Heat Exchanger Duty (Q = U·A·F·LMTD)
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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Where Heat Exchanger Duty (Q = U·A·F·LMTD) fits
- UOverall Heat Transfer Coefficient (U) · Overall U from Total Resistance
- ΔT_lmLog Mean Temperature Difference (Counterflow) · Log Mean Temperature Difference (Parallel Flow)
See also
Number of Transfer Units (NTU): Both use Overall coefficient and heat transfer area.Heat Exchanger Effectiveness (ε = Q/Qmax): Both use exchanger duty.Log Mean Temperature Difference (Counterflow): Both use log mean temperature difference.Log Mean Temperature Difference (Parallel Flow): Both use log mean temperature difference.Stream Duty from Mass Flow (Q = ṁcΔT): Both use exchanger duty.Heat Loss Through an Assembly (Q = A·ΔT/R): Both use heat transfer area.6 more related
Thermal Resistance of a Plane Wall: Both use heat transfer area.Fouled Overall Coefficient: Both use Overall coefficient.Fouling Factor on an Overall Coefficient: Both use Overall coefficient.Overall Heat Transfer Coefficient (U): Both use Overall coefficient.Overall U from Total Resistance: Both use Overall coefficient.U-Factor from Total R-Value (U = 1/R): Both use Overall coefficient.Rating the exchanger →
UniversityThermodynamics & Heat Transfer
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Heat Exchanger Duty (Q = U·A·F·LMTD) explained
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
- = Exchanger duty (kW)
- = Overall coefficient (W/(m²·K))
- = Heat transfer area (m²)
- = LMTD correction factor
- = Log mean temperature difference (C°)
Cite this page
formula.expert. (2026, October 31). Heat Exchanger Duty (Q = U·A·F·LMTD) Formula & Calculator. https://formula.expert/solvers/heat-exchanger-duty-lmtd