Brightness and temperature of a filament

[Ott Rovgeisha]

Hi again.

We have two electric circuits. One is with a thin filament and the other has say 5 times larger filament.
It is clear that the 5 times thicker filament is 5 times more powerful: the energy dissipation from electrons to the atoms is 5 times faster, because the current is higher 5 times...

However, is it safe to say that the thick filament heats up more?
Yes, it's power is greater than that of the thin filament, but it also has more volume.

Yet we do see that 100 W incandescent bulb is brighter and hotter, and it DOES have a thicker filament than 20 W bulb.

Why is that? Does it have something to do with volume to surface area ratio?

Best regards...

 

[CWatters]

Interesting question. I can't claim to be an expert so I might be wrong but..

As I understand it light bulb filaments are designed to run as hot as possible because the hotter they are the more efficient they are. Filament temperature is limited by the properties of tungsten so I would expect the filament designers to try and keep the temperature similar for both?

In equilibrium the power going in and out is the same. The amount of power a filament emits is dependant on it's surface area. See Stefan–Boltzmann law..

http://en.wikipedia.org/wiki/Stefan–Boltzmann_law

So if the temperature is fixed I think they would have to play around with the surface area to suit the power of the bulb. Presumably this is why filaments have a complex "coil of coils" structure.

 

[Philip Wood]

I agree with CWatters's remarks. We have two equations for power, P.

The electrical input power is $P=\frac{\pi r^2 V^2}{\rho L}$ in which $\rho$ is the resistivity.

This power is equal to the emitted power, which, treating the filament surface as a black body, is $P=\sigma 2\pi r L T^4$, in which $\sigma$ is the Stefan-Boltzmann constant.

Re-arranging these equations, $\frac{r^2}{L}=\frac{\rho P}{\pi V^2}$ and $rL =\frac{P}{2 \pi \sigma T^4}$.

These equations enable us to find values for r and L, for a given P and V. For example, multiplying the equations together gives $r^3 = \frac{\rho P^2}{2 \pi^2 \sigma T^4V^2}$. More neatly: $r^3 = \frac{\rho I^2}{2 \pi^2 \sigma T^4}$ for a given current, I..

 

[Vanadium 50]

There is a nice article in The Physics Teacher, reproduced here. There is a temperature dependence with power, but it is small.

 

[Philip Wood]

I should have said in post 3 that the resistivity, $\rho$, is temperature-dependent. It is roughly proportional to kelvin temperature, so, at a tungsten filament's working temperature, will be in the order of 10 times the resistivity of tungsten at room temperature.

Just for interest, I'll put some figures into my equation for r…
For a tungsten filament operating at T = 2500 K, $\rho= 4.65 \times 10^{-8} (1 + 4.5 \times 10^{-3} \times 2200) = 5.1 \times 10^{-7}) \text{\Omega m}$.
$\sigma = 5.7 \times 10^{-8}$ W m^-2 K^-4.
I = 0.43 A for a 100 W lamp operating at 230 V.
This gives r = 0.013 mm, which strikes me as not unreasonable for this lamp. I make the length 0.45 m, which seems rather long, but, as CWatters reminded us, they do use a coiled coil configuration.

 

[Ott Rovgeisha]

So in conclusion, what can we say?

1. 6 Times thicker wire exhibits 6 times more power.
2. But 6 times thicker wire does not mean that the surface area also gets 6 times larger; in fact it gets larger only about 2,4 times if I am correct (for a cylindrical wire)
3. That means that the energy has less surface to leave from, and that means that the thicker wire should get hotter.
4. Since the temperature and the irradiance are dependent of each other through the power 4, that means that the 6 times thicker wire should be hotter some 1,7 times.
5. BUT on the other hand the resistivity of the wire grows with the temperature and that makes the current and therefore the power smaller...

So what does it all add up to exactly?

Intuitively I tried connecting a thick wire to a battery and then a thin one. The thicker did seem to get hotter than the thinner.

Why exactly? (or perhaps a qualitative touch does not do the trick)...

 

[Philip Wood]

1. For a given pd across the filament, and a given length and temperature of filament, the power will be proportional to the cross-sectional area (using $P=\frac{V^2}{R}$ and $R=\frac{\rho L}{A}$), and hence to the square of the radius.

2. If you double the radius of a wire, keeping its length the same, you will double its curved surface area (2 pi r L).

For my taste, there's too much going on here to use this piecemeal, semi-quantitative approach safely. That's why I presented the treatment that I did in post 3.

 

[Ott Rovgeisha]

2. If you double the radius of a wire, keeping its length the same, you will double its curved surface area (2 pi r L).

I tend to reason that making something thicker 2 times does not mean making the radius bigger 2 times, but the cross sectional area. if you double the cross sectional area, you most certainly do not double the curved surface area.

There really IS pretty much going on, that is absolutely true...

Still I would like to come to a conclusive, straightforward answer: is a thicker wire going to be more hot and thus glow brighter or not.
According to what I tried, the thicker wire really is hotter, but the reasons for it, are not so clear to me...

This is interesting, how such important things, like how bright is my bulb going to be and WHY is not even discussed in high school physics classes. It is not hard to imagine why so many people find physics dull, horrible etc... Sad though..

 

[Philip Wood]

I tend to reason that making something thicker 2 times does not mean making the radius bigger 2 times, but the cross sectional area.

Well, you may reason that if you wish, but such reasoning is no good for the purposes of communication, because you won't be understood!

But your basic question is interesting. Are you applying the same pd across the wires of different thickness? Are their lengths always the same? If so, I support your conclusion that the temperature will go up as the radius goes up. I think that the dependency is quite a small one ($T \text{proportional to } r^{\frac{1}{5}}$).

 

[Ott Rovgeisha]

Well, you may reason that if you wish, but such reasoning is no good for the purposes of communication, because you won't be understood!

But your basic question is interesting. Are you applying the same pd across the wires of different thickness? Are their lengths always the same? If so, I support your conclusion that the temperature will go up as the radius goes up. I think that the dependency is quite a small one ($T \text{proportional to } r^{\frac{1}{5}}$).

To be honest, i do not understand what is there so difficult to understand: making a wire thicker 2 times cannot possibly mean making the radius bigger two times, thickness is determined by the cross sectional area.

Yes everything is the same, same electric field, same potential difference, same lengths...

By my calculation, the thicker wire should get almost twice as hot if you make the cross sectional area for example 6 times larger.
But another problem is the resistance, which is growing with temperature...

 

[Ott Rovgeisha]

Interesting:
https://www.physicsforums.com/threads/incandescent-light-bulb-resistance-and-brightness.284581/

 

[Philip Wood]

To be honest, i do not understand what is there so difficult to understand: making a wire thicker 2 times cannot possibly mean making the radius bigger two times, thickness is determined by the cross sectional area.

Just say "double the cross-sectional area'' (if that's what you mean) rather than 'double the thickness' when communicating. Take this as friendly advice.

Yes everything is the same, same electric field, same potential difference, same lengths...

By my calculation, the thicker wire should get almost twice as hot if you make the cross sectional area for example 6 times larger.

May I suggest that you use the two basic equations I gave in post 3? They'll answer your question clearly.

But another problem is the resistance, which is growing with temperature...

This is easily factored in by writing $\rho = \rho_{273} \times \frac{T}{273}$ in which $\rho_{273}$ is the resistivity at 273 K, and so is a constant. This is an approximation, but fit for purpose.

 

[Ott Rovgeisha]

Just say "double the cross-sectional area'' (if that's what you mean) rather than 'double the thickness' when communicating. Take this as friendly advice.
May I suggest that you use the two basic equations I gave in post 3? They'll answer your question clearly.This is easily factored in by writing $\rho = \rho_{273} \times \frac{T}{273}$ in which $\rho_{273}$ is the resistivity at 273 K, and so is a constant. This is an approximation, but fit for purpose.

I appreciate the friendly advice.
I also appreciate the equations...

However, since I teach high-school, I must be able to translate all of this into understandable words. Every teacher's main goal is to word the things that are non-intuitive and even intuitive, because even clear intuition can cause a lot of trouble.

Please do not be offended because of the following:
Since we touched the subject of wording, then let me say this: everyone can throw equations on the screen; I am not saying this to insult anybody here, honestly, but what is very important, is to support the equations with conceptional wording...

Yes you may suggest that I use the equations in post 3... I really am grateful for the response and the equations, but the thing is...I rarely ever post at the forums, partly because equations are findable (is that a word?) without the forums. What is important, and I think you agree, is the presence on subtle substantive concepts that can be quite tricky.

At higher temperatures the brightness seems to be "less dramatically" effected by the current than in lower temperatures. Now what is "less dramatically", is the question that probably some equation will answer :D ( no irony here). So even if the current falls, the brightness will not go down as much.

Cheers!

 

[Philip Wood]

You've now made it clear that you want a qualitative or at most semi-quantitative explanation. I don't want to seem to be saying that my way is the only possible way to understand what's going on, but I do think that the Physics is in these two equations. You must be very careful of arguments based on surface area and volume, lest you oversimplify and get the Physics wrong. Good luck in your quest.

 

[CWatters]

Does it help to look at what happens at lower temperatures? Say a copper wire in air?

Double the cross sectional area -> resistance goes down by half - > Power doubles.

Anyone know what happens to the thermal resistance (wire to air) if the cross section doubles? You need that to work out what happens to the temperature of the wire.