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= 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