Ejector sizing in sewage handling: A cylindrical ejector chamber 2 m high fills every 10 minutes under a peak sewage inflow of 0.0157 m^3/s. Determine the required chamber diameter (assume full usable height).

Introduction / Context:
Sewage ejectors lift wastewater intermittently. Correct chamber sizing prevents excessive cycling and provides surge storage. This problem applies basic volume and geometry to determine a chamber diameter for a given fill time and inflow.

Given Data / Assumptions:

• Cylindrical chamber height, h = 2 m (usable volume assumed full height).
• Peak inflow, Q = 0.0157 m^3/s.
• Fill time between cycles, t = 10 minutes = 600 s.

Concept / Approach:
Volume required V equals inflow rate times time. For a cylinder, V = (π/4) * d^2 * h. Solve for d by equating volumes. This is a straightforward sizing step used prior to mechanical selection and control strategy (on/off levels).

Step-by-Step Solution:
V = Q * t = 0.0157 * 600 = 9.42 m^3V = (π/4) * d^2 * h → d^2 = 4V / (πh)d^2 = 4 * 9.42 / (π * 2) = 37.68 / 6.28318 ≈ 6.00d = sqrt(6.00) ≈ 2.449 m ≈ 2.45 m

Verification / Alternative check:
Back-calculate V using d = 2.45 m: V ≈ (π/4)*(2.45^2)*2 ≈ 9.42 m^3, consistent with Q * t.

Why Other Options Are Wrong:
2.30 m, 2.40 m: Undersized; volume would be lower than 9.42 m^3.2.50 m: Slightly oversized; more volume than required by the target fill time.2.35 m: Also undersized.

Common Pitfalls:

• Forgetting to convert minutes to seconds.
• Not accounting for usable height; in detailed design, high- and low-level set points reduce usable volume.
• Neglecting future peak factors which may require additional storage.

Final Answer:
2.45 m