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

3716942 46599 kgd flow rate of na2s produced

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Unformatted text preview: on of a dilute feed of 3-phenyl-1-propanol (A) and 2-phenyl ethanol (B) in a methanol-water mixture in a 4-section laboratory simulated moving bed. Given: A liquid feed rate of 0.16 mL/min containing 0.091 g/L of A and 0.115 g/L of B, with the balance being a 60/40 wt% methanol-water mixture. Henry’s law applies, in the form, qi = Kici , with KA = 2.36 and KB = 1.40 for the given adsorbent, q and c are concentrations per unit volume. Assume that neither methanol nor water adsorb. External void fraction of the adsorbent beds = εb = 0.572. Switching time = 10 minutes. Find: Using the steady-state, local-composition TMB model for a perfect separation of A from B with a margin, β, of 1.15, estimate initial values for the volumetric flow rates of the extract (E), raffinate (R), desorbent (D), circulation (C), and the solid particles (S). Use those values to determine the recirculation rate for the SMB and the resulting volumetric liquid flow rates in each of the four sections. Analysis: Carry out the calculations using volumetric flow rates as shown in Figure 15.45. Because, KB < KA , B will appear in the raffinate and A will appear in the extract. First make calculations for a TMB. Using (15-149) to (15-151), with QF = 0.16 mL/min, QS = QF KA − K Bβ β = 0.16 2.36 − 1.40 (1.15 ) 1.15 = 0.362 mL/min QE = QS ( K A − K B ) β = 0.362 ( 2.36 − 1.40 )1.15 = 0.400 mL/min QR = QS ( K A − K B ) 0.362 ( 2.36 − 1.40 ) = = 0.302 mL/min β 1.15 Using (15-154), QD = QE + QR − QF = 0.400 + 0.302 − 0.160 = 0.542 mL/min Using (15-153), QC = QS K Aβ − QD = 0.362 ( 2.36 ) (1.15) − 0.542 = 0.442 mL/min Now, make calculations for the SMB, where the solid particles do not flow down through the unit, but are in stationary beds. Referrring to Figure 15.45, the volumetric flow rate in section I of the TMB is, ( QI )TMB = QC + QD = 0.442 + 0.542 = 0.984 mL/min Exercise 15.38 (continued) Analysis: (continued) From (15-175), ( QI )SMB = ( QI )TMB + Then, ( QII )SMB = ( QI )SMB εb 1 − εb ( QS )TMB = 0.984 + 0.572 0.362 = 1.468 mL/min 1 − 0.572 - QE = 1.468 – 0.400 = 1.068 mL/min ( QIII )SMB = ( QII )SMB + QF = 1.068 + 0.160 = 1.228 mL/min ( QIV )SMB = ( QIII )SMB - QR = 1.228 – 0.302 = 0.926 mL/min Exercise 15.39 Subject: Effect of mass-transfer coefficients on the steady-state TMB results of Example 15.17 for the separation of fructose from glucose. Given: Data in Examples 15.16 and 15.17. Find: The compositions of the extract and raffinate. Analysis: In Example 15.17, the component mass-transfer coefficients (MTC) for A and B are 10 min-1. Using Aspen Chromatography, increase the value to 1000 min-1. The following results are obtained, compared to those from Example 15.17. Component Fructose Glucose Water Concentrations in g/L Extract Extract MTC = MTC = 10 min-1 1000 min-1 211.6 8.4 861.7 223.2 2.6 858.9 Raffinate MTC = 10 min-1 Raffinate MTC = 1000 min-1 12.7 295.3 795.8 3.9 303.3 796.1 Some improvement in the separation is noted. Additional improvement may be possible by altering the switching time, the fluid circulation flow rate, and the extract and raffinate flow rates. Exercise 16.1 Subject: Mass balance check on test data...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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