Hello
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 because 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?