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Inert gas pressure in incandescent bulb

  1. Oct 16, 2015 #1
    hi,
    so i want to ask why is the inert gas pressure in an incandescent bulb should be low?

    i thought that when the gas pressure is high, more 'free' tungsten atoms can collide with the gas particles (since the gas particles move at higher speed) and bounce right back towards the filament. hence, the rate of evaporation of atoms are further reduced. so, the bulb will be brighter since the filament can be heated to a higher temperature without disintegrating the filament. (and the bulb would last longer too)

    or my concept is just totally wrong?
    thanks in advance!
     
  2. jcsd
  3. Oct 16, 2015 #2

    Nidum

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  4. Oct 16, 2015 #3

    sophiecentaur

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  5. Oct 16, 2015 #4
    thank you for replying,

    but if you look at this article, why does it says 'The filament is enclosed in a sealed glass jacket (glass bulb) which is filled with a mixture of inert gases at low pressure.'
    http://www.standardpro.com/product-information/incandescent/lamp-construction

    is it possible that the gas pressure must be low so less thermal expansion would occur? since the pressure would be further increased when heated up
     
    Last edited: Oct 16, 2015
  6. Oct 16, 2015 #5

    Nidum

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  7. Oct 16, 2015 #6
  8. Oct 16, 2015 #7

    nsaspook

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    The low pressure INERT gas acts as a barrier for the atoms that boil off the surface of the hot filament. In pure vacuum the detached atoms would quickly coat the inner surfaces of the bulb while prematurely thinning the filament. With a low pressure gas the filament atoms would collide with the heavy gas atoms close to the surface and fall mainly back to the filament surface to be reheated.
    Gas pressure is one of the primary parameters in metal deposition.
     
  9. Oct 16, 2015 #8
    Thank you for replying,
    But how does the gas pressure affect the metal deposition? ( if the same same type of inert gases, hence same molecular mass are used but with different pressures)

    Thank you again
     
  10. Oct 16, 2015 #9

    sophiecentaur

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    That's no problem. When it's filled, the bulb is at room temperature. When it is operating, it is at high temperature. The pressure is then very high. The ratio of the pressures will be roughly the ratio of the temperatures. No one is wrong - different answers apply at different times. Still the basic requirement is that a high gas pressure stops metal evaporation. It is well known that running halogen lamps 'dimmed' harms them.
     
  11. Oct 16, 2015 #10

    nsaspook

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    The gas controls the ballistic regime (straight path) of the evaporated atoms by controlling the pressure around the filament.
    http://users.wfu.edu/ucerkb/Nan242/L06-Vacuum_Evaporation.pdf
     
  12. Oct 16, 2015 #11
  13. Oct 16, 2015 #12
    I think i've got it, thank you very much! :biggrin:
     
  14. Oct 16, 2015 #13

    sophiecentaur

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    It struck me that the reduction of metal loss from the filament when the gas pressure is high, must be analogous to the pressure cooker effect, in which the boiling point of water is raised under high pressure.
     
  15. Oct 16, 2015 #14

    jbriggs444

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    My first guess is that high pressure is important in that it reduces the mean free path of the tungsten vapor and thereby reduces its diffusion rate. Keep the diffusion rate suitably low and the local tungsten vapor pressure may be adequate to re-deposit significant quantities. Instead of evaporation, a ballistic trajectory and deposition on the bulb walls, you get evaporation, a random walk and (probable) re-deposition back on the filament.
     
  16. Oct 16, 2015 #15

    sophiecentaur

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    That makes sense.
    I was trawling around and found these curves. Tungsten is on the very far right of all the elements (The very few liquids at around room temperature are at the far left - as you'd expect), which shows why W is so suitable for filaments. At 3000K, its vapour pressure is still less than 10-7 Atmospheres and there are other elements that aren't too different. But the melting point is also relevant, for a filament and W doesn't melt until 3700K.
    I think the mean free path argument makes more sense than just a high boiling point.
     
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