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

Given feed contains 40 mol p and 60 mol w

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Unformatted text preview: 449 0.3361 0.3277 0.3197 0.3120 0.3048 0.2978 0.2912 0.2848 0.2788 0.2729 0.2673 0.2619 0.2568 0.2518 0.2470 0.2424 0.2379 0.2336 0.2295 0.2255 0.2216 0.2179 0.2143 0.2107 0.2073 0.2040 0.2008 0.1977 0.1947 0.1918 0.1889 0.1862 0.1835 0.1809 0.1783 0.1758 0.1734 0.1711 0.1688 0.1665 0.9816 0.9816 0.9816 0.9816 0.9815 0.9815 0.9815 0.9815 0.9814 0.9814 0.9814 0.9813 0.9813 0.9813 0.9812 0.9812 0.9812 0.9811 0.9811 0.9811 0.9810 0.9810 0.9809 0.9809 0.9808 0.9808 0.9807 0.9807 0.9806 0.9806 0.9805 0.9805 0.9804 0.9803 0.9803 0.9802 0.9802 0.9801 0.9800 0.9799 0.9799 Exercise 13.7 Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene, but with the continual supply of feed of the same composition, while heat is supplied at a rate that maintains the liquid level in the still. Given: Charge of 25 mol of 35 mol% benzene (B) and 65 mol% toluene (T). Continuous feed of 7 mol/h of the same composition as the charge. Constant relative volatility = 2.5. Assumptions: Perfect mixing in the still. Exiting vapor in equilibrium with the liquid. No change in molar density of the mixture as the composition changes. Find: Time for the benzene mole fraction of the instantaneous distillate to fall to 0.45. Analysis: Base the calculations on benzene, which is the more volatile component. Initially, the distillate mole fraction in equilibrium with the initial charge is obtained from Eq. (4-8) for constant relative volatility, α: y= αx 2.5(0.35) = = 0.574 1 + x (α − 1) 1 + 0.35(2.5 − 1) (1) When the benzene mole fraction in the instantaneous distillate = yD = 0.45, the mole fraction of benzene in the liquid in the still is obtained form a rearrangement of Eq. (1), x= 0.45 y = = 0.2466 α + y (1 − α ) 2.5 + 0.45(1 − 2.5) Modify Eq. (13-1) to include the constant feed added to the still, noting that, with the above assumptions, W = total moles in the still = constant = W0 = 25 moles, dW/dt = 0, and the distillate rate, call it D = molar feed rate, F: Fx F − W dx = Dy D = Fy D dt Rearranging, dx F x F − y D = dt W0 (2) Eq. (2) is integrated as follows, noting that yD = y. Substituting Eq. (1) into Eq. (2) gives, F 7 dx dx (1 + 15x )dx . dt = dt = = = 2.5x αx W0 25 0.35 − 1.975x xF − 0.35 − 1 + x (α − 1) 1 + 15x . 7t dt = 0.28t = 25 0 0.2466 0.35 1 + 15x )dx . = 0.599 0.35 − 1.975x Therefore, t = 0.599/0.28 = 2.14 h Exercise 13.8 Subject: Simple differential batch (Rayleigh) distillation of a mixture of A and B, but with a continual supply of feed, while heat is supplied at a rate that maintains the liquid level in the still. Given: Charge of 20 lbmol of 30 mol% A and 70 mol% B. Continuous feed of 10 lbmol/h. Constant relative volatility = 2.5. Assumptions: Perfect mixing in the still. Exiting vapor in equilibrium with the liquid. No change in molar density of the mixture as the composition changes. Composition of the feed is the same as the initial charge. Overhead product is the instantaneous distillate. Find...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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