Heat Exchanger Duty (Q = U·A·F·LMTD)

Q˙=UAF ΔTlm\dot{Q} = U A F \, \Delta T_{lm}

Worked example: U 850, A 24 m2, F 0.95, LMTD 30 K → 581.4 kW — press Try an example to run it live, then adjust anything.

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Heat Exchanger Duty (Q = U·A·F·LMTD) explained

QUAFΔTlm

This is the equation on which the world's exchangers are bought and sold. Compute the counterflow log mean, multiply by U, A and the correction factor F, and you have the duty. F answers a single question: how much worse than pure counterflow is this geometry? A 1-2 shell-and-tube unit — one shell pass, two tube passes — has half its tubes running the wrong way, so F falls below 1; crossflow coils with one or both fluids unmixed sit somewhere between. F comes from charts plotted against the parameters P = (Tc,out − Tc,in)/(Th,in − Tc,in) and R = (Th,in − Th,out)/(Tc,out − Tc,in), and it is a factor, never a bonus: F ≤ 1 always, and true counterflow is F = 1.

The design rule handed down since Bowman, Mueller and Nagle published the F charts in 1940 is: never design below F = 0.80. Not because the physics fails, but because the chart goes vertical there — a one-degree measurement error in a terminal temperature swings F by a tenth, and your exchanger's duty becomes a guess. Cross that line and the answer is more shells in series, not more tubes. Worked example: U = 850 W/(m²·K) on 24 m² with F = 0.95 and a 30 K log mean gives 850 × 24 × 0.95 × 30 = 581 kW. Run it backwards from the observed duty and the F you compute is a fouling alarm — F does not degrade with time, so if the equation only balances at F = 0.6, the real culprit is U.

Heat Exchanger Duty (Q = U·A·F·LMTD) formula

Q˙=UAF ΔTlm\dot{Q} = U A F \, \Delta T_{lm}
Where
  • Q˙\dot{Q}= Exchanger duty (kW)
  • UU= Overall coefficient (W/(m²·K))
  • AA= Heat transfer area (m²)
  • FF= LMTD correction factor
  • ΔTlm\Delta T_{lm}= Log mean temperature difference (C°)