# How do you calculate power at the focal point of a mirror?

I'm a Mechanical Engineering student and I'm working on my senior project, so I need help. My project is about designing a solar dish having a diameter of 1.5 meters and a focal length of 60cm. so at the focal point, a circular coil (copper pipe) will be folded, in order to have a superheated steam as an output. What I'm struggling to know is: How to calculate the Power at the focal point?

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The upper bound is easy. The solar constant through the atmosphere is about $970W \over m^2$ so you're gathering at most $\pi\cdot((0.75m)^2 \cdot 970W / m^2 = 1.7kW$

If your focal point is perfect (it isn't) then you'd have infinite power density.

You will need to measure the actual power produced (and therefor the efficiency) because it depends on a lot of factors like the reflectivity of the coper and so forth.

Edit: fixed solar constant to account for atmospheric absorption

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Aha! Beat me by a few seconds. :) – Chris White Mar 29 '13 at 23:33
Yeah but your answer is better :-) – Brandon Enright Mar 29 '13 at 23:34
@BrandonEnright You have used the AM0 solar spectrum. The AM1 (at one atmosphere absortpion) solar spectrum is only 925 W/m$^2$. – JKL Mar 29 '13 at 23:44
@JKL hmm yeah and latitude seems to really matter too. I adjusted down to 970W and added a link. – Brandon Enright Mar 29 '13 at 23:53

Energy is conserved, and it is presumably not building up anywhere prior to hitting the water. Thus whatever power hits the dish is the same as is delivered to the water, with two catches:

• Nothing is a perfect reflector, especially over a broad range of frequencies, so the dish will only transmit a fraction $\alpha < 1$ of the incoming light. Hopefully $\alpha$ is close to $1$, but it will depend on the material used.
• Some of the light may not be quite focused onto the water-in-copper component. This is due to both the diffraction limit (likely a negligible effect) and just imperfections in the reflective surface. Call the fraction of reflected power that actually reaches the water $\beta$.

Aside from these technical considerations, the power you collect will be the power striking the dish. The Sun gives a flux of about $F = 1.36~\mathrm{kW}/\mathrm{m}^2$ at the Earth's location. This is the yearly averaged solar constant, so it varies depending on time of year (Earth is closer to the Sun around January). Only a fraction $\gamma < 1$ of this light will make it through the atmosphere, where $\gamma$ depends on weather and the Sun's elevation above the horizon.

Your dish has a collecting surface of $A = \pi (0.75~\mathrm{m})^2 = 1.77~\mathrm{m}^2$, so the total power you get is $$P = \alpha \beta \gamma AF = \alpha \beta \gamma \cdot 2.4~\mathrm{kW}.$$ This is averaged over the course of a year and assumes you point the dish directly at the Sun. Of course, $\alpha \beta \gamma$ may be significantly less than $1$.

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