Question

For a more realistic estimate of the maximum coefficient of performance of a heat pump, assume that a heat pump takes in heat from outdoors at $10^{\circ} \mathrm{C}$ below the ambient outdoor temperature, to account for the temperature difference across its heat exchanger. Similarly, assume that the output must be \(10^{\circ} \mathrm{C}\) hotter than the house (which itself might be kept at \(20^{\circ} \mathrm{C}\) ) to make the heat flow into the house. Make a graph of the coefficient of performance of a reversible heat pump under these conditions as a function of outdoor temperature (from $\left.-15^{\circ} \mathrm{C} \text { to }+15^{\circ} \mathrm{C} \text { in } 5^{\circ} \mathrm{C} \text { increments }\right)$

Short answer

Answer: The COP of a reversible heat pump increases as the outdoor temperature rises. From -15°C to 15°C, the COP increases from approximately 0.545 to 1.2.

Step-by-step solution

Definition of Coefficient of Performance

The coefficient of performance (COP) for a heat pump is defined as the ratio of the useful heat output (Q_H) to the work input (W), which can be written as: COP = \(\frac{Q_H}{W}\)

Recall the Relation Between Heat, Work and COP in a Reversible Heat Pump

For a reversible heat pump, we have the following relation: Q_H = Q_C + W Where Q_H is the heat output from the heat pump, Q_C is the heat input taken from the cold reservoir, and W is the work input.

Define Thermal Reservoir Temperatures

According to the problem, the pump takes heat from outdoors at a temperature T_C lower than the ambient outdoor temperature and outputs heat at a temperature T_H higher than the indoor temperature. Let the ambient outdoor temperature be T_A and the desired indoor temperature be T_I. T_C = T_A - 10°C T_H = T_I + 10°C Substituting the indoor temperature T_I = 20°C given in the problem: T_C = T_A - 10°C T_H = 30°C

Relate COP to Reservoir Temperatures

In a reversible heat pump, the COP is related to the reservoir temperatures T_H and T_C through the equation: COP = \(\frac{T_H}{T_H - T_C}\)

Substitute T_H and T_C in COP equation

Substitute T_H = 30°C and T_C = T_A - 10°C into the COP equation: COP = \(\frac{30}{30 - (T_A - 10)}\) Simplify the equation to get: COP = \(\frac{30}{40 - T_A}\)

Create a Graph of COP as a Function of Outdoor Temperature

Now we can use the COP equation to create a graph for different outdoor temperatures from -15°C to 15°C in 5°C increments: - For T_A = -15°C, COP = \(\frac{30}{40 - (-15)} = \frac{30}{55} \approx 0.545\) - For T_A = -10°C, COP = \(\frac{30}{40 - (-10)} = \frac{30}{50} = 0.6\) - For T_A = -5°C, COP = \(\frac{30}{40 - (-5)} = \frac{30}{45} \approx 0.667\) - For T_A = 0°C, COP = \(\frac{30}{40 - 0} = \frac{30}{40} = 0.75\) - For T_A = 5°C, COP = \(\frac{30}{40 - 5} = \frac{30}{35} \approx 0.857\) - For T_A = 10°C, COP = \(\frac{30}{40 - 10} = \frac{30}{30} = 1\) - For T_A = 15°C, COP = \(\frac{30}{40 - 15} = \frac{30}{25} = 1.2\) Using these values, a graph can be plotted with outdoor temperature (T_A) on the x-axis and COP on the y-axis.