Seasonal energy efficiency ratio


عضو جديد
15 يونيو 2006
مجموع الإعجابات
Seasonal energy efficiency ratio
The efficiency of air conditioners are often rated by the Seasonal Energy Efficiency Ratio (SEER) as defined by the Air Conditioning And Refrigeration Institute in its standard 210/240 Performance Rating of Unitary Air-Conditioning and Air-Source Heat Pump Equipment [1] last updated in 2006.

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 (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 BTU/W·h = 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

Relationship of SEER to EER and COP
SEER is related to the Energy Efficiency Ratio (EER) which is the ratio of cooling capacity in Btu/Hr and the input power in watts W at a given operating point and also to the coefficient of performance (COP) commonly used in thermodynamics. COP is a unitless measure of efficiency. The COP of a heat pump is determined by dividing the energy output of the heat pump in watts W by the electrical energy in watts W 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. EER is related to COP by the converting the cooling capacity from watts W to Btu/Hr by multiplying 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). Formulas for the approximate conversion between SEER and EER or COP in California are: [2]

SEER = EER ÷ 0.9
SEER = COP x 3.792
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 removed from indoors per unit of work energy used to run the heat pump.

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.

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 older 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


مواضيع مماثلة