A normal beam of radiation In,b has a flux value of 350 W/m^2. What is the flux reaching a solar surface facing due south and tilted south by 25° at 11 a.m. on the summer solstice at 42° north latitude?

Question

A normal beam of radiation I[latex]_{n,b}[/latex] has a flux value of 350 W/m[latex]^2[/latex]. What is the flux reaching a solar surface facing due south and tilted south by 25° at 11 a.m. on the summer solstice at 42° north latitude?

Accepted Answer

The surface faces south so the value of the surface azimuth is γ = 0. On the summer solstice, the declination is δ = 23.45°, and at 11 a.m. the hour angle is −15°. Solving for solar altitude gives 67.6°. Solving for solar azimuth on the basis of ω, δ, and a gives γ[latex]_s[/latex] = −38.6°. Solving in turn for angle of incidence as a function of α, β, γ, and γ[latex]_s[/latex] gives θ[latex]_i[/latex] = 15.5°. Thus the energy flux onto the device is 350 w/m[latex]^2[/latex] × cos(15.5°) = 337 w/m[latex]^2[/latex].

Question 9.2: A normal beam of radiation In,b has a flux value of 350 W/m^...

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