Inputs
Formula used
This calculator starts from irradiance, which is power per unit area: P = I × A. It then applies practical factors and integrates over time.
I = solar irradiance (W/m^2), A = area (m^2), t = time (s).
Effective irradiance: I_eff = I × (1 + g), where g is albedo gain.
Incident power: P_inc = I_eff × A × f_tilt × τ × (1 − L_shade)
Useful power: P_use = P_inc × η
Energy: E = P × t, and Wh = J / 3600.
How to use this calculator
- Enter the measured or estimated solar irradiance in W/m^2.
- Choose your collector area and its unit.
- Select the exposure time and unit.
- Set efficiency and optional loss or gain factors.
- Pick “Incident” or “Useful” reporting mode.
- Click Calculate to view results above the form.
- Use the CSV or PDF buttons to export the report.
Example data table
| Irradiance (W/m^2) | Area (m^2) | Time (h) | Efficiency (%) | Useful Energy (kWh) |
|---|---|---|---|---|
| 1000 | 1.0 | 1.0 | 20 | 0.2000 |
| 850 | 2.5 | 3.0 | 18 | 1.1475 |
| 600 | 10.0 | 5.0 | 15 | 4.5000 |
Example assumes tilt factor 1.0, transmittance 1.0, and zero shading.
Solar irradiance and energy output overview
Solar irradiance is the instantaneous power from sunlight arriving on a surface, measured in W/m^2. When you multiply irradiance by collector area and exposure time, you obtain energy. This tool also applies practical field factors (tilt, transmittance, shading, albedo, and efficiency) so your estimate reflects real installations, not just ideal laboratory conditions.
1) Typical irradiance ranges you can expect
Under clear skies near solar noon, a horizontal surface often sees 800–1100 W/m^2, while hazy or thin-cloud conditions may drop to 300–700 W/m^2. Early morning and late afternoon values commonly fall below 400 W/m^2 due to a longer atmospheric path and lower incidence angles.
2) From irradiance to incident power on a surface
Incident power is computed as P_inc = I_eff × A × f_tilt × τ × (1 − L_shade). Area is converted to m^2, and time is converted to seconds to keep units consistent. If you select “Incident on surface,” the output ignores conversion efficiency and shows raw solar energy delivered to the collector plane.
3) Practical losses that dominate real output
Shading is often the largest single loss: a 20% shading loss directly reduces energy by 20%. Transmittance models cover glass, dust, and atmospheric clarity; using 0.90 instead of 1.00 represents a 10% reduction. The tilt factor captures orientation, tracking, or mild concentration effects and is kept within a realistic range.
4) Efficiency and “useful” energy conversion
Efficiency (η) converts incident power into useful electrical or thermal output: P_use = P_inc × η. Typical module efficiencies are commonly in the 15–23% range, while some thermal collectors can be higher depending on temperature rise and system design. Set efficiency to 0% to compare purely incident energy.
5) Understanding daily energy with peak-sun-hours
A convenient planning metric is peak sun hours (PSH). For example, 5 PSH means the day’s sunlight is equivalent to 5 hours at 1000 W/m^2. If you have 2 m^2 area and 20% efficiency, an ideal estimate is: 1000 × 2 × 5 h × 0.20 = 2.0 kWh before losses. Apply tilt, transmittance, and shading to refine it.
6) Albedo gain for reflective surroundings
Albedo gain represents extra light reflected from surfaces like snow, light concrete, or bright roofs. A 10% albedo gain increases irradiance by 10% in this model. Use small values unless you have a clearly reflective environment; the gain is capped to prevent unrealistic outputs.
7) Recommended workflow for reliable estimates
Start with measured irradiance when possible. If not, use a conservative average for your location and season, then run multiple scenarios: clear-sky peak, average daytime, and worst-case haze. Document your assumed losses and export CSV/PDF so stakeholders can trace each factor and reproduce results.
8) Example scenario with numbers
Suppose I = 900 W/m^2, A = 3 m^2, time = 2 h, f_tilt = 0.95, τ = 0.92, shading = 10%, η = 18%. Incident power ≈ 900×3×0.95×0.92×0.90 = 2121 W. Useful power ≈ 2121×0.18 = 382 W. Useful energy over 2 h ≈ 0.764 kWh.
FAQs
1) What is the difference between irradiance and energy?
Irradiance is power per area (W/m^2) at a moment. Energy accumulates over time (J or Wh). Multiply irradiance by area and time, then apply system factors to estimate energy.
2) Which reporting mode should I use?
Use “Incident on surface” to see raw sunlight delivered to the collector plane. Use “Useful after efficiency” to estimate output after conversion efficiency, closer to what a system can deliver.
3) What values are reasonable for transmittance?
For clean glazing and clear air, 0.90–1.00 is common. Dust, haze, or older covers can lower it. If you are unsure, run both 0.90 and 0.98 to bracket realistic performance.
4) How do I choose a tilt factor?
Use 1.0 for a reference orientation. Values below 1.0 represent off-angle incidence or poor alignment. Values slightly above 1.0 can represent tracking or mild concentration, but keep it conservative.
5) Why does shading reduce output so strongly?
This model applies shading as a direct fractional loss to incident power, which is a good first-order estimate. In real PV strings, partial shading can cause additional mismatch losses, so conservative inputs are recommended.
6) How can I estimate daily energy quickly?
Use peak sun hours: set time to PSH hours and irradiance near 1000 W/m^2, then apply your factors. This approximates the day’s integrated sunlight and is useful for early feasibility estimates.
7) Why are results shown in both joules and watt-hours?
Joules are the SI unit used in physics. Watt-hours and kilowatt-hours are common in energy billing and system sizing. Showing both helps you compare scientific calculations with practical power-system metrics.