Logarithmic mean temperature difference (LMTD): Given inlet and outlet temperature differences Δt1 and Δt2 between hot and cold streams, the correct expression for the mean temperature difference is

Difficulty: Medium

Correct Answer: tm = (Δt1 - Δt2) / ln(Δt1 / Δt2)

Explanation:


Introduction / Context:
The LMTD method is widely used in heat exchanger analysis to compute heat transfer rate when terminal temperatures are known. The right formula is essential for accurate design.



Given Data / Assumptions:

  • Two fluids exchanging heat with constant overall heat transfer coefficient U and area A.
  • Terminal temperature differences Δt1 and Δt2 at the two ends of the exchanger.
  • No phase change and steady state for simplicity.


Concept / Approach:
Since temperature difference varies exponentially along the length, the appropriate average is the logarithmic mean, not the arithmetic or geometric mean. Heat duty is Q = U * A * tm where tm = (Δt1 - Δt2) / ln(Δt1 / Δt2).



Step-by-Step Solution:
Write local heat transfer: dQ = U * dA * Δt(x).Integrate along length using boundary conditions to derive tm.Obtain tm = (Δt1 - Δt2) / ln(Δt1 / Δt2).Use Q = U * A * tm to compute duty.



Verification / Alternative check:
Limiting case Δt1 → Δt2 gives tm → Δt1 (using L’Hôpital’s rule), matching physical expectation of uniform Δt.



Why Other Options Are Wrong:

  • Arithmetic or geometric means do not satisfy the exponential temperature profile implied by energy balance.
  • ln(Δt1 * Δt2) and (Δt1 - Δt2) * ln(Δt1 / Δt2) are dimensionally or mathematically incorrect.


Common Pitfalls:
Swapping Δt1 and Δt2 positions or using base-10 logarithm incorrectly (natural logarithm is standard in derivation).



Final Answer:
tm = (Δt1 - Δt2) / ln(Δt1 / Δt2)

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