Question

A Carnot heat pump is used to heat and maintain a residential building at \(75^{\circ} \mathrm{F}\). An energy analysis of the house reveals that it loses heat at a rate of \(2500 \mathrm{Btu} / \mathrm{h}\) per \(^{\circ} \mathrm{F}\) temperature difference between the indoors and the outdoors. For an outdoor temperature of \(35^{\circ} \mathrm{F}\), determine (a) the coefficient of performance and ( \(b\) ) the required power input to the heat pump.

Short answer

b) What is the required power input to the heat pump in watts?

Step-by-step solution

List the known variables and convert temperatures to Kelvin

- Indoor temperature, \(T_H = 75^{\circ} \mathrm{F}\) - Outdoor temperature, \(T_C = 35^{\circ} \mathrm{F}\) - Heat loss per temperature difference, \(K = 2500 \mathrm{Btu} / \mathrm{h} ^{\circ} \mathrm{F}\) We need to convert the temperatures to Kelvin in order to work with the Carnot cycle equations: \(T_H(K) = (75 + 459.67) \times (5/9) = \frac{534.67 \times 5}{9} = 297.04\mathrm{K}\) \(T_C(K) = (35 + 459.67) \times (5/9) = \frac{494.67 \times 5}{9} = 274.82\mathrm{K}\)

Calculate the heat transfer rate required to maintain the indoor temperature

Next, we need to find the heat transfer rate required to maintain the indoor temperature. We have the heat loss rate \(K = 2500 \mathrm{Btu} / \mathrm{h} ^{\circ} \mathrm{F}\). We multiply this by the temperature difference (\(T_H - T_C\) in Kelvin) to find the heat transfer rate, \(Q_H\): \(Q_H = K \times (T_H - T_C) = 2500 \times (297.04-274.82) \times \frac{9}{5} = 10000 \ \mathrm{Btu/h}\)

Determine the coefficient of performance (COP) for the heat pump

The coefficient of performance for the Carnot heat pump is given by the formula: \(\mathrm{COP} = \frac{T_H}{T_H - T_C}\) Using the given indoor and outdoor temperatures in Kelvin, we find the COP: \(\mathrm{COP} = \frac{297.04}{297.04 - 274.82} = 12.61\)

Calculate the required power input to the heat pump

The power input, \(W\) is found by using the formula: \(W = \frac{Q_H}{\mathrm{COP}}\) Plugging in the values we found for \(Q_H = 10000 \ \mathrm{Btu/h}\) and \(\mathrm{COP} = 12.61\), we get: \(W = \frac{10000}{12.61} = 792.71 \ \mathrm{Btu/h}\) Note: To convert this power input to watts, we can use the conversion factor: \(1 \ \mathrm{Btu/h} = 0.293071 \ \mathrm{W}\) \(W = 792.71 \ \mathrm{Btu/h} \times 0.293071 \ \mathrm{W/Btu/h} \approx 232.27 \ \mathrm{W}\) #Conclusion: a) The coefficient of performance (COP) for the Carnot heat pump is 12.61 b) The required power input to the heat pump is approximately \(232.27 \ \mathrm{W}\)

Key concepts

Coefficient of Performance

The coefficient of performance (COP) is a vital measure in evaluating the efficiency of a heat pump system. It represents the ratio of heat output to work input for a heat pump. Specifically, for a Carnot heat pump, the COP is determined by the temperatures at which the pump operates.

The formula for COP in the context of a Carnot heat pump is \(\text{COP} = \frac{T_H}{T_H - T_C}\), where \(T_H\) is the higher temperature inside the house and \(T_C\) is the lower outside temperature. Both temperatures should be in Kelvin for accurate computation. The higher the COP value, the more efficient the pump is at transferring heat into the space.

COP is a dimensionless number, meaning it has no units. This assists consumers and technicians in comparing the efficiencies of different heat pump models under consistent conditions without worrying about the effect of units. A high-performance heat pump with a high COP will require less electrical power to transfer a specific amount of thermal energy, hence it's crucial in determining energy cost and environmental impact.

Heat Transfer Rate

The heat transfer rate, often measured in British Thermal Units per hour (BTU/h) in the United States, is a key concept in thermodynamics, particularly when discussing heat pumps and residential heating. It reflects the amount of heat energy transferred over time and can be influenced by factors such as insulation, ambient temperature, and the efficiency of the heat pump.

To maintain a constant indoor temperature, the heat pump must replace the heat lost to the environment. The heat transfer rate required to keep the indoor temperature stable can be calculated by multiplying the rate of heat loss through the building's envelope by the temperature difference between indoors and outdoors. This calculation is crucial for sizing a heat pump correctly and ensuring that it can adequately warm a space.

Carnot Cycle

The Carnot cycle represents an idealized thermodynamic cycle that provides the maximum possible efficiency for a heat engine operating between two temperatures. The cycle consists of four reversible processes: two isothermal (constant temperature) processes and two adiabatic (no heat exchange) processes.

In relation to heat pumps, the Carnot cycle describes the theoretical limit for a heat pump's performance. This ideal cycle is used as a standard against which real heat pump cycles can be compared. However, it's important to note that no real heat pump can achieve the exact performance predicted by the Carnot cycle because of practical limitations like friction and non-reversible processes. But the Carnot cycle serves as an essential benchmark for understanding the potential efficiency of heat pump systems and has key implications in sustainability and energy consumption.