That is very interesting. How do they get 27,000 BTU out of just over 10 amps? That is kind of curious to me. Seems like it would be great for garages, and would give cooling ability too.
Thanks for the reply!
Because it is a heat pump, which has an apparent efficiency of greater than 100%. In other words, you get more heat energy out of it than you put into it in the form of electrical energy, because you are using the electrical energy not to generate heat directly, but to bring in heat energy from the outside air (I think of an analogy of using electrical energy to bring outside coal into a furnace). This "apparent efficiency" is called Coefficient of Performance, abbreviated COP. The following calculations indicate that the specifications show that the COP is in the range that is normally quoted for heat pumps.
Specifications:
1.0 kilowatt (kW) = 3413 Btu/hr
Input current 11.1A(heat)
Input power 2550W(heat) = 2.55kW
Input Voltage
230v/60Hz
Calculations:
11.1Ax230V = 2553w = 2.553kW (which is consistent with their specifications)
2.553kWx 3413 Btu/hr = 8713 Btu/hr
24,000Btu/hr / 8713 Btu/hr = 2.75 COP
"Air source heat pumps are relatively easy (and inexpensive) to install and have therefore historically been the most widely used heat pump type. However, they suffer limitations due to their use of the outside air as a heat source or sink. The higher temperature differential during periods of extreme cold or heat leads to a lower efficiency, as explained above. In mild weather, COP may be around 3.5, while at temperatures below around −5°C (23°F) an air-source heat pump's COP will drop below 2. The average COP over seasonal variation is typically 2.5-2.8,[9] and high efficiency model in Japan over 6.0(2.8kW) written in the IPCC 4th Working Group III report chapter 6 [10]."
http://en.wikipedia.org/wiki/Heat_pump
