Calculate the rate at which a hollow cylinder with a heating element inside will lose heat, specifically through both conduction and radiation.
My goal is to find the temperature inside this cylinder. I realize that the total heat loss per second will be equal to the total amount of power that the heating element consumes.
I used Fourier's Law $q=(\frac{k}{s}) \cdot A \cdot (t1-t2)$, but my understanding is that this law only accounts for heat transfer through conduction and ignores radiation. How can I account for both radiation and conduction to solve this problem?
Can I determine a percent of heat transfer by radiation and a percent of heat transfer through conduction?
Best Answer
If you are neglecting convective heat transfer and conductive heat transfer outside the cylinder, then the radiative heat transfer is in series with the heat transfer through the wall. If the wall is thin, then the heat flow from the heater is related to the temperature difference through the wall by $$Q=\pi D L k\frac{(T_i-T_o)}{W}\tag{1}$$where W is the wall thickness. The radiative heat transfer outside is given by $$Q=\pi DL\epsilon \sigma(T_o^4-T_{\infty}^4)\tag{2}$$where $T_{\infty}$is the surroundings temperature far from the surface, $\sigma$ is the Stefan-Boltzmann constant and $\epsilon$ is the emissivity of the tube surface. So, knowing Q, you can use Eqn. 2 to get $T_0$, and then use Eqn. 1 to get $T_i.$