Heat exchanger analysis – Is LMTD equal to AMTD? State whether the logarithmic mean temperature difference (LMTD) is the same as the arithmetic mean temperature difference for heat exchangers.

Introduction / Context:
The LMTD method is fundamental for sizing and rating heat exchangers. It accounts for the exponential temperature approach of two fluids exchanging heat across a surface. Confusing LMTD with a simple arithmetic mean can lead to serious sizing errors.

Given Data / Assumptions:

• Steady state, no phase change, constant specific heats (typical LMTD assumptions).
• Any common flow arrangement (parallel, counter, cross with correction).

Concept / Approach:
The driving force varies along the exchanger length. Because heat transfer is proportional to local temperature difference, the correct average is the logarithmic mean, not the arithmetic mean, unless the temperature differences at each end are equal.

Step-by-Step Solution:
Define end temperature differences: Δt1 and Δt2.LMTD = (Δt1 − Δt2) / ln(Δt1 / Δt2).AMTD = (Δt1 + Δt2) / 2.Only when Δt1 = Δt2 do both expressions collapse to the same value; otherwise LMTD ≠ AMTD.

Verification / Alternative check:
Consider a parallel-flow exchanger with Δt1 = 30 K and Δt2 = 10 K. LMTD = (20)/ln(3) ≈ 18.2 K, while AMTD = 20 K. They are not equal, confirming the statement is false.

Why Other Options Are Wrong:

• “Only true for counter-flow / equal mass flow / equal cp” are unnecessary constraints; equality depends solely on Δt1 = Δt2.

Common Pitfalls:
Using AMTD in spreadsheets for convenience; forgetting to apply correction factors for crossflow or multipass arrangements; mixing inlet-outlet assignments and getting Δt1, Δt2 swapped.

Final Answer:
Disagree