Arithmetic Logarithmic Mean Temperature

Overview

Engineering reference data for Arithmetic Logarithmic Mean Temperature in heat transfer.

Key Formulas

Fourier's Law

$q = -k \nabla T$

Heat flux proportional to temperature gradient.

Convective Heat Transfer

$Q = hA(T_s - T_\infty)$

Heat transfer between surface and fluid.

Stefan-Boltzmann Law

$q = \varepsilon \sigma T^4$

Radiative heat flux from a surface.

Thermal Resistance

$R_{th} = \frac{L}{kA}$

Resistance to heat conduction.

Variables

References

• Original Source

Additional Flow Configurations

Heat exchangers can be classified by the relative direction of the fluid flows:

• Parallel flow (co-current flow): The primary and secondary fluids enter the exchanger from the same end and flow in the same direction.
• Counter-current flow: The fluids enter from opposite ends and flow in opposite directions.
• Cross flow: The fluids flow perpendicular to each other.

Additional Worked Examples

Example 1: Hot Water Heating Air (Parallel Flow)

Hot water at 80 °C heats air from a temperature of 0 °C to 20 °C in a parallel flow heat exchanger. The water leaves the heat exchanger at 60 °C.

Step 1: Define Inlet and Outlet Temperatures

• Primary inlet temperature (t_{pi}) = 80 °C
• Primary outlet temperature (t_{po}) = 60 °C
• Secondary inlet temperature (t_{si}) = 0 °C
• Secondary outlet temperature (t_{so}) = 20 °C

Step 2: Calculate Arithmetic Mean Temperature Difference (AMTD)

$$\text{AMTD} = \frac{t_{pi} + t_{po}}{2} - \frac{t_{si} + t_{so}}{2} = \frac{80 + 60}{2} - \frac{0 + 20}{2} = 70 - 10 = 60\ ^\circ\text{C}$$

Step 3: Calculate Logarithmic Mean Temperature Difference (LMTD) For parallel flow:

• Inlet temperature difference, dt_i = t_{pi} - t_{si} = 80 - 0 = 80 °C
• Outlet temperature difference, dt_o = t_{po} - t_{so} = 60 - 20 = 40 °C

$$\text{LMTD} = \frac{dt_i - dt_o}{\ln\left(\frac{dt_i}{dt_o}\right)} = \frac{80 - 40}{\ln\left(\frac{80}{40}\right)} = \frac{40}{\ln(2)} \approx 57.7\ ^\circ\text{C}$$

Example 2: Steam Heating Water (Phase Change)

Steam at 2 bar gauge heats water from 20 °C to 50 °C. The saturation temperature of steam at 2 bar gauge is 134 °C.

Note: Steam condenses at a constant temperature (t_{pi} = t_{po} = 134 °C).

Step 1: Define Inlet and Outlet Temperatures

• Primary inlet/outlet temperature (t_{pi} and t_{po}) = 134 °C
• Secondary inlet temperature (t_{si}) = 20 °C
• Secondary outlet temperature (t_{so}) = 50 °C

Step 2: Calculate Arithmetic Mean Temperature Difference (AMTD)

$$\text{AMTD} = \frac{t_{pi} + t_{po}}{2} - \frac{t_{si} + t_{so}}{2} = \frac{134 + 134}{2} - \frac{20 + 50}{2} = 134 - 35 = 99\ ^\circ\text{C}$$

Step 3: Calculate Logarithmic Mean Temperature Difference (LMTD)

• Inlet temperature difference, dt_i = t_{pi} - t_{si} = 134 - 20 = 114 °C
• Outlet temperature difference, dt_o = t_{po} - t_{so} = 134 - 50 = 84 °C

$$\text{LMTD} = \frac{dt_i - dt_o}{\ln\left(\frac{dt_i}{dt_o}\right)} = \frac{114 - 84}{\ln\left(\frac{114}{84}\right)} = \frac{30}{\ln(1.357)} \approx 98.24\ ^\circ\text{C}$$

Practical Application Guidance

The Arithmetic Mean Temperature Difference (AMTD) is an easier but less accurate method. It provides a satisfactory approximation when the smaller of the inlet or outlet temperature differences (min(dt_i, dt_o)) is more than half the larger difference (max(dt_i, dt_o)).

When heat transfer involves a change of phase (condensation or evaporation), the temperature on the phase-change side remains constant. In these cases, the equations can be simplified by setting the inlet and outlet temperatures of that fluid to be equal (e.g., t_{pi} = t_{po} for condensing steam).

A more detailed Logarithmic Mean Temperature Difference Chart is available as a downloadable PDF for common scenarios.

Practical Considerations for AMTD vs. LMTD

The Arithmetic Mean Temperature Difference (AMTD) provides a satisfactory approximation when the smallest inlet or outlet temperature difference is more than half the greatest inlet or outlet temperature difference.

For heat transfer processes involving a change of phase (condensation or evaporation), the temperature of the fluid undergoing the phase change remains constant. The mean temperature difference equations can be simplified by setting the inlet and outlet temperatures of that fluid equal (e.g., tpi = tpo for condensing steam).