Climate Policy

[kandrathe]

You can relax now, though, because Pete has revealed The Truth About Solar Generation Of Electricity, and he has made it clear that he will never choose my side, regardless the amount of nonsense you use for arguments. Not many people will have the guts to question your 'debunking' now, I suspect.

It's not just Pete. Here is a wiki article on photovoltaics. Check the section on disadvantages. Volume of land required, expensive, long pay back period, components wear out, need a storage system(or grid connect) for when power is absent. I actually believe the current concentrator technology will reduce the price of solar below that of grid power, but only in areas where there are many days of good sunshine.

Odd, this is not quite the situation I envisioned, when some people told me this forum was self-regulating, and had zero tolerance for false facts

Which false facts were those? Let's go slow;

1. Energy from the Sun reaches the Earth at a rate of 1,368 W/m2 (watts per square meter). This energy per unit area is called flux. How many MW/km2 (megawatts per square kilometer) is this? The correct answer is 1,368 MW/km2<>
   
2. We will work in W/m2. About 22% of the flux (1,368 W/m2) is absorbed or reflected by the Earth's atmosphere before it reaches the ground (even ignoring clouds). How much power per square meter is left? The correct answer is 1,067 W/m2 <>
   
3. Using the incident flux of 1,067 W/m2 (from Question 2) and the given angles for theta, find the flux available at these a location 26° from the equator (South Texas), 41° from the equator (northern U.S.), 65° from below the Sun (northern U.S. in winter). Answer: South Texas: 959 W/m2, northern U.S.: 805 W/m2, northern U.S. (winter): 451 W/m2<>
   
4. A collector sitting on the Earth's surface as the Earth rotates once in 24 hours (relative to the Sun). The total distance traveled by the collector is the circumference of the circular path, but the cross section of sunlight actually intercepted corresponds to the diameter of the circular path. Therefore, by dividing the above answers by pi (3.14159265...), the results will (approximately) be the average power available over a 24-hour period. One more consideration: the above approximation does not take into account the shorter days during winter (it also would not cover longer days during summer). The result for the northern U.S. in the winter, after dividing by pi, must be multiplied by 0.8 to account for the shorter winter daylight period. Answer: South Texas: 305 W/m2, northern U.S.: 256 W/m2, northern U.S. (winter): 115 W/m2 (remember this is at 100% flux efficiency, so really only about 20% of this power can be captured.)<>
   
5. Using the value of 20% for panel efficiency, find the average power that can be harnessed for each location from above. How do your answers compare to the 1,368 W/m2 you started with? What additional factor have we ignored? Answer: South Texas: 61 W/m2, northern U.S.: 51.2 W/m2, northern U.S. (winter): 23 W/m2. The percentage of usable power is from 4.45% to 1.68%. The factor we ignored is weather.<>
   
6. The usual formula for figuring peak watts needed is to take Watt Hour per Day divided by 4. Let's not focus on the decadent Americans for this question. We'll look at the typical UK home, and assume the Americans will learn to live as frugally as the Brits. Digging around the net, I found 3300 kWh/year was a common number for the UK, or about 9041 Wh / day divided by four gives 2260 Watts. So, now using 23 W/m2 (assuming the flux value is about the same in the UK) for flux to find how much collecting area is needed for a typical UK household. Btw, average US household power needs are about 3000 W. Note also that most home are duel power, with 38% electricity and 62% fossil fuels for heat & hot water (either NG or fuel oil). We are effectively assuming we can store the energy with perfect efficiency in a battery, which is not feasible, and over long periods of time, which is not currently practical. Do you think power demands are high when available sunlight is at its most, or its least?) Answer: 2260 W / ( 23 W/m2) = 98.26 m2<>
   
7. How much solar power collecting area would be required to replace all fossil fuel electrical power supplied in the United Kingdom (the UK has 74,000,000,000 W of electricity generation capacity, of which 5% is already renewable, so use 66,600,000,000 W)? Answer: 66,600,000,000 W / 23 W/m2 = 28956521734 m2 or 11,180.1755 sq. miles (UK has 94,526 sq mi. or about 11.83% of their land area.)<>
   [st]Here are the facts; Solar is currently a very dilute source of power, which is expensive to collect, and still requires a power grid for shoring up during low generation periods. Greater efficiencies in concentrated collectors will help to make it more feasible for places like California, Australia, Egypt and Arizona. It's just not going to work very well in the places where many people live that really need more power.