HKDSE • Elective 3: Energy and Use of Energy • Solar energy

Solar Panel Power & Sun Path

A photovoltaic panel delivers $P = \eta\, I\, A \cos\theta$, where $\theta$ is the angle between the sunlight and the panel's normal. Move the sun across the Hong Kong sky, tilt the panel, and find the tilt that captures the most energy in a day.

Parameters

Season (solar declination $\delta$)

Hong Kong ≈ 22.3° N. Panel faces due south.

Solar constant above the atmosphere ≈ 1370 W/m²; a clear sky at ground level gives ≈ 1000 W/m².

Presets

Live readings

SUNLIT
Sun altitude / azimuth 67.7° / 180°
Angle of incidence $\theta$ 0.3°
Effective intensity $I\cos\theta$ 1000 W/m²
Electrical output $P$ 2.00 kW
Energy today (this tilt) 14.5 kWh
Energy today (flat, $\beta = 0$) 13.9 kWh

HKDSE formula

$$ P = \eta\, I\, A \cos\theta $$
$\eta$: efficiency of the solar cells
$I$: intensity of sunlight at normal incidence (W m⁻²)
$A$: area of the panel (m²)
$\theta$: angle between the rays and the panel's normal

$A\cos\theta$ is the panel's area projected perpendicular to the rays — the only part of the beam it intercepts.

3D rooftop, panel faces south
Drag to rotate • pinch / scroll to zoom
Sunlight $I$ Panel normal & $\theta$ Electrical output

Output power through the day

Physics note

Power peaks when the rays strike the panel head-on ($\theta = 0$). At noon the sun's altitude is $90° - \phi + \delta$, so a south-facing panel tilted at $\beta = \phi - \delta$ is perpendicular to the noon sun. For a fixed panel in Hong Kong a tilt of about 20° gives the best annual yield, while a steeper tilt favours winter and a flatter one favours summer. Efficiency $\eta$ is limited to 15–25% because photons below the band-gap energy cannot free electrons and excess photon energy is lost as heat.