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# THERMAL AND FLUIDS ENGR I ENGR 2250

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This 793 page Class Notes was uploaded by Opal Bernhard on Monday October 19, 2015. The Class Notes belongs to ENGR 2250 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/224822/engr-2250-rensselaer-polytechnic-institute in Engineering and Tech at Rensselaer Polytechnic Institute.

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Date Created: 10/19/15

3 1 An arctic explorer builds a temporary shelter from windpack snow The shelter is roughly hemispherical with an inside radius of 15 meters After completing the shelter the explorer crawls inside and closes off the entrance with a block of snow Assume the shelter is now air tight and loses negligible heat by conduction through the walls If the air temperature when the explorer completes the shelter is lO C how long will it take before the air temperature inside reaches 10 C Assume the explorer does not freeze to death or suffocate but sits patiently waiting for the temperature to rise The explorer generates body heat at a rate of 300 kJh Approach Use the first law to find the change in temperature Assumptions 1 The air behaves like an ideal gas under these conditions 2 The shelter is perfectly insulated and airtight 3 The specific heat of the air is constant Solution Let the system be the air inside the shelter From the first law Q AU mchT The mass of air can be determined from the ideal gas law PW lOlkPajg7cl53m3j2897kgkmol 7 ET 7 8314kJkmolK710273K 93946kg The only heat added to the air is body heat The walls are assumed to be thick and highly insulating The rate of heat transfer is related to the total heat transferred by Q QAt mchT Solving for elapsed time 7 mcv T2 7 Q 946kg0717kJkgK107 710K 300 kJh where CV may be found in Table A8 Evaluating A 045h 27min 4 Answer Comments In actuality there must be some air entering and leaving the shelter or the explorer would be unable to breathe39 therefore the rise in temperature may not be as rapid as calculated 3 2 A wellinsulated room with a volume of 60 m3 contains air initially at 100 kPa and 25 C A lOOW light bulb is turned on for 3 hours Assuming the room is airtight estimate the final temperature Approach Use the first law to find the change in temperature Assumptions 1 The air behaves like an ideal gas under these conditions 2 The room is perfectly insulated and airtight 3 The specific heat of the air is constant Solution Let the system be the air and the light bulb From the first law AU Q 7 W The room is wellinsulated so Q 0 The electrical work done is W 7100W3h lh 7108 X 106 J Work is negative since it is done Q the air Because the air is assumed to be an ideal gas the change in internal energy may be written in terms of the specific heat Using this in the first law produces mCVT2 T 7W To find the mass of air m use the ideal gas law 7 PW T 7 100kPa60m32897 kg 110001381 kmol lkPa 1000 lkJ kI kmolK 8314E j252731lt 702kg where the molecular weight of air from Table Al has been used For CV use Table A8 at 300K Solving the first law for final temperature T27 W T1 mcy 77108gtlt106 10001 0 T2 25 c kl 702kg 07217 kgK 463 C Answer Comments In actuality there would be some air entering and leaving the room through doors windows etc and some conduction through the walls Therefore the room air would not reach this elevated temperature 3 3 An elevator is required to carry 8 people to the top of a 12story building in less than 1 min A counterweight is used to balance the mass of the empty elevator cage Assume that an average person weighs 155 lbf and that each story has a height of 12 ft What is the minimum size of motor in hp that can be used in this application Approach Write the first law in rate form Eliminate all terms except work and potential energy Assumptions 1 The motor speed is constant 2 The elevator is isothermal 3 The elevator is adiabatic Solution Take the elevator to be the system under study From the first law dKE dPE dU 7 W dt dt dt The elevator starts and stops at rest Thus there is no net change in kinetic energy The elevator is isothermal and its internal energy does not change Furthermore there is no heat transfer to or from the elevator With these considerations the first law reduces to E 7W dt For a 1minute time period assuming a constant speed elevator mgAz 7W AI Each person has a mass of 155 lbm So 8x155lbm3221212ft S 60s 3221bmfts2 1min 1 min llbf W72976ft1bf 1gp1bf S 550 7541 hp 4 Answer Comments Work is negative because work is being done Q the elevator cage 3 4 A climate controlled room in a semiconductor factory contains a conveyor belt Electric power is supplied to the motor of the conveyor belt at 220 V and a current which varies linearly with time as I 10 t where I is in amps when lis in minutes An air conditioner removes heat from the room at a constant rate of 2 kW The volume of air in the room is 600 m At I 0 the air is at 25 C and 101 kPa Assume the mass of air is constant during this process and assume constant specific heats a Find the mass of air in the room in kg b Find the air temperature after 30 min in C ignore any temperature change of the motor or conveyor belt Approach Find mass from the ideal gas law Write the first law in rate form Express rate of work as voltage times current and substitute this expression in the first law Integrate over time Assumptions 1 Air behaves like an ideal gas under these conditions 2 The temperature of the conveyor belt and motor is unchanged 3 No air enters or leaves the room 4 The specific heat is constant Solution a From the ideal gas law 8314kJkmoiK298K RT 7 1kg 1 0847m3kg PM lOOOPa 7 ioipam 2897kgkmol The mass of air is then 600m m 0847m3kg b From the first law in rate form dE d QiW PEKEU dl dl The kinetic and potential energy of the air are unchanged and may be eliminated Integrating the first law QEVI39dzjalUU2iU1 Heat is removed by the air conditioner and work is added in the form of electrical work The electrical work is WWW 722201dz mu2 7 ul 709 kg Answer 2 72t220mu2 fulmcv T2 7T1 Solving for T2 use Table A8 at m 300K to get cv 0718kIkgK w 220 2 30 2202 2kW30mmA 6795 721 i 1000 Wkw 2 min T2 T1 1 25 C 1 296 C Answer mcv 709kg0718kJkg K 3 5 An interplanetary probe of volume 300 ft3 contains air at 147 psia and 77 F The heaters fail and the air begins to cool Assume heat is dissipated from the outside of the spacecraft by radiation at a steady rate of 60 Btuh On board electronics generate 12 W on average Estimate the time required for the air to cool to 730 F Approach Use the first law in rate form and the givenvalues of heat and work to find the cooling time Assumptions 1 Air behaves like an ideal gas under these conditions 2 The specific heat of the air is constant Solution Let the system be the air From the first law in rate form LU m Q 7 W 511 dt Assuming constant specific heat dT mc 7 W P dt Q To find the mass of air present use the ideal gas law in the form 147psia300ft32897 lbmol PV M j2221bm quot397 ET 39 f3 1073 w lbmolR Separate variables in preparation for integration 107 460R dT Q Wjdz mcv Both the heat rate and the power are constants therefore this expression integrates to mc AT Q 7 W At Solve for AI and substitute values using the specific heat of air in Table B8 to get Btu 7 mchT 7 lbm R Q W 3 412B tu Btu 39 h 760 7712W h 1W 2221bm0l7l j730777 r AI At213h 4 Answer Comments Note that heat is negative because heat leaves the system the air Work is also negative because electrical work is done on the system 3 6 A fan is installed in a 35 m3 sealed box containing air at 101 kPa and 20 C The exterior of the box is perfectly insulated The fan does 250 W of work in stirring the air and operates for one hour Find the final temperature and pressure of the air Ignore the temperature change of any fan parts Approach Use the first law in rate form to find the change in temperature Then calculate final pressure from the ideal gas law Assumptions 1 The air behaves like an ideal gas under these conditions 2 The box is perfectly insulated and airtight 3 The specific heat of the air is constant 4 The mass of the fan parts is negligible compared to the mass of air in the box Solution Let the system be the air in the box From the first law dt dt Assuming constant specific heat dT mc 7 W p dt Q To find the mass of air present use the ideal gas law in the form lOlkPa35m3 2897 kg 1000133 EVA1 kmol lkPa m ET 1 42kg 1 8314 d 20273K kmolK lkJ Separating variables in preparation for integration dT Q Wjdz mcv Since the rates of work and heat are constants in this case this expression integrates to ATQ WAt mc v Using values och from Table A 8 077250W1hj AT 299 C 42kg07l7ij j kgK IkI ATT27T1 T2720 C T2 499 C 4 Answer To find P2 again use the ideal gas law 42kg83l4 kl j499273K 3313 kmOI39K kg lllkPa 35m22897 kmol P2 111 kPa Answer 3 7 A room contains 4 singlepane Windows of size 5 ft by 25 ft The thickness of the glass is A in If the inside glass surface is at 60 F and the outside surface is at 30 F estimate the heat loss through the Windows Approach Use the onedimensional conduction equation Assumptions 1 The thermal conductivity of the glass is constant S olution The surface area of the glass perpendicular to the direction of heat flow is A 45ft25ft 50ft2 Where all four Windows have been accounted for The rate of heat conduction is given by Fourier s law as W 7T2 Using the thermal conductivity of glass in Table B3 08Btuh ftR50ft260730 F 1f 57 600Btuh Answer 025m F 121n 3 8 An Lshaped extrusion made of Aluminum alloy 2024T6 is well insulated on all sides as shown in the figure Heat flows axially in the extrusion at a rate of 35 W If the cool end is at 25 C find the temperature at the hot end Approach Use the onedimensional conduction equation Assumptions The thermal conductivity of the aluminum is constant Solution Heat flows axially in the bar The surface area of the bar perpendicular to the direction of heat flow is A41542 14 cm2 From the onedimensional conduction equation L Solving for T1 and using thermal conductivity from Table A2 39 35 W 13 T1T2 m 25 C209 C4 Answer kA W 14 2 1777 m K 10000 3 9 The wall of a furnace is a large surface of fire clay brick which is 65 cm thick The outer surface of the brick is measured to be at 35 OC The inner surface receives a heat flux of 23 Wcmz Estimate the temperature of the inner surface of the brick Approach Use the onedimensional conduction equation Assumptions 1 The thermal conductivity of the brick is constant S olution For onedimensional steadystate conduction Q km 7 T2 The wall receives a heat flux of23 Wcmz so 39 k T 7 T 2 23 W2 2 A cm L Solving for T1 T1 5 2 r T k A The thermal conductivity should be evaluated at the average of T1 and T 239 however T1 is unknown We begin by evaluating k at a reasonable temperature and iterating if necessary From Table A3 k 10Wm at T 478 K Using this value 6395cm1dom w 10 2 T1A2HI35 c1530 c18031lt cm 1m 10 mK The average wall temperature is T 7T1T271803308 ave 1056K 2 This is much higher then the temperature at which kwas evaluated and Table A3 shows that kvaries with temperature So find a new estimate for kby interpolating in Table A3 At T 1056 K k 1572 The revised estimate of T1 is then 65 L T1 i23x104308K 1259K 1572 This is much different than the first estimate of T1 1803 K Use the new value of 1259 K to find a new average temperature and a new value of k Iterate until T1 no longer changes The first few iterations are k T1 K 1 000 1803 1 572 1259 1 344 1420 1 435 1350 1 395 1380 For greater accuracy continue the iteration At the last iteration T1380K 4 Answer 3 10 A tungsten filament in a 60 W light bulb has a diameter of 004 mm and an electrical resistivity of 90 pQcm The filament loses heat to the environment which is at 20 C by thermal radiation The emissivity of the filament is 032 and the voltage across it is 115 V Find the length of the filament and the filament surface temperature Electrical resistance equals electrical resistivity times filament length divided by filament crosssectional area Approach Calculate the resistance of the wire from the known Joule heating losses and the applied voltage Use resistance to find the wire length Then equate the Joule heating to the loss by radiation to determine the wire surface temperature Assumptions 1 The only mode of heat transfer is radiation 2 The surface of the filament is gray and diffuse 3 There are no re ecting surfaces near the filament Solution The electrical resistance is related to the electrical resistivity by 911 AL 2 A T 7rD 2f where A is the crosssectional area The power dissipated in the filament by Joule heating is 52 1 Q 5 RE Solving for resistance 2 2 Rgimzzog Q 60W Therefore the filament length is 2 2 2 R 7 2 22097r mmj 1 m L 7 E 2 7 2 1000mm p2 90yQcml ig m 10119 100cm L 0308m 308 cm 4 For radiation from a gray surface to a black environment Q8039A5T54 Ta4 where A5 is surface area TS is surface temperature and Ta is the ambient temperature 1 1 39 Z 39 7 T5 Q Ma Q Ma 80145 goerL T5 60W W 1m 032 567 108 04 0380 f X mZK mm1000mm m Answer 1 20 2734K T5 3040K 4 Answer Comments Tungsten glows brightly at this high temperature The filament is coiled so it fits in the bulb 3 11 On a cold winter day the interior walls of a room are at 55 F A man standing in the room loses heat to the walls by thermal radiation The man s surface area in 16 ftz his clothing has an emissivity of 093 and his surface temperature is 70 F He generates 300 Btuh of body heat What percentage of the man s body heat is transferred by radiation to the walls Approach Use 80247quot 7T 51 to determine radiation heat transfer from the man and compare this to his body heat Assumptions 1 The man s clothing is gray and diffuse 2 There are no re ecting surfaces near the man 3 The man s temperature is uniform Solution The heat lost by radiation is Q 80A T3 7 Tm Converting temperatures to absolute and substituting values Btu 4 7 8 2 7 4 4 de 7 0930l7lgtlt 10 h z R416ft 70 460 55 460 JR Btu 21 8 de h To find the percent transferred by radiation 0726 726 Answer Qbody 300 Comments Typically radiation is an important mode of heat transfer when the only other mode is natural convection in air 312 The sun can be approximated as a spherical black body with a surface temperature of 5762 K The irradiation from the sun as measured by a satellite in earth orbit is 1353 Wmz The distance from rth 39 39 15x1 39 quot inall a directions estimate the diameter of the sun Approach Construct an imaginary sphere around the sun with a radius equal to the sunearth distance The total energy in watts falling on this sphere equals c the total energy emitted by the sun Use this total Q m I 394 energy and the StefanBoltzman law to nd the k e ective solar diameter Assumptions 1 The sun radiates unifome in all directions 2 The sun radiates like a black body Solution Ifthe sun had a radius r then the total heat radiated by the sun would be Q A Q 47571 whereQ is in Wans This heat is spread evenly over an imaginary sphere whose radius is the sunearth distance R Therefore Q 1353 4sz m2 Q 13534Ir15x10 2 83 x 1 026 W The effective radius of the sun is 383x1026W w m2K4 4 N 557x10 8 5752K 7698X108m D27140x109m Answer 3 13 A high torque motor has an approximately cylindrical housing 95 in long and 6 in in diameter The motor delivers 18 hp in steady operation and has an efficiency of 072 All the heat generated by motor losses is removed by natural convection and radiation from the outer surface of the housing The convective coefficient is 168 Btuhft2 F and the housing emissivity is 091 1fthe surroundings are at 58 F what is the housing s outer surface temperature Approach Use the definition of efficiency to calculate the electrical work input to the motor Then determine the rate of heat lost to the environment and set this rate equal to the sum of the radiative and convective heat losses Assumptions 1 The emissivity is not a function of temperature 2 The heat transfer coefficient is not a function of temperature 3 The emissivity is uniform over the surface of the housing 4 The heat transfer coefficient is uniform over the surface of the housing 5 The rate of heat loss is uniform over the surface of the housing 6 The surface of the housing is gray and diffuse Solution Because the motor has an efficiency of 072 72 of the input electric power is converted into mechanical power 072W hp Solve for electric power and convert units to Btuh lhp 2544E Btu 8 h 442 072 1 hp h The energy lost as heat from the outer casing is Q1 17 072442j1238 h h The surface area of the housing is A 27W2 27rrl 27r3in2 27r3in95in 2361112 164 12 The surrounding temperature is Tm 58460518 R Heat is lost by both convection and radiation that is Q Q39md Q39mv SHAW T hAT5 T 5 sun sun Substituting values Btui 8 Btu 2 4 4 Btu 2 124T709101714gtlt10 mjhm ft 7518 168m 164 ft 7 57518 Solving Y54787 F 1 Answer 3 14 A at plate solar collector 6 ft by 12 ft is mounted on the roof of a house The outer surface of the collector is at llO F and its emissivity is 09 The outside air is at 70 F and the sky has an effective temperature for radiation of 45 F The collector transfers heat by natural convection to the air with a heat transfer coefficient of 32 Btuhftz F and also transfers heat by radiation to the sky Calculate the total heat lost from the solar collector Approach Use the fundamental rate equations for convective and radiative heat transfer and add to get the total heat transfer Assumptions 1 The emissivity is not a function of temperature 2 The heat transfer coefficient is not a function of tem erature 3 The emissivity is uniform over the surface of the collector 4 The heat transfer coefficient is uniform over the surface of the collector 5 The surface of the collector is gray and diffuse Solution The surface area of the collector is A 6 ft12 ft 72 ft2 The heat lost by convection is QWhATT gt Btu 32 72ft2 110770 F E x gt 9216E h The heat lost by radiation is QisaAn 7T 39 7 8 Btu 47 4 de7090l7lgtlt10 m110460 45460 4490E h The total heat transferred is QM QM Q39md 9216 4490 13700BT Eu 4 Answer Comments The effective temperature of the sky depends on the cloud cover A clear sky will be much colder than an overcast one 3 15 A CPU chip with a footprint of 3 cm by 2 cm is mounted on a circuit board The chip generates 03l Wcm2 and rejects heat to the environment at 28 C by convection and radiation The outer casing of the chip has an emissivity of 088 and the heat transfer coefficient is 48 WmZK Neglecting the thickness of the chip and any conduction into the circuit board calculate the chip surface temperature Approach Equate the heat generated to the sum of the heat removed by convection and radiation Express the rates of convection and radiation in terms of surface temperature and solve for surface temperature Assumptions 1 No heat is removed from the chip by conduction 2 All heat leaves from the top of the chip the surface area of the sides of the chip is very small and Will be neglected 3 The surface of the chip is gray and diffuse 4 The heat transfer coefficient and emissivity are not functions of temperature 5 The heat transfer coefficient and emissivity are uniform over the surface of the chip Solution The heat generated by the chip is Qgen 031123cm2cml86W cm This heat is removed by convection and radiation 139 e ng va de Q hKTS T 801054 i T MA The area of the top of the chip is lm A 3cm 2cm X E 104 cm 2 The surroundings are at the same temperature as the air The air temperature is T 28 273 301K Substituting values 2 00006m2 NK 00006m2T5 7301K088567x10398 m 186W48 ruffK4 00006m2T 7 3016K By trial and error or using an equationsolving program T5 357K 842 C Answer 3 16 A metal plate 16 cm by 8 cm is placed outside on a clear night The plate which has an emissivity of 07 exchanges heat by radiation with the night sky which is at 7 40 C Air at 710 C flows over the top of the plate cooling it with a heat transfer coefficient of 42 WmZK The plate is insulated on its underside and heated by an electric resistance heater How much electric power must be supplied to maintain the plate at 55 C Approach Calculate the radiation and convection from the fundamental rate equations The sum of these rates equals the heat generated by the first law Assumptions 1 The metal plate is gray and diffuse 2 The plate is perfectly insulated on the back side 3 Emissivity is uniform over the plate 4 Emissivity is independent of temperature 5 Heat transfer coefficient is uniform over the plate 6 Heat transfer coefficient is independent of temperature Solution The heat removed by radiation is Q 4mm 4 where Tp is the plate temperature and T5 is the sky temperature Tp 55 273 328K T5 740273 233K QM 168Cm2 lm W m2K4 2 2 O7567gtlt10398 3284 7 2334 K 104 cm QM 438 w The heat removed by convection is va Tip T where Ta is air temperature Ta 7710273263K w 2 1m2 42 l 8 cm 3287263 K va E m2K6 Wow 349 W QM QM Q39m 438 349 393w By the first law dU 7 W dt Q The plate is in steadystate39 therefore there is no change in internal energy and the first law reduces to Q W The electrical work power to the heater equals the heat lost from the plate W QM 393w Answer 317 A home freezer is 18 m wide 1m high and 12 m deep The interior surface of the freezer must be kept at 710 C The walls of the freezer are made of polystyrene insulation sandwic ed between two thin layers of steel The combined convectiv radiative heat transfer coe icient on the exterior is 82 WmZK and the ambient is at 25 C Ifthe power ofthe refrigeration unit is limitedtolSOW h quot L 39 L 39 39 the thin metal wall panels is very small and can be neglected and that the bottom ofthe freezer is perfectly insulated Approach Use the resistance analogy to find the resistances due to conduction and convection All resistances add in series I J Vi Assumptions 1 Thermal conductivity is independent of temperature 39 2 The resistance ofthe steel is negligible l WI 3 No heat is conducted into the oor V 4 The thickness ofthe insulation is small compared to the size ofthe freezer 39 39 5 The heat transfer coefficient is uniform over the surface of the freezer and independent of temperature Solution To maintain the freezer in steady state the heat transfer into the freezer from the oumide air must be no more than the freezer can remove To find e heat transferred into the freezer use the resistance T T analogy The exposed surface area of the freezer L neglecting the bottom is 2 M rm g a m L M A211821121218 816m 4m am Assume the interior and exterior of the freezer have about the same area since the insulation thickness mva 104 is small compared to the size of the freezer The heat entering the freezer is Solving for L Ta LkA using kfrom Table A4 254710 07 1 250W 52 f 1615quot m K 00481m 481cm Answer L0027i816m2 mK The windshield of an automobile is heated on the inside by a ow ofwarm air Cold air at 45 F ows over the exterior ofthe windshield The heat transfer coefficient on the inside is 16 BtuditWF and the heat transfer coefficient on the outside is 49 BtuhftZ F The glass ofthe windshield has athickness of 025 inch What temperature should the inside air be so that the exterior surface temperature of the windshield is 3 F Approach Use the resistance analogy adding all resistances in Jill series L LI 4 h 3914 15 Assumptions 7 1 Thermal conductivity is independent of temperature 2 The heat transfer coefficient is uniform over the 39T1 surface of the Windshield and independent of K r quot 7 temperature 3 4 LEquot Solution h 1 ll 4 WU F Heat is convected from the inside of the windshield with resistance R3 conducted through the glass with J resistance R2 and convected from the oumide of the 4 A k A Ll A windshield with resistance R All resistances add in A M 1 series Perform the calculation for a unit area of 1 ftz T i The result is independent of area The resistances are 0 l Ru 2 R3 3 a t I V g V RF BL00204 111 5 4 Dd i39SIJE msiclc menm u 2 u r la 7 my m 39 4 5M 4m 0439 OF lzrss 025 in 1 i Bm 12 39 0025 F39h 1 08 1ft B hft F k 15fu 1ft hft F By the resistance analogy heat is AT Q From the resistance circuit the heat through each resistance is the same and 7 1 L R 19 Ri Solving for T3 T3 02 word R1 002500525 37 715 T i M 00204 E81 F Answer 3 319 A copper husbar oflength 40 cm carries electricity and produces 48 W injoule heating The crosssection is square as shown in the gure and is covered with insulation oftherma conductivity 0036 WmK All four sides are cooled by air at 20 C with an average heat transfer coefficient of 18 WmZK 39 L n i i uLiIcuuai 39 39 of the insulation Approach Use the resistance analogy adding all Em resistances in series 40 A ssump ons 1 Thermal conductivity is indep endent of temp erature 2 The heat transfer coefficient is uniform over e surface of the Copper insulation and independent of 392 Gm tern erature 3 Heat transfer is planar and one dimmsionaL Insulation lt gt 12 cm 20m Solution The heat transferred through each side ofthe bar is 4 12 W Q 4 W A a a A A A R r The area ofthe inside of the insulation is 1 mm RI R2 rig Ai 8cm40 cm 320 cm2 The area of the outside of the insulation is A2 12cm40cm480cm2 For conduction through the insulation use the average of the inner and outer areas as an approximation Am 480 320 400m The conduction resistance is R1 2cm 139KW kA 1m 0035w mK 400cm2 100 cm The convection resistance is 1 12 11 6KW 18W m2 K 480cm2 L 10000cm2 The total resistance is the sum ofthe two resistances in series RM 12 122 151KW 39 fthe insulation E n and is givenby TmirTmRmQ Tm 151KW12W20 C Tm 3s1 c 4 Answer 320 A freezer maintains one side of a slab of ice 3 cm thick at 710T The other side exchanges heat with the ambient at 15 C by combined natural convection and radiation In steady state the ice does not melt 39 39 39 fhe 39 39 J exposed to the ambient Approach Use the resistance analogy to relate the I ta surface temperature to the heat transfer I I coe icient R1 R 1 A ssump ons 1 The combined radiativ convective heat transfer coefficient does not 4 depend signi cantly on temperature L 2 The thermal conductivity is constant 3 The heat transfer coefficient is uniform over the surface ofthe ice Solution When L 39 T is0 C 39 39 39 39 Ifthe heat transfer coe icient is higher than this value the ice would begin to melt Conduction through the ice occurs in series with combined convectionradiation at the outer surface as shown in the figure The conduction resistance is R 7 M The heat conducted through the ice is 7T WT 4 For the combined convection and radiation the resistance is 1 R 2 M TL L the iceis equal 39 39 convection and radiation therefore TiT Q 3 2 hAT3 Tz R1 Eliminating Q gives kAT 7T Ma 7T2 Solving for h and using the thermal conductivity of ice from Table A3 T2 Ti 11 issf ijimo ME 7T2 003 m150 418 AIISWBIquot m2 C 321 The door ofa kitchen oven contains awindow made ofa single pane of 14 in thick Pyrex glass The interior oven temperature is 550 F and the room air is at 68 The combined 39 39 39 ancient on L 39 39 is 17BtuhfF F and on the oven exterior it is 088 BtuhftZDF A toddler comes by and touches the window Calculate the temperature ofthe surface that the child39s hand contacts Approach Use the resistance analogy adding all resistances in series 1 I A 9 1 Assumptions R 1 The combined radiativ convective I heat transfer coefficient does no 7 e 4 depend signi cantly on temperature is 2 The thermal conductivity 3 The heat transfer coefficient is uniform over the surface ofthe glass 4 Heat transfer is onedimensional Solution From the resistance network in the figure Q W mom own W 1 12 Substituting values with thermal conductivity from Table B3 Tm n58 7 550TW F 1 1ft 0251n j 088quot u u 17 A 07 A l DJ E Solving for the window temperature TW yields 3 80 F Answer Tm 322 An electronic device may be modeled as three plane layers as shown in the gure The entire package is cooled on both sides by air at 20 C Heat is generated in a very thin layer between two contacting s aces at a rate 0 500 Wmz as shown The heat transfer coef cient on both sides is 87 Wm K Assume the layers are very large in extent in the direction not shown Using data in the gure below calculate the temperature T2 Approach Q m sou wzml Use the resistance analogy adding all resistances in series K 034 WmK A ssump ons 1 The combined radiativ convective heat transfer coe icient does not depend signi cantly on temperature 2 The thermal conductivity is constant 3 The heat transfer coef cient is uniform over the surface ofthe KZ o 16 wt39mK device 4 Heat is generated in a layer of negligible thickness Solution The resistance network that can be used to model this device is 392 A RI 17 2 To TI T2 T3 T4 T0 where I is the air temperature mm In an area of1 mi 39 be calculated as h A L A 03933m 000882E Q LTZ 1 1 0347j1m2 W mK Another way to represent the resistance network is shown in the gure to the right where R4 is given by K R4 7R 122 123 70245W andR is 1251l Hg 01237 W For two resistors in parallel l l L or R 7 Rm R4 R5 Substituting values gives ampamp K Rm 00822 W From the resistance analogy AT QR T2 7 T0 Solving for T2 T2 500 W00822j 20 C T2 61 C 1 Answer 323 A cardboard box is used to ship owers on a summer day when the ambient temperature is 80 F The air inside the box is maintained at 45 F by the use of cold packs The box is lined with a layer of styrofoam k 0015 BtuhftR one half inch thick The cardboard itselfis 18 inch thick and has kc 013 BtuhftR Thebox measures 8 in by 8 in by 25 ft Assume h on the inside is 20 BtuhftZR and h on the oumide is 93 BtuhftzR Calculate the rate of heat transfer into the box Neglect heat transfer on the ends Approach Use the resistance analogy adding all resistances in series A ssump ons 1 The heat transfer coefficient does not depend significantly on tempera re 2 The thermal conductivity ofboth materials is constant 3 The heat transfer coefficient is uniform over the surface ofthe box Solution The area of the inside of the box neglecting the ends is foam cardboard Ai 25 6139 725 tt45525 2 I The area ofthe outside ofthe styrofoam is A2 2 5njn4 54m2 The outer area of the box is A3 25ftj 4 55m inseriesToquot L39 394 1 1 R 0075h F Btu 25A 20Btuh 2 F5625 2 For conduction through the foam the area is different on inside and out Estimate the appropriate area by taking the average of inner and outer areas L R2 12 0417h FBtu kA 0015Bt11h R604 2 Likewise for the cardboard take the average in inner and outer areas to get 11L R R3 L M 013BtuhftR656ft2 The exterior convective resistance is 0012h FBtu i M 013Btuh R656ft2 RM 12 122 123 124 0577h FBtu QAT80745507 Answer Rm 0577 h R1 0012h FBtu 324 A living room oor 3 m by 45 m is constructed ofa layer of oak planks 12 cm thick laid over plywood 20 cm thick In winter the basement air is at 15 C while the living room air is at 20 C 39 quot39 quot39 162nd68 m m WmZK respectively Ifthe home is heated electrically and the cost of electricity is 008 per L L L 4 L r vuLu kWh r ofthe energy walltowall carpeting 16 cm thick k 006 WmK what would the energy cost be Approach se the resistance analogy adding all resistances l in series I v 39 a Assumptions li 1Neglect conduction through the oorjoists not shown 2 Conduction is onedimensional 3 Thermal conductivity is constant Solution First consider the case without the carpet The bike Mil1 al resistances for convection and conduction are using Table A3 and A5 for thermal 2 conductivities 1 0206 0 jams mzK R2 017 000523 0177345m2 W mK 002 K 3 W m 24 012 345m2 W mK 1 001095 w 68 345m2 All these resistances add in series K RM 12 122 123 R4 00205 000523 00124 00109 0049W The heat transfer rate is given by T 7 20715 C lozw Q L Rm 00493 cost1 month M lkw 102h 1month 1day kWh 1000W 587 lt Answer The carpet adds a resistance to the system The resistance of the carpet is 0016 cm R5 5 w 006 j345m2 m K 00198E W Adding this resistance to the previous total resistance gives a revised total resistance of R1 Rm R5 0049 00198 00688 The new heat transfer rate is u 7 20715 7 00688 The ratio of the new cost to the old cost is new cost 7 727W old cost 7 102W Solving for new cost 727 W 727W new cost j5874l9 Answer Comments The carpet may improve appearance or com fort in the room however it does not contribute substantially to energy savings 325 The wall ofa fumace must be designed to transmit no more than 220 Btuhftz Two types of bricks are available for construction one with a thermal conductivity of 038 Btuh R and a maximum allowable temperature of1400 F and the other with a thermal conductivity of 098 Btuh R and a maximum allowable temperature of 2300 F The inside wall ofthe furnace is at 2100 F and the outside wall is at 300 F Both types ofbricks have dimensions of9 by 45 by 3 in and both are e same cost per bric Ifthe bricks can be laid up in any manner determine the most economical arrangement ofbricks Approach Use the resistance analogy to find the temperature drop across the layers ofbrick Use as few high temperature bricks as possible since these have poor insulation properties A ssump ons 1 The thermal conductivity of the bricks is not a function of tempera ure 2 Heat conduction is onedimensional 3 The wall is very large Solution First calculate the necessary thickness ofthe high temperature layer which is made ofhigh conductivity bricks This thickness is found from AT Q T kA Taking A1 2 andQ 220Btuh 220071400 F 098 Bl l z h F 220B h EMT1 312 374in Because ofthe integral size ofthe bricks we cannot design a layer with exactly this dimension The actual layer must be at least 374 in Given the size ofthe bricks the layup which uses the fewest number ofbricks and exceeds 374 in by the least amount is 7 77 7i r quot J ILL which gives an actual thickness of 39 in The twolayer wall may be represented as R102AT21007 F8gt18h F Q 220T Btu L39 39 4 quot eachlayer L2 R M M SolvingforLz Bt h F 39 11 2 L2 sz Rafi 038 quot m2 s1s39 7 2 19A htt F Btu 0981 22 1 5 n The layup ofthe second layer must be a L 7 I 1 llJ J In summary the most economical layup which satis es all constraints is kolqg lto l 42 E 0 A Comments In actual application the mum 39 quot39h b 39 been laid in the other direction with three bricks of edge length 3 in spanning what is shown as one 9in brick in the diagram or two bricks of edge length 45 in spanning what is shown as one 9in brick This 39 UJ mick WHCH me third dimension is considered 326 A chemical reactor is in the shape of a long cylinder as shown in the gure The reactor is covered with a layer of insulation 17 7 cm thick The reactor loses heat through the insulation at a rate of 153 W per meter oflength The thermal conductivity ofthe insulation is 004 WmK On the outside of the insulation air at 26 C removes heat by forced convection with a heat transfer coefficient of 32 WmfK Find the maximum temperature of the insulation Neglect radiation Approach Use the electrical resistivity analogy adding all Reamer resistances in series The maximum temperature is Insulatlon at the inner radius ofthe insulation A ssump ons 1 Heat transfer is onedimensional 2 T e thermal conductivity ofthe insulation is constant 3 The heat transfer coefficient is uniform over the surface of the insulation and independent of temp erature 177 cm gt Solutlo ll The reactor is hot inside and the air is cool outside The maximum temperature of the insulation will be at r 71 3m Q R a r 2 70177m0473m 39 W The outer radius of the insulation is I 1 M 1 L yr0177m055m T W quotM Nth 391 The resistance across the insulation is lnEri 1110155 quotk 039473W 42555 2 2z1m0047 W mK where L1m because the value of Q is given for a onemeter length The resistance for convection is 2z055m1m 7 0007653 W K RM 12 122 1255000755 1273W From the resistance analogy ATQRT T 7T Rearranging Tl Tz QRT K 26 C153W 1273 W 455 C Answer An insulated copper wire with a length of 12 m carries 20 A of current The copper is 1 mm in di ter and the insulation k 013 Wm C has a thickness of 08 cm Air at 25 C blows in cross ow over the 39 39 r mimic of 219 WmZK Assuming the copper is isothermal nd the copper temperature Take the electrical resistivity of copper to be constant at 21 x 10398 Elm Approach L Use the electrical resistivity of the wire to nd the wire s electrical resistance Then calculate the power dissipated by Joule heating Finally add the A conductive resistance across the insulation to the convective resistance and use the total resistance to compute the wire temperature 7 A ssump ons 1 The copper is a perfect conductor and hence is othermal 2 The thermal conductivity and electrical resistivity are constant 3 Heat transfer is onedimensional 4 The heat transfer coef cient is uniform over the surface ofthe wire and independent of temperature 6 Solution The radii of the inner and outer surfaces of the insulation are r i 00005 m 1000mmm 1 72 1 08cm m 10 0 cm The crosssectional area A ofthe copper is A 7n 74000052 785x10397 m2 The electrical resistance ofthe wire is R pL 21x10 3 Qm12m 00321 9 78 X1039 m 00005 0008 00085m The power produced by Joule heating in the wire is Qi R20A200321 2128W The thermal resistances are lnr2 hquot ln0008500005 W 289E 2Ir12m013 W m K L 1 0071 25 Ms 219 2 j2r00085 m12m W m K where A is the surface area ofthe outside ofthe insulation Temperature is found from Q 3quot or znQltRR2 1 12 T 25 c12s W28900712 53m 4 Answer 3 28 The wall of a submarine is 1in thick stainless steel A151 304 insulated on the interior with a 15in layer of polyurethane foam k 0017 Btuhft F The heat transfer coefficient on the interior is 37 Btuhft F At full speed the exterior heat transfer coefficient is 135 Btuhft F The sub is approximately cylindrical with the length 240 ft and outer diameter 30 ft 1f the seawater is a 40 F at what rate must heat be added to the interior air to keep it at 70 F As a first approximation neglect heat transfer through the ends Approach Use the resistance analogy adding all resistances in series Assumptions 1 The combined radiativeconvective heat transfer coefficient does not depend significantly on temperature 2 The thermal conductivity is constant 3 The heat transfer coefficient is uniform over the surface of the submarine 4 Heat transfer is onedim ensional Solution The radii that will be needed in the calculation are r3 15 t r2 15 7112 14917 ft r1 r271512 14792 ft Heat transfer from the 70 F interior air to the 40 F exterior water can be calculated using the following resistance network 7o kl R2 R3 R7 a F The convective resistance on the interior is 1 1 1 1 121x10 5h39 F 1 02 371327r14792ft240ft hft F The conductive resistance through the polyurethane foam is In 12 1n14917 R VI 1439792 328x10quot 2 Lk1 2240 ft0017 Btu Btu hft F The conductive resistance through the stainless steel with thermal conductivity from Table B2 is In g In 15 j 0 R3 72 1439917 430gtlt10 7h39 F 27rLk2 27r24086 Btu Finally the convective resistance on the exterior is R4 327X104h39F hjzmgL 13527r15240 Rm 121 R2 R3 R4 349gtlt10 5 From the resistance analogy L404 879 gtlt104 B mi Answer Rm 341 gtlt 10 h 329 An aluminum wire 25 m long conducts 12 A with an imposed voltage of 15 V The wire which has a diameter of 24 mm is covered with a layer of insulation 2 mm thick The thermal conductivity of the insulation is 015 Wm C Air at 40 C ows over the exterior ofthe wire to give a convective heat transfer coefficient of32 WmZDC Assume the aluminum is isothermal L 394 J 4 r L 394 J 4 e an r insulation Approach Calculate the power dissipated by Joule heating Use the thermal resistance analogy to find the unknon temperatures A ssump ons 1 The aluminum is a perfect conductor and hence isothermal 2 The thermal conductivity is constant 3 Heat transfer is onedimensional 4 The heat transfer coefficient is uniform over the surface ofthe wire and independent of temp erature olu on The radii of the inner and outer surfaces of the insulation are r 12mm 1 00012m 1000 mm 1 m 1000 mm The thermal resistances are In In 00032 R r 00012 72 2mm 00012000200032m 2 M 7 0415 2015 25 mK R77 1 7 7 1 0K hA 112rr2L 32 Y szzo0032m25 m W m The heat generated by Joule heating is Q i 15V12 A 18W From the thermal resistance analogy QEE R R2 K I T QR 122 40 C 18W 04150522W 587 C 4 Answer This is the temperature at the interior of the insulation Similarly at the exterior temperature is K T2 TQR2 40 c18w0522W 512 c 4 Answer 3730 An insulated steel pipe canies hot water at 80 C The outer surface ofthe insulation loses heat to he environment by convection and radiation For convection assume m s 8 wm2 c Th h to e mm of 39 temperature 39L 39 L 39 U data on the gure below Approach 4 cm Use the thermal resistance analogy to ndthe unlmown temperature Assumptions 1 The thermal conductivity is constant 2 Heat transfer is onedimensional 3 The heat transfer coef cient is uniform over the surface ofthe pipe and independent of temperature 4 The insulation is gray and diffuse insulation k0038 VvmK Solution The total neat nan re 39 39 l u A thu hm hm m hm 567 7 73 Tm From the resistance circuit 2277quot quot3 R in ra I3 which becomes K Turf 777quot q 11L ln 17 7 1 2h an I T zuk 1 rij R3 EMA zhhffim It is not possible to solve this equation directly for the unknown temperature T because hm is a function of 7 By trial and error the solution is TSurv z 428 Clt Answer Comme s onsolV quati ing software is Very useful in problems of this type Remember to use absolute temperauture Kelvin in the computations 331 A cylinder ofradius r is covered with a layer of insulation ofthermal conductivity k A uid a E a w E E a E a o E 9 E 3 0 a o m E E a 3 O o a r5 9 E a E E a 9 8 q a a a w 0 combination ofconduction and convection resistances Ifrz is small the conduction resistance is small As 72 increases the conduction resistance increases but the surface area of exposed insulation also increases and this results in a decrease in convective resistance As a result there is anoptimalvalue ofr 39 quot 39 Derive an expression for the optimum value ofrz as a function of r k and h Approach Write an equation for the total resistance due to conduction through the insulation and convection the outer surface To nd the maximum value of resistance take the derivative with respect to the outer radius r and set it equal to zero A ssump ons 1 Thermal conductivity is constant 2 Heat transfer is onedimensional 3 The heat transfer coef cient 39s uniform over the surface of the insulation and independent of temp erature Solution We assume the length L ofthe cylinder is very large compared to the radius so that heat transfer is one dimensional In that case the total resistance is where Rm 39 RmW 39 39 39 quotquot 39 are given by In RM r n erLk Expressing the total resistance explicitly as a function ofrz gives 11172 n r R 2 2sz hZIWZL To find the optimum value ofr2 for whichRm is maximized take the derivative ome2 with respect to 72 and set the result equal to zero Letrn be the optimal value ofr2 then 72 0 U U 2rLk erhL 01L k rnh Solving for 70 k Answer 11 Comments Note thatrn is independent ofr1 but thatrn gt r Ifthe cylinder radius is greater thanrn there is no optimal value ofouter radius and 39 39 39 39 thickne 39 39 A frozen pipe is lled with ice at 0 C A heating tape wrapped around the pipe provides 90 W per meter ofpipe length Insulation is placed over the heating tape The insulation has a thickness of 05 cm and a thermal conductivity of 0082 Wm C Convection and radiation occur from the outside of the insulation to the environment which is at 715 C The heat transfer coefficient is 77 Wm C and the emissivity is 094 The pipe wall remains at 0 C during the heating and the heating tape is very thin T e pipe has as 1D of3 cm and a wa 1 thickness 0 4 mm How much time is required to completely melt the ice heat offusion 334 x105 Jkg density 921 kgmz Approa h C Use a thermal resistance network The heat from the heating tape is a source term added fa T1 a constant temperature of 0 C Determine h much heat conducm through the insulation an subtract this from the heat added by the tape The result is the rate of heat transfer into the ice Using the latent heat offusion the time to melt the ice can be calculated Lilquotm 6 39i39are M 339 a l it l M A ssump ons 1 The pipe wall is a perfect conductor and henc e is a 2 The thermal conductivity of the insulation is constant 3 Heat transfer is onedimensional 4 The heat transfer coefficient is uniform over the surface ofthe insulation and independent of temperature 5 The insulation is gray and diffuse Solution R3 where Q3 is quot hv 39 T is r fthe insulation The heat added by L r rquot 39 quot quot 39 39 Q and the other part Q2 acm to melt the ice First find Q We will need the following radii 3cm 7 T15cm 0015m 4 r 0015m m 0019m 1000 05 73 0019m m 0024m 100 where L lm because the heat rate of the tape is given per meter of pipe length The resistance to conduction is 1113 In 0024 K 22k 039019 W 0453 2 27rlm0082 W m K R1 The resistance to convection is R2 h h W 1 0861E WA wWZWEL 77 2 27r0024mlm W m K The radiative resistance is R L where h 2ch 1T 23 hmdA hmd The surface temperature TS is unknown so had cannot be determined It is necessary to do an iterative calculation We will assume a value of TS 266 K which happens to be the right answer Of course it is very unlikely that a person would pick that exact number as an initial guess however any reasonable guess something between 0 C and 715 C will give a result close to the correct answer The guess can be adjusted and the calculation repeated if greater accuracy is desired With the assumed temperature m2K4 m2 hm 094567gtlt10398 266 258K2662 2582K2 384 REQJ3 384mij2 0024mlm W Define R23 as the parallel combination ofRZ and R3 that is R2R 7 0861173 7 0 5755 39 w 23 T R2R3 T 0861173 T Rm R1 1122 04530575 403 The heat traveling through the insulation is Q T 4 42737258gtK 1 Run 1035 W The heat available to melt the ice is Q392 Q 7Q 907145 755w The total amount of heat needed to melt the ice is 334x105kijn0015m21m921 kg 2l7gtlt105I g l45W Q2 h V p152 hx7W12Lp 1 12 152 E Q2 Q2 AI Therefore 5 At g w 2878s 48min Answer Q2 755w 3 33 Show that the conduction thermal resistance of a spherical shell of inner radius VI and outer radius r2 is given by R 72 71 sphere 47er Approach Start with Fourier s law Write surface area A in terms of the radius of the sphere and integrate from the inner radius to the outer radius of the spherical shell Compare the resulting equation to a resistive circuit Assumptions 1 Thermal conductivity is constant 2 Heat transfer is onedim ensional Solution From Fourier s law dT 7104 Q dr Separate variables and integrate from the inner to the outer radius j 9dr iijz err n A T To perform this integration we needA in terms of r is a constant Substituting the formula for the surface area of a sphere and integrating the right hand side a dr 7k T 7 T Q w lt 2 1 germ Em 7T Compare this result to a resistive circuit to obtain T 7 T 2f2 r1 R quot Answer sphere Q 4 kr1 r2 334 A hollow sphere made ofpure aluminum has an inner radius of3 cm and an outer radius of18 cm The temperature at the inner radius is maintained at 0 C The outer surface is exposed to air at C n cueinciemi 65 Wm K be neglected Calculate the rate ofheat transfer and the temperature ofthe outer surface of the sphere Approach Use the resistance analogy to determine the heat transfer rate an e s ace temperature Calculate the conductive and convective thermal resistances and add them to nd the tota istance A ssump ons 1 The thermal conductivity is constant 2 Heat transfer is onedimensional 3 The heat transfer coef cient is uniform over the surface ofthe sphere and independent of temperature Solution Appludul 39 L R cm 4mm JIIC vaiueiui 39 39 f 39 39 quot 39 TableA2Tquot L39 4 39 values R m1 0187003m 000933 4018003m2 237i mK The convection resistance is Rum 2 4 003785 WW2 55 2 j4r0182m2 W m K 1 IL m H convection occur in series therefore add the resistances to get total resistance The heat transfer b ecomes T T 0725 c Rm Rm Raw 0 00933 00378 Q 71811W Answer a 39 39 owin in t t39 d39 t39 quot To nd T heat is equal hv convection so that T 4 Q7 Rconv Solving for T2 T2 QRW Tm K E71811W 00378 25 C W 495DC Answer A bathosphere of inside diameter 34 m is at an ocean depth where the water temperature is 5 C The wall ofthe bathosphere is made of5 cm thick steel The convective heat transfer coe icient between the air 39 39 is 92Wm 39K the water quot is 860 WmZK After the divers return to the surface they complain to the designer that the bathosphere was chilly Ifthe maximum power of the heater is 25 kW estimate the air temperature inside the bathosphere Approach Use the thermal resistance analogy A ssump ons 1 Neglect body heat generated by the divers 2 The bathosphere is spherical in shape 3 Neglect heat dissipated by any onboard electronic s 4 Heat transfer is onedimensional 5 Thermal conductivity is constant 6 The heat transfer coefficients are uniform ov the sphere and independent of temperature Solution The inside and outside radii ofthe bathosphere are r17m 7217005175m The inside and outside surface areas are respectively Ai 4zrrf 474172 m2 323 m2 2 4742 41752m2 385m2 The thermal conductivity of steel is available in Table A2 The thermal resistances are R L 000299 1 92 2 jam m K R2 ltr2 7quot 175717 221x10395 4 417m175m605 W mK Rj1 W1 302x10 5 hIAZ 860 2 j385m2 W m K K Rm R1 122 123 000305 W quot L L 39 39 The rate ofheat transfer is Q T 1 Solving for the bathosphere air temperature Tm QR Tmm 25kW 1000 W 000305E 5 C 1 kW W 126 C 547 F Answer Comment It is rather chilly in the bathosphere 336 A highpressure chemical reactor contains a gas mixture at 1000T The reactor is made ofAISI 1010 carbon steel and is spherical with an inner diameter of 32 ft and a wallthickness of 075 in The outer wall ofthe reactor is encased in a 25 in thick layer ofinsulation k 003 BtuhftR The convective heat transfer coefficient on the inside wall ofthe reactor is 83 BtuhftzDF and on the outside ofthe insulation the combined convectivdradiative heat transfer coefficient is 117 Btuhft7 T Ifthe ambient is at 80 F find the rate ofheat transfer from the reactor to the surroundings Approach Use the resistance analogy to find the resistances due to conduction and convection All resistances add in series A ssump ons 1 Thermal conductivity is independent of tem era ure 2 The heat transfer coefficient is uniform ov independent of temperature 3 Heat transfer is onedimensional Solution The various radii needed to solve the problem as shown in the figure are 3211 075 7 2 15 721L16 12 166ft r3 7 12 166187 The convective resistance on the inside is R1717 1 7 1 iAAA ADFh 2 M hi4 83 13 416ft2 13 hft F For conduction the resistances are using kfor steel at 1440 R from Table B2 R2 7 r24 7 15571190 JIMM Fh erqrzk 2 Btu Btu 2r16 1 ft 227 lt 6x 6 M F R3 7377 1877166 0356 Fh 2M2ng 2r187166003 Btu The convective resistance on the outside is R4 1 2 1 2 000194 1512 1124 1174r187 As you can see L 39 39 dominates Rm R 122 123 R4 0003740000155 0355 000194 0351 u The heat transferred is T 1000 180 25504 Answer Rm 0351 39 Btu 337 A novelty drink container is made of plastic in the shape of a sphere The container has an outside at the water the outside ofthe container is 94 WmZK Ambient temperature is 18 C The latent heat offusion ofwater is 3337 kJkg and the density ofice is 921 kgmz Neglecting any transient effects estimate the time until all the ice has just melted Approach T3 Use the thermal resistance analogy to determine the heat transfer to the soda an ice mixture Since the surface temperature is unknown and radiation depends on this temperature it will be necessary to T1 iterate As a final step calculate the mass ofice d use the heat offusion ofthe ice to determine T2 an the melting time Radiation A ssump ons 1 Soda has the properties ofwater 2 Heat transfer is onedimensional 3 Thermal conductivity is constant 4 The heat transfer coef cients are uniform over COUVCUUW the sphere and independent of temperature 5 The outer surface is gray and di use 50 Nd ice on Q First find the steadystate rate of heat transfer R R from the ambient to the soda and ice The relevant l 2 radii are o W o M r 6395 0 0325m T T 2 P Convection 2000 as no and 25 Radiation 003m 00 1 The conductive resistance through the plastic is given by R r24 0032570 3 4mer 400300325007 The area of the outside is A 4 00133 m2 The resistance for the convection and radiation on the outside ofthe container is 1 R2 h hmdA A A 291E W Th 39 c v r h mm mm Tgt This cannot be calculated until T2 is knon or assumed Since the heat that is transferred to the outer surface by convection and radiation must conduct through the plastic T T2 T2 Tl R2 R1 The last three equations are solved simultaneously for T2 R2 and hm iteration is necessary All temperatures must be expressed in Kelvin The result is u i m which is hmd 4692l R2 5346E T2 2794K m K W The rate of heat transfer into the soda and ice mixture is thus Q T2 7T1 27947273 R1 2915 The volume of the drink is 218W V gm noo33 ll3gtltlO 4m3 The volume of ice is 30 of total volume so 03V 339gtlt10395m3 me 921k 339x10 5m3 003125kg m The first law for the ice is Q AH mAh QAI mhf Where hfis the heat of fusion of the ice Solving for time 00132kg3337 kg A Q 1k manW 4786s 133h Answer 3 38 A small rod made of pure copper is 05 cm in diameter and 14 cm long The rod is initially at 10 C It is then exposed to a hot air flow at 30 C The heat transfer coefficient between the rod and the air is 25 Wm2 C What Will the rod temperature be after 45 seconds Approach Calculate the Biot number to see if the lumped system approximation is valid Then use the lumped system approximation to determine final temperature Assumptions 1 The lumped system approximation is valid 2 The heat transfer coefficient is uniform over the surface of the rod and independent of temperature 3 The specific heat and thermal conductivity of the copper are constant S olution Check the Biot number to see if the lumpedsystem approximation is valid A representative length is V 7rR2L RL 025cml4cm 106 i i i i cm W A 27rRL 27rR2 2L R 20251 4 cm The thermal conductivity of pure copper is found in Table A2 25 2W 0106cm 1i Biithhm 7 m C 100cm k W 1 66lgtlt10395 401 m K SinceBi ltlt 01 the lumpedsystem approximation is valid The temperature of the rod as a function of time is 7hr T TAT e T Xpch 1 char Using values for the properties of copper from Table A2 25 chmss T10730 Cexp m 39 30 C 893339 385 0106cm m m kgK 100cm T153 C 4 Answer 339 A slab of aluminum 2024T6 which measures 16 cm X 16 cm X 15 cm is initially at 750 K The slab is then annealed by a water spray at 15 C which strikes both sides ofthe slab The convective heat transfer coe 39icient is estimated to be 150 mZK w much time is required to cool the slab to 320 K Neglect convection offthe edges ofthe slab since almost allthe surface area is on the two 16 cmx 16 cm si es Approac Calculate the Biot number to see ifthe lumped system approximation is 4 M valid Then use the lumped s m L I approximation to determine the time required E a Assumptions L 39 a 1 The lumped system approximation 39s valid 2 The heat transfer coefficient is uniform over the surface ofthe slab Ax and independent of temp erature 3 The specific heat and thermal conductivity of the aluminum are constant 4 Heat transfer on the edges is negligible Solution Check the Biot number to see ifthe lumpedsystem approximation is valid A representative length is 15 LhaKSAs i cm075cm A 2A 2 2 where s is the thickness ofthe slab and A is the surface area of one 16 cm X 16 cm side The thermal conductivity of 2024T5 aluminum alloy depends on temperature as shown in Table A2 Assume an average slab temperature of w 750 320K2 535K The Biot number then becomes M 1500 Y Kj075cmj Bi h39 m 39 W cm 505x10 2 k 186 mK SinceBiltlt01the r 39 39 i valid 1 o39 a 1L iL by tympamln NWT 1 rirf Using values for the properties of 2024T5 aluminum alloy from Table A2 at 535 K interpolating for Up and expressing final temperature in Kelvin 7 2770K 1004 075cm 17m 7 m kgK 100cm In 3207288 7507288 1500 Y J m K I 3715 Answer 3 40 Buckshot initially at 450 F is quenched in an oil bath at 85 F The buckshot is spherical with a diameter of 02 in and is made of lead The shot falls through the bath reaching the bottom after 20 s The convective heat transfer coefficient between buckshot and oil is 36 Btuhft2 F Calculate the temperature of the shot just as it reaches the bottom of the bath Approach Calculate the Biot number to see if the lumped system approximation is valid Then use the lumped system approximation to determine final temperature Assumptions 1 The lumped system approximation is valid 2 The temperature of the oil is constant 3 The specific heat and thermal conductivity of the lead is constant 4 The heat transfer coefficient is independent of temperature and uniform over the surface of the shot Solution Check the Biot number to see if the lumpedsystem approximation is valid A representative length is 4 3 lft R 02 in Lchm K 3 2 5A00278 A 47rR 3 2X3 The thermal conductivity of lead is found in Table B2 Btu 36 000278 ft thhar h39ftz39oF Bl T W 204 h ft F Bi 00049 SinceBi ltlt 01 the lumpedsystem approximation is valid The temperature of the buckshot is 7hr pch Using values for the properties of lead from Table B2 36h EEu FJQOSgt 361diJ 39 39 S 85 F 708m T0o3 000278 ft lbm R J 973 F Answer T77Texp Tf T 4507 85 F exp 341 A 39 39 quot quot e ects rm a m in diameter which is suddenly immersed in ice the thermocouple bead should immediately drop to 0 C but in practice there is a 4 two thin wires 01 39 spherical bead Consider a thermocouple 01 water Ideally time delay I WmZDC The density speci c heat and thermal conductivity ofthe bead are 8925 kgm 385 JkgK and 23 WmK respectively Assuming no conduction in the thermocouple wires and an initial 25 C 39 39 39 Approach Calculate the Biot number to see ifthe lumped system approximation is valid Then use the lumped system approximation to determine the unknown time A ssump ons 1 The lumped system approximation is valid 2 There is no conduction in the thermocouple Wires 3 The thermophysical properties of the thermocouple are constant fhcrywcaUJ C Solution Check the Biot number to see if the lumpedsystem approximation is valid The characteristic length is 3 V 3 r Lchav 2 A 4 3 005 1 Lm 7 Hquot i 157x10 5m 3 1000mm The Biot number is given by w 32m2 IK167gtlt10395m 315 W 232x10 5 23 mK Since Bi ltlt 01 the lumped system 39 valid quotquot 39 a H t ipszlnrarnj 89251 385 167x10395m m kg K l 010 250 3 42 A long uninsulatedNichrome Wire of diameter 116 in is cooled convectively by air at 70 F The heat transfer coefficient is 66 Btuhft2 F Current runs through the Wire generating heat at a rate of 19 W per foot a Find the steady state temperature of the Wire b Assume the Wire is initially at 70 F After the current is turned on how long will it take for the Wire temperature to rise to 90 of the difference between its initial temperature and its steadystate temperature Approach For part a set heat generated equal to heat convected and solve for temperature For part b use the lumped system approximation Start from the first law and include both generation and convection in the heat term The first law becomes a differential equation which must be solved to find the required time Assumptions 1 The Wire is very long so that there is no conduction down the Wire 2 The specific heat and thermal conductivity are constant 3 The heat transfer coefficient is uniform over the surface of the andquot1 of 1 1 Solution a 1n steady state the heat generated equals the heat convected from the Wire or Q hATT For a 1ft length of Wire A dL7r im 1ft 00164ft2 16 12 in Solving for surface temperature Btu 341 19E h ft 1w T5in Qg Bt 70 F130 F Answer hA 66 00164ft2 hft F b Check to see if the lumped system approximation is valid by finding the Biot number for the Wire The characteristic length is 1 1ft V 7rr2L r lm 12 in LEM 39 00013ft A 27rrL 2 4 Btu 66 j00013 ft 3 h39 000125 K 69 h ft F Where the thermal conductivity of Nichrome is found in Table B2 The Biot number is less than 0139 therefore the lumped system approximation is reasonable Apply the first law to the Wire dU 7W dl Q By the lumped system approximation the entire Wire is at the same temperature so that internal energy is only a function of time and not a function of space The heat term includes the heat generated and the heat convected from the wire Since we account for the current in the wire as generated heat we set W 0 With these simplifications the first law becomes dT 7 p E 7 Separating variables mc QWihAT7T i QW 7hAT in mcp Integrating both sides t dt Let g QW 7hAT 7Tf d5 ihAdT Using these substitutions on the left hand side and integrating the right hand side produces J T df it T 41145 mop hAI In T 7 IT mcp Evaluating at the limits T MI 111 an hKT T g f ch mQW7hAT7T7 Q39ggphAr7T mop We are asked for the timer at which T 09T5 09130 1 17 F Solving for I ki m ngngTeT hA ng MU Tf lfL is the length of wire L lft If pL rch QggnihAUin hZ rL QW 7 Ma 7 T A 2an 27 iin 1ft 1ft 00164ft2 32 12m using the density and specific heat of Nichrome from Table B2 7 5241b I2n iin 1ft 01 Btu 7 ft 32 12 1n lbm R 66B Eujz hft F 3417Btu Btu h 7 66 2 00164ft2117770 F 1w hft F 19W 1 9341 7 660016470 7 70 Answer 100158 h569 s f 3 43 A copper sphere 3 cm in diameter is painted black so that it has an emissivity very close to l The sphere is heated to 700 C and then placed in a vacuum chamber Whose walls are very cold How long Will it take for the sphere to cool to 300 C Use the lumped system model Approach Calculate the Biot number to see if the lumped system approximation is valid Then use the first law to express the energy balance is terms of temperatures It will be necessary to integrate to find the final temperature Assumptions 1 The lumped system approximation is valid 2 The painted copper is a black body 3 The thermal conductivity and specific heat of the sphere are constant Solution Check the Biot number to see if the lumpedsystem approximation is valid The radiation heat transfer coefficient is with T5 700 273 973K and T 300 273 573K sun h wltTTgt0nxgtlt1gts67xlo m2 IKA 973573K9732 5732K2 W h 112m2 K The representative length for use in the Biot number is 4 3 1m r 30cm K 3 2 17 005m A 47W 3 2X3 The thermal conductivity of copper at the average temperature of 500 C is found in Table A2 With this M 112 YVK j0005m Bi m39 000145 k 3861 m K Since Bi ltlt 01 the lumpedsystem approximation is valid From the first law L char Separating variables J T dT 7 J l O39Adl zT4 0 mcp Integrating both sides T T 3 7 o Al ET 7 mcp I Evaluating at the limits 7 30341 3 3 7 7 T 7 mcp Solving for I 7 4 3 pm LL LL 30A T3 Tf 3047rr2 T3 Tf which simplifies to LL 9039 T 3 73 Using properties of pure copper from Table A2 at the average temperature of 500 K 7 78933Im3s5 1 1 3 3 9567X108 W16 l3oo273 700273 i m2 428s7l3min Answer Comments The radiative heat transfer coefficient changes as the sphere cools since it depends on temperature Evaluating had at the final sphere temperature of 300 C gives a Biot number of 0000553 Thus the Biot number is small enough to justify using the lumped system approximation throughout the entire cooling process 3744 shingles kg 04 Btuh Pj In the attic space the heat transfer coef cient between the air and the plywood is 31 Btuh z The heat transfer coef cient over the snow and the exposed L39 l L 394 L 39 Btuh zDF A 39 r 12 lointi 39 lU in outside ail for the 3011 by 6011 roofpanel shown below Approach parallel and series as necessary Assumptions 1 The Ioofisvery large so edge areisotherrnal V 3 The heat transfer coef cient W is uniform and independent of w temperature 4Thelma1 conductivity is WW constant Auk spine 3 ohm V2quot 3 quot Solution a 39 at 39 39 arrows convcclion snow onsnow VVH N T R0 R1 R2 R3 Rs Tout v NV vVv v R convection plywood shingles 5 inside conviction on shingles The area ofthe roofpanel is 301160 1800112 with properties from Tables 134 and 135 the resistances are 1 1 RDEWwalo h FMBM g 12 4 a 007Btuh R1800 2439xm hi FJBm 5 R7 i04i800djgxm hOFvBm E L R 482 10quot h F Bt 3 kA 00451800064 X u 1 1 4 R 114 10 h F Bt 4 M 761800064 X u R5 1 1 74 o E 761800036 7 23903 h39 FBtu Adding resistances in series and parallel gives R5 R3 R4 R5 R R 3 4 Rm RD 1121 R2 928x10394h FBtu Answer 3 45 A dining area has a glass ceiling built of square unis Each unit consism oftwo glass panes supported by a steel frame as shown below The space between the panes contains a gas The heat transfer coe icienm as shown on the figure are 11 411 WmZDC inside the room 112 363 WmZDC between the panes h 745 WmZDC outside the room The air in the room is at 26 C and the exterior air is at 15 C The glass has a thermal conductivity of14 WmK and the steel has a thermal conductivity of377 WmK Using dimensions on the figure find the total heat loss through one unit Approach 5 9539 I grass analogy quot 39 39 parallel and series as necessary 073 m Assumptions 1 All heat ow is perpendicular to the ceiling 2 39 39 cueurciems Side Vlew of tem erature Top VIeW 3 Thermal conductivity is constant Solution The heat transport may be modeled with the resistance network shown where I T R 1 7 Convection on inside steel surface R R 2 7 Conduction in the steel 4 R3 7 Convection on outside steel surface R R1 7 Conve tion on inside glass surface R gt R5 7 Conduction in the g ass R5 7 Convection in interior gas space R7 7 Convection of outside glass surface 2 The areas needed for the calculation are A 073m2 0533 m2 A2 073 200412 7A1 01254m2 The thermal resistances are R2 1 1 c R 106 3 17312 745O1264 w 1 1 c R 0457 4 11111 4110533 w R5L ooogo4c klAl 14o533 w 1 1 c R 0517 6 17211 3630533 w 12 4 2c thl 7450533 W We now combine the three resistances on the left leg of the circuit into C R8 R1R2 R3 3003W We also combine the resistances on the right leg to get R9 R4 2R5 R6 R7 l76WC The parallel combination of the last two resistances is 3 176 0 RT wm R8 R9 3 176 W Finally the rate of heat loss may be determined as 7 AT 726715 C Q 992wd Answer Rzat w 346 A 39 70 F The are estimated as 08 and 043 Btuh 2 F respectively Model the man s torso as a cylinder ofdiameter 13 ft and height 2 0 Assume the shirt is a layer of cl 39 39 B R J 0th ofthickness 005 in with k 7 012 Btuh R The jacket is 04in 39 rut u 39 are at45 F rut u 39 Approach In Use the thermal resistance 3 mm analogy adding resistances in parallel and series as necessary Assumptions 39 1 The resistance due to trapped gt air layers is small Sh 2 The heat transfer coefficienm are uniform and independent of temperature 3 Thermal conductivity is nstant 4 Thejacket and shirt are gray and diffuse Solution The radiithat will be needed as shown inthe R figure above are r 0 55ft 72 0551 Tl slin R 03925 065416 04 R3 R 4 3 2 3 0396875 R17 conduction in the shirt A mmal resismnce network is Show to me R2 7 convection and radiation from the exposed part right The resistances are39 39 R3 7 conduction in thejacket R1 7 convection and radiation from the jacket In L In 055415 l 065 000424Rh Btu R 2sz 21t2ft012BtuhftR R 4 It w h l T where D is the diameter at radius 72 Substituting values 1 R 0198Rh Bt 2 04308Ir0654162 I H In 39 39 39 halfthebody so the area perpendicular to the direction ofheat ow is only halfthat ofa cylindrical shell The resistance to nduction in thejacket is therefore twice the resistance ofa cylindrical shell less area implies higher resistance This resistance is In 0687 065416 2 7 00842Rh Btu 3 27rLk 27r20094 R7 1 7 1 7 1 4 111 h yrsz 043087r06872 0188RhBtu m 1 hum wad The thermal circuit may now be solved as 122 R3R4 0198008420188 RT R1 1 000404 1 R2R3R4 0198008420188 The rate of heat transfer from the man is Q A T 210Btuhquot Answer R 0119 0119RhBtu 347 The inside wall of a machine is covered with acoustic tile 35 cm thick for noise abatement The tile increases the thermal resistance of the wall and as a result the interior air temperature rises to unacceptable levels An engineer suggests drilling holes in the tile and welding steel rods 35 cm long and 18 cm in diameter to the wall so as to increase its effective thermal conductivity as shown in the figure The rods are in a square array on 10 cm centers The machine dissipates 150 W per square meter of wall area through its outer wall The heat transfer coefficients on the interior and exterior are 46 and 114 Wm2 K respectively If the exterior air temperature is 25 C calculate the interior air temperature with and without the rods Approach 5eel LIqu 39 Use the thermal resistance acoushe 11 la analogy adding resistances in parallel and series as necessary Iquot I CM Assumptions E I 1 The inside wall of the machine is 39OICW very large so that edge effects may be neglected 07 K 09 8w W61 S 2 The steel wall is isothermal 5quot lt 3 Thermal conductivity is constant 3 CM 0 Cm Solution Select a unit cell of wall which contains one rod as shown to the right Defining q as the heat transfer per 3 a i f Ag unit area and A 3 as the surface area of the unit cell the rate of heat transfer through the unit cell is 2 10cm10 cm 150W 10 cm Q qA 2 15w 3 7 1m 1m2 a CM Note that there are 100 of these unit cells in 1 square meter Assuming that the steel wall is isothermal the thermal resistance network is I l I gwa I j l lkg a Ll a uia aliV lulflV T l l39wiv 34 j 39 l D H l M WW R0 inside convective resistance R3 conduction in the steel wall R1 conduction through the tile R4 outside convective resistance R2 conduction through the rod The areas across which heat flows are 1 m2 104 cm 2 A2 7n 2 7r0009m2 254gtlt10394 m2 A A 7A2 000975m2 With these areas the resistances may be calculated as A310cm10cm 001m2 21397 m 3 46m2KO01m2 leg CL1 O35m 610E 11 0058 j000975m2 W mK Rf W0035m 227 2 2 6057j254gtlt10 4m2 mK R kL2 0amp075m 0124 2quot 605 001m2 K h877 cut13 114 2 001m2 W m K Where the thermal conductivity of steel and acoustic tile are found in Tables A2 and A5 respectively For the parallel combination of R1 and R2 619 227 R5 R1R2 lt gtlt gt NE R1R2 619227 This resistance now combines in series with the other three resistances ie Rm RR5 R3 R4 217 21900124877 327 The basic rate equation is 7 7 TiniTlout Rtat Rm Solving for Tm Tm To QR Substituting values 327K W Tm25 C15W 741 c withrods 4 Answer To find the interior temperature Without the steel rods repeat the analysis but set the thermal conductivity of the rods to that of acoustic tile The result is Tm 161 C Without rods Answer Comments The rods make a big difference The engineer might also think about thinner tiles 348 In a certain localized area the earth can be approximately represented by areas of stone soil and iron ore as shown below Using data on the gure and assuming the geometry is twodimensional nd the effective thermal conductivity in the vertical direction This is the conductivity that the earth would have if it were all made of the same materia Approach 4 I analogy adding resistances in T3 parallel and series as 0 5y sums necessaly T K3 l39 Emmi 2 Assumptions 2539 1 All heat ow is in the vertical direction 2 Thermal conductivity is 5 5quot so constant 4 Kl 03 B uhl 5 V TI 15 gt 025 025 Iron ore K2 25 Btuhrl39tR Solution The heat transport may be modeled with the resistance network shown For the soil for a 1 ft depth of earth L 4ft 73 Rl Bt 2421 R 03 quot 152515 1ft B R Slum hftR gt For the iron core R2 i 0 32 25 025025 n 10 MR gt lt gt Forthe stone I R S I run It 2 1 Ul Bto395 0052 u 15 5 ft m MR gt lt gt The total resistance may be calculated as R R R R 0052W0335h39 R 3 Aug 39 42 0 39 Btu The effective thermal conductivity must satisfy L R192 1W1 45 Btu Th 1quot k 224 4 Answer mm 39 RMA 033551 hftR 349 A drinking glass with an oumide diameter of 35 in and a wall thickness of 0125 in is lled to the height of 62 in with a mixture of soda and ice The glass is placed in an insulated soft rubber sleeve 075 inthick cutaway view shown in gure The exposed top surface ofthe drink is at 32 F and gains heat by natural convection and radiation from the surroundings which are at 85 F On the top surface the heat transfer coef cients for convection and radiation are 16 Btuh 2 F and 07 Btuh z F respe 39ve The natural convective heat transfer coef cient between the sodaice mixture and the inside wall ofthe glass is 57 tnh On th ou ide ofthe rubber sleeve the heat transfer coef cients for convection and radiation are 23 BtuhfF F and 085 Btu z F respectively Assume no heat is transferred through the bottom ofthe glass The initial mass ofice in the drink is 009 lbm The latent heat of fusion ofwater is 1435 Btulbm Assuming a steadystate temperature pro le in the glass wall and rubber calculate the time 39 I 39 r L melt 39p fr 1 ss omthega icquu eu JU Approach Use the thermal resistance analogy adding resistances in parallel and series as necessary A ssump ons 1 The heat transfer coef cienm 39 and independent of temp erature 2 Thermal conductivity is cons an 3 The bonom ofthe glass is perfectly insulated 4 Steadystate conditions prevail 5 The glass is a cylin er not slightly conical as shown Solution 39 39 shown above The resistances in this circuit are de ned as R0 7 Convection between the soda and the glass within a I A r R 7 Conduction in the rubber sleeve R3 7 Convection on the outside ofthe rubber sleeve R 7Radiation on the oumide ofthe rubber sleeve R5 7 Convection on the top ofthe soda 5 7 adiation on the top ofthe soda To evaluate these resistances the following radii will be needed see the gure 572125 T 1625 In 35175 in 72 07525 in To calculate RU we need the lateral surface area of the inside ofthe glass A anL z357212552533 in2 2 1144 quot2 71 7 A1 157533 Int 1 big a R i 00285h39 F 2sz 212 08 Btu 12 W E J a R2 7 751455h39 F 2 2z0075 B To nd R3 and R4 we need the outer area ofthe rubber sleeve A2 IrDZL z3507507552 974 in2 1 144 7 1 643 RZ E 2397470 Btu 144 a RA 174 113A 085974 Btu To calculate R5 we need the area ofthe exposed top surface ofthe soda A 7 r15252 830 in R5 i108hF 16830 Btu R6 i 248h39F thz 07830 Btu Combine parallel resistances and rearrange the circuit as shown R7 is the parallel combination ofR and R so I Euda L gt R7 0469 R7 R4 Similarly R3 is the parallel combination ofR5 and R5 so Ra iii gt R8 754 R8 R5 R 9 e L 1 8W 3 1 RMRMR IV R8 Btu 31 AT 857 32 7 F 91 15 h a 335 Rm 15839 h Btu Talr 1435 0091bm lbm Bt 4385 h 231 m 4 Answer 3357quot Therefore 350 A reacting gas is contained in a cubical tank of side length 33 cm The gas is stirred by a pauuiewneei 39 39 a Luiquc o 37 J quotm 39 39 coefficient on the interior wall ofthe tank is 62 WmZK and on the exterior the combined convectiv radiative heat transfer coefficient is 7 WmZK The tank wall is 05 cm thick and is made ofAISI 347 stainless steel The bonom of the tank rests on a highlyinsulating surface Due 39 39 fL Jinthetankf L 39 39 25 C 180 W 0 Find the steadystate temperature of the gases in the tank Approach Apply the first law choosing the gas to be the system under consideration Add the inside ofthe tank conduction through the M O TOR wall of the tank and convection on the oumide ofthe tank Use this total resistance to determine the heat removed 9mm an Assumptions 1 The bottom ofthe tank is perfectly insulated 2 The gas is well stirred and therefore isothermal Solution From the first law Q 7 W t In steady state Taking the system to be the gas in the tank the first law becomes 0 ng Q1952 W W is due to shaft work It is W7 7m wwtso it Wi i min 1 rev 60 s 7233W This is negative because work is being doneg the system ie the gas The heat lost is gm T1 T2 Rm quot 39 T39 39 L 39 andRm is the total thermal resistance Note thatT gtT2 and Q1 mustbe negative sinceQm is leaving the system The convective resistance on the inside wall is 1 R hi The area ofthe five exposed sides is A 5033033m2 0545 m2 4 00296E W W 62 0545 m2 K The conduction resistance is using the thermal conductivity of stainless steel from Table A2 i 39 05m 0000647E w M 142 W j0545m2 K m The exterior convection radiation resistance is R3 L 02625 M 70545m2 W m K Note that the exterior resistance is the important one that is the one that plays the biggest role in determining temperature The same area has been used for both interior and exterior because the wall is thin and these areas and not very different There is no net radiation on the inside walls because all the walls are at the same temperature The total resistance is Rm R1R2 R3 00296 0000647 0262 0293 The heat lost is T 7 25 C Q1 7 K 0293 W Substituting this in the first law 0 Qgen Qlast W T 7 25 C 7 7233 W 0180W7 Solving forT1 results in T1 146 C Answer 3 51 A pistoncylinder assembly is filled with carbon dioxide gas at 250 C 390 kPa The piston and the curved walls of the cylinder are perfectly insulated The bottom wall is maintained at 325 C by an external heater Initially the piston is ll cm above the bottom of the cylinder which has an inside diameter of 6 cm As heat is transferred by convection to the C02 from the cylinder base a control system lifts the piston so as to keep the average COZ temperature constant If the piston rises 3 cm in ll seconds determine the convective heat transfer coefficient Approach Use the first law to calculate the heat transferred during this process Then use the convective rate equation to find the heat transfer coefficient Assumptions 1 The C02 behaves like an ideal gas under these conditions 2 The piston and side walls of the cylinder are perfectly insulated Solution Let the system be the C02 in the cylinder Find the mass of C02 from the ideal gas law 3 390kPa 10 Pa 7r003m20llm 4401 kg PVM l kPa kmol mW kl 000123kg 8314 250273K kmolK The work done in an isothermal expansion of an ideal gas is kJ 000123k 8314 250 273 K imRT V2 f g kmolK 1 7rr2011cm7293J M V1 4401 kg lkJ Lnr2o14cm kmol 1000 I From the first law AU Q 7 W cVAT Q7 W Because there is no temperature change AT 0 and Q W 7293 I Heat is negative because it is being added to the system The convective heat transfer from the base of the cylinder is Q hATW Tg where TW is the base wall temperature andTg is the C02 gas temperature 2 7293J7266W AI 11 s By common practice Q the convective heat transfer is positive Therefore Q 7Q h Q 266 W ATW 7 TE 7r03 m23257250 C h1252l 4 Answer m K 4 1 If atmospheric pressure is 147 lbfin2 What is the pressure at a depth of 10 ft of water Approach Use the equation for pressure as a function of depth in an incompressible uid P Pm pgh Assumptions 1 The density of the water is constant 2 The water is at room temperature 70 F Solution At an assumed temperature of 70 F the density of water from Table B6 is szO 622 lbmft3 The pressure in an incompressible uid as a function of depth is P Pm pgh llbf 1amp2 1471bf 2 6221b ft3 322ft2 10ft m m S 3221bmfts2l44in2 Answer P 190 lbfin2 Hoover dam stands at a height of 725 ft above the Colorado river Assuming atmospheric pressure is 147 lbfin2 calculate the pressure in the reservoir at the base of the dam 4 2 Approach Use the equation for pressure as a function of depth in an incompressible uid P Pm pgh Assumptions 1 The density of the water is constant 2 The water is at room temperature 70 F Solution At an assumed temperature of 70 F the density of water from Table B6 is szO 622 lbmft3 The pressure in an incompressible uid as a function of depth is P Pm pgh 2 147622g1322gk725ft i in ft s 3221bm39ft l441n 752 ii A Answer 4 3 In a manometer containing liquid mercury the height is read as 6 in If atmospheric pressure is 100 kPa What pressure is the manometer reading in kPa 7 Approach Use the equation for pressure as a function of depth in an incompressible fluid P PM pgh Assumptions 1 The density of the mercury is constant 2 The mercury is at room temperature 20 C Solution At an assumed temperature of 20 C the density of mercury from Table B6 is pHg 13579 kgm3 The pressure in an incompressible uid as a function of depth is P Pm pyggh P lOOkPa 13579kgm398ms26inw11 In a P 120 kPa 4 Answer 4 4 A vat in a chemical processing plant contains liquid ethylene glycol at 20 C The air space at the top of the closed vat is maintained at 110 kPa If the depth of the liquid is 08 m What is the pressure at the bottom of the tank Approach Use the equation for pressure as a function of depth in an incompressible fluid P Pgt0 pgh Assumptions 1 The density of the ethylene glycol is constant Solution At a temperature of 20 C the density of water from Table B6 is pl 1 17 kgm3 The pressure in an incompressible uid as a function of depth is P Pu pgh P110 kPaw1117k g39813208 m lkPa m s P118766 Pa119kPa 4 Answer 45 At great ocean depths pressure according to PCln i C2 pa uen it varie wiui 39 Very high where Cl 224 x 109 Pa C 1x105 Pa and p0 1024 kgm3 Assume this relation holds at any depth 2 39 r of3000 m 39 seawater density were A L 1024kgm3 r39 39 pl ulci 1x1 Pa Approach Solve the given equation for density as a function of pres sure Substitute into Eq 47 separate Variables and integra e Assumptions 1 Water density is only a function ofpressure Solution In a compressible uid static pressure is related to depth by dP EPg iven in P Cz P m P exp q Substituting yields d Pic pgexp P 2 dz C Integrate between the two depths shown in the gure a 2 IR TQ L Pgdz exp T Integrating the right hand side and rearranging the left produces I exp QC P wage 2 To make further progress de ne u C2 7 Psothat du7 and 1 Cl u if c expltugtdu go 72 Integrating the lefthand side C 6 7 e p542 7 2 Use the knon boundary condition to evaluate u at 222 1 1X1 Pa 1x105Pa71x105Pa 7 uz 0 Thus C1e pg22 72 Solving for ul produces 1024 kg Igsgjeooom 7 73 u1ln 1 p g22 21 ln 1 m 9 700136 C1 224gtltlO Pa We may now calculate the unknown pressure at station 1 through C2 7P1 ul Cl Solving for P1 P1 7Cu1C2 7224gtlt10970Ol35lgtlt105 P1 3044x107 Pa 4 Answer kg If dens1ty is constant at 1024 2 m P pghP2 10249813000m1x105Pa m s 3024gtlt 107 Pa 4 Answer Comments There is very little difference between the two calculations The compressibility of water due to pressure change is not an important factor in determining hydrostatic pressure in the ocean Note that in reality density varies with depth due to changes in temperature and salinity These effects are generally more significant than the change of density with pressure 4 6 A large tank contains a liquid solution whose density varies with depth as shown in the table below A gas space at the top of the tank contains air at 60 psia Find the pressure at a depth of 30 ft using numerical integration APPT mh depth ft density lbmft3 Integrate Eq 47 from the surface of the l1qu1d to the desired depth of 30 ft For simplicity use the 0 402 trapezoidal rule 5 41 0 Assumptions 10 42397 1 Density varies linearly with depth between given 15 449 data points 20 47 7 25 509 30 546 Solution For a variable density fluid dP 7 2 dz p g First find the pressure at 5 ft by numerical integration 3 Z JR dP 7 pzgdz Use po60g 290 2175ft 1n Approximating the integral with the trapezoid rule AAj 2 Solving for P1 and substituting values 2 Pl pop1 gcliz ai 4024410 111 1311 3223 757M llbf 1ft 2 60 2 2 ft s 322m l441n 2 P1 Po 8Zrzo 1n Now find the pressure at depth 2 p1p2 410427 ft 1 1 j P 7 z 72 P 7 322 7107 75 61416286 2 E 2 jg 1 f 2 s2 l l 322 144 Continue this process using PM p 291144 2 R until you find P5 Intermediate values of pressure are given in the table below A wide variety of software tools can be used to automate this calculation The final result isP6 695 lt 1n Answer 47 A manometer is anached to a rigid tank containing gas at pressure P The manometer uid is mercury at 20 C Using data on the gure below nd the pressure in the t Approach Use the equation for pressure as a function of depth in an incompressible uid P 12m gh Assumptions 1 The density ofthe mercury is constant Solution 16 cm w 3quot PM 101 kPa At an assumed temperature of 20 C the density ofmercury from Table B6 is pHg 13579 kgm The pressure in an incompressible uid as a function of depth is 101kPa 135793 9832 1673cm 1 m m s 100cm 1000Pa P118kPa Answer 48 Write an equation for the mass ofthe piston mp in terms oprzg l 6 andAT See the gure below Approach Perform a force balance on the piston Use the equation for pressure as a function of depth in an incompressible uid to nd the gas pressure A p Assumptions 1 The density ofthe water is constant Solution A force balance on the piston produces Pam 1 quotpg PgasAy The gas pressure is given by g magi sin 6 Pm Substituting the second equation into the first I m1 mpg pyzgglsin EMA Solving for piston mass PnlmAp mpg PguAp mp szglsin 6A7 Answer 49 Liquid water is contained in a pistoncylinder assembly as shown in the gure below An inclined manometer lled with water is attached to the bottom of the cylinder Using data given on the gure calculate the force exerted by the spring on the piston Approach sprng Calculate the pressure at the bottom of the tank A Nquot 30 Iquot 2 starting at the spring and working your way mphm 05 1b downward Derive a second equation for pressure at the bottom of the tank b considering the pressure in the inclined section Equate the two expressions for pressure and h 2 in solve for spring force s 55 in Assumptions 1 The density ofthe water is constant 2 The water is at 70 F Solution The pressure at the bottom ofthe tank is F mum P am g pgh also minquot L r mow Aputon where F is the force exerted by the spring be written Pb pgx sin 6 Pm Equating these two expressions F pgss1n 6 AW Arm Solving for spring force mpmong A put7 F pgssin 67 7 13th Am Using the density ofwater at 70 F from Table B 6 F6221bm 33217 s255insin45 30in2E 2 144m2 n 2 m2 in30m39 144in2 1 12 a 4951bmtts2 1541bflt Answer 705 lbm3217fts276221bmft33217 s22in 4 10 A A umuuulctu iquid mercury also at 80 F low calculate the density of the 39 39 at v F 1 Assuming atmospheric pressure is 142 psia and using data on the gure be air in the tan pproach Calculate the pressure in the tank using the manometer relations Then use the ideal gas law to nd the air density 54 in Air Assumptions 1 The density ofthe mercury is constant 2 Air may be considered as an ideal gas Solut on 2 in First nd the pressure of the air using m ngg Using data for the density ofmercury in Table B6 lbf 8453217154721in P142 s 3 207psia m lbm 12 in 3217 2 3 lbf s 1 it From the ideal gas law PM 207psia2897j p mo 0104 lbm 3 Answer 1073amp 80460R lbmol R 4 11 Two pistoncylinder assemblies are connected by a tube as shown below The diameter of each cylinder is c e m ss of each piston is 04 kg A mass rests on top of each piston The uid in the tube is mercu yyatzooCIY39 A t t 7 Approach Calculate the pressure at the lowest point in the connecting tube by two paths one through the right cylinder and the other through the left cvlinder 39 quot392 solve for the unknown mass Gas Gas Assumptions 1 The density ofthe mercury is constant Solution In The pressure at the lowest point in the tube may be Iquot I calculated using either the right side or the left side This produces quot3918 quot3928 quot3928 quot3918 P P m Rz Rz ght am Rz Rz 13ng where m piston mass andR cylinder radius Simplifying quot391 quot392 2 R294 R2th a quotemtphrh2 R Using the density ofmercury from Table A6 at 20 C kg 1 m 2 1m2 Sk 13579 15745 cm Ir 4cm quot397 g m3 100 cm 10000 cm m1 295 kg 4 Answer 48 412 In the device shown below calculate the gage pressure ofthe gas in the tank Approach Use the expression for Variation of pressure with depth in an incompressible uid 0 Maw Ibmfl3 3 in Water Assumptions Tank 1 The densities ofthe oil and water are constant 86 in Solution a static uid is only a function of horizontal location therefore P 7 h 3 in gasm P quot22 Ig zmwlp ngng h 86m IH 04m The gage pressure is given by Pg E n11 hipigwng 32211152 3221bm s 1n 3 322 1 3 p 3 49lbm 3 8542 522 5 m llbf 12m 32212 Pg 0316psiglt Answer kyrr mm the lengvhl m whmh the Water uses m the rhehheu seeuuh mach Use39he gvenmass and uehsny unhe ml ande mame39er unhe tube m delenmne the length2 Srrhuarly rurme Water delEm39nneLhe surh an and Calculate the pressure anhe luwest pmnl m the tube bytwu paths are ugh the rhehheu seeuuh and the utha39 mugh the mght seeuuh Equals these pressures and salve fur the uhkhuwrr lenng Assum u39 p ans 1 The dmsmes unhe Water and ml are earth 2 The systErms almum tempemlure Sulutiun The mass unhe ml rs grvehhy m1Ju 2J1 11 2 Reamngng 1amp lEIEIEIg 21 kg 1m wk lm 7 USMTUM Thevulumeufwatens gveh hy lEMlm Vwn 7 an Usmgthe de muun uruehsny m mquot V 7131 n 7 11 Sulvmg fur the sum unhe langth and usmg the uehsny ufwala39 at 2w c um Table A 1kg llgm7 UN me1 lEIEIEIg kg 1m 717 99827 ns AL m n Emmnem De heahewmahle 1 1n141rh I141m Thererure Pm gg11pwgk Pm pwg1srh5 M Wquot sure kg 7 0141 m3 In 9982 kg jsin40 9982 kg m3 m3 kg 620 j0041m99827g m l l0101m 10lcm f Answer 414 A manometer connects a large waterth open to the atmosphere to a closed spherical tank of air The ometer contains both oil and water Using data on the gure nd the gage pressure of the air in tank Approach Use the equation for pressure as a function of depth in an incompressible uid to trace pressure through the system Assumptions 1 The densities ofthe water and oil are constant 2 The system is at room temperature Solution The gage pressure at point 1 is Pl Iim Wight 12m 99955 9832 28711cm 1 m m s 100 cm kg m 1 m 7 9995 07 98 1478 cm m7 s2 100cm 1254 Pa Finally relate P2 to 1 using 1g1254 Pa99959824r9 j 12272 kpad Answer um 35 415 The manometer in the gure is designed to measure small changes in pressure Using the data in the table a e rm39 e the initial gage pressure ofthe gas in the sphere b The pressure is increase so at 11 becomes 20 cm During this process none ofthe liquid interfaces change in diameter Find the final gage pressure of the gas Approach Use the equation for pressure as a function of depth in an incompressible uid to find the initial gage pressure For the second part note that the Volume of each liquid must remain constant Assumptions 1 The densities ofthe two liquids are c onstant Solution a From the left side of the manometer the pressure at point 2 is 1 apighrhz quot quot fthem meter amppighrng Equating the last two equations PgaWWIampPighremgr PgarPApighrp2gimim hihz P 086 10001137 98122107746cm 1m 113 100011 9812246722cm 1m m s 100cm m s 100cm 7086 Nook l 98132 91722cm 1 m m s 100cm PM 2590Pa 259kPa b The change in height 112 is Ah20722702cm The Volume of each uid must remain constant Therefore 7rd22A39qrD22Ahl 2 2 7 Ah dDAZh lcm 03922quot 7000247cm 9 cm The height 113 must increase by the same amount as 112 decreases so A113 02cm Similarly AhA 000247cm With these considerations 1 1l 70002479098 11220 l q460248 h4107000247107025 Recomputing the gage pressure with the above Values gives 3003 Pa 300kPa PM Answer 416 A glass tube containing oil is inserted into a tank ofwater as shown in the gure below Using data on the igure calculate the oil density Assume the temperature is 2 Approach Use the equation for pressure as a function of depth in an incompressible uid to trace pressure through the system Assumptions 1 The densities ofthe water and oil are constant 2 The system is at room temperature Solution The pressure at the bottom ofthe oil must equal the pressure at the same depth ofwater that is at L4 m poi qu Pm pyzgglt 4AAA L4 1 171 Substituting L4 and rearranging ipg awr q o Using the density ofwater from Table A6 at 20 C 99821964457115cm m p 445 cm 01 L3 39 7x Ll Mi 1 4o444414s41 L1 115 cm L 96 cm L 445 cm 572k lt Answer m a q mm mass fth an mlbm b Oxhspwuredmla39hemmtu euntdtheantensesl m C cmauthzvalumeufadadded nth m mummpmm 5 mm amepum m m mm m m m m 2 meme mm shave m am pm m swam mm M H mm mm uf m an mm mm E Assumptinms 1 Th demtyaf39he mhs cmmnt 2 Th system is atmam temperature Snllltinn a cmmmgwsmwmmmm Tmmmwemmhm gammy 39quot M m H M samung Mums um cm W mpAhrt1ElE p lkoraD86624Jn15 1047U9 3411bm b E pgwm smmgmw 341 mm p5 7m mm 472m A mw W wlume af ml Addad must be n5xn mn08m lml l xnquot Answer 4713 l M Approach Flnd the resultanthydrostatlc force due to tlne water and lts polnt of apphcatlon Take moments about tlne hmge to determlne tlne unknown force Atmosphere Atmosphere Assump ans 1 The water denslty IS consan Solu on The resultant force FK IS glvenby FK gslne A The locatlon othe centrord othe gate 15 y slnoo 2 Uslng the denslty ofwatar at room temp erature see Table A6 FK 9965319 sl zjsmasoxzt 04m3 46 m8 m m s The polnt of applicationyp ls Front vlew of gate Ixx ya y 4 L 9 LA The moment of lnertla for a rectangular plate 6 L a W 12 a 3 o 2 a ya y 01y 4 7 lZy z 12y 2 amerm 4 29m 124 04m To flnd F take moments about the hlnge an an Ma 7 ya Moment dlagmm 2 Sm so 4 ea F 7 2F 1 7 29 48x105Nl48m 7 7 46m 51 3464Z9148m F812x1 N Answer 4 19 A horizontal pipe 14 m in diameter is half filled with liquid oxygen SGl18 The gas above the liquid is at a pressure of 250 kPa The pipe is closed on both ends by vertical at surfaces Find the magnitude of the resultant force of the fluid and gas acting on one of the end surfaces Approach The total force on the end plate is the sum of the force due to the gas and the liquid The gas force is just pressure times area While the liquid force is 773k H R found us1ng hydrostatic relations 7 39 3 lT Assumptions 4 l The oxygen density is constant lt 7l I M Solution The areaA of half of the end surface is 7 IrR2 7rl4m2 2 A 0770m2 2 The force exerted by the gas is E PA 250kPa077m2 l kPa 192 X 105 N The force exerted by the liquid is F2 PA pg sin chA Where 7 4R 7 407m y 7 g T 37r Substituting F2 250kPa 0297m w077m2 1oook g3118981sin90 0297m077m2 l kPa m s F2 195gtlt105N The total force is FEf72387x105N Answer 4 20 A gravity dam made of concrete p 2200 kgm3 holds back water which is 55 m deep as shown The bottom of the dam rests on the soil and is held in place by friction Calculate the minimum coefficient of friction between the dam and the soil so that the dam does not slide Approach 3 1 In Find the resultant hydrostatic force on the dam Use a H freebody diagram of the dam to do a force balance 143 including the dam s weight the frictional force and I the hydrostatic force The hydrostatic force in the horizontal direction equals the frictional force at the 5 5 m p T I 111 minimum value of friction coefficient Eff Assumptions 1 The water density is constant Sail 9 1 11 Solution The resultant force FR is given by FR pgsinQyEA a2 7 77 ale1 7 91731 6 513 tan 9 b 697 m s1n 9 s1n513 Using the density of water at an assumed temperature of 25 C FR 997g1981gjsin613697mlm m s where the force on a lm deep section of dam has been considered 43 Evaluating FR 188x105N188 kN The resultant force may be resolved into horizontal and vertical components FRJ FR sin 9 188kNsin513 l48kN FRYZ FR cos6 lSSkNcos513 llSkN The volume of the dam is V 11 a2a3 111m 3 l77 05779 l 7 31 470m3 A free body diagram of the dam shows F FuFRzmg L The dam will not slide if the frictional force equals the horizontal force LMI F Fm d F F R 12x Free body diagram 4 FR pVg NOON F u 148 E N 0131 A 4 nswel 1151ltNM2200kg47m3981 S lkN F 4721 shown Caleulate the foree ofthe gate on the stop that holds 1t elosed Appr ach Frndthe resu1tanthyn1rostatre force due to the water andrts pomtof appheauon Take rnornents on the gate about the stop to deterrnrne the unknown force 0 Assumptions 1 The water dens1ty1s eonstant 2 the system 15 at room ternperature Square gate Sxop Soluu39on For a reetangu1ar p1ate the centroxdxs at the eenter so 18 The resu1tant force 15 Fl pgsm nay1 Fl s2 41b m 32 2f t L srn9039181t33tt2 1 01x11 lbf ft s2 lbm 1t 2 272 s Th1s force 15 apphed at L J I 7 y y 1 LA 391 2 13H 12m2 L 3 18 1218 MtJlMi Fwy We have used a large number ofs1gmf1cantdAg1ts because we 111 needthern m the next step Takmg rnornents aboutthe F15Fxyrn Solvmg for F F Fro n 1 5 101x10 1bm18 041mm 15a F 2811bf 4 Answer 4 22 A square plate called apaddle covers apassage in a canal lock as shown The angle o is 15 Find the vertical force F neededto open the p e Approach Winch Find the resultant hydrostatic force due to the vlmter Cable and its point ofapplication on both si es ofthe paddle ake moments about the bottom of the paddle Upn ver D039nrivct to determine the unknown force F 7 Assumptions 40 fl paddle I a l The ther density is constant 2 The system is at room temperature 39 Solution We need the resultant force on the paddle Referring CW8 to the gure below 5 5 40 R 21 Puddle COSZZ 2 s Lib 4 72394 coszz cos15 Atmospheric pressure acts on both side ofthe lock and its effects on the paddle will cancel For simplicity we set PM 0 The centroid is at the middle ofthe paddle therefore the resultant force on the upstream side is Fm Empg wash qjah where 90 d75 Fm 06241T 3223944ra sin 75 2 2ft2 97401bf S tr 394 cu in MM gure to the right 62 180 26g 180 275 115 40734 ft sfgez 1142111 cos15 gt31 2 Em 133 smash 621m x2624322421sin115 22 FM 11791bf note sinolsina2 ctc 11 r39 idFRi y is b b2 pl l 2 12 s 7 r 2 133st 2 2 2 ft yp394 2 4042 12394E0ft Next nd the point of application of the force on the downriver side FRI ypz 52 2 12 b2 b 2 2 pgsln 62 22 2 2 528ft 1242130 421ftft De ne the distance 0 as x yp 7 40427394 10040 Similarly de ne x as x y 7x2 5287421 1054 it Take moments about point A Fnb Fm 1H2 Fm bnxx FR bag 7 F22 bag 9740271004 417924054 b 2 F 4278 lbf To get the force in the Vertical direction cos 6 i F F F7 7 4 7 J 5291bflt Answer cos l cos75 4 23 A cylinder 5 m lon m in diameter is wedged 39nto arectangular opening in the bottom ofatank of water The cylinder seals the opening which is also 5 m long The center ofthe cylinder is lm above the flour m i depth is rn 39 C L m i n the cylinder Approach ecaiise of sym etzry the only net force on the cylinder is in the venical direction The weight of the water above the cylinderprovide downvlmrd force while the water under the cylinder provides an upmrd force Atmospheric pressure cancels out and need not be considered Assumptions water density is constant will cancel Solution B mmehv The force Fv is FH pghA kg rn 2 5 pgh1 7R2RL 997 3 9 8 2 72m 225m 978x10 m S whereL 5m the length ofthe cylinder The mass of water m1 in the region above one halfofthe cylinder is 2 7 1 2 m 7 p R 171R L Jzzi zf 4 m997k m25m2 4279kg m The total downward force is m Fm 2mg 978x105 24279997 106X106 N The force FVZ due to the water pressure at the bottom of the tank is pghA pgwi would FVZ To nd hA note 6sinquot E sinquot 1 30 R 2 hi Rcosa 2cos30 173m tn R411 27173 0268m Fm P h1 52h4 9979817120268 210x105 The mass W5 ofthe vlmter in the region below one half ofthe cylinder is 1 30 997127E1173272 J5 4326kg 1 a 2 th 7 7 L We P2 zha h 360 The net upward force i Fm Fm 72g 21x1057 2433997 201x105 F Fmr Fm 106x106 7 201x105 solxlo5 N 861 kN I Answer 4 22 4724 A semicircular gate hinged at the bottom holds back atank ofwater4 it deep Ifthe gate is 15 it wide what force F is required to keep the gate closed Approach F Find the resultant hydrostatic force due to the A water and its point of application Take moments about the hinge to determine the unknown force 4 4 Assumptions 1 The water densityis constant Hinge Solution Atmosphere forces cancel and will not be considered 16 ST The nonzontal force is n pg sin WA R 7 R2 7 lbm it llbf 4 3 Fyrpgsm90 ZJRL pgTL 5243225 2 W Tjasm 72 s 4881 The point of application ofthe force is I y ycL39 5753 2667 ycA 2 123m 2 o 2 o 2 The mass ofwater in the quarter circle is 2 2 m p R L 524113quot 4 15 117621bm 4 it 4 Taking moments about the hinge 4R FRmgV FAIL R 7 yo R 4 smgr 3n llbf 4 322 3 s 74881bf Answer F 117621bm322 2 s ace A narrow tube partially filled with oil 4 25 Ahemisphere filled with oil so 072 is inverted on a at surf 39 39 39 39 39 39 39 C kg At what height h will the hemisphere L I39Q X L J n Uh 39I I I i i Approach Balance the upward hydmstaticforce against the weight ofthe o and the container When the ydrostatic force is greater the hemisphere will lift h Assumptions k l The oil density is constant 2 The weight ofthe oil in the tube is small Oil 3 the weight ofthe tube is small Solution Atmospheric pressure forces cancel out and will not be considered There is an upmrd force due to static pressure given y 7 pghRR2 This force is balanced by the weight ofthe oil inthe container and the weight ofcontainer itself The mass of oil is 1 4 3 ma 7 p 2 3 71R J M So the force balance is F FFWItg mg quotJV A where m is the mass ofthe container Substituting I Me 3 pgh R2 gp R3mg i R 3 I 1 M03 Solving forh K R 2 3 m 771R h 3 23 R T 71R F 2 3 5kg 701m 3 lOOOlkgyOn m 01m h 2 Il01m 0188 m 4 Answer 4 26 1f the tip of the iceberg that is the volume of the iceberg above the water surface is 79 m3 what is the volume of the submerged iceberg For seawater density use p mm 1027 gcm3 Approach Use the formula for submerged volume of a oating object Assumptions none Solution For a oating body the submerged volume is Vm 9 VW 9 VM WW Pm Pm Gathering terms VS 17 pug Vxp Pm Pm Solving for submerged volume the density of ice is in Table A3 MSW 794 323 quot 3 679m3 Answer sub pm 1 920k g3 1m Pm 17 m 100cm 1027 g amp cm 1000g 4 27 A small boat has a mass of 650 lbm when empty If the volume of the hull is 166 ft3 determine the maximum load the boat can carry in fresh water Approach Draw a free body diagram of the boat Include the buoyant forces and the weight of the boat and the load Assumptions 1 The center of mass of the load is on the same vertical line as the F center of gravity of the hull g 2 Water temperature is near room temperature Solution At the maximum load the weight of the boat plus the weight of the load equals the buoyant force when the hull is submerged as much as possible without flooding The buoyant force is directed along the center of mass of the boat but is offset in the figure for clarity Designating mb as the mass of the boat and m as the mass of the load mtg Wag Fg ng m5 ml pV m4 pV mb 6231jas f yssoibm 96921bm 4 Answer Comments In reality one would need a considerable margin of safety and the boat should carry a much lower load than the maximum A boat submerged to the gunwales is in danger of flooding with any passing wave 4 25 428 A layer of oil 6 cm thick covers a layer ofwater A cylinder made of soft pine oats in this twolayer uid as shown Using data on the gure nd the height x by which the cylinder protrudes from the uid Approach 20 cm Perform a force balance on the cylinder matching weight force with buoyancy lt gt Assumptions air l 14 cm x 1 The density ofthe oil is constant 2 The density ofthe water is constant Solution 5 cm t L 4 mpg must be balanced by the buoyancy forces V due to the oil and water The displaced masses ofthe oil mo and water W can e used to determine their contributions to buoyancy mpg mag quotW mp Irrzhpp m mop mW 72017 67013 where h is the height ofthe cylinder Substituting masses into the force balance Irrzhppg 7 2 61395 72017 6pr 113 63 1176 01 11056pW 6075pW 117 67 xpw h056 5075 117 Six Solving forx x 50757 5 11170 55 5075751417055 x 466 cm Answer 4 29 A hot air balloon has a mass of 250 kg and carries 2 passengers whose average weight is 185 lbf The balloon which has a diameter of 12 m rises through atmospheric air which is at 20 C Find the minimum possible average temperature of the air inside the balloon Atmospheric pressure is 100 kPa Approach Perform a force balance on the balloon using the ideal gas law to find air masses Assumptions 1 The balloon is spherical 2 Air behaves like an ideal gas under these conditions 3 Pressure is the same inside and outside of the balloon Solution The volume of the balloon is V gar 746 905m3 The mass of atmospheric air at 20 C displaced by this volume is 105Pa905m3 2897 kg PVM kmol quot117W mkg 1 8314 20273K kmolK 1kJ The passengers have an average weight of 185 lbf To find their average mass use Fm2g1851bf where m is the average mass of a passenger Solving for m gives lbmft m 7185M 185M 71851bm 2 g ft llbf 7 3227 S2 The buoyancy force is equal to the weight of the air displaced therefore F8 mg The buoyancy force on the balloon must be balanced by the total weight of the balloon hot air and passengers mlg 2m2gm3g m4g where m2 is the average mass of a passenger m3 is the mass of the empty balloon m4 is the mass of hot air in the balloon and m1 is the mass of the 20 C air displaced by the balloon Solving for m m4 m1 72m2 7m 1076kg72185 lbm212k1 m7250kg658kg To find the temperature use the ideal gas law For the displaced air m1 Rn For the hot air in the balloon m4 7 ETA The air inside the balloon is at atmospheric pressure as is the air outside the balloon so we may divide the last two equations to get m4 T Solving for the temperature of the hot air gives T4 202731 m4 658kg T4 479K 206 c Answer 430 A rectangular gate 12 ft high and 3 ft wide is held closed by water pressure as shown in the gure A quot massmis L a quot a top ofthe gate The counterweight which is partially immersed in the water is of 15 ft and a mass of 800 lbm Air 39 39 an gate Calculate the minimum water depth 11 for which the gate will stay closed L pu cylindrical with a diameter Au on L 394 the Approach Draw free body diagrams of the gate the gate counterweight and the cable and do a force balance on the counterweight and cable Perform a moment balance on the gate Express forces and moments in terms of the unknown height h Combine equations and solve for h coumevweigh x Assumptions The water is incompressible 2 The system is at room temperature Pmm Solution rung Begin by considering free body diagrams ofthe gate the cable and the counterweight Atmospheric pressure acts on both sides of the gate so its contribution will cancel and is not shown F ihydrostatic force Gale F2 7 tension in thecable F3 itension in the cable F4 igravity F5 ibouyancy force From a force balance on the cable J 2 7 3 ll L F From a force balance on the counterweight Q 9 I F n F 3 5 Taking momenm about the gate hinge Counterweight Cable x1 i xz 2 To nd the hydrostatic force on the gate F1 use F2 I F Run1 Pgsin9ycA F2 Sincer0 690 and ych2 2 m 2 A where a is the depth ofthe gate into thepage The point of application ofthe hydrostatic force is m Ixx h Substituting the moment of inertia of a rectangular plate from Table 41 y 7 h ah3 P 7 2 12 5 ha 2 By de nition see gure on the previous page Since F3 F2 the free body diagram is 1h 3 XEhiyp Substituting the expressions for xand F into the moment balance gives a XZFZ 3 2 The buoyancy force on the counterweight is F ngn392 h 7 2 The force balance on the counterweight becomes F paw11 2 mg Combine this with the moment balance to obtain 2 E pg a X2 Inge lagr72 11 7 2 Canceling g from each term and substituting values 2 h 524112 3 f0 12ft800 lbm 524112 112017 2m it it Using equation solving software h 545 Answer Approach Express the Velocity ofrainfall in terms of the mass ow rate ofrain falling on the roof Set this mass ow rate equal to the mass ow exiting the gutter and solve for rainfall Velocity Assumptions 1 All rain striking the roofis collected in the gutter 2 The system 3 The rain fal H Rain falling on a roof ows downward over the shingles and is collected in a gutter The gutter which is closed at one end and open at the other is slightly inclined so that water runs out the 0 en end In a h downpour rain falls steadily for several hours and the ow in the gutter reaches steady state Due to the addition of runo from the roof the depth of water in the gutter increases gradually along the length of the gutter in the direction of ow Using data in the gure below calculate the inches per hour ofrainfall that will cause the guner to be completely lled with water at the open end At rainfall rates higher than this the nut 39 39 1 watcl u au w watcl pi over the sides ofthe guner before reaching the end Assume that the exit Velocity of the water from 39n the gutter is 3 M and that all the rain striking the roof is collected in the gutter Note that the home owner has been too cheap to install downspouts 42 ft 85 it a 4 in I rate of rain operates in steady state 4 in ls Vertically Solution b t quot am The mass ow rate of the rain is To find the mass ow rate ofrain on the slanted roof use whereA is the roof surface area projected on the bottom ofthe figure to the right By conservation ofmass Imagine sening a straightsided container of on m surface areaA 39 39 the container would rain is assumed to fall normal to the bottom surface The amount of mass that accumulates in this time I is out in the rain A er time fill up to a height ofL The PLAP A p quot1mm t where ll of rain I h MTM IL A 42 sin 50 85 3092 It quotlvam mglm LAP T A v p wheremwm is the mass ow rate exiting the gutter and AW is the crosssectional area of the gutter Si L 4in4in393 somzll Answer t 1h 3600s 4 32 In a home air infiltrates from the outside through cracks around doors and windows Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 Due to the wind 94 x 1039 kgs of air enters per meter of crack and exits up a chimney Assume air temperature is the same inside and out and air density is constant at 1186 kgm3 If windows and doors are not opened or closed estimate the time required for one complete air change in the building Approach Use the definition of volumetric flow rate and the relation betweenvolumetric and mass flow rates Assumptions 1 Air enters only through the cracks 2 Air temperature is the same inside and outside Solution The volumetric flow rate is related to the volume of the building by KAIV or AIK V Volumetric flow rate may be written in terms of mass flow rate using rh V x p Eliminating volumetric flow rate between the last two equations 210m31186k g3j Vp m AI k 42735 s119 hquot Answer 5 quot 1 94x10395 gj62 m m 4 33 The hull of a vessel develops a leak and takes on water at a rate of 575 galmin When the leak is discovered the lower deck is already submerged to a level of 75 inches At this time a sailor turns on the bilge pump which begins to remove water at a rate of 738 galmin As an approximation the lower deck can be modeled as a atbottomed container with a bottom surface area of 510 ft2 and straight vertical sides How long will it be after the pump is turned on until the deck is clear of water Approach Apply conservation of mass Assumptions 1 The hull is boxshaped Solution From conservation ofmass dm rhx 7m d1 which integrates to 7 m dt J dmcv Since both rates are constant m 7m gtr m where moV is the mass that accumulates in the control volume in time I In our case mass leaves so moV is negative 7510ft2 2 ft 17 mV 7 ipV 7 7V 7 iAx 7 12 7397397 V 397V7V7V7V7 3 m m 9 9V g 5757738ga1 00353ft min 02642gal where Vis a positive number and x is the initial depth of water Evaluating I1464min244h Answer 431 4 34 On April 1 a reservoir has a water depth of 11 m The reservoir is fed by a stream which becomes swollen with snow melt as the month progresses The volumetric flow rate of stream water entering the reservoir during the month of April is V1 25gtlt107 exp0026l Where lis the time in days and the volumetric ow rate has units ofm3day I 0 at 1201 a m on April 1 Water issues from the reservoir through a dam The flow rate of the discharge at the dam is steady at a rate of 04 x 107 m3day for the first 15 days of the month At midnight on April 15 the sluice gates are adjusted to allow a higher flow rate of 635 x 107 m3day This rate remains constant until the end of the month If the surface area of the reservoir is 28 x 106 m3 find the depth on April 30 Assume that the surface area remains unchanged during the month and that the effects of rainfall and evaporation are negligible Approach Apply conservation of mass integrating over the given incoming flow rate Assumptions 1 The surface area of the reservoir is constant 2 There is no evaporation or rainfall and only one stream entering and leaving the reservoir 3 The water is incompressible Solution By conservation of mass dm 1 We have one stream entering with flow rate r39n1 and one stream exiting through the dam with flow rate m2 d i dmw rhldlirh2dt Substituting rh pV and mv pr d M pVdI szdI an zdri w Integrating over the month 30 30 30 de j Vidti j Vzdt 0 0 0 0 day T25gtlt107e dlif04gtlt107dliif635x107dt 0 0 15 V v V 30 days 7 v Defining Ach Vv 30 days 7 Vv 0 day and performing the integrations 630 71 AVw25x107 704gtlt107157635gtlt10715 Noting that 00026 AVw 1235gtlt108m3 The change in reservoir depth is 7 AKV 71235x108 7 AW 7 28x106 The final depth is L2 L1AL11m441m551m 4 Answer AL 4412m 4 35 Hydraulic uid enters a square conduit 2 in on a side at a velocity of 142 fts and a temperature of 60 F The uid leaves the conduit at 100 F Neglecting frictional heating and kinetic energy find the rate of heat addition to the uid in steady state Approach Write the first law for an open system and eliminate all terms except heat and enthalpy change Assumptions 1 Specific heat is constant 2 The system operates in steady state 3 Frictional heating in the uid is negligible 4 There is no change in kinetic or potential energy Solution From the first law 2 2 d5 Q39w Ww Zm Vjgz jizm Eh V78gaj This simplifies to 0 Q WA 41 Assuming constant specific heat Q mchL 1 Where m p A The density of hydraulic uid is taken at the average of the inlet and exit temperatures T 7T 60100 ave 2 80 F Using the density from Table B6 lbm ft 2 1ft2 525 142 2 2 m ft3 S X m 1441M 2071b m S Using cp from Table B6 at80 F Q39w 2071b m 04533 100760 F s lbm F Btu A 375 Answer S 4 36 Air at 20 C and 101 kPa enters a passage between two printed circuit boards inside a desktop computer One board contains 9 chips each dissipating 2 W and 5 chips each dissipating 13 W No heat enters the passage from the other card If the exit temperature is 285 C find the volumetric flow rate of the air Neglect kinetic and potential energy changes Approach Write the first law for an open system and eliminate all terms except heat and enthalpy change Assumptions 1 Specific heat is constant 2 The system operates in steady state 3 Frictional heating in the uid is negligible 4 There is no change in kinetic or potential energy 5 All heat dissipated in the chip enters the air by forced convection no heat is conducted into the card Solution From the first law dE V 2 V 2 d1 Qw Ww Zrh fgz 12m 11 Tegaj This simplifies to 0 Q39w WA 11 Assuming constant specific heat Q WAT T1 The mass flow rate is m Qw cp 7 T1 Using the value of cp for air from Table A7 at 300 K 9 2 5 13 W m EX X 000287k g 1006 J 285720 C S kg C lkJ To find volumetric flow rate use V 7 E p The value of p at the average temperature of the air is needed here The average temperature is 20 285 ave 2425 C 2973K We can either calculate p at 2973K from the ideal gas law or approximate p by the value at 300K in Table A7 We choose the simpler approach and use Table A7 Then m 000287E m So00243 lt Answer 9 118k g3 S m 4 37 Water flows in a pipe 17 cm in diameter and 480 cm long at a velocity of 88 ms The water enters at 100 C and exits at 80 C Calculate the rate of heat removal per square centimeter of pipe wall Neglect frictional losses and kinetic energy Approach Write the first law for an open system and eliminate all terms except heat and enthalpy change Assumptions 1 Specific heat is constant 2 The system operates in steady state 3 Frictional heating in the uid is negligible 4 There is no change in kinetic or potential energy Solution From the first law dEw V 2 Va 2 dl 7 QCV iWw me JrTJrgzx jizmg by Tgzgj This simplifies to 0ch WA 412 Assuming constant specific heat Q WAT T To find mass flow rate use m p A Using the density of water at the average temperature of 90 C from Table A6 2 2 m 965k g3 883 II g cm2 1m 2493 m s 2 104cm s The heat removed is using Table A6 for CF atT 90 C ave Q39wl93k gj 4202 807100 C7162gtlt105W7162kW s kgK To get the heat flux on the wall use 4 Q A 7 5 q 13962X10 W 632 W2 lt Answer 7rl7cm480cm cm 438 In a solar collector air is heated as it ows in a rectangular channel under the collector surface as shown Assume the rate ofheat addition on the top surface is constant and uniform at 400 Wm and that all other sides ofthe channel are insulated The air enters at 1 C an 101 a with a velocity of5 s The eat transfer coefficient inside the channel is 155 WmZK Find the minimum and maximum temperatures of the collector surface Ap roach Write the first law for an open system and eliminate all terms except heat and enthalpy change Use this to calculate the exit temperature of the air Use the convection equation to find the collector temperature at both the inlet where it is a minimum and the outlet where it is a maximum Assumptions 1 Specific heat and density are constant 2 Th tem operates in steady state 3 Frictional heating in the uid is negligible 7f 3 quotr 4 There is no change in kinetic or potential energy 7 V I 36 5 The heat transfer coefficient is constant along the 39 channel and independent of temperature P lo ku 6 The collector is perfectly insulated on all sides except the top Solution From the first law dE V2 7 V2 12 Qr2m h Tgz2m IHTWAJ With the assumptions listed above this simplifies to Assuming constant specific heat QV quot10 T r T To find mass ow rate use In pV A with the density of air at a temperature of15 C and 1 atrn from Table A7 m 1234 kg3j3cm20cm m3 1 2 k k 4m 2 00222 g 193 g 10 cm s s The exit temperature ofthe air is with c from Table A7 at 15 C 40020cm101gm16m 39 quot395 00222k g 100mi w s kgK 1kJ To find the surface temperature of the collector we need QMTrT Ts w 15 C 207 C The surface temperature will vary along the length ofthe collector since Tvaries along the length The ratio QA is given as 400 Wm2 in the problem statement The minimum surface temperature will occur at the entrance where the air temperature is lowest It is 4oo zj155 2N j1sac 175 01 Answer m m K The maximum temperature occurs at the exit Tm 4oo zj155 207 c233 clt Answer 39 lTl W mzK 439 Water is siphoned from a waterbed into a bathtub through a 1 in diameter hose The top surface of the water in the bed is 26 in above the exit of the hose Assume the pressure in the waterbed is atmospheric and that the top surface recedes with negligibly small Velocity Also assume negligible frictional effects in the hose What is the ow rate in galmin of water into the bathtub Approach Bernoulli equation to find the Velocity exiting the hose Use the Velocity to determine the quot ow rate 26 Assumptions 1 Density is constant 2 The ow is frictionless and isothermal 3 The Velocity ofthe surface ofthe water in the waterbed is negligible 4 The pressure in the waterbed is atmospheric Solution Taking station 1 to be the top ofthe water mattress and station 2 to be the exit ofthe hose and applying Bernoulli s equation V 2 1i l V gzl p 2 2 2gzz 2 13 Since the pressure in both the bathtub and the waterbed is atmospheric 1 2 arm Furthermore the surface ofthe water in the mamess is receding at a Very low Velocity V 0 so V 2 0 gr Tziga Solving for V2 The Volumetric ow rate depends on Velocity according to a 11 05 2 2 1g 39 gal V2 VZA118 1r j f 289 4 Answer 5 2228X103937 m s 440 Water ows at 25 kgs through a gradual contraction in a pipe The upstream diameter is 8 cm and the downstream diameter is 56 cm Ifthe exit pressure is 60 kPa nd the entrance pressure Assume frictionless ow Approach Apply the Bernoulli equation Find Velocities from the knon mass ow rate s 39 l Assumptions Density is constant 2 The ow is frictionless and isothermal 3 The system is at room temperature Solution Taking station 1 to be the entrance ofthe contraction and station 2 to be the exit and applying Bernoulli s equation39 Vf 1 2 Since there are no elevation changes 1 V22 gz g22 p 2 Pg Velocities can be determined using the mass ow rate since m i M 25B V k 549 9A1 99757r0042m7 s m 25k g S V21A 1022 p2 9975 jz 039055j m2 s m 2 To nd the pressure rearrange the Bernoulli equation 2 7 V2 1 2 I 50x10313a9975k 1022 4995 m S I 992x103Pa 992kPa lt Answer 441 A constrictionin a 1th 39 39 39 Find the mass ofthe heaviest disk that can be supported Approach Apply the Bernoulli equation Findvelocities from the known mass ow rate 101kPa Pmm V1 3 ms gt Assumptions Water 1 ensity is constant ess and isothermal 3 The system IS at room temper ure 4 The pressure is constant across the throat ofthe constriction 5 There are negligible buoyancy forces on the disk 7 7 cm Solution Taking station 1 to be the entrance ofthe contraction Water and station 2 to be at the throat and applying Bernoulli s equation av5 Z Z 2 g l 2 g 2 p From conservation ofmass WW2 pVArpV2A2 The water is incompressible therefore 2 2 2r 5 V27r 2 2 2 131 2 V2 3 3 39 cm 7662 D2 s 82 cm s 39 L so 21 m 7 m 7 37 7557 2 2 VrV2L1017mnD m ks s s s 2 l k 2 J K ma 2 Let P be the pressure ofthe water in the tube just above the disk The water in the tube is stagnant so PaP2pghrh2 762xlo3 997k g 9813 17 m771x103Pa m3 s2 0 1 1n3 pa 10 Let P4 be the pressure ofthe water on the underside ofthe disk m Po m pg k 37 3 101x103 997 53 98132 m 1014gtlt103 g A m s 100 From a force balance on the disk INN 33 2 0015 2 3 3 2r 7 m 1014x10 r771gtlt10 Pa mimosa E 2 j 3 98132 s m0438 kg Answer 442 39 quot ahum quotquot nd Approach 1 i n at the hole in the tank Then use Newton s second law on a uid particle to nd the time until it hits the surface Wale From the time offall and the horizontal velocity the distance L can be determined As sumptjons 1 Density is constant 3 The system is at room temperature 4 The tankis large so that the water recedes with Very small ve ocity 5 Water issues horizontally from the tank 5 Aerodynamic drag on the water jet is negligible Solution Taking station 1 to be the top sur ce ofthe water and station 2 to be the exit from the tank and applying 391 Bemoulli s equation g2 P g p 2 p 2 2 Since P P Pm and the wter recedes with negligibly small velocity VFW 9815 m 99 5 To nd the time of ight applyNewton s second law to a uid particle at height 22 We only consider forces in the vertical direction Fmamgmd V 9 i L W d x mm M Integrating once at b Emmi Integrating again x 3 ct c2 The boundary conditions are at t0 x0 c2 0 att0 vo jago 2 So that the solutionis xg 7 Solving for time ofdrop 0782 s s The lateral distance traveled during this drop time is Lquot2quot0m Answer K s 443 A re ghter aims a jet of water at a window in a burning building as shown below The jet is horizontal when it enters the window If the nozzle has a diameter of 25 in what is the mass ow rate of the wate l 01 J Approach Apply the Bernoulli equation to the free jet Decompose the Velocities at the nozzl window into horizontal and Vertic the window and that the horizontal component is the same at the nozzle and at the window to determine the Velocity at the nozzle Assumptions Density is constant 2 The ow is frictionless and isothermal 3 The system is at 50 F 4 Water strikes the building horizontally 5 Aerodynamic drag on the waterjet is negligible Solution Taking station 1 to be the exit of the nozzle and station 2 to be at the window and applying Bernoulli s equation R V 2 gz 7P2 V22 gz 2 2 2 In a free jet pressure is atmospheric everywhere so The two Velocities may be decomposed into horizontal and Vertical components v5 v3 v3 V22 V2 V23 Substituting these into the Bernoulli equation th Vt V2 V23 Tum Tag Since thejet is horizontal at point 2 V2 0 There are no forces in the x direction no aerodynamic drag so V V2 and the Bernoulli equationbecomes 0 V2 fgzrzz 1 2gzfz 23217s 22873 401 s Now nd the magnitude of the Velocity at the nozzle exit from 7 sin 70 V L S427 s sin 70 The mass ow rate is lbm ft 25 2 I I 2 4 m pA 624 3 I42 7 sz2lz J 9081bms Answer 444 n 39 39 39nowratein 39 439 DJ Approach PZ101 kPa Apply the Bernoulli equation between poinm 1 and 2 an also between points 1 and 3 Use these two equations and conservation ofmass to nd the unknown diameter D22J cm P1 120 kPa D3 Dl6 cm 39 7 3 Assump 0ns V1 2 m s 1 Density is constant 7 2 The ow is frictionless and isothermal P3 lOl kPa 3 The system is at room temperature Solution Apply Bemoulli s equation between points 1 and 2 to get 1 V12 P2 V22 gz gz2 p 2 p 2 There are no elevation changes soz z2 To nd V J 7073 S 1207101kPa 1000M lkPa m k 939 998 5 5 m The volumetric ow rate is 2 2 V2 VZAZ szr j 939342 cm2 00045113 2 s 2 10 cm x By conservation of mass Marvin 13K3V2pl Since water is incompressible V3 V 4 002700045100154mT3 To nd the diameter apply the Bernoulli equation again between points 1 and 3 to get 1 V12 1 V22 gz 7 gzz p 2 p 2 2 7 7072 7 V3 2V R a 2 120 IOIXIOOO 939EV2 2 p 2 998 s 2 V3 VZA3 V3r 3 001541 S 49393 5 V D 2 3 2 00457m457cm4 Answer I 3 Water at 20 C issues from the bottom of a large tank which is lled to a height of 4 m The air pressure above the water is atmospheric The water ows over an aluminum rod of diameter 11 cm and length 3 cm The heat transfer coefficient between the water and the rod depends on velocity and is given y 11150V where h is in WmZK and V is in ms The rod is initially at 65 C How long will it take for the rod to cool to 30 C ssume the surface ofthe water recedes very slowly so that the depth ofthe water remains at 4 m throughout the process Approach approximation to calculate the time required for the rod to cool to 30 C Assumptions Density is constant Pam 2 The ow In the tank is frictionless and 41mm 4 isothermal 3 The heat transfer coefficient is uniform 3 I over the surface ofthe rod and independent oftemperature I A c a gt 3 3 1 4 The surface ofthe water in the tank recedes very slowly Solution Apply Bernoulli s equation between point 1 the upper surface ofwater in the tank and point 2 the place where water issues from the tank to get 2 2 R V 1 V2 gz gzz p 2 p 2 r 39 r 39 39 39 39 39 mun rut small velocity so V 2 m m gz 4222 7 V2 I2gz7zz 29815 24m886 From the given formula 11150V 8 7150685 915 K Check to see ifthe lumped system approximation can be used find thermal conductivity from Table A2 2 11 cm 3 cm Bi hLquot 39 where I m 7 V 7 I L 2 7 7L 7 7 k A 2rrL2rr 2L2r 23cm211cm 915 Y j0402cm m m K A n 100ch 700155 237i m K Since Bi lt 01 the lumped system approximation is valid and the time of cooling can be calculated as with density and specific heat from Table A2 2702k I903LIWm pcPL ha lTt7TL m kgK 100 h l Bi f 916 0726 s lst Answer L55720j mZK 446 Air at 17 C 100 kPa ows in a duct A stagnation tube connected to a Utube manometer lled w39th mercury is placed in the duct Using data on the gure below nd the air Velocity Assume atmospheric pressure is 100 kPa Approach 1 Apply the Bernoulli equation to nd the i 39 N m 17 C 100 kpa Velocity ofthe air in terms of and P2 1 2 See gure Use hydrostatics to determine 39 39 the relationship between P and P Recognizing that P2 1g and I 11 allows the solution to be completed 3 mm Assumptions 1 The air and the mercury are both incompressible 2 The density ofthe air is much lower than the density ofmercury Solution Apply Bernoulli s equation between point 1 the ow at the entrance upstream from the tube and point 2 the place where the air strikes the stagnation tube to get R gzl P2 822V 22 2 pm 2 Pnint 7 39 39 39 is zero 4 L 39 4 L therefore 4 H Pan 39 7nnrln0lnl3 39 L L 39 lledwithastagnantcolumn ofairthat isI Furthermore L r r 39 439 atmospheric 39 r 39 1thatis Ramp Therefore V 2 P From hydrostatics amppyggh 9 Prawgggh Using mercury density from Table A6 and air density from Table A7 p gh 213586L981 2m V L 80831 Answer Paw 1224 5 m 447 Oil SG 077 ows in a pipe with a sudden contraction as shown in the gure A stagnation tube open e atmosphere is placed in the upstream section Ifthe oil in the stagnation tube rises to a height h 22 cm find the Velocity at the exit ofthe pipe Approach Apply the Bernoulli equation between points 1 and 2 also between poinm 1 and 3 Use hydrostatics to express P3 in terms of P4 Combine equations recognizing that P P4 and solve for the unknon ocity Assumptions 7 1 1 Neglect any elevation difference between points 1 L l gt 2 DJ hfcm The density ofthe oil is constant 3 Frictional effects are sma 1 Solution Apply the Bernoulli equation between points 1 and 2 V 2 2 V2 R gzl t p This simplifies to 1 4 2 V22 2 p 2 the Bernoulli equation between points 1 and 3 1 V a V5 gz gz3 p 2 p 2 p Also apply Because point 3 is a stagnation point V3 0 and iVx2 i p 2 p Combining the two simplified Bernoulli equations V2 2 ilk3541 Combining equations 13 8 p 2 m 1m V 2 2 981 22cm 2 g s2 IOOcm 208E Answer s Comment Frictional effects in a sudden contraction t into accoun 4 48 Hydraulic uid at 80 F flows through a Venturi meter The diameter at the entrance is 81 in and at the throat it is 52 in The pressure at the entrance is 147 psia 1f the pressure at the throat is measured to be 108 psia find the velocity at the entrance Approach Determine the volumetric flow rate from the equation for a Venturi meter Knowing the area and volumetric flow rate calculate the velocity Assumptions 1 The density of the hydraulic uid is constant 2 Frictional effects are small Solution The areas of the entrance and throat are D 2 D 2 4471 515in2 A2n72 212in2 From Table B6 at 80 F p 525 For a Venturi meter 2037132 1 H HAM02 2 2 147 7 1081b2 32171bm2 ft 144 11 2 1ft2 1n 1 lbfs 1 ft 1 V 2121n 2 2 144m 525 lbm 212 73 ft 515 V 4253 ft V1 S119 quot Answer S 449 A block ofmass 4 kg is propelled along a at surface by a waterjet as shown It moves to the right at a constant velocity of 25 ms The coef cient offriction between block and surface is y 033 Lfthe inlet jet area is 67 cm2 find the inlet and outlet velocities of the water Neglect wall shear and elevation changes Approach Determine the relative velocity between the block and the jet Then apply Bernoulli s equation and conservation of momentum Assumptions 1 The density ofthe water is constant 2 Frictional effects are small 3 Neglect the weight ofthe water I Solution Consider a control volume anached to the block as shown in the igure to the right This control volume moves at u Q speed Vb to the right Then 04 CD 39 t V V1 7 Vb J where the r subscript designates relative velocity between jet and block Since the moving control volume is not accelerating we may analyze the system as if it were a stationary block with an impinging jet of velocity Vy Applying conservation ofmass to this system quot391 i quot 1 quot392 mg V M V 13Asz By symmetry if we neglect the weight of the water V V2 A A2 J block Av4v1V22w From the Bernoulli equation P V2 F V2 n 0 JV 744r p 2 t 2 4 P V 2 12 V 2 14 7 J V 7 2 2 07EP 2 11 2 Jag20 sinceljvz13m k V V V2 From conservation of momentum d F B m V 7m V 2 dg m but V 0 and VXV V so umg nij VYVY 0334kg981ms2 9982kgm357cm2 lgfz 44E v V2 4 Answer 5 The inlet velocity is VJ V Vb 442559mlt Answer 5 r39r the velocity quot steady ow nd the 450 Water at 20 C ows through r39r L below is6msquot L 39r 39101kP 39 39 magnitude and direction of the anchoring force Approach Apply the Bernoulli equation to nd the inlet pressure Use conservation oflinear momentum to determine anchoring force Assumptions The water is incompressible 2 The ow is frictionless 3 The system is at room temperature Solution From conservation of mass va 1L 5 922 P MLP V2V A1LZ194mS 2 A2 2452 From the Bernoulli equation 2 P Luge p 2 i 2 2 211 M ego i 2 P V2 12 2 82 1000 Pa 101kPa 194 2 752 j I 9982k lkpa 98103 m 253 kPa m 2 9982 5 5 m Perform a momentum balance in the xdirection 2 FX 2quot V2 4212 quot12 V2 F F FeA2pA2V 92 g Gage pressure must be used in this equation a Opsig Fx 2 2 2 F 99821 I 3 cm 194 39 m 2 10 cm A momentum balance in the ydirection gives 2 V F2IA2pEA 2 2 2 7 2 7 1000 Pa kg m 9 2 1m 7 Fwipiipl 1A1 0537100921 7 9982E 5 2r cm W 7802N lkPa 2 used 2 2 2 F lFx Fy 71090N Answer The direction is F 6arctan quot 1 arctan j 474 Answer F 737 4 4s 4 51 Amterjet moving at galmin mid the ow 1 18 115 strikes avane and is turned through 120 as shown Ifthe owmte is 27 the Vane inp ace Approach Apply conserwtion ofmomentum in both 9 6 th x and y directions Assumptions 1 There are no components ofvelocity in the z direction 2 The system is at room temperature Solution Apply conserwtion ofmomentum in hexdirection F B m V 7m V z in The mass ow me is 3 m m m 272 sum T 0391337 1quotquotquot 374 m min 11 lgal 605 s 2F mVr ViimVir Vii Thexcomponent ofthe exiting velocity is Vx 2 7V2 c056 lsgj os 60 V 5 Substituting values 2 a 374EJ7971812 i 73141bf s s lmy s Apply consemtion ofmomentum in the ydirection d 2 EByevmVm WAVW 1 F 374 18 39 60 1811bf z sm mum Themagnitudeoftheforcemaybecalculated omthe E 131 components F i31421812 363lbf The direction is given by Answer 452 Ajet engine on a commercial aircraft exhausts combustion gases at a rate of8 kgs Upon landing a thrust reverser blocks the exhaust and redirecm the ow forward as sh wn This aids in braking the plane The exhaust gases may be assumed to ow at 200 ms relative to the plane and have properties Very simi ar to those ofair 39 39 on the back ofthe engine Approach Apply conservation of momentum in the x direction P Assumptions H ml rwt 1 Th ere are no components ofvelocity in the z direction 2 The ow is symmetric about thex axis l M 9 Solution Forces in the y direction cancel so we only need to consider thex 7 direction momentum equation 2 Fx 2 m V 2 m V FR quot12 W2 WINK mr V L Apply the Bernoulli equation between points 1 and 2 1 V12 P2 2 gz gz2 p 2 p 2 SincePI and 222 39 V V2 V 200E 1w V 5 Va i2002sin20 39 gt 5 By symmetry v3 7200sin20 3 L The mass ow rates can be found using symmetry 1 kgs 1 mi quota Emr 4kgs Substituting Values FR 43 72003 sin 20 49200 sin 207 8200 s s FR 72147N Answer 453 Ajet ofwater of area 16 in2 and velocity 22 Ms strikes a plate and is de ected into two symmetrical streams as shown If 6 42 nd the anchoring force necessary to hold the plate in place Approach Apply conservation of momentum in the x direction 0 Assumptions 1 There are no components ofvelocity in the z direction 2 The ow is symmetric about thex axis stationary 74 22 fts gt plate Solution Forces in the y direction cancel so we only need to consider the xdirection momentum equation 2 Fx 2 m V 2 quotA Vt FR quote W WM 2m V To nd the velocities apply the Bernoulli equation between points 1 and 2 2 2 j i Ti i gZ jJrTergzz SincePI and z 22 V V2 By symme V 2 By conservation ofmass noting that m2 m1 n11m27i132m2 lbm lft2 lbm V 522 22 1539 2 7152 m p A it s m144in2 Bysymmetry n5m3m 761bm Applying conservation of momentum in thex direction aner mijrmi lbmi22 cos42 llb 5 5 32174 s 1 1 76 722 cos 42 7 152 22 32174 32174 34811bfd Answer 11 1 One approach used to determine the thermal conductivity of metals is to sandwich an electric heater between two identical plates Consider two pieces of a metal each piece is lcm thick lOcm wide and lOcm long All edges are heavily insulated and the exposed faces have the same convective boundary conditions For an applied power input of 173 W to the heater the temperatures of the inner and outer faces of the metal plates are 423 C and 387 C respectively Determine the thermal conductivity of the meta Approach We apply the conduction equation for steady one dimensional conduction Eq 1129 T HE 1111 K 5va Assumptions L 393 010M T1 383 C The processes are steady w 0mm 2 The conduction is one dimensional TI 423 C 3 There is no internal heat generation 3L 4 The thermal conductivity is constant gtl I6 f aol M Solution Assuming steady onedim ensional heat internal conduction with constant thermal conductivity and as heat generation Eq 1129 is k QTAT1T2 Solving fork and recognizing that the heat input goes to both faces because of symmetry k Qt 173WO3901m 240l 4 Answer AT17T2 2010m0lOm4237387K mK 11 2 In an experiment the boiling heat transfer coefficient is to be measured using the apparatus shown in the figure below Condensing steam at 120 C is used to heat the end of the 304stainless steel rod with a 25 mm diameter The outside perimeter of the rod is heavily insulated and the temperature in the rod is measured in two places T4 9119 C and T3 10020 C The boiling uid is at 10 C The condensing heat transfer coefficient is 7500 W mZK Determine a the heat transfer rate in VJ b the temperatures T2 and T5 on the two ends of the rod in C c the heat transfer coefficient on the test specimen end in WmZK Approach D QC 391 We apply the conduction equation for steady one quotMM1h dimensional conduction Eq 1129 T7 k TA ZQHT C a Assumptions T lot 39 T20 C l The processes are stead 939 L BOOK i 2 The conduction is one dimensional k 3 There is no internal heat generation e L 10MM 4 The thermal conductivity is constant Ll gm 3 LLTlth Solution a Assuming steady onedimensional heat conduction constant thermal conductivity and no internal heat generation Eq 1129 is applicable between points 3 and 4 kA Q TU T4 From Appendix AZ for 304 stainless steel k 149Wm K 149wm K7r40025m2 Q 1002079119K 659W 1 Answer 001m b The same equation as used in part a can be used between points 4 and 5 and between points 2 and 3 Solving for the two temperatures 39 659W T5T47QL9119 C7 kA 149wm K 659W 149wm K V 000m 7r40025m2 0020m 2 11822 c 4 Answer 7r40025m 8668 C Answer A T2 T310020 C kA AV c Convective heat transfer is calculated with Q h A T6 Solving for the heat transfer coefficient h Q 659W l75l i AHSWCI AT5Ts Ir40025m28668710K m K 11 3 Large electrical currents are often carried in aluminum conductors Consider a long 2cm diameter cable covered by insulation 2mm thick For a particular application the outside insulation temperature is limited to 35 C and none of the insulation can exceed 50 C The cable is in an environment in which the convective heat transfer coefficient is 25 WmZK the air temperature is 24 C the insulation thermal conductivity is 010 WmK and the electrical resistance per unit length of the wire is 39 X 10394 Qm Determine the maximum current allowed in A Approach L QC This is a composite conduction problem in cylindrical 1 SD lt 0002 coordinates The rate equation ATRm can be used 0 The heat transfer rate current must be calculated from E 3 24 C the given electrical resistance characteristics Define the T g 36C L A 2 w system as the insulation and convective resistance 1 39L 39 39L r Tg TL TI Assumptlons Q lt A A A A Q 1 The processes are steady I 2 The conduction is one dimensional lNDL ID 3 There is no internal heat generation 1quot kL 4 The thermal conductivity is constant Solution For a steady onedimensional constant thermal conductivity system with no internal heat generation Q 7 E 7 T1 7 T R 1 manD h IrD2 L 27r k L The heat transfer rate per unit length is obtained from the electrical characteristics I 2 R Substituting this into the rate equation and solving for 0 5 Hf QM ww1 50724K 39x103949j 1 h100240020 39 m 25wm2 K7r0024m 27r010WmK 1 285A Answer This current ensures that the inside insulation temperature does not exceed50 C To check if the outside surface temperature does not exceed the allowable temperature we can do the same calculation across only the convective resistance 0 5 0 5 T1 7T 357 24 39x10 ml Any current greater than 231 A would cause the surface temperature to exceed the limit so the maximum allowable current is 231 A 231A Answer 11 4 For the design of a chemical processing plant in Hawaii you need to determine the thickness of insulation on several steam lines that would make the most sense economically The steel steam lines have a total length of 350 m have an outside diameter of 25 cm and carry saturated steam at 200 kPa The design air temperature is 27 C We assume we can obtain insulation with thickness in 10 cm increments The insulation thermal conductivity is 005 WmK and costs 000015cm3 From previous studies when you took into account the prevailing wind you developed a simple correlation for the heat transfer coefficient to be h 75D10390 38 where D is the outside diameter in cm and h is in WmZK Natural gas is used in the boiler made 87 and costs 050105 kl For a 20 year life assuming the plant runs 8200 hrsyr determine a the recommended insulation thickness b the maximum savings compared to no insulation Approach I 03g A schematic of the system is shown below k 7S1 D A a gA mMiW 35355133d f i e f iii iii T5212 w 203 Thus beginning with the rate equation ATRm we 16 27 K can integrate it with respect to time to get energy loss 1 Because the pipe is steel we assume its thermal resistance 39 is negligible compared to that of the insulation39 likewise we have saturated steam flowing inside the pipe and also assume that the steam convective thermal resistance is L 739 3500quot negligible compared to the insulation Hence the total resistance Rm consists of the conduction resistance through the insulation and the convective resistance With the cost of energy the boiler efficiency and the total energy loss we can calculate the total cost of the lost energy The thickness of the insulation varies and the cost of it can be calculated with the given information Total cost consists of the sum of the insulation and energy costs over the 20 year life of the processing plant Assumptions 1 The heat transfer is one dimensional 2 The system is steady Solution We assume we have a steady system with constant properties The rate equation is Q AT Rm Integrating this with respect to time where I is the total time period 20 years and 8200hryear and recognizing that both the temperature difference and the thermal resistance are not functions of time we obtain Q I R m The driving temperature difference over the total resistance is from the steam temperature to the air temperature Hence the temperature difference is ATTmmiTw From Appendix All at 200 kPa the saturation temperature is Tmm 1202 C The total resistance is Rm Rmv 11 The convective thermal resistance is l l Rcanv W M 75D225 7rD2L l 7 0000856 K 75Wm2KD2cm2570387rD2cm350mlm100cm chm062 W The insulation thermal resistance is that of a cylinder so ll4 lnD2D17 h1D20m25cm quot5 27rkL 7 27r005WmK350m Energy cost must take into account the energy lost directly from the steam pipes and the inefficiency of the boiler Thus Q 7 Q kl Qbaxler ma a 087 With CWgy 219050105 kJ Q Q k 050 6 Ener cost C C 575gtltlO kl gy Qbaxler Energy ma a Energy 0 87 105 kJ With Cmmman 000015cm3 the cost of the insulation is 0009llnD2cm25cm KnsQnmlanan D12chmzanan Insulation cost 000015 100cm ED2cm2 725cm2350mcm 3JTj 412D2cm2 7 625 Therefore the total cost is Total cost Energy cost Insulation cost cost we can see 65234 10837 54397 Answer 11 5 Consider a solid cylindrical rod with radius R0 a uniform volumetric heat generation qquot a known outside wall temperature TW and constant thermophysical properties Using Example ll3 as a guide and the general conduction equation in cylindrical coordinates develop the following expression for the steady temperature profile in the rod m 2 2 T Tw 1iLj 4k R Approach The conduction equation in cylindrical coordinates TW must be used First we need to simplify the equation by applying what we know about the system being I I 751 analyzed Appropriate boundary conditions need to be 8 R0 stated and used Assumptions 1 The system is steady 2 Conduction is onedimensional 3 All properties are constant Solution The general conduction equation in cylindrical coordinates is l 6 6T 1 6 6T 6 6T 6T kr 2 k x q pcP r6r 6r r 6 6 6x 6x 61 We assume steady onedimensional in the r direction and constant properties Applying these assumptions we obtain li kra Tjq 0 r 6r 6r Because this equation tells us that T is only a function of r we can change this from a partial differential equation to an ordinary one withk constant r qquot 0 r dr dr We require two boundary conditions 1 atrR TTW 2 atr0dTdr0 Separating the variables in the ODE and integrating ir gq r 2 Idr 4M dr dr k dr k 7 m 2 7 q r C1 2 q rig dr 2k dr 2k r Separating the variables again and integrating drq r jdr 2k r 7 m 2 Therefore the general temperature profile is T q r C11nrC2 Applying the two boundary conditions 7 MR2 at rR TW q4k CllnRoC2 7 m 0 ggtcl0 2k 0 Solving for C2 T W quo2 4k Substituting C1 and C2 into the temperature profile and simplifying m 2 2 TT q R 17 L W 4k R0 at r0 0 Answer 116 11 6 Gargantuan Motors has developed a new rear Window defogging system The electric heating element is a thin and transparent film applied to the entire inner surface of the rear Window The glass is 4mm thick with a thermal conductivity of 094 WmK The design operating condition for the defogger is for an outside condition of lO C with a convective heat transfer coefficient of 65 WmZK and an inside condition of 10 C with a convective heat transfer coefficient of 10 WmZK The inner surface of the Window is to be maintained at 10 C Determine a the heat flux that must be supplied to the heater in Wmz b the temperature on the outside surface of the Window in C Approach Glass An energy balance is done on the glass and heater The TEL ldic Kit 4 heatL steady onedimensional heat transfer equation can be used Y 1K o to evaluate the thermal resistances lquot7 39 6 M EA quot3910 C T 40 C Assumptions Tquot I OW LK l The processes are steady t 0004 iquot z w 2 The conduction is one dimensional k t O 94 NM K 3 There is no internal heat generation 39 I 4 The thermal conductivity is constant gwgt l 5 Potential and kinetic energy effects are negligible Tz i Tle 3 Bl Lia Tm IquotLPr IL A 39 Solution a Apply an energy balance to the glassheater combination Assume steady no potential or kinetic effects and a closed system QEW0 Q1Q27W0 W is the power input and is negative due to sign convention The heat transfer rates are obtainedby assuming onedimensional conduction constant properties and no internal heat generation in the glass Conduction across the electric heater is ignored because it is so thin Therefore T 7T T 7T 2 51 1 51 n 7 070 h2A kA hlA Solving for the heat flux T 7T T 7T 107 710 K 107 710 K q 5391 392 5391 391 12181 Answer ii i 1 0004m 1 m2 h2 k h 65Wm2K 094WmK 10Wm2K b The temperature on the outside surface T52 can be found with the onedimensional conduction equation kA Qt q t QTTs1 Ts2 T52TS1 ET517 k 0004 T52 10 C71218 2j 482 c 4 Answer 39 m 094Wm K Comments The surface temperature of the glass on the outside of the Window will melt ice on the outside surface 11 7 An approach to determining the thermal conductivity of a material is to put a known material in series with the unknown material as shown in the figure below A heat input is applied to one end of the assembly and the other end is cooled Temperatures are measured at specific locations in both materials Material A is stainless steel with a thermal conductivity of 152 WmK The specimens are rods 2cm in diameter The rods are heavily insulated In one test the following temperatures were measured T19300 C T28257 C T36921 C T46628 C Determine the thermal conductivity of the unknown material in WmK Approach The same heat transfer rate flows through both materials We assume steady onedimensional conduction with constant properties and use the appropriate conduction quation coubd Assumptions The processes are stead 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant Solution With the heavily insulated circumference heat flows only in the axial direction Assuming steady one dimensional conduction with constant thermal physical properties and no internal heat generation Eq 1 129 is applicable k A AT Q Ax Equating the heat transfer in Material A to that in B kAAATl7T2kBABT37n AxA Ax Noting that AA AB and AxA Ax so that k k TIATZ l930078257 c B ATFTA 39 mK 692176628 C k8 541i Answer m K 118 When natural gas is burned water vapor and other products of combustion are produced These producm can mix with the water vapor to produce a dilute acid To prevent acid attack on the chimney the gases should be kept above a minimum temperature Consider a 304 stainless steel chimney 200mm inside diameter with walls 1mm thick Insulation k 0075 WmK 10mm thick covers the outside ofthe me al The oumide air is 5 C with a convective heat transfer coe icient of25 WmZK The inside convective heat transfer coefficient is 15 W mZK To avoid condensation forming on the inside surface of the chimney an 100 C 39 39 39 39 a temperature in C Approach WW 3 om From the resistance network shown we can calculate the r39 t O 0 heat transfer rate using the rate 39 gt le equation QATRobetween7andf3 Oncewe Sac 39 WWIAm have Q then we use the same equation again but this time TF3 1w t1 001M between I and the only unknown is I which is the kgtzlw t quantity we seek Assumptions k lS WM K 39 The processes are steady Q 12 T3 TL Tl Tng 2 The conduction is one dimensional l I l DiIQ I 3 There is no internal heat generation 94 73H WA39 4 The thermal conductivity is constant 1 4 0quot Solution d39 39 ncm uau m Q H Rm 111D2Di mugD 1 erk 2 k L 111 It D L We can divide by L to obtain the heat transfer per unit length From Appendix A2 for 304 stainless steel k 149Wm K Therefore Q 100475 quotc w L 1n0202020 ln02220202 1 m 2149i 20075 25 2 z0222m mK mK m K Now using the rate equation between Iquot and I T 7T L Q Q TJT Q LMDJ hqu a 407w m a 91100 C 143 C 1 Answer 15wm2 Kr020m 11 9 Consider an infinite plane wall 2L thick with a uniform volumetric heat generation rate 11 The wall surfaces are maintained at T T1 at x 7L and T T2 at x L For constant thermal conductivity k steadystate operating conditions and defining the origin of the coordinate x from the centerline of the plane show that the solution of the general conduction equation for the temperature distribution in the wall IS m 2 2 7 m q L Hi 2k L 2 L 2 Approach We use the general conduction equation in rectangular coordinates With appropriate assumptions we simplify the partial differential equation to an ordinary differential equation that can be solved by integration to give the temperature distribution Assumptions The properties are constant 2 The conduction is onedimensional Solution The general heat conduction equation in rectangular coordinates is 3 3 3 Wopc 6x 6x 6y By 62 62 61 Assuming steady onedim ensional in the xdirection constant thermal conductivity k and a uniform volumetric heat generation rate we reduce the governing equation to 2 ca T q39quot 0 6x 6x This shows thatT is only a function of x so we change this partial differential equation to an ordinary differential equation Removing k from the differential d2T 7 7 gm clx2 k The two boundary conditions are l at x 7L T T1 2at xLTT2 Separating variables and integrating verity 2 7 m 2 Separating variables again and integrating T C1 x C2 This is the general temperature profile Applying the boundary conditions In 2 q LL 1 Tlq c2 m 2 4 L 2 T2TC1LC2 Solving for C1 and C2 T 7T M 2 C1 2 1andC2 T1 72 q L 2L 2 2k Substituting into the general temperature profile and simplifying m 2 2 L T 9 L 1 j j Answer 2k L 2 1110 11 10 A large 2kW electric heater 30cm by 30cm square by 01cm thick can be approximated as an infinite plane wall of thickness 2L 01 cm The heater element is exposed on both sides to air at Tm 25 C and a heat transfer coefficient h 75 WmZK The heater material has a thermal conductivity ofk 05 WmK We need to determine the maximum steadystate temperature in this heater With the equation developed inP llO determine a the maximum temperature inside the heater in C b the location at which it occurs Approach Using the temperature profile given inProblem P 1 19 by inspection we can see that the highest temperature will occur I III at the plate centerline The surface temperature is needed so 73 we will use what we know about the convective heat transfer coefficient to calculate it I X 1 2 Assumptions 1 F The properties are constant I 2 The conduction is twodimensional llt 2L4 Solution The temperature profile in the plane wall with internal heat generation is Tqu2 17 2 JrT277quot1 T1T2 2k L 2 L 2 For the specified conditions this heater is symmetric around the centerline of the plate x 0 and by inspection the highest temperature will occur there To use the temperature profile developed above we need to calculate the surface temperature The rate equation is QhATW7T Note that TW T1 T2 and A 2WD so that Tw T Tf 25 C W ZOOOW 1731 C Answer 75 2 030m 030m m2 K gtlt gtlt gt For the centerline temperature x 0 q39 L ZLWD m 2 39 2 39 TCLq L T1 LTlip 2k 2LWD2k 4WDk 2000w 00005 1731 C1787 C 4 Answer 4o30m030m05l llll 11 11 Water heaters are insulated to minimize heat losses Consider an electric hot water heater that is 60cm inside diameter l75cm tall 4mm wall thickness tank made with 316 stainless steel and covered with a 5 cm layer of fiberglass The basement Where the tank is kept is at 15 C The air side convective heat transfer coefficient is 10 WmZK and that on the water side is 100 WmZK If electricity is 005kWhr determine a the cost required to maintain the water at 60 C for 24 hr when no water is removed or added to the tank b the cost required to maintain the water at 60 C for 24 hr if 100 L of water is removed and replaced with 100 L of 10 C water Approach The cost depends on the amount of energy used Therefore the heat transfer rate can be calculated with the rate equationQ ATRm Multiplying this by the time period will give the total energy used When water flows through the water heater conservation of energy also must be used to determined the total energy 39 arugulath fZOOSM 1 charge t 10 539 3 N K Assumptions I I The processes are steady I I6 The conduction is one dimensional 0004 2 3 There is no internal heat generation 4 The thermal conductivity is constant Solution a Electricity cost is calculated by multiplying the energy used by the cost of energy C Electricity cost WIC Where W is the heat transfer rate andlis the time period of interest We assume steady onedimensional heat transfer and use the rate equation to calculate Heat transfer from the side and two ends must be taken into account Tn TN 1 a TM 7 TM 1 lnD2D1lnD3D2 1 1 II 12 1 h1 A1 27r k5 L 27r kms L h3 A3 h1 1451 k5 1451 km A5 h A5 where A zrD1 L Am an4 A3 7rD3 L and153 nDjz From Appendix A2 and A4 k 134Wm K k 0046Wm K 5 m5 14517r4060m2 0283m2 A 7r407082 0394m2 A1 7r060ml75m 330m2 A3 7r0708ml75m 389m2 L0003035 44 0003535 h1A1 100wm2K330m2 W hlAEJ lOOWm2K0283m2 W In D D 1 0608 060 M 00000903 27rk5L 27rl34Wm K125m w L 002573 L 02543 thg 10389 w thE 100394 w lnD3DZ 7 ln07080608 7 0 301K anWL 2n0046175 39 w 005 0004 2 276 tl00011 A 00460394 w A1 1340283 w 1112 60715 C 60715 C Q 1 39v 0003030000090030l00257 00353000ll2760254 136l48 lSlW Electricity cost 151 0 1 SH Answer w b We use conservation of energy for when water flows through the water heater The control volume is defined as shown Assume the water in the tank remains at 60 C and these are no potential or kinetic energy effects dU Q7Wm h 7m h 7 m w cut out The water temperature in the tank remains constant so dUdt 0 Solving for W W Qmhm 41m The heat transfer rate calculated with the rate equation is always a positive number but with heat transfer from the water for the energy equation it must be negative Assuming an ideal liquid with constant specific heat so that Ah cP AT and multiplying through by time I WI Qtmtcp Tm TauzQH mcp Tin Tm From appendiXA atT lO602 35 C p 9939kgm3 cP 4175kIkgK avg 73 3 Wt7151W24hr9939g100L 41753 10760 C 10 m 1hr 1000i m kgK 1L 1 3600s kJ 73 620W hri 5760W hr 79380W hr 7938kWhr Electricity cost 938kW 047 Answer 1113 11 12 A plane wall has a thickness 01 m and a thermal conductivity of 25 WmK One side is insulated and the other side is exposed to a uid at 92 C with a convective heat transfer coefficient of 500 WmZK The wall has a uniform volumetric heat generation rate of 03 MWm3 The wall is at steady state Using the result given in Problem P 119 determine a the maximum temperature in the wall in C b the location where it occurs Approach Because one side of this wall is insulated this system is equivalent to a symmetric wall with double the given wall thickness That is in the situation presented in Problem P 1 19 a symmetrically cooled wall with uniform heat generation 7 no heat flows across the wall centerline so the centerline acts as an insulted surface identical to that shown above Assumptions 1 The properties are constant 2 The heat transfer is one dimensional Solution The temperature profile given in Problem P ll9 withT1 T2 is Tx if 172T1 By inspection the maximum temperature occurs atx 0 4 Answer M 2 T 1 L T1 2k The surface temperature 1 can be determined from the rate equation Q hAT1 7 T T1 TgTq LATq L hA hA h 300000W m3 0lm 92 c sz c soowm2 K The maximum temperature is 300000wm301m2 152 c212 c 4 Answer 225 Wm K lll4 11 13 The curing of concrete is an exothermic reaction that is the curing process produces heat If a concrete slab is large enough the temperature can rise to the point Where the magnitude of thermal stresses may cause cracking Consider a large slab of concrete lm thick Both sides are maintained at 20 C The curing process produces a uniform internal heat generation rate of 60 Wm If the thermal conductivity of the concrete is 11 WmK and using the results from Problem P ll9 determine the steady temperature at the centerline of the slab Approach Direct application of the equation given in Problem P 1 19 will give us the centerline temperature of this slab Assumptions 1 The system is steady 2 The properties are constant 3 The heat transfer is onedimensional Solution We assume that the heat generation rate is uniform and the thermal conductivity is constant in this curing problem With T1 T2 and x 0 centerline the temperature is qm L2 T1 m 20 c 268 c Answer 1115 11 14 A large steel plate is exposed to 800 C combustion gases on one side the heat transfer coefficient is 300 mZK and the plate is 25mm thick with a thermal conductivity of 40 WmK The other side of the plate is to be insulated so that the insulationls outside temperature does not exceed 35 C To save money two layers of insulation are to be used An expensive hightemperature insulation k 0055 WmK is to be applied to the steel and the second layer is a less expensive insulation k 007l WmK The maximum allowable temperature for the less expensive insulation is 350 C The heat transfer coefficient on the insulation surface is 10 Wm K and the ambient air temperature is 30 C Determine the thickness of each of the layers of insulation Approach I t a T4 4 Dag Using the given information a circuit diagram 7 v can be drawn The basic heat transfer rate TL equation can be usedQ ATRm to determine T TE 4 EEOC the required thicknesses I T 8003 vl7 m t 2an ML CVM14 Assumptlons l 9 I l The system is steady if 0 Gum t 3 2 The properties are constant 39 39 3 The heat transfer is oned1mens10nal 39 1 TL 1 T Tr rI L39 6 1 42 i E J m a ram HA Lama Solution A circuit diagram of the system is shown above We have assumed onedimensional steady heat transfer with perfect contact between the layers The same heat transfer rate flows through all the resistances From point 5 to 6 every thing is known so we can calculate the heat flux qquot Q hAT 7T6 a q QA hT 7T610Wm2 K35730K sowm2 Now equating this known qquot to conduction from point 4 to point 5 k TiT k TiT 0071W K 350735K 61 A 5 3 3m 5 m gtlt2 gt 0447m 13 q 50Wm Likewise from point 1 to point 4 q Ti TA L 1 2 h l kl k2 Solving for 12 12 k2Tl TA Tiff 1 9 I71 k1 7 0055i M7i 39 mK sowm2 300Wm2K 40WmK 2 2 2 0055i 9Km 7000333Km 700006251 0495m 4 Answer mK w w w Comment Note that the resistances due to the convection resistance of the hot gas and conduction resistance through the steel are negligible 1116 11 15 A brass rod 02in in diameter and 3in long with k 64 Btuhrft F connects two plates each of which is at 150 F Air at 75 F flows over the rod with h 40 Btuhrft2 F Determine a the temperature of the rod midway between the two plates in F b the total heat transfer rate from the rod in Btuhr Approach This is a fin problem By inspection because of symmetry no heat flows through the plane midway between the two walls Therefore we analyze this as a pin fin with an insulated tip lL SOAP 21 3M gt 145w gtX Assumptlons l The process is steady 2 The conduction is one dimensional z 7 F 1 3 There is no internal heat generation a 0F h 6 L 4 The thermal conduct1v1ty is constant S olution a Assuming steady onedim ensional conduction with constant properties and a uniform heat transfer coefficient the solution for a fin with an adiabatic tip is given in Table 115 TJC T E ATA COSEEZEZ x 0125ft 3 2x12 ForthetipxLandL hP 5 h D 05 4h 5 mLL L 7 L kA k7r4D2 kD 4 40 hrft2 mL0125ftW 1531 l 141 hIft F 12 cosh0 coshl53l TL75 F15075 F 106 F 4 Answer b The heat transfer from the whole rod is twice that from one fin Again from Tablel 15 Q T 4hpkA tanhmL 2 05 150775 F 40Bm 7r 2ft Btu 1 2ft tanhl53l hrft F 12 hrft F 4 12 ll7 forl fin hr Therefore for whole rod 2 fins Q 2ii7j 234E 4 Answer hr hr lll7 11 16 A solar collector is constructed as shown on the figure below The plate that absorbs the incident solar heat flux is copper and 2mm thick and the space between the absorber plate and the glass cover plate is evacuated so there are no convective heat transfer losses The tubes for the water flow are spaced 20cm apart and the water owing in the tubes is at 50 C Because of excellent conduction the temperature of the absorber plate directly above the tubes is at the same temperature as the water For a net steadystate radiation heat flux of 900 Wm2 absorbed by the plate what is the maximum temperature on the plate Develop the differential equation for this geometry similar to what was done for a convecting fin and then solve the equation Approach thawML As was done for the convecting fin an energy balance 2 7 can be performed on a differential element in this T70 00 Tb l l L l H assembly The resulting differential equation is solved for the temperature profile i quot Assumptions I 6 2L arouel The process is steady muIlia x u 2 The conduction is one dimensional L I 3 There is no internal heat generation 39 I 9 39 v gt Q N 4 The thermal conduct1v1ty is constant chix I m M bxv Solution We assume steady onedimensional conduction in the fin with constant properties We assume the temperature at x 0 is knownTb An energy balance on an element assuming steady no work and negligible potential and kinetic energy effects iQ wndx FqHAxWTFQwndxi x0 7 0 6 x W am With A W t and simplifying we obtain E E I 2 H M q W 0 d q 0 Ax kWt dx kt Separating variables and integrating once Jd 7 qdx g Tq xcl dx kt dx kt Integrating a second time 7 w 7 w 2 jdrj qxC1dx T qxqxc2 kt 2kt The two boundary conditions are lAtthe basex0TTb 2 The centerline between two tubes is a line of symmetry so that plane is an adiabatic surface x L d1quot dx 0 Applying the two boundary conditions and solving for C1 and C2 C1 q Lkt and C2 Tb 7 n 2 w Tx q x q Lx7 2kt kt By inspection the maximum temperature will occur at the midline between two tubes x L Substituting this in and simplifying T L qquotL22kt Tb For the given data with kwppg 40lWm K from Appendix A2 7 900Wm2010m2 50 C556 c 1 Answer 2401wm K0002m 1118 11 17 Baseboard convectors using hot water are used to heat rooms in houses A common design is to attach annular fins to a horizontal tube and use natural convection to add heat to the room Consider a design that has 75mm diameter fins attached to a 25mm outside diameter tube The fins are 1mm thick and are spaced 5mm apart the tube is 3m long with a wall thickness of 2mm The water is at 55 C and has a convective heat transfer coefficient of 1250 WmZK Assume the temperature decrease of the water is small The room air is at 20 C and has a convective heat transfer coefficient of 10 WmZK Determine the heat transfer rate to the air in VJ Approach We assume this is a onedimensional steady heat transfer gt L 001 problem The heat transfer rate with constant water and air temperatures is calculated with the basic rate equation ATRm We need to evaluate all the 3 0009 thermal resistances t QOOIM T Assumptions 0 0 3le 1 1 1 The process is steady 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant Solution Assuming steady onedim ensional conduction with uniform heat transfer coefficients constant properties no radiation constant water temperature and no heat generation the heat transfer rate can be determined from w a Karate Rmbe R n The thermal resistance due to the water is Rm 7 QT 1 111 7 h27rr W 7 1250Wm2 K27r00105m3m The conduction resistance in the tube using from Appendix A2 calm 177 Wm2 K 404gtlt10393 W ln 1 00125 00105 Rm r2r1 n 523gtlt10395 27rkW 27r177Wm W On the fin side NA Rfm Where 77D 17 f 1777 771141 Am We Will use Figure 1117 to obtain the fin efficiency The total area requires the number of fins N over the total length W 7 Wis 7 3m70005m 7 7 4992 SI 0005m 0001m WNStS 2 Therefore N 499 an integer number The total area is taking into account tip area using the corrected length r2 r2 t2 00375m 0001m2 00380m Area of one fin A 27451742 270038m2 00125m2809X10393m2 Am 2an W7N1NA 27r00105m3m 7 4990001m 499809gtlt10393 m2 4202m2 Now for the parameters on Fig 1117 V 00380m 7 362 27 r1 00105m T otal Area 1119 00255m L L 0025m o39oglm AP L r 00255m0001m 255x10 5 m2 05 05 E3 00255m32 0192 kAp 177WmK255gtlt10 sz From the figure 77 z 094 Therefore the overall surface efficiency is 499809x10 3m2 77g 17T07094 0942 11 RM 0033 094210m2 Kj4202m2 55720 K Q 1190w 4 Answer 404x10 3 523gtlt10 5 002525 1120 11 18 In a heat transfer experiment the objective is to determine the convective heat transfer coefficients on three fins Each of the three fins one of which is shown below is a solid rod 15cm long with a diameter of 1 cm The first fin is a pure copper the second is 2024T6 aluminum and the third is 304 stainless steel Small electric heaters are attached to the bases of the fins and the power is adjusted such that the base temperature is 100 C when the room temperature is 25 C The measured powers are copper rod 41 W39 aluminle rod 40 W 304 SS 27 W Estimate a the convective heat transfer coefficient on each of the three fins in WmZK b the tip temperature on each fin in C Approach This is a pin fin Since the heat transfer rate and the base 0 temperature are given if we assume onedimensional 4 39 2 C conduction then we can use the equations given in Table Sam 1139 00 C L 1 15 to calculate the heat transfer coefficient and the tip D M temperature lt L OYMquot I T Assumptions 1 The process is steady 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant S olution a Assuming steady onedim ensional conduction constant properties uniform heat transfer coefficient and no radiation for a fin with an adiabatic tip cosh mL 7 x 1 cosh m L Q39TbiT hpkAxtanhmL 2 where m h pkAX 12 The geometry of all three fins is the same The thermal conductivity andh are different Equation 2 is solved for the heat transfer coefficient and then Eq 1 is solved for the tip temperature e use a corrected length to account for convection from the tip TxTf Tbin p IrD 7r001m003l4m A 7rD247r001m247r 785gtltlO395 m2 E LAXp 015m785x10 5 m2003l4m 01525m From Appendix A2 for copper k 401wm2 K aluminum k 177wm2 K 304 55 k 149wm2 K 12 hW m2 K 00314m 12 200k m391 kW mK 785gtlt105m2 k Q 1007 25Khwm2K00314mkwmK785x10 5m2ltanhm01525m For each rod we solve these two equations for the heat transfer coefficient The results are Copper h37lWm2K Aluminum h 375 Wm2 K Answers 304 ss h414Wm2K a coshm LiL bForthe t1p temperature TL25 C100725K coshm01525m For each rod Copper T L 969 C Aluminum T L 932 C 4 Answers 304 ss TL539 c Comments As you can see the fin material has a significant effect on the temperature distribution llZl 11 19 Electronic equipment is to be encased in an aluminum box Whose temperature must be limited to 60 C Vertical rectangular pure aluminum fins are to be attached to the box top to aid heat removal Air at 25 C will flow over the fins with a convective heat transfer coefficient of 50 WmZK If 10 fins spaced lcm apart are used that are 25mm long 250mm high and 2mm thick What is the heat transfer from the top of the box in W Approach Tb 2 60 C Assuming onedimensional heat conduction in these fins the rate equationQ 77D hAm AT can be used to calculate the heat transfer from this finned surface The overall surface efficiency 77a needs to be determined H1111va T 2r c V i gOWM k Assumptions 1 l The process is steady l is L 0 OZYM I l The conduction is one dimensional t 2M 50 There is no internal heat generation The thermal conductivity is constant There is no radiation 3 39 4 5 Solution Assuming steady onedim ensional constant properties uniform heat transfer coefficient and no radiation the heat transfer from this finned surface can be calculated Q77ah14mTb T We Will assume an adiabatic tip fin and use the corrected fin length E LI2 002500022 0026m The total area is ATot N2LHN1SH 1020026m025m10l001m025m 0158m2 The overall surface efficiency is NA 17 f 17 77 Am 77 The area of one fin is A 2LH 20026m025m 0013m2 From Table 115 the fin efficiency is tanhmL hp hZH 777 Where m mL kAX kIH From Appendix A2 for pure aluminum k 237Wm K 7 250Wm2K 12 7 1452 391 237WmK0002m m mL 1452m3910026m 0378 tanh0378 0954 0378 100Ol3m2 7 17 207 0954 0962 0158m Q 096250Wm2 K0158m260725K 266W 4 Answer 1122 11 20 The failure rate of integrated chips increases rapidly with higher operating temperatures The ability to pack more components into smaller areas results in higher power and therefore more severe cooling requirements One way to improve cooling is to add fins to chips to increase surface area Consider a 125 mm by 125mm chip A 4 X 4 array of pure copper circular pin fins l5mm in diameter and lOmm long is attached to the outer surface of the chip The surface has a maximum operating temperature of 75 C and the heat transfer coefficient with air cooling around the fins is 200 WmZK The design air temperature is 35 C Determine a the maximum power dissipation from the surface if no fins were used in VJ b the maximum power dissipation from the surface if the pin fins are used in W Approach T We 2 200 WM39LK For steady heat transfer the basic rate equation F 3 I L M e D 0001 quot E l M Assumptions W O 0 Z The properties are constant N 2 lug rb 7pc 2 The conduction is steady and onedimensional Q ATRm can be used to determine the maximum chip power The resistance without fins is simply lhA 39 with fins the resistance is 17711 hAm Solution a The basic heat transfer rate equation for the nofin case is 39 7 AT 7 TA T Rm 1 h A Everything is known so 39 125w Answer 200wm2 K00125m2 b With fins assuming steady onedimensional uniform heat transfer coefficient and no radiation the heat transfer rate can be calculated with Q Tb T 177 h Am Assuming an adiabatic tip and using a corrected fin length to account for convection from the tip the overall surface efficiency and fin efficiency are 12 N A tanh m L h I7 i 1 where m t 77 77 mL kA x 77 1 From Appendix A2 for copper k 401 Wm K Evaluating the parameters in the above equations A 7rD24 7r00015m24 1767x10 6 m2 p IrD 7r00015m 4712x10 3 m Lquot LAp LD4 001m00015m4 001038m Area of one fin A 7rDLquot 7r00015m001038m 489x10 5 In2 Total area Am W2 N7rDL 00125m2 167r00015 m001038m 939x10 m2 05 200 2 j4712gtlt10393m m K 71 m 3647m 401l 1767x10396m2 mK mt 3647m391001038m0379 1123 Fin efficiency 7 tanh0379 77 7 0379 Overall surface efficiency 16489x10 5m2 0955 17 170955 0962 77 939x10 m2 Heat transfer rate 75 7 35 K Q 1 723W Answer 0962200Wm2 K939x10 4m2 Comment39 The addition of fins increased the heat transfer by 480 This demonstrates well the effect of extended surfaces 1124 11 21 Air flows over a plane wall with a convective heat transfer coefficient of 40 WmZK Insufficient heat transfer is obtained from this situation so aluminum alloy fins alloy 2024T6 of rectangular profile are attached to the plane wall The fins are 50 mm long 05 mm thick and are equally spaced at a distance of 4 mm 250 finsm With the fins the convective heat transfer coefficient is reduced to 30 WmZK What percentage increase in heat transfer is obtained with the fins compared to the plane wall arrangement without fins Approach We can calculate the relative increase in heat transfer with T t 2 0 OOOS M fins compared to without fins by using the appropriate rate gtl 39amp equations The fin efficiency must be taken into account Only the area associated with one fin needs to be taken into account 0081 Assumptions l The properties are constant l9gtl T S 0004 M 39 b w 39 01W l9 P975 Solution The heat transfer rate for the surface without fins can be obtained for a unit cell39 that is a representative area that encompasses the same base area as associated with one fin is used Qwa hwa ATb 39T hwa SW0 39T For the surface with fins we use Eq 1194 NA Qw77hAmTpT1A1 177hAmET a The relative increase in heat transfer is NA Q if A I39h lth Tb 39T Qwa hwa SW0 39T Note that the temperature different cancels The fin efficiency must be calculated Assuming a rectangular fin with an adiabatic tip using adjusted lengths and obtaining the thermal conductivity from Appendix A2 L L 12 0050m 00005m2 005025m W 12 2 30w m2K W m hp MW 2h 2604m1 a mL1308 kAX kWt kt 177wmK00005m 7 tanhmL 7 tanhl308 77 7 mL 7 308 The other parameters in the equations are N l A 2W Ab WSil Am Ab NA Substituting all these expressions into the heat transfer ratio and canceling terms we obtain 1vlhwltsirgt2fl 066 12L Q 7 S42Lquot Qwa hWDS 2 005025 17A07066 30 2 00047000052005025 7 00047000052005025 m K 40 WKj0004m m2 l3lO or 1210increase Answer 1125 11 22 Motorcycle engines often are aircooled with annular fins attached to the cylinder head Consider a cylinder k 55 WmK with an inside diameter of lOOmm and a wall thickness of 6mm Over the cylinder a fin assembly is interference fit so that there is negligible contact resistance between the cylinder and the fin assembly The fin assembly k 230 WmK has a base thickness of 4mm39 the 6 fins are 25 mm long 2mm thick and spaced 2mm apart A heat flux assumed constant of 105 Wm2 is imposed at the inside surface of the cylinder Air outside the engine is at 30 C with a convective heat transfer coefficient of 100 WmZK Determine a cylinder wall temperature in C b the interface temperature between the cylinder and the base of the fin assembly in C c the fin base temperature in C Approach We assume this is a onedimensional heat transfer problem With the imposed heat flux the temperatures can be calculated with the basic rate equation Q ATRm We need to evaluate all the series thermal resistances Assumptions 1 The properties are constant 2 The conduction is steady and onedimensional 3 Radiation is ignored 3 vam i 1 GOESMa Solution a Assuming steady onedim ensional conduction with uniform heat transfer coefficient constant properties no radiation no resistance between the cylinder wall and fin assembly the rate equation from the inside cylinder wall to the air is Q 7 AT 7 T1 T Rm R4a11Rba52 R ns Solving for the inside wall temperature T1 T Rwa Rba52 R ns Evaluating each term The total heightH is H Nsts 600020002m 0002m 0026m Q q A1 q 27rr1H105Wm227r0050m0026m8168W ln ln 0056 0050 RM r2r1 00126 27mm H 27r55Wm K0026m W ln ln 0060 0056 Rim r3r2 000184 27mg H 27r230Wm K0026m w For the fin side NA RM where 771117 1777 77 Mm Am We will use Figure 1117 to obtain the fin efficiency The total area is Am 2an H 7 NINA rm r4 r2 008500022 0086m Area of one fin A 27rrfc 7 r32 27r0086m2 70060m2 002385m2 Total area Am 27r0060m0026m 760002m6002385m2 01484m2 Now for the parameters in Fig 1117 1126 447 V 0060 L Lt2 00857060000220026m AP m 0026m0002m52gtlt10395 m2 l43 05 05 2 L h 0026m32 w 0469 kAp 230WmK52gtltlO m From Fig lll777 084 The overall surface efficiency is 6002385m2 77a 17 2170840846 01484m 1 K R 00797 0846100wm2 K01484m2 w Finally the inside wall temperature is T1 308168W0012600018400797 1069 C Answer Because the total heat transfer rate is the same through all resistances b At the interface between cylinder and fin assembly T2 30816800018400797 966 c d Answer c At the base of the fins T3 30816800797 951 c4 Answer Commen Note that if no fin assembly were placed over the cylinder wall 1 T1 TQ HUM 1 Zn K H h 27r r2 H l T13O C8168W 00126 933 c W lOOWm2K27r0056m0026m 1127 11 23 A densely populated circuit board has 113 electronic devices attached to it Forty of the devices dissipate 03 W each 30 dissipate 02 W each and the rest dissipate 015 W each Because this is a critical installation the circuit board 3mm thick is constructed of a high thermal conductivity material k 156 WmK so that the heat from the devices is spread out evenly over the 10cm by 20cm area of the board The backside of the board is cooled with air at 25 C with a heat transfer coefficient of 25 WmZK To obtain a worsecase estimate no heat transfer credit is taken for convection from the device side of the circuit board Determine a the temperature of the surface of the circuit board cooled by air in C b the temperature of the surface of the circuit board cooledby air if 500 pin fins 02cm diameter and 2cm long are attached to an aluminum plate 2mm thick k 177 WmK that is epoxied to the circuit board The epoxy k 2 WmK is 01mm thick APPI OMhi w Loaf hawk1 c 1r mic 4 jg 45an 3Q Assum1ng steady onedim ens1onal heat 1 4 3 or 0 transfer the basic rate equation L gt 1 139 093331 Q ATRm can be used to calculate 45 0007 epoxj the surface temperature of the circuit 15139 com boar T CB J13Omml L 001m v MM 2 A t Nz oo s WW Pquot I ssump lOnS 1 The properties are constant 2 Heat transfer is one dimensional S olution Assuming steady onedim ensional heat transfer with uniform heat transfer coefficient constant properties and no radiation the governing equation is Q39 7 AT 7 TUB T T Run Run and Q 4oo3w3002w43015 245w a Without fins the equation to solve is T I 1 Q TC3TQ 1 l r kl ACE 171 ACE k1 AC8 h lACB r 0003m K 000962 k1 AC 156Wm K01m02m w 1 1 2005 171153 25Wm2 K01m02m W TUB 25 C245W000962200 742 C Answer b With fins we have four resistances to take into account 7 TC 7T Q 7 11 12 t3 1 k1 AC8 k2 AC8 k3 AC8 77 hAtot 2 00001m 000255 k2 AC3 2Wm K01m02m W I3 00005653 kEACB 177Wm K01m02m w For the fin resistance assuming an adiabatic tip and using a corrected fin length to account for convection from the tip the overall surface efficiency and fin efficiency are 1128 iNA 12 tanh m L h p l 17 Where and 77 Am 77 77 m L kAX Evaluating the parameters in the above equations 2 0002 2 AK 3142x106 m2 4 4 p IrD 7r0002m 6283gtlt10393 m L LAXp LD4 00200024 00205m Area of one fin A nDLquot 7r0002m00205m 1288x10quot m2 Total area Amt WSN7rDL 01m02m5007r0002m00205m 844gtlt10 2 m2 12 25w m2 K 6283x10 3m gt mm l77Wm K3l42gtlt10396m2 mL 1681mquot00205m 03445 tanh03445 Fin eff1c1ency 09622 03445 5001288X10 m2 Overall surface eff1c1ency 170962 0971 77 844X10392m2 1 1 7 E 77 I240 097125Wm2 K844X102m2 7 0488 W TUB 25 C245w0009620002500005650488373 C Answer Comment As can be seen the addition of fins dramatically reduces the circuit board temperature Care must be taken to minimize contact resistance 1129 11 24 In many new homes hot water pipes are encased in the floors to provide uniform heating in the winter Consider 25cm hot water pipes located at the midplane of a lOcm thick concrete floor k ll WmK The pipes are spaced 20cm apart The air temperature is 23 C and its convective heat transfer coefficient is 10 WmZK The water is at 60 C and assume the heat transfer coefficient inside the pipes is very high Determine a the heat transfer rate per unit length of the pipe in Wm b the surface temperature of the concrete in C This is a steady conduction problem The twodim ensional heat transfer in the concrete will be treated as onedimensional through the use of the conduction shape factor S The basic rate equation ATRm is used to determine the heat transfer rate and the Approach 002 o If W E1JCMl0WHc outside wall temperature CONN 51quot TN ooc F The F Assumptions 0 lt Q l The process is steady rA Emmi T01 NM 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant Solution a Assumin steady constant properties and the water side heat transfer coefficient is very large zero thermal resistance the circuit diagram of this geometry is shown a ove The conduction resistance isR lkS From Table 111 for this geometry the shape factor is Cami 27r L lltajsmhr i This is the conduction resistance for heat transfer out both walls Because of symmetry the heat transfer is the same out of the top and bottom surfaces Therefore T 7T w Q 7 1 1 hA kS where A ZxL is representative of every pipe 27rL 2020m 7r005m 1nH7r0025msmhi 020m Therefore the heat transfer per unit length per tube is 4225L Q 7 60 7 23K 7 E L 7 1 1 7 795 m lowm2 K2o20m 11WmK4225 b The wall temperature T5 can be calculated with 39 39 L 2hAT57T gt T5T L L hAL T5 23 c 79395Wm 429 c 4 Answer lowm2 K2o20m 1130 11 25 Liquid nitrogen is stored in a 3m sphere buried in the earth k 017 WmK with its center 4m below the surface A lOcm thick layer of insulation k 005 WmK covers the sphere The nitrogen is at 7180 C and the surface of the earth is at 10 C Nitrogen vaporizes and the vapor is vented to the surface because of heat transfer from the earth to the tank thus maintaining a constant temperature and pressure in the nitrogen The enthalpy of vaporization of the nitrogen is 1986 kJkg Determine a the heat transfer rate to the nitrogen in VJ b the nitrogen vaporization rate in kghr Approach V N l39 1 00C This is a steady conduction problem The twodimensional L heat transfer through the earth will be treated as one 2 4 M dimensional by using a conduction shape factor The heat transfer rate is determined with the basic rate equation Q ATRt an energy and mass balance on the nitrogen t 3 OJM The vaporization rate can be determined from at39 TNL 480 C Assumptions The process is steady 15 111 2 The conduction is one dimensional gt W 3 There is no internal heat generation Q Raf e walnhaw 4 The thermal conductivity is constant 5 The system does no work and potential and kinetic energy effects are negligible Solution a Assuming steady constant properties and onedim ensional heat transfer except through the earth which is handled by a conduction shape factor the heat transfer rate can be calculated with AT TsrTms TrTms 7 Rm 7 REth Rms 7 1 D2 7 D1 kEmth S 4D1 D2 7 km From Table 111 for a sphere in a semiinfinite solid 7 2713 7 27r320m 7172717 320m 42 44m 1077180K 7 Q 1 320m7300m 711W Answer 017wm K251m l 43m320m7r005Wm K S 25lm b The nitrogen vaporization rate is determined from an energy and mass balance on the nitrogen Defining the control volume as shown and assuming no work and negligible potential and kinetic energy effects the mass and energy balance equations result in q 7 711w h k 358gtlt1073k g 4 Answer g 19861 10001 s kg lkI ll3l 11 26 Hot water is pumped between two buildings in an office complex The 2in inside diameter pipe is buried 15 ft below the earth s surface k 010 Btuhrft F Insulation lin thick k 002 Btuhrft F covers the carbon steel pipe whose wall thickness is 025 in The water has a convective heat transfer coefficient of 2350 Btuhrft2 F lfthe soil surface temperature is 10 F and the water enters the pipe at 170 F determine the initial heat transfer per unit length in Btuhrft when the water enters the pipe Approach E This is a steady conduction problem The tquot quot 39 heat transfer through the earth will be treated as one T 7on 1 dimensional by using a conduction shape factor The heat 2 l L 23mg transfer rate is determined with the basic rate equation 39D 2 N atF QATR V t1 l w Assumptlons R l The process is steady in Z 025 IN 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant Solution Assuming steady constant properties and onedimensional conduction except for that in the earth the governing equation is AT T1 7 T5 Q and Rcand3 mt raw camil The convective resistance is R L l 000081hrft F hA 2350Btuhr ft F7r0167 ftL L Btu For carbon steel pipe from Appendix BZ k1 35Btu hr ft F so across the steel pipe R lnD2D1 ln02080167 000100 hr ft F W 2 k1 L 27r35Btuhr ft FL L Btu Across the insulation lnD3D2 n03750208 7 4690 h ft F R E Z kzL 27002Btuhrft L Btu Through the earth from Table 111 with 2 gt D 2 L 2266 L 7 27rL ln4zD 1n4150375 7 4413 hrft F 1 1 R m kS 010Btuhrft F2266L L Btu Substituting in each resistance and dividing outL 39 170710 F Q 176E d Answer 000081000100469044l3g hIft u Comments The convective resistance and the conduction resistance through the steel pipe are negligible in this case The two other resistances dominate the problem 1132 11 27 A very long electrical conductor is buried in a large trench filled with sand k 003WmK to a centerline depth of 05 m The conductor has an outer diameter of 25 mm and the current flow and resistance of the cable cause a dissipation of l W per meter of length The conductor is covered with an insulating sleeve of thickness 3 mm with k 001 WmK At the surface a 25 C wind blows such that the heat transfer coefficient is 75 WmZK Determine the temperature at the interface between the conductor and the insulating sleeve Note you will need to make an assumption about the convective resistance and you will need to justify it Approach 1 1Pc hamWit This is a multidimensional conduction problem Assuming i we can solve this as a onedimensional problem with the 4 EW gt appropriate conduction shape factor the interface 21039 I t 0003M temperature can be determined with this basic rate equation L DI 0073 QkSAT lLl wM Assumptions 1 The properties are constant 639 T4 2 The conduction is steady and multidimensional amp W 4 39 which 0 MA L5 W 1 Solution The basic heat transfer rate equation is AT L R 1nD2D1 I 1 2 kL km s M For the conduction shape factor from Table 111 with 2 gtgt D S 27rL ln42D For the conduction resistance A LW where W is some width on the surface Because there is no obvious length to use we will use 1 m and check the magnitude of the convective resistance Substituting these expressions into the main equation Q lnD2D1 ln42D l 27rk L 27rkmd L hLW m5 lnD2Dl7 ln003l0025 342m KW 27ch L 7 27r001Wm KL L ln4zD 7 ln405m003lm 7 2211m KW 2mm L 27r003Wm KL L 1 7 1 7 00133m KW hLW 75Wm2KlmL L Note that the convective resistance is negligible compared to the other two resistances so an exact evaluation of W is unimportant Factoring outL and solving for T1 39 ln D D ln 42 D l T1Tg lt 21 lt gt L 27r k 27r k h W In sand W 25 c1 j342 221100133 505 C 4 Answer 11 1133 11 28 Your next door neighbor decides to construct a small underground room in his backyard The room will be 8ft tall and lZft square He will construct it of concrete and bury it under 2ft of earth k 062 Btuhrft F He wants to buy a heater that will maintain the room at 65 F when the outdoor temperature is 0 F and asked you to tell him how big the heater should be Determine a the steadystate heat transfer rate from the room in Btuhr b the steadystate heat transfer rate if 5in of insulation k 002 Btuhrft F is added to the outside of the room in Btuhr Approach T T This is a multidimensional steady conduction problem 2 Using the rate equation ATRm and a conduction shape factor the heat transfer rate can be determined Assumptions 1 The process is steady 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant Solution a Assuming constant properties uniform block temperature and multidim ensional conduction the heat transfer rate is 39 7 T1 7 T2 Q 1 earth From Table 111 the conduction shape factor is Z 70 59 b 0078 2 70 59 8 0078 S2756a lnl j 275612ft lnl j lllft L Z 12 2 6570 F Q 44801Eu 4 Answer 062Btuhr ft F111ft b If insulation is added to all the inside surfaces then the heat transfer rate equation has a second term 39 7 1 2 Q 7 l Ax 77 km S A where Ax 4in 0333ft A 4bL2La 48ft12ft212ft12ft 672ft2 657 0 F Q i 71650Eu Answer 1 033m hr 062Btuhr ft F111ft 002Btuhr ft F672ft2 Comments The thermal resistance of the concrete k 52Btu hr ft T is negligible compared to the other two resistances so that resistance has been ignored 1134 11 29 Molds for plastics and other materials sometimes are heated by the insertion of electrical resistance heaters at appropriate locations on the body Consider a lSOmm long 125 mm diameter 100 W electric heater inserted into a hole drilled perpendicular to the surface of a large mold whose thermal conductivity is 10 WmK The surface temperature of the mold is maintained at 25 C a long way from the heater estimate the steadystate temperature of the heater Approach This is a steady conduction problem in two dimensions We will assume we can treat this as a onedimensional problem through the use of a conduction shape factor kS AT Assumptions 1 The properties are constant 2 The conduction is steady and twodim ensional Solution Assuming steady with constant properties and a mold that is large relative to the heater depth the heater temperature assumed uniform can be determined from QkSTH7Tm 2 HTmkQ S From Table 111 for a vertical cylinder in a semiinfinite solid 27rL 7 27r015m mlliml l The temperature of the heater is TH 25 cvgo 661 C 4 Answer lO j0243m m K 1135 11 30 A thin electronic component 20mm square is epoxied to a large 2024T6 aluminum heat sink The thermal resistance of the epoxy is 035 X 10394 m KW The temperature of the aluminum block far away from the electronic component is 25 C The top of the component is swept by air at 25 C with a convective heat transfer coefficient of 50 WmZK If the maximum component temperature is 75 C what is its maximum allowable operating power in VJ Approach 1 2939 L wM LK An energy balance is performed on the chip to calculate the maximum chip power For the heat transfer rates we W assume the basic rate equationQ ATRm can be used m 01 T 39Y El of The conduction into the heat sink is evaluated through the use of a conduction shape factor or Assumptions quot53965 Ii The properties are constant K T 39 W 2 The conduction is stead and onedimensional y It To Tb WWW a k 39 C9914 QmL lA 1 quxy RS Hk W Solution For a closed system control volume around only the chip and assuming steady state conservation of energy gives gin0 WQQka Using the circuit diagram and assuming onedimensional heat transfer with constant properties ignoring heat transfer from sides of chip T 7T T 7T n 7 r f 7 smk W 1 R hA A kS From Appendix AZ for aluminum k l77 Wm K 39L 7r002m From Table 111 for a thin rectangular plate on the surface S 00453m ln4LL ln4 757 25 c 757 25 c W 1 035gtlt104 m2 KW 1 sowm2 K002m2 oosz 177wm K00453m 7 710W7 2356W 72366W Answer Comment The minus sign is because power is added to the chip which is consistent with the sign convention we adopted for work in the conservation of energy 1136 11 31 You are late preparing for a party39 the soft drinks should have been placed into the refrigerator much sooner It is only 2 2 hours until the party When the party has started will the drinks have reached 5 C The can is 6cm in diameter and lZcm high and is initially at 20 C The air in the refrigerator is at 0 C and you estimate h S 4 WmZK Assume the properties of the drink can be approximated as those of water and the can s contribution to the transient is negligible Determine a the estimated time required for the drink to reach 5 C b a method to speed up the cooling process Approach We will assume initially that the lumped systems analysis method is valid We check the Biot number and then proceed from there Assumptions 1 The properties are constant 2 There is no internal heat generation 3 The metal can contributes little to the process 4 The drink s properties are assumed equal to those of water Solution a To determine if a lumped systems analysis is valid we need to evaluate the Biot number Bi hLEhm k The characteristic length is V 7 H7rD24 7 012m006m2 A 7 IrDH 27rD24 7 4006m012m2006m2 Assuming the drinks properties are the same as water and are constant and the metal can contributes little to the process the water properties from Appendix A6 at 5 202 125 C are p 9993kgm3 0P 4191kJkg K k 059lWm K 4wm2 K0012m Bi 0081 lt 01 059lWmK L char 0012m TxiT T 17 T 7h A The lumped systems method rs valrd The govemrng equatron rs exp I Solving for I and letting VA L I ipcz1Lchm1nTiT hm Yiin 79993k g3j4l9lk1lt K0012mlOOO j 5 0 m g L l7400s484hr Answer 4 w k 70 m2K Based on this time the drinks do not cool down fast enough b If we raise the convective heat transfer coefficient the time required will decrease Placing the cans in an ice water bath does not change the outside temperature but h increases significantly Because time is inversely proportional to h an increase in h going from natural convection in air to water would decrease the time 1137 11 32 You are designing a radiant energy test facility and need to determine how long it will take a test specimen to reach a steady temperature Initially a 3cm thick brass plate is at a uniform temperature of 50 C At time zero one side of the plate is exposed to a radiant heat flux of 6000 Wm2 and the other side is exposed to air at 20 C with h 75 WmZK Determine a the steadystate temperature of the plate in C b the temperature of the plate 15 min after the start of heating in C c the time to reach steady state if steady state is when 99 of the difference between the final plate temperature and the air temperature is achieved in s Approach The first part is solved simply by applying the rate equation hAAT The second part is a transient conduction problem The Biot number should be checked first to determine if the lumped systems method can be used If it cannot then a new analysis will need to be developed for this system Assumptions 1 The properties are constant 2 The heat transfer coefficient is uniform and constant Solution a The steadystate rate equation is QhATS7T T5TQhATq h 2 T5 mamdowc 4 Answer 75Wm K b For the transient problem first evaluate the Biot number laugh Where Lchm KEL k A A From Appendix AZ for brass p 8530kgm3 cP 380 JkgK k llOWm K 75wm2 K0o3m 1 10 Wm K Therefore a lumped systems analysis is valid Using the analysis given in Example ll5 TiT WhA h exp 7 1 T 7 T Wh A m 0P where m pV pLA and W iq A where the minus sign is used because heat is into the system Bi 002 lt 01 Substituting these into the main equation and solving for T T T q Z 7T iiexp 7L h h chP 7 75w m2K 900s 50Zowltexp A 75 8530kgm3003m3801kgK c When T 7T 099T in t 7pLCP ln TTT Tq h h I 7T 7 qquoth 8530kgm3003m380 Ikg K1 f 792 600075 75 wm2 K 507 207600075 750 C Answer 2 20 6000Wm 75Wm2K 099100 7 20K 792K Solving the main equation for time steady 5360s l49hr Answer 1138 11 33 In some lumped systems different parts of the surface may be exposed to different conditions Consider the exhaust pipe of an automobile engine Just before the engine starts the exhaust pipe is at a uniform temperature T When the engine starts at I 0 exhaust gases at T g assumed constant with time flow through the tube with a convective heat transfer coefficient of hg Outside the pipe the air is at T7 with a convective heat transfer coefficient of h Using a lumped system analysis develop an expression for the pipe temperature as a function of time Approach We solve this problem similar to Example ll5 While there iTG Lg is no power addition there are two heat transfer rates that must be taken into account 39 Assumptions f l The properties are constant 2 The heat transfer coefficient is uniform and constant W 3 The lumped systems approach is valid 13 gtTq 0 Solution We define a control volume that encompasses just the exhaust pipe not including the uids For this closed system we assume no potential or kinetic energy effects or work so Assuming the pipe is an ideal solid with constant specific heat so that du 0 d1quot cP dT and describing the heat transfer in terms of convective heat transfer only d1quot mop 7h A T4 7hg A TiTg Rearranging this equation in preparation to separating the variables mpg TH A 7h Agh A Tthg T 7h A thgT 7hATthng dt h A h A Lt hAhA d h A T thng tht W T e an so a mo 7 7 71 a a g g 72 ha A hg Ag P dt 71 72 Separating variables and integrating Vi 71dl A m m 11 ZT772 0mop 7772 ch Exponentiating both sides exp l Answer 7 7 72 m CP Comments Note that ifh A hg Ag andT Tg then 71 ZhA hAm and 72 T TE 2 T Substituting these into the equation and simplifying the equation collapses to the original lumped systems analysis equation T i T h A exp 7 I Tx 7 T c 1139 11 34 Steel balls with a diameter of 2 cm are annealed by heating them uniformly to 950 C and then cooling them to 125 C in air at 35 C The convective heat transfer coefficient is 25 WmZK For 347 stainless steel balls determine the time required for the cooling process in s A roach ITllfis is a transient cooling problem First we check the TOE 113 C Biot number If it is small enough then we will use the 72 3 lumped systems method If lumped systems is not valid 5 then we will use a onedimensional transient solution k T 2 M Assumptions 1 The properties are constant 2 The heat transfer coefficient is uniform Solution The Biot number is 3 Bi hLW39 with LW 1 D 6 2 k A IrD 6 From Appendix AZ for 347 stainless steel at Tavg 9501252 5375 C p 7978kgm3 k 215Wm K 0P 585JkgK E 25Wm2K002m6 1 219WmK So the lumped systems approach is valid and the governing equation is T I 7T 7 L exp t 7 7T chp Solving for time and letting V A L I ipcz1Lchm1nTiT 00038 lt 0l D6 hm T T 7 77978kgm3585 Ikg K002m6 ZSW 2K lnl440s 4 Answer m ll40 11 35 During the startup of a new natural gas furnace the test engineers want to monitor the exhaust gas temperature to assess the unit s performance A 1mm diameter copper constantan thermocouple is used to make the temperature measurement Before they use the thermocouple they want to determine its response characteristics and estimate the heat transfer coefficient on the thermocouple They develop the following experiment Initially the thermocouple is at 25 C They insert it quickly into a gas stream that is at 200 C and has the same velocity as the stream they want to monitor In 83 s the thermocouple reads 199 C Determine the heat transfer coefficient in WmZK Assume the properties of the thermocouple are the same as copper Approach Ignoring radiation and assuming the heat transfer coefficient T 2 100 C is low enough so that the lumped system approach is valid 39c we use the lumped system approach to calculate the heat gt D 0 M transfer coefficient Once it is calculated we need to gt T 39 calculate the Biot number to check the validity of the gt 39 assumption 2 Te3s1l i c Assumptlons 1 The properties are constant Solution The governing equation for the lumped system solution is TltIgt7Tj hA i i h i exp 7 I exp i I 7 T chP pop Lchar 2 where LEM 7 K D 46 Solving forh A IrD 6 h CPDln H 61 7in For copper from Appendix A2 p 8933kgm3 cP 385 Jkg K k 401 Wm K 3 I 78933kgm 385 Jkg K0001mLI199 200 357 W Answer 683s L257200 m2K Now checking the Biot number MW 357Wm2K0001m6 Bl 401WmK Therefore because the Biot number is less than 01 the approach is valid 000015 lt 01 1141 11 36 An electronic device that dissipates 30 W is used infrequently Its maximum allowable operating temperature is limited to 65 C39 as soon as 65 C is reached the device must be shut off The device and an attached heat sink have a combined mass of 025 kg surface area of 563 cm2 and an effective specific heat of 800 JkgK If the device is initially at a uniform temperature of 25 C in air at 25 C with a heat transfer coefficient of 10 WmZK determine a the steadystate operating temperature in C b the time required to reach the maximum operating temperature in s c if a heat sink is to be added to the device so that the operating time is to be doubled what additional mass and area are needed Assume the mass to area ratio of the addedmaterial is the same as that of the original device Approach W 2 30w Because we know little about the shape of this device and we T efJC do not know its effective thermal conductivity we are forced MA to assume that the lumped system approach is valid We use M 1 013139 the equation developed in Example 115 to find the required A 313 CM time c 3303 Assumptions 75 l The properties are constant T IY JC L H WMLK 2 The lumped systems approach is valid 6 Solution Assuming the lumped system analysis is valid the solution to a transient problem with internal heat generation is given in Example ll5 TiTJrWhA hA exp 7 I 7in WhA m cP With 7 T and rearranging the equation T 7 T 1 17 exp 7h A t h A m cP a The steady state temperature occurs when tgt oo so using the above equation 39 30w 100 2 T T 71 25 ww sm c Answer h lowm2 K563cm2 b Solving the above equation for time I 71 TiThA 7025kg800Jkg KlOOcmlm2 hA W lOWmZK563cm2 HP 1 65725K10Wm2i335y63cm2lmlOOcm2 2775 Answer C 563cm2 m2 2 c We want I 2277s 554s From the given data 3 444 kg m Let m CA and substituting into the governing equation T 7T ig i expE hmt m m 0 Solving for m the total mass required to double the operating time 7 WC 4 7 7730W444kgm2 l 7lOWm2 K554s 7 mi liexpC Hilowm2K65725KLliexp 70481kg 444kgm2 800 Ikg K So additional mass is 04727 025kg 0222kg 4 Answer Additional area is 2 444k 2 7563cm2520cm2 4 Answer g m m 1142 11 37 To obtain a hard surface on a metal plate with a somewhat softer core for toughness heat treatment is required Often the hot metal plate is plunged into a much colder bath so that the surface temperature drops very rapidly The rapid temperature drop results in a favorable grain structure that causes a very hard surface material The slower cooling of the core of the plate results in a softer core material Consider a large stainless steel slab initially at a uniform temperature of 1250 F It is to be plunged into a heat treatment bath which is at 150 F The research engineers state that from their studies the correct surface hardness will be reached if the temperature at a depth of 02 in in the slab reaches 700 F in less than 35 sec The steel has a thermal diffusivity of on 0135 ftZhr and a thermal conductivity ofk 7 Btuhrft F Determine a the time required for the temperature to reach 700 F at a depth of 02 in if you assume a very large heat transfer coefficient b the time required for the temperature to reach 700 F at a depth of 02 in if the heat transfer coefficient is 250 BtuhrftZF Approach We are told this is a large slab no thickness is given and the T I500 l1 D F location of interest is close to the surface so this transient conduction problem can be treated as a semiinfinite solid For part a with the very large heat transfer coefficient we have a 5 L gt D step change in the surface temperature Part b must be solved k T3700 Miquot x 011M with a finite heat transfer coefficient Assumptions gt X l The properties are constant 2 The conduction is onedimensional in a semiinfinite solid Solution a The temperature transient in a semiinfinites solid with a step change in surface temperature is T xt 7 T 1 7 erfc x TrT 2J5 7007150 00162ft l 7 erfc 12 1250450 20135ft2hrt The only unknown is time I so solving this equation using information given in Table 114 or appropriate software 1 000226hr 81s 4 Answer b For a step change in surface convection the governing equation is T 7 T 2 W I l7erfc x exp h in erfc x a 7 7 T 2 at k k 2szt k Substituting in what is known x 7 00167ft 7 002273 hx7 250Btuhr ft2 F00167 ft 2J5 20135f121mv2 112 k 7Btuhr ft F Substituting in the known values 0595 Btu ft2 hzm725013mhr ft2 F2 0135 0211 7172 2t ME 7 250m 02 FIO39135Et 112 2 1312 2 k 7Btuhr ft k Btu j hrft F 002273 002273 05l7erfc IV exp05951722zerfc 12 Haw2 Solving with appropriate software t 00132hr 475s Answer 1143 11 38 Heat treating furnaces are heavily insulated on their outer surfaces and are lined with fireclay bricks that are 10cm thick Initially the bricks are at a uniform temperature of 25 C After startup the furnace walls are exposed to 1300 C gases with a combined convectiveradiative heat transfer coefficient of 50 WmZK Determine the temperature in C of a the outer surface of the bricks after 30 min b the outer surface of the bricks after 4 hrs c the temperature next to the insulation after 30 min and 4 hr Approach Because the outer wall is heavily insulated this system is equivalent to a symmetrically heated wall with thickness 2L in the symmetrically heated wall no heat crosses the 1 2 I 300 INSUfA I J centerline of the wall Brick is a poor conductor so we will analyze this transient conduction problem as a one In dimensional transient in an infinite plane wall MLK A lt gt l ssumptlons L O I OM l The properties are constant 39 2 The conduction is onedimensional in an infinite wall Solution The governing equation for onedimensional transient conduction in an infinite plane wall is T x I 7 T 2 x Yin Clexp Al rcos AIL rgt 02 We wantTL I so we need to calculate Bi and obtain A1 and C1 from Table 1172 From Appendix A3 for fireclay brick p 2645kgm3 cP 96OJkgK k l5Wm K ando kpcp 591gtlt10 7 mZs hL sowm2 K010m k 15WmK By interpolation Al 12165 and C1 12164 Bi 333 m 591x10 7m2s1800s a For the outer surface temperature at 30 min I 2 2 01064 L 010m This does not quite satisfy the requirement for the one term approximation but Figure 117 will not give any better results so we will use the oneterm approximation with the understanding of considerable uncertainty TLI1300 C2571300K12165exp712165201064cos12165 840 C 4 Answer 591x10 7 14 400 b For the outer surface temperature at 4 hr 14400 s 1 2 010 08510 for which I gt 02 T 1300257130012165exp7121612 08510cos12165 1147 C d Answel c For the temperature next to the insulation after 30 min T 1300257130012165exp7121652 01064cos0 725 C This is physically impossible The result is a consequence of violating the restriction that the oneterm approximation is valid when r gt 02 In this situation we can conclude that after 30 minutes the wall temperature next to the insulation is still probably about 25 C For the temperature next to the insulation after 4 hours T 1300257130012165exp7121652 08170cos0 837 C Answer Comments The oneterm approximation was used even though parts of the solution did not meet the validity criterion Hence some of the answers must be interpreted and used with caution 1144 11 39 Hot metals are quenched in cold uids to change material properties Consider a long 75cm diameter cylinder of 316 stainless steel that is taken out of a furnace at 500 C and plunged into a cold bath at 25 C The convective heat transfer coefficient is 1000 WmZK Determine a the centerline temperature of the cylinder 90 s after it is quenched in C b the surface temperature of the cylinder 5 min after it is quenched in C c the time required for the centerline temperature to reach 50 C in s Approach 9 This is a transient conduction problem We first check the C D O 39 07 s M Biot number to determine if the lumped systems method is valid If not because the cylinder is very long we can treat this as onedim ensional transient in an infinite cylinder S Assumptions I 3 293C T 3 I b S l The properties are constant 1 C 2 The conduction is onedimensional in an infinite wall k 2 M I Solution L D24 The Biot number is defined as B1 thhm k where LEM VA L D4 From Appendix AZ It for 316 stainless steel assumingTavg m 600Kp 8238kgm2 cP 550Ikg K k 183Wm K or kpcp 404gtlt10396 mZs E 1000wm2K0075m4 1 183 Wm K So the lumped systems approach not valid and a onedim ensional transient analysis is used a For an infinite cylinder the governing equation for the temperature transient is T 1517 T f 2 r cex 7 1 rgt02 T 7 T 1 p 11 a 11 Va 1 l02gt0l We want to determine T 0 90s Recalculating Bi so we can obtain by interpolation Al and C1 from Table 11 2 1000 0075 2 Bih r 205 a 111609 C113424 K 183 404gtlt10396m2 s 90s With JD 0 1 and r if 05 0258 r 0075m2 This is greater than 02 so the oneterm approximation is applicable Trt 25 C5007 25Kl3424exp7l6092 0258 3515 C Answel b Using the same equation for time of 5 min 404x10 6 300s 1 0L 0362 r 00752 From Table 113 by interpolation JD 11 rra JD 11 rDra 0450 Tra300s 25 C500725Kl3424exp7l6092 08620450 558 c 4 Answer c Solving the general equation for r withJa 0 l T liT 1L21n 7 f 1 21ml 50 25 1251 A PTC 1609 50072513424 2 1251 0075 22 11 3 Iiwm Answer r o 404gtlt10 m s 1145 11 40 An 8cmdiameter potato 1100 kgm3 cp 3900 JkgK k 06 WmK initially at a uniform temperature of 25 C is baked in an oven at 170 C until a temperature sensor inserted to the center of the potato indicates a temperature of 70 C The potato is then taken out of the oven and is wrapped into thick towels so that no heat is lost from the potato Assume the heat transfer coefficient in the oven to be 25 WmZK Determine a how long the potato is baked in min b the surface temperature of the potato before it is wrapped in the towel in C c the final equilibrium temperature of the potato in C Approach This is a transient conduction problem We assume the T potato is spherical has constant properties and radiation is 4 not present We check the Biot number to determine if the aw lumped system approach is valid If not we can solve this lquot 2 a as a onedim ensional transient in a sphere h V Ta rw Assumptions 1 The properties are constant 2 Conduction is onedimensional transient in a sphere 3 Moisture is ignored in the potato Solution 2 The Biot number is Bi thhm k where Lchm K 46 2 A IrD 6 25W m2K 008m 6 Bi M 056 gtN 01 06 Wm K So a lumped systems approach is not valid a We will assume that the onedim ensional transient solution for a sphere is applicable Assuming constant properties the governing equation using the oneterm approximation is T rl 7T 39 C1 exp7 l2 am when r gtN 02 7 T 11 77 The criterion 1 gtN 02needs to be checked At the centerline r 0 and m 1 Solving for r 06 A VVa r LJ hk ka To evaluate Al and C1 in Table 112 we need the Biot number BihraW k 06 1667 By interpolation Al 1878 C1 1411 1 70 170 21n 1378 2571701411 So the oneterm approximation is valid Therefore 2 0203 004 2 r i W 2322s 387min 4 Answer a 140gtlt10 7 m s b The surface temperature is 0203 gtN 02 TT 1 Tfclexpi 12 Trim T 170 c25170rlt1411exp18782 0203 I 1878 JMHIQZ C 4 Answer 1146 c The final equilibrium temperature can be calculated by recognizing that if the potato had a uniform temperature throughout its volume the heat transfer to it would be equal to QchT 7T 2 Tavg7 Q avg x m c Because there is a nonuniform temperature we use Eq 1163 to calculate the actual heat transfer Qm 39539 A k 008 3 m pV 1100m jn 0295kg Qm ch T 7 T 0295kg3900j jl707 25K 1668kI kgK 1000 Q70668kl 173707170sinl87871878cos1878 7875M 39 25 170 1878 39 878klj ng 25 C 1011 C 4 Answer 0295k 3900 g kng 1147 11 41 In the manufacturing of laminated wood table tops a more expensive and attractive wood surface layer 1 5mm thick is glued to a less expensive structural wood Heat is applied to the surface of the table to speed the curing of the glue A heater consists of a massive plate maintained at 150 C by an imbedded electrical heater The glue will cure sufficiently if heated above 50 C for at least 2 min but its temperature should not exceed 120 C to avoid deterioration of the glue Assume the laminate and structural wood have an initial temperature of 25 C and that they have equivalent thermophysical properties of k 015 WmK andpcp 15 X 106 Jm3K Determine a how long will it take to heat the glue b the glue temperature at the end of the 2 min curing time c the energy removed from the heater during the time it takes to cure the glue if the heater has a square surface area of 250 mm to the side in kI Approach XO 0 r The table top is very large relative to the depth of L 39 39 M interest so this transient conduction problem can I IMM E be analyzed as onedimensional conduction into 00 a semiinfinite solid with a step change in the 11 I 0 C R surface temperature buck I Wmquot Assumptions 1 The properties are constant and uniform throughout the laminate 2 The heat transfer coefficient is uniform and constant 3 Onedimensional transient conduction in a semiinfinite solid is valid Solution a Assuming the properties of all the materials are the same the glue thickness is negligible compared to the laminate and structural wood and onedim ensional transient conduction into a semiinfinite solid is applicable the temperature transient for a step change in surface temperature is given by Mlierfc 2 on Yin k 7 015WmK pcP 15x1061m3K We want the time required for the glue to reach50 C so substituting in known quantities 2 From the given information or lgtltlO397 s 507150 00015m li erfc 05 25450 21x10 7m2sr 08lierfc237l7t3905 23717t 050906 I 685s 4 Answer b At the end of 2 minutes of curing the glue temperature would be tell Witht260s685s 12685s T T Jj 7Tlierfc 00015m 2lgtlt10 7 mZsl2685s05 This is above the temperature at which the glue deteriorates Because of uncertainties in the thermal physical properties the actual temperature could be higher or lower than the calculated value Therefore a cooler heater should be used T150 C257150K lierfc 135 c4 Answer 1148 c The energy removed from the heater is equal to that which enters the table top We can integrate the power over time to find the energy 1 Qj0dej0qquotAdz The heat ux at the surface is given by k T 7 q 01 Q IkT 4Adt 2kT 4At 2015wm K150725K0250m212685s12 7rlgtlt10 7 mZs 05 If the heater operated continuously during the curing the power required would be approximately Q 471001 4 Answer ngs 47100371l371w 4 Answer t 12685s s 1149 11 42 You have built your dream cabin and now need to lay a water line from the well to the cabin You need to estimate how deep the water line should be placed From historical records you discover that there has never been a cold spell of 710 C weather for longer than 4 weeks but often the temperature is as low as 0 C for 12 weeks Through the fall you note that the ground temperature for a reasonable depth is approximately uniform at about 10 C You assume that if the ground temperature at depth does not reach 0 C in 12 weeks of 710 C weather this includes a factor of safety then the water in the pipes will not freeze The soil on your property has k 21 WmK and 1 7 X 10397 mZs How deep should you lay the water pipe Approach 0 Because the earth s radius is large compared to the depth of 7 0 C the water line we will analyze this problem as transient conduction in a semiinfinite medium with a step change in surface temperature We assume the surface temperature is TL L 7 fixed and the properties constant X 39 39 o T 0 C Assumptions 3 7 W sng 1 The properties are constant 2 The conduction is onedimensional in a semiinfinite E Neglect any freezing of water in the soil Solution The equation governing the temperature transient in a semiinfinite solid with a step change in surface temperature is TiT 17erfc x Tin 2 at We assume properties are constant and neglect any freezing of water in the earth above the water line Substituting in what is known 07710 x 17erfc 05 1040 27x10 7m2s7258x106s Using Table 114 for the erfc x 0477 4508 Solving for x x 215m Answer Comments This depth for the water line is very deep The factor of safety mentioned in the problem statementmay be too large 1150 11 43 When cold weather reaches into Florida if the air temperature remains below freezing 0 C for an extendedperiod of time the orange lime and other citrus fruit crops can be severely damaged Consider an 8cm orange whose properties can be approximated as those of water If the orange is initially at 10 C and the air temperature drops suddenly to 75 C determine how long it will take for any part of the orange to begin to freeze ignore radiation a if it is a relatively still night and the heat transfer coefficient is 10 WmZK b if it is a windy night and the heat transfer coefficient is 40 WmZK Approach The first location that will freeze is the surface We check the Biot number to determine of the lumped systems method 1 quotgoc can be used If not we will use onedimensional transient conduction in a sphere L 1V Assumptions l The properties are constant 0 A39 k L 40 WNL L 2 The conduction is steady and onedimensional Solution a The Biot number is 2 BihLi where L Ki D 62 char 7 A D2 6 Assuming constant properties that are those of water from Appendix A6 at Tavg 10 0 2 5 C p9999kgm3 cP 4203kIkgK k 0578Wm K a 1371x10 7 mZs 10wm2 K008m6 Bi 0230 gt 01 so the lumped system method 1s valid 0578 WmK With the higher h in part b Bi is even larger so we will assume the oneterm approximation for one dimensional transient conductor is valid Assuming constant properties no heat generation and uniform heat transfer coefficient T I 7T 2 39 a z7Crexp 1 Tr when rgt02 Solvingfor 106 17i1ni TTT Air 2 if 42 C1 H sinlt4rrgt Let r ra and recalculate the Biot number so that we can evaluate Al and C1 from Table 112 hrai 100082 k 0578 By interpolation A l3446 and C1 ll954 Bi 0692 1 1 0775 13446 7 2111 0528gt02 13446 11954 10775 s1nl3446 2 008 2 2 0528 So the oneterm approximationis valid IELq26l70sl7lhr 4 Answer 137le0 m s b Withh40Wm2K Bi277 D 112229 and C1l590 07 75 2229 008 2 2 0106 7ln 1 0106 Wl240s034hr Answer 2229f 1590 10775 s1n2229 l37gtlt107 Comment Note that the criterion for using the oneterm approximation is not satisfied for part b Because the figures do not have sufficient detail to provide a better answer and because of other approximations in the problem solution a more involved solution is not justified llSl 11 44 Stainless steel AISI 304 ball bearings which have been uniformly heated in an oven to 850 C are hardened by quenching them in an oil bath that is maintained at about 40 C by removing warm oil from the top of the bath and adding cool oil at the bottom The ball diameter is 20 mm The balls move through the bath on a conveyor belt at a velocity of 015 ms Assume the oil has cp 1960 IkgK and its convective heat transfer coefficient is 1230 WmZK Determine a how long the balls must be in the bath until their surface temperature reaches 100 C b the center temperature at the conclusion of the coolin c what oil flow is required if the oil temperature cannot rise more than 5 C and 10000 balls per hour are to be quenched Approach This is a transient conduction problem The Biot number 1 40 C should be checked first to determine if the lumped systems method can be used If it cannot then we will use a one U dimensional transient solution for a sphere S D k 12131 Assumptlons I MLK 1 The propert1es are constant 2 The heat transfer coefficient is uniform 3 The conduction is onedimensional transient in an 001Jquot 30455 sphere Solution a The Biot number for a sphere is 3 Bi hLW39 where LEM VA 46 2 k IrD 6 From Appendix A2 for 304 stainless steel at 800K p 7900kgm3 k 226Wm K cP 582 Ikg K or kpcp 492gtlt10 6 mZs Bi 7 1230wm2 K0020m6 226Wm K So we cannot use the lumped systems approach so we will use a onedimensional solution Assuming constant properties and heat transfer coefficient the oneterm approximation for a sphere is TiT sin r r f Clexp7 121M r gt 02 T 7 T 1171 2 We will need to check I for the validity of the method To obtain C1 and A from Table 112 we need 7 hra 7 12300022 T k T 226 Solving for r with r ra T 7T 1i21n 71n 1703gt02 A C1s1n 1 7 7T 12093 11561s1n12093 8507 40 So the approach is valid Solving r for the time 2 1703 001 2 1 346s 4 Answer a 492gtlt10 m s 0181gt01 Bi 054 by interpolation Al 12093 and C1 11561 b The center temperature is obtained with same equation Noting that with r 0 W gt1 7 2 T T Yj 7TC1exp7212 r 40 C8507 40K11561exp7120932 1703 1176 C Answer c The heat transfer to the oil can be determined from an energy balance on it Assume steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah CPAT 1152 Q 1 m c T T i m Q Plt 1 CPWTW The heat transfer rate is determined from the number of balls cooled per hour N ballhr and the energy per ball Q Q3 NQ Now that Q1211 iQ baHs For a sphere Q 7 sin 1 711 cos 1 Q 7 1 36115ph2r2 T Qmax chT 7pVCPT Tx 002 3 7900g582m4078501lt 7156001 m kgK 6 7 39 12093 712093 12093 91MJWwW Qm 850740 12093 Q09187156007143201 Q810000balls714320 1h 7398001 hr ball 36005 S 398001 k m406 g 1 Answer 1960 5K S kgK 1153 11 45 For a large party celebrating your graduation from college your parents buy a case of frozen steaks Before you can throw the steaks onto the grill they need to be thawed at room temperature of 77 F with an estimated convective heat transfer coefficient of 5 Btuhrft2 F on both sides of the steak The very large 1 in thick steaks initially are at 10 F and they are thawedwhen the centerline temperature is at 32 F Assume the steaks properties can be approximated as that of water and you can neglect the energy associated with the melting phase change Determine a the time required for the steaks to thaw in min b the time required if you speed up the thawing by hanging the steaks on hooks and blowing air over them with a fan such that the heat transfer coefficient is increased to 50 Btuhrft2 F in min Approach G For this transient conduction problem we first check the Biot P number to determine of the lumped system method is valid If not we will use a onedimensional transient in an infinite T 770p plane wall The steak is large compared to its thickness F 1 k h4LFd F Assumptions 1 The properties are constant DCquot 50 2 The heat transfer coefficient is uniform 3 The conduction is onedimensional transient in an 5 v infinite plane wall 2L l w Solution a The Biot number is Bi hLm k where LW VA 2L AA 2L Assuming the steak properties can be approximated as water evaluate the properties from Appendix B6 at Tm 32772 545 F p 6231bmfti cP 100Btu1bm F k 0336Btuhrft F a 547gtlt10 3 ftZhr SBtuhr ft2 F200833ft 0336Btuhr ft F so the lumped systems approach is not valid With the higher h in part b B1 is even larger so we assume the oneterm approximation for onedim ensional transient conduction in an infinite plane wall is applicable Assuming constant properties no heat generation and a uniform heat transfer coefficient and ignoring the effects of phase change ice to liquid the governing equation is 248gt 01 TiT 777quot C1 exp7 121cosj rgt 02 2 1 1 Tin At the centerline x 0 so cos01 Solv1ng forr oatL a 17 112 C1 7 T T Reevaluate the Biot number so that we can obtain 11 and C1 from Table 112 500833 W124 By interpolation Al 09123 and C1 11336 rln 0712gt02 09123 11336 10777 L2 r 7 00833ft2 0712 So the oneterm approximation is valid t 7 090hr 54min l Answer a 547gtlt10 ft hr b With h50Btuhr ft2 F a Bi 124 a 2 1445 and C1 12639 7 00833 2 0303 1 7111 0303 a t 038hr 23min Q Answer 445 12639 10777 547gtlt10 ft hr Comment F reezingthawing requires much energy so ignoring the change of phase in this problem significantly underestimates the time required to thaw the steak 1154 11 46 One method to experimentally determine thermal conductivity is to measure the temperature response of a thick slab when it is subjected to a step change in surface temperature Consider a solid with p 2500 kgm3 and 0p 630 IkgK The solids temperature initially a uniform 25 C is measured with a thermocouple embedded 6mm from the surface Boiling water at 100 C is brought into contact with the surface and the heat transfer coefficient is very large After 90 s the thermocouple reads 73 C Determine the thermal conductivity of the solid in WmK Approach Zinc This is transient heat conduction in a semiinfinite medium L 3 Because of the very large convective heat transfer 1 3 wc Q Zwol quot coefficient when boiling water is brought into contact with n aw c 650 3 L the surface there is a step change in the surface temperature 39 P quot H Equation 1164 can be used to obtain the thermal conductivity X o 00 6M Assumptions 39gt S MI N ml g l The properties are constant 2 The heat transfer coefficient is uniform 3 The conduction is onedimensional transient in a semi infinite solid Solution The equation governing conduction in a semiinfinite solid subject to a step change in its surface temperature is Txl7T x w li erfc T 7 T Solving for the complementary error function x 17TxtiT 7 737100 1 064 2 at Tin 257100 erfc Tnterpolating in Table 114 at this value of erfc or using appropriate software 2 032 7 0647 06509 0347 032 06306 7 06509 x Z 03308 2401 k With or and solv1ng fork pop 2 kg J 2 0006m 2500 630a W k x 561 2m g 144 Q Answer Z 41 03308 490s m K 1155 11 47 Curing of the epoxy in a laminatedmaterial can be accelerated by the application of heat Consider an electric heater that is pressed tightly against a thick slab of a laminated material Whose properties are estimated to be p 1200 kgm3 cp 1350 JkgK and k l3 WmK The laminate initially is at a uniform 25 C The heater has a heat flux of 350 Wmz Determine a the temperature at the surface of the laminate 3 min after the heat is applied in C and b the temperature 5mm into the laminate 3 min after the heat is applied in C Approach k u L ea lv 3513 Win r 3990 Because the laminate is thick we will analyze this transient conduction problem as a semiinfinite medium with a step change in applied surface heat flux 2Y C Assumptions INSvllffld l The properties are constant 2 The heat flux is uniform 3 The conduction is onedimensional transient in a semi infinite solid Solution a The temperature transient equation for a step change in the surface heat flux into a semiinfinite medium with constant properties is Txl T 2 7x2 x2 x exp 7 erfc q klle It 4061 on 2011 We want to determine T 0180s From the given properties 2 or k 1393WmK 802x10 7m pcP lZOOkgm3l350 JkgK s Because x 0 12 quot 2 350W m2 802gtltlO 7m2 s 180s TO180sZ 21 quotit 25 CM w 287 c4 Answer k 7r l3Wm K II b Using the same general equation Withx 0005m x2 0005m2 00433 4m 4802gtlt10397m2s180s 2 x 400433 01732 on x 0005 1732 E 2802gtlt10397180 H 2 2 v q mquot2m 802x10397m 180s 325K k l3Wm K s Txz 25 C325Kexp7004337 01732quot2 erfcl732 285 C 4 Answer 7 1156 11 48 A consulting engineer is asked to investigate a suspicious fire A room paneled with thick oak planks burned very quickly The insurance company wants an estimate of the time required for the surface of the oak planks to reach their ignition temperature of 400 C Initially the wood was at 25 C The temperature of the hot gases from the fire was estimated to be 850 C with a heat transfer coefficient of 25 WmZK Determine a the time required to reach the ignition temperature in s b the temperature 1 cm inside the wall at this time c comment on the in uence of radiation Approach We will treat this problem as a onedimensional transient conduction problem because the oak is thick PI but not specified and ignition will begin at the surface L 2 1139 2ft m lC 154 400 04 I IONI Assumptlons P 1 The properties are constant 2 The heat transfer coefficient is uniform 3 The conduction is onedimensional transient in a semiinfinite solid Solution a The solution for onedim ensional transient conductor in a semi finfinite medium with a step change in surface convection is given by Eq 1166 T xl 7 T 2 176ch x exp E4112 erfc x 4115 T 7 T 2 k k 2 k From Appendix A3 for oak k 017Wm K p 545kgm3 cP 23851kg K and 2 a L1308X104m pcP 545kgm32385Jkg1 s Calculating what is known at the surface with x 0 T 7 T 7 f m 0545 L 0 E Tin 257850 2 at k hzm ZSWmZK21308X10397mZst k2 7 017wmK2 hm ZSWm211308X10397mZsl05 k T 017WmK Substituting these into the main equation 0545 17erfc0exp00002829terfc0005319t051 Using appropriate software we solve for I 151s Answer b For the temperature 1 cm inside the wall at this time 1151s using the same equation but with 05 x 00m W 25wm2K1308x10397m2s151s 39 1125 0654 N 21308gtlt10397 mZs151s05 k 017WmK hx 25wm2K001m 147 h2g1 25wn12K2 1308X10397 m2s151s 7 017WmK T 39 k2 7 017wmK2 850257 85017 erfc1125exp14704271 erfc11250654 515 C 4 Answer c Because of the high combustion temperature radiation will have an influence and the ignition time will be shorter than that calculated above Answer 0002829t 005319105 04271 1157 11 49 For proper heat treatment of metals the temperature distribution in the metal must be carefully controlled Consider the annealing of a large slab of 304 stainless steel Initially the slab is at 150 C It is placed in an oven in which the air temperature is 1040 C and a heat transfer coefficient of 450 WmZK The average temperature of the slab must be raised to 800 C but the surface temperature should not rise above 900 C Determine a the maximum slab thickness that can be processed in cm b the time required for the annealing process in s and min Approach Assuming that the lumped systems method is not valid we will use the onedimensional transient in an infinite plane wall With neither time nor thickness given an iterative procedure probably will be required Assumptions 1 The system is steady 2 The properties are constant 3 The conduction is onedimensional in a infinite plane wall S olution E mw quotHrSD C k 45033 T5 MLK chs 90m I 39TS 900 C e 2Lquotgt l Assuming constant properties uniform heat transfer coefficient and no internal heat generation the equations governing onedimensional transient heat transfer in an infinite plane wall are Txl7T x W C1 exp7 12 139 cos Q 1 M 5 Qm T T A Note that Qmax ch T 7T and using T 1 avg gt Q chTavg 7T so that x Q TT 717T0gtITsin 1 1040 150 1 800150 7 7 T0t71040 sinil 15071040 A TLt710407 7 2 Wiqex 21 rcos L j 2 T0t710407C exp7 127 3 15071040 7 4 Qmax T 39 7 T i T A For the surface temperature T L I T L of T A L Cex 721 cos 4 I 7T 1 p A L For the centerline temperature T 0 I T 0 r 7 T 2 WClexpi l rcos0 4 The constants Al and C1 from Table 112 depend only on Biot number B LL k The three equations 1 2 and 3 involve 3 unknowns T0I TLI and r The solution procedure is 1 Guess Bi 2 Evaluate Al and C1 3 Solve the three simultaneous equations for the three unknowns 4 Determine if the calculated T L I lt 900 C 5 Return to step 1 if step 4 is not satisfied 1158 For 304 stainless steel from Appendix AZ atTavg 8001502 475 C 748K gt so at 800K p 7900kgm3 k 226JkgK cP SSZJkg K a kpcP 492x10 6 m2s Guessing Bi 20 A 10769 and C1 11795 Solving the equation TOt7465 C TLI9008 C 11099 The surface temperature is close enough to our target so we do not need to iterate Bik 7 20226Wm K h 450 wm2 K So the maximum slab thickness is 2L 0200 m 1 Answer NowL 0100m Calculating the time If 1L2 7 1099o10m2 7W2234s372min Q Answer or X m S 1159 11 50 Quenching of a metal slab in an oil bath requires that the bath temperature does not rise significantly The temperature rise in the bath can be determined if the heat transferred to the oil from the slab is known Consider a plain carbon steel slab 25cm thick and 2m square Initially at 1000 C it is quenched in an oil bath at 100 C The heat transfer coefficient is 450 WmZK Ignoring edge effects determine a the time required for the centerline temperature to reach 425 C b the temperature at a depth of 05 cm from the surface at the same time in C c the heat transfer from the slab to the oil in kl Approach W D 239 Because of the large width and length of this slab compared a to its thickness we will solve this problem as a one 51 moo C dimensional transient in an infinite plane wall We will w calculate the Biot number to see if the lumped systems IA quot K approach is valid but we doubt it is because of the large M heat transfer coefficient 72092 429C Fl 100 Assumptions 1 The system is steady gt 2 The properties are constant I I 2 L 0 01 3 The conduction is oned1mens1onal in a plane wall Solution The Biot number is Bi thhm k where LEM VA 2L AA 2L From Appendix A2 for carbon steel at 1000K p 7854kgm3 k 300Wm K c 1169JkgK a 327x10 6 mZs 450wm2 K0025m 300WmK a Assuming constant properties uniform heat transfer coefficient and no edge effects the equation governing the temperature transient in an infinite plane wall is T xl 7T Clexpi 12 rcos 1xL when 1gtN 02 7 Bi 0375 gt 01 so the lumped systems approach is not valid 1 2 71 Tin For the centerline x 0 and solv1ng for r oatL a r 21n 11 C1 TTf hL 45000125 Evaluating Bl 7 T 01875 gt Al 04213 and C1l0293 T l amp 590gt02 042132 1029310007100 2 590 00125 2 l 2825 Answer a 327gtlt10 m s b The general equation can be used to find the temperature at a depth 05 cm from the surface x 00125m 7 0005m 00075m So the oneterm approximation is valid Solving fort t 04213 00075 T 100 C10007100Kl0293exp7042l32 590cos 4147 c Answer m c The heat transfer is calculate with Q iliQ sin 1 7174257100jsin04213 Qm A 10007100 04213 Qmax chT 7 7854kgm30025m2m2m1169Jkg K10071000K 7826gtlt108 I7826x105kl Q 06497826x105 kJ 7537gtlt105 kJ 4 Answer 0649 1160 11 51 For the cylinder inProblem P 1139 assume it now is only 15cm long For the same conditions as given in that problem determine a the center temperature of the cylinder 1 min after it is quenched in C b the center temperature of the cylinder 5 min after it is quenched in C c the time required for the center temperature to reach 50 C in s Approach T 500 C This is a transient problem in a finite size cylinder We first 39 check the Biot number to see if the lumped systems method D M is valid If not we can combine two onedimensional transient solutions to obtain the transient in this cylinder We can use Example 119 for the procedure to follow T 21155 Assumptions 1 23 C 1 The properties are constant I I OOQWM LK 2 The conduction is twodimensional Solution First check the Biot number Bi hLEhmk where LEM VA L7rD24 V 015m7r0075m24 6627x10 m3 A 7rDL27rD24 7r0075m015m27r0075m24 4418x10 2 m2 LEM 6627X104m24418X10392m 0015m From Appendix A2 for 316 stainless steel at Tavg m 600K p 8238kgm3 op 5501kgK k183Wm K a kpcP 404x10 6m2s Bi 1000wm2 K0015m183wm1lt 082 gt 01 so the lumped systems approach is not valid a Following Example 119 the nondimensional temperature for the infinite wall uses oatL2 404gtlt10396 mZs300s0075m2 0215 gt 02 so the oneterm approximation is okay 132m hLk 10000075183 410 From Table 11 2 2 12695 andC112299 139 wall 6W C1 exp7 12 2 12299exp7126952 0215 0870 404gtlt10396 300 1000 00375 For an infinite cylinder Icy 06 2 0862 Big h r W 205 2 00375 k 183 From Table 112 A 16090 andC113424 6W1 C1 exp7 l2 139 13424exp7160902 0862 0144 The center temperature is T00300s T Tx 7T6avwaugtlt19w 25 C5007 25K08700144 846 C 39 Answer b The same equations are used to determine the time required for the center temperature to reach50 C T0 017 T T T f C1 exp7 12 1wa11 X C1 exp7 12 1 with 15y oatr and r oatL2 we can solve the above equation for time wall 2yl I 71 In 1 T4 waa L2 12y17122 C1y1 CLWaII 7 T 2 71 2 2 1 i13424112299 fji4015668mm Answer 404X106m 12695 16090 U s 0075m 00375m 1161 11 52 Raw clay molded into bricks is fired in a kiln at 1300 C and cooled in air at 25 C with a convective heat transfer coefficient of 50 WmZK The 57cm gtlt 10cm gtlt 20cm brick has the following properties p 2050 kgm339 k 10WmK39 CF 960 JkgK39 or 051gtlt10396 mZs After 50 min of cooling determine a the temperature at the center in C b the temperature of the corners of the brick in C Approach o We will assume lumped heat capacity is not applicable and Boo C 2L 010 450C use the product of three onedrm ensronal trans1ent 2 l conduction solutions for infinite plane walls to determine k 39 502 the two temperatures T 0 0 0 I andTL1 L2 L31 M in ao m Assumptions 2L glam 1 The properties are constant 3 2 The conduction is threedimensional 3 Radiation is ignored Solution As shown on Figure 1113 the intersection of three infinite plane walls is the transient solution for this brick Assuming constant properties uniform heat transfer coefficient and ignoring radiation the very high brick temperature will result in large radiative losses but we will ignore them the equation we need to solve is Tx1x2x3t7Tf 6x1lm72 gtlt9x2tmyg x6x3lmg The oneterm approximation is 6xlp1m C1 exp7 2 1cos 1xL where r gt 02 and r oatL2 wall Each wall has its own values for T x A and C1 and the latter two quantities are obtained from Table 11 2 once the Biot number is evaluated So for the three walls hL 50wm2 K00285m Wall 1 00285m Bi 1425 L1 k lWmK m 051x10 6m2s3000s by rnterpolatron A 09524 andC111448 11 2 21884 L1 00285m 50005 Wall 2 L2 005m 312 250 H 11 11347 and C111949 12 0612 Wall 3 L3010m Big 50010 f500 a 413138 andC112402 130153 a For the centerline temperature x 0 and cos 0 1 so 1 02073 VA 6101 11448exp70952421884 a 01 11949exp7113472 0612 V 1 05434 V a 01 12402exp7131382 0153 1 09524 The center temperature is T0 0 01 25 C1300725K020730543409524 1618 C Answel b For the corner temperature we use 61 L11 610tcos 1LL02073cos09524 01202 62 L2I 05434cos11347 02295 63 L31 09524cos13138 02421 The corner temperature is TL1L2L3t 2513007 25012020229502421 335 C Answel 1162 11 53 Exposed ceiling beams are popular many modern houses However because they are exposed their ignition in the event of a fire is of concern Consider a 15cm by 15cm yellow pine beam initially at 25 C attached to a ceiling thus insulating the base The three other sides are exposed to fire at 600 C with a combined convectiveradiative heat transfer coefficient of 50 WmZK Determine the beam s maximum temperature 5 min after a fire starts Approach Us who By inspection the highest temperature will occur at a corner in this twodimensional transient conduction problem We will assume the lumped heat capacity method is not valid The transient temperature can be a calculated by the intersection of the solutions for two 1 V 900 c infinite plane walls Because the attachment plane to the A S DW ceiling is insulated we need to analyze wall 2 with a length 2L since a line of symmetry acts as an adiabatic gt 1 surface ZLI 019 1 L2 SE39M l t MIH Assumptions 1 The properties are constant 2 The conduction is twodimensional 3 Radiation is ignored Solution As shown on Fig 1113 an infinity long bar is constructed from the intersection of two infinite plane walls Assuming constant properties and uniform heat transfer coefficients the equation we need to solve for the corner TL1L2t7Tf 7 T 6Agt Z7EX6L2gt 5Z7E where 6xlp1m C1 exp7 l2 T cos 1xL if r gt 02 wall temperature TL1L2I is Each wall has its own values for r x A and C1 and the latter two quantities are obtained from Table 11 2 once the Biot number is evaluated For the two walls hL 50w m2 K 0075m Walll Ll 0075m Bi1 lw 250 k 015 Wm K For pine from Appendix A3 p 640kgm3 k 015 WmK cP 2805 Ikg K or kpcp 836gtlt10398 mZs From Table 112 21 15082 and C112708 m 836x10 8m2s300s 1 L 0075m 2 So the oneterm approximation is not valid However Figure 117 is not of any help either because of the very small spacings on the graph so we will continue with this method with the understanding that uncertainties exist Wall 2 using the same approach as used for the first wall L2 015m Bi2 500 2115400 C112727 1000111 Therefore 6L1 t 1 2708exp 7150822 000446 cos15082 00787 000446 lt 02 6L2t 12727exp7154002 000111 cos15400 00391 Solving forTL1L2t TL1L2I T Tx 7T6L1I6L2t 600 C 257600K0078700391 5982 C i Answel Comments This answer suggests that the corner temperature is about the same as the fire Ignition temperature is lower than this Because the oneterm approximation validity certain was not satisfied there is a large uncertainty in the calculation 1163 11 54 For a fire investigation the insurance company for whom you work wants to know how long it would take for an oak beam 2in by 4in to ignite under certain conditions The air temperature is 1000 F with a convective heat transfer coefficient of 18 BtuhrftZOF The initial temperature of the wood is 75 OF and its ignition temperature is 900 OF The wood hasp 45 lbmft3 cp 030 BtulbmOF and k 010 BtuhrftOF Ignore radiation Determine the time required for any of the wood to start burning when suddenly exposed to these operating conditions in s Approach We will assume the lumped heat capacity method is not applicable As shown on Fig 1113 the intersection of 1 OAK 1 L 2 N two infinite plane walls creates an infinite rectangular 11 9 l 39 I bar By inspection we can see that ignition will begin Iquot 3 l lg ll l LL along the corner of the beam We can use these one k egt dimensional solutions to calculate the time required 2 L 4 I N forTL1L2L3900 F 2 Assumptions The process is steady 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant 5 There is no radiation Solution Assuming constant properties uniform heat transfer coefficient and no radiation heat transfer the transient temperature profile for an infinite twodimensional beam can be calculated by T x x I 7T 171 6x1 1 X 6x2 03 where the oneterm approximation is 7 2 x 7 on 6xlm7g 7 C1 exp7 1 rcos l r gt 02 and r 7 F Combining the equations with x1 L1 and x2 L2 C11exp 11 Tl cos 11cl2 exp 122 12 cos 12 Solving for time C11cos 1lcl2 005112 7 T T L Li Each wall has its own values for r x A and C1 and the latter two quantities are obtained from Table 112 once the Biot number is evaluated From the given information 010Btu hrft F 2 or L 000mi pcP 45lbmft3030Btulbm hr For each wall 18Btu hr ft2 F 00833ft E i 150 1 L100833ft Bi1 7 9 k 010Btuhr ft F by interpolation AU 09686 and C11 11493 18 01667 2 L2 01667ft Bi2 30 a in 11925andC1v2 12102 Substituting in numbers 1164 90071000 k 7571000 71 i 1 I L 0 00741ft2 09686 2 11925 2 U11493cos0968612102cos11925 39 E 00833ft 01667ft 0718hI Answer Check the Fourier number 2 000741 071 on hI 0766 gt 02 1 7 1 L 00833ft2 000741 0718 2 016672 This is close enough to r gt 02 to be a good approximation 1165 11 55 A hot dog 20mm in diameter and lScm long at 5 C is placed into boiling 100 C water39 because of the vigorous boiling the heat transfer coefficient on the hot dog is 250 WmZK To be fully cooked its center should be at 80 C The hot dogs properties are p 890 kgm3 cp 3350 JkgK and k 05 WmK Determine how long the hot dog should be in the water in s Approach 0000 L ZED WM39LK For this transient conduction problem we should first check E the Biot number to determine if the lumped system approach J D 0029 can be used If not we will treat this as a multidimensional a conduction problem in a solid with constant properties 11 S C Assumptions l ZL 039 quotwall 1 The Properties are constant T Git 8000 2 The heat transfer coefficient is uniform 3 The conduction is twodimensional transient in short cylinder Solution The Biot number for a short cylinder is Bi hLEhm k where LEM VA Volume V IrD24 2L n0020m24015m 4712x10395 m3 Area A 7rD27rD24 7r0020m015m7r002m22 l005gtlt10392 m2 LW 4712Xl0395m393l005X10392m2 4688x10393 m 250wm2 K4688X10397 m Bi 234gt 01 05 Wm K Because Bi gt 01 the lumped systems approach is not valid Assuming constant properties uniform heat transfer coefficient and no internal heat generation the transient temperature in a short cylinder is obtained by combining the solutions for an infinite plane wall and a infinite cylinder Trxt7T Txt7T X Trt7T 7 7T T T may 7 T m ng mde At the centerline r 0 x 0 the onedimensional transient responses using the oneterm approximation are Plane wall T 01 7 T f 2 vaexp 7 Vwrw Tx 7 T l Cylinder T 0 I 7 T 7 c exp 72 1 Tx 7 T 1 Combining these three equations T 0 01 7T CWexp7 fwrwC1exp7 fr c a emu 1 4 I z 05Wm K 39thIL2dt2k Wl TV 06 an 1 ar 0 MC 890kgm33350JkgK 2 168x10 7m S Solving fort 7 T 0 01 7T I 1n 1 Q Aw 0 AL 0 Cm C1 7 T T 1166 The constants Al and C1 are obtained from Table 112 Plane wall 250w m2 K 0075m Bil Lw375 AW15294 and CW12722 k 05Wm K 39 39 Cylinder 250 001 Bi5o 2 11 19898 and C115029 71 1 1 807100 l52942l68gtlt10397m2s 198982168x107mZs L1272215029K 57100 l 0075m2 001m2 2325386min 4 Answer 1167 11 56 The same heat treating oven used in Problem P 1144 is used to prepare the same geometry balls for a different application If the hot balls are cooled by natural convection in air at 20 C such that the heat transfer coefficient is 15 WmZK determine the time for the surface temperature to reach 100 C in s Approach This is a transient conduction problem The Biot number should be checked first to determine if the lumped system method can be used If it cannot then we Will use a one dimensional solution for a sphere Assumptions 1 The properties are constant 2 The heat transfer coefficient is uniform 3 The conduction is onedimensional transient in a semiinfinite solid Solution The Biot number for this sphere is V 7 7rD36 7 D B39 hL k h L 1 char W ere that A D2 6 From Appendix A2 for 304 stainless steel at 800K p 7900kgm3 k 226Wm K cP 582 Jkg K and or kpcP 492gtltlO396 mZs iswm2 K0020m6 Bi 0013ltOl 226 Wm K So a lumped system analysis is valid The governing equation is T 7 T h A I exp 7 I 7 T f m cP Solving for I with m pV and LEM VA D6 I ipcPDlnFWTTf 6h 77T kgK lt6gt15mY K 77900kg1582 I j0020m ln 2660s 443min 074m 4 Answer 1168 11 57 In the thermal conductivity measurement device described inProblem P ll7 you ask that a new sample of the material be tested and the temperatures have changed to T1 9300 C T2 8497 C T3 7139 C T4 6913 C The same thermal conductivity for the unknown material is obtained However you are puzzled by the different temperatures and a decreased heat transfer rate When you disassemble the test piece you discover that the new operator did not assemble the pieces carefully enough and the two ends have some roughness that should not be there As a result of the roughness there is thermal contact resistance Determine the magnitude of the thermal contact resistance in mZKW Approach The heat transfer rate equation can be used ATRm to determine the contact resistance The same heat transfer flows through both materials We assume steady one dimensional conduction with constant properties and use the appropriate conduction equation Assumptions 1 The process is steady 2 The conduction is one dimensional 3 There is no internal heat generation 4 The thermal conductivity is constant Solution We can use the rate equation between point 2 and point 3 AT 7 T2 7 T3 AxZI Rclimtact AxIE k A A A k8 A where the subscript I represents the interface location Dividing by A 2 q T2 7 T3 A AxZI RN AxIE kA canmct k Solving for Rim T RIM t q kA k3 Heat flux is determined from conduction in Material A k 152 Wm K A T1 42 gt T 12 m The thermal conductivity of Material B is determined by equating the heat flux in A to that in B kA7i T2kBT3 T4 x q 930078497K6100 2 m butAx Ax Ax Ax 12 34 TAT 7 k8kA 1 2 15zljw54oi T344 mK 713976913 mK 849777139 c 2 1ng 2 002m 002m 540gtlt10 4 m K 4 Answer 6100Wm 152WmK 540WmK w 1169 11 58 A device is to be constructed with a 3mm 304 stainless steel plate and a 9mm layer of 2024T6 aluminum The temperature drop across the composite wall will be 150 C Two different manufacturing methods can be used The first method will result in a surface roughness of the contacting parts of 25 um and a contact pressure of 25 MPa The second method will result in a surface roughness of the contacting parts of 15 pm and a contact pressure of7 MPa Estimate a the heat flux for these two manufacturing methods in Wmz b the percent decrease in the heat flux compared to no contact resistance Approach ft 000 The basic heat transfer rate equation ATRm can be gt used to estimate the heat flux The thermal contact resistance 2 1 must be estimated form those given in Table 116 SS AT 50 C k114 w alumnus K miv K Assumptions quot 1 The properties are constant 939 2 The conduction is steady and onedimensional K TamMI Cadm 3 Radiation is ignored 1 0003M KESITHW Solution The heat transfer rate through the composite wall assuming steady onedimensional conduction with constant properties is AT AT Q quot Rm t1 t2 k1 A1 A k2A2 Dividing through by area 39 AT 7 7 R 2 k1 k2 The thermal conductivities are obtained from appendix A and the two thermal contact resistances are estimated to be from Table 116 R1 3x10 4 m2 KW 25ym 25 MPa and R1 05 m21W 1 5 u m 7 MPa ql 150 oz 27160012 Answer 0003m 4 m K 0009m m 3gtlt10 149WmK w 177WmK 7 150 7 w q2 7 07003 0009 7 496 400 4 Answer 705gtlt104 149 177 b If there is no contact resistance 7 150 7 w q 7 W 7 594800F Answer 149 177 594 800 7 271 600 594 800 594 8007 496 400 594 600 X100 543 Answer decrease in case 1 x100165 4 Answer decrease in case 2 Comment The estimate of the contact resistance can have a large impact on the answer Because of uncertainties care should be exercised when choosing a value to use 1170 11 59 Significant manufacturing advances have been made to improve the speed of computer chips The technique is to place more discrete electronic components closer together However this increases the electric power dissipation to such high levels that cooling is becoming a problem To accommodate the high power densities direct cooling with boiling has been investigated Consider the lOmm by lOmm thin chip shown below which is cooled by liquid at 25 C with a boiling heat transfer coefficient of 750 WmZK The chip is attached to the circuit board and the contact resistance between the chip and board is estimated to be 14 X 10394 mZK W The circuit board is 4mm thick with a thermal conductivity of 2 WmK The backside of the board is exposed to air at 25 C with a heat transfer coefficient of 35 WmZK Determine a the thermal circuit of the chip board and cooling uids combinations b the chip temperature if the chip heat dissipation rate is 4 W in C k 73w 5 Agproach d d I 1 WT A row mom out th ssurning stea y one 1mens1ona c l E I conduction the basic rate I 41 L J P Q 39 4 W W uwfmwm q equation ATRm can be used with t 4 M k I 4 x o dln K A M an energy balance on the chip to T k determine the chip temperature T 2m ZwMK QUEW h 2 win 4 Assumptions 9 W 53 l The system is steady T7 L AL 4quot L l A 1T 2 The properties are constant L T 39 3 The heat transfer is one dimensional Solution a Heat is transferred to the liquid and air so the circuit diagram is shown above b Using the circuit diagram in part a T 7 T T 7 T QQLQ 2 1 1 1 h A A h A Solving for the chip temperature I 1 Q39 hz A2 T2 Rmzm L T A h A T 4 Rquot l A h A 14X10394m2KW 1 4W750Wm2K0OOOlm225 C 0000M 35Wm2K00001m2 25 C l4gtlt10394m2 KW 1 71 00001m2 35wm2 K00001m2 750wm2 K00001m2 760 C Answer ll7l 11 60 Annular fins l5mm thick and l5mm long are attached to a 30mm diameter tube Tube and fins are 2024T6 aluminum The thermal contact resistance between the fins and the tube is 25 X 1039 m KW The tube wall is at 75 C the surrounding air temperature is 25 C and the convective heat transfer coefficient is 100 WmZK For a single fin determine a the heat transfer rate Without contact resistance in W b the heat transfer rate with contact resistance in VJ Approach This is steady onedimensional conduction in a constant area fin We can use the basic heat transfer rate Tb 7S C 2quot 2x154M KW equationQ ATRm to calculate the heat transfer from one I t O 0039 3 fin a I L 0 015m Assumptions l The properties are constant T I I 2 The conduction is steady and onedimensional 39 l k IOO M K 3 Radiation is ignored R o my Solution Assuming steady onedim ensional constant properties uniform heat transfer coefficient and no radiation and taking into account both the fin resistance and the contact resistance the heat transfer rate equation is Rm hA Ab We use Figure 1117 for the fin efficiency of this annular fin For aluminum from Appendix A2 kl77WmK r1R0015m 2C r2 I2 RLt200150015000152 00308m rzcr1 003080015 205 L LI2 0015000152 00158m Ap L I 00158m00015m 236gtlt10395 m2 12 12 L32 i 00158m32 M 0307 kA 177wmK236x10 5m2 From Figure 1117 77 2090 A 27rr22C 7132 27r00308m2 7 00151112 1 455gtlt10393 m2 A 27rRt 27r0015m00015ml4lgtlt10 4 m2 Without contact resistance 75725 K i 71 205w 1 Answer 090100wm2 K455x10 3m2 With contact resistance 75725 K Q 7 2 119W 4 Answer 1 25gtlt104m KW 090100Wm2 K455gtlt10 3m2 141x10 4m2 Comment As can be seen in this problem contact resistance can have a significant negative effect on heat transfer 1172 14 1 The tungsten filament of a light bulb is at 2600 K Assuming the filament is black calculate the wavelength at which the maximum power is emitted In what range of the electromagnetic spectrum does 39 wavelength fall Approach Apply Wien s displacement law Assumptions 1 The filament is black Solution From Wien s displacement law Am T 2898 um K 72898umK72898umK7 a A 112m quot Answer M T 2600K From Fig 141 this is in the infrared range Comments Incandescent bulbs are inefficient producing more heat infrared than light visible 14 2 Measurements of the spectrum of the star Vega show that radiation peaks at029 um Neglecting the Doppler shift of the star estimate Vega39s effective surface temperature Approach Apply Wien s displacement law Assumptions 1 The Doppler shift from Vega is small 2 Vega radiates like a black body Solution From Wien s displacement law AMT 2898 um K 7 2898mmK 7 2898mmK A 029 pm max T 9993 K 4 Answer Comments By comparison our sun is cold at only 5800 K 143 The planet Mars has an average radius of3380 km and an average distance from the sun of 228 X 108 km The sun radiates like a black sphere with a radius of 695 x108 m and a temperature of 5800 K Assume 4 L A Mars is black and isothermal 39 39 surface temperature Approach Draw an imaginary sphere around the sun and distribute the energy radiated from the sun uniformly in all directions Apply the first law to I Mars balancing absorbed and emined energy V Assumptions 1 The temperature ofMars does not change with K time 2 The kinetic energy ofMars is constant 3 Mars orbit is circular 4 The sun radiates like a black body 5 The sun is isothermal and spherical 6 Mars does not produce internal heat 7 The sun emim uniformly in all directions 8 The sun s rays are parallel in Mars orbit 9 Mars is spherical 10 Mars radiates like a black body Solution The total energy emitted by the sun in all directions is Q ASGT 4zRfan where R is the radius of the sun Q 4695X108m2 557x10 8 mZVIVKA 5800K 389x1026W 39 L quot and Mars The sun39s Draw an imaginary sphere of d39 h rays are essentially parallel by the time they reach Mars orbit The interaction of Mars and the imaginary sphere t 4 L39 39I I nu mz around the sun is a Luuc ars 0 on ofMars compute the heat ux on the inner surface of the imaginary sphere as 26 Lg Q hf 339 W 2 595x1oX W2 Aw 4 55 4228x108km km The energy absorbed by Mars is QM 1924 Mars emits over its actual surface area not its projected area The energy emitted by Mars is QM AW39TTWA 4 FRm239TTmA Since there is no heat generated by Mars energy absorbed equals energy emitted and Q 4m2 t 2 z 2 quot1 595x1012 W2 T m a m 2 2 4rRm a 4rRm a 4 4567X10 2W 4 m K Tm226K467 C Answer Comments The actual r f 39 L t C There is a f L as a L Mars which 4 L 39 394 39 39 activity 144 Radiation from the sun incident on the earth39s outer atmosphere has been measured as 1353 Wmz The average distance between the sun and the earth is 1510ll m Ifthe planet Pluto is at a distance from the sun of 587x10g km find the solar ux incident on Pluto Approach V lt 0 Draw two imaginary spheres around the sun one to h intersecting the earth and the other intersecting Pluto Equate the total energy impinging on the earth sphere with that impinging on the Pluto sphere l Assumptions l 1 Pluto s orbit is circular 2 Earth s orbit is circular 3 Space is a perfect Vacuum 4 The sun emim uniformly in all directions Solution Draw two imaginary t quot quot Assuming space is a perfect Vacuum the total solar energy impinging on the smaller imaginary sphere equals that impinging on the larger imaginary sphere Therefore quotA q quotA a a p p where qiquot is the solar ux at the earth orbit anqu is the surface area ofthe smaller imaginary sphere around the sun while qp andAp are the corresponding Values in Pluto orbit q quot4rR2 qp quot47er W 15x10 m m 1353 q m2 587x109km 0884 2 Answer m C omments Theorbitof 39 quot 39quot umi 39 394 LL over 14 5 A furnace has an inside area of 320 ft2 and black walls A radiant power of 95 W issues from a rectangular opening in the furnace wall If the opening measures 5 in by 6 in determine the interior wall temperature If the emissivity of the walls is 088 what is the wall temperature Approach Model the furnace as a cavity and treat the opening as a black surface at the temperature of the interior walls Assumptions 1 The radiation issuing from the furnace is black Solution The opening is small compared to the size of the furnace interior therefore the furnace acts like a black cavity The heat flux escaping from the furnace is QA039T 2 g 95WE39i37rn TA mw 542K269 C 4 Answer 039 8 5m 6m 567gtlt10 lt gtlt gt Comments If the emissivity of the furnace walls is 088 the temperature is unchanged The furnace still acts like a black emitter since the interior is so large compared to the opening 14 6 A square window of side length 12 cm is located in the center of an oven wall measuring 35 cm by 45 cm The depth of the oven is 50 cm Considering all the oven walls as one surface and the glass window as a second surface find the view factor from the oven walls to the window Approach The view from the window to the walls is unity by inspection Reciprocity can be used to find the remaining view factor 1 I 1 CM Assumptions 2 l The radiosity is uniform over each surface 35 M D Lm f l39 g 2 All surfaces are diffuse I gt 50 L 39V 3quot m Solutlon Ll L Choose the interior of the oven as the enclosure The view factor from surface 1 the window to surface 2 all the other walls of the enclosure is unity by inspection Therefore Flgt2 1 From reciprocity ArFHz A2F2gt1 FH AIFHZ WWW 9l4gtlt10396 Answer A2 3545W3550245502 Comments The view factor is very small If the window were replacedby an opening the radiation issuing from the opening would be black radiation 147 Two square plates ofsize 8 cm by 8 cm are directly opposite each other in parallel planes What should the 39 39 factor L L is 05 Approach Use the equation in Fig 1416 Iterate to find the correct Value ofplate spacing Assumptions 1 The radiosity is uniform over each surface 2 Both surfaces are di use Solution Use the equation in Fig 1416 with 7 i 8 m 7 L L In HA 2 2 FH205W 1X2Y2 The Value ofL needed so that EA 0 39 39 39 39 39 software be used for this purpose The result is L 304cm 4 Answer Comments LinFi 1446 L L ct inradians A cryogenic dewer consists of two concentric cylinders The inner cylinder has a length of 22 ft and a 39 39 between the 1 48 diameter of 8 In The outer cylinder has a len o 6 ft and a diameter of 10 In The space cyli ders is evacuated Find the view factor from the outer to the inner cylinder Include the end area in the calculation Approach Note that EA lby inspection Find the remaining view factor by reciprocity Assumptions m 1 The radiosity is uniform over each M39 ace 2 All surfaces are diffuse Solution We must find Eavwhere surface 2 is the inner surface of the large cylinder and surface 1 is the outer surface ofthe small cylinder By inspection H2 since all the energy leaving 1 arrives at 2 From reciprocity 8 8 2 F 7 Ana 7 39 22 2m j A2 2 Itg j26ft2z2102 FH 05724 Answer Comments The view factor from the outer cylinder to itselfis FH2 17 F s 0328 149 The inside ofa sphere is divided into two hemispherical surfaces Find the view factor from one hemisphere to the 0 er Approach Construct an imaginary surface bis ecting the sphere Determine view factor using inspection and reciprocity Assumptions 1 The radios39ty is uniform over each ace 2 All surfaces are diffuse Solution Construct ima inary surface 1which divides the sphere in half as shown Surface 1 is an infinitesimally thin disk Surfaces 2 and 3 are the insides ofthe two hemispheres We need Egg All the radiation that leaves 2 and arr39 es at 3 must cross 1 Therefore F gt H Consider an enclosure including the top of surface 1 and surface 2 By inspection H2 since all the energy leaving the top of 1 arrives at 2 From reciprocity lgt2 2 2gtl R2 1 pH A1FH2 05 A2 27 Therefore FH FH3 05 Answer Comment One could also deduce the same result by symmetry 1410 A room in an art gallery has a oor area of 16 ft by 16 ft and 10 ft high ceilings One wall is part ofthe exterior ofthe building and the other three are interior walls In winter the exterior wall is cooler than the 0 er ee an looses heat to the environment by ra iation and convection Calculate the view factor from the oor to the exterior wall and from each interior wall to the exterior wa Approach Use Fig 1417 and 1416 to find some ofthe view factors Then use summation to find the rest Assumptions 1 The radiosity is uniform over each surface 2 All surfaces are diffuse Solution Let surface 1 be the exterior wall surface 2 be the oor surface3 be the ceiling surface 4 be the wall opposite the exterior wal d aces 5 and 6 e the two walls adjacent to the exterior wall as shown First find EM Use Fig 1417 defining X16 Y16 Z10 Then H 0625 14111 5 X 15 15 m 016 By symmetIy F3 1 016 Next find the view factor from the opposite wall to the exterior wall FHP Use Figure 1416 with From the chart in Fig 1417a FH1 at FH1 X 10 Y 16 X 0625 Y 1 L 16 L 16 From the graph F m 014 To find the last view factor note that Figt2Figt3Figt4Figt5Figt l By symmem Fm FM so Fm Fm FM 2FH5 1 Apply reciprocity to each term to get AZFH AZFH AAFH Mfg 1 A1 A A1 Solving for EA F 71510715150157151501571015014 0174 5quot 21510 39 In summary FE 016 F 016 FH 014 1 Answer F51 0174 FH 0174 Comments For greater accuracy use the equations in Figs 1416 and 1417 Note that the output ofthe arc tangent must be in radians 14 11 During a chemicalvapor deposition process a diskshaped silicon substrate 23 cm in diameter is placed in a reaction chamber The chamber is cylindrical with a height of 13 cm and a diameter of 5 cm The silicon substrate rests on the bottom surface as shown Find the view factor from the curved side wall of the chamber surface 1 to the substrate surface 2 Approach Use Figure 1418 to find FEE Then use summation to calculate FHI Finally use Surface 1 side reciprocity to determine FHZ surface Z SUbSt ate Surface S end Assumptions Reaction 1 The radiosity is uniform over each 18 cm Chamber surface 2 All surfaces are diffuse Solution By the summation relation F 241 an 1 substrate Note that surface 2 cannot see any portion of the bottom surface so views to that surface neednot be included in the summation relation To find FEE use the equation in Fig 1418 with 12112 2139156m00885 L 13cm R R 2395 m0192 J L 13cm 2 1 0192 2 S11H23 1 3133 R2 00885 where L is the height of the cylinder The view factor equation is 1 F2gt3S7S274r3r222 Substituting values 1 FM3 1337 1332 7 ZS113212 00354 Thus FH17FH3 1700354 0965 From reciprocity F A2F2gt1 7 VZZF39Zal 7 r22FZgt1 39 cm2 1 L 7 a2 A1 27rrEL 273L 225cml3 cm 1H2 00196 4 Answer Comments Because the view factor FH3 was small it would have been difficult to read it accurately from the chart Using the equation was a better approach 1412 Find the View factors between all the surfaces within a cubical enclosure Approach Use Figure 1416 to find the View factor between two opposite walls Then use summation and symmetry to determine all other View factors Assumptions 1 The radios39ty is uniform over each s ace 2 All surfaces are diffuse Solution Define surfaces 1 and 2 to be two directly opposite sides ofthe cube as shown Define surface 3 to be a side 339 r 1 shown 39 I1 I ure Surfaces 4 5and 6 not shown are the remaining three sides To find EM use the graph in Fig 1416 with 1 and 1 L H L H From the graph Fm m 02 From the summation relation mnFHwEwmwFH 1 Furthermore by symmetry gt4 FHs Fm lazl FH24FH31 17F 1702 F H2 02 Hz 4 4 In summary FH2 FHZ H4 FM FH6 02 lt Answer 1410 1413 A at panel heater is suspended over a warming tray as shown Calculate the View factor from the heater to Approach Create two arti cial surfaces 1b and 1c as shown Break up the warming tray into three surfaces Use shape decomposition to nd the desired View factor in terms 0 e View factor for two aligned rectangles in parallel planes Assumptions 1 The radiosity is uniform over each surface 2 All surfaces are diffuse Solution By shape decomposition m2 Vim Flam Fm2 By symmem Flam Fm therefore Fm Fl a2 Ell1 An additional shape decomposition will be helpful F anagram Fam Ha FE2a2cgtlc By 53quot me Fl2a2cgtla Fl2a2cJgtlUso F2a2tarart ZFhamHa Ell2 By reciprocity 7 AraFiHam Flam AP 2 11 SinceAja Avm 2 Fl 11 2a 2 Fina Substituting into Eq2 yields F F anagram lagt2a2c Again applying shape decomposition F F Fi zancpuam my FM F Wig 7F W Substituting this into Eq1 gives F may Flagt2 F lagt2a 2F 2a2cgtlala F laaza Zfamtpgmt To nd FREQHW useFig 1416 with E 1539quot39 257 710 157 6m 6m F2a2tarart 04 To nd FKHM again use Fig 1416 this time with E E 133 7 10 39 157 6m 6m Flam 03 Substituting EH220470305 4 Answer 1411 1414 Using data on the figure below find the View factor FBI Surfaces 2 and 3 are perpendicular at their common edge Approach Construct surface 12 which is the combination of 1 and 2 The View factor from 12 to 3 can then be calculated using Fig 1417 Fig 1417 can also be used to find FM Apply shape decomposition and reciprocity to find the desired 3 Cm 7 View factor FM 61 K Assumptions 39 Ran 8 m 1 The radiosity is uniform over each surface 2 All surfaces are diffuse Solution Designating 12 as the combination of surfaces 1 and 2 the View factor from 3 to 12 may be decomposed into F3 12 F341 F242 FHi F3412 F3 2 To find ng use Fig 1417 with H 03 W2075 4 4 From the plot 17 z 014 To find VH1 again use Fig 1417 with Hz zojs W23075 4 4 From the plot F 3 12 z 03923 Finally F31 023 014 009 a Answer Comment Greater accuracy could have been obtained by using the equation in Fig 1420 rather than the plot 14 12 1415 quot39 39 fequal 39 39 calculate the View factor from one ring to the other Iauiatiuu Using data on the figure Approach Construct surfaces 3 and 4 as shown Use shape decomposition and the formula for a 2 two disks in parallel planes from Fig 1418 9 5 cm Assumptions quot 52f Cm 1 The radios39ty is uniform over each cm s ace 39 2 All surfaces are diffuse a Solut on Designating 13 as the combination of surfaces 1 and 3 and indicating 24 similarly we may use shape decomposition to obtain F MHUM mm F 12H4 Where the superscript indicates a quantity that can be calculated from Fig 1418 Rearranging FEI3gt2 quawm 7F mp4 From reciprocity 4133155 a FHM sq Again applying shape decomposition Hm Fhi 2 Ell2 We can find the desired View factor F ifwe can calculate FH3 To accomplish that begin with Flam F92 FZgt4 F3gt2 525134 F3gtA By reciprocity 2127H A3F Hm F H Fz eq3 FquotH4 may be calculated using the equation in Fig 1418 with 2 1 1667 1667 S1236 15572 1 1 5 2 g 17254 2367236274 0554 Similarly for FEM 5 11371557 R4 5 8 125572 R3 1557 1pm 2557 S1 2392 3 3 1557 2 1 1 s 2 FHM 3927 392274 0828 2 5 Fromeq3 7 r52 08287 0554 0176 48 75 We must also evaluateF MHM In this case F 2 1413 8 8 126672 Rlt13 2667 RM 2667 s1 2214 3 3 2667 l 1 2 8 2 2 F 13H24 2147 214 74 0689 2 8 The last view factor needed isF Fm3H4 It is calculated using 1 1667 2 Rlt13 2667 R41667 s1 2153 3 3 2667 1 1 2 5 2 2 FMME 1537 153 74 E 0323 Substituting values into eq 1 7r82 06897 0323 FEM 82 7 52 0600 Solving eq 2 for FH1 FH FM 7 FM 0600 7 0176 0424 4 Answer The view factor FH2 is identical to FH1 by symmetry 1414 1416 A tubular enclosure is divided into four surfaces for the purpose ofa radiation analysis Surfaces 1 and 2 are on the inner wall ofthe tube and surfaces 3 and 4 are on the ends Find the view factor EM Approach 1 1 The secret ofthis problem is to construct surface 5 as shown Then consider rst the r enclosure formed from 1 3 and 5 and calculate EA and Ea Next remove L surface 5 and find the remaining view gtr gt factors in the enclosure formed from 1 2 3 l and 4 Assumptions 1 The radios39ty is uniform over each ace 2 All surfaces are diffuse Solution Construct an artificial surface 5 as shown and consider the enclosure formed by surfaces 1 3 and 5 To nd 22 use the equation in Fig 1418 wi L i RTL 05 R5 2 1 05 2 S1 15 R 052 21 1 2 75 2 F srs74 52 l 1 2 af x 741 T01715 The view factor FH may now be found by summation as F 17FH 1701715 08284 Applying reciprocity 2 L 7 7 F F F Fg3m2 H02071 IrLL 4 Now apply summation to surface 1 to get note that surface 1 can see imelf Fm r H Hs Because of symmetry EH 17 25 17 202071 05858 Next remove surface 5 and consider the enclosure formed by surfaces 1 2 3 and 4 The view factor FEM can be found using Fig 1418 M L R 2 2L025RA 2 1025 1 2 s 025 S1 1415 1 2 2 FM s7 S274 7 4 2 r3 1 FM 187 182 7 WE 005573 From summation F3a2 1 F3a4 FHI 17 005573 7 08284 01158 By reciprocity L 2 7r F FZa3 E2 002896 A2 7rLL 4 By symmetry FH2 FH1 05858 FH FH3 002896 By summation FH1FH2 FH3 FH4 1 Finally FHz 17Egt1 7 FH 7 FHA F 17 058587020717 002896 017814 Answer 1gt2 1416 1417 Two rectangular surfaces face each other in parallel planes as shown in the gure The top surface is aligned directly 39 Each r 39 quot quot quot into two equal parts so that A1 A2 A3 A4 Find the Viewfactor EM Approach The View factor from the combination of 1 and 2 to 3 and 4 can be found from Fig 14 16 Repeatedly apply shape decomposition 244 39 symmetry and reciprocity to determine the 51 unknown View factor Assumptions H 1 The radios39ty is uniform over each 1 3 surface q 2 All surfaces are diffuse Solution By shap e decomp osition F MHUM 02 Fu2p4 By symmetry this reduces to FMHM 25m Applying reciprocity and noting that AE Aug 2 2A FH 2 F MHM FHM 12 Again using shape decomposition F i2gt34 F H anz By symmetry Fa Fm Flown F H To find FEMHM use Fig 1416 with X 3075 71Z05 L 4 From the chart Fumwm m 0095 Similarly for Fxhhuse X7 3075 71l025 L 4 L 4 From the chart FE m 005 Therefore H4 00957 005 0045 4 Answer 1417 1418 Three radiation zones are used in an oven analysis Surfaces 1 and 2 are placed on the side wall and surface 3 is placed on the bottom ofthe oven Using dimensions on the gure calculate EA Approach Use shape decomposition symmetIy and reciprocity Assumptions 1 The radiosity is uniform over each urface 2 All surfaces are diffuse Solution By shape decomposition 3gtl2 FH Fm By symmeuy F Bun ZFH Using reciprocity F MIL A5 Hm Solving for FL F A3F1XID 2A H Use Fig 1417 to evaluate YEW with H i 0525 X 4 1 3 075 X 4 From the graph 7quotth m 018 Therefore lt3gtlt4gtlto18gt 7W7 0216 Answer W F a 1418 14719 A planar black surface at 850T is parallel to a second planar black surface at 300 C The two surfaces are separated by a small gap Ifa third black surface is inserted in the gap calculate a quot 39 in outer uuaw wiLilauu 39 39 b the temperature ofthe inserted surface Approach 3 Without the insert use the formula for radiation between two in nite black parallel plates With the insert in place the net radiation from the left 1 surface to the insert equals the net radiation from I the insert to the right surface Equate the heat rates and s 39 olVe for the unknon insert tempera ure Finally calculate the net radiation from one of the plates to the insert Assumptions 1 The surfaces are Very large compared to the size of the a 2 The surfaces are black 3 Radiation is the only mode of heat transfer Solution 39 the small gap Without the insertion of surface 3 the net heat ux is Qlgt2 FMGGA 724 A Since the plates are assumed in nite EA 1 T 850 C 1123 K 557x10 3 2W 4 A m K T2 300 C 573K 10123475734 84066 If surface 3 is inserted in the gap then the net radiation from 1 to 3 must equal the net radiation from 3 to 2 or QH Q3gt2 Mam 7T mam 7T Because plates 1 2 and 3 are all assumed in nite H H2 1 Thus T 7T T 7T I T The net radiative ux with the insert is 9 FHMTFTJ i i T4 T4 7 11234 5734 3 T 2 960K687 C QM 557x10 8 W A 2 A1123 950 42033124 Answer m K m Comment greater ifplate 3 had a lower emissivity 1419 1420 A wire coated with insulation runs through the center ofan evacuated cylindrical channel in a space station Both the wire and the c annel wall may be approximated as black s aces The channel wall is maintained at 50 K and the wire dissipates 75 mW per cm 0 The outer diameter ofthe wire is 03 cm and the inner diameter of the channel is 15 cm Calculate the wire temperature Approach a L iIa VA A i i Use the formula for radiation between two infinitelylong parallel cylinders Assumptions A 1 The wire is Very long compared to the gap between wire an channel 2 The surfaces are black WI My 3 Radiation is the only mode ofheat transfer Solution Ifthe wire and channel 39 39 4 channel end effecm may be ignored For two infinitelylong concentric cylinders from Fig 1435 wt Th 4 QH 1 7 5i Since 5 2 1 this reduces to QH 514103 34 We know the heat generated in a 1cm length ofwire The wire surface area is AI IIDL Ir03cm1 cm 09425cm2 Solving the rate equation for T yields l 4 t i 75mwl7w3vwl i 7 m 344K 1 Answer W J Mimxlorzmz K409425cm21010jzj QHz 1 14 20 1421 A generator is constructed of an inner rotating cylinder the rotor and an outer stationary cylindrical shell the stator During testing the rotor develops a thermallysensitive vibration and the engineer suggests that one side ofthe rotor is hotter than the other To correct the problem a stripe ofblack paint is applied to the hot side The unpainted rotor surface has an emissivity of 02 and the painted surface has an emissivity of 098 The rotor outer surface is as 160quotC an the stator inner surface is at 95 C The diameter ofthe rotor is 12 m and the gap between rotor and stator is 5 cm Calculate the net radiative ux from the rotor on the painted portion and on the unpainted portion Approach Given that the rotorstator gap is small compared to both the rotor size and the width ofthe paint stripe we may use the formula for radiation between two in nitelylong parallel plates Assumptions 1 The rotor diameter and length are large compared to the gap between rotor and s or 2 The paint stripe is wide compared to the ap between rotor and stator 3 All surfaces are gray and diffuse J Twp 39r75 C Solution 39 39 The gap between rotor and stator is very S i mi w small compared to the diameter of the rotor 39Ro i39of therefore we may assume that the rotor and stator act like two in nite at plates The net radiative heat ux is given y A 4 gm 60 42 A L71 Er T160 C433K E95 C368K On the unpainted portion ammo3 2WK4j433473684K4 W QH 7 ml 2 4 Answer A 71 m 02 06 On the painted portion Q 557x10 3 YK4433473584 W m39 565 4 Answer A 1 1 m2 7i71 098 06 Comments T 39 rface can alien dranraricall In this case ifthe radiation is a 39 39 39 oftotal near uau m 39 39 r 39 r then altering the J 39 39 39 couldhaveaL39 39 39 quot 39 uau rer 39 4 r 39 39 unironnit over the entire rotor surface 14 21 1422 Two large at plates in parallel planes are at 60 F and 420 F The cold plate has an emissivity of 06 and e hot plate has an emissivity of 02 Find the net rate ofheat trans er Approach I 1 Use the formula for radiation between two in nite p arallel plate Assumptions 1 Radiation is the only mode ofheat transfer 2 All surfaces are gray and diffuse u 0 Solution 5 F rl 7 20 r For two parallel at plates the net radiative heat ux is given y E E4 7 L4 L71 51 52 T 60 F 520R Substituting values 8 Btu 4 4 A 0171x10 hHillR4sso 420 R Btu 159 lt Answer 2H2 A T2 420 F880R Q1 2 A LL4 h 05 02 Comments The answer is the same if the emissivities are switched as is evident from the rate equation 14 22 1423 A 39 ahnt a 39 39 39 wilucwaiilS lSSOnC A i i i hield 39 r r as shown The shield has an emissiv39ty of 013 The emissivity ofthe thermocouple bead is 068 and the emissivity of the pipe wall is 094 TL cueiiiciem uu mZquotC 39 39 39 39 WC 39 Minimum on the shield is 35 WmZ C he shield has a diameter of35 cm and the pipehas a diameter of6 cm Calculate the actual gas temperature Approach Perform two energy a balancesioneonthe lPc wail lb 5 thermocouple and the other Flu 39 S Lnll T3 6 3 on the shield including i gt w both radiation and gt gt gt co vection T Thzrmo CJv iL T1 51 resulting nonlinear I equations simultaneously Assumptions 1 The thermocouple is small relative to the shield 2 The shield is long relative to is radius 2 All surfaces are gray and diffuse Solution Begin with an energy balance on the thermocouple Designate T1 as the gas temperature T1 as the thermocouple temperature and 739 L 39 39 39s 39 the thermocouple therefore it appears black quot 39 39 r 1 the convective heat transfer to the thermocouple i e AiszaT2tTreAi Ti 7T2 Simplifying T24 T24 2Ti T2 equ N t rm an energy balance on the radiation shield The shield and pipe may be approximated as two in nitelylong cylinders Note that convection occurs on both sides ofthe shield Using Fig 1436 MWJAA 2h3T7T3 eq2 1 1754 7 7 53 54 4 Eq 1 and 2 Ti and T3 39 linear and 39 39 by trial and error Known parameters are w 52068 7500 C773K h270 2 a 7335cm m C W 1013 3350 C623K h335 2 a 7460cm m C a 094 a557x10 8 4 m2K4 Using equation solving software T3 763K 4 Answer Comments en with the shield the measured thermocouple temperature is 10 degrees below the actual gas temperature 39 ll quot L39 quot areused39DJululCl Ev Or r a mm v 14 23 1424 A hotpotato with a surface temperature of 375 F and emissivity of 093 radiates to a large room with walls 0 keep the potato warm someon vers it loosely with a single layer of aluminum foil which has an emissivity of 02 Calculate the net rate ofradiation loss from the potato with and wi out the foil Assume the potato is spherical with a diameter of45 in and the foil forms a spherical shell around the Approach oi Equate radiation from the potato to the foil to radiation from 3 the foil to the room Assumptions d 1 Radiation is the only mode of heat transfer P q o 2 The foil is Ver close to the potato 39 3 All surfaces are gray and di use room 4 The potato is spherical Wequot 1 Solution To be in 39 uom p foil must equal the ux from the foil to the walls ofthe room We are assuming that radiation is the only mode ofheat transfer The potato is assumed tobe spherical and the foil is assumed to form a spherical shell around the potato The net radiative ux between two concentric spheres is 7 46034 Qlgt2 2 LEE 5r 52 2 The foil is very close to the potato so we assume 7 m 72 and QH2 becomes Q MW 42 H2 f 77 5r 52 The net radiative ux from the foil to the walls of the room is Q2 A25239TT24 TEA Equating the net uxes 247T QM 9 a 1 Azsza wafer 52 51 52 Since 7 m 72 A1 m A2 Simplifying and solving for T2 E TATA21752 SSSRV530RA1702 72 7 0 2 39 728R 22 2 2702 g 093 The net radiative ux with the foil is 2 7 47 4 7 45 3 Btu 47 A 47 Btu QHriigz r2 I 74r 212 020171x10 h 2R4728 530 JR 7305 11 The net radiative ux without the foil is Btu A 4 A 7 Btu mj sss 7530 JR 72867 5 ttjmas onnw g Q taadtrx C omments Adding the foil reduced the radiative ux by nearly an order of magnitude In the actual situation natural convection should also be considered 14 24 1425 A motorist leaves a car parked on a driveway with the engine on The driveway is covered with a layer of ice 5 in thick at 32 F The undercarriage ofthe car in the vicinity ofthe catalytic converter is at a temperature 0 190 F 39 quot quot L J 4 L 39 L emissivity of 075 Ignoring heat transfer by convection calculate the time required to melt the ice The latent heat of fusion and density of ice are 1435 Btlilbm and 624 lbmftzl respectively Approach Consider the undercarriage and the ice as two infinite Umitr a nu 3a parallel plates and calculate the net rate of radiative transfer g x I from the undercarriage to the ice Calculate the mass ofice 0 quot it A to be melted per unit area Use the heat transfer rate the e mass of ice and the heat of fusion to determine the melting Lei time Assumptions 1 Radiation is the only mode ofheat transfer 2 Th undercarriage is a black surface 3 The ice is gray and diffuse 4 As the ice melts water drains off and exposes more ice 5 The undercarriage and ice layer are very large compared to the distance between them and compared to the ice thickness therefore they can be treated as infinite parallel plates Solution i i 39 39 39 thenetrate ofradiative transfer is from Fig 1435 QM 6014424 A LL1 5i F1 Using T 19ODF 650R andT 32 7F 492R 0171x103984 550474924 QHL htt R 7154 Btu n 2 A LL71 h 1 075 The mass of1 ft2 ofice is lbm 2 1ft 2 m pV 62405m lzmz 1ft 261bm Designating the latent heat of fusion of ice as L and the rate ofradiative transfer for 1 ft2 of ice as q the time to melt the ice is 14352 lbm 7 Lm 7 lbm 7M h A Answer 1 154 Comments The car would have to be idling a long time for all the ice to melt 14 25 14 26 An astronaut with a surface area of 18 m2 generates 150 W of body heat during a space walk The exterior of the spacesuit is exposed to outer space which is black at 4 K Both exterior and interior surfaces of the suit are silvered with an emissivity of 04 It is necessary to keep the surface of the astronaut s body which has an emissivity of 085 at no less than 16 C in steady state Should the suit be made of one layer of silvered material or two layers Approach The suit follows the contours of the astronaut s body closely39 therefore assume radiation between the body and the suit is like radiation between two infinite parallel plates Equate the generated heat to the net radiation from the body to the inside of the suit Similarly equate the net radiation on the inside to that on the outside of the suit Solve for body temperature Assumptions 1 Conduction in the suit is neglected 2 The suit and body may be represented as two infinite parallel planes 3 The suit and body are gray diffuse surfaces 4 Space is black at 4 K 5 Re ections between locations on the suit exterior are negligible 6 No radiation is incident on the astronaut from the sun or the space vehicle Solution Let the body be surface 1 and the inside of the suit be surface 2 The radiation from the body to the inside of the suit is given by see Fig 1435 aAT 7 T2 Q 1 1 71 1 82 1n steady state the body heat is equal to the heat leaving the outside of the suit 1f the suit has a single layer of silvered material then the net radiation from the exterior of the suit to the environment which is black at temperature T3 is Q A520T24 7 T3 Solving for T24 T 4 T 4 2 A820 3 Substituting this into the first equation gives o A T14 7 7 T34 Ag 039 Q 2 1 1 7 1 1 82 Solving for T1 produces l 1 1 39 Z 150w Q 712A0T 150wLiiijM18M567x10 8 7 8 2 2 1 085 04 04 AU W m2 K4 i 4 WK44K4 m2 18m2567gtlt10398 T1 295K22 C This is above the limit of 16 C There is no need for a second layer of silvered material which would produce a body temperature higher than 22 C Answer 14 26 at 100 K is stored in the In 1427 A cryogenic dewer is constructed oftwo concentric spheres 45 cm and 53 cm in diameter Liquid nitrogen 39 ner sphere and the space between the spheres is evacuated The ou atemperature of220K L 39 39 39 of th 39 If V ter sphere ha 0 p aporization of the nitrogen is 210 kJkg determine the number of kilograms of nitrogen evaporated per hour Approach Use the formula 39 39 39 r39 from Fig 1435 Find the rate of evaporation using the latent heat and the net radiative heat Assumptions 1 1 Radiation is the only mode ofheat transfer 2 Both surfaces are gray and diffuse Solution From Fig 1435 the net rate of radiative transfer between two spheres is A can 7 T24 Q1 L 5i 52 2 Substituting values note that the surface area of a sphere is 472 2 2 4 Ecm 557x10 3 2W 4 100472204K4 2 m K 10 cm 2 7 109w 1 0023 45cm 0023 0023 53cm 39 quot f L Therate ofevaporation ofnitrogen is 1i 35005 7109w 5 1W 1h H 7 A k lt Answer 210 1000J kg lkJ Comments In actuality tsuppons n rL r s t n 4 losses through the structural supports 14 27 1428 During a deposition process a long strip heater is centered above a long wellinsulated plate as shown in the gure The heater is maintained at 400 C and the surroundings are black at 25 C The geometry is effectively twodimensional and radiation is the only mode of heat transfer Using data on the figure find the plate temperature 3 0 cm Approach 391 7 Had Consider the heater the plate and the I I F surroundings to be a threesurface K7 radiation enclosure and apply Eq 1439 a 7 OF 3 and 1440 View factors may be found using Table 142 l a r Assumptions 50 cm 1 Radiation is the only mode ofheat ansfer 2 Both surfaces are gray and diffuse 3 The geometIy is twodimensional 4 Theplate is perfectly insulated T Ln 3 K 7 2178 K Solution 6 I 039 v Let surface 1 be the heater surface 2 be the 3 I E plate and surface 3 be the surroundings 1 0 5 Then apply Eq 1439 to surfaces 1 and 2 I l to get 7 J 7J g 4 3 Eqm RH RH J 7J J 7J Q2 4 Eq 2 R2 R2 Also apply Eq 1440 to surface 1 yielding AUTampQi EqG Surface 2 is perfectly insulated therefore it is a reradiating surface and 2 397 24 Surface 3 is black therefore J 397 3 Substituting these expressions forJz andJ into Eqs 13 and noting that Q2 0 since 2 is insulated gives Ar TAr Q1 EQ 4 RH RH T47 T4 7 o 2 3 J Eq5 R2 RH AU ampQ Ego Once all quot 39 39 quot 39 quot Fn 46 willL L 1 39 39 39 39 39 7 Q and T2 The radiative resistances are Rl ARM 1 a3 Aas 1 amp ampEg By reciprocity 14 28 l l R R 2amp1 A2F2gtl AlFlAZ 1amp2 R1 17 81 A181 The view factor FH2 may be calculated using the first geometry in Table 142 with 1143245 W2 25 L 20 L 20 l l l l W W 2 4 2 W W24 2 15252 4 2 25152 4 2 2 H2111 ll 2W1 215 All the radiation leaving surface 1 arrives either at 2 or at 3 the surroundings Therefore F 17F l0745 0255 1gt3 1gt2 The view factor FH1 may be found using reciprocity 30 0745 FHAF42 0447 A2 50 Finally FH3 is by summation FH3 liFH1 17 0447 0553 All these view factors are now substituted into the radiative resistances to get 121 447m392 o2 03m1m0745 1 3 03m1m0255 1 H 05m1m0553 By reciprocity R R 121 1309m 2 R 362m2 H2 447m392 R1 17 087 03mlm087 These are the resistances for an area of depth 1 m In addition to these resistance values use the given data T1 400 C 673K T3 25 C 298K 2gt1 0498m392 039 567x10 8 2 K m and solve Eqs 46 simultaneously for J1 Q1 and T2 It is recommended that equationsolving software be used for this purpose The result for T2 is T2 545K 272 C lt1 Answer 1429 1429 A chemical reaction chamber is in the shape of a cylinder with a height of 26 ft and a diameter of 06 ft A diskshaped heater with an emissivity of 092 entirely covers the bottom ofthe chamber and generates 13 kW of eat The side wall is at 86 F and the top end is at 65 F Both the side wall and top have an emissivity of 073 The chamber is partially evacuated and the only signi cant mode ofheat transfer present is radiation The back ofthe heater is wellinsulated so that all the heater power is removed by radiation into the chamber Find the heater temperature Approach Model the reaction chamber as a three 0 L lt1 l 3 surface enclosure Apply Eqs 1439 and 5 39 39 z OJ 3 Fig 1418 and the summation relation Assumptions 1 Radiation is the only mode ofheat at 43 transfer 2 All surfaces are gray and di use o 7 Solution C I I Apply Eq 1439 to the three surfaces in Q 2 I 3 k IA the enclosure 39 J 7 J 7 Eq 1 RH Eq 2 Q n R a Also apply Eq 1440 A WampQ E W J2 71229 Eq 5 7 Eq 6 Eq 3 The radiative resistances are R1 2 A15 7 1 MHZ R2 RH A2 Fm By reciprocity 1 1 RI AZFH MHZ 2 1 RH 1 AEFH MHZ 1 1 R2 Asz Asz 3 1 RH R1 R21752 1231753 51 52 The view factor EA may be calculated using the equation in Fig 1418 with R 1 E0115 R2 20115 25 L 14 30 HR 71101152 Rf 01152 FH s 7 s2 7 4r3r1 2 77177712 7 403032 0013 From summation s1 77l FH2 17FH3 170013 0987 To find F z start with F F 0987 37gt2 1a2 From reciprocity 2 7 E 0987 F A3F3 gt2 2 3 A2 7r0626 The radiative resistances may now be calculated as RH 2 358ft392 R 112 0987 2 1 72 R178 2 R3gtl 112 0013 00569 27gt1 l R 359ft392R 3 7r0626ft200569 37gt2 121 172i 0308ft392 112092 2 122 Ljimmssw 7r0626ft 073 121 Zi7131 e 112073 The final step is to solve Eqs l6 simultaneously for the unknowns J1 J2 J3 Tl Q2 Q3 Note that Q1 13kW 4436 Btuh T2 546R andT3 525R Using equationsolving software the result is Q392 74322Btuh Q7114Btuh T11789R1329 F 4 Answer Comments In practice the curved side wall would probably not be isothermal but would be hotter near the heater and cooler farther away A more sophisticated analysis would divide the side wall into a number of separate radiation zones 1431 1430 A Very 39 formed from quot 1 antefl cover plate as shown The crosssection ofthe enclosure is in the shape ofan isosceles right triangle Assuming gray an um Approach Use Eqs 1439 and 1440 for a threesurface enclosure Find View factors from Table 142 10 cm 3 Assumptions 1 Radiation is the only mode ofheat transfer TI 500 C 2 Surfaces 1 and 2 are gray and diffuse 10 cm T2 40090 3 Surface 3 is black T3 1DDoC Solution Apply Eq 1439 to surfaces 1 and 2 J 7J 7 Q Eq 1 7 Q EQ 2 2 R2 From Eq 1440 J 7 W 7amp9 Eq 3 J2 qr 7R2Q2 Eq 4 Surface 3 is black therefore J 7 Eq 5 Eqs 1 7 5 are 5 equations in the 5 unknowns J2J3Q and Q2 The radiative resistances are 1 1 1 RI Rl R2 a 4m 3 MHz a MM 1 17 5 17 RH RH2 R1 R2 52 AqFH A1Fi7gt2 Alf A252 To find EM use Table 142 with 7 10107 102 102 EM W ZW 0293 w 210 By summation Fm 17H2 170293 0707 By symmetry FM 7 Fm 7 070 The radiative resistances may nowbe calculated as 1 4 1 r 341m R 141m R 01m0293 H R 01m0707 1 7 7 5 7 R 141m 10m 3 01m0707 R 01m05 R2 i 333mquot 01m075 The final step is to solve Eqs 15 using I 500 C 773K T2 400 C 673K andT3 100 C 373K Using equationsolving software the result is Answer 1 4 L radiation nuclei of r 14 32 1431 A short thinwalled tube of length 25 in and diameter 075 in is open atboth ends The tube is made 0 copper s 3025 and is maintained at 155 F The surroundings surfaces are at 65 F Including radiation fromboth 394 4 L 39 39J e fr 39 39 Approach I Add two imaginary black surfaces 2 and 3 to reciprocity Find the net heat leaving surface 1 in f the enclosure formed by 1 2 and 3 Add this to the L 1 5 39 net heat leaving the outside ofthe tube Assumptions 1 All surfaces are gray and di use 2 Surrounding surfaces are far from the tube Solution Construct two imaginary black surfaces which cover the ends ofthe tube as shown To calculate FH3 use Fig 1418 with 0375 015 R 015 R 25 J R HR2 10152 s1 2 1 2 464 015 To calculate the View factor use the formula 1 2 i 2 3 FM 1 s7s2741 1 4557454274Mj 00215 2 r 2 0375 By summation FH 17th 1700215 0978 By reciprocity D 2 AF 7 F DF 0 75in0 978 RM 2H 7 H 39 39 rrn Ai 39DL 4L 4251n The thermal circuit for the enclosure is shown to the right The surface resistanceR1 is 17 5 Ir Am m2 2 A1 IrDLr075in25in 144 00409 Therefore Rl 733 ff 00402 025 The space resistance between surfaces 1 and 2 is RH2 L 333 tr2 1 AFH2 00409 200734 14 33 By symmetry Itelirssw Eu Since I I the thermal circuit reduces to the circuit V shown to the right The total resistance is q L R R R 333 333 RH Hz mg gtlt L24 RH2 12 333 333 The net radiative energy leaving surface 1 out the ends ofthe R R tube is calculated as quot 5 A A 3993 Q am gt 1 RM RM 0171x10 2 B2 4615475254R4 Eh 7 h 39n R 7 0 478 240 tr2 39 h From the outside of the tube to the environment QM WA 3442 QM 0250171x1039800409 75154 75254R4 417 h Finally Q Q gm 047s117 155BT lt Answer 14 34 1432 A ceramic cooktop has a heating element 9 in in diameter A person s hands are placed over the heating element at a distance of45 in Model the hands as a disk with an emissivity of 090 and radius of9 in and the heating e ement as a parallel disk with an emissivity of 097 The heating element is at 350 F and the walls ofthe kitchen are at 70 F If the hands are reradiating surfaces what is their temperature Approach 7 Consider the heating element the hands M lt and the walls ofthe room tobeathree 0 H4m5 surface enclosure and apply Eqs 1439 1 T E and 1440 View factors may be found 61 07 3 39 73 F using Fig 1418 LiaM Assumptions T 350 F G 1 Radiation is the only mode ofheat g 077 Hat tr f if 35v rmmlm j A ans er 2 The hands and heating element are gray and diffuse 3 No heat is conducted into the hands they are reradiating 4 The hands are shaped approximately like a disk of 9 in diameter 5 The walls ofthe room are far enough away so that they may be considered black Solution Apply Eq 1439 to the heater and the hands ie Q Eq 1 Q EQ 2 2 R a3 Also apply Eq 1440 to the heater to get J mum Eq 3 The hands are a reradiating surface therefore 4 J an The surroundings may be considered black therefore J 39T 3 Substituting these expressions forJ2 andJ3 into Eqs13 and noting that Q2 0 since 2 is insulated gives JraTZ JraT Q EQ 4 RH RH o 34quot 39 J Eq 5 RH RH Once all quot 39 39 4 39 Fn hS willL L 1 11Q and 7 The radiative resistances are 1 1 l R 2 R 3 R2 3 4 ArFHz 4 ML A A2F2gt3 By reciprocity 1 1 17 5 RI Rl AZFH Arm 2 As The view factor EA may be calculated using the equation in Fig 1418 with 14 35 R1713745in7 1127157451171 L 45m 2 L 45111 2 112 3 11 2 2 3 R1 1 RH2 sis2 74r2r1213732 7 454321 0382 All the radiation leaving surface 1 arrives either at 2 or at 3 the surroundings Therefore FH3 17H2 1703820618 By symmetry F F 0618 2gt3 H3 The areas are 2 A1 A2 12 044m2 Substituting values into the radiative resistances 1 R 593ft392 H2 0442ft20382 1 R 366ft392 H3 0442ft20618 1 R 366ft392 H 0442ft20618 R21 RH2 593 ft392 R1 i 007ft392 0442ft2097 The given data is z350 F810R T370 F530R o 567gtlt10398 m2 K4 Using the radiative resistances and these data in Eqs 35 and solving simultaneously gives T2 677R 217 F It is recommended that the equations be solved using equation solving software Comments Ouch In reality some heat would be conducted into the hands and convected from the hands so this high temperature would not be reached Also hands are usually warmed in a transient process rather than a steadystate one 1436 1433 An attic space is in the shape of an isosceles triangle with dimensions as shown The oor of the attic is at 50 F the two ceiling surfaces under the roof are at 30 F and the air in the attic is at 40 F The convective heat transfer coef cient is 14 Btu hftZ F on the oor and 066 Btu hftZ F on the ceiling surfaces The emissivity of the ceiling surfaces is 093 Calculate the heat transferred from the oor by radiation and by convection per foot of depth if the oor is covered with a unsilvered insulation 5 1 087 b silver backed insulation 5 1 019 Approach The attic space is a three surface enclosure Apply Eqs 14 39 and 14 40 View factors may be found using Table 14 2 T3 30 F 2 30oF Assumptions 53 0 93 1 2 093 l The attic space may be modeled Attic space as two dimensional 30 ft A 2 All surfaces are gray and T1 50 F diffuse Solution Apply Eq 14 39 to the three surfaces in the enclosure 39 J1 T J 2 J1 7 J3 Q Eq 1 1 R1gt2 R1gt3 J 7 J 7 Q2 7 2 3 J2 J1 Eq 2 R293 Rzol J 7 J 7 Q 3 1 3 2 mm 3 R3gt1 R3gt2 Also apply Eq 14 40 J1 0 2114 TRiQi Eq 4 J2 0 2124 TRzQz Eq 5 J3 0T T R3Q3 Eq 6 The radiative resistances are l l l Rt Rt Rb 1 2 A1112 1 A1113 AQFH l l l l l l R R R R RH AQFH A1112 2 H AgFH A1113 3 H A3129 AQFH R 611791 R217 2 631793 A151 A252 A353 To nd FHZ use Table 14 2 F w1w27w3 3O20720 05 1 2 2w1 230 This result could also have been deduced by inspection By symmetry 11 11 2 05 From reciprocity 30 05 FM Ar Loy A2 20 By summation Fzg 1 inl 1 7 075 025 14 37 The radiative resistances may now be calculated as l R H2 30ft05 l o3 30ft 05 1 00667 ft39l R21 R1 00667 ft391 R Eel 02ft391 le2 R H 20ft025 7 17 087 1 30ft087 17 093 R R 000376 ft391 2 3 20ft093 The final step is to solve Eqs l6 using 000498 ft39l Btu h ft2 R4 It is recommended that the calculation be done using equationsolving software The result for the bottom surface with unsilvered insulation 81 087 Q1 426Btuh Answer If the calculation is repeated with silverbacked insulation 81 019 Q1 965Btuh Answer By comparison natural convection from the bottom surface per foot of depth is given by Q hAT1 7T 1430ft1 507 40 F420 1 Answer T150460510R T2T330460490R 0390l7lgtlt10 8 Comments Silverbacked insulation significantly lowers heat loss by radiation from the building Heat transfer by natural convection is comparable in size to heat transfer by radiation when unsilvered insulation is used It is often the case that radiation is important when a surface is cooled or heated by natural convection in air 1438 1434 Two diffuse gray at plates of size 60 cm by 45 cm face each other in parallel planes One plate has an emissivity of 066 and a temperature of460 C while the other has an emissivity of 031 and a temperature of 120 C The spacing between the plates is 14 cm T e surroundings are black at 20 C Calculate the net rate of radiation heat transfer from the hot plate Approach 60 CM 3 v Su ru owd vi 5 e two plates and the HBOMC 6 gt T3 1 20 Consider th 7 surroundings to be a threesurface 0 L 1 radiation enclosure and apply Eqs 394 5 CM 1439 and 1440 View factors may 39 befoundusing Fig 1416 Ham J Jo l V 21 o 3 I Assumptions 1 Radiation is the only mode of ansfer 2 Each plate is gray and di use 3 Only consider radiation from one side of each plateAhe side that faces the opposing plate Solution Apply Eq 1439 to plates 1 and 2 to get Eq 1 Eq 2 R2gt3 R2 r Also apply Eq 1440 to the two plates yielding AUTampQr EqG J2 qr Rig Eq 4 2 Since the surroundings surface 3 are black 3 The radiative resistances are 1 Rl ARM 1 a3 A ArFHz 1 amp 48 By reciprocity 1 1 a amp2 A2F2gtl ArFHz A 1 5 R2 1 5 Am A25 The view factor EA may be calculated using Fig 1416 with 429 Z 321 L 14 L 14 From the plot Fm m 05 By summation Fm 17FH2 17 05 04 14 39 By symmetry F F 2gt3 H3 04 The radiative resistances may now be calculated using A1A2 m jm 027m2 100 100 R 6l7m392 H2 027m206 1 72 R 926m H3 027m204 1 72 R 926m H 027m204 RH RH2 617m2 R102mgii66191m72 R2 02m23l3l 83924m72 We are now prepared to solve Eqsl5 simultaneously with the following additional inputs T1 460 C 733K T2 120 C 393K 039 567gtlt 10398 m2 K4 Using equationsolving software the result is Q12300W 1 Answer 1440 14 35 In the middle of a blizzard a horn eowner s furnace fails In desperation the homeowner turns on the oven of the electric range and opens the oven door The thermostat for the oven is set at 450 F and the inside walls of the oven are black Assume the oven door acts like a reradiating surface ie neglect convection with an emissivity of 089 The walls of the kitchen are at 55 F Calculate the temperature of the oven door and the electric power consumed by the oven Approach The oven cavity may be covered by an imaginary black surface at the temperature of the interior of the oven The door is represented by a gray reradiating surface perpendicular to the imaginary oven surface A threesurface enclosure analysis may then be used taking the environment as the third surface Assumptions 1 Radiation is the only mode of heat transfer 2 The oven door is gray and diffuse 3 The oven is powerful enough so that the interior walls can be kept at 450 F while the door is open 4 The oven walls are black Solution Apply Eq 1439 to plates 1 and 2 to get Q J1 J2 J1 J3 l Rl2 Rl 3 Q J2 J3J2 J1 2 R2gt3 R2al Surfaces l and 3 are black therefore J1 0T1 J UT Surface 2 is a reradiating surface therefore J 2 UT Q2 0 Eqs 1 and 2 become aT 4 aT 7T3 RHz RH 0 0T24 7723 0T24 7714 R2 R2gt1 The radiative resistances are R HZ AIFM 1 R W AIFH l R 2a3 A2F2 3 By reciprocity 39 3 5Uff DJAJIUS 72 55 F Eq 1 Eq 2 Eq 3 Eq 4 1441 l l The view factor FH2 may be calculated using Fig 1417 with 2 08 2ft X 25ft X 25ft From the plot in Fig l4l7 RH2 m 022 By summation RHE 14H2 17022 078 By symmetry RH FM 078 The radiative resistances are 1 72 R 7 225ft2 022 7 090 R 4 0256ft392 HE 225ft2 078 R 0256ft392 l H 7 225ft2078 R RH2 0909ft392 The given temperatures are T1 450 F 910R T3 55 F 515R Solving Eq 4 for T2 2gt1 1 l 0256ft392 910R 7 WM 4 5154R4 T2 RH 0909 ft 2 673R 1 RH 0256ft392 RH 0909 ft392 Substituting values into Eq 3 0l7lx10398j910476734R4 0l7lx10398j910475154R4 Q 7 hft R hft R 0909ft392 0256ft392 Q15007 Answer 1442 1436 An evacuated enclosure is in the shape ofa cube with a side length of 12 cm The top has an emissivity of 0 and 7 quot 39 39 39 o 47aml f210aC The remaining four sides are perfectly insulated Find the side wall temperature Approach llcm 4 39Ti 800 Model the inside ofthe cube as a 9 O 5 threesurface radiation enclosure 3ltA H Li Apply Eqs1439 and 1440 View 17 s I I25 39 39 lCM l fl asctors may befound using Fig 14 l T glove 6 039 m1 gt 3 1 cm 1 04 7 Assumptions 1 Radiation is the only mode ofheat ansfer 2 All surfaces are gray and di use Solution Apply Eq 1439 to the three surfaces in the enclosure Q 1 Eq 1 R172 R172 J 7J J J 92 34 4 Eq R27 RH Q3 g Q Eq 3 3 1 R272 Also apply Eq 1440 A WampQ E W b nampg E16 4a eamp2 qu Because Q3 0 J r7 Eq 7 The radiative resistances are 1 1 1 R12 R13 R22 4 A1572 A A137 A A2F2gt3 1 1 1 1 7 p 7 7 amp tau Aas amp amp 7 7 p 3 141772 1421727 A 52 A151 A252 The view factor EA may be calculated using Fig 1416 with 12cml 12cm1 L 12cm L 12cm Fromtheplot F 02 By summation FHZ 17FHz 1702 08 By symmetry Es s The areas and radiative resistances are now calculated using 12 12 2 m m00144m A A 100 100 14 43 43 4A1 005761112 347m2 12H RHZ 00144m202 7 1 R 7 00144m208 1 R 868m392 R H3 00144m208 H 17047 1251225m392 R2 2783m392 00144m 085 00144m 047 We are now prepared to solve Eqsl5 simultaneously with the following additional inputs T1 800 C 1073K T2 210 C 483K Q30 o 567gtlt10398 868m392 12H W m2 14 Using equationsolving software the result is Q 7Q 349W Answer T3 972K 699 C Answer 1444 1437 A selective surface has a spectral hemispherical absorptivity which Varies with wavelength as shown in the gure Find the total hemispherical absorptivity at 185T Approach Use Eq 1454 obtaining blackbody fractions from Table 143 Assumptions a 04 L 1 The surface is di use Solution 015 I De ne Z 2pm 04 015 2 35 21 7 35 pm a 04 From Eq1454 applied to three bands I aiFW 042 00am T185 C458K 4T 2pm458K 915 pm K 4T 35pm458K1603 pm K By interpolation in Table 143 0001 3 FM 00200 FM FMT 7E 00270000193 00198 FAN kFMT 1700209s Therefore a 0150000193 0400198087098 08614 Answer C omment At this low temperature Very little ofthe radiation is at wavelengths less than 35 pm 14 45 1438 The spectral emissivity ofa polished aluminum plate is 07 for A lt 66pm and 04 for A gt 66pm Calculate the power emitted by this plate at 300 K and at 900 K Approach C Multiply spectral emissivity in each band by the blackb ody fraction for that band Total emissivity is the sum of all the spectral s c ontribution Assumptions 1 The plate is diffuse Solution De ne 1 66pm 5 07 52 04 TA 300K and TL 900K The total hemispherical emissivity ofthe plate at I is 5A FMn EzFu A TA 66um300K1980pmK By interpolation in Table 143 7 54 Farm 17 EMA 17 0054 0935 Ther fore 5A 07 0064 04 0936 0419 At T the total hemispherical emissivity is 55 FMrI EzFu A 66pm900K 5940pmK By interpolation in Table 143 325 Fa rm 17 FM 17 07325 02575 Therefore 55 07073250402575 06198 The emissive power at 300 K is W mZKA EA gAaTA 0419557x103 300K 192124 Answer 1T 14 45 14 39 The tungsten filament of an incandescent light bulb is heated to 2600 K What fraction of the energy emitted is in the visible range 04 to 07 pm 7 What is the wavelength ofmaximum emission Approach Use blackbody fractions to find the total energy emitted in the visible Use Wien s law to find the wavelength of maximum emission Assumptions none Solution The fraction of radiation between A and 2 is F4774 F0457 FUJHT ATO4pm2600K1040 pm K QT 07pm2600K1820 pm K By interpolation in Table 143 raw 0000684 FD 00421 FZMT 0042170000684 00414 4 Answer To find the wavelength of maximum emission apply Wien s law MT 2898 pm K m W112m 1 Answer 2600K Comments Most of the energy emitted by an incandescent light bulb is in the infrared range Eyes are very sensitive instruments and can detect light even at low intensities 1447 14 40 A glowing coal can be seen in a darkened room at the Draper point which is 798 K What fraction of the energy emitted by the coal at this temperature is in the visible range 04 to 07 pm 7 Approach Use blackbody fractions to find the total energy emitted in the visible Assumptions one Solution The fraction of radiation between A and 2 is F F 17427 07127 7 07M AT 04mm798K 319 um K QT 07um798K 559 um K Checking in Table 143 we find that blackbody fractions for these values of AT are virtually zero Effectively we are out of range of the table To make further progress use the equation 15 e 322 62 6 F 23 0717 4 quot2 n n n2 n3 where Z 71438769umK 7 AT The infinite series is best evaluated using appropriate computer software The series converges very rapidly so only one term is needed at these values of AT The results are F 4009gtlt103916 07M F0 l93gtlt10398 The fraction of energy emitted in the visible by the glowing coal is F l93gtlt10 874009gtlt10 16l93gtlt10 8 4 Answer 17427 Comments The eyes are incredibly sensitive organs able to detect radiation even when only 2 parts in 100 million are in the visible range 1448 14 41 A roughly spherical satellite with a diameter of 16 m is in orbit around the sun at an average distance of 67gtlt108 km The sun radiates like a black sphere with a radius of 695 gtlt108 m and a temperature of 5800 K The surface of the satellite is coated with a material having an absorptivity of 088 for it lt 25 pm and an absorptivity of 012 for it gt 25 mm The satellite rotates and may be assumed to be isothermal at 310 K Determine the amount of internal heat generation in the satellite Approach Calculate the heat flux from the sun striking the satellite see example l4l Also determine the total absorptivity and emissivity of the satellite Use these to perform an energy balance on the satellite Assumptions 1 The satellite is isothermal 2 The satellite is not in the shade of a planet or moon 3 The system is in steady state 4 Outer space is at approximately 0 K 5 The satellite is diffuse 6 Solar radiation is emitted uniformly in all directions Solution The total energy emitted by the sun is Q w 4nR3aT where A is the surface area of the sun and R is the radius Substituting values Q5 4695x108 m2 567x10 8 2 4 m K To determine the amount of solar radiation impinging on the satellite draw an imaginary sphere with the sun at the center The radius of the sphere is the distance between the sun and the satellite The heat flux due to solar radiation on the inside surface of the sphere is Q5 Q5 389gtlt1026W 51 7 2 47r67gtlt108km A h T 47rS2 T To find the total absorptivity of the satellite use 5800K 389gtlt1026W 5p 0 D IFlJrATS 0 2 1 Form We use the solar temperature in this calculation because the energy originates at the sun and has the spectrum of the sun ATS 25mm5800Kl4500 um K By interpolation in Table 143 FHA 0966 The total absorptivity becomes 0 088096601217 0966 0854 From Kirchoff s law Therefore the total emissivity may be calculated using 9 glFljrle 8217F071Tm D iForsz 0521 F0717m Where T m is the temperature of the satellite The emission occurs at the satellite temperature so use ATM 25um310K 775 um K From Table 143 1449 Form 0 g 0880012l 012 To find the heat absorbed use the projected area of the satellite in the direction of the sun s rays W 2 m QM 067mth 08547r08m2 69 119w To find the heat emitted use the actual surface area of the satellite QM 84m 2 UT4 01247r08m2 567x10398 sat sat 4 m2K4310K 505w The heat generated is Q39W 7 QM 5057119 387W d Answer 1450 14 42 The Windshield of an automobile is made of glass with a transmissivity of 091 between 03 and 3 pm Outside this range the transmissivity is virtually zero Calculate the total transmissivity for the Windshield for solar radiation Tm m 5800 K and for radiation from the car seats which are at 20 C Approach Use an equation similar to Eq 1454 with or replaced by r Assumptions none Solution efine A 03m 22 3 Hm Also define r 11 it lt A r 12 A lt it lt J r 13 R lt it The total transmissivity for solar radiation is T T1F Hm T2mezzzw 13F Maw From the problem statement 11 13 0 therefore I 12F mmmw 12Foemw Fawn ATM 03m5800K1740 um K um 3um5800K 17 400 pm K By interpolation in Table 143 FUMW 00335 FD 0979 r 091 09797 00335 086 4 Answer For radiation from the car seats ATM 03um293K 879 um K 12TH 3um293K 879 um K By interpolation in Table 143 F0 my m 0 Foimm 0000136 r09l0700001360000124 4 Answer Comment This selective transmissivity is called the greenhouse effect Radiation from the sun can get in but radiation from interior surfaces cannot get out 1451 14 43 An engineer sends an oxidized metal test sample to the laboratory to determine the emissivity The lab reports a value of 072 for the total normal emissivity What value of total hemispherical emissivity can the engineer expect Approach Use Fig l440b Assumptions 1 The surface is diffuse Solution Fig l440b gives the rate of hemispheric to normal emissivity for metals In this case an 39 072 From Fig l440b for metals 4 093 8n Therefore 5 093072 067 4 Answer Comment The oxide layer is a dielectric while the underlying metal is a conductor The surface actually behaves somewhere between these two extremes The insulator plot in Fig 1440 would also give a reasonable estima e 1452 14 44 The normal spectral emissivity of silicon oxide on aluminum is 096 for it lt14 um and 004 for Approach it gt 14 pm Calculate the total hemispherical emissivity at 300 C for this surface emiss1v1ty Use Fig 1440 to determine hemispherical emissivity from normal emissivity Use an equation similar to Eql452 to find total Assumptions 1 Hit lt14um the surface behaves like an insulator since emissivity is high 2 Hit gt 14um the surface behaves like a metal since emissivity is low S olution For it lt14um treat the surface as an insulator From Fig l4 40a if g 39 096 2 m 0945 g g 0945096 0907 s where 81 is the hemispherical emissivity for it lt14um For A gt 14 pm treat the surface as a metal From Fig 14 40b if g 39 004 i 1128 8 g 128004 00512 a where 82 is the hemispherical emissivity for it gt 14 pm The total hemispherical emissivity is g glFllrlT l 52F177w 9 glFllrlT 92 17 F0717 ATl4um573K802 pm K By interpolation in Table 143 Form 0 Therefore g g 00512 4 Answer 1453 S am at 160 psia and 400 F enters a nozzle with a Volumetric ow rate of 6615 cfm cubic feet per 61 te minute The inlet area is 145 in Ifthe steam leaves at 1500 Ms at a pressure of40 psia nd the exit temperature Approach Use the first law for an open system for a nozzle Find properties in bles 39 I 160psia 1g 40 psia ft V2 1500 s specialized the steam ta Assumptions 400ml 7 1 Potential energy change is negligible 2 The system operates in steadystate i 5515 off 3 The nozzle is adiabatic Solution From the rst law specialized for a nozzle 1 2 h 2 hi 2 The inlet Velocity may be found from 3 1min m2 144m 2 llbf Btu lbm 1197E 2 2 5 2 Taking the inlet enthalpy from Table B12 W V22 Btu 1 2 2 n2 7 1218 1095 71500 112 hi 2 2 lbm 2i 1 At P2 40 psia 1 17 1197Bi from Table B12 lbm T2 32an 4 Answer 62 Oxygen at 220 enters a wellinsulated nozzle of inlet diameter 06 it The inlet velocity is 60 Ms The oxygen leaves at 75 F 10 psia The exit area is 001767 z Calculate the pressure at the inlet Approach Use the rst law for an open system specialized for a nozzle Apply conservation ofmass and the ideal gas law to determine gt inlet pressure 7 a T 7 220 F T2 75W Assumptions V 502 122 10 psia 1 Potential energy change is negligible 5 2 The system operates in steadystate 3 The nozzle is adiabatic 4 Speci c heat is constant 5 Oxygen behaves like an ideal gas under these conditions Solution From the rst law specialized for a nozzle V 2 V 2 h 4 h 2 Solving for V2 and noting that for an ideal gas with constant speci c heat Ah CPAT V2 sz mm V3 lzc T 40 V3 Using data for speci c heat from Table B8 2 V2 2 02215 Blquot 220775R M m 50 12593 lbmR lBtu llbfs s s From conservation of mass m quot391 2 pi ViAt p VzAz Using the ideal gas law MR V V RT A RT 2A2 Solving for inlet pressure V2 A2 T 1259001757220450 1 P2 l 10 sia VAiTz p 50 032 63 A wellinsulated nozzle has an entrance area of 028 m2 and an exit area of 0157 ml Air enters at a velocity of 65 ms and leaves at 274 ms The exit pressure is 101 kPa and the exit temperature is 12 C What is the entrance pressure Approach Use the rst law for an open system specialized for a nozzle Apply conservation ofmass and the ideal gas law to determine inlet pressure 2 101 kPa 1 a Assumptions V 7 65 s T2 712 C 1 Potential energy change is negligible V2 27 E 2 The system operates in steadystate s 3 Thenozzle is adiabatic 4 Speci c heat is constant 5 Air behaves like an ideal gas under these conditions Solution From the rst law specialized for a nozzle V 2 V2 hi 2 hi 2 For an ideal gas with constant speci c heat Ah CPAT therefore V2 V2 C 777 2 m 2 2 2 Solving for I V2 V2 n 2 WT 2 Speci c heat depends on temperature and should be evaluated at the average of inlet and outlet temperatures but the inlet temperature is unknown As an approximation evaluate the speci c heat at the exit temperature and correct later ifnecessary From Table A8 At T2 27312285K c m1004L P kgK 2 2 2 m 274 755 15 2 7 k 285K 320K 20004 J 1000 kgK 1k From conservation of mass quot391 quot2 pi ViAt p V2A2 Using the ideal gas law MR MP V 2 V A RT A RT 2 2 Solving for inlet pressure gg g1o1274j j j 268kPa 4 Answer 1 Ai T2 E 028 285 Comment The entrance temperature of320 K is very L 39 0285 K 39r L A signi cantly between r A L 39 A 64 Carbon monoxide enters a nozzle at 520 kPa 100 C with a velocity of 10 ms The gas exits at 120 kPa and 500 ms 39 auu ideal gas behavior quot 4 L 39 Approach Use the rst law for an open system gt specialized for a nozzle Assumptions 1 520kPa a 120 kpa 1 Potential energy change is negligible T 7 1002C z 2 The system operates in steadys te V2 5002 3 The nozzle is adiabatic V 102 5 4 Speci c heat is constant 5 bon monoxide behaves like an ideal gas under these conditions Solution From the rst law specialized for a nozzle 4 v i5 2 2 For an ideal gas with constant speci c heat Ah CPAT therefore V 2 V 2 c r 2 T gt 42 f Speci c heat depends on temperature and should be evaluated at the average of inlet and outlet temperatures but the outlet temperature is unknown As an approximation evaluate the speci c heat at the inlet temperature and J correct later ifnecessary From Table A8 at 1quot1 00 273373 K 5 m1045 k K g m 2 2 107 f 2 2 1 4rf 25 5 k 65 Lowvelocity steam with negligible kinetic energy enters a nozzle at 320 C 3 MPa The steam leaves the nozzle at 2 MPa with a velocity of410 ms The mass ow rate is 037 kgs Determine a the exit state b the exit area Approach Use the rst law for an open gt system specialized for a nozzle Find properties in the steam tables I 320 C 1 2 MPa Assumptions R SMpa 1 Potential energy change is negligible V2 4102 2 e system operates in steadystate 5 3 The nozzle is adiabatic Solution a From the rst law specialized for a nozzle V1 2 V2 h 2 117 2 Determining the inlet enthalpy by interpolation in Table A12 2 2 2 h2 h L7V 2 30mg 1000J 7 410 395 29mm6 i 2950 E 2 2 kg 1k 2 kg kg Interpolate in Table A12 at 2 MPa and h 2960 kJkg to nd the nal temperature of T2 274a C The exit state is superheated vapor at 2 MPa and 274 0C b The mass ow rate is given by m p VzAz 39 L 39 A quotaquot 4 L r 39 39 L interpolation inTableA12 v2 0118m3kg i 43 847k D v2 0118m kg m Solving the mass ow rate equation for exit area P2 107X104m2 Answer sunnundmgs Approach Use the rst law fur an upeh sysLEm allmmaung wk and pummel energy and pmpemes m the steam tables 39 Assumptions 1 zsn c T2 WC 1 Futmnal margy change xsneghgxble 31 S kpa P 15 kpa z The system uperates m steadyrstate Solution a mm the rstlaW 2 7 QVTWHZM 1 gz ZmlhT gzl u wekk m rstlzwbecumes 2 2 UQ m h1L em twi z z The area atthemletxs 2 I25 21an quotq Imam2 z z The velumty atthemletxs usmg data 39um TableArlZ 3 m kg v a a A l 491m2 4 ms The Exit areas 2 Afr L5 Ul m2 Obtammg speci c vulume by mterpulatmg m Table A712 94 198 st swsm A1 196 s Rearrangmg the rst 12w 2 2 em 7 Q h 2 e m 2m 952 646 QN94E 3mg LEE 7 zm LEE 7 5 kg 1k 2 kg 1k 2 Q zem71A wem71kwlt Anwver kFa The est temperature is 1 WC mhe mametez atthe mieth a em a Fmdthe exxtvelun and Assume cunstznt wen c heats nuns 21 t m o Ev u epmp xes T ZXSC Assumptions PI Z 5mg 1 Futennal margy change xsneghgxble V ZSEIms z The system uperates m stead estate 3 The mffusans adiabatic An39behzveshke an ideal gas unda these cundmuns 5 Specx cheats are cunstznt Solution 2 Hum the rst law spemahzed fur a dxffuser 2 2 kn is i Furamdeal gasvnthcunstzntspem cheaL AheM therefure 2 W 97 k IUUUJ 2TrTz V z1u1kg KWxsenu 4 Answer s h By cunservauun ufmass ml 2 W11 BVzAz Usmg Lhexdeal gas law 2 i Mvn e P v RT 2 RT2 P 2 P D v T H 2 on Sulvmg fur em diameter Dzzq ghza sljmwm zsu P2 T V 1m 878Em Answer 85Z73 1m 678 superheated steam enters a well rmsulated diffuser at 14 7 psla 32m saturated vzpur at ayery luw weed Fmd the exrt pressure and temp Approach Use the rst law fur an upeh sys elrrruhauhg vmrk heat and put tern ehu al energy and pmpemes m the steam tahles F and 4mm MS The steam exlts as erature T zzu quotF Assumptions l putehual margy change ls negllglble z The system uperates m steadyrstate 3 The mffusals adlabah Solution a me the rst law speelallzed fur a drffuser 2 2 4 hi V 2 z Usmg yalues ufehthalpy at the mlet 39um Table 312 at 14 7psla 32mquot 2 Em 4mm2 lZUZli 5 lhrh lZEIS KBim 2 Hum s In Table Brl fur h ZUS 3BtulbmLba39e are twu pusslble suluuuns runs at MUDF and 3812 psla and the uLha39 ata uuut WElaF mi 4 l nl H N V has httle effeet uh the exrt state 4 Answer aeyelups 24 5 MW ufpuwer whatrstherhass uwrate7 A pproach Use the rst law fur an upeh system ehrruhauhg heat laheue mar and putehual Energy Fmdpmpemes m the steam tahles quot Itheturbme Tsnnquotc Assumptio l putehual P n a MFa y change ls negllglble e e turbmels adlahaue Solution Assummg an 2642me turhme wllh hegllglhle kmeue and putehual Energy ehahges the rstlawheeurhes W ta 7 wlth values uf enthalpy 39um Tahle A42 24 5 xl sz 348B 67 mm lldkg hr a 610 Air enters an adiabatic turbine at 900 K and 1000 kPa The air exits at 400 K and 100 kPa with a velocity of 30 ms Kinetic and potential energy changes are negligible Ifthe power delivered by the turbine is 1000 kW a nd the mass ow rate b nd the diameter ofthe duct at the exit Approach T 900K Use the rst law for an open system eliminating heat kinetic R loookpa energy and potential energy Find properties using ideal gas relations Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The turbine is adiabatic 5 Air behaves like an ideal gas under these conditions 6 Speci c heat is constant Solution 1 100kPa a Assuming an adiabatic turbine with negligible kinetic and 30ms potential energy changes the rst law becomes 2 W 7 m h 7 l5 For an ideal gas with constant speci c heat Ah CPAT therefore m 7 W 7 W c T 7T2 With values of speci c heat from Table A8 at the average temperature of 650 K 7 kIJOOOkW 7 88k g 4 Answer 1053 9007 400K 5 k K b Exit area is related to velocity and mass ow rate through A2 V2 From the ideal gas law 8314 M 400K 1 RE L 1 1 A 2897 kg 100kPa kg kmol Substituting values 3 1881151115 00721m2 303 5 D2 w0303m 1 Answer I I 611 Saturated steam at 320 C enters a wellinsulated turbine The mass ow rate is 2 kgs and the exit pressure is 50 kPa Determine the nal state ifthe power produce 39 39 0 kW is b 400 kW Approach T 320T Use the rst law for an open system eliminating heat kinetic saturated steam energy and potential energy Find properties in the steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible e system operates in steadystate 4 The turbine is adiabatic Solution a Assuming an adiabatic turbine with negligible kinetic and P 7 SOkPa potential energy changes the rst law becomes 2 7 W m hi 7 la From Table A10 the enthalpy of saturated steam at the initial temperature of 320 C hl 2700kJkg 1000 W 100 kW hfl 1 2700 i lkw 7265x106i 25mg m kg 1 kJ 21 kg kg s From Table A11 at 50 kPa we see that the exit enthalpy is greater than that of a saturated Vapor therefore the 39 39 39 39 andby39 39 39 TableA12 T2834 C 4 Answer v2 50kPa b Recalculating the exit enthalpy for the higher power condition 1000 W 400 kW 2mg 1kw kg 1 25x106 i2500 2 kgs kg From Table A11 at 50 kPa we see that the exit enthalpy is between that ofa saturated liquid and a saturated Vapor therefore the exit state is in the twophase region an hz 11 kg 4 7 h2 7h 7 250073405 7 kg 711 7 25457 3405 Summarizing the nal state is twophase with OkPa x 0937 x0937 Answer 612 Superheated steam at 16 MPa 600 C enters a wellinsulated turbine The exit pressure is 50 kPa T 39 Ifthe exit pipe is 16 m in diameter and carries 11kgs of ow nd gy turbine pro uces 10 MW ofpower the Velocity at the exit Neglect kinetic ener Approach Use the rst law for an open system eliminating heat kinetic energy and potential energy Find properties in the steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The turbine is adiabatic Solution Assuming an adiabatic turbine with negligible kinetic and potential energy changes the rst law be omes 1 SOkPa W mm 45 With Values of enthalpy from Table A12 at 16 MPa 600 IOMW 1000kW 1MW kJ 2784 93 kJ kg W 7 35 7 112 hi Iii kg 11kgs From Table A12 ath 50kPa and 112 2784kJkg I 150 C To ndthe exit Velocity m va2 V2 Solving for Velocity and using Values ofv2 from Table A12 at T2 150 C and 2 50kPa 3 11115118911 V2 7va 7 s 2 g 7 2132 lt Answer A2 15 2 s 1 i m 2 613 Air at 550 C and 900 kPa is expanded through an adiabatic gas turbine to nal conditions of 100 kPa and 00 C The total power output desired is 1 MW Ifthe inlet Velocity is 30 ms what should the inlet pipe 39 energy 3 diameter be Neglect kinetic and potential Approach T 550T Use the rst law for an open system eliminating heat kinetic P gookpa energy and potential energy Find properties using idea gas relations V1 SOWS Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 39 tate 6 Speci c heat is constant Solution 72 300T a Assuming an adiabatic turbine with negligible kinetic and P 7 lookpa potential energy changes the rst law becomes 2 7 W m For an ideal gas with constant specific heat Ah CPAT therefore W W hi 1quot 0 Ti 75 With Values of specific heat from Table A8 at the average temperature of 700 K 1000kW m m 1075i5007300 c 5 k K b Exit area is related to Velocity and mass ow rate through A V From the ideal gas law 8314 M 550273 K 3 RT kmolK m v W k 0252k 2897 g 900kPa g kmol Substituting Values 3 37211510 2524 00325m2 303 S D E W0203m 4 Answer IL39 I 614 Airat510 eal 39 39 39 quot L 39 r 39 101 kPa In steady state the turbine produces 50 kW ofpower Find e e it temperature Hint use Eq 256 b the mass ow rate Approach T 510 C Use Eq 256 to nd the nal temperature Then apply the rst R 450kPa law for an open system eliminating heat kinetic energy and potential energy to obtain mass ow rate Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The turbine is adiaba 5 Air behaves like an ideal gas under these conditions 6 Speci c heat is constant 7 The process is quasistatic 1 101kPa Solution a Sincethe turbine is ideal the process is quasistatic For adiabatic quasistatic process ofan ideal gas with constant speci c heat from Eq 256 I From Table A8 for air k14 E 1 T2 T i 510273 j M 511 K 238 C P 450 b Assuming an adiabatic turbine with negligible kinetic and potential energy changes the rst law becomes h For an ideal gas with constant speci c heat Ah CPAT therefore W W m h n h 0 Tl T2 With Values of speci c heat from Table A8 W 50 kW kg m 0173 4 Answer CP 1 2 Losi 5107238K s kgK 615 Saturated steam at 3 MPa enters a 39 39 39 39 L state quotquot 600 kW ofpower The mass L L39 39 39 4 L the exit temperature quot is 093 Find Approach I 3MPa Use the first law for an open system eliminating heat kinetic saturated steam energy and p t 39 energy To find the nal temperature it is necessary to iterate with data from the steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic ener change is negligible e system operates in steadystate 4 The turbine is adiabatic Solution a Assuming an adiabatic turbine with negligible kinetic and potential energy changes the first law becomes W m hi 7 l5 From Table A11 the enthalpy of saturated steam at the initial pressure of3 M39Pa is h 111 2804 kJkg 9 093 h2Zhl 7 500 kw 2804 E 2375 E m 84kg 1min kg kg min 60 s 39 39 39 39 39 neium 39 39 TL 39 h 7 u found by trial and error Begin by assuming a Value forT then use 9 to comp tel l2 with data from Table A10 If 112 2375 kJkg the iteration is complete ifnot select a new Value off and recompute To begin assume T2 25 C From Table A10 hfhHzlhg hx 10490932547 7 2442 2375 kJkg This is Very close to the calculated Value of 2375 therefore T2 25 C 4 Answer C omment In this example we selected the correct result immediately In reality it would be necessary to try several temperatures before zeroing in on the correct Value 616 In a 3hp compressor carbon dioxide owing at 0023 lbms is compressed to 120 psia The gas enters at 60 F and 147psia The inlet an 39 quot Fi iinai volumetric L 39 39 3min 39 F Approach T 60 F Use the rst law for an open system eliminating heat kinetic R 147psia energy and potential energy Find properties using ideal gas relations Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadysta 4 The compressor is di 1 5 Carbon dioxide behaves like an ideal gas under these conditions E 120psia 6 Speci c heat is constant Solution Assuming an adiabatic compressor with negligible kinetic and potential energy changes the rst law becomes W m hi i h For an ideal gas with constant speci c heat Ah CPAT therefore W mc T 7T2 W T2 I ms with data from Table B8 25441311 7311 p mp T2 60 F7 533 F4 Answer 0 023111 0 195713quot Loos s lbm R 1h From conservation of mass V2 MM quot1 m2 v2 RT 3 ET 00231073533460R n V quotH 2 5 quot1 39 lb quot391quot 7278 Answer gM 120psia 4401 m m lbmol Comments If more accuracy is desired the calculation should be repeated with speci c heat evaluated at the average of the inlet and outlet temperatures ie at 605332 or 2965 F The outlet temperature was not known at the beginning ofthe calculation so speci c heat at a temperature near the inlet temperature was used 617 A wellinsulated compressor is used to raise saturated R134a Vapor at a pressure of 360 kPa to a nal pressure of900 kPa TL u mate in s eady 39 p input U U v I the ow rate r P is 0038 kgs what is the nal temperature Approach 1 360kPa Use the rst law for an open system eliminating heat kinetic saturated vapor energy and potential energy Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The compressor is adiabatic Solution Assuming an adiabatic compressor with negligible kinetic and potential energy changes the rst law becomes 1 900kPa Wm 41 W hz 4 m From Table A15 at 360 kPahl hg 2506kJkg 112 2506E 272 00397g kg kg s From Table A16 at P1 900 kPa andhz 2716 T m 40 C Answer 618 Air owing at 05 m3min enters a compressor at 101 kPa and 25 C The air exits at 600 kPa and 300 C During this process 250 W of 39 Wha 39s 39 input Approach T 25 C Use the rst law for an open system eliminating kinetic energy and P IOIkPa potential energy Find properties using ideal gas relations 3 V 05m min Assumptions 1 Potential en gy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 Air behaves like an ideal gas under these conditions 5 Speci c heat is constant Solution From the rst law for an open system 1 120psia dE V 2 V 2 V 7W m 4 7 m hz Qw V 2 h 2 82 Z 2 539 Assuming steady conditions one stream in and one stream out no change in kinetic or potential energy the rst law becomes 0QWmhrhi For an ideal gas with constant speci c heat AhcpAT therefore WQmcTrT2 Using data in Table A1 3 051 101kPa2897 kg 1mquot V V RM min kmol 60s n A 3kg I s V RT 8314 M 25273K kmol K The average temperature of the air is Tm w152 0 435K Using speci c heat Values from Table A8 interpolated at TM k J W 7250w 984x10393 11018 257 300 DC isooow 73kW 4 Answer s g 619 Refrigerant134a enters a compressor at 0quotF and 10 psia with a Volumetric ow rate of15 3min The refrigerant exim at 70 psia and 140 F If the power input is 2 hp nd the rate of heat transfer in Btuh Approach T 0 F Use the rst law for an open system eliminating kinetic and P 710 7 psia potential energy 3 V 15 mln Assumptions 1 Potential en gy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate Us Solution From the rst law neglecting kinetic and potential energy 0 Q 7 W m h 7 The mass ow rate is T 140 F B 70psia m Vi Taking the inlet speci c Volume v1 from TableB16 3 15L lbm m 3319 4703 m lbm Q WWW 4 1 Again using Table B16 341213 Q2hp m h 31911quot 129171023 M lhp 1W mm lbm 1h Q450 Answer 6 20 A pump is used to raise the pressure of a stream of water from 10 kPa to 07 MPa The temperature of the water is the same at inlet and outlet and equal to 20 C The velocity also does not change across the pump If the mass ow rate is 14 kgs what power is needed to drive the pump Assume frictionless flow and no significant elevation change Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 Water is incompressible Solution For a frictionless pump with no elevation changes W mquot 7P2 so using the density of water at 20 C from Table A6 W m Pl 7P2 p 10kP 7700kP Mk u W 79677W7968kw d Answer 5 99821 lkPa 3 m 6 21 A 2hp pump is used to raise the pressure of saturated liquid water at 5 psia to a higher value Assume the velocity is constant the water is incompressible and the flow is frictionless If the mass flow rate is 6 lbmsec find the final pressure Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 Water is incompressible Solution For a frictionless pump with no elevation changes W MP 7 P P 2 P1 1 From Table Bl l vf 00164 ft3lbm atP 5 psia Substituting values 550ftlbf 1ft2 2 hp S 2 1 hp l441n P2 Sig f2 7 lbm ft3 39 6 00164 s lbm I72 826 psia Answer 619 6 22 Water is pumped at 12 ms through a pipe of diameter 12 cm The inlet pressure is 30 kPa 1f the pump delivers 6 kW find the final pressure Assume frictionless incompressible flow with no elevation or velocity changes Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 Water is incompressible Solution For a frictionless pump with no elevation changes W ma 7P2 P1 Pz p The mass flow rate is 2 m pV A 998k g311z jnwj 135E m s 2 s 76kWMI998k g3 1kW m 445gtlt106 Pa 445MPa kg S P2 445MPa i Answer P2 R 7 30kpa1000Pa7 m 1kPa 135 6 23 A 1hp pump delivers oil at a rate of 10 lbms through a pipe 075 in in diameter There is no elevation change between inlet and exit no velocity change and no oil temperature change The oil density is 56 lbmft3 Find the pressure rise across the pump Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 The oil is incompressible Solution For a frictionless pump with no elevation changes W 7 P2 E 7 I 1 mv m ft lbf 550 2 561b f 1hp 5 If 2 ft 1 hp 144m BAP2 214psia 4 Answer iolb m S 6724 l ace p whlchlsaboullooo lthlgh l p p n edeol lnhp preach Apply the rst law for an open syslecn dropplng lens for Lranslents heat and klneolc mergy Replace the mthalpy dlffamce Wllhh1 eh m e 000 fl Assumptions 1 The syslern ls adlabatlc and lsoLhennal z Kmehc ener change ls negllglble 3 The syslern opaahes ln steadyrsmte 4 The pump lsl ea 5 Water ls lncompresslble l I 1 45 psi1 Solu on The rst law for an open syslen ls dE W W n 7W m 4z mh z dt Q nzlpl Z g2 2 L W 1 l l l l adlabauc operallon to get 07Wm e m z 22 For an lncompresslble llqulol undergolng an lsoLhermal process hi hlVP2 Pl Sub Smung lhls lnLo the rst law and noung that v 1 p Wemfl Pz mgzrzz 2 454mg 144 ln m lhp 621133 550lbf 1 s 2 31b m3217 2 f R 4000a ll amp 5 5 3217i2 5507 s s W 389hp4 Answer ommenl The aclual pump chosen should have mole horsepowa than thls because ln reallty there are some fnctlonal effects 625 Airat150 C quot quot 100kPaquot Lquot quotis36ms quotL 39 I Approach P 40kPa Use the rst law specialized for a thronle T150 C 1100kPa Assumptions V 36ms 1 Potential energy change is negligible 4 Air may be considered an ideal gas under these conditions 5 The throttle is adiabatic 6 The inlet and exit pipe have the same diameter 7 Speci c heat is constant Solution For a throttle with no heat transfer no change in kinetic or potential energy and no work the rst law reduces to For an ideal gas with constant speci c heat AhcpAT therefore oc Trn TH From conservation of mass 2 pViAi p VzAz Assuming the diameter of the inlet and exit pipes are the same and using the ideal gas law ET ET 2 RM PM Solving for exit Velocity V2 V i MEI 144m 4 Answer R s 100 626 Saturated liquid R134a at 24quotC is throttled until the nal quality is 0116 Find the nal temperature and pressure Approach Use the rst law specialized for a thronle I 24a C 39 x2 0115 Assmnptions saturated liquid 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 system operates in steadystate 4 The throttle is adiabatic Solution From Table A14 at 24 C hhf29 kJkg From the rst law applied to a throttle h 19 This problem must be solved iteratively since neither the nal temperature nor the nal pressure is known First assume a nal temperature and determine the corresponding quality X For example assume T1 hen from Table A14 7 111 7h 7 829400 11g 7h 247400 This value ofxz is too high so a different value osz is chosen By trial and error the nal value osz is found to be 0167 X2 T2 m SE C To verify this result calculate the nal quality as 07 X 0116 7 h 252 7 607 f The nal pressure is the saturation pressure at 8 C which is given in Table A14 as 130388MPa Answer 627 Saturated liquid R134a at 80quotF undergoes a throttling process The pressure decreases to 4 of its original Value Find the exit quality Approach Use the rst law specialized for a thronle I SODF P P 4 saturated liquid 2 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 system operates in steadystate 4 The throttle is adiabatic Solution From the rst law applied to a throttle From Table B14 If 7 80 F then P 1014psia sat Bt AtR1014h 11 473113quot 1 P 101 1 252ps1a 4 4 To nd the quality at the exit we need to interpolate in Table B14 From Table B14 I E quot1 539s 5 238 1314 1025 10 267 1466 1032 By interpolation at 252psia 19139 and kg 1028 The quality is 711 2527139 x2 0127 lt Answer hgih 10287139 Appr Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system oper 4 The throttle is adiabatic Solution From the rst law applied to a throttle From Table A12 sup erheated Vap or oach Use the rst law specialized for a thronle A supply line contains a twophase mixture of steam and water at 240 C To determine the quality ofthe mixture a throttling calorimeter is used In this device a small sample ofthe twophase mixture is bled off from L quot 4 ll l in 39 u the L quot ofthe mixture in the Llietluuuliu 125 C L 39 1 main steam line Steam Valer T1 1 240 C Calorimeter ates in steadystate T2125L39 P1 l0 kPn Supply linc h 2725 E From Table A10 at T 240 C h1037 1 1 2804E kg 5 kg Hmd 7h h Mo955 4 Answer 1037 kg 7h 28047 x 6 29 In a heat pump R l34a is throttled through an expansion coil which is a long copper tube of small diameter The tube is bent in a coil both to fit in a compact space and to provide a large pressure drop The refrigerant enters as saturated liquid at 5 C with a flow rate of 0025 kgs and exits as a twophase mixture at a pressure of 200 kPa The wall of the coil may be assumed to be at the average temperature of the inlet and outlet Heat is exchanged by natural convection and radiation from the outer surface of the coil with a combinedheat transfer coefficient of 6 Wm C to the surroundings at 20 C The expansion coil has an outside diameter of 8mm and a length of 22 m Calculate the quality at the exit state Approach Use the first law specialized for an open system eliminating kinetic and potential energy changes work and the transient term Calculate the heat transfer from the convection rate equation Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The heat transfer coefficient is uniform over the surface of the coil and independent of temperature Solution From the first law for an open system 2 2 Q39W 7W5 2 x gzx V gzg Assuming no change in kinetic or potential energy and steady conditions with no work the first law becomes 0 m hz lnteIpolating in Table Al4 gives the enthalpy of saturated liquid at 5 C as h1 567kJkg The heat lost from the surface of the coil is Q WTM Tm The saturation temperature at the exit condition of 200 kPa is from Table Al5 T2 temperature of the outside of the coil is Tm g 5101 72550c 7101 C The average The surface area of the coil is A 7rDL7r8mm l m 22m 00553m2 1000 mm Therefore the rate of heat loss by convectionradiation is 397 7 W 2 o 7 QihA7W7TM76m2oc00553m 7255720 C77748W Solving the first law for exit enthalpy h Q 7748W 6 11000 2 m 0025k g kg 1 S 56396J 564kJ With values at the exit pressure of 200 kPa from Table Al5 the exit quality is x7 h2 h 7 56473684 hg 7h 24137 3684 00956 4 Answer Comment The exit enthalpy was very close to the inlet enthalpy with heat loss having little influence on the final enthalpy This is often the case39 therefore throttles are usually assumed to be adiabatic One way 39 39 39 39 39 39 in steam 40 kgs of subcooledliquid water enters at ls c and50 kPa Superheated steam enters at 200 c and50 kPa 39 39 39 39 39 39 t u a Assume a the tank is wellinsulated Approach Apply conservation ofmass and the rst law for an open system sat liquid Assum tjons 30 kpa 1 Potential energy change is negligible gt0 3 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The tank is adiabatic superheated vapor 200 quotC Solution 50 km From conservation of mass i quot 2 quot 3 From the first law for an open system d5 V 2 V at r Qv 41Vv m h T ngerl kl Tgzlj Neglecting changes in kinetic and potential energy and assuming an adiabatic tank and steady operation the rst law becomes 0 mm quotlth 7 min For state 1 in iiihalp L 39 39 39 inc enthalpy of saturated liquid at 15 C which is from Table A10 k 53 h k From Table AlZ at ZOODC and 50 kPa k 2877 h k From Table A11the enthalpy ofsaturated liquid at 50 kPa is U 3405 k Eliminating ml in the first law by using conservation of mass 0 mlhl m2h2 mi quotEVE 0Mihrhamhrh3 Solving for m2 rm h 7 hi 40 kjgjmr 3405 h ha 2877 7340513 A m2 438kg s Answer 631 In a desuperheater superheated steam is converted to saturated steam by spraying liquid water into the steam Using data on the gure calculate the mass ow rate of liquid water Approach Apply conservation ofmass and the rst LN ii law for an open system q Upg lvmlml 39 gtanv 4 The desuperheater is adiabatic 5 Pressure is constant during the process So t39 ll loll From conservation of mass quot391 quot392 quot392 From the rst law for an open system 2 2 mWw Zm h Lgz Z t 11 Vga a 2 2 Neglecting changes in kinetic and potential energy and assuming an adiabatic tank and steady operation the rst law becomes 0 quotah We From Table A12 the enthalpy of superheated steam at 250 C and 2 MPa is h 2902kJkg For state 2 the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 30 C which is from Table A10 h2 1258 kJkg TL r 1 Y I I L L 39 39 for all three streams The 2 MPa is from Table A11 r e ective enthalpy of saturated vapor at h1 2799 kJkg Eliminating mj in the rst law by using conservation ofmass 0 m quot2112 m nah owmimmzwm Solving for m2 77ml 11171117703290272799 116 I A m2 Id 7 n V Answer 125872799 kg 632 A at JC Water at u oi 39 39 39 39 What are the required mass ow rates ofthe two inlet streams Approach sal slcam Apply conservation ofmass and the rst law for an open system 1 8 A Assumptions T 2 1 Potential energy change is negligible Wm 2 Kinetic energy change is negligible 20 quotC 0 H a 3 The system operates in steadystate I00 kl a 4 The mixing chamber is adiabatic 5 Pressure is constant during the process Solution From conservation ofmass m m2 m3 From the rst law for an open system 2 2 it 94 7W Zm h V7gzezm h V2gz Neglecting changes in kinetic and potential energy and assuming an adiabatic tank and steady operation the rst law becomes 0 mm m a m From Table All for saturated steam at 100 kPa h 26755 kJkg For tate in nme s 2 L me enthalpy 39 39 c which is from Table Alo s For state 3 in nnarp 39 n p 39 39 c which is from Table Alo h 16757 kJkg Using conservation ofmass to eliminate m in the first law 0mrmzhrmhrrnaha 0quot ah m2hi rm2h7Tmaha Gatheringterms ma rkk rh kg k m 01 7h 826755716757k Answer a l kJ g774kgs him 2675578396 3 m 87 m2 026 kgs4 Answer 633 Steam with a quality of 088 and a pressure of 20 kPa enters a condenser The steam ow is divided equally among 20 tubes 21cm in diameter which run in parallel through the condenser The same amount of heat is removed from each tube Liquid water exits each tube with a velocity of 15 s an a temperature of 55 C Find the total amount of heat removed from the entire condenser Approach 1 liquid A e rst law for an open system S earn condenser waler eliminating allterms except heat and gt th 1 h o equot a W c ange x 058 T2 55 C P 20 kPa P2 20 kPa Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is 4 Pressure is constant in each tube Solution From the rst law for an open system 2 2 11 Qr Ww Zm h VfgzjnZ u 11 Vfgaj Neglecting changes in kinetic and potential energy and assuming steady operation with no work the first law becomes h h Using data from Table A11 for saturated steam at 20 kPa h 11 4115 7h2514088261072514 2327kJkg For state 2 the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 55 C which is from Table A10 112 3 The mass ow rate in each tube is with values of density from Table A6 2 9857k j15Er 21ch lm 0512k g m s 2 100cm s The total mass ow rate for all 20 tubes is Lg s mm 92 V m m 20mm 200512 1024 From the rst law Q mh7 7h1024E23072327 721478kw Answer 5 g 634 Saturated steam at 120 F is condensed in a tube as shown Cooling water at 50 F ows in cross ow over xterior ofthe pipe giving a heat transfer coefficient of 200 BtuhfF F Find the exit quality Agpr ag l 1 Cooling Water at 50 F se e erma resistance ana ogy to u 3 0 determine the rate of heat transfer Apply l h 200 Bum n F the first law for an pen system eliminating allterms except heat and enthalpy change Assumptions 2 Kinetic energy chang 4 Pressure is constant in each tube 5 The interior ofthe tube wall is at the Ru R condensing steam temperature T H T I l 1391 391 Solution Apply the thermal resistance analogy to i at o d pipe ZnLlc turated steam The resistance to conduction in the tube wall is 3 Btu 1a Bm 0000212 a 22ft30 h39 F h ft F The resistance to convection on the exterior ofthe tube is 1 A Btu h F 20013 2u 21325m 2ft h ft F 12111 The two resistances add in series The total heat transferred is 7 F Bm41640 Rm 0000212 000147 From the rst law for an open system 2 2 dig Qr 27quot h V7gz2m 11 V7gaj Neglecting changes in kinetic and potential energy and assuming steady operation with no work the first law comes 0 Q mhl r mh2 Solving for exit enthalpy and using data in Table A10 Btu 4139 640 Btu Btu h 11135 735 Holbhm lbm lbm Q h hi m Using data in Table A10 7h x h 1 i o53lt Answer 115711 11135788 635 Superheated R1 39 l v7 MPa 70 C li uid at 0 Pa with a Volumetric ow rate of 6000 cmZmin T e R134a exchanges heat with an air ow which enters at 18 C at a mass ow rate of 195 kgmin Find the exit air temperature Approach air Apply the rst law for an open 8 DC G Assumptions 1 Potential en negligible R134 2 Kinetic energy change is negligible 07 MP3 system operates in steadystate 70 UC 39 V 6000 cm lmin ergy change is r sat liquid air 07 MPa Illa 95 kgmin 5 Pressure is constant during the process 6 Air may be considered an ideal gas under these conditions 7 The speci c heat of air is constant Solution Using data in Table A15 for a saturated liquid at 07MPa the mass ow rate ofthe refrigerant is 3 sooo i 7 ViV 17 min 100cm quot9 p VY V k 72 g lTlln 000083281 3 From the rst law for an open system dE d Qrmzmh V72gz172mhvfzg2 Neglecting changes in kinetic and potential energy and assuming an adiabatic heat exchanger in steady operation the rst law becom 0quot1 445M 414 For an ideal gas with constant speci c heat AhcpAT therefore 0 quot901 40quotan T3 T4 Using data from Tables A16 A15 and A8 kg E TEMUVM Ejnminjcm 8678kg i p 1 Answer quot39 CP 195k g 1 kJ min kgK 636 R134a ows through the evaporator of a refrigeration cycle at a rate of 5 kgs The R134a enters as saturated liquid and leaves as saturated vapor at 12 C Air at 25 C enters the shell side ofthe heat air is required exchanger If the air leaves at 15 C what mass ow rate of Approach Apply the first law for an open system Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate sat vapor sat liquid i ll 0C 4 The evaporator is adiabatic 7 o air L C G 5 Pressure is constant during the process i 5 k IS C 6 Air may be considered an ideal gas under mr g these conditions 7 The speci c heat of air is constant Solution From the rst law for an open system dE V 2 V 2 V 7W m 4zi ihz Qw WELEPA 2 g2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic evaporator in steady operation the first law becomes 0quot1 hirhz ma 414 For an ideal gas with constant speci c heat Ah CPAT therefore 0 quot011 412 an T3 T4 Solving for the mass ow rate ofthe air ma riiy c0344 With values of enthalpy from Table A14 and speci c heat from Table A8 k 2540376618 ma S g g s K k 935 g Answer kJ s 1005 25715 C kgK 637 Superheated steam at 5 psia and 200 F is condensed in a heat exchanger The steam ows at 39 lbms and exim as saturated liquid Cooling water at 45 F is used to condense the steam The water and steam are not mixed in the heat exchanger but enter and leave as separate streams Ifthe maximum allowable water 39 39 F 39 veiucit is 11 s what is the diameter ofthe pipe which carries water to the heat exchanger Approach Apply the rst law for an open system Assumptions 9 1 Potential energy change is negligible 2 Kinetic energy change is negligible Snlummd qum P psm 3 The system operates1n steady state 2 TI 200 F Steam 5 Pressure is constant during the process 6 The cooling water is incompressible 7 The speci c heat of the water is constant 3m P4 147 psin P3 147 psin 45 F Solution From the rst law for an open system dE V 2 V 2 V 7W m zi mh4z Qw V 2 h 2 5n 2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic condenser in steady operation the rst law becomes ht Malaqah mlht Since the streams do not mix m1 m2 and m3 m4 therefore quot391l h2 391h4 h3 Assuming water is an ideal uid quot501145quot551 T4 T2 hi m3 m1 6 T4 2 With 11 from Table B12 112 from Table B11 and c from Table B6 Btu 114871302 mj 39113 ijt lbm 2545113 m s 1 15 F s lbm F To nd the diameter D 2 mjpVA pV 254511 s 221 Answer 11515241sz s it 638 A twophase mixture of steam and water with a quality of 093 and a pressure of5 psia enters a condenser at 143 lbms The mixture exits as saturated liquid River water at 45 F is fed to the condenser through a large pipe The exit temperature of the river water is 70 F less than the exit temperature of the other u h L rr wam stream In I 15 Ms calculate the pipe diameter Approach river water 45 F Apply the rst law for an open system Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible x t h LI 1 3 The system operates 1n steadys wo P ase 5a lqm 4 Th d 39 d39 b t39 39 e con enser 15 a la a 1c x 093 river watel 5 Pressure is constant during the process 6 The river water is incompressible 7 The speci c heat of the river water is c onstant Solution From the rst law for an open system dE V 2 V 2 V 7W m 4zi mhz 12 Q V Z 4 2 5n 2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic condenser in steady operation the rst law becomes quotah rah eh2m4h4 Since the streams do not mix m m2 and m3 m4 therefore MmAFMhrhs Assuming water is an ideal uid ht 49quotan T4 2 quotHM 0p T4 9 State 2 is a saturated liquid at 5 psia since pressure is constant across the condenser From Table B11 T2 1522 F T442770 922 F State 1 is atwophase mixture at 5 psia With data fromTable B11 Btu 11 h 7h 1302093113171302 1061 a f x g x lt gt quotm The enthalpy of state 2 is h 11 1302 Btulbm 39 Table D L 39 I I lClllpClalulC of69nF 14313 105171302 m3s 1bm h Th4 100 Blquot 922745 F lbm F U 282 lbms 42821bl s 0620 Answer 5221137 15 it s 635 639 A heat exchanger is used to cool engine oil The speci c heat ofthe oil is 06 Btulbm F Using data on the gure below nd the exit temperature of the air Approach Air Apply the rst law for an open system T g 00 17 Assumptions 1133 19 lbms 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 Th st 39 te 4 The heat exchanger is adiabatic 5 Pressure is constant during the process 6 The oil is incompressible 7 The speci c heat is constant m 8 Air behaves like an ideal gas under these IAir c onditions Solution From the rst law for an open system dE V 2 V 2 V 7W m 4zi mhz Qw V 2 h 2 5n 2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic heat exchanger in steady operation the rst law becomes quotah We We quot14114 Since the streams do not mix ml m2 and m3 m4 therefore quot5h h2mzhrhz Assuming the oil is an ideal liquid and the air is an ideal gas quotarm Ti 3 MW T4 T2 38113105 Btu 160790 F TiT micpw1TrT2 A s lbm F 7 F Answer 47 3 quot115W 1glbm 024 Btu s lbmR 640 Saturated liquid 9 39 t 4 at 36 C quotquot 39 and exits as saturate vapor The evaporator is used to cool liquid water from 20 C to 10 C If the mass r d ow rate ofrefrigerant is 0013 kgs what is the mass ow rate of the wate Approach D throttle Apply the rst law for an open system to each component R1343 Assumptions Zagiulclqmd 1 Potential energy change is negligible waler wale 2 Kinetic energy change is negligible 20 0C IO 0C 3 The system operates in stea state 4 The throttle and evaporator are adiabatic 5 Pressure is constant across the evaporator 6 The water is incompressible 7 The speci c heat of the water is constant Solution Taking a control volume around the throttle and applying the rst law Using data in Table A14 kl h 1003E kg From Table A14 at 78 C 1 9 3954 kJkg and kg 2425 kJkg Since 112 1003 kJkg falls between these tw values state 2 is a twophase mixture Taking a control volume around the evaporator and applying the rst law dE V 2 V 2 w Qw iWw 27quot h Tgz 727quot h gz Neglecting changes in kinetic and potential energy and assuming an adiabatic evaporator in steady operation the rst law becomes quot392 quot14114 quot3924 mshs The water and R134a do not mix therefore quot392 39z quot quot394 395 w 0 MR h hz mw hrhs mw m h 7m 114 n 175 Assuming the water is an ideal liquid M h 41 c0440 V HMS 2 L 4 39 7i atwo phase mixture p t 39 C 39 L 39 region temperature remains constant Therefore state 3 is a saturated vapor at78 C From Table A14 13913 115 2425kJkg h2 h1 1003kJkg Using values of speci c heat at the average water temperature of 15 C 0013k 242571003 kJ k A g g00441k g Answer 5 W 4187kJkgK20710 C In a ash chamber anre im39 ed quot quot39 iuwei p 39 tw mixture The saturated liquid and vapor streams are removed in separate lines In the gure below liquid R134a at 10 F and 30 psia is thronled to 12 psiaIf2161bmh of saturated vapor exits the ash chamber what is the inlet ow rate Assume the ash chamber is adiabatic 9 A t l Approach rinsii Chamber Apply conservation ofmass and the rst law for an Assumptions 0 open system to the combination of the two Saturach vapor componenm 1 Potential energy change is negligible 1 Ia I 2 Kinetic energy change is negligible Th 39 ta 5 Pressure is constant in the ash chamber 501mmquot Smumlcd liquid In this problem the easiest approach is to define a control volume around the combination of the two components You could also analyze each component separately but that would take mor effort and produce the same final result Applying conservation of mass to the control volume shown I quot391 quot392 quot392 I From the rst law with no kinetic or potential I energy change and no heat transfer or work 39 quotihi We We 39 State 1 is a compressed liquid Approximate the I enthalpy of the liquid as the enthalpy of the I saturated liquid at the same temperature From 11 a 10 F 1466Bt11lbm From Table B15 the enthalpy of states 2 and 3 is 112 kg at 5 psi 9379 Btulbm h lyatS psi 73 Btulbm 39 39 39 value of mass into quot of mass produces l do m1 215113Trr m Substituting m and the values of enthalpy into the rst law 216m3lb m 1455 mom m 9379 mjlb m 7373 h lbm h lbm h lbm Solving for mj gives lbm 929 quot393 n From conservation of mass quotidiomewzame 11 lem 4 Answer 6 42 Saturated liquid water at 40 kPa enters a 140 kW pump The output of the pump is fed into a boiler where heat is added at a rate of 302 MW There is negligible pressure drop across the boiler If the mass ow rate of water is 70 kgs determine the boiler pressure and the state at the exit of the boiler Approach Apply WrhvPliP2to the pump to Sq WW Q 302411 determine the exit pressure form the pump U k 80 I 2 r quot0 kPa This is equal to the pressure at the exit of the boiler since there is negligible pressure drop across the boiler Draw a control volume around both components and apply the first law to determine exit temperature from the boiler Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is adiabatic and ideal 5 Pressure is constant across the boiler 6 The water is incompressible Solution For the pump power is related to pressure rise by quotMR P2 Solving for exit pressure and using the specific volume of saturated liquid water at 40 kPa from Table Al l 1000w P2 12 7 40kPa 1000Pa7 1kwg 199x106 Pa m 20MPa quotN 70 0001027m s kg 7140kw Pressure is constant across the boiler so P3 P2 20MPa To find the state at the exit of the boiler draw a control volume around both components The first law for this control volume is assuming steadystate and no kinetic or potential energy changes 0ch Ww Jrth mghg 0Qermlrhz 39 7W h Q w A m 302000kW77140kW HISE 70kg kg S Answer h3 4634E kg The exit state of the boiler has a pressure of ZMPa and an enthalpy of 4634kJkg From Table Al 2 this is superheated vapor with a temperature of TlOOO C Answer 6743 Air at 2000 R 100 Btu of 39 the air 39 quotb 39 exit ow area of06 01 39quotquot r 39 39 10 lbfinz mt A uii 39 39 lbms Approach D Draw a control volume around each component and apply the first law Assumptions W B 1 Potential energy change is negligible 100 F 2 Kinetic energy change in the turbine is m quot7 negligible W 3 The system operates in steadystate 4 The nozzle and turbine are adiabatic 3 I my 5 Air behaves like an ideal gas under these J in conditions 6 The speci c heat of the air is constant Solution Be inb 39 39 A andpotential energy changes the first law becomes W m s i 2 liming an For an ideal gas with constant specific heat Ah cpAT therefore W quotmy Ti 72 39 39 however make further progress use a of air at the inlet temperature of ZOOOR about L 39 To 1500 F and correct the calculation later ifnecessary So ving for 72 and using values oft from Table 138 at 1500 F W 100m lbm 727 quot ZOOORr u1638R may Btu i Assuming an adiabatic nozzle no work no change in potential energy and steady conditions Wigwagj For an ideal gas with constant specific heat Ah cpAT therefore quot h nwever ect the calculation later ifnecessary u known 39 1m Ror1178 F r For simplicity we assume an average temperature of 1000 F and will corr Using data from Table 138 20228002 112 s2 3 1638R1043R 2 msg 39 32272 lme s From the ideal gas law 2 1545 ft lbf jlO43R 1ft 386ft3lbm ET lbmolR 144m2 v3 MP3 2897 lbm 101bfin2 lbmol The mass flow rate may now be calculated as 2800ft s 06ft2 m V3143 7 W 4351bms 439 Answer v 386ft3lbm 3 Comments For greater accuracy you could recalculate the mass flow rate with improved values of specific heat based on the calculated temperatures The specific heat is not a strong function of temperature over the range considered so this may not be necessary 6 44 A wellinsulated rigid tank of volume 07 m3 is initially evacuated The tank develops a leak and atmospheric air at 20 C 100 kPa enters Eventually the air in the tank reaches a pressure of 100 kPa Find the final temperature Approach Apply the first law for an open system and integrate over time Note that the temperature of the air leaking into the tank is constant with time so the enthalpy of the air is also a constant Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The tank is adiabatic 4 Air may be considered to be an ideal gas 5 The specific heat is constant Solution From the first law for an open system with no change in kinetic or potential energy dU d Q 7 W 2M 7 2mm The tank is adiabatic No work is done and nothing leaves the tank so the first law reduces to v dt Integrating over time dew dt For an ideal gas enthalpy depends only on temperature The air entering the tank is at the same temperature throughout the process so enthalpy is constant and may be removed from the integral U2 7 U1 hjmxdt hxmz Where m2 is the mass in the tank at the end of the process At the beginning of the process the tank is evacuated therefore U1 0 m2u2 hxmz u2 h ux Pvx For an ideal gas with constant specific heat Au chT RT T2 7T 7 Pv 7 V Solving for T2 T 8314101 K20273K 2M 7 k 39 20273K CV 2897 g 0718i kmol kgK T2410K137 C 4 Answer 6 45 Helium at 150 F and 40 psia is contained in a rigid wellinsulated tank of volume 5 ft A valve is cracked open and the helium slowly flows from the tank until the pressure drops to 20 psia During this process the helium in the tank is maintained at 150 F with an electric resistance heater Find a the mass of helium Withdrawn b the energy input to the heater Approach Use a mass balance and the ideal gas law to find the mass of helium Withdrawn Use the first law for an open system to find the energy input to the heater Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The tank is adiabatic 4 Helium may be considered to be an ideal gas 5 The specific heat is constant Solution a From a mass balance my m1 m2 Where mg is the mass of helium exiting Using the ideal gas law PMV PMV M 41bm15 3 mf i j T 2 E Tlazig 40720psia 006llbm 1 2 1073 150460R lbmolR b From the first law with no kinetic or potential energy diw Q 7 W 2m 7 2mm The tank is adiabatic and nothing is entering therefore Uw 7mm dt Integrating over time Ida iijdti dt Since the temperature of the helium in the tank does not change by is constant and U2 7 U1 4V5 WEIquotEd m2u2 T m1u1 Ww T heme m2u2 7 m1u1 Ww 7 a 132V2m2 m2u2 7 m1u1 Tch T 2 Pave ml Tmz Since u is only a function of temperature for an ideal gas and the temperature is constant u1 u2 u u Therefore 0Wchgvgmrm2 ET m T 006llbm 1545 ft39lbf 150460R 7 lbmR lBtu 4 lbm K778ftlbf lbmol WW 7185Btu lt1 Answer 643 646 A residential hot water heater initially contains water at 140 F Someone turns on a shower and draws water from the tank at a rate of 02 lbms Cold makeup water at 50 F is added to the tank at the same rate A burner supplies 5472 Btuh ofheat The water tank which is a cylinder of diameter 18 ft is filled to a height of4ft How long will39 L L L 39 39 reach 100 39 d t Approach Apply the rst law for an open system Use ideal liquid relations to rewrite enthalpy and T 50 quotF internal energy in terms of temperature Separate Variables and integrate In mc m n1CV constant Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 Water is an ideal liquid 4 The tank is wellmixed and all tank contents are at the same temperature 390 Solution The rst law for an open system with one stream in and one stream out neglecting kinetic and potential energy changes is dU V 7W 7 1 dt 9 m m quot L 39 4 quot rate q quot quot quot therefore d quotUM gnmwm The mass in the control Volume does not change with time For an ideal liquidAh CPAT With these ideas the rst law becomes m V mc T 7T w d Q lt gt The tank is wellmixed so the temperature T of the water in the tank equals the temperature of the exiting stream The internal energy of the water in the tank is related to its temperature by w 5 m cPdT since for an ideal liquid CV s CF The rst law takes the form dT quot39pr E Qv quot51 TFT Separating Variables and integrating from initial temperature T to nal temperature T 2 JP 7 22 la QwmcpTrT7I d let Qwch TiT d 7chdT 2 di me I r 1 1 mpg m w 5 7 m 1ng542 whamm z mHQqQ m i imfv 1n ch quot cp T2 12 Qwmc TrT p Find mv 2 mw pV 621bmft37rj ft2 4ft 6311bm 5472Btuh 1h o21bms1 B j507100 F I 776311bm1 26005 lbm F 2 u O39ZIbmS L 5472361W021507140 12 2096535min 4 Answer 6 47 In an industrial process two streams are mixed in a tank and a single stream exits Both streams may be assumed to have the properties of water The volume of uid in the tank is constant A paddle wheel stirs the tank contents doing work W Initially the water in the tank is at temperature T1 At time I 0 stream A at temperature TA enters with mass flow rate r39nA and stream B enters at T3 with rate m3 The quantities TA TB r39nA and ms are all constant with time Assuming a wellmixed tank derive a formula for the time t2 at which the tank water temperature is T2 The tank is well insulated Approach Apply the first law for an open system Use ideal liquid relations to rewrite enthalpy and mA 113 internal energy in terms of temperature TA TB Separate variables and integrate Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The liquid is ideal g 4 The tank is wellmixed and all tank contents are at the same temperature 5 The tank is adiabatic n39i 6 All uids have the properties of water L S olution The first law for an open system neglecting kinetic and potential energy changes is dU d 9 A W 2m 7 2mm Since the tank is wellinsulated d l w 7W rhAhA mghg 7 rhchc The mass of liquid in the tank is constant and conservation of mass requires that rhc rhA m3 therefore d m5 7WrhAhAthhB 7r39nAthhC mcv d TWmAhAThCmB ks Thc The liquid in the tank is ideal and has a temperature T The exiting stream is also at T because the tank is well mixed With these considerations the first law becomes mwcp 1 7W rhAcp TA 7 Tr39nBcp TB 7T m c 7WmAcPTAchpT37mAcPchpT w p dt 7 7WrhAcpTA thCPTB 7 mAmB T dl mwcp mv Define K1 and K2 so that g K1 7 K 2T Separating variables and integrating 13 1 dr Ti K17K2T let 5 K17 KZT d5 7K2dT 7I52 l 51K 2 1 5 iglnflgt2 12Lln2mw K2 51 K2 K17K2Tl 7 EVmATAmBT3mAm3T C 127mw jln Answer mAm3 7 EVmATAm3TBimAmBZ C P 648 A well insulated tank ofvolume 0035 m3 is initially evacuated A valve is opened and the tank is charged with superheated steam from a supply line at 600 kPa 500 C The valve is closed when the pressure reaches 300 kPa How much mass enters Approadli steam 600 kPa 500 quotc Use the first law for an open system Integrate the equation recognizing that the enthalpy entering is constant with time Properties are available inthe steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The tank is adiabatic 4 The state of steam in the supply line is constant Solution The first law for an open system neglecting kinetic and potential energy changes is dU JiZWkth The tank is wellinsulated no mass exits and no work is done therefore aw mh a IdoumaJmhw The state ofthe steam entering remains the same throughout the process so 11 is a constant Ifm2 is the mass in the tank at the end of the process and m is the mass at the start we quotHquot hlm h rm Since the tank is initially evacuated m1 0 and quotwe hm mh From Table A12 h 3483kJkg therefore u2 3483kJkg At the nal pressure of 300 kPa and u2 3483kJkg v2 m15m3 g Therefore the mass in the tank at the end of the process is K M 00233kg lt1 Answer 1 1 V2 649 A wellinsulated pistoncylinder assembly contains 006 kg ofR134a at 715 C with a quality of 092 A supply line Introduces superheated R134a at 10 C 200 kPa into the cylinder Assuming the pressure in the cylinder is constant calculate the volumejust when all the liquid has evaporated Approach Use the rst law for an open system Integrate the equation 5 m recognizing that the enthalpy entering is constant with time P c Assumptions 4 1 Potential energy change is negligible 2 Kinetic energy change is negligible R 3 3 The pistoncylinder assembly is adiabatic a liq R4376 4 The state ofR134a in e supply line is constant 0 C 5 The pressure in the cylinder is constant 200 Jar 6 The expansion is quasistatic Solution From conservation of mass m m2 m1 m2 is the mass in the tank at the end ofthe process In is the mass at the start and m is the total mass introduced during the process The rst law for an open system neglecting kinetic and potential energy changes is W Q 7 W 2 rm 2 mm The tank is wellinsulated and no mass exim therefore w Wvmyh dma uw 7Ww m h Idmwuw 7 Wwal j Mar The state ofthe R134a entering remains the same throughout the process so h is a constant quotwe miui WVh mdtWvhmi quotis139 39 r39 quot 39ofmass quotwe miui PVrVih new we quotHquot PM2V2 7mm quot392 mi quot12111 Prnzvrmui PmiVi h mi 7m Using the de nition of enthalpy h u Pv and recognizing that P I P We WFh quot392 mi Solving for m m MM h h h n The enthalpy ofthe R134a in the cylinder at the start ofthe process is h 11 4115 7h30609223837306 222 where 1 5 and kg are found by interpolation in Table A14 at 715T The contents ofthe cylinder remain at the 34achangesf 39 p A 39 r so the nal enthalpy is the enthalpy of saturated vapor at 715 C that is 112 2383kJkg The enthalpy in the supply line 11 may be found in Table A16 The nal mass may now be calculated as L k 006kg 227 259 worm kg0106k 2 Wi kli 39 g 2 I 2387 259 kg The final volume using data for specific volume interpolated in Table Al4 is 3 V2v2m20120 0106kg00127m3 4 Answer kg 650 The pressure inside a pot is maintained at an elevated level by a steel bob which resm on an open tube of inside diameter 05 cm The bob which has a mass of 0401 kg jiggles whenever the pressure in the pot is high enough to displacei 39 39 f 900 W Heat is lost from the sides and the top ofthepot by natural convection with a heat transfer coef cient of 39 WmZK The pot has a height of 0154 m and a diameter of 0256 m The ambient is at 20 C Assume 39 39 quot is v small 39 39 in the pot only water and steam The pot is half lled with water when the bob rst lifts a Find the temperature inside the ot b Find the net rate of heat addition to the pot c Find the initial mass 0 e twop ase mixture in the ot d Find the fraction ofthe pot which is lled with liquid after one hour Approach W Use the rst law for an open system Integrate the equation recognizing that me en L thalpy time Properties are available in the steam tables F Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The pot contains only steam and 4 The conduction resistance is small you w 5 The convection resistance on the inside ofthe pot is small 6 The work done in displacing the bob is negligible Solution a De ne m as the mass ofthe bob and r as the radius ofthe small tube that the bob resm on By de nition m 0401kg981 2 5 5 2007300132 A M7 025 2 2 739 7 m 100 Therefore throughout the process the pressure in the pot is P 200 3 kPa The saturation temperature at this pressure is from Table A11 T 120 C Answer b The conduction resistance through the walls and lid of the pot is assumed to be small Also the convective heat transfer coef cient on L 39 39 39 VCI L L 39 394 39 39 39 small Therefore the outside of the pot is at the saturation temperature of the water The net heat into the control volume is 2 QMquot hAAT QM an 39 JAT g 900 w739 ij 0255 0154 255 J 1207202C Q 8314 W 4 Answer c The volume of the pot is 2 2 V I L 2 j 0154 000793 m3 The initial volumes of liquid and vapor are VV2 and VgV2 From Table A1 1 v 000106m3 kg and VE 08857 In3 kg at 200 kPa Therefore V V m1 g374 kg Answer V V d The first law for an open system neglecting kinetic and potential energy changes is dU d 7 9 7 W 2m 7 2m No mass enters the pot and no work is done therefore dU W 7 m h dz Qw d mugt Qw m dt I dltmugt7 I 9871 mhdr The state of the steam leaving remains the same throughout the process so he is a constant If m2 is the mass in the pot at the end of the process and m1 is the mass at the start m2u2 T mlul chAt T he Imedt Q stti heme where m is the total mass that escapes Using conservation of mass m2u2 T mlul chAt The quot 2 7 m1 Eq 1 To find ul we need the initial quality of the mixture in the pot Use v1 03900793m3 000212m 3 m1 374kg kg With data from Table All at 200 kPa x V1 V 0002127 000106 vg 7 v 08857 7 000106 u1 u x1ug 7 u 504 0001197 25297504 507 kJkg Let y be the fraction of the liquid at the end by volume so that m2 17yV 0001197 Eq 2 v E Additional equations for the final state are v2 Vm2 Eq 3 v2v x2vg 7vf Eq 4 u2ux2ug7u Eq 5 Equations 1 through 5 are five equations with five unknowns ie m2 1 x2 big and y It is not a linear system of equations because Eq 3 is nonlinear Therefore the solution is iterative One approach is Assume m2 Find v2 from Eq 3 Find x from Eq 4 Find u from Eq 5 Check Ts Eq 1 satisfied lfnot adjusth Once m2 is known findy from Eq 2 The final results are m2 238 kg u2 5098kJkg x2 0002566 v2 0003326 m3kg y 032 Therefore 32 of the pot is filled with water at the end of the process Answer 652 S am at 160 psia and 400 F enters a nozzle with a Volumetric ow rate of 6615 cfm cubic feet per 61 te minute The inlet area is 145 in Ifthe steam leaves at 1500 Ms at a pressure of40 psia nd the exit temperature Approach Use the first law for an open system for a nozzle Find properties in bles 39 I 160psia 1g 40 psia ft V2 1500 s specialized the steam ta Assumptions 400ml 7 1 Potential energy change is negligible 2 The system operates in steadystate i 5515 off 3 The nozzle is adiabatic Solution From the rst law specialized for a nozzle 1 2 h 2 hi 2 The inlet Velocity may be found from 3 1min m2 144m 2 llbf Btu lbm 1197E 2 2 5 2 Taking the inlet enthalpy from Table B12 W V22 Btu 1 2 2 n2 7 1218 1095 71500 112 hi 2 2 lbm 2i 1 At P2 40 psia 1 17 1197Bi from Table B12 lbm T2 32an 4 Answer 62 Oxygen at 220 enters a wellinsulated nozzle of inlet diameter 06 it The inlet velocity is 60 Ms The oxygen leaves at 75 F 10 psia The exit area is 001767 z Calculate the pressure at the inlet Approach Use the rst law for an open system specialized for a nozzle Apply conservation ofmass and the ideal gas law to determine gt inlet pressure 7 a T 7 220 F T2 75W Assumptions V 502 122 10 psia 1 Potential energy change is negligible 5 2 The system operates in steadystate 3 The nozzle is adiabatic 4 Speci c heat is constant 5 Oxygen behaves like an ideal gas under these conditions Solution From the rst law specialized for a nozzle V 2 V 2 h 4 h 2 Solving for V2 and noting that for an ideal gas with constant speci c heat Ah CPAT V2 sz mm V3 lzc T 40 V3 Using data for speci c heat from Table B8 2 V2 2 02215 Blquot 220775R M m 50 12593 lbmR lBtu llbfs s s From conservation of mass m quot391 2 pi ViAt p VzAz Using the ideal gas law MR V V RT A RT 2A2 Solving for inlet pressure V2 A2 T 1259001757220450 1 P2 l 10 sia VAiTz p 50 032 63 A wellinsulated nozzle has an entrance area of 028 m2 and an exit area of 0157 ml Air enters at a velocity of 65 ms and leaves at 274 ms The exit pressure is 101 kPa and the exit temperature is 12 C What is the entrance pressure Approach Use the rst law for an open system specialized for a nozzle Apply conservation ofmass and the ideal gas law to determine inlet pressure 2 101 kPa 1 a Assumptions V 7 65 s T2 712 C 1 Potential energy change is negligible V2 27 E 2 The system operates in steadystate s 3 Thenozzle is adiabatic 4 Speci c heat is constant 5 Air behaves like an ideal gas under these conditions Solution From the rst law specialized for a nozzle V 2 V2 hi 2 hi 2 For an ideal gas with constant speci c heat Ah CPAT therefore V2 V2 C 777 2 m 2 2 2 Solving for I V2 V2 n 2 WT 2 Speci c heat depends on temperature and should be evaluated at the average of inlet and outlet temperatures but the inlet temperature is unknown As an approximation evaluate the speci c heat at the exit temperature and correct later ifnecessary From Table A8 At T2 27312285K c m1004L P kgK 2 2 2 m 274 755 15 2 7 k 285K 320K 20004 J 1000 kgK 1k From conservation of mass quot391 quot2 pi ViAt p V2A2 Using the ideal gas law MR MP V 2 V A RT A RT 2 2 Solving for inlet pressure gg g1o1274j j j 268kPa 4 Answer 1 Ai T2 E 028 285 Comment The entrance temperature of320 K is very L 39 0285 K 39r L A signi cantly between r A L 39 A 64 Carbon monoxide enters a nozzle at 520 kPa 100 C with a velocity of 10 ms The gas exits at 120 kPa and 500 ms 39 auu ideal gas behavior quot 4 L 39 Approach Use the rst law for an open system gt specialized for a nozzle Assumptions 1 520kPa a 120 kpa 1 Potential energy change is negligible T 7 1002C z 2 The system operates in steadys te V2 5002 3 The nozzle is adiabatic V 102 5 4 Speci c heat is constant 5 bon monoxide behaves like an ideal gas under these conditions Solution From the rst law specialized for a nozzle 4 v i5 2 2 For an ideal gas with constant speci c heat Ah CPAT therefore V 2 V 2 c r 2 T gt 42 f Speci c heat depends on temperature and should be evaluated at the average of inlet and outlet temperatures but the outlet temperature is unknown As an approximation evaluate the speci c heat at the inlet temperature and J correct later ifnecessary From Table A8 at 1quot1 00 273373 K 5 m1045 k K g m 2 2 107 f 2 2 1 4rf 25 5 k 65 Lowvelocity steam with negligible kinetic energy enters a nozzle at 320 C 3 MPa The steam leaves the nozzle at 2 MPa with a velocity of410 ms The mass ow rate is 037 kgs Determine a the exit state b the exit area Approach Use the rst law for an open gt system specialized for a nozzle Find properties in the steam tables I 320 C 1 2 MPa Assumptions R SMpa 1 Potential energy change is negligible V2 4102 2 e system operates in steadystate 5 3 The nozzle is adiabatic Solution a From the rst law specialized for a nozzle V1 2 V2 h 2 117 2 Determining the inlet enthalpy by interpolation in Table A12 2 2 2 h2 h L7V 2 30mg 1000J 7 410 395 29mm6 i 2950 E 2 2 kg 1k 2 kg kg Interpolate in Table A12 at 2 MPa and h 2960 kJkg to nd the nal temperature of T2 274a C The exit state is superheated vapor at 2 MPa and 274 0C b The mass ow rate is given by m p VzAz 39 L 39 A quotaquot 4 L r 39 39 L interpolation inTableA12 v2 0118m3kg i 43 847k D v2 0118m kg m Solving the mass ow rate equation for exit area P2 107X104m2 Answer sunnundmgs Approach Use the rst law fur an upeh sysLEm allmmaung wk and pummel energy and pmpemes m the steam tables 39 Assumptions 1 zsn c T2 WC 1 Futmnal margy change xsneghgxble 31 S kpa P 15 kpa z The system uperates m steadyrstate Solution a mm the rstlaW 2 7 QVTWHZM 1 gz ZmlhT gzl u wekk m rstlzwbecumes 2 2 UQ m h1L em twi z z The area atthemletxs 2 I25 21an quotq Imam2 z z The velumty atthemletxs usmg data 39um TableArlZ 3 m kg v a a A l 491m2 4 ms The Exit areas 2 Afr L5 Ul m2 Obtammg speci c vulume by mterpulatmg m Table A712 94 198 st swsm A1 196 s Rearrangmg the rst 12w 2 2 em 7 Q h 2 e m 2m 952 646 QN94E 3mg LEE 7 zm LEE 7 5 kg 1k 2 kg 1k 2 Q zem71A wem71kwlt Anwver kFa The est temperature is 1 WC mhe mametez atthe mieth a em a Fmdthe exxtvelun and Assume cunstznt wen c heats nuns 21 t m o Ev u epmp xes T ZXSC Assumptions PI Z 5mg 1 Futennal margy change xsneghgxble V ZSEIms z The system uperates m stead estate 3 The mffusans adiabatic An39behzveshke an ideal gas unda these cundmuns 5 Specx cheats are cunstznt Solution 2 Hum the rst law spemahzed fur a dxffuser 2 2 kn is i Furamdeal gasvnthcunstzntspem cheaL AheM therefure 2 W 97 k IUUUJ 2TrTz V z1u1kg KWxsenu 4 Answer s h By cunservauun ufmass ml 2 W11 BVzAz Usmg Lhexdeal gas law 2 i Mvn e P v RT 2 RT2 P 2 P D v T H 2 on Sulvmg fur em diameter Dzzq ghza sljmwm zsu P2 T V 1m 878Em Answer 85Z73 1m 678 superheated steam enters a well rmsulated diffuser at 14 7 psla 32m saturated vzpur at ayery luw weed Fmd the exrt pressure and temp Approach Use the rst law fur an upeh sys elrrruhauhg vmrk heat and put tern ehu al energy and pmpemes m the steam tahles F and 4mm MS The steam exlts as erature T zzu quotF Assumptions l putehual margy change ls negllglble z The system uperates m steadyrstate 3 The mffusals adlabah Solution a me the rst law speelallzed fur a drffuser 2 2 4 hi V 2 z Usmg yalues ufehthalpy at the mlet 39um Table 312 at 14 7psla 32mquot 2 Em 4mm2 lZUZli 5 lhrh lZEIS KBim 2 Hum s In Table Brl fur h ZUS 3BtulbmLba39e are twu pusslble suluuuns runs at MUDF and 3812 psla and the uLha39 ata uuut WElaF mi 4 l nl H N V has httle effeet uh the exrt state 4 Answer aeyelups 24 5 MW ufpuwer whatrstherhass uwrate7 A pproach Use the rst law fur an upeh system ehrruhauhg heat laheue mar and putehual Energy Fmdpmpemes m the steam tahles quot Itheturbme Tsnnquotc Assumptio l putehual P n a MFa y change ls negllglble e e turbmels adlahaue Solution Assummg an 2642me turhme wllh hegllglhle kmeue and putehual Energy ehahges the rstlawheeurhes W ta 7 wlth values uf enthalpy 39um Tahle A42 24 5 xl sz 348B 67 mm lldkg hr a 610 Air enters an adiabatic turbine at 900 K and 1000 kPa The air exits at 400 K and 100 kPa with a velocity of 30 ms Kinetic and potential energy changes are negligible Ifthe power delivered by the turbine is 1000 kW a nd the mass ow rate b nd the diameter ofthe duct at the exit Approach T 900K Use the rst law for an open system eliminating heat kinetic R loookpa energy and potential energy Find properties using ideal gas relations Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The turbine is adiabatic 5 Air behaves like an ideal gas under these conditions 6 Speci c heat is constant Solution 1 100kPa a Assuming an adiabatic turbine with negligible kinetic and 30ms potential energy changes the rst law becomes 2 W 7 m h 7 l5 For an ideal gas with constant speci c heat Ah CPAT therefore m 7 W 7 W c T 7T2 With values of speci c heat from Table A8 at the average temperature of 650 K 7 kIJOOOkW 7 88k g 4 Answer 1053 9007 400K 5 k K b Exit area is related to velocity and mass ow rate through A2 V2 From the ideal gas law 8314 M 400K 1 RE L 1 1 A 2897 kg 100kPa kg kmol Substituting values 3 1881151115 00721m2 303 5 D2 w0303m 1 Answer I I 611 Saturated steam at 320 C enters a wellinsulated turbine The mass ow rate is 2 kgs and the exit pressure is 50 kPa Determine the nal state ifthe power produce 39 39 0 kW is b 400 kW Approach T 320T Use the rst law for an open system eliminating heat kinetic saturated steam energy and potential energy Find properties in the steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible e system operates in steadystate 4 The turbine is adiabatic Solution a Assuming an adiabatic turbine with negligible kinetic and P 7 SOkPa potential energy changes the rst law becomes 2 7 W m hi 7 la From Table A10 the enthalpy of saturated steam at the initial temperature of 320 C hl 2700kJkg 1000 W 100 kW hfl 1 2700 i lkw 7265x106i 25mg m kg 1 kJ 21 kg kg s From Table A11 at 50 kPa we see that the exit enthalpy is greater than that of a saturated Vapor therefore the 39 39 39 39 andby39 39 39 TableA12 T2834 C 4 Answer v2 50kPa b Recalculating the exit enthalpy for the higher power condition 1000 W 400 kW 2mg 1kw kg 1 25x106 i2500 2 kgs kg From Table A11 at 50 kPa we see that the exit enthalpy is between that ofa saturated liquid and a saturated Vapor therefore the exit state is in the twophase region an hz 11 kg 4 7 h2 7h 7 250073405 7 kg 711 7 25457 3405 Summarizing the nal state is twophase with OkPa x 0937 x0937 Answer 612 Superheated steam at 16 MPa 600 C enters a wellinsulated turbine The exit pressure is 50 kPa T 39 Ifthe exit pipe is 16 m in diameter and carries 11kgs of ow nd gy turbine pro uces 10 MW ofpower the Velocity at the exit Neglect kinetic ener Approach Use the rst law for an open system eliminating heat kinetic energy and potential energy Find properties in the steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The turbine is adiabatic Solution Assuming an adiabatic turbine with negligible kinetic and potential energy changes the rst law be omes 1 SOkPa W mm 45 With Values of enthalpy from Table A12 at 16 MPa 600 IOMW 1000kW 1MW kJ 2784 93 kJ kg W 7 35 7 112 hi Iii kg 11kgs From Table A12 ath 50kPa and 112 2784kJkg I 150 C To ndthe exit Velocity m va2 V2 Solving for Velocity and using Values ofv2 from Table A12 at T2 150 C and 2 50kPa 3 11115118911 V2 7va 7 s 2 g 7 2132 lt Answer A2 15 2 s 1 i m 2 613 Air at 550 C and 900 kPa is expanded through an adiabatic gas turbine to nal conditions of 100 kPa and 00 C The total power output desired is 1 MW Ifthe inlet Velocity is 30 ms what should the inlet pipe 39 energy 3 diameter be Neglect kinetic and potential Approach T 550T Use the rst law for an open system eliminating heat kinetic P gookpa energy and potential energy Find properties using idea gas relations V1 SOWS Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 39 tate 6 Speci c heat is constant Solution 72 300T a Assuming an adiabatic turbine with negligible kinetic and P 7 lookpa potential energy changes the rst law becomes 2 7 W m For an ideal gas with constant specific heat Ah CPAT therefore W W hi 1quot 0 Ti 75 With Values of specific heat from Table A8 at the average temperature of 700 K 1000kW m m 1075i5007300 c 5 k K b Exit area is related to Velocity and mass ow rate through A V From the ideal gas law 8314 M 550273 K 3 RT kmolK m v W k 0252k 2897 g 900kPa g kmol Substituting Values 3 37211510 2524 00325m2 303 S D E W0203m 4 Answer IL39 I 614 Airat510 eal 39 39 39 quot L 39 r 39 101 kPa In steady state the turbine produces 50 kW ofpower Find e e it temperature Hint use Eq 256 b the mass ow rate Approach T 510 C Use Eq 256 to nd the nal temperature Then apply the rst R 450kPa law for an open system eliminating heat kinetic energy and potential energy to obtain mass ow rate Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The turbine is adiaba 5 Air behaves like an ideal gas under these conditions 6 Speci c heat is constant 7 The process is quasistatic 1 101kPa Solution a Sincethe turbine is ideal the process is quasistatic For adiabatic quasistatic process ofan ideal gas with constant speci c heat from Eq 256 I From Table A8 for air k14 E 1 T2 T i 510273 j M 511 K 238 C P 450 b Assuming an adiabatic turbine with negligible kinetic and potential energy changes the rst law becomes h For an ideal gas with constant speci c heat Ah CPAT therefore W W m h n h 0 Tl T2 With Values of speci c heat from Table A8 W 50 kW kg m 0173 4 Answer CP 1 2 Losi 5107238K s kgK 615 Saturated steam at 3 MPa enters a 39 39 39 39 L state quotquot 600 kW ofpower The mass L L39 39 39 4 L the exit temperature quot is 093 Find Approach I 3MPa Use the first law for an open system eliminating heat kinetic saturated steam energy and p t 39 energy To find the nal temperature it is necessary to iterate with data from the steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic ener change is negligible e system operates in steadystate 4 The turbine is adiabatic Solution a Assuming an adiabatic turbine with negligible kinetic and potential energy changes the first law becomes W m hi 7 l5 From Table A11 the enthalpy of saturated steam at the initial pressure of3 M39Pa is h 111 2804 kJkg 9 093 h2Zhl 7 500 kw 2804 E 2375 E m 84kg 1min kg kg min 60 s 39 39 39 39 39 neium 39 39 TL 39 h 7 u found by trial and error Begin by assuming a Value forT then use 9 to comp tel l2 with data from Table A10 If 112 2375 kJkg the iteration is complete ifnot select a new Value off and recompute To begin assume T2 25 C From Table A10 hfhHzlhg hx 10490932547 7 2442 2375 kJkg This is Very close to the calculated Value of 2375 therefore T2 25 C 4 Answer C omment In this example we selected the correct result immediately In reality it would be necessary to try several temperatures before zeroing in on the correct Value 616 In a 3hp compressor carbon dioxide owing at 0023 lbms is compressed to 120 psia The gas enters at 60 F and 147psia The inlet an 39 quot Fi iinai volumetric L 39 39 3min 39 F Approach T 60 F Use the rst law for an open system eliminating heat kinetic R 147psia energy and potential energy Find properties using ideal gas relations Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadysta 4 The compressor is di 1 5 Carbon dioxide behaves like an ideal gas under these conditions E 120psia 6 Speci c heat is constant Solution Assuming an adiabatic compressor with negligible kinetic and potential energy changes the rst law becomes W m hi i h For an ideal gas with constant speci c heat Ah CPAT therefore W mc T 7T2 W T2 I ms with data from Table B8 25441311 7311 p mp T2 60 F7 533 F4 Answer 0 023111 0 195713quot Loos s lbm R 1h From conservation of mass V2 MM quot1 m2 v2 RT 3 ET 00231073533460R n V quotH 2 5 quot1 39 lb quot391quot 7278 Answer gM 120psia 4401 m m lbmol Comments If more accuracy is desired the calculation should be repeated with speci c heat evaluated at the average of the inlet and outlet temperatures ie at 605332 or 2965 F The outlet temperature was not known at the beginning ofthe calculation so speci c heat at a temperature near the inlet temperature was used 617 A wellinsulated compressor is used to raise saturated R134a Vapor at a pressure of 360 kPa to a nal pressure of900 kPa TL u mate in s eady 39 p input U U v I the ow rate r P is 0038 kgs what is the nal temperature Approach 1 360kPa Use the rst law for an open system eliminating heat kinetic saturated vapor energy and potential energy Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The compressor is adiabatic Solution Assuming an adiabatic compressor with negligible kinetic and potential energy changes the rst law becomes 1 900kPa Wm 41 W hz 4 m From Table A15 at 360 kPahl hg 2506kJkg 112 2506E 272 00397g kg kg s From Table A16 at P1 900 kPa andhz 2716 T m 40 C Answer 618 Air owing at 05 m3min enters a compressor at 101 kPa and 25 C The air exits at 600 kPa and 300 C During this process 250 W of 39 Wha 39s 39 input Approach T 25 C Use the rst law for an open system eliminating kinetic energy and P IOIkPa potential energy Find properties using ideal gas relations 3 V 05m min Assumptions 1 Potential en gy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 Air behaves like an ideal gas under these conditions 5 Speci c heat is constant Solution From the rst law for an open system 1 120psia dE V 2 V 2 V 7W m 4 7 m hz Qw V 2 h 2 82 Z 2 539 Assuming steady conditions one stream in and one stream out no change in kinetic or potential energy the rst law becomes 0QWmhrhi For an ideal gas with constant speci c heat AhcpAT therefore WQmcTrT2 Using data in Table A1 3 051 101kPa2897 kg 1mquot V V RM min kmol 60s n A 3kg I s V RT 8314 M 25273K kmol K The average temperature of the air is Tm w152 0 435K Using speci c heat Values from Table A8 interpolated at TM k J W 7250w 984x10393 11018 257 300 DC isooow 73kW 4 Answer s g 619 Refrigerant134a enters a compressor at 0quotF and 10 psia with a Volumetric ow rate of15 3min The refrigerant exim at 70 psia and 140 F If the power input is 2 hp nd the rate of heat transfer in Btuh Approach T 0 F Use the rst law for an open system eliminating kinetic and P 710 7 psia potential energy 3 V 15 mln Assumptions 1 Potential en gy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate Us Solution From the rst law neglecting kinetic and potential energy 0 Q 7 W m h 7 The mass ow rate is T 140 F B 70psia m Vi Taking the inlet speci c Volume v1 from TableB16 3 15L lbm m 3319 4703 m lbm Q WWW 4 1 Again using Table B16 341213 Q2hp m h 31911quot 129171023 M lhp 1W mm lbm 1h Q450 Answer 6 20 A pump is used to raise the pressure of a stream of water from 10 kPa to 07 MPa The temperature of the water is the same at inlet and outlet and equal to 20 C The velocity also does not change across the pump If the mass ow rate is 14 kgs what power is needed to drive the pump Assume frictionless flow and no significant elevation change Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 Water is incompressible Solution For a frictionless pump with no elevation changes W mquot 7P2 so using the density of water at 20 C from Table A6 W m Pl 7P2 p 10kP 7700kP Mk u W 79677W7968kw d Answer 5 99821 lkPa 3 m 6 21 A 2hp pump is used to raise the pressure of saturated liquid water at 5 psia to a higher value Assume the velocity is constant the water is incompressible and the flow is frictionless If the mass flow rate is 6 lbmsec find the final pressure Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 Water is incompressible Solution For a frictionless pump with no elevation changes W MP 7 P P 2 P1 1 From Table Bl l vf 00164 ft3lbm atP 5 psia Substituting values 550ftlbf 1ft2 2 hp S 2 1 hp l441n P2 Sig f2 7 lbm ft3 39 6 00164 s lbm I72 826 psia Answer 619 6 22 Water is pumped at 12 ms through a pipe of diameter 12 cm The inlet pressure is 30 kPa 1f the pump delivers 6 kW find the final pressure Assume frictionless incompressible flow with no elevation or velocity changes Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 Water is incompressible Solution For a frictionless pump with no elevation changes W ma 7P2 P1 Pz p The mass flow rate is 2 m pV A 998k g311z jnwj 135E m s 2 s 76kWMI998k g3 1kW m 445gtlt106 Pa 445MPa kg S P2 445MPa i Answer P2 R 7 30kpa1000Pa7 m 1kPa 135 6 23 A 1hp pump delivers oil at a rate of 10 lbms through a pipe 075 in in diameter There is no elevation change between inlet and exit no velocity change and no oil temperature change The oil density is 56 lbmft3 Find the pressure rise across the pump Approach Use the equation for pump work in the form W rhvP1 7 132 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is ideal 5 The oil is incompressible Solution For a frictionless pump with no elevation changes W 7 P2 E 7 I 1 mv m ft lbf 550 2 561b f 1hp 5 If 2 ft 1 hp 144m BAP2 214psia 4 Answer iolb m S 6724 l ace p whlchlsaboullooo lthlgh l p p n edeol lnhp preach Apply the rst law for an open syslecn dropplng lens for Lranslents heat and klneolc mergy Replace the mthalpy dlffamce Wllhh1 eh m e 000 fl Assumptions 1 The syslern ls adlabatlc and lsoLhennal z Kmehc ener change ls negllglble 3 The syslern opaahes ln steadyrsmte 4 The pump lsl ea 5 Water ls lncompresslble l I 1 45 psi1 Solu on The rst law for an open syslen ls dE W W n 7W m 4z mh z dt Q nzlpl Z g2 2 L W 1 l l l l adlabauc operallon to get 07Wm e m z 22 For an lncompresslble llqulol undergolng an lsoLhermal process hi hlVP2 Pl Sub Smung lhls lnLo the rst law and noung that v 1 p Wemfl Pz mgzrzz 2 454mg 144 ln m lhp 621133 550lbf 1 s 2 31b m3217 2 f R 4000a ll amp 5 5 3217i2 5507 s s W 389hp4 Answer ommenl The aclual pump chosen should have mole horsepowa than thls because ln reallty there are some fnctlonal effects 625 Airat150 C quot quot 100kPaquot Lquot quotis36ms quotL 39 I Approach P 40kPa Use the rst law specialized for a thronle T150 C 1100kPa Assumptions V 36ms 1 Potential energy change is negligible 4 Air may be considered an ideal gas under these conditions 5 The throttle is adiabatic 6 The inlet and exit pipe have the same diameter 7 Speci c heat is constant Solution For a throttle with no heat transfer no change in kinetic or potential energy and no work the rst law reduces to For an ideal gas with constant speci c heat AhcpAT therefore oc Trn TH From conservation of mass 2 pViAi p VzAz Assuming the diameter of the inlet and exit pipes are the same and using the ideal gas law ET ET 2 RM PM Solving for exit Velocity V2 V i MEI 144m 4 Answer R s 100 626 Saturated liquid R134a at 24quotC is throttled until the nal quality is 0116 Find the nal temperature and pressure Approach Use the rst law specialized for a thronle I 24a C 39 x2 0115 Assmnptions saturated liquid 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 system operates in steadystate 4 The throttle is adiabatic Solution From Table A14 at 24 C hhf29 kJkg From the rst law applied to a throttle h 19 This problem must be solved iteratively since neither the nal temperature nor the nal pressure is known First assume a nal temperature and determine the corresponding quality X For example assume T1 hen from Table A14 7 111 7h 7 829400 11g 7h 247400 This value ofxz is too high so a different value osz is chosen By trial and error the nal value osz is found to be 0167 X2 T2 m SE C To verify this result calculate the nal quality as 07 X 0116 7 h 252 7 607 f The nal pressure is the saturation pressure at 8 C which is given in Table A14 as 130388MPa Answer 627 Saturated liquid R134a at 80quotF undergoes a throttling process The pressure decreases to 4 of its original Value Find the exit quality Approach Use the rst law specialized for a thronle I SODF P P 4 saturated liquid 2 Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 system operates in steadystate 4 The throttle is adiabatic Solution From the rst law applied to a throttle From Table B14 If 7 80 F then P 1014psia sat Bt AtR1014h 11 473113quot 1 P 101 1 252ps1a 4 4 To nd the quality at the exit we need to interpolate in Table B14 From Table B14 I E quot1 539s 5 238 1314 1025 10 267 1466 1032 By interpolation at 252psia 19139 and kg 1028 The quality is 711 2527139 x2 0127 lt Answer hgih 10287139 Appr Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system oper 4 The throttle is adiabatic Solution From the rst law applied to a throttle From Table A12 sup erheated Vap or oach Use the rst law specialized for a thronle A supply line contains a twophase mixture of steam and water at 240 C To determine the quality ofthe mixture a throttling calorimeter is used In this device a small sample ofthe twophase mixture is bled off from L quot 4 ll l in 39 u the L quot ofthe mixture in the Llietluuuliu 125 C L 39 1 main steam line Steam Valer T1 1 240 C Calorimeter ates in steadystate T2125L39 P1 l0 kPn Supply linc h 2725 E From Table A10 at T 240 C h1037 1 1 2804E kg 5 kg Hmd 7h h Mo955 4 Answer 1037 kg 7h 28047 x 6 29 In a heat pump R l34a is throttled through an expansion coil which is a long copper tube of small diameter The tube is bent in a coil both to fit in a compact space and to provide a large pressure drop The refrigerant enters as saturated liquid at 5 C with a flow rate of 0025 kgs and exits as a twophase mixture at a pressure of 200 kPa The wall of the coil may be assumed to be at the average temperature of the inlet and outlet Heat is exchanged by natural convection and radiation from the outer surface of the coil with a combinedheat transfer coefficient of 6 Wm C to the surroundings at 20 C The expansion coil has an outside diameter of 8mm and a length of 22 m Calculate the quality at the exit state Approach Use the first law specialized for an open system eliminating kinetic and potential energy changes work and the transient term Calculate the heat transfer from the convection rate equation Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The heat transfer coefficient is uniform over the surface of the coil and independent of temperature Solution From the first law for an open system 2 2 Q39W 7W5 2 x gzx V gzg Assuming no change in kinetic or potential energy and steady conditions with no work the first law becomes 0 m hz lnteIpolating in Table Al4 gives the enthalpy of saturated liquid at 5 C as h1 567kJkg The heat lost from the surface of the coil is Q WTM Tm The saturation temperature at the exit condition of 200 kPa is from Table Al5 T2 temperature of the outside of the coil is Tm g 5101 72550c 7101 C The average The surface area of the coil is A 7rDL7r8mm l m 22m 00553m2 1000 mm Therefore the rate of heat loss by convectionradiation is 397 7 W 2 o 7 QihA7W7TM76m2oc00553m 7255720 C77748W Solving the first law for exit enthalpy h Q 7748W 6 11000 2 m 0025k g kg 1 S 56396J 564kJ With values at the exit pressure of 200 kPa from Table Al5 the exit quality is x7 h2 h 7 56473684 hg 7h 24137 3684 00956 4 Answer Comment The exit enthalpy was very close to the inlet enthalpy with heat loss having little influence on the final enthalpy This is often the case39 therefore throttles are usually assumed to be adiabatic One way 39 39 39 39 39 39 in steam 40 kgs of subcooledliquid water enters at ls c and50 kPa Superheated steam enters at 200 c and50 kPa 39 39 39 39 39 39 t u a Assume a the tank is wellinsulated Approach Apply conservation ofmass and the rst law for an open system sat liquid Assum tjons 30 kpa 1 Potential energy change is negligible gt0 3 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The tank is adiabatic superheated vapor 200 quotC Solution 50 km From conservation of mass i quot 2 quot 3 From the first law for an open system d5 V 2 V at r Qv 41Vv m h T ngerl kl Tgzlj Neglecting changes in kinetic and potential energy and assuming an adiabatic tank and steady operation the rst law becomes 0 mm quotlth 7 min For state 1 in iiihalp L 39 39 39 inc enthalpy of saturated liquid at 15 C which is from Table A10 k 53 h k From Table AlZ at ZOODC and 50 kPa k 2877 h k From Table A11the enthalpy ofsaturated liquid at 50 kPa is U 3405 k Eliminating ml in the first law by using conservation of mass 0 mlhl m2h2 mi quotEVE 0Mihrhamhrh3 Solving for m2 rm h 7 hi 40 kjgjmr 3405 h ha 2877 7340513 A m2 438kg s Answer 631 In a desuperheater superheated steam is converted to saturated steam by spraying liquid water into the steam Using data on the gure calculate the mass ow rate of liquid water Approach Apply conservation ofmass and the rst LN ii law for an open system q Upg lvmlml 39 gtanv 4 The desuperheater is adiabatic 5 Pressure is constant during the process So t39 ll loll From conservation of mass quot391 quot392 quot392 From the rst law for an open system 2 2 mWw Zm h Lgz Z t 11 Vga a 2 2 Neglecting changes in kinetic and potential energy and assuming an adiabatic tank and steady operation the rst law becomes 0 quotah We From Table A12 the enthalpy of superheated steam at 250 C and 2 MPa is h 2902kJkg For state 2 the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 30 C which is from Table A10 h2 1258 kJkg TL r 1 Y I I L L 39 39 for all three streams The 2 MPa is from Table A11 r e ective enthalpy of saturated vapor at h1 2799 kJkg Eliminating mj in the rst law by using conservation ofmass 0 m quot2112 m nah owmimmzwm Solving for m2 77ml 11171117703290272799 116 I A m2 Id 7 n V Answer 125872799 kg 632 A at JC Water at u oi 39 39 39 39 What are the required mass ow rates ofthe two inlet streams Approach sal slcam Apply conservation ofmass and the rst law for an open system 1 8 A Assumptions T 2 1 Potential energy change is negligible Wm 2 Kinetic energy change is negligible 20 quotC 0 H a 3 The system operates in steadystate I00 kl a 4 The mixing chamber is adiabatic 5 Pressure is constant during the process Solution From conservation ofmass m m2 m3 From the rst law for an open system 2 2 it 94 7W Zm h V7gzezm h V2gz Neglecting changes in kinetic and potential energy and assuming an adiabatic tank and steady operation the rst law becomes 0 mm m a m From Table All for saturated steam at 100 kPa h 26755 kJkg For tate in nme s 2 L me enthalpy 39 39 c which is from Table Alo s For state 3 in nnarp 39 n p 39 39 c which is from Table Alo h 16757 kJkg Using conservation ofmass to eliminate m in the first law 0mrmzhrmhrrnaha 0quot ah m2hi rm2h7Tmaha Gatheringterms ma rkk rh kg k m 01 7h 826755716757k Answer a l kJ g774kgs him 2675578396 3 m 87 m2 026 kgs4 Answer 633 Steam with a quality of 088 and a pressure of 20 kPa enters a condenser The steam ow is divided equally among 20 tubes 21cm in diameter which run in parallel through the condenser The same amount of heat is removed from each tube Liquid water exits each tube with a velocity of 15 s an a temperature of 55 C Find the total amount of heat removed from the entire condenser Approach 1 liquid A e rst law for an open system S earn condenser waler eliminating allterms except heat and gt th 1 h o equot a W c ange x 058 T2 55 C P 20 kPa P2 20 kPa Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is 4 Pressure is constant in each tube Solution From the rst law for an open system 2 2 11 Qr Ww Zm h VfgzjnZ u 11 Vfgaj Neglecting changes in kinetic and potential energy and assuming steady operation with no work the first law becomes h h Using data from Table A11 for saturated steam at 20 kPa h 11 4115 7h2514088261072514 2327kJkg For state 2 the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 55 C which is from Table A10 112 3 The mass ow rate in each tube is with values of density from Table A6 2 9857k j15Er 21ch lm 0512k g m s 2 100cm s The total mass ow rate for all 20 tubes is Lg s mm 92 V m m 20mm 200512 1024 From the rst law Q mh7 7h1024E23072327 721478kw Answer 5 g 634 Saturated steam at 120 F is condensed in a tube as shown Cooling water at 50 F ows in cross ow over xterior ofthe pipe giving a heat transfer coefficient of 200 BtuhfF F Find the exit quality Agpr ag l 1 Cooling Water at 50 F se e erma resistance ana ogy to u 3 0 determine the rate of heat transfer Apply l h 200 Bum n F the first law for an pen system eliminating allterms except heat and enthalpy change Assumptions 2 Kinetic energy chang 4 Pressure is constant in each tube 5 The interior ofthe tube wall is at the Ru R condensing steam temperature T H T I l 1391 391 Solution Apply the thermal resistance analogy to i at o d pipe ZnLlc turated steam The resistance to conduction in the tube wall is 3 Btu 1a Bm 0000212 a 22ft30 h39 F h ft F The resistance to convection on the exterior ofthe tube is 1 A Btu h F 20013 2u 21325m 2ft h ft F 12111 The two resistances add in series The total heat transferred is 7 F Bm41640 Rm 0000212 000147 From the rst law for an open system 2 2 dig Qr 27quot h V7gz2m 11 V7gaj Neglecting changes in kinetic and potential energy and assuming steady operation with no work the first law comes 0 Q mhl r mh2 Solving for exit enthalpy and using data in Table A10 Btu 4139 640 Btu Btu h 11135 735 Holbhm lbm lbm Q h hi m Using data in Table A10 7h x h 1 i o53lt Answer 115711 11135788 635 Superheated R1 39 l v7 MPa 70 C li uid at 0 Pa with a Volumetric ow rate of 6000 cmZmin T e R134a exchanges heat with an air ow which enters at 18 C at a mass ow rate of 195 kgmin Find the exit air temperature Approach air Apply the rst law for an open 8 DC G Assumptions 1 Potential en negligible R134 2 Kinetic energy change is negligible 07 MP3 system operates in steadystate 70 UC 39 V 6000 cm lmin ergy change is r sat liquid air 07 MPa Illa 95 kgmin 5 Pressure is constant during the process 6 Air may be considered an ideal gas under these conditions 7 The speci c heat of air is constant Solution Using data in Table A15 for a saturated liquid at 07MPa the mass ow rate ofthe refrigerant is 3 sooo i 7 ViV 17 min 100cm quot9 p VY V k 72 g lTlln 000083281 3 From the rst law for an open system dE d Qrmzmh V72gz172mhvfzg2 Neglecting changes in kinetic and potential energy and assuming an adiabatic heat exchanger in steady operation the rst law becom 0quot1 445M 414 For an ideal gas with constant speci c heat AhcpAT therefore 0 quot901 40quotan T3 T4 Using data from Tables A16 A15 and A8 kg E TEMUVM Ejnminjcm 8678kg i p 1 Answer quot39 CP 195k g 1 kJ min kgK 636 R134a ows through the evaporator of a refrigeration cycle at a rate of 5 kgs The R134a enters as saturated liquid and leaves as saturated vapor at 12 C Air at 25 C enters the shell side ofthe heat air is required exchanger If the air leaves at 15 C what mass ow rate of Approach Apply the first law for an open system Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate sat vapor sat liquid i ll 0C 4 The evaporator is adiabatic 7 o air L C G 5 Pressure is constant during the process i 5 k IS C 6 Air may be considered an ideal gas under mr g these conditions 7 The speci c heat of air is constant Solution From the rst law for an open system dE V 2 V 2 V 7W m 4zi ihz Qw WELEPA 2 g2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic evaporator in steady operation the first law becomes 0quot1 hirhz ma 414 For an ideal gas with constant speci c heat Ah CPAT therefore 0 quot011 412 an T3 T4 Solving for the mass ow rate ofthe air ma riiy c0344 With values of enthalpy from Table A14 and speci c heat from Table A8 k 2540376618 ma S g g s K k 935 g Answer kJ s 1005 25715 C kgK 637 Superheated steam at 5 psia and 200 F is condensed in a heat exchanger The steam ows at 39 lbms and exim as saturated liquid Cooling water at 45 F is used to condense the steam The water and steam are not mixed in the heat exchanger but enter and leave as separate streams Ifthe maximum allowable water 39 39 F 39 veiucit is 11 s what is the diameter ofthe pipe which carries water to the heat exchanger Approach Apply the rst law for an open system Assumptions 9 1 Potential energy change is negligible 2 Kinetic energy change is negligible Snlummd qum P psm 3 The system operates1n steady state 2 TI 200 F Steam 5 Pressure is constant during the process 6 The cooling water is incompressible 7 The speci c heat of the water is constant 3m P4 147 psin P3 147 psin 45 F Solution From the rst law for an open system dE V 2 V 2 V 7W m zi mh4z Qw V 2 h 2 5n 2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic condenser in steady operation the rst law becomes ht Malaqah mlht Since the streams do not mix m1 m2 and m3 m4 therefore quot391l h2 391h4 h3 Assuming water is an ideal uid quot501145quot551 T4 T2 hi m3 m1 6 T4 2 With 11 from Table B12 112 from Table B11 and c from Table B6 Btu 114871302 mj 39113 ijt lbm 2545113 m s 1 15 F s lbm F To nd the diameter D 2 mjpVA pV 254511 s 221 Answer 11515241sz s it 638 A twophase mixture of steam and water with a quality of 093 and a pressure of5 psia enters a condenser at 143 lbms The mixture exits as saturated liquid River water at 45 F is fed to the condenser through a large pipe The exit temperature of the river water is 70 F less than the exit temperature of the other u h L rr wam stream In I 15 Ms calculate the pipe diameter Approach river water 45 F Apply the rst law for an open system Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible x t h LI 1 3 The system operates 1n steadys wo P ase 5a lqm 4 Th d 39 d39 b t39 39 e con enser 15 a la a 1c x 093 river watel 5 Pressure is constant during the process 6 The river water is incompressible 7 The speci c heat of the river water is c onstant Solution From the rst law for an open system dE V 2 V 2 V 7W m 4zi mhz 12 Q V Z 4 2 5n 2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic condenser in steady operation the rst law becomes quotah rah eh2m4h4 Since the streams do not mix m m2 and m3 m4 therefore MmAFMhrhs Assuming water is an ideal uid ht 49quotan T4 2 quotHM 0p T4 9 State 2 is a saturated liquid at 5 psia since pressure is constant across the condenser From Table B11 T2 1522 F T442770 922 F State 1 is atwophase mixture at 5 psia With data fromTable B11 Btu 11 h 7h 1302093113171302 1061 a f x g x lt gt quotm The enthalpy of state 2 is h 11 1302 Btulbm 39 Table D L 39 I I lClllpClalulC of69nF 14313 105171302 m3s 1bm h Th4 100 Blquot 922745 F lbm F U 282 lbms 42821bl s 0620 Answer 5221137 15 it s 635 639 A heat exchanger is used to cool engine oil The speci c heat ofthe oil is 06 Btulbm F Using data on the gure below nd the exit temperature of the air Approach Air Apply the rst law for an open system T g 00 17 Assumptions 1133 19 lbms 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 Th st 39 te 4 The heat exchanger is adiabatic 5 Pressure is constant during the process 6 The oil is incompressible 7 The speci c heat is constant m 8 Air behaves like an ideal gas under these IAir c onditions Solution From the rst law for an open system dE V 2 V 2 V 7W m 4zi mhz Qw V 2 h 2 5n 2 2 g Neglecting changes in kinetic and potential energy and assuming an adiabatic heat exchanger in steady operation the rst law becomes quotah We We quot14114 Since the streams do not mix ml m2 and m3 m4 therefore quot5h h2mzhrhz Assuming the oil is an ideal liquid and the air is an ideal gas quotarm Ti 3 MW T4 T2 38113105 Btu 160790 F TiT micpw1TrT2 A s lbm F 7 F Answer 47 3 quot115W 1glbm 024 Btu s lbmR 640 Saturated liquid 9 39 t 4 at 36 C quotquot 39 and exits as saturate vapor The evaporator is used to cool liquid water from 20 C to 10 C If the mass r d ow rate ofrefrigerant is 0013 kgs what is the mass ow rate of the wate Approach D throttle Apply the rst law for an open system to each component R1343 Assumptions Zagiulclqmd 1 Potential energy change is negligible waler wale 2 Kinetic energy change is negligible 20 0C IO 0C 3 The system operates in stea state 4 The throttle and evaporator are adiabatic 5 Pressure is constant across the evaporator 6 The water is incompressible 7 The speci c heat of the water is constant Solution Taking a control volume around the throttle and applying the rst law Using data in Table A14 kl h 1003E kg From Table A14 at 78 C 1 9 3954 kJkg and kg 2425 kJkg Since 112 1003 kJkg falls between these tw values state 2 is a twophase mixture Taking a control volume around the evaporator and applying the rst law dE V 2 V 2 w Qw iWw 27quot h Tgz 727quot h gz Neglecting changes in kinetic and potential energy and assuming an adiabatic evaporator in steady operation the rst law becomes quot392 quot14114 quot3924 mshs The water and R134a do not mix therefore quot392 39z quot quot394 395 w 0 MR h hz mw hrhs mw m h 7m 114 n 175 Assuming the water is an ideal liquid M h 41 c0440 V HMS 2 L 4 39 7i atwo phase mixture p t 39 C 39 L 39 region temperature remains constant Therefore state 3 is a saturated vapor at78 C From Table A14 13913 115 2425kJkg h2 h1 1003kJkg Using values of speci c heat at the average water temperature of 15 C 0013k 242571003 kJ k A g g00441k g Answer 5 W 4187kJkgK20710 C In a ash chamber anre im39 ed quot quot39 iuwei p 39 tw mixture The saturated liquid and vapor streams are removed in separate lines In the gure below liquid R134a at 10 F and 30 psia is thronled to 12 psiaIf2161bmh of saturated vapor exits the ash chamber what is the inlet ow rate Assume the ash chamber is adiabatic 9 A t l Approach rinsii Chamber Apply conservation ofmass and the rst law for an Assumptions 0 open system to the combination of the two Saturach vapor componenm 1 Potential energy change is negligible 1 Ia I 2 Kinetic energy change is negligible Th 39 ta 5 Pressure is constant in the ash chamber 501mmquot Smumlcd liquid In this problem the easiest approach is to define a control volume around the combination of the two components You could also analyze each component separately but that would take mor effort and produce the same final result Applying conservation of mass to the control volume shown I quot391 quot392 quot392 I From the rst law with no kinetic or potential I energy change and no heat transfer or work 39 quotihi We We 39 State 1 is a compressed liquid Approximate the I enthalpy of the liquid as the enthalpy of the I saturated liquid at the same temperature From 11 a 10 F 1466Bt11lbm From Table B15 the enthalpy of states 2 and 3 is 112 kg at 5 psi 9379 Btulbm h lyatS psi 73 Btulbm 39 39 39 value of mass into quot of mass produces l do m1 215113Trr m Substituting m and the values of enthalpy into the rst law 216m3lb m 1455 mom m 9379 mjlb m 7373 h lbm h lbm h lbm Solving for mj gives lbm 929 quot393 n From conservation of mass quotidiomewzame 11 lem 4 Answer 6 42 Saturated liquid water at 40 kPa enters a 140 kW pump The output of the pump is fed into a boiler where heat is added at a rate of 302 MW There is negligible pressure drop across the boiler If the mass ow rate of water is 70 kgs determine the boiler pressure and the state at the exit of the boiler Approach Apply WrhvPliP2to the pump to Sq WW Q 302411 determine the exit pressure form the pump U k 80 I 2 r quot0 kPa This is equal to the pressure at the exit of the boiler since there is negligible pressure drop across the boiler Draw a control volume around both components and apply the first law to determine exit temperature from the boiler Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The system operates in steadystate 4 The pump is adiabatic and ideal 5 Pressure is constant across the boiler 6 The water is incompressible Solution For the pump power is related to pressure rise by quotMR P2 Solving for exit pressure and using the specific volume of saturated liquid water at 40 kPa from Table Al l 1000w P2 12 7 40kPa 1000Pa7 1kwg 199x106 Pa m 20MPa quotN 70 0001027m s kg 7140kw Pressure is constant across the boiler so P3 P2 20MPa To find the state at the exit of the boiler draw a control volume around both components The first law for this control volume is assuming steadystate and no kinetic or potential energy changes 0ch Ww Jrth mghg 0Qermlrhz 39 7W h Q w A m 302000kW77140kW HISE 70kg kg S Answer h3 4634E kg The exit state of the boiler has a pressure of ZMPa and an enthalpy of 4634kJkg From Table Al 2 this is superheated vapor with a temperature of TlOOO C Answer 6743 Air at 2000 R 100 Btu of 39 the air 39 quotb 39 exit ow area of06 01 39quotquot r 39 39 10 lbfinz mt A uii 39 39 lbms Approach D Draw a control volume around each component and apply the first law Assumptions W B 1 Potential energy change is negligible 100 F 2 Kinetic energy change in the turbine is m quot7 negligible W 3 The system operates in steadystate 4 The nozzle and turbine are adiabatic 3 I my 5 Air behaves like an ideal gas under these J in conditions 6 The speci c heat of the air is constant Solution Be inb 39 39 A andpotential energy changes the first law becomes W m s i 2 liming an For an ideal gas with constant specific heat Ah cpAT therefore W quotmy Ti 72 39 39 however make further progress use a of air at the inlet temperature of ZOOOR about L 39 To 1500 F and correct the calculation later ifnecessary So ving for 72 and using values oft from Table 138 at 1500 F W 100m lbm 727 quot ZOOORr u1638R may Btu i Assuming an adiabatic nozzle no work no change in potential energy and steady conditions Wigwagj For an ideal gas with constant specific heat Ah cpAT therefore quot h nwever ect the calculation later ifnecessary u known 39 1m Ror1178 F r For simplicity we assume an average temperature of 1000 F and will corr Using data from Table 138 20228002 112 s2 3 1638R1043R 2 msg 39 32272 lme s From the ideal gas law 2 1545 ft lbf jlO43R 1ft 386ft3lbm ET lbmolR 144m2 v3 MP3 2897 lbm 101bfin2 lbmol The mass flow rate may now be calculated as 2800ft s 06ft2 m V3143 7 W 4351bms 439 Answer v 386ft3lbm 3 Comments For greater accuracy you could recalculate the mass flow rate with improved values of specific heat based on the calculated temperatures The specific heat is not a strong function of temperature over the range considered so this may not be necessary 6 44 A wellinsulated rigid tank of volume 07 m3 is initially evacuated The tank develops a leak and atmospheric air at 20 C 100 kPa enters Eventually the air in the tank reaches a pressure of 100 kPa Find the final temperature Approach Apply the first law for an open system and integrate over time Note that the temperature of the air leaking into the tank is constant with time so the enthalpy of the air is also a constant Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The tank is adiabatic 4 Air may be considered to be an ideal gas 5 The specific heat is constant Solution From the first law for an open system with no change in kinetic or potential energy dU d Q 7 W 2M 7 2mm The tank is adiabatic No work is done and nothing leaves the tank so the first law reduces to v dt Integrating over time dew dt For an ideal gas enthalpy depends only on temperature The air entering the tank is at the same temperature throughout the process so enthalpy is constant and may be removed from the integral U2 7 U1 hjmxdt hxmz Where m2 is the mass in the tank at the end of the process At the beginning of the process the tank is evacuated therefore U1 0 m2u2 hxmz u2 h ux Pvx For an ideal gas with constant specific heat Au chT RT T2 7T 7 Pv 7 V Solving for T2 T 8314101 K20273K 2M 7 k 39 20273K CV 2897 g 0718i kmol kgK T2410K137 C 4 Answer 6 45 Helium at 150 F and 40 psia is contained in a rigid wellinsulated tank of volume 5 ft A valve is cracked open and the helium slowly flows from the tank until the pressure drops to 20 psia During this process the helium in the tank is maintained at 150 F with an electric resistance heater Find a the mass of helium Withdrawn b the energy input to the heater Approach Use a mass balance and the ideal gas law to find the mass of helium Withdrawn Use the first law for an open system to find the energy input to the heater Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The tank is adiabatic 4 Helium may be considered to be an ideal gas 5 The specific heat is constant Solution a From a mass balance my m1 m2 Where mg is the mass of helium exiting Using the ideal gas law PMV PMV M 41bm15 3 mf i j T 2 E Tlazig 40720psia 006llbm 1 2 1073 150460R lbmolR b From the first law with no kinetic or potential energy diw Q 7 W 2m 7 2mm The tank is adiabatic and nothing is entering therefore Uw 7mm dt Integrating over time Ida iijdti dt Since the temperature of the helium in the tank does not change by is constant and U2 7 U1 4V5 WEIquotEd m2u2 T m1u1 Ww T heme m2u2 7 m1u1 Ww 7 a 132V2m2 m2u2 7 m1u1 Tch T 2 Pave ml Tmz Since u is only a function of temperature for an ideal gas and the temperature is constant u1 u2 u u Therefore 0Wchgvgmrm2 ET m T 006llbm 1545 ft39lbf 150460R 7 lbmR lBtu 4 lbm K778ftlbf lbmol WW 7185Btu lt1 Answer 643 646 A residential hot water heater initially contains water at 140 F Someone turns on a shower and draws water from the tank at a rate of 02 lbms Cold makeup water at 50 F is added to the tank at the same rate A burner supplies 5472 Btuh ofheat The water tank which is a cylinder of diameter 18 ft is filled to a height of4ft How long will39 L L L 39 39 reach 100 39 d t Approach Apply the rst law for an open system Use ideal liquid relations to rewrite enthalpy and T 50 quotF internal energy in terms of temperature Separate Variables and integrate In mc m n1CV constant Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 Water is an ideal liquid 4 The tank is wellmixed and all tank contents are at the same temperature 390 Solution The rst law for an open system with one stream in and one stream out neglecting kinetic and potential energy changes is dU V 7W 7 1 dt 9 m m quot L 39 4 quot rate q quot quot quot therefore d quotUM gnmwm The mass in the control Volume does not change with time For an ideal liquidAh CPAT With these ideas the rst law becomes m V mc T 7T w d Q lt gt The tank is wellmixed so the temperature T of the water in the tank equals the temperature of the exiting stream The internal energy of the water in the tank is related to its temperature by w 5 m cPdT since for an ideal liquid CV s CF The rst law takes the form dT quot39pr E Qv quot51 TFT Separating Variables and integrating from initial temperature T to nal temperature T 2 JP 7 22 la QwmcpTrT7I d let Qwch TiT d 7chdT 2 di me I r 1 1 mpg m w 5 7 m 1ng542 whamm z mHQqQ m i imfv 1n ch quot cp T2 12 Qwmc TrT p Find mv 2 mw pV 621bmft37rj ft2 4ft 6311bm 5472Btuh 1h o21bms1 B j507100 F I 776311bm1 26005 lbm F 2 u O39ZIbmS L 5472361W021507140 12 2096535min 4 Answer 6 47 In an industrial process two streams are mixed in a tank and a single stream exits Both streams may be assumed to have the properties of water The volume of uid in the tank is constant A paddle wheel stirs the tank contents doing work W Initially the water in the tank is at temperature T1 At time I 0 stream A at temperature TA enters with mass flow rate r39nA and stream B enters at T3 with rate m3 The quantities TA TB r39nA and ms are all constant with time Assuming a wellmixed tank derive a formula for the time t2 at which the tank water temperature is T2 The tank is well insulated Approach Apply the first law for an open system Use ideal liquid relations to rewrite enthalpy and mA 113 internal energy in terms of temperature TA TB Separate variables and integrate Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The liquid is ideal g 4 The tank is wellmixed and all tank contents are at the same temperature 5 The tank is adiabatic n39i 6 All uids have the properties of water L S olution The first law for an open system neglecting kinetic and potential energy changes is dU d 9 A W 2m 7 2mm Since the tank is wellinsulated d l w 7W rhAhA mghg 7 rhchc The mass of liquid in the tank is constant and conservation of mass requires that rhc rhA m3 therefore d m5 7WrhAhAthhB 7r39nAthhC mcv d TWmAhAThCmB ks Thc The liquid in the tank is ideal and has a temperature T The exiting stream is also at T because the tank is well mixed With these considerations the first law becomes mwcp 1 7W rhAcp TA 7 Tr39nBcp TB 7T m c 7WmAcPTAchpT37mAcPchpT w p dt 7 7WrhAcpTA thCPTB 7 mAmB T dl mwcp mv Define K1 and K2 so that g K1 7 K 2T Separating variables and integrating 13 1 dr Ti K17K2T let 5 K17 KZT d5 7K2dT 7I52 l 51K 2 1 5 iglnflgt2 12Lln2mw K2 51 K2 K17K2Tl 7 EVmATAmBT3mAm3T C 127mw jln Answer mAm3 7 EVmATAm3TBimAmBZ C P 648 A well insulated tank ofvolume 0035 m3 is initially evacuated A valve is opened and the tank is charged with superheated steam from a supply line at 600 kPa 500 C The valve is closed when the pressure reaches 300 kPa How much mass enters Approadli steam 600 kPa 500 quotc Use the first law for an open system Integrate the equation recognizing that the enthalpy entering is constant with time Properties are available inthe steam tables Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The tank is adiabatic 4 The state of steam in the supply line is constant Solution The first law for an open system neglecting kinetic and potential energy changes is dU JiZWkth The tank is wellinsulated no mass exits and no work is done therefore aw mh a IdoumaJmhw The state ofthe steam entering remains the same throughout the process so 11 is a constant Ifm2 is the mass in the tank at the end of the process and m is the mass at the start we quotHquot hlm h rm Since the tank is initially evacuated m1 0 and quotwe hm mh From Table A12 h 3483kJkg therefore u2 3483kJkg At the nal pressure of 300 kPa and u2 3483kJkg v2 m15m3 g Therefore the mass in the tank at the end of the process is K M 00233kg lt1 Answer 1 1 V2 649 A wellinsulated pistoncylinder assembly contains 006 kg ofR134a at 715 C with a quality of 092 A supply line Introduces superheated R134a at 10 C 200 kPa into the cylinder Assuming the pressure in the cylinder is constant calculate the volumejust when all the liquid has evaporated Approach Use the rst law for an open system Integrate the equation 5 m recognizing that the enthalpy entering is constant with time P c Assumptions 4 1 Potential energy change is negligible 2 Kinetic energy change is negligible R 3 3 The pistoncylinder assembly is adiabatic a liq R4376 4 The state ofR134a in e supply line is constant 0 C 5 The pressure in the cylinder is constant 200 Jar 6 The expansion is quasistatic Solution From conservation of mass m m2 m1 m2 is the mass in the tank at the end ofthe process In is the mass at the start and m is the total mass introduced during the process The rst law for an open system neglecting kinetic and potential energy changes is W Q 7 W 2 rm 2 mm The tank is wellinsulated and no mass exim therefore w Wvmyh dma uw 7Ww m h Idmwuw 7 Wwal j Mar The state ofthe R134a entering remains the same throughout the process so h is a constant quotwe miui WVh mdtWvhmi quotis139 39 r39 quot 39ofmass quotwe miui PVrVih new we quotHquot PM2V2 7mm quot392 mi quot12111 Prnzvrmui PmiVi h mi 7m Using the de nition of enthalpy h u Pv and recognizing that P I P We WFh quot392 mi Solving for m m MM h h h n The enthalpy ofthe R134a in the cylinder at the start ofthe process is h 11 4115 7h30609223837306 222 where 1 5 and kg are found by interpolation in Table A14 at 715T The contents ofthe cylinder remain at the 34achangesf 39 p A 39 r so the nal enthalpy is the enthalpy of saturated vapor at 715 C that is 112 2383kJkg The enthalpy in the supply line 11 may be found in Table A16 The nal mass may now be calculated as L k 006kg 227 259 worm kg0106k 2 Wi kli 39 g 2 I 2387 259 kg The final volume using data for specific volume interpolated in Table Al4 is 3 V2v2m20120 0106kg00127m3 4 Answer kg 650 The pressure inside a pot is maintained at an elevated level by a steel bob which resm on an open tube of inside diameter 05 cm The bob which has a mass of 0401 kg jiggles whenever the pressure in the pot is high enough to displacei 39 39 f 900 W Heat is lost from the sides and the top ofthepot by natural convection with a heat transfer coef cient of 39 WmZK The pot has a height of 0154 m and a diameter of 0256 m The ambient is at 20 C Assume 39 39 quot is v small 39 39 in the pot only water and steam The pot is half lled with water when the bob rst lifts a Find the temperature inside the ot b Find the net rate of heat addition to the pot c Find the initial mass 0 e twop ase mixture in the ot d Find the fraction ofthe pot which is lled with liquid after one hour Approach W Use the rst law for an open system Integrate the equation recognizing that me en L thalpy time Properties are available in the steam tables F Assumptions 1 Potential energy change is negligible 2 Kinetic energy change is negligible 3 The pot contains only steam and 4 The conduction resistance is small you w 5 The convection resistance on the inside ofthe pot is small 6 The work done in displacing the bob is negligible Solution a De ne m as the mass ofthe bob and r as the radius ofthe small tube that the bob resm on By de nition m 0401kg981 2 5 5 2007300132 A M7 025 2 2 739 7 m 100 Therefore throughout the process the pressure in the pot is P 200 3 kPa The saturation temperature at this pressure is from Table A11 T 120 C Answer b The conduction resistance through the walls and lid of the pot is assumed to be small Also the convective heat transfer coef cient on L 39 39 39 VCI L L 39 394 39 39 39 small Therefore the outside of the pot is at the saturation temperature of the water The net heat into the control volume is 2 QMquot hAAT QM an 39 JAT g 900 w739 ij 0255 0154 255 J 1207202C Q 8314 W 4 Answer c The volume of the pot is 2 2 V I L 2 j 0154 000793 m3 The initial volumes of liquid and vapor are VV2 and VgV2 From Table A1 1 v 000106m3 kg and VE 08857 In3 kg at 200 kPa Therefore V V m1 g374 kg Answer V V d The first law for an open system neglecting kinetic and potential energy changes is dU d 7 9 7 W 2m 7 2m No mass enters the pot and no work is done therefore dU W 7 m h dz Qw d mugt Qw m dt I dltmugt7 I 9871 mhdr The state of the steam leaving remains the same throughout the process so he is a constant If m2 is the mass in the pot at the end of the process and m1 is the mass at the start m2u2 T mlul chAt T he Imedt Q stti heme where m is the total mass that escapes Using conservation of mass m2u2 T mlul chAt The quot 2 7 m1 Eq 1 To find ul we need the initial quality of the mixture in the pot Use v1 03900793m3 000212m 3 m1 374kg kg With data from Table All at 200 kPa x V1 V 0002127 000106 vg 7 v 08857 7 000106 u1 u x1ug 7 u 504 0001197 25297504 507 kJkg Let y be the fraction of the liquid at the end by volume so that m2 17yV 0001197 Eq 2 v E Additional equations for the final state are v2 Vm2 Eq 3 v2v x2vg 7vf Eq 4 u2ux2ug7u Eq 5 Equations 1 through 5 are five equations with five unknowns ie m2 1 x2 big and y It is not a linear system of equations because Eq 3 is nonlinear Therefore the solution is iterative One approach is Assume m2 Find v2 from Eq 3 Find x from Eq 4 Find u from Eq 5 Check Ts Eq 1 satisfied lfnot adjusth Once m2 is known findy from Eq 2 The final results are m2 238 kg u2 5098kJkg x2 0002566 v2 0003326 m3kg y 032 Therefore 32 of the pot is filled with water at the end of the process Answer 652 5 1 At what pressure in kPa does water boil if T 170 C Approach Use the saturated steam tables Assumptions none Solution From Table Al 0 at T 170 C P07917MPa7917kPa 4 Answer Alternatively from Table All at T l7043 C P08MPa800kPa Comments It is easier and more accurate to use Table AlO when temperature is known 5 2 What is the specific volume of saturated water vapor at 600 kPa Approach Use the saturated steam tables Assumptions none Solution From Table All at P 06MPa vg 03157m3nng 4 Answer 5 3 What is the temperature of saturated water vapor with v 03468 m3kg Approach Use the saturated steam tables Assumptions none Solution From Table AlO at T 155 C vg 03468m3kg 4 Answer Alternatively from Table All at T15548 C vg 03427m3kg Comments It is easier and more accurate to use Table AlO when temperature is desired 5 4 Find the temperature in F at which a water boils ifP 35 psia b the specific volume of saturated water vapor is 1207 ftBlbm Approach Use the saturated steam tables Assumptions none Solution a Using Table B1 1 which presents saturated steam data in even increments of pressure T 2593 F lt Answer b Using Table B 10 T 60 F Answer Comments You could also use Table Bll for part b In that case you would have to interpolate 5 5 Find the pressure in kPa at which a water condenses if T 195 C b the specific volume of saturated water vapor is 005 m3kg Approach Use the saturated steam tables Assumptions none Solution a From Table A10 P 13978 MPa 1398 kPa 4 Answer b From Table A1 1 P 4 MPa 4000 kPa Answer Comments You could also use Table A10 for part b In that case you would have to interpolate 5 6 A rigid can contains 090 g of saturated water vapor at 450 kPa Calculate the volume of the can in cubic centimeters Approach Use the saturated steam tables to determine specific volume then compute volume from the known mass Assumptions none Solution From Table All at P 045 MPa vg 0414 m3kg From the definition of specific volume 3 Vmv 090g 04mm 000037m3373cm34 Answer g 1000g kg 5 7 A pistoncylinder assembly contains 012 ft3 of saturated water vapor at 350 F What is the mass of vapor in the tank Approach Use the saturated steam tables Assumptions none Solution From Table B10 at T 350 F vg 3346 ftElbm From the definition of specific volume 7 V 012ft3 A 3 00361bm Answer vg 3346ft lbm 5 8 Find the specific volume of gaseous R 134a at 40 C for P 100 kPa 400 kPa and 800 kPa Use both the ideal gas law and tabulated values Approach Use the superheated vapor tables to determine specific volume compare to results from the ideal gas law Assumptions none Solution From Table All at P 045 MPa vg 0414 m3kg From the definition of specific volume Solution From the ideal gas law using data in Table Al kl 1000 J 8314 40273K j 3 V E H o2551 MP 102 g 100 kPa a g kmol 1 kPa From Table A16 for superheated R134a at 40 C and P 100 kPa 01 MPa 3 v 025076m kg Repeating the calculation at the other pressures and tabulating P v Tdeal gas v Table A16 kPa mSkg m3kg 100 02551 02508 400 006377 005917 Answer 800 003188 002691 Comments The tabulated values are more accurate The ideal gas law is a better approximation at low pressures 5 9 Find the density of steam at 35 MPa and 415 C a using the steam tables b using the ideal gas law Approach Use the superheated vapor tables to determine specific volume which is the reciprocal of density compare to results from the ideal gas law Assumptions none Solution a From Table A12 at 35 MPa I T C v m3kg I 400 008453 450 009196 Interpolating for a temperature of 415 C w 4507400 0091967008453 3 v 008676m kg plll53k g3 4 Answer v m b From the ideal gas law 6 35MPa 1013a 18015 kg 1MP p 1102133 Answer 8314 I m 4152731K m kmolK 1k Comment The ideal gas law works reasonably well under these conditions 5 10 Refrigerant 134a at a pressure of 20 psia and a temperature of 40 F occupies a volume of 05 ft Find the mass a from table values b from the ideal gas law Approach Use the superheated vapor tables to determine specific volume which can be used to find mass compare to results from the ideal gas law Assumptions none Solution a From Table B16 at 20 psia and 40 F v 25244 ft3 lbm 3 m K 01981bm Answer v 25244ft3lbm b From the ideal gas law lbm 20 ps1a05 ft310203 m fwd 0l901bm 4 Answer 1073 40460R lbmolR Comment Not a bad approximation 5 11 A container is filled with 0026 kg of Refrigerant134a at a temperature of 40 C What is the pressure if the vo um i a 364 cm3 b 1560 cm3 Approach Calculate the specific volume and decide whether the substance is a liquid a vapor or a twophase mixture Then use the correct table to determine pressure Assumptions none Solution a First calculate the specific volume V 364cm3j 3 7 7 cm 0014m m 0026kg k From the problem statement it is not clear whether the R l34a is a vapor a twophase mixture or a liquid Consult the saturated table Table Al4 at 40 C and note that 3 3 v 0000871m v 00199m kg g kg The calculated value falls between the saturated vapor and saturated liquid values Therefore the R l34a exists as a two phase mixture and the pressure from Table Al4 is P Pm 10164 MPa b Recalculate the specific volume from the given volume for part b as 1 3 1560 3 7 V 7 100 7 006 m kg m 0026 This is greater than the specific volume for saturated vapor at 40 C so the refrigerant exist as a superheated vapor In Table A16 check values of specific volume at 40 C for various pressures The pressure corresponding to 40 C and 006 m3kg is P 2 04 MPa4 Answer 5 12 A tank contains 005 lbm of water vapor at 20 psia and 500 F Find the volume of the tank in ft3 Approach Use the superheated vapor tables to determine specific volume which can be used to find volume Assumptions one Solution From Table B lZ for superheated water vapor v 2846 ftElbm therefore Vvm 2846ft31bm0051bm1423ft34 Answer 5 13 A container ofvolume 0047 m3 is filled with 67 kg of steam at 600 C Calculate the system pressure Approach Compute the specific volume Determine from the steam tables What regime the substance is in Then read the pressure directly from the table Assumptions none Solution The specific volume is 3 3 VKw 0007m m 67 kg kg From Table A12 at T 600 C and P 40MPa v 0008094 m3kg Also at T 600 C and P SOMPa v 0006112 m3kg Therefore P 2 45 MPa Answer Comment We estimated the answer by eyeballing the table If greater accuracy is needed you could interpolate 5 14 A pistoncylinder assembly with a volume of 400 in3 contains a steamwater mixture at 80 psia If the total mass of the mixture is 0066 lbm find the volume of liquid present in in Approach Compute the specific volume Determine the specific volume of saturated liquid and that of saturated vapor from the steam tables and use this information to compute quality From the definition of quality find the mass of liquid present The specific volume of the saturated liquid can then be used to find the volume of liquid Assumptions none Solution The specific volume is 3 400 i113 7 V 7 12 m 7 3 ft 7 m 7 00661bm 7 39 lbm From Table Bll at 80 psia 3 3 v 001757 f v 5474fL lbm g lbm To determine quality use v v x vg 7 v Viv 3517001757 x 064 vg 7v 5474 7 001757 By definition m xg gt mg xm0640066lbm 004221bm m The mass of liquid present is m m 7 mg 00667 00422 0024 lbm The volume of liquid is 3 39 3 Vmv0024lbm 001757fL 0729in3lt Answer lbm 1ft 5 15 A twophase mixture of steam and water has a quality of 079 and occupies a space of 051 ft3 If the total mass is 00871 m a Find the temperature b Find the volume of liquid present in inf Approach Compute the specific volume Assume a temperature and calculate the specific volume from the saturated steam table at this temperature for a quality of 079 Keep iterating on temperature until the assumed value gives the correct mixture specific volume For part b use the definition of quality to compute the mass of vapor present Subtract from total mass to get the mass of liquid and hence the volume of liquid Assumptions none Solution a The specific volume is 3 VK 5 86f m 0087 lbm Neither the pressure nor the temperature is known so an iterative calculation is needed To begin we assume arbitrarily a temperature of 250 F From Table BlO at 250 F v 00167 ft3 lbm and VE 2782 ft3 lbm To determine specific volume use v v xvg 7 vf 00167 ft31bm 07927827 00167ft3 lbm 2198 ft3 lbm This is not equal to the correct value of 586 ft3lbm so choose a different temperature and repeat the calculation After some exploration the correct value is found to be about T 290 F At this temperature from Table B lO v 0017352ft3lbm and VE 7467ft3 lbm The specific volume is v v xvg ivf 0017352 13 lbm 079746770017352ft3 lbm 5903 13 lbm which is close to the correct value of 586 Therefore T 290 F Answer b By the definition of quality x m Therefore the mass of vapor present is mg xm0790087006873 lbm The mass of liquid is the total mass minus the mass of vapor m mimg001827 lbm The specific volume of the saturated liquid is from part a v 0017352ft3 lbm Volume is related to specific volume by Vmv 00182700174 V 00003172 ft3 0548 11134 Answer 5 16 A tank contains a twophase mixture of steam and water at 40 psia If the volume of the vapor is 10 times that of the liquid What is the quality Approach Express quality as a ratio of mass of vapor to mass of liquid plus vapor Rewrite masses in terms of volumes Use the given fact that vapor volume is 10 times liquid volume and all volumes Will cancel Assumptions None Solution From the given information in the problem statement 10V Vg Using the values of specific volume for saturated liquid and saturated vapor at 40 psia in Table Bll V v 0017146 m V 10V vg g 1o501 m E E By the definition of quality 10V x mg 0016 Answer m mg Lr 0017146 10501 5 17 A tank ofvolurne 530 cm3 contains a twophase mixture of R134a at 712 C The mass of liquid present is four times the mass ofvapor a Find the total mass of R134a in the tank b Find the volume of liquid present Approach Express quality as a ratio of mass of vapor to mass of liquid plus vapor Use the given fact that the mass of liquid is four times the mass of vapor to cancel out masses and determine the quality Then find specific volume of the mixture using table values which leads to mixture mass For part b determine the mass of vapor and subtract from total mass to get the mass of liquid The volume of liquid is then obtained from the specific volume of saturated liquid Assumptions one Solution a From the definition of quality x mvapar mvapar mg m m m m m WP g mm In Since the mass of liquid is four times the mass ofvapor m l x g 02 4m g m g 5 Using values from Table Al4 3 3 3 vv xv ivf 00007498m 0201068700007498 0022m g kg kg kg 3 530 cm 1 m mKM 00241 kg v 0022 m kg b The mass ofvapor is given by mg xmm 0200241 000483 kg The mass of liquid is total mass minus the mass ofvapor so m mm 7mg 002417 000483 00193 kg To find the volume of liquid use 3 6 3 V mv 00193 kg00007498g i m3 1 145 3113 Answer 5 18 A tank with a volume of 48 ft3 contains 6 lbm of liquid water The tank also contains water vapor in equilibrium with the liquid If the pressure in the tank is 30 psia calculate the quality Approach Compute the volume of liquid in the tank from the specific volume in Table Bl 1 Then determine the volume of vapor by subtracting from the known total volume Calculate the mass of vapor and use this in the expression for quality Assumptions none Solution First find the volume of liquid water in the tank Using the specific volume of saturated liquid in Table Bll at P 30 psia ft3 3 V mv 6 lbm 0017004E 0102ft The volume ofvapor is the total volume minus the liquid volume Vg Vm A V 48ft3 A 0102ft3 47ft3 The mass ofvapor is using the specific volume of saturated vapor in Table Bll at P 30 psia V 3 mg iL 3 03421bm V 13748 lbm Finally the quality is x mg 7003 4 Answer m mg 7 60342 5 19 A vial of volume 280 cc contains a twophase mixture of steam and water at 30 C The quality is 045 Find the mass in grams Approach Compute the specific volume using data from the steam tables Determine mass from specific volume Assumptions none Solution First find the specific volume of the mixture in the vial Using values from Table A10 at 30 C v v xvg 7 v 0001004 0453289 7 0001004 148 m3kg 3 lm V7 280cm3100cm 7 0019g 4 Answer v I ij lkg m 39 k g 1000g 5 20 A tank ofvolume 004 m3 contains 06 kg ofR134a at a pressure of 02 MPa a Find the temperature b If the volume is 0068 m3 then What is the temperature Approach Compute the specific volume and then consult the saturated Rl34a tables If the specific volume falls between saturated liquid and saturated vapor then the temperature is the saturation temperature If the specific volume if higher than the saturated vapor specific volume find temperature in the superheated Rl34a tables Assumptions none Solution a The specific volume is 004m 06kg From Table A15 atP 02 MPa vf 00007532 m3kg and vg 00993 ng Sincev lt v lt vg TT 771009 c 4 Answer sat V 0067 m3kg b The new specific volume is 3 3 v 039068m 0114m 06 kg kg As before vg 00993 m3kg Since v gt vg the Rl34a exists as a superheated vapor From Table Al6 T 20 C Answer 5 21 Find the specific volume of H20 in each of the following states a saturated liquid at 160 F b superheated vapor at 80 psia and 440 0F c twophase mixture at a quality of 07 and a pressure of 40 psia d subcooled liquid at 120 F 147 psia Approach Consult the steam tables Assumptions none Solution a From Table B lO v00164 ft3lbm 4 Answer b From Table B12 v654 ftElbm Answer c The specific volume is using data from Table Bl l 3 v v xvg ivf 001714607105017 0017146 7356 Answer 11 d The specific volume of the subcooled liquid is approximated by the specific volume of the saturated liquid at the same temperature thus using data from Table BlO 3 v w v Tm 0016205f Answer lbm 5 22 Determine the volume in m3 of 023 kg of H20 at a temperature of 150 C and a a pressure of 02 MPa b a quality of 06 c a pressure of5 MPa Approach Consult the steam tables Assumptions none Solution a To determine the state consult the saturated steam table at 150 C Table AlO and note that the saturation pressure at this temperature is 04758 MPa The system pressure 02 MPa is less than this so the H20 exists as a superheated vapor From Table AlZ v 09596 m3kg and 3 V mv 023 kg 09596 02207 m34 Answer g b Since quality is specified the substance exists as a twophase mixture With data from Table Al 3 v v xvg7v 0001091 06O3928000109102361i 3 Vmv 023kg02361 543x10 2m3 Answer g c In this case 5 MPa is greater than the saturation pressure of 04758 MPa obtained from Table AlO Therefore the H20 exists as a subcooled liquid Furthermore the pressure is high enough so that the compressed liquid table should be consulted From Table Al3 I T C I v m3kg I 140 00010768 160 00010988 Taking the average of the above two entries v 00010878 at 150 C 3 V mv 023 kg 00010878l 25019gtlt10quot m3 Answer 8 5 23 Find the specific volume of a compressed liquid water at 100 F 1000 psia b saturated liquid water at 100 F c saturated liquid water at 1000 psia Approach Consult the steam tables Assumptions none Solution a From Table B l3 v 0016082ft3 lbm Answer b From Table BlO v 0016130ft3 lbm Answer c From Table Bll v 002159ft3 lbm d Answer Comments The specific volume of compressed liquid water can be approximated by that of saturated liquid water at the same temperature thus the results from a and b are very close If you erroneously evaluate the specific volume at the same pressure rather than the same temperature large differences result as in a and c 5 24 Fill in the values of the specific volume of compressed liquid water at the conditions shown in the table Use scientific notation with four significant figures eg 06216 x 103 Does v depend more on temperature or pressure Approach Consult the steam tables Assumptions none Solution m P 5 MPa 10 MPa 2339 kPa 03613 MPa 20 C 1002 x 10393 0999 x 10393 0997 x 10393 140 C 1080 x 10393 1077 x103 1074 x 10393 J J From Table AlO From Table Al3 Comments The specific volume depends more on temperature than pressure 5 25 Calculate the enthalpy of compressed liquid water at 40 C two ways using the approximate relationship for enthalpy of a compressed liquid and using the compressed liquid tables Perform the calculation at these pressures a 10 MPa b 20 MPa c 50 MPa Approach Consult the steam tables and use the approximate expression for enthalpy of a compressed liquid Assumptions none Solution a The approximate value of enthalpy for a compressed liquid is W hT VTPPsaT With data from Table A10 kJ 3 3 hquot 16757 0001008m 10MPa10 kPa kl kg kg 77384kPa 1776 kg 1 MPa kl From Table A13 h17638 kg b c A table of results at all three pressures is 10 MPa 20 MPa 30 MPa h 1776 1877 2180 176 38 18516 21121 Comment The approximation is good even at very high pressures For this reason compressed liquid tables are rarely needed 526 A rigid tank of volume 06 m3 contains saturated Rl34a vapor at 24 C The contents are cooled until the temperature is 0 C How much heat is removed Show the process on a Pv diagram Approach Since the tank is rigid and mass is constant the speci c volume is constant Find the speci c volume 1 at the initial state from property tables and set it equal P to the nal speci c volume To nd heat use the rst law 2 Assumptions 1 The tank is rigid 2 There are no kinetic energy changes 3 There are no potential energy changes Solution Because the tank is rigid volume is constant Since the mass m is also constant the speci c volume v V m is constant From Table Al4 the speci c volume at the initial state saturated vapor at 24 C is vg 003l7m3kg The mass ofRl34a present is 3 K 39 6m 3 189kg V 003171 kg The rst law assuming no kinetic or potential energy changes is Q AU W Since the volume remains constant no work is done and the rst law becomes Q AU U 2 7 U1 As shown in the P v diagram above the nal state is in the twophase region Because speci c volume is constant vZ vg1 00317 m3kg We can calculate the quality at the nal state from Using values from Table Al4 at 0 C vZ vfZ xvg2 ivfz 00317 0000772 x006897 0000772 0454 Returning to the rst law QUZ 7U1muziul mufzxug2 iufz7 ugl Using data in Table Al4 Q 189 4979 0454227067 49797 240 72074kJ Answer 527 A mixture of steam and water is contained in a rigid tank of volume 3050 cm3 The mixture has a quality of 055 and a temperature of 120 C Heat is added until the temperature is 140 C Find a the nal quality b the amount of heat added Approach Since the tank is rigid and mass is constant the specific volume is constant Find the specific volume at the initial state from property tables and set it equal P 2 to the final specific volume To find heat use the first f law Assumptions V 1 The tank is rigid 2 There are no kinetic energy changes 3 There are no potential energy changes Solution a Because the tank is rigid volume is constant Since the mass m is also constant the specific volume v V m is constant and v1 v2 The initial state is in the twophase region therefore v1vf x1vg ivf With values from Table A10 at 120 C v1000106055089197000106 0491m3kg To find x2 use i22vfx2vg ivf vzivf ivlivf x2 7 M 09654 Answer vgivf vgivf 050897000108 where specific volumes of the saturated liquid and vapor were taken from Table A10 at 140 C b The first law for a closed system is No work is done therefore QAU mu27u1 The mass may be found from 3050 cm3 m Vi m 100cm 3 000621 kg 04911 kg The internal energy of the two states again using data in Table A10 is uluf x1ug7uf 50350552529375o35 1618 kJkg u2ufx2ugiuf588740965255075887 2481kJkg kl Q000621kg248171618F g 536 k1 Answer 528 A rigid tank contains a twophase mixture of water and steam at a quality of 065 and a pressure of 20 psia The mass of the mixture is 026 lbm The mixture is heated until the nal quality is 095 Compute the final pressure and the heat added Approach Since the tank is rigid and mass is constant the specific volume is constant Find the specific volume at the initial state from property tables and set it equal P 2 to the final specific volume It is necessary to iterate f on the final pressure to find the final state To calculate heat use the first law 1 V Assumptlons l The tank is rigid 2 There are no kinetic energy changes 3 There are no potential energy changes Solution Because the tank is rigid volume is constant Since the mass m is also constant the specific volume v Vm is constant and v1 v2 The initial state is in the twophase region therefore with values from Table B1 1 v1 v gvg17v 001683 06520097001683 1306ft3lbm The final state is also in the twophase region with a quality of 095 The specific volume at the final state is v2 va x2 ng iv In this equation v2 and x2 are known and P2 is to be found An iterative approach is necessary Assume a value for P2 and find vzf and v2g in Table Bll Then calculate v2 lfvl v2 you are done if not choose a new P2 and iterate For example assume P2 25 psia Then v2 001692095163067001692 l55ft3 lbm Ts v2 1306 No therefore we must assume a new value for P2 Let s assume P2 30 psia v2 0017004095l374870017 l306ft3 lbm Since v2 1306 the iteration has converged and P2 30 psia i Answer Now calculate the heat transferred from the first law QAUWAUU27U1 Since there is no work done Q mu2 T 1 The internal energy is given by with values from Table B1 1 u1 qu x1ug1 7141 19619065108271962 7719 Btulbm u2 uf2 x2ugfuf2 2188095108872188 1044Btulbm Substituting values Bt Q mu2 7 ul 026 lbm10447 7719 lb u m 7074Btu Answer 529 A twophase mixture of steam and water at 800 kPa x 085 is contained in a rigid wellinsulated tank An electric resistance heater supplies 50 W to the mixture which has a total mass of 13 kg How long must the heater operate to reach a nal temperature of 190T Approach 39 e tank is rigid and mass is constant the speci c Volume is constant Find the speci c Volume at the initial state from property tables and set it equal to the nal speci c Volume Interp olate in the steam tables to nd the nal state To compute time use the Assumptions 1 The tank is rigid 2 There are no kinetic energy changes 3 There are no potential energy changes 4 The system is adiabatic Solution Because the tank is rigid Volume is constant Since the mass m is also constant the speci c Volume v Vm is constant From Table A11 the speci c Volume at the initial state is v v x1vg 7v 0001115 0850240470001115 02045m3 kg The rst law assuming no kinetic or potential energy changes is Q AU W Since the tank is insulated the rst law becomes 7 AU We need the internal energy at the nal state where v2 v 02045m3 kg and T2 190 C It is not clear from quot 4 4 39 p iciunTo swer this question consult Table A10 for saturated steam Note that the speci c Volume of saturated Vapor at 190 C is 015654 mZkg 39 ce v is Luau Lui the nal L r p as shown in the gure above To nd the internal energy at the nal state we need to double interpolate in Table A12 The table below reproduces selected Values from Table A12 at 08 and 10 MPa Interpolated Values are highlighted in bold T C v mZkg u kJkg T C v mZkg u kJkg 170 02402 2577 180 0194 2584 190 02539 2613 190 02000 2603 200 02508 2531 200 02050 2522 A second interpolation between v and u gives 025397 02000 7 2613 7 2603 v2 7 0 2000 u2 7 2503 02539702000 261372603 i 2 ZGOSE 02045 7 02000 112 7 2603 kg Substituting Values in the rst law W mu2 7111 13 kg260372298k 397 k 3 Time is now found from W 397 14 7 m 7 77940 sec 722 lt Answer W 50 w 530 R134a at 40T 15 psia is contained in a rigid tank ofvolume 228 in The tank is cooled at a rate of6 Btuh uuvv 39 39 theRl 4a p 39 39 39 L 39 to nse Approach Since the tank is rigid and mass is constant the 1 speci c Volume is constant Find the speci c Volume P at the initial state from the superheated Vapor table 2 and et it equal to the nal speci c Volume T 134a is a saturated Vapor at the nal state Use the rst law to calculate heat transferred V Assumptions 1 Theth is r1 1 2 There are no kinetic energy changes 3 There are no potential energy changes Solution Because the tank is rigid Volume is constant Since the mass m is also constant the speci c Volume v Vm is constant From Table B 16 the speci c Volume at the initial state is 3 34 lbm The rst law assuming no kinetic or potential energy changes is Q AU W Since the Volume remains constant no work is done and the rst law becomes QAUUrUi maew Find mass from 3 228m i 12in m 3 v ft 003881bm m From Table B16 the initial internal energy is tu 1009 lbm To nd the nal internal energy consult Table B14 You could also use Table B15 The speci c Volume of saturated Vapor at 720 F is 342 Zlbm which is close to v At 720 F u ug 06 Btulbm Substituting Values in the rst law gm my fix 00388 lbm100979061Bb 0403 m At 9 03940B 005711 40min 4 Answer Q 6 Btu h 531 A rigid tank lled with 07 lbm of saturated water vapor at 400 F is cooled at constant volume If the nal temperature is 260 F a Find the nal mass of liquid b Find the heat transferred Approach Since the tank is rigid and mass is constant the speci c volume is constant Find the speci c volume 1 at the initial state from the saturated steam table and P set it equal to the nal speci c volume Use the rst law to calculate heat transferred 2 Assumptions V l The tank is rigid 2 There are no kinetic energy changes 3 There are no potential energy changes Solution a Because the tank is rigid volume is constant Since the mass m is also constant the speci c volume v V m is constant From Table B lO the speci c volume at the initial state is v1 1866ft3lbm vg39 therefore v2 v1 1866ft31bm As shown in the gure above the nal state is in the two phase region vZ varxZ vg ivf x 7 V2 Vf 718667001708 7 7 01573 vg ivf 11777001708 By de nition m x2 g m mg mxZ 0701573 011 lbm mf mimg 077011 0591bm Answer b From the rst law UQ7W No work is done so Q mu2 u1 Using data 39om Table B lO uZ uf x2 ug 7 uf 2286 0157310905 7 2286 3642Btulbm Substituting in the rst law Q 07lbm3642711166Btulbm 75267Btu Answer 532 A wellinsulated piston cylinder assembly ofvolume 0006 m3 contains 625 grams of steam at 150 C The steam expands and during this process 0759 kJ of work is done If the nal temperature is 95 C what is the final volume Approach Apply the first law recognizing that heat is zero Use 1 the state principle and the steam tables to identify property values P Assumptions 1 The system is adiabatic 2 There are no kinetic energy changes 3 There are no potential energy changes Solution From the first law AU Q 7 W Since the assembly is wellinsulated m 112 7 u1 7W We need ul At state 1 the specific volume is V1 0006m3 V 096m3 k 1 m 000625kg g Check Table A10 for the values of specific volume at 150 C The computed value of VI is greater than vg at 150 C therefore the steam is superheated From Table A12 at 150 C and v1 096 P1 02MPa u1 25769E kg Solving the first law for uz and substituting values uZ u1 7K257697L9kl 7 45546E m 000625 kg kg At the final temperature of95 C from Table A10 uf 39788 kJkg and ug 25006kJkg39 since uz is between these two values the final state is a twophase mixture From Table A10 at 95 C vf 000104 m3 kg and vg l982m3 kg To find v interpolate 250067245546 7 19827vZ 25006 739788 19827 000104 v2 194 VZ mvZ 000625194 0012m34 Answer 533 A twophase mixture of steam and water with a temperature of 160 C and a quality of 06 is contained in a pistoncylinder assembly The twophase mixture which has a total mass of 09 kg is compressed slowly and isothermally until only saturated liquid is present What is the work done on the system Approach Evaluate work from W IPd V Property values are found in the saturated steam table P 2 Assumptions 1 The compression is quasistatic 2 The system is isothermal V 3 There are no kinetic energy changes 4 There are no potential energy changes Solution Because the compression is slow we may assume it is quasistatic and the work done is W J39PdV Since the process is isothermal and remains entirely in the twophase region pressure is constant as shown in the figure above Work becomes WPJ39aIVPAVPV2 41 Using data in Table A10 at 160 C v1 vf x1 vg 7 vf 0001102 06030717 0001102 0185m3kg V1 mv1 09 kg0185m3kg 01662 In3 The nal speci c volume is the specific volume of saturated liquid water at 160 C From Table A 10 v20001102 m3kg V2 mv2 09000110299 x10394 m3 The saturation pressure at 160 C is also available in Table A10 It is P 618 kPa Work may now be evaluated as 1000 Pa 1 kPa 7102100 x105 J 7102 kJ4 Answer W618 kPa 99x10 4701662m3 Comments Work is negative because work is done on the system during a compression 534 Refrigerant 134a is contained in a perfectlyinsulated pistoncylinder assembly The refrigerant is initially a saturated vapor at 10 F with a volume of 032 ft3 It is then compressed to a superheated vapor at 120 F and 80 psia Find the work done Approach Evaluate work from the first law Property values are 2 found in the saturated and superheated tables Assumptions 1 The system is adiabatic 2 There are no kinetic energy changes 3 There are no potential energy changes V Solution From the first law Q7 AU W The system is wellinsulated so Q0 and 0AUWmAuW From Table B 14 for saturated R 134a at 10 F vg 1725 ft31bm ug 9468 Btulbm The mass of the system is V 032 m 018561bm vg 1725 From Table B 16 u2 1136 Btulbm W7mAu701856lbml 1367 9468Btulbm 7351Btl1 Answer Comments Work is negative because work is done on the system during a compression 535 Atwophase mixture ofwater and steam with a quality of 063 andT 300 F expands isothermally until only saturated Vapor remains The initial Volume is 0114 N During the process 162 Btu ofheat are added Find the work done Approach Evaluate work from the rst law Property Values are found in the saturated steam tables Assumptions 1 The system is isothermal 2 There are no kinetic energy changes 3 There are no potential energy changes Solution From the rst law QAU AU m 2 i 1 State 1 is in the twophase region so u u xugiuf From Table B10 at 300 F Bt 11 2595 0531 1007 2595 792 quot lbm The final state is saturated Vapor From Table B 10 Bt u2 ug 1100 u lbm To determine the mass we need the speci c Volume At state 1 3 v f xvg ivf 00174 0535472700174 408 g 0114tt3 mK 100279 lbm 408 Rearranging the rst law WQ iAU Q m 2 rui 162Btu700279 lbm110077921Bbi m 75 Btu AIISWBIquot Comments Work is positive because work is done by the system during an expansion 536 Apistoncylinder assembly contains 025 kg ofsatlirated Refrigerant 134a vapor at 16 C The refrigerant is cooled at constant pressure until the volume is one half of its original value Calculate the heat transferred Approach Apply the first law Compute work from WIPdV Property values are found in the saturated R134a P tables Assumptions v 1 The process is quasistatic 2 There are no kinetic energy changes 3 There are no potential energy changes 4 The pressure is constant Solution From the first law AU Q 7 W Because the process is quasistatic PdV P dVPAV Substituting into the first law AU Q 7 PAV mlturuigt7Q7Pmltv7v0 SinceP I Ppwe may rewrite this expression as Q MU V2ui HM quotIUI 41 To find the enthalpy at state 2 we need the qu 39 there The mass ofthe system is constant so when the volume is halved the specific volume is also halved At the initial state with data from Table A14 v 004051 g kg 3 v2 V Z 002025 g v2 v x2 v5 7vf Solving for quality x V2 V 002025 7 00008062 2 v 7v 00405 7 00008062 The final enthalpy is 0490 h2 h x2015 711 7159 04925527 7169162kJkg For state 1 11g 2562kJkg Substituting values Q 0251527 2552 7235kJ 4 Answer 537 A pistoncylinder assembly contains 015 kg ofsatnrated steam at 130 C The piston is held in place by a weight To reach the nal state 8300 J of heat is added Find the nal temperature Approach Apply the rst law Compute work from WIPdV Rewrite the rst law in terms of enthalpy Property P Values are found in the steam tables 2 Assumptions V 1 The process is quasistatic 2 Th are no kinetic energy changes 3 There are no potential energy changes 4 The pressure is constant Solution From the rst law AU Q7 W quot 39 4 4 39 ensures takes r 39 is assumed quasistatic WdeV deVPAV Substituting into the rst law AU Q 7 PAV quot39 2 nut Q7PmV2 Vi SinceP R P2we may rewrite this expression as Qmu2ev27wevi mUrh From Table A10 for saturated steam at 130 C P 027 MPa 270 kPa h 2720 kJkg Solving the last equation for 111 and substituting Values h2 gm 8393 kJ 2720k J 2775 m 015 kg kg kg To nd the nal temperature L A12 at 02 MPa and 03 MPa and g L team table p Valuc num Table ives Values at the nal pressure of 027 MPa obtained by interpolation T 02 MPa 027 MPa 03 MPa 11 h h 150 2769 2763 2761 200 2870 2863 2865 Finally interpolate again to get nal temperature T 150 277 7 763 200 7150 2863 7 2763 T2 155 c 4 Answer 538 quot f 4 C is cooled in quot2kg ofliquid water is present Find the heat transferred Approach Apply the rst law Compute work from WIPdV Rewrite the first law in terms of enthalpy Property P Values are found in the steam tables 1 Assumptions 1 The process is quasistatic 2 There are no kinetic energy changes 3 There are no potential energy changes 4 The pressure is constant Solution From the first law AU Q 7 W 1 WdeV PIdVPAV Substituting into the rst law AU Q 7 PAV 39quotuz iQ P quotV2 Vi SinceP I Ppwe may rewrite this expression as Q quot 2V2 1RV1mU z hr From Table A12 1391l 32671kJkg To nd the enthalpy at state 2 we need the quality By de nition m mim 5 1 0833 12 m m The enthalpy at state 2 is therefore using Values from Table A11 at the nal pressure of 800 kPa kJ h2 11 xhg 71172111083327691772111 2428E Substituting into the rst law Q 12kg2428732671 J710071kJ Answer g 539 Rl34a at 720 C and 200 kPa is heated at constant pressure If the mass of refrigerant present is 62 kg and the heat added is 380 kJ determine the nal state Approach Apply the rst law Compute work from W IPd V Rewrite the first law in terms of enthalpy Property P 1 2 values are found in the steam tables Assumptions 1 The process is quasistatic V 2 There are no kinetic energy changes 3 There are no potential energy changes 4 The pressure is constant 5 The subcooled liquid approximation applies Solution From the first law A Because the process is assumed quasistatic and the pressure is constant W39PdV PJ39dVPAV Substituting into the rst law AU Q 7 PAV mu2 7Q7Pmv2 ivl Since P P1 Pz we may rewrite this expression as Q m uz 132V2 3914r 131V1mh2 7171 From Table Al4 the saturation pressure at 720 C is 133 kPa The pressure of state 1 is higher than this therefore state 1 is a compressed liquid Using data in Table Al4 h hfam vfTP 71100 k 1000 J 3 hl 2426 00007361 2007133kPa kg l k kg Rearranging the first law 24300i 1000 Pa kg lkPa WE 133 J J k 112 2l ll 24300 85590 856 m 62kg kg kg kg From Table AlS this enthalpy lies between the value for hf and 11g at 02MPa therefore the nal state is two phase h2 hfxhg ihf The quality at the nal state is x hi 7 hf M 02394 Answer hg 7 hf 2413 73684 Since pressure is constant the pressure at the nal state is P2 200kPa 391 Answer Saturated liquid water at 70 psia is cooled at constant pressure to 80 F If the Volume of water present is 540 71 ftj nd the heat transferred Approach Apply the first law Compute work from W I PdV 39te e first law in terms of enthalpy Property P m les 2 Rewrl th Values are found in the steam b Assumptions V 1 The process is quasistatic 2 There are no kinetic energy 3 There are no potential energy c 4 The pressure is constant 5 The subcooled liquid approximation applies changes hanges Solution From the first law AU Q 7 W 1 WdeV PIdVPAV Substituting into the first law AU Q 7 PAV m 2ulQPmV2Vx SinceP 1 Ppwe may rewrite this expression as Qmlu2 Izv2ut HM WM To find mass 3 V mm 4061bm m 3 v 00175ft lbm where the specific Volume of saturated liquid at 70 psia from Table B11 has been used From the same table Btu h 2728 h Ibm From Table B1 1 the saturation temperature at 70 psia is 303quotF State 2 has a pressure of 70 psia and a temperature of 80 F therefore it is a subcooled liquid The enthalpy is h2 hT2VT212 12202 2 2 481 00161 7070507 1B 144 quot 2 2 lbm lbm In 778 ftlbf 1ft Btu 483 112 lbm From the first law Btu 79108Btu Answer Q mh2 7h 4061bm483 7 2728 lbm Con uent Heat is negative because the water is being cooled 541 A pistoncylinder assembly of initial volume 06 m3 contains H20 at 500 kPa and 280 C The system is cooled in a twostep process 12 Constant volume cooling until only saturated vapor remains 23 Constant temperature cooling until only saturated liquid remains a Sketch the process on a T V diagram b For process 1 2 calculate the work done and the heat transferred c For process 2 3 calculate the work done Approach For the rst process apply the rst law For the 1 second process recognize that a constant temperature T process in the twophase region is also a constant 3 2 pressure process Compute work from W JPdV Property values are found in the steam tables V Assumptions 1 Both processes are quasistatic 2 There are no kinetic energy changes 3 There are no potential energy changes Solution a From Table A12 the H20 eXists as superheated steam at the initial state The rst step occurs at constant volume Since mass is also constant speci c volume is constant and the process appears as a vertical line on the gure above During the cooling the temperature drops until the saturated vapor line is reached For the second process temperature is constant so the process is represented by a horizontal line on the gure The nal state is on the saturated liquid line b From the rst law AU Q 7 W In the rst process volume is constant so no work is done Qmuziul Find the mass of the system Interpolating in Table A12 3 v1 050331 kg 1 0396 ll9kg v1 05033 Since vZ V1 and v2 is saturated steam from Table AlO P 3613MPa T 140 C u ug 255011 J g Using these values in the rst law QH mu iul 11925507 2771 462k lt1 Answer c The second process is a constant temperature cooling in the twophase region Because temperature is constant pressure is also constant at the saturation pressure of 0361 MPa The work done is WJPdV PJdVPAV Pmv 7v Pmvf ivg 106Pa lMPa W 72183271 72183kl Answer 3 W 03613MPal19kg000117 05089 g 542 A twophase mixture of water and steam at 190 F is contained in a pistoncylinder assembly Initially the piston rests on stops The combined mass of the water and steam is 006 lbm and the initial quality is 03 The piston has a diameter of6 in and a mass 0 12 lbm How much heat must be added to triple the Volume Assume Pm 147 psia Sketch the process on a P v diagram Approach Split the process into two para a process in w ich the pressure rises until the piston lifts and a constant pressure process which continues until the Volume triples For each process apply the first law and use property Values from the steam tables Assumptions 1 Both processes are quasistatic 2 There are no kinetic energy changes 3 There are no potential energy changes Solution As heat is added the pressure will rise At some point the pressure will be enough to just lift the piston Then the process will continue at constant pressure until the Volume triples Let state 2 be the point at which the piston starts to lift offthe stops and state 3 be the nal state Total heat transferred will be the sum ofthe heat from 1 to 2 and from 2 to 3 or or Q Q2 In process 1 to 2 the Volume is constant and the mass is constant therefore the specific Volume is also constant Using data from Table B10 at 190 F the speci c Volume at state 39 v v x v5 7v 00155 034095 7 00155 123 ftZlbm We need the pressure at state 2 We can find this from a force balance on the piston ft mpg 121bm32172 bf I 13 m 147 151ps1a AP 1321n232171bm39ff 139 lbf s From the first law A Q H2 But since the Volume is constant for this part ofthe process WH2 0 and AU QH2 Q1 U2 41 mu2 7 The internal energy at the initial state is found using data from Table B10 at 190 F ul 11 H ug 711 158 0310717158 432Btulbm At state 2 1g 151psia and v2 v 123ft3 lbm From Table B11 at 15 psia we approximate quality using v2 v2 x2 v52 iv V2 sz N1237 00157 v 2 7v 2 253700157 The internal energy at state 2 may be approximated as x2 0457 u2 u H2 52 iufz181 046710787181600Btulbm QH2 mu7 711 0061bm6007432101Btu m TL 4 r 39 39 At state 3 the final state 1 1g 151psia and V3 3V Since mass is constant v3 3v 3123 359E From Table Bll at15psia vg 263 Since v3 gt vg state 3 is a superheated vapor For process 2 to 3 the first law is AU Q2 7W2gt3 Because the process is assumed quasistatic and the pressure is constant W23 deV Pj dVPAV Substituting into the first law AU Q23 iPAV quot 3 712 Q2 Pmv3 Vz Since P P2 P3 we may rewrite this expression as Qm 3 P3V3 2 Jrpzvz mU 15 12 hf2 x2 kg 412 181 046711517181 634Btu1bm From Table Bll at 15 psia vg 2629 Since v3 gt vg state 3 is inthe superheated vapor region To avoid double interpolation we approximate 11 by its value at 147 psia From Table B12 v 11 3631 12588 3877 12873 lnterpolating 143712588 7 36973631 12873712588 387773631 P 3 h3 1266Btu1bm 2 Substituting values 1 QH3 0061bm12667 634E 379Btu lbm V The total heat added for the entire process is Q QH2 QM 1o1 379 480Btu 4 Answer 5 43 A runner whose surface area is 18 m2 generates 650 W of body heat On a hot and cloudy day the air is at 85 F The heat transfer coefficient between runner and air is 14 Btuh F a If the runner is wearing only shorts and does not sweat what would the skin temperature be in F b What volume of sweat in uid oz must be evaporated per hour to keep the skin temperature at 70 F Assume sweat has the properties of water Approach For the first part simply use the convection rate equation For the second part set the body heat equal to the heat of vaporization plus the heat removed by convection Use the saturated steam tables to find the heat of vaporization Assumptions 1 The heat transfer coefficient is uniform over the body and independent of temperature 2 Sweat has the properties of water Solution a For convection QhATT 650 W34112lgttuhj 5 T Btu 328 ft 28 14 hft2 0F18 T5 932 F b Define Q1 as body heat and Q2 as the latent heat from sweat The body heat is removed by both the latent heat from the evaporating sweat and convection to the air so Q1Q2hATT Rearranging Btu 3412 328 ft2 7 7 h Btu 2 o QrQlihATsiT7650W W iatmjhm ijois F Btu 6284 Q h To find the latent heat use the data in Table Bll at atmospheric pressure 147 psia As the sweat evaporates it starts as saturated liquid and ends as saturated vapor Therefore the latent heat is Btu hfg hgih11505718015970E The mass of sweat evaporated per unit time is Btu 7 6284T 7 8 lbm 970 h lbm Consequently the volume of liquid sweat evaporated per unit time is 6481bm 3 7 77010457 lfloz 0104ft3 305ch 7997floz Answer 624ml 39 h 2957cm3 h 1ft 39 h 39 ft3 544 R134a with an initial quality of 073 is contained in a pistoncylinder assembly as shown The curved walls ofthe cylinder are perfectly insulated Initially the R134a occupies a Volume ofheight 20 cm and diameter 75 cm T e pistoncylinder assembly is p ace on a su ace at 10 C and eat conducts upward through the bonom wall and boils the liquid R134a The piston may be assumed to be massless and frictionless The cylinder is constructed of stainless steel AISI 304 with a wall thickness of 08 cm The incatumi m 39 quot quot 4 h 2 take for the piston to rise 5 cm the cylinder is 268 Wm K How long will it Approach Use the thermal resistance analogy to nd the rate ofheat transfer into the R134a adding conduction and convection resistances in series Apply the rst law for a constant pressure heat addition to nd the heat needed to lift the piston 5 cm With the knon rate ofheat transfer and the total heat calculate the time for the process Assumptions 1 The heat transfer coef cient is uniform over the cylinder surface and independent of temperature 2 The conduction is one dimensional 4 Thepiston is perfectly insulated 5 The sides of the cylinder areperfectly insulated 6 The process is quasistatic Solution First nd the rate of heat transfer into the R134a At 100 kPa the saturation temperature is from Table A15 T 725 4 C The area ofthe bottom ofthe cylinder is 2 2 A 4 4 000442m2 Using a thermal conductivity Value from Table A2 the resistance through the stainless steel is L 0008 R 0121W 000442m2 TL 39 39 the Rl the cylinder is R i W 1 084453 1 268 2 000442m2 W m K The total resistance is K Rm R 122 0121 0845 0955W The rate ofheat transfer into the cylinder is T 7 Tw 77 7 107 454 C Q 377W Rm Rm 0955 W Now calculate the total heat needed to raise the piston 5 cm From the rst law for a constant pressure process of a closed system see Eq 236 Q AH m 7 wherel l1 is initial enthalpy andh2 is nal enthalpy Do not confuse enthalpy with heat transfer coe icient To nd the mass ofR134a present use 5 v1 m The initial volume K i V1 AB1 000442m202m 0000884m3 Where B1 is the initial height of the piston The initial specific volume is v1v xxvg 7v With values from Table AlS v1 0000726 07301927 0000726 0140m3kg 3 m 5 000085354m 00063kg v1 014m lltg The mass remains constant Since the piston rises 5 cm the final volume is V2 AB2 000442m2025m 000110m3 The final specific volume is 3 3 V2 5 000110m 0175m m 00063kg kg The final quality is therefore still using Table A lS 3 017570000726m V V x2 2 x kg 0913 V V 01927 0000726mg The initial enthalpy is h1 h x1hg ihf l630732317163 l73kIkg The final enthalpy is h2 h x2hg ihf 163 09132317163 213lltJlltg The total heat which be added to raise the piston 5 cm is Q mh2 7 h 00063kg2137173 1000 2481 g 2 Q AI Ig 2481 659s Answer Q 377w 545 A rigid box made of aluminum with a wall thickness of 025 in contains saturated steam at a pressure of 60 psia The convective heat transfer coefficient on the interior is 19 Btu h a the exterior it is 36 BtuhfF F The ox is a cube with a side length of 1 A heater inside the box maintains the steam at a steadystate temperature The exterior air temperature is 60 F Find the power input to the heater Approach L Use the thermal resistance analogy The r steamtables can be used to findthe ihA kA 434 temperature of the saturated steam in the W box T 151 R1 R1 123 a r Assumptions quot 1 The heat transfer coefficient is uniform over the box and independent of temperature 2 The wall thickness is small compared to the side length ofthe box 3 Thermal conductivity is constant 4 Heat transfer is onedimensional Solution From Table B11 the saturation temperature corresponding to a saturation pressure of 60 psia is 7 293 F t Lullcallluw 101 u e in 39 For a cubewith side length 14 ft the surface area is A 614 14 118 2 u L 394 4 Althou quot fthe box di ers quotquot f 39 small 39 39 The resistance to convection on the interior is 1 1 h 7 R A Bt 039 0446B 1 Hfu 118112 1 hft F quot L 39 39 39 39L 39 Hum Table B2 L 025in 112fl 7 Bt 39 129x10 5 39 137 quota 118112 B h ft F The convection resistance on the outside is 1 h F R3 00235 le 36118 Btu The total resistance is h F R 12 12 12 00445129x10 5 00235 00682 Bm By the resistance analogy 2937 60 a F Q 34191002W1 Answer R 00582 Btu 546 A piston cylinder assembly contains water and steam at a quality of 07 The piston which is made of carbon steel is 15 cm thick and 6 cm in diameter Initially thepiston rests on the steam 9 cm above the bottom ofthe cylinder 39 ii wei IL 39 cylinder are wellinsulated but heat is lost o the top ofthe piston The convective heat transfer coe 39icient on the top ofthe piston is 9 Wm The convective heat transfer coefficient on the bottom ofthe piston is 62 2 K How long will it be before the piston sinks to half is initial height Assume the surroundings are at 20 C Approach Do a force balance on the piston to obtain the system pressure Use the thermal resistance analogy to determine the rate of heat transfer through the piston The steam tables can be used to find the temperature ofthe saturated steam in the box Apply the first law assuming a constant pressure process of a closed tem to determine the final state Assumptions 1 The heat transfer coefficient is uniform over the piston and independent of tempe ermal conductivity is co an W 3 Heat transfer is onedimensional quotm Solution a 39339 h t 39 mixture TL force on thepiston is F my 5 prg 134ng WhereVPis p39 gquot quot 4 quot Tquot andrpis39 39 quot The pressure within the cylinder is 2 F pl N 5 P4 mP z Izmprg Iquot p Using the density of steel from Table A2 P101x105Pa7854 15cm1m 19822 102x105Pa 0102MPa m S 0cm From Table A11 the saturation temperature which corresponds to this pressure is 7 m 100 C To find the heat removed from the steamwater mixture use the thermal resistance circuit shown in the figure above where R1 is m e piston R2 and R3 is convection on the bonom ofthe piston The total resistance may be written 1 L 1 1 1 L 1 Rm Rl R2 1E P My 7047 My AP hi 7 hz Using the thermal conductivity of carbon steel from Table A2 and given values 1 100cm 2 1 0015 m 1 K 2 2 2 964 7E39m 1m 52Wm 605Wm K 9wm K W The heat transfer rate is then 1007 20 C K Rm 083W 96 i W Now apply the first law to the steamwater mixture Since this is a constant pressure process without kinetic or potential energy the form ofthe first law to use is that given by Eq 236 which is Q mAh V m v The specific volume of the twophase mixture at the initial state state 1 with values from Table All is 3 V v x1 vg 7v 0001040716947 0001043 119 2quot Since the piston initially sits 9 cm above the bottom of the cylinder V 7 753cm2 9cm m 7 3 3 2l4gtlt10 4kg v 1 119m 100cm kg lm To evaluate Ah the final quality is needed We know that the volume is reduced by half so V V V 3 V21 and 21 or V2lvl o6m 2 m 2m 2 2 kg v2 v x2 vg ivf Solving for quality and substituting values from Table All x vziv 067000104 0354 2 vgiv 16947000104 39 The enthalpy change Ah may now be calculated as Ah h2 7 hf x2hg 41h x1hg 45 x2 7x1hg 41 kJ k Ah 03547 07267557 41746k 77813 E The total heat rem oval is then QmAh214gtlt10 4kg77813 70167 k kg 70167 k W 1k AI J Q 7083 S AI 201s 336 mi Answer Estimate the specific volume of carbon monoxide at 150 K and 10 MPa using 5 47 a the compressibility chart b the ideal gas law Approach Find the critical temperature and pressure from Table A1 Use the compressibility chart to find Z and the definition of Z to find specific volume Compare with results from the ideal gas law Assumptions none Solution a From Table Al L 133K PE 35MPa TR 113 L 133 P 7 R 2 86 3 5 R P From the compressibility chart at this reduced temperature and pressure By definition Z Ava RT 211 0583l4 j150Kj 3 VW k 0 39 6 00022 lt Answer 280 g j OMPa 10 Pa g kmol 1 MPa b From the ideal gas law 8314 k 150K viRT 7 kmolK lkJ F F 6 MP 280 kg 10MPa kmol 1 MPa m3 v 000445 4 Answer kg Comment Assuming an ideal gas under these circumstances would produce a 50 error 5 48 Steam at 800 F and 5000 psia has a mass of 25 lbm Calculate the volume using a the steam tables b the compressibility chart Approach Find the critical temperature and pressure from Table B1 Use the compressibility chart to find Z and the definition of Z to find specific volume Compare with results from the superheated steam table Assumptions none S olution a From Table B lZ 3 v 00593 ft lbm Therefore 3 Vvm 00593f 251bm148ft3 lbm b The reduced temperature is 7 T 7800460R TE 7 1165R TR108 PR 5 5000psia 214 P 3204 ps1a Where data from Table Bl has been used From the compressibility chart Z 2 03 By definition 2 RT 3 2m 0391073 is 1ftR800460R f 3 v W m0 0059 t ism j ooomm lbm 3 Vvm 0059 ft lbm 251bm 147ft34 Answer Comment an ideal gas Z l The compressibility chart is a good approximation of the superheated steam tables This is especially useful for substances for which a property table is not available The ideal gas law would have produced a large error For 5 49 Does sulfer dioxide at a pressure of 2000 psia behave like an ideal gas at these temperatures b2850 F c350 F Approach Find the critical temperature and pressure from Table B1 Use the compressibility chart to find Z For an ideal gas Z l Assumptions none Solution The reduced pressure is 7 P 7 2000psia R E ummm l 75 Where H is found in Table B1 a The reduced temperature is iTiuwm4wm 7 r 7 T c 5 From the compressibility chart Z m 098 For an ideal gas Z l Using the ideal gas modal gives only 2 error in this case b 85047460 7 2169 7751 Z w 09 The ideal gas model gives a 10 error in this case 0 3507460 7 104 775 Z03 The gas is definitely not ideal in this case giving a 70 error 5 50 A rigid container with a volume of 077 m3 contains 110 kg of gaseous propane at 208 C Using the compressibility chart estimate the pressure of the gas Approach Find the critical temperature from Table A1 Estimate the pressure using the ideal gas law lterate in the compressibility chart to determine the actual pressure Assumptions none Solution The reduced temperature is T TR 3 Using the critical temperature of propane from Table A2 7 208 273K R 7 370K To use the compressibility chart to find the pressure an iterative process is necessary Estimate the pressure with the ideal gas law one could also just select a pressure at random to start with pv E M 280 273K 1 kJ kl 7 T7m T 7 kmolK M W 077m344lk g kmol llOkg83l4 l3gtlt107Pa 13 MPa P Find the reduced pressure PR i 305 P 426MPa From the compressibility chart at TR 13 Z m 066 By definition 7 MPV 7 MPV ET mET Solve this for P to get 066110kg83l4 kJ kmolK MV 441 kg 077m3 kmol This is not close to the value of l3MPa on the last iteration So calculate a new estimate of the reduced pressure as 10001 lkJ j208273K 855gtlt106 Pa 855 MPa From the compressibility chart Z w 0 7 The new estimate for P is P 7 0711083l42082731000 441077 Another iteration will give a value of P of 89MPa Therefore it is safe to conclude that 9lMPa P m 9MPa 4 Answer Comment Tnaccuracies in reading the chart prevent us from giving a more precise value 545 n muhasan r 39 quot L f 39 conditions on heat transfer from a at plate A test section is installed into a wind tunnel The test section consists of 100 thin strip heaters placed on a 2m long at plate Each heater 20mrn long and 250mm wide is lnrated L L 39 p L quot 39 39 i electrically and thermall insulated from the adjacent heaters the backside of the plate is heavily insulated The power to each heater can be individually controlled The freestream air temperature is 25 C and has a velocity of4 ms By t hsmpy lquot t t determine for strips 1 5 25 100 and 200 a the heat transfer rate when the power is adjusted in each heater to maintain a uniform plate temperature of 50 C in W b the wall temperature on strip number 25 when the power is adjusted in each heater to maintain a uniform heat ux equal to that on strip 25 from the first part over the entire plate in C Approach Each heater is individually controlled so that the quot W LONe 2mm temperature or heat ux distribution can be set To 20 M 1nd the total power dissipated by the heaters we need to calculate the eat transfer coefficient for the two a Ie different boundary conditions Assumptions 1 Air 1 at one atmosphere bar it 2 Radiation is ignored Solution a The power from the plate can be calculated from QMR iTm I the Reynolds number Re meLy The properties are evaluated Appendix A7 at the film temperature TM T TW2 25 502 375 C p 1137 kgmz k 00270 WmK u 189 x105 Nsmz Pr 0705 1137kg rn3 4m s 2m 1Ns2 kgm Rg4817000 l89gtlt1039 N sm This is laminar ow The average heat transfer coefficient on a at plate is obtained from N k 410 00270WmK Nu05541221 21 05544s1000 20705 3410 a h rn Q554wm2 K100 002m025m5025 C692W 1 Answer b For the uniform heat ux situation we need to calculate the local heat transfer coe icient on strip number 25 in part a to calculate the local heat ux Then assuming that heat ux is uniform over the whole plate we calculate the heat transfer coe icient with the correlation for the uniform heat ux boundary condition We 39 at the renter ofthe strip 39 39 fthe whole strip 11374240020022 x 5 5 54Wm2 K Re 118 000 The local heat transfer coe icient with a uniform wall temperature is 102 00270 Nu Xx 0332Re l2 Pr 0332118000 2 0705 102 a h w k 24002 0022 So the heat ux on strip number 25 is q hTw Tm559Wm2 K5025 C140Wm2 The heat transfer coe icient for a laminar ow with a uniform heat ux boundary condition is 172 13 00270 Nu0453Re 2Pr 3045311s000 0705 139 a 1154 240020022 559Wm2K 753 Wu2 K So the surface temperature is from qquothTJ iTm Tw Tm qquotIq25 c140wm2 753Wm2 K433 C 1 Answer 12 1 12 2 Many schemes have been proposed to supply arid regions with fresh water One plan involves towing icebergs from the polar regions to dry regions that need fresh water Consider an iceberg lOOOm long and 500m wide that is towed through 10 C water at a velocity of lkmhr The density of ice is 917 kgm3 and the heat of fusion is 3334 kJkg Determine a the average rate at which the at bottom of the iceberg will melt in mmhr b how much ice will melt if the voyage is 1500 km long in kg Approach We define a control volume to encompass the 11 WLA iceberg Heat transfer from the sea water melts the 130 0 DC ice An energy balance will equate the heat transfer A with the change in the iceberg s internal energy wngom The heat transfer rate is calculated once a heat 5 39 transfer coefficient is estimated H I K lt L 000M gt l 62 5 v l 9 6 Assumptlons LV 6 av le so 1 The system is closed with no work or potential or kinetic energy effects 2 The bottom of the iceberg acts as a at plate Solution a The average melting rate dHdl is determined from an energy balance on a closed system that is defined as the iceberg Assuming the bottom of the iceberg acts as a at plate there is no work no mass flow across the control volume boundary and negligible potential and kinetic energy changes Conservation of energy gives us d d d Q7EmuSEpLWHuS dH hLLWTwircehL127752 dl 7 pus LW 7 pusgLW pus The average heat transfer coefficient requires the Reynolds number so for water from Appendix A6 at the T lm Tw T2 10 02 5 C 5 9999 kgm3 k 0578 WmK u 150 x 10394 Nsmz Pr 1091 Re 7 prLi9999kgm3 lkmhr1000mlkmlhr3600s1000m71 85X108 L7 u 7 150x10 Nsm2kgmiNs2 7 39 This is highly turbulent so ignoring the very short laminar contribution the average heat transfer coefficient is NuL 0037Re28Pr13 0037185X10808 1091 337 000 7 NuL k7 337 0000578wm1lt L lOOOm dH l95Wm2K1070 ClIs1W dt 9999kgm33334kJkgIOOOIlkl 585gtlt10396 585gtlt10393 5211 Answer s s hr b The total ice melted for the voyage dH dH L A LW I LW V mp ml p objva 9999kg3 1000m5oom 585x10396 W m s lkmhr lhr3600s l58X10wkg 4 Answer hL 195Wm2K 12 3 Two brothers have rooms side by side in a atroofed mobile home The older brother continually complains that his room is colder than that of the younger brother The older brother decides to add more heating to his room As shown below the first room is 4 m long and the second one is 3m long each is 4 m deep into the plane of the page The roof thickness is 025 m with a thermal conductivity of 12 WmK The outside wind is parallel out of the plane of the page to the roof at a velocity of 20 kmhr at 10 C the inside temperature is to be maintained at 21 C and the inside heat transfer coefficient is 75 WmZK Determine the heat loss from the roof of each of the two rooms in VJ APPI Oach The basic heat transfer rate equation ATRtat can 0 be used to calculate the heat transfer rate The average 46 quot TgtIO C39 heat transfer coefficient for each of the two rooms 9 must be determined Assumptions 1 The heat transfer is one dimensional H LA439WI LB 2 Air is at one atmosphere 3 The roof can be treated as a at plate LI39N 7 YWYM39LK Solution T 7T The heat transfer can be calculated with Q AT 1 mt 1 Run h 4 M h A We need the average heat transfer coefficients for the two rooms which requires the Reynolds number with air properties from Appendix A7 evaluated at T lm 10 212 55 C 2785 K p 1267 kgm3 k 00245 K u169 x10395 Nsmz Pr 0714 7 p1171267kgm3556ms4mNstkgm u 169gtlt10395Nsm2 Similarly ReL22918gtlt106 This is turbulent flow and ignoring the minor laminar contribution for Room A NuL 0037Re 8 Pr 00371667x10608 0714 3140 Nuk 314000245WmK TT For Room B we need to evaluate the average h from L1 4m to L2 7 m Reg 1667x106 hA 192Wm2K hBJ hxd4mIthd if 00296Re28Pr dx 2 2 11 hg Re j iReZB 1PM WU2918XIO SV71667gtlt106080714m145Wm2K Q 1 Piggy 1 1260W t Answer 75Wm2K4m4m 12wmK4m4m 192Wm2K4m4m Q3 1 216210 1 06W 1 Answer WWW 12 4 Rolling mills are used to reduce the thickness of steel plates to create thin steel strips The metal must be at a high temperature so that the power force required to reduce the metal thickness is not excessive and so that the desired material properties are obtained Consider a 304 stainless steel strip 3mm thick leaving a rolling mill at 1000 C at a speed of 20 ms A length of 50 m is exposed to air at 35 C Convective heat transfer occurs on both the top and bottom surfaces of the strip Ignoring radiation and axial conduction in the steel estimate the temperature of the strip when it reaches 50 m from the roller in C Approach T Iagooc We will use conservation of energy on the strip quotquot similar to what was done in Section 125 for a heat 4 3 mm y10m s exchanger with a constant temperature uid Note that this flow of 304 stainless steel is analogous to the flow of a uid When we analyze this metal working process we will obtain the same solution as we obtained for the constant temperature heat exchanger CF 3Y6C Assumptions 1 The heat transfer coefficient is constant 2 Radiation is ignored 3 Air is at one atmosphere 4 The flow can be approximated as flow over a at plate Solution The equation for the outlet temperature is Tm T In 7 T exp 7 hArhcp We have assumed that h is uniform over the length of the metal strip and that radiation can be ignored We assume that flow over a flat plate is a reasonable model of the flow The average heat transfer coefficient requires the Reynolds number with air properties from Appendix A7 evaluated at approximately T lm 1000 352 5175 c 7905 K so p 04465 kgmg u 360 x 10395 Nsmz k 00573 WmK Pr 0688 p L 04465 kgm320ms50m T u T 360x10395Nsm2 This is turbulent flow so ignoring the minor laminar contribution Nu 0037Re 8 Pr 0037124x1070 8 0688 15 450 Re l24gtlt107 15450 00573W K hN k m 177Wm2K L 50m From Appendix A2 for 304 SS at 1200 K cp 640 lkg K p 7900 kgm3 177Wm2K 50m w Tm 35 c 100035Kexp 3 997 c 4 Answer 7900kgm 0003mW20ms6401kgk Comments No radiation is not a good assumption because of the high metal temperature The cooling rate with air is very low Water is more typically used 12 5 The failure rate of computer chips increases with increasing operating temperature Consider a 15mm by l5mm chip that is cooled on its top surface by a 5 ms flow of 25 C air Any heat transfer from its bottom surface to the circuit board is ignored Because of the chip construction the electrical power dissipated in the chip results in a uniform heat flux over the surface of the chip The maximum temperature that any part of the chip can experience is 80 C Determine a the maximum allowable chip power in VJ b the maximum allowable chip power if this chip is the fifth in a column of identical chips all mounted flush to the surface with no space between the chips in W Approach With uniform heat flux on each chip the maximum 1 2 SOC W 2 010 M temperature will occur at the down steam end The TI SMIS L O DIS IR local heat transfer coefficient should be used in the rate equation hA AT to determine the maximum 9 permissible power Assumptions quotI lt 809C IN Sula l EJ 1 Air is at one atmosphere 2 The flow can be treated as flow over a at plate Solution The maximum allowable power to keep all parts of the chip less than 80 C can be calculated with QhXAT5 in where h is the M heat transfer coefficient We assume flow over a at plate with a uniform heat flux is a reasonable model of the flow The Reynolds number must be evaluated at T lm 25 802 525 c 3255 K So from Appendix A7 p 1084 kgm3 k 00281 WmK 1 196 x 10395 Nsmz Pr 0703 ML 1084kgm35ms0015mNszkgm a R 75 2 4150 l96gtltlO Nsm This is laminar flow so Nu 0453Re2 Pr1304534150 20703 259 h Nuk25900281WmK X x 0015m Q486Wm2K0015m0015m80725K 060w 4 Answer b For the fifth chip L 50015 m 0075 m e 7 20750 486Wm2K Nu 580 58000281 0 075 h 217Wm2K x Q 217 0015 0015 80 7 25 027 w 1 Answer Comments Because the heat transfer coefficient decreases with length there is a dramatic decrease in allowable power if the chip is in the fifth place rather than the first This illustrates well why you should be aware of whether a M or an average heat transfer coefficient is needed in a problem 12 6 The walls of a house are constructed of an exterior sheathing insulation framing timber and drywall and their composite resistance is estimated to be 415 m KW A winter wind blows parallel to the 3m high 18 m long wall The wind velocity is 30 kmhr and its temperature is 5 C The heat transfer coefficient at the interior of the house is 5 WmZK For an inside air temperature of 21 C determine the heat transfer rate through the wall in VJ Approach The basic heat transfer rate equation ATRm is used to calculate the heat transfer rate The total thermal resistance is composed of the inside convective resistance the wall resistance and the external convective resistance which m t b determined from appropriate heat transfer coefficient correlations Assumptions 1 The heat transfer is one dimensional 2 The external flow can be treated as flow over a at 3 The air is at one atmosphere Solution AT Tm 7 T The heat transfer rate equation is Q tat Rcaann Rwa11 Rcanvaut 1 1 where R r m 7 000370KW 39 hA 5Wm213m18m Rquot 415m2KVJ R wall 7 111 00769KW A 3m18m 1 Rcanv out 39 ha A We have forced convection over an assumed flat plate and need the Reynolds number For T lm we assume the temperature of the outside surface is T 0 C T lm 052 25 C 2705 K and the air properties from Appendix A7 by interpolation are p 1305 kgm3 u 163 X 10395 Nsmz k 00238 WmK Pr 0716 1305k m3 833ms 18m Ns2 k m Ref311 g g 120gtlt107 u 163x10 5Nsm2 This is turbulent flow Ignoring the minor laminar contribution and using the correlation for the average heat transfer coefficient over the total length NuL 0037Re28Pr13 0037120gtlt1070 80716 15300 00238W K hN k m 202Wm2K L 18m 15300 Rm 4 000092KW 39 202Wm2K3m18m Q 21775K 319w4 Answer 00037000769000092KW 12 7 One wall of an older office building 6m high and 30m long is all glass 7mm thick Wind blows parallel to it at 20 kmhr and 5 C The inside surface temperature of the glass is 20 C Determine a the heat transfer rate from the glass in b the heat transfer rate if the wind velocity is tripled in VJ Approach The basic heat transfer rate equation ATRtat coefficient over the glass Assumptions 1 The heat transfer is one dimensional 2 Air is at one atmosphere 3 Ignore the laminar contribution to the heat transfer coefficient Solution a The heat transfer rate equation is QiAT TwiT 7 TVin T Rm 7 REM Rm 7 IkA 1hA From appendix A for plate glass k 14 WmK The heat transfer coefficient requires the Reynolds number with air properties from Appendix A7 evaluated at the film temperature Assuming TS 310 C T lm 10 52 75 C 2805K so that p 1258 kgm3 u 170 X 10395 Nsmz k 002465 WmK Pr 0714 1258k 3 556 30 Nzk pL gm X m Sgm123x107 170gtlt10395Nsm2 R This is turbulent flow and ignoring the minor laminar contribution the average heat transfer coefficient for flow over a flat plate is NuL 0037 ReESPrm 0037 123 x107 80714 3 15600 h 7 Nuk 7 15600002465 WmK T ml 30m m2K 7 20 7 5K 7 Q 0007m 1 32500 W 4 Answer 14WmK6m30m 128Wm2K6m30m b If V2 3V1 ReL23ReL1 Nu RemRem Nu 37600 712 309 WmZK Q 72300 W d Answer Comments This is a substantial energy loss The use of a double pane window would probably more than halve this energy loss 12 8 The roof of a minivan can be approximated as a at plate 2m wide and 35m long The sun beats down on the roof such that the net solar radiation absorbed is 350 Wmz If the ambient air is 32 C the car is moving at 100 kmhr and the inside surface of the roof is heavily insulated determine the steadystate temperature of the roof in C Ignore radiation emitted from the roof Approach D An energy balance on the roof will equate the T 32 C Tquot 350W v IaowM v 39 v incoming solar energy With the outgomg convective v I heat transfer The convective heat transfer coefficient a VTS 7 must be evaluated to calculate the convective heat 3 39 W 3 2M transfer 396quot L 337M 396 Assumptions 1 The control volume is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the control vo ume 4 The flow is over a at plate 5 Laminar flow can be ignored Solution We define a control volume around the roof and assume steady negligible potential and kinetic energy effects and no work Applying conservation of energy we obtain Q 0 Recognizing that this heat transfer is the net heat transfer Q5111 va 0 qgaimA T Tf 0 T5 T qgalmh The convective heat transfer coefficient requires the Reynolds number with properties evaluated at T lm 7T 2 Assuming T5 m 45 C T m 32452 m 385 C by interpolation the air properties from Appendix A7 are p 1133 kgm339 1 190x10395 Nsmz39 k 00271 WmK39 Pr 0705 p1 ll33kgm3278ms35mNoszkgom 7 u 7 190x10 5 Nosm2 This is turbulent flow Assuming flow over a at plate and ignoring the minor laminar contribution the average heat transfer coefficient is Nu 0037Re28Pr13 0037580x106 08 0705quot 8480 7 NW 7 848000271wmilt 7 L 7 35m ReL 580x106 h 657 Wm 2 K 2 T532 cm373 c 4 Answer 657Wm K Comments Our assumed surface temperature is close enough or we could iterate one time for an improved answer 12 9 Power transformers change the voltage of electricity but the devices are not 100 efficient Dissipated heat must be removed from transformers so that they do not reach a temperature that could damage them Consider a transformer that dissipates 30 W It is 10cm wide and 20cm long with 8 fins 2cm tall 2mm thick and 20cm long evenly distributed across the surface of the transform er Air at 25 C is blown parallel along the length of the fins Assume the fins have a fin efficiency of 100 Ignoring radiation and for a base temperature less than 65 C determine the minimum air velocity required in ms Approach The heat transfer rate equation MA 7 can be used to determine the required heat transfer coefficient h Once that is known the appropriate correlation can be used to calculate the velocity N SQNS Assumptions Air is at one atmosphere Radiation is ignored The fin efficiency is 100 The flow can be approximated as that over a at plate r 39 S olution The heat transfer rate equation is Q hAT5 in 2 hL AT5 4 where A LW ZNHL 02m01m 28002m02m 0084 m2 30 h 893Wm2K 0084m26525K This is the minimum h needed to ensure T lt 65 C We assume flow over a at plate is applicable to this situation and evaluate air properties from Appendix A7 at T lm 65 252 45 C p 1110 kgmz u 193 x 10395 Nsmz k 00276 WmK Pr 0704 An average heat transfer coefficient is needed We assume the ow is laminar This will need to be checked for an isothermal at plate 2 hL N hk L0664R 2P 3 a R quotL Q r 6 0664k Pr 893Wm2K02m 2 e 13 12 000 066400276WmK0704 Assuming transition to turbulence occurs at a Reynolds number of about 500000 this flow is laminar so our assumption is valid Therefore 193x10 5 Nsm2 12000 a V uReL 7104mSEAHSWH ReL u pL T1110kgm202mNs2kgm 12 10 Solarpoweredplanes have been designed to be able to stay aloft for very long times Proposed uses include meteorology and surveillance Photovoltaic PV cells are mounted on the top surface of the Wing The PV panel is 15 m Wide and 8m long The solar energy absorbed but not converted to electricity equals about 850 Wmz and the PV cell conversion efficiency decreases with increasing temperature Determine the temperature of the trailing edge of the panel when the plane flies at 110 kmhr at 5000 m Where the pressure is 54 kPa and temperature is 256 K Approach rU z I D LA The heat transfer rate equation Q hA A T can be used The heat transfer coefficient must be determined Assumptions 1 Laminar effects at the leading edge are ignored 2 Flow behaves as flow over a plate plate 3 Air is an ideal gas Solution An energy balance on the PVcells gives QM Qmv qgglm AhATPV 7 T TPV T qgalmh We assume that flow over a at plate with a uniform heat flux wall condition is appropriate Because we want the PVcell temperature at the trailing edge of the panel the M heat transfer coefficient is used The fluid properties from Appendix A7 are evaluated at T lm We assume TPV 274 K so T lm 274 2562 265 K p 0710 kgm3 u 160 x105 Nsmz k 00235 WmK Pr 0718 D 7 p170710kgm3 306msl5mNs2kgm u l60gtlt10395Nsm2 This is turbulent flow so for the local heat transfer coefficient Nux 00308Re38 Pr 00308204x1060 8 0718 3070 7 Nuk 307000235wmK 204gtlt106 h 482Wm2K L l5m 2 TPV256Km2736K 4 Answer 482Wm K 1210 12 11 For a quick solution to an overheating problem brass rods 6mm in diameter and 5cm long are attached to a surface of a power supply Air at 20 C and 5 ms is blown perpendicular to the tubes If the base temperature must not exceed 75 C how much power can be dissipated by one rod in VJ Approach This is a problem involving conduction heat transfer fly Jig 3 o39oobm in a pin fin The convective heat transfer must be 1 evaluated first and then the fin heat transfer gt T L 0 OS evaluated 7 Assumptions 1 Air is at one atmosphere 2 The fin is one dimensional 3 The flow is as over a circular tube Solution A pin fin was evaluated in Chapter 1 1 Heat transfer from it can be calculated with Q T 7T JthA tanhmL where m JhpkAX This is for an adiabatic tip To account for convection from the tip we use a corrected length L L AXp L D4 We assume the forced convection heat transfer coefficient for flow over a cylinder is applicable and is uniform over the fin The Reynolds number is determined with air properties from Appendix A7 evaluated at T lm Tb Tf2 75 202 475 c 3205 K p 1101 kgm3 1 194 x 10395 Nsmz k 00278 WmK Pr 0704 This temperature is assumed to be a reasonable average over the length of the fin D 7 pD7 1101kgm3 5ms0006mle2kgm u 194x10395Nsm2 From Table 12 1 Nu 0683 Rem Pym 1700 Nuki 19500278WmK Nu 0683 1700 66 0704m 195 a h 902Wm2K D 0006m From Appendix A2 for brass kzllO WmK 902Wm2K 7 0006m W M mmjm m llOWmK7r40006m 4 05 W W 7 2 Q75720K902m2K 0006m110 Z0006m tanhl204 334W 4 Answer lZll 12 12 If the brass rods in P 1211 are replaced by rectangular aluminum alloy fins lOcm wide in the direction of air flow 2mm thick and 5cm long determine how much power can be dissipated by one fin in VJ Approach w Zap This is a problem involving conduction heat transfer a in a rectangular fin The convective heat transfer I coefficient must be evaluated first and then the fin heat transfer evaluated x L 0 351 kt 0439th Assumptions 1 Air is at one atmosphere 7 2 The fin is one dimensional razobc Tb 7330 3 The flow is as over a at plate Solution A fin with a constant crosssectional area was evaluated in Chapter ll Heat transfer from it can be calculated w1 7T JhpkAx tanhmL where m JhpkAX This is for an adiabatic tip To account for convection from the tip we use a corrected length L L AXp L t2 Heat transfer from the two end edges is ignored We assume forced convection heat transfer over a at plate is applicable and is uniform over the fin The Reynolds number is determined with air properties from Appendix A7 evaluated at a film temperature that is reasonable over the length of the fin T lm 75 202 475 c 3205K p 1101 kgm3 u 194 x 10395 Nsmz k 00278 WmK Pr 0704 pVV 1101kgm35ms 0lmNs2kgm 7 7 l94gtlt10 5Nsm2 This is laminar flow so for the length average heat transfer coefficient NuwhTW0664Re2Pr 3 0664 28400 2 0704m 995 28 400 ReW 995 00278 W K h N k m 277Wm2K W 010m From Appendix A2 for aluminum kZ l77 WmK 277Wm2K 2010m mm 177wmK0002m010m 00500022m0638 Q75720K 277m010m1770002111010m05 tanh0638 l37W Answer 1212 1213 In an electric hair drier air at 25 C ows with a Velocity of5 ms perpendicular to a nichrome heating element The heating element is 1mm in diameter and 40cm long with a resistance of 138 Qm The 39 4 A A C 39 39 L and sag Determine a the total power dissipated in W b the electric current in the wire in A Approach This is forced convection perpendicular to a cylinder U gin EZSQC T wrm the heat transfer coe icient geometry and 5 C temperatures we can calculate the maximum power to r not exceed the temperature limit Once the power is 2 D 39 MM known Ohm s Law can be used to calculate current Assumptions 1 L 400 quot 1 Air is at one atmosphere 2 Radiation is ignored Q lsw M 3 The ow is perpendicular to a cylinder Solution quot 39 L 39 39 39 QhAT Tm T rom Appendix A7 at 7quot Tm o obtain 11 39 39 39 39 F m2 25 4302 2275 C 5005K p 07053 kgmz k 004038 WmK y 2571 x 10quot Nsmz Pr 0680 7 3va 7 07053kgm3 5 ms000m1Ns2kgm y 2671X10395Nsm2 From Table 121 Nu0193Re 8 Pr 0193132n8 0580 844 7 Nuk 7 844 00403 SWmK 1m 1 7132 h 341Wm2 K Q341Wm2K M 4 4 2 C173W lt Answer b Electrical power is Q12 R where current and R is resistance r 173W lt I jQR IISSQmXOAm 177A Answer C omments Because ofthe high wall temperature the assumption that radiation can be ignored should be checked 1213 12 14 An existing electric powerline is being examined to determine if a higher current can be used You are asked to calculate the maximum power dissipation per meter of length that is permissible by Ioulean heating such that the inside surface of the cable insulation does not exceed 77 C The copper wire in the cable is 2cm in diameter and the insulation is 0lcm thick with a thermal conductivity of 008 WmK The wind velocity perpendicular to the wire is 5 kmhr and the air temperature is 27 C Neglecting radiation determine the allowable power heat generation rate per unit length Wm 112001 Approach I C M We must take 1nto account two thermal res1stances 1 o conduction and convection to determine the heat I transfer rate from the inner surface of the insulation to m J the air The heat transfer coefficient is calculated with h39gt the appropriate correlation gt Ts Assumptions 1 The heat transfer is one dimensional 1 17 C D D 4 2f 21 cm z n 2 A1r 1s at one atmosphere Solution The heat transfer rate is calculated with A T T RTT at m Where Rm L 1nD2D1 1 lnD2D1 27rkL herzL 27rkL Combining these two equations and solving for heat transfer per unit length T Too 1 lni D2 7 D1 h lj2 erk To obtain h we need the Reynolds number Properties should be evaluated at the film temperature but we do not know it so we estimate T5 N 67 C T lm T ma 27 672 47 C 320 K From Appendix A7 for air at 320 K by interpolation p 1110 kgm3 k 00278 WmK u 194 x 10395 N smz Pr 0704 kg km lOOOm 1h ll1075 O022m e pv Dr m hr 71750 NIIQ lkm 3600s 194x10395 N sm2 From Table 121 for the heat transfer coefficient Nu 06831284 me 06831750 4 0704 3 197 7 Nuk 7 19700278 WmK 7 Y 7 0022m 77 7 27 K W 1 1n0022002 7 649E I Answer 249Wm2K7r0022m 27r008WmK Check on T QhA TSwwwm Tm T T Q 27quotc 64399Wm MD 249Wm2K7r0022m This assumed TS is close enough to the estimated temperature so we do not need to reevaluate the film temperature R h 249 Wm2K 647 c 1214 12 15 The insulation on a 15cm steam pipe deteriorates over time and is to be removed and replaced The outer surface of the steam pipe is at 110 C Air at 6 C blows perpendicular to the pipe at 40 kmhr Determine a the heat transfer rate per unit length of pipe if it is left bare in Wm b the heat transfer rate per unit length of pipe if 4cm insulation k 004 WmK is applied to the pipe in Wm Approach This is forced convection perpendicular to a circular cylinder with and without insulation The basic heat transfer rate equation can be used to find the heat transfer per unit length IIO C s LEINS 7304 Assumptions T jgoc 1 The heat transfer is one dimensional T T 4 T bf MV 2 The air is at one atmosphere I 40 Solution The basic heat transfer rate equation is QAT TSin T5in Rm Rm 11 1h7rDL 1n D2 D127rkL Q Ts T L 1h7rD 1n 132131 27rk a With no insulation D2 D1 D D1 andRW 0 Evaluation of the forced convection heat transfer coefficient requires the Reynolds number with uid properties evaluation at the film temperature T lm 61102 52 C 325 K By interpolation in Appendix A7 p 1086 kgm3 k 00281 WmK u 196 X 10395 Nsmz Pr 7 0703 MD 1086kgm3111ms015mNs2kgm 0139 Re 75 2 92 300 u 196gtlt10 Nsm From Table 12 1 Nu 7 0193 Re 618 Fr Nu 7 0027 923000 8 0703m 7 238 238 00281W K h N k m 447Wm2K D 015m 39 110 6 c g 2440K without insulation Answer L 1447wm21ltn015m m b With insulation the film temperature will be much lower and the uid properties different We should iterate by guessing a temperature on the outside of the insulation evaluating the properties and proceeding as before Let s guess TS 20 C T lm 20 7 62 7 C 280 K p 1261 kgm3 k 002465 WmK u 170 X 10395 Pr 0714 R6W189400 170 gtlt10 5 Nu 7 0027 189400 805 0714m 7 427 h 7 427002465 023 7 458 WmZK Q HO6 c 670 L 1 1n2315 39 m 458Wm2K7r023m 27004WmK We can calculate the outside wall temperature M T 7 Tf hIIDL T 7 Tf 39L TSTQ 6 67390Wm 4 C 4 Answer hirD 458Wm2K7r023m Comments We should recalculate with improved uid properties if we want a more accurate result 1215 12 16 A very long cylinder 25mm in diameter is placed in a large oven whose walls are maintained at 400 C Air at 77 C flows perpendicular to the cylinder at a velocity of 25 ms The emissivity of the cylinder is 065 Determine the steadystate temperature of the cylinder in C Approach Mb WA 5 Two heat transfer modes 7 convection and radiation 7 r Tw 4004 must be taken into account Radiation heats the cylinder and convection cools it At steady state C o by these two heat transfer rates are equal We must 39 39 calculate the heat transfer coefficient to obtain the 39 convective heat transfer Assumptions Ir 4 d l Airisatoneatmosh 11132J39M 5 Fl 77 6 ere 2 The cylinder is small relative to the larger oven Solution At steady state we apply conservation of energy to the cylinder and assume no work and no potential or kinetic energy effects to obtain Q0 7 Q 7 QmV 0 QM sci A ij 7 T5 7 goerL ij 7 T5 hAT5 7T h7rDLT5 7Tf Therefore 8039T 7T54hT5 7 T To evaluate h we need Re with fluid properties from Appendix A7 evaluated at T tm Assuming T 7227 c 7 500 K TM 7 350 5002 7 425 K so by interpolation p7 0831 kgm3 u 7 2385X10395 Nsmz k 7 00354 WmK Pr 7 0686 D 7 pD7 0831kgm325ms0025mNsZkgm u 2385x10395Nsm2 From Table 121 Nu0683Re 66 117r13706832180 66 0686 216 r2180 216 00354W K hN k m 306Wm2K D 0025m Substituting into energy equation W W 4 K4 673K 774306 773501lt Solving this equation T 7 5134K 7 2404 C 4 Answer This is close to the assumed TS so we do not need to reevaluate the uid properties 065567x108 m2 1216 1217 After the extrusion ofa long solid plastic rod it is cooled by a cross ow of30 C air at a Velocity of 20 ms The ro whose i e r is 35cm initially has a uniform temperature of200 C Theplastic s properties are 0 2300 kgmz CF 850 JkgK and k 12 WmK Determine a the time required for the surface temperature of the rod to drop to 100 C in s b the centerline temperature at the same time in C Approach This is a transient conduction problem in an infinite cylinder We 1 t t t t w culLicuL Assumptions 1 Radiation is ignored T T T 2 Air is at one atmosphere 2 lo M R 317 C 3 The oneterm approximation for transient conduction is Valid 1 L Solution a c t u i nvemedby T rt 7T TiTf 6 exp 7 rJu Arrm rgt02 where roztr 2 and C and 1 require B hrDk 1 for 12 1 T004 4 10mm 1 I I z 1 R Wi quot quotum Annenrli A7 evaluated at I39m Note that TM decreases with time so we use TM 150 302 90 c 353 K so 3 0972 kgmz 1 213 x 10quot Nsmz K 00310 WmK Pr 0595 D pw 0972kgm320ms0035mNszkgm 2 213x10395Nsm2 From Table 121 Nu0193Re 8 Prquot3019331900 900 6 8 0595 104 7 Nuk 7 1040031OWm1lt 920Wm2K00352m 7 7 0035m From Table 112 by interpolation 1 1373 C 1252 From Table 113 JD 1 05813 17111 amp 0325 13732 125220073505813 Because rgt 02 the oneterm approximation is okay 2 k 12Wmk1Js1W h 920Wm2K a Bi 1342 1L a 3 614x10397 mis a pep 2300kgm SSOJkgk 0 2 A t 7 2 1625 Answer 614X1039 m s b For the centerline temperature at the same time JD 0 1 H r 4 Ct exp H2 r 35 C 200 e 35 K 1252 exp 13732 0325 1457 C 4 Answer Comments Probably abetterway quot 39 quot quot t I 1 heat transfer coefficient and wall temperature during each interval This would be done until the target wall temperature of 100 C was obtained 4 1217 12 18 Lead shot is made by dropping molten lead p 10600 kgm3 from a drop tower Each pellet a sphere 2 mm in diameter is cooled as it passes through air at 10 C Assume the shot falls at its terminal velocity The lead must be solidified from its molten state at 327 C to a solid state before it reaches the pool of water at the bottom of the drop tower The enthalpy of fusion for lead is 245 kJkg Ignoring radiation determine the minimum required height of the drop tower in m velocity which will need to be calculated too Approach This is a transient cooling problem The lead remains I i 39 at a uniform temperature as it solidifies and we will Ct vwl use an energy balance to develop the governing equation This heat transfer coefficient is needed to T 3 27 C calculate the convective heat transfer That requires a 1 CDC F Sou v1 Assumptions c 1 l The shot falls at its terminal velocity UN AJUM 2 Air is at one atmosphere 3 Radiation is ignore and the heat transfer coefficient is constant 4 Buoyancy force is ignored Solution For a closed system defined as the lead sphere conservation of energy is used assuming no work and no potential or kinetic energy effects Qdd 2 dezde 2 QIAUmu27u1imusf 2 hAT 7T5I7mu We assume the convective heat transfer rate is constant39 heat transfer from the control volume is negative The time I required to completely solidify the lead is used with the terminal velocity to obtain the height H required H V I 5 We assume the sphere moves at the terminal velocity for the whole time so that 7Vmu5 77V pL47rR33usf 7VpLRuS 7 7 VpLDuS hAT7TST h47rR2T7T5 3hT7T576hT7T5 The terminal velocity is determined from a force balance between drag and gravity buoyancy is negligible 2 V2 3 05 degFgmv gt CD E p pL Q g gt V 4 2 6 3 p CD The drag coefficient given on Fig 108 is a function of Reynolds number so evaluating air properties from Appendix A7 at 10 c p 1247 kgm3 1 172x10395 Nsmz k 00249 WmK Pr 0712 at 327 c u 3018X10395 Nsmz The Reynolds number is Re p V Du which requires the velocity that we do not know so we iterate We guess V calculate Re obtain CD and then calculate V When the guessed V matches the calculated value then we have a solution V 10600981ms20002mHI2 1mS llze1247kgm3 V mS0002m 145 V 3 1247 CD l72gtlt10395Nsm2 Estimate V 25 ms Re 3630 CDZ 04 V 236 ms 6 V 236ms Re 3420 CDZ 04 V 236 m s The heat transfer coefficient is determined from 14 7 12 23 04 147 12 23 04 l72gtltlO395 7 Nu7204Re 006Re 1P7 uys 72043420 0063420 10712 73018X1075 7299 hiNuki 29900249WmK 372Wm3K D 0002m 1218 7 7236ms10600kgm30002m24500 Jkg 3461 1 1 Answer 3372Wm2K107327KIJsIW 1219 12 19 False temperature readings can be obtained from temperature sensors if they are used incorrectly or if the effects of radiation are not taken into account For example if a thermometer were used in direct sunlight or in shade to measure air temperature significantly different readings would be obtained Consider a thermocouple that is a l5mm sphere which is used to measure the temperature of an air stream in a large duct The air velocity is 4 ms The walls of the duct are at 150 C The thermocouple indicates an air temperature of 300 C and has an emissivity of 05 Determine the actual air temperature in C Approach Two heat transfer modesconvection and radiation E 513 C must be taken into account Radiation cools the TN quot thermocouple and convection heats it At steady 4M 5 state the two heat transfer rates are equal We must U I D t 000 3 M calculate the heat transfer coefficient before we can a 9 g obtain the convective heat transfer 5 9 t 0 Assumptlons 1 5 300 C 1 Air is at one atmosphere 2 The thermocouple is small relative to the duct size Solution At steady state we apply conservation of energy to the thermocouple and assume no work and no potential or kinetic energy effects to obtain Q0 gt QmV 7 Q 0 With Qmv hAT 7T5 and Q39md gaA T54 7T4 so that h Tfi TS 817014 7 TWA Solving for the uid temperature TTntva To evaluate h for a sphere we use Eq 1211 which uses air properties from Appendix A7 evaluated at the bulk uid temperature We assume Tbulk 300 C so that p 0616 kgm3 u 2926 X 10395 Nsmz k 00450 WmK Pr 0680 We assume the thermocouple is a sphere so that D 7 pD 706l6kgm3 4ms00015mNs2kgm T u T 2926x10395Nsm2 For a sphere NuD 204Re22 006Reg31Pr04 ufys Assume z 320 C y 2994 x 10395 Nsmz Nu204126m 006126230680042926x103952994x10395W711 126 14 7 Nuk771100450WmK h 213Wm2K D 00015m Therefore 05 567x10398W m2K T 573KMBSBKY7423K4583K310 C Answer 213Wm K Comments For a better estimate the uid properties should be reevaluated at this new uid temperature and the calculations repeated 1220 1220 39 39 39 with L 39 In such a device the temperature ofa small diameter cylinder is maintained constant by varying the electric current through it in response to varying uid velocity a wheatstone bridge is used to control the current A typical hot wire is constructed of a 02mm diameter polished platinum wire 10mm long Air at 23 C ows over the hot wire maintained at 200 C The electrical resistivity ofplatinum is 17 uQcm Determine the electric current required for a velocity of a 1 and ms b 10 ms Approach Ujlmls 0L Ions This is forced convection perpendicular to a cylinder With the heat transfer coe icient geometry and hr 7 30C 3 temperatures we can calculate the power to reach the X 5 operating temperature Once the power it known 3 Ohm s Law can be used to calculate the current DO0001M J Assumptions 67 Hits 17 rm 1 Radiation is ignored HES ldouc 2 Air is at one atmosphere Solution urewireis QhAT7 whereA IrDL From Ohm s Law QIZR where is the current andR is the electrical resistance R p5 LAX whereA Ir Ad39h quot quotquot quot 39 39andsolvingforI 12 17 11731730 4 4p To obtain 11 we need the Reynolds number with air properties from Appendix A7 evaluated at 71 23 2002 1115 C by interpolation ofproperties in the appendix 3 0918 kgmz y 223 gtlt10395 Nsmz k 00327 WmK Pr 0691 pw 0918kgm11ms00002m7 7 u 7223X10395NXsm2kgmNs2 From Table 121 Nu 09mmn Spy 0911823n 3850591 1813 7Nuk 7 181300327WmK 7 296 w 7 7 m 7 m2K Re h 3 I 7 295 WrrBK12 00002m 2007 23K100cmm 417x10396 ncm b For V 10ms Re823 Nu 0683Re 697 0683823 4660591m 472 m 247A 1 Answer 472 00327 kw 771Wm2K 00002 771 7 000023 200723 100 X 1221 12 21 After a heat treating process a 2024T6 aluminum sphere 20mm in diameter is removed from an oven which is at 85 C The sphere is placed in an air stream at 27 C that has a velocity of 10 ms Determine the time required for the sphere s temperature to cool to 40 C in s Approach This is a transient conduction problem We need to U 10 W4 5 determine if lumped systems approach Will work or if a A um Nd quot7 a onedimensional transient solution is required The 3911 83 C heat transfer coefficient is determined from the gt appropriate correlation 72 400C gt Assumptions o 1 All properties are constant 1 T 27 C Solution To determine if the lumped systems approach is valid we need to calculate Bi h LCMk Lm VA r08 001m3 000333m For aluminle from Appendix A2 k 177 WmK cp 875 JkgK p 2770 kgm3 The heat transfer coefficient requires the Reynolds number The sphere heat transfer coefficient correlation uses properties at the bulk temperature so from Appendix A7 at 300K p 1774 kgm3 k 002624 WmK u 1846 x 10395 Nsmz Pr 0708 at 350 K u 2075 x 10395 Nsmz D 7 pD 7 11774kgm310ms002mNs2kgm 1846 X105 NXsm2 The heat transfer coefficient is determined from Nu 204Re12 006Re231Pr04 Wis 12800 14 2 0412800 2 00612800 070804 1846X103952075x10395 679 hi Nuk 7 679002624WmK 891Wm2K 002m 891Wm2K000333m 000168 lt 01 l77WmK so the lumped systems approach is valid TAT 711A Am J exp t exp T T T mop chhar CP Solving for time I ichhmcp In Tin 72770kgm3 000333m8751lltgK 40 27 7 h 7 7T 7 891Wm2K1Js1W 85 7 27 136 s Answer 1222 12 22 In an oil refinery a steam pipe k 15 WmK with an inside diameter of 10 cm and an outside diameter of 11cm is coveredwith 35cm of insulation k 003 WmK The steam is at 300 C Air and surrounding surfaces are at 17 C and the air flows perpendicular to the pipe at a velocity of 2 ms The heat transfer coefficient of the steam is 100 WmZK Determine the heat transfer rate per unit length of pipe in Wm Approach 1 15003 wk The heat transfer rate equation ATRtat can be lt 003 used directly The heattransfer coefficient on the Stem 39 outs1de must be determ1ned D Dz 011M WOW u Y Assum tions 11L 1 Thepair is at one atmosphere 1 d T T 1 T U ZZmlj T ll 2 The heat transfer is one dimensional Solution The governing equation is Q7 AT 7 Rm 1 1nD2D11nD3Dz hmA 271k L 27rkm5L haAa For flow perpendicular to the pipe the Reynolds number with air properties from Appendix A7 at T lm is required to calculate h Assuming TSme E 27 C T lm Z 17 272 22 C 295K p 1197 kgmz k 00258 WmK u 181 x105 Nsmz Pr 0709 pD 1197kgm32ms01120035m 23 800 7 u 7 181x10395Nsm2mszng 7 From Table 121 Nu 0193128618137m 019323800 5180709 3 872 1 h7Nuk 87200258WmKL125th D 018m 1 7 1 7000318 h A 100Wm2K 110m L L W 1nD2D57 1n011010 7000101 27rk L 2 1SWmKL L 5221 1nD3D2 1n018011 2613 27rkm5L 27r003WmKL7 L 1 7 1 7 01415 ha AG 7 1257r018L 7 L 39 300717 K Qi 103Wm Answer L 00031800010126l301415KmN Comments The insulation dominants the total thermal resistance Inaccuracies in the other resistances would have little effect 1223 12 23 If you have ever changed a hot incandescent light bulb then you know that much of the power going into the bulb is converted to heat about 90 rather than to light about 10 Fluorescent light bulbs are much more efficient All the heat is dissipated from the glass bulb Consider an 8cm diameter 100W light bulb cooled by air at 30 C Both convection and radiation 3 085 cool the glass Assuming the surroundings are at 30 C for radiation purposes determine the temperature of the glass bulb if air at 2 ms flows across it in C Approach An energy balance on the light bulbs is needed The power into the glass 90 of 100 VJ is removed by convection and radiation The heat transfer coefficient must be evaluated Assumptions 1 Air is at one atmosphere 2 Radiation is from a small body to a large area 3 The light bulb is a sphere Solution A steady energy balance on the glass gives Qm Qmd Qmv Q39m go AT 7ThAT5 4 The light bulb is assumed to be a sphere and the surroundings are very large The heat transfer coefficient correlation for a sphere is based on air properties from Appendix A7 evaluated at the bulk temperature 300K p 11774 kgmz 1 1846 x 10395 Nsmz k 002624WmK Pr 0708 assume T 450K y 2484 x 10395 Nsmz RC MD ll774kgm32ms008m u 1846x10395Nsm2kgmNs2 The heat transfer coefficient on a sphere is determined from Nu 204Re12 006Re23Pr04 fl15TH 10200 1 20410200m00610200mO708041846x103952484x10395 4 585 7Nuki 585002624WmK D 008m 09100w085567x10 8wmZK 47r004m2 7147300K4 226Wm2K47r004m2717300K 90969x103910714781x10903867 7300 Solving the equation for the surface temperature T 4503K 1773 C Answer h l92Wm2K 1224 1224 After preheating long 316 stainless steel rods 50mm in diameter to a uniform temperature of 1000 C they L AuLaLi in theplauuur 39 39 39 e conveyor moves the rods perpendicular to the direction oftravel at a velocity of3 ms The rod emissivity is 05 and th 39 and quot are at 27 C quot L quot r Luau w p for the next processing step Determine onvective heat transfer coefficient at the start ofthe travel in WmZK b the radiation heat transfer coefficient at the start ofthe travel between one processing station and another in WmzK L 39 uau it uctwccu L quot heat transfer coefficient radiation and convection is the sum of those calculated in para a and b in s Approach The convective heat transfer coefficient for forced 12039 2 0C convection perpendicular to a cylinder and the o radiative heat transfer coe icient both can be 1 500 C evaluated with the given information Because the diameter is small relative to the cylinder length we will analyze it as an in mite linder The Biot number should be checked with the total heat E transfer coe icient to determine if lumped systems approach is applicable Ifnot the one rm approximation for onedimensional transient conduction in an infinite cylinder will be used of M39t 1mm quotrpm Assumptions Air is at one atmosphere Solution a To obtain L 39 cueincieut 39 r r 39 at 71 I T2 Because we want 11 at the start of travel T 1000 C 7 787 K From Appendix A7 by interpolation p 0449 kgm y 359 x 10quot Nsmz k 0 0571 WmK Pr 0688 pw 0449kgm33ms0050m Re QSSO y 359x1039 Nsm From Table 121 Nu 0583R20 55Pr 0683 1880 4 0588m 202 kg Nuk 20200571WmK m 231Wm2K b Assuming the surroundings are very long and are at T the radiative heat transfer coefficient is defined as a ma was 22 With T 1000 273 1273 K and T 27 273 300 K 05557x108 W 1273K300K1273K2300K2763Wm3K mzK4 11 L c The Biot number with a combined coefficient is Bi m where LCM VA D4 From Appendix A2 for 316 stainless steel at 1000 K p 8238 kgmz k 242 WmK c 602 Jkg K 231763Wm2K005m4 Bl 0051lt01 242WmK The lumped systems approach is valid and the equation governing it is T T M ht 7 L wc TiT exp 7 3 exp 7 a kbln ref ch pLc h TrT 575 S AIISWBIquot 78238kgm3005 m4602JkgK1 90027 1 231 763Wm2K quot 100027 1225 12 25 Ventilation ducts are often uninsulated when they run through attics and other uninhabited spaces Consider an air ow at 70 C and 15 m3min which enters a 20m long 30cm square duct The duct runs through a space that is at 10 C Ignoring the temperature drop across the metal duct determine a the outlet temperature of the air in C b the heat transfer rate from the hot air in VJ A roach This is a heat exchanger a constant wall temperature T i 20 0C gt W 039 3nquot so we can use the equation given in Section 125 g amp The appropriate heat transfer coefficient correlation i 0 30A is used once the Reynolds number is determined Assumptions L 39 20quot 1 Air is at one atmosphere vL 2 The duct wall temperature is the same as the air v IS Mquot mud m temperature outside of the duct 3 The system is steady with negligible potential and kinetic energy effects no work and air an ideal gas with constant specific heat Solution a An equation for a constant wall temperature heat exchanger with only one convective thermal resistance is Tm T Tm 7T exp ihAr39ncp The mass flow rate is evaluated with the air density at the inlet p 1029 kgm3 3 mpV39 1029kgm315 m min 1min 60 s For the heat transfer coefficient we evaluate the air properties from Appendix A7 at the average air temperature Because we do not know the exit temperature assume TM 60 C TaVg 70 602 2 65 C so p 1044 kgm3 u 202 x 10395 Nsmz k 00291 WmK Pr 0700 CF 1009 kJkgK This is a square duct so the hydraulic diameter is needed 4AX 4030m030m 0257kgs D 030 H Fwd 4030m m R pDH 1044kgm315m3min03Jm1min60s41100 202x10 5Nsm2030m kgmNs2 This is turbulent flow so using the DittusBoelter equation with an exponent of 03 on Pr since the air is being cooled Nu 0023Re0 8Pr03 00234310008 0700 105 hiNuK 710500291Wm1lt 7 DH 7 030m 102Wm2K 102Wm2K4030m20m T 0257kgs1009kkgK1000Jkl 20 c7020 c exp 394 c Answer b The heat transfer rate can be determined from conservation of energy Assuming steady negligible potential and kinetic energy effects no work and air an ideal gas with constant specific heat kill 009 70394K793kw Answer 5 g out Q ch Tm 7T 0257 Comments A better estimate can be obtained by reevaluating the air properties at the average temperature 1226 12 26 Because of a shallow ocean floor and deep draft an oil tanker must use an offshore oil depot to unload The depot is connected to a shore installation by a llOOm long 45cm pipe In the winter the ocean water temperature is 5 C The oil properties equivalent to unused engine oil initially at a temperature of 20 C is pumped from the tanker at a flow rate of 008 m3min Ignoring the water and pipe thermal resistances determine a the outlet temperature of the oil in C b the heat transfer rate in VJ Approach This is a heat exchanger with a constant wall temperature OT IN 7 29 C V 008M1 mlrquot so we can use the analysis given in Section 125 The I appropriate heat transfer coefficient correlation is used 1 a once the Reynolds number is determined 7 e 1 lo M a Assumptions o l The water and pipe thermal resistances are ignored TEE 5 C 2 The system is steady with no work negligible potential and kinetic energy effects and oil is an ideal liquid Solution An equation for a constant wall temperature heat exchanger with one convective thermal resistance is 42A T T T 7T ex out m pmcp The heat transfer coefficient requires the Reynolds number with properties evaluated at the average oil temperature Because we do not know TM we assume TM N 15 C so TaVg 15 202 l75 C For unused engine oil from Appendix A6 p 890 kgm3 1 9990x10394 Nsmz k 0145 WmK Pr 12900 cp 1868 kJkgK Re pw but u m m 2 Re p7r 4D 7ruD 4890kgm3008m3minlmin6OS 336 7r9990gtlt10394Nsm2045mms2ng 7 39 This is laminar flow so check the entrance length Lem 0037 Re PrD 0037 336 12900 045 m 722 In More than half of the llOOm pipe is in the entrance region so we take the entrance length into account noting that Gz RePrDL 1 Nu3 00668Gz 7 00668336129000451100 1004Gz 39 100433612900045110023 l48Wm2K 459 hiNuki 4590145Wm1lt T D 7 045m o l48Wm2K7r045ml100m Tm 5 C 205 C exp 890kgm3 008m3min1868kJkgK 1min60s1000 JkJ 103 C Answer b For the heat transfer rate using the energy equation Q mop Tau Tm chP Tam Tm 890kgm3008m3minlmin6OS1868kJkgK10320K 21500 w 4 Answer 1227 12 27 A steam condenser downstream of a turbine in a Rankine cycle power plant has 5000 tubes each with an internal diameter of 075in The steam condenses at 120 F on the outside of the tubes The total cooling water flow rate is 3500 lbms enters the tubes at 54 F and leaves at 85 F Because the condensation heat transfer coefficient is very high ignore the steam and tube wall thermal resistances Determine a the heat transfer rate in Btuhr b the tube length required in ft Approach o F With the flow rating and inlet and outlet temperatures i 20 N Svoohb j given for the cooling water the heat transfer rate can be calculated with conservation of energy Because NAN 39 this is a constant uid temperature heat exchanger the 9 DA 3 O ITIN 9 analysis given in Section 125 is used to calculate the tube length This requires the heat transfer coefficient 174 1 51quot F Tm gde 1030 3500 mes Assumptions The system is steady with no work and negligible potential and kinetic energy effects 2 The liquid water is an ideal liquid with constant properties Solution a Applying the conservation of energy to the water and assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat Q WP Tm Tm For water at an average temperature of 54 852 695 F cp l00 Btulme Q 35001bmsl00Btulme8554R3600 shr3906X108 Btuhr b For a heat exchanger with one constant temperature uid and Rm lhA we ignore the condensation and wall thermal resistances T M Tm Tm 7 T f exp hArh cp where A NIIDL and m mm Solving for L L 7ch In Tm 7T NIrDh Tm A T We need h which requires the Reynolds number for ow in a single tube with the cooling water properties from Appendix B6 evaluated at TaVg 695 F u 658 X 10395 lbmfts k 0347 Btuhrft F Pr 682 Re p D but V u Re4m NilJD Np7r4D2 435001bms 50007r658x10395 lbmfts00625 ft This is turbulent flow so using the DittusBoelter correlation Nu 0023Re08PrD 4 002326700 8682 4 146 7 Nuk 7 1460347Btuhrft0F D 00625ft 7735001bmsl00Btulbm F3600slhr1 85120 50007r00625 ft810Btuhrft2 L547120 2l700 h 810Btuhrft0F 100ft4 Answer l228 12 28 An air compressor used in a large car body shop is located in an inside equipment room Fresh air at 5 C 96 kPa is conveyed to the compressor from outside through a 30cm circular duct that is l5m long The duct runs along the ceiling of the facility where the temperature is 34 C If the air flow rate is 035 m3s determine a the temperature of the air when it reaches the compressor in C b the heat transfer rate in VJ Approach 3 10 This is a heat exchanger with a constant wall 7 39 C temperature We can use the analysis given in Section 125 with Rm lhA to calculate the outlet temperature To do so we need to determine the A m 39 convective heat transfer coefficient Once TOUT is a determined the heat transfer rate is calculated with gt conservation of energy TN Assumptions 7 5 L 1 04 9 l 1 Air is at one atmosphere 3 2 The system is steady with no work and negligible V OI 35 M 5 potential and kinetic energy effects 3 Air is an ideal gas with constant specific heat Solution a An equation for a constant wall temperature heat exchanger with Rm lhA is Tom Tf Tm 7 Tf exp hArh CF The mass flow rate is in pV Assuming air is an ideal gas FM 96kNm22897kgkmol1kIkNm 7 ET 8314kJkmolK5273K rhl20kgm3 035m3s042kgs The heat transfer coefficient requires the Reynolds number with properties from Appendix A7evaluated at TWg Assume Tam Z10 C Tng S 75 C Z 2805 K u l70X10395 Nsmz k 00246 WmK Pr 0714 CF 1005 kJkgK RepD but V u l20kgm3 Re 4m 7ruD p7r 4D2 Re 403942kgS 105000 7rl 70x10 5Nsm2 030m This is turbulent flow so using the DittusBoelter equation Nu 0023Re0 Pr 4 0023105 0000 8 0714 209 7 Nuk 7 20900246 WmK D 030m h l7lWm2K l7lWm2K7r030ml5mlle 042kgs1005kJkgK10001kl b The heat transfer rate is determined from the energy equation Assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cp AT Qr39ncp71m77 042kgsl005kkgKl7675K1000Wslkl 5320 W Answer T 176 C 4 Answer out 34 C534K exp 1229 12 29 A proposed cooling technique for high power computer chips is to machine microchannels into the backside of the silicon chip Because Nusselt number is defined as Nu hkLchm for a given value of Nu as the characteristic length decreases the heat transfer coefficient increases Consider a l5cm by l5cm computer chip that dissipates 50 W Water at 20 C is used as the coolant and its outlet temperature is limited to 25 C Heat transfer is primarily from the base of the channel that is ignore heat transfer to the water from the channel sides Assume the distance between the sides of two channels is 0lmm Determine a the number of 025mm deep and 025mm wide microchannels that are on a chip and the average surface temperature of the base of the microchannels in C b the number of lmm deep and lmm wide microchannels that are on a chip and the average surface temperature of the base of the microchannels in C Approach Wok WW 39 52gtw If we assume the chip is at a 4 5949quot L gtD o Mmlm Q uniform temperature T S we can 039 58quot use the analysis given in Section 125 for a heat exchanger with a v constant temperature wall We H 39 39 w 1 O39OIYM need to determine the water ow 39 i L WM 9 rate and the heat transfer M orolfm l coefficient The water flow rate can be determined from conservation of energy tM Nb39 born 6 0 0 Assumptions 1 The system is steady with no work and negligible potential and kinetic energy effects 2 Water is an ideal liquid with constant specific heat S olution a For a constant wall temperature heat exchanger with Rm lhA the water exit temperature can be calculated from TM T Tm 7 TS exp hA m CF Tm iTm exp 7 hArhcp 1 exp 7 hArhcp Therefore we must evaluate A h and rh The heat transfer area depends on the number of channels The number of channels N on a chip is based solely on geometry Solving for T5 WNLEN1D 2 NW D 0015m 00001m 425 LE D 000025m 00001m This must be an integer so N 42 Area for heat transfer is only the base of the channel A NLCL 42 000025 m 0015 m 1575 x104 m2 The water flow rate is determined from conservation of energy heat plus the heat transfer rate and temperatures Assuming steady no work no potential or kinetic energy effects and an ideal liquid with constant specific heat so that Ah CF AT QrhcpTirT i m L m 0p Tm Tm For evaluating CF and eventually the heat transfer coefficient we need water properties from Appendix A6 at TaVg 20 252 225 C so by interpolation p 9976 kgm3 cp 4180 kJkgK k 0607 WmK u 929 X 10394 Nsmz Pr 640 50WlWs 4180kJkgK1000JkJ2520K 000239kgs 1230 4 000025 0000025 pVDH where DH MK 4HL mx m000025m w PW 2HL 2000025m000025m The Reynolds number is Re 211201 The velocity is for the flow in one channel 7 rpm 7 000239kgs pAX 4299761ltgn 000025m2 R 9976kgm309lms000025m e 925gtlt10 Nsm2 This is laminar flow so check the entrance length Lem 0037 Re Pr D 0037 245 640 000025 m 00145 In Because the entrance length is about the same length as the channel we will take into account entrance conditions Assuming T N 25 C 71 872 x 10394 Nsm2 245640000025 3 929x10 0 N 186G 3 014186 557 Z fl5 015 876x10 091 ms 557 0607W K hNukMl3SOOWmzllt DH 000025m hA 135OOWm2K1575x10quot m21JWs 0213 rhcp 7000239kgs4180kJkgK1000Jkl 7 39 T W3190K460 c 4 Answer 5 lexp0213 b For a channel with LCH 0001 m N13 A 195 x104 m2 DH 0001m v 0184ms Re 199 Lem 0047 m Nu 824 h 5000 WmZK T 3467 K 737 c4 Answer Comments Increasing the size of the channels has a detrimental effect on the wall temperature Even though the heat transfer area increased the decrease in the heat transfer coefficient was larger so that the hA product decreased and it s the hA product that is most important 1231 12 30 Consider turbulent ow of a uid through a tube maintained at constant temperature The mass ow rate is 032 kgsec and the heat transfer coefficient is 250 WmZK Now the freestream velocity of the uid is doubled Assume the ow regime remains unchanged Estimate a the percent change in the pressure drop of the uid between the old and new ow rates b the percent change in the M heat ux between the uid and the walls of the channel c the percent change in the M heat transfer rate over the length of the channel if the heat transfer area is 37 m2 specific heat is 2200 JkgK inlet temperature is 25 C and wall temperature is 75 C Approach This is a comparison study of changes in performance AP h when a turbulent ow is varied inside a tube Basic expressions for AP f h and are used Assumptions 1 The system is steady with no work or potential or kinetic energy effects Solution L V 2 a Let V2 2 V1 m22m1 An express1on for pressure drop is AP f3 p7 The tube geometric and 7 f2LDpV222 igE I AR fLDpvf2 1 v1 7 0184 7 0184 R602 pD 02V02 Substituting this into the AP expression and canceling like terms 18 218348d Answer 1 uid properties remain fixed Therefore For turbulent ow in a tube f AP 1 gm Twin h2 b The change in the local heat ux 2H 11 h1Tw Tf h1 0 8 Using the DittusBoelter equation h0023 Pr04 C V 0 8 u Substituting this into the local heat ux comparison and canceling like terms 0 8 n V 1 2 20 8 174 Answer 41 V1 c For the total transfer rate we can use conservation of energy Assuming steady no work negligible potential energy and kinetic energy effects and ideal uid so that Ah cp m2 cp Tom 7 m2 Q1 m1 cp 711111117 mi We can obtain the outlet temperatures for this constant wall temperature heat transfer situation from 2 2 71mTW7TW77ZKeXp7hArhc 4 Tml75 c75725 cexp w 616oc 39 032kgs2200 JkgK nut275775725exp 591 c 39 20322200 39 m c T 7 7 Therefore M259125186 Answer Q1 m1 Cp Tami Tmi 616725 1232 1231 Glycerin is pumped through a 15 cm diameter tube that is 5 m long The inlet temperature is 32 C the required outlet temperature is 22 C and the ow rate is 100 kghr Determine the wall temperature required to obtain this outlet temperature in C Approach s This is a constant wall temperature heat exchanter so I L SM gt we can use Eq 12 3 The appropriate heat transfer coe icient correlation is used once the Reynolds gt D number is determined lc yw p 00 UL 3m 395 Assumptions 1quot 1 The heat transfer is one dimensional To in we Solution An equation for a constant wall temperature heat exchanger is Tom Tw LTw 7T 6X WAmt Solving for TW T T er exp elmmap w liexthAmcp To calculate h we need Re 39 39 L 39 4 Annendi L 32222 27 C C 2 27 kJkgK k 0286 WmK y 7990 X 10394 Nsmz Pr 6780 4m with TinK temp erature 2 Q 1m4quot39 prD 39 D 4100kghr1hr3600s 72 95 M7990X10ANsm20015m7 39 Because this is laminar ow we must check entrance length W 0 037 Re Pr D 003729567800015m111m This is much ion 1 L Using Eq 12 42 and evaluating the Viscosity at an assumed wall temperature of7 0 MW 53400x10394Nsm2 Nu185Gz 3 WWW 185Re PrDL yyw 12 A an 482955700015 7990x10 558 Rep but V 11 Re i i i 39oLooI 53400x10 hM5580286Wmk106Wm2K D 015m 5 M a a l 106Wm2K1t0015m5m l T VU100kghr1hr3600s2427kJkgK1000JkJj W 1 ex 145Wm2K1t0015m5m p 100kghr1hr3600s2427kJkgK1000JkJ 2727K03 C 4 Answer Comments Because the Viscosity changes so dramatically with temperature we will iterate one time At 273 K MW106000X104 Nsmz Nu 507 h 967 WmZK TW 2701 K 29 C 1233 12 32 In a pharmaceutical application the product is subjected to a final sterilization by heating it from 32 C to 0 C A flow of 60 cm3s is passed through a lOmm tube that is heated with a uniform heat ux produced by wrapping the tube with an electric resistance heater If product properties can be approximated by those of ethylene glycol and the tube is 25m long determine a the required power in VJ b the wall temperature at the tube exit in C Approach u M The required heat transfer rate can be obtained from w c 35 J conservation of energy and the given information V l The wall temperature at the tube outlet can be t calculatedusing the basic rate equation q hAT 0 Tm 32 C 13 ED C Assumptions 4 L39 23 M quotgt 1 The system is steady with no work or potential or kinetic energy effects 2 The liquid is ideal with constant specific heat Solution a For conservation of energy on the liquid assume steady no work no potential or kinetic energy effects and an ideal liquid with constant specific heat so that Ah cp AT Qme Tm Tm The properties required to calculate the heat transfer coefficient are evaluated from Appendix A6 at ng 32 802 56 c N 330 K p10895 kgm3 cp 2549 kIkgK 1 561x10394 Nsmz k 0260 WmK Pr 550 Q10895kgm3 60cm3 s1m 1 00cm3 2549kIkgK807 32K1000W1ks 8000w b For the wall temperature at the tube outlet q hT5 7TH 2 T5 Tm qquoth T T QAh T QIrDLh To calculate the heat transfer coefficient we need the Reynolds number p D 2 V L2 2 Re 4 u p7r4D 7ruD 4 10895kgm3 60 cmEs 1m1 00cm 3 75561 X101 Nsm20OlmkgmNs2 This is laminar flow so we will check the entrance length Lem N 0053 Re PrD 0053 1480 550 001 m 432 m This is significantly longer than the tube length so entrance effects must be taken into account We do not have a correlation that gives the heat transfer coefficient for laminar developing flow with a constant wall heat flux boundary condition So we will use the Seider and Tate correlation Assuming T N 350 Re re 1480 K 11 342x10394 Nsmz 13 014 1480 550 001 4 Nu186Gz s2186in 63937 637 0260W K hNI k 001 m 166Wm2K m T580 c1413 c 7r001m25m166Wm2K Reevaluating us at 373 K highest temperatures for which we have properties us 215 gtltlO 4 Nsm2 Nu680 h 177Wm2K T5 1375 c 4 Answer Comments The wall temperature will be lower than this since us should be smaller than with what we have available 1234 12 33 Air enters a compressor operating at steady state with a volumetric flow rate of 37 m3min at 105 kPa and 30 C and exits with a pressure of 690 kPa and temperature of 240 OC The compressor is cooled with 40 kgmin of water that circulates in a water jacket enclosing the compressor The water jacket can be approximated as 25 3cm diameter 2m long pipes The water enters at 20 C and the wall temperature of the pipes can be approximated as being constant at 135 OC Assume fully developed flow Determine a heat transfer rate from the compressor to the water in kW b the mass flowrate of air in kgsec c the power input to the compressor in kaf Approach The heat transfer rate from the compressor to the 2 s7 N water jacket can be determined from a combination of a 2v I 136 690 L an energy balance on the water and the use of the A constant wall temperature heat exchanger equation 9 403C given in Section 125 Assuming onedimensional 39PA Dyna T Tg 1 flow the air mass flow rate can be determined from its T l 11quot definition Finally an energy balance on the g J compressor is used to determine the required input 1 duw power N ZY ka J ne405 CV3 D 003m Assumptions To L 7 Z M 1 The system is steady with negligible potential and kinetic energy effects 2 Air and water are ideal fluids 3 No work is done in the heat exchanger Solution a Using control volume CVH defined in the schematic we apply conservation of energy to the water Assume steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah cp AT Q11 m c TD ATS The water outlet temperature TD is obtained with the analysis for a constant wall temperature heat exchanger with Rm IhA TD TM To 7 Tm exp hArh CF The heat transfer coefficient is calculated with the Reynolds number for flow in one tube evaluated at TaVg TC TD2 From Appendix A7 and assuming TD Z 30 c ng 20 302 25 c so 1 872 x 10394Nsm2 k 0611 WmK cp 4178 kJkgK Pr 597 pD rhN 2 a R 4m 440kgmin1min60s 1300 p7r4D 7ruDN 77872x104Nsm2oo3m25 This is laminar flow so assuming fully developed flow for constant wall temperature boundary condition Nuk73660611WmK D 7 003m 745Wm2K7r003m2m25 40kgmin 1min60s4178kJkgK1000 JkJ Q11 40kgmin1min60s4178kIkgK 3367 20K 379kw 4 Answer b For the air mass flow rate 105kN m2 2897k kmol 37m3 min 1min 60s mp4pVP V g 0745 4 Answer RT 8314kgkmolK30273K s c Applying the energy equation to C W and assuming steady negligible potential and kinetic energy effects ideal gas with constant specific heat so Ah CF AT Q1 7WmhA ihg0 WQ1 ch TA 7TB Note that Q 7Q39H gt W7Q39H rhcp TA 7TB At the average air temperature c 1013 JkgK W7379kW0745kgs1013JkgK 307 240K1kJ10007196kW Answer Re Nu366 gt h 745Wm2K TD135 C20135Kexp 336 C 1235 12 34 The condenser downstream of the turbine in a large Rankine cycle power plant is constructed of 30000 25 mm tubes The steam condenses at 50 C with a heat transfer coefficient of 9000 WmZK on the outside the tubes The cooling water enters the tube side of the condenser at 20 C at a flow rate of 17000 kgs For a 100011W net power output and a cycle thermal efficiency of 42 determine a the cooling rate required in W b the outlet temperature of the cooling water in C c the length of tubing required in m Approach itquot The Rankine cycle thermal efficiency definition is x 3 CM used to calculate the cooling rate The energy j k 39 CidOOWMl k equation applied to the cooling water is used to a T 7 SD06 calculate the outlet temperature Condensing steam is a constant uid temperature so the analysis in Section cosh mm N 1 30000 leS 125 is used to calculate the required length 17H 10 C 0 0 0182M n39A modems Assumptlons 1 The system is steady with no work or potential or kinetic energy effects WN LJ L do MW V a 41 2 Water is an ideal liquid with constant specific heat 10 Solution a The definition of Rankine cycle thermal efficiency is 77m Wm Q39m Also Wm an 7Q gt Qm Wm Qm Substituting this into the definition of 776W QM 17 WW 1 17042042 iooon1381Mw 4 Answer b The outlet cooling water temperature is calculated with conservation of energy appliedto a control volume around the cooling water Assuming steady no work no potential or kinetic energy effects and an ideal liquid with constant specific heat so Ah cp AT Q r39ncp Tm 7 7 gt Tm Tm Qirhcp Evaluating water properties from Appendix A6 at the average temperature assume TM Z 40 C TaVg Z 30 C CF 4176 kJkgK k 0618 WmK u 779 gtlt10394 Nsmz Pr 526 1381000kW lkJs lkW Tm ammwagyc 4 Answer 17000kgs4176kkgK c For a heat exchange with one constant fluid temperature Tm T In 7 T exp 7 1rhcp Rm 1 man an 1 where Rm Because no information is given about the tube material or wall thickness M 2m Iqu 1 1 1 1 1 wall res1stance must be ignored andA Ag NIIDL 4 Rm hNirDL hDN DL N7rDL 1 ha W 1 1 Tm T Solv1ng for the tube length L P 1n N IrD 1 ha Tmin The Reynolds number in one tube is Re pw V 2 4m u p7r4D N 7r uD 4 17000k s Ns2 k m I g g 37050 300007779x10 Nsm20025m Turbulent flow so using DittusBoelter equation Nu 0023Re0 8Pr0 4 0023 370500 8 52604 202 hi Nuk 72020618WmK D 4990Wm2K 0025m 17000kgs4176kJkgK10001klf 1 2 1 1 2 1n 39550 985m Answer 300007r0025m L4990wm K 9000Wm K 2050 1236 12 35 Air at a mass flow rate of 00015 lbms and an inlet temperature of 80 F enters a rectangular duct 35ft long 015in high and 060in wide A uniform heat flux of 50 Btuhrft2 is imposed on the duct surface Determine a the outlet temperature of the air in F b the highest wall temperature and its location in F and ft Approach e SD kMGR The outlet temperature can be determined from an energy balance on the air flow combined with the x J given information The highest wall temperature V V V will occur at the end of the duct where the uid L 3 319 temperature is highest and can be found by using the K rate equation q 54 hAAT LHLTDJSM Assumptions 1 Air is an ideal gas at one atmosphere 2 The system is steady with no work or potential 39 AM W 039 so iN A 6F 39 or kinetic energy effects 1quotquot 39 80 m O 39 00 I S1de Solution a For an energy balance on the air flow assume steady no work no potential or kinetic energy effects and an ideal gas with constant specific heat so that Ah CF AT QmcpT 7T a T Tm Qn39wp Tm q Amcp The fluid properties from Appendix B7 are evaluated at TWg so assuming TM Z 100 F TaVg Z 90 F p 0007251bmft3 CF 0240 Btulbm F k 00152 Btuhr ft F u 1267 X 103951bmfts Pr 072 50Btuhrft2200125005ft35ft Tm 80 F 969 F Answer 000151bms3600slhr0240Btulbm b Using the rate equation Qq AhATW 7 Tm gt TW Tm q h The Reynolds number is Re pDH u For the noncircular duct we need the hydraulic diameter D 7 4A 7 400125ft005ft H 7PM 7 200125005ft m 000151bms ft 331 pA 007251bmft300125ft005ft s R 007251bmft3331fts002ft 6 1267gtlt103951bmfts 0020ft The velocity is V 3790 8 R 71000 P Turbulent flow so using the Gnielinski correlation Nu w 1127f8 Prmil f07901nRe71643920791n3790716439200422 004228379071000072 Nu m m 130 1127004228 072 71 130 00152Btuhrft F hNuk 989 Btu DH 002ft hrft F 2 Tw969 Fm1020 F 4 Answer 989Btuhrft F 1237 12 36 Water enters at 40 F and flows at a rate of 025 ft3s inside a 20ft long annulus whose inner and outer radii at l in and 2 in respectively The inner surface is maintained at 150 F and the outer surface is heavily insulated Determine a the outlet temperature in F b the heat transfer rate in Btuhr Approach This is a constant wall temperature heat exchanger lt L 109 gt Using the analysis from section 125 we can determine the outlet temperature Once that is known conservation of energy is used to calculate the heat 2 W NitL V r 939 13 G1 13 transfer rate D if Assumptions l The system is steady with no work and negligible V DL 3 2w potential and kinetic energy effects 2 The water properties are constant Solution a For a constant wall temperature heat exchanger with Rm lhA the exit temperature can be obtained from Tom T Tm 7 TS exp hAr39n CF The heat transfer coefficient requires the Reynolds number with uid properties evaluation at the average temperature Because we do not know the outlet temperature we assume TM l20 F TaVg 40 l202 80 F and the properties from Appendix B6 are p 622 lbmft3 cp 100 Btulbm F k 0353 Btuhr ft F u 578 x 103951bmftsPr 589 ipVDH 4AX 747r4D227D12 Re where DH 7 D2 7D1 00833ft Pwetmd D2 D1 39 39 4 025ft3s v KV 153fts A 7r4D227D12 7r01672 7008332ft2 622lbmft3153 fts00833 ft Re 137 000 578X10395 lbmft s This is turbulent flow so using the DittusBoelter correlation Nu 002312881 04 0023 13700008 58904 601 601 0353Btuhrft F hN k 2550 Bil DH 00833ft hrft F T 2550Btuhrft2R7t00833ft20ft out l50 F 40 150 F ex p62211111183025ft3s1ooBtulme3600s1hr 634 F 4 Answer Because the calculated TM is significantly different than the assumed Tom the uid properties should be reevaluated and TM calculated again b The heat transfer rate can be determined from conservation of energy applied to the water Assuming steady no work no potential or kinetic energy effects and an ideal liquid with constant specific heat so that Ah cp AT Q m cp AT 3 622lb I3n 025 100 Btu 63440 F 1310X106 Btuhr Answer ft s lme lhr 1238 12 37 Unused engine oil is to be heated from 20 C to 65 C using condensing steam at 100 C The oil flows inside a 1cm diameter tube at a flow rate of 01 kgs The resistance of the condensing steam and the tube wall can be ignored Determine the length of tube required in m Approach any 24 10000 This is a constant uid temperature heat exchanger Using the analysis given in Section 12xx we can 9 0 45 I determine the required length The heat transfer 39 coefficient must be determined 0 39 gt Assumptions T 210 1 95 0c 1 The condensing and wall thermal resistances are L T 9 1 ignored Solution For a heat exchanger with one constant temperature uid and ignoring the condensing coefficient and the wall resistance thean IhA Therefore T Tst Tm 7 T5 exp 7 hArhcp where A IIDL Substituting this out into the main equation and solving for L me In T 7T L 17T IrDh The heat transfer coefficient requires the Reynolds number with oil properties from Appendix A6 evaluated at the average temperature 65 202 425 C39 by interpolation u 1970x10394 Nsmz k 0144 WmK cp 1972 kJkgK Pr 2683 pVD V 7 m 4 401kgsNs2kgm u 7 p7r4D2 Ir D 7r1970gtlt10ANsm2001m Re a R 46 This is laminar flow so check the entrance length Lm 0037RePrD00376462683001m642m We assume the tube length will be shorter than this so we use the SeiderTate correlation Nu186Gr13uys014 186RePrDL 3 uys 014 But this requires the tube length Therefore an iterative solution is required The viscosity at the wall temperature of about 100 C is approximately 180 X 10394 Nsmz Assume a tube length of 10 m 13 014 646 2683 001 A 145 0144W K Nu186 M M 145 4 hN kM209wm21 10m 180gtlt104 D 001m 701kgs1972kJkgKlnEZ SSJOOOOJkl L 39 248m 7r001m209Wm2K1Js1W This is longer than our assumed length so using this L 13 014 646 2683 001 394 107 0144 Nu1g m w 107 4 hw154wm21 248 180gtlt10 001 65100 7 01 1972 1 1000 L lt gtlt gtnlzoqoollt gt 337 m 7r001154 Additional iterations give Nu966 h139Wm3K L373m Nu9333 h134Wm2K L386m Answer 1239 12 38 In a small ship with limited space water must be heated from 10 C to 50 C with condensing steam at 100 C The water flow rate is 15 kgs Either one 4cm diameter tube two 3cm tubes or three 2cm tubes can be used in parallel Determine which configuration will yield the shortest length Approach a Whether there is 1 2 or 3 tubes the flow is evenly lawF1 00 a distributed among the tubes This is a constant fluid J temperature heat exchanger Using the analysis given V 4 2 9 SC in Section 125 and the appropriate flow in each tube 1 M which depends on the diameter the required length 8 can be determined The heat transfer coefficient for W 5 each configuration must be evaluated D 4CM Nf l n39nl k 5 Base Nz Assumptions TN ION D Z ZCM N 73 1 The condensing and wall thermal resistances are ignored Solution For a heat exchanger with one constant temperature uid and ignoring the condensing heat transfer coefficient and the wall resistance then Rm IhA Tm 7 4771 7T5 exp7hAr39ncp where A IrDL for one tube Substituting this into the equation recognizing that the flow in one tube is m N me In T 7 T L T 7 7 N7rDH The heat transfer coefficient requires the Reynolds number for flow in one tube with fluid properties from Appendix A6 evaluated at the average temperature 10 502 300C so cp 4176 kJkgK u 779 X 10394 Nsmz k7 0618 WmK Pr 7 526 pVD VmN Re 4m y 7ruD Nil1D ForD4cm andN1 4 15k s Ns2 k m R9M 61 300 1 7r004m779gtlt10ANsm2 Similarly for D 3 cm andN 2 Re 40860 D 2 cm andN 3 Re 40860 These are all turbulent flow so using the DittusBoelter correlation ForD 7 4 cm Nu 7 00231288137 7 00236130008 52604 7 302 7 Nuk 73020618WmK Re h 4 670Wm2K D 004m Similarly for D3cm Nu7218 h4500Wm21 D2cm Nu7218 h6750Wm21 ForD 4 cm 15kgs4176MltgKm g ll ij uoooykl L7 7 27m 17r004m4670Wm2K Similarly for D3cmandN2 L868m D2cmandN3 L868m The single 4 cm tube yields the shortest length Answer 1240 12 39 A thick stainless steel AIST 316 pipe with inside and outside diameters of 20 mm and 40 mm respectively is heated electrically to provide a uniform heat generation rate of 107 Wm3 This pipe is encased within a larger concentric tube with an inside diameter of 50 mm whose outer surface is heavily insulated Pressurized water flows through the annular region between the two tubes with a flow rate of 06 kgs The water inlet temperature is 20 C Determine a the required pipe length if the desired outlet temperature is 40 C in m b highest surface temperature and its location in C and m APPmach 4 1 gt1 This is an internal forced convection problem With the known heat generation rate and water flow we can determine the length using conservation of energy M L I The highest surface temperature can be determined quot9 WA I m 39 aquot 7 S from the heat transfer rate and the heat transfer 5 coefficient 39 Do 1 40wa Assumptions 391 l The heat transfer is one dimensional 03950quot 2 The system is steady with no work or potential or Em kinetic energy effects 3 Water is an ideal liquid with constant specific heat S olution a Assume a steady flow with no work negligible potential and kinetic energy effects and water is an ideal liquid with constant specific heat so that Ah cpAT Applying these to the conservation of energy and mass Q mhwhwo a WWW mcpmiTm0 7 quotWP Tm TM 4 quot39 4 D3 432 Evaluating the water specific heat at the average temperature TaVg Tm 7 Tag2 20 402 30 C From Appendix A6 cp 4176 kJkgK 7 06kgs4176kJkgK40720KlOOOJlkJ 7 107 Wm37r4 004m27002m2 11wS b The location of the highest surface temperature is at the tube outlet There the bulk water temperature is 40 C so from Appendix A6 p 9922 kgm3 k 0631 WmK 1 634 x 10394 N smz Pr 419 m q Dfinlumcparniko a 5 32m Answer q W gm 111271121 The heat flux of that location is 1 hTS TM a T5 Tm Tm W Tm h hAS h 4DDL Because this is flow in an annulus we need the hydraulic diameter for Re and Nu 4A Dh x DuiDx 50 mm 7 40 mm 10mm 001m pwetted m m 06kgs 855 9A p 4D2 D3l 9922kgm3 4005m27004m2 S Re pDh 9922kgm30855ms001m1Ns2kgm 2 13 400 634gtlt10A N sm This is turbulent flow so using the DittusBoelter equation 817 0631WmK Nu 0023Re08Pr04 0023 1340008 41904 817 a h N k 5160 Dh 001m m K 1 o7 wmE 004m27002m2 T5 40 C 40 C273 C427 C1 Answer 5160 Wm2K4004m532m 1241 12 40 Pasteurization is the sterilization of milk to ensure no diseases are transmitted with the milk Consider a flow of 15 kgs ofmilk whose temperature must be raised from 35 C to 75 C in a 2cm tube The wall temperature is 100 C 1Lilk properties are p 1030 kgm3 u 212 X 10393 Nsmz cp 3850 JkgK and k 06 WmK Determine the required tube length in m Approach With the given information the analysis in Section M t 191 IS Tl 125 for a constant wall temperature heat exchanger can be used to calculate the required length The 39 heat transfer coefficient must be evaluated gt gt TM7 c Assumptions 1 Only the milk thermal resistance is taken into TH Sidc account Solution For a constant wall temperature heat exchanger the governing equation from Section 125 is Tm TmTmitexpi mCP where A IrDL Substituting this in and solving for L rhc 7 L p IHET Tw her 7 7 Tw To calculate the heat transfer coefficient we need the Reynolds number Re pD V m Re 4 4 pA 7WD Using the given uid properties 415k s Nos2 k om g g 45040 g 7r212gtlt10393Nosm2002m This is turbulent flow so using the DittusBoelter correlation 212x10 3 Nos m2 38501 k K Pramp g 136 k 06 WmoK Nu 0023Re0 8W 00234504008 13604 345 345 06W K hMM10360Wm21 L 002m 15k 38501koK 7 L7wm75 100g4gm4 Answer 10360wm2K7r002m 357100 1242 12 41 Pressurized liquid water enters a 2cm diameter 6m long tube at 20 C at a flow rate of 05 kgs The tube surface temperature is constant and the total power transferred to the water is 150 kW Determine the surface temperature in C Approach This is a constant wall temperature heat exchanger Using the analysis given in Section 125 we can determine the required surface temperature Assumptions 1 The system is steady with no work or potential or m 033 L1 M Q kinetic energy effects 5 2 Liquid water is an ideal liquid with constant specific heat Solution For a constant wall temperature heat exchanger with Rm 1 M the exit temperature can be calculated from T T Tm 7T expthArhcp Tm Tm exp hIllma liexp7hAr39ncp The outlet temperature can be obtained from an energy balance on the water Assuming steady no work no potential or kinetic energy effects and an ideal liquid with constant specific heat so that Ah cp AT QEWICp Term Tm Tm Qimcp All the properties should be evaluated at the average water temperature From Appendix A6 and assuming TM Z100 C Ta 20 1002 60 C CF 4181 kIkg K p 9832 kgm3 k 0653 WmK u 452X10394 89 Solving for T5 Nsmz Pr f2 150kW1000s1kW 05kgs4181kJkgK1000 JkJ To calculate h we need the Reynolds number p D 39 39 u T 918 C out 20 C m m 4 2 gt Re pA p7r4D 7ruD 405kgsNs2kgm e n452x10 Nsm2002m This is turbulent flow so using the DittusBoelter equation Nu 002312881 0 4 002370400 8289 4 266 266 0653W k NWM867owm21 D 002m M 8670Wm2K7r002m6m1Is1W rho 05kgs4181kIkgK1000Tk p 918 273 7 7 nw3g3gK 1108 c1 Answer liexp71564 Re 3 70 400 h 1243 12 42 Parts of the Alaskan oil pipeline 1 m in diameter are buried 3m below the surface of the earth k 065 WmK and covered with 20cm of insulation k 005 WmK Pumping stations are 60km apart To decrease pumping power the oil is heated to about 100 C to reduce its viscosity before it enters the pipeline at a pumping station In the winter the surface temperature of the earth is 30 C Assume the oil properties can be approximatedwith those of unused oil given in the appendix For a flow rate of 05 m3s and using properties evaluated at the average temperature of the oil determine a the oil temperature when it reaches the next pumping station in C b the heat transfer required at the pumping station to raise the oil temperature back to 100 C in VJ c the pumping power required in VJ d the pumping power if the inlet oil temperature is 50 C instead of 100 C in W Approach This is a heat exchanger with one fluidwith constant 39 temperature so that analysis given in Section 125 can t be used to calculate the outlet temperature Once that temperature is known conservation of energy is used to calculate the heat transfer rate The pumping power 2 T 339 D l M also is calculated with conservation of energy but first V L I 60 LA the pressure drop is needed D I 4m Assumptlons 1 2 1 The heat transfer is one dimensional 20 M o 3 2 The heat transfer coefficient is uniform o M 3 The system is steady with no work or potential or 0 FED 39 02 C V 039 S kinetic energy effects Oil is an ideal liquid with constant specific heat r e Solution a For a constant wall temperature heat exchanger the exit temperature is calculated with T TltTTexp4 meRm Three resistances contribute to Rm 1 lnD2D1 s 27rkm5L hA First evaluating the ground resistance we assume twodimensional conduction in the ground with constant kgmund and from Table 111 for this configuration 2 ND s 27rLcosh 12zD 27r60000mcosh 123m1m152100m Evaluating the oil properties from Appendix A6 at TaVg Z 370 K p 8418 kgm3 cp 2206 kJkgK u 186 X 10394Nsm2 k 0137 WmK Pr 300 mpl48418kgm3 05m3s421kgs 39 4 421k For a circular tube Re4 m 7ruD 7r186 X 101 Nsm2 This is turbulent flow so using the Dittus Boelter correlation Nu 0023 Re 8 Fr0 3 002328 8000 8 3000 3 470 470 0137W K hN k m 645Wm2K D 1m 7 1 1n141 1 065 WmK 152100m 27r005WmK60000m 645 Wm2K7r1m60000m Tm 7300c100 7730Kexp71421kgs2206kTkgK 280x10395KW10001k1 951 c 3681 K d Answer Using this temperature TaVg 3705 K This is close enough to our assumed value so no iteration is required Rm k ground 28800 280gtlt10395KV RID 12 44 b Applying conservation of energy to the oil and assuming steady no work negligible potential or kinetic energy effects and ideal liquid with constant specific heat so that Ah cp AT Q mop Tm ATM 421kgs2206kJkgK100795lKlOOOJkl455gtlt106W4550kW l Answel c For pumping power we define a control volume around the pum Apply conservation of energy and assume steady adiabatic negligible potential and kinetic energy effects isothermal and incompressible 39 39AP 9V2 D 2 3 1 7 4V 7 405m s A 7rD2 1m2 For the friction factor we need the Reynolds number which is the same as we used above f 0790 ln28 8007164392 00239 0637 2 2 AP00239 M g41gkg3 NS 244700 2 lm m 2 kgm m Pressure drop is APf V 0637ms Finally pumping power is 3 W 05m 24470032 i M 7122300w71223kw 4 Answer s m lNm lJs d Using oil properties from Appendix A6 for Tm Z 50 C 4 use ng 1410 x 10quot Nsm2 4421kgs e n1410x10 Nsm21m f 00421 2320K p8718 kgm3 u 3800 0637 2 AP00421 87187 447000Nm2 W 05 447000 223500 w 2235 kW 4 Answer 60000m 1 Comments Because of the steep change in viscosity with changing temperature the pumping power increases dramatically when the oil temperature is decreased 1245 1243 Pressurized liquid water owing inside a tube at a rate of1 kgs is to be heated from 25 C to 90 C using condensing steam The 304 stainless steel tube has an inside diam ter of 25mm a wall thickness of 1mm and a length of 6m The c densing coefficient on the oumide ofthe tube is 6500 WmZK Determ39ne39 a the steam temperature and pressure required in C and kPa L 1 4 L a saturated vapor and exits as a saturated liquid in kgs Approach 519m Lo GlsuowMIK This is a constant uid temperature heat exchanger t J L J Using the analysis given in Section 125 we c lt 1m determine the required steam temperature Assumptions 1 The heat transfer is one dimensional Solution For a heat exchanger with one constant temperature uid T T TFT29XP 1 quotCPRM where Rm 1 1nDoD 1 T 7T e ilmc Solvingfor T 7 m m M 11 quot liexp71chRM a A 2sz 11vo From Appendix A2 for 304 stainless steel k 149 WmK The inside heat transfer coe icient requires Re with properties from Appendix A6 evaluated at the average temperature ng 90 252 565 so y interpolation 2 470 x 10394 Nsmz k 0550 WmK Pr 02 c 4179 kJkg K 4 1k s N s2 k m ampM a V L2 1 4 quot MIOS4OO 2 pr4D MD 7r470x104N smz 0025m This is turbulent ow so using the Gnielinski correction fS 1m 71000 u 1127fs 2 1 7quot 71 f07901n 1227154 2 07901n 1084001542 00177 00177810836071000302 A h Nuk 4350650Wmk a m 1127001778 2 302 71 D 0025 1300Wm2K 1 111 2725 R102 7 1 11300Wm2KIr0025m6m 2r149wmK6m 6500Wm2Kr0027m6m 0000188 0000137 0000302 0000627 KW 1 7 1 1Js1W in 382 chRm 1kgs4179kJkgK0000627KW1000JkJ 39 90273 25273 0382 n w owkaswc t Answer 1exp 0382 From the saturated water table Appendix A10 at 230 C P 2795 kPa b For the condensation rate we apply conservation of energy to a control volume aron the steam and liquid water Assume steady adiabatic no work no potential or kinetic energy effects and an ideal liquid water quot394 1 mw CPWT12 T m C T 7T 1k 4179ka K 90725 K m w M m gsgtlt g gtlt gt 715011 Answer 18138kJkg s with lyg from the steam table 1246 1244 The oil from a large Diesel engine ows through an oil cooler before it is returned to the engine Consider a ow rate of 01 kgs that must be cooled from 90 C to 40 C by passing through a thinwalled tube with a diameter of 127 mm Air at 30 C is in cross ow outside the tubes with a Velocity of 10 ms Determine the required tube length in m Approach a This is a constant uid temperature heat exchanger Am PIAF 3009 U 3 MS Using the analysis given in Section 125 we can determine the required length The outside heat transfer coe icient must be determined Assumptions a 1 The heat transfer is one dimensional Tquot To C 2 Air is at one atmosphere M 0M9 3 Ignore wall resistance Solution For a heat exchanger with one constant temperature uid 1 7 In i uexp71mcp Rm Tom 1 1 imcp 1n T T 77 1 1 1 1 1 m T a an where Rm Solving forL k 1101 IrDL hm IrDL IrDL 110 km The oil hear uau 11 39 39 wiur nuiu properties at 7ng 90 402 65 C338 K From Appendix A6 at 340 K 1 531x104 Nsmz k 0139 WmK Pr 793 c 2076 kJkg K 4 01k s Nsz k m RepD 4m 7 g g 189 a V 2 amp 1 pIr4D IwD Ir531x10394Nsm200127m This is laminar ow so the entrance length is Lem m 0037 1 Pr D 0037 189 793 00127 m 704 m We assume the tube length will be shorter than this so we will use the SeiderTate equation Nu 186 Gz m M1 4 186 Re Pr DLm M1 4 39 39 39 39 39 39 39 39 39 39 AssumingTSZ330K15836X10394 Nsmz and a tube length of40 m 172 189 394 NFLSG 7930 0127 531x10 40 836x10 59 2Wm2 K iNuk 7 6330139WmK 4 7 D 7 00127 11 j 533 a hm Th wiur pruperrie nom Appendix A7 at e air TM T T2 340 273 302 322 K p 1095 kgm 1 195 x10quot Nsmz k 00279 WmK Pr 0703 pw 1096kgm310ms00127mNszkgm y 195x10395Nsm2 From Table 121 for cross ow over a cylinder Nu 0193 1 5 Pr 0193 7140 6 0703m 413 km NukD413002790 127908Wm2K 2 0 112g 2075 1000i In 40 3930 LL m K s kgK kJ 90 30 592 908 w 1 00127m This is much longer than the calculated entrance length so we assume fully developed laminar ow with constant properties for the constant wall temperature boundary condition 366 0139 Nu366 2PM p 7140 40 1Wm2 K The air side 11 is the same as before so calculating the new length L 335 m Answer Comments This is an impractical length Multiple tubes in parallel should be used 1247 12 45 For some applications enhanced cooling capabilities are obtained by attaching a heat generating system to a cold plate which is maintained at a cold temperature by passing water through it Consider the copper cold plate shown below that has heat generating equipment attached to its top and bottom surfaces Each of the six channels is 6mm square and 100mm long and the walls of each channel is 4mm thick 1f chilled water at 10 C is pumped through the channels at a velocity of 05 ms and the surfaces of the cold plate must stay below 45 C determine a the maximum allowable power top and bottom surfaces to the cold plate in VJ b the water outlet temperature in C Approach Assumptions Heat enters the cold plate from the top and bottom surfaces An 1 The heat transfer is one dimensional energy balance on the water can give the total heat transfer but 2 The system is steady with no work and the exit temperature is unknown With the top and bottom no potential or kinetic energy effects surfaces at a constant temperature we have a heat exchanger with 3 Water is an ideal liquid with constant one constant temperature fluid so the analysis from Section 125 specific heat is applicable Examining the figure a unit cell shown can be analyzed that is representative of all 12 unit cells We have heat transfer from a fin The heat transfer coefficient must be determined once one unit cell is analyzed we multiply by 12 to determine the total performance of the cold plate S 4M v a WC j nlwm L WW1 L21 2 2 El 12 121 MW A 491 e all e 4quot 42 wb m 3 49m Solution An energy balance is applied to the water flow Defining a control volume that encompasses only the water and assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cp AT Qrhcp Tm 77 The outlet temperature can be determined by applying the analysis from Section 125 for a heat exchanger with a T Tm 7Tbexp71r39nchm The resistances that must be taken into account are the conduction through the plane wall with thickness 5 and through the fin as described by the unit cell Therefore the total resistance is NA mi where 77017 1777 k A 77 Mm Am For a unit cell N 1 Am s WL 0004 m 0006 m 01 m 0001 m2 Af 2W2L 0006 m 01 m 00006 m2 Ab WL 00006 m2 Am Ab NAf 00006 m2 100006 m2 00012 mi The convective resistance and the fin efficiency require the heat transfer coefficient From Appendix A6 assuming TM Z 30 C TaVg Z 10 302 20 C p 9982 kgm3 cp 4182 kJkg K k 0603 WmK u 985 X 10394 Nsmz Pr 683 In addition for copper k 401 WmK For the heat transfer coefficient we need the Reynolds number Re pDH u with DH 4AXPWEM 4W2 2WW W0006m 9982kgm3 05ms0006m Re 985gtlt10 Nsm This is turbulent flow so using the Gnielinski correlation f07901nRe716472 0791n30407164392 00454 constant wall temperature T out 3040 1248 A f8R671000Pr 7 004548304071000683 7227 M 1127f8 2Pr2371 1127004548 26832371 39 227 0603Wm3K hN k 2280Wm2K DH 0006m The fin is a straight rectangular fin with an adiabatic tip so fin efficiency is tnh W 2 12 12 a m gt when M1 hp 1 m mW2 2 2 kAX 2 kSL 0006m 2280Wm2K2 7 2 401wmK0004m 77tanh016001600992 100006 01 12 0160 77D 17 1709920996 For one unit cell 039004m l 0377m2KW 401WmK0001m2 09962280Wm2K00012m2 rhpV AX 9982kgm3 05ms0006m0006m00180kgs Tm45 C wt 10 45K exp 1001 8kgs4182kJkgKlOOOJk0377m2KW ll2 C Answer We have a total of 12 unit cells so the total heat transfer rate is Q120018kgs4182kJkgK112710KlOOOJkl1095W Answer Comments Because the assumed T a was higher than the calculated Tom the uid properties probably should be reevaluated and the calculations repeated 1249 12 46 The total thermal resistance between the outside and inside of a home consists of the external convective resistance the wall resistance and the internal convective resistance Adding additional insulation reduces the heat transfer but it also changes the inside wall temperature Compare the average natural convection heat transfer coefficient on a 25m tall wall for two situations a inside air temperature of 22 C and wall temperature of 10 C b inside air temperature of 22 C and wall temperature of 17 C Approach I T 21 C This 1s a natural convection problem We assume that Ag the wall can be approximated as a vertical at plate H Z M In 39 Assumptions V 1 Air is at one atmosphere 7 la c OIL 7 C Solution The natural convection heat transfer coefficient requires the Rayleigh number with properties from Appendix A7 evaluated at T lm TW Tf2 1 TM 10 222 16 0c 289 K so 5 1221 kgm339 1 177x10395 Nsmz39 k 00254 Wmk Pr 0711 2 TM 17 222 195 0c 2925 K so 5 1207 kgm3 1 179x105 Nsmz39 k 00256 WmK39 Pr 0710 ig pz TW 763Pr7981ms2 l289Kl22lkgm32 22710K25m3 0711 7 2 7 177X10395Nsm22 From Table 123 for a vertical wall Nu0lRa 30l215gtlt10w13278 iNuk727800254WmK H 25m 9811292512072227172530710 179x103952 Nu 01 846 x199 3 204 204 00257 h 2 209Wm2K N 26 smaller 4 Answer 1 PM 215x1010 h 283Wm2K Answer 946gtlt109 2 PM Comments As can be seen a small change in the driving temperature difference reduces the inside convective heat transfer coefficient So adding insulation helps in two ways increased conduction resistance and increased convection resistance 1250 1247 Consider again the light bulb in Problem P 1223 For all the same conditions determine the temperature of the glass bulb if it is cooled by natural convection in 3 Approach An energy balance on the light bulbs is needed The power into the glass 90 of 100W is removed by Hi 527 convection and rad t on The natural convection heat w 39 d 1a 1 transfer coef cient must be evaluate Q 9 W Ts Assumptions Air is at one atmosphere 2 Radiation is from a small 3 The light bulb is a sphere i1 083 body to a large area Solution A steady energy balance on the glass gives 2 Q QM wATf 7T MA T T The light bulb is assumed to be a sphere and the surroundings are very large The natural convection heat transfer coefficient correlation for a sphere is 0589 Ra 4 m 10469Pr9 6 gJ IJ2 TfD3 P7 2 where Ra Evaluate the uid properties from Appendix A7 at 71 I 12 because we seek 7 we must first estimate 7 2000 TM 200 272 1135 C 3855K p 0913kgm3 y 223 x 10quot Nsmz k 00327 WmK Pr 0591 D 981ms2 13855K09131qgm32 2007 27K008m3 0593 2 61 106 7 7 x 223x10395N smz 2 0589261x105m Nu2 m202 1045905919 6 202 00327WmK hN k 826Wm2K 008m Q 09 100 W 90 w QM 085567x10 3 Wm2 KA 4r004m2 If 7300K4 9590x10quot If 781x109 QM 826Wm2K 47 004m2 I 7 300K 015510 7 300 Combining these expressions 90 959 gtlt10quot T4 781x109 015510 7300 Solving R5061K 233 C 4 Answer This is reasonably closed to our assumed value off so we will not iterate 1251 12 48 In any design process decisions have to be made about placement of components Cost performance and maintainability are some of the criteria used Consider the placement of a 12 W 60mm by 60mm electric component in a larger device The component s surface temperature must not exceed 85 C The air is quiescent at 25 C Neglecting radiation determine if the component can be located facing downward or facing upward In other words what is the component s surface temperature if it is facing upward or downward Approach This is a natural convection heat transfer problem JigSo c The maximum power transferred from a 85 C surface N to 25 C air should be calculated and compared to the H r oSSI x 12W specification Appropriate heat transfer quot r coefficient correlations for the two geometrics must be 39TaD 2st Oca h u use 0 N1 Assumptlons quotOMMK M 1 Radiation is neglected 2 The plates match the geometry for the correlations in Table 123 3 Air is at one atmosphere Solution Neglecting radiation the heat transfer rate is calculated with hA TW Tm We need the Rayleigh number to calculate h Evaluating air properties from Appendix A7 at T lm 85 252 55 C 328K by interpolation p 1076 kgm3 k 00283 WmK 1 199 x 10395 Nsmz v 185 x106 mZs Pr 0702 85 Tw To 5 T Ra Pr From at plates the characteristic length is 5Ap 006m 006 4 006m 00l5m 981ms21328K85725K0015m3 0702 a 2 12400 l85gtlt10396 m2s From Table 123 for ahorizontal plate 1 Hot plate facing upward Nu 054Ratl4 05412400 4 570 hi Nuk 57000283WmK 10 w R a 00l5m m2K Q 108Wm2K 006m2 85725K 233w 4 Answer 2 Hot plate facing downward Nu 0271211 4 027 12400 4 285 h 285 002830015 538 WmZK Q 538 006 006 85 725116W d Answer The safe choice is to use the location with the component facing upward However because facing downward is close to satisfying the design specification it might usable if other design considerations had to taken into account Comment A second approach would be to estimate the wall temperature for the 12W power input This would require an iterative solution 1252 12 49 To lower the viscosity of an oil before it is used in a process an electric resistance heater l5cm in diameter and 30cm long is immersed horizontally in a vat of unused engine oil which is at 20 C If the heater surface should not rise above 150 C so that the oil does not smoke determine the maximum power that can be dissipated in the heater in VJ Approach This is a natural convection problem Because of the large length compared to the diameter we will ignore heat transfer from the two ends The basic heat transfer rate equation hAAT is used to calculate L0 5Mgtl T3950 the maximum power The heat transfer coefficient must be evaluated 11 710 C I Oll Assumptions 1 The system is steady 2 All properties are constant and are evaluated at the appropriate temperatures 3 The ends of the heater are ignored Solution The heat transfer rate is calculated with hAT5 7T Ignoring the ends A IrDL From Table 123 the heat transfer coefficient correlation is 0387Ra16 8m 1o559Pr9 g T5 7TD3Pr 2 Nu 06 where Ra Evaluating the oil properties from Appendix A6 at TM 150202 85 c358K v 321X10396m2s k 0138 WmK39 Pr 425 8 07X10393K 981ms07X10393K15020K0015m3 425 Ra 2 124x106 321x10396 mZs 0387l24gtlt10616 105594259 T27 7 NW 7 2100138wm1lt 7 7 0015m Q l93Wm2K7r0015m030m15020K355 w 4 Answer Nu 06 210 h 193wm2K Comments A factor of safety might be used to derate the power to ensure the surface temperature remains below 150 C In addition as the oil temperature increases lower power would be used to ensure the surface temperature remains below 150 C 1253 12 50 An electric resistance heater lOmm in diameter and 300mm long is rated at 550 W If the heater is horizontally positioned in a large tank of water that is at 20 OC estimate the surface temperature of the heater in C Approach This is a natural convection heat transfer problem 7 1 WA We will ignore heat transfer from this ends of this cylinder and assume that the tank is large enough to Q 139 S30 w not influence this heat transfer coefficient Because the driving force is the temperature difference and we do not know it an iterative solution is required Assumptions 1 The ends of the heater are ignored L 03 M D l 2 Only natural convection occurs Solution The surface temperature can be determined from the basic heat transfer rate equation QhATW7Tw TWT QhA We ignore the cylinder ends so thatA IIDL The natural convection heat transfer coefficient for a horizontal cylinder is used Nu 06 where Ra 2 0387Ra 85Tw 7T8 P 10559Pr 16 V2 r The properties should be evaluated at the film temperature T lm TW Tw2 so as first approximation estimate TW 40 C Therefore T lm 30 C and the properties from Appendix A6 are 8 306 X lO39AK u 779 X 10394 Nsmz 5 9956 kgm3 Pr 526 v 0782 8106 mZs k 0618 WmK 2 74 7 3 D 7981ms 306gtlt10 K40 20K001m 526 2 516000 0782gtlt1039 m2s 2 0387 516000 6 146 0618W K Nu 06 W 146 a hN uk m902Wm2K 105595269 6 D 001m TW 20 C847 c 902Wm2K7r001m03m This temperature is significantly about what we first estimated so we iterate T lm 847202 523 c a at 50 c the uid properties are 8 495 x 10 1 488 x 15394 Nsmz 5 9857 kgm3 Pr 315 k 0648 WmK v 495 x 10395 mZs 981 495x10 4 847720 001 3 315 Ra 2 H 7404x106 0495x10 0387404gtlt10 16 2 m 254 h 254 0648001 1650 Wm K Nu 06 10559315W T 20554 c W 16507r001030 One more iteration based on trends let s assume TW 70 C 3 T lm 145 C 8 423 X 10394K u 577 X 10394 mZs 5 9902 kgm3 Pr 423 k 0637 WmK v 0583 x 10395 mZs a Ra 258X10 Nu 228 h 1450 WmZK a TW 602 c Another iteration gives TW N 62 C Answer 1254 1251 Apa ive ulai wall a and then 39 when 111 aii quot waii is lower than that ofthe wall Consider a long 3m tall wall well insulated on its backside that has a net radiant solar energy ux into the wall of 150 Wmz The air temperature is 21 C Assuming that the temperature ofthe wall changes very slowly and the wall operation canL 39 1 39 t d 4 39 L J temperature of the wall in C Approach This is a natural convection problem that can be solved with the basic rate equation Q q 14 11A AT The heat transfer coef cient must be calculated Assumptions 1 The system is quasisteady 2 Air is at one atmosphere Solution Assuming a steady state system the convective rate equation is 29340 9 Solving for T T T qw Using an inn will L wan temperature oh Rayleigh number at I I T2 Because I is what we seek we must rst estimate 7 evaluate h and then calculate 7 If L 39 A value are close L 4 39 u 1 no From Appendix A7 and assuming I 73 gt 71 21 732 47 C 320 K p 1103 kgmz 2 194 x10quot Nsmz k 0027 WmK Pr 0703 m 1 kg 2 z 981 1103 73721K3m 0703 7 TiT HzPr 2132 3 D sg gp 2 s s m 2 78x10 14 194x10395 N smz FromTable123 Nu011m 301978x10 3451 451 00278WmK hN k 427Wm2K H 3m 2 T21 C 561 C 42 Wm K Assuming I 554 C gives TM 312 K so thatJ 1131 kgmz 2 190 x 10quot Nsmz k 00272 WmK Pr0704 981 1312 11312 554721 33 0704 Mg gtlt gtlt lt 2 gtlt gt750x10 190x10395 Nu 01 750 x10 m 421 421 00272 h382Wm2K T21150382602 C lt Answer This is close enough and another iteration is not needed 1255 1252 A steamheated cooking vat in a food processing plant has a bottom that is 15 m by 15 m The vat is lled with water initially at 25 C and the bottom is heated with condensing steam at 105 C Determine a the initial heat transfer rate from the bottom of the vat to the water in b how long it would take for the water temperature to rise to 30 C if the water depth is 60cm in min Approach This is a natural convection heat transfer problem We will assume that the heat transfer coe icient from an upward facing heated plate is applicable The time to heat the water from 25 C to 30 C can be determined with conservation of energy H 3cm Assumptions w 2quot 1 The heat transfer coefficient correlationefor an L 4 L LY upward facing heated plate is applicab 2 The system is cl 39 With no work or potential or Too 5 ZWC J T kinetic energy e ects 3 The condensing resistance is ignored 4 The heat transfer rate is constant Solution a The initial heat transfer rate is calculated with QI 1ATV iTm L 39 cuemciem i VCI high 4 L 39 39 bottom is very small so that the wall temperature is the same as the condensing temperature TM T TW2 25 1052 55 C From Appendix A6 p 9805 kgmz k 055s WmK 1 421 x 10394Nsm2 v 0429 x 10396 mZs 6553 x 10394K Pr 257 The characteristic length is 5Ap 15 m 15 m 4 X 15 m 0 7 m 3 D 7g TLJ 77953 p 7981ms2553x10AK105725K0375m 2577 32m v2 0429x10396 mzs 39 From Table 123 for an upward facing heated plateNu 015th We are outside the range of the correlation s applicability but with nothing else available we are forced to use this Nu 015 332 x10 m 1040 hi Nuk 7 10400658WmK 7 5 7 0375m Q 1820 WmZK 15 m 15 m 105 7 25 K 328000 w Answer b The time to reach 30 quotC can be determined by conservation of energy applied to a closed system Assume no work and no potential or kinetic energy changes dU d mu Q dt dt For an ideal liquid du c dT c N c N CV the vat 4 1 820Wm2 K mc Q 1 dt 39 constant 39 quot ME AT IthIchdT gt t At the averagebulk temperature 25 302 275 C c 4177 kJkg K p 996 kgm3 996k m3 15m 15m 06m 4177kJ k K 1000 k 5K 7 g lt gtlt gtlt x g gtlt gtlt gt 443mm Answer 328000w1Js1w 1256 12 53 You devise a transient heat transfer experiment to measure the natural convection heat transfer coefficient on a 2024T6 aluminum sphere Initially the 3cm sphere is at a uniform temperature of 90 C as measured by a thermocouple inserted into the sphere s center You plunge the sphere into 10 C water and record the center temperature as it decreases with time The center temperature reaches 80 C after 083 s 50 C after 566 s and 40 C after 978 s Determine a the average heat transfer coefficient as the sphere changes temperature from 90 to 80 C in WmZK b the average heat transfer coefficient as the sphere changes temperature from 50 to 40 C in WmZK c Whether or not this approach is valid Approach o This is a transient heat transfer problem Because the quot 0 C sphere is aluminle and natural convection has a relatively low heat transfer coefficient h we Will assume a lumped systems analysis is applicable to this problem and the recorded sphere temperatures are uniform We also assume h is relatively constant over the time intervals AiUMINUM I 9 0893 805 A t 5 Has 93 C ssump lOnS l The lumped systems analysis is valid S 2 The heat transfer coefficient is constant over the time intervals Solution TeT ihAl exp 7 7T chp 7 Vc TiT Solving for h hLln A 7 7T For a sphere V47rR33 7rD36 and A 4er2 IrDZ Combining these with the main equation h7pDcp 1n Tin 61 I 7T From Appendix AZ for 2023T6 aluminum p 2770 kgm3 k l77 WmK cp 875 JkgK a For the average heat transfer coefficient as the sphere changes temperature from 90 to 80 C 7 2770kg m3 003m 875JkgK hwh m l950Wm2K lt Answer 6083s 90 10 b For this time interval we must reset T 50 C T 40 C and the time interval is I 9785 7 566s 4125 7 2770 003 875 hwln 846Wm2K 394 Answer 6412 50 10 c To determine if this approach is valid check the Blot number D MW 7 hay6 7 1950Wm2K003m k 6177Wmk So the approach is valid The second condition does not need to be checked explicitly because that h is smaller than the one used The lumped systems equation is 0055lt0l 1257 12 54 Farmer Brown installs an electric resistance heater in the watering trough for his cows so that the water will not freeze during the long cold winter He does not want a cow to burn its tongue if it accidentally touches the heater He places a 25mm diameter 30cm long 100W heater horizontally in the water which is maintained at 5 C Determine a the surface temperature of the heater in C b the surface temperature of the heater if the water trough develops a leak all the water drains out and the air temperature is 15 C in C Approach c This is a natural convection problem We ignore the heat Q 100 W TS transfer from the heater ends and use the rate equation hAAT to calculate the surface temperature The appropriate heat transfer coefficient must be evaluated Assumptions lt39 L 03 C 1 Ignore heat transfer from the ends 2 Air is at one atmosphere 3 There is no radiation in air Solution a The surface temperature can be determined from the basic convective heat transfer rate equation ignoring heat transfer from the ends Qh7rDLT5 7 T gt T5 T Qh7rDL The natural convection heat transfer coefficient for a horizontal cylinder is 2 us 2 T 7T D3Pr Nu om w when 10559Pr9 6 Evaluating the properties of water from Appendix A6 at T lm T 7 Tf2 we first assume T Z 30 C a TM 2 290 K p 999 kgm3 u 108 x 10 Nsmz k 0598 WmK Pr 756 8 174 810 981ms2 174gtlt10AK 999k m3 2 3075 K 0025m 3 756 M g lt 2 gt gtlt 1431st 108x104Nsm2 R 0387 431x106 6 275 a hNuk275O598WmK Nu 06 w l05597569 6 D 0025m 657Wm2K T5 5 C100W657Wm2K n0025m030m 115 c The first estimated TS is significantly larger than the calculated temperature The uid properties should be re evaluated and a new TS calculated Using TS 115 C and properties evaluated at T lm 280 K Ra t233gtlt105 gt Nu 121 gt h282Wm2K gt Ts 201 C Guessing T Z 16 C and T lm Z 285KRa 113gtlt106 gt Nu188 gt h443Wm2K gt T5 1460C Answer This is close enough so additional iterations are not required b Now with air we also assume no radiation Following the same procedure as above guess T N 500 C TM 2 515K 5 0685 kgm3 u 275 810395 Nsmz k 00417 WmK Pr 0680 Ra98115150685250015002530680275gtlt10395264600 Nu691 a h116Wm2K a T5715 C1001l67r0025030 353 C lterating a second time with T Z 353 C and evaluating properties at T lm Z 480 K Ra77650 a Nu725 h114Wm2K 7357 C Answer Comment At this temperature radiation is significant so more calculations with only convection are not justified 1258 12 55 The manufacturer of the electric resistance heater described in Problem P 1254 wants to expand her sales and considers using the heater for fuel oil tanks too Concern about a possible fire hazard if the oil anywhere in an oil tank reaches too high a temperature makes her contact a consulting engineer for an analysis If the oil has properties of unused oil and the oil is at 0 C determine the surface temperature of the heater in C A roach ITllfis is a natural convection problem We will ignore Qw w TS heat transfer from the two ends The basic convective heat transfer equation hA AT is used to calculate the surface temperature The appropriate heat transfer coefficient must be evaluated 5 L 013 M 39gt Assumptions T quot 0 C 0 n l Ignore heat transfer from the ends Solution Ignoring heat transfer from the two ends the basic convective heat transfer rate equation is used to calculate the surface temperature Qh7rDLT5 7T gt T5T Qh7rDL The natural convection heat transfer coefficient for a horizontal cylinder is 2 16 2 T 7T D3 Pr Nu 06 when 10559Pr9 6 Evaluating the oil properties from Appendix A6 at T lm TS Tf2 we first must estimate TS Z 35 C a TM 2 2905 K p 890 kgm3 1 9990x10394 Nsmz k 0145 WmK Pr 12900 8 07 x103 K 981m s2 07gtltlO393K 890k m3 2 3570 K 0025m 3 12900 RM g lt 2 lt gtlt L385 9990x10 4Nsm2 038738500 6 Nu 06 m7 807 10559129009 6 807 0145W K hN kM46gwm21 D 0025m TS0 C90 C 468Wm2K7r0025m030m Because this calculated TS is significantly different than the estimated TS we must iterate Guessing T S 35 902 625 c 2 TM 305K p 881 kgm3 1 3700 x 10394 Nsmz k 0145 WmK Pr 4900 8 07 x 10 981 07gtlt10393 8812 62570 00253 4900 RM 2 Jim 3700x104 Nu124 a h720Wm2K a 759 C lt Answer Close enough so no more iterations are required 1259 12 56 The coils in electric power transformers mounted on telephone poles in every neighborhood are cooled by oil If the coils reach too high a temperature the transformer can fail To prevent this problem the transformer is externally cooled by air The worstcase scenario occurs on hot still summer days Consider a transformer that is 55cm in diameter and l5m tall on a day when the temperature is 40 C Assume the heat transfer coefficients on the ends are the same as on the cylinder sides If 250 W must be dissipated determine a the surface temperature of the transform er in C when ignoring radiation b the surface temperature if radiation is included with an emissivity of 06 in C Approach This is a natural convection problem Using the basic heat transfer rate equation hA AT the surface quotr H l quotM temperature can be determined An iterative solution f 39 must be used to find the natural convective heat Q V transfer coefficient Assumptions T5 1 Air is at one atmosphere 2 The same heat transfer coefficient is used on sides and ends 3 The transformer is small relative to its surroundings Solution a The rate equation is QhAAT hAT5 7 T Solving for the surface temperature T5 T whereA IrDH 27rD24 7r055 m 15 m 27r055 m24 3070 m2 We assume that h can be approximated by that on a vertical wall and this also is assumed to be valid for the two ends The Rayleigh number is evaluated with properties from Appendix A7 at T lm T Tf2 We assume T Z 65 C so T lm Z 65 402 525 C Z 325 K and p 1086 kgm3 1 196 x 10395 Nsmz k 00282 WmK Pr 0703 g p2 eTH3Pr 7981ms2 1325K1086kgm3 265e4015m30703 R 2 550x109 4 196x10395Nsm2 From Table 12 3 Nu0lRa1301550gtlt109ml76 176 00281W K hNWM33lwm2llt H l5m T54OUC6470C lt Answer 33lWm2K3070m2 This is reasonably close to our assumed TS so iteration is not required b If we include radiation then the energy balance is QQcanv Qmd hATe TTSUAT54 TA The surface temperature will be lower so h should be reevaluated at a lower TS Assume T Z 55 C 328 K TM 2 475 C 2 320 K p 1103 kgm3 1 194 x105 Nsmz k 00278 WmK Pr 0703 7 981132011032 55740153 0703 194x10395Nsm22 152 00278 Nu01353x109 152 h 15 D 353gtlt109 282Wm2K Substituting into the energy balance 250W282Wm2K 307m2T5 313 C06567X10 3 wm2K4307m2Tj 313K4 Solving for T N 3243 K N 513 c d Answer l260 1257 39 dc ui ued in Problem P 1256 16 fins made of plain carbon steel are attached Each fin has the same length as the transformer is 4mm thick and extends from the surface 100mm Using only natural convection determine the surface temperature of the transformer in C Approach This is a natural convection problem from fins Using the convective heat transfer rate equation applied to D 339 0 I5 m the finned surface and to the ends the surface N 392 l 9 66 temperature can be determined An iterative solution may be required since the calculation ofthe natural T H quot y convection heat transfer coefficient depends on the L surface temperature gtII 0 0M Assumptions t 39 O39OO tM 1 The heat transfer is one dimensional Solution The rate equation for convective heat transfer from a finned surface and from the ends is Q74 M T 7T 2114 T 7T Solving 1m I 7 i 1 i la M 2441 quot 4 to the finned surface Assuming a straight rectangular profile fin with an adiabatic tip and using a corrected length NA qUI7 A 17 and AOA5NA N16 A2LH2Lt2H 201000042m15 m031m2 AbrDHNIH r055 m 15 m7160004 m 15 m250m2 Am 250 15 031 740 m2 Am 7004 7r055 m24 024 m2 tank ml 12 2 Thefin efficiency is q where m h p m H M M ktH We assume that 11 can be approximated by that on a vertical wall and this is also valid for the ends The Rayleigh number is evaluated with properties from Appendix A7 at 71 I T2 We assume I Z 54 C gt TM 2 320 K p 1103 kgmz 1134 x10quot Nsmz k 00278 WmK Pr 0703 2 1 1 2 3 D 1320575 981ms 1103kgm 54740Kl5m 0703 29 109 7 7 7 x 2 194x10395Nsm22 149 00278WmK FromTable123 Nu011m 301329x109ml49 7 N k276wmZK H 15m 275wm7K2 2 For pla1n carbon steelk605WmK mf 01000042m0487 605WmK0004m 2 itanh04877 HM quot01 15031m h Aim EA 1 f 0487 740m2 T40 C 250W i 1 520 C 1 Answer 275Wm2 K L0951740m2 2024m2 This is close enough to our estimate so we do not need to iterate Comments 39 ation r 647 C Jhi 39 wen i of fins in increasing heat transfer 1261 12 58 Two 15m in diameter 2m tall tanks connected to a common piping header are used to store propane for use in an isolated cabin in the mountains Unknown to the owner the spring in the pressure relief valve on the system weakens and allows the pressure in the two tanks to drop to atmospheric The temperature of the propane falls to 42 C when the pressure inside the two tanks reaches one atmosphere The still ambient air is at 15 C Heat transfer from the air to the propane causes it to vaporize and the vapor is vented from the tank Properties of propane are vf 0001755 m3kg vg 04127 m3kg hfg 425 kJkg and hg 493 kJkg Ignoring the wall resistance of the tank and radiation determine how long it will take for the tank to empty in days Approach The rate of the conservation of energy and mass are applied to the control volume shown These equations are integrated with respect to time The heat transfer coefficient is required to calculate the heat transfer rate Assumptions The air is at one atmosphere 1 2 Radiation and wall resistance is ignored 3 The system has no work and potential and kinetic energy effects are negligible Solution For the control volume shown assume no work constant pressure and no potential or kinetic energy effects Conservation of mass and energy give 7mm dmdl and 7 in hm dUdt where how is the enthalpy of the leaving vapor and is equal to hg saturated vapor Integrating both equations with respect to time dzjdm dtm2 7m 17ij kg dzde With hg constant and assuming is constant Im2 2 m1u17m2 Tml hg The initial internal energy is saturated liquid u1 uf The final internal energy is saturated vapor u ug From the given information we can determine the internal energies and masses m1 Vv1 Vvf V7 211D2H4 7 21 15 ml 2 m4 7 707 m3 m1707 m3 0001755 ng 7 4030 kg and m 7 Vvg 7 707 mg 04127 m3kg 171 kg The final mass is negligible compared to the initial mass so one will ignore m2 that is use m2 7 0 The initial internal energy is 111711 hf 7Pv hg7 hg 7Pv 4937 424kIkg 71013kNm2 0001755m3kg1kI1kNm688kkg This is a natural convection problem ignore radiation We assume the heat transfer coefficient can be calculated for a vertical wall the convection from the top of the tanks can use the same heat transfer coefficient and there is no heat transfer from the tank bottoms they sit on the ground Evaluating air properties from Appendix A7 at TM 7 T7TP2260K p7 1359 kgm3 1 7156 x105 Nsmz k7 00230 WmK Pr 0719 2 3 RM g p2T7TPH3PV 981ms21260K1359kgm3 1542K2m 0719 2 156x10395Nsm22 Q17m2 7 m1 hg m2 u2 7 m1 u1 Solving for time 39gtlt 1010 454 00230W K From Table 12 3 Nu 7 011211 3 7 01939 x1010 3 7 454 7 hN7 k2 m m The surface area is insulated bottoms A 27rDH 27rD24 Zn 1 5m 2m 1 5m 24 236m2 The heat transfer rate is Q hAT 7TP 523wm2K236m2157 7 42K 77040w The time requires for the tanks to empty 7 4030k 688k k 4030k 493ka gx gm ggtlt g243gtlt105s 675 hrs 7 281 daysquot Answer 7040W1kW1000W1ks1kW 523Wm 2K Comments Note that radiation heat transfer will have an influence and will accelerate vaporization 1262 12 59 An experiment is performed to determine the heat transfer coefficient on a horizontal circular cylinder Radiation effects are minimized by polishing the cylinder s surface The 30cm long 25cm diameter cylinder has well insulated ends Measurements show that 30 W are dissipated when the cylinder surface temperature is 95 C and the surrounding air and surfaces are at 20 C Determine a the natural convection heat transfer coefficient from the data b the natural convection heat transfer coefficient if radiation is taken into account and the surface emissivity is estimated to be 007 in WmZK c compare the calculated heat transfer coefficient to one calculated with the appropriate correlation in Wm K Comment on the accuracy of the experimental results Approach T 201C The convective heat transfer rate equation hAAT TS ag is used to determine the convective heat transfer coefficient If radiation is included the heat transfer rate due to convection is reduced so h increases is L camN Nsuulw Assumptions 30W 51 l The air is at one atmosphere Solution a The rate equation is M T 7 Tf Solving for h and noting that the ends are well insulated 1 Q 30W 170Wm2K4 Answer AT57T 7r0025m030m95720K b If radiation is included Q QM hA T5 T gaA T54 7 T 39 780A TtT Solving for h hQm 5 AT5 in 30W7007567gtlt10398 wmZK n0025m030m95273 720273 1K 7 7r0025m030m95720K 164Wm21lt4 Answer c To calculate h from a correlation use properties from Appendix A7 T lm 95 202 575 C 3305 K p 1068 kgm3 k 00285 WmK 1 198 x 10395 Nsmz Pr 0701 85920 T DEPr T 981ms2 133051lt 1068 m3 2 95720 K 0025m 3 0701 lt gt kg lt 2gtlt gtlt L70 198x10395Nsm2 Ra From Table 123 6 0387 70900 6 Nu 0 1 0387Ra 0 1 710 10559Prm Tm 105590701 16 Tm 7 71000285Wm2K h 81Wm2K 4 Answer 0025m Comments This is poor accuracy Radiation plays a small role reducing the experimental heat transfer coefficient by about 35 However that is minor compared to the factor of 2 differences between the experimental h and the h calculated with the correlation The experimental is flawed and the measurement sensors must be checked as well as the experimental procedures For example the smallest air draft can raise the experimentth significantly so care must be exercised to enclose the test section with panels to minimize drafts 1263 12 60 Arrays of vertical fins are often attached to equipment to aid passive 139 6 natural convection cooling of the device Consider the assembly shown below that is located in air at 20 C Each fin has a length of 25 mm a thickness of 15 mm and a height of 100 mm Assume the fin has a fin efficiency of 100 the base temperature is 75 C and each fin operates as if it is independent of all other surfaces nearby For a fin spacing of 8 mm determine the heat transfer rate from an array of fins that covers 150 mm of wall in W This is a natural convection between fins We assume that it is similar to natural convection in vertical parallel plate channels and the heat transfer coefficient calculated from the appropriate correlation quot is applicable to all the surfaces T e number of fins must be determined and then the basic rate equation is used to calculate the heat transfer rate Approach w 2 0 39WM 4 H Tb 7r c 7 lamp092 Tg 10 c 1 9025 p gtl lt 000137 L 010m Assumptions 1 Air is at one atmosphere 3 o oogm 2 The heat transfer is one dimensional Solution The rate equation for a finned surface is Q 770 17A Tb in Because 77f 1 the overall surface efficiency also is unity that is 770 1 The total area is Am NZLH WL For thenumber offins N WNSt gt NL o39lsom 158 SI0008m00015m Because there must be an integer number of fins N 15 Am 15 2 010 m 0025 m 015 m 01 m 0090 m2 For the average heat transfer coefficient we assume isothermal surfaces and use the BarCohen and Rohsenow correlation 712 q s 576 287 5 W t A Ra5 j Ra5 j L L 2 T 7T S3Pr Rafg p 2 u From Appendix A7 at T lm 74 202 465 c 3205 K p 1101 kgm3 1 194x10 5 Nsmz k 00278 WmK Pr0704 981ms2 13205K1101kgm3 2 75720K0008m3 0704 194X10395Nsm22 1950 712 us 576 2 1 287 m 199 19500008010 19500008010 iNukTT719900278Wm1lt757201lt s 0008m Qq Am 380wm2 0090m2 342w1 Answer q u 380Wm2 1264 12 61 A Window 30cm tall and 45cm Wide is centered in an oven door that is 50cm tall and 75cm Wide During operation when the room temperature is 24 C the Window reaches a temperature of 45 C and the door surface reaches 33 C Assume that both the door and Window have an emissivity of 10 and the surroundings also are at 24 C Estimate a the heat transfer from the door and Window in W b the heat transfer if the door did not have a Window in W Approach We Will need to use approximations for part A because of the natural convection over two surfaces at different temperatures Heat loss is due to radiation and convection We Will assume that the Window is flush with the door and we have the same natural H1 convection heat transfer coefficient over both door and Window gt 1 mm Ito33M L1 97SM HL 030M Assumptions 0 T 2 C l The heat transfer coefficient for he Window and T 1 43 0 C Tl quot C id door is the same 2 Air is at one atmosphere Solution a The total heat loss flow the door and Window combination is sz QDOOR QWINDOW QDOOR gcrAD T4 7T hAD T2 77 where AD L2 H2 711 N1 QWINDOW 50 1484 T14 7 T4 hAw TTf Where Aw L1 H1 The natural convection heat transfer coefficient requires the Rayleigh number with properties from Appendix A7 evaluated at T lm T Tf2 Because the door and Window are at different temperatures we assume T N TS T22 45 332 39 C 312K so that T lm 312 2972 3045 K The properties are p 1159 kgm3 1 1887 810395 Nsmz k 00266 WmK Pr 0707 D 7 g p2 eTH Pr 7 981ms2l3045Kll59kgm3 2 3127297KO5m3 0707 2 161x108 1887x10395Nsm2 From Table 123 the vertical plate correlation is 665 00266W K Nu059Ra4059161gtlt108m665 a hN WM354Wm2llt H2 05m Aw 045 m 030 m 0135 m2 AD 050 m 075 m 0135 m2 0240 m2 39 8 W 2 4 4 W 2 QDOOR1567X10 m2K4024m 306K 7297K 354F024m 3067297K 13476210W QWINDOW1567x103980135318472974135401353187297187100287w Q39m 210 287 497w lt1 Answer b For a door with no Window assuming the heat transfer coefficient is the same as in part a Q1567gtlt10398050753064 72974354050753067297 210119329w 1 Answer 1265 12 62 The heat loss situation described in Problem P 1215 changes when the wind stops and the air is calm For this new condition determine a the heat transfer rate per unit length of pipe in Wm with no insulation and b the heat transfer rate per unit length of pipe if 4cm insulation k 004 WmK is applied to the pipe in Wm Approach This is a natural convection problem on a circular 03c cylinder The basic heat transfer rate equation ATRm can be used to find the heat transfer per unit length The heat transfer coefficient must be DZ evaluated Assumptions 1 The heat transfer is one dimensional 2 Air is at one atmosphere Solution The basic heat transfer rate equation is Q AT Ts T Ts T Ts T or R222 R222 1222 lhIrDL 1nD2 D127rkL L 11mD1nD2 D1 27rk a With no insulation D2 D1 D D1 and Row 0 Evaluation of the natural convection h requires the Rayleigh number with uid properties evaluated of T lm 6 llO2 52 C 325K From Appendix A7 p 1086 lltgm3 1 196 x105 Nsmz k 00281WmK Pr 0703 2 g p2 T2 7TD3 Pr 2 981ms2l325Kl086kgm32 110476015111 0703 D 2 2 255 x 107 196x105Nsm2 From Table 123 for a horizontal cylinder 0387B 6 2 0387255x10716 a Nu 0 1 W 2222 0 1 W 2222 373 10559Pr 105590703 373 00281W K hN k X m 699Wm2K D 015m 39 1107 76 K 2 382E without insulation Answer L 1699Wm21ltn015m m b Once we add the insulation the outside insulation temperature will be low Assuming TS 0 C T lm 2 p 1307 lltgm3 1 163 x 10395 Nsmz k 00239WmK Pr 0716 7981127013072 1107760233 0716 PM 2 236x108 163x10 5 738 00239 Nu738 ww mwmk 023 1107 76 E 1 617Wm Q L 1767r0231n23152n004 The outside wall temperature can be calculated with hIIDL T2 7 Tf 2 T2 Tf T5 T QLh7rD 0 617Wm 7176 c 51 Clt Answer 767Wm2K7r023m This is close to what we assumed so no iteration is required 1266 12 63 A power amplifier is mounted vertically in air that is at 27 C The case is made of anodized aluminum with a surface area of 3800 mm2 and a height of 40 mm If the amplifier operates at 127 C estimate the total power dissipation natural convection and radiation from the unit in VJ Assume a surface emissivity of 076 heat transfer problem With the surface temperature known this is a straight forward calculation The appropriate heat transfer coefficient correlation must be chosen and we assume the surface is small relative Approach This is a combined natural convection plus radiation H 443M g to its surroundings i 0quot0 Assumptions d 1 Air is at one atmosphere R C 2 The power amp is small relative to the large surrounding area Solution The total heat transfer from the surface by convection and radiation is Q QWQW hATweTwmAT 7T3 To evaluate h we need the Rayleigh number with properties from Appendix A7 evaluated at T lm TW T2 400 3002 350 K p 0998 kgm3 v 2076 x 10395 mZs k 003003 WmK Pr 0697 gMTFTw H3pr 981ms2l350K1277 27K004m3 070 V2 2076gtlt10396m2s2 From Table 12 3 Nu 05911211 4 059 290 x 105 4 137 hi Nuk 137003003WmK T H 7 004m W 00038m2 4007300K076 567gtlt10398 K D 290gtltlO5 lO3Wm2K W K4 2 2 m m Q103 Q391w287w678w 4 Answer 00038m34001lt 73OOK 1267 12 64 In car paint shops and other drying applications radiant heaters are often used because the radiant thermal energy heats the surface directly with minimal heating of the surrounding air Consider a vertical at panel lm tall and 4m long with an emissivity of 085 mounted on the wall of a large room The panel is maintained at a uniform temperature of 330 C and the walls and air in the room are maintained at 25 C Determine the heat transfer rate from the panel to the room in W Approach 4M Power input to the heater must equal radiation plus LT convection from panel We assume the panel is 0 located in a very large room for radiation purposes l 1 1 C and we use hA AT for the convection The E 1 0 natural convection heat transfer coefficient must be evaluated Hquot M T quot luv ch Assumptlons l 1 Air is at one atmosphere 2 The heater is small relative to the large room Solution An energy balance on the heater gives Qm Qad Qmv SO39AT54 7 T4 hAT5 in where A HL The natural convection heat transfer coefficient requires the Rayleigh number with properties from Appendix A7 evaluated at T lm 330 252 1775 c 4505 K p 0784 kgm3 7 2484 x 10395 Nsmz k 003707 WmK Pr 0683 m kg 2 3 g p2T 7T HEPV 9815 214505KI0784EJ 6037298K1m 0683 PM 5 2 2 452x109 2484x10395Nsm2 From Table 123 for a vertical wall Nu Ra 01 452 x 109 3 165 7 Nuk 7 165003707WmK 7 H 7 1m h 613Wm2K W K4 de 085567x10398 m2 j1m4m603K47298K423970W 6l31m4m6037298K 7480w m Q39m 23970w7480w31450w 1 Answer 1268 12 65 A home hobbyist builds a kiln to fire her ceramic pots Plans obtained from the Internet state that because of the thick fireclay bricks used to construct the kiln insulation on the outside surface are not required When she uses the lm by lm by lm kiln for the first time in a room at 30 C the kiln s outside wall temperature is 90 C Assuming that there is heat loss from the four sides and the top only and that these surfaces have an emissivity of 08 determine a the total heat loss from the kiln in VJ b the total heat loss from the kiln if 4 cm thick insulation with k 004 WmK and s 01 assume the outside brick temperature remains at 90 C c the simple payback time for the insulation if the insulation costs 400 the cost of natural gas is 5050105 kI the furnace has an efficiency of 84 and the furnace operates 2000 hryr Approach This is a combined convection and radiation problem 39 c in part a and conduction convection and radiation E T T Ci BC in part b The natural convection heat transfer 5 coefficient must be evaluated for the two different situations E Assumptions H 39gt D W l M 1 Air is at one atmosphere 2 The kiln is small relative to the large room size Solution a The heat transfer rate is calculated with the convective rate equation for the sides and top and the radiation rate equation assuming the kiln is in a very large room Qh 4 177 ATT JAM1142442 The heat transfer coefficient on the sides is assumed to be that on a vertical wall39 for the top we use a correlation for a heated plate facing upward From Appendix A7 at T lm Z 90 302 60 C 333 K p 1060 kgmg 1 200 x 10395 Nsmz k 00288 WmK Pr 0702 For vertical wall sides 981ms2l333Kl 060kgm32 90730Klm 3 0702 Rs 2 349x109 200X10395Nsm2 From Table 12 3 Nu 011211 3 01 349 x 109 3 152 152 00288Wm2K h5N k 438Wm2K H lm A 4HW m For the top LCM AP HD2H D 025m R117 545 x107 Nu 0151211 3 015 545 x 107 3 569 117 569 00288025 655 WmZK 2 AT lm 39 7 W 2 W 2 78 W 2 4 4 Q7H438 m2K 4m 655m2K lm 90730K08567x10 m2K 5m 90273K 7302731lt 1051 393 2026 w 3470 w 4 Answer b Draw a circuit diagram to evaluate the heat loss when insulation is added From the diagram QM QW Qmd T i T I T 5 1 2041 Tm 7T Am hsAs 6 We want the outside temperature of the insulation but the heat transfer coefficient depends on that temperature so assuming s 44 C TM 2 310 K p 1139 kgm3 1 189 x 10395 Nsmz k 00270 WmK Pr 0706 1269 981131011392 4473013 0706 R115 5 114x109 189x10 NuS 01 114 x109m 104 a 15282 WmZK R117 178x107 NuT 015 178 x107m 392 a hT 423 WmZK Now substituting these into the main energy balance may 117303K 4 4 4 004m 7 1 101567gtlt10 5Tm303 004WmK5m2 28244231 Solving for Tm 3153 K 423 C and Q 0 239 w lt Answer Because the calculated Tm is close to the assumed value no iteration is required 31 W c The energy savings with the insulation is 3470 7 239 32 For 2000 hrsyr in a 84 efficient furnace this is results in an energy savings of 2000hr 323l lk 1000W lkJ lkW 3600 hr lt M W w084 gtlt s gtlt s gt2771Wyr Gas costs 050105 k so total fuel expense savings is 277gtlt107 kJyr050105kl 138yr If the insulation costs 400 the simple payback is 400 29yrs d Answer 138yr l270 12 66 In oil refineries and chemical process plants insulation on pipes is wrapped in thin aluminum metal sheaths to protect the insulation from the weather After weathering and exposure to harsh air borne chemicals around the plants the surface of the metal sheath corrodes and the emissivity is about 04 Consider a 30 cm TD 40cm OD carbon steel pipe k 60 WmK carrying saturated steam at 350 C covered with 75cm of fiberglass insulation k 0036 WmK The steam convective heat transfer coefficient is 600 WmZK With an air temperature of 0 C determine the heat transfer rate per unit meter of pipe length when a a crossflow at 30 kmhr is on the outside of the pipe in Wm b natural convection is on the outside of the pipe in Wm use an approximation based on the results of part a and justify Approach 07 r M Assuming onedimensional heat transfer the base heat g 534 It T 0 transfer rate equation ATRm can be used to m it 0 calculate the heat transfer rate The heat transfer 39 coefficient on the outside must be evaluated and the DI 03939 0 outside surface temperature is needed to determine the D 0 MM radiation heat transfer 7 Assumptions 1 The heat transfer is one dimensional 2 Air is at one atmosphere Solution For steady onedim ensional heat transfer the total heat transfer from steam to outer surface is Q T143 Rm 1 JrlnD2D1lnD3D2 h IrDlL 27rksL 27rkw L where the outside surface temperatures T3 is unknown This heat transfer rate must equal that from the surface to the air QQ39m de h nDLT 7Tgoer3LT34 7 Tj The length L can be factored out of both equations We have assumed that the surroundings are very large and are at the same temperature as the air We have two equations and two unknowns QL and T3 that can be solved simultaneously We can recast the radiation contribution in terms of a radiative heat transfer coefficient T3 7 T Q where h 8039T3 TT32 T2 her3 L h IrD3 L For the forced convection heat transfer coefficient from Appendix A7 and assuming T3 6 7 C T lm 0 DZ 35 C 2 275 K p 1284 kgm3 1 167 x105 Nsmz k 00243 WmK Pr 0715 MD 1284kgm330000mhrlhr3600s055m 1 7 l67gtlt10395Nsm2 R 348gtlt105 0806 13 5 0806 13 From Table 121 Nu 0027Re Pr 0027348gtlt10 0713 706 7 Nuk 7 70600243WmK 7 D 7 055m The resistances are Nu 312Wm2K 7 1 7000177 hstirDlL 600Wm2K7r030mL L 1nD2D17 1n040030 70000763 27rk5L 2n60WmKL7 L 1271 1nD3D27 1n055040 71408 27rkm5L 27r0036WmKL L 1 7 1 7 00185 hnD3L 312wm21ltn 055mL 7 L With a first approximation of T3 7 C 280 K h 04567gtlt10398Wm2K4ZSOK 273K280K2 273K219Wm2K 1 7 1 7 0302 Solving the two rate equations we obtain T 3 3379 K This does not match our assumed value lterating on T3 we obtain h 246Wm2K 1 7 02351 h 7rD3L L QL 210 Wm 4 Answer b F or the natural convection case if we assume hm 0 lhfmgd as a first approximation then solving for the heat loss as above T3350K77 C h 279 WmZK 1 7 02079 h 7rD3L 7 L QL194 Wm d Answer Comments Because the heat loss per unit length does not change much with a factor of ten change in the air side heat transfer coefficient h a more accurate estimate of h is not needed 1272 12 67 The power cables for an electric welding rig are suspended above the floor of a factory so that a tripping hazard is not created The cables are 25m long the copper is 10mm in diameter and is covered by a 2 mm thick rubberized cover k 026 WmK which is black has an emissivity of 09 and cannot have a temperature greater than 65 C The cable resistance is 4 X 1039 ohms 1f the cable is suspended in calm air at 27 C determine the maximum allowable current in A Approach T Soc Because the cable is long we assume onedimensional 1 5 quot 95 heat transfer This is a combined natural convection and radiation problem We will need to estimate the surface tquot oOOZ39 temperature to obtain the natural convection heat transfer coefficient and the radiation An iterative solution may be D1 D 4 required Once the heat transfer rate is determined the maximum allowable current can be calculated 8 o la Assumptions 394Q X10 1 The air is at one atmosphere L 25 M R 439 2 The heat transfer is one dimensional Solution The total heat transfer rate is calculated with Q Qmd Qmv Qmd Assuming the surroundings are large and at the same temperature as the air and all properties are constant 27rkLT 7T 4 4 5 hA T crA T 7T 1nD2D1 T f 5 f The convective heat transfer coefficient must be determined with fluid properties from Appendix A 7 evaluated at the film temperature T lm T Tf2 We assume T Z 45 C 318K T lm Z 27 452 36 C m 310 K so p 1143 kgm3 k 00268 WmK u 190 X 10395 Nsmz Pr 0711 When calculating the Rayleigh number we will keep these properties constant and only change TS during the iteration This is for simplicity only g 32TTD Pr 2 981ms2 T 3001143kgm3 2 T3000014m2 0711 1138510 5 190x10 5Nsm22 X For T 318 K Ra 4035 so from Table 123 Ra 717300 T5300 2 6 0387 4035 6 Nu 0 1 0387B 0 1 354 827 2 10559Pr9 6 1055907119 6 hiNuKi 35400268Wm1lt T D T 0014m Therefore the right hand side of the heat transfer rate equation is Q 678 WmZK 1 0014 m 25 m 318 7 300 K 09567gtlt10398Wm2K4 n0014m25m3181lt 7300K4134211932535W 27r026WmK25m ln00140010 These two heat transfer rates do not match so an iteration is required Doing so we obtain T5 3336K606 C Ra7350 Nu406 h778Wm2K Q529W 678Wm2K The left hand side is 338318K2428W Electric power is 12R where I is current Solving for current I QR05 529w4x10 205 1150A 4 Answer 1273 12 68 A manufacturer of prefabricated buildings is considering using the same structure shown below for walls and roofs for quickly and cheaply assembled buildings The inner and outer surfaces are l27cm thick plywood k 0115 WmK and the airfilled gap is lOcm wide the panels are 25m long and 125m deep out of the plane of the page For temperatures on the outside of the two plywood sheets of lO C and 15 C determine a the heat transfer rate for both horizontal and vertical orientations in VJ b the effect of inserting a baffle at midheight for when the panel would be used in a vertical orientation in VJ Approach lo an Assuming steady onedim ensional heat transfer the basic rate equation ATRmt can be used to calculate the heat transfer rate The natural convection heat transfer coefficient inside the wall must be estimated 5 w n 5 1C pr 112mb Assumptions 1 The heat transfer is one dimensional 2 Air is at one atmosphere T 40 3 Radiation is ignored l ltm7 CM 4 Heat transfer from the ends is ignored Solution Assuming adiabatic ends and that radiation can be ignored the rate equation is Q 5 1 Run RWALLJ RWALL2 Rcanv The wall resistances are I 00127m R R 00353KW W W kA 0115WmK125m250m a The convective resistance in a vertical enclosure requires the natural convective heat transfer coefficient with properties evaluated at the average of the two wall temperatures We first estimate the heat transfer rate b assuming the convective resistance is equal to twice that contributed by the plywood and then use this estimate to calculate the inside wall temperatures 157 710 K Q2 M177W 40035kW The temperature drop across both sheets of plywood is the same so TAT QM 5110 a T27T2VQmtkA177W00353KW63K So Tm Z 10 63 37 C and T2 S15 7 63 87 C Therefore from Appendix A7 at TM 737872 25 c2751lt p 1284 kgm3 1 167 x105 Nsmz k 00243 WmK Pr 0715 R g p2T2Vm7TlvmSgPr 981ms21275Kl284kgm328737010m30715 187106 a 1 X 5 2 167x10395Nsm2 From Table 126 Nu0073Ra3 HSfm 0073187gtlt106m 250010 9606 606 00243W K hN kM147wm31 s 010m 1 mmmw 147Wm2Kl25m250m Therefore for vertical configuration 1274 157710K Q 868W 200353KW0217KW Calculating a new Tm 69 C and Tzfm 119 C Ra5284gtlt106 Nu723 h176 WmZK Q 989w i Answer We could iterate again for an improved estimate For the horizontal configuration from Table 126 and using the same Ra as for the vertical configuration Nu0061Ra 3 0061284x106m 864 864 00243 h N k 210Wm2K s 010 1 1 0152KW hA 210Wm2K125m250m 157710 112W Answer Q 2003530152 b If a baf e is inserted in the vertical wall the only change in the calculation of the heat transfer coefficient is in the term HS391 9 kW 7HSiji 125 quot1108 39 mama 108 176 WmZK 190 WmZK Q 105 W Answer 1275 12 69 A thermal pane window often is constructed of two panes of plate glass separated by a short distance this arrangement provides increased thermal resistance compared to a single pane of glass Consider a 4 ft wide and 6ft high window whose inside wall temperature is 75 F and outside wall temperature is 45 F It is constructed of glass that is 025in thick Determine the heat transfer rate a for a single pane of glass in Btuhr b for two panes separated by a lin air gap in Btuhr c for two panes of glass with if a thin 0lin sheet of glass is inserted between the other two panes that are still separated by l in in Btuhr Approach Assuming onedimensional heat transfer the heat transfer rate is determined by the rate equation Q ATRm This thermal resistance can be found for FIT l reels 2 gem combinations of IkA and lhA Radiation is ignored T1 WP Assumptions 1 L 1 Air is at one atmos here between the two anes of glass p p gt l gtl lr 2 Radiation is ignored t l 39 03925 N39 S I quot 39 H L l w 49 Solution a The thermal resistance of a single pane of glass is REM IkA From Appendix B3 for plate glass k 08 Btuhr ft F and 02512ft 0001085hr FBtu R g1 08Btuhrft F6ft4ft The heat transfer rate is Q ATRglm 75 e 45 F0001085 hr mam 27650 Btuhr Answer b Resistance of two panes of glass and an air gap is Rm ZREW lhA For air properties from Appendix B7 at TaVg 75452 60 F p 0077 lbmft3 k 00146 Btuhrft F u 1214x10 5 lbmfts Pr 072 MM Tm T2m 5313r 2 The temperatures to use are the inside wall temperatures We could estimate the temperatures from the glass resistance Let s assume for the air gap that resistance is approximation equal to that contribution by both panes of glass Therefore 7 75 45 F 6912 4Rg1m 40001085hr mam hr Now calculate the AT across the glass Ra ATQWW Rm 6912Btuhr 0001085hr FBtu 75 F So Tm 75 775 675 F and Tm 45 75 525 F 322fst 1520R 0077lbm ft3 2 6757525 R 112 ft 3 072 D lt gt lt 2 gtlt 5570 1214x103951bmfts From Table 12 5 Nu0197Ra HSfm 01971557o 611239 9 1369 1369 00146Btu hrft F hN kM 024Btuhlft2 F 112ft 1 1 o1738hr0FBtu hA 024Btuhrft2 F6ft4ft 1276 75 45 F m 0001085hr FBtu0l738hr FBtu hr We need to reestimate h Because of the symmetric system the uid properties are the same so ATglm 170 0001085 0185 F T1quot 748 F T2quot 452 F Ra 30720 Nu 1623 h 028 Btuhr ftz F lhA 0147 hr FBtu and Q 2017 Btuhr Answer Therefore Q 2 c Adding the thin glass between the two panes changes the Rayleigh number because the characteristic length is changed as is the AT We will keep the same fluid properties Because of symmetry half the temperature drop occurs on each side of the thin glass partition so D 322152000772 757 6017012123 072 1419 1214x10 52 From the Table 125 Nu l h100146 039 Btu hrftUF 212 0 L 1 0107 hI hA 039Btuhrft0F6ft4ft Btu AT Q 2Rg1a55 2Rcanv Rpamtum 75745 F 138 1 Answer 0112ft hr 2 0001085hr 0F Btu 2 0107hr 0F Btu 08Btuhrft0F6ft4ft Comments Using a double pane window reduces the heat loss substantially Inserting a thin glass partition between the two other panes of glass further reduces the heat loss by 32 Another technique to reduce heat loss is to lower the interpane pressure and to use a gas such as argon with a lower thermal conductivity 1277 12 70 Flat plate solar collectors have their best efficiency if they are tilted toward the sun at an angle that equals the latitude of the location of the collector Consider a 3m wide and 2m high solar collector The solar absorber plate maintained at 65 C is separated from the glass cover plate which is at 30 C by a distance of 5cm Determine a the heat loss from the collector if it horizontal in W b the heat loss from the collector if it is tilted at an angle of 33 from the horizontal in VJ Approach mini b Assum1ng steady onedimens10nal heat transfer the Tquot ZTGC 1 Ti 6311 bas1c rate equat1on Q ATRmt can be used to h Dwnqx p 5 me calculate the heat transfer rate The natural a convective heat transfer coefficient in the enclosure 1 30 must be determined mm M w 3M mil op I c Assumptlons P M 1 Air is at one atmosphere Solution a Assuming constant properties no radiation and no heat loss from the ends on back the heat transfer from the absorber plate is hA Tp To The natural convective heat transfer coefficient is calculated with properties from Appendix A 7 evaluated at T T 712 475 c 3205 K p 1103 kgm3 k 00278 WmK 1 194 810395 Nsmz Pr 0704 avg PM g p2 T 483Pr 981ms2 l3205KllO3kgm3 2 65730K005m3 0704 3 05 105 gtlt 2 194x10395Nsm22 From Table 126 we see that we are near the dividing Ra for two correlations We use the one in our range but understand that there may be a larger uncertainty than if we are not near the end of the range of applicability Nu 02121211025 0212 305 8105025 498 hi Nuk 7 49800278 WmK T s 7 005m Q 277 WmZK 3 m 2 m 65 4 30 K 582 w 4 Answer b For an angle 330 from the horizontal we can use gcos in the Rayleigh number rather than just g Note 9 90 39 57o Ra Ratcos6305X105cos57 166X105 Nu 0212 166 8105025 428 428 00275 rpm L 238Wm2K 005 Q 238 3 2 65 7 30 500 w lt Answer 277Wm2K 1278 1 1 For the following systems define a control volume and state whether the system is open or closed and steady or unsteady Identify any and all heat transfer energy flows mass flows and energy transformations flows cket b Pot of boiling water with no lid c Portable space heater with fan d The jet airplane in Figure 11 c e The house in Figure 14 Approach For each system identify a control volume with a dotted line and then describe the various processes occurring to the system Solution a Define the control volume to surround the rocket Because the hot exhaust gases cross the boundary this is an open system As the propellant burns a mass flow exits through the rocket s nozzle and the mass of the rocket decreases changes with time so the system is unsteady The nozzle is very hot compared to its surroundings so there is heat transfer across the boundary but it is negligible compared to the energy flow associated with the hot exhaust gases The stored energy in the rocket propellant is converted to kinetic energy in the rocket s nozzle b Define the control volume as shown in the figure to the right Heat transfer crosses the boundary from the gas ame to the pot bottom Because the water is boiling water vapor crosses a mass flow rate the boundary at the top of the pot so the mass of the system decreases with time so this is an unsteady system The vapor leaving the system removes energy from the system Note that if the water was not boiling and heat added then the mass of the water would remain the same but the temperature of the water would increase with time so the system would still be unsteady c Define the control volume as shown in the figure Electricity crosses the boundary in the wire39 electricity is considered work Air is sucked into and pushed out of the heater by the fan39 this is the same mass flow rate crossing the boundary in two locations Because neither air nor energy is stored in the heater and the air flow is constant this is a steady system Note that once the electricity is inside the heater some of the electricity is converted to mechanical work to drive the fan and some electricity is dissipated in the heater to heat the air Because neither of these conversion processes results in work or heat crossing the boundary we have defined we do not have heat transfer or a second work process Electricity mechanical work is converted to thermal energy in the flowing air and kinetic energy in the air d Define a control volume to encompass the airplane The processes are identical to those identified in part a for the rocket e Define a control volume as shown in the figure to the right Heat transfer enters the house due to the solar radiation Assuming the air outside the house is warmer than the air inside the house there also is heat transfer from the surroundings into the house Electricity enters the house to drive the air conditioner electricity is considered mechanical work A mass flow of refrigerant enters and leaves the house at different conditions where the control volume cuts through the two pipes Assuming the air temperatures solar radiation and flow rate of refrigerant are constant this is a steady system 1 2 Describe some of the thermalfluid systems in a typical residence define a boundary and describe the energy andor mass flows associated with them Approach Chose several systems identify a control volume with a dotted line and then describe the various processes occurring to the system Solution a Define a control volume to surround a refrigerator as shown in the first diagram39 the control volume passes along the surface of the heat exchanger at the back of the refrigerator Assume the cooling unit is operating Electricity crosses the boundary39 electricity is considered work There is heat transfer from the warmer room to the colder interior of the refrigerator As the cooling unit is running for example to freeze ice cubes the total energy level in the refrigerator drops so that the system is unsteady If the cooling unit is not running there is no work crossing the boundary but heat transfer still occurs from the warm er room to the cooler refrigerator interior The energy level in the refrigerator increases so the system is unsteady Note that no mass flows across the boundary so this is a closed system Note that the cooling unit has a heat exchanger often on the back of the refrigerator that transfers energy from the hot refrigerant to the cooler room Depending on where the control volume surface is drawn the processes occurring will change Defining the control volume to pass along the surface of the heat exchanger then this there is a heat transfer process with heat flowing from the surface of the heat exchanger to the room air but no mass flow so this is a closed system If the control volume is drawn farther away from the heat exchange surface to include the air then there is a mass flow of air at different temperatures into and out of the control volume but there is no heat transfer39 this results in an open system If the control volume is drawn to cut the tubes conveying the refrigerant then there is a mass flow of refrigerant at different condition into and out of the control volume but no heat transfer or mass flow of air but again this is an open system b Define a control volume around a gas oven as shown on the figure to the right and assume the oven has been just turned on Gas flows mass flow rate across the boundary39 so this is an open system Electricity is also flows across the boundary so there is a work process Air is needed to burn the gas so cold air enters and hot exhaust gases exit the control volume so these are two more mass flows carrying energy that need to be taken into account The surface of the oven is warmer than the room so there is a heat transfer process With time the oven heats so this is an unsteady system If the oven has been on for a long time and the desired oven temperature has been reached the oven thermostat maintains the fixed temperature The gas and air flows remain constant and the temperature is fixed so this then would be a steady open system c Define a control volume to encompass the lightbulb No mass crosses the boundary so it is a closed system Electricity flows to the bulb so there is a work power term The bulb is hot so there is heat transfer from the bulb to the surrounding air Likewise some energy leaves the systems as visible light When the lightbulb is first turned on it takes a little while before it reaches a steady temperature so it would be an unsteady system After a long time operating the temperature would stabilize and the system would be steady side v1qu 01C La uf u 4 1 3 For the following four systems define a control volume state if the system is steady or unsteady open or closed constant volume or changing volume constant uid density or changing uid density Also identify all heat transfer energy flows and mass flows a Swimming pool being filled Choose one control volume as the whole pool then choose a second control volume one surface of which follows the surface of the rising water b Helium tank being filled c Helium balloon being filled Approach For the given systems identify a control volume with a dotted line and then describe the various processes occurring to the system Solution a Define a control volume to surround the physical pool envelope the concrete tile etc and cutting across the top opening as shown Water flows in so mass crosses the boundary The total mass of the system is increasing so this is an unsteady open system The rising water level 39 displaces air so air leaves the system However the air mass flow rate is very small compared to the water flow rate and can be assumed to be negligible We assume the water has the same temperature as the pool envelope so that there is no heat transfer process The total volume of the system remains constant but the mass increases The density of the water is constant Mechanical work crosses the boundary because the increasing volume pushes against atmospheric pressure39 force is pressure times area and work is force times distance but this work would be very small Define the control volume to include only the water in the pool During the filling water crosses this boundary so this is an unsteady open system Unlike the previous control volume air does not play a roll in the present defined control volume The volume of the system is changin with time39 both the mass and the energy in it increase Mechanical work crosses the boundary because the increasing volume pushes against atmospheric pressure39 force is pressure times area and work is force times distance but this work would be very small The density of the water in the control volume remains constant b Define a control volume to follow the inside surface of the helium tank Mass enters the control volume from the filling port so this is an unsteady open system Energy flows along with this mass The volume of the system is fixed The density of helium increases with time With fixed volume no mechanical work occurs As the gas is inside the tank is compressed the gas temperature will rise If the temperature of the tank becomes greater than that of the surroundings then there will be heat transfer c Define a control volume to follow the inside surface of the balloon Mass enters the control volume at the balloon s opening so this is an unsteady open system The volume of the system increases The density of the helium also increases slightly with time With the increasing volume mechanical work occurs across the boundary but this would be very small The gas compression is small39 any heating of the gas inside the balloon due to the compression would be small so heat transfer would be negligibly small 14 A thermal solar energy system consists ofa solar collector on the roof ofa house a hot water storage tank to store hot water a heat exchanger through which the hot water passes a fan that blows air through the heat exchanger to heat the house and a pump to circulate water through the complete system Define several 439 quot 39 quot 39 39 fe uipment or collections of equipment and identify if the control volume is steady or unsteady open or closed what heat transfer energy ows mass ows and energy transformations occur and if constant or varying volume Approach Identify several control volumes in this thermal solar energy system with dotted lines and then describe the various processes occurring to the systems Solution a Define a control volume to surround the solar collector on the roof Solar energy enters the system this is a heat transfer process Assuming the collector surface is a temperature greater an the surrounding air there is a second heat transfer process The pump circulates water through the collector Assuming the solar radiation the water ow the surrounding air temperature an the entering water temperature are constant this is a steady open system The volume is xed the water entering the solar collector is not constant that is we are trying to increase the water temperature in the storage tank and everything else is constant as in the first part ofthis answer then the system would be unsteady because the overall r level ofthe cuneuur 39 39 b Define a control volume to encompass the pump and is electric motor Assume they are running at a constant speed Electricity crosses the boundary electricity is considered mechanical work Water enters and leaves across the boundary mass ow rates This is an open steady system c Define a control volume to encompass the storage tank Assume it is well insulated so there is no heat transfer from it u t I 1 temperature This is an unsteady open system Volume is fixed No mechanical work occurs d Define a control volume to surround the fan and heat exchanger Air and water ow into and out of the control volume Electricity work crosses the boundary to drive the fan This is an open steady system 15 In hydroelectric plants electric power is generated from the ow ofwater from a reservoir such as shown in Figure 118 The water ows continuously with a seemingly endless supply How is the wa er replenished Where does the energy in the water come from that is converted to electrical power Approach The question can be answered with either an open system analysis or a closed system analysis For the open system choose a control volume that includes the dam the reservoir and the river waterjust downstream ofthe dam For the closed system choose a control volume that includes all the water on the planet Solution a De ne a control volume that includes the dam the reservoir and the river waterjust downstream f the dam 39 39s sy m w39 leaves via the down en the control volume at a higher ow rate than it leaves water level falls Lfthe ow rates are equal e sy te is in steady state This choice ofcontrol kinetic energy of the downstream river water b De ne a control volume that includes all the water on the earth No water enters or leaves this control volume so it is a closed system The water at the base of the dam ows down the river and empties into an ocean Water evaporates from the ocean surface and is suspended in the air as water vapor and clouds The water is transported over the land by air currents and eventually falls to the earth as rain snow or hail The water runs over the land and soaks into the soil Some water is st snow or ice and later melts into liquid form Both location upstream ofthe dam and acts to replenish the supply of water An energy balance on the control volume reveals that heat is added to the earth s water by the sun and by intemal heat generation within the arth s crust ssion reactions Work is done by the hydroturbine in the power 1 shown in the gure driving water circulation on the earth There are of course many more eat and work interactions with the earth s water than the two shown 1 6 The radiator of a car is a heat exchanger Energy from the hot water that flows through the heat exchanger is transferred to the cooler air that also flows through the radiator For the three control volumes defined below state if the system is steady or unsteady open or closed and what heat transfer energy flows mass flows and energy transformations occur a Water b Air c Complete heat exchanger Approach For the three systems specified clearly identify the control volumes with dotted lines and then describe the various processes occurring to the systems Solution a Define a control volume CV l to surround the water flowing through the car radiator Assume the air flow rate and inlet temperature are constant39 also assume the water flow rate and inlet temperature are constant Mass water flows across the boundary so this is an open system Heat flows from the water across the boundary to the air so there is a heat transfer process Everything is constant so this is a steady system No mechanical work occurs b Define a control volume CV H to surround the air flowing through the car radiator Assume the air flow rate and inlet temperature are constant39 also assume the water flow rate and inlet temperature are constant Mass air flows across the boundary so this is an open system Heat flows from the water across the boundary to the air so there is a heat transfer process Everything is constant so this is a steady system No mechanical work occurs Note that Q 7 7Q because of the sign convention c Define a control volume CV 1 CV H to encompass the complete heat exchanger and assume the outside surface of the heat exchanger is well insulated Assume the air flow rate and inlet temperature are constant also assume the water flow rate and inlet temperature are constant Mass air and water flows across the boundary so this is an open system Everything is constant so this is a steady system No mechanical work occurs Note tha there is no heat transfer Heat transfer occurs only across boundaries The energy flowing from the water to the air is internal to the control volume and not across a boundary so by definition there is no heat transfer 39 v39v cod mL QJZZ I CV IE Iquot A 1 17 An acom is planted in the ground After many years the acom grows into a mighty oak tree De ne a 5y 7 1 39rlxxL I39AL 39L 0 Approach n v transformations Solution n c 4 L r u 39rL r 7 since mass in the form of air water and organic materials crosses the control Volume over time The tree grows so the system is transient Sunlight enters the control Volume and provides energy for photosynthesis This energy is stored in the tree as chemical energy in the molecules The tree is composed of organic molecules with four main constituenm carbon hydrogen oxygen and nitrogen The tree absorbs carbon dioxide C02 from the air reduces it and returns gen to the atmosphere The tree also draws water H20 and dissolved nutrients which contain nitrogen through the room trunk and branches Much ofthe mass in the tree comes from the carbon in the surrounding air 39 and oil Discuss mass ow and energy t 4 l rowsW dull I NANA l Fcrlxly u l EQ Wafer below state ifthe system is ows and energy tran rmations occur 9 n c Complete turbine d All the equipment shown Approach A Rankine cycle power plant is shown schematically in Figure 113 For the control volumes de ned 39 39 steady or unsteady open or closed and what heat transfer energy ows mass For the four systems speci ed clearly identify the control volumes with dotted lines and then describe the various processes occurring to the systems For all the systems we assume that they operate at steady sta e Solution a De ne a control volume surrounding the electric generator Mechanical energy crosses the system from the turbine that drives the electric generator mechanical energy leaves the controlvolume in the form of electricity Not all ofthe mechanical energy from the turbine is 4 quot moto39 39 the electricity is converted to mechanical energy so the generator is hotter e surrounding air and there is heat transfer Generators also are actively cooled by passing some sort of uid water or air or other gases through them so m often a generator is an open system Mechanical energy is converted to electricity another form of mechanical work and heat b De ne a controlvolume aron the steam generator Fuel and air enter the steam enerator and exhaust gases leave Water steam enters the system at two locations and leave at two locations Hence this is an open steady system Ifthe oumide surface ofthe steam generator is heavily insulated then there is not heat transfer if it is not insulated there would be heat transfer from the hot steam generator to the colder surroun ing air No mec anical work occurs The owin ermal energy in the hot gases formed by combusting the fuel and air is converted to owing thermal energy in the steam c De ne a control volume surrounding the steam turbine Assume the turbine is heavily insulated so that there is no heat transfer Steam enters and leaves the turbine at di erent conditions so this is an open system Mechanical energy leaves the turbine where is shaft crosses the boundary Th owing energy in the steam is converted to mechanical energy in the turbine d De ne a control volume around all the equipment shown Mass crosses the boundary in three locations air the and exhaust gases Electricity mechanical energy crosses the boundary in one location eat transfer crosses the bound at the steam generator and steam turbine if not well insulated at the electrical generator and at the condenser The owing thermal energy in the hot gases formed by combusting the fuel and air ultimately is converted to electricity mechanical work in the generator and heat 5533 w i eleci vur39iw gt var biogeuw r E 5355 A56 0 4 19 A vaporcompression refrigeration cycle similar to what is used in air conditioning systems is shown schematically in Figure 15 For the control volumes de ned below state ifthe system is stea or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations 0 a Electric motor b Refrigerant owing through condenser e All the equipment shown Approach Forth f stems specified clearly identify the control volumes with do ed lines and then describe the various 39 t e ive sy processes occurring to the systems For all the systems we assume that they operate at steady sta e Solution a Define a control volume surrounding the electric motor Mechanical energy crosses the system from the turbine that drives the electric g tor mechanical energy leaves the control volume in the form of electricity Not all of the mechanical energy from the turbine is converted to electricity feel a motor it gets hot since not all the electricity is 39 energy L i IIULLCI uiaii e ui undin aii 39 Generators quot cooled by passing some sort of uid water or air or other gases through them so most often a generator is an open sys em b Define a control volume around the refrigerant owing through the condenser The w rate is constant and crosses the boundary so this is a steady open system There is heat transfer from the refrigerant to the surrounding environment No work occurs c Define a control volume around the complete condenser Refrigerant enters and quot f 439 39 and at a steady ow rate Air ows through the condenser crosses the control volume boundary and removes energy from the refrigerant at a steady rate Hence this is an open steady system Assume the condenser is insulated By definition there is no heat transfer between the a 1 L I nu vv volume b oundary d Define a control volume surrounding the throttling valve Assume the valve is heavily insulated so that there is no heat transfer Refrigerant enters and leaves the valve at different conditions so this is an open system No mechanical work is done e Define a control volume around all the equipment shown Mass air crosses the boundary in four locations in and out at both the evaporator and condenser There is no heat transfer Electricity work crosses the boundary at the electric motor 110 A hot cup of coffee is placed on a tabletop to cool De ne a control volume and state ifthe system is steady or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations occur Approach 39 the cup p quot and one Solution De ne a control volume around the coffee in the cup as shown Ifwe take a relatively short time period such as one hour we may assume quot quot f 39 closed system The system is unsteady because the coffee cools during the hour 39 r ro quot quot convection and heat conducm into the sides and bottom ofthe coffee cup No wor is done on or by the coffee The internal energy ofthe coffee decreases as heat is transferred from it Ifwe choose a relatively long time period such as a week then we need an open system analysis During the week the coffee will not only cool but also evaporate This is an unsteady open system since mass crosses the control volume as water vapor Energy leaves by heat transfer as before and also is removed by evaporation 1 11 The water in a canal lock is at the downstream river level and the gates are opened A boat enters the locks and the downstream gates are closed A valve is opened and water from upstream flows into the lock raising the boat After the water reaches the upstream river level the upstream gates are opened and the boat travels upstream Finally the first valve is closed and a second valve is opened allowing the water in the lock to drain to the downstream river level Another boat arrives from downstream and the process is repeated Neglect the energy required to open and close the gates and valves Where does the energy come from to raise the boat Approach Define a control volume around the canal lock Alternatively draw a control volume around the entire earth to show the primary source of the energy used to raise the boat Solution Define a control volume around the canal lock This is an open transient system with water and boats entering and leaving In the context of this control volume the energy to raise the boat com es from the potential energy stored in the water upstream of the canal We may also choose a control volume that encloses all the water on the earth This is a closed transient system Heat enters this control volume from the sun and also from nuclear reactions within the crust of the earth The heat energy evaporates water from the surface of all the bodies of water on the earth The water is transported throughout the earth by wind and returns to the land as rain snow or hail Water runs into the river upstream of the canal and acts to raise the boat Thus the ultimate source of the power to lift the boat is sunlight and geothermal energy 112 A closed pan of cold water is placed on a burner of an electric stove which is already turned on For the control volumes de ned below state ifthe system is steady or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations occur b Burner c Pan ofwater plus burner Approach For the three systems specified clearly identify the control volumes with dotted lines and then describe the 39 s various processes occurring to the system Solution a Define a control volume CV I surrounding the pan of water Heat crosses the boundary into the control volume from the burner beneath the pan Because the cold water does not boil initially no vapor leaves the pan so the mass is constant but as the temperature ofthe water increases the energy level increases Thus this is a closed unsteady system No mechanical work occurs As the temperature of the waterpan combination increases there is heat transfer from the pan to the surrounding air The heat transfer is converted to internal energy b Define a control volume CV 11 around the burner on the electric stove Electricity e ectricity 39 39 39 WUIK Heat transfer crosses the boundary from the heater to the bonom of the pan TL 39 L 39 burner o the is attained quickly so this is a closed steady system The electricity mechanical energy is converted to heat Note that Q Q because of the sign convention c Define a control volume CV HI around the pan ofwater and burner l Electrici e ec ici WUIK ecause the cold water does not boil initially no vapor leaves the pan so the mass is constant but as e temperature of the water increases the energy level increases Thus this is a closed unsteady system As the temperature of L39 39 increa 39 39 JIUIII the pan to the surrounding air The electricity mechanical energy is converted to internal energy 113 Water from 39 a nliei p If 39 the water in the pool and filter is this an open or closed system Lfthe system is just the water in the filter is this an open or closed system Approach Ifmass crosses the control volume the system is open otherwise it is closed Solution For the first control volume which is all the water in the pool we have a closed system We are assuming that no mass enters or leaves the pool by evaporation rainfall or addition of city water from a hose e second control volume which is the water in the filter we have an open system Water enters and leaves the filter crossing the control volume as it does so 114 Wind turbine systems such as shown in Figure 11b consist ofa wind turbine an electric generator 39 39 and eiLh r power quot 39 L 39 grid or to banery storage In a steady wind for the control volumes de ned below state if the system is steady or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations occur a Wind turbine B b a ery c Electric generator d Wind turbine electric generator and electrical grid e Wind turbine electric generator and battery Approach For the ve systems speci ed clearly identify the control volumes with doned lines and then describe the various processes occurring to the systems Solution a De ne a control volume surrounding the wind turbine as shown in the gure to the right Air enters and leaves the control volume at different conditions 39 i 39 39 generator For the steady wind this is a closed steady system No heat transfer occurs The kinetic energy in the wind is converted to mechanical energy in the turbine b De ne a control volume around the battery Electricity enters or leaves the battery depending on the operation of the system so the stored energy level changes No heat transfer occurs No mass crosses the boundary so this is closed unsteady system The electricity mechanical energy is converted to stored energy in the anery c De ne a control volume around the outside surface of the electric generator Mechanical energy is transferred to the generator from the wind turbine Not all the mechanical energy is converted to electricity touch an electric motor not all of the electricity is converted to mechanical work so the generator heats up and there is heat transfer from the generator to the surrounding air No mass ows through the generator control volume so this is a steady closed system d De ne a control volume around the wind turbine electric generator and electrical grid For a steady wind a steady ow of electricity mechanical work ows from the generator into the grid but since we include the grid in the control volume no mechanicalwork occurs Air ows into and out ofthe control volume Assuming all the electricity is used no energy is stored in the grid likewise the mass and energy level ofthe control volume are constant so this is a closed steady system If all the electric energy is not used but some is stored or used from the battery then this would be a closed unsteady system The kinetic energy in the wind is converted to mechanical energy in the turbine the mechanical energy is converted to electrical energy another form of mechanical energy in the generator e De ne a control volume around the wind turbine electric generator and battery Air ows into and out ofthe control vo ume The mass ofthe system s constant Electricity may ow out of the system if the grid demands it or the energy may be stored in the battery if the grid does not require it Thus this is a closed system but it could be either steady or unsteady depending on the operation The kinetic energy in the wind is converted to mechanical energy in the turbine the mechanical energy is converted to electrical energy another form of mechanical energy in the generator 1 15 Global warming has been in the news much in recent years Define an appropriate control volume to study this system and state whether it is steady or unsteady open or closed and what heat transfer energy flows mass flows and energy transformations occur Approach If mass crosses the control volume the system is open otherwise it is closed Solution Choose the entire planet earth as the control volume Neglecting the small amount of mass that enters via meteorites and solar wind and the small amount of mass that leaves as interplanetary probes this is a closed system Sunlight enters the control volume is transmitted through the atmosphere and absorbed at the surface of the earth The earth reemits radiation39 however the reemitted radiation has a much longer wavelength than the incident solar energy The atmosphere is largely transparent to solar radiation but partially opaque to the longer infrared wavelengths As a result reemitted radiation is absorbed in the atmosphere and global warming occurs Global warming is caused by the socalled greenhouse gases which include water vapor carbon dioxide methane ozone and certain refrigerants These are gases that are transparent to incident solar energy but absorb infrared radiation reemitted by the earth The increase in concentration of some of these gases over time leads to global warming Especially important is the concentration of carbon dioxide which has increased substantially since the industrial revolution due to the burning of fossil fuels and deforestation Water vapor is also an important factor since as atmospheric temperature rises the concentration of water vapor in the atmosphere increases Increases in methane concentration result from increases in the cattle population Volcanic eruptions also added gases and particulates to the atmosphere that can be important in the global energy balance Since the earth s temperature changes over time scientists estimate an average rise of 05 C over the last century this is an unsteady system The fundamental energy transformation is the conversion of incident solar radiation to stored internal energy in the earth s atmosphere Question 1 EXAMPLE 14 Chaim ul 30mm Volume m Mnmenlum Analysis Water fmm a Minnan 321139 strikes lm plane as sham The water leaves the nozle at 5 mfs the IMJZTJE area is 001 m Assuming the walm is directed normal to ma ptm mm ows along the man d ermme the h mn39a lm 0 be quot392 Nggzte support EmuFm Panatm M GWEN Walt um 21 slmimmy DUDE in inchd mornl In Ihr plane submuent aw is parallel in plate kt velocity V liftW39s Noam sum 4 00 m1 HM Hurizonul farce cm mg swoon SULUTIOM We chose a mordmase sywmn m d ning the problem abawe We must now change a sumblc cnmml 139 Two paysibln chuices an show by the dashnd 11mm beiuw uh1mmmmmuw uwualkwnmln smuhmhnunndzqmlulh mk wi umnudwluwl lmawhdm n bh xu numyuquothm Wm mmmwamauamu 1 A A 2 F F vauvLv M m1 h pawL JIa m mumuhmmm mww Warmmmymmm mm mm munme hum EFII yaw m Uni o a hennw lhdnhmmlwmmummmm um mumomwnuunnMEnmammamuwm aunt mmmm mmm mmmwMMmmrmnh nun aw Kwnwmmmmkmvamlnmeulmwmlwlm hQNMuthRHdNHM mmm4mmzmllk k mm Chmalum Mummymuumuuu run1 j m a 5 mm ummuhhulnmulmumumemam Pu h 39 IA 39 Iv h mMmW mm mmquot mum mm Ammw mquot Kw quotm A V lt a Mwwmm M Ly m MQDV M mammme 39 LquotH39 quot mmme 1 mm dmmm m m mu Ill mmnmmuummawmmx n mn fummwnzumnnnnk 1 mm um HaremI ramsm mmmnmwunm mummmummumnmqmma murmMMMMhu mwmvnm nmmmulmmmplm mammmp kdn w Whamnk nnmmnumlaummmlthaunrl IraIncu We mamnrmmlvulmx mmm ThMylonuumsauwlidmmmtmmkxm Mnlmllmmmuumm yua n M n my 1 I ulVn39uAD mm y4 735 uumwmmundlknwyawulmm 1 M u Mf 275m zLm gummuymmmux a Lulu Mmmm f nest on 1 EXAMPLE 45 Tank nn Scale army Force A mm canumcl z n mm mm in msid cmssseclimu ma uf 1 n1 weigh 5 IN when amply The mnmm 1 mu m a an and mm ows In llmmgh an opening in m mp and mu mmugn mu m aquz rucl upcmngs in m we as x F L9 1 Damnun me rudmg on m mic Ax A3Dl 1 EXAMPLE PRO LEM nww Mull wnlninn 1mm 2 n m mxSsccmnnl m A 1 hquot mm 5 mm mm um ulmvulsnnml Undarslurlyllawwn llwmunlrrdzpmhn m Wageunlknwmu y u xwm nnd knubmmuny thugh mum und mm 5mm wumnu um vnlunu mwmnwm m1 bummd yum mandamus pmwwmuummmwummuWW Am zmlmu m yw mukmm hmhum Mmmhmn nlwhlm utbmmlmmwmmlwlmmumnwmgym mm 01 mi 1 muf m DUI h vmq w u lumrmludnnlpmm Mummmmmummmmwwuu mdumwmwlmwnmmMWumMnHJK Ikm kmmm ngmmunhnwhmmnwmnwmkmhku a u mm1 mam1mmumummmmammuw mmnmm mum m sway mm m uraniumWNW 3 Ummum mmm mmk mu Wewmmymomntwmmummm n 4 me39 n x mrzismnclreruduelannlmhdiuv n 7w wn mmimm 39 y wlun NV VM I 7454 I39pV axf vlwmn u m 4 V mm Mmmanumummpnhn Subununnnnmhl HM l W m IKDIVIAI u 1 1 is n Bil 53m xix asl uumzrz an ai l atlllrlnlrinl Illllliltrn 9 Lil ITIn rIIIIIYIuiilL rtlulrlninlliil Jul 52 IIII39IIEIKIIIIIIIIIIILIIIIISIIJ Question 3 A duct carries air from the furnace of a residence along the basement ceiling and then executes a 90 turn to travel up through a wall The basement portion of the duct is 20 ft long and loses heat to the room air at an average rate of 70 Btuh per foot of duct length Air leaves the furnace at 110 F 17 psia with a velocity of 13 fts The duct is 7 in in diameter Calculate the air temperature at the 90 turn Assume kinetic energy is negligible and use constant specific heat at 110 F Solution Apply the first law to the air in the 20 ft section of duct For an open system dE v2 v2 cvch chmihigzi mehe 5 gZe dt ln steady operation the left hand side is zero No work is done and there is no change in elevation With negligible kinetic energy the first law becomes 0 ch mihl mehe 2ch mhi he 0Q mcT T To find mass flow rate mszA lbm 3 p densityAir T 110P 17 00806 ft 2 m 008061br3nj 13 7 2 ft2 ft s 12 m 028113 m S Btu lbm R Solving the first law for exit temperature ch mcp 707Btu 20ft 71h h ft 3600s 0 7 1b B 110 F 028m102405 t j S cp CPAir T 110 02405 T 6 2 lbmR Te 104 F Inkma Viscous Flow P7P zwhama Problcmlv Wk quot N39fw How JenPu a Psz39 Hum is a gym in which 9 40 391 anrfDrmcd FVOM ah rro39fat onal ca 4 a ro0diond viscous rimht Lg H dudmmznf 143 Le L l L2H 9 mg Um bud o My Ha rife Viscous lHla I g zwwudzahly r1de uid 0439 50 wall 6 Molgcwlar di ucon 603332 dude 1 M Hw LM 491 9 1 NM rayon who ll ONCFAMHV V 1erf 54 IVOJtlI 0 HQ who Vegan Visuus a Hedo quotquot41 391 cuthr OF M 71 low thOMgr u dtva921dU all owed ParkaJar m veD Hy Fag an i VArSAnI wHA 56 la Dlvzlalbm c En h39j la 1 W absfarm 790 7 Milaf vimf 5 r 7urod or A How 0 Fully d Vdgf Far I Mnov flow Lt 30 ReD 39 W Mrz 2 ltzgpo plum rm kc anold flanon NHNL it 096 2 D P 1 Au RC 2 D Dialwill 195M 50 Hlvwf r4 row a Fb I M b l r J ff MW 3274000 Reynold Numbcr fl Dgfg rat39a of inc e rm 4 wrcous fares fVLdquot dcush y Rt J v Veo leiw a xaagwsnz M39 dynamig can First Ic39f r cvqluqI 1k Mv39h al ferns E9 cube alv L FluWA M Mn 0 AV 391tg 0 T quotquot MHz 2 w I V 5 ME M b 3930 dIs quot N0 rewrih V 7 Fi39w39 run 7 far0 f 79 law I fhlyz 73h Now avaluctz 51 v I ou wees she F 395 If vituu Qu EX 12 win zuq f I vh A h Taking m rat390 of inc11 1 0 I iION tat 42 1 iv Mvh 39 M a R Sam ELIE Deal LlM Mr n has answer an indch Prod arxuut 42an n ovumu FMr Veloaly pro le u aulr AV 4 c Vclo tl 39 M f 9 6 3 MGKMJ IOQ Prawn Inf AP 5 a fund 7 flow rah Shear 3 abshibu 39ons39 Lc39lquot Ar fro Hun Conxcrvn on of M43 alumina fV2A vtVz Haw valua fz Ixc mun4mm o NOMINI39MM 1ua bu X diuc on a e um Fsif ixsl lsusdA u tux V PVuAe V lflet o t 5 ll oucrv on of Mamaham guru s info 61quot5 quot 0 From orcg balladquot hefty 0 CV EA PA Z A mjsnao whzu A 39 an can pro 44 4ch 44 Side and Wc ud area g 2 I gt I Ag A s Su uft39 j Ac Try It A 239er smo P w P2 hrr I2mWquot LgtL Dawn 1 9 7Tquot P Pzt3 z7 quot9 5 Recall m skew Iresx 4 74 Sohdw wall I O Zlu thxicrrm We Would likt 0 wrth sitar 473 as a func39 on hf radms MS IuM 39 06hrquot Tk r 2 47 Jr d Thnforz39 hear chats am 6 trusted as d I 5447 vu39uou was ad M u a Inc an allyquot SuLs ruz In quot0 8192 2ij5 54 Rearrange x rewrzh as 7 2 0 fjO a 397 g Nutf Parfng hfara39h bn of varia u and ialcjra wN A Br 41 rsk use nosir Ac Pzfj 09 39 rdr 4 2L 2 391 fas quot95 l 2f 31an R f2 shag ran 6 r veloa y raftg hams PaPz ur 1 J39sma 1 Natquot 1 Putnam dro P f 0 5 AP when 4f 70 Also dmn k MalAP by 339 Mr 35 r quot D39QW a OBSIVV H wl I u is 1n tf n r 1 Paraboch whetl Pro le 2 WM 90 Dryowfg39 FP39AGO 910 vcl hcal fir gn91 New Wt would kn 0 altferme H averagc WO f7 7 LurdA u A fub 39 f u n air Mh a 7rz Snarkfymz yieldg mg39H m 3 4MP oE jd39 439 dh ZUr l39 Inhjmh and cvaIMac a Hc Isms5 a IAP397 quot9 EE I ZAL 2 43 S mpli qm g yields 2 a 3 4p wgtn9k Ml gol ng 1 AP 9 L5 WW intll39Md 7 9 39 039 muquot W os u Ar 8amp2 hOHonql rare NexlI Icl39 3nd 2 maxmum velar7 N041 Th4 a maximum choc l MM du 70 1 4 an Far A Abf3ln F39F u did I a 9773quot MaxMum Vclu v i a no N 220 TAU fl 1quot9um 5 ALP kor yawh39 rl 019th ka A 1 u max Zu XML Neatf leiquot inVexlija39fe 9 volume flow in Q afar a Aon39jawfal 0 fA 3917 fufr21rrdr f 4 L wir 0 ZTTrdr 7114 AP Recall 39HM39Iquot 212134 m HWX r d39an Thufare 3 Nu run fauna of fluid 0 flow Finally skew Insr abrhibut ong I A I LAP Lorijol 39 rCPz IO 13 1 Hot exhaust gases are used in the reheat section of a Rankine cycle Consider a commercial steel tube 5cm outside diameter 45cm inside diameter used to convey the steam The air side heat transfer coefficient is 85 WmZK and that of the steam side is 200 WmZK Determine a the overall heat transfer coefficient based on the inside tube area in WmZK b the overall heat transfer coefficient based on the outside tube area in WmZK c the overall heat transfer coefficient based on the inside area if air side fouling is 00015 mZKW and steam side fouling is 00005 mZKW in WmZK Approach 39 Use Eq 133 and Eq 134 which give the overall heat Lo XSWMIK transfer coefficient in terms of its component parts Assumptions 1 There are no fins y 2 There 1s no foul1ng L loowM L L Solution The overall heat transfer coefficient is calculated with Eq 133 and Eq 134 1 1 1 R R 1 U14 UGAD 771114 77114 W 77114 7711114 No fins are used so 77 77w l For a circular tube 11 M M 27rkL 27rkL From Appendix AZ the thermal conductivity of commercial steel is k 605 WmK The areas are A IerL and AG DDL a Incorporating the above expressions into the equation for the overall heat transfer coefficient based on the inside area noting that the length cancels and ignoring fouling L iw Q 147 27rkL haeraL L 1 0045mln00500045 1 0045m Q 200Wm2K 2605WmK SSWm2K 0050m Q 640Wm2K Answer b The overall heat transfer coefficient based on the outside area is U U i 640 W ij l W 4 Answer a 1 AD m2K 0050 m2K c With fouling on both sides of the heat exchanger we obtain 0005392gtlt10 5 00106 2 0045 1 0050 0045 2 i 1 2 000 mK mn 00005 mK 0045m 1 2 0045m Q 200wm K w 2605WmK w 0050m 85Wm K 0050m q603wm2K d Answer Comment The overall heat transfer coefficient is low so the effect of fouling is small If the overall heat transfer coefficient had been large then the addition of fouling would have had a much greater effect 13 2 A two shell pass eight tube pass heat exchanger with a surface area of 8300 ft2 is used to heat l700 lbmmin of water from 75 F to 210 F Hot exhaust gases enter at 570 F and exit at 255 F Assuming the exhaust gases have the same properties as air determine a the overall heat transfer coefficient in BtuhrftzOF b the overall heat transfer coefficient if fouling on both sides equivalent to 0005 hrftZOFBtu is present in the heat exchanger in BtuhrftZOF A roach PAIith all temperatures known along with the area and HUN 2 513 one flow rate sufficient information is given to evaluate the heat transfer rate and the LM TD 8 lvb posse l Therefore Eq 1321 can be used to calculate the overall heat transfer coefficient A39 3309 gt12 139 z ooF I 3 Assumptions waitLia 7P 1 The system is steady 2 Potential and kinetic energy effects are negligible MN nook Tum zsx M 3 No work is done on or by the system 4 Water is an ideal uid with constant specific heat Solution a The governing equation for the LMTD method is Q UAFAT a U Q M AFATLW The heat transfer rate can be obtained from conservation of energy applied to the water flow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Q rhcpAT rhcp Tm 7 Tm From Appendix B6 at TaVg 752102 1425 F cp l00 Btulme Q l7001bmminl 00Btulme21075R230gtlt105 Btumin The LMTD is AT1 TH 7ch 5707 210 360 F AT2 THVW ichm 2557 75 180 F AT1 7 AT2 360 7 7 2597 F M lnAT1AT2 111360180 Using Figure l37b to determine F R M 233 P 210775 0273 210 7 75 7 7 So from Figure l37b FZ 097 Finally U 230gtlt105Btumin60minlhr 6 59 Btu 7 8300ft20972597R 39 2 d Answer hrftR b Using Eq 133 and rearranging it ignoring fouling and wall resistance since no wall thickness is given 1 1 2 R Looosjhr R UM U 659 Btu clean U 638 Btu duty Answer hrft2R Comments larger effect Because the overall heat transfer coefficient is small the addition of the fouling does not have a significant effect If the overall heat transfer coefficient had been large then the addition of fouling would have had a 13 3 In a desalination plant salt water is used to create pure water Salt water is boiled and salt concentrates in the boiler the salt water solution is drained from the boiler and the pure water vapor is condensed for use Condensing vapor at a high pressure is used to boil salt water at a lower pressure Consider an experiment on a single tube condensing steam at 105 C inside the tube is used to boil salt water at 85 C The 304 stainless steel tube is 3m long 25cm inside diameter and 2mm thick The overall heat transfer coefficient based on the inside area is 830 WmZK and the condensing coefficient is 1500 WmZK Determine the heat transfer coefficient of the boiling salt water in Wm K Approach Tc gfdc We can use Eq 133 to determine the overall heat T ogoC transfer coefficient No information is given about k 39 fouling so we ignore those resistances Likewise no fins are used so the overall surface efficiencies are unity 1700 17011 00001 1 114 lt L 3M gt 1 U 1 There is no fouling and no fins J 39 853 WMLV 39quotH be 2 The heat transfer is one dimensional Assumptions Solution Using Eq 133 1 7 1 11 R 1 ILA 77114 nA wt nhA There is no fouling 1 R 0 and 1700 17051 For 304 stainless steel from Appendix AZ km 149 WmK The wall resistance is R lnDa Dx 27rkL The areas are 4 IerL and AD IIDDL Substituting these expressions into the overall heat transfer coefficient equation simplifying and solving for the outside heat transfer coefficient 1 1 A 1A A 1nltDDgt 1 D 1D D 1nltDDgt ha UIA hA 27rkL UID 1113 2k i R4411 i 7 1 0029m 1 0029m 0029mln0029m0025m 830Wm2K 0025m 1500wm2K 0025m 2149WmK ha 2080Wm2K 4 Answer 134 L 39 39 of a finned L are determined in a laboratory The heat exchanger has 100 tubes that have inside diameters of 12 mm and lengths of 26 m a dense arr continuou 39 quot fthe tubes 39 op 39 439 39 L L L m iiui water 39 174 C liui waver 121 C nuiwauci mZs cold air 39 7 C cold air 39 r kPa cold air inlet ow rate 22 mls Determ39 e the overall heat transfer coe 39icient based on the inside tube area in WmZK Approach AIL TQM 25quot Sufficient information is given to evaluate the heat PC I qj kPR transfer rate the LMTD and the inside surface area 1 2 1 V Therefore Eq 1322 can be used to calculate the V V 39 39 M 5 overall heat transfer coefficient W m a Assumptions TMquot 392 gtTuw 1 C 1 The system is steady 39 17 2 Potential and kinetic energy effects are negligible W Glows bakfimd 3 No work is done on or by the stern N x a 4 Air and water are ideal uids with constant specific heats Solution The governing equation for the LMTD method is Q U FAT gt U Q 4 um AFATUM I l i b ined 39 ofenergy ppquot 39 nnwork negligible potential and kinetic energy e ects and ideal liquid with constant specific heat so that Ah CPAT Q chAT We Th 7TH From Appendix A6 at 174 C 0 8931 kgmz at Tzwg 1741212 1475 C c 4267 kJkgK Q 8931kgm3000051m3s4267kJkgK174121K103kW The area is A NrDL 10010012m25m98m2 The LMTD requires four temperatures We can calculate the air outlet temperature from an energy balance similar to that done with the water Therefore Q Q Q mac Tacit TCM 9 Tacit TQM TCM mc We 97kN m2 2897 kg kmol Using the ideal gas equation p M RT 8314kJkmolK25273K The air specific heat should be evaluated from Appendix A7 at the air average temperature Because we are calcu ating the air outlet temperature to evaluate CW we assume Tam 100 C and ng 251002 625 C c 1008 kJkgK 1134k m 103 kW1kJ1kWs 1134kgm322m3s1008kJkgK AT TH from 1747 550 108 C AT2 TM 40 1217 25 950 Tm 25 c 550 c AT 7 AT 1 08 7 96 ATM f 1019 C 39 1nAT AT2 ln10896 257660 1217174 Using Figure 137c to determine F R 077 P 036 So from Figure 137c FE 085 1217174 257174 103k 1000 k many U M it 4 Answer 98m20851019K m K 13 5 Water at 200 F flows inside a lin inside diameter 304 stainless steel tube with wall thickness of 005 in Air flows over the outside surface of the tube The water side heat transfer coefficient is 80 BtuhrftzOF while that of the air side is 40 BtuhrftzOF Determine a the overall heat transfer coefficient based on the inside surface area in BtuhrftZOF b the overall heat transfer coefficient if the air side fouling factor is 00007 hrftzOFBtu and that on the water side is 00003 hrftzOFBtu in BtuhrftZOF Approach 39 OOYm Because of the given information we can use Eq 13 WM d termine the overall heat transfer coefficient 39 3 to e gt gt A t 1 quot yogih ssump Ions 1 91 l The heat transfer is one dimensional x P T 1 1 All I ha 4 1F 2 The heat transfer coefficients are uniform over the 2 heat exchanger Solution Using Eq 133 1111 R391 7 Rwy U14 771114 7714 77111 7711114 Multiplying through by AD incorporating A IerL A IIDDL and recognizing there are no fins 77 771 1 D In D D LLRMRg ii U h 271k DD h DD a With no fouling RE RE 0 and from Appendix B2 the thermal conductivity of 304 stainless steel of 86 BtuhrftR so that 1 1 l12ftlnlll 1 1 U 80Btuhrft2R 286BtuhrftR 40Btuhrft2R k11 x U 280Btuhrft2R f Answer b With fouling obtain estimates of the fouling factors from Table 131 D 2 2 2 1 1 R R 2OOOO3 hm R 00007 hm R L 00366 hm R UW Um DD 280Btuhrft R Btu Btu 11 Btu UM 273Btuhrft2R 4 Answer Comments Because the overall heat transfer coefficient is low the addition of the fouling did not have much of an effect on the overall heat transfer coefficient If the clean overall heat transfer coefficient were high then the addition of fouling would have had a much larger effect 13 6 A heat exchanger tube 25mm outside diam eter has 20 longitudinal fins with rectangular cross sections equally spaced around the circumference of the tube The fins are 25mm from base to tip and l6mm thick The tube has a 2mm wall thickness and tube and fins are both made of plain carbon steel k 605 WmK The inside and outside convective heat transfer coefficients are 1000 WmZK and 200 WmZK respectively Determine a the overall heat transfer coefficient based on the inside surface area in WmZK b the overall heat transfer coefficient based on the outside area in Wm K Approach J With the given information we can use Eq 133 to P Nib determine the overall heat transfer coefficient First g w I though the overall surface efficiency for the fins must P 7 be determined I 4 LEMquot Assumptions I l The system is steady 2 Potential and kinetic energy effects are negligible 9 N T 20 S 3 No work is done on or by the system 9 e by 100 MHLK 4 Hydrogen and water are ideal fluids with constant 1quot I h loo 3 WMLL specific heats Solution a We use Eq 133 to determine the overall heat transfer coefficient 1 l R R l l l Ila111 l l U14 771114 7714 77114 7711114 There are no fins on the inside Va1 I no fouling 11 R 0 and the wall resistance is R lnD Dl 27rkWN The areas are A7 IrDW and A IIDDW2MW Substituting these expressions into the equation for the overall heat transfer coefficient and simplifying we obtain 171 D1nDD 1 7rD U h 2k 77111 MD 2NL The given wall thermal conductivity is k 605 WmK For the fins we use the equation for a fin with an adiabatic tip and correct the length to account for NA convection at the fin tip The overall surface efficiency is Eq 1194 77 li TKO 7 77 1 With a corrected fin length L LI2 0025000162 000258m Am A and A ZLW From Table 115 Case A with mL hp104 05 77 tanhmLmL ML 05 2200Wm2K 05 Therefore mL L 00258m 166 ktL 605WmK00016m t nh 166 20 2 00258 w 0561 and 7 17170561 0592 39 7r002522000258 1 7 1 0021m1n0025m0021m 1 7r0021m U 71000Wm2K 2605 WmK 0592200wm21ltLno025m22000258m U 653Wm2K Answer b For the overall heat transfer coefficient based on the outside area 0021 U U i653 A 388 2 4 Answer A m K 7r0025m22000258m m K Comments Because of the significantly different areas on the inside and outside surfaces the magnitudes of the overall heat transfer coefficients are similarly significantly different 13 6 13 7 Very thinwalled low chromium steel k 37 WmK tubes of diameter 10 mm are used in a condenser A convection coefficient of h 5000 WmZK is associated with condensation on the inner surface of the tubes while a coefficient of he 100 WmZK is maintained by air ow over the tubes For a 1m long section of tube with 286 fins determine a the overall heat transfer coefficient if the tubes are unfinned in WmZK b the fin efficiency and overall heat transfer coefficient based on inner area if low chromium steel annular fins of thickness I 15 mm outer diameter D0 20 mm and axial spacing S 35 mm are addedto the outer tube surface in WmZK Approach deli1mm I snow M We can use Eq 133 to determine the overall heat 39LD loo WALK transfer coefficient The overall surface efficiency L m WIMIK 70 must be determined av 02 3 0 m A m Assumptions L w 15442 J th 1 The heat transfer is one dimensional N 154 539 Solution R Us1ng Eq 133 1 i 1 1 R 1 Rm 1 D 1 1 UL milL 77M ag1 nacho1 There is no fouling 11quot R 0 and wall thickness is not specified so D D0 and RW 0 a With no fins and D D0 1 1 1 1 1 1 2 a U 980Wm oK Answer Q UD 1 kg 5000 100 b With annular fins we can determine the fin efficiency using Figure 1118 and the overall surface efficiency NA w1tth 1194 77 177 Am 12 V2 2 5 L32 h Lt2 j 0005m00015m2M 0244 kAp kl 37Wm2K00015m From Figure 1118 r2 I2r1 00100001520005 215 so that nfz 092 Total area is AnnWM A n Area of one fin A 27rr22 7 rf 27rr2l 20012 7 0005227r00100015 0000566m2fin Total fin area A 0000566m22860162m2 Unfinned area AnnWM NSit7rD1 28600035m00015m7r0010m00180m2 Total area Am 001800162 0180m2 0162 Overall surface efficiency 77D a 17 17 092 0928 39 0180 l l l The overall heat transfer coefficient is U7rDW WW awn1 1 1 2 7r001m1m Q 5000Wm K 0928100Wm2K0180m2 U 481Wm2K 4 Answer Comments Because of the addition of the fins the overall heat transfer coefficient increases dramatically 13 8 A double pipe heat exchanger consists of a 4cm pipe inside a 6cm pipe the heat exchanger is 2m long The water inside the inner pipe has an average temperature of 40 C and a flow rate of 0016 m3s In the annulus between the inner and outer pipes unused engine oil has an average temperature of 147 C and a flow rate of 001 m3s The inner tube has a wall thickness of 1mm and is made of 304 stainless steel Determine a the overall heat transfer coefficient based on the outside area of the inner tube in WmZK b the overall heat transfer coefficient based on the outside surface area of the inner tube if the water and oil sides are fouled choose representative fouling factors from Table 131 in WmZK Approach We can use Eq 133 to determine the overall heat L 2M gt Dz39o gk transfer coefficient Sufficient information is given lt to calculate the two heat transfer coefficients With 7 If D 0 04 M hm 400C gt appropriate correlations aquot 7 C It v 39 3 I Assumptlons YW o39olej t quot 0 00 m 1 The overall heat transfer coefficient is uniform VD 001 9035 over the heat exchanger 2 The flows are fully developed Solution Using Eq 133 to calculate the overall heat transfer coefficient 1 7 1 R lnDaDx R 1 UaAa 7711th 77aan Z kL 771111141 771ahaAa There are no fins so 1700 17051 The two areas are A 7rDL and AD IIDDL 7rQ ZIL Substituting in these expressions and simplifying D 21 D 2 Di211n Dx21 Dx L L W L MW UD 1 D D 2k ha The therm a1 conductivity of 304 stainless steel from Appendix A2 is 149 WmK The heat transfer coefficient requires the Reynolds number x 4m 7 4pV39 gt Re 7 7ruDx 7ruDx Water RepD lL 4 p4 37r4Dx2 For water from Appendix A6 with TW 40 C u 634 X10394 Nsmz k 0631 WmKPr 419 p 9922 1gm3 R 7 49922kgm30016m3s1Ns2kgm e T This is turbulent flow so using the DittusBoelter equation Nu 002312e0 Pr 00237970000 8 419 2150 7 Nuk 7 21500631WmK 733 900 w m2K 797 000 hw D 004m For the oil Reynolds number and Nusselt number we need to use the hydraulic diameter as the characteristic length 7 4A7 747r4D327D1212 7 7rD37rD121 P DE 71 21 00670042gtlt0001 00180111 Wm 7139 V 001m3s AX 2 2 6933 4D32 7D121 n40062 0040002 jmz s For 011 from Appendix A6 with TD 147 C 11 564 xio394 Nsmz k 0133 WmKPr103p 8121 kgm3 Ref 8121kgm3693ms0018m1Ns2kgm 564X10 Nsm2 This is turbulent flow so again using the DittusBoelter equation Nu 0023Re08Pr03 00231800008 103 372 18000 ha Nuk 3720l33WmK27501 I 0018m m2K The wall resistance is D 21 In D 21 D 0040002 In 0042 004 2 RA gt lt gtlt gt lt gt0000069m1lt 2k 2149WmK W a Without fouling the overall heat transfer coefficient is with consistent units omitted for brevity L 1 0000069 Jr 0000031 0000069 0000364 U 33 900 004 2750 UH 2160Wm2K Answer b With fouling using fouling factors estirn ated from Table 131 of R5 m 0000175m2KW and R m 00004m2KW 1 0042 000003 1 00004 U 0 040 0000069 0000175 0000364 000106m2KW a Ug944Wm2l 1 Answer Comments The oil side heat transfer coefficient dominates the situation When there is no fouling Adding fouling has a significant effect on the overall heat transfer coefficient E lene glycol enters a double pipe heat exchanger at 17 C with ow rate of 15 kgs It is heated with water that en er the heat exchanger at 100 C with a ow rate of 004 kgs The inner pipe is 25cm in di meter the outer pipe is 375cm in diameter and the length is 3 m Determine the overall heat transfer coef cient in WmzK if a the wa ows in the inner tube b the water ows in the annular space between the two tubes Approach We can use Eq 133 to determine the overall heat transfer coe icient Suf cient information is given to calculate the two heat transfer coef cients with appropriate correlations 4 L3m gt Assumptions d 1 The overall heat transfer coef cient is uniform 5 law 7 394 I YLfST HDC over the heat exchanger quot8197 M 03904 LiS IT 0 be 2 The water ow ls fully developed 3 Wall resistance is negligible Solution a Us39 g Eq 133 to calculate the overall heat transfer coef cient U914 7 09414 quotWA 2 quotMA There are no flns so 7100 710 1 nd we assume negligible fouling With no wall thickness given the two areas are A IrDL and A IrDL Substituting inthese expressions simplifying and solving for the overall heat transfer coe icient U 71 11811 The 39 39 Annmd L n r 39 Ethylene glycol M 247x10394 Nsmz k 0248 WmK Pr 235 0 11188 kgmz Water M 2 x 0quot Nsmz k 0681 WmK Pr 171 0 9583 kgmz 11 ware quot 39 39 39 Re pm l 03904kgs4 2 00853 u M 9583kgm3n0025m s 9583kgm3 0085 ms0 025m f 73800 276gtlt10 Nsm This is turbulent ow so using the DittusBoelter equation Nu 0023amp Pr 002301800 171 212 212 0681W mK 2 11 MM 770 1558X10 4m K D 025m m K h w For ethylene glycol in the annulus 15kgs4 V 1 2 2 2185 M 11188kgm3n00375m 70025m Jul ule annulu to calculate the D 4APmm D2 7D 0037570025 00125m pwh 11188kgm32185ms00125m Re 1240 247x10 Nsm This is laminar ow so we need to check the entranc l Lm 00371er 0037124023600125m135m Therefore entrance effecm must be taken into account Using the SeiderTate correlation 1310 Nu 186Gr 3 Ia115014 186RePrDLV3 uys014 Assuming the wall temperature is Ts m 171002 585 C us 542810394 Nsm2 13 7 014 1240 236 00125 4 Nu186 246 542gtlt107 h 7Nuki2460248WmK488 w 7 D 7 00125m m2K a 71 U1 LL 450 l 4 Answer 5770 488 m K b Now putting the water on the annulus side V 7 m 7 004kgs4 pA 9583kgm37t00375m270025m2 R pDh 9583kgm30068lms00125m e 276gtlt10395 Nsm2 This is still turbulent flow so Nu 0023Re08Pr03 00232950008 171 102 102 0681W K kw gtlt m L550 v D 00125m m K x 006802 S 29500 For the ethylene glycol 39 15 k 4 l gS 2 2733 p4 lll88kgm37r0025m s MD 1 ll88kgm3273ms0025m 7 7 247x104 Nsm2 The ethylene glycol is now turbulent but this is a low Reynolds number so we will use the Gnielinski correlation 72 f 079lnRe7164 079ln3090716472 00451 f8Re71000Pr 7 004518309071000236 7 Re 3090 7 763 1127f81 2Pr2371 1127oo4518V2236234 763 0248W K So thathNukM756 Z D 0025m m K x 71 U1 LJFL 666 Answer 5550 756 m K Comments Just by putting the two uids on different sides of the heat exchanger the overall heat transfer coefficient increased by 48 which would translate into a significantly higher heat duty l3ll 13 10 A heat exchanger used to heat air with hot water is constructed of individually finned tubes as shown in the figure below The tube lm long 10mm inside diameter l3mm outside diameter and fins lZmm long 05mm thick spaced on 5mm centers are constructed of brass Air flows over the tubes with a heat transfer coefficient of 100 WmZK Water with a velocity of 2 ms enters the tube at 80 OC Determine the overall heat transfer coefficients based on the inside area U and the outside area U0 in WmZK Approach I We can use Eq 133 to determine the overall heat transfer coefficient The water side heat transfer coefficients and the fin efficiency must be determined q o OITM N l T IS 393 000 M Assumptions l l The overall heat transfer coefficient is uniform wzl39 I 7 39 03900 M over the heat exchanger L 00 in MW 7k 2 The water flow is fully developed 3 Fouling is negligible 1 0390069quot 0 000 Solution Using Eq 133 to calculate the overall heat transfer coefficient 11 R 1nr1ra R 1 UGA milL m4 Z kW min1 nigh1m There are no fins on the inside so 17W 1 and we assume negligible fouling The inside area is 4 IIQL For the wall resistance the thermal conductivity of brass from Appendix A2 is 110 WmK so that ln ln 00065 0005 Vire 380X104m2KW 27rkW 27rllOWmKlm With annular fins we can determine the fin efficiency using Figure 111 8 and the overall surface efficiency with Eq 1194 77 17NAAmli 77 The total area requires the number of fins N over the total length W WNSts NW7S lm0005m SI 0005m00005m Therefore we use N 180 an integer number of fins The total area will take into account the tip area by using the corrected length so we need the corrected l809 radius r2 r2 r2 00185m000052001875m Area of one fin A 27rr 7 r12 27r0018752 7000652l943gtlt10393 m2fin Total area Am 27rr1W iNzNA 27r00065mlml8000005ml80l943X10393m2 0387m2 Now for the parameters inFig lll8 r2r1 00187500065 288 L LI2 0012m00005m2 001225m A Lt 001225m00005m6125X10396m2 12 12 2 Li h 001225m32 w 0522 kAp llOWm2K6125X10396m2 From Figure lll8 so that 17fZ 078 l80l943X10393m2 Overall surface ef ciency 77 17 l 7 078 0801 0387m2 The finned side resistance is wonmqw navahaAm 0801100Wm2K0387m2 1312 The water side heat transfer coefficient requires the Reynolds number Water properties from Appendix A 6 at T 80 C 1 348 810394 Nsmz k 0670 WmK Pr 218 p 9718 kgm3 MD 971Skgm32ms0010m 7 348x104Nsm2 This is turbulent flow so using the DittusBoelter equation Nu 0023Re08Pr03 00235590008 218 182 Re 55 900 182 0670W K 11 N k m 12200 iv Dx 001m m K 2 L261x104 m K M 12200Wm2K7r001mlm w The inside overall heat transfer coefficient is with consistent units omitted for brevity U 261x10393 380x10 00322514 903 Wm2K 4 Answer 7500lmlm The outside overall heat transfer coefficient is 001 1 UD U i 903Wm2K mg m 733wm2K4 Answer AD 0387m Comments Because the surface areas are significantly different on both sides of the tube the overall heat transfer coefficients are similarly different This illustrates well that the side on which the overall heat transfer coefficient is based must be specified 1313 13 11 Hot water at 100 C flows at a rate of 45 X 10394 m3s through a horizontal 316 stainless steel pipe with a 5 cm inside diameter and a 5mm wall thickness Outside of the pipe is still air at 25 C and 1 atmosphere Determine the overall heat transfer coefficient based on the inside and outside surface areas of the pipe in WmZK Approach We can use Eq 133 to determine the overall heat M transfer coefficient Sufficient information is given to WA calculate the two heat transfer coefficients with 4gt 394 appropriate correlations V345 g B Assumptions T m C If If 4 IT Allquot Tc zwc The overall heat transfer coefficient is uniform over the heat exchanger The water flow is fully developed Fouling is negligible 1 WP S olution Using Eq 133 to calculate the overall heat transfer coefficient 1 1 I1 1nDaD R 1 UGA 7 771114 771A erkL 77114 nigh1 There are no fins so 1700 17W 1 and we assume negligible fouling The two areas are A IerL and AD IIDDL Substituting in these expressions simplifying and solving for the overall heat transfer coefficient U U 1 12 2k hD a a For the wall resistance the thermal conductivity of 316 stainless steel from Appendix A2 is 134 WmK so that Dx lnDaDx 005mln006005 340x10 m2KW 2k 2134WmK The water side heat transfer coefficient requires the Reynolds number Water Re p D 4m 7 4pV rh m a Re p4 pvt4W 7WD 7WD For water from Appendix A6 with TW 100 C u 276 XlO394 Nsmz39 k 0681 WmK39 Pr 171 p 9583 kgm3 4 9583k 3 45x10 3 1N 2 k Re gm m W W m 29800 7276X10 Nsm2005m This is turbulent flow so using the DittusBoelter equation Nu 0023Re08Pr03 00233980008 171 1292 l292 068lW K 2 h NWMl760 i56gxlor4 D 005m m K h w The air side is natural convection We must assume a surface temperature to calculate a Rayleigh number We know natural convection heat transfer coefficients are low compared for example to forced convection of water Hence the wall temperature will be closer to the water temperature than the air temperature so we assume an outside wall temperature that is only slightly lower than the water temperature T Z 95 C Therefore T lm 95 25 2 60 C and the air properties from Appendix A7 at this temperature are approximately 1 199 x105 Nsmz39 k 00288 WmK39 Pr 0700p 1060 kgm3 Ra g p2 72TD Pr 981ms213331lt10601ltgm32 9525K006m3 0700 2 885000 199X10395Nsm2 l3l4 Using Table 123 for natural convection on a horizontal circular cylinder 2 2 0387885 000 6 140 0387Ra16 Nu 916 827 0 916 827 l0559Pr 105590700 140 00288W K 2 NWM673 i E 239j0124 m K ha DD 673Wm K 006 w ha DD 006m The inside overall heat transfer coefficient is with consistent units omitted for brevity U 568x104 34gtlt10394 012414 802Wm2K Answer The outside overall heat transfer coefficient is UH U i 802Wm2K 668wm2K4 Answer A 006 Comments Note that the water and wall resistances are negligible and our assumed wall surface temperature will be close to the actual wall temperature so no iteration is required 1315 13 12 Ethylene glycol flows inside a copper tube that has a 05in inside diameter and a 065in outside diameter The heat transfer coefficient for the ethylene glycol is 300 BtuhrftzOF Water flows outside the tube and has a heat transfer coefficient of 550 BtuhrftzOF Determine a the overall heat transfer coefficient based on the outside tube area in BtuhrftzOF b the overall heat transfer coefficient based on the outside tube area if fouling is present on both the water and ethylene glycol sides in BtuhrftzOF CEstimate fouling factors from Table 13 c Discuss how much the overall heat transfer coefficient can vary depending on the choice of fouling factor Approach I I I 8 AD 2 550 Gav 11 Because of the g1ven1nformat1on we can use Eq 133 me to determine the overall heat transfer coefficient Fouling factors are estimated from Table 131 quot39 DJ 003119 TD 0041791 if v v l The heat transfer is one dimensional l 5 1 cat L 39 3 30x35h 49 lt 2 The heat transfer coefficients are uniform over em 9 i Y A the heat exchanger Assumptions Solution Using Eq 133 1 1 1quot 1nltDDgt UAnhA A 2m 111 1 RDquot l 77A 77hA Multiplying through by A0 incorporating A IerL A IIDDL and recognizing there are no fins 77 77 1 Dl D D L D Rg n 0 U h D D k ha I 2 a With no fouling RE RDquot 0 and the thermal conductivity of copper from Appendix B 2 of 2317 BtuhrftR so that 1 7 1 wj00542ft1n0054200417 1 UD 300 BtuhrftZR R 00417 22317BtuhrftR 550Btuhrft2R U0 0004330000031000182f1 162Btuhrft2R Answer b With fouling obtain estimates of the fouling factors from Table 131 Water 12 00009 000053m2KW OR 000051 00030hrft2RBtu Ethylene glycol 11 000035 mZKW 0002hr ft2RBtu Using the maximum fouling resistance 71 U0 0004330002g39gi j000003l0003000182 849Btuhrft2Ri Answer Using the minimum fouling resistance 71 U0 000433 0002 000003l 000051 000182 108Btuhrft2R Answer Comments The choice of the magnitude of the fouling resistance can have a large effect on a heat exchanger design or rating problem Care must be exercised and the effect of the choice must be evaluated 1316 13 13 To use as much energy as possible from the combustion of natural gas heat exchangers are often placed in exhaust stacks to recover waste energy Consider a single pass crossflow heat exchanger Exhaust gases assume air properties enter at 180 F with a flow rate of 031 lbms and exit at 130 F Fresh air enters at 70 F with a flow rate of 062 lbms The heat exchanger construction is such that both uids are unmixed and the overall heat transfer coefficient is estimated to be 35 BtuhrftZOF Determine the required area of the heat exchanger in ftz Approach stamens Because we want to determ1ne the heat exchanger T W a area this is a design problem We use the LMTD 39 H i 0 method for the analySIS The fresh an outlet C C V m Y U MS temperature needs to be calculated so that the log 1 ML 175 fwd mean temperature difference can be determined 1214170 7 UNNliea Assumptions ML a The system is steady h Potential and kinetic energy effects are negligible 1 30 C No work is done on or by the system Tum The fresh air and exhaust gases are ideal with constant specific heats uww Solution The governing equation for the LMTD method is Q U FAT A Q 1A LM UFATLM39E The heat transfer rate can be obtained from conservation of energy applied to the exhaust gas flow Assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cpAT We also assume that the exhaust gases can be approximated as air Therefore evaluating the specific heat from Appendix B7 at the average temperature 1801302 155 F cp 0241 Btuhrft F Q m AT 0311b m 0241B tu 180 130 F374 E p s hrftR s To calculate the LMTD we need to calculate TQM Applying conservation of energy to the fresh air as before Q mop Tam TCJK Tam Tcm mcp Assuming TM 100 F so that ng 85 F cp 0240 Btuhrft F and TC 7WQ5I F 39 062lbms0240Btulme AT 7 AT ATLM Wgt ATI THm Tcump ATz Tm FTC AT11807951849 F AT213077060 F 8 7 0 AT 717 F M ln84960 Using Figure 137c to determine F R180 130199 lD951770OI228 951770 180770 ThereforeFZ 096 Finally 374Btus3600s1hr 35Btuhrft2R096717R 559ft2 Answer 1317 13 14 A shellandtube heat exchanger has tubes with 18mm outside diameter and a wall thickness of 12 mm Cold water outside the tubes with a flow rate of 250 kgmin is heated from 30 C to 50 C with hot water that enters the heat exchanger at 105 C with a flow rate of 150 kgmin The company design specification is to use a fluid velocity inside the tubes of about 04 ms From previous designs the overall heat transfer coefficient based on the inside surface area is estimated to be 1800 WmZK Determine the number of tubes and the required tube length if the heat exchanger is a counterflow b parallel ow c one shell pass and two tube passes d two shell passes and four tube passes Approach D 0 0 87 M Sufficient information is given to evaluate the heat 601 M transfer rate Because we are seeking the area this L quot5 1s a des1gn problem We use the LMTD method for 4164 1 the analysis 1 U 1203 K gt 1114 109 er 331C Assumptions 50 odd wn18 Tc IN 1 The system is steady hequot I EN two 3 zwbiNH 2 Potential and kinetic energy effects are negligible T m 3 No work is done on or by the system C M 4 Water is an ideal uid with constant specific heat Solution The number of tubes is obtained from conservation of mass and the definition of mass flow rate mNp4 NrhpV4X For the hot water assuming an outlet temperature of 65 C Tavg 105 652 85 C so that the hot water properties from Appendix A6 at this temperature are CF Z 4198 kJkgK and p 9686 kgm3 Therefore 50kgmin1min60s 733 8 9686kgm304ms7400156m2 So use N 38 and V 0397 ms The governing equation for the LMTD method is Q U FAT wzth NIrDL a L Q A W A U1 Nm F MW The heat transfer rate can be obtained from conservation of energy applied to the cold water flow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT From Appendix A6 for water at an assumed average temperature of 40 C cp Z 4175 kJkgK kg kJ 1min 39 T 7T 250 4175 5030 K 348kW Q a kgK gt 605 The hot water outlet temperature is obtained with an energy balance on the water using the same assumptions as used as before and using the heat transfer rate calculated above Therefore QwmwcpwTHpTH THTH7 mwcpw TH 150 C 348kw 1168 C 39 150kgm1n4198kJkgK1m1n60s The LMTD for counterflow is An TH 7ch 105750 55 C AT2 TM 7T5 1168730 868 C AT 55 86398 698 C M ln55868 The LMTD for parallel flow is 1318 AT1 Tm 7ch 105730 75 C AT2 Tm 7 757668 M ln75668 7T AT L 708 c Finally 7 348kW1000 WkW 7 116mK 7 1SOOWmZK347rO0156mFATLM 7 FATLM The value of F depends on the heat exchanger configuration Caut 1168750668 C 4 Answers 169 FMo157 11687105 307105 Using Figure 137 for the F factors and using the appropriate LMTD we calculate the tube lengths Counter ow F 1 ATM L 1662 m Parallel ow F1 ATMVP L 1638 m 1 shell2 tube F2 095 ATLMVE L 1696 m 2 shell4 tube FS 1 ATM L 1662 m Comments For this combination of conditions the variation in length with different heat exchanger configurations is not large However with other conditions the length can vary significantly 1319 13 15 A counterflow concentric tube heat exchanger is designed to heat water from 20 C to 80 C using hot oil which is supplied to the annulus at 160 C and discharged at 140 OC The thinwalled inner tube has a diameter of D 20 mm and the overall heat transfer coefficient is 500 WmZK The design condition calls for a total heat transfer rate of 3000 W Determine the length of the heat exchanger in m After 3 years of operation performance is degraded by fouling on the water side of the exchanger and the water outlet temperature is only 65 C for the same uid flow rates and inlet temperatures What are the corresponding values of the heat transfer rate outlet temperature of the oil the overall heat transfer coefficient and the water side fouling factor Approach W L Because we want to determ1ne the tube length which U 500 V4 K I Q is equivalent to finding the heat exchanger area this is a design problem We use the LMTD method for the ml MegAim Em 10 C 4 a Assumptlons Tum Hose THIIM 90 C l The system is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 The oil and water are ideal liquids with constant specific heats Solution a The governing equation for the LMTD method is UAFATLM 5 gt L 39 UnDFATWy AT1 7 AT Because this is counterflow F l and ATM 1n AT T 7T AT T 7T ATlATZ gt 1 Hm Cautgt 2 Haut Cm AT1 160780 80 C AT 140720120 C ATUM 987 C 11180120 3000W Finally L A 097m 4 Answer 500Wm2K7r0020m987K b With the fouling resistance 1 LR gt R 1 derty U512 derty U512 We need to determine the dirty overall heat transfer coefficient and to do that we need the heat transfer rate The new heat transfer rate after three years of operation can be obtained from conservation of energy applied to the water flow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Qnew rhCPATW mop Tam 7 Tam 39 7 T For the original ngd rhcpAT r39ncp T Gym DM 0101 Cant Taking the ratio of these two heat transfer rates 39 rhc T 7 T T 7 T A QW 65 20 075 A QM o753ooow2250w Qazd mop Tam TCM M T 7T M 80 20 Cum Cm For the new oil outlet temperature we use the same approach Q Tm W QW 39 THautn2w THm T 39 Qazd THm T THautD1d Qazd The overall heat transfer coefficient is U QAATLMVE TH 4AM 16070751607140 145 C AT1 160765 95 C AT 145720125 C gt ATM dwyc 1n95125 U ZZSOW 38 iv a R 1 1 0000955 Answer 70020m097m1093K m K 338 500 w 1320 13 16 In a cogeneration plant the exhaust from the turbine in a Brayton cycle is used in a crossflow heat exchanger to heat pressurized liquid water inside tubes from 300 F to 400 OF The exhaust gas flow enters the heat exchanger at 850 F with a flow rate of 18 lbms and is considered unmixed The overall heat transfer coefficient is 80 BtuhrftzOF The tubes are lin diameter and 16ft long If the heat exchanger effectiveness must be at least 75 determine a the water flow rate in lbms b the number of tubes Approach Using the definition of effectiveness the given 6 703 UNM EH information to calculate the heat transfer rate and W sI 1quot LbMS assuming the hot gas has the minimum heat capacity u w 81 F rate the hot outlet temperature and the water flow rate Nessaum can be obtained Finding the number of tubes is WHEEL U gig sizzp equivalent to determining the area so this is a design quot problem and the preferred approach is the LMTD 131 300 H D T 0 0933 method L r E Q Q Tam 4m F J s 75 Assumptlons l The system is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 Both uids are ideal with constant specific heats Solution a The definition of effectiveness is g QQmaz Where me Cmm THm 7 Tam The heat transfer rate is obtained from conservation of energy applied to either of the flows Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Q rhcpAT lfwe assume the hot exhaust gas has the minimum heat capacity rate then mgcpvg THVW 7 THVM Combining this expression with the definition of effectiveness and the maximum heat transfer rate we obtain TM TM 7 5THVM 4H 850 F075850200 F4375 F To find the water mass flow rate equate the heat transfer rate expressions for the gas and water mgcpg THm 7TH mwcpmz TCaut 7 Tam m mgcpg THm 771112 W cpw TCaut Tc m Assuming the hot gas can be approximated with air properties and evaluating specific heats from Appendix B7 at average temperatures Gas Tavg 85043752 644 F cpg 0251Btulbm F Water Tavg 300 4002 350 F cw 105Btulbm F mw 181b m 039251 85039437395 17751b m 1 Answer s 105 400300 s Checking the heat capacity rate as a check on the assumption above CW l775lbmsl05Btulbm F188Btus F Cg lSlbms025lBtulbm F452Btus F Therefore our assumption above that the gas had the minimum heat capacity rate is valid b The governing equation for the LMTD method is UAFATLMVE 1321 With A N DL solving for the number of tubes N 4 UIrDLFATLMVE 39 Using the energy balance on the gas from above Q mgcpg THm T T Hm 1 8lbms0251Btulbm F8504375 F1864Btus The LMTD is 7 8507400 7 43757300 MUM Ml HZ 2636 F 39 lnAT1AT2 ln8507 40043757300 Using Figure l37c to determine F for cross ow with both uids unmixed 850 7 7 8507300 39 Therefore FZ 087 Finally 1864Btu 3600 1hr S 874 tubes 80Btuhrft2R700833ft16ft0872636 F Therefore choose N 88 tubes an integer number Answer 1322 13 17 In a refrigeration unit R134a at 018 MPa is evaporated inside in a long thinwalled tube The refrigerant W ose flow rate is 0001 kgs enters the tube as a saturated liquid and exits as a saturated vapor and its heat transfer coefficient is 500 WmZK Air at 27 C flows with velocity of 6 ms perpendicular to the outside the tube Shown in the figure below is the aluminle tube k l77 WmK that has eight rectangular fins inside the tube Each fin is 5mm long and lmm thick The tube diameter is 3 cm Determine the required tube length in m Approach Both uids have constant temperatures so the basic heat transfer rate equation N1 Kinks D 3am LT fob l wi Fe Q ATRm can be used to determine L MM 5 Amp 4 34 the required length since each of the 392 P 011 9 IVWA resistances involve area and length is 7 h goo NFl kr used to calculate area The outside air PM 5 27 I eU GMIS heat transfer coefficient must be evaluated from the given information Assumptions The heat transfer is one dimensional The system is steady No work is done or by the control volume Potential and kinetic energy effects are negligible WN 4 Solution Assuming steady onedim ensional heat transfer the rate equation is Q E T TR Rm Ran R ns Note that no wall thickness is given so we ignore the wall resistance The air side resistance is Rm L hDAD haerLT The air side heat transfer coefficient requires the Reynolds number with uid properties evaluated from Appendix A7 at TM Assuming TM 2 300 K p 11774 kgm3 1 1846X10395 Nsmz k 002624 WmK Pr 0708 pD ll774kgms6ms003m e R 11500 1 1846X10395 Nsm2 From Table 12 1 Nu 0193Re0618Pr13 0193115000618 0708 555 Nuk 55502624wmk 7 w h 7 486 D 003m m2K l 7 0284 486wmZK 003mr7 LT w The refrigerant side resistance is Rf l77uvxhxz47 an NA where 77 17 17 77 Assuming a rectangular fin with an adiabatic tip and using a corrected length to a 147 f 1 account for convection at the tip 77 tanhmLmL and m hpkAX 0 5 Evaluating the parameters in the equations A 1LT 0001LT p 2LT L LAXp LI2 0005 00012 000505m Area of one fin A LT 2Lr LT 20005m0001m 0011LT Total area Am IrDLT NZLLT LT n003m820005m 0174LT 1323 05 500w m2K 2L m A 7516m391 and mL7516mquot000505m03796 177WmK0001LT 8 0011L 77tanh03796037960955 and 77a17 1709550977 R 1 1 7001176mKW nahJ 0977500Wm2K0174L LT The heat transfer rate can be obtained from an energy balance on the refrigerant Assuming steady no work and negligible potential and kinetic energy effects we obtain Q rhhg From the saturated liquid table Tm 1273 C and hfg 20625 kIkg Q 0001kgs20626kJkg0206kJs Substituting this into the main equation to determine the tube length 271273 c 0206 1 s 02184mKW001176mKW 1 1 0206kJs02184mKW001 176mKw10001k1 ll4m Answer 27 1273 c11WS Comments Note that the fin resistance is essentially negligible compared to the air side resistance 1324 1218 A one shell pass four tube pass heat exchanger contains 20 thinwalled 25mm tubes It must be designed to heat 25 kgs ofwater from 15 C to 85 C The heating is to be accomplished with hot oil CF 235 kJkgK which enters the shellside ofthe heat exchanger at 160 C The oil heat transfer coef cient is 400 WmZK The oil leaves the heat exchanger at 100 C Determine the length ofthe shell in m Approach The length of the shell is required which is equivalent to requiring the tube length and heat transfer area Q I This is a design problem and the preferred analysis TQW NlDivlatS approach is the LMI D method The water side heat D 0075 transfer coefficient must be determined quot9 4 PM LT Mt MK Oil Tm 00 C ho 7 wow Ln Assumptions 1 The ow is steady and there is no work Taw Ir c LP gt otential and kinetic energy e ects are negligible M 1 l5 with c 2 P L l nou W0 3 Water is an ideal liquid onstant specific heat Solution The governing equation for the LMTD method is Q UAFATMquot No tube thickness is given so A NrDL NrDN L where N number of tubes andM number of tube passes Substituting into the heat p1 transfer rate equation and solving for the pass length L 1 QNzDNPF 71M iaie can be obtained 39 ofenergy ppquot ow 39 no work negligible potential and kinetic energy effecm and ideal liquid with constant specific heat so that AhcPAT 39 piupeiiie iiomAppendix J T 7 v L M 529 X104 Nsm k 0643 WmK Pr 344 CI 4178 kJk K Q mcpAT 25kgmin4178kJkgK8515K1kJskW732kW AT 7 AT 75 7 85 MUM AT 160785 750 AT21001585 C MUM 799 0 39 1nAT AT2 39 ln7585 1607100 85 715 Using Figure 137 to determine F R 0857 P 0483 So from Figure 137a FZ 087 85715 150715 We use Eq 133 to determine the overall heat transfer coe 39icient 1 LRmai UA rigAA TINA 719914 nacho1 There are no fins 7191 71951 no fouling R R0quot 0 the wall resistance is RW 0 and no wall thickness so that simplifying this equation we obtain hihi The water side heat transfer coefficient requires the Reynolds number Re pm V1L2 Rg4 m H M Mir4m M7 Remember that the given mass ow rate is for 20 tubes and the Reynolds number must be calculated for one tube 425kgs Re 12030 20n529x10 Nsm20025m This is turbulent ow so using the DinusBoelter equation Nu 0023RemPr 002312030 8 344 593 h Nuk 6930643WmK w i7so 2 D 0025m mK Therefore L t 7 U327 U 400 1780 mK L 7 732kw1000Wkw 7 P 327WmzK201t0025m4087799K 512m Q Answer 1325 13 19 Because of its construction the heat transfer area of a plate heat exchanger can be changed easily by adding or removing plates in addition counter ow can be achieved which results in good performance Consider the counter ow plate heat exchanger shown below The plates are 304 stainless steel lmm thick and 2m wide and 3m long The channels on the hot and cold sides have 5mm gaps Engine oil enters at 80 C with a ow rate of 003 m3s and should leave at 55 OC Water in counter ow enters at 20 C and should leave no hotter than 30 OC Determine a the required water mass ow rate in kgs b the number of channels required Approach The water ow rate can be determined from an t 20m lmx energy balance Because we are seeking the number 39 03 w 39 of channels and we know the channel dimensions ml 3 5 5pm T anon gt ST C this is equivalent to determining the area which 1quot 9M makes this a design problem We use the LMTD lt l S MM c Tf quot44 39 C method for the analys1s Jwm CM 2 l P L 3M ml Assumptions 1 The system is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 Both uids are ideal with constant specific heats Solution a Apply energy balances to both uids Assuming steady no work negligible potential and kinetic energy effects and ideal liquids with constant specific heats so that Ah cpAT Q mwcpmz THm 77111214 macpa TCaut 7 Tam Evaluating all the uid properties from Appendix A 6 for the oil at Tavg 80552 675 C p 8599 kgm3 1 531 x104 Nsmz39 k 0139 WmK39 Pr 793 c 2076 kJkgK For water at ng 20302 25 c p 997 kgm3 1 872 x104 Nsmz39 k 0611 WmK39Pr 597 cp 4178 JkgK Q macwM 8599kgm3003m3 s2076kJkg1lt80 55K1340kw m L wan kgs 4 Answer CMATW 4178kJkgK30 20K b The governing equation for the LMTD method is Q UAFATLMVE where A NLW Assuming equal number of channels for the oil and water and solving for the number of channels N 1 ULWFATLMVE Since this is a counter ow heat exchanger F l The LMTD is 7 807 307 55 7 20 M 1n807 30557 20 We use Eq 133 to determine the overall heat transfer coefficient assuming no fins 770 lye01 no fouling R R 0 and the wall resistance is R IkA Substituting these expressions into the equation for the overall heat transfer coefficient and simplifying we obtain 71 l I l U 2 ih ml 0 From Appendix A for 304 stainless steel k 149 WmK To calculate both heat transfer coefficients we need the Reynolds number based on the hydraulic diameter ATL 421 C 1326 Re pvjh where D i 2s 20005m001m 17mm 2SW The velocity depends on the number of channels V 1 PV p4 pNWS So writing the velocity for both uids in terms of what is known 3 Oil VD 003m s 33 N2m0005m N s 8599kgm33N ms001m 486 7 531x10ANsm2 7 For any N the oil flow is laminar Water V 320kgs 7321m w N997kgm32m0005m77 997 kgm3321N ms001m 36 700 872x104Nsm2 N For any N lt 18 the flow is turbulent so we will assume turbulent Checking the oil entrance length even if N is reasonably large the entrance length is much longer than the heat exchanger length We don not have a constant wall temperature or a constant wall heat flux boundary condition so we will assume the SeiderTate equation is valid Nu 186Gzl3uus014 186RePrDL13uys014 Assuming T5 80552203022 45 C yS1630X10394 Nsmz 13 4 014 haNuk0139WmK186 j793001j 531gtlt1074 2410 f 3 Dh 001m N 3 1630x10 N m K For the water also use the Seider Tate equation Solving for the heat transfer coefficient 13 014 ha Nuk 0611WmK 186 36900 597 001 872 109 V3 4 D 001m N 3 597 NV m K Now iterating Equations 1 through 4 with the known information N 1404 therefore use N 140 parallel channels Answer ha 462 WmZK hw 210 WmZK U 377 WmZK 1327 13 20 In a counter ow heat exchanger 36 kgs of an organic uid ows with a specific heat of 850 JkgK39 it enters the heat exchanger at 12 C and leaves at 340 0C A high temperature oil with a specific heat of 1900 JkgK enters at 650 C with a ow rate of 3 kgs If the outlet temperature of the cooler uid must be increased to 450 0C with everything else remaining constant determine the percentage increase in the heat transfer area required Approach T 3 400 Sufficient information is given to evaluate the heat C Wr a ENI1 C transfer rate and we want to determine area so this is 0quot lt 0453quot Ffu d 5 345 a design problem and the preferred approach is the 5 gt 7 LMTD method T o u 50 C x 39 amp C Assumptions m 7 9 qt l The system is steady Cp l i03 Jkft 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 Both uids are ideal with constant specific heats Solution The governing equation for the LMTD method is Q UAFATLMVE a A For a counter ow heat exchanger F 1 Between the two cases the overall heat transfer coefficient and the ow rates remain constant The heat transfer rate can be obtained from conservation of energy applied to the organic uid ow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Q rhcpAT rhcp TM 7 Tam For the old case QM 36kgs850JkgK340 12Kl004X106W For the new case QM 36kgs850 JlltgK45012Kl34X106W For the LMTD the oil outlet temperature is obtained by conservation of energy on the oil With the same assumptions as before Q rhcpAT mop TM 4H a TM TH eQmcp For the old case Tm 650 C1004X106W 3kgs19001kg1lt4739 c For the new case TH 650 Cl34X106W3kgsl900IkgK 4149 C The LMTD is ATM M AT1 7AT2 650734074739712 39 39 1nAT1AT2 ln65073404739712 MW 650745074149712 39 W 11136507 4504149712 Therefore the ratio of areas is AW 7 QUFATLMVE 1W 7 QATUM 1W 7 134x1062897W 71 75 AM QUFATLMVE 1M QATUM 1M 1004x10638091M 39 Therefore a 75 increase in area is required Answer 3809 C 2897 C Comments Note that while the heat transfer rate increased by about 34 the required area increased by 75 This is a consequence of the decreasing average temperature different between the two uids 1328 13 21 The water flow inProblem P 135 is 130 lbmmin The air enters the heat exchanger at 700 F and leaves at 500 F with a flow rate of 300 lbmmin If no fouling is present determine the inside heat transfer area in ftz if the heat exchanger is a counterflow b parallel flow Approach I A a 1LT 70 l 5001 We can evaluate the heat transfer rate from the given a uquot I CV33 LbsMl information We want to determine area so for this a J I J J T 2 o 004 design problem the LMTD approach is preferred we Assumptions T l The system is steady with no work Fri439 1m 1 I 6L 2 Potential and kinetic energy effects are negligible Mw13095 33 L my Incl F 3 Both uids are ideal with constant specific heats Solution Using the LMTD method F l for both counterflow and parallel flow Hence the governing equations for the LMTD method are Q4 MAIATM and Q UAWATM The LMTD is ATLM amp honM2 Where for counterflow AT1 THVW 7 TQM and AT2 TH ichm and for parallel flow AT1 THVW 7ch and AT2 TH 7 T you Cput The heat transfer rate is obtained from conservation of energy applied to the air flow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Q macpaAT macpa THm 7THaut 7005002 600 F CM 0250Btuhrft2R Q 3001bmmin0250Btulme700500R15000Btumin The water outlet temperature is obtained by conservation of energy on the water With the same assumptions as before mwcpwAT mwcpyv Tam 7 Tom Tam Tam Qmwcpw For water from Appendix B6 assuming Tam Z 300 F Tavg 200 3002 250 F gt 0va l01Btuhrft2R Therefore TC 200 F15000Btumin 130 lbmminl 01 Btulme 3142 F For air from Appendix B7 at T avg 0111 38587300 For counterflow AT1 7007 3142 3858 F and AT2 5007 200 300 F ATLM 341 F 39 ln3858300 For parallel flow AT1 7007 200 500 F and AT2 500731421858 F 5001858 AT 3l74 F W 1n5001858 D l D D The inside heat transfer coefficient with no fouling or fins using Eq 133 is UL hi 39 ngka 39 From Appendix BZ for 304 stainless steel k 86 BtuhrftR U 7 1 00833ft1110091700833 1 00833 1 7 Btu 7 80Btuhrft2R 286BtuhrftR 40Btuhrft2RK00917 39hrftZR 4 5 Q5 15000Btumin60minlhr 943 2 39 LATUM 28Btuhrft2R341 F Therefore 4 Answers Q5 15000Btumin60minlhr 2 A pf 1013ft 39 MATH 28Btuhrft2R3174 F 1329 13 22 The counterflow heat exchanger in Problem P 1321 is operated for a year All flows and inlet temperatures remain constant but the hot uid exits at 550 OF Determine the magnitude of the fouling factor in hrftzOFBtu Approach Using the information given in Problem 1320 we can use the governing equation for the LMTD method to calculate the overall heat transfer coefficient Comparison of the dirty to the clean overall heat transfer coefficient will give the fouling factor Assumptions 1 The system is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 Air and water are ideal uids with constant specific heats Solution The governing equation for the LMTD method is Q U F AT gt U Q 1A LM 1 AIFATLM39E For counterflow F l The heat transfer rate can be obtained from conservation of energy appliedto the air flow Assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cpAT Q macpa THm 7 THaut For air from Appendix B7 at an average temperature of 625 F cp Z 0250 Btulme Q 300ml0250700550R1130 min lme min To calculate the LMTD we need the cold water outlet temperature which is obtained with an energy balance on the water using the same assumptions as used for the air and the heat transfer rate calculated above Q mwcpmz Tam TCJK Tam Tam JV mwcpw For water from Appendix B6 at an assumed average temperature of 240 F CF Z 101 Btulme TM 200 FWas57 r 39 l30lbmm1nlOlBtulme The LMTD is AT1 TH 7T5 70072857 4143 F AT2 TM 7ch 5507 200 350 F ATLME M 3812 F 39 1n4143350 Therefore 11250Bt 39 60 39 lhr U was Btg 4 Answer 943ft23812R hrft R When fouling has occurred R 1 a R 1 1 derty U512 derty U512 2 R 42 420175 hm R 4 Answer 188 Btuhrft R 280Btuhrft R Btu 1330 13 23 A closed feedwater heater is used in a Rankine cycle power plant The feedwater 150 kgs is to be heated from 30 C to 90 C using steam extracted from the turbine at 200 kPa at a quality of 0987 and the condensate should leave as a saturated liquid at 200 kPa The overall heat transfer coefficient is estimated to be 2000 WmZK Determine a the required heat transfer area in mi b the condensate flow rate in kgs Approach I P 1 2 LR Enough information is given to calculate the heat J x o 937 U wow transfer rate We want to find the required area so 39 a this is a design problem and the preferred approach is wantL the LM TD method The condensation rate and the gt gt E w C heat transfer rate are determined using conservation of 13901 got energy MwI uk S ng fUABEJ L81 Assumptions 1 The system is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 Water is an ideal uid with constant specific heat Solution a The governing equation for the LMTD method is Q UAF AT a A Q M UFATLMVE Because of the constant temperature condensation process F l The heat transfer rate can be obtained from conservation of energy appliedto the feedwater flow Assuming steady no work negligible potential and kinetic energy effects and ideal fluid with constant specific heat so that Ah cpAT where the specific heat from Appendix A6 is evaluated at the average temperature 30902 60 C CW 4181 kIkgK Q rhcpAT lSOkgs4l8lkJkgK9030K376gtlt104kW The LMTD is calculated with the given feedwater temperatures and the saturation temperature at 200 kPa Tm 12023 C AT 7 AT AT 1 2 1nAT1AT2 gt AT1 12023790 3023 C 3023 7 9023 ln30239023 376gtlt 104kw1000 WkW 343m2 Answer 2000Wm2K549K ATI THm T TCautgt ATz Thu 7 Tom AT2 12023730 9023 C MUM 549 C b An energy balance on the steam with assumptions similar to those used on the feedwater gives us Q ham 7 From the steam table Appendix Al l at 200 kPa the inlet enthalpy is hmme hf xhg 5047098722019 26780kJkg For the outlet enthalpy h 5047kJkg 376x10 kW1kJ1kWs mm 173kgs4 Answer 26780 5047kIkg 1331 1324 A shellandtube heat exchanger is to be constructed with 075in outside diameter 003in thick tubes Cold water inside the tubes has a ow rate of 500 lbmmin and is to be heated from 80 F to 110 F Hot ater wi a ow rate of350 lbmmin enters the heat exchanger at 210 F T e overall heat transfer coefficientbased on the oumide area is 300 BtuhrftZ F For one shell pass tube side water velocity of1 s and a maximum tube length of8 ft determine a the number oftubes per pass b the number oftube passes c the length ofthe tubes in ft Approach 110 who M u 350 The number oftubes N is obtained from application I l L TH 04 2390 F ofconservation of mass For the number oftube V asses assume onepass and then analyze to see ifthe WA gth of8 i p heat exchanger is less than the maximum en TD DObZYH Sufficient information is given to evaluate the heat Tc I n M transfer rate Because we seek the tube length which T39 on L a 3H 9 U0 303 is equivalent to area this is a design problem We use 94 a v4 l R H471 the LMTD method for the analysis quot 0 50 MW e l5 Assumptions The system is steady with no work 2 Potential and kinetic energy e ects are negligible 3 Water is an ideal uid with constant speci c heat Solution a The numbero b 39 b 39 quot f 39 fmass quot quot quot quot39 ofmass ow rate mNpI4X gt Nmpl4 For the cold water at T vg 95 F the roperties from Appendix B6 are c S 100 BtuhrfLR and1 620 lbmft 5001bmmin 1min60s 7 51 6 520lbmn3msii4oos75n2 So useN 52 Answer b c The governing equation for the LMTD method is Therefore N Q UOAOFATLM L with A0 NrDDL gt L 4 39 UoNrDoFATLMV f i t atcl ow Ass min rate can be obtained f 39 of energy nquot w u steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that r lbm Btu 60min Btu Ah AT T 7T 500 100 11080 R 900000 c Q q min M gt W hr For the LMTD we need Tam The hot water outlet temperature is obtained with an energy balance on the 39 39 a II ed L f 39 lav calculated above Therefore wam Qw quot1wa Ta 7TH Tim TH Qmwcw From water at an assumed average temperature of about 190 F c S 100 BtuhrfLR 900 000Bt h THO 210 F39 ur1671 F 39 3501bmmin10Btulme60min1 hr The LMTD for counterflow is AT 739va irem 2107110 100 F AT2 TM 40 1671780 871 F 100 7 871 AT 934 F M 1n100871 The value ofF depends on the heat exchanger configuration We assume 1 shellpass 1 tube pass in counter ow so F 1 Finally 900 000Bmhr 315 Answer 300Btuhr 2R521r00625ft1934 F Because this is less than 8 ft the assumed con guration works and the length is 315 ft 1332 1325 A cross ow heat exchanger is to be designed to heat hydrogen gas with hot water The water is on the tube side and enters at 150 C at a ow rate of3 gs with a heat transfer coefficient of 1250 W mZK The hydrogen c 144 kJkgK is on the shell side and enters at 30 C at a ow rate of 120 kgmin with a heat transfer coefficient of 1800 WmZK The required hydrogen outlet temperature is 60 C The heat exchanger has 100 15cm inside diameter 25mm thick tubes made of 347 stainless steel Determine a the overall heat transfer coefficient based on the inside area in WmzK b the required tube length in m Approach Because we are seeking the tube length this is Tb gc N 00 equivalent to determining the area which makes this a 39M dc O O o D 0 Ol flu design problem We use D method for e Tinsquot D a 0131 analysis Sufficient information is given to evaluate Q the heat transfer rate and overall heat tran f s er a O O O coef c ient 39 gt Assumptions gt The system is steady 39 Potential and kinetic energy e ects are negligible Me 37 quot 4 3 W5 No work is done on or by the sys em h WOWnLK LI WWWquotHQ Hydrogen and Water are ideal uids with constant a specific heam AwwH Solution a We use Eq 133 to determine the overall heat transfer coefficient 1 1 r1 Rwli V4 144 am UNA 09911vo There are no fins 710 71001 no fouling R R0quot 0 and the wall resistance is RW ln DoD l27rkLN The areas are A IrDL and AD IrDoL Substituting these expressions into the equation for the overall heat transfer coefficient and simplifying we obtain U 4 2k 11 Do From Appendix A2 for 347 stainless steel k 142 WmK 1 7 1 015mln0020m0015m 1 015m U 1250Wm2K 2142WmK 1800wm7K koozom U 731Wm2K 1 Answer b The governing equation for the LMTD method is Q U FAT gt L Q 4 M QNMFMLMV I lav can be obtained 39 of energy nquot ow Assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah CPATWlLl1 the hydrogen specific heat from Appendix A7 Qmc AT wok g 144i 6030K 1mm lkw 864kW P min kgK 60s 1kJs The LMTD is TiAT AT 1 2 ATT 7T ATT 7T rm lnATAT2 1 HM can 2 H912 Cm The wave 39 39 quot balms U the L for the hydrogen and using the heat transfer rate calculated abo a e 39 39 as used ve Note that because we defined heat transfer in is positive QM 7me Therefore 1333 Qwaler thd Qw mwcpyw THaut 7TH TH THm THm 7 mwcpw mwcpmz From Appendix A6 for water at an assumed average temperature of 120 C cp Z 4232 kJkgK o 864kW1kIlkWs 0 TH 150 C 819 C 39 3kgs4232kJkgK ATI15076090 C AT2819730519 C 907 519 AT 692 C M 1n90519 Using Figure 137 to determine F R 30760 044 P8217150057 8217150 307150 The tube side is unmixed the shellside is mixed so from Figure 137d FZ 093 Finally 7 864kW1000WkW 7 731Wm2K1007r0015m093692K L 390 m 4 Answer 1334 1326 A small oil re nery uses river water to cool some ofthe uid streams in the re nery Consider a two shell pass four tube pass heat exchanger that uses 25 kgs ofriver water at 10 C on the shell side to cool 20 kgs process uid UP 2300 JkgK from 80 C to 25 C Ifthe overall heat transfer coef cient is 600 i e a the outlet temperature of the coolant in C b the heat transfer area required in n3 Approach The water outlet temperature can be obtained from gitL a conservation ofenergy Suf cient information is U wowAlk 123m 0 C gi en to evaluate the heat transfer rate We want to NA 2 ti539 determine area so this is a esign pro lem and the T a 1be preferred approach is the LMTD method 39 w Assumptions stem is steady Messrs u14 2 Potential and kinetic energy effects are negligible 5 20 5 Tu RFC 3 No work is done on or by the stem P CP 2 13 3kg 4 Both uids are ideal with constant speci c heats Solution a TL p 39 quot from 39 fenergy pquot n w ssuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant speci c heat so that i 5 AT p Q mcpAT ms Tam Tcw Tm Tc Qmcp The heat transfer rate is obtained by conservation of energy on the process uid With the same assumptions as before Q chAT ch TM 7T H10 20kgs2300 JkgK8025K253gtlt106W For water from Appendix A6 at an assumed avera e tem er ture of 25 C c 4178 kJkgK Therefore p a Tm 10 C253gtlt106W25kgs4178kJkgK1000JkJ342 C Answer b The governing equation for the LMTD method is QUAFATUM gt AL 39 UFATLW TheLMTDis 7 807342 7 25710 AT AT 275 c AT M 7 1nAT AT2 ln 80734225 710 Using Figure 137b to determine Ffor 2 shell passes and 4 tube passes 107342044 25780079 25 7 80 10 7 80 Therefore F S 099 Finally 6 A warm 4 Answer 600Wm2K099276 C 1335 13 27 Water is heated from 25 C to 80 C in a one shell pass two tube pass shellandtube heat exchanger The hot uid is oil cp 1750 JkgK with a ow rate of l kgs that enters the tubeside of the heat exchanger at 175 C and exits at 145 C With an overall heat transfer coefficient of 350 WmZK determine a the heat transfer rate in VJ b the water ow rate in kgs c the required heat transfer area in mi Approach u L The heat transfer rate and the water ow rate can be TC m 25 determined using conservation of energy Because we 39 want to determine the area this is a design problem Tm We U350W K and the preferred approach is the LMTD method Assumptions 1 The system is steady 1 w r7936 2 Potential and kinetic energy effects are negligible 39 S 3 No work is done on or by the system m 4 The uids are an ideal uid with constant specific C 7 0 Illa K heats 0 ll lth We S olution a The heat transfer rate can be obtained from conservation of energy applied to the oil ow Assuming steady no work negligible potential and kinetic energy effects and ideal uid with constant specific heat so that Ah cpAT Q mac TM 7TH 10 kgsl750 JkgK175 145K52500wlt Answer b The water ow rate is obtained with an energy balance on the water using the same assumptions as used for the oil and using the heat transfer rate calculated above Being careful with signs on the heat transfer rate Qw QW m c T ichm gt mw cpw Tom 7 Tom From Appendix A6 for water at the average temperature of 80252 525 C cp Z 4179 kJkgK 52500W 0228k Answer 4179kJkgK80 25K1000kl gS c The governing equation for the LMTD method is Q UAFAT a A Q M UFAT LM w pw cm mw The LMTD is calculated with ATM AT1 iATZ ATT 7T AT T ATM l Hm Caut 2 H t C AT117578095 C AT2145725120 C A 957120 7 107 c M 1n95120 Using Figure l37a to determine F 1833 p ozo 1457175 257175 From Figure l37a F S 097 w mzi Answer 350Wm2K097107K 1336 13 28 Car radiators are singlepass cross ow heat exchangers with both uids unmixed Water at 005 kgs enters the tubes at 125 C and leaves at 55 C Air enters the heat exchanger at 35 m3min 25 C and 97 kPa The overall heat transfer coefficient is 225 Wm2 2 m K Determine the required heat transfer area in Approach EA Because we want to determine the area this is a W design problem We use the LMTD method for the TH 191C analysis We need to determine the heat transfer rate MN 00 using conservation of energy on the water ow L ML Assumptions quot lm 25 C U 275 4 le l The system is steady L s ball Plum 2 No work is done or by the control volume P n UN MIXED 3 Potential and kinetic energy effects are negligible 7 58 l L 4 Air is an ideal gas Tw H I mNT 550C Solution The governing equation for the LMTD method is UAFATLM 5 gt A L 39 UFATLMYE The heat transfer rate can be obtained from conservation of energy applied to the water ow Assuming steady no work negligible potential and kinetic energy effects and ideal uid with constant specific heat so that Ah 0 AT P Q rhwcpvw TH 7T Hauz From Appendix A6 for water at an average temperature of l25552 90 C CW Z 4206 kJkgK Q 005kgs4206kJkgK12555KlkWlkJsl472kW For the LMTD we need Tam The air outlet temperature is obtained with an energy balance on the air using the same assumptions as used for the water and using the heat transfer rate calculated above Therefore Q macpa Tam TCJK TCDIA TCM Qmacpv From Appendix A7 for air with an estimated TM Z 45 C TWg 25452 35 C cp Z 1006 kJkgK For the mass ow rate ma V v Assuming air is an ideal gas 7 ET 7 8314kJkmolK25273K 0 882m 7 PM 797kNm22897kgkmol1kI1kNm 39 kg 35m3minlmin60s 0882m3kg 147kW1kJ1kWs TM 25 c 4 1 c 39 066kgsl006kJkgK a 0661 g s The LMTD is AT1 Tva 7ch 1257471 779 C AT2 THVM 7ch 55725 30 C WE amp 5011 lnAT1AT2 ln77930 Using Figure l37c to determine F 2574717032 11755712507O 557125 25 7125 From Figure l37c for cross ow with both uids unmixed F S 099 Finally 147kW1000WkW 132m2 I Answer 225Wm2K099502K 1337 13 29 A small Rankine cycle power plant is used in a ship The condenser is cooled by seawater Consider a one shell pass steam side four tube pass seawater side shelland tube heat exchanger Steam enters the condenser at 50 C with a quality of 95 and a flow rate of 075 kgs and exits as a saturated liquid its condensing heat transfer coefficient is approxim ately 7500 WmZK Seawater enters the condenser at 18 C and its temperature at the exit should be no higher than 40 C Assume seawater properties can be approximated with fresh water properties The heat exchanger has 20 brass tubes 25cm inside diameter 28cm outside diameter Determine a the water side heat transfer coefficient in WmZK b the overall heat transfer coefficient based on the inside area in WmZK c the tube length required in m d the tube length required if after a long time in service both sides of the heat exchanger have been fouled in m Approach Because we are seeking the tube length this is Tquot m gooc x 045quot equivalent to determining the area which makes this a 7V 5 A z w 1 design problem We use the LMTD method for the nMs 40b quot 9 039 Li 539 15 00 M k analysis Sufficient information is given to evaluate U the heat transfer rate and the waterside flow rate using N w 5 an energy balance 54on 00171 l 00181 Assumptions Tc IXOC Soiunolbxl Db The system is steady IIBV Potential and kinetic energy effects are negligible No work is done on or by the s stem Seawater is an ideal uid with constant specific eat i rams S olution a We need the water flow rate to calculate the water side heat transfer coefficient So applying conservation of energy to the seawater assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat 7 Tam mw cpwTCm1 Team The heat transfer rate can be obtained from conservation of energy applied to the steam flow Assuming steady no work and negligible potential and kinetic energy effects Q msleamthg From Appendix A10 at 50 C hfg 23827 kIkg Therefore Q 075kgs09523827kJkg1698kW The water side heat transfer coefficient requires the Reynolds number for one tube with properties evaluated at the average temperature For water from Appendix A6 at Tavg 18 402 29 C u 798 810394 Nsmz39 k 0617 WmK39 Pr 54039 CF 4176 kJkgK 1698kW 185kg 4176kJkgK4018K s Q rhwcpvw T Cw mw Re m 4m m 2 Re NM Np7r4IZ Nit1D 4185kgs 207798X10A Nsm2 0025m This is turbulent flow so using the DittusBoelter equation Nu 0023Re08Pr04 00235900008 540 296 296 0617W K hNwltgtlt mgt73oo lt Answer D 0025m mK 59000 1338 b We use Eq 133 to determine the overall heat transfer coefficient 1 R i Ra 7 l Ila111 UA milL 77M min1a ash1a There are no fins 170 lye01 no fouling R R 0 and the wall resistance is lg lnDa Q 27rkLN The areas are A QL and Au IIDDL Substituting these expressions into the equation for the overall heat transfer coefficient and simplifying we obtain Dl D D 14 n Lg 7 h 2k h D From Appendix AZ for brass k llO WmK 1 7 1 0025min0028mo025m 1 002m U 7 7300Wm2K 2110wmK 7500wm21ltk0028m U 3720Wm2K lt Answer c Combining the expression for A N IerL and the governing equation for the LMTD method Q U F AT a L Q 1147 LM JIN DxFATLMVEf Because of the condensing steam F l The LMTD is gt ATI THm Tc mm munM2 39 39 AT15074010 C AT25071832 C ATLM ATz Thu 7 Tam 107 32 AT 189 C M 1n1032 Finally 1698kw1000wkw 3720Wm2K2070025m189K d With fouling on both sides of the heat exchanger we estimate the fouling factors from Table 131 154m Answer Seawater ms 00003m2KW Steam R m 00003m2KW 2 2 1 Rj R 3 2oooo3 00003 M UW Um D 3720Wm K w w 0028 UM lZOOWm2K 1698kW 1000w1w L 479mlt Answer 1200wm2K20n0025m189K Comments Because the clean overall heat transfer coefficient was very large the additional of any fouling will have a large effect 1339 1330 The oil cooler in a large Diesel engine is a oneshell pass fourtube pass shellandtlibe heat exchanger with 15 brass tubes 10mm outside diameter 1mm wall thi kness Oil enters the tubes at 135 C and 05 kgs and leaves at 95 C Water enters the shell at 15 C with a ow rate of2 kgs and a heat transfer coe 39icient of 1100 WmZK Determine the shell length in m Approach w Because we are seeking the tube length this is T Po c m equivalent to determining the area which makes this a a it design problem We use the LMTD method for the w v 1 1 an analysis Suff ient information is given to evaluate ENT39 PE h I ic the heat transfer rate and overall heat transfer N I S Mou coefficient Do o clam Oil Assumptions DA meow tem is steady libel Hyde lt L 2 Potential and kinetic energy e ects are negligible 39 0 115 5 3 No work is done on or by the stem 0 39 4 The uids are ideal with constant specific heats Solution a The governing equation for the LMTD method is Q UAFATlMV f Using the inside area A NrDI7 NrD 4L where L7 is the tube length and L5 is the shell length Combining this with the governing equation and solving for the shell length L Q 4U NrD FATLW r r r quotn iauc can b of ener n 4 39 ow 39 work negligible potential and kinetic energy e ects and ideal liquid with constant specific heat so that Ah CPAT Q moo Tam 4a From Appendix A6 for oil at an average temperature of135952 115 BC 5 m 2294 kJkgK u 110gtlt10394 Nsmz k 0135 WmK Pr 187 Q mcpAT 05kgs2294kJkgK13595K1kW1kJs t59kW For the LMTD we need T rm Th water outlet temperature is obtained with an energy balance on the water using the same assumptions as used for the oil and using the heat transfer rate calculated above Therefore Q mwcm T em Tep Te Tem Qmwc w From Appendix A6 for water at an assumed outlet temperature of 20 C 7ng 15202 175 C c Z4185 kJkgK T 7159 459kW1kJ1kWs 7 Com 7 7205 C 2kgs4185kJkgK The LMTD is AT Tm ergo 135 7 205 1145 C AT2 TM era 95 715 80 C AT 1 14 5 7 8 AT 0 96 2 C M 1nAT AT2 1n1 14580 Using Figure 137 to determine F R1572050138 P 957135 95 7135 7135 From Figure 137a FZ 10 0333 e use Eq 133 to determine the overall heat transfer coefficient Rm MA 144 am novo mahvo There are no fins 710 71001 no fouling R Ro39 0 and the wall resistance is RW ln DoD 2rkLN 1340 For a single tube the areas are A IerL and AD IIDDL Substituting these expressions into the equation for the overall heat transfer coefficient and simplifying we obtain 1 1 D1nDD 1 D U h 2k h D From Appendix AZ for brass k llO WmK To calculate the oil side heat transfer coefficient we need the Reynolds number for one tube 3 V L 2 4 N M N P W 4 D 405 kg s 15n110x10 Nsm20008m This is laminar flow so check the entrance len th LEW m 0037RePrD 00374821870008m267m We will assume that the tube length is shorter than this and that entrance effects must be taken into account Using the SeiderTate equation and assuming the wall temperature is approximately equal to the average water temperature because the water side heat transfer coefficient is so large for the oil u W 9990XlO39 Nsmz For a first estimate of LT m 10m 4L5 4 Re N 7ruDx gtRe 014 Nu 186RePrDLV3 1111 0 4 1864821870008101 3110x10 49990x104 412 hNuk412O135WmK695 w D 0008m m2K 71 0008 1 0010 0008 U 1 2 mn m m 1 039008 661Wm2K 695Wm K 2110wmK 1100wm2Kk0010m Finally the first estimates of the shell and tube lengths are 459kW1000WkW L 5 466lWm2K1570008m1962K This is much longer than our assumed value Using the calculated value in the SeiderTate equation gives a value of the Nu lt 366 so using the fully developed and minimum value Nu 366 h 618 WmZK U 59 l WmZK L 536 m and LT 214 m 4 Answer 479m a LT4L5191m Comments Note that the water side resistance is very small compared to the oil side resistance 1341 13 31 The regenerator in a small Brayton cycle is a singlepass crossflow heat exchanger with both uids unmixed Compressed air enters the exchanger at 300 C at 15 kgs Hot exhaust gases enter the exchanger at 850 C at 16 kgs assume the properties can be estimated as air The overall heat transfer coefficient is 250 WmZK If we want a heat exchanger effectiveness of 75 determine the surface area required in mi Approach H5455 Because we want to determine the area this is a T N 8500C design problem We use the LMTD method for the 1 l QkfS analysis We need to determine the heat transfer rate quot 5 7 39 using the given heat exchanger effectiveness Cam 4cer M Ham a N Assumptlons gt big ade 1 The system is steady 12quot 303 C Uplle ed 2 The overall heat transfer coefficient is uniform 39 L I m LY 5 over the heat exchanger 5 i a on r 3 The hot gas behaves as air V Solution The governing equation for the LMTD method is UAFATLMVE gt Q UFATLMVE The LMTD requires all four temperatures We have two temperatures given and the information that the heat exchanger effectiveness is 075 The governing equation for the s NTUmethod is Q ngax 8rhcp W THm TCJK To determine the minimum heat capacity rate evaluate the specific heats at the average temperature Assume THO Z 350 C so Tavg 8503502 600 C so from Appendix A7 cmg Z 1115 kJkgK For the cold air assume Tam Z 800 C so TaVg 3008002 550 C so cm Z 1104 kJkgK The heat capacity rates are Cg 16kgs1115kJkgK1784kJsK and Ca 15kgs1104kJkgK1656kJsK Therefore Ca CW and Q 0751656kWK850300K683kW To obtain the actual outlet temperatures apply conservation of energy to the air and hot gas flows Assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cpAT rhcpAT Solving for the outlet temperatures Gas TH m THm 7 Q 850 C 4671 c 39 mgchg 1784kWK Air TC TCm Q 300 c 7125 c 39 39 mac 1656 kWK The LMTD is AT1 Tva 7T3 850771251375 C ATMCf AT AT2 137571671 ISIISOC 39 lnAT1AT2 ln13751671 Using Figure 137c to determine F AT2 Tm rm 467173001671 C 785074671093 lD71257300OI75 71257300 8507300 From Figure 137c for crossflow with both uids unmixed F S 067 Finally 683kW1000WkW 269m2 Answer 250Wm2K0671518K 1342 13 32 A boiler is constructed as an unfinned crossflow heat exchanger Hot gases at 1200 C enter the heat exchanger and flow over 400 25mm diameter tubes at 12 kgs assume the hot gas properties as those of air Saturated liquid water enters the tubes at 8 MPa with a flow of 35 kgs and leaves as saturated vapor The overall heat transfer coefficient is 75 WmZK Determine a the gas outlet temperature in C b the required tube length in m Approach The heat transfer rate and the gas outlet temperature can be determined from energy balances on the two uids Because we are seeking the tube length this is equivalent to determining the area which makes this a design problem We use the LMTD method for the gglwrolol ML MM 7 N 400 Twaich O D 0w M analysis a O O O U Ma 11H 5 M Assumptrons gt O O 1 The system is steady l 2 Potential and kinetic energy effects are negligible Shh Mkquot 8 3 No work is done on or by the system 1 8 MP in 41 4 Hydrogen and water are ideal fluids with constant specific heats S olution a The gas outlet temperature is obtained with an energy balance on the gas assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cpAT TH T Q macpa THm T THaut Hm T Qmacpa For an assumed outlet temperature of 800 C T W 7 1200 8002 1000 C and cwz Z 1185 kJkgK The heat transfer rate is determined by applying conservation of energy to the vaporizing water with the same assumptions as before except for ideal gas Q mwhfg From the saturated steam table Appendix Al 1 at 8 MPa Tm 29506 C and hfg 14413kJkg Q 35kgs14413kJkg5050kW 5050kW T 1200 c Hm 12kgs1185kJkgK 845 C Answer b With an area of 4 NIIQL incorporated into the governing equation for the LMTD method Q UAFATLM a L UNIerFATLMVE Because the water is evaporating constant temperature F 1 The LMTD is AT1 Tva 7ch 12007 29506 9049 C AT2 THVW 7ch 8457 29506 5499 C AT AT AT2 904975499 L 713 C 394 lnAT1AT2 1n90495499 Finally 5050kW1000W1kW 300m Answer 75Wm2K4007r0025m713K 1343 13 33 To condense 3 kgs of saturated steam at 40 C a shellandtube heat exchanger with one shell pass steam side and several tube passes is used The condensing heat transfer coefficient is 11000 WmZK Cooling water enters the 19mm thin wall tubes at 15 C and exits at 24 C the maximum velocity allowable is 15 ms Determine a the number of tubes required b the number of passes required if the maximum shell length is 2 m c the actual length per pass in m d the percent increase in heat transfer rate if the water velocity is increased to 175 ms and all other conditions remain the same as in parts a b and c Approach The number of tubes depends on the total water mass WITH WC l flow rate and the maximum allowable velocity The Mgr 3 1955 mass flow can be obtained from energy balances on L H 0 o o WINL the steam and liquid water Finding the required tube 391 211 length is equivalent to finding the area so this is a M lt design problem and the preferred approach is the D 1quot 00 LMTD method For part d if the water mass flow IAEN rate is increased holding everything else constant gt then we have a rating problem and the preferred TOW ve e L5 lt be t approach is the s NTUmethod We need to calculate U 1 T I the water side heat transfer coefficient so that the MK 39 M 5 overall heat transfer coefficient can be evaluated Assumptions 1 The overall heat transfer coefficient is uniform over the heat exchanger 2 The water side flow is fully developed Solution a The heat transfer rate can be obtained from conservation of energy applied to the steam flow Assuming steady no work and negligible potential and kinetic energy effects Q rhhg From the saturated steam table Appendix A10 at 40 C hfg 24067 kJkg Therefore Q 3kgs24067kJkg7220kW To determine the number of tubes required apply an energy balance on the liquid water using the same assumptions as used for steam plus ideal liquid with constant specific heat and using the heat transfer rate calculated above and the maximum velocity Therefore Q cpw Tam T Tam From the definition ofmass ow rate r39nw NpV lx NpWD24 gt N 4mWp WD2 For water from Appendix A6 at ng 15242 195 c u 998 x10394 Nsmz39 k 0602 WmK39 Pr 694 cp 4182 kJkgK p 9985 kgm3 Therefore Q mwcpw Tam T Tam mw 7220kW 7 k mw 7192 4182kJkgK2415 s NLkgS45l7tubes 9985kgm315ms7t0019m2 So we use N 454 tubes b If the maximum tube length is 2 m we need to evaluate the tube length The governing equation for the LMTD method is Q UAFATUM a L 4 39 UN IrDF ATLMVE Because of the condensing steam F 1 The LMTD is 13 44 AT1THVM7TCVW ichm407l525 C ATLM AT1 TATZ 16 25 202 C 39 lnAT1AT2 1111625 For the overall heat transfer coefficient using Eq 133 there are no fins 1700 lyeFl no fouling or wall resistance so 1 l l U h ha The water side heat transfer coefficient requires the Reynolds number MD 9985kgm3l5ms0019m 7 7 998x10 Nsm2 This is turbulent flow so using the DittusBoelter equation Nu 0023Re08Pr04 00232850008 694 183 4072416 C AT2T Hauz Re 28500 183 0602W K h N k m 5800 D 0019m mK 71 U 4 3800Wm2K 11000Wm2K 5800Wm2K 7220kW 1000W kW L X 349m 3800Wm2K45270019ml202K Because L gt 2 In two tube passes are required c The actual pass length is 3492 1745 m d If the maximum velocity is increased to 175 ms this now becomes a rating problem The governing equation for the s N TU method is Q st 2ch m Tm 4W Because of the condensing steam CW ch g l 7 exp7NTU and NTU UACm The new overall Wm heat transfer coefficient requires a new water side heat transfer coefficient Taking the ratio of the DittusBoelter equation for the new and old conditions 08 08 hm ham VW iSOO ZV 1 175 6560 VW m K 1 0 m K 71 U 422 4llOWm2K ll000Wm K 6560Wm K 7 UA 741lOWmZK4527r0019m349m NTU 7 0482 CW 4182kJkgKl92kgs1000JkJ g 17 exp70482 0382 Therefore Q 03824182kJkgK192kgs4oo15K7680kw increase wx100 64 4 Answer 7 220 1345 13 34 The regenerator in a Brayton cycle power plant is a crossflow heat exchanger Air enters the regenerator at 200 C and exits at 380 C with a flow rate of 10 kgs Exhaust gases enter at 580 C and leave at 325 C their properties can be approximated with those of air The overall heat transfer coefficient is estimated to be 150 WmZK Determine a the required heat transfer area if both uids are unmixed in mi b the heat transfer area if the air is unmixed and the exhaust gas is mixed in mi c the heat exchanger effectiveness in parts a and b Approach I I I fowvs l39 T4 565 Sufficient information is given to evaluate the heat 11 m 530 C transfer rate and we want to determrne the required 39 area which makes this a design problem We use the 1 h w LMTD method for the analysis On glob5 re501 33 C A A Jgt gt Assumptions w l The system is steady U SOan 2 Potential and kinetic energy effects are negligible 4 3 No work is done on or by the system Tuldur 32YJC 4 Air is an ideal gas with constant specific heat Solution a The governing equation for the LMTD method is Q UAFATLMVE a A QUFATUM The heat transfer rate can be obtained from conservation of energy applied to the cold air flow Assume steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cpAT From Appendix A7 for the cold air at its average temperature of 290 C CF Z 10056 kJkgK Q rhcp TM ichm lOkgsl 0056kJkgK380 200K 1000 JkJ 181x106W The LMTD is M 7 AT1 7AT2 7 580738073257 200 M 7 1nAT1AT2 7 ln5807380325 A 200 Using Figure 137 for crossflow both fluids unmixed to determine F 1596 C 5807325142 133807200O47 3807200 5807200 From the figure FZ 087 6 181x10 W 869m2 Answer lSOWm2K0871596K b Using Figure 137 for cross ow one uid mixed and one unmixed to determine F R andP do not change From the figure F E 077 6 A w SQ 4 Answer 150wm2K0771596K c The definition of effectiveness is g Qm where QVW CW THVW 7 Tam We need the hot gas heat capacity rate so applying an energy balance to the complete heat exchanger using the same assumptions as before mHCpH THm 7 TM chpC TCaut 7 Tam F mH cp chpC Tcmiz 7 Tom THm 7 Thu chva i0kgsi0056kJkgilt380 200580 325 10056 JsK07067098kJsilt Because r39nH 0va rhccpvc 0706 the exhaust gas has the minimum heat capacity rate and the effectiveness is g chp THm T THaut THm 7 Thu i 5807325 06714 Answer chpH THm Tcumz THm Tcumz 5807200 1346 13 35 An oil cooler operates in counterflow mode Oil cp 05 BtulbmOF enters the heat exchanger at 195 F and leaves at 125 F with a flow rate of 400 lbmmin Water enters at 80 OF The overall heat transfer coefficient is 100 BtuhrftZOF and the heat transfer area is 360 ftz Determine the water flow rate in min lbm Approach I l inquot Several approaches could be use We will use the THMI 0 if Lb LM TD method Three of the four temperatures d Morwg needed to calculate the LMTD are given Sufficient w 911 4 cf faSg information is given to calculate the heat transfer rate a awe gt Assumptions 1 30 F 6 1 The system is steady F L 2 Potential and kinetic energy effects are negligible U 3 10031 A 7 36w 3 No work is done on or by the system UNIQ LF 4 Oil and water are ideal fluids with constant specific heats Solution The governing equation for the LMTD method is Q UIAFATLMpf For counterflow F 1 The heat transfer rate can be obtained from conservation of energy applied to the oil flow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Q macpa THm 7 THaut Q 4001b m 050 Btu 195 125R 60mm 840000 min lb R 1hr hr m The LMTD is ATI THm 7 Tam ATz TH 7Tan AT iAT AT 1 2 M lnAT1AT2 We do not know Tam However we can calculate the LMTD from the governing equation ATM 5 L MZ333 F 39 UAF 100Btuhr ft R360ft 1 However AT1 1957chm AT2 125780 1957Tcm7125780 AT 39 2333 F a TM 185 F M 7 In 195 7 TM 125 7 80 Applying an energy balance on the water using the same assumptions as used for the air and the heat transfer rate calculated above 39 Q T 7 T Q mwcpw cm Cm mw cpvw Tam 7 Tam For water at an average temperature of 185802 1325 F CF Z 100 Btulme 840000Bt hr 1hr 60 39 mww134mmlt Answer 100Btulme18580R mm 1347 13 36 A single pass shellandtube heat exchanger in counterflow is to be used to heat 5000 galmin of water from 50 F to 90 F using condensing steam on the shell side at one atmosphere The condensing heat transfer coefficient is 2000 BtuhrftzOF The tubes are carbon steel with a l32in outside diameter and a l05in inside diameter The maximum pressure drop through the tubes is 5 lbfinz Determine the required number of tubes in parallel and the tube length in ft Approach I cb mmj T2 2 There are heat transfer requ1rem ents and a pressure 59 Lo 2 100 what drop limitation For the heat transfer part we are l l 4 l J seeking both the tube length and the number of tubes w n45quot V V V This is equivalent to determining the area which makes this a design problem We use the LMTD method for the analysis The basic pressure drop equation can be used to address the pressure drop 12 IN 3 50 restriction Assumptions 1 The system is steady 2 Potential and kinetic energy effects are negligible 3 No work is done on or by the system 4 Water is ideal uid with constant specific heat Solution The governing equation for the LMTD method is Q UxAFATLMpf 1 Because the steam is condensing F l Sufficient information is given to calculate U and ATM The area is A N IrDL where the number of tubes N and the tube length L are unknown For pressure drop we can use L v 2 AP 2 f D p 2 For N tubes the velocity is V pAX 4rhNp7rD2 and the friction factor depends on the Reynolds number Re pDu Again we have two unknowns N and L Thus the problem requires the simultaneous solution of the two main equations We must evaluate the terms in the equations The heat transfer rate can be obtained from conservation of energy applied to the liquid water flow Assuming steady no work negligible potential and kinetic energy effects and ideal liquid with constant specific heat so that Ah cpAT Q rhcpAT rhcp TM 7 Tm For water from Appendix B6 at TaVg 50902 70 F the properties are p 6221bmft3 u 658 X10395 lbmfts k 0347 BtuhrftR Pr 682 cp l00 Btulme 3 Q622mmISOOOEIOHWLILOO Btu 90 50 F6j998x107 ft3 m1n gal lme The LMTD is M 7 2127 9072127 50 M 7 ln212790212750 The overall heat transfer coefficient is evaluated with Eq 133 There are no fins 170 lye01 no fouling 11quot R 0 and the wall resistance is lg h1DaQ2 kLN The areas are A7 NIIQL and Au N DDL Therefore 1 D D 11n1 3 UGAD hA 27rkLN hDAD From Bppendix BZ for carbon steel k 35 BtuhrftR The water side heat transfer coefficient requires the Reynolds number l4ll F 1348 Repr V1 quot 4 r34 Nan4m Re 439quot s Nirny We will assume that the flow is turbulent so using the DittusBoelter equation Nu 0023Re0 8Pro4 a h gjwommeoxprm 6 For the friction factor we will use f 079011102071642 7 The above seven equations can be solved with the given information for N L V Re f h and UA Using appropriate software to do the calculations we obtain N 1403 tubes 6 N 140 tubes L 212 ft V 132 fts Re 109200 Answer f 00177 h 2110 BtuhrftZR UA 705880 BtuhrR Comments Iterative solutions are not uncommon This is a good example to show how all the parts work together While such an iteration can be solved by hand appropriate software makes the task much easier 1349 13 37 A counterflow heat exchanger is designed to cool 20 kgs of air from 70 C to 40 OC Cold air at 10 C enters on the other side with a flow rate of 26 kgs For a modified application the basic design of the heat exchanger will remain the same as will the two air flow rates and the cold air inlet temperature However the hot air now enters at 67 C and must leave at 25 OC Assume the uid properties of the air are constant and equal on both sides Determine the ratio of the length of the new heat exchanger to the length of the original heat exchanger Approach ac T a The flow and operating conditions are given and the Tu39m 40 M 30 tube length is sought This is equivalent to finding 5 Z quot 39 the are so this is a design problem We use the LM TD method for the analysis 59 T 0 C Assumptions c w 6quot l The system is steady quotAc235 C 2 Potential and kinetic energy effects are negligible 5 bu My TilH 3 No work is done on or by the system I WWII Mot 4 Air is an ideal gas w1th constant specific heat Solution The governing equation for the LMTD method is UAFATLMVE with A IrDL Combining and solving for area L QU DFATLMVE For counterflow F 1 With the uid properties and flow rates the same in the old and new operating conditions UW UM Taking the ratio of the lengths Lnew U DFATWvE W QM ATLMzmzd L 7 QU DFATLMvcf 7 39 AT Qazd LMn2w old The heat transfer rate can be obtained from conservation of energy applied to the hot air flow Assuming steady no work negligible potential and kinetic energy effects and ideal gas with constant specific heat so that Ah cpAT Q chp THm TTngz 39 m c T 7T Therefore 70 40 0714 Qazd mH Cp THm 7 Thu 1 To calculate the LMTD we need TQM Using an energy balance on the hot and cold air streams with the same assumptions as above plus an adiabatic heat exchanger chp THm 771112 chp Tam TCM Tam Tam chp chpTHm 7THaut Old condition TM 10 C226707 40 c 3307 c 7 7073307740710 M T h1707330740710 New condition Tam lO C22667725 C 423l C 677 4239425710 ATL 3335 C AT 1944 c M ln677 nan25710 Therefore LW 0714 1224 4 Answer Lam 1944 1350 13 38 Saturated steam at 100 C condenses in a shellandtube heat exchanger one shell pass two tube passes with a surface area of 05 m and an overall heat transfer coefficient of 2000 Wm K Water enters at 05 kgsec and 15 C Determine a the outlet temperature of the water in C b the rate of steam condensation in kgsec Approach We seek the water outlet temperature which is equivalent to seeking the heat transfer rate Therefore V T m IODBC this is a rating problem and the preferred approach is I w Ah the s NTUmethod The condensation rate can be T S39OC determined from an energy balance on the steam once CWquot S the heat transfer rate is calculated gt MN Assumptions 1 amp Trc39m The overall heat transfer coefficient is uniform A 2 0 SM over the heat exchanger 1 W The system is steady U LL No work is done on or by the steam or liquid water Potential and kinetic energy effects are negligible Liquid water is an ideal liquid with a constant specific heat 1 V HeP N Solution a The governing equation for the s NTUmethod is Q ngax 8rhcp m THm TCJK Because the steam is condensing constant temperature Cm is on the cooling water side and g 17 exp7NTU and QVW rhwcpvw THVW 7 Tam The cooling water outlet temperature can be obtained from conservation of mass and energy applied to the water Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah cpAT rhwcpyw Tam 7 Tam st mwcpw Combining the above equations and solving for the cooling water outlet temperature Tam Tam From Appendix A6 for water assuming Tm 45 C T avg H 30 C CpW 4176 kJkgK Qm 05kgs4176kJsK10015K1775kJs1775kW To calculate the NTU we use the given information UA 2000 wm2K05m2 CW 05kgs4176kkgK1000 JkJ 817exp70479 0381 TC 15 c 039381177395kJS 474 c 4 Answer 39 05kgs4176kJkgK b The steam condensation rate is obtained from an energy balance on the steam Using the same assumptions as was used for the water except for ideal liquid Q mhg a m Qhg From Appendix A10 at 100 C hfg 2257 kIkg Q gQ39m 03811775kJs676kIs rh M0030k g lt Answer 2257kJkg s NTU 0479 1351 13 39 A crossflow condenser for a twospeed air conditioning system has both uids unmixed At the highest fan speed the heat transfer rate is 35 kW and the refrigerant condenses at 65 OC The air inlet temperature is 40 C and the air cannot have more than a 5 C temperature rise the overall heat transfer coefficient is 150 WmZK At the lower fan speed the air velocity is half of that at the high speed and the overall heat transfer coefficient is 125 Wm K Determine the percentage decrease in heat transfer rate at the low fan speed compared to the high fan speed Approach M 7 6c At high speed we have sufficient information to A Frail 4o evaluate WINS and the areaA Because of the W condensing steam g 17 exp7NTU and air has the LeCui led u l SDpr minimum heat capacity rate From the original Ta 4 03 0 Q7 3 LW condition we can determine 8 and N TU Once I M LN 77 those are known the new operating heat transfer rate Vquot canbe determined 7amp5 39 39 a 12 4 c LS Assumptions TQMquot 451 It 1 The overall heat transfer coefficient is uniform over the heat exchanger The water side flow is fully developed The system is steady Potential and kinetic energy effects are negligible No work is done by the system ldeal gas with constant specific heat QEJ HeP N Solution The governing equation for the 87 NTU method is Q ngax 5mcp m THm TCJK For the highspeed case we can determine the effectiveness from its defining equation 7 Q me Tam TW 7 Ta 7T5 7 457 40 11 Qmax mop THm 7 Tam THm 7 Tam 65 7 40 Applying conservation of energy to the air assuming steady negligible potential and kinetic energy effects no work and constant specific heat QmcpTcrTCm Cm 391cTLW7000E can 7 Tam 45 40K K Because of the condensing we know that g 17 exp7NTU and NTU UACm equations and solving for area A 7 7cm hlt1777 7 7000wk U 150Wm K For the lowspeed operation which has half the flow rate as the highspeed operation 125w m2K 104m2 NTUWO37Z CW 057000 WK gm 17exp70372 0311 Because the two inlet temperatures are the same Q 7 i Qf m39m g mcp m 39m Ob3il 0776 Answer 0200 m so comblning these ln1702104m2 QHS gHS me s gHS mop W HS 2 1352 13 40 A shellandtube heat exchanger with single shell and tube passes in counterflow is used to cool the oil of a large marine engine Lake water shell side uid enters the heat exchanger at 20 kgs and 15 C while the oil enters at 10 kgs and 100 C The oil flows through 100 brass tubes each 500mm long and having inner and outer diameters of 6 and 8mm respectively The shell side heat transfer coefficient is 500 WmZK Determine the oil outlet temperature in C Approach We seek the oil outlet temperature which is Ah led quot 00 equivalent to seeking the heat transfer rate Therefore i 930 wKLK this is a rating problem and the preferred approach is M w L lads the s NTUmethod We need to calculate the oil side heat transfer coefficient so that the overall heat 4 transfer coefficient can be evaluated 1 Assumptions N 39 Ioo hbbs 3 Mo I l39JilS The overall heat transfer coefficient is uniform 39 Q 50 M over the heat exchanger 0 1 00010 The system is steady Do 3 0 00 No work is done on or by the oil Potential and kinetic energy effects are negligible Oil is an ideal liquid with a constant specific heat 1 V HeP N Solution The governing equation for the s NTUmethod is Q ngax 8rhcp m THm TCJK The oil outlet temperature can be obtained from conservation of mass and energy applied to the oil Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat Q macp THm 71111214 Combining the two equations and solving for the outlet temperature SQW macpa lfwe determine Cm rhop m CW Cm and NTU UAC From Appendix A6 for water assuming TM 30 C TaVg 20 C c 4182 kJkgK For oil assuming TM 90 C Tavg 370K the properties are u 186X10394 Nsmz39 k 0137 WmK39 Pr 30039 cp 2206 kIkgK The heat capacity rates are CW 2kgs4182kJkgK8364kJSK and C 1 kgs2206kJkgK2206kJsK Therefore C Cm CW Cm 22068364 0253 Qm 2206kJsK100 15K1875kJs1875kw To calculate the NTU we need the overall heat transfer coefficient So using Eq 133 assuming no fouling and no fins so 1700 lyeFl L L R L UA M W M Wall resistance uses the metal thermal conductivity From Appendix A2 for brass k 110 WmK 7 lnDa Q 7 ln00080006 T 27rkLN T 27r110WmK050m100 The water side resistance is 1 7 1 m7 500Wm2K7r0008m050m100 The oil side heat transfer coefficient requires the Reynolds number for one tube with properties evaluated at the average oil temperature THaut THm then we can evaluate s 832gtlt10396 KW 159X10393KW 1353 4llltgs100 e 7r186x10 Nsm20006m This is laminar flow so checking the entrance length L m 0037RePrQ 00371l43000006m76m gnu R Because the tubes are only 05 m long entrance effects must be taken into account Using the SeiderTate correlation Nu 186RePrDLV3 uyw014 Assuming the wall temperature is Tm m T T 2 m 330K uT W W W 836x1074 Nsm2 Nu 186114300000605v3186x104836x10394014 112 7 Nuk 7 ll20l37WmK 256 w 7 D 7 0006m m2K x h 1 1 I14 7 256Wm2K7r0006m050m100 The overall heat transfer coefficient is UA 415x1073 832x10396 159x1073171WK 174WK The oil side resistance is 415X10393KW NTUM 00789 CW 2206WK For a counterflow heat exchanger from Table 133 liex iNTU 17C liex 7 00789 170253 8 p1 lt gt1 p1lt gtlt gt1 0075 717Cexp7NTUliC 717 0253exp700789li 0253 Therefore 00752 1875kW THW100 C936 C 4 Answer 2206kWK Comment Note that with this very low effectiveness this is a poorly designed heat exchanger 1354 13 41 Water enters a heat exchanger at 70 C with a flow rate of 2 kgs On the other side air enters at 25 C with a flow rate of 3 kgs The heat transfer area is 15 m and the overall heat transfer coefficient is 200 WmZK Determine the heat transfer rate in kVJ if the heat exchanger is a counterflow b parallel flow c crossflow with one flow the air flow mixed d crossflow with both flows unmixed Approach 6quot d The uids and geometry are given and we want the w a T l T 70 C heat transfer rate so this is a rating problem The V MW 2 L 5 preferred approach is the s NTUmethod The appropriate relation for the heat exchanger effectiveness for the specific geometry must be used Assumptions 1 The overall heat transfer coefficient is uniform over the heat exchanger Solution The governing equation for the s NTUmethod is SQVW Where Qm CW THVW ichm The effectiveness 3 depends on CW Cm and NTU UACmm The heat capacity rates use the specific heat from Appendix A6 and A7 evaluated at the average temperature Assuming TQM Z 50 C TaVg S 375 C cm Z 10064 kJkgK and THWZ 60 C TaVg Z 65 C CFW Z 4184 kJkgK Ca rh c 3kgs10064kJkgK3019kWK and Cw rhwcva 2kgs4184kJkgK8368kWK a M Therefore Ca CW and CW Cm Therefore 200W m2K 15m2 C 3390190361 and NTU 0993 8368 Cm 3019kWK1000Wkw Qm 3019kwK70 25K1359kw a From Table 133 for counterflow liexp7NTU1iC liexp70993170361 0 581 5 717Cexp7NTU17C T170361exp70993170361 Q05811359kw790kw4 Answer b From Table 133 for parallel flow 7 liexpENTU1C 7 liexp7099310361 8 1C T 10361 Q05451359kw740kwi Answer c From Table 133 for crossflow with air flow mixed 817exp717 exp7NTU 01 17 exp7li exp70993l 036m 0566 Q 0566135900w76900wlt1 Answer d From Table 133 for crossflow with both uids unmixed NTU022 078 0993022 817exp c exp7CNTU 171 17exp 0361 Q 0566135900W76900W Answer 0545 exp70361 0993078 171 0566 1355 1342 A two shell pass eight tube pass heat exchanger uses liquid water at 100 C to heat 24 kgs ofa uid UP 7 kJkgK from 25 C to 50 C The water exits the heat exchanger at 50 C The overall heat transfer coefficient is 700 Wm K Determine the heat transfer area in mi using LMTD method b the a NTU method Approach 17quot Itch The basic equations for the LMTD method and the a wait 1 NTU method are used The water ow rate needs to U 903 WNLK 7 a be determined using conservation of energy gtT u r 5 b C Assumptions 1 The system is steady 2 Potential and kinetic energy e ects are negligible 3 No work is done on or by the stern THM u ci 4 Water is an ideal uid with constant speci c heat Solution a The governing equation for the LMTD method is Q AAFATZMM gt A QMFATLMV I The LMTD is calculated with ATLMV I AT iATZ 1nATAT2 AT TH 7T cow AT Tim TcM AT 1007 50 50 C AT2 507 25 25 C MUM 507 251n5025 351 0 Using Figure 137 to determine F R 100750507 25 2 P507 251007 25033 From Figure 137b FZ 095 b 39 quot from quot fenergy quot 39 ow Assuming steady no work negligible potential and kinetic energy effects and ideal uid with constant speci c heat so that M 7 CPAT Q mcpAT 24kgs27kJkgK5025K1kW1kJs162kW 162kW 1000W IkW A 2 l 75m 4 Answer 700Wm K095351K b The governing equation for the a NTUmethod is SQW mcp m Tam ichm Ifwe determine CW mcp W then we can calculate a This can be used with Cm Cm to determine NTU UACm which would give us A ow rate is obtained with an energy balance on the water using the same assumptions as used for the tubeside uid and using the heat transfer rate calculated above Being careful with signs on the heat transfer rate Q a era a m From Appendix A6 for water at the average temperature of 75 C c 5 4190 kJkgK 7 162kW1kJ1kWs m 7 4190kJkgK10050K The heat capacity rates H 7 077kgs are 0 7kgs 4190kJkgK324kJsK and CC 24kgs27kJkgK648kJsK Therefore CH Cm gt Cm Cm 324648 05 7 Q 7 162kW1kJkWs From Figure 138d NTU m 15 Therefore A 7 NTUCW 7 15324kJsK1000JkJ7 U 700Wm2K1J1Ws 067 5 94m Answer Comments TL 1356 1343 Hot air at 250 C 100 kPa with a ow rate of 08 kgs leaves a counterflow heat exchanger at 100 C On the other side ofthe heat exchanger oil CF 2100 JkgK enters at 35 C and leaves at 110 C The overall heat transfer coe icient is estimated to be 85 WmZK Determine a the required heat transfer area in m2 air outlet temperatures if the area is increased to 25 m2 in C b the oil and Approach a m0 With given information we can calculate the heat T nwf 39 w 0 Z T transfer rate We want the area in part a which makes v Liquot ad this a design problem and the LMTD method is used 0 E Part b is a rating problem since the heat transfer rate is gt 39 U gt sought and the preferred approach is the a NTU method t 39 Term ZPC lt TW Assumptions C9 003 The system is steady with no work V 2 Potential and kinetic energy e ects are negligible 3 Hydrogen and water are ideal uids with constant specific heats Solution a The governing equation for the LMTD method is Q UAFATlMV f gt A QUFATLMY I Because this is counterflow F heat transfer rate can be obtained from conservation of energy applied to the air ow Assumin steady no work negligible potential and kinetic energy effects and ideal as with constant specific heat so that Air CPAT For air from Appendix A7 at an average temperature of 2501002 175 C c 2 10207 kJkgK Q mcpAT 8kgs10207kJkgK250 100K122kW AT TH ere 2507110 140 C AT2 TM 7ch 100735 550 7 7 122kW 1000W kW MUM 7 AT ME 7 140 5 in 8 C a A r 7m2 39Answer 39 1nATAT2 m14055 SSWmzK1978K b The governing equation for the a NTUmethod is Q SQW mcp Tam 7T0 We need to determine which uid is CW The oil mass ow rate can be obtained from conservation ofmass and quot 4 L quot 39 nnwnrk quot quot 39 and kinetic energy effects and an ideal energy pp liquid with constant specific heat 7 122kW1000WkW 7 T 7T gt 7 Q m CPv Cv 39 C cpvochmichm 2100JkgK11035K 0775k g s Heat capacity rates Ca 08kgs10207 kJkgK0817kJsK and Co 0778kgs2100 kJkgK1631kJsK Therefore C CW gt C CW 08171631 050 QM 0817kJsK25035K176kJF176kW NTU UACW 85 wrrBK25m2 817wK 250 For a counterflow heat exchanger from Table 133 liexp7NTU17C lrexp726017050 17C expENTrkc 17 050exp726017050 Q 0842176kW148kW Therefore using the energy balances above for the air and water OilTC Tcni35 c 1256 c Answer 39 39 mcp 1631kWK in An Air THomTHm7 Q awe amwc 4 Answer 39 39 mcp 0817kWK 1357 1344 A shellandtube heat exchanger has 135 tubes 125mm ID 04mm wall thickness in a doublepass arrangement Each tube pass is 448 m long Total inside surface area is 4 aust gas UP 102 kJkgK at 250 C ows outside ofthe tubes at 10 kgs the gas side heat transfer coef cient is 700 WmZK Boiler feedwater enters the tubes at 65 C and ows at a total ow rate of5 kgs The fouling factor on the water side is 00002 mZKW The air side fouling factor has the same value Ignoring wall resistance determine the heat transfer rate in kW Approach Because the heat transfer rate is sought this is a rating problem and the preferred approach is the a NI U method We need to calculate the water side heat transfer coe icient so that the overall heat transfer coe icient can be evaluated h CMLIQZWKIK Assumptions T gt 1 The overall heat transfer coef cient is uniform c I I n KW over e heat exchan er meg 17 S 4 g ea 0130on WA 2 The water side ow is fully developed Solution The governing equation for the a NTUmethod is Q SQW mcp m Tam 7T0 Ifwe determine CW mcp W CmCm and NTU UACm then we can evaluate a The exhaust gas specific heat is given We need the Water properties at the average temperature and estimate the outlet temperature as 120 C and leg 120 652 925 C and the properties from Appendix A 6 are H 299X10394 Nsmz k 06775 WmK Pr 186c 4204 kJkgK The heat capacity rates are Cw 5 kgs4204kJkgK210kJsK and CW 10kgs102kJkgK102kJsK Therefore Cg Cm gt CW CW 102210 0485 Using Eq 133 and ignoring Wall resistance since the type oftube material is not given 1 1 R39 o UA m A quotWA MA molt1 There are no ns so 7100 01 1h walcl 39 1m amp V 1 L2 5 Re 439quot 13A Mir4m M7 Remember that the given mass ow rate is for 135 tubes and Reynolds number must be calculated for one tube 7 45 kgs135 7 1r299x10quotNsm200125m This is turbulent ow so using the DittusBoelter equation Re 12600 Nuk 56306775WmK w 3050 2 00125m mK 1 7 1 00002m2KW 00002m2KW L 1 UA 3050Wm2K475m2 475m2 475m2133125 700Wrr K475m2 133125 UA 23070WK a NTUEamp226 Cm 102kWK1000W1kW From Figure 138c 5 059 Therefore Q 059102kwK25055K1302kw4 Answer Comments Using the appropriate effectiveness equation from the table the e ectiveness could be evaluated more accurately that what canberead from the gure However 39 39 in 39 39 and uid properties so the additional effort probably is not justi ed 1358 13 45 An engine oil cooler is made from a single tube lOmm diameter 3m long laid out in a serpentine path with fins on the tube outside surface both uids are unmixed in this crossflow heat exchanger The air side effective area is twelve times the inside area Air at 35 C blows perpendicular to the plane of the serpentine fincovered tube with a flow rate of 06 kgs and a heat transfer coefficient of 120 WmZK Oil enters the tube at 75 C with a flow rate of 0025 kgs Determine a the overall heat transfer coefficient based on the inside surface area assuming fully developed flow in WmZK b the oil exit temperature in C c the oil exit temperature if entrance effects are taken into account in C Approach The overall heat transfer coefficient can be determined L 7 3 with Eq 133 once the inside heat transfer coefficient D T 00 IO M is evaluated Finding the oil outlet temperature is or I equivalent to determining the heat transfer rate so this a a is a rating problem and the preferred approach is the quotr 7Yd s NTUmethod quotM quot C 4 It I 39r Mn MOZODZA39 3SJC Assumptlons 5 TE 14 l The overall heat transfer coefficient is uniform n 0J0 5 over the heat exchanger ho mi pqut K Oil and water have constant specific heats The system is steady No work is done or by the control volume Potential and kinetic energy effects are negligible V HeP N Solution a Using Eq 133 to calculate the overall heat transfer coefficient with no fins on the inside 1700 l fouling 1 1quot RD 0 or wall resistance since we have no information about the tube thickness or material 1 l l a 7 M nigh1 We are told that the the outside effective area is 12 times that of the inside so U h 12h To calculate the inside heat transfer coefficient we need the Reynolds number using the oil properties from Appendix A 6 evaluated at the average temperature Assuming the outlet temperature is 50 C ng 75502 625 c 1 623 XlO39ANsmz39 k 0140 WmK Pr 917 op 2055 kJkgK i V 1 L2 4 p4 P 4D R 40025kgs e 75623x10 Nsm20010m This is laminar flow Initially assuming fully developed flow and noting that for this crossflow heat exchanger the boundary condition is equivalent to a constant wall temperature then Nuk 73660140WmK7512 w D 7 001m 39 m2K Re a Re 4m 7WD Nu366 ah 71 1 1 U 495w Zki Answer 512Wm2K 12120Wm2K m b The governing equation for the s NTUmethod is Q ngax 5mcp W THm T Tam First we will evaluate Cm From Appendix A7 for air at 40 C 0 1007 kJkgK The heat capacity rates are 1359 Cm rhcp 06kgs1007 kIkgK 0604kwK Cm rhcp 0025 kgs2055 kIkgK00514kWK Therefore Cm CW and CyanCm 005140604 00851 Qm 00514kwK75 35K2055kw NTU 495Wm2K7r001m3m C 00514kWK1000WkW mm For both uids unmixed in crossflow 022 817expN7VU022 exp7CNTUU7871 17 expwexpi00851009l078171 00863 0091 00851 Q 008632055kW 0177kW Using an energy balance on the oil assuming steady negligible potential and kinetic energy effects no work and ideal liquid with constant specific heat and solving for the oil outlet temperature TM va Q 75 c715 c Answer mcp 0025kgs2055 JkgK c If entrance effects are taken into account use the SeiderTate equation Assuming the wall temperature is about 35 c 11W 3000 x10394Nsm Nu 186Gr13uuw014 186511917001313 623x10 43000x103940 4 804 Nuk 8040140wm1lt w h ll3 001m m2K x 1 104Wm2K 1 1 U 113Wm2K 12120Wm2K UA 7 104Wm2K7r001m3m NTU 0191 C 514WK mm 00851 g liexpmexpi00851019107871 0172 Q 01722055kW 0354kW TH m 75 c 039354kw 819 c f Answer 39 0025kgs2055kIkgK Comments Taking into account entrance effects doubled the heat transfer rate This problem is a good illustration of the need to check entrance effects in lam1nar flows 1360 13 46 A shellandtube heat exchanger with one shell pass and 20 tube passes uses hot water on the tube side to heat unused engine oil on the shell side The single 304stainless steel tube has inner and outer diameters of 20 and 24 mm respectively and a length per pass of 3 m The water enters at 87 C and 02 kgs The oil enters at 7 C and 09 kgs The shellside oil heat transfer coefficient is 1880 WmZK and the tubeside water heat transfer coefficent is 3250 WmZK Determine a the outlet temperature of the oil in C b the new outlet temperature of the oil if over time the oil fouls the surface such that a fouling factor of 0003 mZKW can be assumed in C Approach 0 Il Lo WkV944 The outlet temperature is sought which is equivalent T 70C I SZeU my to determining the heat transfer rate This means that c f mt 09 10 3 5quot 1 001 Do 2 0024quot L9 SMPnn this is a rating problem and the preferred approach is the s NTUmethod The overall heat transfer coefficient must be evaluated from the given information Assumptions The overall heat transfer coefficient is uniform m M 01 is V over the heat exchanger h 5 zmw LK The water side flow is fully developed quot M The system is steady No work is done or by the control volume Potential and kinetic energy effects are negligible Oil is an ideal liquid with constant specific heat QEJ HeP N Solution a The governing equation for the s NTUmethod is ng 5mcp THVW 7 Tam Ifwe determine Cm ch CW C max gt and NTU UAC mm then we can evaluate 3 Once we have we apply conservation of energy to the oil flow to obtain the outlet temperature Assume steady no work negligible potential and kinetic energy and an ideal liquid with constant specific heat so that Ah cpAT QC rhccpvc TM 7 Tm Equating the energy equation and the rate equation and solving TC m TC QW rhcpvc We evaluate the oil specific heat at its average temperature We estimate the oil outlet temperature to be 33 C and Tavg 7 332 20 C so that from Appendix A6 CF l880 kJkgK Likewise estimate for water ng 55 C and 0p 4l79 kJkgK The heat capacity rates are CH 02kgs4l79kJkgKlkWslkl0835kWK and CC 09kgsl880kIkgKlkWslkll692kWK Therefore CH Cm a CW Cm 08351692 0494 Qm mop Tva 4W 0835kWK877K668kW Using Eq 133 to calculate the overall heat transfer coefficient with no fins so 1700 lyeFl or fouling 11quot R 0 and the thermal conductivity of 304 stainless steel from Appendix A2 is k 149 WmK 1 1 1nltDDgt 1 am 27rkNL thg 1 ln00240020 1 3250Wm2K7r002m203m 27l49WmK203m l880Wm2K7r0024m203m 00000816000003250000118 UA4320WK l36l NTU 4320WK 517 35WK From Figure l3 8c 8 Z 076 T 770c076668kw C out 370 C Answer 1692kWK b With the addition of fouling the overall heat transfer coefficient changes Therefore 2 L 0000081600000325 0000118 M UA 70024m203m UA 1120 WK NT U 134 835 g m 060 060 668kW TC 7 CL307 C Answer 1692kWK Comments As calculated fouling can have a significant detrimental effect on the heat transfer in a heat exchanger 1362 13 47 Air at 27 C 100 kPa approaches a crossflow heat exchanger with a velocity of 34 ms Hot water enters the tubes at 93 C mass flow rate is 166 kgs The heat exchanger mixed on the shell side has 70 3cm diameter 2m long tubes Neglect wall resistance The tubes are placed five deep in an inline array with longitudinal and transverse distances between tube centers of 375 cm The air side heat transfer coefficient is 125 WmZK Determine the heat transfer rate in VJ Approach 50 037 Tn We are given the heat exchanger geometry and the inlet gm M1154 I HI N 1 v96 conditions and we want to determine the heat transfer rate Tc1 17 C 0 0 3m Therefore this is a rating problem and the preferred 3 O O D 39 039 approach is the s NTUmethod We need to calculate the P 0quot K L 2quot water side heat transfer coefficient so that the overall heat 7 transfer coefficient can be evaluated U 3394 0amp3 o39ornm gt O Assumptions wAiISc Tu 734C 1 The overall heat transfer coefficient is uniform over the n nw I 1 heat exchanger The water side flow is fully developed The system is steady Potential and kinetic energy effects are ignored No work is done on or by the pressurized water The water is an ideal liquid with constant specific heat Air is at one atmosphere SQEJ HeP N Solution The governing equation for the s NTUmethod is Q so 4414 Tm 7 TM The evaluation of the heat exchanger effectiveness requires CmmCm and NTU UA CW The air mass flow rate and the water side heat transfer coefficient need to be calculated We need water and air properties to calculate these two quantities For air its density should be evaluated at the inlet temperature while its specific heat should be evaluated at its average temperature For air from Appendix A7 at 27 C p 1177 kgm3 Assuming Tam Z 47 C TaVg m 27472 m 37 C cm 1007 IkgK For the water from Appendix A6 assuming THO Z 60 C Tavg m 93602 m 765 C u 363 X10394 Nsmz39 k 0667 WmK39 Pr 228 cp 4191 kJkgK With 70 tubes in 5 columns there are 14 tubes in each column so that tThe air mass flow rate is pV4 pV 14SL 1177kgm334ms1400375m2m420kgs The heat capacity rates are CW 166kgs4191kJkgK696kJsK and Cm 420kgs1007kIkgK423ksK So the air has the minimum heat capacity rate and C CW Cm 423696 0608 The heat transfer rate can be evaluated once NTU UACW is determined To calculate the overall heat transfer coefficient we use Eq 133 There are no fins 1700 17051 we ignore fouling and no wall thickness is l l l l l l g1ven so that U4 114 114 A h h The water side the heat transfer coefficient requires the Reynolds number for a single tube Rp a VmNmN2 a Re 439quot x4 M p7r4IZ Nit4D 4166k e 2770 707363X10 Nsm2003m This is turbulent flow but a low Reynolds number so using the Gnielinski correlation f 079111012071642 0791n27707164392 00468 1363 f8Re71000Pr 7 004688277071000228 Nu 7 138 1127f81 2 PM 71 1127004688V2 22823 71 138 0667W K so that hNukM307 Dx 003m mK 2 2 1 4 1170WK NTUM0277 707E003m2m125W 307w CW 4230WK From Table 133 for cross ow heat exchanger with Cmm mixed 817exp717 exp7NTUoC 1 17 exp7 liexp7027700608 0225 Q ng 0225423 kwK93 27K628kw 4 Answer 1364 13 48 A lowpressure boiler is a shellandtube heat exchanger with one shell pass and two tube passes with 100 thinwalled tubes each with a diameter of 20 mm and a length per pass of 2 m Pressurized liquid water enters the tubes at 10 kgs and 185 C and is cooled by boiling the water at 1 atm on the outer surface of the tubes The heat transfer coefficient of the boiling water is 4000 WmZK Determine the liquid water outlet temperature in OC Approach Slam We want to determine the hot pressurized water outlet T p wllg LP temperature which is equivalent to finding the heat In 4 00 Int K transfer rate Hence this is a rating problem and the Haw I 5 9 w kl preferred approach is the s NTUmethod We need to myoc calculate the water side heat transfer coefficient so D 0 01 Tuquot gt mwlolcj5 that the overall heat transfer coeff1c1ent can be evaluated Assumptions 1 The overall heat transfer coefficient is uniform over the heat exchanger The water side flow is fully developed The system is steady with no work and potential and kinetic energy effects are ignored Water is an ideal liquid with constant specific heat 9 He Solution The governing equation for the s NTUmethod is gQm rho n THVW 7 Tam Because the steam is boiling constant temperature Cm is on the hot pressurized water side and g 17 exp 7NTU and Q rhwcpvw THVW 7 Tam The hot pressurized water outlet temperature can be obtained from conservation of mass and energy applied to the water Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah cpAT rhwcpvw TM 7THW Combining the above equations and solving for the cooling water outlet temperature THVW TM 7 Qrhwcva The heat transfer rate can be evaluated once NT U UACW is determined To calculate the overall heat transfer coefficient we use Eq 133 No wall thickness is given so ignore wall resistance There are no fins 1700 701 and we ignore fouling so that i UA 11A hA The hot pressurized water side heat transfer coefficient requires the Reynolds number for a single tube with properties evaluated at the average temperature Assuming Tm m 130 C so that Tm m 1575 C the properties from Appendix A6 are 1 7 171 x10 4 Nsmz39 k 7 0685 WmK39 Pr 7 107 cp 7 4282 kJkgK 39 39 39 4 10k Rw a V N quotN2 a Re 4 quot 7 gS 737200 p4 p7r4D Nit1D 1007r171X10A Nsm2002m This is turbulent flow so using DittusBoelter equation Nu 0023Re0 8Pr0 3 0023372000 8 1070 3 106 So thath NukQ 1 060685wm1lt 002m 3650Wm2K L ii 1 f 1 2 1 2 7 UA479OOWK UA NirDZLP h h 100n002m22m13650wm K 4000Wm K NTU UA 47900WK 1120 d 17 71120 0674 CW 10kgs4282kIkgK10001kJ an 8 pl Q ng 7 067410kgs4282kJkgK1000Jk1185 100K72452x106w 6 THm7185 C71277 C Answer 39 10kgs4282kJkgK10001kJ 1365 13 49 After the lowpressure boiler described in Problem P 1348 has operated for six months fouling occurs such that the fouling factor is 00005 mZKW Determine a the new outlet temperature in 0C b the percent decrease in heat transfer rate Approach Fouling affects the overall heat transfer coefficient U Changes in Ucascade through all the calculations reducing the effectiveness and the heat transfer rate and raising the pressurized water outlet temperature Once the dirty U is calculated then the same calculations done in Problem Pl348 can be repeated here Assumptions These are the same as in Problem Pl348 Solution a Using Eq 133 and the information given in Problem Pl343 we can determine the overall heat transfer coefficient 1 l Rquot l l R UA hlA A haAa UALW A Using data from Problem Pl343 1 7 1 00005m2KW UAW 47900w1lt 1007E002m22m UA W 24500Wm2K 2 NTU UA 24500Wm K 0573 CW 10kgs4282kJkgK1000JkJ g liexp7NTU l iexp70573 0436 Q ng 043610kgs4282kJkgK1000Jk1185100K159gtlt106W 6 ut185 C14T9 C 4 Answer lOkgs4282kJkgKIOOOJkl du b The decrease in the heat transfer rate is 2 452 000 7 l 590 000 gtlt100352 Answer 2 452000 decrease inQ 1366 1350 Liquid R134a CF 1260 JkgK ows inside the inner tube ofa double pipe heat exchanger at 720 C 39 rate of 0265 kgs the heat transfer coefficient is 800 W mz n unterflow water at 25 C has a ow rate of 014 kgs The thinwall inner tube has a diameter of2 cm and the outer tube has diameter of3 cm both are 8 m long Determine a the heat transfer rate in W b the water and refrigerant outlet temperatures in C c ifice will form Hint calculate walltemperatures Approach We are given the heat exchanger geometry and want to determine the heat transfer rate Therefore this is a rating problem and the preferred approach is the a NTUmethod We need to calculate the water side heat transfer coefficient so that the overall heat transfer coef cienth be evaluated Assumptions The overall heat transfer coefficient is uniform over the heat exchanger The water side ow is fully developed The system is steady Potential and kinetic energy effects are ignored No work is done on or by the pressurized Wat The water is an ideal liquid with constant specific heat 95quotwa Solution The governing equation for the a NTU method is Q me 5mcm TH from 4 L L 39 F and NTU UA Cm We need Water 39 39 so assuming THWZST r r 39 Annmdi 39 39 q TM a 255215 C M 112 x104 Nsmz k 0595 WmK Pr 790 c 4184 kJkgK 0 999 kgm The heat capacity rates are Cw 0140kgs4187 JkgK586JsK and Cquot 0255 kgs1260JkgK334JsK So the refrigerant has the minimum heat capacity rate and Cquot CWCm 334586 057 To calculate the overall heat transfer coefficient for use in NTU we use Eq 133 There are no fins 711 1 1 1 1 1 UA M 1w A h h quot as the 7E1 We ignore fouling and no wall thickness is given so that The wave 39 39 39 characteristic length Re pWh y The hydraulic diameter is 4 z 4 D2 7D D 4Ax M D2 in 0037002 001m 13mm D2 D1 7L7 0140kgs 7 PA 7999kgm114003m2 002m2 999kgm30357ms001m 7 3180 112X10394Nsm2 7 UuLalOW 39 03573 S f 07911112045414 0791n31807154 2 00447 7 f8amp71000Pr 7 004478318071000790 7 25 3 1127f8v2 Prm 71 1127004478quot2 79023 71 39 39 correlation 1367 Nuk 2530595wmK w So thath 1500 D1 001m m2K 1 2 2 391 UAA ii 7r002m8m m K 262E h ha 1500W 800W K CW 334WK From Table 131 for counterflow liexp7NTUliC liexp70786li 057 g 17Cexp7NTUl 70 17 057exp7078617057 Q gQ39m 0483334wK25 20K7260w 4 Answer 0483 b The water outlet temperature can be obtained from conservation of mass and energy applied to the water Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah cpAT r39nwcpyw THVW 7 THW Solving the above equation 7260W 0140kgs4184kIkgK10001kl 7260W 0265kgs1260kgK Tm Tm 7Qmc 25 c 126 c Answer Likewise TCW chm Qrhwcmf 720 C 174 C Answer 1368 13 51 Saturated steam at 015 bar is condensed in a shellandtube heat exchanger with one shell pass and two tube passes with 130 brass tubes k 114 WmK each with a length per pass of 2 m The tubes have inner and outer diameters of 134 and 159 mm respectively Cooling water enters the tubes at 20 C with total flow rate of 230 kgsec The heat transfer coefficient for condensation on the outer surfaces of the tubes is 10000 WmZK Determine a the overall heat transfer coefficient in WmZK based on the outside surface area b the cooling water outlet temperature in C c the steam condensation rate in kgs Approach The overall heat transfer coefficient can be calculated SEAN P 39 lb LPC with Eq 133 once the water side heat transfer a 3000 WIMLK coefficient is evaluated Because we seek the outlet N b lns cooling water temperature that is equivalent to L 2M finding the heat transfer rate This is a rating problem P 39 and the preferred approach is the s NTUmethod lav 0 08 ODIWM Assumptions 1 kb I 4V1mlc 1 The overall heat transfer coefficient is uniform who 2 k5 35 over the heat exchanger The water side flow is fully developed The system is steady No work done on or by the control volume Potential and kinetic energy effects are negligible V HeP N S olution The overall heat transfer coefficient is calculated using Eq 133 There are no fins so 1700 17051 or fouling 11quot RDquot 0 so that 1 1nDD 1 UDAD hJ 27rkNL thg iiampDa1nDaD 1 UG hx Dx 2k h To calculate the inside heat transfer coefficient we need the Reynolds number From Appendix A6 and assuming the cooling water average temperature is 30 C u 779 X10394 Nsm k 0618 WmK Pr 526 cp 4176 kJkgK 5 1L2 a Re4m 4 p4 P 4Dx 7WD Remember that the given mass flow rate is for 130 tubes and the Reynolds number must be calculated for one tu e Re 4 23k 130 Re gS n779x10 Nsm200134m This is turbulent flow so using the DittusBoelter equation Nu 0023Re08Pr04 0023216000 8 526 131 21600 131 0618W K hNuk m 6040 V Q 00134m m K 1 7 1 15900159mln159134 1 UD 7 10000wm2KK134 2114wmK 6040Wm2K UD3380Wm2K 4 Answer b The cooling water outlet temperature can be found once we have the heat transfer rate which is calculated with the governing equation for the s NT U method 1369 Q ngax 5mcp m THm 7 Tam The condensing steam has Cm gt 00 SO that Cm mopcaa1mg my Cm 23kgs4176kJkgK960kIs1lt The hot inlet temperature is the saturation temperature evaluated at 15 kPa so that from Appendix A11 TH Tm 15kPa5397 C Jn Qm 960kWK539720K3260kW For a heat exchanger with one constant uid temperature 8 17 exp iNTU 3380W m2K 130 7 00159m 2 2m NTU UA L095 mm 960kWK1000W1kW 817exp70915 0599 Q 05993260kW1950kW From an energy balance on the cooling water assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat Q mop Tam TCJK Solving for the outlet temperature 1950kW1kJ1kWs 23kgs4176kJkgK c An energy balance on the steam with assumptions similar to those used on the cooling water gives us T 403 C Answer Caut Tam 20 c mc Q msleamhfg From the steam table Appendix A11 at 15 kPa hfg 2373 kJkg so that 39 1950kW 1kl 1kW msm wo82kgs Answer h g 2373 kJkg Comments Using the appropriate effectiveness equation from the table the effectiveness could be evaluated more accurately that What can be read from the figure However there are uncertainties in the heat transfer coefficient correlation and uid properties so additional effort probably is not justified 1370 13 52 Milk is pasteurized in a plate heat exchanger with hot water The two parallel passages in the heat exchanger are formed with lmm thick 304 stainless steel plates that are 3m high and l2m wide The gap between the plates for both the water and milk flows is 5 mm Water at 85 C enters the heat exchanger at a flow rate of 4 kgs The milk enters the heat exchanger at 5 C with a flow rate of 3 kgs The milk properties are p 1040 kgm3 u 00021 Nsmz CF 3900 JkgK k 065 WmK Determine the exit temperature in C of the milk if the heat exchange is a counterflow b parallel flow Approach We are given the geometry and the inlet conditions 7w Il 2M and we want to find the milk s exit temperature which T Soc is equivalent to finding the heat transfer rate This is a 6 L 3 gt 9 quot k1 rating problem and the preferred approach is the s 2 M c NT Umethod We need to determ1ne both heat transfer coefficients 3 0003 e gt Assumptions way gt 5 i f 000 l The overall heat transfer coefficient is uniform o K over the heat exchanger THWquot 39 ys c M W 39 S 2 The system is steady with negligible potential and kinetic energy effects no work and ideal gas with constant specific heat Solution The milk outlet temperature can be obtained from conservation of mass and energy applied to the milk Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah cpAT r39nmcp Tam 7 Tam Solving the above equation Tam Tam Qmmcpm The heat transfer rate is unknown Using the governing equation for the s NTUmethod Q ngax 5mcp m THm TCJK Combining these two equations Tam Tam 9 mop m THm T Tam JmmcpJn Tam 8THm T Tam We need to evaluate the effectiveness and for that we need the NTU and CWTCm From Appendix A6 and assuming the water exits at THO Z 35 C TaVg H 8535260 C u 452 x10 4 Nsmz39 k 0653 WmK39 Pr 289 cp 4181 kJkgK p 9832 kgm3 The heat capacity rates are Cm mmcpvm 3lltgs3900 JkgKl l700 JsK CW mychW 4kgs4l8lkJkgK10001kl16724JsK Therefore Cw CW a Cm Cm 1170016724 070 We need to evaluate NTU UACW Using Eq 133 to calculate the overall heat transfer coefficient with no fins so 1700 17051 or fouling 11quot R 0 and the thermal conductivity of 304 stainless steel from Appendix AZ is k 149 WmK l l I l UA 147 kL hDAD For the areas 4 AD LW The heat transfer coefficients on both sides need the Reynolds number based on the hydraulic diameter as the characteristic length Re pDh u The hydraulic diameter for both sides is 4 12 0005 D 4A 4WS m m000996m pwmd 2WS 2l20005m The velocities are l37l Mjlk V lio4gl pAX lO40kgm30005ml2m 5 Water vw l06783 pAX 9832kgm30005ml2m s The Reynolds numbers are 1040k m3 0481m s 000996m Milk Rem M 2370 00021Nsm 9832 m3 0678m s 000996m water Rew M14700 452gtlt10394 Nsm2 The milk Reynolds number is just above the transition With entrance effects the flow is probably turbulent so we Will use the Gnielinski correlation for both milk and water Milk f 07911102071642 079111237071642 00494 Pr 7 pop 7 00021Nsm23900IkgK 7 k 065 WmK u 7 f8Re71000Pr 7 004948237071000126 71977 1127f812Pr2371 11270049481 2126234 7 Nuk 7 197065 WmK 71290 w 7 7 m2K h m Dh 000996m Water f0791n12e7164392 079lnl47007l642 00283 7 f8Re71000Pr 700283814700710002897 1127f8V2 Pr23 71 1127002838quot2 28923 71 789 0653W K hw Nuk X m 5170 Dh 000996m m K The overall heat transfer coefficientarea product is 789 r1 7 1 0001m 1 7 UAi 1290Wm2K42m2 l496WmK42m2 5170Wm2K42m2 4050WK NTU M41347 CW 117000WK x For a counterflow heat exchanger 7 7 liexp iNTUUiC 7 17 exp70347li 070 7 0 099 17Cexp7NTUliC l070exp70347li 070 39 For a parallel flow heat exchanger 17 exp7NTUliC 17 exp7034717070 Pf 1 C 1 070 Outlet milk temperature Counterflow T 5 c009985 5 c129 c Answer 5 c0058855 c965 c1 Answer 0058 Caut Parallel flow T Caut Comments This heat exchanger does not raise the milk temperature sufficiently so additional area needs to be added 1372 13 53 For the heat exchanger described in Problem P 1352 if the ow length is doubled for example by having a 1800 bend at the end of one pass so that the overall length of the heat exchanger remains 3 m determine the milk exit temperature for a counter ow arrangement in OC Approach If the heat exchanger in Problem P 1352 has its area doubled the the N TU s and Tao Will change Everything else remains the same so Assumptions These are the same as in Problem Pl352 Solution Using the results from Problem P 1349 the new NTU is NTU W 2NTUW 20347 0694 Me For counter ow 17 exp7NTUliC 17 exp706941707 7 0436 817CexpiNTUliC71707exp706941707 TM5 co436855 c398 c 4 Answer For parallel ow 717exp7NTUliC 7 liexp706941707 7 0 m 8 1C 107 39 TM5 c0111855 c138 c 4 Answer 1373 1354 In 39 39 45000 lb of air from 70 F to 170 F The owners ofth so the aii 4 L 4 If remain at 170 F heat transfer 39 39 W 39 determine the new required steam pressure in psia L n in tn heat e plant want to increase production and to do 39 se L ow A roach pp CMMNK slew For the original condition we can evaluate Q and I P 20 Psi UA1 where the subscript 1 indicates the original v I condition Also from the given information we know AIL 39T 0339 cm 397 Q2 and UQAZ We can use the siNTU method to gt Twat find the required temperature because With the I W A z condensing steam 5 17 exp iNTU Using the M 431000 u 7 Maquot 39 TIL 1 1U definition of QM we can calculate the hot UL I temperature from which the corresponding saturation pressure can be found Assumptions 1 The overall heat transfer coef cient is uniform over the heat exchanger The system is steady with no Work Potential and kinetic energy e ects are negligible Ideal gas with constant specific heat bww Solution The definition ofeffectiveness is 5 QQmaxv and 2 Qzvaz a Q QM 7Q m0pmm2TamrTcMLi gjm a Q 2 Q m TH 7T k2 1 Solving for the new required temperature Tva2 Tam 5 2 Tva1 7 TC Because of L 39 we know that M 7quotquot so we need to evaluate NTU UACW For the first condition applying conservation of energy to the air assuming steady negligible potential and kinetic energy e ects no work and constant specific heat Q mac Tc TCM Qmacw Tam TCM From the governing equation for the LMTD method UA QFA UA T Tc C Tim Dividing by CW NTU mac aFATLMc l mm For condensing steam F 1 and from the saturated steam table Appendix B11 7 20 psia 22796 F 227957170 422735770 ln22796717022796770 T 7 T 17070 R M 1003 and 5 liexp710030633 FATlMV f 1997R NTUZ JlZWM 2cm ATM 997 F NTU 06NTU 0602 and C 52 liexp 70502 0452 mm 0633 TH2 70 F 22796770 F 2912 F 39 39 0452 Therefore L 4 team L H 439 D quot interpolation P2596psia 4 Answer 1374 1355 A space heater used in a university gymnasium is constructed of 60 brass tubes with 063in outside 39ameter 048in inside diameter and 3ft length The air blower provides 2000 Zmin of air at 65 F and the heat transfer coefficient is 50 BtuhrftZ F Inside the tubes 10 psig saturated steam is condensed with a heat transfer coefficient of 750BtuhrftZ F Determine a the heat transfer rate in Btuhr b the air exit temperature in F c the steam condensation rate in lbmmin Approach We are given the geometry and the inlet conditions and we want to find the heat transfer rate This is a rating problem and the preferred approach is the 5 od Db Mser N53Mm 135 a L 150133 1 T 1 39r T 1 he NIL Tclw GPFI L g F Va 103 RVMW Assumptions 1 The overall heat transfer coefficient is uniform over the heat exchanger 2 The system is steady with negligible potential and kinetic energy e ects no work and ideal gas with constant specific heat 3 Air is at one atmosphere Solution a The governing equation for the a NTUmethod is Q SQW mcp m Tam iTcym Because ofthe condensing steam E 17 exp NTU where CW ms 1 and NTU UACm Using Eq 133 to calculate the overall heat transfer coefficient with no fins so 7100 noy1 or fouling R R 0 and the thermal conductivity ofbrass from Appendix B2 is k 636 BtuhrfLR lrDoD 1 UA 114 2rkNL 11vo We estimate the air outlet temperature to be 95 F and 12 65 952 80 F so from Appendix B7 5 0240 Btulme Also from the saturated steam table Appendix B11 at 247 psia 1 2394 F and lyg 9527 Btulbm For the heat capacity rate we need the air mass ow rate In pV Assuming an ideal gas p 7 pM 147lbfin 2897lbmlbmol144in2ft2 4 1b RT A 154m lbflbmolR 55450 R 39 n3 m 00756lbmft32000 3min151lbmmin CW 1511bmmin0 240Btulme60minhr2180Bt11hrR The areas are A NrDL 60004011 311 226ft2 and A NrDoL 60z00525 3ft297ft2 The overall heat transfer coefficientarea product is 1 ln0 05250040 1350BtuhrR 1 1 750Btuhr 2R226 2 21t636BtuhrftR603 50Btuhr 2R297 2 NTUUACW 1360BtuhrR2180BtuhrR0625 a 517exp706250465 Q 0455 2180BtuhrR2394 65 F177x105Btuhr 4 Answer b Using an energy balance on the air assuming steady negligible potential and kinetic energy effects no work and ideal gas with constant specific heat and solving for the air outlet temperature Tm TCYmQCW 55 F177x105 Btuhr2180BtuhrR146 F Answer c For the steam condensation rate apply conservation of energy to the steam and using the same assumptions as before except the ideal gas assumption Q mhfg gt m Qhfg 177gtlt105Btuhr1hr60min 9527Btulbm3101bmmin 0 Answer 1375 13 56 A proposed Ocean Thermal Energy Conversion OTEC power plant uses ammonia as the working uid in a Rankine cycle Warm water from the surface of the ocean 80 0F is the heat source used to vaporize the ammonia cold water 45 OF pumped from low ocean depths is used to condense the ammonia Because of the small temperature difference between the warm and cold water the cycle thermal efficiency is very low The cycle has four identical evaporators Each evaporator has 120000 tubes made of aluminum k 92 BtuhrftOF that is 20in outside diameter 004in wall thickness and 552 ftlong Water enters the evaporator at 80 F with a velocity in each tube of 53 fts The ammonia evaporates at 72 F with a heat transfer coefficient of 1500 BtuhrftZOF The water side fouling factor is 00003 hrftzOFBtu Seawater properties are p 641 lbmft3 u 232 lbmhrft cp 094 BtulbmOF k 0340 BtuhrftOF Determine a the heat transfer rate in one evaporator in Btuhr b the maximum theoretical power output from a plant using four evaporators in kaf Approach We are given the geometry and the uid inlet temperatures and the heat transfer rate is sought AmMon 121 710 1 1015008MALQ P same This is a rating problem and the preferred approach D39 910013 is the s NT U method For part b the maximum theoretical cycle thermal eff1c1ency can be quotTil N 3 goal D o I 01 determined assumrng a Carnot cycle Usrng that and 0 the evaporator heat transfer rate the maximum 1 5 3M5 L1 557pf theoretical power output can be calculated 239 00003 MWm N 110000 A Assumptions 1 There is no fouling on the ammonia side 2 The water side flow is fully developed Solution a The governing equation for the s NTUmethod is gQm 5mcp THVW 7 Tam Because of the evaporating ammonia constant temperature CW rhcp and g 17 exp7NTU where NTU UACW Using the given seawater properties the mass flow rate is m vazr 12000064llbmft353fts7r40160ft2 8197X105 lbms Cm 8197 gtlt105lbms094Btulme7705gtlt105 BtuSR Using Eq 133 with no fins so 170a 17051 and no fouling on the ammonia side ln D D L M UA 111A A 27rkLN haAa The water side heat transfer coefficient requires the Reynolds number MD 6411bmft353fts0160ft 84 400 11 232lbmhrftlhr3600s This is turbulent flow so using the DittusBoelter equation m The heat capacity rate is Re Nu 002312e0 81w 0023844000 8 641 351 and h N k W445i D 016ft hrft R A N71131 1200007r016ft552ft 333gtlt106ft2 SimilarlyAa 347x105 f 1 7 1 00003hrft2RBtu UA 7 745Btuhrft2R333gtlt106ft2 333gtlt106ft2 ln01670160 1 27r92BtuhrftR552ft120000 lSOOBtuhrftzR347gtlt106ft2 UA l44gtlt109 BtuhrR 1376 9 NTUW0518 and 817exp705180404 CW 7705gtlt 105 BtusR3600slhr Q 04047705 x105 BtusR8072R249gtlt106 Btus 897X109 Btuhr 4 Answer b F or a Carnot cycle with four evaporators W Q39m liTLTH 1745 46080460 00648 775mg net W 4897X109Btuhr006481W3412Btuhr 681gtlt108W Answer m Comments While the cycle efficiency will be low the energy source is free Hence economics the trade off between capital costs interest on borrowed money maintenance etc and the sale of electricity using the free fuel will dictate whether or not an OTEC plant would be built 1377 1357 A one shell pass two tube pass heat exchanger uses condensing steam on the shell side with a heat transfer coefficient of 3000 WmZK to heat liquid water from 27 C to 68 The water ow rate is 5 kgs The 2 m long heat exchanger has 25 tubes of304 stainless steel each 2cm inside diameter with 1mm wall thickness Determine the required steam pressure in kPa Approach 52 Enough information is given to calculate the heat transfer at 3000 wN39LK rate and the geometIy is given The steam pressure can be evaluated if we determine the required steam temperature sztht AM to produce the given heat transfer for the given geometIy Tall 27 Several approaches can be used and we will use the 5 9 03901quot law ya5 method We need to calculate the water side heat D D QMM gt transfer coe icient so that the overall heat transfer 0 12M 580C coefficient can be evaluated LP zMFA 1 Assumptions The overall heat transfer coefficient is uniform over the heat exchanger The water side ow is fully developed The system is steady Potential and kinetic energy effects are ignored No work is done on or by the pressurized water The water is an ideal liquid with constant specific heat 95quotwa Solution The governing equation for the a NTU method is Q me 4mm TH from Because the steam is condensing constant temperature CM is on the liquid water side and 517 exp NTU and QM quot1wa Tim Tc The heat transfer rate can be obtained from conservation ofmass and energy applied to the water Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah CPAT Q mwcw Tm 7 THW quot 39 39 Tm Tcm Tc Tcm5 The heat transfer rate can be evaluated once NTU UACW is determined To calculate the overall heat transfer coefficient we use Eq 133 There are no fins 7100 mfl and we ignore fouling so that R UA M m For the wall resistance the 304 stainless steel thermal conductivity from Appendix A2 is 149 WmK n o D 111 00240020 75 R l95gtlt10 KW W erkLN 21t149WmK2m225 The condensing resistance is 1 u 1 7115 WKW 11vo 3000Wm2Kn0024m2m225 The liquid wate 39 39 39 Jul 1 39 quot p p 39 evaluated at the average temperature With Tzwg m 27 682 475 C the properties from Appendix A6 are H 553 x104 Nsmz k 0540 WmK Pr 351 c 4177 kJkgK Reamp a VLquot39N2 a Re4m u M pIr4D NitJD 45k s g 23000 1 251553x10A Nsm2002m 1378 This is turbulent ow so using the Dittus Boelter equation Nu 0023Re08Pr04 00232300008 36104 119 Nuk 119O640WmK 3800 Dx 002m m K 1 7 1 H7 3800Wm2K7r002m2m225 1 So thath 419x10395 KW UA 419x10 5 195gtlt10395 442x10395 WK 9470wK NT UA 9470WK 0453 and 816Xp04530364 CW 5kgs4177kkgK1000kJ Tm 27 c68 27 c0364140 c From the saturated liquid table Pm140 C 0361 MPa 361kPa Answer 1379 13 58 Hot exhaust gases are used on the shellside of a two shell pass four tube pass shellandtube heat exchanger to heat 25 kgsec of liquid water from 35 C to 85 C The gases assumed to have the properties of air enter at 200 C and leave at 100 C The overall heat transfer coefficient when the exchanger is clean is 180 WmZK If a fouling factor of 00006 mZKW is known to exist after operating for a period of time determine the additional area required in the heat exchanger to have the same heat transfer rate in mi Approach 3de 11 20305 Because areas are sought this is a design problem and 7655 39 the preferred approach is the LMTD method We need U sowMIK to determine the required area with and without the 1 gt 1c m S s C in uence of fouling gray 000931 A w malty ssumptlons 1 The system is steady Tam 3m 2 No work is done on or by the control volume mt 231115 3 Potential and kinetic energy effects are negligible T39 M le C 4 Water is an ideal liquid with constant specific heat Solution The governing equation for the LMTD method is Q UAF AT a A Q LM UFATLMVE The LMTD is calculated with AT 7 AT ATLM Wgt ATI THm Tcump ATz Tm FTC AT1200785115 C AT210073565 C 115 7 65 AT 7 876 C M ln11565 Using Figure 137b to determine F RM2 p 85735 0303 85735 200735 From Figure 137b FZ 097 The heat transfer rate can be obtained from conservation of energy appliedto the water flow Assuming steady no work negligible potential and kinetic energy effects and ideal uid with constant specific heat so that Ah cpAT From Appendix A7 at the average water temperature of 60 C cp 4181 kJkgK Qnquotlc AT25k g 41813 8535K IOOOW 522600W p s kgK 1kJs A 7 522600W 12 7 W When fouling has occurred 1 342m2 Answer R derty clean 522600W AM 2 3 9m 162wm K097876K Therefore the additional area required because of fouling A 379 342 37 m2 4 Answer Comments Because the clean overall heat transfer coefficient was low the addition of the fouling did not change the required area very much However if the clean overall heat transfer coefficient is high even a small amount of fouling can have a dramatic effect 00006L gt U 180 dirty 162Wm2K lt Answ er 1380 13 59 A crossflow heat exchanger has 50 tubes made from 302 stainless steel and each tube is 25cm inside diameter 25mm wall thickness and 4m long Water enters the tubes at 27 C with a total flow rate of 150 kgmin Air flow on the shell side mixed enters at 260 C with a ow rate of 100 kgmin39 the shell side heat transfer coefficient is 525 WmZK Determine a the heat transfer rate in VJ b the water outlet temperature in C A roach pp A447mxeo1 NSDUl7eS We are given the heat exchanger geometry and want to determine the heat transfer rate Therefore this is a gt O O O clown rating problem and the preferred approach is the s NTUmethod We need to calculate the water side 131quot 39 mac Do 0 0 39 heat transfer coefficient so that the overall heat a O O L 4quot transfer coefficient can be evaluated Ma DOMN gt O O Assumptlons k 75 1 The overall heat transfer coefficient is uniform 4 M1 Weltvfl39 M 17 C over the heat exchanger Ismail The water side flow is fully developed M W quot NH The system is steady Potential and kinetic energy effects are ignored No work is done on or by the pressurized water The water is an ideal liquid with constant specific heat QEJ HeP N Solution a The governing equation for the s NTUmethod is Q st 2ch m T1 46 The evaluation of the heat exchanger effectiveness requires CmmCm and NTU UA CW We need water properties from Appendix A6 to calculate these two quantities so assuming Tam Z 60 C TaVg m 2760j2 m 44 C u 588 810394 Nsmz39 k 0636 WmK39 Pr 38639 cp 4176 kJkgK Likewise from Appendix A7 assuming THO Z 100 C Tavg m 1002602 m 180 C cm 10207 kJkgK The heat capacity rates are CW 150 kgmin4176kJkgK1min60s1044kJsK CW 100kgmin10207kJkgK1min60sl700kJsK So the air has the minimum heat capacity rate and C CW Cm 17001044 0163 The heat transfer rate can be evaluated once NT U UACW is determined To calculate the overall heat transfer coefficient we use Eq 133 There are no fins 1700 17051 and we ignore fouling so that l l l RW UA M hA For the wall resistance the 302 stainless steel thermal conductivity from Appendix A2 is 151 WmK ln D D 111 0030 0025 lt gt lt gt 96110KW 27rkLN 27151WmK4m50 The air side resistance is 1 11111 7 525Wm2K7r0030m4m50 The water side the heat transfer coefficient requires the Reynolds number for a single tube 1018104 KW Rp a mNmN2 a Re4m 4 p4 POI0D Nit1D 4150 39 1 39 60 Re kgmmmm S 4330 507r588X104 Nsm20025m This is turbulent flow but a low Reynolds number so using the Gnielinski correlation 1381 72 f 079111012071642 07911143307164 00404 f8Re71000Pr 7 004048433071000386 7 7 280 1127f812Pr2371 1127004048123862371 280 0636W K W kwwwg D 0025m mK The water side resistance is l 7 X 75 a712Wm2K7r0025m4m507894 10 Km UA 7 5000 WK UA 894x105 96lgtlt10396 101x104 1 wK5000wK NTU 7 1700WK 294 From Table 133 for crossflow heat exchanger with Cm mixed 1 l 817ex 7 liex iNTUoC k liex 7 liex 729400163 0903 pl C4 pl 1 pl 016 pl 1 Q ng 09031700kwK260 27K358kw4 Answer b The water outlet temperature can be obtained from conservation of mass and energy applied to the water Assuming steady no work negligible potential and kinetic energy effects and an ideal liquid with constant specific heat so that Ah cpAT r39nwcpvw THVW 7TH T Qmc 27 c 613 c 4 Answer 5quot W W 1044kWK Solving the above equation T Caut 1382 13 60 In a cross ow heat exchanger the hot and cold sides are separated by a plate l0mm thick The hot side of the plate has straight rectangular cross section fins 5mm long and 0 lmm thick spaced 4mm on center The cold side of the plate also has straight rectangular cross section fins 5mm long 0lmm thick and spaced 3mm on center The hot side uid 05 13 kJkgK has a ow rate of 70 kghr enters at 250 C and has a heat transfer coefficient of 80 Wm K The cold side uid cp 21 kJkgK has a ow rate of 90 kghr enters at 70 C and has a heat transfer coefficient of 80 WmZK The height of both the hot and cold side ow passages is 5 mm The heat exchanger length in the direction of hot ow is l m and that in the cold ow direction is 075 m The separating plate and fins are 2024T6 aluminum Determine a the overall heat transfer coefficient based on the hot side area in Wm K b the uid outlet temperatures in C Approach 7 We can use Eq 133 to vb mmv oblate Tu10 C determine the overall heat 9 I Su z 44Wquot M R 7Q killw transfer coefficient The overall I lm gt Lquot x WnHL surface efficiencies on both sides g I c quot13 31qu of the heat exchanger need to be 3 M P 39 evaluated Finding the uid quot t g 1 outlet temperature is equivalent Dc39 m 5 an COM 3 to finding the heat transfer rate h c sowquotLL so this is a rating problem and 9 lg 5 3m C c 1 2 K1 g the s NTUmethod is preferred t mum P c 144 e IT Assumptions 1 The overall heat transfer coefficient is uniform over the heat exchanger 2 Fouling is negligible Solution a Using Eq 133 to calculate the overall heat transfer coefficient and assuming negligible fouling l l Ax l UH AH 771HhH AH kAwaH 771ChCAC Multiplying through by A H and rearranging 71 1 Ax AH 1 A 77th k UH H Awall 77DChC AC For the wall resistance the thermal conductivity of aluminle from Appendix AZ is 177 WmK The overall surface efficiency is evaluated with Eq 1194 77 17NAAmli 77 We will assume an adiabatic tip fin and take into account the tip area by using a corrected length LH LC LH I2 0005m00001m2000505m The area of one fin is AN 2LDH 2000505mlm00101m2 and A C 2LDC 2000505m075m000758m2 The Width W of each side is equal to the length of the other side so the number of fins is WCDHNCSCIS a NC D H 250fins SC 1 0003m0001m DC 7 075m SH 1 0004m0001m The total area AW NHAVH NHSHDH 150o0101m20004m1m2115m2 AW NCAVC NCSCDC 250o00758m2ooo3m075m 2456m2 AW DHDC 1m075m075m2 NH 150fins 1383 The fin efficiency is 77 tanhmLmL Where m hPkAX 0 5 2hkt0 5 05 280Wm2K 05 Hotside mLL2hkt 000505mW 01518 In 1 1 1 tanh0l518 1500OlOlm2 77M W0992 77M 17W1709920995 0 250Wm2K 0 Colds1de mLL2hkt 000505mW 0l20 1 1 1 1 1 1 tanh0l20 250000758m2 gm W 0995 a 77w 17W070995 0996 71 U 7 1 0001m 2115 1 2115 335 w Answer H 099580Wm2K l77WmK 075 o9965owm2K 2456 39 m2K b The governing equation for the s NTUmethod is Q 2ch m TH in The heat capacity rates are CH r39ncp 70kghrl300 JkgKlhr3600slOOOJlkl253WK CC rhcp 90kghr2100 JkgKlhr3600slOOOJlkl525WK Therefore CH CW and CyanCm 253525 0482 Qm 253WK250 70K4550w NTU 335Wm2K2ll5m2 W 253 WK For both uids unmixed in crossflow NTUU 22 8176xpi Cg expCNTU0787117expexpi0482280078Jig 0820 Q 08204550W 3730w Using an energy balance on the oil assuming steady negligible potential and kinetic energy effects no work and ideal liquids with constant specific heat and solving for the outlet temperatures Tm va 250 c m1025 c Answer chva 253WK TamTCVIKL70 C 14l1 C 4 Answer mccpvc 525WK 1384 13 61 Your supervisor assigns you the task of purchasing a heat exchanger to cool 85 galmin of oil 01 139 X 10395 lbmfts k 0074 BtuhrftOF cp 052 BtulbmOF p 53 lbmft3 from 250 F using 18750 ft3min of air at one atmosphere and 70 OF However in the storage building you find a new singlepass crossflow heat exchanger that has a 25 X 25 array of 2ft long 05in outside diameter 0025in thick 304 stainless steel tubes k 94 BtuhrftOF The air side outside the tubes heat transfer coefficient is 80 tuhrftZOF Determine a the possible heat transfer rate in Btuhr b the oil outlet temperature in 0F c the new heat transfer temperature and oil outlet temperature if an oil fouling factor of 0005 hrftz FBtu and an air fouling factor of 0002 hrftZOFBtu are used Approach Am The heat transfer rate is sought so this is a rating 39 Q yum 1 problem and the preferred approach is the s NTU Va 39 750 0 M 2 JimJo MW method Once the heat transfer rate is determined T O O L 29 t D then the 011 outlet temperature can be calculated 1 H t DI O I 0mg 39th b l w1 an energy a ance Pa l 41 quot O O O Assumptlons in 3310396 Q The overall heat transfer coefficient is uniform over the heat exchanger Oil and water have constant specific heats ml V quot gfj lMw THIN 2580 The system is steady No work is done or by the control volume Potential and kinetic energy effects are negligible Q 004179 V HeP N Solution a The governing equation for the s NTUmethod is Q ngax 5mcp m THm T Tam lfwe determine Cm ch m CW C m and NT U UAC then we can evaluate s First we will evaluate Cm Oil properties are given At the air inlet temperature from AppendixB7 p 0075 lbmftj We estimate the air outlet temperature to be 130 F and Tavg 70 1302 100 F so that CF 024 Btulme The heat capacity rates are pV00751bmft318750ft3min 1406lbmmin CW rh c l406lbmmin024 Btulme338BtuminR an p mm pV39 53lbmft385galmin0l337ft3gal 602lbmmin Cm rhmlcp 6021bmmin052 Btulme 3l3BtuminR Therefore Cm CW Qm 313BtuminR25070R56340Btumin For the effectiveness we need NTU UACmm Using Eq 133 to calculate the overall heat transfer coefficient with no fins so 1700 lyeFl or fouling Hiquot R 0 and the thermal conductivity of 304 stainless steel is k 94 BtuhrftR 1 1 1nDaD 1 UA th 27rkNL haAa To calculate the inside heat transfer coefficient we need the Reynolds number using the given oil properties Rp a 1L2 a Re4m 4 p4 POI4 7WD 1385 Remember that the given mass flow rate is for 625 tubes and the Reynolds number must be calculated for one tube Re 7 46021bmminlmin60s625 7 392 71139x10395 lbmfts00375 ft This is laminar flow so check the entrance length Pr 7 ucp 7 139x105 lbmfts3600shr052Btulme k 0074BtuhrftR L s 0037RePrD 003739235200375ft191ft gnu Because the tubes are only 2 ft long entrance effects must be taken into account Using the SeiderTate equation and ignoring the in uence of the wall to bulkviscosity difference since no data are given Nu 186GrV311W 186RePrDLV3 13 392 352 00375 Nul86 ll9 h7 Nuk 71190074BtuhrftR723 4 Btu Q 7 0037511 39 hrft2R 1 1 ln004l700375 1 UA234Btuhrft2R700375ft2ft625 2794BtuhrftR6252ft 80Btuhrft2R7004l7ft2ft625 000029100000014400000763 UA 2720BtuhrR U 7 2720BtuhrR I 7 0144 313Btum1nR60slm1n Using Figure 138f Cm mm Cm le 338313 108 s z 01 Q 0l56340Btumin5630Btumin Answer b Using an energy balance on the oil and solving for the oil outlet temperatureTHm Hm Q 250 F560B232 F Answer 39 39 mcp 602lbmm1n052Btulme n c With the addition of fouling the overall heat transfer coefficient changes Therefore 1 00 2 l m A l A l U A 7r003752675 l 7r004l72675 l 2720 duty clean UdWA 2410BtuhrR U O 128 31360 From the figure 5 m 01 which is about the same as before so both the heat transfer rate and the outlet temperature would be about the same However in reality they both would be slightly lower but because the reading of the accuracy of the figure we cannot determine how much lower with any accuracy Comments Because the overall heat transfer coefficient is low adding fouling did not have much affect on the heat transfer rate If the overall heat transfer coefficient is large fouling can have a significant detrimental effect on the heat transfer in a heat exchanger 1386 13 62 A steam condenser is constructed of 100 brass tubes each 125in outside diameter 10in inside diameter and 8ft long Water enters the tubes at 60 F with a total flow rate of 800 galmin Saturated steam at 5 psia is condensed on the shell side of the heat exchanger and has a heat transfer coefficient of 1250 BtuhrftzOF Determine a the heat transfer rate in Btuhr b the outlet temperature of the water in 017 Approach slum P 9an 10 mm 31 We are given the geometry and we want to find the l k 91 heat transfer rate so this is a rating problem and the V V preferred approach is the s NTUmethod We need to wAlElL calculate the water side heat transfer coefficient so gt 39 9 that the overall heat transfer coefficient can be 12 m 60 evaluated quot 39 W a 9 sz goo NW N goofyth Assumptions 90 o OM 4 1 The overall heat transfer coefficient is uniform over the heat exchanger 2 The water side flow is fully developed 3 The system is steady with negligible potential and kinetic energy effects no work and ideal liquid with constant specific heat Solution The governing equation for the s NTUmethod is Q ngax 8rhcp m THm TCJK Because of the condensing steam constant temperature 8 17exp7NTU and CW ch We need to evaluate NTU UACmm Using Eq 133 to calculate the overall heat transfer coefficient with no fins so 1700 17051 or fouling 11quot RDquot 0 and the thermal conductivity of brass from Appendix B2 is k 636 BtuhrftR l D D L LM UA 174 27rkNL hDAD The water side heat transfer coefficient requires the Reynolds number for one tube with properties evaluated at its average temperature We estimate the liquid water outlet temperature to be 120 F and ng 601202 90 F so that from Appendix B 6 op 100 Btu1me p 621 lbmft3 1 514 x105 lbmfts k 0359 BtuhrftR39 Pr 513 Also from Appendix B11 at 5 psia Tm 16221 F Re7pD 7 m 7 m 67 4m 7 4pV39 11 NpA Np7r4D2 N7ruD N7ruD 4 621lbm ft 800 al min 01337 ft al 1min 60s Re lt g gt g lt gt732920 1007t515x10395 lbmfts00833ft This is turbulent flow so using the DittusBoelter equation Nu 0023Re08Pr 00233292008 513 182 9 h 7 Nuk 7 1820359BtuhrftR 7784 Btu 7 D 00833ft hrft2R L 609x10 6 hA 784Btuhrft2R100700833ft8ft Btu The wall resistance is lnDaDx ln0104200833 700x10397 27rkNL 27636BtuhrftR1008ft Btu The condensing steam resistance is 1387 l l haA 1250Btuhrft2R1007L01042ft8ft The overall heat transfer coefficientarea product is UA 609x10396 700gtlt10 7 3055 gtlt1039671 BtuhrR 101 500BtuhrR The heat capacity rate is CW 6211bmft3800 galmin0l337 ftEgal60minlhrl 00 Btulme 399gtlt105BtuhrR UA 7 101500 BtuhrR NTU o255 C 399gtlt10 BtuhrR g 17 exp70255 0225 Q o225399x105BtuhIR16221 60 F9l6gtlt106 Btuhr Answer Using an energy balance on the water assuming steady negligible potential and kinetic energy effects no work and ideal liquid with constant specific heat and solving for the water outlet temperature 6 TCautTCm Q 60 FW830 F Answer 39 39 Cm 399gtlt10 BtuhrR 73055x10 6 Bt 1388 13 63 The condenser tubes described in Problem P 1362 are retrofitted with annular exterior fins The brass fins are 025in long 0125in thick and spaced 0375in apart All other dimensions remain the same as does the heat transfer coefficients on both sides Determine a the heat transfer rate in Btuhr b the outlet temperature of the water in 0F Approach 1 The addition of fins to the tubes in Problem Pl362 I T gt changes the overall heat transfer coefficient We use Eq 133 to determine it The fin efficiency must be determined 3 CC 31 H Assumptions k t t C L 0 1 l The overall heat transfer coefficient is uniform I 7 over the heat exchanger F lt3 3574 If 2 The water flow is fully developed r o t 417 H 3 Fouling is negligible 1 L t O O Lk 51 I Solution Following the solution of Problem Pl362 we use Eq 133 to calculate the overall heat transfer coefficient but now incorporate the overall surface efficiency on the outside surface 1 1 1nDD 1 mm 27rkWN nigh1 1 ml The thermal conductivity of brass from Appendix BZ is 636 BtuhrftR and the wall resistance is the same lnDaDx ln0104200833 7oo3x10397 27rkWN 27636BtuhrftR8ft100 Btu The inside thermal resistance is the same 1 7 1 H7 784Btuhrft2R1007r00833ft8ft With annular fins on the outside we can determine the fin efficiency using Figure 1118 and the overall surface efficiency with Eq 1194 77 17NAfAm 17 77 The total area requires the number of fins N over the total length W 609lx10396 Btu WWW a NHMtgtzm SI 003l3ft00104ft Therefore we use N 191 an integer number of fins The total area will take into account the tip area by using the corrected length so we need the corrected radius r2 r2 l2 00729ft00104ft2 00781ft Area of one fin A 27152 r12 27r007812 7 005212 00213ft2fin Total area Am 27rr1W7NtNA 200521ft8ft 19100104ft19100213ft2 6031ft2 Now for the parameters in Fig 1118 r2r1 0078100521 15 L L I2 00208ft00104ft20026ft A Lt0026ft00104ft000027ft2 12 12 L3 Oioz ftpr 1250Btuhrft2R 113 kA 636BtuhrftR000027ft2 From Figure 1118 nfz 057 19100213ft2 Overall surface efficiency 77D a 17 2 39 6013ft 17057 071 1389 The finned side resistance is 1 1 man1m 7 0711250Btuhrft2R6013ft2100 The overall heat transfer coefficientarea product is 1868x10396 Btu 71 UA 6091X10 6 7003gtlt10 7 hI tR4r1868gtlt10 6 115000B tu u Btu NTU UA 115000BtuhrR 0290 CW BtuhrR 816Xp0290 0251 Q gQ39m 0251399gtlt10516221601024gtlt106Btuhr 4 Answer T 6 W TCVM Q 600F1024gtlt10 Btuhr m 5 857 F 1 Answer Wow 399gtlt10 BtuhrR 1390 1 1 For the following systems define a control volume and state whether the system is open or closed and steady or unsteady Identify any and all heat transfer energy flows mass flows and energy transformations flows cket b Pot of boiling water with no lid c Portable space heater with fan d The jet airplane in Figure 11 c e The house in Figure 14 Approach For each system identify a control volume with a dotted line and then describe the various processes occurring to the system Solution a Define the control volume to surround the rocket Because the hot exhaust gases cross the boundary this is an open system As the propellant burns a mass flow exits through the rocket s nozzle and the mass of the rocket decreases changes with time so the system is unsteady The nozzle is very hot compared to its surroundings so there is heat transfer across the boundary but it is negligible compared to the energy flow associated with the hot exhaust gases The stored energy in the rocket propellant is converted to kinetic energy in the rocket s nozzle b Define the control volume as shown in the figure to the right Heat transfer crosses the boundary from the gas ame to the pot bottom Because the water is boiling water vapor crosses a mass flow rate the boundary at the top of the pot so the mass of the system decreases with time so this is an unsteady system The vapor leaving the system removes energy from the system Note that if the water was not boiling and heat added then the mass of the water would remain the same but the temperature of the water would increase with time so the system would still be unsteady c Define the control volume as shown in the figure Electricity crosses the boundary in the wire39 electricity is considered work Air is sucked into and pushed out of the heater by the fan39 this is the same mass flow rate crossing the boundary in two locations Because neither air nor energy is stored in the heater and the air flow is constant this is a steady system Note that once the electricity is inside the heater some of the electricity is converted to mechanical work to drive the fan and some electricity is dissipated in the heater to heat the air Because neither of these conversion processes results in work or heat crossing the boundary we have defined we do not have heat transfer or a second work process Electricity mechanical work is converted to thermal energy in the flowing air and kinetic energy in the air d Define a control volume to encompass the airplane The processes are identical to those identified in part a for the rocket e Define a control volume as shown in the figure to the right Heat transfer enters the house due to the solar radiation Assuming the air outside the house is warmer than the air inside the house there also is heat transfer from the surroundings into the house Electricity enters the house to drive the air conditioner electricity is considered mechanical work A mass flow of refrigerant enters and leaves the house at different conditions where the control volume cuts through the two pipes Assuming the air temperatures solar radiation and flow rate of refrigerant are constant this is a steady system 1 2 Describe some of the thermalfluid systems in a typical residence define a boundary and describe the energy andor mass flows associated with them Approach Chose several systems identify a control volume with a dotted line and then describe the various processes occurring to the system Solution a Define a control volume to surround a refrigerator as shown in the first diagram39 the control volume passes along the surface of the heat exchanger at the back of the refrigerator Assume the cooling unit is operating Electricity crosses the boundary39 electricity is considered work There is heat transfer from the warmer room to the colder interior of the refrigerator As the cooling unit is running for example to freeze ice cubes the total energy level in the refrigerator drops so that the system is unsteady If the cooling unit is not running there is no work crossing the boundary but heat transfer still occurs from the warm er room to the cooler refrigerator interior The energy level in the refrigerator increases so the system is unsteady Note that no mass flows across the boundary so this is a closed system Note that the cooling unit has a heat exchanger often on the back of the refrigerator that transfers energy from the hot refrigerant to the cooler room Depending on where the control volume surface is drawn the processes occurring will change Defining the control volume to pass along the surface of the heat exchanger then this there is a heat transfer process with heat flowing from the surface of the heat exchanger to the room air but no mass flow so this is a closed system If the control volume is drawn farther away from the heat exchange surface to include the air then there is a mass flow of air at different temperatures into and out of the control volume but there is no heat transfer39 this results in an open system If the control volume is drawn to cut the tubes conveying the refrigerant then there is a mass flow of refrigerant at different condition into and out of the control volume but no heat transfer or mass flow of air but again this is an open system b Define a control volume around a gas oven as shown on the figure to the right and assume the oven has been just turned on Gas flows mass flow rate across the boundary39 so this is an open system Electricity is also flows across the boundary so there is a work process Air is needed to burn the gas so cold air enters and hot exhaust gases exit the control volume so these are two more mass flows carrying energy that need to be taken into account The surface of the oven is warmer than the room so there is a heat transfer process With time the oven heats so this is an unsteady system If the oven has been on for a long time and the desired oven temperature has been reached the oven thermostat maintains the fixed temperature The gas and air flows remain constant and the temperature is fixed so this then would be a steady open system c Define a control volume to encompass the lightbulb No mass crosses the boundary so it is a closed system Electricity flows to the bulb so there is a work power term The bulb is hot so there is heat transfer from the bulb to the surrounding air Likewise some energy leaves the systems as visible light When the lightbulb is first turned on it takes a little while before it reaches a steady temperature so it would be an unsteady system After a long time operating the temperature would stabilize and the system would be steady side v1qu 01C La uf u 4 1 3 For the following four systems define a control volume state if the system is steady or unsteady open or closed constant volume or changing volume constant uid density or changing uid density Also identify all heat transfer energy flows and mass flows a Swimming pool being filled Choose one control volume as the whole pool then choose a second control volume one surface of which follows the surface of the rising water b Helium tank being filled c Helium balloon being filled Approach For the given systems identify a control volume with a dotted line and then describe the various processes occurring to the system Solution a Define a control volume to surround the physical pool envelope the concrete tile etc and cutting across the top opening as shown Water flows in so mass crosses the boundary The total mass of the system is increasing so this is an unsteady open system The rising water level 39 displaces air so air leaves the system However the air mass flow rate is very small compared to the water flow rate and can be assumed to be negligible We assume the water has the same temperature as the pool envelope so that there is no heat transfer process The total volume of the system remains constant but the mass increases The density of the water is constant Mechanical work crosses the boundary because the increasing volume pushes against atmospheric pressure39 force is pressure times area and work is force times distance but this work would be very small Define the control volume to include only the water in the pool During the filling water crosses this boundary so this is an unsteady open system Unlike the previous control volume air does not play a roll in the present defined control volume The volume of the system is changin with time39 both the mass and the energy in it increase Mechanical work crosses the boundary because the increasing volume pushes against atmospheric pressure39 force is pressure times area and work is force times distance but this work would be very small The density of the water in the control volume remains constant b Define a control volume to follow the inside surface of the helium tank Mass enters the control volume from the filling port so this is an unsteady open system Energy flows along with this mass The volume of the system is fixed The density of helium increases with time With fixed volume no mechanical work occurs As the gas is inside the tank is compressed the gas temperature will rise If the temperature of the tank becomes greater than that of the surroundings then there will be heat transfer c Define a control volume to follow the inside surface of the balloon Mass enters the control volume at the balloon s opening so this is an unsteady open system The volume of the system increases The density of the helium also increases slightly with time With the increasing volume mechanical work occurs across the boundary but this would be very small The gas compression is small39 any heating of the gas inside the balloon due to the compression would be small so heat transfer would be negligibly small 14 A thermal solar energy system consists ofa solar collector on the roof ofa house a hot water storage tank to store hot water a heat exchanger through which the hot water passes a fan that blows air through the heat exchanger to heat the house and a pump to circulate water through the complete system Define several 439 quot 39 quot 39 39 fe uipment or collections of equipment and identify if the control volume is steady or unsteady open or closed what heat transfer energy ows mass ows and energy transformations occur and if constant or varying volume Approach Identify several control volumes in this thermal solar energy system with dotted lines and then describe the various processes occurring to the systems Solution a Define a control volume to surround the solar collector on the roof Solar energy enters the system this is a heat transfer process Assuming the collector surface is a temperature greater an the surrounding air there is a second heat transfer process The pump circulates water through the collector Assuming the solar radiation the water ow the surrounding air temperature an the entering water temperature are constant this is a steady open system The volume is xed the water entering the solar collector is not constant that is we are trying to increase the water temperature in the storage tank and everything else is constant as in the first part ofthis answer then the system would be unsteady because the overall r level ofthe cuneuur 39 39 b Define a control volume to encompass the pump and is electric motor Assume they are running at a constant speed Electricity crosses the boundary electricity is considered mechanical work Water enters and leaves across the boundary mass ow rates This is an open steady system c Define a control volume to encompass the storage tank Assume it is well insulated so there is no heat transfer from it u t I 1 temperature This is an unsteady open system Volume is fixed No mechanical work occurs d Define a control volume to surround the fan and heat exchanger Air and water ow into and out of the control volume Electricity work crosses the boundary to drive the fan This is an open steady system 15 In hydroelectric plants electric power is generated from the ow ofwater from a reservoir such as shown in Figure 118 The water ows continuously with a seemingly endless supply How is the wa er replenished Where does the energy in the water come from that is converted to electrical power Approach The question can be answered with either an open system analysis or a closed system analysis For the open system choose a control volume that includes the dam the reservoir and the river waterjust downstream ofthe dam For the closed system choose a control volume that includes all the water on the planet Solution a De ne a control volume that includes the dam the reservoir and the river waterjust downstream f the dam 39 39s sy m w39 leaves via the down en the control volume at a higher ow rate than it leaves water level falls Lfthe ow rates are equal e sy te is in steady state This choice ofcontrol kinetic energy of the downstream river water b De ne a control volume that includes all the water on the earth No water enters or leaves this control volume so it is a closed system The water at the base of the dam ows down the river and empties into an ocean Water evaporates from the ocean surface and is suspended in the air as water vapor and clouds The water is transported over the land by air currents and eventually falls to the earth as rain snow or hail The water runs over the land and soaks into the soil Some water is st snow or ice and later melts into liquid form Both location upstream ofthe dam and acts to replenish the supply of water An energy balance on the control volume reveals that heat is added to the earth s water by the sun and by intemal heat generation within the arth s crust ssion reactions Work is done by the hydroturbine in the power 1 shown in the gure driving water circulation on the earth There are of course many more eat and work interactions with the earth s water than the two shown 1 6 The radiator of a car is a heat exchanger Energy from the hot water that flows through the heat exchanger is transferred to the cooler air that also flows through the radiator For the three control volumes defined below state if the system is steady or unsteady open or closed and what heat transfer energy flows mass flows and energy transformations occur a Water b Air c Complete heat exchanger Approach For the three systems specified clearly identify the control volumes with dotted lines and then describe the various processes occurring to the systems Solution a Define a control volume CV l to surround the water flowing through the car radiator Assume the air flow rate and inlet temperature are constant39 also assume the water flow rate and inlet temperature are constant Mass water flows across the boundary so this is an open system Heat flows from the water across the boundary to the air so there is a heat transfer process Everything is constant so this is a steady system No mechanical work occurs b Define a control volume CV H to surround the air flowing through the car radiator Assume the air flow rate and inlet temperature are constant39 also assume the water flow rate and inlet temperature are constant Mass air flows across the boundary so this is an open system Heat flows from the water across the boundary to the air so there is a heat transfer process Everything is constant so this is a steady system No mechanical work occurs Note that Q 7 7Q because of the sign convention c Define a control volume CV 1 CV H to encompass the complete heat exchanger and assume the outside surface of the heat exchanger is well insulated Assume the air flow rate and inlet temperature are constant also assume the water flow rate and inlet temperature are constant Mass air and water flows across the boundary so this is an open system Everything is constant so this is a steady system No mechanical work occurs Note tha there is no heat transfer Heat transfer occurs only across boundaries The energy flowing from the water to the air is internal to the control volume and not across a boundary so by definition there is no heat transfer 39 v39v cod mL QJZZ I CV IE Iquot A 1 17 An acom is planted in the ground After many years the acom grows into a mighty oak tree De ne a 5y 7 1 39rlxxL I39AL 39L 0 Approach n v transformations Solution n c 4 L r u 39rL r 7 since mass in the form of air water and organic materials crosses the control Volume over time The tree grows so the system is transient Sunlight enters the control Volume and provides energy for photosynthesis This energy is stored in the tree as chemical energy in the molecules The tree is composed of organic molecules with four main constituenm carbon hydrogen oxygen and nitrogen The tree absorbs carbon dioxide C02 from the air reduces it and returns gen to the atmosphere The tree also draws water H20 and dissolved nutrients which contain nitrogen through the room trunk and branches Much ofthe mass in the tree comes from the carbon in the surrounding air 39 and oil Discuss mass ow and energy t 4 l rowsW dull I NANA l Fcrlxly u l EQ Wafer below state ifthe system is ows and energy tran rmations occur 9 n c Complete turbine d All the equipment shown Approach A Rankine cycle power plant is shown schematically in Figure 113 For the control volumes de ned 39 39 steady or unsteady open or closed and what heat transfer energy ows mass For the four systems speci ed clearly identify the control volumes with dotted lines and then describe the various processes occurring to the systems For all the systems we assume that they operate at steady sta e Solution a De ne a control volume surrounding the electric generator Mechanical energy crosses the system from the turbine that drives the electric generator mechanical energy leaves the controlvolume in the form of electricity Not all ofthe mechanical energy from the turbine is 4 quot moto39 39 the electricity is converted to mechanical energy so the generator is hotter e surrounding air and there is heat transfer Generators also are actively cooled by passing some sort of uid water or air or other gases through them so m often a generator is an open system Mechanical energy is converted to electricity another form of mechanical work and heat b De ne a controlvolume aron the steam generator Fuel and air enter the steam enerator and exhaust gases leave Water steam enters the system at two locations and leave at two locations Hence this is an open steady system Ifthe oumide surface ofthe steam generator is heavily insulated then there is not heat transfer if it is not insulated there would be heat transfer from the hot steam generator to the colder surroun ing air No mec anical work occurs The owin ermal energy in the hot gases formed by combusting the fuel and air is converted to owing thermal energy in the steam c De ne a control volume surrounding the steam turbine Assume the turbine is heavily insulated so that there is no heat transfer Steam enters and leaves the turbine at di erent conditions so this is an open system Mechanical energy leaves the turbine where is shaft crosses the boundary Th owing energy in the steam is converted to mechanical energy in the turbine d De ne a control volume around all the equipment shown Mass crosses the boundary in three locations air the and exhaust gases Electricity mechanical energy crosses the boundary in one location eat transfer crosses the bound at the steam generator and steam turbine if not well insulated at the electrical generator and at the condenser The owing thermal energy in the hot gases formed by combusting the fuel and air ultimately is converted to electricity mechanical work in the generator and heat 5533 w i eleci vur39iw gt var biogeuw r E 5355 A56 0 4 19 A vaporcompression refrigeration cycle similar to what is used in air conditioning systems is shown schematically in Figure 15 For the control volumes de ned below state ifthe system is stea or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations 0 a Electric motor b Refrigerant owing through condenser e All the equipment shown Approach Forth f stems specified clearly identify the control volumes with do ed lines and then describe the various 39 t e ive sy processes occurring to the systems For all the systems we assume that they operate at steady sta e Solution a Define a control volume surrounding the electric motor Mechanical energy crosses the system from the turbine that drives the electric g tor mechanical energy leaves the control volume in the form of electricity Not all of the mechanical energy from the turbine is converted to electricity feel a motor it gets hot since not all the electricity is 39 energy L i IIULLCI uiaii e ui undin aii 39 Generators quot cooled by passing some sort of uid water or air or other gases through them so most often a generator is an open sys em b Define a control volume around the refrigerant owing through the condenser The w rate is constant and crosses the boundary so this is a steady open system There is heat transfer from the refrigerant to the surrounding environment No work occurs c Define a control volume around the complete condenser Refrigerant enters and quot f 439 39 and at a steady ow rate Air ows through the condenser crosses the control volume boundary and removes energy from the refrigerant at a steady rate Hence this is an open steady system Assume the condenser is insulated By definition there is no heat transfer between the a 1 L I nu vv volume b oundary d Define a control volume surrounding the throttling valve Assume the valve is heavily insulated so that there is no heat transfer Refrigerant enters and leaves the valve at different conditions so this is an open system No mechanical work is done e Define a control volume around all the equipment shown Mass air crosses the boundary in four locations in and out at both the evaporator and condenser There is no heat transfer Electricity work crosses the boundary at the electric motor 110 A hot cup of coffee is placed on a tabletop to cool De ne a control volume and state ifthe system is steady or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations occur Approach 39 the cup p quot and one Solution De ne a control volume around the coffee in the cup as shown Ifwe take a relatively short time period such as one hour we may assume quot quot f 39 closed system The system is unsteady because the coffee cools during the hour 39 r ro quot quot convection and heat conducm into the sides and bottom ofthe coffee cup No wor is done on or by the coffee The internal energy ofthe coffee decreases as heat is transferred from it Ifwe choose a relatively long time period such as a week then we need an open system analysis During the week the coffee will not only cool but also evaporate This is an unsteady open system since mass crosses the control volume as water vapor Energy leaves by heat transfer as before and also is removed by evaporation 1 11 The water in a canal lock is at the downstream river level and the gates are opened A boat enters the locks and the downstream gates are closed A valve is opened and water from upstream flows into the lock raising the boat After the water reaches the upstream river level the upstream gates are opened and the boat travels upstream Finally the first valve is closed and a second valve is opened allowing the water in the lock to drain to the downstream river level Another boat arrives from downstream and the process is repeated Neglect the energy required to open and close the gates and valves Where does the energy come from to raise the boat Approach Define a control volume around the canal lock Alternatively draw a control volume around the entire earth to show the primary source of the energy used to raise the boat Solution Define a control volume around the canal lock This is an open transient system with water and boats entering and leaving In the context of this control volume the energy to raise the boat com es from the potential energy stored in the water upstream of the canal We may also choose a control volume that encloses all the water on the earth This is a closed transient system Heat enters this control volume from the sun and also from nuclear reactions within the crust of the earth The heat energy evaporates water from the surface of all the bodies of water on the earth The water is transported throughout the earth by wind and returns to the land as rain snow or hail Water runs into the river upstream of the canal and acts to raise the boat Thus the ultimate source of the power to lift the boat is sunlight and geothermal energy 112 A closed pan of cold water is placed on a burner of an electric stove which is already turned on For the control volumes de ned below state ifthe system is steady or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations occur b Burner c Pan ofwater plus burner Approach For the three systems specified clearly identify the control volumes with dotted lines and then describe the 39 s various processes occurring to the system Solution a Define a control volume CV I surrounding the pan of water Heat crosses the boundary into the control volume from the burner beneath the pan Because the cold water does not boil initially no vapor leaves the pan so the mass is constant but as the temperature ofthe water increases the energy level increases Thus this is a closed unsteady system No mechanical work occurs As the temperature of the waterpan combination increases there is heat transfer from the pan to the surrounding air The heat transfer is converted to internal energy b Define a control volume CV 11 around the burner on the electric stove Electricity e ectricity 39 39 39 WUIK Heat transfer crosses the boundary from the heater to the bonom of the pan TL 39 L 39 burner o the is attained quickly so this is a closed steady system The electricity mechanical energy is converted to heat Note that Q Q because of the sign convention c Define a control volume CV HI around the pan ofwater and burner l Electrici e ec ici WUIK ecause the cold water does not boil initially no vapor leaves the pan so the mass is constant but as e temperature of the water increases the energy level increases Thus this is a closed unsteady system As the temperature of L39 39 increa 39 39 JIUIII the pan to the surrounding air The electricity mechanical energy is converted to internal energy 113 Water from 39 a nliei p If 39 the water in the pool and filter is this an open or closed system Lfthe system is just the water in the filter is this an open or closed system Approach Ifmass crosses the control volume the system is open otherwise it is closed Solution For the first control volume which is all the water in the pool we have a closed system We are assuming that no mass enters or leaves the pool by evaporation rainfall or addition of city water from a hose e second control volume which is the water in the filter we have an open system Water enters and leaves the filter crossing the control volume as it does so 114 Wind turbine systems such as shown in Figure 11b consist ofa wind turbine an electric generator 39 39 and eiLh r power quot 39 L 39 grid or to banery storage In a steady wind for the control volumes de ned below state if the system is steady or unsteady open or closed and what heat transfer energy ows mass ows and energy transformations occur a Wind turbine B b a ery c Electric generator d Wind turbine electric generator and electrical grid e Wind turbine electric generator and battery Approach For the ve systems speci ed clearly identify the control volumes with doned lines and then describe the various processes occurring to the systems Solution a De ne a control volume surrounding the wind turbine as shown in the gure to the right Air enters and leaves the control volume at different conditions 39 i 39 39 generator For the steady wind this is a closed steady system No heat transfer occurs The kinetic energy in the wind is converted to mechanical energy in the turbine b De ne a control volume around the battery Electricity enters or leaves the battery depending on the operation of the system so the stored energy level changes No heat transfer occurs No mass crosses the boundary so this is closed unsteady system The electricity mechanical energy is converted to stored energy in the anery c De ne a control volume around the outside surface of the electric generator Mechanical energy is transferred to the generator from the wind turbine Not all the mechanical energy is converted to electricity touch an electric motor not all of the electricity is converted to mechanical work so the generator heats up and there is heat transfer from the generator to the surrounding air No mass ows through the generator control volume so this is a steady closed system d De ne a control volume around the wind turbine electric generator and electrical grid For a steady wind a steady ow of electricity mechanical work ows from the generator into the grid but since we include the grid in the control volume no mechanicalwork occurs Air ows into and out ofthe control volume Assuming all the electricity is used no energy is stored in the grid likewise the mass and energy level ofthe control volume are constant so this is a closed steady system If all the electric energy is not used but some is stored or used from the battery then this would be a closed unsteady system The kinetic energy in the wind is converted to mechanical energy in the turbine the mechanical energy is converted to electrical energy another form of mechanical energy in the generator e De ne a control volume around the wind turbine electric generator and battery Air ows into and out ofthe control vo ume The mass ofthe system s constant Electricity may ow out of the system if the grid demands it or the energy may be stored in the battery if the grid does not require it Thus this is a closed system but it could be either steady or unsteady depending on the operation The kinetic energy in the wind is converted to mechanical energy in the turbine the mechanical energy is converted to electrical energy another form of mechanical energy in the generator 1 15 Global warming has been in the news much in recent years Define an appropriate control volume to study this system and state whether it is steady or unsteady open or closed and what heat transfer energy flows mass flows and energy transformations occur Approach If mass crosses the control volume the system is open otherwise it is closed Solution Choose the entire planet earth as the control volume Neglecting the small amount of mass that enters via meteorites and solar wind and the small amount of mass that leaves as interplanetary probes this is a closed system Sunlight enters the control volume is transmitted through the atmosphere and absorbed at the surface of the earth The earth reemits radiation39 however the reemitted radiation has a much longer wavelength than the incident solar energy The atmosphere is largely transparent to solar radiation but partially opaque to the longer infrared wavelengths As a result reemitted radiation is absorbed in the atmosphere and global warming occurs Global warming is caused by the socalled greenhouse gases which include water vapor carbon dioxide methane ozone and certain refrigerants These are gases that are transparent to incident solar energy but absorb infrared radiation reemitted by the earth The increase in concentration of some of these gases over time leads to global warming Especially important is the concentration of carbon dioxide which has increased substantially since the industrial revolution due to the burning of fossil fuels and deforestation Water vapor is also an important factor since as atmospheric temperature rises the concentration of water vapor in the atmosphere increases Increases in methane concentration result from increases in the cattle population Volcanic eruptions also added gases and particulates to the atmosphere that can be important in the global energy balance Since the earth s temperature changes over time scientists estimate an average rise of 05 C over the last century this is an unsteady system The fundamental energy transformation is the conversion of incident solar radiation to stored internal energy in the earth s atmosphere 21 A 2000kg car accelerates from 20 to 60 kmh on an uphill road The car travels 120 m and the slope of the road from the horizontal is 25 Determine the work done by the engine Approach Apply the first law choosing the car to be the system under consideration Eliminate all terms except gt kinetic energy potential energy and work Assumptions 1 The car is isothermal 2 No heat is transferred to or from the car 25 Solution Consider the car to be the system From the first law AEzQ W or AKEAPEAUQ W The car does not change temperature so AU 2 O No heat is transferred so Q 0 and the first law becomes AKE APE W KE2 KE1 PE2 PE 2 W Using expressions for kinetic and potential energy 1 W mV12 V22mgz1 ZZ Let 21 O 22 120msin250 2 507m 2 2 1 km 2 lOOOm 1h m W 2000k 202 602 20001lt 981 507m 2 g h lkm 36005 g s2 l24gtlt106 J 1240 k The system is the car Work done Q a system is negative The work done b1 the engine is W 1240 k I Answer 2 2 A missile is launched vertically upward from the surface of the earth with an initial velocity of 350 ms If the missile mass is 1200 kg calculate the maximum height the missile will attain Assume no aerodynamic drag or other work during the ight and no heat transfer Approach Write the first law for a closed system and eliminate all terms except kinetic and potential energy changes Assumptions 1 The missile does not exchange heat with the surroundings 2 Aerodynamic drag is negligible 3 The missile is isothermal Solution From the first law AKEAPEAUQ7W With no work heat transfer or change in internal energy this becomes AKE APE 0 1 V 2 7 V 2 7 7 Em 2 1 mgzz Zi 0 Setting z1 0 and solving for 22 2 Wivz 3503 70 22 12 2 S 6244m 4 Answer 8 29813 S 2 Comment Mass is not needed in the calculation 23 A system of conveyor belts is used to transport a box of 30 lbm as shown in the gure Note quot Ucll that the uni uavcl at a m lCl done by the motor which drives the inclined belt Neglect all friction Approach Apply the first law choosing the box to be the system under consideration Eliminate all 1 terms except kinetic energy potential energy va 7 3 4 and work 5 i 1 Assumptions s 39 1 The box is isothermal quot1 5 7 15quotquot 2 No heat is transferred to or from the Zf2 3 The conveyor belts operate in steady state 4 Frictional e ects are negligi e ffs 59 gt Solution Consider the box to be the system From the first law AE Q7 W or AKEAPEAUQ7W APEii The box does not change temperature thus AU 0 No heat is transferred so Q 0 and the first law reduces to AKE KEQ KEiPEz PEi W 1 2 2 W m V 7V m z 72 2 i 2 g i 2 where 1 is the bottom ofthe belt and 2 is the top Letz 0 Then 22 8ft sin20 274ft 2 Wl301bm2524332 301bm3217 474m 2 5 3217 5 3217 s s W 834 lbf The system is the box Work done on a system is negative The work done by the motor is W 834ftlbf Answer 2 4 In a frontWheeldrive car 60 of the braking energy is dissipated in the front Wheels and 40 is dissipated in the rear If a car with a mass of 2650 lbm is decelerated from 60 mph to 15 mph on level ground by braking calculate the energy dissipated in each front Wheel in Btu Neglect aerodynamic drag and rolling resistance Approach Write the first law for a closed system and eliminate all terms except kinetic and internal energy changes The energy dissipated in the brakes may be expressed as a rise in internal energy Assumptions l The car does not exchange heat with the surroundi s 2 Aerodynamic drag is negligible 3 Rolling resistance is negligible Solution The system is the car From the first law AKEAPEAUQ7W The car moves on level ground so APE 0 There is no drag or rolling resistance so W 0 The work of the brake pads on the Wheels is internal to the system and is not included Assume that in the short time frame no heat Q leaves the vehicle This is a conservative assumption Which Will result in the highest estimate of brake pad temperature Therefore AKE AU 0 AU7AKEm12 7 v22 2 2 147E S llbf 113m 1 3217 lmeft 778 ftlbf S 1 2 2 I AU E26501bm60 715 AU 386 Btu Of this internal energy 60 is dissipated in the front Wheels therefore each front Wheel receives 30 The final solution is AUWl 03386Btu 116Btu 4 Answer 2 5 A mass of 1200 kg of fish at 20 C is to be frozen solid at 720 C The freezing point of the fish is 722 C and the specific heats above and below the freezing point are 32 and 17 kJkgK respectively The heat of fusion the amount of heat needed to freeze 1 kg of fish is 235 kJkg Find the heat transferred Approach Write the first law for a closed system and eliminate all terms except heat and internal energy Calculate the rise in internal energy as the sensible heat needed to reach the freezing point the latent heat needed to freeze the fish and the sensible heat needed to lower the fish to the final temperature Assumptions 1 Kinetic and potential energy are negligible 2 No work is done on the fish 3 The specific heats are constant Solution The first law is AKEAPEAUQ7W In this case there is no kinetic energy potential energy or work Therefore AU Q The change in internal energy is calculated in 3 parts from 20 C to 22 C from unfrozen to frozen and from 22 C to 20 C Thus mclAT1 mhfs mczAT2 Q where hf is the latent heat of fusion of the fish Substituting values Q 1200kg32 1le 20 22 c235k117L7227720 c g kg kg K 404x105 kJ 4 Answer Comments This would take a big freezer 2 6 A steel bar initially at 1000 F is quenched by immersion in a bath of liquid water initially at 70 F The mass of the bar is 25 lbm and the volume of the water is 7 ft3 Heat is transferred from the bath to the surroundings which are at 70 F After some time the bar and water reach an equilibrium temperature of 70 F Find the heat transferred For the steel use c 0106 BtulbmR Approach Select the combination of the bar and the liquid as the closed system and apply the first law All terms except heat and internal energy may be eliminated Assumptions 1 Kinetic and potential energy are negligible 2 No work is done on the bar or water 3 The specific heats are constant Solution Take the bar and water together to be the system From the first law AKEAPEAUQ7W In this process there is no change in kinetic energy or potential energy and no work is done Thus mbcb T52 T TbimwcwTw2 wl Q where 1 refers to the bar and w refers to the water The water begins at 70 F rises in temperature due to heat transfer from the hot bar and then cools back to 70 F As a result there is no net change in the water temperature and TW1 Tw2 Then Q mbcb T52 Tbi Btu 251b 0106 7071000 F Q m MR Q 7 7247 Btu 4 Answer Comments Q is negative because the system water and bar is being cooled 27 A 014 lbm aluminumball at 400 F is dropped into awater bath at 70 F The bath contains 052 t3 of Water and is Wellinsulated What is the nal tempemture of the ball a er the ball and Water reach equilibrium Ap proach Apply the rst law choosing the ball and Water to be the system under consideration Eliminate all terms except enthalpy change and heat Assumptions 1 The process occurs at constant pressure 2 The container is Wellinsulated 3 Speci c heats are constant 4 Kinetic and potential energy changes are negligible 5 No Work is done on or by the system Solution Take the ball and water together to be the system The process occurs at constant pressure therefore from the rst law Q AH Since the bath is Wellinsulated on the outside 0 The total enthalpy change is the sum of the enthalpy changes of the ball and Water 0 AHb AHW mbAhb mWAhW Assuming constant speci c heat 0 rnbcpbATb WCWATW mt Tt 7T1 mwcpw T2 7T1 Both the ball and the Water reach the same nal temperature T2 at equilibrium so 0 quot1101502 7Tb mch T2 7T The mass of Water is calculated from the density and volume so that 0 mbcpb T2 Tbt PWVWCPW T2 Ttw Rearmnging min anew mtch pWVWcPW Using speci c heat data in Tables B2 and B6 014lbm0216 Btu 400 F6221 Tj052 30398 Bl 70 F T2 T lbmR lbm F 2 0141bm 0216 Bl 622113quot 0521 0998 Blquot lbmR ft lbm F T2 703 F d Answer 2 8 In a new process a thin metal film is produced when very high velocity particles strike a surface melt and adhere to the surface Imagine an aluminum particle with a diameter of 40 pm 1 pm 10396 m at a temperature of 20 C The particle strikes a cold aluminum surface also at 20 C The particle energy is just high enough so that the particle and a portion of the surface with the same mass as the particle completely melt What is the velocity of the particle Assume pure aluminum with a constant specific heat of 1146 JkgK The latent heat of fusion the amount of heat needed to melt 1 kg of aluminum is 404 kJkg Approach Write the first law for a constant pressure process of a closed system and eliminate all terms except the kinetic energy change and the enthalpy change Include both sensible and latent heat of fusion in evaluating the enthalpy term Assumptions 1 There is no potential energy change 2 There is no heat transfer from the surface during the process 3 Specific heat is constant Solution Choose the particle and the part of the surface which melts as the control volume The process begins just before the particle hits the surface and ends as the particle is brought to a complete halt and the aluminum has melted The process occurs at constant pressure therefore the first law may be written in the form AKE APE AH Q We assume no change in potential energy and no heat transfer to the air from the surface therefore AKE AH 0 The enthalpy change has two components the enthalpy rise as the particle temperature changes from its initial value to the melting point of aluminum and the enthalpy rise as the solid aluminum and the particle melt Thus m22 i me 2mcp T2 7 T1 th 0 where hf is the latent heat of fusion of aluminum The final velocity 2 is zero Solving for 1 V1 quot40 T2 7T14h5 2J0 T2 7 T1h5 Substituting the melting temperature of aluminum from Table A2 V1 2 1146 9337 2737 20K 404 m kg K kg 1k V1 2133ms Answer 2 9 An object weights 40 N on a space station which has an artificial gravitational acceleration of 5 msz What is the weight of the object on Earth Approach Use Newton s second law to determine the mass of the object knowing its weight on the space station Use this mass to find the weight on the surface of the earth Assumptions none S olution On the space station by Newton s second law Fl quot 81 Mass is the same on the space station and on earth Therefore on Earth F2 mg 2 Eliminating mass between the last two equations E F2 g1 82 m 981 40N ingi 52 127 7 g1 52 S2 F2785N Answer 210 A mass of 5 lbm is acted on by an upward force of 16 lbf The only additional force on the mass is the force of gravity Find the acceleration in Ms Is this acceleration up or own Approach Draw a free body diagram of the mass included both upward and downward forces Find the net force and F H 5 39t in Newton s second law to determine the use 1 acceleration Assumptions None Fl ma Solution The weight force on the object is by Newton s second law F2mg 51bm3217 Slbf 5 3217 S The net force is Fm F 7F 1675111bf Since F is greater than F2 the net force is upward To nd acceleration use Newton s second law Fma Rearranging 1 1 lbf a m 5 lbm 1 lbf 2 3217lbmfLs ft 11708 2 upward Answer s 2 11 An airplane of mass 18300 kg travels at 500 mph through the atmosphere Calculate the kinetic energy of the plane in k Approach Use the formula for the kinetic energy of a moving object and convert units Assumptions none Solution Kinetic energy is given by KE 1m V 2 2 Substitute values and convertmiles per hour to ms to obtain m 1 KE 18300k 500m h 2 S 2 g p 2237mph Since kg m and s are all SI units the result will be in J KE 457gtlt108 J KE 457gtlt105 kJ 4 Answer 2 12 A gas is contained in a pistoncylinder assembly as shown in the figure below A compressed spring exerts a force of 60 N on the top of the piston The mass of the piston is 4 kg and the surface area is 35 cmz If atmospheric pressure is 95 kPa what is the pressure of the gas in the cylinder Approach Draw a free body diagram for the piston including atmospheric pressure forces the pressure force due to sprmg the gas in the cylinder the force exerted by the spring and the weight of the piston In equilibrium these forces sum to zero Assumptions none Solution A force balance on the piston gives PuunA PA PmAFS mpg Solving for gas pressure F m P P 5 i A A Inpg Substituting values PA 60N IlOOcmT lkPa 4kg98mSZ100cm2 lkPa 35cm2 lOOOPa 35 cm2 lOOOPa P1233kPa 4 Answer P 95kPa lm lm 213 A gas is contained in a pistoncylinder assembly as shown below A compressed spring exerts a downward L r39 quot L r 39 39 r 4 in 39 39 l fin The piston is made of steel with a density of 490 lbmft3 and a thickness of 05 in The cylinder has a 7 in diameter Calculate the gage pressure ofthe gas in the tank Approach Perform a force balance on thepiston including atmospheric pressure forces the pressure force due to the gas in the cylinder the force exerted by the spring Spring and the weight ofthe piston In equilibrium these forces sum to zero Piston Gus Assumptions 110116 Solution A force balance on the piston gives A IEWA kx mg where Pub is the absolute gas pressure x is the length of compression ofthe spring and k is the spring constant The gage pressure is the difference between absolute and atmospheric pressure so rearrange the above equation in the form ab 10 at 7 Pm Using the fact that the mass ofthe piston is the density times the Volume P 7 kx mg 7 10 phAg g A A where h is the thickness of the piston and PE is gage pressure Simplifying Pg E pkg Substituting Values 39 in 2m lbm I 490 05in 322 llbf 1 ft sum m 12 in MI 2 14 Find the density of hydrogen at a pressure of 150 kPa and a temperature of 50 C Approach Use the ideal gas law Assumptions 1 Hydrogen behaves like an ideal gas under these conditions Solution From the ideal gas law p The value of the molecular weight M is found in Table A l Substituting values and making unit conversions lOOOPa lSOkPa 2016k kmol lt gtlt g gt Pa 8314kkmolK50273K1 p 0ll3kgm3 4 Answer 2 15 A pressurized nitrogen tank used on a paintball gun has a volume of 88 in3 If the pressure of nitrogen is 4500 psia calculate the mass of nitrogen in the tank Assume a temperature of 70 F Approach Use the ideal gas law Assumptions 1 Nitrogen behaves like an ideal gas under these conditions Solution From the ideal gas law MP V m RT The value of the molecular weight M is found in Table B1 Substituting values and making unit conversions 3 280 lbm 4500psia88in3 lbmol 12 m3 sia ft3 1073 M 70460R lbmol R mll3lbm 1 Answer 2 16 Air is pumped from a vacuum chamber until the pressure drops to 3 torr If the air temperature at the end of the pumping process is 5 C calculate the air density Eventually the air temperature in the vacuum chamber rises to 20 C because of heat transfer with the surroundings Assuming the volume is constant find the final pressure in torr Approach Use the ideal gas law to find the air density Density remains constant during the heating process The ideal gas law can be used again to find the final pressure Assumptions 1 Air behaves like an ideal gas under these conditions Solution From the ideal gas law ET B L M Solving for density PM P ETI From Table A1 M 2897 for air Substituting values 3ton0l33kPaj lOOOPa 2897 kmol ltorr lkPa 8314i 5273K kmolK lkI p 0005k g3 m Since all units have been converted to SI the result for density is in the appropriate SI unit kgm3 At the end of the heating P2 pRT2 M Density does not change during the heating so i 7 3 P2 T2 P2 Plg you 20237K T1 5 237K P2 316 torr 39 Answer 2717 m gure Apprnzch W Thewuzk dmexsthe Amman clxve ufFvexsus V v Us geamekym hummus area W Assnmptinns in 1 m placessxs quarequlhbnum Snlutin The wk m a quarethbmxn pmcess xs The area we the clxve rm state 1 W State 2 5 represented 5 W onnm24cm mmmjmjcm mnnh 9 1m 1 W um cm WEIISJ Answer u 2 18 In a certain quasiequilibrium process pressure increases from 200 kPa to 350 kPa The initial gas volume is 025 m3 During the process pressure varies with volume according to V701105 11771002 where Vis in m3 andP is in kPa Calculate the work done Approach Calculate the work as the integral of PdV Use the initial and final volumes as the limits of the integral Assumptions 1 The process proceeds through a set of equilibrium states Solution In a quasiequilibrium process W PdV V Solve the expression given in the problem statement for pressure to get 1000 Pa P V7013105 100 lkPa 1 11 P V7012102 105 Substitute this expression into the equation for work y 1 WJ VV701 102105 dV where V1 and V2 are the initial and final volumes Performing the integration V2 105V2Vl V Z 2 2 W10 3V 01 3 3 W 102 VZ70137V70l3105V2 7V1 To find V2 use the equation in the problem statement bearing in mind that V is in m3 and P is in kPa V P7100210395 01 V2 3507100210 5 01 0725m2 Work may now be evaluated as 11 3 3 W 102 07257 012 7 025701210507257 025 139gtlt105 1139kJ Answer Comments In the evaluation of work the SI units for P and Vwere used Therefore the SI unit for work I will result 219 Air is contained in apistoncylinder assembly as shown in the gure The piston which is assumed massless is held in place by a spring Initially the spring is not compressed and exerts no force on the piston Then e ail i catcu ulth Lu 39 25 TL f 4 spring on the piston is F 10 where k 130 Ncm The piston diameter is 6 cm and the initial height ofthe piston is 8 L fwork done by 39 quot r 39 pie mei cm 101 kPa Approach Find an expression for the pressure ofthe gas as a function ofx the amount by which the spring is compressed Use this expression to find work from W deV Assumptions 1 The process is quasiequilibrium 2 The piston is massless 3 There is no friction between the piston and the cylinder Solution The pressure of the gas is kx PM Patm I Because the process is quasiequilibrium 7 7 X 10 W 7 IPdVijPAdx 7 PM 7Adx Performing the integration and simplifying 2 7 2 W PmAxrxk 2 2 3 2 101kPa132cm72cm l m lOOOPa 1301 2 cm l m 100cm lkPa cm 2 100cm W831J lt Answer 2 20 A propeller operating at 85 rpm applies a torque of 61 Nm If the propeller has been rotating for 30 minutes find the work done in kWh kilowatthours Approach Use the expression for shaft work integrating over time Since both torque and speed are constant they may be removed from the integral and the integration becomes trivial Assumptions 1 Torque does not vary with time 2 Rotational speed does not vary with time Solution For shaft work W ISwdl 5ade 5am Substituting values W 611853j 27 ad 30min 1 kvfh 1 kl m1n lrev 36X10 k 1000 W 02715 kWh 4 Answer 2 21 A resistance heater is being used to heat a tank of nitrogen If 3 amps are supplied to the resistor which has a resistance of 60 Q how long will it take for 1200 J of work to be don 7 Approach Use the expression for electrical work integrating over time Since both voltage and current are constant they may be removed from the integral and the integration becomes trivial Assumptions 1 Voltage does not vary with time 2 Current does not vary with time Solution For electrical work W Ifidt gijdz a From Ohm s law 5 iR Substituting gives W i2RI Solving for time I W 12001 1392R 3A26OQ Answer 1222s 2 22 An electric motor operates in steady state at 1000 rpm for 45 minutes The motor draws 8 amps at 110 volts and delivers a torque of 76 Nm Find the total electrical energy input in kWh kilowatthours and the total shaft work produced in both kWh and Btu Approach Use the formula for electrical work to find the electrical energy input and the formula for shaft work to find shaft work Assumptions 1 Current and voltage to the motor are constant Solution Electrical work is given by W 5 dt Because this is a DC motor current and voltage are constant and the integral is just W am 110V8A45minij lkw 60 min 1000 W 066 kWh Recall that volts times amps equals watts For shaft work W 5 rodt Since speed and torque are constant W Scam 76Nm1000 45minZ rad U lkWh j min rev lNm 36gtlt10 0597kWh 3 0597kvfh 36gtltlO kl lBtu lkWh 1055kl 2037 Btu Answer 223 Nitrogen at 28 OC and 100 kPa is heated in a pistoncylinder assembly Initially the spring shown is uncompressed and exerts no force on the piston which is massless If 45 J of work is done by the N2 a how far does the piston rise b what is the final temperature Approach The work done is the sum of the expansion work PM I 00 Ha against the atmosphere and the work done on the spring Knowing the total work done one can calculate the amount of compression of the spring Find the final temperature from the ideal gas law 6 quot4 444414 Assumptions 1 The mass of the piston is negligible 2 Nitrogen behaves like an ideal gas under these conditions 3cm Solution a The piston does expansion work against the atmosphere and work against the spring The sum is W Waxy erring Defining x1 as the imtial piston location and x2 as the final piston location the expansion work is Waxy BYIMAVZBYDW x2 xlA Arbitrarily set x1 0 so that Waxy BithZA The spring work is 1V1 l ltd ID I s rovv x22 x12 x 2 CM 95 H39IIM Wpringzk T 2k s The total work done is x2 W 28msz 1672 tot Substituting values N 100 2 45J100000Pa x2 7z0025 m2 28 cmjx Z cm 1 m 2 Solving for x2 x2 2002 m2 cmi Answer b From the ideal gas law In T1 ml T2 PM M M and PZV2 Dividing gives PM T1 Psz T2 The final pressure is kx P2 2811177 72 Substituting values cm 28 2 cm P2 100 000 Pa 2 129gtlt103 Pa129 kPa 7r 0 025 m The initial and final volumes are x lt8 cmgtltAgt V2 10 cmA Substituting these expressions into the ideal gas law produces 1001ltPa8 cmA 7 28273 129 kPa10 cmA T2 273 T2 212 C 4 Answer 2 24 A pistoncylinder assembly contains 049 g of air at a pressure of 150 kPa The initial volume is 425 cm3 The air is then compressed while 164 J of work are done and 32 I of heat are transferred to the surroundings Calculate the final air temperature Approach Apply the first law Write the change in internal energy in terms of the specific heat at constant volume and the temperature difference Find the initial temperature from the ideal gas law Assumptions 1 Specific heat is constant 2 Air behaves like an ideal gas under these conditions Solution From the first law QAUWmAuW Writing internal energy in terms of specific heat Q mchTW mcv T2771 W What is T1 From the ideal gas law 3 150kPa425cm3 M m 2897 kg j PM lkP km 1 T 7 a 106cm3 01 17 mR 049g 1 kg 8314 kJ 000 lOOOg kmolK lkJ T1 453K 180 C Solving the first law for T2 T2 Q W mc v tTl The preferred practice is to evaluate specific heat at the average of the initial and final temperatures however we do not know the final temperature Instead we will evaluate specific heat at the known initial temperature We can make a correction later if more accuracy is desired With values of CV from Table A8 at 450 K lkJ 32 7 16 J lt gt lt gt 1000 0 T2 0733kJ 1k 180 C 049g 39 g kgK 1000g T2 215 C 4 Answer Comments For more accuracy repeat the calculation evaluating specific heat at the average of 215 C and 180 C Specific heat does not vary substantially between these two temperatures so the estimate of final temperature will change very little 225 In the gure below a piston is resting on a set of stops The cylinder contains CO initially at 730 C and kPa The mass ofthepiston is quot quot 39 39 39 39 39 101 45 kPa how much heat must be add in 17 I vvu ed tojust lift the piston offthe stops Approach A force balance on the piston can be used to calculate the gas pressure at piston li o Th 39 gas law is used to determine the nal temperature PM As a last step apply the rst law to nd heat added r Assumptions 1 Speci c heat is constant 2 Carbon dioxide behaves like an ideal gas under these conditions 101 Gas L 0042 m Solution The piston will just li o when the pressure inside equals the pressure outside plus the weight of the piston per unit area 12kg982 1003m2 1051 kPa m g Pm Pg 101x103Pa AP To nd T2 use the ideal gas law noting that mRT mET gV and 12212 2 M M Dividing the last two equations and realizing that V V2 1 Tl P2 T2 P T2 T 2 730273K 568K I 45 From the rst law QAU Since there is no volume change W 0 and U mAu chT27739 To evaluate CV use average temperature ieTw T I 2 243568 2 405K From Table A8 cv m 075 kJkg K To nd m use the ideal gas law k 451033 0042 003 2 4401 1W x alt mm mgt km m RT 0116kg 8314 243K kmol From the rst law Q mcv T2 4 0116kg075568243K g Q 283 kJ Answer 2 26 A closed tank of volume 28 ft3 contains oxygen at 70 F and an absolute pressure of 143 lbfinz The gas is then heated until the pressure becomes 45 lbfinz Treating oxygen as an ideal gas a find the final temperature b find the total change in enthalpy H in Btu for this process Approach Use the ideal gas law to find the final temperature T2 The change is enthalpy is found using the specific heat at constant pressure Assumptions 1 Oxygen is an ideal gas at these conditions 2 Specific heat is constant Solution a From the ideal gas law mzRT2 RT EVI and Psz Neither the volume nor the mass changes during the heating process so V1 V2 and m1 m2 Dividing the last two equations gives H T1 P2 T2 70460 R 45 39 T2 w 1668R1208 0F P l43ps1a l b The total enthalpy change is related to specific enthalpy change by AHmAhmlthfmmcpltrfngt Evaluate specific heat at the average temperature T 7012082 639 F From Table B8 ave 0p m 024Btulbm R To find mass use the ideal gas law 39 2 14323 ft332 lbm j 144 11 1311 j EKM 7 1n lbmol 1 ft 77821bfft quot7 Bt 02251bm RTI 1986711 70460R lbmolR Substituting values AH02251bm 024 Btu 1208770 lbmR AH 6l5Btu Answer 227 A rigid tank ofvolume 026 m1 contains hydrogen at 15 a C and 101 kPa A paddlewheel stirs the tank adding 178 kJ ofwork Over the same time period the tank loses 93 kJ ofheat to the environment 39 39 39 hy gen uoe not w 39 nd the nal temperature Approach Ideal Gas H Use the rst law to relate heat work and 39 internal energy Apply the ideal gas law to nd the total mass of hydrogen present Express e internal energy is terms oftemperature and solve for the a Q nal temp erature Assumptions 1 Hydrogen is an ideal gas at these conditions 2 Speci c heat is constant Solution From the rst law AU mAu Q7 W Writing internal energy in terms of speci c heat mam TtQW Mass may be found from the ideal gas law PVM 101kPa026m32016IITEEPZIJ m T k 0 a 00221kg 8314 J 15273K kmolK 1kJ where molecular weight was obtained from Table A1 Solving the rst law for nal temperature 7 Tt ms The nal temperature is unknown As an approximation evaluate speci c heat at the initial temperature Lfthe r L 39 arge L r quot L L initial and mm 39 perform the calculation again Using c from Table A8 at 15 C 7937 7178 kJ T 1 k 15 00221kg 10127 kg K T2 530 CltI Answer Comments For more arrnmrv 15 C and 53 C Speci c heat L does not vary substantially between these two temperatures so the estimate of nal temperature will change very little LII39 u 39L 2 28 A chamber is divided equally in two parts by a membrane One side contains H2 at a pressure of 130 kPa and the other side is evacuated The total chamber volume is 0004 m3 At time I 0 the membrane ruptures and the hydrogen expands freely into the evacuated side If the chamber is considered adiabatic find the final pressure Approach Use the first law to relate heat work and internal energy There is no work done and no heat transferred therefore internal energy does not change and the process is isothermal Apply the ideal gas law to find the final pressure Assumptions 1 Hydrogen is an ideal gas at these conditions 2 The chamber is perfectly insulated Solution From the first law Q AU W No work is done in a free expansion and no heat is transferred since the chamber is adiabatic so AU 0 Writing internal energy in terms of temperature mchT mcv T2 7T1 0 which implies T1 T2 From the ideal gas law mET mRT M 1 and PZV2 M 2 Dividing these two equations and remembering that T1 T2 PIVT P2V2 Rearranging 3 P2 130kPa 039002m3 V2 0004m 5 kPa lt Answer PM P2 0 2 29 Air at 20 C 250 kPa is contained in a pistoncylinder assembly Initially the piston is held in place by a pin Then the pin is removed and the gas expands rapidly During the expansion there is no time for any heat transfer to occur The final air temperature and pressure are 716 0C and 100 kPa The mass of air in the cylinder is 04 kg Find the work done on the atmosphere Approach Use the first law for a constant pressure process of a closed system Calculate mass from the ideal gas law Assumptions 1 Air behaves like an ideal gas under these conditions 2 Specific heat is constant 3 The process is adiabatic 4 There is negligible friction between the piston and cylinder walls 5 The mass of the piston is negligible Solution Choose the air to be the system under consideration From the first law Since the process is adiabatic Assuming constant specific heat W mo T1 7 T2 The specific heat may be found in Table A8 Substituting values 7 k 7 7 o W704kg07l7kg Kj20 16 c W 103kJ Answer Comments The work is positive because work is done by the gas on the atmosphere 2 30 Nitrogen at 50 psia and 650 F is contained in a pistoncylinder assembly The initial volume is 25 ft1 The nitrogen is cooled slowly While the pressure stays constant until the temperature drops to 150 F Find the heat transferred Approach Use the first law for a constant pressure process of a closed system Calculate mass from the ideal gas law Assumptions 1 Nitrogen behaves like an ideal gas under these conditions 2 Specific heat is constant Solution From the first law for a constant pressure process Q AH chAT From the ideal gas law PW ETI 50psia25 ft3280j m 294 lbm 10736504r460 lbmol R From Table B8 at an average temperature of 400 F cP 025113 Eu 7150650 2 1me Using these values in the first law gives Q 294 lbm 0251B tu 1507650 F lbm R 7369Btu Answer Comments The heat transfer is negative because the system is being cooled Air at 30 C is contained in a pistoncylinder assembly as shown in the figure The piston has a weight of 15 N and a crosssectional area of 012 m2 The initial volume of air is 35 m3 Heat is added until the 231 volume of the air becomes 65 m3 Atmospheric pressure is 100 kPa a Find the final air temperature b Determine the work done by the air on both the piston and the atmosphere Approach Use the ideal gas law to find the final temperature and 0 a 4 m zffs le W JPdV to find the work Assumptions 1 Air behaves like an ideal gas under these conditions 2 There is no friction between the piston and cylinder walls 0 V 3 The process proceeds through a succession of quasi equilibrium states Solution a Select the air to be the system under consideration This is a constant pressure process therefore Pl 2 P2 From the ideal gas law RT RT PlVl m 1 Psz m 2 The last two equations imply that E 3 V2 T2 Solving for T 2 V 65mg TT 2 30273K 2 VJ 356 l T 563K290 C and Answer b Since this is a quasiequilibrium process V2 V2 W V1 PdV 13le dV PV2 V1 where pressure can be removed from the integral because in this process it is a constant The pressure is due to atmospheric pressure plus the weight of the piston distributed over the piston area mg PM 100kPa UN lkPa 012m 1000Pa 1001kPa Work may now be calculated as W 1001kPa 1000 Pa 65 35m2 lkPa Answer W 300300J 300k Hi Comments Note that the weight of the piston is insignificant compared to atmospheric pressure in this case 2 32 An ideal gas with CF 07 kJkgK and a molecular weight of 256 is initially at 75 kPa and 40 C First the gas is expanded at constant pressure until its volume doubles Then it is heated at constant volume until the pressure doubles If the mass of gas is 45 kg find a the total work for the entire process b the heat transferred for the entire process Approach Consider this as a twostep process Calculate work for each step and add to get total work Also calculate heat for each step and add Find the work from the integral of PdV and the heat from the first law after work is known Assumptions 1 Both processes proceed through a succession of quasiequilibrium states 2 Specific heat is constant Solution a For a quasiequilibrium process work is W j PdV Applying this to the first step of the process which proceeds from state 1 to state 2 7 V2 7 V2 W12 7 K PdV 7 Pjy dV where P has been removed from the integral because pressure is constant in the first step Performing the integral and noting that the final volume is twice the initial volume W12 PV2 7V1P2V7V Using the ideal gas law k 7 45kg8314 4o2731lt W12 PV1 PIV kmOll K 457k M 256 g kmol For the second step of the process work done is zero because volume does not change39 therefore the total work for the process is W W12 457kJ lt Answer b To find the heat transferred use the first law AUiQiW For the first step of the process U2 U1 Q12 7 Wi2 Q12 mcvT2 F TlVVIZ From the ideal gas law 1le1 sz2 E M M but P1 P2 and V2 2Kso 7 E 7 L 7 1 V2 T 2 2V1 2 The temperature at state 2 becomes T2 2T1 240 273K626K To find cvuse kJ 8314 cv Oji7M 0375i kg39K 256g kg39 kmol Substituting values Q12 45kg0375kk IK6267 313K457 kl 985k g For the second step of the process apply the first law AU Q 7 W U3 U2 Q23 7W2 Volume is constant during the second step therefore W23 0 Using the relation for the specific heat of an ideal gas Q23 mcv 7 T2 From the ideal gas law Mf p2V2 M M SinceP3 2P2 and V2 V3 amp2 P2 T2 P2 T3 2T2 2626K1252K The heat transferred becomes Q23 45kg0375i1252 7 626K kgK 1057 k The total heat transferred is the sum of the heat in each step that is QT Q12 Q23 9851057 2043M 4 Answer 2 33 A pistoncylinder assembly contains 02 kg of argon at 200 K and 50 kPa If the argon is expanded isothermally to 30 kPa find the work done Approach Apply the formula for work done during an isothermal expansion of an ideal gas Use the ideal gas law to rewrite the formula in terms of pressure rather than volume Assumptions 1 Argon is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states Solution For an ideal gas undergoing an isothermal expansion W ln 1 From the ideal gas law ply and p2 V2 E M M Solving for V1 and V2 and substituting 111 mm h5 M MP2 ME M P2 Using molecular weight from Table A1 02kg8314km1iKj200Klnj W 399 kmol 4 W 426k Answer 2 34 An ideal gas with a molecular weight of 372 is contained in a pistoncylinder assembly The gas is initially at 130 kPa 25 C and has a mass of 234 x 10394 kg The gas expands slowly and isothermally until the final pressure is 100 kPa Calculate the work done Approach Apply the formula for work done during an isothermal expansion of an ideal gas Use the ideal gas law to determine initial and final volumes Assumptions 1 The process proceeds through a succession of equilibrium states Solution For an ideal gas undergoing an isothermal expansion W ln M V 1 From the ideal gas law 3 mm 234x10 4kg831411almK25273K Vl M Pl k 0 l2gtlt10 4m3 1 372 g jl30kPa kmol 3 gtlt 39 g mm 234 104k 83141K 25 273K V2 MP2 1 l56gtlt10394m3 2 372ijaoo kPa kmol Finally the work done is 3 234X10394kg 831411 1 m 25273K W7 kmolK 1 156x10394 7 372 kg lkPa l12x104 39kmol lOOOPa W409J Answer 2 35 An ideal gas with a volume of 05 ft3 and an absolute pressure of 15 lbfin2 is contained in a piston cylinder assembly The gas is compressed isothermally until the pressure doubles Calculate the heat transferred in Btu Is the heat moving from the gas to the surroundings or viceversa Approach Apply the formula for work done during an isothermal compression of an ideal gas Use the first law to determine heat transferred Assumptions 1 The process proceeds through a succession of equilibrium states Solution From the ideal gas law PM and PZV Since temperature is unchanged in this process PM P2V2 Rearranging 15 sia 05ft3 V3 p 025ft3 P2 30 psia For an isothermal compression of an ideal gas the work done is 39 2 W 131V1 m5 15039 ln 77486ft1bf V1 in ft 05 From the first law AU Q W But since the process is isothermal AU 0 Q W 77486ftlbflBi 70962Btui Answer 778 ftlbf Comment Since Q is negative heat moves from the gas to the surroundings 2 36 Air in a pistoncylinder assembly is compressed slowly and isothermally from an initial volume of 350 cm3 to a final volume of 200 cm3 The air is initially at 100 kPa a Find the work done b Find the heat transferred Approach Apply the formula for work done during an isothermal compression of an ideal gas Assumptions 1 Air is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states Solution a For an ideal gas undergoing an isothermal compression V W PM ln 2 V1 3 W100kPa 350cm3 1nj lkPa 10 cm 350 W7196J 4 Answer b From the first law QAUWchATW but the process is isothermal and AT 0 therefore Q W Q7196J 4 Answer Comments Work is negative because during a compression work is done on the system Heat is negative because the system is cooled during the process One must cool the system while compressing it so that temperature will remain constant 2 37 A pistoncylinder assembly contains 04 kg of C02 The gas expands at constant temperature from an initial state of 250 kPa 100 C to a final pressure of 100 kPa Calculate the heat transferred during the process Approach Apply the formula for work done during an isothermal expansion of an ideal gas Use the ideal gas law to rewrite the formula in terms of pressure rather than volume Finally use the first law to find heat transferred Assumptions 1 Carbon dioxide is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states Solution For an ideal gas undergoing an isothermal expansion W ln M V1 From the ideal gas law ply and p2 V2 E M M Solving for V1 and V2 and substituting P W mRT1n mRTmRT mRT 1 M MP2 ME M P2 Using molecular weight from Table A1 8314 kJ km 1K W 041ltg A 37315Kln 4401kgkmol 100 W 258kI From the first law recognizing that temperature is constant AUQWCVATO Therefore QW258kJ Answer Comments Work is positive because during an expansion work is done by the system Heat is positive because the system is heated during the process One must heat the system during expansion so that temperature will remain constant 2 38 Air at 180 F and 25 psia is compressed slowly and isothermally to 86 psia If the initial mass of air is 00043 lbm find a the work done b the heat transferred Approach Apply the formula for work done during an isothermal compression of an ideal gas Use the ideal gas law to rewrite the formula in terms of pressure rather than volume Apply the first law to find heat transferred Assumptions 1 Air is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states Solution a For an ideal gas undergoing an isothermal compression W ln M V1 From the ideal gas law ply and p2 V2 E M M Solving for V1 and V2 and substituting W mRT1n mRTmRT mRTn i M MP2 ME M P2 Using molecular weight from Table A1 1545 ftlbf 1b lR W00043lbm A 180460R1n 2897lbmlbmol 86 W7181ftlbf Answer b From the first law recognizing that temperature is constant AU QiW CVAT 0 Therefore Q W 7181ft1bf Converting to Btu l Btu 7181ftlbf Q 7781bfft 70233 Btu 4 Answer Comments Work is negative because during a compression work is done on the system Heat is negative because the system is cooled during the process One must cool the system while compressing it so that temperature will remain constant pre ssure 2 39 A piston cylinder assembly of initial volume 150 cm3 contains 03 g of oxygen at 120 kPa The oxygen is then compressed slowly isothermally and frictionlessly while 59 J of heat is removed Find the final Approach Use the first law to determine the work done From the formula for work done during an isothermal compression of an ideal gas you can determine the final pressure Use the ideal gas law to find properties Assumptions 1 Oxygen is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states S olution From the first law recognizing that temperature is constant AU QiWcVAT Therefore 7591 For an ideal gas undergoing an isothermal compression W ln 1 To find temperature use the ideal gas law the molecular weight of oxygen is found in Table A1 3 120kP3150cm33l99 kg 1m jElOOOPa T1P Vly kmol 100cm lkPa 231K mR osg gsm kJ lkg 1000 kmolK lOOOg 1k Rearranging the work equation 3199 101 1759Jj o 1 k1 lkg 03 8314 7 231K ggt kmolKIIOOOg V2108cm3 To find the final pressure apply the ideal gas law 03g83l4 kl 231K 1kg PimRT27 kmolK 2 1000 g 1k MV2 3 3199 kg 108cm3 1m kmol 100cm P2 166700Pa 1667kPa 4 Answer 2 40 Carbon dioxide is expanded slowly and isothermally in a pistoncylinder assembly from 337 psia to 147 psia The initial volume is 39 in3 and the temperature is 100 F Calculate the work done Approach Apply the formula for work done during an isothermal expansion of an ideal gas Use the ideal gas law to rewrite the formula in terms of pressure rather than volume Assumptions 1 Carbon dioxide is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states Solution For an ideal gas undergoing an isothermal expansion V W PIV 111 2 V1 From the ideal gas law PM mRT mRT M and PZV2 M Solving for V1 and V2 and substituting WPlzln mRTmRT Hzln MP2 ME P2 Substituting values W 33739m3m 1 1f j in 147 12m W 909ft1bf u Answer Comment The problem can be solved without knowing the temperature 2 41 Fifteen grams of nitrogen in a pistoncylinder assembly is compressed slowly and isothermally from 100 kPa 25 C to 2500 kPa Calculate the heat transferred and the work done Approach Apply the formula for work done during an isothermal expansion of an ideal gas Use the ideal gas law to rewrite the formula in terms of pressure rather than volume Use the first law to find heat transferred Assumptions 1 Nitrogen is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states S olution For an ideal gas undergoing an isothermal expansion W ln 1 From the ideal gas law RV and 1 Solving for V1 and V2 and substituting W mRT1n mRTmRT mRTn 5 M MP2 ME M P2 Using molecular weight from Table A1 kI 100 0015kg 8314 25273K1n 7 kmol K 2500 W kg 2801 kmol W 7 7427kJ Answer From the first law recognizing that temperature is constant cyAT 0 Therefore Q W 7427kJ 4 Answer Comments Work is negative because during a compression work is done on the system Heat is negative because the system is cooled during the process One must cool the system while compressing it so that temperature will remain constant 2 42 Air in a pistoncylinder assembly is slowly compressed from 100 kPa to 300 kPa The mass of the air isl5 gtltlO394 kg and its initial temperature is 20 C During the entire process pressure is related to volume as PV1 4 a constant Calculate the work done Approach Apply the formula for work done during a polytropic process of an ideal gas Use the ideal gas law to determine volume Assumptions 1 Air is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium sta 3 Specific heat is constant Solution For a polytropic process of an ideal gas work done is W Psz A BK 17 n where n 14 in this case To find V1 apply the ideal gas law 15X104kg 8314 lt gt km 1 K 20273K PM 39 k l26gtlt10quot m3 1 lOOkPa 2897 g1 0 To find V2 note that PIX P2V214 therefore L 1 14 V2 5 V1 jm 126x10quot 575x10 5 m3 P2 300 300kPa W F gt 575x10 5m37100kPal26gtlt10394m3 lOOOPa 1714 lkPa W 7116 J Comments Answer Work is negative because work is being done on the air in this compression process polytropic with PV constant The initial volume of air is 1 m3 Find a the value ofn b the work c the heat transfer Air is compressed from 150 kPa to 600 kPa while the temperature rises from 20 C to 100 C The process is Approach Use the defining equation for a polytropic process and the ideal gas law to find n After work is calculated using the expression for work done in a polytropic process heat may be calculated from the first law Assumptions 1 Air is an ideal gas under these conditions 2 The process proceeds through a succession of equilibrium states 3 Specific heat is constant Solution a For a polytropic process by definition PrVrquot Pszquot To find V2 apply the ideal gas law at each state prI p2V2 E M M Dividing one equation by the other mlTT1 PiVi M P2V2 mm M Solving for V2 yields V g 150kpax1m3 100273K 2 P T 600kPa 20273K 2 1 To find n return to the defining equation for a polytropic process PrVrquot Pszquot Rearranging 5 2 P2 V1 Taking the natural logarithm of both sides In E nln E P2 V1 Solving for n l 11 121 0318m3 1n 5 P2 J n 0318 q Answer 11 V 1 b As long as n i l the work done in a polytropic process is Psz PM 7 17 n Substituting values W 1000Pa W 6001ltPa0318m37150kPalm3 lkpa jl95gtlt1051 1121 W71951d 4 Answer c From the first law AUQ7W Q WAU WmcyT2 7T1 The average temperature is 201002 60 C 333K From Table A8 for air at 333K CV m 072 kJkgK The molecular weight of air is available in Table A1 Find the mass of air from the ideal gas law PW 15 7 lSOkPa1m32897k gjlooopaj kmol l kPa 1000 W m kJ kmolK 8314 2o273rlt m 178kg Substituting values in the first law Q 7195kll78kg 0723 100720 C kg K 4 Q 7923k Answer Comments The heat calculated is negative and this indicates that the system is losing heat during this process A pistoncylinder assembly oftotal mass 16 lbm is free to move within a housing as shown in the gure Initially the cylinder contains gas at an absolute pressure of 20 lbfin2 and 007 if and is at rest The piston 39 39 L 39 nal Velocity of75 s During this process the gas is compressed to a nal pressure of35 lbinn2 The process is adiabatic and the pressure is related to the Volume by PV 4 constant Calculate the change in internal energy for this process in Btu Approach Apply the formula for work done during a polytropic process of an ideal gas Use the rst law including both kinetic energy chan e and compression work to determine the internal energy change Assumptions 1 The gas is ideal 2 gigfzzt hsermass ofthe gas compared to the mass pislo mylmdcr assembly 3 Speci c heat is constant 4 The process is adiabatic 5 The compression is a quasiequilibrium process Solution p39 39 a cum 4 L 39 39 run 39 an ideal gas work done is W Psz 71m where n 14 in this case To nd V1 note that RV1 A sly2 therefore L l 14 V2 1 V1 Q 007 oo459tt2 g 35 lbf lbf 35 0045 7 20 007m 2 gt 2lt J 7 1714 m2 W 7874 lbf Next ndthe change in kinetic energy 2 AKE1mV2 1151bm 755 amp 1398 lbf 2 2 s n 322ng S From the rst law AUAKEAPEQ7W is L rlenergyy AU7AKE7W713988741bf 1B 778 lbf AU 0994 Btu Answer 2 45 Natural gas is a mixture of methane ethane propane and butane as well as other components Composition varies by point of origin of the gas Consider natural gas with an equivalent molecular weight of 236 and an equivalent specific heat CF of 201 kIKg K The gas is slowly compressed in a frictionless adiabatic process from an initial volume of 212 cm3 to a final volume of 98 cm3 1f the initial pressure is 39 kPa and the initial temperature is 15 C find the final temperature and pressure Assume the mixture can be modeled as an ideal gas Approach39 To find the final temperature use It 71 2 7 V1 in v2 The ratio of specific heat k can be determined after CV is calculated with cp cV 11 M Assumptions 1 Natural gas is ideal under these conditions 2 The process is a quasiequilibrium process 3 Specific heat is constant 4 The process is adiabatic Solution For an ideal adiabatic process of an ideal gas kil if V1 T1 V2 5 where k p To find CV use 0 k 314 cvcp7 201 7 kI1147166kJ M kg K 236 g kg K kmol The ratio of specific heats may now be determined from c P 23901 121 c 166 Solving for T2 and substituting values k4 12171 T2 T1 V l 15273 v2 98 T2 3387K 657 C lt Answer For final pressure use k 5711 H V2 Rearranging 1 21 F2 39 E 98 P2 992 kPa Answer 2 46 Carbon monoxide is expanded slowly in a wellinsulated frictionless pistoncylinder assembly from 300 cm3 25 C to 400 cm3 Find the final temperature Approach To find the final temperature use a 7 l k4 T1 V2 Assumptions 1 Carbon monoxide behaves like an ideal gas under these conditions 2 The process is a quasiequilibrium process 3 Specific heat is constant 4 The process is adiabatic Solution Since the process is slow and frictionless it may be assumed quasistatic Since it is also wellinsulated it maybe be assumed adiabatic For an ideal adiabatic process of an ideal gas k4 T 1 2 Values for the specific heat ratio k are available in Table A8 The final temperature is not known but it is expected to be less than the initial temperature As shown in the table k is very insensitive to temperature being almost the same at 250 300 and 350 K Therefore k 14 will do and 300 1471 0 T2 25273 m 266K 77 cd Answer 2 47 Hydrogen with a mass of 11 kg is compressed slowly and adiabatically from 100 kPa 25 C to 450 kPa in a pistoncylinder assembly Assuming constant specific heat calculate the final temperature and the work done Approach 17K 7 Find the final temperature from g Then apply the first 2 law to determine work Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Specific heat is constant 3 Hydrogen behaves like an ideal gas under these conditions 4 The process is adiabatic Solution For an adiabatic compression of an ideal gas with constant specific heat 17K T2 E 7 Eli Solving for final temperature and substituting values 171 405 T2 25 273K 05 460K 450 kPa where k 1405 at 300K from Table A 8 From the first law Since the process is adiabatic Q 0 and the first law becomes W iAU imcvT2 7T1 Use values for cv from Table A8 at 300 K to get 7 7 W 11kg102kgK460 25273K W71818kJ 4 Answer 2 48 Air at 147 psia and lOO F is contained in a wellinsulated pistoncylinder assembly of initial volume 06 ft3 The air is slowly expanded by applying 560 ftlbf of work What is the final pressure Assume constant specific heats Approach Use the formula for work done in a polytropic process with n k Solve this simultaneously with k PLVl E V2 Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Specific heat is constant 3 Air behaves like an ideal gas under these conditions 4 The system is adiabatic Solution This is an adiabatic expansion of an ideal gas with constant specific heat so it is polytropic withn k The work done is WBKEK 17 k Also use 5 5 k E V2 Rewriting the numerator of the work equation P H JKEK l 17 k Replacing the pressure ratio with a volume ratio k V P l V iPV W 2 11P1V1kV21k7P1V1 l 7 k 1 7 k Solving for V2 Waimaz V 17k 2 ERIC Simplifying and substituting the value of k for air at 100 F from Table B8 1 2 m L 560ftlbf114 147 06ft3 k 7 1n lft 0975 ft WlikPIV l Vz PVk 1 1 7 lbf 144m2 3 m l47 06ft in2 lft2 Finally the final pressure is k 2 l4 P2 5 12147E 0396 3 V2 1n 0975ft P2 7451bf lt Answer 2 49 Oxygen at 147 psia and 70 F is contained in a pistoncylinder assembly with an initial volume of 150 in3 The oxygen is compressed slowly and adiabatically to a final volume of 50 in Assume constant specific heat Find a the final temperature b the final pressure c the work done in ftlbf Approach kil T V Find the final temperature from F2 j and the l 2 k final pressure from P2 Then apply the first 2 law to determine work Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Specific heat is constant 3 Oxygen behaves like an ideal gas under these conditions 4 The process is adiabatic Solution a For an adiabatic compression of an ideal gas with constant specific heat g 7 5 1H T1 V2 Solving for T2 k l T2 T1 L V2 The final temperature is unknown For simplicity we evaluate the ratio of specific heats k at the given initial temperature Once final temperature is estimated we can correct the value of k if necessary For oxygen at 70 F k 139 from Table B8 Using this and given values 3 1 3971 T2 70 460R150i 50in3 T2 813R 353 F Answer b To find the final pressure use 2 5 E V2 Solving for P k 139 V 39 3 P2 P1 1 147ps1a 159m V2 50m P2 677psia lt Answer c From the first law AU Q 7 W Since the process is adiabatic Q 0 and AU 7 7W The oxygen is assumed to be an ideal gas therefore 2 49 W mcv T1 T2 To find the mass use RKM m Rn Substituting values l47psia150 in332 lbm W 31bm 1 0007181bm 1073 70 460R lbmolR lft The work may now be evaluated from W 0007181bm0157 Btu 707353 M lbm R lBtu W 7248ft1bf Answer Comments The final temperature is 353 F39 therefore the average temperature for the process is Tm 703532 211 F The ratio of specific heats k at the average temperature is close to the value at the initial temperature of 70 F as seen in Table B8 As a result there is no need to iterate 2 50 Nitrogen at 850 K 2 MPa expands slowly and adiabatically until the final temperature is 300 K Assuming constant specific heat find the final pressure and the ratio of final to initial volume Approach k l T2 P2 7 Find the final pressure from Then apply T1 P1 k l T2 V1 to find the percent increase in volume T1 V2 Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Specific heat is constant 3 Nitrogen behaves like an ideal gas under these conditions 4 The process is adiabatic Solution For an adiabatic expansion of an ideal gas with constant specific heat 71 2 7 i 7 T1 P1 Solving for P2 k T 3 p2 p1 1 T2 Evaluate k for nitrogen at the average temperature of 575 K From Table A8 k 138 Substituting values 1 38 P2 2000 kPa H 38 300 P2 456kPa Answer For percent increase in volume use kil 7 5 T1 V2 Rearranging VLTiE V1 T2 Substituting values l jlm155 Answer V1 300 Comments For this process to occur the volume must increase by more than 15 times 2 51 Air with a mass of 017 lbm is slowly compressed in a wellinsulated frictionless pistoncylinder assembly from 147 psia to 68 psia If the air is initially at 60 F a find the final temperature b find the work done in ftlbf Approach k l 7 Find the final pressure from Then apply 1 1 the first law to find the work done Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Specific heat is constant 3 Air behaves like an ideal gas under these conditions 4 The process is adiabatic S olution a For an adiabatic compression of an ideal gas with constant specific heat k l PT T2T12 E The final temperature is unknown As an approximation we use the ratio of specific heats at the initial temperature We will iterate later if necessary From Table B8 for air at 60 F k 14 Therefore 71 T2 60460mj1 805R 345 F 4 Answer b From the first law QAU The process is adiabatic the gas is ideal and the specific heats are constant so U 7 mAu imchT imcv T2 7 T1 To evaluate specific heat we take advantage of having a good estimate of the final temperature from part a and evaluate specific heat at the average of initial and final temperatures 0 m M 202 F cv m 0173 Btu 2 lbm R where Table B8 has again been used Substituting values W70171bm0173 Btu j345760R lbmR 7838311 778ft1bf 1Btu W 6521ftlbf Answer 2 52 Air is slowly expanded at constant pressure from an initial temperature of 300 K to a final temperature of 700 K in a pistoncylinder assembly The initial volume of air is 250 cm3 and the pressure is 150 kPa Calculate the work done and the heat transferred a using variable specific heats b using constant specific heats Approach Find work from W P V2 7 K and then use the first law to find heat In part a use values of internal energy from Table A 9 and in part b rewrite internal energy in terms of temperature and specific heat from Table A8 Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Air behaves like an ideal gas under these conditions Solution a From the first law Since the process is slow and pressure is constant the work is obtained from W IPdVPIdVPV2 7V3 The ideal gas law may be used to find the final volume V2 Start with Dividing one equation by the other and noting that l31 132 E 7 T1 V2 T 2 Solving for V2 T V2 V1 2 250cm3 E 583cm3 T1 300K Work may now be calculated as 3 W 150kPaj5837 250cm31010m 501 a cm Find heat from the first law in the form Q WAU Wmu2 ful To find mass use the ideal gas law lm 3 PVM 150kPa250cm32897le quotFL 1 0 cm 435gtlt104kg 1 8314 300K kmolK In part a variable specific heats are used Taking values of ul and uz from Table A9 Q 50J435x10 kg5123 2141E kg lkJ Q 1801 Answer b If constant specific heats are used the first law may be written Q WAUWmu2iul WmcvT2 7T1 The value of CV is chosen at the average of T1 and T2 or Tm 500 K Using CV for air from Table A8 and noting that work mass and temperature are unchanged Q 50J435X104kg0742i700300KE 1000 kgK lkI 2 53 gem lt Anwer Cnmment gaseswhzn me temperatu39e change xsnatlaa large 2 53 A rigid tank ofvolume 42 ft3 contains air initially at 100 F and 147 psia Heat is added until the final pressure is 709 psia Assuming variable specific heat find the heat added Approach Find heat from the first law noting that work is zero in a rigid tank Use values of internal energy from Table B9 and calculate mass with the ideal gas law Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Air behaves like an ideal gas under these conditions Solution From the first law AU Q 7 W The tank is rigid so W 0 and m u2 7 u1 Q To find uz we need the final temperature T2 From the ideal gas law mRT mRT 1le1 1 and 11sz2 2 M M Dividing one equation by the other and noting that K V2 3 T1 P2 T2 Rearranging T2 T1 5 100 460RE 2701R E 14 7 We also need the mass m of air in the tank From the ideal gas law lbm l47ps1a42 ft32897 quot1 M l 02981bm 1 1073 100460R lbmolR Taking values of ul and uz from Table B9 interpolating for uz Btu 02981bm 518579547 Q lt gtlt gt1bm Q 126Btu lt Answer 2 54 A rigid tank contains 005 kg of air at 800 K and 300 kPa The tank is cooled While 635 kI of heat are transferred Find the final air temperature and pressure assuming variable specific heat Approach Find heat from the first law noting that work is zero in a rigid tank Use values of internal energy from Table B9 to determine the final air temperature The pressure can then be found from the ideal gas law Assumptions 1 The process proceeds through a succession of quasiequilibrium states 2 Air behaves like an ideal gas under these conditions Solution From the first law AU Q 7 W The tank is rigid so W 0 and m 2 7 1 Q Using the value of uz at 800 K from Table A9 and noting that Q is negative because the air is cooled u2 u1 63935kl5923 4653E m 005kg kg kg From Table A9 the temperature corresponding to uz is T2 640K lt Answer To find pressure use the ideal gas law mRT mRT Hz 1 and Psz 2 M Dividing one equation by the other and noting that K V2 3 T1 P2 T2 Rearranging P2 E g 300kPa 240kPa 439 Answer T1 800 10 1 Air at 23 C 100 kPa with a freestream velocity of 100 kmhr flows along a at plate How long does the plate have to be to obtain a boundary layer thickness of 8 mm Approach For the distance x from the leading edge to where the boundary layer thickness equals 8 mm if we assume a turbulent ow we can use Eq 108 to calculate the distance We must check the turbulent flow assumption Assumptions 1 The boundary layer is turbulent from the leading e ge Solution Assuming a turbulent boundary layer from the leading edge the boundary layer thickness can be calculated with Eq 1 5 037 7 037 x 7 Re5 7 prxu15 Solving for the distance x 5 025 x 5 pi 037 u From Appendix A7 the air properties at 23 C are p ll77 kgm3 and u l817gtlt10395 Nsm2 02 0304m Answer 7 0008m5ll77kgm3100000mhrlhr3600s x7 037 1817X10395Nsm2kgmNs2 Now we check the Reynolds number 7 pv x 7 ll77kgm3100000mhrlhr3600s0304m u 1817X10395Nsm2 This is turbulent ow as assumed if turbulence begins at 500000 Re i 547 000 10 l 10 2 A large cruise ship has a length L 250 m beam width W 65 m and draft depth D 20 m It cruises at 25 kmhr Assume the flow over the hull can be approximated as that over a at plate Estimate a the total skin friction drag in N and the power in kVJ to propel the ship on a voyage in the Caribbean where the water temperature is 28 C b the total skin friction drag inN and the power in kVJ to propel the ship on a voyage to Alaska where the water temperature is 4 Approach P 1 5 Assuming we can treat the sides and bottom as flow D 7f 2 kM M q 1 over a flat plate and ignoring the form drag at the front the skin friction can be calculated with the basic drag equation and the appropriate drag coefficient Power is force times velocity Q Assumptlons W H M 4 T u 0C on C l The s1des and bottom can be treated as flat plates 2 Seawater properties can be approximated with pure water properties Solution a The friction drag force is defined by Eq 106 FD CDprZAZ where A is the planform area A LW 2LD 250m65m220m 26250m2 We need the Reynolds number to evaluate the drag coefficient We assume we can treat this area as flow over a at plate Assuming seawater properties can be approximated by those of pure water from Appendix A6 at 28 C p 9961 kgm3 and1 816X10394 Nsmz The length Reynolds number is 3va 9961kgm3962 ms250m e L 1 816x10 Nsm2 This is turbulent flow so ignoring the laminar contribution and using Eq 108 even though the Reynolds number is larger than the applicable range D 0455 258 0455 258 000138 10gReL log 2936x109 FD 000138996lkgm3962ms2 26250m2NsZkgm21670X106N Power is obtained from W FD v 1670x106N962ms1kw10001s16060kw4 Answer b The only difference i part b is the temperature at which properties are evaluated At 4 C p 1000 kgm3 and1 155x10394 Nsmz 3va lOOOkgm3962ms250m T 1 T l55XlO Nsm2 CD 0455 258 0455 258 000149 logReL 1og1552x109 FD 0001491000kgm3962ms2 26250m2Ns2kgm21809X106N W FD vw 1 809x106N962ms1kw10001s17410kw Answer 2936X109 ReL 1552X109 Comments Just by changing the water temperature the power increase was 84 This is almost entirely due to the change in the water viscosity 10 2 10 3 A 5mm X 15m X 4m plastic panel SG 175 is lowered from a ship to a construction site on a lake floor at a rate of 15 ms Determine the tension in the cable lowering the pane a assuming the panel descends vertically with its wide end down inN b assuming the panel descends vertically with its narrow end down in N Approach This problem requires a force balance on the plastic panel The buoyancy drag tension and weight forces must be evaluated We will assume that we can treat the flow as if it is over a flat plate Assumptions 1 The plastic panel is treated as a at plate 2 The properties can be approximated with pure water properties Solution a A force balance on the plate is ZF0TFDFAW7W a TW7FD7FA For the weight W ppWVg prSGPWVg 1000kgm31750005m4m15m981ms2 515N Buoyancy force is Fm pWVg 1000kgm30005m4m15m981ms2294N The drag force is defined by Eq 106 Accounting for drag on both sides FD 2CDpV 2A2 where A is the planform area A LH 15m4m60m2 We need the Reynolds number to evaluate the drag coefficient We assume we can treat this area as flow over a at plate Assuming properties can be approximated by those of pure water from Appendix A6 at 10 C p 9996 kgm3 and1 129X10394 Nsmz The length Reynolds number is nay Re 1744x106 L u 129x10 Nsm2 This is turbulent flow but taking into account the laminar contribution at the leading edge 7 0074 1740 7 0074 1740 7 0 00318 D 7 Ref ReL 1744X106V5 1744gtlt106 FD 20003189996kgm315ms2 6m2Ns2kgm2429N Therefore T 515N 429N 294N178N Answer b For the narrow end down the only change is in the drag force pVL 9996kgm315ms4m Re 4649X106 L y 129x10 Nsm2 CD 7 00354 1740 7 0074 vs 0 6 000306 ReL ReL 4649x106 4649 FD 2000306996kgm315ms2 6m2Ns2kgm2413N T 515N413N2943N1797N 4 Answer Comments Drag plays a minor role in this problem compared to weight and buoyancy 10 3 10 4 Your car has broken down on an Interstate highway and you are on the medium strip between the lanes By the time you decide you need to cross to the other side of the empty road a stream of cars moving bumper to bumper at 120 kmhr passes only 1 m away from you For an air temperature of 25 C determine the velocity inkmhr of the wind that will hit you 10 seconds after the first car has passed Approach C M quotII IZOLLW Assuming the bumpertobumper traffic acts as a flat vy W plate moving through air at 120 kmhr a boundary 1 k l x l gt layer forms We can calculate the distance the first Xlt car travels in 10 s We can check the Reynolds l M3 number to determine if the flow is laminar or S d turbulent and using the appropriate equation we can g V amp bod 9 calculate the boundary layer thickness Finally using IA 1 64 an expression for the velocity profile we can then determine the velocity Assumptions 1 The traffic acts as a at plate moving through still 2 Pressure is one atmosphere Solution The distance x the first car travels is x v Il20000mhrlhr3600s10s333m For air from Appendix A7 at 25 C u l832gtlt10 5 Nsm2 and p 1181kgm3 e 7 MD 7 1181kgm3120000mhrlhr3600s333m u 1832x10395Nsm2 This is turbulent flow Assuming turbulence begins at the start of the first car by 10 s later the boundary layer thickness is 037 333 037 a V r owns x Rex 715500 Assuming a 1711 power law velocity profile V 17 17 l a v 1209 ll IOSE 4 Answer V 5 hr 209m hr Comments The main assumption traffic acts as flat plate is somewhat weak because the cars act as blunt bodies and the flow around them are different than over a flat plate Nevertheless we needed an answer however approximate so we needed to make assumptions that would let us determine an answer R 715500 10 4 10 5 A atbottomed river barge 60m long and lZm wide is towed through still water at 25 C at 10 kmhr Determine a the force required to overcome the drag in N b the power required by the towboat in kW c the boundary layer thickness at the end of the barge in mm Approach Assuming the bottom can be treated as flow over a at w A amp 1200 W 39 plate and the skin friction can be calculated with the gt basic drag equation and the appropriate drag coefficient Power is force times velocity Depending on whether the flow is laminar or turbulent at the end w of the plate the appropriate boundary layer thickness M equation can be used Assumptions 1 Pf 7 54 l The plate can be treated as flow over a flat plate VA Solution a The drag force is defined by Eq 106 FD CDprZAZ where A is the planform area A LW 12m 60m 720m2 We need the Reynolds number to evaluate the drag coefficient We assume we can treat this area as flow over a at plate For water from Appendix A6 at 12 C p 1000 kgm3 andz 1225X10393 Nsmz The length Reynolds number at 10 kmhr 278 ms is 3va 1000kgm3278ms60m T T 1225x10 3 Nsm2 This is turbulent flow so ignoring the laminar contribution and using Eq 108 0455 0455 000204 D 10gReL ng logl36gtlt108258 FD0002041000kgm3278ms2720m2Ns2kgm25674N 4 Answer ReL 136x108 b Power is obtained from W FD Vw 5674N278mslkW10001sl58kW Answer c For the turbulent boundary layer thickness at the end of the plate 60 037 i 00327 a m 15 050m AnSWer L ReL 138X108 10 5 106 Outboard racing boas are designed for part ofthe hull to rise completely out of the water when a high speed is reached Then the boat planes on the remainder ofthe hull At 50 mihr on 60 F water the area ofthe hull plani g is G long and S wide 39 r a 39 39 quot quot 39 hp Approach Assume the boat bottom can be treated as ow over a at plate and the skin friction can be calculated with the basic drag equation and e appropriate drag coe icient Power is force times Velocity SUML 473391 Assumptions 1 The plate can be treated as ow over a at plate Solution The drag force is de ned by Eq 106 C 1 2 whereA is the planform area A LW 6ft5 30 2 I u treat this area as ow over a at plate Note that V 50 mihr 733 s For water from Appendix B6 at 60 F0 623 lbm 3 and M 75x10quot lbm s 623lbm 733ft s an Rel 3me 5 lt 361x10 75x10 lbmfts This is turbulent ow so ignoring the laminar contribution and using Eq 108 CD imm 000247 10g1 log361x107 FD 0002475231bm 3733 s2 30 2lbfs2 322 n 1bm23s4 lbf Power is obtained from W FD Vm 3851bf733 s1 hp 5550 lbf512hp 4 Answer 107 A ceiling fan has ve thin blades each 55cm long and 15cm wide Assume the blades can be approximated as at plates Air temperature is 27 C For rotational speeds of 50 RPM 100 M and 50 RP etermine the power in W needed to overcome the drag force Hint Because Velocity Varies with distance from the center of rotation you must integrate the drag coef cient Approach 6 P wer is force time Velocity or torque times angular 727ac Velocity Velocity Varies from zero at the center of V rotation to a maximum at the blade tip Drag force depends on velocity and drag coef cient both of lt L 0WM gtl which Vary along the length ofthe blade Hence we must integrate to obtain the total power Assumptions C 23 n 53913O0113 DL10M 1 Flow over the blade behaves as ow over a at plate 2 Air is at one atmosphere Solution Using the differential element shown on 1 blade 5 where the factor of2 is used to take into account both sides of the blade The Velocity is V rm The differential drag force is dFD CDpV 2Wd72 Because the drag coef cient is a function ofvelocity and hence distance along the blade we must determine if e ow is laminar or turbulent The maximum Reynolds number will occur at the blade tip at the m 39 rotational speed The air properties from App ix 7 1173 kgm3 and 1846X10395 Nsmz 3 Re 7 pw 7 mm 7 1173 kgm 150reVmin1min60s2madrev055m015m 7 M W quotm y y 1846X10395 Nsrn2 Because this is less than 500000 we assume the whole blade is in laminar ow and we will use the laminar ow drag coe icient for ow over a at plate C 1328 1328 8 132 D U5 ns n 5 Rew P W FEWy Combining all the expressions 5W 2rm polar Wdr me y r 3 1 Simpifying and integrating W I5W I 1328pr 52 g 2 1328me5 L y mW 7 mWMn Substituting in Values t Mm 7 1173kgm3mrads015m kgm Nm lJs 1846X10395 Nsm2kgmNs2 This is for one blade Finally At 50reVmin1min60s2madrev524rads W 5000008445245 2 0 0255w 4 Answer At 1 00reVmin 1min60s21nquotadrev 1047rads W 50000084410475 2 015w Answer At 150reVmin1min60s21nquotadrev1571rads W 50000084415715 2 0413W Answer 10 8 The highspeed trains in France and Japan are streamlined to reduce drag forces Consider a 120m long train whose outer surface can be approximated by a flat plate with a width of 10 m At 101 kPa 20 C determine the drag inN due to skin friction only and the power required in kVJ to overcome this drag at a 100 kmhr 200 kmhr and 300 kmhr b The front of the train can be approximated by a hemisphere facing forward with a circumference of 15 m Estimate the drag force caused by the front of the train and the power required to overcome it Approach This is a straightforward application of the drag force P 39 I0 I ICPC E w lo M equation We assume we can treat the plate as flow T ana gt over a at plate and the skin friction can be calculated with the basic drag equation and the appropriate drag iblzo M coefficient Power is force times velocity Assumptions 0 00 2col 3 30 AL 1 The plate can be treated as flow over a flat plate Solution a The drag force is defined by Eq 106 FD CD p v 2A2 where A is the planform area A LW 120m 10m1200m2 We need the Reynolds number to evaluate the drag coefficient We assume we can treat this area as flow over a at plate For air from Appendix A7 at 20 C p 1206 kgm3 andz 180810395 Nsm2 The length Reynolds number at 100 kmhr 278 ms is pv L 1206kgm3278ms120m 7 7 180X10395Nsm2 This is turbulent flow so ignoring the laminar contribution and using Eq 1010 0 ReL 224x108 55 000191 D logReL258 1og224x108258 FD 0001911206kgm3278ms2 1200m2Ns2kgm21070N Power is obtained from W FD v 1070N278ms1 Jl Nm29750w1 Answer b At velocity of 200 kmhr 556 ms ReL 448X108 CD 000174 FD 3890N W 216300W Answer At velocity of 300 kmhr 834 ms ReL 672X108 CD 000165 FD 8300N W 692 200W quot Answer c The front of the train can be 1 J 39 as a 39 p with a 39 f of 15m Hence the diameter is IrD 5m gt D 477 From Table 102 CD 04 so at 100 kmhr 278 ms FD 041206kgm3278ms2 Ir4477m2 stkgm23330N WFDV 3330N278ms1T1Nm92600W Answer At 200 kmhr FD 13320N and W 740 600W At 300 kmhr FD 29 970N and W 2500000w 4 Answer Comments The blunt nose of the train increases the power requirement by a dramatic amount even though its area is relatively small compared to the at areas of the train This illustrates well the effect of different shapes on drag 10 8 10 9 In large electric power plants cool water flows through condensers downstream of the steam turbines which drive the electric generators This water is recirculated so it is often cooled in cooling towers such as shown on the figure below The design specification is that the tower must withstand a 100 mihr wind at 70 F Approximating the drag coefficient from information given in one of the tables determine a the drag force on the tower in lbf b the moment that must be resisted by the foundation of the tower in ftlbf Drag force is calculatedwith the basic drag We will need to approximate the drag coefficient because we do not have one specifically for this shape The moment is force times distance we will assume the drag force acts through the center of the cooling tower Approach D 7 Z ZDPf l Assumptions The flow over the cooling tower can be approximated as flow perpendicular to a cylinder with a small LD ratio 2 The flow is uniform over the tower 3 The air is at one atmosphere Solution a The drag force is defined by Eq 106 FD CDprZAZ We assume the velocity is uniform over the whole tower We assume this shape can be approximated as flow over a cylinder with a small LD ratio Therefore the area is A 7rL24 and DW Dm Dmm 2 2302002 215ft From Table 102 LDavg 300215 1395 so we let CD 064 The velocity is 100 mihr 1467 fts For air at 70 F p 00754 lbmf FD 06400754lbmft31467 fts2 7r4300ft2 11be2 322 ft lbm2 228gtlt1061bf 4 Answer b The moment around the base of the tower is ZM Fd FD L210ft 228x106 lbf300ft210ft 365gtlt108 ftlbf 1 Answer Comments Several approximations were made These were necessary since we needed an answer and we did not have a drag coefficient for this specific shape 10 9 1010 A child releases a helium lled balloon that is spherical in shape into 80 F 147 psia air Ifthe balloon weighs 001 lbf and has a diameter of 1 determine its terminal Veloci Approach Terminal Velocity occurs when there is a balance of forces among weight buoyancy and drag Drag force T T depends on Velocity and is calculated with the basic F drag force equation 50ml 611 Assumptions T H0 1 The balloon is spherical and smooth 1 V Jillii 2 Helium and air are ideal gases at the same 39 conditions w whalbdr Solution A force balance on the balloon is F 0 Fm FD WH Mallow Theweight ofthe heliumin L J 4 r 4 on quot quot quot 39 quot ideal gas 2 147ng 4003 mm 144 2 In lbmol lbm 1hr 7 3 RT 1545 lbf 80460R lbmolR WM pMVg 322 s2001021bm 31t1 3 lbst322 lbm60005321bf The buoyancy force is the weight ofthe displaced air Air density is pa 001021bm 3 28974003007381bm 3 Fm 3an 322 s2007381bm 31r1 3lbfs2322ftlbm600387lbf The drag force is de ned by Eq 106 0 2A 2 where ArD24 TL 39 quot quot the 39 39 velocir For air from Appendix B7 at 80 F u 125x103951bffts Re 7 pm 7 007381bmft3V s1 y 125x103951bf s Substituting the knon quantities into the force balance and solving for velocity 013 2 Fm WH Walloon CD 007381bmtt3 v 2 a 1ft2 4 2 003877 0005327 001 00234lbf n5 200234lbf322 lbmlbfs2 250 11 CD 007381bm 3 sz 4 CD From Figure 1010 guess CD m 04 Therefore as V g 806g Re 5900806 47600 s s 5900v s s At this Reynolds number again from Figure 1010 CD m 05 US V 721 Re590072142600 05 s s At this Reynolds number again from Figure 1010 CD m 05 which is the same as previously calculated so the Velocity has been determined V 721 s Answer 1010 1011 You hike to the top of a mountain and climb the re tower The wind is blowing at 80 kmhr The air temperature is 17 C and the pressure is 94 kPa Estimate the wind force in N that would act on you Approach The basic drag force equation is used We need to U W k estimate the frontal area of a person 711 M Assumptions 1 An average person is 18 m tall and 40cm wide T W C 2 Air is an ideal gas 3 Solution The drag force is de ned by Eq 106 DpV 2A2 From Table 102 for a standing person CD m 12 For the frontal area assume an average person is 18m tall and 40cm wi e A 18m040m072m2 For the air density use the ideal gas equation 942s97 kg 42 m k km 1129k 8314 17273K m kmolK FD 121129kgm3222ms2 072m7NsZkgm2241Nt Answer 1011 10 12 In the westem United States empty boxcars sometimes are blown over by strong crosswinds Shown in the quot L 439 39 one type ofa V J N b 39 39 39 39 velocity in m and 39 39quot to the side ofihe uu m ll over 39 at 22 C 101 kPa Approach Assume the drag force acm through the center ofthe llt L I A Mgt side of the boxcar and the weight acts through the center of mass The boxcar would just begin to tip Pb when the total moment around a wheel is zero lw Assumptions 1 Dra force acts throu h the center ofa side ofthe T7 exit g 303 lt gtI 39 Sid Vl w D4Q M m 20000 kg cm mew Solution As shown in the schematic taking a moment balance around the left hand wheel in the end View M 0 FD SH2 7WD2 The drag force is de ned by Eq 106 FD CDpV 2A 2 From Figure 105 for a rectangle with an aspect ratio of 12532 39 CD m 12 For the air density at one atmosphere at 22 C p 1193kgm3 FD 121193kgm3v ms2 125m32m22853kgmV ms2 Substituting into the moment equation and solving for Velocity WW2 V 7 2853 kgmSH2 20000kg981m52144m2 ill429932 Answer 2863kgm09m32m2 s hr 1012 10 13 In a bicycle race a bicyclist coasts down a hill with a 7 grade to save energy The mass of the bicycle and rider is 85 kg the projected area is 022 ml and the drag coefficient is 09 Air temperature is 17 C Neglecting rolling friction and bearing friction determine a the maximum velocity if the air is still inms b the maximum speed if there is a head wind of 5 ms in ms c the maximum speed if there is a tail wind of 5 ms in ms Approach I m gfk For a steady speed there is a force balance between A 0 L M1 the drag force and the weight in the direction of travel R I 397 Assumptions 1 Rolling resistance is ignored 2 Air is at one atmosphere 70 644915 Wm Solution As shown in the schematic taking a force balance on the bikerider combination 2F 0 Wsin iFD The drag force is defined by Eq 106 FD CD p v 2A2 The weight is W mg Combining equations and solving for velocity V 2mg sin605 C D pA The angle for a 7 grade is 6 tan l 007 400 For the air density at one atmosphere at 17 C p l2l4kgm3 V 285kg981ms2sin40 05 22 0 m 7 09l2l4kgm3022m2 39 s The velocity calculated is the relative velocity between the object and the air Therefore the rider s velocity relative to the ground is In still air 220 ms With a 5 ms headwind 220 5170 ms Answers With a 5 ms tailwind 220 5 270 ms 10 13 10 14 A 25cm sphere with a specific gravity of 025 is released into a uid with a specific gravity of 071 The sphere rises at a terminal velocity of 05 cms Determine the dynamic viscosity of the uid in Nsmz Approach Terminal velocity occurs when there is a balance of 3915 O S CMIS forces among weight buoyancy and drag Drag force depends on velocity and is calculated with the basic FD drag force equation J B U01 Assumptions I l The sphere rises at a steady velocity Fluid 2 The Reynolds number is lt N 1 SG 01739 V w Solution A force balance on the sphere is 2F0FbpFD7W E E The buoyancy force is the weight of the displaced liquid Fm pLVg gprG D36 981ms2lOOOkgm307l7r0025m3 stkgm6 0057N The weight of the sphere is W psphng gprG D36 981ms2lOOOkgm30257t0025m3 Ns2kgm6o020N The drag force is defined by Eq 106 FD CD p v 2A2 where A 7rD24 From Table 102 we assume the Reynolds number is small so that CD 24Re 24pDu Substituting this into the drag equation FD24pDupV2A237rDu a u3f D Combining the equations for force balance drag force weight and buoyancy and simplifying we obtain FD 005770020 0037N 7 0037N T 37r0005ms0025m Now checking the Reynolds number 0711000 kgm30005 ms0025m 313 Nsm 2 This Reynolds number is well within the valid range of the assumed drag coefficient equation so the calculated viscosity is correct FD nay 7 u 313N 4 Answer 111 00028 1014 1015 Many sports cars are convertibles The air ow over such a car is significantly different depending o w ether the convertible top is up or down The engine ofthe 1000kg car delivers 135 kW 0 the wheels the car frontal area is 19 m2 and rolling resistance is 25 of the car weight The drag coefficient when the top is down is 043 and 031 when it is up For 20 C air at one atmosphere determine a the maximum speed with the top up in ms b the maximum speed with the top down in ms Approach Power is force times velocity The total force consists R ofrolling resistance Fm and drag force FD The 7 39gt L basic equation can be used to calculate the drag force w HS w Assumptions M I 10003 1 Frontal area does not change with top up or down W 5quot A 39 Him CD om for VP CD 04 4w WW Solution Power is force times velocity W Fm V where Ea FD FRR The drag force is defined by Eq 106 FD CDpV 2A2 For the air density at one atmosphere at 20 C p 1201kgm3 Because we have two different drag quot 39 we will L 4 39 39 4 39 In t FD CD 1201kgm3v rns2 19m2Nsikgm21141CD V ms2 The drag force is in terms ofN he rolling resistance is Fm 0025W 0025mg 00251000kg981ms2Ns2kgrn245N L L 1 fordra force andpower 39 quot 39 weobtainthe following equation 135000w WsNm245N1141CD v ms2 v ms Solving this equation for the two conditions Top up v 693ms155mihr Answer Top down v 625ms139mihr Answer 1015 10 16 Wind speed is measured with an Ah m m 391 can be J from a thin plate hinged on one end when the plate is hung from the hinge wind impinging on the plate will cause the plate to rotate around the hinge The angular de ection is a measure of the wind speed For a brass plate 20 mm wide and 50mm long derive a relationship between wind speed and angular de ection 0 Assume the drag force on the plate depends only on the velocity component normal to the surface for angles less that about 40 and the air temperature is 25 C Determine a the relationship between wind speed and angular de ection b the thickness of brass needed for 0 30 at a wind speed of 60 kmhr in mm Approach L 2 ODCM In a steady wind the angle 0 is set by a moment k39Nfe balance between the wind force and the weight of the plate The appropriate component of each force must 3 be used Assumptions 9 Tt 02W 1 Air is at one atmosphere Solution a A moment balance around the hinge using the normal component of the drag force and the weight is 2M 0 FDVNLZiWNLZ The friction drag force is defined by Eq 106 and using the normal component of velocity FDVN Dpa V cos62A2 where A is the area A LD The normal component of the weight is WN Wsin 9 mgsin6 pBLDIgsinQ Combining the expressions and solving for velocity 12 V 2pglgsng 139 Answer C D pa cos 6 b Solving the above equation for thickness I CDpa cos2 9 V 2 ZpBgsinQ The drag coefficient is obtained from Figure 105 With an aspect ratio of 005002 25 gt CD 115 For air at 25 C pa 1184 kgm3 For brasspB 8530 kgm3 7 1151184 0082 30 60000m j981ms2sin30 3600s 2 7 2 j 000421m421mm Answer Comments If the angle becomes too large then this analysis will become invalid since the ow over the plate will become too different than ow perpendicular to a plate and the drag coefficient will be affected 10 16 10 17 A 70kg bicycle racer in the Tour de France can maintain about 40 kmhr on a calm day over level ground The bike has a mass of about 10 kg and has a rolling resistance of 1 of the weight of bicycle and rider The drag coefficient of the bike and rider is 11 and their frontal area is 024 ml The air temperature is 25 C Determine a the power output by the rider in kVJ on level ground b the velocity the rider could attain going up a hill that has a slope of 6 in kmhr Approach Power is force times velocity The forces in the x direction are drag rolling resistance and a component of the weight Each of these forces can be evaluated with the given information Assumptions 1 Air properties are at 25 C M 2 70quot lo 43m lilMIS Um M I Solution a Power is W Fm V The total force is Fm FD FRR Wsin6 Drag force is FD CDpV 2A2 For air at 25 C pa 1184 kgm3 Also V 40000mihr1hr3600s 111ms FD 111184kgm3111ms2 024m2Ns2kgm2 193N The force due to the rolling resistance and weight on horizontal ground 6 0 FRR Wsin6 FRR 001mg 00180kg981msNs2kgm785N Combining these expressions and solving for the power W 193N785N111ms301W Answer b When traveling uphill 6 6 the weight term is no longer zero W Fm v CDpvjA2o01mg Wsin6 v Substituting in known values and using consistent units 301W 111184kgm3 v 2 024m22785N80kg981ms2sin6 v 301 01564V 3786V 820V Solving for velocity V329ms118kmhr Answer Comments Using the same power going up a hill as can be produced on a horizontal road results in a significantly lower velocity as anyone who has ridden a bicycle knows 10 17 10 18 Assume the bicycle rider in Problem P lOl7 adds a fairing to streamline his bike and body The drag coefficient is reduced to 024 but the frontal area is increased to 029 m From the power the rider can produce estimate the new speed in kmhr the rider can maintain on level ground Approach We use the equation developed in part b of Problem PlOl7 to obtain the new speed Assumptions 1 Air properties are at 25 C 1 atm Solution We developed the equation W Fmv CDpV 2A2001mgWsin6 v On horizontal ground 6 0 Using the variable values obtained in Problem PlOl7 with the new drag coefficient and frontal area 301W 024l184kgm3 v 2029n122785N801ltg981ms2sino v 301 00412v 3786V v 162ms 582kmhr Answer Comments This velocity is a significant increase from the 40 kmhr obtained without the fairing The effect of streamlining is well illustrated by this comparison 10 18 10 19 A parachutist controls her freefall speed by falling spreadeagle CD 5 12 to slow down or head down CD S 04 to speed up The frontal areas in the two positions are about 070 m2 and 025 ml respectively For a 55kg skydiver at 3000 m assume the density and temperature are approximately constant at this elevation determine a the terminal speed in each position in kmhr b the time in s and distance in m to reach 95 of the terminal speed Approach F Terminal velocity is achieved when the drag force 391 D equals the weight of the parachutist so a simple force 39 m Sgt balance is used When terminal velocity has not been V g a achieved an additional term for the acceleration must be included in the force balance gtlt V 1 Assumptions CD L L W CD 04 1 Air properties are evaluated at 3000 m and L 0 lYM39L assumed constant A 073 A 2 Drag coefficients are constant Solution a A force balance on the parachutist in free fall is ZF0FD7W FDW Drag force is FD CDpV 2242 12 Combining equation and solving for velocity V i 2mg CD pA CD p14 From Table 105 for air at 3000 m pa 0909 kgm3 Using the drag coefficients given in the problem statement 12 Spread eagle V 376 4 Answer l20909kgm3070m2 s 12 2 55k 981 Head down V w 1093 Answer 040909kgm 025m s b The time required to obtain 95 of the terminal velocity is determined by another force balance but this time another term involving the acceleration is included dV C v 2A ZFymayFD7W a FDW 7m7img Rearranging this equation and separating the variables 2 dV 7CDpV AgCDpA V2 2mg JV dV JtCDpAdI dl 2m 2m CDpA 0 V2 2mg 0 2m CDpA Let a2 2mg CDpA Note that a2 gt V 2 and that a is the terminal velocity Performing the integration Ami1 a a 2m Re arranging this equation 12 12 metal W Cw1 a 2m CDpA 2m 2m 12 I 2m tanh 1 C D pAg a To reach 95 of terminal velocity 10 19 Spread eagle 12 255kg a 1 S a 1 S If1I20909kgm3070m2981mS tnh 095 3832 tnh 095 702 4 Answer Head down 12 255kg a 1 S a 1 S z040909kgm3025m2981mSJ tnh 095 1111 tnh 095 203 t Answer For the distance traveled we solve the equation above for velocity and integrate with respect to time 12 V mm w 1 2m Because V dydt 12 12 7 y 7 t 7 t C pAg 7 a C pAg yejodyijSWtijoatanh DZ m IW10g005h2 m 1 D 2m Spread eagle y 373961 mSlog cosh 3 81328 702s 728m Answer 383 I Head down y 10gcoshll111203s 610m Answer llllsj I Comments The constant drag coefficient assumption is reasonable since for blunt bodies after a certain Reynolds number the drag coefficient does not vary much 10 20 10 20 In the United States the Bonneville Salt Flats in Utah are used by individuals trying to set land speed records in various class vehicles One challenger has developed a 1750 lbf car that has a 675hp engine a streamlined body with a drag coefficient CD Z 029 a frontal area of 135 ftz and rolling resistance of only 3 of the body weight The car s transmission has an efficiency of 88 that is 88 of the engine power is transferred to the tires On a day when the air temperature is 95 OF determine the maximum speed of the car in mihr Approach 0 Power is force times velocity For this car force W urkpw consists of drag force and rolling resistance both of F0 gt t 0 N which can be evaluated from the given information Fm 003 Assumptions D l The air is at one atmosphere 2 The drag coefficient is constant CD 1073 W WWWf A 35 Solution Power is obtained from WFmV FDFRRV The rolling resistance is FRR 00317501bf5251bf The drag force is definedby Eq 106 FD CDpV 2Az For air at 95 F pa 00715 lbmft3 FD 029007151bmft3V fts2 135ft2lbfs2322ftlbm2 ooo435v 21bf Combining these equations and using the given information 2 lhps 088675hp7 000435v 5251bf v ftsmj Solving for velocity 326 700 000435 V 3 525 V v 428 fts 292mihr 4 Answer Comments Note that if we ignore the rolling resistance the top speed would be 437 fts 10 21 10 21 A BMW 520 has a drag coefficient of 031 and a frontal area of 225 ftz It weighs 3500 lbf If rolling resistance is 15 of the weight determine a the speed at which drag resistance becomes larger than the rolling resistance in mihr b the power in kW and hp required to cruise at 45 mihr and 75 mihr Approach Power is force times velocity For this car force consists of drag force and rolling resistance both of which can be evaluated from the given information Assumptions 1 The air is at 1 atm 77 OF A22T6 C9203 Solution a The rolling resistance is FRR 0015W 001535001bf5251bf The drag force is defined by Eq 106 FD CD p v 2A2 For air at 77 F pa 0074 lbmftj Equating the drag and rolling resistance and solving for velocity 12 2 12 2 525lbf 322ftlbf lbm s V ZFRR 809 ft 552E Answer CDpA 03100741bmft3225ft2 E hr b Power is obtained from WFmV FDFRRV FD 0310074lbmft3 v fts2 225ft2lbfs2322ftlbm2000801V 2lbf Combining these equations and using the given information lhp s 550ftlbf At 45 mihr 66 fts W 105hp7822kw lt Answer At 75 mihr 110 fts W 299hp223kw 4 Answer W 000801v 2 5251bfv fts Comments Note that the power required for the cruising speed is much small that the typical engine installed in cars More power is required to accelerate a car than is required to maintain a constant speed Power requirements increase as the time to reach a given speed decreases 10 22 Some military jets deploy parachutes when they land to reduce the distance required to stop Suppose a 14500kg jet uses two 6m diameter parachutes and lands at 300 kmhr in 20 C air Determ39 a the total force the cables connecting the parachutes to the plane must withstand inN b the time in s and distance m required to decelerate the plane to 150 kmhr without using brakes and ignoring drag from the plane 10 22 Approach 3d Stu1 93 ZMIS L For part a the drag force can be determined from the Y basic drag equation For part b all forces including T a deceleration forces must be included in a force lt X balance Assum tions 7 muLa 4L7MIS p i m l ANDOE 1 Air is at one atmosphere 2 There is no interaction between the two parachutes Solution a Drag force is defined by Eq 106 FD CD p v 2A2 where A is the projected area of the parachutes A 7rD24 The drag coefficient is obtained from Table 102 CD 13 For air at 20 C p 1205 kgm3 FD 131205kgm305833ms2 76m24153800N For two parachutes Fm ZFD 307 600N Answer b Using Newton s second law of motion in the xdirection d V ZFX 72F max m dt Drag force is as given above Substituting that expression into the force balance separating variables and integrating from the initial velocity to a second ve o it V v C A C A dV27JquotLdI a 7i Lii i V V 0 m V V VI V m Solving for t 7 m Liii l4500kg 1 1 Cap1 V V 7131205kgm37546m2K4l66ms 8333ms I 393s Answer To find the distance we revisit the original force balance 72FDmax gt 7 DpVZAmax Note that ax d V dt Using the chain rule on the derivative d V dx d V ax V dx dt dx Substituting this expression into the force balance separating variables and integrating DpVZAm I J V dx 0 m V V x1n a xLln m V CDpA VI 1 14500kg 227m 1 Answer 131205kgm3746m2 300 10 23 1023 In Lquot quot quot One day on a long drive a bored engineering student realizes that his gas mileage is 20 lower traveling into a head wind than w en ere was no head wind The road is level e temperature is 7 C and his speed is 120 kmhr The driver a car enthusiast knows that the drag coef cient ofhis car is 035 frontal area is 21 m2 mass i 950 kg and rolling resistance is 3 ofthe body weight To pass the time he uses this information to calculate the head wind velocity What is it in kmhr 21quotch 4439 I A L39 Approach U 10 lamL4 33 3M5 We assume gas mileage is inversely proportionalto power which is calculated with force times velocity The total force consists ofrolling resistance Fm gt and drag force FD The velocity in the drag force is the relative velocity between the car velocity and the air velocity The basic equation can be used to calculate the drag force CD 2 V w W A 1iM FMEOAD Assumptions I m Cl gu 1 Mileage is inversely proportional to power Air is at one atrnosp ere 3 The quantity C is constant Solution Power is force times velocity W Fm V where F 0 FD FRR Using a constant ofproportionality C to convert power to gas mileage GM GM cW CM v The drag force is defined by Eq 106 C 2A 2 For the air density at one atmosphere at 7 C p 1257 kgm3 Therefore the drag force is FD 0351257kgm3v ms2 21mNsZkgm20452V ms2 The drag force is in terms ofN The rolling resistance is Fm 003W 003mg 003950kg981ms2Ns2kgm280N Combining the equations for gas mileage drag force rolling resistance andpower and simplifying we obtain the following equations In still air GM 2 1 280N0462V ms 1w ms C With a head Wind 2 2 2 280N 045212 ms 1 ms We also know that 080 3 GM And V2 V1 thdwmd 333 thdwmd 4 We have iuui 39 unmiuwn 39 39 V2 3667m s The head wind is vmm 35577 333 333ms12kmhr 4 Answer 10 24 1024 Ifyou have ever been hit by a hailstone you know it can hurtbecause of its high speed Consider a 4c hailstone falling in 17 C 96 kPa air Assume the hailstone has a speci c gravity of 084 Determine its terminal velocity in ms and mi for a smooth hailstone b for a hailstone with a surface roughness similar to that ofa golf ball Approach Terminal velocity occurs when there is a balance of forces among weight buoyancy and drag The drag force must be determined iteratively Assumptions 1 Hailstone is a sphere Solution a A force balance on the hailstone is F0Fm FDmW gt FDW7Fmy Assuming the hailstone is a sphere its weight is W pSPW39Vg 1000kgm3084981ms21t004m3 stkgrn60276N The buoyancy force is the weight of the displaced air For the air density use the ideal gas equation 95E2 2897 kg PM in krnol 7 kg 3 kJ e1153 3 8314 17273K m kmolK Fmy pVg gprDZG 981ms2 1153kgm3 1r004m3 NsZkgrn638x1039AN The drag force is de ned by Eq 106 FD CDpV 2A2 where A IrD74 The drag force depends on the Reynolds number From Appendix A7 for air at 17 c y 177x10 5 Nsm2 Combining the equations for force balance drag force weight and buoyancy and simplifying we obtain V 8WFiwns 80276 38gtlt10 N TLEEJME CDpIrD2 CD11153kgm3 004m2 CD 5 Using Figure 1010 for a smooth sphere guess CD m 05 Calculating the velocity v 38105 276ms Now checking the Reynolds number 1153kgm3 275ms004m 177x10395 Nsm2 Atthi CD e 051 39 v 381051 273ms 6 Answer i n I I i A 71900 39 39 39 39 velocit c A b Using Fig 1014 for the rough hailstone at the above Reynolds number CD m 023 v 381023 407ms the Reynolds number is 105000 Reevaluating the drag coef cient CD m 024 V 381024IS 398ms Answer Comments n i n i i in u u a A wnl uu loss in accuracy 10 25 1025 A beginning bicyclist can produce 84W for short periods of time On a hot day 32 C how fast can the bicyclist travel ifthe projected area ofthe bike and cyclist is 05 m2 and the drag coefficient is 11 Approach Power is force times Velocity Ignoring rolling resistance the only force is drag force and the basic equation can be used Assumptions 1 Rolling resistance is ignored 2 Air is at one atmosphere Solution Power is force times Velocity W F V The drag force is de ned by Eq 106 3 2A 2 00 FD Ignoring rolling resistance combining equations and solving for Velocity Wm V CDpA For the air density at one atmosphere at 32 C p 1154kgm3 V i284WkgmNs21NmWs 7 111154kgm305m2 4ZE231 s hr 10 26 km Answer 1026 A copper sphere 10mm in diameter is dropped into a 1m deep drum of asphalt The asphalt has a density 0 1 m3 and a Viscosity of 10S Nsmz Estimate the time in hours it takes for the sphere to reach the bonom of the dmm Approach Ignoring the time required for the sphere to accelerate F If F to its terminal Velocity the time required for the D Co ppUt 8W1 particle to settle back to earth would be LV Terminal Velocity occurs when there is a balance of A SPQH forces among weight buoyancy and drag Sukim Q TL M Assumptions M IogNxM1 l iW l U H D00ltv 1 Ignore time required to accelerate to terminal Velocity 2 The sphere is smooth Solution The time required for the sphere to fall 39 wiLh t LV a umiu 39 39 39 accelerate to terminal Velocity which we can determine from a force balance force balance on the sphere is F0F FDeW FD WiFtw With a density ofcopper of 8933 kgm3 the weight ofthe p 39 39 W Psphng 981ms2 sgsskgrrE 1001m3 stkgm600459N The buoyancy force is the weight ofthe displaced liquid Fm pVg gJIIDZ6 981ms71150kgm31t001m3 stkgm6591x10393N The drag force is de ned by Eq 106 FD CDpV 2A2 where A D24 From Table 102 because ofthe Very high Viscosity we assume the Reynolds number is small so that CD 24Re 24pWy Substituting this into the drag equation FD 24pWypV2A237rmy a v Combining the equations for force balance drag force weight and buoyancy and simplifying we obtain 00459591gtlt10393N 7 31t105Nsm2001m With such a low Velocity and high Viscosity there is no need to che number will be well within L quot4 the t39 124x10 3 5 ck the Reynolds number since the Reynolds assumed w 39 39 Now 39 L quot i2358x105s655 ht273 days 4 Answer V 424x10 ms 10 27 1027 A meteorological balloon is to be lled with helium at 0 C 100 kPa The surrounding air is at the same 39I L i i i i 4 r 0 kg and the balloon material has a mass of 015 kgmz Ifan upward vertical velocity of3 ms is desired what diameter in m balloon is require Ap roach At terminal velocity there is a balance offorces among weight buoyancy and drag The drag d F l buoyancy forces depend on the balloon diameter D which is unknown A force balance must be wrinen and each force evaluated Assumptions V 1 The balloon is spherical and smooth Wham 2 Air and helium are ideal gases at the same conditions Solution 7W msmm Each term except the weight of balloon A force balance on the balloon is 2 F 0 Fmy 7 FD 7 W 7 W the instruments is wrinen in terms of the unknown diameter e instrumentation package wei t is Wmm mg 30kg981ms2 NsZkgm7294N The balloon weight is WWW mg mAAg mAIrD2g 015kgm2 1Drn2 981rns2 stkgm 4623D2N The buoyancy force is the weight ofthe displaced air Assuming air is an ideal gas 7 7100kNm22897kgkmol1 2761i ET 7 8314kJkrnolK0273K 39 m3 Fmy 3an 981nis2 1275kgrn3 1D 3 N szkg m55555D3N The helium weight is pH R g 1275 j 0175 WM pMVg 981ms20176kgm31tD 3 N szkg m50905D3N The drag force is de ned by Eq 106 FD CDpV 2A2 where ArD24 TL 39 quot quot the 39 39 velocir For air from Appendix A7 at 0 C y 1653x10395 Nsrn2 7 pw 7 1275kgrn33rnsDrn Pa Re 231500 Dm y 1553x10 5 Nsrn2 The drag force is FD CD 1275kgrn33rns2 z4Drn2 Ns2kgrn2L510CDD2 N Combining the force expressions 0 5555D3 74510CDD2 e 0905D3 74623D2 7294 The procedure to obtain a solution is 1 guess a value of D 2 calculate Re from the above equation 3 at this Re obtain CD from Figure 1010 and 4 calculate D from the cubic equation and compare to the guessed value Continue the iteration until convergence is obtained Performing the iteration CD 2020 a Dz409rn a 1222910 X105 1 n 10 28 1028 In dry regions wind storms can entrain much dust into the air For a particle 005 mm in diameter with a density of 18 gcm3 raised to a height of 100 m in such a storm estimate how long it will take the particle to settle back to earth Assume the air is at 27 C 100 kPa and that the time required to reach the terminal Velocity is negligible Approach FD Ignoring the time required for the particle to accelerate to its terminal Velocity the time required for the particle to settle back to earth would be t HV Terminal Velocity occurs when there is a balance of forces among weight buoyancy and drag Assumptions W 70 1 Ignore time required to accelerate to terminal H11 1 T 2 C Velocity 39U o mu H a 2 The particle is a smooth sphere 3 Reynolds number is lt 1 Solution A force balance on the particle is F0F5WFDW FDF5WW The weight ofthe sphere is W p51quotng 981ms218gcm3 15x10395m3 100cm1mf stkgm1kg1000g61156X10399N The buoyancy force is the weight of the displaced air For the air density use the ideal gas equation 100k Ij 2897 kg PM m kmol illszkg 7 7 3 RT 8314 N 27273K m kmolK Fm PVg gpIrDZG 981ms21162kgm31r5x10395m3 stkgm67458x10quot3N The drag force is de ned by Eq 106 FD CDpV 2A2 where A IrD24 From Table 102 we assume the Reynolds number is small lt 1 so that CD 24Re 24pWy Substituting this into the drag equation FD 24pWypV2A237rmy a v From the appendix for air from Appendix A7 at 27 c y 1s45x10 5 Nsm2 Combining the equations for force balance drag force weight and buoyancy and simplifying we obtain 1155x1039g 7458x1039 N 31r1846x10395 Nsm25gtlt10395m Now checking the Reynolds number 3 5 Re 1162kgm 013ms25X10 m 0 42 1846x10 Nsm 01333 S Thi i we so we cannow calculate the settling time H t 752512 5min 4 Answer V 0133ms 10 29 10 29 A 5mm iron sphere is dropped into a tank of 17 C unused engine oil Determine the sphere s terminal velocity in cms Approach VD Terminal velocity occurs when there is a balance of T forces among weight buoyancy and drag The drag f rce can be determined iterative y T we 94w ml J JAN Assumptions Sphere is smooth v Solution a A force balance on the sphere is ZF0F5WF07W gt FDW7Fmy With a density of iron of 7870 kgmz the weight ofthe sphere W p51quotng 7870kgm3981ms210005m3 stkgrn65053x10393N The buoyancy force is the weight ofthe displaced oil From Appendix A6 for oil at 20 C p 8882kgm3 is y 8450x10 A Nsm2 Fm pVg gprDZG 981ms2 8882kgm31t0005m3 Nsikgm55703x10394N T e drag force is de ned by Eq 106 FD CDpV 2A2 where ArD24 The drag force depends on the Reynolds num er Combining the equations for force balance drag force weight and buoyancy and simplifying we obtain n5 n5 V 8W7me s5053x103935703x10394N 0752 SE CDpIrD2 CD1E8882kgm30005m3 CD 5 The Reynolds number is 8882k m3 Vm s 0005m g 74 H2 5256Vm 8450x10 Nsm quot39 39 39 velocity r 39 339 108 2 calculatethe velocity 4 t 39 Figure quot 4 39 39 39 guess 50 repeat the proceeding steps until converged Guess CD m 05 calculate velocity V 075205D5 123ms calculate Reynolds number Re 5256123 ms 646 reevaluate CD m 6 recalculate velocity V 07527IS 0354ms Reynolds number Re 186 At this Reynolds number CD 24Re 24186 129 and V 0752129IS 0241 ms Continuing several more steps in the iteration we obtain Dz 275 v 0165ms amp0868 Answer Comm Note that ifwe had initially assumed that the Reynolds number would be lt 1 then we could have used CD 24Re o obtain an analytic solution One the terminal velocity had been determined we would have 10 30 1030 The military landing strips r p quot39 quot auunuuer 39 39 anno trike the ground at a Velocity greater than 10 ms Determine how many 20m diameter parachutes are required when the air is at 17 C 9 kPa Approach a Terminal Velocity occurs when there is a balance of leb l quot quotl forces among wei t buoyancy and drag Buoyancy is negligible compared to the drag and weight We F U4 IO M need to determine the drag force on a parachute D T 17 C Assumption P 2 LE 1 Buoyancy is ignored V 2 Air is an ideal gas W 4Y1 Solution A force balance on the bulldozer is 2 F NFD gt N WFD where N 39 Th dr s the number of parachutes and FD is the drag force on one parachute e ag force is de ned by Eq 106 FD CDpV 2A2 where A IrD24 Using Table 102 the drag coe 39lcient for a parachute is CD 13 For the air density use the ideal gas equat39 PM 9513211289742 1 p 11415 3 8314 290K m kmolK kg The drag force i FD 13 11415 kgmz 10rns2 Ir420m2 st kgm 223300N Solving for the number of parachutes 4 5000N 193 23300N Therefore we use two parachutes and the terminal Velocity would be slight under 10 ms Answer 1031 1031 A heliumfilled spherical balloon is released into air at 40 F 140 psia The combined weight ofthe balloon and its payload is 300 lbf If a vertical velocity of 10 Ms is desired what diameter balloon is required in ft 7 39 39 the air Ifthis balloon is tethered to the ground in a 10 mihr wind what angle does the restraining cable make with the ground Approach F TU IOQj At terminal velocity there is a balance offorces D among wei L b ancy and drag The drag and T buoyancy forces depend on the balloon diameter 9 Em which is unknown A force balance must be wrinen vro IF and each force evaluated For part b there is a M T 40 horizontal velocity only Again a force balance must P H39OF 4 be valuated which includes the tension 39n the tethering cable T w 300 U Assumptions 1 The balloon is spherical and smooth 2 Air is an ideal gas 3 The cable acts as a rigid link Solution a A force balance on the balloon is unknon diameter The buoyancy force is the weight ofthe displaced air Assuming air is an ideal gas pM 14lbfin2 2s97Ham113mm144irBtt2 M A 1b ET 154m lbflbmolR40460R 39 ft Fmy 3an 322tts2 007561bm 3 Mm 113st 322ft lbm600396D31bf The drag force is de ned by Eq 106 FD CDpV 2A 2 where ArD24 TL from Appendix B7 at 40 F u 1179x103951bm s 007561bm ft 10ft s m Re pW 5 64100Dm y 1179x1039 lbmfts FD CD 007561bm 310 s2 Ir4D 2 113st 322tt lbm 200922CDD71bf Combining the force expressions 0 00396D3 7 00922CDD2 7 300 The procedure to obtain a solution is 1 guess a value of D 2 calculate Re from the above equation 3 at this Re obtain CD from Figure 1010 and 4 calculate D from the cubic equation and compare to the guessed value Continue the iteration until convergence is obtained Performing the iteration z 20 7 D219 7 Res 127 x106 Answ r b For the stationary balloon tethered by a cable two force balances are require We need to calculate a new drag force for the 10 mihr 147 s wind 127x10 5 14710 1s7x106 From the Figure 1010 CD 2025 FD 025007561bm 3147 s2 Ir4198 2 lbfs2322 1bm21951bf From above the buoyancy force is Fmy 003961983 307 lbf Using the expression given in Example 104 and adapting it to this problem F 7 W 7 6 canquot W tanquot M 197 Answer FD 195 ZF0FW7FD7W Eachtermiswrinenintermsofthe A A u 1 Forair The drag force is e d as was done in Example 104 Comment One assumption used in Example 104 was that the connecting cable acted as a rigid link This is reasonable ifthe quot n 39 innt nnrlL L quot quot 10 32 1032 A 40mm ping pong ball weighing 0025 N is released from the bottom ofa 4m eep wimm39ng poo whose 39 39 d s 1 1 temperature is 20 C I noring the time to reach terminal velocity how long does the ball take to reach the pool surface in s Approach 7 The time required for the ball to reach the surface of q the swimming pole is obtained by dividing the dep T l 4M m drag The drag force must be determined iteratively Assumptions 1 Ignore the time required to accelerate to terminal V w O 02y velocity Solution Ignoring the time required to accelerate the ping pong ball to is terminal velocity the time required for the sphere to reach the surface is t H V Hence we need to determine the terminal velocity A force balance on the ping pong ball is 2F0F5WFDW FDF5WW The buoyancy force is the weight ofthe displaced liquid For water from Appendix A6 at 20 C 9982 kgm and y 985x10394 Nsm2 Fmy pVg gprDZ 5 981ms2 9982kgm31t004m3 stkgm60328N The drag force is de ned by Eq 106 FD CDpV 2AZ where A D24 Combining the equations for force balance drag force weight and buoyancy and simplifying we obtain n5 5 8Fm W 7 803280025N 7 0483 5 m IDurn2 CD19982kgm3004m2 CD 5 The drag coef cient depends on Reynolds number 9982kg m3 v s 004m Re warm 985x 10 Nsm The procedure to use is Guess a drag coef cient CD Calculate the velocity Calculate Reynolds number 1 I u 4 V Figure 1 A 1 A L 4 I Continue until converged Guess CD m 4 US v w 4105 amp405401144550 04 s From the gure CD m 05 US v j 0983 Re 40540098 39800 s From the fgur D m 05 Because ofthe dif culty in reading the gure more closely another iteration is not justi ed Therefore the time for the ball to reach the surface is 4m L085 t Answer H t V 098ms 10 33 1033 A 50 mihr 60 F wind blows perpendicular to an outdoor movie screen that is 70 wide and 35 tall the screen is supported on 10 tall pilings Estima a the dr force on the screen in b b the moment at the base ofthe pilings in lbt Approach The basic drag force equation is used to calculate the force The drag coe icient must be evaluated The moment is calculated by assuming the force acts through the center ofthe sc ee Assumptions 1 The air is at one atmosphere 2 Drag force on supportpoles is ignored Solution a The drag force is de ned by Eq 106 FD CDpV 2A2 whereA is the area A Mi 701t35 2450ft2 As uming the Reynolds number is greater than 1000 we use Figure 107 with bH 7035 2 Therefore CD 5 12 From the append39x for ai 39 B7 at 60 F0 00 FD 120077lbm 3733 s2 245m21bst322 lbm218900 lbf Answer b Assuming the force acts through the center of area ofthe screen 7 F 39 Answer 10 34 10 34 A hotdog company decides to create a giant heliumfilled balloon of a hotdog to oat in parades for advertising purposes It will oat 75 ft above the street and will be controlled by people holding onto tethering lines The balloon is 50ft long and 10ft in diameter and can be approximated as a cylinder Air at 70 F 147 psia is funneled down the street between the buildings at a velocity of 15 mihr Determine the drag force in lbf Approach The drag force with a uniform approach velocity is calculated with the basic equation Assumptions 1 The balloon can be treated as a short cylinder 2 Air is an ideal gas Solution The drag force is FD CDpV 2A2 For air at one atmosphere and 70 F p 00749 lbmftj From Table 102 for a short cylinder with its axis parallel to the flow and DL 5010 5 CD Z 085 2 2 2 FD o85o0749m mj2s j h j 10ft2lbfs2322ftlbm2 ft h mi 3600s 2091bf Answer 10 35 A telephone wire 5mm in diameter is suspended between telephone poles spaced 50 m apart 1f the wind velocity is 100 kmhr and the air is at 2 C 1 atm determine the horizontal force in N the wire exerts on the poles Approach The basic drag force equation is used to calculate the force The drag coefficient must be evaluated T 1 C P Assumptions 1 The cable acts as a long cylinder 2 Flow is perpendicular to the cable Solution The drag force is definedby Eq 106 FD CD p v 2A2 where A is the area A LD From Appendix A7 the air properties 2 C are p 1280 kgm3 and u 167gtlt10 5 Nsm2 To evaluate the drag coefficient we need the Reynolds number 7 3va 7 1280kgm3100000mhr1hr3600s0005m Re 167X10395Nsm2 10650 u From Figure 1010 CDZ 14 FD 141280kgm3100000mhr2 1hr3600s2 50m0005mNs2kgm2173N Answer 10 35 1036 When parachuting an Army Ranger and his gear may weigh as much as 2501bf To prevent injury the Ve ica 39 quot 39 39 39 39 15 s If L r L 39 39 a an nnen hemisphere and the air is at 70 F 147 psia what diameter in ft parachute is required Ap roach z k D gt I Terminal Velocity occurs when there is a balance between the weight of the parachutist and drag 7 Buoyancy is negligible compared to the drag an Us I EQ weight We need to determine the drag force on a F T 7011 5 parachute D P HMML Assumptions lW 250M 1 Buoyancy is ignored 2 Air is an ideal gas Solution A force balance on the parachutist is F0F07W gt FDW The drag force is de ned by Eq 106 FD CDpV 2A2 where A IrD24 Combining the three equations and solving for diameter 8W 5 D 2 CDprV Using Table 102 the drag coef cient for a parachute is CD 14 For the air density use the ideal gas equation 2 1472897l bm111 1 2339 lbf 77 mi lbf m0 10749 1545 530R lbmolR 7 82501bf322 lbmlbfs7 5729 Answer 7 14007491bmt n15 s2 39 10 36 10 37 An office building approximately 90 m wide and 150 m tall is to be built in a new development far from any other building Its drag coefficient is 14 Determine a the drag force inN if the wind at 17 C is uniform at 15 ms b the drag force inN if the velocity profile can be approximated with the 1739h power law VX Vw y 7 with a boundary layer thickness of 100 m and a freestream velocity of 15 ms Hint integrate to obtain the total drag force Approach w 70 The drag force with a uniform velocity can be calculated with the basic equation 1f the velocity 110 1 Mi T profile is taken into account then a differential analysis must be perform ed and the resulting equation 5 integrated X H I 0M Assumptions 1 The air is at one atmosphere 2 The drag coefficient for a building is the same for 1 II the whole structure and for a differential element Solution a Drag force is FD CDpV 2A2 For air at 17 C pa 1214 kgm3 FD 141224kgm315ms2 90m150mNs2kgm2 260x106N b When the velocity profile is taken into account begin with the differential force on a differential area as shown on the schematic dFD CD3v 22 4 CD3v 22Wdy Substituting in the expression for the velocity profile we recognize that we must break this into two parts 0 g y lt a dFD 05CDpVw yaWT Wdy a y H dFD 05CDpVw2Wdy With the drag coefficient density and freestream velocity all constant both expressions can be integrated 0 g y lt a FD deD 050pr 5quot2 Wdy 0579CDpv3W5 FD 0579141214kgm315ms2 90m100mNs2kgm1338x106 N 5g y H FD deD joscpp vm2 Wdy 05CDpr2WHi 39 FD 05141214kgm315ms2 90m150710Omstkgm0860gtlt106 N The total drag force is Fm1338gtlt106N0860gtlt106N2198gtlt106N Answer Comments Note that by taking into account the changing velocity over the height of the building the total drag force is reduced by 148 10 37 1038 The superintendent ofa national cemetery wants to erect a larger than usual agpole and ag The ag pole is 125 ft tall The ag has a heightH 20 ft and a length L 38 ft A e the ag pole must wi a wind of 60 mihr at 32 F when the ag is ying The drag coef cient ofthe ag can be estimated by CD 005LH If the pole has a diameter of9 in determine a the total force exerted on the pole in lbf b the moment at the base ofthe pole in ftlbf Approach This is a composite body We assume the combined drag force is the sum ofthe individual drag forces Moment is force time distance and we assume each individual drag force acts through the center of is object Assumptions 1 The air is an ideal gas at one atmosphere 2 The pole is a smooth cylinder 3 The drag on each component is calculated as if it stands alone Solution a The drag force is de ned by Eq 106 FD CDpV 2142 The drag coe icient depends on the Reynolds number For air from Appendix B7 at 40 F p 00811bf 3 y 1179X103951bffts For the ag CD 005 LH 00538ft20ft 0095 FDM SSfts2 rommszmnhr 232 I For the pole we need the Reynolds number to evaluate the drag coef cient 00811b 03 sstt s 0750 amppm m VS gtlt 5370 y 1179x10 lbffts Assuming the pole is a smooth cylinder from Figure 1010 CD a 022 F010 02200811bmft3 sstts2 125 075 lbst322 lbm2 2011bf Therefore the total force is 27 703 201 9041bf Answer b The bending moment at the base is M Fuzzm Hpgt1 2FDIag Hpm ag2 2011bf125 27031bf125ft 2002 93400 lbf Answer Comments at the dip in L 4 39 cm W should be performed at a lower velocity and a high velocity to see ifthe resulting force is much different For example ifthe velocity were 50 mihr then Re 380000 and CD 5 05 which would give a drag on the pole of 317 lbf which is 50 greater than what was calculated previously However the ag drag force would decrease to 488 lbf and the total would be 805 lbf which is 11 less than what was previously calculated 10 38 10 39 Antennas on old cars are vertical circular cylinders 025 in in diameter and 4ft long Some people attach objects to the top of their antenna so that their car is more easily found in crowded parking lots If the car is driven at 65 mihr and the air is at 80 F 147 lbfin2 determine a the bending moment in ftlbf at the base of the antenna without the object attached b the bending moment in ftlbf at the base of the antenna if an object shaped like a sphere 3in in diameter is attached to the top of the antenna A moment is force times distance Assume the drag forces on the antenna and the sphere act through their centers For this composite body assume the drag on the individual components is calculated as if the other part were not present Approach U25 933 D 3 IN T 70 p 1 47 iiw D 0251 Assumptions 1 The drag force on each part is calculated as if the other part were not present Solution a For the antenna the drag force is defined by Eq 106 FD CD p v 2A2 where A is the area A Ld 4ft025inlft12 in00833ft2 For the drag coefficient we need the Reynolds number so from Appendix B 7 for air at 80 F p 0074 lbmft3 andz 125X10395 lbmfts pv d 0074lbmft3953fts00208ft Re 75 125X10 lbmfts ll740 u Using Figure 1010 CDZ 12 FD 1200741bmft3953fts200833ft21bfs2322ft1bm2104lbf Assuming the force acts through the center of the antenna M FDLZ 1041bf4ft2208ft1bf 4 Answer b For the composite body we need to calculate the force and moment created by the sphere For the sphere pv D 00741bmft3953fts025ft T 125X103951bmfts From Figure 1010 for a smooth sphere CD S 05 FD 050074lbmft3953fts2 7r4025ft2 lbfs2322ft lbm2o2561bf M 208ft1bf 02561bf40252ft314ft1bf 4 Answer Re 141000 10 39 10 40 The external rearview mirrors two each on old cars were circular disks lOcm in diameter New cars use streamlined rearview mirrors two each to reduce drag losses these mirrors can be approximated as hemispheres facing upstream A car without mirrors has a drag coefficient of 036 a frontal area of 15 m2 and rolling resistance can be ignored For a car speed of 125 kmhr in air at 23 C 100 kPa what percent increase in gas mileage could be obtained by replacing the oldmirrors with two new ones of the same diameter Approach The car plus mirror combination is a composite body For this composite body the individual forcepower contributions from the car and the mirrors need to be calculated The total power with and without the mirrors then can be compared Power is force times velocity Assumptions 1 The drag on a composite body can be calculated as if the separate parts act independently 2 Gas mileage is inversely proportional to power Solution The drag force is defined by Eq 106 FD CDprZAZ For composite bodies we assume that the drag force contribution from each part is simply additive Power is W FD v0 chvjAz For a composite body we add the contributions from the parts With velocity air conditions and engine efficiency the same for the two mirror types on the car two mirrors on the ca Therefore the ratio of power with new and old mirrors is CDA2CDVAWWpvjz C A 2C A W Dmr m DJVRND mum Wild 7 CDarAmr 2CDmxnarAmxnm M p Va 2 CDmr Ara 2CDmnmAmmma1d From Table 102 For a disk CDVWWVDM m 11 and A 7rD24 For a hemisphere facing into the velocity CDVWWVW m 04 and A 7rD24 Therefore 03615m22047E401m2 Wald 036l5m22ll740lm2 Therefore the new mirrors reduce the fuel consumption by about 20 Answer 0980 Comments By itself the changing of the mirror configuration does not affect the gas mileage dramatically But many small changes in the aerodynamics of a car plus decreasing the car weight can significantly increase the gas mileage 10 40 engine cle th al ef ciency is 25 A reasonable average air condition is 10 C 100 kPa The fuel costs 040L its specific gravity is 082 and its energy content is 40000 kJkg 39LM mach W quot Dle Increased fuel cost is directly proportionalto a increased power The sign causes a drag force and L OJm power is force times Velocity The total time the taxi T 0 C p loo LR total energy used which is related to the gasoline consumed sowa 13 qms Assumptions 1 The sign can be treated as ow over a long rectangular ro 2 Air is at one atmosphere Solution The drag force on the sign is defined by Eq 106 FD ch v 2Az From Table 101 we assume the sign acm as a long rectangular rod with DL 7 1203 4 so CD m 13 For air at 10 C 1244 kgm3 FD 131244kgm3139ms2 03m12mNs7kgm2562N Power is obtained from W FDV 562N139ms1W1Js781W The total energy used for the 100000 kmyr requires the time t distance 100000kmyr1000mkm 7194X106 syr Velocity 139ms Total energy used including engine thermal efficiency is W 789W7194gtlt106syr1kJ1000J 025 The mass offuel used is m 2247gtlt107 kJyr40000kJkg562kgyr Volume of fuel used m 7 562kgyr1L10393m3 F7 0821000kgm3 Cost offuel 040L685 Lyr274yr Answer E 2247x107 kJyr V GSSL yr Comments The drag coefficient is approximate so this answer has a fair amount of uncertainty The revenue generated by canying the sign would be balanced against the additional fuel cost 1041 10 42 A thin at plate 10ft long and 2ft Wide is mounted horizontally on a 10ft long 3in diameter pole Air flows at 60 F 147 psia along the 10ft length of the plate The velocity profile of the air flow varies from 0 at the base of the pole to 50 fts at 10 ft along the top of the plate Taking into account the variation in velocity determine the total drag force in lbf acting on the composite body Approach This is a drag force on a composite body The 1U T 5081 contribution from each part must be evaluated gt Because of the velocity variation on the pole that V V 7H force will need to be obtained by integration OIA 39 9 H 139 1061 Assumptions 1 The contributions from each part of the composite D glbgt 639 body are add1t1ve I I I I 2 The pole drag coeff1c1ent 1s constant over 1ts length Solution The drag force is on the at plate both sides is given by FD 2CDpV 242 Where the plan area is A LW The drag force depends on the Reynolds number From Appendix B7 for air at 60 F p 0077 lbmft3 and1 1214X10395 lbmfts The Reynolds number is 0077lbmft350 fts10ft e L 1214gtlt103951bmfts Assuming transition to turbulent occurs at a Reynolds number of about 500000 0074 1740 7 0074 1740 7 0 00316 DRe5 ReL 3171X106V5 3171gtlt106 200031600771bmft350fts2 10ft2ft1bf s2322ftlbm2 0381bf For the drag on the pole begin with a differential area dFD 05CDpV 251A 05CDpV 2Ddy The velocity varies linearly with distance from the ground Using the given information V Vw yH For the drag coefficient the pole Reynolds number varies from zero at the ground to a maximum atH 10 ft Therefore 0077lbmft350 fts025ft Re 7 1214gtlt103951bmfts From Figure 1010 we assume a reasonable value of the drag coefficient over the complete pole CD 13 Therefore integrating the expression for the pole drag force 2 H 1an 1 OSCDp v Ddy CppngD6 130077lbmft350fts2 10ft025ftlbf s2 322ftlbm 63241bf Finally the total force is FD 324038 3621bf 4 Answer 3171gtlt106 F Bylaw 79280 10 42 1043 A lar m pecan cu ge family is going on a vacation in their minivan that has a drag coef cient of 044 and a 5 2 frontal area 15m wide 30cm h39 the car top carrier compared to without it Approach For this composite body the individual forc power contributions from the car and the car top carr39 need to be calculated The total power with and without the carrier then can be compared Power is force times velocity Assumptions 1 The drag on a composite body can be calculated as ifthe separate pars act independ l 2 The car top carrier is approximated as a at rectangular plate perpendicular to the velocity Solution The drag force is defined by Eq 106 chv 2 2 run r 39 d Power is W FDvm chngz Ac Jul ureu Au a e they w1 car top carrier that is gh and 2m long Estimate the increase in power required to drive at 10 39 0kmhrw1th 2 W 304 L K H 030M 7 A KTML k L cm 11 IOO A CW 0 44 from each part is simply additive Therefore the ratio of power with and without the carrier is Wm CDmpVAy2 W CDACDApv2ic A c am am Dmvmv AcIZYYMY DcavAcav Assuming we can treat the car top carrier as a at rectangular plate from Fig 107 Camme N 1 2 WW 7044351203015 W 5 Wo 135 or a 10 43 Answer 10 44 A small aircraft has a wing area of 27 m2 a takeoff mass of 2500 kg a lift coefficient at takeoff of 049 and a drag coefficient at takeoff of 00074 For standard atmospheric conditions determine a the takeoff speed at sea level in kmhr b the power required at takeoff in kVJ c the maximum mass in kg possible at takeoff speed using the power from part b if the airport is at 500 m Approach The takeoff speed is obtained by knowing that lift force equals weight at takeoff The basic equations are used and a standard atmosphere is assumed Assumptions 1 The air conditions equal the standard atmosphere Solution a At takeoff speed lift force equals weight FL W mg The lift force is calculated with FL C L pV 2142 Combining these two expressions and solving for velocity 2 05 v mg CLpA From Table 10 5 p 1225kgm3 22500kg981ms2 7 049l225kgm327m2 b Power is obtained from W FD V CDpV 2142 Note that FD CDpV 2142 Dividing the drag force by the lift force FDFL CDCL Therefore W CDCL mgv 000740492500kg981ms2550msle2kgmleNm20370W Answer c At higher elevation we have two equations FL W CLpVZA2mg FDV W a CDpVEA2W From the second equation we can obtain the maximum velocity at the higher elevation using the power from part b Solving for velocity 2W W V CDpA From Table 105 by interpolation p b 07821225kgm3 0958kgm3 05 550319s Answer S 13 7 220370WNmWskgmNs2 7 000740958kgm327m2 From the first equation 2 m CLpV 2A 0490958kgm3597ms 27m2 2g 2981ms2 5973 215 s hr 2300kg Answer 10 44 10 45 For a small plane the lift coefficient at the landing speed is 115 and the maximum lift coefficient at the stall speed is 142 The landing speed of the airplane is 8 ms faster than its stall speed Determine both the landing and stalling speeds in ms Approach At landing and stall the lift force just equals the weight of the plane We can use the given information and the basic definitions of lift force to determine the two speeds Assumptions 1 The air is an ideal gas 2 The pole is a smooth cylinder Solution Let V L the speed at landing and V s the speed at stall These two speeds are related by VL VS 8ms At landing and at stall lift force equals weight Landing FL CLVLpVLzA2mg Stall FL CLVSpVSZA2mg Equating the two equations and simplifying cps V32 CAL VL2 Substituting in the relationship between stall and landing speed CLS V32 CAL VS 8mS2 142 v 115 VS 8ms2 Solving for the velocities VS 719ms Answer vL 7198798ms 4 Answer 10 45 1046 A 250kg glider with a wing area of 22 m2 has a minimum glide angle of 17 Its lift coef cient is 11 For a still day at 15 C 10 Pa 39 a the 0 k determine total horizontal distance for the glider to descend from 1500 m to sea level in km b the time required in min Approach Part a is a geometry problem for a right triangle of p heightHand angle 639 the use ofthe tangent will give us K D the base length L For part b we must determine the H plane speed by recognizing that at steady ight lift force equals weight Once we have the velocity then t39 39 i E L gt determined noting that velocity is distance divided by time Assumptions 1 The air is an ideal gas at one atmosphere Solution a From geometry tan HL gt L Htan 1500rrtan17 50540m505km 4 Answer b We can ulc glider to reach L 4 39 L D V where D the distance traveled is obtained from geometry D Hsin 1500rrsin17 50553m Velocity is determined from the fact that in steady ight weight equals li FL W CLpV 2AZ mg For air at 15 C p 1222kgm3 Solving for velocity V T 2250kg981ms2 1292 CLPA 111222kgm322m2 s IM3907s51min Answer 129ms 1046 1047 When quot 39 lift drag in equilibrium Show that the glide slope angle 639 is given by 6 tan39l CDCL Approach At equilibrium for unpowered ight lift drag and weight forces balance out Force balances in the x and y directions must be written Solution a Force balances in the two coordinate directions are 2 Fx FD 7 39 39 F FL 7 W cos 6 gt Dividing these two expressions FD 7 Wsin 6 7L 7 W cos 6 Drag force is FD CDpV 2142 Li force is FL CLpV 2A 2 Combining these two expressions FL Wcos tan FDCo Cl Substituting this into the ratio of the force balances D D an a F C Therefore 6 tanquot 4 Answer I 10 47 10 48 A hydrofoil is a watercraft that rides above the surface of the water on foils which are essentially wings attached to the bottom of struts connecting the foils to the hull of the boat Suppose the area of the foils in contact with the water on a 2000kg hydrofoil is 11 m2 Their lift and drag coefficients are 172 and 045 respectively Determine a the minimum speed required for the foils to support the hydrofoil in kmhr b the power required to propel the hydrofoil at the speed calculated in part a in kVJ c the top speed if the boat has a 175 kW engine in kmhr Note that at higher speeds the hydrofoil rises further out of the water and the lifting area is decreased Approach F FL The minimum speed for the foils to support the V hydrofoil occurs when the lift force matches the weight We have sufficient information given to evaluate the speed Power is drag force times velocity W hyLW W M l looquot C L7 Assumptlons A k L d 1 Water density is 1000 kgm3 m V we 3 CD 0quot Solution a At the minimum speed FL W mg where liftforce is FL CLszAZ 2mg U2 CombLmng equations and solv1ng for velocity memm C L pA We assume for water p 1000 kgm3 22000kg981ms l721000kgm3llm2 mmxmm 12 4331 4 Answer S b The power required at the minimum speed is W FD v chv 31420451000kgm3455ms3llm2223310W Answer c At the maximum speed FL W mg Comparing the lift and drag equations we can see that C C FD CFL Cmg CL CL In addition power is Wm F D Vm Combining these equations l75000W lJ lW vm E amp Wm jwam m 4 Answer FD CD mg 045 2000kgm3981ms2 s hr 10 48 1049 Consider a U 39 at 35000 ft m er 39 39 If its lift and drag characteristics are similar to those given in Figure 1019 determine a the optimum angle of anack for maximum glide distance b if it can make it to an airport 425 miles away Ignore initial velocity Approach F At equilibrium for unpowered ight lift drag and L weight forces balance out Force balances in the x and H 330w 1 y directions must be written 056 i H eL39Qal Solution a Force balances in the two coordinate directions are ZFXFDiWsin gt FDWsin6 F FliWcosS gt FLWcos6 Dividing these two expressions Fo m an 9 FL W cos 6 Drag force is FD CDpV 2AZ Lift force is FL CLpV 2AZ Combining these two expressions FD Co C l 1 Substituting this into the ratio of the force balances F C D D an a Cl Therefore 6 tanquot 4 Answer L b From Figure 1016 we can obtain CD and C We want the minimum angle 639 to maximize the distance traveled Choosing values of drag and lift coef cients for different angles of anack So using an angle of attack of about 8 the maximum glide distance will be obtained Answer c Using the given geometry H H H 35000ft lmi tan6 i L L tan a CDCL 00082 52800 Theplane can reach the airport Answer 808mi C omments This is an optimistic answer In reality the plane could not glide this far 10 49 10 50 If you have ever own a kite in a strong wind you know that the pull on the string can be quite strong Consider a 12m by 08m kite that has a mass of 05 kg Its lift coefficient can be approximated by C L 27r sino where 11 is the angle of attack In a 45 kmhr wind the kite s string has an angle of 500 to the horizontal and the kite has an angle of attack of 5 Determine the force on the string in N Approach The tension in the string can be obtained from a force balance on the kite Assumptions 1 Air is an ideal gas at one atmosphere S olution A force balance on the kite in the ydirection is ZE0FL7Tsin67Wm kg Fz mg s1n 9 sin 9 The lift force is defined by FL Cva 2A2 Assuming air is an ideal gas at 25 C one atmosphere p 1181 kgm3 For the drag coefficient CL 27rsin5 0548 The lift force is FL 05481181kgm3125ms212m08mNs21ltgm2485N The tension force is 2 2 T 485N 05kg981ms Ns kgm 569N Answer sm50 10 51 A small experimental plane has a mass of 750 kg a drag coefficient of 0063 and a lift coefficient of 04 In level flight it is own at 175 kmhr For standard conditions 1 atm 25 C determine the effective lift area of the plane in mi Approach The basic equation for lift force is applied directly to this problem For steady flight weight equals lift Sufficient information is given to solve for the plan area Assumptions 1 Air properties are at 25 C 1 atm Solution The lift force is calculated with the following equation FL CL pV 2A2 In steady flight lift equals weight FL W mg Combining the equations and solving for effective area A 2mg Cm V 2 For air at 25 C p 1184 kgm3 2750kg981ms23600shr2 7 041184kgm3175000mhr2 10 52 Because of the decrease in density and temperature with increasing elevation in the atmosphere lift and drag forces change Consider a plane flying at velocity V at sea level For the same lift and drag coefficients determine a the speed required at 10000 m to generate the same lift force b the change in drag force Approach At steady ight lift equals weight We can compare the velocity requirements at two different elevations using the basic definition of lift force Assumptions 1 The lift coefficient is constant Solution a In steady flight F L W The lift force is defined by FLCLpV2A2 CLpV2A2W For two different elevations we can find the relative velocities required by taking a ratio of the equation above evaluated at the two conditions CL1P1V12A12 E Cmpz V221422 7 W2 Because the weight area and lift coefficient are the same at the two conditions 0 5 0 5 V22 amp a Lamp LlPo V12 92 V1 92 pzpo Obtaining the values of density at different elevations from Table 105 V2 1 0 j 172 Answer v1 03376 b Drag force is FD CDp V 2142 Taking the ratio of the forces at the two conditions FD2 CD2p2 V22A22 7 p2 V22 FDA CD1pl V1214l2 7 pl V12 Using the relation given above for the ratio of velocities FDv2 amp 1 Answer FDi p1 p2 10 51 10 53 When planes takeoff and land at airports at higher elevations the lower density air due to reduced atmospheric pressure must be taken into account because of its effect on lift and drag If a plane requires 15 s to reach its takeoff speed of 220 kmhr at sea level in a distance of 500m estimate for an airport at 2000 m a the takeoff speed in kmhr b the takeoff time in s c the additional runway length required in m Assume the same constant acceleration for both cases Approach The takeoff speed is obtained by knowing that lift force equals weight at takeoff the takeoff speed at the second elevation can be obtained by comparison The additional time and distance required at the second elevation can be approximated by a force balance that now takes into account the acceleration Use Newton s second law of motion Assumptions 1 Lift coefficient is the same at both elevations and speeds Solution a At takeoff speed lift force equals weight FL W mg The lift force is calculated with FL C L pV 214 2 Combining these two expressions and solving for velocity V 2mg 05 CL pA Assuming the lift coefficient is the same at both elevations and speeds the ratio of the speeds is 0 5 0 5 Lamp We V1 92 p2po From Table 105 at 2000 m p2p0 08217 6 V2 220kmhr10821705 243kmhr Answer b To find the additional time we need to take into account the acceleration of the plane Using Newton s second law ofmotion 2F FD7max7dedt x Drag force is FD C D pV 2142 Substituting this into the force balance separating variables and integrating from zero the initial velocity to the takeoff velocity W fwd i i V LL Capra i t 2mii v v2 0 2m v V v v 2m CDpA v v If we let the initial velocity go to zero then its inverse goes to infinity Therefore assume the initial velocity is 1 kmhr and the ratio of times is 2 t2jl24ij15s183s Answer t1 1p11V171 08217 122071 c For the distance we revisit the original force balance 7F imax gt CDpV 2142 max Note that ax dV dl and using the chain rule on the derivative 11 dv dv dx i V dt dx E 7 dx Substituting this expression into the force balance separating variables and integrating K v v CD2V2A2mv a agpjv a x1n f x 239quot ln f dx 2m Vx 0 2m V V CDpA VI 1 1n V V Therefore w Assuming the initial velocity is 1 kmhr x1 1p1nV1Vx 1n V l 243 x2x1p1p2 2500m 1 j 620m 4 Answer lnV1 08217 ln220 10 52 10 54 An airplane is to be designed that will fly at 650 kmhr at an altitude where the density is 0655 kgm3 and the kinematic viscosity is 2 X 1039 m s A 11511 scale model that is the model is geometrically similar to the prototype but is 11511 its size is built for use in a wind tunnel whose velocity is 650 kmhr The air in the wind tunnel is at 55 C and viscosity is independent of pressure Determine a the test section pressure in kPa so that the model data are useful in designing the prototype Hint match Reynolds numbers b what is the relation between the drag on the prototype and that on the model Approach To match the flow conditions the Reynolds number of the prototype and model should be equal For the relationship between the drag of the prototype and the drag of the model we assume because the Reynolds numbers are the same the drag coefficients are equal We then can use the drag coefficient to determine the relationship Assumptions Flow similarity occurs when Reynolds numbers match Drag coefficient depends only on Reynolds number 3 Air is an ideal gas N Solution a For flow similarity we set the Reynolds numbers equal to each other Re Re p m Pp Vpr pm VmLm p m The velocities and viscosities are the same and LmLp ll5 Therefore pm pp LPLyn 0655kgm3159825kgm3 For the pressure we use the ideal gas law ET 9825kgm3s3i4kl lt552731lt 112 k 0 925kPa Answer 2897 g kmol b To find the relationship between the drag forces on the prototype and model we take the ratio of the drag force expressions FDVP CDvppp VPZAP 2 FDm CDmpm Vm2A m2 Canceling like terms and noting that AWAp LynLp 2 1152 F L 2 amp P E 15215 Answer FDvm pm Lm 9825 10 53 10 55 A 1m by 15m plate moves through still water at 3 ms and is at an angle of 12 to the Velocity Vector For 39 39 39 4 f 39 39 17 4 L quot f 39 39 72 Determine a the resultant force on the plate inNL L L39 L this L p w the puwcl 39 in kW and d the drag force inN and power in kW ifthe plate moves through 20 C air instead of water Approach F The resultant forced is due to the Vector sum ofthe lift L F2 and drag forces The angle ifmakes with the horizontal can be determined by geometry U Zmls I 9R 5 PD Assumptions 9 73910 f 1 The water density is 1000 kgmz 2 Air is at one atmosphere CL CD of W 3 I TM L x i M Solution a 39 L vcuul sum of L quot A A F F F5 The lift force is FL CLpV 2A2 With awater density of1000 kgmz FL 0721000kgm33ms2 15m1mstkgm24860N The drag force is FD CDpV 2A2 FD 0171000kgrr 3ms2 15m1mNs2kgm21150N The resultant force is us FR 11502 48602 4994N 4 Answer b The angle the resultant force makes with the horizontal is obtained with tan 6R i gt 6R tanquot i tanquot 133 lt Answer FR FR 4860 c Power is force times Velocity The force is th