Pump upgrade at a cooling-water transfer station
A cooling-water transfer pump runs at 1450 rpm and is rated 20 L/s at 30 m head, with a published efficiency of 68%. It is driven by an 11 kW, 415 V, three-phase squirrel-cage motor. The plant adds a new heat exchanger and the process now asks for 25 L/s at a system head of about 25 m. The operator has throttled the discharge valve to raise flow, and the motor current has risen to 22 A per phase at 415 V with a power factor of 0.84. The pump suction is flooded, the suction pipe is 100 mm diameter, and the suction gauge reads 2.5 m absolute. The motor is a TEFC unit in a 40°C ambient. Maintenance proposes increasing the pump speed to 1800 rpm by changing the pulley to meet the new flow.
1. Calculate the hydraulic power and shaft power at the rated duty, then use the measured motor current to decide whether the 11 kW motor is overloaded. State any assumption.
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Hydraulic power: P_h = rho g Q H = 1000 x 9.81 x 0.020 x 30 = 5.89 kW. Shaft power at 68% efficiency: P_s = 5.89/0.68 = 8.66 kW, so at the design duty the 11 kW motor has about 2.3 kW margin. Measured motor input: P_in = sqrt(3) x 415 x 22 x 0.84 = 13.30 kW. Even before applying motor efficiency, this exceeds the 11 kW nameplate. Taking a typical 4-pole motor efficiency of 0.89, shaft output is about 11.8 kW, which is above rating. Conclusion: the motor is overloaded in the present throttled condition; throttling has moved the pump to a higher-head, lower-efficiency region and increased power. Assumption: supply is balanced at 415 V and current is a true RMS value.
2. Using affinity laws, estimate flow, head and shaft power if speed is raised to 1800 rpm. Is the pulley change acceptable?
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Speed ratio = 1800/1450 = 1.241. Q2 = 20 x 1.241 = 24.8 L/s, close to the 25 L/s target. H2 = 30 x 1.241² = 46.2 m, far above the 25 m system head. The pump will not operate at that point; it will run out on its curve, so the affinity estimates are upper-bound indicators, not the actual duty. P2 is about 8.66 x 1.241³ = 16.5 kW at unchanged efficiency. Even with a 5% efficiency gain, shaft power is about 15.7 kW, well above 11 kW. The pulley change is therefore not acceptable without a larger motor, and actual flow will be set by the intersection with the system curve. Also check NPSH: a suction pressure of 2.5 m absolute leaves only about 2.3 m after vapour pressure, and NPSHr rises roughly with N², so cavitation risk increases.
3. Recommend a practical fix with approximate numbers.
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Do not simply speed up the existing unit. First measure the actual system curve and pump curve. For 25 L/s at 25 m, hydraulic power = 1000 x 9.81 x 0.025 x 25 = 6.13 kW. At a realistic 65% efficiency, shaft power = 9.4 kW, so an 11 kW motor is adequate only if the pump is trimmed or speed-controlled to that duty and efficiency is maintained. Options:
- Trim the impeller to meet 25 L/s at 25 m; this reduces head and power and keeps the existing motor.
- Fit a VFD and limit speed to about 1550 rpm, verifying motor current and NPSHa; a VFD also avoids throttling losses.
- If 1800 rpm operation is essential, replace with a 15 kW motor and verify NPSHr and pipe velocities. Check suction: 100 mm pipe at 25 L/s gives 3.2 m/s, which is high and will increase suction loss; consider 125 mm suction piping.