### Solar Power vs. Coal

He forgot that was the power and energy required to manufacture the cells. Now they need to be installed for another half that amount.

Why don't we just say that Solar PV panels are 10,000% efficient and it makes it so much easier?

"Mauried" wrote in message

On Wed, 17 May 2006 05:55:34 GMT, "JoeSP" wrote:

The Solar Resource

Could solar energy power the United States? To answer this question, let's compute how much land area would realistically be required to produce that much energy from Solar.

Then, for comparison, lets compare that to the land area we would need to mine coal to supply just our electricity (about 1/3 of our total energy usage - the rest being from oil and natural gas) over a period of fifty years (fifty years being roughly the time needed to make a complete transition to renewable energy sources).

The simple calculations done here are done explicitly, so that everyone may see the details and reproduce the results themselves.

Let BTU represent British Thermal Units, m2 represent square meters, kw represent kilowatts, hr represent hours, d represent days, yr represent years. Therefore, for example, the quantity "1 kilowatt hour per day per square meter" would be written as 1 kw-hr/d-m2. It is also handy to know that "kilo" means 1000 or 103 in exponential notation, one million is 106 , and one billion is 109.

Fact: The total energy usage of the US, including electricity, oil, natural gas, nuclear, and renewables, is currently approximately 1017 BTUs/year.

Let's assume that the efficiency of the solar collectors is 30% (nowadays a very realistic figure - recent photovoltaic cells are achieving 34% for example), and that we receive eight hours of sunlight per day at about 1 kilo-watt per square meter. Thus we can capture .3 kilowatts per square meter for eight hours, which means we can collect 8 hours x .3 kilowatts = 2.4 kw-hr/d-m2.

(Incidentally, an energy efficient home uses about 10 kw-hr/d or less of electricity. This implies that we would need 4 square meters or less of 30% efficient photovoltaic panels. This is much much less than the roof area of a home, so simply from this one can see that solar should be sufficient. But lets continue on anyways) Over the whole year we capture 365 d/yr x 2.4 kw-hr/d-m2 = 876 kw-hr/yr-m2.

Now, one square kilometer is equal to one million (106) square meters. By multiplying the figure above by one million and adjusting the exponents correctly, we find that over one square kilometer we capture 876 million kw-hr/yr-km2. If we know how to related BTUs and kilowatt-hours, then we can finish our calculation. One BTU (British Thermal Unit) is equal to 1055 Joules. A Joule is the amount of energy delivered by a 1 watt over one second. The term kilowatt-hour means the energy delivered by 1 kilowatt over 3600 seconds, or 3600 kilo-Joules. Dividing this by the factor 1055 Joules/BTU, we find that one kilowatt-hour is about 3.41 kilo-BTUs.

So, the yearly energy usage of the US, in kilo-watt hours, is 1017 BTUs divided by 3.41 kilo-BTUs, or 2.93 x 1013 kw-hr/yr.

Dividing this by 876 million kw-hr/yr-km2, we find that we need 3.34 x 104 km2. This is an area of 334 by 100 kilometers.

One square mile is equal to 2.58 square kilometers, so:

We find that we would need an area of 1.29 x 104 square miles, which is an area 100 miles by 129 miles, to completely power the US.

How does this compare to coal?

Facts: The average thickness of a coal seam is about 1 meter, the density of coal is about 1.1 gram per cubic centimeter, and the energy contained in one gram of coal is about 30 kilo-Joules/gram.

The average thickness of 1 meter means that we have about 1 cubic meter of coal per square meter of land area. One cubic meter is one million cubic centimeters. Therefore, the density of 1.1 gram per cubic centimeters implies that we have about 1.1 106 grams of coal per square meter of land area. Multiplying this by the energy density of 30 kilo-Joules/gram then implies that we obtain about 33 billion Joules per square meter of land area from coal.

Because a kilo-watt hour is 3600 kilo-Joules, dividing this into the previous figure means that coal yields about 9100 kilo-watt hours per square meter, or about 9100 million kilo-watt hours per square kilometer.

Dividing this into 1/3 of the US usage of 2.93 x 1013 kw-hr/yr (the 1/3 coming from the fact that we just consider electricity here), we find that we need to mine 1066 square kilometers per year. Dividing this by 2.58, we find that we must mine 413 square miles per year.

This in turn implies that over 50 years we must mine 20,670 square miles.

In conclusion, we find that over a fifty year time period, to produce just our electricity from coal, we must mine over twice the land area needed to completely power the US with solar alone!

Stolen from: NMSEA is a chapter of the American Solar Energy Society NM Solar Energy Association 1999

Unless Im missing something , if you are going to completely power the US from solar , then where does the power come from at night and when the sun isnt shining?