# Will Solar WORK For You?

Here’s how to calculate the size of a grid-tied photovoltaic system by assessing your electrical use, your roof and your yard.

Your household electrical needs will help determine the size of your solar electric system. Photo: Dan Bihn

By DAN CHIRAS

Before you calculate what a solar system will cost to install, and whether it’s a good economic investment, you need to know how much electricity you consume and how much electricity a solar electric system can provide.

Household electrical needs vary. My own house, designed for efficiency, consumes 75 to 80 percent less electricity than a standard home of similar size, so we get all the electricity we need from a modest 1.1-kilowatt (kW) photovoltaic (PV) array.

In a typical home using natural gas for cooking, water heating and space heat, electrical consumption may run about 500 kilowatt- hours (kWh) per month. A 3- to 4-kW PV system might cover most household needs in a good sunny region. A larger system might be required in a cloudier area.

In an all-electric home equipped with the usual assortment of appliances — especially heavy hitters like a central air conditioner, a water heater, a stove and space heaters — average monthly electrical consumption can run between 2,000 and 3,000 kWh per month. It might take a system 10 kW or larger to meet that demand.

Assessing the electrical needs in an existing house is different from planning the needs for a new building. For your existing house, start by gathering your electric utility bills for the past two or three years. If you don’t have them, you can usually obtain the information by phoning your utility company. The company may even make the information available to you via its website.

When studying your utility bills, look for the kilowatt-hours consumed each month. Calculate the total for each year, and then calculate a yearly average. Look for trends in energy use. Are you using more electricity now than two or three years ago? Or less? Is consumption steady? Perhaps you’ve installed a big-screen TV or an air conditioner. Perhaps you’ve replaced an inefficient furnace or refrigerator with a new, more frugal model. Plan for energy use going forward, with the equipment you have now or expect to obtain in the future. That way, you’ll avoid oversizing or undersizing your PV system.

Determining electrical demand in a new home is more difficult. There’s no track record, just some design assumptions. If the new home is of roughly the same size as your old home, you plan to use similar appliances and you don’t expect dramatic changes in lifestyle, electrical consumption may be similar to that in the old home. But that’s not often the case. You may be building in a new climate or with a different orientation to the sun and wind. When building a new home, it’s customary to improve efficiency by installing more energy-efficient appliances and lighting. You may choose passive solar heating and cooling designs. If you follow these strategies, electrical consumption could easily be 50 or 75 percent lower. You can adjust electrical demand to reflect efficiency upgrades.

Another way to estimate electrical consumption is to perform a load analysis. Begin by listing all the appliances, lights and electronic devices in the new home or office. Rather than try to list each light bulb separately, just lump them together room by room.

Average Electrical Consumption of Common Appliances in Watts

The power consumed by any device can be estimated from the chart shown to the left. For more accurate data, check out the nameplate on the appliance or on its power supply. The sticker or engraved plate lists the unit’s rated voltage (typically 120) and power draw in either amps or watts. If it’s given in watts, simply enter that number in the watts column. If it’s in amps, enter the number in the amps column, and the spreadsheet will calculate watts. If you’re doing this manually, on a hand calculator, the formula is: Volts x Amps = Watts.

For appliances with large induction motors — fans, washing machines, dryers, dishwashers, pumps and furnace blowers — the rated wattage is called the apparent power. To calculate the real power drawn by these devices, multiply the rated wattage by the power factor 0.6.

Be sure, when determining the run wattage of laptop computers, cordless drills, cordless phones and other rechargeable devices, to use the wattage listed on the charger, not the device.

A more accurate way to determine the wattage of any electrical device is to measure it directly with a watt-hour meter. Plug the meter into an electrical outlet and then plug the appliance into the outlet on the face of the meter. A digital readout indicates the instantaneous power in watts. When measuring a device that cycles on and off, such as a refrigerator, leave the watt-hour meter connected for a day or two. The meter will record the total energy used during this period as watt-hours. Divide by the number of hours to get the average wattage for the period.

(An example is the Kill-A-Watt.  They are sometimes available for checkout at the local library.)

Estimate the number of hours each device is used daily and the number of days
it’s used during a typical week. Remember that some devices run seasonally: Electric lights burn longer in winter, and furnace blowers work only in winter; air conditioners often run only in summer months. Calculate the weekly energy consumption of all devices in the home. Divide this number by seven to determine the average daily consumption of the home in watt-hours. You will use this number to determine the size of a PV system.

If your home is energy efficient, you can get by with a smaller photovoltaic array. Photo: Dan Bihn

Total up your daily use of electrical power — and look for ways to reduce it. The lower your energy demand, the smaller your solar electric system needs to be, and the less it will cost to install. It’s been estimated that every dollar invested in energy efficiency saves \$3 to \$5 on the cost of a photovoltaic system. Make sure the house is tightly sealed and well-insulated. Turn off appliances when they’re not in use. Many electronic devices draw power even when they’re turned off. We call these phantom loads. They include television sets, VCRs, satellite receivers, cell phone and computer chargers and a host of other common household devices. A few, like satellite receivers, draw nearly as much power when they’re off as when they’re on. Phantom loads typically account for 5 to 10 percent of the electrical consumption in U.S. homes.

Recalculate your projected electricity use based on improved habits and equipment — and then it’s time to figure out the size of a solar electric system. A utility-connected or grid-tied system can be designed to meet all or some of the electrical needs in a home or business. Excess electricity is fed back onto the grid, running the electric meter backward.

To meet all your needs and virtually eliminate your electric bill, divide your average daily electrical demand (in kilowatt-hours) by the average peak sun hours per day for your area. Find sun hours at nrel.gov/rredc, and select the options labeled average, annual, and flat plate tilted south.

As an example, let’s suppose that you and your family consume 6,000 kWh of electricity per year, or 500 kWh per month. This is about 17 kWh per day. Let’s suppose you live in Lexington, Ky., where there are, on average, 4.5 peak sun hours per day. Dividing 17 by 4.5 gives the array size, or capacity, of 3.8 kilowatts. But don’t run out and order a system based on that number.

The American Solar Energy Society’s National Solar Tour can help you get started with solar.

This calculation yields a usable system size only if the system were 100 percent efficient and unshaded throughout the year. Unfortunately, no PV system is 100 percent efficient. As a result, most solar installers de-rate grid-connected PV systems by 22 to 25 percent. This loss is due to voltage drop through cables; resistance at fuses, breakers and connections; dust on the array; inefficiencies of system components such as the inverter and other factors. If your PV array is not shaded by trees or buildings, you’d need a 22 to 25 percent larger system to provide 17 kWh per day. Divide 3.8 kW by 0.77 (using a 23 percent de-rate factor) to get 4.94 kW. Let’s round it up to 5 kW to be on the safe side. That will work provided the array is not shaded.

Will the array fit on the roof? A top-quality silicon PV module can produce roughly 140 watts per square meter, or about 12 watts per square foot. As an example, a 20-by-22 foot garage with a pitched roof might offer about 230 square feet (21 square meters) of usable south-facing roof surface — enough to accommodate two rows of modules that are each 20 feet (6 meters) long. That ought to accommodate a 2.6-kW PV array, depending on how the modules are tilted up from the roof surface. The rack needs to be designed to prevent the lower tier of modules from shading the bot- tom of the upper tier when the sun is low in the sky. If your roof won’t accommodate a south-facing array, consider a ground-mount system in the yard.

Now you need to get out on the roof with a solar resource evaluation tool (SRET), like the Solar Pathfinder, SunEye or Acme ASSET. The SRET is designed to predict how much sun will fall on your site and pinpoint when the shade will fall on it. You may be able to borrow or rent an SRET, but this may also be the time to call in an expert who already has one. (Call your local American Solar Energy Society chapter to see if a member has an SRET. Or ask a local installer to come out for a site evaluation.)

5-kW PV Array Output Based on Shading

The SRET data is entered into a computer program that determines the output of the array by month and annually, based on shading. The result is a table like the one above, which shows the output of a 5-kW array based on a set of shading data, month by month through the year, with an annual total.

According to this table, if the family required 6,000 kWh per year, the array size would need to be increased. To adjust the array size, divide 6,000 kWh by 5,632 (AC energy produced for the year with shading). That equals 1.06. Multiply the result by 5 kW to find that the owner needs a 5.3-kW PV system to meet the house- hold needs. In today’s market, that system might cost about \$26,000, including the 30 percent federal tax credit and depending on the local price of modules, inverters and labor.

Alternatively, you could size the system to your budget. If you have \$15,000 to invest, with the federal tax credit, you might be able to purchase a 3.1-kW system that would, in this case, reduce the annual electric bill by about 70 percent. Don’t forget to factor in state and local incentives, which can be considerable. See dsireusa.org.