Conduction heat transfer is a fundamental concept in thermodynamics and heat transfer, where heat is transferred through a material without the involvement of any bulk motion of the material. This process is governed by Fourier’s law of heat conduction, which relates the heat flux to the temperature gradient within the material. In this comprehensive guide, we will delve into the various aspects of conduction heat transfer, providing a detailed and technical exploration of the topic.

## Thermal Conductivity: The Key to Conduction

Thermal conductivity, denoted by the symbol `k`

, `λ`

, or `κ`

, is a measure of a material’s ability to conduct heat. It is a fundamental property that determines the rate of heat transfer through a material and is measured in watts per meter-kelvin (W/mK). The higher the thermal conductivity, the more efficiently the material can conduct heat.

The thermal conductivity of common materials can vary significantly:

Material | Thermal Conductivity (W/mK) |
---|---|

Copper | 386 |

Aluminum | 205 |

Steel | 50 |

Glass | 0.8 |

Wood | 0.1 – 0.5 |

Air | 0.026 |

These values highlight the wide range of thermal conductivities found in different materials, which is crucial in understanding and predicting conduction heat transfer.

## Temperature Gradient: The Driver of Conduction

The temperature gradient, denoted by `∇T`

, is the rate of change of temperature with respect to distance. It is measured in kelvin per meter (K/m) and is a crucial quantity in conduction heat transfer, as it determines the rate of heat flow.

The temperature gradient can be expressed mathematically as:

`∇T = dT/dx`

where `dT`

is the change in temperature and `dx`

is the change in distance.

The temperature gradient is directly proportional to the heat flux, as described by Fourier’s law of heat conduction.

## Fourier’s Law of Heat Conduction

Fourier’s law of heat conduction is a fundamental principle that relates the heat flux, `q`

, to the temperature gradient, `∇T`

, and the thermal conductivity, `k`

. The law states that the heat flux is proportional to the negative of the temperature gradient, and is expressed as:

`q = -k ∇T`

where:

– `q`

is the heat flux, measured in watts per square meter (W/m²)

– `k`

is the thermal conductivity of the material, measured in watts per meter-kelvin (W/mK)

– `∇T`

is the temperature gradient, measured in kelvin per meter (K/m)

The negative sign in the equation indicates that heat flows from a region of higher temperature to a region of lower temperature.

## Thermal Resistance: Quantifying Resistance to Heat Flow

Thermal resistance, denoted by the symbol `R`

, is a measure of a material’s resistance to heat flow. It is measured in kelvin-meters per watt (K·m/W) and is given by the equation:

`R = L/kA`

where:

– `L`

is the thickness of the material, measured in meters (m)

– `k`

is the thermal conductivity of the material, measured in watts per meter-kelvin (W/mK)

– `A`

is the cross-sectional area of the material, measured in square meters (m²)

The thermal resistance is inversely proportional to the thermal conductivity, meaning that materials with lower thermal conductivity have higher thermal resistance and vice versa.

## Steady-State Heat Transfer

In steady-state heat transfer, the temperature distribution in a system does not change with time. The rate of heat transfer in steady-state heat transfer is given by the equation:

`Q = (kA/L)ΔT`

where:

– `Q`

is the rate of heat transfer, measured in watts (W)

– `k`

is the thermal conductivity of the material, measured in watts per meter-kelvin (W/mK)

– `A`

is the cross-sectional area of the material, measured in square meters (m²)

– `L`

is the thickness of the material, measured in meters (m)

– `ΔT`

is the temperature difference across the material, measured in kelvin (K)

This equation is derived directly from Fourier’s law of heat conduction and the definition of thermal resistance.

## Transient Heat Transfer

In transient heat transfer, the temperature distribution in a system changes with time. The rate of heat transfer in transient heat transfer is given by the equation:

`Q = ρcAdT/dt`

where:

– `Q`

is the rate of heat transfer, measured in watts (W)

– `ρ`

is the density of the material, measured in kilograms per cubic meter (kg/m³)

– `c`

is the specific heat capacity of the material, measured in joules per kilogram-kelvin (J/kg·K)

– `A`

is the cross-sectional area of the material, measured in square meters (m²)

– `dT/dt`

is the rate of change of temperature with respect to time, measured in kelvin per second (K/s)

This equation takes into account the thermal inertia of the material, as represented by the product of density and specific heat capacity, which determines the material’s ability to store and release heat.

## Numerical Examples and Problem-Solving

To further illustrate the concepts of conduction heat transfer, let’s consider some numerical examples and problem-solving exercises.

**Example 1: Steady-State Heat Transfer through a Wall**

Suppose a wall made of brick (k = 0.72 W/mK) has a thickness of 0.2 m and a cross-sectional area of 10 m². The temperature difference across the wall is 20 K. Calculate the rate of heat transfer through the wall.

Using the steady-state heat transfer equation:

`Q = (kA/L)ΔT`

Substituting the values:

`Q = (0.72 W/mK × 10 m² / 0.2 m) × 20 K`

`Q = 720 W`

**Example 2: Transient Heat Transfer in a Metal Plate**

Consider a metal plate with a thickness of 0.05 m, a cross-sectional area of 1 m², a density of 2700 kg/m³, and a specific heat capacity of 900 J/kg·K. The plate is initially at a uniform temperature of 20°C, and its surface is suddenly exposed to a heat flux of 1000 W/m². Calculate the temperature of the plate after 10 seconds.

Using the transient heat transfer equation:

`Q = ρcAdT/dt`

Rearranging the equation to solve for `dT/dt`

:

`dT/dt = Q / (ρcA)`

Substituting the values:

`dT/dt = 1000 W/m² / (2700 kg/m³ × 900 J/kg·K × 1 m²)`

`dT/dt = 0.4 K/s`

Integrating the equation to find the temperature change after 10 seconds:

`ΔT = dT/dt × Δt`

`ΔT = 0.4 K/s × 10 s`

`ΔT = 4 K`

Therefore, the temperature of the plate after 10 seconds is 20°C + 4°C = 24°C.

These examples demonstrate the application of the equations and principles discussed earlier, allowing you to solve practical problems in conduction heat transfer.

## Conclusion

Conduction heat transfer is a fundamental concept in thermodynamics and heat transfer, with a wide range of applications in various fields, from engineering to materials science. By understanding the key principles, such as thermal conductivity, temperature gradient, Fourier’s law, thermal resistance, and the differences between steady-state and transient heat transfer, you can effectively analyze and solve problems related to conduction heat transfer.

This comprehensive guide has provided you with the necessary tools and knowledge to delve deeper into the subject of conduction heat transfer. Remember to practice solving numerical problems and applying the concepts to real-world scenarios to solidify your understanding of this important topic.

## Reference:

- Thermtest – Thermal Conductivity – What It Is and It’s Formula: https://thermtest.com/what-is-thermal-conductivity
- ScienceDirect – Quantitative analysis on the heat transfer modes in the process of thermal runaway propagation in lithium-ion battery pack: https://www.sciencedirect.com/science/article/abs/pii/S001793102100586X
- C-Therm – What is Thermal Conductivity? How is it Measured?: https://ctherm.com/resources/newsroom/blog/what-is-thermal-conductivity/

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