Logarithmic mean temperature difference for heat transfer

  • #1

Homework Statement


[Update: just realized that the LMTD is a temperature difference, so my question was not valid] :P

Calculate the logarithmic mean temperature difference (LMTD) to heat water flowing through a tube from 21 C (Ti) to 40 C (Te) if the tube has a fixed temperature of 45 C (Ts).
I should then calculate the length needed, which I can do. I am stuck with calculating the logarithmic mean temperature difference as I get a number that is less than the initial temperature of the water, which I know should be wrong!

Homework Equations


LMTD = (Ti - Te)/(ln[(Ts-Te)/(Ts-Ti)])

The Attempt at a Solution


For the given parameters:
LMTD = (21-40)/(ln[(45-40)/(45-21)]) = 12.1 C
 
Last edited:

Answers and Replies

  • #2
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