Solar concentration required for 700 deg f. ?

[cujet]

OK, I want to achieve 700 degree temperatures in linear fashion. In other words, I want to heat oil in a tube.

What solar concentration do I need? I have a Microsharp solar Fresnel Lens with a 10 to 1 (or so) concentration. I can't get more than about 120 degrees F. However, the plastic lens is horribly inefficient in my setup. It's certainly not able to achieve the temperatures of a good magnifying lens.

So, I was wondering what exact number is required. Would it be 60 to 1?

[rbodell]

OK, I want to achieve 700 degree temperatures in linear fashion. In other words, I want to heat oil in a tube.

What solar concentration do I need? I have a Microsharp solar Fresnel Lens with a 10 to 1 (or so) concentration. I can't get more than about 120 degrees F. However, the plastic lens is horribly inefficient in my setup. It's certainly not able to achieve the temperatures of a good magnifying lens.

So, I was wondering what exact number is required. Would it be 60 to 1?

I think multiple mirrors would probably be your best bet, but that would still be a real hassle with aiming them and you would probably have to have a tracker to follow the sun too. I built one once with something like 1200 or 1300 mirrors that would turn a soda can to molten aluminum in about 30 to 45 seconds, but the aiming all those mirrors was nearly impossible. I have often thought about trying a big satellite dish like the first ones that came out, that were like 8 or 10 feet across and lining that with mirrors. It wouldn't focus as close, but you could get a lot more mirrors on it.

[mike90045]

You are looking at a "trough" collector.
either E-W non tracking, or N-S with active tracking.

what sort of oil do you belive will last @ 700F ?

You will HAVE to have a vacuum collector tube, double wall glass, and a dark tint to your oil.



Check this "hot plate" cooker, 1 tube  http://solarcooking.wikia.com/wiki/Evacuated_Tube_Indirect_Solar_Cooker


http://greenterrafirma.com/evacuated_tube_collector.html

[Curbie]

cujet,

You're looking at three parameters here:
1)   The area of the collector dictates the volume of heat potential.
2)   The area of the receiver (relative to the area of the collector) dictates the concentration ratio and maximum temperature potential.
3)   The temperature of fluid is dictated by receiver temperature potential, the rate of fluid flow, and thermal dynamics of the receiver.

So you have to work the math backwards, starting with how much temperature AND volume your fluid will need, then use that to calculate the areas of both a collector and receiver necessary to produce it.

You can use the math here:
Link to an engineering student's did his thesis on a 10' satellite dish he converted to a solar concentrator for a steam turbine, the idea fell-short, but he has good math on solar concentration:
http://www.redrok.com/NewtonSolarSteamManuscript.pdf

I also agree with mike90045 about the oil at those temperatures, you may need to look at a phase change material like aluminum or draw-salt and a solar furnace like a Scheffler Reflector:
http://www.solare-bruecke.org/

Good luck,

Curbie

[cujet]

Well,

Let's look at the basics first. Assume I simply want to heat a small quantity of oil to 700F. No flow rate, just heat. (yes, there are oils used for this). Also, evacuated glass tube is a simple matter. I can use Neon glass and easily evacuate it. Tracking is not an issue. That's a simple matter in my case. I may use my telescope clock drive for the experiment.

What level of concentration is required? There has got to be some level of experience on this. I could do this with one or more optics. That's not an issue.

Anybody want to take a guess and give me a place to start?

[mobile_bob]

cujet:

you don't mention two important factors

1. how much oil do you want to heat, and

2. how long do you want to wait for it to get to 700 degree's F

as an example

a 4" magnifier will certainly heat a drop of oil within 30 seconds to well over 700 degree's F, i would think

now obviously it would take a much larger lens to heat a gallon of the same oil in the same amount of time.

so it is hard to say how big or what you might need to do the job, without understanding exactly what the job is
and how long you have to do it in.

bob g

[mike90045]

You need a double wall tube, very inner has the dark oil,  and is surrounded by vacuum, so the air does not cool the tube.  How much depends on other losses in the system, the less losses, the smaller the collector can be.

[Curbie]

cujet:

The "flow" I was referring to was heat flow not oil flow and it comes in two forms, non-useful (losses) and useful (work). In my view, flow is the starting not ending point and a system needs to be designed to run at equilibrium a point at which the heat collected is equal (or greater) to the heat flow (work plus losses).

Once you determine flow (which is the hard part) it's pretty easy to determine collector area and concentration ratio. This notion seems to gum up a lot of people playing with solar, but it's really no different than any other type of fuel; as you use it, the fuel needs to be replenished.

Good luck,

Curbie

[Ronmar]

Well put curbie. 

As you increase the temps, the losses also increase till you reach a point where you cannot gain any temp without increasing your energy input.  For temps that high, you are going to need some carefully thought out insulation to keep the losses manageable.  That is where the vacume outer tube comes in.  Evacuating the outer envelope cuts your conduction and convection losses out of the picture and you are only left with re-radiation losses.  Without removing the first two, getting to 700 is going to require a lot of energy(a lot more than it has to)...

[cujet]

OK, so the quantity is very small. Say less than one ounce. Quite simply, a quantity large enough to place a termocouple in. With the thermocouple full submerged. In fact, let's make it 1cc for the sake of this discussion.

As I mentioned, an evacuated tube is easy. It could be simple neon tube or something larger. I don't care.

Also, as I mentioned, I used a Microsharp solar concentrator and the temps achievable were quite limited. No where near my needs.

I don't care if it takes 10 minutes or more to heat up. Not a problem. However, it has to be a reasonable length of time for obvious reasons.

[Curbie]

cujet,

According to the engineering student's thesis I posted the link to, temperature at the receiver depends on the following values which in turn are themselves dependent on yet other values.

Aperture Area of Receiver
Aperture Area of Collector
Concentration Ratio (geometric)
Concentration Ratio (ideal)
Absorbance of Receiver for Solar Radiation (a)
Emittance of the Receiver in Infrared Region (e)

Once receiver temperature is calculated, thermal dynamics come into play to determine heat flow into and out of whatever the receiver is heating. Sometimes questions that are simple to ask are very difficult to answer, I have no way to answer your question as currently posed.

In my view, if you're intent on a answer, short of hiring an engineer you have two ways to go; a mathematical/theoretical route OR an experimental route, maybe a combination of the two. I went the mathematical/theoretical route and put the entire engineering student's thesis in a spread-sheet, but even with sound math (remember, he had only partial success) I would expect the need for some experimentation.

Good luck,

Curbie

[mike90045]

I don't have a thermocouple, but I have a glass fresnel lens from an overhead projector, about 14"x14", that is really dangerous, it will light anything flammable, and spall brick & concrete in 10 seconds.  A drop of oil, it would boil, I'm guessing.  I wear welding goggles when using it.  I first thought you wanted to use moving oil to accumulate heat.