It's been said that almost all forms of energy on earth, be they fossil fuel, renewable biofuels, tidal, wind or solar are all really solar energy. I won't get into the details of solar power too much, but I want to get into the sun itself. Well not literally! I'll spend this installment on describing some aspects of the sun and astrophysics.
We are of the sun.
Everything we grow, everything we burn, everything we see is solar in nature. With the small exception of geothermal energy, everything else comes from solar energy, or stored solar energy in biochemical processes. But how do we get more efficient forms of solar energy into our energy system to take advantage of this almost free source of energy?
The sun gets most of its energy from converting atoms of hydrogen (really hydrogen nuclei) into helium. This process of nuclear fusion liberates a tiny amount of mass, the mass of the helium atom is very slightly less than the mass of 4 hydrogens. The nuclear force holding the helium nucleus together is smaller than that of the constituent hydrogen atoms, and that slight loss of mass (IIRC, its about 1 part in 400) generates a lot of random heat. Photons emitted in nuclear fusion today will take about a million years after bouncing around in the hot core of the sun to finally make their way to the surface, being absorbed and emitted trillions and trillions of times along the way. When they finally make their way to earth, the total amount striking the equator at high noon is about (doing this from memory...) 1366 Watts per square meter, or on average 342 watts per square meter for the entire surface, for the entire year.
The sun does it's little dance in the sky as the earth spins around. Our planet does not "circle" around the sun, rather it travels in an elliptical path. Because this ellipse isn't "round" there are times when the earth travels slightly faster, slightly slower and the day will also be slightly longer or shorter. It is because of this imbalance, that we have the Equation of Time to describe how the sun "moves" across the earth's tropics during the year. When I say "moves" I really mean the point at which the sun appears to be directly on top of, when there is no shadow cast at all.
Remind me to do a diary on Eratosthenes one of these days...
If you were to add up all of the time the sun shines every day, and find the median time of the sunshine (the time where you have the same amount before and after) that median time is 12:00 PM. And the median time when the sun is set (the "middle of the night") is you guessed it, midnight.
The earth, at least last time we've checked is a oblate spheroid. The circumference at the equator is slightly more than at the poles, all rotating bodies eventually get this "pot belly". But at the scale we're interested in, let's consider the earth to be "flat".
The angle any point on earth makes with the sun determines how much energy can be derived from solar power at that time. Now at night, that energy is essentially zero. At that rare time when there is no shadow, that power is 1366 W/m2. So how do you calculate how much light/power is available?
Take the cosine of that angle, and multiply it by 1366 watts per meter. Now calculating that angle gets tricky, when you have to deal not only with the Equation of Time, your latitude, and the time of day. You can balance the absolute maximum by subtracting the sun's current position in the Analemma (figure "8" of the Equation of Time) with your latitude to determine the sun's altitude. Let's use NY and Miami as an example- NY is roughly 41°N latitude, Miami is roughly 26°N.
For these three days, the first day of summer, the first day of autumn, and the first day of winter, NY will see at the height of the day:
Cos(41°-23.5°) = .954 * 1366 W/M2 = 1302 W/M2
Cos(41°-0°) = .755 * 1366 W/M2 = 1030 W/M2
Cos(41°+23.5°) = .431 * 1366 W/M2 = 588 W/M2
And for Miami
Cos(26°-23.5°) = .998 * 1366 W/M2 = 1364 W/M2
Cos(26°-0°) = .899 * 1366 W/M2 = 1228 W/M2
Cos(26°+23.5°) = .649 * 1366 W/M2 = 887 W/M2
Now you know why people who want to stay warm go south for the winter!
But wait, there's more.
The further you go away from the sun in the winter, the shorter the day is. Go far enough away (above 66.5°N) and you are in the Arctic Circle. You know, where the Eskimos live, and the home of the very long night. Go up far enough, and you not only have a few days without any sun at all, but you might never see the sun at all for a few months at a time.
Similarly the further up in latitude you go in the summer, the more hours of daylight you have. There are some days above the arctic circle where they never see night, and go up to the pole and you have six months of weak sunlight each year, followed by six months of darkness.
This is just a general rule of thumb, but in order to calculate the total power on a given meter of earth in a day you need to find out how much time that spot will have sunlight in the day (consult your sunrise and sunset tables) and the previous maximum energy calculations for that day. Divide the maximum by two and multiply that out by time. This "fudge" factor is based on the Cross product of the two angles that the sun makes with you, the altitude, and the right ascension (how much "east or west" it is. When the sun is at sunrise or sunset at the equator, it is still much weaker than at noon. That is the second component of the instantaneous power equation.
There is one other serious intangible, and that is cloud cover and atmospheric scattering! Since you can't predict or control either, I've taken the liberty of setting them both to zero. So for the purpose of this little thought experiment, there are no clouds, and no atmosphere.
Again using the NY and Miami locations for those three days, we get the following:
1302 W/M2 / 2 * 15 Hours = 9765 WattHours
1030 W/M2 / 2 * 12 Hours = 6180 WattHours
588 W/M2 / 2 * 9.2 Hours = 2705 WattHours
Again, this graphically displays how much more solar energy we receive in the summer than the winter.
Now for Miami
1364 W/M2 / *13.75Hours= 9377 WattHours
1228 W/M2 /2 *12 Hours = 7370 WattHours
887 W/M2 2 *10.5 Hours = 4655 WattHours
Hard to imagine, but NYC gets more photons in the dead of summer than Miami! Only problem is, that's just for a few days a year, and it is spread out over a longer period of time. And in the cold of the winter, NY gets less than 60% of the solar energy. Luckily we have the residual heat of the atmosphere, ground, and water to balance things out, or else things would be much worse.
Now, there is a way to CHEAT! I wish I knew the exact math, but if we were to take this square meter and put it on a mechanized table that could track the sun, it would no longer have as the "fudge factor" .5 (which is the sum of all cosines, as you have to multiply the first angle of altitude with the second of right ascension) but you'd be able to get much more bang out of the buck, nearly double the earlier calculations. More sophisticated solar power generators do this, but it isn't cheap.
Just to finish off the diary, the best we can do right now is less than 15% efficiency for commercial solar electric power. We might be able to go up a notch or two, but it's going to be hard. Using the sun to heat a working fluid such as water is slightly more efficient, but unless you get a really big difference in temperature even the most efficient engines will not give you much more efficiency than the electrical ones. I'm sure we will be able to boost this as well, in time.