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Energy efficiency ratio, a measure to asses heat pumps;
Relationship of SEER to EER and COP
SEER is related to the Energy Efficiency Ratio (EER) and also to the coefficient
of performance (COP) commonly used in thermodynamics. COP is a measure of
efficiency. The COP of a heat pump is determined by dividing the energy output
of the heat pump by the electrical energy needed to run the heat pump. The
higher the COP, the more efficient the heat pump. For example resistivie heat
has a COP = 1. The EER is the efficiency rating for the equipment at a
particular pair of external and internal temperatures, while SEER is calculated
over a range of expected external temperatures (i.e., the temperature
distribution for the geographical location of the SEER test). Formulas for the
approximate conversion between SEER and EER or COP in California are: [2]
(1) SEER = EER ÷ 0.9
(2) SEER = COP x 3.792
(3) EER = COP x 3.413
From equation (2) above, a SEER of 13 is approximately equivalent to a COP of
3.43, which means that 3.43 units of heat energy are moved indoors to out per
unit of work energy.
The relationship between SEER and EER is relative depending on where you live
because equipment performance is dependent of air temperatures, humidities, and
pressures. The relationship stated above is typical if you live in the
lower-elevation portions of California; however, if you live in Georgia it is
better approximated by
SEER = EER ÷ 0.80
due to the much higher humidities. A similar relationship exists in relating
SEER and COP, also depending on where you live.
US Government SEER Standards
Today, it is rare to see systems rated below SEER 9 in the United States because
aging, existing units are being replaced with new, higher efficiency units. The
United States now requires that residential systems manufactured after 2005 have
a minimum SEER rating of 13, although window units are exempt from this law so
their SEERs are still around 10.[3] Substantial energy savings can be obtained
from more efficient systems. For example by upgrading from SEER 9 to SEER 13,
the power consumption is reduced by 30% (equal to 1 - 9/13). It is claimed that
this can result in an energy savings valued at up to US$300 per year depending
on the usage rate and the cost of electricity.
With existing units that are still functional and when the time value of money
is considered, most often retaining existing units rather than proactively
replacing them is the most cost effective. Maintenance should be performed
regularly to keep their efficiencies as high as possible.
But when either replacing equipment, or specifying new installations, a variety
of SEERs are available. For most applications, the minimum or near-minimum SEER
units are most cost effective, but the longer the cooling seasons, the higher
the electricity costs, and the longer the purchasers will own the systems,
incrementally higher SEER units are justified. Residential split-system ACs of
SEER 18 or more are now available, but at substantial cost premiums over the
standard SEER 13 units.
Calculating the annual cost of power for an air conditioner
Air conditioner sizes are often given as "tons" of cooling where 1 ton of
cooling is defined as being equivalent to 12,000 BTU/h. The annual cost of
electric power consumed by a 72,000 BTU/h (6 ton) air conditioning unit
operating for 1000 hours per year with a SEER rating of 10 and a power cost of
$0.12 per kilowatt-hour (kW·h) may be calculated as follows:
unit size, BTU/h × hours per year, h × power cost, $/kW·h ÷ SEER, BTU/W·h ÷ 1000
W/kW
(72,000 BTU/h) × (1000 h) × ($0.12/kW·h) ÷ (10 BTU/W·h) ÷ (1000 W/kW) = $864
annual cost
As another example, a 2000 ft2 residential unit near Chicago would require a 4
ton air conditioner based on a location-specific rule-of-thumb that 1 ton is
required for each 500 ft2 for a typical house: [4]
(2000 ft2) ÷ (500 ft2/ton) = 4 tons.
(4 tons) × (12,000 BTU/h/ton) = 48,000 BTU/h.
The estimated cost of electrical power for the 4 ton unit with a SEER rating of
10 and a power cost of $0.10 per kilowatt-hour, using 120 days of 8 hours/day
operation, would be:
(48,000 Btu/h) × (960 h/year) × ($0.10/kW·h) ÷ (10 BTU/W·h) ÷ (1000 W/kW) = $461
annual cost
