Question

Calculate the average power per unit area that is incident on a 5-cm diameter focal area for a 10-m diameter parabolic dish (see photo of dish). Midday insolation is 650 W/m² on a sunny day.

a. Dish Area =
b. Total insolation on the dish =
c. Focused on an area of: 5 cm =
d. Power per unit area =

Answer 1

To calculate each of the points, we will first calculate the circular area with the given data, then we will multiply the area by the given intensity to find the insolation. The focal area will be determined by the focal area, and finally the power per unit area that will be determined by the insolation and the focal area. So we will have:

A ) Dish Area

`A_D = (\pi d^2)/(4)`

`A_D = (\pi (10)^2)/(4)`

`A_D = 78.54 m^2`

B) Total insolation n disc

`\alpha = I * A_D`

`\alpha = 650 *78.54`

`\alpha = 51051 W`

C) Focused area of disc

`A_F = (\pi d^2)/(4)`

`A_F = (\pi 0.05^2)/(4)`

`A_F = 0.00196 m^2`

D) Power per unit area

`P = (\alpha)/(A_F)`

`P = (51051)/(0.00196)`

`P = 2.60*10^7 W/m^2`