## Equating solar power to usable heat

I am hoping to find how to determine the heat available from focused and concentrated sunlight. I am not confident in my math skills or understanding of some terms, so will describe what I want to do and how I expect to accomplish it. I mostly need an overview and some specific details or suggestions.

I want to use sunlight, focused by some number, or area, of arrayed Fresnel lenses, - each designed for a position in an array and tracked to the sun, with the ability to move the focal point across a surface.

I would like to be able to fuse layers of sand particles within a roughly 4" to 6" swath at a rate of around one square foot per minute - that is to bring the surface layer to around 1450 C as it travels across the surface layer.

Any details or thoughts will be appreciated. Thank you

 Quote by ThoughtRay I am hoping to find how to determine the heat available from focused and concentrated sunlight. I want to use sunlight, focused by some number, or area, of arrayed Fresnel lenses, - each designed for a position in an array and tracked to the sun, with the ability to move the focal point across a surface. I would like to be able to fuse layers of sand particles within a roughly 4" to 6" swath at a rate of around one square foot per minute - that is to bring the surface layer to around 1450 C as it travels across the surface layer. Any details or thoughts will be appreciated. Thank you
Mean insolation at the outside of the Earth's atmosphere averages some 342 watts per square meter, measured normal to the Sun's rays. The atmosphere absorbs roughly 102 watts, and scatters another 134 watts. This leaves approximately 106 watts per square meter of direct beam insolation at the Earth's surface. These are world average figures, and vary with cloud cover and thickness, humidity, dust and particulate concentrations, and the like. There is another 57 watts of diffuse solar radiation (skylight), but I don't know whether your array system can focus this radiation.

It seems to me that you would need a damn tight focus to raise surface temperatures to 1450°C. A square foot fused each minute??? I don't think so. You're expecting more energy than simple sunshine can provide.
 klimatos: I'm just going from memory here, but I think your numbers are off. I think the solar flux at the outside of the atmosphere is more like 1.3 kW/m2, not 340 watts/m2.

## Equating solar power to usable heat

 Quote by cjl klimatos: I'm just going from memory here, but I think your numbers are off. I think the solar flux at the outside of the atmosphere is more like 1.3 kW/m2, not 340 watts/m2.
Yes. That is the solar constant for the flux through a square meter of space. However, to calculate the insolation on the surface, we have to divide by 4 (the surface of a sphere is four times the surface of a circle of the same diameter). The figure of 342 watts per square meter is pretty universally agreed upon by climatologists.

 Quote by klimatos Mean insolation at the outside of the Earth's atmosphere averages some 342 watts per square meter, measured normal to the Sun's rays. The atmosphere absorbs roughly 102 watts, and scatters another 134 watts. This leaves approximately 106 watts per square meter of direct beam insolation at the Earth's surface. These are world average figures, and vary with cloud cover and thickness, humidity, dust and particulate concentrations, and the like. There is another 57 watts of diffuse solar radiation (skylight), but I don't know whether your array system can focus this radiation.
Sorry, ThoughtRay, I got carried away trying to give a quick answer and did not give your question proper thought. Since you will not be trying this procedure at night, we don't want long-term averages. The current Solar Constant is 1366 watts per square meter at the outside of the Earth's atmosphere. Scattering, reflection, and absorption by the atmosphere varies tremendously from place to place and time to time. If you pick your location, season, and time of day very carefully, you might end up with as much as 70% of this value at the surface. From here on, its an engineering calculation that depends on the nature of your array mechanism.

 Quote by klimatos Yes. That is the solar constant for the flux through a square meter of space. However, to calculate the insolation on the surface, we have to divide by 4 (the surface of a sphere is four times the surface of a circle of the same diameter). The figure of 342 watts per square meter is pretty universally agreed upon by climatologists.
Well, then you shouldn't specify "normal to the sun's rays". You should say that the mean insolation averaged over the entire surface of the earth is 342 watts per square meter. It's a very different statement.

 Quote by cjl Well, then you shouldn't specify "normal to the sun's rays". You should say that the mean insolation averaged over the entire surface of the earth is 342 watts per square meter. It's a very different statement.
You are absolutely right. That was careless of me, and I do know better. I shall try to avoid quick posts in the future; and to give my responses more thorough consideration.