Preheat Train Fouling and a Tempting Flow Increase
A crude preheat exchanger (TEMA AES, 400 ft2, one shell pass, two tube passes) uses 80,000 lb/h of 400 F vacuum residue, cp 0.65 Btu/lb-F, to heat 120,000 lb/h of crude, cp 0.55 Btu/lb-F, entering at 150 F. Immediately after the last cleaning, the overall coefficient was 88 Btu/h-ft2-F against a design clean value of 90 and a design dirty value of 60, which includes the fouling allowance. Ten months later the residue leaves at 340 F while the crude leaves at 197 F, short of the 215 F target needed to hit the desalter temperature. Shell-side pressure drop has risen from 5 psi to 12 psi against a 15 psi TEMA allowable. To recover duty, the unit engineer proposes raising residue flow by 20 percent, since the residue is available and the crude side cannot be increased without cutting the main unit rate. The exchanger has no spare shell, and the next window is four months out.
1. Compute the current duty and overall coefficient, and determine whether the shortfall is fouling or underflow.
Read the worked answer
Take duty from the hot side, which is fully specified: Q = 80,000 x 0.65 x (400 - 340) = 3.12 x 10^6 Btu/h.
Log mean temperature difference, countercurrent: hot in 400, hot out 340, cold out 197, cold in 150. dT1 = 400 - 197 = 203 F; dT2 = 340 - 150 = 190 F. LMTD = (203 - 190) / ln(203/190) = 13 / 0.0661 = 197 F.
U = Q / (A x LMTD) = 3.12 x 10^6 / (400 x 197) = 39.6 Btu/h-ft2-F.
Now separate the resistances. The design dirty allowance implies a fouling resistance of 1/60 - 1/90 = 0.0056 h-ft2-F/Btu. The current operating point gives 1/39.6 - 1/90 = 0.0253 - 0.0111 = 0.0142 h-ft2-F/Btu, which is 2.5 times the design allowance.
So the crude outlet is low because the exchanger is fouled, not because the hot side is short of flow. The hot side is delivering exactly the duty its current U allows, and more residue flow would only raise the hot outlet temperature and reduce LMTD.
2. Quantify the 20 percent residue flow increase against both the thermal and mechanical limits.
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Mechanically, shell-side pressure drop scales with the square of flow: 12 x (1.2)^2 = 17.3 psi. That exceeds the 15 psi TEMA allowable, and it exceeds it starting from a shell that is already at 12 psi. A 20% flow increase also raises the risk of tube vibration and baffle damage in an AES shell, and this shell has already lost crossflow area to fouling.
Thermally, the gain is much smaller than 20%. With U essentially unchanged at 39.6, adding 20% to the residue flow means more heat is available but the hot outlet rises, and the LMTD falls as the two streams converge. A rough check: at 1.2 times flow with the same Q, the hot outlet would rise, so Q must rise only enough to keep the new LMTD consistent. The realistic recovery is well under 10% of duty, not 20%.
So the proposal spends the entire mechanical margin, including the vibration margin, to recover a fraction of the thermal shortfall. It should be rejected.
3. Recommend a path to the 215 F target over the next four months.
Read the worked answer
Cleaning is the only measure that restores the resistance you have actually lost, but the window is four months out, so manage the interim.
Interim: hold residue flow at design and accept a reduced crude outlet, then make up the desalter preheat elsewhere, for example by trimming the fired heater duty or rerouting a hotter side stream. Do not run above 12 psi shell-side; add a dP alarm at 13 psi and a trip-level alarm at 14.5 psi. Trend U monthly, since dP alone lags the thermal loss.
At the window, clean the shell and tubes and verify the gaskets and baffles. If U recovers to 88 but falls by more than about 15% within six months, the fouling is chronic and cleaning frequency, not cleaning, is the problem, so consider a larger shell, a rod insert in the tubes, or a feed-side antifoulant. Be explicit with operations that the 20% flow idea is off the table until dP headroom exists, and that running at 12 psi already leaves only 3 psi of margin.