There is this video ("R5800 Solar Death Ray") where teenager built a 5000 mirror device which concetrates the solar rays which is showing the potency of the mosaic method of concentrating sunlight using elliptical dish. It can fire the wood, boil the water in seconds, melt an aluminum, glass or even burn the whole in concrere or rock.
What is the approximate maximum temperature (could be a range) is achieved by this mirror device in its focus point on sunny day?
The R5800 solar concentrator, replaced by the new R23k, is made from an ordinary fiberglass satellite dish. It is covered in about 5800 3/8" (~1cm) mirror tiles. When properly aligned, it can generate a spot the size of a dime with an intensity of 5000 times normal daylight. This intensity of light is more than enough to melt steel, vaporize aluminum, boil concrete, turn dirt into lava, and obliterate any organic material in an instant. It stands at 5'9" and is 42" across.
Technical details & calculations:
Calculations: Area of dish aperture (elliptical)= (pi x 102cm x 73cm)/4 = 5848 sq cm
Area of focal point (circular)= pi(0.6cm)^2 = 1.14 sq cm
Dish Type= 0.7m x 1m offset focus elliptical.
Concentration Power= 5000x
Output Power Estimate = 560 watts
Answer
There are several ways to approach this problem. If we can estimate the power density achieved in $W/m^2$, then the temperature that can be reached follows from the Stefan-Boltzmann law.
First method:
1) Take the total power collected, and see the size it got focused down to. You state the area of the mirror array is 0.6 m$^2$ (roughly), and with power incident on surface of earth about 1 kW, you get 600 W (consistent with the claimed 580 W from your question). If this is incident on an area of 1.14 cm$^2$, the power density is $\frac{580}{0.000114}\approx 5 MW/m^2$. A perfect black body of 1.14 cm$^2$, with this power incident on one side, and insulated perfectly from the other side, would be able to reach a temperature $T$ such that
$$\Phi = \sigma T^4$$
so:
$$T = \sqrt[4]{\frac{5 \cdot 10^6}{5.67\cdot 10^{-8}}} = 3000 K$$
(round numbers...)
However, if you tried to heat a disk (twice the area - no insulation on the back) the temperature reached would drop by $\sqrt[4]{2}$, to T = 2600 K (~2330°C). Note that the temperature doesn't drop all the way to 1500 K (~1230°C) - this is that 4th power in the Stefan-Boltzmann law shows its - ahem - power...
2) A second method would look at "how big The Sun looks" from the vantage point of the focus. When you are at the focal point of the mirrors, you "see" that it is as large as the satellite dish. That means that the heat flux increases, compared to the flux from The Sun, by the ratio of apparent areas. Which is actually the same thing as saying "you appear to be a lot closer to The Sun and can use the inverse square law to determine how much more power per unit area you experience".
Now The Sun looks like a disk that is 0.5° diameter; and with the dimensions given, your dish is equivalent to a disk with a diameter of 86 cm ($\sqrt{102\cdot73}=86.3$) and the focal distance is 138 cm (which I derived from the size of the focal spot, which is really an "image" of The Sun).
At a distance of 138 cm, a disk of 86 cm diameter "looks" 69x larger than The Sun - so it has an apparent area that is 4800x larger than The Sun - it therefore "feels the heat of 4800 suns". That is remarkably similar to the answer we got before, despite a different approach (but not really, if you look closely). So we will again get the same estimate for the temperature you can reach.
This second approach helps us understand that to get to higher temperatures, we need "The Sun to look even bigger" - that is, we need a larger dish or a smaller focal length. Making the focal length smaller will only work if the individual mirrors in the dish are small compared to the size of the focal spot - otherwise they will result in significant blurring of the focus and thus lower the power density. Simply increasing the size of the mirror array doesn't increase the power density - only the quality of the focus does. In principle the best you can do is create a giant 3D array of mirrors that make it look like The Sun is "everywhere" - a full $4\pi$ array would in principle give you the light of 50,000 suns (10x more than this mirror). Such a device would illuminate an object from all sides, and the temperature of that object would be (by the same equation as above) $\sqrt[4][10]$ x greater, or 5500 K (~5230°C). This is very close to the temperature of the surface of The Sun - and that is no surprise. If I had not used rounded values along the way (since there is a lot of estimating going on) I might have expected the answer to be 5776 K (~5500°C) - the surface temperature of The Sun, and the theoretical limit of such a device. So 5500 K (~5230°C) is "close enough for estimating".
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