Wanderport Corp (WDRP)

[n1tr08urner]

You say 12,000 Btu/hr of cooling energy. If this is achieved the traditional way, with a compressor, in an enclosed space without the help of the outside environment, then by definition you have to put that exact amount of energy (or more to account for losses) into the compressor. Heat pumps that just move heat from one environment to another are different.

It is like saying that the fan (or water pump or whatever) in a car engine is single-handedly cooling the engine removing a huge amount of heat with only a tiny amount of current. Technically yes you are putting a little 12V current in and removing thousands of btus of heat. But it is obvious that the entire process is dependent on a very specific setup where you have a radiator AWAY from the engine with a lot of COOLER air running over it. So the environment is doing the work, not your little electric motor or little pump. In the case of the water heater the environment is obviously cooler than the water so it would actually work AGAINST you. It is the equivalent of you trying to cool your car engine if the outside temperature is 300 F. All of a sudden to achieve the same temperatures you need to actually honestly spend the full amount of energy, you will need a giant AC system with a compressor etc to keep the engine at a reasonable temperature. If your radiator goes to 300 F normally and surrounding air is 300F then you get no cooling, you need to raise your radiator temperature to 500F or whatever, which takes a lot of energy.

The bottom line of the problem with your reasoning is that you are interchanging the terms efficiency and COP. COP only applies to a heat pumps and does not apply here. See here, the efficiency section. http://en.wikipedia.org/wiki/Heat_pump#Efficiency

In our specific case unless there is a heat pump setup, where heat can be transferred AWAY from something and into the water, then the efficiency will be under 100%.