Question

A thermoelectric device is made from a material with resistivity 5 . \(10^{-8} \Omega \cdot \mathrm{m}\) and Seebeck coefficient \(8.5 \cdot 10^{-5} \frac{\mathrm{V}}{\mathrm{K}}\). A cube, \(1 \mathrm{~cm}\) on each side, was used to determine the thermal conductivity. One side of the cube was heated. At a steady state, the rate of energy transfer by conduction through the cube is \(1.8 \mathrm{~W}\). The temperature distribution through the material is linear, and a temperature difference across is measured to be \(20 \mathrm{~K}\) across the cube. Find the thermal conductivity \(\kappa,\) and find the figure of merit \(Z\) for the material.

Short answer

The thermal conductivity \( \kappa = 0.9 \mathrm{~W/m \cdot K} \), and the figure of merit \( Z \approx 0.1606 \mathrm{~K}^{-1} \).

Step-by-step solution

Identify the formula

The rate of energy transfer by conduction can be described by Fourier's law of thermal conduction: \[ P = \kappa \cdot A \cdot \frac{\Delta T}{L} \] Where \( P \) is the power (or rate of energy transfer), \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference, and \( L \) is the thickness of the material.

Calculate the cross-sectional area and thickness

Since the cube has sides of \( 1 \mathrm{~cm} \), we have:- The area \( A = (0.01 \mathrm{~m})^2 = 0.0001 \mathrm{~m}^2 \)- The thickness \( L = 0.01 \mathrm{~m} \)

Substitute values to find the thermal conductivity

Using Fourier's law with \( P = 1.8 \mathrm{~W} \), \( \Delta T = 20 \mathrm{~K} \), \( A = 0.0001 \mathrm{~m}^2 \), and \( L = 0.01 \mathrm{~m} \):\[ 1.8 = \kappa \cdot 0.0001 \cdot \frac{20}{0.01} \] Solve for \( \kappa \):\[ \kappa = \frac{1.8 \times 0.01}{0.0001 \times 20} = 0.9 \mathrm{~W/m \cdot K} \]

Determine the figure of merit Z

The figure of merit \( Z \) is given by the formula: \[ Z = \frac{S^2}{\rho \cdot \kappa} \] Where \( S \) is the Seebeck coefficient, \( \rho \) is the resistivity, and \( \kappa \) is the thermal conductivity calculated earlier.

Calculate Z using the given values

Substituting the provided values:- \( S = 8.5 \times 10^{-5} \mathrm{~V/K} \)- \( \rho = 5 \times 10^{-8} \Omega \cdot \mathrm{m} \)- \( \kappa = 0.9 \mathrm{~W/m \cdot K} \)Substitute into the formula:\[ Z = \frac{(8.5 \times 10^{-5})^2}{5 \times 10^{-8} \times 0.9} \]Calculate \( Z \):\[ Z = \frac{7.225 \times 10^{-9}}{4.5 \times 10^{-8}} \approx 0.1606 \mathrm{~K}^{-1} \]

Key concepts

Fourier's law of thermal conduction

Fourier's law of thermal conduction is a fundamental principle that describes how heat energy is transferred through a material. It states that the rate at which heat is conducted through a material is proportional to the negative gradient of temperatures and the area perpendicular to that gradient. This is represented mathematically as: \[ P = \kappa \cdot A \cdot \frac{\Delta T}{L} \] Here, \( P \) stands for the power or the rate of energy transfer, \( \kappa \) is the thermal conductivity of the material, \( A \) is the cross-sectional area, \( \Delta T \) is the temperature difference across the material, and \( L \) is the thickness of the material. In practical terms, Fourier's law helps us understand how quickly or efficiently a material can transfer heat from a hotter side to a colder one.

• Materials with high thermal conductivity are better at conducting heat.
• If a material has low thermal conductivity, it acts more like an insulator.

Understanding Fourier’s law is crucial in designing systems where thermal management is important, such as in thermoelectric materials used for power generation or cooling. This law was utilized in the exercise to find the rate of energy transfer through a cubic material heated on one side, knowing its dimensions and temperature difference.

thermal conductivity

Thermal conductivity, denoted as \( \kappa \), is a material-specific property that indicates how well a material can conduct heat. The higher the thermal conductivity, the more efficient the material is at transferring heat. For the cube discussed in the exercise, thermal conductivity was determined using Fourier's law by rearranging the formula to: \[ \kappa = \frac{P \times L}{A \times \Delta T} \] From the values provided, including power, area, thickness, and temperature difference, the thermal conductivity was calculated as \( 0.9 \, \mathrm{W/m \cdot K} \).

• Metals usually have high thermal conductivity and are good heat conductors.
• Insulators, such as plastics and rubber, have low thermal conductivity.

Thermal conductivity is essential when choosing materials for heat sinks, insulation, and thermoelectric devices. Where applications demand efficient heat removal or containment, knowing the thermal conductivity helps engineers and scientists make informed decisions. In the context of thermoelectrics, a balance between thermal conductivity and electrical properties is vital for efficiency.

Seebeck coefficient

The Seebeck coefficient, often represented by the symbol \( S \), measures the magnitude of an induced thermoelectric voltage in response to a temperature difference across that material. It is expressed in \( \mathrm{V/K} \). The Seebeck effect is one of the three main effects in thermoelectrics, where a voltage is generated in a circuit composed of two different or similar materials with a temperature difference between their junctions. In practical terms:

• The Seebeck coefficient tells us how much voltage will be generated by a given temperature difference.
• Metals typically have low Seebeck coefficients, while semiconductors can have high values.

In the exercise given, the Seebeck coefficient was provided as \( 8.5 \times 10^{-5} \, \mathrm{V/K} \). This value is crucial in evaluating the efficiency of the material in a thermoelectric application. Higher Seebeck coefficients mean that a material can produce more voltage—and therefore more power—with the same amount of heat input, which is crucial for creating efficient thermoelectric generators.

figure of merit in thermoelectrics

The figure of merit, commonly represented by the symbol \( Z \), is a dimensionless measure that describes the efficiency of a thermoelectric material. It integrates the electrical and thermal properties, using the formula: \[ Z = \frac{S^2}{\rho \cdot \kappa} \] Where \( S \) is the Seebeck coefficient, \( \rho \) is the electrical resistivity, and \( \kappa \) is the thermal conductivity. The higher the figure of merit, the better the material is at converting temperature differences into electrical energy. This makes \( Z \) a critical value when assessing materials for thermoelectric applications.

• An ideal thermoelectric material has a high Seebeck coefficient, low resistivity, and low thermal conductivity.
• Higher \( Z \) values indicate higher efficiency in solid-state energy conversion devices.

For the material in question, the exercise calculated \( Z \) as approximately \( 0.1606 \, \mathrm{K}^{-1} \), which provides insight into its potential efficiency as a thermoelectric material. This metric is crucial in developing new thermoelectric materials for energy harvesting, cooling technologies, and waste heat recovery.