Case 1
- Panel or array wattage
- 400W
- Peak sun hours per day
- 5h
- Performance ratio
- 0.75
1.500kWh
Open with these valuesSolar Panel Output Calculator
1.500kWh/day
Result: 1.500 kWh/dayA 400 W panel at a site with 5 peak sun hours and a performance ratio of 0.75 makes 1.5 kWh on an average day. Peak sun hours are not daylight hours — they count only sunshine at full strength. Both inputs are local estimates, so treat the result as a planning figure, not a meter reading.
The numbers at a glance
Held fixed: Panel or array wattage 400 W, Peak sun hours per day 5.00 h.
| Performance ratio | Result (kWh) |
|---|---|
| 0.60 | 1.200 |
| 0.70 | 1.400 |
| 0.75Your value | 1.500 |
| 0.80 | 1.600 |
| 0.90 | 1.800 |
Worked examples
Case 2
- Panel or array wattage
- 300W
- Peak sun hours per day
- 4h
- Performance ratio
- 0.8
0.960kWh
Open with these valuesCase 3
- Panel or array wattage
- 6000W
- Peak sun hours per day
- 4.5h
- Performance ratio
- 0.78
21.060kWh
Open with these valuesHow it's calculated
kWh per day = W × peak sun hours × ratio ÷ 1000
- StepEnter the rated wattage from the spec sheet — the whole array, if several.
- StepEnter the peak sun hours for your location as a year-round average.
- StepEnter a performance ratio: about 0.75 for a good rooftop system.
- ResultRead the daily energy; multiply by 365 for a rough year.
Reference table
| Wattage, sun hours, ratio | Situation | Daily energy |
|---|---|---|
| 1, 1, 1 | Unit check: one watt for one hour | 0.001 |
| 400, 0, 0.75 | No sun at all | 0 |
| 300, 4, 0.8 | Older panel, northern Europe | 0.96 |
| 400, 5, 0.75 | Typical rooftop panel | 1.5 |
| 400, 5, 1 | The same panel with no losses at all | 2 |
| 6000, 4.5, 0.78 | A 6 kWp home array | 21.06 |
Questions
How do I calculate a solar panel's daily output?
Multiply the panel wattage by the peak sun hours and a performance ratio, then divide by 1000. A 400 W panel with 5 peak sun hours and a ratio of 0.75 produces 1.5 kWh per day.
What are peak sun hours?
Peak sun hours are the equivalent number of hours per day at which sunlight reaches full strength, 1000 W/m². They are not daylight hours — a long hazy day may still deliver only a few. Typical values run from about 2.5 to 3 in northern climates up to 5 to 6 in sunny regions, and they are a local estimate rather than a constant.
What performance ratio should I use?
It is a factor between 0 and 1 that absorbs real-world losses: inverter and wiring losses, heat, dust and shading. It is an empirical rule of thumb, not a measured constant — 0.75 is the usual figure for a well-installed rooftop system, while modern solar parks are expected to exceed 0.8. Use a lower value where heat or shading is severe.
How do I estimate a whole array?
Enter the combined wattage of all the panels. Ten 400 W panels make a 4000 W array, so at 5 peak sun hours and a ratio of 0.75 the daily output is 15 kWh. The formula scales directly with total wattage.
How do I turn the daily figure into a year?
Multiply by 365 for a rough annual estimate — 1.5 kWh a day is about 548 kWh a year. Because peak sun hours swing with the season, feed in a year-round average rather than a summer day.
Why is real output lower than the rated wattage suggests?
Rated wattage is measured in the lab under full sun at 25 °C. Panels rarely see those conditions all day, and the performance ratio trims the result for heat, dust, shading and losses in the wiring and inverter. That is why a 400 W panel yields far less than 400 Wh in any single hour of normal operation.
Sources and last check
Information, not professional advice.