Design sewage flow for a separate system: A township has 40 hectares at a density of 250 persons/hectare. Water demand is 200 L/person/day and sewage flow is 80% of water supply. What peak sewage discharge (assuming peak factor ≈ 3 for separate sanitary sewers) should be used for design?

Introduction / Context:
Sanitary sewers in a separate system are sized for peak sanitary flows, often using a peaking factor to account for diurnal variations. Converting population and per-capita water demand to peak sewage discharge is a standard design exercise.

Given Data / Assumptions:

• Area = 40 ha; density = 250 persons/ha → population = 10000.
• Water demand = 200 L/person/day.
• Sewage flow = 80% of water use (average).
• Use peak factor ≈ 3 for separate sanitary sewer design.

Concept / Approach:

Compute average sewage flow from population and per-capita demand, then apply a peaking factor to obtain design peak discharge. Finally, convert daily volume to flow rate in cubic meters per second.

Step-by-Step Solution:

Verification / Alternative check:

Back-calculation: 0.0556 * 86400 ≈ 4800 m^3/day → matches peak volume computed.

Why Other Options Are Wrong:

0.05552 and 0.05554 are rounding variations but 0.0556 is the conventional rounded value; 0.0558 overstates; 0.0450 underestimates (would correspond to a smaller peaking factor).

Common Pitfalls:

Forgetting to multiply by the peaking factor; mixing average and peak flows; omitting the liters-to-cubic-meters or day-to-seconds conversions.

Final Answer:

0.0556 cumec