Membranes - Principles of Chemical Separations with Environmental Applications - page 253

Determine the membrane area required.
Solution:
The total water flux is given by:

(m 2 N w = 0 . 2L / · day · kPa)(2500 kPa − 300 kPa)

m 2 at 25 ◦ C

= 440 L / day · .

The membrane area at 10 ◦ C is given by:

m 2 ) 2690 m 2 .

A = (750,000 L / day)[(day · / (440 L)](1.58) =

Example 9.4

Problem:
Experiments at 25 ◦ C were performed to determine the water permeability and
salt (NaCl) rejection of a cellulose acetate membrane. The membrane area is A

=

10 − 3 m 2 . The inlet salt concentration is C s1 m 3 2.00 × = 10.0 kg NaCl / solution

(10.0 g NaCl L). The water recovery is low so that the salt concentration in the

entering and exit feed solutions are assumed to be equal. The product solution contains
C s2

/

m 3 10 − 9 m 3 s. A

pressure differential of 55 atm is used. Calculate (a) the solute rejection R , and (b) the
permeability coefficient of the membrane.
Solution:
The total salt flux across the membrane ( N s ) is:

= 0.4 kg NaCl / solution and the flowrate is 2 × solution /

2 0.4 kg salt

m 3

10 − 8 m 3

soln
s

×

kg salt
m 2

soln

10 − 6

N s = = 4 ×

s . 2 × 10 − 3 m 2 ·

(a) The rejection R is:

C s1 − C s2 10 g / L − 0 . 4g / L

R = = = 0 . 96 .

c s1 10 g / L

Thus, we can assume that the permeate flowrate is predominantly water.
The total water flux ( N w ) is:

2 1

10 3 kg H 2 O
m 3

10 − 8 m 3 × ×

s kg H 2 O

m 2

10 − 2

N w = = 1 ×

s . 2 × 10 − 3 m 2 ·

(b) To calculate the permeability coefficient, we first need to determine the osmotic
pressure on each side of the membrane.