Question

Consider a spherical shell of inner radius \(r_{1}\), outer radius \(r_{2}\), thermal conductivity \(k\), and emissivity \(\varepsilon\). The outer surface of the shell is subjected to radiation to surrounding surfaces at $T_{\text {sarr }}$, but the direction of heat transfer is not known. Express the radiation boundary condition on the outer surface of the shell.

Short answer

The radiation boundary condition for the heat transfer on the outer surface of the spherical shell is expressed as: $$\frac{dT}{dr} = -\frac{\varepsilon\sigma}{k} (T^4 - T_{sarr}^4)$$ Where: - \(\frac{dT}{dr}\) is the temperature gradient in the radial direction - \(\varepsilon\) is the emissivity of the shell - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times 10^{-8} W m^{-2}K^{-4}\)) - \(k\) is the thermal conductivity of the shell - \(T\) is the temperature of the outer surface of the shell - \(T_{sarr}\) is the temperature of the surrounding surfaces

Step-by-step solution

Identify the heat transfer mechanisms

In this problem, there are two modes of heat transfer involved: conduction within the spherical shell and radiation between the outer surface of the shell and the surrounding surfaces.

Express the conduction heat transfer rate

The rate of conduction heat transfer through the shell can be expressed using Fourier's law of heat conduction. For a spherical shell, the heat transfer rate (Q_cond) is given by: $$Q_{cond}=-kA\frac{dT}{dr}$$ Here, - \(k\) is the thermal conductivity of the shell - \(A\) is the surface area of the shell at radius \(r\), which can be calculated as \(A = 4\pi r^2\) - \(\frac{dT}{dr}\) is the temperature gradient in the radial direction

Express the radiation heat transfer rate

To calculate the rate of heat transfer between the outer surface and the surroundings due to radiation, we need to apply the Stefan-Boltzmann law. The heat transfer rate (Q_rad) can be expressed as: $$Q_{rad}=\varepsilon A\sigma (T^4 - T_{sarr}^4)$$ Here, - \(\varepsilon\) is the emissivity of the shell - \(A\) is the surface area of the outer surface of the shell - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times 10^{-8} W m^{-2}K^{-4}\)) - \(T\) is the temperature of the outer surface of the shell - \(T_{sarr}\) is the temperature of the surrounding surfaces

Equivalent conduction and radiation heat transfer rates

At the steady-state condition, the rate of heat transfer through conduction within the shell must be equal to the rate of heat transfer due to radiation between the outer surface and its surroundings. Therefore, we can equate the two expressions to get the radiation boundary condition: $$-kA\frac{dT}{dr} = \varepsilon A\sigma (T^4 - T_{sarr}^4)$$

Express the boundary condition for radiation heat transfer

Finally, we can solve for the temperature gradient at the outer surface to obtain the radiation boundary condition: $$\frac{dT}{dr} = -\frac{\varepsilon\sigma}{k} (T^4 - T_{sarr}^4)$$ This expression represents the radiation boundary condition on the outer surface of the spherical shell.