Title 130 Heat Eng_Pumps Hvac Temperature Heat Engines Heat Pump 1.2 MB 4
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Heat Engines and Heat Pumps
www.curriculum-press.co.uk Number 130

FactsheetPhysics

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

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.

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:

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.

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).

High Temperature
Energy Reservoir

Useful Work

Q (Heat Flow)

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.

High Temperature
Energy Reservoir

Heat Engine

Q
1

W

Q
2

Low Temperature
Energy Reservoir

The useful work done: W = Q
1
- Q

2

The efficiency:

Eff =
W × 100 %

Q
1

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.

Eff =
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.

Heat Engine
It would be helpful if heat flow could be transformed completely
into useful work.

Energy transfer
(Heat Flow)

Higher Temperature Lower Temperature

Heat Flow

Lower TemperatureHigher Temperature

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130. Heat Engines and Heat Pumps Physics Factsheet

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This is a simplistic engine cycle:

1

2

3

4

A

B

Volume

Pressure

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.

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

The table shows some examples of common heat engines:

Efficiency calculations
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.

Example 2: Find the high temperature to low temperature ratio
of two reservoirs at 175oC and 25oC.

Answer: All temperatures must be in Kelvin.

Heat
engine

Power
station

Petrol
engine

Steam
engine

Working
fluid

water/steam

air plus
combustion
gases

water/steam

High temp
reservoir

burning fuel

burning
petrol

burning
coal/oil

Waste
heat

low pressure
steam

exhaust
gases

low pressure
steam

Low temp
reservoir

river/lake

atmosphere

atmosphere

Ratio =
175 + 273

= 1.50
25 + 273

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

225 + 273

= 1.43
75 + 273

The theoretical maximum efficiency would be less.

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.

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

where T
C
and T

H
are the temperatures of the low and high temperature

reservoirs.

Example 4: For the reservoirs just mentioned at 175oC and
25oC, find the theoretical maximum efficiency.

Answer: Change the temperatures to Kelvin.

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 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.

100Cmax
H

T
Eff %

T
= − ×

max

298
Eff 1 100 33%

448
= − × =

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).

The heat transferred into the room (Q
1
) is always greater

than the heat removed from the refrigerator (Q
2
), as the

electrical energy supplied will also end up as heat.

High Temperature
Energy Reservoir

Heat Pump

Q
1

W

Q
2

Low Temperature
Energy Reservoir

1

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Physics Factsheet

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130. Heat Engines and Heat Pumps

Coefficient of Performance
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.

C
max

H C

T
COP

T T
=

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

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?

(a) COP =
283

= 28.3
10

(b) COP =
263

= 8.8
30

Outside house

Warmer
air in

Cooler
air out

Cooler
air in

Warmer
air out

Inside house

circulating
liquid/gas

Electric pump to
circulate liquid/gas

Air Source

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.

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.

Ground Source : Closed loop system