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Heat Exchanger LMTD

Log mean temperature difference used as the effective driving force for heat transfer along a heat exchanger.

Chemical EngineeringHeat TransferProcess Design

Log Mean Temperature Difference CalculatorLMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)

LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂)
ΔT₁ = hot fluid temperature difference  ·  ΔT₂ = cold fluid temperature difference
⟹ LMTDΔT₁, ΔT₂
°C
°C
°C
Common:
Solve for:
LMTD
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LMTD Level
Low (<20°C) Medium (20–50°C) High (50–80°C) Very High (>80°C)
LMTD vs. ΔT₁fixed ΔT₂
LMTD(ΔT₁) for fixed ΔT₂ Computed point
LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂)  ·  valid for ΔT₁ ≠ ΔT₂; if equal, LMTD = ΔT₁

Interpretation

LMTD = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂) for heat exchangers. It gives the average temperature driving force. Example: ΔT₁=60, ΔT₂=20 → LMTD = 40/ln3 ≈ 36.4°C.

LMTD = (dT1 - dT2) / ln(dT1/dT2)
Heat Exchanger LMTD

Variables

SymbolQuantityUnit
LMTDLog mean temperature difference°C
dT1Temperature difference at one end°C
dT2Temperature difference at the other end°C

What it means

The Log Mean Temperature Difference (LMTD) is a key parameter in heat exchanger design. It represents the effective average temperature difference between the hot and cold fluids over the length of the exchanger, assuming constant specific heats and overall heat transfer coefficient. The formula is LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁ / ΔT₂), where ΔT₁ is the temperature difference at one end and ΔT₂ at the other. For counter‑current or co‑current flow, the LMTD is used in the heat exchanger equation Q = U A LMTD to calculate the required heat transfer area A. It is applicable to single‑pass exchangers with no phase change. For more complex configurations (multi‑pass, cross‑flow), a correction factor is applied. The LMTD is a fundamental concept in thermal engineering and is taught in all heat transfer courses. It allows engineers to size exchangers for a given heat duty and to analyse existing equipment performance.

Worked example

Heat Exchanger LMTD – Two Examples

Real‑World
Scenario: Hot fluid cools from 100 °C to 60 °C (ΔT₁=40), cold fluid warms from 20 °C to 50 °C (ΔT₂=10). Find LMTD.
ParameterValue
ΔT₁40 °C
ΔT₂10 °C
1LMTD = (40-10)/ln(40/10) = 30/ln(4) ≈ 21.6 °C
Result ≈ 21.6 °C ✓ Effective ΔT
Scenario: ΔT₁ = 50 °C, ΔT₂ = 5 °C. Calculate LMTD.
ParameterValue
ΔT₁50 °C
ΔT₂5 °C
1LMTD = (50-5)/ln(50/5) = 45/ln(10) ≈ 19.5 °C
Result ≈ 19.5 °C ✓ Lower driving force
Key insight: LMTD is the effective temperature difference for heat exchangers.

Common mistakes

  • Temperature differences: ΔT₁ and ΔT₂ are the terminal temperature differences at the two ends of the exchanger (hot‑cold). Ensure they are positive.
  • Logarithm: Use natural log (ln), not log₁₀.
  • Counter‑flow vs. parallel‑flow: LMTD applies to both configurations, but the expression for ΔT₁ and ΔT₂ differs; for counter‑flow, the larger ΔT is at one end.
  • Assumptions: LMTD assumes constant specific heats, no phase change, and no heat losses; for variable properties, use the ε‑NTU method.
  • Units: Temperature differences can be in °C or K (same magnitude).

Applications

The log‑mean temperature difference (LMTD) is the driving force for heat transfer in heat exchangers with constant wall temperature or co‑current/counter‑current flow. It is calculated as LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂). This value is used in the equation Q = U·A·LMTD to size heat exchangers. Engineers apply LMTD in the design of shell‑and‑tube, plate, and air‑cooled heat exchangers, ensuring that the required heat duty is achieved with a reasonable area. The LMTD accounts for the changing temperature difference along the exchanger length. For complex flow arrangements (e.g., cross‑flow), correction factors are applied. By correctly computing LMTD, engineers can select appropriate exchanger geometry and thermal performance, balancing capital cost against energy efficiency.

  • Sizing of shell‑and‑tube and plate heat exchangers
  • Design of condensers and reboilers
  • Performance analysis of existing heat exchangers
  • Optimisation of heat recovery networks
  • Thermal design of HVAC systems and refrigeration units

Frequently Asked Questions

Q01What is the Log Mean Temperature Difference (LMTD) and why is it used?
A01

The LMTD is the effective average temperature difference driving heat transfer in a heat exchanger. It is defined as LMTD = (ΔT₁ – ΔT₂) / ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the temperature differences at the two ends of the exchanger. It accounts for the fact that the temperature difference varies along the length.

Q02What are the common mistakes when using the LMTD?
A02

  • Using the arithmetic mean instead of the log mean – this overestimates the driving force.
  • Applying the LMTD to a cross‑flow or multi‑pass exchanger without the appropriate correction factor (F).
  • Using the wrong temperature values (e.g., mixing up hot and cold streams).
  • Forgetting to check if the flow is counter‑current or parallel.

Q03How do you calculate the LMTD for a counter‑current heat exchanger?
A03

For counter‑current flow, ΔT₁ = T_h,in – T_c,out (hot inlet minus cold outlet), and ΔT₂ = T_h,out – T_c,in (hot outlet minus cold inlet). Then apply the LMTD formula. For parallel flow, both differences are at the same ends.

Q04What is the correction factor F for LMTD?
A04

For cross‑flow or multi‑pass exchangers, the true mean temperature difference is LMTD × F, where F is a correction factor (≤ 1). F depends on the P‑R diagram (temperature effectiveness and heat capacity rate ratio). It is used to account for the non‑ideal flow pattern.

Q05What are the assumptions of the LMTD method?
A05

  • Steady state.
  • Constant specific heats.
  • No heat losses to the surroundings.
  • No phase change.
  • Overall heat transfer coefficient U is constant.

Q06How does the LMTD relate to the total heat transfer rate?
A06

The heat transfer rate is Q = U·A·LMTD, where U is the overall heat transfer coefficient and A is the heat transfer area. This is the fundamental equation for heat exchanger design.

Q07What is the LMTD for a condenser or evaporator?
A07

In a condenser or evaporator, one stream is isothermal (phase change). The LMTD simplifies to ΔT₁ – ΔT₂ divided by ln(ΔT₁/ΔT₂), where one ΔT is zero? Actually, if one stream is isothermal, the LMTD still works; it is often used with the appropriate temperature differences.

Q08How do you choose between the LMTD method and the NTU method?
A08

The LMTD method is used when the inlet and outlet temperatures are known (rating problems). The NTU method is used when only the inlet temperatures and exchanger size are known (sizing problems). Both are equivalent; they can be converted using relationships.