Heat Exchanger Duty (Q = U·A·F·LMTD)

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

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

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)
Q˙=UAFΔTlm\dot{Q} = U A F \, \Delta T_{lm}
Where
  • Q˙\dot{Q}= Exchanger duty
  • UU= Overall coefficient
  • AA= Heat transfer area
  • FF= LMTD correction factor
  • ΔTlm\Delta T_{lm}= Log mean temperature difference