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Membrane Separation: Basics and Applications 295
R ¼ 1
DAmKARTcBP
cBmDBmvBðDp DPÞ
1
: (29)
Example 7.1
The following numerical values were given for reverse osmosis of aqueous sodium chloride solution at
25 C(47):
cBmDBm ¼ 2:7 10 8 kg=ms;
DAmKA ¼ 4:2 10 14 m 2 =s:
Calculate the solute separation and water flux when the feed sodium chloride molality is 0.1 mol/kg
and the operating pressure is 4.134 10 6 Pa (gauge). The thickness of the selective layer is 10 7 m.
Use the following numerical values:
From Eq. (10),
cBP ¼ 10 3 kg=m 3
RT ¼ 2:479 10 3 J=mol for 25 C;
and vB ¼ 18:02 10 6 m 3 =mol:
P ¼ 7;720 ð25 þ 273Þ 0:2 ¼ 0:460 10 6 Pa:
As P = 0 for pure water, firstly, we can assume pure water on the permeate side. Approximating
DP ¼ð0 0:460 10 6 Þ¼ 0:460 10 6 Pa, from Eq. (29)
R ¼ 1
ð4:2 10 14Þð2:479 103Þð103 ð2:7 10
Þ
8Þð18:02 10 6Þ½ð 4:134 106Þ ð 0:460 106ÞŠ Then, the sodium chloride molality in the permeate will be
0:1 ð1 0:945Þ ¼0:0055 mol=kg:
From Eq. (10), the osmotic pressure of the permeate will be
P ¼ 7;720 ð25 þ 273Þ 2 0:0055 ¼ 0:0253 10 6 Pa:
1
¼ 0:945:
Then, the osmotic pressure difference, DP, between the permeate and feed will be
DP ¼ð0:0253 0:460Þ 10 6 ¼ 0:4347 10 6 Pa;
ð4:2 10
R ¼ 1
14Þð2:479 103Þð103Þ ð2:7 10 8Þð18:02 10 6Þ½ð 4:134 106Þ ð 0:4347 106ÞŠ Hence, R ¼ 0.945 is an accurate answer.
The flux (kg/m 2 s) is calculated from Eq. (18) as
JB ¼ ð2:7 10 8 Þð18:02 10 6 Þ½ð 4:314 10 6 Þ ð 0:4347 10 6 ÞŠ
ð2:479 10 3 Þð10 7 Þ
¼ 76:14 10 4 kg=m 2 s:
¼ 0:945: