LMTD Calculator

Log Mean Temperature Difference for heat exchangers, with full step-by-step working

°C
°C
°C
°C
LMTD:
--
--

What is LMTD?

The Log Mean Temperature Difference (LMTD) is the correct way to average the temperature difference between two fluids exchanging heat along the length of a heat exchanger, since that difference changes continuously from one end to the other — a simple arithmetic average would overstate it.

LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂)

where ΔT₁ and ΔT₂ are the temperature differences between the hot and cold streams at each end of the exchanger. Which end is which depends on the flow arrangement:

ArrangementΔT₁ΔT₂
Counter-currentTh,in − Tc,outTh,out − Tc,in
Co-current (parallel)Th,in − Tc,inTh,out − Tc,out

Why It Matters

LMTD is used directly in the core heat exchanger design equation: Q = U × A × LMTD, where Q is heat duty, U is the overall heat transfer coefficient, and A is heat transfer area. Getting LMTD wrong means the calculated exchanger area (and therefore cost and physical size) is wrong too.

Counter-Current vs Co-Current

For the same inlet and outlet temperatures, counter-current flow always gives an LMTD equal to or greater than co-current flow — which is why counter-current arrangements are preferred whenever possible; they need a smaller heat transfer area for the same duty.

Solved Examples (Practice Problems)

Click "Try This Example" to auto-fill the calculator above and see the full step-by-step working.

Example 1 — Counter-current heat exchanger

Hot oil enters at 150°C and leaves at 90°C. Cold water enters at 30°C and leaves at 80°C, in a counter-current arrangement. Find the LMTD.

Example 2 — Co-current (parallel flow) heat exchanger

The same hot oil (150°C → 90°C) and cold water (30°C → 80°C), but now in a co-current (parallel flow) arrangement. Find the LMTD and compare it to Example 1.

Example 3 — Steam condenser (counter-current)

Steam condenses at a near-constant 120°C. Cooling water enters at 25°C and leaves at 60°C, counter-current arrangement. Find the LMTD.

Worked Solutions in Full

The three examples above, solved in full. Example 2 also shows why the arithmetic average of the two end temperature differences cannot be used.

Example 1 — Counter-current

Given: Hot oil 150 → 90 °C, cold water 30 → 80 °C

Step 1 — Pair the ends for counter-current flow: The hot inlet meets the cold outlet at one end, and the hot outlet meets the cold inlet at the other.

Step 2 — ΔT₁: 150 − 80 = 70 °C

Step 3 — ΔT₂: 90 − 30 = 60 °C

Step 4 — Apply the formula: LMTD = (70 − 60) / ln(70/60) = 10 / 0.1542 = 64.87 °C

Answer: LMTD ≈ 64.87 °C

Example 2 — Same temperatures, co-current

Given: Hot oil 150 → 90 °C, cold water 30 → 80 °C, but now flowing in the same direction

Step 1 — Pair the ends for co-current flow: Both inlets meet at one end, and both outlets meet at the other.

Step 2 — ΔT₁: 150 − 30 = 120 °C

Step 3 — ΔT₂: 90 − 80 = 10 °C

Step 4 — Apply the formula: LMTD = (120 − 10) / ln(120/10) = 110 / 2.4849 = 44.27 °C

Answer: LMTD ≈ 44.27 °C. Counter-current gave 64.87 °C from identical temperatures, so it needs less area for the same duty.

Example 3 — Steam condenser

Given: Steam condensing at 120 °C, cooling water 25 → 60 °C, counter-current

Step 1 — ΔT₁: 120 − 60 = 60 °C

Step 2 — ΔT₂: 120 − 25 = 95 °C

Step 3 — Apply the formula: LMTD = (60 − 95) / ln(60/95) = -35 / -0.4595 = 76.16 °C

Answer: LMTD ≈ 76.16 °C. Because the steam temperature is constant, counter-current and co-current give the same result here.

Why not just average the two temperature differences?

CaseArithmetic meanLMTDArithmetic mean too high by
Example 1 (ratio 1.17)65.00 °C64.87 °C0.2%
Example 2 (ratio 12)65.00 °C44.27 °C46.8%
Example 3 (ratio 1.58)77.50 °C76.16 °C1.8%

When the two end differences are close, the arithmetic mean is nearly right. When they differ a lot, as in Example 2, it overestimates the driving force badly, and an exchanger sized from it would be far too small.

Practical Notes and Common Mistakes

LMTD is the driving-force term in the basic exchanger equation Q = U × A × LMTD. It is used for double-pipe and shell-and-tube exchangers, condensers, coolers and heaters. Once you have it, the Heat Exchanger Area Calculator converts it into the surface area you need.

Mistakes that give wrong answers

Reading the result

A larger LMTD means a stronger driving force and therefore a smaller, cheaper exchanger. A small LMTD (a close temperature approach) means a large exchanger, which is why very tight approach temperatures are expensive.

Frequently Asked Questions

Why not just use the arithmetic mean temperature difference instead?

Because the temperature difference between the two streams doesn't change linearly along the exchanger length — it decays exponentially. Using a simple arithmetic mean overestimates the true driving force, which leads to an undersized heat exchanger. LMTD accounts for this correctly.

What if ΔT₁ equals ΔT₂?

When ΔT₁ and ΔT₂ are equal (or very close), the LMTD formula divides zero by zero (ln(1) = 0), which is mathematically undefined. In that specific case, the LMTD simply equals ΔT₁ (or ΔT₂, since they're the same) — this calculator detects that case and handles it automatically.

What does a negative or undefined result mean?

If ΔT₁ or ΔT₂ comes out zero or negative, it usually means a "temperature cross" — the cold outlet temperature is higher than the hot outlet temperature in an arrangement where that shouldn't happen. Double-check which temperature is inlet vs outlet, and which flow arrangement you actually have.

Does LMTD work for exchangers with more than one shell or tube pass?

Not directly. Multi-pass and cross-flow exchangers need an additional correction factor (F), applied as LMTD × F, because the flow isn't purely counter-current or co-current. This calculator gives the true counter-current/co-current LMTD; the correction factor is a separate step.

Related Tool

Already have your LMTD? Use it directly to size the exchanger with the Heat Exchanger Area Calculator — solves Q = U × A × LMTD for the required area.