"Catching energy from the Sun is pretty simple. Sit in the sun on a cool day and benefit from its warmth. Situate your house so that south-facing windows can swallow sunlight and offset (or obviate) conventional sources of heat. Use thermal collectors for domestic hot water and/or interior heating. Stick a PV panel outside and it will generate electricity provided it is not placed face-down. Concentrate sunlight to heat a fluid and/or create steam for electricity production in a heat engine (possibly combined with thermal storage). Lots of options, when the sun shines."
'via Blog this'Solar vs. Wind - Who Wins? at Oil Price:
Of the 1370 W/m² incident on the upper atmosphere from the Sun, 30% is reflected straight away without pausing long enough to say hello. About 20% is absorbed in the atmosphere and clouds, and 50% gets absorbed at ground level. Note that 7% of the energy budget goes into conduction and rising air (separate phenomena; the latter relating to wind). Virtually no heat is able to conduct through the thick atmosphere, so really this figure is all about convection, or moving air. For comparison, the energy consumption (conversion) rate of the human race is about 13 TW (13 trillion Watts), which works out to an average of about 2,000 W per person on the globe (Americans are 10 kW). We can also divide by the area of the globe to get a power density of 0.025 W per square meter, or 0.09 W/m² if we just count land area.
If 50% of the incoming solar radiation makes its way to the ground, then we have about 700 W/m² for the average terrestrial square meter facing the Sun. But the Sun puts this onto the projected ?R² area of the Earth (the disk of the Earth as seen from the Sun), while the actual 3-D globe has an area of 4?R². So we must divide by four to get the flux per unit area of actual terra firma, yielding 170 W/m². We can think of this factor of four as being made up of a factor of two for day and night, plus a factor of two because the Sun is not overhead all the time, resulting in a loss of intensity per square meter at the ground. A panel tilted to the site latitude can compensate for some of the slanted-sun-angle loss (for high latitudes, the ground always suffers from this geometric dilution, even at “high” noon), so that the ½ factor becomes 2/?, or 0.64, representing a global 30% boost over horizontal panels. In this scheme, we get 220 W/m² for our latitude-tilted panel (nearly independent of latitude, weather notwithstanding). The tilted panels will require more land to avoid self-shadowing, so that the amount of land area needed is stuck with the pre-adjusted value of 170 W/m².Note how much bigger the solar potential is than our demand of 0.09 W/m² of land area. This implies that we need only 0.05% of the land to capture adequate sunlight, or that enough sunlight strikes land (the entire Earth) in 4.5 hours (1.25 hours) to satisfy our needs for a year. That’s a powerful resource!
But once we factor in efficiency—say 10% for simplicity and conservatism—we need ten times the land area computed above. Still, it’s a pittance. I have used the following graphic before to illustrate how much land would be occupied by solar photovoltaics (PV) at 8% efficiency to produce 18 TW of electrical output (note that about half of the 13 TW consumption today is lost in heat engines, so 18 TW of electricity more than satisfies our current demand)
