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

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Unformatted text preview: t3. Holdup in the column is 0.10 ft3 for each stage. Rectification campaign: Step Reflux ratio Stop criterion 1 5 98 mol% C3 in accumulator 2 20 95 mol% C4 in instantaneous distillate 3 25 99.8 mol% C4 in accumulator 4 15 80 mol% C5 in instantaneous distillate 5 25 99 mol% C6 in final residue Assumptions: SRK thermodynamics. Total reflux condition is established by time zero. Product compositions are in mol%. Find: (a) Time for rectification. (b) Total amount of distillate produced. A campaign that will get larger amounts of the main products (Cuts 1, 3, and final residue) Analysis: Use the Batch distillation program of Chemcad to make the calculations. The flowsheet for Chemcad is shown below. Note that it includes the batch column (1), a time switch (2), and 5 tanks (3, 4, 5, 6, and 7). The input data for Chemcad are as follows: Batch column: No. of stages = 10 (8 in column + reboiler + condenser) No. of operation steps = 5 Total condenser Condenser pressure =50 psia Column pressure drop = 2 psia (52 psia in the boiler) Holdup units in ft3 Condenser holdup = 1 ft3 Stage holdup = 0.1 ft3 per stage (total of 0.8 ft3) Method of calculation = Inside-out Exercise 13.27 (continued) Analysis: Exercise 13.27 (continued) Analysis: Operation parameters: Step 1 Startup option Total reflux Tank no. 3 Reflux ratio 5 Boilup,lbmol/h 40 Step size, hr 0.02 Record frequency 3 Stop criterion x of C3 in accum. = 0.98 Stop tolerance 0.0001 2 Column content 4 20 40 0.02 3 x of C4 in instant. distillate = 0.95 0.0001 3 Column content 5 25 40 0.05 3 x of C4 in accum. = 0.998 0.0001 4 Column content 6 15 40 0.02 3 x of C5 in instant. distillate = 0.80 0.0001 5 Column content 7 25 40 0.02 3 x of C6 in final residue = 0.99 0.0001 The results are as follows for the contents of tanks 3, 4, 5, 6, and 7 after Steps 1, 2, 3, 4, and 5, respectively, and the final residue: Vessel Amount, lbmol Mole fractions: C3 C4 C5 C6 Operation time, hr Tank 3 5.70 Tank 4 7.18 Tank 5 24.54 Tank 6 3.87 Tank 7 Final residue 9.29 48.65 0.9766 0.0234 0.0000 0.0000 0.62 0.6103 0.3897 0.0000 0.0000 2.88 0.0022 0.9976 0.0002 0.0000 12.3 0.0000 0.6031 0.3945 0.0024 1.34 0.0000 0.0273 0.8399 0.1328 5.65 0.0000 0.0000 0.0100 0.9900 22.79 (a) The total time for rectification = 22.79 hours. (b) The total amount of distillates = 5.70+7.18+24.54+3.87+9.29 = 50.58 lbmol. The above data indicate that the column and condenser holdup at the end of the campaign = 100 - 5.70 - 7.18 - 24.54 - 3.87 - 9.29 - 48.65 = 0.77 lbmol, which is relatively small. The amount of C3 product (Step 1) is only 5.70/10.0 = 57% of the C3 in the feed. The amount of C4 product (Step 3) is 24.54/30 = 81.8% of the C4 in the feed. The amount of C6 (Step 5) in the final residue product = 48.65/50 = 97.3% of C6 in the feed. Thus, there is room for improvement in the campaign, especially in the recovery of C3. The very high purity requirement for C4 probably makes it difficult to significantly increase the recovery of C4. Exercise 13.27 (c...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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