Solar Irradiance Calculator (Clear-Sky, Tilted Panel)
Estimate clear-sky solar irradiance in W/m² on a horizontal surface and on a tilted solar panel for any latitude, date and time — with sun elevation, air mass, angle of incidence and panel output.
How it works
δ = 23.45° sin[360° (284 + n) ÷ 365] · ω = 15° × (solar time − 12) · cos θz = sin φ sin δ + cos φ cos δ cos ω · AM = 1 ÷ [cos θz + 0.50572 (96.07995° − θz)^−1.6364] · DNI = 1353 × [(1 − 0.14h) 0.7^(AM^0.678) + 0.14h] W/m² (h in km) · diffuse ≈ 0.1 DNI · panel = DNI cos θ + diffuse (1 + cos β) ÷ 2, cos θ from Duffie & Beckman eq. 1.6.2
On a clear day the sunlight reaching the ground depends mainly on how much air it passes through — the air mass, which is 1 with the sun overhead and grows as the sun gets lower. The Meinel model used by PVEducation estimates the direct beam from the air mass and adds about 10% for diffuse skylight; the Kasten–Young formula gives the air mass. The angle between the sun and the panel’s normal (angle of incidence) then sets how much of the beam the panel catches. Real irradiance is lower with haze, cloud and pollution, so these are clear-sky upper values; use solar time (local clock time corrected for longitude and the equation of time).
Worked example
At 40° N on 21 June (day 172) at solar noon, the declination is 23.45°, the sun is 73.4° high and the air mass is 1.043. DNI = 1353 × 0.7^(1.043^0.678) = 937 W/m². A panel tilted 30° facing south meets the beam at 13.4°, receiving 911 W/m² direct plus 87 W/m² diffuse — about 999 W/m² in total. A 2 m², 20% efficient panel would then produce about 400 W.
Sources: PVEducation — Air mass and solar radiation intensity (Meinel model) · Duffie & Beckman — Solar Engineering of Thermal Processes, chapter 1 (angle of incidence)
Frequently asked questions
What is solar irradiance?
The power of sunlight falling on a surface, in watts per square metre. Outside the atmosphere it averages 1,361 W/m² (the solar constant); at the ground on a clear day it is typically up to about 1,000 W/m².
Why do solar panels use 1,000 W/m² as standard?
Standard Test Conditions rate panels at 1,000 W/m², 25 °C cell temperature and air mass 1.5 — roughly a clear midday sun at mid-latitudes.
What is air mass?
The length of the sunlight’s path through the atmosphere relative to the vertical path. With the sun 48° above the horizon the air mass is 1.5; near sunrise it can exceed 10.
What tilt gives the most irradiance?
At noon a panel facing the equator and tilted so the sun hits it perpendicularly (tilt = latitude − declination) catches the most. Over a year a tilt close to the latitude is a good compromise.
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This calculator is for educational purposes. Double-check important results before acting on them.