Toddsterpatriot
Diamond Member
If that were true you wouldn't still be chasing me around.
Pointing and laughing.
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If that were true you wouldn't still be chasing me around.
still proving my point.Pointing and laughing.
still proving my point.
You sure are worked up.Only if your point was your faulty understanding of physics.
What percentage of solar radiation striking solar panels is in the visible light spectrum.
The answer is 40 to 44%.
What percentage of visible light striking solar panels is converted into electricity?
The answer is 15 to 20%.
So per square meter solar panels effectively reduce the 240 W/m² striking it by:
240 W/m² x 0.44 x 0.2 = 21.12 W/m²
What is the total land surface of the planet?
The answer is 149 million square kilometers or 1.49x10^14 m²
What is the solar radiation striking the total land surface of the planet?
The answer is 240 W/m² x 1.49x10^14 m² = 3.5760x10^16 W
What is the reduction needed to make the planet net cooling?
The answer is 0.7 W/m² x 1.49x10^14 m² = 1.0430x10^14 W
How many square meters of solar panels are needed to change the planet from net warming to net cooling?
The answer is 1.0430x10^14 W / 21.12 W/m² = 4.9384x10^12 m²
What percentage of the total land surface of the planet would need to be covered by solar panels to change the planet from net warming to net cooling?
The answer is 4.9384x10^12 m² / 1.49x10^14 m²= 0.0331 or 3.31% of the total land surface of the planet.
I sure did wind you up.I just bought a new LG refrigerator.
The freezer section is 21 in. X 30 in.
Every time a satellite goes over Chicago it measures my freezer at 40C cooler than my home.
How many of those freezers would it take to change the planet from net warming to net cooling?
ReinyDays
Nothing in these calculations about waste heat or fossil fuels.
What is the reduction needed to make the planet net cooling?
The answer is 0.7 W/m² x 1.49x10^14 m² = 1.0430x10^14 W
How many square meters of solar panels are needed to change the planet from net warming to net cooling?
The answer is 1.0430x10^14 W / 21.12 W/m² = 4.9384x10^12 m²
OMG!
You think solar panels somehow reducing the warming caused by solar radiation.
That's hilarious!
Moving that energy from a solar panel to my coffee maker can cool the planet.
DURR
The answer is 1.0430x10^14 W / 21.12 W/m² = 4.9384x10^12 m²
If you absorb 95% of the solar radiation hitting 4.9384x10^12 m², instead of absorbing 70%,
wouldn't you be warming the planet?
Why do you think there's exhaust gases from solar panels?
CH4 + 2O2 --> CO2 + 2H2O ...
∆E = - 55.7 kJ per gram of methane ... on average only half that energy is converted to electricity ... the other half is wasted on the heat in the exhaust products ... especially keeping water in her vapor state ... so my question is what exactly are solar panels emitting that wastes heat energy ...
Your calculations assume solar panels work at peak efficiency at night? ... where are you getting your numbers from? ... or are you just following ding-a-ling's incorrect logic? ... radiative forcing is generally considered to be 1.8 W/m^2 and Earth's surface area is closer to 5.1 X 10^14 m^2 ... but that's meaningless because solar panel convert SOLAR energy into electricity, not terrestrial longwave radiation ... your vectors are wrong ... and yes, power is a vector value, direction matters ..
Using solar power vectors ... we find that solar panels are net warmers in all circumstances ... and the last sentence in the above quoted post is SPOT-ON CORRECT: "If you absorb 95% of the solar radiation hitting 4.9384x10^12 m², instead of absorbing 70%, wouldn't you be warming the planet?" ... except you forgot to divide by two root two, irradiation follows a half-wave cosine function, so we use the appropriate RMS formula for locations along the equator, otherwise we still have to multiply by the cosine of latitude ...
Solar panels reduce the solar radiation striking the planet!!!!!!
LOL!