Seasonal energy efficiency ratio

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The efficiency of air conditioners are often, but not always, rated by the Seasonal Energy Efficiency Ratio (SEER). The higher the SEER rating, the more energy efficient is the air conditioner. The SEER rating is the Btu of cooling output during a simulated, typical cooling-season divided by the total electric energy input in watt-hours (W·h) during the same period. [1]

SEER = BTU ÷ W·h

For example, a 5000 Btu/h air-conditioning unit, with a SEER of 10, operating for a total of 1000 hours during an annual cooling season (e.g., 8 hours per day for 125 days) would provide an annual total cooling output of:

5000 Btu/h × 1000 h = 5,000,000 Btu

With a SEER of 10, the annual electrical energy usage would be about:

5,000,000 Btu ÷ 10 = 500,000 W·h

This is equivalent to an average power usage during the cooling season of:

500,000 W·h ÷ 1000 h = 500 W

The average power usage may also be calculated more simply by:

Average power = (Btu/h) ÷ (SEER, Btu/W·h) = 5000 ÷ 10 = 500 W


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[edit] 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 resistive 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.

[edit] 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.

[edit] 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

[edit] References

  1. ^ Definition of SEER (scroll down to "Seasonal energy efficiency ratio")
  2. ^ SEER conversion formulas from Pacific Gas and Electric
  3. ^ Minimum SEER ratings required in the US
  4. ^ How Contractors Really Size Air Conditioning Systems

[edit] See also

[edit] External links

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