Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
07-Bld-A5 Building Science — National Exam, December 2019. Six problems of 20 marks each were printed; per the paper's own NOTES only the first five in the answer book are graded, but all six are answered below as a complete study resource.
Reference texts: ASHRAE Handbook — Fundamentals (Chapters 1 Psychrometrics, 4 Heat Transfer, 14 Climatic Design Information, 16 Ventilation and Infiltration, 25 Thermal and Water Vapor Transmission Data, 26 Heat, Air, and Moisture Control in Building Assemblies); McQuiston, Parker & Spitler, Heating, Ventilating, and Air Conditioning: Analysis and Design; National Building Code of Canada (NBCC).
Check: two inputs are assumed since they were not in the printed data — (1) clock time is taken as solar time (no equation-of-time/longitude correction, the standard simplification at this exam level); (2) ground reflectance ρg = 0.2, the ASHRAE default for ordinary (non-snow) ground.
Find. Total solar irradiation (direct + diffuse + ground-reflected) on the tilted, east-facing collector at the stated time, using the ASHRAE clear-sky model, for tilt 45° and again for tilt 30°.
Lowering the tilt from 45° to 30° turns the east-facing collector further toward the low afternoon sun, cutting its angle of incidence.
Approach. Find the solar position (declination, hour angle, altitude, azimuth) for the stated place/date/time — this is independent of collector tilt — then resolve the angle of incidence and the ASHRAE clear-sky direct, diffuse and ground-reflected components separately for each tilt.
Solar declination for day n = 202 (July 21):
$$\delta = 23.45\sin\left(\frac{360(284+n)}{365}\right) = 23.45\sin(479.3^\circ) = 20.44^\circ$$
Hour angle at 1:00 PM, one hour past solar noon (15°/hour): H = 15°.
Solar altitude β, from sinβ = cosL·cosδ·cosH + sinL·sinδ:
$$\sin\beta = \cos(44^\circ)\cos(20.44^\circ)\cos(15^\circ) + \sin(44^\circ)\sin(20.44^\circ) = 0.8937 \Rightarrow \boxed{\beta = 63.3^\circ}$$
Solar azimuth φ (from south, +west), from sinφ = cosδ·sinH/cosβ:
$$\sin\phi = \frac{\cos(20.44^\circ)\sin(15^\circ)}{\cos(63.3^\circ)} = 0.540 \Rightarrow \phi = 32.7^\circ\ \text{(west of south, afternoon)}$$
Direct-normal irradiation (tilt-independent), ASHRAE clear-sky model, July constants A = 1085 W/m², B = 0.207:
$$I_{DN} = \frac{A}{\exp(B/\sin\beta)} = \frac{1085}{\exp(0.207/0.894)} = 861\ \text{W/m}^2$$
(i) Tilt Σ=45°: angle of incidence on the east-facing surface (surface azimuth γ = −90° from south), using
$$\cos\theta = \cos\beta\cos(\phi-\gamma)\sin\Sigma + \sin\beta\cos\Sigma$$
$$\cos\theta = \cos(63.3^\circ)\cos(122.7^\circ)\sin(45^\circ) + \sin(63.3^\circ)\cos(45^\circ) = 0.460 \Rightarrow \boxed{\theta = 62.6^\circ}$$
(ii) Repeat at Σ=30°. Solar position (β, φ, IDN) is unchanged — only the surface geometry terms change:
$$\cos\theta = \cos(63.3^\circ)\cos(122.7^\circ)\sin(30^\circ) + \sin(63.3^\circ)\cos(30^\circ) = 0.653 \Rightarrow \theta = 49.3^\circ$$
$$I_{DT}=861\times0.653=562\ \text{W/m}^2,\quad I_{dT}=0.136\times861\times\frac{1+0.866}{2}=109\ \text{W/m}^2,\quad I_{rT}=861\times1.030\times0.2\times\frac{1-0.866}{2}=12\ \text{W/m}^2$$
$$\boxed{I_{T,30^\circ} = 562+109+12 = 683\ \text{W/m}^2}$$
Reducing the tilt from 45° to 30° INCREASES total irradiation by roughly 161 W/m² (≈31%): the shallower tilt turns the collector's face further from the high overhead sky and more directly toward the still-fairly-low (63.3°-altitude) early-afternoon sun, shrinking θ enough that the gain in direct-beam capture outweighs the small loss in sky-diffuse view factor.
