PV Array Sizing for kWh

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Residential Solar-Electric Array
The first step in going solar is not sizing the PV system, but reducing your electricity usage through conservation and efficiency measures.

While your budget may dictate how big your PV array will be, you likely have a kWh production goal that you’d like to meet. Most folks want to have a grid-tied PV system designed to offset most or all of their annual electricity usage. 

The first step in going solar is not sizing the PV system, but reducing your electricity usage through conservation and efficiency measures. Once these measures have been implemented, you’re ready to size a PV system to offset the remaining energy usage. To determine the PV array size needed, you’ll need to know the average daily peak sun-hours for your location. Once you’ve determined your annual kWh production goal and the peak sun-hour value, use an overall system efficiency factor to calculate the required PV array size.

Let’s say we have a home near Eagle, Colorado. After implementing energy-efficiency strategies, this home consumes an average of 300 kWh per month. However, we would also like to use an electric car, which will use an estimated 250 kWh per month for charging. Now we have a total annual energy consumption estimate of 6,600 kWh per year. Using solar data for Eagle, supplied by the National Renewable Energy Laboratory (bit.ly/SolarData), you’ll find the average peak sun-hours per day for a south-facing array, mounted with tilt equal to latitude (in this case, 40°) is 5.5. Let’s use an overall average system efficiency factor of 72% (see below).

To calculate the array size needed to meet our predicted annual energy consumption, divide the annual kWh consumption by 365. This gives an average daily consumption in kWh. Divide this amount by average daily peak sun-hours to get the approximate array size in kW. That value is then divided by the system’s efficiency factor:

6,600 kWh/yr. ÷ 365 days/yr. = 18.1 kWh/day

18.1 kWh/day ÷ 5.5 sun-hours/day = 3.3 kW

3.3 kW ÷ 0.72 efficiency factor = 4.6 kW array

To offset 100% of the home’s (and EV’s) electricity consumption over the course of a year, a 4.6 kW system is needed. 

The 0.72 efficiency factor is based on the following assumptions: average solar access of 95% (shading derate factor); modules with a positive-only production tolerance; inverter efficiency of 96%; module temperature derate factor of 0.88; DC and AC wiring derate of 0.98 and 0.99; module soiling derate = 0.95; module mismatch derate = 0.98; system availability derate = 0.99:

0.95 × 0.96 × 0.88 × 0.98 × 0.99 × 0.95 × 0.98 × 0.99 = 0.72

Comments (2)

Justine Sanchez's picture

Hi Calvert,
thanks for your post! That's great you found it helpful.
Best,
Justine Sanchez
Home Power Magazine

Calvert Bowen's picture

Your example is dead on, It really simplified the process for me.

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