Is there a better way to calculate the LMTD factor? 2

[renatoab]

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?

 

[25362]

I wonder how can you get negative results. For example, take a temperature crossing:

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

 

[FredRosse]

LMTD is defined as follows: LMTD = (GTD - LTD) / Loge (GTD / LTD), where GTD is the Greater Temperature Difference, and LTD is the Lower Temperature Difference. The only case where the LMTD equation falls apart is when the temperature differences are both equal, and in that case the LMTD is equal to exactly that Temperature Difference, which is the same on both ends of the exchanger. As far as I know, there are no other computational difficulties.

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.

 

[electricpete]

I'm not a big hx guy but I think the responses apply to single pass hx?

Maybe op can clarify more details about the heat exchanger and the equation used.

=====================================
(2B)+(2B)' ?

 

[renatoab]

No. I think I was not clear. I don't have problem with de LMTD.
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.

 

[davefitz]

The root equations for determining the F factor in the LMTD method are contained in the e-NTU method ( ie compact heat exchangers by Kays + London) . It is suggested the effectiveness of teh HX be comuted using the e-NTU method directly.

"Whom the gods would destroy, they first make mad "