工程热力学课后答案英文版电子工业出版社

1、1-11 A barometer is used to measure the height of a build ing by record ing read ing at the bottom and atthe top of the build ing. The height of the build ing is to be determ ined.Assumptions The variation of air density with altitude is n egligible.730 mmHg3Properties The density of air is given to

2、 beP = 1.18 kg/m . The3den sity of mercury is 13,600 kg/m .Analysis Atmospheric pressures at the top and at the bottom of the build ing areRop =( Pgh)top=(13,600 kg/m3 )(9.807 m/s2 )(0.730 m)1 kPa7.55. .mmHg=97.36 kPaPbottom =( gh)bottom=(13,600 kg/m3)(9.807 m/s2 )(0.755 m)=100.70 kPa1N,1kg m/s21 kP

3、a2XJ000 N/mTaking an air colu mn betwee n the top and the bottom of the buildi ng and writi ng a force bala nee per un it base area, we obta inWair / A = Pbottom - Rop (;gh) air = Pbottom Ptop(1.18kg/m3)(9.807 m/s2)(h)Y1 kg -m/s2 j1000 N/m2 丿1 kPa=(100.70-97.36) kPaIt yieldsh = 288.6 mwhich is also

4、the height of the buildi ng.1-21 The air pressure in a duct is measured by an in cli ned mano meter. For a give n vertical level differe nee, the gage pressure in the duct and the len gth of the differe ntial fluid colum n are to bedeterm ined.AssumptionsThe mano meter fluid is an in compressible su

5、bsta nee.Properties The den sity of the liquid is give n to be = 0.81 kg/L = 810 kg/m 3.Analysis The gage pressure in the duct is determ ined fromPgage二 Pabs Patm =动32= (810 kg/m )(9.81 m/s )(0.08 m)=636 Pa1 NQ kg m/s 21 Pa21 N/mThe len gth of the differe ntial fluid colu mn isL 二h/sin j-(8cm)/sin 3

6、5 =13.9 cmDiscussion Note that the length of the differential fluid column is extended considerably by inclining the mano meter arm for better readability.2-4 No. This is the case for adiabatic systems on ly.2-6 A classroom is to be air-c on diti oned using win dow air-c on diti oning un its. The co

7、oli ng load is due to people, lights, and heat tra nsfer through the walls and the win dows. The n umber of 5-kW win dow air con diti oning un its required is to be determ in ed.AssumptionsThere are no heat dissipat ing equipme nt (such as computers, TVs, or ran ges) in the room.Analysis The total c

8、ooling load of the room is determined fromQcooling QlightsQaeopleOneatgainwhereQlights =10 100 W =1 kWQpeople =40 360kJ/h =4kWQheatgain = 15,000 kJ / h = 4.17 kWSubstitut ing,Qcooling =14 417 =9.17kWThus the n umber of air-c on diti oning un its required is9.17 kW1.83 '2 units5 kW/unit2-18 The f

9、low of air through a flow cha nnel is con sidered. The diameter of the wind cha nnel dow nstream from the rotor and the power produced by the wi ndmill are to be determ in ed.Analysis The specific volume of the air isRT(0.287 kPa m3/kg K)(293 K)-1 0 k P a3=0.8 4 0m / k gThe diameter of the wi nd cha

10、 nnel dow nstream from the rotor is2 2A1VA2V2 C -D12/4)VC ：D2/4)V2 D2= (7m)10m/s- 9 m/s= 7.38 mThe mass flow rate through the wind mill is2= 457.7 kg/s-A1V1n(7 m) (10 m/s)m3v 4(0.8409 m /kg)The power produced is the nW=m字=(457.7kg/s)22(10 m/s)2 -(9 m/s)21 ""g 2=4.35 kW1000 m2/s22-19 The av

11、ailable head, flow rate, and efficiency of a hydroelectric turbine are given. The electric power output is to be determ in ed.Assumptionsl The flow is steady. 2 Water levels at the reservoir and the discharge site remaincon sta nt. 3 Fricti on al losses in pip ing are n egligible.Properties We take

12、the density of water to be3"=1000 kg/m = 1 kg/L.Analysis The total mecha ni cal en ergy the water in a dam possesses is equivale nt tothe pote ntial en ergy of water at the free surface of the dam (relative to free surface of discharge water), and it can be con verted to work en tirely. Therefo

13、re, the power potential of water is its potential energy, which is gz per unit mass, and mgz for agive n mass flow rate.2emech 二 pe 二 gz 二(9.81m/s )(120m)盘需 “177 kJ/kgThe mass flow rate ism = -V =(1000kg/m3)(100m3/s) =100,000kg/sThe n the maximum and actual electric power gen erati on become1 MW117.

14、7MW1000 kJ/sWmax 二 E mech memech 二(100 ,000 kg/s)(1.17 7 kJ/kg)Welectric 二 overallWma0.80(117.7MW) =94.2 MWDiscussion Note that the power gen erati on would in crease by more tha n 1 MW for each perce ntagepoint improveme nt in the efficie ncy of the turbi ne-gen erator un it.3-9 A rigid container t

15、hat is filled with R-134a is heated. The temperature and total enthalpy are to be determ ined at the in itial and final states.Analysis This is a constant volume process. The specific volume isV 0.014 m33v 1 v 20.0014 m /kgm 10 kgVThe initial state is determined to be a mixture, and thus the tempera

16、ture is the saturati on temperature at the give n pressure. From Table A-12 by in terpolati onT1 Tsat 300 kPa =0.61 Candv1 -v fX:Vfg(0.00140.0007736) m3/kg3(0.067978-0.0007736) m3 /kg= 0.009321hi =hf X1hfg =52.67 (0.009321)(198.13) =54.52kJ/kgThe total en thalpy is the nH 1 =mh1 =(10 kg)(54.52 kJ/kg

17、 ) =545.2 kJThe final state is also saturated mixture. Repeating the calculations at this state,T2 =Tsat 600 kPa =21.55 CV2 fX2Vfg(0.00140.000819S) m3/kg(0.034295-0.0008199) m3/kg= 0.017332h2 =hfx2hfg =81.51 (0.01733)(180.90) =84.64kJ/kgH 2 =mh2 =(10 kg)(84.64 kJ/kg) =846.4 kJ3-22 rigid tank contain

18、s an ideal gas at a specified state. The final temperature is to be determined for two differe nt processes.Analysis (a) The first case is a constant volume process. When half of the gas is withdrawn from the100 kPa300 kPa600 K) =400 Ktank, the final temperature may be determined from the ideal gas

19、relation asm1 P2(b) The sec ond case is a con sta nt volume and con sta nt massprocess. The ideal gas relati on for this case yieldsT2 p = 400 KT1 1 一 600 K300 kP a) =200 kPaIn sufficie nt in formatio n3-32 Complete the following table for H 2 O:P, kPaT, C3v, m / kgu, kJ/kgPhase description200300.00

20、1004125.71Compressed liquid270.31302004001.54932967.2Superheated steam300133.520.5002196.4Saturated mixture,x=0.825500473.10.68583084Superheated steam4-14 Oxyge n is heated to experie nee a specified temperature cha nge. The heat tra nsfer is to be determ ined for two cases.Assumptionsl Oxygen is an

21、 ideal gas since it is at a high temperature and low pressure relative to its critical point values of 154.8 K and 5.08 MPa. 2 The kinetic and potential energy changes are negligible, . ke 三.pe 三0 . 3 Constant specific heats can be used for oxygen.Properties The specific heats of oxyge n at the aver

22、age temperature of (25+300)/2=162.5C=436 K areCp = 0.952 kJ/kg K and c = 0.692 kJ/kg K (Table A-2 b).:-EsystemChangein intern al, kin etic,potential, etc.energiesAnalysis We take the oxygen as the system. This is a closed system since no mass crosses the bou ndaries of the system. The en ergy bala n

23、ee for a con sta nt-volume process can be expressed asQEin - EoutNet en ergy tran sferby heat,work, and massQin 二 U 二 mq(T2 -Ti)The en ergy bala nee duri ng a con sta nt-pressure process (such as in a pist on-cyli nder device) can be expressed asEin Eout=- ：Esystem'W''71Net energy transf

24、erChangein internal, kinetic,by heat,work, andmasspotential, etc.energiesQin -Wb,out = ：UQin 二 Wbg ；UQin =：H = mcp (T2 _Ti)since =U + Wb = UH during a constant pressure quasi-equilibrium process. Substitut ing for both cases,Qin,v 攻nst =mCv(T2 TJ =(1kg)(0.692 kJ/kg K)(300 25)K =190.3 kJQin,p 攻nst =m

25、Cp(T2 Ti) =(1kg)(0.952 kJ/kg K)(300 25)K =261.8 kJ4-25 A rigid tank filled with air is connected to a cylinder with zero clearanee. The valve isopened, and air is allowed to flow into the cylinder. The temperature is maintained at 30 C at all times. The amou nt of heat tra nsfer with the surrou ndin

26、gs is to be determ in ed.Assumptionsl Air is an ideal gas. 2 The kinetic and potential energy changes are negligible,.；ke 三.:pe 三0 . 3 There are no work interactions involved other than the boundary work.3Properties The gas constant of air is R = 0.287 kPa.m /kg.K (Table A-1).Ein - EoutNet en ergy t

27、ran sferby heat,work, andmassAnalysis We take the entire air in the tank and the cyli nder to be the system. This is a closed system since no mass crosses the bou ndary of the system. The en ergy bala nee for this closed system can be expressed asi_EsystemChange in internal, kinetic,potential, etc.e

28、nergiesQin -Wb,out = .：U = m(U2 -Ul) =0Qn Wb, outsince u = u (T) for ideal gases, and thus u2 = uiPV,P2V2Ti 一 T2whenTi = T 2. The initial volume of air is400 kPa 1(0.4 m3) =0.80 m3200 kPaThe pressure at the pist on face always rema ins con sta nt at 200 kPa. Thus the bou ndary work done duri ng this

29、 process is21 kJ 1Wbout = PdV =P2(V2 M) =(200 kPa)(0.8 0.4)m3 =80 kJ' 山Ji kPa m3 丿Therefore, the heat transfer is determined from the energy balance to beWb,out =Qin =80 kJ4-27 An in sulated cyli nder is divided into two parts. One side of the cyli nder contains N 2 gas and the other side contai

30、ns He gas at different states. The final equilibrium temperature in the cylinder when thermal equilibrium is established is to be determined for the cases of the piston being fixed and movi ng freely.Assumptionsl Both N 2 and He are ideal gases with constant specificheats. 2 The energy stored inthe

31、container itself is negligible.3 The cylinder is well-insulated and thus heat transfer is negligible.3Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m /kg.K iscv3=0.743 kJ/kg °C for N 2, and R = 2.0769 kPa.m /kg.K is c = 3.1156 kJ/kg °C -for He (Tabl

32、es A-1 and A-2)Analysis The mass of each gas in the cyli nder ism“2PV1 帀2_(500 kPa « m3 )=4 77 kg0.2968 kPa m3/kg K 353 KmHe.ie(500 kPa 1 m3 ) 2.0769kPa m3/kg K 298 K=0.808 kgN2彳31 m500 kPa80CHe1 m3500 kPa25 °CTaking the entire contents of the cylinder as our system, the 1st law relation c

33、an be written asEin - Eout.:EsystemNet energy transferChangein internal, kinetic,by heat, work, andmasspotential, etc. energies0 wUN U He0 二mQ(T2 -Ti)N2mcv(T2 _T)HeSubstituti ng,It givesTf = 57.2 Cwhere Tf is the final equilibrium temperature in the cylinder.The answer would be the same if the pisto

34、n were not free to move since it would effect only pressure, and not the specific heats.Discussion Using the relation PV = NRuT, it can be shown that the total number of moles in the cyli nder is 0.170 + 0.202 = 0.372 kmol, and the fin al pressure is 510.6 kPa.6-9 An inven tor claims to have develop

35、ed a heat engine. The inven tor reports temperature, heat tran sfer, and work output measureme nts. The claim is to be evaluated.Analysis The highest thermal efficie ncy a heat engine operat ing betwee n two specified temperaturelimits can have is the Carnot efficiency, which is determined from290 K

36、0.42 or 42%500 KThe actual thermal efficie ncy of the heat engine in questi on isWnet300 kJ一 700 kJ= 0.429 or 42.9%which is greater tha n the maximum possible thermal efficie ncy. Therefore, this heat engine is a PMM2 and the claim is false.6-11 A heat pump maintains a house at a specified temperatu

37、re. 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 in put to a heat pump will be a mini mum whe n the heat pump operates in a reversible ma

38、nn er. The coefficie nt of performa nee of a reversible heat pump depe nds on the temperature limits in the cycle only, and is determined fromC°PhP® 一 1 - Tl /Th2 273 K / 22 273 K= 14.75The required power in put to this reversible heat pump is determined from the definition of the coeffici

39、ent of performa nee to beHouse22 C110,000 kJ/hHP5 kW4.77 kg 0.743kJ/kg C Tf 80 C 0.808kg 3.1156 kJ/kg C Tf 25 C =0W _ qh net,in, min -COPhp14.75110，000 亠=2.07 kW3600 sThis heat pump is powerful enough since 5 kW > 2.07 kW.7-8 A reversible heat pump with specified reservoir temperatures is con sid

40、ered. The en tropy cha nge of two reservoirs is to be calculated and it is to be determ ined if this heat pump satisfies the in crease in en tropy pr in ciple.Assumptions The heat pump operates steadily.Analysis Since the heat pump is completely reversible, the comb in ati on of the coefficie nt of

41、performa nee expressi on, first Law, and thermod yn amic temperature scale givesC°PHp,rev11 -几 /Th11(283 K)/(294 K)= 26.73The power required to drive this heat pump, accord ing to the coefficie nt of performa nee, is the nWnet,in =QHJ" =3.741 kWC°PHP,rev 26.73According to the first la

42、w, the rate at which heat is removed from the low-temperature energy reservoir isQl =Qh -Wnet,in =100kW 3.741kW =96.26kWThe rate at which the entropy of the high temperature reservoir changes, according to the definition ofthe en tropy, is and that of the low-temperature reservoir isSh_ Qh_Th100 kW2

43、94 K=0.340 kW/K. SL =鱼二 -96.26 kW _ _0.340 kW/KTl283 KThe net rate of entropy change of everything in this system isStotal = ；ShSl =0.340-0.340=0 kW/Kas it must be since the heat pump is completely reversible.7-11 Steam is expa nded in an ise ntropic turbi ne. The work produced is to be determ in ed

44、.Assumptionsl This is a steady-flow process since there is no cha nge with time. 2 The process is isen tropic (i.e., reversible-adiabatic).Analysis There is one in let and two exits. We take the turbi ne as the system, which is a con trol volume since mass crosses the bou ndary. The en ergy bala nee

45、 for this steady-flow system can be expressed in the rate form asEin - Eout' -= -Rateof net energy transfer by heat,work, and masssystem0 (steady)Rateof cha ngei n intern al, kin etic,potential, etc.energiesEin = Eo u tm1 h-i =m2h2 m3h3 WoutWout =m1hm2h2 -m3h3From a mass bala nee,sm2 =0.05mi =(0

46、.05)(5kg/s) =0.25 kg/sm3 =0.95m1 =(0.95)(5 kg/s) =4.75 kg/sNoti ng that the expa nsion process is ise ntropic, the en thalpies at three states are determ ined as follows:P3 =50 kPaT3 =100 C(Table A -6)Kh3 =2682.4 kJ/kgs3 =7.6953 kJ/kgP-i =4 MPa：> hi =3979.3kJ/kg (Table A -6)si =S3 =7.6953 kJ/kg K

47、P2 =700 kPah2 =3309.1 kJ/kg (Table A - 6)S2 =S3 =7.6953 kJ/kg K |Substitut ing, Wout =m1h -m2h -m3h3=(5kg/s)(3979.3kJ/kg) -(0.25kg/s)(3309.1 kJ/kg) -(4.75 kg/s)(2682.4 kJ/kg) =6328 kW7-13 The en tropy cha nge relati ons of an ideal gas simplify to-s = cp ln( T2/T1) for a con sta nt pressure process

48、and -s = cv ln(T2/T1) for a constant volume process.Noting that cp > cv, the entropy change will be larger for a constant pressure process.7-22 Air is compressed in a piston-cylinder device. It is to be determined if this process is possible.Assumptionsl Changes in the kinetic and potential energ

49、ies are negligible. 4 Air is an ideal gas with con sta nt specific heats. 3 The compressi on process is reversible.3Properties The properties of air at room temperature are R = 0.287 kPa m /kg K, Cp = 1.005 kJ/kg K (Table A-2a).Analysis We take the contents of the cylinder as the system. This is a c

50、losed system since no mass en ters or leaves. The en ergy bala nee for this stati onary closed system can be expressed asEin iE°ut =- EsystemNet energy transfer Changein internal, kinetic, by heat, work, andmass potential, etc. energiesWb,in Qout = U 二 m(U2-Ui)Vb,in -Qout = mcp (T2 -Ti)Wb,in -Q

51、out =0(sinceT2 =Ti)Qout =Wb,inThe work in put for this isothermal, reversible process isP2250 kP aWin =RT ln 2 =(0.287 kJ/kg K)(300 K)ln78.89 kJ/kgPi100 kP aThat is,qout =win =78.89 kJ/kgThe en tropy cha nge of air duri ng this isothermal process isT2P2P2250 kP aLSair =Cpln R lnR ln(0.287 kJ/kg K)ln

52、0.2630 kJ/kg KT1P-iP-)100 kPaThe en tropy cha nge of the reservoir isSr =亚=78.89 kJ/kg =0.2630 kJ/kg KR Tr 300 KNote that the sig n of heat tra nsfer is take n with respect to the reservoir.The total en tropy cha nge (i.e.,en tropy gen erati on) is the sum of the en tropy cha nges of air and the res

53、ervoir:stotal = *sR =-0.2630 0.263 = 0 kJ/kg KNot only this process is possible but also completely reversible.&1 The four processes of an air-sta ndard cycle are described. The cycle is to be show n onP-v andT-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are t

54、o be determ ined.Assumptionsl The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are n egligible. 3 Air is an ideal gas with con sta nt specific heats.Properties The properties of air at room temperature areCp= 1.005 kJ/kg.K,Cv = 0.718 kJ/kg K,and k = 1.4 (Table A-2)

55、.Analysis (b) From the ideal gas ise ntropic relati ons and en ergy bala nee,=3 0 K0 . 4 / 1 .41 0 0I0P a1 0 (k P a1 q414qin =h3二Cp T3 T22800 kJ/kg = 1.005 kJ/kg K T3 _579.2 Tmax =T3 = 3360 K(C)P3V3P4V4可> T4 =2丁3100 kPa 3360 K ；=336 KP31000 kPa'厂qout q34,out q41,out u3 _u F"l.h4 - h1 I=cv T3 _T4 cp T4 _T1=0.7 1k8J/kK 0 3 63 3g<+(1 .0 0k5J/kK3 3 63 0(K=2 2 1I2J/ kg4許亠14器210%qiq34Discussion The assumption of constant specific heats at room temperature is not realistic in this case the tempera