calc_LMTD¶

pychemengg.heattransfer.heatcommonmethods.calc_LMTD(deltaT1=None, deltaT2=None)[source]¶

To compute log mean temperature difference (LMTD)

Parameters

deltaT1int or float: Temperature difference.
deltaT2int or float: Temperature difference.

Returns

LMTDint or float: Reynolds number for a flowing fluid.

Notes

The following formula is used.

\[LMTD = \cfrac {\Delta T_1 - \Delta T_2} {ln[ \Delta T_1 / \Delta T_2 }\]

where:

\(\Delta T_1\) = temperature difference #1

\(\Delta T_2\) = temperature difference #2

\(\Delta T_1\) and \(\Delta T_2\) vary on the type of problem under consideration

Internal to this function, the temperature differences are converted into complex numbers to allow for the use of this function in situations where temperature must be iteratively solved using solvers such as ‘fsolve’ from scipy. The value that is returned is the real part.

References

[1] G. F. Nellis and S. A. Klein, “Introduction to Engineering Heat Transfer”, 1st Edition. Cambridge University Press, 2021.

[2] Y. A. Cengel and A. J. Ghajar, “Heat And Mass Transfer Fundamentals and Applications”, 6th Edition. New York, McGraw Hill Education, 2020.

[3] T.L. Bergman, A. S. Lavine, F. P. Incropera, D. P. Dewitt, “Fundamentals of Heat and Mass Transfer”, 7th Edition, John Wiley, 2011.

Examples

First import the module heatcommonmethods.

>>> from pychemengg.heattransfer import heatcommonmethods as hcm 
>>> hcm.calc_LMTD(deltaT1=90-40, deltaT2=65-20)
47.456107905149494