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

Column pressure of 300 psia kvalues in figs 28 and 29

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Unformatted text preview: = 5 1+ α 10.3 = 195 i ,C3 fi (4) 3 Using the above K-values to compute geometric mean values of αi, C3= , followed by use of Eqs. (3) and (4) with the material balance, fi = di + bi , the following results are obtained: Component Methane Ethylene Ethane Propylene Propane n-Butane αi, C3= at top αi, C3= at bottom 27.8 5.83 3.83 1.00 0.78 0.18 8.70 3.22 2.35 1.00 0.88 0.34 αi, C3= average 15.55 4.33 3.00 1.00 0.83 0.25 di , lbmol/h bi , lbmol/h 1000 2499.973 1999 5 0.375 8.1x10-7 2.1x10-8 0.027 1 195 99.625 50 Exercise 9.9 Subject: Recovery of a key component as a function of distillate flow rate for the distillation of a paraffin hydrocarbon mixture when the minimum number of equilibrium stages is fixed. Given: Feed composition and flow rate of 1000 lbmol/h. Column pressure of 250 psia. Minimum equilibrium stages = 15. K-values from Figs. 2.8 and 2.9. Find: Percent recovery of propane in the distillate as a function of distillate flow rate. Analysis: Assume that the average relative volatility between propane and n-butane is not sensitive to the separation. If a perfect separation were made between propane and n-butane at 250 psia, the distillate temperature would be approximately 110oF and the bottoms temperature would be approximately 270oF. Assume an average temperature of 190oF. From Fig. 2.8, the Kvalues and relative volatilities at 250 psia and 190oF, with the light key as propane and the heavy key as n-butane are: Component Ethane Propane n-Butane n-Pentane n-Hexane Feed rate, lbmol/h 30 200 370 350 50 1000 Total: K-value 4.3 1.74 0.72 0.30 0.133 α referred to nC4 5.97 2.42 1.00 0.417 0.185 Based on these relative volatility values, assume that little of the nonkey components will distribute. Therefore, D = d C2 + d C3 + d nC 4 + d nC5 + d nC6 = 30 + d C 3 + d nC 4 (1) The Fenske equation, (9-12) becomes: log N min = 15 = Rearranging, 370 − d nC4 d nC4 = d C3 bnC4 d nC4 bC 3 log α C3 , nC4 200 − d C 3 d C3 avg log = dC3 370 − d nC4 d nC 4 200 - d C3 log 2.42 2.4215 = 571800 200 − d C 3 d C3 (2) Assume a value of d for C3. Calculate d for nC4 from Eq. (2). Then compute D from Eq. (1). Then compute recovery of C3 = d for C3/200 and make plot. Exercise 9.9 (continued) Analysis: (continued) d of C3, lbmol/h % recovery of C3 d of nC4, lbmol/h 198 99 0.064 190 95 0.012 180 90 0.006 150 75 0.002 100 50 0.00065 D, lbmol/h 228.1 220.0 210.0 180.0 130.0 We see that with 15 minimum stages, the distribution of nC4 to the distillate is almost negligible. Exercise 9.10 Subject: Minimum reflux ratio by the Underwood equation for the separation of a binary mixture as a function of feed vaporization. Given: Binary feed of 30 mol% propane and 70 mol% propylene. Distillate to contain 99 mol% propylene, and bottoms to contain 98 mol% propane. Column pressure of 300 psia. Kvalues in Figs. 2.8 and 2.9. Find: Minimum reflux ratio by the Underwood equation for: (a) Bubble-point liquid feed. (b) Feed of 50 mol% vapor. (c) Dew-point vapor feed. Analysis:...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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