## Potential temperature of a fresnel lens focal point

Hi,

I have a fresnel lens from a old big screen tv and it is roughly 1 meter square. The focal point comes down to roughly 2.5cm square (1"). On a sunny day in Hawaii I can melt copper wire after about a minute. Copper has a melting point of 1083°C (1981.4 °F) . That's some serious fun! I can also re-melt lava rocks back into obsidian.

My question is: What is the general equation to figure out the maximum temperature that can be achieved assuming no losses or distortion when an area of sunlight is focused down to another area?

For example,

Taking 1 m^2 of sunlight and focusing it to 2.5 cm^2 the potential temperature is?

Taking 1m^2 of sunlight and focusing it to .5cm^2 the potential temperature is?

Taking 10m^2 of sunlight and focusing it to 1m^2 the potential temperature is?

Thanks.
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 Recognitions: Science Advisor Say the solar power is 1 kW/m^2 in your location. That sets the theoretical upper limit of achievable temperature. The power radiated by a perfect absorber is given by the Stefan-Boltzmann law, so setting the two powers equal: 1 kW = sAT^4 (s = 5.67*10^-8 W/m^2*K^4, A = 6.3*10^-4 m^2) T = 2300K In principle, the actual achievable temperature is much lower, but be careful, obviously. Substitute any numbers you like.

 Quote by Andy Resnick Say the solar power is 1 kW/m^2 in your location. That sets the theoretical upper limit of achievable temperature. [..] 1 kW = sAT^4 (s = 5.67*10^-8 W/m^2*K^4, A = 6.3*10^-4 m^2) [..] Substitute any numbers you like.
You're saying the limit is when the focal spot radiates onto one side of the lens with the same power as the sun radiates back?

But what if the lens is good enough that most of that sunlight is actually focussed on a smaller area? Does moving to a location of far higher solar power really raise the limit? Neglecting to account for the solid angle of the sun at your location may give results contradicted by thermodynamics; in principle the upper limit is the surface temperature of the sun.

Recognitions:

## Potential temperature of a fresnel lens focal point

No... at least I don't mean to.

I mean that the maximum equilibrium temperature is reached when the absorbed power is the same as the emitted power.

The surface temperature of the sun is in excess of 10^4 K, well above the number I calculated. There is no contradiction.

 Quote by munglet My question is: What is the general equation to figure out the maximum temperature that can be achieved assuming no losses or distortion when an area of sunlight is focused down to another area? For example, [..] Taking 1m^2 of sunlight and focusing it to .5cm[ square ..]
 Quote by Andy Resnick The surface temperature of the sun is in excess of 10^4 K, well above the number I calculated. There is no contradiction.
If the equation you answered with is used for the OP's examples, the second result (5200K) is comparable to the sun's surface temperature (5800K). How about a 10m x 10m lens focusing to a 0.5cm x 0.5cm area?
 The sun's surface temperature is not an upper limit. The upper limit on power derived from the sun is the power output of the sun. In the case of the lens, the upper limit of power applied to the piece of area in question is the power passing through the lens(actually, it's probably less than that since Fresnel lenses have some inherent 'unideal' behavior). The upper limit of the temperature of that slice then, is the equilibrium point where the power absorbed by the piece of area is equal to the dissipation of power from that sheet, which I think will come in the form of heat dissipation and blackbody radiation.

Recognitions: