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Radiant (all wavelength) flux emitted by a light source

  1. Feb 23, 2014 #1

    If I were to hang a traditional 100W clear glass light bulb from the ceiling, would it radiate 80W (radiant flux), the other 20W lost by air convection/conduction up the wires, or would it radiate closer to 60% or 40% of consumed energy?

    I can find plenty of figures for the eye-sensitive flux, but no figures for the all-wavelength flux.

  2. jcsd
  3. Mar 4, 2014 #2
    As nobody came up wiith a reply, I spent quiite a lot of time browsing the Net for a figure. Much to my surprise no manufacturer, or State body quotes figures for total radiant flux emitted by light sources. There are hoards of graphs giving spectal irradiance (watts/m2 arriving on a surface often placed at 50 cm from the bulb), but only at the brightest position with regards to the shape of the bulb. So that's of no use.

    I found just one spectral graph giving total watts emitted per nm of wavelength. By summing the area under the curve, this gave 60W radiated and therefore 40W conduction/convection.

    I also found some theretical calculations, working up from the the tunsten filament. One gave 61 % radiated. Here it is for what it's worth :

    "The IFS-66 uses an internal 150W Tungsten-filament Quartz Halogen Lamp.  Assume that the filament has a color temperature of 3000K, a surface area of 50mm2, and an emissivity of 40%. From Stefan’s Law, the total radiance is 0.5E-04*0.4*5.67E-08*30004 = 92 W  The remaining 58 W is lost by convection and conduction through the base."

    Theoreticaly, if you know the luminous efficiency lumens/watt, and the spectral distribution from 0 to 1, assumed to be the same all around the bulb which is probably true, you can determine the total watts radiated. I haven't pursued this approach.
  4. Mar 4, 2014 #3


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  5. Mar 4, 2014 #4


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    That's because Radiant Flux, measured in watts, is for optical wavelengths from Ultra Violet to Infra Red (.01 to 1000 μm) and all other wavelengths produce a negligible amount of energy. So the marked wattage on a incandescent lamp IS the Radiant Flux.

    The flux seen by the human eye is called Luminous Flux, measured in lumens, and the eye doesn't see every wavelength equally, so your summing won't give you the correct answer. i.e. you must factor the Radiant Flux by the sensitivity of the human eye (683 lumens/watt).


    Image compliments of http://hyperphysics.phy-astr.gsu.edu
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  6. Mar 5, 2014 #5


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    I should have said "visible light"
  7. Mar 5, 2014 #6
    1. "So the marked wattage on a incandescent lamp IS the Radiant Flux." This is false. Marked wattage is power consumption.
    "For years, people have chosen light bulbs by the watt, learning over time about how bright a typical 40-watt or 60-watt bulb is. But wattage tells you only how much energy a bulb uses — not how bright it is." Source amongst others http://www.consumer.ftc.gov/articles/0164-shopping-light-bulbs

    2. "That's because Radiant Flux, measured in watts, is for optical wavelengths" This is false Radiant flux is for all wavelengths, luminous flux is for eye-visible wavelegths. In general -RAD- and -LUM- are dedicated to ALL and VISIBLE wavelengths respectively For example irradiance and illuminance, radiance and luminance

    3. "That is, a 60 W incandescent lamp would have a radiant flux of 60 W, and a 250 W Mercury vapor lamp would have a radiant flux of 250 W." This is false. The person who wrote this should be notified of his mistake, but maybe he just wants to make it simple for the layman. A 60W lamp consumes 60W of electricity and its radiant (all wavelength) flux would, I guess very very approximately, be around 36W, the rest lost in heat conduction and convection.
  8. Mar 5, 2014 #7


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    Your reference is confusing you. Brightness is a photometric measurement dealing with the wavelengths the human eye sees. If you put all/most of the bulbs energy in a few visible colors (what the eye sees), like a CFL does, sure you can get the same "brightness" in lumens while consuming less energy.

    Isn't that what I said?
    .01 to 1000 μm pretty much covers all emission wavelengths.

    So you must agree.

    You need to study the difference between Radiometry and Photometry.

    So if you know everything about the subject, why did you start this thread? Actually most of the radiant flux is from the Infra Red wavelengths which you can't see and is the least energetic part of the bulbs spectrum; hence is a smaller contributor to the total radiant flux.

    https://www.palagems.com/Images/spd_incandescent.jpg [Broken]
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  9. Mar 6, 2014 #8
    I found time (many hours) to work back from the lumen count to estimate the total radiant flux of a bulb, as nobody publishes this figure. I chose an FEL 1000W. See page 21 of the Newport Oriel product training to be found at http://assets.newport.com/webDocuments-EN/images/Light_Sources.PDF.. Newport doesn't give a lumen count for their bulb but as it's used by professionals in laboratory analysis (image on page 16 of the Newport document), I felt that the 25000 lumens quoted on vendor sites wouldn't be grossly overinflated. There is also very useful information at the top of page 21. Unfortunately their Y axis is logarithmic, which is ghastly for manually numerising a graph, so I hunted elsewhere.

    I found the numerical values on page A2 and a drawing of the lamp on page A7 , about half way down a very long document issued by the US Dep't of Commerce, National bureau of Standards in 1986 to be found at http://www.nist.gov/calibrations/upload/sp250-20.pdf. Unfornunately the wavelengths at which they made their there measurements back in 1986 aren't pratical to use, so I used polynomial fitting to fill the gaps. I also exponentialy extended the values from 1600nm down to 2400nm, which is the limit on the Newport graph. .The result gave 12,2% of toal radiant flux (from 2400nm up) is in the visible region between 400 an 700 nm, which compares favorably with 12% flux visible mentioned on the top of page 21 of the Newport graph.

    I then multiplied by the 1931 photopic values (every 5nm) to get a weighted value at each 5nm wavelength, and as you can imagine, I summed in order to find the constant which would bring the lumen count up to 25000. Applying this constant to my polynomial curves, and summing, gave a total radiant flux of 805 watts, so 80% of consumed energy radiated and 20% conduction and convection.

    I suppose a 1000 FEL bulb doen't have ten times more surface than a 100W bulb, so direct heat loss by convection is proportionately less, and I think you would agree with me that convection rather than conduction is the main direct cooling mechanism.

    In fact it's probably more than 80%. Newport meausures from 2400 nm up, probably beacause quartz transparency sharply drops from 90% to next to nothing below this range. We should add the much longer wavelength radiation from the surface of the quartz bulb at temperatures approaching 900°C. If we know the sq cms of surface of a body at 900°C, and supposing an ambient temp of 20°C, the Stefan-Bolzmann T4 equation gives the watts radiated

    But I don't know the sq cms of a FEL 1000, I don't know the emissivity of quartz at this temperature as it could well be different from room temperature, and furtheremore at lower wavelengths it's surely different from visible wavelengths. Too much. I'm not prepared to spend many more hours on the Net finding them so that refinement will have to wait.

    To reply to your comments, my original question was would radiant power be 20 or 40 or 60% or more. We always hope that somebody knowlegeable from the industry, or an enlightened amateur who has already gleaned the net, will give a set of figures the next morning. Alas that was not the case, so I had to hunt them down myself, which proved difficult as they aren't there, so I had to figure out how to calculatre them. Which is a good thing as things you thought you knew turn out to be more complex than you thought. You learn an awful lot.
    One thing I dicovered to my amazement was that the photopic 1931 lumen weightings are practically a normal (Gaussian) distribution. Now why on earth should that be so?
  10. Mar 6, 2014 #9


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    Awesome post Roger44. The devil is always in the details. But the ironic part of understanding the complexities is when in comes to actually putting it to use in specifications or testing; Measured radiance or intensities by good meters with calibrations traceable to NIST.
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