# Surface area of a bulb filament

Does anyone know what it is, or where I can find the surface area of a typical 60W incandescent bulb filament? (if possible, by GE)

I'm working on a research project and we're applying the stefan-boltzmann law. I'm calculating an area around 3e-5 m^2.
I figure I'm in the ballpark, but I'd like something to use as a reference to check it.

Anyone got any ideas?
Thanks for any help

## Answers and Replies

berkeman
Mentor
Does anyone know what it is, or where I can find the surface area of a typical 60W incandescent bulb filament? (if possible, by GE)

I'm working on a research project and we're applying the stefan-boltzmann law. I'm calculating an area around 3e-5 m^2.
I figure I'm in the ballpark, but I'd like something to use as a reference to check it.

Anyone got any ideas?
Thanks for any help

Can you just break open a bulp (carefully) and measure it?

I really doubt I could do something like that accurately. But that's worth a shot if I get desperate. heh.

berkeman
Mentor
I really doubt I could do something like that accurately. But that's worth a shot if I get desperate. heh.

Why would it be hard? If you cut the filament(s) off of their posts, stretch them out, measure their lengths and use calipers to measure their diameters, it seems like you could get pretty close.

Of course if they are too brittle to straighten out, then nevermind...

berkeman
Mentor
Oh wait, I guess I had a simpler filament structure in mind. Is it just the exterior surface area of the filament that you need to use?

Does anyone know what it is, or where I can find the surface area of a typical 60W incandescent bulb filament? (if possible, by GE)

I'm working on a research project and we're applying the stefan-boltzmann law. I'm calculating an area around 3e-5 m^2.
I figure I'm in the ballpark, but I'd like something to use as a reference to check it.

Anyone got any ideas?
Thanks for any help

Unless you do some complex calculations what you'll get will be some effective surface are not the actual surface area. That happens because the bent shape of the filament allows many of the photons emitted by the filament to be absorbed by the filament itself. That effect must be taken into account when calculating the actual area of the filament. Did you take that into account?

K^2
I'm pretty sure they just want to estimate the effective area. It should be within an order of magnitude of the physical surface area, at any rate.

Wikipedia said:
For a 60-watt 120-volt lamp, the uncoiled length of the tungsten filament is usually 22.8 inches (580 mm),[40] and the filament diameter is 0.0018 inches (0.046 mm).
That's 8x10-5m². So the value you are getting is in the right ballpark.

Keep in mind that what you want to keep constant when changing wattage of the bulb is the temperature of the filament. And since that means energy in = energy out, a 40W bulb is going to have 2/3 of the surface area of the 60W bulb. So you should have no trouble converting the above number to whatever wattage that applies to your computation.

As dauto pointed out, you shouldn't expect it to be exactly the same. But they should be within a small multiple of each other. I'd be surprised if it's more than 2-3 times off.

NascentOxygen
Staff Emeritus
The area probably almost doubles when white hot, though it's difficult to gauge by eye because of an optical illusion that glowing objects seem larger than they really are. (The coefficient of thermal expansion of tungsten should tell us unequivocally.)

Thanks for the replies everyone, we've been working on modeling the light bulb spectrum at various currents with the planck spectral distribution (planck's law), and from that we calculate the filament temperature. We then took the stefan-boltzmann law and fit $P=IV=a\sigma T^{4}$ and found a value for a, which should represent the surface area of the filament. Trouble is, we really don't have a viable way to check it. And I hadn't thought about an 'effective' surface area, so that really throws us off. We're at a dead end for now, but thanks for the help.

berkeman
Mentor
Thanks for the replies everyone, we've been working on modeling the light bulb spectrum at various currents with the planck spectral distribution (planck's law), and from that we calculate the filament temperature. We then took the stefan-boltzmann law and fit $P=IV=a\sigma T^{4}$ and found a value for a, which should represent the surface area of the filament. Trouble is, we really don't have a viable way to check it. And I hadn't thought about an 'effective' surface area, so that really throws us off. We're at a dead end for now, but thanks for the help.

What do you mean that you don't have a way to check it? That's what experiments are all about...

tell me a value for 'a' and I will calculate our error. See what I mean? I have no way to say anything other than that we're in the ballpark, and that's not good enough.
Even if I were to take apart the filament and measure it, I have no way to know the 'effective' surface area.

Andy Resnick
Thanks for the replies everyone, we've been working on modeling the light bulb spectrum at various currents with the planck spectral distribution (planck's law), and from that we calculate the filament temperature. We then took the stefan-boltzmann law and fit $P=IV=a\sigma T^{4}$ and found a value for a, which should represent the surface area of the filament. Trouble is, we really don't have a viable way to check it. And I hadn't thought about an 'effective' surface area, so that really throws us off. We're at a dead end for now, but thanks for the help.

This seems like an odd thing to check- the Stefan-Boltzmann also incorporates material emissivity which varies with both wavelength and temperature. Unless you know the filament's emissivity, you are not going to be able to accurately calculate either T or a. If you've assumed e = 1 (blackbody), which is common enough, then you shouldn't be too concerned about accurately determining 'a' or 'T'.

This seems like an odd thing to check- the Stefan-Boltzmann also incorporates material emissivity which varies with both wavelength and temperature. Unless you know the filament's emissivity, you are not going to be able to accurately calculate either T or a. If you've assumed e = 1 (blackbody), which is common enough, then you shouldn't be too concerned about accurately determining 'a' or 'T'.

We did regression with matlab and fit both the emmissivity and temperature in planck's law, that's where we got our value for T, and our value for P came from the voltage and current data.

We assumed emmissivity was a constant in planck's law, which got us pretty close, but we are now searching for a better approximation.
I have seen an approximation of the form e = aT + bλ + cλT
does anyone know of a reference regarding the emmissivity of a tungsten filament?
thanks

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There are special lamps for precision measurements in photometric, for example this one:
http://www.pyrometer.com/ribbon_lamp.html

I´ve found one source for emissivity of tungsten here:
http://www.monarchserver.com/TableofEmissivity.pdf [Broken]

If you expect high accuracy in photometric measurements you may be in for a disappointment. Within ballpark is probably good enough (at least in an educational context)

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thanks maimonides, that's close, but I'm looking for a model of emmissivity as a function of wavelength (particularly) but also temperature. I suspect I will end up modeling it.

For those who are interested in the finer details of why:
we have two operating modes with our spectrometer, and relative irradiance mode (RIM) uses a specific formula to calculate the values it outputs. The trouble is that when you work out that formula you will find it outputs the spectrum we want, divided by the emmissivity of the calibration source. The calibrated lamp is a tungsten source, So theoretically, if we knew the emmissivity of the tungsten calibration source as a function of wavelength, we could correct the RIM data to match the AIM data. The plot shows reference data, data from absolute irradiance mode (AIM) and data from RIM, normalized. This is the sun's spectrum - It was cloudy today when I took these data sets, so there is some error with the AIM data, but that is irrelevant for now.

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