Volume reduction during sludge dewatering: In a digestion/thickening tank, if the sludge volume is V1 at moisture content p1% and is reduced to moisture content p2% (solids mass constant), what is the resulting volume V2 ?

Introduction / Context:
Sludge handling often involves thickening or dewatering to reduce volume. When moisture content changes while the mass of dry solids remains the same, we can relate initial and final volumes using simple mass-fraction bookkeeping.

Given Data / Assumptions:

• Initial sludge volume V1 at moisture content p1% (water by mass).
• Final moisture content p2% after thickening/dewatering.
• Dry solids mass is constant; density approximated close to water for volume–mass proportionality.

Concept / Approach:
Let Ms be the dry solids mass. Initially, solids fraction by mass is (100 - p1)%. Finally, solids fraction is (100 - p2)%. With Ms constant, the total mass (and hence volume) scales inversely with the solids fraction, yielding a direct relation between V1 and V2.

Step-by-Step Solution:
Let Ms = constant.Initial total mass ∝ V1, with solids fraction (100 - p1)/100.Final total mass ∝ V2, with solids fraction (100 - p2)/100.Equal solids mass gives: V2 / V1 = (100 - p1) / (100 - p2).Therefore: V2 = V1 * (100 - p1) / (100 - p2).

Verification / Alternative check:
Edge check: If p2 < p1 (drier sludge), denominator increases, so V2 < V1, which is physically correct. If p2 → 100%, volume would tend to infinity for constant Ms, which also aligns with the limiting behavior (pure water with vanishing solids fraction is not meaningful for fixed Ms).

Why Other Options Are Wrong:
(p2/p1) or (p1/p2) assume proportionality to moisture, not solids fraction.Expressions with (100 + p) mix up solids and water fractions.Linear difference (p1 - p2)/100 lacks mass balance basis.

Common Pitfalls:

• Using moisture %, instead of solids % = (100 - moisture %).
• Forgetting that solids mass, not total mass, remains constant.

Final Answer:
V2 = V1 * (100 - p1) / (100 - p2)