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

000 ic 4 15 15000 nc 4 25 23125 ic 5 20 3600 nc 5 35

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Unformatted text preview: .27, bnC5 = 0.2605 mol and dnC5 = 0.0095. Therefore, B = 0.0057 + 0.2605 + 0.20 + 0.10 = 0.5662. Repeating the calculations with this value of B, the result is bnC4 = 0.00566, dnC4 = 0.19434, bnC5 = 0.2604, and dnC5 = 0.0096. Using Eq. (1) for each of the nonkey components shows that they do not distribute to any extent. Therefore the distillate and bottoms compositions at total reflux are: xD xB Component f, mol d, mol b, mol C2 0.08 0.0800 0.184 0.0000 0.000 C3 0.15 0.1500 0.346 0.0000 0.000 nC 4 0.20 0.1943 0.448 0.0057 0.010 nC 5 0.27 0.0096 0.022 0.2604 0.460 nC 6 0.20 0.0000 0.000 0.2000 0.353 nC 7 0.10 0.0000 0.000 0.1000 0.177 Total 1.00 0.4339 1.000 0.5661 1.000 Exercise 9.16 (continued) Analysis: (c) Because the split of the two key components is quite sharp, assume a Class 2 separation for the purpose of minimum reflux. Thus, only components methane to n-pentane will distribute to the distillate. To determine the phase condition of the feed, perform an adiabatic flash with bubble-point conditions at 300 psia upstream of the feed valve to the column, and 120 psia downstream of the valve, using Chemcad. The result is a V/F ratio for the feed of 0.334. From Eq. (7-19), q = 1 - (V/F) = 1 - 0.334 = 0.666 or 1 - q = 0.334. From Underwood Eq. (9-28), α i ,nC5 zi , F i α i ,nC5 − θ = 1 − q = 0.334 = 14.6(0.08) 5.9(015) 2.35(0.20) 1.00(0.27) 0.424(0.20) 0.187(010) . . + + + + + 14.6 − θ 015 − θ . 2.35 − θ 100 − θ . 0.424 − θ 0187 − θ . Solving this equation for the root between 2.35 and 1.00 gives θ = 1.5962 From Underwood Eq. ((9-29), α i ,nC5 xi , D 14.6(0184) . 5.9(0.346) 2.35(0.448) 1.00(0.022) = + + + = 2.04 = 1 + Rmin 14.6 − 15962 5.9 − 15962 2.35 − 15962 1.00 − 15962 . . . . i α i , nC5 − θ Therefore, the internal minimum reflux ratio = 2.04 - 1.00 = 1.04 Assume the external minimum reflux ratio = 1.04 (d) For the number of theoretical stages, use the Gilliland equation, (9-34), where for this exercise, R = 1.5R = 1.5(1.04) = 1.56. X = (R - Rmin)/(R + 1) = (1.56 - 1.04)/(1.56 + 1) = 0.203. Using Eq. (9.34), Y = 0.458 = (N - Nmin)/(N + 1). Solving, N = 15.6 equilibrium stages. Exercise 9.17 Subject: mixture. Use of the FUG method for the distillation of a light paraffin hydrocarbon Given: Feed mixture of: Component C 3 i C 4 nC 4 i C 5 nC 5 lbmol/h 5 15 25 20 35 Column pressure = 120 psia. Liquid distillate contains 92.5% of the nC4 in the feed. Bottoms contains 82 mol% of iC5 in the feed. Use Chemcad with SRK instead of Figs. 2.8 and 2.9 for Kvalues. Find: (a) Minimum number of equilibrium stages. (b) Distribution of nonkey components. (c) Minimum reflux ratio for a bubble-point liquid feed. (d) Number of equilibrium stages for R = 1.2 Rmin. (e) Feed-stage location. Analysis: (a) To obtain the average relative volatilities, run bubble points on the following assumed distillate and bottoms compositions, using the given key-component recoveries: lbmol/h: Component Feed Distillate C3 5 5.000 iC 4 15 15.000 nC 4 25 23.125 iC...
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