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?

  • $\begingroup$ Are you considering natural convection heat transfer from the cylinder also? $\endgroup$ – Chet Miller Nov 8 '18 at 1:14
  • $\begingroup$ I don't think that convection is really relevant to this problem, I'm just assuming that the entire space inside the cylinder is the same temperature. $\endgroup$ – William Norton Nov 8 '18 at 1:45
  • $\begingroup$ I’m talking about convection outside the cylinder. Are you asking about conductive heat transfer outside the cylinder? $\endgroup$ – Chet Miller Nov 8 '18 at 1:55
  • $\begingroup$ I'm not really sure what you mean by convection outside the cylinder, but I am asking about conductive heat transfer to the outside of the cylinder. $\endgroup$ – William Norton Nov 8 '18 at 2:12
  • $\begingroup$ Conductive heat transfer through the tube wall? $\endgroup$ – Chet Miller Nov 8 '18 at 2:13

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 $T_i.$

  • $\begingroup$ " the radiative heat transfer is in series with the heat transfer through the wall." What did you mean by this? $\endgroup$ – Buraian Oct 9 '20 at 9:56
  • $\begingroup$ @Buraian This is an analogy to an electrical circuit involving resistors in series, where T is analogous to voltage and heat flow Q is analogous to current. $\endgroup$ – Chet Miller Oct 9 '20 at 11:27
  • $\begingroup$ So, the sentence says " the radiative current flow is in series with the current flow through the wall"... I still don't understand what it means for the current flow through the wall to be in series with the radiative flow $\endgroup$ – Buraian Oct 9 '20 at 11:31
  • $\begingroup$ It means that Q in the two equations is the same, and equal to the heat load of the heating element. This allows you to solve for To in the 2nd equation, and substitute that into the first equation to get Ti. $\endgroup$ – Chet Miller Oct 9 '20 at 11:34
  • 1
    $\begingroup$ They are happening at the same time. But heat flow can be regarded in a way as analogous to a flowing fluid. So it first flows through the wall and then through the radiative resistance. $\endgroup$ – Chet Miller Oct 9 '20 at 11:41

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