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# ThermodynamicAnalysisI MEM310

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This 38 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM310 at Drexel University taught by MinjunKim in Fall. Since its upload, it has received 60 views. For similar materials see /class/212388/mem310-drexel-university in Mechanical Engineering at Drexel University.

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Date Created: 09/23/15

MEM 310 Thermodynamic Analysis 1 Fall 2010 Homework 5 Solutions Wednesday December 1 2010 11 The Clapeyron equation is Q 7 7 hfg dT sat T vfg T vag The derivative dPdTsa can be computed using values from the water saturation table A4 E 145 C 713500 145 C 713500 1025 kPaK dP N Psa 145oc 7 Psa 135oc 415168 kPa 7 31322 kPa lt gtsatl40 C N From this table we also have vfgmoc 7 vgymoc 7 mime 7 0150850 mSkg7 01001080 mSkg 7 0150742 mSkgi For our computations we will also need the temperature in Kelvins T 140 C 27315 K 41315 K Using these results gives hfgmoc 7 vagmoc g 7 41315 K 0150742 mSkg1025 kPaK 7 2148198 kJkg sat140 C 0 2148198 M k sfgmoc 7 7 WK 7 511990 kJkg Ky These results match the tabulated values to three decimal places The tabulated values are hfg140 C 214413 lltJlltg7 ng14ooc 511901 kJkg 2 The molar masses and gas constants for the two species are from table A 1 a A 0 V MN2 28013 kgkmol7 MO2 31999 kgkmol7 RN2 02968 kJkgK7 R02 02598 kJkg K The number of moles of each component is 7an2 04 kg 7 0014279 k 1 nN2 MN2 23013 kgkmol m0 06 k n02 W32 7 g 0013751 kmol Veg 7 Wgkmol So the total number of moles of gas is n nN2 7202 0014279 kmol 0018751 kmol 0033030 kmol7 While the total mass of the gas is m 7an2 72202 04 kg 00 kg 10 kg The average molar mass of this mixture is therefore 10 k 7 i 3 lM n 0033030 kmol 30276 kgkmol The average gas constant can be computed from the average molar mass 7 331447 kJkmolK lR M 30276 kgkmol 027462 kJkg K The partial pressure of each gas can be computed by ignoring the presence of the other gas and simply applying the ideal gas law PNZ mNZ IENzT 04 kg02968 kJkg K300 K 71232 kPa7 m P02 72202530211 06 kg02598 kJkg K300 K 93528 kPa Ideal gas case Under the ideal gas assumptions the total pressure is simply the sum of the individual partial pressures of the gases Pideal PN2 Po2 71232 kPa 93528 kPa 16476 kPa Kay s rule case To do the Kay s rule computation we need the critical properties of the two gases TmN2 1262 K7 Tarp2 1548 K7 PmN2 339 MPa7 Pmo2 508 MPa7 We also need the molar fractions of the two gases 7th2 7 0014279 kmol n 7 0033030 kmol n02 7 0018751 kmol n 7 0033030 kmol W2 043200 902 056770 From these we can calculate the pseudocritical temperature and pseudocritical pressure of the gas Tcr yNzTcer yOZTcrpg 0432001202 K 0567701543 K 14307 K PCr yNZPmNZ yOZPmOZ 043200339 MPa 056770508 MPa 43434 MPa The actual molar speci c volume is K 7 05 m3 n 7 0033030 kmol f 13133 mSkmol and so the reduced temperature and reduced molar speci c volume are T 300 K i 7 20260 TCr 14307 K 7 15138 mSkmol 7 v R 7 RumPCr 331447 kJmol K14807 K 43434 kPa 53468 Since the reduced molar speci c volume is so large in other words we are far from the critical point a glance at the generalized compressibility chart reveals that the compressibility factor is essentially Z 10 Thus according to Kay s ruley the pressure is ideal 164 kPa 7 nRuT 7 nRuT mwafiv Van der Waals case The van der Waals equation of state is P7gt17713RuT Using this equation of state requires the determination of effective Wholemixture values for the constants a and b The values of these constants for the individual gases are 27R3T M 7 27331447 Jmol K21262 K am 7 64mm 64330106 Pa 7 013702 P a mB 10127 3N2 W 3369010 5 mSmol 02 W 31670 105 mSmol The averaged parameters are 2 7 1 2 1 2 a yNgaNg yo o 2 lt043200013702 Pa mBm01212 056770013757 Pa mBm01212gt 013725 Pa m5 m012 3 yNJIN2 M32302 04320038690 105 mSmol 05677031670 105 mSmol 3469310 5 mSmol Using these values we can solve for the pressure in the van der Waals equation of state T a 7 831447 Jmol K300 K 013725 Pam5m012 7 0015138 mSmoli 3469310 5 meol 7 0015138 mSmol2 16514 kPa This closely matches the other results as expected for this comparatively lowdensity gas Under the assumption of constant speci c heats and using the fact that the volume of the gas cannot change in its rigid container we can see that the heat added in warming the gas from 300 K to 330 K A 39D V must be Q vasooKAT The average value of the constantvolume speci c heat may be computed by averaging the speci c heats of the individual gases Cv300K yNgcvN2300K yOQCVOg300K 0432000743 kJkg K 0567700657 kJkg K 0694 kJkg A K where the CV values for the individual gases are derived from table A2 Using this result we have Q mcv300KAT 10 kg 0694 kJkg K330 K 7 300 K 203 10 The change in pressure is MELAT 0033030 kmol331447 giksmol K330 K 7 300 K 1648 mm m AP 3 a The given pressure dry bulb and wet bulb temperatures are P 110 kPa7 Tdb 20 C7 T wb 15 C Since we are operating near atmospheric pressure the speci c humidity may be determined from the equation w 7 Cpairdbwa 7 Tdb Whfgwaterwb s hgwaterdb 7 hfwaterwb 7 Where 0622me OJ P 7 Psatwaterwb The constant pressure speci c heat of air at the drybulb temperature here 293 K may be obtained from table A2 Cpair293K N Cpair300K 11005 kJkg The saturation pressure and speci c enthalpy values for saturated water at various temperatures may be obtained from table A4 17057 kPa 712281 kPa Psatwaterwb 12281 kPa 15 C 710 C 14 C 710 C 16102 kPa hgmatendb 2537A kJkg7 24654 kJkg7 24772 kJkg 15 C 7 10 C 62982 kJkg 7 42022 kJkg 1 0 C hfgmmwb 7 24772 kJkg 14 C 7 10 C 7 24678 kJkg hawmnwb 7 42022 kJkg 14 C 7 10 C 7 58790 kJkg Using these values gives 062216102 kPa m 00092402 kg HQOkg dry air7 and inserting this into the equation for the speci c humidity gives 7 1005 kJkg K15 C 7 20 C 7 00092402 kg H20kg dry 31124678 kJkg 5 25374 kJkg 7 58790 kJkg 0011227 kg HQOkg dry air b To compute the relative humidity we also need the saturation pressure at the dry bulb temperature Psakwatendb Z3392 kPa The relative humidity is ab 1 L 0622 wsPsatwaterdb 7 0011227 kg HQOkg dry air 110 kPa 7 0622 0011227 kg HQOkg dry air23392 kPa 083376 c The vapor pressure of water under the conditions described above is PV Psatwa erdb 08337623392 kPa 195 kPa The deW point Will be the temperature at Which this vapor pressure is equal to the saturation pressure We can interpolate the data in the saturation pressure table to determine this temperature po satwaterl9503kPa 17500C 713020C 1302 0 20 kPa715 kPa 19503 kPai 15 kPa 1705500 As the air cools the relative humidity Will rst increase to 100 and then water Will begin to condense out of the air While the relative humidity remains at 45 1 Thus the vapor pressure When the air temperature reaches 5 C Will be the saturation pressure of water at this temperature Which is Pv satwater5 C 08725 kPaA Assuming the overall pressure remains constant we can solve for the speci c humidity of the air w 0 622 Psatwater5 C s P Psacwacer5oc 06221 08725 kPa 110 kPa7 108725 kPa 00049730 kg HQOkg dry air Since the mass fraction of water vapor in the air is very small the amount of water Which condenses per kg of original air sample is essentially equal to the change in the speci c humidity Which is Awg wsgooc 7 ws5oc 0011227 kg HQOkg dry air 7 00049730 kg HQOkg dry air 0006254 kg HQOkg dry air 1 MEM 310 Thermodynamic Analysis I Fall 2010 Homework 1 Solutions Available Wednesday September 29 2010 a The gauge pressure of the air in the tank Pgauge is the amount by which the tank air pressure is greater than the external air pressure Pext the pressure at the mouth of the manometer tube We can solve this problem by using the principle that in equilibrium any two points at the same height in a uid system will have the same pressure The pressure at the airwater interface inside the tank is P1 Pgauge Pext39 This must be identical to the pressure at the same height inside the leftmost manometer tube We may then use the fact that the pressure due to a column of uid with density p and height h is P pgh to compute the pressure at the level of the oilmercury interface in the middle manometer tube as P2 P1 poilghl pwaterghQ Pgauge Pext poilghl pwaterghQ39 This pressure must be equal to the pressure at the same height in the rightmost manometer tube This pressure can also be computed as due to the external air pressure and the pressure from the column of mercury of height hg above it P2 Pext pmercuryghBA Comparing these expressions for P2 gives Pgauge pmercuryghB poilghl pwatergh2gt 104 Pa atm39 b Since the gauge pressure is now zero we can use the equation for the gauge pressure from part a and set it equal to zero to get 0 pmercuryghg poilghl pwatergh2gt39 where h l HQ and hg are the new values for these distances with the tank lid removed In order to compute the change in the mercury level at the mouth of the manometer we must see how each of the heights h1 h2 and hg changes when the pressure is released Suppose that the mercury level at the manometer mouth goes down by some distance Aactube Then assuming the liquids are incompressible the level of the oilmercury interface Will rise by Azmbe so that hg decreases to he ha ZAItube Similarly the oilwater interface Will go down by Azmbe so that hg becomes h hg 7 2Azmbei When the oilwater interface moves down by Azmbe a volume AV Ambe Azmbe of water Will move from the tube into the tank Where Ambe is the crosssectional area of the manometer tuber So the water level in the tank Will rise by Aztank AVAtank Where Atank is the crosssectional area of the water tank and hl Will become A i hl Artube Arcank hi Artube lt1 Lube Atank Inserting all of this into the gauge pressure equation gives 0 7 memuryg ha 7 2 Azmbe A 7 pong lthl AItube lt1 pwaterg M 7 2 Alma tank A b Pgaugeoriginal 7 Artubeg lt2 memury 1 Am egt nal 7 2pwatergt tank Solving for Azmbe gives P 2 a nri am Azmbe m 241 cmi g lt2pmercuty lt1 23 pan 7 2pm 2 a The pressure at the bottom of the building is equal to the pressure at the top of the building plus the pressure of a column of air With height equal to that of the building We can Write Pbottom Hop pairghbuilding The air pressure measurements in this problem are given in terms of the height of a column of mercury Which would produce an equivalent pressure To convert them to the SI unit of Pascals we therefore simply multiply by the density of mercury and the gravitational acceleration near Earth s surface PLop 13600kgm39i81ms20i683m s 911104 Pa Phenom 13600kgm39i81ms20i755m s 101105 Pa Solving for hbuilding gives Pbottom Hop aky hbuilding m 829 mi b The building in question is the newlybuilt Burj Khalifa skyscraper in Dubai c I ll let you know if anybody comes up With anything good 9 3 a The heat current J transmitted through an object is the temperature difference AT between the two sides of the object divided by the thermal resistance Rm of the object J ATRth The thermal resistance is computed as Rm LHA where L is the length of the object along the direction of the heat ow A is its crosssectional area and H 1s its thermal conductivity For the window l m Rimwmdow W N 000521 Kw For the wall 01 m Ema N 001 KW So the ratio of the two heat currents is Jwall 7 ATRthwall iRthwindow 7 a 0467 Jwindow ATRthwindow Rthwall The heat current passing through a thermal resistor is J ATRm The heat currents through the window and the wall ow in parallel so the total heat current is simply their sum 1 JoaJwinow Jwa AT 7 c c 1 d 11 lt Emma 7 s 2820 W Rthwindow gt 4 a We will proceed step by step through the gas cycle The results are summarized in two tables at the end i 2 kg of nitrogen corresponds to 2000 g 7 7142 1 n 28 gmol m0 7 or N nNA 7142 mol 6022 1023 moleculesmol 430 1025 molecules Using Boltzmann s constant 103 138 10 23 Jmolecule K and the ideal gas law PV N kBT gives P1V1 T1 NkB 168K for the initial temperature As the expansion in stage ii is isothermal T2 T1 and we are given V2 3V1 3 m3 We can determine the new value of the pressure using the ideal gas law iNkBTgiNkBTlil 71 5 P27 V2 7 3V1 7 P17310Pa As a consequence the work done on the gas during the expansion is V2 Won 7 PdV Vi WN 77w AV gt V1 E V2 7110105 J N mlt PlVl ln lt Here we have used the fact that P NkBTlV is true throughout an isothermal expansion Conservation of energy requires that AU Won Qin Since the internal energy of an ideal gas is U aNkBT where a for a diatomic gas is 52 a constant temperature means that AU 0 and therefore 7o dlel Cooling the gas at constant pressure gives P3 P2 and we are given T3 T22 842 K The new volume is therefore 7 NkBTg 7 NkBT12 7 3NkBT1 a 7 Rm 2 H The change in the internal energy of the gas is 3 V3 EVl 15 m3 AU7MM wWRaNltRgt7H7m3m Since the pressure is constant the work done on the gas is simply 1 Won 7P2 AV 7P2V3 7 V2 7 lt P1gt 3 7 3V1 7 EPlVl 5 104 J By conservation of energy we therefore have in AU 7 Won 7173105 J ivi We are given that V4 V1 1 m3 and we know that during an adiabatic process the lt quantities VT and PV397 remain constant where 7 75 for a diatomic gas Thus we can compute V3 7 4 P4 P3 i s58810 Pa V4 V3 10 T4 7 T3 9910K The heat which ows into the gas during an adiabatic process is by de nition Qin 0 so by conservation of energy we have Won AU 1ng AT s 22000 J i We are given that V5 V4 V1 1 m3 and that P5 P1 105 Pa and therefore T5 T1 168 L as well No work can be done during an isobaric constant volume process and therefore Won 0 and by conservation of energy Qan AU 1ng AT s1103105 Jr Stage After stage i l Pressure Pa Volume m3 Temperature 105 1 168 After stage ii 105 3 168 After stage iii 2 105 115 8412 After stage iv 5188 104 1 9910 After stage v 105 1 168 Stage 1 AU J Won J Qin J ii 0 7110105 110105 iii 7123105 5 104 7173105 iv 220104 220 104 0 v 103 105 0 103 105 Computing the total change in internal energy across steps ii v con rms that the gas has indeed returned to its initial state The total work done on the gas was 7318 104 J negative and the total heat put into the gas was 318 104 J positive and thus the total energy of the system was conserve r 1f the gas goes through one cycle every ten seconds and generates 38104 J of energy every cycle as we computed in part a then the rate of power generation during this process is Powerom 318 kJ AWby 7 38104 J At 7 10 s 1 MEM 310 Thermodynamic Analysis I Fall 2010 Homework 4 Solutions Wednesday November 3 2010 a As the process is isothermal the working uid and the heat sink are both at the same constant A 0quot V C temperature 300 K and therefore this is the temperature at which the heat transfer takes place For the working uid 7 Ash Q uid uid and so Qin uid T uid AS uid 300 K71 5 kJK Since all of the heat rejected by the working uid goes into the heat sink at the same temperature as that of the working uid the heat sink receives no other heat and there are no other sources of entropy generation it must be the case that ASheat sink AS uid 1 5 kJK The total entropy change is MSW As md Mm sink 0 kJK l 2 a We can assume that this is a steady state process so the mass in ow and out ow must balance The Thinf 7hin 5 kgs 025 kgs 525 kgs No heat leaks are mentioned which means that the total energy ow carried by the incoming steam and water must equal the total energy ow carried by the outgoing water Since no velocity or elevation changes are described we can also assume that the changes in kinetic and potential energy are negligible so the only energy that must be considered is the internal energy of the water and the ow work in other words the enthalpy The energy balance equation is therefore minfhinf minyhiny mouthoum It can be seen from table A7 that the enthalpy of compressed liquid water varies slowly with pres sure consequently we can assume that the speci c enthalpies of the incoming and outgoing liquid streams are approximately equal to the speci c enthalpy of liquid water at the same temperature and saturation pressure So from table A4 we have hinf hfsat200C 83915 kJkg hum hf75a 75ooc 20934 kJkg Solving for the speci c enthalpy of the incoming steam gives h mouthout minfhinf m v Thimv 7 525 kgs20934 kJkg 7 50 kgs83915 kJkg 7 25 kg s 27178 kJkg According to table A 4 this is the enthalpy of saturated steam at a temperature somewhere between 125713000 Linear interpolation gives Tin 12836 C e exergy must e ca cu ate w1t respect to some equ1 1 r1um state w 1c 1n t 1s case w1 e bTh b ll d 39h 39l39b39 h39h39 h39 39llb the state of the water when it is at the temperature and pressure of the external environment 20 C The incoming stream of liquid water is already at the environment temperature so its speci c exergy is dim 0 The speci c exergies of the other two streams can be calculated relative to this state inv hiny hinf Tinf3inv Sam 27178 kJkg 7 83915 kJkg 7 20 C 27315 K 70431 kJkg K 7 02965 kJkg K 65612 kJkg out hout hinf Iinf30ut Sinf 20934 kJkg 7 83915 kJkg 7 20 C 27315 K 7038 kJkg r K 7 02965 kJkg K 60250 kJkg The change in the exergy of the system is the outgoing exergy minus the incoming exergy and so AX mou bout minf39LJinf mimv39LJiny 525 kgs60250 kJkg 7 025 kgs65612 kJkg 50 kgs0 kJkg The secondlaw ef ciency for this process is the ratio of the outgoing exergy to the incoming exergy 7 mom20m 7 525 kgs60250 kJkg 7 7 minimquot miny m 025 kgs65612 kJkg 50 kgs0 kJkg A 0 V 3 a Table A 1 gives M 002897 kgmol and R 02870 kJkg K for air while table A 17 provides the idealgas properties of air across a wide range of temperatures The calculations for each stage of the cycle are described below The results are Stage T P kPa 11 mgkg u kJkg h kJkg PT 1 310 200 044485 22125 31024 15546 2 42315 600 020263 30332 42493 46638 3 2000 28328 020263 16787 22521 2068 4 10837 200 15551 81383 11191 14600 Process Au kJkg qm kJkg won klkg 1 H 2 8207 0 8207 2 3 13754 13754 0 3 A 4 786487 0 786487 4 A 1 759258 780886 22205 totals 0 5665 75608 Stage 1 Initial state The initial temperature must of course be converted to Kelvins T1 37 C 37 273 K 310 K Using the ideal gas equation the initial speci c volume is therefore RT1 02870 kJkg K310 K 3 7 044485 k 7 1 P1 2 100 kPa m g Since internal energy and enthalpy are a function only of temperature for ideal gases the speci c internal energy and enthalpy may be read from table A17 in the text ul 22125 kJkg hl 31024 kJkg 1 7gt 2 Isentropic compression from P1 200 kPa and T1 37 C to P2 600 kPa For an ideal gas undergoing an isentropic process under nonconstant speci c heat assump tions the ratio of the initial and nal pressures must equal the ratio of the initial and nal relative pressures The initial and nal pressures are known and the initial relative pressure from table A17 is Pm 15546 so the nal relative pressure is P2 600 kPa P P 7 15546 7 46638 2 Pl l 200 kPagt This falls between 4522 and 4915 which are the relative pressures at 420 K and 430 K respectively lnterpolation gives 430 K 7 420 K T2 420 K lt 4915 7 4522 gt 46638 7 4522 42361 K The new speci c volume internal energy and enthalpy are therefore 7 RT 7 02870 kJkgK42361 K 7 3 7 j 7 W 7 020263 In lltg 30799 kJkg 7 30069 kJkg 3902 W 7 30069 kJkg lt 46633 7 4522 7 30332 kJkg 4915 7 4522 7 43142 kJkg 7 42126 kJkg 7 h2 7 42126 kJkg 7 W 46633 7 4522 7 42493 kJkg Since this is an isentropic compression the heat input must be zero and therefore the work done on the system per unit mass is equal to the change in internal energy per unit mass womlag Aul g ug 7 ul 30332 kJkg 7 22125 kJkg 8207 kJkg 2 7gt 3 Constantvolume heat addition to T3 2000 K The nal temperature is given and the nal speci c volume must be the same as the initial Value 63 7 62 7 020263 mSkg The nal relative pressure speci c internal energy and speci c enthalpy may be read from table A17 ng 2068 ug 16787 kJkg hg 22521 kJkg The nal pressure is 02870 kJkgK2000 K 7 28328 kP 63 020263 mSkg a P3 In a constantvolume expansion no work is done on or by the air so the speci c heat input is equal to the change in speci c internal energy qing g Augng ug 7 ug 16787 kJkg 30332 kJkg 13754 kJkg 3 7gt 4 Isentropic expansion to P1 200 kPa We can calculate the nal relative pressure as we did for the expansion 1 7gt 2 to be P 200 kP PM 7 Pm 7 2068 7 14600 This falls between 1439 and 1552 which are the relative pressures at 1060 K and 1080 K respectively and interpolation gives 1080 K 7 1060 K T4 1080 K 1 lt 1552 7 1439 RT4 02870 kJkg K10837 K 114 7 gt 14600 71439 10837 K 715551 mSkg P4 2100 kPa 7 82788 kJkg 7 81062 kJkg 7 u4 7 81062 kJkg 7 W 1460071439 7 81383 kJkg h4 7 111486 kJkg 7 W 14600 71439 11191kJkg As in the isentropic expansion 1 7gt 2 the heat input is zero so the work input per unit mass is equal to the change in the speci c internal energy womgn4 Au344 u4 7 ug 81383 kJkg 7 16787 kJkg 786487 kJkg 4 7gt 1 Constantpressure heat rejection until the working uid returns to its initial state The change in the speci c internal energy can be computed from the table Au421 ul 7 u4 22125 kJkg7 81383 kJkg 759258 kJkg We also have for a constantpressure process wonixl 7 7P4 Av 7 P4014 7 v1 7 200 kPa15551 mSkg 7 044485 mSkg 7 22205 kJkg gm 7 Ah 7 hl 7 h4 7 31024 kJkg 7111911 kJkg 7 780886 kJkg Consequently the P v and T s graphs are as follows 3000 2500 2000 1500 P kPa 1000 500 2000 1500 T K 1000 500 Pv diagram 3000 7 2500 7 2000 7 1500 7 1000 0 2 0 4 0 6 0 8 1 1 2 1 4 1 6 vhfkg Ts diagram 7 2000 7 1500 7 1000 8 kJkg K Note that the area enclosed by each cycle is equal to the work done by the system during that cycle b The net work output per unit mass is the sum of the work output over all steps so wbymet 710mm 56018 kJkgi c The thermal ef ciency is the ratio of the net work done by the air wbymet to the the heat absorbed from the hot reservoir qing gi whynet 56018 kJkg 0540775 77 qingug 137554 kJkg 4 As in problem 3 we summarize our results in a table rst and then show the work below Stage T P kPa 1 mgkg 1 323 100 092701 2 87272 32423 0077251 3 26917 10000 0077251 4 99621 30842 092701 Process Au klkg qm kJkg won kJkg 1 A 2 45847 0 45847 2 A 3 15170 15170 0 3 A 4 714140 0 714140 4 A 1 756145 756145 0 totals 0 9555 79555 Stage 1 Initial state s before we use R 02870 kJ kg K as the gas constant of air We are told to use the speci c heat of air at 900 K which according to table A2 is 00 0834 kJkg The initial temperature must of course be converted to Kelvins T1 37 C 50 273 K 323 K Using the ideal gas equation the initial speci c volume is therefore RTl 02870 kJkg 323 K 3 7 092701 k 7 1 P1 1100kPa m g 1 A 2 Isentropic compression to minimum volume he compression ratio is 7 12 so the nal speci c volume is 3 v2 U71 W 0077251 mSkg The compression is assumed to be polytropic Pun constant Tvn l constant with n 14 so the nal pressure and volume will be n P2 Pl lt11 PIT 100 kPa121394 32423 kPa v2 v n71 T2 T1 A lesl 323 K1214 1 37272 K 3902 No heat is added to the gas during this process since it is isentropic so the work done will be equal to the change in internal energy woml g A1213 T2 7 T1 0834 kJkg K87272 K A 323 K 45847 kJkg 2 A 3 Constant volume heat addition to P3 10 MPa The volume is constant so v3 v2 0077251 mSkg The nal temperature is therefore ng3 104 kPa 000077251 10 5 mSkg T3 f 02370 kJkg K 26917 K No work is done during a constantvolume process so the heat added to the gas is equal to its change in internal energy gimgg A0233 Mug 7 ug 0334 kJkg K26917 K 7 37272 K 15170 kJkg 3 7gt 4 Isentropic expansion to original volume We are given that v4 v1 092701 mSkg and as this process is again polytropic with n 4 we have P4 7 P3 7 Pgrn 7 104 kPa12 1394 7 30642 kPa v4 n71 T4 7 T3 T 1 7 26917 K121 14 7 99621K v4 Since the process is isentropic no heat is added to or removed from the gas and therefore wowH4 7 AuH4 7 cum 7 T3 7 0834 kJkg K99621 K 7 26917 K 7 714140 kJkg 4 7gt 1 Constant volume cooling to original state As this process takes place at constant volume no work is done so the heat added to the gas 756145 kJkg must be the same as the change in its internal energy qin4 gt1 All4A1 cVT1 7 T4 0834 kJkg K323 K 7 99621 K a From our data table above the temperature at the end of the expansion process is T4 9962 K b From the data table the net work output is wbymet 710mm 9555 kJkg wbymet 9555 kJkg 06299 gingag 15170 kJkg c The thermal ef ciency is 77th Note that this result could also have been obtained from the equation 1 l 7 l 7 W 06299 77W 1 7 T7171 d The mean effective pressure is wbymet 9555 kJkg MEP 11244 lltP i 61 7 62 092701 mSkg7 0077251 mSkg a 5 The properties of air at room temperature N 300 K can be found in table AZ The speci c heats are up 1005 kJkg K and CV 0718 kJkg K so that k 1400 As before R 02870 kJkg K for air An ideal Brayton cycle consists of two isentropic processes compression in a compressor and expansion in a turbine and two isobaric processes heating and cooling in a heat exchanger the given turbine and compressor ef ciencies modify the amount of work which is producedconsumed by the working uid during the two isobaric processes We will rst calculate the data for an ideal cycle and then take the nonidealities into account afterwards For an ideal Brayton cycle operating between the given temperatures and pressures Stage T P kPa 1 mgkg 1 310 100 088970 2 67203 1500 012858 3 1030 1500 019707 4 47513 100 13636 Process Au kJkg gm kJkg won kJkg 1 A 2 25994 0 36384 2 A 3 25702 35976 0 3 A 4 739840 0 755765 4 A 1 711856 716595 0 totals 0 19381 719381 Stage 1 Initial state The air begins at a temperature of T1 37 C 37273 K310 K and a pressure of P1 100 kPa The initial speci c volume is therefore 7 RT1 7 02870 kJkg 310 K 7 3 i i i W 7 088970 m kg v1 1 A 2 Isentropic compression to the maximum pressure P2 1500 kPa The nal pressure for this compression is given so that we know the pressure ratio is P2 1500 kPa 15 T i 7 P P1 100 kPa and we may treat the compression as a polytropic process Pun constant Tvn l con stantwith n k 1400 The nal speci c volume and temperature are therefore 1 v2 v1 g my 08897 mSkg15 11394 012858 mSkg 2 kil 71 T2 T1 T1 TIEk mamk 310 K1514 114 67203 K As this is an isentropic compression there is no heat input Since it is an open system the work done on the gas must be the source of not only the change in the internal energy of the gas but also the ow work and so we have Aulag cVT2 A T1 0718 kJkg K67203 K A 310 K 25994 kJkg won1A2 AulAQ P2112 P1111 25994 kJkg 1500 kPa012858 mSkg 7 100 kPa088970 mSkg 36384 kJkg 2 7gt 3 Isobaric heating to the maximum temperature T3 757 C 757 273 K 1030 K The nal temperature is given and the nal pressure is the same as the initial pressure P3 P2 1500 kPa So the nal speci c volume is 7 RTg 7 02870 kJkg 1030 K 7 P3 7 15 100 kPa 3903 7 019707 mSkg Because this process is isobaric and because no work is done on the working uid by a heat exchanger the heat added to the gas during this process is equal to the change in internal energy plus the ow work qingng up T3 7 T2 1005 kJkg K1030 K 7 67203 K 35976 kJkg and the change in the internal energy of the gas is Augng cVT3 7 T2 0718 kJkg K1030 K 7 67203 K 25702 kJkg 3 7gt 4 Isentropic expansion back to the minimum pressure P4 100 kPa As with the isentropic compression 1 7gt 2 this process may be treated as a polytropic process with n k 1400 With the given pressure ratio we have as before P 1115 v4 7 US 7 53513 7 019707 mSkg1511394 713636 mSkg P4 kink 7k71k 7147114 T4 7 T3 7 TSTP 7 1030 K15 7 47513 K 3 The heat input to the working uid during this process is again zero so the work must again include the ow work AuH4 7 aT4 7 T3 7 0718 kJkg K47513 K 71030 K 7 739840 kJkg won37gt4 Au374 P4114 Psvs 7 739840 kJkg 7 100 kPa13636 mSkg 7 1500 kPa 019707 mSkg 755765 kJkg 4 7gt 1 Isobaric cooling back to the initial state As with the isobaric heating we have qual 7 515Tl 7 T4 7 1005 kJkg K310 K 7 47513 K 7 716595 kJkg M431 7 cVT1 7 T4 7 0718 kJkg K310 K 7 47513 K 7 711856 kJkg won47gt1 0 a The net work done by the actual cycle as opposed to the ideal cycle must take the thermal ef ciencies of the compressor and the turbine into account w 39d l 776 M 0857 wcactual we c l m 0 95 wtideal The compressor operates during the process 1 7gt 2 while the turbine operates during the process 3 7gt 4 The actual work done by the compressor and turbine is therefore why17gt2ideal 736384 kJkg 776 085 why m cma ntwby nm dea 095 55765 kJkg 52986 kJkg why17gt2actual 742805 kJkg The actual net work done during one cycle is therefore lwbymetaml 7 whyH2acua1 wbyHWal 7 7428905 kJkg 52986 kJkg 7 10181 kJkg l b The thermal ef ciency of the cycle is whynetactual 77th 1inaccual qin is the thermal energy added during the isobaric heating process 2 7 3 The states 1 and 3 in the actual process are the same as in the ideal process maximum temperature and pressurey but the states 2 and 4 are different and their enthalpies must be recalculated based on the actual ef ciencies of the compressor and turbine in order to recompute the heat transfers for the processes 2 7gt 3 and 4 7gt 1 We can recompute hg using won1gt2actual h2actual h17 won3gt4actual h4actual h37 and hl CPTl 1005 kJkgK310 K 31155 kJkg7 hg CPTg 1005 kJkg K1030 K 103515 kJkg so that hg ctua hl womlnzacmal 31155 kJkg 42805 kJkg 73960 kJkg7 magma hg womsu l cmal 103515 kJkg 752986 kJkg 50529 kJkg This gives gimml 7 4223 7 53 7 ham 7 103515 kJkg 7 73960 kJkg 7 29555 kJkg quuhmal 7 74421 7 7h1 7 Manual 7 731155 kJkg 7 50529 kJkg 7 19374 kJkg W 10181 kJkg 7 034447 am 29555 kJkg 77th 6 As usualy we present the results table rsty followed by the calculations Stage T P kPa 1t 1 mgkg 5 klkg h kJkg u kJkg 1 33103 1600 1 0012123 090784 27786 25847 2 29470 600 097730 0034295 090784 25830 23818 3 29470 600 0 00008199 030799 8151 8102 4 1600 0 00008199 030799 8233 8102 Process Au klkg qm kJkg won kJkg 1 A 2 0 956 2 A 3 715716 717679 0 3 A 4 0 082 4 A 1 17745 19553 0 totals 0 1874 71874 Stage 1 Initial state Initially the refrigerant is a saturated vapor at P1 16 MPa Table A12 gives the other initial properties T 3901 T15Mpa 7 5788 7 5788 7 27315 K 7 33103 K vga15Mpa 7 012123 mSkg ul ugsat16MPa 25847 kJkg7 hl hgsat15Mpa 7 27786 kJkg7 81 8gsat15Mpa 7 090784 kJkg K 1 7gt 2 Isentropic expansion from P1 16 MPa to P2 06 MPa The nal pressure is given andy as the refrigerant is under the saturation curve and coolingy it will remain in a saturated statey albeit with reduced quality So the nal temperature can be read from table A lZ T2 Tsa 05MPa 21550C 2155 27315 K 29470 K As the compression was isentropicy we can compute the nal quality by setting 32 81 where So we have Noting that we can then conclude that I 5fg2 81 A 8m 82 1 ISf2 18g 2 Isfgjy 3f Sfsat05MPa 030799 kJkg r K7 ngg ngsa 05Mpa 061378 7 090784 kJkg K 7 030799 kJkg K 7 061378 kJkg K 097730 um 7 vfsa05Mpa 7 00008199 mSkg 71g 7 vga05Mpa 7 0034295 mSkg um ufsat06MPa 8102 kJkg7 ugg ug5a 05Mpa 24183 lJlg7 hm hfsat06MPa 8151 kJkg7 hfg2 hfgsato5Mpa 7 18090 kJkg7 v2 s pg 7 0034295 mSkg W 7 1 7 zuf2 7 rug 7 17 0977308102 kJkg 7 09773024183 kJkg 7 23818 kJkg h2 7 hm 7 mg 7 8151 kJkg 09773018090 kJkg 7 25830 kJkg 11 No heat is added during this process so since the Rankine cycle does not operate as a closed system the work done on the refrigerant is simply the change in the internal energy plus the flow work that is the change in the enthalpy womldg hg 7 hl 25830 kJkg 7 27786 kJkg 71956 kJkg The change in internal energy is Auldg ug 7 ul 23818 kJkg7 25847 kJkg 72029 kJkg 3 Isobaric cooling to a saturated liquid state The nal state in this case is a saturated liquid at the same pressure and therefore temperature as in stage 2 so T3 T2 29470 K P3 P2 06 MPa The other nal parameters may be read from table Al2 v3 vgsamsMpa 00008199 mSkg us ufsat06MPa 8102 kJkg7 ha hfsat06MPa 8151 kJkg7 83 Sfsat05MPa 030799 kJkg The change in internal energy is Augag ug 7 ug 8102 kJkg 7 23818 kJkg 715716 kJkg and since no work is done by the heat exchanger the heat added must include both the change in internal energy and the flow work in other words the change in the enthalpy qingdg hg 7 hg 8151 kJkg 7 25830 kJkg 717679 kJkg 4 Isentropic compression from P3 06 MPa to P4 16 MPa As the refrigerant is a saturated liquid at stage 3 isentropic compression will bring it to a com pressed liquid state The nal pressure is given and the speci c volume of the liquid can be assumed not to change signi cantly over the relatively small pressure difference so we have v4 e vs 00008199 mSkg As the process is isentropic we also have 84 83 030799 kJkg K and 405354 0 Consequently the work done on the system must equal the change in internal energy plus the flow work The change in internal energy may also be assumed to be negligibly small so that u4 m ug 8102 kJkg and Au37gt4 07 and the only work done on the refrigerant will be the flow work Since the volume does not change appreciably this flow work is womgd4 v3P4 7 P3 00008199 mSkgl6106 Pa 7 06106 Pa 08199 kJkg and so the new enthalpy is h4 h3 womgn 8151kJkg 08199 kJkg 8233 kJkg The readily available data do not allow us to calculate T4 but it is not needed for our analysis 12 4 7gt 1 Isobaric heating from compressed liquid to saturated vapor T e change in internal energy is Au4nl ul 7 u4 25847 kJkg 7 8102 kJkg 17745 kJkg7 andy since no work is done by the heat exchangery the heat added must include both the change in internal energy and the ow worky in other wordsy the change in the enthalpy qiwag hl 7 h4 27786 kJkg 7 8233 kJkg 19553 kJkg a From the data tabley the Ts graph is Ts diagram 380 380 360 7 l 7 360 340 7 7 340 g a 320 7 7 320 300 7 7 300 280 i i i i i 280 02 03 04 05 06 07 08 09 1 8kJkgK The saturation curve is plotted from the data in table Ally which gives the entropies of the saturated liquid and vapor states of refrigerant134a across a range of temperatures We can see that almost the entire process takes place under the saturation curvey except in the neighborhood of stage 4 b The thermal ef ciency is whymet why17gt2 why3 4 1956 kJkg 7 708199 kJkg 01042 qin gin27gt3 19553 kJkg 77th c The power output is simply the speci c work times the rate of mass ow Power mwbymet 4 kgs1874 kJkg 7496 kW H a MEM 310 Thermodynamic Analysis I Fall 2010 Homework 3 Solutions Wednesday October 13 2010 We know the velocity and crosssectional area of the incoming stream of steam so the mass ow rate can be found simply by using the speci c volume of steam at 4 MPa and 450 C from the steam table table A6 in the text 1 WMPMWC 003004 mSkg MA 70 ms 00060 m2 m 7 5247 k 1 003004 mSkg gS Since this is a steady ow the total energy and mass of the uid inside the inlet must not vary over time Thus both the net energy in ow rate and the net mass in ow rate must be zero The net mass in ow rate is 752 752 7 mo 07 where 752 and 7520 are the mass in ow rate via the inlet and out ow rate via the outlet respectively Therefore 7520 7h 5247 kgs The net energy in ow rate is t t t 39 EEi Eo Qout 07 where E and E0 are the energy in ow rate via the inlet and out ow rate via the outlet respec tively and Q0 is the rate of heat ow out the control volume given as Qom 100 kJs The energy ow terms include the internal energy ow rate of the steam the rate of ow work the kinetic energy ow rate of the steam and the potential energy of the steam We have already assumed that the potential energy is negligible The internal energy and the ow work PV can both be related to the enthalpy via H U PV and we can therefore use the speci c enthalpy values from the steam table A 6 in the text h thMpamoC 33312 MJkg ho hggMpa oooC 30242 MJkg Using these data Hip miohio The rate of kinetic energy ow through the inlet or outlet is of course simply 127i2i0Vi20 where V is the velocity of the steam at the inlet or outlet So we have 10 1 2 1 2 E mihi Emivi moho Emovo Qout 0 Using the fact that 7520 752 we can simplify this to E39 an 7 he 0 7 23gt 7 QM 0 Solving for V0 the only unknown 1 2 v0 72 2h i he V 7621 183 mi c To nd the crosssectional area of the outlet we can use m0 mi and mo Vvovot The speci c volume at the outlet is v0 Ug2Mpa3oooC 012551 Inskg So the crosssectional area is A 864210 4 m2 8642 cm move V 2 Because the capillary tube is adiabatic there is no heat ow in or out of the systemi Since the tube is horizontal there is no signi cant change in the potential energy of the refrigerant along the tube Since the ow velocity is constant there is no signi cant change in kinetic energy along the tube So the only remaining type of energy which can vary along the tube is the enthalpy internal energy plus ow work However since this is a steady state system the energy in ow must equal the out ow and so the enthalpy in ow and out ow rates must balance as well Moreover because this is a steady state system the mass in ow and out ow rates must also be the same and consequently the speci c enthalpy of the refrigerant at either end of the tube must be the same hi her The frictional drag due to the capillary tube walls causes the uid pressure to drop as the refrigerant goes through the tube causing a certain amount of the refrigerant to use some of the uids thermal energy to vaporize which causes the temperature to drop as well At the inlet the refrigerant enthalpy is the enthalpy of saturated refrigerant134a liquid at 60 C h haw ooc 13936 kJkgi table A 11i Assuming a fraction 1g of the uid mass turns into vapor we can compute the enthalpy at the outlet in terms of the enthalpies of saturated liquid and gaseous refrigerant134a at 716 C hof hfsat716 C 3069 kJkg hog hg at soc 24087 kJkg ho 17 Ighof zghog Inserting this into the condition that enthalpy be constant and solving for 1g gives hi hof I 7 05170 g hog hof This quantity is the quality of the output refrigeranti 3 a Because the ow is steady mass conservation requires that the mass in ow and out ow rates must e ba ance 751m Thu 7516 where the subscripts c and h refer to the cold and hot streams respectively We also know that the mass ow rate of the cold stream is 15 times greater than that of the hot stream so 752 157i2ih and therefore m0 257i21h As we have assumed all other forms of energy ow in this system are negligible only the enthalpy ow rate must be balanced The inlet enthalpy ow is a combination of the enthalpy in ows from the cold and hot streams according to table Al3 and the enthalpy out ow will be a combination of enthalpy out ows of refrigerant in liquid and vapor form If we let 1g be the fraction of the output refrigerant which is in vapor form 6 the quality then Hi michic mghhgh Ho moho 75101 Ighof Ighogl Substituting in the relationships amongst the mass ow rates 157hihhic Tthhul 257711th 257511 1 7 Ighof Jr Ighog which may be simpli ed to 15 C Jr th 25h0 251 7 Ighof Ighog We can use hm m hfsat100j 6543 kJkg7 since table A ll clearly shows that a pressure of 16 MPa is above the saturation pressure for refrigerant134a at 10 C and so the refrigerant should be in its uid state The hot stream is in the vapor phase as we can see from table Al3 h hg15Mpa7ooC 29325 kJkg So the outlet speci c enthalpy h0 is l h0 15hic hm 1566 kJkg Examining the saturation table for refrigerant134a table A ll we can see by interpolation that at 16 MPa the nal pressure as well as the initial pressure the speci c enthalpies of saturated liquid and vapor are hfo hf16MPa 7 13936 kJkgi 13296 kJkg 13291 kJkg m 16 MPa715291MPa 1359 kJkg hgo hg16MPa 7 27346 kJkg 7 27730 kJkg 27730 kJkg m 16 MPa7 15291 MPa 2773 kJkg Since the speci c enthalpy at the outlet is in between these two values the output refrigerant will be a saturated liquidvapor combination whose temperature may be found by interpolation of the pressuretemperature values from the saturation table 60 C 7 56 C To 56 C m 16 MPa7 15291 MPa 5785 O b The quality 1g may then be found by solving h0 1 7 Ighof zghog We get he i he f 7 0 1458 Ig hog 7 he 4 a Treating the air as an ideal gas we can use Mair 002895 kgmol from table A l to nd the initial and nal mass of air in the tan MPZV 7 002895 kgmol2 105 Pa 3 m3 m1 831447 Jmol K300 K 6964 k RTi E7 7 MPiV 7 002895 kgmol8105 Pa3 m3 m2 RTZ 831447 Jmol K350 K So the amount of air added to the tank through the inlet is 2388 kg mi m2 7m1 1691 kg As we have assumed that kinetic energy potential energy and work are not important effects in this system the change in the energy stored in the tank is due both to the heat added and the enthalpy in ow from the supply line Although the ow is unsteady the conditions at the inlet are uniform throughout the process matching those of the supply line so we can use these conditions together with the air ideal gas table table A l7 to determine the speci c enthalpy of the in owing air to be h haingoo K 30019 kJkg The total amount of energy added must be equal to the change in the energy inside the tank so Qin mihi m2u2 mlulv where ul and ug are also read from table A17 ul uair oo K 21407 kJkg7 ug uair350 K 25002 kJkg Solving for Qin Qin 7nqu 7 mlul 7mm 75981 kl So in this case we can see that heat was actually removed from the air 5 The Carnot ef ciency Which is the ef ciency of a reversible heat engine is the maximum possible ef ciency for any realworld heat engine Any heat engine With higher ef ciency could be used to construct a perpetual motion machine and consequently is impossible The Carnot ef ciency for the reservoir temperatures given for this machine is TL 300 K 80 717 7TH 7 77600K 705 The claimed ef ciency of the inventors machine is Wont 510 kJ 7 051i 5 Qquot 1000 Id Since this is greater than the Carnot ef ciency this machine violates the second law of thermodynamics and would allow one to construct a perpetual motion machine of the second kind So the claim is not reasonable H MEM 310 Thermodynamic Analysis I Fall 2010 Homework 2 Solutions Wednesday October 6 2010 a The speci c volume of saturated liquid water at 250 C is v1 7 5125000 7 0001252 mSkg So from table A4 in the text the initial volume of the water tank is V1 7 mvl 7 2 kg 0001252 mSkg 7 0002504 m3 b The nal volume of the water tank is V2 7 3Vl 7 30002504 m3 7 0007512 m3 so the nal speci c volume of the water is 7 E 7 0007512 m3 70003756 3 k m 2 kg m g Since the nal state of the water is saturated vapor this must be the speci c volume of water vapor at some temperature Consulting table A 4 we can see that the nal temperature is therefore somewhere between 370 C and 37395 C and linear interpolation between these two values gives 37395 C 7 370 C T 7 370 C 2 l H 0003106 mSkg 7 0004953 mSkg 3726 C gt 0003756 mSkg 7 0004953 mSkg The nal pressure will be the saturation pressure at this temperature and so can similarly be esti mated by linearly interpolating between table A 47s saturation pressures at 370 C and 37395 C 22064107 Pa 7 21044107 Pa 7 7 P2 7 21044 10 Pallt 3739500 7 370 C gt 372600737000 7 2171107 Pa The initial speci c internal energy of the water was the speci c energy of liquid water at 250 C which according to table A 4 is A o ul 7 11135000 7103077106 Jkg and the nal internal energy may be found by interpolation of the speci c internal energy of water vapor 20843 7105 Jkg 7 22301106 Jkg 373950C 7 370 C u m 7 22301106 Jkg lt gt 3726 C7370 C 7 20917106 Jkg Therefore the change in internal energy of the water is AU 7 u2 7 u1m 7 2091106 Jkg710807106 Jkg2 kg 7 2021106 J d Now we return to the initial condition and repeat our computations for an isobaric expansion A 39D V A l h V In this case we know that the pressure remains constant at the initial saturation pressure from table A4 P2 P1 ngsooc 39762105 Pa Since both the initial and nal states are saturated states the water must remain at the saturation temperature for this pressure which is the same as the initial temperature T2 T1 250 C Since the nal state of the water is saturated vapor its nal speci c volume will be the speci c volume of saturated vapor at this temperature from table A4 52 5325000 0050085 mSkg giving a nal volume of V2 mv2 2 kg 0050085 mSkg 0100170 mg The initial and nal speci c internal energies are table A 4 ul 11135000 10807105 Jkg u2 11325000 26018105 Jkg Consequently the change in internal energy is AU W 7 u1m 3042106 J The work done by the water on its surroundings is V2 P dV Vi P1 AV P1V2 V1 09762 106 Pa 0100170 m3 7 0002504 m3 3883105 J Why The heat energy added to the water may be calculated using energy conservation We have AU Won Qin Wby Qin and consequently in AU Why 3042106 J3883105 J 3431106 J 2 a A 0quot V Treating the carbon dioxide as an ideal gas means using the equation of state PV nRuT so we ave We can compute n the number of moles of C02 gas to be m m 2kg M 7 M 7 004401 kgmol 7 4544 101 n where we have used the molar mass M of C02 from table Al M 004401 kgmol Therefore the inital volume of the gas is 07219 m3 7 nRuTl V1 7 P1 For a polytropic process PlVfl PQVQ so we have P 17 V2 V1 4 1733 m3 P2 The nal temperature is therefore 4601 K 1869OC PV T2 Treating the carbon dioxide as a van der Waals gas requires us to use the van der Waals equation of state RuT a 13 U712 7M where the van der Waals constants a and b are derived from the critical temperature and pressure of carbon dioxide table Al TC 3042 K PC 739106 Pa according to 7 27R3Tf 7 27331447 JmolK23O42 K2 a 7 64M2PC 7 64004401 kgmol2739105 Pa b7 BJC 7 331447 Jmol1lt3042 K 7 8MB 7 8004401 kgmol739 106 Pa Rearranging the van der Waals equation of state into the form 1885 mF Pang 972110 4 mSkg P11 7 ltP1b R3213 v av1 fab 0 allows us to plug in the values of all the known quantities 3 105 P207137 1086 105 mSPakgv 1885 mBPakg2vl 7 01833 mgPakgs 0 and solve this cubic equation analytically or with a numerical root nder resulting in v1 03602 mSkg So the initial volume of the van der Waals carbon dioxide is V1 m1 07203 m3 almost but not quite the same as for the ideal carbon dioxide As before we do not really need to know the total mass or volume of the carbon dioxide as the polytrope equation equation may be Written either as PlVl l P2V2 l or P11 P21 The nal speci c volume and volume are therefore P2 V2 mvg L735 ms7 P1 17 v2 v1 i 08674 mSkg again close but not identical to the results for the ideal gas We can then use the van der Waals equation of state to compute the nal temperature M a 0 T2 E P2 v2 7 b 459 K 1866 C Which nearly matches the ideal gas resulti 3 a A 0quot V We will assume that the neon behaves as an ideal gas The ideal gas equation of state is PV nRuT Using V mv and n mM we may divide both sides by m to obtain RuT Pv 7 Therefore the speci c volume is v 7 iRuT 7 MP Initially 7 RuTl 7 831447 Jmol K20 C 27315 K 7 MP1 7 002018 kgmol105 Pa Since neon is a monatomic an adiabatic compression of ideal neon is a polytropic process with 7 53 The quantity PV V is constant as is Pu and using the fact that v RuTMP we can see that PRuTMP39Y is also constant Since Ru and M do not change PTP39Y T YIW 1 is therefore constant Consequently we may write T 7 T3 P971 v1 1208 mSkg and therefore P 1 17 T2 T1 lt1 20 C 27315 K lt 5 lamas P w 5104 K 1 105 Pa The nal value of v is therefore RuTZ 2 MP2 05257 mSkg The change in speci c volume is therefore Av v2 7 v1 706821 kag The enthalpy is de ned as H U PV and the speci c enthalpy is therefore h u Pv For an ideal gas the internal energy is U anRuT where a 32 for a monatomic gas like neon The speci c internal energy is therefore 7 aRuT u 7 M We can compute ozth 5 7 l81210 J k m M g7 u2 3154105 Jkg which give hl ul Pm 3020105 Jkg hg ug ngg 5257 r 105 Jkg Therefore the change in speci c enthalpy of the neon is Ah h2 7 h1 2237105 Jkg 4 The speci c volume of saturated water vapor at 710 C is table AS v1 vg ooc 46717 mSkg Therefore the container initially holds V i 7 m3 7214110 3k v1 7 46717 mSkg g ml of water vapor The speci c volume of saturated water vapor at 720 C is table A S v2 7134000 11313 mSkg So after cooling the container holds V 1m3 4 mgigim7883910 kg of water vapor The missing water vapor has frozen onto the surface of the ice blocky so the mass of the block has increased by Am m1 7 m2 125710 3 kg

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