Lube-Oil Cooler Shortfall: Fouling, Plugged Tubes, or Water Flow?
A shell-and-tube lube-oil cooler is specified to cool 120 gpm of oil (SG 0.87, cp 0.45 Btu/lb·°F) from 150°F to 120°F using 70 gpm of cooling water entering at 85°F. The unit is 1-2 pass with F = 0.93, 295 ft² of area, a clean overall coefficient of 65 Btu/h·ft²·°F, and a design fouling allowance of 0.0015 h·ft²·°F/Btu. After 18 months the oil leaves at 132°F instead of 120°F, and the bearing temperatures on the machine it serves have risen 12°F. Measured conditions: oil in 150°F, water in 85°F, water out 97°F, water flow 70 gpm by ultrasonic meter, oil flow steady at 120 gpm, water control valve wide open. Tube-side pressure drop is 4 psi lower than the clean commissioning value at the same flow; shell-side pressure drop is unchanged. The plant can spare a 6-hour outage this quarter and wants to know whether cleaning alone will restore duty or whether capital is needed.
1. From the measured data, compute the actual duty, the actual UA, and the implied fouling factor. Check the energy balance first.
Read the worked answer
Flows and heat capacity rates:
- Oil: 120 gpm × 8.34 lb/gal × 0.87 = 871 lb/min = 52,200 lb/h; C_oil = 52,200 × 0.45 = 23,490 Btu/h·°F
- Water: 70 gpm × 500 = 35,000 lb/h; C_water = 35,000 Btu/h·°F
Energy balance check:
- From the oil side: Q = 23,490 × (150 − 132) = 422,800 Btu/h
- From the water side: Q = 35,000 × (97 − 85) = 420,000 Btu/h
Agreement within 0.7%, so the readings are trustworthy and there is no significant ambient loss or unmetered bypass.
LMTD (counterflow): ΔT₁ = 150 − 97 = 53°F; ΔT₂ = 132 − 85 = 47°F. LMTD = (53 − 47)/ln(53/47) = 6/0.1198 = 50.1°F. Corrected: 50.1 × 0.93 = 46.6°F.
Actual UA = 422,800/46.6 = 9,070 Btu/h·°F, so U_actual = 9,070/295 = 30.7 Btu/h·ft²·°F against 65 clean.
Fouling: 1/30.7 = 1/65 + R_f → 0.0326 − 0.0154 = R_f ≈ 0.017, more than ten times the 0.0015 allowance. The unit has lost roughly half its UA. This is not measurement noise; something has substantially degraded the surface.
2. The tube-side pressure drop is 4 psi below the clean baseline at the same indicated flow. What does that imply, and is cleaning alone sufficient to restore 120°F oil outlet?
Read the worked answer
Interpreting the low tube-side ΔP. At identical flow, ΔP should rise with fouling (smaller bore, rougher surface), not fall. A drop of 4 psi at the same total flow means the water is passing through fewer effective tubes - typically tubes plugged after leak repair, or a pass-partition/tube-sheet bypass - or that the ultrasonic meter is reading a flow that is partly recirculated. Fewer active tubes means less area and higher velocity in the remainder, which partially offsets the loss in U but shrinks the 295 ft² actually available. That is a mechanical/capital issue, not a cleaning issue.
Is cleaning enough? Test the design case at clean coefficient. Design duty Q = 23,490 × 30 = 704,700 Btu/h. With 70 gpm water, outlet would be 85 + 704,700/35,000 = 105°F. ΔT₁ = 150 − 105 = 45; ΔT₂ = 120 − 85 = 35. LMTD = 10/ln(45/35) = 39.8°F; corrected 39.8 × 0.93 = 37.0°F. Required UA = 704,700/37.0 = 19,050, i.e. required U = 19,050/295 = 64.6 - essentially identical to the 65 clean coefficient.
So even a perfectly clean exchanger has zero margin: the design fouling allowance was effectively never available as excess area. If some tubes are plugged, required U exceeds 65 and cleaning cannot succeed. Cleaning is necessary but not sufficient; the outage must include tube counting/eddy-current inspection and a plan for added surface.
3. Recommend the scope of work for the 6-hour outage and the interim operating strategy, with justification.
Read the worked answer
Outage scope (6 hours):
- Hydro-jet or brush the tube side and inspect the water-side fouling type (scale suggests water chemistry; silt suggests low velocity).
- Count plugged tubes and eddy-current test the rest. If more than about 5% are plugged, retubing or a new bundle is required.
- Record clean ΔP at a known flow immediately after cleaning; that becomes the new baseline for trending.
- Verify the water control valve stroke and that no bypass around the exchanger is open.
Interim operation:
- Keep the oil outlet under about 130°F by confirming that the 132°F reading reflects steady state, and review bearing temperatures against alarm limits with the higher-viscosity oil. Do not simply accept elevated bearing temperature.
- Keep the water valve fully open and confirm flow daily; raising water flow above 70 gpm is only a partial help because more flow lowers LMTD. At 85 gpm the water outlet falls to about 101.6°F and LMTD falls to roughly 35.6°F, so the required U rises to about 67 - above clean. Flow alone cannot solve a lost-UA problem.
- Do not run the machine above its normal load if bearing temperatures trend upward; be ready to reduce load.
Capital recommendation: add a plate-and-frame trim cooler or replace the bundle with more area, sized for the 704,700 Btu/h duty plus a genuine 0.0015 fouling factor (target U about 55 with area ≈ 345 ft² minimum, preferably larger). Also investigate the fouling source: 0.017 in 18 months usually means low tube velocity or poor water treatment, so add water-side monitoring and a corrosion/scale program.