Is the Hostel Pump Really Undersized?
A polytechnic hostel lifts water from a ground-level sump into a 2000 litre overhead tank on the roof. The water surface in the tank stands 12 m above the sump surface, and the pipework, bends, foot valve and strainer together add friction equivalent to 3 m of head at the design flow. The existing pump is a 0.75 kW (1 hp) single-phase monoblock set, and the electrician's clamp meter and flow test put the pump efficiency at about 60 per cent. The original design promised that the tank would fill in roughly 25 minutes. It now takes nearly 45 minutes, and the caretaker wants a 1.5 kW pump bought at once, blaming 'more students than the pump was designed for'. Before sanctioning the money, the principal asks the workshop instructor to establish from the numbers whether the motor is genuinely too small or whether something else has gone wrong with the installation.
1. Using the design flow (2000 L in 25 minutes), calculate the hydraulic power and the required motor output. Is 0.75 kW adequate?
Read the worked answer
Flow Q = 2000 L / (25 × 60 s) = 1.333 L/s = 1.333 × 10^-3 m³/s.
Total head H = static 12 m + friction 3 m = 15 m.
Hydraulic power = ρgQH = 1000 × 9.81 × 1.333×10^-3 × 15 ≈ 196 W.
Shaft power required = 196 / 0.60 ≈ 327 W ≈ 0.33 kW.
The 0.75 kW motor is asked for only about 44 per cent of its rating - a healthy loading for a small monoblock set. At the design duty the motor is therefore not undersized; it has roughly 2.3 times the necessary output. A 1.5 kW replacement would cost money and raise starting current without adding a litre per minute if the fault lies elsewhere, so the request must be refused until the installation is checked.
2. Estimate the overall wire-to-water efficiency at the observed 45 minutes and say what the figure tells you.
Read the worked answer
At 45 minutes, Q = 2000 / 2700 = 0.741 L/s = 7.41 × 10^-4 m³/s.
Hydraulic power = 1000 × 9.81 × 7.41×10^-4 × 15 ≈ 109 W.
Taking motor input as roughly its 0.75 kW rating (a lightly loaded induction motor still draws close to full current), overall efficiency = 109 / 750 ≈ 14.5 per cent, against about 26 per cent (196/750) at design duty.
The set delivers less than half its design hydraulic output while the motor still runs. That is lost head or lost flow, not a small motor: a partly closed or furred gate valve, a choked foot valve or strainer, an air leak on the suction side, a worn impeller or wear-ring clearance, or low terminal voltage reducing speed. None of these is cured by a bigger motor; each raises losses or lowers delivered head.
3. Which single measurement would you take first, and how would the result separate the possible causes?
Read the worked answer
Fit a pressure gauge on the delivery side close to the pump and read it with the discharge valve fully open; convert using head (m) = pressure (bar) × 10.2 for water.
- If the gauge reads about 1.5 bar (≈15 m), the pump is still generating full head but flow is restricted - look for a throttled or furred valve, choked strainer or scaled pipe.
- If it reads well below 1.5 bar, the pump is not generating head - look for air drawn in at a suction joint, a worn impeller or wear ring, or a blocked foot valve acting as a throttle.
Either way the diagnosis is narrowed in ten minutes with one gauge, before any purchase order is raised. A suction vacuum reading and a check of terminal voltage under load would complete the picture and confirm whether the supply is also sagging.