About the LMTD Calculator
This LMTD calculator works out the log mean temperature difference between the hot and cold streams of a heat exchanger from their four terminal temperatures. Choose counterflow or parallel (co-current) flow, enter the inlet and outlet temperatures of each fluid, and it returns the two end temperature differences, the LMTD and the simpler arithmetic mean for comparison.
Add the heat duty, an overall heat transfer coefficient U and, for shell-and-tube or cross-flow units, a correction factor F, and the calculator also sizes the required heat transfer area from Q = U·A·F·LMTD. It works in metric (°C, kW, W/m²·K, m²) or US units (°F, Btu/h, Btu/h·ft²·°F, ft²), which suits process, HVAC and mechanical engineers as well as students learning heat exchanger design.
The LMTD method assumes steady flow, constant specific heats and a constant U along the exchanger, with no heat loss to the surroundings. If the temperatures cross — the cold outlet hotter than the hot outlet in parallel flow, or an end difference of zero or less — the arrangement cannot achieve that duty and the calculator says so.
With the default inputs, the log mean temperature difference is 69.52 °. Change any value above to recalculate instantly.
How to use the lmtd calculator
- 1Choose metric or US units and the flow arrangement.
- 2Enter the hot fluid inlet and outlet temperatures.
- 3Enter the cold fluid inlet and outlet temperatures.
- 4Optionally enter the heat duty, U value and correction factor F.
- 5Read the LMTD and the required heat transfer area.
Formula and method
The temperature difference between the two fluids changes along a heat exchanger, and with constant U and specific heats it varies exponentially with heat transfer area. Integrating the local heat flux over the area gives the logarithmic mean of the two end differences, which is the single driving temperature difference that makes Q = U·A·LMTD exact. When ΔT₁ equals ΔT₂ the LMTD simply equals that difference.
For multi-pass shell-and-tube and cross-flow exchangers the true mean difference is smaller than the counterflow LMTD, so it is multiplied by a correction factor F read from charts using the P and R ratios. Area is Q ÷ (U·F·LMTD); in metric units Q in kW is converted to watts, while in US units Btu/h, Btu/h·ft²·°F and °F are already consistent. Temperature differences are the same in °C and K.
- ΔT₁, ΔT₂
- Temperature differences at the two ends of the exchanger
- LMTD
- Log mean temperature difference
- Q
- Heat duty (kW or Btu/h)
- U
- Overall heat transfer coefficient
- F
- LMTD correction factor for multi-pass or cross-flow
- A
- Required heat transfer area
Worked examples
Counterflow: hot 150→90 °C, cold 30→70 °C, 500 kW
In counterflow ΔT₁ = 150 − 70 = 80 °C and ΔT₂ = 90 − 30 = 60 °C, so LMTD = 20 ÷ ln(80/60) ≈ 69.52 °C. Transferring 500 kW with U = 800 W/m²·K needs 500,000 ÷ (800 × 69.52) ≈ 8.99 m².
Same temperatures in parallel flow
In parallel flow the ends see 150 − 30 = 120 °C and 90 − 70 = 20 °C, giving LMTD = 100 ÷ ln 6 ≈ 55.81 °C. The same duty now needs about 11.2 m² — 25% more surface than counterflow.
US units, shell-and-tube with F = 0.9
ΔT₁ = 200 − 100 = 100 °F and ΔT₂ = 120 − 60 = 60 °F, so LMTD = 40 ÷ ln(100/60) ≈ 78.30 °F. With F = 0.9 the effective difference is 70.47 °F, and 250,000 Btu/h at U = 50 needs about 70.9 ft².
Frequently asked questions
What is LMTD?+
The log mean temperature difference is the effective average temperature difference driving heat transfer in a heat exchanger. It is (ΔT₁ − ΔT₂) ÷ ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the hot-to-cold differences at each end.
Why use the log mean instead of the arithmetic mean?+
The temperature difference falls exponentially along the exchanger, not linearly, so the arithmetic mean overstates the driving force. The error is small when the end differences are close but grows quickly when one end difference is much smaller than the other.
Is counterflow better than parallel flow?+
For the same terminal temperatures, counterflow always gives a higher LMTD and so needs less area. It also allows the cold fluid to leave hotter than the hot fluid’s outlet, which parallel flow can never achieve.
What is the LMTD correction factor F?+
F adjusts the counterflow LMTD for shell-and-tube exchangers with multiple passes and for cross-flow units, where the flow is not purely counter-current. It is read from charts using the temperature ratios P and R and is usually kept above about 0.75 for a good design.
Do I use °C or K for LMTD?+
Either. LMTD is a temperature difference, and a difference of 1 °C equals a difference of 1 K. In US units use °F, which matches U values in Btu/h·ft²·°F.