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First, a bit about my thoughts. I believe we have the capability today to provide energy, water, food, education, and transportation to every man woman and child on the planet. To that end, I would like to become a force that brings about this change.

In trying to meet the first goal, which is to provide energy, I have come across two technologies which greatly interest me, the first of which must be in place to begin the second.

The first is the high efficiency solar cells developed by Patrick Pinhero at the University of Missouri. Assuming that said solar cell captures 80% of available light, how much energy can I expect them to produce per meter of cell? How would this vary betweeen environments such as the Nevada desert and central Florida, how did you come to these conclusions, and is there any formula I can use to calculate an expected energy output?

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It's about solar cell efficiency, so it's not clearly off topic, but I certainly do understand why one might consider it off topic. Let's see what the rest of the community thinks. (By the way, welcome to the site, JG214!) –  David Z Mar 2 '12 at 9:48
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See the excellent answers below, but your question reminds me of when I had this very problem assigned as homework for an electromagnetic fields class. The instructor informed us later that everyone in the class arrived at different answers and that his answer and the book's answer were different still. He gave credit to most of us and explained that in such a complex situation, there can be more than one right answer depending on the assumptions of the model. –  AdamRedwine Mar 2 '12 at 15:15
    
If there was just one right answer, that would be great. I work with financing companies on multi MW PV system production analysis all the time, and the industry can only barely agree on certain best practices. NREL's meta analysis of simulation packages showed that the biggest differences between simulations and reality was the accuracy of the meteorological data, and the choice of derate efficiencies. The real problem, mathematically, is that we are trying to give an "answer" to something that is actually a distribution of answers. P50/P90 statistical analysis is where the process is headed. –  phidauex Mar 2 '12 at 16:15
    
Hang on, hang on!! There is a world of difference between "solar cell captures 80% of available light" and "solar cell converts 80% of available light energy into usable electric power". The latter is extremely unlikely. Now I'm not intimately familiar with this work, I just read the press release and brief paper. I see no indication that power conversion efficiency will be anywhere near 80%. I'm certainly willing to believe that antennas –  Steve B Mar 5 '12 at 4:41
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4 Answers

A fantastic, free, book for calculations of this kind is provided by David McKay in Renewable energy without the hot air.

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That doesn't really answer the question, does it? And the book has more than a couple of flaws, so while the text is elegantly written, it's not always the most reliable source. Perhaps you could copy over the relevant calculation from the book's website, as it has a Creative Commons licence? –  EnergyNumbers May 17 '12 at 19:07
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On PV efficiency

A recent study by Vandenbosch & Ma in Nano Engineering looked at the theoretical efficiencies obtainable from the sort of nano antennae that Pinhero and others are looking at, and found that maximum theoretical figures to be:

radiation efficiencies near or slightly above 90%, and a total solar power harvesting efficiency of about 60–70%

So the reported radiation efficiency in the question of 80% will translate to a lower solar power harvesting efficiency. For the calculation below, I'll assume it translates to 50% conversion of sunlight to electricity.

From efficiency to energy yield

Here's the calculation in sketch form: $$annual~energy~per~square~metre (kWh/y) = panel~efficiency~{\times}~peak~sun~hours~per~year~{\times}~electrical~system~efficiency$$

Average insolation at sea-level is around $1kW/m^2$, so a panel efficiency of 50% (I don't believe it, but let's run with it) would give a mean power of $0.5kW/m^2$ in peak sun.

The quick way to do the calculation is to then look at the number of peak sun hours in an average year, and multiply that by the peak power per unit area, to give you an estimate of mean energy per year. Miami has about 1670 peak sun hours per year; Las Vegas has around 2120 peak sun hours per year. Figures are from the PVWatts 2.0 calculator, and assume we've picked a static tilt to maximise mean annual output.

So, for Miami we'd have $0.5\times1670 kWh/m^2/y$, and Las Vegas would have $0.5\times2120 kWh/m^2/y$. Then multiply by the efficiency of your electrical system (inverter, etc), probably around 85% or so. That gives final figures of $713kWh/m^2/y$ in Miami and $900kWh/m^2/y$ for Las Vegas.

Now, h/y is a constant ($\frac{1}{8760}$; let's ignore leap-years), so we can drop that out of those figures, to express it in $kW/m^2$ rather than $kWh/m^2/y$, and we find that these panels would yield a mean power (averaged over an average year) of $$0.081kW/m^2~in~Miami,~and~0.102kW/m^2~in~Las~Vegas$$

That's a gross simplification, because panel output will depend on whether the mounting is static, 1-d tracking or 2-d tracking; and what the combined profile is of ambient temperature, direct sunlight, and ambient light. But it's probably going to be within 10% of a near-enough number.

Closing thoughts on efficiency

But efficiency really isn't an issue for PV: there's no shortage of sunny area. It's all about cost per unit energy for terrestrial applications. Extremely-high efficiencies are likely to be of interest only for rooftop applications, where available space is strictly confined; and space applications, where power per unit mass is critical.

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Definitely a valid way of approaching the problem - just don't forget that 1 sq.m of module will require 2-3 sq.m of ground area for inter-row shading, infrastructure, roads, etc., which is a factor I often see left out of "back of the envelope" energy calcs. Entirely agree with you about the sunny area issue, and the "value" of high efficiency, we look for lowest LCOE (Levelized cost of energy), and ignore almost all other specifications. –  phidauex Mar 2 '12 at 16:20
    
As an aside, I clearly need to read up on formatting equations. I spend most of my time on the Cooking.SE, and equations don't come up there quite as often. ;) –  phidauex Mar 2 '12 at 16:23
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there are some easy guides to LaTex (the equation-formatting language used here) around the web, and there is some help somewhere on meta I think on how LaTeX is done here (this site only uses a small subset of it). But please don't use my LaTeX as a good example to learn from - I'm rubbish at it. –  EnergyNumbers Mar 2 '12 at 17:01
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Have a look at this innovative solar panel technology from CERN.

The ultra-high vacuum provides the panels' heat chambers with exceptional insulation, vastly reducing heat loss and greatly improving efficiency. "We've had temperatures of 80°C inside the panel when the panels were covered in snow", says Benvenuti.The panels also recover the energy produced by diffuse light more efficiently than traditional panels. The two technologies make them particularly suited to colder, less sunny climates where classic solar panels are less efficient.

In Geneva, the vacuum-sealed solar panels can supply heat at over 300 degrees Fahrenheit with 50 percent efficiency, meaning each square meter of panel generates a half kilowatt of power. Moreover, unlike other solar panels that use parabolic mirrors, Benvenuti’s do not rely only on direct sunlight. Instead, they capture diffuse light, which makes up more than 50 percent of the light in central European climates like Switzerland.

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This is an incredibly complex question, but we can get you going by ignoring a few of the trickier issues, and get you to an "order of magnitude" sort of result.

PV cells output a given power (Watts) that is roughly proportional to the irradiance (Watts/Sq. Meter) they are exposed to, corrected for the angle of incidence against their surface. Temperature and spectral content of the light affect the "curve" describing the relationship between output power and irradiance.

PV modules made for actual utility power generation are tested at 1000 W/sq.m, at 25C cell temperature. This roughly corresponds to the peak expected irradiance at the surface of the Earth in sunny places. Actual daily peaks in the summer tend to be in the 800-1100 W/sq.m in the central US. Horizontal irradiance at the top of Earth's atmosphere averages 1350 W/sq.m, though is a bit higher in January, when the Earth is at it's closest to the sun.

An 80% efficient solar module, exposed to 1000 W/sq. meter, would output approximately 800W instantaneously. This oversimplifies things a bit, but is a truthy way of describing the result.

The problem for your calculations is that irradiance is constantly changing, as is ambient temperature, cloud cover, and other things that have a huge impact on your output. Knowing the peak wattage at any given time isn't helpful, because peak watts don't run our lives - we need to know the energy produced over time, by integrating the changing irradiance, and accounting for efficiency losses and weather conditions, in order to determine the amount of energy produced (typically measured in Kilowatt-hours (kWh) in the US).

This would be very hard, except that NREL (US National Renewable Energy Labs) have done most of that work for you, in their package SAM (System Advisor Model) which is a free application (registration required) that is very powerful, but relatively easy to use.

Below are the results of two quick simulations using a hypothetical 80% efficient module:

The system as simulated:

  • Module: 80% efficient, 1.0 m^2, 800W-DC output at standard test conditions.
  • Inverter: 97% efficient
  • DC and AC losses: Standard losses for DC and AC conversion, NOT including utility transmission losses, ~0.88 including soiling, DC losses and AC losses.
  • 1 acre of space (~4000 sq. meters)
  • Ground Coverage Ratio of 0.33 (33% of the ground area is occupied by modules)
  • Module tilt: 30 degrees

This gives us 1320 modules, for a DC system size of 1.056 MegaWatts (nice round number), using commonly used values for GCR and tilt. The layout is a bit conservative, and assumes we have a large amount of space available, and that maximum annual production is our goal, rather than production any given month.

The cost to install a system like this, using today's inverter, racking and labor rates, and a little margin for the installers and financiers would be around 500,000USD + module cost. Crystalline PV modules today cost about 0.85USD/watt on the utility-scale market. I won't speculate about the cost of hypothetical 80% efficient modules (but I can promise you it won't be as cheap as any researcher claims).

Simulation Results:

Site 1: Nevada Desert. Meteorological source, NREL TMY2, Las Vegas Station
Net annual energy produced: 1,989,000 kWhs

Site 2: Florida. Meteorological source, NREL TMY2, Tallahassee Station
Net annual energy produced: 1,586,000 kWhs

Site 3: Alaska. Meteorological source, NREL TMY2, Anchorage Station
Net annual energy produced: 1,010,000 kWhs

Boiling that down to a usable "formula" would be simple - each acre of land dedicated to this technology would generate the above values annually. You can see that regional conditions contribute to nearly a 2:1 ratio in expected output. Globally, solar has been the most popular in locations that are politically and financially favorable, rather than environmentally favorable, but long-term plans would favor regions with good environmental conditions.

NOTES:

This simulation uses accepted parameters for estimating the production of realistic systems, but these hypothetical modules make huge assumptions. Experience has shown us that new technologies never perform in the real world like they do in the lab. Thin-film technologies have been in development for over 30 years and are still barely pushing 12.5%, while crystalline silicon is at ~15%, and costs less per watt (due to the enormous manufacturing infrastructure).

I've ignored a HUGE issue with supplying vast quantities of renewable energy to a grid. When you get above 15% distributed penetration on a grid, things get unstable fast. Storage methods need to be in place to ensure that periods of excess production aren't wasted, and energy is still available during times of reduced production. This is a non-trivial problem for large systems.

I have also ignored the transmission problem. Even if you could make all the world's energy in Nevada, you couldn't get it where it needs to go. One of distributed generation's benefits is the ability to generate close to the loads they serve, reducing transmission losses.

I don't think I can attach files, but let me know if it is possible for me to send or attach the .zsam simulation file I used, and I will be happy to provide it.

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