EGEE 102
Energy Conservation and Environmental Protection

Air-Source Heat Pump or Air-to-Air Heat Pump Examples

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Example 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:

T hot =( 70-32 )× 5 9 = 21 o C T cold =( 30-32 )× 5 9 =- 1 o C

Next, convert the Celsius temperatures to Kelvin temperatures by adding 273.

T hot = 21 o  C + 273 = 294K

T cold =- 1 o C+273 = 272K

Finally, use the formula from the previous screen to solve for the COP.

COP=( T hot T hot - T cold )

COP=( 294K 294K-272K )= 294 22 =13.3

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.

Comparison of the same heat pump installed in State College, PA and Ann Arbor, MI
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
COP= T hot T hot - T cold ( 294 294-272 )
COP= T hot T hot - T cold ( 294 294-264 )
=17.3 = 9.8
Important Point!

During a heating season, the heat pump's efficiency increases on mild days and decreases on cold days.