What is Extraterrestrial solar radiation? Explain.

Extraterrestrial solar radiation refers to the solar radiation that reaches the Earth's atmosphere before it is affected by atmospheric absorption, scattering, or reflection. It is the total amount of solar energy incident on a plane perpendicular to the Sun's rays at the mean distance of the Earth from the Sun. This radiation is essentially constant, though it varies slightly due to the Earth's elliptical orbit. It is often referred to as the solar constant, though its value is not perfectly constant. The extraterrestrial solar radiation is the source of all solar energy that eventually reaches the Earth's surface and is crucial for understanding and predicting solar energy availability.

The intensity of extraterrestrial solar radiation is typically measured in watts per square meter (W/m²) or joules per second per square meter (J/s/m²). The accepted value for the solar constant (Gsc) is approximately

The amount of solar radiation received at the top of the atmosphere depends on several factors:

• Earth's orbital position: The Earth's orbit is elliptical, meaning the distance from the Sun varies throughout the year. This variation causes the solar irradiance at the top of the atmosphere to fluctuate by about ±3.4%.
• Solar activity: The Sun's output is not perfectly constant and can vary slightly due to solar cycles.

The extraterrestrial solar radiation is a fundamental parameter used in:

• Solar energy system design: It provides the maximum possible solar energy input for calculations.
• Atmospheric modeling: It serves as the baseline for understanding atmospheric effects on solar radiation.
• Climatology: It is used in studies of Earth's energy balance.

At Suva (ϕ = −180) at 9 a.m. on 20 May, the irradiance measured on a horizontal plane was Gh = 1.0 MJh−1 m−2. Determine the angle θz between the beam radiation and the vertical, and hence find the irradiance G∗ = (Gb + Gd)∗ measured in the beam direction.

(Assume that Gd << Gb, as might be the case on a clear day).

Given Data:

• Location: Suva
• Latitude (ϕ): −18° (Note: The provided ϕ = -180 is likely a typo and should be -18° for Suva, Fiji. Assuming ϕ = -18° for calculation.)
• Time: 9 a.m.
• Date: May 20
• Irradiance on horizontal plane (Gh): 1.0 MJh⁻¹m⁻²
• Assumption: Gd << Gb (diffuse radiation is negligible compared to beam radiation)

Calculations:

1. Convert Gh to W/m²:

• 1 MJ = 10⁶ J
• 1 hour = 3600 seconds
• Gh = 1.0 MJh⁻¹m⁻² = (1.0 × 10⁶ J) / (3600 s × m²) = 277.78 J/s/m² = 277.78 W/m²

2. Calculate the Hour Angle (ω):

• Standard time is usually 12:00 p.m. (solar noon). 9 a.m. is 3 hours before solar noon.
• Hour angle ω = (Time difference from solar noon) × 15°/hour
• ω = 3 hours × 15°/hour = 45°

3. Calculate the Declination Angle (δ):

• The declination angle can be approximated using the formula:
δ ≈ 23.45° sin(360/365 × (n + 284))
• Where 'n' is the day number of the year.
• May 20th: January (31) + February (28) + March (31) + April (30) + May (20) = 140. So, n = 140.
• δ ≈ 23.45° sin(360/365 × (140 + 284))
• δ ≈ 23.45° sin(360/365 × 424)
• δ ≈ 23.45° sin(1.095 × 424)
• δ ≈ 23.45° sin(420.9°)
• δ ≈ 23.45° × (-0.485)
• δ ≈ -11.37°

4. Calculate the Angle of Incidence (θ):

• The angle of incidence (θ) between the beam radiation and the surface normal is given by:
cos(θ) = sin(ϕ)sin(δ) + cos(ϕ)cos(δ)cos(ω)
• Using ϕ = -18°, δ = -11.37°, ω = 45°:
• cos(θ) = sin(-18°)sin(-11.37°) + cos(-18°)cos(-11.37°)cos(45°)
• cos(θ) = (-0.3090)(-0.1974) + (0.9511)(0.9803)(0.7071)
• cos(θ) = 0.0609 + 0.6592
• cos(θ) = 0.7201
• θ = arccos(0.7201) ≈ 43.95°

5. Determine the Angle θz between the beam radiation and the vertical:

• The angle θz between the beam radiation and the vertical is the zenith angle. This is exactly the angle of incidence (θ) when the surface is horizontal.
• Therefore, θz = θ ≈ 43.95°

6. Find the irradiance G∗ measured in the beam direction:

• The irradiance on a horizontal surface (Gh) is related to the beam irradiance on a surface normal to the sun (Gb) by:
Gh = Gb cos(θ) + Gd
• Given that Gd << Gb, we can approximate Gh ≈ Gb cos(θ).
• We are asked to find G∗ = (Gb + Gd)∗, which represents the total irradiance in the beam direction. On a clear day, Gd is small, and the direct beam component (Gb) dominates. The irradiance measured in the beam direction is essentially the beam irradiance itself.
• From Gh ≈ Gb cos(θ), we can find Gb:
• Gb = Gh / cos(θ)
• Gb = 277.78 W/m² / cos(43.95°)
• Gb = 277.78 W/m² / 0.7201
• Gb ≈ 385.75 W/m²
• Since Gd << Gb, G∗ ≈ Gb.
• Therefore, G∗ ≈ 385.75 W/m².

Summary:

• The angle θz between the beam radiation and the vertical is approximately 43.95°.
• The irradiance G∗ measured in the beam direction is approximately 385.75 W/m².