---
title: "Solar angles for a site in Nepal"
subject: "Energy Resources"
chapter: 2
key: "semester_6/energy_resources:solar-angles-kathmandu-calculation"
source: https://www.bctengineer.com/q/semester_6/energy_resources/solar-angles-kathmandu-calculation
site: BCT Engineering
---

# Solar angles for a site in Nepal

Energy Resources · Chapter 2: Solar energy

- Asked in: year not given, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/energy_resources/solar-angles-kathmandu-calculation
- Whole chapter: https://www.bctengineer.com/md/semester_6/energy_resources/chapter-2

## Question

Calculate the declination, hour angle, solar altitude angle, zenith angle, solar azimuth angle, sunrise hour angle and day length at a place of latitude 27.7° N at 10:00 hours solar time on 15 May.

## Short answer

For 15 May (n=135) at latitude 27.7°N and solar time 10:00: declination δ = 18.79°, hour angle ω = -30°. Solar zenith angle θz = 28.88°, altitude α = 61.12°. Solar azimuth γs = 78.5° east of south. Sunrise hour angle ωs = 100.3°, day length = 13.37 hours.

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the day number n for 15 May?
   A. 135
   B. 120
   C. 105
   D. 145
   Answer: A. 31+28+31+30+15 = 135. (Key point: Day number)

2. What is the hour angle at 10:00 solar time?
   A. +30°
   B. -15°
   C. +150°
   D. -30°
   Answer: D. ω = 15(10-12) = -30°, negative in the morning. (Key point: Hour angle)

3. Given cos θz = 0.8756, how are zenith and altitude related?
   A. Altitude = 28.88° and zenith = 61.12°
   B. Altitude = zenith = 28.88°
   C. θz = 28.88° and altitude = 90° - θz = 61.12°
   D. Altitude = 180° - 28.88°
   Answer: C. Altitude and zenith are complementary. (Key point: Zenith and altitude)

4. How is day length found?
   A. N = 2ω/15 using the hour angle at 10:00
   B. cos ωs = -tanφ tanδ, then N = 2ωs/15
   C. N = 24 cos δ
   D. N = 12 + δ
   Answer: B. With ωs = 100.29°, N = 13.37 h. (Key point: Day length)

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