Unformatted text preview: on of a dilute feed of 3phenyl1propanol (A) and 2phenyl ethanol (B) in a
methanolwater mixture in a 4section 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% methanolwater 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 steadystate, localcomposition 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 (15149) to (15151), 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 (15154), QD = QE + QR − QF = 0.400 + 0.302 − 0.160 = 0.542 mL/min Using (15153), 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 (15175), ( 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 masstransfer coefficients on the steadystate 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 masstransfer coefficients (MTC) for A and B are
10 min1. Using Aspen Chromatography, increase the value to 1000 min1. 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 min1
1000 min1
211.6
8.4
861.7 223.2
2.6
858.9 Raffinate
MTC =
10 min1 Raffinate
MTC =
1000 min1 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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 Spring '11
 Levicky
 The Land

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