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1-1、已知B=105Pa=0.1MPa=100kPa p = pg+ B = 1.5 + 0.1 = 1.6MPa p = B H = 100 4 = 96kPa H = B p = 100 90 = 10kPa pg= p B = 1 0.1 = 0.9MPa第一章1-4、H=gL sin 30=0.81039.806 0.2 0.5 =784.48kPa= 105 784.48 =p = BHV99215.52Pa99.22kPa1-5、W = 2pdV = p(VV21) = 0.7 106103(0.05 =kJV10.02)21= V2lnV2=630.05=(图略)WV1pdV= p V11V10.7 10100.02ln0.0212.83kJ1-7、需由热泵向室内提供的热量Q1=7000010003600=2 4010019264.4W故热泵的功率为Q1N =519264.4=3852.9W 3.853kW1-9、Q12=W105736003.97=Q= QP105 73600=/21 =1=PQ27.78kW74800kJh第二章2-3、 p= pBkPa, p= pB 1g1+=131.3252g2+=401.325kPaR=/RM= 8314 / 44 =188.95kJ / kgKCO20CO2Vp1p23401.325131.325=m = mm() =21RCO2T1T2188.95(70 + 273)(45 + 273)=18.577 6.557 = 12.02kgp V1105 32-6、吸气速率 m=1RT1=287 (15 + 273)=3.63kg / minm = mm=21p V22RT2p V11RT=VRT(pp) =218.52872886 (0.8 0.1) 10=kg71.99= m=71.99=119.83 min2-13、mpV3.63T0(431.2 +100100) 683.72733V0= 638.58mTpVp0T0(431.2 +290100100) 683.71013252733V02=T2p02-14、按平均比热计算250c=,30325=101325= 611.18mp01.0155kJ / kg Kcp01.0045kJ / kgKV0V028.973500 28.97m =v025022.4=2522.4=4526.6/kgh6Q = m cp0 250 cp025 = 1035516/kJh=1.035516 10/kJh按真实比热经验公式计算由和是空气的定压摩尔比热与温度的关系为P26表23McP=3+29328.106 + 1.966510T4.802310T1.966110TT2=T23+293Q = nMcdTn28.106 + 1.966510T4.802310T1.966110T1pT11TdT= T ) +322n28.106(1.9665 10(TT)1T216233121944+TTTT34.802310(21)41.966110(21)=V022.41128.106(523 298) +263321.9665 103(5232 298)1944+34.8023 10(523 298) 461.966110(523 298)=1.04050 10/kJh按定值比热计算由P24表2 2得空气的定压摩尔比热为:McP=kJkmol K29.099/V3500Qn Mc(TT=0McTT=P21)22.4P(21)22.429.099 (523 298)= 1.02301106kJ / h2-15、(3) 按平均比热容计算RR =0= 259.8kJ / kgKMVT，3m =p V11RT1=608103 0.5259.8293=3.9934kg(V0=p11T0=p60810 0.5273293101325=3 )2.795Nm10 由表查得氧的平均比热容：640cp= 0.9978c640=0.7380kJ / kg Kc020=0.9166kJ / kgKv020c=0.6568kJ / kgKp0640kJ / kgK20p0Q = m cv0640cp0202-17、=3.9934 0.7380 640 0.6568 20 =831483141833.7kJR = giRi=0.232 32+ 0.768 28=288.32J / kgKR08314MM=/;=RrgM288.3228.84kgkmol28.84griiMio2=o2Mo2=0.232 32=0.209rN2= gNRTM2MN2= 0.76828.8428= 0.791v0=0p01=288.322731013251=3mkg0.7768/30=v00.7768=kgm1.287/2-19、(1) 理想气体状态方程mRTmR T0.5 8314373p =0=19382012.5P=19.38MP或p =VmRT=MV0.5 16 0.005518.2373= 19328860P=a19.33MPaV0.005aa362(2) 范氏方程：由表6-1查得：a =228.50 10(MPa m/ kmol33)b =42.69 10(/mkmol)由( p +aVVM=nMVm=16 0.0050.5，得=30.16/mkmol)(VM b) = R T20VMR T36p =0a=8314373228.5101020.16 32VMbVM42.69100.16= 17509492.8Pa 17.51MPa2-20、已知：氮，V0=1Nm3, p = 20MPa,t = 70 ；求V解：(1) 理想气体状态方程TV =pVp00T=(273 70) 0.1013251=0.00377m30(2) 通用压缩因子图20273由表2-5查得T=126.2,=3.39456;对比参数值cKpc27370MPap20Tr=T=Tc126.21.61, pr=pc3.394565.89由通用压缩因子图查得压缩因子 z = 0.935代入状态方程,得zR T0.93583142033VM=0=6=/0.0789mkmol=V = VMnVMpV022.4=201010.078922.4=0.00352m33-3、 Q = W = m gz 第三章= 450 9.806 30103=132.38kJW =0;(U= Q = mc t2 t1)tQ132.38=+ t=+=2mc15 4.18682026.323-4、W = QU = 12 20 = 8kJ ，是压缩过程，外界对气体作功8kJ。3-5、UQWkJ21=2b1 2b1=7 (4) = 3a UkJW1a2=Q12U12=Q1a2(21)=103 = 7;Q= UW=kJ3-6、c12过程1-22-33-44-112+1c2=3 + 25Q（kJ）W（kJ）110000100-9500050E（kJ）1100-100-950-503-8、取整个容器内的气体为热力系，这是一个闭口系统能是方程为： Q = U + WW= 0(气体向真空膨胀绝热), Q = 0()U = 0,即U左+U右=0 T2= T左=p VpV300K=27由22左左质量相同=(m),=又TTT2T左左2 p2=pV左左V21= 600 =6100kPa3-10（1）风机出口温度Nt2=st+=1000+ 0 = 1.78mcp10.561005（2）空气在加热器中的吸热量) = 0.56 1.005 (250 1.78) =Q = m cp(tt139.7/kJh=55.0310/kJh22（3）过程不可逆时，以上结果仍然正确。3-13、CO2，p1= 800kPa,t1= 900 ，p1=120kPa,t2= 600 。解：（1）用定值比热容CO2为多原子气体，Mcp=99R= 8.314 =37.413kJ / kmolK202=u = ctMcvt =McpR0tvMM=37.413 8.31444 (600 900) =198.4kJ / kg=h = ctMcpt =37.413 (600 900) =/255.1kJkgpM44（2）用平均比热容由表查得CO2的平均比热容：900c= 1.104kJ / kg K ,c900=0.915kJ / kgKcp0600=1.040kJ / kg Kv0600c=0.851kJ / kgKp0600900p0u = cv0 600 cv0=900 = 0.851 600 0.915900312.9kJ / kg600900 =hcp0 600 cp0=900 = 1.040 600 1.104900369.6/kJkgwt= h =/255.1kJkg/或369.6kJkg3-15、解：因空气预热器无热损，故有 Q烟气=Q空气即烟所放出的热量等于空气吸收的热量。（1）又，对应于1kg 空气，V0,烟=1.09RT烟0=1.09 286.45273=0.8412Nm3从而 Q烟气=0.8412 245 =Q空气=1 cP(t2 t1)代入式（1），得：Q烟206.094t=t+=+=P0206.094kJ1013252cp11.004510m215.171215.2 或，对应于1Nm3烟气，m空气=烟气1.09=1.09p V00RT烟气0=10132511.09 286.45273=1.1887kg从而 Q空气=m空气代入式（1），得：Q烟t=t(cPt2 t1245)，而Q烟气=245kJ2+=+ 10 = 215.18 215.2mcp11.1887 1.00453-17、解：将混合过程视为绝热过程，则混合前后的焓值相等， H = 0又将空气看作定比热理想气体，则有：(tmctt) +(=)0m1cpt312p32t3=+ mtm1t122+ mm12=120 500 + 210200+120210=309.13-19、解：本题为空气的定压加热过程cv= cp R = 1.01 0.287 =由理想气体状态方程，可得：p V2068.4 103 0.030.723kJ / kg.KT=11=216.2K1T=Rp V22287= 21= 432.42RTK所以：U= mcv(TT21) = 1 0.723 216.2 =156.3kJH = mcp(TT21) = 1 1.01 216.2 =218.4kJQ = HWt+= H= 218.4kJ补充题:W = QU = 218.4 156.3 = 62.1kJ图示燃气轮机装置. 已知h1=300kJ/kg, c1=20m/s的燃料与空气的混合物在燃烧室内定压燃烧,吸热q=850kJ/kg. 燃烧后的燃气在喷管内绝热膨胀到状态3, 焓值为h3，流速达到c3 =1000m/s. 然后进入动叶,推动转轮作功. 若燃气在动叶中的热力状态不变,离开燃气轮机的速度c4=150m/s.燃气流量为4.0kg/s。求:(1) 燃气在喷管出口的焓h3；(2) 每千克燃气在燃气轮机中所作的功Ws,3-4；(3) 燃气轮机的功率N.解: (1)取热力系 “1-3”：能量方程：122(h+c cq13=h31)2(31)燃气在喷管出口的焓h3为: 1221223h= qh+31312(c3 c1+) = 8503002(1000 20)10=kJkg650.2/(2)取热力系 “3-4:能量方程:1220 =( c4 c3) + ws2故每千克燃气在燃气轮机中所作的功:ws=12(cc2122JkgkJkg234) =2(1000 150) = 488750/=488.75/(3)燃气轮机的功率:N = m ws= 4.0 =kW488.751955第四章思考题4-3、 q0,w0,u0; q0,w0,u0或：思考题4-4、题4-2、解：(1)可逆绝热膨胀RT1287 (150 + 273)0.11.41w =1 11 (p2) =p1(kJkg)1.4= 111.88/s= 011.410.5(2)不可逆绝热膨胀，T2= 300Kw = q u = u = c(T T T2vp212) = 0.718 (423 300) =3000.188.31kJ / kg=scplnT1Rlnp1=1.005ln4230.287 ln0.5=0.117.kJ / kg K(3)可逆定温膨胀p10.5=wRT1lnp2p2=0.287 423ln0.10.1= 195/.40kJkg= sRlnp1=0.287 ln=0.50.462.kJ / kg K(4)可逆多变膨胀， n = 2n1RT1p20.287 (150 + 273)2 1w=n11 () n =p1n121211(0.520.1)=/67.11kJkg或：=TTp2()n=423(0.1)2=189.17K21Rw =p1=0.2870.5=TT 423189.17/n112T221p2189.1767.11kJkg0.1=scplnT1Rlnp1=1.005ln4230.287 ln0.5=0.347.kJ / kg K4-7、解：由多变过程初终参数间关系式p2=p1v1()v2n，得过程的多变指数为：n =ln(p2vp1)=ln(0.12ln(0.2360.6) 1.30ln(1)0.815)p v11v260.6 10 0.236T1=Rp v=2876=493.38KT2=w =22R1(=0.12 10 0.8152871 p v) = 340.77K(0.6 0.236 0.12 6n 1p1v1221.3010.815)10= 0.146106J / kg = 1.30 146 =wsnwu = c(TT/146kJkg189.8/kJkgkJkgv) = 0.718 (340.77 493.38) =109.57/21h = c(TTkJkgp21) = 1.005 (340.77 493.38) =153.37/q = wu+ =+146(p2v2109.57)=kJkg36.43/0.120.815scc=vlnp1+plnv1=0.718 ln0.6+1.005 ln0.236= 1.1556 + 1.2456 =4-8、由多变过程初终参数间关系式T2T0.0900kJ / kg.Kn1p2= ()np，得过程的多变指数为：Tln(2)1ln(3731n= 1 pT11= 1)673ln( 111.49由ln(2p1)()6)q w40200u = q w = cvT2 T1，得： cv=又： wt= nw = 1.49 200 = 298kJ / kg ，TT21=400100(= 0.)533kJ / kg.K得：cP=qwtTT21=40298400100=h = q wt= cpT2 T10.860kJ / kg.K注意：本题中未指明气体种类，故不能认为其绝热指数 = 1.4 ；同理，气体常数亦不能按 R = 287.J / kg K处理。4-14 （一）按V2=3/600mh计算时（注意：此时，三种压缩过程的质量流量不相等）：定温压缩：m =pV22RT=0.6 106 600 / 3600=/1.189kgs2p12872930.1NT= m RT1ln() = 1.189 0.287 293lnp20.6=kW = N179.15max定熵压缩：k11.41=TTp2()k=293(0.6)1.4=489.87K21p10.1m =pV22RT2k=0.6 106 600 /3600287 489.87=0.7113/kgsNs=k 1m R TT12=1.41.410.7113 0.287 293489.87=140.66kWNmin多变压缩：1=TTp2()nn=293(0.6)1.2211.22=404.75K21p10.1m =pV22RT2n=0.6 106 600 / 3600287 404.75=0.8609/kgsNs=n 1m RTT12=1.221.2210.8609 0.287 293 404.75 =153.13kW（二）按Vmh 计算时，1= 6003/Nmin= NT= p V11pln(1) = 0.1103p2k16003600 ln0.10.6=29.86kWNNkpVp2kmax=111 ()1.4sk 16600p10.61.4 13=1.41 0.11036001(n1-1601.4)-66100.1=kW38.99Nn=nn 1p V11p1 (2)np11.22 1=1.226 0.1106000.63=kW1.221（三）按 m = 600/kgh3600计算时，p1(6000.1)1.22100.135.254Nmin=NT= m RT11ln() =p236000.287 293ln0.6=25.11kWNNkm RTp2k1kmax=11 ()=1.4sk 16000.287 p10.61.4 13=1.4136002931() 1.4100.132.79kWnp2n1nNn=n 1m RT11 ()p11.221=1.226000.63=1.22136000.287 293 1()1.22100.129.64kW4-15 压缩过程的总升压比为p终p16= 600.1，采用二级压缩时，各级的升压比为opt=60 = 7.746 ，故应采用二级压缩（也可根据压缩终温的高低来确定压缩级数）最佳中间压力为： p= p1= 7.746 0.1 = 0.7746MPa2,optoptn11.251压缩终温为：T= T ()n= 293 0.7746 1.25=441.25K=168.25终T21opt4-18 用下标“1”表示空气，下标“2”表示冷却水，则冷却水吸收的热量： Q2= c2mT2 2= 4.1868465360014 = 7.571kW空气的质量流量： m1=p V11RT1=0.1106 250 / 3600287293=/0.08258kgs由热平衡关系可知： Q1= Q2，即m1nk= cTQkc+ Q/ mn 1v112由此可解得：n =v1T121cv1Tq+ Q/ m21n=1.4 0.718130 + 7.571/ 0.08258+ 7.571/ 0.082580.7181301.2021= 1.202因此：p=p(T2,max)n1=0.1(423)1.202=0.889MPa2,max1nT1pn1293N =p Vnn 1111 (2)p11.2021=1.2026 0.1102500.8891.2023=1.202136001(0.1)1018.33kW或： = N = QH=Q2 m1cT=7.571 0.082581.00513018.36kW11p11补充题(多级压缩过程计算)某三级压缩的活塞式压气机，每小时吸入p1=0.1MPa，t1=20的空气90m3，压缩到终压p4=34.3MPa。设三级压缩过程均为n=1.25的可逆多变压缩。求:(1)最佳分级增压比及各中间压力；(2)各气缸的出口温度；(3)压缩机的总耗功率；(4)压缩过程中空气的总放热量；(5)中间冷却器中空气的总放热量解：(1)最佳分级增压比：= 3p4= 7各中间压力：12optp1p=p= 7 0.1 = 0.7,= = 7 0.7 = 4.9211(2)各气缸的出口温度：MPapp322n1=MPa1.25 129371.25T2= T3= TTn=432.4K(3)压缩机的总耗功率：411m =p V11RT=0.1 106 90287293=/107.03kgh=0.0297/kgs1= 3= 3n)NW t= 3m wt1.25mn 1R(TT21=107.0364230.52/1.251=17.840.287 (432.4 293)kJhkW(4)压缩过程中空气的总放热量：= 3n =1.25 1.4Qmc(TT )3107.030.718 (432.4 293)n1v211.251=19282.58/kJh=5.356kW(5)中间冷却器中空气的总放热量：Q= 2 m c(TT=中p22/=29989.16kJh)28.330kW107.03 1.005 (293第五章432.4)5-2、解:该循环的热效率为:W700t=Q1相应卡诺循环的热效率为:1000T2= 70%400c= 1 = 1T1000=60%1tc,不符合卡诺定理.故循环不能实现.5-3、解：循环的T s 图 q1= c(T T) = 1.01 (1000 300) =707kJ / kgp21T2q= T (s s ) = T (s s ) = T cln23313211p1000= 3001.01 ln=/364.8kJkg300q2364.8T1t= 1= 1q707=48.40%1=w = q1q2707364.8=/342.2kJkg5-6、已知：逆向卡诺循环，Q1 =1200kJ / K h求：（1）按热泵循环运行， t1= 20 ，t2= 0 时的功率N；（2）按制冷循环运行时，T1的极限值N值与（1）相同。解：（1） QkJh1=20 1200 = 2400/供暖系数2=T1TT12=293293273=14.65耗功率Q1N =22400014.65=/1638.23kJh=0.455kW（2）此时，T2= 293K制冷系数 =T2Q2=1200(T1 T2)1 TT12T NN293 1638.231638.23T T22(12)=1200=1200= 400T T() = 20,T=+K=5-12、已知：12=T1220 = 31340，定温压缩，p10.5MPa T,1400Kp2= 1MPa ；实际耗功 ws=,1.25wsT；环境温度T0= 300K求：压缩每kg气体的总熵变解：对应于1kg气体，w= 1.25w= 1.25lnp2=s,s TRT1p1= 1.25 0.287 400lnq = h + wsws0.1kJkg330.42/kJkgq环=q =/330.42/总熵变为：330.42kJkgS总=+ SS环境=Rlnp1p2+q环境T0=0.287ln0.11+330.42300=0.6608 + 1.1014 = 0.4406kJ / kg.K5-14、推导循环热效率计算式。解：循环的 p v 图与T s 图：v2 (s s= RTq1= T121=)1lnv1=+v3qc()(scTRTv)()ln2T1 T2+ T2s34vT122c(v4+TRTv3qcv()( s)vT12)2lnv2=T1 T2+ T2s34=4t1q11(ss )T1211RT lnv2（注意：q1, q2均为绝对值）1v15-16、解:由已知条件可得:m1=p V11RT=0.4453kg,m2=p V22RT=kg0.23781本题为定容绝热混合过程,故有 U = 02设混合后的温度为 Tm,则有: m cv(Tm T+ m cvTmT11)2(2=)0因此,Tm=+ m Tm1T122+ m=0.4453 313 + 0.2378293=0.4453 + 0.2378306.04K =33.04m12混合过程的熵变化则为:S = S1+ S2= m (clnTmV+ R ln) + mc(lnTmV+ R ln)1v306.04T1V12vT2V2306.04=0.4453 (0.718ln313+0.287ln 2) + 0.2378 (0.718ln293+0.287 ln 2)=0.4453 (0.01624 +0.136kJ / K0.19893) + 0.2378 (0.03117 + 0.19893)注：本题中，因过程绝热，故过程的熵变也就是过程的熵产；如把题目改为定压绝热混合过程，则能量平衡式变为 H= 0 ，混合过程的熵变化则为:S = S1+ S2= m1cplnTmT1+ m2cplnTmT25-18、已知： m =/0.5kgs（1）不可逆绝热膨胀， t1= 25 ， p2=1p1， N = 20kW ，求 t2 ；2.5（2）定熵膨胀，p=1p ，求 N理论22.51解：
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