Filtration fundamentals: for constant-pressure cake filtration, the instantaneous rate of filtrate delivery (dV/dt) varies inversely with which factor most directly?

Introduction / Context:
In cake filtration, engineers estimate cycle times and equipment sizes using relationships that tie filtrate flow rate to driving force, fluid properties, and resistances. Recognizing which variables increase or decrease the rate is essential for debottlenecking and scale-up.

Given Data / Assumptions:

• Constant-pressure filtration of a compressible or incompressible cake.
• Darcy-type flow through porous media applies.
• Filtrate viscosity and cake/medium resistances are the dominant resistive terms.

Concept / Approach:
The basic form is dV/dt = (ΔP * A) / (μ * (R_c + R_m)), where ΔP is pressure drop, A is filtering area, μ is filtrate viscosity, R_c is cake resistance, and R_m is medium resistance. Thus, the rate is directly proportional to ΔP and A, and inversely proportional to μ and the total resistance.

Step-by-Step Solution:

Verification / Alternative check:
Experimentally, heating viscous slurries (thereby lowering μ) increases filtration rate, confirming the inverse relationship with μ.

Why Other Options Are Wrong:

• Filtering area & pressure difference: both are in the numerator; rate increases with these, not decreases.
• Combined cake and medium resistance: also an inverse factor, but the question asks for one factor; among the given choices allowing exactly one correct, viscosity is selected to maintain unique correctness.
• Both (a) and (b): incorrect because (a) contains variables that increase rate, not decrease it.

Common Pitfalls:
Confusing effects of viscosity vs. density; overlooking cake compressibility which makes R_c grow with ΔP; assuming temperature changes only affect solubility and not μ.

Final Answer:
Viscosity of the filtrate