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I'm trying to see what percentage of total electricity demand will be met by solar power in some year x. Let's say I have data saying that the total electricity demand in country A is 1000 billion units (BU) in 2025. Let's also say that the installed solar power capacity in that country in 2025 is 200 GW.

How much electricity will these power plants with a cumulative capacity of 200 GW produce? I also have information that solar power plants in these countries run with an efficiency of 20%, but I'm just confused. What does a 200 GW power plant actually mean (how much electricity will it produce, say, in an hour?) and if the efficiency is 20%, how does it change my analysis?

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    \$\begingroup\$ Plants are rated by their output power. Please note that solar plants only put out power when the sun shines. In order to meet demand when the sun is not shining, plants must be accompanied by some type of energy storage system. \$\endgroup\$ – mkeith Feb 28 '18 at 7:24
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    \$\begingroup\$ The important number for renewables is not "efficiency" but "capacity factor". That tells you how to derate their "nameplate" (maximum) capacity to a reasonable estimate of how much will be produced over a year. \$\endgroup\$ – pjc50 Feb 28 '18 at 12:26
  • \$\begingroup\$ Thanks for the note on capacity factor. I now understand efficiency is irrelevant to knowing just how much energy will be generated given capacity. \$\endgroup\$ – WorldGov Feb 28 '18 at 18:57
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A power plant rated at 1GW can produce 1GW of power, at the rated conditions.

If it has an efficiency of 20%, then it will be consuming 5GW of energy in some form to do that.

If the power plant is (say) thermal steam, then the calculations are fairly easy, because we can assume that it can do this continuously, as long as fuel arrives. It will generate 1GWh of energy in 1 hour. Note that 20% efficiency is pretty poor, archaic even, for a combustion based plant, but might be reasonable for geothermal (low temperature) sources.

If the power plant is solar, then

(A) it's weather dependent (don't work well in cloudy weather)

(B) time of day dependent (don't work as well when the sun is low in the sky) and

(C) the efficiency is less relevant than for fuel burning plant, as it gets whatever sun it does for free all the time (subject to a and b above), though cost of plant, installation and real estate to deploy it will increase with lower intercept efficiency, so that will continue to be improved by manufacturers

We come back to the rating conditions. Is that 1GW at solar max, best time of day? Or is it average between the hours of (say) 10am and 4pm? For what the 1GW means, you will need to read the fine print for what it really means.

Once you know what it means, as a function of time of day, and as a function of seasonal weather, then you can integrate the amount of energy that the solar plant will produce over a day, or a year.

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  • \$\begingroup\$ Just a side note: Solar efficiency is not a useless spec. Based on irradiation charts combined with efficiency specs, you can make production predictions. Even with a simple spec such as solar efficiency, which is oversimplified, it's useful on a higher level. \$\endgroup\$ – gommer Feb 28 '18 at 9:07
  • \$\begingroup\$ @gommer I don't think I said it was useless, just implied it was less relevant than for fuel burning plant, though may update to make that more explicit. However, I just read an article that said that with the Chinese reduction in cost of cells, half of a typical plant cost was physical installation, so higher efficiency was more important to cost of plant and real estate than I thought. I might update to reflect that as well. \$\endgroup\$ – Neil_UK Feb 28 '18 at 9:25
  • \$\begingroup\$ especially don't work well when the sun is below the horizon. \$\endgroup\$ – Jasen Feb 28 '18 at 10:40
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    \$\begingroup\$ @Jasen - we just need to launch some biiiig mirrors ... and hope that the neighbors don't want to sleep at night. \$\endgroup\$ – brhans Feb 28 '18 at 12:56
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    \$\begingroup\$ Or just launch the panels into space and hope the microwave beam doesn't miss the receiver. \$\endgroup\$ – Phil Frost Feb 28 '18 at 14:48
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It depends on the letter(s) after GW.

A 1GWe plant produces 1GW of electrical power. At 20% efficiency, it will have to get rid of 4 GW heat.

You will sometimes see 1GWth - that produces 1GW of thermal power; as you have told us its efficiency is 20%, it'll produce 200MW electrical power (200MWe).

But no practical thermal plant of that size will have efficiency below 30% - some (like AGR) are around 40% as opposed to PWR in the 30-35% range, and CCGT can easily exceed 50%.

Without a suffix, you can assume it's rated by its electrical output.

Solar power is rated a little differently, but again its rating is its electrical output under optimum conditions, so a 1 GW plant (with 20% efficient solar cells) is intercepting 5GW of sunlight and producing 1 GW of power. That means, 200GW capacity will produce 200GWh in one really good hour.

But that's not the whole story, because that 200GW capacity doesn't reflect the power you'll get all day every day. Factoring in night-time, cloudy days, weaker morning and evening sunshine etc, a solar plant may only operate at something like 20% of its actual capacity, averaged throughout the whole year.

And I have to wonder if this (more properly called 20% "availability" or "capacity factor" as pjc50 says) is what your "20% efficiency" is really referring to? I suspect it is.

This means that a country with a 200GW power demand cannot meet it with a 200GW solar installation alone.

Grid planning is a fascinating problem anyway thanks to demand variability during the day; adding variable supply to the mix doesn't fundamentally change this but modifies the constraints. Currently, solar (in the UK) is helping, as peak sunlight coincides with daytime demand, leaving two smaller peaks morning and evening, doubling the value of storage (pumped hydro storage, and batteries in future).

More advanced solar users (Germany) sometimes see the daytime spot price go negative, as supply exceeds demand. This is more of an opportunity than a problem ... watch solutions emerge (not only storage, but also time-shifting big loads) to utilise a free or negative price resource.


Now, what can we get from your numbers?

1000 billion units (kWh) = 10^12 kWh over 8000 hours (a very approximate year!) gives us a mean demand of 10^12/8 W = 120GW.

Demand will be lower at night and higher during the day - say, 80GW and 160GW respectively. Therefore there is enough solar capacity alone to exceed likely daytime demand during (probably rare) peak solar periods.

There will normally be some storage capacity to consume some of the excess, helping fill the morning/evening demand peaks, and some other renewable power generation (wind, hydro) to cover part of the night demand. And probably thermal generation to fill in otherwise.

Best case : assuming enough storage to consume peak solar... solar provides 200GW * 20% capacity factor * 8000 hours = 320 billion units or 32% of your annual demand.

Close to worst case : assume half the operating hours are limited by demand to 160GW, so 80% of nameplate capacity. One way to account for this would be factor this in to the capacity factor as (10% + (10% * 0.8)) = 18% capacity factor.

Then solar provides 200GW * 18% capacity factor * 8000 hours = 288 billion units or 29% of your annual demand.


Better inputs (like actual demand curves, storage capacity, and knowledge of other sources) will refine these crude estimates, of course. But hopefully this is good enough to get you started.

You can make some conclusions even from a simple analysis like this : for example, if you have this much solar power, you need charging points wherever you park your car during daylight hours. Which means, for many people, at the workplace.

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How much electricity will these power plants with a cumulative capacity of 200 GW produce?

200 GW will be the peak output power and for solar this will be at optimum solar conditions.

I also have information that solar power plants in these countries run with an efficiency of 20%, but I'm just confused.

So they will take 1000 GW of solar energy and give out 200 GW of electrical energy.

What does a 200 GW power plant actually mean (how much electricity will it produce, say, in an hour?) ...

200 GW solar power plant will give 200 GWh in one hour if all the available supply is consumed.

... and if the efficiency is 20%, how does it change my analysis?

I don't see your analysis but the energy provider can only charge for what is produced and consumed no matter what the efficiency is.

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The power of a plant is its maximum power, what it can deliver to the grid. You have nothing to do with the efficiency. It just means that with technological improvement in the future a solar plant with the same area can produce more electrical power under the same circumstances.

Note, that solar systems does not always operate at their maximum efficiency.

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Capacity is peak sun while Efficiency may include Solarity.

Make sure someone dont confuse Availability with Efficiency or Average Monthly capacity.

Solar power peak is capacity at full sun. Geographical solarity figures factor weather that block sun power and useable daylight time. On average 12hr is 50% efficiency of peak capacity. Then there are losses for directional losses as efficiency using a curve called Lambertian for radiant or absorbant efficiency vs angle. Or the cost of motor energy to reposition panels when useable sun exists.
Then there are storage and conversion losses if that exists.

Nuclear reactors are about 50% efficient becuase they always run at 100% power and most is lost in heat transfer efficiency (steam) to turn turbines.

In Australia, most of the power is needed to stablize the grid with Tesla batteries and only 10% is sold, but that avoids frequency errors, surge failures, downtime from insufficient capacity or worse burnt transformers in power substations... like the huge number of high quality transformers that burn out or totally destroyed last month in Nigeria from insufficient capacity and poor instability protection management skills.

If solar power cost is 4 cents per kWh and if coal power cost is 4.1 cents /kWh

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