The total power used by humans worldwide is approximately 15 TW (terawatts). Sunlight striking Earth provides 1.336 kW per square meter (assuming no clouds). The surface area of Earth is approximately 197,000,000 square miles. What percentage of the Earth's surface would we need to cover with solar energy collectors to power the planet for use by all humans? Assume that the solar energy collectors can only convert 10 % of the available sunlight into useful power.

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joetheelite joetheelite · Answer 1 · 2019-10-08T00:02:19+02:00

Answer:

0,022%

Explanation:

Basically, the first thing you need to do is to convert all the units to the same. You can't work with TW and kW at the same time.

So, the first thing we'll do is the unit conversion.

There are 1,000,000,000 kW in a TW so:

15 TW * 1,000,000,000 = 15,000,000,000 kW

Now that we have done this, let's begin with the exercise:

Sinlight provides 1.336 kW/m2, and convert the 10% so this would be at the end:

1.336 * (10/100) = 0.1336 kW/m2

Using this number to get only the miles:

15,000,000,000 kW / 0.1336 kW/m2 = 1.122754x 10 elevated to 17 m2

Now, we need to convert square meter to square mile, the conversion is the following:

1.122754 x 10 elevated 17 m2 * 1 sq mile/ 2,589,999 sq meters = 43349 mi2

Finally the percentage is:

% = (43,349 / 197,000,000) * 100 = 0,022%