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

Explanation:

The given data is as follows.

           Water flux, $J_{w}$ = 25 $m^{3}/m^{2}h$

                                       = $\frac{25}{3600} m^{3}/m^{2}h$

So, let velocity (u) = $\frac{25}{3600}$ m/s = $6.9 \times 10^{-3}$

             $\rho$ = 998 $kg/m^{3}$

             Pore size, d = $0.8 \times 10^{-6} m$

             $\mu$ = 0.9 cP = $9 \times 10^{-4}$ Pa.s

Hence, calculate the reynold number as follows.

                 $R_{e} = \frac{\rho \times u \times d}{\mu}$            

                        = $\frac{998 kg/m^{3} \times 6.9 \times 10^{-3} \times 0.8 \times 10^{-6} m}{9 \times 10^{-4} Pa.s}$    

                        = $612.1 \times 10^{-5}$

                        = 0.006

This means that the flow is laminar.

Now, we use Hagen-Poiseuille equation as follows.

             $J_{w} = \frac{\varepsilon \times d^{2}}{32 \times \mu \times \tau} \times \frac{\Delta P}{L_{m}}$

where,     $\varepsilon$ = membrane porosity = 0.35

                              d = $0.8 \times 10^{-6}$ m

                       $\Delta P$ = $2 \times 10^{5} Pa$

                      $\mu$ = $9 \times 10^{-4}$

                      $\tau$ = tortuosity

                      $L_{m}$ = membrane thickness = $75 \times 10^{-6} m$

                    $\frac{25}{3600} = \frac{0.35 \times (0.8 \times 10^{-6})}{32 \times (9 \times 10^{-4}) \times \tau}$

                            $\tau$ = 3.73

Hence, the tortuosity factor of the pores is 3.73.

As flow resistance = $R_{m}$

               $J_{w} = \frac{\Delta P}{r \times R_{m}}$

               $R_{m} = 3.2 \times 10^{10} m^{-1}$

Water permeability is represented by $L_{p}$.

                    $J_{w} = L_{p} \times \Delta P$  

             $6.9 \times 10^{-3}$ = $L_{p} \times 2 \times 10^{5} Pa$  

              $L_{p} = 3.45 \times 10^{-8} m^{3}/m^{2}s Pa$

Therefore, the resistance to flow is $3.2 \times 10^{10} m^{-1}$ and its water permeability is $3.45 \times 10^{-8} m^{3}/m^{2}s Pa$.