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

# Column 2 packed column with a total liquid holdup of

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Unformatted text preview: = 4, αB,C = 2, and αC,C = 1. Assumptions: No condenser and column holdups. Constant molar overflow. No pressure drop. Startup under total reflux. Find: Variation of instantaneous residue and distillate compositions with time following startup. Analysis: For startup conditions of total reflux, use Eq. (13-27) to compute the distillate 0.333 ( composition using N = 4, x D0C) = = 0.00366 4 4 0.333 4 + 0.333(2) 4 + 0.333 1 0.00366 4 ( 4 = 0.9370 and x D0B) = 1 − 0.0037 − 0.9370 = 0.0593 0.333 Take a time increment = ∆t = 0.25 h. 100 From Eq. (13-22), W (1) = 100 − 0.25 = 9583 kmol . 1+ 5 9583 − 100 . ( From Eq. (13-23), x W1) = 0.3333 + (0.9370 − 0.3333) = 0.3081 A 100 95.83 − 100 ( x W1) = 0.3333 + (0.0593 − 0.3333) = 0.3447 B 100 ( From Eq. (13-25), x D0A) = 0.333 ( x W1) = 1 − 0.3081 − 0.3448 = 0.3471 C 4 − N min 5 − Rmin From Eq. (13-28), = 0.75 1 − 4 +1 5 +1 0.5668 (1) 4 N min − 4 From Eq. (13-30), Rmin = (2) (4 − 1)[0.3081(4) N min + 0.3448(2) N min + 0.3471(1) N min ] When Eqs. (1) and (2) are solved simultaneously, the results are Rmin = 0.917 and Nmin = 3.265. Using the latter value in Eq. (13-27), 0.3471 x DC = = 0.0108 3.265 3.265 0.3081 4 + 0.3447 2 + 0.3471(1) 3.265 From Eq. (13-25), 0.0108 3.265 0.0108 3.265 x DA = 0.3081 4 = 0.8861 and x DB = 0.3447 2 = 0.1031 0.3471 0.3471 Exercise 13.24 (continued) Analysis: The calculations above for the first time increment are repeated for subsequent time increments, using a spread sheet, up to 5 hours. The results are as follows. On the next page, the mole fraction compositions are plotted as a function of time. Analysis: (continued) Exercise 13.24 (continued) Exercise 13.25 Subject: Calculation of batch rectification of a ternary mixture by the shortcut method of Sandaram and Evans. Given: Charge of 200 kmol of 40 mol% A, 50 mol% B, and 10 mol% C. Reflux ratio, R = 10 and vapor rate, V = 100 kmol/h. Column has two equilibrium plates, a total condenser, and a partial reboiler. Thus, N = 3. αA,C = 2, αB,C = 1.5, and αC,C = 1. Assumptions: No condenser and column holdups. Constant molar overflow. No pressure drop. Startup under total reflux. Find: Variation of instantaneous residue and distillate compositions for 1hour following startup. Analysis: For startup conditions of total reflux, use Eq. (13-27) to compute the distillate 0100 . ( composition using N = 3, x D0C) = = 0.0199 3 3 0.4 2 + 0.5(15) 3 + 0.1333 1 . 0.0199 3 ( 2 = 0.6368 and x D0B) = 1 − 0.0199 − 0.6368 = 0.3433 01 . Take a time increment = ∆t = 0.10 h. 100 From Eq. (13-22), W (1) = 200 − 0.10 = 199.09 kmol 1 + 10 199.09 − 200 ( From Eq. (13-23), x W1) = 0.4 + (0.6368 − 0.4) = 0.3989 A 200 199.09 − 200 ( x W1) = 0.5 + (0.3433 − 0.5) = 0.5007 B 200 ( From Eq. (13-25), x D0A) = 0.4 ( x W1) = 1 − 0.3989 − 0.5007 = 01004 . C 3 − N min 10 − Rmin From Eq. (13-28), = 0.75 1 − 3 +1 10 +1 0.5668 (1) 2 N min − 2 From Eq. (13-30), Rmin = (2) (2 − 1)[0.3989(2) N min + 0.5007(15) N min + 0.1004(1) N min ] . Whe...
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