Question

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

Hint:

In ultrafiltration, the rate of filtration is directly proportional to the pressure difference and the permeability constant, and inversely proportional to the length of the membrane.

Answer 1

We can calculate the length of the membrane tube using the formula for ultrafiltration rate: \[ Q = K \cdot A \cdot \Delta P \] Where:
- \( Q \) is the volumetric flow rate (kg/s),
- \( K \) is the permeability constant,
- \( A \) is the membrane area,
- \( \Delta P \) is the pressure difference.
First, we calculate the flow rate \( Q \) using the mass flow rate and the change in total solids: \[ Q = 10 \, {kg/min} = \frac{10}{60} = 0.167 \, {kg/s} \] Now, using the formula for the membrane area \( A \) (where \( A = \pi d L \), with \( d \) as the diameter of the tube and \( L \) as the length): \[ A = \pi \times (0.10)^2 \times L = 0.0314 \times L \] Substitute values into the equation: \[ 0.167 = 5 \times 10^{-5} \times 0.0314 \times L \times 2000 \] Solving for \( L \): \[ L = \frac{0.167}{5 \times 10^{-5} \times 0.0314 \times 2000} = 3.7 \, {m} \] Thus, the length of the membrane tube lies between 3.65 to 3.75 m.