We know that conservation is important, but exactly how many dollars should be invested to save a KWH? We answer this question by comparing the investment necessary to produce one KWH/day with the cost of conserving one KWH/day. Even though a KWH is relatively expensive, with effective investments in conservation you can substantially reduce the total cost of operating a home with a stand alone power system.

### The Value Of Energy Efficiency In Stand Alone PV Systems: A Simple Approach

John Richter and Larry Schlussler | Sun Frost (owner-Larry Schlussler)

The Value of Energy Efficiency in Stand Alone PV Systems: A Simple Approach

We know that conservation is important, but exactly how many dollars should be invested to save a KWH? We answer this question by comparing the investment necessary to produce one KWH/day with the cost of conserving one KWH/day. Even though a KWH is relatively expensive, with effective investments in conservation you can substantially reduce the total cost of operating a home with a stand alone power system.

Larry Schlussler is the owner of the Sun Frost company. of Arcata, CA.

This article is excerpted from the Spring 2003 newsletter of EORenew, the Eastern Oregon Renewable Energies Non-profit, which is the sponsor of SolWest Renewable Energy Fair.

Numerous studies have been done to determine the cost of generating electricity from solar photovoltaic (PV) systems. Such computations typically attempt to compare two quantities, the cost of a Kilowatt-hour (KWH) of electricity and the investment in a PV system necessary to produce a KWH of electrical energy. These studies are important to financial institutions considering large-scale investments in grid-connected PV systems. But these financial computations are complicated and involve a number of fuzzy assumptions, so this approach has severe limitations when considered by the typical homeowner.

Our Approach

Because PVs have high capital costs and lower operating costs, the financial analyst must find a way to compare the initial capital cost of the PV system to the monthly expense of a utility bill. This method is relatively complicated and requires some representation of the time value of money, or &quot;discount rate". Based on the discount rate, the future costs of electricity, adjusted for projected increases and reduced to a &quot;Net Present Value" or NPV are compared to the capital cost of the PV system. While this is old hat for corporate accountants, it often doesn't seem meaningful to the average homeowner. Corporations often choose the interest rate on their loans as a discount rate. But if you can't get a bank to finance an off-grid home, there is no interest rate to use.

What we present here is a simpler and more direct method of figuring the cost of the PV required to produce one additional KWH/day vs. the cost of conserving that KWH/day using energy efficient appliances in the home. In a stand-alone system, calculating the cost of a KWH of electricity doesn't give the homeowner much useful information, especially when the many assumptions made in this calculation are considered. There are no financial decisions that can be readily made based on this calculated cost of a KWH. More useful would be a measure that would let the homeowner know when conservation is a better investment than increased power generation. This article presents a straightforward method for making this decision by comparing the cost of generating an additional KWH/day with the cost of conserving one KWH/day.

The Cost of Generating One KWH/day

&quot;CFs (compact fluorescents) on the ceiling before PVs on the roof!" is a Home Power adage. But just how much is efficiency worth? How much should an off-grid homeowner be willing to invest in more efficient appliances (refrigerator, washer/dryer, lighting, well pumps etc.) to save a KWH? The answer depends on how much capital it takes to generate off-grid power. We will consider the investment necessary to produce an additional 1 KWH/day from an existing PV system.

The cost of the PV system will be divided into two parts, the initial investment and operating costs. Initial costs will include PV panels, mounting hardware, original battery set etc. Operating costs will include battery replacement and the cost of backup generator power during the winter months. The system will be sized so that it can provide 1 KWH/day of AC power with 4.2 hours of insolation, the average yearly insolation for Arcata, a Northern California coastal town. It will be assumed that the system is being increased in size and that the inverter, charge controller, load center, and battery housing are large enough to handle the increase in size.

Battery Mismatch Losses

Unfortunately, a 1 KW solar array sitting in direct sun for one hour will not produce 1 KWH of useable AC power. There are a number of ways energy is lost in this system. A major loss occurs because the panels produce their peak rated power at 17 Volts and most of the energy input into the battery occurs at a voltage closer to 13 volts. The energy is then taken out at about 12.4 Volts. The voltage mismatch alone results in a loss of over 25%.

When analyzing battery powered PV systems, it is relatively easy to take these mismatches into consideration by using a current-based analysis. With this type of analysis voltage mismatches are automatically taken into account. To get a quantitative picture of the losses in a stand-alone system, let's calculate the size array necessary to produce 1KWH with one hour of full sun. In a loss-less system, it would of course take a 1000-watt array to produce 1000 watts AC power. In a real system there will be a number of losses along the way. The input to the inverter will have to be about 1100 watts due to the inefficiency of the inverter. If this power is supplied at an average voltage of 12.4 Volts, then the batteries must supply 1100 watts/12.4 volts or 88.7 amps.

For the inverter to produce 1 KWH the batteries must then supply 88.7 amps for one hour, or 88.7 amp hrs. (An amp hour is a unit of energy like a KWH). Assuming a 10% loss in the batteries the PV panels will have to supply an additional 10% more energy or 97.6 amp/hrs. To produce 1KWH with 1 hour of insolation will then require a 97.6 amp solar array. For each 17 watts of solar panels the panel will produce one amp; i.e., a 34-watt panel will produce 2 amps. The energy required to produce 1KWH of AC power with 1 hour of direct sun then be 17 watts/amp X 97.6 amps or 1659 watts which is a lot more than the 1000 watt array required in a loss-less situation. These large losses make stand-alone power more expensive than a grid-tied system. In a grid-tied system the inverter loads the panels so that they are putting out their peak power. In addition, there are no battery losses and no backup system required during cloudy winter months.

The following table summarizes the assumptions made in calculating the initial capital costs and the operating costs of the PV system.

 System Assumptions: Location, Arcata - North Coast of CaliforniaYearly average sunlight hours = 4.2 hrs/day†System sized so that with 4.2 hours of sun the system will supply 1 average KWH/day of AC power. Requirement: 393 watts of additional solar at \$6/watt, including rack. Misc. costs and installation add \$550 to this.During the winter, when insolation averages are less than 4.2 hrs/day, a back-up generator will be required to provide 50 KWH/year during overcast weather at \$1/KWH.Inverter and battery losses are both 10%. Batteries are \$1/amp-hour with an expected life of 8 years. 500 amp-hours is added to the battery bank for the extra storage.No tracking, array fixed at 45° tilt.† If you live in an area where there are 5 average sun hours per day, you may multiply the solar cost by .86

The investment for increasing the capacity of the system by 1 AC KWH/day is \$5606.00: \$3406.00 for the initial cost of the equipment and \$2200.00 for the operating costs (battery replacement and generator operation). If you include the cost of all the equipment to set up a system, the inverter, load center, charge controller, and battery housing add about \$1000.00, the total cost to produce 1KWH/day is \$6600.00.

Conservation will be a good investment if you can purchase a device that can save 1KWH/day for less than \$5606.00. If the device will last less than 24 years, its replacement cost should be considered. If you spend \$1000.00 on an energy-saving device that saves 1KWH/day and lasts 12 years, you'll buy 2 over the life of your PV system and you're investing \$2000.00 that will save you \$3606.00.

Cost per KWH - A Minimum Value

If you want to get an estimate of how much a KWH of electricity costs from your PV system, you could divide the total investment needed to produce 1KWH/day by the number of KWHs those PVs will produce in 24 years ( \$6606.00 / 8760 KWH=\$.75/KWH). If a more detailed financial analysis is carried out considering interest rates and Net Present Value of money, etc., this cost will increase, \$.75 is as low as it will get.

How Good an Investment is a Compact Fluorescent?

If you replace a 100-watt light bulb with a 20-watt compact fluorescent and the bulb is used 4 hours per day your energy savings will be .32 KWH/day. If you have to expand your system to produce this energy, choosing instead to reduce your load by .32 KWH/day will result in a savings of \$1794.00: \$1090.00 initially and \$704.00 in operating costs for the life of your system. If a compact fluorescent has a life of 10,000 hours and an incandescent bulb a life of 750 hours, over 24 years 3.5 compact fluorescents or 46 incandescent lamps will be required. The 3.5 compact fluorescents will cost roughly as much as the 46 incandescent lamps; so for essentially no additional investment you can reduce the total cost of your power system by \$1794 - that's a good deal.

Refrigeration

What would the savings be in purchasing a DC Sun Frost rather than an energy efficient AC refrigerator such as the one reviewed in HP#84: Aug/Sept. 2001, page 84? Sun Frost's refrigerators are produced in both AC and DC models. If a DC model can be conveniently hooked up it has the advantage of eliminating inverter losses and increased reliability because an inverter failure will not leave you without refrigeration. In addition, there will be more inverter capacity left for other applications. A DC Sun Frost unit will typically consume about .5 KWH/day at 70°. With 10% inverter losses, the fridge reviewed in HP#84 (page 84) consumed about 1.5 KWH/day including inverter losses. The savings with a Sun Frost are then 1KWH/day. This will result in an initial saving of \$3406 in cost of the PV system and an additional saving of \$2200 in operating costs for the life of the system. This savings of \$5606 will be approximately three times the cost difference between the two refrigerators, making the Sun Frost an excellent investment. If the expected life of the appliance is factored in the savings will be even greater. The expected life of a Sun Frost is over 24 years and the expected life of a conventional refrigerator is only 15 years. Over the life of a PV system this will increase the savings by an additional factor of 1.6

Modifying this approach to your Personal System

We know that conservation is important, but exactly how many dollars should be invested to save a KWH? We answer this question by comparing the investment necessary to produce one KWH/day with the cost of conserving one KWH/day. Even though a KWH is relatively expensive, with effective investments in conservation you can substantially reduce the total cost of operating a home with a stand alone power system.

You can also modify our results to fit your personal situation. For example, if you live in the Southwest and receive 20% more sunlight you could reduce the cost of producing a KWH by about 20%, or you may choose to do your own installation and deduct that cost from your solar investment. As for rebates, we have found that unfortunately they are seldom applicable to stand alone systems (except in Oregon, where a \$3 per watt tax credit with a \$1500 maximum applies to both on-grid and off-grid systems).

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