Announcement

Collapse
No announcement yet.

What are the limits....

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • What are the limits....

    ... of a solar furnace? What's the maximum amount of power per unit area you can get if you're willing to build an array of mirrors of arbitrary size? If you're willing to make a 100x100 grid, for example?

    For the uninitiated: Wiki Link
    The first google link

  • #2
    What would you use this for? Frying Pakistan? Won't work, it's already a desert as it is.

    Comment


    • #3
      Frying some.... stuff, shall we say?

      I'd like to try it just because it's cool (and relatively cheap, for a hobby). With the help of the scorching Indian sun, I could easily fry metal in under three seconds with this thing, if done right. Or I could sell it to the college workshop, who could use it for their blacksmithy work - from ice-cold to white-hot in under three seconds!

      But the uses are secondary. What are its practical limits? What's the maximum power that I could get if I'm willing to build, for example, a 100x100 array?

      Comment


      • #4


        The problem would IMHO be finding concave mirrors. Do they even sell something like that?

        Comment


        • #5
          An image, for reference:

          Comment


          • #6
            You could use polished metal OTOH, but presumably it would reflect a narrower spectrum.

            Comment


            • #7
              Originally posted by VetLegion


              The problem would IMHO be finding concave mirrors. Do they even sell something like that?
              That's the beauty of it - you don't need concave mirrors!

              You arrange a large number of small flat mirrors so that light coming from a point source infinitely far away is reflected from each mirror to an area equivalent to the area of one mirror. You have to align each mirror like that.

              The beauty is the scalability - it is possible to build a 1000x1000 array, but I don't want to anywhere NEAR that mofo when the sun rises

              I built a small one (10x10, 100 mirrors) in school, but because I used plaster to align the mirrors, and because it deformed slightly while drying, I never got a completely clear focus, because of this slight misalignment of each mirror. But it could heat black objects to very high temperatures (melted plastic in a few minutes).

              Comment


              • #8
                How much metal would you want to melt? You can calculate the power needed from that, and from that the area you need.

                Comment


                • #9
                  Originally posted by VetLegion
                  How much metal would you want to melt? You can calculate the power needed from that, and from that the area you need.
                  As much as I reasonably can.

                  Comment


                  • #10
                    Well, if you have some figures we can do some calculations.

                    BTW. you asked about limiting factors. Heated metal would lose temperature through radiation, convection and conduction. They are proportional to the size of the thing you are heating and its temperature. I'm kind of curious if this can be done now too. Maybe tommorow I'll do a calculation and let you know what I got.

                    Comment


                    • #11
                      OK. Assume I want to soften 1 kg of iron to a white-hot state within less than fifteen seconds.

                      A quick calculation shows me that a 1 kg ball of iron has a radius of ~ 3.12 cms, giving me a surface area of ~ 30.58 cm^2. For a mass of 1 kg of iron, the heat required to bring it from 23 C to 1538 C is (1538-23)*1*450 = 681750 J (450 is the sp. heat in J/kg/C). I need that much heat in under fifteen seconds - so my power requirement is 45,450 Watts.

                      The energy density of sunlight I'm taking to be 250 W/m^2. Unfortunately, at this rate, I'll need roughly 181 m^2 of area for my little toy - completely unfeasible. So from 1 KG, we reduce the requirement to 10 grams. That brings down the required area to 1.81 m^2 - an achievable goal.


                      Unfortunately, that also brings down the area available - to 1.419 cm^2.

                      Convert 1.81 m^2 to cm^2, getting 18100. Divide by 1.419 to get the number of mirrors: ~12752. This is far too many. So we expand the estimate - instead of waiting 15 seconds, we are willing to wait for a full minute. This reduces the size requirement to 0.4525 m^2, and the number of mirrors to 3188. Still far too many. So we're now willing to wait for a full three minutes. This reduces the area requirement to a further 0.150833333 m^2, and the mirrors requirement to 1062. Let's sacrifice a bit of performance and make that 1024, and make this a 32x32 array.

                      Why, however, restrict ourselves to 0.151 m^2? Why not simply build a more powerful array of 40x40, fully 1 m^2? That'll give us the requisite power to melt iron in under three minutes. That would satisfy my requirements, too - it could heat iron to smithy temperatures in under one minute, hopefully.

                      Comment


                      • #12
                        You're completely ignoring the heat loss from the metal.
                        12-17-10 Mohamed Bouazizi NEVER FORGET
                        Stadtluft Macht Frei
                        Killing it is the new killing it
                        Ultima Ratio Regum

                        Comment


                        • #13
                          Yes, I am. Gimme a rough idea of how much of a factor that would be.

                          Comment


                          • #14
                            You're also ignoring the fact that generally even "black" materials are not particularly good blackbodies. They generally have an emissivity of ~0.4 (this is spectrum dependent, however). I'd be willing to accept that in the visible range (i.e. the solar spectrum) your black piece of metal has an absorption coefficient of ~0.7. So any energy you hit it with, only 70% of it gets absorbed

                            Secondly, the heat loss goes as the stefan-boltzmann law. Your true blackbody emissivity is still probably ~0.4 at temperatures under 2000K (this is good, meaning that you're a more efficient heat absorber than you are heat emitter) and at temps above 500K radiative cooling is probably dominant (so we ignore convection, since most of the important cooling will take place in this range or higher anyway). So heat loss = 0.4*sigma*(T^4). sigma = 5.67*10^-8 W/(m^2*K^4). At 500K your heat loss is 1400 W/m^2. At 1000 K it's 23000 W/m^2. At 1500K it's 115000 W/m^2
                            12-17-10 Mohamed Bouazizi NEVER FORGET
                            Stadtluft Macht Frei
                            Killing it is the new killing it
                            Ultima Ratio Regum

                            Comment


                            • #15
                              Hmm...

                              At the peak, that would result in a loss of roughly 64 W, whereas the input is around 150 W. Not good. But even if it takes five minutes instead of three, it's good enough. And the temperatures needed to justify the investment (temps. required for smithy work) are far below 1500 C.

                              Comment

                              Working...
                              X