Question

Explain how it is possible for a heat pump like that in Example 20.2 to operate with a power of only \(6.28 \mathrm{~kW}\) and heat a house that is losing thermal energy at a rate of \(21.98 \mathrm{~kW}\).

Short answer

Short Answer: The heat pump can effectively heat the house because it has a high Coefficient of Performance (COP) of 3.5, meaning that for every unit of electrical power consumed, it transfers 3.5 units of heat energy to the house. As a result, the heat pump can output 21.98 kW of thermal energy, compensating for the heat loss and maintaining the house's temperature, with a power consumption of only 6.28 kW.

Step-by-step solution

Understand the concept of a heat pump

A heat pump is a device that transfers heat energy from a lower temperature reservoir (source) to a higher temperature reservoir (sink) using mechanical work or a high-temperature heat source. In the case of heating a house, the heat pump extracts heat from the external environment (source) and releases it into the house (sink) to maintain the desired temperature.

Introduce the Coefficient of Performance (COP)

The efficiency of a heat pump is measured by its Coefficient of Performance (COP). It is defined as the ratio of the heat energy transferred to the sink (house) to the amount of work done by the heat pump (consumed electrical power): COP = \frac{Q_\textrm{output}}{W} where \(Q_\textrm{output}\) is the thermal energy transferred to the house per unit time, and \(W\) is the work done per unit time or power consumed by the heat pump.

Solve for the required heat output

In this exercise, the house loses thermal energy at a rate of 21.98 kW. Therefore, the desired heat output (\(Q_\textrm{output}\)) should be equal to this value to maintain the house's temperature: \(Q_\textrm{output} = 21.98 \mathrm{~kW}\).

Solve for the COP of the heat pump

The heat pump is said to operate with a power of 6.28 kW, which is the work done (\(W\)) by the heat pump per unit time. Using the formula for COP, we can find the COP value: COP = \frac{Q_\textrm{output}}{W} = \frac{21.98}{6.28} = 3.5

Explain how the heat pump can maintain the house's temperature

The heat pump has a COP of 3.5, which means that for every unit of electrical power consumed by the heat pump, it transfers 3.5 units of heat energy to the house. Since the heat pump has an output power of 21.98 kW, it can compensate for the house's heat loss (also 21.98 kW) and maintain the house's temperature, despite consuming only 6.28 kW of electrical power. This is due to the efficient energy transfer process of the heat pump, which allows it to provide more heat energy than the consumed electrical power.