Solar angles for a site in Nepal

Data

$\phi = 27.7°$ N; 15 May is day number $n = 31+28+31+30+15 = 135$. Solar time = 10:00.

Declination

$$\begin{aligned} \delta &= 23.45 \sin\left[\frac{360(284+n)}{365}\right] \\ &= 23.45 \sin\left[\frac{360 \times 419}{365}\right] = 23.45 \sin(413.3°) \\ &= 23.45 \sin(53.3°) = 18.79° \end{aligned}$$

Hour angle

15° per hour from solar noon, negative in the morning:

$$\omega = 15(10 - 12) = -30°$$

Zenith and altitude

$$\begin{aligned} \cos\theta_z &= \cos\phi\cos\delta\cos\omega + \sin\phi\sin\delta \\ &= (0.8854)(0.9467)(0.8660) + (0.4648)(0.3221) \\ &= 0.7259 + 0.1497 = 0.8756 \end{aligned}$$ $$\theta_z = 28.88°, \qquad \alpha = 90° - \theta_z = 61.12°$$

Solar azimuth

$$\cos\gamma_s = \frac{\sin\alpha\sin\phi - \sin\delta}{\cos\alpha\cos\phi} = \frac{(0.8756)(0.4648) - 0.3221}{(0.4830)(0.8854)} = 0.1985$$

$\gamma_s = 78.5°$, measured from south towards the east (morning). That is, the azimuth is 78.5° east of south.

Sunrise hour angle and day length

$$\begin{aligned} \cos\omega_s &= -\tan\phi\tan\delta = -(0.5250)(0.3402) = -0.1786 \\ \omega_s &= 100.29° \\ N &= \frac{2\omega_s}{15} = \frac{2(100.29)}{15} = 13.37\ \text{h} \end{aligned}$$

Sunrise is at $12 - 6.69 \approx 5{:}19$ solar time.

Answer: $\delta = 18.79°$, $\omega = -30°$, $\theta_z = 28.88°$, altitude $= 61.12°$, azimuth $= 78.5°$ east of south, $\omega_s = 100.3°$, day length $= 13.37$ h.