工程热力学习题详解

《工程热力学》习题详解及简要提示(2009.9-任课教师：，助教：胡第一1-（法一）按照式（1-10）P物理大气压定义为：1atm760mmHg133.3P故有hgP

133.3133.32hHO1039.812

133.3789

133.3760860（法二）直接用公HgghHggh故有h

Hg

13.610376010336mm10.336m同理

76013100mmh

13.610376012019mm12.019m1-2PEP0HOghmlg2EP= 9.810.54-9009.81E1.499105Pa1-解：PP ghsin15gh g 9.810.770.2588136009.810.0081039.8197405Pa1-(pdp)SgSdhdp

p和的关系式pc11.4c04dgdh 式（3）中c可由题目中所给条件求得。H1.4c06dH H=5607（法二）由dpgdh有：1.4c04d 1.4c06d1.4c0

21

c8.06104Pa/21

p21

0.711kg/3 3 0 0 1.48.06 0 4h

2)

0.4

)dp1-PcdPbPv76175011mmHg1-1-

PPAPv3601.999362kPaPPPB362170PcPPv1921.999 FPPS1.01310521030.22521.58104 说明Pv表示真空室压Pb表示大气压力P表示水银柱压力力为1.58x104N。1-解：设该温标为tnew，摄氏温标为t/t/newkt/oc75100k200ktnew0.55tc

1-解：假定新温标N与摄氏温标Ct at [N [C400100ab从以上二式得：a3 故关系式为：t 3t [N [C氧气处于600Nt[C

600100166.67(3[CTt 273.15[Ct=540解 按关系式

32如 9 F9则 t155 9

t53241-解：由已知条件有：0aln6100aln36联立上二式得：a

bt*100lnZlnZlnZ当t*10当t*90

1.1ln6ln61Z611 lnZ190ln61.9ln6ln61 Z6191027.887

1100A1041446A1.989105R0B6.83107(1/(C)2R10(14.32103t6.83107t2解：按照功的定义WV2则有WV2PdV0.5106(0.80.4)2105J 解：由PV13 VV2PV1VW PdV 11

111 030dV

对外作出膨胀功85.7kJ解：本可逆过程对外做功量为：W pap

10PadV

D3

WpdV04aDD2dD510504D3dD510510.440.343436J0 0 4 A-B-C-A的净功量为：W0.5(v2v1p2A-C-B-A的净功量为：W0.5(v2v1p2据做功和积分的定义WpdvA-B-C-的净功量就是该循环所包围的三角形面积。解：由图可见，由做功定义WW1ACCB1vvpp 如果循环为ACBA，则（被做功W'1vvpp 解：(1)1kWh13.6106J3.61033.610324 3.610324故B 276000千克/q0 29000

b24250000.46kgkWh)0.46kg解：(1)qwuuqwrp(v22250 22502071.5kJ/(2)焓的变化为hupvqr2250kJ/0--0-0-5解：

Q12U12W1213900Q23U23W23U23Q34U34W3410000 由Q12Q23Q34Q41W12W23W34W41139001000003950解：如维持一稳定室温，则说明车间 系的U由UQ则WQU50 3105103/2.833104W1.02105kJ/为维持合适室温，必须外加采暖设备，其热量为1.02105kJh解：（1）Q1U1 Q1 W1 U1

Q2U2U2

Q239.1kJ(已知条件 U2U1上升5cm，换热量为98J解：设空气为理想气体 T1T2300K VP1VP1LA 又VLA2 2LLP112 195

P 1.013101.911101.028102.9410 P959.87711.0131050.9311051.0281051.96105 2.94 L2L11.96105110 QU T1T2，U1U2，U2QWPAL1.9610510051062上升了3.57cm，气体终温271.2Kxcm P2.94105Pa，P1.96105Pa，T1

T222T1 300 2010xP 2.94 1001061Q

U其中WPAx1.96105100x104 P而UmcTT 11cTT

11由cv1 （cv,mcv

24502x101960xx3.571cm代回计算得出T2Q

(h2

gz)outmin(h2gz)inW 2按题设条件：Q0,Wnet0,mout0,( gz)in2

dEc,v

minE0dEc,vE

minhdmu'0u'h

1 1QWs

mout(2

gzh)outmin(2cgz1而Q

Ws

mout0(2

)in0(gz)in1EdEc,v0minhdmm0)hE2E1(mm0h1mu'mu(mm0 解：m 0,m0RT RTmumu(mm)h0 0cT00cT0( )cRT'

RT'v

RT

P'cPc( 0)cTT 0 TT0P'

P'TPTkT T' 若原为真空P00，则TkT，分析从略

1 1QWs

mout(

gzh)outmin(2cgzmout0Ws又因与大气充分换热，故T' u' 10QdEdEc,vh0mind01Q(E2E1)(mm0)h(mu'm0u0)(mm0)hp(V'V(mu'm0u0)(mm0)hp(mv'p'

T'

v'故Qmm0)(h0 1 1 (QWs out gzh)outmin(2cgzWs mout 0Qd00

dEc,v

minQ(E2E1)(mm0(mu'm0u0)(mm0又T' u'0 故Qmu'mu(mm)h(0 (PV0P0V0)(cTcT

vV0(PP)c(kTT 解：v10.001m3W理想状态下：

3.988105kJ/h m7q(m1m7)h6(m1h1m7h7) Pm7q(m1m7)h6(m1h1m7h7 [70041800(500.7)103418(5010312700

wsh1 220103483.4 轴功率WsDh1h2 2201031402（2）

Dc2

449.20.0067

G2202386.5146.5411770t/ n 53.5 （1）已知q50kJkgu146.5kJkgquwqu50146.5196.5(kJ/由qhwtwtqhququp2v2p1v150146.50.80.1751030.10.845103252kJ/1kg压缩空气需轴功252kJkg55 h

1c2h

1 2 2而h1h2cp(T1T2)1.005(573T2 c1

)v1260mAm

0.278

85.5m/ c (A)v22602105100.373T2m/ (d T20.0696103T 2 T2555Kc2207.1mq01c20gz020hwthh1h2290580290kJNc

1002902.9104 Bmq1006701.52kg/ q(hh)1(c2c2) wnet02[q(2[q(hh)]32222[6702[670-(800-580)]103qh1(c2c2) 由于q0h 1(c2c2)1(94921002)445.3kJ/ 故Ntmwnet4.453104 NNN4.4531042.91041.553104 3-

（1）T2T12293 7（2）mmmmmV1(p1p3)13 R 133-P1105PTT V12

m112m2压入气量mmm

(PP1 1mPVm 1 PV1

(7.84561818.2513- UAB即(U2U1)AU2U1)BT1A 即U1AU2AU2B0.5cvT2AT2B20.5又因完全平衡时T2AT2 T2A即T2AT2BT1AT1BT1A

∴P

P

1A

1B 1.2 41A1B 5 V1A 4 又∵T1AT2A,m1Am2 V1A0.3V1B4 2 2 2 V1AV2A 2

0.3V1B241051.8461053- 解：电机发热功率W 0.05qcp(t2t11552035kJmq

315.7899.0kg/s32400kg/3-t360C ct10.732 300(0.7410.732)0.7374kJ/kg v 400t1700C，ct20.857kJ/kg v故quct2tct1t 1191.44kJ/ v v 5按定比热计算（双原子气体cv2R2.50.28710.71775kJkgq2ucv(t2q2q2 20%，这是因为当t21700C时，已远远偏离cv0.71775kJkgK，从表中所查得ct20.857kJ/kgK，也可以看出与定比热相去甚远。vq1q23-解：折算出在P2T2下的空气流VP0T2V760270273.151.08105 m3/ P2T0

1.013105 m 0 1.395110kg/

/ ttt1 p020.71.0191.0121.0121.0169kJ/kgtpp p Qhm(ct2tp p (1.01692701.0044355.02105kJ/pp2 0.28711.00485kJ/kg5Qhcptt1.0048527020350.4710kg/5 3-

1.0131059解：计算冷空气质量流量1

1.1626105kg/h 287.1273.15由于绝热、定压条件，则HHconstm2m1h1m2h2(m1m2 m2(h2h3)m1(h3p由t120℃ct11.0044kJkgptt2350℃c21.0235kJkgtpt40℃ct31.0048kJkgpct3

ct1

则m p p 1 kg/p p 1ct2tct3p p m2RT0 287.1273.15

5689.4m/h .80665 V 287.1(350 V

1.2959104m3/h3.5997m3/由VD2 )23.14159D2 3.141593-T2T12PV1P16.01052105V V 3、Sms

P2mRln m 1 0.0050.208ln21.14103kJ/63- h1932.93kJ/ u1674.38kJ/ PP2P0.175.29 1

PrPr

hh

uuv2

4807.5297.26810 2287484.71.391m3/ u344.704.7352.08344.70348.17kJ/ h482.494.7492.74482.49487.31kJ/ Wmu1u23674.58348.17979.2kJUW HWtRlnP2S PT2P1T由S°2.84856kJ/kg RlnP20.287ln10.661kJ/kgT1 1S°2.848560.6612.18756kJ/kgS由T2查得：h2487.1kJ/ u2348kJ/由T1h1932.93kJkgu1674.58kJkgwth1h2932.93487.1445.83kJ/kgWtmwt3445.83Wmw3674.38348 k 10如用kconst，则T2 2

900()14由T2查得：h2468.152kJ/kg u2334.38kJ/kgWm(u1u2)3674.58334.381020.6kJ

s2S

Rlnp2 ppe(s2s1)/ 3-1 V(p)k0.1(1)141

1(p

pV) 1(0.98070.11.96140.0612)105H k 1 2 V(V )20.10.0612

pa 2

a1p 0.9807105a1QUa 又WaWH R而Umc(TTR

V,a(p pV a

a2aa

a1代入可得：空气终温799.5K，加入的热量为W

p2pQUW0573.22573.22kJT1T2303KWmRT1[1p2 287

11 k

p

]6

[1

143

Q

1 1P 3

1 nWmRT1[1p2

287

11 n

p

]

[1

123

1 1p 3

Q

1nk

T)650.2871.21.4(252.3303)rn

T2240K，放出的质量为等温膨胀：放出的质量为 k 73.551（1）T2T1(2)

p

)14mmm

11p2V2V1(p1p2)

R p

T373.5510529389.79105T2 T2 mmm'p1V1p2V1V1(pp 259.778

105(147.173.55)3.87kg解：由题有：V2 T1 T2

T2

(V1所以n1In(T2T1)1In(333573)R

In(1/膨胀功Wmn1(T1T2R418.68(1.4941)0.4309kJ/kgn n 吸热量Qm c(TT)m

T)n1 n1k kc R1.61750.43091.1287kJ/kg k c 0.43090.6978kJ/kg k nssclnT2RlnTp Tp n 而2(2) 故In2 In (n n 所以

n

(c

k又cpk1k

代入整理得：p

nk

RIn 由于

2

上式可写成：ss n

2(n

n1In (n1)(k TppssclnT4Rlnp4Rlnp4const（其中TTTpp p

p

ln

1V2 所以nln

1 ln

2

p2

V1

（2）W (pVpV) (0.11060.0323.21060.021) np

1 2

13.21（3）T2T1(2)

323.15

0.1QHWsWsmcp(T2T1mp1V13.45102kg由于是理想气体，且为双原子，故i5 i2R 8.31429.099kJ/kmolK ccpnm1.01kJ/kg Q

5.8861] 287.1283.15[1 wt

n

)

13]178.034kJ/100 m 0.39318kg/s1415.46kg/0vmRT01415.46287.1273.151095.8m3/0 k 5.8861wt

k

R[T1T1(2)p

] 287.1283.15[1 )1

]187.612kJ/100 m 0.37311kg/s1343.198kg/0vmRT01343.198287.1273.151039.8m3/0

1105m11 297.3kg/ Qwater=mwatercwatert4654.1871427260kJ/h7.57kW=Qwater故c 0.7025kJ/kg m(TT 297.3(423 又cnk 而c

R 287717.5J/kg n1 有0.7025n1.4n 由22)n

nppT)1

( )1219.0510

T

n1 n1 n

T2T1(2)

29360125 2n 4231 1 w mR(TT) 1.2297.3287(293423) n 1.2 利用c nkcnk5R或者WQH nmR(T 1p2 p1p2 0.110661067.75105 )，计算T3T1(2)n故采用二级压缩，中间压力7.75105Pa，终温设余隙容积为V3，开始进气的容积为V4，进气终了为余隙比 1所 V1V3(0.061 3p V1VVp21 V

pp VV1VV p 31 v10.065141 v10.0651251 v10.065101 4-abOca、bA、B，则三条线构成一个循环。若某工质依循环ABOA正向循环工作，则是单热源热机循环，了热力学第二定律4-故作功WQ(1T0)1对于热机有Q1W1Q2，对于制冷机有Q1W1Q2'Q'WW QQ Q 2T 那么总效率 1 1 1 210，故并无提高Q QT2T2注意两种典型错误：一是直接说加入制冷机会额外消耗功，效率降低；二是错误应用overall，在几个T之间倒腾，用不等式证明——事实上是不能导出正确结论的。4- )QQt,AQQ1

t, t, t11Tm1T2[(1Tm)(1T2)]1t

即这种串联工作的卡诺热机总效率与工作于同一T1、 热源间的单个卡诺机效率相同4-由热一律QHQLW1000250 SSS1250100000.17[kJ/K] SSS240020000 由热一律QLQHW30002502750[kJ SSS3000275001[kJ/K

4-

1、设循环自T2200KQQ1Q2Q3

Q3 解得：Q3WQ1Q2Q340080015002、设循环自T2200KQQ1Q2Q3

Q3 解得：Q3WQ1Q2Q34008009004-

2580001.7kJ/K04-HHbHcnb(cp)(tbta)nc(cp)(tctanb(cp)tbnc(cp)tcna(cp174.1868(15)174.186860274.186825只有向环境放热146.538kJSairSbScSaSbaScanb(SbSa)nc(ScSa ln n(clnTb)nclnTc174.1868(ln258.15 ln 0.9689kJ/(其中PP lnPblnPca a

Q0

0.5365kJ/SisoSairSsurr0.4292kJK为使装置成为可能，必须满足Siso0Sairnb(SbSa)nc(ScSa

lnTbRlnPb)(c

lnTcRlnPc Q0Hcp(2TaTbTcTTTTTT SisoSairc(lnTbTc2TaTbTc)RlnPbPcT T Ta,Tb,TcPaPbPcPa是最切实可行的，代入得：4-

0.0694kJ/kgTT21T127320293KT2273200473K对于理想气体 0psclnT2Rlnp 1.003ln4738.3143ln 0.0694kJ/kg4-

Q Q(11)100(11 0.1789kJ/做功能力损失：T0Siso2780.1789

U1m1c(tmt1) U2m2c(tmt2m1c(tmt1)m1rm2c(tmt2)tm Sm1rTmm1cdTTmm2cdT 533354.18ln28.4273254.18ln28.4 2730.9343kJ/

Sm1rTmm1cdTTmm2cdT 4-

W pdvpvInp1nRTInp118.314400In100pp pp SnRInp218.314In100019.14kJ T Tr

17657.7519.14kJ/a，即Q2W2aS19.14kJ而热源温度为300K，故热源熵变： Q27657.7525.53kJ/TrTr

S2SSr,225.5319.146.39kJ摩擦损耗要多消耗20%的功，则Q3W3W1(120%)aS19.14kJ而热源温度为300K，故热源熵变：

Q3

30.63kJ/TrTr

S3SSr,330.6319.1411.49kJ即QA

0 TmQATm TA TB T

Tm

InTmIn T

（2）WQAQBmc[(TATm)(TmTB)]mc(TATB

TATBmc(TTmc(T'T 则T'

(TT

T' T' T' T'S A Bmc( m TA TB TA TB (TTmc(InTmInTAInTmInTB)mc

设左右两边气体摩尔数为nn，由于活塞导热，达到新平衡时终温相同为T' UU1U2即nMc(TTnMc(TTp p因pVnRT，pVnRT，VV，TT n1 1np1 2 np 设终态左右两边容积分别为V，VpVp'VpVpV 又活塞平衡时p'p'，且有V V

1 1 2 2则：1 1 又由于V1V2V 于是可解得：V

0.001333m3，V'

2V 3 3V p V 2p 左侧气体熵变：S1n1MRIn111In1 22 V p 右侧气体熵变：S2n2MRIn2 22 pV 42 1.0105 总熵变：SS1S222In[()] In 0.0566J/ （1）QH故Q1HmcpT若使 最小，则必须为可逆过程，由孤立系熵增原理有而

T2T T

T

dT7.964kJ/2p1 2pT Q2 T大 273

7.964Q2（3）WQ1Q241162389解：由连续过程m1m2 V10V20PVmR

PV

90.123168273 0

22

/ 0 故V302Nm3/由能量方程QHWs,绝热Q0,无功Ws 则HVh'Vh'V t h'1h'1h'21h'1h'2h'tt 3 如为可逆绝热过程，则P3为最大，此Sm1(s3s1)m2(s3s2)V10V20,且均为空 故有m1 s3s1s3s2 T1T2 ClnT3RlnP3RlnTP TP C

3R

3R P 则s

3

3) 2lnPln(PP)P2P9P39

1 0.981051.10990.98105VT1P0V01.109m3/1只有T3，求不出P3，因至少需两个状态参

SisoS水+S+S大气+S

S=280

mc

1091J/水295

2 =Q1Q2W54180(295280)W313500TTTT0 0

故Siso 可逆时所用功最小，令Siso0Wmin8345J

WminQ1Q2mc(TgTimc(TiTf)而对可逆过程，孤立系熵增SisoSSSSTfmcdTTgmcdT

mc(InTfInTg)故

TT 将

mc(iTf2Ti 1、先求W1，为使功最小，必为可逆Q2mcpt14.1983(250)

273

mcln2731

2730.3678kJ/ TT Q1Q1|W1|104.96|W1TT大 0.3678104.96|W1|0 解得|W1|0.36780.3522)2982、求出W2（也为可逆过程Q'mr16013.5 Q'334.1S水

1.224kJ/冰 Q'|W |W 1 2 T0大 T0SisoS水SS热泵故1.224334.1|W2|0 |W2|3、总功量WW1W235.297kJ0.01kW最小电费为0.160.010.0016 T T'T

2W1cm(01)dTcmT0ln2

ln273(273 W2 30.59kJ

ppT20.098(273170) 1

p m(cInT2RInp2)0.474kJ/pTp Tp 所以，此不可逆过程中作功能力损失为T0Siso27318)0.474（1）wsh1h2cp(t1t2)190.88kJ/1（2）进口ex,h(h1h0T0s1s01 (hh)T(ss)190.88T(c

T2

p2)159.6kJ/Tp Tp wminex,hex,h159.6kJ/ wswmin190.88159.631.28kJ/ ex,h2

1kg氮定压下由300C升高到850Cqcp(850300)2856kJ/ 273 )qqTc

1814kJ/x, 0 wmaxexp1814kJwmax

该循环的最大火用效率为 wmax x,第五5-p2

1 1 )p2p1108.5 210PaT2(v1)k1

Tk13008.504 P3241 R(TT

k

k1T9200.4706 2 p3 2T3 19885.6310 p4v3k

v1

pp

6 2.8110 (v

(v

3k

8.51 T 1 3k 8.50 t1k118.5040.575 9200.5756.96106 v 0.861 其中

RT10.861m3/Pv1vv1

0.101m3/

6.96106Pa其中m v1v25-（1）k=1.3p2 ppk9.81047.5131.35106PaTTk12857.503 t k（2）k=1.4

7.50

0.454p2p1k9.81047.5141.65106PaTTk12857.504638K t k

7.50

0.5335-余隙比

V1

故有 ppk10512.5143.43106 TTk129312.504 Q1420000.2/28300kJ/TT

1804.7

m

P32T3 28251.210 PP11.2107 3.495105 3 4T 1 3k

12.50Q2mcv(T4T1)0.207140.717(1028.6293)109.3kJ/ （2）WQ1Q2300109.3190.7kJ/（3）

2

0.636 1mRT10.207142872930.17419m3/pV1p1

11 11V2

0.013935m/PW

1.19106V V

5-TTk1300604 TTq1614540

TT 137513k 60 Wmq1mcv(T4T1)100[5400.71(671300)]27659kJ/h P 3.85105 1 V1111

100287

其中：V1

86.1m/h，V2 14.35m/1 110.512 k 45-

232341 323241 pppk910418145.15106 vRT3 0.132m3/3 34vRT12872830.902m3/4 9vv/0.0501m3/12p v312

0.1321 v p3 v4

5.1510(

3.4910 k

0.1321

k1k( 18041.4(

5-

Tk1288160414 vTk 114T2v

24

1

Tk Tk1 Tk1 k1k1 0 0 4故T3kT2kTk161487314940142.216.9213341cv(T4T1) 940 c(TT 1.4(2034

20341 k4 kk(4

1 3.271

940K1因为此处T4是指T32034K

5- （1）由题TTk13001404 TTq1862500

q T4T311559 vRT12873000.861m3/ 1vv

v1

0.0615m3/ vv3T0.061520560.0811m3/

3T 3

kv T4 v5

故放热量为：q2cv(T5T10.717799300)358kJ循环热效率：wnet642t 5- clnT50.717ln5730.327kJ/(kgK 1

clnT30.717ln8630.178kJ/(kgKT T2

TclnT4 0.3270.1780.149kJ/(kgKT 30T4T3e10058631.16循环热效率： T5 (TT)k(TT (863673)1.4(1001 求T4也可用下述方法

13631 v 1 1 v 1 TT 4 T3(2)k14T3(2)k1(5)k 3

1 v(1)k 1T k 673 0 T4(2)kT3kT5

( 41000.95K5-

p2k 0解：压气机耗功量wc [1 )k]4 ( )14)0k 1000

4 ( )14)1 11循环热效率为：t k110 4

kRT1[1(p2)k1]kRT4[1(p2)k1 k k 即T1T4T2T3 773 此时压缩比为2(2)k1(3)k1( )04 降低后的压缩比为 )'0.5(2)0.535.8 1.4287 0w' 1[1(2)'k] [117.914]357kJ/1 k 1P1)'k 4P1)'k 4

k]1.4287773[10.035914]436kJ/2 k 02 w'w'79kJ/ 5-wkRT1(1k1)1.6672078317(130667)909kJ/kck

1w'wc

9091069kJ/ 1.6672078 10w 3(1 ) [1()1667]1816kJ/ k k

1.667 wt'wtt18160.91634kJ/w'ww'16341069565kJ wmw

590.1044kg/循环增温比T3983 1最佳增压比opt0.90.853.1)20667wkRT1(1k1)1.6672078317(12.940667)888kJ/kck

1w'wc

8881044kJ/ 1.6672078 0kwt 3(1 ) [1( )1667]1789kJkk

1.667 wt'wtt17890.91610kJ/w'ww'16101044566kJ wmw

590.1042kg/解：现需计算T2323241 k

5000T2T1(2)

293( )14 k pk 1000T4T3(4)

T3(1)

1173( )14741Kqc(TT')kR(1173741)434kJ/ kqc(T'T)1.4287(464293)172kJ/ t1q21172t 1 (

)2(k1)(0.80.851300)08wkRT1[1k1]1.4287291

217kJ/0 6.99140k w'

271kJ/

0wkRT 0k

k

]1.42871300

]537kJ/ 6.991w'w5370.85473kJ/ T w'w'473271202kJ/

m 14.85kg/w'w

解：需计算T'和T2w'

T'T1 c291

T'T3T1300

' Q节省mq节省14.85248.3rT2Ar

r 2典型错误为 cm(TT)，忽略了因不r 2T 3 3 21TS576857684231Sp2k 4T2T4 )

2932.714TTT(6)k11073( )04 p p5

11wc2cp(T2T1)21.004(389293)192.8kJ/kgwT2cp(T5T6)21.004(1073807.9)532kJ/kgwnetwTwc339.2kJ/kgq11cp(T5T4)1.004(1073389)686.7kJ/kgq12cp(T2T3T7T6)266kJ/kgwnet339.2 q12266532kJt339.2t 11

T(p)k

623(1)04 p6 p64TT(5)k1740604 4 6wc2cp(T4T3)21.004(623373)502kJ/wT2cp(T5T6)21.004(1235740)993.96kJ/kgwnetwTwc993.96502491.96kJ/kgpk T2T12p1

288314wclcp(T2T1)1.004(394288)106kJ/w'

123kJ/wchcp(T4T3)1.004(459309.2)150kJ/kgT3T2(T2T1)0.839484.8309.2KpkT4T34p3

0309.2414w wc

177kJ/ c(TT)c[TT(8)k1]1.004(13731003)311kJ/ p p7w'w

3710.86319kJ/

6kwc(TT)c[TT(p)k]1.004(1373924)451kJ/p p5w'w

4510.86388kJ/ w'w'w'w'410kJ/ 质量流率m 14.6kg/3点温度，“中间冷却使空气温度下降为低压级压气机温升80%应理解为T3T2a0.8(T2aT1，有同学错误认为T30.8T2，另外很多同学忽略了因不可逆造成的温度变动，处理成T3T20.8(T2T1，但第六6-P91Tx--vhs6-（91温度t——v[m3/h[kJ/s[kJ/kgK6-

h1129.75kJ/kgh23393.35kJ/QG(hh)20103(3393.35129.75) kJ/ G'Q/0.86.5272/0.81072.7103kg/h2.7t/ 6-

（1）4.04（2）4.012饱和水 s0.4224kJ/kg h121.41kJ/ v0.001004m3/

s20.4224kJkg h2125.45kJkg（插值

h2h1125.45121.414.04kJ/Wpvdpvp0.001004(400.04)1054.012kJ/4.044.012100%6-203029s[kJ/kgKh[kJ/p20.06MPas1s26.7656kJkgKs'0.5209[kJ/kgKh'151.5[kJ/

s''8.3305[kJ/kgKh''2567.1[kJ/x6.76560.5209 h20.9962567.1(10.7996)151.52083.1kJ/NG|ws|G(h1h2)3.4(3117.92083.1)35186-p14MPat14500Cs16.9379kJ/(kgK)，h13330.7kJ/p20.005MPa2s2'0.4762kJ/(kgK)，s''8.3952kJ/(kgK2h'137.77kJ/kg，h''2561.2kJ/ x2s8.39520.4762h2s0.8162561.2(10.816)137.772115.2kJ/x20.9h20.92561.210.9)137.772318.9kJ汽轮机相对内效率oih1h1

6-解：乏汽绝对压力：ppbp40.006MPa，x0.9可得：v121. m3/kg，h12325.11kJ/kg对于饱和水：ppbp40.006MPa m3/kg，h2151.48kJ/21232倍，1kg2174kJ。提示：注意真空度=大气压力—绝对压力6-

p

v11.0017103m3/kg，h184kJ/ p214MPa，h2179.8kJ/（1）Qm(u2u1)20[(h2p2v2)(h120[(h2h1)v(p2p16-解：查图做（饱和线附近查图更方便p13MPa和t1300℃查图可v10.079m3/kg，s16.545kJ/kgK，h12983kJ/由

v10.0263m3kg和232

p27.5MPa，h22812kJ/kg，s25.89kJ/kg则放热qTs(300273)(6.5455.89)5730.655375.3kJwquq[(h2p2v2h1p1v1375.3[(2812751020.0263)(298350102375.3[(2812197.25)(2983280.1kJ/6745kJ热量；放汽的情况，需要加入5.73105kJ放汽：QE mh mhW mumumh' out in 2 1 放水：Q m mh mumum out in 2 1 放汽：m*v''(300m*)v'第七7-P49d[kg/(kWh)7-p24kPah'121.41kJ/kg，h''2554.1kJ/kg，s'0.4224kJ/kgK，s''8.4747kJ/kgt1=4000t1=5000h1(kJ/kg (kJ/kgKxh2(kJ/kgwnet(kJ/kgq1(kJ/kg (kg/kWh7-p13.5MPat1440℃查表可有：h13314.8kJkgs16.9797kJkgK如查h-s图，则有h13302kJ/kgp2p2h1[kJ/s1[kJ/kgKh'[kJ/h''[kJ/s'[kJ/kgKs'[kJ/kgKh2[kJ/w净[kJq1[kJ/d[kg汽kW7-p12.6MPa和t1380℃h13193.08kJ/kg，s16.9373kJ/kgp20.007MPah'163.38kJ/kg，h'2572.2kJ/kg，s'0.5591kJ/kgK，s''8.276kJ/kg s2s16.9373kJkgKs x2 2 0.8265kJ/kg s'' hh'x(h''h')163.380.8265(2572.2163.38)2154.27kJ/ h'h(hh)3193.080.8(3193.082154.27)2362kJ/ hx 2 h'' ss'x(s''s')0.55910.9127(8.2760.5591)7.6023kJ/kg wneth1h2'831.05kJ/t't'

27.43%3193.08d36004.3319kg/kWK7-p1 1点：t5350 s6.4109kJ/kgK p

hb2954.43kJ/kgbsb

291.450C pa ha3529.95kJ/kga点：t5350 s7.2628kJ/kgK pe he3534.9kJ/kget5350Cs7.3338kJkgKe2

h'2564.2kJ/kgp0.0055MPa s'0.4995kJ/kgKs''8.3613kJ/kgK s2sa7.2628kJ/kg x2hxh''1xh'2225.5kJ/ 2 2’

2s2'se7.3338kJ/kg x'2h2'2248kJ/h1hbhah2 无阻力时：thh

h'有阻力时： hh'h h13378kJ/kg hb2957kJ/kg ha3515kJ/kg he3520kJ/kgh22228kJ/kg h2'2252kJ/kg

tt7-p1 h13219.18kJ/kg(3209kJ/1点：t3900 s7.01296kJ/kg2点：p 2 1h'137.77kJ/kg 2点：p 2 s'0.4762kJ/kgKs''8.3952kJ/kgK 又s2s17.01296kJ/kg x2h22138.2kJ/kg(2141'kJ/

439.36kJ/kg

2683.8kJ/kgap0.12MPa

'1.3609kJ/kgKs

7.2996kJ/kgK a’点：hh'439.36kJ/ 3点：hh'137.77kJ/a haha'1ha'h30.1237net,回＝h1ha1hah21026.9kJkg net,回36.94% t

h1h2h

36003.51kg汽/kWh

d3600

3.33kg汽/kWh

h 7-11223S1点：p3.5MPa，t440℃ h13314.8kJ/kg a1点：pa

s16.9797kJ/kg1s6.9797kJ/kg1 压力（MPas（kJ/kgKh（kJ/kgh6.97976.9424(3040.383018.18)3018.183039.4kJ/kg 6.9815a2点：pa

6.9797kJ/kg s'1.6717kJ/kgKh'561.4kJ/kga a2

s''6.9930kJ/kgKh''2725.5kJ/kg 2720.1kJ/kg2p2h'151.5kJ/ h''2567.1kJ/ s''8.3005kJ/kg x2 h22149.2kJ/kg3hh'151.5kJ a'点：h 830.1kJ/1 1a'点： h'561.4kJ/2 2由于1(hah')(11)(h'h' 10.1084 2(hah')(112)(h'h3 20.1423 a a ha)(121)(hah2)986.7kJ/kg aq1h1ha1

2484.7kJ/kgd36003.648kg/kWhw净39.7%t7-p1 h13396.4kJ/

5350

6.4304kJ/kg b点：pb sbs16.4304kJ/kgK(270tbh6.43046.3974(2941.82914.2)2914.22932.2kJ/ 6.4477

5350

s7.3338kJ/kg a点：pa sases''（为过热蒸汽

ha2846.4kJ/kg

h''2561.2kJ/kgs''8.3952kJ/kgKx2 hh'xh'h' 3h3h2'137.77kJa’haha'561.4kJ抽汽率： 0.1564ha比功：neth1hbheha1hah21667.2kJ/ 热效率：tnetq148.5%(

汽耗率： 2.159(kg汽/kW抽汽率为0.156，加热量为3437.7kJ/kg，净功为1667.2kJ/kg，热效率为第八8-解：同温限卡诺逆循环的制冷系数cc

4.450.75c0.754.45 Q1Q229

WQ1Q211.69993.86kW，故制冷量Q23.862.338-

T

W c c Q2Q1W25018.78-解：

T15 Q2W4624kW6.22

T

1 Q'W568-

输入功率为1“冷吨”需要2.0kW，那么 TT

T2

'

0.563kW/8-解：制冷系数(COP)R

1kk k

304/14 k

604/148-T T T0 s为200K（新版为220K其次，不要受同温限间工作的回热非回热误导，直接用非回热循环的COP

kk

T)(TT T) T 2 2 8-2424351S查氟利昂-12phh1137kcal/kg573.6kJ/kgh4107kcal/kg448kJ/kgh2141kcal/kg590.3kJ/h1h4h2

Q2W7.523.526.32kW6.82 Q1Q2W26.323.5h1400kJ/ h4243kJ/ h2419kJ/4002432 2W

Q1Q2W8-44141532hp-h图得：h1385kJkgh2420kJh4245kJ/故COPh1h5h1h4 COPT01

Qm h1

0.00694kg/8-p10.261MPa2.66kgf/cm2 p20.9MPa9.189kgf/cm2h1136.5kcal/kg571kJ/kg h2142kcal/kg595kJ/kgh4109kcal/kg456kJ/kg t037℃ t15℃故

h1h4h2

5714565953ex(2681) m5714560.101kg/CFC12lnphh1136.5kcal/kg571.5kJ/kgh2141kcal/kg590.3kJ/kgh4107kcal/kg448kJ/kgh2ah1（h2h1）/c571.5(590.3571.5)/0.75596.6kJ/h1h2a

571.5448wh2ah125.1kJ/ 4.92(3031)如用HFC134a，则查图h1396kJ/ h2420kJ/ h4243kJ/h2ah1（h2h1）/c396(420396)/0.75428kJ/kgwh2ah132kJ/kg153ex8-4解：查图得h1398kJkgh2418kJkgh240kJ4则COP

h1h4h24

第九9- 5105解：R 265.176J/kg3 32RN28103296.939Jkg2

8.3143

188.961J/kg22RiRiN2296.9391N222N

CO9-iPi9-iPi9- 7 915.9J/kg

77

1012J/kg1046.7J/kg mOcp,O(300275)macp,a(400275)mCOcp,CO(275