Logarithmic Mean Temperature Difference

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SUMMARY

The discussion centers on the calculation of the Logarithmic Mean Temperature Difference (LMTD) in a counter flow heat exchanger. The correct formula involves taking the temperature differences at both ends of the exchanger: (700-450) and (1100-960), and applying the LMTD formula: ΔT1 - ΔT2 / ln(ΔT1/ΔT2). Participants clarified that the confusion arose from the definition of which temperature differences to use, ultimately confirming that the method of calculation was correct. The provided resource further elucidates the definitions and calculations involved.

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  • Understanding of heat exchanger principles
  • Familiarity with the LMTD formula
  • Basic knowledge of logarithmic functions
  • Experience with counter flow heat exchangers
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  • Study the detailed workings of the LMTD formula in heat exchangers
  • Learn about different types of heat exchangers, focusing on counter flow designs
  • Explore practical applications of LMTD in thermal engineering
  • Review logarithmic functions and their applications in engineering calculations
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Students and professionals in thermal engineering, mechanical engineers, and anyone involved in the design or analysis of heat exchangers will benefit from this discussion.

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Homework Statement


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The Attempt at a Solution



I am confused as to which streams I should use for the log mean temperature difference. I am pretty sure it should be the difference in temperature between the two streams on the same side, so I think it's

(700-450)-(1100-960) / ln [(700-450)/(1100-960)], but I am not sure.

The term wasn't well defined, and my search on wikipedia gave the definition as ''The LMTD is a logarithmic average of the temperature difference between the hot and cold streams at each end of the exchanger.''

This is too vague, which end of the exchanger??
 

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The way you calculated it is correct. You are not confused about it after all.

Chet
 
I am confused, because the only reason I know it had to be that was because I would get a negative logarithm if I did it the other way. Suppose it was such that I didn't get a negative number in the logarithm either way that I subtracted it, then how would I do it??
 
Last edited by a moderator:
Read the attachment to Post #2.
 
Woopydalan said:
I am confused, because the only reason I know it had to be that was because I would get a negative logarithm if I did it the other way. Suppose it was such that I didn't get a negative number in the logarithm either way that I subtracted it, then how would I do it??

You take hot temperature minus cold temperature at one end of the heat exchanger, and hot temperature minus cold temperature at the other end of the heat exchanger. How you do things after that doesn't matter.
 
Okay great, I have convinced myself of that now. Thank you!
 

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