# What temperature is achieved in focus point by 5000 flat 1x1cm mirrors onto a satellite dish?

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

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".

• @kenorb - I had rounded the numbers in my answer since I was doing a fair bit of estimating. Converting a round number in K to a "precise" number in Celsius defeats the purpose of this rounding - it creates a false impression of precision with a lot of significant digits. – Floris Mar 5 '15 at 13:44

A rough estimate can be obtained as follows. At the extremely high intensity radiation the temperature will be determined by radiative equilibrium. Now, in the case of purely radiative equilibrium, an object that receives the solar radiation that reaches the Earth would reach the a similar temperature as the lunar surface. At the equator, the maximum temperature there is 390 K. The flux of solar radiation is then bit higher than on the surface of the Earth, it is 1361 Watt/m^2 instead of the 1050 W/m^2 because on the Moon there is no atmosphere.

Since the radiated power is proportional to the fourth power of the temperature, we can estimate the maximum temperature using the above figures. The device concentrates sunlight by a factor of 5000 but this then yields 5000*1050/1361 = 3860 times the intensity outside the Earth's atmosphere while with the normal solar intensity you could reach a maximum temperature of 390 K. The maximum temperature should thus be roughly 390*3860^1/4 K = 3100 K.

• It's in Kelvin? So it would be around 2800°C? – kenorb Oct 25 '14 at 16:55
• Yes around 2800 C according to this estimate. – Count Iblis Oct 25 '14 at 17:00
• Thank you. Do you think more mirrors could generate more heat? – kenorb Oct 25 '14 at 17:12
• Yes, the higher the total surface area of the mirrors, the more solar energy is intercepted. The maximum power that one can extract per unit surface area is limited. – Count Iblis Oct 25 '14 at 17:56