Re: Filament resistance numbers

[brian whatcott <inet@intellisys.net>]

At 16:46 4/7/98 -0700, Dan MacIsaac wrote:
> >
> .. It'd still be nice to
> have a geometric determination of the length of that wire spiral as well.
>
> ...  I'd feel happier if I knew the WHOLE
> picture, including a correct temp for the filament (is 3000K the photo
> standard?) and a better value on the diameter of the filament.  As well, 
> the composition of the filament -- is it really just W?

> Dan M

I'd like to help with measuring a 150 watt lamp's coiled-coil filament -
but after destroying a perfectly good heavy duty 40W lamp to measure its
filament diameter, my wife will encourage no further conspicuous waste in
the cause of science. She says that 150 W lamps can cost 3 or 4 dollars.

 You might try a micrometer or a microscope with calibrated graticule if
you can get spousal permission. I can provide a couple of values for alpha
- the linear expansivity for tungsten - again the property is somewhat
non-linear.

Temperature K    Alpha x 10^-6   (Yarwood & Castle, Physical & Math Tables
300              4.44              Macmillan)
2300             7.26

The alloy you expect to find in a coiled-coil lamp is a 1% thoria tungsten.
The thoria provides increased resistance to hot-sagging apparently.
(Enc Brit 15th ed.)

With a value of 3125K for the filament temperature (it varies...), and the
alpha and rho values I provided you could expect to reasonably estimate the
total active length. But a note to a manufacturer might provide a sensible
alternative and cross-check. It's easy to see that the suggested answer
from the text is possibly twenty times too long.

 Here's what I calculated - ignoring the alloying:

temp K    Resistivity Rho (micro ohm cm)
300       5.65
1000     24.93
2000     56.67
3000     92.04

Extrapolate to 3125K:
92.04 + 125/1000 * (92.04 - 56.67) = 96.5 uohm.cm


Temperature K    Alpha x 10^-6   (Yarwood & Castle, Physical & Math Tables
300              4.44              Macmillan)
2300             7.26

Extrapolate to 3125K:
7.26 + 825/2000 * (7.26 - 4.44) = 8.4 * 10^-6


For a 150W lamp at 115 V, filament radius 0.045 mm.
power = E^2/R(hot)   R(hot) = E^2/W = 88.2 ohm

Terms:
R resistance, E line voltage, W power dissipated, L length, 
A cross-section area, r filament radius


R(hot) = Rho(hot)*L(hot)/A(hot) = Rho(hot)*L(hot)/pi*r(hot)^2
So:
L(hot) = R(hot) * pi * r(hot)^2/Rho(hot)


But:
L(cold) = L(hot)/(1 + alpha*T)
     
r(cold) = r(hot)/(1 + alpha*T) 

So:
L(cold) = (R(hot) * pi * [r(cold)/(1 + alpha*T)]^2/Rho(hot))/(1 + alpha*T)

Substituting the best numeric values we have estimated:

L(cold) = (88.2 * 3.142 * [4.5E-5/(1 + 8.4E-6 *3125)]^2/96.5E-8)
            /(1+8.4E-6*3125)

 = (277.09 * [4.5E-5/(1.0263)]^2/96.5E-8)/(1.0263)

 = (277.09 * 1.92E-9 / 96.5E-8) / (1.0263)

 = 0.54 meter   ("my estimate")

cf: 11 meters    ("text estimate")

 The thermal expansivity correction was hardly worth the effort in view 
of the probable error of the calculation, wouldn't you say?

Whatcott   Altus OK