Question

5. a. A vertical wall faces south and is located at latitude 40 deg. At solar noon on January 1, insolation on a horizontal surface is 320 W/m*. a. Determine the intensity of beam radiation on the wall. Assume the insolation on the horizontal is 75% beam. A two-dimensional sketch may help with the trigonometry. (answer -471 Wm') b. Repeat for the same location but at solar noon on July 1. Insolation on the horizontal is now 600 W/m, and 70% beam. (answer =128 W/m2).

Answer 1

5 (a)
We know,
E = E₀ Cos(A-B)
Where E= Insolation
A = Latitude angle
B = Solar Declination
320= E₀Cos (40 - B)

We know, B= 23.45*cos(2*π*(JD-172)/365)
On 1st Jan, Julian Day (JD) = 0
Thus, B= 22
Now, E₀ = 320 / cos (18)
E₀ = 337

Given, I= 4/3 E₀
Where, I = Intensity
I = 450.5

We know,
I= I₀ Cos (A-B)
450.5 = I₀ Cos (18)
Thus, I₀ = 471 W/m²

5 (b)
On July 1 J.D = 182
Thus B= 23.45*cos(2*π*(182-172)/365)
B= 23.44
E₀ = 600 / cos (16.56)
E₀ = 626

Given, I= 10/7 E₀
Where, I = Intensity
I = 895
K I₀ = 895/ cos (16.56)
Where K= constant, On 1 July, K = 7.3
I₀ = 933 / 7.3
I₀ = 128 W/m²