Question

A fruit juice is concentrated using an ultrafiltration membrane. A feed stream at \(10~\mathrm{kg/min}\) with \(6%\) total solids (by weight) is increased to \(20%\) total solids (by weight). The membrane tube has \(10~\mathrm{cm}\) inside diameter and the pressure difference across the membrane is \(2000~\mathrm{kPa}\). If the permeability constant of the membrane is \(5 \times 10^{-6}\ \mathrm{kg\ water/(m^2\ kPa\ s)\), the length of membrane tube (in m) is \underline{\hspace{2cm}}. (rounded off to 2 decimal places)}

Hint:

In tubular membranes, \(A=\pi D L\). Get permeate flow from mass balance, flux from \(J=K\Delta P\), then \(L=A/(\pi D)\).

Answer 1

Correct Answer: 3.65

Step 1: Overall mass balance and solids balance.
Feed \(F=10~\mathrm{kg/min}\) at \(6%\) solids \(\Rightarrow\) solids \(=0.06F=0.6~\mathrm{kg/min}\).
Retentate concentration \(=20%\) \(\Rightarrow 0.20R = 0.6 \Rightarrow R=3~\mathrm{kg/min}\).
Permeate (assumed water only) \(P = F-R = 7~\mathrm{kg/min} = \dfrac{7}{60}=0.1167~\mathrm{kg/s}\).


Step 2: Permeate flux from permeability.
Flux \(J = (\text{permeability})\times \Delta P\) \[ J = \left(5\times 10^{-6}\ \frac{\mathrm{kg}}{\mathrm{m^2\,kPa\,s}}\right)(2000~\mathrm{kPa}) = 0.01~\frac{\mathrm{kg}}{\mathrm{m^2\,s}}. \]

Step 3: Required membrane area.
\[ A = \frac{\dot m_{\text{perm}}}{J} = \frac{0.1167}{0.01}=11.6667~\mathrm{m^2}. \]

Step 4: Convert area to tube length.
For a tubular membrane of inside diameter \(D=0.10~\mathrm{m}\), area \(A=\pi D L\). Thus \[ L=\frac{A}{\pi D}=\frac{11.6667}{\pi \times 0.10}=37.134~\mathrm{m}. \] Rounded to two decimals: \(\boxed{37.13~\mathrm{m}}\).