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Finding the temperture of the glass on an incandescent light bulb

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fang54

fang54

Aerospace
Nov 11, 2003
5
Hi,

I have a problem at work that is similar to estimating the temperture of the glass on an incandescent light bulb. The glass(quartz) is heated(P1) uniformly by some resistivity filament(T1=1700 deg-C) some distance away in vacuum, but cooled from the other side by natural convection, this(IR light bulb) is used to heat a sample material in the open air oven.

How would I estimate(rough) the temperture of the quartz? Quartz's emissivity is listed as 0.93, so seems I can apply the Stefan-Boltzmann Law P = ε’σ(T1^4-T2^4) to find the temperture of the glass(T2)? and how would I estimate how much heat(or temp) my sample will see from the Transmitted radiation? How would I account for the transmission of infrared radiation, Quart supposedly being very efficient for the transmission of infrared radiation in the some Short-wave infrared (SWIR) range?


 
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Is the filament heated to incandescence? If so, the window is heated by the amount it absorbs while passing the rest of the radiation.
Complicating it is that the oven will radiate back from its operating temperature and the sample will be directly heated by radiation from the filament.

Edit: It occurs to me that few lamps are evacuated - that tends to allow the filament to evaporate and deposit on the envelope. Instead they use a halogen gas that chemically combines with the tungsten and only breaks down at high temps, allowing the metal to replate onto the filament and not onto the far too cold (relatively) envelope.
 
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Hi,
Did you consider a pyrometer or IR camera?
My 2 cents
Pierre
 
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Several degrees below the melt temperature of glass used in light bulbs. a 100 watt light bulb could have a surface temperature of less than 500 dF as a guess.
 
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Another way of looking at this is by using a nominal 10% (approx) light emitting eff of an incandescent bulb. So 90% of the rated power of the bulb goes past the quartz layer as heat, while 10% is transmitted out as visible light. Knowing the thermal conductivity of quartz, and the surface total emissivity, one can estimate both the inside temp of the quartz and the outside surface temp. Heat emitted out would consist of radiated heat plus natural convective heat.
 
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Heat output of the bulb cannot be more than the wattage input to the bulb (if that is what you are after).
 
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Light bulb temperatures
25W = 75F
40W = 115F
150W= 250F

Spectrum analysis
Screenshot_20230310-075923_Brave_kski6k.jpg


--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
 
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Thermocouple, no point in theoretical calculations if you can directly measure it... Also no idea why anyone would recommend an IR camera, its a polished surface, you'll need to apply a surface treatment to get an accurate measure, change the camera setup to cut the signal, recalibrate to the cut signal... etc...
 
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