Lube-Oil Cooler Fouling: Clean, Upsize or Increase Water Flow
A shell-and-tube lube-oil cooler on a 2 MW diesel generator set is designed to remove 150 kW from the oil: 5.0 kg/s of oil (cp 2.0 kJ/kg·K) entering at 80 °C and leaving at 65 °C, cooled by 3.0 kg/s of water (cp 4.18 kJ/kg·K) entering at 30 °C. The unit has 8.65 m² of area and a clean overall coefficient of 500 W/m²·K; the design fouling allowance is 0.0005 m²·K/W. The correction factor F is 0.97 in both design and current operation. After 14 months, with the same oil flow, the same water flow and the same oil inlet temperature, the oil leaves at 70 °C and the water leaves at 38 °C. Maintenance records show the last cleaning was 14 months ago, and the cooling water is a once-through river supply. Operators want to raise the cooling water flow to 4.5 kg/s instead of cleaning the exchanger.
1. Verify the field data, calculate the current overall coefficient and quantify the fouling resistance. Is the exchanger actually fouled?
Read the worked answer
Data consistency check - do this first, always.
- Oil side: 5.0 × 2.0 × (80 − 70) = 100 kW
- Water side: 3.0 × 4.18 × (38 − 30) = 100.3 kW
The two independent energy balances agree within 0.3%, so the flow and temperature readings can be trusted. If they had disagreed by more than about 5%, the first action would be to calibrate the sensors and check for a fouled flow meter, not to open the exchanger.
Current LMTD and UA. Counterflow terminal differences: ΔT₁ = 80 − 38 = 42 K; ΔT₂ = 70 − 30 = 40 K. LMTD = (42 − 40)/ln(42/40) = 41.0 K. Corrected: 41.0 × 0.97 = 39.8 K.
UA_actual = 100 000/39.8 = 2513 W/K, so U_actual = 2513/8.65 = 290 W/m²·K.
Fouling resistance. 1/U_f = 1/U_clean + R_f, so R_f = 1/290 − 1/500 = 0.003442 − 0.002000 = 0.00144 m²·K/W, nearly three times the 0.0005 design allowance.
Design case for comparison. With hot 80→65 and cold 42←30: ΔT₁ = 38, ΔT₂ = 35, LMTD = 36.5 K, corrected 35.4 K, UA_design = 150 000/35.4 = 4237 W/K, U = 490 W/m²·K - consistent with the stated clean coefficient once the design fouling allowance is included.
Conclusion. The exchanger is heavily fouled. Note how the duty fell from 150 kW to 100 kW with the same inlet conditions - a third of the capacity lost. Operators often see only 'the oil is hotter than it should be' and miss that the exchanger has effectively been derated by 40%.
2. Evaluate the operators' proposal to raise the water flow to 4.5 kg/s. Will it restore the duty, and what are the side effects?
Read the worked answer
What raising water flow does. Tube-side heat transfer coefficient scales roughly with velocity to the 0.8 power. Raising flow from 3.0 to 4.5 kg/s is a factor of 1.5, so h_i rises by about 1.5^0.8 = 1.38.
Estimate the clean-side split. The clean overall resistance is 1/500 = 0.0020 m²·K/W. Assume the tube side is roughly 40% of that, 0.0008, and the shell side and wall make up the rest. At the higher flow the tube-side resistance falls to 0.0008/1.38 = 0.00058, saving 0.00022 m²·K/W.
New total resistance. 0.003442 − 0.00022 = 0.00322, so U rises from 290 to 310 W/m²·K - about 7%. Duty rises to roughly 107 kW and the oil outlet falls only from 70 °C to about 69.3 °C. The target of 65 °C is not remotely achieved.
Side effects. Pumping power rises roughly with the cube of flow: 1.5³ = 3.4 times. On a raw-water once-through supply this is a real operating cost and may exceed the pump motor rating. Higher velocity also increases erosion-corrosion risk on the tube inlets, particularly with river water carrying silt, and raises the tube-side pressure drop, which may lift the water pressure above the shell design if the return is restricted.
Judgement. The proposal treats a symptom. Increasing flow is a legitimate temporary measure on a hot day - it buys about 7% - but it cannot substitute for cleaning, and it adds pump power at the exact moment the plant is trying to save it. The correct answer is to reject the proposal as a cure but to note that the current 3.0 kg/s should be checked against the design value, since a low water flow would compound the fouling.
3. Q3. Set out the corrective and preventive actions, including how you would decide between cleaning and adding area, and how you would monitor the exchanger afterwards.
Read the worked answer
Decision: clean, do not add area. To achieve 150 kW with the fouled coefficient you would need UA = 150 000/35.4 = 4237 W/K, hence A = 4237/290 = 14.6 m² against 8.65 m² installed - a 70% increase. That is a new exchanger, not a modification. Cleaning restores 500 W/m²·K for the cost of a day's labour and a set of gaskets.
Corrective work.
- Open both ends, inspect and record tube condition. The oil side is the shell; expect varnish and degraded oil sludge. The water side is river water and should be checked for silt and both general and under-deposit corrosion.
- Clean the water side chemically (mild acid or EDTA, with a neutralisation and passivation step) rather than by high-pressure jet alone if the deposit is hard; use mechanical cleaning only after confirming tube wall thickness, since river-water tubes are often thinned.
- Renew gaskets, hydrotest, and re-commission with a fresh oil charge if the oil analysis shows oxidation or a drop in viscosity.
- Address the root cause: check whether the oil is running hot enough to oxidise, whether the river water needs filtration or a strainer upstream, and whether a dosing regime (antiscalant or biocide) is justified.
Monitoring.
- Install or use the existing differential pressure across the water side and trend it weekly; a rising ΔP at constant flow is the first sign.
- Log oil inlet and outlet temperatures, water flow and water outlet temperature every shift, and compute U weekly from the energy balance. A 10% fall from the clean baseline should trigger inspection.
- Record a clean-baseline U after every overhaul so there is a reference, and set the cleaning interval from the historical trend rather than calendar.
Judgement on adding area. Only if the fouling is intrinsic and unavoidable - for example a chronic silt load that no filtration can fix - should the exchanger be replaced with a larger one, and even then the added area should be justified against the cost of a more frequent cleaning regime.