Pump Efficiency on a Smallholding
A smallholding needs to lift water from a well 12 m below ground level into a storage tank whose inlet is 4.0 m above ground level. A pump rated at 1.5 kW is used. In a test, the pump delivers 4.5 kg of water per second. The water leaves the pipe at negligible speed. The farmer wants to fill a 2000-litre tank that is initially empty. The pipe has a narrow section, and the pump casing becomes noticeably warm during use. The mains supply is 230 V and the pump current is 6.8 A. The farmer claims that the pump is wasting more than half of the energy supplied. The density of water is 1.0 kg/L and g = 10 N/kg.
1. Calculate the useful power delivered to the water and the overall efficiency of the pump, and assess the farmer's claim.
Read the worked answer
The useful power is the rate of increase of gravitational potential energy:
useful power = mgh per second.
The total lift is 12 + 4.0 = 16 m, so:
useful power = 4.5 × 10 × 16 = 720 W.
The electrical input power is:
P = VI = 230 × 6.8 = 1564 W.
Efficiency = 720/1564 × 100% = 46%.
The farmer is correct: about 54% of the supplied energy is not doing useful lifting, so more than half is wasted. The rated 1.5 kW is close to the measured input, so the pump is drawing roughly its rated power; the low efficiency is not simply because it is underloaded.
2. Estimate the time needed to fill the 2000-litre tank, stating any assumptions.
Read the worked answer
A 2000-litre tank holds 2000 kg of water. The energy needed to raise this mass through 16 m is:
E = mgh = 2000 × 10 × 16 = 3.2 × 10^5 J.
At a useful power of 720 W:
t = E/P = 3.2 × 10^5/720 = 444 s,
which is about 7.4 minutes. This assumes a steady flow of 4.5 kg/s and that the useful power remains 720 W throughout. In practice the time will be longer because the flow rate may fall as the tank fills and because the pump may not deliver 4.5 kg/s at the same head.
3. Explain where the wasted energy goes, using evidence from the description.
Read the worked answer
The missing input power is about 1564 − 720 = 844 W. It appears as internal energy in the pump casing, pipe and water. The main causes are friction in the narrow pipe (pressure drop), turbulence, friction in the pump bearings and impeller, electrical resistance heating in the motor windings, and sound. The warm casing is direct evidence that internal energy is increasing. Since the water leaves at negligible speed, kinetic energy is not a significant useful output. Reducing the narrow section would reduce the frictional pressure loss and increase the efficiency.