How big would a wind farm need to be to power London?

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I get rather frequent traffic to a piece I once wrote called “How many wind turbines would it take to power the UK?”

There are multiple varients of the google search that have led people to reading the piece. Most strange are those who interested in finding out how big a wind farm would need to be to power a fun fair. Don’t ask.

Others are clearly searching google to find the answers to a question asked by a high school teacher. The sudden influx of almost identically worded google referrals is the giveaway.

But one question that people appear to want to know the answer to is of genuine interest, and the answer is very uninformative of the deep challenges we would face if we attempted to transition to a be a predominantly wind powered society.

How many wind turbines would it take to power London? Or perhaps more revealing: how big would a wind farm need to be to power London? When I mean I power here, I use the word colloquially, meaning electricity, and not scientifically. Enough people seem to want to know that it would be worth doing the basic calculations.

So, here goes.  Roughly eight million people in Greater London, which covers an area of around 1.5 thousand square kilometres, or 600 square miles if you prefer long defunct units of measure.

According to DECC, annual electricity consumption in Greater London is just under 40, 000 GWh. This is, on average, 4.5 gigawatts.

This could be supplied by a couple of big coal or nuclear power plants. A typical nuclear power plant is made up of two 1.3 or so GW units, which have capacity factors of around 90%. So, London could be supplied by a couple of regular sized nuclear power plants.

But what about wind farms? Here we have the problem of low power density. Supplying London’s electricity with a nuclear or coal power plant requires you to build a big centralised unit that can be seen by more or less no one, except travellers on the occasional mainline railway. They just need a few square kilometres of land and can be tucked out of sight.

Not so with wind farms, which take up vast amounts of land. Powering London with a wind farm would require more land than London itself. The numbers are quite clear, and they are best understood by thinking in terms of power density.

Power density is a much under-discussed issue. Put simply, power density is a measure of the spatial requirements of a source of energy. We measure this in watts per square. It is rather like the way we measure crop yields. “Bushels per acre” or “tonnes per square kilometre” or whatever takes your taste. When it comes to coal, oil, natural gas or nuclear power, we don’t think to much about these things. They don’t take up too much space. Not so with some renewables.

What is the power density of London’s electricity consumption? On average it is just under 3 watts per square metre. It will, of course, vary a bit during the year. But that is a detail?

Wind farms however supply less than this throughout Britain. Using data for wind farms throughout Britain David Mackay has calculated that the power density of the average British wind farm is 2.5 watts per square metre.

In other words, per unit of land, London consumes more electricity than can be provided by wind farms; providing all of London’s electricity with wind farms will require an area greater than London itself to be covered in wind turbines.

london100

This is the true reality of so-called “community renewables” in Britain. An essentially urban society cannot be powered by local, community energy. It is hard physical reality, and no amount of wishful thinking can result in the laws of physics being cast aside.

And we must ask what kind of society the various promoters of community renewables think we live in. A Google image search of “community renewables” is revealing. Pictures of wind turbines and flowing green fields. Everyone  is white, happy and appearing to live a traditional British life. One must ask why, if these images are accurate,  Ukip dislike wind farms so much.

And I have news for these people. We stopped living in villages more than a century ago, and there are no signs we are going to start moving back. The city is here to stay and it is time to make that work. London, Manchester, Glasgow, Birmingham, Leeds, Liverpool; these are the communities we now live in, not the villages of community renewables enthusiasts imaginations.

Powering humanity, then, means powering cities. This will require big scale things. Big wind farms, big nuclear power plants, big carbon capture and storage facilities. Big scale things may be out of fashion, but necessity and fashion are separate things.

And this stuff has already begun. The London Array wind farm covers 100 square kilometres of sea in the English Channel. 100 square kilometres. This seems a lot, yet it only provides the equivalent of 6% of London’s electricity demand.

So, next time you hear someone getting enthusiastic about community renewables, think about the communities we actually live in, and how big a wind farm would need to be to power one of them.

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19 thoughts on “How big would a wind farm need to be to power London?

    michaelroberts4004 said:
    December 11, 2014 at 10:31 pm

    Reblogged this on Peddling and Scaling God and Darwin and commented:
    That will change the face of London

    Like

    Mark Miller said:
    December 12, 2014 at 2:11 pm

    Robert excellent visualization of how much land is needed to power London’s electrical energy needs via wind turbines.

    Out here in normally sunny CA, the 5” of rain in my rain gauge for the last 24 hours will help in filling up Folsom dam, our transportation sector of the economy is the sector that is causing the most trouble in terms of decarbonizing our economy.

    We are in the early phases of adapting EV’s in CA. The UK’s automobile and truck fleet are more efficient than CA’s and the miles driven per vehicle are less hence I was wondering if you have estimated the land area needed for turbines, or PV farms, to power say 20% of the transportation sector of the greater London area?

    Like

      Robert Wilson said:
      December 12, 2014 at 2:53 pm

      Thanks Mark,

      I haven’t done that calculation. A quick mental calculation tells that the area figure for 100% EVs in London would be around half of the area quoted in the piece. But don’t quote me on that.

      I have looked into the land requirements of solar in California.

      http://theenergycollective.com/robertwilson190/364526/how-big-worlds-biggest-solar-farm

      I had been planning on writing a longer piece for the Energy Collective into land requirements of 100% wind around the world. But I’ve decided to stop writing there. Though I might see if I can write it up as an academic paper.

      Like

        Mark Miller said:
        December 12, 2014 at 8:13 pm

        Robert

        Thanks for the link to your post at the “collective”. I flew over the concentrating solar farm back in August. It was a Salvador Dali moment for me. The shadows from some white fluffy clouds were kind of cool to see as they progressed across different sections of the mirrors below.

        I was rather surprised to learn, from a nuclear physicist friend, that the output of the facility would be affected more than my little PV system (6.12 STS rated) by the atmospheric conditions I saw back in August. I really should of ask a couple of follow up questions….. was it the humidity in the air that was effecting the output of the facility or was it the specific clouds that were passing over the mirrors that were the primary causes of the drop off of the output.

        By chance have you reviewed the way that Barnhart, et al, evaluated the storage vs curtailment question?

        http://www.cpuc.ca.gov/NR/rdonlyres/1521FE3B-2FB5-4A6A-A93B-45125D6EF895/0/Barnhart20140116CPUCPHSWorkshop.pdf

        I am in favor of SMUD’s approach to stabilizing the grid by the way- a pumped hydro facility is being evaluated for construction about 10 miles from my location.

        Given their real world experience with some of the other alternative energy storage options (See page 95+ of a public report- https://www.smud.org/en/about-smud/company-information/board-of-directors/documents/documents-meetings/board-packet-09-04-2014.pdf ) I can see why they are holding off on installing any additional energy storage technologies at this time.

        Mark

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        Robert Wilson said:
        December 13, 2014 at 10:52 am

        Mark

        I have read the work of Barnhart et al. and it raises major questions about whether non-geological forms of energy storage are a good idea without huge improvements in the embodied energy of storage.

        I keep meaning to write about their work, but I’ve never had the time. I try to stick to less than a thousand words, and their stuff would likely to take more or I would need to simplify it.

        But the problem is that we can easily end up with a situation where we buy batteries from China. The emissions from that are on China’s books, but they are still real. Such things should be accounted for, but currently they are not. The same goes with solar panels. In Britain the emissions are around 100 g CO2/kWh if you import the panel from China. But most of that is not contributing to Britain’s official emissions. So, states such as California should probably make provisions for imported emissions in their energy storage mandate.

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        Fred Golden said:
        June 19, 2015 at 8:11 am

        I was also running some ‘rough’ numbers. I think that your estimate are a bit off. I figure it this way, and am open to discussion on the results.

        You quoted as needing about 40,000 GWH per year to run London. Lets say that is about 120 GWH daily or around 50 GWH per hour on average (though we all know that afternoon peaks are much higher, and overnight is much lower). So a ‘average’ Westinghouse 1,100 MW Nuclear machine it would take about 50 of them running at 90% capacity to generate 50 GW per hour, or 40,000 GW in the 8,700 (+/-) hour long year (deducting for maintenance, and other downtime).

        Then it comes to the Wind Machines. At say 30 MPH it might be at it’s full 1.5 MW rating, but lets say that winds ‘average 15 MPH’ and we know that is only 25% of the energy that is found at 30 MPH, so say 400 KW per machine, or about 5 MW per day per machine. 5 MW X 365 days = 1,825 MW per machine.

        Lets say that it was windy that year, and each machine makes 2 GW per year. 40,000 GW would be produced by installing about 20,000 each 1.5 MW rated machines.

        Now figure how many acres is required spacing for each machine, and you have a simple answer. You can disperse all of these wind machines over a farm, it actually only takes about 200 square feet of the 10 or so acres spaced between the machines. So while a wind farm might cover a total of 200,000 acres, the actual area required for the post that it sits on is only a few square feet, the remaining land can be planted, or cows use it, ect.

        It would also take a ‘few’ acres to located the 50 nuclear plants too. AS well as space for the nuclear fuel to be deposited ‘in someone’s back yard’ – if you can get them to agree to accept the spent fuel. . .

        Not many people want to live within 25 miles of Fujisaki anymore. So figure that one plant is using up about 25 miles square in Japan, or 25X25 square miles of land, that will probably never be lived in again! That is about 625 square miles, if I did my math right. Of course 1/2 of that amount is at sea, in the case of the coastal Japan nuclear plants. . .

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        Robert Wilson said:
        June 19, 2015 at 8:29 am

        Fred

        These aren’t ‘rough’ numbers. They are an order of magnitude out. 120 GWh daily does not correspond to 50 GWh hourly.

        Like

        Fred Golden said:
        June 19, 2015 at 8:51 am

        The Eight paragraph stated:

        ‘This could be supplied by a couple of big coal or nuclear power plants. A typical nuclear power plant is made up of two 1.3 or so GW units, which have capacity factors of around 90%. So, London could be supplied by a couple of regular sized nuclear power plants.’

        What I am stating is that a 1,100 MW nuclear plant can only produce about 8,000 hours X 1.1 GW yearly, or less, so say about 8,000 GW per plant. 40,000 GW total would require about 50 nuclear plants, not just ‘a couple of regular sized nuclear power plants’. In America, we typically have a 800 MW steam turbine at both gas fired and coal fired ‘large plants’, and they might group 5 – 8 steam turbines at one site, for easy maintenance. A recent natural gas plant opened with a 500 MW capacity, and the cost to build it was around $250,000,000 or about 50 cents per rated watt.

        Large nuclear plants seem to cost around $8 per rated watt (but don’t quote me on that amount).

        Hydroelectric can cost a lot to install a 200 MW system, but fuel is much less expensive!

        Wind has ‘free’ fuel, but somewhat expensive maintenance – considering the need to maintain 20,000 of the large wind machines to power a large town.

        Nuclear has very expensive waste disposal, and a long time before it ‘becomes safe to live near it’. While damage from a coal power plant may exceed millions of lives per year, in coal dust, mine safety accidents, and lost productivity due to smog and health issues – even 100 miles away from the actual mine or power plant.

        Each power system has it’s advantages and dis-advantage.

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        Fred Golden said:
        June 19, 2015 at 9:00 am

        Robert Wilson,

        Now I see the mistake in my math. It is 1 am local time, forgive me.

        1.1 GW per nuclear plant per hour, X roughly 8,000 hours per year = something close to 8,000 GW per large plant. 5 plants = the required 40,000 GW per year.

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        Robert Wilson said:
        June 19, 2015 at 9:01 am

        Perhaps if you could get your units straight it might help.

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    Todd said:
    December 19, 2014 at 7:35 pm

    This article provides a great understanding about power density, thank you. Was there an assumed capacity factor used in the calculations?

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      Robert Wilson said:
      December 20, 2014 at 2:10 pm

      Todd,

      I’m not sure if I understand your question. Power density has little to do with capacity factor. One is a measure of average kinetic energy per unit area, with the efficiency of wind turbines and turbine configuration factored in. The other is a measure of the variability of that kinetic energy after it has been converted to electricity.

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        Todd De Ryck said:
        December 20, 2014 at 2:27 pm

        Sorry I wasn’t clear, what capacity factor did you use in this scenario? For example, if 100% was used, I assume that would require less land than using 30% capacity factor with some form of storage or backup? For example, a capacity factor comparison is done in this video at 9:15 and I am essentially wondering how that might effect land use, or are you saying power density accounts for capacity factor? https://www.youtube.com/watch?v=IzbI0UPwQHg

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        Robert Wilson said:
        December 20, 2014 at 4:19 pm

        Todd,

        a) I provided a link to where David MacKay did the calculation. If you want to find this out read that, instead of asking me to take time out of my day to explain.

        b) Take some time to understand what a capacity factor actually is. Your question implies you don’t understand the concept.

        Like

      Mark Miller said:
      December 20, 2014 at 4:08 pm

      Morning Todd-

      Lincoln Wolverton posted a comment, from a 2010 paper, entitled “Working With Wind”
      http://judithcurry.com/2014/12/11/all-megawatts-are-not-equal/#comment-656769

      that covers some of the nuts and bolts of keeping the grid up at all times with intermittent generation in the Pacific Northwest (CA’s is included in his evaluation).

      Like

        Todd De Ryck said:
        December 20, 2014 at 4:35 pm

        Excellent, thanks Mark, I have been reading some of “Planning Engineer” ‘s posts on that site

        Like

    Chris Vernon said:
    December 28, 2014 at 10:40 am

    I’m not a fan of the power density unit when it comes to wind farms – as the land the wind farm occupies isn’t ‘used up’. Most of it is still available for crops to the grown under, animals to be raised under etc. MacKay’s calculations do not consider this mixed use or layering.

    That said, it’s an interesting question to ask. A few hundred years ago folk may also have asked how much area it takes to feed a city. Today we totally accept, unquestioningly, that large cities can’t feed themselves and instead rely on imported food. No reason why the same can’t be true for renewable electricity.

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    Fred Golden said:
    June 19, 2015 at 8:18 am

    I don’t spell well, so here is a google copy.

    Fukushima Daiichi nuclear disaster

    Lets say that nobody lives within 50 miles of this site for the next 100 years. You could s

    Like

    Fred Golden said:
    June 19, 2015 at 8:35 am

    Fukushima Daiichi nuclear disaster

    Lets say that nobody lives within 50 miles of this site for the next 100 years. You could say that 50X 50 square miles is taken out of being productive farming area, and not the ‘best place’ to go fishing either. That is 2,500 square miles!

    Even installing 20,000 of the 1.5 MW wind machine or say 5,000 of the much larger modern 6 MW wind machines will take up less space than that!

    Somehow I think the 40,000 GWH per year estimate is a bit off, as Japan has only a few dozen nuclear sights, and I can not imagine that London would need 50 of that size power plant to run only about 600 square miles of the country.

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