I'm exploring a DIY project about solar power and have found very few resources that can explain, in a not too simple but not too advanced way, how to go about calculating important values in solar power, specifically concentrated solar power. So maybe someone here could help me get started or show me where I can find answers to this question.

I am wondering how I can calculate the temperature that a receiving point of focus can attain from concentrated sunlight. The following example shows the basic variable factors that I am aware of. Suppose:

  • There is a material in the shape of a dish with a surface area $x$ and a reflectivity factor $r$
  • There is a clear cloudless sky with sun directly above the dish and the current air temperature is $y$ (e.g. $25^\circ\mathrm{C}$ or $77^\circ\mathrm{F}$)
  • The dish reflects light onto a receiver which is at a perfect focal point
  • The receiver has heat absorption efficiency of $a$ and insulation efficiency of $i$.

Is there a simple formula that can put these factors (and probably others I am unaware of) together to calculate or approximate the temperature $q$ the receiver can reach after a time $t$. For example, something like $$q = rxyt + ai \tag{no doubt this is wrong}$$

Ignore for simplicity (if you like) some of the details for achieving a geometrically optimal shape, and the different amounts of heat depending on the location of the dish on the earth and the time of the day.

I am most interested in the qualities needed by the reflecting material. Besides geometric efficiency, is reflectivity the main factor in the performance of the reflecting material?

I have fairly beginner/amateur level knowledge of physics. In this area I have read chapters about optics in a first year level college physics book. But optics seems like such a vast area, at the moment I am just interested in finding out the details of just a small area of it.

  • $\begingroup$ It can't get any hotter than the surface of the sun, about 10^4 Fahrenheit. Other than that, it's all a matter of how efficiently you can draw energy away from the spot. $\endgroup$ – Mike Dunlavey Dec 3 '11 at 15:17
  • $\begingroup$ @MikeDunlavey: Why should it not get hotter than the surface of the sun? $\endgroup$ – Alexander Dec 3 '11 at 17:35
  • $\begingroup$ Welcome to Physics.SE! Notice the nicely rendered math in the answer. That comes to us courtesy of MathJax which renders a subset of LaTeX. There are some notes in the FAQ. $\endgroup$ – dmckee --- ex-moderator kitten Dec 3 '11 at 17:51
  • 2
    $\begingroup$ @Alexander: because then the sun would be cooler than it, so it would radiate backward. $\endgroup$ – Mike Dunlavey Dec 3 '11 at 18:09
  • 1
    $\begingroup$ @Alexander: The temperature matters. Check out the Stefan-Boltzman Law. Black-body radiative heat emission is proportional to T^4, and radiation goes both ways between two bodies. As a result, they will seek the same temperature if there's nothing else going on. If the receiver is at a higher temperature than the sun, it will cool off because the sun is colder than it is, no matter how much sun-coldness is focussed on it. $\endgroup$ – Mike Dunlavey Dec 4 '11 at 3:36

The first thing you should be aware of, is the difference between temperature and heat. Heat is thermal energy (energy related to the mean speed of the molecules/atoms in your material) that a body has, and is useful because energy is conserved. So if you have heat transfer from a body to another, if you know that the first body gives the second 10 Joules, the second receives the same amount. Temperature is related to the amount of heat a body has, but it is not conserved, because different materials have different heat capacities. This means that different materials will have different temperatures when they have the same thermal energy.

In your solar heater, the receiver gets heat from the sun ($Q_\mathrm{sun}$), and loses heat to the atmosphere ($Q_\mathrm{atm}$). The steady state will be achieved when the energy flux to the body is equal to the energy flux from the body. That is, when $Q_\mathrm{sun} = Q_\mathrm{atm}$. Note that the units of $Q$ are Energy/time (e.g., Watts)

From the parameters you give, the power the receiver gets is $Q_\mathrm{sun}=P_\mathrm{sun}\,x\,r\,a$, that is, the power density of solar radiation at sea level (about 1.0 kW/m${}^2$) times the area of your dish times its reflectivity times the fraction of that that your receiver absorbs. Notice that this is (essentially) independent of the temperature of the receiver. I believe $a$ might depend a little on that, but not very much.

On the other hand, the heat your receiver loses is not so easy to calculate. It will lose heat by three different processes: radiation, conduction and convection. I think (though I'm not very familiar with this) that convection will be the most important process, but this requires that your receiver is thermally isolated form the rest of the structure of your solar condenser.

So lets say that $Q_\mathrm{atm} = Q_\mathrm{convection}$. If we use Newton's law of convection (not always valid, but a reasonable approximation here, I think) we have $Q_\mathrm{convection}= h\,S(T-y)$, where $h$ is the heat transfer coefficient and $S$ is the surface of the receiver. Notice that this does depend on the temperature of the receiver($T$). I'm not sure what your parameter $i$ is, but I think is related to the conduction of heat.

From the above equations, you can solve for $T$: $$P_\mathrm{sun}\,x\,r\,a = h\,S(T-y)$$ $$T = y + \frac{P_\mathrm{sun}\,x\,r\,a}{h\,S}$$

Regarding the most important factors in the performance, it's clear that the reflectivity is very important, not only in the visible, but in the infrared too, sine a lot of the energy of the sun's power is infrared. A metallic mirror is good for this. The absorption efficiency will be equally important, too. I think tis will be related to the reflectivity of your receiver: you want it as pitch black as possible (again, in the visible and infrared). You might want to paint it with soot, for example. I'm sure there is a lot of literature on the subject, but again, I'm not an expert on this.


I would encourage you to get familiar with the concept of a Heat Engine, of which a Stirling Engine is my favorite example. It has a region of high temperature, and a region of low temperature. Heat energy flows away from the high-temperature region (thus cooling it), and is split into two components, 1) useful work, and 2) waste heat in the low-temperature region (warming it up).

It can work in reverse as well. If you put work in, rather than take it out, it can move heat from the low-temperature region (cooling it) to the high-temperature region (warming it). That's what a refrigerator does. (In fact, if you power a stirling engine with an electric motor, its cooling fins will get colder.)

So in the case of your solar collector, it's important to know what your cold sink is, how efficiently you are moving heat from the hot to the cold, and how you are harnessing the useful work that results.

  • 1
    $\begingroup$ I appreciate the input, but I think your answer deals with how to efficiently utilise the heat generated by a solar concentrator, whereas I am asking in this case how to calculate how hot it could get. $\endgroup$ – dbjohn Dec 3 '11 at 19:23
  • $\begingroup$ @dbjohn: It's tough to calculate how hot it would get because that depends on how efficiently you can extract the heat. Just consider some extremes. If it were perfectly insulated and surrounded by mirrors so it could not radiate to the sides, it would just go up to sun temp. OTOH, if it were immersed in water, it couldn't get any hotter than 212F. If it were in tight contact with dry ice or liquid nitrogen, that would determine its temp. It all depends on how you carry away the heat. $\endgroup$ – Mike Dunlavey Dec 4 '11 at 2:42

Lets say your receiver is a copper plate, if you leave it in sunshine in the conditions you describe you get certain temperature, lets say 50 c, lets say you calculate the area of your dish an it is 1 square meter, now you place the piece of copper close to the focal point, and you can see a concentrated bright dot, and you measure the area of that dot, lets say it was .0025 square meters this means you concentrate the sun 400 times, I would say if ones sun gives you 50 degrees you are going to get 400 times 50 in that spot if the heat can't go any were else (which not be the case). That is 20000 c approximate 4 times the temperature at the surface of the sun.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.