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

70 068 066 064 62 08550 08428 08302 08172 08037 6452

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Unformatted text preview: aterial balance on P gives Eq. (13-6) for the mole fraction of C7 in the cumulative distillate, Therefore, for a basis of W0 = 1 mole, yD avg = W0 x0 − Wx W0 − W ( yD )avg = (3) 1(0.4) − 0.3(0.0575) = 0.547 1 − 0.3 Exercise 13.2 (continued) x 0.40 0.39 0.38 0.37 0.36 0.35 0.34 0.33 0.32 0.31 0.30 0.29 0.28 0.27 0.26 0.25 0.24 0.23 0.22 0.21 0.20 0.19 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.11 0.10 0.09 0.08 0.07 0.06 0.05 0.04 y f =1/(y-x) 0.580 0.578 0.575 0.573 0.571 0.569 0.566 0.564 0.562 0.559 0.557 0.554 0.552 0.549 0.546 0.543 0.540 0.536 0.533 0.529 0.525 0.521 0.517 0.512 0.507 0.502 0.497 0.491 0.486 0.479 0.473 0.466 0.459 0.452 0.444 0.436 0.427 5.558 5.329 5.117 4.922 4.741 4.573 4.417 4.272 4.137 4.011 3.894 3.785 3.683 3.588 3.499 3.416 3.338 3.266 3.198 3.135 3.076 3.021 2.970 2.923 2.879 2.838 2.801 2.767 2.735 2.707 2.681 2.658 2.638 2.621 2.606 2.594 2.584 Value of Integral 0.054 0.107 0.157 0.205 0.252 0.297 0.340 0.382 0.423 0.462 0.501 0.538 0.575 0.610 0.645 0.678 0.711 0.744 0.775 0.806 0.837 0.867 0.896 0.925 0.954 0.982 1.010 1.037 1.065 1.092 1.118 1.145 1.171 1.197 1.223 1.249 Exercise 13.3 Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene. Given: Charge of 100 moles of 30 mol% benzene (B) and 70 mol% toluene (T). Distillation until the cumulative distillate contains 45 mol% benzene. Assumptions: Relative volatility = α = αB-T = 2.5. Perfect mixing in the still. Exiting vapor in equilibrium with the liquid. Find: Moles of residue. Analysis: For constant relative volatility, Eq. (13-5) applies with α = 2.5, W0 = 100, x0 = 0.30. Thus, ln 100 2 0.30 1− x = ln + 2.5 ln W 3 x 0.70 (1) From Eq. (13-6), the mole fraction of benzene in the cumulative distillate is given by: yD avg = 0.45 = 30 − Wx 100 − W (2) Eqs. (1) and (2) are solved by first solving Eq. (2) for x in terms of W to obtain x = 0.45 − 15 W (3) Substituting Eq. (3) into Eq. (1) and solving the resulting nonlinear equation gives W = 57.9 moles. In lieu of a nonlinear equation solver, a spreadsheet can be used by assuming a value of W and solving Eq. (3) for x. Then substitute this value into Eq. (1) and solve for W. Repeat until the assumed value of W equals the calculated value. This is readily done by setting up a table in W, working in increments from W = 100 moles down to a low value of W. Exercise 13.4 Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene. Given: Charge of 250 lb of 70 mol% benzene (B) and 30 mol% toluene (T). Distillation at 1 atm until 1/3 of the original charge is distilled. Assumptions: Ideal K-values. Perfect mixing in still. Exiting vapor in equilibrium with liquid. Find: Compositions of the residue and distillate. Analysis: Using molecular weights of 78.11 for benzene and 92.14 for toluene, the average molecular weight of the charge = 0.7(78.11) + 0.3(92.14) = 82.32. Therefore, the charge contains 250/82.32 = 3.037 lbmol. Final mass of t...
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