Centrifugal Pump Cavitation in Vacuum Column Service
A centrifugal pump takes 120 m3/h of a toluene-like liquid (density 830 kg/m3, vapor pressure 18 kPa abs at 60°C) from the bottom of a vacuum column operating at 30 kPa abs. The pump is located 3.5 m below the column liquid level. The suction line is 20 m long with 100 mm inside diameter, two elbows, and a fully open gate valve; estimated friction loss at design flow is 2.2 m of liquid. The pump curve at 120 m3/h gives NPSHr = 3.8 m. The plant requires at least 0.6 m NPSH margin. The column skirt cannot be raised without a long shutdown, but the suction line can be replaced with 150 mm pipe in a short outage. A variable-speed drive could reduce flow to 80%, where NPSHr falls to about 3.0 m, but production would be lost. The suction valve has been found partly closed in the past.
1. Calculate the available NPSH and assess the cavitation risk.
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NPSHa = (P_vessel - P_vapor)/(rho g) + static head - suction friction loss. Pressure difference = 30 - 18 = 12 kPa = 12,000 Pa. Head = 12,000/(830 x 9.81) = 1.47 m. Thus NPSHa = 1.47 + 3.5 - 2.2 = 2.77 m. The pump needs 3.8 m, and the required margin is 0.6 m, so the required available head is 4.4 m. The deficit is 4.4 - 2.77 = 1.63 m. The pump is therefore operating in certain cavitation, not marginal cavitation. The low suction pressure and high line velocity make it likely that the pump is already suffering capacity loss, noise, and impeller damage. The suction valve should be checked first for partial closure, because that alone could explain the friction loss.
2. Compare raising the column skirt with replacing the suction line by 150 mm pipe, and recommend one.
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Raising the skirt must add 1.63 m of static head to reach NPSHa = 4.4 m, so the skirt would need to rise by at least 1.7 m, preferably 2 m. That requires a long shutdown and structural work. Replacing the 100 mm line with 150 mm reduces velocity from 4.24 m/s to 1.89 m/s. Friction loss scales roughly with diameter to the fifth power, so the new loss is about 2.2 x (100/150)^5 = 2.2 x 0.132 = 0.29 m. New NPSHa = 1.47 + 3.5 - 0.29 = 4.68 m, giving a margin of 4.68 - 3.8 = 0.88 m, above the required 0.6 m. The line replacement is the better recommendation: it solves the problem in a short outage, avoids structural work, and also lowers operating energy loss. Confirm the 150 mm line routing and check the suction valve is fully open.
3. What immediate operating change would give relief without capital, and how much flow reduction is needed?
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The immediate action is to open the suction valve fully and reduce pump flow by variable-speed drive or throttling. At 80% flow, friction loss falls roughly as Q^1.8: 2.2 x 0.8^1.8 = 2.2 x 0.67 = 1.47 m. NPSHa becomes 1.47 + 3.5 - 1.47 = 3.50 m, while NPSHr at 80% is about 3.0 m, giving only 0.50 m margin, just below the 0.6 m requirement. Reducing to about 75% flow gives loss 2.2 x 0.75^1.8 = 1.31 m and NPSHa = 3.66 m, with NPSHr perhaps 2.9 m, giving 0.76 m margin. That is acceptable as a temporary measure, but it costs production and does not repair a partly closed valve or a fouled strainer. The permanent fix remains the larger suction line.