Equating solar power to usable heat

  • Thread starter ThoughtRay
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  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #2
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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.
 
  • #3
cjl
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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.
 
  • #4
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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.
 
  • #5
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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.
 
  • #6
cjl
Science Advisor
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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.
 
  • #7
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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.
 
  • #8
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This sounds like a fascinating project, and I can certainly help you with the solar part of the question. I will probably make some mistakes, so let me know if anything is unclear. This will be quite a simplistic, back of the envelope type calculation, but hopefully it helps to clarify some things.

For now, let's just assume you know the flux (power per unit area), q_out [W/m2], due to conduction, convection, phase change, chemical reaction, etc. (i.e. anything that is not radiation), required to maintain the flow of sand at the specified temperature of Ts ~ 1700 K. This could be calculated a priori but would require knowledge of how the apparatus is insulated, etc. Once you know this, it just comes down to doing an energy balance on the surface. For radiation heat transfer, you will need to consider both the flux absorbed by the surface, and the flux emitted by the surface. The absorbed flux is equal to: alpha*q_solar, where alpha is the solar absorptance (how much of the incident sunlight is absorbed) and q_solar is the incident solar flux. The emitted flux is equal to: epsilon*sigma*Ts^4, where epsilon is the total emissivity (how much energy the surface emits relative to a perfect emitter (blackbody) at Ts); sigma is the Stephen-Boltzmann constant = 5.67 e-8 W/m2-K; and Ts is the surface temperature of the sand. Both alpha and epsilon are always between 0 and 1.

We can now do an energy balance on the surface:

q_in = q_out

alpha*q_solar = q_out + epsilon*sigma*Ts^4

This can be easily solved for q_solar:

q_solar = (q_out + epsilon*sigma*Ts^4)/alpha

The most difficult part is to evaluate the q_out term. For a back of the envelope calculation, I would do the following: the radiative loss term, epsilon*sigma*Ts^4, is proportional to T^4, whereas the q_out term is *probably* proportional to T. Therefore at very high temperatures, and 1700 K is indeed high, the radiative term dominates. A *probably* conservative approach is to assume q_out = epsilon*sigma*Ts^4. Then we get:

q_solar = 2*epsilon*sigma*Ts^4/alpha

Let's plug in some numbers. I am just guessing here, but I can look up values for alpha and epsilon for sand tomorrow at my office. Both epsilon and alpha must be between 0 and 1, so let's take 0.5 for both. We get:

q_solar = 2*0.5*5.67e-8*(1700)^4/0.5 ~ 1 000 000 W/m2

Although this number seems high, it is achievable depending on the type of concentrator you use. Usually in the solar concentration field, we express things in terms of "suns" where 1 sun = 1000 W/m2, which is about the maximum direct normal solar irradiance that you can get on Earth. When expressed in terms of suns, we usually call this the "solar concentration ratio", or simply "concentration". Then:

Concentration = q_solar = 1000 suns

Therefore we need to concentrate the sunlight, at least 1000 times. Of course, depending on where you live, you may expect values closer to 500 - 800 W/m2 for the direct normal solar irradiance on a clear day. Then to be safe, you need a concentration of around 2000 suns.

In order to get a concentration this high, you will need a point-focus two-axis tracking system. I don't think you'll be able to get a Fresnel lens system to achieve these kinds of concentration, so you may want to consider parabolic dish shaped mirrored reflectors. If you are still interested, I can provide some more info on potential concentrators.
 
  • #9
Thank you all for your responses. I have tried responding back but each time it has disallowed my posting.

I will study and try to work through your notes, but am not very adept with math. By the way, I am located in WA State at Lat: 47° 19' 45" N.

If choosing to, I can be contacted more directly at this e-mail: husom-teo@india.com.

Again thanks, (Thought)Ray
 

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