UNIT 9: SECOND LAW OF THERMODYNAMICS –

PART 2

Unit Structure

9.0 Overview
9.1 Learning objectives
9.2 Reversible and Irreversible Processes
9.2.1 Internally Reversible Processes
9.3 The Carnot Cycle
9.3.1 Introduction
9.3.2 The Carnot Cycle
9.3.3 The Carnot Principles
9.4 The Thermodynamics Temperature Scale or the Kelvin Scale
9.5 The Carnot Heat Engine and Thermal Efficiency
9.6 The Carnot Refrigeration Cycle
9.7 The Carnot Refrigerator, Heat Pump and Coefficient of Performance
9.8 Tutorial Sheet
9.9 Summary
9.10 Answers to Activities and Tutorial Sheet

9.0 OVERVIEW

The second law of Thermodynamics states that no heat engine can have an efficiency
of 100 percent. Therefore, we have to determine the highest efficiency that a heat
engine can possibly have.

We will first start by defining idealised processes and then introduce the Carnot Cycle
and the Carnot Principles. These will lead towards defining the Thermodynamics
Temperature scale called the Kelvin Scale.

Finally, the Carnot Heat Engine, the Carnot efficiency, the Carnot Refrigerator and
Heat Pump (HP) and their coefficients of performance will be examined.

Unit 9 1
9.1 LEARNING OBJECTIVES

By the end of this unit, you will be able to do the following:

1. Distinguish between reversible and irreversible processes.

2. Determine the highest efficiency engine, the Carnot engine.
3. Explain the basis of the Thermodynamics Temperature Scale.
4. Calculate the Carnot efficiency and Coefficient of Performance.
5. Assess the performance of heat engines.

9.2 REVERSIBLE AND IRREVERSIBLE PROCESSES

The second law of Thermodynamics states that no heat engine can have an efficiency
of 100%. However, we can try to determine the highest efficiency that a heat engine
can possibly have.

To determine this, we first have to define an idealised process, also known as the
reversible process since reversible processes form the basis of heat engines with the
highest efficiency.

A reversible process is defined as a process which can be reversed without having

any trace on the surroundings; that is, both the system and the surroundings are
returned to their initial states at the end of the reverse process. For examples of
reversible processes, refer to section 5.4 in Unit 5.

Processes that are not reversible are called irreversible processes. One example of an
irreversible process is the case of the cooling of the cup of coffee illustrated in Section
8.2 in Unit 8. The hot cup of coffee lost heat to the cooler surroundings. However, the
cup will not heat up by retrieving the heat lost from the surroundings.

The factors that make a process irreversible are termed irreversibilities, for example
friction, free expansion, compression and chemical reactions.

Unit 9 2
Note that a system can be restored to its initial state following a process, regardless of
whether the process is reversible or irreversible. However, for reversible processes,
this restoration is made without leaving any net change on the surroundings, whereas
for irreversible processes, the surroundings usually do some work on the system and
therefore they do not return to their original state.

9.2.1 Internally Reversible Processes

Reversible processes do not occur in nature and can never be achieved. All processes
occurring in practice, are irreversible. However, reversible processes are easy to
analyse and serve as idealised models to which actual processes can be compared.
The more closely a process approximates a reversible process, the more work will be
delivered by a work producing device and the less work required by a work
consuming device.

For that purpose, idealised actual processes can be approximated by “Internally

Reversible Processes”.

A process is called internally reversible if no irreversibilities occur within the

boundaries of the system during the process. During an internally reversible process, a
system proceeds through a series of equilibrium states and when the process is
reversed, the system passes through exactly the same equilibrium states while
returning to its initial state, that is the same path is followed for the forward and
reversed processes.

The quasi-equilibrium process explained in section 1.8 in Unit 1 is an example of an

internally reversible process.

Unit 9 3
9.3 THE CARNOT CYCLE

9.3.1 Introduction

In the last unit, we have seen that the net work delivered by the heat engine is the
difference between the heat supplied during one part of the cycle and the heat rejected
during another part. We have also noted that heat engines are cyclic devices and the
working fluid of the heat engine returns to its initial state at the end of each cycle.

The net work as well as the cycle efficiency can be maximised by using processes that
require the least amount of work and deliver the most, that is, by using reversible
processes. Heat engines and refrigerators that work on reversible processes serve as
models to which actual heat engines and refrigerators can be compared.

Hence, the most efficient cycle will be the one consisting entirely of reversible
processes. The best known reversible cycle is the Carnot Cycle, first proposed in
1824 by a French engineer, Sadi Carnot.

9.3.2 The Carnot Cycle

The Carnot Cycle is composed of four reversible processes – two isothermal and two
adiabatic. It can be executed either in a closed or a steady flow system. All the heat
supplied is at a fixed temperature and all heat rejected at a lower fixed temperature.
Let us consider the steady flow Carnot Cycle.

Unit 9 4
 HIGH TEMPERATURE SOURCE

QH

Liquid Vapour
4 Boiler 1

Pump
Turbine
Work Output

3 CONDENSER
2
Liquid Vapour + Liquid
QL

LOW TEMPERATURE SINK

Figure 9.1: The Carnot Heat Engine

PROCESSES:

4 1: Reversible Isothermal Process: Change of phase from liquid to vapour at

constant temperature in Boiler with heat supplied QH from high
temperature source.
1 2: Reversible adiabatic Process: Fluid expands in turbine, temperature and
pressure are lowered.
2 3: Reversible Isothermal Process: The fluid condenses at constant
temperature, liberating heat (QL) to the low temperature reservoir.
3 4: Fluid is pumped and pressure and temperature increase.

Unit 9 5
Figure 9.2 shows the representation of the Carnot Cycle, that is, the four reversible
processes on a pressure-volume diagram.

P
QH
4 1

TH = cst
Wnet
3 TL = cst
2
QL

Figure 9.2: Representation of Carnot Cycle V

4 1: Reversible Isothermal Evaporation

1 2: Reversible adiabatic expansion, also referred to isentropic expansion
2 3: Reversible Isothermal Condensation
3 4: Reversible adiabatic compression also referred to isentropic Compression

On a P-V diagram, the area under the process curve represents the boundary work for
internally reversible processes. Hence, the area under the curve 4-1-2 is the work done
by the gas during the expansion part of the cycle ( that is turbine) and the area below
the curve 2-3-4 is the work done on the gas during compression part of the cycle
(pump). The area enclosed by the path of the cycle is the difference between these two
and represents the net work done during the cycle.

9.3.3 The Carnot Principles

Based on Kelvin Planck and Clausius statements, two conclusions pertaining to the
thermal efficiencies of reversible and irreversible heat engines, known as Carnot
Principles, can be drawn.

1. The efficiency of an irreversible heat engine is always less than the efficiency
of a reversible one operating between the same two reservoirs.

Unit 9 6
2. The efficiency of all reversible heat engines operating between the same two
reservoirs are the same.

9.4 THE THERMODYNAMICS TEMPERATURE SCALE

A temperature scale that is independent of the properties of substances that is used to

measure temperature is called an absolute temperature scale. The Carnot Principles
form the basis for establishing an absolute temperature scale, also called the Kelvin
Scale. The latter is related to the heat transfer between a reversible device and the
high and low temperature reservoirs.

For a reversible heat engine operating between one reservoir at high temperature TH
and another reservoir at low temperatureTL, Lord Kelvin proposed the following
relationship:

 QH  T
  = H Equation 9.1
 Q L  rev TL

The temperatures on the Kelvin scale are called absolute temperatures. The
1
magnitude of a Kelvin is defined as of the temperature interval between
27316
.
absolute zero and the triple point temperature of water.

9.5 THE CARNOT HEAT ENGINE AND THERMAL

EFFICIENCY

The hypothetical heat engine that operates on the reversible Carnot Cycle is called the
“Carnot Heat Engine”. The thermal efficiency of any heat engine, reversible or
irreversible is given by equation 8.2.
QL
nth = 1 −
QH

Unit 9 7
 QL T
For reversible heat engines, the heat transfer ratio can be replaced with L so
QH TH
that the efficiency of any reversible heat engine or the Carnot heat engine is

TL
nth , rev = 1 − Equation 9.2
TH

This relation is often referred to as the Carnot thermal efficiency or Carnot efficiency.

The Carnot efficiency is the highest efficiency a heat engine operating between two
thermal energy reservoirs at temperatures TH and TL can have. An actual heat engine
cannot achieve this maximum theoretical efficiency because it is impossible to
completely eliminate all irreversibilities associated with the actual cycle.

Note that TL and TH are absolute temperatures, and TL and TH have to be used in
Kelvin.

Hence, the following can be deduced

< nth, rev : irreversible heat engine

nth = nth, rev : reversible heat engine
> nth, rev : Impossible heat engine

The thermal efficiency of actual heat engines can be maximised by supplying heat to
the engine at the highest possible temperature (but the latter is limited by material
strength) and rejecting heat from the engine at the lowest possible temperature which
is limited by the temperature of the cooling medium.

Activity 1

a. What are the four processes that make up the Carnot Cycle?

Unit 9 8
b. A Carnot heat engine receives 500 KJ of heat per cycle from a high
temperature source at 6520C and rejects heat to a low temperature sink at
300C.

Determine: (i) the thermal efficiency

(ii) the amount of heat rejected to the sink.

9.6 THE CARNOT REFRIGERATION CYCLE

The Carnot heat engine cycle is a totally reversible cycle. Therefore, all the processes
that comprise it, can be reversed, in which case it becomes the Carnot Refrigeration
cycle. The cycle remains exactly the same, except that the directions of all heat and
work interactions are reversed.

HIGH TEMPERATURE TH

QH

4 Condenser 3

Winput
TURBINE
COMPRESSOR

EVAPORATOR
1 2

QL

LOW TEMPERATURE TL

Figure 9.3: The Carnot Refrigeration Cycle

Unit 9 9
PROCESSES:
1 2: Evaporation in evaporator, taking heat from low temperature reservoir, TL.
2 3: Reversible adiabatic compression in pump.
3 4: Condensation from vapour to liquid state rejecting heat QH to high
temperature reservoir, TH.
4 1: Reversible adiabatic expansion in turbine.

9.7 THE COP’s OF CARNOT REFRIGERATOR AND HEAT

PUMP

A refrigerator or a heat pump that operates on the reversed Carnot Cycle is called a
Carnot refrigerator or a Carnot heat pump.

As seen in sections 8.5 in the previous unit, the efficiency of a Carnot refrigerator or
heat pump, also called Coefficient Of Performance, (COP) is defined as
desired output
.
required input
For a refrigerator, the desired output is keeping the cold space cool and the required
input is the compressors’s work

QL QL 1
COPR = = = Equation 9.3
W QH − QL QH
−1
Ql

TH

QH

QL

TL

Unit 9 10
For a heat pump, the desired output is keeping the hot space hot and the required input
is the pump’s work.

QH QH 1
COPHP = = = Equation 9.4
W QH − QL Q
1− L
QH

The COPs’ of any refrigerator or heat pump, reversible or irreversible are given by

1 1
COPR = COPHP =
QH QL − 1 1 − QL QH

QH T
For reversible refrigerators or heat pumps, the quantity can be replaced by H ,
QL TL
the highest coefficients of performance that a refrigerator or a heat pump operating
between the temperature limits of TL and TH can have, are as follows :

1 1
COPR ,rev = COPHP ,rev = Equation 9.5
TH T
−1 1− L
TL TH

The coefficients of performance of actual and reversible refrigerators operations

between the same temperature limits can be compared as follows:-

< COPR, rev : irreversible refrigerator

COPR = COPR, rev : reversible refrigerator
> COPR, rev : impossible refrigerator

A similar relation can be obtained for heat pumps by replacing all values of COPR by
COPHP.

Unit 9 11
 < COPHP, rev : irreversible heat pump
COPHP = COPHP, rev : reversible heat pump
> COPHP, rev : impossible heat pump

Activity 2

a. What are the four processes that make up the reversed Carnot Cycle?
b. A refrigerator is to remove heat from the cooled space at a rate of 300 KJ/min
to maintain its temperature at –8 0C. If the air surrounding the refrigerator is
at 250C, determine the minimum power input required for this refrigerator.
c. Draw the reversed Carnot Cycle on a P-V diagram.

9.8 TUTORIAL SHEET

1. A cyclic machine is used to transfer heat from a higher to a lower temperature

reservoir, as shown in Figure 9.4 below. Determine whether this machine,
with energy transfer values is reversible, irreversible or impossible.

TH =1000K

QH = 325 Btu

Cyclic
Machine W = 200 Btu

QL = 125 Btu

TL = 400K

Figure 9.4

Unit 9 12
2. A heat pump is to be used to heat a house in the winter and then reversed to
cool the house in summer. The interior temperature is to be maintained at
200C in the winter and 250C in the summer. Heat transfer through the walls
and roof is estimated to be 2400 kJ per hour.
(a) If the outside temperature in the winter is 00C, what is the minimum
power required to drive the heat pump?

(b) If the power input is the same as in part (a), what is the maximum
outside summer temperature for which the inside of the house can be
maintained at 250C?

9.9 SUMMARY

A process is said to be reversible if both the system and the surroundings can be
restored to their original conditions. But this holds only for ideal processes. Any
other process is irreversible.

The Carnot Cycle is a reversible cycle that is composed of four reversible processes,
two isothermal and two adiabatic. The Carnot Principles state that the thermal
efficiencies of all reversible heat engines between the same two reservoirs are the
same, and that no heat engine is more efficient than a reversible one operating
between the same two reservoirs. These statements form the basis for establishing the
Q  T
absolute temperature scale, the Kelvin Scale, where  H  = H
 Q L  rev TL

A heat engine that operates on the reversible Carnot Cycle is called the Carnot heat
engine and its thermal efficiency is given by

TL
nth rev = 1−
TH

Unit 9 13
This is the maximum efficiency a heat engine operating between two reservoirs at
temperature TH and TL can have.

The processes in the Carnot heat engine can be reversed to form the Carnot
refrigerator cycle and the COP’s of reversible refrigerators and heat pumps are

1
COPR , rev =
TH TL − 1

1
COPHP , rev =
1 − TL TH

These are the highest COPs’ a refrigerator and a heat pump can have.

In the next unit, you will see the application of the 2nd law of Thermodynamics to
processes.

9.10 ANSWERS TO ACTIVITIES AND TUTORIAL SHEET

Activity 1

a) Four processes that make up the Carnot Cycle are:

• Reversible Isothermal expansion
• Isentropic expansion
• Reversible Isothermal Condensation
• Reversible adiabatic compression

b) nth = 0.672
QL = 163.82 kJ

Unit 9 14
Activity 2

a) Four processes that make up the reversed Carnot Cycle are:

1. Isothermal condensation
2. Reversible adiabatic compression
3. Isothermal evaporation
4. Reversible adiabatic expansion

b) COPR = 8.03
Power input = 0.62 kW

c) P
QH
4 3

1
2
QL

Tutorial No 1

QL 125
nth = 1 − = 1− = 0.615
QH 325

TL 400
nth , rev = 1 − = 1− = 0. 6
TH 1000

Since nth > nth,rev; the machine is an impossible machine.

Unit 9 15
Tutorial No 2

Heat Pump
Winter Summer

House 20°C Hot Air

2400 kJ/hr

HP

2400 kJ/hr

Cold Air 0°C House

Heat Pump
QH QH QL
(COP ) HP = = =
1
(COP )R =
Wnet QH − QL Q Wne t
1− L
QH
1 2400
= = = 14.64
T 163.9
1− L
TH
QL
=
1
= 14.64. (COP )R =
273 Q H − Q1
1−
293
QH
(COP )HP = = 14.64 =
1
= 14.64
Wnet TH
−1
TL
2400 1 T
Wne t = = 163.9kJ / hr. = H −1
14.64 14.64 TL
TH
= −1
298
TH = 318 K = 45°C ← .

Unit 9 16