A very-rough calculation Comparing Solar Energy Output to the Gulf Oil Slick.

Oil is leaking out at 5,000 Barrels / Day.

Apparently the most efficient 4-stroke engines run at about 43% efficient, as well, so the previous estimation of 43% will still work.

Previous Calculations give around 14.25 kWh/Gallon of Gas.

I'm going to be very rough, because gas is only a bit less than half of the stuff that comes out of a barrel of oil ( http://tonto.eia.doe.gov/dnav/pet/pet_pnp_pct_dc_nus_pct_m.htm ).

So a Barrel of oil is worth about 21 Gallons of gas, 12 Gallons of Fuel Oil, and 4 Gallons of Jet Fuel.

I've not got it in me to look up the details of the fractions of different types of fuel oil and jet fuel, so I'm just going to stick with 21 Gallons of Gas per Barrel of Oil right now.

So, 5000 Barrels of oil per day times 21 Gallons of Gas / Barrel of Oil gives 105,000 Gallons of Gas per Day draining into the Gulf.

At 14.25 kWh/Gallon of Gas, the leak is equivalent to 1,496,250 kWH / Day in Energy output.



19 Solar Panels = 1 Gallon Gas / Day @ 200W / Panel (In much of the USA).

So, it takes roughly 1,995,000 Solar Panels to produce the equivalent in Energy to 105,000 Gallons of Gas per Day.

Multiply this times 200W per Panel gives a total of 399 MW of Solar Installation in order to equal the gas content of this single spill. If we assume that the remainder of the oil doubles the energy content of the barrel of oil (which I bet is giving false advantage to Oil), then we need 798 MW of Solar to replace it. This seems like alot. Just a few years ago there wasn't this much Solar Capacity in the World. Last year the World Installed about 7 GW of Solar Energy, or a bit over 8.7 offshore platforms, and this year we're looking at the possibility of 13 GW in new Installation, or the equivalent of 16 Platforms. It goes wild from there.

Now, at $4/W Installed, 798MW of Solar Energy would cost around $3.2 Billion Dollars, which is surely much higher than the cost to produce a single oil platform. On the other hand, it's surely alot less than what this disaster is going to cost BP for this particular Rig. In any case, the Price of Solar will keep coming down, while the price of fossils is going to keep going up.


Anyway, just sayin'...

Unit Idea... PowerPeak-Lifetime25 (or somesuch).

A Solar Cell performance degrades over time. The effective lifetime of Solar Panels are right now considered to be in the area of 25 years.

The rate at which a collection of cells degrades, on average, over the lifetime may not be a nice line.

I can only guess what the curve looks like for a standard Silicon, Wafer-based, Cell but I imagine that it's close to linear. If you have good data, I'd love to see it.

On the other hand, particularly when future cells have their efficiency enhanced by things like coatings, the curve could get alot more complicated. You might have a Silicon Solar Cell that will degrade linearly over 25 years, but the cell might be coated with a product that will increase its initial efficiency by a very significant amount, but which might degrade in effect completely after only 15 years.

Depending on the cost to add the coating, etc, it might very well be financially advantageous to buy this panel with a rapidly degrading initial phase, and a slowly degrading long term component.

The way I see it is that if, for example, a 100W panel were to degrade at 1% per year over its 25 year rated lifetime, then the effective Peak Power is really 87.5W. That's the effective peak power over the panel's lifetime, of Power_Peak-Lifetime₂₅.

In the same way, if you had a panel of the same surface area area that was 160W, but it degraded at 2% per year for 15 years, and then 1% after that till 25 years, it would have an effective Power_Peak-Lifetime₂₅ of 140W.

This would allow the customer to know what they're really buying over 25 years, even if it would take some estimations and tricky modeling on the part of manufacturers. They'd have to try to figure out with great care exactly what the FUTURE degradation curve of their product is going to look like. No worries, they'll appreciate the challenge. :)

In any case, if you're not reflecting rates of lifetime degradation in the cost per Watt, then there is trouble on the horizon for everyone involved.

EDIT: I suppose I spoke prematurely. I assumed that there wasn't a standard name for this. There must be. I ask then, what is it?

So, you want to buy a solar plant, part 2.

I thought I'd extend my calculations from So, you want to buy a solar plant, part 1, to make them a bit more realistic. You'll find the basic logic for this post at that location. Here's a recap:

(1) Outgoing$Monthly = TotalPeakPower (kW) * Cost/Wp * C1

(2) Incoming$Monthly = TotalPeakPower (kW) * Rate ($/kWh) * C2

(3) C1 = i / (1 - (1+ i)^-n)

Note: i = .0042, n = 300. Previously, I had set n to 240 for a 20 year loan, but in this case, I'll be setting n to 300 for a 25 year payoff period.

(4) C2 = Insolation Ratio * 1year * 365days/year * 24hours/day / 12months

Note: Insolation Ratio = .1875. Previously I had used .20 for the Solar Plant's local Insolation Ratio. The particular location that's being considered is in the area of Bangalore. I see from this Insolation Map that that area seems to be recieving between 4kWh/day/m2 and 5kWh/day/m2 in Solar Energy. For this estimation's sake, I'll pick the middle, or 4.5kWh/day/m2. Using previous calculations, this works out to an Insolation Ratio of 18.75%.


(5) Setting Outgoing$Monthly = Incoming$Monthly gives the break-even long term average energy price per kWh with respect to the cost per Watt of the Installation:

(6) Cost/kWp ($/Wp) = Rate ($/kWh) * C2/C1

Or conversely:

(7) Cost/kWp ($/Wp) * C1/C2 = Rate ($/kWh)

I want to take this one step further, and take into account degradation of the system over time, as well as various losses like Inverter / Substation losses. I'll try to be aggressive on this correction, to err towards the worse case. So, over the 25 years over which I'm considering, I'll say that the modules lose 1% per year, and Inverter losses are 8%. Averaging the degradation over 25 years gives a loss of 12.5%, plus the 8% Inverter loss, gives a total of a 20.5% loss.

I'll reflect this in my equation by going back to another equation from Part 1 (one of the roots of the (6), above):

(8) TotalPeakPower (kW) * Cost/kWp * C1 = TotalPeakPower (kW) * Rate ($/kWh) * C2 (Hours/Year)

Just prior to (6), I had canceled TotalPeakPower out of my equation. Essentially, these equations should give a rough estimate of the Levelized Cost of Energy irrespective of the total size of the Installation. Of course, there are alot of variables involved when changing scale, and these aren't reflected here. Consider this to be more along the lines of an ideal solar farm. It costs the same per Watt to add a thousand Watts of capacity, or a hundred thousand Watts.

Remember that on the lefthand side of the equation is the initial cost per Watt of the system, while on the righthand side of (8) is the income per unit energy. So, essentially, the "TotalPeakPower" on the left is the ideal TotalPeakPower, while the TotalPeakPower on the right is the effective TotalPeakPower as reflected in the Energy produced over the lifetime of the system, by which income is derived.

So, I'll throw in some subscripts.

(9) TotalPeakPower_ideal * Cost/kWp * C1 = TotalPeakPower_effective * Rate ($/kWh) * C2 (Hours/Year)

In the ideal case, TotalPeakPower_effective = TotalPeakPower_ideal, but in this case, there's a 20.5% loss, so:

(10) TotalPeakPower_effective = .795 * TotalPeakPower_ideal

Plugging (10) back into (9) gives:

(11) TotalPeakPower_ideal * Cost/kWp * C1 = TotalPeakPower_ideal * .795 * Rate ($/kWh) * C2 (Hours/Year)

So, I can now still cancel out the TotalPeakPower_ideal, and I'm left with a correction to the income side of the equation, which reflects system degredation losses and Inverter losses.

The final result is:

(12) Cost/kWp * C1 = (Rate ($/kWh) * C2 (Hours/Year)) * .795

Taking a case, let's say your installation is going to cost $5.25/W, or $5250/kW.

Solving (12) for Rate gives:

(13) Rate ($/kWh) = Cost/kWp * C1/.795C2

C1 = i / (1 - (1+ i)^-n) = .0058 (i = .0042, n = 300 (25 years))

C2 = .1875 * 8760 Hours/Year / 12 Months/Year = 137 Hours/Month

C1/.795C2 = .00005325

Rate ($/kWh) = $5250/kWp * .00005325 = $.28/kWh

Calculation Central.

One of the fun things that I have been enjoying doing for this blog is to run various calculations that in some way apply to Solar Energy, or Energy in general. Since these things tend to get lost down the blog history, so I've decided to keep a reference to them here.

Note: I would consider all of these to be rough. Since I am typically looking at very general cases, I make assumptions. As much as possible, I try to point out the assumptions in the detail of the article.

Also, I'm typically choosing some specific variable or variables to focus on, so I am probably not calculating exactly what you are looking to know. However, hopefully that doesn't mean that my work isn't some value in giving direction in possible ways to look at a problem and to work out a reasonable solution.


Selected Calculations:


So, you want to buy a solar plant, part 2.

Description: Extends Part 1 by correcting for panel degradation and inverter losses.


So, you want to buy a solar plant.

Description: A rough start at a look of Calculating a Levelized Cost of Energy (LCOE) for Solar.


Part II : Percentage Land Area required for 100% Replacement of 2006 Energy Demand.

Description: State-by-State comparison of Total Land Area required in order to replace 100% of that State's Electricity Consumption.


Percentage Land Area required for 100% Replacement of 2006 Energy Demand.

Description: State-by-State comparison of Total Land Area required in order to replace 100% of that State's Energy Consumption.


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

Description: Comparison between the energy output of typical modern Solar Panels to the energy contained in a Gallon of Gas.


How much is 1% in efficiency worth in Solar?

Description: Discussion of Diminishing Returns in Increasing Conversion Efficiency.


Does Solar Tracking make sense?

Description: Comparison between a Stationary and a Tracking Installation, discussion of potential advantage of tracking in Energy Output.


Real World Estimation of Land Use per Watt - Sunpower

Description: Land Use Scenario using Tracked Sunpower Modules. Expands upon Solar vs Coal, Land Area Comparison, below.


Solar vs Coal, Land Area Comparison, below.

Description: Comparison between Kentucky Coal Output to potential Kentyucky Solar Resource. Ideal Situation. Expanded on in Real World Estimation of Land Use per Watt - Sunpower , above.


A Note on Units of Energy and Insolation.

Description: Description of the concept of Insolation, and description of a rough way to use Insolation as a basis for Annual Solar Energy Output.


Looking forward to 2020.

Description: Scenario for 2020 making assumptions based on 15% replacement of Fossils by Renewables by 2020. Assumptions are very optimistic.


Coal / Solar Cost Comparison - Final Draft.

Description: Comparison between Coal and Solar Costs over the long term, assuming various rates of Inflation. Written prior to Economic Crash, so certain Assumptions should be reworked.

FSLR; The Betamax of Solar?

FSLR announced their earnings today. The results were great, particularly considering the overall economy.

I've given FSLR considerable thought in the last couple years, and I remain convinced that they have unspeakable future problems. On their investor relations page, they link to the pdf associated with their Q1 conference call. In it, they mention what they consider to be risks to their business, but nowhere do they mention the risk associated with availability of their critical Tellurium supply. Ok, so maybe they have it all figured out; but nobody's asking, and nobody's telling.

Ok, I don't know, but I want to get an idea of what kind of supply issue they're up against, so I've gathered some info.

Per Greentech Media, FSLR uses 6 grams of Tellurium per square meter. (See.)

Per First Solar, the FS-277 Module is .72m2 and has a peak power of 77.5W. (See.)

77.5W / .72m2 = 107.64W/m2

So, at 6 grams / m2, the amount of Tellurium required per Watt works out to be 6g / 107.64W = .056 g/W.

Well, we know that FSLR is aiming for a bit over a GW in annual production for '09 and '10, so rounding to 1GW gives roughly 55.7 Metric Tons of Tellurium required to produce that GW of modules.

The question, then, is how much Tellurium is out there, and what does it cost?

According to the USGS, the price has ranged from $41,800/MT in 2004 to
$82,000/MT in 2007. The World Supply of Tellurium according to US Geological Survey was 132MT in 2006.

Ah, no problem. If they're using 55MT to produce 1W worth of modules, and they're paying even the high price of $82,000/MT for their supply, then they're only paying a total of $4.5 Million for their entire yearly supply of Tellurium. That's less than a penny per Watt. In fact, during the CC, Jesse Peechel stated, quite possibly accurately, that First Solar's largest cost was glass.

Wait, a problem. Solar is big. A sensible look at the required future scale of Solar Energy puts the annual Global installation rate to be around 30GWp per year by just 2012. What if FSLR wants to maintain a significant share in this market?

Well, as it is today, it appears that over a third of the World's Tellurium supply is required for the production of a single Gigawatt of First Solar modules.
If FSLR were to take 10% of that market, they'd have to produce 3GW of modules, which by today's efficiencies would require 165MT of Tellurium, or more Tellurium than the World produced in 2006! Well, maybe the price of Tellurium is a pittance when the company is demanding only a third of the World supply of material, but I can guarantee that it won't remain so when that company is demanding 33MT MORE than the World's annual supply.

A big part of this problem is that there's no such thing as a Tellurium mine. Tellurium is only produced as a byproduct of mining other commodities, such as Copper. This means that it's very difficult to increase the World Supply independently of the supply of those other materials. If you were to mine Tellurium alone, the cost would be astronomical, and yet if you were to drive up the mining activity in Tellurium's sister elements, then you'd have the affect of driving down the prices of those materials, thus making them into less desirable targets for mining.

What about efficiency gains? Sure, if FSLR is able to pull off a tripling, or even just a doubling of their efficiency, then they could make do with dramatically less material. I can imagine several possible ways that they could do this, but I suspect that it will be a tough path. As it stands, per the CC pdf, FSLR has increased the conversion efficiency of their product by .3% since Q1 of '08. That's simply not going to cut it, particularly if you look out past 2012 when the market gets even larger.

I don't know. They have some very smart people there, and they're working hard in an exciting industry. The particular technology just doesn't seem to stack up to me, though, and like I said, nobody is asking questions and nobody is volunteering answers.

Ah well, in the short term, I'm quite certain that they are going to do great. Wall Street loves them, and they have excellent margins for the time being. They very well might be able to leverage some of that temporary financial advantage in order to open up new technologies to their benefit, so we'll see.

All that said, I'm not short FSLR, and I suspect that to go short FSLR would be a very bad plan.

Also, a final note, it's pretty obvious that I think that the strongest players at this time are out of China, but it's not that I don't like some US Companies. I really like Applied Materials, and Sunpower to name a couple of domestic players.

So, you want to buy a solar plant.

A Scenario.

Note: A follow-up scenario includes accounting for system degradation and inverter losses.

The cost of the install + Interest will equal some amount of money to be paid out per month. I'll call this Outgoing$Monthly.

Power generated per month will be sold on the market for some amount of money. I'll call this Incoming$Monthly.

Set Incoming$Monthly = Outgoing$Monthly.

This would be the point at which your investment broke even on a monthly basis (not including maintenance cost at the moment, this is just to include interest expense into the equation). It's not going to be quite right, because of seasonal variation, as mentioned below, but I'm not looking for anything exact, just a rough way to start gauging cost / benefits.

The end result will be a relationship between the Installation Cost per Watt, Interest Rate, and Required Sales Price of Energy produced in order to break even.

I'll skip to the chase, for those that don't want to read through the whole thing.

Cost/kWp ($/Wp) = Rate ($/kWh) * C2/C1

Note that the assumed interest rate (5%) for purposes of this post has been set and absorbed by C1, and the Insolation Ratio has been absorbed into C2.. Other assumptions are pointed out below.

To give an example of what this tries to point out, let's say you can sell the energy produced by the power plant for $.25/kWh (equal to the low range of this estimate of costs for future nuclear power plants).

Cost/kWp ($/Wp) = $.25/kWh * 146 Hours/Year / .0066 = $5,530/kWp, or $5.53/Wp.

So, if you can sell your power for $.25/kWh, then you break even (roughly) if you can complete the installation for $5.53/Wp or less. Note that the equations below DO NOT include the existing 30% Federal Tax Credit for Solar Installation. That's icing (of course, it also doesn't include lifetime performance degradation or inverter losses).

Fact: this is very much in the range of possibility in TODAY's market. Particularly in the case of mid-large scale installations.

The basis follows.

If there's one thing that I've learned being on the Internet this many years, it's that if you're wrong, somebody will point it out. Have at it with my thanks!

Here goes:



First, find the Monthly Payment required to make the loan payment for an installation of some total cost.


(1) Outgoing$Monthly = (Principle * i) / (1 - (1+ i)^-n) See http://en.wikipedia.org/wiki/Amortization_calculator.

This is the Monthly Payment on the loan for the power plant with the below assumptions.

Principle = Total Original Loan amount used to finance the entire plant = the Total Peak Power of the plant * the overall Cost per Watt of the system.
i = periodic interest rate (Monthly. Assume 5% APR, so i = .05 / 12 = .0042).
n = total number of payments (Months. Assume 20 Year Loan, so n = 240).


(2) Principle = TotalPeakPower * Cost/Wp

The Principle is the amount of the loan, where the total cost of the installation is given by the Total Peak Power * Cost per Watt. Substituting for "Principle," from (2) into (1) gives:


(3) Outgoing$Monthly = (TotalPeakPower * Cost/Wp * i) / (1 - (1+ i)^-n)

For simplicity, and ease of double-checking results, I'm going to treat n and i as constants (they are part of the assumptions above), and will pull a constant out of the above equation (3):


(4) Set C1 = i / (1 - (1+ i)^-n) and substitute into (3).


(5) Outgoing$Monthly = TotalPeakPower * Cost/Wp * C1



Now, to figure out what's coming in every month on the sale of the Energy.


This doesn't include seasonal variations. On thinking about it, though, in an Energy market where consumers are paying based on momentary supply and demand, wintertime prices could actually go up based on decreased supply, and so help to balance out the annual cycle for the energy supplier. Then, in the summer where supplies were higher, the prices to the consumer would decrease to offset some winter costs.

In any case, following similar logic to my note on Insolation, the Annual Energy output of the plant can be written as below.


(6) Annual Energy (kWh) = TotalPeakPower (kW) * 20% * 8760 Hours/Year * 1 Year

Start by writing down an equation to relate the Installation's Total Peak Power, to it's Annual Energy Output. I'm plugging in an assumption of a 20% Insolation Ratio, which would include a broad swath of non-sunbelt States. The Insolation Ratio Assumption for this post applies to such shady states as Tennessee, Missouri, and even North Dakota.


(7) Incoming$Yearly = Annual Energy (kWh) * Rate ($/kWh)

Multiplying the Annual Energy Output by the Rate at which it sells for, gives the Total Income for the year. Divide by 12 (below) and you have the Average Monthly Income.


(8) Incoming$Monthly = Incoming$Yearly / 12 Months


(9) Set C2 = .2 * 365 * 24 / 12

Once again, I'm going to pull all of the Constants out of the equation (6) to come up with C2.


(10)Incoming$Monthly = TotalPeakPower (kW) * Rate ($/kWh) * C2


Ok, so now we have the Monthly Outlay required for loan payments, and we have the Monthly Income from energy sales.


To break even - let's set them equal to each other.


(11) Set Outgoing$Monthly = Incoming$Monthly


(12) TotalPeakPower (kW) * Cost/kWp * C1 = TotalPeakPower (kW) * Rate ($/kWh) * C2 (Hours/Year)


(13) Cost/kWp ($/kWp) * C1 = Rate ($/kWh) * C2


Canceling out TotalPeakPower (kW) from both sides of the equation, gives a very simple equation relating the Rate at which the energy is sold, to the Cost/kWp of the initial plant installation.

Neat.



Ok, so to an example and a factcheck.


First, Calculate out C1 and C2.

(14) C1 = i / (1 - (1+ i)^-n) = .0066 (i = .0042, n = 240)

(15) C2 = .20 * 8760 Hours/Year / 12 Months/Year = 146 Hours/Month

Then, pick a target Sale Price for the power that is produced by the Installation, and solve (13) for Cost/kWp. I'm using $.25 in this case, so:

(16) Cost/kWp = Rate * C2/C1 = $.25/kWh * 146 Hours/Month / .0066 = $5,530/kW

Now to check it, or at least check the Interest Calculations:

Since TotalPeakPower was canceled out of the above equation, I'll pick a value to use for the factcheck, say, 1000kW.

So, using (3), Outgoing$Monthly = (TotalPeakPower * Cost/Wp * i) / (1 - (1+ i)^-n) = 1000kW * $5,530/kW * .0042 / (1 - (1+ .0042)^-240) = $36,617/Month.

Then, using (6), Annual Energy (kWh) = TotalPeakPower (kW) * 20% * 8760 Hours/Year * 1 Year = 1,752,000kWh/Year and Dividing by 12 to get a monthly Energy Output, gives 146,000kWh/Month.

Multiplying this by $.25/kWh gives $36,500/Month

Pretty Close. Exponentials are subject to rounding errors. Another way to check would be to put the total cost, or Principle (in this case, $5,530,000) into any number of online mortgage calculators.


Fin

Part II : Percentage Land Area required for 100% Replacement of 2006 Energy Demand.

Yesterday I posted a chart showing a rough estimate of how much land area would be required by each State in order for that State to replace 100% of its Energy Demand (per DOE numbers).

I posted it at DailyKos, and on the LDK board for comments.

Apsmith of DailyKos makes a good point that there are generator losses, etc., which should be used to reduce the overall total energy required to be replaced, and China_s2 of Yahoo agrees, and points out a different set of data, which is based on retail electricity use, so should closely represent actual electricity delivered, as opposed to total Energy Input.

So, I copied over the old data to a new sheet, plugged in the new data, and came up with a rough estimate of the total land are required to replace 100% of US 2007 Electricity demand.

Percentage Land Area required for 100% Replacement of 2006 Energy Demand.

The following chart represents the percentage of land for each State, and the USA as a whole (without Alaska), that would be required to replace 100% of that State's Annual Energy Demand.

Make no mistake, the numbers are huge. Then again, nobody is actually talking about 100% replacement by Solar, Ever. This is just to give an idea that it is physically possible, at all.

Assumptions and references follow.




here's the spreadsheet.


References:


State Energy Data.
State Land Area Data.
State Insolation Estimates.
Sunpower Power/Area Claim.


Assumptions / Notes:


The percentages reflected in the Graph are based on a Stationary system, though the value for Power/Area is based on a Sunpower claim related to their tracking system. This should be irrelevant, as Power is independent of whether the system tracks or not. Since these are Sunpower numbers, the Panel's Conversion Efficiency should be around 22%.

The Demand cited is irrespective of source, and so includes existing production of renewables such as Hydropower. Here's a very interesting page from the DOE giving detailed map-based information on US Energy sources. There's a "Select a State" dropdown that will take you to a close-up of the individual State including facts and demographics.

In order to work out an the Area, I used the equation:

Annual Energy Output = 1 Year * Power/UnitArea * Insolation Ratio * TotalSolarArea * 8760.

For more info, see A Note on Units of Energy and Insolation. Solve for TotalSolarArea, and divide by the State's Total Land Area, and you will get the percentage. Most of the trouble here is just in the conversion of units. On a political note, can we just all go metric please?

The Insolation values were eyeballed from the map. If anybody's got some better data on State Average Insolations, I'd love to see!

The base data does not seem to include Transportation Energy, though it didn't specify.

Of course, this assumes nice flat areas of land, on which to set up installations, and it also assumes that each state takes care of its own needs irrespective of local conditions or capacity. It's a brief look from 1000 miles up above. It's not exhaustive, but it's fun, and maybe interesting.

By all means, if my basic math is way off, let me know.

This post is followed by Part II, which calculates the same area percentage, but only for the replacement of Electricity End Use.

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.

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.

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_Efficiency_final / Conversion_Efficiency_initial - 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 Power_Peak (W) = Insolation_Peak (W/m2) * Area (m2) * Conversion_Efficiency (%).

Setting Power_Peak and Insolation_Peak 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.

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.

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.

Real World Estimation of Land Use per Watt - Sunpower

Sunpower gives us an idea of a realistic value for Land Area per Watt of 2-3 Hectares per MW.

Convert Units:

2.5 hectare = 6.2 acres per MW.

Determine Total Peak Power Output:

1.5 Million Acres / 6.2 Acres/MWp = 241935 MWp = 241,935,000,000 Wp = 242 GWp

Calculate Average Annual Energy Output:

242 GWp * 18.75% Average Annual Insolation = 45.4 GW*Year = 3974 GWh = 397 Billion kWh

This is about 4 times less than the ideal number calculated in this Ideal Situation. Not a problem at all, IMO, considering that you don't actually want to cover every square inch of a chunk of land with flat panels. Tracking is sure a nice option.

Follows: Solar vs Coal, Land Area Comparison.

Solar vs Coal, Land Area Comparison.

Thirty-eight years of Kentucky strip mining have, at one time or another, destroyed 1.5 Million Acres (6,070,284,633 m2) of land, and torn down 470 Mountains.

Let's roughly compare the land impact of Coal to the potential land impact of Solar, with this 1.5 Million Acres as a basis.


First, let's see how much Energy could be produced by all the Coal that was mined in Kentucky in 2007*. Answer: 553 Billion kWh.


Per Salon, 158 Million Tons of Coal were produced in Kentucky 2007. I know from previous calculations that a very efficient coal plant can produce around 3.5 MWh/Ton, so burning the entire 158 Million Tons of coal produced in Kentucky in 2007 gives, 158 Million Tons * 3.5 MWh/Ton = 553 Million MWh, or 553 Billion kWh.


Now lets look at how much Solar Energy is available to an equivalent area of Kentucky. Answer: 10,000 Billion kWh.


Take a look at the US Insolation Map and note that the State of Kentucky is almost entirely bright yellow. Look at the legend, and note the value of 4.5-5.0 kWh/m2/day for this color. This is the Average amount of Solar Energy striking a 1 m2 panel on some single Day of the year (Averaged over the whole Year). Let's take the low estimate, and get the total Solar Energy incoming onto a 1 m2 panel, over an entire year, by multiplying 4.5 kWh/m²/day * 365 days = 1642.5 kWh/m²**. Multiplying this by 6.07 * 10⁹ m² gives us the total Solar Energy Striking the mined area of Kentucky in 2007, or 10 * 10¹² kWh, or 10,000 Billion kWh.


So, how much of this Energy could actually be converted to Electricity using modern Solar Panels? Answer: 1,600 Billion kWh.


Of course, Photovoltaic Panels don't convert all of the Energy Striking them into Electricity. At this time, it would be fair to use 16% as a rough Average Conversion Efficiency***. So, if you were to cover that 1.5 Million Acres with Solar Panels, and each panel was 16% efficient at converting that light to Energy, that installation would produce 10,000 Billion kWh * .16 = 1,600 Billion kWh of electricity.


The Conclusion?


If you covered the 1.5 Million Acre area of Kentucky that has been affected by Strip Mining and Mountaintop Removal with Solar Panels like those commonly manufactured today, then you would produce 2.9 times the energy every year from that Installation than you would from mining the coal. In addition, unlike in coal mining, where once you've mined out an area, you have to move on to another, in the case of Solar, the Installation would produce Energy Year after Year from the same pieces of land. If you just wanted to produce the same amount of Energy as the 2007 Coal Production, you would only have to set up solar panels on 517 Thousand Acres of land. Of course, the Installation doesn't have to be all in one place, the Panels could be distributed among small Installations all across the State (2% of the total land area of Kentucky).

Note that this post does not attempt to address price. Those calculations are elsewhere, and ongoing. However, just in terms of land use, it becomes clear that the energy content of Coal could, in fact, be replaced by an Installation of Solar, while distrupting a Third of the land area of ongoing Coal Mining Operations.

Also note that this does not address the availability of Solar Panels. This will be addressed incrementally by a growing industry.



* Depends on numerous factors, including Coal Chemistry, and Power Plant Design. The values that I used assume very high Energy Content Coal, and highly efficient, state of the art, Power Plants.

** Note that this isn't quite correct, as the Insolation values on the map are not based on a panel laying flat on the ground, but are based on a panel tilted to the South. Within the scope of these calculations, though, this should be negligible.

*** Expect Conversion Efficiencies of low cost Crystalline Silicon Solar Panels to increase significantly within the next 3 years.

Conversions:
1.5 Million Acres = 6,070,284,633 m2
Kentucky total area = 104 658 829 550 m2 = 40409 mile2

For more information on "Insolation," see "A Note on Units of Energy and Insolation".

This article is followed by http://americansolareconomy.blogspot.com/2009/02/real-world-estimation-of-land-use-per.html, which estimates the Solar Output of 1.5 Million Acres of Kentucky Land, using real Land Use data from Sunpower Corporation.

A Note on Units of Energy and Insolation.

This post is in reference to my use of units like "Watt*1Year," or "Watt*25Years," etc, in posts such as This, This, and This, and might just be useful in laying out some of the basic math behind Solar Energy Output. Feel free to critique.


In Physics, Power is described in Watts. Energy is described by Power * Time. Typically when we think Electrical Energy, we think in terms of Kilowatt*Hours, but the actual units used for Time are arbitrary, it's just a matter of the increment of time over which you are considering the flow of Power.

The Quantity of Energy streaming down on the planet can be measured in terms of its Insolation. The problem that I'm seeing out there is that it's not firmly decided what units we should be using for Solar Insolation, and there is little way to translate at a glance quantities from one choice of Unit to the next. Now, maybe there's a reason that somebody would want to use "kW·h/(m²·day)" or "kWh/kWp•y" for practical applications to Solar Energy, but the rationale certainly escapes me. What I do know is that a Solar Module is rated in Watts Peak (W_p), which is the Power Generated when the Panel is exposed to an Insolation of 1000W/m². So, to match this, I want my units to be comparable to W/m².

Following is a map of US Annual Insolation in kW/m²*.



By taking the given values in terms of kWh/m²/day, converting from KiloWatts to Watts, and multiplying each by 1day/24h to cancel out the elements of time, we get the Annual Average Power, in W/m². Once we know this Annual Average Power, then by dividing it by the 1000W/m² rated Peak Power used by the Photovoltaic Industry, we get a very useful percentage.

Example: Looking at the map, let's take a spot on one of the bright yellow areas, like is found in most of Virginia. The legend shows an Insolation Value of 4.5-5kWh/m²/day. Converting to Watts, and taking the range's lowest value of 4500Wh/m²/day, multiplying by 1day/24h, and canceling out the hours and days, gives 187.5W/m².

So, now that we have the average Rated Insolation for the location, then we divide this number by 1000W/m² in order to get the Actual Insolation as a percentage of Rated Peak Insolation, in this case, for Virginia, at 18.75%.

I've run this calculation for the various brackets in the map legend, and have added these percentages to the graphic. The spreadsheet is here.

Lets say that you want to know roughly how much actual Energy some Solar Installation will produce over a year. You just take the Peak Power rating of the Installation, and multiply by the Percentage that was calculated above, and then multiply by 1Year in order to get the Energy produced on average over that Year.

Example: You want to know how much Energy is going to be produced over the year by a 5kW_p Installation in Virginia where the expected average Insolation is 18.75% as calculated, above. Simply take the Peak Rated Power of the Installation, and multiply by the Percentage and 1Year, in this case, 5kW*18.75%*1Year = 937W*1Year.

This is the Average Energy Produced over a Year for this Installation, even though it's not in the usual units. To convert to kWh, just convert the Year to Hours using the factor of 8760Hours/Year and 1kW/1000W to get 8208kWh.


Now let's say that you want to make a comparison in Cost per Watt between a Solar Installation and a Coal Plant, or a Natural Gas Turbine, or any other conventional Electrical Generator running at a Constant Output over the year. Just remember that the total Energy Produced by a constant generator over a year, in W*1Year (or kW*1Year, or GW*1Year), is roughly it's Rated Output * 1Year, so a 100MW Coal Plant should produce in the area of 100MW*1Year in Energy over the year. We could convert this to kWh just like was done above for the Solar Installation, but there's no need to do so if we're just using it for comparisons-sake.


* This map measures Insolation assuming optimally angled panels, so for flat-roof installations, particularly at higher Latitudes, will over-estimate output. For a European Map and Insolation Values that assume flat placement of Panels, see Lightbucket.

Looking forward to 2020.

Per Wikipedia, Total World Consumption of Energy in 2005 was somewhere in the area of 15TW. It's suggested that 86.5% of this total is derived from Fossil Fuels, which amounts to a total World rate of Energy Consumption of 13TW. Note, this value includes the energy content of Oil and Liquid Fuels, so is not just the Electricity component.

Using 2005 as a rough basis:

The Total Energy Consumed over the whole Year could be written as 13TW*1Year.

There is a growing consensus in the World that the first International Targets will be in the area of 20% production of Energy from Renewables by 2020. Let's be conservative, and suggest that this target will be missed, at least on a Worldwide scale. There are several major Economies that might not play along, particularly among the heavy coal users.

We'll go with just a 15% target. 15% of 13TW*1Year = 1.95TW*1Year of renewables needed for, say, the year 2020.

Pulling a number out of my butt, let's say that Solar PV will provide just 10% of this amount of Energy by 2020. That makes for a Solar PV contribution of 195GW*1Year in 2020.

Since a Solar Panel doesn't provide constant Energy, we can take some averages, and assume that over an entire year, the panel will have provided a total Energy of about about 20% of its Peak Power Rating, so in order to provide 195GW*1Year, you'd need to install 971GWp of Solar Panels.

Hmm, 971GW of installed Solar Panels by 2020. Sounds crazy.

2007 Total Installation was in the area of 8.7 GWp. That leaves 962GWp to produce over the next 12 years.

What would this look like?

Photon Consulting put out some numbers quite some time ago suggesting what the growth curve in Solar would look like up to 2012. I took 10% off of the top from each of their yearly estimates to reflect the effects of the present slowdown, and came up with the following path to 962GWp by 2020.

The Spreadsheet is here.

For sake of completeness, I also made a more conservative scenario where the present downturn caused the Photon Numbers to be slashed by 30% over the next 3 Years. See the "Scenario 2" tab.

In the end, it makes little difference whether we slow down a bit for the moment, as the long term goal is largely set, and will almost certainly be acted on with great vigor by the Obama Administration.

Is it any wonder that I look with some scorn at the short-sighted calculations regularly drawn up by Yahoo bashers who suggest that today's Solar Manufacturers will wither due to lack of future demand for their products? The market that we're talking about is simply larger by orders of magnitude than most people can visualize, and it follows that so is the opportunity at a time when wholesale replacement of Existing Technology and Energy Sources are the order of the day.

Coal / Solar Cost Comparison - Final Draft

Note: This article is under revision, considering current fluctuations in price. The Concept is sound (IMO) as a way to make rough comparisons in cost, but the Prices are presently off.

Also note, the 33% Insolation that is used as a basis for comparison, is very high for a stationary system, but well within the range of a tracking system. For more information on Insolation, see "A Note on Units of Energy and Insolation."

I'll set up two equivalent scenarios using Coal and Solar, and will then make comparisons.

Note: I make numerous assumptions, and will mention these where appropriate.




For this thought experiment, we'll imagine that both of these industries are starting from scratch with equal Energy Production Capacities. In reality, of course, Coal has tremendous existing Scale Advantage over Solar.

This will be a demonstration of how fuel costs could affect the long term comparative cost of the Coal Energy vs. Solar Energy.

First, imagine two industries; Solar and Coal. The goal of both of these industries is to produce Energy.

Now, divide each of the industries into three groups.

Group One:

Group One is made up of those segments of the Industries that produce the actual Electrical Generation Facilities.

In the case of Coal, this is the industry that produces the actual Power Plant. It would include everything from the ground up, like the steelworks that made the metal, to the quarries that produced the Concrete. It would include the Engineers, Managers, and Laborers for the Plant Construction, as well as the Lawyers and Lobbyists required to work with the Government and Public to support the plant's construction.

In the case of Solar, this would include all of the players from TCS, Wafers, Crucibles, and Modules, through the final Solar Power Plant Installation. Once again, it would include all of the extraneous support required for the project.


Group Two:

This group is made up of everyone associated with supplying the fuel for the Power Plants that were produced by Group One, above.

- For Coal, this would include everything from the actual Mining of the Coal; the engineers, geologists, equipment operators, supervisors, etc. This group would also include the Transportation of the Coal to the Power Plant.

- For Solar, there is no Second Group. There is no Delivery of fuel to the Solar Plant.


Group Three:

This is all of those involved in the upkeep over time of the power plants. I'll ignore this group, and give Coal a freebie. I think it's safe to say that Solar will beat Coal on Upkeep Costs over time.

Notes and Assumptions:

Note: The Solar Plant is going to have to be rather larger in peak rating than the Coal Plant, since Sunlight isn't constant. 33% is a fair conversion for a very sunny place, so our Solar Plant has to be three times the rated output of the Coal Plant (Say, 350MW Coal = 1050MW Solar). Whatever actual output we settle on, we just want the total yearly output of Energy from both plants to be the same for comparison purposes.

Note: Solar does not provide a base load like Coal. We're just looking at total Energy Output, not the convenience or timing of the final product. Ultimately, for future base-loads, we'll need a heck of a grid, plus some other provider like Sequestered Coal, Geothermal, or something else like that.

Note: I have Silicon-based Photovoltaic Solar in mind in writing this.

Note: PV Solar lends itself to a decentralized solution. Therefore, when talking about a 1050 MW Installation, we don't have to assume that some company has bought 1050 MW worth of Panels and Installed them as a single project. Instead, we can talk about a total of 1050 MW of Panels installed anywhere, in any distribution. Whether Centralized, or not, a Watt of Solar Energy offsets a Watt of Fossil-based Production.

Note: The referred-to Spreadsheet is likely not entirely clear to anyone but myself. I did try to add descriptions to help, but there are a lot of numbers involved. Feel free to counter my numbers with your own if you think that I'm off on anything.

Assumption: I've worked out several Cost scenarios involving guesses on future Inflation / Coal Price Increases. Of course there's no telling how the price of Coal will vary over the next 25 Years. There are numerous reasons to suggest, however, that the price of Coal will not remain static, particularly in the face of Peak Fossil and US Dollar Depreciation. Even if Coal is not near Peak, Peak Oil will put increasing upward price pressure on Fuel to support Coal Deliveries. Nearly all of the price pressures in the foreseeable future point towards a continued Increase in the Price of Coal, particularly in the price of non-local Coal that requires long distance transport.

Assumption: No cost factors related to future Climate Change Regulations are included in this Document. This gives a huge Freebie to Coal, as Sequestration and Carbon Credits will add greatly to the cost of Energy Production from Coal Sources over the next 25 Years.

Assumption: No cost factors related to Increased Healthcare Costs due to the Burning of Coal. This is another Freebie for Coal as far as this paper is concerned.

Assumption: I assume for the initial calculations that the lifespan of the Coal Plant and the Solar Plant are equal to 25 Years. The lifespan of either a Coal Plant or a Solar Plant is certainly greater than 25 Years. I'll look back at this in a later section.

Assumption: I assume for the initial calculations that the Conversion Efficiency of the Solar Panels stay constant throughout the life of the plant. Again, I'll look back at this in a later section.

Imagine both a Coal Industry and a Solar Industry, each capable of producing a single Power Plant per year (or arbitrary unit of time, really).

Year One: Both a Coal and a Solar Plant are built.

By the end of year One, both the Coal Plant and the Solar Plant have produced one Yearly Energy Unit. The Coal Plant has consumed it's required yearly supply of Coal.

Year Two: Both a Coal and a Solar Plant are built.

By the end of this year, the Plants that were built last year, each produce their total yearly capacity in Energy. In addition, the new plants being constructed this year have each produced a Yearly Energy Unit. The two Coal Plants consume a total of 2 Units of Coal for the year.

The Total Amount of Coal burned since the first Year is 3 Units.

Year Three: Both a Coal and a Solar Plant are built.

By the end of this year, the Plants that were built in the two previous years, each produce their total yearly capacities in Energy, for a total of 2 Units of Energy from Solar and Coal Plants. In addition, the new plants from this year have each produced a Yearly Energy Unit. The three Coal Plants consume a total of 3 Units of Coal for the year.

The Total Amount of Coal burned since the first Year is 6 Units.

Now, to make some Comparisons between Coal and Solar based on the above setup.

Comparison One: Side-by-Side – Energy Output

Take a look at this spreadsheet, I'll take it out 25 Years.

See Sheet 1.

This first set just shows that over 25 years, the total Energy Output of both our Coal and Our Solar Industries are the same. Easy enough, that was part of the basic assumption.


Comparison Two: Side-by-Side – Feedstock Demand

This next set shows how the total demand for Coal Feedstock grows over time.

See Sheet 2.

Per Plant, of course, it's linear; just One Unit of Coal Fuel per Year per Plant; however, as the number of plants increases, the Total Yearly Demand for Coal for the Industry increases exponentially based on the rate of increase of demand. This is a recipe for increased cost of that fuel over time, particularly since the Coal is utterly destroyed in the process of burning; there is no recycling or conservation of raw materials.

In fact, over the first 25 years of the scenario, the yearly demand for Coal from the Power Plants has increased 25 times. Unless supply increases similarly, prices will have to increase due to the additional demand.

This is where our assumption that the Coal Industry isn't actually a behemoth in comparison to Solar comes in. Of course, the Industry is so large that an extra 25 Plants worth of Demand isn't going to stress out the Suppliers too much. However, the ability of the Coal Industry to increase supply to meet demand is not infinite, particularly since, once the coal is gone from a site, it's gone and the total production from that site has to be replaced by production from a new site. Finding new sites gets more difficult over time, particularly as International Politics and Dependence on Support from Sovereign Governments creates Long Term Complications and various forms of Blowback.

Looking at some actual Coal Consumption Numbers (See P.35), we see that in 10 years between '97 and '07, consumption of Coal increased from 2317 to 3177 (Millions of Tons of Oil Equivalent), or by 37%. According to The World Coal Institute, "at current production levels coal will be available for at least the next 147 years." They specify at "current production rates," which says to me that they are not taking into account increases in Demand / Production, as production rates would either have to increase to meet demand, or else price would have to go through the roof to take into account the discrepancy. Oddly enough, at the beginning of the writing of this paper, the World Coal Institute estimation was that we had 155 years worth of Coal remaining, but now, having confirmed my links, I see that they've updated this number to 147 Years, which means that in about two weeks of time, the World Coal Institute revised their estimate down by eight years*. For a counter opinion on the timing of Peak Coal, see this article which concludes that it could be in as soon as 15 years.


Comparison Three: Costs – Inflation Scenarios

Looking at a specific example, I'll take a look at some samples of Coal Plants, to see how much coal they each go through in a year. I've grabbed a couple of examples from the web, which gives some idea of how much coal a plant will go through, compared to its rated power output. It looks like Milliken Station on Cayuga Lake is quoted as the most efficient plant of the four that I found (in Energy per ton of Coal), so I'll use that plant as an example, and support it as within a reasonable estimation with some averages from www.powerofcoal.com.

See Sheet 3.

In fact, it appears that the fuel cost that I derive for Milliken Station is slightly above the average in the Industry. Per PowerofCoal. Working out the Cost per Watt from Milliken Station over 25 Years at $100 / ton gives $6.26/Watt*25 Years. This compares to the National Average, which works out to $5.18/Watt*25 Years. Note that since this “PowerofCoal” reference was dated, most Coal Prices have increased quite dramatically, so the national average costs have probably increased by 25% or more.

Note: Per “Checking my Numbers,” below, it appears that Milliken Station is very close to the theoretical maximum in terms of Energy Production / Ton of Coal. Therefore the PowerofCoal Numbers are likely skewed in some way, likely due to the Particularly large amount of easily recoverable Coal in the Powder River Basin in Wyoming, and possibly also due to Government incentives at some stage of the Coal Energy Production Cycle.

Over the first 25 years of this plant's life, it costs a total of around $2.6 Billion in initial Construction Costs and Yearly Deliveries of Coal Fuel. Of course, this assumes that the price of Coal doesn't increase over this 25 years, and it assumes that the plant costs nothing in maintenance. As shown on Sheet 3, if Inflationary factors are considered, total cost for this near average plant over 25 years could actually approach $6+ Billion.

For Fuel Cost Estimation for other Coal Plants, see Sheet 4.

Ok, now to look at an equivalent Solar Installation (1050MW @ 33% of Peak in Total Energy Output). There are alot of different ways to work out sample prices for equivalent Solar Installations. The first, and ugliest example would be to use the retail price data from Solarbuzz.

According to Solarbuzz, the average US Retail Price for Panels is $4.82 Watt, and the Total Cost of the Project is about Twice the cost of the PV Modules. Using this method arrives at an end resulting cost of between about 2 and 5 times the cost of an equivalent Coal Plant over 25 years (Depending on Future Inflation). At this price, the total cost of the Installation would be $4.82/W * 1MillionW/MW * 1050MW * 2 = $10.12 Billion (compared to $2-$6 Billion for an equivalent Coal Plant). Wow! Ok, but this number reflects the many inefficiencies of small-scale retail distribution and installation. It also represents the current high demand / low supply that we see in the World PV Market, reflected among other things by a cost of Polysilicon of 5-10 times (or more) the cost of its production (Polysilicon costs are around 40% of the total cost of producing Solar Panels at this time).

So, with a 25 Year window, it's tough to compare the Best-case scenario for Coal to the Worst-case scenario for Solar at present Solar Prices. We'll get back to this one a bit later.

Instead, I'll try to gauge the cost of some existing large scale PV Solar Installations. Attached you'll see a few price references to indicate the Cost / Peak Power that is currently available for mid-size Installation sizes.

See Sheet 5.

This spreadsheet shows some examples of Solar Power Plants in the real World, their output, and their projected costs. Remember, that I've chosen a 1050MW Solar Plant to be equivalent to a 350MW Coal Plant in annual Energy Output.

The Price per Watt ranges from $5.33 -$8.05. So, using this range of prices to construct a theoretical Solar Plant of 1050MW would give us costs ranging from $5.8-$8.5 Billion. Remember, this is compared to a cost for coal of my just slightly above US average Coal Power Production Costs of $2.6-$6.5 Billion.

Personally, I think that assuming future inflation to be zero is ludicrous, and can't help but think that the much safer bet is that Coal Fuel Prices will increase significantly faster in the near and mid-term future than we're used to thinking about. If this is the case, then there are cases in this scenario in which Solar Installation would be the best economic choice for installation RIGHT NOW.


Comparison Three A: Costs – Inflation Scenarios – Extended to 50 Years

We know that a Coal Plant Lifespan isn't limited to 25 Years. We know that Solar Panels are typically warranted out to 25 Years. We also know, however, that Solar Panels can last significantly longer than 25 Years. Sheet 7 gives some idea of what kind of useful lifespan we are looking at as far as Solar Panels, based on a .5% degradation in output per Year. Considering this degradation would certainly throw off the previous Calculations, so I'll consider it for this scenario. I'll also cut down the total output of the Panels by 5% due to Inverter Losses, and by 10% for High Temperature Loss. In addition, I'll take into account the Panel Output loss over that 50 Years using the Chart on Sheet 7 by reducing the overall Output by an extra 12.5%. All of this means that now, instead of needing 1050MW to equal the 350MW Coal Plant, we're going to need a 1364MW Solar Plant.

As before, using Solarbuzz, $4.82/W * 1MillionW/MW * 1364MW * 2 = $13.15 Billion for the entire Solar Installation.

Now, for the Coal Plant, we'll figure out the cost over 50 Years assuming some inflation rate. This time I'll assume a rate of 4%. See Sheet 8. It seems that assuming 4% Inflation in the price of Coal over these 50 Years, even with all of the negative offsets that I've just added to the cost of the Solar Installation, the Coal Plant LOSES with a total fuel cost of $13.4 Billion.

Remember, Solarbuzz Numbers are Retail. How much money can we save for a utility-scale operation by buying bulk? I'm going to take a wild guess.

In the real World, Trina Solar recently reported an ASP, or Average Selling Price, of $3.85 / Watt, which is relatively high relative to other Solar Manufacturers, but well below Retail. Given a direct relationship with a Modulemaker such as Trina, and the ability to buy at around $3.85 / Watt, brings the cost of our 1364MW Solar Plant cost down by $2.6 Billion to $10.50 Billion.

We can do more. Solarbuzz says that the total cost of the Installation is twice the cost of the Modules. Well, clearly this reflects the cost of Installation on the Retail Level, which will certainly be higher than the cost of Installation on a Utility Scale. It is much more challenging to do thousands of Individual Installations on unique rooftops all over a region, than it is to take a piece of land and set up a large scale array of panels. Another Efficiency factor to be found in Large-scale installations will be the savings due to an efficiently engineered wiring and electrical design. For instance, a large scale system won't need the vast number of small inverters that would be required for an equally powered Residential Distribution. I think it's pretty safe to assume that 20% in efficiencies could be found in this situation, so we work out a Installation cost per Watt of $3.85, or $2.6 Billion Dollars off of the cost of the 1364MW Installation, leading to a total cost of $7.9 Billion Dollars.

So, the results of this scenario show that over 50 Years, our 1364W Installation should compare very favorably with Coal. The Total Installation Cost of $7.9 Billion is much lower than the Coal Plant's 50 Year Cost of $13.4 Billion assuming a low low inflation rate of 4%. Is fact, just considering a low 4% Inflation Rate, the Solar Plant breaks even with the Coal Plant at 39 Years. Anything beyond this time is Icing.

A Note on Scale

So far I've been assuming that 1050MW or 1364MW of Solar panels could be even bought on the Open Market. This is a questionable assumption.

According to the Chart on Sheet 5, the total annual installation for 2007 was 2.2 GW. However, as can be seen on the same chart, the rate of increase of installation capacity (limited by production capacity) is taking off, and is expected to increase by Eighteen Hundred Percent, to 37GW Annually, in the next Four Years.

This is when things will start to get interesting, because Utility-scale Developers will for the first time ever, have the opportunity to supply large-scale plants with decreasing lead times, and at the prices that I have talked about in this document, or less.

Conclusion

Well, so far, what I've shown is that there is overlap in the long term price of a Solar Installation and Coal Installations. Much depends on the future rate of Inflation, or at least Inflation in terms of increased Price of Coal. However, given that Future Inflation is not knowable, but in today's World Economic Climate could be explosive, Solar, even at today's high prices, fills a lucrative Energy Niche as a hedge against increasing Coal prices.

As it is, Solar Producers have more than enough Customers to easily sell all the product that they can make at today's prices. Industry Production Capacity is increasing incredibly fast, though, and will likely soon outstrip demand. However, long term Coal Generation costs would indicate that a price bottom for Solar Products will arrive as defined by projections of long-term production costs from Fossil Fuels similar to what I've shown above.

In a future Post I will look at Comparisons between Solar and Natural Gas Electricity Production, which is really a much closer fit to the particular niche that Solar fills, but in this first case I wanted to compare the costs to Coal, which is typically acknowledged as the cheapest current source of Electricity.

Checking my Numbers

Energy Capacity per Ton of Coal:


Is it reasonable to assume that a Coal Plant like Milliken Station actually consumes 876000 Tons of Coal per Year in order to produce its 350MW of Power?

Per Wikipedia, Coal Plants produce approx. 2KW*Hour/KG of Coal.

I want to solve the equation (2KW*Hour/KG)*X Tons of Coal Burned / Year = 350 MW * Year.

I'll convert to Years because because I want the Annual Average to make Comparisons to. As for the Mass, I want Long Tons, which are equal to 1016 KG, and for Power I want Megawatts.

So, X Tons / Year = (350MW * Year)/(2KW*Hour/KG)

Then, X Tons / Year = (350MW * Year)/(2KW*Hour/KG*1MW/1000KW*1Year/8760Hours*1016KG/1Ton)

Finally, X Tons / Year = (350MW * Year)/(.000232MW*Year/Ton) = 1.5 Million Tons of Coal / Year. Wow, this is rather a lot higher than my estimated Coal Fuel Demand for a 350MW Plant, which makes the Solar Plant considerably cheaper in Comparison.

Let's try another estimation. A physicist friend of mine, who works in Coal, estimated for me that a Ton of Thermal Coal, when burned, produces 26 GJ of Energy (Wikipedia has it at 24 GJ/Ton (after some conversions)). Using an Online Converter, 26 GJ works out to 7.22 MW*Hour. Not all of that Energy is converted into Electricity at the Coal Plant, only between 30%-35% is typically converted with a theoretical limit at about 45%.

Using 35% Efficiency would put the Energy / Ton of Coal at 9.1 GJ/Ton, or 2.52 MW*Hour/Ton.

Using the same process as above, for a 350 MW Power Plant, this works out to 1.22 Million Tons of Coal / Year, also higher than my earlier Estimation for Milliken Station.

Let's go one step better for Coal. I've seen reference to 30 GJ per Ton and 42% Conversion Efficiency at a particular plant. I'll work out the Tons of Coal / Year for a 350MW Coal Plant under these Conditions.

30 GJ per Ton = 8.33 MW*Hour/Ton.

At 42% Efficiency in converting this energy to Electricity at a Coal Plant, we get 3.5 MW*Hour/Ton.

Calculating as above, at this incredibly efficient example we come up with Total Tons per Year = (350MW * Year)/(3.5 MW*Hour/Ton*1Year/8760Hours) = 877,000 Tons per Year. This almost exactly matches our estimation for Milliken Station. Nice!



PowerofCoal Data Check:


PowerofCoal Claim: The Average Cost of Production of all US Coal Plants (as of Jan '08) = $23.68 per MW*Hour

In Comparison Three I used this number to calculate a Cost / Watt over 25 Years of $5.18/Watt*25 Years. To do this, I did the following conversion:

Average Cost / Watt*Year = ($23.68/MW*Hour)(1MW/1,000,000W)(365Days/1Year)(24Hours/1Day) = $0.20 / Watt*Year = $5.18 / W*25Years or $10.36 / W*50Years.

Note: These numbers for PowerofCoal.com include all of the cost of production, including presumably, maintenance and upkeep. So, they should be slightly more representative of the actual costs to produce Energy with Coal in the US, however, as shown above, Milliken Station is close to the peak of Efficiency in terms of Energy Output to Coal Consumed, so in order to arrive at a lower average cost than Milliken Station, the average cost of Coal to these US Coal Plants must be much lower than $100 / Ton, or else the cost to produce Coal Energy in the US must be Subsidized. We can see from the Chart that the US does indeed have access to very cheap Coal from Powder River Basin, though from the same Chart we can also see that other Coal Sources are increasing their prices dramatically.

Using the above numbers as a starting place, and then calculating in 4% in Inflation Increases per year, gives $8.63 / W*25Years or $31.63 / W*50Years.

Additional References

Commodity Price Data (Pink Sheets)

PV Costs to Decrease 40% by 2010

China Spurs Coal-Price Surge -WSJ



* Note on World Coal Institute Archives. Based on Archived Reports :

2008 Estimated Reserves: 147 Years
2007 Estimated Reserves: 147 Years
2006 Estimated Reserves: 155 Years
2005 Estimated Reserves: 164 Years
2004 Estimated Reserves: 190 Years
2003 Estimated Reserves: 200 Years
2001 Estimated Reserves: 200 Years

Conclusion, since 2001, we've used 53 Years worth of Coal. LOL!