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I am trying to understand the challenge of decarbonising the power grid, and the equivalence between different generation technologies.

If a nuclear power plant has a capacity of 4GW and a capacity factor of 90%, how much installed capacity of wind turbines with a capacity factor of 40% would be required to match that output?

How about the same with solar pv with a capacity factor of 25%?

What area of land/sea would be required by that installed capacity of wind or pv to match that 4GW power output?

These are the numbers I get but I have been told they are ridiculously wrong. One of the worlds largest offshore sites is the London Array - 630 MW 122 sq km (47 sq mile) That works out to 194 sq km per GW installed capacity. (75 sq mile) The London Array capacity factor 36.8% So to produce 1GW average power you need to overbuild by a factor of nearly three, and would need about 200 sq miles. A large nuclear power station puts out 4GW, so equivalent power to about 800 square miles of offshore turbines.

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    \$\begingroup\$ It's not clear what you're asking – is it really "percentages" that are a mathematical hurdle for you? As it is now, this feels like homework where you didn't even make the slightest attempt. \$\endgroup\$ – Marcus Müller Jun 24 '19 at 21:35
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    \$\begingroup\$ I'm voting to close this question as off-topic because it appears to be a homework or study question with no effort to solve demonstrated. \$\endgroup\$ – Charles Cowie Jun 24 '19 at 21:41
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    \$\begingroup\$ The answers to your questions can be found in David McKay's excellent and free to download Sustainable Energy - without the hot air. It is specific to the UK but the general numbers for wind farm land usage will be applicable anywhere. Most of your question you can answer yourself if you do the maths. \$\endgroup\$ – Transistor Jun 24 '19 at 21:42
  • \$\begingroup\$ I've tried the maths myself but have been accused of being out by several orders of magnitude. David McKay is now somewhat out of date. I'm 67 years old and don't get homework any more. \$\endgroup\$ – Mike H Jun 24 '19 at 21:46
  • \$\begingroup\$ This question may be a better fit on Sustainability.SE. \$\endgroup\$ – LShaver Jun 24 '19 at 21:57
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Approximately 1,600 square miles of solar panels, perhaps in west Texas to shade the grass and cattle during midday, is enough to supply all the USA with 1,000 watts per person.

Then you need about 4 cubic miles of pumped storage, lifted up 1,000 feet, to store energy for the next 24 hours. Some states (Florida) do not have 1,000 feet or even 500 feet or even 300 feet, of convenient adjacent elevation-change, and must depend o other states for their pumped-storage.

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  • \$\begingroup\$ "all the USA with 1,000 watts per person." What's that in MW of output after factoring in capacity factor, from these 1,600 sq miles of hypothetical solar panels in Texas?. It's not clear if capacity factor was incorporated. \$\endgroup\$ – Mike H Jun 25 '19 at 3:22
  • \$\begingroup\$ I believe a 5-sun-hour/day factor was used in the calculation. Thus capacity factor for average day in USA; in USA when Air Conditioning loads hit the power-grid, the west Texas days become10 or 12 hours long based on whether the panels are tracking. \$\endgroup\$ – analogsystemsrf Jun 25 '19 at 10:15
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    \$\begingroup\$ 1000 Watts per person continuous ( = 24 kWh/day) or for one hour per day or ... . ... Expressing the energy in kWh/day seems more useful. || Lessee - SAY 50% area coverage and 15% efficiency and 4h/day annual mean and 350 million population (all assumptions are wrong :-) ) = ~~~ 7 kWh/day/person E&OE. [1600 *1.6^2 km^2 x E6 m^2/km^2 x 0.15 kW/m^2 x 4h/d /350e6 people =~7] \$\endgroup\$ – Russell McMahon Jun 25 '19 at 10:22
  • \$\begingroup\$ What does this have to do with the OPs question? \$\endgroup\$ – Voltage Spike Jul 31 '19 at 6:04

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