### Can the UK get 100% of its electricity from wind farms?

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For some reason the most popular blog post I have written is one that looked at the number of wind turbines it would take to power the UK. There is clearly a market for this kind of thing, and based on my google referrals there is an audience for a post on whether the UK could get 100% of its electricity from wind farms. So, here goes.

If the UK was not full of NIMBYs or if wind power had a much higher power density, then plastering the UK with enough wind turbines to provide the equivalent of 100% of UK electricity or energy demands might be doable. It probably isn’t. But here, I will just assume that we can stick wind turbines wherever we want.

Instead I’ll consider a rather obvious problem: the wind does not always blow, and whether this means 100% wind power is impossible. Of course some people imagine that if it’s not windy in one place, it will be windy some place else. Well, the data indicates that such talking points need to be reconsidered.

You can see this simply be picking a random month of output from UK wind farms. For example December 2011:

So, if we want 100% wind we will need to find a way of storing it, but can we find enough. David MacKay has convincingly argued that with existing technologies there is not enough storage available. Things may change, so for our purposes I will start of by assuming that we have access to one rather large battery that is capable of storing and outputting a day’s worth of UK peak electricity demand, i.e. 60 GW. So, we have 1440 GWh of storage. I will also assume very generously that this battery is 100% efficient. So, 1 GWh of excess wind farm output can be stored and then fed to the grid later with no losses.

Here is my simple electricity grid: the only source of power is wind farms and batteries. If wind farm output is greater than electricity demand then it is stored in the battery. Of course once the battery has 1440 GWh of juice stored, it will take no more. And the second wind power output goes below electricity demand and the battery goes flat you get a blackout.

How would this work? I’ve used historic data from October 2012 to see how many days we could go without the battery going flat.

There was roughly 5 GW of wind farm capacity visible to the grid in October 2012. UK average annual electricity demand is 40 GW. So, again I’ll be generous and have a total of 250 GW of wind farms on the grid. And we can use the output of the 5 GW of wind farms in October 2012 to see what the output of 250 GW would look like. Their total output should be almost double total electricity demand. Our wind farm output in October will then look like this:

In contrast electricity demand in October 2012 looked like this:

Given that demand never goes much below 25 GW, it appears that things might get rather dicey around the 9th of October. And this is exactly what happens: our battery runs flat in the middle of October the 9th. Here is what happens to the amount of energy stored in the battery throughout October:

So, 1 day of storage and a vast excess of wind power is not enough to get us to 100% wind. What if we were able to store 2 days of peak electricity demand? This looks a bit better. It can get us through the final three months of 2012 without our battery running flat:

What about the entire year? Unfortunately things come undone sometime towards the end of summer:

2 days of storage did not work. How about 3? Well, this manages to give us a system that will work all year long:

So, 100% wind powered electricity is possible in the UK. But we got there by building wind farms that produce two times more electricity than is consumed within the UK. We also need the ability to store three days of peak electricity demand. And that 100% efficiency assumption for storage is not going to reflect reality.

None of these things have any chances of happening. On the storage side we actually need the ability to store more electricity than probably exists on the entire planet currently. The technology breakthroughs will need to be profound. And the idea that we can cover the required amount of land in turbines is fanciful at best. Remember these simple statistics: the UK consumes on average 1.2 watts per square metre of land. Wind farms can provide us with 2.5 watts per square metre, maybe significantly less. For whatever reason the British like to use the area of Wales as a unit, “a region of rainforest the size of Wales is lost every year.” Well, 100% wind energy will require something like six Wales’ worth of land to be plastered by wind turbines. These land considerations should make what I have written above seem rather academic.

So, our original question: Can the UK get 100% of its electricity from wind power? Well, there only seems to be one answer, and that is no.

## 16 thoughts on “Can the UK get 100% of its electricity from wind farms?”

roddycampbell said:
July 14, 2013 at 1:31 pm

‘UK consumes on average 1.2 watts per square metre of land. Wind farms can provide us with 2.5 watts per square metre’. roughly 1:2. Which matches your 6 x Wales, Wales being 1/12 of UK.

At the moment wind is doing c. 6% of UK electricity? Which would imply 3% of surface covered with wind farms. Is that what we have, including offshore surface area?

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Thomas Gerke said:
July 15, 2013 at 5:15 pm

Modern wind turbines easily produce 6 watts per square meter.

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Robert Wilson said:
July 15, 2013 at 5:18 pm

Thomas

Can you please some evidence to back up this assertion? And if it is true, then why are the offshore wind farms being built in the North Sea so large?

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Thomas Gerke said:
July 16, 2013 at 7:01 am

Simply use the “power per windmill / land area per windmill” formular from McKays book, but enter real world values instead of simplistic theorethical half-knowledge.

A modern 3 MW low-wind optimized wind turbine at a capacity factor of 40% generates 1200000W of power. It has a rotor diamitor of 114m and requires a circular spacing of 4d. (160k m²)

The result is 7,5 W/m².
(It more than 6 W/m² in Germany due to the less ideal wind conditions compared to the UK)

Even with Mr. McKays theoratical 5d square spacing it would still be 3.7 W/m² – that’s 50% more than what you call the “maximum”.

Or do you want to tell me that a moden 3 MW windmill generates just 3.5 GWh per year? (2.5 W/m²)
That would be a capacity factor of 13%.

I am not anti-arithmetic, I am just pro-reality.

So could you please stop sourcing most of your wisdom from a popular science book that seems to be stuck in the faulty 1990s conventional (academic) wisdom?

————-
In addition there’s the revised onshore wind energy potential study recently released by the Federal Enviromental Agency. It’s conclusion was that 13.8% of the German land area are theoretically suited for wind power generation (taking landuse & minmum distance regulation into account).

According to that study (based on a detail GIS & simulation) the current potential for wind energy in Germany is 1.19 TW generating 2,898 TWh of electricity (capacity factor 28%). Again, on “just” 13.8% of the land area.

When one would take only the best locations (capacity factor above 30%) into account, only 5.2% of the land area would be enough to install 450 GW generating 1,119 TWh of energy.

It’s obviously a potential study that can not be directly translated into reality, but it still shows what’s possible today.

UBA Study (german)
http://www.umweltdaten.de/publikationen/fpdf-l/4467.pdf

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Robert Wilson said:
July 16, 2013 at 9:39 am

Thomas

Your arithmetic here is incorrect and I am not going to explain why because I do not believe it is worth my time.

As for MacKay’s calculation, no it is not theoretical. He based it on data from a large number of existing UK wind farms.

And try reading some scientific papers on the power density of wind farms. You will find that no mainstream paper claims it is close to 6 W/m2.

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Thomas Gerke said:
July 16, 2013 at 10:40 am

You are a chicken Robert.
I gave you a calculation and a source and you say “it’s wrong because my book told me”.

Would you at least recognize that the result of McKays power-density calculation heavily depends on the size of the spacing area?

In his book McKay states this about the neccessary spacing:
“Experts say that windmills can’t be spaced closer than 5 times their diameter without losing significant power. ” No Source given, nothing… scientific hear-say?

I used information from a recent scientific study (based on current real world data) that shows that his assertion for what is a minimum spacing area is utterly wrong.

So are you telling me that you choose the information of a small section in a popular science book over the results of an extensive study that looked into the potential of wind energy? Simply because you prefer somebody telling you that wind power can not supply all the energy in the UK?

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Robert Wilson said:
July 16, 2013 at 11:04 am

Thomas

I am not a chicken. It’s just that the idea the UK can get a power density of 7.5 W/m2 from wind farms is not worth arguing against. MacKay’s calculation is based on actual wind farms. He found the area of the wind farm and the output. And divided one by the other. Also just google some big offshore wind farms. Look at their areas. If they are getting 7.5 W/m2 then they should have capacity factors of over 100%.

But again, I have better thing to do with a “pro-reality” ideologue.

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roddycampbell said:
July 16, 2013 at 3:17 pm

Is 2.5W what the UK is achieving? London Array is 630MW x 30%(?) = 189MW / 100sqkm = 1.89MW psk = 1.89 wpm. Maybe they spread it too thin, or I’ve got CF wrong, or both, but that’s a lot closer to 2.5w than 6w?

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Robert Wilson said:
July 16, 2013 at 3:19 pm

London Array is expected to have a capacity factor of 39%, slightly higher than the current offshore average of 35%.

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Roddy Campbell said:
July 16, 2013 at 3:22 pm

sorry delete my repetition. thanks. so very close to the 2.5 W/m2, in fact bang on.

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Roddy Campbell said:
July 16, 2013 at 3:20 pm

London Array http://www.londonarray.com/the-project if it gets a capacity factor of 30% is 1.89 w/m2.

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Thomas Gerke said:
July 16, 2013 at 5:26 pm

Look, the formular used by David MacKay has two significant modificators that are relevant.
Average Windspeed and the required spacing. Both are variables and not constants.

For example average windspeeds of 9m/s aren’t that uncommon at 120m in the UK.
Do the very same calculation with 9m/s and you get 7.3 W/m².

Same goes for using a different spacing area.(5d*5d not that common for onshore wind)

Absolutly no rocket science.

[hope this gets posted, my other replies were not]

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roddycampbell said:
July 19, 2013 at 3:04 pm

Thomas, if the London Array with an expected Capacity Factor of 39% (according to Robert) does 2.5 W/M2, and you maintain 6W+ is do-able, then the Array have done something terribly stupid or inefficient? Or are you saying they could in fact have built over double the density of turbines, but chose not to for some reason?

I’d like to know why they seem to have missed what you say isn’t rocket science.

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[…] in winter with very low wind speeds where does No Dash for Gas expect we get our electricity from? Energy storage is almost certainly not going to help us, while interconnectors appear to off some, but not much help. There can be occasions when there […]

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John Ashcroft said:
July 13, 2014 at 4:21 pm

I have never been pro wind power, the variability looks a big problem, and it is. However I was interested in your calculations. It doesn’t actually look like you need to store that much power… Even if the round trip efficiency was 75%, the size would be what ?.. 5-6days?..

This doesn’t sound enough, .. And though it is huge by current technology there is probably an answer… PHES , by isentropic ltd looks like it may work.

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john ashcroft said:
November 2, 2014 at 6:07 pm

looking back over previous comments, I suspect the amount of wind power available/ m2 of area will be higher today than in the 1990s, and will be greater in the future.
The reason is that tubine blades are bigger, and they are being placed on higher towers. Both will increase the output per tower, but also per unit area.

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