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

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Unformatted text preview: ,D 6 5 75 0.01 = = 2.86 log(3.4 / 0.15) log For component distribution, use the following rearrangements of the Fenske equation: xD xB xD xB = C2 H 4 HCl xD xB x =D xB N α C2min4 ,C2 H5Cl = 0.01 H C 2 H 5Cl α C 2 H 5Cl N min HCl,C 2 H 5Cl 51 . 015 . 3.8 = 0.01 015 . 2 .86 = 240 2 .86 = 103 Because the separation between the two key components is quite sharp, assume that the two nonkeys do not distribute at minimum reflux. The relative volatilities with respect to the heavy key are: 34 for A, 25.3 for B, and 22.7 for C. Now, use the Class 2 Underwood equation, Eq. (928),which for a bubble-point feed is: Exercise 9.18 (continued) Analysis: (continued) α i , D zi , F i αi ,D − θ = 1− q = 0 = 34(0.05) 25.3(0.05) 22.7(0.10) 1.00(0.80) + + + (1) . 34 − θ 25.3 − θ 22.7 − θ 100 − θ There are three roots to Eq. (1). The one of interest is the one between 22.7 and 1.0, which from a spreadsheet calculation is θ = 4.318. The composition of the distillate by material balances is: Component distillate, mole fraction C2H4 , A 0.25405 HCl , B 0.25405 C2H6 , C 0.48195 C2H5Cl , D 0.00995 Apply Underwood Eq. (9-29): α i , D xi , D 34(0.25405) 25.3(0.25405) 22.7(0.48195) 100(0.00995) . = + + + = 1 + Rmin 34 − 4.318 25.3 − 4.318 22.7 − 4.318 1.00 − 4.318 i αi ,D − θ = 0.291 + 0.306 + 0.595 − 0.003 = 1189 = 1 + Rmin . Therefore, Rmin = 0.189. Assume this is equal to the external minimum reflux ratio. Operating reflux ratio = 1.5(0.189) = 0.284. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (0.284 - 0.189)/0.284 + 1) = 0.074. Using Eq. (9.34), Y = 0.581 = (N - Nmin)/(N + 1). Solving, N = 8.2 stages. For feed stage location, apply the Kirkbride equation (9-36): From above, using the Fenske distribution with a material balance, D/B = 0.245. Also, mole fractions in the bottoms product are: xC = 0.0064 and xD = 0.9936 NR = NS zD , F xC , B zC , F xD , D 2 0.206 B D = 0.80 0.10 0.0064 0.00995 2 1 0.245 0.206 = 171 . Therefore, of 8.2 equilibrium stages, (1.71/2.71)(8.2) = 5.2 stages are in the rectifying section. With a partial condenser acting an equilibrium stage, the feed is about stage 4 from the top stage in the column. Exercise 9.19 Subject: Use of the FUG method for the distillation of a ficticious ternary mixture. Given: 100 kmol/h of a bubble-point feed of: Component A B C Feed rate, kmol/h 40 20 40 αi,C 5 3 1 Find: (a) Product distribution for a distillate rate of 60 kmol/h and 5 minimum theoretical stages. (b) Minimum reflux and boilup ratios for separation of part (a). (c) Number of equilibrium stages and feed-stage location for R = 1.2Rmin. Analysis: (a) With a distillate rate of 60 kmol/h, the LK is B and the HK is C. From the rearranged form of the Fenske equation, given by Eq. (9-14), using component C as the reference, dA d d = C 55 = 3125 C bA bC bC dB d d = C 35 = 243 C bB bC bC The following 4 material balances also apply: dA + bA = 40 dB + bB = 20 dC + bC = 40 dA + dB + dC = 60 Solving these 6 equations, tw...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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