Question

The daily average insolation on the earth surface is approximately 6 kWh/m2 (ignoring cloudy weather). If a commercially available solar panel can achieve an energy conversion efficiency of 20%, what is the area that is required to supply the electricity demand of the US in 2014 (the population was about 324 million and the average per capita power was 1378 watts). What is the fraction (or multitude) of Tennessee (109,250 km2) that is required to power the entire US.

Answer 1

The area is 15.502 km².

The fraction is
`(31)/(218500)`

Step-by-step explanation:

Given that,

Daily average insolation = 6 kWh/m²

Efficiency = 20%

Population = 324 million

Per capita power = 1378 watts

Area of tennessee = 109250 km²

We need to calculate the daily power required

Using given data

`daily\ power=population* Per\ capita\ power`

Put the value into the formula

`daily\ power = 324*10^(6)*1378`

`daily\ power=4.4647*10^(11)\ watts`

`daily\ power=4.4647*10^(8)\ kW`

We need to calculate the total energy per day

Using given data

`E=6\tiimes24`

`E=144\ kW/m^2`

We need to calculate the out of total energy

Using energy and efficiency

`E'=E*\eta`

`E'=144*(20)/(100)`

`E'= 28.8\ kW/m^2`

We need to calculate the area

Using formula of area

`A=(daily\ power)/(out\ of\ energy)`

Put the value into the formula

`A=(4.4647*10^(8))/(28.8)`

`A=15.502\ km^2`

We need to calculate the fraction of Tennessee that is required to power

Using given data

`fraction =(Area)/(tennessee)`

Put the value into the formula

`fraction =(15.502)/(109250)`

`fraction=(31)/(218500)`

Hence, The area is 15.502 km².

The fraction is
`(31)/(218500)`