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

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Unformatted text preview: antaneous distillate accumulator = 0.04 = 0.80 Stop tolerance 0.0001 0.0001 0.0001 Case 2: Plate column: Chemcad input data same as for Case 1 except: Condenser holdup = 1 lbmol Stage holdup = 0.8 lbmol per stage (total of 8 lbmol for 10 stages) Case 3: Packed column: Chemcad input data same as for Case 1 except: Condenser holdup = 1 lbmol Stage holdup = 0.2 lbmol per stage (total of 2 lbmol for 10 stages) Exercise 13.26 (continued) Analysis: All 3 cases converged. The results are as follows for the contents of tanks 3, 4, and 5 after Steps 1, 2, and 3, respectively, and the final residue: Case 1: No holdups: Vessel Tank 3 Tank 4 Tank 5 Final residue Amount, lbmol 32.9987 7.6664 27.1655 32.1696 Mole fractions: C6 0.9487 0.3537 0.0362 0.0000 C7 0.0513 0.6460 0.9231 0.1019 C8 0.0000 0.0003 0.0407 0.8981 Operation time, hr 3.96 0.92 3.26 8.14 Case 2: Plate column: Vessel Tank 3 Amount, lbmol 34.9987 Mole fractions: C6 0.9478 C7 .0.0522 C8 0.0000 Operation time, hr 4.20 Case 3: Packed column: Vessel Tank 3 Amount, lbmol 33.4988 Mole fractions: C6 0.9497 C7 .0.0503 C8 0.0000 Operation time, hr 4.02 Tank 4 3.4999 Tank 5 30.9987 Final residue 21.5030 0.2859 0.7136 0.0005 0.42 0.0267 0.9136 0.0417 3.72 0.0000 0.0244 0.9756 8.34 Tank 4 6.6664 Tank 5 28.1655 Final residue 28.6694 0.3334 0.6662 0.0004 0.80 0.0342 0.9255 0.0403 3.38 0.0000 0.0677 0.9323 8.20 The above results show a significant effect of holdup. If the 4 amounts are totaled, Case 1 gives 100.0002 lbmol (since no holdup), Case 2 gives 91.0003 lbmol (since 1+ 9 lbmol of holdup), and Case 3 gives 97.0001 lbmol (since 1+2 lbmol of holdup). The plate column gives the most product from Step 1, by far the least from Step 2, and the most from Step 3. The compositions for Tank 3 (Step 1) are all close. The compositions for Tank 4 (Step 2) are all different, with the best cumulative purity for C7 being the plate column. The compositions for Tank 5 (Step 3) are fairly close, while the compositions of the final residue are different, with the best purity for C8 given by the plate column. Total operation times are close, but the time for Step 2 is much lower for the plate column. This exercise shows the great difficulty in predicting the effect of holdup. Composition plots for the no holdup case are shown on the following pages. Exercise 13.26 (continued) Analysis: (continued) No holdup case Exercise 13.26 (continued) Analysis: (continued) No holdup case Exercise 13.26 (continued) Analysis: (continued) No holdup case Exercise 13.27 Subject: Calculation of batch rectification of a four-component mixture by a method that accounts for holdup. Given: Charge of 100 lbmol of 10 mol% propane (C3), 30 mol% n-butane (C4), 10 mol% npentane (C5), and 50 mol% n-hexane (C6). Batch rectification with 50 psia in condenser a column pressure drop of 2 psia. Column with the equivalent of 8 theoretical stages, plus a partial reboiler and a total condenser. Vapor boilup rate, V = 40 lbmol/h. Holdup of condenser-reflux drum = 1 f...
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