Question

A spacecraft orbiting Earth relies on solar panels that have an efficiency of 26% (percent of incoming energy the panels can convert) for power. The spacecraft has a battery to store energy for the half of isorbi the spacecraft is on the night side of the planet and incoming solar irradiance is 1367 W/m2. How big do the panels have to be at minimum if the spacecraft require 1000 W of power? Show your work.

Answer 1

The solar panel should have a surface area with a size of 2.8 m^2

Step-by-step explanation:

efficiency of solar panel = 26% = 0.26

solar irradiance = 1367 W/m^2

required power = 1000 W

area of panel = ?

The power that need to fall on the panel = 1000/0.26 = 3846.15 W

This power is gotten from the solar irridiance incident on the plate

The minimum area the plate must have to absorb this power must be

From

A = E/I

where

A is the area of the solar panel

E is the energy needed by the panel

I is the irridiance from the sun

substituting values, we have

A = 3846.15/1367 = 2.8 m^2