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

01 286 log34 015 log for component distribution

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Unformatted text preview: 47.872 − 9.9057(d nC5 ) = 1.644(d iC4 ) + 93.386 − 9.9057(d nC5 ) . . D + Lmin = 0.84 (d nC5 ) 2.36 (d iC 4 ) 1.88 (23125) 1.00 (3.6) 4.36 (5) . + + + + . . . . 4.36 − 1317 2.36 − 1.317 188 − 1317 100 − 1317 0.84 − 1317 . . . . = 7.164 + 2.263(d iC 4 ) + 77.22 − 1136 − 1.761(d nC5 ) = 2.263(d iC4 ) + 73.02 − 1761(d nC5 ) D + Lmin = 2.36 (d iC4 ) 1.88 (23125) 1.00 (3.6) 0.84 (d nC5 ) 4.36 (5) . + + + + 4.36 − 2.173 2.36 − 2.173 188 − 2.173 100 − 2.173 0.84 − 2.173 . . = 9.968 + 12.62(d iC4 ) − 148.38 − 3.07 − 0.63(d nC5 ) = 12.62(d iC4 ) − 14148 − 0.63(d nC5 ) . Because the total distillate rate, D, is also an unknown, the following equation is also needed: D = dC3 + diC4 + dnC4 + diC5 + dnC5 = 5 + diC4 +23.125 + 3.6 + dnC5 = 31.725 + diC4 + dnC5 Solving these 4 linear equations, the following results are obtained: D = 53.27 lbmol/h, Lmin = 64.74 lbmol/h, diC4 = 20.61 lbmol/h, dnC5 = 0.93 lbmol/h Note that the distillate rate for iC4 is impossible because it exceeds the feed rate of 15 lbmol/h. Therefore, iC4 does not distribute, and the calculations must be repeated, eliminating diC4 as an unknown by replacing it with a value of 15 lbmol/h. Also, the θ root of 2.173 is not used. The following three equations result: Exercise 9.17 (continued) Analyze: (c) (continued) D + Lmin = 106.96 − 1761d nC5 . D + Lmin = 118.05 − 9.9057d nC5 D = 46.725 + d nC5 Solving these three equations: D = 48.085 lbmol/h, Lmin = 56.49 lbmol/h, and dnC5 = 1.36 lbmol/h This is the correct value of internal minimum reflux rate. Assume the external minimum reflux ratio = internal minimum reflux ratio = 56.49/48.085 = 1.175. Note that this value is about 15% less than that obtained by assuming a Class 1 separation. (d) The operating reflux ratio = 1.2(1.175) = 1.41. In the Gilliland equation, (9-34), X = (R Rmin)/(R + 1) = (1.41 - 1.175)/1.41+1) = 0.0975. Using Eq. (9.34), Y = 0.556 = (N - Nmin)/(N + 1). Solving, N = 15.6 stages. (e) Apply the Kirkbride equation (9-36): From above, using the Fenske distribution, D = 48.799 lbmol/h and B = 51.201 lbmol/h NR = NS ziC5 , F xnC4 , B znC4 , F xiC5 , D 2 0.206 B D = 0.20 0.25 0.0366 0.0738 2 51.201 48.799 0.206 = 0.723 Therefore, of 15.6 equilibrium stages, (0.723/1.723)(15.6) = 6.5 stages are in the rectifying section. Thus, the feed stage is equilibrium stage 6 or 7 from the top. Exercise 9.18 Subject: Use of the FUG method for the distillation of a chlorination effluent. Given: Bubble-point liquid feed with the following composition and average K-values: Component C2H4 , A HCl , B C2H6 , C C2H5Cl , D Mole fraction 0.05 0.05 0.10 0.80 Aver. K-value 5.10 3.80 3.40 0.15 Partial condenser and partial reboiler. Column pressure = 240 psia. (xD/xB) = 0.01 for C2H5Cl and 75 for C2H6. Find: Minimum equilibrium stages Component distribution Minimum reflux ratio Number of equilibrium stages for R = 1.5 Rmin Feed stage location Analysis: For minimum equilibrium stages, use the Fenske equation (9-11): log N min = ( xD / xB ) C H ( xD / xB )C H Cl 2 2 log α C...
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