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

57 mol then bnc4 001057 00057 therefore dnc4 020

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Unformatted text preview: be favored over the Gilliland correlation, which emphasizes rectification, rather than stripping. Subject: mixture. Exercise 9.16 Use of the FUG method for the distillation of a normal paraffin hydrocarbon Given: Bubble-point liquid feed at 300 psia, with composition in mole fractions: C3 nC 4 nC 5 nC 6 nC 7 Component C2 Mole fraction 0.08 0.15 0.20 0.27 0.20 0.10 Find: (a) For a sharp separation between nC4 and nC5, determine column pressure and type condenser for condenser outlet temperature of 120oF. (b) At total reflux, determine the separation for 8 equilibrium stages, with 0.01 mole fraction of nC4 in the bottoms. (c) Minimum reflux ratio for separation in Part (b). (d) Number of equilibrium stages for R/Rmin = 1.5 using Gilliland correlation. Analysis: (a) For a sharp split, the distillate composition is as follows: Component C2 C3 nC 4 Total: xd 0.186 0.349 0.465 1.000 K at 120oF, 250 psia 2.8 1.13 0.34 K xd 0.520 0.394 0.158 1.072 In this table, a distillate bubble-point pressure of 250 psia is assumed and the bubble-point equation (4-12) is applied, using K-values in Fig. 2.8. The summation is 1.072. In order for the sum to be 1.0, an even higher pressure is needed, since the K-value is almost inversely proportional to pressure. But, in Fig. 7.16, we are already > 215 psia, therefore use a partial condenser. To determine the pressure, run a dew point using Eq. (4-13). Component C2 C3 nC 4 Total: yD 0.186 0.349 0.465 1.000 K at 120oF, 140 psia 4.7 1.58 0.53 yD/K 0.040 0.221 0.877 1.138 K at 120oF, 120 psia 5.4 1.80 0.60 yD/K 0.030 0.194 0.775 0.999 On the second trial in the above table, the dew-point pressure is found to be 120 psia. Set the column pressure at this value with a partial condenser. Exercise 9.16 (continued) Analysis: (continued) (b) Run a bubble point on an assumed bottoms composition, using Eq. (4-12): xB K at 320oF, xB K K at 290oF, xB K Component 120 psia 120 psia nC 4 0.010 3.3 0.033 2.75 0.028 nC 5 0.469 1.7 0.797 1.38 0.647 nC 6 0.347 0.92 0.319 0.72 0.250 nC 7 0.174 0.48 0.084 0.36 0.063 Total: 1.000 1.233 0.988 o A third trial gives a bottoms temperature of 295 F The K-values and relative volatilities referred to the nC5, the heavy key, at distillate and bottoms conditions are as follows for a pressure of 120 psia: Component K at 120oF, αI, nC5 at 120oF, K at 295oF, αI, nC5 at 295oF, Geometric 120 psia 120 psia 120 psia 120 psia average α C2 5.40 25.7 12.0 8.28 14.6 C3 1.80 8.6 5.8 4.0 5.9 nC 4 0.60 2.86 2.8 1.93 2.35 nC 5 0.21 1.00 1.45 1.00 1.00 nC 6 0.074 0.352 0.74 0.51 0.424 nC 7 0.028 0.133 0.38 0.262 0.187 log From the Fenske Eq. (9-12), N min = 8 = d nC4 bnC5 d nC5 bnC4 log α nC4 ,nC5 avg log = d nC 4 bnC5 d nC5 bnC4 log 2.35 (1) Take as a basis, 1 mol of feed. Assume that the distribution of the nonkeys is negligible. That is, only the LK and the HK are found in both the distillate and bottoms. Assume the total bottoms, B, is 0.57 mol. Then bnC4 = 0.01(0.57) = 0.0057. Therefore, dnC4 = 0.20 - 0.0057 = 0.1943 mol. Solving Eq. (1), bnC5/dnC5 = 27.3. Since bnC5 + dnC5 = fnC5 = 0...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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