The seasonal energy efficiency ratio (SEER) is a meaurement used to rate the efficiency of air conditioners. The SEER rating of a device is the cooling output during a typical cooling-season divided by the total electric energy input during the same period. The higher a device's SEER rating, the more energy efficient it is. SEER is defined for the United States by the Air Conditioning, Heating, and Refrigeration Institute, a trade association, in its 2008 standard AHRI 210/240, Performance Rating of Unitary Air-Conditioning and Air-Source Heat Pump Equipment. This defines SEER as the ratio of cooling in British thermal units (BTUs) to the energy consumed in watt-hours. The device testing protocol was changed in 2023 to include more accurately the energy used when cooled air is pushed through ducts. The revised protocol produces smaller numbers identified in documentation as SEER2 measurements. European legislation defines SEER as the reference annual cooling demand divided by the annual electricity consumption for cooling of a device. The legislation requires that many air conditioners offered for sale in the EU must bear labels including SEER ratings measured in a standardized way. This is often referred to as the European seasonal energy efficiency ratio (ESEER) to distinguish it from the US standard. SEER (ESEER) values are calculated in Europe using Europe-specific load conditions, outdoor temperatures, and weightings, and SI units rather than BTU, making the numbers not directly comparable with those seen in the U.S. The rest of this article discusses SEER as defined by the AHRI 210/240 standard.
Example For example, consider an air-conditioning unit with a cooling capacity of 5,000 BTU/h and a SEER of 10 BTU/Wh, operating for 8 hours daily during a 125-day annual cooling season (1,000 hours each year). The annual total cooling output would be:
5 , 000 B T U 1 h × 8 h 1 d a y × 125 d a y 1 y e a r = 5 , 000 , 000 B T U y e a r . {\displaystyle {\frac {\mathrm {5,000~BTU} }{\mathrm {1~{\cancel {h}}} }}\times {\frac {\mathrm {8~{\cancel {h}}} }{\mathrm {1~{\cancel {day}}} }}\times {\frac {\mathrm {125~{\cancel {day}}} }{\mathrm {1~year} }}=\mathrm {5,000,000~{\frac {BTU}{year}}} .}
The annual electrical energy usage would be:
A n n u a l t o t a l c o o l i n g o u t p u t S E E R = 5 , 000 , 000 B T U 10 B T U p e r W h = 500 , 000 W h = 500 k W h {\displaystyle {\frac {\mathrm {Annual~total~cooling~output} }{\mathrm {SEER} }}={\frac {\mathrm {5,000,000~BTU} }{\mathrm {10~BTU~per~Wh} }}=500,000~\mathrm {Wh} =500~\mathrm {kWh} }
The average power usage may be calculated simply as:
C o o l i n g c a p a c i t y S E E R = 5 , 000 B T U p e r h o u r 10 B T U p e r W h = 500 W = 0.5 k W {\displaystyle {\frac {\mathrm {Cooling~capacity} }{\mathrm {SEER} }}={\frac {\mathrm {5,000~BTU~per~hour} }{\mathrm {10~BTU~per~Wh} }}=500~\mathrm {W} =0.5~\mathrm {kW} }
If the electricity cost is $0.20/(kW·h), then the cost per operating hour is:
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