Heat pumps and refrigerators

State what is meant by the coefficient of performance of a heat pump.

A jeweller hopes to reduce workshop heating costs by replacing a gas-fired boiler with an electric heat pump. The heat pump will provide underfloor heating using water at a temperature of 40 °C.

The air outside the building provides the low-temperature reservoir for the heat pump. Over the winter period the outside temperature will vary between 12 °C and −8 °C.

The salesperson for a heat pump company claims that:

• the coefficient of performance will vary greatly, depending on the outside temperature

• the colder the outside temperature, the lower the energy costs will be for running the heat pump.

Comment on the salesperson’s claims.

A heat pump is installed. The average COP_hp over the winter period is 7.5 In this period the heat pump delivers 2880 kW h of underfloor heating.

What is the electrical energy used by the heat pump in this period?

Tick () one box.

Figure 6 shows a low-voltage solid-state thermoelectric cooling element. The element is a square of side 40 mm and is 4 mm thick.

Figure 7 shows how the element is used as part of a thermoelectric refrigerator to keep small quantities of medicine at a low temperature.

The manufacturer’s data for the element show that when the temperature of the hot side is 35 °C and the temperature of the cold side is 5 °C:

• the rate at which energy is dissipated from the hot side is 65 W

• the electrical power supplied is 28 W.

It is claimed that the coefficient of performance (COP) of a thermoelectric refrigerator is much less than the COP of an ideal refrigerator.

Discuss whether the claim is valid for the thermoelectric refrigerator in this question.

Suggest why a small value of the COP might be acceptable for this particular application of a thermoelectric cooling element.

Which is a correct statement about an ideal heat engine? Tick (✓) one box.

An ideal heat engine has an efficiency of 0.33 The same engine works in reverse as an ideal refrigerator between the same hot and cold spaces.

Determine the coefficient of performance COP_ref of the refrigerator.

Tumble-dryers blow hot air over wet clothes that are moving in a rotating drum. Conventional tumble-dryers heat the air in the drum electrically; other dryers use a heat pump to heat the air.

A typical conventional tumble-dryer uses about 0.6 kW h per kg of clothes. A heat pump tumble-dryer uses about 0.25 kW h per kg.

Explain why the heat pump tumble-dryer uses less electrical energy than the conventional tumble-dryer to dry the same load.

The cold space of the heat pump is the room in which the tumble-dryer is placed. The hot space is the air in the tumble-dryer and is at a temperature of 160 °C.

A heat pump tumble-dryer can be placed in a kitchen at a temperature of 20 °C, or in a garage at around 5 °C.

Deduce which place would result in lower running costs for the tumble-dryer. Support your answer with calculations.

An ideal heat pump and an ideal refrigerator operate between the same hot and cold spaces.

Which statement relating to the coefficient of performance (COP) is correct? Tick (✓) the correct answer.

An ideal refrigerator operates between a cold space at a temperature of ̶ 1 °C and a hot space at a temperature of 70 °C.

Calculate the input power to the refrigerator if the rate of transfer of energy to the hot space is 100 W.