Last week I meet with sustainable design & renewable energy expert Troy Van Beek, of Ideal Energy in Fairfield Iowa to discuss what size solar array will be needed to meet the Khare family’s energy needs.

Troy informed me that the first step in determining the size of array needed is to find out how many kilowatt hours a one kilowatt panel will produce at a maximal angle.  If the array is going on a roof, it should–if possible–be placed so that it faces south.  West and east facing roofs can also be used, but at an efficiency loss of about 14%.

If the roof is pitched, then the panels should be tilted to the same degree as the roof pitch.  This is because –in order to maximize roof space, and also avoid shading out surrounding panels–all panels should lay flush on the roof.

Once you’ve determined the direction and angle at which to place the panels, you can use this information– along with the latitude of the your site–to determine how many kilowatt hours a one kilowatt system will produce per year at your specific location .  *Simply plug this information into a solar calculator, like the one provided by Iowa Energy Center:

Using this calculator, I determined that in Farifield Iowa, a one kilowatt system at a 40 degree tilt will produce  1,611 kilowatt hours per year.

The next step is to find out how many kilowatt hours the household consumes per year–which can be done by referencing utility bills.   Then simply divide the total kilowatt hours used per year by the number of kilowatt hours produced per year by a one kilowatt system, and this will tell you how large of a system (in kilowatts) you will need.

For example, lets say the total household consumption is 35,362 kilowatt hours a year, and a one kilowatt system will produce 1611 kilowatt hours per year.  The calculation for determining the size of system needed looks like this:

35,362 kwh per year / 1611 kwh per year, per kilowatt  = 21.9 kilowatt system

The next step is to determine how much space such a system will take up, and also how much it will cost.  According to Troy, the best price right now, is a 240 Watt panel measuring 3.5 ft x 5.5 ft.

First determine how many square feet one of these panels takes up:

3.5 ft x 5.5 ft = 19.25 sq ft.

*So, it takes 19.25 square feet for one of these panels to produce 240 watts.

Next we need to know how many watts are produced by a 21.9 kilowatt system–which we will round to 22 kilowatts.  *Kilo means thousand, so there are 1,000 watts per kilowatt.

1,000 watts per kilowatt  x  22 kilowatts =  22,000 watts.

To determine how many 240 watt panels you need, simply divide total watts by 240 watts:

22,000 total watts / 240 watts per panel = 9.1 panels (which we will round to 9 for this exercise)

We know that each panel requires 19.25 square feet, so to calculate the total number of square feet that will be taken up by the system, simply multiply the number of  panels in the system by 19.25 ft.

9 panels x 19.25 sq ft per panel = 173.25 total square feet.

To calculate the cost of the system, simply multiply the cost per watt by the the total number of watts.  –According to Troy, the going rate is approximately $5 per watt.

22,000 total watts   $5 per watt = $110,000 system (not including installation).

*Combining the Federal tax credit with state incentives you qualify for may significantly reduce the cost of a solar pv system.

Solar Incentives by state:

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  1. Good review of solar array sizing considerations.
    But 35 MWh is a huge number. Average household consumption varies by state from 6 MWh to a maximum of 16 MWh.

    • Thank you. And yes, you are absolutely right– 35 MWH is a huge number and not by any means the average for a typical sized family. This was the number of MWH used by a family comprised of 6 adults and 2 children (for which I did a consultation). Granted, this is still a large number and I definitely had some recommendations of how they could decrease their usage. Thanks for the actual stats on the average home. Very good info to have.

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