Question 3 of 6

Question 3 (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. Total solar irradiation (direct + diffuse + ground-reflected) on the tilted, east-facing collector at the stated time, using the ASHRAE clear-sky model.

Approach. Find the solar position (declination, hour angle, altitude, azimuth) for the stated place/date/time, resolve the angle of incidence on the tilted east-facing surface, then evaluate the ASHRAE clear-sky direct, diffuse and ground-reflected components and sum them.

1. Solar declination for day n = 202 (July 21): $$\delta = 23.45\sin\left(\frac{360(284+n)}{365}\right) = 23.45\sin(479.3^\circ) = 20.44^\circ$$
2. Hour angle at 2:00 PM, two hours past solar noon (15°/hour): H = 30°.
3. Solar altitude β, from sinβ = cosL·cosδ·cosH + sinL·sinδ: $$\sin\beta = \cos(44^\circ)\cos(20.44^\circ)\cos(30^\circ) + \sin(44^\circ)\sin(20.44^\circ) = 0.8262 \Rightarrow \boxed{\beta = 55.7^\circ}$$
4. Solar azimuth φ (from south, +west), from sinφ = cosδ·sinH/cosβ: $$\sin\phi = \frac{\cos(20.44^\circ)\sin(30^\circ)}{\cos(55.7^\circ)} = 0.832 \Rightarrow \phi = 56.3^\circ\ \text{(west of south, as expected mid-afternoon)}$$
5. Angle of incidence θ on the tilted east-facing surface (surface azimuth γ = −90° from south), using $$\cos\theta = \cos\beta\cos(\phi-\gamma)\sin\Sigma + \sin\beta\cos\Sigma$$ $$\cos\theta = \cos(55.7^\circ)\cos(146.3^\circ)\sin(60^\circ) + \sin(55.7^\circ)\cos(60^\circ) = 0.0074 \Rightarrow \boxed{\theta = 89.6^\circ}$$ An angle of incidence approaching 90° is exactly what is expected: by 2 PM the east-facing collector is nearly edge-on to the sun, which now sits well into the southwest sky.
6. Direct-normal irradiation, ASHRAE clear-sky model, July constants A = 1085 W/m², B = 0.207: $$I_{DN} = \frac{A}{\exp(B/\sin\beta)} = \frac{1085}{\exp(0.207/0.826)} = 845\ \text{W/m}^2$$
7. Direct component on the tilted surface: $$I_{DT} = I_{DN}\cos\theta = 845\times 0.0074 = 6\ \text{W/m}^2$$ Nearly zero, because the collector is almost edge-on to the beam.
8. Diffuse component (C = 0.136 for July): $$I_{dT} = C\,I_{DN}\frac{1+\cos\Sigma}{2} = 0.136\times 845\times\frac{1+0.5}{2} = 86\ \text{W/m}^2$$
9. Ground-reflected component (ρᵣ = 0.2): $$I_{rT} = I_{DN}(\sin\beta+C)\,\rho_g\,\frac{1-\cos\Sigma}{2} = 845\times(0.826+0.136)\times 0.2\times\frac{1-0.5}{2} = 41\ \text{W/m}^2$$
10. Total irradiation on the collector: $$\boxed{I_T = I_{DT}+I_{dT}+I_{rT} = 6+86+41 = 133\ \text{W/m}^2}$$