Solar: Jean Luc Perrier, solar hydrogen production

[chatelot16]

it's not the instantaneous power but the average power over a day ... which explains the difference in 1000W / m2 in maximum instantaneous power and 342W / m2 in average power over a day

the ratio of 1000 to 342 is easy to understand ... we divide by 2 because of the night ... and we reduce a little more because in the morning and in the evening the sun is low and must cross a greater thickness of 'atmosphere

the figure in average power per day makes it possible to calculate the production of a system having a constant yield whatever the power received

the maximum power figure is useful for scaling so that the maximum power is used without limitation ... it would be a big mistake to limit the power to 342W / m2 by insufficient hardware ... to reach the 342W / m2 means the 1000W / m2 must pass lossless

[izentrop]

If I follow you it is h24 and 365 days so?

Yes!
The CNRS definition

The solar energy that reaches the top of the atmosphere per apparent square meter (ie on a surface perpendicular to the solar rays) is 1367W / m2, which, averaged over the year and on the whole of Earth, represents 342 W / m2. http://www.cnrs.fr/cw/dossiers/dosclim1 ... mentS.html

That of wikipedia

The total power received by the earth is calculated simply by making the product of the average power irradiated by the calculated total surface area:
P = 1 361 * 1,274 1014 watts = 1,734 * 1017 watts = 173 petawatts (1015W).

That is an average power of 340 W / m² on the ground (1361 W / m² / 4, see the surface of a sphere) https://fr.wikipedia.org/wiki/Irradiation_solaire

[Remundo]

Here is a finer analysis according to the geography of solar power average received per m²:

If I follow you it is h24 and 365 days so?

yes, this is the "full" average,

even in Africa, instantaneous flows are very variable, from 0 at night, to more than 1000 W / m² at the zenith.

[A.D. 44]

Hello,

I do not think you understand what the solar constant is

1000w per m² is 1368 w / m² out of atmosphere and before absorption by this one

rest in the best case 340w / m² whether you like it or not ...

awesome, amazing !!!

my solar cells pass very regularly 50% yield!

As for my thermal insulators they they patatent to more than 200%, it is simply fabulous !!!

[izentrop]

Hello,

I do not think you understand what the solar constant is
1000w per m² is 1368 w / m² out of atmosphere and before absorption by this one
rest in the best case 340w / m² whether you like it or not ...

awesome, amazing !!!
my solar cells pass very regularly 50% yield!
As for my thermal insulators they they patatent to more than 200%, it is simply fabulous !!!

The yield is always relative to a reference. 340 W / m² is the average power on earth divided by the number of m² ensolleiled.

What is the reference of your returns?

[A.D. 44]

Hello,

I do not think you understand what the solar constant is
1000w per m² is 1368 w / m² out of atmosphere and before absorption by this one
rest in the best case 340w / m² whether you like it or not ...

awesome, amazing !!!
my solar cells pass very regularly 50% yield!
As for my thermal insulators they they patatent to more than 200%, it is simply fabulous !!!

The yield is always relative to a reference. 340 W / m² is the average power on earth divided by the number of m² ensolleiled.

What is the reference of your returns?

but I do not know

sorry if I was taken seriously

of course, that all my various sensors do not display such efficiencies ... they correspond, of course, to the classical performances: about 17% for PV and 80 for thermal.

I was just reacting to the so-called 340W / m² max advertised on our floor. What saw the power delivered to m² by my sensors, would have given me crazy yields ...

Uh ... I do not know if I was very clear there for once ...

in short ... it was a joke.

[chatelot16]

if we want to calculate the efficiency of a solar collector we have to measure either the instantaneous yield with the instantaneous solar power and the power produced at the moment of the measurement ... and when the sun is a is maximum it is closer to 1000W / m2 only from 340W / m2

you can also choose to measure the average yield over a day, and you have to compare the average power produced by the sensor over a whole day, with the average power received from the sun over a day ... which is the type of 340W / m2

performance measurement with average power is complicated because you have to measure over a whole day, with a sun that is only in its head and every day is different

Measuring instantaneous power is much easier because it only takes a few minutes of full sun

[izentrop]

of course, that all my various sensors do not display such efficiencies ... they correspond, of course, to the classical performances: about 17% for PV and 80 for thermal.

It's also a joke.

The yield is always relative to a reference. 340 W / m² is the average power on earth divided by the number of m² ensolleiled.

The best is to make a calibrated measurement of instantaneous or cumulative illumination, as do the researchers and compare it to the electrical power actually supplied to the installation or the cumulative energy / cumulative illumination.

[Christophe]

This sacred JULES VERNE was right again!

Bluntly!

[Christophe]

Jean-Pierre Barrault, who died in 2014, was a mechanic, an inventor. With his friend Jean-Luc Perrier, they dreamed of developing hydrogen fuel. They proved that it worked, and the car still exists, exposed in the family garage, at the gates of Poitiers. Tribute to an inventor and a humanist.