Question

It is desired to build a heat pump that has an output temperature of \(23.0^{\circ} \mathrm{C}\). Calculate the maximum coefficient of performance for the pump when the input source is (a) outdoor air on a cold winter day at \(-10.0^{\circ} \mathrm{C}\) and (b) groundwater at \(9.00^{\circ} \mathrm{C}\).

Short answer

Answer: The maximum coefficients of performance are approximately 6.51 for outdoor air at -10.0°C and 17.6 for groundwater at 9.00°C.

Step-by-step solution

Convert temperatures to Kelvin

Before starting, we need to convert the temperatures to absolute scale (Kelvin). To do this, we will add 273.15 to each temperature in Celsius. (a) \(T_{source}=-10.0^{\circ} \mathrm{C}= -10.0 + 273.15 = 263.15\,\mathrm{K}\) (b) \(T_{source}=9.00^{\circ} \mathrm{C} = 9.00 + 273.15= 282.15\,\mathrm{K}\) \(T_{output}=23.0^{\circ} \mathrm{C}=23.0 + 273.15= 296.15\,\mathrm{K}\)

Calculate the maximum coefficient of performance (COP)

According to the Carnot cycle, the maximum coefficient of performance for a heat pump can be calculated by the formula: \(\mathrm{COP}_{\mathrm{max}}=\frac{T_{output}}{T_{output} - T_{source}}\) Now, we will apply this formula for both cases. (a) For outdoor air at \(-10.0^{\circ} \mathrm{C}\) \(\mathrm{COP}_{\mathrm{max}}=\frac{296.15\,\mathrm{K}}{296.15\,\mathrm{K} - 263.15\,\mathrm{K}} \approx 6.51\) (b) For groundwater at \(9.00^{\circ} \mathrm{C}\) \(\mathrm{COP}_{\mathrm{max}}=\frac{296.15\,\mathrm{K}}{296.15\,\mathrm{K} - 282.15\,\mathrm{K}} \approx 17.6\)

Results

The maximum coefficient of performance for the heat pump is approximately: (a) 6.51 when the input source is outdoor air at \(-10.0^{\circ} \mathrm{C}\); (b) 17.6 when the input source is groundwater at \(9.00^{\circ} \mathrm{C}\).

Key concepts

Maximum Coefficient of Performance

Understanding the maximum coefficient of performance (COP) is key to evaluating the efficiency of a heat pump. The COP represents the ratio of heating or cooling provided to the energy consumed. Higher COP values indicate a more efficient heat pump. For the maximum COP, we refer to the idealized scenario described by the Carnot cycle. This theoretical value assumes there are no losses in the system due to friction, resistance, or other inefficiencies.

The formula for the maximum COP of a heat pump is \[ \mathrm{COP}_{\mathrm{max}} = \frac{T_{output}}{T_{output} - T_{source}} \] where \( T_{output} \) and \( T_{source} \) must be in absolute temperature units (Kelvin). The exercise provided showcases how substantial variances in the source temperature can greatly affect the heat pump's COP, revealing that the COP is highest when the temperature difference between the source and output is smallest. As seen in the solution, a warmer source such as groundwater greatly increases the COP when compared to colder outdoor air.

Carnot Cycle

The Carnot cycle serves as the gold standard for the most efficient heat engine cycle that is theoretically possible. In the context of heat pumps, the cycle also defines the limits of efficiency for transferring heat from a cooler place (the source) to a warmer place (the output). The cycle is composed of two isothermal processes – where the temperature remains constant, and two adiabatic processes – where no heat is exchanged with the environment.

For an ideal heat pump operating in a reverse Carnot cycle, the COP is determined by the temperatures of the hot and cold reservoirs. However, real-world heat pumps cannot achieve this maximum performance due to practical limitations such as non-reversible processes and other losses. When students learn about the Carnot cycle, understanding that it is a theoretical model helps them appreciate why actual heat pumps have lower efficiencies than the calculated maximum COP.

Temperature Conversion to Kelvin

Temperature conversion to Kelvin is crucial because thermodynamic calculations, including those for the COP of a heat pump, require absolute temperature measurements. Kelvin is the SI unit for thermodynamic temperature and is one of the seven base SI units. To convert Celsius to Kelvin, we add 273.15 to the Celsius temperature as per the formula:\[ T(\text{K}) = T(^\circ\text{C}) + 273.15 \]

This conversion is necessary because the Kelvin scale aligns with the laws of thermodynamics and allows for the correct interpretation of temperature differences. In Kelvin, zero represents absolute zero, the point at which molecular energy is at a minimum. Kelvin allows for consistent and accurate calculations across various thermodynamic equations. The use of Kelvin in the exercise ensures that the computation of COP relates directly to the absolute thermodynamic temperatures, eliminating the confusion that could arise from temperature scales that start at points other than absolute zero, like the Celsius or Fahrenheit scales.