Answer :
Explanation:
The given data is as follows.
Water flux, [tex]J_{w}[/tex] = 25 [tex]m^{3}/m^{2}h[/tex]
= [tex]\frac{25}{3600} m^{3}/m^{2}h[/tex]
So, let velocity (u) = [tex]\frac{25}{3600}[/tex] m/s = [tex]6.9 \times 10^{-3}[/tex]
[tex]\rho[/tex] = 998 [tex]kg/m^{3}[/tex]
Pore size, d = [tex]0.8 \times 10^{-6} m[/tex]
[tex]\mu[/tex] = 0.9 cP = [tex]9 \times 10^{-4}[/tex] Pa.s
Hence, calculate the reynold number as follows.
[tex]R_{e} = \frac{\rho \times u \times d}{\mu}[/tex]
= [tex]\frac{998 kg/m^{3} \times 6.9 \times 10^{-3} \times 0.8 \times 10^{-6} m}{9 \times 10^{-4} Pa.s}[/tex]
= [tex]612.1 \times 10^{-5}[/tex]
= 0.006
This means that the flow is laminar.
Now, we use Hagen-Poiseuille equation as follows.
[tex]J_{w} = \frac{\varepsilon \times d^{2}}{32 \times \mu \times \tau} \times \frac{\Delta P}{L_{m}}[/tex]
where, [tex]\varepsilon[/tex] = membrane porosity = 0.35
d = [tex]0.8 \times 10^{-6}[/tex] m
[tex]\Delta P[/tex] = [tex]2 \times 10^{5} Pa[/tex]
[tex]\mu[/tex] = [tex]9 \times 10^{-4}[/tex]
[tex]\tau[/tex] = tortuosity
[tex]L_{m}[/tex] = membrane thickness = [tex]75 \times 10^{-6} m[/tex]
[tex]\frac{25}{3600} = \frac{0.35 \times (0.8 \times 10^{-6})}{32 \times (9 \times 10^{-4}) \times \tau}[/tex]
[tex]\tau[/tex] = 3.73
Hence, the tortuosity factor of the pores is 3.73.
As flow resistance = [tex]R_{m}[/tex]
[tex]J_{w} = \frac{\Delta P}{r \times R_{m}}[/tex]
[tex]R_{m} = 3.2 \times 10^{10} m^{-1}[/tex]
Water permeability is represented by [tex]L_{p}[/tex].
[tex]J_{w} = L_{p} \times \Delta P[/tex]
[tex]6.9 \times 10^{-3}[/tex] = [tex]L_{p} \times 2 \times 10^{5} Pa[/tex]
[tex]L_{p} = 3.45 \times 10^{-8} m^{3}/m^{2}s Pa[/tex]
Therefore, the resistance to flow is [tex]3.2 \times 10^{10} m^{-1}[/tex] and its water permeability is [tex]3.45 \times 10^{-8} m^{3}/m^{2}s Pa[/tex].