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B Stupid Light Bulbs

  1. Mar 11, 2017 #1
    Ok, according to my in depth research, wiki, the filament of standard incandescent light bulb will reach a temperature of roughly 2500 K. According to Stefan-Boltzmann, the power radiated from an object at temperature T (K) is given by P= A ε σ T^4 . Suppose the filament has a radius of 1mm and is 25.4mm long. The filament has a surface area, A, of 1.6E-4 m^2. Further assume emissivity, e, is 1.0.
    The radiant power is then calculated is then 367 watts.
    According to wiki, a 100 W bulb has a resistance of 144 ohms when it is lit. If the bulb is plugged into a standard 120V outlet, the electrical power delivered to the bulb is P = V^2/R = 100 W.
    For a 1 second time period 100 J of electrical energy power the bulb, but somehow 367 J comes out.
    Where does the extra 267 J come from?

    Is the 267 J stored in the filament material and then released from the filament as it burns?
     
  2. jcsd
  3. Mar 11, 2017 #2

    DaveC426913

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    No. The filament in a bulb is not a consumable; it does not burn.
     
  4. Mar 11, 2017 #3

    DaveC426913

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    Perhaps these are poor assumptions.
     
  5. Mar 11, 2017 #4

    Vanadium 50

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    1 mm? Yowza! :oldsurprised:
     
  6. Mar 11, 2017 #5
    Match the power in = to power out and solve for surface area assuming 2500 K ? something like this ? 100 = A (1) σ (2500^4) --> A = 4.5E-5 m^2 if L = 25.4mm then R = 0.54mm.. this would make everything balance...?
     
  7. Mar 11, 2017 #6

    DaveC426913

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    "For a 60-watt 120-volt lamp, the uncoiled length of the tungsten filament is usually22.8 inches (580 mm), and the filament diameter is 0.0018 inches (0.046 mm)."
    https://en.wikipedia.org/wiki/Incandescent_light_bulb

    This makes for a surface area of 84mm2, or 8.4E-5 m2.

    Hmph. That's only a factor of 2 from your estimation of 16E-5 m2.
     
  8. Mar 11, 2017 #7
    Appreciate the comments, I did notice in the wiki that the tungsten does evaporate so it is a consumable? correct?
    I think I can make things balance now... thanks again for the comments.
     
  9. Mar 11, 2017 #8

    Vanadium 50

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    Tungsten does evaporate but this is a weakness, not part of the design. It evaporates at hot points, not throughout the filament.
     
  10. Mar 11, 2017 #9

    russ_watters

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    Does this calculation include the radiation of the bulb? It isn't radiating into space near absolute zero, right...?
     
  11. Mar 11, 2017 #10

    Drakkith

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    Also note that the coiled filament will undoubtedly absorb much of its own radiation (since different parts of the coiled filament will block the LOS of other areas) whereas a straight filament will not. Looking at pictures of the filament from wiki, it looks like a reasonable assumption to ignore most of the "inside" surface of the coiled filament when calculating the surface area.
     
  12. Mar 11, 2017 #11
    The point (I take it) is that the calculated output doesn't match the electrical power input; the fact that room temperature isn't at absolute zero seems irrelevant: 25004 vs 3004 is a factor of nearly 5000.

    I don't suppose wiki specified the wattage of the bulb?

    Yes you might be a bit generous in your estimate of the filament size. After a quick look at a low-wattage bulb, I'd guesstimate 0.5 mm for the diameter, which would reduce the area by a factor of 0.25. I think the working temperature may be quite variable too (Wiki again, from memory). If it was as low as 2000 K, your power estimate would drop by another factor of 0.4, bringing the radiated power below 40 W. Care to guess at the emissivity?

    I think there's enough intrinsic slop (technical term) in the parameter estimates to conclude that incandescent light bulbs were probably not gathering power out of the vacuum fluctuations. (Unless of course they are now being phased out to prevent us from Discovering the Truth.)
     
  13. Mar 11, 2017 #12
    If it was tightly coiled, wouldn't the effective area be less than this?
     
  14. Mar 11, 2017 #13

    Drakkith

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    Using these values: ##A = 5x10^{-5}m^2, T=2500K, σ=5.67x10^{-8}, ε=1.0##

    I get ##P= A ε σ T^4 = 110.7522W##

    So about 111 watts.

    I took Dave's calculated surface area and cut it almost in half under the assumption that much of the radiation from the interior of the coil would be re-absorbed by the filament.
     
  15. Mar 12, 2017 #14

    DaveC426913

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    Seems to me, if that were true, it would defeat the purpose of making it that way.

    Not really a compelling argument, I agree. But why are they building it tightly coiled (and using excess tungsten) if it isn't effective?
     
  16. Mar 12, 2017 #15
  17. Mar 12, 2017 #16

    OmCheeto

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    Excellent document!
    Though I find some of that information hard to comprehend.

    "... Another result of coiling is to produce a black body effect due to multiple reflections of the radiation within the coil. This increases the brightness of the interior of the coil to about 175% of the exterior brightness.
    ...
    It should be noted that the temperature of a coiled coil filament is no greater than for the single coil type, so life is the same. ... "

    How on earth does that work? I guess I still don't understand what "temperature" means, nor how this all works.

    ps. Just discovered that the light bulb filament "temperature", per wiki: 2550K, is the same as that star in the news: Trappist-1. Weird!
     
  18. Mar 12, 2017 #17

    Drakkith

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    A triple-coiled filament just has a third coiling done to reduce its length inside the bulb. A coiled-coil has two coilings done and is the standard filament in most bulbs. The article is just saying that neither type runs hotter than the other.

    As for the interior vs exterior brightness, I'm not sure how that works either. I assume the interior isn't much hotter than the exterior, else the bulb would suffer early burnout, but I don't really know what's going on.
     
  19. Mar 12, 2017 #18

    OmCheeto

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    I rest my case. :biggrin:
     
  20. Mar 12, 2017 #19
    The first part seems okay: the coil isn't at equilibrium--its inside is exposed to temperatures of around 2500 K, while its outside sees temperatures an order of magnitude lower. (The interior of the coil, being somewhat enclosed, might also have a high enough density of tungsten vapour to mitigate evaporation.) I can't visualise a coiled-coil well enough to know whether or not I should be surprised that it behaves the same as a simple coil (which is not necessarily the same as an uncoiled filament).

    It occurred to me, after I posted, that this emphasis on coils is probably British. I'd come across the terms "coiled filament" and "coiled coil", but that must have been when I lived in England. Because the power output is V2/R, working at 240 V would mean they could use a filament nearly five times as long to get the same power as in a North American bulb.

    Yes, I noticed that too--a reminder that the light of "red dwarf" stars isn't all that red.
     
  21. Mar 12, 2017 #20
    Afterthought: maybe the coiled coil has the same temperature but greater surface area.
     
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