Thursday, March 26, 2009

Chinese Solar Rocking today!

It looks like this is the reason.

LDK up 30% at the moment. STP up 35%+.

Wow!

Wednesday, March 18, 2009

For the Survivalists: How much gasoline is one Solar Panel worth?

Ok, first, what is the Kilowatt*Hour equivalent of a gallon of gas?

A Gallon of gas contains 114,000 BTU/gallon per Wikipedia (and other sources).

So, 1kWh is ideally equal to 3412 BTU, but no Generator is ideal. The generator's conversion efficiency is measured by its "heat rate," and the common range seems to be centered around 8,000-11,000 BTU/kWh. For this estimation I took a very efficient generator and used 8000 BTU/kWh (about 43% Efficiency).

Using these numbers gives a Total Energy Output/Gallon of 114,000 BTU/Gallon * 1kWh/8000BTU, or 14.25 kWh/Gallon.

Cost: $2.50/Gallon. This gives Cost/kWh = $2.5/14.25kWh = $.18/kWh


Now, let's look at a single 200Wp Solar Panel over one year at a 17% Insolation location (like in Massachusetts).

200Wp * .17 * 1Year = 34W*Year = 34W*Year*365Days/Year*24Hours/Day = 297.8kWh
Cost: $800/Panel. This gives Cost / kWh = $800/297.8kWh = $2.68/kWh


Woah! Ok, so obviously the Solar System doesn't pay off in a year. Going out 25 years, though, (assuming 10% average degradation over that time) gives a total of 6700.5kWh produced over that time for a total 25 Year Cost/kWh of $0.12/kWh.
For another comparison, over 25 years this single solar panel will produce the equivalent of 470 Gallons of Gas, or at this rate, 19 Solar Panels (3800Wp) will produce the equivalent of a gallon of gas per day.


Of course, this isn't exhaustive. I didn't compare costs of the generator involved, or of the installation and inverter costs for the Solar (this will at least double the cost for Solar Energy, but Government Incentives will bring it back down quite a bit). The focus here is a comparison between energy output over time. The point being, it's a potentially valid hedge for those that might be worried about future disruptions in such things like the supply of gasoline for generators. Prior to such a time, there are choices to be made, and in the case of a very long term potential outage, Solar Panels will provide much more energy than a person could even safely store in the form of Gas for an extended period of time. I also didn't account for such things as Interest on debt, because a Survivalist isn't necessarily going to care about that. If the time comes that they are preparing for, they know that money just might not worth what it is at the moment, and a working light bulb may be worth alot more.

Of course, remember that if you're one of these people, the neighbors will know that you have Solar Panels (or a Generator), and they'll want in on it. Therefore, the best thing we can all do now, is to do everything possible to make sure that not just "we" have a system, but to make sure that as many of our neighbors have them, too. Desperate people are dangerous.

Tuesday, March 17, 2009

Converting Energy to Peak Watts.

I put this out on the LDK board today. I figured I'd keep it here for posterity.

The debate starts with a claim that a company's product can put out 500MWh / acre / year, and that this is a good thing.
Well, it may be a good thing, but I can't really compare it to anything without converting it to Peak Power. So, that's what I do.


500MWh is energy, not power. So, we need to convert to Peak Power in order to compare to other systems.

Energy = Power * Time, so Power = Energy / Time.

Average Power per Acre = 500,000kWh/Year/Acre / Time (1 Year) = 500,000kWh*1day/24h*1year/365days*1/acres*1/year.

Do some cancelling and division:

The Average Power required to produce 500,000kWh in a year per acre is 57kW/acre.

Ok, so the company didn't give any idea of what assumed insolation ratio they are using here, but if it were set up in, say Arizona, and was on a dual axis tracker, 33% insolation would be a reasonable guess.

Start with Peak Power * Insolation Ratio = Actual Average Power.

Solve for Peak Power = Actual Average Power / Insolation Ratio = 57kW / .33 = 173kWp

This is the Peak Power Rating of their 500MWh/acre/year system assuming dual axis tracking, and 33% Insolation Ratio.

Comparing to a real world scenario (see).

Per Sunpower Tracker Advertising, their system works out to 161kWp/acre, which is just slightly less peak power than this reflecting system, which makes sense if the reflecting system gets a 28% conversion efficiency.

Tuesday, March 10, 2009

Great article on Black Silicon.

http://www.xconomy.com/boston/2008/10/12/sionyx-brings-black-silicon-into-the-light-material-could-upend-solar-imaging-industries/


"Black silicon is between 100 and 500 times more sensitive to light than untreated silicon."


"The company won’t build semiconductors or even semiconductor fabrication equipment, but will instead work with as-yet-unnamed partners to develop specifications for machines that can treat isolated areas of silicon wafers to create black silicon."


They're either going to sell the capacity to produce black silicon to one, or to many companies. This will be fun to see.

Heck of a day today in Solar.

Sunday, March 8, 2009

More Mathematical Mumbo Jumbo.

The other day I pointed out the diminishing retrurns of increasing a Module's Conversion Efficiency. The folks on Daily Kos nicely pointed out how trivial the results really were. Well, I can live with that. I think the post still serves to make very clear that the percentage change in output Energy is, in fact, proportional to the percentage change in Conversion Efficiency (I don't know, I guess I just had to see it for myself).

As usual, if there are errors, please let me have it; though please point out a specific or two rather than just saying "check your math."

So, turned into a simple equation, increasing a module's Conversion Efficiency increases the total energy panel output per unit time and per unit area by (Conversion_Efficiencyfinal / Conversion_Efficiencyinitial - 1) * 100%.

For example, the percentage difference between the Annual Energy Output of a 16% Efficient Panel and a 20% Efficient Panel would be (20/16 - 1) * 100% = 25% (assuming constant Area).


Following from this, I'd like to get a few more bits of information from these variables.


Effects on Surface Area of Improving Conversion Efficiency:


It could be said that PowerPeak (W) = InsolationPeak (W/m2) * Area (m2) * Conversion_Efficiency (%).

Setting PowerPeak and InsolationPeak as Constants, then we can say that C = Area * Conversion_Efficiency.

Take two scenarios, say, Case 1 and Case 2.

C1 = Area1 * Conversion_Efficiency1.

C2 = Area2 * Conversion_Efficiency2.

C1 = C2

Area2 / Area1 = Conversion_Efficiency1/Conversion_Efficiency2

Let's imagine a Solar Manufacturer and set today's average Conversion Efficiency at 16%, and let's say that by 2012 the average Conversion Efficiency will be 22% for some company.

Area2 / Area1 = .16/.22 = .72 = 72%

So, in order to generate the same amount of Peak Power at 22% Conversion Efficiency vs. 16% Conversion Efficiency, the manufacturer need produce only 72% as much area of PV material. Nice.


I'm not sure how "deep" this thought is, but I'm putting it out here, at the very least as a future resource for myself.

Friday, March 6, 2009

How much is 1% in efficiency worth in Solar?

Ok, so say you start with a Solar Panel that's 15% efficient (like today's low-end Crystalline Silicon Panels). For simplicity's sake, lets say that the panel has an area of 1 M^2.

So, at 1000W/m^2 Insolation, the panel will produce 150W, so this would be called its Peak Power Rating.

Let's say that you set that panel in an area with an Insolation Ratio of 20%.

In one year, that panel will produce 30W*Year = 262.8kWh [150W * .20 * 1Year * 365 Days/Year * 24 Hours/Day]



Now, say that the solar panel is 16% efficient.

At 1000W/m^2, the panel will produce 160W, so this is its peak rating.

You set that panel in an area with an Insolation Ratio of 20%.

In one year, that panel will produce 32W*Year = 280.3kWh



What is the percentage difference in the Energy Produced by the two panels in one year?


280.3kWh/262.8kWh = 1.067, so the 16% efficient panel will produce 6.7% more energy in a year than a 15% efficient panel.



Now, say that the solar panel is 22% efficient.

At 1000W/m^2, the panel will produce 220W, so this is its peak rating.

You set that panel in an area with an Insolation Ratio of 20%.

In one year, that panel will produce 44W*Year = 385.44kWh

This panel prouces 46.7% more energy in a year than the 15% panel.


See http://spreadsheets.google.com/pub?key=pNlmSU6te4mhtvbGWKl6KCA for a Spreadsheet that shows the interesting, but maybe obvious results.


So, let's say I have a choice between a 14% Module and a 15% Module. Well, the 15% module produces 7.14% more Energy per year than the 14% one. So, I had better look at the prices, and if the 15% module is more that 7.14% more costly, then you're better off sticking with the 14% one. This is assuming that space, quality, etc, aren't factors, of course. This is "all things being the same."

What if I has a choice between a 45% module and a 46% module? Well, the 46% Efficient Panel will produce just 2.22% more Energy per year than the 45% Efficient one. So, once again assuming that space isn't a factor, the 46% efficient panel had better be no more than 2.22% more costly.


I'm thinking that this is something that manufacturers have to be thinking about, too. Of course, there could be marketing reasons why a panel of a higher percentage efficiency might sell for more, and there are certainly applications that put surface area at a premium, but from a basic cost perspective at the very least, if a manufacturer of 50% efficient modules thinks that they have some technology that will take that efficiency up to 51%, then they'd better be able to manufacture those panels for less than 2% more than it costs them to make their 50% Modules. If the additional materials and manufacturing operations are going to add more than 2% to the cost of manufacture, then they very well might not have gained anything by the "breakthough."


As usual, if my thinking is wrong, by all means, let me have it.

Monday, March 2, 2009

Does Solar Tracking make sense?

I want to know, so I'm going to try to work out a rough scenario.


Looking at Wattsun Tracker Datasheets, I've decided to use 12 175W Suntech Panels. See http://www.wattsun.com/prices/Wattsun_Tracker_Prices.pdf

Cost of Tracker Equipment: $6250.

Additional Installation Costs (Rough Guess): $3000-$4000 (lower costs if you can put together an out-of-work electrician, welder, and some laborers).

Panel Total cost at $4.50/W = $9450; Total Peak Watts: 2100W

Inverter Cost: $2500 (small inverter, for just this application).


Cost of Tracking System:

Using these rough estimates, the total cost of the Tracking System with Panels would range from $21,200 - $22,200. Just to assume the worst, I'll stick with $22,200, or $10.57/Watt.

The cost of JUST the Tracker and Installation ($4000), runs $10,250, or $4.88/Watt.


Cost of Stationary System:

Calculating a rough cost of an Installed Stationary System, I'll go with the above Panel Cost of $4.50/Watt, and using the Solarbuzz estimation, which suggests that the total installed cost of the system will be twice the cost of the panels (I believe that this would include the Inverter). So, for comparison purposes, I'll set the Installed Stationary system at a total of $18,900, or $9/Watt.


Insolation Comparison:

In a normal stationary scenario, the Installation would produce energy according to the usual local Insolation values. However, the fact that it's a tracker, leads to an INCREASE in the effective Insolation value. Using a US Government Insolation Reference, it looks safe to say that for at least a very large portion of the US, there's a 2 kWh/M2 difference in Annual Insolation between a "Flat Plate Tilted South at Latitude," and a "Two Axis Tracking Flat Plate." I know from previous calculations that 2 kWh/m2 is equivalent to an insolation ratio of 8.33%.

Let's put this percentage in terms of our original 2.1 kW System. Assume that the Stationary Installation is on a roof angled at latitude, in a region that recieves an average of 20% Insolation over the course of the year. In ideal conditions, this system will produce 2.1 kW*Year * 20% = 0.42 kW*Year = 3679kWh.

Now, let's put that same system on a tracker, thus increasing the effective Insolation Value by 8.33%. This system will produce 2.1 kW*Year * 28.33% = 0.59 kW*Year = 5212kWh.

We can see that an 8.33% increase of in the effective Insolation Ratio has increased the total Annual Energy Output by 29.5%!


Does the Tracker pay off?

To start out with, let's find out how much Energy each system will produce in 25 years. To be a bit more accurate to the real World, I'll take off 25% from each value to reflect Inverter losses, efficiency degredation over the 25 year lifespan, and variation from the Manufacturers Test Conditions that went into the initial rating of the Panels.

Stationary: 3679kWh/Year * 25 Years * .75 = 68,961kWh.

Tracking: 5212kWh/Year * 25 Years * .75 = 97,725kWh.

So, over the course of 25 Years, the Tracking System produces 28764kWh more than the Stationary System.

Since the Tracking System cost $3300 more than the Stationary System, this is our target to beat.

Taking the difference between the two outputs, and multiplying by a reasonable energy selling price ($.12/kWh) gives 28,764kWh * $.12/kWh = $3451, which, compared to the additional cost of the Tracking System ($3300) is a win over 25 Years, just barely.


Conclusion:

Yes, the tracker pays off slightly over 25 years, using rough estimations. Much would depend on the specific local conditions, and the Electricity Costs.


Final Comparison:

The Stationary Roof Installation had a Total Cost of $18900, or $9/Wp, and produced 68,961kWh over 25 Years.

$18900 / 68,961kWh = $.27 / kWh.

The Tracking Installation had a Total Cost of $22,200, or $10.57/Wp, and produced 97,725kWh over 25 Years.

$22,200 / 97,725kWh = $.23 / kWh.

From this, we can see quite clearly how, though the price per Peak Watt for a Tracking System is higher than for a Stationary System, the actual cost per unit of Energy of a Tracking System is lower.


Note:
Of course, there are many variables unaccounted for in these basic Calculations, including Government Subsidies, Interest on Loans, and Insurance Considerations. More detailed Calculations would have to be done on a specific case-by-case basis. I think this is good for a start.