Potential temperature of a fresnel lens focal point


by munglet
Tags: fresnel, lens, sunlight
munglet
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#1
May22-09, 05:01 PM
P: 1
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 1083C (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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Andy Resnick
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#2
May24-09, 01:04 PM
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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.
cesiumfrog
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#3
May24-09, 07:29 PM
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Quote Quote by Andy Resnick View Post
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.

Andy Resnick
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#4
May24-09, 09:52 PM
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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.
cesiumfrog
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#5
May25-09, 12:59 AM
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Quote Quote by munglet View Post
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 Quote by Andy Resnick View Post
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?
Bobbadillio
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#6
May25-09, 01:06 AM
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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.
Andy Resnick
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#7
May25-09, 08:07 AM
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Quote Quote by cesiumfrog View Post
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?
I'm not sure what you are getting at. Why not dream up a 1-mile wide lens focused down to a square micron? Do you have a different approach?
Dr Duck
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#8
Aug17-11, 03:31 AM
P: 1
I'm new at this, but I'll give this question a go.
Firstly, the Power coming in from the sun is ~1kWatt, if the Fresnel lens is 1m2.
Secondly you need to define what you are heating. Say it is a block of copper (as this is what you stated you were melting earlier. Say that it is 10cm square. (That is length = width = height = 10 cm).Also the bottom side is perfectly insulated. Also this block is sitting in air wich a slight breeze, which is 25 deg Celcius.
Since we are now assuming, lets say that ALL the heat from the sun is focused perfectly onto the block, and that the block absorbs all the light energy.
The volume of the block will begin to heat, and it will be hottest on the top, where the light is focused, and become cooler as you get further into the block. The block will also be hotter on the edges than it is in the centre.
Now that the problem is defined, you can find the temperature point, where the energy lost to the air = 1kWatt coming from the sun.
q_in = power from sun_per second = 1kW
q_out = power lost to air_per_second = q in =1kW
The heat equation in its simplest form is q = u.A.Delta_T, that is
q_out = conductance * surface area * Change in Temperature
conductance of copper = 400(W/m.K), this will change with temperature, but I'll assume itis constant.
surface area = 0.05 m2. (Remember we assumed the bottom was on a perfectly insulating surface)
And that is it...
Delta_T = 1000 / (400*0.05) = 50.
Temperature of edge of copper = 75 deg C.

Lets assume that you are using a flat rectagular sheet of copper.
length = 5cm, width = 5cm, height = 1 mm.
surface area = 0.0025.
The temperature of the sheet of copper would be around 950 degC.

The temperature of the copper would get higher than that given above because the air around the copper would heat up, and thus reduce the amount of energy escaping across the boundary.

In relation to the copper wire you were melting, the volume of the copper is small, and if the air is still, then the temperature loss from the surface would be lower than that assumed above.

Hope this helps... even if it is a little long winded.
Tutupoopy
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#9
Sep22-11, 04:52 AM
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Continuing further on that topic of using lense to heat up the objects,

I want to find out the rate of evaportion of the water.

say, there is a swimming pool and i put a 1 m2 lense on top in a way that light is concentrated on surface of the water over the area of 0.05cm2. How long will it take to empty the pool?
sophiecentaur
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#10
Sep22-11, 05:01 AM
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All you need is the 'Latent Heat' of vaporisation of water. That tells you the energy needed to vaporise 1kg of water. The sums are easy after that. (Ignoring any other heat loss, of course)


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