Heat Exchanger Duty & Area Calculator

Understanding the heat exchanger duty and area calculator — concept and principle

Sizing a heat exchanger starts with two questions: how much heat must move, and how much surface is needed to move it. This calculator finds the duty from the hot stream's flow and temperature drop, the log-mean temperature difference (LMTD) for counter-flow, and the area required for a chosen overall heat-transfer coefficient U. It is used for preliminary sizing, checking vendor quotes and teaching process heat transfer.

The two end temperature gaps set the LMTD; a bigger LMTD needs less area.

How it is calculated

Duty: Q = ṁ · c_p · (T_h,in − T_h,out), in kW when ṁ is kg/s and c_p is kJ/(kg·K).
End differences (counter-flow): ΔT₁ = T_h,in − T_c,out, ΔT₂ = T_h,out − T_c,in.
LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂); if ΔT₁ = ΔT₂ it equals that value.
Area: A = Q × 1000 / (U · LMTD), with U in W/(m²·K) and A in m².

The principle

Heat flows through a wall at a rate U·A·ΔT, but ΔT changes along the exchanger as the hot stream cools and the cold one warms. Integrating the local rate over the length, assuming constant U and c_p, gives exactly the logarithmic mean of the two end differences. The arithmetic mean always overestimates the driving force and would undersize the unit. Counter-flow keeps a more even ΔT than parallel flow, so it gives the largest LMTD and the smallest area for the same temperatures.

Worked example

Inputs: ṁ = 2 kg/s, c_p = 4.18 kJ/(kg·K), hot 90 → 50 °C, cold 20 → 45 °C, U = 800 W/(m²·K).
Q = 2 × 4.18 × 40 = 334.4 kW.
ΔT₁ = 90 − 45 = 45 K; ΔT₂ = 50 − 20 = 30 K.
LMTD = (45 − 30) / ln(1.5) = 15 / 0.405465 = 36.995 K.
A = 334,400 / (800 × 36.995) = 11.299 m².

Each point is the calculator re-run with only “Hot in (°C)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Hot in (°C)	Duty (kW)
22.5	-229.9
45	-41.8
67.5	146.3
90	334.4
112	522.5
135	710.6
180	1,086.8

How to read the result

Area scales directly with duty and inversely with U and LMTD. A small LMTD (a close temperature approach) saves energy but needs much more surface. U depends strongly on the fluids and fouling: water-to-water units are often in the hundreds to low thousands of W/(m²·K), while gas-side duties can be far lower. Designers usually add margin for fouling. Check that the cold stream can actually absorb Q: its flow must be Q / (c_p,c × ΔT_c).

Common mistakes

Pairing the wrong temperatures for ΔT₁ and ΔT₂ (counter-flow pairs hot inlet with cold outlet).
Mixing kW and W, giving an area 1000 times wrong.
Using the counter-flow LMTD for multi-pass shell-and-tube units without the correction factor F.
Entering temperatures that cross (cold outlet above hot inlet), which makes the log undefined.

Keep in mind: This is a preliminary estimate assuming constant U and c_p and no phase change. Real exchanger design needs fouling factors, pressure-drop checks, mechanical codes and review by a qualified engineer.

Frequently asked questions

What is a typical U value?

It varies widely with fluids, velocity and fouling, from tens of W/(m²·K) for gases to a few thousand for condensing steam; use data for your service.

Why not use the average temperature difference?

The arithmetic mean overstates the driving force when end differences differ, which would give too small an area.

What happens if the two ends have the same ΔT?

Then LMTD equals that ΔT; the calculator handles this case directly.

Sources: F. P. Incropera et al., Fundamentals of Heat and Mass Transfer, Wiley · Perry's Chemical Engineers' Handbook, McGraw-Hill

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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