串级萃取工艺设计实例.pdf

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1、串级萃取工艺设计实例Er/Tm分组工艺设计（水相进料）一、萃取体系的确定，1.5mol/lP507+煤油，36%皂化，β=β'=3.5，(M)F=1.5mol/l，pH=3~4，(M)n+1=0.9mol/l，余酸为1.1mol/lHCl。二、分析原料组成，确定fA',fB'La2O3CeO2Pr6O11Nd2O3Sm2O3Eu2O3Gd2O3Tb4O7W%0.060.100.411.831.050.455.206.60M162.91172.12170.24168.24174.4175.96181.25186.92Mol0.00040.0006

2、0.00240.01090.00600.00260.02870.0353Mol%0.00070.00100.00430.01960.01090.00460.05180.0637Dy2O3Ho2O3Er2O3Tm2O3Yb2O3Lu2O3Y2O3W%37.076.6217.602.2912.481.806.44100M186.50188.93191.26192.93197.04198.97112.90Mol0.19880.03500.09200.01190.06330.00900.0570T=0.5539Mol%0.35880.06330.166

3、10.02140.11430.01630.1030fA=0.1521,fB=0.8479M=180.54已知：PAn+m=0.995,PB1=0.995PAn+m/(1−PAn+m)0.995/0.005a===1109.35fA/fB0.1521/0.8479P/(1−P)0.995/0.005B1B1b===35.70f/f0.8479/0.1521BAa(b−1)b(a−1)YA==0.9720YB==0.9991ab−1ab−1f'=fA⋅YA=0.1486'fB⋅YB0.8514AfB==PAn+mPB1三、确定EM,EM'，S、W和

4、nmf0.8479B1.0>EM>(EM)≈f+=0.1521+=0.3944minAβ3.51

5、'/EM'−13.5/1.2086−1PB*≈==0.7584β'−13.5−1PfBnBPBn===0.6143β‘−(β’−1)Pβ−(β−1)⋅fBnBlgaPB*−PBn+mm=+2.303lg−1=7.25lg(β'/EM')PB*−PBn其它K值的工艺参数KWSnmR0.60.89241.04105.976.8413.330.70.71250.86117.007.2512.270.80.57990.72858.417.8011.810.90.47900.627610.558.6412.04选K=0.8一组为最优四、流比S0.7285

6、MF1Vs===4.0472，V===0.6667有效F(M)S0.18(M)F1.53WW0.5799Vw====0.6443C+(M)0.9Hn+1VsVw4.04720.6443归一化流比：Vs:VF:Vw⇒:1:=:1:=6.0705:1:0.9664VFVF0.66670.6667MfB'0.85141(M)====0.59281VsVs4.0472×3%VF+Vw+×3%VF+Vw+×3%0.6667+0.6443+97%97%97%Mn+mfA'0.1486(M)====0.0367n+mVsVs4.04722W0.5799(M)

7、===0.9n+1Vw0.6443W+11.5799(M)===1.2052−nVF+Vw0.6667+0.6443洗酸酸度：3×0.9+1.1=3.8mol/ln理8.41m理7.8实际级数：n===10.51，取11级m===9.75，取10级实实η80%η80%五、工艺框图Vs有效=4.0472VF=0.667Vw=0.6443MF=1.0级数1………..1125M0.72850.72850.72850.72850.1486M0.85141.57991.57990.57990.5799(M)0.180.180.180.180.0367(M

8、)0.59281.2051.2050.90.93