# Arithmetic and Logarithmic Mean Temperature Difference

According to **Newton's Law of Cooling** heat transfer rate is related to the instantaneous temperature difference between hot and cold media

- in a heat transfer process the temperature difference vary with position and time

### Mean Temperature Difference

The mean temperature difference in a heat transfer process depends on the direction of fluid flows involved in the process. The primary and secondary fluid in an heat exchanger process may

- flow in the same direction -
**parallel flow or co-current flow** - in the opposite direction -
**counter-current flow** - or perpendicular to each other -
**cross flow**

With saturation steam as the primary fluid the primary temperature can be taken as a constant since the heat is transferred as a result of a change of phase only. The temperature profile in the primary fluid is not dependent on the direction of flow.

### Logarithmic Mean Temperature Difference -* LMTD*

The rise in secondary temperature is non-linear and can best be represented by a logarithmic calculation. A logarithmic mean temperature difference is termed

*LMTD**(or DT*- Logarithmic Mean Temperature Difference_{LM})

LMTD can be expressed as

LMTD = (dt_{o}- dt_{i}) / ln(dt_{o}/ dt_{i})(1)

where

LMTD =Logarithmic Mean Temperature Difference ()^{o}F,^{o}CFor parallel flow:

dt_{i}= t_{pi}- t_{si}=inletprimary and secondary fluid temperature difference (^{o}F,^{o}C)

dt_{o}= t_{po}- t_{so}=outletprimary and secondary fluid temperature difference (^{o}F,^{o}C)For counter flow:

dt_{i}= t_{pi}- t_{so}=inletprimary and outlet secondary fluid temperature difference (^{o}F,^{o}C)

dt_{o}= t_{po}- t_{si}=outletprimary and inlet secondary fluid temperature difference (^{o}F,^{o}C)

The Logarithmic Mean Temperature Difference is always less than the Arithmetic Mean Temperature Difference.

### Arithmetic Mean Temperature Difference - *AMTD*

An easier but less accurate way to calculate the mean temperature difference is the

*AMTD (or DT*- Arithmetic Mean Temperature Difference_{AM})

AMTD can be expressed as:

AMTD = (t_{pi}+ t_{po}) / 2 - (t_{si}+ t_{so}) / 2(2)

where

AMTD= Arithmetic Mean Temperature Difference (^{o}F,^{o}C)

t_{pi}= primary inlet temperature (^{o}F,^{o}C)

t_{po}= primary outlet temperature (^{o}F,^{o}C)

t_{si}= secondary inlet temperature (^{o}F,^{o}C)

t_{so}= secondary outlet temperature (^{o}F,^{o}C)

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

t_{p1}= t_{p2}

or

t_{s1}= t_{s2}

### Arithmetic and Logarithmic Mean Temperature Difference Calculator

The calculator below can be used to calculate Arithmetic and Logarithmic Mean Temperature Difference of counter-flow an parallel-flow heat exchangers.

* t _{pi} - primary flow - inlet temperature (^{o}F, ^{o}C)*

*t _{po} - primary flow - outlet temperature (^{o}F, ^{o}C)*

*t _{si} - secondary flow - inlet temperature (^{o}F, ^{o}C)*

*t _{so} - secondary flow - outlet temperature (^{o}F, ^{o}C)*

Counter-flow Parallel-flow

### Logarithmic Mean Temperature Difference Chart

### Example - Arithmetic and Logarithmic Mean Temperature, Hot Water Heating Air

Hot water at *80 ^{ o}C* heats air from from a temperature of

*0*to

^{ o}C*20*in a parallel flow heat exchanger. The water leaves the heat exchanger at

^{ o}C*60*.

^{ o}CArithmetic Mean Temperature Difference can be calculated as

AMTD= ((80^{ o}C) + (60^{ o}C)) / 2 - ((0^{ o}C) + (20^{ o}C)) / 2

= 60^{o}C

Logarithmic Mean Temperature Difference can be calculated as

LMTD= ((60^{ o}C) - (20^{ o}C)) - ((80^{ o}C) - (0^{ o}C))) / ln(((60^{ o}C) - (20^{ o}C)) / ((80^{ o}C) - (0^{ o}C)))

= 57.7^{o}C

### Example - Arithmetic and Logarithmic Mean Temperature, Steam Heating Water

Steam at *2 bar* gauge heats water from *20 ^{ o}C to 50^{ o}C*. The saturation temperature of steam at

*2 bar*gauge is

*134*.

^{ o}C**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^{ o}C) + (134^{ o}C)) / 2 - ((20^{ o}C) + (50^{ o}C)) / 2

= 99^{o}C

Log Mean Temperature Difference can be calculated like

LMTD= ((134^{ o}C) - (20^{ o}C) - ((134^{ o}C) - (50^{ o}C))) / ln(((134^{ o}C) - (20^{ o}C)) / ((134^{ o}C) - (50^{ o}C)))

= 98.24^{o}C

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