Solar Thermal Optics - Concentrated Light

AI Thread Summary
The discussion focuses on calculating the heating energy potential of solar thermal systems using parabolic mirrors. The typical solar energy received is about 1000W/m2, translating to 0.1W/cm2, but this value varies based on location and conditions. To estimate the energy increase from concentrated sunlight, one should multiply the area of the reflector facing the sun by its efficiency and the target area receiving sunlight. The efficiency of the mirror and the absorption characteristics of the target material also play crucial roles in the calculations. Understanding these factors is essential for optimizing the design and performance of solar thermal devices.
Steven B
Hi,
Can someone start me out with creating predictable temperatures and inputs for calculations.
Here is a miniature example of the type of parabolic circulated water heating devices I have been building.



My first questions would be what is the potential heating energy of the sun per cm2?
And can I simply multiply that, by the cm2 area of the heating target (copper tube) receiving sunlight, and then by the surface area of the parabolic mirror, to calculate the increase in energy due to concentrated sunlight?
I definitely have additional questions to follow!
Thanks in advance,
Steven
 
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Steven B said:
My first questions would be what is the potential heating energy of the sun per cm2?
The usual figure quoted is 1000W/m2 but it depends on where you live, time of day, time of year, the weather so will probably be lower.

1000W/m2 = 0.1W/cm2

Steven B said:
And can I simply multiply that, by the cm2 area of the heating target (copper tube) receiving sunlight, and then by the surface area of the parabolic mirror, to calculate the increase in energy due to concentrated sunlight?

First multiply by the area of the reflector facing the sun. That's not the same as the curved surface area of the reflector. It also changes if the reflector isn't pointed correctly at the sun.

Then multiply by a number between 0 and 1 to account for the efficiency of the mirror (the mirror absorbs some energy). I think a reasonable figure for a good mirror might be 90% or 0.9.

If the reflector is well designed (right curve) then all of that energy arrives at the target. If not well designed then some will miss the target. Assume well designed or pick your own figure?

Not all of the energy that arrives at the target will be absorbed by it. So again multiply by another number between 0 and 1. This depends on how reflective it is. Something like a matt black anodised pipe would be quite good (perhaps assume x0.9). A polished chrome plated pipe much lower (0.1?).
 
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PS If you know the flow rate for the water going through the pipe and can measure the temperature going in and coming out then you can calculate the power captured.
 
I'm going to re-read your second part to absorb it (no pun intended) better.
But I thought 1kw per square meter was just the potential photovoltaic electrical energy. Interesting.
The reflective surface is mirror grade stainless. The target pipe is 1/2 copper with black stove paint coating. I'll have to look up the gps of the pump.
I do know that the amount of heat gathered by a parabolic is related to it's diameter, " ... Not it's surface length. Photons do not compress, but they can be culmative on a focal point... Yes?
 
Steven B said:
I'm going to re-read your second part to absorb it (no pun intended) better.
But I thought 1kw per square meter was just the potential photovoltaic electrical energy. Interesting.

It's the amount of energy from the sun that arrives at the Earth's surface. See..

http://www.reuk.co.uk/wordpress/solar/solar-insolation/

I think most PV panels are only about 20% efficient so at best produce around 200W per sqm.
 
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