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Solar panel wattage calculator

If you know how much electricity you use and how much sun your roof gets, the array size follows from one equation. Enter any two of annual consumption, peak sun hours and array wattage, and the calculator returns the third.

Calculator

Array wattage needed (kW)
9.01
Annual consumption covered (kWh)
10,000
Solar hours per day needed
3.80
Average generation per day (kWh)
27.40
Panels of 400 W
23
What this means
At 3.8 peak sun hours a day and a 80 percent derate, a 9.0 kW array produces about 27.4 kWh a day, or 10000 kWh a year. That covers 100 percent of an annual consumption of 10000 kWh.

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How this calculator works

Solar array sizing rests on one relationship: the energy an array produces in a year equals its rated power, multiplied by peak sun hours, by 365 days and by a system derate. Rearranging it answers the three questions people actually ask. If you know your consumption and your local sun, divide to get the kilowatts. If you have a quote for a given array, multiply to see what it can cover. If the array is fixed and the coverage is short, solve for the sun hours you would need, which tells you whether the shortfall is the weather or the array size. Peak sun hours are the daily average of hours at 1,000 watts per square metre, not the hours of daylight: a long summer day at lower intensity can equal fewer peak sun hours than you might expect. The derate of 0.8 covers the inverter, wiring, soiling, snow and heat, and setting the offset below 100 percent lets you size for a share of consumption rather than all of it.

Formula

array kW = consumption (kWh per year) x offset / (365 x peak sun hours x derate). Annual generation (kWh) = array kW x 365 x peak sun hours x derate. Sun hours needed = consumption x offset / (365 x array kW x derate). Daily generation = array kW x peak sun hours x derate.

Worked example

A household using 10,000 kWh a year, covering all of it, with 3.8 peak sun hours and an 80 percent derate needs 10,000 / (365 x 3.8 x 0.8) = 9.01 kW. That is 23 panels of 400 W. Running the calculation the other way, a 7 kW array in the same place covers 7 x 365 x 3.8 x 0.8 = 7,767 kWh a year, about 78 percent of the consumption.

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Frequently asked questions

What sun hours should I enter if I do not know them?

Use 3.0 to 3.5 for coastal British Columbia, Newfoundland and the Maritimes, 3.5 to 3.8 for southern Ontario and Quebec, and 3.8 to 4.2 for the Prairies. Natural Resources Canada publishes city tables and PVWatts computes them from weather files.

Why does my installer quote a different size?

Installers use hourly simulations with your roof pitch, azimuth, shading and the inverter clipping, not an annual average. They may also size around a net metering cap or the roof area rather than annual consumption. Use this as a first estimate and compare the quote with a simulation.

Is a bigger array always better?

No. Beyond the point where you cover your own use, extra output is exported, often at a lower rate, and may be curtailed by the utility. Batteries or a time-of-use tariff can change that, but they add cost and losses.

Does the derate include snow?

Partly. A generic 0.8 derate includes soiling and some snow losses, but in a snowy province panels can be covered for days. A steeper tilt helps the snow slide off; a winter-cleared array produces far more than a neglected one.

Sources

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