4

A stiffer, more formal (form, schmform!) way to ask this is:

Is a solar panel field the size of Spain enough to cover the worlds energy needs?

(This assumes that everything in the world runs on electricity. This of course, can be technically true, but we would need much smaller ships and different planes etc.)

There's this Businessinsider article that states that the world energy consumption can be covered by a solar panel area the size of Spain? (Source seems to be TechInsider) Here's a picture too. Since Spain is only 0.335% of total global land area, then, if every country dedicated one third of one percent of their area to solar panels, we would be able to supply the worlds energy, aside from rockets, (according to Elon Musk in 2nd paragraph)

But doing the math told me that:

  • a. the world consumes 157,5 PetaWh/year (1.575 × 10^17 Wh)
  • b. about 1000 Watt of sunlight hits one square meter,
  • c. at 100% efficiency, 157.000.000.000.000 m2 is needed or 157 million km2
  • d. but actual average efficiency/effectiveness is 16,6% efficiency so we would need six times that much solar panels or 942 million km2
  • e. Even at recent Japanese claims of 26.6% efficiency we would need about 590 million km2

World area 510.million km2 of which land is less than 150 million km2

So, where do I or the article go wrong?

EDIT: sorry to edit so much.

GwenKillerby
  • 140
  • 2
    Looks very much on topic to me as it stands now... – kkm -still wary of SE promises Mar 16 '19 at 13:38
  • 1
    Gwen, welcome, and certainly do not be sorry about edits. The clearer you make the question, the better! It would be good if you provided the sources for some numbers under the points a-e. – kkm -still wary of SE promises Mar 16 '19 at 13:43
  • Also, your Tech Insider link doesn't lead to an article. I'd love to see what they have to say about their own estimate. – flevinBombastus Mar 16 '19 at 13:51
  • Since 3 people think it's on-topic, can the off topic thing be removed? Thanks. – GwenKillerby Mar 16 '19 at 13:51
  • 2
    I closed this question as off-topic because I think it's I think it's clearly engineering - the problem is supplying the world with power, and it's asking how to design a solar panel to do that. If someone asked us "how large does a solar panel need to be to power my house?", it would be the same type of question, just on a smaller scale. Also: If a question is closed, you should not offer answers in comments, but vote to reopen it if you truly think it's on-topic. – ACuriousMind Mar 16 '19 at 13:59
  • If you don't have the reopen vote privilege (3k rep), such requests for reopening are welcome on the meta site. –  Mar 16 '19 at 13:59
  • 2
    You should write 1.57 PetaWh/year, not just 1.57 PetaWh. – Thomas Fritsch Mar 16 '19 at 14:05
  • I haven't been able to figure out how to use the meta site, but... it's really asking how a numerical estimate went wrong, isn't it? The basic conceptual misunderstanding is one of units, not of how to engineer a solar panel. – flevinBombastus Mar 16 '19 at 14:06
  • 4
    I agree, this is on-topic. Voted to re-open. We are talking about the distribution of solar radiation that hits Earth and of energy conversion efficiencies. This is within the scope of physics in my view. – Steeven Mar 16 '19 at 14:11
  • 1
    And it's INCONSISTENT to close it, since there are load of engineering or engineering adjacent question on physics. for example:

    https://physics.stackexchange.com/questions/301626/what-is-the-difference-of-engineering-and-true-stress

    https://physics.stackexchange.com/questions/443694/true-strain-engineering-strain-strain-gauges

    https://physics.stackexchange.com/questions/52617/does-graphene-actually-remain-strong-for-macroworld-engineering

    There's EVEN a whole engineering TAG on this site, so I call Bees. .... https://physics.stackexchange.com/questions/tagged/nuclear-engineering

     – GwenKillerby Mar 16 '19 at 14:15
  • @GwenKillerby By the way, I don't think 1000 W/m^2 is correct for Spain (maybe it is a global average). See here: http://www.adrase.com/en/. – Steeven Mar 16 '19 at 14:22
  • 1
    @Steeven I think the idea is that when you compute using the global average, apparently the original article gets out a number that is roughly equal to the area of Spain. Nobody suggests actually covering Spain in solar panels. – flevinBombastus Mar 16 '19 at 14:26
  • @GwenKillerby In any case, just be careful about the units between (a) and (b), and be sure to use the correct number in (a). Those are the main errors I was able to find. – flevinBombastus Mar 16 '19 at 14:27
  • @flevinBombastus Well, since the numbers don't add up, I am wondering of this may actually be the case. Maybe the calculation is based on both Spain's area and Spain's typical solar irradiation. – Steeven Mar 16 '19 at 14:29
  • @flevinBombastus "Nobody suggests actually covering Spain in solar panels." My first LOL for the day, so thanks for that? But how are the numbers at a and b wrong? This might explain it. – GwenKillerby Mar 16 '19 at 14:31
  • 1
    Your Wikipedia link says the power supply in year 2013 was 157,482 TWh, which is 157 PetaWh, but not 1.57 PetaWh as you have written. – Thomas Fritsch Mar 16 '19 at 14:31
  • 1
    @ThomasFritsch ah, yes, it's the European-French comma thing vs Anglo comma use. You're right. Wikipedia doesn't use international standards, but American ones. – GwenKillerby Mar 16 '19 at 14:34
  • 1
    I've already tried to comment on how (a) and (b) are wrong, but those comments were removed because apparently you're not supposed to post "answers" to questions on hold in the comment section. Use @Steeven's comment above, and note that in (b), you have a power, but in (a) you have an energy. To get to a total area needed you need to get rid of all the time units, so you need to calculate the total energy, not just the power, collected by a square meter of solar panel in a year. – flevinBombastus Mar 16 '19 at 14:59
  • 1
    @Steeven Using Spain's irradiance will change the answer by less than an order of magnitude, but as it stands Gwen is off by many orders of magnitude. See my comment above. – flevinBombastus Mar 16 '19 at 15:00
  • @flevinBombastus yes, a. and b. are sloppy. I stand corrected. Would changing b. to "about 1000 Watt of sunlight hits one square meter per year/hour/second" fix it? Please advise on how to change this. Thanks. – GwenKillerby Mar 16 '19 at 15:24
  • 1
    Way aside, rockets powered by hydrogen and oxygen also run off electricity in the end; $H_2$ is produced by electrolysing water. But this is the fuel cycle Elon's competitors (wink, wink) use in Delta IV; Falcon runs on RP-1/LOX, and RP-1 comes from hydrocarbs. Having said that, all rocket launches in the world use an entirely insignificant amount of energy, compared to overall use. – kkm -still wary of SE promises Mar 16 '19 at 15:30
  • More engineering questions, even on solar cells! Like mine. Which makes ur on-hold decision even more INCONSISTENT & by your own rules, very premature.

    1st even admits it's about ENGINEERING! https://physics.stackexchange.com/questions/78846/are-fresnel-lenses-widely-used-for-solar-electricity-if-not-why-not/78848#78848 https://physics.stackexchange.com/questions/36041/how-far-from-the-sun-is-a-photovoltaic-effective/36063?r=SearchResults&s=2|23.6376#36063 https://physics.stackexchange.com/questions/69652/concentrating-sunlight-to-initiate-fusion-reaction/69705#69705

     – GwenKillerby Mar 16 '19 at 15:33
  • 2
    @ACuriousMind, I respectfully disagree. Your purported reading of the question as if being about in fact engineering a solar panel to power the whole world is strained at best. To me, the question is how much radiant energy the Earth gets from the Sun on the planet's surface. The panel efficiency, transmission losses etc. are secondary to this; “but we can use only X% in the best current engineering scenario, so multiply the required area by $1/X$ in the end.” Understood this way, the question is chiefly about how nature works, thus in scope. And it's very, very educational, really! – kkm -still wary of SE promises Mar 16 '19 at 15:48
  • @GwenKillerby No, because that's incorrect. "1000 $W/m^2$" is correct. The issue is that you then take the number in (a) (which was the wrong number) and just divide by 1000. That is nonsensical as far as the units are concerned. Instead, take the 1000 $W/m^2$, which is 1000 J/s per square meter, and multiply it by the number of seconds in a year to get the total energy collected by one sq. m. of solar panel in a year. Convert (a) to Joules also, and only THEN can you safely divide to get an area. – flevinBombastus Mar 17 '19 at 12:14
  • 1
    I used the calculator at the below site to calculate the Joules but it was a bit large:

    number under (a): 1,575 × 10^17 Wh resulted in: 4.951.152.000.000.000.000.000.000.000.000.000 joules ==================== And the number under (b) would be: One square meter would collect energy to the sum of: 31.536.000 x 1.000 = 31.536.000.000 joules ================ This seems wrong? Especially since dividing it results in the same thing: 157.000.000.000.000.000.000.000

    ==================== https://www.rapidtables.com/calc/electric/Watt_to_Joule_Calculator.html

     – GwenKillerby Mar 19 '19 at 23:59

0 Answers0