Introduction to solar energy/Equations

Solar resources and availability

$T S T = U T C + \frac{l o n g i t u d e}{15} + E o T$

Precise method

$E o T = Δ t_{e y} = - 7.659 \sin (D) + 9.863 \sin (2 D + 3.5932)$ (minutes)

where: $D = 6.240 040 77 + 0.017 201 97 (365.25 (y - 2000) + d)$

where $d$ represents the number of days since 1 January of the current year, $y$ .

Rough approximation

For $E o T$ in minutes, see the following equation. Ensure that the trigonometry relations are calculated in radians mode.

$E o T (m i n u t e s) = Δ t_{e y} = 7.678 \sin (B + 1.374) - 9.87 \sin (2 B)$

Where B is:

$B = 2 π \frac{(n - 81)}{365}$

Where $n$ is the day of the year.

For $E o T$ in degrees, see the following equation. Ensure that the trigonometry relations are calculated in degrees mode.

$E o T (d e g r e e s) = 0.0072 c o s (J) - 0.0528 c o s (2 J) - 0.0012 c o s (3 J) - 0.1229 s i n (J) - 0.1565 s i n (2 J) - 0.0041 s i n (3 J)$ Where J is:

$J = n \times \frac{360}{365}$

Where $n$ is the day of the year.

Solar declination rough approximation, in degrees

$δ_{d e g r e e s} = 23.45 s i n (360 (n + 284) / 365)$

Solar declination finer result, in radians

$δ_{r a d i a n s} = a r c s i n (0.369 s i n (360 (n - 82) / 365 + 2 s i n (360 (n - 2) / 365))$

Solar declination finer result, in degrees

$δ_{r a d i a n s} = 0.33281 - 22.984 c o s (J) - 0.3499 c o s (2 J) - 0.1398 c o s (3 J) + 3.7872 s i n (J) + 0.03205 s i n (2 J) + 0.07187 (3 J 0$

Where J is the day of the year, with January 1st being n=1.

$J = n \times (360 / 365)$ , in degrees.

Half hour angle, ω₀

Calculate the hour angle when the altitude angle is equal to 0.

$ω_{0} (degrees) = a r c c o s (- (\tan δ) * (\tan L a t))$

Using this equation, you can find true solar time (TST) at sunrise and at sunset.

$T S T_{s u n r i s e} = 12 - (\frac{1}{15}) ω_{0}$ , make sure that $ω_{0}$ is in degrees.

$T S T_{s u n s e t} = 12 + (\frac{1}{15}) ω_{0}$ , make sure that $ω_{0}$ is in degrees.

$S_{0} = (\frac{2}{15}) ω_{0}$ , make sure that $ω_{0}$ is in degrees.

if: $ω_{0} > 90$ , then the day is longer than the night, and the time that the sun is in the sky is at least 12 hours.

if: $ω_{0} < 90$ , then the day is shorter than the night, and the time that the sun is in the sky is less than 12 hours.

Elevation angle, h

$h = a r c s i n [c o s (δ - L a t)] = 9 0^{\circ} - L a t + δ$

$h = a r c s i n (\cos δ \cos H A \cos L a t + \sin δ \sin L a t)$