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Chapter 18 Membranes page 18-1
Chapter 18 Membranes
1. Permeability and diffusion coefficient
P = DH/RT = 63 barrer = 6310-10 cm3(STP)-cm/cm2/cmHg/s
2. N2 permeation from a hollow fiber module
Following Example 17.2-2,
a = 4(1 - )/d = 4*0.59/0.034 = 69.41 cm-1
k
1 = 0.0076 cm/s,
3. Diffusion in a glove bag
a)
0.21
Chapter 18 Membranes page 18-2
dt
c
4. Ammonia recovery
Following Example 17.2-2,
5. Acetic acid diffusion through membrane
(a) j1 = (D/l)(c10 – c1l) = (DH/l)(C10 – C1l) where H = c/C
(b) Ka = cH2/CHA => cH = KaCHA
6. Properties of an ultrafiltration membrane
From eq. 17.3-15,
Lp = jv
p = 0.62 m/day
3.4 atm = 0.182 m/day-atm = 0.000211 cm/s-atm
In the diffusion experiment, from eq. 5.5-1,
10
7. Polyethylene glycol concentration
From eq. 17.3-15,
-12
8.
First we find the osmotic pressure at the beginning and end of the run:
010
1
1
8 3145 293 0 16 390
1
8 3145 293 0 62 1460
ff
JmolLkPa
R
Tc . K . kPa
Kmol L J
JmolLkPa
R
Tc . K . kPa
Kmol L J
The average osmotic pressure is therefore:
2
avg
For reverse osmosis, we know that:
where jv is a volumetric flux. Since the membrane rejects nearly all the salt, σ is nearly one.
We can find the area from:
3
2
7
026
10 136
1 3 10 2400 926
vpavg
.m
QQ min
m
jLp .k
kPa min
Pa
9. Facilitated diffusion
Because no reaction occurs in the membrane, the total flux is
j1 = D
l(c - c + c - c)
10 1l1s0 1sl
dz2 = 0, d2c1s
dz2 = 0 => c + c = const =
-
10. Cu separation using a liquid membrane
(a) The form of Cu in the membrane is predominantly R2Cu (2), thus
The average carrier concentration is c
-R = 0
l(c1 + 2c2)dz/l
jCu = (D/l)c
(b) Measure jCu vs c
-R, and check the linearity.
(c) Could be attribute to the formation of carrier clusters as inverted micells.
11. Nitric oxide pump
at z = l, FeNO2+
Fe + NO
Chapter 18 Membranes page 18-5
Fe2+ Fe3+ + e-
In the liquid film, let r = k[Fe2+][NO]
0 = Dd2[NO]
dz2 - r (4)
B. C.
[FeNO2+] = K[Fe2+][NO] (8)
The total flux of NO is obtained by integrating twice eq. (2) + (4),
jNO = D
l([NO] - [NO] + [FeNO ]) (9)
0l2+ 0
jNO = D
l Kc
1 + K[NO]0 (11)
The electrode current
i = DF
3d[Fe3+]
= -DFc
l (12)
