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

3 fi fi thus for the llk i bi 3 5 d c n 103 3 min 1

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Unformatted text preview: sults are obtained: Component Propane Isobutane n-Butane Isopentane n-Pentane n-Hexane n-Heptane n-Octane αi, iC5 at top 9.06 3.55 1.00 0.783 0.272 0.094 0.038 αi, iC5 at bottom αi, iC5 average 3.93 5.97 2.24 2.82 1.00 0.848 0.448 0.228 0.128 0.815 0.349 0.146 0.070 di , kmol/h 2499.996 399.63 594 15 6.12 0.001 8.2x10-7 1.92x10-10 bi , kmol/h 4.2x10-3 0.37 6 85 193.88 39.999 50 40 Exercise 9.8 Subject: Type condenser and operating pressure. Minimum equilibrium stages and distribution of nonkey components using the Fenske equation for distillation of a light hydrocarbon mixture. Given: Feed composition and split of two key components in Fig. 9.24. K-values from Figs. 2.8 and 2.9. Find: Type condenser. Operating pressure. Minimum equilibrium stages. Distribution of nonkey components at total reflux. Analysis: To determine the type condenser and operating pressure, follow the algorithm in Fig. 7.16. The estimated distillate and bottoms compositions from the given data are: Component Feed, lbmol/h Methane Ethylene Ethane Propylene Propane n-Butane Distillate, lbmol/h 1000 2500 1999 5 0 0 1000 2500 2000 200 100 50 Bottoms, lbmol/h 0 0 1 195 100 50 Using Fig. 2.8, it is quickly shown that at 120oF, both the bubble-point and the dew-point pressures for the estimated distillate composition are greater than 1000 psia because the K-value for ethane is greater than 1.0 at 120oF and 1000 psia. Therefore, select the operating pressure as 415 psia and use a partial condenser. Run a dew point on the distillate at 415 psia. Use Eq. (4-13) in the following form: i di 1000 2500 1999 5 =D= + + + = 5504 Ki K C1 KC = KC 2 KC = 2 (1) 3 By trial and error, using Fig. 2.9 for K-values, the temperature that satisfies Eq. (1) is 0oF, which gives: di 1000 2500 1999 5 =D= + + + = 5505 5.0 105 0.69 018 . . i Ki The relative volatility of the two key components is αC2, C3= = 0.69/0.18 = 3.83 Exercise 9.8 (continued) Analysis: (continued) Now run a bubble-point on the estimated bottoms composition at 415 psia. Use Eq. (4-12) in the following form: bi Ki = B = KC 2 + 195KC= + 100 KC3 + 50 KnC4 = 346 (2) 3 i By trial and error, using Fig. 2.8 for K-values, the temperature that satisfies Eq. (1) is 155oF, . . which gives: bi Ki = B = 2.7 + 195(115) + 100(101) + 50(0.39) = 347 i The relative volatility of the two key components is αC2, C3= =2.7/1.15 = 2.35 The geometric mean relative volatility = [(3.83)(2.35)]1/2 = 3.0 From the Fenske equation, (9-12), d C2 N min = bC= d C= log bC2 3 3 log α C log = = 2 , C3 1999 195 5 1 log 3.0 = 10. 3 To compute the distribution of nonkey components at total reflux, use for the lighter than light key components, LLK, Eq. (9-15), and use Eq. (9-16) for the heavier than heavy key, HHK, with C= as the reference component, r, and the above value of Nmin. 3 fi fi = Thus, for the LLK, i, bi = (3) 5 d C= N 10.3 3 min 1+ α i,C= 1 + 195 α i,C= 3 3 b C= 3 fi For the HHK, i, di = d C= 3 bC = α iN min ,C = 3 3 1+ d C= 3 bC= α iN min ,C = 3 5 α 10.3 = 195 i ,C3...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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