Heat Pumps for Heating and Cooling

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Heat Pumps for Heating and Cooling ( heat-pumps-heating-and-cooling )

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Heat Pumps for Heating and Cooling 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 3 to 4 is typical. Even with this range, it is an excellent choice, because for every watt of power that we use, we transfer 2 to 3 additional watts from outside. What this means is that they will always cost less to operate than electric resistance heat. If it cost $30 per week to heat a space with electric resistance heat, it would cost $10 per week to heat the space with a heat pump. Unfortunately, this coefficient of performance varies with the outdoor temperature. This makes sense when you think about it, because it is going to be a lot easier to remove heat from 50oF outdoor air than it is to remove heat from 10oF outdoor air. Let’s check this with an example. Example 2 Compare the ideal coefficients of performance of the same heat pump installed in two different locations with average outdoor temperatures of 40°F and 15°F respectively. Assume the inside temperatures in both the cases are maintained at 70°F. Location #1 Thot =(70−32)×5/9=21°C Tcold =(40−32)×5/9=4°C Next, convert the Celsius temperatures to Kelvin temperatures by adding 273. T hot = 21°C + 273 = 294K T cold = 4°C+273=277K Finally, use the formula to solve for the COP. COP= 294K / (294K−277K) = 294 / 17 =17.3 Location #2 Thot =(70−32)×5/9=21°C T cold = (15−32) ×5 / 9 = -9.4°C 24

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