Question

An experimental RO cellulose acetate membrane with an active layer 6 microns thick achieveda flux of 0.33x10^-4 g/cm2 sec. for a 40atmosphere pressure difference with 0.1N NaCl in water (pi=4.9atm) what was the water permeability and list the units if:

1.M=1

2.M=3

also this membrane had R=0.96. what was the salt permeability for

1. M=1

2. M=3

D7 An experimental RO cellulose-acetate membrane with an active layer 6 microns thick achieved a flux of 0.33 x 10 g/cm2 sec for a 40 atmo- sphere pressure difference with 0.1 N NaCl in water (-4.9 atm). What was the water permeability and list the units if: al) M Cw/CB 1 a2) M cw/CB 3 b) This membrane had Ro 0.96. What was the salt permeability for b2) M cw/CB = 3 bl) M Cw/CB1

Answer 1

Hi In this case of the reverse osmosis, letus draw the diagram first to understand the process C2 Cb Here Cbis the bulk concentration Cw is the wall concentration C2 is the concentration on the other side)

J=Pm(Pd-Opd)

Here

Pm is the permeability.

Pd is the pressure difference

Opd is the osmotic pressure difference

Pm is the permeability

J is the flux

Please note that osmotic pressure difference is proportional to the concentration

Osmotic pressure difference =3.14*8.314*(Cw-C2)

Part a

When m =1

Cw= cb = 0.1 N

Opd = 4,9

Pd = 40 atm

Assuming 100 % rejection

0.33x10^-4 g/cm2= pm*(40-4.9)

Pm= 9.4*10 ^-7 g/cm2.atm

When m = 3

Cw= 3*0.1

Opd = 3*4.9 = 14.7 atm

Assuming 100 % rejection

0.33x10^-4 g/cm2= pm*(40-14.7)

1.3*10 ^-6 g/cm2.atm

Part b

When m =1

Cw= cb = 0.1 N

Opd = 4.9 for 0.1 N

Pd = 40 atm

Given rejection 96 %.

So C2= 0.04*0.1 =0.004

OPd= (4.9/0.1)*(0.1-0.004)= 4.7

0.33x10^-4 g/cm2= pm*(40-4.7)

Pm= 9.34*10 ^-7 g/cm2.atm

When m = 3

Cw= 3*0.1

Given rejection 96 %.

So C2= 0.04*0.3 =0.012

OPd= (4.9/0.1)*(0.3-0.012)= 14.1

Assuming 100 % rejection

0.33x10^-4 g/cm2= pm*(40-14.1)

1.27*10 ^-6 g/cm2.atm

Hope this helps

Feel free to comment in the case of any doubt.

Thanks