2
$\begingroup$

This refers to a previous question re: Solar Cell Systems and as part of the answer the statement:

If we concentrated sunlight to [the] maximum amount 42600× was given.

My questions are:

By what physical method is this achieved?

How do we know that figure of 42600x is the theoretical (or practically achievable) maximum limit of concentration?

$\endgroup$
  • $\begingroup$ This (p. 7) looks close to the number given by you. $\endgroup$ – Keep these mind Apr 12 '15 at 18:23
  • $\begingroup$ Maybe a question for engineering SE $\endgroup$ – Steeven Apr 12 '15 at 21:07
  • $\begingroup$ @Steeven no prob. at doing that, just please forget the Solar Cell part for a sec, I think the rest is pure physics when you look at Glen's answer, it seems all maths and physics/astronomy more than fancy mirror systems...regards $\endgroup$ – user74893 Apr 12 '15 at 21:22
4
$\begingroup$

The sun is an extended source. This means that it occupies a definite solid angle in the sky $\omega = 6.8\times 10^{-5} Sr$.

To visualise this (not to scale), let say that the black area in the following diagram is the angular extend of the sun as seen from the surface of the Earth (ignore the other labels),

Angular extend of the sun.

What happens when we concentrate sunlight from the perspective of the white plate in the diagram? Answer: the sun looks bigger! It beings to fill more of the "sky".

When light rays are travelling from all angles towards the plate then we have reached maximum concentration.

Full concentration.

So maximum concentration is defined as when the true angular size of the sun is made to fill a hemisphere solid angles. The angular size of the sun is approx. $\theta_s = 0.2666^{\circ}$, note this is the half-angle, therefore performing the solid angle integration yields,

$$ X_{2D} = \frac{\iint_{2\pi}\cos\theta\ d\omega}{\iint_{\omega_s}\cos\theta\ d\omega} = \frac{\int_0^{2\pi}d \phi \int_0^{\pi/2} \cos\theta\sin\theta\ d \theta }{\int_0^{2\pi}d \phi \int_0^{\theta_s} \cos\theta\sin\theta\ d \theta} = \frac{\frac{2\pi}{2}}{\frac{2\pi\sin^2\theta_s}{2}} = \frac{\pi}{6.8\times 10^{-5} Sr} = 46200\times $$

This assumes that we can change the angular extend of light in both $x$ and $y$ and focus to a point. If instead you can only focus to a line then the limit is much smaller,

$$ X_{1D} = 220\times $$

$\endgroup$
  • $\begingroup$ I was thinking, before Glen's comment, of the most complicated system of mirrors! Simplest & most straightforward answers are the hardest to think of sometimes. thanks and regards $\endgroup$ – user74893 Apr 12 '15 at 21:34
  • $\begingroup$ If you like this answer, you may be interested in a similar question that asks about the temperature that can be achieved at the focus of a given mirror arrangement. $\endgroup$ – Floris Apr 12 '15 at 21:42
  • 1
    $\begingroup$ The real limit is that you see the sun in every direction, not just the upper hemisphere of directions. Mirrors do exist! The upper hemisphere alone should only give you 25000x, iirc. You must have made an error but I'm not sure what. $\endgroup$ – Steve Byrnes Apr 13 '15 at 0:23
  • $\begingroup$ This is correct and gives the general maximum solar concentration limit. Write an alternative answer that backs up your point if you have something to contribute. $\endgroup$ – boyfarrell Apr 13 '15 at 7:35
  • $\begingroup$ The factor of two is cancelled when performing the angular integrations. $\endgroup$ – boyfarrell Apr 13 '15 at 17:41
1
$\begingroup$

You can actually do this a bit more simply (or at least without integrations).

Luminance is invariant in geometrical optics. That is, the brightness of an image cannot be brighter than the source.

The radius of the sun is 0.6958 x 10^6 meters. The radius of the earth's orbit is a mean of 149.6 x 10^6. Then the brightness at the surface of the sun is the square of the ratios, or (149.6 /.6958)^2, or 46,228 as bright as it is at earth orbit, and this represents the best performance of a solar concentrator.

$\endgroup$

Your Answer

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