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

4 8 applies ya 1 xa ab 1 solving this equation for

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Unformatted text preview: achford-Rice equation to be satisfied. Given: Eq. (3), Table 4.4, which is the Rachford-Rice equation. Find: Conditions under which the equation can be satisfied for 0 ≤ V ≤ 1. F Analysis: A necessary, but not sufficient, condition is that at least one K-value is < 1 and at least one K-value is > 1. If all K-values are > 1, the sum: zi 1 − Ki will be negative and can not be zero. i =1 1 + Ψ Ki − 1 C If all K-values are < 1, the numerator in the sum will be positive for each term. With Ψ between 0 and 1, the term Ψ(Ki - 1) will always be < 1. Therefore, the denominator will be positive also and the sum will be positive and can not be zero. Exercise 4.19 Subject: Flash vaporization of a benzene (A) - toluene (B) mixture for αA-B = 2.3. Given: Feed is 40 mol% A and 60 mol% B. Find: Percent of A in the equilibrium vapor if 90% of the toluene leaves in the liquid by graphical means. α A,B xA Analysis: For constant relative volatility, Eq. (4-8) applies, yA = 1+ xA α A,B − 1 Solving this equation for yA as a function of xA , xA 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 yA 0.2035 0.3651 0.4964 0.6053 0.6970 0.7753 0.8429 0.9020 0.9539 A plot of the calculated equilibrium curve is given below. To use this plot for a graphical solution of the equilibrium, draw a q-line, using the following equation above Eq. (4-6), for an assumed value of Ψ = V/F and check the resulting % recovery of toluene in the liquid. Vary Ψ until the % recovery = 90%. Then compute, for the corresponding Ψ , the % recovery of benzene in the vapor. yA = Ψ −1 1 Ψ −1 1 xA + zA = xA + 0.40 Ψ Ψ Ψ Ψ (1) Analysis: (continued) Exercise 4.19 (continued) Basis: F = 100 moles, 60 moles toluene (B). Want 0.9(60) = 54 moles B in liquid. Therefore, 60 - 54 = 6 moles B in vapor. Therefore, want (nB)V = yBV = (1 - yA)100Ψ = 6. Then compute % recovery of benzene in vapor = (nA)V/40 x 100% = yAV/40 x 100% = 2.5 yAΨ x 100%. The following are typical values for the trial and error procedure, with the final result at the bottom. Assumed Ψ yA 0.3 0.2 0.15 0.142 0.54 0.56 0.575 0.58 xA (nB)V , moles 0.35 13.8 0.36 8.8 0.37 6.4 0.375 6.0 % recovery of A in vapor 40.5 28.0 21.6 20.6 Exercise 4.20 Subject: Flash vaporization of a benzene (A) - toluene (B) mixture. Given: Feed is 40 mol% A and 60 mol% B. Vapor pressure data. Assumptions: Raoult's law (ideal solutions). Pressure = 1 atm. Find: Percent of A in the equilibrium vapor if 90% of the toluene leaves in the liquid. Analysis: Basis: F = 100 mole with 60 moles B and 40 moles A. Want 0.9(60) = 54 moles B in liquid. Therefore, 60 - 54 = 6 moles B in vapor. Therefore, want (nB)V = yBV = (1 yA)100Ψ = 6. Then compute % recovery of benzene in vapor = (nA)V/40 x 100% = yAV/40 x 100% = 2.5 yAΨ x 100%. The following trial and error procedure can be used, based on material balance and equilibrium equations: (1) Guess a temperature. (2) Read vapor pressures from Fig. 2.4 and compute K-values from Raoult's law (Eq. (3), Table 2.3), Ki = Pi s / P . (3) Solve for Ψ = V/F using the fifth equation in Exercise 4.17, zA KA − KB / 1 − KB − 1 0.40 KA − KB / 1 − KB − 1 Ψ= =...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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