Underperforming Lube-Oil Cooler After Plant Turndown
A lube-oil cooler is a 1 shell-pass, 2 tube-pass TEMA BEM unit with 17.2 m² of area, designed to cool 3.0 kg/s of oil (cp 2.0 kJ/kg·K) from 85 °C to 55 °C using 2.0 kg/s of cooling water at 24 °C. The design overall coefficient is 340 W/m²·K with F = 0.90. Since a plant turndown, the oil leaves at 62 °C, and the water leaves at 49.5 °C at a measured flow of 1.3 kg/s. There are no obvious tube leaks. The shell vent was never opened after the last maintenance, and the water valve is 40% open. Operations proposes to raise water flow to design and, if that fails, to chemically clean the tubes. The design oil-side fouling factor was 0.0002 m²·K/W.
1. Check the energy balance and calculate the actual overall coefficient. What does the comparison with the design value tell you?
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Energy balance
- Oil duty: Q = 3.0 × 2.0 × (85 − 62) = 138 kW.
- Water duty: Q = 1.3 × 4.18 × (49.5 − 24) = 138.6 kW.
The two sides agree within 0.5%, so the flow and temperature measurements are credible and there is no large heat loss or unaccounted stream. This matters: a large mismatch would point to instrument error before fouling.
LMTD and U
Counterflow terminal differences: ΔT1 = 85 − 49.5 = 35.5 K; ΔT2 = 62 − 24 = 38.0 K.
LMTD = (38.0 − 35.5)/ln(38.0/35.5) ≈ 36.7 K. With F = 0.90, the effective mean difference is 33.1 K.
U = Q/(A·F·LMTD) = 138,600/(17.2 × 33.1) ≈ 244 W/m²·K versus 340 W/m²·K design, a 28% shortfall. At the design U and area, the unit would transfer about 180 kW, matching the original duty. So the area is not the problem; the coefficient is.
Interpretation
The measured U is an aggregate. It cannot by itself separate low water velocity, fouling, and non-condensable gas. The next step is to test the competing explanations, not to clean immediately.
2. Separate the effect of low water flow from fouling using heat-transfer reasoning. How much of the 28% loss is plausibly flow, and how much is fouling?
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Flow effect
Tube velocity scales with volumetric flow: 1.3/2.0 = 0.65. For turbulent tube-side flow, h ∝ v^0.8, so h falls to 0.65^0.8 ≈ 0.71, a 29% reduction in the tube-side coefficient.
If tube-side resistance is roughly half of the clean overall resistance, a 29% drop in h_i lowers U by about 15%: 340 → approximately 289 W/m²·K. The measured 244 W/m²·K is below that, so low flow alone does not explain the shortfall.
Fouling estimate
Using the remaining gap: 1/244 − 1/289 ≈ 0.00410 − 0.00346 = 0.00064 m²·K/W. That is about three times the design fouling allowance of 0.0002 m²·K/W. The estimate is sensitive to the assumed split of tube-side and shell-side resistance, so treat it as an order-of-magnitude diagnosis, not a precise fouling factor.
Non-condensables
A shell that was never vented can hold air or vapour, which lowers U and often makes outlet temperature unstable. That mechanism would not disappear with higher water flow, and it is cheap to test. The conclusion is: low water flow is a real but partial cause; fouling or gas is needed to explain the rest.
3. Recommend a diagnostic and corrective sequence. Justify why chemical cleaning should not be the first action.
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Sequence
- Vent the shell at the high point and confirm that the vent is clear. Record whether gas or air is released. This costs almost nothing and removes one explanation.
- Restore water flow to design and let the unit reach steady state. Recalculate U with the same method. If U rises to about 290 W/m²·K, much of the problem was flow; if it stays near 244, fouling or gas remains.
- Check the water strainer and valve position, and confirm the measured flow with a second method if possible. A 40% open valve may be limiting flow even if the pump can deliver more.
- Inspect the oil side. Oil-side deposits are often varnish or sludge that chemical cleaning of the water side will not remove. Sample the oil and check its temperature and viscosity; higher viscosity lowers the shell-side coefficient.
- Clean only after the diagnosis. Select chemical or mechanical cleaning based on the deposit type, and verify with a U test before and after.
Why not clean first
Chemical cleaning costs downtime, generates waste, and can damage tubes if the deposit and metallurgy are not matched. It also treats only one of at least three credible causes. A vent and a flow test are faster and cheaper, and they produce evidence. Log the baseline U, flows, and temperatures so future degradation can be trended rather than rediscovered.