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Separation Process Principles- 2n - Seader & Henley - Solutions Manual

102 abch ln1079 1 03247 0545ln1079 achb 0545 ln1079

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Unformatted text preview: ar constants, use Eqs. (2-73) and (2-74) with 1= benzene and 2 = cyclohexane: 2 0.455ln1.102 AB,CH = ln1.079 1 − = 0.3247 0.545ln1.079 ACH,B 0.545 ln1.079 = ln1.102 1 − 0.455 ln1.102 2 = 0.3247 Compute with a spreadsheet values of activity coefficients, using these values for the binary interaction parameters with the van Laar equations, (3), Table 2.9: 1 ln γ B = 2 1 + 0.3247 xB / 0.3647 xCH ln γ CH = 1 1 + 0.3647 xCH / 0.3247 xB 2 Note that because the activity coefficients are provided, the vapor pressure data are not needed. Exercise 2.22 Analysis: (b) (continued) Temp., oC 79.7 79.1 78.5 78.0 77.7 77.6 77.6 77.6 77.8 78.0 78.3 78.9 79.5 xB 0.088 0.156 0.231 0.308 0.400 0.470 0.545 0.625 0.701 0.757 0.822 0.891 0.953 Experimental γΒ 1.300 1.256 1.219 1.189 1.136 1.108 1.079 1.058 1.039 1.025 1.018 1.005 1.003 γCH 1.003 1.008 1.019 1.032 1.056 1.075 1.102 1.138 1.178 1.221 1.263 1.328 1.369 van Laar______ γB 1.317 1.271 1.224 1.181 1.136 1.107 1.079 1.054 1.035 1.023 1.013 1.005 1.001 It is seen that the van Laar equation fits the experimental data quite well. γCH 1.002 1.007 1.016 1.030 1.052 1.074 1.102 1.139 1.181 1.218 1.266 1.326 1.387 Exercise 2.23 Subject: Activity coefficients from the Wilson equation for the ethanol-benzene system at 45oC. Given: Wilson constants and experimental activity coefficient data. Find: Wilson activity coefficients and comparison with experimental data. Analysis: Let: 1 = ethanol and 2 = benzene The Wilson constants are Λ12 = 0.124 and Λ21 = 0.523 From Eqs. (4), Table 2.9, ln γ1 = − ln ( x1 + 0.124 x2 ) + x2 0.124 0.523 − x1 + 0.124 x2 x2 + 0.523 x1 ln γ 2 = − ln ( x2 + 0.523 x1 ) − x1 0.124 0.523 − x1 + 0.124 x2 x2 + 0.523 x1 Using a spreadsheet and noting that γ = exp(ln γ), the following values are obtained, x1 0.0374 0.0972 0.3141 0.5199 0.7087 0.9193 0.9591 Experimental γ1 8.142 5.029 2.032 1.368 1.140 1.000 0.992 γ2 1.022 1.053 1.297 1.715 2.374 3.735 4.055 It is seen that the Wilson equation fits the data very well. γ1 8.182 4.977 2.033 1.370 1.120 1.009 1.002 Wilson_______ γ2 1.008 1.044 1.294 1.708 2.350 3.709 4.108 Exercise 2.23 (continued) Exercise 2.24 Subject: Activity coefficients for the ethanol (1) - isooctane (2) system at 50oC. Given: Infinite-dilution activity coefficients for the liquid phase. Find: (a) (b) (c) (d) (e) splitting van Laar constants Wilson constants Activity coefficients from van Laar and Wilson equations Comparison to azeotropic point y-x curve from van Laar equation to show erroneous prediction of phase Analysis: (a) From van Laar Eqs. (2-72) for infinite dilution, ∞ A12 = ln γ1 = ln(21.17) = 3.053 A21 = ln γ ∞ = ln(9.84) = 2.286 2 (b) From Wilson Eqs. (2-80) and (2-81) for infinite dilution, ∞ ln γ 1 = 3.053 = 1 − ln Λ 12 − Λ 21 (1) ln γ ∞ = 2.286 = 1 − ln Λ 21 − Λ 12 2 (2) Solving simultaneous, nonlinear Eqs. (1) and (2) using Newton's method, Λ12 =0.1004 and Λ21 = 0.2493 (c) Activity coefficients can be calculated with the above constants,...
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