Constant-pressure cake filtration: For filtration operated at a constant pressure drop, how does the instantaneous filtrate flow rate vary with time (t)?

Introduction / Context:
In incompressible-cake filtration at constant pressure, resistance increases as cake thickness grows. Understanding how filtrate flow rate declines with time is fundamental for cycle time estimates and area sizing in process filtration.

Given Data / Assumptions:

• Constant pressure drop across the filter and cake.
• Incompressible cake; viscosity and temperature constant.
• Filter medium resistance is small or absorbed into constants.

Concept / Approach:
For constant-pressure cake filtration, the classical equation gives t = α * V^2 + β * V (α relates to specific cake resistance and solids concentration, β to medium resistance). Differentiating with respect to time gives the instantaneous flow rate dV/dt, which is inversely proportional to a linear function of V. Since V ∝ √t (for large t when cake resistance dominates), dV/dt ∝ 1/√t.

Step-by-Step Solution:
Start with t = a V^2 + b V.Differentiate: dt/dV = 2 a V + b ⇒ dV/dt = 1 / (2 a V + b).For cake-dominated regime (b negligible): dV/dt ≈ 1 / (2 a V).But V ≈ k √t ⇒ dV/dt ∝ 1 / √t.

Verification / Alternative check:
Plotting V versus √t typically yields a straight line under constant-pressure filtration; slope relates to cake parameters. This confirms the √t dependence and the 1/√t decay in rate.

Why Other Options Are Wrong:

• 1/t: too steep a decay and not supported by the filtration equation.
• Constant with time or increasing: contradicts the growing resistance of the cake.
• Zero after a few seconds: unrealistic; flow declines gradually, not abruptly.

Common Pitfalls:
Ignoring medium resistance at early times (b term) or cake compressibility, which can further modify the time dependence. Always check regime validity.

Final Answer:
Varies inversely with the square root of time, i.e., dV/dt ∝ 1/√t