A quick production estimate closes more deals than a simulation nobody reads. System kW times peak sun hours, times 30 days, times a real-world derate — monthly kWh.
An 8 kW system at 5 sun hours with a 0.8 derate produces about 960 kWh a month — the number to set against the utility bill.
The example
Three systems, each with a size, local sun hours, and derate factor.
| A | B | C | D | E | |
|---|---|---|---|---|---|
| 1 | System | Size (kW) | Sun hrs / day | Derate | kWh / month |
| 2 | 8 kW system | 8 | 5 | 0.80 | 960 |
| 3 | 6 kW shaded | 6 | 4.2 | 0.75 | 567 |
| 4 | 12 kW ranch | 12 | 5.5 | 0.80 | 1,584 |
The formula
Four factors, multiplied straight through:
How it works
Each factor scales the ideal output toward reality:
B2is the DC system size in kilowatts.C2is average peak sun hours per day for the location.*30scales a day to a month, and the derateD2(typically 0.75–0.85) accounts for inverter losses, temperature, and soiling.
Use a month-by-month sun-hours table for a full-year model; this single-line version is the sales-call estimate.
Try it: interactive demo
Enter the system size, sun hours, and derate factor.
Variations
Annual estimate
Swap the 30 for 365 to see yearly production.
Dollar value of production
Multiply the kWh by the utility rate for monthly savings.
Pitfalls & errors
Skipping the derate overstates production by 20–25% — a promise the first utility bill breaks loudly.
Sun hours vary more by month than customers expect; quote the annual average, not July's.
Practice workbook
Frequently asked questions
What derate should I use?
Where do sun hours come from?
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