SolarGraphene

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

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?

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. https://www.youtube.com/watch?v=02v-XJf1pPg https://www.youtube.com/watch?v=LONKzXSfGjE 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?

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.

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?