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# Solutions to Cengel and Boles Thermodynamics textbook problems ENGR 222

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L m SA CE GEL I gemgjgmzx 39 BOLES Eighth Edition ll Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Cengel Michael A Boles McGrawHill 2015 Chapter 1 INTRODUCTION AND BASIC CONCEPTS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 12 Thermodynamics 11C Classical thermodynamics is based on experimental observations whereas statistical thermodynamics is based on the average behavior of large groups of particles 12C On a downhill road the potential energy of the bicyclist is being converted to kinetic energy and thus the bicyclist picks up speed There is no creation of energy and thus no violation of the conservation of energy principle 13C A car going uphill without the engine running would increase the energy of the car and thus it would be a violation of the first law of thermodynamics Therefore this cannot happen Using a level meter a device with an air bubble between two marks of a horizontal water tube it can shown that the road that looks uphill to the eye is actually downhill 14C There is no truth to his claim It violates the second law of thermodynamics Mass Force and Units 15C Kgmass is the mass unit in the SI system whereas kgforce is a force unit 1kgforce is the force required to accelerate a 1kg mass by 9807 ms2 In other words the weight of 1kg mass at sea level is 1 kgforce 16C In this unit the word light refers to the speed of light The lightyear unit is then the product of a velocity and time Hence this product forms a distance dimension and unit 17 C There is no acceleration thus the net force is zero in both cases 18 The variation of gravitational acceleration above the sea level is given as a function of altitude The height at which the weight of a body will decrease by 03 is to be determined Analysis The weight of a body at the elevation 2 can be expressed as Z W mg 2 m9807 332 x 10 In our case W 1 03 100Ws O997Ws O997mgs O997m9807 Substituting 0 09979807 9807 332 gtlt106 z z 8862 m SeZe PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 13 19 The mass of an object is given Its weight is to be determined Analysis Applying Newton39s second law the weight is determined to be W mg 200 kg96 msz 1920N 110 A plastic tank is filled with water The weight of the combined system is to be determined Assumptions The density of water is constant throughout Properties The density of water is given to be p 1000 kgm3 Analysis The mass of the water in the tank and the total mass are mtank 3 kg V02 m3 mw pV1000 kgm302 m3 200 kg H20 mtotal mw mmk 200 3 203 kg Thus W mg 203 kg981ms2 1 N2 1991 N 1kgms 111E The constantpressure speci c heat of air given in a speci ed unit is to be expressed in various units Analysis Using proper unit conversions the constantpressure specific heat is determined in various units to be lkJkg K 1 kJkg C Mx j 1005Jg C cp 1005kJkg C j1005kJkgK 1005kJk C 6 g lkJ 1000g cp 1005kJkg C amp 0240kcakg C 41868kJ M 024OBtulbm F 0 1005kJk C P g 41868kJkg C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 112 A rock is thrown upward with a specified force The acceleration of the rock is to be determined Analysis The weight of the rock is 1N W mg 2 3 kg979 ms2 2 1kgms 2 2937 N Then the net force that acts on the rock is FM 2 Fup Fdown 200 2937 1706 N Stone From the Newton39s second law the acceleration of the rock becomes 1N F 1706N lkgmsz m 3kg J 569 ms2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 113 proper units Analysis The problem is solved using EES and the solution is given below quotThe weight of the rock isquot Wmg m3 kg g979 ms2 quotThe force balance on the rock yields the net force acting on the rock asquot Fup200 N Fnet Fup Fdown FdownW quotThe acceleration of the rock is determined from Newton39s second lawquot Fnetma quotTo Run the program press F2 or select Solve from the Calculate menuquot SOLUTION a5688 msAZ Fdown2937 N Fnet1706 N Fup200 N g979ms2 m3 kg W2937 N m kg a msz 200 1 1902 2 9021 3 5688 160 4 4021 5 3021 712039 6 2354 1 3 7 1878 g 8 1521 a 80 9 1243 10 1021 40 Z 0 1 2 3 10 15 339 Problem 112 is reconsidered The entire EES solution is to be printed out including the numerical results with PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 16 114 A resistance heater is used to heat water to desired temperature The amount of electric energy used in kWh and kJ are to be determined Analysis The resistance heater consumes electric energy at a rate of 4 kW or 4 kJs Then the total amount of electric energy used in 3 hours becomes Total energy Energy per unit timeTime interval 4 kW3 h 12 kWh Noting that 1 kWh 1 kJs3600 s 3600 kJ Total energy 12 kWh3600 kJkWh 43200 M Discussion Note kW is a unit for power whereas kWh is a unit for energy 115E An astronaut took his scales with him to space It is to be determined how much he will weigh on the spring and beam scales in space Analysis a A spring scale measures weight which is the local gravitational force applied on a body 1 lbf W mg 1501bm548 S 2 2 322 lbm s 255bf b A beam scale compares masses and thus is not affected by the variations in gravitational acceleration The beam scale will read what it reads on earth W 1501bf 116 A gas tank is being filled with gasoline at a specified ow rate Based on unit considerations alone a relation is to be obtained for the lling time Assumptions Gasoline is an incompressible substance and the ow rate is constant Analysis The lling time depends on the volume of the tank and the discharge rate of gasoline Also we know that the unit of time is seconds Therefore the independent quantities should be arranged such that we end up with the unit of seconds Putting the given information into perspective we have ts lt gt VL and V Ls CC 99 S It is obvious that the only way to end up with the unit for time is to divide the tank volume by the discharge rate Therefore the desired relation is t V Discussion Note that this approach may not work for cases that involve dimensionless and thus unitless quantities PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 17 117 A pool is to be filled with water using a hose Based on unit considerations a relation is to be obtained for the volume of the pool Assumptions Water is an incompressible substance and the average ow velocity is constant Analysis The pool volume depends on the filling time the crosssectional area which depends on hose diameter and ow velocity Also we know that the unit of volume is m3 Therefore the independent quantities should be arranged such that we end up with the unit of seconds Putting the given information into perspective we have V m3 is a function of t s D m and V ms It is obvious that the only way to end up with the unit m3 D Therefore the desired relation is v CDZVt for volume is to multiply the quantities t and V with the square of where the constant of proportionality is obtained for a round hose namely C 7r4 so that V 71924 Vt Discussion Note that the values of dimensionless constants of proportionality cannot be determined with this approach Systems Properties State and Processes 118C The radiator should be analyzed as an open system since mass is crossing the boundaries of the system 119C The system is taken as the air contained in the pistoncylinder device This system is a closed or fixed mass system since no mass enters or leaves it 120C A can of soft drink should be analyzed as a closed system since no mass is crossing the boundaries of the system 121C Intensive properties do not depend on the size extent of the system but extensive properties do 122C If we were to divide the system into smaller portions the weight of each portion would also be smaller Hence the weight is an extensive property 123C Yes because temperature and pressure are two independent properties and the air in an isolated room is a simple compressible system PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 18 124C If we were to divide this system in half both the volume and the number of moles contained in each half would be onehalf that of the original system The molar specific volume of the original system is V 7 N and the molar specific volume of one of the smaller systems is 7 V2 K N2 N which is the same as that of the original system The molar specific volume is then an intensive property 125C A process during which a system remains almost in equilibrium at all times is called a quasiequilibrium process Many engineering processes can be approximated as being quasiequilibrium The work output of a device is maXimum and the work input to a device is minimum when quasiequilibrium processes are used instead of nonquasiequilibrium processes 126C A process during which the temperature remains constant is called isothermal a process during which the pressure remains constant is called isobaric and a process during which the volume remains constant is called isochoric 127 C The pressure and temperature of the water are normally used to describe the state Chemical composition surface tension coefficient and other properties may be required in some cases As the water cools its pressure remains fixed This cooling process is then an isobaric process 128C When analyzing the acceleration of gases as they ow through a nozzle the proper choice for the system is the volume within the nozzle bounded by the entire inner surface of the nozzle and the inlet and outlet crosssections This is a control volume since mass crosses the boundary 129C The speci c gravity or relative density and is de ned as the ratio of the density of a substance to the density of some standard substance at a speci ed temperature usually water at 4 C for which pHZO 1000 kgm3 That is SG 2 0 pHZO When speci c gravity is known density is determined from p SGx pHZO PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 19 130 239 The variation of density of atmospheric air with elevation is given in tabular form A relation for the variation of density with elevation is to be obtained the density at 7 km elevation is to be calculated and the mass of the atmosphere using the correlation is to be estimated Assumptions 1 Atmospheric air behaves as an ideal gas 2 The earth is perfectly sphere with a radius of 6377 km and the thickness of the atmosphere is 25 km Properties The density data are given in tabular form as r km z km p kgm3 6377 0 1225 6378 1 1112 6379 2 1007 6380 3 09093 6381 4 08194 6382 5 07364 6383 6 06601 6385 8 05258 6387 10 04135 6392 15 01948 6397 20 008891 6402 25 004008 Analysis Using EES 1 Define a trivial function rho az in equation window 2 select new parametric table from Tables and type the data in a twocolumn table 3 select Plot and plot the data and 4 select plot and click on curve t to get curve fit window Then specify 211d order polynomial and enteredit equation The results are pz a bz cz2 120252 0101674z 00022375z2 for the unit of kgm3 or pz 120252 0101674z 00022375z2gtlt109 for the unit of kgkm3 where z is the vertical distance from the earth surface at sea level At z 7 km the equation would give p 060 kgm3 b The mass of atmosphere can be evaluated by integration to be h h mdeVI abzcz247zr0z2dz47z abzcz2r022r0zz2dz z0 V 47 ar02hr02abr0h2 2a2br0 cr02h3 3b2cr0h4 4ch5 5 where r0 6377 km is the radius of the earth h 25 km is the thickness of the atmosphere and a 120252 9 0101674 and c 00022375 are the constants in the density function Substituting and multiplying by the factor 109 for the density unity kgkm3 the mass of the atmosphere is determined to be m 5092x1018 kg Discussion Performing the analysis with excel would yield exactly the same results EES Solution for final result a12025166 b010167 C00022375 r6377 h25 m4piarquot2hr2abrhquot22a2brcrquot2hquot33b2crhquot44chquot551 E9 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Temperature 131C They are Celsius C and kelvin K in the SI and fahrenheit F and rankine R in the English system 132C Probably but not necessarily The operation of these two thermometers is based on the thermal expansion of a uid If the thermal expansion coefficients of both uids vary linearly with temperature then both uids will expand at the same rate with temperature and both thermometers will always give identical readings Otherwise the two readings may deviate 133C Two systems having different temperatures and energy contents are brought in contact The direction of heat transfer is to be determined Analysis Heat transfer occurs from warmer to cooler objects Therefore heat will be transferred from system B to system A until both systems reach the same temperature 134 A temperature is given in C It is to be expressed in K Analysis The Kelvin scale is related to Celsius scale by TK T C 273 Thus TK 37 C 273 310 K 135E The temperature of air given in C unit is to be converted to F and R unit Analysis Using the conversion relations between the various temperature scales T F 18T C 32 18150 32 302 F TR T F 460 302 460 762 R 136 A temperature change is given in C It is to be expressed in K Analysis This problem deals with temperature changes which are identical in Kelvin and Celsius scales Thus ATK AT C 70 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 137E The ash point temperature of engine oil given in F unit is to be converted to K and R units Analysis Using the conversion relations between the various temperature scales TR T F 460 2 363 460 2 823R LR 2 457K T K 18 18 138E The temperature of ambient air given in C unit is to be converted to F K and R units Analysis Using the conversion relations between the various temperature scales T 40 C 2 4018 32 2 40 F T 40 27315 23315K T 40 45967 41967R 139E A temperature change is given in F It is to be expressed in C K and R Analysis This problem deals with temperature changes which are identical in Rankine and Fahrenheit scales Thus ATR AT F 45 R The temperature changes in Celsius and Kelvin scales are also identical and are related to the changes in Fahrenheit and Rankine scales by ATK ATR18 4518 25 K and AT C ATK 25 C Pressure Manometer and Barometer 140C The atmospheric pressure which is the external pressure exerted on the skin decreases with increasing elevation Therefore the pressure is lower at higher elevations As a result the difference between the blood pressure in the veins and the air pressure outside increases This pressure imbalance may cause some thinwalled veins such as the ones in the nose to burst causing bleeding The shortness of breath is caused by the lower air density at higher elevations and thus lower amount of oxygen per unit volume 141C The blood vessels are more restricted when the arm is parallel to the body than when the arm is perpendicular to the body For a constant volume of blood to be discharged by the heart the blood pressure must increase to overcome the increased resistance to ow PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 112 142C No the absolute pressure in a liquid of constant density does not double when the depth is doubled It is the gage pressure that doubles when the depth is doubled 143C Pascal s principle states that the pressure applied to a con ned uid increases the pressure throughout by the same amount This is a consequence of the pressure in a uid remaining constant in the horizontal direction An example of Pascal s principle is the operation of the hydraulic car jack 144C The density of air at sea level is higher than the density of air on top of a high mountain Therefore the volume ow rates of the two fans running at identical speeds will be the same but the mass ow rate of the fan at sea level will be higher 145 The pressure in a vacuum chamber is measured by a vacuum gage The absolute pressure in the chamber is to be determined Analysis The absolute pressure in the chamber is determined from Pabs 2 Pa Pvac 92 35 57 kPa Pabs 35 kPa Patm 92 kPa 146 The pressure in a tank is given The tank39s pressure in various units are to be determined Analysis Using appropriate conversion factors we obtain 2 M 1200kNm2 a P 1200 kPa lkPa lkNm2 1000 kg msz j1200000kgm s2 lkPa lkN b P 1200 kPa l 2 1 k 2 1 c P1200kpa kNm 000 g ms 000m 1200000900kgkmsz lkPa lkN 1km PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 147E The pressure in a tank in SI unit is given The tank39s pressure in various English units are to be determined Analysis Using appropriate conversion factors we obtain 208861bf7 2 a P 1500 kPa a J 31 330lbfft2 2 2 I 1 I b P1500kPa 20 8861mm If 2 pSIaZJ2176pSIa 1kPa 144in llbf in 148E The pressure given in mm Hg unit is to be converted to psia Analysis Using the mm Hg to kPa and kPa to psia units conversion factors 2 290psia 6895 kPa P 1500mmH g 1mmHg 01333kPa 1psia 149E The pressure in a tank is measured with a manometer by measuring the differential height of the manometer uid The absolute pressure in the tank is to be determined for the cases of the manometer arm with the higher and lower uid level being attached to the tank Assumptions The uid in the manometer is incompressible Properties The specific gravity of the uid is given to be SG 125 The density of water at 32 F is 624 lbmft3 Table A3E Analysis The density of the uid is obtained by multiplying its speci c gravity by the density of water p SGgtlt p 125624lbmit3 7801bmit3 H20 28 in SG 125 The pressure difference corresponding to a differential height of 28 in between the two arms of the manometer is Patm 127 P31a 2 AP pgh 78lbm 332174 sz2812 1 lbf 2 m 2 126psia 32174 lbm s 144in Then the absolute pressures in the tank for the two cases become a The uid level in the arm attached to the tank is higher vacuum Pabs 2 Pa Pvac 127 126 1144psia b The uid level in the arm attached to the tank is lower Pabs Pgage Paltm 127 126 2 1396 psia Discussion Note that we can determine whether the pressure in a tank is above or below atmospheric pressure by simply observing the side of the manometer arm with the higher uid level PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 114 150 The pressure in a pressurized water tank is measured by a multi uid manometer The gage pressure of air in the tank is to be determined Assumptions The air pressure in the tank is uniform ie its variation with elevation is negligible due to its low density and thus we can determine the pressure at the airwater interface Properties The densities of mercury water and oil are given to be 13600 1000 and 850 kgm3 respectively Analysis Starting with the pressure at point 1 at the airwater interface and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach point 2 and setting the result equal to Patm since the tube is open to the atmosphere gives P1 pwater ghl poilth pmencury gh3 Patm Solving for P1 HEB J P1 Patm pwaterghl poilth pmercury 8113 T i ia39hTER or P1 Patm gpmencuryh3 pwaterhl poith Noting that PLgage P1 Pam and substituting 981 ms2 136OO kgm3 04 m 1000 kgm3 02 m 850 kgm3 03 m IN I a 2 2 1kg ms 1000 Nm P1gage 489kPa Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same uid simpli es the analysis greatly 151 The barometric reading at a location is given in height of mercury column The atmospheric pressure is to be determined Properties The density of mercury is given to be 13600 kgm3 Analysis The atmospheric pressure is determined directly from Patm pgh 1 lkP 13600 kgm3 981 ms2 0750 m N a 2 2 1kgms 1000 Nm 2 1001 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 115 152E The weight and the foot imprint area of a person are given The pressures this man exerts on the ground when he stands on one and on both feet are to be determined Assumptions The weight of the person is distributed uniformly on foot imprint area 39g Analysis The weight of the man is given to be 200 lbf Noting that pressure is force per unit area the pressure this man exerts on the ground is a On both feet K 2781bf in 2 278psi 2A 2x 36 in b On one foot P K 200 M 556 1bf7in 2 556psi A 36in2 Discussion Note that the pressure exerted on the ground and on the feet is reduced by half when the person stands on both feet 153 The gage pressure in a liquid at a certain depth is given The gage pressure in the same liquid at a different depth is to be determined Assumptions The variation of the density of the liquid with depth is negligible Analysis The gage pressure at two different depths of a liquid can be expressed as P1 Pghl and P 2 pghz Taking their ratio hl i pghz 11 2 V Pl pgh1 h 1 h V Solving for P2 and substituting gives h2 P 2 hl P1 9 m42 kPa126 kPa 3m Discussion Note that the gage pressure in a given uid is proportional to depth PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 116 154 The absolute pressure in water at a specified depth is given The local atmospheric pressure and the absolute pressure at the same depth in a different liquid are to be determined Assumptions The liquid and water are incompressible Properties The speci c gravity of the uid is given to be SG 085 We take the density of water to be 1000 kgm3 Then density of the liquid is obtained by multiplying its speci c gravity by the density of water 0 SGgtlt szO 0851000 kgm3 850 kgm3 Analysis a Knowing the absolute pressure the atmospheric pressure can be determined from P atm Patm P pgh 3 2 1 kPa 185 kPa 1000 kgm 981ms 9 m 2 h 1000 Nm 2 967 kPa v P b The absolute pressure at a depth of 5 m in the other liquid is P Patm pgh 3 2 1 kPa 967 kPa 850 kgm 981ms 9 m 2 1000 Nm 2 1718 kPa Discussion Note that at a given depth the pressure in the lighter uid is lower as expected 155E A submarine is cruising at a speci ed depth from the water surface The pressure exerted on the surface of the submarine by water is to be determined Assumptions The variation of the density of water with depth is negligible P atm Properties The speci c gravity of seawater is given to be SG 103 The density of water at 32 F is 624 lbmft3 Table A3E Analysis The density of the seawater is obtained by multiplying its speci c gravity by h the density of water p SGX szO 103624 lbmit 3 6427 lbmit3 Sea I1 F The pressure exerted on the surface of the submarine cruising 300 ft below the free surface of the sea is the absolute pressure at that location P Patm pgh 3 2 1 lbf 1 it 2 147 ps1a 6427 lbmft 322 fts 175 it 2 2 322 lbmfts 144 in 928 pSIa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 117 156 The mass of a woman is given The minimum imprint area per shoe needed to enable her to walk on the snow without sinking is to be determined Assumptions 1 The weight of the person is distributed uniformly on the imprint area of the shoes 2 One foot carries the entire weight of a person during walking and the shoe is sized for walking conditions rather than standing 3 The weight of the shoes is negligible Analysis The mass of the woman is given to be 70 kg For a pressure of 05 kPa on the snow the imprint area of one shoe must be AZKZE P P 2 70kg981ms 1N 2 lkPa 2137m2 05kPa lkgms lOOONm Discussion This is a very large area for a shoe and such shoes would be impractical to use Therefore some sinking of the snow should be allowed to have shoes of reasonable size 157E The vacuum pressure given in kPa unit is to be converted to various units Analysis Using the definition of vacuum pressure Pgage not applicable for pressures below atmospheric pressure Pabs 2 Pa Pvac 98 8018kPa Then using the conversion factors 2 Pabs 18 kPaM 18 kNm2 lkPa 2 Pabs 18 kPa 261Ibfin2 6895 kPa 6895 kPa 1mmHg j135mmHg 01333kPa Pabs 18 kPaamp 261psi Pabs 18 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 118 158 A mountain hiker records the barometric reading before and after a hiking trip The vertical distance climbed is to be determined 650 mbar Assumptions The variation of air density and the gravitational acceleration with altitude is negligible Properties The density of air is given to be p 120 kgm3 Analysis Taking an air column between the top and the bottom of the mountain and writing a force balance per unit base area we obtain Wair lA Pbottom Ptop 750 mbar pgh air Pbottom Ptop 1 1 120 kgm3 981 ms2 h N 2 bar 2 0750 0650 bar 1 kg ms 100000 Nm It yields h 850 m which is also the distance climbed 159 A barometer is used to measure the height of a building by recording reading at the bottom and at the top of the building The height of the building is to be determined Assumptions The variation of air density with altitude is negligible Properties The density of air is given to be p 118 kgm3 The density of mercury is 13600 kgm3 675 mmHg Analysis Atmospheric pressures at the top and at the bottom of the building are Ptop pghtop 1 1 kP 13600 kgm3981ms20675 m N a 1 kg ms2 1000 Nm2 h 9006 kPa Pbottom pgh bottom 1N 1 Id 2 13600 kgm3981ms20695 m 2 a 2 695 mmHg 1kg ms 1000 Nm 9272 kPa Taking an air column between the top and the bottom of the building and writing a force balance per unit base area we obtain Wair lA Pbottom Ptop pgh air Pbottom Ptop 1 18 kgm3981 ms2h1k 1 11 210010kpj 2 j 9272 9006 kPa g s N m It yields h 231 m which is also the height of the building PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 119 160 0 539 Problem 159 is reconsidered The entire EES solution is to be printed out including the numerical results with proper units Analysis The problem is solved using EES and the solution is given below Pbottom695 mmHg Ptop675 mmHg g981 msquot2 quotlocal acceleration of gravity at sea levelquot rho118 kgmquot3 DELTAPabsPbottomPtopCONVERTmmHg kPa quotkPaquot quotDelta P reading from the barometers converted from mmHg to kPaquot DELTAPh rhoghConvertPa kPa quotDelta P due to the air fluid column height h between the top and bottom of the buildingquot DELTAPabsDELTAPh SOLUTION DELTAPabs2666 kPa DELTAPh2666 kPa g981 msAZ h2303 m Pbottom695 mmHg Ptop675 mmHg rho118 kgmA3 161 The hydraulic lift in a car repair shop is to lift cars The uid gage pressure that must be maintained in the reservoir is to be determined Assumptions The weight of the piston of the lift is negligible W mg Analysis Pressure is force per unit area and thus the gage pressure required is simply the ratio of the weight of the car to the area of the lift Pm W P gage ng A 72D 4 2 1 2000 kgxg 821mm kN 2 278 kNm2 278 kPa P 7r030 m 4 1000 kg ms Discussion Note that the pressure level in the reservoir can be reduced by using a piston with a larger area PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 120 162 A gas contained in a vertical pistoncylinder device is pressurized by a spring and by the weight of the piston The pressure of the gas is to be determined Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield F spring PAzPathWF spring Patm Thus m F n pzpmw A P 2k 1 2 1 lkP 95 kPaJr3 g98 m4s 2 SON a 2 Wmg 35x10 m 1000 Nm 147kPa 163 39 Problem 162 is reconsidered The effect of the spring force in the range of 0 to 500 N on the pressure inside the cylinder is to be investigated The pressure against the spring force is to be plotted and results are to be discussed Analysis The problem is solved using EES and the solution is given below g981 msquot2 Patm 95 kPa mpiston32 kg Fspring150 N A35CONVERTcmquot2 mquot2 Wpistonmpistong FatmPatmACONVERTkPa NmA2 quotFrom the free body diagram of the piston the balancing vertical forces yieldquot Fgas FatmFspringWpiston PgasFgasACONVERTNm 2 kPa Fspring Pgas I l I I I I l I I I l l I I l l l I I l N kPa i 0 104 240 50 1183 220 100 1325 a 150 1468 39 l 200 1611 39339 20 250 175 4 D 39 quot 39 39 x 180 300 1897 3939 6 350 204 g 160 I 39 400 2183 ofquot 450 2325 140 0 500 2468 0 120 I l I I I I l I I I l l I I l l I I I l o 100 200 300 400 500 spring N PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 121 164 39 Both a gage and a manometer are attached to a gas to measure its pressure For a speci ed reading of gage pressure the difference between the uid levels of the two arms of the manometer is to be determined for mercury and water Properties The densities of water and mercury are given to be pwater 1000 kgm3 and be pHg 13600 kgm3 Analysis The gage pressure is related to the vertical distance h between the F 33 Wit two uid levels by P Pgagengh hz 08 a For mercury h 2 P gage pHg g 80 kP 1 2 1000 k 2 3 a 2 kNm gms 060m 13600 kgm 981ms 1kPa 1 kN b For water P kP 2 1 k 2 h gage 803 a 2 lkNm 000 gm s 2 816m PHzog 1000 kgm 981ms lkPa lkN PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 122 39239 Problem 164 is reconsidered The effect of the manometer uid density in the range of 800 to 13000 kgm3 on the differential uid height of the manometer is to be investigated Differential uid height against the density is to be plotted and the results are to be discussed 165 Analysis The problem is solved using EES and the solution is given below quotLet39s modify this problem to also calculate the absolute pressure in the tank by supplying the atmospheric pressure Use the relationship between the pressure gage reading and the manometer fluid column height quot Function fluiddensityFluid quotThis function is needed since ifthenelse logic can only be used in functions or procedures The underscore displays whatever follows as subscripts in the Formatted Equations Windowquot lf fluid39Mercury39 then fluiddensity13600 else fluiddensity1000 end Input from the diagram window If the diagram window is hidden then all of the input must come from the equations window Also note that brackets can also denote comments but these comments do not appear in the formatted equations window Fluid39Mercury39 Patm 101325 kPa DELTAP80 kPa quotNote how DELTAP is displayed on the Formatted Equations Windowquot g9807 msquot2 quotlocal acceleration of gravity at sea levelquot rhoFluiddensityFluid quotGet the fluid density either Hg or H20 from the functionquot quotTo plot fluid height against density place around the above equation Then set up the parametric table and solvequot DELTAP RHOgh1000 quotInstead of dividiing by 1000 PakPa we could have multiplied by the EES function CONVERTPakPaquot hmmhconvertm mm quotThe fluid height in mm is found using the builtin CONVERT functionquot Pabs Patm DELTAP quotTo make the graph hide the diagram window and remove the brackets from Fluid and from Patm Select New Parametric Table from the Tables menu Choose Pabs DELTAP and h to be in the table Choose Alter Values from the Tables menu Set values of h to range from 0 to 1 in steps of 02 Choose Solve Table or press F3 from the Calculate menu Choose New Plot Window from the Plot menu Choose to plot Pabs vs h and then choose Overlay Plot from the Plot menu and plot DELTAP on the same scalequot Manometer Fluid Height vs Manometer Fluid Density p hm 11000 kgm3 mm 800 10197 2156 3784 8800 3511 2323 4867 1676 I 6600 6222 1311 E 7578 1076 8933 9131 E 4400 10289 7928 E 11644 7005 2200 13000 6275 0 I I l I I I l I I 1 l 0 2000 4000 6000 8000 10000 12000 14000 p kgmA3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 123 166 The air pressure in a tank is measured by an oil manometer For a given oillevel difference between the two columns the absolute pressure in the tank is to be determined Properties The density of oil is given to be p 850 kgm3 Analysis The absolute pressure in the tank is determined from P Patm lkP 98 kPa 850 kgm3981ms2080 m az AIR 080 m lOOONm 21047 kPa 10tm 98 kPa 167 The air pressure in a duct is measured by a mercury manometer For a given mercurylevel difference between the two columns the absolute pressure in the duct is to be determined Properties The density of mercury is given to be p 13600 kgm3 T Analysis a The pressure in the duct is above atmospheric pressure since the uid column on the duct side is at a lower level AIR 5 b The absolute pressure in the duct is determined from P Patm pgh 1 N 1 kP 100 kPa 13600 kgm3981mSZ0015 111 2 a 2 1 kgms 1000 Nm 2102 kPa 168 The air pressure in a duct is measured by a mercury manometer For a given mercurylevel difference between the two columns the absolute pressure in the duct is to be determined Properties The density of mercury is given to be p 13600 kgm3 T Analysis a The pressure in the duct is above atmospheric pressure since the uid m column on the duct side is at a lower level AIR 5 b The absolute pressure in the duct is determined from P PZPatmpgh 100 kPa 13600 kgm3981ms20045 m 2 a 2 1 kg ms 1000 Nm 2 106 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 124 169E The pressure in a natural gas pipeline is measured by a double Utube manometer with one of the arms open to the atmosphere The absolute pressure in the pipeline is to be determined Assumptions 1 All the liquids are incompressible 2 The effect of air column on pressure is negligible 3 The pressure throughout the natural gas including the tube is r uniform since its density is low 1039 Properties We take the density of water to be pW 624 In Water lbmft3 The specific gravity of mercury is given to be 136 V and thus its density is pHg 136gtlt624 8486 lbmft3 hw Analysis Starting with the pressure at point 1 in the natural gas pipeline and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until Natural I gas we reach the free surface of oil where the oil tube is eXposed to the atmosphere and setting the result equal to Patm gives P1 pHgthg pwaterghwater Patm Solving for P1 P1 Patm pHg thg pwaterghl Mercury Substituting llbf 1 it 2 P 142 psia 322 lts 2 8486 lbm 3 612 it 624 lbm 3 2712 it 2 2 322 lbm s 144 in 181p51a Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same uid simplifies the analysis greatly Also it can be shown that the 15in high air column with a density of 0075 lbmft3 corresponds to a pressure difference of 000065 psi Therefore its effect on the pressure difference between the two pipes is negligible PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 125 170E The pressure in a natural gas pipeline is measured by a double Utube manometer with one of the arms open to the atmosphere The absolute pressure in the pipeline is to be determined Assumptions 1 All the liquids are incompressible 2 The pressure throughout the natural gas including the tube is A uniform since its density is low 011 Properties We take the density of water to be p W 624 lbmft3 The specific gravity of mercury is given to be 136 water and thus its density is pHg 136gtlt624 8486 lbmft3 The specific gravity of oil is given to be 069 and thus its A quot39 7V density is p011 069x624 431 lbmft3 Analysis Starting with the pressure at point 1 in the natural gas pipeline and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until Natural I gas we reach the free surface of oil where the oil tube is eXposed to the atmosphere and setting the result equal to Patm gives P1 pHg thg poilghoil pwater ghwater 2 Farm Solving for P1 Mercury P1 Pm pHg thg pwaterghl ponghon Substituting P1 142 psia 322 fts 2 8486 lbmft3 612 it 624 lbmft3 2712 it 1 lbf 1 2 4311bm 31512 2 2 322 lbmfts 144 in 177pSla Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same uid simpli es the analysis greatly PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 126 171E The systolic and diastolic pressures of a healthy person are given in mmHg These pressures are to be expressed in kPa psi and meter water column Assumptions Both mercury and water are incompressible substances Properties We take the densities of water and mercury to be 1000 kgm3 and 13600 kgm3 respectively Analysis Using the relation P pgh for gage pressure the high and low pressures are expressed as 1kPa 1000 Nmz 1N 10i p h i 13600k m3981ms2012 m 160 kPa hgh 8 hgh g gm82 1N 1kPa P w p h w 13600 kgm3981ms2008 m 107 kPa 1quot g 1quot 1kg ms2 1000 Nmz Noting that 1 psi 6895 kPa 1 psi 1 pSi P 160 Pa 232 Si and P 107 Pa 155 5 h1g1 6895kPa p I 6895kPa p For a given pressure the relation P pgh can be expressed for mercury and water as P owatelg1lzwalter and P pmury ghmemury Setting these two relations equal to each other and solving for water height gives A pmercury 339 P 2 I0 water ghwater I0 mercury ghmercury gt hwater hmercury water h Therefore 39Z39 10mm 13600 kgm3 II hwater high j hmercuryhigh W 012 m 1 63m V ommy 13600 kgm3 W hwater low pwater hmercurylow W m 1 09m Discussion Note that measuring blood pressure with a water monometer would involve differential uid heights higher than the person and thus it is impractical This problem shows why mercury is a suitable uid for blood pressure measurement devices PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 127 172 A vertical tube open to the atmosphere is connected to the vein in the arm of a person The height that the blood will rise in the tube is to be determined Assumptions 1 The density of blood is constant 2 The gage pressure of blood is 120 mmHg Properties The density of blood is given to be p 1050 kgm3 Analysis For a given gage pressure the relation P pgh can be expressed for mercury and blood as P pbloodghblood and P pmemury ghmemury Setting these two relations equal to each other we get P pbloodghblood pmemury ghmemury Solving for blood height and substituting gives pmercury h hblood mercury 012 m 2 155m pblood 1050 kgm3 Discussion Note that the blood can rise about one and a half meters in a tube connected to the vein This explains why IV tubes must be placed high to force a uid into the vein of a patient 173 A diver is moving at a speci ed depth from the water surface The pressure exerted on the surface of the diver by water is to be determined Assumptions The variation of the density of water with depth is negligible Properties The speci c gravity of seawater is given to be SG 103 We take the density of water to be 1000 kgm3 Analysis The density of the seawater is obtained by multiplying its speci c gravity by the density of water which is taken to be P atm 1000 kgm3 Sea 0 SGx pHZO 1031000 kgm3 1030 kgm3 h The pressure exerted on a diver at 45 m below the free surface 439 of the sea is the absolute pressure at that location P Patm pgh 1 kP 101kPa1030 kgm39807 ms245 m a 2 1000 Nm 556kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 128 174 Water is poured into the Utube from one arm and oil from the other arm The water column height in one arm and the ratio of the heights of the two uids in the other arm are given The height of each uid in that arm is to be determined Assumptions Both water and oil are incompressible substances Properties The density of oil is given to be p 790 kgm3 We take water on the density of water to be p 1000 kgm3 A A Analysis The height of water column in the left arm of the monometer is given to be hwl 070 m We let the height of water and oil in the right arm to be hwz and ha respectively Then ha 4hw2 Noting that hwl both arms are open to the atmosphere the pressure at the bottom of the hw2 Utube can be expressed as V Pbottom Patm pw ghwl and Pbottom Patm pw ghwz pa gha Setting them equal to each other and simplifying pw ghwl pw ghwz paella gt pw hwl pw hwz paha gt hwl hwz pa pwha Noting that ha 4hW2 the water and oil column heights in the second arm are determined to be 07mhw2 79010004hw2 gt kw 0168m 07m0168m7901000ha gt ha 0673m Discussion Note that the uid height in the arm that contains oil is higher This is eXpected since oil is lighter than water 175 A double uid manometer attached to an air pipe is considered The specific gravity of one uid is known and the specific gravity of the other uid is to be determined Assumptions 1 Densities of liquids are constant 2 The air pressure in the tank is uniform ie its variation with Alf elevation is negligible due to its low density and thus the P 76 kPa pressure at the airwater interface is the same as the indicated gage pressure A 40 cm Properties The specific gravity of one uid is given to be 1355 We take the standard density of water to be 1000 CHI Fl id 1 Analysis Starting with the pressure of air in the tank and G1 moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach the Fluid 2 SG2 free surface where the oil tube is eXposed to the atmosphere and setting the result equal to Patm give Pair Prgh1 Pzghz Patm gt Pair Patm SGzpwghz SGlpwghl Rearranging and solving for 8G2 h P P 2 SG223G11m21355022nn 736 100kPa2 1000 kg m2s 2134 h2 ngh2 040m 1000 kgm 981ms 040 m lkPam Discussion Note that the right uid column is higher than the left and this would imply above atmospheric pressure in the pipe for a single uid manometer PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 129 17 6 Fresh and seawater owing in parallel horizontal pipelines are connected to each other by a double Utube manometer The pressure difference between the two pipelines is to be determined Assumptions 1 All the liquids are incompressible 2 The effect of air column on pressure is negligible Properties The densities of seawater and mercury are given to be esea 1035 kgm3 and pHg 13600 kgm3 We take the density of water to be p W 1000 kgm3 Analysis Starting with the pressure in the fresh water pipe point 1 and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach the sea water pipe point 2 and setting the result equal to P2 gives P1 pwghw pHgthg pairghair pseaghsea P2 Rearranging and neglecting the effect of air column on pressure Mercu P1 P2 pwghw pHgthg pseaghsea gpthHg pwhw pseahsea ry Substituting P1 P2 981ms213600 kgm301 m 1000 kgm 06 m 1035 kgm 04 m 2 1000 kg ms 339 kNmz 339kPa Therefore the pressure in the fresh water pipe is 339 kPa higher than the pressure in the sea water pipe Discussion A 070m high air column with a density of 12 kgm3 corresponds to a pressure difference of 0008 kPa Therefore its effect on the pressure difference between the two pipes is negligible PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 130 177 Fresh and seawater owing in parallel horizontal pipelines are connected to each other by a double Utube manometer The pressure difference between the two pipelines is to be determined Assumptions All the liquids are incompressible Properties The densities of seawater and mercury are given to be esea 1035 kgm3 and pHg 13600 kgm3 We take the density of water to be p W 1000 kgm3 The speci c gravity of oil is given to be 072 and thus its density is 720 kgm3 Analysis Starting with the pressure in the fresh water pipe point 1 and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach the sea water pipe point 2 and setting the result equal to P2 gives Pl pw ghw pHgthg poilghoil pseaghsea P2 Rearranging P1 P2 pw ghw pHg thg poilghoil pseaghsea gpHg hHg poilhoil pw hw pseahsea Substituting 101 P2 981 ms2 136OO kgm3 01 m 720 kgm3 07 m 1000 kgm3 06 m 1035 kgm3 04 m 2 1000 kg ms 834 kNmz 834kPa Therefore the pressure in the fresh water pipe is 834 kPa higher than the pressure in the sea water pipe 17 8 The pressure indicated by a manometer is to be determined Properties The specific weights of uid A and uid B are AT EEEEEEIC given to be 10 kNm3 and 8 kNm3 respectively Pl HE Analysis The absolute pressure P1 is determined from P1 Patm pghA pghB 13 L1 39I i 15 13111 hB Patm7AhA7BhB 758 mm Hg wj hA CE 1 Fluid E1 mm Hg f e kan ll 10 kNm3 005 m 8 kNm3 0 15 m 31 em 1027kPa Flujd Note that 1 kPa 1 kNmz m WT r PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 131 179 The pressure indicated by a manometer is to be determined Properties The specific weights of uid A and uid B are given to be 100 kNm3 and 8 kNm3 respectively P 39 39 in Analysis The absolute pressure P1 is determined from P1 Patm Patm 7AhA 7BhB 15 cm hB 90 kPa 100 kNm3 005 m 8 kNm3 0 15 m t 962kPa Fluid Er Note that 1 kPa 1 kNmz Iquot t Wm Rial 3E cm Fluid Po 180 The pressure indicated by a manometer is to be determined ATMUSFHEMC Properties The specific weights of uid A and uid B are PRESSWE given to be 10 kNm3 and 20 kNm3 respectively P Analysis The absolute pressure P1 is determined from P1 Patm pghA pghB 1 cm h Patm7AhA73hB B 01333kPa 1mm Hg Fluid E xquot 20 kNm3 720 mm Hg 10 kNm3 005 m 20 kNm3 0 15 m m m 995kPa quot Fluid 35 Note that 1 kPa 1 kNmz 1t WIDE PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 132 181 The uid levels in a multi uid Utube manometer change as a result of a pressure drop in the trapped air space For a given pressure drop and brine level change the area ratio is to be determined Assumptions 1 All the liquids are incompressible 2 Pressure in the brine pipe remains constant 3 The variation of pressure in the trapped air space is A negligible Air Properties The specific gravities are given to be 1356 for mercury and 11 for brine We take the standard density of water to be pw 1000 kgm3 Area A1 Analysis It is clear from the problem statement and the gure that the brine pressure is much higher than the air SG11 pressure and when the air pressure drops by 07 kPa the quotquotquotquotquotquotquotquotquot 39 39 pressure difference between the brine and the air space increases also by the same amount Mercury Ahb 5 mm SG1356 4L Starting with the air pressure point A and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach the brine Area A2 pipe point B and setting the result equal to PB before and after the pressure change of air give Bef07 33 PAl Pw ghw pHg thg1 pbrghbr1 2 PB After P A2 nghw pHg thg2 pbrghbr2 2 PB Subtracting P P M SGHgAhHg SGbrAhbr 0 1 PA2 PA1 pHggAhHg pbrgAhbr 0 gt ng where AhHg and Ath are the changes in the differential mercury and brine column heights respectively due to the drop in air pressure Both of these are positive quantities since as the mercurybrine interface drops the differential uid heights for both mercury and brine increase Noting also that the volume of mercury is constant we have AlAhHgJe AzAhHg ght and PA2 PA1 07 kPa 700 Nm2 700 kgm s2 Ahbl 0005 m AhHg MHgJight MHgJeft Ahbr AhbrAz A1 Ahbr 1 142 A1 Substituting 700 kgmsz 1000 kgm3981 ms2 1356x00051 A2A1 1 1x0005 m It gives A2A1 01 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 133 Solving Engineering Problems and EES 182C Despite the convenience and capability the engineering software packages offer they are still just tools and they will not replace the traditional engineering courses They will simply cause a shift in emphasis in the course material from mathematics to physics They are of great value in engineering practice however as engineers today rely on software packages for solving large and compleX problems in a short time and perform optimization studies efficiently quot Determine a positive real root of the following equation using EES 2x3 10x05 3x 3 Solution by EES Software Copy the following line and paste on a blank EES screen to verify solution 2Xquot310Xquot053X 3 Answer X 2063 using an initial guess of x2 r 39 Solve the following system of 2 equations with 2 unknowns using EES X3 y2 775 3xy y 35 Solution by EES Software Copy the following lines and paste on a blank EES screen to verify solution Xquot3yquot2775 3xyy35 Answer X2 y05 I Solve the following system of 3 equations with 3 unknowns using EES x2y z 1 x 3y 395xz 2 xy z2 Solution by EES Software Copy the following lines and paste on a blank EES screen to verify solution xquot2yz1 x3yquot05xz2 xy z2 Answer X1 y1 z0 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 1 34 Solve the following system of 3 equations with 3 unknowns using EES 2x y z 7 3x2 2y z 3 xy 22 4 Solution by EES Software Copy the following lines and paste on a blank EES screen to verify solution 2xyz7 3xquot22yz3 xy2z4 Answer X1609 y09872 22794 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 135 187E 39 Specific heat of water is to be expressed at various units using unit conversion capability of EES Analysis The problem is solved using EES and the solution is given below EQUATION WINDOW quotGIVENquot Cp418 kJkgC quotANALYSISquot Cp1CpConvertkJkgC kJkgK Cp2CpConvertkJkgC BtuIbmF Cp3CpConvertkJkgC BtuIbmR Cp4CpConvertkJkgC kCaIkgC FORMATTED EOUATIONS WINDOW IGIVEN Cp 418 kJkgC ANALYSIS C C kJkg K p 39 p kJkg C c c 0 238846 Btwmm F p2 p 39 kJkg C c c 0 238846 Btwmm R p3 39 p 39 kJkg C c 0 0238846 W p394 39 p 39 kJkg C SOLUTION Cp418 kJkgC Cp1418 kJkgK Cp209984 BtuIbmF Cp309984 BtuIbmR Cp409984 kCaIkgC PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 136 Review Problems 188 The gravitational acceleration changes with altitude Accounting for this variation the weights of a body at different locations are to be determined Analysis The weight of an 80kg man at various locations is obtained by substituting the altitude 2 values in m into the relation 1N W mg 80kg9807 332gtlt106 zms2 1kg ms2 Sea level 2 0 m w 80x9807332X10396gtlt0 80x9807 7846 N Denver 2 1610 m w 80x9807332X10396gtlt1610 80x9802 7842 N Mt Ev z 8848 m w 80x9807332X10396gtlt8848 80x9778 7822 N 189E A man is considering buying a 1202 steak for 550 or a 300g steak for 520 The steak that is a better buy is to be determined Assumptions The steaks are of identical quality Analysis To make a comparison possible we need to express the cost of each steak on a common basis Let us choose 1 kg as the basis for comparison Using proper conversion factors the unit cost of each steak is determined to be 12 ounce steak Unit Cost 550 16oz llbm 1617kg 12 oz llbm 045359 kg 300 gram steak Unit Cost M w 1733 rg 300 g 1 kg Therefore the steak at the traditional market is a better buy PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 137 190E The mass of a substance is given Its weight is to be determined in various units Analysis Applying Newton39s second law the weight is determined in various units to be W mg 2 1 kg981 ms2 IN 2 J 981N 1kgms W mg 2 1 kg981 ms2i2 000981kN 1000 kgms W mg 1kg981msz1kgmsz Wmg1kg981ms2 1N 1kgf1kgf 1kgm32 981N 22 1 Wmg 1 kg 32211s271lbmfts2 1kg 22 1 11 f Wmg1 kg 3221032 bz 221bf 1kg 3221bmfts PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 138 191 A hydraulic lift is used to lift a weight The diameter of the piston on which the weight to be placed is to be determined Weight Assumptions 1 The cylinders of the lift are vertical 2 I1 2500 kg There are no leaks 3 Atmospheric pressure act on both sides and thus it can be disregarded 25 kg I I Analysis Noting that pressure is force per unit area the F2 1 pressure on the smaller piston is determined from D P quot118 1 A 2 I x 10 m 1 7ZD1 4 K C D2 a 7r010m24 1000 kg ms2 2 3123 kNm2 3123 kPa From Pascal s principle the pressure on the greater piston is equal to that in the smaller piston Then the needed diameter is determined from Z 25 kg981ms2 lkN F 2 gt31231ltNm 2 k 1 2 Plzpr 2 2500 g8ms 1kN Dzzwm A2 702 4 702 4 1000 kgmsz Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal s principle 192E The efficiency of a refrigerator increases by 3 per C rise in the minimum temperature This increase is to be expressed per F K and R rise in the minimum temperature Analysis The magnitudes of 1 K and 1 C are identical so are the magnitudes of 1 R and 1 F Also a change of 1 K or 1 C in temperature corresponds to a change of 18 R or 18 F Therefore the increase in efficiency is a 3 for each K rise in temperature and b c 3 18 167 for each R or F rise in temperature 193E Hyperthermia of 5 C is considered fatal This fatal level temperature change of body temperature is to be expressed in F K and R Analysis The magnitudes of 1 K and 1 C are identical so are the magnitudes of 1 R and 1 F Also a change of 1 K or 1 C in temperature corresponds to a change of 18 R or 18 F Therefore the fatal level of hypothermia is a 5K b 5x189 F c 5x189R PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 139 194E A house is losing heat at a rate of 1800 kJh per C temperature difference between the indoor and the outdoor temperatures The rate of heat loss is to be expressed per F K and R of temperature difference between the indoor and the outdoor temperatures Analysis The magnitudes of 1 K and 1 C are identical so are the magnitudes of 1 R and 1 F Also a change of 1 K or 1 C in temperature corresponds to a change of 18 R or 18 F Therefore the rate of heat loss from the house is a 1800 kJh per K difference in temperature and b c 180018 1000 kJh per R or F rise in temperature 195 The average temperature of the atmosphere is expressed as Tltm 28815 65z where z is altitude in km The temperature outside an airplane cruising at 12000 m is to be determined Analysis Using the relation given the average temperature of the atmosphere at an altitude of 12000 m is determined to be Tatm 28815 652 28815 65gtlt12 21015 K 63 C Discussion This is the average temperature The actual temperature at different times can be different 196 A new Smith absolute temperature scale is proposed and a value of 1000 S is assigned to the boiling point of water The ice point on this scale and its relation to the Kelvin scale are to be determined Analysis All linear absolute temperature scales read zero at absolute zero pressure and are K A S constant multiples of each other For example TR 18 TK That is multiplying a temperature value in K by 18 will give the same temperature in R 37315 1000 The proposed temperature scale is an acceptable absolute temperature scale since it differs from the other absolute temperature scales by a constant only The boiling temperature of water in the Kelvin and the Smith scales are 31515 K and 1000 K respectively Therefore these two temperature scales are related to each other by 1000 37315 TS TK 26799TK The ice point of water on the Smith scale is mm 26799 mom 26799x27315 7320 s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 140 197E An expression for the equivalent wind chill temperature is given in English units It is to be converted to SI units Analysis The required conversion relations are 1 mph 1609 kmh and T F 18T C 32 The first thought that comes to mind is to replace T F in the equation by its equivalent 18T C 32 and V in mph by 1609 kmh which is the regular way of converting units However the equation we have is not a regular dimensionally homogeneous equation and thus the regular rules do not apply The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph Therefore if V is given in kmh we should divide it by 1609 to convert it to the desired unit of mph That 1s Tequiv F 914 914 Tambiarlt H 0475 00203V 1609 03044 V 1609 0139 T equiv o F 914 914 Tambiant P 0475 00126V 0240JV where V is in kmh Now the problem reduces to converting a temperature in F to a temperature in C using the proper convection relation 18Tequi C 32 914 914 18Tmbient C 32 0475 00126V 0240JV which simpli es to Tequiv C 330 330 Tmbient 0475 00126V 024M where the ambient air temperature is in C 198E quot Problem 197E is reconsidered The equivalent windchill temperatures in F as a function of wind velocity in the range of 4 mph to 40 mph for the ambient temperatures of 20 40 and 60 F are to be plotted and the results are to be discussed Analysis The problem is solved using EES and the solution is given below Tambient20 IIV20H Tequiv914914Tambient0475 00203V 0304sqrtV V Tequiv mph F 4 5994 50 8 5459 40 12 5107 16 485 30 20 4654 E 24 4502 20 28 4382 32 4288 g 10 36 4216 I O 40 4161 The table is for Tmbiem60 F 10 20 0 5 10 15 20 25 30 35 40 V mph PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 141 199 A vertical pistoncylinder device contains a gas Some weights are to be placed on the piston to increase the gas pressure The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined Assumptions Friction between the piston and the cylinder is negligible Analysis The gas pressure in the pistoncylinder device initially depends on the local WEIGTHS atmospheric pressure and the weight of the piston Balancing the vertical forces yield I m 5 k 81 m 2 1 PM P M2100kPa gxg 2 S W 2 9566kNm2 957kPa A 7r012m 4 1000 kgms The force balance when the weights are placed is used to determine the mass of the weights GAS mpiston mweightsg P Patm A 5 kg mweights9 81 msz 1kN 7r012 m2 4 1000 kg ms2 200 kPa 9566kPa gt mweights 115 kg A large mass is needed to double the pressure 1100 One section of the duct of an airconditioning system is laid underwater The upward force the water will exert on the duct is to be determined Assumptions 1 The diameter given is the outer diameter of the duct or the thickness of the duct material is negligible 2 The weight of the duct and the air in is negligible Properties The density of air is given to be p 130 kgm3 We take the A density of water to be 1000 kgm3 P D 15 cm 6 v Analysis Noting that the weight of the duct and the air in it is negligible L 35 m the net upward force acting on the duct is the buoyancy force exerted by FB water The volume of the underground section of the duct is V AL 702 4L W015 m2435 m 06185 m3 Then the buoyancy force becomes 1kN FB ng 1000 kgm3981ms206185 m3 2 1000 kg ms j607kN Discussion The upward force exerted by water on the duct is 607 kN which is equivalent to the weight of a mass of 619 kg Therefore this force must be treated seriously PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 142 1101E The average body temperature of a person rises by about 2 C during strenuous exercise This increase in temperature is to be expressed in F K and R Analysis The magnitudes of 1 K and 1 C are identical so are the magnitudes of 1 R and 1 F Also a change of 1 K or 1 C in temperature corresponds to a change of 18 R or 18 F Therefore the rise in the body temperature during strenuous exercise is a 2 K b 2x18 36 F c 2x18 36 R 1102 A helium balloon tied to the ground carries 2 people The acceleration of the balloon when it is rst released is to be determined Assumptions The weight of the cage and the ropes of the balloon is negligible Properties The density of air is given to be p 116 kgm3 The density of helium gas is 17th of this Analysis The buoyancy force acting on the balloon is Vballoon 4751373 2 m3 Z m3 FB pairgvballoon D121m 116 kgm3981ms29048 m3 2 210296 N we 7r39air 1 kg ms The total mass is mHe pHeV 1397 16kgm3 9048 m3 1499 kg mtotal mHe mpeople 21499 2X 85 3199 kg The total weight is J 1 N I W mtota1g 3199kg981ms2 2 23138 N J 1 kg ms Thus the net force acting on the balloon is m 170 kg Fnet 2 FB W 210296 3138 7157 N Then the acceleration becomes F 7157 N 1k m 2 a net g S 224 ms2 mtotal 3199 kg 1N PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 143 1103 339 Problem 1102 is reconsidered The effect of the number of people carried in the balloon on acceleration is to be investigated Acceleration is to be plotted against the number of people and the results are to be discussed Analysis The problem is solved using EES and the solution is given below quotGivenquot D12 m Nperson2 mperson85 kg rhoair1 16 kgmquot3 rhoHerhoair7 quotAnalysisquot g981 msA2 VbaonpiDquot36 FBrhoairgVbaon mHerhoHeVbaon mpeopleNpersonmperson mtotalmHempeople Wmtotalg FnetFBW a Fnetmtota Nperson a 39 m2 1 34 3o 2 2236 3 1561 25 4 112 0 5 8096 6 579 1 2 7 401 g 39 0 8 2595 m 15 9 1443 10 04865 10 ID 0 5 0 0 quot0 1 2 3 4 5 6 7 8 9 10 Nperson PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 144 1104 A balloon is lled with helium gas The maximum amount of load the balloon can carry is to be determined Assumptions The weight of the cage and the ropes of the balloon is negligible Properties The density of air is given to be p 116 kgm3 The density of helium gas is 17th of this Analysis The buoyancy force acting on the balloon is vballoon 4m33 47t6 m3 3 9048 m3 FB pairgvballoon 116 kgm3981ms29048 m3 2 j 10296 N 1kgms The mass of helium is mHe pHeV 1397 16kgm3 9048 m3 1499 kg In the limiting case the net force acting on the balloon will be zero That is the buoyancy force and the weight will balance each other W mg 2 FB F 10296N mtotal 2 32 221050 kg 8 981ms Thus mpeople mtotal mHe 1050 1499 900 kg HELIUM D 12 m 939 39He 71bit 1105 A 6m high cylindrical container is lled with equal volumes of water and oil The pressure difference between the top and the bottom of the container is to be determined Properties The density of water is given to be p 1000 kgm3 The specific gravity of oil is given to be 085 Analysis The density of the oil is obtained by multiplying its speci c gravity by the density of water 0 SGgtlt pHZO 0 851000 kgm3 850 kgm3 The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two uids APtotal AP oil AP water pgh oil pgh water 850 kgm3981ms23 m 1000 kgm3981ms23 m 544 kPa Oil SG 085 1 kPa 1000 Nm2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 145 1106 The pressure of a gas contained in a vertical pistoncylinder device is measured to be 180 kPa The mass of the piston is to be determined Assumptions There is no friction between the piston and the cylinder Pm Analysis Drawing the free body diagram of the piston and balancing the l l 1 vertical forces yield I A A A I W 2 PA Palth m8 P Patm A P m981ms2 180 100 kPa25 gtltlO4m2 J V W2 mg It yields m 204 kg 1107 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock The mass of the petcock is to be determined Assumptions There is no blockage of the pressure release valve P atm Analysis Atmospheric pressure is acting on all surfaces of the petcock which balances itself out Therefore it can be disregarded in calculations if we use the gage pressure as the cooker pressure A force balance on the petcock 2Fy 0 yields W PgageA W mg m PgageA 100 kPa4 x106m2 1000 kgmsz g 981 ms2 lkPa 00408kg 1108 A glass tube open to the atmosphere is attached to a water pipe and the pressure at the bottom of the tube is measured It is to be determined how high the water will rise in the tube Properties The density of water is given to be p 1000 kgm3 Analysis The pressure at the bottom of the tube can be expressed as P Patm pghtube Solving for h atm pg 110 99 kPa 1kg ms2 1000 Nm2 1000 kgm3981ms2 1 N lkPa 2112 m P P h PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 146 1109E Equal volumes of water and oil are poured into a Utube from different arms and the oil side is pressurized until the contact surface of the two uids moves to the bottom and the liquid levels in both arms become the same The excess pressure applied on the oil side is to be determined Assumptions 1 Both water and oil are incompressible substances 2 Oil does not miX with water 3 The crosssectional area of the Utube is constant 39 4511 Properties The density of oil is given to be p011 493 lbmft3 We take the If density of water to be pw 624 lbmft3 Analysis Noting that the pressure of both the water and the oil is the same at the contact surface the pressure at this surface can be expressed as P I 11113quot 13 METquotTr E39iIII m PcontathPblowpagha Patmpwghw V Noting that ha hw and rearranging Pgageblow Pblow Patm pw p0ilgh 3 2 1 lbf 1 it 2 624493 lbmit 322 s 3012 it 2 2 322 lbm s 144 in 0227 pSl Discussion When the person stops blowing the oil will rise and some water will ow into the right arm It can be shown that when the curvature effects of the tube are disregarded the differential height of water will be 237 in to balance 30in of oil 1110 A barometer is used to measure the altitude of a plane relative to the ground The barometric readings at the ground and in the plane are given The altitude of the plane is to be determined Assumptions The variation of air density with altitude is negligible Properties The densities of air and mercury are given to be p 120 kgm3 and p 13600 kgm3 Analysis Atmospheric pressures at the location of the plane and the ground level are Pplane pghplane 3 2 1 N 1 kPa 13600 kgm 981ms 0690 m 2 2 1 kgms 1000 Nm 2 9206 kPa Pground pghground 3 2 1 N 1 kPa 13600 kgm 981ms 0753 m 1kgms2 1000 Nm2 2 10046 kPa Taking an air column between the airplane and the ground and writing a force balance per unit base area we obtain h Wair A Pground Pplane pghair Pground Pplane 3 2 1 N 1 kPa 120 kgm 981ms h 2 2 10046 9206 kPa 0 Sea 1 kgms 1000 Nm It yields h 714 m which is also the altitude of the airplane PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 147 1111E A water pipe is connected to a doubleU manometer whose free arm is open to the atmosphere The absolute pressure at the center of the pipe is to be determined Assumptions 1 All the liquids are incompressible 2 The solubility of the liquids in each other is negligible 1311513 LED Properties The speci c gravities of mercury and oil are given to be 136 and 080 respectively We take the density of water to be pw 6241bmft3 131i SE LSD a Analysis Starting with the pressure at the center of the water pipe and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach the free El in surface of oil where the oil tube is eXposed to the atmosphere and 25 in setting the result equal to Patm gives O p n D P water pipe p water ghwater poilghoil pHg thg poilghoil P am hi f l 1 SE 135 Solving for Pwater pipe P water pipe 2 P am p water 8 hwater Soilhoil SHg hHg Soilhoil Substituting 142psia 624lbmit 3 322 s 2 2012 it 086012 it 1362512 it 11 f 1 2 O83012 x b 2 2 3221bmits 144 in 264 pSla P water pipe Therefore the absolute pressure in the water pipe is 264 psia Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same uid simpli es the analysis greatly PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 148 1112 A gasoline line is connected to a pressure gage through a doubleU manometer For a given reading of the pressure gage the gage pressure of the gasoline line is to be determined Assumptions 1 All the liquids are incompressible 2 The effect of air Pgage 370 kPa column on pressure is negligible Properties The speci c gravities of oil mercury and gasoline are Oil given to be 079 136 and 070 respectively We take the density of water to be pw 1000 kgm3 Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding as we go down or subtracting as Air we go up the pgh terms until we reach the gasoline pipe and setting the result equal to Pgasoline gives Pgage pw ghw poilghoil pHg thg pgasolineghgasoline Pgasoline Rearranging Water Pgasoline Pgage pw ghw SCloilhoil SCng hHg SClgasolinehgasoline V V Mercury Substituting Pgasolinez 370 kPa 1000 kgm3981ms2045 m 07905 m 13601 m 070022 m X lkN lkPa 1000 kg ms2 lkNmz 3546 kPa Therefore the pressure in the gasoline pipe is 154 kPa lower than the pressure reading of the pressure gage Discussion Note that sometimes the use of speci c gravity offers great convenience in the solution of problems that involve several uids PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 149 1113 A gasoline line is connected to a pressure gage through a doubleU manometer For a given reading of the pressure gage the gage pressure of the gasoline line is to be determined A 10gage 180 kPa Oil Assumptions 1 All the liquids are incompressible 2 The effect of air column on pressure is negligible 1 Properties The specific gravities of oil mercury and gasoline are 45 given to be 079 136 and 070 respectively We take the density of water to be pW 1000 kgm3 Air cm Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding as we go down or subtracting as we go up the pgh terms until we reach the 50 cm gasoline pipe and setting the result equal to Pgasoline gives Water 139 V V M Rearranging Pgasoline Pgage pw ghw SCoilhoil SCHg hHg SCgasolinehgasoline Substituting 2 180 kPa 1000 kgm3 9 807 ms2 045 m 07905 m 13601 m 070022 m X lkN lkPa 1000kgmsz lkNmz 1646kPa P gasoline Therefore the pressure in the gasoline pipe is 154 kPa lower than the pressure reading of the pressure gage Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several uids 1114 The average atmospheric pressure is given as Patnfl 1013251 002256z539256 where z is the altitude in km The atmospheric pressures at various locations are to be determined Analysis The atmospheric pressures at various locations are obtained by substituting the altitude 2 values in km into the relation 11 1013251 002256z5256 Atlanta z 0306 km Pam 1013251 002256gtlt03065256 977 kPa Denver z 1610 km Pam 1013251 002256gtlt16105256 834 kPa M City z 2309 km Pam 1013251 002256gtlt23095256 765 kPa Mt Ev z 8848 km Pam 1013251 002256gtlt88485256 314 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 150 1115 The temperature of the atmosphere varies with altitude z as T T0 z while the gravitational acceleration varies by gz 2 go 1 z 63703202 Relations for the variation of pressure in atmosphere are to be obtained a by ignoring and b by considering the variation of g with altitude Assumptions The air in the troposphere behaves as an ideal gas Analysis a Pressure change across a differential uid layer of thickness dz in the vertical z direction is dP 0gdz P Then RT RT0 Bz From the ideal gas relation the air density can be eXpressed as 0 gdz RTo Z Separating variables and integrating from z 0 where P P0 to z z where P P Pd PJIZ gdz PO P 0 RT0 Bz Performing the integrations T 1n 3 2 i 1n 0 6Z P0 R T0 Rearranging the desired relation for atmospheric pressure for the case of constant g becomes 8 0 b When the variation of g with altitude is considered the procedure remains the same but the eXpressions become more complicated P g0 dz RT0 8z 1 z 63703202 Separating variables and integrating from z 0 where P P0 to z z where P P P d P Jz godz P0 P o RT0 330 z63703202 Performing the integrations go 1 1 1 lkz lnP IP n Po RB1kT0B1kz 1kT0B2 TO z 0 where R 287 JkgK 287 m2s2K is the gas constant of air After some manipulations we obtain go 1 1 1n lkz RBkT0 llkz 1kT0B 1 BzT0 P P0 exp where T0 28815 K 8 00065 Km g0 9807 ms2 k 16370320 ml and z is the elevation in m Discussion When performing the integration in part b the following eXpression from integral tables is used together with a transformation of variable x T0 z xal9x2 aal9x a2 x J dx 1 11 abx Also for z 11000 m for example the relations in a and 9 give 2262 and 2269 kPa respectively PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 151 1116 The variation of pressure with density in a thick gas layer is given A relation is to be obtained for pressure as a function of elevation z Assumptions The property relation P Cp is valid over the entire region considered Analysis The pressure change across a differential uid layer of thickness dz in the vertical z direction is given as dP 0gdz Also the relation P Cp can be expressed as C P 0 P0 p3 and thus ppoltPPogt quot Substituting Separating variables and integrating from z 0 where P P0 2 C08 to z z where P P P 1n z I PPO dpz pogj39dz P0 0 Performing the integrations P n 1n lnl P0 n P0 0 Solving for P nn l p p 1 0 n P0 which is the desired relation Discussion The nal result could be expressed in various forms The form given is very convenient for calculations as it facilitates unit cancellations and reduces the chance of error PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 152 1117 The ow of air through a wind turbine is considered Based on unit considerations a proportionality relation is to be obtained for the mass ow rate of air through the blades Assumptions Wind approaches the turbine blades with a uniform velocity Analysis The mass ow rate depends on the air density average wind velocity and the crosssectional area which depends on hose diameter Also the unit of mass ow rate rh is kgs Therefore the independent quantities should be arranged such that we end up with the proper unit Putting the given information into perspective we have rh kgs is a function of p kgm3 D m and V ms It is obvious that the only way to end up with the unit kgs for mass ow rate is to multiply the quantities p and V with the square of D Therefore the desired proportionality relation is m is proportional to pDZV or m CpDZV where the constant of proportionality is C 7r4 so that m 07ZD2 4V Discussion Note that the dimensionless constants of proportionality cannot be determined with this approach 1118 A relation for the air drag exerted on a car is to be obtained in terms of on the drag coefficient the air density the car velocity and the frontal area of the car Analysis The drag force depends on a dimensionless drag coef cient the air density the car velocity and the frontal area Also the unit of force F is newton N which is equivalent to kgmsz Therefore the independent quantities should be arranged such that we end up with the unit kgms2 for the drag force Putting the given information into perspective we have FD kgmSZ e CDgAfmmm21 prkgm and Vrms It is obvious that the only way to end up with the unit kgms2 for drag force is to multiply mass with the square of the velocity and the fontal area with the drag coef cient serving as the constant of proportionality Therefore the desired relation is FD CDrag pA ont l2 Discussion Note that this approach is not sensitive to dimensionless quantities and thus a strong reasoning is required PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 153 Fundamentals of Engineering FE Exam Problems 1119 An apple loses 45 kJ of heat as it cools per C drop in its temperature The amount of heat loss from the apple per F drop in its temperature is a 125 k b 250 k c 50 k d 81 k e 41 k Answer b 250 k Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values QperC45 quotkJquot QperFQperC18 quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1QQperC18 quotmultiplying instead of dividingquot W2QQperC quotsetting them equal to each otherquot 1120 Consider a sh swimming 5 m below the free surface of water The increase in the pressure exerted on the sh when it dives to a depth of 25 m below the free surface is a 196 Pa b 5400 Pa c 30000 Pa d 196000 Pa e 294000 Pa Answer d 196000 Pa Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho1000 quotkgm3quot g981 quotm52quot z1 5 quotmquot 2225 quotmquot DELTAPrhog22z1 quotPaquot quotSome Wrong Solutions with Common Mistakesquot W1Prhog22z11 000 quotdividing by 1000quot W2Prhogz1 22 quotadding depts instead of subtractingquot W3Prhoz1 22 quotnot using gquot W4Prhog022 quotignoring 21 quot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 154 1121 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 960 and 980 kPa If the density of air is 10 kgm3 the height of the building is a 17 m b 20 m c 170 m d 204 m e 252 m Answer d 204 m Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho10 quotkgm3quot g981 quotmsZquot P1 96 quotkPaquot P298 quotkPaquot DELTAPP2 P1 quotkPaquot DELTAPrhogh1000 quotkPaquot quotSome Wrong Solutions with Common Mistakesquot DELTAPrhoW1h1000 quotnot using gquot DELTAPgW2h1000 quotnot using rhoquot P2rhogW3h1000 quotignoring P1quot P1rhogW4h1000 quotignoring P2quot 1122 Consider a 2m deep swimming pool The pressure difference between the top and bottom of the pool is a 120 kPa b 196 kPa c 381 kPa d 508 kPa e 200 kPa Answer b 196 kPa Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho1000 quotkgmquot3quot g981 quotmsZquot z1 0 quotmquot 222 quotmquot DELTAPrhog22z11000 quotkPaquot quotSome Wrong Solutions with Common Mistakesquot W1Prhoz1221000 quotnot using gquot W2Prhog22z12000 quottaking half of zquot W3Prhog22 z1 quotnot dividing by 1000quot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 155 1123 During a heating process the temperature of an object rises by 10 C This temperature rise is equivalent to a temperature rise of a 10 F b 42 F c 18 K d 18 R e 283 K Answer d 18 R Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TinC10 quotCquot TinRTinC1 8 quotRquot quotSome Wrong Solutions with Common Mistakesquot W1TinFTinC quotF setting C and F equal to each otherquot W2TinFTinC1832 quotF converting to F quot W3TinK18TinC quotK wrong conversion from C to Kquot W4TinKTinC273 quotK converting to Kquot 1124 At sea level the weight of 1 kg mass in SI units is 981 N The weight of 1 lbm mass in English units is a 1 lbf b 981 lbf c 322 lbf d 01 lbf e 0031 lbf Answer a 1 lbf Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m1 quotlbmquot g322 quotft2quot Wmg322 quotlbfquot quotSome Wrong Solutions with Common Mistakesquot gSI981 quotmsZquot W1W mgSI quotUsing wrong conversionquot W2W mg quotUsing wrong conversionquot W3W mgSI quotUsing wrong conversionquot W4W mg quotUsing wrong conversionquot 1125 1126 Design and Essay Problems PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 1 Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Cengel Michael A Boles McGrawHill 2015 Chapter 2 ENERGY ENERGY TRANSFER AND GENERAL ENERGY ANALYSIS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 2 Forms of Energy 21C The sum of all forms of the energy a system possesses is called total energy In the absence of magnetic electrical and surface tension effects the total energy of a system consists of the kinetic potential and internal energies 22C The internal energy of a system is made up of sensible latent chemical and nuclear energies The sensible internal energy is due to translational rotational and vibrational effects 23C Thermal energy is the sensible and latent forms of internal energy and it is referred to as heat in daily life 24C The mechanical energy is the form of energy that can be converted to mechanical work completely and directly by a mechanical device such as a propeller It differs from thermal energy in that thermal energy cannot be converted to work directly and completely The forms of mechanical energy of a uid stream are kinetic potential and ow energies 25C Hydrogen is also a fuel since it can be burned but it is not an energy source since there are no hydrogen reserves in the world Hydrogen can be obtained from water by using another energy source such as solar or nuclear energy and then the hydrogen obtained can be used as a fuel to power cars or generators Therefore it is more proper to view hydrogen is an energy carrier than an energy source 26C In electric heaters electrical energy is converted to sensible internal energy 27 C The forms of energy involved are electrical energy and sensible internal energy Electrical energy is converted to sensible internal energy which is transferred to the water as heat 28E The total kinetic energy of an object is given is to be determined Analysis The total kinetic energy of the object is given by V2 50 rrs2 lBtulbm KEm lOlbm 2 2 0499Btu20503tu 2 2 25037lt s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 3 29E The total potential energy of an object is to be determined Analysis Substituting the given data into the potential energy expression gives 1 Btulbm pE mgz 2001bm322 s 2 10 it W 25037 it s 2 257Btu 210 A person with his suitcase goes up to the 10th oor in an elevator The part of the energy of the elevator stored in the suitcase is to be determined Assumptions 1 The vibrational effects in the elevator are negligible Analysis The energy stored in the suitcase is stored in the form of potential energy which is mgz Therefore 1 kJkg AE APE m Az 30 k 981ms2 35 m g g 1000m2 2 suitcase S j103kJ Therefore the suitcase on 10th oor has 103 kJ more energy compared to an identical suitcase on the lobby level Discussion Noting that 1 kWh 3600 k the energy transferred to the suitcase is 1033600 00029 kWh which is very small 211 A hydraulic turbinegenerator is to generate electricity from the water of a large reservoir The power generation potential is to be determined Assumptions 1 The elevation of the reservoir remains constant 2 The mechanical energy of water at the turbine eXit is negligible Analysis The total mechanical energy water in a reservoir possesses is equivalent to the potential energy of water at the free surface and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per unit mass and rngz for a given mass ow rate lkJk Generator g e pe gz 981ms2120 m 1177 kJkg mech 1000 m2s2 Then the power generation potential becomes 1kW 2400k s 1177kJk g 291leS Wmax Emech memech 2 2825kW Therefore the reservoir has the potential to generate 2825 kW of power Discussion This problem can also be solved by considering a point at the turbine inlet and using ow energy instead of potential energy It would give the same result since the ow energy at the turbine inlet is equal to the potential energy at the free surface of the reservoir PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 24 212 Wind is blowing steadily at a certain velocity The mechanical energy of air per unit mass and the power generation potential are to be determined Assumptions The wind is blowing steadily at a constant uniform velocity 9 Properties The density of air is given to be p 125 kgm3 Wind Wind turbine Analysis Kinetic energy is the only form of mechanical 10 MS energy the wind possesses and it can be converted to work a 60 m entirely Therefore the power potential of the wind is its g kinetic energy which is V22 per unit mass and mvz 2 for e a given mass ow rate 2 2 H 10 m 1 kJk emhzkeV S 2g20050kJkg 2 2 1000 m s 702 2 m pVA pV T 125 kgm310 nusme 35340 kgs Wmax mech n39wmech 35340 kgs0050 kJkg 1770 kW Therefore 1770 kW of actual power can be generated by this wind turbine at the stated conditions Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity and thus the power generation will change strongly with the wind conditions 213 A water jet strikes the buckets located on the perimeter of a wheel at a speci ed velocity and ow rate The power generation potential of this system is to be determined Assumptions Water jet ows steadily at the speci ed speed and ow rate Analysis Kinetic energy is the only form of harvestable mechanical energy the water jet possesses and it can be converted to work entirely Therefore the power potential of the water jet is its kinetic energy which is V22 per unit mass and mvz 2 for a given mass ow rate V2 60 ms2 lkJkg e ke 18 kJk m1 2 2 1000 m2s2 g Wmax Emech memech 120 kgs18kJkg W 216kW 1kJs Therefore 216 kW of power can be generated by this water jet at the stated conditions Discussion An actual hydroelectric turbine such as the Pelton wheel can convert over 90 of this potential to actual electric power PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 25 214 Two sites with specified wind data are being considered for wind power generation The site better suited for wind power generation is to be determined Assumptions 1The wind is blowing steadily at speci ed velocity during speci ed times 2 The wind power generation is negligible during other times Properties We take the density of air to be p 125 kgm3 it does not affect the nal answer Analysis Kinetic energy is the only form of mechanical energy the wind possesses and it can be converted to work entirely Therefore the power potential of the wind is its kinetic energy which is V22 per unit mass and mv2 2 for a given mass ow rate Considering a unit ow area A 1 m2 the maximum wind power and power generation becomes V12 7 ms2 1kJkg e ke 00245 kJk mh l 1 2 2 1000m2s2 g V22 10 ms2 1kJkg e ke 0050kJk mm 2 2 2 1000 m2s2 g ViHm1 13th Inleth leAkel 125 kgm3 7 ms1m200245 kJkg 02144 kW Vilmax2 15th 71226th pVZAkeZ 125 kgm3 10 ms1m20050 kJkg 0625 kW since 1 kW 1 kJs Then the maximum electric power generations per year become E WmJAtl 02144 kW3000 hyr 643 kWhyr per m2 ow area max1 E ma 2 Wm2Ar2 0625 kW1500 hyr 938 kWhyr per m2 ow area Therefore second site is a better one for wind generation Discussion Note the power generation of a wind turbine is proportional to the cube of the wind velocity and thus the average wind velocity is the primary consideration in wind power generation decisions PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 26 215 A river owing steadily at a speci ed ow rate is considered for hydroelectric power generation by collecting the water in a dam For a specified water height the power generation potential is to be determined Assumptions 1 The elevation given is the elevation of the free surface of the river 2 The mechanical energy of water at the turbine eXit is negligible Properties We take the density of water to be p 1000 kgm3 River Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam relative to free surface of discharge water and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per unit mass and mgz for a given mass ow rate 1kJkg eInec pe gz 981ms2 80 m 07848 kJkg h 10 0m2s2 The mass ow rate is m pt 1000 kgm3175 m3s 175000kgs Then the power generation potential becomes 1MW 1000 kJs Wmax 39mh memh 175000 kgs07848 kJkg j 1 37 MW Therefore 137 MW of power can be generated from this river if its power potential can be recovered completely Discussion Note that the power output of an actual turbine will be less than 137 MW because of losses and inefficiencies 216 A river is owing at a specified velocity ow rate and elevation The total mechanical energy of the river water per unit mass and the power generation potential of the entire river are to be determined Assumptions 1 The elevation given is the elevation of the free surface of the river 2 The velocity given is the average velocity 3 The mechanical energy of water at the turbine eXit is negligible Properties We take the density of water to be p 1000 kgm3 Analysis Noting that the sum of the ow energy and the z potential energy is constant for a given uid body we can IF take the elevation of the entire river water to be the elevation of the free surface and ignore the ow energy Then the total mechanical energy of the river water per unit mass becomes V2 2 3ms2 1kJkg emec eke h 981ms 90m 0887kJk h p g 2 x gt 2 10mm IS 9 The power generation potential of the river water is obtained by multiplying the total mechanical energy by the mass ow rate m pi 1000 kgm3 500 m3s 500000 kgs W E me 500000 kgs0887 kJkg 444000 kW 444 MW max mech mech Therefore 444 MW of power can be generated from this river as it discharges into the lake if its power potential can be recovered completely Discussion Note that the kinetic energy of water is negligible compared to the potential energy and it can be ignored in the analysis Also the power output of an actual turbine will be less than 444 MW because of losses and inef ciencies PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 7 Energy Transfer by Heat and Work 217 C The form of energy that crosses the boundary of a closed system because of a temperature difference is heat all other forms are work 218C a The car39s radiator transfers heat from the hot engine cooling uid to the cooler air No work interaction occurs in the radiator 19 The hot engine transfers heat to cooling uid and ambient air while delivering work to the transmission 0 The warm tires transfer heat to the cooler air and to some degree to the cooler road while no work is produced No work is produced since there is no motion of the forces acting at the interface between the tire and road d There is minor amount of heat transfer between the tires and road Presuming that the tires are hotter than the road the heat transfer is from the tires to the road There is no work exchange associated with the road since it cannot move 6 Heat is being added to the atmospheric air by the hotter components of the car Work is being done on the air as it passes over and through the car 219C a From the perspective of the contents heat must be removed in order to reduce and maintain the content39s temperature Heat is also being added to the contents from the room air since the room air is hotter than the contents 9 Considering the system formed by the refrigerator box when the doors are closed there are three interactions electrical work and two heat transfers There is a transfer of heat from the room air to the refrigerator through its walls There is also a transfer of heat from the hot portions of the refrigerator ie back of the compressor where condenser is placed system to the room air Finally electrical work is being added to the refrigerator through the refrigeration system 0 Heat is transferred through the walls of the room from the warm room air to the cold winter air Electrical work is being done on the room through the electrical wiring leading into the room 220C It is a work interaction 221C It is a work interaction since the electrons are crossing the system boundary thus doing electrical work 222C It is a heat interaction since it is due to the temperature difference between the sun and the room 223C This is neither a heat nor a work interaction since no energy is crossing the system boundary This is simply the conversion of one form of internal energy chemical energy to another form sensible energy PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 224 The power produced by an electrical motor is to be expressed in different units Analysis Using appropriate conversion factors we obtain a W5W j1N39mj5Nms 1w 1 2 b W5W1N mjpkg mS j5kgm2s3 1w 1J 1N 225E The power produced by a model aircraft engine is to be expressed in different units Analysis Using appropriate conversion factors we obtain a W Z 10 W lBtus j7781691bfits 738bf s 1055056 w lBtus lhp b W10W 00134h 7457 w p Mechanical Forms of Work 226C The work done is the same but the power is different 227E A construction crane lifting a concrete beam is considered The amount of work is to be determined considering a the beam and b the crane as the system Analysis a The work is done on the beam and it is determined from 1 lbf W mgAz 3X 20001bm32174 fts 2 2 32174 lbm s 24 ft 144000Ibfft lBtu 1440001bf 7781691bf it 2185Btu 9 Since the crane must produce the same amount of work as is required to lift the beam the work done by the crane is W 144000Ibf ft 2185 Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 9 228E A man is pushing a cart with its contents up a ramp that is inclined at an angle of 10 from the horizontal The work needed to move along this ramp is to be determined considering a the man and b the cart and its contents as the system Analysis a Considering the man as the system letting l be the displacement along the ramp and letting 9 be the inclination angle of the ramp W 2 Fl sin6 100180lbf1001tsin10 4862Ibf ft lBtu 4862 lbf it 778169 lbf it 2 624BBtu This is work that the man must do to raise the weight of the cart and contents plus his own weight a distance of lsinl9 9 Applying the same logic to the cart and its contents gives W 2 Fl sin 6 1001bf100 1tsin10 1 736bf ft lBtu 1736 lbf it 778169 lbf it 2231Btu 229E The work required to compress a spring is to be determined Analysis Since there is no preload F kx Substituting this into the work expression gives 2 2 2 k WzIFdszIkxdxsz39xdxzamg xlz 1F 1 1 1 T whim2 02i833b1 2 x 2 12m s331bf wj2001073 778169 lbf it 230 A car is accelerated from 10 to 60 kmh on an uphill road The work needed to achieve this is to be determined Analysis The total work required is the sum of the changes in potential and kinetic energies 2 2 1 1 1 Wu 2 V12 1300 kg kJ22 21755 k 2 2 3600 s 3600 s 1000 kg m s and Wg mgz2 z1 1300 kg981msz40 m 22 5100 kJ 1000 kgm s VVt O ta 2 W Wg 1755 5100 686 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 10 231E The engine of a car develops 450 hp at 3000 rpm The torque transmitted through the shaft is to be determined Analysis The torque is determined from amp 450 hp 5501bf s 27m 27r300060s lhp 2788 Ibfft 232E The work required to expand a soap bubble is to be determined Analysis Noting that there are two gasliquid interfaces in a soap bubble the surface tension work is determined from 2 W ZIO39SdA 039A1 A2 200051b17ft 7r312ft2 0512ft2 1 1Btu 0001909 lbf ft 0001909 lbf ft 77821bf ft 1245gtlt106 Btu 233 A linear spring is elongated by 20 cm from its rest position The work done is to be determined Analysis The spring work can be determined from W spring 2 gm x12 g 70 kNm022 0 m2 14 kNm 14 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 11 234 A ski lift is operating steadily at 10 kmh The power required to operate and also to accelerate this ski lift from rest to the operating speed are to be determined Assumptions 1 Air drag and friction are negligible 2 The average mass of each loaded chair is 250 kg 3 The mass of chairs is small relative to the mass of people and thus the contribution of returning empty chairs to the motion is disregarded this provides a safety factor Analysis The lift is 1000 m long and the chairs are spaced 20 m apart Thus at any given time there are 100020 50 chairs being lifted Considering that the mass of each chair is 250 kg the load of the lift at any given time is Load 50 chairs250 kgchair 12500 kg Neglecting the work done on the system by the returning empty chairs the work needed to raise this mass by 200 m is lkJ Wg mgz2 z1 12500 kg981ms2200 m 22 1000 kgm s j 24525 kJ At 10 kmh it will take distance lkrn velocity 10 km h At 0lh360s to do this work Thus the power needed is W 24525 kJ g g 681kW At 360 s The velocity of the lift during steady operation and the acceleration during start up are V 10 kmh j 2778 ms 36 kmh aAV W0556m32 At 5 s During acceleration the power needed is l 2 mV2 V2At l 12 500 kg2 778 ms2 0 lkJkg 2 1 2 39 1000 m2s2 W js s 96 kW Assuming the power applied is constant the acceleration will also be constant and the vertical distance traveled during acceleration will be 2 1 h latzsina la 0556 ms25 s202 139 m 2 2 1000 m 2 and lkJk Wg mgz2 z1At 12500 kg981ms2139 m g2 2 5 s 341 kW 1000 kgm s Thus Wm1 W Wg 96 341 437 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 212 235 The engine of a car develops 75 kW of power The acceleration time of this car from rest to 100 kmh on a level road is to be determined Analysis The work needed to accelerate a body is the change in its kinetic energy a 5 2 1 1 100 000 1kJ W mv v12 1500 kg 0 3600 s 1000 kg 1112 2 5787 k Thus the time required is W 5787 k W 75 kJs a 772 s This answer is not realistic because part of the power will be used against the air drag friction and rolling resistance 236 A car is to climb a hill in 12 s The power needed is to be determined for three different cases Assumptions Air drag friction and rolling resistance are negligible Analysis The total power required for each case is the sum of the rates of changes in potential and kinetic energies That is Wtotal Wa a 0 since the velocity is constant Also the vertical rise is h 100 msin 30 50 m Thus Wg mgz2 z1At 1150 kg981ms250 m 2 j12 s 470 kw 1000 kg m2s and Wm1 W Wg 0 470 470 kW 9 The power needed to accelerate is 1 2 2 1 2 Waz mV V At 1150kg 30ms 0 2 2 1 2 k i 1000 kgm2s 1kJ 2 j12 s431kW and Wm1 W Wg 470 431 901 kW 0 The power needed to decelerate is Wu 2 gmax V12At g 1150 kg5 ms2 35 ms2 1kJ 22 12 s 575 kW 1000 kgm s and Wtotal 2 Wu Wg 575 471 105 kW breaking power PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 13 The First Law of Thermodynamics 237 C Energy can be transferred to or from a control volume as heat various forms of work and by mass transport 238C Warmer Because energy is added to the room air in the form of electrical work 239 Water is heated in a pan on top of a range While being stirred The energy of the water at the end of the process is to be determined Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible Analysis We take the water in the pan as our system This is a closed system since no mass enters or leaves Applying the energy balance on this system gives Ein Eout AE system V W Net en ergy tranSfer Chan gem 1ntemal krn et1c by heat W0rkand mass potential etc energies Qin Wshin Qout 2 U2 U1 30 kJO5 kJ 5kJU2 lOkJ U2 2355 Therefore the nal internal energy of the system is 355 k 240E Water is heated in a cylinder on top of a range The change in the energy of the water during this process is to be determined Assumptions The pan is stationary and thus the changes in kinetic and potential energies are negligible Analysis We take the water in the cylinder as the system This is a closed system since no mass enters or leaves Applying the energy balance on this system gives Ein Eout AEsystem Net energy Hanger Chan gein internal kinetic by heat W0rkandmaSS potentirl etc energies Qin Wout Qout AU 2 U2 U1 65Btu 5Btu 8Btu AU Therefore the energy content of the system increases by 52 Btu during this process PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 2 14 241E The heat loss from a house is to be made up by heat gain from people lights appliances and resistance heaters For a specified rate of heat loss the required rated power of resistance heaters is to be determined Assumptions 1 The house is wellsealed so no air enters or heaves the house 2 All the lights and appliances are kept on 3 The house temperature remains constant Analysis Taking the house as the system the energy balance can be written as lt90 ad Ein Eout dEsystem dt Ste y 2 O Ein Eout k V n J n atEOftnet eiergi l manger Rate of changern 1ntemalk1net1c y 63 a W01 aan mass potentialetc energies where Eout Qlout 60000 Btuh and Lights Qout Ein Epeople Elights E appliance Eheater 6000 Bullh Eheater Energy gt Appliance 39 gt Substituting the required power rating of the heaters becomes Heaters 1 kW Eheater 60000 6000 54000 Btuh 158kW 3412 Btuh Discussion When the energy gain of the house equals the energy loss the temperature of the house remains constant But when the energy supplied drops below the heat loss the house temperature starts dropping 242E A water pump increases water pressure The power input is to be determined Analysis The power input is determined from WVP2 Pl 08ft3s70 15psia 3m lhp 5404 psia it 3 07068 Btus 115hp The water temperature at the inlet does not have any signi cant effect on the required power PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 215 243 A water pump is claimed to raise water to a speci ed elevation at a specified rate while consuming electric power at a specified rate The validity of this claim is to be investigated Assumptions 1 The water pump operates steadily 2 Both the lake and the pool are open to the atmosphere and the ow velocities in them are negligible Properties We take the density of water to be p 1000 kgm3 1 kgL Q Analysis For a control volume that encloses the pumpmotor unit the energy balance can be written as 39 lt90 steady Ein Eout dEsystemdt O ar J k V J Rate 0f net energy tranSfel Rate of changein internal kinetic by heat W0 and mass potentialetc energies E in 2 E out Win mpel Z mpez Win MAI mgZ2 Z1 since the changes in kinetic and ow energies of water are negligible Also m 2 pi 1 kgL50 Us 2 50 kgs Substituting the minimum power input required is determined to be 1 kJkg Wmm z z 50k s981ms230m gm 1 g 1000m2s2 2 147 kJs 147kW which is much greater than 2 kW Therefore the claim is false Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the power required will be considerably higher than 147 kW because of the losses associated with the conversion of electricaltomechanical shaft and mechanical shafttopotential energy of water 244 A classroom is to be airconditioned using window airconditioning units The cooling load is due to people lights and heat transfer through the walls and the windows The number of 5kW window air conditioning units required is to be determined Assumptions There are no heat dissipating equipment such as computers TVs or ranges in the room Analysis The total cooling load of the room is determined from Qcooling Qlights Qpeople Qheat gain where Q39lights 10 X100 w 1 kW QMple 40 x 360 kJ h 4 kW Room Qheat gain 2 15000 M h 2 417 kW 15000 kJh gt 40 people gt Qcool I 10 bulbs Substituting Q0001ng 2 14 417 917 kW Thus the number of airconditioning units required is 917 kW 183 gt2 units 5 kWunrt PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 16 245 The classrooms and faculty of ces of a university campus are not occupied an average of 4 hours a day but the lights are kept on The amounts of electricity and money the campus will save per year if the lights are turned off during unoccupied periods are to be determined Analysis The total electric power consumed by the lights in the classrooms and faculty offices is Elighting classmom 2 Power consumed per lamp x No of lamps 200 x12 x110 W 264000 2 264 kW Elightingpf ces 2 Power consumed per lamp x No of lamps 400 x 6 x110 W 264000 2 264 kW Elightingtotal E lightingclassroom Elightingpf ces 264 264 Z 5 28 kW Noting that the campus is open 240 days a year the total number of unoccupied work hours per year is Unoccupied hours 4 hoursday240 daysyear 960 hyr Then the amount of electrical energy consumed per year during unoccupied work period and its cost are Energy savings Eligh ngmtal Unoccupied hours 2 528 kW960 hyr 506880 kWh Cost savings Energy savingsU nit cost of energy 506880kWhyr011kWh 55757yr Discussion Note that simple conservation measures can result in signi cant energy and cost savings 246 An industrial facility is to replace its 40W standard uorescent lamps by their 35W high ef ciency counterparts The amount of energy and money that will be saved a year as well as the simple payback period are to be determined Analysis The reduction in the total electric power consumed by the lighting as a result of switching to the high efficiency uorescent is Wattage reduction Wattage reduction per lampNumber of lamps 40 34 Wlamp700 lamps 4200 W Then using the relations given earlier the energy and cost savings associated with the replacement of the high efficiency uorescent lamps are determined to be Energy Savings Total wattage reductionB allast factorOperating hours 42 kW112800 hyear 12936 kWhyear Cost Savings Energy savingsUnit electricity cost 39 12936 kWhyear0105kWh 1358year I The implementation cost of this measure is simply the extra cost of the energy efficient uorescent bulbs relative to standard ones and is determined to be Implementation Cost Cost difference of lampsNumber of lamps 226177lamp700 lamps 343 This gives a simple payback period of Implementation cost 343 Simplepayback period 025 year 30 months Annual cost savings 1358 year Discussion Note that if all the lamps were burned out today and are replaced by highef ciency lamps instead of the conventional ones the savings from electricity cost would pay for the cost differential in about 4 months The electricity saved will also help the environment by reducing the amount of C02 CO NOX etc associated with the generation of electricity in a power plant PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 17 247 A room contains a light bulb a TV set a refrigerator and an iron The rate of increase of the energy content of the room when all of these electric devices are on is to be determined Assumptions 1 The room is well sealed and heat loss from the room is negligible 2 All the appliances are kept on Analysis Taking the room as the system the rate form of the energy balance can be written as Ein Eout dEsystemdt dEroomdZL Ein Rate 0f net energy tranSfel Rate of changein internal kinetic by heat workand maSS ptentialetc energies ROO since no energy is leaving the room in any form and thus Eom 0 Also Ein Elights ETV Erefrig Eiron Lights 401103001200w TV Electr1c1ty gt Refrig 1650 W Iron Substituting the rate of increase in the energy content of the room becomes dE room dtEin 1650W Discussion Note that some appliances such as refrigerators and irons operate intermittently switching on and off as controlled by a thermostat Therefore the rate of energy transfer to the room in general will be less 248E A fan accelerates air to a specified velocity in a square duct The minimum electric power that must be supplied to the fan motor is to be determined Assumptions 1 The fan operates steadily 2 There are no conversion losses Properties The density of air is given to be p 0075 lbmft3 Analysis A fan motor converts electrical energy to mechanical shaft energy and the fan transmits the mechanical energy of the shaft shaft power to mechanical energy of air kinetic energy For a control volume that encloses the fanmotor unit the energy balance can be written as 39 39 lt90 steady 39 39 Ein Eout dEsystemdt O Ein E ar J k V J Rate 0f net energy tranSfel Rate of changein internal kinetic by heat WOIku smd mass ptentialetc energies Iquot I V I 1quot out Welectin mairkeout mair 2 Hill F II where quot II pVA 00751bm 33 X 3 it 222 113 14851bms Substituting the minimum power input required is determined to be V2 22 2 Win 2 mm A 14851bms S lBtullbzm 2 01435 Btus 151 w 2 2 25037it s since 1 Btu 1055 kJ and 1 kJs 1000 W Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the power required will be considerably higher because of the losses associated with the conversion of electricaltomechanical shaft and mechanical shafttokinetic energy of air PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 218 249 The fan of a central heating system circulates air through the ducts For a speci ed pressure rise the highest possible average ow velocity is to be determined Assumptions 1 The fan operates steadily 2 The changes in kinetic and potential energies across the fan are negligible Analysis For a control volume that encloses the fan unit the energy balance can be written as 39 39 lt90 steady 39 39 Ein Eout dEsystemdt 0 Ein Eout ar J k V J Rate 0f net energy tranSfer Rate of changein internal kinetic by heat WOIk Ild mass potentialetc energies W mltPvgt1mPv2 a Win mltP2 PlgtvVAP Ap50pa since m Vv and the changes in kinetic and potential energies of Air gasoline are negligible Solving for volume ow rate and G VmS substituting the maXimum ow rate and velocity are determined to be 39 3 Vmax W1n 60Js lPa m 1392m3S AP 50Pa l V V 3 V m m lzmS 170ms max AC 702 4 7r030m24 Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the velocity will be less because of the losses associated with the conversion of electricaltomechanical shaft and mechanical shaftto ow energy 250 A gasoline pump raises the pressure to a specified value while consuming electric power at a specified rate The maximum volume ow rate of gasoline is to be determined Assumptions 1 The gasoline pump operates steadily 2 The changes in kinetic and potential energies across the pump are negligible Analysis For a control volume that encloses the pumpmotor unit the energy balance can be written as Ein Eout dE dtlt90ltsteadygt 0 gt E39in E39out t sys em V J RateOf net energy manger Rate of changein intemalkinetic 38 kW byheata workaandmass potentialetc energies Win man1 man2 gt Win ma P1v VAP s1nce m Vv and the changes 1n k1net1c and potent1al energ1es of M otor gasoline are negligible Solving for volume ow rate and substituting the maXimum ow rate is determined to be V Wi 38kJs1kPam3 max AP7kPa 1kJ 0543m3s Discussion The conservation of energy principle requires the energy to be conserved as it is converted from one form to another and it does not allow any energy to be created or destroyed during a process In reality the volume ow rate will be less because of the losses associated with the conversion of electricaltomechanical shaft and mechanical shaftto ow energy PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 251 An inclined escalator is to move a certain number of people upstairs at a constant velocity The minimum power required to drive this escalator is to be determined Assumptions 1 Air drag and friction are negligible 2 The average mass of each person is 75 kg 3 The escalator operates steadily with no acceleration or breaking 4 The mass of escalator itself is negligible Analysis At design conditions the total mass moved by the escalator at any given time is Mass 50 persons75 kgperson 3750 kg The vertical component of escalator velocity is V V sin 45 2 06 rnssin45O vert Under stated assumptions the power supplied is used to increase the potential energy of people Taking the people on elevator as the closed system the energy balance in the rate form can be written as E E dE Id 0 E dE d sys in out system t in sys t Z J At Rate 0f net energy manger Rate of changein intemalkinetic byheatW0rkandmaSS potentialetc energies APE mgAz in mgvm At At That is under stated assumptions the power input to the escalator must be equal to the rate of increase of the potential energy of people Substituting the required power input becomes 1 kJkg Win 2 mngt 3750 kg981ms206 mssin45 2 2 1000 m s 2 125 kJs 2156 kW When the escalator velocity is doubled to V 12 ms the power needed to drive the escalator becomes 1 kJkg Win 2 mgVVeIt 3750 kg981ms212 mssin45 2 2 250 kJs 312 kW 1000 m s Discussion Note that the power needed to drive an escalator is proportional to the escalator velocity PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 20 252 A car cruising at a constant speed to accelerate to a speci ed speed within a speci ed time The additional power needed to achieve this acceleration is to be determined Assumptions 1 The additional air drag friction and rolling resistance are not considered 2 The road is a level road Analysis We consider the entire car as the system except that let s assume the power is supplied to the engine externally for simplicity rather that internally by the combustion of a fuel and the associated energy conversion processes The energy balance for the entire mass of the car can be written in the rate form as Ein Eout dEsystem f J f J Rate 0f net energy manger Rate of changein intemalkinetic by heat workaand mass potentialetc energies AESyS dt 0 gt 13in dESySdt E At AKE mV22 V122 At At since we are considering the change in the energy content of the car due to a change in its kinetic energy acceleration Substituting the required additional power input to achieve the indicated acceleration becomes Win V22 V12 2 2 1 kJk W 2 m M Z 1400 kg 1103 6ms 703 6ms g 1 25 s 1000 m2s2 778 kJs 778 kW 1n since 1 ms 36 kmh If the total mass of the car were 700 kg only the power needed would be V22 V2 11036ms2 7036ms2 1 kJkg Winm 1 700k At g 25 s 1000 m2s2 2389 kW Discussion Note that the power needed to accelerate a car is inversely proportional to the acceleration time Therefore the short acceleration times are indicative of powerful engines Energy Conversion Efficiencies 253C The combined pumpmotor e iciency of a pumpmotor system is defined as the ratio of the increase in the mechanical energy of the uid to the electrical power consumption of the motor Emechout Emechin AEmech uid Wpump npump motor npumpnmotor Welectin Welectin Welectin The combined pumpmotor efficiency cannot be greater than either of the pump or motor ef ciency since both pump and motor efficiencies are less than 1 and the product of two numbers that are less than one is less than either of the numbers 254C The turbine ef ciency generator ef ciency and combined turbinegenerator ef ciency are de ned as follows n Mechanical energy output Wshaftout turbme Mechanical energy extracted om the uid AEmech uid Electrical power output Welectout 77 generator 2 Z 39 Mechanical power input Wsha in Welectout VVelectout nturbine gen nturbinengenerator 39 39 Emechin Emechout I AErrrgtch uid I PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 221 255C No the combined pumpmotor ef ciency cannot be greater that either of the pump ef ciency of the motor ef ciency This is because npumpmotor npumpnmotor and both npump and 77mtor are less than one and a number gets smaller when multiplied by a number smaller than one 256 A hooded electric open burner and a gas burner are considered The amount of the electrical energy used directly for cooking and the cost of energy per utilized kWh are to be determined Analysis The ef ciency of the electric heater is given to be 73 percent Therefore a burner that consumes 3kW of electrical energy will supply ngas 38 nelectric 73 Qutiumd Energy input gtlt E icienc y 24 kW07 3 1 75 kW of useful energy The unit cost of utilized energy is inversely proportional to the ef ciency and iS determined from Cost of energy input 010kWh 0137kWh Cost of utilized energy E iciency 073 Noting that the ef ciency of a gas burner is 38 percent the energy input to a gas burner that supplies utilized energy at the same rate 175 kW is Qutilized 175 kW 461kW 15700Btuh E icrency 038 Qinputgas since 1 kW 3412 Btuh Therefore a gas burner should have a rating of at least 15700 Btuh to perform as well as the electric unit Noting that 1 therm 293 kWh the unit cost of utilized energy in the case of gas burner is determined the same way to be Cost of energy input 120 293 kWh 01 08 kWh E icrency 0 38 Cost of utilized energy PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 22 257 A worn out standard motor is to be replaced by a high ef ciency one The amount of electrical energy and money savings as a result of installing the high ef ciency motor instead of the standard one as well as the simple payback period are to be determined Assumptions The load factor of the motor remains constant at 075 Analysis The electric power drawn by each motor and their difference can be eXpressed as Welectricinstandard Z Wshaft nstandard 2 Power ratingLoad 13101 nstandard Welectric inef cient Wshaft 77 ef cient Power ratingLoad factor 77 ef cient 770M 2 Power saVngS Welectric instandard electric inef cient 2 Power ratingLoad factor1 nstandard 1 nef cient where nstandard is the ef ciency of the standard motor and nef cient is the efficiency of the comparable high ef ciency motor Then the annual energy and cost savings associated with the installation of the high ef ciency motor are determined to be Energy Savings Power savingsOperating Hours Power RatingOperating HoursLoad Factor1nstandard 1n ef cient 75 hp0746 kWhp4368 hoursyear0751091 10954 9290 kWhyear Cost Savings Energy savingsUnit cost of energy 9290 kWhyear012kWh 11 14year The implementation cost of this measure consists of the excess cost the high efficiency motor over the standard one That is Implementation Cost Cost differential 5520 5449 71 This gives a simple payback period of Implementation cost 71 Simplepayback period 00637year or 076 month Annual cost savings 1114 year Therefore the highef ciency motor will pay for its cost differential in less than one month 258 An electric motor with a speci ed ef ciency operates in a room The rate at which the motor dissipates heat to the room it is in when operating at full load and if this heat dissipation is adequate to heat the room in winter are to be determined Assumptions The motor operates at full load Analysis The motor efficiency represents the fraction of electrical energy consumed by the motor that is converted to mechanical work The remaining part of electrical energy is converted to thermal energy and is dissipated as heat Q39dissipated 1 77mm Wigwamc 1 08820 kW 24 kW which is larger than the rating of the heater Therefore the heat dissipated by the motor alone is suf cient to heat the room in winter and there is no need to turn the heater on Discussion Note that the heat generated by electric motors is signi cant and it should be considered in the determination of heating and cooling loads PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 23 259E The combustion efficiency of a furnace is raised from 07 to 08 by tuning it up The annual energy and cost savings as a result of tuning up the boiler are to be determined Assumptions The boiler operates at full load while operating Analysis The heat output of boiler is related to the fuel energy input to the boiler by Boiler output Boiler inputCombustion ef ciency Boiler 70 0139 Qout Qin nfumace 5 5X 106 The current rate of heat input to the boiler is given to be Qlimumem 55 gtlt106 Btuh Then the rate of useful heat output of the boiler becomes Q out Q39inn nmmnt 55 gtlt106 Btuh07 385 gtlt106 Btuh The boiler must supply useful heat at the same rate after the tune up Therefore the rate of heat input to the boiler after the tune up and the rate of energy savings become Qimew gout77mmnew 385gtlt106 Btuh08 481gtlt106 Btuh Qimaved Q39mment Qimew 55 gtlt106 481gtlt106 069 gtlt106 Btuh Then the annual energy and cost savings associated with tuning up the boiler become Energy Savings Qlimsaved Operation hours 069x106 Btuh4200 hyear 289x109 Btuyr Cost Savings Energy SavingsUnit cost of energy 289x109 Btuyr435106 Btu 12600year Discussion Notice that tuning up the boiler will save 12600 a year which is a significant amount The implementation cost of this measure is negligible if the adjustment can be made by inhouse personnel Otherwise it is worthwhile to have an authorized representative of the boiler manufacturer to service the boiler twice a year PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 24 260E 39 quot Problem 259E is reconsidered The effects of the unit cost of energy and combustion efficiency on the annual energy used and the cost savings as the ef ciency varies from 07 to 09 and the unit cost varies from 4 to 6 per million Btu are the investigated The annual energy saved and the cost savings are to be plotted against the ef ciency for unit costs of 4 5 and 6 per million Btu Analysis The problem is solved using EES and the solution is given below quotGivenquot Qdotincurrent55E6 Btuh etafurnacecurrent07 etafurnacenew08 Hours4200 hyear UnitCost435E6 Btu quotAnalysisquot QdotoutQdotincurrentetafurnacecurrent QdotinnewQdotoutetafurnacenew QdotinsavedQdotincurrentQdotinnew EnergysavingsQdotinsavedHours CostSavingsEnergySavingsUnitCost 6x109 nfumacemw EnergySavings CostSavings Btuyear year 39E39 5x109 07 000E00 0 339 072 642E08 3208 3 4x109 074 125E09 6243 e 076 182E09 9118 a 939 078 237E09 11846 g 3 10 08 289E09 14437 g 939 082 338E09 16902 9 2X10 084 385E09 19250 g 086 430E09 21488 c 109 088 473E09 23625 L 09 513E09 25667 0x10o 068 072 076 08 084 088 092 139lfurnacenew 30000 Fl 25000 5 g 20000 2 39 6 15000 6x10 Btu 5x106 Btu 03 10000 8 4x10396 Btu o 5000 0 l l I l I I I 068 072 076 08 084 088 092 Tlfurnacemew Table values are for UnitCost 5E5 Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 2 25 261 Geothermal water is raised from a given depth by a pump at a specified rate For a given pump ef ciency the required power input to the pump is to be determined Assumptions 1 The pump operates steadily 2 Frictional losses in the pipes are negligible 3 The changes in kinetic energy are negligible 4 The geothermal water is exposed to the atmosphere and thus its free surface is at atmospheric pressure Properties The density of geothermal water is given to be p 1050 kgm3 200 Analysis The elevation of geothermal water and thus its potential energy In changes but it experiences no changes in its velocity and pressure Therefore the change in the total mechanical energy of geothermal water is equal to the G V change in its potential energy which is gz per unit mass and mgz for a given mass ow rate That is Pump AE mAemch n39tApe mgAz pVgAz mech 1 1k 1050kgm303m3s981ms2200m N 2 W gmS 1000Nms 6180kW Then the required power input to the pump becomes W 415mb 618 kW pumpelect pumpmotor 74 835 kW Discussion The frictional losses in piping systems are usually significant and thus a larger pump will be needed to overcome these frictional losses 262 Several people are working out in an exercise room The rate of heat gain from people and the equipment is to be determined Assumptions The average rate of heat dissipated by people in an exercise room is 600 W Analysis The 6 weight lifting machines do not have any motors and thus they do not contribute to the internal heat gain directly The usage factors of the motors of the treadmills are taken to be unity since they are used constantly during peak periods Noting that 1 hp 7457 W the total heat generated by the motors is Qmotors NO39 Of meters X Wmotor X fload X usageUnlotor 7 x 25 x 746 Wgtlt 070gtlt10077 11870 W The heat gain from 14 people is Qpeople 14 x 600 W 8400 w Then the total rate of heat gain of the exercise room during peak period becomes Q39total thors Qpeople 11870 8400 20270 w PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 26 263 A room is cooled by circulating chilled water through a heat exchanger and the air is circulated through the heat exchanger by a fan The contribution of the fanmotor assembly to the cooling load of the room is to be determined Assumptions The fan motor operates at full load so that fload 1 Analysis The entire electrical energy consumed by the motor including the shaft power delivered to the fan is eventually dissipated as heat Therefore the contribution of the fanmotor assembly to the cooling load of the room is equal to the electrical energy it consumes Qintemalgeneration ine1ectric shaft nmotor 025 hp054 0463 hp 345 w since 1 hp 746 W 264 A hydraulic turbinegenerator is to generate electricity from the water of a lake The overall ef ciency the turbine ef ciency and the shaft power are to be determined Assumptions 1 The elevation of the lake and that of the discharge site remains constant 2 Irreversible losses in the pipes are negligible Properties The density of water can be taken to be p 1000 kgm3 The gravitational acceleration is g 981 msz Analysis a We take the bottom of the lake as the reference level for convenience Then kinetic and potential energies of water are zero and the mechanical energy of water consists of pressure energy only which is P emechin emechout 2 8h Generator p WigwamJr 1kJk 9 81 ms2 50 m 2g2 1000m s 0491 kJkg Then the rate at which mechanical energy of uid supplied to the turbine and the overall ef ciency become I AEmech uid I n39Kemechdn emhjn 5000 kgs0491kJkg 2455 kW Welecmut 1862 kw 0760 AEmh uid 2455 kW noverall nturbine gen 9 Knowing the overall and generator ef ciencies the mechanical ef ciency of the turbine is determined from W 2 076 095 nturbine gen nturbinengenerator gt nturbine Z generator c The shaft power output is determined from the definition of mechanical ef ciency Wsha m mum I AE39mh uid I 0800x2455 kWgt1964 kw z 1 960 kW Therefore the lake supplies 2455 kW of mechanical energy to the turbine which converts 1964 kW of it to shaft work that drives the generator which generates 1862 kW of electric power PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 227 265 A pump with a specified shaft power and efficiency is used to raise water to a higher elevation The maximum ow rate of water is to be determined Assumptions 1 The ow is steady and incompressible 2 The elevation difference between the reservoirs is constant 3 We assume the ow in the pipes to be frictionless since the maximum ow rate is to be determined 39 39 Properties We take the density of water to be p 1000 kgm3 Analysis The useful pumping power the part converted to mechanical energy of water is Wpumu npunpwpumsha 0827 hp 574 hp The elevation of water and thus its potential energy changes during pumping but it experiences no changes in its velocity and pressure Therefore the change in the total mechanical energy of water is equal to the change in its potential energy which is gz T per unit mass and mgz for a given mass ow rate That is Q 415m mew mm mgAz pVgAz 33333333333 Noting that AEmech Wpumpu the volume ow rate of water is determined to be W 4h 4 1 1k 2 v 2 WW 2 35 7 P 2 7 5 7 W N In g 1 5 00291m3s pgzz 1000 kgm 981ms 15 m lhp 1W 1N Discussion This is the maXimum ow rate since the frictional effects are ignored In an actual system the ow rate of water will be less because of friction in pipes 266 Wind is blowing steadily at a certain velocity The mechanical energy of air per unit mass the power generation potential and the actual electric power generation are to be determined Assumptions 1 The wind is blowing steadily at a constant uniform velocity 2 The efficiency of the wind turbine is independent of the wind speed Wind 3 Wlnd turbine Properties The dens1ty of an 1s g1ven to be p 125 kgm a l o 7 AnalySts K1net1c energy 1s the only form of mechanrcal energy 80 m the w1nd possesses and 1t can be converted to work ent1rely a Therefore the power potential of the wind is its kinetic energy E which is V22 per unit mass and sz 2 for a given mass ow rate H 2 2 7 m lkJk emhzkeV S 2g200245kJkg 2 2 1000 m s 7rD2 2 m pVA pV T 125 kgm3 7 wow 43982 kgs Wm E me 43982 kgs00245 kJkg1078kW mech mech The actual electric power generation is determined by multiplying the power generation potential by the efficiency W 16c nmdtumewm 0301078 kW 323 kW e Therefore 323 kW of actual power can be generated by this wind turbine at the stated conditions Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity and thus the power generation will change strongly with the wind conditions PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 28 267 Problem 266 is reconsidered The effect of wind velocity and the blade span diameter on wind power generation as the velocity varies from 5 ms to 20 ms in increments of 5 ms and the diameter varies from 20 m to 120 m in increments of 20 m is to be investigated Analysis The problem is solved using EES and the solution is given below Parametrii Table quotGivenquot T H 1 1 a E V7 1 I12 I13 I14 II D80 154 D quotVquot m welect eta overa030 quotquot1 WE WE kw A 33111 23 3 1333 2333 rho125 kgm 3 33112 23 13 3322 333 3311 3 23 13 3333 1333 quotAnalysisquot 3311 4 23 23 2334 4212 g981 mSAQ 33113 43 3 2334 2343 34334 2 3 1 3 3 un mdotrho A V 33113 43 23 31413 1333 WdotmaxmdotVA22ConvertmA2sA2 kJkg Run 9 ED 5 14541 5524 WdoteectetaoveraWdotmax 331113 33 13 33343 3331 331111 33 13 33314 1233 3311 12 33 23 23333 4241 331113 33 3 31413 1123 3311 14 33 13 32332 3423 331113 33 13 34243 3131 331113 33 23 123334 2343 331112 133 3 43332 1341 331113 133 13 33123 1423 331113 133 13 142232 4323 3311 23 133 23 133333 11231 3311 21 123 3 23333 2331 331122 123 13 141322 2121 3311 23 123 13 212333 2132 21 1 6000 12000 I 10000 8000 120 m D1094 39 380 39 6000 r 4000 39 D GOWIE 2000 xgt II39 n m 11 quot o E 39 U H welect kw PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 229 268 Water is pumped from a lake to a storage tank at a specified rate The overall ef ciency of the pumpmotor unit and the pressure difference between the inlet and the eXit of the pump are to be determined Assumptions 1 The elevations of the tank and the lake remain constant 2 Frictional losses in the pipes are negligible 3 The changes in kinetic energy are negligible 4 The elevation difference across the pump is negligible Properties We take the density of water to be p 1000 kgm3 Analysis a We take the free surface of the lake to be point 1 Q and the free surfaces of the storage tank to be point 2 We also take the lake surface as the reference level zl 0 and thus the A Storage potential energy at points 1 and 2 are pe1 0 and pe2 gz2 tank The ow energy at both points is zero since both 1 and 2 are open to the atmosphere P1 P2 Pm Further the kinetic energy at both points is zero ke1 ke2 0 since the water at both locations is essentially stationary The mass ow rate of water and its potential energy at point 2 are m p1 1000 kgm30070 m3s 70 kgs 1 kJkg pe gz 981ms215m 2 2 1000 m2s2 01472 kJkg Then the rate of increase of the mechanical energy of water becomes AE39mh uid Memo emhmgt mltpe2 0 mpez 70 kgsgtlt01472 kJkggt103kW The overall efficiency of the combined pumpmotor unit is determined from its definition AEmech uid 103 kW W 154kW electin npumpfnotor Z 01 9 Now we consider the pump The change in the mechanical energy of water as it ows through the pump consists of the change in the ow energy only since the elevation difference across the pump and the change in the kinetic energy are negligible Also this change must be equal to the useful mechanical energy supplied by the pump which is 103 kW 39 P P AEmech uid memechout emhjn m Solving for AP and substituting AE 3 AP mech urd 103kJ3s lkPa m 147kpa V 0070 m s 1 k Therefore the pump must boost the pressure of water by 147 kPa in order to raise its elevation by 15 m Discussion Note that only twothirds of the electric energy consumed by the pumpmotor is converted to the mechanical energy of water the remaining onethird is wasted because of the inefficiencies of the pump and the motor PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 30 269 A large wind turbine is installed at a location where the wind is blowing steadily at a certain velocity The electric power generation the daily electricity production and the monetary value of this electricity are to be determined Assumptions 1 The wind is blowing steadily at a constant uniform velocity 2 The ef ciency of the wind turbine is independent of the wind speed 6 H Wind Properties The density of air is given to be p 125 kgm3 Wind turbine Analysis Kinetic energy is the only form of mechanical 8 mS energy the wind possesses and it can be converted to work a 100 m entirely Therefore the power potential of the wind is its EJ kinetic energy which is V22 per unit mass and mv2 2 e for a given mass ow rate V2 8 ms2 lkJkg emhzkez z 2 2j0032kJkg 2 2 1000m s 7rD2 1 2 m pVA pV T 125 kgm38 ywm 78540 kgs Wmax mech memh 78540 kgs0032 kJkg 2513 kW The actual electric power generation is determined from W lee 77in tumwmax 0322513 kW 8042kW 6 Then the amount of electricity generated per day and its monetary value become Amount of electricity Wind powerOperating hours8042 kW24 h 19300 kWh Revenues Amount of electricityUnit price 19300 kWh009kWh 1737 per day Discussion Note that a single wind turbine can generate several thousand dollars worth of electricity every day at a reasonable cost which eXplains the overwhelming popularity of wind turbines in recent years 270 The available head of a hydraulic turbine and its overall ef ciency are given The electric power output of this turbine is to be determined Assumptions 1 The ow is steady and incompressible 2 The elevation of the reservoir remains constant Properties We take the density of water to be p 1000 kgm3 Analysis The total mechanical energy the water in a reservoir possesses is equivalent to the potential energy of water at the free surface and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per Eff91 unit mass and rngz for a given mass ow rate Therefore the actual power produced by the turbine can be expressed as Generator Wturbine nturbinemghturbine nturbinepvghturbine Substituting 1N lkW W 091 1000k m3 025m3s 981msz 85m 190kW turb1ne g I mSZ N I mS Discussion Note that the power output of a hydraulic turbine is proportional to the available elevation difference turbine head and the ow rate PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 231 271E A water pump raises the pressure of water by a speci ed amount at a speci ed ow rate while consuming a known amount of electric power The mechanical efficiency of the pump is to be determined Assumptions 1 The pump operates steadily 2 The changes in velocity and elevation across the pump are negligible 3 Water is incompressible Analysis To determine the mechanical efficiency of the pump we need to know the increase in the mechanical energy of the uid as it ows through the pump which is 39 AP 12 si AEmech uid memechout emechin p quot391PV2 PV1 P1V VP2 P1 15ft3s12psi L113 5404psi 333Btus471hp since 1 hp 07068 Btus m 2 pi V v and there is no change in kinetic and potential energies of the uid Then the mechanical ef ciency of the pump becomes AE 7 pump 2 39 Wpumpshaft mem uid 43971hp 20786 or 786 6hp Discussion The overall ef ciency of this pump will be lower than 838 because of the inef ciency of the electric motor that drives the pump PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 232 272 Water is pumped from a lower reservoir to a higher reservoir at a specified rate For a specified shaft power input the power that is converted to thermal energy is to be determined Assumptions 1 The pump operates steadily 2 The elevations of the reservoirs remain constant 3 The changes Q in kinetic energy are negligible Reservoir Properties We take the density of water to be p 1000 kgm3 Analysis The elevation of water and thus its potential energy changes during pumping but it experiences no changes in its velocity and pressure Therefore the change in the total mechanical energy of water is equal to the Reservoir change 1n 1ts potent1al energy wh1ch 1s gz per un1t mass and mgz for a given mass ow rate That is AEmech n39iAemech n39iApe IngAz pVgAz 1 1k 2 1000 kgm3 003 m3s981ms2 45 m N 2 W 2 132 kW 1kgms 1000Nms Then the mechanical power lost because of frictional effects becomes W ict W AE pumpin mach 20 132 kW 2 Discussion The 68 kW of power is used to overcome the friction in the piping system The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy which results in a slight rise in uid temperature Note that this pumping process could be accomplished by a 132 kW pump rather than 20 kW if there were no frictional losses in the system In this ideal case the pump would function as a turbine when the water is allowed to ow from the upper reservoir to the lower reservoir and extract 132 kW of power from the water 273 The mass ow rate of water through the hydraulic turbines of a dam is to be determined Analysis The mass ow rate is determined from W 100000 kJs gltZ2 Z1 W rngz2 z1 gtn391 49500kgs 98 ms2206 0 m1kJk2g2 1000 m s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 233 274 A pump is pumping oil at a speci ed rate The pressure rise of oil in the pump is measured and the motor efficiency is specified The mechanical efficiency of the pump is to be determined Assumptions 1 The ow is steady and incompressible 2 The elevation difference across the pump is negligible Properties The density of oil is given to be p 860 kgm3 Analysis Then the total mechanical energy of a uid is the sum of the potential ow and kinetic energies and is eXpressed per unit mass as emh gh Pv V2 2 To determine the mechanical ef ciency of the pump we need to know the increase in the mechanical energy of the uid as it ows through the pump which is 39 V22 V12 39 V22 V12 AEmech uid memechout emechin m P102 TPV1T V P2 HHpT since m 2 pi V v and there is no change in the potential energy of the uid Also V V 1 3 A1 701 4 7r008m 4 V V 1 3 2 0 13 884ms A 2 no 4 7r012m2 4 Substituting the useful pumping power is determined to be Wpumpu AEmech uid 01m3s SOOkNmz 860kgm3 884mS2 199mS2 lkN lkW 2 lOOOkgmsz lkNms 363 kW Then the shaft power and the mechanical ef ciency of the pump become Wpumpshaft UmtorWelecmc 09044 kW 396 kW Pump villoumwmft 396kW 0918 918 Discussion The overall ef ciency of this pumpmotor unit is the product of the mechanical and motor efficiencies which is 09x0918 0826 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 34 Energy and Environment 275C Energy conversion pollutes the soil the water and the air and the environmental pollution is a serious threat to vegetation wild life and human health The emissions emitted during the combustion of fossil fuels are responsible for smog acid rain and global warming and climate change The primary chemicals that pollute the air are hydrocarbons HC also referred to as volatile organic compounds VOC nitrogen oxides NOx and carbon monoxide CO The primary source of these pollutants is the motor vehicles 27 6C Fossil fuels include small amounts of sulfur The sulfur in the fuel reacts with oxygen to form sulfur dioxide 802 which is an air pollutant The sulfur oxides and nitric oxides react with water vapor and other chemicals high in the atmosphere in the presence of sunlight to form sulfuric and nitric acids The acids formed usually dissolve in the suspended water droplets in clouds or fog These acidladen droplets are washed from the air on to the soil by rain or snow This is known as acid rain It is called rain since it comes down with rain droplets As a result of acid rain many lakes and rivers in industrial areas have become too acidic for sh to grow Forests in those areas also experience a slow death due to absorbing the acids through their leaves needles and roots Even marble structures deteriorate due to acid rain 277 C Carbon monoxide which is a colorless odorless poisonous gas that deprives the body39s organs from getting enough oxygen by binding with the red blood cells that would otherwise carry oxygen At low levels carbon monoxide decreases the amount of oxygen supplied to the brain and other organs and muscles slows body reactions and re exes and impairs judgment It poses a serious threat to people with heart disease because of the fragile condition of the circulatory system and to fetuses because of the oxygen needs of the developing brain At high levels it can be fatal as evidenced by numerous deaths caused by cars that are warmed up in closed garages or by exhaust gases leaking into the cars 278C Carbon dioxide C02 water vapor and trace amounts of some other gases such as methane and nitrogen oxides act like a blanket and keep the earth warm at night by blocking the heat radiated from the earth This is known as the greenhouse e ect The greenhouse effect makes life on earth possible by keeping the earth warm But excessive amounts of these gases disturb the delicate balance by trapping too much energy which causes the average temperature of the earth to rise and the climate at some localities to change These undesirable consequences of the greenhouse effect are referred to as global warming or global climate change The greenhouse effect can be reduced by reducing the net production of C02 by consuming less energy for example by buying energy ef cient cars and appliances and planting trees 279C Smog is the brown haze that builds up in a large stagnant air mass and hangs over populated areas on calm hot summer days Smog is made up mostly of groundlevel ozone 03 but it also contains numerous other chemicals including carbon monoxide CO particulate matter such as soot and dust volatile organic compounds VOC such as benzene butane and other hydrocarbons Groundlevel ozone is formed when hydrocarbons and nitrogen oxides react in the presence of sunlight in hot calm days Ozone irritates eyes and damage the air sacs in the lungs where oxygen and carbon dioxide are exchanged causing eventual hardening of this soft and spongy tissue It also causes shortness of breath wheezing fatigue headaches nausea and aggravate respiratory problems such as asthma PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 35 280E A person trades in his Ford Taurus for a Ford Explorer The extra amount of C02 emitted by the Explorer Within 5 years is to be determined Assumptions The Explorer is assumed to use 850 gallons of gasoline a year compared to 650 gallons for Taurus Analysis The extra amount of gasoline the Explorer will use Within 5 years is Extra Gasoline Extra per yearNo of years 850 650 galyr5 yr 1000 gal Extra C02 produced Extra gallons of gasoline usedC02 emission per gallon 1000 gal197 lbmgal 19700 Ibm C02 Discussion Note that the car we choose to drive has a significant effect on the amount of greenhouse gases produced 281E A household uses fuel oil for heating and electricity for other energy needs Now the household reduces its energy use by 15 The reduction in the C02 production this household is responsible for is to be determined Properties The amount of C02 produced is 154 lbm per kWh and 264 lbm per gallon of fuel oil given Analysis Noting that this household consumes 14000 kWh of electricity and 900 gallons of fuel oil per year the amount of C02 production this household is responsible for is Amount of CO2 produced 2 Amount of electricity consumedAmount of CO2 per kWh Amount of fuel oil consumedAmount of C02 per gallon 14000 kWhyr154 lbmkWh 900 galyr264 Ibmgal 45320 C02 Ibmyear Then reducing the electricity and fuel oil usage by 15 will reduce the annual amount of C02 production by this household by Reduction in CO2 produced 2 015Current amount of CO2 production 2 01545320 C02 kgyear 6798 602 Ibmyear Therefore any measure that saves energy also reduces the amount of pollution emitted to the environment PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 2 36 282 A power plant that burns natural gas produces 059 kg of carbon dioxide C02 per kWh The amount of C02 production that is due to the refrigerators in a city is to be determined Assumptions The city uses electricity produced by a natural gas power plant Properties 059 kg of C02 is produced per kWh of electricity generated given Analysis Noting that there are 300000 households in the city and each household consumes 700 kWh of electricity for refrigeration the total amount of C02 produced is Amount of C02 produced 2 Amount of electricity consumedAmount of C02 per kWh 300000 household700 kWhyear household05 9 kgkWh 123 gtlt108 co2 kgyear 2123000 602 tonyear Therefore the refrigerators in this city are responsible for the production of 123000 tons of C02 283 A power plant that burns coal produces 11 kg of carbon dioxide C02 per kWh The amount of C02 production that is due to the refrigerators in a city is to be determined Assumptions The city uses electricity produced by a coal power plant 1 f 1 Properties 11 kg of C02 is produced per kWh of electricity generated given Analysis Noting that there are 300000 households in the city and each household consumes 700 kWh of electricity for refrigeration the total amount of C02 produced is Amount of C02 produced 2 Amount of electricity consumedAmount of C02 per kWh 300000 household700 kWhhouseholdl 1kgkWh 231 x 108 co2 kgyear 231000 602 tonyear Therefore the refrigerators in this city are responsible for the production of 231000 tons of C02 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 237 284 A household has 2 cars a natural gas furnace for heating and uses electricity for other energy needs The annual amount of NOX emission to the atmosphere this household is responsible for is to be determined Properties The amount of NOX produced is 71 g per kWh 43 g per therm of natural gas and 11 kg per car given 1 Jig HUI per year Analysis Noting that this household has 2 cars consumes 1200 therms of natural gas and 9000 kWh of electricity per year the amount of NOX production this household is responsible for is Amount of NOX produced 2 No of carsAmou nt of NOX produced per car Amount of electricity consumedAmount of NOX per kWh Amount of gas consumedAmount of NOX per gallon 2 cars11 kgcar 9000 kWhyr0 0071 kgkWh 1200 thermsyr00043 kgtherm 9106 NOx kgyear Discussion Any measure that saves energy will also reduce the amount of pollution emitted to the atmosphere Special Topic Mechanisms of Heat Transfer 285C The three mechanisms of heat transfer are conduction convection and radiation 286C Diamond has a higher thermal conductivity than silver and thus diamond is a better conductor of heat 287 C In forced convection the uid is forced to move by external means such as a fan pump or the Wind The uid motion in natural convection is due to buoyancy effects only 288C A blackbody is an idealized body that emits the maXimum amount of radiation at a given temperature and that absorbs all the radiation incident on it Real bodies emit and absorb less radiation than a blackbody at the same temperature 289C Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface The Kirchhoff39s law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength 290C No It is purely by radiation PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 238 291 The inner and outer surfaces of a brick wall are maintained at speci ed temperatures The rate of heat transfer through the wall is to be determined Assumptions 1 Steady operating conditions eXist since the surface temperatures of the wall remain constant at the speci ed values 2 Thermal properties of the wall are constant Brink Properties The thermal conductivity of the wall is given to be k 069 Wmo wall C Analysis Under steady conditions the rate of heat transfer through the wall is anis 30 mam I AT 20 5 0C k QcondzkA 069 Wm C5x6m21035w L 03 m BEIGE PC 292 The inner and outer surfaces of a window glass are maintained at specified temperatures The amount of heat transferred through the glass in 5 h is to be determined Assumptions 1 Steady operating conditions eXist since the surface temperatures of the glass remain constant at the speci ed values 2 Thermal properties of the glass are constant Properties The thermal conductivity of the glass is given to be k 078 Wm C Glass Analysis Under steady conditions the rate of heat transfer through the glass by conduction is Q39com1 kAA LT 078 Wm C2 x 2 m2 5616 w Then the amount of heat transferred over a period of 10 h becomes 15 C J 6 C Q Qcondm 5616 kJs10 gtlt 3600s 202200kJ x 05 cm If the thickness of the glass is doubled to 1 cm then the amount of heat transferred will go down by half to 101100 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 239 293 339 Reconsider Prob 292 Using EES or other software investigate the effect of glass thickness on heat loss for the speci ed glass surface temperatures Let the glass thickness vary from 02 cm to 2 cm Plot the heat loss versus the glass thickness and discuss the results Analysis The problem is solved using EES and the solution is given below FUNCTION klookupmaterial If material39Gass39 then klookup078 If materia39Brick39 then klookup072 If material39Fiber Glass39 then klookup0043 If materia39Air39 then klookup0026 If material39Woodoak39 then klookup01 END L2 m W2 m material39Gass39 Tin15 C Tout6 C k078 WmC t10 hr thickness05 cm ALW QdotossAkTinToutthicknessconvertcm m QosstotaQdotosstconverthrsconvertJkJ ThiCkneSS Qlosstota1 600000 cm kJ 02 505440 500000 04 252720 06 168480 08 126360 2 400000 1 101088 E 39 12 84240 3 300000 14 72206 g 16 63180 2 200000 18 56160 0 2 50544 100000 39 0 0 o 02 04 06 08 1 12 14 16 18 2 thickness cm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 240 294 Heat is transferred steadily to boiling water in the pan through its bottom The inner surface temperature of the bottom of the pan is given The temperature of the outer surface is to be determined Assumptions 1 Steady operating conditions eXist since the surface temperatures of the pan remain constant at the specified values 2 Thermal properties of the aluminum pan are constant Properties The thermal conductivity of the aluminum is given to be k 237 Wm C Analysis The heat transfer surface area is A 7H2 n01 m 2 00314 m2 Under steady conditions the rate of heat transfer through the bottom of the pan by conduction is 2 Mg 2 L L Substituting T 105 C 700W 2 237Wm C00314m2 2 0006m which gives T2 15 C 700w x 06 cm 295 The inner and outer glasses of a double pane window with a 1cm air space are at specified temperatures The rate of heat transfer through the window is to be determined Assumptions 1 Steady operating conditions eXist since the surface temperatures of the glass remain constant at the specified values 2 Heat transfer through the window is onedimensional 3 Thermal properties of the air are constant 4 The air trapped between the two glasses is still and thus heat transfer is by conduction only Properties The thermal conductivity of air at room temperature is k 0026 Wm C Table 23 Analysis Under steady conditions the rate of heat transfer through the window by conduction is AT Qcond M T 0026 Wm C2gtlt 2 mam m 125W0125kW 18 C J Air 6 C 1cm V PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 41 296 Two surfaces of a at plate are maintained at speci ed temperatures and the rate of heat transfer through the plate is measured The thermal conductivity of the plate material is to be determined Assumptions 1 Steady operating conditions eXist since the surface temperatures of the plate remain constant at the specified values 2 Heat transfer through the plate is onedimensional 3 Thermal properties of the plate are constant Plate 2 cm A V Analysis The thermal conductivity is determined directly from the steady 1000C 0 C onedimensional heat conduction relation to be E 500 Wmz L QAL 500 Wm2002 m T1 T2 100 0 C QkA k 01 Wm C 297 Hot air is blown over a at surface at a specified temperature The rate of heat transfer from the air to the plate is to be determined Assumptions 1 Steady operating conditions eXist 2 Heat transfer by radiation E 80 C is not considered 3 The convection heat transfer coefficient is constant and gt A1r un1form over the surface gt AnalySts Under steady cond1trons the rate of heat transfer by convectron 1s gt QCODV 55 Wm2 C2 X 4 m280 30 C 22ooow 22 kW 298 A person is standing in a room at a speci ed temperature The rate of heat transfer between a person and the surrounding air by convection is to be determined Assumptions 1 Steady operating conditions eXist 2 Heat transfer by radiation is not considered 3 The environment is at a uniform temperature Analysis The heat transfer surface area of the person is A nDL 703 m175 m 1649 m2 Under steady conditions the rate of heat transfer by convection is Qconv hAAT 10 Wm2 C1649 m2 34 20 c 231 w PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 242 299 A spherical ball whose surface is maintained at a temperature of 110 C is suspended in the middle of a room at 20 C The total rate of heat transfer from the ball is to be determined Assumptions 1 Steady operating conditions eXist since the ball surface and the surrounding air and surfaces remain at constant temperatures 2 The thermal properties of the ball and the convection heat transfer coef cient are constant and uniform Properties The emissivity of the ball surface is given to be 8 08 Analysis The heat transfer surface area is A 702 7r009 m2 002545 m2 Under steady conditions the rates of convection and radiation heat transfer are Qconv hAAT 15 Wm2 C002545 m2110 20 C 3435 W Q39rad 80 ATS4 T04 08002545 m2567 gtlt108Wm2 K4383 K4 293 104 1633 W Therefore Q39total Qconv Q39rad 3435 1633 507 w PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 243 2100 quot Reconsider Prob 299 Using EES or other software investigate the effect of the convection heat transfer coef cient and surface emissivity on the heat transfer rate from the ball Let the heat transfer coef cient vary from 5 Wm2 C to 30 Wm2 C Plot the rate of heat transfer against the convection heat transfer coef cient for the surface emissivities of 01 05 08 and 1 and discuss the results Analysis The problem is solved using EES and the solution is given below quotGivenquot D009 m TsConvertTempCK1 10 TfConvertTempCK20 h15 WmA2C epsilon08 quotPropertiesquot sigma567E8 WmA2KA4 quotAnalysisquot ApiD 2 QdotconvhATsTf QdotradepsilonsigmaATsquot4Tf 4 QdottotaQdotconvQdotrad h Qtotal 90 I WmzC W 5 278 80 1 8 AZ 125 4498 7 15 507 a 39 gzy quot39 60 quot I Q r 175 5643 a y 20 6216 g 32V 225 6788 0 50 A 332 30 8506 30 5 1O 15 20 25 3O hWm2 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 244 2101 A 1000W iron is left on the iron board With its base exposed to the air at 23 C The temperature of the base of the iron is to be determined in steady operation Assumptions 1 Steady operating conditions eXist 2 The thermal properties of the iron base and the convection heat transfer coefficient Iron are constant and uniform 3 The temperature of the surrounding surfaces 1000 W is the same as the temperature of the surrounding air a All 23 C Properties The emissivity of the base surface is given to be 8 04 gt Analysis At steady conditions the 1000 W of energy supplied to the iron Will be dissipated to the surroundings by convection and radiation heat transfer Therefore Qtotal Qconv Qrad 1000 W where Qconv hAAT 20 Wm2 K002 m2 Ts 296 K 04Ts 296 K W and Qquotrad 80ATs4 T04 04002m2567 gtlt108Wm2 K4Ts4 296 K4 004536 gtlt108Ts4 296 K4 W Substituting 1000 W 04Ts 296 K 004536 gtlt108Ts4 296K4 Solving by trial and error gives Ts 1106 K 833 C Discussion We note that the iron Will dissipate all the energy it receives by convection and radiation when its surface temperature reaches 1106 K 2102 A hot water pipe at 80 C is losing heat to the surrounding air at 5 C by natural convection With a heat transfer coef cient of 25 W m2 C The rate of heat loss from the pipe by convection is to be determined Assumptions 1 Steady operating conditions eXist 2 Heat transfer by radiation is not considered 3 The convection heat 80 C transfer coef cient is constant and uniform over the surface Analysis The heat transfer surface area is E gt L D 7 cm 3 2 A IZ39DL 7007 m18 m 3958m L18m Q Under steady conditions the rate of heat transfer by convection is Ail 50C Qconv hAAT 25 Wm2 C3958 m2 80 5 c 7422 W 742 kW PROPRIETARY MATERIAL 2015 McGraWHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 2 45 2103 The backside of the thin metal plate is insulated and the front side is exposed to solar radiation The surface temperature of the plate is to be determined when it stabilizes Assumptions 1 Steady operating conditions eXist 2 Heat transfer through the insulated side of the plate is negligible 3 The heat transfer coefficient is constant and uniform over the plate 4 Heat loss by radiation is negligible Properties The solar absorptivity of the plate is given to be OL 08 Analysis When the heat loss from the plate by convection equals the solar radiation absorbed the surface temperature of the plate can be determined from 450 Wmz Qsolarabsorbed Qconv aQsolm hATs To 2 2 06 08 08 x A x 450 Wm 2 50 Wm CATs 25 250C Canceling the surface area A and solving for T gives Ts 32200 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 2104 2 46 quot Reconsider Prob 2103 Using EES or other software investigate the effect of the convection heat transfer coef cient on the surface temperature of the plate Let the heat transfer coefficient vary from 10 Wm2 C to 90 Wm2 C Plot the surface temperature against the convection heat transfer coefficient and discuss the results Analysis The problem is solved using EES and the solution is given below quotGivenquot alpha08 qdotsoar450 WmA2 Tf25 C h50 WmA2C quotAnalysisquot qdotsoarabsorbedalphaqdotsolar qdotconvhTsTf qdotsolarabsorbedqdotconv h T8 65 WmzC C i 10 61 60 15 49 20 43 55 25 394 50 30 37 A 39 35 3329 g 45 m 45 33 l 40 l 50 322 35 I 55 3155 l 60 31 30 quotII 65 3054 quotI g 70 3014 25 75 298 10 20 30 40 50 60 70 80 90 80 295 h Imace 85 2924 90 29 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 247 2105 A spacecraft in space absorbs solar radiation While losing heat to deep space by thermal radiation The surface temperature of the spacecraft is to be determined when steady conditions are reached Assumptions 1 Steady operating conditions eXist since the surface temperatures of the wall remain constant at the speci ed values 2 Thermal properties of the spacecraft are constant Properties The outer surface of a spacecraft has an emissivity of 06 and an absorptivity of 02 Analysis When the heat loss from the outer surface of the spacecraft by radiation equals the solar radiation absorbed the surface temperature can be determlned from 1000 W m2 Qsolarabsorbed Qrad 4 4 atholar 0 ATs Tspace Ot 02 8 06 02x Ax 1000 Wmz 06x A x 567 gtlt108 Wm2 K4TS4 0 K4 Canceling the surface area A and solving for T gives Ts 2769K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 248 2106 339 Reconsider Prob 2105 Using EES or other software investigate the effect of the surface emissivity and absorptivity of the spacecraft on the equilibrium surface temperature Plot the surface temperature against emissivity for solar absorptivities of 01 05 08 and 1 and discuss the results Analysis The problem is solved using EES and the solution is given below quotGivenquot epsilon02 alpha06 qdotsoar1000 WmA2 Tf0 K quotspace temperaturequot quotPropertiesquot sigma567E8 WmA2KA4 quotAnalysisquot qdotsoarabsorbedalphaqdotsoar qdotradepsilonsigmaTsquot4Tf 4 qdotsoIarabsorbedqdotrad 8 Ts 650 K 3 01 648 600 02 5449 550 03 4924 g 04 4582 500 05 4334 g 450 06 4141 7 3 07 3984 I 400 08 3853 09 3741 350 1 3644 3003 Table for e 1 250 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 2 49 2107 A hollow spherical iron container is lled with iced water at 0 C The rate of heat loss from the sphere and the rate at which ice melts in the container are to be determined Assumptions 1 Steady operating conditions eXist since the surface temperatures of the wall remain constant at the speci ed values 2 Heat transfer through the shell is onedimensional 3 Thermal properties of the iron shell are constant 4 The inner surface of the shell is at the same temperature as the iced water 0 C Properties The thermal conductivity of iron is k 802 Wm C Table 23 The heat of fusion of water is at 1 atm is 3337 kJkg Analysis This spherical shell can be approximated as a plate of thickness 04 cm and surface area A 702 n 04 m2 05027 m2 Then the rate of heat transfer through the shell by conduction is AT 3 0 C kA 802 Wm c 05027 m2 30 235w Qcond L 0 9 m Considering that it takes 3337 kJ of energy to melt 1 kg of ice at 0 C the rate at which ice melts in the container can be determined from 39 30235le m Q S 00906k Is 106 hi 3337 kJkg 9 Discussion We should point out that this result is slightly in error for approximating a curved wall as a plain wall The error in this case is very small because of the large diameter to thickness ratio For better accuracy we could use the inner surface area D 392 cm or the mean surface area D 396 cm in the calculations PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 50 Review Problems 2108 The weight of the cabin of an elevator is balanced by a counterweight The power needed when the fully loaded cabin is rising and when the empty cabin is descending at a constant speed are to be determined Assumptions 1 The weight of the cables is negligible 2 The guide rails and pulleys are frictionless 3 Air drag is negligible Analysis a When the cabin is fully loaded half of the weight is balanced by the counterweight The power required to raise the cabin at a constant speed of 12 ms is 39 E mgV 400 kg981ms212 ms IN 2 lkw 471 kW At 1kgms 1000Nms If no counterweight is used the mass would double to 800 kg and the power would be 2x471 942 kW 9 When the empty cabin is descending and the counterweight is ascending there is mass imbalance of 400150 250 kg The power required to raise this mass at a constant speed of 12 ms is m Z 1 N lkw Counter Cabin W g mgV 250 kg981ms212 ms1k j 294 kW weight At gms2 1000 Nms If a friction force of 800 N develops between the cabin and the guide rails we will need 39 F 39 1 Hz IkW W iction IZ lf F ic onv 800 N12 msm 096 kW of additional power to combat friction which always acts in the opposite direction to motion Therefore the total power needed in this case is Wm1 W Wmc on 294 096 390 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 25 1 2109 A decision is to be made between a cheaper but inef cient natural gas heater and an expensive but ef cient natural gas heater for a house Assumptions The two heaters are comparable in all aspects other than the initial cost and efficiency Analysis Other things being equal the logical choice is the heater that will cost less during its lifetime The total cost of a system during its lifetime the initial operation maintenance etc can be determined by performing a life cycle cost analysis A simpler alternative is to determine the simple payback period The annual heating cost is given to be 1200 Noting that the existing heater is 55 ef cient only 55 of that energy and thus money is delivered to the house and the Gas Heater rest is wasted due to the inef ciency of the heater Therefore the monetary value of the 771 82 heating load of the house is 772 95 Cost of useful heat 55Current annual heating cost 055gtlt1200yr660yr This is how much it would cost to heat this house with a heater that is 100 efficient For heaters that are less efficient the annual heating cost is determined by dividing 660 by the ef ciency 82 heater Annual cost of heating Cost of useful heatEf ciency 660yr082 805yr 95 heater Annual cost of heating Cost of useful heatEf ciency 660yr095 695yr Annual cost savings with the efficient heater 805 695 110 Excess initial cost of the ef cient heater 2700 1600 1100 The simple payback period becomes Excess initial cost 1100 Annaul cost savings 110 yr Simple payback period 2 1 0 years Therefore the more efficient heater will pay for the 1100 cost differential in this case in 10 years which is more than the 8 year limit Therefore the purchase of the cheaper and less ef cient heater is a better buy in this case PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 52 2110E The energy contents unit costs and typical conversion efficiencies of various energy sources for use in water heaters are given The lowest cost energy source is to be determined Assumptions The differences in installation costs of different water heaters are not considered Properties The energy contents unit costs and typical conversion efficiencies of different systems are given in the problem statement Analysis The unit cost of each Btu of useful energy supplied to the water heater by each system can be determined from Unit cost of energy supplied Unit cost of useful energy Convers1on ef c1ency Substituting 3 3 Natural gas heater Unit cost of use il energy 039012 Ht 2 138 gtlt106 Btu 085 1025 Btu 22 1 Heating by oil heater Unit cost of 118le energy lgal gal 211 x 10 6 Btu 075 138700 Btu 1 1 h Electric heater Unit cost of use il energy 0 kW IkWh 358 x 10 6 Btu 090 3412 Btu Therefore the lowest cost energy source for hot water heaters in this case is natural gas 2111 A home owner is considering three different heating systems for heating his house The system with the lowest energy cost is to be determined Assumptions The differences in installation costs of different heating systems are not considered Properties The energy contents unit costs and typical conversion efficiencies of different systems are given in the problem statement Analysis The unit cost of each Btu of useful energy supplied to the house by each system can be determined from Unit cost of energy supplied Unit cost of useful energy Convers1on ef c1ency Substituting Natural gas heater Unit cost of 118le energy 13924therm 1 them 2 135 gtlt106 kJ 087 105500kJ Heating oil heater Unit cost of 118le energy 2393gal lgal 191 x 10 6 kJ 087 138500kJ Electric heater Unit cost of use il energy 03912kWh IkWh 333 gtlt106 kJ 10 3600 k Therefore the system with the lowest energy cost for heating the house is the natural gas heater PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 5 3 2112 It is estimated that 570000 barrels of oil would be saved per day if the thermostat setting in residences in winter were lowered by 6 F 33 C The amount of money that would be saved per year is to be determined Assumptions The average heating season is given to be 180 days and the cost of oil to be 40barrel Analysis The amount of money that would be saved per year is determined directly from 570000 barrelday180 daysyear110barrel 1 1 290001000 Therefore the proposed measure will save more than 11billion dollars a year in energy costs 2113 Caulking and weatherstripping doors and windows to reduce air leaks can reduce the energy use of a house by up to 10 percent The time it will take for the caulking and weatherstripping to pay for itself from the energy it saves is to be determined Assumptions It is given that the annual energy usage of a house is 1100 a year and the cost of caulking and weather stripping a house is 90 Analysis The amount of money that would be saved per year is determined directly from Money saved 1 100year0 10 110yr Then the simple payback period becomes Cost 90 0818yr Money saved 110yr Payback period Therefore the proposed measure will pay for itself in less than a year 2114E The work required to compress a gas in a gas spring is to be determined Assumptions All forces except that generated by the gas spring will be neglected Analysis When the expression given in the problem statement is substituted into the work integral relation and advantage is taken of the fact that the force and displacement vectors are collinear the result is 2 2 WZIFdSZJIQ n Skmmdx 1 1 x F Constant 1k 1k 1 x x2 x1 i 2001bfin1394 04 04 1r 7m 39 21n 39 1 14 112m 212451bfft 12451bfft 13 2001603tu 7781691bf it PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 2 54 2115E A man pushes a block along a horizontal plane The work required to move the block is to be determined considering a the man and b the block as the system Analysis The work applied to the block to overcome the friction is found by using the work integral 2 2 WIFdsIfWx2 x1 gtx 1 1 021001bf100 ft fW 20001bf 1 Btu 20001bfft 257Btu lt fW v The man must then produce the amount of work W W 257 Btu 2116 A diesel engine burning light diesel fuel that contains sulfur is considered The rate of sulfur that ends up in the eXhaust and the rate of sulfurous acid given off to the environment are to be determined Assumptions 1 All of the sulfur in the fuel ends up in the eXhaust 2 For one kmol of sulfur in the eXhaust one kmol of sulfurous acid is added to the environment Properties The molar mass of sulfur is 32 kgkmol Analysis The mass ow rates of fuel and the sulfur in the eXhaust are mm 336 kg airh AF 18 kg airkg iel mmel 1867 kg ielh mSulfur 750 x106n391fuel 750 gtlt106 1867 kgh 0014 kgh The rate of sulfurous acid given off to the environment is M H2503 mSulfur W 0014 kgh 0036kgh mH2SO3 Sulfur Discussion This problem shows why the sulfur percentage in diesel fuel must be below certain value to satisfy regulations 2117 Lead is a very toxic engine emission Leaded gasoline contains lead that ends up in the eXhaust The amount of lead put out to the atmosphere per year for a given city is to be determined Assumptions Entire lead is exhausted to the environment Analysis The gasoline consumption and the lead emission are Gasoline Consumption 70000 cars15 0 00 kmcar year85 L 100 km 2 8925 gtlt107 Lyear Lead Emission GaolineConsumptionmlead flead 8925 x 107 Lyear0 15 X 10393 kgL050 6694kgyear Discussion Note that a huge amounts of lead emission is avoided by the use of unleaded gasoline PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 255 2118 A TV set is kept on a specified number of hours per day The cost of electricity this TV set consumes per month is to be determined Assumptions 1 The month is 30 days 2 The TV set consumes its rated power when on Analysis The total number of hours the TV is on per month is Operating hours 6 hday30 days 180 h Then the amount of electricity consumed per month and its cost become Amount of electricity Power consumedOperating hours0 120 kW180 h 216 kWh Cost of electricity Amount of electricityUnit cost 216 kWh012kWh 259 per month Properties Note that an ordinary TV consumes more electricity that a large light bulb and there should be a conscious effort to turn it off when not in use to save energy 2119E The power required to pump a speci ed rate of water to a speci ed elevation is to be determined Properties The density of water is taken to be 624 lbmft3 Table A3E Analysis The required power is determined from W mgZ2 Z1 pvgZ2 Z1 3 11 f 624 lbmft 3 200 galmin 32174 fts 2 300 ft 15850 galmrn 32174lbmfts lkW 8342 lbf fts 8342 lbf fts 73756lbf s 113kW 2120 The power that could be produced by a water wheel is to be determined Properties The density of water is taken to be 1000 m3kg Table A3 Analysis The power production is determined from W n1gz2 Z1 pVgZ2 Z1 1kJkg 1000 kgm3 O3206O m3s981ms214 m 2 2 1000m s 2 0732kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 256 2121 The ow of air through a ow channel is considered The diameter of the wind channel downstream from the rotor and the power produced by the windmill are to be determined Analysis The speci c volume of the air is v 0287 kPam3kgK293 K P 100 kPa 08409 m3kg channel The diameter of the wind channel downstream from the rotor is AlVl AZVZ gt7le2 4V1 7022 4V2 L l 1 V I D22D1 17m 8mS 777m quot39 I V2 65 ms i The mass ow rate through the wind mill is 39 2 m AIVI W m 8 315 3661kgs v 408409 m Ikg The power produced is then 2 2 W 3661 kgs 8 ms2 65 ms2 lkJkg 398 kW 2 1000 m2s2 2122 The available head ow rate and efficiency of a hydroelectric turbine are given The electric power output is to be determined Assumptions 1 The ow is steady 2 Water levels at the reservoir and the discharge site remain constant 3 Frictional losses in piping are negligible Properties We take the density of water to be p 1000 kgm3 1 kgL Analysis The total mechanical energy the water in a dam possesses is equivalent to the potential energy of water at the free surface of the dam relative to free surface of discharge water and it can be converted to work entirely Therefore the power potential of water is its potential energy which is gz per unit mass and rhgz for a given mass ow rate 90 m J 1 novem emh pe gz 981 ms2 90 m 08829 kJkg quot 1000 m s l The mass ow rate is Generator 1 pV 1000 kgm365 m3s 65000kgs Then the maximum and actual electric power generation become W E39 me 65 000 kgs0 8829 kJk lM W 57 39 Mw max mech mach 39 g IOOOkJs 39 Welectric novemuwm 0845739 MW 482 MW Discussion Note that the power generation would increase by more than 1 MW for each percentage point improvement in the ef ciency of the turbine generator unit PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 257 2123 An entrepreneur is to build a large reservoir above the lake level and pump water from the lake to the reservoir at night using cheap power and let the water ow from the reservoir back to the lake during the day producing power The potential revenue this system can generate per year is to be determined Assumptions 1 The ow in each direction is steady and incompressible 2 The elevation difference between the lake and the reservoir can be taken to be constant and the elevation change of reservoir during charging and discharging is disregarded 3 Frictional losses in piping are negligible 4 The system operates every day of the year for 10 hours in each mode Properties We take the density of water to be p 1000 kgm3 Analysis The total mechanical energy of water in an upper reservoir relative to water in a lower reservoir is equivalent to the potential energy of water at the free surface of this reservoir relative to free surface of the lower reservoir Therefore the power potential of water is its potential energy which is gz per unit mass and rhgz for a given mass ow rate This also represents the minimum power required to pump water from the lower reservoir to the higher reservoir Wmaxturbine Wminpump Wideal mech mAemech mApe pVgAZ 1000 kgm32 m3s981ms240 m IN 2 lkw 1kgmS 1000NmS 7848 kW The actual pump and turbine electric powers are W Wideal 7848 kW u l t p mp 6 6C pumpmotor 1046 kW erbme nmrbmge mdeal O757848 kW 5886 kW Then the power consumption cost of the pump the revenue generated by the turbine and the net income revenue minus cost per year become Cost WP At gtlt Unit price 1046 kW365 X 10 hyear0 05kWh 190968year umpelect Revenue 2 WturbineAt gtlt Unit price 2 5886 kW365 x 10 hyear012kWh 257807year Net income Revenue Cost 257807 190968 66839year Discussion It appears that this pumpturbine system has a potential to generate net revenues of about 67000 per year A decision on such a system will depend on the initial cost of the system its life the operating and maintenance costs the interest rate and the length of the contract period among other things PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 258 2124 The pump of a water distribution system is pumping water at a speci ed ow rate The pressure rise of water in the pump is measured and the motor efficiency is specified The mechanical efficiency of the pump is to be determined Assumptions 1 The ow is steady 2 The elevation difference across the pump is negligible 3 Water is incompressible Analysis From the de nition of motor ef ciency the mechanical shaft power delivered by the he motor is W pumpshaft nmotorW electric 09015 kW 135 kW To determine the mechanical efficiency of the pump we need to know the increase in the mechanical energy of the uid as it ows through the pump which is AEmechluid Mensch ennui 164th Poll 17391er Pl gtv 0er Pl 0050 m3s300 100 kPa1i kP 3210kJs210kW am since m 2 pi V v and there is no change in kinetic and potential energies of the uid Then the pump efficiency becomes AE39mh uid 10 kw 77 Pump Wpumshaft 135kW 0741 or 741 Discussion The overall ef ciency of this pumpmotor unit is the product of the mechanical and motor efficiencies which is 09gtlt0741 0667 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 259 Fundamentals of Engineering FE Exam Problems 2125 In a hot summer day the air in a wellsealed room is circulated by a 050hp shaft fan driven by a 65 efficient motor Note that the motor delivers 050 hp of net shaft power to the fan The rate of energy supply from the fanmotor assembly to the room is a 0769 kJs b 0325 kJs c 0574 kJs d 0373 kJs e 0242 kJs Answer c 0574 kJs Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Eff065 Wfan05007457 quotkWquot EWfanEff quotkJsquot quotSome Wrong Solutions with Common Mistakesquot W1EWfanEff quotMultiplying by efficiencyquot W2EWfan quotIgnoring efficiencyquot W3EWfanEff07457 quotUsing hp instead of kWquot 2126 A fan is to accelerate quiescent air to a velocity to 12 ms at a rate of 3 m3min If the density of air is 115 kgm3 the minimum power that must be supplied to the fan is a 248 w b 72 w c 497 w d 216 w e 162 w Answer a 248 W Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values rho115 V12 Vdot3 quotm3squot mdotrhoVdot quotkgsquot WemdotVquot22 quotSome Wrong Solutions with Common Mistakesquot W1WeVdotVquot22 quotUsing volume flow ratequot W2WemdotVquot2 quotforgetting the 2quot W3WeVquot22 quotnot using mass flow ratequot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 260 2127 A 2kW electric resistance heater in a room is turned on and kept on for 50 min The amount of energy transferred to the room by the heater is a 2 k b 100 k c 3000 k d 6000 k e 12000 kJ Answer d 6000 k Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values We 2 quotkJsquot time5060 quotsquot WetotaWetime quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1EtotaWetime60 quotusing minutes instead of squot W2EtotaWe quotignoring timequot 2128 A 900kg car cruising at a constant speed of 60 kmh is to accelerate to 100 kmh in 4 s The additional power needed to achieve this acceleration is a 56 kW b 222 kW c 25 kW d 62 kW e 90 kW Answer a 56 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m900 quotkgquot V1 60 quotkmhquot V2100 quotkmhquot Dt4 quotSquot WamV236quot2 V136quot22000Dt quotkWquot quotSome Wrong Solutions with Common Mistakesquot W1WaV236quot2 V136quot22Dt quotNot using massquot W2WamV2quot2 V1quot22000Dt quotNot using conversion factorquot W3WamV236quot2 V136quot22000 quotNot using time intervalquot W4WamV236V1361000Dt quotUsing velocitiesquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 261 2129 The elevator of a large building is to raise a net mass of 400 kg at a constant speed of 12 ms using an electric motor Minimum power rating of the motor should be a 0 kW b 48 kW c 47 kW d 12 kW e 36 kW Answer c 47 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m400 quotkgquot V12 quotmsquot g981 quotmsZquot WgmgV1000 quotkWquot quotSome Wrong Solutions with Common Mistakesquot W1WgmV quotNot using gquot W2WgmgVquot22000 quotUsing kinetic energyquot W3WgmgV quotUsing wrong relationquot 2130 Electric power is to be generated in a hydroelectric power plant that receives water at a rate of 70 m3s from an elevation of 65 m using a turbine generator with an efficiency of 85 percent When frictional losses in piping are disregarded the electric power output of this plant is a 39 MW b 38 MW c 45 MW d 53 MW e 65 MW Answer b 38 MW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vdot70 quotm3squot z65 quotmquot g981 quotmsZquot Eff085 rho1000 quotkgm3quot WerhoVdotgzEff10A6 quotMWquot quotSome Wrong Solutions with Common Mistakesquot W1WerhoVdotzEff10A6 quotNot using gquot W2WerhoVdotg Eff10A6 quotDividing by efficiencyquot W3WerhoVdotgz10quot6 quotNot using efficiencyquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 262 2131 Consider a refrigerator that consumes 320 W of electric power when it is running If the refrigerator runs only one quarter of the time and the unit cost of electricity is 009kWh the electricity cost of this refrigerator per month 30 days is a 356 b 518 c 854 d 928 e 2074 Answer b 518 Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values We0320 quotkWquot Hours0252430 quothyearquot price009 quotkWhquot CostWehoursprice quotSome Wrong Solutions with Common Mistakesquot W1cost We2430price quotrunning continuouslyquot 2132 A 2kW pump is used to pump kerosene p 0820 kgL from a tank on the ground to a tank at a higher elevation Both tanks are open to the atmosphere and the elevation difference between the free surfaces of the tanks is 30 m The maximum volume ow rate of kerosene is a 83 US b 72 US c 68 US d 121 US e 178 US Answer a 83 US Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values W2 quotkWquot rho0820 quotkgLquot z30 quotmquot g981 quotmsZquot WrhoVdotg 1000 quotSome Wrong Solutions with Common Mistakesquot WW1Vdotg 1000 quotNot using densityquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 263 2133 A glycerin pump is powered by a 5 kW electric motor The pressure differential between the outlet and the inlet of the pump at full load is measured to be 211 kPa If the ow rate through the pump is 18 US and the changes in elevation and the ow velocity across the pump are negligible the overall ef ciency of the pump is a 69 b 72 c 76 d 79 e 82 Answer c 76 Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values We5 quotkWquot Vdot 0018 quotm3squot DP21 1 quotkPaquot EmechVdotDP EmechEffWe 2134 A 75 hp shaft compressor in a facility that operates at full load for 2500 hours a year is powered by an electric motor that has an efficiency of 93 percent If the unit cost of electricity is 006kWh the annual electricity cost of this compressor is a 7802 b 9021 c 12100 d 8389 e 10460 Answer b 9021 Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Wcomp75 quothpquot Hours2500 hyear Eff093 price006 lkWh WeWcomp07457HoursEff CostWeprice quotSome Wrong Solutions with Common Mistakesquot W1cost Wcomp07457HourspriceEff multiplying by efficiency W2cost WcompHourspriceEff not using conversion W3cost WcompHourspriceEff multiplying by efficiency and not using conversion W4cost Wcomp07457Hoursprice Not using efficiency PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 264 The following problems are based on the optional special topic of heat transfer 2135 A 10cm high and 20cm wide circuit board houses on its surface 100 closely spaced chips each generating heat at a rate of 008 W and transferring it by convection to the surrounding air at 25 C Heat transfer from the back surface of the board is negligible If the convection heat transfer coefficient on the surface of the board is 10 Wm2 C and radiation heat transfer is negligible the average surface temperature of the chips is a 26 C b 45 C c 15 C d 80 C e 65 C Answer e 65 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values A010020 quotmA2quot Q 100008 quotWquot Tair25 quotCquot h10 quotWmquot2Cquot Q hATsTair quotWquot quotSome Wrong Solutions with Common Mistakesquot Q hW1TsTair quotNot using areaquot Q h2AW2TsTair quotUsing both sides of surfacesquot Q hAW3TsTair quotAdding temperatures instead of subtractingquot Q100 hAW4TsTair quotConsidering 1 chip onlyquot 2136 A 50cm long 02cm diameter electric resistance wire submerged in water is used to determine the boiling heat transfer coef cient in water at 1 atm experimentally The surface temperature of the wire is measured to be 130 C when a wattmeter indicates the electric power consumption to be 41 kW Then the heat transfer coef cient is a 43500 Wm2 C b 137 Wm2 C c 68330 Wm2 C d 10038 Wm2 C e 37540 Wm2 C Answer a 43500 Wm2 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values L05 quotmquot D0002 quotmquot ApiDL quotmA2quot We41 quotkWquot Ts130 quotCquot Tf100 quotC Boiling temperature of water at 1 atmquot We hATsTf quotWquot quotSome Wrong Solutions with Common Mistakesquot We W1hTsTf quotNot using areaquot We W2hLpiDquot24TsTf quotUsing volume instead of areaquot We W3hATs quotUsing Ts instead of temp differencequot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 265 2137 A 3m2 hot black surface at 80 C is losing heat to the surrounding air at 25 C by convection with a convection heat transfer coef cient of 12 Wm2 C and by radiation to the surrounding surfaces at 15 C The total rate of heat loss from the surface is a 1987 W b 2239 W c 2348 W d 3451 W e 3811 W Answer d 3451 W Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values sigma567E8 quotWmquot2Kquot4quot eps1 A3 quotmA2quot hconv12 quotWm A20quot Ts80 quotCquot Tf25 quotCquot Tsurr15 quotCquot QconvhconvATsTf quotWquot QradepssigmaATs273quot4Tsurr273quot4 quotWquot QtotaQconvQrad quotWquot quotSome Wrong Solutions with Common Mistakesquot W1QQconv quotIgnoring radiationquot W2QQrad quotignoring convectionquot W3QQconvepssigmaATsquot4Tsurr 4 quotUsing C in radiation calculationsquot W4QQtotaA quotnot using areaquot 2138 Heat is transferred steadily through a 02m thick 8 m by 4 m wall at a rate of 24 kW The inner and outer surface temperatures of the wall are measured to be 15 C to 5 C The average thermal conductivity of the wall is a 0002 Wm C b 075 Wm C c 10 Wm C d 15 Wm C e 30 Wm C Answer d 15 Wm C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values A84 quotmquot2quot L02 quotmquot T1 1 5 quotCquot T25 quotCquot Q2400 quotWquot QkAT1 T2 L quotWquot quotSome Wrong Solutions with Common Mistakesquot QW1kT1T2L quotNot using areaquot QW2k2AT1T2L quotUsing areas of both surfacesquot QW3kAT1T2L quotAdding temperatures instead of subtractingquot QW4kALT1T2 quotMultiplying by thickness instead of dividing by itquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 266 2139 The roof of an electrically heated house is 7 m long 10 m wide and 025 m thick It is made of a at layer of concrete whose thermal conductivity is 092 Wm C During a certain winter night the temperatures of the inner and outer surfaces of the roof are measured to be 15 C and 4 C respectively The average rate of heat loss through the roof that night was a 41 w b 177 w c 4894 w d 5567 w e 2834 w Answer e 2834 W Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values A710 quotmquot2quot L025 quotmquot k092 quotWmCquot T1 1 5 quotCquot T24 quotCquot QcondkAT1 T2 L quotWquot quotSome Wrong Solutions with Common Mistakesquot W1QkT1T2L quotNot using areaquot W2Qk2AT1T2L quotUsing areas of both surfacesquot W3QkAT1T2L quotAdding temperatures instead of subtractingquot W4QkALT1T2 quotMultiplying by thickness instead of dividing by itquot 2140 2147 Design and Essay Problems PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 31 Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Cengel Michael A Boles McGrawHill 2015 Chapter 3 PROPERTIES OF PURE SUBSTANCES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 32 Pure Substances Phase Change Processes Property Diagrams 31C Yes Because it has the same chemical composition throughout 32C A vapor that is about to condense is saturated vapor otherwise it is superheated vapor 33C No 34C Because one cannot be varied while holding the other constant In other words when one changes so does the other one 35C Yes The saturation temperature of a pure substance depends on pressure The higher the pressure the higher the saturation or boiling temperature 36C At critical point the saturated liquid and the saturated vapor states are identical At triple point the three phases of a pure substance coexist in equilibrium 37C Yes 38C Case c when the pan is covered with a heavy lid Because the heavier the lid the greater the pressure in the pan and thus the greater the cooking temperature Property Tables 39C A given volume of water will boil at a higher temperature in a tall and narrow pot since the pressure at the bottom and thus the corresponding saturation pressure will be higher in that case 310C The molar mass of gasoline C8H18 is 114 kgkmol which is much larger than the molar mass of air that is 29 kgkmol Therefore the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air As a result the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 33 311C Yes Otherwise we can create energy by alternately vaporizing and condensing a substance 312C No Because in the thermodynamic analysis we deal with the changes in properties and the changes are independent of the selected reference state 313C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure It can be determined from hfg hg hf 314C Yes It decreases with increasing pressure and becomes zero at the critical pressure 315C Yes the higher the temperature the lower the hfg value 316C Quality is the fraction of vapor in a saturated liquidvapor mixture It has no meaning in the superheated vapor region 317 C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure the lower the hfg 318C The compressed liquid can be approximated as a saturated liquid at the given temperature Thus UT P E If T 319C Ice can be made by evacuating the air in a water tank During evacuation vapor is also thrown out and thus the vapor pressure in the tank drops causing a difference between the vapor pressures at the water surface and in the tank This pressure difference is the driving force of vaporization and forces the liquid to evaporate But the liquid must absorb the heat of vaporization before it can vaporize and it absorbs it from the liquid and the air in the neighborhood causing the temperature in the tank to drop The process continues until water starts freezing The process can be made more ef cient by insulating the tank well so that the entire heat of vaporization comes essentially from the water PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 320 Complete the following table for H2 0 T C P kPa u kJ kg Phase description 14361 400 1450 Saturated mixture 220 23196 26013 Saturated vapor 190 2500 80515 Compressed liquid 46621 4000 3040 Superheated vapor 321 Complete the following table for H 2 0 T C P kPa v m3 kg Phase description 50 1235 772 Saturated mixture 1436 400 04624 Saturated vapor 250 500 04744 Superheated vapor 1 10 350 0 001051 Compressed liquid PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 322 is to be repeated for refrigerant134a refrigerant22 and ammonia Analysis The problem is solved using EES and the solution is given below quotGivenquot T150 C v1772 mA3kg P2400 kPa x21 T3250 C P3500 kPa T41 10 C P4350 kPa quotAnalysisquot Fluid39steamiapws39 quotChange the Fluid to R134a R22 and Ammonia and solvequot P1pressureFluid TT1 vv1 x1qualityFuid TT1 vv1 T2temperatureFuid PP2 xx2 v2voumeFuid PP2 xx2 v3voumeFuid PP3 TT3 x3qualityFuid PP3 TT3 v4voumeFuid PP4 TT4 x4qualityFuid PP4 TT4 quotx 100 for superheated vapor and x 100 for compressed liquidquot SOLUTION for water T C P kPa x v kgm3 5000 1235 06419 772 14361 40000 1 04624 25000 50000 100 04744 11000 35000 100 0001051 SOLUTION for R134a T C P kPa x v kgm3 5000 341 100 772 891 40000 1 005127 25000 50000 11000 35000 100 008666 35 39539 Problem 321 is reconsidered The missing properties of water are to be determined using EES and the solution PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission SOLUTION for R22 TC PkPa x vkgm3 5000 402 100 772 656 40000 1 005817 25000 50000 100 009959 11000 35000 100 0103 SOLUTION for Ammonia T C P kPa X v kgm3 5000 2040 100 772 189 40000 1 03094 25000 50000 100 05076 11000 35000 100 05269 700 Steam 600 500 400 T C 300 200 100 700 40 50 s kJkgK Steam 600 500 400 T C 300 200 100 8600 kPa 2600 kPa 500 kPa 45 kPa o I I I I 10394 10393 10392 10391 100 101 102 v m3kg 36 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 37 39339 n E a 100 I 10393 10392 10391 100 101 102 v makg 5 Steam 10 I I l I I I I I I I I 104 250 C 3 10 170 C N O n i 2 110 C n 10 75 C 101 A I I I I I I I I I I I 0 500 1000 1500 2000 2500 3000 h kJkg Steam l I I I I I l I I I I I I I I I I I I 8600 kPa 3500 2600 kPa 39 3000 500 kPa 2500 39339 i 2000 i 1500 1000 500 0 I 00 10 20 30 40 50 60 70 80 90 100 s kJkgK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 323 Complete the following table for H2 0 38 T C P kPa v m3 kg Phase description 140 36153 005 Saturated mixture 15546 550 0001097 Saturated liquid 125 750 0 001065 Compressed liquid 500 2500 0140 Superheated vapor 324E Complete the following table for H2 0 T F P psia u Btu lbm Phase description 300 6703 782 Saturated mixture 26722 40 23602 Saturated liquid 500 120 11744 Superheated vapor 400 400 37384 Compressed liquid PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 325E A 39 Problem 324E is reconsidered The missing properties of water are to be determined using EES and the solution is to be repeated for refrigerant134a refrigerant22 and ammonia Analysis The problem is solved using EES and the solution is given below quotGivenquot T1300 F u1782 BtuIbm P240 psia x20 T3500 F P3120 psia T4400 F P4420 psia quotAnalysisquot Fuid39steamiapws39 P1pressureFluid TT1 uu1 x1qualityFuid TT1 uu1 T2temperatureFuid PP2 xx2 u2intenergyFluid PP2 xx2 u3intenergyFuid PP3 TT3 x3quaityFuid PP3 TT3 u4intenergyFluid PP4 TT4 x4qualityFuid PP4 TT4 quotx 100 for superheated vapor and x 100 for compressed liquidquot Solution for steam T F P psia x u BtuIbm 300 67028 06173 782 2672 40 0 236 500 120 100 1 174 400 400 100 3738 Solution for R134a T F P psia x u BtuIbm 300 782 2901 40 0 2124 500 120 100 2035 400 400 100 1736 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission Solution for R22 T F P psia x u BtuIbm 300 782 1534 40 0 7808 500 120 100 2401 400 400 100 2187 Solution for ammonia T F P psia x u BtuIbm 300 782 1167 40 0 6347 500 120 100 7855 400 400 100 727 310 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 326 Complete the following table for Refrigerant134a 311 T C P kPa v m3 kg Phase description 4 320 0000764 Compressed liquid 10 41489 00065 Saturated mixture 3345 850 002409 Saturated vapor 60 600 004632 Superheated vapor 327E Complete the following table for Refrigerant134a T F P psia h Btu lbm x Phase description 6589 80 78 0566 Saturated mixture 15 29759 6992 06 Saturated mixture 10 70 1536 Compressed liquid 160 180 12946 Superheated vapor 110 16116 11725 10 Saturated vapor 328 A rigid tank contains steam at a speci ed state The pressure quality and density of steam are to be determined Properties At 220 C uf 0001190 m3kg and u 008609 m3kg Table A4 Analysis a Two phases coexist in equilibrium thus we have a saturated liquidvapor mixture The pressure of the steam is the saturation pressure at the given temperature Then the pressure in the tank must be the saturation pressure at the specified temperature 9 The total mass and the quality are determined as Vf 13gtlt18 m3 uf 0001190m kg Vg 23gtlt18 m3 m 1394k 8 ug 008609 m3kg g ml 2 mf mg 50421394 5181kg 5042 kg xampo0269 mt 5181 c The density is determined from Steam 18 m3 220 C v If xvg If 2 0001190 00269008609 0003474 m3kg 1 1 3 2878k m p 1 0003474 9 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 312 329 A pistoncylinder device contains R134a at a speci ed state Heat is transferred to R134a The final pressure the volume change of the cylinder and the enthalpy change are to be determined Analysis a The final pressure is equal to the initial pressure which is determined from m 12k 81 m 2 pf 88kPa gxg 2 S N 2 904kPa 70 4 7r025 m 4 1000kgms PZZPIZPatm b The speci c volume and enthalpy of R134a at the initial state of 904 kPa and 10 C and at the nal state of 904 kPa and 15 C are from EES v1 02302 m3kg h1 24777 kJkg u2 02544 m3kg h2 26818 kJkg The initial and the nal volumes and the volume change are R134a r V1 2 mv1 085 kg02302 m3kg 01957 m3 085 kg Q v2 mu2 085 kg02544 m3kg 02162 m3 10 C AV 02162 01957 2 00205m3 c The total enthalpy change is determined from AH mh2 h1085 kg268 18 24777 kJkg 174kJkg 330E The temperature of R134a at a speci ed state is to be determined Analysis Since the speci ed speci c volume is higher than 18 for 80 psia this is a superheated vapor state From R134a tables P 80 psia 3 T80 F TableA 13E v 06243 it lbm 331 A rigid container that is filled with R134a is heated The nal temperature and initial pressure are to be determined Analysis This is a constant volume process The speci c volume is v 139 4 3 R134a v1v2 2 z 01348m3kg 4ooc m 10 kg 10 kg 3 The initial state is determined to be a mixture and thus the pressure is the 1348 m saturation pressure at the given temperature P1 PS40C 5125kPa TableAll P A The final state is superheated vapor and the temperature is determined by 2 interpolation to be 3 T2 663 C TableA 13 1 12 01348m kg gt V PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 313 332 The enthalpy of R134a at a speci ed state is to be determined Analysis The speci c volume is v 9m3 V m 300kg 003 m3kg Inspection of Table A11 indicates that this is a mixture of liquid and vapor Using the properties at 10 C line the quality and the enthalpy are determined to be v vf 003 00007929 m3kg 3 06001 Vfg 0049466 00007929 m kg x h h f thg 6542 0600119080 1799kJkg 333 The specific volume of R134a at a specified state is to be determined Analysis Since the given temperature is higher than the saturation temperature for 200 kPa this is a superheated vapor state The speci c volume is then P 200 kPa 11 47 3k b1A13 TZZSOC v 0 6 m gCTae 334 The average atmospheric pressure in Denver is 834 kPa The boiling temperature of water in Denver is to be determined Analysis The boiling temperature of water in Denver is the saturation temperature corresponding to the atmospheric pressure in Denver which is 834 kPa 335E The temperature in a pressure cooker during cooking at sea level is measured to be 250 F The absolute pressure inside the cooker and the effect of elevation on the answer are to be determined Assumptions Properties of pure water can be used to approximate the properties of juicy water in the cooker Properties The saturation pressure of water at 250 F is 2984 psia Table A4E The standard atmospheric pressure at sea level is 1 atm 147 psia Analysis The absolute pressure in the cooker is simply the saturation pressure at the cooking temperature P abs Psat25PF 2984 PSia H20 It 1s equ1valent to 25001 1 Pabs 2984 psiaa tm 203 atm 7 ps1a The elevation has no effect on the absolute pressure inside when the temperature is maintained constant at 250 F PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 314 336E A springloaded pistoncylinder device is filled with R134a The water now undergoes a process until its volume increases by 50 The final temperature and the enthalpy are to be determined Analysis From Table A11E the initial specific volume is v1 If xlvfg 001143 08044286 001143 2 35452 it 3lbm and the initial volume will be V1 2 mvl 013 lbm35452 ft 3lbm 04609ft 3 P A With a 50 increase in the volume the final volume will be 2 v2 141 1504609ft3 206913ft3 V C The distance that the piston moves between the initial and final conditions is Ax V2 V1 06813 04609rt3 2 2 202934ft AI 70 4 7r1 4 As a result of the compression of the spring the pressure difference between the initial and final states is AP kAx 437lbf7in02934gtlt12in Ap Ap 7zD24 7r12in2 1152lbf in2 1152psia The initial pressure is P1 Psat 30 F 2 9869 psia TableA 11E The final pressure is then P2 P1 AP 9869 1152 1102psia and the nal specific volume is 06813 3 m 013lbm v2 25318 31bm At this final state the temperature and enthalpy are P2 1102 psia T2 1047 F om EES v2 25318 3lbm h 1247Btulbm Note that it is very dif cult to get the temperature and enthalpy readings from Table A13E accurately PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 315 337E A pistoncylinder device that is lled with water is cooled The nal pressure and volume of the water are to be determined Analysis The initial specific volume is v1 24264 it 3 3 u 24264 lbm 1 m llbm H20 600 F This is a constantpressure process The initial state is determined to be superheated 1 lbm vapor and thus the pressure is determined to be 24264 ft3 T1 600 F 3 P1 P2 250pSIa TableA 6E v1 24264 it lbm P A The saturation temperature at 250 psia is 4001 F Since the nal temperature is less than this temperature the final state is compressed liquid Using the 2 incompressible liquid approximation 2 V 2001 001663 it 3lbm TableA4E V C The final volume is then v2 mu2 11bm001663it 3lbm 001663ft3 338 The volume of a container that contains water at a specified state is to be determined Analysis The speci c volume is determined from steam tables by interpolation to be P 100 kPa 3 TZISOOC v19367m lkg TableA6 Water 3 kg The volume of the container is then 100 kPa 3 3 150 C V mu 2 3 kg19367 m kg 2 581m 339 Water is boiled at sea level 1 atm pressure in a pan placed on top of a 3kW electric burner that transfers 60 of the heat generated to the water The rate of evaporation of water is to be determined Properties The properties of water at 1 atm and thus at the saturation temperature of 100 C are hfg 22564 kJkg Table A 4 Analysis The net rate of heat transfer to the water is Q060gtlt3kW18kW H20 100 C Noting that it takes 22564 k of energy to vaporize 1 kg of saturated liquid water the rate of evaporation of water is determined to be Q 18 kJs m 080gtlt10 3 k s2872k h evapmu hfg 22564 kJkg g g PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 316 340 Water is boiled at 1500 m 845 kPa pressure in a pan placed on top of a 3kW electric burner that transfers 60 of the heat generated to the water The rate of evaporation of water is to be determined Properties The properties of water at 845 kPa and thus at the saturation temperature of 95 C are hfg 22696 kJkg Table A4 Analysis The net rate of heat transfer to the water is Q060gtlt3kW18 kW H20 95 C Noting that it takes 22696 kJ of energy to vaporize 1 kg of saturated liquid water the rate of evaporation of water is determined to be Q 18 kJs m 0793x103 s 2855k h evaporatrcn hfg 226 kJkg kg 9 341 A rigid container that is lled with R134a is heated The temperature and total enthalpy are to be determined at the initial and final states Analysis This is a constant volume process The speci c volume is V 0 014 3 R134a v1 2 v2 A 00014 m3kg 300 kPa m 10 kg 10 kg 14 L The 1n1t1al state 1s deterrmned to be a mrxture and thus the temperature 1s the saturation temperature at the given pressure From Table A12 by interpolation T1 Tsat300kPa 051 C P A Using EES we would get 065 C Then 2 v v 00014 00007735 3k x1 1 f m3g 20009351 V fg 0067776 00007735 m kg 1 h h x h 5271 000935119817 5456 kJkg 1 f 1 f8 V C The total enthalpy is then H1 2 mh1 10 kg5456 kJkg 5456kJ The final state is also saturated mixture Repeating the calculations at this state T2 Tsat600kPa 21 55 C V2 Vf 00014 00008198 m3kg 3 001731 vfg 0034335 00008198 m kg x2 h h x h 8150 00173118095 8464 kJkg 2 f 2 f8 H2 2 mh2 10 kg8464 kJkg 8464kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 317 342 A pistoncylinder device that is filled with R134a is cooled at constant pressure The final temperature and the change of total internal energy are to be determined Analysis The initial specific volume is 1 12322 m3 3 v 012322m k 200 kPa The initial state is superheated and the internal energy at this state is 100 kg 12322 m3 P1 200 kPa 3 ul 2 26308 kJkg TableA 13 v1 012322 m lkg P A The final specific volume is 012322 3 k 1 v2 i 006161 m3kg 2 2 2 This is a constant pressure process The final state is determined to be saturated V C mixture Whose temperature is The internal energy at the nal state is Table A12 V2 Vf OO6161 OOOO7532m3kg O 6135 vfg 0099951 00007532m3kg 39 x2 u2 uf x214 fg 3826 O613518625 15252 kJkg Hence the change in the internal energy is Au uz u1 15252 26308 1106kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 318 343 A pistoncylinder device tted with stops contains water at a specified state Now the water is cooled until a nal pressure The process is to be indicated on the T vdiagram and the change in internal energy is to be determined Analysis The process is shown on T Vdiagram The internal energy at the initial state is P1 200 kPa ul 2 28088 kJkg Table A 6 T1 300 C Water State 2 is saturated vapor at the initial pressure Then 200 kPa 300 C v2 08858m3kg TableA5 x2 I sat vapor Process 23 is a constantvolume process Thus it 15086kJk ableA5 3 2 08858m3kg 3 g T The overall change in internal energy is An M1 u3 28088 15086 1300kJkg 344 Saturated steam at Tsat 40 C condenses on the outer surface of a cooling tube at a rate of 130 kgh The rate of heat transfer from the steam to the cooling water is to be determined Assumptions 1 Steady operating conditions eXist 2 The condensate leaves the condenser as a saturated liquid at 30 C Properties The properties of water at the saturation temperature of 40 C are hfg 24060 kJkg Table A4 Analysis Noting that 24060 kJ of heat is released as 1 kg of saturated vapor at 40 C condenses the rate of heat transfer from the steam to the 40 C cooling water in the tube is determined directly from A D Q mevaphfg PL 35 m v 3 cm 3 130 kgh24060 kJkg 312780 kJh 86 9 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 319 345 The boiling temperature of water in a 5cm deep pan is given The boiling temperature in a 40cm deep pan is to be determined Assumptions Both pans are full of water Properties The density of liquid water is approximately p 1000 kgm3 Analysis The pressure at the bottom of the 5cm pan is the saturation pressure corresponding to the boiling temperature of 98 C 40 cm 5 cm P Psat98C 9439 kPa Table A 4 PI PI The pressure difference between the bottoms of two pans is 3 2 1 kPa AP pgh 1000 kgm 9807 ms 035 m 2 343 kPa 1000 kgm s Then the pressure at the bottom of the 40cm deep pan is P 9439 343 9782 kPa Then the boiling temperature becomes Tboiling sat9782 kPa Table A39s 346 A cooking pan is filled with water and covered with a 4kg lid The boiling temperature of water is to be determined Analysis The pressure in the pan is determined from a force balance on the lid PA P6th W Patm or P Patm m f 2 2101kpa4kg981nzs lkPa 2 P Wmg 7r0l m 1000 kgms 10225 kPa The boiling temperature is the saturation temperature corresponding to this pressure T I Tsat10225 kPa 1 Table PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 320 347 Prob 346 is reconsidered Using EES or other software the effect of the mass of the lid on the boiling temperature of water in the pan is to be investigated The mass is to vary from 1 kg to 10 kg and the boiling temperature is to be plotted against the mass of the lid Analysis The problem is solved using EES and the solution is given below quotGiven dataquot Patm101kPa Did20 cm mid4 kg quotSolutionquot quotThe atmospheric pressure in kPa varies with altitude in km by the approximate functionquot Patm101 3251002256zquot5256 quotThe local acceleration of gravity at 45 degrees latitude as a function of altitude in m is given byquot g980733210quot6zconvertkmm quotAt sea levelquot z0 quotkmquot AlidpiDidquot24convertcmquot2mquot2 Wlidmlidgconvertkgmsquot2N PlidWlidAlidconvertNmquot2kPa PwaterPlidPatm TwatertemperaturesteamiapwsPPwaterx0 mlid Twater 1001 1001 1002 1003 1004 1005 1006 1007 1007 1008 COCDVCDO IAOON L L O 1009 1008 1007 1006 1005 1004 Twater C 1003 1002 1001 100 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 321 110 100 mass of lid 4 kg I3water kPa O 1 2 3 4 5 6 7 8 9 2 km Effect of altitude on boiling pressure of water in pan with lid 105 100 masls of Ilidl 394 lg 90 85 80 75 O 1 2 3 4 5 6 7 8 9 2 km Twater C Effect of altitude on boiling temperature of water in pan with lid PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 322 348 A vertical pistoncylinder device is lled with water and covered with a 40kg piston that serves as the lid The boiling temperature of water is to be determined Analysis The pressure in the cylinder is determined from a force balance on the piston PA PmA W Patm l l 1 II I 4 k 1 2 100kPa 0 98 11215 lkpa 2 P 00150 m 1000 kgm s W mg 212615kPa The boiling temperature is the saturation temperature corresponding to this pressure T Tsat12615kPa 210620C Table A5 349 Water is boiled in a pan by supplying electrical heat The local atmospheric pressure is to be estimated Assumptions 75 percent of electricity consumed by the heater is transferred to the water Analysis The amount of heat transfer to the water during this period is Q fEelecttime 0752 kJs30 x 60 s 2700 kJ The enthalpy of vaporization is determined from Q 2700 k mbo 119 kg 2269 kJkg Using the data by a trialerror approach in saturation table of water Table A5 or using EES as we did the saturation pressure that corresponds to an enthalpy of vaporization value of 2269 kJkg is Psat which is the local atmospheric pressure 350 A rigid tank that is filled with saturated liquidvapor mixture is heated The temperature at which the liquid in the tank is completely vaporized is to be determined and the T vdiagram is to be drawn Analysis This is a constant volume process 1 Vm constant and the speci c volume is determined to be H20 3 90 C 3 1398m 2012m3kg m 15 kg When the liquid is completely vaporized the tank will contain saturated vapor only Thus T A 12 zug 2012 m3kg 2 The temperature at this point is the temperature that corresponds to I this 18 value 1 T 20290c Table A4 V C Tsatug012m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 323 351 A pistoncylinder device contains a saturated liquidvapor mixture of water at 800 kPa pressure The mixture is heated at constant pressure until the temperature rises to 200 C The initial temperature the total mass of water the nal volume are to be determined and the P vdiagram is to be drawn Analysis a Initially two phases coexist in equilibrium thus we have a saturated liquidvapor mixture Then the temperature in the tank must be the saturation temperature at the speci ed pressure T Tsat600kPa 215880C b The total mass in this case can easily be determined by adding the mass of each phase Vf 0005 m3 m f 4543 kg vf 0001101 m3kg P A V 0 3 mg g9 m32852kg Vg 03156 m kg m mf mg 2 4543 2852 7395 kg 1 2 c At the nal state water is superheated vapor and its speci c volume is V P2 600 kPa 3 03521 k T bl A6 T22200 C quot2 m gm Then v2 my 7395 kg03521 m3kg 2604 m3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 324 352 339 Prob 351 is reconsidered The effect of pressure on the total mass of water in the tank as the pressure varies from 01 MPa to 1 MPa is to be investigated The total mass of water is to be plotted against pressure and results are to be discussed Analysis The problem is solved using EES and the solution is given below P1600 kPa P2P1 T2200 C Vf1 0005 mquot3 Vg109 mquot3 spvsatf1voumeSteamiapws PP1x0 quotsat liq specific volume mA3kgquot spvsatg1voumeSteamiapwsPP1x1 quotsat vap specific volume mA3kgquot mf1Vf1spvsatf1 quotsat liq mass kgquot mg1Vg1spvsatg1 quotsat vap mass kgquot mtotmf1 mg1 V1Vf1 Vg1 spvol1V1mtot quotspecific volume1 mquot3quot T1temperatureSteamiapws PP1vspvol1quotCquot quotThe final volume is calculated from the specific volume at the final T and Pquot spvol2volumeSteamiapws PP2 TT2 quotspecific volume2 mA3kgquot V2mtotspvol2 1o6 Steam39APWS P1 kPa mtot kg 105 100 5324 200 5731 104 300 6145 400 6561 3 200 C 500 6978 g 0 133 39 600 7395 E P600 kPa 700 7812 102 800 823 900 8648 101 1000 9066 10 39 39 39 39 10393 10392 1o1 100 1o1 102 Vm3kg 95 9 85 39 I 8 I 39639 75 E 39 E 7 l E 65 6 39 55 5 I I I I I I I I I 100 200 300 400 500 600 700 800 900 1000 P1 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 325 353E A rigid tank contains water at a specified pressure The temperature total enthalpy and the mass of each phase are to be determined Analysis a The speci c volume of the water is g 5 F m 5 lbm At 20 psia vf 001683 ft3lbm and 8 20093 ft3lbm Table A12E Thus the tank contains saturated liquidvapor mixture since vf lt vlt vg and the temperature must be the saturation temperature at the speci ed pressure T Tsat20psia 10 3lbm b The quality of the water and its total enthalpy are determined from H20 V V x f 23 3930315186383 2 004897 5 lbm Vfg 39 39 20 psia h h f thg 19627 004897 X95993 24328 Btulbm H mh 5 lbm24328 Btulbm 12164 Btu c The mass of each phase is determined from mg 2 mt 004897X5 0245Ibm mf 2 mt mg 5 0245 4755Ibm 354E A rigid tank contains saturated liquidvapor mixture of R134a The quality and total mass of the refrigerant are to be determined Analysis At 50 psia Vf 001252 ft3lbm and vg 094791 ft3lbm Table A12E The volume occupied by the liquid and the vapor phases are 3 3 Vf lft and lg 4ft R134a Thus the mass of each phase is 5 ft3 V lit 3 50 psia mf f 2 3279881bm Vf 001252ft lbm V 3 m g L4221bm g vg 094909ft3lbm Then the total mass and the quality of the refrigerant are m mf mg 7988 422 8410 lbm m 4221b xz gz m005018 mt 84101bm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 326 355E Superheated water vapor cools at constant volume until the temperature drops to 25 0 F At the nal state the pressure the quality and the enthalpy are to be determined Analysis This is a constant volume process v m constant and the initial speci c volume is determined to be P1 18035121 v1 30433 it 3lbm Table A 6E T1 500 F H20 At 250 F uf 001700 ft3lbm and ug 13816 ft3lbm Thus at the 180 psia nal state the tank will contain saturated liquidvapor mixture since o If lt vlt vg and the nal pressure must be the saturation pressure at 500 F the final temperature P ll satmsoF 2984 ps1a T A 1 b The quality at the nal state is determined from V2 Vf 30433 001700 219 ufg 13816 001700 x2 c The enthalpy at the final state is determined from h hf thg 21863 0219x94541 4260 Btulbm V c PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 327 356E 39 quot Problem 355E is reconsidered The effect of initial pressure on the quality of water at the final state as the pressure varies from 100 psi to 300 psi is to be investigated The quality is to be plotted against initial pressure and the results are to be discussed Analysis The problem is solved using EES and the solution is given below T1500 F P1180 psia T2250 F v 1volumesteamiapwsTT1PP1 V2V1 P2pressuresteamiapwsTT2vv2 h2enthalpysteamiapwsTT2vv2 x2quaitysteamiapwsTT2vv2 P1 X2 Steam psia 1400 I I I 100 04037 1200 1222 03283 1444 02761 1000 1667 02378 1889 02084 I 800 2111 01853 9 2333 01665 39 600 2556 0151 39 2778 01379 4 f 300 01268 200 O 102 104 v 1131me 045 04 035 03 quot39 N x 025 02 015 01 I I I I I 1 00 1 4O 1 80 220 260 300 P11 psia PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 328 357 Superheated steam in a pistoncylinder device is cooled at constant pressure until half of the mass condenses The final temperature and the volume change are to be determined and the process should be shown on a T vdiagram Analysis 9 At the final state the cylinder contains saturated liquid vapor mixture and thus the nal temperature must be the saturation temperature at the final pressure H20 0 The quality at the final state is speci ed to be x2 05 The specific 2000C volumes at the initial and the nal states are 05 MPa P 05 MP 1 o a u 042503 m3kg Table A 6 T1 200 C T A VZZVf szfg 0001093 05 x 037483 0001093 2 01880 m3kg 2 Thus AV 2 mv2 v1 06 kg0 1880 042503m3kg 014222m3 gt V 358 The water in a rigid tank is cooled until the vapor starts condensing The initial pressure in the tank is to be determined Analysis This is a constant volume process 1 Vm constant and the T 0C A 1n1t1al spec1f1c volume 1s equal to the nal spec1 c volume that 1s 1 v1 2 v2 ug124oc 079270 m3kg Table A 4 25 quot H20 39 th t t d 39 t 150 C Th f o s1nce e vapor s ar s con ens1ng a en rom T1 250 C 15 Table A6 101 2 T1 250 C V 3 P1 030 MPa 11 079270 m lkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 329 359 Heat is supplied to a pistoncylinder device that contains water at a speci ed state The volume of the tank the nal temperature and pressure and the internal energy change of water are to be determined Properties The saturated liquid properties of water at 200 C are If 0001157 m3kg and uf 85046 kJkg Table A4 Analysis a The cylinder initially contains saturated liquid water The volume of the cylinder at the initial state is v1 mu1 14 kg0001157 m3kg 0001619 m3 The volume at the nal state is v 40001619 0006476m3 Water 9 The nal state properties are 14 kg 200 C Q U 64 6 3 sat liq v2 M 0004626 m3 kg m 14kg 0 004626 3k T2 2 371300 quot2 139 m gP2 21 367kPa Table A4 or A5 or EES x2 u2 22015 kJkg c The total internal energy change is determined from AU mu2 u1 14 kg22015 85046kJkg 1892kJ 360E The error involved in using the enthalpy of water by the incompressible liquid approximation is to be determined Analysis The state of water is compressed liquid From the steam tables P 3000 psia h 37841 Btulbm Table A 7E T 400 F Based upon the incompressible liquid approximation P 3000 psia h g h 37504 Btulbm Table A 4E T 400 F f 400 F The error involved is Percent Error 2 x100 089 37841 which is quite acceptable in most engineering calculations PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 330 361 A pistoncylinder device that is filled with R134a is heated The volume change is to be determined Analysis The initial specific volume is P1 60 kPa 3 T 20 C 1 033608 m Ikg TableAl3 1 R 134a and the initial volume is 60 kPa 20 C v1 mu1 0100 kg033608 m3kg 0033608 m3 100 g At the final state we have 2 050410m Ikg TableA l3 T2 100 C v 0100k 050410 31lt 0050410 3 1 2 2 quotW2 g m g m The volume change is then AV 2 V2 v1 0050410 0033608 00168m3 gt v 362 A rigid vessel is lled with refrigerant134a The total volume and the total internal energy are to be determined Properties The properties of R134a at the given state are Table A13 P2500 kPa u32991kJkg R134 T2120 C 120061687 m3kg a 8 kg Analysis The total volume and internal energy are determined from 500 kPa V mu 2 8 kg0061687 m3kg 0494 m3 120 C U mu 2 8 kg3299l kJkg 2639 k PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 331 363 Heat is supplied to a rigid tank that contains water at a speci ed state The volume of the tank the final temperature and pressure and the internal energy change of water are to be determined Properties The saturated liquid properties of water at 200 C are If 0001157 m3kg and uf 85046 kJkg Table A4 Analysis 61 The tank initially contains saturated liquid water and air The volume occupied by water is V1 2 mvl 14 kg0001157 m3kg 0001619 m3 which is the 25 percent of total volume Then the total volume is determined from 1 v 0001619 2 0006476m3 025 9 Properties after the heat addition process are g 0006476 m3 2 0004626 m3 kg m 14kg 0 004626 3k T2 2 371306 V2 1 m gP2 21 367kPa Table A4 or A5 or EES x2 u2 22015 kJkg c The total internal energy change is determined from AU 2 mu2 u114 kg22015 85046 kJkg 1892kJ 364 A pistoncylinder device that is initially lled with water is heated at constant pressure until all the liquid has vaporized The mass of water the nal temperature and the total enthalpy change are to be determined and the T vdiagram is to be drawn Analysis Initially the cylinder contains compressed liquid since P gt Psat40oc that can be approximated as a saturated liquid at the specified temperature Table A4 v1 uf4ooc 0001008 m3kg 11 E Z H20 61 The mass 1s deterrmned from 40 C V 0050 3 quot12 1 m3 24961kg 200kPa V1 0001008m kg T b At the nal state the cylinder contains saturated vapor and thus I the nal temperature must be the saturation temperature at the nal pressure T Tsat200kPa 1 2021OC c The final enthalpy is h2 hg 200 kpa 27063 kJkg Thus AH mh2 h1 4961 kg27063 16753kJkg 125950 kJ V t PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 332 Ideal Gas 365C A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to its critical temperature and pressure 366C Ru is the universal gas constant that is the same for all gases whereas R is the specific gas constant that is different for different gases These two are related to each other by R Ru M where M is the molar mass of the gas 367 C Propane molar mass 441 kgkmol poses a greater fire danger than methane molar mass 16 kgkmol since propane is heavier than air molar mass 29 kgkmol and it will settle near the oor Methane on the other hand is lighter than air and thus it will rise and leak out 368 A rigid tank contains air at a speci ed state The gage pressure of the gas in the tank is to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table A1 P Analysis Treating air as an ideal gas the absolute pressure in the tank is determined from 2 5k 0287kP 3kK2 K g szRT g a 13 g X 98 210691kPa Air Thus the gage pressure is 25 C Pg 2 P rgm 210691 97 9721 kPa 369E The temperature in a container that is lled with oxygen is to be determined Assumptions At specified conditions oxygen behaves as an ideal gas Properties The gas constant of oxygen is R 03353 psiaft3lbmR Table A1E Analysis The definition of the speci c volume gives V3 3 15 3lbm m 21bm Using the ideal gas equation of state the temperature is T 80 psia15 it 3lbm 3 358R R 03353psiaft lbmR PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 333 370 The volume of a container that is filled with helium at a specified state is to be determined Assumptions At specified conditions helium behaves as an ideal gas Properties The gas constant of helium is R 20769 kJkgK Table A1 Analysis According to the ideal gas equation of state mRT 2 kg20769 kPa m3kg K27 273 K P 300 kPa 4154m3 V 371 The pressure and temperature of oxygen gas in a storage tank are given The mass of oxygen in the tank is to be determined Assumptions At specified conditions oxygen behaves as an ideal gas Pg 500 kPa Properties The gas constant of oxygen is R 02598 kPam3kgK Table A1 Analysis The absolute pressure of 02 is j PPgPaun50097597kPa 02 3 Treating 02 as an ideal gas the mass of 02 in tank is determined to be 597 kPa25 m3 1908kg m RT 02598 kPa m3kg K28 273K 372 A balloon is filled with helium gas The mole number and the mass of helium in the balloon are to be determined Assumptions At specified conditions helium behaves as an ideal gas Properties The universal gas constant is Ru 8314 kPam3kmolK The molar mass of helium is 40 kgkmol Table A1 Analysis The volume of the sphere is v gym gyms m3 3817 m3 Assuming ideal gas behavior the mole numbers of He is determined from PV 200 kPa3817 m3 N 3 3061 kmol RuT 8314 kPa m kmol K300 K Then the mass of He can be determined from m NM 3061 kmol40 kgkmol 123 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 334 quot Problem 372 is to be reconsidered The effect of the balloon diameter on the mass of helium contained in the balloon is to be determined for the pressures of a 100 kPa and b 200 kPa as the diameter varies from 5 m to 15 m The mass of helium is to be plotted against the diameter for both cases 373 If Analysis The problem is solved using EES and the solution is given below quotGiven Dataquot D9 m T27 C P200kPa Ru8314 kJkmoIK quotSolutionquot PVNRuT273 V4piD2quot33 mNMOLARMASSHeIium Dm mkg 5 2101 a 6111 3835 500 7222 6331 8333 9725 400 9444 1416 1056 1976 1167 2669 g 300 P 200 W 1278 3506 a 1389 4502 E 200 i 39 15 5672 P2oo kPa 100 39 P1OO kPa l 39 339 5 7 11 13 15 D m PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 335 374E An automobile tire is under in ated with air The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined Assumptions 1 At speci ed conditions air behaves as an ideal gas 2 The volume of the tire remains constant Properties The gas constant of air is R 03704 psiaft3lbmR Table AlE Analysis The initial and final absolute pressures in the tire are 0 13113 P1 P81 Patm 20 146 346 psia 90 F P2 P82 Rtm 30 146 446 psia 20 psig Treating air as an ideal gas the initial mass in the tire is Pv 4 39 3 m1 1 3 6 pSIaXO 53 009001bm RT1 03704 psia 3lbm R550 R Noting that the temperature and the volume of the tire remain constant the final mass in the tire becomes 2 446 psia053 3 0 1160mm m RT2 03704 psiait3lbm R550 R Thus the amount of air that needs to be added is Am 2 1112 m1 01160 00900 2 002601bm 375 Two rigid tanks connected by a valve to each other contain air at specified conditions The volume of the second tank and the nal equilibrium pressure when the valve is opened are to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table A1 Analysis Let39s call the first and the second tanks A and B Treating air as an ideal gas the volume of the second tank and the mass of air in the rst tank are determined to be RT 3 VB 2 m1 1 Z 3 kg0 287 kPa m kg K308 K 21326m3 A B P1 B 200 kPa Air 39 P v 350 kP 10 3 All mA 1 X m 4309kg V1m3 I m3kg RTl A kPa 39 m 39 T 10 C T 35 C Thus P 350 kPa P 200 kPa v VA VB 101326 2326 m3 mmA mB 430937309 kg Then the final equilibrium pressure becomes RT 3 P2 2 m 2 Z 7 309 kg0 287 kPa m kg K293 K 2 264 kPa V 2326 m3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 336 37 6 One side of a twosided tank contains an ideal gas While the other side is evacuated The partition is removed and the gas lls the entire tank The gas is also heated to a nal pressure The nal temperature is to be determined Assumptions The gas is speci ed as an ideal gas so that ideal gas relation can be used Analysis According to the ideal gas equation of state P 2 P1 V V 2V 2 3V 2 1 1 1 Ideal gas Eva2011ated Applying these 927 C 1 1 Q quot 1 2 m1 P1V1 P2V2 T1 T2 V1 V2 T1 T2 V 3V T2 T172 T171 3T1 3927 273 K 3600 K 332700 1 1 377 A pistoncylinder device containing argon undergoes an isothermal process The nal pressure is to be determined Assumptions At specified conditions argon behaves as an ideal gas Properties The gas constant of argon is R 02081 kJkgK Table Al Analysis Since the temperature remains constant the ideal gas equation gives Argon PM P2V2 15 kg m RT RT 39Plvl P2V2 004 m3 550 kPa wh1ch when solved for nal pressure becomes V1 V1 P2 P1 P1 05P1 05550 kPa 275 kPa V2 V1 378E A rigid tank contains slightly pressurized air The amount of air that needs to be added to the tank to raise its pressure and temperature to the recommended values is to be determined Assumptions 1 At speci ed conditions air behaves as an ideal gas 2 The volume of the tank remains constant Properties The gas constant of air is R 03704 psiaft3lbmR Table AlE Analysis Treating air as an ideal gas the initial volume and the nal mass in the tank are determined to be V mlRTl 201bm03704 psiait 3lbm R530 R 21963 3 P1 20 ps1a 39 1 1021 35 psia1963 it 3 Alr 20 bm m2 2 3 33731bm 20 psia RTZ 03704 ps1ait lbm R550 R 70 F Thus the amount of air added is Am m2 m1 3373 200 1373 lbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 337 Compressibility Factor 379C All gases have the same compressibility factor Z at the same reduced temperature and pressure 380C Reduced pressure is the pressure normalized with respect to the critical pressure and reduced temperature is the temperature normalized with respect to the critical temperature 381E The temperature of R134a is to be determined using the ideal gas relation the compressibility chart and the R134a tables Properties The gas constant the critical pressure and the critical temperature of refrigerant134a are from Table AlE R 010517 psia 31bmR Tc 6736 R PC 5887 psia Analysis a From the ideal gas equation of state T 400 psia0 1384 it 3lbm 3 526 R R 010517 psia1t lbm R 9 From the compressibility chart Fig A15a P 400 39 PR ps 1quotf 0679 PC 5887 ps1a 3 TR 2103 vactual 01384 it lbm5887 ps1a 1 115 RTC Pc 010517 psia 11 3lbm R6736 R Thus T TRTcr 103 x 6736 2 694 R c From the superheated refrigerant table Table A13E P 400 psia v 01384 11 3lbm T Z 240 F 700 R PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 338 382 The specific volume of steam is to be determined using the ideal gas relation the compressibility chart and the steam tables The errors involved in the rst two approaches are also to be determined Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kgK Tcr 6471 K Pcr 2206 MPa Analysis 61 From the ideal gas equation of state 3 WE 04615 kPa m kg 62315 K oo1917 m3kg 670error P 15000 kPa 9 From the compressibility chart Fig A15 10 MP PR i a 0453 H20 P6 2206 MPa Z Z O 65 15 MPa K 39 0 TR i 673 104 350 C T 6471 K Thus I 20161 065001917 m3kg 001246 m3kg 85 error 0 From the superheated steam table Table A6 5 3235153 1 2001148 m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 339 383 quot Problem 382 is reconsidered The problem is to be solved using the general compressibility factor feature of EES or other software The speci c volume of water for the three cases at 15 MPa over the temperature range of 350 C to 600 C in 25 C intervals is to be compared and the error involved in the ideal gas approximation is to be plotted against temperature Analysis The problem is solved using EES and the solution is given below P15 M PaConvertMPakPa TCesius 350 C TTCesius273 quotKquot TcriticalTCRTSteamiapws PcriticaPCRTSteamiapws vVom PtableP PcompPPideaIgasP TtabeT TcompTTideagasT vtabevoumeSteamiapwsPPtabeTTtabe quotEES data for steam as a real gasquot PtabepressureSteamiapws TTtablevv TsattemperatureSteamiapwsPPtablevv MMMOLARMASSwater Ru8314 kJkmoIK quotUniversal gas constantquot RRuMM quotkJkgK Particular gas constantquot PideagasvideagasRTideagas quotIdeal gas equationquot 2 COM PRESSTcompTcriticaPcompPcritica PcompvcompzRTcomp quotgeneralized Compressibility factorquot ErrorideaIgasAbsvtabIevideagasvtabeConvert ErrorcompAbsvtabevcompvtabeConvert Errorcomp 00 Errorideal gas 00 TCeIcius C 9447 6722 350 2725 4353 375 04344 3221 400 05995 2523 425 1101 2044 450 1337 1692 475 1428 1422 500 1437 121 525 1397 1039 550 1329 8976 575 1245 7802 600 70 I I I I I I I 60 0 Ideal Gas q Io I 50 EI Compressibiity Factor E h g 9 4o h gt m 39 Steam at 15 MPa 39 30 E C 8 8 20 h D h a D n 10 o I I I I I I I I I I I 300 350 400 450 500 550 600 TCelsius C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 340 384 The specific volume of steam is to be determined using the ideal gas relation the compressibility chart and the steam tables The errors involved in the rst two approaches are also to be determined Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kgK Tcr 6471 K 10cr 2206 MPa Analysis 61 From the ideal gas equation of state I 3 I v E 04615 kPa m lkg me K 009533m3kg 37 error P 3500 kPa 9 From the compressibility chart Fig A15 PRziz ZOJSQ H20 PC 2206 MPa Z 2 0961 35 MPa Tc 6471 K Thus I Zvidaall 0961009533 m3kg 009161m3kg 04 error 0 From the superheated steam table Table A6 5 22513g1 a v 009196m3kg 385 Somebody claims that oxygen gas at a specified state can be treated as an ideal gas with an error less than 10 The validity of this claim is to be determined Properties The critical pressure and the critical temperature of oxygen are from Table A1 Tc 1548 K and PC 508 MPa Analysis From the compressibility chart Fig A15 3 MP PRi a0591 02 PC 508 MPa T 160 K 2 079 3 MPa TR 2 21034 160K Tc 1548 K Then the error involved can be determined from Error 2 m 1 i 21 266 v Z 079 Thus the claim is false PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 341 386E Ethane in a rigid vessel is heated The nal pressure is to be determined using the compressibility chart Properties The gas constant the critical pressure and the critical temperature of ethane are from Table A1E R 03574 psia 3lbmR Tc 5498 R Rcr 708 psia Analysis From the compressibility chart at the initial state from Fig A 15 or EES We used EES throughout the solution T TRl 1 560R 1019 Tcr 5498R Zl 0977 i 50 ps1a O 0706 Ethane R1 n Pcr 708 PSla 50 psia Q The speci c volume does not change during the process Then 10001 Z RT 39 3 V1 2 V2 2 1 1 Z 0 9770 3574 ps1a lbm R560 R 39m 31bm P1 50 ps1a At the final state T R2 2 2 lOOOR 21819 Tcr 5498R V2actua1 3911 lbm VR2 Z 14 1 RTcrPcr 03574 psia it 3lbm R549 8 R708 psia Thus 2 RT 39 3 P2 2 2 2 100 3574ps1a R3 lbm RlOOO R 914p5Ia 2 39111 lbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 342 387 Ethylene is heated at constant pressure The speci c volume change of ethylene is to be determined using the compressibility chart Properties The gas constant the critical pressure and the critical temperature of ethane are from Table A1 R 02964 kPam3kgK Tcr 2824 K Pcr 512 MPa Analysis From the compressibility chart at the initial and final states Fig A15 1 T TR1 1 293K 2 1038 Tcr 2824K P 5 MP gt Z1 2 056 I PR1 1 a0977 Pcr 512MPa Ethylene T TR2 2 473K 21675 5MPa Q Tcr 2824KR gt 21 0961 20cc The speci c volume change is R 02964kPam3kgK 5000 kPa 00172m3kg 0961473 K 056293 K 388 The error involved in treating C02 at a speci ed state as an ideal gas is to be determined Properties The critical pressure and the critical temperature of C02 are from Table A1 Tel 2 3042 K and Per 2 739 MPa Analysis From the compressibility chart Fig A15 P MP PR 0947 Pcr 739 MPa C02 2 084 l 380 K 7 MPa R Tcr 3042 K 39 380 K Then the error involved in treating C02 as an ideal gas is Error 2m 21 21 2 0190 or 190 v Z 084 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 343 389 Water vapor is heated at constant pressure The final temperature is to be determined using ideal gas equation the compressibility charts and the steam tables Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kgK Tcr 6471 K Analysis 61 From the ideal gas equation PC 2206 MPa T2 2 T1 quot 2 2 350 273 K2 1246K 1 b The pressure of the steam is Water 350 C P1 P2 Psat350gtC 216529 kPa sat vapor From the compressibility chart at the initial state Fig A15 T TR1 1 2 2 0963 Tcr 6471KR P 16 529 MP Z1 2 0593 le 075 PR1 1 39 a 0749 Pcr 2206 MPa At the final state P P 0749 R2 R1 22 088 um 2le 2075 150 Thus P2412 2 szTc 16529kPa 1506471K 2 Z2R 22 P 088 22060kPa CI39 2826K c From the superheated steam table T1 350 C 3 v1 0008806 m Ikg Table A4 x1 1 P2 16529 kPa 3 T2 477 C 2 750 K from Table A6 or EES PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 3 44 390 Methane is heated at constant pressure The nal temperature is to be determined using ideal gas equation and the compressibility charts Properties The gas constant the critical pressure and the critical temperature of methane are from Table A1 R 05182 kPam3kgK Tcr 1911 K 1Dcr 464 MPa Analysis From the ideal gas equation T2 T1 V 2 300 K18 2 540K V 1 I I From the compressibility chart at the initial state Fig A15 Methane Q TR1 7711 1 1393011 2 157 1301341123 Cr 39 Z1 2 086 um 063 PI 10 MPa PR1 P 4 64 MPa CI At the final state 22 042 Fig A 15 UR2 218le 18063 21134 Thus P2112 1R2Tcr 10000kPa 11341911K 22R 22 P 042 4640 kPa CI39 1112K 2 Of these two results the accuracy of the second result is limited by the accuracy with which the charts may be read Accepting the error associated with reading charts the second temperature is the more accurate PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 345 391 C02 gas ows through a pipe The volume ow rate and the density at the inlet and the volume ow rate at the eXit of the pipe are to be determined Properties The gas constant the critical pressure and the critical temperature of C02 are Table A1 R 01889 kPam3kgK Tc 3042 K Pcr 739 MPa Analysis 3 MPa 500 K C02 gt 450 K 2 kgs a From the ideal gas equation of state 3 V mRT1 2 kgs01889 kPa m lkg K500 K 2006297m3kg 21error 1 P1 3000 kPa p1 i 3000 kPa 31 76 kgm3 21 error RT1 01889 kPa m3kg K500 K 39 RT 3 v m 2 2kgSX01889 kPa m lkg K450 K 005667m3kg 36 error P2 3000 kPa 9 From the compressibility chart EES function for compressibility factor is used pR Ziz zmm PC 739 MPa T 500 K Z1 2 09791 Tc 3042 K PR i 0407 PC 739 MPa T 450 K Z2 2 09656 2 48 T I T 3042K CT Z 39RT 3 Thus V1 2 1m 1 Z 0 97912kgs01889kPa m lkg K500 K 006165m3kg p1 3000 kPa 01 P1 3000 a 23244 kgm3 3 ZlRTl 0979101889 kPam lkg K500 K Z 39RT 3 V2 2 2m 2 Z 0 96562 kgs0 1889 kPa m Ikg K450 K 005472m3kg P2 3000 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 346 392 The pressure of R134a is to be determined using the ideal gas relation the compressibility chart and the R134a tables Properties The gas constant the critical pressure and the critical temperature of refrigerant134a are from Table A1 R 008149 kPam3kgK Tcr 3742 K Pcr 4059 MPa Analysis The speci c volume of the refrigerant is 0016773 3 v K m 0016773 m3kg m 1 kg R134a a From the 1deal gas equatlon of state 0016773 Inskg 3 o Pz z 008149 kPa m lkg3 K383 K Z1861 kPa 110 C 1 0016773 m Ikg 9 From the compressibility chart Fig A15 fl 383K 1023 Tcr 3742K PRO39 UR yam 0016773 m3kg 2 RTcrPcr 008149 kPa m3kg K3742 K4059 kPa Thus P PRPcr O394059 kPa 1583 kPa c From the superheated refrigerant table Table A13 T 110 C v 0016773 m3kgP 1600kpa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 347 Other Equations of State 393C The constant 61 represents the increase in pressure as a result of intermolecular forces the constant 9 represents the volume occupied by the molecules They are determined from the requirement that the critical isotherm has an in ection point at the critical point 394 The pressure of nitrogen in a tank at a speci ed state is to be determined using the ideal gas van der Waals and Beattie Bridgeman equations The error involved in each case is to be determined Properties The gas constant molar mass critical pressure and critical temperature of nitrogen are Table A1 R 02968 kPam3kgK M 28013 kgkmol Tcr 1262 K Rcr 339 MPa Analysis The speci c volume of nitrogen is v 2 3 3 7 m 00327 m3kg N2 m 100 kg 3 00327 m kg a From the 1deal gas equatron of state 175 K RT 02968 kP 3k K 175 K P a m f X 21588 kPa 55error V 00327 m kg 9 The van der Waals constants for nitrogen are determined from a 27R2TCZ 2702968 kPam3 kgK21262 K2 0175 m6 kPang 64Pc 643390 kPa RT 3 b c 2 0 2968 kPa m kg K1262K 2000138 m3kg 8P0 8 X 3390 kPa Then RT or 02968x175 0175 P 2 1495 kPa 07 error 11 2 00327 000138 003272 c The constants in the BeattieBridgeman equation are AA0 1 2 13623151 03902617 v 09160 B B0 1 2 005046 PM 005084 1 09160 2 132339 0 42x104m3K3km61 since 7 Mu 28013 kgkmol00327 m3kg 09160 m3kmol Substituting RT A P 32 1 L3VB 2 1 7T 1 14 1 4 2 83 X 725 1 4 2X10 3 09160005084 1323392 09160 09160x175 09160 2 1504 kPa 007 error PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 348 395 Methane is heated in a rigid container The nal pressure of the methane is to be determined using the ideal gas equation and the BenedictWebbRubin equation of state Analysis a From the ideal gas equation of state T P2 P1 2 80 kpa 573 K 1565kpa Methane T1 293 K 80 kPa Q 20 C 9 The speci c molar volume of the methane is 7 Run 8314kPam3kmolK293 K 11 2 3045 m3kmol P1 80 kPa Using the coef cients of Table 34 for methane and the given data the BenedictWebbRubin equation of state for state 2 gives RT C 1 bRuT a aa c 7 2 P Z Z BRuT 0 2 l jex lv 2 v2 0 2 A0 T3 73 76 WT 72 p 7 8314573 3045 6 004260x8314x573 18791 239286X10 J 1 0003380X8314x573 500 5732 30452 30453 500x1244x10 4 2578x105 00060 3 2 2 30456 3045 573 3045 1565kPa jexp 00060 30452 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 349 396E The temperature of R134a in a tank at a specified state is to be determined using the ideal gas relation the van der Waals equation and the refrigerant tables Properties The gas constant critical pressure and critical temperature of R134a are Table AlE R 01052 psia ft31bmR Tc 6736 R PC 5887 psia Analysis a From the ideal gas equation of state T 400 psia0 1144 11 3lbm 3 435 R R 01052 psiaft lbmR b The van der Waals constants for the refrigerant are determined from a 27R2T 270 1052 psia 11 3lbm R26736 R2 64P 645887 psia RT 01052 psia ft 3lbm R6736 R b 2 CT 8P 8x5887 psia CT 3596 it 6 psialme 001504 it 3lbm Then TlPiv b 1 400 01144 001504638R R 62 2 01052 034792 c From the superheated refrigerant table Table A13E P 400 psia v 01144 11 3lbmT 210001 660 R PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 350 397 39 The pressure of nitrogen in a tank at a speci ed state is to be determined using the ideal gas relation and the Beattie Bridgeman equation The error involved in each case is to be determined Properties The gas constant and molar mass of nitrogen are Table A1 R 02968 kPam3kgK and M 28013 kgkmol N2 Analysis a From the ideal gas equation of state 0041884 m3kg 02968 kPam3kgK150 K 150 K P 3 1063 kPa 63 error 1 0041884 m lkg b The constants in the BeattieBridgeman equation are AA0 1 2 13623151 03902617 v 1 1733 B B0 1 2 0050461 M 005076 2133193 1 11733 c 42x104m3 K3kmol since 7 MU 28013 kgkmol004l884 m3kg 211733 m3kmol Substituting 133193 117332 4 RT c3 B A8314x1501 42x10 V 1 u 3 11733005076 T 11733x150 172 117332 210004 kPa negligible error I PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 351 398 quot Problem 397 is reconsidered Using EES or other software the pressure results of the ideal gas and Beattie Bridgeman equations with nitrogen data supplied by EES are to be compared The temperature is to be plotted versus specific volume for a pressure of 1000 kPa with respect to the saturated liquid and saturated vapor lines of nitrogen over the range of 110 K lt Tlt 150 K Analysis The problem is solved using EES and the solution is given below Function BeattBridgTvMRu vbarvM quotConversion from mquot3kg to mquot3kmolquot quotThe constants for the BeattieBridgeman equation of state are found in textquot Ao1362315 aa002617 Bo005046 bb000691 cc4201 E4 BBo1 bbvbar AAo1 aavbar quotThe BeattieBridgeman equation of state isquot BeattBridg RuTvbar21 ccvbarT3vbarBAvbar2 End T150 K v0041884 mquot3kg Pexper1000 kPa TtabeT TBBTTideagasT PtablePRESSURENitrogenTTtabevv quotEES data for nitrogen as a real gasquot TtabetemperatureNitrogen PPtablevv MMOLARMASSNitrogen Ru8314 kJkmoIK quotUniversal gas constantquot RRuM quotParticular gas constantquot PidealgasRTideagasv quotIdeal gas equationquot PBBBeattBridgTBBvMRu quotBeattieBridgeman equation of state Functionquot PBB kPa Ptable kPa Pidilqaj kPa V m3kg TBB K Tum K Ttable K 1000 1000 1000 001 9123 3369 1038 1000 1000 1000 002 9552 6739 1038 1000 1000 1000 0025 105 8423 1061 1000 1000 1000 003 1168 1011 1172 1000 1000 1000 0035 1301 1179 1301 1000 1000 1000 004 1444 1348 1443 1000 1000 1000 005 1746 1685 1745 Nitrogen T vs v for P1000 kPa Ei Idea Gas I o BeattieBridgeman 14o 160 150 o EES Table Value 130 120 T K 110 1000kPa 100 90 80 70 I 10393 10392 10391 v m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 399 Carbon dioxide is compressed in a pistoncylinder device in a polytropic process The nal temperature is to be determined using the ideal gas and van der Waals equations Properties The gas constant molar mass critical pressure and critical temperature of carbon dioxide are Table A1 R 01889 kPam3kgK M 4401 kgkmol Tcr 3042 K Pcr 739 MPa Analysis a The speci c volume at the initial state is 01889 kPam3kgK473 K 1 008935 m3kg P1 1000 kPa C02 Accord1ng to process spec1 catron 1 Mpa 200 C 1000 kPa 112 003577 m3kg 3000 kPa 1n P 12 2 v1 008935 m3kg 2 The final temperature is then P2112 3000 kPa003577 m3kg 3 2568K R 01889kPam IkgK T2 b The van der Waals constants for carbon dioxide are determined from 27R2T2 2 1 kP 3k K2 42K2 cr 70 889 a m g 30 01885m6kPakg2 a 6410cr 64739O kPa RT 1 kP 3k K 42K b r 20 889 a m l g X30 200009720m3kg 8Pcr 8x7390kPa Applying the van der Waals equation to the initial state Pizju b 2 RT V 01885 V2 1000 V 00009720 01889473 Solving this equation by trialerror or by EES gives 11 008821m3kg According to process speci cation 1000 kPa 112 003531 m3kg 3000 kPa P 1n 92 2 VI 008821m3kg 2 Applying the van der Waals equation to the final state a P v b 2 RT V2 85200353 1 00009720 01889T 3000 003531 Solving for the nal temperature gives T2 2 573K 352 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 353 3100 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation van der Waals equation and the steam tables Properties The gas constant critical pressure and critical temperature of steam are Table A1 R 04615 kPam3kgK Tcr 6471 K 1Dcr 2206 MPa Analysis The speci c volume of steam is 1 3 15 2 203520 m3kg H20 m 2841 kg 3 1 m a From the ideal gas equation of state 2841 kg 3 06 MPa T 2 2 600 kPa0352 m lkg 4576K R 04615 kPa m3kg K b The van der Waals constants for steam are determined from 27R2T2 3 2 2 a c 270 4615 kPa m kg K 6471K 21705 ng 64Pc 642206O kPa RT 41kP3 K 41K 9 Cr O 6 5 a m kg 6 7 000l69 m3kg 8P 8x22060kPa CT Then T Piv b 1 600 0352 000169 4659K 2 04615 035202 c From the superheated steam table Tables A6 P06MPa V03520m3kgT200 c 473 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 354 3101 39 Problem 3100 is reconsidered The problem is to be solved using EES or other software The temperature of water is to be compared for the three cases at constant speci c volume over the pressure range of 01 MPa to 1 MPa in 01 MPa increments The error involved in the ideal gas approximation is to be plotted against pressure Analysis The problem is solved using EES and the solution is given below Function vanderWaasTvMRuTcrPcr vbarvM quotConversion from mquot3kg to mA3kmolquot quotThe constants for the van der Waals equation of state are given by equation 324quot a27Ruquot2Tcr 264Pcr bRuTcr8Pcr quotThe van der Waals equation of state gives the pressure asquot vanderWaaIsRuTvbarbavbar2 End m2841 kg VoI1 mA3 P6convertMPakPa TcrTCRTSteamiapws Pcr PCRITSteamiapws vVolm PtableP PvdWPPideagasP TtabetemperatureSteamiapwsPPtabevv quotEES data for steam as a real gasquot PtablepressureSteamiapws TTtabevv TsattemperatureSteamiapwsPPtabevv MMMOLARMASSwater Ru8314 kJkmoIK quotUniversal gas constantquot RRuMM quotParticular gas constantquot PidealgasRTideagasv quotIdeal gas equationquot quotThe value of PvdW is found from van der Waals equation of state Functionquot PvdWvanderWaasTvdWvMMRuTcrPcr ErroridealgasAbsTtabeTideagasTtabeConvert ErrorvdWAbsTtabeTvdWTtabeConvert P kPa Tideagas K Ttable K TvdW K Errorideal gas IKI 100 7627 3728 8635 7954 200 1525 3934 1623 6122 300 2288 4067 2382 4374 400 3051 4168 3141 268 500 3814 425 390 1027 600 4576 473 4659 3249 700 5339 5453 5418 2087 800 6102 6191 6177 1442 900 6864 6937 6936 1041 1000 7627 7686 7695 07725 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 355 100 90 80 39 van der Waals 7o Ideal gas 60 50 40 30 20 Erro rvdw 10 800 900 1000 L C C l C C 0 C C 4 C C Cl C C C C C l C C T vs v for Steam at 600 kPa I I 900 800 Steam Table 7 00 0 Ideal Gas 500 0 van der Waals 500 103 102 101 100 1o1 102 103 v m3kg T K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission T vs v for Steam at 6000 kPa 1000 900 800 7001 ldea Gas T K 500 6000 kPa 400 300 Steam Table 500 ovan der Waals 103 102 101 100 101 102 v m3kg 800 750 700 Steam table 650 van der Waals 600 Ideal gas 550 500 450 400 350 300 250 200 150 100 50 Ttable K 100 200 300 400 500 600 700 800 900 1000 356 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 357 Special Topic Vapor Pressure and Phase Equilibrium 3102 The vapor pressure in the air at the beach when the air temperature is 30 C is claimed to be 52 kPa The validity of this claim is to be evaluated Properties The saturation pressure of water at 30 C is 4247 kPa Table A4 30 C Analysis The maximum vapor pressure in the air is the saturation pressure of water at the given temperature which is WATER P P vmax satTair sat30 C 24247 kPa which is less than the claimed value of 52 kPa Therefore the claim is false 3103 A glass of water is left in a room The vapor pressures at the free surface of the water and in the room far from the glass are to be determined Assumptions The water in the glass is at a uniform temperature Properties The saturation pressure of water is 2339 kPa at 20 C and 1706 kPa at 15 C Table A4 Analysis The vapor pressure at the water surface is the saturation pressure of water at the water temperature Pv water surface PsatTwaner Psat15 C kPa Not1ng that the an 1n the room 1s not saturated the vapor pressure 1n the room far from the 150C glass is 3104 The temperature and relative humidity of air over a swimming pool are given The water temperature of the swimming pool when phase equilibrium conditions are established is to be determined Assumptions The temperature and relative humidity of air over the pool remain constant Properties The saturation pressure of water at 25 C is 317 kPa Table A4 Analysis The vapor pressure of air over the swimming pool is Patm 25 C Pv PsatTair Psat250C Z POOL Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface Therefore Pv water surface Pv air 2 L902 kPa and Twater Tsat PV Tsat 1902kPa 167 C Discussion Note that the water temperature drops to 167 C in an environment at 25 C when phase equilibrium is established PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 358 3105 A person buys a supposedly cold drink in a hot and humid summer day yet no condensation occurs on the drink The claim that the temperature of the drink is below 10 C is to be evaluated Properties The saturation pressure of water at 35 C is 5629 kPa Table A4 AnalySts The vapor pressure of a1r 1s 35 C Pv air PsatTair Psat350C 075629 kPa 3940 kPa 70 The saturation temperature corresponding to this pressure called the dewpoint temperature is T sat Tsat PV 2 Tsat3940 kPa 283 C That is the vapor in the air will condense at temperatures below 287 C Noting that no condensation is observed on the can the claim that the drink is at 10 C is false 3106 Two rooms are identical except that they are maintained at different temperatures and relative humidities The room that contains more moisture is to be determined Properties The saturation pressure of water is 2339 kPa at 20 C and 317 kPa at 25 C Table A4 Analysis The vapor pressures in the two rooms are ROWquot 13 P v1 11Dsatr1 1Psat25 c 04317 kPa 127 kPa Therefore room 1 at 30 C and 40 relative humidity contains more moisture 3107E A thermos bottle halffilled with water is left open to air in a room at a speci ed temperature and pressure The temperature of water when phase equilibrium is established is to be determined Assumptions The temperature and relative humidity of air over the bottle remain constant Properties The saturation pressure of water at 60 F is 02564 psia Table A4E Analysis The vapor pressure of air in the room is Thermos Pv air mama Psat7ooF 03502564 psia 008973 psia bottle Phase equilibrium will be established when the vapor pressure at the water surface equals the vapor pressure of air far from the surface Therefore 60 F 35 P1 watersurface P1ai1 008973ps1a and Twater TsatPV Tsat008973psia 323039 Discussion Note that the water temperature drops to 323 F in an environment at 60 F when phase equilibrium is established PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 359 Review Problems 3108E Water in a pressure cooker boils at 260 F The absolute pressure in the pressure cooker is to be determined Analysis The absolute pressure in the pressure cooker is the saturation pressure that corresponds to the boiling temperature P llamaF 3545 psia 26001 3109 Carbon dioxide ows through a pipe at a given state The volume and mass ow rates and the density of C02 at the given state and the volume ow rate at the eXit of the pipe are to be determined Analysis 3 MPa 500 K C02 gt 450 K 04 krnols a The volume and mass ow rates may be determined from ideal gas relation as NR T 3 V1 2 u 1 Z 0 4krnols8 314kPa rn kmol K500 K 0 5543m3S P 3000 kPa 39 3 m1 P1V1 3000 kPa035543m ls 21760kgS RTl 01889 kPam kgK500 K The density is p1 2 31 76kgm3 V1 05543m s b The volume ow rate at the eXit is NR T 3 V2 2 u 2 Z 0 4krnols8 314kPa rn kmol K450 K 0 4988m3S P 3000 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 360 3110 A tank contains argon at a specified state Heat is transferred from argon until it reaches a speci ed temperature The nal gage pressure of the argon is to be determined Assumptions 1 Argon is an ideal gas Properties The local atmospheric pressure is given to be 100 kPa Q Analysis Noting that the speci c volume of argon in the tank remains constant Argon from ideal gas relation we have 600 C T 300 273 K 200 kPa gage P2 P1 2 200100 kPa 1969kPa T1 600 273K Then the gage pressure becomes Pgage2 P2 Pml 1969 100 969 kPa 3111 The cylinder conditions before the heat addition process is speci ed The pressure after the heat addition process is to be determined Assumptions 1 The contents of cylinder are approximated by the air properties 2 Air is an ideal gas combustlon chamber Analysis The nal pressure may be determined from the ideal gas relation 12 MP3 450 C T 1 2 K P2 2 P1 750 731200 kPa 3358kPa T1 450 273 K 3112 A rigid container that is lled with R13a is heated The initial pressure and the nal temperature are to be determined Analysis The initial specific volume is 0090 m3kg Using this with R134a the initial temperature reveals that the initial state is a mixture The 400C initial pressure is then the saturation pressure 1 kg 3 T1 40 C P P 51 25kPa 0 b1 A 11 0090 m Z a 6 39 V1 2 m3kg I sat 40 C P This is a constant volume cooling process 1 Vm constant The final A state is superheated vapor and the nal temperature is then 2 T 250 C ableAl3 v2 1 20090 m3kg 2 T V C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 361 3113 The refrigerant in a rigid tank is allowed to cool The pressure at which the refrigerant starts condensing is to be determined and the process is to be shown on a Pv diagram Analysis This is a constant volume process 1 V m constant and the specific volume is determined to be R134a v 0 117 3 240 km 12 2A 20117 m3kg m 1kg P A When the refrigerant starts condensing the tank will contain saturated 1 vapor only Thus 1 v2 2 ug 0117 m3kg The pressure at this point is the pressure that corresponds to this 8 value P2 Psatug0ll7m3kg 170kpa 3114E A pistoncylinder device that is lled with water is cooled The nal pressure and volume of the water are to be determined Analysis The initial specific volume is v 2649 it 3 3 v 2649ft lbm 1 m llbm H20 400 F This is a constantpressure process The initial state is determined to be 1 lbm superheated vapor and thus the pressure is determined to be 2649 ft3 T1 400 F P P 180 sia T b1 A 6E a e v1 2649 1t 3lbm 1 2 p P A The saturation temperature at 180 psia is 3731 F Since the nal 2 1 temperature is less than this temperature the nal state is compressed liquid Using the incompressible liquid approximation u2 of mF 001613 1t 3lbm TableA4E V C The final volume is then v2 mu2 11bm0016131t 3lbm 001613ft3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 362 3115 Ethane is heated at constant pressure The nal temperature is to be determined using ideal gas equation and the compressibility charts Properties The gas constant the critical pressure and the critical temperature of ethane are from Table A1 R 02765 kPam3kgK Tc 3055 K PC 448 MPa Analysis From the ideal gas equation T2 T1 2 373 K16 5968K 1 I I From the compressibility chart at the initial state Fig A15 Ethane Q T 2 T1 2 373K 21221 10 MPa R1 Tcr 3055 K 100 C P IOMP z1 061 um 035 PR1 1 a 2232 P 448 MPa CI At the final state uR 16UR1 16035 056 2 39 Thus P2112 i vaor 10000kPa 0563055K 460K 2 22R 22 Per 083 4480 kPa Of these two results the accuracy of the second result is limited by the accuracy with which the charts may be read Accepting the error associated with reading charts the second temperature is the more accurate 3116 A large tank contains nitrogen at a speci ed temperature and pressure Now some nitrogen is allowed to escape and the temperature and pressure of nitrogen drop to new values The amount of nitrogen that has escaped is to be determined Properties The gas constant for nitrogen is 02968 kPam3kgK Table A l Analysis Treating N2 as an ideal gas the initial and the final masses in the tank are determined to be P v kP 1 3 m1 2 1 2 600 X 3 m 9062 kg RTI 02968kPa m kg K29O K m sz 400 kPa13 m3 6O 84 kg N39 2 RT2 02968 kPa m3kg K278 K 600 113 Thus the amount of N2 that escaped is 170C 13 m3 Am m1 m2 9062 6084 298 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 363 3117 Superheated refrigerant134a is cooled at constant pressure until it exists as a compressed liquid The changes in total volume and internal energy are to be determined and the process is to be shown on a T v diagram Analysis The refrigerant is a superheated vapor at the initial state and a compressed liquid at the final state From Tables A 13 and A11 Pl 1ZMPa u1 27723 kJkg T A T1 70 C 01 0019502 m3kg 1 P2 12MPa u2 Euf20c 27885 kJkg 111343 2 T2 20 C 2 z vf20c 00008160 m3kg 70 C 12 MPa Thus 9 AV 2 mv2 v1 10 kg00008160 0019502 m3kg 0187 m3 c AU 2 mu2 ul 2 10 kg7885 27723 kJkg 1984 kJ 3118 The rigid tank contains saturated liquidvapor mixture of water The mixture is heated until it exists in a single phase For a given tank volume it is to be determined if the nal phase is a liquid or a vapor Analysis This is a constant volume process 1 Vm constant and thus the nal speci c volume will be equal to the initial specific volume V2 2V1 H20 The critical speci c volume of water is 0003106 m3kg Thus if the nal specific V 4 L volume is smaller than this value the water will exist as a liquid otherwise as a vapor m 2 kg T 50 C v 3 v 4L w m 0002 m3kg lt ac Thus liquid m 2 kg v 04 3 v 400L w m 02 m3kggt 10 Thus vapor m 2 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 364 3119 The pressure in an automobile tire increases during a trip While its volume remains constant The percent increase in the absolute temperature of the air in the tire is to be determined Assumptions 1 The volume of the tire remains constant 2 Air is an ideal gas Properties The local atmospheric pressure is 90 kPa Tire 200 kPa Analysis The absolute pressures in the tire before and after the trip are 003 5 1113 P1 Pgage1 Patm 200 90 290 kPa Noting that air is an ideal gas and the volume is constant the ratio of absolute temperatures after and before the trip are 101V1 PZV2 T2 310kPa gt 1069 T1 T2 T1 P1 290kPa Therefore the absolute temperature of air in the tire will increase by 69 during this trip 3120 A pistoncylinder device contains steam at a specified state Steam is cooled at constant pressure The volume change is to be determined using compressibility factor Properties The gas constant the critical pressure and the critical temperature of steam are R 04615 kPam3kgK Tc 6471 K PC 2206 MPa Analysis The exact solution is given by the following PZZOOkPa 131623 31lt V m Tr 300 c 1 g Steam gt P 200 1d Table A6 02 kg r 2 a v2 095986 m3kg 200 kPa Q T2 150 C 3000C AV mv1 2 02kg131623 095986m3kg 007128m3 ex act Using compressibility chart EES function for compressibility factor is used P 02 MP PC 2206 MPa T 300 273 K Zl 09956 Tc 6471 K P 2 MP pR 2 2 2 u 00091 pc 2206 MPa T 150 273 K 22 2 09897 TR22 065 Tc 6471 K 3 V1 ZlmRTl O9956O2kg04615 kPa m kg K300 273 K 2 02633 m3 Pl 200 kPa 3 V2 22mm2 O9897O2kg04615 kPa m kg K150 273 K 2 019321113 P2 200 kPa Avchm 11 v2 02633 01932 007006m3 Error 17 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 365 3121 The temperature of steam in a tank at a specified state is to be determined using the ideal gas relation the generalized chart and the steam tables Properties The gas constant the critical pressure and the critical temperature of water are from Table A1 R 04615 kPam3kg K Tel 2 6471 K Pa 2206 MPa Analysis a From the ideal gas equation of state 3 P 2 E 2 04615 kPa m lkg K673 K 215529kpa H20 1 002 m kg 002 m3kg 9 From the compressibility chart Fig A15a 400 C TR 2 l 673 K 21040 Tcr 6471 K vactual 002 m kg22060 kPa v R RTcrPcr 04615 kPam3kgK6471K Thus P PRPcr 057 x 22060 12574kPa c From the superheated steam table T 400 C U20 2 m3kg P12515kPa from EES 3122 One section of a tank is lled with saturated liquid R134a While the other side is evacuated The partition is removed and the temperature and pressure in the tank are measured The volume of the tank is to be determined Analysis The mass of the refrigerant contained in the tank is v 3 m 1 003m 3 3496kg V1 00008580 m kg R134a Sll lCC P 09 MPa Evacuated At the final state Table A13 2 007997 m3kg T2 20 C Thus vtank v2 mu2 3496 kg007997 m3kg 280 m3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 366 3123 539 Problem 3122 is reconsidered The effect of the initial pressure of refrigerant134 on the volume of the tank is to be investigated as the initial pressure varies from 05 MPa to 15 MPa The volume of the tank is to be plotted versus the initial pressure and the results are to be discussed Analysis The problem is solved using EES and the solution is given below quotGiven Dataquot X100 Vo1003 mquot3 P11200 kPa T23O C P2400 kPa quotSolutionquot v1voumeR134aPP1xx1 Vo1mv1 v2voumeR134aPP2TT2 Vo2mv2 P1 Vol2 3 39 39 39 39 kPa m3 500 2977 600 2926 29 700 288 800 2837 I 900 2796 1000 2757 39739 28 1100 2721 g 1200 2685 N 1300 2651 3 2 1400 2617 gt 1500 2584 26 500 700 900 1100 1300 1500 P1 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 367 3124 A propane tank contains 5 L of liquid propane at the ambient temperature Now a leak develops at the top of the tank and propane starts to leak out The temperature of propane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire propane in the tank is vaporized are to be determined Properties The properties of propane at 1 atm are Tsat 421 C 0 581 kg m3 and hfg 4278 kJkg Table A3 Analysis The temperature of propane when the pressure drops to 1 atm is simply the saturation pressure at that temperature T Tsat1atm 421 C Propane The initial mass of liquid propane is m pv 581kgm30005 m3 2905 kg The amount of heat absorbed is simply the total heat of vaporization Q mh 2905 kg4278 kJkg 1243 kJ absorbed fg 3125 An isobutane tank contains 5 L of liquid isobutane at the ambient temperature Now a leak develops at the top of the tank and isobutane starts to leak out The temperature of isobutane when the pressure drops to 1 atm and the amount of heat transferred to the tank by the time the entire isobutane in the tank is vaporized are to be determined Properties The properties of isobutane at 1 atm are Tsat 117 C p 5938 kgm3 and hfg 3671 kJkg Table A3 Analysis The temperature of isobutane when the pressure drops to 1 atm is simply the saturation pressure at that temperature T Tsat1atm 11397OC Isobutane 5 L 20 C The initial mass of liquid isobutane is m pv 5938 kgm3 0005 m3 2969kg The amount of heat absorbed is simply the total heat of vaporization Qabsorbed mhfg 2969 kg3 671 k kg 1090 kJ Leak 3126 A tank contains helium at a speci ed state Heat is transferred to helium until it reaches a speci ed temperature The nal gage pressure of the helium is to be determined Assumptions 1 Helium is an ideal gas Properties The local atmospheric pressure is given to be 100 kPa Analysis Noting that the speci c volume of helium in the tank remains constant from ideal gas relation we have Q Helium 4 T 273 K P2 P1 2 140 100 kPaM 3662 kPa 37 C T1 37 273K 140 kPa gage Then the gage pressure becomes PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 3127 The table is completed as follows P kPa T C 1 m3kg u kJkg Condition description and quality if applicable 300 250 07921 27289 Superheated vapor 300 13352 03058 15600 x 0504 Twophase mixture 10142 100 Insuf cient information 3000 180 0001127 76192 Compressed liquid Approximated as saturated liquid at the given temperature of 180 C 368 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 369 3128 Water at a speci ed state is contained in a pistoncylinder device tted with stops Water is now heated until a nal pressure The process will be indicated on the Pvand T vdiagrams Analysis The properties at the three states are P1 300 kPa I 3 T1 1335 C TableA 5 v1 05 m Ikg P2 300 kPa 3 Water v2 06058 m lkg T2 1335 C TableA 5 Q 300 kPa x2 I sat vap 05 Inskg P2 600 kPa T 5178 C ableA6 V3 206058m3kg 2 T Using Property Plot feature of EES and by adding state points we obtain following diagrams SteamlAPWS 106 105 104 39E39 n 3 5 10 300 kPa D 102 300 kPa 101 100 I I I I I I 104 103 102 10391 05 100 101 102 v m3kg SteamlAPWS I I 600 39 600 kPa 39 300 kPa 500 400 9 l 300 200 15880C 39 100 1335 C 1 2 0 I I I I I 103 102 10391 05 100 101 102 v m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 370 3129E Argon contained in a pistoncylinder device at a given state undergoes a polytropic process The nal temperature is to be determined using the ideal gas relation and the BeattieBridgeman equation Analysis a The polytropic relations for an ideal gas give P n ln 0616 T2T1 2 300460R ps1a 986R P1 1000 ps1a b The constants in the BeattieBridgeman equation are expressed as Argon A A01 2 13078021 003328 1000 pm V v 300 F B BO1 3 2 003931 2 V V c 599gtlt104m3 K3kmol Substituting these coefficients into the BeattieBridgeman equation and using data in SI units P 1000 psia 6895 kPa T760 R 4222 K Ru 8314 kJkmolK P R T 1 C 7B i 72 7T3 72 and solving using an equation solver such as EES gives 7 05120 m3kmol 8201 ft 3lbmol From the polytropic equation P1 1 n 3 116 3 v2 2 v1 05120m kmol 03319 m kmol Substituting this value into the BeattieBridgeman equation and using data in SI units P 2000 psia 13790 kPa and RM 8314 kJkmolK R T c A P 7 2 l UT3VB 7 2 and solving using an equation solver such as EES gives PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 371 3130E The speci c volume of nitrogen at a given state is to be determined using the ideal gas relation the BenedictWebb Rubin equation and the compressibility factor Properties The properties of nitrogen are Table AlE R 03830 psia 31bmR M 28013 lbmlbmol Tc 2271 R PC 492 psia Analysis a From the ideal gas equation of state V 03830 psiait 3lbm R360 R P 400 psia 03447ft3lbm Nitrogen 400 psia 100 F b Using the coef cients of Table 34 for nitrogen and the given data in SI units the BenedictWebbRubin equation of state is RT CO 1 bRuT a aa c y 2 P 7 B0RuT Ao FETJ3T2 l l ngXIK yV 8314200 72 2758 5 004074 X 8314 X 200 10673 m 1 0002328 x 8314 x 200 254 002 E 3 V 254x1272gtlt104 7379x104 00053 eX 00053 72 76 732002 72 j p The solution of this equation by an equation solver such as EES gives 17 05666 m3kmol Then 1 05666 m3kmol1602 3lbm 03240ft3lbm M 28013kgkmol 1m3kg c From the compressibility chart Fig A15 TR 2 i 360R 21585 Tcr 2271R P 400 Z 094 R 0813 Pcr 492 ps1a Thus I 21411 0940 3447 11 3lbm 03240ft3lbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 372 3131 A hot air balloon with 3 people in its cage is hanging still in the air The average temperature of the air in the balloon for two environment temperatures is to be determined Assumptions Air is an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table Al Analysis The buoyancy force acting on the balloon is vbauoon 47 r3 3 47r10m33 41891113 P 90 kPa 2 3 1089 kgm3 RT 0287 kPam kgK288 K pcoolair FB pcoolairgvballoon 1 N 1089 kgm398 ms24189 m3 2 44700 N 1kg ms The vertical force balance on the balloon gives FB Whotair Wcage Wpeop1e mhotair mc e m as people3 Substituting 1N 44700 N 2 m r 80 k 195 k 98 ms2 howl g g 1kg ms2 which gives mhotair 4287 kg Therefore the average temperature of the air in the balloon is T 90 kPa4189 m3 3 3065 K mR 4287 kg0287 kPa m lkg K Hot air balloon P atm Al T 15 C u u I uuuuuuu 39 u I l u u u I u mcage r I u u u I u IIIIIIII I l I 39 u n u u Repeating the solution above for an atmospheric air temperature of 30 C gives 3236 K for the average air temperature in the balloon PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 373 3132 39 Problem 3131 is to be reconsidered The effect of the environment temperature on the average air temperature in the balloon when the balloon is suspended in the air is to be investigated as the environment temperature varies from 10 C to 30 C The average air temperature in the balloon is to be plotted versus the environment temperature Analysis The problem is solved using EES and the solution is given below quotGiven Dataquot quotatmatmosphere about balloonquot quotgasheated air inside balloonquot g9807 msquot2 dbaoon20 m mcage 80 kg m1person65 kg NoPeopIe 6 TatmCesius 15 C Tatm TatmCesius273 quotKquot Patm 90 kPa R0287 kJkgK Pgas Patm TgasCesiusTgas 273 quotCquot quotCalculated valuesquot Patm rhoatmRTatm quotrhoatm density of air outside balloonquot Pgas rhogasRTgas quotrhogas density of gas inside balloonquot rbaoondbaoon2 Vbaoon4pirbaoonquot33 mpeopleNoPeopem1 person mgasrhogasVbaoon mtotamgasmpeopemcage quotThe total weight of balloon people and cage isquot Wtotamtotag quotThe buoyancy force acting on the balloon Fb is equal to the weight of the air displaced by the balloonquot FbrhoatmVbaoong quotFrom the free body diagram of the balloon the balancing vertical forces must equal the product of the total mass and the vertical accelerationquot Fb Wtotamtotaaup aup 0 quotThe balloon is hanging still in the airquot 100 I I I I I I I 90 Tatm Celcius C T as Celcius C 80 10 1732 70 5 2342 539 0 2955 T 50 9 people 5 3571 10 4189 15 4809 a 20 5431 I 25 6057 30 6684 0 l I I I I I l I l I I 1o 5 o 5 1o 15 20 25 30 TatmCelsius c PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 374 3133 A hot air balloon with 2 people in its cage is about to take off The average temperature of the air in the balloon for two environment temperatures is to be determined Assumptions Air is an ideal gas Properties The gas constant of air is R 0287 kPam3kgK HOt air balloon D 18 m Analysis The buoyancy force acting on the balloon is Pa 93 kP ubauoon 47rr3 3 47r9 m33 3054 m3 g 12 C a P 93 kP pcoolair 3 a 1137 kgm3 RT 0287 kPa m kg K285 K FB pcoolairgvballoon 1 N 1137 kgm398 ms23054 m3 2 34029 N 1 kg ms The vertical force balance on the balloon gives meage FB Whotair Wcage Wpeop1e mhotair mcage mpeopleg Substituting 1N 34029 N m r 120 k 140 k 981ms2 howl g g 1kgms2 which gives mhotair Z 3212 kg Therefore the average temperature of the air in the balloon is 3 T PV 93 kPa3054 m 2 308K mR 3212 kg0287 kPa m3kg K Repeating the solution above for an atmospheric air temperature of 25 C gives 323 K for the average air temperature in the balloon PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 375 Fundamentals of Engineering FE Exam Problems 3134 A 300m3 rigid tank is filled with saturated liquidvapor mixture of water at 200 kPa If 25 of the mass is liquid and the 75 of the mass is vapor the total mass in the tank is a 451 kg b 556 kg c 300 kg d 331 kg e 195 kg Answer a 451 kg Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vtank300 quotm3quot P1 200 quotkPaquot x075 vfVOLUMESteamIAPWS x0PP1 vgVOLUMESteamIAPWS x1 PP1 vvfxvgvf mVtankv quotkgquot quotSome Wrong Solutions with Common Mistakesquot R04615 quotkJkgKquot TTEMPERATURESteamIAPWSx0PP1 P1 VtankW1mRT273 quotTreating steam as ideal gasquot P1 VtankW2mRT quotTreating steam as ideal gas and using degCquot W3mVtank quotTaking the density to be 1 kgmquot3quot 3135 Water is boiled at 1 atm pressure in a coffee maker equipped with an immersiontype electric heating element The coffee maker initially contains 1 kg of water Once boiling started it is observed that half of the water in the coffee maker evaporated in 10 minutes If the heat loss from the coffee maker is negligible the power rating of the heating element is a 38 kw b 22 kw c 19 kw d 16 kw e 08 kw Answer c 19 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values m11 quotkgquot P101325 quotkPaquot time1060 quotSquot mevap05m1 Powertimemevaphfg quotkJquot hfENTHALPYSteamIAPWS X0PP hgENTHALPYSteamIAPWS x1 PP hfghghf quotSome Wrong Solutions with Common Mistakesquot W1Powertimemevaphg quotUsing hgquot W2Powertime60mevaphg quotUsing minutes instead of seconds for timequot W3Power2Power quotAssuming all the water evaporatesquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 376 3136 A 1m3 rigid tank contains 10 kg of water in any phase or phases at 160 C The pressure in the tank is a 738 kPa b 618 kPa c 370 kPa d 2000 kPa e 1618 kPa Answer b 618 kPa Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vtank1 quotmquot3quot m10 quotkgquot vVtanldm T160 quotCquot PPRESSURESteamlAPWSvvTT quotSome Wrong Solutions with Common Mistakesquot R04615 quotkJkgKquot W1PVtankmRT273 quotTreating steam as ideal gasquot W2PVtankmRT quotTreating steam as ideal gas and using degCquot 3137 Water is boiling at 1 atm pressure in a stainless steel pan on an electric range It is observed that 2 kg of liquid water evaporates in 30 minutes The rate of heat transfer to the water is a 251 kW b 232 kW c 297 kW d 047 kW e 312 kW Answer a 251 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values mevap2 quotkgquot P101325 quotkPaquot time3060 quotSquot Qtimemevaphfg quotkJquot hfENTHALPYSteamIAPWS X0PP hgENTHALPYSteamIAPWS X1 PP hfghghf quotSome Wrong Solutions with Common Mistakesquot W1Qtimemevaphg quotUsing hgquot W2Qtime60mevaphg quotUsing minutes instead of seconds for timequot W3Qtimemevaphf quotUsing hfquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 377 3138 Water is boiled in a pan on a stove at sea level During 10 min of boiling its is observed that 200 g of water has evaporated Then the rate of heat transfer to the water is a 084 kJmin b 451 kJmin c 418 kJmin d 535 kJmin e 2257 kJmin Answer b 451 kJmin Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values mevap02 quotkgquot P101325 quotkPaquot time10 quotminquot Qtimemevaphfg quotkJquot hfENTHALPYSteamIAPWS X0PP hgENTHALPYSteamIAPWS X1 PP hfghghf quotSome Wrong Solutions with Common Mistakesquot W1Qtimemevaphg quotUsing hgquot W2Qtime60mevaphg quotUsing seconds instead of minutes for timequot W3Qtimemevaphf quotUsing hfquot 3139 A rigid 3m3 rigid vessel contains steam at 4 MPa and 500 C The mass of the steam is a 3 kg b 9 kg c 26 kg d 35 kg e 52 kg Answer d 35 kg Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values V3 quotmquot3quot mVv1 quotmquot3kgquot P1 4000 quotkPaquot T1 500 quotCquot V1 VOLUMESteamIAPWSTT1 PP1 quotSome Wrong Solutions with Common Mistakesquot R04615 quotkJkgKquot P1 VW1mRT1 273 quotTreating steam as ideal gasquot P1VW2mRT1 quotTreating steam as ideal gas and using degCquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 378 3140 Consider a sealed can that is filled with refrigerant134a The contents of the can are at the room temperature of 25 C Now a leak developes and the pressure in the can drops to the local atmospheric pressure of 90 kPa The temperature of the refrigerant in the can is expected to drop to rounded to the nearest integer a 0 C b 29 C c 16 C d 5 C e 25 C Answer b 29 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T1 25 quotCquot P290 quotkPaquot T2TEMPERATURER134ax0PP2 quotSome Wrong Solutions with Common Mistakesquot W1T2T1 quotAssuming temperature remains constantquot 3141 A rigid tank contains 2 kg of an ideal gas at 4 atm and 40 C Now a valve is opened and half of mass of the gas is allowed to escape If the final pressure in the tank is 22 atm the nal temperature in the tank is a 71 C b 44 C c 100 C d 20 C e 172 C Answer a 71 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values quotWhen Rconstant and V constant P1P2m1T1m2T2quot m1 2 quotkgquot P1 4 quotatmquot P222 quotatmquot T1 40273 quotKquot m205m1 quotkgquot P1P2m1T1m2T2 T2CT2273 quotCquot quotSome Wrong Solutions with Common Mistakesquot P1P2m1T1273m2W1T2 quotUsing C instead of Kquot P1P2m1T1m1W2T2273 quotDisregarding the decrease in massquot P1P2m1T1m1W3T2 quotDisregarding the decrease in mass and not converting to deg Cquot W4T2T12732 quotTaking T2 to be half of T1 since half of the mass is dischargedquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 379 3142 The pressure of an automobile tire is measured to be 190 kPa gage before a trip and 215 kPa gage after the trip at a location where the atmospheric pressure is 95 kPa If the temperature of air in the tire before the trip is 25 C the air temperature after the trip is a 511 C b 642 C c 272 C d 283 C e 250 C Answer a 511 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values quotWhen R V and m are constant P1P2T1T2quot Patm95 P1 190Patm quotkPaquot P2215Patm quotkPaquot T1 25273 quotKquot P1P2T1T 2 T2CT2273 quotCquot quotSome Wrong Solutions with Common Mistakesquot P1P2T1273W1T2 quotUsing C instead of Kquot P1PatmP2Patm T1W2T2273 quotUsing gage pressure instead of absolute pressurequot P1PatmP2PatmT1273W3T2 quotMaking both of the mistakes abovequot W4T2T1273 quotAssuming the temperature to remain constantquot 3143 3145 Design and Essay Problems 3145 It is helium PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 1 Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Cengel Michael A Boles McGrawHill 2015 Chapter 4 ENERGY ANALYSIS OF CLOSED SYSTEMS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 2 Moving Boundary Work 41C The area under the process curve and thus the boundary work done is greater in the constant pressure case 42 Nitrogen gas in a cylinder is compressed at constant temperature until its pressure rises to a specified value The boundary work done during this process is to be determined Assumptions 1 The process is quasiequilibrium 2 Nitrogen is an ideal gas Analysis The boundary work is determined from its de nition to be P A 2 V P Wout 2 PdV PIVIIn Z P1V11n 1 2 1 V1 P 2 T 300 K 150 kPa02 m3 111150 kPa J 3 800 kPa 1 kPa m 1 502 H gt v Discussion The negative sign indicates that work is done on the system work input 43 Helium is compressed in a pistoncylinder device The initial and final temperatures of helium and the work required to compress it are to be determined Assumptions The process is quasiequilibrium Properties The gas constant of helium is R 20769 kJkgK Table Al Analysis The initial specific volume is V1 5m3 P Using the ideal gas equation 180 2 A 1 P 3 T1 lvl 2 180 kPa5 m kg 433I3K R 20769 kJkg K Since the ressure stays constant p 3 7 v m3 V 3 T2 2T1 2amp3 4333 K 1733K V1 5 m and the work integral expression gives 2 WboutzI PdVP2 1180kPa2 5m3 3 540kJ 1 kPa m That is Whirl PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 44E The boundary work done during the process shown in the figure is to be determined Assumptions The process is quasiequilibrium Analysis The work done is equal to the the sumof the areas under the process lines 12 and 23 A ps1a P P 3 2 Who Area 1 2 v2 v1P2ltv3 v2 300 1 39 2 300 5ps1a 331 3 1Btu 3 2 5404 ps1aft 1 15 1Btu 300 psia2 33rt 3 3 5 404 psia it 1 2 33 v ft3 514Btu The negative sign shows that the work is done on the system 45 A pistoncylinder deVice contains nitrogen gas at a specified state The boundary work is to be determined for the polytropic eXpansion of nitrogen Properties The gas constant for nitrogen is 02968 kJkgK Table A2 Analysis The mass and volume of nitrogen at the final state are 3 m 101 130 kPaO39m m 007802 kg N2 RT1 02968 kJkgK120 273 K 130 kPa 120 C mRT2 007802 kg02968 kPam3kgK100 273 K P2 100 kPa 008637 m3 V2 The polytropic index is determined from Flt1 PZVZ 130 kPa007 m3 100 kPa008637 m3 n 1249 The boundary work is determined from 102V2 101V1 100 kPa008637 m3 130 kPa007 m3 1 n 1 1249 Wb 1 86 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 46 A pistoncylinder device with a set of stops contains steam at a specified state Now the steam is cooled The compression work for two cases and the final temperature are to be determined Analysis a The speci c volumes for the initial and nal states are Table A6 Pl1MPa 3 P21MPa 3 v1 030661 m kg 12 023275 m kg T1 400 C T2 250 C Steam II E Noting that pressure is constant during the process the boundary work is determined from 400 C r Q Wb mPv1 v2 06 kg1000 kPa030661 023275m3kg 443kJ b The volume of the cylinder at the nal state is 40 of initial volume Then the boundary work becomes Wb mPv1 040v1 06 kg1000 kPa030661 040 X 030661m3kg 1 104kJ The temperature at the nal state is 3 T2 151 8 C Table A5 v2 040 x 030661 2 01226 m kg 47 A pistoncylinder device contains nitrogen gas at a specified state The final temperature and the boundary work are to be determined for the isentropic eXpansion of nitrogen Properties The properties of nitrogen are R 02968 kJkgK k 1395 Tables A2a A2b Analysis The mass and the nal volume of nitrogen are P v 1 kP 3 N m 1 1 30 am 07 m 006768 kg 130 1133 RT1 02968 kJkgK180 273 K 1800C Flt1k P2V2k gt130 kPa007 m3 1395 80 kPa2139395 gtv2 009914 m3 The final temperature and the boundary work are determined as P v 3 T2 2 2 2 Z 80 kPa0 09914m 3 2395K mR 006768 kg02968 kPam kgK Wb 102V2 101V1 80 kPa009914m 130 kPa007 m 2236 l k 1 1395 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 5 48 Saturated water vapor in a cylinder is heated at constant pressure until its temperature rises to a speci ed value The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Properties Noting that the pressure remains constant during this process the speci c volumes at the initial and the nal states are Table A4 through A6 P P1 300 kPa 3 A V1 Vg 300kPa m Sat vapor P2 2 300 kPa 0 71643 31lt 1 2 v m T2 2 200 C 2 g 300 Analysis The boundary work is determined from its de nition to be V C 2 WboutJ1Pdv PV2 V1 mPV2 v1 3 1k 2 5 kg300 kPa071643 060582m kg 3 lkPam 2166 kJ Discussion The positive sign indicates that work is done by the system work output 49 Water is eXpanded isothermally in a closed system The work produced is to be determined Assumptions The process is quasiequilibrium Analysis From water table P1 P2 Psat20PC 215549 kPa P A 3 am VZZVf i39XVfg 1 gt 2 0001157 0800 12721 0001157 2 010200 m3kg The definition of specific volume gives 1 88 16 V g3 010200 31lt v2 v1quot 21m3gt m g 3 8816 m3 1 0001157 m kg The work done during the process is determined from lkJ 2 Whom L PdV 1002 1 15549 kPa88 16 1m3 3 1355x105 kJ kPam PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 46 410 A gas in a cylinder is compressed to a speci ed volume in a process during which the pressure changes linearly with volume The boundary work done during this process is to be determined by plotting the process on a P V diagram and also by integration Assumptions The process is quasiequilibrium Analysis a The pressure of the gas changes linearly with volume and thus the process curve on a P V diagram will be a straight line The boundary work during this process is simply the area under the process curve which is a trapezoidal Thus P1 2 W1 b 2 1200 kPam3042 m3 600 kPa 96 kPa P P2 W b 1200 kPam3012 m3 600 kPa 456 kPa kPa A and P2 P1 P 2 Wbout 2 Area T V2 V1 01 1 W012042m3 kJ3 x 2 1 kPa m i gt V 2 828 k 012 042 m3 9 The boundary work can also be determined by integration to be WboutI PdVIaVbdVabVz 1 1 0122 0422m6 GAS 1200 kPam3 2 600 kPaO 12 042m3 828kJ PaVb Discussion The negative sign indicates that work is done on the system work input 411 Air in a cylinder is compressed at constant temperature until its pressure rises to a specified value The boundary work done during this process is to be determined Assumptions 1 The process is quasiequilibrium 2 Air is an ideal gas Properties The gas constant of air is R 0287 kJkgK Table A1 P A Analysis The boundary work is determined from its de nition to be 2 2 V2 Pr Whom L PdV PIVIInVI mRTlnP Z T240C 120 kPa 1 15 k 0287 kJk K 297 K In g g 600 kPa gtV 206 kJ Discussion The negative sign indicates that work is done on the system work input PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 7 412 V A gas in a cylinder expands polytropically to a specified volume The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis The boundary work for this polytropic process can be determined directly from n 15 v 003 3 10211 1 350 kPa m3 2033kPa P and 15 n 2 PV PV Wboutzj pdvz 1 l n 2033x02 350gtlt003kPam3 lkJ 1 15 1kPam3 i gt V 00 02 m3 129kJ Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 48 413 A quot Problem 412 is reconsidered The process described in the problem is to be plotted on a P V diagram and the effect of the polytropic exponent n on the boundary work as the polytropic eXponent varies from 11 to 16 is to be plotted Analysis The problem is solved using EES and the solution is given below Function BoundWorkP1V1P2V2n quotThis function returns the Boundary Work for the polytropic process This function is required since the expression for boundary work depens on whether n1 or nltgt1quot If nltgt1 then BoundWorkP2V2P1V11nquotUse Equation 322 when n1quot else BoundWork P1V1nV2V1 quotUse Equation 320 when n1quot endif end quotInputs from the diagram windowquot n15 PM 350 kPa V1 003 mA3 V2 02 mA3 Gas39AR39 quotSystem The gas enclosed in the pistoncylinder devicequot quotProcess Polytropic expansion or compression PVquotn Cquot P2V2quotnP1V1 n quotn 13quot quotPolytropic exponentquot quotInput Dataquot Wb BoundWorkP1V1P2V2nquotkJquot quotIf we modify this problem and specify the mass then we can calculate the final temperature of the fluid for compression or expansionquot m1 m2 quotConservation of mass for the closed systemquot quotLet39s solve the problem for m1 005 kgquot m1 005 kg quotFind the temperatures from the pressure and specific volumequot T1temperaturegasPP1vV1m1 T2temperaturegasPP2vV2m2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 160 120 100 5 D 5 80 a 60 40 20 o I I I I I I I 002 004 01 014 018 02 v m3 n WbkJ 19 11 1814 18 1156 1725 1211 1641 1267 1563 17 1322 149 1378 1422 16 1433 1358 E39 1489 1298 E 15 1544 1242 a 16 1189 B 14 I 13 I l 12 4 11 11 12 13 14 15 16 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 410 414 Nitrogen gas in a cylinder is compressed polytropically until the temperature rises to a speci ed value The boundary work done during this process is to be determined Assumptions 1 The process is quasiequilibrium 2 Nitrogen is an ideal gas Properties The gas constant for nitrogen is R 02968 kJkgK Table A2a Analysis The boundary work for this polytropic process can be determined from P A 2 P V P V R T T WI sz PdV 22 11m2 1 1 l n l n 5 kg02968 kJkg K36O 250K 1 14 1 Discussion The negative sign indicates that work is done on the system work input 408 kJ V C 415 The unit of the quantity 10 and the boundary work done during this process are to be determined A gas whose equation of state is 7 P 10 7 2 RuT eXpands in a cylinder isothermally to a specified volume Assumptions The process is quasiequilibrium Analysis a The term 107 2 must have pressure units since it is P A added to P Thus the quantity 10 must have the unit kPam6kmolz b The boundary work for this process can be determined from P RuT 10 RuT 10 NRuT10N2 v 72 VN N2 V v2 gtV and 2 2 NRT 2 v WboutZIPdVZI 10i VNRuT1n 210N2ii 1 1 V V2 V1 V2 V1 3 4 02 kmol83 14 kJkmol K350 Kln m m 1 10 kPam6kmolz05kmol2 1 3 1 1d 3 403kJ 4 lkPam 3 m 2m3 Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 416 A quot Problem 415 is reconsidered Using the integration feature the work done is to be calculated and compared and the process is to be plotted on a P V diagram Analysis The problem is solved using EES and the solution is given below quotInput Dataquot N02 kmol v1bar2N quotmquot3kmolquot v2bar4N quotmquot3kmolquot T350 K Ru8314 kJkmoIK quotThe quation of state isquot vbarP10vbarquot2RuT quotP is in kPaquot quotusing the EES integral function the boundary work WbEES isquot WbEESNintegraIPvbar v1bar v2bar001 quotWe can show that Wbhand integeral of Pdvbar is one should solve for PFvbar and do the integral 39by hand39 for practicequot Wbhand NRuTnv2barv1bar 1 01v2bar1v1bar quotTo plot P vs vbar define Ppot fvbarpot T asquot vbarpotPpot1 0vbarpotquot2RuT quot PPplot and vbarvbarpot just to generate the parametric table for plotting purposes To plot P vs vbar for a new temperature or vbarpot range remove the 3939 and 3939 from the above equation and reset the vbarpot values in the Parametric Table Then press F3 or select Solve Table from the Calculate menu Next select New Plot Window under the Plot menu to plot the new dataquot PDIot Vplot 320 39 2909 10 1 2618 1111 280 238 1222 39 2182 1333 240 2014 1444 39 I T 350 K 187 1556 7 20 39 1746 1667 g 16039 39 1637 1778 1 V 2 154 1889 2 39 39 1455 20 If 120 80 Area Wboundary 40 0 9 11 13 15 17 19 21 17pm mskmol PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 412 417E A gas in a cylinder is heated and is allowed to expand to a specified pressure in a process during which the pressure changes linearly with volume The boundary work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis a The pressure of the gas changes linearly with volume and thus the process curve on a PV diagram will be a straight line The boundary work during this process is simply the area under the process curve which is a trapezoidal Thus At state 1 P1 al1 b P 15 psia5 psia 37 3b Psia A Pavb b 20 psia 100 2 At state 2 1 P2 aVZ 19 15 100 psia 5 psiait 3 2 20 psia v2 24 it 3 7 gtfi and P P 1 1 39 113 Who tArea 1 2 tz 1 00 5pSIa24 7 3 tu u 3 2 2 54039 ps1a it 181 Btu Discussion The positive sign indicates that work is done by the system work output 418 A pistoncylinder device contains nitrogen gas at a specified state The boundary work is to be determined for the isothermal eXpansion of nitrogen Properties The properties of nitrogen are R 02968 kJkgK k 14 Table A2a Analysis We rst determine initial and nal volumes from ideal gas relation and nd the boundary work using the relation for isothermal eXpansion of an ideal gas V1 mRT 04 kg02968 kJkgK140 273 K 2 03064 m3 P1 160 kPa V2 mRT 04 kg02968 kJkgK140 273 K 2 04903 m3 N2 P2 100 kPa 160 kPa 140 C v 3 W 2 PM ln 2 160 kPa03064m31n 230kJ V1 03064 m PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 413 419E Hydrogen gas in a cylinder equipped with a spring is heated The gas eXpands and compresses the spring until its volume doubles The final pressure the boundary work done by the gas and the work done against the spring are to be determined and a P V diagram is to be drawn Assumptions 1 The process is quasiequilibrium 2 Hydrogen is an ideal gas Analysis a When the volume doubles the spring force and the nal pressure of H2 becomes 15113 F kx k H 150001191711 750001bf s A 3 1t 2 P I F 75 OOOlbf 111 2 P2 P1 s 147 psia 2 2 1883 psia A 3 r 144 in b The pressure of H2 changes linearly with volume during this process and thus the process curve on a P V diagram will be a straight line Then the boundary work during this process is simply the area under the process curve which is a trapezoid Thus 15 30 P1 P2 2 gt V 1131 Wm Area 02 v1 1 Btu 1883147psia 540395 psiait 3 2 30 151t 3 1 2817 Btu c If there were no spring we would have a constant pressure process at P 147 psia The work done during this process would be 2 Wboutno spring 2 I1 Pdv PV2 1 Bt 147 psia3O 15it3 u3 408 Btu 540395 psia it Thus Wspn39ng Wb Wbn0 Spring 2 Z Discussion The positive sign for boundary work indicates that work is done by the system work output PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 420 A pistoncylinder device contains air gas at a specified state The air undergoes a cycle with three processes The boundary work for each process and the net work of the cycle are to be determined Properties The properties of air are R 0287 kJkgK k 14 Table A2a Analysis For the isothermal eXpansion process V1 mRT 015 kg0287 kJkgK350273 K 001341m3 P1 2000 kPa Air 2 MPa 2 K 2 K V2 mRT 0 15 kg0 87 kJkg 350 73 2005364 m3 3500C P2 500 kPa 3 Wb12 Plllln 2000 kPa001341m3ln m 3718kJ V1 001341m For the polytropic compression process PM P3V3 gt500 kPa005364 m3 12 2000 kPa31392 gtv3 001690 m3 103v3 102V2 2000 kPa001690 m3 500 kPa005364 m3 1 n 1 12 3486kJ For the constant pressure compression process Wb3l P3 V1 V3 2 2000 kPa001341 001690m3 697 kJ The net work for the cycle is the sum of the works for each process W Wb12 Wb23 Wb3l Z net 421 A saturated water mixture contained in a springloaded pistoncylinder device is heated until the pressure and temperature rises to speci ed values The work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis The initial state is saturated mixture at 90 C The pressure and the specific volume at this state are Table A4 P A P1 70183 kPa 800 kPa A 2 v1 If xvfg 0001036 01023593 0001036 1 023686 m3kg v The nal speci c volume at 800 kPa and 250 C is Table A6 22 029321m3kg Since this is a linear process the work done is equal to the area under the process line 12 P1P2 WboutArea 70183 800kPa 2 quot102 V1 1kg029321 023686m3 1kPa m 2452kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 15 422 A saturated water mixture contained in a springloaded pistoncylinder device is cooled until it is saturated liquid at a specified temperature The work done during this process is to be determined Assumptions The process is quasiequilibrium Analysis The initial state is saturated mixture at 05 MPa The speci c volume at this state is Table A5 P V1 Vf XVfg 0001093 030037483 0001093 011321 m3kg 0395 MPa 1 The nal state is saturated liquid at 100 C Table A4 2 a P2 210142 kPa 12 uf 0001043 m3kg CV Since this is a linear process the work done is equal to the area under the process line 12 P1P2 WboutArea 500 10142kPa mv2 v1 075 kg0001043 01 1321m3 lkPa m 253kJ The negative sign shows that the work is done on the system in the amount of 253 k PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 416 423 An ideal gas undergoes two processes in a pistoncylinder device The process is to be sketched on a P V diagram an expression for the ratio of the compression to eXpansion work is to be obtained and this ratio is to be calculated for given values of n and r Assumptions The process is quasiequilibrium Analysis a The processes on a P V diagram is as follows P 2 p const b The ratio of the compressiontoeXpansion work is called the 3 backwork ratio BWR 2 V 1 Process 12 Wb12 I Pd V P COHSt 1 constant V The process is PVl constant P n and the integration results in V W P2V2 PrV1 1RT2 T1 b l39z l n l n where the ideal gas law has been used However the compression work is mRT2 T1 Wcomp Wb1 2 Z n 1 3 Process 23 Wb 23 I Pd V 2 The process is P constant and the integration gives Wb2 3 PV3 42 where P P2 P3 Using the ideal gas law the eXpansion work is Wexp Z Wb2 3 quot1RT3 T2 The backwork ratio is de ned as mRT2 T1 BWRWcomp n1 1 TZ T1 1 T21 T1T2 1 1 T1T2 Since process 12 is polytropic the temperaturevolume relation for the ideal gas is quot 1 n l T2 V1 r where r is the compression ratio V1 V2 Since process 23 is constant pressure the combined ideal gas law gives P V P V T V V T3 T2 T2 V2 V2 The backwork ratio becomes 1 1 r1quot n 1 r 1 BWR c For n 14 and r 6 the value of the BWR is 1 14 BWR 1 1 6 14 1 6 1 0256 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 417 424 Water in a cylinder equipped with a spring is heated and evaporated The vapor expands until it compresses the spring 20 cm The nal pressure and temperature and the boundary work done are to be determined and the process is to be shown on a P V diagram Assumptions The process is quasiequilibrium Analysis a The nal pressure is determined from F P3 2P2 SP2 E250 kPa 100 kNm 02m lkpaz 450kPa A A 01m lkNm The speci c and total volumes at the three states are T1 25 C 3 P1 2 250 kpav12 vf25C 0001003 m kg VI mu1 50 kg0001003 m3kg 005 m3 V2 2 02 m3 3 2 V2 x23Ap 02 m3 02 m01m2 022 m3 U v v3 022 m3 3 m 50 kg At 450 kPa vf 0001088 m3kg and 8 041392 m3kg Noting that If lt 3 lt 8 the final state is a saturated mixture and thus the final temperature is T3 Tsat450kPa 14790C 00044 m3kg b The pressure remains constant during process 12 and changes linearly a straight line during process 23 Then the boundary work during this process is simply the total area under the process curve P2 P3 Wbout Area Fla2 V1 250 kPa02 005m3 W 022 02m3 i 2 1 kPam3 V3 V2 2 445 H Discussion The positive sign indicates that work is done by the system work output PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 418 425 x V 39 Problem 424 is reconsidered The effect of the spring constant on the final pressure in the cylinder and the boundary work done as the spring constant varies from 50 kNm to 500 kNm is to be investigated The nal pressure and the boundary work are to be plotted against the spring constant Analysis The problem is solved using EES and the solution is given below P3P2SpringconstV3 V2 quotP3 is a linear function of V3quot quotwhere Springconst WAAZ the actual spring constant divided by the piston face area squaredquot quotInput Dataquot P1150 kPa m50 kg T125 C P2P1 V202 mA3 A01mA2 k100 kNm DELTAx20 cm SpringconstldAquot2 quotkNmquot5quot V1mspvo1 spvo1volumeSteamiapwsPP1TT1 V2mspvo2 V3V2ADELTAxconvertcmm V3mspvo3 quotThe temperature at state 2 isquot T2temperatureSteamiapwsPP2vspvo2 quotThe temperature at state 3 isquot T3temperatureSteamiapwsPP3vspvol3 Wnetother 0 WoutWnetother Wb12Wb23 Wb12P1V2V1 quotWb23 integral of P3dV3 for Deltax 20 cm and is given byquot Wb23P2V3V2Springconst2V3V2 2 k kNm P3 kPa wout kJ 50 350 4346 1 00 450 4446 150 550 4546 200 650 4646 250 750 4746 300 850 4846 350 950 4946 400 1 050 5046 450 1 150 51 46 500 1 250 5246 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 105 104 103 IO 1389 C I 102 1 1114 C 2 P kPa 25 C 100 I I I I I 104 103 102 101 100 1o1 102 v mskg 1 39 39 l 39 I 39 39 39 39 I 39 I 1100 900 700 P3 kPa 500 I I I I I I I I I I I I I I 50 100 150 200 250 300 350 400 450 500 k kNm Tl Igt wout kJ 42 I I I I I I I I I I I I I I I 50 100 150 200 250 300 350 400 450 500 k kNm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 420 Closed System Energy Analysis 426E The table is to be completed using conservation of energy principle for a closed system Analysis The energy balance for a closed system can be expressed as Ein E out AE system r J r J Net energy traleel Chan gein intemalkinetic by heat workaand mass potentialetc energies Qin Wout 2 E2 E1 2 quotK32 31 Application of this equation gives the following completed table Qin Wout E1 E2 m 62 el Btu Btu Btu Btu lbm Btulbm 350 510 1020 860 3 533 350 130 550 770 5 440 560 260 600 900 2 150 500 0 1400 900 7 71 4 650 50 1 000 400 3 200 427E The heat transfer during a process that a closed system undergoes Without any internal energy change is to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 The compression or expansion process is quasiequilibrium Analysis The energy balance for this stationary closed system can be expressed as Ein Eout AE system r J Net energy thfer Chan gein intemalkinetic byheata workaandmass potentialetc energies Qin Wout 2 AU 2 0 since KE PE O Qin Wout Then 1 Bt Qin 11gtlt1061bfit u 1414Btu 77817lbf if PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 421 428 Motor oil is contained in a rigid container that is equipped with a stirring device The rate of speci c energy increase is to be determined Analysis This is a closed system since no mass enters or leaves The energy balance for closed system can be eXpressed as E in E out AE system EK J W Net energy t1 Sfer Chan gein intemalkinetic byheata workaand mass potentialetc energies Qin Wshjn Then AEQmWS 11525w hin Dividing this by the mass in the system gives g 25Js m 25kg A 10Jkgs 429 A rigid tank is initially lled with superheated R134a Heat is transferred to the tank until the pressure inside rises to a specified value The mass of the refrigerant and the amount of heat transfer are to be determined and the process is to be shown on a Pvdiagram Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions Analysis a We take the tank as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as Ein Eout AEsystem f J r J Net energy traleel Chan gein internalkinetic R39134a by heat workaand mass potentialetc energies 160 kPa Qin AU mu2 u1 sinceW KE PE0 Using data from the refrigerant tables Tables A11 through A13 the properties of R134a are determined to be P1 160 kPa V 00007435 ug 012355 m3kg x1 04 uf 3106 ufg 19031kJkg v1 uf xlufg 00007435 04012355 00007435 004987 m3kg n1 J xlufg 3106 0409031 2 10719 kJkg P2 700 kPa n2 2 37723 kJkg Superheated vapor V2 2 V1 Then the mass of the refrigerant is determined to be u 3 m242L1113 21o03kg v1 004987 m kg 1 9 Then the heat transfer to the tank becomes gt u Qin quot1W2 1 1003 kg37723 10719 kJkg 2708 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 422 430E A rigid tank is initially lled with saturated R134a vapor Heat is transferred from the refrigerant until the pressure inside drops to a speci ed value The final temperature the mass of the refrigerant that has condensed and the amount of heat transfer are to be determined Also the process is to be shown on a P vdiagram Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions Analysis a We take the tank as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as E in E out AEsy stem r J af J Net energy thfer Chan gein intemalkinetic by heat workaand mass potentialetc energies Qout 2 AU 2 mu2 u1 sinceW 2 KB PE 0 Qout Z quot1W1 2 Using data from the refrigerant tables Tables A11E through A13E the properties of R134a are determined to be P1 160 psiavl ug160psia 029339 rt 3lbm sat vapor ul 2 ug160psia 10851 Btulbm R 134a 160 sia 3 P P2 50 ps1a vf 001252 vg 094909 ft lbm Sat vapor v2 2 v1 uf 24824 ufg 75228 Btulbm The final state is saturated mixture Thus T2 Tsat 50 psia 40 23 F P b The total mass and the amount of refrigerant that has condensed are I v 20 it 3 1 m 1 36817lbm V1 029339 ft lbm V V 2 12 2 x2 2 f0 9339 00 5 202999 2 vfg 094909 001252 V C mf 1 x2 m 1 029996817 lbm 4773 Ibm Also L12 2 uf xzufg 24824 0299975228 4738 Btulbm c Substituting Qout mu1 2 6817lbm10851 4738 Btulbm 4167Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 423 431 Water contained in a rigid vessel is heated The heat transfer is to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved 3 The thermal energy stored in the vessel itself is negligible Analysis We take water as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem Water Q r J W Net energy traleel Changein internalkinetic 10 L A byheat W0FkandmaSS potentialetc energies 1 000 C Qin AUmu2 u1 sinceKEPE0 x0123 The properties at the initial and final states are Table A4 T1 100 C 11 If xvfg 0001043 012316720 0001043 2 02066 In3 kg x1 0123 ul 2 uf xu fg 41906 012320870 67576 kJkg V V 2 f 02066 0001091 05250 T o x Z T2 150 C 2 ufg 0392480001091 A 12 v102066m3kg 2 fx2 fg 63166 0525019274 16435 kJkg 2 The mass in the system is Cv v 3 mZAZmzamgmkg v1 02066 m3kg Substituting Qin quot1042 ul 2 004841 kg16435 67576 kJkg 469kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 424 432 Saturated vapor water is cooled at constant temperature and pressure to a saturated liquid The heat rejected is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be eXpressed as Ein Eout AEsystem ar J Net energy thfer Changein intemalkinetic byheata workaand mass potentialetc energies Q W AU 39 KEPEO Water quot39 Qout bout m 2 1 Slnce 400 kPa Qout Z Wbout m 2 1 sat vap Qout Z quot1012 hr Qout Z mall hz T A qout 2 hr hz since AU Wb AH during a constant pressure quasiequilibrium process 2 1 Since water changes from saturated liquid to saturated vapor we have qout hg hfg 400kPa Table Note that the temperature also remains constant during the process and it is the saturation temperature at 400 kPa which is 1436 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 425 433 A cylinder is initially filled with steam at a specified state The steam is cooled at constant pressure The mass of the steam the final temperature and the amount of heat transfer are to be determined and the process is to be shown on a T v diagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem ar J Net energy thfer Chan gein intemalkinetic byheataworkaandmass potentialetc energies W 2 AU 2 39 KE PE 0 Qout bout mu2 1 Slnce H20 Q Qout quot1012 hl 1 MPa since AU Wb AH during a constant pressure quasiequilibrium 450 C process The properties of water are Tables A4 through A6 P1 1MPa u1 033045 m3kg T A T2 450 C h1 33713 kJkg 1 v 25 3 m 1 m37565kg V1 033045 m kg 2 b The final temperature is determined from sat vapor P2 1MPa T2 Tsat1MPa 1799 C gt V 0 Substituting the energy balance gives Qout 7565 kg27771 33713 kJkg 4495 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 426 434 A cylinder is initially lled with saturated liquid water at a speci ed pressure The water is heated electrically as it is stirred by a paddleWheel at constant pressure The voltage of the current source is to be determined and the process is to be shown on a Pvdiagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The cylinder is well insulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AE system ar J W Net energy MSfer Chan gein intemalkinetic by heat WOIkaandmaSS potentialetc energies Wein pr in Wbout 2 AU s1nce Q KE PE 0 Wein prin mh2 hl H20 VIAt prjn mh2 h P const A 7 since AU Wb AH during a constant pressure quasiequilibrium W e process The properties of water are Tables A4 through A6 I kPa hl hf175kPa v1 uf175kPa 0001057 m3kg satliquid 0 5 h hf xzhfg 48701 05x 22131 2 15936 kJkg xz 3 m2 2 0005m3 4731kg 2 11 0001057 m kg 1 Substituting VIAt 400kJ 4731 kg15936 48701kJkg gt v VIAt 4835 kJ 4 1 A 835kJ 000V 2239V 8 A45 x 60 s 1 kJs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 427 435 V A cylinder equipped with an external spring is initially filled with steam at a specified state Heat is transferred to the steam and both the temperature and pressure rise The final temperature the boundary work done by the steam and the amount of heat transfer are to be determined and the process is to be shown on a Pvdiagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The thermal energy stored in the cylinder itself is negligible 3 The compression or expansion process is quasiequilibrium 4 The spring is a linear spring Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves Noting that the spring is not part of the system it is external the energy balance for this stationary closed system can be expressed as Ein E out 2 AE system Net energy tl meer Changein intemalkinetic byheata workaandmass potentialetc energies Q H O Qm Wbout 2 AU 2 mu2 ul s1nce KE PE 0 20011332 Qin mu2 1 Whom 200 C The properties of steam are Tables A4 through A6 P1 200 kPa v1 108049 m3kg p T1 200 C u 26546 kJkg A 2 v 04 3 mz lz m3203702kg 11 108049 m kg 1 v 3 v2 2 06 1 16207 m3kg m 03702 kg gt V 2 216207 m3kg u 33120 kJkg b The pressure of the gas changes linearly with volume and thus the process curve on a PV diagram will be a straight line The boundary work during this process is simply the area under the process curve which is a trapezoidal Thus lkJ P P 1 2 06 04m3 1 3 200 250kPa oe 7 Wb 2 Area kPam jz4skd c From the energy balance we have Qin 03702 kg33120 26546kJkg 45 kJ 288 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 428 436 5 39539 Problem 435 is reconsidered The effect of the initial temperature of steam on the final temperature the work done and the total heat transfer as the initial temperature varies from 150 C to 250 C is to be investigated The nal results are to be plotted against the initial temperature Analysis The problem is solved using EES and the solution is given below quotThe process is given byquot quotP2P1kxNA and as the spring moves 39x39 amount the volume changes by V2V1quot P2P1SpringconstV2 V1 quotP2 is a linear function of V2quot quotwhere Springconst ldA the actual spring constant divided by the piston face areaquot quotConservation of mass for the closed system isquot m21mlll quotThe conservation of energy for the closed system isquot quotEin Eout DeltaE neglect DeltaKE and DeltaPE for the systemquot Qin Wout m1u2u1 DELTAUm1u2u1 quotInput Dataquot P1200 kPa V104 mA3 T1200 C P2250 kPa V206 mA3 Fluid39SteamlAPWS39 m1V1spvol1 spvol1volumeFluidTT1 PP1 u1intenergyFluid TT1 PP1 spvol2V2m 2 quotThe final temperature isquot T2temperatureFluidPP2vspvol2 u2intenergyFluid PP2 TT2 Wnetother 0 WoutWnetother Wb quotWb integral of P2dV2 for 05ltV2lt06 and is given byquot WbP1V2V1Springconst2V2V1quot2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 106 I I SteamlAPws I I 105 104 606 C E 103 200 C i D 102 1 10 39 AreaWb 100 I I I 10393 10392 10391 100 1o1 102 vm3kg 725 100 T1 Qin T2 Wout C N C N 150 2815 5082 45 680 39 39 80 160 2828 5279 45 39 170 2842 5475 45 39 180 2855 567 45 a 635quot 47 3960 190 2869 5864 45 I I I I 4 I I i 200 2883 6058 45 E 590 40 210 2898 625 45 220 291 2 6443 45 230 2927 6634 45 545 20 240 2942 6826 45 0 Z 250 2957 7017 45 500 I I I I I I I I 0 150 170 190 210 230 250 TH C 296 29439 292 29o E 288 E 0 286 284 282 a 280 150 170 230 250 19039 I I 210 TH 01 429 Wout kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 430 437 A cylinder equipped with a set of stops for the piston to rest on is initially lled with saturated water vapor at a specified pressure Heat is transferred to water until the volume doubles The final temperature the boundary work done by the steam and the amount of heat transfer are to be determined and the process is to be shown on a P vdiagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein E out 2 AE system Net energy thfer Changein intemalkinetic 300 kPa byheata workaandmass potentialetc energies Qin Wbout 2 AU 2 mu3 ul s1nce KE PE 0 H20 Qin quot1W3 1 l Wb0ut 250 kPa Sat vapor The propert1es of steam are Tables A4 through A6 P1 250 kPa v1 ug250kPa 071873 m3kg satvap0r ul 2 ug250kPa 25368 kJkg P 3 11 071873 m kg 3 v3 1396 m 14375 m3kg m 1113 kg CV 103 300 kPa T3 662 C V3 214375 m3kg u3 34114 kJkg b The work done during process 12 is zero since V const and the work done during the constant pressure process 23 is 1kJ 3 3 Whom LPdV Pa3 42 300 kPa16 08m3 am 2240 kJ 0 Heat transfer is determined from the energy balance Qin mu3 1 Wbout 1113 kg34114 2536 8 kJkg 240 kJ 1213 k PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 431 438 A room is heated by an electrical radiator containing heating oil Heat is lost from the room The time period during which the heater is on is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141 C and 377 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant speci c heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 The local atmospheric pressure is 100 kPa 5 The room is airtight so that no air leaks in and out during the process Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cu 0718 kJkgK for air at room temperature Table A2 Oil properties are given to be 0 950 kgm3 and CI 22 kJkg C Analysis We take the air in the room and the oil in the radiator to be the system This is a closed system since no mass crosses the system boundary The energy Room balance for th1s statlonary constantvolume closed system can be expressed as 10 C Ein Eout AEsystem R d39 J W a 1ator Net energy t1 Sfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies Viin QoutAt AJair AJoil E mcu T2 T1air mcp T2 T1o since KE 2 PE 2 0 The mass of air and oil are Pv 3 m 2 a1 100 kPa50m Z 6232 kg m RTI 0287kPa m3kg K1O 273 K moil po voil 950 kgm3 0040 m3 38 kg Substituting 24 035 kJsAt 6232 kg0718 kJkg C20 10 C 38 kg22 kJkg C50 10 C gtAt 218505 308min Discussion In practice the pressure in the room will remain constant during this process rather than the volume and some air will leak out as the air eXpands As a result the air in the room will undergo a constant pressure eXpansion process Therefore it is more proper to be conservative and to using AH instead of use AU in heating and airconditioning applications PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 432 439 A saturated water mixture contained in a springloaded pistoncylinder device is heated until the pressure and volume rise to speci ed values The heat transfer and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as P A A 2 Ein Eout AEsystem 225 kPa f J W Net energy thfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies 1 Qin Wb0ut 2 AU 2 mu2 ul since KE PE 0 75 kPa Qin Wbout m 2 1 gt v The initial state is saturated mixture at 75 kPa The speci c volume and internal energy at this state are Table A5 v1 If 3wfg 0001037 00822172 0001037 01783 m3kg ul 2 uf xufg 38436 00821118 55330 kJkg The mass of water is V 3 m 1 L31r22kg V1 01783 m kg The final specific volume is V 3 2 2 2 5i 2 04458 m3kg m 1122 kg The final state is now xed The internal energy at this speci c volume and 225 kPa pressure is Table A6 u2 16504 kJkg Since this is a linear process the work done is equal to the area under the process line 12 lkJ P P WboutZArea 1 2 2 lkPam3 75225 kPa v21 5 2m3 j 450kJ Substituting into energy balance equation gives Qin Wb out mu2 ul 2 450 kJ 1122 kg16504 55330 kJkg 12750kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 433 440E Saturated R134a vapor is condensed at constant pressure to a saturated liquid in a pistoncylinder device The heat transfer and the work done are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be eXpressed as Ein Eout AE system r J W Net energy thfer Chan gein intemalkinetic byheat W0rkand mass potentialetc energies Whirl Qout 2 AU 2 mn2 ul since KE PE 0 R134a7 O Qout Wbin mu2 1 F The properties at the initial and final states are Table A11E T1 100 F v1 2 ug 034074rt 3 lbm TA x1 1 ul 2 big 2 10746 Btulbm T2 100 F v2 V 2001386ft3 lbm x2 0 u2 uf 44774 Btulbm 2 1 Also from Table A11E P1 P2 13893 psia ufg 62690 Btulbm hfg 71094 Btulbm tV The work done during this process is 1Btu 2 wb out J39 Pdv Pv2 v1 13893 psia001386 034074 31b 3 1 5404 psia 11 j 8404 Btulbm That is Whirl Substituting into energy balance equation gives qout Whirl 142 M1 wb in ufg 840 62690 2 71 OQBtulbm Discussion The heat transfer may also be determined from q out hz h1 qout h fg 7109 Btulbm since AU Wb AH during a constant pressure quasiequilibrium process PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 434 441 One part of an insulated tank contains compressed liquid while the other side is evacuated The partition is then removed and water is allowed to expand into the entire tank The final temperature and the volume of the tank are to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 The tank is insulated and thus heat transfer is negligible 3 There are no work interactions Analysis We take the entire contents of the tank as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as Ein Eout AE system Net energy traleer Chan gein intemalkinetic byheatworkandmass potentialetc energies Evacuated O 2 AU 2 mu2 ul since W Q 2 KB PE 0 Partition 1 Z 2 The properties of water are Tables A4 through A6 H20 P1 600 kPa u uf6ooc 0001017 m3kg T1 60 C u uf6ooc 25116 kJkg We now assume the final state in the tank is saturated liquidvapor mixture and determine quality This assumption will be verified if we get a quality between 0 and 1 P2 10 kPa V 0001010 ug 14670 m3kg u ul 17 19179 ufg 22454 kJkg 2 uf 25116 19179 x2 002644 ufg 22454 Thus T2 Tsat 10 kPa 4581 C 02 uf xzufg 0001010 002644x 14670 0001010 038886 m3kg and V mug 25 kg038886 m3kg 0972 m3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 435 442 x 39 Problem 441 is reconsidered The effect of the initial pressure of water on the final temperature in the tank as the initial pressure varies from 100 kPa to 600 kPa is to be investigated The final temperature is to be plotted against the initial pressure Analysis The problem is solved using EES and the solution is given below quotKnownsquot m25 kg P1600 kPa T160 C P210 kPa quotSolutionquot FIuid39SteamIAPWS39 quotConservation of Energy for the closed tankquot EinEoutDELTAE Ein0 Eout0 DELTAEmu2 u1 5o u1INTENERGYFIuidPP1 TT1 v1voumeFuidPP1 TT1 T2temperatureFuidPP2 uu2 40 T2T2 v2voumeFuidPP2 uu2 30 Vtotamv2 539 P1 T2 N 20 kPa C l 100 4579 200 4579 10 300 4579 400 4579 0 I I I I I 500 4579 600 4579 100 200 300 400 500 600 PH kPa Steam l 600 500 quot39 400 g l 300 200 600 kPa 100 1 1o kP 2 o I la v I I 10394 10393 10392 10391 10 101 102 103 v m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 436 Specific Heats Au and Ah of Ideal Gases 443C It can be used for any kind of process of an ideal gas 444C It can be used for any kind of process of an ideal gas 445C Very close but no Because the heat transfer during this process is Q mcpAT and CI varies with temperature 446C The energy required is mcpAT which will be the same in both cases This is because the CI of an ideal gas does not vary with pressure 447 C The energy required is mcpAT which will be the same in both cases This is because the CI of an ideal gas does not vary with volume 448C For the constant pressure case This is because the heat transfer to an ideal gas is mcpAT at constant pressure chT at constant volume and CI is always greater than 0 449 The desired result is obtained by multiplying the rst relation by the molar mass M Mc p Mcu MR or cp cURu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 437 450 The enthalpy change of oxygen is to be determined for two cases of speci ed temperature changes Assumptions At specified conditions oxygen behaves as an ideal gas Properties The constantpressure specific heat of oxygen at room temperature is 0 0918 kJkgK Table A 2a Analysis Using the specific heat at constant pressure Ah cpAT 0918kJkg K250 150K 91 8kJkg If we use the same room temperature specific heat value the enthalpy change will be the same for the second case However if we consider the variation of speci c heat with temperature and use the speci c heat values from Table A 2b we have 0 0964 kJkgK at 200 C 473 K and CI 0934 kJkgK at 100 C E 373 K Then Ahl cpATl 0964 kJkg K250 150K 964 kJkg M2 cpxrl 0934 kJkg 19100 0K 934 kJkg The two results differ from each other by about 3 The pressure has no in uence on the enthalpy of an ideal gas 451E Air is compressed isothermally in a compressor The change in the specific volume of air is to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E Analysis At the compressor inlet the speci c volume is 1 2 03704ps1a it lbrn R70460 R 981mi 3lbm P1 20 ps1a Similarly at the compressor exit 1 4 3 1 R 4 R V2 Z zm 370 ps1a it brn 70 60 le309 3lbm P2 150 ps1a f The change in the speci c volume caused by the compressor is Av 12 91 1309 9816 851ft3lbm 452 The total internal energy changes for neon and argon are to be determined for a given temperature change Assumptions At specified conditions neon and argon behave as an ideal gas Properties The constantvolume speci c heats of neon and argon are 06179 kJkgK and 03122 kJkgK respectively Table A2a Analysis The total internal energy changes are AU 2 mcUAT 2 kg06179 kJkg 19180 20K 1977kJ 116011 AUargon mcUAT 2 kg03122 kJkg K180 20K 999kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 438 453 The enthalpy changes for neon and argon are to be determined for a given temperature change Assumptions At specified conditions neon and argon behave as an ideal gas Properties The constantpressure specific heats of argon and neon are 05203 kJkgK and 10299 kJkgK respectively Table A2a Analysis The enthalpy changes are Ahargon cpAT 05203 kJkg K25 75K 260kJ kg Ahneon cpAT 10299 kJkg K25 75K 515kJkg 454 The internal energy change of hydrogen gas during a heating process is to be determined using an empirical specific heat relation constant specific heat at average temperature and constant specific heat at room temperature Analysis a Using the empirical relation for 5 T from Table A2c and relating it to EU T EUTEP R a RubTcT2 dT3 Where a 2911 b 01916gtlt10392 c 04003x10395 and d 08704gtlt10399 Then 2 2 M 2 EVTdT 2 a RubT cT2 dT3dT 1 1 a RuT2 T119T22 T12 cT23 T13 dT24 T14 2911 8314800 200 01961gtlt10 28002 2002 04003 gtlt1058003 2003 08704 gtlt1098004 2004 12487 kJkmol ALT 12487 kJkmol A 6194 kJkg M 2016 kgkmol b Using a constant cp value from Table A2b at the average temperature of 500 K 0an CU500K 39 K Au cmvga T1 10389 kJkg K800 200K 6233 kJkg c Using a constant 0 value from Table A 2a at room temperature 0an CU300K 39 K Au 0 T2 T1 10183 kJkg K800 200K 6110 kJkg uav g PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 439 455 The enthalpy change of nitrogen gas during a heating process is to be determined using an empirical speci c heat relation constant speci c heat at average temperature and constant speci c heat at room temperature Analysis a Using the empirical relation for El T from Table A2c Ep 2 abTcT2 dT3 where a 2890 b 01571gtlt10392 c 08081x10395 and d 2873gtlt10399 Then 2 2 2 3 Ah 2 cpTdT 2 abTcT dT ldT 1 1 4 T1 9022 T cT T MT T14 28900000 600 01571gtlt 10 210002 6002 08081gtlt10 510003 6003 2873 gtlt10910004 6004 12544 kJkmol Ah A l 12544 kJkmol M 28013 kgkmol b Using the constant 0 value from Table A2b at the average temperature of 800 K Cpavg Z 1 39 K Ah 2 0 T2 T1 1121 kJkg K1000 600K 4484 kJkg 4478 kJkg pavg c Using the constant 0 value from Table A2a at room temperature Cpavg Cp300K 39 K Ah cpavgT2 T1 1039 kJkg K1000 600K 4156 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 440 456E A springloaded pistoncylinder device is filled with air The air is now cooled until its volume decreases by 50 The changes in the internal energy and enthalpy of the air are to be determined Assumptions At specified conditions air behaves as an ideal gas Properties The gas constant of air is R 03704 psiaft3lbmR Table A1E The speci c heats of air at room temperature are cu 0171 BtulbmR and CI 0240 BtulbmR Table A2Ea Analysis The mass of the air in this system is m 10111 250 psia1 3 3 07336 lbm RTl 03704 psia1t lbm R460 460 R The final specific volume is then u 1 3 2 2 2 1681611 31bm m 07336lbm As the volume of the air decreased the length of the spring will increase by AV 405113 2 2 2091671t 1100in AI 7zD 4 7r1012 The final pressure is then Pzzpr Apzpr A szr Ezpr kgx A A 71D 4 4 1 39 11 39 250 psia 5 b mx 200 m 249311917111 2 mom 2493psia Employing the ideal gas equation of state the final temperature will be 10212 2493psia05 3 458 R quot1R 07336 lbm0 3704 psia1t 3lbm R T2 Using the specific heats Au 2 CUAT 0171BtulbmR4587 920R 78QBtulbm Ah 2 cpAT 0240 Btulbm R4587 920R 1107Btulbm Pl 2 CV PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 441 Closed System Energy Analysis Ideal Gases 457 C No it isn39t This is because the rst law relation Q W AU reduces to W 0 in this case since the system is adiabatic Q 0 and AU 0 for the isothermal processes of ideal gases Therefore this adiabatic system cannot receive any net work at constant temperature 458 The hydrogen gas in a rigid tank is cooled until its temperature drops to 300 K The nal pressure in the tank and the amount of heat transfer are to be determined Assumptions 1 Hydrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 240 C and 130 MPa 2 The tank is stationary and thus the kinetic and potential energy changes are negligible Ake E Ape E 0 Properties The gas constant of hydrogen is R 4124 kPam3kgK Table A1 The constant volume specific heat of hydrogen at the average temperature of 450 K is c avg 10377 kJkgK Table A2 Analysis a The nal pressure of hydrogen can be determined from the ideal gas relation P l P l T 350 K 1 2 gt P2 2 2101 2 250 kPa 1591kPa T1 T2 T1 550 K b We take the hydrogen in the tank as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein E out 2 AE system r J Net energy thfer Changein intemalkinetic byheata workaandmass potentialetc energies H2 Qout AU 250 kPa Qout AU 2 mu2 ul E va T1 T2 550 K where Q P v 250 kP 30 3 m 1 ax m 03307kg RT1 4124 kPam3kgK550 K Substituting into the energy balance Qout 033307 kg10377 kJkgK550 350K 6862 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 442 459E A paddle wheel in an oxygen tank is rotated until the pressure inside rises to 20 psia while some heat is lost to the surroundings The paddle wheel work done is to be determined Assumptions 1 Oxygen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 181 F and 736 psia 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 The energy stored in the paddle wheel is negligible 4 This is a rigid tank and thus its volume remains constant Properties The gas constant and molar mass of oxygen are R 03353 psiaft3lbmR and M 32 lbmlbmol Table A1E The specific heat of oxygen at the average temperature of Tavg 7355402 638 R is cuavg 0160 BtulbmR Table A 2E Analysis We take the oxygen in the tank as our system This is a closed system since no mass enters or leaves The energy balance for this system can be expressed as Ein E0ut AE system Net energy thfer Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies 02 prjn Qom AU 147 psia 80 F 20 Btu prin Qout mu2 1 a E Qout mcu T2 T1 l The final temperature and the mass of oxygen are ivz Tzzinzm W54OR735R T1 T2 P1 147 ps1a 3 m 147 ps1a10 it 08121bm RTI 03353 psia 3lbmol R54O R Substituting Wm 20 Btu 0812 lbm0160 BtulbmR735 540 R 453 Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 443 460E The air in a rigid tank is heated until its pressure doubles The volume of the tank and the amount of heat transfer are to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 221 F and 547 psia 2 The kinetic and potential energy changes are negligible Ape E Ake E 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications Properties The gas constant of air is R 03704 psiaft3lbmR Table AlE Analysis a The volume of the tank can be determined from the ideal gas relation Air 3 10 lbm V mRTl 10 lbm03704 ps1a lbm R525 R 2 648113 30 psia P1 30 pm 65 F b We take the air in the tank as our system The energy balance for this stationary closed system can be expressed as E in E out 2 AE system ar J EH Net energy thfer Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies Qin quot1012 1 E mcu T2 T1 The final temperature of air is PV P V P 1 2 2 T2 2T1 2gtlt525 R1050R T1 T2 P1 The internal energies are Table A17E ul U525R Substituting Qin 10 lbm18148 8948Btulbm 920 Btu Alternative solutions The specific heat of air at the average temperature of Tavg 52510502 788 R 328 F is from Table A2Eb cuavg 0175 BtulbmR Substituting Qin 101bm 0175 BtulbmR1050 525 R 919 Btu Discussion Both approaches resulted in almost the same solution in this case PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 444 461E Paddle Wheel work is applied to nitrogen in a rigid container The nal temperature is to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K 2271 R and 339 MPa 492 psia 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant speci c heats at room temperature can be used for nitrogen Properties For nitrogen cu 0177 BtulbmR at room temperature and R 03830 psia ft 3lbm R Table A1E and A 2Ea Analysis We take the nitrogen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Ein Eout AE system af J 1E enerng traaner Chan gein intemalkinetic by eat W01 aan mass potent1aletcenerg1es Nitrogen prin 2 AU 2 mcu T2 T1 since KE PE 0 20 psia lOO F The mass of nitrogen is pr P v 2 39 1 3 m 1 0 psi 009325 lbm RT1 03830 psia it lbm R560 R Substituting 1 B 5000 lbf it A 0093251bm0177 Btulbm RT2 560R 77817 lbf T 2 949B 489 F 462 One part of an insulated rigid tank contains an ideal gas While the other side is evacuated The nal temperature and pressure in the tank are to be determined when the partition is removed Assumptions 1 The kinetic and potential energy changes are negligible Ake E Ape E 0 2 The tank is insulated and thus heat transfer is negligible Analysis We take the entire tank as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be eXpressed as Ein Eout AEsystem b zgfneggri tangirass Change in internal kinetic W 39 39 y potent1al etc energles IDEAL 0 AUmu2 u1 GAS 2 1 Evacuated Th f 800 kPa ere ore 50 C T2 T1 Since it uT for an ideal gas Then PV P V V 1 A A P2 1101 800 kPa 400kPa T1 T2 v2 2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4 45 463 A resistance heater is to raise the air temperature in the room from 5 to 25 C Within 11 min The required power rating of the resistance heater is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141 C and 377 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 Heat losses from the room are negligible 5 The room is airtight so that no air leaks in and out during the process Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cu 0718 kJkgK for air at room temperature Table A2 Analysis We take the air in the room to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary constantvolume closed system can be eXpressed as Ein E out 2 AE system ar J W Net energy ttmsfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies Wein AU 3 mcmvg T2 T1 s1nce Q 2 KE 2 PE 2 0 4gtlt5gtlt6 m3 or 5 C gt VVeinAt mcuavgT2 Tl AIR We The mass of air is 6 V4gtlt5gtlt6120 m3 P v 3 m 2 1 100 kPa120 m 1506 kg RTI 0287 kPam kg K278 K Substituting the power rating of the heater becomes 1506 kg0718 kJkg C25 5 C ein 11x60s Discussion In practice the pressure in the room will remain constant during this process rather than the volume and some air will leak out as the air eXpands As a result the air in the room will undergo a constant pressure eXpansion process Therefore it is more proper to be conservative and to use AH instead of using AU in heating and airconditioning applications PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 446 464 A student living in a room turns her 150W fan on in the morning The temperature in the room when she comes back 10 h later is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141 C and 377 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 All the doors and windows are tightly closed and heat transfer through the walls and the windows is disregarded Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cu 0718 kJkgK for air at room temperature Table A2 Analysis We take the room as the system This is a closed system since the doors and the windows are said to be tightly closed and thus no mass crosses the system boundary during the process The energy balance for this system can be expressed as Ein Eout AE N f system t V t W by CEZIiegne aif i iil fg im ROOM W 2 AU 61quot 3 m x 4 m x 4 m We ln mu2 u1E mcu T2 T1 Fan The mass of air is V3gtlt4gtlt448 m3 P v 100 kP 48 3 m 1 ax m 5708kg I RT1 0287 kPa m3kg K293 K The electrical work done by the fan is We 2 WeAt 0100kJs8 x 3600 s 2880 k Substituting and using the cu value at room temperature 2880 kJ 5708 kg0718 kJkg CT2 20 C T2 903 C Discussion Note that a fan actually causes the internal temperature of a con ned space to rise In fact a 100W fan supplies a room with as much energy as a 100W resistance heater PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 447 465 A room is heated by a radiator and the warm air is distributed by a fan Heat is lost from the room The time it takes for the air temperature to rise to 20 C is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 141 C and 377 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant specific heats at room temperature can be used for air This assumption results in negligible error in heating and airconditioning applications 4 The local atmospheric pressure is 100 kPa 5 The room is airtight so that no air leaks in and out during the process Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also cu 0718 kJkgK for air at room temperature Table A2 Analysis We take the air in the room to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary constantvolume closed system can be expressed as Ein Eout AEsystem 5000 kJh Net energy Sfer Chan gein intemalkinetic byheataworkaandmass potentialetc energies Qin Wfanin Qout E mcuavgT2 Since KE 2 PE 2 or 4m x 5m x 7m Qin VVfaan QoutAt mcuavgT2 The mass of air is SE 4gtlt5gtlt714Om3 10000kJh 14 3 W m 100 km 0 m 1724 kg RT1 0287 kPam kgK283 K lt Using the cu value at room temperature 10000 50003600 kJs 01kJsAt 1724 kg0718 kJkg C20 10 C It yields At 831 s Discussion In practice the pressure in the room will remain constant during this process rather than the volume and some air will leak out as the air expands As a result the air in the room will undergo a constant pressure eXpansion process Therefore it is more proper to be conservative and to using AH instead of use AU in heating and airconditioning applications PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 448 466 Argon is compressed in a polytropic process The work done and the heat transfer are to be determined Assumptions 1 Argon is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 151 K and 486 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 Properties The properties of argon are R 02081kJkgK and C 03122 kJkgK Table A2a Analysis We take argon as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as E in E out AE system r 4 W Argon Net energy thfer Chan gein intemalkinetic 1 2 kP by heat workaand mass potentialetc energies O a Q l 0 C W AUmc T T Qm bout V 2 1 Puquot constant Using the boundary work relation for the polytropic process of an ideal gas gives n 1n 0212 RT1 i 1 02081 kJkg K283 K 800 I 109 SkJkg w 2 b m 1 n P1 1 12 120 Thus whim 1095kJkg The temperature at the nal state is 111 n 0212 P T2 T1 2 283 19800 kPa 3882K P1 120 kPa From the energy balance equation qin wb out cu T2 T1 1095 kJkg 03122 kJkg 106882 283K 766 kJkg Thus qout 766 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4 49 467 An insulated cylinder is initially filled with air at a speci ed state A paddlewheel in the cylinder stirs the air at constant pressure The final temperature of air is to be determined Assumptions 1 Air is an ideal gas with variable speci c heats 2 The cylinder is stationary and thus the kinetic and potential energy changes are zero 3 There are no work interactions involved other than the boundary work 4 The cylinder is well insulated and thus heat transfer is negligible 5 The thermal energy stored in the cylinder itself and the paddlewheel is negligible 6 The compression or eXpansion process is quasiequilibrium Properties The gas constant of air is R 0287 kPam3kgK Table A1 Also 0 1005 kJkgK for air at room temperature Table A2 The enthalpy of air at the initial temperature is 11 h298 K Table Analysis We take the air in the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AE system Net energy quot31 Sfel Chan gein intemalkinetic 7 byheatW0rkandmaSS potentialetc energies prin Wbout AU prin hl AIR s1nce AU Wb AH dur1ng a constant pressure quasrequrlrbrrum process P const The mass of air is m 2 2 3 0468 kg 39 RT1 0287 kPam kgK298 K pr PIV 400 kPa01m3 lg Substituting into the energy balance 15 k 0468 kgh2 29818 kJkg h2 33023 kJkg From Table A17 T2 3299 K Alternative solution Using speci c heats at room temperature cp 1005 kJkg C the nal temperature is determined to be prjn mm hl mp0 71 a 15 kJ 0468 kg1005 kJkg CT2 25 c which gives T2 563 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 450 468 Carbon dioxide contained in a springloaded pistoncylinder device is heated The work done and the heat transfer are to be determined Assumptions 1 C02 is an ideal gas since it is at a high temperature relative to its critical temperature of 3042 K 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 Properties The properties of C02 are R 01889 kJkgK and P C 0657 kJkgK Table AZa kPa A Analysis We take C02 as the system This is a closed system 1000 2 since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Ein E out 2 AE system ar J W Net energy thfer Changein intemalkinetic 100 1 by heat workand mass potentialetc energies Qin Wbout AU mcu T2 T1 V m3 The initial and nal speci c volumes are ka1 1 kg0 1889 kPam3kg K298 K P1 100 kPa v1 05629 m3 mRT2 1 kg0 1889 kPam3kg K573 K P2 1000 kPa 01082 m3 V2 Pressure changes linearly with volume and the work done is equal to the area under the process line 12 P P Wm Area 04 v1 lkJ W01082 05629m3 3 2 lkPa m 2501kJ Thus Whin 2501kJ Using the energy balance equation Qin Wb out mcu T2 T1 2501kJ 1 kg0657 kJkg K3OO 25K 694 kJ Thus Qout 694 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 451 469E A cylinder is initially lled with nitrogen gas at a speci ed state The gas is cooled by transferring heat from it The amount of heat transfer is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium 5 Nitrogen is an ideal gas with constant specific heats Properties The gas constant of nitrogen is 03830 psiaft3lbmR The speci c heat of nitrogen at the average temperature of Tavg 7002002 450 F is cpavg 02525 Btulbm F Table A2Eb Analysis We take the nitrogen gas in the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be expressed as Ein Eout AEsystem J W Net en ergy traleel Chan gein intemalkin etic by heat workaand mass potentialetc energies Qout Wb0ut mu2 ulQout hl mcpT2 Tl since AU Wb AH during a constant pressure quasiequilibrium process The mass of N2 40 ps1a n1trogen 1s 700 F I Q Plv 40 psia25 it 3 m 3 2251lbm RT1 03830 psia lbm R1160 R Substituting Qout 2251 lbm02525 Btulbm F700 200 F 2842 Btu 470 A pistoncylinder device contains air A paddle wheel supplies a given amount of work to the air The heat transfer is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1325 K and 377 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant specific heats can be used for air Analysis We take the air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Ein Eout AEsystem Net energy MSfer Changein intemalkinetic by heat W01kandmaSS potentialetc energies Air I prin Wb0ut Qin ch T2 T1 kPa I Q prin Wbout Qin 0 Slnce T1 2 T2 170C Qin WbOUt Using the boundary work relation on a unit mass basis for the isothermal W pw process of an 1deal gas g1ves wout 2 RT ln 2 2 RT ln 3 0287 kJkg K290 Kln3 914 kJkg 1 Substituting into the energy balance equation eXpressed on a unit mass basis gives qin Wbput pr in Discussion Note that the energy content of the system remains constant in this case and thus the total energy transfer output via boundary work must equal the total energy input via shaft work and heat transfer PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 452 471 A cylinder is initially filled with air at a specified state Air is heated electrically at constant pressure and some heat is lost in the process The amount of electrical energy supplied is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 Air is an ideal gas with variable speci c heats 3 The thermal energy stored in the cylinder itself and the resistance wires is negligible 4 The compression or eXpansion process is quasiequilibrium Properties The initial and final enthalpies of air are Table A17 hl h298K AIR P const Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this closed A system can be expressed as We 1 Ein Eout AEsystem V J E 4 Net energy tralsfer Changeinintemalkinetic by heat W0rkand mass potentialetc energies Wein Qout Wbout AU Wein hl Qout since AU Wb AH during a constant pressure quasiequilibrium process Substituting Wein 15 kg35049 29818kJkg 60 kJ 845 k or 1 kWh 3600 k We 845kJ 0235 kWh Alternative solution The speci c heat of air at the average temperature of Tavg 25 772 51 C 324 K is from Table A2b cpavg 10065 kJkg C Substituting Wein mcp T2 T1 Qout 15 kg10065 kJkg C77 25 C 60 k 845 kJ or Wein 845 kJ 0235 kWh 3600 k Discussion Note that for small temperature differences both approaches give the same result PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 45 3 472 A cylinder initially contains nitrogen gas at a speci ed state The gas is compressed polytropically until the volume is reduced by onehalf The work done and the heat transfer are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The N2 is an ideal gas with constant speci c heats 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or expansion process is quasiequilibrium Properties The gas constant of N2 is R 02968 kPam3kgK Table Al The cu value of N2 at the anticipated average temperature of 350 K is 0744 kJkgK Table A2b Analysis We take the contents of the cylinder as the system This is a closed system since no mass crosses the system boundary The energy balance for this closed system can be expressed as Ein Eout AEsystem r J W Net energy thfer Chan gein intemalkinetic by heat W0rkand mass potentialetc energies Wbin Qout AU mu2 1 Wbin Qout mcu T2 N2 100 kPa The nal pressure and temperature of n1trogen are o 25 C Q P 13 C M v Plug3 Flt113 gt P2 1 P1 2 213000 kPa 2462 kPa M2 P2 T 2462kPa Prvr szz gt T2 1 100 kPa XO5gtlt 298 K 3669K T1 T2 P1 V1 Then the boundary work for this polytropic process can be determined from Wbin I2Pdvzmzm 1 l n 1n 22kg02968 kJkg K3669 298K 149 9 kJ 1 13 I Substituting into the energy balance gives Qout Wbin mcu T2 T1 1499 k 22 kg0744 kJkgK3669 298K 372 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4 54 473 39539 Problem 472 is reconsidered The process is to be plotted on a P V diagram and the effect of the polytropic exponent n on the boundary work and heat transfer as the polytropic exponent varies from 10 to 14 is to be investigated The boundary work and the heat transfer are to be plotted versus the polytropic eXponent Analysis The problem is solved using EES and the solution is given below Procedure WorkP2V2P1V1nWin If n1 then WinP1V1InV2V1 Else WinP2V2P1V11n endif End quotInput Dataquot Vratio05 quotV2V1 Vratioquot quotn13quot quotPolytropic exponentquot PM 100 kPa T1 25273 K m22 kg MMmolarmassnitrogen Ru8314 kJkmoIK RRuMM V1mRT1P1 quotProcess equationsquot V2VratioV1 P2V2T2P1V1T1quotThe combined ideal gas law for states 1 and 2 plus the polytropic process relation give P2 and T2quot P2V2 nP1V1 n quotConservation of Energy for the closed systemquot quotEin Eout DeltaE we neglect Delta KE and Delta PE for the system the nitrogenquot Qout Winmu2u1 u1intenergyN2 TT1 quotinternal energy for nitrogen as an ideal gas kJkgquot u2intenergyN2 TT2 Call WorkP2V2P1V1nWin quotThe following is required for the Pv plotsquot PpotspvpotTpotP1V1mT1 quotThe combined ideal gas law for states 1 and 2 plus the polytropic process relation give P2 and T2quot PpotspvpotquotnP1V1mquotn spVpotRTpotPpotquotmquot3quot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Pressure vs specific volume as function of polytropic exponent 1800 4500 1600 4000 I110 1400 1 3500 1200 2 3000 E 1000 2500 5 n39 39 a E 800 2000 n 46 600 1500 13 400 39 1000 200 500 0 I I I 0 0 02 04 06 08 1 spvplot mA3kg n Qout Win kJ K1 1 1349 1349 1044 1218 137 1089 1082 1391 1133 9425 1413 1178 798 1435 1222 6486 1458 1267 4942 1481 1311 3345 1505 1356 1693 1529 14 01588 1554 160 140 Win kJ 120 c 100 x v g 80 Qout kJ o 0 60 40 20 O I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I 1 105 11 115 12 125 13 135 14 n 455 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 456 474E A cylinder initially contains air at a speci ed state Heat is transferred to the air and air expands isotherrnally The boundary work done is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The air is an ideal gas with constant speci c heats 3 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass crosses the system boundary The energy balance for this closed system can be expressed as Ein Eout AEsystem L r J W Net energy thfer Chan gein intemalkinetic byheataworkaandmass potentialetc energies QinWboutAU m 2 1mcuT2 T10 A1r since u uT for ideal gases and thus u2 ul when T1 T2 Therefore 40 Btu T const Wbout Qin 40 B111 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 457 475 V A cylinder equipped with a set of stops on the top is initially filled with air at a specified state Heat is transferred to the air until the piston hits the stops and then the pressure doubles The work done by the air and the amount of heat transfer are to be determined and the process is to be shown on a Pv diagram Assumptions 1 Air is an ideal gas with variable speci c heats 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 The thermal energy stored in the cylinder itself is negligible Properties The gas constant of air is R 0287 kPam3kgK Table Al Analysis We take the air in the cylinder to be the system This is a closed system since no mass crosses the boundary of the system The energy balance for this closed system can be expressed as Ein Eout AE system r J W Net en ergy tr Sfer Chan gein internal kinetic Q by heat W0rkand mass potentialetc energies AIR 1 Qin Wbout AU mu3 1 200 kPa Qin mu3 1 Wbout The initial and the nal volumes and the nal temperature of air are determined from A mRTl 3 kg0287 kPam3kg K300 K P1 200 kPa v3 2v1 2x129 258 m3 rel v1 P3V3 P3 3 400 kPa T1 T3 Flt1 200kPa V1 129 m3 x2gtlt300 K 1200K 1 2 V No work is done during process 23 since 2 3 The pressure remains constant during process 12 and the work done during this process is 2 W1 devP2ltv2 v1 l 200 kPa258 129 mfi 1kPam312258kJ The initial and nal internal energies of air are Table A17 Substituting Qin 3 kg93333 21407kJkg 258 k 2416 kJ Alternative solution The specific heat of air at the average temperature of Tavg 300 12002 750 K is from Table A 2b c Wg 0800 kJkgK Substituting Qin mu3 1 Wbout E mcu Wbout 3 kg0 800 kJkgK1200 300 K 258 k 2418 k PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 458 476 Air at a speci ed state contained in a pistoncylinder device with a set of stops is heated until a final temperature The amount of heat transfer is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 3042 K 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 Properties The properties of air are R 0287 kJkgK and C 0718 kJkgK Table AZa Analysis We take air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be eXpressed as E in E out AEsy stem I I ar J W Net energy thfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies Qin Whom 2 AU 2 mcu T2 T1 100 kPa 23 C The volume will be constant until the pressure is 200 kPa 025 m3 P T2 T1 2296 K ZOOkPa 592K P1 100 kPa P The mass of the air is kPa 3 m 101v1 100 kPa025 m 2 02943 kg 200 2 g RT1 0287 kPa m Ikg K296 K A The boundary work done during process 23 is 100 1 Wbout P2 V3 V2 mRT3 T2 V 02943 kg0287 kPa m3kg K7OO 592K 912 kJ Substituting these values into energy balance equation Qin Wb out mcu T3 T1 912 kJ 02943 kg0718 kJkg K700 296K 945 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4 59 477 Air at a speci ed state contained in a pistoncylinder device undergoes an isothermal and constant volume process until a nal temperature The process is to be sketched on the P V diagram and the amount of heat transfer is to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 3042 K 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 Properties The properties of air are R 0287 kJkgK and C 0718 kJkgK Table AZa Analysis a The processes 12 isothermal and 23 constantvolume are sketched on the P V diagram as shown b We take air as the system This is a closed system since no mass crosses h the boundaries of the system The energy balance for this system fort he process 13 can be expressed as Air Ein Eout AESyStem 100 kPa Net energy traleer Chan gem 1ntemalk1net1c 3 byheata workaandmass potentialetc energies 04 111 Wbout1 2 Qin AU mcu T3 T1 The mass of the air is P A 3 kPa P V m 1 1 Z 600 kPa0 8m 139394kg 600 RT1 0287 kPa m3kg K1200 K 300 The work during process 12 is determined from boundary work relation for an isothermal process to be V2 P1 Wb0m12 mRTl ln 7 mRT1 In P 1 2 600 kPa 1394k 0287 kPa m3k K 1200 K In g g 300 kPa 3328 kJ V P since 2 1 for an isothermal process 1 P 2 Substituting these values into energy balance equation Qin Wbout1 2 mcu T3 T1 3328 kJ 1394 kg0718 kJkg K3OO 1200K 568 kJ Thus Qout 568 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 460 478 A cylinder initially contains argon gas at a speci ed state The gas is stirred While being heated and expanding isothermally The amount of heat transfer is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 The air is an ideal gas with constant speci c heats 3 The compression or expansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass crosses the system boundary The energy balance for this closed system can be expressed as 15 kJ Ein Eout AEsystem V W 1 Net energy thfer Changein intemalkinetic byheataworkaandmass potentialetc energies Qin prin Wbout mu2 1 mcu T2 Z 0 Ar T const s1nce u uT for 1deal gases and thus u2 ul when T1 T2 Therefore 3 1d 3 Qin Wbout Vpr in 15 3 12kJ Q Closed System Energy Analysis Solids and Liquids 479E Liquid water experiences a process from one state to another The internal energy and enthalpy changes are to be determined under different assumptions Analysis a Using the properties from compressed liquid tables h1 hf50 FvfPPsatT 1Btu 1807 Btulbm 001602 ft 3lbm50 017812 psia 3 5404 psia ft 1 1821 Btulbm P2 2000 psia L12 2 6736 Btulbm T able A 7E T2 100 F h 7330 Btulbm Au 2 uz ul 2 6736 1807 49293tulbm Ah 2 h2 h1 7330 1821 5508Btulbm b Using incompressible substance approximation and property tables Table A4E ul E uf 50 F 21807 Btulbm h1 E hf 50 F 21807 Btulbm uz E uf1003F 6803 Btulbm h2 E hf1003F 6803 Btulbm Au 2 uz ul 2 6803 1807 49963tulbm Ah 2 h2 h1 6803 1807 49963tulbm c Using speci c heats and taking the speci c heat of water to be 100 BtulbmR Table A3Ea Ah 2 Au 2 CAT 2 100 Btulbm R100 50R 50 Btulbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 461 480E A person shakes a canned of drink in a iced water to cool it The mass of the ice that will melt by the time the canned drink is cooled to a specified temperature is to be determined Assumptions 1 The thermal properties of the drink are constant and are taken to be the same as those of water 2 The effect of agitation on the amount of ice melting is negligible 3 The thermal energy capacity of the can itself is negligible and thus it does not need to be considered in the analysis Properties The density and speci c heat of water at the average temperature of 85372 61 F are p 623 lbmft3 and CI 10 Btulbm F Table A3E The heat of fusion of water is 1435 Btulbm Analysis We take a canned drink as the system The energy balance for this closed system can be expressed as Ein Eout AE system ar J W Net energy traleel Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies Cola Qout AUcanneddrink mu2 1 Qout mCT1 T2 85 F Noting that 1 gal 128 oz and 1 ft3 748 gal 9575 oz the total amount of heat transfer from a ball is 3 1 m pV 623 lbmft 3 12 ozcan l gill 07811bmcan 748 gal 128 urd oz Qout mcT1 T2 07811bmcan10 Btulbm F85 37 F 375 Btucan Noting that the heat of fusion of water is 1435 Btulbm the amount of ice that will melt to cool the drink is Qout 375 Btucan m Ice 0261 lbm per can of drink hif 1435Btulbm since heat transfer to the ice must be equal to heat transfer from the can Discussion The actual amount of ice melted will be greater since agitation will also cause some ice to melt PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 462 481 An iron Whose base plate is made of an aluminum alloy is turned on The minimum time for the plate to reach a specified temperature is to be determined Assumptions 1 It is given that 85 percent of the heat generated in the resistance wires is transferred to the plate 2 The thermal properties of the plate are constant 3 Heat loss from the plate during heating is disregarded since the minimum heating time is to be determined 4 There are no changes in kinetic and potential energies 5 The plate is at a uniform temperature at the end of the process Properties The density and speci c heat of the aluminum alloy plate are given to be p 2770 kgm3 and CI 875 kJkg C Analysis The mass of the iron39s base plate is m pv pLA 2770 kgm3 0005 m003 m2 04155 kg Air Noting that only 85 percent of the heat generated is transferred to the plate 22 C the rate of heat transfer to the 1ron s base plate 1s IR ON Qin O90gtlt 1000 W 900 W 1000 W gt We take plate to be the system The energy balance for this closed system can be expressed as gt Ein E out 2 AE system gt af J Net energy traleel Chan gein intemalkinetic byheata workaandmass potentialetc energies Qin AUPIate mu2 ul gtQin At mcT2 T1 Solving for At and substituting chTplate 04155 kg875 Jkg C200 22 C Q39in 900 Js At 7195 which is the time required for the plate temperature to reach the speci ed temperature PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 463 482 Stainless steel ball bearings leaving the oven at a specified uniform temperature at a specified rate are exposed to air and are cooled before they are dropped into the water for quenching The rate of heat transfer from the ball bearing to the air is to be determined Assumptions 1 The thermal properties of the bearing balls are constant 2 The kinetic and potential energy changes of the balls are negligible 3 The balls are at a uniform temperature at the end of the process Properties The density and speci c heat of the ball bearings are given to be p 8085 kgm3 and CI 0480 kJkg C Analysis We take a single bearing ball as the system The energy balance for this closed system can be expressed as Furnace Ein Eout AE sy stem Water 25 C f J W Net en ergy thfer Chan gein intemalkin etic by heat workaandmass potentialetc energies Qout Z AUbau 2 quotKM 1 Qout 171001 T2 The total amount of heat transfer from a ball is Steel balls 900 C ooo oo 3 3 70 12 m pv pT 8085 kgm w 0007315 kg Qout mcT1 T2 0007315 kg0480 kJkg C900 850 C 01756 kJball Then the rate of heat transfer from the balls to the air becomes Q total nbaugoutwerbam 800 ballsmingtlt 01756 kJball 1405 kJmin 234 kW Therefore heat is lost to the air at a rate of 234 kW 483E Large brass plates are heated in an oven at a rate of 300min The rate of heat transfer to the plates in the oven is to be determined Assumptions 1 The thermal properties of the plates are constant 2 The changes in kinetic and potential energies are negligible Properties The density and speci c heat of the brass are given to be p 5325 lbmft3 and CI 0091 Btulbm F Analysis We take the plate to be the system The energy balance for this closed system can be eXpressed as a E Plates Ein Eout system 0 Net energy t1 Sfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies Qin AJplate mu2 1 mCT2 The mass of each plate and the amount of heat transfer to each plate is m pV pLA 53251bm 3 1612 it 2 it2 11 284 lbm Qin mcT2 T1 284lbmplate0091Btulbmf F900 75 F 21320 Btuplate Then the total rate of heat transfer to the plates becomes Q total nplategmerplate 300 platesmin X 21320 Btuplate 6396000Btumin 106600 Btus PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 464 484 Long cylindrical steel rods are heattreated in an oven The rate of heat transfer to the rods in the oven is to be determined Assumptions 1 The thermal properties of the rods are constant 2 The changes in kinetic and potential energies are negligible Properties The density and speci c heat of the steel rods are given to be p 7833 kgm3 and CI 0465 kJkg C Analysis Noting that the rods enter the oven at a velocity of 2 mmin and eXit at the same velocity we can say that a 3m long section of the rod is heated in the oven in 1 min Then the mass of the rod heated in 1 minute is m pv pLA 1002 4 7833 kgm3 2 m7r008 m2 4 7875 kg We take the 2m section of the rod in the oven as the system The energy balance for this closed system can be expressed as Ein Eout AEsystem 39 39 39 2037911 39 39 r J I I I I I I I I I I I I Net energy traleer intemaLkinetiC I I I I I I I I I I I I byheata workaandmass potentialetc energies gt O QmAvmdmltu2 u1gtmcltT2 ngt Steer C Substituting Qin mcT2 T1 7875 kg0465 kJkg C700 30 C 24530 kJ Noting that this much heat is transferred in 1 min the rate of heat transfer to the rod becomes Qin Qin At 24530 kJl min 24530kJmin 409 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 465 485 An electronic device is on for 5 minutes and off for several hours The temperature of the device at the end of the 5 min operating period is to be determined for the cases of operation with and Without a heat sink Assumptions 1 The device and the heat sink are isothermal 2 The thermal properties of the device and of the sink are constant 3 Heat loss from the device during on time is disregarded since the highest possible temperature is to be determined Properties The speci c heat of the device is given to be 0 850 Jkg C The speci c heat of aluminum at room temperature of 300 K is 902 Jkg C Table A3 Analysis We take the device to be the system Noting that electrical energy is supplied the energy balance for this closed system can be eXpressed as Electronic device 25 C Ein Eout AEsystem ar J W Net energy transfer Changein intemalkinetic by heat W0rkandmaSS potentialetc energies Wein AUdevice mu2 1 W ein At mcT2 T1 Substituting the temperature of the device at the end of the process is determined to be 25 Js5 x 60 s 0020 kg850 Jkg CT2 25 C gt T2 466 C Without the heat sink Case 2 When a heat sink is attached the energy balance can be eXpressed as W ein AUdevice AUheat sink W einAt mCT2 Tldevice mCT2 Tlheat sink Substituting the temperature of the deviceheat sink combination is determined to be 25 Js5 x 60 s 0020 kg850 Jkg CT2 25 C 0500 kg902 Jkg CT2 25 C T2 41 0 C With heat sink Discussion These are the maXimum temperatures In reality the temperatures will be lower because of the heat losses to the surroundings PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 466 486 quot Problem 485 is reconsidered The effect of the mass of the heat sink on the maximum device temperature as the mass of heat sink varies from 0 kg to 1 kg is to be investigated The maximum temperature is to be plotted against the mass of heat sink Analysis The problem is solved using EES and the solution is given below quotKnownsquot quotT1 is the maximum temperature of the devicequot Qdotout 25 W mdevice20 g Cpdevice850 JkgC A5 cm 2 DELTAt5 min Tamb25 C msink05 kg quotCpal taken from Table A3b at 300Kquot Cpa0902 kJkgC T2Tamb quotSolutionquot quotThe device without the heat sink is considered to be a closed systemquot quotConservation of Energy for the closed systemquot quotEdotin Edotout DELTAEdot we neglect DELTA KE and DELTA PE for the system the devicequot Edotin Edotout DELTAEdot Edotin 0 Edotout Qdotout quotUse the solid material approximation to find the energy change of the devicequot DELTAEdot mdeviceconvertgkgCpdeviceT2T1deviceDELTAtconvertmins quotThe device with the heat sink is considered to be a closed systemquot quotConservation of Energy for the closed systemquot quotEdotin Edotout DELTAEdot we neglect DELTA KE and DELTA PE for the device with the heat sinkquot Edotin Edotout DELTAEdotcombined quotUse the solid material approximation to find the energy change of the devicequot DELTAEdotcombined mdeviceconvertgkgCpdeviceT2 T1deviceampsinkmsinkCpaT2 T1deviceampsinkconvertkJJDELTAtconvertmins msink T1 deviceampsink 500 kg C 39 0 4662 01 9496 400 02 6299 03 5108 5 I 04 4485 3OO 05 4103 E 06 3844 3953 07 3657 6 200 08 3515 5 I 09 3405 If 3 1 3316 100 I I 3939l I I I I I 0 02 04 08 1 06 msink kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 467 487 The face of a person is slapped For the speci ed temperature rise of the affected part the impact velocity of the hand is to be determined Assumptions 1 The hand is brought to a complete stop after the impact 2 The face takes the blow without signi cant movement 3 No heat is transferred from the affected area to the surroundings and thus the process is adiabatic 4 No work is done on or by the system 5 The potential energy change is zero APE 0 and AE AU AKE Analysis We analyze this incident in a professional manner without involving any emotions First we identify the system draw a sketch of it and state our observations about the specifics of the problem We take the hand and the affected portion of the face as the system This is a closed system since it involves a xed amount of mass no mass transfer We observe that the kinetic energy of the hand decreases during the process as evidenced by a decrease in velocity from initial value to zero while the internal energy of the affected area increases as evidenced by an increase in the temperature There seems to be no significant energy transfer between the system and its surroundings during this process Under the stated assumptions and observations the energy balance on the system can be expressed as Ein Eout AE system Wr J W Net energy thfer Chan gein intemalkinetic by heat workaandmass potentialetc energies O AUaffected tissue AKE hand 0 mCATaffected tissue V 2 2 hand That is the decrease in the kinetic energy of the hand must be equal to the increase in the internal energy of the affected area Solving for the velocity and substituting the given quantities the impact velocity of the hand is determined to be V 2mCAT affected tissue hand mhand 20 15 kg38 kJkg C24 C 1000 m2s2 Vhand 09 kg 1kJkg 551ms or 198 kmh Discussion Reconstruction of events such as this by making appropriate assumptions are commonly used in forensic engineering PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 468 488 A number of brass balls are to be quenched in a water bath at a speci ed rate The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined Assumptions 1 The thermal properties of the balls are constant 2 The balls are at a uniform temperature before and after quenching 3 The changes in kinetic and potential energies are negligible Properties The density and speci c heat of the brass balls are given to be p 8522 kgm3 and CI 0385 kJkg C Analysis We take a single ball as the system The energy balance for this closed system can be expressed as Ein Eout AEsystem r J Brass balls 120 C Net energy thfer Chan gem 1ntemalk1net1c gt byheat W0rkandmaSS potentialetc energies Qout Z AUbau quot1012 1 out mCTr T2 The total amount of heat transfer from a ball is 3 3 m pv 5 8522 kgm w 0558 kg Qout mcT1 T2 0558 kg0385 kJkg Cl20 74 C 988 kJball Then the rate of heat transfer from the balls to the water becomes Qtotal baHQbaH 100 ballsmingtlt 988 kJball 988 kJrnin Therefore heat must be removed from the water at a rate of 988 kJmin in order to keep its temperature constant at 50 C since energy input must be equal to energy output for a system Whose energy level remains constant That is Em 2 EM when AE 0 system PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 469 489 A number of aluminum balls are to be quenched in a water bath at a specified rate The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined Assumptions 1 The thermal properties of the balls are constant 2 The balls are at a uniform temperature before and after quenching 3 The changes in kinetic and potential energies are negligible Properties The density and specific heat of aluminum at the average temperature of 120742 97 C 370 K are p 2700 kgm3 and CI 0937 kJkg C Table A3 Analysis We take a single ball as the system The energy balance for this closed system can be expressed as Ein E0ut AE system Aluminum balls Net energy thfer Chan gein intemalkinetic I by heat W0rkand mass potentialetc energies Qout Z AUbau quot1W2 1 Qout Z mCT1 T2 The total amount of heat transfer from a ball is 3 3 m 2 pl 2 p72 2 2700 kgm m 01767 kg Qout mcT1 T2 01767 kg0937 kJkg C120 74 C 762 kJball Then the rate of heat transfer from the balls to the water becomes Qtotal baHQbau 100 ballsmingtlt 762 kJball 762 kJmin Therefore heat must be removed from the water at a rate of 762 kJmin in order to keep its temperature constant at 50 C since energy input must be equal to energy output for a system whose energy level remains constant That is Em Eout when AE 0 system Special Topic Biological Systems 490C The food we eat is not entirely metabolized in the human body The fraction of metabolizable energy contents are 955 for carbohydrates 775 for proteins and 977 for fats Therefore the metabolizable energy content of a food is not the same as the energy released when it is burned in a bomb calorimeter 491C Yes Each body rejects the heat generated during metabolism and thus serves as a heat source For an average adult male it ranges from 84 W at rest to over 1000 W during heavy physical activity Classrooms are designed for a large number of occupants and thus the total heat dissipated by the occupants must be considered in the design of heating and cooling systems of classrooms 492C 1 kg of natural fat contains almost 8 times the metabolizable energy of 1 kg of natural carbohydrates Therefore a person who lls his stomach with carbohydrates will satisfy his hunger without consuming too many calories PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 470 493 The average body temperature of the human body rises by 2 C during strenuous exercise The increase in the thermal energy content of the body as a result is to be determined Properties The average specific heat of the human body is given to be 36 kJkg C Analysis The change in the sensible internal energy of the body is AU chT 80 kg36 kJkg C2 C 576 k as a result of body temperature rising 2 C during strenuous exercise 494 Two men are identical except one jogs for 30 min while the other watches TV The weight difference between these two people in one month is to be determined Assumptions The two people have identical metabolism rates and are identical in every other aspect Properties An average 68kg person consumes 540 Calh while jogging and 72 Calh while watching TV Table 42 Analysis An 80kg person who jogs 05 h a day will have jogged a total of 15 h a month and will consume AE C onsumed 540 72 Ca1h15 h4391868 I 80 kg 34578 kJ 1 Cal 68 kg more calories than the person watching TV The metabolizable energy content of 1 kg of fat is 33100 kJ Therefore the weight difference between these two people in 1month will be AEconsumed 34578 kJ 104kg Energy content of fat 33100 kJkg AWifat 495 A bicycling woman is to meet her entire energy needs by eating 30g candy bars The number of candy bars she needs to eat to bicycle for 1h is to be determined Assumptions The woman meets her entire calorie needs from candy bars while bicycling Properties An average 68kg person consumes 639 Calh while bicycling and the energy content of a 20g candy bar is 105 Cal Tables 41 and 42 Analysis Noting that a 20g candy bar contains 105 Calories of metabolizable energy a 30g candy bar will contain Ecandy 105 ma a g 1575 Cal g of energy If this woman is to meet her entire energy needs by eating 30g candy bars she will need to eat 639Calh N 4cand bars l candy 1575Ca1 y PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 471 496 A 90kg man eats 1L of ice cream The length of time this man needs to jog to burn off these calories is to be determined Assumptions The man meets his entire calorie needs from the ice cream while jogging Properties An average 68kg person consumes 540 Calh while jogging and the energy content of a 100ml of ice cream is 110 Cal Tables 41 and 42 Analysis The rate of energy consumption of a 55kg person while jogging is k Econmd 540 Carh kg 715 Calh g Noting that a 100ml serving of ice cream has 110 Cal of metabolizable energy a 1liter box of ice cream will have 1100 Calories Therefore it will take 1100 Cal 1 54h 715 Calh of jogging to burn off the calories from the ice cream 497 A man switches from an apple a day to 2001111 of ice cream and 20min walk every day The amount of weight the person will gain or lose with the new diet is to be determined Assumptions All the extra calories are converted to body fat Properties The metabolizable energy contents are 70 Cal for a an apple and 220 Cal for a 200ml serving of ice cream Table 41 An average 68kg man consumes 432 Calh while walking Table 42 The metabolizable energy content of 1 kg of body fat is 33100 kJ Analysis The person who switches from the apple to ice cream increases his calorie intake by Eextm 220 70 150Cal The amount of energy a 60kg person uses during a 20min walk is E CODSU med 2 432 Calh20 min 1h w 127 Cal 60 mm 68 kg Therefore the man now has a net gain of 150 127 23 Cal per day which corresponds to 23gtlt30 690 Cal per month Therefore the man will gain 690Ca1 41868kJ mfat 33100kJkg 1Cal 0087 kg of body fat per month with the new diet Without the exercise the man would gain 0569 kg per month PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 472 498 A man with 20kg of body fat goes on a hunger strike The number of days this man can survive on the body fat alone is to be determined Assumptions 1 The person is an average male who remains in resting position at all times 2 The man meets his entire calorie needs from the body fat alone Properties The metabolizable energy content of fat is 33100 Calkg An average resting person burns calories at a rate of 72 Calh Table 42 Analysis The metabolizable energy content of 20 kg of body fat is Efat 33100 kJkg20 kg 662000 kJ The person will consume E C 41 onsumed 72 CalhX24 h 7235 kJday Therefore this person can survive 662000 k 915 days 7235 k day on his body fat alone This result is not surprising since people are known to survive over 100 days without any food intake 499 Two 50kg women are identical except one eats her baked potato with 4 teaspoons of butter while the other eats hers plain every evening The weight difference between these two woman in one year is to be determined Assumptions 1 These two people have identical metabolism rates and are identical in every other aspect 2 All the calories from the butter are converted to body fat Properties The metabolizable energy content of 1 kg of body fat is 33100 k The metabolizable energy content of 1 teaspoon of butter is 35 Calories Table 41 Analysis A person who eats 4 teaspoons of butter a day will consume 365 days 1 year E CO mum 35 Calteaspoon4 teaspoonsday 51100 Calyear Therefore the woman who eats her potato with butter will gain 65kg 51100 Cal 41868 kJ mfat 33100 kJkg 1 Cal of additional body fat that year PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 473 4100 A woman switches from 1L of regular cola a day to diet cola and 2 slices of apple pie It is to be determined if she is now consuming more or less calories Properties The metabolizable energy contents are 300 Cal for a slice of apple pie 87 Cal for a 200ml regular cola and 0 for the diet drink Table 43 Analysis The energy contents of 2 slices of apple pie and 1L of cola are Epie 2x 300 Cal 2 600 Cal Ecola 5 X 87 Cal 2 435 Cal Therefore the woman is now consuming more calories 4101E A man and a woman have lunch at Burger King and then shovel snow The shoveling time it will take to burn off the lunch calories is to be determined for both Assumptions The food intake during lunch is completely metabolized by the body Properties The metabolizable energy contents of different foods are as given in the problem statement Shoveling snow burns calories at a rate of 420 Calh for the woman and 610 Calh for the man given Analysis The total calories consumed during lunch and the time it will take to burn them are determined for both the man and woman as follows Man Lunch calories 720400225 1345 Cal 1345 Cal shovelingman m Woman Lunch calories 3304000 730 Cal 730 Cal Shovelin time At g shovelrngwoman 420 C 31 h Shoveling time At 2201 174h 4102 Three people have different lunches The person who consumed the most calories from lunch is to be determined Properties The metabolizable energy contents of different foods are 5 30 Cal for the Big Mac 640 Cal for the whopper 350 Cal for french fries and 5 Cal for each olive given Analysis The total calories consumed by each person during lunch are Person 1 Lunch calories 530 Cal Person 2 Lunch calories 640 Cal Person 3 Lunch calories 3505gtlt50 600 Cal Therefore the person with the Whopper will consume the most calories PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 74 4103 A 75kg man decides to lose 5 kg by exercising without reducing his calorie intake The number of days it will take for this man to lose 5 kg is to be determined Assumptions 1 The diet and exercise habits of the person remain the same other than the new daily exercise program 2 The entire calorie de ciency is met by burning body fat Properties The metabolizable energy content of body fat is 33100 Calkg text Analysis The energy consumed by an average 68kg adult during fastswimming fast dancing jogging biking and relaxing are 860 600 540 639 and 72 Calh respectively Table 42 The daily energy consumption of this 75kg man is 860 600 540 639 Ca1h1h 72 Calh23 h 4737 Cal g Therefore this person burns 4737 4000 737 more Calories than he takes in which corresponds to mfat 737 Cal 41868 k 009324 kg 33100 kJkg 1 Cal of body fat per day Thus it will take only 5kg 2 536 days 009324 kgday for this man to lose 5 kg 4104E The range of healthy weight for adults is usually expressed in terms of the body mass index BMI in SI units as Wkg BMI H2m2 This formula is to be converted to English units such that the weight is in pounds and the height in inches Analysis Noting that 1 kg 22 lbm and 1 m 3937 in the weight in lbm must be divided by 22 to convert it to kg and the height in inches must be divided by 3937 to convert it to m before inserting them into the formula Therefore Wkg W1bm22 7 Wlbm H2m2 H2in239372 H2in2 Every person can calculate their own BMI using either SI or English units and determine if it is in the healthy range PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 475 4105 A person changes hisher diet to lose weight The time it will take for the body mass index BMI of the person to drop from 30 to 20 is to be determined Assumptions The de cit in the calori intake is made up by burning body fat Properties The metabolizable energy contents are 350 Cal for a slice of pizza and 87 Cal for a 200ml regular cola The metabolizable energy content of 1 kg of body fat is 33100 kJ Analysis The lunch calories before the diet is E 2gtltecoke 3gtlt350 Cal2gtlt87 Cal 1224 Cal 6163 pizza The lunch calories after the diet is E M 2xe 1xecoke 2gtlt350Cal1gtlt87 Cal 787 Cal 0 pizza The calorie reduction is Emduction 21224 787 2 437 Cal The corresponding reduction in the body fat mass is mfat 437 Cal 41868kJ 005528 kg 33100kJkg 1Cal The weight of the person before and after the diet is W1 13Mr1 x mm 30 x 16 m2 768 kg W2 BMI2 gtlt thim 20 x 16 m2 512 kg Then it will take W1 W2 2 768 512 kg mfat 005528 kgday Time 463 days for the BMI of this person to drop to 20 4106E An average American adult switches from drinking alcoholic beverages to drinking diet soda The amount of weight the person will lose per year as a result of this switch is to be determined Assumptions 1 The diet and exercise habits of the person remain the same other than switching from alcoholic beverages to diet drinks 2 All the excess calories from alcohol are converted to body fat Properties The metabolizable energy content of body fat is 33100 Calkg text Analysis When the person switches to diet drinks he will consume 210 fewer Calories a day Then the annual reduction in the calories consumed by the person becomes Reduction in energy intake E duced 210 Cal day365 days year 76650 Cal year 1396 Therefore assuming all the calories from the alcohol would be converted to body fat the person who switches to diet drinks will lose R 39 39 39 E 41 Reduction in weight eductron 1n energy 1ntake Ieduced 76650 Calyr 868 kJ 970 kgyr Enegy content of fat efat 33100 kJkg 1 Cal or about 21 pounds of body fat that year PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 476 Review Problems 4107 For a 10 C temperature change of air the final velocity and final elevation of air are to be determined so that the internal kinetic and potential energy changes are equal Properties The constantvolume speci c heat of air at room temperature is 0718 Air kJkg C Table A2 00C 100C Analysis The internal energy change is determined from 0 mSV612 Au 2 CUAT 0718 kJkg C10 C 718 kJkg 0 m gt Z2 AUAKEAPE Equating kinetic and potential energy changes to internal energy change the nal velocity and elevation are determined from 1kJkg 1000 m2s2 AuzAke 2sz V12 gt718kJkgV22 0m2s2 gtV2 120ms 1kJkg 1000 m2s2 AuApegz2 z1 gt718kJkg981ms2z2 0m gtz2 732m 4108 Heat is transferred to a pistoncylinder device containing air The expansion work is to be determined Assumptions 1 There is no friction between piston and cylinder 2 Air is an ideal gas Properties The gas contant for air is 0287 kJkgK Table A2a Analysis Noting that the gas constant represents the boundary work for a unit mass and a unit temperature change the expansion work is simply determined Alf r om AT5 C W mATR 05 kg5 K0287 kJkgro 07175kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 477 4109E An insulated rigid vessel contains air A paddle wheel supplies work to the air The work supplied and final temperature are to be determined Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 2385 R and 547 psia 2 The kinetic and potential energy changes are negligible Ake E Ape E 0 3 Constant specific heats can be used for air Properties The speci c heats of air at room temperature are 0 0240 BtulbmR and C 0171 BtulbmR Table A2Ea Analysis We take the air as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as E in E out 2 AE system ar J W Net energy thfer Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies As the speci c volume remains constant during this process the ideal gas equation gives P T2 T1 2 520 R 40 psla 6933 R 2333 F P1 30 psra Substituting pr in mcu T2 T1 21bm0171Btulbm R6933 520R 593 Btu 4110 Air at a given state is eXpanded polytropically in a pistoncylinder device to a speci ed pressure The nal temperature is to be determined Analysis The polytropic relations for an ideal gas give P 1 1 kPa 0212 T T 2 400273K 466K193 C 2 11 1000kPa 3135 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 478 4111 Nitrogen in a rigid vessel is heated The work done and heat transfer are to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K and 339 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E O 3 Constant speci c heats at room temperature can be used for nitrogen Properties For nitrogen cu 0743 kJkgK at room temperature Table AZa Analysis We take the nitrogen as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as Ein Eout AEsystem f J W Net energy transfer Changein intemalkinetic NltI OgCIl byheat W0rkandmaSS potentialetc energies 100 kPa Q Qm AU metT2 T1 25 C There is no work done since the vessel is rigid w 0 kJkg Since the specific volume is constant during the process the nal temperature is determined from ideal gas equation to be P T2T12298KW894K 1 100 psia Substituting qin 2 cu T1 T2 0743 kJkg K894 298K 4428le kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 479 4112 A wellinsulated rigid vessel contains saturated liquid water Electrical work is done on water The nal temperature is to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 The thermal energy stored in the tank itself is negligible Analysis We take the contents of the tank as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein E out AEsystem ar J W Net energy thfer Changein intemalkinetic P A byheat workaandmass potentialetc energies 2 Wain 2 AU 2 mu2 u1 The amount of electrical work added during 30 minutes period is 1W lVA Wain VIAt 50 V10 A30 X 60 s j 900000J 900 kJ Cv The properties at the initial state are Table A4 ul 2 uf 400C 2 16753 kJkg v1 V 400C 2 0001008 m3kg Substituting W 6 1 16753kJkg 93 W e m mu2 u1 gtu2 u1 46753 kJkg m The final state is compressed liquid Noting that the speci c volume is constant during the process the final temperature is determined using EES to be u 46753 kJkg v2 1 0001008m kg Discussion We cannot find this temperature directly from steam tables at the end of the book Approximating the final compressed liquid state as saturated liquid at the given internal energy the nal temperature is determined from Table A4 to be T2 5 Tsat u46753kJkg 15 C Note that this approximation resulted in an answer which is not very close to the exact answer determined using EES PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 480 4113 A 1 ton 1000 kg of water is to be cooled in a tank by pouring ice into it The nal equilibrium temperature in the tank is to be determined Assumptions 1 Thermal properties of the ice and water are constant 2 Heat transfer to the water tank is negligible 3 There is no stirring by hand or a mechanical deVice it will add energy Properties The specific heat of water at room temperature is c 418 kJkg C and the speci c heat of ice at about 0 C is c 211 kJkg C Table A 3 The melting temperature and the heat of fusion of ice at 1 atm are 0 C and 3337 kJkg Analysis We take the ice and the water as our system and disregard any heat transfer between the system and the surroundings Then the energy balance for this process can be written as Em Eout AEsystem af J Net energy tranSfer Change in internal kinetic by heat work and mass potential etc energies 0 2 AU 0AU AU rce water 711600 C T1Soljd mhif mCT2 0 C liquidice 7776012 Tl water 2 0 Substituting 80 kg2l 1 kJkg C0 5 c 3337 kJkg 418 kJkg CT2 0 C 1000 kg418 kJkg CT2 20 C 0 It gives T2 124OC which is the nal equilibrium temperature in the tank 1 WATER 1 ton ice 5 C 80 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 481 4114 A cylinder equipped with a set of stops for the piston is initially lled with saturated liquidvapor mixture of water a specified pressure Heat is transferred to the water until the volume increases by 20 The initial and nal temperature the mass of the liquid when the piston starts moving and the work done during the process are to be determined and the process is to be shown on a Pv diagram Assumptions The process is quasiequilibrium Analysis a Initially the system is a saturated mixture at 160 kPa pressure and thus the initial temperature is T1 T at160kPa Z 1133OC S The total initial volume is v1 mfuf mgvg 1gtlt00010542x109152184 m3 H 0 Then the total and speci c volumes at the final state are 3 1g v3 121 12gtlt 2184 26211113 v 3 3 3 08736m3kg m 3 kg Thus A P3 500 kPa 2 3 T3 675 C 3 08736 m3kg b When the piston rst starts moving P2 500 kPa and 2 1 2184 m3 The speci c volume at this state is 1 v 2 184 3 gt V m 2 2 07280 m3kg m 3 kg which is greater than 8 03748 m3kg at 500 kPa Thus no liquid is left in the cylinder when the piston starts moving 0 No work is done during process 12 since 1 2 The pressure remains constant during process 23 and the work done during this process is lkJ W J39 3Pdv 10203 v2 500 kPa2621 2184m3 3 2 lkPam J218kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 482 4115 A cylinder is initially filled with saturated R134a vapor at a speci ed pressure The refrigerant is heated both electrically and by heat transfer at constant pressure for 6 min The electric current is to be determined and the process is to be shown on a T vdiagram Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible 2 The thermal energy stored in the cylinder itself and the wires is negligible 3 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be eXpressed as Ein Eout AEsystem T Net energy transfer Ch 39nint mal kin ti byheatworkandmass pggneiial tzen rgie C Rl34a Qin Wein Wb0ut Since Q KE PE A P conSt39 Qin Wein hl W since AU Wb AH during a constant pressure quasiequilibrium process The properties of R134a are Tables All through A13 T A P1 240 kPa 2 11 hg24kPa sat vapor P2 240 kPa 1 h2 31453 kJkg T1 70 C Substituting gt v 1 A 300000 VAs 110 VI6 x 60 s 12 kg31453 24732kJkg s I128A PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 483 4116 Saturated water vapor contained in a springloaded pistoncylinder device is condensed until it is saturated liquid at a specified temperature The heat transfer is to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem ar J W Net energy thfer Chan gein intemalkinetic P byheat W0rkandmaSS potentialetc energies A m 2 Wbin Qout AU 111 1 Slnce KE PE 0 1235 kPa Qout Wbin m 2 1 qout Z wbin 2 1 1555 kPa The properties at the initial and final states are Table A4 at 200 C 21555 kPa tV P1 2 Ps Since this is a linear process the work done is equal to the area under the process line 12 PrP2 wbout Area 2 V2 V1 1555 1235kPa 2 0001012 01272 m3 kgi3 989kJkg lkPam Thus Whirl Substituting into energy balance equation gives qout wan 142 ul 2 989 kJkg 20933 25942 kJkg 2484kJ kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 484 4117 The ideal gas in a pistoncylinder device is cooled at constant pressure The gas constant and the molar mass of this gas are to be determined Assumptions There is no friction between piston and cylinder Properties The speci c heat ratio is given to be 1667 Analysis Noting that the gas constant represents the boundary work for a unit mass and a unit temperature change the gas constant is simply determined from Ideal gas r gt R W 16396 k1 2075kJkgK 0398 1 Q mAT 08 kg10 C AT 10 C The molar mass of the gas is M R 839314kJm 139K 4007kgkmol R 2075 kJkgK The speci c heats are determined as C R 2075 kJkgK quot k 1 1667 1 cp 2 cu R 3111 kJkgK2075 kJkgK 5186 kJkg C 3111 kJkg C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 485 4118 A cylinder is initially lled with helium gas at a specified state Helium is compressed polytropically to a speci ed temperature and pressure The heat transfer during the process is to be determined Assumptions 1 Helium is an ideal gas with constant specific heats 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Properties The gas constant of helium is R 20769 kPam3kgK Table A1 Also cu 31156 kJkgK Table A2 Analysis The mass of helium and the eXponent n are determined to be Pv 1 kP 2 3 m 11 00 330 m 003403kg RT1 20769 kPa m kg K283 K PV PV TP 1 kP 11 2 2 gtv2 2142563KgtltMx02m3005684m3 RT1 RT2 T1P2 283K 700 kPa He P v P2V2 2P1V1 2 1 Hm 02 j n1547 P1 2 100 005684 Q Then the boundary work for this polytropic process can be determined from Wb in J 12PdVP2V2 P1V1 mRT2 71 1 n 1 n 2 003403 kg207169 ids1 K563 283K 3619 k We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves Taking the direction of heat transfer to be to the cylinder the energy balance for this stationary closed system can be expressed as Ein Eout AEsystem V Net energy thfer Chan gem 1ntemalkrnetrc by heat W0rkand mass potentialetc energies Qin Wb ir1 2 AU 2 mu2 ul Qin m 2 1 Wbin mCUT2 Tr Wbin Substituting Qin 003403 kg31156 kJkgK563 283K 3619 kJ 651 k The negative sign indicates that heat is lost from the system PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 486 4119 An insulated cylinder initially contains a saturated liquidvapor mixture of water at a speci ed temperature The entire vapor in the cylinder is to be condensed isothermally by adding ice inside the cylinder The amount of ice that needs to be added is to be determined Assumptions 1 Thermal properties of the ice are constant 2 The cylinder is wellinsulated and thus heat transfer is negligible 3 There is no stirring by hand or a mechanical device it will add energy Properties The specific heat of ice at about 0 C is c 211 kJkg C Table A 3 The melting temperature and the heat of fusion of ice at 1 atm are given to be 0 C and 3337 kJkg Analysis We take the contents of the cylinder ice and saturated water as our system which is a closed system Noting that the temperature and thus the pressure remains constant during this phase change process and thus Wb AU AH the energy balance for this system can be written as Ein Eout AE system f J W Net en ergy thfer Chan gein intemalkin etic byheatW0rkandmaSS potentialetc energies ijnZAU gtAH 0 WATER AHiceAH 0 x 001 m3 120 C water mC0 C T1 d quotICT2 Oo Cliquidice hl water 2 0 soli The properties of water at 120 C are Table A4 uf 20001060 ug 2089133 m3kg hf 250381 hfg 22021kJkg Then 11 uf xlufg 000106002gtlt 089133 0001060 017911 m3kg h1 h f xlhfg 5038102gtlt 22021 94424 kJkg 3 m amp 005583kg cam 11 017911m3kg Noting that T1 ice 0 C and T2 120 C and substituting gives m0 3337 kJkg 418 kJkg C1200 C 005583 kg50381 94424 kJkg 0 m 00294 kg 294 g ice PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 487 4120 Nitrogen gas is expanded in a polytropic process The work done and the heat transfer are to be determined Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of 1262 K and 339 MPa 2 The kinetic and potential energy changes are negligible Ake E Ape E O 3 Constant specific heats can be used Properties The properties of nitrogen are R 02968 kJkgK and C 0743 kJkgK Table A2a Analysis We take nitrogen in the cylinder as the system This is a closed system since no mass crosses the boundaries of the system The energy balance for this system can be expressed as E in E out AEsy stem N2 Q J E J 2 MP3 Net en ergy thfer Chan gein intemalkin etic by heat W0rkand mass potentialetc energies 1200 K Qin Wbout AU mcu T2 T1 Using the boundary work relation for the polytropic process of an ideal gas gives n 1n 025125 RT1 i 1 2 02968 kJkg K1200 K 200 j I 526kJkg w b t 1 n P1 1 125 2000 The temperature at the final state is quot391 yn 025125 P T2 2 T1 2 1200 KMj 7571 K P1 2000 kPa Substituting into the energy balance equation qin wb out c T2 T1 526 kJkg 0743 kJkg K757 1 1200K 1 97 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 488 4121 The heating of a passive solar house at night is to be assisted by solar heated water The length of time that the electric heating system would run that night with or without solar heating are to be determined Assumptions 1 Water is an incompressible substance with constant speci c heats 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water 3 The house is maintained at 22 C at all times Properties The density and speci c heat of water at room temperature are p 1 kgL and c 418 kJkg C Table A 3 Analysis a The total mass of water is 50000 kJh mw pv 1 kgLXSOX 20 L 1000 kg Taking the contents of the house including the water as our system the energy balance relation can be written as Ein Eout AE sy stem f J W Net en ergy thfer Chan gein intemalkin etic byheata workaand mass potentialetc energies Wein Qout AU AUwater AUair AUwater mCT2 water or einAt Qout mag 11water Substituting 15 kJsAt 50000 kJh10 h 1000 kg4 18 kJkg C22 80 C It gives At 17170 s 477 h b If the house incorporated no solar heating the energy balance relation above would simplify further to einAt Qout 0 Substituting 15 kJsAt 50000 kJh10 h 0 It gives At 33333 s 926 h PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 489 4122 One ton of liquid water at 50 C is brought into a room The nal equilibrium temperature in the room is to be determined Assumptions 1 The room is well insulated and well sealed 2 The thermal properties of water and air are constant Properties The gas constant of air is R 0287 kPam3kgK Table Al The specific heat of water at room temperature is c 418 kJkg C Table A3 Analysis The volume and the mass of the air in the room are 4gtlt5gtlt6120m3 4mgtlt5mgtlt6m 101v1 95 kPa120 m3 ROOM mair 39 3 1379kg o RT1 02870 kPa m Ikg KX288 K 15 C 95 kPa Taking the contents of the room including the water as our system the energy balance can be written as 2 Heat Ein E out 2 AE system water N ct energy t1 Sfer Chan gem 1ntemalk1net1c byheata workaandmass potentialetc energies 0 2 AU 2 AUwater AUair or mCT2 Talus m T2 T0 0 Substituting 1000 kg4 18 kJkg CTf 50 c 1379 kg0718 kJkg CTf 15 c 0 It gives 7 492 C where I is the nal equilibrium temperature in the room PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 490 4123 Water is boiled at sea level 1 atm pressure in a coffee maker and half of the water evaporates in 25 min The power rating of the electric heating element and the time it takes to heat the cold water to the boiling temperature are to be determined Assumptions 1 The electric power consumption by the heater is constant 2 Heat losses from the coffee maker are negligible Properties The enthalpy of vaporization of water at the saturation temperature of 100 C is hfg 22564 kJkg Table A4 At an average temperature of 100182 59 C the specific heat of water is c 418 kJkg C and the density is about 1 kgL Table A3 Analysis The density of water at room temperature is very nearly 1 kgL and A W V Water 100 C Heater Noting that the specific heat of water at the average temperature of 181002 59 C is c 418 kJkg C the time it takes for the entire water to be heated from 18 C to 100 C is determined to be chT 1 kg4 18 kJkg C100 18 C I 0752 kJs thus the mass of 1 L water at 18 C is nearly 1 kg Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a liquid at a specified temperature the amount of electrical energy needed to vaporize 05 kg of water in 25 min is mhfg 05 kg22564 kJkg At 25 X 60 s 0752kW We WeAtmhfg gt W 2 Therefore the electric heater consumes and transfers to water 0752 kW of electric power 2 456 s 760min We WeAtchT gt At 6 Discussion We can also solve this problem using vf data instead of density and hf data instead of specific heat At 100 C we have If 0001043 m3kg and hf 41917 kJkg At 18 C we have hf 7554 kJkg Table A4 The two results will be practically the same PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 491 4124 A room is to be heated by 1 ton of hot water contained in a tank placed in the room The minimum initial temperature of the water is to be determined if it to meet the heating requirements of this room for a 24h period Assumptions 1 Water is an incompressible substance with constant speci c heats 2 Air is an ideal gas with constant specific heats 3 The energy stored in the container itself is negligible relative to the energy stored in water 4 The room is maintained at 20 C at all times 5 The hot water is to meet the heating requirements of this room for a 24h period Properties The speci c heat of water at room temperature is c 418 kJkg C Table A 3 Analysis Heat loss from the room during a 24h period is Qloss 6000 kJh24 h 144000 kJ 6000 kJh Tak1ng the contents of the room 1nclud1ng the water as our system the energy balance can be written as Ein Eout AE system Net energy thfer Chan gein intemalkinetic 20 C byheata workaandmass potentialetc energies Qout AUwater Aairltj0 01 water 39Qout mCT2 T1water Substituting 144000 kJ 1000 kg4 18 kJkg C20 T1 It gives T1 5450C where T1 is the temperature of the water when it is first brought into the room PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 92 4125 A sample of a food is burned in a bomb calorimeter and the water temperature rises by 32 C when equilibrium is established The energy content of the food is to be determined Assumptions 1 Water is an incompressible substance with constant speci c heats 2 Air is an ideal gas with constant specific heats 3 The energy stored in the reaction chamber is negligible relative to the energy stored in water 4 The energy supplied by the mixer is negligible Properties The speci c heat of water at room temperature is c 418 kJkg C Table A3 The constant volume speci c heat of air at room temperature is cv 0718 kJkg C Table A2 Analysis The chemical energy released during the combustion of the sample is transferred to the water as heat Therefore disregarding the change in the sensible energy of the reaction chamber the energy content of the food is simply the heat transferred to the water Taking the water as our system the energy balance can be written as Ein Eout AEsystem Qin AU r J EH Net energy thfer Chan gein intemalkinetic byheat WO i ndmaSS potentialetc energies Water or Reactlon Qin AUwater mc T2 T1water Chamber Substituting Foojd Q 3 kg418 kJkg C32 C 4013 kJ AT 32 C for a 2g sample Then the energy content of the food per unit mass is 4013 kJ 1000 g 20060kJ 2g 1kg quotg To make a rough estimate of the error involved in neglecting the thermal energy stored in the reaction chamber we treat the entire mass within the chamber as air and determine the change in sensible internal energy AUchamber mcu T2 T1chamber 0102 kg0718 kJkg CX32 C 023 kJ which is less than 1 of the internal energy change of water Therefore it is reasonable to disregard the change in the sensible energy content of the reaction chamber in the analysis PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 493 4126 A man drinks one liter of cold water at 3 C in an effort to cool down The drop in the average body temperature of the person under the in uence of this cold water is to be determined Assumptions 1 Thermal properties of the body and water are constant 2 The effect of metabolic heat generation and the heat loss from the body during that time period are negligible Properties The density of water is very nearly 1 kgL and the specific heat of water at room temperature is c 418 kJkg C Table A 3 The average speci c heat of human body is given to be 36 kJkg C Analysis The mass of the water is mw pVl kgL1 L1kg We take the man and the water as our system and disregard any heat and mass transfer and chemical reactions Of course these assumptions may be acceptable only for very short time periods such as the time it takes to drink the water Then the energy balance can be written as Ein Eout AEsystem V W V Net en ergy MSfer Chan gein intem a1kin etic by heat workaand mass potentia1etc energies I water or mCT2 T1body mCTz T1water 0 Substituting 68 kg36 kJkg CTf 39 C 1 kg4 18 kJkg CTf 3 C 0 It gives 7 384 C Then AT39 384 06 C Therefore the average body temperature of this person should drop about half a degree celsius PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4 94 4127 An insulated rigid tank initially contains saturated liquid water and air An electric resistor placed in the tank is turned on until the tank contains saturated water vapor The volume of the tank the nal temperature and the power rating of the resistor are to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions 3 Energy added to the air is neglected Properties The initial properties of steam are Table A4 T1 200 C u1 0001157 m3kg x1 0 u1 85046 kJkg Analysis a We take the contents of the tank as the system We neglect energy added to the air in our analysis based on the problem statement This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as Ein Eout AE sy stem EK J W Net en ergy t1 Sfel Chan gein intemalkin etic Air by heat workaand mass potentialetc energies Wain 2 AU 2 mu2 ul since Q 2 KB PE O We The initial water volume and the tank volume are Water VI mu1 14 kg0001157 m3kg 0001619 m3 14 kg 200 C 1 1 3 vtank 0006476 m3 9 Now the final state can be fixed by calculating specific volume 0006476 m3 m 14 kg 12 0004626 m3kg The final state properties are 12 0004626 m3kg F 371 3 c x2 1 u2 22015 kJkg c Substituting Wain 14 kg22015 85046kJkg 1892 kJ Finally the power rating of the resistor is W Wein 1892 kJ em 1576kW At 20x60s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 495 4128 A 03L glass of water at 20 C is to be cooled with ice to 5 C The amount of ice or cold water that needs to be added to the water is to be determined Assumptions 1 Thermal properties of the ice and water are constant 2 Heat transfer to the glass is negligible 3 There is no stirring by hand or a mechanical device it will add energy Properties The density of water is l kgL and the speci c heat of water at room temperature is c 418 kJkg C Table A 3 The specific heat of ice at about 0 C is c 211 kJkg C Table A 3 The melting temperature and the heat of fusion of ice at 1 atm are 0 C and 3337 kJkg Analysis a The mass of the water is mw pV 1 kgL03 L 03 kg Ice cubes We take the ice and the water as our system and disregard any heat and mass transfer This is a reasonable assumption since the time period of the process is very short Then the energy balance can be written as Ein Eout AEsystem r J W Net energy thfer Chan gein intema1kinetic by heat W0rkandmaSS potentia1etc energies 0 2 AU 0 AUice AUwater 711600 C T1 de mhif mCT2 0 C1iquidioe mCT2 T1 water 2 0 Noting that T1 ice 0 C and T2 5 C and substituting gives m0 3337 kJkg 418 kJkg C5 0 C 03 kg418 kJkg C5 20 C 0 m 00546 kg 546 g b When T1 ice 20 C instead of 0 C substituting gives m211 kJkg C0 20 C 3337 kJkg 418 kJkg C5 0 C 03 kg418 kJkg C5 20 C 0 m 00487 kg 487 g Cooling with cold water can be handled the same way All we need to do is replace the terms for ice by a term for cold water at 0 C AU coldwater AU Z 0 mCT2 T1coldwater mCT2 T1water Z 0 Substituting mcoldwm 418 kJkg C5 0 C 03 kg4 18 kJkg C5 20 C 0 water It gives m09kg900g Discussion Note that this is about 16 times the amount of ice needed and it explains why we use ice instead of water to cool drinks Also the temperature of ice does not seem to make a significant difference PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 496 4129 0 quot Problem 4128 is reconsidered The effect of the initial temperature of the ice on the final mass of ice required as the ice temperature varies from 26 C to 0 C is to be investigated The mass of ice is to be plotted against the initial temperature of ice Analysis The problem is solved using EES and the solution is given below quotKnownsquot rhowater 1 kgL V 03 L T1ice 0 C T1 20 C T2 5 C Cice 211 kJkgC Cwater 418 kJkgC h if 3337 kJkg T 1CodWater 0 C mwater rhowaterV quotkgquot quotThe mass of the waterquot quotThe system is the water plus the ice Assume a short time period and neglect any heat and mass transfer The energy balance becomesquot Ein Eout DELTAEsys quotkJquot Ein 0 quotkJquot Eout 0quotkJquot DELTAEsys DELTAUwaterDELTAUicequotkJquot DELTAUwater mwaterCwaterT2 T1quotkJquot DELTAUice DELTAUsoidiceDELTAUmetedicequotkJquot DELTAUsoIidice miceCice0T1ic micehifquotkJquot DELTAUmeltedicemiceCwaterT2 0quotkJquot micegramsmiceconvertkggquotgquot quotCooling with Cold Waterquot DELTAEsys DELTAUwaterDELTAUCoIdWaterquotkJquot DELTAUwater mwaterCwaterT2CodWater T1quotkJquot DELTAUCoIdWater mColdWaterCwaterT2CodWater T1ColdWaterquotkJquot mColdWatergramsmCodWaterconvertkggquotgquot T1 ice micegrams 54 C 9 5339 39 26 4594 24 4642 52 22 4691 quot 20 474 51 39 18 4791 a 0 16 4843 quot39 50 14 48 97 E 39 0 39 39 a 49 12 4951 a 0 10 5007 g 48 8 5064 E 5 5122 47 4 5181 0 46 2 5242 0 5305 45 24 20 16 12 8 4 0 T1ice C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 497 4130 A wellinsulated room is heated by a steam radiator and the warm air is distributed by a fan The average temperature in the room after 45 min is to be determined Assumptions 1 Air is an ideal gas with constant speci c heats at room temperature 2 The kinetic and potential energy changes are negligible 3 The air pressure in the room remains constant and thus the air eXpands as it is heated and some warm air escapes Properties The gas constant of air is R 0287 kPam3kgK Table Al Also cp 1005 kJkgK for air at room temperature Table A2 Analysis We first take the radiator as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be eXpressed as Ein Eout AE system r J EH Net energy thfer Changein intemalkinetic byheatW0rkaI1dmaSS potentialetc energies 3 m x 4 m x 6 m Qout m 1 2 Using data from the steam tables Tables A4 through A6 some properties are determined to be gt 5 Steam P1 200 kPa 11 2108049 m3kg 3 radiator T1 200 C u 26546 kJkg E P2 100 kPa V 0001043 ug 16941 m3kg lt 8 v2 2 v1 uf 41740 ufg 20882 kJkg V2 Vf 108049 0001043 ufg 16941 0001043 u uf xzufg 41740 06376x 20882 17487 kJkg 06376 x2 v 1 3 m1amp001388kg v1 108049 m3kg Substituting Qout 001388 kg 26546 17487kJkg 1258 kJ The volume and the mass of the air in the room are V 3 x 4 x 6 72 m3 and 2 10111 I 100 kPa72 m3 2 8960 kg RT1 02870 kPa m3kg KX280 K mair The amount of fan work done in 45 min is W mr1 WfaanAt 0120 kJs45 x 60 s 324 k We now take the air in the room as the system The energy balance for this closed system is eXpressed as Ein Eout AEsystem Qin Wfanin Wbout 2 AU Qin Wfanin 2 AH E meT2 T1 since the boundary work and AU combine into AH for a constant pressure eXpansion or compression process It can also be eXpressed as Qin VVfaninAt mcpavgT2 Tl Substituting 1258 kJ 324 kJ 8960 kg1005 kJkg CT2 7 C which yields T2 1070C Therefore the air temperature in the room rises from 7 C to 107 C in 45 min PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 498 4131 Two rigid tanks that contain water at different states are connected by a valve The valve is opened and the two tanks come to the same state at the temperature of the surroundings The nal pressure and the amount of heat transfer are to be determined Assumptions 1 The tanks are stationary and thus the kinetic and potential energy changes are zero 2 The tank is insulated and thus heat transfer is negligible 3 There are no work interactions Analysis We take the entire contents of the tank as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as Q H20 H20 400 kPa g 200 kPa Ein Eout AE af J af J Net energy thfer Chan gein intemalkinetic by heat workaand mass potentialetc energies Qout AUAUA AUB sinceWKEPE0 Qout Z U2AB U1A U1B Z m2tota1 2 m1 1A m1u1B sy stem The properties of water in each tank are Tables A4 through A6 Tank A P1 400 kPa V 0001084 ug 046242 m3kg x1 080 uf 60422 ufg 19489 kJkg ILA uf xlufg 0001084 08x 046242 0001084 037015 m3kg uLA uf xlufg 60422 08x19489 21633 kJkg Tank B 191 200 kPa uLB 11989 m3kg T1 250 C LB 2 27314 kJkg v 02 3 m1A A 2 m3 05403kg VLA 037015 m kg v 05 3 mle B 2 m3 04170kg V13 11989 m kg ml 2 mLA mLB 05403 04170 2 09573 kg v 07 3 v2 quot m 073117 m3kg mt 09573 kg T2 25 C V 0001003 ug 43340 m3kg u2 073117 m3kg uf 10483 ufg 23043 kJkg Thus at the final state the system will be a saturated liquidvapor mixture since vf lt 2 lt Vg Then the final pressure must be P2 Psalt 25 cc 317 kPa Also V2 Vf 073117 0001 001685 vfg 43340 0001 142 uf xzufg 10483001685gtlt 23043 2 14365 kJkg x2 Substituting Qout 0957314365 0540321633 0417027314 2170 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 499 4132 39 Problem 4131 is reconsidered The effect of the environment temperature on the nal pressure and the heat transfer as the environment temperature varies from 0 C to 50 C is to be investigated The nal results are to be plotted against the environment temperature Analysis The problem is solved using EES and the solution is given below quotKnownsquot VoA02 mA3 PA1400 kPa xA108 TB1250 C PB1200 kPa VoB05 mA3 Tfina25 C quotTfinal Tsurroundings To do the parametric study or to solve the problem when Qout 0 place this statement in quot Qout0 kJ quotTo determine the surroundings temperature that makes Qout 0 remove the and resolve the problemquot quotSolutionquot quotConservation of Energy for the combined tanksquot EinEoutDELTAE Ein0 EoutQout DELTAEmAuA2 uA1mBuB2 uB1 mAVoAvA1 mBVoBvB1 Fluid39SteamIAPWS39 uA1INTENERGYFuidPPA1 xxA1 vA1voumeFuidPPA1 xxA1 TA1temperatureFuidPPA1 xxA1 uB1INTENERGYFuidPPB1TTB1 vB1voumeFuidPPB1TTB1 quotAt the final state the steam has uniform properties through out the entire systemquot uB2ufina uA2ufina mfinamAmB VofinaVoAVoB vfinalVofinamfina ufinalINTENERGYFuidTTfina vvfina PfinapressureFuidTTfina vvfina Pfinal Qout Tfinal 06112 2300 0 09069 2274 5556 1323 2247 1111 1898 2218 1667 2681 2187 2222 3734 2153 2778 513 2116 3333 6959 2075 3889 9325 2030 4444 1235 1978 50 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4100 I I I I 39 39 2250 2200 2150 2100 Gout le 2050 2000 39 1950 I I I I T nalc 15IIIIIIIIIIIIIIIIII 1 PfInal kPa O 3 0 0 o I I I I I I I I I I I I I o 5 1o 15 20 25 30 35 4o 45 5o 39T nal quot31 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4101 4133 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The nal equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible Properties The gas constants and the constant volume speci c heats are R 02968 kPam3kgK is cu 0743 kJkg C for N2 and R 20769 kPam3kgK is cu 31156 kJkg C for He Tables A1 and A2 Analysis The mass of each gas in the cylinder is 3 mN2 500 kPa1 m 4287 kg N2 He N2 RT1 02968 kPam31g39Kx393 K 5301 0111 a a m 2 L 101 11 Z 500 kPa1m3 07691kg 120 C 40 C He RT1 He 20769 kPam3kg KX313 K Taking the entire contents of the cylinder as our system the 1st law relation can be written as Ein Eout AE system V V Net energy thfer Chan gem 1ntemalkrnetrc byheat W0rkandmaSS potentialetc energies 0 2 AU 2 AUN2 AUHe 0 mCUT2 T1N2 mCUT2 T1He Substituting 4287 kg0743 kJkg CXTf 120Pc 07691 kg3 1 156 kJkg CTf 40c 0 It gives 7 857 C where I is the nal equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats Discussion Using the relation P V NRuT it can be shown that the total number of moles in the cylinder is 0153 0192 0345 kmol and the final pressure is 515 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4102 4134 An insulated cylinder is divided into two parts One side of the cylinder contains N2 gas and the other side contains He gas at different states The nal equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and moving freely Assumptions 1 Both N2 and He are ideal gases with constant specific heats 2 The energy stored in the container itself except the piston is negligible 3 The cylinder is wellinsulated and thus heat transfer is negligible 4 Initially the piston is at the average temperature of the two gases Properties The gas constants and the constant volume speci c heats are R 02968 kPam3kgK is cu 0743 kJkg C for N2 and R 20769 kPam3kgK is cu 31156 kJkg C for He Tables A1 and A2 The speci c heat of copper piston is c 0386 kJkg C Table A3 Analysis The mass of each gas in the cylinder is 3 N H mN2 PI V1 2 500 111 m i 4287 kg 1 11213 1 13 RT1 N2 02968 kPa m kg K393 K 500 kpa 500 kpa 3 120 C 40 C mHe PI V1 500 kPa1 m 07691 kg 7 RT1 He 20769 kPam3kgKX313 K Copper Taking the entire contents of the cylinder as our system the 1st law relation can be written as Ein Eout AE system J E 4 Net en ergy thfer Chan gein intern alkin etic byheat W0rkand mass potentialetc energies 0 2 AU 2 AUN2 AUHe AUCu 0 mCuT2 T1lN2 mCuT2 TlHe quot1002 T1lCu where TLCu 120 40 2 80 C Substituting 4287 kg0743 kJkg CXTf 120 C 07691 kg3 1156 kJkg Cfo 40 C 8 kg0386 kJkg Cfo 80 C 0 It gives 7 837 C where I is the nal equilibrium temperature in the cylinder The answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4103 4135 A quot Problem 4134 is reconsidered The effect of the mass of the copper piston on the final equilibrium temperature as the mass of piston varies from 1 kg to 10 kg is to be investigated The final temperature is to be plotted against the mass of piston Analysis The problem is solved using EES and the solution is given below quotKnownsquot Ru8314 kJkmoIK VN211 mA3 CvN20743 kJkgK quotFrom Table A2a at 27Cquot RN202968 kJkgK quotFrom Table A2aquot TN21120 C PN21500 kPa VHe11 mquot3 CvHe31156 kJkgK quotFrom Table A2a at 27Cquot THe140 C PHe1500 kPa RHe20769 kJkgK quotFrom Table A2aquot mPist8 kg CvPist0386 kJkgK quotUse Cp for Copper from Table A3b at 27Cquot quotSolutionquot quotmass calculationsquot PN21VN21mN2RN2TN21273 PHe1VHe1mHeRHeTHe1273 quotThe entire cylinder is considered to be a closed system neglecting the pistonquot quotConservation of Energy for the closed systemquot quotEin Eout DELTAEnegPist we neglect DELTA KE and DELTA PE for the cylinderquot Ein Eout DELTAEnegIPist Ein 0 kJ Eout 0 kJ quotAt the final equilibrium state N2 and He will have a common temperaturequot DELTAEnegIPist mN2CvN2T2negIPistTN21mHeCvHeT2negPistTHe1 quotThe entire cylinder is considered to be a closed system including the pistonquot quotConservation of Energy for the closed systemquot quotEin Eout DELTAEwithPist we neglect DELTA KE and DELTA PE for the cylinderquot Ein Eout DELTAEwithPist quotAt the final equilibrium state N2 and He will have a common temperaturequot DELTAEwithPist mN2CvN2T2withPistTN21mHeCvHeT2withPist THe1mPistCvPistT2withFistTPist1 TPist1TN21THe12 quotTotal volume of gasesquot VtotaVN21VHe1 quotFinal pressure at equilibriumquot quotNeglecting effect of piston P2 isquot P2negPistVtotaNtotaRuT2negPist273 quotIncluding effect of piston P2 isquot NtotamN2molarmassnitrogenmHemoarmassHeium P2withPistVtotaNtotaRuT2withPist273 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4 104 mPist T2negPist T2withPist 86 kg 0 C g 855 H H H H W ithout piston h I 3 8565 8468 4 8565 8443 85 5 8565 842 3 6 8565 8399 g 845 I I 7 8565 8381 393 IWlth plston 8 8565 8364 I 9 8565 8348 84 39 10 8565 8334 I I I 835 li 8339 1 2 3 7 9 10 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4105 4136 A pistoncylinder device initially contains saturated liquid water An electric resistor placed in the tank is turned on until the tank contains saturated water vapor The volume of the tank the nal temperature and the power rating of the resistor are to be determined Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions Properties The initial properties of steam are Table A4 u 0001060 m3kg h1 50381 kJkg T1 120 C P1 19867 kPa x1 Analysis a We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be eXpressed as Ein E out 2 AE system ar J W Net energy thfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies Wain Wbout 2 AU 2 mu2 ul since Q 2 KB PE 0 Water 18 k 120 C W Wein Wbout AU AH mh2 hl g e sat liq since Wb out AU 2 AH for a constantpressure process The initial and nal volumes are v1 mu1 18 kg0001060 m3kg 0001909 m3 v2 40001909 m3 0007634m3 b Now the nal state can be fixed by calculating specific volume v 3 v2 2 m 0004241 m3 kg m 18kg The final state is saturated mixture and both pressure and temperature remain constant during the process Other properties are P2 2 P1 219867 Pa Table A 4 or EES v2 0004241 m kg x2 2 000357 c Substituting Wain 18 kg51168 50381kJkg 1416 kJ Finally the power rating of the resistor is W Wein 1416kJ em At 10x60 s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4106 4137 A pistoncylinder device contains an ideal gas An external shaft connected to the piston exerts a force For an isothermal process of the ideal gas the amount of heat transfer the nal pressure and the distance that the piston is displaced are to be determined Assumptions 1 The kinetic and potential energy changes are W negligible Ake E Ape E 0 2 The friction between the piston and the cylinder is negligible Analysis a We take the ideal gas in the cylinder to be the system This is a closed system since no mass crosses the system boundary 5 The energy balance for this stationary closed system can be eXpressed GAS f a 1 bar 39 Q as 240C l EinEout AEsystem 13333333333 f J W I Net energy traleer Chan gein intemalkinetic byheatW0rkandmaSS potentialetc energies Wbin Qout AUidealgas E mcu T2 Tlidea1gas O Slnce T2 2 TI and KE 2 PE 2 O Wbin Qout Thus the amount of heat transfer is equal to the boundary work input Qout Wbin 01 b The relation for the isothermal work of an ideal gas may be used to determine the nal volume in the cylinder But we rst calculate initial volume 72132 7r012m2 V1 lq 02 m 0002262 m3 4 4 Then V2 Wbin Z P1V11n 71 V2 0002262 m3 01kJ 100 kPa0002262 m3ln j w 0001454 m3 The final pressure can be determined from ideal gas relation applied for an isothermal process Flt1 sz2 gt100 kPa0002262 m3 P2 0001454 m3 gtP2 1556kPa c The final position of the piston and the distance that the piston is displaced are 702 7z012m2 V2 L2 gt0001454 m3 AL 2 11 L2 020 01285 007146 m 71cm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4107 4138 A pistoncylinder device contains an ideal gas An external shaft connected to the piston exerts a force For an isothermal process of the ideal gas the amount of heat transfer the final pressure and the distance that the piston is displaced are to be determined Assumptions 1 The kinetic and potential energy changes are W negligible Ake E Ape E 0 2 The friction between the piston and the cylinder is negligible Analysis a We take the ideal gas in the cylinder to be the system This Z Z Z Z Z Z Z Z Z Z Z is a closed system since no mass crosses the system boundary The Z Z Z 39 39 Z Z Z energy balance for this stationary closed system can be expressed as E Z 5 Z 5 Z 1 bar 5 Z 5 Q Ein Eout AEsystem af J I I I I I I I I I I I Net energy traleer Changein intemalkinetic Z 39 Z 39 Z 39 39 39 39 39 39 39 39 39 39 39 byheata workaandmass potentialetc energies Wb iIl Qout AUidealgaS E mcu T2 T1idea1gas 0 since T2 2 T1 and KE 2 PE 2 O Wbin Qout Thus the amount of heat transfer is equal to the boundary work input Qout Wbin 01 b The relation for the isothermal work of an ideal gas may be used to determine the nal volume in the cylinder But we rst calculate initial volume 72132 7r012m2 V1 lq 02 m 0002262 m3 4 4 Then V2 Wbin Z P1V1 IDLE V2 0002262 m3 01kJ 100 kPa0002262 m3ln w 0001454 m3 The final pressure can be determined from ideal gas relation applied for an isothermal process Flt1 sz2 gt100 kPa0002262 m3 P2 0001454 m3 gtP2 1556kPa c The final position of the piston and the distance that the piston is displaced are 702 7z012m2 V2 L2 gt0001454 m3 L2 gtL2 01285m AL 2 L1 L2 020 01285 007146 m 71cm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4108 4139 A pistoncylinder device with a set of stops contains superheated steam Heat is lost from the steam The pressure and quality if mixture the boundary work and the heat transfer until the piston first hits the stops and the total heat transfer are to be determined Assumptions 1 The kinetic and potential energy changes are negligible Ake E Ape E 0 2 The friction between the piston and the cylinder is negligible Analysis a We take the steam in the cylinder to be the system This is a closed system since no mass crosses the system boundary The energy balance for this stationary closed system can be eXpressed as Ein Eout AE system aka E 4 Net energy trmsfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies Wb iIl Qout 2 AU since KE 2 PE 2 0 Denoting when piston first hits the stops as state 2 and the final state as 3 the energy balance relations may be written as Wbin Qout12 mu2 39 111 Wbin Qout13 Z m 3 39 111 The properties of steam at various states are Tables A4 through A6 T at35MPa 24256 C S T1 2 T1 ATsat 24256 74 250 C P1 35 MPa I1 005875 m3kg T1 250 C ul 2 26239 kJkg P2 P1 35 MPa F 0001235 m3kg x2 2 0 u 10454 kJkg 3 x3 2000062 V3 12 0001235 m kgP3 Z1555kpa T3 200 C 3 85155 kJkg Steam 035 kg 35 MPa tel b Noting that the pressure is constant until the piston hits the stops during which the boundary work is done it can be determined from its definition as Wb in mP1v1 v2 035 kg3500 kPa005875 0001235m3 7045kJ c Substituting into energy balance relations Qout m 7045 kJ 035 kg10454 26239kJkg 6229kJ d Qoum 7045 kJ 035 kg85155 26239kJkg 6908kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4109 4140 An insulated rigid tank is divided into two compartments each compartment containing the same ideal gas at different states The two gases are allowed to mix The simplest expression for the mixture temperature in a speci ed format is to be obtained Analysis We take the both compartments together as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem ar J W Net energy thfer Chan gein intemalkinetic byheatW0rkandmaSS potentialetc energies 0AU s1nceQWKEPE0 m1 m2 0 mlcu mzcu T2 T1 T2 m1 m2T3 mlji szz Solving for final temperature we nd m1 m2 m3 m3 4141 Carbon dioxide contained in a springloaded pistoncylinder device is compressed in a polytropic process The nal temperature is to be determined using the compressibility factor Properties The gas constant the critical pressure and the critical temperature of C02 are from Table A1 R 01889 kPam3kgK Tcr 3042 K 10cr 739 MPa Analysis From the compressibility chart at the initial state Fig A15 or EES we used EES T 4 K TR1 1 73 155 Tcr 3042 K P O 5MP Z1 2 09953 a CO P 1 00677 2 R1 10cr 739 MPa 05 MPa 200 C The speci c volume at the initial state is ZlRTl 0995301889kPa m3kg K473 K 01779 m3kg P1 500 kPa V1V2 The speci c volume at the nal state can be determined from the polytropic process to be Flt1 qug gt 500 kPa01779 m3kg1393 3000 kPau 3 gt 02 004483 m3kg At the final state P PR2 2 0406 Pcr 739 MPa v 0 04483 3k ZZ 210 VRZ 2actual m g RTcrPcr 01889kPa m3kg K3042 K7390 kPa Thus T P2v2 3000 kPa004483 m3kg 2 3 712K439 C ZZR 1001889 kPa m lkg K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4110 4142 Solar energy is to be stored as sensible heat using phasechange materials granite rocks and water The amount of heat that can be stored in a 5m3 5000 L space using these materials as the storage medium is to be determined Assumptions 1 The materials have constant properties at the specified values 2 No allowance is made for voids and thus the values calculated are the upper limits Analysis The amount of energy stored in a medium is simply equal to the increase in its internal energy which for incompressible substances can be determined from AU 2 mcT2 T1 a The latent heat of glaubers salts is given to be 329 kJL Disregarding the sensible heat storage in this case the amount of energy stored is becomes AUsalt mhif 5000 L329 kJL 1645000 kJ This value would be even larger if the sensible heat storage due to temperature rise is considered 9 The density of granite is 2700 kgm3 Table A3 and its speci c heat is given to be c 232 kJkg C Then the amount of energy that can be stored in the rocks when the temperature rises by 20 C becomes AUmck chAT 2700 kgm3 5 m3232 kJkg C20 C 626400 kJ c The density of water is about 1000 kgm3 Table A3 and its speci c heat is given to be c 40 kJkg C Then the amount of energy that can be stored in the water when the temperature rises by 20 C becomes AUmck chAT 1000 kgm3 5 m340 kJkg C20 C 40000 kJ Discussion Note that the greatest amount of heat can be stored in phasechange materials essentially at constant temperature Such materials are not without problems however and thus they are not widely used PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4111 4143 The cylinder of a steam engine initially contains saturated vapor of water at 100 kPa The cylinder is cooled by pouring cold water outside of it and some of the steam inside condenses If the piston is stuck at its initial position the friction force acting on the piston and the amount of heat transfer are to be determined Assumptions The device is airtight so that no air leaks into the cylinder as the pressure drops Analysis We take the contents of the cylinder the saturated liquidvapor mixture as the system which is a closed system Noting that the volume remains constant during this phase change process the energy balance for this system can be expressed as Ein Eout AE system f J W Net en ergy t1 Sfer Chan gein internal kinetic by heat workaand mass potentialetc energies Q01 AU 2 mu2 ul The saturation properties of water at 100 kPa and at 30 C are Tables A4 and A5 101 100 kPa gt uf 20001043 m3kg ug 16941m3kg uf 41740 kJkg ug 25056 kJkg T2 30 C gt uf 0001004 m3kg ug 32879 m3kg 17 12573 kJkg ufg 22902kJkg P 42469 kPa sat Then Cold water 192 Pm30c 42469 kPa V1 Vg100kPa 1 ug100kPa and v 005 3 m 1 In3 Z Steam V U 2 f 2 16941 0001 2 005150 100 kPa ufg 32879 0001 u 11 xzufg 12573 005150x 22902 24367 kJkg V2V1 x2 The friction force that develops at the pistoncylinder interface balances the force acting on the piston and is equal to 1000 Nmz F 141 P2 01m2100 42469kPa a 29575 N The heat transfer is determined from the energy balance to be Qout mu1 2 002951 kg25056 24367kJkg 668 k PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 4112 Fundamentals of Engineering FE Exam Problems 4144 The speci c heat of a material is given in a strange unit to be C 360 kJkg F The specific heat of this material in the SI units of kJkg C is a 200 kJkg C b 320 kJkg C c 360 kJkg C d 480 kJkg C e 648 kJkg C Answer e 648 kJkg C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C360 quotkJkgFquot CSC18 quotkJkgCquot quotSome Wrong Solutions with Common Mistakesquot W1CC quotAssuming they are the samequot W2CC18 quotDividing by 18 instead of multiplyingquot 4145 A 3m3 rigid tank contains nitrogen gas at 500 kPa and 300 K Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800 kPa The work done during this process is a 500 k b 1500 k c 0 k d 900 k e 2400 k Answer b 0 k Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values V3 quotmquot3quot P1 500 quotkPaquot T1 300 quotKquot P2800 quotkPaquot W0 quotsince constant volumequot quotSome Wrong Solutions with Common Mistakesquot R0297 W1WVP2 P1 quotUsing WVDELTAPquot W2WVP1 W3WVP2 W4WRT1nP1P2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4113 4146 A 05m3 cylinder contains nitrogen gas at 600 kPa and 300 K Now the gas is compressed isotherrnally to a volume of 01 m3 The work done on the gas during this compression process is a 720 k b 483 k c 240 k d 175 k e 143 kJ Answer b 483 kJ Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R831428 V1 05 quotmA3quot V201 quotmA3quot P1 600 quotkPaquot T1 300 quotKquot P1V1mRT1 WmRT1 nV2V1 quotconstant temperaturequot quotSome Wrong Solutions with Common Mistakesquot W1WRT1 nV2V1 quotForgetting mquot W2WP1V1V2 quotUsing VDetaPquot P1V1T1P2V2T1 W3WV1V2P1P22 quotUsing PaveDeta Vquot W4WP1V1P2V2 quotUsing WP1 V1 P2V2quot 4147 A wellsealed room contains 60 kg of air at 200 kPa and 25 C Now solar energy enters the room at an average rate of 08 kJs while a 120W fan is turned on to circulate the air in the room If heat transfer through the walls is negligible the air temperature in the room in 30 min will be a 256 C b 498 C c 534 C d 525 C e 634 C Answer e 634 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 quotkJkgKquot Cv0718 quotkJkgKquot m60 quotkgquot P1200quotkPaquot T1 25 quotCquot Qsol08 quotkJsquot time3060 quotsquot Wfan012 quotkJsquot quotApplying energy balance EinEoutdEsystem givesquot timeWfanQsomCvT2 T1 quotSome Wrong Solutions with Common Mistakesquot Cp1005 quotkJkgKquot timeWfanQsomCpW1T2T1 quotUsing Cp instead of Cv quot timeWfanQsomCvW2T2 T1 quotSubtracting Wfan instead of addingquot timeQsomCvW3T2 T1 quotIgnoring Wfanquot timeWfanQso60mCvW4T2T1 quotUsing min for time instead of squot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4114 4148 A 2kW baseboard electric resistance heater in a vacant room is turned on and kept on for 15 min The mass of the air in the room is 75 kg and the room is tightly sealed so that no air can leak in or out The temperature rise of air at the end of 15 min is a 85 C b 124 C c 240 C d 334 C e 548 C Answer d 334 C Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 quotkJkgKquot Cv0718 quotkJkgKquot m75 quotkgquot time1560 quotsquot We2 quotkJsquot quotApplying energy balance EinEoutdEsystem givesquot timeWemCvDELTAT quotkJquot quotSome Wrong Solutions with Common Mistakesquot Cp1005 quotkJkgKquot timeWemCpW1DELTAT quotUsing Cp instead of Cvquot timeWe60mCvW2DELTAT quotUsing min for time instead of 5quot 4149 A room contains 75 kg of air at 100 kPa and 15 C The room has a 250W refrigerator the refrigerator consumes 250 W of electricity when running a 120W TV a 18kW electric resistance heater and a 50W fan During a cold winter day it is observed that the refrigerator the TV the fan and the electric resistance heater are running continuously but the air temperature in the room remains constant The rate of heat loss from the room that day is a 5832 kJh b 6192 kJh c 7560 kJh d 7632 kJh e 7992 kJh Answer e 7992 kJh Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 quotkJkgKquot Cv0718 quotkJkgKquot m75 quotkgquot P1100 quotkPaquot T1 15 quotCquot time3060 quotsquot Wref0250 quotkJsquot WTV0120 quotkJsquot Wheater18 quotkJsquot Wfan005 quotkJsquot quotApplying energy balance EinEoutdEsystem gives EoutEin since Tconstant and dE0quot EgainWrefWTVWheaterWfan QossEgain3600 quotkJhquot quotSome Wrong Solutions with Common Mistakesquot Egain1WrefWTVWheaterWfan quotSubtracting Wrefrig instead of addingquot W1QossEgain13600 quotkJhquot Egain2WrefWTVWheaterWfan quotSubtracting Wfan instead of addingquot W2QossEgain23600 quotkJhquot Egain3WrefWTVWheaterWfan quotSubtracting Wrefrig and Wfan instead of addingquot W3QossEgain33600 quotkJhquot Egain4WrefWheaterWfan quotIgnoring the TVquot W4QossEgain43600 quotkJhquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4115 4150 A pistoncylinder device contains 5 kg of air at 400 kPa and 30 C During a quasiequilibrium isothermal expansion process 15 kJ of boundary work is done by the system and 3 kJ of paddlewheel work is done on the system The heat transfer during this process is a 12 k b 18 k c 24 k d 35 k e 60 kJ Answer a 12 kJ Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R0287 quotkJkgKquot Cv0718 quotkJkgKquot m5 quotkgquot P1400 quotkPaquot T30 quotCquot Woutb15 quotkJquot Winpw3 quotkJquot quotNoting that Tconstant and thus dEsystem0 applying energy balance EinEoutdEsystem givesquot QinWinpwWoutb0 quotSome Wrong Solutions with Common Mistakesquot W1QinQinCv quotDividing by CVquot W2QinWinpwWoutb quotAdding both quantitiesquot W3QinWinpw quotSetting it equal to paddlewheel workquot W4QinWoutb quotSetting it equal to boundaru workquot 4151 A 6pack canned drink is to be cooled from 18 C to 3 C The mass of each canned drink is 0355 kg The drinks can be treated as water and the energy stored in the aluminum can itself is negligible The amount of heat transfer from the 6 canned drinks is a 22 k b 32 k c 134 k d 187 k e 223 kJ Answer c 134 kJ Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values c418 quotkJkgKquot m60355 quotkgquot T1 1 8 quotCquot T23 quotCquot DELTATT2 T1 quotCquot quotApplying energy balance EinEoutdEsystem and noting that dUsystemmCDELTAT givesquot QoutmCDELTAT quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1QoutmCDELTAT6 quotUsing one can onlyquot W2QoutmCT1T2 quotAdding temperatures instead of subtractingquot W3Qoutm10DELTAT quotUsing specific heat of air or forgetting specific heatquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4116 4152 A glass of water with a mass of 045 kg at 20 C is to be cooled to 0 C by dropping ice cubes at 0 C into it The latent heat of fusion of ice is 334 kJkg and the speci c heat of water is 418 kJkg C The amount of ice that needs to be added is a 56 g b 113 g c 124 g d 224 g e 450 g Answer b 113 g Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 quotkJkgKquot hmeting334 quotkJkgKquot mw045 quotkgquot T1 20 quotCquot T20 quotCquot DELTATT2 T1 quotCquot quotNoting that there is no energy transfer with the surroundings and the latent heat of melting of ice is transferred form the water and applying energy balance EinEoutdEsystem to icewater givesquot dEiced Ew0 dEicemicehmelting dEwmwCDELTAT quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1micehmeltingT1T2mwCDELTAT0 quotMultiplying hlatent by temperature differencequot W2micemw quottaking mass of water to be equal to the mass of icequot 4153 A 2kW electric resistance heater submerged in 5kg water is turned on and kept on for 10 min During the process 300 kJ of heat is lost from the water The temperature rise of water is a 04 C b 431 C c 574 C d 718 C e 1800 C Answer b 431 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 quotkJkgKquot m5 quotkgquot Qloss300 quotkJquot time1060 quotsquot We2 quotkJsquot quotApplying energy balance EinEoutdEsystem givesquot timeWeQloss dUsystem dUsystemmCDELTAT quotkJquot quotSome Wrong Solutions with Common Mistakesquot timeWe mCW1T quotIgnoring heat lossquot timeWeQloss mCW2T quotAdding heat loss instead of subtractingquot timeWeQloss m10W3T quotUsing specific heat of air or not using specific heatquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4117 4154 15 kg of liquid water initially at 12 C is to be heated to 95 C in a teapot equipped with a 800 W electric heating element inside The speci c heat of water can be taken to be 418 kJkg C and the heat loss from the water during heating can be neglected The time it takes to heat the water to the desired temperature is a 59 min b 73 min c 108 min d 140 min e 170 min Answer c 108 min Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C418 quotkJkgKquot m15 quotkgquot T1 1 2 quotCquot T295 quotCquot Qloss0 quotkJquot We08 quotkJsquot quotApplying energy balance EinEoutdEsystem givesquot time60WeQloss dUsystem quottime in minutesquot dUsystemmCT2 T1 quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1time60WeQloss mCT2T1 quotAdding temperatures instead of subtractingquot W2time60WeQloss CT2T1 quotNot using massquot 4155 An ordinary egg with a mass of 01 kg and a specific heat of 332 kJkg C is dropped into boiling water at 95 C If the initial temperature of the egg is 5 C the maXimum amount of heat transfer to the egg is a 12 k b 30 k c 24 k d 18 k e infinity Answer b 30 kJ Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C332 quotkJkgKquot m01 quotkgquot T1 5 quotCquot T295 quotCquot quotApplying energy balance EinEoutdEsystem givesquot Ein dUsystem dUsystemmCT2 T1 quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1Ein mCT2 quotUsing T2 onlyquot W2EinmENTHALPYSteamIAPWSTT2x1ENTHALPYSteamIAPWSTT2x0 quotUsing hfgquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4118 4156 An apple with an average mass of 018 kg and average specific heat of 365 kJkg C is cooled from 22 C to 5 C The amount of heat transferred from the apple is a 085 k b 621 k c 177 k d 112 k e 71 k Answer d 112 k Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values C365 quotkJkgKquot m018 quotkgquot T1 22 quotCquot T25 quotCquot quotApplying energy balance EinEoutdEsystem givesquot Qout dUsystem dUsystemmCT2T1 quotkJquot quotSome Wrong Solutions with Common Mistakesquot W1Qout CT2T1 quotNot using massquot W2Qout mCT2T1 quotadding temperaturesquot 4 157 The specific heat at constant volume for an ideal gas is given by cv 0727gtlt10394T kJkg K where T is in kelvin The change in the enthalpy for this ideal gas undergoing a process in which the temperature changes from 27 to 127 C is most nearly a 70 kJkg b 721 kJkg c 795 kJkg d 821 kJkg e 840 kJkg Answer c 795 kJkg Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T1 27273 K T2127273 K quotPerforming the necessary integration we obtainquot DELTAh07T2T127E42T2 2T1A2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4119 4158 An ideal gas has a gas constant R 03 kJkgK and a constantvolume speci c heat cu 07 kJkgK If the gas has a temperature change of 100 C choose the correct answer for each of the following 1 The change in enthalpy is in kJkg a 30 b 70 c 100 Answer c 100 2 The change in internal energy is in kJkg a 30 b 70 c 100 Answer b 70 3 The work done is in kJkg a 30 b 70 c 100 Answer d insuf cient information to determine 4 The heat transfer is in kJkg a 30 b 70 c 100 Answer d insuf cient information to determine 5 The change in the pressurevolume product is in kJkg a 30 b 70 c 100 Answer a 30 d insuf cient information to determine d insuf cient information to determine d insuf cient information to determine d insuf cient information to determine d insuf cient information to determine Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values R03 kJkgK cv07 kJkgK DELTAT100 K quot1 ll cpRcv DELTAhcpDELTAT quot2quot DELTAucvDELTAT quot5quot PVRDELTAT PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 4120 4159 An ideal gas undergoes a constant temperature isothermal process in a closed system The heat transfer and work are respectively a 0 cAT b cuAT 0 c cpAT RAT d R lnT2T1 R lnT2T1 Answer d R lnT2T1 R lnT2T1 4160 An ideal gas under goes a constant volume isochoric process in a closed system The heat transfer and work are respectively a 0 cAT b CVAT 0 c cpAT RAT d R lnT2T1 R lnT2T1 Answer 9 CVAT 0 4161 An ideal gas under goes a constant pressure isobaric process in a closed system The heat transfer and work are respectively a 0 cAT b CVAT 0 c cpAT RAT d R lnT2T1 R lnT2T1 Answer c cpAT RAT 4162 4167 Design and Essay Problems 4165 A claim that fruits and vegetables are cooled by 6 C for each percentage point of weight loss as moisture during vacuum cooling is to be evaluated Analysis Assuming the fruits and vegetables are cooled from 30 C and 0 C the average heat of vaporization can be taken to be 2466 kJkg which is the value at 15 C and the speci c heat of products can be taken to be 4 kJkg C Then the vaporization of 001 kg water will lower the temperature of 1 kg of produce by 24664 6 C Therefore the vacuum cooled products will lose 1 percent moisture for each 6 C drop in temperature Thus the claim is reasonable ewe PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 51 Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Cengel Michael A Boles McGrawHill 2015 Chapter 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 52 Conservation of Mass 51C Flow through a control volume is steady when it involves no changes with time at any specified position 52C Mass ow rate is the amount of mass owing through a crosssection per unit time whereas the volume ow rate is the amount of volume owing through a crosssection per unit time 53C The amount of mass or energy entering a control volume does not have to be equal to the amount of mass or energy leaving during an unsteady ow process 54C No a ow with the same volume ow rate at the inlet and the eXit is not necessarily steady unless the density is constant To be steady the mass ow rate through the device must remain constant 55 The ventilating fan of the bathroom of a building runs continuously The mass of air vented out per day is to be determined Assumptions Flow through the fan is steady Properties The density of air in the building is given to be 120 kgm3 Analysis The mass ow rate of air vented out is lt pVail 120 kgm3 0030 m3s 0036 kgs Then the mass of air vented out in 24 h becomes m mmm 0036 kgs24x3600 s 3110kg Discussion Note that more than 3 tons of air is vented out by a bathroom fan in one day 56E The ducts of an airconditioning system pass through an open area The inlet velocity and the mass ow rate of air are to be determined AIR gt Assumptions Flow through the air conditioning duct is steady Properties The density of air is given to be 0078 lbmft3 at the inlet 450 ft3min I D 10 in Analysis The inlet velocity of air and the mass ow rate through the duct are 2 V 450 it 3 39 2 1 Z 21 1 825 ftmin 138fts A1 no 4 7r1012 it 4 V1 m plvl 00781bm 3450 it 3 min 3511bmmin 05851bms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 53 57 Air ows through a pipe Heat is supplied to the air The volume ow rates of air at the inlet and eXit the velocity at the eXit and the mass ow rate are to be determined QN Air 200 kPa 180 kPa 20 C 40 C 5 ms Properties The gas constant for air is 0287 kJkgK Table A2 Analysis a b The volume ow rate at the inlet and the mass ow rate are 202 7r028 m2 V1 f P 2 kP 2 2 m plACVl 1 D 1 200 a quotO 8 m 5 ms 07318kgs R71 4 0287 kJkgK20273 K 4 VI ACVl 5 ms 2 03079m3s c Noting that mass ow rate is constant the volume ow rate and the velocity at the eXit of the pipe are determined from I I 1k v2 m 073 8 gS 203654m3s p2 i 180 kPa R72 0287 kJkgK40 273 K 39 3 V2 2amp2 03654m s 2594mm Ac 7r028 m2 4 58E Helium at a specified state is compressed to another speci ed state The mass ow rate and the inlet area are to be determined Assumptions Flow through the compressor is steady 200 psia F Properties The gas cosntant of helium is R 26809 psiaft3lbmR Table AlE ft2 Analysis The mass ow rate is determined from Compressor 2 m A2V2 A2V2P2 001 if 102 SXZOO psra 007038Ibms V2 RT2 26809 psia it lbm R1060 R 15 psia The inlet area is determined from 70 F 50 fts mvl 77397RT1 007038lbms26809 psia it 3lbm R530 R 0 1333112 A 1 V1 V1101 50 s15psia PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 54 59 A rigid tank initially contains air at atmospheric conditions The tank is connected to a supply line and air is allowed to enter the tank until the density rises to a speci ed level The mass of air that entered the tank is to be determined Properties The density of air is given to be 118 kgm3 at the beginning and 530 kgm3 at the end Analysis We take the tank as the system which is a control volume since mass crosses the boundary The mass balance for this system can be expressed as Mass balance 4 min m0ut Z Amsystem gt mi 2 m2 quotquot1 Z pzv prv 3 V1 2 m Substituting p1 118 kgm3 m p2 p1V530 118kgm32m3 824kg Therefore 824 kg of mass entered the tank 510 A cyclone separator is used to remove ne solid particles that are suspended in a gas stream The mass ow rates at the two outlets and the amount of y ash collected per year are to be determined Assumptions Flow through the separator is steady Analysis Since the ash particles cannot be converted into the gas and viceversa the mass ow rate of ash into the control volume must equal that going out and the mass ow rate of ue gas into the control volume must equal that going out Hence the mass ow rate of ash leaving is mash yashn39iin 000110kgs 001kgs The mass ow rate of ue gas leaving the separator is then m uegas 2 mm n391ash 10 00l 999kgs The amount of y ash collected per year is mash mashm 001kgs365 x 24x 3600 syear 31 5400kgyear PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 511 A spherical hotair balloon is considered The time it takes to in ate the balloon is to be determined Assumptions 1 Air is an ideal gas Properties The gas constant of air is R 0287 kPam3kgK Table Al Analysis The speci c volume of air entering the balloon is 3 v 2 E 2 0287 kPa m Ikg K20 273 K 207008 m3kg P 120 kPa The mass ow rate at this entrance is A V 2 2 mz C D K 7 0 0m 3mS 3362kgs v 4 v 4 07008 m3kg The initial mass of the air in the balloon is 71D3 7r5 m3 u 6v 607008 m3kg 9339 kg i Similarly the nal mass of air in the balloon is 703 7r15m3 3 2522 kg 1 6v 607008 m Ikg mf The time it takes to in ate the balloon is determined from m m 2 f z 2 2522 9339kg Z7228 m 3362kgs At 120min 512 A desktop computer is to be cooled by a fan at a high elevation where the air density is low The mass ow rate of air through the fan and the diameter of the casing for a given velocity are to be determined Assumptions Flow through the fan is steady Properties The density of air at a high elevation is given to be 07 kgm3 Analysis The mass ow rate of air is mir pVail 07 kgm3 034 m3min 0238 kgmin 00040 kgs If the mean velocity is 110 mmin the diameter of the casing is 2 3 4 4 vavz i V D Z Mzaomnn Therefore the diameter of the casing must be at least 63 cm to ensure that the mean velocity does not exceed 110 mmin 7r110mmin Discussion This problem shows that engineering systems are sized to satisfy certain constraints imposed by certain considerations PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 56 513 A water pump increases water pressure The diameters of the inlet and eXit openings are given The velocity of the water at the inlet and outlet are to be determined Assumptions 1 Flow through the pump is steady 2 The specific volume remains constant Properties The inlet state of water is compressed liquid We approximate it as a saturated liquid at the given temperature Then at 15 C and 40 C we have Table A4 T 15 c 3 v1 0001001m lkg x 0 T 40 C 3 0 v1 0001008m kg x Analysis The velocity of the water at the inlet is n39w1 4n39w1 405 kgs0001001m3kg A1 7012 7z001m2 637ms VI Since the mass ow rate and the specific volume remains constant the velocity at the pump eXit is 2 2 V2 Vl Vl 637ms 001m 283ms A2 D2 0015m Using the speci c volume at 40 C the water velocity at the inlet becomes n39w1 4n39w1 405 kgs0001008m3kg 642mS A1 7012 7r001m2 V1 which is a 08 increase in velocity PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 57 514 Refrigerant134a ows through a pipe Heat is supplied to R134a The volume ow rates of air at the inlet and eXit the mass ow rate and the velocity at the eXit are to be determined Properties The speci c volumes of R134a at the inlet and eXit are Table A13 P1 200kPa 3 P1 180kPa 3 1 01142 m kg 2 01374 m kg T1 20 C T1 40 C Analysis Q R134a 200 kPa 180 kPa 20 C I 40 C 5 ms a b The volume ow rate at the inlet and the mass ow rate are 2 2 11 Ach 711 we ms 2 03079m3s 1 1 702 1 7r028 m2 m Ach 1 3 v1 v1 4 01142m kg 4 5 ms 2 2696kgs c Noting that mass ow rate is constant the volume ow rate and the velocity at the eXit of the pipe are determined from V n39w2 2696 kgs0 1374 m3kg 03705m3s 39 3 V2 2 03705m s 2602ms Ac 7r028 m2 4 515 A smoking lounge that can accommodate 15 smokers is considered The required minimum ow rate of air that needs to be supplied to the lounge and the diameter of the duct are to be determined Assumptions In ltration of air into the smoking lounge is negligible Properties The minimum fresh air requirements for a smoking lounge is given to be 30 US per person Analysis The required minimum ow rate of air that needs to be supplied to the lounge is determined directly from Vair Vairperpason No of persons 30 US person15 persons 450 US 045 m3s The volume ow rate of fresh air can be expressed as Smoking Lounge VVAV7zD2 4 15 smokers Solv1ng for the d1ameter D and subst1tut1ng 30 US person 3 D 2 2 0263111 7rV 7r8 ms Therefore the diameter of the fresh air duct should be at least 268 cm if the velocity of air is not to exceed 8 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 58 516 Warm water is withdrawn from a solar water storage tank while cold water enters the tank The amount of water in the tank in a 20minute period is to be determined Properties The density of water is taken to be 1000 kgm3 for both cold and warm water C 01d water Analysis The initial mass in the tank is first determined from 5213 m1 thank 1000 kgm3 03 m3 300 kg quot132 The amount of warm water leaving the tank during a 20min period is 7r002 m2 m pACVAt 1000 kgm3 05 ms20gtlt 60 s 1885 kg The amount of cold water entering the tank during a 20min period is m pVCAt 1000 kgm30005 m3min20 min 100 kg The final mass in the tank can be determined from a mass balance as ml memQ m1 gtmQm1mi me300100 18852115kg Flow Work and Energy Transfer by Mass Warm water 45 C 05 ms 517 C Flow energy or ow work is the energy needed to push a uid into or out of a control volume Fluids at rest do not possess any ow energy 518C Flowing uids possess ow energy in addition to the forms of energy a uid at rest possesses The total energy of a uid at rest consists of internal kinetic and potential energies The total energy of a owing uid consists of internal kinetic potential and ow energies PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 59 519 Warm air in a house is forced to leave by the in ltrating cold outside air at a specified rate The net energy loss due to mass transfer is to be determined Assumptions 1 The ow of the air into and out of the house through the cracks is steady 2 The kinetic and potential energies are negligible 3 Air is an ideal gas with constant specific heats at room temperature Properties The gas constant of air is R 0287 kPam3kgK Table A1 The constant pressure speci c heat of air at room temperature is cp 1005 kJkg C Table A2 Analysis The density of air at the indoor conditions and its mass ow rate are P 101325 kPa p 1189kgm3 3 RT 0287 kPa m kg K24 273K m p1 1189 kgm3 150 m3h 17835kgh 00495kgs Noting that the total energy of a owing uid is equal to its enthalpy when the kinetic and potential energies are negligible and that the rate of energy transfer by mass is equal to the product of the mass ow rate Warm and the total energy of the uid per unit mass the rates of energy C01 d air gt air gt Warm air transfer by mass into and out of the house by air are 5 C 24 C 24 C Emassin Z quot1111 Emassout mgout th The net energy loss by air in ltration is equal to the difference between the outgoing and incoming energy ow rates which is AE Emsmut E n39th2 h1 2 mp T2 T1 00495 kgs1005 kJkg C24 5 C 0945 kJs 0945kW mass mass in This quantity represents the rate of energy transfer to the refrigerant in the compressor Discussion The rate of energy loss by infiltration will be less in reality since some air will leave the house before it is fully heated to 24 C 520E A water pump increases water pressure The ow work required by the pump is to be determined Assumptions 1 Flow through the pump is steady 2 The state of water at the pump inlet is saturated liquid 3 The specific volume remains constant Properties The speci c volume of saturated liquid water at 15 psia is v uf 15pm 001672 11 3lbm Table A5E Then the ow work relation gives W ow 2P2V2 P1V1VP2 P1 Btu xK i I 1 39 5 ps1a 5404psia1t 3 001672 ft 3lbm80 15psia 0201Btulbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 510 521 Refrigerant134a enters a compressor as a saturated vapor at a speci ed pressure and leaves as superheated vapor at a specified rate The rates of energy transfer by mass into and out of the compressor are to be determined Assumptions 1 The ow of the refrigerant through the compressor is steady 2 The kinetic and potential energies are negligible and thus they are not considered Properties The enthalpy of refrigerant134a at the inlet and the eXit are Tables 2 A12 and A13 h h 23919 kJk 1 g014MPa g T2 600C h2 29682 kJkg compressor Analysis Noting that the total energy of a owing uid is equal to its enthalpy when the kinetic and potential energies are negligible and that the rate of energy transfer by mass is equal to the product of the mass ow rate and the total energy of the uid per T unit mass the rates of energy transfer by mass into and out of the compressor are 1 E mom nah1 006 kgs239 19 kJkg 1435 kJs 1435kW 014 MPa massin E mass out moo n39zh2 006 kgs29682 kJkg 1781kJs 1781kW Discussion The numerical values of the energy entering or leaving a device by mass alone does not mean much since this value depends on the reference point selected for enthalpy it could even be negative The significant quantity here is the difference between the outgoing and incoming energy ow rates which is E massin AEmass Emassout 1781 1435 346 kW This quantity represents the rate of energy transfer to the refrigerant in the compressor PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 511 522E Steam is leaving a pressure cooker at a specified pressure The velocity ow rate the total and ow energies and the rate of energy transfer by mass are to be determined Assumptions 1 The ow is steady and the initial startup period is disregarded 2 The kinetic and potential energies are negligible and thus they are not considered 3 Saturation conditions eXist within the cooker at all times so that steam leaves the cooker as a saturated vapor at 20 psia Properties The properties of saturated liquid water and water vapor at 20 psia are If 001683 ft3lbm 8 20093 ft3lbm ug 10818 Btulbm and kg 11562 Btulbm Table A5E Analysis 61 Saturation conditions eXist in a pressure cooker at all times after the steady operating conditions are established Therefore the liquid has the properties of saturated liquid and the eXiting steam has the properties of saturated vapor at the operating pressure The amount of liquid that has evaporated the mass ow rate of the eXiting steam and the eXit velocity are AV 3 m 2 liquid 2 06 gal 0133681t 4766119111 vf 001683it lbm lgal H20 m m 01059lbmmin1765gtlt10393 lbms Sit Vapo Q At 45 mm P 20 ps1a n39w 3 3 2 E5E5E5555555555555555555555552 V 2 m 2 g 2 1765 x 10 lbms30093 1t lbm 144121 2 341 pgAc Ac 015ih 1t 9 Noting that h u Pvand that the kinetic and potential energies are disregarded the ow and total energies of the exiting steam are em Pu h u 11562 10818 744 Btulbm 6hkepezh11562 Btulbm Note that the kinetic energy in this case is ke V22 341 fts2 2 581 ft2s2 00232 Btulbm which is very small compared to enthalpy c The rate at which energy is leaving the cooker by mass is simply the product of the mass ow rate and the total energy of the eXiting steam per unit mass Ems 2 me 1765 x 10 3 lbms11562 Btulbm 204 Btus Discussion The numerical value of the energy leaving the cooker with steam alone does not mean much since this value depends on the reference point selected for enthalpy it could even be negative The signi cant quantity is the difference between the enthalpies of the eXiting vapor and the liquid inside which is hfg since it relates directly to the amount of energy supplied to the cooker Steady Flow Energy Balance Nozzles and Diffusers 523C It is mostly converted to internal energy as shown by a rise in the uid temperature 524C The kinetic energy of a uid increases at the eXpense of the internal energy as evidenced by a decrease in the uid temperature 525C Heat transfer to the uid as it ows through a nozzle is desirable since it will probably increase the kinetic energy of the uid Heat transfer from the uid will decrease the eXit velocity PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 512 526E Air is accelerated in a nozzle from 150 fts to 900 fts The eXit temperature of air and the eXit area of the nozzle are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with variable speci c heats 3 Potential energy changes are negligible 4 There are no work interactions Properties The enthalpy of air at the inlet is h1 14347 Btulbm Table A17E Analysis a There is only one inlet and one eXit and thus m1 1512 m We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady E E AE I 0 1n 0 t s stem ug f J 65 Btulbm Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies 1 AIR 2 Ein Eout mm V12 2 Q39out mm v 2 since W Ape 0 V2V2 Qout mLhz hl or V22 V12 h2 qout hl 2 2 2 65 Btulbm 14347 Btulbm 900 it s 150 it s 1 Btu1192111 2 2 25037 it s 1212 Btulbm Thus from Table A17E T2 507 R b The eXit area is determined from the conservation of mass relation 1 1 RT P AZV2 AIV1 A2 ZQEA1 EA V2 V1 V1 V2 RT1P1 V2 508147150 its0 1amp2 o 048 z 2 600509OO its PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 513 527E Air is accelerated in an adiabatic nozzle The velocity at the eXit is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with constant speci c heats 3 Potential energy changes are negligible 4 There are no work interactions 5 The nozzle is adiabatic Properties The speci c heat of air at the average temperature of 70064526725 F is cp 0253 BtulbmR Table A 2Eb Analysis There is only one inlet and one eXit and thus m1 n3912 m We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as Em E01 2 A 70 steady O sy stem r J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies 300 pSIa 250 psia AIR Ein Eout 700 F 64501 gt 2 2 ftS mh1V1 2 mh2 V2 2 h1V122 h2 V222 Solving for eXit velocity 5 5 V2 2 V12 20114120 2 V12 201T1 T20 05 25037 it 2s2 1 Btulbm 80 its 2 20253 Btulbm R7OO 645R 8386st PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 514 528 Air is decelerated in an adiabatic diffuser The velocity at the eXit is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with constant specific heats 3 Potential energy changes are negligible 4 There are no work interactions 5 The diffuser is adiabatic Properties The speci c heat of air at the average temperature of 30902 60 C 333 K is cp 1007 kJkgK Table A 219 Analysis There is only one inlet and one eXit and thus m1 n3912 m We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as Em Emu A 70 steady O sy stem r J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies 100 kPa Ein Eout 300C AIR 200 kPa 350 m 90 C 2 2 S mh1V1 2 mh2 V2 2 h1V122 h2 V222 Solving for eXit velocity 5 5 V2 2 V12 2h1 h20 VIZ 20pT1 T20 05 1000 m2s2 350ms221007kJk K 30 90K g lkJkg 407ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 515 529 Air is accelerated in a nozzle from 120 ms to 380 ms The eXit temperature and pressure of air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with variable speci c heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The enthalpy of air at the inlet temperature of 500 K is h1 50302 kJkg Table A17 Analysis a There is only one inlet and one eXit and thus m1 n3912 m We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies E39 2 E 1 AIR 2 1n out 1 mm V22n391 v22 since QEWEApeEO 1 1 2 2 2 or 2 2 2 2 m 12 m 1 k hzzhl M250302kJkg 380 S 0 S W E 2 243802kJkg 2 2 1000 m s Then from Table A17 we read T2 4365 K E 437 K b The eXit pressure is determined from the conservation of mass relation 1 1 1 1 AV AV gt AV 2 2 ll 2 2 A1V1 v2 1 RT2 P2 Thus P A1T2V1 P 3 4365 K120 ms 2 1 600 kPa 331 kPa AZTle 1 500 K380 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 516 530 Heat is lost from the steam owing in a nozzle The velocity and the volume ow rate at the nozzle eXit are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy change is negligible 3 There are no work 1nteractrons 4000C STE AM 3000C Analysis We take the steam as the system which is a control 800 kPa gt 200 kPa volume since mass crosses the boundary The energy balance 10 W3 for this steady ow system can be expressed in the rate form as Q Energy balance 39 70 steady Ein Eout AEsystem 0 r J Rate 0f net energy MSfer Rate of changein intemalkinetic byheat W0rkandmaSS potentialetc energies Ein Eout V2 V2 ml n391 1122 Qout sinceWEApeEO V2 V2 39 or hlLh2 2 Q ut 2 2 m The properties of steam at the inlet and eXit are Table A6 P1 800 kPa 1 038429 m3kg T1 400 C h1 32677 kJkg P2 200 kPa 2 131623 m3kg T1 300 C 72 30721 kJkg The mass ow rate of the steam is 1 mZ Arvr 3008 m210 ms 2 2082kgs v1 038429m s Substituting 2 32677 kJkg 10 ms 1k kg 2 1000 m2s2 2 j30721kJkgV 2 Jkg 25km 2 1000m2s2 2082kgs The volume ow rate at the eXit of the nozzle is 2 mu 2082 kgs131623 m3kg 274 m3s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 517 531 39 Steam is accelerated in a nozzle from a velocity of 40 ms to 300 ms The eXit temperature and the ratio of the inlettoeXit area of the nozzle are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From the steam tables Table A6 P1 3 MPa 1 009938 m3kg T1 400 C h1 32317 kJkg Analysis a There is only one inlet and one eXit and thus 51 n3912 m We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic byheatW0rkaIldmaSS potentialetc energies P1 3 MPa Steam P2 25 MPa n o E 15 T1 400 C 9 V2300mls 1 m VI 40 ms mm V12 2 2 mm v222 since Q 2 W E Ape g 0 VZ VZ or hzzhl V2 V2 300ml 2 40m 2 lkJk 232317 kJkg S S f 2 2 2 lOOOm s J 31875 kJkg Thus h2 31875 kJkg v2 011533 m3kg b The ratio of the inlet to eXit area is determined from the conservation of mass relation 1 l A V 009938 3k 300 m 2 1 A2 2 V1 011533 m kg40 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 518 532E Air is decelerated in a diffuser from 750 fts to a low velocity The eXit temperature and the eXit velocity of air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with variable speci c heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The enthalpy of air at the inlet temperature of 65 F or 520 R is h1 12540 Btulbm Table A17E Analysis a There is only one inlet and one eXit and thus m1 n3912 m We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat workaand mass potentialetc energies AIR E39 Eout 1 2 111 1 mm V12 2 2 mm V222 since Q 2 W E Ape 0 2 2 or V22 V2 0 750 s 2 1 Btulbm h2 h1 1 212540 Btu1bm 2 2 13663 Btulbm 2 2 25037 it s From Table A17E T2 5716 R 112 F b The eXit velocity of air is determined from the conservation of mass relation 1 1 1 1 A2V2 AlVl gt A2V2 AlVl v2 v1 RT2 no RTl no1 Thus ATP 1 5716R 13 39 V2 2 VI X psla 750 s 244 fts A2T1P2 3 525 R145 ps1a PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 519 533 C02 gas is accelerated in a nozzle to 450 ms The inlet velocity and the eXit temperature are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 C02 is an ideal gas with variable speci c heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The gas constant and molar mass of C02 are 01889 kPam3kgK and 44 kgkmol Table A1 The enthalpy of C02 at 500 C is 171 30797 kJkmol Table A20 Analysis a There is only one inlet and one eXit and thus m1 1512 m Using the ideal gas relation the speci c volume is determined to be RTl 01889 kPam3kgKX773 K 11 0146 m3kg Thus 1 608 ms 1 39 k 14 3 m A1Vl VI mu1 60003600 gs 0 2 6m lkg v1 A1 40x10 m b We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout mm V12 2 2 mm v222 since Q 2 W E Ape g 0 VZ VZ Substituting V2 V2 h2 hl M 2 450 2 608m 2 lkJk 30797 kJkmol m S S f 2 44 kgkmol 2 1000 m s 26423 kJkmol Then the eXit temperature of C02 from Table A 20 is obtained to be T2 6858 K E 686 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 520 534 R134a is accelerated in a nozzle from a velocity of 20 ms The eXit velocity of the refrigerant and the ratio of the inlettoeXit area of the nozzle are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From the refrigerant tables Table A13 P1 700 kPa v1 0043358 m3kg T1 120 C h1 35892 kJkg and W P2 400 kPa v2 0056796 m3kg T2 30 C h 27509 kJkg Analysis a There is only one inlet and one eXit and thus m1 1512 m We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat workaand mass potentialetc energies 2E Ein out mm 1132 2 mm v222 since Q 2 W E Ape g 0 2 2 Substituting 2 2 0 27509 35892kJkg V2 20 mS 1 kJkg 2 1000 m2s2 It yields V 4099 ms 9 The ratio of the inlet to eXit area is determined from the conservation of mass relation 1 1 A V 0043358 3k 40 m A2V2 A1V1 1 2 m 3gx 99 S1565 V2 V1 A2 12 V1 0056796 m ng20 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 521 535 Nitrogen is decelerated in a diffuser from 275 ms to a lower velocity The eXit velocity of nitrogen and the ratio of the inlettoeXit area are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Nitrogen is an ideal gas with variable specific heats 3 Potential energy changes are negligible 4 The device is adiabatic and thus heat transfer is negligible 5 There are no work interactions Properties The molar mass of nitrogen is M 28 kgkmol Table Al The enthalpies are Table A18 T1 7 C 280 K gt I7 8141 kJkmol T2 27 C 300 K gt 712 8723 kJkmol Analysis a There is only one inlet and one eXit and thus 51 n3912 m We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 1n out 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies 1 N2 2 E E gt mm V12 2 2 mm V222 since Q E w E Ape 0 V22 V12 712 r V22 V12 2 M 2 O hz h1 Substituting 8723 8141 kJkmol V22 275 ms2 1 kJkg 28 kgkmol 2 1000 m2SZ It yields V2 185 ms 9 The ratio of the inlet to eXit area is determined from the conservation of mass relation A RT P LAZVZ iA1V11 V2 V2 12 11 A2 12 V1 RT2 P2 V1 or 0887 i T1P1 V2 280 K60 kPa185 ms A2 T2 no2 V1 300 K85 kPa200 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 522 536 quot Problem 535 is reconsidered The effect of the inlet velocity on the eXit velocity and the ratio of the inletto eXit area as the inlet velocity varies from 210 ms to 350 ms is to be investigated The final results are to be plotted against the inlet velocity Analysis The problem is solved using EES and the solution is given below Function HCaIWorkFuid TX PX quotFunction to calculate the enthalpy of an ideal gas or real gasquot If 39N239 WorkFluid then HCaIENTHALPYWorkFuidTTx quotIdeal gas eququot else HCaIENTHALPYWorkFuidTTx PPxquotRea gas eququot endif end HCaI quotSystem control volume for the nozzlequot quotProperty relation Nitrogen is an ideal gasquot quotProcess Steady state steady flow adiabatic no workquot quotKnownsquot WorkFluid 39N239 T1 7 C P1 60 kPa Ve1 275 ms P2 85 kPa T2 27 C quotProperty Data since the Enthalpy function has different parameters for ideal gas and real fluids a function was used to determine hquot h1HCaWorkFuidT1P1 h2HCaWorkFluidT2P2 quotThe Volume function has the same form for an ideal gas as for a real fluidquot v1voumeworkFuidTT1pP1 v2voumeWorkFuidTT2pP2 quotFrom the definition of mass flow rate mdot AVellv and conservation of mass the area ratio ARatio A1A2 isquot ARatioVe1v1 Ve2v2 quotConservation of Energy SSSF energy balancequot h1Ve1quot221000 h2Ve2quot221000 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission Ve1 VGIZ ARatio 300 ms ms 39 210 5001 03149 1 224 9261 05467 250 238 1227 06815 39 252 148 07766 266 1708 08488 3 200 280 1918 09059 g 294 2117 09523 H 39 308 2308 09908 g 150 322 2492 1023 g I 336 267 1051 350 2844 1075 100 I 5039 200 220 240 260 280 300 320 340 360 Vel1 ms 11 1 09 08 3 07 E f lt 06 i 03 J 200 220 240 260 280 300 320 340 360 Vel1 ms 523 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5 24 537 R134a is decelerated in a diffuser from a velocity of 160 ms The eXit velocity of R134a and the mass ow rate of the R134a are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions Properties From the R134a tables Tables All through A13 P1 600 kPa v1 0034335 m3kg 2 kJs h1 26246 kJkg 1 R134lia 2 and P2 700 kPa v2 0031696 m3kg T2 40 C h2 27859 kJkg sat vapor Analysis 61 There is only one inlet and one eXit and thus m1 1512 m Then the eXit velocity of R134a is determined from the steady ow mass balance to be 1 1 A 1 0031696 3k A2V2 A1V1V2v 2 1V1 m 3 160 ms 8206 ms b We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies E E 1n out Q39in mm V12 2 2 mm vzzz since W Ape g 0 V2 V2 Qin mLhz h1 Substituting the mass ow rate of the refrigerant is determined to be 2 2 kJs m27859 26246kJkg 8206 mS 160 ms2 lkJkg 2 1000 m2s2 It yields nquot 02984kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 525 538 Steam is accelerated in a nozzle from a velocity of 60 ms The mass ow rate the eXit velocity and the eXit area of the nozzle are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 There are no work interactions Properties From the steam tables Table A6 3 75 kJs P1 4 MPa 1 007343 m lkg BK T1 400 C h1 32145 kJkg 1 Steam 2 and P2 2 MPa 2 012551m3kg T2 300 C h 30242 kJkg Analysis a There is only one inlet and one eXit and thus m1 n3912 m The mass ow rate of steam is m im 43 60 ms50 X104m2 4085 kgs v1 007343 m kg 9 We take nozzle as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies E E 1n out mm V12 2 Q39out mm vzzz since W Ape 0 VZ VZ Qout Z mLhz hr Substituting the eXit velocity of the steam is determined to be 2 2 75 kJs 4085 kgs30242 32145 V2 60 mS 1kJkg 2 1000 m2s2 It yields V2 5895 ms 0 The eXit area of the nozzle is determined from 3 m LVZAZ A2 mu2 4085 kgs012551m Ikg OX 1041112 2 V2 5895 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 526 539 Air is decelerated in a diffuser from 220 ms The exit velocity and the eXit pressure of air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with variable speci c heats 3 Potential energy changes are negligible 4 There are no work interactions Properties The gas constant of air is 0287 kPam3kgK Table Al The enthalpies are Table A17 7 27 C300K gt hl 230019 kJkg T2242 C315K gt h231527kJkg Analysis a There is only one inlet and one eXit and thus m1 1512 m We take diffuser as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 stead Ein Eout AE y 0 system f J r J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat workand mass potentialetc energies Ein Eout MAR mm VIZ2 Q out mm V222 since w Ape 0 2 2 Substituting the eXit velocity of the air is determined to be sz 220 ms2 lkJkg 18kJs 25k s 31527 30019kJk g E g 2 1000 m2s2 It yields V 620 ms 9 The eXit pressure of air is determined from the conservation of mass and the ideal gas relations 1 A V 004 2 62 m m 2 A2V2 gt 2 Z 2 2 Z m x S 2 2 m 25 kgs and RT 0287kP 3k K 315K 2 0992m lkg Turbines and Compressors 540C The volume ow rate at the compressor inlet will be greater than that at the compressor eXit 541C Yes Because energy in the form of shaft work is being added to the air 542C No PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 527 543E Air expands in a turbine The mass ow rate of air and the power output of the turbine are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible 4 Air is an ideal gas with constant speci c heats Properties The gas constant of air is R 03704 psiaft3lbmR The constant pressure speci c heat of air at the average temperature of 900 3002 600 F is cp 025 Btulbm F Table A2a Analysis a There is only one inlet and one eXit and thus m1 1512 m The inlet speci c volume of air and its mass ow rate are 3 V1 2 03704 ps1a 11 0an RX1360 R 3358 31bm 1 P1 150 ps1a 1 1 l m AlVl 3 0 1 11 2 X350 its 10421bms V1 3358 lbm AIR 6 b We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow J system can be eXpressed in the rate form as N 39 39 39 70 steady Ein Eout AEsystem O 2 f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic byheata workaandmass potentialetc energies E E 1n out mall V12 2 Wout n392h2 V222 since Q g Ape 0 V 2 V 2 V 2 V 2 Wout mh2 h1 mcp T2 T1 Substituting 2 2 W 10421bms0250 Btu1bm FX300 900 F700 its 350 its lBtulbm 22 2 25037 it 2s2 14865 Btus 1568 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 528 544 Refrigerant134a is compressed steadily by a compressor The power input to the compressor and the volume ow rate of the refrigerant at the compressor inlet are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible Properties From the refrigerant tables Tables All through 13 T1 24 C u1 017398 m3kg P2 08 MPa h2 296 82 kJkg 2 h1 23594 kJkg T2 60 C satvap0r Analysis a There is only one inlet and one eXit and thus 51 1512 m We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as a 39 39 39 70 steady Ein Eout AEsystem O f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic byheata workaandmass potentialetc energies Ein Eout 1 Win nah1 2112 since Q Ake Ape 0 Win 2 quot1072 171 Substituting Wm 12 kgs296 82 23594kJkg 7306 kJs b The volume ow rate of the refrigerant at the compressor inlet is V mu 12 kgs0 17398 m3kg 0209 m3s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 529 545 Saturated R134a vapor is compressed to a specified state The power input is given The eXit temperature is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer with the surroundings is negligible Analysis We take the compressor as the system which is a control volume since mass crosses the boundary Noting that one uid stream enters and leaves the compressor the energy balance for this steady ow system can be eXpressed in the rate form as u u d Ein Eout AEsystem Stea y 2 O f r J RateOf netenergy thfer Rateof changein intemalkinetic kPa by heat W0fkand mass potentialetc energies Em E01 Compressor 3 Win 1139th 1139th since Ake Ape 0 Wm Win mh2 hz From R134a tables Table A12 180 kPa sat vap Pl 2180 kPa h1 24290 kJkg 035 1113ij x1 0 v1 01105 m3kg The mass ow rate is rh 005283 kgs v1 01104m3kg Substituting for the eXit enthalpy Wm mazz ho 235 kW 2 005283 kgsh2 24290kJkg gth2 28741 kJkg From Table A13 P2 700 kPa h2 28741 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 530 546 Steam expands in a turbine The change in kinetic energy the power output and the turbine inlet area are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 The deVice is adiabatic and thus heat transfer is negligible Properties From the steam tables Tables A4 through 6 P1 4 MPa T1 500 C P1 4 MPa 1 0086442m3kg VI 80 ms T1 500 C h1 34460kJkg 21 P2 30 kPa STEAM x2 092 m 12 kgs w Analysis a The change in kinetic energy is determined from 2 2 2 2 Akezvz V150ms 80ms 1kJkg 195kJkg h2 h f xzhfg 28927 092gtlt 23353 24377 kJkg 2 2 1000 m2s2 P2 30 kPa x2 b There is only one inlet and one eXit and thus m 1512 m We take the turbine as the V2 50 ms system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat workaand mass potentialetc energies 2E E39 out mm V12 2 Wout mm V222 since Q g Ape 0 V2 V2 Wout 2 mLh2 hr Then the power output of the turbine is determined by substitution to be Wout 12 kgs24377 34460 195kJkg 12 123 kW 2 121 MW c The inlet area of the turbine is determined from the mass ow rate relation 39 12k 42 3k m 2 i AlVl A1 mt1 gs0 0864 m g 1 V1 80 ms 00130 m2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 531 547 1 quot Problem 546 is reconsidered The effect of the turbine eXit pressure on the power output of the turbine as the eXit pressure varies from 10 kPa to 200 kPa is to be investigated The power output is to be plotted against the eXit pressure Analysis The problem is solved using EES and the solution is given below quotKnowns quot T1 500 C P1 4000 kPa Ve1 80 ms P2 30 kPa X2092 13o Ve2 50 ms mdot112 kgs 120 Fluid39SteamlAPWS39 11 39 O quotProperty Dataquot 100 h1enthapyFuidTT1PP1 h2enthapyFuidPP2xx2 90 T2temperatureFuidPP2xx2 a 8039 v1voumeFuidTT1pP1 39 v2voumeFuidPP2xx2 70 quotConservation of mass quot 60 mdot1 mdot2 50 quotMassflowratequot 40 mdot1A1Ve1v1 o 40 80 120 160 200 mdot2 A2Ve2v2 P2 kPa quotConservation of Energy Steady Flow energy balancequot mdot1h1Ve1quot22Convertmquot2squot2 kJkg mdot2h2Ve2 22Convertmquot2squot2 kJkgWdotturbconvertMWkJs DELTAkeVe2 22Convertmquot2squot2 kJkgVe1quot22Convertmquot2squot2 kJkg P2 erb T2 3 kPa MW Cl 1 28 10 1267 4581 12 6 3111 121 6993 39 5222 1182 824 124 7333 1163 9116 9444 1148 9802 E 122 1156 1136 1037 E 12 1367 1125 1086 g 118 1578 1116 1129 1789 1108 1167 116 200 1101 1202 114 112 1 1 1 60 200 80 120 P2 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 532 548 Steam expands in a turbine The mass ow rate of steam for a power output of 5 MW is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The deVice is adiabatic and thus heat transfer is negligible Properties From the steam tables Tables A4 through 6 1 P1 10 MPa h1 33751kJkg T1 500 C 0 90 112 hf xzhfg 19l81090gtlt 23921 23447 kJkg x2 2 Analysis There is only one inlet and one eXit and thus m1 1512 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance 2 for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout nah1 Wout mh2 since Q Ake Ape 0 W n391h2 hl out 2 Substituting the required mass ow rate of the steam is determined to be 5000 kJs n39123447 33751 kJkg m39i 4852 kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 533 549E Steam expands in a turbine The rate of heat loss from the steam for a power output of 4 MW is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties From the steam tables Tables A4E through 6E P1 1000 psia h1 14486 Btulbm T1 900 F 6 P2 5 psia h2 11307 Btulbm satvap0r Analysis There is only one inlet and one eXit and thus m1 1512 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout m1 Qquotout Wout m2 since Ake Ape 0 Qout mh2 h1 Iiut Substituting 1 B Qout 450003600 lbmsl 1307 14486Btulbm 4000 kJs j 1820 Btus PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 534 550 Helium is compressed by a compressor For a mass ow rate of 60 kgmin the power input required is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas with constant specific heats Properties The constant pressure specific heat of helium is 0 51926 kJkgK Table A2a Analysis There is only one inlet and one eXit and thus m1 1512 m We take the compressor as the system which is a control volume since mass crosses the boundary 102 700 kPa The energy balance for this steady ow system can be eXpressed in the rate form as T2 460 K 39 39 39 70 steady Ein Eout AEsystem O RateOf net energy MSfer Rate of changein intemalkinetic 39 by heat WOIkaIldmaSS potentialetc energies Q H6 3 Em Eout 60 kgmin Win nhh1 Qlout n39mz since Ake E Ape E 0 W VVinQoutmh2hlmcpT2Tl T Thus P1 105 kPa T1 K Win Qout mcp T2 T1 606 0 kgS15 kJkg 6060 kgS51926 kJkg K460 295K 872 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 535 551 C02 is compressed by a compressor The volume ow rate of C02 at the compressor inlet and the power input to the compressor are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas with variable speci c heats 4 The device is adiabatic and thus heat transfer is negligible Properties The gas constant of C02 is R 01889 kPam3kgK and its molar mass is M 44 kgkmol Table Al The inlet and eXit enthalpies of C02 are Table A20 T1 300 K gt 131 9431kJkmol 2 T2 450 K gt 132 15483 kJkmol Analysis a There is only one inlet and one eXit and thus m1 1512 m The inlet specific volume of air and its volume ow rate are C02 3 RT 0188 kP 3k K 300K v1 1 9 am g x 205667m3kg P1 100 kPa T V n39w1 05 kgs05667 m3kg 0283 m3s 1 b We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy t1 Sfer Rate of changein intemalkinetic byheat W0rkand mass potentialetc energies E E 1n out Win 1111 2112 since E Ake E Ape E 0 W 42012 hl 2032 43044 Substituting Win 2 05 kgsX15483 9431 kJkmol 688 kw 44 kgkmol PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 536 552E Air is compressed by a compressor The mass ow rate of air through the compressor and the eXit temperature of air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas with variable speci c heats Properties The gas constant of air is R 03704 psiaft3lbmR Table AlE The inlet enthalpy of air is Table A17E T1 60 F R 11 h 520R Btulbm Analysis a There is only one inlet and one eXit and thus 51 1512 m The inlet speci c volume of air and its mass ow rate are 3 1 2 2 03704 ps1a it lbm RX520 R Z 131 gllbm P1 147 ps1a 2 5000 13111111 3 3817lbmmin636lbms 1 13111 lbm AIR 3 b We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as T 10 Btulbm 39 39 39 70 steady Ein Eout AE 0 system p x f J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat workaand mass potentialetc energies Ein Eout Win m1 Qout m2 since Ake Ape 0 Win Qout Z quot1072 171 Substituting 700 hpWj 636 lbmsgtlt 10 Btulbm 6361bmsh2 12427 Btulbm p h 19206 Btulbm Then the eXit temperature is determined from Table A17E to be T2 801 R 341 F PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 537 553E l i quot Problem 552E is reconsidered The effect of the rate of cooling of the compressor on the eXit temperature of air as the cooling rate varies from 0 to 100 Btulbm is to be investigated The air eXit temperature is to be plotted against the rate of cooling Analysis The problem is solved using EES and the solution is given below quotKnowns quot T1 60 F P1 147 psia Vdot1 5000 ftquot3min P2 150 psia qout10 BtuIbm Wdotin700 hp quotProperty Dataquot h1enthalpyAirTT1 h2enthapyAirTT2 TR2T2460 quotRquot v1volumeAirTT1pP1 v2voumeAirTT2pP2 quotConservation of mass quot mdot1 mdot2 quotMass flow ratequot mdot1Vdot1v1 convertftquot3 mi n ftquot3s mdot2 Vdot2v2convertftquot3 min ftquot3s quotConservation of Energy Steady Flow energy balancequot WdotinconverthpBtusmdot1h1 mdot1qoutmdot1h2 qout T2 400 BtuIbm F 350 0 382 10 3409 3 20 2997 250 30 2583 L 200 40 2159 H 50 1754 E 15 60 1338 100 70 9225 50 80 5067 0 90 9053 100 3263 50 o 20 4o 60 80 100 qout Btulbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 538 554 Air is expanded in an adiabatic turbine The mass ow rate of the air and the power produced are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 The turbine is wellinsulated and thus there is no heat transfer 3 Air is an ideal gas with constant specific heats Properties The constant pressure specific heat of air at the average temperature of 5001272314 C5 87 K is 0 1048 kJkgK Table A 2b The gas constant of air is R 0287 kPam3kgK Table A1 Analysis a There is only one inlet and one eXit and thus m1 2 m2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 710 t d Ein Eout AEsystem S ea y 2 O V 13 MPa Rate 0f net energy MSfer Rate of chan gem 1ntemalk1net1c byheata workaandmass potentialetc energies 5000C 39 39 40 ms Ein Eout V12 V22 quotlth I mLhz 7 Wout VZ VZ Vz VZ 100kPa Wont quot1th h2 1 m cp T1 T2 127 C The speci c volume of air at the inlet and the mass ow rate are 0287 kPa m3kg K500 273 K 1 01707m3kg 1D1 1300kPa AV 2 2 4 m 1 1 m X 2m4688kgs v1 01707m lkg Similarly at the outlet RT 2 kP 3 K 12 2 K V2 22m 87 amkg 7 73 1148mgkg 1D2 100kPa 39 4 k 114 3 k V2 mv2 688 gs 8m g533982mS A2 1m2 9 Substituting into the energy balance equation gives Vz VZ 2 2 4688 kgs1048 kJkg K500 127K 40 mm 5382 mS Jkg 2 1000111282 18300kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 539 555 Steam expands in a twostage adiabatic turbine from a specified state to another state Some steam is extracted at the end of the first stage The power output of the turbine is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The 8 MPa turbine is adiabatic and thus heat transfer is negligible 600 C Properties From the steam tables Tables A5 and A6 13 kg S P 8 MPa l 1 o h1 36424 kJkg L T1 2 600 C STEAM 6 p O3MPa 13 kgs x2 085 19181 O8523921 22251 kJkg 1001i X Analysis We take the entire turbine including the connection part 03 MP3 between the two stages as the system which is a control volume since 13 kgS mass crosses the boundary Noting that one uid stream enters the sat Vap turbine and two uid streams leave the energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 lt90 steady Ein Eout AEsystem O r J f J Rate 0f net energy WSfer Rate of changein intemalkinetic byheats worksandmass potentialetc energies Ein E out W 1411011 01h2 0911 out Substituting the power output of the turbine is W out 13 kgs36424 01 x 27249 09 X 22251 kJkg 17776 kW 2 1 78 MW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 540 556 Steam is expanded in a turbine The power output is given The rate of heat transfer is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties From the steam tables Table A4 A5 A6 P1 6 MPa T1 600 C P2 05 MPa T2 200 C h1 36588kJkg h2 28558 kJkg Analysis We take the turbine as the system which is a control volume since mass crosses the boundary Noting that one uid stream enters and leaves the compressor the energy balance for this steady ow system can be eXpressed in the rate form as lt90 steady Ein Eout AEsystem O V r J RateOfnetenel39gy WSfel39 Rate of changein intemalkinetic S byheatsworksandmass potentialetc energies 6 MPa Ein Eout C V2 V2 mth mLh2 Wout Qout s1nce Ape E 0 39 39 V12 V22 Qout Wout quot1th h2 T MPa 200 C Substituting V2 V2 Qout Wout quotlth hZ J 0 180ms2 lkJkg J 2 1000m2s2 20000 kW 26 kgs36588 28558kJkg 455 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 541 557 Air is compressed by a compressor The mass ow rate of air through the compressor is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy changes are negligible 3 Air is an ideal gas with variable specific heats Properties The inlet and eXit enthalpies of air are Table A17 T1 25 C K 21 h 298 K T2 347 C K 12 h 620K 2 Analysis We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as AIR 5 7I0 t ad Ein Eout AEsystem S e y 2 O RateOf net energy MSfer Rate of chan ein internal kinetic by heat WOI39kaIldl39naSS potentigletc energi s T Ein Eout 1 Win mm V12 2 Qquotout mm V222 since Ape 0 VZ VZ Win Qout mLhZ hl Substituting the mass ow rate is determined to be 250 kJs 150060 kJs m 62807 2982 2 2 1000 m 90 ms2 0 1 kJkg S gt m 0674 kgs Throttling Valves 558C Because usually there is a large temperature drop associated with the throttling process 559C No Because air is an ideal gas and h hT for ideal gases Thus if h remains constant so does the temperature 560C If it remains in the liquid phase no But if some of the liquid vaporizes during throttling then yes 561C Yes PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 542 562 Refrigerant134a is throttled by a valve The temperature drop of the refrigerant and the specific volume after expansion are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the uid is negligible 4 There are no work interactions involved Properties The inlet enthalpy of R134a is from the refrigerant tables Tables A11 through 13 P1 07MPa T1 Tsat 2669 C P1 700 kPa sat llqu1d h1 hf 8882 kJkg Sat liquid Analysis There is only one inlet and one eXit and thus 51 1512 m no We take the throttling valve as the system which is a control volume 139 since mass crosses the boundary The energy balance for this steady A ow system can be eXpressed in the rate form as R4343 70 ad Ein Eout AEsystem Ste y 2 O Ein out mhl th h 2 72 1 Cu since Q2 WAkezApezOThen P2 160 kPa P2 160kPa hf 3118kJkg Tsat 1560 C h2 h1 hg 224114 kJkg Obviously hf lth2 lthg thus the refrigerant eXists as a saturated mixture at the eXit state and thus T2 Tsalt 1560 C Then the temperature drop becomes AT 2 T2 T1 1560 2669 423 C The quality at this state is determined from h2 hf 8882 3118 h fg 20996 x2 02745 Thus 2 If xzvfg 00007435 02745 x 012355 00007435 00345 m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 543 563 Steam is throttled from a specified pressure to a specified state The quality at the inlet is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the uid is negligible 4 There are no work interactions involved Analysis There is only one inlet and one eXit and thus m1 n3912 m We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 7390 d Ein Eout AEsystem Stea y 2 O Em Eont Throttling valve n39zh1 n39zh2 Steam 69 50 kPa hl hz MP3 1000C since QEWZAkeEApCEO The enthalpy of steam at the eXit is Table A6 h2 26824 kJkg T2 100 C The quality of the steam at the inlet is Table A5 x 12 h f 26824 84455 1 h fg 19464 2 0944 P1 1500 kPa h1 I12 26824 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 544 564 Refrigerant134a is throttled by a valve The pressure and internal energy after expansion are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the uid is negligible 4 There are no work interactions involved Properties The inlet enthalpy of R134a is from the refrigerant tables Tables A11 through 13 P1 08MPa T1 25 C h15 hf25C 8641kJkg Analysis There is only one inlet and one exit and thus m1 1512 m We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 7390 d Ein Eout AEsystem Stea y 2 O 51 358031321 1 Ein Eout l i lhl 11 h 1 2 C R134a since QEWAk EAp EOThen 1 T2 20 c hf 2547 kJkg uf 2537 kJkg T2 20 C I12 hl hg 23843 kJkg ug 21886 kJkg Obviously hf lt hz lthg thus the refrigerant exists as a saturated mixture at the exit state and thus P2 Psat 200C kPa Also 12 hf 8641 2547 hfg 21296 02862 x2 Thus u2 uf xzufg 2537 02862 x 19349 807 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 545 565 Steam is throttled by a wellinsulated valve The temperature drop of the steam after the expansion is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the uid is negligible 4 There are no work interactions involved Properties The inlet enthalpy of steam is Tables A6 P1 8 MPa h1 29881 kJkg T1 350 C Analysis There is only one inlet and one eXit and thus m1 1512 m We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O Ein Eout l i lhl h1 hz since E W Ake E Ape E 0 Then the eXit temperature of steam becomes T 285 C hzzhl 2 P1 8 MPa T1 350 C 139 C H20 P22MPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 546 566 quot Problem 565 is reconsidered The effect of the eXit pressure of steam on the eXit temperature after throttling as the eXit pressure varies from 6 MPa to 1 MPa is to be investigated The eXit temperature of steam is to be plotted against the eXit pressure Analysis The problem is solved using EES and the solution is given below WorkingFIuid39Steamiapws39 quotWorkingFIuid can be changed to ammonia or other fluidsquot Pin8000 kPa Tin350 C Pout2000 kPa quotAnalysisquot mdotinmdotout quotsteadystate mass balancequot mdotin1 quotmass flow rate is arbitraryquot mdotinhinQdotWdotmdotouthout0 quotsteadystate energy balancequot Qdot0 quotassume the throttle to operate adiabaticallyquot Wdot0 quotthrottles do not have any means of producing powerquot hinenthapyWorkingFIuidTTinPPin quotproperty table lookupquot TouttemperatureWorkingFluidPPouthhout quotproperty table lookupquot xoutquaityWorkingFluidPP0uthhout quotxout is the quality at the outletquot P1Pin P2Pout h1hin h2hout quotuse arrays to place points on property plotquot Pout Tout 340 39 39 39 39 kPa C 330 39 1000 2705 1500 2777 32039 39 2000 2846 2500 2912 g 310 J 3000 2975 H D 3500 3037 300 4000 3095 I 4500 3152 290 5000 3207 U 5500 3259 280 5000 331 1000 2000 3000 4000 5000 6000 Pout kPa 1 39 l 39 I 39 IStean39IlAPwlS I 105 4 350 C 10 28500 39 G g 103 2 D 102 101 l I I I I I I I I I I 0 500 1000 1500 2000 2500 3000 3500 hkJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 547 567E Refrigerant134a is throttled by a valve The temperature and internal energy change are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the uid is negligible 4 There are no work interactions involved Analysis There is only one inlet and one eXit and thus m 1512 m We take the throttling valve as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as Em Emu 70 steady O system 1 T 2 Ein Eout gt gt n39zhl 1139th hr hz since E W Ake E Ape E 0 The properties are Tables A11E through 13E P1 120 psia ul 2 4149 Btulbm x1 2 O h1 4179 Btulbm T1 9049 F P2 20 psia T2 243 F h2 h1 4179 Btulbm u2 3896 Btulbm AT 2 T2 T1 243 9049 929 F Au M2 ul 2 3896 4149 253 Btulbm That is the temperature drops by 929 F and internal energy drops by 253 Btulbm Mixing Chambers and Heat Exchangers 568C Under the conditions of no heat and work interactions between the miXing chamber and the surrounding medium 569C Under the conditions of no heat and work interactions between the heat exchanger and the surrounding medium 570C Yes if the mixing chamber is losing heat to the surrounding medium PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 548 571 Liquid water is heated in a chamber by mixing it with superheated steam For a speci ed mixing temperature the mass ow rate of the steam is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties Noting that T lt Tsat 300 kpa 13352 C the cold water stream and the mixture exist as a compressed liquid which can be approximated as a saturated liquid at the given temperature Thus from steam tables Tables A4 through A 6 111 E hf20 C 25118 kJkg h3 hf 60 C and P2 300 kPa 30696 kJk T2 300 C i172 g Analysis We take the mixing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steady ow system can be expressed in the rate form as M 7I0 steady Mass balance Inin mout system O Inin mout m1 quot 2 quot 3 Energy balance 39 39 39 lt90 steady Ein Eout AEsystem O RateOf net energy thfer Rate of chan gem 1ntemalk1net1c 39 byheatworkandmass potentialetc energies m1 18 kgS Ein Eout H20 P 300 kPa mlhl mzhz m3h3 s1nce Q W Ake E Ape E O T 60 C 3 Combining the two T2 300 C W 1712 mlhl quotth 2 11 Solving for Substituting 8391 25118kJkg quot12 25118 30696kJkg 18 kgs 0107 kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 549 572 Feedwater is heated in a chamber by mixing it with superheated steam If the mixture is saturated liquid the ratio of the mass ow rates of the feedwater and the superheated vapor is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties Noting that T lt Tsalt 1MPa 17988 C the cold water stream and the mixture exist as a compressed liquid which can be approximated as a saturated liquid at the given temperature Thus from steam tables Tables A4 through A6 hl E hf 50 C 20934 kJkg 13 E hf1Mpa and h2 28283kJkg T2 200 C Analysis We take the mixing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steady ow system can be expressed in the rate form as 7I0 steady Mass balance Inin mout AInsystem O min mout m1 quot12 quot 3 Energy balance I n n t d Ein Eout AEsystemCy0 S ea y 2 O o RateOf net energy thfer Rate of changein intemalkinetic 39 byheat W0rkandmaSS potentialetc energies m1 Ein Eout H20 P 1 MPa mlh1 m22 m33 s1nce Q W Ake E Ape E O Sat liquid Combining the two 12 200 C W mlhl quot72172 2 11 Dividing by yields yhl 12 y 1h3 Solving for y h3 h2 y hlh3 where y 1511 1512 is the desired mass ow rate ratio Substituting 76251 28283 y 373 20934 76251 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 550 573E Liquid water is heated in a chamber by mixing it with saturated water vapor If both streams enter at the same rate the temperature and quality if saturated of the eXit stream is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible Properties From steam tables Tables A5E through A6E hl E hf 65 F 3308 Btulbm 12 18 20 psia Analysis We take the miXing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steady ow system can be eXpressed in the rate form as Mass balance 7I0 stead T 65 F min mout An tsystem y O 1 min mout H20 n39zl n39zz 2 m3 2n391 P 20 psia quot11 Z quot12 Z T3 x3 Sat va or Energy balance p m2 m1 39 39 39 lt90 steady Ein Eout AEsystem O r J r J RateOf net energy thfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies Ein Eout m1h1 mth 1311 since Q39 W Ake Ape 0 Combining the two gives mh1 th 2mh3 or h3 h1 h22 Substituting h3 3308 115622 5946 Btulbm At 20 psia hf 19627 Btulbm and kg 11562 Btulbm Thus the eXit stream is a saturated mixture since hf lt h3 lt hg Therefore T3 Tsat 20 psia 22801 and h3 hf 5946 19627 x 3 hfg 11562 19627 0415 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 551 574 Two streams of refrigerant134a are mixed in a chamber If the cold stream enters at twice the rate of the hot stream the temperature and quality if saturated of the exit stream are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The deVice is adiabatic and thus heat transfer is negligible Properties From R134a tables Tables A11 through A13 11 E hf 20 C 12 h 1Mpa 80 C Analysis We take the mixing chamber as the system which is a control volume since mass crosses the boundary The mass and energy balances for this steady ow system can be expressed in the rate form as Mass balance 7l0 steady T 20 C min mout A lsystem O 1 I m1 21712 min mout R134a m1 Z Z SiIlCe m1 2 1 T3 x3 Energy balance T2 80 C Em Emit AE lt90 steady 0 system J J Rate 0f net energy thfer Rate of changein intemalkinetic byheat W0rkand maSS potentialetc energies Ein Eout 141111 mth 1413113 since Q39 W Ake Ape 0 Combining the two gives 214127 11392th 31712713 or 173 2h1 h23 Substituting h3 2x7932 314273 15764 kJkg At 1 MPa hf 10734 kJkg and kg 27104 kJkg Thus the exit stream is a saturated mixture since hf lt 123 lt hg Therefore T3 Tsat 1MPa 3937 C and hs hf 15764 10734 hfg 27104 10734 0307 x3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 552 575 A quot Problem 574 is reconsidered The effect of the mass ow rate of the cold stream of R134a on the temperature and the quality of the eXit stream as the ratio of the mass ow rate of the cold stream to that of the hot stream varies from 1 to 4 is to be investigated The mixture temperature and quality are to be plotted against the coldtohot mass ow rate ratio Analysis The problem is solved using EES and the solution is given below quotInput Dataquot mfrac 2 quotmfrac mdotcodmdothot mdot1mdot2quot T120 C P11000 kPa T280 C P21000 kPa mdot1 mfracmdot2 P31000 kPa mdot1 1 quotConservation of mass for the R134a Sum of mdotinmdotoutquot mdot1 mdot2 mdot3 quotConservation of Energy for steadyflow neglect changes in KE and PEquot quotWe assume no heat transfer and no work occur across the control surfacequot Edotin Edotout DELTAEdotcv DELTAEdotcvO quotSteadyflowrequirementquot Edotinmdot1h1 mdot2h2 Edotoutmdot3h3 quotProperty data are given byquot h1 enthapyR134aTT1PP1 h2 enthapyR134aTT2PP2 T3 temperatureR134aPP3hh3 x3QUALITYR134ahh3PP3 mfrac T3 X3 055 C 05 1 3937 05467 125 3937 0467 045 15 3937 04032 0 4 175 3937 0351 39 2 3937 03075 035 lt2 225 3937 02707 gt3 25 3937 02392 03 g 275 3937 02119 025 3 3937 0188 325 3937 01668 02 35 3937 01481 0 15 375 3937 01313 39 4 3937 01162 01 mfrac PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 55 3 57 6 Water is heated in a heat exchanger by geothermal water The rate of heat transfer to the water and the exit temperature of the geothermal water is to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 3 Changes in the kinetic and potential energies of uid streams are negligible 4 Fluid properties are constant Properties The speci c heats of water and geothermal uid are given to be 418 and 431 kJkg C respectively Analysis We take the cold water tubes as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 60 C 39 70 steady Ein Eout AEsystem O f 112311610 ftn Ct eiergi l 31 Sfer Rate 0 f chan gein intemalkin etic ea 01 an mass 39 39 y w I I potent1aletcenerg1es Bnne q Ein Eout 140 C gt gt E gt 8 gt Qin mhl th s1nce Ake E Ape E O gt I 2 mcpltT2 T1gt Water Then the rate of heat transfer to the cold water 1n the heat 25 C exchanger becomes Q icpT0ut Tin 1Water 02 kgs4 18 kJkg C60 C 25 C 2926 kW Noting that heat transfer to the cold water is equal to the heat loss from the geothermal water the outlet temperature of the geothermal water is determined from 39 2 2 k Q mcp Toutgeotwater gtTout i 2 140 C 9 6 W 117394OC n39wp 03 kgs431kJkg C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 554 577E Steam is condensed by cooling water in a condenser The rate of heat transfer in the heat exchanger and the rate of condensation of steam are to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 3 Changes in the kinetic and potential energies of uid streams are negligible 4 Fluid properties are constant Properties The speci c heat of water is 10 Btulbm F Table A3E The enthalpy of vaporization of water at 75 F is 10509 Btulbm Table A4E Steam 75 F Analyszs We take the tubes1de of the heat exchanger where cold water 1s m o owing as the system which is a control volume The energy balance for r d 65 F this steady ow system can be expressed in the rate form as Q G J 39 39 39 70 steady k Ein Eout AEsystem O V Q Rate 0f net energy thfer Rate of changein intemalkinetic byheata workaandmass potentialetc energies J k D Em Eout 1 J 50 F k n391hn391 sinceAkeEAeEO Qm I 1 ha p Lo Water Qin mcpT2 Then the rate of heat transfer to the cold water in this heat exchanger becomes 75 F i Q mop T0 Tm water 45 lbms10Btulbm F65 F 50 F 675 Btus Noting that heat gain by the water is equal to the heat loss by the condensing steam the rate of condensation of the steam in the heat exchanger is determined from Q 675 Btus h o642lbms Q m fg steam msteam hfg 10509 Btulbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 555 57 8 Oil is to be cooled by water in a thinwalled heat exchanger The rate of heat transfer in the heat exchanger and the exit temperature of water is to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 3 Changes in the kinetic and potential energies of uid streams are negligible 4 Fluid properties are constant Properties The speci c heats of water and oil are given to be 418 and 220 kJkg C respectively Hot oil Analys1s We take the 011 tubes as the system wh1ch 1s a control 150 C volume The energy balance for this steady ow system can be 2 k ls expressed in the rate form as C old I I g Em Emu AESyStemzo steady 0 water I ar J J 22 C RateOfnet energy thfer Rate of changein intemalkinetic 1 5 k byheataworkaandmass potentialetc energies g S r Ein Eout 40 C 1quotth1 2 Qlout n391h2 since Ake E Ape E O Q mop T1 T2 Then the rate of heat transfer from the oil becomes Q 7139ch Tin Tout0i1 2 kgs22 kJkg C150 C 40 C 484 kW Noting that the heat lost by the oil is gained by the water the outlet temperature of the water is determined from 39 4 4 Q mcp Tout Tin water gtTout Tin 2 22 C 8 kJS 9920C m c 15 kgs4 18 kJkg C water p PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 556 579 Air is preheated by hot exhaust gases in a cross ow heat exchanger The rate of heat transfer and the outlet temperature of the air are to be determined Assumptions 1 Steady operating conditions exist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 3 Changes in the kinetic and potential energies of uid streams are negligible 4 Fluid properties are constant Properties The speci c heats of air and combustion gases are given to be 1005 and 110 kJkg C respectively Analysis We take the exhaust pipes as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady 39 39 Ein Eout AEsystem O E39 E af J r J Rate 0f net energy thfer Rate of changein intemalkinetic byheat workandmaSS potentialetc energies I n39zhl Qlout 1139th since Ake E Ape E 0 I I Air I Qout mcp T2 kPa b 5 Then the rate of heat transfer from the exhaust gases becomes 20 3 lt5 06 m s Q E Q mcp Tout gas b 095 kgs11kJkg C160 C 95 C R k 7 kW 6 93 Exhaust gases The mass ow rate of air is 095 kgs 95 C 39 3 m PV 95 kPa06m Is 2 06778 kgS RT 0287 kPam3kgK X 293 K Noting that heat loss by the exhaust gases is equal to the heat gain by the air the outlet temperature of the air becomes Q mcp Tc out Tc ingtTc out Tc in 20 C kW leooc mcp 06778 kgS1005 kJkg C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 557 580E determined Air is heated in a steam heating system For speci ed ow rates the volume ow rate of air at the inlet is to be Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 5 Air is an ideal gas with constant speci c heats at room temperature Properties The gas constant of air is 03704 psiaft3lbmR Table AlE The constant pressure specific heat of air is cp 0240 Btulbm F Table A2E The enthalpies of steam at the inlet and the exit states are Tables A4E through A6E P3 30 psia 113 12379 Btulbm T3 400 F P4 25 pm h 180 21B 1b T4 21201 4 2 WW 39 t m Analysis We take the entire heat exchanger as the system which is a control volume The mass and energy balances for this steady ow system can be expressed in the rate form as Mass balance for each uid stream AIR lt90 steady i i min mout AInsystem 0 1391 min mom Steam mlzmzzma and m3m4ms 3 7 Energy balance for the entire heat exchanger lt Q9 39 39 39 70 steady i 4 Ein Eout AEsystem 0 1 2 ar J f J Rate 0f net energy thfer Rate of changein intemalkinetic byheat workiandmass potentialetc energies 2 2 Ein Eout llhl 1123113 I llth me since Q W Ake Ape 0 Combining the two nean h1 ms h3 h4 Solving for ma Substituting m 12379 18021Btulbm 0240 Btulbm F130 80 F 15 lbmmin 21322 lbmmin 22041bms Also RTl 03704 psia1t 3lbmR540 R P1 147 psia 21361 31bin 1 Then the volume ow rate of air at the inlet becomes 1 maul 22041bms1361 ft 31bin 300 ft3s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 558 581 Refrigerant134a is to be cooled by air in the condenser For a speci ed volume ow rate of air the mass ow rate of the refrigerant is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 5 Air is an ideal gas with constant speci c heats at room temperature Properties The gas constant of air is 0287 kPam3kgK Table A1 The constant pressure speci c heat of air is cp 1005 kJkg C Table A 2 The enthalpies of the R134a at the inlet and the eXit states are Tables A11 through A13 P3 ZIMPa h 324 66 kJk T3 90 C 3 39 g AIR P421MPa h h 9358kJk i l1 7 T4 230 C 4 2 f 307 39 g R134a Analysis The inlet speci c volume and the mass ow rate of air are 1 RT1 0287 kPam3kgKX300 K O 861m3kg iegt 4 P1 100 kPa l2 V1 and 600 m3min m 3 6969 kgmin V1 0861 m kg We take the entire heat exchanger as the system which is a control volume The mass and energy balances for this steady ow system can be eXpressed in the rate form as Mass balance for each uid stream 70steady Inin mout A nsystrem O Inin mout m1 quot 2 ma m4 mR Energy balance for the entire heat exchanger 39 70 steady Ein Eout AEsystem O r J r J Rate 0f net energy thfer Rate of changein intemalkinetic byheat W0rkandmaSS potentialetc energies Ein Eout mlhl 13113 mth m4h4 since Q W Ake Ape 0 Combining the two 161012 h1 mRh3 h4 Solving for 141 mR ma E ma n m n n Substituting 1 2 mR 005 kJkg CW 7 C 6969kgmin 1000 kgmin 32466 9358 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 559 582E Refrigerant134a is vaporized by air in the evaporator of an airconditioner For specified ow rates the exit temperature of the air and the rate of heat transfer from the air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the device to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 5 Air is an ideal gas with constant speci c heats at room temperature Properties The gas constant of air is 03704 psiaft3lbmR Table A1E The constant pressure specific heat of air is cp 0240 Btulbm F Table AZE The enthalpies of the R134a at the inlet and the exit states are Tables A11E through A 13E P3 20 psia AIR l1 i P4 20 psia R134a h4 hg20psia satvapor 3 Analysis a The inlet speci c volume and the mass ow rate of air are h3 hf x3hfg 11436 03x 91302 3883 Btulbm 03704 psia 1t 3lbm R550 R P1 147 psia 21386ft31bm 2 V1 and V 2 3 39 m 1 L 1443lbmmin v1 1386ftlbm We take the entire heat exchanger as the system which is a control volume The mass and energy balances for this steady ow system can be expressed in the rate form as Mass balance for each uid stream Am 70 steady lm mom system Ornin out mIZrnQZma andr a n4 nR Energy balance for the entire heat exchanger 39 70 steady Ein Eout AEsystem O r J r J Rate 0f net energy thfer Rate of changein intemalkinetic byheat W0rkand mass potentialetc energies E E out mlhl 1123113 mth m4h4 since Q W Ake Ape 0 1n Combining the two 77112013 h4 77161012 h1 macp T2 T1 mR h3 h4 macp Solving for T2 T2 2 T1 4 lbmmin38 83 10274Btulbm 162 F 1443 Btumin024 Btulbm F Substituting T2 2 90 F b The rate of heat transfer from the air to the refrigerant is determined from the steady ow energy balance applied to the air only It yields Qair0ut ma 012 hl quot1an T2 T1 QWW 1443 lbmmin024 Btulbm F162 90 F 2556 Btumin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 560 583 Two streams of cold and warm air are mixed in a chamber If the ratio of hot to cold air is 16 the mixture temperature and the rate of heat gain of the room are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 The deVice is adiabatic and thus heat transfer is negligible Properties The gas constant of air is R 0287 kPam3kgK The enthalpies of air are obtained from air table Table A17 as Cold 11 1 280K 12 h 307 K 24 C 1 hroom h 297 K 297 8 kJkg Warm Analysis a We take the mixing chamber as the air 7 system which is a control volume since mass 34 C crosses the boundary The mass and energy balances for this steady ow system can be expressed in the rate form as Mass balance 7I0 t ad min mout Amsystem S e y 0 min 2 out m1 16m1 2 m3 26m1 srnce me 16m1 Energy balance 39 39 39 70 steady Ein Eout AEsystem O ar J f J Rate 0f net energy thfer Rate of changein intemalkinetic byheata workaandmass potentialetc energies Ein Eout n39zlhl 172th 1723113 since E W E Ake E Ape E 0 Combining the two gives mlh 16141111 261411113 or 723 h1 16h226 Substituting h3 28013 16 x 3072326 29681 kJkg From air table at this enthalpy the mixture temperature is T3 T h29681k1kg 2966 K 236 C b The mass ow rates are determined as follows 0287 kPa m3kg K7 273 K P 105 kPa VI 055 m3s m12 2 3O7186kgs V1 07654m kg m3 261121 2 2607186 kgs 1868 kgs 07654 m3 kg V1 The rate of heat gain of the room is determined from anin 2 m3 h100m ha 2 1868 kgs29718 29681 kJkg 0691kW Therefore the room gains heat at a rate of 0691 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 561 584 A heat exchanger that is not insulated is used to produce steam from the heat given up by the exhaust gases of an internal combustion engine The temperature of exhaust gases at the heat exchanger exit and the rate of heat transfer to the water are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Exhaust gases are assumed to have air properties with constant speci c heats Properties The constant pressure specific heat of the exhaust gases is taken to be cp 1045 kJkg C Table A2 The inlet and exit enthalpies of water are Tables A4 and A5 Twin 15 C hW in 6298 kJkg x O sat llq Exh gas f Q PW out 2 MPa 400 C hW out 27983 kJkg gt x I sat vap Heat exchanger Analysrs We take the ent1re heat exchanger as the system wh1ch is a control volume The mass and energy balances for this 2 MP3 Water steady ow system can be expressed 1n the rate form as O sat vap 15 C Mass balance for each uid stream 7I0 steady Inin mout AInsystem O Inin mout Energy balance for the entire heat exchanger 39 39 39 70 steady Ein Eout AEsystem O r J f J Rate 0f net energy MSfer Rate of chan gein intern a1 kinetic byheata workaandmass potentialetc energies Ein Eout mexhhexhjn mwhwin mexhhexhput mwhwout Qout Slnce W Aka E Ape E O or mexthTexhjn mwhwjn mexthTexhput mwhwput Qout Noting that the mass ow rate of exhaust gases is 15 times that of the water substituting gives 1511quotW 1045 kJkg C400 C mw 6298 kJkg 21511quotW 1 045 kJkg CTemout mw 27983 kJkg Q0 1 The heat given up by the exhaust gases and heat picked up by the water are QM meth Text Tam 15mWlt1045 kJkg cgtlt400 Texhputrc 2 Qw mw hwout thn In 27983 6298kJkg 3 The heat loss is Qout fheatlossQexh O39lQexh 4 The solution may be obtained by a trialerror approach Or solving the above equations simultaneously using EES software we obtain Texhput 2061 C QW 9726kW mw 003556 kgs mexh 05333 kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 562 585 Refrigerant22 is evaporated in an evaporator by air The rate of heat transfer from the air and the temperature change of air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions between the evaporator and the surroundings Analysis We take the condenser as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 70 steady Ein Eout AEsystem 0 A 4 f J RateOfnet energy MSfer Rate of changein intemalkinetic mw byheataworkaandmss potentialetc energies gt 3 Em Eout Evaporator 1 mRhl 2171ha l i1ah4 l mR hz hr 2 ma 0 3 h4 manATa 2 Refrigerant mR If we take the refrigerant as the system the energy balance can be written as 39 70 steady Ein Eout AEsystem 0 EK J r J Rate 0f net energy MSfer Rate of changein intemalkinetic byheat W01kandmaSS potentialetc energies Ein Eout mRhl Qin quot1ha Q m hz hl a The mass ow rate of the refrigerant is 2653600m3s v1 00253m3kg mR 002910 kgs The rate of heat absorbed from the air is Qin 11gtng h1 002910 kgs3980 2202kJkg 517kW b The temperature change of air can be determined from an energy balance on the evaporator QL MR 013 hZ quot1an Tal TaZ 517 kW 2 075 kgs1005 kJkg CATa AT 2 69 C The speci c heat of air is taken as 1005 kJkg C Table A2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 563 586 Steam is condensed by cooling water in the condenser of a power plant The rate of condensation of steam is to be determined Assumptions 1 Steady operating conditions eXist 2 The heat exchanger is wellinsulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 3 Changes in the kinetic and potential energies of uid streams are negligible 4 Fluid properties are constant Properties The heat of vaporization of water at 50 C is hfg 23820 kJkg and speci c heat of cold water is cp 418 kJkg C Tables A3 and A4 Analysis We take the cold water tubes as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as Steam 50 C 27 C 39 39 39 70 steady Ein Eout AEsystem O 6 gt E J t J u RateOfnetenergy thfer Rate of changein intemalkinetic K J by heat workaandmass potentialetc energies Q 3 r J Ein Eout Q J Qin mh1 th s1nce Ake E Ape E O G r 18 C Qin mcpT2 L a lt Then the heat transfer rate to the cooling water in the condenser becomes Lo water Q mcp Tout Tin coolingwater 50 C 101 kgs4 18 kJkg C27 C 18 C 3800 kJs The rate of condensation of steam is determined to be Q 3800mm 39 mh gtm Q fgsteam Steam hfg 23820kJkg 160 kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 564 587 x i 39 Problem 586 is reconsidered The effect of the inlet temperature of cooling water on the rate of condensation of steam as the inlet temperature varies from 10 C to 20 C at constant exit temperature is to be investigated The rate of condensation of steam is to be plotted against the inlet temperature of the cooling water Analysis The problem is solved using EES and the solution is given below quotInput Dataquot Ts150 C Ts250 C mdotwater101 kgs Twater118 C Twater227 C CPwater 420 kJkg C quotConservation of mass for the steam mdotsinmdotsoutmdotsquot quotConservation of mass for the water mdotwaterinmdotwateroutmdotwaterquot quotConservation of Energy for steadyflow neglect changes in KE and PEquot quotWe assume no heat transfer and no work occur across the control surfacequot Edotin Edotout DELTAEdotcv DELTAEdotcvO quotSteadyflow requirementquot Edotinmdotshs1 mdotwaterhwater1 Edotoutmdotshs2 mdotwaterhwater2 quotProperty data are given byquot hs1 enthapysteamiapwsTTs1x1 quotsteam dataquot hs2 enthapysteamiapwsTTs2xO hwater1 CPwaterTwater1 quotwater dataquot hwater2 CPwaterTwater2 hfgshs1hs2 quothfg is found from the EES functions rather than using hfg 2305 kJkgquot ms Twater1 39 I 39 I 39 I I kgS C 3028 10 2671 12 2315 14 1959 16 1603 18 1247 20 1 I I I I I I I 1O 12 14 16 18 20 Twater1 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 565 588 Two mass streams of the same idela gas are mixed in a mixing chamber Heat is transferred to the chamber Three expressions as functions of other parameters are to be obtained Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis a We take the mixing deVice as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem O ar J r J Rate 0f net energy thfer Rate of changein intemalkinetic Q byheat W0rkand mass potentialetc energies Em Emu Cold gas 39 mrhr mzhz Qin quot13113 gt From a mass balance gt Q m m m2 Hot gas Since h cpT Then m m Q T3 1T1 2T2 4 m3 m3 m3cp b Expression for volume ow rate RT 2 7quot2V2 7 l3 3 3 n39zR m m 39 32 1T12T2Qi P3 m3 m3 m3cp eaeP n391RT n391RT R39 V3 1 1 2 2 an P1 P2 P30 R39 Pc P c If the process is adiabatic then PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 566 Pipe and duct Flow 589E Saturated liquid water is heated in a steam boiler The heat transfer per unit mass is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis We take the pipe in which the water is heated as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 70 steady Ein Eout AEsystem 0 ar J RateOfnet energy MSfer Rate of changein intemalkinetic byheat W0rkandmaSS potentialetc energies qm E E 1n out 500 pSIa Water 500 psra mhr Qin mhz sat liq 600 F Qin Z quot1012 hr 1m 2 hz hr The enthalpies of water at the inlet and eXit of the boiler are Table ASE A6E P1 500 psia x Z 0 hr 5 hf 500psia 44951 Btulbm P2 500 psia h2 12986 Btulbm T2 60001 Substituting qin 12986 44951 8491Btulbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 567 590 Air at a speci ed rate is heated by an electrical heater The current is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The heat losses from the air is negligible Properties The gas constant of air is 0287 kPam3kgK Table Al The constant pressure specific heat of air at room temperature is cp 1005 kJkg C Table AZa Analysis We take the pipe in which the air is heated as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 70 steady Ein Eout AEsystem 0 RateOf net energy thfer Rateofchangein intemalkinetic W byheata workaandmass potentialetc energies 6 1 E E g m t 100 kPa 15 C Air 100 kPa mhl Wen mhz 03 m3s gt 30 C Wein hl The inlet speci c volume and the mass ow rate of air are RT 0287 kP 31lt K 288 K 1 a m g X 208266m3kg 1 1 100 kPa v 03 3 m 1 38 03629 kgs v1 08266m kg Substituting into the energy balance equation and solving for the current gives 1126 T T I 2 p 2 1 2 03629 kgs1005 kJkg K30 15K 1000VI 497 Amperes V 110V lkJs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 568 591E The cooling fan of a computer draws air which is heated in the computer by absorbing the heat of PC circuits The electrical power dissipated by the circuits is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 All the heat dissipated by the circuits are picked up by the air drawn by the fan Properties The gas constant of air is 03704 psiaft3lbmR Table A lE The constant pressure speci c heat of air at room temperature is 0 0240 Btulbm F Table A2Ea Analysis We take the pipe in which the air is heated as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 ar J J RateOfnet energy MSfer Rate of changein intemalkinetic W byheat W0rkandmaSS potentialetc energies 6111 Ein Eout w O mhl Wein mhz 147 psrag 70 F Arr 147 psra 03 ft s gt 83 F Wein quot1012 11 Wein mcp T2 T1 The inlet speci c volume and the mass ow rate of air are 03704psia it 3lbm R530 R P1 147 psia V 03 it 3 m 1 38 002246 lbms v1 1335 lbm v1 1335 ft 3lbm Substituting We out 002246 lbms0240 Btulbm R83 70Btulb lk W 00740kW 094782 Btus PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5 69 592E Electronic devices mounted on a cold plate are cooled by water The amount of heat generated by the electronic devices is to be determined Assumptions 1 Steady operating conditions eXist 2 About 15 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation 3 Kinetic and potential energy changes are negligible Properties The properties of water at room temperature are p 621 lbmft3 and CI 100 Btulbm F Table A3E Analysis We take the tubes of the cold plate to be the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 70 steady Em Eout AEsystem 0 Cold plate Water inlet J Rate of net energy thfer Rate of changein intemalkinetic by heat W0 rkand mas S p otentialetc energies Ein Eout Qin n39th1 n39th2 since Ake E Ape E 0 Q mcpltT2 T1 Then mass ow rate of water and the rate of heat removal by the water are determined to be 7202 7r02512 102 V 621lbmit 3 40 itmin 0 8483 lbmmin 509 lbmh 1 51 2 pAV 0 Q n39icp Tout Tin 509 lbmh100 Btulbm F105 70 F 1781 Btuh which is 85 percent of the heat generated by the electronic devices Then the total amount of heat generated by the electronic devices becomes 1781Btuh 085 2096 Btuh 614 W PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 570 593 A sealed electronic box is to be cooled by tap water owing through channels on two of its sides The mass ow rate of water and the amount of water used per year are to be determined Assumptions 1 Steady operating conditions exist 2 Entire heat generated is dissipated by water 3 Water is an incompressible substance with constant specific heats at room temperature 4 Kinetic and potential energy changes are negligible Properties The speci c heat of water at room temperature is l W at 61 1 cp 418 kJkg C Table A3 inlet Analysis We take the water channels on the sides to be the system 1 which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 stead 39 Ein Eout AEsystem y 0 Electronic box E J J 2 kW RateOf net energy thfer Rate of changein intemalkinetic byheataworkaandmass potentialetc energies Ein Eout Qin n391h1 11quotch since Ake E Ape E O Water Qin 2 quot10 p T2 Tl exit Then the mass ow rate of tap water owing through the electronic box becomes l 2 1 Q 2kJs cpAT 418kJkg C4 C szcpAT gtm 201196kgs Therefore 01196 kg of water is needed per second to cool this electronic box Then the amount of cooling water used per year becomes m m 01196 kgs365 daysyr x 24 hday x 3600 sh 3772000kgyr 3772 tonsyr PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 571 594 A sealed electronic box is to be cooled by tap water owing through channels on two of its sides The mass ow rate of water and the amount of water used per year are to be determined Assumptions 1 Steady operating conditions exist 2 Entire heat generated is dissipated by water 3 Water is an incompressible substance with constant specific heats at room temperature 4 Kinetic and potential energy changes are negligible Properties The specific heat of water at room temperature is cp 418 kJkg C Table A3 Analysis We take the water channels on the sides to be the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 stead Bin 150ut AE y 0 system J r J Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0 rkand mas S p otentialetc energies Ein Eout Qin n391h1 11quotch since Ake E Ape E O Q T2 T1 Then the mass ow rate of tap water owing through the electronic box becomes Q 4kJs l Water inlet 1 Electronic box 4 kW Q mcpAT gtm cpAT 418 kJkg C4 C 02392kgs Water exit Therefore 02392 kg of water is needed per second to cool this electronic box Then the amount of cooling water used per year becomes m mm 023923 kgs365 daysyrx 24 hday x 3600 sh 7544400kgyr 7544 tonsyr PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 572 595 39 The components of an electronic device located in a horizontal duct of rectangular cross section are cooled by forced air The heat transfer from the outer surfaces of the duct is to be determined Assumptions 1 Steady operating conditions eXist 2 Air is an ideal gas with constant speci c heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The gas constant of air is R 0287 kJkg C Table A1 The specific heat of air at room temperature is 0 1005 kJkg C Table A2 Analysis The density of air entering the duct and the mass ow rate are P 101325 kPa p 3 RT 0287 kPam kgK30 273K m pi 1165 kgm3 O6 m3 min 0700 kgmin 40 C 1165 kgm3 We take the channel excluding the electronic components to be the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AE 0 system J r J Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout Qin n39th1 n39m2 since Ake E Ape E 0 Q T2 T1 Then the rate of heat transfer to the air passing through the duct becomes Q air n39icpT0ut Tin air 070060 kgs1005 kJkg C40 30 C 0117 kW 117 W The rest of the 180 W heat generated must be dissipated through the outer surfaces of the duct by natural convection and radiation Qexternal Qtotal Qinternal 2180 117 2 63 W PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 573 596 The components of an electronic device located in a horizontal duct of circular cross section is cooled by forced air The heat transfer from the outer surfaces of the duct is to be determined Assumptions 1 Steady operating conditions eXist 2 Air is an ideal gas with constant speci c heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The gas constant of air is R 0287 kJkg C Table A1 The specific heat of air at room temperature is 0 1005 kJkg C Table A2 Analysis The density of air entering the duct and the mass ow rate are P 101325 kPa p 3 RT 0287 kPam kgK30 273K m p1 1165 kgm3 O6 m3 min 0700 kgmin 1165 kgm3 We take the channel excluding the electronic components to be the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem 0 R 7quot f 1 2 3160 net energy 313 er Rate of chan gem 1ntemalk1net1c byheatW0rkandmaSS potentialetc energies Air gt Ein Eout o h Ak A 0 30 C m I m SlIlCC C E C E Qm 1 7 2 p 1113 S Qin mcpT2 Then the rate of heat transfer to the air passing through the duct becomes Qair n39icpT0ut Tin air 070060 kgs1005 kJkg C40 30 C 0117 kW 117 W The rest of the 180 W heat generated must be dissipated through the outer surfaces of the duct by natural convection and radiation Qexternal Qtotal Qinternal 2180 117 2 63 W PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 574 597 Air enters a hollowcore printed circuit board The eXit temperature of the air is to be determined Assumptions 1 Steady operating conditions eXist 2 Air is an ideal gas with constant speci c heats at room temperature 3 The local atmospheric pressure is 1 atm 4 Kinetic and potential energy changes are negligible Properties The gas constant of air is R 0287 kJkg C Table Al The speci c heat of air at room temperature is cp 1005 kJkg C Table A2 Analysis The density of air entering the duct and the mass ow rate are i 101325 kPa RT 0287 kPam3kgK25 273K m pi 1185 kgm3 00008 m3 Is 2 00009477 kgs p 21185 kgm3 We take the hollow core to be the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout Qin n39th1 n39th2 since Ake E Ape E 0 Qin mcpT2 Then the eXit temperature of air leaving the hollow core becomes Q39in 225 15Js mop 00009477 kgs1005 Jkg C 460 C Q39in mCPT2 T1 gtT2 T1 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5 75 598 A computer is cooled by a fan blowing air through the case of the computer The required ow rate of the air and the fraction of the temperature rise of air that is due to heat generated by the fan are to be determined Assumptions 1 Steady ow conditions eXist 2 Air is an ideal gas with constant speci c heats 3 The pressure of air is 1 atm 4 Kinetic and potential energy changes are negligible Properties The speci c heat of air at room temperature is 0 1005 kJkg C Table A2 Analysis a We take the air space in the computer as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O RateOfnetenergy thfer Rate of changein intemalkinetic by heat workaandmass potentialetc energies Ein Eout Qin Win n39th1 n39m2 since Ake E Ape E O Qin VVin mcpT2Tl a Not1ng that the fan power 1s 25 W and the 8 PCBs transfer a total of 80 WEE Q W of heat to air the mass ow rate of air is determined to be Q39m Win 8XI0W25W 00104kgs CPU 1 1005 Jkg C10 C Qin Win mcpa Tl gtn391 b The fraction of temperature rise of air that is due to the heat generated by the fan and its motor can be determined from Q39zmcpATaAT Q 25W 24 C mcp 00104 kgs1005 Jkg C f 2394 C 2024 24 10 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 576 599 A room is to be heated by an electric resistance heater placed in a duct in the room The power rating of the electric heater and the temperature rise of air as it passes through the heater are to be determined Assumptions 1 Steady operating conditions eXist 2 Air is an ideal gas with constant speci c heats at room temperature 3 Kinetic and potential energy changes are negligible 4 The heating duct is adiabatic and thus heat transfer through it is negligible 5 No air leaks in and out of the room Properties The gas constant of air is 0287 kPam3kgK Table A1 The speci c heats of air at room temperature are cp 1005 and cu 0718 kJkgK Table A2 Analysis a The total mass of air in the room is 3 3 150kJmin V4gtlt5gtlt6m 120m m PIV 98 kPa120 m3 142 3 kg 4x5gtlt6 m3 RTI 0287 kPa m3kg K288 K 39 lt We rst take the entire room as our system which is a closed system We since no mass leaks in or out The power rating of the electric heater is determined by applying the conservation of energy relation to this W W constant volume closed system Ein Eout AE system Net energy thfer Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies Wain Wmn Q0 2 AU since AKE APE 0 AtVVeirl Wfanin Qout mcuavgT2 Solving for the electrical work input gives Wein Qout Wfanin mcu T2 T1 At 15060 kJs 02 kJs 1423 kg0718 kJkg C25 15 C20 x 60 s 3151 kW 9 We now take the heating duct as the system which is a control volume since mass crosses the boundary There is only one inlet and one eXit and thus m1 2 1512 m The energy balance for this adiabatic steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r J Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies Ein Eout Wain Wfan n n391h1 nah since Ake E Ape E 0 Wein Wfanin hl mcp T2 Thus WmW in 11 2 ATZTFTIZ 6 fan 3 5 0 kJs 5ooc mcp 4060 kgs1005 kJkg K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 577 5100 A long roll of large lMn manganese steel plate is to be quenched in an oil bath at a speci ed rate The rate at which heat needs to be removed from the oil to keep its temperature constant is to be determined Assumptions 1 Steady operating conditions eXist 2 The thermal properties of the roll are constant 3 Kinetic and potential energy changes are negligible Properties The properties of the steel plate are given to be p 7854 kgm3 and CI 0434 kJkg C Analysis The mass ow rate of the sheet metal through the oil bath is m pi 0th 7854 kgm3 2 m0005 m10 mmin 7854 kgmin We take the volume occupied by the sheet metal in the oil bath to be the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem O ar J r J RateOf net energy thfer Rate of changein intemalkinetic gt Oil bath byheat W01kandmaSS potentialetc energies i 45 C Ein Z Eout Steel plD r n39th1 Qlout n391h2 since Ake E Ape E O 10 IIImin Q mop T1 T2 Then the rate of heat transfer from the sheet metal to the oil bath becomes Qlout n39icpTir1 Toutmeta1 7854 kgmin0434 kJkg C820 511 C 262090 kJmin 4368 kW This is the rate of heat transfer from the metal sheet to the oil which is equal to the rate of heat removal from the oil since the oil temperature is maintained constant PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 578 5101 39 Problem 5100 is reconsidered The effect of the moving velocity of the steel plate on the rate of heat transfer from the oil bath as the velocity varies from 5 to 50 mmin is to be investigated Tate of heat transfer is to be plotted against the plate velocity Analysis The problem is solved using EES and the solution is given below quotKnownsquot Vel 10 mmin Tbath 45 C T1 820 C T2 511 C rho 785 kgmA3 CP 0434 kJkgC width 2 m thick 05 cm quotAnalysis The mass flow rate of the sheet metal through the oil bath isquot Vodot widththickconvertcmmVelconvertmins mdot rhoVodot quotWe take the volume occupied by the sheet metal in the oil bath to be the system which is a control volume The energy balance for this steadyflow systemthe metal can be expressed in the rate form asquot Edotmetain Edotmetaout Edotmetainmdoth1 Edotmetaoutmdoth2Qdotmetaout Qdotoiout Qdotmetaout Qoilout VGI 250039I39I39I39l39l39l39l39l kW mmin 2183 5 4366 10 200 6549 15 E 8732 20 x 1500 1091 25 3939 1310 30 5 1528 35 1000 1746 40 o 1965 45 C 2183 50 50 0 1 1 l 1 1 l 1 1 l 1 1 l 51015 20 25 30 35 40 45 50 Vel mmin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 579 5102E Water is heated in a parabolic solar collector The required length of parabolic collector is to be determined Assumptions 1 Steady operating conditions eXist 2 Heat loss from the tube is negligible so that the entire solar energy incident on the tube is transferred to the water 3 Kinetic and potential energy changes are negligible Properties The speci c heat of water at room temperature is 0 100 Btulbm F Table A3E Analysis We take the thin aluminum tube to be the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 70 steady Ein Eout AEsystem 0 J J 1 2 Rate 0f net energy thfer Rate of changein intemalkinetic byheatW0rkandmaSS potentialetc energies Water gt Ein Eout 0 Q mh n391h2 sinceAkeA e0 55 F 1801 m 1 P 4 lbms Qi mwatercp T2 Then the total rate of heat transfer to the water owing through the tube becomes Q total mop Te Ti 4 lbms100 Btulbmf PXl 80 55 F 500 Btus 1800000Btuh The length of the tube required is QM 1800000 Btuh Q 400 Btuhit L 450011 5103 A house is heated by an electric resistance heater placed in a duct The power rating of the electric heater is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible Properties The constant pressure specific heat of air at room temperature is 0 1005 kJkgK Table A 2 Analysis We take the heating duct as the system which is a control volume since mass crosses the boundary There is only one inlet and one eXit and thus m1 2 1512 m The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 R t f t tl al f a 90 He energy S er Rate of changein intemalkinetic byheat W0rkandmaSS potentialetc energies 300 W E Ein Eout I I N W gin Wfanin mhl Qout mhz s1nce Ake Ape O T e Wein Wfanin Qout hl Qout mcp T2 300 W Substituting the power rating of the heating element is determined to be Wein gout rncpAT mm 03 kJs 06 kgs1005 kJkg C7 C 03 kW 422 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 580 5104 Steam pipes pass through an unheated area and the temperature of steam drops as a result of heat losses The mass ow rate of steam and the rate of heat loss from are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 4 There are no work interactions involved Properties From the steam tables Table A6 p1 2 MPa u 012551 m3kg 2 MPa STEAM 18 Mpa T1 300 C h 30242 kJkg 300 C gt 250 C P2 1800 kPa h2 29117 kJkg T2 250 C Q Analysis a The mass ow rate of steam is determined directly from 1 1 2 AV 008 25m 04005k ls m u 11 012551m3kg7 In S g b We take the steam pipe as the system which is a control volume since mass crosses the boundary There is only one inlet and one eXit and thus m1 n3912 m The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem 0 r J r Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout 111111 2 Qlout n39m2 since W E Ake E Ape E 0 Q mall hz Substituting the rate of heat loss is determined to be Q1088 04005 kgs30242 29117 kJkg 451 kJS PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 58 1 5105 R 134a is condensed in a condenser The heat transfer per unit mass is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis We take the pipe in which R134a is condensed as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 ar J f J Rate 0f net energy MSfer Rate of changein intemalkinetic qout by heat workaandmass potentialetc energies Em Eout 900 kPa R134a I 900 kPa 60 C gt sat liq Z Qout Q quotth hz qout 2 hr hz The enthalpies of R134a at the inlet and eXit of the condenser are Table A12 A13 P1 900 kPa h1 29515 kJkg T1 60 C P2 900 kPa x20 hz hf900kpa 10162kJkg Substituting qout 29515 10162 1935kJkg 5106 Water is heated at constant pressure so that it changes a state from saturated liquid to saturated vapor The heat transfer per unit mass is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions Analysis We take the pipe in which water is heated as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 70 steady Ein Eout AEsystem 0 q ar J r J in Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies V Em Eout 500 kPa Water 500 kPa Sat Liq sat vap mhr Qin mhz Qin Z quot1012 hr 4m 2 h2 h1 hfg where hfg SOOkPa Table Thus qin 2108kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 582 5107 Water is heated by a 7kW resistance heater as it ows through an insulated tube The mass ow rate of water is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Water is an incompressible substance with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible 4 The tube is adiabatic and thus heat losses are negligible Properties The speci c heat of water at room temperature is c 418 kJkg C Table A3 Analysis We take the water pipe as the system which is a control volume since mass crosses the boundary There is only one inlet and one eXit and thus m1 1512 m The energy balance for this steady ow system can be eXpressed in the rate form as E lt90 steady 0 in out system f r J Rate 0f net energy thfer Rate of changein intemalkinetic WATER byheata workaandmass potentialetc energies 20 C a 75 C Ein Eout nth1 nth2 since Qlout E Ake E Ape E O W n391h2 h1 n391cT2 T1 uAPt O mcT2 T1 ein W ein 4x 7kW V Substituting the mass ow rates of water is determined to be m mm 7 kJs cT2 T1 4184 kJkg C75 20 C 00304 kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 583 5108 Electrical work is supplied to the air as it ows in a hair dryer The mass ow rate of air and the volume ow rate at the eXit are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Heat loss from the dryer is negligible Properties The gas constant of argon is 0287 kPam3kgK The constant pressure specific heat of air at room temperature is cp 1005 kJkg C Table A2a Analysis a We take the pipe as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 7I0 d Ein Eout AEsystem Stea y 2 0 Rateof netenergy transfer R t f h mint mal kin ti 6 byheatworkandmass aegoie gfewegergi es e C T2 80 C P1 100 kPa V221ms VWW T1300K Ein Eout 9 V2 V2 11391th LJWin 2 t h2 2 4 V 2 2 K 1500 W V22 V12 We W11l m 2 K K 2 2 V V K Substituting and solving for the mass ow rate m W1 V22 V12 cp T2 T1 T 150 kW 21 m 2 0 lkJk 1005 kJkg K353 300K S 2g 2 j 2 1000m s 00280kgs b The eXit speci c volume and the volume ow rate are RT2 0287 kPa m3kg K353 K P2 100 kPa v2 n39w2 002793 kgs1013 m3kg 00284m3s 2 1013 m3kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5109 rate is to be investiagted Analysis The problem is solved using EES and the solution is given below quotGivenquot T1 300 K P100kPa Ve1 0 ms Wdotein15 kW T280273 K Ve221 ms quotPropertiesquot cp1005 kJkgK R0287 kJkgK quotAnalysisquot 584 A 539 Problem 5108 is reconsidered The effect of the eXit velocity on the mass ow rate and the eXit volume ow WdoteinmdotcpT2T1mdotVe2quot2Vel1A2Convertmquot2squot2kJkg quotenergy balance on hair dryerquot v2RT2P Vodot2mdotv2 v61 m V012 281 5 ms kgs m3s o 0281 EK 5 002815 002852 39 75 002813 00285 oo2805 10 002811 002848 125 002808 002845 0028 a 15 002804 002841 5 175 0028 002837 5 0mm 20 002795 002832 225 00279 002826 00279 25 002783 00282 002785 00278 5 9 13 17 21 Vel2 ms 002855 00285 002845 u E 00284 no 39 E D 002835 N 6 u gt 00283 4 002825 00282 002815 39 39 39 39 5 13 Vel2 ms 17 21 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 585 5110E The ducts of an airconditioning system pass through an unconditioned area The inlet velocity and the eXit temperature of air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible 4 There are no work interactions involved Properties The gas constant of air is 03704 psiaft3lbmR Table AlE The constant pressure specific heat of air at room temperature is cp 0240 BtulbmR Table A2E Analysis a The inlet velocity of air through the duct is V V 450 it 3 39 V1 1 12 m1 825 ftmin 3 quot A1 m 7z512 it 450 ft min AIR D 10 in R 9 v Then the mass ow rate of air becomes 2 Btus 3 V1 2 2 03704 pSIa it lbm RX510 R 2 126 3lbm P1 15 psia 3 m w 357 lbmmin 05951bms 1 126 it lbm b We take the airconditioning duct as the system which is a control volume since mass crosses the boundary There is only one inlet and one eXit and thus m1 1512 m The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O J r J Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Eout Qin n39th1 n39th2 since W E Ake E Ape E 0 Q 1072 h0 mcpltT2 T1gt E in Then the eXit temperature of air becomes Qin 500F 2 6400F T2 2 T1 o mcp 0595 lbmS024 Btulbm F PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 586 Charging and Discharging Processes 5111 Steam in a supply line is allowed to enter an initially evacuated tank The temperature of the steam in the supply line and the ow work are to be determined Analysis Flow work of the steam in the supply line is converted to sensible internal energy in the tank That is hline tank where Steam gt 4 MPa utank 31895 kJkg Table A6 Ttank 550 C Now the properties of steam in the line can be calculated Inrtrally P 4 MPa T 3895 c evacuated lrne lrne Table A The ow work per unit mass is the difference between enthalpy and internal energy of the steam in the line w ow hum u ne 31895 29015 288 kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 587 5112 Steam owing in a supply line is allowed to enter into an insulated tank until a speci ed state is achieved in the tank The mass of the steam that has entered and the pressure of the steam in the supply line are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid entering the tank remains constant 2 Kinetic and potential energies are negligible Properties The initial and nal properties of steam in the tank are Tables A5 and A6 P1 1MPa v1 019436 m3kg x1 1sat vap I 225828 kJkg Steam gt 400 C p22Mpa012551m3kg I T1 300 C u2 27732 kJkg Sat vapor Analysis We take the tank as the system which is a control volume since mass crosses 2 m3 the boundary Noting that the microscopic energies of owing and non owing uids 1 MP3 are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be eXpressed as Mass balance min mout Amsystem ml quoth m1 Energy balance Ein Eout AESystem V V Net energy Sfer Chan gem 1ntemalk1net1c by heat W0rkand mass potentialetc energies mihl mza2 mla1 since Q E ke E pe E 0 The initial and nal masses and the mass that has entered are u 2 3 v1 019436m kg V 2 m3 V2 012551m kg m m2 m1 1594 1029 5645kg 1594 kg Substituting 5645 kghi 15 94 kg27732 kJkg 1029 kg2582 8 kJkg gthl 31203 kJkg The pressure in the supply line is h 31203 kJkg P 8931kPa determined from EES T 400 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 588 5113 Helium ows from a supply line to an initially evacuated tank The ow work of the helium in the supply line and the nal temperature of the helium in the tank are to be determined Properties The properties of helium are R 20769 kJkgK cp 51926 kJkgK cu 31156 kJkgK Table A2a Analysis The ow work is determined from its definition but we first determine the speci c volume RT 2 K 12 2 K Helium gt200 kP 120 C V 1me 0769kJkg 0 73 40811m3kg a P 200 kPa 3 E w ow Pu 200 kPa40811 m3kg 8162kJkg Noting that the ow work in the supply line is converted to sensible internal Initially energy in the tank the nal helium temperature in the tank is determined as evacuated follows tank 2 hline hline cpTline 51926 kJkgK120 273 K 20407 kJkg wank curtank gt 20407 kJkg 31156 kJkgKTtank gtTank 6550K Alternative Solution Noting the definition of specific heat ratio the final temperature in the tank can also be determined from Ttank kTline 1667120 273 K 6551K which is practically the same result PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 589 5114 An evacuated bottle is surrounded by atmospheric air A valve is opened and air is allowed to ll the bottle The amount of heat transfer through the wall of the bottle when thermal and mechanical equilibrium is established is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 Air is an ideal gas with variable specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The direction of heat transfer is to the air in the bottle will be veri ed Properties The gas constant of air is 0287 kPam3kgK Table Al Analysis We take the bottle as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance ruin mout An lsystem mi n72 Slnce mout quotlinitial Energy balance Ein Eout AE system Net energy traleel Chan gein intemalkinetic by heat W0FkandmaSS potentialetc energies Evacuated Qin mihl mguz since W E Eout Einitial ke E pe E O Combining the two balances Qin quot12u2 where P v 1 kP 3 m2 2 00 a 035 m 004134 kg RTz 0287 kPam kg K295 K T Z T Z K TableA17 hi 2 2 u 21049 kJkg Substituting Qin 004134 kg21049 29517 kJkg 350 kJ 0139 Qout 350 kJ Discussion The negative sign for heat transfer indicates that the assumed direction is wrong Therefore we reverse the direction PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 590 5115 A rigid tank initially contains superheated steam A valve at the top of the tank is opened and vapor is allowed to escape at constant pressure until the temperature rises to 500 C The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process by using constant average properties for the steam leaving the tank 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be veri ed Properties The properties of water are Tables A4 through A6 P1 2 MPa 1 012551 m3kg K T1 300 C ul 2 27732 kJkg h1 30242 kJkg STEAM T Q P2 2 MPa 2 017568 m3kg 2 MPa T2 500 C u2 31169 kJkg h2 34683 kJkg Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be eXpressed as Mass balance ruin mout A nsystem me 2 m1 quot72 Energy balance Ein Eout AE sy stem Net energy traleel Chan gein intemalkinetic by heat W01kandmass potentialetc energies Qin mehe mza2 mla1 since W E ke E pe E 0 The state and thus the enthalpy of the steam leaving the tank is changing during this process But for simplicity we assume constant properties for the eXiting steam at the average values Thus 2112 3024234683kJkg 3246 kag he 2 2 III The initial and the nal masses in the tank are u 02 3 V1 012551m kg v 02 3 m2 2 2 m 1138kg v2 017568 m3kg Then from the mass and energy balance relations me 2 m1 m2 1594 1138 20456 kg Qin Z mehe m2 2 quotW1 2 0456 kg32462 kJkg 1 138 ng31169 kJkg 1594 kg27732 kJkg 6068 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 591 5116E A rigid tank initially contains saturated water vapor The tank is connected to a supply line and water vapor is allowed to enter the tank until onehalf of the tank is filled with liquid water The nal pressure in the tank the mass of steam that entered and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniforrn ow process since the state of uid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be veri ed Properties The properties of water are Tables A4E through A6E Steam 200 psia o 3 a 400 F 5 Ii 2 F 1 Vg3moF sat vapor ul 2 ug3000F 210998Btulbm g T2 300 F If 001745 lg 64663 ft3lbm sat mixture uf 26951 a 10998 Btulbm Wat g 3 ft P 200 sia p h 12109 Btulbm 300 F Q Tl 2 400 Sat vapor Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mm Amsystem gt mi 2 m2 m1 Energy balance Em Eout AESystem Net energy Sfer Chan gein intemalkinetic by heat W0rkand mass potentialetc energies Qin mihl mga2 mla1 since W E ke E pe E O a The tank contains saturated mixture at the final state at 250 F and thus the eXit pressure is the saturation pressure at this temperature P2 Psat 300B 2 6703 PSia b The initial and the final masses in the tank are u 31t3 m1 v1 646631t1bm V V 151t3 15rt3 f g 8597 0232 8620 lbm m m m 2 f 539 vf vg 0017451t3lbm 646631t3lbm Then from the mass balance ml m2 m1 8620 0464 28574 Ibm c The heat transfer during this process is determined from the energy balance to be Qin Z mihi m2 2 quotW1 2 85741bm12109 BtuIbm 23425 Btu O4641bm10998 BtuIbm 80900 Btu gt Qout 80900 Btu since U2 2 m2u2 mfuf mgug 8597 x 26951 0232x10998 23425 Btu Discussion A negative result for heat transfer indicates that the assumed direction is wrong and should be reversed PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 592 5117 A pressure cooker is initially half lled with liquid water If the pressure cooker is not to run out of liquid water for 1 h the highest rate of heat transfer allowed is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Properties The properties of water are Tables A4 through A6 P1 175 kPa gt V 0001057 m3kg ug 10037 m3kg uf 48682 kJkg ug 25245 kJkg 102 175 kPa v2 ug175kPa 10036 m3kg sat vapor a2 ug175kPa 25245 kJkg 13533111 P6 175 kPa 4 L he 2 hg175kPa 27002kJkg 175 kPa sat vapor Analysis We take the cooker as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout 2 Am system me 2 m1 quot72 Energy balance Ein Eout 2 AE system Net energy Sfer Chan gein intemalkinetic by heat W0rkand mass potentialetc energies Qin mehe mauz mlul since W E ke E pe E 0 The initial mass initial internal energy and final mass in the tank are Vf Vg 0002 m3 0002 m3 mlszmg 3 3 uf ug 0001057 m kg 10036 m kg U1 2 m1 mfuf mgug 189348682 000225245 9266 kJ v 0004 m3 0004kg m2 u2 10037 m3kg 1893 0002 1895 kg Then from the mass and energy balances me 2 m1 m2 21895 0004 1891 kg Qin Z mehe quot6 m1u1 1891kg27002 kJkg 0004 kg25245 kJkg 9266 kJ 4188 k Thus 4188kJ Q2 1 163kW At 3600 s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 593 5118 A cylinder initially contains saturated liquidvapor mixture of water The cylinder is connected to a supply line and the steam is allowed to enter the cylinder until all the liquid is vaporized The nal temperature in the cylinder and the mass of the steam that entered are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 The eXpansion process is quasiequilibrium 3 Kinetic and potential energies are negligible 3 There are no work interactions involved other than boundary work 4 The device is insulated and thus heat transfer is negligible Properties The properties of steam are Tables A4 through A6 7 P1 200 kPa h h h x W x1 06 1 f 1 fg P 200 kPa 5047106gtlt 2201618256kJkg m110kg P 200 kPa H O 2 h2 hg200kpa 27063 kJkg 2 P 05 MPa sat vapor T Ti 350 C h 31681 kJkg T 350 C Analysis a The cylinder contains saturated vapor at the nal state at a pressure of 200 kPa thus the final temperature in the cylinder must be T2 Tsat 200 kPa 12020C b We take the cylinder as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout 2 Am system mi 2 quot72 m1 Energy balance Ein Eout AEsystem f J W Net energy traleel Chan gein intemalkinetic by heat W0Fkand mass potentialetc energies mihl Wbout mza2 mla1 since Q E ke E pe E O Combining the two relations gives 0 Z VVbout ma m1hi quot72112 mil 1 or 0 2 ma m1hi maha mrhr since the boundary work and AU combine into AH for constant pressure eXpansion and compression processes Solving for m2 and substituting h h1 m 31681 18256kJkg h h 1 31681 27063kJkg l 10 kg 2907 kg Thus m m2 m1 2907 10 1907 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 594 5119E A scuba diver39s air tank is to be filled with air from a compressed air line The temperature and mass in the tank at the final state are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 Air is an ideal gas with constant specific heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The tank is well insulated and thus there is no heat transfer Properties The gas constant of air is 03704 psiaft31bmR Table A 1E The speci c heats of air at room temperature are cp 0240 BtulbmR and cu 0171 BtulbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be eXpressed as Mass balance min mout Am system mi 2 m2 m1 Alf 39 0 Energy balance gt 120 13813 85 F EK J W Net energy transfer Chan gein intemalkinetic byheat W0rkandmaSS potentialetc energies 20 psia mihi Z m2 2 m1u1 6001 mc Tch ch 2ft 1 p z 2 U 2 1 I 1 Combining the two balances m2 m1cpTi Z m2CuT2 mlcuTl The initial and nal masses are given by P v 39 3 m1 1 2 20 131am 02077 lbm RTl 03704 p31a it lbm R60 460 R sz 120 psia2 3 6479 RT2 03704 psia it 3lbm RT2 T2 quot 2 Substituting 6479 2 6479 11735 020770171520 02077024545 2 whose solution is The final mass is then 6479 6479 T2 709 2 0914bm m2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 595 5120 R 134a from a tank is discharged to an airconditioning line in an isothermal process The final quality of the R134a in the tank and the total heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the eXit remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be eXpressed as Mass balance min mout AInsystem AC line gt me 2 m2 m1 me m1m2 Energy balance Liquid R134a Ein E out 2 AE system 5 af J W Net energy thfer Chan gein intemalkinetic 24 C by heat workaand mass potentialetc energies Qin mehe mzuz m1 1 Qin mzuz m1 1 mehe Combining the two balances Qin quot12142 mil 1 m1 m2 he The initial state properties of R134a in the tank are T1 24 C 0 ul 2 8444 kJkg Table A11 x u 00008260 m3kg he 8498 kJkg Note that we assumed that the refrigerant leaving the tank is at saturated liquid state and found the eXiting enthalpy accordingly The volume of the tank is v mlvl 5 kg00008260 m3kg 0004130 m3 The final specific volume in the container is 3 92 l 2 101652 m3kg m2 025 kg The final state is now xed The properties at this state are Table A11 T2 24 C 12 001652 m3kg x2 05056 ufg 0031869 00008260 V2 Vf 001652 00008260 u2 uf xzufg 8444 kJkg 0505615868 kJkg 16467 kJkg Substituting into the energy balance equation Q1 2 mzuz mlul m1 m2 he 2 025 kg16467 kJkg 5 kg8444 kJkg 475 kg8498 kJkg 226kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 596 5121E Oxygen is supplied to a medical facility from 10 compressed oxygen tanks in an isothermal process The mass of oxygen used and the total heat transfer to the tanks are to be determined Assumptions 1 This is an unsteady process but it can be analyzed as a uniform ow process 2 Oxygen is an ideal gas with constant speci c heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved Properties The gas constant of oxygen is 03353 psiaft31bmR Table AlE The speci c heats of oxygen at room temperature are cp 0219 BtulbmR and cu 0157 BtulbmR Table A2Ea Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout Am system 39 me m2 m1 me 2 m1 m2 Energy balance Oxygen Ein Eout AEsystem 1500 psia Net energy thfer Changein intemalkinetic 80 F 15 ft3 by heat W0rkandmaSS potentialetc energies Qin mehe Z mzuz mrur Qin Z mzuz mrur mehe Qin meCpTe Combining the two balances Qin Z m2cuT2 mlcuTl m1 m2 cpTe The initial and nal masses and the mass used are m Jul V 15OOPSiaX15 3 124 3lbm 1 RTl 03353psia it 3lbm R80 460 R sz 300 psia15 3 Z 24 85mm m 2 RT2 03353psia it 3lbm R80 460 R me 2 m1 m2 1243 2485 9941Ibm Substituting into the energy balance equation Qin m2CUT2 mlcuTl meCpTe 2485O157540 1243O157540 9941O219540 3328 Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 597 5122 A rigid tank initially contains saturated R134a vapor The tank is connected to a supply line and R134a is allowed to enter the tank The mass of the R134a that entered and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of refrigerant are Tables A11 through A13 P1 08 MPa F ug08 MP 002565 m3kg R134a 12 MPa satvapor ul 2 u g 08 MPa 2 24682 kJkg gt 360C gt 102 12 MPa 2 uf12MPa 00008935 m3kg sat 2 Mf12NIPa Pl 12MPa R4343 h h o 10234kJk Tl 36 C f36 C g 006 m3 08 MPa Q Analysis We take the tank as the system which is a control volume since mass Sat vapor crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniforrn ow system can be expressed as Mass balance ruin mout A nsystem mi n72 m1 Energy balance Ein Eout AEsystem r J W Net energy traleel Chan gein intemalkinetic by heat W01kandmass potentialetc energies Qin mihl mga2 mla1 since W E ke E pe E O a The initial and the nal masses in the tank are u 006 3 v1 002565 m kg v 006 3 V2 00008935 m3kg Then from the mass balance ml m2 m1 6716 2340 6482 kg c The heat transfer during this process is determined from the energy balance to be Qin Z mihi m2u2 mrur 64 82 ng102 34 kJkg 6716 kg11672 kJkg 2 340 kg246 82 kJkg 627kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 598 5123 A rigid tank initially contains saturated liquid water A valve at the bottom of the tank is opened and half of the mass in liquid form is withdrawn from the tank The temperature in the tank is maintained constant The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be verified Properties The properties of water are Tables A4 through A6 o 3 T1 200 C u1 vf200C 0001157 m kg sat liquid u1 uf200C 85046 kJkg Te 2 200 C he 2 h mm 85226 kJkg sat liquid f Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniforrn ow system can be expressed as Mass balance min mout 2 Am me 2 m1 mg system Energy balance H20 Ein Eout AEsystem sat thud f T 200 C Net energy traleel Chan gem 1ntemalk1net1c 3 Q byheatW01 kandmaSS potentialetc energies v 03 m Qin mehe mga2 mla1 since W E ke E pe E O The initial and the nal masses in the tank are 11 03 m3 v1 0001157m3kg m2 m1 2594 kg 1297 kg m1 2594 kg Then from the mass balance me 2 m1 m2 2594 1297 21297 kg Now we determine the nal internal energy v 03 3 v2 m 0002313 m3kg 1112 1297 kg V V x2 2 2 f 2 0002313 0001157 2 0009171 ufg 012721 0001157 T2 2 200 C 850 46 0 0091711743 7 866 46 kJk ll 2 ll x ll 2 x2 0009171 2 f 2 fg g Then the heat transfer during this process is determined from the energy balance by substitution to be Q 1297 ng85226 kJkg 1297 kg86646 kJkg 2594 ng85046 kJkg 2308 k PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 599 5124E A rigid tank initially contains saturated liquidvapor mixture of R134a A valve at the top of the tank is opened and vapor is allowed to escape at constant pressure until all the liquid in the tank disappears The amount of heat transfer is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid leaving the device remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Properties The properties of R134a are Tables A1 1E through A13E P1 160 psia gtuf 001413 1t 3lbm ug 0293391t 3lbm u 4811 Btulbmu 10851 Btulbm R1 4a f g 3 Sat vapor 102 160 ps1a v2 vg160psia 029339 1t lbm P 160 psia Q sat vapor a2 ug160psia 10851Btulbm V 2 ft3 Pe 160 psia 23232323232321EIEIEIEIEIEIEIEIEIEIE he 2 h g 160psia 117 20 Btulbm sat vapor Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance ruin mout A nsystem me 2 m1 quot72 Energy balance Ein Eout AE sy stem V Net energy traleel Chan gem 1ntemalk1net1c by heat W01kandmass potentialetc energies Qin mehe mza2 mla1 since W E ke E pe E 0 The initial mass initial internal energy and final mass in the tank are Vf V 2gtlt021t3 2gtlt081t3 g 3 3 Vf Vg 001413 it lbm 029339 it lbm U1 2 mlul mfuf mgug 70774811 647610851 10432 Btu v 2 1t 3 m 2 2 v2 0293391t 3lbm m1 mf mg 27077 647613553 lbm 26817 lbm Then from the mass and energy balances me 2 m1 m2 13553 6817 6736 lbm Qin Z mehe m2u2 mrur 67361bm11720 Btulbm 68171bm10851Btu1bm 10432 Btu 486 Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5100 5125 A rigid tank initially contains saturated R134a liquidvapor mixture The tank is connected to a supply line and R 134a is allowed to enter the tank The final temperature in the tank the mass of R134a that entered and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The direction of heat transfer is to the tank will be veri ed Properties The properties of refrigerant are Tables A11 through A13 T1 14 C 1 If xlvfg 00008018 055gtlt 004347 00008018 002427 m3kg x1 055 ul 2 uf xlufg 7056 055gtlt 16730 16257 kJkg P2 1MPav2 ug1MPa 002033 m3kg sat vapor u2 ug1MPa 25071 kJkg P 14 MPa 1 o h 33032 kJkg R134a 14 MPa T 00 C 100 C gt Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and 0 3 m3 internal energy a respectively the mass and energy balances for this RL134a uniforrn ow system can be expressed as Mass balance min mout 2 Am system mi 2 quot72 m1 Energy balance Ein Eout AE sy stem r J W Net energy traleel Chan gein intemalkinetic by heat W01kandmass potentialetc energies Qin mihl mga2 mla1 since W E ke E pe E 0 a The tank contains saturated vapor at the nal state at 800 kPa and thus the final temperature is the saturation temperature at this pressure T2 Tsat 1MPa 394 C 9 The initial and the final masses in the tank are 3 mIZKZLm321236kg V1 002427m kg v 03 m3 m2 1476 kg V2 2 002033 m3kg Then from the mass balance ml m2 m1 21476 1236 2396kg c The heat transfer during this process is determined from the energy balance to be Qin Z mihi mzuz mrur 2396 kg33032 kJkg 1476 kg25071kJkg 1236 kg16257 kJkg 899kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission P1 100kPa 3 1 19367m Ikg Table A6 T1 150 C P 1 kP 12 211 50 a50m375m3 P1 IOOkPa V 3 m1 150 m2582kg v1 19367 m3kg The final temperature may be determined if we rst calculate speci c volume at the nal state 2 75m3 m2 2m1 2gtlt2582kg v2 14525 m3kg P2 150kPa 3 T2 2025 C Table A6 2 214525 m lkg P1 P2 100 150kPa Wb 2 12 41 f 75 50m3 3125kJ 5101 5126 A large reservoir supplies steam to a balloon whose initial state is speci ed The nal temperature in the balloon and the boundary work are to be determined Analysis Noting that the volume changes linearly with the pressure the final volume and the initial mass are determined Steam 150 kPa Steam 50 m3 100 kPa 150 C Noting again that the volume changes linearly with the pressure the boundary work can be determined from PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5102 5127 A hotair balloon is considered The nal volume of the balloon and work produced by the air inside the balloon as it expands the balloon skin are to be determined Assumptions 1 This is an unsteady process since the conditions Within the device are changing during the process but it can be analyzed as a uniform ow process 2 Air is an ideal gas with constant speci c heats 3 Kinetic and potential energies are negligible 4 There is no heat transfer Properties The gas constant of air is 0287 kPam3kgK Table A 1 Analysis The speci c volume of the air at the entrance and eXit and in the balloon is V 0287kPam3kgK35273 K P 100kPa 08840 m3kg The mass ow rate at the entrance is then AV 1m22 ms 3 2262 kgs V 08840m kg While that at the outlet is A V 2 1 e e e 0 5m X 13 05656kgs v 08840 m3kg Applying a mass balance to the balloon min mout An lsystem mi me 2 m2 m1 m2 m1 m me At 2262 05656 kgs2 x 60 s 2036 kg The volume in the balloon then changes by the amount AV m2 m1v 2036 kg08840 m3kg 180 m3 and the nal volume of the balloon is v2 V1 AV75180255m3 In order to push back the boundary of the balloon against the surrounding atmosphere the amount of work that must be done is is 18000kJ Wmut PAV 100 kPa180 m3 kPa m PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5103 5128 An insulated rigid tank initially contains helium gas at high pressure A valve is opened and half of the mass of helium is allowed to escape The final temperature and pressure in the tank are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process by using constant average properties for the helium leaving the tank 2 Kinetic and potential energies are negligible 3 There are no work interactions involved 4 The tank is insulated and thus heat transfer is negligible 5 Helium is an ideal gas with constant speci c heats Properties The speci c heat ratio of helium is k 1667 Table A2 Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout 2 Am me 2 m1 mg system 1 39 1 m2 m1g1ven gt me m2 m1 IF He Energy balance 015 m3 Ein Eout AEs 3 MPa ystem 0 Net energy tfaleel Chan gein intemalkinetic by heat W0Fkand mass potentialetc energies mehe mgaz m1a1 since W E Q E ke E pe E 0 Note that the state and thus the enthalpy of helium leaving the tank is changing during this process But for simplicity we assume constant properties for the eXiting steam at the average values Combining the mass and energy balances O 1 mlhe m1a2 mla1 a T1 T2 D1v1d1ng by m12 0 2 he u2 2u1 0r 0 2 cp CUTZ ZCUTI Dividing by c 0 kT1 T2 2T2 4T1 since k cp c Solving for T2 T2 4 k T m 403 K 257 K 2 k 1 2 1667 The final pressure in the tank is PV RT T 1 2 1 2 m1 1 gt P2 2 m2 2 P1 30001ltPn956 kPa sz mZRTZ mlTl 5 403 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5104 5129E An insulated rigid tank equipped with an electric heater initially contains pressurized air A valve is opened and air is allowed to escape at constant temperature until the pressure inside drops to 25 psia The amount of electrical work transferred is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the eXit temperature and enthalpy of air remains constant 2 Kinetic and potential energies are negligible 3 The tank is insulated and thus heat transfer is negligible 4 Air is an ideal gas with variable speci c heats Properties The gas constant of air is R 03704 psiaft3lbmR Table AlE The properties of air are Table A17E r 580 R gt h 13866 Btulbm T1 580 R gt u1 9890 Btulbm T2 580 R gt u2 9890 Btulbm Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance quotin mout An lsystem me 2 m1 m Energy balance AIR 40 f 3 V t Ein Eout AEsystem 9 W 50 ps1a W Net energy t1 Sfer Chan gein intemalkinetic o e byheatWOIkandmaSS potentialetc energies 120 F I I Wein mehe mzu2 mla1 s1nce Q ke pe O J T The initial and the nal masses of air in the tank are 3 m1 50 pSIa4O 9310 lbm RT1 03704 psia rt 3lbm RX580 R 3 m2 sz 25 ps1a40 1t 2 4655119111 RTZ 03704 psia 1t 3lbm RX580 R Then from the mass and energy balances me 2 m1 m2 9310 4655 4655kg Wein Z mehe mzuz m1u1 4655 1me13866 Btulbm 4655 lbm9890 Btulbm 93101me9890 Btulbm 185 Btu PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5105 5130 A vertical cylinder initially contains air at room temperature Now a valve is opened and air is allowed to escape at constant pressure and temperature until the volume of the cylinder goes down by half The amount air that left the cylinder and the amount of heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the eXit temperature and enthalpy of air remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions other than boundary work 4 Air is an ideal gas with constant speci c heats 5 The direction of heat transfer is to the cylinder will be verified Properties The gas constant of air is R 0287 kPam3kgK Table Al Analysis a We take the cylinder as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance ruin mout A nsystem me 2 m1 quot72 Energy balance AIR Ein Eout AEsystem 300 kP3a N t tral f f J 39 39 0392 m 6 energy S er Changern 1ntemalk1net1c byheata WOIkaandmass potentialetc energies 20 C Qin Wb iIl mehe mguz mlul since ke E pe E 0 gt The initial and the nal masses of air in the cylinder are 3 m 101v1 300 kPa02m 207l4kg RT1 0287 kPa m3kg KX293 K 3 m2 2 szz 2 300 kPa01m 2 0357 kg 2 ml RTZ 0287 kPam3kg KX293 K Then from the mass balance me 2 m1 m2 0714 0357 0357 kg 9 This is a constant pressure process and thus the W and the AU terms can be combined into AH to yield Q mehe quot12172 mlhl Noting that the temperature of the air remains constant during this process we have h h 1 h2 h Also m1 NIH me 2 quot72 2 Thus Qm1m1 m1 10 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5106 5131 A vertical pistoncylinder device contains air at a speci ed state Air is allowed to escape from the cylinder by a valve connected to the cylinder The nal temperature and the boundary work are to be determined Properties The gas constant of air is R 0287 kJkgK Table Al Analysis The initial and final masses in the cylinder are PV kP 2 3 m1 1 600 am 5m 09121m3 RT1 0287 kJkgK300 273 K Air m2 025m1 02509121 kg 02280 kg 025 m3 600 kPa Air Then the final temperature becomes 3000C gt PV 3 T 2 600kPa0 05m 458I4K 2 m2R 02280 kg0287 kJkgK Noting that pressure remains constant during the process the boundary work is determined from Wb PV1 42 600 kPa025 005m3 120 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5107 5132 A cylinder initially contains superheated steam The cylinder is connected to a supply line and is superheated steam is allowed to enter the cylinder until the volume doubles at constant pressure The nal temperature in the cylinder and the mass of the steam that entered are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 The eXpansion process is quasiequilibrium 3 Kinetic and potential energies are negligible 3 There are no work interactions involved other than boundary work 4 The device is insulated and thus heat transfer is negligible Properties The properties of steam are Tables A4 through A6 P1 500 kPa v1 042503 m3kg T1 200 C u 26433 kJkg P 1 MPa T 350 C hi 31582 kJkg P 500 kPa T1 200 C Analysis a We take the cylinder as the system which is a control VI 001 m3 volume since mass crosses the boundary Noting that the microscopic Steam Pi 1 MPa energies of owing and non owing uids are represented by enthalpy T Ti 350 C h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as I Olt Mass balance min mom Amsystem gt mi 2 quot72 m1 Energy balance Ein Eout AESystem f J W Net energy traleel Chan gein intemalkinetic byheatW0FkandeS potentialetc energies mihl Wbput mza2 mla1 s1nce Q E ke E pe E 0 Combining the two relations gives 0 Z VVbout m2 m1hi quot72112 mil 1 The boundary work done during this process is Wb 0 2 IdeV P2 41 500 kPa002 001m3 i 5 k 1 lkPam3 The initial and the nal masses in the cylinder are u 1 3 m1 1 2 L1 00235 kg 11 042503 m kg V2 002 m3 m2 V2 V2 2 2 Substituting 0 5 0023531582 u 0023526433 v2 12 Then by trial and error or using EES program T2 2617 C and v2 04858 m3kg b The nal mass in the cylinder is V2 002 m3 m2 3 22 04858m kg 00412 kg Then m mg m1 00412 00235 00176 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5108 5133 The air in an insulated rigid compressedair tank is released until the pressure in the tank reduces to a speci ed value The final temperature of the air in the tank is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process 2 Air is an ideal gas with constant speci c heats 3 Kinetic and potential energies are negligible 4 There are no work interactions involved 5 The tank is wellinsulated and thus there is no heat transfer Properties The gas constant of air is 0287 kPam3kgK Table A l The specific heats of air at room temperature are cp 1005 kJkgK and cu 0718 kJkgK Table AZa Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout An tsystem me m2 m1 3 Energy balance Air Ein Eout AEsystem 4000 kPa Net energy thfer Changein intemalkinetic 20 C m3 by heat workaandmass potentialetc energies mehe Z mzuz m1u1 0 mza2 m1u1 mehe 0 Z meCpTe Combining the two balances 0 Z mZCUTZ mlcuTl m1 m2 CpTe The initial and nal masses are given by P v 4000kP 05 3 quot1121 3 ax m 2378kg RT1 0287 kPa m kg K20 273 K sz 2000 kPa05 m3 3484 RT2 0287kPam3kgKT2 T2 m2 The temperature of air leaving the tank changes from the initial temperature in the tank to the nal temperature during the discharging process We assume that the temperature of the air leaving the tank is the average of initial and final temperatures in the tank Substituting into the energy balance equation gives 0 mzcuT2 mlcUT1m1 m2cpTe 3484 2 0 0718T2 23780718293 2378 3841005 2932 T2 2 whose solution by trialerror or by an equation solver such as EES is T2 241 K 32 c PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5109 5134 An insulated pistoncylinder device with a linear spring is applying force to the piston A valve at the bottom of the cylinder is opened and refrigerant is allowed to escape The amount of refrigerant that escapes and the final temperature of the refrigerant are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process assuming that the state of uid leaving the device remains constant 2 Kinetic and potential energies are negligible Properties The initial properties of R134a are Tables A11 through A13 3 P1 14MPa V1 002039 m Ikg u1 32357 kJkg T1 120 C h1 35211 kJkg Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout Amsystem me 2 m1 mg Energy balance Ein Eout AEsystem f J Net energy thfer Changein intemalkinetic byheata workaandmass potentialetc energies Wb in mehe mza2 mla1 since Q E ke E pe E 0 R134a 3 The initial mass and the relations for the final and eXiting masses are 1012 a v 08 3 120 C m1 m33924kg gt V1 002039 m kg v 05 m3 m2 V2 V2 05 3 me m1 m223924 m 2 Noting that the spring is linear the boundary work can be determined from 1400700 kPa 2 v V P1P2 Wbin 0805m3 315kJ Substituting the energy balance 3 3 315 3924 0395 m 0395 m Ju 3924 kg32357 kJkg Eq 1 V2 2 where the enthalpy of eXiting uid is assumed to be the average of initial and nal enthalpies of the refrigerant in the cylinder That is h1 112 35211kJkg 112 he 2 2 Final state properties of the refrigerant hz uz and V2 are all functions of final pressure known and temperature unknown The solution may be obtained by a trialerror approach by trying different final state temperatures until Eq 1 is satis ed Or solving the above equations simultaneously using an equation solver with builtin thermodynamic functions such as EES we obtain T2 962 C me 268 kg h2 33451 kJkg a2 30643 kJkg u2 004011 m3kg m2 1247 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5110 Review Problems 5135 The air in a hospital room is to be replaced every 15 minutes The minimum diameter of the duct is to be determined if the air velocity is not to exceed a certain value Assumptions 1 The volume occupied by the furniture etc in the room is negligible 2 The incoming conditioned air does not mix with the air in the room Analysis The volume of the room is V 6 m5 m4 m 120 m3 To empty this air in 20 min the volume ow rate must be 3 Hospital 2 K 120 m 01333 m3s Room At 15 X 60 S If the mean velocity is 5 ms the diameter of the duct is 6X5gtlt4 1113 10 bulbs 2 39 3 v 2 AV D V gt D 1 40391333m IS 0184m 4 7rV 7r5 ms Therefore the diameter of the duct must be at least 0184 m to ensure that the air in the room is exchanged completely within 20 min while the mean velocity does not exceed 5 ms Discussion This problem shows that engineering systems are sized to satisfy certain constraints imposed by certain considerations 5136 A long roll of large 1Mn manganese steel plate is to be quenched in an oil bath at a specified rate The mass ow rate of the plate is to be determined 0 I Assumptions The plate moves through the bath stead11y 7 Pro erties The densit of steel late is iven to be 7854 k m3 Steel plate U Analysis The mass ow rate of the sheet metal through the oil bath is m pi 0th 7854 kgm31m0005 m10 mmin 393 kgmin 655kgs Therefore steel plate can be treated conveniently as a owing uid in calculations PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 51 1 1 5137 Air is accelerated in a nozzle The density of air at the nozzle eXit is to be determined Assumptions Flow through the nozzle is steady Properties The density of air is given to be 418 kgm3 at the inlet Analysis There is only one inlet and one eXit and thus m1 n3912 m Then plAIVI Pzszz A V 12 11p1 2 OIIlS A2 V2 380 ms 02 418 kgm3 264 kgm3 Discussion Note that the density of air decreases considerably despite a decrease in the crosssectional area of the nozzle 5138 An air compressor consumes 62 kW of power to compress a specified rate of air The ow work required by the compressor is to be compared to the power used to increase the pressure of the air Assumptions 1 Flow through the compressor is steady 2 Air is an ideal gas Properties The gas constant of air is 0287 kPam3kgK Table A 1 08 MPa Analysis The specific volume of the air at the inlet is RTl 0287 kPa m3kg K20 273 K P1 120 kPa Compressor v1 07008 m3kg The mass ow rate of the air is 3 20 C V1 0015 m s m 3 002140 kgs v1 07008 m kg Combining the ow work expression with the ideal gas equation of state gives the ow work as w ow P2u2 Plul RT2 T1 0287 kJkg K300 20K 8036 kJkg The ow power is W ow n39iw 0W 002140 kgs8036 kJkg 1 72 kW The remainder of compressor power input is used to increase the pressure of the air W Wtotal in W ow I PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5112 5139 Saturated refrigerant134a vapor at a saturation temperature of Tsat 34 C condenses inside a tube The rate of heat transfer from the refrigerant for the condensate eXit temperatures of 34 C and 20 C are to be determined Assumptions 1 Steady ow conditions eXist 2 Kinetic and potential energy changes are negligible 3 There are no work interactions involved Properties The properties of saturated refrigerant134a at 34 C are hf 9941 kJkg hg 26861 kJkg and hfg 16921 kJkg The enthalpy of saturated liquid refrigerant at 20 C is hf 7932 kJkg Table A11 Analysis We take the tube and the refrigerant in it as the system This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one eXit and thus m1 n3912 m Noting that heat is lost from the system the energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem 0 KJ Qout Rate 0f net energy thfer Rate of changern 1ntemalkrnet1c by heat W0rkand mass potentialetc energies E39in E39Ollt R134a n39thl Qlout 1139ch since Ake E Ape E O 34 C Q 1011 ha where at the inlet state h1 kg 26861 kJkg Then the rates of heat transfer during this condensation process for both cases become Case 1 T2 34 C h2 hf34 C 9941 kJkg Qout 01 kgmin26861 9941 kJkg 169kgmin Case 2 T2 20 C 12 E hf20 C Qout 01 kgmin26861 7932 kJkg 189 kgmin Discussion Note that the rate of heat removal is greater in the second case since the liquid is subcooled in that case PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5113 5140 Steam expands in a turbine Whose power production is 12350 kW The rate of heat lost from the turbine is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties From the steam tables Tables A6 and A4 P1 16 MPa h1 31460kJkg T1 350 C T2 30 C h2 25556kJkg xz Analysis We take the turbine as the system which is a control volume since mass crosses the boundary Noting that there is one inlet and one eXiti the energy balance for this steady ow system can be eXpressed in the rate form as lt90 steady 0 system Ein Eout Wr J Wf J Rate 0f net energy thfer Rate of chan g ein intemalkinetic by heat workaand mass potentialetc energies Ein Eout mlhl mzhz Wout Qout Qout h2 Wout Substituting Qout 22 kgs31460 25556 kJkg 12350 kW 2 640 kW 16 MPa 350 C Heat 22 kgs 30 C sat vap PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5114 5141E Nitrogen gas ows through a long constantdiameter adiabatic pipe The velocities at the inlet and eXit are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Nitrogen is an ideal gas with constant specific heats 3 Potential energy changes are negligible 4 There are no work interactions 5 The re is no heat transfer from the nitrogen Properties The speci c heat of nitrogen at the room temperature iss cp 0248 BtulbmR Table A2Ea Analysis There is only one inlet and one eXit and thus 1 12 m We take the pipe as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O Ef J f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Em E out 100 psia N2 50 psia 120 F 70 F mm V12 2 2 mm V222 h1V122 h2 V222 V12 V22 2 Combining the mass balance and ideal gas equation of state yields m1 21712 A1V1 A2V2 V1 2 A v u T P V2 12V1 2V1 21V1 A2 1 1 T1 P2 Substituting this expression for V2 into the energy balance equation gives 05 05 20102 T1 2024870 120 25037 it 2s2 1 T2 P1 2 3 10 2 580 50 The velocity at the eXit is T P 21V1530515941 s T1 P2 580 50 j 2515st V2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5115 5142 Water at a speci ed rate is heated by an electrical heater The current is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The heat losses from the water is negligible Properties The speci c heat and the density of water are taken to be cp 418 kJkg C and p 1 kgL Table A3 Analysis We take the pipe in which water is heated as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 70 steady Ein Eout AEsystem 0 af J RateOfnet energy MSfer Rate of changein intemalkinetic by heat WOTkandmaSS potentialetc energies Ein out e m mhl Wein mhz 18 C water 30 C Wain n391h2 hl 01 US gt The mass ow rate of the water is m pi 1 kgLO 1113 01kgs Substituting into the energy balance equation and solVing for the current gives I mcpT2 T1 01kgs418kJkgK30 18K 1000VI 45 6A V 110V lkJs 39 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5116 5143 Steam ows in an insulated pipe The mass ow rate of the steam and the speed of the steam at the pipe outlet are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions Analysis We take the pipe in which steam ows as the system which is a control volume The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 ar J Rate 0f net energy MSfer Rate of changein intemalkinetic byheata workaandmass potentialetc energies E Z t 1200 kPa 250 C Water 1000 kPa in on DO15m4ms DOlm hr h2 The properties of the steam at the inlet and eXit are Table A6 P 1200 kPa 1 019241 m3kg T 250 C h1 29356 kJkg P2 21000 kPa v2 023099 m3kg h2 h1 29356 kJkg The mass ow rate is 2 2 m AlVl 7le 7r015 m 4ms3 203674kgS V1 4 V1 4 01924lm kg The outlet velocity will then be 4 I I 3 quotW2 mvz 4 3674 kgSXO 230991 kg 108ms V 2 A2 7022 7r010m2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 51 17 5144 Air ows through a nonconstant crosssection pipe The inlet and exit velocities of the air are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy change is negligible 3 There are no work interactions 4 The device is adiabatic and thus heat transfer is negligible 5 Air is an ideal gas with constant speci c heats Properties The gas constant of air is R 0287 kPam3kgK Also 0 1005 kJkgK for air at room temperature Table A 2 Analysis We take the pipe as the system which is a N control volume since mass crosses the boundary The D1 D2 mass and energy balances for this steady ow system 200 kpa 175 kPa can be expressed in the rate form as 650C gt Alf 600C Mass balance V1 4 V2 70 steady min mout AInsystem O 2 2 i v i P P min mout gt101Arvr Z P2A2V2 gt 1 V2 gt 1D12Vr Z 2D22V2 RT1 4 RT2 4 T1 T2 Energy balance 39 39 39 70 steady 39 Ein Eout AESystem 0 s1nceW Ape 0 r J r Rate 0f net en ergy t1 Sfer Rate of chan gein intemalkin etic byheatW0rkandmaSS potentialetc energies V2 2 1 2 EinEout 3h1 2 hQ 2qout 2 2 T V T 2 0139 C p 1 2 Cp 2 2 q out Assuming inlet diameter to be 14 m and the eXit diameter to be 10 m and substituting given values into mass and energy balance equations 200kPa 2 175 kPa 2 14m V 10m V 1 338K 1 333K 2 2 2 1005 kJkgK338 K V 1005 kJkgK333 K V 2 33 kJkg 2 2 1000m s 2 1000m Is There are two equations and two unknowns Solving equations 1 and 2 simultaneously using an equation solver such as EES the velocities are determined to be V1 299 ms V2 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 51 18 5145 Heat is lost from the steam owing in a nozzle The eXit velocity and the mass ow rate are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Potential energy change is negligible 3 There are no work interactions Anal sis We take the steam as the s stem hich is a control y a y W 150 C STEAM 75 kPa volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 200 kPa gt sat Vap Energy balance q 39 70 steady Ein Eout AEsystem 0 r J r J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies Ein Eout V2 V2 m 11 n391 2 Q since WEApeEO 1 2 2 out 0139 V2 2 V2011 h2 qout The properties of steam at the inlet and eXit are Table A6 101 200 kPa h1 27691 kJkg T1 150 C 102 75 kPa u2 22172 m3kg sat vap h2 26624 kJkg Substituting lkJk V2 2 42071 I72 qout 227691 26624 26kJkg 2gzj 401 7ms 1000m s b The mass ow rate of the steam is 1 1 m A2V2 3 00011112 4017 ms 2 0181kgs V2 22172m lkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5119 5146 Water is boiled at a speci ed temperature by hot gases owing through a stainless steel pipe submerged in water The rate of evaporation of is to be determined Assumptions 1 Steady operating conditions eXist 2 Heat losses from the outer surfaces of the boiler are negligible Properties The enthalpy of vaporization of water at 180 C is 2 142 T l A 4 hfg 0 kJkg ab e A Water Analysis The rate of heat transfer to water is given to be 48 kJs M 180 C Noting that the enthalpy of vaporization represents the amount of energy needed to vaporize a unit mass of a liquid at a speci ed Heater temperature the rate of evaporation of water is determined to be HOt gases Q gt0L boiling 48 kJs m 00238 s evapmu h fg 20142kJkg kg 5147 Saturated steam at 1 atm pressure and thus at a saturation temperature of Tsat 100 C condenses on a vertical plate maintained at 90 C by circulating cooling water through the other side The rate of condensation of steam is to be determined Assumptions 1 Steady operating conditions eXist 2 The steam condenses and the condensate drips off at 100 C In reality the condensate temperature will be between 90 and 100 and the cooling of the condensate a few C should be considered if better accuracy is desired Plate 9 2 Properties The enthalpy of vaporization of water at 1 atm 101325 kPa p Steam is hfg 22565 kJkg Table A5 3 Analysis The rate of heat transfer during this condensation process is Q lt given to be 180 kJs Noting that the heat of vaporization of water 0 391 represents the amount of heat released as a unit mass of vapor at a 90 C 100 C specified temperature condenses the rate of condensation of steam is determined from m 2 00793kgs condensatbn hf 22565kJkg i l Condensate g PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5120 5148E Saturated steam at a saturation pressure of 095 psia and thus at a saturation temperature of Tsat 100 F Table A 4E condenses on the outer surfaces of 144 horizontal tubes by circulating cooling water arranged in a 12 x 12 square array The rate of heat transfer to the cooling water and the average velocity of the cooling water are to be determined Assumptions 1 Steady operating conditions eXist 2 The tubes are isothermal 3 Water is an incompressible substance with constant properties at room temperature 4 The changes in kinetic and potential energies are negligible 03995 pSIa Saturated steam Properties The properties of water at room temperature are p 621 lbmft3 and CI 100 Btulbm F Table A3E The enthalpy of vaporization of water at a saturation pressure of 095 psia is hfg 10367 Btulbm Table A4E Analysis a The rate of heat transfer from the steam to the cooling water is equal to the heat of vaporization released as the vapor condenses at the speci ed temperature Q X VJ C A A A FA A 6 Cooling Q mhfg 68001bmh10367 Btulbm 7049560 Btuh 1958 Btus LC water 9 All of this energy is transferred to the cold water Therefore the mass ow rate of cold water must be Q 1958 Btus Q water P Water cpAT 100Btulbm F8 F Then the average velocity of the cooling water through the 144 tubes becomes m pAV V 2 l m 24481bms IDA 2 3 2 5021118 0n7zD 4 6211bmlt 1447z112lt 4 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5121 5149 The feedwater of a steam power plant is preheated using steam extracted from the turbine The ratio of the mass ow rates of the extracted seam the feedwater are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the deVice to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid Properties The enthalpies of steam and feedwater at are Tables A4 through A6 Pl 2 1 Mpa h1 28283 kJkg T1 200 C STEAM a 1MPa h2 hf1MPa 76251kJkg l1 sat liquid T2 1799 C Feedwater 3 9 gt D and 4 T3 2500C h3 E f50C 20934 kJkg Analysis We take the heat exchanger as the system which is a control volume The mass and energy balances for this steady ow system can be expressed in the rate form as Mass balance for each uid stream 70 steady quotEn moutZAmsystem 0rnin out mIZmQZmS and quot732m42n1fw Energy balance for the heat exchanger 39 39 39 70 steady Ein Eout AEsystem O r J f J Rate 0f net energy thfer Rate of changein intemalkinetic byheat W0rkandmaSS potentialetc energies E E 1n out m1h1 1413113 mth m4h4 since Q W Ake Ape 0 Combining the two ms h2 h1 mfwh3 h4 Dividing by mfw and substituting m h6 h4 71855 20934kJkg S 0246 mw h2 h1 28283 76251kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5122 5150 Cold water enters a steam generator at 20 C and leaves as saturated vapor at Tsat 200 C The fraction of heat used to preheat the liquid water from 20 C to saturation temperature of 200 C is to be determined Assumptions 1 Steady operating conditions eXist 2 Heat losses from the steam generator are negligible 3 The speci c heat of water is constant at the average temperature Properties The heat of vaporization of water at 200 C is hfg 19398 kJkg Table A4 and the specific heat of liquid water is c 418 kJkg C Table A3 Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a specified temperature Using the average speci c heat the amount of heat transfer needed to preheat a unit mass of water from 20 C to 200 C is CAT 418 kJkg C200 20 C 7524 kJkg Cold water 20 C q preheatin g Heater and q total 2 q boiling q preheating 219398 3641 26922 kJkg Therefore the fraction of heat used to preheat the water is qpreheating 7524 qtotal 26922 Fraction to preheat 02795 or 280 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5123 5151 Cold water enters a steam generator at 20 C and is boiled and leaves as saturated vapor at boiler pressure The boiler pressure at which the amount of heat needed to preheat the water to saturation temperature that is equal to the heat of vaporization is to be determined Assumptions Heat losses from the steam generator are negligible Properties The enthalpy of liquid water at 20 C is 8391 kJkg Other properties needed to solve this problem are the heat of vaporization hfg and the enthalpy of saturated liquid at the speci ed temperatures and they can be obtained from Table A4 Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a specified temperature and Ah represents the amount of heat needed to preheat a unit mass of water from 20 C to the saturation temperature Therefore q preheating q boiling hfTsa hf20 C hfg Ta The solution of this problem requires choosing a boiling temperature reading hf and hfg at that temperature and substituting the values into the relation above to see if it is satisfied By trial and error Table A4 C01 d water Heater At 310 C hfTsat hfgTsat 14020 13259 761 kJkg 20 C At 315 C hfTsat hfgTsat 14316 12834 1482 kJkg The temperature that satis es this condition is determined from the two values above by interpolation to be 3106 C The saturation pressure corresponding to this temperature is 994 MPa 5152 An ideal gas eXpands in a turbine The volume ow rate at the inlet for a power output of 650 kW is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible Properties The properties of the ideal gas are given as R 030 kPam3kgK cp 113 kJkg C C 083 kJkg C Analysis We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 P1 900 kPa W f J RateOf net energy thfel Rate of changein intemalkinetic T1 1200 K byheat W0rkand mass potentialetc energies 41 Em Bout Wk Wont nilIQ since E Ake Ape E 0 which can be rearranged to solve for mass ow rate Ideal gas 8 650 kW Wout Wout 650 kW m 2 1438 kgs h1 h2 cpT1 T2 113kJkgK1200 800K The 1nletspec1 c volume and the volume ow rate are T2 800 K V 03 kPam3kgKX1200 K 1 P1 900kPa 04 m3kg Thus V mu 1438 kgs04 m3kg 0575 m3s PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5124 5153 Chickens are to be cooled by chilled water in an immersion chiller The rate of heat removal from the chicken and the mass ow rate of water are to be determined Assumptions 1 Steady operating conditions eXist 2 The thermal properties of chickens and water are constant Properties The speci c heat of chicken are given to be 354 kJkg C The speci c heat of water is 418 kJkg C Table A 3 Analysis a Chickens are dropped into the chiller at a rate of 500 per hour Therefore chickens can be considered to ow steadily through the chiller at a mass ow rate of mcmcken 500 chicken h22 kg chicken 2 1100 kg h 03056kg s Taking the chicken ow stream in the chiller as the system the energy balance for steadily owing chickens can be Immersion expressed in the rate form as chilling 05 C 39 70 steady Ein Eout AEsystem 0 Rate 0f net en ergy tr Sfer Rate of chan gein internal kinetic by heat W0rkandmaSS potentialetc energies Ein Eout n39th1 Q0 n39th2 since Ake E Ape E O Qout Qchicken I chickencp T2 Then the rate of heat removal from the chickens as they are cooled from 15 C to 3 C becomes Qchicken n391cpAT chicken 03056 kgs354 kJkg C15 3 C 130 kW The chiller gains heat from the surroundings at a rate of 200 kJh 00556 kJs Then the total rate of heat gain by the water is Q39Water Q39cmcken Q39heat gain 2 130 0056 13056 kW Noting that the temperature rise of water is not to exceed 2 C as it ows through the chiller the mass ow rate of water must be at least 39 13056 kW mwater Qwater 156 kgS cpAT 418 kJkg C2 C water If the mass ow rate of water is less than this value then the temperature rise of water will have to be more than 2 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5125 5154 Chickens are to be cooled by chilled water in an immersion chiller The rate of heat removal from the chicken and the mass ow rate of water are to be determined Assumptions 1 Steady operating conditions eXist 2 The thermal properties of chickens and water are constant 3 Heat gain of the chiller is negligible Properties The speci c heat of chicken are given to be 354 kJkg C The speci c heat of water is 418 kJkg C Table A 3 Analysis a Chickens are dropped into the chiller at a rate of 500 per hour Therefore chickens can be considered to ow steadily through the chiller at a mass ow rate of mcmcken 500 chicken h22 kg chicken 2 1100 kg h 03056kg s Taking the chicken ow stream in the chiller as the system the energy Immersion balance for steadily owing chickens can be eXpressed in the rate form Chilling 0 5 C as d iCken 5 C I w r V Em Emu A 70 steady 0 system J r J Rate 0 f net en ergy tr Sfer Rate 0 f chan gein intern a1 kinetic by heat W0 rk and H138 S p otentialetc energies Ein Eout mill 2 Qlout n391hQ since Ake E Ape E 0 Qout Qchicken I chickencp T2 Then the rate of heat removal from the chickens as they are cooled from 15 C to 3 C becomes Qchicken n3910pATchick6r1 03056 kgs354 kJkg C15 30 C 130 kW Heat gain of the chiller from the surroundings is negligible Then the total rate of heat gain by the water is Qwater Qchicken kW Noting that the temperature rise of water is not to exceed 2 C as it ows through the chiller the mass ow rate of water must be at least Qwater 130 kW m 2 Z 156 s water CPAT 418 kJkg C2 C kg water If the mass ow rate of water is less than this value then the temperature rise of water will have to be more than 2 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5126 5155E A refrigeration system is to cool eggs by chilled air at a rate of 10000 eggs per hour The rate of heat removal from the eggs the required volume ow rate of air and the size of the compressor of the refrigeration system are to be determined Assumptions 1 Steady operating conditions eXist 2 The eggs are at uniform temperatures before and after cooling 3 The cooling section is wellinsulated 4 The properties of eggs are constant 5 The local atmospheric pressure is 1 atm Properties The properties of the eggs are given to p 674 lbmft3 and CI 080 Btulbm F The specific heat of air at room temperature 0 024 Btulbm F Table A2E The gas constant of air is R 03704 psiaft3lbmR Table A1E Analysis a Noting that eggs are cooled at a rate of 10000 eggs per hour eggs can be considered to ow steadily through the cooling section at a mass ow rate of megg 10000 eggsh014 lbmegg 1400 lbmh03889 lbms Taking the egg ow stream in the cooler as the system the energy balance for steadrly owrng eggs can be expressed 1n the rate form as 0 14glgbm 39 39 39 70 steady 44 m 1n out system Alr RateOf net energy thfer Rate of changein intemalkinetic 34 F by heat workaandmass potentialetc energies gt Ein Eout gt n39th1 Qlout 11391h2 since Ake E Ape E 0 Q Q megch T1 T2 Then the rate of heat removal from the eggs as they are cooled from 90 F to 50 F at this rate becomes Qegg n3910pATegg 14001bmh080 Btulbmf F90 50 F 44800 Btuh b All the heat released by the eggs is absorbed by the refrigerated air since heat transfer through he walls of cooler is negligible and the temperature rise of air is not to exceed 10 F The minimum mass ow and volume ow rates of air are determined to be mair Qair 44 800 Btu h 2186671bmh cpATair 024Btulbm FlO F pair 2 i Z 17 pm 2 00803lbmft3 RT 03704 ps1aft lbmR34 460R V mair W 2 232500 ftslh a pair 00803lbrnft3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5127 5156 Glass bottles are washed in hot water in an uncovered rectangular glass washing bath The rates of heat and water mass that need to be supplied to the water are to be determined Assumptions 1 Steady operating conditions eXist 2 The entire water body is maintained at a uniform temperature of 50 C 3 Heat losses from the outer surfaces of the bath are negligible 4 Water is an incompressible substance with constant properties Properties The speci c heat of water at room temperature is cp 418 kJkg C Also the speci c heat of glass is 080 kJkg C Table A3 Analysis a The mass ow rate of glass bottles through the water bath in steady operation is rnbot e mbot e x Bottle ow rate 0150 kgbottle4506O bottless 1125 kgs Taking the bottle ow section as the system which is a steady ow control volume the energy balance for this steady ow system can be eXpressed in the rate form as Water bath 55 C Em Emu 70 steady 0 system ar J r J Rate 0 f net en ergy tMSfer Rate 0 f chan g ein internal kinetic byheat W01kand mass potentialetc energies Ein Eout Qin n391h1 nah since Ake E Ape E O Qin Qbottle mwatercp T2 o Q Then the rate of heat removal by the bottles as they are heated from 20 to 55 C is Qboule Z mbottlecp AT 1125 kgSXO8 kJ1ltg C50 20 C 27 kW The amount of water removed by the bottles is m Flow rate of bottlesXWater removed per bottle waterout 450 bottlesminX00002 gbottle 90 gmin 00015kgs Noting that the water removed by the bottles is made up by fresh water entering at 15 C the rate of heat removal by the water that sticks to the bottles is edcp AT 00015 kgs418 kJkg C50 15 C 02195 kW Qwater removed mwater remov Therefore the total amount of heat removed by the wet bottles is Qtotalremoved leassmnnved Qwater removed 2 27 0392195 Discussion In practice the rates of heat and water removal will be much larger since the heat losses from the tank and the moisture loss from the open surface are not considered PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5128 5157 Dough is made with refrigerated water in order to absorb the heat of hydration and thus to control the temperature rise during kneading The temperature to which the city water must be cooled before mixing with our is to be determined to avoid temperature rise during kneading Assumptions 1 Steady operating conditions exist 2 The dough is at uniform temperatures before and after cooling 3 The kneading section is wellinsulated 4 The properties of water and dough are constant Properties The specific heats of the our and the water are given to be 176 and 418 kJkg C respectively The heat of hydration of dough is given to be 15 kJkg Analysis It is stated that 2 kg of our is mixed with 1 kg of water and thus 3 kg of dough is obtained from each kg of water Also 15 kJ of heat is released for each kg of dough kneaded and thus 3x15 45 kJ of heat is released from the dough made using 1 kg of water Flour Tak1ng the coollng sectlon of water as the system wh1ch is a steady ow control volume the energy balance for this I steady ow system can be expressed in the rate form as l 39 39 39 70 steady Em Eout AEsystem 0 Water 15 kJkg 1 5 C Rate 0f net en ergy tr Sfer Rate of chan gein internal kinetic byheataworkaandmass potentialetc energies Ein Eout Dough n39th1 Q0 n391h2 since Ake E Ape E O Qout Qwater mwatercp T2 In order for water to absorb all the heat of hydration and end up at a temperature of 15 C its temperature before entering the mixing section must be reduced to Qm Qdoughm6pT2 T1gt T1 T2 i 15 C 45k 42 C mcp l kg4 18 kJkg C That is the water must be precooled to 42 C before mixing with the our in order to absorb the entire heat of hydration PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5129 5158 Long aluminum wires are extruded at a velocity of 8 mmin and are exposed to atmospheric air The rate of heat transfer from the wire is to be determined Assumptions 1 Steady operating conditions exist 2 The thermal properties of the wire are constant Properties The properties of aluminum are given to be p 2702 kgm3 and CI 0896 kJkg C Analysis The mass ow rate of the extruded wire through the air is m pi p717quot02 v 2702 kgm3 7r00025 m2 8 mmin 04244 kgmin 0007074 kgs Taking the volume occupied by the extruded wire as the system which is a steady ow control volume the energy balance for this steady ow system can be expressed in the rate form as Em Emu AESyStemZO steady 0 511172 3 V V 3 lelgaftnggrtnris fer Rategfoil g egtl Clinetregrggail lginet1c 5 4 qquot 1 7 350 C 8 mmin n w H Ein jout 2 7 I l I gt nth1 Qout n39ihQ since Ake E Ape E 0 57 5 3394 Alumlnum Wll e Qout Qwine mwirecp T2 Then the rate of heat transfer from the wire to the air becomes Q mop Tt Too 0007074 kgs0896 kJkg C350 50 C 190 kW 5159 Long copper wires are extruded at a velocity of 8 mmin and are exposed to atmospheric air The rate of heat transfer from the wire is to be determined Assumptions 1 Steady operating conditions exist 2 The thermal properties of the wire are constant Properties The properties of copper are given to be p 8950 kgm3 and CI 0383 kJkg C Analysis The mass ow rate of the extruded wire through the air is m pt p717quot02 v 8950 kgm3 700025 m2 8 60 ms 002343 kgs Taking the volume occupied by the extruded wire as the system which is a steady ow control volume the energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 LV 17 quot KJ an f y7 1 RateOf net energy thfer Rateof chan ein intemalkinetic 39 39 39i V39 1 byheataworkaandmass potenti 1etc energies 5 u 392 39 7 350 C 8 mmln u 9 39 7quot39r I I 39 39 0 d o n I 7 9 39 39 39 7 mh1 Qout th s1nce Ake E Ape E 0 f g f 39 a 39Ei Copper Wire in 5 39 39 a 0 0f 1 Qout Qwine mwirecpal T2 5r quot Then the rate of heat transfer from the wire to the air becomes Q mop Tt Too 002343 kgs0383 kJkg C350 50 C 269 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5130 5160E Steam is mixed with water steadily in an adiabatic device The temperature of the water leaving this device is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions 4 There is no heat transfer between the mixing device and the surroundings Analysis We take the mixing device as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 7I0 ad Ein Eout AE system Ste y 2 0 S ar J r J team RateOf net energy MSfer Rate of changein intemalkinetic 80 39 byheat Workaandmass potentialetc energies pSIa E E 400 F in out 005 lbms 1111111 l i12h2 G From a mass balance gt 80 psia m3 m m2 005 1 1051bms Water 80 39 The enthalp1es of steam and water are Table A6E and A4E 60281281 1 Z psia 1 h1 12308 Btulbm T1 400 F P2 2 80 pm h h 28 08 Bt 1b E u m T2 2 60 F 2 f60 F Substituting into the energy balance equation solving for the exit enthalpy gives m1 h1 mth 0051bms12308 Btulbm 11bms2808 Btulbm m 1051sz h3 8535 Btulbm At the exit state P3 80 psia and h3 8535 kJkg An investigation of Table A5E reveals that this is compressed liquid state Approximating this state as saturated liquid at the speci ed temperature the temperature may be determined from Table A4E to be P3 80 psia h 85 35 Btulbm T3 E Tfh8535Btu1bm 1173 F 3 Discussion The exact answer is determined at the compressed liquid state using EES to be 1171 F indicating that the saturated liquid approximation is a reasonable one PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5131 5161 A constantpressure R134a vapor separation unit separates the liquid and vapor portions of a saturated mixture into two separate outlet streams The ow power needed to operate this unit and the mass ow rate of the two outlet streams are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work and heat interactions Analysis The speci c volume at the inlet is Table A12 P1 320 kPa 3 0 55 v1 If x 1vg If 2 00007771 055006368 00007771 003537 m kg x1 2 The mass ow rate at the inlet is then R134a VI 0006 m3s 320 kPa Saturated m 01696kgs gt gt 1 V1 003537 m3kg x 05 5 Vapor separatron vapor 6 Us unlt For each kg of mixture processed 055 kg of vapor are processed Therefore l m 07m1 055 X 01696 009329kgs Saturated liquid The ow power for this unit is W ow Z m2P2V2 m3P3V3 mrprvr 009329 kgs320 kPa006368 m3kg 007633 kgs320 kPa00007771m3kg 01696 kgs320 kPa003537 m3kg 0 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5132 5162 Two identical buildings in Los Angeles and Denver have the same infiltration rate The ratio of the heat losses by in ltration at the two cities under identical conditions is to be determined Assumptions 1 Both buildings are identical and both are subjected to the same conditions except the atmospheric conditions 2 Air is an ideal gas with constant speci c heats at room temperature 3 Steady ow conditions exist Analysis We can view in ltration as a steady stream of air that is heated as it ows in an imaginary duct passing through the building The energy balance for this imaginary steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 R rm f r J Los Angeles 101 kPa 3160 He energy S er Rate of chan ein intemalkinetic byheatworkandmass potentigl ta energies DCIIVCI39 83 kPa Ein Eout Qin rth 1139ch since Ake E Ape E O Q n39wT2 T1 chpT2 T1 Then the sensible infiltration heat loss heat gain for the infiltrating air can be expressed Qin ltration maircp To 2 p0 air ACPDGbuilding Cp To where ACH is the in ltration volume rate in air changes per hour Therefore the in ltration heat loss is proportional to the density of air and thus the ratio of in ltration heat losses at the two cities is simply the densities of outdoor air at those cities Qin ltrationLos An eles p0 air Los An eles In ltratr on heat loss ratro g g in ltrationDenver p 0 air Denver PORTOLosAngeles PoLosAngeles 101kPa P0 lRT0Denver P0 Denver 83 kPa 122 Therefore the infiltration heat loss in Los Angeles will be 22 higher than that in Denver under identical conditions 5163E A study quanti es the cost and benefits of enhancing IAQ by increasing the building ventilation The net monetary benefit of installing an enhanced IAQ system to the employer per year is to be determined Assumptions The analysis in the report is applicable to this work place Analysis The report states that enhancing IAQ increases the productivity of a person by 90 per year and decreases the cost of the respiratory illnesses by 39 a year while increasing the annual energy consumption by 6 and the equipment cost by about 4 a year The net monetary benefit of installing an enhanced IAQ system to the employer per year is determined by adding the bene ts and subtracting the costs to be Net benefit Total bene ts total cost 9039 64 119year per person The total bene t is determined by multiplying the bene t per person by the number of employees Total net bene t No of employees gtlt Net bene t per person 120gtlt119year 14280year Discussion Note that the unseen savings in productivity and reduced illnesses can be very significant when they are properly quantified PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5133 5164 The ventilating fan of the bathroom of an electrically heated building in San Francisco runs continuously The amount and cost of the heat vented out per month in Winter are to be determined Assumptions 1 We take the atmospheric pressure to be 1 atm 1013 kPa since San Francisco is at sea level 2 The building is maintained at 22 C at all times 3 The in ltrating air is heated to 22 C before it eX ltrates 4 Air is an ideal gas with constant speci c heats at room temperature 5 Steady ow conditions eXist Properties The gas constant of air is R 0287 kPam3kgK Table A1 The speci c heat of air at room temperature is cp 1005 kJkg C Table A 2 Analysis The density of air at the indoor conditions of 1 atm and 22 C is P0 1013 kPa p Z 2 21197 kgm3 quot RTO 0287 kPam3kgK22 273 K 30 US F 122 C Then the mass ow rate of air vented out becomes mail 2 pVail 1197 kgm3 0030 m3s 003590 kgs T T We can view in ltration as a steady stream of air that is heated as it ows in an imaginary duct passing through the house The energy balance for this imaginary steady ow system can be expressed in the rate form as g 710 t d Ein Eout AEsystem S ea y 2 0 220C RateOfnetenergy thfer Rate of changein intemalkinetic by heat workaandmass potentialetc energies E Ein Eout Qin Ink1 nth2 since Ake E Ape E 0 Q mcpltT2 T1 Noting that the indoor air vented out at 22 C is replaced by in ltrating outdoor air at 122 C the sensible in ltration heat loss heat gain for the in ltrating air due to venting by fans can be expressed Qlossbyfan mair Cp Tindoors Toutdoors 00359 kgs1005 kJkg C22 122 C 03536 kJs 03536 kW Then the amount and cost of the heat vented out per month 1 month 30x24 720 h becomes Energy loss 2 Qlossbyfan At 03536kW720 hmonth 2546kWhmonth Money loss 2 Energy lossUnit cost of energy 2546 kWhmonth012kWh 229month Discussion Note that the energy and money loss associated with ventilating fans can be very signi cant Therefore ventilating fans should be used with care PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5134 5165E The relationships between the mass ow rate and the time for the in ation and de ation of an air bag are given The volume of this bag as a function of time are to be plotted Assumptions Uniform ow eXists at the inlet and outlet Properties The speci c volume of air during in ation and de ation are given to be 15 and 13 ft3lbm respectively Analysis The volume of the airbag at any time is given by 0 J39 n39win dt J39 mmoutd in ow time out ow time Applying at different time periods as given in problem statement give t 2 1 1 Vt151131bm 0 W3 S 0 10ms 1000ms tdt OStSIOms t Vt J39 00151 2 11 3ms2 0 s t s 10 ms 0 t 1 Vt 10 ms J39 15 11 3lbm201bms10008 10ms tdt 10 lt t S 12 ms ms 0 V10 ms 00311 3011320 10 ms 10 lt t g 12 ms t 12 ms 00311 3ms2t 12 ms t J39 1311 3lbm 12ms 16lbms ls t 12msdt 12ltt325ms 3012ms 1000ms v0 12 ms 00311 3ms2t 12 ms t I0011556ft3mszt 12msdt 12lttg25ms 12ms t 12 ms 00311 3ms2t 12 ms O390121556 t2 144 ms2013867t 12 ms 12 lt t S 25 ms 0011556 r 25 ms 12 625 ms2 013867t 25 ms 25 lt r g 30 ms ls 1000 ms t 30 ms I 13 11 3lbml6lbms 12ms dt 30ltt 50ms v0 30 ms 0208 11 3mst 30 ms 30 lt t g 50 ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5135 The results with some suitable time intervals are 6 Time ms V ft3 0 0 2 006 5 4 024 6 m 4 I 8 096 t 10 150 a 39 12 210 g 3 15 295 O 20 413 gt 25 502 2 27 470 30 413 1 40 205 46 080 4985 0 o 0 1O 20 30 4O 50 Time ms Alternative solution The net volume ow rate is obtained from V migt111 mvout which is sketched on the figure below The volume of the airbag is given by v J39 W The results of a graphical interpretation of the volume is also given in the gure below Note that the evaluation of the above integral is simply the area under the process curve 400 6 300 5 503 200 4 414 39I 100 mg m E o E 3 gt gt 100 2 210 200 1 300 400 0 39 39 39 39 39 39 0 1o 20 30 4o 50 60 1 2 3 4 5 60 time mlllisec time millisec PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5136 5166 Outdoors air at 5 C and 95 kPa enters the building at a rate of 60 US While the indoors is maintained at 25 C The rate of sensible heat loss from the building due to in ltration is to be determined Assumptions 1 The house is maintained at 25 C at all times 2 The latent heat load is negligible 3 The in ltrating air is heated to 25 C before it eX ltrates 4 Air is an ideal gas with constant specific heats at room temperature 5 The changes in kinetic and potential energies are negligible 6 Steady ow conditions eXist Properties The gas constant of air is R 0287 kPam3kgK The speci c heat of air at room temperature is cp 1005 kJkg C Table A 2 Analysis The density of air at the outdoor conditions is P 95 kP p0 o 3 a 1235 kgm3 RTO 0287 kPam kgK 5 273 K We can View infiltration as a steady stream of air that is heated as it ows in an imaginary duct passing through the building Warm The energy balance for this imaginary steady ow system can C01 d air gt all gt Warm air be expressed in the rate form as o o 5 C 25 C 25 C E Z 70 steady O kPa in out system LS RateOfnetenergy thfer Rate of changein intemalkinetic by heat WOTkandmaSS potentialetc energies Ein Eout Qin n391h1 n39mz since Ake E Ape E 0 Qin mcpT2 Then the sensible infiltration heat load corresponding to an in ltration rate of 60 US becomes Qin ltration p0 Van39GC To 1235 kgm3 0060 m3sl005 kJkg C25 5 C 223 kW Therefore sensible heat will be lost at a rate of 223 kJs due to infiltration PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5137 5167 The average air velocity in the circular duct of an airconditioning system is not to exceed 8 ms If the fan converts 80 percent of the electrical energy into kinetic energy the size of the fan motor needed and the diameter of the main duct are to be determined Assumptions 1 This is a steady ow process since there is no change with time at any point and thus AmCV 0 and AECV 0 2 The inlet velocity is negligible V1 E 0 3 There are no heat and work interactions other than the electrical power consumed by the fan motor 4 Air is an ideal gas with constant specific heats at room temperature Properties The density of air is given to be p 120 kgm3 The constant pressure speci c heat of air at room temperature is cp 1005 kJkg C Table A2 Analysis We take the fanmotor assembly as the system This is a control volume since mass crosses the system boundary during the process We note that there is only one inlet and one eXit and thus m1 2 m2 m The change in the kinetic energy of air as it is accelerated from zero to 8 ms at a rate of 130 m3min is m 2 pt 2 120 kgm3 130 m3min 156 kgmin 26kgs AKE m 26 kgs 8 mS 0 km 2 j 00832 kW T T 2 1000 m s It is stated that this represents 80 of the electrical energy consumed by the motor Then the total electrical power consumed by the motor is determined to be 0 W 2 ME gt W AKE 00832kW 39 motor motor 08 08 0104kW 8 ms 130 m3min The diameter of the main duct is 39 3 VVAV7ZD2 4 gt DM 2 403mm lmln1mm 0587m 7rV 7r8 ms 60s Therefore the motor should have a rated power of at least 0104 kW and the diameter of the duct should be at least 587 cm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5138 5168 The maximum ow rate of a standard shower head can be reduced from 133 to 105 Lmin by switching to low ow shower heads The ratio of the hottocold water ow rates and the amount of electricity saved by a family of four per year by replacing the standard shower heads by the low ow ones are to be determined Assumptions 1 This is a steady ow process since there is no change with time at any point and thus AmCV 0 and AECV 0 2 The kinetic and potential energies are negligible ke E pe E 0 3 Heat losses from the system are negligible and thus E 0 4 There are no work interactions involved 5 Showers operate at maXimum ow conditions during the entire shower 6 Each member of the household takes a 5min shower every day 7 Water is an incompressible substance with constant properties 8 The efficiency of the electric water heater is 100 Properties The density and specific heat of water at room temperature are p 1 kgL and c 418 kJkg C Table A3 Analysis a We take the mixing chamber as the system This is a control volume since mass crosses the system boundary during the process We note that there are two inlets and one eXit The mass and energy balances for this steady ow system can be expressed in the rate form as follows Mass balance 7390 steady 39 min mout AInsystem O Ml ture mianoutamlInQZIn3 T Cold Hot Energy balance E E AE 7390 steady 0 1n out system water water W W RateOf net energy thfer Rate of changein intemalkinetic 1 2 byheat W0rkandmaSS potentialetc energies out I m1h1m2h2 n3913h3 since QEOW 0ke peEO E 2E 111 Combining the mass and energy balances and rearranging mlhl mzhz 2 11 M2 M3 72012 h3 7771013 h1 Then the ratio of the mass ow rates of the hot water to cold water becomes E 113 h1 cT3 T1 T3 T1 42 15 C 2 08 m1 112 113 CTZ T3 T2 T3 55 42 c b The low ow heads will save water at a rate of Vsaved 133 10 5 Lmin5 minperson day4 persons3 65 daysyr 20440 Lyear msaved staved 1 kgm20440 Lyear 20440 kgyear Then the energy saved per year becomes savechT 20440 kgyear4 18 kJkg C42 15 C 2307000kJyear 641 kWh since 1 kWh 3600 kJ Energy saved m Therefore switching to low ow shower heads will save about 641 kWh of electricity per year PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5139 5169 39 Problem 5168 is reconsidered The effect of the inlet temperature of cold water on the energy saved by using the low ow showerhead as the inlet temperature varies from 10 C to 20 C is to be investigated The electric energy savings is to be plotted against the water inlet temperature Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot cp 418 kJkgK density1 kgL T1 15 C T2 55 C T3 42 C Vdotold 133 Lmin Vdotnew 105 Lmin mdot11kgs quotWe can set mdot1 1 without loss of generalityquot quotAnalysisquot quota We take the mixing chamber as the system This is a control volume since mass crosses the system boundary during the process We note that there are two inlets and one exit The mass and energy balances for this steadyflow system can be expressed in the rate form as followsquot quotMass balancequot mdotin mdotout DELTAmdotsys DELTAmdotsys0 mdotin mdot1 mdot2 mdotout mdot3 quotThe ratio of the mass flow rates of the hot water to cold water is obtained by setting mdot11kgs Then mdot2 represents the ratio of mdot2mdot1quot quotEnergy balancequot Edotin Edotout DELTAEdotsys DELTAEdotsys0 Edotin mdot1h1 mdot2h2 Edotout mdot3h3 1 cpT1 cpT 2 C3 h h h T3 2 3 quotb The lowflow heads will save water at a rate of quot Vdotsaved Vdotold VdotnewquotLminquot5quotminpersondayquot4quotpersonsquot365quotdaysyearquot quotLyearquot mdotsaveddensityVdotsaved quotkgyearquot quotThen the energy saved per year becomesquot EdotsavedmdotsavedCPT3 T1quotkJyearquotconvertkJkWh quotkWhyearquot quotTherefore switching to lowflow shower heads will save about 641 kWh of electricity per year quot mratio mdot2mdot1 quotRatio of hottocold water flow ratesquot 800 Esaved kWhyear T1 C 7595 10 750 712 12 a g 700 5171 15 E 5595 18 E 650 5221 20 II 39U g 600 3 quot39 550 500 I I I I I 10 12 14 16 18 20 T1 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5140 5170 An adiabatic air compressor is powered by a directcoupled steam turbine which is also driving a generator The net power delivered to the generator is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The devices are adiabatic and thus heat transfer is negligible 4 Air is an ideal gas with variable specific heats Properties From the steam tables Tables A4 through 6 193 125 MPa h3 33436 kJkg T3 500 C and O 92 h4 hf x4hfg 1918109223921 23925 kJkg X4 2 From the air table Table A17 T1 295 K gt h1 29517 kJkg 2 125 MP3 3 1 MPa 500 C T2 620 K gt h2 62807 kJkg 620 K Analysis There is only one inlet and one eXit for either device and Air Steam thus min mom 2 m We take either the turbine or the comp turbine compressor as the system which is a control volume since mass crosses the boundary The energy balance for either steady ow system can be expressed in the rate form as 4 1 98 kPa 39 39 70 t ad Ein Eout AEsystem S e y 2 O 295 K 10 kPa Rate 0f net energy MSfer Rate of changein intemalkinetic byheat W0rkandmaSS potentialetc energies Ein Eout For the turbine and the compressor it becomes Compressor Wcomp in mairhl mairh2 gt Wcomp in mair hz 11 Turbine msteamh3 Wturb out msteamh4 Wturb out msteam h3 h4 Substituting vicompin 10 kgsX62807 295 17 kJkg 3329 kW Wmout 25 kgs33436 23925 kJkg 23777 kW Therefore Wnetout Wturbout W comp in 23777 3329 20448kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5141 5171 Helium is compressed by a compressor The power required is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Helium is an ideal gas with constant specific heats 4 The compressor is adiabatic Properties The constant pressure specific heat of helium is cp 51926 kJkgK The gas constant is R 20769 kJkgK Table A2a Analysis There is only one inlet and one eXit and thus m1 n3912 m We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O ar J r Rate 0f net energy MSfer Rate of changein intemalkinetic 43900 kPa by heat W0rkand mass potentialetc energies E E 1n out Compressor Win mhl rhhz since Ake E Ape E 0 Win mh2 h1 mcpqz T1 Helium The speci c volume of air at the inlet and the mass ow rate are 110 kpa 3 20 C V 20769 kPa m lkg K20 273 K 1 P1 110kPa 5532 m3kg m AlVl 01m29 ms V1 5532m3kg 01627 kgs Then the power input is determined from the energy balance equation to be Win mop T2 T1 01627 kgs5 1926 kJkg K200 20K 152kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5142 5172 Saturated R134a vapor is compressed to a specified state The power input is given The rate of heat transfer is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible Properties The enthalpy at the compressor eXit is given to be h2 28139 kJkg From the R134a tables Table A11 T1 10 C h1 25622 kJkg Analysis We take the compressor as the system which is a control volume since mass crosses the boundary Noting that one uid stream enters and leaves the compressor the energy balance for this steady ow system can be expressed in the rate form as 39 39 39 lt90 steady Ein Eout AEsystem O kPa EK J r J RateOfnetenel39gy thfel39 Rate of changein intemalkinetic IIIS byheats worksandmass potentialetc energies Ein Eout V2 V2 Win Qout n391 h1i n391 h2 72 since APCEO 2 5 kgs V o QoutZWin mLhr hz 2 J 10 C 2 sat vap Substituting V2 QoutZVVin m h1h272 2 1324 kW 5 kgs25622 28139kJkg 50 ms lkJkg 2 1000 m2s2 03 kW which is very small and therefore the process is nearly adiabatic PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5143 5173 A submarine that has an airballast tank originally partially lled with air is considered Air is pumped into the ballast tank until it is entirely filled with air The final temperature and mass of the air in the ballast tank are to be determined Assumptions 1 Air is an ideal gas with constant speci c heats 2 The process is adiabatic 3 There are no work interactions Properties The gas constant of air is R 0287 kPam3kgK Table Al The specific heat ratio of air at room temperature is k 14 Table AZa The speci c volume of water is taken 0001 m3kg Analysis The conservation of mass principle applied to the air gives dma 2 min dt and as applied to the water becomes dm W dt mout The first law for the ballast tank produces d d 0 2 Wu mmw hwmw mama dt dt Combining this with the conservation of mass eXpressions rearranging and canceling the common dt term produces dmua dmuw hadma hwdmw Integrating this result from the beginning to the end of the process gives ltmugt2 ltmugt1 ltmugt2 ltmugt1 haltm2 m1gt hwltm2 m1gtw Substituting the ideal gas equation of state and the speci c heat models for the enthalpies and internal energies eXpands this to Pl2 Pl1 cuT2 cT1 mwlchw cpT RT2 RT1 in PV2 Pt1 T m c RT2 RT1 1 W W When the common terms are cancelled this result becomes 2 700 v1 1 100 1 v2 v1 T1 kT 288 14293 T2 3868K 700 100 in The final mass from the ideal gas relation is P V2 1500 kPa700 m3 RT2 0287 kPam kgK3868 K m2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5 144 5174 A submarine that has an airballast tank originally partially lled with air is considered Air is pumped into the ballast tank in an isothermal manner until it is entirely lled with air The nal mass of the air in the ballast tank and the total heat transfer are to be determined Assumptions 1 Air is an ideal gas with constant speci c heats 2 There are no work interactions Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heats of air at room temperature are cp 1005 kJkgK and cu 0718 kJkgK Table A2a The specific volume of water is taken 0001 m3kg Analysis The initial air mass is P v 3 m1 1 1 1500kla100m 21814kg RT1 0287 kPa m kg K288 15 K and the initial water mass is V 3 w 1 600 n 600000 kg V1 0001 m kg and the nal mass of air in the tank is P v 1 kP 3 m2 2 2 2 Z 500 a700 m 212697kg RT2 0287 kPam3kg K288 15 K The first law when adapted to this system gives Qin mihz mehe Z mzuz mrur Qin Z mzuz mrur mehe mihi Qin mchT malcuTmwuw mwhw m2 m1cpT Noting that it hw 6298 kJkg Substituting Qin 12697 X 0718 x 288 1814 x 0718 x 288 600000 X 6298 600000 X 6298 12697 1814 x 1005 x 288 0 kJ The process is adiabatic PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5145 5175 Water is heated from 16 C to 43 C by an electric resistance heater placed in the water pipe as it ows through a showerhead steadily at a rate of 10 Lmin The electric power input to the heater and the money that will be saved during a 10min shower by installing a heat exchanger with an effectiveness of 050 are to be determined Assumptions 1 This is a steady ow process since there is no change with time at any point within the system and thus AmCV O and AECV O 2 Water is an incompressible substance with constant speci c heats 3 The kinetic and potential energy changes are negligible Ake E Ape E 0 4 Heat losses from the pipe are negligible Properties The density and speci c heat of water at room temperature are p 1 kgL and c 418 kJkg C Table A 3 Analysis We take the pipe as the system This is a control volume since mass crosses the system boundary during the process We observe that there is only one inlet and one exit and thus 51 m m Then the energy balance for this steady ow system can be expressed in the rate form as gin Eon 2 AE quotOswaldv 0 gt E sy stem J r J Rate 0f net energy thfer Rate of changein intemalkinetic by heat W0rkand mass potentia1etc energies Wain n39zh1 1139th since Ake E Ape E O Wein mm2 h1 n391cT2 T1 vP2 Pl7390 2 mar T1 where m pi 1 kgL10 Lmin 10 kgmin m WATER Substituting 16 C gt 43 C Wein 1060 kgs4 18 kJkg CX43 16 C 188 kW v The energy recovered by the heat exchanger is A Qsaved MCTmax Tmin 051060 kgs4 18 kJkg 0X39 16 C 80 kJs 80kW Therefore 80 kW less energy is needed in this case and the required electric power in this case reduces to W new Q saved 188 80 108 kW 1nnew The money saved during a 10min shower as a result of installing this heat exchanger is 80 kWX106O h115 centskWh 1 53 cents PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5146 5176 Problem 5175 is reconsidered The effect of the heat exchanger effectiveness on the money saved as the effectiveness ranges from 20 percent to 90 percent is to be investigated and the money saved is to be plotted against the effectiveness Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot density 1 kgL Vdot 10 Lmin C 418 kJkgC T1 16 C T2 43 C Tmax 39 C Tmin T1 epsilon 05 quotheat exchanger effectiveness quot EIeRate 115 centskWh quotFor entrance one exit steady flow mdotin mdotout mdotwat erquot mdotwater densityVdot convertmins quotEnergy balance for the pipequot Wdoteein mdotwaterh1mdotwaterh2 quotNeglect ke and pequot quotFor incompressible fluid in a constant pressure process the enthalpy isquot h1 CT1 h2 CT2 quotThe energy recovered by the heat exchanger isquot QdotsavedepsilonQdotmax Qdotmax mdotwaterCTmax Tmin quotTherefore 80 kW less energy is needed in this case and the required electric power in this case reduces toquot Wdoteenew Wdoteein Qdotsaved quotThe money saved during a 10min shower as a result of installing this heat exchanger isquot Costssaved QdotsavedtimeconvertminhEleRate time10 min Costssaved g 30 cents 6142 02 26 9213 03 Pg 1228 04 3 l 1536 05 g 22 1843 06 o 215 07 18 l 2457 08 393 2754 09 gt 3 14 lt 9 m l 3 1o 6 I I I I I I I 02 03 04 05 05 DJ 03 03 Heatexchangere ec veness PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5 147 5177 The air in a tank is released until the pressure in the tank reduces to a speci ed value The mass withdrawn from the tank is to be determined for three methods of analysis Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process 2 Air is an ideal gas with constant speci c heats 3 Kinetic and potential energies are negligible 4 There are no work or heat interactions involved Properties The gas constant of air is 0287 kPam3kgK Table A l The specific heats of air at room temperature are cp 1005 kJkgK and cu 0718 kJkgK Also k 14 Table A2a Analysis a We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout Am system me 2 m2 m1 me 2 m1 m2 Energy balance Alf 800 kPa Ein Eout AEsystem 25 C 1 In3 KJ W 7 Net energy thfer Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies mehe mzu2 m1u1 0 mzu2 m1u1 mehe 0 mchTz m1cUT1 mecpTe Combining the two balances O m2cUT2 m1cUTl m1 m2 cpTe The initial and nal masses are given by P v kP 1 3 m1 1 8003 ax m 9354 kg RT1 0287 kPa m kg K25 273 K RT2 0287kPam3kgKT2 T2 m sz 150kPa1m3 5226 2 The temperature of air leaving the tank changes from the initial temperature in the tank to the nal temperature during the discharging process We assume that the temperature of the air leaving the tank is the average of initial and final temperatures in the tank Substituting into the energy balance equation gives 0 mzcuT2 mlcUT1m1 m2cpTe 5226 2 0 0718T2 93540718298 9354 52T261005 2982 T2 2 whose solution is T2 1910K Substituting the nal mass is 5226 m 2736 k 2 191 g and the mass withdrawn is me 2 m1 m2 9354 2736 6618kg b Considering the process in two parts rst from 800 kPa to 400 kPa and from 400 kPa to 150 kPa the solution will be as follows From 800 kPa to 400 kPa 1021 400 kPa1 m3 1394 RT2 0287kPam3kgKT2 T2 m2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5148 298 T 0 94 OJISWZ 93540718298 9354 1941005 2 2 T2 2451K m2 5687 kg 2451 From 400 kPa to 150 kPa 2451 T 0 5sz 6 0718T2 568707182451 5687 5sz 61005 2 2 22 m2 u 2803 kg 1865 me 2 m1 m2 5687 2803 2884 kg The total mass Withdrawn is me 2 ma me 3667 2884 6551kg c The mass balance may be written as dm E Z e When this is combined with the ideal gas equation of state it becomes V d P T R dt 2 6 since the tank volume remains constant during the process An energy balance on the tank gives cu M cpT d m dt dt V dP V dPT cu cpT R dt R dt dP dP P dT cu 2 cp dt dt T dt dP dT 0p 0v p E When this result is integrated it gives k lk 0414 P T2 2T1 2 298K150kpa 1847K P1 800 kPa The final mass is sz 150 kPalm3 2 830 kg m 2 RT2 0287kPam3kgK1847K and the mass Withdrawn is me 2 m1 m2 9354 2830 6524kg Discussion The result in first method is in error by 14 While that in the second method is in error by 04 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5149 517 8 A tank initially contains saturated mixture of R134a A valve is opened and R134a vapor only is allowed to escape slowly such that temperature remains constant The heat transfer necessary with the surroundings to maintain the temperature and pressure of the R134a constant is to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the eXit remains constant 2 Kinetic and potential energies are negligible 3 There are no work interactions involved Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance min mout Amsystem me m2 m1 me m1 m2 Energy balance R134a 055 kg 1 5 L Ein Eout AES St IIl 250C ar J EH Net energy thfer Chan gein intemalkinetic by heat W0rkandmaSS potentialetc energies Qin mehe m2u2 m1u1 Qin Z m2 2 m1 1 mehe Combining the two balances Qin quot12142 quot11141 m1 m2 he The speci c volume at the initial state is l 00015m3 0002727 m3kg ml 055 kg 1 The initial state properties of R134a in the tank are u 0002727 m3kg x 2 006499 1 ufg 0030008 00008312 Table A11 V1 V f 0002727 00008312 ul 2 uf xlufg 8726 00649915689 9745 kJkg The enthalpy of saturated vapor refrigerant leaving the bottle is h hg mm 26473 kJkg e The speci c volume at the nal state is l 00015 m3 m2 015 kg 12 001m3kg The internal energy at the nal state is V2 Vf 001 00008312 O 3143 ufg 0030008 00008312 39 Table A11 u 17 xlufg 8726 0314315689 13656 kJkg 12 001 m3kg Substituting into the energy balance equation Qin Z m2u2 m1u1 m1 m2 he 2 015 kg13656 kJkg 055 kg9745 kJkg 055 015 kg26473 kJkg 728 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5150 5179 39 Steam expands in a turbine steadily The mass ow rate of the steam the eXit velocity and the power output are to be determined Assumptions 1 This is a steady ow process since there is no change with time 1 20 Mk 2 Potent1al energy changes are neg11g1ble g Properties From the steam tables Tables A4 through 6 l P1 7 MPa v1 005567 m3kg H20 3 T1 600 C h1 36506 kJkg and 102 25 kPa v2 2 VJ xzufg 000102 09562034 000102 58933 m3kg x2 095 h2 h f xzhfg 27196 09523455 25002 kJkg 2 Analysis 61 The mass ow rate of the steam is l 1 360 ms0015 m21617kgs 1 005567 m kg 9 There is only one inlet and one eXit and thus m 1512 m Then the eXit velocity is determined from 39 1617k 58 33 3k rapiva2 gt V2 quotw2 gSX 92 m g 6806ms 2 A2 014m c We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem O r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout mm V12 2 2 went Q39out mm V222 since Ape 0 VZ VZ Wout Qout mLhZ hl 1 Then the power output of the turbine is determined by substituting to be Wout 16 17 X 20 kJs 1617 kgs25002 36506 6806 ms2 60 ms2 1 kJkg 2 1000 m2s2 214560kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5151 339 Problem 5179 is reconsidered The effects of turbine eXit area and turbine eXit pressure on the eXit velocity and power output of the turbine as the eXit pressure varies from 10 kPa to 50 kPa with the same quality and the eXit area to varies from 1000 cm2 to 3000 cm2 is to be investigated The eXit velocity and the power output are to be plotted against the eXit pressure for the eXit areas of 1000 2000 and 3000 cm2 5180 0 Analysis The problem is solved using EES and the results are tabulated and plotted below FIuid39SteamIAPWS39 A1150 cm 2 T1600 C P17000 kPa Ve1 60 ms A21400 cm 2 P225 kPa qout 20 kJkg mdot A1Ve1v1convertcmquot2m 2 v1voumeFuid TT1 PP1 quotspecific volume of steam at state 1quot Vel2mdotv2A2convertcmquot2mquot2 v2voumeFuid x095 PP2 quotspecific volume of steam at state 2quot T2temperatureFuid PP2 vv2 quotCquot quotnot required but good to knowquot quotconservation of Energy for steadyflowquot quotEindot Eoutdot DeltaEdotquot quotFor steadyflow DeltaEdot 0quot DELTAEdot0 quotFor the turbine as the control volume neglecting the PE of each flow steamquot EdotinEdotout h1enthalpyFIuidTT1 PP1 Edotinmdoth1 Ve1A22Convertmquot2squot2 kJkg h2enthapyFuidx095 PP2 Edotoutmdoth2 Ve2 22ConvertmA2squot2 kJkg mdot qout Wdotout PowerWdotout Qdotoutmdotqout P2 Power Ve2 kPa kW ms 10 22158 2253 1444 1895 1595 1889 6071 1239 2333 9998 1017 2778 12212 8632 3222 13573 751 1 3667 14464 6654 4111 15075 5978 4556 15507 543 50 15821 4977 Table values are for A21000 cm 2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5152 18000 j D r F I 16000 A230Q90 F A210000nF M 14000 A220000nF zd1 a f a 12000 N 5 10000 8000 6000 r 10 15 20 25 30 35 40 45 50 P2 kPa 2200 2000 1800 1600 1400 1200 Ve2 ms A21OOO cm 000 CO CO CD 0 O A22000 cm2 400 200 A 3000 cm 10 15 20 25 30 35 40 45 50 P2 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5153 5181 Air is preheated by the exhaust gases of a gas turbine in a regenerator For a speci ed heat transfer rate the exit temperature of air and the mass ow rate of exhaust gases are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 There are no work interactions 4 Heat loss from the regenerator to the surroundings is negligible and thus heat transfer from the hot uid is equal to the heat transfer to the cold uid 5 Exhaust gases can be treated as air 6 Air is an ideal gas with variable specific heats Properties The gas constant of air is 0287 kPam3kgK Table Al The enthalpies of air are Table A17 T1 550 K gt h1 55574 kJkg T3 800 K gt lg 82195 kJkg AIR T4 600 K gt h4 60702 kJkg F 1 Exhaust 1 Analysrs a We take the cur szde of the heat exchanger as the system G ases 5 wh1ch 1s a control volume s1nce mass crosses the boundary There 1s only 0 one inlet and one exit and thus 51 n3912 m The energy balance for this 3 J steady ow system can be expressed 1n the rate form as 9 a 7I0 t d 7 4 Ein Eout AEsystem S ea y 2 0 12 RateOfnet energy thfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies 4 J Ein Eout Qin math 2 117th since W Ake E Ape E 0 Qin mairU Q h1 Substituting 3200 kJs 80060 kgsh2 55471 kJkg gt h2 79471 kJkg Then from Table A 17 we read T2 7751 K b Treating the exhaust gases as an ideal gas the mass ow rate of the exhaust gases is determined from the steady ow energy relation applied only to the exhaust gases Ein Eout mexhausthG Qout mexhausth4 Since W Aka 5 AP6 5 0 Q mexhausrhg hp 3200 kJs mexhaust82195 60702 kJkg It yields mexhaust 149 kgs PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5154 5182 Water is to be heated steadily from 20 C to 48 C by an electrical resistor inside an insulated pipe The power rating of the resistance heater and the average velocity of the water are to be determined Assumptions 1 This is a steady ow process since there is no change with time at any point Within the system and thus AmCV 0 and AECV 0 2 Water is an incompressible substance with constant speci c heats 3 The kinetic and potential energy changes are negligible Ake E Ape E 0 4 The pipe is insulated and thus the heat losses are negligible Properties The density and speci c heat of water at room temperature are p 1000 kgm3 and c 418 kJkg C Table A 3 Analysis a We take the pipe as the system This is a control volume since mass crosses the system boundary during the process Also there is only one inlet and one eXit and thus m1 n3912 m The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 lt90 steady Ein Eout AEsystem 0 WATER f r J RateOfnet energy me Rate of changein intema1kinetic 24 Lmln D 75 cm by heat workaandmass potentia1etc energies VNVV Ein Eout 39 39 39 Wain mhl mhz s1nce Qout Ake Ape 0 We W n391h2 h1 n391cT2 T1 uAPt O mcT2 T1 ein The mass ow rate of water through the pipe is m p0 1000 kgm3 o024 m3min 24 kgmin Therefore W ein n39tcT2 T1 2460 kgs418kJkg C48 20 C 468kW b The average velocity of water through the pipe is determined from V Lquot 0024 m3min 2 22543mmin A or n00375m V PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5155 5183 39 An insulated cylinder equipped with an external spring initially contains air The tank is connected to a supply line and air is allowed to enter the cylinder until its volume doubles The mass of the air that entered and the nal temperature in the cylinder are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid at the inlet remains constant 2 The eXpansion process is quasiequilibrium 3 Kinetic and potential energies are negligible 4 The spring is a linear spring 5 The device is insulated and thus heat transfer is negligible 6 Air is an ideal gas with constant specific heats Properties The gas constant of air is R 0287 kJkgK Table A 1 The specific heats of air at room temperature are cu 0718 and CI 1005 kJkgK Table A2a Also a cVT and h cpT Analysis We take the cylinder as the system which is a control volume since mass crosses the boundary Not1ng that the mrcroscop1c energ1es Spring of owing and non owing uids are represented by enthalpy h and p 7 internal energy a respectively the mass and energy balances for this uniforrn ow system can be expressed as Mass balance Air P 150 kPa ruin mout A nsystem mi n72 m1 T1 22 C 0 7 MP 3 P a Ener balance 1 011 m 1 83 T T 22 C Ein Eout AEsystem lt O lt f J Net energy traleel Changein intemalkinetic 7 r by heat workaandmass potentialetc energies mihl Wbput mguz mlul since Q E ke E pe E 0 Combining the two relations m2 m1hl Whom m2u2 m1 ul or m2 m1cpTi Wb0mmzcuT2 mlcUTl The initial and the nal masses in the tank are 2 PM 2 150 kPa011m3 RT1 0287 kPa m3kg K295 K m2 2 szz 600 kPa022 m3 2 4599 m1 01949 kg RTZ 0287kPam3kgKT2 T2 4599 2 Then from the mass balance becomes m 2 m2 m1 01949 The spring is a linear spring and thus the boundary work for this process can be determined from P P 1 W Area 212 awo2z o11m3 4125 kJ Substituting into the energy balance the final temperature of air T2 is determined to be 4125 45T9399 019491005295 45T93990718T2 019490718295 2 2 It yields T2 351 K Thus m2 4599 1309 kg T2 3514 and mi m2 m11309 01949 111 kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5156 5184 R134a is allowed to leave a pistoncylinder device with a pair of stops The work done and the heat transfer are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid leaving the device is assumed to be constant 2 Kinetic and potential energies are negligible Properties The properties of R134a at various states are Tables All through A13 v1 0032659 m3kg P 800 kP 1 a u 29086 kJkg T1 80 C h1 31699 kJkg P 2 500 kPa 12 0042115 m3kg 2 u 24242 kJkg T2 20 C h2 26348 kJkg Analysis a We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniforrn ow system can be eXpressed as Mass balance n lin mout A nsystem me 2 m1 quot72 Energy balance Ein Eout AEsystem V Net energy t1 Sfer Chan gem 1ntemalk1net1c by heat W0rkandmaSS potentialetc energies Wb iIl Qout mehe mza2 mla1 since ke E pe E 0 The volumes at the initial and nal states and the mass that has left the cylinder are V1 2 mlv1 2 kg0032659 m3kg 006532 m3 V2 mzvz 1 2m1v2 122 kg0042115 m3kg 004212 m3 me m1 m2 2 llkg The enthalpy of the refrigerant withdrawn from the cylinder is assumed to be the average of initial and nal enthalpies of the refrigerant in the cylinder he 2 1 2h1 h2 1 231699 26348 29023 kJkg Noting that the pressure remains constant after the piston starts moving the boundary work is determined from Wb in P2 1 v2 500 kPa006532 004212m3 1 1 6kJ 9 Substituting 1 16 kJ Q01t 1 kg290 23 kJkg 1 kg24242 kJkg 2 kg290 86 kJkg Qout 607 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5157 5185 Air is allowed to leave a pistoncylinder device with a pair of stops Heat is lost from the cylinder The amount of mass that has escaped and the work done are to be determined Assumptions 1 This is an unsteady process since the conditions within the device are changing during the process but it can be analyzed as a uniform ow process since the state of uid leaving the device is assumed to be constant 2 Kinetic and potential energies are negligible 3 Air is an ideal gas with constant speci c heats at the average temperature Properties The properties of air are R 0287 kPam3kgK Table A1 cu 0733 kJkgK cp 1020 kJkgK at the anticipated average temperature of 450 K Table A2b Analysis We take the tank as the system which is a control volume since mass crosses the boundary Noting that the microscopic energies of owing and non owing uids are represented by enthalpy h and internal energy a respectively the mass and energy balances for this uniform ow system can be expressed as Mass balance mm mom 2 Am me 2 m1 m2 system Energy Ein Eout AEsystem o f af J balance39 Net energy thfer Changein intemalkinetic byheataworkaandmass potentialetc energies Wb iIl Qout mehe mguz mlul since ke E pe E 0 Air r quotV 12 kg Q or Wbin Qout meije quot720uT2 mlcuTl kPa 200 C The temperature of the an w1thdrawn from the cylrnder 1s assumed to be the average of initial and final temperatures of the air in the cylinder That is Te 2 1 2T1 T2 12473 T2 The volumes and the masses at the initial and final states and the mass that has escaped from the cylinder are given by mlRTl 12 kg0287 kPam3kgK200 273 K P1 700 kPa V2 2 08011 08002327 01862 m3 PZVZ 600 kPa01862m3 38918 kg RT2 0287 kPam3kgKT2 T2 me 2 m1 m2 12 3893918kg T2 02327 m3 V1 2 Noting that the pressure remains constant after the piston starts moving the boundary work is determined from Wb ir1 P2 V1 V2 2 600 kPa02327 01862m3 279kJ Substituting 38918 2 279 k 40 kJ 12 10020 kJkgK12473 T2 1 389 8 O733 kJkgKT2 12 kg0733 kJkgK473 K 2 The final temperature may be obtained from this equation by a trialerror approach or using EES to be T2 50 K Then the amount of mass that has escaped becomes m 12 38918 0262kg 4150 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5158 5186 Geothermal water ows through a ash chamber a separator and a turbine in a geothermal power plant The temperature of the steam after the ashing process and the power output from the turbine are to be determined for different ash chamber eXit pressures Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The deVices are insulated so that there are no heat losses to the surroundings 4 Properties of steam are used for geothermal water Analysis For all components we take the steam as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as Energy balance 39 39 39 70 steady Ein Eout AEsystem O R f t tral f ateo 3 energy S er Rateof changein intemalkinetic se arator byheat workandmass potentialetc energies FlaSh p chamber 15 E steam lIl out 1 gt turbrne For each component the energy balance reduces to 230 C 2 Flash chamber h1 h2 sat liq I I I liquid Separator m2h2 m3h3 mhquidhhquid Turbine WT 13073 h4 20 kPa x095 a For a ash chamber eXit pressure of P2 1 MPa The properties of geothermal water are S h2 2 hr h 14 2 1 x2 hz f1000kPa 990 76 5 2 0113 hfg 1000kPa 20146 T2 Tsat1000kPa 179900 P3 1000 kPa 113 27771kJkg O 95 h4 hf x4hfg 25142 00523575 kJkg 24911 kJkg X4 2 The mass ow rate of vapor after the ashing process is 13 x21152 011350 kgs 5649kgs Then the power output from the turbine becomes WT 5649 kgs2777 1 24911 1616kW Repeating the similar calculations for other pressures we obtain 9 For P2 500 kPa T2 1518 C WT 2134kW c For P2 100 kPa T2 996 C WT 2333kW d For P2 50 kPa T2 81 3 c WT 2173kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5159 5187 The turbocharger of an internal combustion engine consisting of a turbine a compressor and an aftercooler is considered The temperature of the air at the compressor outlet and the minimum ow rate of ambient air are to be determined Air Assumptions 1 All processes are steady since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Air properties are used for eXhaust gases 4 Air is an ideal gas with constant speci c heats 5 The mechanical Turbine Compressor ef ciency between the turbine and the compressor is 100 6 All devices are adiabatic 7 The local atmospheric pressure is 100 kPa Exhaust Properties The constant pressure specific heats of eXhaust gases Aftercooler gt gt gases warm a1r and cold amb1ent a1r are taken to be 0 C 01d air 1063 1008 and 1005 kJkgK respectively Table A2b Analysis a An energy balance on turbine gives l WT mexhcp xh Tm1 Tam 002 kgs1063 kJkg K400 350K 1063 kW This is also the power input to the compressor since the mechanical efficiency between the turbine and the compressor is assumed to be 100 An energy balance on the compressor gives the air temperature at the compressor outlet WC 2 1an3 1063 kW 2 0018 kgs1008 kJkg KTa2 50K gtTa2 1086 C b An energy balance on the aftercooler gives the mass ow rate of cold ambient air macp Ta2 7213 mcacppa Tca2 Tca1 0018 kgs1008 kJkg OC1086 80 C 2 mm 1005 kJkg C40 30 C mm 005161 kgs The volume ow rate may be determined if we first calculate speci c volume of cold ambient air at the inlet of aftercooler That is 0287 kJkg K30 273 K v 08696 m3k ca P 100 kPa g 4 mac 005161kgs08696 m3kg 00449m3s 449 US PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5160 5188 A building is to be heated by a 30kW electric resistance heater placed in a duct inside The time it takes to raise the interior temperature from 14 C to 24 C and the average mass ow rate of air as it passes through the heater in the duct are to be determined Assumptions 1 Steady operating conditions eXist 2 Air is an ideal gas with constant specific heats at room temperature 3 Kinetic and potential energy changes are negligible 4 The heating duct is adiabatic and thus heat transfer through it is negligible 5 No air leaks in and out of the building Properties The gas constant of air is 0287 kPam3kgK Table A1 The speci c heats of air at room temperature are 0 1005 and cu 0718 kJkgK Table A2 Analysis a The total mass of air in the building is 101v1 I 95 kpa40 m3 450 kJmin RTl 0287 kPa m3kg KX287 K 4613kg T2 T1 5 C We first take the entire building as our system which is a closed system since no mass leaks in or out The time required V 400 3 j to raise the air temperature to 24 C is determined by applying P 95 1 the energy balance to this constant volume closed system a E Ein Eout AEsystem T We Netenergy transfer Chan einintemal kinetic 14 24 byheataworkaandmass potegntialetc energies C C Wein Wfanin Qout Since AKE APE Fl T1 7 250 W At Wein Wfanin Qout mcuavgT2 Solving for At gives quot10 T T 41 k 1 k 24 14 At WC 1 lt6 3 ggtlt078kJIgCgtlt gtC146S e1n Mam Qout 30 kJs 025 kJs 45060 kJs b We now take the heating duct as the system which is a control volume since mass crosses the boundary There is only one inlet and one eXit and thus 51 2 1512 m The energy balance for this adiabatic steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O r J r J Rate 0f net energy thfer Rate of changein intemalkinetic by heat workand mass potentialetc energies Ein Eout Wein Wfanin mhl th Slnce Q Ake E Ape E 0 Wein Wfanin hl mcp T2 Thus W W m e1n fan1n 0 kJS kgS cpAT 1005 kJkg C5 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 5161 5189 A water tank open to the atmosphere is initially lled with water The tank discharges to the atmosphere through a long pipe connected to a valve The initial discharge velocity from the tank and the time required to empty the tank are to be determined Assumptions 1 The ow is incompressible 2 The draining pipe is horizontal 3 The tank is considered to be empty when the water level drops to the center of the valve Analysis a Substituting the known quantities the discharge velocity can be expressed as 2 2 V gZ gz 01212gz 15 fLD 150015100m010m Then the initial discharge velocity becomes VI 01212gz1 J01212981ms2 2 m 154 ms where z is the water height relative to the center of the ori ce at that time b The ow rate of water from the tank can be obtained by multiplying the discharge velocity by the pipe crosssectional area EDZ v Apipe T 01212gz Then the amount of water that ows through the pipe during a differential time interval dt is 7102 dV th 01212gzdt 1 which from conservation of mass must be equal to the decrease in the volume of water in the tank 7202 dV Amk dz T dz 2 where dz is the change in the water level in the tank during dt Note that dz is a negative quantity since the positive direction of z is upwards Therefore we used dz to get a positive quantity for the amount of water discharged Setting Eqs 1 and 2 equal to each other and rearranging 2 2 2 2 D D D r w 01212gzdt 0 dz gt dt 0 dz 0 z zdz D2 01212gz D201212g The last relation can be integrated easily since the variables are separated Letting tfbe the discharge time and integrating it from t 0 when z zl to t if when z 0 completely drained tank gives 2 0 2 0 2 1 H D201212g zz1 f D201212g D201212g 1 z1 Simplifying and substituting the values given the draining time is determined to be 2D2 21 2 o Z1 0m 2 25940s721h t f D2 01212g 01m2 01212981ms2 Discussion The draining time can be shortened considerably by installing a pump in the pipe PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5162 5190 A uid is owing in a circular pipe A relation is to be obtained for the average uid velocity in therrns of Vr R and r Analysis Choosing a circular ring of area dA Zardr as our differential area the mass ow rate through a crosssectional area can be expressed as R dr m IpVrdA IpVr27rrdr A 0 Solving for Van 2 R VaVg FIOVrrdr 5191 Two streams of same ideal gas at different states are mixed in a mixing chamber The simplest expression for the mixture temperature in a speci ed format is to be obtained Analysis The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem 0 r J Rate of netenergy transfer r J Rate of chan gein intern alkinetic by heat workand mass potentialetc energies quot11 T1 Ein Eout gt T h 0 Mixing m3 3 m1 1 mzhz make 511103 Q W device gt quotI llcpTl mchTz m3CpT3 W and quot13 2 11 m2 Solving for final temperature we nd m1 m2 T3 T1 T2 m3 m3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5163 Fundamentals of Engineering FE Exam Problems 5192 Steam is compressed by an adiabatic compressor from 02 MPa and 150 C to 2500 MPa and 250 C at a rate of 130 kgs The power input to the compressor is a 144 kW b 234 kW c 438 kW d 717 kW e 901 kW Answer a 144 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values quotNote This compressor violates the 2nd law Changing State 2 to 800 kPa and 3500 will correct this problem it would give 511 kWquot P1 200 quotkPaquot T1 1 50 quotCquot P22500 quotkPaquot T2250 quotCquot mdot130 quotkgsquot Qdotoss0 quotkJsquot h1ENTHALPYSteamIAPWSTT1 PP1 h2ENTHALPYSteamIAPWSTT2PP2 quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot WdotinQdotossmdoth2 h1 quotSome Wrong Solutions with Common Mistakesquot W1WinQdotossh2 h1mdot quotDividing by mass flow rate instead of multiplyingquot W2WinQdotossh2 h1 quotNot considering mass flow ratequot u1NTENERGYSteamIAPWSTT1PP1 u2NTENERGYSteamIAPWSTT2PP2 W3WinQdotossmdotu2 u1 quotUsing internal energy instead of enthalpyquot W4WinQdotossu2 u1 quotUsing internal energy and ignoring mass flow ratequot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5164 5193 Steam enters a diffuser steadily at 05 MPa 300 C and 122 ms at a rate of 35 kgs The inlet area of the diffuser is a 15 cm2 b 50 cm2 c 105 cm2 d 150 cm2 e 190 cm2 Answer d 150 cm2 Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Vel1122 quotmsquot m35 quotkgsquot T1300 quotCquot P1 500 quotkPaquot quotThe rate form of energy balance is Edotin Edotout DELTAEdotcvquot v1 VOLUMESteamIAPWSTT1 PP1 m1v1AVe1 quotA in mquot2quot quotSome Wrong Solutions with Common Mistakesquot R04615 quotkJkgKquot P1 v1 ideaIRT1 273 m1v1ideaW1AVe1 quotassuming ideal gasquot P1 v2ideaRT1 m1v2ideaW2AVeI1 quotassuming ideal gas and using Cquot mW3AVeI1 quotnot using specific volumequot 5194 An adiabatic heat exchanger is used to heat cold water at 15 C entering at a rate of 5 kgs by hot air at 90 C entering also at rate of 5 kgs If the eXit temperature of hot air is 20 C the eXit temperature of cold water is a 27 C b 32 C c 52 C d 85 C e 90 C Answer b 32 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 quotkJkgCquot Cpair1005 quotkJkgCquot Tw115 quotCquot mdotw5 quotkgsquot Tair1 90 quotCquot Tair220 quotCquot mdotair5 quotkgsquot quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot mdotairCpairTair1 Tair2mdotwCwTw2Tw1 quotSome Wrong Solutions with Common Mistakesquot Tair1Tair2W1Tw2Tw1 quotEquating temperature changes of fluidsquot Cvair0718 quotkJkgKquot mdotairCvairTair1Tair2mdotwCwW2Tw2Tw1 quotUsing Cv for airquot W3Tw2Tair1 quotSetting inlet temperature of hot fluid exit temperature of cold fluidquot W4Tw2Tair2 quotSetting exit temperature of hot fluid exit temperature of cold fluidquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5165 5195 A heat exchanger is used to heat cold water at 15 C entering at a rate of 2 kgs by hot air at 85 C entering at rate of 3 kgs The heat exchanger is not insulated and is loosing heat at a rate of 25 kJs If the exit temperature of hot air is 20 C the exit temperature of cold water is a 28 C b 35 C c 38 C d 41 C e 80 C Answer b 35 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 quotkJkgCquot Cpair1005 quotkJkgCquot Tw115 quotCquot mdotw2 quotkgsquot Tair1 85 quotCquot Tair220 quotCquot mdotair3 quotkgsquot Qoss25 quotkJsquot quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot mdotairCpairTair1 Tair2mdotwCwTw2Tw1 Qoss quotSome Wrong Solutions with Common Mistakesquot mdotairCpairTair1Tair2mdotwCwW1Tw2Tw1 quotNot considering Qossquot mdotairCpairTair1Tair2mdotwCwW2Tw2Tw1Qoss quotTaking heat loss as heat gainquot Tair1Tair2W3Tw2Tw1 quotEquating temperature changes of fluidsquot Cvair0718 quotkJkgKquot mdotairCvairTair1Tair2mdotwCwW4Tw2Tw1Qoss quotUsing Cv for airquot 5196 An adiabatic heat exchanger is used to heat cold water at 15 C entering at a rate of 5 kgs by hot water at 90 C entering at rate of 4 kgs If the exit temperature of hot water is 50 C the exit temperature of cold water is a 42 C b 47 C c 55 C d 78 C e 90 C Answer b 47 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 quotkJkgCquot Tcold115 quotCquot mdotcod5 quotkgsquot Thot1 90 quotCquot Thot250 quotCquot mdothot4 quotkgsquot Qoss0 quotkJsquot quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot mdothotCwThot1 Thot2mdotcodCwTcod2Tcod1 Qoss quotSome Wrong Solutions with Common Mistakesquot Thot1Thot2W1Tcod2Tcod1 quotEquating temperature changes of fluidsquot W2Tcod290 quotTaking exit temp of cold fluidinet temp of hot fluidquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5166 5197 In a shower cold water at 10 C owing at a rate of 5 kgmin is mixed with hot water at 60 C owing at a rate of 2 kgmin The exit temperature of the mixture will be a 243 C b 350 C c 400 C d 443 C e 552 C Answer a 243 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cw418 quotkJkgCquot Tcold1 1 0 quotCquot mdotcod5 quotkgminquot Thot1 60 quotCquot mdothot2 quotkgminquot quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot mdothotCwThot1 mdotcodCwTcold1 mdothotmdotcoldCwTmix quotSome Wrong Solutions with Common Mistakesquot W1TmixTcod1Thot12 quotTaking the average temperature of inlet fluidsquot 5198 In a heating system cold outdoor air at 7 C owing at a rate of 4 kgmin is mixed adiabatically with heated air at 70 C owing at a rate of 3 kgmin The exit temperature of the mixture is a 34 C b 39 C c 45 C d 63 C e 77 C Answer a 34 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cair1005 quotkJkgCquot Tcold1 7 quotCquot mdotcod4 quotkgminquot Thot1 70 quotCquot mdothot3 quotkgminquot quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot mdothotCairThot1 mdotcodCairTcold1 mdothotmdotcoldCairTm ix quotSome Wrong Solutions with Common Mistakesquot W1TmixTcod1Thot12 quotTaking the average temperature of inlet fluidsquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5167 5199 Hot combustion gases assumed to have the properties of air at room temperature enter a gas turbine at 1 MPa and 1500 K at a rate of 01 kgs and eXit at 02 MPa and 900 K If heat is lost from the turbine to the surroundings at a rate of 15 kJs the power output of the gas turbine is a 15 kW b 30 kW c 45 kW d 60 kW e 75 kW Answer c 45 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cpair1005 quotkJkgCquot T11500 quotKquot T2900 quotKquot mdot01 quotkgsquot Qdotloss15 quotkJsquot quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot WdotoutQdotlossmdotCpairT1 T2 quotAlternative Variable specific heats using EES dataquot WdotoutvariableQdotlossmdotENTHALPYAirTT1ENTHALPYAirTT2 quotSome Wrong Solutions with Common Mistakesquot W1WoutmdotCpairT1T2 quotDisregarding heat lossquot W2WoutQdotlossmdotCpairT1T2 quotAssuming heat gain instead of lossquot 5200 Steam eXpands in a turbine from 4 MPa and 500 C to 05 MPa and 250 C at a rate of 1350 kgh Heat is lost from the turbine at a rate of 25 kJs during the process The power output of the turbine is a 157 kW b 207 kW c 182 kW d 287 kW e 246 kW Answer a 157 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values T1 500 quotCquot P1 4000 quotkPaquot T2250 quotCquot P2500 quotkPaquot mdot13503600 quotkgsquot Qdotloss25 quotkJsquot h1ENTHALPYSteamIAPWSTT1 PP1 h2ENTHALPYSteamIAPWSTT2PP2 quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot WdotoutQdotlossmdoth1 h2 quotSome Wrong Solutions with Common Mistakesquot W1Woutmdoth1h2 quotDisregarding heat lossquot W2WoutQdotlossmdoth1h2 quotAssuming heat gain instead of lossquot u1NTENERGYSteamIAPWSTT1PP1 u2NTENERGYSteamIAPWSTT2PP2 W3WoutQdotlossmdotu1u2 quotUsing internal energy instead of enthalpyquot W4WoutQdotlossmdotu1u2 quotUsing internal energy and wrong direction for heatquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5168 5201 Steam is compressed by an adiabatic compressor from 02 MPa and 150 C to 08 MPa and 350 C at a rate of 130 kgs The power input to the compressor is a 511 kW b 393 kW c 302 kW d 717 kW e 901 kW Answer a 511 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1 200 quotkPaquot T1 1 50 quotCquot P2800 quotkPaquot T2350 quotCquot mdot130 quotkgsquot Qdotoss0 quotkJsquot h1ENTHALPYSteamIAPWSTT1 PP1 h2ENTHALPYSteamIAPWSTT2PP2 quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot WdotinQdotossmdoth2 h1 quotSome Wrong Solutions with Common Mistakesquot W1WinQdotossh2 h1mdot quotDividing by mass flow rate instead of multiplyingquot W2WinQdotossh2 h1 quotNot considering mass flow ratequot u1NTENERGYSteamIAPWSTT1PP1 u2NTENERGYSteamIAPWSTT2PP2 W3WinQdotossmdotu2 u1 quotUsing internal energy instead of enthalpyquot W4WinQdotossu2 u1 quotUsing internal energy and ignoring mass flow ratequot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5169 5202 Refrigerant134a is compressed by a compressor from the saturated vapor state at 014 MPa to 09 MPa and 60 C at a rate of 0108 kgs The refrigerant is cooled at a rate of 110 kJs during compression The power input to the compressor is a 494 kW b 604 kW c 714 kW d 750 kW e 813 kW Answer c 714 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P1140quotkPaquot x11 P2900 quotkPaquot T260 quotCquot mdot01 08 quotkgsquot Qdotoss1 10 quotkJsquot h1ENTHALPYR134axx1PP1 h2ENTHALPYR134aTT2PP2 quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot WdotinQdotossmdoth2 h1 quotSome Wrong Solutions with Common Mistakesquot W1WinQdotossmdoth2 h1 quotWrong direction for heat transferquot W2Win mdoth2h1 quotNot considering heat lossquot u1NTENERGYR134axx1PP1 u2NTENERGYR134aTT2PP2 W3WinQdotossmdotu2 u1 quotUsing internal energy instead of enthalpyquot W4WinQdotossu2 u1 quotUsing internal energy and wrong direction for heat transferquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5170 5203 Refrigerant134a expands in an adiabatic turbine from 12 MPa and 100 C to 018 MPa and 50 C at a rate of 125 kgs The power output of the turbine is a 447 kW b 664 kW c 727 kW d 892 kW e 1120 kW Answer a 447 kW Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11200quotkPaquot T1 1 00 quotCquot P2180quotkPaquot T250 quotCquot mdot125 quotkgsquot Qdotloss0 quotkJsquot h1ENTHALPYR134aTT1PP1 h2ENTHALPYR134aTT2PP2 quotThe rate form of energy balance for a steadyflow system is Edotin Edotoutquot WdotoutQdotlossmdoth2 h 1 quotSome Wrong Solutions with Common Mistakesquot W1WoutQdotlossh2h1mdot quotDividing by mass flow rate instead of multiplyingquot W2WoutQdotlossh2h1 quotNot considering mass flow ratequot u1NTENERGYR134aTT1PP1 u2NTENERGYR134aTT2PP2 W3WoutQdotlossmdotu2 u1 quotUsing internal energy instead of enthalpyquot W4WoutQdotlossu2u1 quotUsing internal energy and ignoring mass flow ratequot 5204 Refrigerant134a at 14 MPa and 90 C is throttled to a pressure of 06 MPa The temperature of the refrigerant after throttling is a 22 C b 56 C c 82 C d 80 C e 900 C Answer d 80 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11400quotkPaquot T1 90 quotCquot P2600quotkPaquot h1ENTHALPYR134aTT1PP1 T2TEMPERATURER134ahh1PP2 quotSome Wrong Solutions with Common Mistakesquot W1T2T1 quotAssuming the temperature to remain constantquot W2T2TEMPERATURER134ax0PP2 quotTaking the temperature to be the saturation temperature at P2quot u1NTENERGYR134aTT1PP1 W3T2TEMPERATURER134auu1PP2 quotAssuming uconstantquot v1 VOLUMER134aTT1 PP1 W4T2TEMPERATURER134avv1PP2 quotAssuming vconstantquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5171 5205 Air at 27 C and 5 atm is throttled by a valve to 1 atm If the valve is adiabatic and the change in kinetic energy is negligible the eXit temperature of air will be a 10 C b 15 C c 20 C d 23 C e 27 C Answer e 27 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values quotThe temperature of an ideal gas remains constant during throttling and thus T2T1quot T1 27 quotCquot P1 5 quotatmquot P21 quotatmquot T2T1 quotCquot quotSome Wrong Solutions with Common Mistakesquot W1T2T1P1P2 quotAssuming vconstant and using Cquot W2T2T1273P1P2 273 quotAssuming vconstant and using Kquot W3T2T1P2P1 quotAssuming vconstant and pressures backwards and using Cquot W4T2T1273P2P1 quotAssuming vconstant and pressures backwards and using Kquot 5206 Steam at 1 MPa and 300 C is throttled adiabatically to a pressure of 04 MPa If the change in kinetic energy is negligible the specific volume of the steam after throttling will be a 0358 m3kg b 0233 m3kg c 0375 m3kg d 0646 m3kg e 0655 m3kg Answer d 0646 m3kg Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values P11000quotkPaquot T1 300 quotCquot P2400quotkPaquot h1ENTHALPYSteamIAPWSTT1 PP1 v2VOLUMESteamIAPWShh1PP2 quotSome Wrong Solutions with Common Mistakesquot W1v2VOLUMESteamIAPWSTT1PP2 quotAssuming the volume to remain constantquot u1NTENERGYSteamTT1PP1 W2v2VOLUMESteamIAPWSuu1PP2 quotAssuming uconstantquot W3v2VOLUMESteamIAPWSTT1PP2 quotAssuming Tconstantquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 5172 5207 Air is to be heated steadily by an 8kW electric resistance heater as it ows through an insulated duct If the air enters at 50 C at a rate of 2 kgs the eXit temperature of air will be a 460 C b 500 C c 540 C d 554 C e 580 C Answer c 540 C Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Cp1005 quotkJkgCquot T1 50 quotCquot mdot2 quotkgsquot Wdote8 quotkJsquot WdotemdotCpT2 T1 quotChecking using data from EES tablequot WdotemdotENTHALPYAirTT2tabIeENTHALPYAirTT1 quotSome Wrong Solutions with Common Mistakesquot Cv0718 quotkJkgKquot WdoteCpW1T2 T1 quotNot using mass flow ratequot WdotemdotCvW2T2T1 quotUsing Cvquot WdotemdotCpW3T2 quotIgnoring T1 quot 5208 5212 Design and Essay Problems PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 1 Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Qengel Michael A Boles McGrawHill 2015 Chapter 6 THE SECOND LAW OF THERMODYNAMICS PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission The Second Law of Thermodynamics and Thermal Energy Reservoirs 61C Transferring 5 kWh of heat to an electric resistance wire and producing 6 kWh of electricity 62C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity 63C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room 64C No Heat cannot ow from a lowtemperature medium to a higher temperature medium 65C A thermalenergy reservoir is a body that can supply or absorb finite quantities of heat isothermally Some examples are the oceans the lakes and the atmosphere 66C Yes Because the temperature of the oven remains constant no matter how much heat is transferred to the potatoes Heat Engines and Thermal Efficiency 67 C Heat engines are cyclic devices that receive heat from a source convert some of it to work and reject the rest to a sink 68C It is expressed as quotNo heat engine can exchange heat with a single reservoir and produce an equivalent amount of wor quot 69C No Such an engine violates the KelvinPlanck statement of the second law of thermodynamics 610C No Because 100 of the work can be converted to heat 611C a No b Yes According to the second law no heat engine can have and ef ciency of 100 612C No Such an engine violates the KelvinPlanck statement of the second law of thermodynamics 613C No The KelvinPlank limitation applies only to heat engines engines that receive heat and convert some of it to work 614C Method b With the heating element in the water heat losses to the surrounding air are minimized and thus the desired heating can be achieved with less electrical energy input PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 3 615 The rates of heat supply and heat rejection of a power plant are given The power output and the thermal efficiency of this power plant are to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are taken into consideration Analysis a The total heat rejected by this power plant is QL 1458 153 GJh Then the net power output of the plant becomes QH 280 GJh W QH QL 2230 1532127 GJh353MW netout b The thermal efficiency of the plant is determined from its definition Wnetout 0454 454 QH 280 GJh 77th 616E The power output and thermal ef ciency of a car engine are given The rate of fuel consumption is to be determined Assumptions The car operates steadily Properties The heating value of the fuel is given to be 19000 Btulbm Analysis This car engine is converting 28 of the chemical energy released during the combustion process into work The amount of energy input required to produce a power output of 110 hp is determined from the definition of thermal efficiency to be Fuel QH new 2 p u 999598 Btuh 39 77m 028 1hP 19 000 Btulbm 110 hP To supply energy at th1s rate the eng1ne must burn fuel at a rate of 5 8Btuh 28 7 msz lbmh 0 19000 Btulbm since 19000 Btu of thermal energy is released for each lbm of fuel burned 617E The rate of heat input and thermal efficiency of a heat engine are given The power output of the heat engine is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal ef ciency to the heat engine 3gtlt104 Btuh 77th 40 Wnet 77th Wnet 1 hp 04 3gtlt104 Btuh X 25445 Btuh 472 hp PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 64 618 The power output and thermal ef ciency of a heat engine are given The rate of heat input is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal ef ciency to the heat engine W 60 hp 07457 kJs 39 net QH nth 035 lhp 128kJs 619 The power output and thermal efficiency of a power plant are given The rate of heat rejection is to be determined and the result is to be compared to the actual case in practice Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis The rate of heat supply to the power plant is determined from the thermal efficiency relation Wnetout MW 77th The rate of heat transfer to the river water is determined from the first law relation nth 40 for a heat engine 600 MW Q QH Wigwamt 1500 600 900 MW 2 1500 MW QH In reality the amount of heat rejected to the river will be lower since part of the heat will be lost to the surrounding air from the working uid as it passes through the pipes and other components at 620 The work output and heat input of a heat engine are given The heat rejection is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the first law to the heat engine gives QH QL QH Wnet 700kJ 250kJ 450kJ Wnet 6W3 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 5 621 The heat rejection and thermal ef ciency of a heat engine are given The heat input to the engine is to be determined Assumptions 1 The plant operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis According to the definition of the thermal efficiency as applied to the heat engine IH wnet 77 thq H IL 4H 4L 77thQH which when rearranged gives qH qL 500 mg 909kJkg 1 77th 1 045 wnet 622 The power output and fuel consumption rate of a power plant are given The thermal efficiency is to be determined Assumptions The plant operates steadily Properties The heating value of coal is given to be 30000 kJkg 60 Uh Analysis The rate of heat supply to this power plant is QH mcoaquVcoal coal 60000 kgh30000 kJkg18gtlt109 kJh 150 MW 2 500 MW Then the thermal ef ciency of the plant becomes Wnetout MW nth 0300 300 QH 500 Mw 623 The power output and fuel consumption rate of a car engine are given The thermal ef ciency of the engine is to be determined Assumptions The car operates steadily Properties The heating value of the fuel is given to be 44000 kJkg Analysis The mass consumption rate of the fuel is me Mm 08kgLgtlt22 Lh 176kgh Fuel The rate of heat supply to the car is EEEE QH mcoaquVcoal kW 2 176 kgh44000 kJkg 774400 kJh 2151 kW Then the thermal ef ciency of the car becomes Wnetout 55 kW QH 2151kW nth 0256 256 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 66 624 The United States produces about 51 percent of its electricity from coal at a conversion efficiency of about 34 percent The amount of heat rejected by the coalfired power plants per year is to be determined Analysis Noting that the conversion ef ciency is 34 the amount of heat rejected by the coal plants per year is Wcoal Wcoal 77th Qin Qout W00a1 fffffffff W 1 IEIEIEIEIEZEIEIEZE Qout Wcoa1 1878x1012 kWh th 1878gtlt1012 kWh 034 3646x1012 kWh 1878gtlt 1012 kWh 625E The power output and thermal ef ciency of a solar pond power plant are given The rate of solar energy collection is to be determined Assumptions The plant operates steadily Analysis The rate of solar energy collection or the rate of heat supply to the power plant is determined from the thermal efficiency relation to be Wnemut 180 kW 1Btu 3600s nth 003 1055kJ 1h QH j 205x107Btuh 626 A coalbuming power plant produces 300 MW of power The amount of coal consumed during a oneday period and the rate of air owing through the furnace are to be determined Assumptions 1 The power plant operates steadily 2 The kinetic and potential energy changes are zero Properties The heating value of the coal is given to be 28000 kJkg Analysis a The rate and the amount of heat inputs to the power plant are W MW Q em 300 9375 Mw 77th 032 Qin QinAt 9375 MJs24 X 3600 s 81107 M The amount and rate of coal consumed during this period are Qin 81x 107 M qHV 28 MJkg 6 mcoal mC a1 239893 X10 kg 3348 kgs At 24 x 3600 s b Noting that the airfuel ratio is 12 the rate of air owing through the furnace is mail 2 AFn391coal 12kg airkg iel 3348 kgs 401 8kgs J 2893x106 kg mcoal PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 7 627E An OTEC power plant operates between the temperature limits of 86 F and 41 F The cooling water experiences a temperature rise of 6 F in the condenser The amount of power that can be generated by this OTEC plans is to be determined Assumptions 1 Steady operating conditions eXist 2 Water is an 1 g f incompressible substance with constant properties Properties The density and speci c heat of water are taken p 640 lbmft3 and c 10 Btulbm F respectively Analysis The mass ow rate of the cooling water is 3 74804 gal mwater rimr 6401bmit313300 galmin 1137901bmmin1897lbms The rate of heat rejection to the cooling water is Qlout mwatercagmt n 1897 lbms10Btulbm F6 F 11380 Btus Noting that the thermal efficiency of this plant is 25 the power generation is determined to be nzlzi 40025 W gt W292Btuls308 kW Qin W Qout W 11380 Btus since 1 kW 09478 Btus Refrigerators and Heat Pumps 628C The difference between the two devices is one of purpose The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium 629C The difference between the two devices is one of purpose The purpose of a refrigerator is to remove heat from a refrigerated space whereas the purpose of an airconditioner is remove heat from a living space 630C No Because the refrigerator consumes work to accomplish this task 631C No Because the heat pump consumes work to accomplish this task 632C The coef cient of performance of a refrigerator represents the amount of heat removed from the refrigerated space for each unit of work supplied It can be greater than unity 633C The coef cient of performance of a heat pump represents the amount of heat supplied to the heated space for each unit of work supplied It can be greater than unity 634C No The heat pump captures energy from a cold medium and carries it to a warm medium It does not create it PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 68 635C No The refrigerator captures energy from a cold medium and carries it to a warm medium It does not create it 636C No device can transfer heat from a cold medium to a warm medium without requiring a heat or work input from the surroundings 637 C The violation of one statement leads to the violation of the other one as shown in Sec 64 and thus we conclude that the two statements are equivalent 638 The heat removal rate of a refrigerator per kW of power it consumes is given The COP and the rate of heat rejection are to be determined Assumptions The refrigerator operates steadily Analysis The coefficient of performance of the refrigerator is determined from its definition 39 4 h 1k COPR QL 500kJ W 1kW Wnet in 1kW 3600 kJh 5040 mm The rate of heat rejection to the surrounding air per kW of power consumed is determined from the energy balance QH QL Wnetjn 5040 kJh 1 X 3600 kJh 8640 kJh Refrigerator 639 The rate of heat supply of a heat pump per kW of power it consumes is given The COP and the rate of heat absorption from the cold environment are to be determined Assumptions The heat pump operates steadily Analysis The coefficient of performance of the refrigerator is determined from its definition I House QH 8000 kJh 1 kW COP 222 HP Wmin 1 kW 3600 kJh 8000 kJh The rate of heat absorption from the surrounding air per kW of power consumed is I 1kW determined from the energy balance Q QH Wnein 8000kJh 13600 kJh 4400kJh PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 9 640E The COP and the power input of a residential heat pump are given The rate of heating effect is to be determined Reservoir QH 5 hp COP 24 Reservoir Assumptions The heat pump operates steadily Analysis Applying the definition of the heat pump coefficient of performance to this heat pump gives 3053OBtuh 25445 Btuh Q H COPHPWnet in 245 hp 641 The power input and the COP of a refrigerator are given The cooling effect of the refrigerator is to be determined Assumptions The refrigerator operates steadily Analysis Rearranging the definition of the refrigerator coefficient of performance and applying the result to this refrigerator gives 39 COP18 Q COPRWnet in 1812 kW 2 216kW Reservoir Reservoir 642 The power consumption and the cooling rate of an air conditioner are given The COP and the rate of heat rejection are to be determined Assumptions The air conditioner operates steadily Analysis a The coefficient of performance of the airconditioner or refrigerator is determined from its de nition QL 750 kJmin lkW 108 6 kW 60 kJmin Outdoors COPR 6 kW netin 5 Id 2 7 O 39 b The rate of heat discharge to the outside air is determined from the QL mm energy balance QH 2 QL Wnet in 750 kJmin 6 x 60 kJmin 1110 kJmin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 10 643 A refrigerator is used to keep a food department at a specified temperature The heat gain to the food department and the heat rejection in the condenser are given The power input and the COP are to be determined Assumptions The refrigerator operates steadily Analysis The power input is determined from Wen QL 4800 3300 1500 kJh 1500 kJh W 0417kw 3600 kJh The COP is COP a 2 M 2 W 1500 kJh 1 644 The COP and the power consumption of a refrigerator are given The time it will take to cool 5 watermelons is to be determined Assumptions 1 The refrigerator operates steadily 2 The heat gain of the refrigerator through its walls door etc is negligible 3 The watermelons are the only items in the refrigerator to be cooled Properties The specific heat of watermelons is given to be c 42 kJkg C Analysis The total amount of heat that needs to be removed from the watermelons is QL WATwaterrrBlons 5 X 39 OCX28 k The rate at which this refrigerator removes heat is QL corR Wnetjn 1 5045 kW 0675 kW 450 W That is this refrigerator can remove 0675 kJ of heat per second Thus the time COP 1 5 required to remove 4200 kJ of heat is 39 42 zgz z622232104mm Q L 0675 kJs This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air which will increase the work load Thus in reality it will take longer to cool the watermelons PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 11 645 An air conditioner with a known COP cools a house to desired temperature in 15 min The power consumption of the air conditioner is to be determined Assumptions 1 The air conditioner operates steadily 2 The house is wellsealed so that no air leaks in or out during cooling 3 Air is an ideal gas with constant speci c heats at room temperature Properties The constant volume speci c heat of air is given to be cu 072 kJkg C Analysis Since the house is wellsealed constant volume the total amount of heat that needs to be removed from the house is QL chATHouse 800 kg072 kJkg C35 20 C 8640 k Outside This heat is removed in 30 minutes Thus the average rate of heat removal from the house is QL 8640 k COP 28 48 kW QL At 30 X 60 s AC Using the definition of the coef cient of performance the power input to the air conditioner is determined to be QL 1 House 35 20 C Wu n 6 COPR 28 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 12 E 646 Problem 645 is reconsidered The rate of power drawn by the air conditioner required to cool the house as a function for air conditioner EER ratings in the range 5 to 15 is to be investigated Representative costs of air conditioning units in the BER rating range are to be included Analysis The problem is solved using EES and the results are tabulated and plotted below quotSince it is well sealed we treat the house as a closed system constant volume to determine the rate of heat transfer required to cool the house Apply the first law closed system on a rate basis to the housequot quotInput Dataquot T1 35 C T220 C cv 072 kJkgC mhouse800 kg DELTAtime30 min EER5 COPEER3412 quotAssuming no work done on the house and no heat energy added to the house in the time period with no change in KE and PE the first law applied to the house isquot Edotin Edotout DELTAEdot Edotin 0 Edotout QdotL DELTAEdot mhouseDELTAuhouseDELTAtime DELTAuhouse cvT2T1 quotUsing the definition of the coefficient of performance of the NCquot Wdotin QdotLCOP quotkJminquotconvert39kJmin3939kW39 quotkWquot QdotH Wdotinconvert39KW3939kJmin39 QdotL quotkJminquot EER win 35 kW 39 5 3276 6 273 3 7 234 8 2047 9 182 H 25 10 1638 E I 11 1489 c 12 1365 gquot 2 13 126 r 14 117 u 15 1539 LR EER PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 613 647E The rate of heat supply and the COP of a heat pump are given The power consumption and the rate of heat absorption from the outside air are to be determined Assumptions The heat pump operates steadily 60000 Analysis a The power consumed by this heat pump can be determined House Bmh from the definition of the coef cient of performance of a heat pump to be QH 60000 Btuh Wnetin COPHP 25 b The rate of heat transfer from the outdoor air is determined from the conservation of energy principle QL QH Wnen 60000 24000Btuh 36000Btuh 24000 Btuh 943 hp 648 A refrigerator is used to cool bananas to a speci ed temperature The power input is given The rate of cooling and the COP are to be determined Assumptions The refrigerator operates steadily Properties The speci c heat of banana is 335 kJkg C Analysis The rate of cooling is determined from QL mop T1 T2 21560 kgmin335 kJkg C24 13 C 132kJmin The COP is COP2ZWZL57 win 14kW 649 The rate of heat loss the rate of internal heat gain and the COP of a heat pump are given The power input to the heat pump is to be determined Assumptions The heat pump operates steadily Analysis The heating load of this heat pump system is the difference between the heat 83930 lost to the outdoors and the heat generated in the house from the people lights and House appliances QH 85000 4000 81000 kJh Using the definition of COP the power input to the heat pump is determined to be QH 81000 kJh lkW 3600 kJh W 6 COPHP 32 2 703kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 14 650E The COP and the refrigeration rate of an ice machine are given The power consumption is to be determined Assumptions The ice machine operates steadily Analysis The cooling load of this ice machine is QL qu 281bmhX169 Btulbm 4732 Btuh Using the definition of the coef cient of performance the power input to the ice machine system is determined to be 39 4 2 B 1h Wnetin QL 73 tuh 1 0775 hp 156 COPR 24 2545 Btuh Water 39 1ce 55 F Mach1ne 25 F 651 The COP and the refrigeration rate of a refrigerator are given The power consumption of the refrigerator is to be determined Assumptions The refrigerator operates steadily Analysis Since the refrigerator runs onefourth of the time and removes heat from the food compartment at an average rate of 800 kJh the refrigerator removes heat at a rate of Q 4 x 800 kJh 3200 kJh when running Thus the power the refrigerator draws when it is running is Refrigerator QL 3200 kJh 1455 kJh 040 kW W new COPR 22 652 The rate of heat loss from a house and the COP of the heat pump are given The power consumption of the heat pump when it is running is to be determined Assumptions The heat pump operates onethird of the time 22 000 kJh Analysis Since the heat pump runs onethird of the time and must supply heat to the house at an average rate of 22000 kJh the heat HOUS6 pump supplies heat at a rate of QH 3 x 22000 kJh 66000 kJh when running Thus the power the heat pump draws when it is running is QH 66000 kJh lkW W 6 COPHP 28 655 kW 3600 kJh PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 15 653E An office that is being cooled adequately by a 12000 Btuh window airconditioner is converted to a computer room The number of additional airconditioners that need to be installed is to be determined Assumptions 1 The computers are operated by 7 adult men 2 The computers consume 40 percent of their rated power at any given time Properties The average rate of heat generation from a person seated in a roomof ce is 100 W given Analysis The amount of heat dissipated by the computers is equal to the amount of electrical energy they consume Therefore Qconlputem Rated power x Usage factor 84 kW04 336 kW Q39people No of people gtlt Qpemon 7 x 100 W 700 W Qtotal QlcomputeIS Qpeople 3360 700 4060 W 13853 Btuh w since 1 W 3412 Btuh Then noting that each available air conditioner provides T 7000 Bt h 7000 Btuh cooling the number of airconditioners needed becomes u Coolingload 13853 Btuh Computer No of air cond1t10ners Coollng capac1ty of NC 7000 Btuh room 198 z 2 Air conditioners 654 A decision is to be made between a cheaper but inef cient airconditioner and an eXpensive but efficient air conditioner for a building The better buy is to be determined Assumptions The two air conditioners are comparable in all aspects other than the initial cost and the efficiency Analysis The unit that will cost less during its lifetime is a better buy The total cost of a system during its lifetime the initial operation maintenance etc can be determined by performing a life cycle cost analysis A simpler alternative is to determine the simple payback period The energy and cost savings of the more ef cient air conditioner in this case is Energy savings 2 Annual energy usage of A Annual energy usage of B 2 Annual cooling load1 COPA 1 COPB 40000 kWhyear123 136 Air Cond A 6280 kWhyear COP 23 Cost savings Energy savingsUnit cost of energy 6280 kWhyear0 10kWh 628year The installation cost difference between the two airconditioners is Cost difference Cost of B cost of A 7000 5500 1500 A ggongg Therefore the more efficient airconditioner B will pay for the 1500 cost differential in this case in about 1500628 239 years Discussion A cost conscious consumer will have no dif culty in deciding that the more eXpensive but more ef cient air conditioner B is clearly the better buy in this case since air conditioners last at least 15 years But the decision would not be so easy if the unit cost of electricity at that location was much less than 010kWh or if the annual airconditioning load of the house was much less than 40000 kWh PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 16 655 Refrigerant134a ows through the condenser of a residential heat pump unit For a given compressor power consumption the COP of the heat pump and the rate of heat absorbed from the outside air are to be determined Assumptions 1 The heat pump operates steadily 2 The Q kinetic and potential energy changes are zero 800 kPa H 800 kpa 35 C Properties The enthalpies of R134a at the condenser 950 inlet and eXit are Condenser l P1 800 kPa T1 35 C P2 800 kPa h1 27124 kJkg Expansion W valve Compressor K h2 9548 kJkg Analysis a An energy balance on the condenser gives the heat rejected in the condenser A QH 2 mm ha 2 0018 kgs27124 9548 kJkg 3164 kW K QL The COP of the heat pump is CQPZQJZMZZM W 12 kW 9 The rate of heat absorbed from the outside air QL QH Win 3164 12196kW Evaporator PerpetualMotion Machines 656C This device creates energy and thus it is a PMMl 657 C This device creates energy and thus it is a PMMl Reversible and Irreversible Processes 658C No Because it involves heat transfer through a finite temperature difference 659C This process is irreversible As the block slides down the plane two things happen a the potential energy of the block decreases and b the block and plane warm up because of the friction between them The potential energy that has been released can be stored in some form in the surroundings e g perhaps in a spring When we restore the system to its original condition we must a restore the potential energy by lifting the block back to its original elevation and b cool the block and plane back to their original temperatures The potential energy may be restored by returning the energy that was stored during the original process as the block decreased its elevation and released potential energy The portion of the surroundings in which this energy had been stored would then return to its original condition as the elevation of the block is restored to its original condition In order to cool the block and plane to their original temperatures we have to remove heat from the block and plane When this heat is transferred to the surroundings something in the surroundings has to change its state eg perhaps we warm up some water in the surroundings This change in the surroundings is permanent and cannot be undone Hence the original process is irreversible PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 17 660C The chemical reactions of combustion processes of a natural gas and air mixture will generate carbon dioxide water and other compounds and will release heat energy to a lower temperature surroundings It is unlikely that the surroundings will return this energy to the reacting system and the products of combustion react spontaneously to reproduce the natural gas and air mixture 661C Adiabatic stirring processes are irreversible because the energy stored within the system can not be spontaneously released in a manor to cause the mass of the system to turn the paddle wheel in the opposite direction to do work on the surroundings 662C When the compression process is nonquasi equilibrium the molecules before the piston face cannot escape fast enough forming a high pressure region in front of the piston It takes more work to move the piston against this high pressure region 663C When an expansion process is nonquasiequilibrium the molecules before the piston face cannot follow the piston fast enough forming a low pressure region behind the piston The lower pressure that pushes the piston produces less work 664C The irreversibilities that occur within the system boundaries are internal irreversibilities those which occur outside the system boundaries are external irreversibilities 665C A reversible expansion or compression process cannot involve unrestrained expansion or sudden compression and thus it is quasiequilibrium A quasiequilibrium expansion or compression process on the other hand may involve external irreversibilities such as heat transfer through a nite temperature difference and thus is not necessarily reversible 666C Because reversible processes can be approached in reality and they form the limiting cases Work producing devices that operate on reversible processes deliver the most work and work consuming devices that operate on reversible processes consume the least work The Carnot Cycle and Carnot39s Principle 667 C The four processes that make up the Carnot cycle are isothermal expansion reversible adiabatic expansion isothermal compression and reversible adiabatic compression 668C They are 1 the thermal efficiency of an irreversible heat engine is lower than the efficiency of a reversible heat engine operating between the same two reservoirs and 2 the thermal efficiency of all the reversible heat engines operating between the same two reservoirs are equal 669C a No b No They would violate the Carnot principle PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 618 670C False The second Carnot principle states that no heat engine cycle can have a higher thermal efficiency than the Carnot cycle operating between the same temperature limits 671C Yes The second Carnot principle states that all reversible heat engine cycles operating between the same temperature limits have the same thermal ef ciency Carnot Heat Engines 672C No 673C The one that has a source temperature of 600 C This is true because the higher the temperature at which heat is supplied to the working uid of a heat engine the higher the thermal ef ciency 674 Two pairs of thermal energy reservoirs are to be compared from a workproduction perspective Assumptions The heat engine operates steadily Analysis For the maXimum production of work a heat engine operating between the energy reservoirs would have to be completely reversible Then for the rst pair of reservoirs T 325 K Q 1 L1 o519 H 77mm TH 675 K For the second pair of reservoirs QL Wnet T 275 K 1 L1 0560 0 77mm TH 625 K The second pair is then capable of producing more work for each unit of heat extracted from the hot reservoir PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 19 675E The sink temperature of a Carnot heat engine the rate of heat rejection and the thermal efficiency are given The power output of the engine and the source temperature are to be determined Assumptions The Carnot heat engine operates steadily Analysis a The rate of heat input to this heat engine is determined from the definition of thermal ef ciency B 39 nth 1 gt 055 1 800 mmm gt QH 17778 Btumin H H Then the power output of this heat engine can be determined from Wmout 77tth 05517778 Btumin 9778 Btumin 231 hp 9 For reversible cyclic devices we have L mv TL 800 Btumin Thus the temperature of the source TH must be 39 1 39 TH Q H TL m1 520 R 11556R QL rev 800 Btumrn 676 The sink temperature of a Carnot heat engine and the rates of heat supply and heat rejection are given The source temperature and the thermal ef ciency of the engine are to be determined Assumptions The Carnot heat engine operates steadily T Analysis a For reversible cyclic devices we have 650 kJ L 6V TL Thus the temperature of the source TH must be 250 kJ QH 650k T T 297K7722K H 9L L 250le 9 The thermal ef ciency of a Carnot heat engine depends on the source and the sink temperatures only and is determined from TL 1 297K 77thc 1 0615 or 615 TH 7722 K 677 The source and sink temperatures of a Carnot heat engine and the rate of heat supply are given The thermal ef ciency and the power output are to be determined Assumptions The Carnot heat engine operates steadily Analysis a The thermal ef ciency of a Carnot heat engine depends on the source and the sink temperatures only and is determined from T K 800 kJmin nthC1 L1 300 0700r 70 TH 1000 K b The power output of this heat engine is determined from the definition of thermal ef ciency w W 2 ntth 2 070800 kJmin 560 kJmin 933 kW netout PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 620 67 8 The source and sink temperatures of a heat engine and the rate of heat supply are given The maximum possible power output of this engine is to be determined Assumptions The heat engine operates steadily Analysis The highest thermal efficiency a heat engine operating between two speci ed temperature limits can have is the Carnot ef ciency which is determined from 65000 kJmin T 298 K 1 L1 06000r600 7 77mm 77 TH 477 273 K 0 Then the maximum power output of this heat engine is determined from the definition of thermal efficiency to be W nth QH 060065000 kJmin 39000 kJmin 653 kW netout PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 21 1 39 679 Problem 678 is reconsidered The effects of the temperatures of the heat source and the heat sink on the power produced and the cycle thermal ef ciency as the source temperature varies from 300 C to 1000 C and the sink temperature varies from 0 C to 50 C are to be studied The power produced and the cycle efficiency against the source temperature for sink temperatures of 0 C 25 C and 50 C are to be plotted Analysis The problem is solved using EES and the results are tabulated and plotted below TH 477 C TL 25 C QdotH 65000 kJmin quotFirst Law applied to the heat enginequot QdotH QdotL Wdotnet 0 WdotnetKWWdotnetconvertkJminkW quotCycle Thermal Efficiency Temperatures must be absolutequot etath 1 TL 273TH 273 quotDefinition of cycle efficiencyquot etathWdotnet QdotH 850 TH WnetkW nth qw C kW 800 I It 300 5672 05236 W 400 6439 05944 750 TL0 C 500 7007 06468 600 7446 06873 E 700 700 7794 07194 5 65039 M T25 C 800 8077 07456 1 MTFWLC 900 8312 07673 g 600 1000 851 07855 AL Values forTL0 C 55 500 450 300 400 500 600 700 800 900 1000 TH C 08 075 065 T TL25 C 06 1 Ith 055 05 045 04 I 300 400 500 600 700 800 900 1000 TH C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 22 680E The claim of an inventor about the operation of a heat engine is to be evaluated Assumptions The heat engine operates steadily Analysis If this engine were completely reversible the thermal efficiency would be T 490R 1 L1 0467 77mm TH 920R QH 45 hp When the first law is applied to the engine above 15000 Bmh 25445 Btuh QH Wnet Q 45 hp 15000 Btuh 26450 Btuh The actual thermal ef ciency of the proposed heat engine is then Wnet 45hp 25445Btuh QH 26450Btuh lhp nth 0433 Since the thermal ef ciency of the proposed heat engine is smaller than that of a completely reversible heat engine which uses the same isothermal energy reservoirs the inventor39s claim is valid but not probable with these values 681 The work output and thermal ef ciency of a Carnot heat engine are given The heat supplied to the heat engine the heat rejected and the temperature of heat sink are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal ef ciency and an energy QH 77th 40 balance to the heat engine the unknown values are determined as follows 500 H W 500 kJ QL 77th QL QH Wnet 1250 500750kJ TL TL 77mm 1 gt0401 gtTL8838K611 C TH 1200 273 K 682 The source and sink temperatures of an OTEC Ocean Thermal Energy Conversion power plant are given The maximum thermal efficiency is to be determined Assumptions The power plant operates steadily Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot ef ciency which is determined from W T 276 K 1 L1 007l or 717 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 23 683 The source and sink temperatures of a geothermal power plant are given The maximum thermal efficiency is to be determined Assumptions The power plant operates steadily Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot ef ciency which is determined from W TL 1 20273K 77mm mac 1 0291 or 291 E 140273K Carnot Refrigerators and Heat Pumps 684C The difference between the temperature limits is typically much higher for a refrigerator than it is for an air conditioner The smaller the difference between the temperature limits a refrigerator operates on the higher is the COP Therefore an airconditioner should have a higher COP 685C The deep freezer should have a lower COP since it operates at a much lower temperature and in a given environment the COP decreases with decreasing refrigeration temperature 686C By increasing TL or by decreasing TH 687 C No At best when everything is reversible the increase in the work produced will be equal to the work consumed by the refrigerator In reality the work consumed by the refrigerator will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 688C No At best when everything is reversible the increase in the work produced will be equal to the work consumed by the refrigerator In reality the work consumed by the refrigerator will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant 689C Bad idea At best when everything is reversible the increase in the work produced will be equal to the work consumed by the heat pump In reality the work consumed by the heat pump will always be greater than the additional work produced resulting in a decrease in the thermal efficiency of the power plant PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 24 690 An experimentalist claims to have developed a refrigerator The experimentalist reports temperature heat transfer and work input measurements The claim is to be evaluated Analysis The highest coef cient of performance a refrigerator can have when removing heat from a cool medium at 30 C to a warmer medium at 25 C is 1 1 2 442 TH TL 1 25 273 K 30 273 K 1 COPRJnaX COPRJBV The work consumed by the actual refrigerator during this experiment is W At 2 kJs20 x 60 s 2400 k netin W netin 2 kW Then the coefficient of performance of this refrigerator becomes 30000 kJ QL 30000kJ COPR 125 w Wnet in 2400M which is above the maXimum value Therefore these measurements are not reasonable 691 The refrigerated space and the environment temperatures of a Carnot refrigerator and the power consumption are given The rate of heat removal from the refrigerated space is to be determined Assumptions The Carnot refrigerator operates steadily Analysis The coef cient of performance of a Carnot refrigerator depends on the temperature limits in the cycle only and is determined from 1 1 TH TL 1 22 273K3 273K 1 a The rate of heat removal from the refrigerated space is determined from the 2 kW definition of the coefficient of performance of a refrigerator QL COPR 39netdn 1452 kW 2 290 kW 1740 kJmin 692 The cooled space and the outdoors temperatures for a Carnot airconditioner and the rate of heat removal from the air conditioned room are given The power input required is to be determined Assumptions The airconditioner operates steadily Analysis The COP of a Carnot air conditioner or Carnot refrigerator depends on the temperature limits in the cycle only and is determined from 1 1 COP Z 270 RC TH lTL1 35 273 K24 273 K 1 The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator W QL 750kJmin new COPRmax 270 House 27 8 kJmin 0463 kW 24 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 25 693 The validity of a claim by an inventor related to the operation of a heat pump is to be evaluated Assumptions The heat pump operates steadily Analysis Applying the definition of the heat pump coef cient of performance COPHP QH 200 kW W 75 kW netin 267 The maximum COP of a heat pump operating between the same temperature limits is 1 1 1 TL TH 1 273 K293 K M Since the actual COP is less than the maXimum COP the claim is valid 694 The power input and the COP of a Carnot heat pump are given The temperature of the lowtemperature reservoir and the heating load are to be determined Assumptions The heat pump operates steadily Analysis The temperature of the lowtemperature reservoir is TH 8 7 297 K COP gt gtT 2 263K The heating load is COPHPmaX Q H gt87 187kW 215 kW 695 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat removal from the refrigerated space are given The minimum power input required is to be determined Assumptions The refrigerator operates steadily Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner The coef cient of performance of a reversible refrigerator depends on the temperature limits in the cycle only and is determined from COPR rev 2 1 1 803 TH TL 1 25 273 K 8 273 K 1 The power input to this refrigerator is determined from the de nition of a the coefficient of performance of a refrigerator QL 300 kJmin 300 kJmin Wnetin min 3736 kJmin 0623 kW COPRM 803 Q PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 626 696 An inventor claims to have developed a refrigerator The inventor reports temperature and COP measurements The claim is to be evaluated Analysis The highest coef cient of performance a refrigerator can have when w removing heat from a cool medium at 12 C to a warmer medium at 25 C is 1 1 COP COP 71 Rm Rm TH TL 1 25 273 K 12 273 K 1 lt COP 65 The COP claimed by the inventor is 65 which is below this maXimum value thus the claim is reasonable However it is not probable 697 A heat pump maintains a house at a speci ed temperature The rate of heat loss of the house and the power consumption of the heat pump are given It is to be determined if this heat pump can do the job Assumptions The heat pump operates steadily Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner The coef cient of performance of a reversible heat pump depends on the temperature limits in the cycle only and is determined from 1 1 COP 141 HR 1 TL TH 1 4 273 K25 273 K 9 4 110000 11 House The required power input to this reversible heat pump is determined from the 25 C definition of the coefficient of performance to be QH 110000kJh 1h 215 kW 3600 s 475 kW This heat pump is powerful enough since 475 kW gt 215 kW W 2 mm COPHP 1419 698E The power required by a reversible refrigerator with specified reservoir temperatures is to be determined Assumptions The refrigerator operates steadily Analysis The COP of this reversible refrigerator is TL 450R COP 2 RM TH TL 540R 450R Using this result in the coef cient of performance expression yields Wnet in Q 15 000 B uh lkW 15 000 Bmh Wnetjn L 2 t 0879kw COPRJW 5 341214 Btuh PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 27 699E The cooled space and the outdoors temperatures for an airconditioner and the power consumption are given The maximum rate of heat removal from the airconditioned space is to be determined Assumptions The airconditioner operates steadily Analysis The rate of heat removal from a house will be a maximum when the airconditioning system operates in a reversible manner The coef cient of performance of a reversible airconditioner or refrigerator depends on the temperature limits in the cycle only and is determined from 1 1 COP 296 Rm TH TL 1 90 460 R72 460 R 1 The rate of heat removal from the house is determined from the definition of the coefficient of performance of a refrigerator H 5 hp ouse 4241 B 39 o QL COPR new 2965 hplj 6277 Btumin 72 F P 6100 The power input and heat rejection of a reversed Carnot cycle are given The cooling load and the source temperature are to be determined Assumptions The refrigerator operates steadily Analysis Applying the definition of the refrigerator coefficient of performance Q QH Wmin 2000 200 1800kW Applying the definition of the heat pump coefficient of performance COPR QL 1800kW Wmin 200 kW The temperature of the heat source is determined from T R n aquot TH TL 300 TL PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 628 6101 A commercial refrigerator with R134a as the working uid is considered The condenser inlet and eXit states are specified The mass ow rate of the refrigerant the refrigeration load the COP and the minimum power input to the compressor are to be determined Assumptions 1 The refrigerator operates steadily 2 The kinetic and potential energy changes are zero Properties The properties of R134a and water are Steam and R134a tables 0 Water P1 212 MPa 26 C 18 C T 50 C h1 27828 kJkg 1 o 12 MPa QH 12 MPa T2 Tsat12MPa ATsubcoolz 46393 5 Z 413 C 5 C SUbCOOl 500C P2 2 12 MPa Condenser I h2 11019 kJkg T2 413 C Expansion Win Tw1 18 C valve th 7554 kJkg Compressor K xw1 Tw 2 26 C hw 2 10901 kJkg xw 2 Evaporator A Analysis a The rate of heat transferred to the water is the K Q L energy change of the water from inlet to eXit QH mw hw2 hwl 025 kgs10901 7554 kJkg 8367 kW The energy decrease of the refrigerant is equal to the energy increase of the water in the condenser That is Q 8367 kW QH mRh1 h2 gtmR H h 112 27828 11019kJkg 00498kgs b The refrigeration load is Q QH Win 837 330 507kW c The COP of the refrigerator is determined from its definition copz zm W 33kW 1 2154 d The COP of a reversible refrigerator operating between the same temperature limits is COPmax 1 1 2 TH TL 1 18 273 35 273 1 449 Then the minimum power input to the compressor for the same refrigeration load would be QL 507 kW VVianin COP max 113kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 29 6102 A heat pump maintains a house at a speci ed temperature in winter The maximum COPs of the heat pump for different outdoor temperatures are to be determined Analysis The coefficient of performance of a heat pump will be a maXimum when the heat pump operates in a reversible manner The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only and is determined for all three cases above to be COPHPM 1 1 293 1 TL TH 1 10 273K20 273K 1 1 COP Z 117 IIPreV 1 TL TH 1 5 273K 20 273K 1 1 COP 53986 a Ham 1 TL TH 1 30 273K20 273K 6103E A heat pump maintains a house at a speci ed temperature The rate of heat loss of the house is given The minimum power inputs required for different source temperatures are to be determined Assumptions The heat pump operates steadily Analysis a The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner If the outdoor air at 25 F is used as the heat source the COP of the heat pump and the required power input are determined to be 1 COPHP COPHP V maX ale 1 TL TH 70000 BtUh 1 1015 House 1 25 460 R78 460 R 7801 and W QH 70000 Btuh lhp 2 71h i Wm C013Pme 1015 2545 Btuh 39 p b If the wellwater at 50 F is used as the heat source the COP of the heat pump and the required power input are determined to be 25 F or 1 1 50 F COP COP Z 19392 HPmax HPrev 1 TL TH 1 50 460 R78 460 R and W QH 70000 Btuh 1 hp 1 43 h neuan CQPHPM 192 2545 Btuh 39 p PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 630 6104 A Carnot heat pump consumes 48kW of power when operating and maintains a house at a specified temperature The average rate of heat loss of the house in a particular day is given The actual running time of the heat pump that day the heating cost and the cost if resistance heating is used instead are to be determined Analysis a The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only and is determined from 1 1 COP 2 21296 HP 1 TL TH 1 2 273 K25 273 K 55000 kJh The amount of heat the house lost that day is House QH QH 1 day 55000 kJh24 h 1320000kJ 25 C Then the required work input to this Carnot heat pump is determined from the definition of the coefficient of performance to be lt 48kw QH 21320000k1210L880kJ W 6 COPHP 1296 Thus the length of time the heat pump ran that day is W At et in 2101 880kJ 21225s590h W 48kJs netin b The total heating cost that day is Cost 2 W X price 2 vimtin gtlt AtXpriee 48 kWX590 h0 11 kWh 31 1 c If resistance heating were used the entire heating load for that day would have to be met by electrical energy Therefore the heating system would consume 1320000 kJ of electricity that would cost 1 k h New Cost QH gtlt price 1320000ltJ36OVKJ0 11 kWh 403 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 3 1 6105 A Carnot heat engine is used to drive a Carnot refrigerator The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined Assumptions The heat engine and the refrigerator operate steadily Analysis a The highest thermal ef ciency a heat engine operating between two speci ed temperature limits can have is the Carnot efficiency which is determined from TL 300 K l l 0744 77thmax 77thC TH 1173 K Then the maximum power output of this heat engine is determined from the de nition of thermal ef ciency to be W 77tth 0744800 kJmin 5952 kJmin netout which is also the power input to the refrigerator Wneun The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used The COP of the Carnot refrigerator is 1 1 TH TL 1 27 273 K 5 273 K 1 Then the rate of heat removal from the refrigerated space becomes QLR COPRM XviHem 8375952 kJmin 4982 kJmin 837 C OPRJBV b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine QILHE and the heat discarded by the refrigerator QHR Q39LHE Q39HHE Wmtout 800 5952 2048 kJmin QHR Q39LR 3911th 4982 5952 55772 kJmin and Qarnbient QLHE Q39HR 2048 55772 5782 kJmin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 32 6106E A Carnot heat engine is used to drive a Carnot refrigerator The maximum rate of heat removal from the refrigerated space and the total rate of heat rejection to the ambient air are to be determined Assumptions The heat engine and the refrigerator operate steadily Analysis a The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot ef ciency which is determined from T 540 R 77thmax 77thc 1 L 1 075 TH 2160 R 700 Then the maximum power output of this heat engine is Btumin determined from the de nition of thermal ef ciency to be a Wnemut 77tth 075700 Btumin 525 Btumin which is also the power 1nput to the refr1gerator Wnet in C The rate of heat removal from the refrigerated space will be a maximum if a Carnot refrigerator is used The COP of the Carnot refrigerator is 1 1 COP Z 8390 1W TH TL l 80 460 R20 460 R 1 Then the rate of heat removal from the refrigerated space becomes Q39LR COPRMXWWH 80525 Btumin 4200 Btumin b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine QILHE and the heat discarded by the refrigerator QIHR Q39LHE Q39HHE Wmout 700 525 175 Btumin QHR QLR 39netjn 4200 525 4725 Btumin and Qambient Q39LHE Q39HR 175 4725 4900 Btumin 6107 A heat pump that consumes 4kW of power when operating maintains a house at a speci ed temperature The house is losing heat in proportion to the temperature difference between the indoors and the outdoors The lowest outdoor temperature for which this heat pump can do the job is to be determined Assumptions The heat pump operates steadily Analysis Denoting the outdoor temperature by TL the heating load of this house can be expressed as 39 3800kJhK 297 T 1056kWK 297 T K QH X L X L 3800kJhK The coef cient of performance of a Carnot heat pump depends on the temperature limits in the cycle only and can be expressed as 1 1 1 TLTH1 TL297 K or 4 kW COPHP QH 1056 kWK297 TL K 4w 0 Equating the two relations above and solving for TL we obtain TL 2635 K 95 C House 24 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 633 6108 An airconditioner with R134a as the working uid is considered The compressor inlet and eXit states are speci ed The actual and maximum COPs and the minimum volume ow rate of the refrigerant at the compressor inlet are to be determined Assumptions 1 The airconditioner operates steadily 2 The kinetic and potential energy changes are zero Properties The properties of R134a at the compressor inlet and eXit states are Tables A11 through A13 QH t P1 400 kPa h1 25561 kJkg x1 1 v1 005127 m3kg Condenser 12 MPa P2 12 MPa 70 C T2 700C h2 30063 kJkg Exlpansion Win va ve Analysis a The mass ow rate of the refrigerant and the compressor l power consumption of the compressor are 3 400 kPa 80 Lmi 1m 1mm Evaporator sat39 vap39 V1 1000 L 60 s A mR 3 002601kgs V1 005127 m kg QL Win 10 k2 bl 002601 kgs30063 25561 kJkg 1171 kW The heat gains to the room must be rejected by the airconditioner That is 1min QL Qheat Qequipment 250 kJm1n 6O 09 kW 2 5067 kW 3 Then the actual COP becomes COP2ZMZ433 Wi 1171kW b The COP of a reversible refrigerator operating between the same temperature limits is COP 1 1 max 22691 TH TL 1 34 273 23 273 1 c The minimum power input to the compressor for the same refrigeration load would be QL 5067 kW viian 201883 kW COPmax 2691 The minimum mass ow rate is VVin min mm O 1883 kW 0004182 kgs I12 h1 30063 25561 kJkg Finally the minimum volume ow rate at the compressor inlet is V 2 mm 11 0004182 kgs005127 m3kg 00002144 m3s 129 Lmin nnn1 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 34 6109 An expression for the COP of a completely reversible refrigerator in terms of the thermalenergy reservoir temperatures TL and TH is to be derived Assumptions The refrigerator operates steadily Analysis Application of the first law to the completely reversible refrigerator yields Wnetin QH QL This result may be used to reduce the coef cient of performance a 1 COPRJBV QL QL Wnetin QH QL QH IQL 1 QH Since this refrigerator is completely reversible the thermodynamic G definition of temperature tells us that Q Wnetin L 9 H TH When this is substituted into the COP eXpression the result is 1 TL COP R THTL l TH TL Special Topic Household Refrigerators 6110C Today s refrigerators are much more ef cient than those built in the past as a result of using smaller and higher efficiency motors and compressors better insulation materials larger coil surface areas and better door seals 6111C The energy consumption of a household refrigerator can be reduced by practicing good conservation measures such as 1 opening the refrigerator door the fewest times possible and for the shortest duration possible 2 cooling the hot foods to room temperature first before putting them into the refrigerator 3 cleaning the condenser coils behind the refrigerator 4 checking the door gasket for air leaks 5 avoiding unnecessarily low temperature settings 6 avoiding excessive ice buildup on the interior surfaces of the evaporator 7 using the powersaver switch that controls the heating coils that prevent condensation on the outside surfaces in humid environments and 8 not blocking the air ow passages to and from the condenser coils of the refrigerator 6112C It is important to clean the condenser coils of a household refrigerator a few times a year since the dust that collects on them serves as insulation and slows down heat transfer Also it is important not to block air ow through the condenser coils since heat is rejected through them by natural convection and blocking the air ow will interfere with this heat rejection process A refrigerator cannot work unless it can reject the waste heat 6113C It is a bad idea to overdesign the refrigeration system of a supermarket so that the entire airconditioning needs of the store can be met by refrigerated air without installing any airconditioning system This is because the refrigerators cool the air to a much lower temperature than needed for air conditioning and thus their efficiency is much lower and their operating cost is much higher 6114C It is a bad idea to meet the entire refrigeratorfreezer requirements of a store by using a large freezer that supplies suf cient cold air at 20 C instead of installing separate refrigerators and freezers This is because the freezers cool the air to a much lower temperature than needed for refrigeration and thus their ef ciency is much lower and their operating cost is much higher PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 35 6115 A refrigerator consumes 400 W when running and 170 worth of electricity per year under normal use The fraction of the time the refrigerator will run in a year is to be determined Assumptions The electricity consumed by the light bulb is negligible Analysis The total amount of electricity the refrigerator uses a year is Totalcost of energy 170year Unit cost of energy 0125kWh Total electric energy used e otal 1360 kWhyear The number of hours the refrigerator is on per year is W eitotal 1360 kWhyear 3400 hyear We 0400 kW Noting that there are 365 x 24 8760 hours in a year the fraction of the time the refrigerator is on during a year is determined to be Total operating hours 2 At Total operatrng hours 3400year 0388 Time action on Total hours per year 8760 hyear Therefore the refrigerator remained on 388 of the time 6116 The light bulb of a refrigerator is to be replaced by a 25 energy ef cient bulb that consumes less than half the electricity It is to be determined if the energy savings of the ef cient light bulb justify its cost Assumptions The new light bulb remains on the same number of hours a year Analysis The lighting energy saved a year by the energy ef cient bulb is Lighting energy saved 2 Lighting power savedOpe rating hours 2 40 18W60 hyear 1320 Wh132kWh I This means 132 kWh less heat is supplied to the refrigerated space by the light bulb which must be removed from the refrigerated space This corresponds to a refrigeration savings of Lighting energy saved 132 kWh COP 13 Then the total electrical energy and money saved by the energy efficient light bulb become Total energy saved 2 Lighting Refrigeration energy saved 2132 102 2 234 kWh year Money saved Total energy savedUnit cost of energy 234 kWh year008 kWh 019 year That is the light bulb will save only 19 cents a year in energy costs and it will take 25019 132 years for it to pay for itself from the energy it saves Therefore it is not justified in this case 102 kWh Refrigeration energy saved 2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 36 6117 A person cooks three times a week and places the food into the refrigerator before cooling it rst The amount of money this person will save a year by cooling the hot foods to room temperature before refrigerating them is to be determined Assumptions 1 The heat stored in the pan itself is negligible 2 The specific heat of the food is constant Properties The specific heat of food is c 390 kJkg C given Analysis The amount of hot food refrigerated per year is mfood 5 kgpan3 pansweek 52 weeksyear 780 kgyear The amount of energy removed from food as it is unnecessarily cooled to room temperature in the refrigerator is Energy removed Qout mfoochT 780 kgyear3 90 kJkg C95 23 C 219024 kJyear Energy removed 219024kJyear lkWh 4056 kWhyear COP 15 3600kJ Money saved 2 Energy savedUnit cost of energy 4056 kWhyear010kWh 406year Energy saved Esaved 2 Therefore cooling the food to room temperature before putting it into the refrigerator will save about four dollars a year PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 37 6118 The door of a refrigerator is opened 8 times a day and half of the cool air inside is replaced by the warmer room air The cost of the energy wasted per year as a result of opening the refrigerator door is to be determined for the cases of moist and dry air in the room Assumptions 1 The room is maintained at 20 C and 95 kPa at all times 2 Air is an ideal gas with constant specific heats at room temperature 3 The moisture is condensed at an average temperature of 4 C 4 Half of the air volume in the refrigerator is replaced by the warmer kitchen air each time the door is opened Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of air at room temperature is 0 1005 kJkg C Table A2a The heat of vaporization of water at 4 C is hfg 24914 kJkg Table A4 Analysis The volume of the refrigerated air replaced each time the refrigerator is opened is 03 m3 half of the 06 m3 air volume in the refrigerator Then the total volume of refrigerated air replaced by room air per year is v replaced 2 03 m3 20day365 daysyear 2190 m3year 31 The density of air at the refrigerated space conditions of 95 kPa and 4 C and the mass of air replaced per year are P0 95 kPa RTO 0287 kPam3kgK4 273 K I 1195 kgm3 ll 00 man pVair 1195kgm32190 m3year 2617 kgyear The amount of moisture condensed and removed by the refrigerator is m mair moisture removed per kg air 2617 kg airyear0006 kgkg air 1570 kgyear The sensible latent and total heat gains of the refrigerated space become moisture aninsensible maircp Troom Tre ig 2617 kgyear1 005 kJkg C20 4 C 42082 kJyear hfg 1570 kgyear24914kJkg 39121kJyear anintotal aninsensible aninlatent 42082 39121 2 81202 kJyear anin latent mmoistune For a COP of 14 the amount of electrical energy the refrigerator will consume to remove this heat from the refrigerated space and its cost are Electrical energy used total anin total 81 202 kJyear 1kWh COP 14 3600 k Cost of energy used total Energy usedUnit cost of energy 1611kWhyear0115kWh 1 85year If the room air is very dry and thus latent heat gain is negligible then the amount of electrical energy the refrigerator will consume to remove the sensible heat from the refrigerated space and its cost become aninsensible 42082kJyear 1kWh COP 14 3600M Cost of energy used sensible Energy usedUnit cost of energy 8 350 kWhyear0 115kWh 096year 1611kWhyear Electrical energy used sensible 8350 kWhyear PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 38 Review Problems 6119 An airconditioning system maintains a house at a specified temperature The rate of heat gain of the house the rate of internal heat generation and the COP are given The required power input is to be determined Assumptions Steady operating conditions eXist Analysis The cooling load of this airconditioning system is the sum of the heat gain from the outdoors and the heat generated in the house from the people lights and appliances QL 20000 8000 28000 kJ h COP 25 Using the de nition of the coef cient of performance the power input House to the airconditioning system is determined to be 1 Q W 6 COPR 25 QL 28000 kJh lkW 3600 kJh 2 311 kW 6120E A Carnot heat pump maintains a house at a specified temperature The rate of heat loss from the house and the outdoor temperature are given The COP and the power input are to be determined Analysis a The coef cient of performance of this Carnot heat pump depends on the temperature limits in the cycle only and is determined from C0 1 1 13 4 2500 P 0 HR 1 TL TH 1 35 460 R75 460 R Honse Btuh F b The heating load of the house is 75 F QH 2500 Btuh F75 35 F 100000 Btuh Then the required power input to this Carnot heat pump is determined a from the definition of the coef cient of performance to be Q 100000 Btuh 1 hp Wmin H 2 293 hp a COPHP 134 2545 Btuh 6121 The work output and the source and sink temperatures of a Carnot heat engine are given The heat supplied to and rejected from the heat engine are to be determined Assumptions 1 The heat engine operates steadily 2 Heat losses from the working uid at the pipes and other components are negligible Analysis Applying the definition of the thermal ef ciency and an energy QH balance to the heat engine the unknown parameters are determined as follows 500 k T 50 273 K QL 1 L1 0781 77mm TH 1200 273 K W QH quott 500kJ 640kJ 77th 0781 net PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 39 6122E The operating conditions of a heat pump are given The minimum temperature of the source that satisfies the second law of thermodynamics is to be determined Assumptions The heat pump operates steadily Analysis Applying the rst law to this heat pump gives 341214 Btuh QL QH Wnetin 32000 Btuh 18kW lkw 25860 Btuh In the reversible case we have TL amp TH QH Then the minimum temperature may be determined to be 39 2 B h TL 2TH 2 530mm 2 zen Q H 32000Btuh 6123E A refrigerator with a watercooled condenser is considered The cooling load and the COP of a refrigerator are given The power input the eXit temperature of water and the maXimum possible COP of the refrigerator are to be determined Assumptions The refrigerator operates steadily Analysis a The power input is Water 39 65 F W QL 24000Btuh 1055kJ 1h 23702kw 1 COP 19 lBtu 3600 s b The rate of heat rejected in the condenser is QH Z QL Win 2 24000 Btuh 3702 kW 13m 3600 S 1055 kJ 1 h 36632 Btuh The eXit temperature of the water is mcp 65 F 3366332 Btuh 7202 F 1451bms S 10 Btulbm F c Taking the temperature of hightemperature medium to be the average temperature of water in the condenser TL 25 460 Z 1 2 TH TL 0565 7202 25 cova PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 640 6124 A heat pump with a specified COP and power consumption is used to heat a house The time it takes for this heat pump to raise the temperature of a cold house to the desired level is to be determined Assumptions 1 Air is an ideal gas with constant specific heats at room temperature 2 The heat loss of the house during the warpup period is negligible 3 The house is wellsealed so that no air leaks in or out Properties The constant volume specific heat of air at room temperature is cV 0718 kJkg C Analysis Since the house is wellsealed constant volume the total amount of heat that needs to be supplied to the house is QH nichTh 1500 kg0718 kJkg C22 7 C 16 155 kJ ouse The rate at which this heat pump supplies heat is QH COPmWnem 285 kW 14 kW House That is this heat pump can supply 14 k of heat per second Thus the time required to supply 16155 kJ of heat is 5 kW QH 16155 kJ At 1154s192 rmn QH 14 kJs 6125 A solar pond power plant operates by absorbing heat from the hot region near the bottom and rejecting waste heat to the cold region near the top The maXimum thermal ef ciency that the power plant can have is to be determined Analysis The highest thermal efficiency a heat engine operating between two specified temperature limits can have is the Carnot ef ciency which is determined from 80 C TL 308 K 1 1 0127 or 127 7 77thmax 77thC TH 353 K In reality the temperature of the working uid must be above 35 C in the W condenser and below 80 C in the boiler to allow for any effective heat transfer Therefore the maXimum ef ciency of the actual heat engine will be w lower than the value calculated above PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 41 6126 A Carnot refrigeration cycle is executed in a closed system with a fixed mass of R134a The net work input and the ratio of maximumtominimum temperatures are given The minimum pressure in the cycle is to be determined Assumptions The refrigerator is said to operate on the reversed Carnot cycle which is totally reversible Analysis The coefficient of performance of the cycle is Also COPR 2 gt QL COPR me 522 kJ110kJ QH QL W11022132kJ and QH 132k gt V 1375kJk 11 1quot m 096kg g fgTH since the enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P Therefore TH is the temperature that corresponds to the hfg value of 1375 kJkg and is determined from the R134a tables to be TH E 613 C 2 3343 K Then T 4 K TL H 33 3 2786K 56 C 12 12 Therefore P mi 2 Psat56 C Z 356kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 42 Eli 6127 Problem 6126 is reconsidered The effect of the net work input on the minimum pressure as the work input varies from 10 k to 30 kJ is to be investigated The minimum pressure in the refrigeration cycle is to be plotted as a function of net work input Analysis The problem is solved using EES and the results are tabulated and plotted below Analysis The coefficient of performance of the cycle is given byquot mR134a 096 kg THtoTLRatio 12 quotTH 12TLquot quotWin 22 kJquot quotDepending on the value of Win adjust the guess value of THquot COPR 1 THtoTLRatio 1 QL WinCOPR quotFirst law applied to the refrigeration cycle yieldsquot QL Win QH quotSteadyflow analysis of the condenser yields mR134ah3 mR134ah4 QH QH mR134ah3h4 and hfg h3 h4 also THT3T4quot QHmR1 34ahfg hfgenthapyR134aTTHx1 enthalpyR134aTTHx0 THTHtoTLRatioTL quotThe minimum pressure is the saturation pressure corresponding to TLquot Pmin pressureR134aTTLx0convertkPaMPa TLC TL 273 VVm Pmm TH TL TLE kJ MPa K K C 10 08673 3688 3073 3432 15 06837 3589 299 2605 20 045 3427 2856 1261 25 02251 3193 2661 6907 30 006978 2871 2392 3378 Pmin MPa 1O 14 18 22 26 30 wiquot le w kJ in PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 43 6128 Two Carnot heat engines operate in series between speci ed temperature limits If the thermal efficiencies of both engines are the same the temperature of the intermediate medium between the two engines is to be determined Assumptions The engines are said to operate on the Carnot cycle which is totally reversible 0 Analysis The thermal ef ciency of the two Carnot heat engines can be eXpressed as a 77thI1 and 77thH 1 Equating o 1 l 2 6 T 6 Solving for T T JTHT J1400 K300 K 648 K 0 6129 A Carnot heat engine drives a Carnot refrigerator that removes heat from a cold medium at a speci ed rate The rate of heat supply to the heat engine and the total rate of heat rejection to the environment are to be determined Analysis a The coefficient of performance of the Carnot refrigerator is COPR C 1 1 6143 TH TL 1 300 K258 K 1 Then power input to the refrigerator becomes QL 250 kJmin Wnetin COPKC 6143 250 kJmin 407 kJmin QH R which is equal to the power output of the heat engine Wnemut The thermal ef ciency of the Carnot heat engine is determined from T 300K nthc 1 L1 06667 TH 900K Then the rate of heat input to this heat engine is determined from the definition of thermal efficiency to be Wnetput 40397 kJmin HHE nthHE 06667 39 b The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine QAHE and the heat discarded by the refrigerator QIHR QAHE Q39HHE vimmt 611 407 204 kJmin QHR Q39LR 39nem 250 407 2907 kJmin and QAmbient Q39LHE Q39HR 204 2907 311kJmin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 44 5 6130 Problem 6129 is reconsidered The effects of the heat engine source temperature the environment temperature and the cooled space temperature on the required heat supply to the heat engine and the total rate of heat rejection to the environment as the source temperature varies from 500 K to 1000 K the environment temperature varies from 275 K to 325 K and the cooled space temperature varies from 20 C to 0 C are to be investigated The required heat supply is to be plotted against the source temperature for the cooled space temperature of 15 C and environment temperatures of 275 300 and 325 K Analysis The problem is solved using EES and the results are tabulated and plotted below QdotLR 250 kJmin Tsurr 300 K TH 900 K TLC 15 C TL TLC 273 quotCoefficient of performance of the Carnot refrigeratorquot THR Tsurr COPR 1THRTL1 quotPower input to the refrigeratorquot WdotinR QdotLRCOPR quotPower output from heat engine must bequot WdotoutHE WdotinR quotThe efficiency of the heat engine isquot TLHE Tsurr etaHE 1 TLH ETH quotThe rate of heat input to the heat engine isquot QdotHHE WdotoutH EetaHE quotFirst law applied to the heat engine and refrigeratorquot QdotLHE QdotHHE WdotoutHE QdotHR QdotLR WdotinR TH QHHE qurr K kJmin kJmin 500 3661 2866 200 600 3041 2804 700 2713 2771 180 800 251 2751 160 900 2372 2737 140 1000 2272 2727 g g 120 5 100 TLC QHHE qurr I C kJmin kJmin 80 20 313 2813 6 60 18 2824 2782 16 2521 2752 40 14 2224 2722 20 12 1931 2693 0 3910 163943 266394 500 600 700 800 900 1000 8 1358 2636 TH K 6 1079 2608 4 803 258 2 5314 2553 0 2637 2526 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 645 440 420 400 380 360 340 ism kJmin 320 300 280 260 500 600 70 0 900 1000 0 80 TH K 35 25 20 DEI 15 10 2H HE kJmin 20 16 12 8 4 0 285 280 275 270 qurr N 8 250 20 16 12 8 TLc C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 46 6131 Half of the work output of a Carnot heat engine is used to drive a Carnot heat pump that is heating a house The minimum rate of heat supply to the heat engine is to be determined Assumptions Steady operating conditions eXist Analysis The coef cient of performance of the Carnot heat pump is 1 l COP Z 1475 HPC 1TLTH 12 273 K22 273 K 62000 kJh Then power input to the heat pump which is supplying heat to the house at the same rate as the rate of heat loss becomes W QH 262000kJh 6 C013leC 1475 4203 kJh which is half the power produced by the heat engine Thus the power output of the heat engine is W 2Wn netout 24203 kJh 8406 kJh etin To minimize the rate of heat supply we must use a Carnot heat engine whose thermal efficiency is determined from T 23K nthc1 L1 9 0727 TH 1073K Then the rate of heat supply to this heat engine is determined from the de nition of thermal efficiency to be WWW 8406 kJh 11560 kJh 77mm3 0727 6132E An extraordinary claim made for the performance of a refrigerator is to be evaluated Assumptions Steady operating conditions eXist Analysis The performance of this refrigerator can be evaluated by comparing it with a reversible refrigerator operating between the same temperature limits 1 l TH TL 1 85 460 40 460 1 COPRJnaX COPRIBV Discussion This is the highest COP a refrigerator can have when absorbing heat from a cool medium at 40 F and rejecting it to a warmer medium at 85 F Since the COP claimed by the inventor is above this maXimum value the claim is false PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 47 6133 The thermal efficiency and power output of a gas turbine are given The rate of fuel consumption of the gas turbine is to be determined Assumptions Steady operating conditions eXist Properties The density and heating value of the fuel are given to be 08 gcm3 and 42000 kJkg respectively Analysis This gas turbine is converting 21 of the chemical energy released during the combustion process into work The amount of energy input required to produce a power output of 6000 kW is determined from the de nition of thermal efficiency to be WWW 6000 kJs QH 0 21 28570 kJ S Combustion 77th 39 chamber To supply energy at this rate the engine must burn fuel at a rate of 28 570 kJ m S 06803 kgs 42000 kJkg since 42000 kJ of thermal energy is released for each kg of fuel burned Then the volume ow rate of the fuel becomes 0 06803 kgs 0 08kgm 0850 Ls 6134 A performance of a refrigerator declines as the temperature of the refrigerated space decreases The minimum amount of work needed to remove 1 k of heat from liquid helium at 3 K is to the determined Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner The coef cient of performance of a reversible refrigerator depends on the temperature limits in the cycle only and is determined from 1 1 COP 00101 Rm THTL 1 300 K3 K 1 The power input to this refrigerator is determined from the de nition of the coef cient of performance of a refrigerator QR 1 kJ Wnetin min 2 Z Z kJ COPKmaX 00101 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 648 6135 A Carnot heat pump cycle is executed in a steady ow system with R134a owing at a speci ed rate The net power input and the ratio of the maximumtominimum temperatures are given The ratio of the maximum to minimum pressures is to be determined Analysis The coefficient of performance of the cycle is 1 1 6O Z Z HP 1 TLTH 1 112 and QH COPHP xWin 605 kW 2 300 kJs QH 300 kJs m 022 kgs since the enthalpy of vaporization hfg at a given T or P represents the amount of heat transfer per unit mass as a substance is converted from saturated liquid to saturated vapor at that T or P Therefore TH is the temperature that corresponds to the hfg value of 13636 kJkg and is determined from the R134a tables to be TH 2 620 C 335 K qH 13636 kJkg h fgTH and Pmax sat620 C 21764 kPa T 3351K TL H 22914K2183 C 125 12 Also P 39 mr S Then the ratio of the maximum to minimum pressures in the cycle is P m 1764kPa 2325 Pmi 542kPa V PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 649 6136 Switching to energy ef cient lighting reduces the electricity consumed for lighting as well as the cooling load in summer but increases the heating load in winter It is to be determined if switching to ef cient lighting will increase or decrease the total energy cost of a building Assumptions The light escaping through the windows is negligible so that the entire lighting energy becomes part of the internal heat generation Analysis a Ef cient lighting reduces the amount of electrical energy used for lighting yeararound as well as the amount of heat generation in the house since light is eventually converted to heat As a result the electrical energy needed to air condition the house is also reduced Therefore in summer the total cost of energy use of the household definitely decreases b In winter the heating system must make up for the reduction in the heat generation due to reduced energy used for lighting The total cost of energy used in this case will still decrease if the cost of unit heat energy supplied by the heating system is less than the cost of unit energy provided by lighting The cost of 1 kWh heat supplied from lighting is 012 since all the energy consumed by lamps is eventually converted to thermal energy Noting that 1 therm 105500 k 293 kWh and the furnace is 80 efficient the cost of 1 kWh heat supplied by the heater is Cost of 1 kWh heat supplied by fumace 2 Amount of useful energy nfumace Price 1 kWhO 80140therm m 293kWh 0060 per kWh heat which is less than 012 Thus we conclude that switching to energy ef cient lighting will reduce the total energy cost of this building both in summer and in winter g Discussion To determine the amount of cost savings due to switching to energy efficient lighting consider 10 h of operation of lighting in summer and in winter for 1 kW rated power for lighting Current lighting Lighting cost Energy usedUnit cost 1 kW10 h012kWh 120 Increase in air conditioning cost Heat from lightingCOPunit cost 10 kWh35012kWh 034 Decrease in the heating cost Heat from lightingEff unit cost1008 kWh140293kWh 060 Total cost in summer 120034 154 Total cost in winter 120 060 060 Energy ef cient lighting Lighting cost Energy usedUnit cost 025 kW10 h012kWh 030 Increase in air conditioning cost Heat from lightingCOPunit cost 25 kWh35012kWh 0086 Decrease in the heating cost Heat from lightingEff unit cost2508 kWh140293kWh 015 Total cost in summer 0300086 039 Total cost in winter 030015 015 Note that during a day with 10 h of operation the total energy cost decreases from 154 to 039 in summer and from 060 to 015 in winter when efficient lighting is used PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 50 6137 A heat pump is used to heat a house The maximum money saved by using the lake water instead of outside air as the heat source is to be determined Assumptions 1 Steady operating conditions eXist 2 The kinetic and potential energy changes are zero Analysis When outside air is used as the heat source the cost of energy is calculated considering a reversible heat pump as follows 1 1 COPmax 1192 1 TL TH 1 0 273 25 273 Wmmm QH 1400003600 kW 2 3262 kW COPmax 1192 Cost air 3262 kW100 h0 105kWh 3426 Repeating calculations for lake water 1 1 COPmaX 1987 1 TL TH 1 10 27325 273 39 14 k Wm min QH 00003600 W 21957 kW COPmax 1987 Costlake 1957 kW100 h0105kWh 2055 Then the money saved becomes Money Saved 2 Costair Costlake 3426 2055 137 6138 The cargo space of a refrigerated truck is to be cooled from 25 C to an average temperature of 5 C The time it will take for an 11kW refrigeration system to precool the truck is to be determined Assumptions 1 The ambient conditions remain constant during precooling 2 The doors of the truck are tightly closed so that the in ltration heat gain is negligible 3 The air inside is sufficiently dry so that the latent heat load on the refrigeration system is negligible 4 Air is an ideal gas with constant speci c heats Properties The density of air is taken 12 kgm3 and its speci c heat at the average temperature of 15 C is cp 10 kJkg C Table A2 Analysis The mass of air in the truck is mair oairVtIuck 12 kgm312 mx 23 mgtlt 35 m 116 kg Truck The amount of heat removed as the air is cooled from 25 to 5 C Qwoungai MAT 116 kggtlt10 kJkg cgtlt25 5gt C T1 25 C T2 5 C 2320 kJ Noting that UA is given to be 80 W C and the average air temperature in the truck during precooling is 2552 15 C the average rate of heat gain by transmission is Q determined to be 2 UAAT 120 WPC25 15 C 1200 W 12kJs Qtransmissionavg Therefore the time required to cool the truck from 25 to 5 C is determined to be QnefrigAt Qcoolingair QtransmissionAt Qcoolingair 2320K 237s395min 11 12kJs gtAt nefrig Qtransmission PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 51 6139 The maximum ow rate of a standard shower head can be reduced from 133 to 105 Lmin by switching to low ow shower heads The amount of oil and money a family of four will save per year by replacing the standard shower heads by the low ow ones are to be determined Assumptions 1 Steady operating conditions eXist 2 Showers operate at maximum ow conditions during the entire Sh shower 3 Each member of the household takes a 5min ower shower every day Head 133 Lmin Properties The specific heat of water is c 418 kJkg C I and heating value of heating oil is 146300 kJgal given The density of water is p 1 kgL Analysis The low ow heads will save water at a rate of Vsaved 133 10 5 Lmin6 minperson day4 persons3 65 daysyr 24528 Lyear pvsaved 1 kgL24528 Lyear 24528 kgyear Then the energy fuel and money saved per year becomes Energy saved n39isavechT 24528 kgyear418kJkg C42 15 C 2768000kJyear Energy saved 2768000kJyear E icienc yHeating value of fuel 065146300kJgal Money saved 2 Fuel savedUnit cost of fuel 29 1galyear280 gal 815year Therefore switching to low ow shower heads will save about 80 per year in energy costs Fuel saved 2 291 galyear PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 52 6140 The maximum work that can be extracted from a pond containing 105 kg of water at 350 K when the temperature of the surroundings is 300 K is to be determined Temperature intervals of a 5 K b 2 K and c 1 K until the pond temperature drops to 300 K are to be used Analysis The problem is solved using EES and the solution is given below quotKnownsquot TL 300 K mpond 1E5 kg Cpond 418 kJkgK quotTable A3quot THhigh 350 K THIow 300 K deltaTH 1 K quotdeltaTH is the stepsize for the EES integral functionquot quotThe maximum work will be obtained if a Carnot heat pump is used The sink temperature of this heat engine will remain constant at 300 K but the source temperature will be decreasing from 350 K to 300 K Then the thermal efficiency of the Carnot heat engine operating between pond and the ambient air can be expressed asquot etathC 1 TLTH quotwhere TH is a variable The conservation of energy relation for the pond can be written in the differential form asquot deltaQpond mpondCponddeItaTH quotHeat transferred to the heat enginequot deltaQH detaQpond ntegrandWout etathCmpondCpond quotExact Solution by integration from TH 350 K to 300 Kquot Woutexact mpondCpondTHIow THhigh TLnTHIowTHhigh quotEES integral function where the stepsize is an input to the solutionquot WEES1 integrantegrandWoutTH THIow THhighdetaTH WEES2 integrantegrandWoutTH THIow THhigh2detaTH WEES5 integraintegrandWoutTH THIow THhigh5detaTH SOLUTION Cpond418 kJkgK deltaQH418000 kJ deltaQpond418000 kJ deltaTH1 K etathC0 1 429 ntegrandWout59714 kJ mpond100000 kg TH350 K THhigh350 K THIow300 K L300 K EES1 1 569E06 kJ EES21569E06 kJ EES51569E06 kJ outexact1570E06 kJ T w w w w PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 653 This problem can also be solved exactly by integration as follows The maximum work will be obtained if a Carnot heat engine is used The sink temperature of this heat engine will remain constant at 300 K but the source temperature will be decreasing from 350 K to 300 K Then the thermal efficiency of the Carnot heat engine operating between pond and the ambient air can be expressed as TL 300 77m 1 1 H TH where TH is a variable The conservation of energy relation for the pond can be written in the differential form as onnd mchH and g anond mchH 105 ng418kJkg KdTH Also aWne 2 77M aQH 1 105 ng4 18 kJkg K TH H The total work output is obtained by integration 350 350 300 5 Wnet 300nmC QH 2 I300 1 Ej o ng4 18 kJkg KdTH 350 418x105I 1 deH 157gtlt 105 kJ 300 TH which is the exact result The values obtained by computer solution will approach this value as the temperature interval is decreased PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 54 6141 A refrigeration system is to cool bread loaves at a rate of 1200 per hour by refrigerated air at 30 C The rate of heat removal from the breads the required volume ow rate of air and the size of the compressor of the refrigeration system are to be determined Assumptions 1 Steady operating conditions eXist 2 The thermal properties of the bread loaves are constant 3 The cooling section is wellinsulated so that heat gain through its walls is negligible Properties The average specific and latent heats of bread are given to be 293 kJkg C and 1093 kJkg respectively The gas constant of air is 0287 kPam3kgK Table A1 and the speci c heat of air at the average temperature of 30 222 26 C z 250 K is cp 10 kJkg C Table A2 Analysis a Noting that the breads are cooled at a rate of 500 loaves per hour breads can be considered to ow steadily through the cooling section at a mass ow rate of n39zbmad 1200 breadsh0350 kgbread 420 kgh 01167 kgs Then the rate of heat removal from the breads as they are cooled from 30 C to 10 C and frozen becomes Qbread n391cp Ahmad 420 kgh293 kJkg C30 10 C 49224 kJh Q eezing mamamd 420 kgh1093kJkg 45906kJh and Qtotal Qbmad Q eezing 49224 45906 95130 kJh b All the heat released by the breads is absorbed by the refrigerated air and the temperature rise of air is not to exceed 8 C The minimum mass ow and volume ow rates of air are determined to be Q39air 95130kJh mm 2 211891kgh cpATair 10 kJkg C8 C p i 13 3 1453 kgm3 RT 0287 kPam kgK 30 273 K V mair 11891kgh 8185mm arr pair 1453kgm3 c For a COP of 12 the size of the compressor of the refrigeration system must be Qm ig 95130 kJh Wm ig 79275 kJh 2202 kW COP 12 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 55 6142 The drinking water needs of a production facility with 20 employees is to be met by a bobbler type water fountain The size of compressor of the refrigeration system of this water cooler is to be determined Assumptions 1 Steady operating conditions eXist 2 Water is an incompressible substance with constant properties at room temperature 3 The cold water requirement is 04 Lh per person Properties The density and speci c heat of water at room temperature are p 10 kgL and c 418 kJkg CC Table A 3 Analysis The refrigeration load in this case consists of the heat gain of the reservoir and the cooling of the incoming water The water fountain must be able to provide water at a rate of mm pvwm 1 kgL04Lhperson20 persons 80 kgh Water in To cool th1s water from 22 C to 8 C heat must removed from the water at 22 C a rate of m Qcooling mcp Tout 80 kgh4 18 kJkg C22 8 C J 468kJh130W since1W36kJh i Then total refrigeration load becomes Qrefrig total Qcooling Qtransfer Q 6 Noting that the coefficient of performance of the refrigeration Refrig system is 29 the required power input is Water out L Q 175 W 8 C i W 603 W e ig COP 29 Therefore the power rating of the compressor of this refrigeration system must be at least 603 W to meet the cold water requirements of this of ce PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 656 6143 A typical heat pump powered water heater costs about 800 more to install than a typical electric water heater The number of years it will take for the heat pump water heater to pay for its cost differential from the energy it saves is to be determined Assumptions 1 The price of electricity remains constant 2 Water is Hot an incompressible substance with constant properties at room temperature 3 Time value of money interest in ation is not considered Analysis The amount of electricity used to heat the water and the net amount transferred to water are Total energy used electrical Total COSt Of energy Un1t cost of energy Cold 350year Water a 0 1 1kWh 3182 kWhyear Total energy transfer to water Em E icienc yTota1 energy used 095gtlt 3182 kWhyear 3023 kWhyear The amount of electricity consumed by the heat pump and its cost are Energy transfer to water 3023 kWhyear COPHP 33 Energy cost of heat pump Energy usageUnit cost of energy 9160 kWhyear011kWh 100 8year Then the money saved per year by the heat pump and the simple payback period become Energy usage of heat pump 9160 kWhyear Money saved Energy cost of electric heater Energy cost of heat pump 350 1008 2492 Simplepayback period Addrtronal 1nstallatron cost 800 321years Money saved 2492year Discussion The economics of heat pump water heater will be even better if the air in the house is used as the heat source for the heat pump in summer and thus also serving as an airconditioner PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 57 6144 V I Problem 6143 is reconsidered The effect of the heat pump COP on the yearly operation costs and the number of years required to break even are to be considered Analysis The problem is solved using EES and the results are tabulated and plotted below quotEnergy supplied by the water heater to the water per year is EEecHeaterquot quotCost per year to operate electric water heater for one year isquot CostEectHeater 350 year quotEnergy supplied to the water by electric heater is 90 of energy purchasedquot eta095 EEectHeater etaCostEectHeater UnitCost quotkWhyearquot UnitCost011 kWh quotFor the same amont of heated water and assuming that all the heat energy leaving the heat pump goes into the water thenquot quotEnergy supplied by heat pump heater Energy supplied by electric heaterquot EHeatPump EEectHeater quotkWhyearquot quotElectrical Work enegy supplied to heat pump Heat added to waterCOPquot COP33 WHeatPump EHeatPumpCOP quotkWhyearquot quotCost per year to operate the heat pump isquot CostHeatPumpWHeatPumpUnitCost quotLet NBrkEven be the number of years to break evenquot quotAt the break even point the total cost difference between the two water heaters is zeroquot quotYears to break even neglecting the cost to borrow the extra 800 to install heat pumpquot CostDifftota 0 CostDifftotaAddCostNBrkEvenCostHeatPumpCostEectHeater quotquot AddCost800 BBrkEven COStHeatPump COStEIektHeater years year year 2 4354 1663 350 23 3894 1446 350 26 3602 1279 350 29 3399 1147 350 32 3251 1039 350 35 3137 95 350 38 3048 875 350 41 2975 811 350 44 2915 7557 350 47 2865 7074 350 5 2822 665 350 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 658 45 4 39639 h m a a 35 IIJ E m z 3 25 2 2 5 3 35 4 4 5 5 COP 400 350 Electric 300 T 8 250 2 a 5 200 8 150 Heat pump 50 2 2 5 3 35 4 4 5 5 COP PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 59 6145 A home owner is to choose between a highefficiency natural gas furnace and a groundsource heat pump The system with the lower energy cost is to be determined Assumptions The two heaters are comparable in all aspects other than the cost of energy Analysis The unit cost of each k of useful energy supplied to the house by each system is Natural gas furnace Unit cost of useful energy 13942therm 1 them 139 gtlt106 k 097 105 500 kJ Heat Pump System Unit cost of useful energy O39115kWh 1kWh 913 gtlt106 kJ 35 3600 kJ The energy cost of groundsource heat pump system will be lower 6146 A washing machine uses 85year worth of hot water heated by an electric water heater The amount of hot water an average family uses per week is to be determined Assumptions 1 The electricity consumed by the motor of the washer is negligible 2 Water is an incompressible substance with constant properties at room temperature Properties The density and speci c heat of water at room temperature are p 10 kgL and c 418 kJkg C Table A3 Analysis The amount of electricity used to heat the water and the net amount transferred to water are Total cost of energy 85year Unit cost of energy 0113kWh Total energy transfer to water Em E icienc yTotal energy used 091gtlt 7522 kWhyear 7522 kWhyear Total energy used electrical 6845 kWhyear 6845 kWhyear 3600 1d 1 Year j lkWh 52 weeks 47390 kJweek Then the mass and the volume of hot water used per week become Em mcTout Tin gt m m 47 390 kJWCek 2637 kgweek CTout Ti 418 kJkg C55 12 C and V 263397 kgWCCK 264 Lweek water p 1 IL Therefore an average family uses 264 liters of hot water per week for washing clothes PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 60 6147 The ventilating fans of a house discharge a houseful of warmed air in one hour ACH 1 For an average outdoor temperature of 5 C during the heating season the cost of energy vented out by the fans in 1 h is to be determined Assumptions 1 Steady operating conditions eXist 2 The house is maintained at 22 C and 92 kPa at all times 3 The in ltrating air is heated to 22 C before it is vented out 4 Air is an ideal gas with constant speci c heats at room temperature 5 The volume occupied by the people furniture etc is negligible Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of air at room temperature is cp 10 kJkg C Table A 2a Analysis The density of air at the indoor conditions of 92 kPa and 22 C is P 2 kp 5 C 00 0 2 a 1087 kgm3 92 kpa 7 RT 0287 kPam kgK22 273 K Noting that the interior volume of the house is 200 x 28 560 m3 the Bathroom T T mass ow rate of air vented out becomes fan p1ir 1087 kgm3 560 m3h 6087 kgh 0169kgs Noting that the indoor air vented out at 22 C is replaced by infiltrating outdoor air at 5 C this corresponds to energy loss at a rate of Si Qlossfan 771er p Tindoors Toutdoors 0169 kgs10 kJkg C22 5 C 2874 kJs 2874 kW 22 C Then the amount and cost of the heat vented out per hour becomes 7 7 Fuel energy loss 2 Qlloss fanAt nfumace 2874 kW1h096 2994 kWh Money loss 2 Fuel energy lossUnit cost of energy 1 therm 29 3 kWh Discussion Note that the energy and money loss associated with ventilating fans can be very signi cant Therefore ventilating fans should be used sparingly 2994 kWh120therm 0123 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 61 6148 The ventilating fans of a house discharge a houseful of airconditioned air in one hour ACH 1 For an average outdoor temperature of 28 C during the cooling season the cost of energy vented out by the fans in 1 h is to be determined Assumptions 1 Steady operating conditions eXist 2 The house is maintained at 22 C and 92 kPa at all times 3 The in ltrating air is cooled to 22 C before it is vented out 4 Air is an ideal gas with constant speci c heats at room temperature 5 The volume occupied by the people furniture etc is negligible 6 Latent heat load is negligible Properties The gas constant of air is R 0287 kPam3kgK Table A1 The specific heat of air at room temperature is cp 10 kJkg C Table A 2a Analysis The density of air at the indoor conditions of 92 kPa and 22 C is 33 C P p0 0 92 kpa 1087 kgm3 92 kPa 7 RTO 0287 kPam3kgK22 273 K Noting that the interior volume of the house is 200 x 28 560 m3 the mass Bathroom T T ow rate of air vented out becomes fan pVair 1087 kgm3 560 m3h 6087 kgh 01690kgs Noting that the indoor air vented out at 22 C is replaced by infiltrating outdoor air at 33 C this corresponds to energy loss at a rate of Qlossfan 772er p Toutdoors Tindoors 01690 kgs10 kJkg C33 22 C 1859 kJs 1859 kW 22 C Then the amount and cost of the electric energy vented out per hour becomes 7 7 Electric energy loss 2 Q1088 fan At COP 1859 kW1h21 0 8854 kWh Money loss 2 Fuel energy lossUnit cost of energy 08854 kWh012 kWh 0106 Discussion Note that the energy and money loss associated with ventilating fans can be very signi cant Therefore ventilating fans should be used sparingly PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 6 62 6149 A geothermal heat pump with R134a as the working uid is considered The evaporator inlet and eXit states are specified The mass ow rate of the refrigerant the heating load the COP and the minimum power input to the compressor are to be determined Assumptions 1 The heat pump operates steadily 2 The kinetic and potential energy changes are zero 3 Steam properties are used for geothermal water Properties The properties of R134a and water are t QH Steam and R134a tables Condenser T1 12 C h1 9654 kJkg x1 015 P1 4433 kPa Expansmn W P2 2 P1 2 kPa valve In 1 h2 25733 kJkg Compressor K xz Tw1 60 C wa th VH Evapirator p Tw 2 40 C 12 C A Sat vap hw 2 16753 kJkg X 015 QL xw2 0 Geo water Analysis a The rate of heat transferred from the water is the 600C 40 C energy change of the water from inlet to eXit QL 141me hwg 0065 kgs25118 16753 kJkg 5437 kW The energy increase of the refrigerant is equal to the energy decrease of the water in the evaporator That is QL 5437 kW 39 m h h gtm 00338k ls QL R 2 1 R h h 25733 9654 kJkg g b The heating load is QH Q Win 5437 16 704kW c The COP of the heat pump is determined from its definition CQPZQJZM4AO Wi 16 kW d The COP of a reversible heat pump operating between the same temperature limits is COP 1 1 max 951 1 TL TH 1 25 273 60 273 Then the minimum power input to the compressor for the same refrigeration load would be W I QH 2704kW mm COP 951 max 0740kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 663 6150 A heat pump is used as the heat source for a water heater The rate of heat supplied to the water and the minimum power supplied to the heat pump are to be determined Assumptions 1 Steady operating conditions eXist 2 The kinetic and potential energy changes are zero Properties The speci c heat and specific volume of water at room temperature are cp 418 kJkgK and v0001 m3kg Table A3 Analysis a An energy balance on the water heater gives the rate of heat supplied to the water QH meT2 T1 V CpT2 T1 v 00260 3 418 kJkg C50 10 C 0001m kg 5573kW b The COP of a reversible heat pump operating between the specified temperature limits is 1 TL TH 1 0 273 30 273 Surroundings WC COPmaX Then the minimum power input would be QH 5573 kW VVianin COP 101 max 552kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 64 6151 A heat pump receiving heat from a lake is used to heat a house The minimum power supplied to the heat pump and the mass ow rate of lake water are to be determined Assumptions 1 Steady operating conditions eXist 2 The kinetic and potential energy changes are zero Properties The speci c heat of water at room temperature is cp 418 kJkgK Table A3 Luke water inlet Lake water pump Lake water n tei HP heat quot quotquot339 eeehunger quotLukegi c39 He nee water 1 min e111 Analysis a The COP of a reversible heat pump operating between the specified temperature limits is COP 1 1 max 1 TLTH 1 627323273 1741 Then the minimum power input would be W QH 520003600kW1444kW mm COP 1741 1741 max 9 The rate of heat absorbed from the lake is Q QH Wimn 1444 0830 1361kW 0830kW An energy balance on the heat exchanger gives the mass ow rate of lake water mwater QL 1361kJS CpAT 418 kJkg C5 C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 65 6152 It is to be proven that the COP of all completely reversible refrigerators must be the same when the reservoir temperatures are the same Assumptions The refrigerators operate steadily Analysis We begin by assuming that COPA lt COPB When this is the case a rearrangement of the coef cient of performance eXpression yields QL QL gt B A TH That is the magnitude of the work required to drive QH A QH B refrigerator A is greater than that needed to drive refrigerator B Applying the rst law to both refrigerators yields a A QHA gt QHB QL QL since the work supplied to refrigerator A is greater than that supplied to refrigerator B and both have the same cooling effect QL Since A is a completely reversible refrigerator we can reverse it without changing the magnitude of the heat and work transfers This is illustrated in the figure below The heat QL which is rejected by the reversed refrigerator A can now be routed directly to refrigerator B The net effect when this is done is that no heat is exchanged with the TL reservoir The magnitude of the WA WB heat supplied to the reversed refrigerator A QHA has been shown to be larger than that rejected by refrigerator B There is then a net heat transfer from the TH reservoir to the combined device in the dashed lines of the gure whose magnitude is given by QHA QHB Similarly there is a net work production by the combined device whose magnitude is given by WA WB TL The combined cyclic device then exchanges heat with a reservoir at a single temperature and produces work which is clearly a violation of the KelvinPlanck statement of the second law Our assumption the COPA lt COPB must then be wrong If we interchange A and B in the previous argument we would conclude that the COP 3 cannot be less than COPA The only alternative left is that COPA COPB 6153 A Carnot heat engine is operating between speci ed temperature limits The source temperature that will double the efficiency is to be determined Analysis Denoting the new source temperature by TH the thermal ef ciency of the Carnot heat engine for both cases can be eXpressed as T T l L and 1 2 77thc TH 77thc TH 77thc Substituting 0 0 1 TL 2 T L T TH 77th 277th Solving for TH T which is the desired relation PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 666 6154 A Carnot cycle is analyzed for the case of temperature differences in the boiler and condenser The ratio of overall temperatures for which the power output will be maximum and an eXpression for the maXimum net power output are to be determined Analysis It is given that MltTT v Therefore W nth l hAHTH 4 1 hAH TH H H TH 0139 T39H lt L 1 TH1rx 1 hAH TH TH TH where we defined r and x as r TLTH and x 1 THTH W For a reversible cycle we also have gzgi1hAHTH TZ hAHrH1 TTH TL QL r hAL T TL hAL TH T TH TL TH but TLTLTH1 T TH T TH Substituting into above relation yields 1 H x 7 hALr1xTL lTH Solving for X r TLTH hAH hAL 1 Substitute 2 into 1 x 739 r TLTH W hAgtHTH1 rrhAHhAL1 6W Taking the partial derivative a holding everything else constant and sett1ng 1t equal to zero g1ves r 4 TH TH which is the desired relation The maXimum net power output in this case is determined by substituting 4 into 3 It simpli es to 6155 It is to be shown that COPHP COPR 1 for the same temperature and heat transfer terms Analysis Using the definitions of COPs the desired relation is obtained to be W QH QL net1n QL 1 2 1 Wnetin Wnetin netin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 67 Fundamentals of Engineering FE Exam Problems 6156 A 24m high 200m2 house is maintained at 22 C by an airconditioning system whose COP is 32 It is estimated that the kitchen bath and other ventilating fans of the house discharge a houseful of conditioned air once every hour If the average outdoor temperature is 32 C the density of air is 120 kgm3 and the unit cost of electricity is 010kWh the amount of money vented out by the fans in 10 hours is a 050 b 160 c 500 cl 1100 e 1600 Answer a 050 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP32 T1 22 quotCquot T232 quotCquot Price010 quotkWhquot Cp1005 quotkJkgCquot rho120 quotkgmquot3quot V24200 quotmA3quot mrhoV mtotam1 0 EinmtotaCpT2T1COP quotkJquot CostEin3600Price quotSome Wrong Solutions with Common Mistakesquot W1CostPrice3600mtotaCpT2T1COP quotMultiplying by Eff instead of dividingquot W2CostPrice3600mtotaCpT2T1 quotIgnoring efficiencyquot W3CostPrice3600mCpT2T1COP quotUsing m instead of mtotalquot W4CostPrice3600mtotaICpT2T1COP quotAdding temperaturesquot 6157 The drinking water needs of an of ce are met by cooling tab water in a refrigerated water fountain from 23 C to 6 C at an average rate of 10 kgh If the COP of this refrigerator is 31 the required power input to this refrigerator is a 197 w b 612 w c 64 w d 109 w e 403 w Answer c 64 W Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP31 Cp418 quotkJkgCquot T1 23 quotCquot T26 quotCquot mdot103600 quotkgsquot QLmdotCpT1T2 quotkWquot WinQL1000COP quotWquot quotSome Wrong Solutions with Common Mistakesquot W1WinmdotCpT1T2 1000COP quotMultiplying by COP instead of dividingquot W2WinmdotCpT1T2 1000 quotNot using COPquot W3WinmdotT1T2 1000COP quotNot using specific heatquot W4WinmdotCpT1T2 1000COP quotAdding temperaturesquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 668 6158 The label on a washing machine indicates that the washer will use 85 worth of hot water if the water is heated by a 90 ef ciency electric heater at an electricity rate of 009kWh If the water is heated from 18 C to 45 C the amount of hot water an average family uses per year in metric tons is a 116 tons b 158 tons c 271 tons d 301 tons e 335 tons Answer b 271 tons Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Eff090 C418 quotkJkgCquot Hmur T245quotCquot Cost85 quotsquot Price009 quotkWhquot Enercosvpncer3eooquotkw39 EinmCT2T1Eff quotkJquot quotSome Wrong Solutions with Common Mistakesquot EinW1mCT2T1Eff quotMultiplying by Eff instead of dividingquot EinW2mCT2T1 quotIgnoring efficiencyquot EinW3mT2T1Eff quotNot using specific heatquot EinW4mCT2T1Eff quotAdding temperaturesquot 6159 A heat pump is absorbing heat from the cold outdoors at 5 C and supplying heat to a house at 25 C at a rate of 18000 kJh If the power consumed by the heat pump is 19 kW the coef cient of performance of the heat pump is a 13 b 26 c 30 d 38 e 139 Answer b 26 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL5 quotCquot TH25 quotCquot QH180003600 quotkJSquot Win19 quotkWquot COPQHW in quotSome Wrong Solutions with Common Mistakesquot W1COPWinQH quotDoing it backwardsquot W2COPTHTHTL quotUsing temperatures in Cquot W3COPTH273THTL quotUsing temperatures in Kquot W4COPTL273THTL quotFinding COP of refrigerator using temperatures in Kquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 69 6160 A heat engine cycle is executed with steam in the saturation dome The pressure of steam is 1 MPa during heat addition and 04 MPa during heat rejection The highest possible efficiency of this heat engine is a 80 b 156 c 202 d 798 e 100 Answer a 80 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1000 quotkPaquot PL400 quotkPaquot THTEMPERATURESteamIAPWSx0PPH TLTEMPERATURESteamIAPWSx0PPL EtaCarnot1 TL273TH273 quotSome Wrong Solutions with Common Mistakesquot W1EtaCarnot1PLPH quotUsing pressuresquot W2EtaCarnot1TUTH quotUsing temperatures in Cquot W3EtaCarnotTLTH quotUsing temperatures ratioquot 6161 A heat engine receives heat from a source at 1000 C and rejects the waste heat to a sink at 50 C If heat is supplied to this engine at a rate of 100 kJs the maXimum power this heat engine can produce is a 254 kW b 554 kW c 746 kW d 950 kW e 1000 kW Answer c 746 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH1000 quotCquot TL50 quotCquot Qin100 quotkWquot Eta1TL273TH273 Wout EtaQin quotSome Wrong Solutions with Common Mistakesquot W1Wout1TLTHQin quotUsing temperatures in Cquot W2WoutQin quotSetting work equal to heat inputquot W3WoutQinEta quotDividing by efficiency instead of multiplyingquot W4WoutTL273TH273Qin quotUsing temperature ratioquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 70 6162 A heat pump cycle is executed with R134a under the saturation dome between the pressure limits of 14 MPa and 016 MPa The maximum coefficient of performance of this heat pump is a 11 b 38 c 48 d 53 e 29 Answer c 48 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1400 quotkPaquot PL160quotkPaquot THTEMPERATURER134ax0PPH quotCquot TLTEMPERATURER134ax0PPL quotCquot COPH PTH273THTL quotSome Wrong Solutions with Common Mistakesquot W1COPPHPHPL quotUsing pressuresquot W2COPTHTHTL quotUsing temperatures in Cquot W3COPTLTHTL quotRefrigeration COP using temperatures in Cquot W4COPTL273THTL quotRefrigeration COP using temperatures in Kquot 6163 A refrigeration cycle is executed with R134a under the saturation dome between the pressure limits of 16 MPa and 02 MPa If the power consumption of the refrigerator is 3 kW the maximum rate of heat removal from the cooled space of this refrigerator is a 045 kJs b 078 kJs c 30 kJs d 116 kJs e 146 kJs Answer d 116 kJs Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH1600 quotkPaquot PL200quotkPaquot Win3 quotkWquot THTEMPERATURER134ax0PPH quotCquot TLTEMPERATURER134ax0PPL quotCquot COPTL273THTL QLWinCOP quotkWquot quotSome Wrong Solutions with Common Mistakesquot W1QLWinTLTHTL quotUsing temperatures in Cquot W2QLWin quotSetting heat removal equal to power inputquot W3QLWinCOP quotDividing by COP instead of multiplyingquot W4QLWinTH273THTL quotUsing COP definition for Heat pumpquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 71 6164 A heat pump with a COP of 32 is used to heat a perfectly sealed house no air leaks The entire mass within the house air furniture etc is equivalent to 1200 kg of air When running the heat pump consumes electric power at a rate of 5 kW The temperature of the house was 7 C when the heat pump was turned on If heat transfer through the envelope of the house walls roof etc is negligible the length of time the heat pump must run to raise the temperature of the entire contents of the house to 22 C is a 135 min b 431 min c 138 min d 188 min e 808 min Answer a 135 min Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COP32 Cv0718 quotkJkgCquot m1200 quotkgquot T1 7 quotCquot T222 quotCquot QHmCvT2T1 Win5 quotkWquot WintimeQHCOP60 quotSome Wrong Solutions with Common Mistakesquot WinW1time60mCvT2T1 COP quotMultiplying by COP instead of dividingquot WinW2time60mCvT2T1 quotIgnoring COPquot WinW3timemCvT2T1 COP quotFinding time in seconds instead of minutesquot WinW4time60mCpT2T1 COP quotUsing Cp instead of Cvquot Cp1005 quotkJkgKquot 6165 A heat engine cycle is executed with steam in the saturation dome between the pressure limits of 7 MPa and 2 MPa If heat is supplied to the heat engine at a rate of 150 kJs the maXimum power output of this heat engine is a 81 kW b 197 kW c 386 kW d 107 kW e 130 kW Answer b 197 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values PH7000 quotkPaquot PL2000 quotkPaquot Qin150 quotkWquot THTEMPERATURESteamIAPWSX0PPH quotCquot TLTEMPERATURESteamIAPWSX0PPL quotCquot Eta1TL273TH273 WoutEtaQin quotSome Wrong Solutions with Common Mistakesquot W1Wout1TLTHQin quotUsing temperatures in Cquot W2Wout1PUPHQin quotUsing pressuresquot W3WoutQinEta quotDividing by efficiency instead of multiplyingquot W4WoutTL273TH273Qin quotUsing temperature ratioquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 72 6166 An airconditioning system operating on the reversed Carnot cycle is required to remove heat from the house at a rate of 32 kJs to maintain its temperature constant at 20 C If the temperature of the outdoors is 35 C the power required to operate this airconditioning system is a 058 kW b 320 kW c 156 kW d 226 kW e 164 kW Answer e 164 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL20 quotCquot TH35 quotCquot QL32 quotkJsquot COPTL273THTL COPQLWin quotSome Wrong Solutions with Common Mistakesquot QLW1WinTLTHTL quotUsing temperatures in Cquot QLW2Win quotSetting work equal to heat inputquot QLW3WinCOP quotDividing by COP instead of multiplyingquot QLW4WinTH273THTL quotUsing COP of HPquot 6167 A refrigerator is removing heat from a cold medium at 3 C at a rate of 7200 kJh and rejecting the waste heat to a medium at 30 C If the coef cient of performance of the refrigerator is 2 the power consumed by the refrigerator is a 01 kW b 05 kW c 10 kW d 20 kW e 50 kW Answer c 10 kW Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TL3 quotCquot TH30 quotCquot QL72003600 quotkJsquot COP2 QLWinCOP quotSome Wrong Solutions with Common Mistakesquot QLW1WinTL273THTL quotUsing Carnot COPquot QLW2Win quotSetting work equal to heat inputquot QLW3WinCOP quotDividing by COP instead of multiplyingquot QLW4WinTLTHTL quotUsing Carnot COP using Cquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 73 6168 Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one If the source temperature of the first engine is 1300 K and the sink temperature of the second engine is 300 K and the thermal ef ciencies of both engines are the same the temperature of the intermediate reservoir is a 625 K b 800 K c 860 K d 453 K e 758 K Answer a 625 K Solution Solved by EES Software Solutions can be veri ed by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values TH1300 quotKquot TL300 quotKquot quotSetting thermal efficiencies equal to each otherquot 1TmidTH1TLTmid quotSome Wrong Solutions with Common Mistakesquot W1TmidTLTH2 quotUsing average temperaturequot 6169 Consider a Carnot refrigerator and a Carnot heat pump operating between the same two thermal energy reservoirs If the COP of the refrigerator is 34 the COP of the heat pump is a 17 b 24 c 34 d 44 e 50 Answer d 44 Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values COPR34 COPH PCOPR1 quotSome Wrong Solutions with Common Mistakesquot W1COPCOPR1 quotSubtracting 1 instead of adding 1quot W2COPCOPR quotSetting COPs equal to each otherquot 6170 A typical new household refrigerator consumes about 680 kWh of electricity per year and has a coefficient of performance of 14 The amount of heat removed by this refrigerator from the refrigerated space per year is a 952 MJyr b 1749 MJyr c 2448 MJyr d 3427 MJyr e 4048 MJyr Answer d 3427 MJyr Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Win68036 quotMJquot COPR1 4 QLWinCOPR quotMJquot quotSome Wrong Solutions with Common Mistakesquot W1QLWinCOPR36 quotNot using the conversion factorquot W2QLWin quotIgnoring COPquot W3QLWinCOPR quotDividing by COP instead of multiplyingquot PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 6 74 6171 A window air conditioner that consumes 1 kW of electricity when running and has a coef cient of performance of 3 is placed in the middle of a room and is plugged in The rate of cooling or heating this air conditioner will provide to the air in the room when running is a 3 kJs cooling 9 1 kJs cooling 0 033 kJs heating d 1 kJs heating 6 3 kJs heating Answer d 1 kJs heating Solution Solved by EES Software Solutions can be verified by copyingandpasting the following lines on a blank EES screen Similar problems and their solutions can be obtained easily by modifying numerical values Win1 quotkWquot COP3 quotFrom energy balance heat supplied to the room is equal to electricity consumedquot EsuppiedWin quotkJs heatingquot quotSome Wrong Solutions with Common Mistakesquot W1EWin quotkJs coolingquot W2ECOPWin quotkJs coolingquot W3EWinCOP quotkJs heatingquot W4ECOPWin quotkJs heatingquot 6172 6178 Design and Essay Problems PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 1 Solutions Manual for Thermodynamics An Engineering Approach 8th Edition Yunus A Qengel Michael A Boles McGrawHill 2015 Chapter 7 ENTROPY PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of McGrawHill Education and protected by copyright and other state and federal laws By opening and using this Manual the user agrees to the following restrictions and if the recipient does not agree to these restrictions the Manual should be promptly returned unopened to McGrawHill Education This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the af liated textbook No other use or distribution of this Manual is permitted This Manual may not be sold and may not be distributed to or used by any student or other third party No part of this Manual may be reproduced displayed or distributed in any form or by any means electronic or otherwise without the prior written permission of McGrawHill Education PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Entropy and the Increase of Entropy Principle 71C Yes Because we used the relation QHTH QLTL in the proof which is the defining relation of absolute temperature 72C No A system may reject more or less heat than it receives during a cycle The steam in a steam power plant for example receives more heat than it rejects during a cycle 7 3C Yes 7 4C That integral should be performed along a reversible path to determine the entropy change 7 5C No An isothermal process can be irreversible Example A system that involves paddlewheel work while losing an equivalent amount of heat 7 6C The value of this integral is always larger for reversible processes 7 7 C No Because the entropy of the surrounding air increases even more during that process making the total entropy change positive 7 8C If the system undergoes a reversible process the entropy of the system cannot change without a heat transfer Otherwise the entropy must increase since there are no offsetting entropy changes associated with reservoirs exchanging heat with the system 7 9C The claim that work will not change the entropy of a uid passing through an adiabatic steady ow system with a single inlet and outlet is true only if the process is also reversible Since no real process is reversible there will be an entropy increase in the uid during the adiabatic process in devices such as pumps compressors and turbines 710C Sometimes 7 11C Never PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 12C Always 7 13C Increase 7 14C Increases 7 15C Decreases 7 16C Sometimes 7 17 C Greater than 7 18C They are heat transfer irreversibilities and entropy transport with mass 7 19E The source and sink temperatures and the entropy change of the sink for a completely reversible heat engine are given The entropy decrease of the source and the amount of heat transfer from the source are to be determined Assumptions The heat engine operates steadily Analysis According to the increase in entropy principle the entropy change of the source a must be equal and opposite to that of the sink Hence QH ASH ASL 1 o BtuR Applying the definition of the entropy to the source gives QH 2 TH ASH 1500 R 1O BtuR 15000Btu a which is the heat transfer with respect to the source not the device PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 20 Air is compressed steadily by a compressor The air temperature is maintained constant by heat rejection to the surroundings The rate of entropy change of air is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 Air is an ideal gas 4 The process involves no internal irreversibilities such as friction and thus it is an isothermal internally reversible process Properties Noting that h hT for ideal gases we have h1 hz since T1 T2 25 C Analysis We take the compressor as the system Noting that the enthalpy of air remains constant the energy balance for this steady ow system can be eXpressed in the rate form as 39 70 steady Ein Eout AEsystem 0 P2 r J r Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkandmaSS potentialetc energies Ein Eout Q AIR VVin Qout T COIlSt 3 Qout Win 15 kW T Noting that the process is assumed to be an isothermal and internally reversible process the rate of entropy change of air is determined to be P1 AS QM 00503kWK a Tsys 298K 7 21 Heat is transferred directly from an energysource reservoir to an energysink The entropy change of the two reservoirs is to be calculated and it is to be determined if the increase of entropy principle is satisfied Assumptions The reservoirs operate steadily 1200 K Analysis The entropy change of the source and sink is given by 100 kJ 100 kJ AS Q Hamp 200833kJK 100k TH TL 1200 K 600 K v Since the entropy of everything involved in this process has increased this transfer of heat is possible 600 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 5 7 22 It is assumed that heat is transferred from a cold reservoir to the hot reservoir contrary to the Clausius statement of the second law It is to be proven that this violates the increase in entropy principle Assumptions The reservoirs operate steadily Analysis According to the de nition of the entropy the entropy change of the hightemperature reservoir shown below is ASH g 100 k1 008333 kJK TH 1200 K and the entropy change of the lowtemperature reservoir is Q 100 kJ ASL g 4001 01667 kJK TL 600 K The total entropy change of everything involved with this system is then Asm1 ASH ASL 008333 O1667 00833kJK which violates the increase in entropy principle since the total entropy change is negative 7 23 A reversible heat pump with speci ed reservoir temperatures is considered The entropy change of two reservoirs is to be calculated and it is to be determined if this heat pump satis es the increase in entropy principle Assumptions The heat pump operates steadily Analysis Since the heat pump is completely reversible the combination of the coef cient of performance eXpression first Law and thermodynamic temperature scale gives 1 1 17 47 1 TL TH 1 280 K 297 K COPPIPJBV The power required to drive this heat pump according to the coefficient of performance is then W QH 300 kW 6 COPHPJBV 1747 21717 kW According to the rst law the rate at which heat is removed from the lowtemperature energy reservoir is QL QH Wnetjn 300 kW 1717 kW 2 2828 kW The rate at which the entropy of the high temperature reservoir changes according to the definition of the entropy is ASH 9 H 300 kW 101kWK TH 297 K and that of the lowtemperature reservoir is ASL zgz 1o1kwK TL 280K The net rate of entropy change of everything in this system is AS39WI ASH ASL 101 101 0 kWK as it must be since the heat pump is completely reversible PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 76 7 24 Heat is transferred isotherrnally from a source to the working uid of a Carnot engine The entropy change of the working uid the entropy change of the source and the total entropy change during this process are to be determined Analysis a This is a reversible isothermal process and the entropy change during such a process is given by A52 T Noting that heat transferred from the source is equal to the heat transferred to the working uid the entropy changes of the uid and of the source become As uid Q uid Qm uid 900 1 1337 kJK T uid T uid 673 K 400 C 9 AS Q t s e 900 kJ 1337 kJK source Tsource Tsource 673 K A 900 k 0 Thus the total entropy change of the process is 5gen AStotal As uid ASsoume 1337 1337 0 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 7 quot Problem 724 is reconsidered The effects of the varying the heat transferred to the working uid and the source temperature on the entropy change of the working uid the entropy change of the source and the total entropy change for the process as the source temperature varies from 100 C to 1000 C are to be investigated The entropy changes of the source and of the working uid are to be plotted against the source temperature for heat transfer amounts of 500 k 900 kJ and1300 kJ Analysis The problem is solved using EES and the results are tabulated and plotted below 725 quotKnownsquot TH 400 C QH 1300 kJ TSys TH quotAnalysis a amp b This is a reversible isothermal process and the entropy change during such a process is given by DELTAS QTquot quotNoting that heat transferred from the source is equal to the heat transferred to the working fluid the entropy changes of the fluid and of the source become quot DELTASsource QHTH273 DELTASquid QHTSys273 quot0 entropy generation for the processquot Sgen DELTASsource DELTASquid ASfluid ASsource Sgen TH kJK kJK kJK C 3485 3485 0 100 2748 2748 0 200 2269 2269 0 300 1932 1932 0 400 1682 1682 0 500 1489 1489 0 600 1336 1336 0 700 1212 1212 0 800 1108 1108 0 900 1021 1021 0 1000 4 I I I I I I I I I I I I I I I I 3395 ASsource 39Asfluid L 3 39J I E 2 5 I g 2x Q 1300 kJ A I 5 I lt1 1 5 I o 906 kJ 39u 05j QH500 kJ O I I I I I I I I I I 100 200 300 400 500 600 700 800 9001000 TH C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 8 7 26E Heat is transferred isotherrnally from the working uid of a Carnot engine to a heat sink The entropy change of the working uid is given The amount of heat transfer the entropy change of the sink and the total entropy change during the process are to be determined Analysis a This is a reversible isothermal process and the entropy change Heat during such a process is given by 3 4 AS 2 95 F Noting that heat transferred from the working uid is equal to the heat camOt heat engine transferred to the sink the heat transfer become Q uid T uidAS uid Z gt Q uid out b The entropy change of the sink is determined from Sink Q5111ij 3885 Btu 07 BtuR Tsmk 555 R AS 0 Thus the total entropy change of the process is Sgen total A S uid A Ssink 0397 Z 0 This is eXpected since all processes of the Carnot cycle are reversible processes and no entropy is generated during a reversible process 7 27 R 134a enters an evaporator as a saturated liquidvapor at a speci ed pressure Heat is transferred to the refrigerant from the cooled space and the liquid is vaporized The entropy change of the refrigerant the entropy change of the cooled space and the total entropy change for this process are to be determined Assumptions 1 Both the refrigerant and the cooled space involve no internal irreversibilities such as friction 2 Any temperature change occurs within the wall of the tube and thus both the refrigerant and the cooled space remain isothermal during this process Thus it is an isothermal internally reversible process Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes the entropy change for them can be determined from AS 2 T a The pressure of the refrigerant is maintained constant Therefore the temperature of the refrigerant also remains constant at the saturation value A Q i tin 180k R39134a Then Asm igem 8 gem 07076kJK 140 kPa 71 Tm igerant 2544 K 180 H 9 Similarly Qspaceput 180 kJ Asspace T 263 K 06844kJK space 0 The total entropy change of the process is S en 2 AStotal 2 AS AS 07076 06844 2 00232kJK g re igerant space PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission Entropy Changes of Pure Substances 7 28C Yes because an internally reversible adiabatic process involves no irreversibilities or heat transfer 7 29E A pistoncylinder device that is lled with water is heated The total entropy change is to be determined Analysis The initial specific volume is 25113 v1 125 11 3lbm m 21bm which is between If and V8 for 300 psia The initial quality and the entropy are H20 then Table ASE 300 pm 2 lbm V1 Vf 125 0018901t 3lbm 25 13 08075 x1 Vfg 15435 001890 it 3lbm s1 sf xlsfg 058818 Btulbm R 08075092289 Btulbm R 13334 Btulbm R The final state is superheated vapor and P A T2 500 F s2 15706 Btulbm R Table A 6E P2 P1 300 ps1a 1 2 Hence the change in the total entropy is ASms2 s1 gt V 21bm15706 13334 Btulbm R 04744BtuR 7 30 An insulated rigid tank contains a saturated liquidvapor mixture of water at a specified pressure An electric heater inside is turned on and kept on until all the liquid vaporized The entropy change of the water during this process is to be determined Analysis From the steam tables Tables A4 through A6 101 200 kPa v1 uf xlufg 0001061 O25O88578 0001061 022224 m3kg 1340 x1 025 s1 sf xlsfg 15302 O2555968 29294 kJkg K 200 kgpa 22224 3 We quot2 quot1 0 m Ikg s2 66335 kJkg K x2 1 sat vapor Then the entropy change of the steam becomes AS ms2 s1 3 kg66335 29294 kJkg K 111kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 10 731 The radiator of a steam heating system is initially filled with superheated steam The valves are closed and steam is allowed to cool until the temperature drops to a specified value by transferring heat to the room The entropy change of the steam during this process is to be determined Analysis From the steam tables Tables A4 through A6 P1 200 kPa v1 095986 m3kg T1 150 c sl 272810 kJkg K g H20 3233 200 kPa T2 240 C x2 2 2 If 095986 0001008 2 004914 150 C gtQ ufg 19515 0001008 s2 sf xzsfg 05724 00491476832 09499 kJkg K The mass of the steam is 0020 3 m K 2 m3 002084 kg v1 095986 m kg Then the entropy change of the steam during this process becomes AS ms2 s1 002084 kg09499 72810 kJkg K 0132kJK 7 32 r A rigid tank is divided into two equal parts by a partition One part is filled with compressed liquid water While the other side is evacuated The partition is removed and water eXpands into the entire tank The entropy change of the water during this process is to be determined Analysis The properties of the water are Table A4 25 kg P1 400 kPa 1 E uf6ooc 0001017m3kg compressed T1 60 C s1 sf6ooc 08313 kJkg K liquid Vacuum Noting that 400 kPa 3 60 C 2 201 20001017 0002034 m kg P2 2 40 kPa x2 2 V2 Vf 0002034 0001026 00002524 2 4 3 k ufg 3993 0001026 2 000 03 m l g s2 sf xzsfg 210261 000025246643010278 kJkg K Then the entropy change of the water becomes AS ms2 s1 25 kg10278 08313kJkg K 20492 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 11 7 33 An insulated cylinder is initially lled with saturated liquid water at a speci ed pressure The water is heated electrically at constant pressure The entropy change of the water during this process is to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The cylinder is wellinsulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis From the steam tables Tables A4 through A6 v v 0001053 m3k H P1 150 kPa 1 f150kPa g H O 39 39 hl hf150kpa 2 satlzqmd 150 kPa Sr 2 Sf150kPa 214337 kJkg 39 K 2200 k Sat liquid Also 7 v 0005 m3 m 3475kg V1 0001053m Ikg A We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AE system f J W Net en ergy tr Sfer Chan gein internal kinetic by heat workaand mass potentialetc energies Wein Wbout Wein hl since AU Wb AH during a constant pressure quasiequilibrium process Solving for hz w 112 h1 e m 46713 2200M m 475 kg 93033 kJkg Thus hz hf 93033 46713 hfg 22260 s s x s 14337 02081 57894 226384kJkgK 2 f 2 fg P2 150 kPa x2 202081 h2 93033 kJkg Then the entropy change of the water becomes AS ms2 s1 475 kg26384 14337kJkg K 572 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 12 7 34E R134a is compressed in a compressor during which the entropy remains constant The final temperature and enthalpy change are to be determined Analysis The initial state is saturated vapor and the properties are Table A11E hl hg 60F S1 Sg 60F R T A The final state is superheated vapor and the properties are Table A13E 2 P2 80 psia T2 757 F 52 51 022477 Btulbm R h2 11452 Btulbm 1 The change in the enthalpy across the compressor is then Ah I12 hl 11452 10398 105BtuIbm V 9 7 35 Water vapor is eXpanded in a turbine during which the entropy remains constant The enthalpy difference is to be determined Analysis The initial state is superheated vapor and thus P1 6 MPa h1 31783 kJkg T Table A 6 A T1 400 C 51 65432 kJkg K 1 The entropy is constant during the process The nal state is a mixture since the entropy is between sf and 58 for 100 kPa The properties at this state are Table A5 s s 42 12 kK 2 2 f653 30 8kJg 208653 x2 V 9 sfg 60562 kJkg K h2 h f x211 fg 41751 O865322575 23709 kJkg The change in the enthalpy across the turbine is then Ah I22 h1 23709 31783 8074kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 13 7 36 R 134a undergoes a process during which the entropy is kept constant The nal temperature and internal energy are to be determined T Analysis The initial entropy is I T1 25 C 1 sl 2 09335 kJkg K TableA 13 P1 600 kPa The entropy is constant during the process The nal state is a mixture since the entropy 2 V 9 is between sf and s8 for 100 kPa The properties at this state are Table A12 T2 2 Tsat100kPa Z 26370C Sz S f 09335 007182 kJkg K sfg 08801 kJkg K u2 uf x214 fg 1719 O979119801 2111kJkg 09791 x2 7 37 Refrigerant134a is is expanded in a turbine during which the entropy remains constant The inlet and outlet velocities are to be determined Analysis The initial state is superheated vapor and thus 1 0 v1 004307m g TableA13 1 T1 70 C 51 10706 kJkg K The entropy is constant during the process The properties at the eXit state are 2 v2 02274 m Ikg TableA 13 s2 s1 10706 kJkg K V 9 The inlet and outlet veloicites are n39w1 075 kgs004307 m3kg A 05 2 00646ms 1 m V1 n w2 075 kgs02274 m3kg A 10 2 0171ms 2 m V2 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 14 7 38 A cylinder is initially filled with saturated water vapor at a specified temperature Heat is transferred to the steam and it eXpands in a reversible and isothermal manner until the pressure drops to a speci ed value The heat transfer and the work output for this process are to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The cylinder is wellinsulated and thus heat transfer is negligible 3 The thermal energy stored in the cylinder itself is negligible 4 The process is stated to be reversible and isothermal Analysis From the steam tables Tables A4 through A6 T1 200 C 1 g20rc 2 25942 kJkg H Satvap0r S1 Sg2003C 39 K H20 P2 800 kPa u2 26311kJkg 200 C sat vapor The heat transfer for this reversible isothermal process can be determined from R Q Q TAS Tms2 s1 473 K12 kg6 8177 64302kJkg K 2199 H We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be eXpressed as Ein Eout AE system f J W Net en ergy tr Sfer Chan gein internal kinetic by heat workaand mass potentialetc energies Qin Wbout AU mu2 1 Wbout Qin mu2 1 Substituting the work done during this process is determined to be Wbout 2199 kJ 12 kg26311 25942 kJkg 1756 kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 15 quot Problem 738 is reconsidered The heat transferred to the steam and the work done are to be determined and plotted as a function of nal pressure as the pressure varies from the initial value to the final value of 800 kPa 739 If Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot T1 200 C x1 10 msys 12 kg P2 800quotkPaquot quotAnalysis quot FIuid39SteamIAPWS39 quot Treat the pistoncylinder as a closed system neglect changes in KE and PE of the Steam The process is reversible and isothermal quot T2 T1 Ein Eout DELTAEsys Ein Qin Eout Workout DELTAEsys msysu2 u1 P1 pressureFuidTT1x10 u1 NTENERGYFuidTT1x10 v1 volumeFuidTT1x10 s1 entropyFuidTT1x10 Vsys msysv1 200 1 I quot The process is reversible and isothermal Then P2 and T2 specify state 2quot u2 NTENERGYFuidPP2TT2 O 39 I V s2 entropyFIuidPP2TT2 800 1000 P 1330 1400 1600 Qin T1273msyss2s1 2 a P2 Qin Workout I 39 39 kPa kJ kJ 200 800 2199 1757 39 900 1837 1447 160 1000 1506 117 H 1100 120 9184 3 120 1200 9123 6885 1300 6408 4765 E 80 1400 382 2798 0 1500 1332 9605 1553 04645 03319 40 0 39 39 I I I 800 1000 1200 1400 1600 P2 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 16 7 40 R 134a undergoes an isothermal process in a closed system The work and heat transfer are to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved other than the boundary work 3 The thermal energy stored in the cylinder itself is negligible 4 The compression or eXpansion process is quasiequilibrium Analysis The energy balance for this system can be eXpressed as T A E in E out 2 AE system r J W Net energy t1 Sfel Chan gein intemalkinetic R39134a byheat W0Fkandfnass potentialetc energies 320 kpa 1 Win Qout 2 AU 2 mu2 ul T1T240 C 2 The initial state properties are gt s P12320kPa u126162kJkg T M A 13 a e T1 40 C s1 10452 kJkg K For this isothermal process the nal state properties are Table A11 T2 2 T1 40 C u2 uf xzufg 10739 04514361 17202 kJkg x2 045 s2 sf xzsfg 039493 045052059 062920 kJkg K The heat transfer is determined from qin 2 T0 52 sl 2 313 K062920 10452 kJkg K 1302kJkg The negative sign shows that the heat is actually transferred from the system That is qout 1302kJkg The work required is determined from the energy balance to be win qout u2 ul 2 1302 kJkg 17202 26162 kJkg 406kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 17 7 41 A rigid tank contains saturated water vapor at a speci ed temperature Steam is cooled to ambient temperature The process is to be sketched and entropy changes for the steam and for the process are to be determined Assumptions 1 The kinetic and potential energy changes are negligible Analysis b From the steam tables Tables A4 through A6 v1 vg 16720 kJkg T1 100 C x 1 ul 2 ug 25060 kJkg 51 73542 kJkg K T2 250C x2 003856 u2 19367 kJkg 52 06830 kJkg K The entropy change of steam is determined from ASW ms2 s1 5 kg06830 73542ng K 3336kJK c We take the contents of the tank as the system This is a closed system since balance for this closed system can be expressed as Ein Eout AEsystem f J W Net en ergy traleel Chan gein intemalkin etic by heat workaand mass potentialetc energies Qout AU 2 m 2 1 That is Qout mu1 u2 5 kg25060 19367kJkg 11562kJ The total entropy change for the process is 11562kJ Sgen ASW 3336 kJK 544 kJK SUIT 700 SteaT39APWS 600 500 400 Si 39 300 200 1 100 1014 kPa 39 317 kPa l 2 39 0 I l l l I 103 1o2 101 100 1o1 102 103 v mskg H2O 100 C x 1 no mass enters or leaves The energy PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 18 7 42 A rigid tank is initially lled with a saturated mixture of R134a Heat is transferred to the tank from a source until the pressure inside rises to a specified value The entropy change of the refrigerant entropy change of the source and the total entropy change for this process are to be determined Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions Analysis 61 From the refrigerant tables Tables All through A13 u 11 xlufg 3826 0418625 11276 kJkg s1 sf xlsfg 015449 04078339 04678 kJkg K 01 2 VJ xlufg 00007532 040099951 00007532 004043 m3kg P1 200 kPa x1 2 04 x V2 V f 004043 00007905 2 07853 vfg 0051266 00007905 P2 400 kPa u uf xzufg 6361 0785317149 19829 kJkg V2 2 V1 32 sf xzsfg 024757 07853067954 07813 kJkg K The mass of the refrigerant is R 134a 05 3 quot1212 1113 1237kg 200kPa v1 004043m kg Q 4 Then the entropy change of the refrigerant becomes AS ms2 s1 1237 kg07813 04678 kJkg K 3876kJK system 2 b We take the tank as the system This is a closed system since no mass enters or leaves Noting that the volume of the system is constant and thus there is no boundary work the energy balance for this stationary closed system can be expressed as Ein Eout AEsystem J W Net energy MSfer Chan gein intemalkinetic by heat workaand mass potentialetc energies Qin 2 AU 2 mu2 ul Substituting Qin mu2 111 1237 kg19829 11276 1058 kJ The heat transfer for the source is equal in magnitude but opposite in direction Therefore Qsource out 39 Qtank in 39 8 H and Q u ut 1058kJ ASsourvce 11069 2 308K SOUICC 3434 k K c The total entropy change for this process is Asml ASSystem ASsoume 3876 3434 0441 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 43 temperature for nal pressures of 250 kPa 400 kPa and 500 kPa Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot P1 200 kPa x1 04 Vsys 05 mquot3 P2 400 kPa Tsource 35 C quotAnalysis quot quot Treat the rigid tank as a closed system with no work in neglect changes in KE and PE of the R134aquot Ein Eout DELTAEsys Eout 0 kJ Ein Q DELTAEsys msysu2 u1 u1 NTENERGYR134aPP1xx1 v1 volumeR134aPP1xx1 Vsys msysv1 quotRigid Tank The process is constant volume Then P2 and v2 specify state 2quot v2 v1 u2 NTENERGYR134aPP2vv2 quotEntropy cacuationsquot s1 entropyR134aPP1xx1 s2 entropYR134aPP2vv2 DELTASsys msyss2 s1 quotHeat is leaving the source thusquot DELTASsource QTsource 273 quotTotal Entropy Changequot DELTAStotaI DELTASsource DELTASsys AStotal Tsource kJK C 03848 30 06997 60 09626 90 1185 120 1376 150 1 542 180 1687 210 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation Astota kJK Problem 742 is reconsidered The effects of the source temperature and nal pressure on the total entropy change for the process as the source temperature varies from 30 C to 210 C and the nal pressure varies from 250 kPa to 500 kPa are to be investigated The total entropy change for the process is to be plotted as a function of the source 3 25 P2 250 kPa I 400 kPa 2 500 kPa O 15 0 1 I I I I O I I I I I I 25 65 105 145 185 225 TSOUI CB If you are a student using this Manual you are using it Without permission 7 20 7 44 The heat transfer during the process shown in the gure is to be determined Assumptions The process is reversible T Analysis No heat is transferred during the process 23 since the 0C 1 area under process line is zero Then the heat transfer is equal to 600 3 the area under the process line 12 2 T T c112 zITdszArea 1 2 52 s1 1 2 200 2 2 K 2 2 K 600 73 OO 73 1003kJkgIK 03 l gt 1 0 471 kJkg s klkg K 7 45E The heat transfer during the process shown in the gure is to be determined Assumptions The process is reversible Analysis Heat transfer is equal to the sum of the areas under the process 12 and 23 2 3 T T q12ITd5ITds 12 2sz slT2s3 sz A 1 2 lt 3 2 360 4 R 4 R 2 55 60 2660 60 30 10Btulbm R 1 360 460R 20 30Btulbm R 55 515 Btulbm l 2 3 S BtulbmR PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 21 7 46 Steam enters a diffuser at a speci ed state and leaves at a specified pressure The minimum outlet velocity is to be determined Analysis The inlet state is superheated vapor and thus A P1 150kPa h127114kJkg 2 T able A 6 T1 120 C 51 72699 kJkg K For minimum velocity at the eXit the entropy will be constant during the process The 1 eXit state enthalpy is Table A6 gt 5 P2 300 kPa h 228457 kJk ableA6 s2s172699kJkgK 2 g T We take the diffuser as the system which is a control volume since mass crosses the boundary Noting that one uid stream enters and leaves the diffuser the energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 lt90 steady Ein Eout AEsystem O af J r J Rate 0f net energy thfer Rate of changein intemalkinetic byheats workaand mass potentialetc energies Ein E out V2 V2 1mm 1h272 sinceWEQEApeEO 2 2 h1 h2 Solving for the eXit velocity and substituting 2 2 h1 hz 05 1000 m2s2 V2 2 V12 2h1 h20395 550 ms2 227114 28457 kJkgL 1ka g 1841ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 22 7 47E Steam expands in an adiabatic turbine The maximum amount of work that can be done by the turbine is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 Kinetic and potential energy changes are negligible 3 The device is adiabatic and thus heat transfer is negligible Analysis The work output of an adiabatic turbine is maXimum when the eXpansion process is reversible For the reversible adiabatic process we have s2 s1 From the steam tables Tables A4E through A6E P1 800 psia h1 14560 Btulbm T1 900 F s1 16413 Btulbm R S S P2 40 pm x2 2 2 f 2 16413 039213 209725 sfg 128448 32 Sl h2 h f xzhfg 23614 0972593369 11442 Btulbm There is only one inlet and one eXit and thus m1 n3912 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be expressed in the rate form as 39 70 steady Ein Eout AEsystem O f J Rate 0f net energy MSfer Rate of changein intemalkinetic by heat W0rkand mass potentialetc energies Ein Eout mhl Z Wont quot1172 Wout Z 1071 I72 Dividing by mass ow rate and substituting wout h h2 14560 11442 2 3118 Btulbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 23 748E 39 Problem 747E is reconsidered The work done by the steam is to be calculated and plotted as a function of nal pressure as the pressure varies from 800 psia to 40 psia Also the effect of varying the turbine inlet temperature from the saturation temperature at 800 psia to 900 F on the turbine work is to be investigated Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot P1 800 psia T1 900 F P2 40 psia TsatP1 temperatureFuidPP1x10 FIuid39SteamIAPWS39 quotAnalysis quot quot Treat theturbine as a steadyflow control volume with no heat transfer in neglect changes in KE and PE of the Steamquot quotThe isentropic work is determined from the steadyflow energy equation written per unit massquot ein eout DELTAesys Eout Workouth2 quotBtulbmquot ein h1 quotBtuIbmquot 350 DELTAesys 0 quotBtulbmquot h1 enthalpyFIuidPP1TT1 300 lsentroplc Process s1 entropyFuidPP1TT1 E 250 p1 300 psia quotThe process is reversible E 200 T1 900 F and adiabatic or isentropic E Then P2 and s2 specify state 2quot 5 150 s2 s1 quotBtuIbmRquot x0 h2 enthalpyFIuidPP2ss2 3 100 T2isentemperatureFuidPP2ss2 50 39 T W k 0 I I I I l I I I l I I I I 1 or out o 100 200 300 400 500 600 700 800 F BtuIbm P 520 2193 2 ps39a 560 2296 600 2391 650 2507 320 39 39 39 39 39 39 39 39 39 39 39 39 39 39 690 260 lsentroplc Process 730 2694 E 298 P1 800 psia 770 279 Q 820 2913 g 276 P2 40 Ps39a 860 3015 E 900 3119 quot39 g 254 i o g 232 21 O I I I I I I I I I I I I I I 500 550 600 650 700 750 800 850 900 T1 F PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 24 7 49 Steam is expanded in an isentropic turbine The work produced is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 The process is isentropic ie reversible adiabatic Analysis There is one inlet and two eXits We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 steady Ein Eout AEsystem O Er J f f r 3 MPa Rate 0 net energy trans er Rate of changern 1ntemalk1net1c by heat workaand mass potentialetc energies 2 kg S Ein Eout Steam turbine Wout mlhl m2h2 mshs 500 kPa 50 kPa From a mass balance 100 C m2 005n3911 0055 kgs 025 kgs m3 095n3911 0955 kgs 475 kgs Noting that the eXpansion process is isentropic the enthalpies at three states are determined as follows P3 50 kPa h3 26824 kJkg Table A 6 T3 100 C 53 76953 kJkg K P1 3MPa h38512kJk ableA6 s1s376953kJkgK 1 g T P2 500 kPa h 32065 kJk ableA 6 52 s3 276953kJkgK 2 g T Substituting Wout mrhr m2h2 quot13113 2 kgs38512kJkg 01 kgs32065 kJkg 19 kgs26824kJkg 2285 kW PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 25 7 50 Water is compressed in a closed system during which the entropy remains constant The final temperature and the work required are to be determined Analysis The initial state is superheated vapor and thus T P1 70 kPa u 25094 kJkg A 1 Table A 6 T1 100 C 51 75344 kJkg K The entropy is constant during the process The properties at the eXit state are 2 P2 4000 kPa u2 33965 kJkg Table A 6 gt S S2 2 S1 2 39 K T2 2 To determine the work done we take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem ar J r J Net energy WSfel Chan gein intemalkinetic by heat workaandmass potentialetc energies win Auu2 u1 sinceQKEPE0 Substituting wi uz u1 33965 25094kJkg 8871le kg 7 51 Refrigerant134a is eXpanded in a closed system during which the entropy remains constant The heat transfer and the work production are to be determined Analysis The initial state is superheated vapor and thus P1 800kPa u126387kJkg A 1 Table A 13 T1 50 C s1 09803 kJkg K The entropy is constant during the process The properties at the eXit state are 2 P2 140 kPa u2 22777 kJkg Table A 13 gt S 52 2 s1 09803 kJkg K Since the process is isentropic and thus the heat transfer is zero Q 0 kJ To determine the work done we take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be eXpressed as Ein Eout AEsystem ar J r J Net energy WSfel Chan gein intemalkinetic by heat WOIkaandmaSS potentialetc energies Wout 2 AU mu2 ul since Q 2 KB PE 0 Substituting Wow 2 mu1 u2 07 kg26387 22777kJkg 253kJ PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 26 7 52 Water vapor is expanded adiabatically in a pistoncylinder device The entropy change is to be determined and it is to be discussed if this process is realistic Analysis a The properties at the initial state are P16OOkPa u125668kJkg T M A 5 a 6 x1 1 S1 67593 kJkg K We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem Net energy transfer Ch 39 39 t all k39 t by heat workandmass p31 aifgtiingrglgilc T I 600 kPa Wout AUmu2 u1 sinceQKEPEO 100 kPa Solving for the nal state internal energy 1 W u2 2 ul out 25668 kJkg 700k 2 22168 kJkg 2 2s m 2 kg gt s The entropy at the final state is from Table A5 M2 uf 22168 41740 ufg 20882 52 sf xsfg 13028 08617 x 60562 65215 kJkg K P2 100kPa x2 208617 u2 22168 kJkg The entropy change is b The process is not realistic since entropy cannot decrease during an adiabatic process In the limiting case of a reversible and adiabatic process the entropy would remain constant PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 27 7 53 Steam enters a nozzle at a speci ed state and leaves at a speci ed pressure The process is to be sketched on the T s diagram and the maximum outlet velocity is to be determined Analysis b The inlet state properties are T A 6000 kPa P1 6000 kPa h1 27846 kJkg TableA 5 1200 kPa x1 1 S1 58902kJkgK 1 v For the maXimum velocity at the eXit the entropy will be constant during the process The eXit state enthalpy is Table A6 2 52 Sf 58902 22159 gt 5 P2 1200 kPa x2 sf 4 3058 418533 g s2 2 s1 58902 kJkg K h2 hf thg 79833 08533 x 19854 2 24925 kJkg We take the nozzle as the system which is a control volume since mass crosses the boundary Noting that one uid stream enters and leaves the nozzle the energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 lt90 steady Ein Eout AEsystem O Er J r J RateOf net energy thfel39 Rate of changein intemalkinetic by heat W0 rksand ms 5 p otentialetc energies Ein Eout V2 V2 mh1i n391h272 Si CBWEQEApEEO 2 2 h1h2 Solving for the eXit velocity and substituting 2 2 h1 h2 05 1000 m2s2 5 V V22h h 39 Oms2227846 24925 kJk 2 1 1 20 lt g1kJkg 7643ms PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 28 7 54 Heat is added to a pressure cooker that is maintained at a speci ed pressure The minimum entropy change of the thermalenergy reservoir supplying this heat is to be determined Assumptions 1 Only water vapor escapes through the pressure relief valve Analysis According to the conservation of mass principle d out dm E mout An entropy balance adapted to this system becomes dS SUIT It 11 outs 2 0 When this is combined with the mass balance it becomes dSsurf dms S dm gt 0 dt dt dt Multiplying by dt and integrating the result yields AS surr m252 quotquot151 Soutm2 m1 20 The properties at the initial and final states are from Table A5 at P1 175 kPa and P2 150 kPa v1 V xufg 0001057 01010037 0001057 01013 m3kg s sf xsfg 14850 O1056865 20537 kJkg K 12 V xufg 0001053 04011594 0001053 04644 m3kg s2 sf xsfg 14337 O4057894 37494 kJkg K The initial and nal masses are 3 m1 1 amp01974kg v1 001013 m3kg v 0020 m3 m2 2 3 004307 kg 12 04644 m kg The entropy of escaping water vapor is Sout Sg 150kPa 72231kJkg 39 K Substituting ASsurf mzs2 mls1 soutm2 m1 2 0 ASsurr 00430737494 0197420537 72231004307 01974 2 0 ASsurr 08708 2 0 The entropy change of the thermal energy reservoir must then satisfy AS gt 08708kJ K SLII39I PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 29 7 55 Heat is added to a pressure cooker that is maintained at a speci ed pressure Work is also done on water The minimum entropy change of the thermalenergy reservoir supplying this heat is to be determined Assumptions 1 Only water vapor escapes through the pressure relief valve Analysis According to the conservation of mass principle d out dm E mout An entropy balance adapted to this system becomes dS SUIT It 11 outs 2 0 When this is combined with the mass balance it becomes dSsurf dms S dm gt 0 dt dt dt Multiplying by dt and integrating the result yields AS surr m252 quotquot151 Soutm2 m1 20 Note that work done on the water has no effect on this entropy balance since work transfer does not involve any entropy transfer The properties at the initial and final states are from Table A5 at P1 175 kPa and P2 150 kPa v1 V xufg 0001057 01010037 0001057 01013 m3kg s sf xsfg 14850 O1056865 20537 kJkg K 12 V xufg 0001053 04011594 0001053 04644 m3kg s2 sf xsfg 14337 O4057894 37494 kJkg K The initial and nal masses are 3 m1 12w201974kg v1 001013m kg 3 m2 1 039020 m 004307 kg V2 04644 m3kg The entropy of escaping water vapor is S Sg150kPa out Substituting A SSUIT AS surf 00430737494 0197420537 72231004307 01974 2 0 AS 08708 2 0 mzsz mlsl Soutm2 m1 2 0 SUIT The entropy change of the thermal energy reservoir must then satisfy AS gt 08708kJ K SUIT PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 30 7 56 A cylinder is initially filled with saturated water vapor mixture at a speci ed temperature Steam undergoes a reversible heat addition and an isentropic process The processes are to be sketched and heat transfer for the rst process and work done during the second process are to be determined Assumptions 1 The kinetic and potential energy changes are negligible 2 The thermal energy stored in the cylinder itself is negligible 3 Both processes are reversible Analysis b From the steam tables Tables A4 through A6 T1 100 C 0 5 h1 h f thg 41917 0522564 15474 kJkg x h2 kg 26756 kJkg u2 ug 25060 kJkg T2 100 C 52 2 5g 2 73542 kJkg K x221 u3 22479 kJkg 53 52 We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this closed system can be eXpressed as by heat workand mass Ein Eout AE system V Chan gem 1ntemalk1n et1c potentialetc energies Qin Wbout AU mu2 1 f J Net energy trarsfer For process 12 it reduces to Q12 in mlb h1 5 kg26756 15474kJkg 5641kJ c For process 23 it reduces to T C 300 200 W23 hm mu2 173 5 kg25060 22479kJkg 1291kJ 700 StgaT39APVquot 600 500 400 10142 kPa 15 kPa 100 00 11 22 33 44 55 66 77 88 99 110 s kJkgK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 31 7 57E An insulated rigid can initially contains R134a at a specified state A crack develops and refrigerant escapes slowly The final mass in the can is to be determined when the pressure inside drops to a speci ed value Assumptions 1 The can is wellinsulated and thus heat transfer is negligible 2 The refrigerant that remains in the can underwent a reversible adiabatic process Analysis Noting that for a reversible adiabatic ie isentropic process sl s2 the properties of the refrigerant in the can are Tables A1 1E through A13E P 90 sia 1 p S1 E Z R S S 004750 002603 P2 20 psia x2 2S f 01997 01075 S2 2 S1 fg v2 2 VJ xzufg 001181 0107522781 001181 0255511 3lbm Thus the nal mass of the refrigerant in the can is 3 mzl 2151bm 12 02555 it 3lbm PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 32 7 58E An electric windshield defroster used to remove ice is considered The electrical energy required and the minimum temperature of the defroster are to be determined Assumptions No no heat is transferred from the defroster or ice to the surroundings Analysis The conservation of mass principle is dmcv m m d 2 2 m out which reduces to dt mout while the the rst law reduces to d W mm m out 2 out dt h out Combining these two eXpressions yield dm dmu W h CV CV out out dt dt When this is multiplied by dt and integrated from time when the ice layer is present until it is removed m 0 gives Wout houtmi The original mass of the ice layer is The work required per unit of windshield area is then W t 02512 it 1 2 ul 4w 2 1u uf in 144Btulbm 1873Btuit2 A u v v 001602 lbm That is Win 21873 Btuft2 The second law as stated by Clasius tells us that the temperature of the defroster cannot be lower than the temperature of the ice being melted Then Trnin PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 33 Entropy Change of lncompressible Substances 7 59C No because entropy is not a conserved property 7 60 A hot copper block is dropped into water in an insulated tank The final equilibrium temperature of the tank and the total entropy change are to be determined Assumptions 1 Both the water and the copper block are incompressible substances with constant specific heats at room temperature 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The tank is well insulated and thus there is no heat transfer Properties The density and speci c heat of water at 25 C are p 997 kgm3 and CI 418 kJkg C The specific heat of copper at 27 C is 0 0386 kJkg C Table A3 Analysis We take the entire contents of the tank water copper block as the system This is a closed system since no mass crosses the system boundary during the process The energy balance for this system can be expressed as Ein Eout AEsystem r J Net energy traleel Chan gein intemalkinetic byheata workaandmass potentialetc energies O 2 AU WATER or Copper AJCu Aljwater 0 50 kg mCT2 TlCu quot3602 Tlwater O 90 L where mwm W 997 kgm3 0090 m3 8973 kg Using speci c heat values for copper and liquid water at room temperature and substituting 50 kg0386 kJkg OCT2 140 C 8973 kg4 18 kJkg CT2 10 C 0 T2 164 C 2894 K The entropy generated during this process is determined from T 2 4 K AScopper mcavg ln 2 50 kg0386 kJkg Kln L 6864 kJK T1 413 K T 2 4 K 45Wr 21man ln 2 8973kg418kJkg Kln L 8388 kJK T1 283 K Thus 45m AS AS 6864 8388 2152 kJK copper water PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 34 7 61 Computer chips are cooled by placing them in saturated liquid R134a The entropy changes of the chips R134a and the entire system are to be determined Assumptions 1 The system is stationary and thus the kinetic and potential energy changes are zero 2 There are no work interactions involved 3 There is no heat transfer between the system and the surroundings Analysis a The energy balance for this system can be expressed as E in E out AE system ar J W Net en ergy thfer Chan gein intern alkinetic by heat workaand mass potentialetc energies 0 2 AU 2 mu2 ul Chips mu2 u1R134a m 1 2lchips m 2 1R 134a The heat released by the chips is Qchips mcTl T2 0010 kg03 kJkg K20 40 K 018kJ The mass of the refrigerant vaporized during this heat exchange process is QR 134a QR 134a 018kJ g z ug uf ufg4ooc 20740kJkg m 00008679 kg Only a small fraction of R134a is vaporized during the process Therefore the temperature of R134a remains constant during the process The change in the entropy of the R134a is at 40 F from Table A1 1 A511 13431 Z mg2Sg2 mf2Sf2 mf1Sf1 00008679O96866 0005 000086790 00050 0000841kJK b The entropy change of the chips is mc lnT 2 0010 kg03 kJkg Kln 0000687kJK T1 20 273K AS chips c The total entropy change is AStotal S gen 2 ASR134a ASchips 0000841 0000687 2 0000154kJK The positive result for the total entropy change ie entropy generation indicates that this process is possible PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 35 7 62 A hot iron block is dropped into water in an insulated tank The total entropy change during this process is to be determined Assumptions 1 Both the water and the iron block are incompressible substances with constant speci c heats at room temperature 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The tank is well insulated and thus there is no heat transfer 4 The water that evaporates condenses back Properties The speci c heat of water at 25 C is 0 418 kJkg C The speci c heat of iron at room temperature is 0 045 kJkg C Table A3 Analysis We take the entire contents of the tank water iron block as the system This is a closed system since no mass crosses the system boundary during the process The energy balance for this system can be expressed as Ein Eout AEsystem J W Net energy traleel Chan gein intemalkinetic byheata workaandmass potentialetc energies WATER O 0 2 AU 18 C or AJinon Aljwater 0 mCT2 Tlinon mCT2 Tlwater 0 Substituting 25 kg045 kJkg KxT2 350 C 100 kg4 18 kJkg KxT2 18 C 0 gzmmc The entropy generated during this process is determined from T 2997 K AS ln 2 2 25k 045 kJk K In 8232 kJK 1m mCavg T1 g g 623K T 2997 K AS mc ln 2 100k 418kJk K In 12314kJK g g 291K Thus sgem Astotal ASim ASWm 8232 12314 2 408 kJHlt Discussion The results can be improved somewhat by using specific heats at average temperature PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 36 7 63 An aluminum block is brought into contact with an iron block in an insulated enclosure The nal equilibrium temperature and the total entropy change for this process are to be determined Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specific heats 2 The system is stationary and thus the kinetic and potential energies are negligible 3 The system is wellinsulated and thus there is no heat transfer Properties The speci c heat of aluminum at the anticipated average temperature of 400 K is 0 0949 kJkg C The specific heat of iron at room temperature the only value available in the tables is 0 045 kJkg C Table A3 Analysis We take the ironaluminum blocks as the system which is a closed system The energy balance for this system can be expressed as Ein Eout AEsystem r J W Net energy traleel Chan gein intemalkinetic byheata workaandmass potentialetc energies Iron Aluminum 0 2 AU 40 kg 30 kg 60 C 140 C or AUalum AUimn 0 mCT2 T1a1um mCT2 T1iron 0 Substituting 30 kg0949 kJkg KT2 140 C 40 kg045 kJkg KT2 60 C 2 0 The total entropy change for this process is determined from T2 382 K Asim mcavg lnL T1 1 40 kg045 kJkg K 111E 333 K 2 2472 kJK 382 K 413 K T ASalum mcavg 111Lsz 30 kg0949 kJkg K 1n 2 2221kJK 1 Thus AStotal ASiIon ASalum 2472 2221 0251 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 37 7 64 i quot Problem 763 is reconsidered The effect of the mass of the iron block on the final equilibrium temperature and the total entropy change for the process is to be studied The mass of the iron is to vary from 10 to 100 kg The equilibrium temperature and the total entropy change are to be plotted as a function of iron mass Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot T1iron 60 C miron 40 kg T1al 140 C mal 30 kg Cal 0949 kJkgK quotFromTable A 3 at the anticipated average temperature of 450 Kquot Ciron 045 kJkgK quotFromTable A 3 at room temperature the only value availablequot quotAnalysis quot quot Treat the iron plus aluminum as a closed system with no heat transfer in no work out neglect changes in KE and PE of the system quot quotThe final temperature is found from the energy balancequot Ein Eout DELTAEsys Eout 0 Ein 0 DELTAEsys mironDELTAuiron maDELTAua DELTAuiron CironT2iron T1iron DELTAua CaT2a T1al quotthe iron and aluminum reach thermal equilibriumquot T2iron T2 T2a T2 DELTASiron mironCironnT2iron273 T1iron273 DELTASa maCanT2a273 T1a273 DELTAStota DELTASiron DELTASa 045 Astotal kJkg miron k9 T2 C 0 4 39 6 008547 10 1291 31 01525 20 1208 035 r 02066 30 1143 g a 02511 40 109 a 03 1 02883 50 1047 i 02539 032 60 1011 g 39 03472 70 9798 g 02 x 03709 80 9533 r 03916 90 9302 5 041 100 91 0 005 130 10 20 30 40 50 60 70 80 90 100 miron kg 125 120 115 E110 105 100 95 90 10 20 30 40 50 60 70 80 90 100 miron kg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 38 7 65 An iron block and a copper block are dropped into a large lake The total amount of entropy change when both blocks cool to the lake temperature is to be determined Assumptions 1 The water the iron block and the copper block are incompressible substances with constant specific heats at room temperature 2 Kinetic and potential energies are negligible Properties The speci c heats of iron and copper at room temperature are cm 045 kJkg C and ccopper 0386 kJkg C Table A3 Analysis The thermalenergy capacity of the lake is very large and thus the temperatures of both the iron and the copper blocks will drop to the lake temperature 15 C when the thermal equilibrium is established Then the entropy changes of the blocks become T2 288 K Asim mcavg 1n T1 1 30 kg045 kJkg Kln 353 K 2 2746 kJK T2 288 K AScopper mcavg IHLFJ 40 kg0386 kJkg Kln 353 K 2 3141kJK 1 We take both the iron and the copper blocks as the system This is a closed system since no mass crosses the system boundary during the process The energy balance for this system can be eXpressed as Ein Eout AE system Net energy MSfer Changein intemalkinetic byheata WOIkaandmaSS potentialetc energies Qout AU AUiron AUcopper or Qout T2inon T2copper Substituting Q01t 30 kg045 kJkg K353 288K 40 kg0386 kJkg K353 288K 1881 kl Thus Tlake 288 K ASIake Then the total entropy change for this process is Astotal Asiron Ascopper Asme 2746 3 141 6528 0642 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 39 7 66 An adiabatic pump is used to compress saturated liquid water in a reversible manner The work input is to be determined by different approaches Assumptions 1 Steady operating conditions eXist 2 Kinetic and potential energy changes are negligible 3 Heat transfer to or from the uid is negligible Analysis The properties of water at the inlet and eXit of the pump are Tables A4 through A6 h1 19181 kJkg P1 10 kPa 0 s1 06492 kJkg x 1 v1 0001010 m3kg 15 MPa T P2 15 MPa h2 20690 kJkg 52 2 s1 2 0001004 m3kg a Using the entropy data from the compressed liquid water table 10 kPa gt WP 2 I22 h1 20690 191811510kJkg b Using inlet speci c volume and pressure values WP 11P2 Pl 2 0001010 m3kg15000 10kPa 1514kJkg Error 03 c Using average speci c volume and pressure values wP vavgP2 P1 l120001010 0001004 m3kg115000 10kPa 1510kJkg Error 0 Discussion The results show that any of the method may be used to calculate reversible pump work Entropy Changes of Ideal Gases 7 67 C No The entropy of an ideal gas depends on the pressure as well as the temperature 7 68C The entropy of a gas can change during an isothermal process since entropy of an ideal gas depends on the pressure as well as the temperature 7 69C The entropy change relations of an ideal gas simplify to As CI lnT2T1 for a constant pressure process and As chnT2 T1 for a constant volume process Noting that cp gt cu the entropy change will be larger for a constant pressure process PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 740 770 For ideal gases 0 cv R and P2V2PrV1 3 szpr T2 T1 V1 T1P2 Thus T V 52 51 culn F2 Rln K 1 K 1 2 cu ln Rlnf T211 Tl T1P2 T T P culn 2 Rln zj Rl 2 E Kg 3 771 For an ideal gas dh CI dT and v RTP From the second Tds relation dp 0 IF T T T PTpT P Integrating T2 P2 52 51 cp ln F Rln F 1 1 Since 0 is assumed to be constant PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 41 7 72 The entropy changes of helium and nitrogen is to be compared for the same initial and final states Assumptions Helium and nitrogen are ideal gases with constant specific heats Properties The properties of helium are 0 51926 kJkgK R 20769 kJkgK Table A2a The speci c heat of nitrogen at the average temperature of 427272227 C500 K is 0 1056 kJkgK Table A2b The gas constant of nitrogen is R 02968 kJkgK Table A2a Analysis From the entropy change relation of an ideal gas T2 P2 ASHe cp ln Rln 1 1 51926 kJkg Kln w 20769 kJkg Kln m 427 273K 2000 kPa 03826kJkg K T2 P2 As c ln Rln N2 p 1 P1 1056 kJkg Kln w 02968 kJkg Kln m 427 273K 2000 kPa 02113kJkgK Hence helium undergoes the largest change in entropy 7 73 The entropy difference between the two states of air is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The speci c heat of air at the average temperature of 500502275 C 548 K E 550 K is cp 1040 kJkgK Table A2b The gas constant of air is R 0287 kJkgK Table A2a Analysis From the entropy change relation of an ideal gas T P Asair cp lnFZ Rln2 1 1 50 273K 0287 kJkg K n 100 kPa 500 273K 2000 kPa 00478kJkg K 1040 kJkg Kln PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 42 7 74E The entropy difference between the two states of air is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The speci c heat of air at the average temperature of 902102150 F is cp 0241 BtulbmR Table A2Eb The gas constant of air is R 006855 BtulbmR Table A2Ea Analysis From the entropy change relation of an ideal gas T P Asair cp lnFZ Rln2 1 1 21 4 R 39 006855 Btulbm Rln m 90 460R 15 Psla 001973Btulbm R 0241 Btulbm Rln 7 75 Oxygen gas is compressed from a speci ed initial state to a specified nal state The entropy change of oxygen during this process is to be determined for the case of constant speci c heats Assumptions At specified conditions oxygen can be treated as an ideal gas Properties The gas constant and molar mass of oxygen are R 02598 kJkgK and M 32 kgkmol Table A1 Analysis The constant volume speci c heat of oxygen at the average temperature is Table A2 2 avg 98560 429 K gt cuan 0690 kJkgK Thus v T V s s 0 ln 2Rln 2 2 1 1an Tl VI 0 8 023k 0690 kJkg K In 560 02598 kJkg 101110g 25 C 298 K 08 m kg 0105 kJkg K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 43 7 76 An insulated tank contains C02 gas at a speci ed pressure and volume A paddlewheel in the tank stirs the gas and the pressure and temperature of C02 rises The entropy change of C02 during this process is to be determined using constant specific heats Assumptions At specified conditions C02 can be treated as an ideal gas with constant specific heats at room temperature Properties The speci c heat of C02 is cV 0657 kJkgK Table A2 Analysis Using the ideal gas relation the entropy change is determined to be P2VP1V amp150amp15 023 T2 T1 T1 P1 100 kPa 15 m 100 kPa Thus 27 kg lt90 T V T l 4 AS msz S1 m cuvavgln2 Ran2 j mow g In2 T 1 1 1 27 kg0657 kJkg Kln15 0719 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 44 7 77 An insulated cylinder initially contains air at a speci ed state A resistance heater inside the cylinder is turned on and air is heated for 15 min at constant pressure The entropy change of air during this process is to be determined for the cases of constant and variable specific heats Assumptions At specified conditions air can be treated as an ideal gas Properties The gas constant of air is R 0287 kJkgK Table A1 Analysis The mass of the air and the electrical work done during this process are IV 12 kP 3 quot12 O 04325kg 1 RTI 0287 kPa m Ikg KX290 K Wein Weinm 02 kJs15 x 60 s 180 kJ 021113 m The energy balance for this stationary closed system can be expressed as A 120 kPa 7 17 C Ein Eout AEsystem W r J af J e Net energy traleel Chan gein intemalkinetic I byheataworkaandmass potentialetc energies Wein Wbout A11 Wein hl E Cp T2 since AU Wb AH during a constant pressure quasiequilibrium process a Using a constant cp value at the anticipated average temperature of 450 K the final temperature becomes Thus w e m 290 K 180 Id 2 698 K 04325 kg102 kJkg K p Then the entropy change becomes lt90 T P T AS msz slmc V ln z Rln 2 mcV ln 2 sys pa g T1 P1 P a g T1 04325 kg1020 kJkg Kln 698 K 0387 kJK 290 K 9 Assuming variable specific heats W Wein mh2 h1 gt h2 h1 6 1 29016 kJkg 70634 kJkg m 04325 kg From the air table Table A17 we read s3 25628 kJkgK corresponding to this hz value Then lt90 ASSyS ms sf Rln J ms sf 04325 kg25628 166802kJkg K 0387 kJK 1 PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 45 7 7 8 A cylinder contains N2 gas at a specified pressure and temperature The gas is compressed polytropically until the volume is reduced by half The entropy change of nitrogen during this process is to be determined Assumptions 1 At speci ed conditions N2 can be treated as an ideal gas 2 Nitrogen has constant speci c heats at room temperature Properties The gas constant of nitrogen is R 02968 kJkgK Table Al The constant volume speci c heat of nitrogen at room temperature is cV 0743 kJkgK Table A2 Analysis From the polytropic relation n l 111 L T2 21 T2 41 310 K213 1 3817 K T1 12 V2 N2 Then the entropy change of nitrogen becomes Pvrs C T V ASN2 m c lnFZRln 2 uav g l l 3817K 075 kg 0743 kJkg K In 310 K 02968 kJkg Kln05 00384 kJK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 46 7 79 A 339 Problem 778 is reconsidered The effect of varying the polytropic exponent from 1 to 14 on the entropy change of the nitrogen is to be investigated and the processes are to be shown on a common Pvdiagram Analysis The problem is solved using EES and the results are tabulated and plotted below quotGivenquot m075 kg P1140kPa T1 37273 K n1 3 Rati0V05 quotRatioVV2V1quot quotPropertiesquot cv0743 kJkgK R0297 kJkgK quotAnalysisquot T2T11Rati0Vquotn1 quotfrom polytropic relationquot DELTASmcvInT2T1RInRati0V P1 V1 mRT1 002 n AS 0 kJkg r 1 01544 39 39 2 105 01351 004 r 11 01158 115 009545 006 12 007715 5 008 125 005783 0 13 003852 lt1 01 135 001921 39 39 14 0000104 012 I J 014 016 I I I 1 105 11 115 12 125 13 135 14 1000 900 800 700 600 500 400 300 200 100 0 01 02 03 04 05 06 07 v m3kg P kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 47 780 Air is compressed steadily by a 5kW compressor from one speci ed state to another speci ed state The rate of entropy change of air is to be determined Assumptions At specified conditions air can be treated as an ideal gas 2 Air has variable speci c heats Properties The gas constant of air is R 0287 kJkgK Table Al Analysis From the air table Table A17 P2 600 kPa T 290 K 1 S 166802 kJkg K T2 440 K T2 440 K 0 s2 20887 kJkgK P2 600 kPa A1r compressor 3 Then the rate of entropy change of air becomes 5 kW 39 o o ASSyS ms2 s1 RlnE T P1 kPa 1660 kgs20887 166802 0287 kJkg Kln f 8 T1 290 K a 000250 kWK PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 48 7 81 Air is accelerated in an nozzle and some heat is lost in the process The eXit temperature of air and the total entropy change during the process are to be determined Assumptions 1 Air is an ideal gas with variable speci c heats 2 The nozzle operates steadily Analysis a Assuming variable speci c heats the inlet properties are determined to be h 35049 kJkg T12350K gt 0 s1 185708kJkgK Table A 17 We take the nozzle as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as 39 39 39 lt90 d Ein Eout AEsystem Stea y 2 O r J r J RateOfnetenergy thfer Rate of changein intemalkinetic 3392 kJS by heat workaandmass potentialetc energies k Ein Eout I 1 AIR 2 mm V12 2 2 mm V222 Qout gt V2 V2 quouthQhl Therefore 2 2 2 2 2 l k hfh1qomvz V1 235049323 0ms 50ms kJzg 2 2 1000 m s 29734 kJkg At this hz value we read from Table A17 T2 2972 K s 16924 kJ kg K b The total entropy change is the sum of the entropy changes of the air and of the surroundings and is determined from AStotal ASair ASsurr where O O P kP Asa 2 s2 s1 Rln 2 16924 185708 0287 kJkg Klnamp 01775 kJkg K 101 280 kPa and Assurr qsumi 3392kJkg 00109 kJkg K Tsurr 293 K Thus Astotal 01775 00109 01884kJkg K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 49 7 82 l Problem 781 is reconsidered The effect of varying the surrounding medium temperature from 10 C to 40 C on the eXit temperature and the total entropy change for this process is to be studied and the results are to be plotted Analysis The problem is solved using EES and the results are tabulated and plotted below Function HCaIWorkFuid TX PX quotFunction to calculate the enthalpy of an ideal gas or real gasquot If 39Air39 WorkFluid then HCaIENTHALPY39Air39TTx quotIdeal gas eququot else HCaIENTHALPYWorkFuidTTx PPxquotRea gas eququot endif end HCaI quotSystem control volume for the nozzlequot quotProperty relation Air is an ideal gasquot quotProcess Steady state steady flow adiabatic no workquot quotKnowns obtain from the input diagramquot WorkFluid 39Air39 T1 77 C P1 280 kPa Ve1 50 ms P2 85 kPa Ve2 320 ms qout 32 kJkg quotTsurr 20 Cquot quotProperty Data since the Enthalpy function has different parameters for ideal gas and real fluids a function was used to determine hquot h1HCaWorkFuidT1P1 h2HCaWorkFluidT2P2 quotThe Volume function has the same form for an ideal gas as for a real fluidquot v1voumeworkFuidTT1pP1 v2voumeWorkFuidTT2pP2 quotIf we knew the inlet or exit area we could calculate the mass flow rate Since we don39t know these areas we write the conservation of energy per unit massquot quotConservation of mass mdot1 mdot2quot quotConservation of Energy SSSF energy balance for neglecting the change in potential energy no work but heat transfer out isquot h1Ve1quot22Convertmquot2squot2 kJkg h2Ve2 22Convertmquot2squot2 kJkgqout s1entropyworkFuidTT1pP1 s2entropyWorkFuidTT2pP2 quotEntropy change of the air and the surroundings arequot DELTAsair s2 s1 qinsurr qout DELTAssurr qinsurrTsurr273 DELTAstotal DELTAsair DELTAssurr PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 50 AStotal Tsurr T2 kJkgK C C 0189 10 2422 01888 15 2422 01886 20 2422 01884 25 2422 01882 30 2422 0188 35 2422 01879 40 2422 0189 01888 xI g 01886 I I 01884 9 39 2 01882 m 0188 01878 I I I I I I 10 15 20 25 30 35 40 Tsurr C PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 5 1 7 83E A xed mass of helium undergoes a process from one speci ed state to another specified state The entropy change of helium is to be determined for the cases of reversible and irreversible processes Assumptions 1 At speci ed conditions helium can be treated as an ideal gas 2 Helium has constant specific heats at room temperature Properties The gas constant of helium is R 04961 BtulbmR Table A1E The constant volume speci c heat of helium is cu 0753 BtulbmR Table A2E i Analysis From the idealgas entropy change relation T He ASHezm cuaveln 2Ran 2 T1520R T1 V1 T2 700 R 1 3 1 25 lbm 0753 Btulbm R In 700 R 04961 Btu119m R1n 0 3 bm 520 R 50 1t lbm 144BtuR The entropy change will be the same for both cases 7 84 Air is expanded in a pistoncylinder device isothermally until a final pressure The amount of heat transfer is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The speci c heat of air at the given temperature of 127 C 400 K is 0 1013 kJkgK Table A2b The gas constant of air is R 0287 kJkgK Table A2a Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be eXpressed as Ein Eout AE system EK J W Net energy MSfer Chan gein intemalkinetic byheata workaandmass potentialetc energies Qin Wout 2 AU 2 mu2 ul since KE PE 0 Qin Wout 2 AU 2 mu2 ul 2 0 since T1 2 T1 Qin Wout The boundary work output during this isothermal process is P 2 kP W mRT 11110 1 1 kg0287 kJkg K127 273 Kln M 796kJ m 2 100 kPa Thus Qin Wout PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 52 7 85 The final temperature of nitrogen when it is compressed isentropically is to be determined Assumptions Nitrogen is an ideal gas with constant specific heats Properties The speci c heat ratio of nitrogen at an anticipated average temperature of 450 K is k 1395 Table A2b Analysis From the isentropic relation of an ideal gas under constant speci c heat assumption P W 1000kP T2T1 2 27273K a P1 100kPa 03951395 576 K Discussion The average air temperature is 3005762438 K which is sufficiently close to the assumed average temperature of 450 K 7 86 Air is expanded in an adiabatic turbine The maXimum work output is to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 The process is adiabatic and thus there is no heat transfer 3 Air is an ideal gas with constant speci c heats Properties The properties of air at an anticipated average temperature of 550 K are 0 1040 kJkgK and k 1381 Table AZb Analysis There is only one inlet and one eXit and thus 1 ma 2 m We take the turbine as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 35 MPa 500 C 39 39 39 70 steady Ein Eout AE 0 system EK J r J Rate 0 f net en ergy tun Sfer Rate 0 f chan g ein intern a1 kinetic by heat W0rkand mass potentialetc energies Ein Eout Air turbine Z Wont Wout Z quot1011 hz ou T For the minimum work input to the compressor the process must be reversible as well as adiabatic ie isentropic This being the case the 35 MPa 1 eXit temperature will be k 1k 03811381 P T2 T1 2 500 273 K M 351 K 02 MPa P1 3500 kPa 2 V m Substituting into the energy balance equation gives wout hl h2 CI T1 T2 1040 kJkg K773 351K 439kJkg Discussion The average air temperature is 7733512562 K which is suf ciently close to the assumed average temperature of 550 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 53 7 87E Air is compressed in an isentropic compressor The outlet temperature and the work input are to be determined Assumptions 1 This is a steady ow process since there is no change with time 2 The process is adiabatic and thus there is no heat transfer 3 Air is an ideal gas with constant speci c heats Properties The properties of air at an anticipated average temperature of 400 F are 0 0245 BtulbmR and k 1389 Table A2Eb Analysis There is only one inlet and one eXit and thus m1 2 m2 m We take the compressor as the system which is a control volume since mass crosses the boundary The energy balance for this steady ow system can be eXpressed in the rate form as 39 39 39 70 stead Ein Eout AEsystem y O 13 net eiergz i MSfer Rate of changein intemalkinetic y eat WOT aan masS I potent1aletcenerg1es compressor Ein E out mhl Win nfhz 70 F A The process 1s revers1ble as well as ad1abat1c 1e 1sentrop1c Th1s being the case the eXit temperature will be 200 psia 2 k 1k 03891389 P 200 sia T2T1 2 70460R p 1095R P1 15 ps1a 15 P31a 1 gt s Substituting into the energy balance equation gives win h2 h1 cp T2 T1 0245 Btulbm R1095 530R 2138 Btulbm Discussion The average air temperature is 53010952813 K 353 F which is suf ciently close to the assumed average temperature of 400 F PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 54 7 88 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature and pressure While the other side is evacuated The partition is removed and the gas lls the entire tank The total entropy change during this process is to be determined Assumptions The gas in the tank is given to be an ideal gas and thus ideal gas relations apply Analysis Taking the entire rigid tank as the system the energy balance can be expressed as Ein Eout AE system ar J W Net energy thfer Changein intemalkinetic byheatW0rkandmaSS potentialetc energies IDEAL 0 2 AU 2 mu2 ul GAS 2 Z I 12 kmol 50 C T2 2 T1 since u uT for an ideal gas Then the entropy change of the gas becomes lt90 T V V ASN Euavgln Z R 1n 2 NRu 1n 2 T1 V1 V1 2 12 kmol8314 kJkmol K 1n2 692kJK This also represents the total entropy change since the tank does not contain anything else and there are no interactions with the surroundings 7 89 An insulated rigid tank contains argon gas at a specified pressure and temperature A valve is opened and argon escapes until the pressure drops to a speci ed value The nal mass in the tank is to be determined Assumptions 1 At speci ed conditions argon can be treated as an ideal gas 2 The process is given to be reversible and adiabatic and thus isentropic Therefore isentropic relations of ideal gases apply Properties The speci c heat ratio of argon is k 1667 Table A2 Analysis From the ideal gas isentropic relations r k 1k 06671667 ARGON P 200 kP T2 4 2 303 2190 K 4 kg P1 450 kPa 450 kPa 30 C The nal mass 1n the tank 1s deterrmned from the 1deal gas relatron P V RT P T 200 kP 303 K PZV mZRTZ Psz 450 kPa219 K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 55 7 90 i quot Problem 789 is reconsidered The effect of the final pressure on the nal mass in the tank is to be investigated as the pressure varies from 450 kPa to 150 kPa and the results are to be plotted Analysis The problem is solved using EES and the results are tabulated and plotted below quotKnownsquot cp 05203 kJkgK cv 03122 kJkgK R02081 kPamA3kgK P1 450 kPa T1 30 C m1 4 kg P2 150 kPa quotAnalysis We assume the mass that stays in the tank undergoes an isentropic expansion process This allows us to determine the final temperature of that gas at the final pressure in the tank by using the isentropic relationquot k cpcv T2 T1273P2P1Ak1k273 v2 v P1V1m1RT1 273 P2V2m2RT2273 4 P2 m2 kPa kg 150 2069 36 200 2459 250 2811 300 3136 31 32 x 350 344 400 3727 N 2 8 450 4 E 24 l 150 200 250 300 350 400 450 P2 kPa PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 56 7 91E Air is accelerated in an adiabatic nozzle Disregarding irreversibilities the eXit velocity of air is to be determined Assumptions 1 Air is an ideal gas with variable speci c heats 2 The process is given to be reversible and adiabatic and thus isentropic Therefore isentropic relations of ideal gases apply 2 The nozzle operates steadily Analysis Assuming variable speci c heats the inlet and eXit properties are determined to be P 1230 T1 1000 R gt 1 h1 24098 Btulbm and 1 AA 2 P 2 P pS1510230 246 gt 2 2 Pl 1 60 ps1a h2 15211 Btulbm We take the nozzle as the system which is a control volume The energy balance for this steady ow system can be expressed in the rate form as d Em ECllt AEsystem stea Y Z 0 r J r RateOfnetenergy thfer Rate of changein intemalkinetic byheatW0rkandmaSS potentialetc energies Ein Eout mm V12 2 2 mm v222 2 2 hZLlL0 2 Therefore 25037 it 2 2 V2 2 2h1 h2V12 224098 15211Btulb S 200 rm2 1 Btulbm 2119 fts Alternative Solution The enthalpy at the nozzle eXit for this isentropic process can also be determined as follows From the air table Table A 17E T1 21000 R sf 2 075042 Btulbm R For the isentropic process the entropy change is zero Then 0Ass sf RlnP2Pl0 or s3 2 s10 RlnP2 Pl 2 075042 Btulbm R 006855 Btulbm Rln1260 06401 Btulbm R From Table A17E at this value the enthalpy is obtained as S 06401 Btulbm R h2 15215 Btulbm which is practically identical to the value obtained above PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 57 792 Air is expanded in a pistoncylinder device until a nal pressure The maximum work input is given The mass of air in the device is to be determined Assumptions Air is an ideal gas with constant specific heats Properties The properties of air at 300 K is cu 0718 kJkgK and k 14 Table A2a Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AE system Net energy WSfel Chan gein intemalkinetic T by heat WOIkaandmaSS potentialetc energies I Wout 2 AU 2 mu2 ul since Q 2 KB PE O 400 kPa 2 For the minimum work input to the compressor the process must be reversible as well as adiabatic ie isentropic This being the case the 100 kPa ex1t temperature wrll be 1 gt s k 1k 0414 P 100 kP T2 T1 2 257 273 K a 3567 K P1 400 kPa Substituting into the energy balance equation gives W 1000 kJ Wout mu1 u2 mcv T1 T2 gtm out 804kg c T1 T2 0718 kJkg K530 3567K 7 93 Air is compressed in a pistoncylinder device until a nal pressure The minimum work input is given The mass of air in the device is to be determined Assumptions Air is an ideal gas with constant speci c heats Properties The properties of air at 300 K is C 0718 kJkgK and k 14 Table A2a Analysis We take the contents of the cylinder as the system This is a closed system since no mass enters or leaves The energy balance for this stationary closed system can be expressed as Ein Eout AEsystem r J Net energy traleel Chan gein intemalkinetic byheataworkaandmass potentialetc energies T A Win AU mu2 u1 sinceQ 2 KB PE0 600 kPa 2 For the minimum work input to the compressor the process must be reversible as well as adiabatic ie isentropic This being the case the exit temperature will be 100 kPa k 1k 0414 gt s P 600 kP T2 T1 2 27 273 K a 5005K P1 100 kPa Substituting into the energy balance equation gives W 1000 k Win mu2 u1mcpT2 T1 gtm m 694kJkg c T2 T1 2 0718 kJkg K5005 300K PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it without permission 7 58 7 94 Air is compressed in a pistoncylinder device in a reversible and isothermal manner The entropy change of air and the work done are to be determined Assumptions 1 At speci ed conditions air can be treated as an ideal gas 2 The process is speci ed to be reversible Properties The gas constant of air is R 0287 kJkgK Table Al Analysis a Noting that the temperature remains constant the entropy change of air is determined from lt90 T ASair 2 cp avgln z Rlni Rlnamp T1 PI PI i O287 kJkg Kln M 0428 kJkg K 90 kPa AIR Q Also for a reversible isothermal process T const q TAs 293 K 0428 kJkg K 1254 kJkg gt qout 1254 kJkg b The work done during this process is determined from the closed system energy balance Ein E out AE system ar J W Net energy thfer Chan gein intemalkinetic by heat workaandmass potentialetc energies Win Qout AU mcuT2 T1 0 win I qout 1254kJkg PROPRIETARY MATERIAL 2015 McGrawHill Education Limited distribution permitted only to teachers and educators for course preparation If you are a student using this Manual you are using it Without permission 7 59 7 95 Helium gas is compressed in a pistoncylinder device in a reversi