Question

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.

Answer 1

To meet global power demand using 10% efficient solar collectors, covering 0.022% of Earth's surface with solar panels is necessary, which amounts to around 112,200 square kilometers.

Step-by-step explanation:

To calculate what percentage of the Earth's surface needs to be covered with solar energy collectors to meet the global power demand, we first need to find out how much power these collectors can generate. Given that sunlight provides 1.336 kW per square meter and the solar panels have a 10% conversion efficiency, each square meter of panels will generate 0.1336 kW of power.

The total power demand is 15 TW, which is 15,000,000 MW (since 1 TW = 1,000,000 MW). To find how many square meters are needed to generate this power, we divide the total demand by the power per square meter:

Total area required (in square meters) = Total power demand (in MW) / Power per square meter (in kW)

Total area required = 15,000,000 MW / 0.1336 kW

Therefore, the total area required is approximately 112,200,000,000 square meters or 112,200 km2.

The Earth's surface area needs to be converted to square meters to match the units. The surface area of Earth in square miles is approximately 197,000,000 which is about 510,100,000,000,000 m2 when converted.

To find the percentage of the Earth's surface required, we divide the total area required for solar panels by the Earth's surface area and multiply by 100:

Percentage of Earth's surface required = (Total area required / Earth's surface area) × 100

Percentage of Earth's surface required = (112,200,000,000 m2 / 510,100,000,000,000 m2) × 100 ≈ 0.022%

Thus, to power the planet using solar energy collectors with 10% efficiency, we would need to cover approximately 0.022% of the Earth's surface.

Answer 2

0,022%

Step-by-step 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%