130 Heat Eng_Pumps

Physics Factsheet www.curriculum-press.co.uk

Number 130

Heat Engines and Heat Pumps Heat Engine

A simple starting point for Thermal Physics is the fact that heat naturally flows from a warmer object to a cooler object.

It would be helpful if heat flow could be transformed completely into useful work. High Temperature Energy Reservoir

Heat Flow

Q (Heat Flow) Useful Work

Higher Temperature

Lower Temperature

However, the Second Law of Thermodynamics says that heat cannot be completely transformed into work. Some of the heat must be transferred into a lower temperature energy reservoir.

And heat, of course, is a form of energy. Heat is a form of energy. So heat flow involves energy transfer. And by definition, energy can do work.

High Temperature Energy Reservoir

At the molecular level, the temperature of an object is linked to the average energy of the molecules. In a solid, the energy is in the form of oscillations:

Q1 Heat Engine

W

Q2 Low Temperature Energy Reservoir Higher Temperature

Lower Temperature

The useful work done: W = Q1 - Q2 When these objects touch each other, and the molecules at the junction are basically in contact, energy will transfer from the larger amplitude oscillations to the smaller amplitude oscillations. We recognise this as heat flow.

The efficiency: Eff =

W × 100 % Q1

The efficiency of a heat engine is always less than 100%. Example 1 : In one cycle of a heat engine, 5000J of heat energy flow from the high temperature reservoir, and 2000J flow into the low temperature reservoir. Calculate the efficiency of the engine. Answer:

Energy transfer (Heat Flow)

Eff =

In this Factsheet we will look at how heat flow can be used to do work (heat engine), and how an energy input can be used to obtain heat flow (heat pump).

5000 - 2000 × 100 = 60 % 5000

In practise, useful heat engines operate in repeating cycles. In each cycle, heat is received from a high temperature reservoir, part of this heat is converted into work, and the remaining (waste) heat is rejected to a low temperature reservoir e.g. the atmosphere or a lake, perhaps. Usually a fluid (gas or liquid) is used to transfer the heat. This is called the working fluid.

1

Physics Factsheet

130. Heat Engines and Heat Pumps This is a simplistic engine cycle:

Efficiency can be improved by lowering the temperature of the low temperature reservoir. However this is often determined by outside factors e.g. the temperature of a river or of the atmosphere.

2 Pressure 3

A

1

The actual equation for the theoretical maximum efficiency in terms of the reservoir temperatures is:

4 B

Eff max = 1− Volume

where TC and TH are the temperatures of the low and high temperature reservoirs.

1) The working fluid (a gas) is heated at fixed volume in a cylinder. Pressure increases. 2) The gas is still being heated. It forces the piston forward, doing work, at constant pressure. 3) Heat is extracted from the gas at fixed volume. The pressure drops. 4) More heat is extracted. The piston pushes the gas back, doing work on the gas.

Example 4: For the reservoirs just mentioned at 175oC and 25oC, find the theoretical maximum efficiency. Answer: Change the temperatures to Kelvin. 298 Eff max = 1− × 100 = 33% 448

Area A+B is the input heat energy. Area A is the useful work done. Area B is the waste heat.

Heat pump A heat pump is often referred to as a “reverse heat engine”. Its usual purpose is to transfer heat energy from a cold reservoir to a hot reservoir. (Normally heat would flow in the opposite direction.)

The table shows some examples of common heat engines: High temp reservoir

Waste heat

Low temp reservoir

Heat engine

Working fluid

Power station

water/steam burning fuel low pressure river/lake steam

Petrol engine

air plus combustion gases

Steam engine

water/steam burning coal/oil

burning petrol

exhaust gases

The Second Law of Thermodynamics tells us that it is impossible to transfer heat from a cooler body to a warmer body without any work input. High Temperature Energy Reservoir

atmosphere Q1

low pressure atmosphere steam

Heat Pump

The theoretical maximum efficiency of a heat engine is defined by the Carnot theorem. This is investigated in factsheet 125 – Energy efficiency in power generation. A key conclusion of this theorem is that the greater the temperature ratio of the high temperature reservoir to the low temperature reservoir, the greater the maximum possible efficiency.

Low Temperature Energy Reservoir

We can see now why it is called a reverse heat engine. A refrigerator is an example of a heat pump. Input electrical energy drives a pump which circulates the working fluid around a circuit. The result is heat being removed from the inside of the refrigerator (the cold reservoir) and transferred into the room (the hot reservoir).

Example 2: Find the high temperature to low temperature ratio of two reservoirs at 175oC and 25oC. Answer: All temperatures must be in Kelvin. 175 + 273 Ratio = = 1.50 25 + 273

The heat transferred into the room (Q1) is always greater than the heat removed from the refrigerator (Q 2), as the electrical energy supplied will also end up as heat.

Example 3: If the temperature of both reservoirs was increased by 50K, would the theoretical maximum efficiency be greater or smaller? Ratio =

W

Q2

Efficiency calculations

Answer:

TC ×100% TH

225 + 273 = 1.43 75 + 273

The theoretical maximum efficiency would be less.

2

Physics Factsheet

130. Heat Engines and Heat Pumps

Coefficient of Performance

Example 5: (a) A heat pump is used to warm a house in winter. The required inside temperature is 20oC. the outside temperature is 10oC. Find the COP. (b) How does the COP change if the outside temperature drops to –10oC?

If you consider the meaning of the term “efficiency” involving useful work done, it doesn’t fit in with heat pumps. Instead we define the Coefficient of Performance (COP). The larger the value of the COP, the more useful the heat pump is. For instance, a COP of 6 would mean that 1 unit of electrical energy supplied to the pump would result in 6 units of heat energy being transferred into the hot reservoir. TC COPmax = TH − TC

Answer:

Using heat pumps to warm buildings is least effective in the winter (when it would be most useful).

(a) COP =

283 = 28.3 10

(b) COP =

263 = 8.8 30

Heating and cooling We have mentioned that a refrigerator is a type of heat pump, where heat is removed from the inside of the refrigerator (and expelled into the room). Heat pumps can also be used as heating systems, drawing heat from a low temperature reservoir and pumping it into the house (the high temperature reservoir). In domestic heating systems, the low temperature reservoir can be the air outside or the ground.

Air source and ground source The air source heating system is much cheaper to install, but its effectiveness drops in the winter. At –18oC the COP drops to about one. This means that the heat provided to the house is only the same as the electrical energy required to drive the pump. The ground source system is much more expensive to install, requiring pipes containing the working fluid to be buried underground. However, as the average below ground temperature remains between 8oC and 13oC, the system is effective year round. Ground source systems can have a COP of five. This means that five units of heat energy are transferred into the house for one unit of electrical input energy.

Ground Source : Closed loop system

Air Source

Inside house

Outside house

Cooler air in

Warmer air in

Warmer air out

Cooler air out circulating liquid/gas Electric pump to circulate liquid/gas

3

130. Heat Engines and Heat Pumps

Practice Questions 5. The table shows some approximate comparative figures for gas and electric heating systems and a ground source heat pump system (driven by electricity) for a small, new house. The values include the hot water system as well as domestic heating.

1. Determine (a) the input energy for the simple energy cycle shown below (b) the useful work done (c) the waste energy emitted (d) the efficiency

Gas costs about 3p per kWh. Electricity costs about 12p per kWh.

Pressure/Nm-2 6.6 × 105

Value

System

2.4 × 105

0.10

0.30

Volume/m3

2. For a diesel engine, state the working fluid, the high and low temperature reservoirs, and the form of the waste heat.

Gas heating

Capital cost Annual consumption

£3500 12000kWh / yr

Electric heating

Capital cost Annual consumption

£3500 9000kWh / yr

Ground heat pump Capital cost Annual consumption

£6000 5000kWh / yr

(a) Find the annual running costs of each system. (b) How long is the payback time for using the heat pump rather than electric heating? (c) What is the advantage of using the heat pump compared to the gas heating system?

3. State the likely temperature of the low temperature reservoir for most heat engines. 4. For a ground source heating system, the heat pump produces about 100W of heat for each metre of trench containing buried pipework. Estimate the length of trench required for an average house with no alternative heating system.

Answers 1. (a) E = 6.6×105 × 0.20 = 1.32×105J (b) W = 4.2×105 × 0.2 = 8.4×104J (c) E = (13.2 - 8.4)×104 = 4.8×104J (d) Eff = 64% 2. Virtually the same as for the petrol engine. 3. Air or water temperature e.g. 10oC or 20oC. 4. My gas boiler has a maximum output of 18kW. A ground source heating system would require 180m of trenches. 5. (a) Gas: £360

Electric: £1080 Heat pump: £600

(b) Extra capital cost £2500 Annual saving £480 Payback time 5.2yr (c) Because it uses electricity, it doesn’t save on cost, but it uses considerably less energy, conserving fossil fuel stocks and reducing emissions of greenhouse gases.

Acknowledgements: This Physics Factsheet was researched and written by Paul Freeman The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered subscriber. No part of these Factsheets may be reproduced, stored in a retrieval system, or transmitted, in any other form or by any other means, without the prior permission of the publisher. ISSN 1351-5136

4