PrintExample 1
Calculate the ideal coefficient of performance (COP) for an air-to-air heat pump used to maintain the temperature of a house at 70 °F when the outside temperature is 30 °F.Solution:
First, convert the Fahrenheit temperatures to Celsius temperatures using this formula:
Next, convert the Celsius temperatures to Kelvin temperatures by adding 273.
Finally, use the formula from the previous screen to solve for the COP.
The example above shows that for every watt of power we use (and pay for) to drive this ideal heat pump, 13.3 W is delivered to the interior of the house and 12.3 from the outside (we don’t pay for this). This seems to be a deal that one cannot refuse. However, the theoretical maximum is never achieved in the real world. In practice, a COP in the range of 2 to 6 is typical. Even with this range, it is an excellent choice, because for every watt of power that we use, we transfer 1 to 5 additional watts from outside.
Example 2
Compare the ideal coefficients of performance of the same heat pump installed in State College, PA and Ann Arbor, MI when the inside temperature of a house is maintained at 70°F at both locations and the outside temperatures on a given day were 40°F and 15°F at State College and Ann Arbor, respectively.
| State College, PA | Ann Arbor, MI |
|---|---|
| Thot= 70 ºF = 21 ºC = 294 K | Thot= 70 ºF = 21ºC = 294 K |
| Tcold = 40 ºF = 4 ºC = 277 K | Tcold = 15 ºF = -9.4 ºC = 264 K |
| =17.3 | = 9.8 |
During a heating season, the heat pump's efficiency increases on mild days and decreases on cold days.