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Cavitation on the 80 °C Cooling Loop

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Cavitation on the 80 °C Cooling Loop

A cooling loop draws 80 °C water from an open atmospheric tank whose surface is 2.5 m below the centrifugal pump centreline. At the duty flow of 45 L/s the 150 mm ID suction line, its two 90° bends and a basket strainer account for 1.8 m of head loss, and the pump's NPSHr is 3.0 m. For water at 80 °C: density 972 kg/m³, vapour pressure 47.4 kPa; barometric pressure is 101.3 kPa. After two weeks of running, the pump rattles, the discharge pressure fluctuates by about 1 bar, and the flow is 15% below design. Maintenance proposes closing the discharge valve to about 60% of design flow to quieten it. The loop holds 15 m³ and the pump absorbs 30 kW at duty.

1. Calculate the net positive suction head available at the pump and compare it with the requirement.

Read the worked answer

Step 1 - pressure head available at the suction source

NPSHa = (p_atm − p_vap) / (ρ·g) − static lift − suction losses

(p_atm − p_vap) = 101,300 − 47,400 = 53,900 Pa

ρ·g = 972 × 9.81 = 9,535 N/m³

53,900 / 9,535 = 5.65 m

Step 2 - subtract the physical penalties

5.65 − 2.5 (lift to the pump centreline) − 1.8 (friction, fittings, strainer) = 1.35 m NPSHa

Step 3 - compare

Margin = 1.35 − 3.0 = −1.65 m

The pump is roughly 1.65 m short of its own requirement, which is why it is cavitating. The symptoms line up: rattling, a fluctuating discharge pressure (vapour bubbles collapsing), and a loss of head and therefore flow. The 15% flow loss is the consequence, not a separate fault.

The useful insight is that of the 5.65 m of head the pressure available gives you, the vapour pressure has already eaten nearly half. At 80 °C you cannot afford any suction lift at all - a pump taking water at that temperature from an open tank needs a flooded suction, not a 2.5 m lift.

2. Evaluate the proposal to throttle the discharge to 60% flow. Show what happens numerically.

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Throttling reduces flow, and suction losses scale roughly with the square of flow:

  • New suction loss = 1.8 × 0.60² = 0.65 m
  • New NPSHa = 5.65 − 2.5 − 0.65 = 2.50 m

So you recover about 1.15 m and end up with a margin of roughly zero against a pump requirement that, at 60% flow, may be anywhere between about 1.1 m and 2.5 m depending on the pump's low-flow behaviour. NPSHr generally falls with flow, but on the left side of the curve it often rises again because of inlet recirculation - which is precisely the region you would be forcing the pump into.

That makes the proposal a mask, not a fix:

  1. The strainer that contributed to the 1.8 m will keep fouling, so the margin keeps shrinking from an already zero starting point.
  2. Running at 60% of duty on a fixed-speed pump puts it left of its best efficiency point: recirculation, shaft deflection, and accelerated wear on bearings and seals.
  3. In a closed loop, all 30 kW of shaft power (less what leaves with any make-up or losses) ends up in 15 m³ of water. At about 25 kW absorbed, the bulk temperature rises at roughly 1.4 K per hour, which raises the vapour pressure, which cuts NPSHa further - the fix feeds back on itself.

Verdict: acceptable for a few hours as an emergency measure with continuous monitoring, not acceptable as a repair.

3. Recommend remedies, ranked, with the numbers that support each.

Read the worked answer

1. Restore the suction path (first, because it is cheap and diagnostic). The strainer is part of the 1.8 m. Clean or replace it, verify the tank level, check for a vortex or air entrainment, and confirm the isolation valve is fully open. If the clean-system loss is 1.0 m rather than 1.8 m, NPSHa rises to 5.65 − 2.5 − 1.0 = 2.15 m - better, still short.

2. Reduce the suction losses by re-sizing. Going from 150 mm to 200 mm ID cuts the friction loss by roughly (150/200)⁵ ≈ 0.24, so 1.8 m becomes about 0.43 m: NPSHa = 5.65 − 2.5 − 0.43 = 2.72 m. Necessary but not sufficient on its own, because the lift still dominates.

3. Remove the lift - flooded suction. Move the pump below the tank or raise the tank so the level is above the pump centreline. With a 1.0 m static head and clean losses: NPSHa = 5.65 + 1.0 − 0.5 = 6.15 m. This is the robust solution and the one that matches the fluid temperature.

4. Lower the temperature if the process allows. At 65 °C the vapour pressure is about 25 kPa, giving 7.9 m of pressure head and NPSHa = 7.9 − 2.5 − 1.8 = 3.6 m. It works, but it changes the process and may increase the heat load elsewhere.

5. Change the pump. A low-NPSH impeller or an inducer rated at 1.0 m would fit the existing installation, but it is the most expensive option and does not address the air-entrainment risk at the tank.

Recommended sequence: clean and verify first (it may restore the margin enough to run safely while the permanent fix is engineered), then flooded suction plus the larger suction line. Record NPSHa in the operating manual, because this pump will cavitate again the day someone raises the tank temperature or drops the level.