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Surface of the lense and temperature

  1. Sep 19, 2006 #1
    What is the formula that relates surface (m^2) of a circular convex lense with temperature (Celsius) in a light focus point of a lense.
    (calculating temperature in the light focus according to size of a lense...)

    (Glass lense, regular... (so, no coefficients, just assume it's some nice common average colourless glass lense... Some simple forumula, something practical - you get what I mean...))

    (I guess it relates [power of sunlight]per[unit of surface]...)

    Example, please... :smile:

    (...and formula for mirror used instead of lense to concentrate light (I guess it's a better solution, but nevertheless I prefere lense))
    Link if there is a place you know where I could find it (it's a too complex concept, google gives me all sorts of irrelevant results)
    Thank you in advance.
    Last edited: Sep 19, 2006
  2. jcsd
  3. Sep 19, 2006 #2


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    Good guess --- a perfectly transparent lens, or perfectly reflecting mirror, concentrates power only from the area it intercepts. That means, power is a function of source power per unit area at the interception distance. Temperature? Can only equal the source temperature, never exceed it.
  4. Sep 19, 2006 #3
    O', sorry - maybe I've used a wrong term.

    What I've ment to ask is: how to achive some silly high temperature such as e.g. 3000 Celsius degrees at the light focus spot using a lense?
    (how huge the lense must be in order to do that?)
  5. Sep 19, 2006 #4


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    Any size converging lens focusing sunlight will do that. Heat loss from the point at which the light is focused reduces the temperature you actually achieve; reduce the heat losses, and you raise the temperature.
  6. Sep 20, 2006 #5
    ...And I've thought I was asking a simple question...
    I was not asking a theoretical question; I was asking a practical question.

    ...If we imagine (for ilustration purposes) that Sun-light shining on ground consists of lines of some density running into ground (density of those hypotetic lines define how strong it's shining), then putting a bigger (say 0.5m diameter) lense "catches" more of those lines (and concentrates energy of all of them into a single point), than some small (say 1cm diameter) lense would...

    ...And you are telling me that a 1cm diameter this-world plain-old glass lense can heat some object (natural magnet for instance) in the focal point on it to 3000 Celsius degrees...

    I don't know, but somehow two words (for some strange reason) come to my mind... leg and pull...
    Last edited: Sep 20, 2006
  7. Sep 20, 2006 #6


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    It's a heat transfer problem: supply heat to an object at a fixed rate; restrict heat loss from the object to a rate less than that of the supply; the object increases in T. Boil the earth's oceans with a flashlight --- it's gonna take on the order of 1021 to 1022 years.

    You want lens sizes? You're going to have to come up with time constraints, methods to control heat loss, heat capacities for whatever you're cooking, and a couple other parameters.
  8. Sep 21, 2006 #7


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    It is very dependent upon things like thermal conductivity, coefficient of expansion, and thickness gradients. This is a pretty complicated question with a number of variables. Your garden variety engineer would field test this application in a heartbeat. The company would go bankrupt waiting for the scientist to perfect the math. Engineers assume Ms. nature will invariably toss you a curve ball on a full count. . . e.g., spare part kits tend not to include more than one of the least expensive part most likely to be lost at the job site.

    I went to a customer facility once to replace a 50 hp motor. We took every precaution to guarantee the safe arrival of the motor. The plan worked to perfection. The motor mount bolts - 4 required - was a different story. My parts kit included exactly 4 bolt assemblies - bolts, washers, nuts, etc., which cost a total of 5 bucks. After spending half a day unloading and transporting the motor to the machine, I dropped one of the bolts, which fell about 5 stories before making a faint, but distinct, splashing sound. Not good.
    Last edited: Sep 21, 2006
  9. Sep 29, 2006 #8
    OK, you have a perfectly sunny summer day on Earth (exactly halfway between northpole and equator) at noon, and exactly 5 seconds (not more not less) to melt 1kg of some hypotetic pitch-black material (90% of all rays are absorbed; conducts heat like concrete; dimnsions 1x1x1dm) that melts at 3000 C degrees. What is the diameter of a convex glass lense you would use? THINK FAST!
    Last edited: Sep 29, 2006
  10. Sep 29, 2006 #9


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    What is the heat capacity of the material in question? That is, how many joules or calories does it take to raise one gram of it by one degree in temperature?

    Once you know that, you can calculate how much energy in joules you need, as a minimum (not taking radiation losses into account); the required time then gives you the required joules per second.

    Sunlight at the earth's surface carries about 1400 or 1500 joules of energy per second, per square meter perpendicular to the sun's rays. Comparing this with the result from the preceding paragraph gives you the number of square meters needed.
  11. Oct 1, 2006 #10
    Well, as I've mentioned that it "conducts heat like concrete", then let's say same as concrete.
    So what is the diameter for this particular example?
  12. Oct 1, 2006 #11


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    You haven't given us a heat of fusion for your "hypothetical material."
  13. Oct 4, 2006 #12
    What fusion now?
    You mean melting?
    What do you mean by 'fusion'?
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