Question

A typical home may require a total of 2.00×10^3 kWh of energy per month. Suppose you would like to obtain this energy from sunlight, which has an average daylight intensity of 1500 W/m^2 . - Assuming that sunlight is available 8.0 h per day, 25 d per month (accounting for cloudy days), and that you have a way to store energy from your collector when the Sun isn't shining, determine the smallest collector size that will provide the needed energy, given a conversion efficiency of 26 % .

Answer 1

Answer:

The the smallest size of the collector is 25.64 m²

Explanation:

Given that,

Total energy $E=2.00\times10^{3}\ kWh$

Intensity $I= 1500 W/m^2$

Efficiency = 26%

The intensity of light can be transformed to the required energy = Available intensity of light

$I=1500\times\dfrac{26}{100}$

$I=390\ W/m^2$

We need to calculate the smallest size of the collector

Using formula of energy related to the intensity through area and time

$E=IA\Delta t$

$A=\dfrac{E}{I\Delta t}$

Where, E= energy

I = intensity

$\Delta t$ = time

Put the value into the formula

$A=\dfrac{2.00\times10^{6}}{390\times25\times8}$

$A=25.64\ m^2$

Hence, The the smallest size of the collector is 25.64 m²