華理化工原理課件下8章

平衡觀(guān)點(diǎn)： 氣體吸苯水1.4 氣液液氣氣液液pyxp0ei iipyiipii mpHMS101.3kPa20ye<y

101.3kPa20101.3kPa20由y=ey=0.05或xe>xyy2xy1x1Fick定律：JBBA壓力溫度不變,總濃度 ccp dcA J BA雙組分?jǐn)U散速率JA11）等分子反向擴(kuò)散：NANAJAAONDA c ccNJ JcAc m ANAMMN(1cc)D AMc dzAcc1ANAM(cA1cA2DccB2clncMcMcDT1液體Dp量級(jí)10-9~-pc=k(y-y)=k(x-界面 xkkGf(,,u,d,無(wú)因次化ShkdReduScD k∝D氣液k∝D0.5gk∝D0.5yx 界面相平衡 界面濃度難以獲得，設(shè)法消去xi,NAyyixixyyim(xi1y1m 11m K(yy Ky11m Kx11 y討論：Ky 1 11m 1)當(dāng)m很小時(shí)，1m Ky≈ky 如HCl1Ky≈kx/mKx≈kx, H2CO2Ky與kxky兩者有關(guān) 常數(shù)m=2,總壓為試求:1)推動(dòng)力y-yexe-液氣y總傳質(zhì)系數(shù)Ky,Kx,傳質(zhì)速率xK11y 2.2104kmol/(m2112K11x 4.4104kmol/(m21 1 m pm'pm101.32k0.023D(dG)083()1/3m3/ D1,pk yD2/3p0與py-ye=0.05-↑K1y1k 5 82.8104kmol/1 1 1.25↑↑流率G,L沿塔為常量G,L單位為kmol/(m2.s)y2Lxyxkxkyy1Gx10 得-Gdy=NAadhyHHyKy yy-Gdy=Kya(y-HdhHG0KyayyHOG N含義y1 HdhHLx0KxaxeHOLN討論：H與H xGy1+Lx=LxyyLxyLG Gyy1xG 1xe-y2yex xxxGy1得yyeyL1ymGymGy d(yye)1mGΔy1 Ly1dyy1 yd(yye yy1 yd(yyeyy y1yy1y2lny1y1y1 my1 (y1mx1)(y2mx2y1 HGy1Kyay2同樣HLx1Kxa y xGy+Lx=yyLxyLG2Gy2或Gy+Lx=yy LL12GG 對(duì)數(shù)平均推動(dòng)力法(適用于 d(yye1 y Ly1y1y2lny1 lny1y L2 y y21 xxG(yy L y1mx1(1mG)y1mx2y2 y2 N1mG)y1mx2mGL Ly2 LA 111A1)y1mx21Ay2 A越大1/A解吸因數(shù)越大 相平衡H KyaGyGLy1y2x1x2HmKya共9個(gè)給定條件：Gy1m分離要求：y或回收率 y2y1選擇：流向(并流或逆流)x2 2xxx逆流Δym大泛點(diǎn)流率限制,y2并流Δym小氣液流率可很大,y2經(jīng)濟(jì)上L大,吸收容易H小 ,H大,再生費(fèi)用小技術(shù)限制,L>Lmin(L(LG)min yx1ex通常取L1.1~2(L(LGGGminLL，GG y2=0.001相平衡服從 L=1.2Lmin解: K G0.0318yLminy1 1y210.001 L1.2LminGGmG 1.21N ln(1 1L1 L1GL ln(10.877)1 11HHOGNOG0.619.86.0已知：HG求：y2y1m 由H (1mG)y1mx2mGKy1mG y LL得ymxexp(1mG)KyaH1 G y21L 1) 已知：HGy1y2x2mKya流向求：Lx1(需試差) 1mGymG1) 2 y L解：1)2)LG mη0.90.93)L20.81G1mG0.9 mG11mG1 11mGLL10.55611 14)例4一逆流操作吸收塔，相平衡 試求：1)因解吸不良，x’=0.00721y2Lxy1GxHKy11LmG)y1mx2mG y Ly 2 1 y'y2mx y y'20.120.0030.122y'2y1η'120.732y12’yx2x’x1x xxG(yy)0.0030.5(0.10.02) Lxx (yy)0.0070.5(0.10.0268) G'L1解吸塔計(jì)算解吸塔計(jì)算 yyyx2x22)有 mx x12實(shí)際3) Ny1y2m y x1 HGKay1y2OGN 吸收時(shí) y1 lny1mx1(ymx)(ymx y 1 lny1y2解吸時(shí)L y1 lnmx1y1(mx1y1)(mx2y2)mx2y lnmx1mGLmx2例5例5已知：HOG=0.5m相平衡 x2=0.005Lmx1y23x1 0.05L 3.33 yyL(xx G yy1mx1y130.050.125y2mx2y230.0050my1y20.0250.015lnlnNy1y20.125my2xLrLx2LxL(1)1Gyxx'x2x1G(y1)L(1θ)(xx' yxyx2y1y2Gyx2x2x1x xLGyx1yx2xH=H上+HGyxLxxx'x1極限問(wèn)題：極限問(wèn)題：H→∞時(shí)，y2=?技術(shù)上兩點(diǎn)限制a相平 b物HHOGy1yΔy→0ym yy1 y2 yLx11GLG 2逆流LGy=mx 1yL x1myy L xx 2y L1G1G2y1yyA22逆流當(dāng)逆流當(dāng)Lm 1L xxGyyxGymx L L yyLm呢？A x1x逆流操作，HyL2LGm yLGG12Lm時(shí)，yG GLA.L↑,y2y1∴NOGm若y2Δym↑,y1-y2不變NOG2 若x1Δym↓y1-y2↑NOGL↑L↑,y2↓x1y12 流量增加使本身的出口濃度變差(氣體↑吸收液↓),另一流體的出口濃度變好(氣↓,液↑)B.t↑,m↑y2↑x1↓Δymy2C.yC.y↑,m ————2 1 yx↑↑1x=(ε/m)y+(1-ε 2y1y2y2x21？N mG 1mGymxmG1Ly LLy1y 當(dāng)x時(shí)，不變，1 y1y2當(dāng)x20時(shí)，令B 2，且Byy2 By2(BBB1y22y1↑y2↑y2/y1↓η ABSLGA+BL↑x2↓tpA.當(dāng)LGLGLH=BG時(shí)，x2↓t↓p 20℃,壓力為101.3KPaye0.9x。當(dāng)混合物的流率為22.4kmol/(hm2),水的流率為44.8kmol/(hm2)時(shí),氨的吸收率可達(dá)99.5％。問(wèn)KaG0.8y解y20.08(10.9951mG0.922.4 ln10.450.080.458.551ΔymQ GKaKyyHOG1mG N1ln10.90.080.91y y2'ηy1y2'0.080.0055 Δym2)HOGH'1.15N'NOG1mG 1y2y2'ηy1y2'0.080.000194 Δym——用GBLSkDyRT1k1gmy1ym)全塔分段計(jì)算數(shù)值解yn-yn-1=(yN-Gn x B1 1L n nn1s1 1nn1t xnxn1CL yi=f(xiN k'ayka(x mΔHGynyn11k'yyymiGGyxt12GyxtLka'.G,. n.nkxa2G,NGyxtL,CfxLnGxt五個(gè)式子解五個(gè)未知數(shù)：xntnΔHyi B溶解度=∞B相平衡物理平 CClH pKecHOC