Uncovering the Source of Extra 267 J in Standard Incandescent Light Bulb

[Vector1962]

Ok, according to my in depth research, wiki, the filament of standard incandescent light bulb will reach a temperature of roughly 2500 K. According to Stefan-Boltzmann, the power radiated from an object at temperature T (K) is given by P= A ε σ T^4 . Suppose the filament has a radius of 1mm and is 25.4mm long. The filament has a surface area, A, of 1.6E-4 m^2. Further assume emissivity, e, is 1.0.
The radiant power is then calculated is then 367 watts.
According to wiki, a 100 W bulb has a resistance of 144 ohms when it is lit. If the bulb is plugged into a standard 120V outlet, the electrical power delivered to the bulb is P = V^2/R = 100 W.
For a 1 second time period 100 J of electrical energy power the bulb, but somehow 367 J comes out.
Where does the extra 267 J come from?

Is the 267 J stored in the filament material and then released from the filament as it burns?

 

[DaveC426913]

Is the 267 J stored in the filament material and then released from the filament as it burns?

No. The filament in a bulb is not a consumable; it does not burn.

 

[DaveC426913]

Suppose the filament has a radius of 1mm and is 25.4mm long. The filament has a surface area, A, of 1.6E-4 m^2. Further assume emissivity, e, is 1.0.

Perhaps these are poor assumptions.

 

[Vanadium 50]

. Suppose the filament has a radius of 1mm

1 mm? Yowza!

 

[Vector1962]

Match the power in = to power out and solve for surface area assuming 2500 K ? something like this ? 100 = A (1) σ (2500^4) --> A = 4.5E-5 m^2 if L = 25.4mm then R = 0.54mm.. this would make everything balance...?

 

[DaveC426913]

"For a 60-watt 120-volt lamp, the uncoiled length of the tungsten filament is usually22.8 inches (580 mm), and the filament diameter is 0.0018 inches (0.046 mm)."
https://en.wikipedia.org/wiki/Incandescent_light_bulb

This makes for a surface area of 84mm², or 8.4E-5 m².

Hmph. That's only a factor of 2 from your estimation of 16E-5 m².

 

[Vector1962]

Appreciate the comments, I did notice in the wiki that the tungsten does evaporate so it is a consumable? correct?
I think I can make things balance now... thanks again for the comments.

 

[Vanadium 50]

Tungsten does evaporate but this is a weakness, not part of the design. It evaporates at hot points, not throughout the filament.

 

[russ_watters]

Ok, according to my in depth research, wiki, the filament of standard incandescent light bulb will reach a temperature of roughly 2500 K. According to Stefan-Boltzmann, the power radiated from an object at temperature T (K) is given by P= A ε σ T^4 . Suppose the filament has a radius of 1mm and is 25.4mm long. The filament has a surface area, A, of 1.6E-4 m^2. Further assume emissivity, e, is 1.0.
The radiant power is then calculated is then 367 watts.

Does this calculation include the radiation of the bulb? It isn't radiating into space near absolute zero, right...?

 

[Drakkith]

"For a 60-watt 120-volt lamp, the uncoiled length of the tungsten filament is usually22.8 inches (580 mm), and the filament diameter is 0.0018 inches (0.046 mm)."
https://en.wikipedia.org/wiki/Incandescent_light_bulb

This makes for a surface area of 84mm², or 8.4E-5 m².

Hmph. That's only a factor of 2 from your estimation of 16E-5 m².

Also note that the coiled filament will undoubtedly absorb much of its own radiation (since different parts of the coiled filament will block the LOS of other areas) whereas a straight filament will not. Looking at pictures of the filament from wiki, it looks like a reasonable assumption to ignore most of the "inside" surface of the coiled filament when calculating the surface area.

 

[John Park]

The point (I take it) is that the calculated output doesn't match the electrical power input; the fact that room temperature isn't at absolute zero seems irrelevant: 2500⁴ vs 300⁴ is a factor of nearly 5000.

I don't suppose wiki specified the wattage of the bulb?

Yes you might be a bit generous in your estimate of the filament size. After a quick look at a low-wattage bulb, I'd guesstimate 0.5 mm for the diameter, which would reduce the area by a factor of 0.25. I think the working temperature may be quite variable too (Wiki again, from memory). If it was as low as 2000 K, your power estimate would drop by another factor of 0.4, bringing the radiated power below 40 W. Care to guess at the emissivity?

I think there's enough intrinsic slop (technical term) in the parameter estimates to conclude that incandescent light bulbs were probably not gathering power out of the vacuum fluctuations. (Unless of course they are now being phased out to prevent us from Discovering the Truth.)

 

[John Park]

"For a 60-watt 120-volt lamp, the uncoiled length of the tungsten filament is usually22.8 inches (580 mm), and the filament diameter is 0.0018 inches (0.046 mm)."
https://en.wikipedia.org/wiki/Incandescent_light_bulb

This makes for a surface area of 84mm², or 8.4E-5 m².

Hmph. That's only a factor of 2 from your estimation of 16E-5 m².

If it was tightly coiled, wouldn't the effective area be less than this?

 

[Drakkith]

Using these values: $A = 5x10^{-5}m^2, T=2500K, σ=5.67x10^{-8}, ε=1.0$

I get $P= A ε σ T^4 = 110.7522W$

So about 111 watts.

I took Dave's calculated surface area and cut it almost in half under the assumption that much of the radiation from the interior of the coil would be re-absorbed by the filament.

 

[DaveC426913]

If it was tightly coiled, wouldn't the effective area be less than this?

Seems to me, if that were true, it would defeat the purpose of making it that way.

Not really a compelling argument, I agree. But why are they building it tightly coiled (and using excess tungsten) if it isn't effective?

 

[John Park]

Seems to me, if that were true, it would defeat the purpose of making it that way.

Reference https://www.physicsforums.com/threads/stupid-light-bulbs.907265/#post-5714783

A fair point, and I can't claim to have thought of it. But this seems to provide the answer:

http://lamptech.co.uk/Documents/IN Coiling.htm

Looks as though it's to do with convection in the gas. You can find complicated physics everywhere.

 

[OmCheeto]

A fair point, and I can't claim to have thought of it. But this seems to provide the answer:

http://lamptech.co.uk/Documents/IN Coiling.htm

Looks as though it's to do with convection in the gas. You can find complicated physics everywhere.

Excellent document!
Though I find some of that information hard to comprehend.

"... Another result of coiling is to produce a black body effect due to multiple reflections of the radiation within the coil. This increases the brightness of the interior of the coil to about 175% of the exterior brightness.
...
It should be noted that the temperature of a coiled coil filament is no greater than for the single coil type, so life is the same. ... "​

How on Earth does that work? I guess I still don't understand what "temperature" means, nor how this all works.

ps. Just discovered that the light bulb filament "temperature", per wiki: 2550K, is the same as that star in the news: Trappist-1. Weird!

 

[Drakkith]

How on Earth does that work? I guess I still don't understand what "temperature" means, nor how this all works.

A triple-coiled filament just has a third coiling done to reduce its length inside the bulb. A coiled-coil has two coilings done and is the standard filament in most bulbs. The article is just saying that neither type runs hotter than the other.

As for the interior vs exterior brightness, I'm not sure how that works either. I assume the interior isn't much hotter than the exterior, else the bulb would suffer early burnout, but I don't really know what's going on.

 

[OmCheeto]

...I'm not sure how that works either ... I don't really know what's going on.

I rest my case.

 

[John Park]

Excellent document! Though I find some of that information hard to comprehend. "... Another result of coiling is to produce a black body effect due to multiple reflections of the radiation within the coil. This increases the brightness of the interior of the coil to about 175% of the exterior brightness. ... It should be noted that the temperature of a coiled coil filament is no greater than for the single coil type, so life is the same. ... "

Reference https://www.physicsforums.com/threads/stupid-light-bulbs.907265/#post-5714897

The first part seems okay: the coil isn't at equilibrium--its inside is exposed to temperatures of around 2500 K, while its outside sees temperatures an order of magnitude lower. (The interior of the coil, being somewhat enclosed, might also have a high enough density of tungsten vapour to mitigate evaporation.) I can't visualise a coiled-coil well enough to know whether or not I should be surprised that it behaves the same as a simple coil (which is not necessarily the same as an uncoiled filament).

It occurred to me, after I posted, that this emphasis on coils is probably British. I'd come across the terms "coiled filament" and "coiled coil", but that must have been when I lived in England. Because the power output is V²/R, working at 240 V would mean they could use a filament nearly five times as long to get the same power as in a North American bulb.

Just discovered that the light bulb filament "temperature", per wiki: 2550K, is the same as that star in the news: Trappist-1. Weird!

Reference https://www.physicsforums.com/threads/stupid-light-bulbs.907265/#post-5714897

Yes, I noticed that too--a reminder that the light of "red dwarf" stars isn't all that red.

 

[John Park]

I can't visualise a coiled-coil well enough to know whether or not I should be surprised that it behaves the same as a simple coil

Afterthought: maybe the coiled coil has the same temperature but greater surface area.

 

[OmCheeto]

Afterthought: maybe the coiled coil has the same temperature but greater surface area.

I would try and analyze this off the top of my head, but the last time I tried that, I had a stroke^[3].

The document is both word-wise and situationally^[1] "busy", so it may take me some time to extract the data into something I can comprehend^[2].

[1] vacuum vs inert gas, coil count: 0, 1, 2, 3
[2] a spreadsheet has been my latest tool of choice
[3] it's all fun and games, until a PF HW problem tries to kill you...

 

[John Park]

[3] it's all fun and games, until a PF HW problem tries to kill you...

Don't forget the changes in geometry. There may be more chance to stretch a coiled-coil into a (fat) cylinder, giving more surface area than a blob of single coil.

 

[Vector1962]

Seems like the current through a coil would develop some type of magnetic field? Seems like this field would be capable of storing either energy or charge?

 

[Drakkith]

Seems like the current through a coil would develop some type of magnetic field? Seems like this field would be capable of storing either energy or charge?

It would store and release energy at the mains frequency (60 hz here in the states), but that has no effect on the lightbulb except to increase the impedance compared to a non-coiled filament.

 

[OmCheeto]

Ugh. There is just TMI for a days worth of analysis...

Triple coil incandescent filament [US Patent 4499401A, 1985]

Google claims there are no hits for: "quadruple coil incandescent filament".

Thank god.

 

[Drakkith]

Google claims there are no hits for: "quadruple coil incandescent filament".

Thank god.

Quick, Om! To the patent office!

 

[John Park]

It would store and release energy at the mains frequency (60 hz here in the states), but that has no effect on the lightbulb except to increase the impedance compared to a non-coiled filament.

Reference https://www.physicsforums.com/threads/stupid-light-bulbs.907265/page-2#post-5715768

I was told (in Britain) that when a bulb burned out, it was the back-emf in the coil that was responsible. Have't tried to verify this.

 

[Vanadium 50]

What is this thread going on about now?

The original question was why, using some badly estimated numbers (one more than an order of magnitude off), and an equation with fourth powers in it, the power calculated is a factor of 3.5 off. That's been answered - use more accurate numbers and the power calculated gets much closer to the number on the box. Hasn't that been answered adequately?

So what are we talking about now?

 

[Drakkith]

So what are we talking about now?

We're talking about why filaments are coiled and not straight lines and how that affects the operation of the light bulb. If you have a problem with something in the thread, please report it.

 

[Vanadium 50]

We're talking about why filaments are coiled and not straight lines and how that affects the operation of the light bulb

Thank you.

Filaments are coiled to get the resistance up. To get the resistance where you want it, you have two options - lengthening the filament (which requires coiling to make it fit) or narrowing it. If you narrow it, you are producing more heat per unit of filament (however you choose to measure it) and shorten the bulb life.

 

[Drakkith]

I see. So then if you want to dissipate 100 watts you have to balance length and thickness to get the resistance at the right amount and also keep the temperature from either being too low (which will make a dim, inefficient bulb) or too high (reducing filament life).

Also, do you know if my assumption is post #13 about using less surface area in the calculation is accurate at all?

 

[Vanadium 50]

I said heat, not temperature. The thing that destroys bulbs is heat.

I am not sure how important the shadowing effect is. If you model a coil as a hollow tube, the light's dependence on the inner radius is only through the thickness of the tube and thus its resistivity.

 

[Drakkith]

I said heat, not temperature. The thing that destroys bulbs is heat.

I don't know what you mean by this. Could you elaborate?

 

[Paul Colby]

I don't have a reference but I recall coiling a filament reduces failure due to thermal shock. Time lapse movies of lights turning on show considerable vibration due to sudden temperature changes.

 

[Vanadium 50]

I don't know what you mean by this. Could you elaborate?

Heat and temperature are two different quantities - I can stick my hand in a 350 degree oven without injury, but I can't stick it in a 212 degree pot of boiling water. I wrote heat because I meant heat. That's all I meant.