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Chapter 20: Problem 18

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

Expert verified
Answer: A heat pump can achieve this by having an efficient coefficient of performance (COP), which indicates the ability to transfer more thermal energy than the work it consumes. In this example, the heat pump has a COP of 3.50, allowing it to transfer 3.50 units of heat for every 1 unit of work it consumes, effectively compensating for the heat loss in the house.

Step by step solution

01

Understand the concept of a heat pump

A heat pump is a device that transfers thermal energy from a cold reservoir (usually outside air, ground, or water) to a warm reservoir (usually the inside of a building) by doing work on the system. The main working principle of a heat pump is based on the refrigeration cycle, which consists of compression, condensation, expansion, and evaporation processes. A refrigerant fluid circulates through these processes, absorbing heat from the cold reservoir and releasing it into the warm reservoir.
02

Define the coefficient of performance for a heat pump

The efficiency of a heat pump is measured using the coefficient of performance (COP), which is defined as the ratio of the heat transferred (Q) to the work done (W) by the heat pump: COP = (Q_Warm_reservoir) / W The COP is an indicator of how effectively the heat pump can transfer thermal energy from the cold reservoir to the warm reservoir. The higher the COP, the more efficient the heat pump is at transferring heat.
03

Calculate the coefficient of performance (COP)

In the given exercise, we know that the heat pump operates at a power of 6.28 kW while heating a house that is losing thermal energy at a rate of 21.98 kW. We can use these values to calculate the COP: COP = (Q_Warm_reservoir) / W = (21.98 \mathrm{~kW}) / (6.28 \mathrm{~kW}) = 3.50
04

Explain how a heat pump can heat a house losing energy at a higher rate

Based on our calculation, the COP of the heat pump in the given example is 3.50, which indicates that it can transfer 3.50 units of heat for every 1 unit of work it consumes. In other words, for every 1 kW of electrical power supplied to the heat pump, it can transfer 3.50 kW of heat from the cold reservoir to the warm reservoir. So, even though the heat pump is only operating at a power of 6.28 kW, it can effectively heat a house that is losing energy at a rate of 21.98 kW because it can transfer a total of 21.98 kW of heat, which compensates for the heat loss.
05

Conclusion

A heat pump can heat a house that is losing thermal energy at a rate of 21.98 kW while operating at a power of only 6.28 kW due to its ability to transfer more thermal energy than the work it consumes. It can achieve this through its working principle, the refrigeration cycle, and an efficient coefficient of performance (COP). In this example, the heat pump has a COP of 3.50, meaning that it can transfer 3.50 units of heat for every 1 unit of work it consumes. This allows the heat pump to effectively compensate for the heat loss in the house.

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