Your roof receives daylight for much longer than your panel delivers its rated power. Using daylight duration as solar hours per day can therefore leave a battery short of energy overnight. Start with the solar energy reaching the panel, then account for the electrical path between the panel and the load.
What does solar hours per day mean for panel sizing?
A peak sun hour is an equivalent-duration measure of accumulated solar irradiation. For the engineering calculations in this article, use a reference irradiance of 1 kW/m² (1,000 W/m²): equivalent hours = daily irradiation in kWh/m² ÷ reference irradiance in kW/m². With that stated calculation convention, 1 kWh/m² ÷ 1 kW/m² = 1 peak sun hour, and several hours of weaker sunlight can contribute that amount.
The governing relation is daily energy in Wh = panel rating in W × peak sun hours × retained-output factor. The retained-output factor is the fraction remaining after the losses included in your calculation. Its boundary matters: panel output, energy entering a battery, and energy delivered to an AC appliance are different measurement points.
For an assumed 100 W panel and 5 peak sun hours, the ideal calculation is 100 × 5 = 500 Wh per day. If an assumed 80% of that energy reaches the chosen measurement point, the estimate becomes 100 × 5 × 0.80 = 400 Wh. Neither number is a measured result or a location forecast.
A watt-hour is the energy transferred by one watt of power operating for one hour. Watts describe the rate of energy transfer; watt-hours describe the accumulated amount. For an assumed constant 20 W load operating for an assumed 5 hours, energy = power × duration = 20 W × 5 h = 100 Wh, even though the load never draws 100 W.
Which hours are considered solar hours?
All daylight intervals with solar irradiation contribute to peak sun hours according to their intensity; there is no universal clock-time window. Calculate equivalent hours by adding each interval's irradiation and dividing the daily total by the stated reference irradiance of 1 kW/m².
For a hypothetical interval receiving 500 W/m² for 2 hours, the contribution is 0.5 kW/m² × 2 h = 1 kWh/m², or 1 peak sun hour. An interval receiving 250 W/m² for 4 hours contributes the same amount. Morning and afternoon light count, although their contribution per clock hour differs.
The irradiation input should describe the panel's orientation. A horizontal weather measurement and irradiation on a tilted roof are different inputs, especially when the sun is low. Shade from a nearby wall also changes what reaches the panel, even when the surrounding location remains sunny.
A shadow and solar-yield calculator is useful for sizing when its output represents energy reaching the intended panel surface over the relevant season. A shadow image alone does not provide peak sun hours. Check whether the calculation accounts for the duration of shading and the irradiation during those intervals.
Choose the period that matches the job. An annual daily average helps estimate annual energy, while a year-round standalone load needs attention to the demanding season and sequences of low-output days. A camper that changes parking locations has a different exposure pattern from a fixed installation; the camper solar sizing guide connects that operating pattern to panel and battery decisions.
How much solar do I need for 70 kWh per day?
Supplying 70 kWh per day requires 70 ÷ peak sun hours kilowatts of panel capacity before losses, or about 16.96 kW at an assumed 5 peak sun hours when using a 0.8256 retained-output factor calculated from defaults in the NLR (formerly NREL) PVWatts Version 5 Manual by Dobos, published in 2014. That factor combines the same manual's 14% system losses with its 96% inverter efficiency for this simplified AC energy calculation.
The same PVWatts manual specifies default total system losses of 14% and default inverter efficiency of 96%. Multiplying the remaining fractions gives (1 − 0.14) × 0.96 = 0.8256. These are model defaults, rather than a site-specific performance prediction.
The calculation is array capacity in kW = 70 kWh ÷ (peak sun hours × 0.8256). Each irradiation input in the table is an assumption for comparison, not a regional solar resource value. Capacity results are rounded to two decimal places.
| Assumed peak sun hours per day | Capacity before losses | Capacity with 0.8256 factor |
|---|---|---|
| 3 h | 23.33 kW | 28.26 kW |
| 4 h | 17.50 kW | 21.20 kW |
| 5 h | 14.00 kW | 16.96 kW |
| 6 h | 11.67 kW | 14.13 kW |
A daily energy target does not determine inverter power. Equipment operating simultaneously sets the required instantaneous output, and nighttime consumption determines the storage requirement. A grid-connected system and a standalone system can therefore need different supporting hardware despite sharing the same daily energy target.
Worked example: how do you calculate a 100 W panel's daily output step by step?
A 100 W panel produces an estimated 400 Wh per day with an assumed 5 peak sun hours and an assumed 0.80 retained-output factor. The calculation below treats 400 Wh as energy available at the selected output boundary, rather than promising battery charging performance.
- Set the panel rating. Use 100 W as the hypothetical panel's nameplate power, rather than treating 100 W as continuous daylight output.
- Set the solar input. Assume 5 peak sun hours per day for this example; substitute the relevant site and season input for an actual design.
- Calculate energy before losses. Multiply 100 W × 5 h = 500 Wh.
- Apply the stated loss allowance. Multiply 500 Wh × 0.80 = 400 Wh, using an illustrative factor rather than the AC model defaults.
- Compare energy with the load. An assumed 20 W device running for 10 hours consumes 200 Wh, leaving an arithmetic difference of 200 Wh before any losses outside the chosen boundary.
If the same panel receives 3 assumed peak sun hours, the estimate falls to 100 × 3 × 0.80 = 240 Wh. The panel rating has stayed constant; the daily energy input has changed. That is why a wattage label alone cannot answer how many watt-hours a 100 watt solar panel produces.
Amp-hours express electric charge and need a voltage to become an energy estimate. For an assumed 200 W array, 5 peak sun hours, and a 0.80 factor, energy is 800 Wh; dividing by an assumed nominal 12 V gives 66.7 Ah, while dividing by 24 V gives 33.3 Ah. Actual battery charging also depends on changing battery voltage and charge-control behavior.
What most guides leave out: how much extra solar is needed after cloudy days?
Cloudy-day recovery requires panel output sufficient for the current daily load plus the stored energy deficit divided by the desired recovery period. A design that balances average daily consumption exactly has no calculated surplus for replenishing that deficit.
Assume a CCTV camera draws 12 W continuously. Consumption is 12 × 24 = 288 Wh per day, or 12 × 24 × 30 ÷ 1,000 = 8.64 kWh over 30 days. Measure the complete installation's consumption, including any separate recorder or communications equipment, before substituting a real power value.
Two assumed days with no usable solar create a 288 × 2 = 576 Wh deficit. Recovering that deficit over 3 suitable solar days requires 576 ÷ 3 = 192 Wh of extra energy each day. The array must then supply 288 + 192 = 480 Wh daily during recovery.
At an assumed 4 peak sun hours and a 0.80 retained-output factor, the recovery target requires 480 ÷ (4 × 0.80) = 150 W. Covering daily operation alone requires 288 ÷ (4 × 0.80) = 90 W. Battery capacity, allowable charging current, and the actual weather sequence still need separate checks.
This recovery calculation adds a procurement decision beyond a location's average solar hours: specify the desired time to restore the reserve. The same energy accounting applies to overnight lighting, where the street-light panel sizing guide addresses the panel and battery relationship.
Choosing hardware: which specifications follow from solar hours?
Solar hours determine the required energy-producing panel capacity, while controller compatibility and installation geometry determine whether that capacity can be used. Specify the design-period irradiation input alongside the daily load energy.
Check panel operating voltage against the controller's charging requirements and check open-circuit voltage against the controller's input limit across the expected temperature conditions. Connector polarity and cable sizing belong in the same electrical specification. Increasing panel watts cannot correct an incompatible charging path.
Choosing a mini solar panel for IoT in a cold climate starts with the demanding season's solar input and the complete device's energy consumption. Check the controller and battery operating limits for the expected temperatures separately. An annual average alone cannot establish whether the installation will replenish its reserve during winter.
The same PVWatts manual defaults array tilt to site latitude, but the actual mounting arrangement must reflect the available surface and shading. For a constrained mobile roof, the flexible-panel guide for RV and boat roofs provides relevant context for fitting the calculated wattage to the installation.
For a fixed enclosure or unusual mounting footprint, carry the energy target, electrical requirements, and dimensional drawing into a custom panel specification. LinkSolar, a China-based B2B supplier, handles specification and quality checking through partner factories. The solar-hours calculation establishes the required capacity before a panel format is selected.
FAQ
Which hours are considered solar hours?
Every daylight interval contributes to peak sun hours in proportion to the solar irradiation received at the panel. Using a reference irradiance of 1 kW/m² for the calculation, two hypothetical hours at 500 W/m² contribute (0.5 kW/m² × 2 h) ÷ 1 kW/m² = one peak sun hour. A fixed midday clock window does not capture the whole day's energy.
How much solar do I need for 70 kWh per day?
A 70 kWh daily target requires about 16.96 kW of panels at an assumed 5 peak sun hours and a 0.8256 retained-output factor. Before losses, the result is 14 kW. Seasonal irradiation and the required storage arrangement can change the design.
What appliances will 100 watt-hours of solar energy run?
Available energy of 100 Wh runs an assumed 10 W load for 10 hours or a 50 W load for 2 hours before further losses. Runtime equals available watt-hours divided by load watts. Starting power and inverter capacity require separate checks.
What is amp-hours in batteries?
Amp-hours measure battery charge capacity, while watt-hours estimate energy by multiplying amp-hours by nominal voltage. An assumed 10 Ah battery at 12 V represents approximately 120 Wh. Usable energy depends on permitted discharge and delivery losses.
Do the 33% rule or the 20 rule determine solar hours per day?
Neither the 33% rule nor the 20 rule defines solar hours per day; calculate peak sun hours as accumulated irradiation divided by the stated reference irradiance of 1 kW/m². A percentage rule needs its own stated meaning and application before it can inform a design. Use site irradiation, daily load energy, explicit losses, and the required recovery period for the panel-capacity calculation.
Next step
To get started, compare your calculated daily energy requirement and installation dimensions with the options on the custom solar panels page.
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