Over the last two decades, the cost of solar energy systems has come down drastically. As the demand for solar power systems continues to rise, the costs are further decreasing. In an NREL report it has been noted that from 2006 to 2016, the average cost of a system dropped from $9/watt to approximately $4/watt and even to $2/watt when self-installed. An important parameter for profitability assessment of a solar installation is to evaluate its Return on investment or the ROI. It is important to take advantage of the RECs, the federal and state tax incentives to reduce the initial costs of the system, and thus increase the returns on investment (ROI) of the system
Follow these steps while calculating ROI of a solar installation:
- It is important to review the electricity bills to determine the average electricity consumption in a year. Local utility statements can help figure out the amount.
- Determine the total installed cost of a solar energy system. This includes expenses such as one spent on PV panels, installation costs, permits, etc.
- From the utility bill determine how much you pay per kilowatt hour (kWh).
- Next figure out how much electricity your solar panels will generate. Factor snow load, panel degradation factor etc that negatively impact the energy production.
- Identify the types of federal, state, and local government incentives you can take advantage of towards installation of your solar panels. Calculate the total financial benefit that your system can avail. This includes any RECs you can sell to the utility company. All these benefits will reduce the initial cost of your system and increase your ROI.
- Find out available financing options.
- Calculate your return on investment (ROI) using the following steps shown in example below. Here is an example:
Let's say we have a 5040-watt system
Assume 5 $/watt. So, initial cost of system = $ 5x 5040 = $25200
5040 W system is about 26.1 m^2 in area (using a sun power type of panel)
Let's say the place of interest is in North America and has insolation of 5.5 Kwh/m^2/day
So per day intensity = 26.1 m^2 x5.5KWh/m^2/day = 144 DC KWh/day (hitting the 26.1 m^2 panels)
Now find out the efficiency of solar panels – In the current case sunpower panels have efficiency of 19.3 %
So per day intensity = 144KWh/day x0.193 =27.8 DC KWh/day
Let's say the area experiences snow, so above figure will get reduced = 27.8x 0.95 =26.4 DC/KWh/day
Assume Panel derate factor of 82 % (derate factor is efficiency of DC to AC conversion)
Revised intensity =26.4 DC KWh /day x 0.82 =21.6 AC KWh/day
Convert this to per year = 21.6 AC KWh/day x365 =7884 KWh /year ----àEq. (1)
But panels degrade over time, hence Equation (1) minus degradation @ 0.8% compounded annually for 25 years = Average electricity generation over 25 years = 7095 KWh per year -----àEq. (2)
Assume utility electricity is priced @ 10 c/KWh = Equation (2) ÷10 = 7095 ÷ 10 = $709.5 = $710
We had spent @25200 as initial investment.
So ROI = (what we earned) ÷ (what we invested)= 710 ÷ 25200 =2.8% return on investment
Note that we haven’t applied any incentives/rebates/RECs as yet. If we apply federal incentives, rebates and REC payments to initial cost of the system, then ROI will significantly improve, as initial cost will come down by a big margin
For example: Initial cost after application of incentives in above example will be:
$25200 – (federal tax credit @30% lets say) – (local incentives) = $Y
So new ROI = (710 + REC payments) ÷ ($ Y) = $ Z
Note that $Z will be a lot more and can be as close to 10 -20 % as new ROI is based on rebates/incentives and RECs availed by the customer.