In a digestion tank, the sludge moisture content decreases from 90% to 80%. Assuming solids mass remains constant, what is the percentage decrease in sludge volume?

Introduction / Context:
Sludge conditioning aims to reduce water content. When solids mass is constant, volume change is governed by moisture change. This type of computation is common in sizing digesters, thickeners, and storage.

Given Data / Assumptions:

• Initial moisture = 90% → solids fraction = 10%.
• Final moisture = 80% → solids fraction = 20%.
• Solids mass remains constant; sludge density approximated uniform for mass–volume proportionality.

Concept / Approach:

Let solids mass be S. Total mass (and volume, by proportionality) initially Mi = S/(solids fraction) = S/0.10 = 10S. Finally Mf = S/0.20 = 5S. The percentage decrease is (Mi − Mf)/Mi * 100.

Step-by-Step Solution:

Verification / Alternative check:

Doubling the solids fraction from 10% to 20% halves the volume for constant solids mass, confirming 50% reduction.

Why Other Options Are Wrong:

25%, 10%, and 5% underestimate the strong effect of halving the water fraction; 40% still underestimates.

Common Pitfalls:

Using moisture percentages directly on water mass rather than on total mass; forgetting that constant solids mass means volume is inversely proportional to solids fraction.

Final Answer:

50%