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The efficiency of air conditioners is often rated by the Seasonal Energy Efficiency Ratio (SEER) which is defined by the [[Air Conditioning, Heating, and Refrigeration Institute]] in its standard ARI 210=240, Performance Rating of Unitary Air-Conditioning and Air-Source Heat Pump Equipment.
The higher the SEER rating of a unit, the more energy efficient it is. The SEER rating is the Btu of cooling output during a typical cooling-season divided by the total electric energy input in watt-hours during the same period. ## Relationship of SEER to EER and COP

## Theoretical maximum

## US government SEER standards

SEER rating more accurately reflects overall system efficiency on a seasonal basis and EER reflects the system’s energy efficiency at peak day operations. Both ratings are important when choosing products. As of January 2006, all residential air conditioners sold in the United States must have a SEER of at least 13. ENERGY STAR qualified Central Air Conditioners must have a SEER of at least 14.## 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 12¢ per kilowatt-hour (kW·h) may be calculated as follows:

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 BTU/W·h = 500,000 W·h

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

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

If your electricity cost is 20¢/kWh, then your operating cost is:

- 0.5 kW × 20¢/kWh = 10¢/h

SEER is related to the Energy Efficiency Ratio (EER), which is the ratio of output cooling in Btu/Hr and the input power in watts W at a given operating point and also to the COP commonly used in thermodynamics. Performance ratios can be a unitless output over input ratio, never to exceed one, and it is also proper to state what kind of energy is in the numerator and denominator. The COP of a heat pump is determined by dividing the power output of the heat pump by the electrical power needed to run the heat pump, with both powers measured using the same units, e.g. watts. 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. EER ((Btu/(W*hr)) is converted to COP (Btu/Btu) by dividing by 3.413 Btu/(Hr*W).

The SEER is calculated over a range of expected external temperatures (i.e., the temperature distribution for the geographical location of the SEER test).

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 removed from indoors per unit of work energy used to run the heat pump.

The SEER and EER of an air conditioner are limited by the laws of thermodynamics. The refrigeration process with the maximum possible efficiency is the Carnot cycle. The COP of an air conditioner using the Carnot cycle is:

- $COP\_\{Carnot\}=frac\{T\_\{C\}\}\{T\_\{H\}-T\_\{C\}\}$

where $T\_C$ is the indoor temperature and $T\_H$ is the outdoor temperature. Both temperatures must be measured using a thermodynamic temperature scale such as Kelvin or Rankine. The EER is calculated by multiplying the COP by 3.413:

- $EER\_\{Carnot\}=3.413\; frac\{T\_\{C\}\}\{T\_\{H\}-T\_\{C\}\}$

For an indoor temperature of 80F (299.81 K) and an outdoor temperature of 95F (308.15 K), the above equation gives an COP of 36.0. This is about 3 times more efficient than a typical home air conditioner available today.

The maximum COP decreases as the difference between the inside and outside air temperature increases, and vice versa. In desert climates, where the temperature may be as high as 120F, the maximum COP drops to 13.5 (assuming an outdoor temperature of 120F and an indoor temperature of 80F).

The maximum SEER can be calculated by averaging the maximum EER over the range of expected temperatures for the season.

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.

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 20 or more are now available, but at substantial cost premiums over the standard SEER 13 units.

- 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) × (12¢/kW·h) ÷ (10 BTU/W·h) ÷ (1000 W/kW) = $864 annual cost

As another example, a 2000 ft^{2} 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 ft^{2} for a typical older house:

- (2000 ft
^{2}) ÷ (500 ft^{2}/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 10¢ per kilowatt-hour, using 120 days of 8 hours/day operation, would be:

- (48,000 Btu/h) × (960 h/year) × (10¢/kW·h) ÷ (10 BTU/W·h) ÷ (1000 W/kW) = $461 annual cost

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Last updated on Wednesday September 24, 2008 at 11:10:01 PDT (GMT -0700)

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This article is licensed under the GNU Free Documentation License.

Last updated on Wednesday September 24, 2008 at 11:10:01 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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