LMTD Heat Exchanger Calculator
Compute the log mean temperature difference (LMTD) — the effective driving force for heat transfer in a heat exchanger, given the four end-of-fluid temperatures. The standard input to the design equation Q = U·A·LMTD for sizing or rating exchangers.
Last updated: May 2026
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About this calculator
In a heat exchanger, the temperature difference between hot and cold streams changes along the exchanger's length. The log mean temperature difference (LMTD) is the appropriately weighted average that lets you use the simple design equation Q = U · A · LMTD, where Q is total heat duty, U is the overall heat transfer coefficient, and A is heat transfer area. For two end temperature differences ΔT₁ and ΔT₂ (defined consistently for either counter-current or co-current flow), LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂). When ΔT₁ = ΔT₂ the formula is undefined as written (0/0), and LMTD just equals that common value. This calculator uses the counter-current convention: ΔT₁ = t_hot_in − t_cold_out (the "hot end" of the exchanger) and ΔT₂ = t_hot_out − t_cold_in (the "cold end"). Variables: t_hot_in, t_hot_out are the hot fluid inlet and outlet temperatures; t_cold_in, t_cold_out are the cold fluid inlet and outlet temperatures, all in consistent units (°C or K — LMTD is a difference so the choice doesn't matter as long as it's consistent). Edge cases: temperature crossover (cold outlet hotter than hot outlet) is physically impossible in a simple co-current exchanger but allowed in counter-current; if ΔT₁ and ΔT₂ have opposite signs the formula breaks down and the geometry needs reconsideration. For multi-pass or cross-flow exchangers, multiply LMTD by a correction factor F (typically 0.8–1.0) read from charts that depend on the inlet/outlet temperatures of both fluids; F = 1 for pure counter-current. Counter-current always gives higher LMTD than co-current for the same end temperatures, which is why most exchangers are designed counter-current — it maximises driving force and minimises area.
How to use
Example 1 — Counter-current water-water exchanger. Hot water enters at 120 °C and leaves at 80 °C; cold water enters at 25 °C and leaves at 65 °C. Enter t_hot_in = 120, t_hot_out = 80, t_cold_in = 25, t_cold_out = 65. ΔT₁ = 120 − 65 = 55 °C; ΔT₂ = 80 − 25 = 55 °C. Since ΔT₁ = ΔT₂, LMTD = 55 °C. ✓ When the two end ΔTs are equal (constant ΔT along the exchanger), LMTD just equals that constant value — a degenerate case but mathematically clean. Example 2 — Standard unequal ΔT case. Hot fluid 150 → 90 °C; cold fluid 30 → 70 °C. Enter 150, 90, 30, 70. ΔT₁ = 150 − 70 = 80 °C; ΔT₂ = 90 − 30 = 60 °C. LMTD = (80 − 60) / ln(80/60) = 20 / ln(1.333) = 20 / 0.2877 ≈ 69.5 °C. ✓ The LMTD lies between the two end ΔTs (60 and 80) but closer to the geometric mean (√(80·60) ≈ 69.3) than the arithmetic mean (70). The log-mean weights toward the smaller end ΔT — appropriate because that's where the driving force is lowest and the most exchanger area is needed.
Frequently asked questions
Why log mean and not arithmetic mean?
The temperature difference along the exchanger doesn't change linearly with area — it changes exponentially, because the heat transfer rate at any point is proportional to the local ΔT. The differential equation dQ = U·dA·ΔT(x), combined with the fact that ΔT(x) decays exponentially in counter-current and co-current exchangers (for constant U and constant heat capacity flows), yields the log mean as the exact average that lets you replace the integral with Q = U·A·LMTD. The arithmetic mean would overstate the driving force; the geometric mean would understate it slightly. For most engineering ΔT ratios (1.5 < ΔT₁/ΔT₂ < 4) the LMTD is between the arithmetic and geometric mean, closer to the geometric. When ΔT₁/ΔT₂ approaches 1, all three means converge.
What's the difference between counter-current and co-current flow?
In counter-current flow, the two streams flow in opposite directions — hot in and cold out at one end, cold in and hot out at the other. This maintains a relatively uniform ΔT along the entire length, maximising LMTD. In co-current (parallel) flow, both streams flow in the same direction — hot in and cold in at one end, both out at the other. The ΔT starts very large at the inlet end and approaches some minimum (the streams can never crossover) at the outlet end, giving a smaller LMTD for the same end temperatures. Counter-current is almost always preferred in design — for the same end temperatures it requires less area and can achieve closer approach (cold outlet can exceed hot outlet temperature in counter-current, impossible in co-current). Co-current is occasionally used when the higher initial driving force is desirable (rapid initial heat-up) or when fluid mechanics requires it.
What is the F-correction factor and when do I need it?
The simple LMTD formula assumes pure counter-current (or co-current) flow with no cross-flow components. Real exchangers — shell-and-tube with multiple tube passes, cross-flow with one or both fluids mixed/unmixed — don't match these idealisations exactly. The F-correction factor accounts for the geometric difference: Q = U·A·F·LMTD, where F < 1 reflects the loss of driving force compared to pure counter-current. F depends on the temperature parameters P = (t_cold_out − t_cold_in)/(t_hot_in − t_cold_in) and R = (t_hot_in − t_hot_out)/(t_cold_out − t_cold_in), and is read from published charts for each exchanger geometry. Typical F values: 0.85–0.95 for shell-and-tube with 1 shell pass and 2 tube passes; 0.75–0.85 for cross-flow; below 0.75 is poor design — increase the number of shells or change geometry. For pure counter-current, F = 1.
What are the most common mistakes computing LMTD?
The first is mixing up which end ΔT is which — for counter-current, ΔT₁ pairs hot inlet with cold outlet, ΔT₂ pairs hot outlet with cold inlet. Reversing them produces an undefined or wrongly-signed LMTD (though the symmetry of the formula often forgives this). The second is using the simple LMTD for non-counter-current geometries without applying the F-correction factor; for shell-and-tube exchangers this can overstate the driving force by 10–25%. The third is mixing units — Celsius and Kelvin produce the same ΔT, but mixing °F into a metric calculation is a recipe for confusion. The fourth is forgetting that LMTD assumes constant fluid properties and constant U along the exchanger; when properties change significantly (large temperature swings, phase changes), divide the exchanger into segments and compute LMTD for each. The fifth is misinterpreting LMTD = ΔT₁ = ΔT₂ as an error; this is the legitimate degenerate case of constant ΔT throughout the exchanger.
When should I not use this calculator?
Skip it for non-counter-current geometries without an F-correction factor; cross-flow, multi-pass shell-and-tube, and complex flow patterns all need geometry-specific corrections. Don't use it for exchangers involving phase change (condensers, evaporators, reboilers) without recognising that the constant-temperature side gives ΔT_LMTD = (T_hot − T_sat) − (T_cold − T_sat) / ln(...) but with one stream at constant temperature it simplifies; even so, segment-by-segment analysis is often needed when superheat or subcooling is involved. It's the wrong tool for exchangers with significantly varying fluid properties (high-viscosity oils, supercritical fluids near the critical point) where the constant-U, constant-Cp assumption fails. Avoid it for highly fouled exchangers where the local U varies dramatically along the length. Finally, for design from scratch, the LMTD method assumes you know both inlet and both outlet temperatures; if some outlets are unknown, use the ε-NTU (effectiveness-NTU) method which works directly from inlets and exchanger size.