Pump Upsizing at a Hilltop Water Scheme
A hilltop community water scheme serves 1,800 residents through a 4.2 km distribution network. The existing centrifugal pump is rated 11 kW and, at present demand, delivers 18 L/s against a total head of 32 m. Of that head, 18 m is static lift and 14 m is pipe friction at 18 L/s. During the evening peak, demand has grown to 24 L/s, and pressure at the farthest standpipe falls to about 9 m, below the required minimum of 15 m. The operator proposes raising the variable-frequency drive speed by 33% so that flow rises from 18 to 24 L/s, arguing that the affinity laws will take care of head. The motor has only a 10% short-term service margin, and pipe friction varies approximately with the square of flow. The pump curve is not available, but the duty point above is reliable. The scheme has a 20 m³ elevated tank that currently cycles between 40% and 80% full.
1. Calculate the total head the system requires at 24 L/s and explain why the proposed 33% speed increase does not simply 'take care of head'.
Read the worked answer
At 18 L/s, friction is 14 m. Friction scales with Q², so at 24 L/s: 14 × (24/18)² = 14 × 1.777 = 24.9 m. Required total head = static 18 m + 24.9 m = 42.9 m.
The affinity law would map the old duty point (18 L/s, 32 m) to (24 L/s, 32 × 1.333² = 56.9 m) at 33% higher speed. That is not the head the system needs at 24 L/s; it is about 14 m too high. Because pump head would exceed system head at 24 L/s, the operating point would move to a higher flow, not sit exactly at 24 L/s. The correct control target is pressure or flow, not a fixed speed ratio. A VFD should be set to hold the required pressure, and the pump curve should be checked to confirm it can reach 42.9 m at 24 L/s.
2. Using affinity laws, estimate the power demand at 33% overspeed and assess whether the 11 kW motor with 10% margin is adequate. Recommend a practical alternative.
Read the worked answer
Affinity laws give power ∝ N³, so a 1.333 speed ratio multiplies power by 1.333³ = 2.37. Even if the existing pump drew only 8 kW at the old duty, it would draw about 19 kW at the new speed; a full 11 kW would become 26 kW. The 11 kW motor with a 10% margin can safely deliver only about 12.1 kW.
Check hydraulically: P_h = ρgQH = 1000 × 9.81 × 0.024 × 42.9 ≈ 10.1 kW. At a generous 70% pump efficiency, shaft power is 14.4 kW, already above 12.1 kW. So the proposal is not viable.
Practical options: reduce peak demand by using the 20 m³ tank for equalization and limiting pump cycling; install a higher-head, higher-power pump/motor, around 18.5 kW with a matched impeller; or keep the existing pump and add a small booster near the critical standpipe if demand can be managed. I would first model tank operation, then select a pump against the 24 L/s, 42.9 m duty with at least 15% motor margin.