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  1. #1
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    surface area calculation for cooling

    I apologize for not being able to find this in a search of the newsgroup but I need to calculate the cooling needs for a proposed light (single xml, to be driven at 2800mA).
    I plan on using a simple square tubular aluminum stock and know I will need to add some additional cooling. I know there is a formula that gives you the area required for leds driven at (?) amperage. I don't know the formula or even the variables. Any help?
    thanks in advance.

  2. #2
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    2 sq. Inches per watt works pretty well. In rounded numbers your comnfig will be 9 watts. Led Vf x A = W

  3. #3
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    Here's what i found,

    Quote Originally Posted by Vancbiker
    Somewhere, several years ago, I heard 2 sq. in. per watt was a good rule of thumb to follow. I have used that or near that on my lights with good results. Yes, at room temp in still air they will reach 50-60C in about 5 minutes on max. A walking pace will keep them easily below 50C in warm night air of ~18-20C. ~8C air keeps them just lukewarm. You can get a pretty small housing to work if you can get a decent amount of fins on it. Look at this thread for a pretty small light that still manages heat well.

    Micro helmet light
    I think if you can use this, and adjust it by +/- 10%, depending on how warm it is and youre riding speed. So you will need 22"-26" square of surface to cool you're 12 watt LED, plus more if you use an inefficient driver.

  4. #4
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    if my maths are right for 9 watts that would be 9/6=1.5inch? or 38.10mm square cube?

    edit its length of each side would be 1.23 inches/31.24mm

    SA=6a2 where a = length of each side
    Last edited by Goldigger; 01-16-2011 at 09:57 AM.

  5. #5
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    Quote Originally Posted by Goldigger
    if my maths are right for 9 watts that would be 9/6=1.5inch? or 38.10mm square cube?

    edit its length of each side would be 1.23 inches/31.24mm

    SA=6a2 where a = length of each side
    Er, no.

    9w x 2 sq in = 18 sq in of cooling.

    Assuming all 6 sides are open to the air, each side needs 18/6 = 3 sq in of surface.

    3 sq in has sides of 1.73 in, or 44mm.

    However, the front will presumably contain an optic, the bottom will have a mount and there will be switches, cables etc reducing the o/a surface area. So you will probably need to go even bigger, or used fins to increase the area.

    I don't know where the 2 sq in/w figure comes from and whether it has been scientifically tested or is just a rough rule of thumb. The finish of your light will have a big effect too - matt black radiates heat far more effectively than a polished surface which is why air cooled motorbike engine fins are matt black. Also a rough surface (eg shot peened or sand cast) has a much greater surface area.

  6. #6
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    Quote Originally Posted by JezV
    Er, no.

    9w x 2 sq in = 18 sq in of cooling.

    Assuming all 6 sides are open to the air, each side needs 18/6 = 3 sq in of surface.

    3 sq in has sides of 1.73 in, or 44mm.

    However, the front will presumably contain an optic, the bottom will have a mount and there will be switches, cables etc reducing the o/a surface area. So you will probably need to go even bigger, or used fins to increase the area.

    I don't know where the 2 sq in/w figure comes from and whether it has been scientifically tested or is just a rough rule of thumb. The finish of your light will have a big effect too - matt black radiates heat far more effectively than a polished surface which is why air cooled motorbike engine fins are matt black. Also a rough surface (eg shot peened or sand cast) has a much greater surface area.
    So my tripple cuboid is fine at 36mmx36mmx36mm
    3xXPG=10w
    ?

  7. #7
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    Quote Originally Posted by Goldigger
    So my tripple cuboid is fine at 36mmx36mmx36mm
    3xXPG=10w
    ?
    Well, based only on the size of the cube you've provided - no, it should not be fine. Each side of your cube is approx 2 square inch. Let's assume that front and back side of the cube are not really contributing much at emitting heat away from the light, so that leaves you with 4 "useful" sides. That gives you about 8 square inches.

    However, if I remember correctly, those four sides of your light's body are finned. Probably those fins more than double your cooling surface area. So actually, yes, your light satisfies that rule of thumb criteria about required cooling area.
    Last edited by ortelius; 01-16-2011 at 02:21 PM.

  8. #8
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    Quote Originally Posted by Goldigger
    So my tripple cuboid is fine at 36mmx36mmx36mm
    3xXPG=10w
    ?
    I was only checking your maths If it runs at full power without tripping the thermal cutout then it doesn't matter.

    I think I read somewhere in another thread that the 2 sq in rule is pretty conservative, maybe more relevant to the Californian summer than the British winter. Here we get the benefit of water cooling too!

  9. #9
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    The 2 sq. in. per Watt figure that I have promoted and been quoted on is a "guideline" value that I heard a few years ago. It works for a quick and easy method to see if a housing and led combo is feasible. Stay within 10% of that you"ll have a light that will be suited for the majority of users in most conditions.

    Newer LEDs will get by with a little less area as the efficiency is improving (greater lumens per watt)

  10. #10
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    Thank you all for the many responses. Very helpful information.

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