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 1:00 PM, one hour past solar noon (15°/hour): H = 15°.
3. Solar altitude β, from sinβ = cosL·cosδ·cosH + sinL·sinδ: $$\sin\beta = \cos(44^\circ)\cos(20.44^\circ)\cos(15^\circ) + \sin(44^\circ)\sin(20.44^\circ) = 0.8937 \Rightarrow \boxed{\beta = 63.3^\circ}$$
4. Solar azimuth φ (from south, +west), from sinφ = cosδ·sinH/cosβ: $$\sin\phi = \frac{\cos(20.44^\circ)\sin(15^\circ)}{\cos(63.3^\circ)} = 0.540 \Rightarrow \phi = 32.7^\circ\ \text{(west of south, afternoon)}$$
5. Angle of incidence θ on the tilted east-facing surface (surface azimuth γ = −90° from south), using $$\cos\theta = \cos\beta\cos(\phi-\gamma)\sin\Sigma + \sin\beta\cos\Sigma$$ $$\cos\theta = \cos(63.3^\circ)\cos(122.7^\circ)\sin(45^\circ) + \sin(63.3^\circ)\cos(45^\circ) = 0.460 \Rightarrow \boxed{\theta = 62.6^\circ}$$ The shallower 45° tilt (vs. a steeper collector) points more toward the zenith and less toward the low afternoon sun, but still catches it at a moderate incidence angle rather than a grazing one.
6. Direct-normal irradiation, ASHRAE clear-sky model, July constants A = 1085 W/m², B = 0.207: $$I_{DN} = \frac{A}{\exp(B/\sin\beta)} = \frac{1085}{\exp(0.207/0.894)} = 861\ \text{W/m}^2$$
7. Direct component on the tilted surface: $$I_{DT} = I_{DN}\cos\theta = 861\times 0.460 = 396\ \text{W/m}^2$$
8. Diffuse component (C = 0.136 for July): $$I_{dT} = C\,I_{DN}\frac{1+\cos\Sigma}{2} = 0.136\times 861\times\frac{1+0.707}{2} = 100\ \text{W/m}^2$$
9. Ground-reflected component (ρᵣ = 0.2): $$I_{rT} = I_{DN}(\sin\beta+C)\,\rho_g\,\frac{1-\cos\Sigma}{2} = 861\times(0.894+0.136)\times 0.2\times\frac{1-0.707}{2} = 26\ \text{W/m}^2$$
10. Total irradiation on the collector: $$\boxed{I_T = I_{DT}+I_{dT}+I_{rT} = 396+100+26 = 522\ \text{W/m}^2}$$