Heat exchangers and LMTD, explained

Why a log mean

The temperature difference between the hot and cold streams changes along the exchanger, and the heat flux follows it. Integrating over the length gives a logarithmic average, not an arithmetic one. ΔT₁ and ΔT₂ are the differences at the two ends; which end is which does not matter.

For a sizing problem, where you know all four temperatures and need the area, LMTD is direct: get Q from an energy balance on either stream, then A = Q/(U·ΔT_lm).

Counterflow, parallel flow, and the ε–NTU method

In counterflow the streams enter at opposite ends. The temperature difference stays more uniform, the LMTD is higher, and you need less area for the same duty. The cold outlet can even end up hotter than the hot outlet, which parallel flow can never do. Cross-flow and shell-and-tube exchangers use the counterflow LMTD multiplied by a correction factor F.

When you know the exchanger and need the outlet temperatures, a rating problem, LMTD needs iteration. The effectiveness–NTU method solves it directly: effectiveness is the actual duty over the maximum possible, and charts or formulas give it from NTU and the heat capacity ratio.

A worked number

Oil cools from 150 °C to 90 °C while water heats from 30 °C to 70 °C, with a duty of 100 kW and U = 300 W/m²·K. In counterflow, the end differences are 80 and 60 K, the LMTD is 20/ln(80/60) ≈ 69.5 K, and the area is 100,000/(300 × 69.5) ≈ 4.8 m².

In parallel flow the end differences are 120 and 20 K, the LMTD falls to about 55.8 K, and the same duty needs about 6.0 m², a quarter more.

Common mistakes

• Using the arithmetic mean of ΔT₁ and ΔT₂, which overstates the driving force.
• Pairing the wrong temperatures at each end in counterflow.
• Applying the counterflow LMTD to a shell-and-tube exchanger without the correction factor.