Re: What are the factors that cause a nonmoving incand. light bulb to burn out?

Greetings:

You have taken on a very difficult experiment and I’ll share
with you some of the information that I have about incandescent light bulbs
and the types of measurement problems that you are going to encounter.
You topic is very interesting, however, you might want to measure less difficult life limiting parameters such as the Tungsten (Wolfram in Europe) is element number 74 in the
Periodic Table of Elements. It has a chemical symbol W and is metal of many
superlatives. Not only does it have the highest melting point of all elements except
carbon - sources in scientific literature vary between 3,387°C and 3,422°C - but it
also has excellent high temperature mechanical properties and the lowest
expansion coefficient of all metals. A temperature of about 5,700°C is needed to
bring tungsten to boil - which corresponds approximately to the temperature of the
sun’s surface. With a density of 19.25 grams/cubic cm, tungsten is also among the
heaviest metals. Its electrical conductivity at 0°C is about 28% of that of silver
which itself has the highest conductivity of all metals.

Incandescent lamp failure is interesting but difficult topic and theoretically a lamp
with a perfectly shaped filament would never burn out. A perfect filament would
gradually evaporate, equally from all points on its surface. Thus it would get
thinner, and the resistance would go up. With higher resistance, and the same
voltage, the current goes down, and the power (P=V^2/R) also goes down. Thus
the filament would run cooler, and evaporate more slowly. Because the
evaporation is exponential with temperature, theoretically the filament would never
disappear, but just gradually get thinner and dimmer.

In real life, the filament has defects--spots that are a bit thinner than the rest of the
filament. These spots must carry the same current as the rest of the filament, but
the spots have a higher resistance per unit length. So the local power at these
points, (P=I^2*R) is higher, and the spots have a higher temperature. This makes
the tungsten evaporate faster, which makes the spot get thinner, which makes it
run hotter and so on....until the filament breaks at one of the hot spots.
When a spot in the filament breaks, the entire line voltage (plus some inductive
kick, perhaps) appears across a tiny gap between the filament wires. This ionizes
the fill gas in the bulb (usually argon - Ar). The resulting arc is an avalanche
process, and more and more gas is ionized, resulting in lower arc resistance,
resulting in more ionization, etc. Finally, the arc is between the two lead-in wires,
not just across the break in the tungsten, and because the resistance of the tungsten
is out of the circuit, there can be a much higher current causing a bright flash of light.

The cold resistance of the
tungsten is about ten times lower than the resistance at normal operating
temperature (this is important for your experiments). So at turn-on there is a large
current surge until the filament has heated up (actually only 16 milliseconds ~10
cycles of 60 Hz voltage). The defect spots will not only be heated up more by
having a high resistance, but they will also get heated up faster, because they have
less thermal mass (actually thermal mass to surface area ratio is what matters).
This fast heating continues not until the hot spots reach their operating
temperature, but until the rest of the filament reaches its operating temperature.
Remember, the rest of the filament has most of the resistance, and so controls the
current. Thus the hot spots will overshoot their steady-state operating
temperature, which is already hotter than the proper operating temperature. This
overshoot is only for ~16 milliseconds, so the extra evaporation during the
overshoot is not significant for accelerating the aging of the bulb. But if the bulb is
very close to failure, and the hot spots are really bad, they can over heat to the
point where the tungsten softens or melts, and the filament can break in a flash.

Does turning lamps on and off result in shorter life? In general experts such as
Charles Sullivan of the University of California at Berkeley indicate that the lamp
must be within about 4-8 hours of failure for the turn-on to push it over the edge
and make it fail. Cycling the lamp will only decrease the life by a maximum of 4-8
hours, no matter how often it is cycled.

You write: “One part of my experiment is observing 2 clear glass light bulbs and 2
soft white light bulbs (all 60 watt) and seeing if one type burns out sooner than the
other.”
A typical light bulb at temperatures below 3000 C emits about 85% of its energy
at infrared wavelengths as heat and only 15%
as visible light. It would be interesting to measure the transmission
of clear bulbs and soft white bulbs at infrared wavelengths. I’m not sure what the
difference would be; however, it probably is a small difference because of the
thinness of the glass and you would need many hundreds, perhaps thousands, of
bulbs to be tested to statistically determine the differences between the two types
of bulbs for the failures caued by the thin spot problem in the filaments would
mask the data that you are trying to measure.

You also write: “The second part of my experiment is measuring the resistance of
each type over a period of time to see if there is a difference in how fast the
resistance changes.”

The hot resistance is what you need to measure and this cannot be accomplished
with an Ohm meter. You need to measure the voltage and current through the bulb
very accurately to determine the resistance by Ohm’s Law (Resistance = Volts
divided by current). Making these measurements is difficult and DANGEROUS
and they should be performed under the guidance of a person knowledgeable in
making electrical measurements. In the place of making current measurements I
would suggest that you place a very small value, precision resistor, such as a one
ohm one watt resistor, in series with the light bulb and measure the voltage across
the resistor to determine the current through the bulb. The current through a
typical 60 watt bulb will be; I (amps) = Power divided by voltage (R = 60
watts/110 volts ac = 0.56 amps. The hot resistance would be about R = V/I =110
vac/0.56 amps = 196 ohms, so that a one ohm resistor would not have a large effect on
your measurements. However, the power line voltage can change 10 % or more
over the day and this will effect your data unless you maintain a constant voltage.

You ask: “Do you know of the length of a regular 60 watt bulb can live?”
The "Big Three" light bulb manufacturers, are General Electric, Osram/Sylvania, and
Philips. They are the main producers of good light bulbs in the USA. Please note that store
brand light bulbs with lumen light output figures and hour life expectancy figures similar to
those of "Big Three" actually are specially branded "Big Three" lightbulbs.
The life average life of "Big Three" "regular" (A19) lightbulbs will usually be:
100 watt bulbs with 750 hour average life and 1670-1750 lumens average light output
75 watt bulbs with 750 hour average life and 1150-1210 lumens average light output
60 watt bulbs with 1000 hour average life and 840-890 lumens average light output
40 watt bulbs with 1000-1500 hours and 440-505 lumens average light output

If you search the Mad Science archives for “light bulb” you will find a number of helpful
answers to questions similar to yours.

Best regards and good luck with your experiments, Adrian Popa