Unformatted text preview: .3 mol / h
Therefore, the material balance in kmol/h is:
Component
Feed Sweep Retentate Permeate
HCl
60
0
6.0
54.0
NaCl
30
0
6.7
23.3
H2 O
16,650 16,650
16,650.0
16,650.0
Total
16,740 16,650
16,662.7
16,727.3
Case 2 (95% transfer of HCl:
Transfer of HCl to the permeate = 0.95(60) = 57 mol/h
Neglect any transfer of water. Then the concentrations of HCl are:
cHCl P = 57 / 300,000 = 0.00019 mol / cm3
= 3 / 300,000 = 0.00001 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  0.00019 = 0.00001 mol/cm3 nHCl
(57 / 60)
=
= 1, 730, 000 cm 2 or 173.0 m 2
K HCl ( ∆cHCl )LM 0.055(0.00001)
Now determine the rate of permeation of NaCl.
From Eq. (14.57,
nNaCl = KNaCl AM ∆cNaCl LM = 0.021(1,730,000) ∆cNaCl LM = 36,300 ∆cNaCl LM
(1) From Eq. (1457), AM = But, nNaCl = cNaCl ∆cNaCl LM P = cNaCl 300,000
= 5,000 cNaCl P
60
− cNaCl P = 0.0001  cNaCl
F (2)
P (3) Exercise 14.14 (continued) Analysis: Case 2 (continued)
Substituting Eqs. (2) and (3) into (1),
5,000 cNaCl P = 36,300 0.0001 − cNaCl P Solving, cNaCl P = 0.0000879 and the transfer of NaCl = 30(0.0000879 / 0.0001) = 26.4 mol / h
Therefore, the material balance in kmol/h is:
Component
Feed Sweep Retentate Permeate
HCl
60
0
3.0
57.0
NaCl
30
0
3.6
26.4
H2 O
16,650 16,650
16,650.0
16,650.0
Total
16,740 16,650
16,656.6
16,733.4
Case 3 (98% transfer of HCl:
Transfer of HCl to the permeate = 0.98(60) = 58.8 mol/h
Neglect any transfer of water. Then the concentrations of HCl are:
cHCl P = 58.8 / 300,000 = 0.000196 mol / cm3
= 12 / 300,000 = 0.000004 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  0.000196 = 0.000004 mol/cm3 nHCl
(58.8 / 60)
=
= 4, 450, 000 cm 2 or 445.0 m 2
K HCl ( ∆cHCl )LM 0.055(0.000004)
Now determine the rate of permeation of NaCl. From Eq. (14.57),
nNaCl = KNaCl AM ∆cNaCl LM = 0.021(4,450,000) ∆cNaCl LM = 93,500 ∆cNaCl LM
(1) From Eq. (1457), AM = 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 = 93,500 0.0001 − cNaCl (2)
P (3) P Solving, cNaCl P = 0.0000949 and the transfer of NaCl = 30(0.0000949 / 0.0001) = 28.5 mol / h
Therefore, the material balance in kmol/h is:
Component
Feed Sweep Retentate Permeate
HCl
60
0
1.2
58.8
NaCl
30
0
1.5
28.5
H2 O
16,650 16,650
16,650.0
16,650.0
Total
16,740 16,650
16,652.7
16,737.3 Exercise 14.15
Subject: Desalinization of an aqueous solution by electrodialysis.
Given: 86,000 gal/day of an aqueous solution of 3,000...
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