Heat Exchanger (LMTD)

A hot fluid warms a cold one through a wall. How much heat passes, how much wall area you need, or how good the heat transfer must be — with the log mean temperature difference, for counterflow or parallel flow.

Heat transferHeat exchangersEngineering

Inputs

Input examples

Choose the flow arrangement and the two fluids, enter the four temperatures, then click Solve next to the heat duty, U or the area. The flow rate of each fluid is worked out from CoolProp enthalpies at 1 bar.

Hot fluid

Cold fluid

Hot fluid in T_h,in

Hot fluid out T_h,out

Cold fluid in T_c,in

Cold fluid out T_c,out

Heat duty Q

Overall coefficient U

Area A

What is the LMTD?

In a heat exchanger a hot fluid passes heat through a wall to a cold fluid. The temperature difference between them changes along the way, so no single ΔT describes it. The log mean temperature difference ΔT_lm is the exact average for this: Q = U·A·ΔT_lm.

U says how easily heat gets through the wall, including the fluid films on both sides; A is the wall area. In counterflow the fluids run in opposite directions. That keeps the temperature difference more even, so it needs less area than parallel flow — and the cold fluid can leave warmer than the hot fluid does.

hotgives off Q

coldtakes up Q

ΔT₁ = ΔT₂ΔT_lm = ΔT

c_pCoolProp, 1 bar

Temperature Profile

counterflow

hot fluid

cold fluid

arrows: direction of flow

Result

Enter the values above

Specific Heat vs Temperature

1 bar

Fluids at 1 bar

CoolProp data, rounded

Fluid	c_p at 20 °C, J/(g·K)	Boils at 1 bar	Data range

Worked Techniques

Sizing a Heat Exchanger

Work out the two end temperature differences, take their log mean, and divide the heat duty by U·ΔT_lm.

Example:

Flow Rate from the Energy Balance

Each fluid carries the same Q: the hot one gives it off, the cold one takes it up. The calculator uses the enthalpy change at 1 bar, which also counts any boiling.

Example:

Counterflow vs Parallel Flow

Same temperatures, different ends: the first form is for counterflow, the second for parallel flow. Counterflow gives the larger ΔT_lm.

Example: oil 120 → 60 °C, water 25 → 45 °C: 52.5 K vs 43.3 K, so parallel flow needs 21% more area.

Sources & Verification

References

F. P. Incropera, D. P. DeWitt, T. L. Bergman & A. S. Lavine, Fundamentals of Heat and Mass Transfer, 7th ed., Wiley, 2011 — Ch. 11 (Heat Exchangers).
I. H. Bell, J. Wronski, S. Quoilin & V. Lemort, “Pure and Pseudo-pure Fluid Thermophysical Property Evaluation and the Open-Source Thermophysical Property Library CoolProp”, Ind. Eng. Chem. Res. 53 (6), 2498–2508, 2014.