Question:
In a constant-rate cake filtration operation, the collected filtrate volumes are 120 m$^{3}$ and 240 m$^{3}$ at 1 min and 2 min, respectively. Assume the cake resistance to be constant and the filter medium resistance to be negligible. If the pressure-drop across the cake is 10 kPa at 1 min, its value at 2 min is \(\underline{\hspace{2cm}}\) kPa (rounded off to the nearest integer).
In a constant-rate cake filtration operation, the collected filtrate volumes are 120 m$^{3}$ and 240 m$^{3}$ at 1 min and 2 min, respectively. Assume the cake resistance to be constant and the filter medium resistance to be negligible. If the pressure-drop across the cake is 10 kPa at 1 min, its value at 2 min is \(\underline{\hspace{2cm}}\) kPa (rounded off to the nearest integer).
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In constant-rate filtration, pressure drop increases linearly with filtrate volume because cake resistance grows proportionally.
Updated On: Jan 13, 2026
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Correct Answer: 20
Solution and Explanation
In constant-rate filtration, the flow rate is constant: \[ \frac{V}{t} = \text{constant} \] Given: \[ V_1 = 120,\; t_1 = 1\ \text{min} \] \[ V_2 = 240,\; t_2 = 2\ \text{min} \] Thus, \[ \frac{V_1}{t_1} = \frac{V_2}{t_2} \] Cake resistance is proportional to cake mass, which is proportional to filtrate volume: \[ R_c \propto V \] At constant rate, pressure drop must increase proportionally to cake resistance: \[ \frac{\Delta P_2}{\Delta P_1} = \frac{R_{c2}}{R_{c1}} = \frac{V_2}{V_1} \] \[ \Delta P_2 = 10\ \text{kPa} \times \frac{240}{120} = 20\ \text{kPa} \]
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