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LMTD Calculator for Heat Exchangers (Counter-flow / Parallel Flow)

Enter the flows, specific heats and temperatures of both streams to get the heat duty, cold outlet temperature and log-mean temperature difference ΔTlm; add the overall coefficient U to get the required area. Defaults are the benchmark from Incropera Example 11.1.

Inputs

Units
300 W/(m²·K) is only an example; enter your own value, or leave empty to skip the area

Results

Log-mean temperature difference ΔTlm43.2 K
Heat duty Q
8.524 kW
Cold outlet temperature
40.2 °C
Terminal difference ΔT₁
59.8 K
Terminal difference ΔT₂
30 K
UA = Q / ΔTlm
197.3 W/K
Required area A
0.6577 m²
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Results are for reference; check them against the applicable engineering codes. Everything is calculated in your browser and your inputs are not uploaded.

Formula and basis

  • Q = ṁₕ · cₚ,ₕ · (Tₕ,ᵢ − Tₕ,ₒ), T_c,o = T_c,i + Q / (ṁ_c · cₚ,c)
  • ΔTlm = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂)
  • Counter-flow: ΔT₁ = Tₕ,ᵢ − T_c,o, ΔT₂ = Tₕ,ₒ − T_c,i
  • Parallel flow: ΔT₁ = Tₕ,ᵢ − T_c,i, ΔT₂ = Tₕ,ₒ − T_c,o
  • A = Q / (U · ΔTlm)

Standards and sources

  • Incropera, DeWitt, Bergman, Lavine, Fundamentals of Heat and Mass Transfer (6th ed.), Eq. 11.15 and Example 11.1 (counter-flow concentric tube exchanger)
  • Multi-pass and cross-flow exchangers need a correction factor F on the LMTD (TEMA, Bowman et al.); this tool covers pure counter-flow and parallel flow only

Worked example (industry benchmark)

Counter-flow double-pipe exchanger: oil at 0.1 kg/s cooled from 100 °C to 60 °C by 0.2 kg/s of water entering at 30 °C

Inputs: Hot (oil) 0.1 kg/s, cₚ 2131 J/(kg·K), 100 → 60 °C; cold (water) 0.2 kg/s, cₚ 4178 J/(kg·K), inlet 30 °C; counter-flow

Log-mean temperature difference ΔTlm43.2 K
Heat duty Q8.524 kW
Cold outlet temperature40.2 °C
Terminal difference ΔT₁59.8 K
Terminal difference ΔT₂30 K
UA = Q / ΔTlm197.3 W/K

The example matches industry benchmark GC-IND-HX-LMTD-001 in our repository (expected values derived independently with 50-digit decimals and checked by automated tests).

FAQ

Why the log-mean and not the arithmetic mean temperature difference?

The temperature difference varies exponentially along the surface; integrating it gives the log-mean. When both terminal differences are close the two are similar, but the arithmetic mean becomes increasingly wrong as they diverge.

Counter-flow or parallel flow?

For the same terminal temperatures counter-flow gives a larger LMTD and needs less area, and the cold outlet can exceed the hot outlet. In parallel flow the cold outlet can never exceed the hot outlet.

What is a temperature cross?

A terminal difference of zero or less, so heat cannot flow from the hot to the cold stream. The page tells you when it happens; switch to counter-flow or change the flows or temperatures.

How do I handle multi-pass shell-and-tube exchangers?

Compute the counter-flow LMTD and multiply by the correction factor F from TEMA charts or equations. This tool does not include F.

What value of U should I use?

It depends on the fluids, velocities and fouling and should come from heat transfer correlations or handbooks. The 300 W/(m²·K) default is only an example.

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