Constant-pressure cake filtration (incompressible cake): 5 m³ of filtrate are produced in 20 minutes. How much additional time is needed to produce the next 5 m³ under the same constant pressure? (Repaired options.)

Introduction / Context:
In constant-pressure filtration with an incompressible cake and negligible medium resistance, filtrate volume V grows with time t according to t = K * V^2, where K bundles viscosity, specific cake resistance, solids concentration, pressure, and area. This quadratic relation implies progressively longer times for equal volume increments.

Given Data / Assumptions:

• Constant pressure operation throughout.
• Incompressible cake; medium resistance neglected.
• First 5 m³ obtained in 20 minutes.
• Find time for V = 5 m³ → 10 m³ increment.

Concept / Approach:
With t = K * V^2, evaluate K from the first data point, then compute incremental time for the next equal volume.

Step-by-Step Solution:
Use t = K * V^2.At V = 5 m³, t = 20 min ⇒ K = 20 / 25 = 0.8 min·m^-6.Time to reach V = 10 m³: t_10 = K * 10^2 = 0.8 * 100 = 80 min.Additional time for next 5 m³: Δt = t_10 − t_5 = 80 − 20 = 60 min.

Verification / Alternative check:
For constant pressure, plotting t versus V^2 yields a straight line through the origin. Doubling volume quadruples time; thus, from 20 min at 5 m³, four-fold gives 80 min at 10 m³ → +60 min more.

Why Other Options Are Wrong:
20/30/40 minutes underestimate because filtration rate falls as cake thickens; equal volumes take longer at later stages.

Common Pitfalls:
Using a linear t ∝ V model (constant rate) instead of quadratic t ∝ V^2 for constant-pressure cake filtration.

Final Answer:
60 minutes