Incident Solar Radiation W per m2

Incident solar radiation (W/m²)

Beyond Earth's atmosphere the solar constant averages about 1,361 W/m². Once that beam reaches the ground, irradiance on a clear-sky day at solar noon drops to roughly 1,000 W/m² — and that happens to be the reference value behind the STC (Standard Test Conditions) rating used for PV panels. What actually arrives depends on the solar zenith angle, how transparent the air is, your altitude, and the cloud cover overhead. A simplified formula handles the direct beam: I = I₀ · cos(θz) · τ, where θz is the zenith angle and τ the atmospheric transmittance. Plug in θz = 30° and τ = 0.75 and you get I ≈ 1,361 · cos(30°) · 0.75 ≈ 884 W/m². In Brazil the peak-sun energy (HSP) runs 4.5-5.5 kWh/m²/day on average, and in the Northeastern sertão it goes past 6 kWh/m²/day, which ranks among the highest figures recorded anywhere.

Applications

Sizing photovoltaic and solar thermal systems. Crop modelling in agriculture, where it feeds evapotranspiration and photosynthesis estimates. Thermal comfort and HVAC load calculations in buildings. And, on the research side, climate studies and the calibration of remote sensing instruments.

FAQ

Why is the solar constant 1,361 W/m² and not 1,000? Because 1,361 W/m² is the value measured at the top of the atmosphere. On the way down, air, water vapour and aerosols absorb and scatter part of the beam, so at sea level on a clear day only about 1,000 W/m² is left.

How does the zenith angle affect irradiance? Irradiance scales with cos(θz), so at sunrise and sunset, when θz sits near 90°, the beam component drops toward zero. That is why HSP counts only the hours equivalent to 1,000 W/m².

Does altitude increase incident radiation? Yes. The higher you climb, the shorter the path the light travels through the atmosphere, so transmittance goes up and UV/visible irradiance can exceed 1,100 W/m² on clear mountain days.