化工原理(下册)(第三版陈敏恒)习题解答.docx
= 1.95x10-3= 7.28xW43、（D-f =0.1×101.34.0×106= 2495×10' = - -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=1879MXeCeCeMCSMS2.857×102×182、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=1.21×102W（绝）'：p=EXrp=FQy-y1-1.08×10解：20。C 时，5 = 4.06×10 = 4.06XlO6×0.3l1.661×105y2-1.21×102×1.0-F-1.661×105对稀溶液，其比质量分率X=Me-195×10-3X1=44×=4.78×10-3CO22O1C'72R×13X*=44×=1.78×103CO2O1芯解得=1.38x10-3w=1.38×10-8×32×103=44.27wgw3（2）tn=17.7wgw34、解：为0Cb系统，20。C时，E1=0.537×10Wa=0.537×05MPa-f=o三=-Ex0.537×105×0.8×10-5101.3=4.24×10"(勺-X)=I.09x10<J_M)=o.0057640：E2=020×02MPa=OSxW5KPapyl%×101.3E0.8×105=1.27×1Q-50.8×105×0,8×10'5101.3一X=0.47X10"y-ye=0.003685、解：yet=w1x=50×2×10-4=0.01O-)1=0.025-0.01=0.015(一)=(5-2)xl0=3xlC)Y2?1V>¾=50X=25y,=W2X=25×2×10'=50x10L="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_8.0×10(Xe-X)I3.0×10-46、解：查20。C水，pja=2.3346KPat=101.3-2.3346=98.9654½P32=101.3-1.33=99.97O°CZ)o=0.220XW4w2s20CD=0.252×104w2s“DP.Pb2D.P.、NA-In=(-RT1RTP.NAAr=2ABMr=2137s=0.59Ar7、解：查300C,D=O.268cw2sfp=995.7kg/w3,饱和蒸汽压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=22×l"KrnoIf(SW)KykykX,M=:e8-”)=2.2x1(*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'-=-+->,三2.81×10Klk'yHyy-y,e=0.05-.25×Q.0VK2-心=1.05x10"N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、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) /( - 0.0004) = 1.75 Gx1 = y(-Kl) + 2 = y×(0.02-0,001) + 0,0004 = 0,0113(2) y1 =,y1-wx1 = 0.02-1.2×0.0113 = 6.44×103Ay2=Jz2-2 =0,001-1.2 × 0.0004 = 5.2 ×10-4_ 曲一电 _ 6.44x10-3 -5.2XlCT4Iny鸣5.2×10= 2.35x10-3 =¾=s'°9G,=-=-=0,033bnolsM29/o=-=0.052=0.265w的aH=HQGNC)G=0.265×80,9=2.14w16、解：(1)=0.9=1,3277=1.3×0.9w=1.17w吧=-L=O.855Z1.17w1.17NCfG=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=1.3×0.99w=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=1.76x10，f545x2.5Xi。-、。Gm=(j)ttm0i=176XlQ-3×278=0.489?/w227R(2)-=695>w=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=545×2,5×10-5=0,0136x=x-y=2.5x10"-×0.0136=5.43×1021Zzi69518、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=1.43×0.7w=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./Hl=HLH=HaGNOG=1.2×2.33=22m19、解：(I)M廿58再=60/58+lW18=°0183G=U=2250x273=920批.打22.4722.4×298畀蕊=造低r”861=2×0.386=0,772ALNoq=-rln(l-)-+-=-ln(l-0.772)+0.772=7.1911-0.7720.0026Ac4G,4x92.2,G=T=1VfkmolItnhD23.14×12Hqg -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)=4.36bnolh=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-2×0.0002CUCcln(l-0.667)×+0.667=5.380.00l-2×0.0002当为上升时，由于H不变，HoG不变，U.NaG=也不变，即HOGCCIIr/ICCf0.01-2×0.000355.38=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=3.2×1031.3当W=0.00035时,xj=2(>1-ey-2)+x'2=l×(0.01-0.0013)+0.00035=3.25×103L322、解：vZ G=四口 0y1-次.N0Q = In - = In为一尔20.0145-1.05×i4×0.80.000322-1.05×10×0.8= 4.10当流率增加一倍时，凡酎=£ccSocg因为塔高不变，则H=HogNoqi=Jj1¾=l=410×(0-2=357"GN
