3 square miles is 7.77 square km. Just over 1 kW of sunlight falls per square meter, so I will round up and say that there are 8 GW falling on the plant total.
So 280 MW of production is 3.5% efficient of peak irradiance.
It looks like 5 kWh/m2 for Arizona, so that's 38.85 GW per day irradiance on 7.77 sq km. Assuming the plant can put out its maximum capacity for the 18 hours it runs each day, that's 280 times 18 = 5 GW per day.
So it's operating at almost 13% efficiency of average irradiance.
That is very good, considering that they are able to store heat much more cheaply and environmentally friendly than storing electricity in batteries.
The most important thing of all is that it doesn't have to wait for innovations in photovoltaics. Anyone can buy land, set up parabolic mirrors, and tinker with forgotten heat engines like Stirling cycle engines or Tesla turbine engines that easily achieve 30, 40, 50% efficiency and approach the Carnot limit. Conversion of motion to electricity is a solved problem at 95% efficiency.
Oh and these plants can be supplemented by biomass, say biodiesel from algae or fuel pellets made of hemp. If that's too granola for the fossil fuel industry, they can also use natural gas.
In fact if you study this long enough, you find that there are only two real hurdles: energy storage and connection to the grid (made difficult because of resistance from established utilities). Generation turns out to be relatively inexpensive because there's a sea of free energy all around us. To put it in perspective, Grand Coulee dam puts out less power than the irradiance falling on the solar plant. It's just more efficient at converting the motion of falling water to electricity:
It puts out 6.8 times 24 = 163 GW/day. That's 33 of these solar plants. So every 100 sq miles (260 sq km) of desert is equal to the 324 sq km of area flooded by Grand Coulee.
Solar thermal is the hydroelectric of the future and uses less land, which will only improve going forward. IMHO this will someday dwarf wind and nearly eliminate intermittency issues.
> (...) that's 280 times 18 = 5 GW per day. (...)
> (...) It puts out 6.8 times 24 = 163 GW/day. (...)
GW (GigaWatt) is already energy over time (Joules per second), you don't multiply it by the number of hours a day (why hours, why not seconds or anything else for example). Unless you're counting multiple plants you're adding, but that's not the case here.
edit after finishing this post: actually, you seem to be using W(att) for both W (watt, a unit of power) and Wh (Watt-hour, a unit of energy), which, besides being wrong, leads to confusion. Your post is very coherent under this new light.
So this :
> It looks like 5 kWh/m2 for Arizona, so that's 38.85 GW per day irradiance on 7.77 sq km. Assuming the plant can put out its maximum capacity for the 18 hours it runs each day, that's 280 times 18 = 5 GW per day.
Actually becomes :
5 kWh/m²/day -> 5/24 kW/m² (since there are 24 hours a day and a kWh is the energy of 1 kW power over 1 hour), so 7.77 x 1000000 x 1000 x 5 / 24 Watt = about 1.6 GW for 7.77 km².
And assuming the plant can run at 280 MW for 18 hours per day, its average power output is 280 x 18 / 24 = 210 MW.
Unfortunately I'm unable to edit my post now because it's too old. For anyone finding this, cataflam is partially right, I had some typos but the math is the same. I stand by using total power output per day to calculate efficiencies. cataflam's right that over a 24 hour period, you could say the average output would be 210 MW. Here is the corrected version:
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3 square miles is 7.77 square km. Just over 1 kW of sunlight falls per square meter, so I will round up and say that there are 8 GW falling on the plant total. 280 MW over 8 GW = 0.035.
So 280 MW of production is 3.5% efficient of peak irradiance.
It looks like 5 kWh per m2 per day for Arizona, so that's 5 kWh per m2 times 7,700,000 sq m = 38.85 GWh per day irradiance on 7.77 sq km. Assuming the plant can put out its maximum capacity for the 18 hours it runs each day, that's 280 MW times 18 hrs = 5 GWh per day. 5 GWh over 38.85 GWh = 0.1287.
So it's operating at almost 13% efficiency of average irradiance.
That is very good, considering that they are able to store heat much more cheaply and environmentally friendly than storing electricity in batteries.
The most important thing of all is that it doesn't have to wait for innovations in photovoltaics. Anyone can buy land, set up parabolic mirrors, and tinker with forgotten heat engines like Stirling cycle engines or Tesla turbine engines that easily achieve 30, 40, 50% efficiency and approach the Carnot limit. Conversion of rotational motion from a turbine to electricity with a generator is a solved problem at 95% efficiency.
Oh and these plants can be supplemented by biomass, say biodiesel from algae or fuel pellets made of hemp. If that's too granola for the fossil fuel industry, they can also use natural gas.
In fact if you study this long enough, you find that there are only two real hurdles: energy storage and connection to the grid (made difficult because of resistance from established utilities). Generation turns out to be relatively inexpensive because there's a sea of free energy all around us. To put it in perspective, Grand Coulee dam puts out less power than the irradiance falling on the solar plant. It's just more efficient at converting the motion of falling water to electricity:
It puts out 6.8 GW times 24 hrs = 163 GWh/day. That's 33 of these solar plants. So every 100 sq miles (260 sq km) of desert is equal to the 324 sq km of area flooded by Grand Coulee.
Solar thermal is the hydroelectric of the future and uses less land, which will only improve going forward. IMHO this will someday dwarf wind and nearly eliminate intermittency issues.
So 280 MW of production is 3.5% efficient of peak irradiance.
BUT average irradiance is smaller than peak:
http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/redbook...
It looks like 5 kWh/m2 for Arizona, so that's 38.85 GW per day irradiance on 7.77 sq km. Assuming the plant can put out its maximum capacity for the 18 hours it runs each day, that's 280 times 18 = 5 GW per day.
So it's operating at almost 13% efficiency of average irradiance.
That is very good, considering that they are able to store heat much more cheaply and environmentally friendly than storing electricity in batteries.
The most important thing of all is that it doesn't have to wait for innovations in photovoltaics. Anyone can buy land, set up parabolic mirrors, and tinker with forgotten heat engines like Stirling cycle engines or Tesla turbine engines that easily achieve 30, 40, 50% efficiency and approach the Carnot limit. Conversion of motion to electricity is a solved problem at 95% efficiency.
Oh and these plants can be supplemented by biomass, say biodiesel from algae or fuel pellets made of hemp. If that's too granola for the fossil fuel industry, they can also use natural gas.
In fact if you study this long enough, you find that there are only two real hurdles: energy storage and connection to the grid (made difficult because of resistance from established utilities). Generation turns out to be relatively inexpensive because there's a sea of free energy all around us. To put it in perspective, Grand Coulee dam puts out less power than the irradiance falling on the solar plant. It's just more efficient at converting the motion of falling water to electricity:
http://en.wikipedia.org/wiki/List_of_largest_hydroelectric_p...
It puts out 6.8 times 24 = 163 GW/day. That's 33 of these solar plants. So every 100 sq miles (260 sq km) of desert is equal to the 324 sq km of area flooded by Grand Coulee.
Solar thermal is the hydroelectric of the future and uses less land, which will only improve going forward. IMHO this will someday dwarf wind and nearly eliminate intermittency issues.