Centrifugal Pump Cavitates at Low Tank Level
A centrifugal pump transfers 60 m³/h of 30 wt% NaOH at 60°C from an atmospheric storage tank to a reactor. The tank liquid level is normally 4.0 m above the pump centerline, but it can fall to 2.0 m before the transfer is complete. The suction line is 100 mm ID with an equivalent length of 22 m and fitting loss coefficient K = 4.5; Darcy friction factor is 0.021. At 60°C the solution density is 1310 kg/m³ and vapor pressure is 3.5 kPa abs; atmospheric pressure is 101.3 kPa. The pump manufacturer gives NPSHr of 4.0 m at 60 m³/h, 5.5 m at 75 m³/h, and 6.0 m at 78 m³/h. At the low tank level the pump cavitates once flow reaches about 78 m³/h, but not at 60 m³/h. Production wants 75 m³/h. The discharge valve is currently throttled to hold 60 m³/h.
1. Calculate NPSHa at 60 m³/h and 78 m³/h with the tank at its low level, and compare each with NPSHr.
Read the worked answer
Available static head from pressure:
(Patm − Pv)/(ρg) = (101.3 − 3.5) × 1000/(1310 × 9.81) = 7.61 m.
Suction pipe area = π(0.100)²/4 = 0.007854 m². Total loss coefficient = fL/D + K = 0.021 × 22/0.100 + 4.5 = 4.62 + 4.5 = 9.12. Thus hf = 9.12 v²/(2g).
At 60 m³/h: v = (60/3600)/0.007854 = 2.12 m/s; hf = 9.12 × 2.12²/19.62 = 2.09 m. NPSHa = 7.61 + 2.0 − 2.09 = 7.52 m. NPSHr = 4.0 m, so margin = 3.52 m.
At 78 m³/h: v = 2.76 m/s; hf = 9.12 × 2.76²/19.62 = 3.54 m. NPSHa = 7.61 + 2.0 − 3.54 = 6.07 m. NPSHr = 6.0 m, so margin = only 0.07 m.
This explains the observation: at 78 m³/h the pump is at essentially zero NPSH margin and cavitates. At 60 m³/h it has ample margin.
2. Determine the maximum flow that keeps at least 1.0 m of NPSH margin at the low tank level, and assess whether 75 m³/h is acceptable.
Read the worked answer
At 75 m³/h, v = (75/3600)/0.007854 = 2.65 m/s; hf = 9.12 × 2.65²/19.62 = 3.27 m. NPSHa = 7.61 + 2.0 − 3.27 = 6.34 m. NPSHr = 5.5 m, so the margin is 0.84 m - less than the 1.0 m criterion and vulnerable to level variation, dissolved gas, or a partially blocked suction strainer. So 75 m³/h is not reliably acceptable.
For a 1.0 m margin, NPSHa must be at least NPSHr + 1.0. Around 73 m³/h, NPSHr is about 5.4 m, so NPSHa must be at least 6.4 m. With 7.61 m pressure head and 2.0 m static head, allowable hf is 7.61 + 2.0 − 6.4 = 3.21 m. Using hf = 9.12 v²/19.62, v = √(3.21 × 19.62/9.12) = 2.63 m/s, giving Q = 2.63 × 0.007854 × 3600 = 74 m³/h. The practical limit is therefore about 73–74 m³/h, not 78, and 75 m³/h has insufficient margin at low level.
3. Evaluate three ways to reach 75 m³/h reliably, including the operator's idea of throttling the discharge.
Read the worked answer
Throttling the discharge reduces flow and therefore reduces both suction-line loss and NPSHr, so it can stop cavitation - but it does not achieve 75 m³/h and it wastes energy. It is a temporary protection, not a debottleneck.
Three real options:
- Raise the tank level or lower the pump. Adding 1 m of static head raises NPSHa from 6.34 to 7.34 m at 75 m³/h, giving about 1.84 m margin over NPSHr. This is simple but may require tank or foundation changes.
- Enlarge the suction line to 150 mm ID. At 75 m³/h the velocity falls to about 1.18 m/s, and hf falls to roughly 0.5–0.6 m, giving NPSHa near 9 m. This is the most robust hydraulic fix.
- Reduce the liquid temperature or use a low-NPSH pump/inducer. Lower temperature reduces vapor pressure; a low-NPSH impeller reduces NPSHr. Either must be verified at the minimum tank level, not just at normal level.
Do not rely on the normal 4.0 m level. The design case is the low-level condition, and a 1.0 m margin should be demonstrated with the strainer dirty, not clean.