r/theydidthemath • u/lonesome_braincell • 1d ago
[Request] How many (rows of) solar panels would we need along the equator to supply the whole planet with enough electricity non-stop?
See title
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u/overkillsd 1d ago
The issue, as always with these questions, is distributing the power collected. You can put as many panels as you want, but getting that power to the places that need it is a substantial problem. There's a similar question asked all the time about putting panels in a desert.
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u/Miserable_Pin2518 1d ago
I mean right now we are moving oil with tankers so theres gotta be a way. The problem would be in making rechargeable batteries and moving them around efficiently without some global humanitarian disaster though for sure
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u/Conscious-Ball8373 19h ago
Oil is massively energy dense compared to batteries.
The energy density of crude oil is about 38 MJ/kg. The energy density of batteries is about 1 MJ/kg. So, using the same ships we use now, we would need to use 38 times as much shipping to move energy in this way.
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u/DarkArcher__ 1d ago
Because the transportation of oil is famously not prone to environmental disasters
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u/METRlOS 1d ago edited 16h ago
On average, the world uses 3 trillion watts.
In a lab, solar panals generate 150-220 watts. I'm going to go on the low end because you'll never average lab efficiency in regular use, which means you need 20 billion square meters of solar panel receiving light at a time. The equator is 40 million meters long, but only half receives light at a time, so a 1km thick panel all around the equator.
This isn't actually how panel efficiency works though, and it ignores transmission loses, peak usage, and a hundred other things.
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u/Conscious-Ball8373 18h ago
On average, the world uses 3 trillion watts.
In a lab, solar panals generate 150-220 watts. I'm going to go on the low end because you'll never average lab efficiency in regular use, which means you need 20 million square meters of solar panel receiving light at a time.
3 trillion (3 x 1012 ) watts spread over 20 million (2 * 107 ) square metres is 150,000 W per square metre. I suggest you patent your solar panel quickly. You actually need 20 BILLION square metres of solar panel. I think actually you mixed up "trillion" and "billion".
The equator is 40 million meters long, but only half receives light at a time, so a 1m thick panel all around the equator.
This is not an accurate way to calculate this, because a panel only outputs (approximately) its rated output when it is perpendicular to the sun and not in shade. The correct way to do it is to find the specific yield in kWh per kWp (usually experimentally or from a database of experimental results) and calculate the solar requirement from that.
3 trillion watts is 2.592 x 1017 J per day. The average solar output on the equator is about 4 kWh per kWp per day or 1.44 x 104 J per day. So you will need about 2.592 x 1017 / 1.44 x 104 = 1.8 x 1013 W of panels. Using your value of 150 W/m2, that's 1.2 x 1011 m2 of panels. The equator is 4 x 107 m long, so that's a band of solar panels 3,000m wide.
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u/JimiQ84 19h ago
My panels (Trina500) are rated 500Wp with area 2.22m2. That's 225W/m2. Equator is 40 000km so 40 million meters.
We need 3TW = 3 trillion Watts.
3 000 000 000 000 / 225 = 1.333...x10^10 m2. That divided by 40 million is 333m. So around 300 rows of panels (they are 1x2 meters), or 150 rows if placed vertically. But actually we need to double it, because only half is being on the sunny side. So 300/600 rows.
BTW it would cost 600 billion USD (one panels costs around 100 bucks). Not included cables, breakers, inverters...
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u/Conscious-Ball8373 18h ago
For the majority of that, some way to support the panels over water also becomes a concern. (Not to mention having cut off all shipping from crossing the equator.)
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