Hot-Oil Exchanger Cannot Reach the Required Water Outlet
A 1–2 shell-and-tube exchanger with 30 m² of area cools 10 kg/s of hot oil (cp = 2.2 kJ/kg·K) from 200°C to 145.5°C. The cooling water flow is 8.0 kg/s (cp = 4.18 kJ/kg·K), entering at 25°C and leaving at 60.9°C. The exchanger was designed with U_clean = 450 W/m²·K and a fouling allowance Rf = 0.00035 m²·K/W. The 1–2 correction factor is F = 0.95 at these conditions. Shell-side pressure drop has risen from 35 kPa at start-up to 48 kPa, and the cooling-water control valve is 90% open at design flow. The last mechanical cleaning was 14 months ago, and the unit has been on cooling-water service with occasional suspended solids. Production now needs the water outlet at 65°C to supply a reactor preheater. A colleague suggests increasing cooling-water flow by 20% to raise the outlet temperature.
1. Calculate the duty, corrected mean temperature difference, actual overall coefficient, and fouling resistance. Compare the fouling resistance with the design allowance.
Read the worked answer
Heat duty from the hot oil:
Q = 10 × 2.2 × (200 − 145.5) = 1199 kW.
Check with water: 8.0 × 4.18 × (60.9 − 25.0) = 1199 kW, so the heat balance closes.
Countercurrent LMTD:
- Hot end: 200 − 60.9 = 139.1 K
- Cold end: 145.5 − 25.0 = 120.5 K
- LMTD = (139.1 − 120.5)/ln(139.1/120.5) = 129.4 K
Corrected ΔTm = 0.95 × 129.4 = 122.9 K.
Actual U = 1,199,000/(30 × 122.9) = 325 W/m²·K.
Fouling resistance:
Rf,actual = 1/325 − 1/450 = 0.003077 − 0.002222 = 0.000855 m²·K/W.
This is about 2.4 times the design allowance of 0.00035. The exchanger is heavily fouled, so the measured outlet temperatures are consistent with a lost-capacity problem rather than a flow-measurement error.
2. Determine whether cleaning alone can restore a 65°C water outlet, assuming clean U = 450 W/m²·K and design fouling.
Read the worked answer
Required duty at 8.0 kg/s water and 25 → 65°C:
Q = 8.0 × 4.18 × 40 = 1338 kW.
At 10 kg/s hot oil, the hot-oil outlet would be 200 − 1338/(10 × 2.2) = 139.2°C.
New LMTD:
- Hot end: 200 − 65 = 135.0 K
- Cold end: 139.2 − 25 = 114.2 K
- LMTD = 20.8/ln(135/114.2) = 124.5 K
With F ≈ 0.95, corrected ΔTm ≈ 118.3 K.
Required U = 1,338,000/(30 × 118.3) = 377 W/m²·K.
At design fouling, U = 1/(1/450 + 0.00035) = 390 W/m²·K. That is only about 3% above the required 377 W/m²·K, so cleaning can just meet 65°C - but with very little margin. If the exchanger fouls rapidly, the target will be missed again. Cleaning is necessary, and the operating plan should include a fouling monitor or additional area.
3. Evaluate the colleague's suggestion to raise cooling-water flow by 20%. What does the shell-side pressure drop indicate?
Read the worked answer
Increasing cooling-water flow will not raise the water outlet to 65°C at fixed hot-oil conditions. With a fixed heat source and fixed area, adding water flow increases the water-side heat-transfer coefficient somewhat, but it also lowers the water outlet temperature because the same duty is spread over more mass. The hot-oil outlet would fall slightly and the water outlet would move below 60.9°C, not above it. To raise the water outlet, the exchanger must transfer more duty or the hot-oil supply must be hotter; more cooling water alone is the wrong direction.
The shell-side pressure drop rising from 35 to 48 kPa points to shell-side fouling, baffle deposits, or flow maldistribution. The water valve being 90% open at design flow also suggests increased tube-side pressure drop. Both sides need inspection. The heat balance confirms that water flow is about 8.0 kg/s, so the problem is not a hidden low-flow condition. Clean both sides, inspect baffles and tube bundle, verify the F factor at the new conditions, and then retest before considering a larger exchanger.