Quick calculations for radiation heat transfer

In certain cases, radiation heat transfer is important to include in one’s calculations. Radiation heat transfer is nonlinear because the heat flux is proportional to the temperature to the fourth power. I find two frequent simplifications for radiation problems quite useful. The first approximation is to linearize and create an effective convection coefficient. The second approximation is the effective emissivity for two bodies which are transferring heat between each other via radiation.

Linearization of radiation heat transfer for effective convection coefficient

Heat transfer by radiation is nominally given by

$\dot{q}=\sigma \epsilon (T^4-T_0^4)$

$\dot{q}$ is the heat flux (W/m $^2$ in SI units), $\sigma$ is the Stefan-Boltzmann constant (5.670 x 10 $^-8$ W/m $^2$ K $^4$ ), and $\epsilon$ is the emissivity, and $T$ and $T_0$ are the absolute temperatures of the surfaces. Note: Another consideration that one must also include is the view factor which describes how well the two bodies are facing each other, but we will discuss that in a future post.

When the temperatures $T$ and $T_0$ are not too far apart, then we can linearize the nonlinear relation above to obtain a linear convection relation of the form

$\dot{q}=h (T-T_0)$

where $h$ is the convection coefficient (W/m $^2$ K in SI units) to obtain

$\dot{q}=4\sigma \epsilon T_{char}^3 (T-T_0)$

Therefore, the effective heat transfer coefficient is

$h_{rad}=4\sigma \epsilon T_{char}^3$

where $T_{char}$ is a characteristic temperature. As the characteristic temperature, one may use the nominal value $T$ , $T_0$ , or some value in between, such as the mean. Depending on the problem one may choose the value based in order to make a conservative approximation.

This relation can be derived in two ways. First, one can factor the original relation into

$\dot{q}=\sigma \epsilon (T-T_0) (T+T_0) (T^2+T_0^2)$

If one defines that mean temperature as $T_{mean}=(T+T_0)/2$ and assumes that T and T_0 are not too far apart, then $T^2+T_0^2 \approx 2T_{mean}^2$ and one finds

$\dot{q}=4\sigma \epsilon T_{mean}^3 (T-T_0)$

A second way involves using a Taylor series expansion with $T=T_0+\delta T$ . Then the heat flux is given by

$\dot{q}=\sigma \epsilon \left( (\delta T+T_0)^4-T_0^4 \right)$

Expanding and dropping higher order terms of $\delta T^2$ and above, we obtain

$\dot{q}= 4 \sigma \epsilon T_0^3 \delta T = 4 \sigma \epsilon T_0^3 (T-T_0)$

Above, I wrote that a few different characteristic temperatures can be used. Those can be obtained here depending on the expansion. One could write $T=T_{mean}+\delta T/2$ and $T_0=T_{mean}-\delta_T/2$ for example.

Effective emissivity between two surfaces

Suppose we have two bodies with emissivities $\epsilon_1$ and $\epsilon_2$ at temperatures $T_1$ and $T_2$ transferring heat between each other as sketched below. Each body emits radiation as $\dot{q}_i=\sigma \epsilon_i T_i^4$ but the net heat transfer is not the simple difference because the incoming and outgoing radiation between the two bodies are coupled. The net heat transfer is given by