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.
Residential Solar-Electric Array

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 (9)

Dustin's picture

Hi Justine,
I agree with Calvert: this is wonderfully written. Plain, simple, yet informative, and down to the facts. There's even an example! Homepower is really lucky to have you.

Would you mind if I ask just one more question? How does one calculate the "average solar access (shading derate factor)"? Neither PVWatts nor SAM applications have this ability, and they recommend using Solar Pathfinder hardware tool for calculating shading derate factor. Would you recommend some other way? Can shading derate factor be calculated with some regular insolation analysis application, by calculating Sky view factor?
Thank you for the reply.

Justine Sanchez's picture
Hi Dustin, Thanks so much for your post! So if you don't have access to a solar site analysis tool, you can do a shade analysis by hand. This will require you to generate a sun path chart for your location and then use this chart to plot any obstructions (objects that will shade your potential array location) on the horizon on to the chart. (This means you will have to figure out the azimuth (angle east or west of south) and altitude angle (angle off the horizon) of each object, which you can do with a compass and protractor). I found this research paper from the University of Oregon that you might find helpful, Take a look: http://solardat.uoregon.edu/download/Papers/UsingSunPathChartstoEstimate... And here is the link to generate a chart for your area. http://solardat.uoregon.edu/SunChartProgram.html Best wishes! Justine Sanchez Home Power Magazine
Dustin's picture

Hi Justine,

Thank you for the quick reply.
I must say I tried to follow the University of Oregon paper instructions and got a bit lost.
Furthermore the paper also uses Solar Pathfinder hardware tool for determination of horizon.

I was wondering is some other solution, like this one could be used:
It's an annual Shading mask preview of a particular point (TP in upper image) surrounded by buildings. Basically the software calculates the annual preview of the portion of the sky dome that is masked by surrounding objects and its percentage , and its percentage (that's the black part). It also calculates the percentage of the visible sky dome (non masked part of the sky dome)

visible sky dome equals 72 % -> annual shading derate factor = 0.72
Do you think this method is valid for calculating annual shading derate factor?

Thank you for the reply.

Michael Welch's picture
Hello Dustin. I recommend spending some time trying to find a Solar Pathfinder to borrow. They're not uncommon. For example, I know of three in the area I live in (around Humboldt Bay, CA). Call local energy related nonprofits, then give a call to local solar installers to see if they know of any available.
Dustin's picture

Hi Michael,
Thank you for the help.
Sadly I do not reside in USA, so borrowing a Solar Pathfinder is not an option.
Also I am really interested in doing this by hand or some application.

Justine Sanchez's picture
Hi Dustin, I am not familiar with that particular software, also looks like it is accounting for objects on all sides, rather than just to the southern portion of the proposed area. I did however find another resource, that is a little easier to read and describes step-by-step on how to determine your shading…check it out, but you will see it is a labor intensive process…so if you can find a Solar Pathfinder to borrow, that would be the easiest. http://www.thesolarplanner.com/array_placement3.html Best, Justine Home Power Magazine
Dustin's picture

Once again, thank you Justine.
The part on assigning particular shading percentage to squares in the solar window is a bit confusing, but I guess that's the question for the author of the article. Will try to contact him.
Thank you once again for the help.

Justine Sanchez's picture

Hi Calvert,
thanks for your post! That's great you found it helpful.
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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