Oil-Cooler Fouling: Clean, Add Area, or Increase Water Flow
A counterflow shell-and-tube oil cooler must cool 5.0 kg/s of oil (cp = 2.10 kJ/kg·K) from 120°C to 70°C using 6.0 kg/s of cooling water (cp = 4.18 kJ/kg·K) entering at 30°C. The installed area is 30 m² and the clean overall heat-transfer coefficient is 450 W/m²·K. After 18 months, the measured overall coefficient is 280 W/m²·K. The oil outlet is now about 75°C, and the operations team wants to increase the cooling-water flow rather than clean the exchanger. The minimum acceptable oil outlet is 70°C; a 3°C short-term excursion is tolerated, but sustained operation above 73°C will affect downstream seals. You must decide between cleaning, adding area, and increasing water flow, using the current fouling level.
1. Compute the design duty, LMTD, required U, and fouling factor.
Read the worked answer
Design duty Q=5.0×2.10×(120−70)=525 kW. Water outlet=30+525/(6.0×4.18)=30+20.9=50.9°C. Counterflow ΔT1=120−50.9=69.1 K, ΔT2=70−30=40 K; LMTD=(69.1−40)/ln(69.1/40)=53.5 K. Required U=525,000/(30×53.5)=327 W/m²K. Fouling resistance Rf=1/327−1/450=0.00306−0.00222=0.000836 m²K/W.
2. Using U=280 W/m²K, calculate the oil outlet at the same flows and compare it with the limit.
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C_h=5×2.1=10.5 kW/K; C_c=6×4.18=25.08; Cmin=10.5; Cr=0.419. At U=280, NTU=UA/Cmin=280×30/10,500=0.800. Counterflow ε=[1−exp(−NTU(1−Cr))]/[1−Cr exp(−NTU(1−Cr))]. NTU(1−Cr)=0.800×0.581=0.465; exp=0.628; ε=(1−0.628)/(1−0.419×0.628)=0.372/0.737=0.505. Qmax=Cmin(120−30)=945 kW; Q=0.505×945=477 kW. Oil outlet=120−477/10.5=74.6°C. This is above the 73°C sustained limit, so current operation is not acceptable.
3. Evaluate increasing water flow, determine whether it can restore 70°C, and recommend an action.
Read the worked answer
Increasing water flow to 8 kg/s gives C_c=33.4, Cr=0.314. At NTU=0.8, ε≈0.515, Q=487 kW, oil outlet=73.6°C-still above 73°C. Even infinite water flow gives ε=1−exp(−0.8)=0.551, Q=521 kW, oil outlet=70.4°C, so flow alone cannot guarantee 70°C at U=280. Required NTU for Q=525 kW: ε=525/945=0.556; E=(1−ε)/(1−εCr)=0.444/(1−0.233)=0.579; NTU=−lnE/(1−Cr)=0.546/0.581=0.939; required UA=0.939×10.5=9.86 kW/K. At U=280, area=9.86/0.280=35.2 m², i.e. add about 5.2 m². Cleaning restores U=450 and gives ample capacity. Recommend clean now, investigate water-side fouling, and add area only if fouling returns quickly.