Question

The US currently produces about 27 GW of electrical power from solar installations. Natural gas, coal, and oil powered installations produce about 740 GW of electrical power. The average intensity of electromagnetic radiation from the sun on the surface of the earth is 1000 W/m2 . If solar panels are 30% efficient at converting this incident radiation into electrical power, what is the total surface area of solar panels responsible for the 27 GW of power currently produced

Answer 1

The total surface area is "90 km²".

Step-by-step explanation:

Given:

Power from solar installations,

= 27 GW

Other natural installations,

= 740 GW

Intensity,

`(F)/(At)=(P)/(A)=1000 \ W/m^2`

%n,

= 30%

Now,

⇒ %n =
`(out.)/(Inp.)* 100`

then,

⇒
`Inp.=(27)/(30)* 100`

`=90 \ GW`

As we know,

⇒
`I=(P)/(A)`

by substituting the values, we get

`1000=(90* 10^9)/(A)`

`A = (90* 10^9)/(10^3)`

`=90* 10^6`

`=90 \ km^2`