Log Mean Temperature Difference (Counterflow)

Also known as log mean temperature difference · LMTD

ΔTlm=ΔT1−ΔT2ln⁡(ΔT1/ΔT2)\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}
sig figs

Worked example: 150 → 90 C against 30 → 70 C counterflow → 69.52 K. Press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.

Where Log Mean Temperature Difference (Counterflow) fits

Your answer feeds

See also

Log Mean Temperature Difference (Parallel Flow): Both use log mean temperature difference and hot stream inlet.Maximum Possible Heat Transfer (Qmax): Both use hot stream inlet and cold stream inlet.Heat Exchanger Duty (Q = U·A·F·LMTD): Both use log mean temperature difference.
Here the solver did the work. Could you?

Log mean temperature difference →

UniversityThermodynamics & Heat Transfer

Test your skills in the Exam Room: new numbers every attempt. Free lessons for students, no sign-up, just pure learning. Find 1 more lesson on this formula.

See your Report Card
Learning zone

Log Mean Temperature Difference (Counterflow) explained

Th,inTh,outTc,outTc,inΔTlm

The driving ΔT in an exchanger is not constant along its length, so you cannot use the arithmetic mean: integrating Q̇ = UAΔT along the tube produces the logarithmic mean of the two terminal differences instead. In counterflow the streams run opposite ways, so the hot end pairs the hot inlet with the cold outlet: ΔT₁ = Th,in − Tc,out and ΔT₂ = Th,out − Tc,in. Cool 150 °C oil to 90 °C against water warming from 30 °C to 70 °C and the terminals are 80 K and 60 K, giving ΔTlm=(80−60)/ln⁡(80/60)=69.5\Delta T_{\mathrm{lm}} = (80 - 60)/\ln(80/60) = 69.5 K, not the 70 K an average would suggest. The log mean is always the smaller of the two, and the gap widens fast as the terminals diverge.

Counterflow is the reason so many exchangers are plumbed the way they are: it permits a temperature cross, where the cold stream leaves hotter than the hot stream leaves, which parallel flow can never do. Two traps. First, a genuine cross at either terminal makes the logarithm undefined: the calculator refuses, because the arrangement you described cannot exist. Second, the four terminal temperatures sit inside a logarithm and a difference at once, so recovering an inlet temperature from a known ΔTlm\Delta T_{\mathrm{lm}} is transcendental: it takes the Lambert W function, which this page uses when you leave a temperature blank.

Log Mean Temperature Difference (Counterflow) formula

ΔTlm=ΔT1−ΔT2ln⁡(ΔT1/ΔT2)\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}
Where
  • ΔTlm\Delta T_{lm}= Log mean temperature difference (C°)
  • Th,inT_{h,in}= Hot stream inlet (°C)
  • Th,outT_{h,out}= Hot stream outlet (°C)
  • Tc,inT_{c,in}= Cold stream inlet (°C)
  • Tc,outT_{c,out}= Cold stream outlet (°C)
Revised 1 November 2026
Cite this page

formula.expert. (2026, November 1). Log Mean Temperature Difference (Counterflow) Calculator. https://formula.expert/solvers/lmtd-counterflow