高等化工热力学部分课后习题答案.docx

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1、第一次课后作业答案:选择题：3-1m:=2.77涯4吗=2.77小3600ssu=12-5O2=109124J-mu=1.6510-2sgzm=81.724I26Jh+-mu-gzm=2,56710-w=I2.5671(A1I=25671gW2.583-2.56710(%2.567wc=0.6233-31 2h-u+gz=q-wq2 sh2:=30.23kJkg1h1:=188.45kJkg-,z=61+18.：9879.3-130.23-188.45f+3.168=-154.275kJkgIOOO105页参考答案4-1：b4-2:c4-4:a4-5:a105页4-7绝热稳流过程，M=m+m2

2、,AH=0所以Mg=mhj+叱力？-1-120kgs(90+273.15K+30kgs(50+273.1AK50kg-s1T3=339.1K339.15-273.15=66ASg=Zmjsj-ZSi=mCpms1n+咚CpmsInjiI2查表可得,h1=376.92h2=20933用内插法求得h390-50_90-66376.92-209.33376.92-h3=0.239h3=276.5Cn3n1-100.42，CDmh=4.184PmnT3-T166-90,Cms(339、(339、-1-1So=204.1841d+304.184IN=0.345kJKsgV363；323j不同温度的S值也

3、可以直接用饱和水表查得。计算结果是0.3365-116801000I836OO=25.926n:=25.926mo1sT0:=298KH(T)r=n8.314(3.47+1.45163T+O.1211O5T2)dTJ703rTS(T,p):=n-J7038.314(3.47+1.451(3T+O.121105T2)i,t,CeJP-dT-n83141n-T13.72Zt=100.964100-t=10Q-95=o298101.325-104.9101.325-84.55Wid=(-H(374.1i+T0S(374.110.104Wid=3.60衣JW:=-AH(374.11HWsnQ/1/1v

4、n.,.f5S:=a=0.844WC=3.0461(Wid(b)Wid：=(-AH(33+T0S(330.0147)Wid=4.9261(W=-H(333WS.,仙或r5S=-=0692Ws=3.4084WidASg=ASSySH=h2-h1=292.98-376.92=-83.94=0.044(-8304+Sr=0.9549-1.1925surI298)根据热力学第一定律，热损失为Q=H=-83.94kJkg-1或Q=-1.5I1x10?Jmo1功损失为Wt=T0ASg=BJkJ.kg或W1=235.8Jio15-6理想气体节流膨胀过程温度不变，焰变为零。Wid=-H+T0S=T0S=88.

5、34()：；：)=7.42103Jmo1W1=Wid=TSg=12988.314In(嘿既)=7.421(？-JmoF15-13HQ+Q20Qni(h3-h)Q2=1吸(卜3-卜2)m1(h3-h1)+m2(h3-h2)=072000,-110800011kgs1,36002.3600h1:=376.92kJkg1S1:=1.1925kJkg1Kh2:=20933kJkg1S2:=0.7038kJkg1Kh3:=276.366t=66.03CS3-0.8935Q,9549-o,893566.03-65-70-65S3=0.906kJkg1K,mhh,+r2,(h-h2)expand,h3-5O

6、h13818.3013818.3-1hq=276.366kJkg35090t_90-50376.92-276.366376.92-209.33To：=298.15S3:=0.906W1=T0Sg=T0Ssys=T0(S3-S1)+m2(S3-S2)=100.178kJh0:=104.89S0:=0.3674M:=m1+m2eX1-m1(h0-h1)+m1T0(s0-S1)E2=-m2(h0-h2)+m2T0(s0-S2)E3:=-M(h0-h3)+MT(s0-S3)W1j=E+E2-E3W1=100.178kJV1p=2.64x5.28x+92.6x1=1V1p=V1=89.96x=0V2p=

7、N2-I。9.4V=V-V前=109416.8x1-2.64xj2-(x1V1+x2V2)=2.64xf(1-xj6-4根据吉布斯-杜亥姆方程2a匕=O可推导出如下关系式iZxidVi=2bxx2+(Z?-a)x12bx+(b-a)x2=0in=j+2表明白/不可能为常数。2b且当K=1X2=1时，1-丫=3工0也不满足热力学的要求。所以上述方程不合理在总压101.33kPa、温度350.8K下，苯(1)-正己烷(2)形成x=0.525的恒沸混合物。此温度下两组分的蒸汽压分别是99.4kPa和97.27kPa,液相活度系数模型选用MargU1eS方程,气相服从理想气体,求350.8K下的气液平衡关系Pf和X七的函数式。解：将低压下的二元气液平衡条件与共沸点条件结合可以得到:QZPm101.331A/1=1.042P5297.27将此式代入Margu1es方程：In%=A122(A1-12)x1x;Iny2=A1+2(12-A1)x2xi2得1n1.02=12+2(A1-12)。，5250.4752In1.04=A1+2(4-A1)0.4750.5252解出12=0.1495,A1=0.0879由此得新条件下的气液平衡关系：P=PK%+，；七九=994dM59-06XM1F)2+9727(O,O879+O,1I6(1-X1)J.PZx=99.4VET)2PP