Crude Preheat Exchanger Fouling and Cleaning Economics
A crude preheat exchanger is designed to heat 20 kg/s of crude oil (cp = 2.0 kJ/kg K) from 30°C to 110°C using 20 kg/s of hot residue (cp = 2.4 kJ/kg K) entering at 200°C. The unit is a single shell-and-tube exchanger with 100 m2 of area. At the last test, crude outlet was 100°C and residue outlet was 142°C. Both flow rates were confirmed by calibrated meters. Design clean overall coefficient is 500 W/m2 K, and design fouled coefficient is 300 W/m2 K. The fired heater downstream has 85% efficiency, and fuel costs $8/GJ. Production loses $2,000 per hour when the exchanger is offline. Cleaning costs $30,000 and takes 12 hours. The next planned shutdown is in seven months. The crude unit runs 8,000 hours per year.
1. Calculate the actual overall heat-transfer coefficient and the fouling resistance. Compare with the design fouled condition.
Read the worked answer
Crude duty = 20 x 2.0 x (100 - 30) = 2,800 kW. Check with residue: 20 x 2.4 x (200 - 142) = 2,784 kW, close enough given roundoff. For countercurrent flow, delta T1 = 200 - 100 = 100 K and delta T2 = 142 - 30 = 112 K. LMTD = (112 - 100)/ln(112/100) = 12/0.1133 = 106 K. Actual UA = 2.8 x 10^6/106 = 26,415 W/K, so U = 26,415/100 = 264 W/m2 K. Actual fouling resistance Rf = 1/264 - 1/500 = 0.00379 - 0.00200 = 0.00179 m2 K/W. Design fouled Rf = 1/300 - 1/500 = 0.00333 - 0.00200 = 0.00133 m2 K/W. The exchanger has about 0.00046 m2 K/W more fouling resistance than the design allowance, so it is fouled beyond its design margin, but only moderately.
2. Estimate the annual fuel penalty of running at 100°C instead of 110°C, and decide whether to clean now.
Read the worked answer
The missing preheat duty is 20 x 2.0 x 10 = 400 kW. At 85% fired-heater efficiency, extra fuel energy = 400/0.85 = 471 kW. Over 8,000 h/year, that is 471 x 8,000 x 3,600/10^6 = 13,560 GJ/year. At $8/GJ, the penalty is about $108,500/year, or roughly $9,000/month. Cleaning costs $30,000 plus 12 h x $2,000/h = $24,000, for a total of $54,000. Payback is about 54,000/108,500 = 0.50 year, or six months, before the next planned shutdown in seven months. Cleaning now is economically justified if the fouling trend is confirmed and the cleaning restores at least the design fouled coefficient. If cleaning only restores 300 W/m2 K, the recovered duty is still substantial; if it restores the clean coefficient, the benefit is larger.
3. Before authorizing the shutdown, what checks could show that low duty is not caused by fouling?
Read the worked answer
Verify that the crude and residue flows match the design and that neither control valve is bypassing or partly shut. Check exchanger pressure drop: a rising shell-side or tube-side pressure drop supports fouling, while a normal pressure drop with low duty suggests bypassing, an air pocket, or maldistribution. Confirm temperature transmitters are calibrated and not reading a stratified outlet. Check for a closed vent or drain that could leave vapor in the shell. Compare the calculated U trend with the last two inspections; a gradual decline supports fouling, while a sudden drop points to a mechanical or instrumentation cause. Only after these checks should the shutdown be scheduled.