# How to calculate solar energy produced by a space station?

I have designed the "blueprint" of an spaceport. I know that in Earth solar energy is 1.200W/m^2 approx., but my spaceport is in the space (no atmosphere).

Information
- My blueprint has 60 solar panel blocks. Each block has 100m^2 area, for 60 * 100m^2 = 6,000m^2 total area of solar panels.
- The spaceport is at 1 AU from the sun (149,597,870.7 km).
- The sun is the same as our sun.
- Spaceport always gets sunlight.
- Solar panels are high-tech level, 80% efficiency.

Question
How can I calculate energy output?

I would put a hard-science tag, but I don't think it is necessary. I only want a calculation formula.

• This question does not look like world building. As it is now it's "just" about calculating a solar plant yield... – L.Dutch - Reinstate Monica Jun 1 '17 at 13:46
• @L.Dutch In world building there are plenty of hard science calculations, so I think my question is fine – Ender Look Jun 1 '17 at 13:47
• Again, you are not looking for world building when googling "how to calculate the yield of a solar plant" provides you an answer photovoltaic-software.com/PV-solar-energy-calculation.php – L.Dutch - Reinstate Monica Jun 1 '17 at 13:49
• @L.Dutch this link not answer how to calculate solar output in space. – Ender Look Jun 1 '17 at 13:55
• "Each block have 100m^2, so 60 * 200m^2 = 1,200m^2 of solar panels." You might want to take another look at your maths... – F1Krazy Jun 1 '17 at 13:56

Question How can I calculate energy output?

Not too reliably.

You start from the energy actually incoming on the panels. This is the total solar irradiance and is the black body radiation given off by the Sun. It actually depends on several factors, so a star "identical" to the Sun might be off by up to 5% either direction (the Sun is itself a weakly variable star, by about 0.09-0.15% depending on the source).

For the Sun that value is around 1365 W/m^2.

Now, the problems are the following:

• your solar panels (any solar panel) will only capture a fraction of that energy, if we're talking photovoltaics, because the photovoltaic effect requires photon of a precise wavelength or, at the very least, some specific frequency bands. To improve the yield you can re-radiate some waste heat, mainly due to the longer wavelengths in the infrared, through some Seebeck converters. But that adds to the mass and complexity of the station.

• That energy will not be converted with 100% efficiency. You're probably looking at around 20-40% if using photovoltaics.

There is talk of "quantum traps" rectennas made up of microscopic crystal "hairs" (a previous version employed carbon nanotube 'hairs') pointed towards the Sun, organized in such a way that an incoming photon would always pass through a hair-gap "tuned" to steal its energy, no matter what frequency it was. But even the theoretical efficiency for this setup is not 100%, but around 90-95%, because it depends on the re-radiation temperature of the panels; to achieve 100% you would need to cool them to absolute zero, and that would take energy... .

• Conversion and storage efficiency. The panels will produce electricity at some optimum voltage, then you'll have to smooth, buffer and convert this energy to "pair" it to whatever uses it. Possibly you'll have to store it into batteries or capacitors and retrieve it later. All these operations will waste a percentage of the available energy.

• Wear and tear. The panels are exposed to hard radiation and charged particles from the sun, and this is hard on many materials. Glass will slowly become more opaque, semiconductor junctions will slowly "corrupt". The decay rate for the ISS Solar Array Wings is around 3% each year.

A reasonable, back-of-a-napkin calculation for energy yield would be around 400 W/m2. Having 6000 m2 available will then give you around 2.4 megawatt of power, and I think you can rely on two solid megawatts for most purposes for any reasonable time frame.

Of course, inconveniences like severe coronal mass ejections and charged particle onslaught, high-speed meteoric dust or even, Lord forbid, Kessler syndrome could easily reduce available power either gradually or even catastrophically.

A way to increase the yield and durability of the setup could be to set up the panels facing away from the Sun, and use a micrometer-thick Fresnel mirror to concentrate the radiation on the panels. The mirror would protect the panels, and it would be easier to replace; possibly it could even be self-healing (a corroded area of the mirror could be re-melted to erase pocks and scratches). Of course, the mirror performances in a vacuum then might become a problem.

To calculate the irradiance you need to know the star's luminosity. Since energy is conserved, the total flux of it is constant through any ideal sphere with the Sun inside, and the flux through a given area of such a sphere only depends on the sphere's surface; the constant flux is "diluted" on a larger and larger surface getting farther and farther out from the star.

The surface of the sphere is an easy thing - 4*pi*r2. So you divide the luminosity flux by this surface, multiply again for the surface of your panels, and get the fraction of energy that they intercept. Multiply again by the conversion efficiency and you get the power output.

The difficult thing is calculating the star's luminosity. Usually this is done by approximating the Sun with an object known as a perfect radiator or black body (which stars, while close enough, really aren't).

There is an equation known as Stefan-Boltzmann's equation that gives the total unit radiative power of a black body, which only depends on its temperature:

$W = k T{^4}$

with $k = {5.67} \cdot {10^{-8}} W m^{-2} K^{-4}$ and T, the temperature, is expressed in Kelvin.

Supposing this is constantly radiated by the Sun's whole surface, the equation giving irradiance of a star of radius R and temperature K at a distance D is

$w = {5.67} \cdot 10^{-8} \cdot {T^4}({\frac{R}{D}})^2$

(we get rid of the PI's)

Solving for the Sun, where T = 5778, R = 695700 and D = 149600000 (1AU), gives us the value of 1366.7 W/m2.

This is the mathematical answer for the Universe where cows have spherical symmetry and uniform density. In the real world, things are not so simple.

Plugging in the surface area of the panels A (in meters squared) and the efficiency factor (from 0 to 1), you would have a power output P, when the panels are new, of

$P = {5.67} \cdot 10^{-8} \cdot {T^4}({\frac{R}{D}})^2 \cdot A \cdot \epsilon$

• Great info with "Wear and tear" and the Fresnel mirror. If you could explain how to calculate the 1365 W/m^2 (my real question) I'll accept it grateful. – Ender Look Jun 1 '17 at 23:31
• It is not calculated - you measure it using bolometric methods. Several attempts have been made using satellites with special sensors; the Wikipedia link on irradiance has the details. – LSerni Jun 1 '17 at 23:41
• b.Lorez answer have used a very interesting formula: P = (0/(4.π.r^2))*A.π.e. If you insert that in your answer it would be complete for me. – Ender Look Jun 1 '17 at 23:46
• That equation depends on the luminosity L0, and you would still have the problem of knowing that. I've updated the answer with Stefan-Boltzmann's equation which gives an approximate answer for any reasonable distance and temperature, under some assumptions about the star (won't work very well for variable stars except at maxima and minima). BTW, I suspect that the equation you quoted has an extra PI before the efficiency factor. – LSerni Jun 3 '17 at 9:58
• k is the Stefan–Boltzmann constant?. Also, great information! Excelent. – Ender Look Jun 3 '17 at 16:01

The general answer is the following:

Take the full power output of the Sun (in watts). Divide it by $4 \pi$ and divide it by the distance from the sun squared (in metres). They multiply it by the effective area of your solar panels, (if they block each other or are not directly facing the sun, they produce less.), and multiply with the efficiency. You get the power output in watts. Of course, your panels can't convert all the wavelengths the Sun emits, so you might would like to use the luminosity of the Sun only in visible light.

So: $p = \frac{L_0}{4 \pi r^2} A \pi e$ where $L_0$ is solar luminosity, $A$ is the area of the panels, and $e$ is the efficiency of the panels.

But of course, if you only need it at 1 AU, the solar contant there is 1368 W/m2.

Well first off your number for solar irradiance or insolation of 1200 W/m^2 is for the surface, as you noted; in space it would be higher around 1361 W/m^2 without the atmosphere blocking some energy.

Then this is fairly straightforward, to get the power output you just multiply the total area by the solar input times the efficiency.

There were some math errors in you question 60 blocks at 100m^2 would be 6000 m^2 or if they are double sided as you seem to assume (which doesn't make much sense for a solar collector because one side would always be in the shade not producing energy) 12,000 m^2. For my maths I assumed single sided area of 6,000 m^2.

6000 m^2 * 1361 W/m^2 * 0.8 = 6,532,800 Watts or 6.5 MW

To find energy you just multiply by how long it is operating, so assuming it gets constant sunlight, again fairly straightforward depending on what units you want. Super easy in the metric system where 1 watt = 1 Joule / second.

24 hours = 86400 seconds

1.15 MW * 86400 s = 561 GigaJoules per day

or for a more common energy unit the kWh that would be ~156,000 kWh per day.

Note: I find that Google is very helpful for converting units, just type in the starting number then "to" whatever unit you want and it will bring up their unit converter, i.e. search for "561 Gigajoules to kWh" and it will do the conversion and bring up the tool which has dropdowns for other units.

• Isn't 80% efficiency absurdly high? – M i ech Jun 1 '17 at 14:07
• @Miech Current research cell efficiency maxes out at 46%. The Asker specified the 80% number which is far from the most egregious case of handwaving on this site. – sphennings Jun 1 '17 at 14:09
• Oh, right. 80% was OP's idea. Never mind then. – M i ech Jun 1 '17 at 14:12
• @Miech efficiency is defined by spectrum and energy gap in semiconductor and resistivity. Multilayer panels with different gaps can have high efficiency. one of the problems is to vary those gaps. Just as fun fact. – MolbOrg Jun 1 '17 at 14:20
• @Miech, yes, in my setting humans have developed high-tech solar panel, 80%. – Ender Look Jun 1 '17 at 14:35

This shouldn't be very complicated. The atmosphere reduces useful solar flux by ~20%. So correct for that. And make a better estimate about PV efficiency. 40% would be more realistic.

So if you start with 1200W/m^2 on the earth's surface, multiply by 1.25 to reclaim the 20% and you get 1500, and multiply by 0.4 for your PV efficiency and you get 600W/m^2.

If it's constantly exposed, you should see that consistently.