I am solving the following problem:

A liquid-to-liquid counterflow heat exchanger is used to heat a cold fluid from 120 °F to 310 °F. Assuming that the hot fluid enters at 500 °F and leaves at 400 °F, calculate the log mean temperature difference for the heat exchanger.

The given answer is 232 °F

I used:

$${ LMTD = \frac{\triangle max - \triangle min }{ ln( \frac {\triangle max}{\triangle min} ) } }$$

$${ \triangle max = 500-120 = 380 }$$ $${ \triangle min = 400-310= 90}$$

I plug them all together and I got 201.3383331.

What am I doing wrong?

  • 4
    $\begingroup$ Algo is right BUT what you are doing wrong is the same as in the two questions yesterday - slow down!. Look at the question. Delta_max is the delta of the maxima. Delta_min is the delta of the minima. IF that is not clear at 1st try then at lest try it out when you get the wrong answer., Occam says however that that is what you should have tried 1st in the abswence of more info.' $\endgroup$ Commented Jul 18, 2015 at 5:17

1 Answer 1


$\triangle max$ and $\triangle min$ are not defined as maximum and minimum temperature differences in a heat exchanger. Quoting from Wikipedia:

The LMTD is a logarithmic average of the temperature difference between the hot and cold feeds at each end of the double pipe exchanger.

Temperature distribution of HEX inlets and outlets

$${ \triangle T_{max} = 400-120 = 280 \text{ F}}$$ $${ \triangle T_{min} = 500-310= 190\text{ F}}$$ $${ LMTD = \frac{\triangle T_{max} - \triangle T_{min} }{ ln( \frac {\triangle T_{max}}{\triangle T_{min}} ) } } = \frac{280-190}{ln( \frac {280}{190})}= 232\text{ F}$$


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