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Suppose we have a light bulb, for which we know its power rating, like voltage of $12\mathrm V$, and power consumption of $10\mathrm W$. We also know it's a halogen bulb with a tungsten filament inside. Suppose we also know temperature of the surrounding air.

Is this data enough to compute temperature of the filament? If not, what should also be included? And anyway, how do we find the temperature?

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You really are asking two questions.

First - how do we calculate the temperature:

At the typical temperatures of a halogen bulb, the large majority of heat loss is due to thermal radiation (although there is some conductive loss in a halogen bulb as the bulb is not evacuated). Because of this, the most important factor is the "apparent size" of the filament. I say "apparent" because when you have a tightly wound coil, the parts of the coil facing other parts of the coil don't contribute to a net heat loss.

If you took for example a 5 mm long, tightly wound filament with a mean diameter (after winding) of 0.5 mm, you would have a surface area of approximately 5 * π * 0.5 ~ 8 mm2. If you had 10 W of emission, you would use the Stefan Boltzmann law to get the power per unit area:

$$I = \sigma T^4$$

from which we get a temperature of

$$T = \sqrt[4]{\frac{10}{8\cdot 10^{-6}\cdot5.67\cdot 10^{-8}}} \approx 2100 K$$

Getting more accurate numbers is quite hard - there are lots of subtle effects (conduction down the support wires, heat lost to the filler gas, and "true effective area" to name just three).

Second - how do we measure the temperature. For such high temperatures, the disappearing filament pyrometer is very effective: you send a current through a calibrated filament, and compare its color against the color of the object of interest. When the filament "disappears" against the background, the temperatures are matched. Often, filters are used to do the comparison in a narrower range of wavelengths; the result can give resolution down to 10C according to the above linked article. There are obvious problems with this is the emissivity of the object of interest is very different than that of the filament - but if you are trying to determine the temperature of a filament that is not likely to be a problem.

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  • $\begingroup$ the surface of a cylinder is 2πrh not πrh, in addition, you need to multiply the term of emissivity of a body, for tungsten is 0.3 $\endgroup$ – Ricardo Casimiro Jun 26 '20 at 19:39
  • $\begingroup$ @RicardoCasimiro I did say “ diameter” not “radius”... fair point about emissivity. Note that by the time you take the fourth root, 0.3^(0.25)=0.74 so the effect is not huge ( but also not insignificant) $\endgroup$ – Floris Jun 26 '20 at 21:39
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The filament will be a reasonable approximation to a black body emitter, so it's spectrum will be given by Planck's law:

$$ B = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{k\lambda T}} - 1} $$

So just measure the radiance of the light from the filament for a range of wavelengths and do a fit to Planck's law by varying $T$. This will give you an excellent approximation to the temperature.

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    $\begingroup$ There is a wonderful instrument called a "disappearing filament bolometer (or pyrometer)" that superposes the image of a (calibrated) glowing filament over the image of a hot surface. You adjust the current through the filament until it "disappears" against the background (the color matches). That is in essence what this method does. $\endgroup$ – Floris Apr 18 '15 at 18:05
  • $\begingroup$ Too bad this formula is what I was going to use to calibrate spectral response of my spectrometer. So for me, this answer appears circular :) $\endgroup$ – Ruslan Apr 18 '15 at 18:07
  • $\begingroup$ @Ruslan - it sounds to me like you actually intended to ask a different question than the one you posed, then? $\endgroup$ – Floris Apr 18 '15 at 18:13
  • $\begingroup$ @Floris well, if I asked the whole question about calibration, this would be "too many questions in one", so I just asked what I thought to be first. I was wondering about this particular question I actually asked for a long time though, so it's also interesting on its own. $\endgroup$ – Ruslan Apr 18 '15 at 18:14

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