how do you do coefficient of performance on heatpumps?

Answer 1

εk=Q/W
Q- is the heat given off at the higher temperature
W- is only the work of the process : W=ΣQall
Check your material on Carnot processes. Hope that gives you a direction to look into.

Answer 2

COP is the ratio of the heat pump's BTU heat output to the BTU electrical input, as shown in this equation:

COP = Heat Delivered (BTU per hour)/(Electrical Input(Watts) x 3.413BTU per Watt-Hour)

Conventional electric resistance heaters have a COP of 1.0. This means it takes one watt of electricity to deliver the heat equivalent of one watt.

Air-source heat pumps generally have COPs ranging from 2 to 4; they deliver two to four times more energy than they consume. Water and ground source heat pumps normally have COPs somewhere between 3 and 5.

The COP of air-source heat pumps decrease as the outside temperature drops. Therefore, two COP ratings are usually given for a system: one at 47 and the other at 17. When comparing COPs, make sure ratings are based on the same outside air temperature.

COPs for ground- and water-source heat pumps don't vary as much because ground and water temperatures are more constant than air temperatures.

While comparing COPs is helpful, it doesn't tell the whole story. When the outside temperature drops below 40F, the outdoor coils of a heat pump must be defrosted periodically. It's actually possible for the outdoor coil temperature to be below freezing when a heat pump is in the heating cycle.

Under these conditions, any moisture in the air will freeze on the surface of the cold coil. Eventually the frost could build up enough to keep air from passing over the coil and the coil would then lose efficiency. When the coil efficiency is reduced enough to appreciably affect system capacity, the frost must be eliminated. To defrost the coils, the heat pump reverses its cycle and moves heat from the house to the outdoor coil to melt the ice. This reduces the average COP significantly.

Some units have an energy saving feature that will allow the unit to defrost only when necessary. Others will go into a defrost cycle at set intervals whenever the unit is in the heating mode.

Another factor which lowers the overall efficiency of air-to-air heat pumps is their inability to provide enough heat on the coldest days of the winter. This means a back-up heating system is required. This back-up is often electric resistance heat, which only has a COP of one. Whenever the temperature drops into the 25 to 30F range, or whatever its balance point is, and this electric resistance heat kicks in, overall system efficiency drops.

Answer 3

The industry standard is 1.5 horse power per ton according to the ARI. this is equated to CoP of 2.89 which is the reciprocal of Hp times .23 times