mattr888 Posted October 16, 2010 Share Posted October 16, 2010 Is there a sum to work out how hot the point of focusing could get with a parabolic mirror? thanks, matt Link to comment Share on other sites More sharing options...
swansont Posted October 16, 2010 Share Posted October 16, 2010 The brightness theorem tells you it can't be any hotter than the source of the light. That's a manifestation of the second law of thermodynamics. Link to comment Share on other sites More sharing options...
mattr888 Posted October 16, 2010 Author Share Posted October 16, 2010 so if the souse of light is the sun it can't be any hotter than that Link to comment Share on other sites More sharing options...
CaptainPanic Posted October 18, 2010 Share Posted October 18, 2010 (edited) First you must find out how much power you are focussing on an object (that is not a temperature, but an energy flow). As I mentioned in a reply in a previous post, you have to find the insolation for where you live. That needs to be converted to Watt/m2s (power per area). Then you focus that energy on to a much smaller area (same power, smaller area). Then you have to look up some parameters of the material you try to heat up (probably water). You need to know the Cp value (specific heat). That's 4180 J/kgK for water. You need to know the reflectivity of your materials (tubes, and the mirror itself)... but you can also just assume 50% efficiency (the rest of the light reflects, and is not absorbed by the water)... probably a safe guess. Then you must decide on a flow of water (kg/s, or liter/s)... because if you have no flow, you will heat up the water more and more and more... Now, you have an amount of energy, a flow, a material that needs to be heated, and an efficiency. Your formula now is: [math]P\cdot{\epsilon} = F\cdot{C_P\cdot{\Delta{T}}}[/math] [math]P[/math] = power (energy per second, in Watts) [math]\epsilon[/math] = efficiency (%) [math]F[/math] = flow (kg/s) [math]C_P[/math] = specific heat (J/kgK) [math]\Delta{T}[/math] = difference between temperature of incoming cold water, and water after heating up Please note that you will probably reduce your efficiency as well because of heate losses. In short, because some parts get hot, they will start heating up the air around it, which means you lose energy. If you have no flow of any material (such as water), but instead have just a single sample which is heated up, then the temperature will rise and rise until the heat losses to the air from the sample are equal to the energy from the sunlight. And depending on the mirror, the sample, the air flow, and some other parameters, that can be really hot. On the other end of the scale, if you have a massive flow of water (say, a complete river), then the temperature rise will be very small... You've entered the realm of engineering, sir. And I congratulate you. What you're trying to do (and what I just helped you with) is an energy balance. Edited October 18, 2010 by CaptainPanic Link to comment Share on other sites More sharing options...
swansont Posted October 18, 2010 Share Posted October 18, 2010 so if the souse of light is the sun it can't be any hotter than that Yes. You would be limited to ~6000 K. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted October 18, 2010 Share Posted October 18, 2010 Yes. You would be limited to ~6000 K. Hardly makes it worth while. Link to comment Share on other sites More sharing options...
swansont Posted October 18, 2010 Share Posted October 18, 2010 Hardly makes it worth while. Tell that to the ants. Link to comment Share on other sites More sharing options...
CaptainPanic Posted October 18, 2010 Share Posted October 18, 2010 Why exactly wouldn't you be able to make it hotter than 6000 K? If you would have a theoretical lens of a square kilometer, and focus all the sunlight of that square kilometer (approx. 200 MW of insolation) onto a square centimeter, I'd say that higher temperatures can be reached (assuming also a theoretical way of immobilizing the vaporized material that is being heated)? Are the individual photons not energetic enough or something? Where is the limitation? Link to comment Share on other sites More sharing options...
mattr888 Posted October 18, 2010 Author Share Posted October 18, 2010 (edited) well the parabolic mirror i think should be 11.43cm diameter. The attachment has all the info on that you should need except the 15 degrees c should be 11 as that is the yearly average temp Edited October 18, 2010 by mattr888 Link to comment Share on other sites More sharing options...
swansont Posted October 18, 2010 Share Posted October 18, 2010 Why exactly wouldn't you be able to make it hotter than 6000 K? If you would have a theoretical lens of a square kilometer, and focus all the sunlight of that square kilometer (approx. 200 MW of insolation) onto a square centimeter, I'd say that higher temperatures can be reached (assuming also a theoretical way of immobilizing the vaporized material that is being heated)? Are the individual photons not energetic enough or something? Where is the limitation? There are practical considerations: you can't collect all the light, you can't focus the light down to an arbitrarily small size, and as the target heats up, it will radiate more strongly. Every ray that comes in must have an identical path leading away, and in the best-case scenario, where the target "sees" the sun everywhere, due to the massive mirror coverage, these will equilibrate when the temperatures are the same. In the less-than-ideal case, the target "sees" a colder reservoir somewhere, and radiates more energy wherever that happens, and keeps the target temperature below the source temperature. The overall principle at play here is the second law of thermodynamics, which tells us that you cannot spontaneously transfer heat from a colder object to a warmer object. Link to comment Share on other sites More sharing options...
John Cuthber Posted October 18, 2010 Share Posted October 18, 2010 "well the parabolic mirror i think should be 11.43cm diameter." why? Link to comment Share on other sites More sharing options...
mattr888 Posted October 18, 2010 Author Share Posted October 18, 2010 (edited) "well the parabolic mirror i think should be 11.43cm diameter."why? well it is that at the moment but it could be bigger. 11.43cm is 4.5" and i plan to use that one Edited October 18, 2010 by mattr888 Link to comment Share on other sites More sharing options...
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