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How to use
- Enter System size kW.
- Enter Peak sun hours.
- Enter Efficiency %.
- Enter Days.
- Calculate, then review power, price and period before comparing.
How it is calculated
Output = system kW × sun hours × efficiency × days
Example
A 4.5 kW system with 5.5 sun hours and 80% efficiency averages 19.8 kWh daily and about 7,227 kWh over 365 days.
Important notes
Using one sun-hours value for the whole year is simplified. For better annual estimates, use monthly site data or a simulation that accounts for season and shade.
Worked examples and interpreting results
| Case | Calculation | Result |
|---|---|---|
| Daily average | 4.5 × 5.5 × 0.8 | 19.8 |
| Annual output | 19.8 × 365 | 7,227 |
Frequently asked questions
Does 365 days mean full sun every day?
No. Peak sun hours are intended to summarize average irradiance, though this remains a simplified model.
Is the result an actual utility bill?
No. It is a calculation from your usage, price and inputs; real tariffs, taxes and fixed fees may add other items.
How can I improve accuracy?
Use measured or device-specific power and usage plus the actual electricity price for the same period.