Considerthe picture.

It can be shown that the filtration time is equal to:

$t=\frac{V_o{C}_w}{kA{C}_f}[f(-ln(\frac{C_w}{C_o}))-f(-ln(\frac{C_w}{C_f}))]$

Where,

$V_o$ : Volume of solution in the holding tank

$k$ : Mass transfer coefficient

A: Area of the membrane

$C_W$ : Concentration of solute at the surface of the membrane

$C_f$ : Concentration of solute in the feed

$C_0$ : Initial concentration of solute in the holding tank

$f(x)=-{\displaystyle \underset{-x}{\overset{}{}}}\frac{e^{-t}}{t}dt=0\mathrm{.}577216+ln|x|+{\displaystyle \underset{n=1}{\overset{}{}}}\frac{x^n}{x!x}$

$($a$)200L$ of an antibody solution $(10\frac{g}{L})$ needs to be ultrafiltered in a fed batch process in which $50L$ of the solution is placed in the holding tank and the remainder of the feed is fed at a flowrate equal to the permeate one. $C_W$ was determined in other experiment and found equal to

$250\frac{g}{L}.$ The mass transfer coefficient, $k,$ is equal to $3{10}^{-6}\frac{m}{s}$ and the membrane area is $1\mathrm{.}5m^2.$ Find the time needed to produce $140L$ of permeate.

$($b$)$ Compare the time obtained in $($a$)$ with the time needed to reduce $200L$ of the same solution down to $60L$ by a conventional batch process.