Light bulb efficiency.......I'm so lost

[Guest A_question]

I'm trying to figure out how efficient/inefficient a certain light bulb is. I'm not sure where to post this, but I'm not sure which area this applies to. I don't know where to start on this and was hoping someone could point me out in the right direction. Thanks.

[Sayonara]

Just to clarify for anyone who can help, do you mean in terms of ratio of visible light and waste heat produced, or something else?

[Guest A_question]

yeah - ive figured out the luminous efficiency of a 100 watt, 120 volt incandescent light bulb to be about 17.5 lumens per watt. But i was wondering if anyone knew how to figure the percentage of useful, visible light of the total radiant energy. I dont know if any of this makes sense. I'm kinda new to this.

[Phospholipid]

Efficiency = (useful energy output/energy input) x100

 

Is that 17.5 lumens per watt a give value or an experimental value? If it's experimental then you should have a given value to compare it to. If it's a given value you need to do an experiment (somehow) to determine how many lumens per watt are actually being used.

[Lance]

You need to convert the lumen to watts.

 

On second thought, that may not even be possible. You need to somehow find the total energy of visible light emitted and compare it with the energy input.

[Martin]

You need to convert the lumen to watts.

 

On second thought' date=' that may not even be possible. You need to somehow find the total energy of visible light emitted and compare it with the energy input.[/quote']

 

I just saw this thread, I remember reading that the temperature of the tungsten filament of a 100 watt litebulb is about 3000 kelvin

 

by comparison the sun is about 5800 kelvin

 

You can use the blackbody curve to find out what fraction of the radiation from an object at 3000 kelvin is in the visible

 

My guess is that it will be on the order of 10 percent efficiency

 

that is, most of the radiant energy output from the filament will be in the invisible infrared part of the spectrum and only a small amount of the radiation out in the high energy short wavelength tail of the distribution will be be visible.

 

Off the top of my head I cant say anything for sure, but I am pretty sure of the 3000 kelvin for the filament, if that is any help

[Martin]

You need to convert the lumen to watts.

 

my handbook gives a conversion

it says that one watt (in the maximum visibility 555 nanometer part of the spectrum) is 680 lumens

 

the illumination value of a watt of light depends on the light being in a good part of the spectrum for the eye, the eye is most sensitive to 555 nanometers.

if you have a watt of light that is all in that very good wavelength then it is worth 680 lumens, my handbook says

 

the poster says the litebulb produces 17.5 lumens for each watt it uses

 

what this says to me is 2.6 percent efficiency

 

because 100 percent efficiency would be that the entire watt of electricity was converted to a watt of perfect 555 nanometer light and that would be 680 lumens

 

from a purely illumination standpoint, permitting the eye to see things, that would be the ideal

but it would sure look funny to have your room illuminated by 555 nm.

[Guest A_question]

hey thanks - i actually did the math myself earlier and got 2.56%. So being roughly 2.6% efficient, where would the remaining energy go to? I think "invisible" light and heat, but i may be wrong

[Martin]

hey thanks - i actually did the math myself earlier and got 2.56%. So being roughly 2.6% efficient, where would the remaining energy go to? I think "invisible" light and heat, but i may be wrong

 

 

I'd guess some heat goes to the glass bulb and heats the surrounding air and some heat flows into the metal socket, so convection and conduction take some away

and a lot of it is radiated in the invisible infrared of thermal radiation

which I guess could be considered light even tho it is no good for illumination

I cant estimate the percentages of where the waste goes

 

white LED light is much more efficient. does anybody know how much more, and how they work (must be based on a blue LED that stimultes reradiations from dyes in other colors to get the mix that looks white, no one LED would do)

[YT2095]

I`de say the easiest way would be to put an Ameter in line with the bulbs and take the current reading from each, a simple calculation will tell you how many watts of electricity is being used.

 

then using a simple light meter as used in photography (maybe borrow one or hire one from a shop or school/college) and put this at a set distance in the same room and then take a reading from each bulb.

 

from there you`ll be able work out a percentile of efficiency for each

[swansont]

You can revisit a previous discussion on a related topic.

[boxhead]

usually it is about 40 %. i/p to o/p ratio

[Dave]

What? More like a tenth of that.

 

I point you to: http://www.wordiq.com/definition/Light_bulb#Efficiency

 

(From the other thread).

[Lance]

usually it is about 40 %. i/p to o/p ratio

 

I'm pretty sure he was talking about an incandescent bulb which as dave pointed out is nowhere near that.

[boxhead]

I'm pretty sure he was talking about an incandescent bulb which as dave pointed out is nowhere near that.

 

oh i am sorry.

[Lance]

oh i am sorry.

 

Don't be sorry, we all make mistakes. I have no Idea what the efficiency of other light sources are though. I'm sure you're right on some level or other.