Is there a better way to calculate the LMTD factor?
Is there a better way to calculate the LMTD factor?
(OP)
I'm using the equation from the Kern's book to calculate the correctiojn factor for LMTD (log mean temperature difference), of a heat exchanger with one pass in the shell, and 2 in the tubes.
In some cases, a numeric error occurs, leading to a logarithm of a negative number. I tested some values, and it seems that when there is a temperature cross in the exchanger (assuming that temperature changes in a linear way, or other simplified way), the formula crashes.
I understand that there should always exist some value for that factor, am I wrong?
I've seen other formulas for this factor, but with the same error. Do you know any other formula for the LMTD factor, that can handle this situation?
In some cases, a numeric error occurs, leading to a logarithm of a negative number. I tested some values, and it seems that when there is a temperature cross in the exchanger (assuming that temperature changes in a linear way, or other simplified way), the formula crashes.
I understand that there should always exist some value for that factor, am I wrong?
I've seen other formulas for this factor, but with the same error. Do you know any other formula for the LMTD factor, that can handle this situation?





RE: Is there a better way to calculate the LMTD factor?
Hot fluid 100 ---> 50
Cold fluid 90 <--- 20
Delta T: 10 & 30
LMTD: (10-30)/ln(10/30) = -20/ln(1/3) = -20/-1.0986 = 18.2
RE: Is there a better way to calculate the LMTD factor?
The effectiveness equations are more simple to use, and only require the inlet temperatures of both fluids for solution. I find these relationships far more easy to use in heat exchanger calculations.
RE: Is there a better way to calculate the LMTD factor?
Maybe op can clarify more details about the heat exchanger and the equation used.
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(2B)+(2B)' ?
RE: Is there a better way to calculate the LMTD factor?
The Kern's formula is for the factor f, from Q = U * A * f * LMTD, for a heat exchanger with an even number of passes on the tubes.
Using the values of 25362:
Hot fluid 100°C ---> 50°C (tubes)
Cold fluid 90°C <--- 20°C (shell)
We can have, for the tubes:
100 -------> 75
50 <------- 75
And, for the shell, this is possible:
90 <------- 20
But, this is not (considering the supposed temperature for the tubes):
20 -------> 90
This kind of values of temperature lead to error on the formula for the "f" factor, although they are possible to exist in a real heat exchanger.
I don't have the formula now, but I can paste it here later.
RE: Is there a better way to calculate the LMTD factor?
"Whom the gods would destroy, they first make mad "