## Friday, November 6, 2009

### Power to the People (FPOTW)

Incandescent light bulbs are very inefficient. Use a google search engine to find out just how inefficient.

Compact florescent bulbs are much more efficient in the percentage of watts producing visible light.

This information is also on google.

www.google.com "how to use google" is a good place to start. =)

---- Start Fermi Problem ----

Estimate how many incandescent 100 Watt light bulbs there are in the United States.

Estimate how much power is used to power the light bulbs.

Estimate how much power is saved by switching all the light bulbs to equivalent lumen compact florescent.

Assume that the old light bulbs are thrown away. How much landfill space is taken up if the bulbs are not broken?

How about if the bulbs are broken?

What is the mass of the broken bulbs?

Some Physics: What is the weight of the broken bulbs compared to the weight of the unbroken bulbs? Remember, I said "weight" here.

---- End Fermi Problem ----

Estimate how many Poles it takes to screw in all those light bulbs. (old POLEitically incorrect ethnic joke here)

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What about the mercury in all the CF bulbs?

ReplyDeleteOf course the weight would be the same, unless you are saying the force of gravity is somehow changed by crushing the glass? Compressing the earth under the broken shards? Mass is always concerned. You need to go back and study your freshman physics.

ReplyDeleteconcerned = conserved

ReplyDelete"Some Physics: What is the weight of the broken bulbs compared to the weight of the unbroken bulbs? Remember, I said "weight" here."

ReplyDeleteThe measured weight F of a bulb in the air is the

difference between the true weight w (weight in the vacuum) and the buoyant force B:

F=w-B (for unbroken bulb)

F'=w'-B' (for broken bulb)

Obviously w>w' because unbroken bulb contains some amount of an inert gas and w-w'=g*di*V

(The most common is probably argon)

Obviously B>B' and B-B'=g*da*V

where V is glass-enclosed volume

di is the density of the inert gas in the unbroken bulb

da is the density of the surrounded air

g is the acceleration of gravity

The difference between the weights is:

F-F'=w-w'-(B-B')=g*(di-da)*V

So we need to estimate the density of the inert gas in the bulb.

Let's take the argon.

From ideal gas law we have:

di=P*M/(R*T)

where P is the pressure inside the bulb

(Typically 70kPa or so)

M is molar mass of the argon

(40g/mol)

R =8.31 J/(K*mol)is gas constant

T is temperature

Assume that temperatures are the same inside and outside the bulb and let's take T=293K (20C)

di=7*10^4*4*10^(-2)/(8.31*293)=1.15kg/m^3

This is very close to air density at 20C and P=10^5Pa which is 1.2kg/m^3

So i assume that in reality there may be so these light bulbs that weigh more at broken state, as well as those that weigh more at unbroken state.

Totally, the weight of the broken bulbs should be approximately equal to the weight of the unbroken bulbs, i think.

Thanks for interesting problem(s)!

Some Physics yet:

Does an ordinary light bulb explode in vacuum?

Oh, sneaky. But Varney should still go back and study his freshman physics.

ReplyDelete@Martin

ReplyDeleteCool analysis. What may be interesting to consider is that in Colorado, the air density is 1.05 kg/m^3. =)

Fun question on the lightbulb in a vacuum.

Lets see,

Assume a constant pressure inside a spherical shell. For the shell to rupture, the internal pressure must exert a force greater than the yield point stress (or tensile stress) of the material.

We can use dimensional analysis to get a relationship:

Any stress has units equivalent to pressure: sigma == [N] m^-2

Lets assume the only variables we need to consider are the pressure inside the shell (P), the radius of the shell (r), and the thickness of the shell (h).

We can use Buckingham Pi theory to determine the combination of pressure, r and h to give us the dimensions of stress, or we can do it by inspection, for which I get:

stress = constant (p r)/h

For a light-bulb of radius r = 2.5 cm, and an internal pressure of 70kPa, and a thickness of 1 mm and assuming the constant is of order unity we get a stress the gas exerts of a bit less than 2MPa. This is order of magnitude.

The tensile strength of glass is greater than 25MPa, so the pressure inside the glass bulb is not sufficient to rupture the light-bulb.

Now, since the light-bulb is not a perfect sphere, we can have less yield strength at the neck. Or if there is a bonding agent between the bulb and the metal base, that may rupture.

As an experiment today, I took a 100 W incandescent bulb and put in a chamber and took the pressure down with a roughing pump to a few milli torr. The bulb lived.

Empirical evidence wins again! ;)

I think you have inspired me to write a short blog on dimensional analysis and Buckingham Pi, using the light-bulb as an example.

light-bulb.

Oh and Trade... bite me! ;)

Population of the US: about 300 million. Assuming 10 % ( I have no idea if this no is even practical or not) use incandescent light bulbs, at a rate of 5 blulbs per person per yr, that 150 million.

ReplyDeleteIf each bulb is 100 W, thats 150*10^8 W

Assuming the surface area of each bulb is 10 cm^ (slight underestimation for the sake of simpler calculations), thats 1500 million square cm or .15 million square meters

The combined mass would be 150 millions * 10 gms or something like that

I would assume that there are at least one light-bulb per person in the USA. =)

ReplyDelete