Thermal Diffusivity

Under steady-state conditions, thermal conductivity indicates how rapidly heat moves through a substance from the hot side to the cold side, thermal diffusivity of a substance is how well a heat may be dispersed by material, taking into consideration both how rapidly the heat may be transmitted through and the speed at which its own temperature may fluctuate.

A measurement of a substance's or energy's ability to diffuse or to permit something to pass through diffusion define diffusivity.

What is Thermal Diffusivity?

The rate at which temperature spreads through a material is referred to as thermal diffusivity. Thermal conductivity and heat thermal capacity are used to compute thermal diffusivity. We use D or α (alpha) to denote thermal diffusivity. The SI unit of thermal diffusivity is m2/s.

Thermal Conductivity

It is the process by which heat is transferred from hotter to cooler regions of a body, bringing the temperature closer to equilibrium. Thermal conduction, as opposed to heat transmission through convection, results from a direct energy transfer between particles, such as molecules, atoms, and electrons, with higher and lower energies. Thermal conduction is unrelated to macroscopic displacements in the body. We use ‘k’ or ‘λ’ to represent it. Thermal resistivity is the reciprocal of thermal conductivity.

Thermal Diffusivity Formula

Thermal diffusivity is the ratio of thermal conductivity to the product of material density and specific heat capacity when the pressure is constant.

$\alpha ={\displaystyle \frac{k}{\rho c_p}}$

Where

k – Thermal conductivity

$\rho$- Density of the material

$$c_p$$- Specific heat capacity

$\rho c_p$- volumetric heat capacity

In other words, we can also say that thermal diffusivity is the ratio of the heat allowed to travel through the material to the heat held in the material per unit volume.

Because thermal diffusivity is a ratio between thermal conductivity and volumetric heat capacity at constant pressure, a material with a high value of thermal diffusivity doesn't always disperse heat better than a material with a low value of thermal diffusivity.

Measurement of Thermal Diffusivity

Laser Flash Method: The thermal diffusivity of several different materials is measured using the laser flash analysis or laser flash technique. A plane-parallel sample has one side heated by an energy pulse, and the time-dependent temperature increase caused by the energy input is seen on the opposite side. Energy moves toward the backside more quickly the higher the sample's thermal diffusivity.

Infrared Thermography: The science of collecting and analysing thermal data from non-contact thermal imaging sensors is known as infrared (IR) thermography. IR thermography uses the electromagnetic spectrum's infrared region to find the energy that has been released. This translates to wavelengths outside of the range of visible light.

Thermal Diffusivity of Different Materials

Interesting Facts

• A heat sink is a device created to transfer heat from one piece of equipment to another. To permit the rapid transfer of heat, a heat sink must have a very high thermal diffusivity. These are used in electronic devices and computers.

• Used in electric devices and machines like refrigeration systems, heating machines, machining, and other industrial purposes.

Solved Problem

1. Calculate the thermal diffusivity of the water at 25 degrees Celsius with a density of 997 kg/m3, a specific heat capacity of 4182 J/kg.K, and thermal conductivity of 0.6071 W/m.K.

Sol:

Given

k – 0.6071 W/m.K

$\rho$- 997 kg/m3

$$c_p$$- 4182 J/kg.K

The formula for thermal diffusivity:

$\alpha ={\displaystyle \frac{k}{\rho c_p}}$

$\Rightarrow \alpha ={\displaystyle \frac{0.6071}{997\times 4184}}$

$$\therefore \alpha =1455$$ mm2/s

The thermal diffusivity of water is $$\alpha =1455$$ mm2/s

2. Calculate the thermal diffusivity of the Gold with a density of 19300 kg/m3, a specific heat capacity of 130 J/kg.K, and thermal conductivity of 318 W/m.K.

Sol:

Given

k – 318 W/m.K

$\rho$- 19300 kg/m3

$$c_p$$- 130 J/kg.K

The formula for thermal diffusivity:

$\alpha ={\displaystyle \frac{k}{\rho c_p}}$

$\Rightarrow \alpha ={\displaystyle \frac{318}{19300\times 130}}$

$$\therefore \alpha =126.74$$ mm2/s

The thermal diffusivity of Gold is $$\alpha =126.74$$ mm2/s

Summary

In many sectors, thermal diffusivity is crucial for choosing the best materials that can transfer heat efficiently without overheating themselves. In a real-world application, it is also very useful where heat transfer takes place in computers and electronic devices. It is very important that this process happens easily otherwise creating heating problems.