OFF-GRID DESIGN
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SYSTEM SIZING INFORMATION
The size of a solar electric system depends on the amount of
power that is required (watts), the amount of time it is used (hours)
and the amount of energy available from the sun in a particular area
(sun-hours per day). The user has control of the first two of these
variables, while the third depends on the location. |
Conservation
Conservation plays an important
role in keeping down the cost of a photovoltaic system. The use of
energy efficient appliances and lighting, as well as non-electric
alternatives wherever possible, can make solar electricity a cost
competitive alternative to gasoline generators and, in some cases, utility
power.
Cooking, Heating, & Cooling
Conventional electric cooking,
space heating and water heating equipment use a prohibitive amount of
electricity. Electric ranges use 1500 watts or more per burner, so bottled
propane or natural gas is a popular alternative to electricity for cooking.
A microwave oven has about the same power draw, but since food cooks more
quickly, the amount of kilowatt hours used may not be large. Propane and
wood are better alternatives for space heating. Good passive solar design
and proper insulation can reduce the need for winter heating. Evaporative
cooling is a more reasonable load than air conditioning and in locations
with low humidity, the results are almost as good. One plus for cooling—the
largest amount of solar energy is usually available when the temperature is
the highest.
Lighting
Lighting requires the most study
since many options exist in type, size, voltage and placement. The type
of lighting that is best for one system may not be right for another. The
first decision is whether your lights will be run on low voltage direct
current (DC) or conventional 110 volt alternating current (AC). In a small
home, an RV, or a boat, low voltage DC lighting is often the best choice. DC
wiring runs can be kept short, allowing the use of fairly small gauge wire.
Since an inverter is not required, the system cost is lower. When an
inverter is part of the system, a home will not be dark if the inverter
fails and the lights are powered directly by the battery. In addition to
conventional-size medium-base low voltage bulbs, the user can choose from a
large selection of DC fluorescent lights, which have 3 to 4 times the light
output per watt of power used compared with incandescent types. Halogen
bulbs are 30% more efficient and actually seem almost twice as bright as
similar wattage incandescent bulbs given the spectrum of light they produce.
High quality fluorescent lights are available for 12 and 24 volt systems.
In a large installation or one with many lights,
the use of an inverter to supply AC power for conventional lighting is cost
effective. AC compact fluorescent lights will save a tremendous amount of
energy. It is a good idea to have a DC-powered light in the room where the
inverter and batteries are in case there is a problem. AC light dimmers will
only function properly on AC power from inverters that have pure sine wave
output.
Refrigeration
Gas powered absorption
refrigerators are a good choice in small systems if bottled gas is
available. Modern absorption refrigerators consume 5-10 gallons of LP
gas/month. If an electric refrigerator will be used in a stand-alone system,
it should be a high-efficiency type. Some high-efficiency conventional AC
refrigerators use as little as 1200 watt-hours of electricity/day at a 70º
average air temperature. A comparably sized Sun Frost refrigerator/freezer
uses half that amount of energy and a Sundanzer refrigerator (without a
freezer) uses less than 100 watt-hours per day. The higher cost of good
quality DC refrigerators is made up by savings in the number of solar
modules and batteries required.
Major Appliances
Standard AC
electric motors in washing machines, larger shop machinery and tools,
swamp coolers, pumps, etc. (usually 1/4 to 3/4 horsepower) require a large
inverter. Often, a 2000 watt or larger inverter will be required. These
electric motors are sometimes hard to start on inverter power, they consume
relatively large amounts of electricity, and they are very wasteful compared
to high-efficiency motors, which use 50% to 75% less electricity. A standard
washing machine uses between 300 and 500 watt-hours per load, but new
front-loading models use less than 1/2 as much power. If the appliance is
used more than a few hours per week, it is often cheaper to pay more for a
high-efficiency appliance rather than make your electrical system larger to
support a low-efficiency load. Vacuum cleaners usually consume 600 to 1,000
watts, depending on how powerful they are, about twice what a washer uses,
but most vacuum cleaners will operate on inverters larger than 1,000 watts
since they have low-surge motors.
Small Appliances
Many small
appliances such as irons, toasters and hair dryers consume a very large
amount of power when they are used but by their nature require very short or
infrequent use periods. If the system inverter and batteries are large
enough, they will be usable. Electronic equipment, such as stereos,
televisions, VCR’s and computers have a fairly small power draw. Many of
these are available in low voltage DC as well as conventional AC versions.
In general, DC models use less power than their AC counterparts.
OFF-GRID LOADS WORKSHEET
Use the below worksheet to determine the total energy in amp-hours per day
used by all the AC and DC loads in your system.
Calculate your AC loads
If there are no AC loads, skip to Step 5
1. List all AC loads, wattage and hours of use per week in the spaces
provided. Multiply watts by hours/week to get watt-hours per week (WH/Wk).
Add up all the watt hours per week to determine AC watt-hours per week. Use
a separate sheet of paper if you need to list more loads than the space
below allows
NOTE:
Wattage of appliances can usually be determined from tags on the back of the
appliance or from the owner’s manual. If an appliance is rated in amps,
multiply amps by operating voltage (120 or 240) to find watts.
2. Convert to DC watt-hours per week. Multiply line 1 by 1.15 to correct
for inverter loss. ______________
3. Inverter DC input voltage; usually 12, 24 or 48 volts. This is DC
system voltage. ______________
4. Divide line 2 by line 3. This is total DC amp-hours per week used by
AC loads. ______________
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Use this worksheet to determine what size battery
is required for your system. Battery size is measured in amp-hours. This
is a measure of battery capacity. Battery voltage is determined by the
number of "cells" in series. All lead-acid battery cells have a nominal
output of 2 volts. Actual cell voltage varies from about 1.7 volts at
full discharge to 2.4 volts at full charge. 12 volt lead-acid batteries
are made of 6 separate cells in one case. 6 volt batteries are made of 3
cells in one case. Putting battery cells in parallel increases amp-hour
capacity, but does not change voltage.
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Battery Temperature |
Multiplier |
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80ºF/26.7ºC |
1 |
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70ºF/21.2ºC |
1.04 |
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60ºF/15.6ºC |
1.11 |
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50ºF/10.0ºC |
1.19 |
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40ºF/4.4ºC |
1.3 |
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30ºF/-1.1ºC |
1.4 |
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20ºF/-6.7ºC |
1.59 |
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Step 1 |
Total average amp-hours per day
required from the Systems Load Worksheet, line 9: |
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Step 2 |
Maximum number of continuous cloudy
days expected in your area : |
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Step 3 |
Multiply line 1 by line 2: |
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Step 4 |
Divide line 3 by 0.8 to maintain a 20%
reserve after deep discharge period. (Divinding line 3 by a more
conservative 0.5 will maintain a 50% reserve and increase battery life):
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If no special conditions below
apply, skip to line 9: |
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Special Condition #1: Heavy
Electrical Load |
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Step 5 |
Maximum amperage that will be drawn by
the loads for 10 minutes or more : |
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Step 6 |
Multiply line 5 by line 5.0 |
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Special Condition #2: High
Charge Current |
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Step 7 |
Maximum output amperage of PV array or
other battery charger : |
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Step 8 |
Multiply line 7 by 5.0 hours: |
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Step 9 |
Amp hours from line 4, 6 or 8,
whichever is largest : |
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Step 10 |
If you are using a lead acid battery,
select the multiplier from the Battery Temperature Chart above which
corresponds to the battery’s wintertime average ambient temperature |
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Step 11 |
Multiply line 9 by line 10. This is
your optimum battery size in amp-hours: |
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Step 12 |
Amp-hours of battery chosen.
(Industrial Cell, T105=220, L16=350, etc.): |
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Step 13 |
Divide line 11 by line 12. This is the
total number of batteries in parallel required |
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Step 14 |
Round off to the next highest whole
number. This is the number of parallel strings required. |
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Step 15 |
To determine the number of batteries
required in series, divide the system voltage (12, 24,48) by the voltage
of the chosen battery (2V, 6V or 12V). |
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Step 16 |
Multiply line 14 by line 15. This is
the total number of the chosen battery needed for the system |
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