Solar power is one of the most accessible renewable energy sources in our planet. Also it is the most abundant source of energy, delivering about 173,000 Terrawats on Earth every second. This power is more than 10,000 times the total demand in our plant. Primary objective of this article is to estimate the net monthly insolation for any place on earth. Designing the solar system is published in a separate article.
Amount of Solar irradiance falling on a unit surface area (W/m^2) is depending Sun’s position at that time.
Sun’s position
Sun’s position in any time is given by two angles Altitude angle (β) and Azimuth angle (φs)
Altitude angle
Is the angle between the sun and the local horizon directly beneath the sun.
Azimuth angle
Is the angle between local horizon beneath the sun and due South. Azimuth angle towards East are taken positive while towards West are negative.
Calculation of Azimuth and Altitude angles
In order to calculate Altitude and azimuth angles, we need to know solar declination angle, latitude angle and hour angle.
Solar declination angle δ
is the angle between the light rays from the Sun and the Earth’s equator.
Solar declination angle can be calculated using the following equation
δ=23.45*Sin(360/365 (n-81))
Where n is the number of days from the beginning of the year. Jan 1st – n=1, Dec 31st – n= 365
Solar declination angle for 21st of Each month
|
Date |
δ |
| 21st of Jan | -20.1 |
| 21st of Feb | -11.2 |
| 21st of Mar | 0 (equinox) |
| 21st of Apr | 11.6 |
| 21st of May | 20.1 |
| 21st of Jun | 23.4 |
| 21st of Jul | 20.4 |
| 21st of Aug | 11.8 |
| 21st of Sep | 0(equinox) |
| 21st of Oct | -11.8 |
| 21st of Nov | -20.4 |
| 21st of Dec | -23.3 |
In this calculation, solar declination angle is assumed to be constant throughout the month. Although for the actual case, it changes each and every day.
Latitude angle
Is the line joining any position on surface of earth to the center of the earth is tilted above the equator.
Hour angle
The angle by which the earth must rotate before the Sun will be over the local meridian. Earth is rotating 15 degrees in every hour. Therefore hour angle can be calculates as,
H=15˚×hours before solar noon.
Here solar noon time can be assumed as 12 noon. But the actual value is a little more or less.
Following example shows real values of Solar noon time for the month of September 2019 for Colombo, Sri Lanka.
| Date | Time (hours) |
| 0 | 12:10 (88.6°) |
| 1 | 12:10 (88.9°) |
| 2 | 12:10 (89.3°) |
| 3 | 12:09 (89.7°) |
| 4 | 12:09 (90.0°) |
| 5 | 12:09 (89.6°) |
| 6 | 12:08 (89.2°) |
| 7 | 12:08 (88.8°) |
| 8 | 12:08 (88.5°) |
| 9 | 12:07 (88.1°) |
| 10 | 12:07 (87.7°) |
| 11 | 12:07 (87.3°) |
| 12 | 12:06 (86.9°) |
| 13 | 12:06 (86.6°) |
| 14 | 12:05 (86.2°) |
| 15 | 12:05 (85.8°) |
| 16 | 12:05 (85.4°) |
| 17 | 12:04 (85.0°) |
| 18 | 12:04 (84.6°) |
| 19 | 12:04 (84.3°) |
| 20 | 12:03 (83.9°) |
| 21 | 12:03 (83.5°) |
| 22 | 12:03 (83.1°) |
| 23 | 12:02 (82.7°) |
| 24 | 12:02 (82.3°) |
| 25 | 12:02 (81.9°) |
| 26 | 12:01 (81.5°) |
| 27 | 12:01 (81.1°) |
| 28 | 12:01 (80.8°) |
| 29 | 12:00 (80.4°) |
Calculation of hour angle from 6 a.m to 6 p.m assuming solar noon is at 12 noon.
| Time | Hours before solar noon | Hour angle |
| 6 a.m | 6 | 90 |
| 7 a.m | 5 | 75 |
| 8 a.m | 4 | 60 |
| 9 a.m | 3 | 45 |
| 10 a.m | 2 | 30 |
| 11 a.m | 1 | 15 |
| 12 noon | 0 | 0 |
| 1 p.m. | 1 | -15 |
| 2 p.m. | 2 | -30 |
| 3 p.m. | 3 | -45 |
| 4 p.m. | 4 | -60 |
| 5 p.m. | 5 | -75 |
| 6 p.m. | 6 | -90 |
Altitude angle and azimuth angle can be calculated using Solar declination angle, latitude angle and hour angle.
Sinβ=CosLCosδCosH+SinLSinδ
Sin(φ)=(CosδSinH)/Cosβ
Since there are two values for inverse of sin, respective angle is selected from the following condition,
if CosH > tanδ/tanL, then |∅| < 90 , if not |∅|>90
Following tables shows calculated values for altitude angle and azimuth angle for a location of 6.9 degrees latitude (a place in Sri Lanka).
| Date | Time | Altitude angle (β) | Azimuth angle (φs) |
| 21st of Jan | 6 a.m. | -2.37 | 69.99623435 |
| 7 a.m. | 11.53 | 67.75164235 | |
| 8 a.m | 25.13 | 63.90791742 | |
| 9 a.m. | 38.15 | 57.58552483 | |
| 10 a.m. | 49.98 | 46.88812 | |
| 11 a.m. | 59.20 | 28.32944323 | |
| 12 a.m. | 62.96 | 0 | |
| 1 p.m. | 59.20 | -28.32944323 | |
| 2 p.m. | 49.98 | -46.88812 | |
| 3 p.m. | 38.15 | -57.58552483 | |
| 4 p.m. | 25.13 | -63.90791742 | |
| 5 p.m. | 11.53 | -67.75164235 | |
| 6 p.m. | -2.37 | -69.99623435 |
Accordingly we can calculate values of latitude and azimuth for 21st of each month.
Sunpath diagram
When we plot altitude against azimuth for different days of month, we get a diagram called Sunpath diagram.
Estimation of total clear sky insolation (W/m^2)
Total clear sky insolation on a collector is depending upon three main radiation factors.
- Direct beam radiation (Ibh)
- Diffuse beam radiation (Idh)
- Reflected beam radiation (Irh)
Total clear sky insolation on a tilted collector is given by following equation.
I(horizontal) = Idh + Ibh + Irh
Total clear sky insolation on a horizontal collector is given by following equation.
I(horizontal) = Idh + Ibh
In order to estimate these three components, we need to know following factors.
Extraterrestrial flux (A)
The extra-terrestrial flux is the value of solar power ‘intensity’ expressed in watts per meter square which is incident on the earth but does not face any disturbance from the Earth’s atmosphere.
A= 1160 + 75*sin((360/365)*(n-275))
Value of A for 21st of Each Month
| Date | A (W/m^2) |
| 21st of Jan | 1230 |
| 21st of Feb | 1215 |
| 21st of Mar | 1186 |
| 21st of Apr | 1136 |
| 21st of May | 1104 |
| 21st of Jun | 1088 |
| 21st of Jul | 1085 |
| 21st of Aug | 1107 |
| 21st of Sep | 1151 |
| 21st of Oct | 1192 |
| 21st of Nov | 1221 |
| 21st of Dec | 1233 |
Optical depth (k)
Is the absorption coefficient of solar radiation by the atmosphere. K is a dimensionless constant
k = 0.174+0.035((360/365)*(n-100))
Value of k for 21st of Each Month
| Date | k |
| 21st of Jan | 0.142 |
| 21st of Feb | 0.144 |
| 21st of Mar | 0.156 |
| 21st of Apr | 0.18 |
| 21st of May | 0.196 |
| 21st of Jun | 0.205 |
| 21st of Jul | 0.207 |
| 21st of Aug | 0.201 |
| 21st of Sep | 0.177 |
| 21st of Oct | 0.16 |
| 21st of Nov | 0.149 |
| 21st of Dec | 0.142 |
Air mass ration (m)
The air mass coefficient defines the direct optical path length through the Earth’s atmosphere, expressed as a ratio relative to the path length vertically upwards, i.e. at the zenith.
m = (1/sinβ)
Sky diffusion factor (C)
Solar radiation is diffused by molecules in the atmosphere. This is a dimensionless constant.
C=0.095+0.04*sin((360/365)*(n-100))
Value of C for 21st of Each Month
| Date | c |
| 21st of Jan | 0.058 |
| 21st of Feb | 0.06 |
| 21st of Mar | 0.071 |
| 21st of Apr | 0.097 |
| 21st of May | 0.121 |
| 21st of Jun | 0.134 |
| 21st of Jul | 0.136 |
| 21st of Aug | 0.122 |
| 21st of Sep | 0.092 |
| 21st of Oct | 0.073 |
| 21st of Nov | 0.063 |
| 21st of Dec | 0.057 |
Total insolation on a tilted collector (W/m^2)
When the collector surface is inclined by an angle Σ, faced towards an azimuth angle of Φc, total insolation (direct + diffuse + reflected) on the collector is given by the following equation.
I = Ae^(-km) {(Cosβ*Cos(φs-φc)*SinΣ+Sinβ*CosΣ+C*((1+CosΣ)/2)+ρ*(sinβ+C)*((1-CosΣ)/2)}
Here ρ is the reflectance coefficient of the surface which the solar panels are mounted. It can vary from 0.1 to 0.8.
For an asbestos roof, ρ is approximately 0.1. normal ground/grass = 0.2, snow = 0.8
Total insolation on a horizontal collector (W/m^2)
Direct beam radiation for a horizontal collector is given by
Idh = A*exp(-k*m) * Sinβ
Diffuse beam radiation for a horizontal collector is given by
Idh = A*exp(-k*m) * C
Therefore total insolation (Ibh+Idh) on a horizontal collector is given by
I = A*exp(-k*m) ( Sinβ + C)
Daily insolation (kWh/m^2/day)
Daily insolation is calculated by taking the sum of hour by hour insolation throughout the day.
Sample calculation – Following example shows sample calculation of daily insolation for 21st of January for a tilted surface of 6.9 degrees and oriented towards due south (0 orientation).
Accordingly, daily insolation was calculated for 21st of each month for 6.9 degrees tilt angle and 0 degrees orientation. Results are shown in the following table.
Estimation of net monthly insolation by considering shading factor (kWh/m^2/month)
For this approximation, daily insolation is assumed to be constant throughout the month. Also, by considering average number of rainy days per month, insolation is assumed to be dropped by 50% on a rainy day.
| Month | Average rainy days |
| Jan | 7 |
| Feb | 4 |
| Mar | 8 |
| Apr | 14 |
| May | 16 |
| Jun | 14 |
| Jul | 11 |
| Aug | 9 |
| Sep | 13 |
| Oct | 19 |
| Nov | 16 |
| Dec | 10 |
Following example shows a sample calculation for January.
For a tilt of 7 degrees and facing due south.
Daily clear sky insolation = 7.9 kWh/m^2/day
Average number of rainy days = 7
Number of clear sky days = 31-7 = 24
Net monthly insolation = 7.9 * 24 + 7.9*7*0.5 = 217.27 kWh/m^2/month
Accordingly, net monthly insolation was approximated for each month. Results are
shown below.
Sri Lankan Dream 