Learn To Make Your Own Solar Panels And SAVE $$$
Click Here!!!
Solar Power Images
Using mirrors, what is the maximum temperature you can get focusing the sun's rays.?
Solar power is coming. One of the techniques is to use a mirror to focus the sun's rays. What is the maximum theoretical temperature you could get the collector?
Here is a picture: http://en.wikipedia.org/wiki/Image:Power_Tower.jpg
Please disregard potential material failure.
********
And please explain your reasoning
*********
Scythian:
It is not that hard to prove optically. The sun is NOT a point source. The minimum size of its image will be proportional to the size of the collection lens/mirror.
Interesting Thermo = Optics +Blackbody
*******
Thanks Remember.Kelly for the optical analysis. You posted it while I was writing the above.
********
There are several answers worth reading including gtnurse, edward, scythian and remember.kelly to get a full understanding of this problem.
The maxium temperature is limited to the temperature of the surface of the sun for both optical/blackbody and thermodymic reasoning.
Thank you everyone.
Suppose we use a lens of focal length f and radius a to project an image of the sun. The power of the sunlight that is collected over the area of the image is equal to
P = I A Ω,
where I is the intensity of all radiation emitted by the Sun (in watts per unit surface area per steradian), A is the area of the sun's disk, and Ω≈(a/D)^2 is the solid angle subtended by the lens as seen from the Sun. D is the distance of the Sun from the Earth.
Likewise, the total power received by the Sun that is emitted by the image is given by
P' = I' A' Ω',
where I' is the intensity of the radiation emitted by the image, A' is the area of the image, and Ω' ≈ (a/f)^2 is the solid angle subtended by the lens at the image.
In thermal equilibrium
P'=P,
so that
I A Ω = I' A' Ω'.
Or
I' / I = A/A' Ω /Ω'
= (D/f)^2 (a/D)^2 / (a/f)^2 = 1.
Hence:
I =I'.
But, in thermal equilibrium I and I' are given by the black-body radiation law
I= σT^4
I'= σT'^4,
where T=temperature of Sun's surface, T' =temperature of the terrestial image. Hence:
T'=T.
In case I bored somebody to death, Yahoo will pay for the funeral expenses.
*********************************
Edit:
The above result, T'=T is a limiting case based on the assumption that thermal equilibrium exists (expressed by the assumption P'=P), which is unlikely In practice a certain asymmetry exitsts, since the Sun's temperature remains steady whether or not it receives heat from the from Earth, while this is not true of the heated spot in the focus of the lens, which will also lose heat to its surroundings in addition to the Sun. Thus in reality P'<
*****************
To Deep Sky:
Adding more lenses in parallel will increase the rate of heating, but will not alter the final condition that T' can not be larger than T. This is because as you add lenses, it merely increases both the solid angles Ω and Ω' proportionally, so that the result is the same.
Note the subtle point made in the previous addition.
************
To Scythian:
Your statement:
"Also, thermodynamic efficiency of ANY machine goes up with greater temperature differences"
may be true, but to have a thermodynamic machine it has to go through cycles in which it performs work. The contraption under discussion is in thermodynamic equilibrium under static conditions, so it can not do work. With W=0, the efficiency is zero no matter what the temperature difference. May be you had something else in mind, but if so, I would be at a loss to understand what you are driving at.
**********************
I agree with Scythian's observation that my derivation is not "exhaustive". It can be generalized by noting first that corresponding to the product dA dΩ in the object space of an arbitrary optical system there corresponds dA' dΩ' in the image space, defined by the mapping (x,y,px,py→x',y',px',py') , where dA=dxdy, dΩ=dpx dpy, dA'=dx'dy', dΩ'=dpx' dpy' . Conservation of energy, when satisfied, yields
∫ ∫ I'(x',y',px',py') dA' dΩ' = ∫ ∫ I (x,y,px,py). dA dΩ
Since dA dΩ is a Poincaré invariant, it follows that
dA' dΩ' = dA dΩ.
Hence,
I'(x',y',px',py') = I (x,y,px,py).
The rest follows. Note that in thermal equilibrium the intensity functions I'(x',y',px',py') and I (x,y,px,py) are actually constants.
***************************
I intend to pose the following question in engineering:
What is the key scientific or technological challenge that needs to be overcome in solar cell technology? What solutions have been tried or proposed, and what solution would you propose?
If interested, turn there.
New and Improved Solar Powered Path Lights
If you’re looking for an easy and inexpensive way to add some outdoor lighting to your front or backyard, then some solar powered path lights just might be what you need. Most of these great lights are mounted on narrow poles that easily stick in most terrains, from soil to rock. These eco-friendly lighting options require no wiring and use absolutely no electricity – just mount them anywhere they’ll receive direct sunlight and watch ‘em glow when the sun sets.
If the phrase “solar powered light” puts an image of a cheap-looking plastic fixture with a faint, dim glow in your head, you’re probably not alone. These older, cheaper models don’t light up much of anything and they don’t add much to the look of your yard in the daylight. Newer versions are available in multiple colors and metallic finishes and come in a variety of styles to suit just about any decorating taste. Choose from sleek modern designs or add a touch of country charm with hanging or carriage-style path lights.
Many of these solar powered lights are referred to as “path lights,” as that is the common use for them. But these accent lights work great in other outdoor areas as well. Some other uses include flower beds, raised planters and in pots, large and small. Anywhere you want to add some accent light that receives direct sunlight during a good portion of the day will work. The best part is since there are no wires and they are easy to install and remove, you can try out different spots until you find the look you want.
Even though these newer solar powered models greatly outperform their older counterparts, they are still only meant for accent lighting. The energy-efficient LED bulbs cannot illuminate large areas, so we recommend using other lighting options as your primary light source and then using multiple solar lights to augment and enhance your outdoor areas. And again, since these great little lights use absolutely no electricity, there are no costs to worry about after purchase. Check out all the new styles at the Eco-Lights website today.
About the Author
For a HUGE selection of lighting, including outdoor lighting, pendant lights, and wall sconces, check out the Eco-Lights website today.
