Hong Life CalculatorsUnderstanding the LMTD heat exchanger calculator — concept and principle
The log mean temperature difference (LMTD) is the effective average temperature difference that drives heat between the hot and cold streams of a heat exchanger. Engineers use it to size exchangers: once the LMTD and the overall heat transfer coefficient U are known, the required area follows from the heat duty.
How it is calculated
LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂), and Q = U · A · LMTD, so A = Q / (U · LMTD).
- Counter-flow (streams run in opposite directions): ΔT₁ = T_hot,in − T_cold,out and ΔT₂ = T_hot,out − T_cold,in.
- Parallel flow (same direction): ΔT₁ = T_hot,in − T_cold,in and ΔT₂ = T_hot,out − T_cold,out.
- U (W/m²K) is the overall coefficient including both film resistances, the wall and fouling.
If ΔT₁ = ΔT₂ the formula becomes 0/0; the limit is simply that common value, which the calculator uses. The second output, U × LMTD, is the heat transferred per square metre of area.
The principle
Along the exchanger the temperature gap between the streams changes continuously, so the local heat flux changes too. Assuming constant U and constant specific heats, each stream's temperature varies such that the gap changes exponentially with area. Integrating the local heat flux over the whole area gives exactly the logarithmic mean of the end differences, which is always smaller than the arithmetic mean.
Worked example
- Inputs: hot 90 → 55 °C, cold 20 → 50 °C, U = 500 W/m²K, counter-flow.
- ΔT₁ = 90 − 50 = 40 K; ΔT₂ = 55 − 20 = 35 K.
- LMTD = (40 − 35) ÷ ln(40 ÷ 35) = 5 ÷ 0.133531 = 37.44437845 K, a little below the arithmetic mean of 37.5 K.
- Heat per m²: 500 × 37.444 = 18,722.18922 W/m².
- For a duty of, say, 100 kW, the area would be 100,000 / 18,722 ≈ 5.34 m².
| Cold in (°C) | LMTD (K) |
|---|---|
| 5 | 44.81420118 |
| 10 | 42.45093508 |
| 15 | 40 |
| 20 | 37.44437845 |
| 25 | 34.76059497 |
| 30 | 31.91464718 |
| 40 | 25.4886362 |
How to read the result
A higher LMTD means a smaller exchanger for the same duty. With the same terminal temperatures, parallel flow gives ΔT₁ = 70 K and ΔT₂ = 5 K, so LMTD ≈ 24.63 K, which needs about 52% more area than counter-flow. In parallel flow the cold outlet can never exceed the hot outlet, whereas counter-flow allows the cold stream to leave hotter than the hot stream leaves.
Common mistakes
- Pairing the wrong terminal temperatures for the flow arrangement.
- Using LMTD directly for shell-and-tube or cross-flow units without the correction factor F (Q = U·A·F·LMTD).
- Entering temperatures that cross (a negative ΔT), which is physically impossible for that arrangement.
Frequently asked questions
When should I use ε-NTU instead of LMTD?
When the outlet temperatures are unknown. LMTD is easiest when all four temperatures are known and you need the area.
Why not simply use the arithmetic mean?
It overestimates the driving force when the end differences are very different, leading to an undersized exchanger.
What is a typical U?
It depends strongly on fluids and fouling, from tens of W/m²K for gas-to-gas units to a few thousand for water-to-water or condensing service.
Sources: Incropera, DeWitt, Bergman and Lavine, Fundamentals of Heat and Mass Transfer, Wiley · Kern, Process Heat Transfer, McGraw-Hill