Unformatted text preview: From a modification of Eq. (1455), AM = 90.4 − 65.0 − 22.9 − 0.0
= 24.2 kg / m3
90.4 − 65.0
ln
22.9 − 0.0 mNa 2SO 4
KNa 2SO 4 ∆cNa 2SO 4 = 56.5(0.4536)
= 692 m2
−3
153 × 10 24.2
. Because the membrane area for Na2SO4 is larger than for A, the % transfer of Na2SO4 is less than
the assumed 75%. Repeat the calculations assuming 70% transfer of Na2SO4, but use the same
liquid densities. Now the permeate contains 0.7(75.3) = 52.7 lb/h Na2SO4 and the retentate
contains 22.6 lb/h Na2SO4. The new concentrations are:
Concentrations in kg/m3:
Component
Feed
Sweep Retentate Permeate
Na2SO4
90.4
0.0
28.1
65.0
A
67.8
0.0
63.1
7.0
3
3
∆c of Na2SO4 = 26.8 kg/m and membrane area = 52.7(0.4536)/[1.53x10 (26.8)] = 583 m2
∆c of Na2SO4 = 62.0 kg/m3 and membrane area = 5.7(0.4536)/[7.63x105 (62.0)] = 547 m2
By extrapolation, membrane area = 545 m2 and the % transfer of Na2SO4 = 65%. Exercise 14.14
Subject: Removal of HCl from an aqueous solution by dialysis.
Given: 300 L/h of an aqueous solution of 0.1 M NaCl and 0.1 M HCl. Sweep of 300 L/h of
H2O. Microporous membrane for which lab experiments give the following overall masstransfer coefficients:
Component K, cm/min
H2 O
0.0025
NaCl
0.021
HCl
0.055
Assumptions: Log mean concentration driving force. Negligible change in volume flow rate of
feed and sweep with permeation.
Find: Membrane area in m2 and material balance for 90, 95, and 98% transfer of HCl.
Analysis: Determine feed conditions, noting that it is very dilute in NaCl and HCl. Therefore,
assume that it contains 55.5 mol of H2O per liter. For 300 L/h,
nNaCl F = 0.1(300) = 30 mol / h
nHCl F nH 2 O F = 0.2(300) = 60 mol / h
= 55.5(300) = 16,700 mol / h The feed concentrations are:
cNaCl F = 0.1 / 1000 = 0.0001 mol / cm3 cHCl F cH 2 O F = 0.2 / 1000 = 0.0002 mol / cm3
= 55.5 / 1000 = 0.0555 mol / cm3 The sweep flow rate = 55.5(300) = 16,700 mol/h of pure water.
Case 1 (90% transfer of HCl:
Transfer of HCl to the permeate = 0.9(60) = 54 mol/h
Neglect any transfer of water. Then the concentrations of HCl are:
cHCl P = 54 / 300,000 = 0.00018 mol / cm3
= 6 / 300,000 = 0.00002 mol / cm3 cHCl R cHCl Sweep = 0.0 mol / cm3 Note that because the volumetric flow rates of feed and sweep are equal to each other and to the
permeate and retentate, by material balance, the concentration driving forces for HCl and NaCl
are equal at both ends and equal to ci F − ci P
Therefore, ∆cHCl LM = cHCl F − cHCl P = 0.0002  0.00018 = 0.00002 mol/cm3 Exercise 14.14 (continued) Analysis: Case 1 (continued)
nHCl
(54 / 60)
=
From Eq. (1457), AM =
= 818, 000 cm 2 or 81.8 m 2
K HCl ( ∆cHCl )LM 0.055(0.00002)
Now determine the rate of permeation of NaCl.
From Eq. (14.57,
nNaCl = KNaCl AM ∆cNaCl LM = 0.021(818,000) ∆cNaCl LM = 17,200 ∆cNaCl LM
(1)
300,000
= 5,000 cNaCl P
P
60
= cNaCl F − cNaCl P = 0.0001  cNaCl
LM But, nNaCl = cNaCl ∆cNaCl Substituting Eqs. (2) and (3) into (1),
5,000 cNaCl P = 17,200 0.0001 − cNaCl (2)
P (3) P Solving, cNaCl P = 0.0000775 and the transfer of NaCl = 30(0.0000775 / 0.0001) = 23...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details