# Solar experiment that gets 7KW per square meter of sunlight??

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I have been doing experiments with solar thermal technology trying to get the most efficient system possible. I keep on getting crazy figures and I’m not sure if its my calculations, an oversight or if I am correct. I did the experiment with the superconductor Graphene, so it should be possible, but i'm not sure.

So the experiment is a simple setup: (In video)

I use a Fresnel lens that is 83 x 57 cm2, so the amount of sunlight is known.

I heat the superconductor Graphene with the lens, the graphene in in a Pyrex beaker that has 800ml of water in it being mixed by a magnetic mixer, so the temperature readings are accurate and give an overall temperature rise over time.

The thermometer is in the back of the beaker about half way down and the LCD display is glued to the front of the magnetic mixer, so you can see the temperature rise.

Have a lux meter to give the lux reading just for clarity.

Video:

I calculated the temperature rise from 49-50 degrees C, 50-51C and 51-52C

Formula used:

Watts = 3.1 x Gallons x ΔT (in °F) / Heat-Up Time (in hrs)

15 seconds

0.8 litres water used is = 0.175975 imperial gallons

1 degree C temperature rise in °F = 33.8°F

In video temperature rise calculations:

49 to 50 degrees = 13 seconds = 0.00361111 hours

Watts = 3.1 x 0.175975 x 33.8 / 0.00361111

= 5103 watt (5.1KW)

50-51 degrees = 21 seconds = 0.00583333 hours

Watts = 3.1 x 0.175975 x 33.8 / 0.00583333

= 3161 watts (3.16KW)

51-52 degrees = 15 seconds = 0.00416667 hours

Watts= 3.1 x 0.175975 x 33.8 / 0.00416667

= 4424wats (4.42KW)

My question is am I correct that 1 square meter of sunlight has 7KW+ in energy or did I make a mistake in my calculations?

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"1 degree C temperature rise in °F = 33.8°F"

No it isn't

1 degree C change is equivalent to a 1.8 degree F change.

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The word is insolation and Wikipedia suggests the annual max average insolation at the equator is 0.25kW/m2, making a nonsense of your claim ofnearly 30 times that.

http://en.wikipedia.org/wiki/Insolation

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Studiot, I suspect that SolarGraphene did his experiment in day time.

In that case the figure you cite (0.25KW/m^2) is the wrong value.

The actual value depends on the season and time.

It might be as high as about 1Kw/m^2

As I explained his value will be wrong by a factor of 18.8 or so.

Correcting for that gives about 0.4KW/m^2

Without knowing the time, etc when he did the experiment, you can't rule out the idea that his measured value (0.4 Kw/m^2) is better than your cited value of 0.25 Kw/m^2.

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I was done in the UK the 19th July 2014 at 12:20. Lux reading is 1035 set to times 100x setting, 103 500 lux as I understand. There is more videos on the day, got the same results.

I can see the C to F conversion was wrong.

New calculations:

49 to 50 degrees = 13 seconds = 0.00361111 hours

Watts = 3.1 x 0.175975 x 1.8 / 0.00361111

= 270 watt (0.2KW)

50-51 degrees = 21 seconds = 0.00583333 hours

Watts = 3.1 x 0.175975 x 1.8 / 0.00583333

= 166 watts (0.16KW)

51-52 degrees = 15 seconds = 0.00416667 hours

Watts= 3.1 x 0.175975 x 1.8 / 0.00416667

= 232 watts (0.23KW)

How efficient would this system then be at converting the energy, available on that day, into usable energy? Guessing it’s about 60% for the high number and 40% for the low one, but what is the exact figures.

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http://en.wikipedia.org/wiki/Lux

suggests that 100,000 lux is pretty much as bright as sunlight gets and the page Studiot cited earlier says that's about 1KW/m^2.

Without a lot of complicated measurements and calculations (like how much infrared gets through the lens) you are not going to get a more accurate figure

Full sunlight would deliver something like 470 Watts to your system.

You should also take account of how much heat is delivered to the system by the stirrer and how much is lost or gained from the surroundings.

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To heat 1 gram of water for 1 C, you need to use approximately 4.1855 J energy (XIX century calorie unit)

So for 0.8 L it's ~800 grams (better measure mass, it'll be more precise than measuring volume) = 800 * 4.1855 J = 3348.4 J needed for 1 C increase.

If it takes your device 13 seconds to receive such increase in temperature of water, divide it by time:

3348.4 J / 13 s = 257 W (= 257 J/s)

My question is am I correct that 1 square meter of sunlight has 7KW+ in energy or did I make a mistake in my calculations?

No. As you can read on

http://en.wikipedia.org/wiki/Sunlight

it's max 1050 W/m^2

at Zenith.

Your Fresnel lens have area 0.4731 m^2

0.4731 m^2 * 1050 W/m^2 = 496 W

Receiving 257 W from max 496 W possible is not bad result IMHO for home made experiment.

(and we don't know at what hour you did experiment).

Fix video subtitles..

If I were you I would surround water container by mirrors from all sides (buy little aquarium and cut mirrors perfectly to match its borders), and leave just little hole on top for incoming light from Fresnel lens.

You might also consider using vacuum flask, or surround it by Styrofoam.

Edited by Sensei
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I understand how I can boost the output, but I deliberately didn't want to make it look fake or complicated, in the video. there should be no doubt that it is real.

Here is two other videos made on the same day. Same thing, but a little different to show other angles, etc.

Possible boosters:

mirrors to reflect back sunlight reflecting off glass beaker / solar graphene < 3% increase

Insulate beaker <2%

Optimal focal point <20%

Highest lux reading and light intensity <15% (have to be in a different country closer to equator)

Angle of the beaker to the lens to minimize reflection <5%

More layers of solar graphene as this version is porous and lets about 10-20% of light through.

I calculate 51.8% efficiency in the 13 second, 1 degree temperature rise. I would expect to be able to boost it with ideal conditions to 60-70%. I think that currently the most efficient solar systems only manage 43%. Unless someknows of ones that can do more?

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I did the experiment with the superconductor Graphene [...]

Did you?

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I would expect to be able to boost it with ideal conditions to 60-70%. I think that currently the most efficient solar systems only manage 43%. Unless someknows of ones that can do more?

But solar panels generate electricity. Your setup is just heating water.

For now you can use it to make a cup of tea or coffee in 6 minutes or so.

Edited by Sensei
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Yes, its solar thermal.

I have built a version that makes electricity (small scale) from the hot water. So you get electricity + hot water at the same time. The 51% if for the hot water only, and I don't know how to work out or add the electricity part.

Another experiment, and have a question about it:

There is a 75 watt light bulb hanging over 400 ml of water that has solar graphene inside. The light shines on the graphene in the water, raising the temperature, so you can do the calculations. The calculated efficiency comes in around 44%, but that is using all 75 watt from the bulb. Not all the light is falling on the graphene.

The question is: How much of the 75 watt light actually hit the foam? So that the real figure can be calculated. I’m guessing < 50 watt that would make it > 67% efficient. Is there a way of getting a better idea, over guessing?

Edited by SolarGraphene
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I did the experiment with the superconductor Graphene,

I'm willing to bet that you didn't.

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Don't get your insinuation. That statement is a fact.

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As far as I am aware there is no known material which is a superconductor at (or anywhere near) 50C.

So your statement is not a fact at all, unless you are about to announce a Nobel Prize winning discovery.

Incidentally, even if it were superconductive, it wouldn't make any real difference to the experiment. You could just as well use soot.

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• 3 weeks later...

And if the liquid is black? This black liquid will heat the water.

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