There's a ton of information online about Solar Electricity, not all of it useful. I hope that you find the information on this website to be an exception, because my intent really isn't to sell you anything. I really just want to share the things that I've learned from building and designing systems. I want you to feel confident that PV will work.
If you are interested in assessing your own home, and you think that you're able to take measurements, here's what you can do:
Collect Data
- Determine your home's orientation, using a compass. (south, or 180°, is optimal).
- Find the southernmost facing roof surface, and determine its pitch with an inclinometer.
- Measure the area of that roof surface. Be safe. You can also measure the house perimeter.
- Pick a type of solar panels, and use those dimensions to determine how many panels will fit
Additional Data:
- What is the busbar rating of your main service panel? (100, 200, 250 are common).
- What is the main breaker rating of your service panel? (The top breaker that shuts all power off)
- What is the spacing of rafters in your roof? Is it a truss system? (structural system)
- How much shading is in your environment? (a pathfinder might be useful).
Estimate Power
Once you can determine how many panels will fit, you can then multiply the number of panels by their STC wattage to determine what the wattage of your system will be.
Panels x Peak Power = "Max Rating" (the laboratory rating of your system at maximum power).
Panels x Peak Power = "Max Rating" (the laboratory rating of your system at maximum power).
For example, if you can fit 30 x SPR-230 panels, that's technically 6900 watts, or a 6.9 KW system.
Next, take your 6.9 KW number and perform the following calculations:
- Subtract the percentage that will be shaded. If no shade, skip this step.
- Subtract the percentage that derates the system by orientation and angle.
- Take 75% of whatever's left.
Let's say that your 6.9 KW system has 10% shading, it faces Southeast, and the roof is 30°.
Southeast at 30° is 96% efficient.
6.9 x .90 x .96 x .75 = 4.4 kw
6.9 x .90 x .96 x .75 = 4.4 kw
The 75% factor is an industry-standard practice recommended by Sunpower and other agencies, to keep your estimates from being too high. Typically, other factors such as dust and debris on the panels, as well as voltage drop in the wiring and inverter efficiency, will subtract from the production of the system. By taking a conservative estimate, you allow yourself some room.
Your new rating is 4.4 KW
Next: Factor Your Insolation
Depending on the part of the country you're in, this system will perform differently. You can find this information from the National Renewable Energy Laboratory Red Book. Most of the country is between 4 and 6 w/m². In this case, let's say we are in an area that has a rating of 5 w/m².
4.4 kW multiplied by 5 w/m² is 22
That number essentially represents the average amount of KWH per day that you can expect from your system on an annual basis.
Of course, that's going to fluctuate depending on what time of year it is, but that's why NREL provides annual averages for irradiance, because it's the best way of estimating your system's potential, with all the varying factors of sunlight energy potential throughout the year. Annually, you can expect to see:
22 kWh x 365 = 8,030 kWh. Therefore you have estimated your home to be capable of producing 8,000 kWh, or 8 megawatt-hours, of power, annually. At 22¢, that's $1766 in savings per year.
This formula is used standard by professionals in estimating the annual production of the systems that they sell to people across the US. You can try it yourself if you'd like. Remember to account for your own factors, as listed by the information that you gather about your home's orientation, angle of your roof, the area shading, your region's irradiation, plus the type and number of solar panels that you wish to install.
See more site surveys for more detailed information about estimating energy potential.
4.4 kW multiplied by 5 w/m² is 22
That number essentially represents the average amount of KWH per day that you can expect from your system on an annual basis.
Of course, that's going to fluctuate depending on what time of year it is, but that's why NREL provides annual averages for irradiance, because it's the best way of estimating your system's potential, with all the varying factors of sunlight energy potential throughout the year. Annually, you can expect to see:
22 kWh x 365 = 8,030 kWh. Therefore you have estimated your home to be capable of producing 8,000 kWh, or 8 megawatt-hours, of power, annually. At 22¢, that's $1766 in savings per year.
This formula is used standard by professionals in estimating the annual production of the systems that they sell to people across the US. You can try it yourself if you'd like. Remember to account for your own factors, as listed by the information that you gather about your home's orientation, angle of your roof, the area shading, your region's irradiation, plus the type and number of solar panels that you wish to install.
See more site surveys for more detailed information about estimating energy potential.
