化工原理(下册)(第三版陈敏恒)习题解答.docx

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1、= 1.95x10-3= 7.28xW43、（D-f =0.1101.34.0106= 249510 = - -IS r n第八章气体吸收1、解：查30。C水的Ps=424KpaP=P-PS=IoI.3-4.24=97.06KPa一=2=&.=起_=卫_=97.06厚XIO-3=1879MXeCeCeMCSMS2.857102182、解：查25。，8厂出。系统=l661xlSV/，设当地大气压为Iatm即1.033at,且不计溶剂分压。P1=10+1,033=11.033aZ=1.08xl0w2（绝）P2=0.2+1.033=1.233=1.21102W（绝）：p=EXrp=FQy-y1-1.

2、0810解：20。C 时，5 = 4.0610 = 4.06XlO60.3l1.661105y2-1.211021.0-F-1.661105对稀溶液，其比质量分率X=Me-19510-3X1=44=4.7810-3CO22O1C72R13X*=44=1.78103CO2O1芯解得=1.38x10-3w=1.3810-832103=44.27wgw3（2）tn=17.7wgw34、解：为0Cb系统，20。C时，E1=0.53710Wa=0.53705MPa-f=o三=-Ex0.5371050.810-5101.3=4.2410(勺-X)=I.09x10=50X=25y,=W2X=25210=50

3、x10L=竺=LoXIO_325O-M)2=0.025-0.005=0.02。=QO-2)x10Y=8x10”vO-)2O)i0.020.015= 1.33 倍= 2.67 倍(xe-x)2_8.010(Xe-X)I3.010-46、解：查20。C水，pja=2.3346KPat=101.3-2.3346=98.9654P32=101.3-1.33=99.97OCZ)o=0.220XW4w2s20CD=0.252104w2s“DP.Pb2D.P.、NA-In=(-RT1RTP.NAAr=2ABMr=2137s=0.59Ar7、解：查300C,D=O.268cw2sfp=995.7kg/w3,饱

4、和蒸汽压a=31.82阳wPBI=760-31.82=728=IAAmfnHgp12=760fnfnHgPA17601nIn州728二=迎=1.02外744纥d=处，b=001+/MM=念凡-加W=MDPf0(0.01 +力)劭=L(p,0H+-H2)MDplP2号需鬻04+Rw)=.44(fs=6.7day51HPH101.34.62”小8、解：(1)=+=+=+=0.542Kakak1akyak1a64.616.6Kya=PKGa=549KjnOiKh)HoG -G 16=0.29mKya 54.99、解：(1)-=+K=22lKrnoIf(SW)KykykX,M=:e8-”)=2.2x1

5、(*x(005-2x001)=6.6XIOY%H(s)降膜塔勺=0.023(也)083(g)0,：p0P,Ds上=kCCLnk=上产=号与P无关PbPySyEpm=m=W=I.25PP-=-+-,三2.8110KlkyHyy-y,e=0.05-.25Q.0VK2-心=1.05x10N广纺=577%NA10、证明：.-1=1_+二Ky匕kxKvakvaLakYZZ即HQG=挺LC：y=mx:.x=mKy而H=HaoG=HmoL两式相除得Noq=ANol即Na=工坊G证毕A11、证明：由物料衡算得x=x2+-(y-y2)L低浓度吸收ye=wxmG”=+万一3一乃)1.Nog=上=1dyy-y外加G

6、、kCtnGmGz入0-2)Q-)+(丁乃一叩2)1.LaL,4mG、fnG1(1一丁1+-；-乃一缈2=.InLL-2乃一1.L(x1-xz)=G(yl-y2)mGmG-,2=wx11.aZj得维Gfln1 -Ly1-nx11y2-2mG1LIn必证毕乃12、解：（万M =叼1 mG m 1A L m 乃二1乃 IFNJG = 7ln (1 - 7)及 十=一1ln(l- -)+= In的 1一n l- (1-7)15、解：(1)=1(l-7) = 0.02x(1-0.95) = 0.001 = 1.5(y1 - 2) - X2) = 1.5 (0.02 - O. OO1) /( - 0.0

7、004) = 1.75 Gx1 = y(-Kl) + 2 = y(0.02-0,001) + 0,0004 = 0,0113(2) y1 =,y1-wx1 = 0.02-1.20.0113 = 6.44103Ay2=Jz2-2 =0,001-1.2 0.0004 = 5.2 10-4_ 曲一电 _ 6.44x10-3 -5.2XlCT4Iny鸣5.210= 2.35x10-3 =s9G,=-=-=0,033bnolsM29/o=-=0.052=0.265w的aH=HQGNC)G=0.26580,9=2.14w16、解：(1)=0.9=1,3277=1.30.9w=1.17w吧=-L=O.855

8、Z1.17w1.17NCfG=L-111(1-)+=5ln(l-0.855)+0.855=5.761-fnLlFL1-0.8551-0.9TN=Ho%=0.8x5.76=461(2)=099=1,37=1.30.99w=1.29wG=-=-L=0.7771.1.29加1.29Nog=5ln(1-0.777)+0.777=14.0。1-0.7771-0.99H=HoGN0G=0.8x14.0=11.2w(3)比较(1)、(2)可知，当回收率提高，则吸收剂利用量为原来的口竺二1倍1.17w由W=1.筋q可知,吸收剂利用量与回收率成正比。17、解：(1)Z=278bwow2-M18(=上生=丝TH曾

9、=1.76x10，f545x2.5Xi。-、。Gm=(j)ttm0i=176XlQ-3278=0.489?/w227R(2)-=695w=545G0.4此时，塔高不受限制，则可在塔顶达到汽液平衡，贝J:%二痛=5452,510-5=0,0136x=x-y=2.5x10-0.0136=5.431021Zzi69518、解：*)Eia-他哈卜=哈)b=143(刍Eia=%十=*=/+(1FM=2nT%乃0.91=勿-得”0.77mia=E卦=自=1.4哈.=1.43(刍Ein=1437=1.430.7w=1用说明操作线与平衡线斜率相等，即推动力处处相等.由心X2 -加叼 1-707=t=2331U

10、./Hl=HLH=HaGNOG=1.22.33=22m19、解：(I)M廿58再=60/58+lW18=0183G=U=2250x273=920批.打22.4722.4298畀蕊=造低r”861=20.386=0,772ALNoq=-rln(l-)-+-=-ln(l-0.772)+0.772=7.1911-0.7720.0026Ac4G,4x92.2,G=T=1VfkmolItnhD23.1412Hqg -HNOG5719=0.659也C11z7Kya=168bwoZ/w3D=0.0467bnol/w2sHOQ0.659(2)回收丙酮量町=G-乃)=92x(0.05-0.0026)=4.36b

11、nolh=253影/h20s解：W=g=6匕=0.592P101.3内=竺=06郃W=O592即操作线与平衡线平行，此时,G0.03必=AyI=AXz=乃一加=必,G0.03ACHog=0.3wOGKya0.1NaG =27=9.00.3.%=滋=丝也得：72=0.00221、解：(1)%=yl(l-瑜=O.01x(1-0.9)=0.0011.四3o.667AL3(l-l)2i+l)r _ 1 1 NQQ -n1- AAy2fnx2工y2-21-0.6671r/1、0.0l-20.0002CUCcln(l-0.667)+0.667=5.380.00l-20.0002当为上升时，由于H不变，Ho

12、G不变，U.NaG=也不变，即HOGCCIIr/ICCf0.01-20.000355.38=ln(l-0.667)X-+0.6671-0.667乃-2x0.00035得乃=OoOI3001-q013 = 0.870.01(2)当Xz=OOoO2时,C1x1=(y1-2)+x2=(0.01-0.001)+0.0002=3.21031.3当W=0.00035时,xj=2(1-ey-2)+x2=l(0.01-0.0013)+0.00035=3.25103L322、解：vZ G=四口 0y1-次.N0Q = In - = In为一尔20.0145-1.05i40.80.000322-1.05100.8

13、= 4.10当流率增加一倍时，凡酎=ccSocg因为塔高不变，则H=HogNoqi=Jj1=l=410(0-2=357GN.3.57=ln竺3R=41。-1.05100.8解得其=0.000489第九章液体精储1、解：查安托因方程log *6.031-1211甲恭 IOg 耳=6.080-t + 220.81345 + 219.5IOIIIogrf=6.031-=2.3479(1) 1080C:108+220.8收=222.80kPaIogrf=6.080-=1.9731108+219.5=94.00IOIIlog招=6.031=2.018481C：81+220.8W=IO4.33kPaIog

14、rf=6.080-1345-=1.604181+219.5片=40.19%4=2.5962.484xax(2) %=Q+/)/2=(2.596+2.371)/2=2.484X0.0880.200.300.3970.489计算y0.1930.3830.5150.6210.704实嗡y0.2120.3700.5000.6180.710X0.5920.700.8030.9030.9301.00计算y0.7830.8530.9100.9590.9791.00实验y0.7890.8530.9140.9570.9791.00由y=求得下表l+(a-l)x1+(2.484-I)X取大庆差0.212-0.19

15、30.212=9.0%2、解:=4设温度为65.35Clog*6.08240-1424.22565.35+213.206耳=9.312kPaIOg 月=6.08232-1445.5865.35+209.43琉=6.626熠X8-6626=o.595893l2-6.626原式正确，2=65.35C母二总33、解:(1)设温度为81.37OC10g=6824-8三6收二17.6769kPa1POq0o01445.58logPk6.0823281.37+209.43耳=12.915313.6-12.91531yljl=0.14417.6769-12.9153原式正确，Z=8137C(2)=0144x

16、17.6769=q187JlP13.64、解：因40C时，Z5=373.3kPa=117.1kPaX尸琮.303.9-1171729，邦-以373.3-117.1xaP2以=07293733=0,896303.9jx+j=0.8按总物料为I摩尔计：一；得n1,-0-8_0.896-0,8_情nf0.8-xx0.8-0.72915、解：(1)山吸二一InJaln匕2W2c-x2l-x1由In吧=LUn+3xlnUW23-10.31-0.4得名=6&7丘/K=100-68.7=31.3bwo-Wy68.7y=X+(五一必)=04+(0.4-0.3)=0.61931.3ax2l+(a-l)x2l+(

17、3-l)0.3=0,563由（阳一练）+综J=跖得券=%Xi-XoICC0.40.3-oa.-=100x=38.Obnoly-20.563-0.36、解：（1）2=F(2)r-i-i+2=0.667.g=17=RD+F-W=(R+l)D0.470(jR+1)Z)F-(2+1)0.228I-RD!F+20.228+l(3)1._R_4F-T+-i+4=0.81.68RDfF+_40.228(+1)DF-(4+l)0.2287、解：由教材附录查得：料液的泡点4=986C由Z=,+zf)=(98.6+40)=69.30C查得：c=35.5kcal!blfiC,cp=4.9kcal/bnolqC由泡点

18、温度查得：=79567a7/hnol,7甲本=9384衣R/hnolcp=CP享xr+Cj(l-XF)=35.50.3+41.90,7=40kcalfhnoly=y*Xf+/甲矛(I-XF)=79560.3+93840,7=8956aZ/hnolJ40q=l+-2-lf)=l+-2-(98.6-40)=1.26CXF-X耐a0.30.03.co,tD=F-竺二IoX=2.93bnollhXDXW0.950.03=(1+A)Z)=2.93(13)=Wlbnollh=-(1-(7)F=11.7-(1-1.26)XlO=14.3wZ8、解：(1)D=FXfXw=IOx0105=0.53Sbnolfh

19、Xp-Xffr0.9-0.05V=F=IOhnOHh(2)由，=(1+R)Z)得A=-1=-1=16D0.588Z=16=0941V(A+1)DR+179、解：(1)D=7aF=0.9-100=60bwo6XD09W=F-D=00-60=40bnolfh_Fxf-Dxd_1000.6-600.9_n.W40对11塔作全塔物料衡算，得T=印JFJ400315=30bnolhxr-x30.5-0.3,.(7=1Vn叫=(+I)T=(3+l)30=l20hnoUh(2)对I塔塔顶与F和T之间塔段截面作物料衡算:FXF+Vly=DXD+LiXy = -x+%Dxd - Fxfi%,.(7=1:.%=%

20、=(&+DZ)=(3+1)X60=2406H/h1.1=Lj+qFL)R卢F=603+100=28ObnOljh= 1.17x-0.025280600.9-1000.6y=XH24024010、解：由精福段任一截面与塔顶(包括分瀚器在内)作物料衡算:X=LX+Dy若回流比为RlLDR1贝IJy=-+-y=X+yQV，AlA+D1对于全凝器时精窗段作操作线y=x+xDR+1五+1可知：当选用的回流比一致，且XD=yo时，两种情况的操作线完全一致。在六X图上重合。分凝器相当于一块理论板。11、证明：(1)对塔顶第一块板作物料衡算得卬O+Ul=叫O+Li忽略V与VO的温度变化，则I0=I又：对该板作

21、物料衡算得匕-七将此两式代入能量衡算式，整理可得：若以WC为烙的基准，则/=CW+Ai=与，JO=TU尸代入上式得=4左四二Z2r两边同除以DR=Rf十0式笃-T)得证r(2)若对精福段任一截面与塔顶作物料衡算:7=L+DVy=Lx+Dxd-=RD1.DRxd.V=X+X=X+VVR+火+1故冷回流下精圈段操作线方程形式不变。12、解：(1).q=lq=工尸=0.5乂=255=0,714l+(c-l) l+(2.5-l)0.5.,a=Const,Rmm=RJ96-叫5兀0.714-0.5=1.2mn=1.21.15=1.3精修段操作线方程:+=RXdxn+lR+1.380.96X”+1.38+

22、11.38+1=0.58+0.403提留段操作线方程:0.96-0.050.5-0.05=2.021.38+2.021.38+1l-2.02-11.38+1X0.05=1.43与0.0214平衡方程，yn=2541+1%25-1.5/自塔顶往下计算：y1=xd=0.96XI=为2.5-1.5y10.962.5-1.5x0.96=0.906为=0.58x0.906+0.403=0.9280.9282.5-1.50.928=0.S38依次反复计算，可得:NOyX10.960.90620.9280.83830.8890.76240.8450.68650.8010.61760.7610.560NO78

23、9101112y0.7280.7030.6740.6250.5500.451X0.5170.4860.4530.4000.3290.246NO13141516y0.3300.2150.1200.0518X0.1640.09810.05120.0214由上计算可知,第十六块理论板xxw=0.05,所以全塔需理论板为16块（包括筌）。以上计算，当第八块板时，xD=F上-=15=5AlbnoIfhxd-xw0.95-0.04(7=1.0.塔釜蒸发量G=(R+l)D=(l5+l)x5.11=128hwH415、解：因为XD丽不变,=(R+T)D-(-q)F=(R+T)D%+F=D+印 2.5D + 1

24、5 = D + J7FXF=DXD+%1.50.35=0.95D+0.0W解得D=4.6kmoVhrW=21.QkmoVhr=(A+1)D=2.54.=ll.5bwor因是常压下的饱和蒸汽直接通入筌内加热，故加热蒸汽消耗量为:=11.518=207回收率= 83.3%Dxd4.6x0.95Fx150.35改用直接蒸汽加热后，精濯段操作线方程没变提留段：%+工=沅+用1.-x=Vy-Wxw(1)用作图法可得N=9块(包括筌)(即在点0.42,0.6和点点2,0.叫连线上作梯级)图中相平衡数据由V=计算而得.数据如下:l+(-l)xX0.10.20.30.40.50.60.70.80.9Y0.21

25、70.3850.5170.6250.7140.7890.8540.9090.957(2)在设计条件下，若板数不限，则NL时。塔顶产物均可达XDlMX=0.644axF_2.50,421+(C-DXF-1+L5xO.4217、解：该塔精修段操作线方程形式不变。y=RX+XD=x+-=0.667x+0.3R+R+2+12+1由Dl出料口以下至加料口以上任取截面与塔顶作物料衡篝:Vy=Lx+Dxd+DlXny=因DI为液相，.7=(R+1)Z),Z=L-Z?代人操作线方程得:R-DJDxd-DJDxd2-0.50.9+0.50.7cA+lX+=X+=0.5x+0.417qXFl.050,42Cq方程

26、y=XHX-=21x-8q-q-1.05-11.05-1提圈段操作线方程形式亦不变.相平衡数据由y=求出。l+（-l）xX0.10.20.30.40.50.60.70.80.9y0.2110.3750.5070.6150.7060.7830.8480.9060.956用作图法求理论板数，得NT=I5块（包括釜）18、解：（1）由全塔物料衡菖F+S=D+用Fxf+Sxs=Dxd+Wxw_S0.35-0.02CC056-0.02Cg+一X-=+0.2=0.456Fxzf-xffr0.98-0.020.98-0.02DXDSxs + Fxf F (0.2xs + xf)= 0.4560.980.20

27、.56+0.35=96.7%(2)两股加料口之间的操作线方程可由该段任一塔截面与塔顶作物料衡算而得:y+02%=Zx+%LV = KDxd-0.2Fxsq=1.0,=(+1)D1.=L-qs=L-0.2F=RD+0,2FFA+0.2 -y=-& + 10 _ 0.2 泉S R+0.2/0.4560.98-0.2 0.56/0.456X +2- =X +即+A+1K + 10.735R+1&+1(1)精留段、提圈段操作与常规塔类同.作图可知，在最小回流比下可能出现的有A点或B点成为挟点。当A点挟紧时，xa=xf=0.35Zi =axA2.40.35 .= 0.564l+(a-1)以 1+1.4

28、0.35将yA.Xa代入方程(D得0564 = = + 4380.35+江。+1V=l51(ii)当B点挟紧时，XB=XS=56324x056S31+1.4x0.56代人方程(1)得in=1.18)i)三故A点光挟紧，&in=L5119、解：由全塔物料衡算可得=受二2+DIFw先设“0.194,得怒受+。精播段操作线方程：Rxd30.889_y=x+=Xd0.75x+0.222及+1R+3+13+1提溜段操作线方程：（q=l）R + Fi D FD-y =XXW =A + lA+1平衡线方程：3+1/0441/044-1X-0.194 = 1.32x-0.06173+131y_ya-(a-l)

29、y2.47-1.47y自塔顶往下作逐板计算：（计算时由X3X4起用提留段操作线方程）12345X0.8890.7950.6800.5480.373y0.7640.6110.4620.3290.194由计算结果可知，XWe=X5=0194,即原设正确。得XD=O.889Xw=O.19420、解：先设了4=0.207贝G=导+=若泮+。2。7=。873精溜段操作线方程:2OTQ=0.75x+021844提圈段操作线方程：0J4+0J4o.2O7=1.32x-0.065744相平衡方程:2.47-1.4712345y0.8730.7700.6930.5650.390X0.7360.5750.4780

30、.3450.206逐板计算得：计算时由为起用提留段操作线方程）由结果求得：xffr = x5 = 0.206 0.207,与原设相符。.xd = 0.873 xw = 0.20721、解：设塔仅有精溜段，其操作线方程Rxd 4y =X+ = x+A + l l 55XD=0.8x+0.2xzj作平行线y = kx与操作线yax+b图可得:现上=6,4,a = O.Stb = G2xd,N= 3 得:/OS6 =（R）立 +0.2和（O.8/6.4）3-16.40.8/6.4-1=0.0376xp及 0.002 = 0.8Xwr+ 0.2 %= 0.00869”3.27 xl（422、解：=-2

31、 = 0.8Fxf.=0.8=0.8-=0.571FXD0.28由物料衡算得：X=XTXJo2-0571u028=00935W1-1-0.571塔筌为理论板，即axw2.50.0935yw=0.205l+(a-l)xwl+1.50.0935由怛摩尔流得：F=L,V三D对实际板作物料衡算：Jyw+Fxf=Dxd+Lx1：.1=(.yw-X1,)+xf=0.571x(0.205-0.28)+0.2=0,157FM=XQ=0.28；=ax1=25157=03181+(a-l)x1l+1.50,157/W=三=664%若以液相板效率表示，则xI =0.282.5-1.50.28= 0.135= 66.

32、2%xf-x1_0.2-0.157xf-Xi0.2-0.13523、解：（1）Nr=8,设居，%达相平衡，则K=&ix=08q=1Xq=XF=0.5= 0.667a%_2x0.5l+(-l)为1+0.58=三得xd=0.80_D验物料衡售，XfFXF一万XQ0.5-0.60.8c=Us1_21-0.6F原假设成立.（2）仍设为,达平衡,11Xq-0.6670.667-0.5验物料衡算，Dxf-xd05-06x0.9JO5(1-My)“/慧=三Koi夬4、解：Fdt=F(-)5=2_x(2Z)05=0,0449吸29003.85查费尔泛点关联图，得：Crao=0.083uf=cx(3严(包二立严=0.083()02(7738V=l.Mw/s20就203.85=0.8uy=0.81.14=0.912ws4=V=36OoX12=O88W由=桀=0.7渣得2=0.124_二3=UW1-0.121-0.12力-(44严(4x1严-113.圆整取D=12m3.14此时，A=-D2=0.785l.22=1.1344=4O-12)=1.13x0.88=0.994w-S=2900%-3600x0994=0.81ws泛点百分率为:u_0.81=0.715、解：由芬斯克方程得:= 23.5块