Cavitating Pump Suction: Throttling the Discharge Is Not a Cure
A chemical plant circulates water at 85 °C from an open atmospheric tank to a pump. The pump centreline is 2.0 m above the tank water level. Suction piping is 4.0 m long, 100 mm internal diameter, with a foot valve (K = 4.5) and two 90° elbows (K = 0.75 each). The Darcy friction factor is 0.020. Design flow is 20 L/s. At this flow the pump curve gives NPSHr = 4.0 m. Local atmospheric pressure is 101.3 kPa; at 85 °C the vapour pressure of water is 57.8 kPa and density is 968 kg/m³. The operators report rattling noise and erratic discharge pressure, especially in summer. To reduce the noise, the night shift throttles the discharge valve. A proposal is to raise the tank by 1.5 m; another is to install a variable-frequency drive and run at 85% of design flow. The pump manufacturer states that NPSHr varies approximately with Q².
1. Calculate the available net positive suction head (NPSHa) at 20 L/s and determine whether the pump is cavitating. Show the velocity, friction and minor losses.
Read the worked answer
Pipe area: A = π(0.100)²/4 = 7.854×10^-3 m². Velocity: v = Q/A = 0.020 / 7.854×10^-3 = 2.546 m/s. Velocity head: v²/2g = (2.546)²/(2 × 9.81) = 0.3305 m.
Friction loss: hf = f(L/D)(v²/2g) = 0.020 × (4.0/0.1) × 0.3305 = 0.264 m. Minor-loss coefficient: ΣK = 4.5 + 2(0.75) = 6.0. Minor loss: hm = 6.0 × 0.3305 = 1.983 m. Total suction loss = 0.264 + 1.983 = 2.247 m.
Pressure head difference: (p_atm − p_v)/(ρg) = (101.3 − 57.8)×10³ / (968 × 9.81) = 4.581 m.
NPSHa = 4.581 − 2.0 − 2.247 = 0.334 m.
NPSHr is 4.0 m, so the margin is 0.334 − 4.0 = −3.67 m. The pump is cavitating severely; the rattling noise and erratic pressure are consistent with this.
2. Evaluate quantitatively the two proposals: raising the tank by 1.5 m, and running at 85% of design flow with a VFD. State whether either alone restores a 0.5 m margin.
Read the worked answer
Raising the tank by 1.5 m reduces the static lift from 2.0 m to 0.5 m, so NPSHa becomes 0.334 + 1.5 = 1.834 m. The requirement remains 4.0 m, so the margin is still −2.17 m. It is insufficient.
At 85% flow, Q = 0.017 m³/s. Suction losses scale with Q²: 2.247 × 0.85² = 1.624 m. Static lift remains 2.0 m, so NPSHa = 4.581 − 2.0 − 1.624 = 0.957 m. NPSHr scales as Q²: 4.0 × 0.85² = 2.89 m. The margin is 0.957 − 2.89 = −1.93 m. The VFD alone is also insufficient.
To restore a 0.5 m margin at design flow, the static lift must satisfy 4.581 − z − 2.247 ≥ 4.5, so z ≤ −2.17 m. The pump centreline must be at least about 2.2 m below the tank water level, or the suction line must be enlarged and the pump replaced with a lower-NPSHr unit. A combination of a larger suction line, lower pump elevation and reduced flow is the practical route.
3. The night shift throttles the discharge valve and the noise disappears. Explain the mechanism and why this is not a cure. Estimate the cooling needed to restore a 0.5 m NPSH margin at design flow and comment on practicality.
Read the worked answer
Throttling reduces flow, so suction losses fall and NPSHr falls approximately with Q². NPSHa rises while the requirement drops, which can eliminate audible cavitation. But the pump is then operating off its design point at lower efficiency, the process may not receive design flow, and the root cause remains. The noise may return with warmer water, a fouled suction strainer or a partly closed valve.
For a 0.5 m margin, required NPSHa = 4.0 + 0.5 = 4.5 m. At design flow the losses and static lift total 4.247 m, so the pressure head term must be 8.747 m. That requires p_atm − p_v = 8.747 × 9.496 = 83.1 kPa, so p_v = 101.3 − 83.1 = 18.2 kPa. The corresponding saturation temperature is roughly 58 °C.
Cooling the water from 85 °C to about 58 °C is not a practical plant fix. The suction system and pump elevation must be changed, or a booster/low-NPSH pump installed.