In the previous articles of this series, we examined whether sufficient shadow-free rooftop area is available and how household electricity consumption can help determine the required capacity of a solar PV system. But knowing how large a system we need raises another equally important question: how much electricity will that system actually generate at a particular location? The answer determines not only the technical performance of the system but also its financial viability. After all, a solar power plant is an investment, and its economic value ultimately depends on how much money it can save—or earn—over its operating life.
The amount of energy generated by a solar PV plant is highly location-specific. It depends on several factors, including the length of the day, the intensity of solar radiation, and the orientation and tilt of the PV modules. The orientation and inclination of the modules determine the angle at which the Sun’s rays strike their surface; the closer the solar rays are to being perpendicular to the module surface, the greater the irradiance intercepted by the module. Therefore, before estimating the energy output of a solar PV system, it is important to understand the solar conditions prevailing at a particular location. Let us begin with one of the most fundamental parameters—the length of the day.
Earth–Sun Geometry and Its Importance in Solar Energy
Day and night occur because the Earth rotates on its axis from west to east. As a result of this rotation, the Sun appears to move across the sky in the opposite direction—rising in the east, reaching its highest position around solar noon, and setting in the west. This apparent movement continuously changes the position of the Sun relative to a particular location on the Earth’s surface. Understanding this position is important in solar-energy calculations because the amount and angle of solar radiation received by a surface vary throughout the day.
At the same time, the Earth revolves around the Sun in a slightly elliptical orbit, completing one revolution in approximately one year. The Earth’s rotational axis is tilted by approximately 23.450 from the perpendicular to the plane of its orbit around the Sun. As is familiar from basic geography, the combined effect of this axial tilt and the Earth’s revolution around the Sun produces the seasonal variation experienced during the year. In solar-energy studies, however, this phenomenon has another important consequence: it changes both the length of the day and the angle at which solar radiation reaches the Earth’s surface. Consequently, the solar energy available at a particular location varies not only during the course of a day but also from one season to another, directly influencing the potential energy generation of a solar PV system.
This seasonal change in the apparent position of the Sun is represented by the solar declination angle (δ). Solar declination is the angle between the Sun’s rays and the Earth’s equatorial plane. It changes continuously throughout the year, ranging approximately from +23.450 at the June solstice to −23.450 at the December solstice, while becoming approximately 00 at the March and September equinoxes. For any particular day of the year, the declination angle can be estimated as
where n is the day number of the year, with 1 January taken as n = 1.
While declination describes the seasonal north–south position of the Sun, its apparent east–west movement during a day is represented by the solar hour angle (ω). Since the Earth completes a rotation of 3600 in approximately 24 hours, it rotates through 150 per hour, or 10 every four minutes. This relationship between angular displacement and time provides the basis for defining the solar hour angle.
The solar hour angle should not be confused with geographic longitude. It represents the angular displacement of the Sun from the local meridian due to the Earth’s rotation. At solar noon, when the Sun crosses the local meridian and reaches approximately its highest position in the sky, the hour angle is defined as
Since the apparent solar position changes by approximately 150 for every hour of solar time, the hour angle can be calculated from
where tsolar is the local solar time expressed in hours. Using the standard convention, the hour angle is negative before solar noon and positive after solar noon. Thus, at 11:00 solar time ω = −150, at 10:00 it is −300, at 12:00 noon it is 00, at 1:00 p.m. it becomes +150, and at 2:00 p.m. it is +300.
The combined effects of latitude (φ), solar declination (δ), and the Earth’s rotation determine how long the Sun remains above the horizon at a particular location. At sunrise and sunset, the corresponding angular displacement from solar noon is represented by the sunrise/sunset hour angle. Its magnitude, commonly expressed as the sunset hour angle (ωₛ), can be calculated as:
where φ is the latitude of the location and δ is the solar declination for the selected day. Once ωₛ is known, the theoretical length of the day ‘N’ can be calculated directly:
where N is the day length in hours. The factor 15 appears because the Earth rotates through approximately 150 in one hour.
These simple relationships provide a powerful way of understanding solar availability at any location. Latitude tells us where the location lies on the Earth, declination tells us where the Sun appears to be seasonally, and the hour angle describes its apparent movement during the day. Together, these parameters allow us to estimate day length and solar sunrise and sunset times for any selected location and date—an important first step toward estimating the solar radiation available to a PV system.
Calculating Day Length, Sunrise and Sunset for Solar Energy
Let us now apply these relationships to a real location. Consider Nainital, situated at approximately 29.380 N latitude, and determine its theoretical day length on 23 December, close to the winter solstice. For this date, the solar declination calculated from the preceding equation is approximately −23.40. Substituting the latitude φ = 29.380 and declination δ = −23.40 into the sunset-hour-angle equation gives:
This angle represents the interval from solar noon to sunset. Since the Sun follows the corresponding interval before solar noon, the total theoretical day length N is
Thus, on 23 December, Nainital receives approximately 10.12 hours of theoretical daylight, equivalent to about 10 hours 7 minutes. In other words, the Sun remains above the horizon for approximately 10 hours and 7 minutes. Since solar noon lies at the midpoint of this theoretical daylight period, half of this duration occurs before solar noon and the other half after it. Because solar noon divides this period into two equal halves, half of the day length is approximately 5 hours 3 minutes 36 seconds. Subtracting this interval from 12:00 solar noon gives the sunrise time,
while adding it to solar noon gives the sunset time,
Therefore, for Nainital on 23 December, the calculated sunrise is approximately at 6:56 a.m., sunset approximately 5:04 p.m., and theoretical day length approximately 10 hours 7 minutes. The same procedure can be followed for any location and any date: first find the location’s latitude, calculate the solar declination for the required date, determine the sunset hour angle, and then convert that angle into day length and sunrise/sunset times.
One important distinction should be kept in mind when applying this calculation to a solar PV system. The value calculated above is more accurately called astronomical or theoretical day length, rather than actual bright sunshine hours. Clouds, fog, mountains, buildings, trees and other obstructions can reduce the period during which useful direct solar radiation reaches a PV array. Also, the calculated sunrise and sunset are in local solar time; they will not necessarily be identical to the times shown on a clock. Conversion to standard clock time requires corrections for the location’s longitude and the equation of time. Nevertheless, this calculation provides an excellent first step for understanding the solar availability of a location before moving on to solar radiation and expected PV energy generation.
Thus, day length tells us how long the Sun is theoretically above the horizon, but it does not by itself tell us how much solar energy is available. For that, we must next consider the intensity of solar radiation received at the location—a parameter that ultimately determines how many units of electricity a solar PV system can generate.
