Lightening Energy Spectrum

All energy and matter decays to heat - eventually. In 10¹⁰⁰ years, there will be nothing else left.

In solar industry maximum solar radiation received on the surface of the Earth is estimated to be 800W - 1000W per m2. If we use value of 1000W/m2 and your answer of 54% being visible light, which would indicate that every m2 of surface receive visible light of 540W.

To produce same amount of visible light, incandescent bulbs that emit 10% of consumed power as light would need to be 5400W in power.

Does that seem plausible?

Yes, very. I've used a 'Sun Gun' made by Osram that was rated at 1000 watts, which produced a focused beam of incandescent light over a circle less than 1 foot in diameter. This lamp more-or-less re-created the luminous intensity of the Sun. A more efficient arrangement of lamps to light an area of 1 square meter could probably be done with 5400 Watts.

I'm not sure where your "800W - 1000W per m2" figure comes from- a more common figure is on the order of 300 W/m², at the earth's surface at the equator- less at higher latitudes (more like 100 W/m² in England, for example).

The Sun's rays are attenuated as they pass though the atmosphere, thus reducing the insolation at the Earth's surface to approximately 1,000 watts per square meter for a surface perpendicular to the Sun's rays at sea level on a clear day.

http://en.wikipedia.org/wiki/Insolation

Quoting from your reference:

The actual figure varies with the Sun angle at different times of year, according to the distance the sunlight travels through the air, and depending on the extent of atmospheric haze and cloud cover. Ignoring clouds, the average insolation for the Earth is approximately 250 watts per square meter (6 (kW·h/m²)/day), taking into account the lower radiation intensity in early morning and evening, and its near-absence at night.

The same article mentions the 1000 W/m² figure, but read closely- "the average insolation for the Earth is approximately 250 watts per square meter".

Other sources are more generous with this number, citing 300 W/m² (if you happen to be close to the equator). Your reference also explains the effect of latitude on the actual insolation at a particular location, and the illustrations on the same wiki page provide good information for insolation for various locations.

Taking this divergent thread one step closer to the actual answer, let us not forget that unless there is some form of tracking involved then the actual amount collected by, as opposed to incident to, the collector surface will be a fraction of what passes through the atmosphere. Here's the next paragraph in the wikipedia article:

When calculating the output of, for example, a photovoltaic panel, the angle of the sun relative to the panel needs to be taken into account as well as the insolation. (The insolation, taking into account the attenuation of the atmosphere, should be multiplied by the cosine of the angle between the normal to the panel and the direction of the sun from it).

I can't believe that this is still a source of confusion.

Did you include conduction and convection heat loss along with the radiated heat loss for an incandescent bulb?

I think you've mixed up the numbers for efficiency and efficacy. The wiki text is somewhat ambiguous, but the table data (table1 and table2) clearly shows <3 [%] for luminous efficiency and ~15 [lumen/watt] for luminous efficacy. Although percentage is dimensionless, it is important to keep track of units during these types of comparisons.

The average insolation figure must be from pole to pole.

Yesterday (15 may 11) insolation at my home in Izmir, Turkey was 7,564 watts/m2 over a 15 hour period - with about 85% of that being useful for solar panels in a 9 hour period.