抽水成本：一台表现不佳的潜水泵
A farmer irrigates from an open well using a 5 hp (3.73 kW) three-phase submersible pump rated at 400 V. A clamp meter shows a line current of 7.2 A on each phase and the power factor is 0.82. The pump discharges 28,000 litres per hour into a storage tank. A survey shows that the water level in the tank is 18 m above the water level in the well while pumping, and friction and fitting losses in the 65 mm delivery line are estimated at 4 m of head. The pump runs 6 hours a day for 25 days a month and the electricity tariff is ₹7.20 per kWh. The motor-pump set was installed nine years ago, and the farmer says the same pump used to fill the tank in 4 hours. A mechanic offers to overhaul the pump and replace the worn impeller and bushes for ₹18,000, after which an overall efficiency of about 60% is expected.
1. Calculate the hydraulic power, the electrical input power and the overall pump efficiency. State each formula and the assumptions you use.
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Discharge: Q = 28,000 L/h = 28,000 ÷ 3,600 = 7.78 × 10⁻³ m³/s.
Total head: H = static lift + friction = 18 + 4 = 22 m (velocity head neglected).
Hydraulic (water) power: P_h = ρgQH = 1000 × 9.81 × 7.78 × 10⁻³ × 22 = 1,679 W ≈ 1.68 kW.
Electrical input power (three-phase): P_in = √3 × V × I × cos φ = 1.732 × 400 × 7.2 × 0.82 = 4,090 W ≈ 4.09 kW.
Overall efficiency: η = P_h ÷ P_in = 1.68 ÷ 4.09 = 0.41, i.e. about 41%.
A healthy small submersible set should reach 55–65%, so roughly one-third of the billed energy is being lost. Note that 4.09 kW input against a 3.73 kW rated output corresponds to about 88% loading if motor efficiency is taken as 0.8 - the motor itself is not overloaded, so the fault lies in the pump end or the installation.
2. List three likely causes of the low efficiency, and describe one measurement that would tell whether the problem is hydraulic or electrical.
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Likely causes:
- Worn impeller and wear rings after nine years - water recirculates inside the casing instead of being thrown into the delivery; discharge falls while current stays high, which matches the farmer's observation that the tank now takes far longer to fill.
- Excessive friction in the delivery line - scale, silt or a partly closed valve or a corroded bend adds head; here friction is already 4 m, about 18% of total head, and would rise further if scaled.
- Wrong duty/trimmed impeller or a worn foot valve - the set may be running far from its best-efficiency point, or leaking back through a defective foot valve so it re-pumps the same water.
- Low supply voltage causing the motor to draw more current for less speed is a further possibility.
Deciding measurement: close the delivery and read the shut-off head with a pressure gauge, then measure actual discharge by a volumetric or ultrasonic method while recording head, voltage, current and power factor together. If the measured head and discharge are well below the duty point at normal current, the fault is hydraulic (impeller/wear rings); if the hydraulic output is right but input is high with low voltage, the fault is electrical.
3. Compute the present monthly electricity cost, the cost after the overhaul, and the simple payback period of the ₹18,000 repair. State any assumption that makes the payback conservative.
Read the worked answer
Present: energy = 4.09 kW × 6 h × 25 days = 613.5 kWh/month; cost = 613.5 × ₹7.20 = ₹4,417 per month.
After overhaul: input power needed for the same hydraulic output = 1.68 ÷ 0.60 = 2.80 kW; energy = 2.80 × 150 = 420 kWh; cost = 420 × ₹7.20 = ₹3,024 per month.
Saving: 4,417 − 3,024 = ₹1,393 per month (about 193 kWh).
Payback: 18,000 ÷ 1,393 = ≈ 12.9 months, i.e. about 13 months of running.
This is a good investment because the pump runs for many months each year. The payback estimate is conservative: it assumes the overhaul only restores efficiency and ignores the extra water delivered, which shortens pumping hours and pays back faster. If the pump also delivers more discharge after the impeller is replaced, the real saving will be larger than ₹1,393 per month.