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

V hcs ki hcs ki using the above table this sum is 1

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Unformatted text preview: carbons in the entering steam simplifies to: l1 = lN+1 φS (1) where, from Eq. (5-50), where, from Eq. (5-53), By overall material balance, φS = S −1 S −1 =4 N +1 S −1 S −1 (2) S = KV/L = KV0/LN+1 = K(1,000)/(1,000) = K υN = lN+1 + υ0 − l1 (3) (4) The results from applying the above equations with a spreadsheet are as follows: Flow rates, kmol/h: Component Ki = S φS lN+1 = l in li = l out υN = υ out C1 60 0 0.3 0.0 0.3 C2 28 0 2.2 0.0 2.2 C3 14 0 18.2 0.0 18.2 nC 4 6.5 0.003 44.7 0.1 44.6 nC 5 3.5 0.0168 85.9 1.4 84.5 nC10 0.20 0.8013 848.7 680.0 168.7 Steam 1,000.0 Total 1,000.0 681.5 1,318.5 Now check the primary dew point. Partial pressure of steam in the exit vapor = yi P = (1,000)/(1,318.5) x 50 = 37.9 psia. Vapor pressure of water at 300oF = 67 psia. Therefore, steam will not condense. Now check the possibility of the HCs condensing. yi 1 υi No condensation if, for the HCs, from Eq. (4-13), = < 10 . V HCs Ki HCs Ki Using the above table, this sum is: 1 0.3 2.2 18.2 44.6 84.5 168.7 + + + + + = 0.0 + 0.0 + 0.001 + 0.005 + 0.018 + 0.640 = 0.664 1,318.5 60 28 14 6.5 3.5 0.2 Therefore, no condensation occurs. CHEMCAD computes a primary dew point of 287oF. Exercise 5.20 Subject: Given: Successive flash distillation as in Fig. 5.12 Two multiple flash arrangements, one with no recycle and one with recycle. Assumptions: Equilibrium at each flash stage. Find: Cooling requirements if heaters in Stages 2 and 3 are removed, and if so whether system will produce the same products. Analysis: Nothing is gained by totally condensing the vapor leaving each stage, followed by partial vaporization in the stage because equilibrium compositions depend only on the temperature and pressure. Therefore, remove a heater at a stage (2 or 3) and reduce the condenser duty on the vapor feed to that stage by the amount of the heater. Product compositions will be the same. In Fig. 12(a), new duty of condenser feeding stage 2 = 734 - 487 = 247 MBH. New duty of condenser feeding stage 3 = 483 - 376 = 107 MBH. Save two heat exchangers and reduce the size of the two condensers. Save total heating and cooling duties of 247 + 107 = 364 MBH. In Fig. 12(b), new duty of condenser feeding stage 2 = 883 - 665 = 218 MBH. New duty of condenser feeding stage 3 = 646 - 503 = 143 MBH. Save two heat exchangers and reduce the size of the two condensers. Save total heating and cooling duties of 218 + 143 = 361 MBH. Exercise 5.21 Subject: Effect of reflux rate on key-component distribution in multicomponent distillation by the group method. Given: Results in Example 5.4 for distillation at 400 psia of a superheated vapor feed at 105oF and containing, lbmol/h of 160 C1, 370 C2, 240 C3, 25 nC4, and 5 nC5 in a column with a partial condenser, 5 equilibrium rectification stages, equilibrium feed stage, 5 equilibrium stripping stages, and a partial reboiler, for an external reflux rate of 1,000 lbmol/h. Assumptions: No changes in stage temperatures when reflux rate is changed. Find: Molar ratio of C3 flow rate in the distillate to that in the bottoms, with a plot, for reflux rates of (a) 1,500, (b) 2,000, and (c) 2,500 lbmol/h. Analysis: Using the Edmister group method,...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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