What is LMTD in a heat exchanger?

The log mean temperature difference: (ΔT₁ − ΔT₂) ÷ ln(ΔT₁/ΔT₂), using the temperature differences between the fluids at each end. It is the correct average driving force for Q = U·A·LMTD.

What is a fouling factor?

An allowance for the extra resistance of deposits that build up in service, added to 1/U in design so the exchanger still meets its duty when dirty.

Module 09 · Heat Transfer3 min read

The Heat Exchanger Equation: U-value, LMTD and Fouling

Every heat exchanger, from a boiler economizer to a process heater, follows one equation: heat transferred equals an overall coefficient, times area, times a mean temperature difference. Knowing how each term behaves explains how exchangers are sized and why they lose output as they foul.

ByShikhar Singh·Steam Consultant & IT Consultant·Published 13 October 2026

You'll learn

Use Q = U A LMTD
Calculate a log mean temperature difference
Understand what the overall U-value includes
Allow for fouling

Before this: Conduction, convection and radiation, Sensible heat

BeginnerThe idea in plain words, and where you meet it. No maths.

Three things decide how much heat moves

How easily heat crosses the wall between the fluids — the overall coefficient, U.
How much wall there is — the area, A.
How big the temperature difference is between the two fluids — the driving force.

Double any one of them and, other things equal, the heat transferred doubles. A dirty exchanger loses U; a small one lacks A; one fed with steam at too low a pressure lacks temperature difference.

IntermediateThe formula, its units, and a worked example on real plant numbers.

Q = U A LMTD

Heat exchanger equation

Q = U A Δ T_{l m}

Log mean temperature difference

Δ T_{l m} = \frac{Δ T_{1} - Δ T_{2}}{\ln (Δ T_{1} / Δ T_{2})}

Symbols in the formula
Symbol	Meaning	Unit
$Δ T_{1}$ , $Δ T_{2}$	Temperature differences between the fluids at the two ends	K

Worked example

A steam water heater

Given

5,000 kg/h of water heated from 30 to 80 °C by steam at 4 kg/cm²(g) (151.4 °C).
Overall U assumed at 1,500 W/(m²·K) for illustration.

End differences: 121.4 K and 71.4 K.
LMTD:
$\frac{121.4 - 71.4}{\ln (121.4 / 71.4)} = 94.1 K$
Heat duty: 290 kW.
Area:
$A = \frac{2,90,434}{1,500 \times 94.1} = 2.1 m^{2}$

Answer: About 2.1 m² of heating surface, before any allowance for fouling.

ExpertThe fine print: where the simple model breaks, and what it costs.

What U contains, and fouling

Overall coefficient, resistances in series

\frac{1}{U} = \frac{1}{h_{1}} + \frac{t}{k} + \frac{1}{h_{2}} + R_{f}

Symbols in the formula
Symbol	Meaning	Unit
$h_{1}$ , $h_{2}$	Film coefficients on each side	W/(m²·K)
$t / k$	Wall resistance	m²·K/W
$R_{f}$	Fouling resistance allowance	m²·K/W

Fouling — scale, sludge, product deposits, soot — adds resistance over time. Designers add a fouling allowance so that the exchanger still meets duty when dirty. Adding 0.0002 m²·K/W to the example drops U from 1,500 to 1,154 and raises the required area from 2.1 to 2.7 m².

Arithmetic mean temperature difference overstates the driving force whenever the end differences are very unequal; LMTD is the correct mean for exchangers with one side at constant temperature, as condensing steam is. For other flow arrangements, a correction factor is applied to the counterflow LMTD.

Key takeaways

Q = U × A × LMTD.
LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂).
1/U is the sum of the film, wall and fouling resistances.
Fouling allowances add area; cleaning restores U.

The Heat Exchanger Equation: U-value, LMTD and Fouling

Three things decide how much heat moves

Q = U A LMTD

What U contains, and fouling

Key takeaways

Check your understanding

Frequently asked

Related equipment

Training a boiler team?

Sources

About the author