contestada

What area is needed for a solar collector to absorb 47.0 kW of power from the Sun's radiation if the collector is 84.0 % efficient? (At the surface of Earth, sunlight has an average intensity of 1.00×103W/m2. )

Respuesta :

MathPhys

Answer:

56 m²

Explanation:

The efficiency of the solar collector is equal to the power absorbed divided by the power radiated from the Sun, which is equal to the intensity times the area.

e = P / (IA)

A = P / (Ie)

A = (47×10³ W) / (1×10³ W/m² × 0.84)

A = 56 m²

Otras preguntas