The Maury Equation, redux March 17, 2012Posted by Maury Markowitz in solar power satellites.
Tags: bolognium, solar power satellites
It seems that every time a space-based solar power supporter reads my post on The Maury Equation, they pick one point of the many and try to attack the entire argument on that. That point normally has to do with launch costs, which isn’t the real argument at all.
So, let’s make this simple:
A solar cell in space will deliver less energy to the grid than the same cell on Earth.
The amount of energy delivered to the grid is easily estimated. It looks like this: E = R x I x L x T Where:
- E is the total energy delivered over the lifetime of the system
- R is the “nameplate rating” of the cell (or panel), the amount of power it produces under optimum conditions
- I is the “insolation”, the amount of sunlight hitting the cell in a year
- L is the system lifetime, in years
- T is the loss in transmission between the supply and demand
R is simply a measure of the system’s size. In order to avoid apples-to-oranges comparisons, we want our systems to be the same size when comparing. Generally we use standard size systems, 1 kW, so this number is 1 in both equations and you can ignore it.
I, for insolation, can be measured in a number of ways. The simplest, and the one I use here, is in “hours of bright direct sunlight per year”. In space that is simply 365 x 24, but down here on Earth we have night and weather to contend with. There’s no way to guess it, the only way to get an accurate number is us an insolation calculator.
That equation is normally used for ground-based applications, but we’re talking about space. So I’m going to change it very slightly to make this part of the argument clearer. Instead of a single term for T, I’m going to use two (and R has been eliminated): E = I x L x Tg x Ts Where:
- Tg is the loss in transmission between the supply and demand (g for ground)
- Ts is the loss between the two antennas (s for space)
Since Tg is the same for a field of solar panels in a field or a field of rectennas in a field, it too can be eliminated from both sides of the equation.
There are some other effects, like the slightly strong sunlight in space and the slightly higher efficiency of panels when on Earth (yes, that’s correct), but they are minor considerations that have no material effect on the outcome.
So let’s put in the numbers. For solar panels on mountings facing south in the Nevada desert, the industry-standard PVWatts tools and real-world panel lifetime produces uses these input numbers:
- I = 2300 (using a tracker)
- L = 40 years
- Ts = 1
So every 1 kW of panels in Nevada will produce 92,000 kWh of power over its lifetime. In space, the numbers look like this:
- I = 8760
- L = 12 years
- Ts = 0.50 (see this article for reasons why)
So every 1 kW of panels in space 52,560 kWh over its lifetime. 52,560 < 92,000.
The long and short of it is that space is an extremely nasty place. Putting anything up there seriously shortens its life, and that includes humans.
If you’re wondering where those lifetime numbers come from, it’s the measured real-world 20% degradation rate. That is, it’s the measured time it takes for the panels to drop their output by 20% from their original power. The numbers for ground-mounted panels are widely available, and you can read NASA’s experience, NATO papers and this lengthy report (page 422) for the space examples.
Why that number, 20%? It’s just the number the industry uses. It’s widely measured and available, as opposed to some other number, say the time to 50% reduction which you will not easily find.
End of story
If you take a panel and put it in Nevada it will deliver some kWh. If you put that same panel in space, it will make less kWh.
What else needs to be said?
If previous history is any guide, lots. So I’ll reiterate my previous note: if you’re going to take issue with the basic gist of this article, post your math.