I would like to calculate the heating of a polyethylene solar collector as a result of the sun's radiation. the collectors are uncovered collectors.

Suppose I have a certain intensity $I$ of solar radiation in $\mathrm{W/m}^2$, how can I determine the heating of the collector?

I myself think through the following formula: $$ I = \sigma * (T_c ^ 4 - T_o ^ 4) $$ where $T_c$ is the temperature of the collector and $T_o$ is the outside temperature. The collector is assumed to be a black body (stefan-boltzmann).

What is your opinion about this calculation?

I apologize for my English but I am from the Netherlands


This assumes your solar panels are blackbody absorbers. This is definitively false, since solar panels can give you an efficiency rating for converting incident radiation to electrical energy, which means it is not transformed immediately to heat. For example, if the efficiency were $17\%$ and the reflectance was $0$, then the absoptivity would be $0.83$ instead of $1$. However, solar panels also reflect quite a bit, which means it's likely that your panel is far from a perfect blackbody absorber.

Furthermore, to calculate the ideal equilibrium temperature of the panel, you would not assume the solar irradiance is simply the difference in the total irradiance from the collector and environment. If $I$ is the irradiance at the solar panel, then at equilibrium we find that

$$\sigma(\epsilon T_c^4-\alpha T_o^4)=\frac{A_{Cc}}{A_{Ct}}\alpha I$$ where $\epsilon$ is the emissivity of the collector, $\alpha$ is the absorptivity, $A_{Cc}$ is the cross-sectional area of the collector exposed to sunlight, and $A_{Ct}$ is the total surface area of the collector.

The solar irradiance, $I$, already includes the solar temperature and accounts for your distance from the surface of the Sun. This equation asserts equilibrium by ensuring that the energy absorbed by the sun-facing side as well as from the surroundings is completely re-emitted across all sides of the collector. It also accounts for the fact that solar collectors often have very difference values of absorptivity and emissivity. If we were to exclude the ratio of areas, it would be like having the sun shine uniformly on all sides of the collector, which would heat up quite a bit more.

However, I want to caution you from using this approximation in a practical setting. Unless your solar panel is in space, this is a poor approximation because cooling mechanisms like convection and conduction will drastically lower the equilibrium temperature.

If this isn't for practical considerations or the solar panel is in space, then feel free to use this. Values for the absorptivity and emissivity of most commercially available solar panels can be found online with a bit of searching.

  • $\begingroup$ I'm using PE-tubes to absorb the sun's radiation and directly use it to heat up the temperature of the liquid in the collector. I don't use solar panels to make electricity from it. The emissivity of the material is 1 so all the radiance is used to heat up the liquid $\endgroup$ May 10 '18 at 15:28
  • $\begingroup$ @MischaUrlings Great, so then the only correction would be for the cross-sectional area exposed to the sun. Also, you'd change the total area to simply be the area exposed to the liquid $\endgroup$
    – Jim
    May 10 '18 at 17:51
  • $\begingroup$ I don't really know the area that is exposed to the sun but let's say it's half of the absorber area. What would my formula look like in my situation? $\endgroup$ May 11 '18 at 8:05
  • $\begingroup$ @MischaUrlings Usually, the area exposed to the sun is what you'd get if you projected the object into 2D. For a sphere like Earth, it's 1/4 the surface area. For a cylinder side-on, its $rh$ instead of $2\pi r(h+ r)$. A good rule of thumb is take a picture of it, trace the outline of the object in the picture and find the area of that shape, that's what you use. $\endgroup$
    – Jim
    May 11 '18 at 12:17
  • $\begingroup$ If we actually say it's half, then you still need the area exposed to the liquid. I daresay it's not the entire surface area, unless the collector is free-floating in the liquid. But, as I said, just replace the $A_{Ct}$ in the equation I gave with $A_{Cl}$ (area of collector exposed to liquid) and then that is exactly what your equation looks like $\endgroup$
    – Jim
    May 11 '18 at 12:20

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