Re: [Phys-l] light bulbs

["Karshner, Gary" <gkarshner@stmarytx.edu>]

Brian,
	Well argued with one small problem, above 3700K Tungsten is a liquid and would not make a very good filament material.  Photofloods that have a short lifetime because the burn very hot run at 2870 K. The low pressure gas in "modern" lamps is there to slowdown filament evaporation. Initially argon was used because it was chemically inert. Last I heard they were using nitrogen gas, which works a little better and is much cheaper. 
	Also with that much temperature change could you still get away with a simple linear term for the temperature correction. I would expect the higher order derivative terms in the Taylor would be getting important. 
Gary

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of brian whatcott
Sent: Tuesday, March 08, 2011 12:17 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] light bulbs

On 3/8/2011 7:06 AM, Anthony Lapinski wrote:
> I have some incandescent bulbs from Europe. In America, the voltage is
> half as much, so the current is half as much. Thus, the power should be
> one quarter as much (P = IV)
>
> A 60-W European bulb should behave like a 15-W American bulb (4X less
> power). However, when I plug it in, it looks more like the glowing
> filament of a 7.5-W bulb (8X less power). So I am a bit puzzled with this.
> I am wondering if it has something to do with the filament specs. I
> believe the electrical frequency in Europe is 50 Hz, but that shouldn't
> really matter with 60 Hz in the US.
>
> Anyone have experience with this, or a similar demo?
>
> _
The resistance of incandescent light bulbs varies with filament temperature.
Resistance = Rref*( 1 + alpha (T - Tref))

Rref is  the resistance at reference temperature T (Tref)
Alpha the thermal coefficient for tungsten is 0.004403 /K at Tref = 293K
T is the temperature of interest.

A 120volt 60 watt lamp has an R(hot)  = 240 ohms
a 240 volt 60 watt lamp has an R(hot) = 960 ohms

Let's suppose that a 240 volt, 60 watt lamp that would burn at 5000K
  only heats to 3000K at the lower voltage.
What would its effective resistance be for this condition?
First, its reference resistance = Hot Resistance/(1 + alpha(Thot - Tref))
= 960 ohms/(1 + 0.00444403*(5000K - 293K)) for the 240 volt bulb at 240 
volt.
= X ohms/(1 + 0.004403*(3000K - 293K)    for the 240 volt bulb at 120 volts.

so X ohms = 960 ohms * (1 + alpha(3000K- 293K) / (1 + alpha(5000K - 293K))
say 550 ohms that's the 240 volt filament resistance when burning dimly 
at 3000K

That leads to a power estimate of V^2/R = 120^2 volts*volts/550ohms = 26 
watts,
mostly in the infra red....
26 watts to produce what looks like 7.5 watts of light equivalence? How 
inefficient!
That's why some economists argued for 1000 hour bulbs - as more efficient
when burning the candle at both ends.

Brian W


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