Glossary of Terms
Q: Appliances and Equipment
A: When referring to the efficiency of an appliance or energy system, we are actually talking about how much energy that system must use to perform a certain amount of work. The higher its energy consumption per unit of output, the less efficient the system is. For example, an air conditioner that requires 750 watts of electricity to provide 6,000 Btu of cooling will be less efficient than one that can provide the same amount of cooling for only 500 watts. The most common ratings applied to energy systems are EER and SEER for most central cooling systems; COP for some heat pumps and chillers; HSPF for heat pumps in their heating modes; and AFUE for gas furnaces and boilers.
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Q: EER
A: EER (energy efficiency ratio) is a measure of how efficiently a cooling system will operate when the outdoor temperature is at a specific level (usually 95° F). A higher EER means the system is more efficient. The term EER is most commonly used when referring to window and unitary air conditioners and heat pumps, as well as water-source and geothermal heat pumps. The formula for calculating EER is:
For instance, if you have a window air conditioner that draws 1500 watts of electricity to produce 12,000 Btu per hour of cooling when the outdoor temperature is 95°, it would have an EER of 8.0 (12,000 divided by 1500). A unit drawing 1200 watts to produce the same amount of cooling would have an EER of 10 and would be more energy efficient.
Using this same example, you can see how efficiency can affect a system’s operating economy. First, you’ll need to determine the total amount of electricity — measured in kilowatt-hours — the unit will consume over a period of time. (A kilowatt-hour is defined as 1,000 watts used for one hour. This is the measure by which your monthly utility bills are calculated.) To do this, let’s assume you operate your 8 EER window air conditioner — drawing 1500 watts at any given moment — for an average of 12 hours every day during the summer. At this rate, it will use 18,000 watt-hours, or 18 kilowatt-hours (kWh) each day, leading to a total consumption of 540 kWh over the course of a 30-day month. At a summer electric rate of 6.34¢ per kWh, it would cost you $34.24 to operate that window air conditioner each month. At the same time, a 1200-watt, 10 EER system, consuming 14.4 kilowatt-hours per day and 432 kWh per month, would cost you $27.39, a 20% savings over the less efficient model.
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Q: SEER
A: SEER (seasonal energy efficiency ratio) measures how efficiently a residential central cooling system (air conditioner or heat pump) will operate over an entire cooling season, as opposed to a single outdoor temperature. As with EER, a higher SEER reflects a more efficient cooling system. SEER is calculated based on the total amount of cooling (in Btu) the system will provide over the entire season divided by the total number of watt-hours it will consume:
By federal law, every central split cooling system manufactured in the U.S. today must have a seasonal energy efficiency ratio of at least 13.0.
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Q: COP
A: COP (coefficient of performance) is the measurement of how efficiently a heating or cooling system (particularly a heat pump in its heating mode and a chiller for cooling) will operate at a single outdoor temperature condition. When applied to the heating modes of heat pumps, that temperature condition is usually 47°F. The higher the COP, the more efficient the system. COP can be calculated by two different methods. In the first, you divide the Btu of heat produced by the heat pump by the Btu equivalent of electricity that is required to produce that heat. This formula is stated:
For instance, let’s assume a heat pump uses 4000 watts of electricity to produce 42,000 Btu per hour (Btu/hr) of heat when it is 47°F outside. To determine its COP, you would first convert the 4000 watts of electrical consumption into its Btu/hr equivalent by multiplying 4000 times 3.413 ( the number of Btu in one watt-hour of electricity). Then you would divide your answer — 13,648 Btu/hr — into the 42,000 Btu/hr heat output. This would show your heat pump to have a 47°F COP of 3.08. This means that, for every Btu of electricity the system uses, it will produce a little more than three Btu of heat when the outdoor temperature is 47°F.
The second formula is most frequently used to determine chiller efficiency. Using this calculation method, you would divide 3.516 by the number of kilowatts (kW) per ton used by the system. This formula is stated:
For example, a chiller that consumes 0.8 kW per ton of capacity would have a COP of 4.4 (3.516 divided by 0.8). On the other hand, the COP of a new, more efficient chiller, using as little as 0.5 kW per ton, would be greater than 7 (3.516 divided by 0.5).
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Q: HSPF
A: HSPF (heating seasonal performance factor) is the measurement of how efficiently all residential and some commercial heat pumps will operate in their heating mode over an entire normal heating season. The higher the HSPF, the more efficient the system. HSPF is determined by dividing the total number of Btu of heat produced over the heating season by the total number of watt-hours of electricity that is required to produce that heat. The formula is written:
Most heat pumps installed in Springfield today have HSPFs in the 7.0 to 8.0 range, meaning they operate with seasonal efficiencies of anywhere from 205% to 234%. (To convert the HSPF number into a percentage, you just divide the HSPF by 3.414, the number of Btu in one watt-hour of electricity.) That means that, for every Btu-worth of energy they use over the entire heating season, these systems will put out anywhere from 2.05 to 2.34 Btu of heat. Compare this to electric furnaces, which have nominal efficiencies of 100% (for each Btu worth of electricity, they put out one Btu of heat), or new gas furnaces, which have efficiency ratings of about 80% to 97% (for each Btu worth of gas, they put out 0.8 to 0.97 Btu of heat).
NOTE: When comparing energy systems that use different primary fuel sources with different costs per Btu, it is important that you understand that higher operating efficiency is not necessarily equivalent to better operating economy. Although an electric heat pump might work with greater efficiency than a gas furnace, it won’t necessarily be more economical to run due to the pricing difference between the two fuel sources.
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Q: AFUE
A: AFUE (annual fuel utilization efficiency) is the measurement of how efficiently a gas furnace or boiler will operate over an entire heating season. The AFUE is expressed as a percentage of the amount of energy consumed by the system that is actually converted to useful heat. For instance, a 90% AFUE means that for every Btu worth of gas used over the heating season, the system will provide 0.9 Btu of heat. The higher the AFUE, the more efficient the system.
When comparing efficiencies of various gas furnaces, it is important to consider the AFUE, not the steady state efficiency. Steady state refers to the efficiency of the unit when the system is running continuously, without cycling on and off. Since cycling is natural for the system over the course of the heating season, steady state doesn’t give a true measurement of the system’s seasonal efficiency. For instance, gas furnaces with pilot lights have steady-state efficiencies of 78% to 80%, but seasonal efficiencies B AFUEs B closer to 65%. Virtually all gas forced-air furnaces installed in this area from the 1950s through the early 1980s had AFUEs of around 65%. Today, federal law requires most gas furnaces manufactured and sold in the U.S. to have minimum AFUEs of 78%. (Mobile home furnaces and units with capacities under 45,000 Btu are permitted somewhat lower AFUEs.) Gas furnaces and boilers now on the market have AFUEs as high as 97%.
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