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

For each method and compare analysis calculations

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Unformatted text preview: KA − 1 KA − 1 (4) Solve for yA from the third equation in Exercise 4.17, y A = KA K B − KA / K B − KA (5) Compute (nB)V = (1 - yA)100Ψ . If the value is 6, then temperature guess is correct. Otherwise, guess another T, and repeat steps (1) to (5). If 6, compute % recovery of benzene in the vapor from 2.5 yAΨ x 100%. Guess T, oF 195 205 205.5 Ps of A, psia 20.0 23.4 23.6 Ps of B, psia 8.0 9.5 9.6 KA KB Ψ yA 1.36 1.59 1.605 0.544 0.646 0.653 -0.79 0.113 0.161 Moles B in vapor 0.596 0.585 4.6 6.7 By interpolation, T = 205.3oF to obtain 6 moles of B in the vapor. This corresponds to Ψ = 0.145 and yA = 0.588. From above, % recovery of benzene in the vapor = 2.5(0.588)(0.145)100% = 21.3% Exercise 4.21 Subject: Equilibrium flash of a seven-component mixture. Given: Feed mole fractions and K-values. Find: Ψ = V/F by: zi 1 − Ki (a) Rachford- Rice equation, f 1 {Ψ} = = i =1 1 + Ψ Ki − 1 C C C i =1 zi Ki (b) Alternative flash equation, f 2 {Ψ} = = i =1 1 + Ψ Ki − 1 g1 {i , Ψ} C i =1 g2 {i , Ψ} Make plots of f{Ψ} vs. Ψ for each method and compare. Analysis: Calculations with a spreadsheet, for values of Ψ from 0 to 1.0 in intervals of 0.1: (a) I zF K Ψ=0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 0.0079 16.2 g{i,Ψ -0.120 -0.048 -0.030 -0.022 -0.017 -0.014 -0.012 -0.010 -0.009 -0.008 -0.007 2 0.1321 5.2 -0.555 -0.391 -0.302 -0.245 -0.207 -0.179 -0.158 -0.141 -0.127 -0.116 -0.107 3 0.0849 2.6 -0.136 -0.117 -0.103 -0.092 -0.083 -0.075 -0.069 -0.064 -0.060 -0.056 -0.052 4 0.2690 1.98 -0.264 -0.240 -0.220 -0.204 -0.189 -0.177 -0.166 -0.156 -0.148 -0.140 -0.133 5 0.0589 0.91 0.005 0.005 0.005 0.005 0.005 0.006 0.006 0.006 0.006 0.006 0.006 6 0.1321 0.72 0.037 0.038 0.039 0.040 0.042 0.043 0.044 0.046 0.048 0.049 0.051 7 0.3151 0.28 0.227 0.244 0.265 0.289 0.319 0.354 0.399 0.457 0.535 0.645 0.810 f{Ψ}: -0.805 -0.508 -0.345 -0.227 -0.130 -0.042 0.045 0.137 0.245 0.380 0.568 K Ψ=0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 (b) I zF 1 0.0079 16.2 g{i,Ψ 0.128 0.051 0.032 0.023 0.018 0.015 0.013 0.011 0.010 0.009 0.008 2 0.1321 5.2 0.687 0.484 0.373 0.304 0.256 0.222 0.195 0.174 0.158 0.144 0.132 3 0.0849 2.6 0.221 0.190 0.167 0.149 0.135 0.123 0.113 0.104 0.097 0.090 0.085 4 0.2690 1.98 0.533 0.485 0.445 0.412 0.383 0.357 0.335 0.316 0.299 0.283 0.269 5 0.0589 0.91 0.054 0.054 0.055 0.055 0.056 0.056 0.057 0.057 0.058 0.058 0.059 6 0.1321 0.72 0.095 0.098 0.101 0.104 0.107 0.111 0.114 0.118 0.123 0.127 0.132 7 0.3151 0.28 0.088 0.095 0.103 0.113 0.124 0.138 0.155 0.178 0.208 0.251 0.315 f{Ψ}: 0.805 0.457 0.276 0.159 0.078 0.021 -0.018 -0.041 -0.049 -0.038 0.000 The values of f{Ψ} are plotted on the next page, where it is observed that the Rachford-Rice and Alternative equations give the same result of Ψ = 0.55. However, the alternative equation also has a trivial root at Ψ = 1.0. With a Newton procedure, the alternative equation may converge to the trivial root. Therefore, the Rachford-Rice equation is preferred because of its uniqueness. Analysis: (continued) Exercise 4.21 (continuous...
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