0

I would like to calculate the steady state temperature of a small cube of any material heated by electromagnetic radiation. Suppose the cube floats in air and doesn't touch anything. The air is at 20 °C.

What I have understood so far

By knowing the complex dielectric constant of the material, which is $$\underline{\varepsilon_r}=\varepsilon_r'-j\varepsilon_r''$$ and the electric field strength $E_\mathrm{rms}$, I can calculate the power dissipated in the cube like $$\frac{P}{V}=\omega\varepsilon_0\varepsilon_r''E_\mathrm{rms}^2$$ where $V$ is the volume of the cube. From there I can calculate the heat rise rate $\Delta T/t$ together with the density $\rho$ and the specific heat capacity $c_\mathrm{p}$ of the cubes material like $$\Delta T/t=\frac{P}{V}\cdot\frac{1}{\rho c_\mathrm{p}}$$ I know that these formulas are only valid for a homogenous electrical field inside the cube, which isnt the case in reality. But this simplification is good enough for me.

What I would like to know

How can i calculate the steady state temperature of the cube from the heat rise rate? What additional Information do I need to know (like emission coefficient? or something like that?)

I have read a few things about newtons law of cooling but didnt really understand how to use it in my case. I would have somehow incorporated the heat rise rate in the law of cooling but i dont even know how to find the coefficients required to use the law of cooling alone (like in the case of an already heated cube floating in the air that cools down)...

I would be very thankful for any help!

Edit: What I have learned from the answers

So from the answer of @Newbie I have found the following solution for a time dependent description of the temperature increase $$T(t)=T_\mathrm{air}+\frac{\omega\varepsilon_0\varepsilon_\mathrm{r}''E_\mathrm{rms}^2a}{6k}\left[1-e^{-\frac{6ka^2}{mc_\mathrm{p}}t}\right]$$ which looks like it makes sense. (It behaves similar to the charges on a capacitor over time when charging it).

x3b7z99
  • 165

2 Answers2

0

I wouldn't recommend calculating the rate of temperature variation $\Delta T/t$ as you attempted. Instead just start with the definition of steady state; that the input and output heat to the cube should cancel out. As you already mentioned the power per unit volume dissipated in the cube is $$P^{\rm in}_{V}=\omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}$$ On the other hand, the heat transferred to the surrounding air may be calculated via Newton's law of cooling via $$P^{\rm out}=kA(T_{\rm ss}-T_{\rm air})$$ Note that this power is not per unit volume and will be lost through each 6 surfaces of the cube; there is a surface area $A$ in $P^{\rm out}$. Assuming the cube has dimension $a$ we have $$P^{in}_{V}a^{3}=P^{\rm out}\rightarrow \omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}a^{3}=6ka^{2}(T_{\rm ss}-T_{\rm air})$$ Thus, $$T_{\rm ss}=T_{\rm air}+\frac{1}{6}\frac{\omega\epsilon_{0}\epsilon_{\rm r}^{''}E^{2}_{\rm rms}a}{k}$$

Considering the convection, conduction, and radiation mechanisms of heat transfer and the context of the problem, i.e., the fact that the cube is in direct contact with air, I think the process of heat transfer should be conduction through the cube surfaces.

Newbie
  • 703
-1

The question is that of heat balance: the object obtains energy at a rate $R_{in}$ and loses it at rate $R_{out}(T)$, which typically depends on the temperature. One then could find the steady state temperature by solving $$R_{in}=R_{out}(T)$$ for $T$.

The cooling can proceed via different mechanisms (see, e.g., the discussion in this answer, and particularly the calculations in the linked article): radiation, convection, heat conductance. Heat conductance is usually approximated as proportional to the temperature difference (which is essentially a discrete form of the Fourier's law): $$R_{conductance}=q(T_{object} - T_{environment}),$$ whereas the heat lost via radiation is well described by the Stefan-Boltzmann law: $$R_{radiation}=\alpha T^4.$$

Radiation is probably the best first try in your case, judging by how you try to address the problem via dielectric constant - here it is perhaps possible estimate $\alpha$ in terms of the dielectric constant.

Roger V.
  • 68,984