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Solar Geometry & Irradiation Modelling

60 min Solar Thermal Systems — Collectors, Geometry & Storage

Solar Geometry & Irradiation Modelling

Accurate solar resource assessment requires computing the sun's position in the sky (altitude and azimuth angles) as a function of latitude, day-of-year, and time of day, then decomposing the global horizontal irradiance (GHI) into beam (direct), diffuse, and ground-reflected components on a tilted collector surface. These calculations underpin simulation tools (SAM, Polysun, TRNSYS) used for system sizing and bankable energy yield assessments.

1. Solar Declination & Hour Angle

The solar declination $\delta$ (deg) varies seasonally as the Earth orbits the sun:

$$\delta = 23.45° \sin\left(\frac{360(284 + N)}{365}\right)$$

where $N$ is the day-of-year (1 = 1 Jan). Solar hour angle $\omega$ (deg):

$$\omega = 15°(t_{solar} - 12)$$

where $t_{solar}$ is solar time in hours (not local clock time; the equation of time and longitude correction apply in practice, but we use solar time here for clarity).

2. Solar Altitude and Azimuth

For latitude $\phi$ (deg, positive North), the solar altitude angle $\alpha_s$ (elevation above horizon):

$$\sin\alpha_s = \sin\phi\sin\delta + \cos\phi\cos\delta\cos\omega$$

Solar azimuth $\gamma_s$ (deg from South, positive West):

$$\cos\gamma_s = \frac{\sin\delta\cos\phi - \cos\delta\sin\phi\cos\omega}{\cos\alpha_s}$$

Sunrise and sunset occur when $\alpha_s = 0$: $\cos\omega_{ss} = -\tan\phi\tan\delta$. Day length = $\frac{2}{15}\arccos(-\tan\phi\tan\delta)$ hours.

3. Irradiance on a Tilted Surface

The angle of incidence $\theta$ of beam radiation on a surface tilted at $\beta$ from horizontal, facing azimuth $\gamma$ (from South):

$$\cos\theta = \sin\delta\sin\phi\cos\beta - \sin\delta\cos\phi\sin\beta\cos\gamma + \cos\delta\cos\phi\cos\beta\cos\omega + \cos\delta\sin\phi\sin\beta\cos\gamma\cos\omega + \cos\delta\sin\beta\sin\gamma\sin\omega$$

For a south-facing surface in the Northern Hemisphere ($\gamma = 0$) this simplifies significantly. The beam irradiance component on the tilted surface $G_b = G_{b,h}/\sin\alpha_s \cdot \cos\theta = G_{b,h} \cdot R_b$ where $R_b = \cos\theta/\cos\theta_z$ ($\theta_z$ is the solar zenith angle).

4. Diffuse & Reflected Components — Perez Model

The isotropic diffuse model (Liu–Jordan) for diffuse irradiance on a tilted surface:

$$G_{d,T} = G_d \frac{1+\cos\beta}{2}$$

Ground-reflected component (albedo $\rho \approx 0.2$ for grass, 0.6–0.8 for snow):

$$G_{r,T} = G_H \rho \frac{1-\cos\beta}{2}$$

Total: $G_T = G_{b,T} + G_{d,T} + G_{r,T}$. The more accurate Perez (1990) model replaces the isotropic diffuse with circumsolar and horizon-brightening terms, reducing RMSE from ~15% to ~8%.

Worked Example — Solar Position Calculation

Latitude 40°N ($\phi = 40°$), Day 172 (21 June, summer solstice), $t_{solar} = 10:00$. $\delta = 23.45° \sin(360\times456/365) = 23.45°\sin(449.6°) = 23.45°\sin(89.6°) = 23.44°$. $\omega = 15(10-12) = -30°$. $\sin\alpha_s = \sin40°\sin23.44° + \cos40°\cos23.44°\cos(-30°) = 0.6428\times0.3979 + 0.7660\times0.9173\times0.8660 = 0.2558 + 0.6083 = 0.8641$. $\alpha_s = 59.9° \approx 60°$. ✓

Solar Position & Day Length Calculator
Sunrise =?h solar
Sunset =?h solar
Max elevation =?° (at solar noon)

Practice Problems

1. For Berlin (latitude 52.5°N) on the winter solstice (N=355): compute solar declination, day length, and maximum solar elevation at noon. Comment on the implications for solar collector tilt angle selection.
2. A south-facing collector at latitude 35°N is tilted at $\beta = 35°$. At solar noon on the equinox (N=80), compute the angle of incidence $\theta$ and the beam irradiance ratio $R_b = \cos\theta/\sin\alpha_s$.
3. The global horizontal irradiance (GHI) at a site is 700 W/m², with measured diffuse horizontal 250 W/m² and direct normal 600 W/m². For a surface tilted at 30° from horizontal facing South at latitude 40°N on day 180 at solar noon: compute $G_T$ using the three-component model.