LMTD Heat Exchanger Calculator
Calculate the log mean temperature difference (LMTD) of a counter-flow or parallel-flow heat exchanger from the inlet and outlet temperatures of both fluids.
Last updated: October 2026
Log mean temperature difference (LMTD)
69.52 °C
Formula below · 3 sources (web.mit.edu, engineeringlibrary.org, workforce.libretexts.org) · Updated Oct 2026
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About this calculator
The log mean temperature difference (LMTD) is the average temperature difference that drives heat from the hot fluid to the cold fluid along a heat exchanger. It goes into the design equation Q = U × A × LMTD, where U is the overall heat-transfer coefficient and A the heat-transfer area.
The formula
LMTD = (ΔT₁ − ΔT₂) ÷ ln(ΔT₁ ÷ ΔT₂), where ΔT₁ and ΔT₂ are the temperature differences between the two fluids at the two ends of the exchanger. • Counter-flow: ΔT₁ = hot inlet − cold outlet, ΔT₂ = hot outlet − cold inlet • Parallel flow: ΔT₁ = hot inlet − cold inlet, ΔT₂ = hot outlet − cold outlet When ΔT₁ = ΔT₂ the formula becomes 0 ÷ 0, and the LMTD simply equals that constant difference.
Example (the defaults): hot fluid 150 → 90 °C, cold fluid 30 → 70 °C. Counter-flow: ΔT₁ = 80, ΔT₂ = 60, LMTD = 20 ÷ ln(1.333) = 69.52 °C. The same temperatures in parallel flow give ΔT₁ = 120, ΔT₂ = 20 and an LMTD of 55.81 °C.
Why counter-flow is usually preferred
The DOE heat-transfer handbook notes that counter-flow keeps a more uniform temperature difference between the fluids, lets the cold outlet approach the hot inlet temperature, and needs less surface area for the same duty, because its mean temperature difference is higher. In parallel flow the cold outlet can never exceed the hot outlet; the calculator flags temperatures that would require that.
Multi-pass and cross-flow exchangers
Shell-and-tube exchangers with several tube passes, and cross-flow units, have mixed flow patterns. Their effective driving force is F × LMTD(counter-flow), where the correction factor F (at most 1) is read from charts for the geometry. Use the counter-flow result here and multiply by F. The method assumes constant specific heats and U; for condensers, boilers or strongly varying properties, split the exchanger into zones or use the effectiveness–NTU method.
Temperature differences in °C and K are the same size, so the result is valid in kelvin too; for °F inputs the LMTD is in °F.
How to use
Enter the inlet and outlet temperatures of both fluids and choose the flow arrangement. The defaults (hot 150 → 90 °C, cold 30 → 70 °C, counter-flow) give an LMTD of 69.52 °C; switch to parallel flow and it drops to 55.81 °C. To size an exchanger, divide the heat duty by U × LMTD: 200 kW with U = 500 W/m²·K needs 200,000 ÷ (500 × 69.52) = 5.75 m² of area in counter-flow.
Frequently asked questions
How do you calculate LMTD?
Find the temperature difference between the fluids at each end, ΔT₁ and ΔT₂, then compute (ΔT₁ − ΔT₂) ÷ ln(ΔT₁ ÷ ΔT₂). For end differences of 80 °C and 60 °C the LMTD is 69.52 °C.
Why use a log mean instead of an arithmetic mean?
The temperature difference changes exponentially, not linearly, along the exchanger when U and the fluid heat capacities are constant. Integrating the local heat flow gives the logarithmic mean. The arithmetic mean overstates the driving force, slightly when the end differences are similar and a lot when they are not.
Why is counter-flow LMTD higher than parallel flow?
In counter-flow the hottest hot fluid meets the hottest cold fluid, so the difference stays more even along the length. In parallel flow the difference starts large and collapses toward the outlet, lowering the average.
What is the F correction factor?
For multi-pass shell-and-tube and cross-flow exchangers, the true mean driving force is F × the counter-flow LMTD. F depends on the geometry and two temperature ratios and is read from published charts; values well below 0.8 usually mean the configuration should be changed.
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Sources & references
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