Answer:
The membrane diffusivity would be 3.968 x [tex]10^{-10}[/tex] [tex]m^{2}[/tex]/s
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.
[tex]V = \frac{(P_{1}-P_{2})AD }{T}[/tex] ..............................1
Where
V is the rate of purified product = 2 kg/hr = 2 kg/hr x 1 hr / 3600 sec = 1/1800 kg/sec
[tex]P_{1}[/tex] is the pressure at the supply gas point = 0.75 kg/[tex]m^{2}[/tex]
[tex]P_{2}[/tex] is the pressure at the take-off side = 0.05 kg/[tex]m^{2}[/tex]
A is the area of the membrane = 100 [tex]m^{2}[/tex]
T is the thickness of the membrane = 0.05 mm = 0.05/ 1000 = 5 x [tex]10^{-5}[/tex] m
Substituting the values into equation 1 we have;
[tex]\frac{1}{1800}= \frac{(0.075 - 0.05)*100*D}{5*10^{-5} }[/tex]
5 x [tex]10^{-5}[/tex] = 126000 x D
D = 5 x [tex]10^{-5}[/tex] / 126000
D = 3.968 x [tex]10^{-10}[/tex] [tex]m^{2}[/tex]/s
Therefore the membrane diffusivity would be 3.968 x [tex]10^{-10}[/tex] [tex]m^{2}[/tex]/s
