化工热力学(第三版)课后答案完整版-朱自强.docx

其次章流体的压力、体积、浓度关系：状态方程式2-1试分别用下述方法求出4O0C.4.053MPa下甲烷气体的摩尔体积.（1）志向气体'：（2）RK方程：（3PR方程：（4维里裁断式（2-7）.其中B用PitZCr的普Jfi化关联法计算。解1（I）依据志向气体状态方程,可求出甲烷气体在志向状况下的摩尔体积VM为V嗅工业姆*组1.381x103，。尸p4.053×1（T（2）用RK方程求摩尔体枳格RK方程稍加变形,可写为V=-+b-Y-©I)pV(V+b)其中0.42748/?-7；25P(,0.08664RFO=-Pr从附表1连得甲烷的临界程度和乐力分别为q=190.6K.P,=4.60MPa,将它们代入a.b表达式得0.42748X8.3142X1906,4.6()XIO6=3.221.7m6Pamo1.2K50.08664x8.314x190.64.60×IO6=2.9M6×105m-w1.以志向气体状态方程求得的VW为初值，代入式（E1.）中迭代求解，第一次迭代得到K值为8.314x673.154.053XIOh+2.9K46×1O53.2217x(1.38I×10-3-2.9846×105)673.155X4.053×106×1.381×10"j×(1.3«1×10,+2.9K46×105)1.381X10、'+2.9g46×10-,-2J246×IoT=1.3896x10'm'-7Mr>1.其次次迭代得匕为V,=1.381x1.0-+2.9846x1.0-府337。38处史二2.9846冷673.1SuX4.053XIO6X1.3896×I0×(1.3896×I0+2.98460s)=1.38xO+2.98461.s-2.1.1.2O×IOs=1.3897XIOHm1.匕和匕己经相爰很小，可终止迭代“故用RK方程求得的摩尔体枳近似为V=1.39O×IOm'w,（3）用PR方程求摩尔体积将PRzf程稍加变形，可写为(E2)RT、卜”(丫一切ppV(V+b)+pb(V-b)式中«=0.45724-P.?T=0.07780工P.a*s=1.+(0.37464+1.54226-0.26992疗X1.一般S)从附表I直为甲烷的0=0.008。将工与。代入上式ft5=1+(0.37464+1.54226×0.008-0.26992XO.()O82)(1-严)=0.659747=0.435266用P,、工和“求a和b,«=0.457248.314?X1.90.624.60x(ys×0.435266=0.10864，/Pamo1b=O.O778O8314-16=2.68012105my-n>1.,4.60XIO6以RK方程求得的V值代入式（E2）.同时将a和b的值也代入该式的右边,藉此求式（E2>左边的Vf,得v=i三÷2,2×1°0.1064X(1.39()XI(尸-2.6»)12X10)4.053x1(×I390x10-,x(I390x10-,+2.680I2x0-5)+2.68012×10-s×(1.390×10'-2.680120-5)=I.38I×IO'+2.68OI2×IO,-1.821.7x1.0,=1.3896×10'm'>no1.临界点也符合式<E1).得黑一声不(E5)式(E3)(E5)三个方程中共有a、b、p,、7；和，五个常数，由于匕的试蕤值误差较大,通常将其消去，用p和Tc来表达a和b解法步臊如下：令必=Z,(临界压缩因子)，即匕=空工,P.同理，令.=也立,b=典PePe代入式(E3>(E5>,J1.整理得C,和C(I为两个待定常数,将a、b、匕的友达式u(2Z÷)1Z(21+J2NS(E6)Q(3Z+3+白二_1z：(4+5)3-(Zr-*<(Zr+1.1.)Z<-Qt1.式(E6)除以式(E7.武(E6)除以式(ES)得Z-3ftZr2-3fc,Zr-i'=0(E7)(E8)(E9)-2Z+Z；+3Z-2Z.-?=0对式(E8>整理后，得Z(2i+.)(1-Z4+a)(E9)减去(EIO),如(1.-3Z<rV+2"Z,-Zj)=0由式(EI2)解得Z,=g,或=(2-1)Z.(此解不肯定为最小正根)，或(E1.O)(EI1.)<E12>,=-(0+1.)Z,(Qt,不能为负曲宣握弃)再将Z,=g代入式（E9）或式（EIO）,得'+16-=o解式（E13）.得最小正根为(EI3)=0.08664将Z<=j和Q.=0.08664代入式(E1.1.)，得Q1.r=O.42748,故a三O4274(E!4)P.,0.08664RTb=*Pf<E15>式（EI4）和式（EIS即为导出的a、b与临界常数的关系式,（2）SRK方程立方型状态方程中的a、b与临界常数间的通用关系式可写为aR2TeiCa=-1=aaP.戒生RPeSRK方程的是7；与。的函数.而RK方程的=Tfas,两者有所区分.至于C“与。卜的来算方法对RK和SRK方程样.因此就可顺当地写出SRK方程中a、b与够界常数间的关系式为(EI6)P.卜().()8664K7(EI7)P.(3)PR方程I1.1.fPR方程也属于立方型方程.a,b与修界常数间的通用关系式仍IH适用,但Q.、a的他却与方程的形式有关,须要堂新推导PR方程由下式表达p=V-bV(V-¥b)+b(V-b)(EI8)(EI9)禹+2qN_7=0WE(Vf-h)ieVc(Vf+)+XV.-b)f经简化，上式可写为RT=¼(V÷)(V,-b)2-(匕2+护)2+W,匕(匕2-)把匕=4旺、QRz-/,=2i四代入式(E19)中，化简得出PtPePe一!一=Ra+=)(E，。)(Zi-Q1.1.)2(Z+)-4Z,Z-)对式(EI8)再求导，得曲二竺7：I2J(I-+)2+4>K(1./-6)-(K+/以4匕"+4匕+12M1.-毋“而T)E=(,-)'+(Vei+b')2+4bVf(Ve2-bi)2=0(E2I)将上式化筒后得出2RT2«(3V+1劝匕3+14-K2+二-5/)(V.-b)iV.x+SfcV+2WV,f'+Hb-Vi-2hbiVi'+2(V?V2-8/?½,7+/(E22)再带匕=空工、QR工2、，)=也三代入式(E22)',化简得出PtPcPe1Q,(3Z,"+I211Z'+14C/Z：+4%Z-5)(Z1.-J,-Zs+8f,Z.7+2()Z46+8C(IZS-26Z44-8ft5Z?+20n,1Z.2-8Zr+6(E23)PR方程的Z<=03074,将其分别代入式(E2I)和(E23)后,就可联立价出Q“与C1.i,得到tf=0.45724和f=0.0778.烛终料到”0.4572吟，和P.,0.0778/?!b=-Pe2-4反应器的容积为1.213加内有45.4Okg乙酥蒸气，温度为227C。试用卜列四种方法求算反应零的压力,1.1.知试蛤值为2.75Mpa.（I）RK方程：（2SRK方程：（3）PR方程：(4)三参数普遍化关联法.储(1)用R-K方程法计编从附表1查得乙静的化和T分别为6.38MPa和516.2K.W1.RK方程参数a.b为=042748isP10.42748×8.3142×5I6.22-s6.38XIO6=28.O39MP山尸KU10.08664K7：0.08664x8.314x516.2CO,八、16.38×1O'bs-75.828×10m>>1.P,再求乙醇在该状态下的摩尔体枳，V1.213(45.4046)×10,=I.229x0w-1按R-K方程求豫压力，有RTP=V-bT,V(V+?)28.0398.314x(227+273.15)1.229×1Oj-5.828×IO5500.15°$×1.229*1.()-,×(1.229×10'+5.828×105)=(3.5519-0.7925)×106=2.759X106Pa=2.759.WY/(2)用SRK方程计算从附表I查得乙胖的为0.63SSRK方程中的a和b分别计算如卜:小迎£=0.89'516.205=I+(0.480+1.574×0.635-0.176×0.635?X1-0.9689"-5)=1.022=1.0222=1.04460.42748x8,3I42×5I6.226.38X1.(Tx.0446=1.2891.w6P11w2人。.086田834x516.2=5.828x1。-".“'6.38O'在给定条件下乙醇,摩尔体枳为2291.(尸符上述有关数值代入SRK方程，得8.314x500.151.2891pI.229×1()3-5.828×1O5-1.229×103×(1.229×103+5.828×105)=(3.5519-0,8148)X10"Pa=2.131MPa(3)用PR方程计克"=1+(0.37464+1.542260.635-0.26992×O.6352X1-0.9689"3)=1.0195=1.0952=1.0394C,a=35.03-ww'Kf*W体枳流连为：V=-111(<2)j=3.14-30.0752=0.0132w,r'摩尔潦只为I=4.015mo1.sVV0.0132一Vw-RT/p-8.314593.I515(XXXX)依据热力学第肯定律,绝热时W1.-ZXH,所以H=nCpm(T2-T1)+n-(HR2-HR1)Ws=-4.015-8.40SIO3+(-56.91+576.771=3.167xI(AW梃二，依据过热蒸汽表，内插法应用可查得35kPa.NOC的乏汽处在过热蒸汽区,其恰值h2=2645.6kJkg:1500kPa,32(C的水蒸汽在过热蒸汽区，其培值h=3081.5kJkg,:-W=Ah-i(U22-U12)=2645.6-3081.5-4.464103=-435.9(>4Jkg1按理妞气体体积泞厚的体枳xrRTV=P8.314593.15,强。-33-1500000Nr=4.01.SO.O132nWs-I_$(“型3.28<1.(3m3mo'w=435.90418N=3.15×1.'W4-6解：二巩化碳T1:=303.15R:=8.314P:-1,51.O6PaP2>O.1O133O6PaTc304.2Pc:-7.357106Pa3=0.225Cp(T)=45.369+8.68103T-9.6!9105T2,、RTCP，h2r(t2)p-O.O83*0.139-1097j-°894sf)I2Cp(T)d+H2r(T2)-Hir=-8.32725xIO3JmoJJT1.146页第五章54:a5-5:a-5解：可逆过程烦产为零，即鼠=AS,"-AS=5g-/=OnASw,V0,5-2：解：不行逆过程熠产大于零，即5,=5r,-Sz=S,->O=>S,1.,>U.即系统燧变可小于零也可大于W：不行逆绝热过程燧产大于零，即S=AS,”-1.5f=Sty,>O.所以海体燧变大于解:不行逆过程墙产大于零，即ASe=AS,-5,=Sty,-*°=A%,>£,5-3：解：电阻滞作为系统,温度维持1(X)C,W373.15K,属干放热:环境温度298.15K.蟠于吸热，依据孤立体系的埔变为系统烟变加环境雌变，可计尊如下:7R5-(20A)-2-360s=1.44×10JVV-1.44×IOJ1.44×IOJia411+=9.707IO-J373.I5K298.15KK5-6：蟀：志向气体节流过程即是等焰改变，温度不变，而且过程绝热,所以系统的烯变等于焰产，计算如下：-8.31.4Jmo1.K1.1.0.098071.96.24.901JKmo1.所以过程不行逆,5-7：解：绝热Ia流过程M=m+n2.H=0所以Mg=ri"h+吗与T_20kgs(9O+273.I9K+30kgs(50273.19K5(Kkgs-1T3=339.15K339.15-273.15=66直表可用h1.=376.9Zh2=209.3-376.92-209.3390-50=4.1.9kJkg1KASg=2。4.18可言）-30418414言）=0.345kJK-'s1.不同温度的S做也可以干脆用饱和水衣自汨，计算结果是0.336,5-12（2）水泵功与透平功之比rh=解：（1）耐环的热效率QHH1.HtH2=3562,38Ukg1.,Hj=2409.3kJkg,.Ht=162.60kJkg,H5=2572J4kJkgH4-VpHt=162.60+(14-0.007)10jO.(X)I=176.6kJ-kgi帆J=V>=0.001(I4-O.()()7)1(P=()()17jwjH2-H3-3562.38-2409.3=X口二生=0.345(3)供应1kw电功的蒸气循环量，"幽=g=0857*1.Wn1167.085-15三:4-15,0W>WTo-T1.I)T=(76ff=2Wv(,-)W0555-0.60.2I_21+273.15V6+273.15118+273.1521-6-0.12=0.555194页第六章解：水蒸气的摩尔流业为:m16801000“=25.926mo1.sM36(X)1836AH(T)I=J8.314(3.47+1.4S103T+O.I21.IO5TBdT1.J703S(T.p):=n.T8.314(3.47+I.4S!O3T+O.I2I1O5T2)tt.。，id-n8.31.41.703（八）通过内插法求出0UM9MPa时对应的温度.加下=°°5=029,101.325-104.9101.325-84.551=100.96,WU1.=-AH(374.114+T0S(374J1.40.104=3.607×IO5J1.V=-H(374.II4=3.046×IO5Js-1WSa=°844wMWid=-AH(333)+ToS(333,O.OI4-J=4.926×IO5Js-1Ws=-H(333)=3.408×IO5Js1-=0.692idH=h2-h1=292.98-376.92=-83.940.9549-1.1925-荔=0.044Ukg-1.K根抠热力学第一定律，热损失为Q=8H=-83.94kJkgT成Q=-1.51Ix10'Jmof'功损失为W1.=T0S1j=13.1kJkg't&W=235.8Jmof166:解：志向气体经一模孔降味过程为节流过程A"=0.且。=0.故卬s=0过程恒温.则绝热膨胀过程的志向功和物髭功计算如下：Wid=-H+T0S=TS=（-2988.314I吊零7）=7.42×IO3JmoJW1.=Wid=ToASg=（-29&8.31叫可翳I）=7.42×IdjmO1.6-8：解：1产品是纯制和纯毓时.Wu=RT0(y1.Iny1+y2bi,y2)=8.314298.15-(0.21In0.21+().79In().79)=-1.274w,（2）产品是98%Nz和50%Oz的空气时,设计计算流程如下:%»98%N20.21O:小沏d纯（工?今.0.98N2WH纯N2W：50%O2总的功W=W,i+VV1+W2W1=FR(0.98In0.98+0.02In0.02)=-298.15-8.314-(-0.098)=242.924-wo,W2=-7/(0.5In().5+0.5In().5)=-298.158.314(-0.693)=I.7I8m<,：.W=-1.274+242.924-10j+I.718=0.687W三,6-12：解：包衣将H2H20NH302CH30H002N2H0-285.84-46.190-238.64-393.510S0130.5969.94192.51205.03126.8213.64191.4!H2H2+IO2H2O(I)H=-285.84kJ-mo1S=69.94-130.59-0.5I205.03-8.3141J1.1.02cxsi05II=-69.785Jmof'KI0.10i33)Ec(H2)=-H+T0AS=285.84-2三罂=235.244kJmo'NH3-N2+-H2NH322Exc(NbU)=31.17.61+)0.335-116.63=336.535kJmo'CH30f1.EXC(CH30H)=4-117.61+1.1.966+410.54-166.31=716.636kJ11o16I3解：Q2=11b(h3-)AH=Q+Q2=QI=叫(h3f)m,:-j13600IOSOOOi11b>Kgs36EH376.92kJkgS-1.1925kJkgK与K209.33kJkgS2：=0.7038kJkgKh3=276.366kJkg"'cpm4,3-1.)+m2cpm3-t2)=0运用内插法可求汨66.03C时的焙位，可得t3=6603CS3-0.89350.9549-0.893566.03-65S3=0.906kJkgK()利用墙分析法计算损耗功,W1.=TOASg=T0Ssys=Tom(S3-S1)+m2(S3-S2)=100.178kJs(2)利用Wi分析法Nh=J04.89S:=0.3674M:=m+n11E-n-(h0-h)+m1T0(S0-S1)E2-11b(h()-112)+n(s0-s2)eX3z=-M(h0-h3)+MS0-S3)W1.:=eX1+eX2-eX3W1.=100.78Us,或者W1.=3.606×IO5kJh1241页第七章7-2解：假设需水mkg,则100o96%1()0()+w=56%=>/?/=714.3小产品酒中含水(100O+7I4.3)(1-56%)=754.产品油中含醇(KXX)+714.3)56%=9604g所以酒的体枳V=Z町Vi=754.30.953+9601.243=191210,cm3=1.912m,7-3解：V=(109.4-16.SxI-2.64-xjcm'1mo1Vip=2.64-x1.2-5.28x1+92.6X1Vp=VI-89.96cn?mo1.x=0V,p=V,=1.09.4cn/mo1.1V=V-V符=109.4-I6.8x-2.64xJ-xV+即?)=2.64xI-X1.)7-4解：依堀吉布斯杜亥姆公式，恒温恒压时>,函=0f则有ZXWE=0,所以.V1JV1.+A1JV1=X1(/?-«)+2bx1.pA1+x,(>-)+Ibx2Jrh,.dX=-dx1：.1JV1+X1JV1=(i-X,-)+2b(x-x1.).v1.把的=1-七代入上式，得.r1.JVI+JVi=(动-«)+2(«+bxit1.x1.0所以i殳计的方程不合理.