Question

In a constant pressure cake filtration with an incompressible cake layer, the volume of filtrate (V) is measured as a function of time, t. The plot of \( \frac{t}{V} \) versus \( V \) gives a straight line with slope \( 10^4 \, \text{sm}^{-3} \). Filter area is \( 0.005 \, \text{m}^2 \), viscosity of filtrate is \( 10^{-3} \, \text{Pa.s} \), and the overall pressure drop across filter is \( 200 \, \text{kPa} \). What is the value of filter medium resistance in \( \text{m}^{-1} \)?

• A. \( 1 \times 10^9 \)
• B. \( 1 \times 10^{10} \)
• C. \( 1 \times 10^{11} \)
• D. \( 1 \times 10^{12} \)

Hint:

For filtration problems, apply Darcy's law and remember to account for all units correctly when calculating resistance.

Answer 1

The Correct Option is A

Step 1: Darcy's law for filtration.
The equation for constant pressure filtration is given by Darcy's law: \[ \frac{t}{V} = \frac{\mu A}{\Delta P} R \] Where: - \( \mu \) is the viscosity of the filtrate (\(10^{-3} \, \text{Pa.s}\)), - \( A \) is the filter area (\(0.005 \, \text{m}^2\)), - \( \Delta P \) is the pressure drop (\(200 \, \text{kPa} = 200 \times 10^3 \, \text{Pa}\)), - \( R \) is the filter medium resistance (in \( \text{m}^{-1} \)).

Step 2: Rearranging the equation.
We can solve for \( R \): \[ R = \frac{1}{A} \times \frac{\Delta P}{\mu} \times \frac{t}{V} \] Substitute the known values: \[ R = \frac{1}{0.005} \times \frac{200 \times 10^3}{10^{-3}} \times 10^4 \]

Step 3: Calculation.
Solving the above equation: \[ R = 200 \times 10^9 \, \text{m}^{-1} \] Thus, the filter medium resistance is \( 1 \times 10^9 \, \text{m}^{-1} \).

Final Answer: \[ \boxed{1 \times 10^9} \]