Sea-Water Cooling Pump Fouling, Flow and NPSH
A sea-water cooling pump serves the central coolers of a 12,000 kW diesel installation. At rated speed its characteristic is H = 42 − 0.00035Q², where H is head in m and Q is flow in m³/h. With clean coolers and strainers the system curve is H = 12 + 0.00022Q²; after two months of operation, biofouling and debris change it to H = 25 + 0.00045Q². The pump motor is rated for the original duty. The HT cooling-water outlet is normally 34°C with an 8 K rise at the clean operating point, and the alarm trips at 40°C. The sea temperature is 32°C, atmospheric pressure is 10.3 m of water, the suction static head is 2.5 m, and the suction line loss at the fouled flow is 3.2 m. The pump NPSHr is 7.0 m. The second engineer suggests closing the discharge valve a little to stop the pump running out on its curve, while the chief engineer wants to increase pump speed to restore the original flow.
1. Determine the clean and fouled operating flows and heads. Quantify the flow reduction.
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The operating point is where pump head equals system head.
Clean system: 42 − 0.00035Q² = 12 + 0.00022Q² 30 = 0.00057Q² Q² = 52,632, so Q = 229.4 m³/h. H = 42 − 0.00035 × 52,632 = 42 − 18.42 = 23.6 m.
Fouled system: 42 − 0.00035Q² = 25 + 0.00045Q² 17 = 0.00080Q² Q² = 21,250, so Q = 145.8 m³/h. H = 42 − 0.00035 × 21,250 = 42 − 7.44 = 34.6 m.
The flow falls from 229.4 to 145.8 m³/h, a reduction of 83.6 m³/h or about 36%. The pump has moved up its curve to a higher head and lower flow. That is the signature of increased system resistance, not of the pump losing its own capability.
2. Estimate the HT outlet temperature at the fouled flow, assess the alarm margin, and check whether cavitation is a credible cause.
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For the same heat load, temperature rise is inversely proportional to flow: ΔT_fouled = 8 × (229.4 / 145.8) ≈ 12.6 K.
If the inlet remains at 34 − 8 = 26°C, the fouled outlet is about 26 + 12.6 = 38.6°C. The 40°C alarm has not tripped, but the margin has fallen from 6 K to about 1.4 K. Any further fouling or increase in engine load will trip the alarm, and the HT water is already hot enough to reduce cooling performance and increase liner/charge-air temperatures.
Cavitation check: NPSHa = p_atm + static head − suction loss − vapour pressure = 10.3 + 2.5 − 3.2 − 0.48 ≈ 9.1 m. The vapour pressure of sea water at 32°C is about 0.48 m. NPSHa is about 9.1 m against NPSHr = 7.0 m, leaving roughly 2.1 m margin. Cavitation is therefore not the primary explanation for the reduced flow; the fouled system curve and higher head are. NPSH margin should still be monitored if sea temperature rises or suction strainers block further.
3. Evaluate the two proposals: throttling the discharge valve versus increasing pump speed. What action should be taken?
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Closing the discharge valve adds system resistance. The operating point moves to a lower flow and higher head, which makes the cooling water even hotter. It may reduce pump power slightly, but it does not remove the fouling and it directly worsens the HT temperature problem. It is not an acceptable remedy.
Increasing speed is also unattractive. To restore 229.4 m³/h against the fouled system, the pump would need H = 25 + 0.00045 × 52,632 ≈ 48.7 m. At the original speed the pump would produce only 23.6 m at that flow, so the affinity ratio is n2/n1 = sqrt(48.7/23.6) ≈ 1.44, a 44% overspeed. Pump power varies roughly as n³, so the motor would be asked for about 3 times its original duty and would likely trip or overload the shaft. The motor is rated only for the original duty.
The correct action is to clean the sea strainers and heat exchangers, inspect and renew anodes, and recheck the operating point. If cooling is still marginal, reduce engine load until cleaning is complete. Do not use throttling or overspeed as a substitute for maintenance.