Question

A gas purification system needs to produce 2 kg/hour of purified product. If the purification membrane is 100 m2 and 0.05mm thick, and the supply gas is at 0.75 kg/m3 and the take-off side maintains a pressure of 0.05 kg/m3 find required membrane Diffusivity in sq m/sec.

Answer 1

The membrane diffusivity would be 3.968 x
`10^(-10)`
`m^(2)`/s

Step-by-step explanation:

According to Fick's law of diffusion, the rate of the purified product is related with to membrane diffusivity with the expression in equation 1.

`V = ((P_(1)-P_(2))AD )/(T)` ..............................1

Where

V is the rate of purified product = 2 kg/hr = 2 kg/hr x 1 hr / 3600 sec = 1/1800 kg/sec

`P_(1)` is the pressure at the supply gas point = 0.75 kg/
`m^(2)`

`P_(2)` is the pressure at the take-off side = 0.05 kg/
`m^(2)`

A is the area of the membrane = 100
`m^(2)`

T is the thickness of the membrane = 0.05 mm = 0.05/ 1000 = 5 x
`10^(-5)` m

Substituting the values into equation 1 we have;

`(1)/(1800)= ((0.075 - 0.05)*100*D)/(5*10^(-5) )`

5 x
`10^(-5)` = 126000 x D

D = 5 x
`10^(-5)` / 126000

D = 3.968 x
`10^(-10)`
`m^(2)`/s

Therefore the membrane diffusivity would be 3.968 x
`10^(-10)`
`m^(2)`/s