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It depends on what you mean by efficiency.
Suppose you want to heat your house. An electric heater like you're considering would do this by converting electrical energy directly into heat. Pretty much all the electrical energy does get converted to heat, as you suggest. The energy used to get a certain amount of heat into the house is simply equal to that amount of heat. In that sense, the electric heater is 100% efficient, since energy not directly turned into heat will be turned into heat soon. That isn't a very useful way of thinking about efficiency, though, because any form of energy in your house will probably decay into heat energy pretty quickly. Your computer, television, and refrigerator are 100% efficient at heating your house from this point of view, because although they do things other than generate heat, the energy they use to do those things becomes heat in short order.
By contrast, a heat pump would heat your house by taking heat from the outside and moving it inside. The energy it needs to do this depends on the outside and inside temperatures. If the temperatures inside and outside are $T_i$ and $T_o$, an ideal heat pump (i.e. a Carnot engine) would require
$(1-\frac{T_o}{T_i})*dH$
Joules of work energy to move $dH$ Joules of heat energy from outside to inside (if the outside temperature were greater, this number is negative, meaning the heat pump can extract energy).
The efficiency of the electric heater, compared to the idealized heat pump, is
$1-\frac{T_o}{T_i}$
for given inside and outside temperatures. When the inside and outside temperatures are the same, the electric heater is zero percent efficient. If it's 0C outside and 25C inside, the electric heater is about 8% efficient.
One of the things many people don’t understand is there is little or no difference between the efficiencies of resistance-style electric heaters.
That’s because they all put out 3,414 Btu per watt, per hour. A 1,000-watt wall baseboard heater or an oil-filled electric radiator puts out exactly the same amount of heat per hour (3,414 Btu/hr.) as a 1,000-watt electric hair drier, a 1,000-watt electric range element, a 1,000-watt motor or even 1,000 watts of efficient light-emitting diodes (LEDs).
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And the only way to get anything more out of an electric device beyond just heat is to also put it to work making light, sound or magnetism to run motors. In other words, electric-resistance heaters of any kind probably are the least efficient way to use electricity.
On the other hand, when it comes to heating, heat pumps also use electricity’s power of magnetism to very efficiently drive compressors and motors in order to take the heat out of even very cold outdoor air and then move it inside. How efficient are heat pumps at doing this? Well, that’s what one of the ratings tells us.
I’m currently looking at a news release Carrier Corp. sent me about a new product it says has a Heating Seasonal Performance Factor (HSPF) up to 10.5. What this means is the particular unit will average putting out 10.5 Btu of heat for every watt of electricity it uses per hour of operation during a normal heating season, which is about three times more efficient that an electric resistance heater — or you can get the same amount of heat using a third of the electricity.
On the air-conditioning side, what the industry always has used as a measurement of comparison is the amount of cooling you can get out of a one-ton block of ice if it were defrosted over exactly 24 hours. That is, 12,000 Btu/hr., or a one-ton air conditioner.
But when it comes to cooling efficiency, the unit of measure is the Seasonal Energy Efficiency Ratio (SEER), which is roughly how many Btu of cooling the unit will squeeze out of every watt of electricity per hour of operation across a cooling season. The rating for the unit Carrier advertised in its press release is pretty impressive — 23.8 SEER. Considering the current “builder model” air conditioner usually is rated at
13-SEER, you can see this system, theoretically, probably will use about a third less energy. Why theoretically? Well, that’s the rest of the story.
Understand that these advertised efficiencies are reached on test stands in laboratories under the rated conditions with the proper refrigerant charge, with perfectly matched components (piping, etc.), with the exact amount of airflow distributed evenly across the coils and located in areas that provide the optimal environment. As you can see, whether the units actually achieve their ratings out in the field is heavily influenced by the quality of the installation, the commissioning and the servicing, which is more “real world.”
Nevertheless, assuming that a higher-efficiency unit is being installed to replace a lower-efficiency unit in the same environment, the net result still should produce an impressive energy savings.
This article was originally titled “What does it really mean” in the November 2016 print edition of Supply House Times.
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