dti = tpi - tsi = inlet primary and secondary fluid temperature difference (oF, oC)
dto = tpo - tso = outlet primary and secondary fluid temperature difference (oF, oC)
For counter flow:
dti = tpi - tso = inlet primary and outlet secondary fluid temperature difference (oF, oC)
dto = tpo - tsi = outlet primary and inlet secondary fluid temperature difference (oF, oC)
The Logarithmic Mean Temperature Difference is always less than the Arithmetic Mean Temperature Difference.
An easier but less accurate way to calculate the mean temperature difference is the
AMTD can be expressed as:
AMTD = (tpi + tpo) / 2 - (tsi + tso) / 2 (2)
where
AMTD = Arithmetic Mean Temperature Difference (oF, oC)
tpi = primary inlet temperature (oF, oC)
tpo = primary outlet temperature (oF, oC)
tsi = secondary inlet temperature (oF, oC)
tso = secondary outlet temperature (oF, oC)
A linear increase in the secondary fluid temperature makes it more easy to do manual calculations. AMTD will in general give a satisfactory approximation for the mean temperature difference when the smallest of the inlet or outlet temperature differences is more than half the greatest of the inlet or outlet temperature differences.
When heat is transferred as a result of a change of phase like condensation or evaporation the temperature of the primary or secondary fluid remains constant. The equations can then be simplified by setting
tp1 = tp2
or
ts1 = ts2
The calculator below can be used to calculate Arithmetic and Logarithmic Mean Temperature Difference of counter-flow an parallel-flow heat exchangers.
tpi - primary flow - inlet temperature (oF, oC)
tpo - primary flow - outlet temperature (oF, oC)
tsi - secondary flow - inlet temperature (oF, oC)
tso - secondary flow - outlet temperature (oF, oC)
Counter-flow Parallel-flow
Hot water at 80 oC heats air from from a temperature of 0 oC to 20 oC in a parallel flow heat exchanger. The water leaves the heat exchanger at 60 oC.
Arithmetic Mean Temperature Difference can be calculated as
AMTD = ((80 oC) + (60 oC)) / 2 - ((0 oC) + (20 oC)) / 2
= 60 oC
Logarithmic Mean Temperature Difference can be calculated as
LMTD = ((60 oC) - (20 oC)) - ((80 oC) - (0 oC))) / ln(((60 oC) - (20 oC)) / ((80 oC) - (0 oC)))
= 57.7 oC
Steam at 2 bar gauge heats water from 20 oC to 50 oC. The saturation temperature of steam at 2 bar gauge is 134 oC.
Note! that steam condenses at a constant temperature. The temperature on the heat exchangers surface on the steam side is constant and determined by the steam pressure.
Arithmetic Mean Temperature Difference can be calculated like
AMTD = ((134 oC) + (134 oC)) / 2 - ((20 oC) + (50 oC)) / 2
= 99 oC
Log Mean Temperature Difference can be calculated like
LMTD = ((134 oC) - (20 oC) - ((134 oC) - (50 oC))) / ln(((134 oC) - (20 oC)) / ((134 oC) - (50 oC)))
= 98.24 oC
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