Question

How do you calculate the radiative heat transfer coefficient?

Answer 1

Hint: You could firstly give a brief description on what radiative heat transfer is. Then you could give a brief account on Stefan-Boltzmann law and then relate it with the above coefficient. Thereby, you could recall the required expression used in the calculation. Also mention the significance of the coefficient.

Complete answer:
In the question, we are asked how we calculate the radiative heat transfer coefficient. These Heat transfer coefficients are normally used in characterizing the thermal energy in terms of the temperature difference between the two participating media.

From the Stefan-Boltzmann law, we could relate the total amount of emitted radiation in terms of the object’s temperature as,
$\varepsilon =\sigma {{T}^{4}}$
Here, $\varepsilon $ is the total amount of radiation emitted per meter square (emissivity), $\sigma $ is the Stefan-Boltzmann constant and T is the temperature of the radiating object in Kelvin.

We could define Radiative heat transfer as the comparison of the emissivity of the other materials to that of a blackbody.

Let,${{h}_{rad}}$ be the radiative heat transfer coefficient, T be the temperature of the media that are participating in a given radiative exchange, $\varepsilon $ the emissivity of the emitting surface, $\sigma $ Stefan- Boltzmann constant and numbers 1 and 2 denotes the radiation of the emitting and absorbing surfaces respectively, then, for the radiative exchange between two gray diffuse surfaces, the radiative heat transfer coefficient can be calculated by,
${{h}_{rad}}=\varepsilon \sigma \left( {{T}_{1}}^{2}+{{T}_{2}}^{2} \right)\left( {{T}_{1}}+{{T}_{2}} \right)$

Therefore, we found that we could calculate the radiative heat transfer coefficient by substituting accordingly in the above equation.

Note:
We could also find the thermal resistance to the radiative transfer by using the following expression,
${{R}_{rad}}=\dfrac{1}{{{h}_{rad}}{{A}_{1}}}$
Where, ${{A}_{1}}$ is the area of the emitting surface. Also, the value of Stefan-Boltzmann constant is given by,
$\sigma =5.67\times {{10}^{-8}}W/{{m}^{2}}{{K}^{4}}$
Also, gray surface could be defined as that reflects or absorbs the thermal radiation that would have been absorbed by a blackbody surface.