Solar PV / system sizing

Solar panel sizing

Size a PV array from a daily energy demand, then get the panel count, battery bank and simple payback. System losses are applied before sizing, which is what separates a real design from a nameplate calculation.

Performance ratio82%CO₂ avoided2.0 t/yrAnnual saving$760.27

Array size

3.78 kW

9 × 420 W panels

Annual yield

4752 kWh

13.0 kWh/day average

Battery bank

22.5 kWh

1.5 days autonomy

Simple payback

4.2 yr

Against $3,213 installed cost

Array and storage sizing

The energy balance from daily demand through to panel count, battery bank and payback.

9 panels

Array and storage sizing
Item	Value	Basis
Daily demand	12.00 kWh	Entered
Peak sun hours	4.20 h	Site-specific, worst-month or annual average
System losses	18.0%	Inverter, wiring, soiling, temperature
Required array	3.484 kW	E/(PSH × (1 − losses))
Panel rating	420 W	Nameplate at STC
Panels needed	9	Rounded up to whole modules
Installed capacity	3.780 kW	9 × 420 W
Annual yield	4752 kWh	13.02 kWh/day average
Battery bank	22.50 kWh	1.5 days autonomy at 80% DoD
Installed cost	$3,213	$0.85 per watt
Simple payback	4.2 yr	At $0.16/kWh

Sizing on the annual average under-builds for winter. Size on the worst-month PSH if the system must be self-sufficient year-round.

Monthly yield

Seasonal variation around the annual average.

Array size vs peak sun hours

Better insolation needs proportionally less installed capacity for the same demand.

Canonical equations

The exact relationships implemented by this engine.

P = E_daily/(PSH·(1−losses)) · energy balance over one day

N = ⌈P·1000/P_panel⌉ · rounded up to whole modules

C = E_daily·autonomy/DoD · depth of discharge derating

E_year = P·PSH·365·(1−losses) · annual energy production

Show model notation and computationOpen the symbolic form, the variable definitions, and the step-by-step derivation when you need the detail.

A clear path from inputs to results.

Solar sizing is an energy balance, not a wattage guess. Daily demand divided by effective sun hours gives the array; autonomy divided by depth of discharge gives the battery. The derate factor is where most optimistic designs go wrong, so it is applied before the array size is calculated.

Step-by-step methodWorked exampleAssumptions explained

How to use

Enter the daily energy demand in kilowatt-hours. Use a real bill, not an estimate.

Set the peak sun hours for the site — the worst month if the system must be self-sufficient.

Apply a system loss factor: inverter, wiring, soiling and temperature together are typically 15–25%.

Choose the panel rating.

Set battery autonomy days and depth of discharge. Use zero autonomy for a grid-tied system.

Read the panel count, annual yield and simple payback.

The formula

P_{\text{array}} = \frac{E_{\text{daily}}}{\mathrm{PSH}\,(1-\eta_{\text{loss}})}, \qquad N = \left\lceil \frac{P_{\text{array}}\cdot 1000}{P_{\text{panel}}} \right\rceil, \qquad C_{\text{batt}} = \frac{E_{\text{daily}}\cdot \text{autonomy}}{\mathrm{DoD}}

Reference: IEC 62548 (PV array design); NREL PVWatts derate conventions.

Worked example

12 kWh/day at 4.2 peak sun hours (default inputs)

Inputs

Daily demand 12 kWh.
Peak sun hours 4.2.
System losses 18%.
Panels 420 W.
Autonomy 1.5 days at 80% depth of discharge.

Calculation path

1
Effective sun hours: 4.2 × 0.82 = 3.444.
2
Required array: 12 / 3.444 = 3.484 kW.
3
Panels: ⌈3,484 / 420⌉ = 9 panels, giving 3.78 kW installed.
4
Annual yield: 3.78 × 4.2 × 365 × 0.82 = 4,752 kWh.
5
Battery: 12 × 1.5 / 0.80 = 22.5 kWh of nominal capacity.

Result

3.48 kW required · 9 panels (3.78 kW installed) · 4,752 kWh a year · 22.5 kWh battery.

Interpreting the result

The 18% derate turns a 3.48 kW requirement into 3.78 kW of installed panels — a 9% oversize that is not waste but physics. Annual yield of 4,752 kWh against a 4,380 kWh demand leaves a small surplus, which is the right side to err on. The battery is the expensive part: 22.5 kWh of nominal capacity at 80% DoD delivers 18 kWh usable, covering one and a half days.

Assumptions & limitations

Peak sun hours are an annual or worst-month average, not a peak instantaneous value.

Losses are lumped into one derate. A detailed design separates inverter, wiring, soiling, mismatch and temperature coefficients.

Panels face the optimal direction at the optimal tilt with no shading.

Battery round-trip efficiency is folded into the depth-of-discharge figure; in practice it is a separate 5–10% loss.

No degradation over time; panels lose about 0.5% of output a year.

Frequently asked questions

What are peak sun hours?

The number of hours per day at an irradiance of 1,000 W/m² that would deliver the same total energy as the actual varying sun. It is an energy measure, not a count of daylight hours.

How do I size for winter?

Use the worst-month peak sun hours rather than the annual average. In northern latitudes that can be half the summer value, which roughly doubles the array.

Is the payback calculation realistic?

It is a simple payback with no discounting, no degradation and no maintenance. Treat it as an order-of-magnitude figure; a full model would use levelised cost of energy.

Why apply losses before sizing?

Because the array must produce the demand after losses, not before. Sizing on nameplate output under-builds by the derate factor, which is how systems end up 20% short in winter.

What depth of discharge should I use?

80% for lithium iron phosphate, 50% for lead-acid if you want reasonable cycle life. Deeper discharge shortens battery life, so the nominal capacity must be larger.

Do I need a battery for a grid-tied system?

No. A grid-tied system without storage exports surplus and draws at night, and is considerably cheaper. Batteries are for autonomy, not for saving money in most markets.