LMTD & Heat Exchanger Duty

The temperature difference that drives heat transfer is not the same at both ends of an exchanger, and it does not change linearly between them — it decays exponentially. Averaging the two end differences arithmetically therefore overstates the driving force every time. The log mean, (ΔT₁ − ΔT₂) ÷ ln(ΔT₁ ÷ ΔT₂), is the correct average for a constant coefficient and constant specific heats, and the page shows the arithmetic mean beside it so the gap is visible. When the two ends are equal the formula is indeterminate, and the limiting value is used instead.

In counter-flow the ends pair hot-in with cold-out and hot-out with cold-in; in parallel flow both streams enter together, so the ends pair inlet with inlet. For the same four temperatures counter-flow always gives the larger log mean and therefore the smaller exchanger. A shell-and-tube or cross-flow unit sits between the two, and the correction factor F from the chart for that exact number of shell and tube passes is your input — it depends on the configuration, so it is not computed here. Q = U·A·ΔT_lm·F is then solved for whichever term you left out.

Clean-surface, steady-state, one-dimensional: no fouling resistance, no wall resistance, no phase change, no pressure drop and no variation of properties with temperature. U is your input and spans three orders of magnitude between a gas-to-gas service and condensing steam, so no default would be honest; c_p is yours too. The process-side cross-check ṁ·c_p·ΔT is there because when the two sides disagree, the assumption that is wrong is usually U. Nothing leaves the browser; the same four anonymous usage counts as the rest of the site apply.