Question

A 75 um thick polysulphone microporous membrane has an average porosity of E 0.35. Pure water flux through the membrane is 35 m'/m h at a pressure drop of 2 bar at 25°C. The average pore size is estimated to be 0.8 um. Calculate the tortuosity factor of the pores, the resistance to flow offered by the membrane and its water permeability. The viscosity of water at 25°C is 0.9 cP

Answer 1

Step-by-step explanation:

The given data is as follows.

Water flux,
`J_(w)` = 25
`m^(3)/m^(2)h`

=
`(25)/(3600) m^(3)/m^(2)h`

So, let velocity (u) =
`(25)/(3600)` m/s =
`6.9 * 10^(-3)`

`\rho` = 998
`kg/m^(3)`

Pore size, d =
`0.8 * 10^(-6) m`

`\mu` = 0.9 cP =
`9 * 10^(-4)` Pa.s

Hence, calculate the reynold number as follows.

`R_(e) = (\rho * u * d)/(\mu)`

=
`(998 kg/m^(3) * 6.9 * 10^(-3) * 0.8 * 10^(-6) m)/(9 * 10^(-4) Pa.s)`

=
`612.1 * 10^(-5)`

= 0.006

This means that the flow is laminar.

Now, we use Hagen-Poiseuille equation as follows.

`J_(w) = (\varepsilon * d^(2))/(32 * \mu * \tau) * (\Delta P)/(L_(m))`

where,
`\varepsilon` = membrane porosity = 0.35

d =
`0.8 * 10^(-6)` m

`\Delta P` =
`2 * 10^(5) Pa`

`\mu` =
`9 * 10^(-4)`

`\tau` = tortuosity

`L_(m)` = membrane thickness =
`75 * 10^(-6) m`

`(25)/(3600) = (0.35 * (0.8 * 10^(-6)))/(32 * (9 * 10^(-4)) * \tau)`

`\tau` = 3.73

Hence, the tortuosity factor of the pores is 3.73.

As flow resistance =
`R_(m)`

`J_(w) = (\Delta P)/(r * R_(m))`

`R_(m) = 3.2 * 10^(10) m^(-1)`

Water permeability is represented by
`L_(p)`.

`J_(w) = L_(p) * \Delta P`

`6.9 * 10^(-3)` =
`L_(p) * 2 * 10^(5) Pa`

`L_(p) = 3.45 * 10^(-8) m^(3)/m^(2)s Pa`

Therefore, the resistance to flow is
`3.2 * 10^(10) m^(-1)` and its water permeability is
`3.45 * 10^(-8) m^(3)/m^(2)s Pa`.