Question

The location of an array of photovoltaic panels on the roof of a residence is shown in the figure above. The orientation of each of the exterior walls of the residence is given in the figure. Each of the rows of photovoltaic panels has the same separation between rows. The height of the roof roof of the residence is 17 inches and the photovoltaic panels have an inclination of 10 degrees facing southeast. It is desired to place the panels at a distance from the parapet so that there is no shade of 8:00 a.m. at 4:00 p.m. Likewise, it is desired that the space between the rows of panels meets the requirement of not having a shadow of 8:00 a.m. at 4:00 p.m. The panel to be specified is a SWA 295W and its specifications are included with the exam.

a) Calculate the separation distance between the panels and the parapet for the SE, SW and NE sides of the residence.

b) Calculate the separation distance between the rows of panels.

Answer 1

a) At 8 am and 4 pm, the sun is at an angle of 360*4/24 = 60^o from the vertical.

As shown in the figure, the length of the shadow is given by

o sun

d= h * tan(60)

Here, h = 17 inches.

So,

d=17* tan (60) = 29.445 inches

But due to the alignment of the house, this distance is the distance from a corner of house to a corner of the panel.

So, distance from edge =

d/V2 = 29.445/V2 = 20.821 in

dse = dne = dsw = 20.821 in

The panel faces south east direction. So, the panel is elevated in the NW side, and need not have this much distance.

The length of SWA 295 W panel is 66 inches. So, for a slope of 10 degrees, the elevation is

H = 66 * sin(10) = 11.461 in

So, the effective height of the wall in this side = 17-11.461 = 5.539 inches.

So,

d=5.539 * tan(60) = 9.594 inches

duw = d/V2=9.594/V2 = 6.784 inches

b) Since the height of the panel is 11.461 inches, its shadow must not fall on the panel above it.

So, Here, distance of the shadow =

d = 11.461 * tan (60) = 19.851 inches

So,

dpanel = d/V2 = 19.851/V2 = 14.037 inches