get started sizing your solar electric system
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.
The lower your energy
demand, the smaller your
solar electric system needs
to be, and the less it
will cost to install.
redbook/atlas, 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 k Wh of electricity per year, or 500 k Wh per month. This is
about 17 k Wh 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
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 k Wh per
day. Divide 3. 8 k W by 0.77 (using a 23 percent
de-rate factor) to get 4.94 k W. Let’s round it up
to 5 k W to be on the safe side. That will work
provided the array is not shaded.
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
Month Insolation ACenergy Percent ACenergy
kWh/m2/day kWh sun with
5-k W PV Array Output Based on Shading
January 3. 23
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 rredc.nrel.gov/solar/old_data/nsrdb/
Copyright © 2010 by the American Solar Energy Society Inc. All rights reserved.
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 — see page 54 for listings — to see if a
member has an SRET. Or ask a local installer
to come out for a site evaluation.)
The SRET data is entered into a computer
program that determines the ouput 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-k W 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 k Wh per year, the array size would need
to be increased. To adjust the array size, divide
6,000 k Wh by 5,632 (AC energy produced for
the year with shading). That equals 1.06. Multiply the result by 5 k W to find that the owner
needs a 5.3-k W PV system to meet the household 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-k W 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. ST
The American Solar Energy Society’s National
Solar Tour can help you get started with solar.