bLAGI HOME · Total Surface Area Required to Fuel the World With Solar

Total Surface Area Required to Fuel the World With Solar

August 13, 2009 | 105 comments



click for larger image

According to the US Department of Energy (Energy Information Administration), the world consumption of energy in all of its forms (barrels of petroleum, cubic meters of natural gas, watts of hydro power, etc.) is projected to reach 678 quadrillion Btu (or 7.15 exajoules) by 2030 – a 44% increase over 2008 levels (levels for 1980 were 283 quadrillion Btu and we stand at around 500 quadrillion Btu today).

I wonder what surface area would be required and what type of infrastructural investment would be required to supply that amount of power by using only solar panels. To create fuel that can be used in vehicles and equipment I am assuming that some of the electricity generated would be used to create hydrogen. We should all start wondering about these things since we will have really no other choice* by the turn of the next century.

So to find this out we start with the big number 678,000,000,000,000,000 Btu.

Converting this to KW•h [1 Btu = .0002931 kW•h (kilowatt hours)] makes 198,721,800,000,000 kW•h (199,721 TW•h). This is for an entire year. As a comparison, the average household uses approximately 18,000 kW•h per year (1/11 billion of the total world usage).

We can figure a capacity of .2KW per SM of land (an efficiency of 20% of the 1000 watts that strikes the surface in each SM of land).

So now we know the capacity of each square meter and what our goal is. We have our capacity in KW so in order to figure out how much area we’ll need, we have to multiply it by the number of hours that we can expect each of those square meters of photovoltaic panel to be outputting the .2KW capacity (kilowatts x hours = kW•h).

Using 70% as the average sunshine days per year (large parts of the world like upper Africa and the Arabian peninsula see 90-95% – so this number is more than fair), we can say that there will be 250 sun days per year at 8 hours of daylight on average. That’s 2,000 hours per year of direct sunlight.

Therefore, we can multiply each square meter by 2,000 to arrive at a yearly kW•h capacity per square meter of 400 kW•h.

Dividing the global yearly demand by 400 kW•h per square meter (198,721,800,000,000 / 400) and we arrive at 496,804,500,000 square meters or 496,805 square kilometers (191,817 square miles) as the area required to power the world with solar panels. This is roughly equal to the area of Spain. At first that sounds like a lot and it is. But we should put this in perspective.

If divided into 5,000 super-site installations around the world (average of 25 per country), it would measure less than 10km a side for each. The UAE has plans to construct 1,500MW of capacity by 2020 which will require a space of 3 km per side. If the UAE constructed the other 7 km per side of that area, it would be able to power itself as a nation completely with solar energy. The USA would require a much larger area and approximately 1,000 of these super-sites.

According to the United Nations 170,000 square kilometers of forest is destroyed each year. If we constructed solar farms at the same rate, we would be finished in 3 years.

There are 1.2 million square kilometers of farmland in China. This is 2 1/2 times the area of solar farm required to power the world in 2030.

Compare it to the Saharan Desert:

The Saharan Desert is 9,064,958 square kilometers, or 18 times the total required area to fuel the world.

By another measure, “the unpopulated area of the Sahara desert is over 9 million km², which if covered with solar panels would provide 630 terawatts total power. The Earth’s current energy consumption rate is around 13.5 TW at any given moment (including oil, gas, coal, nuclear, and hydroelectric).” This measure arrives at a multiplier of 46 times the area needed and shows that my numbers are very conservative.

Compare it to highways:

At a density ratio of 800km per 1000 square kilometers and a total length of 75,440km, the overall area of the US interstate highway system (constructed entirely between 1956 and 1991 – 35 years) is 94,000 square kilometers, or 20% of the overall required area for the world. The US also consumes about 20% of the world’s energy. (if the efficiency of conversion from solar to electricity was 100%, the area of USA highway would be equal to exactly that required to run the world). Indeed if every nation were to embark on a state program of the scale of the US highway system we could be finished with the required infrastructure in 20-40 years.

Compare it to golf courses:

The typical golf course covers about a square kilometer. We have 40,000 of them around the world being meticulously maintained. If the same could be said for solar farms we would be almost 10% of the way there.

Also remember that we are working here with a worst case scenario based on projections for the year 2030 that assume a lot about growth. What could we do to lower the overall Btu load? And what other sources of clean energy could contribute to lower the area needed for solar panels?

Wave:

World wave energy potential = 2,100,000,000,000 KW•h (2,100 TW•h) or 1% of the required load.

Wind:

A 5 MW turbine can be expected to produce 17 GWh per year (they are 40% effective from their peak rated capacity – 5 MW x 365 x 24 = 43.8 GWh). Therefore, it would require 11,748,294 of the 5 MW capacity turbines to create the same yearly output. There are 500 million cars in the world so it’s not like that’s an unattainable goal from a manufacturing standpoint. And each 5 MW turbine is a 30 year lifespan money making machine for whoever buys it. The same can not be said for my car. But if we can build 90,000 Cape Wind size installations, we would be there on wind alone. Based on that installation, each turbine requires 1/2 square mile of area for offshore sites. This would require 5.85 million square kilometers for 2030 world energy needs.

Here is a graphic for wind based on the notes above. The area in the North Sea is taken directly from the OMA proposal by Rem Koolhaas the pdf of which can be seen here.


click for larger image

Existing Hydroelectric:

I say existing hydroelectric because it would be damaging to the environment to construct more dams on rivers. Such designs have been shown conclusively to have a deleterious effects on the ecosystems of the watersheds that are fed by the existing river.

As of 2004, hydroelectric power accounted for 6% of the energy production in the world. A conversion of this percentage into energy capacity makes 28 quadrillion Btu (492 quadrillion Btu x 6%). As a percentage of 2030 levels and accounting, this would be more like 4% and accounting for a hopeful decommissioning of existing dams, let’s assume 2%.

So these other sources together have the potential to reduce the area required by 5% – 25% based on the amount of wind power we tap into. Solar panels are really going to have to do the vast majority of the work but a sustainable solution is going to require a great mix of solutions that are diversified as much as possible.

The technologies are improving and the efficiencies are getting greater. We must make it our goal to by the end of this century construct the area required by at the same time reducing our demand and by starting the necessary infrastructure projects today everywhere around the world. Otherwise the consequences are unthinkable.

*As for nuclear power, it currently produces 2.5% of the world’s energy or 10 quadrillion Btu per year. In 2008, the International Atomic Energy Agency (IAEA) predicted that nuclear power capacity could double by 2030, though that would not be enough to increase nuclear’s share of electricity generation. As for the non-renewable resource of uranium, according to the nuclear industry’s own estimation:

Current usage is about 65,000 tU/yr. Thus the world’s present measured resources of uranium (5.5 Mt) in the cost category somewhat below present spot prices and used only in conventional reactors, are enough to last for over 80 years.

80 years does not equal sustainable. And this is only assuming current use rates (the 5% of world energy needs).

An average plant puts out 3 cubic meters of spent fuel each year. Assuming 1000 plants operating around the world (there are 500 today), that would makes 3,000 cubic meters per year. Over those 80 years this would create a volume of 240,000 cubic meters or a cube of 60 meters on each side (bigger than the Pantheon and roughly equivalent to the volume of the Gol Gumbaz Mausoleum. What do we do with that amount of dangerous radioactive material that has a half life of 2 million years?

Update 1: some comments being posted here:
reddit
digg

Update 2: Many comments have to do with the distribution of energy. I reiterate that I am in favor of a maximizing of diversity of clean energy technologies and of points of generation. For example, if we use the figure of 6 billion people in the world, and if over the course of each person’s lifetime they would be responsible for creating a panel to use their equal share of the worldwide demand (never mind the non-equal distribution) then we would each be in for a 9m x 9m square, or something that gives off 33,000 kW•h per year. With a typical home roof installation that assumes 15 kW capacity. Obviously this extreme localization is also not ideal — what is needed is a plan that captures the best balance of centralized/localized and best mix of renewable and clean resources.

Update 3: SES technology would bring down the solar area required to 315,000 square kilometers (based on the 629 kW•h per square meter listed on the site sourced as from Southern California Edison and Sandia National Laboratories). This is a 40% reduction just on efficiency of the capturing device. The technology will continue to get better and better…

Tags: , , , ,

  14. David on September 16, 2009 at 9:25 pm

    Where are all of these solar panels going to come from? What is the environmental cost of creating all of these solar panels and the other infrastructure needed to create these solar farms? How are you going to get the panels and the other infrastructure to the sites? Where are you going to find 1000 10km square sites in the US to build these farms that won’t be objected to by environmentalist based on the environmental impact?

    Reply

  15. JD on September 17, 2009 at 10:03 am

    Although the calculation is stunning when visualized like that, solar power does not produce energy in a particularly convenient form. Batteries are inefficient, heavy, and expensive. The current infrastructure is built upon an energy in liquid form. This is why the idea of biofuels is so attractive since it could potentially convert sunlight to a diesel form with high efficiency.

    Reply

  18. Rich Hessler Solar on September 17, 2009 at 11:11 pm

    We could put it in the New Mexico salt mines! The mines contain what remains of the Cold War and are thousands of feet underground.

    This picture is amazing. If we create solar plants of this size, image how the view from space will change in the coming years. One other idea that is in development is generating electricity via satellite in space. The energy is transferred to earth through electronic waves and converted back into electricity. In space, we can generate electricity night or day and do not have to worry about the environmentalists.

    Reply

    1. Joseph on October 19, 2009 at 5:04 pm

      space is an environment, too.. everything everywhere impacts everything everywhere. any exceptions you can think of?

      Reply

  20. willhorspool on September 19, 2009 at 5:44 pm

    Fuel of the future could be found in a fuel from the past!

    Early cars were duel fuel, and could run on petrol or alcohol. Due to the energy used to distil beer or wine into 85% alcohol or more for fuel, it was cheaper to buy oil-based fuel, despite these burning with more impurities than alcohol does.

    (Alcohol is already part oxidized, so produces less btu of energy when burned. The extra oxygen released however makes it less likely that carbon released will form poisonous carbon monoxide. Also, as alcohol mixes with water, in a combustion engine the steam produced makes a boost of pressure against the pistons, which compensates for lower btu. This has been used for years in indy car racing with run on methanol)

    Brazil has excess sugar cane to ferment, and has now been running many of its vehicles on ethanol for years. Buses, cars, even planes!

    So! My point is. Why build solar panels to catch the suns energy, when all plant life on earth has evolved to do just that?

    All green plants on earth are on average 33% cellulose. Cellulose is a polymer of sugar. To humans, it is indigestible fibre, but many forms of life produce the enzymes that break it down into fermentable sugars.

    if you google for “cellulosic ethanol”, you’ll find that NREL researchers in Colorado are working on improving the efficiency of the process. Improvements in the Methods of seperation, production of enzymes, or yields of yeasts could make fueling the world from waste plants a viable option.

    Reply

  21. jay on September 21, 2009 at 7:54 am

    i don’t understand why environmentalists get so upset about putting solar panels in certain locations when it would save many other environments from destruction by pollution and possibly prevent the extinction of certain species from climate change due to excessive emissions of several “greenhouse” gasses. i believe it would be much better than relying on combustion for out energy needs. The bio diesel is a good idea except for the fact that it still emits harmful gasses. The solution will be a combination of energy resources, but i don’t think any type of burning, other than hydrogen, should be part of it.

    Reply

  27. Visio Guy on October 8, 2009 at 4:18 pm

    A thought:

    If you create a huge solar farm, you create a bunch of shade, which means any water you might leave laying around in that shade won’t evaporate as quickly.

    That means you could possibly grow stuff in the shade of a solar farm and slowly reclaim desert areas (and gum up the solar panels)

    Has anybody heard of such a strategy being studied?

    Reply

    1. admin on October 8, 2009 at 4:41 pm

      Great idea! I’ve had similar thoughts about this. If the array is designed to allow 25% of the light through, it may well start to simulate a temperate ecology over time and function as a sort of green house especially if the space below is sealed and allowed to accumulate humidity.

      Reply

      1. Rich on October 30, 2009 at 11:46 pm

        But, if you’ve got solar panels covering the area, how does the water get to the ground? Also, you would have to make sure that the light that gets through are the red and blue wavelengths if you want plants to be able to grow — do we have solar cells that are this selective? And do we want to give up these wavelengths?

        Why not focus solar generation on areas that are already built up and covered with impermeable structures? For example, houses, government buildings, parking garages, airport hangars, etc, which are also the places where electricity is needed. Adding in wind, etc, we will then greatly reduce the areas of currently undisturbed land that we would need to cover (areas that are often far from where the electricity demand is which means we have to factor in energy losses for transporting to markets via wires).

        Reply

        1. admin on October 31, 2009 at 11:06 am

          Not either or, but both and. Use all of the area of existing unused impermeable surfaces. You will perhaps still require arrays in the landscape. If these are heliostatic which they should be to maximize efficiencies, then during the day, they will move and allow reflected and ambient light sufficient to use the “understory” for some agricultural purposes, especially if they are place apart and at a sufficient elevation. In a desert area, a proper design could have the added benefit of reclaiming moisture to the soil.

          Reply

  28. Ewan MacLachlan on October 10, 2009 at 2:21 pm

    There is surely the possibility of using the sea, lake, loch, harbour and river surfaces to host the solar panels?

    Reply

  31. Rich on October 30, 2009 at 9:58 pm

    As best I can tell, and as much as I’d like your area calculations to be correct, your calculations for the area needed to provide all energy via solar is way off.

    First, the number you are using for solar input is an approximation of the solar constant (no problem here). However, this is based on the earth being a disc when the earth is actually a sphere. Thus, to account for angles of incidence and being on the wrong side, etc, the average solar input needs to be divided by 4 = 340w/m2.

    Now this 340w/m2 is the solar energy incident at the top of the atmosphere. Only approximately half of this energy makes it to the surface of the earth.

    Thus, just looking at your values for incoming solar energy, you have underestimated the area needed by a factor of 6. I haven’t looked at any of your other numbers at this point.

    However, increasing the area by a factor of 6 still occupies only a small amount of the total land area. Have you done an estimate of the area of roof space and parking lot space in the US & Europe? This might be a good comparison figure so that we could evaluate just how much new land would need to be covered by panels vs using current structures to generate some of our energy needs.

    Reply

    1. admin on October 30, 2009 at 10:19 pm

      In 2009, companies like Sunrgi (http://www.sunrgi.com/pv-efficiencty.html) are converting 375W per square meter of surface area with heliostatic XCPV fresnel lens systems. By the year 2030, I imagine that we will be able to go well beyond the 200W per square meter that I used in my calculation. In fact I was extremely conservative because I expected people to have the reaction that you had. If we apply the same rate of technological progress that has occurred to the microchip over the past 20 years, I think hitting 500W is something to aim for by 2030.

      Reply

      1. Rich on October 30, 2009 at 11:10 pm

        Thanks for your reply, but I guess I’m still stuck on if the incident solar energy at the earth’s surface is less than 200W/m2, it’s kind of difficult to get more than that.

        Even the SUNRGI system (of which I was unaware and of which I still don’t fully understand) appears to only reduce the size of the panel at which light is converted — it still needs a large unobstructed area from which to gather light and focus it onto a smaller area. Just like a magnifying glass or mirrors, you can’t just calculate the area of the spot that you’ve focused light onto — you must include the area occupied by the glass or mirrors.

        I’m not coming at this from complete ignorance nor hostility. I have solar PV and solar hot water and have spent some time in my past measuring photosynthetic rates in plants, etc.

        Even following your link above for Incoming Solar Energy, it shows only about 50% of the incoming energy reaching the earth’s surface. I then followed up on this to get figures for the Earth’s energy balance which contrasts the solar constant that you have used to the average energy per square meter of earth’s atmosphere.

        So, perhaps I’ve got this all wrong. I’d be happy to be wrong! And I have no doubt that technological improvements (like what we’ve seen with computing and cell phones) will result in unimaginable improvements in our ability to convert sun and wind into usable electricity.

        Thanks again. Your graphics are great. But I’m not going to show these energy area visuals to my students until I have more confidence in where the numbers come from.

        Reply

        1. admin on October 31, 2009 at 12:37 pm

          Rich,

          The incident solar energy striking the earth’s surface is more than 200KW/m2. See this US Department of Energy white paper page 215: http://www.science.doe.gov/bes/reports/files/SEU_rpt.pdf. It references the research of Fletcher and Moen 1977 regarding direct solar insolation (DNI). Perhaps you are thinking about the reflected or radiated value (http://asd-www.larc.nasa.gov/erbe/). But that is what is reflected – not what strikes the surface. Every source I can find agrees with 800-1300w per m2 at between 40 and 0 degrees latitude. Or perhaps you are relying on an equation that averages the solar constant over the entire surface area of the globe which is not practically useful since we will never be placing collection devices above 60 degrees latitude. I’m relying on measured surface data from http://www.nrel.gov/.

          Perhaps you may be convinced another way. Look at the graphic regarding solar incidence at http://re.jrc.ec.europa.eu/pvgis/countries/europe/EU-Glob_opta_presentation.png. This data is expressed not in KW/m2 but in KWh per year for optimally inclined south facing (not heliostatic) m2 surface. Depending on the latitude it ranges but let’s assume 2000kwh/m2/year. Now let’s say the efficiency of the system is 20% so that the output is 400kwh per year per m2. I’ve allotted a surface area of 496,804,500,000 in my study so that comes to 198,721,800,000,000Kwh. This is exactly the number we are expected to have to produce in 2030.

          How about this graphic: http://en.wikipedia.org/wiki/File:Us_pv_annual_may2004.jpg. It figures the KWh per m2 per day which ranges from 5-8. Let’s use 6 as the value as it covers the entire southwest quarter of the USA. x 365 = 2190 Kwh per year. x 496,804,500,000m2 = 1,088,001,855,000,000. We’ve exceeded the amount required in 2030 by 889,280,055,000,000 KWh or 22%. Assuming that there is 22% complete cloud cover effect, we have it exactly right. Again this is assuming that we are not using tracking technology.

          I hope this has convinced you. With these incident values of for example 6KWh/m2/day like on the second graphic it is easy to see that in a day with just 6 hours of direct sunlight the incident value is at 1000w per m2. If you were correct that the incident value was less than 200w/m2 then the KWh/m2/day number would never exceed 1.6-2.

          Reply

        2. admin on October 31, 2009 at 1:49 pm

          Oh…I GET IT!

          You are looking at: “Ignoring clouds, the average insolation for the Earth is approximately 250 watts per square meter (6 (kW·h/m2)/day), taking into account the lower radiation intensity in early morning and evening, and its near-absence at night.” on the Wikipedia entry for Insolation.

          This is averaged over a 24 hour period! I’m using the actual 1000w per m2 hitting during the day and multiplying it times the expected hours of direct sunlight per year.

          Reply

          1. Rich on October 31, 2009 at 11:45 pm

            Aaah!
            Thanks for taking so much time to figure out where our differences were — what a difference to be trying to compare global averages (as I was) to what you were doing (essentially practical application). That you were limiting the parts of the globe you were considering, and taking expected instantaneous inputs and multiplying by hours of max light was not obvious to me.

            Thanks again for your efforts to educate me. I am, at least for a short while (!), reassured and confident in your numbers.

            I am a fervent supporter of solar power, but I do not like getting caught using estimates that I can’t explain or justify. This all makes a lot more sense. Keep it up!

  33. Kyle on January 7, 2010 at 6:06 am

    Hope your still answering posts! I was trying to investigate the answer to this question that I have heard before but wanted to be very clear before I go and advertise this…

    Does the sun shine enough in 8 hours on America to power the world for 1 day?I know if that isnt right, it was somewhere around that. Just looking for a definite answer.

    Thanks
    Kyle

    Reply

    1. admin on January 7, 2010 at 6:52 am

      Using the US area of 9,629,091 square kilometers or 9,629,091,000,000 square meters:
      .2KW per square meter of land x 9,629,091,000,000 square meters = 1,925,818,200,000KW peak capacity.
      Assume that during the 8 hours, the efficiency of the entire system is only 25% due to weather conditions: 1,925,818,200,000KW x .25 = 481,454,550,000KW
      Now multiply by the 8 hours (the above is the capacity and now we are multiplying to make KWh): 481,454,550,000KW x 8 hours = 3,851,636,400,000KWh

      The world consumed about 140,000,000,000,000KWh in 2009. /365 days in a year = 384,000,000,000KWh per day.
      In fact you have ten times more than enough. Or in other words, you could reduce all of your operating efficiencies by a factor of 10 and it still works out.

      Reply

  38. Paul Crowley on January 28, 2010 at 6:19 pm

    Er, don’t you have to use an equal-area map to get a meaningful result here? Looking at the hugely disproportionate size of Canada and Africa, this blatantly isn’t one.

    Or is it deliberate, so that the solar panel placed close to the equator looks disproportionately small?

    If you do an accurate version of this map using an appropriate projection, I look forward to linking to it!

    Reply

    1. admin on January 30, 2010 at 8:12 pm

      Map projection is an equal area Mollweide projection: http://en.wikipedia.org/wiki/Mollweide_projection
      In any case, the scale of the areas is founded on a three measure cross-check of the longest East-West measure of three continents: Australia, North America, and Africa.

      Reply

      1. Paul Crowley on February 1, 2010 at 3:49 pm

        Oops! Sorry to have wronged you, and thanks!

        Reply

· 1 · 2