Heat Exchanger LMTD Calculator

Find the log mean temperature difference for a heat exchanger and the heat it moves, for parallel flow or counter flow.

Physics and EnergyUpdated 2026-10-215 min read

Heat Exchanger LMTD Calculator

First end temperature difference

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Second end temperature difference...

Log mean temperature difference...

Heat transferred...

What it does

A heat exchanger moves heat from one stream of fluid to another without mixing them. The design question is how much heat it can shift, and the answer comes from three numbers: how well the surface conducts, how big the surface is, and how large the temperature difference is that drives the flow of heat. The last of those is not the plain average of the two end differences, and the reason why is the point of this page.

Heat flows faster where the temperature difference is bigger, so the average difference that matters is weighted towards the wide end of the exchanger. The correct average is the logarithmic mean, which is what the engineering texts use. The University of California San Diego plate exchanger notes derive it from the differential balance, and Purdue's heat exchanger lecture notes write the design equation as Q equals U A LMTD F, where F is a correction factor for the flow pattern.

That gives the procedure. Work out the temperature difference at each end of the exchanger, take the logarithmic mean of the two, multiply by the overall heat transfer coefficient and by the area, and you have the heat duty. Which end differences you use depends on how the fluids run: in counter flow the cold outlet sits opposite the hot inlet, and in parallel flow the two inlets sit opposite each other. Counter flow gives the larger differences, so it moves more heat for the same surface.

This page runs that calculation for both arrangements. Give it the four temperatures, the overall heat transfer coefficient and the area, and it returns both end differences, the log mean temperature difference and the heat transferred, in kilowatts or in BTU per hour.

Two things are left out on purpose. The correction factor F is not applied, so the counter flow answer is the best case for a given set of temperatures, while a real shell and tube exchanger with a one shell and two tube passes needs a factor read off a chart. And the page does not work backwards from a required heat duty to the area needed, because that needs the mass flow rates and the specific heat of each stream. The geometry and the duty are the easy half of a heat exchanger design.

How to use

Choose metric or imperial units for the temperatures, the coefficient and the area.
Pick the flow arrangement: counter flow for the best performance, parallel flow for the simplest piping.
Enter the hot fluid inlet and outlet temperatures, with the inlet always the hotter one.
Enter the cold fluid inlet and outlet temperatures, with the outlet always the warmer one.
Enter the overall heat transfer coefficient and the heat transfer area.
Read the log mean temperature difference and the heat transferred, then check the end differences make sense for the arrangement you picked.

Formula

Counter flow: end differences are Th,in minus Tc,out and Th,out minus Tc,in. Parallel flow: end differences are Th,in minus Tc,in and Th,out minus Tc,out. Log mean temperature difference LMTD = (dT1 - dT2) / ln(dT1 / dT2), which equals dT1 when the two differences are equal. Heat transferred Q = U x A x LMTD.

Th,in and Th,out: Hot fluid temperatures at the two ends of the exchanger.
Tc,in and Tc,out: Cold fluid temperatures at the two ends of the exchanger.
LMTD: Log mean temperature difference, the driving force for heat transfer.
U: Overall heat transfer coefficient, in W per m2 per K.
A: Heat transfer area in m2.

When the two end differences are equal the logarithm is undefined, and the log mean is simply that shared difference, which is what the page returns. The 1 BTU per hour per ft2 per °F equals 5.678263 W per m2 per K, and a temperature difference of 1 K equals 1.8 °F, so the imperial numbers on this page describe exactly the same exchanger as the metric ones. In practice a shell and tube exchanger also carries a correction factor F below one, and a plate exchanger with a different number of passes carries one too.

Worked example

Example: a counter flow exchanger heating water from 20 to 50 degrees C

Defaults: metric units, counter flow, a hot stream from 95 to 60 degrees C, a cold stream from 20 to 50 degrees C, U = 350 W per m2 per K and 25 m2 of area.

Counter flow end differences: dT1 = 95 - 50 = 45 K and dT2 = 60 - 20 = 40 K.
Log mean temperature difference: (45 - 40) / ln(45 / 40) = 5 / 0.117783 = 42.4509 K, so 42.45 K.
Heat transferred: Q = 350 W per m2 per K x 25 m2 x 42.4509 K = 371,445.7 W, so 371.45 kW.
In BTU per hour: 371,445.7 W x 3.412141633 = 1,267,465 BTU/h.
In imperial units the same exchanger has U = 61.64 BTU per hour per ft2 per °F, an area of 269.10 ft2 and an LMTD of 76.41 °F, which gives the same 1,267,465 BTU/h.
The tool shows 45.00 K, 40.00 K, 42.45 K and 371.45 kW.

Reference table

Log mean temperature difference and heat transferred for different flow arrangements and temperature splits, at U = 350 W per m2 per K and 25 m2
Flow and temperatures	End differences	LMTD	Heat transferred
Counter flow, hot 95 to 60 °C, cold 20 to 50 °C	45 K and 40 K	42.45 K	371.45 kW
Parallel flow, hot 95 to 60 °C, cold 20 to 50 °C	75 K and 10 K	32.26 K	282.27 kW
Counter flow, hot 80 to 60 °C, cold 20 to 50 °C	30 K and 40 K	34.76 K	304.16 kW
Counter flow, hot 120 to 80 °C, cold 20 to 100 °C	20 K and 60 K	36.41 K	318.58 kW

Limits

No correction factor is applied, so the counter flow figure is the best case for the temperatures given rather than the answer for a real shell and tube pass arrangement.
The duty is calculated from the geometry, so the four temperatures have to be consistent with the flow rates and the specific heats of the two streams.
The overall heat transfer coefficient is treated as constant over the exchanger, which ignores the fact that it changes with temperature and with fouling.
Heat lost to the surroundings is not accounted for, so a real exchanger moves slightly less than the calculation shows.

Common mistakes

Using the plain average of the two end differences instead of the logarithmic mean. For the worked example the plain average is 42.50 K and the log mean is 42.45 K, and the gap grows as the differences spread apart.
Mixing up the end differences for parallel flow and counter flow. In counter flow the cold outlet pairs with the hot inlet, in parallel flow the inlets pair together.
Reading the LMTD in degrees Fahrenheit as if it were the same number as in kelvin. A temperature difference of 42.45 K is 76.41 °F.
Forgetting that the hot fluid has to enter hotter than it leaves and the cold fluid has to leave hotter than it enters, otherwise the direction of the heat flow is wrong.
Forgetting the correction factor for a real shell and tube design, which makes the calculated area too small.

Try it yourself

Quick checks. Answers sit tucked under each question.

What is the LMTD for the counter flow example?

42.45 K, from (45 - 40) divided by the natural log of 45 over 40, which is 5 / 0.117783 = 42.4509 K.

What happens if the same four temperatures are arranged in parallel flow?

The end differences become 75 K and 10 K, so the LMTD falls to 32.26 K and the exchanger moves 282.27 kW instead of 371.45 kW.

How much heat does the example exchanger move?

371.45 kW, from 350 W per m2 per K times 25 m2 times 42.4509 K, which is 371,445.7 W or 1,267,465 BTU per hour.

Questions

What is the log mean temperature difference?

The average temperature difference between two fluids across a heat exchanger, weighted so that the wide end counts more. It replaces the plain arithmetic average because heat flows faster where the difference is larger.

Why does counter flow perform better?

Because the cold outlet meets the hot inlet, so the difference stays large across the whole exchanger. With the same inlet and outlet temperatures in parallel flow the two ends nearly equalise and the LMTD drops.

What is the correction factor F?

A factor below one that accounts for a flow pattern that is not pure counter flow, such as a shell and tube exchanger with one shell and two tube passes. It is read from a chart or an equation for the geometry, and the Purdue notes show a worked example with F = 0.87.

What is the overall heat transfer coefficient?

A single number that lumps together the conductivity of the wall, the films on both sides of the surface and the fouling on it. Typical values run from a few hundred for a shell and tube exchanger up to a few thousand for a plate exchanger.

How do I size a heat exchanger?

Work out the duty from the mass flow rates and the temperature change, then set A = Q divided by U F LMTD. This page covers the second half of that job, so you need the duty from an energy balance first.

Why do my numbers differ slightly from a textbook?

Most textbooks round the constants and apply a correction factor for the pass arrangement, and some use the plain average rather than the logarithmic mean. Those differences are small, but they account for a percent or so.

Sources

How this page was checked. The formula above was matched against the cited sources, and the worked example was run through the tool on this page before publishing. If you spot an error, contact us so we can fix it and note the correction.