Solar Panel Size Calculator

Background.

The question usually starts with a power bill. A household that used 900 kilowatt-hours in a 30-day billing period used 30 kWh a day, and an array meant to cover all of it has to produce 30 kWh on an average day. Where the panels get the equivalent of 5.2 hours of full sunshine daily and lose 14 percent of their output to heat, wiring, dust and downtime, each kilowatt of panels yields about 4.47 kWh a day. The home therefore needs about 6.71 kW of panels: seventeen 400-watt modules, a 6.8 kW array covering roughly 357 square feet of roof.

Panel wattage is the figure shoppers fixate on, and it changes the answer least. Switching to 350-watt panels turns 17 panels into 20, but the system still has to be about 6.7 kW. The sunshine figure is different. Move the same house to a site with 4 peak sun hours and it needs 8.72 kW, 22 panels and 462 square feet. Required size scales inversely with the sun-hours input, so a guess there is a guess in the result. NREL's PVWatts tool and its National Solar Radiation Database give the solar resource for a specific location, tilt and orientation, and are worth the few minutes they take.

If you know the roof space you have, enter it and the result checks whether the panels fit, and if they do not, how many will and what that smaller array amounts to. Keep in mind what the numbers are: energy, in kilowatt-hours and kilowatts of panel capacity, worked out from an average day. They say nothing about money, about which month the panels outproduce the house, or about how your utility credits what you export.

What is solar panel size calculator?

A solar system's size is quoted in kilowatts DC: the combined nameplate rating of its panels, each measured under standard test conditions of 1,000 watts of sunlight per square metre with the cells at 25 °C. On a real roof the panels rarely see those conditions for long, so sizing works through peak sun hours. A peak sun hour is one kilowatt-hour of sunlight falling on a square metre, the amount delivered by one hour at that test intensity, so a site receiving 5.2 kWh per square metre per day has 5.2 peak sun hours however long the day actually is. In principle, one kilowatt of panels then produces one kilowatt-hour per peak sun hour. In practice some is lost: panels run hot, wires and connections have resistance, modules are mismatched, dust and shade cut the light, and systems are occasionally down. NREL's PVWatts model uses 14 percent for these system losses by default, which leaves a performance ratio of 0.86. The offset is the share of your annual electricity use the array is meant to generate. The calculation here sizes for an average day across the year, with no seasonal or hourly modelling.

How to use this calculator.

1. 01Enter your electricity use in kWh and the number of days it covers. One bill works, but a full year is better: enter the 12-month total with 365 days, which irons out summer air conditioning or winter heating. Add the expected use of anything you plan to install, such as an electric car.
2. 02Enter peak sun hours for your location: the average daily solar radiation, in kWh per square metre, on panels at your roof's tilt and direction. NREL's PVWatts and the National Solar Radiation Database provide it. Do not use hours of daylight.
3. 03Enter the system loss percentage. The 14 percent default follows PVWatts; raise it for partial shade, a hot climate or panels that will rarely be cleaned.
4. 04Enter the rated power of one panel from its datasheet, in watts, and the share of your use you want to cover. The offset is capped at 100 percent; to plan for more generation, add the extra load to the usage instead.
5. 05Enter the area of one panel in square feet (its length times width) and, if you want a roof-fit check, the usable roof area. Leaving the roof area at 0 skips the check.

The formula.

The required size is kW = (E ⁄ d × o) ⁄ (H × (1 − L)). E ⁄ d turns the bill into average daily use in kWh, E being the energy used over d days. Multiplying by the offset o, as a fraction, gives the daily energy the array must produce. H is peak sun hours and L the system loss as a fraction, so H × (1 − L) is what one kilowatt of panels delivers on an average day: 5.2 × 0.86 = 4.472 kWh in the default case. Dividing the target by that yield gives the DC capacity needed.

The rest is counting. Panels = ⌈1,000 × kW ⁄ P⌉ for a panel rating P in watts, rounded up because you cannot install part of a module, and the installed capacity is panels × P ⁄ 1,000. Roof area is panels × the area of one panel. The annual target is the daily target × 365. When you enter an available roof area, the margin is that area minus the area required, and the number of panels that fit is the available area divided by the area per panel, rounded down and never more than the panel count.

Everything rests on one average day, so the result cannot show that a system matched to your annual use will overproduce in summer and fall short in winter, and it does not consider shading by the hour, temperature by the season, or panel degradation over the years. In PVWatts the inverter's conversion loss is modelled separately from the 14 percent, so if you want a comparable result, fold an allowance for the inverter into the loss figure. The roof check counts panel area only. Setbacks, fire access paths, vents and the shape of the roof all reduce what actually fits.

A worked example.

Frequently asked questions.

References& sources.

1. [1]National Renewable Energy Laboratory (2014). PVWatts Version 5 Manual. NREL/TP-6A20-62641.
2. [2]National Renewable Energy Laboratory. PVWatts API. NREL Developer Network.
3. [3]National Renewable Energy Laboratory. National Solar Radiation Database.
4. [4]U.S. Department of Energy. Planning a Home Renewable Energy System. Energy Saver.
5. [5]National Institute of Standards and Technology (2008). Guide for the Use of the International System of Units (SI), NIST Special Publication 811.