Question

The performance of a heat pump degrades (i.e., its COP decreases) as the temperature of the heat source decreases. This makes using heat pumps at locations with severe weather conditions unattractive. Consider a house that is heated and maintained at \(20^{\circ} \mathrm{C}\) by a heat pump during the winter. What is the maximum COP for this heat pump if heat is extracted from the outdoor air at \((a) 10^{\circ} \mathrm{C},(b)-5^{\circ} \mathrm{C},\) and \((c)-30^{\circ} \mathrm{C} ?\)

Short answer

Question: Calculate the maximum Coefficient of Performance (COP) of a heat pump under different outdoor temperature conditions, given that the indoor temperature is maintained at 20°C. Find the COP values for outdoor temperatures of 10°C, -5°C, and -30°C. Answer: The maximum Coefficient of Performance (COP) for the heat pump under different outdoor temperature conditions is as follows: a) At 10°C, the maximum COP is 29.315. b) At -5°C, the maximum COP is 11.726. c) At -30°C, the maximum COP is 5.863.

Step-by-step solution

Convert temperatures to Kelvin

To convert temperatures from Celsius to Kelvin, we will use the formula: \(T_K = T_C + 273.15\). We need to convert both indoor and outdoor temperatures: - Convert indoor temperature: \(T_H = 20^{\circ} \mathrm{C} + 273.15 = 293.15 \mathrm{K}\). - For each part of the problem, convert outdoor temperature, \(T_L\), to Kelvin: a) \(T_L = 10^{\circ} \mathrm{C} + 273.15 = 283.15 \mathrm{K}\), b) \(T_L = -5^{\circ} \mathrm{C} + 273.15 = 268.15 \mathrm{K}\), c) \(T_L = -30^{\circ} \mathrm{C} + 273.15 = 243.15 \mathrm{K}\).

Calculate the maximum COP using the formula

Now that we've converted all temperatures to Kelvin, let's use the formula: \(COP_{HP}=\frac{T_H}{T_H-T_L}\) to find the maximum COP for each outdoor temperature condition: a) For \(T_L = 283.15 \mathrm{K}\): \(COP_{HP}=\frac{293.15}{293.15-283.15} = \frac{293.15}{10} = 29.315\) b) For \(T_L = 268.15 \mathrm{K}\): \(COP_{HP}=\frac{293.15}{293.15-268.15} = \frac{293.15}{25} = 11.726\) c) For \(T_L = 243.15 \mathrm{K}\): \(COP_{HP}=\frac{293.15}{293.15-243.15} = \frac{293.15}{50} = 5.863\)

Report the results

Finally, report the maximum COP for the heat pump under each outdoor temperature condition: a) At \(10^{\circ} \mathrm{C}\), the maximum COP is 29.315. b) At \(-5^{\circ} \mathrm{C}\), the maximum COP is 11.726. c) At \(-30^{\circ} \mathrm{C}\), the maximum COP is 5.863.