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J2006 THERMODYNAMICS 1 ( 0 )

KAMARUZZAMAN BIN DAUD (PUO)

ROSLAN BIN HASHIM (PUO)

MODULE J2006

THERMODYNAMICS 1

MALAYSIA

POLYTECHNICSMINISTRY OF EDUCATION

http://modul2poli.blogspot.com/


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Name : Kamaruzzaman b. Daud

Address : Mechanical Engineering Department,

Ungku Omar Polytechnic,

Jln. Raja Musa Mahadi,

31400 Ipoh, Perak.

Telephone No. : 05-5457622 ext.1041

e-mail : [email protected]

Position : Polytechnic Lecturer

Name : Roslan b. HashimAddress : Mechanical Engineering Department,



31400 Ipoh, Perak.

Telephone No. : 05-5457622 ext.1041


EditorName : Nor Resom bt . Buyong

Address : English Language Unit,

General Studies Department,



31400 Ipoh, Perak.

Telephone No. : 05-5457622 ext. 2225


BIODATA OF MODULE WRITERS

J2006 THERMODYNAMICS 1



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What Do You Think Of This Module?

Title of Module: _______________________ Module Code : ___________

Students Name: _______________________ Registration No. : ___________

Course : ____________________________________

Module Writers: ______________________________

Please use the following scale for your evaluation:

4 Strongly Agree3 Agree2 Disagree1 Strongly Disagree

Instruction: Please on the space provided.

No. How much do you agree with the follow ing statements? SCALE

A. FORMAT1 2 3 4

1 The pages are organized in an interesting manner.

2 The font size makes it easy for me to read the module.

3The size and types of pictures and charts used are suitable forthe input.

4 The pictures and charts are easy to read and understand.

5 The tables used are well-organised and easy to understand.

6 The arrangement of the Input makes it easy for me to follow.

7 All the instructions are displayed clearly.

B. CONTENTS1 2 3 4

8 I understand all the objectives clearly.

9 I understand the ideas conveyed.

10 The ideas are presented in an interesting manner.

11 All the instructions are easy to understand.

12 I can carry out the instructions in this module.

13 I can answer the questions in the activities easily.

14 I can answer the questions in the self-assessment.

15 The feedback section can help me identify my mistakes.

16 The language used is easy to understand.

17 The way the module is written makes it interesting to read.18 I can follow this module easily.

19 Each unit helps me understand the topic better.

20I have become more interested in the subject after using thismodule.



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CURRICULUM GRID

The curriculum grid of this module is based on the curriculum used by Malaysian

Polytechnics.

No. TOPIC UNIT TotalHours

1 Basic

Thermodynamics

1

( 1 H)

2

( 2 H )

3

( 2 H ) 5 Hours

2 Non-Flow

Process

4

( 3 H)

5

( 2 H)

5 Hours

3 Flow Process 6

( 2 H )

7

( 2 H )

4 Hours

4 Properties of

Steam

8

( 5 H )

5 Hours

5 The Second

Law of

Thermodynamics

9

( 3 H )

10

( 3 H )

6 Hours

6 The Steam Power

Cycle

11

( 5 H )

5 Hours



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UNIT 1 BASIC THERMODYNAMICS

1.0 Introduction

1.1 Fundamental and derived quantities

1.1.1 Force

1.1.2 Energy

1.1.3 Power1.1.4 Pressure

1.1.5 Density

1.2 Unit conversions


2.0 Introduction

2.1 Definitions of system, boundary, surrounding, open system and close system

2.2 Property, state and process

2.3 The First Law of thermodynamics

2.4 Work and heat transfer

2.5 Sign convention for work transfer

2.6 Sign convention for heat transfer

2.7 Internal energy


3.0 Definition of perfect gas

3.1 Boyles Law

3.2 Charles Law

3.3 Universal Gases Law

3.4 Specific heat capacity at constant volume

3.5 Specific heat capacity at constant pressure

3.6 Relationship between the specific heats

3.7 Specific heat ratio

UNIT 4 NON-FLOW PROCESS

4.0 Introduction

4.1 Differences between the Flow and Non-Flow Process

4.1.1 Flow Process4.1.2 Non-Flow Process

4.2 Constant temperature (isothermal) process

4.3 Adiabatic process



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UNIT 5 NON-FLOW PROCESS

5.0 Non-flow process

5.1 Polytropic process

5.2 Constant volume process

5.3 Constant pressure process

UNIT 6 FLOW PROCESS

6.0 Steady flow processes

6.1 Steady flow energy equation

6.2 Application of steady flow equation

6.2.1 Boilers

6.2.2 Condensers

UNIT 7 FLOW PROCESS

7.0 Application of steady flow equation

7.0.1 Turbine7.0.2 Nozzle

7.0.3 Throttle

7.0.4 Pump

7.1 Equation of continuity

UNIT 8 PROPERTIES OF STEAM

8.0 Introduction

8.1 Phase-change process

8.2 Saturated and Superheated Steam

8.3 Properties of a Wet Mixture

8.3.1 Specific volume

8.3.2 Specific enthalpy

8.3.3 Specific internal energy

8.3.4 Specific entropy

8.4 The use of Steam Tables

8.4.1 Saturated Water and Steam Tables

8.4.2 Superheated Steam Tables

8.5 Interpolation

8.5.1 Single Interpolation

8.5.2 Double Interpolation



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UNIT 9 THE SECOND LAW OF THE THERMODYNAMICS

9.0 Introduction to the Second Law of Thermodynamics

9.1 The heat engine and heat pump

9.2 Entropy

9.3 The T-sdiagram for a steam

9.4 To show that Q = h2 h19.5 Reversible processes on the T-sdiagram for steam

9.5.1 Constant pressure process

9.5.2 Constant volume process

9.5.3 Constant temperature (or isothermal) process

9.5.4 Adiabatic (or isentropic) process

9.5.5 Polytropic process

UNIT 10 THE SECOND LAW OF THE THERMODYNAMICS

10.0 The P-Vand T-sdiagram for a perfect gas

10.1 Reversible processes on the T-sdiagram for a perfect gas10.1.1 Constant pressure process

10.1.2 Constant volume process

10.1.3 Constant temperature (or isothermal) process

10.1.4 Adiabatic (or isentropic) process

10.1.5 Polytropic process

UNIT 11 THE STEAM POWER CYCLE

11.0 Introduction

11.1 The Carnot cycle

11.1.1 Thermal efficiency of Carnot cycle

11.1.2 The work ratio for Carnot cycle

11.2 Rankine cycle

11.2.1 Thermal efficiency of Rankine cycle

11.2.2 The work ratio for Rankine cycle

11.3 Specific steam consumption



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MODULE GUIDELINES

To achieve maximum benefits in using this module, students must follow the instructions

carefully and complete all the activities.

1. This module is divided into 11 units.

2. Each page is numbered according to the subject code, unit and page number.

J2006 / 1 / 5

Subject Page Number 5

Unit 1

3. The general and specific objectives are given at the beginning of each unit.

4. The activities in each unit are arranged in a sequential order and the following

symbols are given:

OBJECTIVES

The general and specific objectives for each learning topic are stated

in this section.

INPUT

This section introduces the subject matter that you are going to learn.

ACTIVITIES

The activities in this section test your understanding of the subject matter.

You have to complete this section by following the instructions carefully.

FEEDBACKAnswers to the questions in the activity section are given here



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FEEDBACK TO SELF-ASSESSMENT

This section contains answers to the activities in the self-assessment.

5. You have to follow the units in sequence.

6. You may proceed to the next unit after successfully completing the unit and you are

confident of your achievement.

SELF-ASSESSMENT

Self-assessment evaluates your understanding of

each unit.



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GENERAL AIMS

This module is prepared for the second semester students who are undergoing

Certificate/Diploma programmes in Malaysian Polytechnics. In each unit, the aim is to

expose the students to the concepts of Thermodynamics and to lead them towards self-

directed learning with guidance from their lecturers.

PREREQUISITE SKILLS AND KNOWLEDGE

At least a pass in Mathematics and Science at SPM level

GENERAL OBJECTIVES

At the end of this module, students should be able to:

1. understand the principles and concepts of units and dimensions2. define the fundamental concepts of system, boundary, surrounding, open system and

close system

3. understand the state of working fluid, its example and the definition of the first law

of thermodynamics

4. describe the differences between work and heat transfer

5. define and show the definition and application of internal energy, the Boyles law,

Charles law and universal gases law

6. define and apply the principle of specific heat capacity of constant pressure and

constant volume

7. provide definition, differences and give examples of the flow process and the nonflow process

8. provide definitions of heat and work in reversible processes

9. define and calculate the following non-flow processes :

9.1 constant temperature (isothermal) process

9.2 adiabatic process

9.3 polytropic process

9.4 pressure and volume constant processes

10. derive the meaning and interpret the steady-flow energy equations

11. apply the steady-flow energy equations on :

11.1 boiler

11.2 condenser

11.3 turbine

11.4 nozzle

11.5 throttle

11.6 pump

12. define the following phases:

12.1 solid

12.2 liquid



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J2006 THERMODYNAMICS 1 ( 10)

12.3 steam

13. define steam at constant pressure

14. differentiate the wet steam, dry saturated steam and superheated steam

15. define dryness fraction (x) and internal energy

16. derive enthalpy from the energy equations

17. apply of the wet steam equations18. define and evaluate the properties of steam using u = h + pvand the steam tables

19. calculate the specific volume, enthalpy, internal energy and entropy using the steam

table, equation and interpolation

20. define The Second Law of Thermodynamics

21. define, give examples and state the differences between the efficiency of heat engine

and heat pump

22. define entropy, s

23. derive Q =T. dsequations for reversible process

24. use T sdiagram to show the changes in entropy and its properties for steam

25. define constant pressure using T sdiagram

26. define and calculate Q = h2- h1

27. draw volume and pressure properties using T sdiagram

28. use equations and calculate the changes in entropy for constant pressure and

constant volume

29. draw reversible isothermal, isentropic and polytropic process using T-sdiagram for

vapour and perfect gas

30. calculate heat, work done and the changes in entropy for isothermal, isentropic and

polytropic process

31. draw Carnot cycle using the T sdiagram

32. define and calculate the Carnot cycle efficiency

33. differentiate between Carnot and Rankine cycle

34. describe the two characteristics of Rankine cycle

35. draw Rankine cycle using the T sdiagram

36. define the processes in Rankine cycle

37. draw block diagrams for Rankine cycle

38. derive equation for turbine, condenser, pump and boiler

39. calculate heat, work done and efficiency of Rankine cycle.



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J2006 THERMODYNAMICS 1 ( 11)

TEACHING AIDS AND RESOURCES NEEDED

1. Calculator

2. Mayhew,Y.R. & Rogers, G.F.C. Steam tables

3. Plant laboratory

REFERENCES

1. Ahmad Taufek Mohd Tiblawi, (1990). Haba dan Bendalir II; IBS Buku Sdn. Bhd.

2. Dr. Yunus A. Cengel & Boles, M.A.(1994). Thermodynamics: An Engineering

Approach; McGraw-Hill, Inc

3. Eastop, T.D. and Mc Conkey, A.(1978). Applied Thermodynamics for

Engineering Technologists; Longman

4. Irving Granet & Maurice Bluestein, (2000). Thermodynamics and Heat Power

(6th

Edition); Prentice Hall Inc.

5. K. Iynkaran & David J. Tandy, (1993). Basic Thermodynamics: Applications and

Pollution Control; Prentice Hall, Simon & Schuster (Asia) Pte. Ltd.

6. Mayhew,Y.R. & Rogers, G.F.C., (1981). Thermodynamic and Transport

Properties of Fluids in SI Units(3rd

Edition); Oxford, Basil Blackwell.

7. Metcalfe, F. (1972). Heat Engines and Applied Heat; Cassell & Company Ltd

8. Moran,M.J. and Shapiro, H.N.(1988). Fundamentals of Engineering

Thermodynamics; John Wiley & Sons, Inc

9. Rayner Joel, (1971). Basic Engineering Thermodynamics in SI Units

(3rd

Edition); Longman Group Ltd.

10. Thomas,T.H. and Hunt, R. (1987). Applied Heat; Heinemann Educational Books

11. http://www.engr.lousiana.edu

12. http://www.mme.tcd.ie



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BASIC THERMODYNAMICS J2006/1/1

BASIC THERMODYNAMICS

OBJECTIVES

General Objective : To understand the concept of units and dimensions

Specific Objectives : At the end of the unit you will be able to:

state the difference between fundamentals and derived

quantities

describe the physical quantities of thermodynamics

understand the conversion units of thermodynamics

calculate the examples of conversion factors

UNIT 1



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1.0 INTRODUCTION

id you realize that the work of an engineer is limited unless he has a source

of power to drive his machines or tools? However, before such a study can

begin, it is necessary to be sure of the number of definitions and units, whichare essential for a proper understanding of the subject. We are familiar with most ofthese items in our everyday lives, but science demands that we have to be exact in

our understanding if real progress is to be made.

When engineering calculations are performed, it is necessary to be concerned with

the units of the physical quantities involved. A unit is any specified amount of aquantity by comparison with which any other quantity of the same kind is measured.

For example, meters, centimeters and millimeters are all units of length. Seconds,

minutes and hours are alternative time units.

D

10 Kilometer + 5 Feet +

25 Yard + 100 Inches

= ? Meter

Could you give me an

answer?

INPUT



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1.1 Fundamental and Derived Quantities

In the present discussion, we consider the system of units called SI (International

System of Units) and it is a legally accepted system in many countries. SI units will

be used throughout this module.

Length, mass, time, electric current, thermodynamic temperature and luminous

intensity are the six fundamental physical quantities. These six quantities are

absolutely independent of one another. They are also called the Indefinables of

mechanics. The SI base units are listed in Table 1.1-1.

Table 1.1-1Fundamental units

Quantity Unit Symbol

Mass kilogram kgTime second s

Length meter m

Thermodynamic temperature degree Kelvin K

Electric current ampere A

Luminous intensity candela cd

All other physical quantities, which can be expressed in terms of one or more of

these, are known as derived quantities. The unit of length, mass, time, electric

current, thermodynamic temperature and luminous intensity are known as

fundamental units. Physical quantities like area, volume, density, velocity,acceleration, force, energy, power, torque etc. are called derived quantities since

they depend on one or more of these fundamental quantities. The units of the derived

quantities are called derived units as shown in Table 1.1-2.

Table 1.1-2 Derived units

Quantity Unit Symbol Notes

Area meter square m2

Volume meter cube m3 1 m

3= 1 x 10

3litre

Velocity meter per second m/s

Acceleration Meter per secondsquared

m/s2

Density kilogram / meter cube kg/m3

Force Newton N 1 N = 1 kgm/s2

Pressure Newton/meter square N/m2 1 N/m

2= 1 Pascal

1 bar = 105N/m

2= 10

2kN/m

2



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1.1.1 Force

Newtons second law may be written as force (mass x acceleration), for a

body of a constant mass.

i.e. F = kma (1.1)

(where m is the mass of a body accelerated with an acceleration a, by a force

F, k is constant)In a coherent system of units such as SI, k = 1, hence:

F = ma (1.2)

The SI unit of force is therefore kgm/s2. This composite unit is called the

Newton, N.

i.e. 1 N = 1 kg.m/s2

1.1.2 Energy

Heat and work are both forms of energy. The work done by a force is the

product of the force and the distance moved in the same direction.

The SI unit of work = force x distance in the Newton meter, Nm.

A general unit for energy is introduced by giving the Newton meter the name

Joule, J.

i.e. 1 Joule = 1 Newton x 1 meteror 1 J = 1 Nm

A more common unit for energy in SI is the kilo joule (1 kJ = 103J)

1.1.3 Power

The use of an additional name for composite units is extended further by

introducing the Watt, W as the unit of power. Power is the rate of energy

transfer (or work done) by or to a system.

i.e. 1 Watt, W = 1 J/s = 1 N m/s



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1.1.4 Pressure

Pressure is the force exerted by a fluid per unit area. We speak of pressure

only when we deal with gas or liquid. The pressure on a surface due to

forces from another surface or from a fluid is the force acting at 90o to the

unit area of the surface.

i.e. pressure = force/ area

P = F/A (1.3)

The unit of pressure, is N/m2and this unit is sometimes called the Pascal, Pa.

For most cases occurring in thermodynamics the pressure expressed in Pascal

will be a very small number. This new unit is defined as follows:

1 bar = 105N/m

2= 10

5Pa

1.1.5. Density

Density is the mass of a substance per unit volume.

The unit of density is kg/m3

.

Force, F = ma

Pressure, P = F/A

Work, W = F x L

Density, = m/V

(1.4)

volume

massDensity

V

m=

=



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Calculate the pressure of gas underneath the piston in equilibrium for a 50 kg

mass that reacts to a piston with a surface area of 100 cm2

.

A density of = 850 kg/m3of oil is filled to a tank. Determine the amount of

mass min the tank if the volume of the tank is V= 2 m3.

Example 1.1

Solution to Example 1.1

2N/m05.49

0.01

9.81x50

area

force(P)Pressure

=

=

=

Example 1.2


We should end up with the unit of kilograms. Putting the given information into

perspective, we have

= 850 kg/m3and V= 2 m

3

It is obvious that we can eliminate m3and end up with kg by multiplying these

two quantities. Therefore, the formula we are looking for is

V

m=

Thus, m = V

=(850 kg/m3)(2 m

3)

= 1700 kg

Force = mass x acceleration

Pressure = force/area



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TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE

NEXT INPUT!

1.1 What is the work done by an expanding gas if the force resisting the motion

of the piston is 700 N and the length of the stroke is 0.5 m ?

1.2 What is the force required to accelerate a mass of 30 kg at a rate of 15 m/s2?

1.3 The fuel tank of a large truck measures 1.2m x 0.9m x 0.6m. How many litres

of fuel are contained in the tank when it is full?

1.4 A weather research instrument is suspended below a helium filled balloon

which is a 3.8m diameter sphere. If the specific volume of helium is

5.6m3/kg, what is the weight of helium in the balloon? Explain briefly why

the balloon rises in the atmosphere.

Activity 1A



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Feedback to Activity 1A

1.1 Work = Force x Distance

= (700 N)(0.5 m)

= 350 Nm or J

1.2 Force = mass x acceleration

F =ma

= (30 kg)(15 m/s2)

= 450 kg.m/s2or N

1.3 Volume = 1.2 x 0.9 x 0.6 = 0.648 m3

Since 1m3= 1000 litres

Then, contents of full tank = 0.648 x 1000

= 648 litres

1.4 Radius of volume, r =2

d

=2

3.3= 1.9 m

Volume of balloon, V= 33

4r

= 3)9.1(3

4

= 28.73 m3

Mass of helium in balloon, m=v

V

= 28.73/5.6

= 5.13 kg



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w = mg

= 5.13 x 9.81

= 50.3 N

Density of helium, =v

1

=6.5

1

= 0.1786 kg/m3

The balloon rises in the atmosphere because the density of helium is less than

the density of atmosphere.



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1.2 Unit Conversions

We all know from experience that conversion of units can give terrible

headaches if they are not used carefully in solving a problem. But with some

attention and skill, conversion of units can be used to our advantage.

Measurements that describe physical quantities may be expressed in a variety of

different units. As a result, one often has to convert a quantity from one unit to

another. For example, we would like to convert, say, 49 days into weeks. One

approach is to multiply the value by ratios of the equivalent units. The ratios are

formed such that the old units are cancelled, leaving the new units.

The Dimensional Homogeneity

Despite their causing us errors, units/dimensions can be our friends.

All terms in an equation must be dimensionally homogeneous.

That is, we cant add apples to

oranges

Neither can we add J/mol to J/kg s.

By keeping track of our units/dimensions,

we can automatically do a reality check

on our equations.

But the fun doesnt stop there

A dimensional analysis can help to determine the form of an equation

that we may have forgotten.

The example of unit conversions are:

1 kg = 1000 g

1 m = 100 cm = 1000 mm 1 km = 1000 m = (100 000 cm @ 10

5cm) = (1 000 000 mm @ 10

6mm)

1 hour = 60 minutes = 3600 seconds

1 m3= 1000 litre, or 1 litre = 1 x 10

-3m

3

1 bar = 1 x 105N/m

2= 1 x 10

2kN/m

2

INPUT



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Multiple and sub-multiple of the basic units are formed by means of prefixes, and the

ones most commonly used are shown in the following table:

Table 1.2 Multiplying factors

Multiplying Factor Prefix Symbol

1 000 000 000 000 1012

tera T

1 000 000 000 109 giga G

1 000 000 106 mega M

1 000 103 kilo k

100 102 hector h

10 101 deca da

0.1 10-1

desi d

0.01 10-2

centi c

0.001 10-3

milli m

0.000 001 10-6 micro

0.000 000 001 10-9

nano n

0.000 000 000 001 10-12

pico p

Example 1.3

Convert 1 km/h to m/s.


m/s278.0

s3600

m1000

s3600

j1x

km1

m1000x

j

km1

j

km1

=

=

=



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Example 1.4

Convert 25 g/mm3to kg/m

3.


1 kg = 1000 g

1 m = 1000 mm

1 m3= 1000 x 1000 x 1000 mm

3

= 109m

3

36

3

9

3

39

33

kg/m10x25

m1000kg1x10x25

g1000

kg1x

m1

mm10x

mm

g25

mm

g25

=

=

=

How could I convertg/mm3to kg/m3?



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NEXT INPUT!

1.5 Convert the following data:

a) 3 N/cm2 to kN/m

2

b) 15 MN/m2to N/m

2

1.6 Convert 15 milligram per litre to kg/m3.

I hope youve learnt somethingfrom this unit. Lets move on tothe next topic.

Activity 1B



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Feedback To Activity 1B

1.5 a) 1 kN = 1000 N

1 m2= 100 x 100 = 10

4cm

2

b) 1 MN = 106N/m

2

1.6 1 kg = 1 000 000 mg

1 m3= 1000 litre

2

2

4

2

24

22

kN/m30

m1000

kN10x3

N1000

kN1x

m1

cm10x

cm

N3

cm

N3

=

=

=

26

6

22

N/m10x15

MN1

N10x

m

MN15

m

MN15

=

=

33-

3

kg/m10x15

m1

litre1000 x

mg0000001

kg1x

litre

mg15

litre

mg15

=

=



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You are approaching success. Try all the questionsin this self-assessment section

and check your answers with those given in the Feedback to Self-Assessment on the

next page. If you face any problem, discuss it with your lecturer. Good luck.

1. A gas is contained in a vertical frictionless piston-cylinder device. The

piston has a mass of 4 kg and a cross-sectional area of 35 cm2. A compressed

spring above the piston exerts a force of 60 N onto the piston. If theatmospheric pressure is 95 kPa, determine the pressure inside the cylinder.

2. A force of 8 N is applied continuously at an angle of 30oto a certain mass.

Find the work done when the mass moves through a distance of 6 m.

3. A man weighing 60 kg goes up a staircase of 5 m in height in 20 secs.

Calculate his rate of doing work and power in watts.

4. The density of water at room temperature and atmospheric pressure is

1.0 g/cm3. Convert this to kg/m3. Find also the specific volume of water.

SELF-ASSESSMENT



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Have you tried the questions????? If YES, check your answers now.

1. 123.4 kPa

2. 41.57 J

3. 147 J, 147 watt.

4. 1000 kg/m3; 0.001 m

3/kg

CONGRATULATIONS!!!!..

May success be with you

always.

Feedback to Self-Assessment



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OBJECTIVES

General Objective : To understand the basic concept and the First Law of

Thermodynamics


Define the fundamental concepts of system, boundary,surrounding, open system and close system

explain the property, state and process of the working fluid

and provide example

state the definitions of the First Law of Thermodynamics

describe the differences between work and heat transfer

define the definitions and show the application of internal

energy

UNIT 2



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2.0 Introduction

Every science has a unique vocabulary associated with it, and thermodynamics is no

exception. Precise definition of the basic concepts forms a sound foundation for the

development of science and prevents possible misunderstandings. In this unit, the

systems that will be used are reviewed, and the basic concepts of thermodynamics

such as system, energy, property, state, process, cycle, pressure and temperature are

explained. Careful study of these concepts is essential for a good understanding of

the topics in the following units.

2.1 Definitions of system, boundary, surrounding, open system and close system

A thermodynamic system, or simply a system, is defined as a quantity of matter or

a region in space chosen for study. The fluid contained by the cylinder head,

cylinder walls and the piston may be said to be the system.

The mass or region outside the system is called the surroundings. The surroundings

may be affected by changes within the system.

The boundaryis the surface of separation between the system and its surroundings.

It may be the cylinder and the piston or an imaginary surface drawn as in Fig. 2.1-1,

so as to enable an analysis of the problem under consideration to be made.

Boundary

Surrounding

System

Figure 2.1-1 System, surroundings and boundary

INPUT



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A system can either to be close or open, depending on whether a fixed mass or a

fixed volume in space is chosen for study. A close system(also known as a control

mass) consists of a fixed amount of mass, and no mass can cross its boundary. That

is, no mass can enter or leave a close system, as shown in Fig. 2.1-2. But energy,

in the form of heat or work can cross the boundary, and the volume of a close system

does not have to be fixed.

SURROUNDINGS

BOUNDARY

Fig. 2.1-2 A closed system with a moving boundary

An open system, or a control volume, as it is often called, is a properly selected

region in space. It usually encloses a device, which involves mass flow such as a

boiler, compressor, turbine or nozzle. Flow through these devices is best studied by

selecting the region within the device as the control volume. Both mass and energy

can cross the boundary of a control volume, as shown in Fig. 2.1-3.

QOUT

WOUT

Fig 2.1-3 Open system in boiler

2.2 Property, State and Process

SYSTEM

Fluid Inlet

Fluid Outlet

SYSTEM

SURROUNDINGS

BOUNDARY



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Properties are macroscopic characteristics of a system such as mass,

volume, energy, pressure, and temperature to which numerical values can be

assigned at a given time without knowledge of the history of the system.

Many other properties are considered during the course of our study of

engineering thermodynamics. Thermodynamics also deals with quantities

that are not properties, such as mass flow rates and energy transfers by work

and heat. Properties are considered to be either intensiveor extensive.

Intensive properties are those which are independent of the size of the

system such as temperature, pressure and density.

Extensiveproperties are those whose values depend on the size or extent of

the system. Mass, volume and total energy are some examples of extensive

properties.

The word state refers to the condition of system as described by its

properties. Since there are normally relations among the properties of a

system, the state often can be specified by providing the values of a subset of

the properties.

When there is a change in any of the properties of a system, the state changes

and the system are said to have undergone a process. A process is a

transformation from one state to another. However, if a system exhibits thesame values of its properties at two different times, the state remains the

same at these times. A system is said to be at a steady state if none of its

properties changes with time. A process occurs when a systems state (as

measured by its properties) changes for any reason. Processes may be

reversible or actual (irreversible). In this context the word reversible has a

special meaning. A reversible process is one that is wholly theoretical, but

can be imagined as one which occurs without incurring friction, turbulence,

leakage or anything which causes unrecoverable energy losses. All of the

processes considered below are reversible and the actual processes will bedealt with later.

Processes may be constrained to occur at constant temperature (isothermal),

constant pressure, constant volume, polytropic and adiabatic (with no heat

transfer to the surroundings).



33/132



NEXT INPUT!

2.1 Fill in the blanks with suitable names for the close system in the diagram

below.

2.2 Study the statements in the table below and decide if the statements are

TRUE(T)or FALSE (F).

STATEMENT TRUE or FALSE

i. The mass or region inside the system is called

the surroundings.

ii. In a close system, no mass can enter or leave

a system.

iii. Intensive properties are those which are

independent of the size of the system

iv. Mass, volume and total energy are some

examples of intensive properties.

Activity 2A

ii. _________

i. _____________

iii. _____________



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Feedback To Activity 2A

2.1 i. Surroundings

ii. System

iii. Boundary

2.2 i. False

ii. True

iii. True

iv. False

CONGRATULATIONS, IF YOUR ANSWERS ARE CORRECT YOU CAN

PROCEED TO THE NEXT INPUT..



35/132


2.3 The First Law of Thermodynamics

Figure 2.3 Pictures showing types of energy

The first law of thermodynamics is simply a statement of conservation of

energy principle and it asserts that total energy is a thermodynamic property.

Energy can neither be created nor destroyed; it can only change forms. This

principle is based on experimental observations and is known as the First

Law of Thermodynamics.The First Law of Thermodynamics can therefore be

stated as follows:

When a system undergoes a thermodynamic cycle then the

net heat supplied to the system from its surroundings is

equal to the net work done by the systems on its surroundings.

The First Law of Thermodynamics

In symbols,

dQ = dW (2.1)

where represents the sum of a complete cycle.

Energy can exist inmany forms such asthermal, kinetic,potential, electric,chemical,

INPUT



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2.4 Work and Heat Transfer

Worktransfer is defined as a product of the force and the distance moved in

the direction of the force. When a boundary of a close system moves in the

direction of the force acting on it, then the system does work on its

surroundings. When the boundary is moved inwards the work is done on the

system by its surroundings. The units of work are, for example, Nm or J. If

work is done on unit mass of a fluid, then the work done per kg of fluid has

the units of Nm/kg or J/kg. Consider the fluid expanding behind the piston

of an engine. The force F (in the absence of friction) will be given by

F = pA (2.2)

where

pis the pressure exerted on the piston and

Ais the area of the piston

If dx is the displacement of the piston and pcan be assumed constant

over this displacement, then the work done W will be given by,

W = Fx dx

= pA xdx

= p xAdx

= p xdV

= p(V2 V1) (2.3)

where dV = Adx = change in volume.

Figure 2.4 Work transfer

When two systems at different temperatures are in contact with each other,

energy will transfer between them until they reach the same temperature (thatis, when they are in equilibrium with each other). This energy is called heat,

or thermal energy, and the term "heat flow" refers to an energy transfer as a

consequence of a temperature difference.

Heat is a form of energy which crosses the boundary of a system during a

change of state produced by the difference in temperature between the system

FPRESSURE

dx



37/132


and its surroundings. The unit of heat is taken as the amount of heat energy

equivalent to one joule or Nm. The joule is defined as the work done when

the point of application of a force of one newton is displaced through a

distance of one meter in the direction of the force.

2.5 Sign Convention for Work Transfer

It is convenient to consider a convention of sign in connection with work

transfer and the usual convention adopted is:

if work energy is transferred from the system to the surroundings, it is

donated as positive.

if work energy is transferred from the surroundings to the system, it is

donated as negative.

WORK W2

+ veSYSTEM

SURROUNDINGS BOUNDARY

Figure 2.5 Sign Convention for work transfer

WORK W1-ve



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2.6 Sign Convention for Heat Transfer

The sign convention usually adopted for heat energy transfer is such that :

if heat energy flows into the system from the surroundings it is said to

be positive.

if heat energy flows from the system to the surroundings it is said to be

negative. It is incorrect to speak of heat in a system since heat energy

exists only when it flows across the boundary. Once in the system, it is

converted to other types of energy.

Figure 2.6 Sign convention for heat transfer

2.7 Internal Energy

Internal energy is the sum of all the energies a fluid possesses and stores

within itself. The molecules of a fluid may be imagined to be in motion

thereby possessing kinetic energy of translation and rotation as well as the

energy of vibration of the atoms within the molecules. In addition, the fluid

also possesses internal potential energy due to inter-molecular forces.

Suppose we have 1 kg of gas in a closed container as shown in Figure 2.7.For simplicity, we shallassume that the vessel is at rest with respect to the

earth and is located on a base horizon. The gas in the vessel has neither

macro kinetic energy nor potential energy. However, the molecules of thegas

are in motion and possess a molecular or 'internal' kinetic energy. The term is

usually shortenedto internal energy. If we are to study thermal effectsthen

HEAT ENERGYQ2

-ve

SURROUNDINGS

HEAT

ENERGYQ1

+ ve

SYSTEM

BOUNDARY



39/132


we can no longer ignore this form of energy. We shall denote the specific

(per kg) internal energy as u J/kg.

Now suppose that by rotation of an impeller within the vessel, we add work

dW to the closed system and we also introduce an amount of heat dQ. The

gas in the vessel still has zero macro kinetic energy and zero potential

energy. The energy that has been added has simply caused an increasein the

internal energy.

The change in internal energy is determined only by the net energy that has

been transferred across the boundary and is independent of the form of that

energy (work or heat) or the process path of theenergy transfer. In molecular

simulations, molecules can of course be seen, so the changes occurring as a

system gains or loses internal energy are apparent in the changes in the

motion of the molecules. It can be observed that the molecules move faster

when the internal energy is increased. Internal energy is, therefore, a

thermodynamic property of state. Equation 2.4 is sometimes known as the

non-flow energy equation and is a statement of the First Law of

Thermodynamics.

or,

Figure 2.7 Added work and heat raise the internal energy of a close system

dQ

dW

121212

d-d

WQUU

WQdU

=

=

(2.4)



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The figure above shows a certain process, which undergoes a complete cycle

of operations. Determine the value of the work output for a complete cycle,Wout.

Example 2.1


Q = Qin+ Qout

= (10) + (-3)

= 7 kJ

W = Win+ Wout

= (-2) + (Wout)

Hence Q- W= 0

W= Q

(-2) + (Wout) = 7

Wout= 9 kJ

Qin= +10 kJ Wout= (+) ?

Win= -2 kJQout= -3 kJ

SYSTEM

Qin is +10 kJ

Qout is 3 kJ

Winis 2 kJ

Woutis +ve



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A system is allowed to do work amounting to 500 kNm whilst heat energy

amounting to 800 kJ is transferred into it. Find the change of internal energy

and state whether it is an increase or decrease.

Example 2.2


U2 U1= Q12 W12

now,W12= +500 kNm = 500 kJ

Q12= +800 kJ

U2 U1= 800 500= 300 kJ

Since U2>U1, the internal energy has increased.



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NEXT INPUT!

2.3 During a complete cycle operation, a system is subjected to the following:

Heat transfer is 800 kJ supplied and 150 kJ rejected.

Work done by the system is 200 kJ.Calculate the work transferred from the surrounding to the system.

2.4 Each line in Table 2.4 gives information about a process of a closed system.

Every entry has the same energy unit i.e. kJ. Fill in the empty spaces in the

table with the correct answers.

PROCESS Q12 W12 (U2 U1)

a. +50 -20 i. ________

b. +100 ii. _______ -30

c. iii. _______ -70 +130

d. -50 +20 iv. _______

2.5 A close system undergoes a process in which there is a heat transfer of 200 kJ fromthe system to the surroundings. The work done from the system to the surroundings

is 75 kJ. Calculate the change of internal energy and state whether it is an increase

or decrease.

Activity 2B



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2.3 Q = Qin+ Qout= (800) + (-150) = 650 kJ

W = Win+ Wout= (Win) + (200)

HenceQ

-W

= 0W= Q

(Win) + (200) = 650

Win= 450 kJ

2.4 i. 70

ii. 130

iii. 60

iv. -70

2.5 U2 U1= Q12 W12

now,Q12= -200 kJW12= 75 kJ

U2 U1= (-200) (75) = -275 kJ

(Since U2-U1 = -ve, the internal energy is decreased)

CONGRATULATIONS, IF YOUR ANSWERS ARE CORRECT THEN YOU

ARE SUCCESSFUL.



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Problem Solving MethodologyThere are several correct and effective steps to problem solving. There are many

variations as to what various authors give for their problem solving strategy.

Some of these steps are:

Read the ENTIREproblem carefully and all the way through before starting work on theproblem. Make sure that you understand what is being asked.

List the data based on the figures given in the question. This will include both explicitand implicit data items. Note that not all of the explicitly given data are alwaysnecessarily involved in the problem solution. Be wary of introducing implicit conditionsthat may be unnecessary for the problem solution.

Draw a diagram of the physical situation. The type of drawing will depend upon theproblem.

Determine the physical principles involved in the particular problem. What are thepertinent equations and how can they be used to determine either the solution orintermediate results that can be further used to determine the solution. Often oneequation will be insufficient to solve a particular problem.

Simplify the equations as much as possible through algebraic manipulation beforeplugging numbers into the equations. The fewer times numbers are entered intoequations, the less likely numerical mistakes will be made.

Check the units on the quantities involved. Make sure that all of the given quantities arein a consistent set of units.

Insert the given data into the equations and perform the calculations. In doing thecalculations, also manipulate the units. In doing the calculations, follow the rules forsignificant figures.

Is the result reasonable and are the final units correct?

GOOD LUCK, TRY YOUR BEST.

!Tips



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You are approaching success. Try all the questions in this self-assessment

section and check your answers with those given in the Feedback to Self-

Assessment on the next page. If you face any problem, discuss it with your

lecturer. Good luck.

1. A thermodynamic system undergoes a process in which its internal energy decreases

by 300 kJ. If at the same time, 120 kJ of work is done on the system, find the heat

transferred to or from the system.

2. The internal energy of a system increases by 70 kJ when 180 kJ of heat is transferred

to the system. How much work is done by the gas?

3. During a certain process, 1000 kJ of heat is added to the working fluid while 750 kJ

is extracted as work. Determine the change in internal energy and state whether it is

increased of decreased.

4. If the internal energy of a system is increased by 90 kJ while the system does 125 kJ

of work to the surroundings, determine the heat transfer to or from the system.

SELF-ASSESSMENT



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1. Q= - 420 kJ

2. W= 110 kJ

3. U2 U1= 250 kJ (Since U2-U1 = +ve, the internal energy is increased)

4. Q= 215 kJ



always.

Feedback to Self-Assessment



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OBJECTIVES

General Objective : To understand the laws of thermodynamics and its constants.


define the definitions of Boyles Law, Charles Law andUniversal Gases Law

define and show the application of the specific heat capacity at

constant pressure

define and apply the specific heat capacity at constant volume

UNIT 3



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3.0 Definition Of Perfect Gases

Did you know, one important type of fluid that has many applications in

thermodynamics is the type in which the working temperature of the fluid remains

well above the critical temperature of the fluid? In this case, the fluid cannot be

liquefied by an isothermal compression, i.e. if it is required to condense the fluid,then cooling of the fluid must first be carried out. In the simple treatment of such

fluids, their behavior is likened to that a perfect gas. Although, strictly speaking, a

perfect gas is an ideal which can never be realized in practice. The behavior of many

permanent gases, e.g. hydrogen, oxygen, air etc is very similar to the behavior of a

perfect gas to a first approximation.

A perfect gas is a collection of particles that:

are in constant, random motion,

have no intermolecular attractions (which leads to elastic collisions in which

no energy is exchanged or lost),

are considered to be volume-less points.

You are more familiar with the term ideal gas. There is actually a distinction

between these two terms but for our purposes, you may consider them

interchangeable. The principle properties used to define the state of a gaseous system

are pressure (P), volume (V) and temperature (T). SI units (Systems International)

for these properties are Pascal (Pa) for pressure, m3for volume (although liters and

cm3are often substituted), and the absolute scale of temperature or Kelvin (K).

Two of the laws describing the behavior of a perfect gas are Boyles Law and

Charles Law.

INPUT



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3.1 Boyles Law

The Boyles Law may be stated as follows:

Provided the temperature Tof a perfect gas remains constant, then volume, Vof a

given mass of gas is inversely proportional to the pressure Pof the gas, i.e. P1/V(as shown in Fig. 3.1-1), or P x V= constant if temperature remains constant.

Figure 3.1-1 Graph P1/V

If a gas changes from state 1 to state 2 during an isothermal process, then

P1V1= P2V2= constant (3.1)

If the process is represented on a graph having axes of pressure Pand volume V, the

results will be as shown in Fig. 3.1-2. The curve is known as a rectangularhyperbola, having the mathematical equationxy= constant.

P

P1 1

P2 23

P3

V1 V2 V3 V

Figure 3.1-2 P-Vgraph for constant temperature

P

1/V

P1/V

PV= constant



50/132


A quantity of a certain perfect gas is heated at a constant temperature from an

initial state of 0.22 m3

and325 kN/m

2to a final state of 170 kN/m

2. Calculate

the final pressure of the gas.

Example 3.1


From equation P1V1= P2V2

3.2 Charles Law

The Charless Law may be stated as follows:

Provided the pressure Pof a given mass of gas remains constant, then the volume V

of the gas will be directly proportional to the absolute temperature Tof the gas, i.e.

VT, or V= constant x T. Therefore V/T= constant, for constant pressure P.

If gas changes from state 1 to state 2 during a constant pressure process, then

If the process is represented on a P Vdiagram as before, the result will be as shown

in Fig. 3.2.

( ) 32

23

2

112 m0.421

kN/m170

kN/m325m0.22x =

==

P

PVV

constant2

2

1

1 ==T

V

T

V(3.2)

1 2

P

V0V1 V2

Figure 3.2 P-Vgraph for constant pressure process



51/132


A quantity of gas at 0.54 m3and 345

oC undergoes a constant pressure process

that causes the volume of the gas to decreases to 0.32 m3. Calculate the

temperature of the gas at the end of the process.

Example 3.2


From the questionV1= 0.54 m

3

T1= 345 + 273 K = 618 KV2= 0.32 m

3

3.3 Universal Gases Law

Charles Law gives us the change in volume of a gas with temperature when the

pressure remains constant. Boyles Law gives us the change in volume of a gas with

pressure if the temperature remains constant.

The relation which gives the volume of a gas when both temperature and the

pressure are changed is stated as equation 3.3 below.

i.e. (3.4)

( )

K366

m0.54

m0.32K618

x

3

3

1

212

2

2

1

1

=

=

=

=

V

VTT

T

V

T

V

RT

PV== constant (3.3)

2

22

1

11

T

VP

T

VP=



52/132


No gases in practice obey this law rigidly, but many gases tend towards it. An

imaginary ideal that obeys the law is called aperfect gas, and the equation

is called the characteristic equation of state of a perfect gas.

The constant, R, is called the gas constant. The unit of R is Nm/kg K or J/kg K.

Each perfect gas has a different gas constant.

The characteristic equation is usually written

PV = RT (3.5)

or for mkg, occupying Vm3,

PV = mRT (3.6)

Another form of the characteristic equation can be derived using the kilogram-moleas a unit. The kilogram-moleis defined as a quantity of a gas equivalent to mkg of

the gas, whereMis the molecular weight of the gas (e.g. since the molecular weight

of oxygen is 32, then 1 kg mole of oxygen is equivalent to 32 kg of oxygen).

From the definition of the kilogram-mole, for mkg of a gas we have,

m = nM (3.7)

(where nis the number of moles).

Note: Since the standard of mass is the kg, kilogram-mole will be written simply asmole.

Substituting for mfrom equation 3.7 in equation 3.6,

PV = nMRT or (3.8)

RT

PV=

nT

PVMR=



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0.046 m3of gas are contained in a sealed cylinder at a pressure of 300 kN/m

2

and a temperature of 45oC. The gas is compressed until the pressure reaches

1.27 MN/m2and the temperature is 83

oC. If the gas is assumed to be a perfect

gas, determine:

d) the mass of gas (kg)e) the final volume of gas (m3)

Given:

R= 0.29 kJ/kg K

NowAvogadros hypothesisstates that the volume of 1 mole of any gas is the same

as the volume of 1 mole of any other gas, when the gases are at the same temperature

and pressure. Therefore V/nis the same for all gases at the same value of P and T.

That is the quantity PV/nT is constant for all gases. This constant is called the

universal gas constant, and is given the symbolRo.

i.e. (3.9)

or sinceMR = Ro then,

RR

M

o= (3.10)

Experiment has shown that the volume of 1 mole of any perfect gas at 1 bar and 1o

Cis approximately 22.71 m

3. Therefore from equation 3.8

From equation 3.10 the gas constant for any gas can be found when the molecular

weight is known, e.g. for oxygen of molecular weight 32, the gas constant is

KJ/kg8.25932

4.8314

=== M

R

Ro

Example 3.3

TnRPVnT

PVRMR

oo === or

KJ/mole8314.4273.15x1

22.71x10x1 5

0 ===nT

PVR



54/132



From the questionV1= 0.046 m

3

P1= 300 kN/m2

T1= 45 + 273 K = 318 KP2= 1.27 MN/m

2= 1.27 x 10

3kN/m

2

T2= 83 + 273 K = 356 KR= 0.29 kJ/kg K

From equation 3.6PV = mRT

From equation 3.4, the constant volume process i.e. V1= V2

kg0.1496318x0.29

0.046x300

1

11 ===RT

VPm

( ) ( ) K1346300

10x1.27318

3

1

212

2

2

1

1

=

=

=

=

P

PTT

T

P

T

P



55/132



NEXT INPUT!

3.1 Study the statements in the table below. Mark the answers as TRUE or

FALSE.

STATEMENT TRUE or FALSE

i. Charles Law gives us the change in volumeof a gas with temperature when the

temperature remains constant.

ii. Boyles Law gives us the change in volume of

a gas with pressure if the pressure remains

constant.

iii. The characteristic equation of state of a

perfect gas is .

iv. Rois the symbol for universal gas constant.v. The constant Ris called the gas constant.

vi. The unit ofRis Nm/kg or J/kg.

3.2 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3at a pressure

6.76 bar and a temperature of 127oC. Calculate the molecular weight of the

gas (M). When the gas is allowed to expand until the pressure is 2.12 bar the

final volume is 0.065 m3. Calculate the final temperature.

Activity 3A

RT

PV=



56/132


Feedback To Activity 3A

3.1 i. False

ii. False

iii. True

iv. True

v. Truevi. False

3.2 From the question,

m= 0.04 kg

V1= 0.072 m3 V2= 0.072 m

3

P1= 6.76 bar = 6.76 x 102kN/m

2 P2= 2.12 bar = 2.12 x 10

2kN/m

2

T1= 127 + 273 K = 400 K

From equation 3.6

P1V1= mRT1

Then from equation 3.10

RR

M

o=

i.e. Molecular weight = 27

KkJ/kg3042.0400x0.04

0.0072x10x6.76 2

1

11 ===mT

VPR

kg/kmol273042.0

3144.8===

R

RM o



57/132


From equation 3.6P2V2= mRT2

i.e. Final temperature = 1132.5 273 = 859.5oC.

CONGRATULATIONS, IF YOUR ANSWERS ARE CORRECT YOU CAN

PROCEED TO THE NEXT INPUT..

K1132.5

0.3042x0.04

0.065x10x2.12 2222 ===

mR

VPT



58/132


3.4 Specific Heat Capacity at Constant Volume (Cv)

The specific heat capacities of any substance is defined as the amount of heat energyrequired to raise the unit mass through one degree temperature raise. In

thermodynamics, two specified conditions are used, those of constant volume and

constant pressure. The two specific heat capacities do not have the same value and itis essential to distinguish them.

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise the

temperature of the gas by 1 degree whilst the volume of the gas remains constant,then the amount of heat energy supplied is known as the specific heat capacity at

constant volume, and is denoted by Cv. The unit of Cvis J/kg K or kJ/kg K.

For a reversible non-flow process at constant volume, we have

dQ= mCvdT (3.11)

For a perfect gas the values of Cvare constant for any one gas at all pressures andtemperatures. Equations (3.11) can then be expanded as follows :

Heat flow in a constant volume process, Q12= mCv(T2 T1) (3.12)

Also, from the non-flow energy equationQ W = (U2 U1)

mcv(T2 T1) 0 = (U2 U1)

(U2 U1) = mCv(T2 T1) (3.13)

i.e. dU = Q

Note:

In a reversible constant volume process, no work energy transfer can take

place since the piston will be unable to move i.e.W

= 0.

INPUT



59/132


3.4 kg of gas is heated at a constant volume of 0.92 m3and temperature 17

oC

until the temperature rose to 147oC. If the gas is assumed to be a perfect gas,

determine:

a) the heat flow during the processb) the beginning pressure of gasc) the final pressure of gas

GivenCv= 0.72 kJ/kg KR= 0.287 kJ/kg K

The reversible constant volume process is shown on a P-Vdiagram in Fig. 3.4.

Figure 3.4 P-Vdiagram for reversible constant volume process

Example 3.4


From the questionm= 3.4 kgV1= V2= 0.92 m

3

T1= 17 + 273 K = 290 KT2= 147 + 273 K = 420 KCv= 0.72 kJ/kg KR= 0.287 kJ/kg K

P2

P1 1

2

P

VV1= V2



60/132


a) From equation 3.13,

Q12= mCv(T2 T1)

= 3.4 x 0.72(420 290)

= 318.24 kJ

b) From equation 3.6,

PV = mRT

Hence for state 1,

P1V1= mRT1

2

3

1

11 kN/m6.307

m92.0

K290kJ/kgK x287.0xkg4.3===

V

mRTP

c) For state 2,

P2V2= mRT2

2

3

2

22 kN/m5.445

m92.0

K042kJ/kgK x287.0xkg4.3===

V

mRTP

3.5 Specific Heat Capacity at Constant Pressure (Cp)

If 1 kg of a gas is supplied with an amount of heat energy sufficient to raise thetemperature of the gas by 1 degree whilst the pressure of the gas remains constant,

then the amount of heat energy supplied is known as the specific heat capacity at

constant pressure, and is denoted by Cp. The unit of Cp is J/kg K or kJ/kg K.

For a reversible non-flow process at constant pressure, we have

dQ= mCpdT (3.14)

For a perfect gas the values of Cpare constant for any one gas at all pressures andtemperatures. Equation (3.14) can then be expanded as follows:

Heat flow in a reversible constant pressure process Q= mCp(T2 T1) (3.15)



61/132


3.6 Relationship Between The Specific Heats

Let a perfect gas be heated at constant pressure from T1to T2. With reference to the

non-flow equation Q= U2 U1+ W, and the equation for a perfect gas

U2 U1= mCv(T2 T1), hence,

Q= mCv(T2 T1) + W

In a constant pressure process, the work done by the fluid is given by the pressure

times the change in volume, i.e. W = P(V2 V1). Thenusing equationPV = mRT,

we have

W = mR(T2 T1)

Therefore substituting,

Q= mCv(T2 T1) + mR(T2 T1) = m(Cv+ R)(T2 T1)

But for a constant pressure process from equation 3.15,

Q= mCp(T2 T1)

Hence, by equating the two expressions for the heat flow Q, we havemCp(T2 T1) = m(Cv+ R)(T2 T1)

Cp= Cv+ RAlternatively, it is usually written as

R = Cp- Cv 3.16

3.7 Specific Heat Ratio ()

The ratio of the specific heat at constant pressure to the specific heat at constant

volume is given the symbol (gamma),

i.e. =v

p

C

C (3.17)

Note that since Cp - Cv=R, from equation 3.16, it is clear that Cp must be greater

than Cv for any perfect gas. It follows therefore that the ratio Cp/Cv= , is always

greater than unity. In general, is about 1.4 for diatomic gases such as carbonmonoxide (CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic



62/132


gases such as argon (A), and helium (He), is about 1.6, and for triatomic gases such

as carbon dioxide (CO2), and sulphur dioxide (SO2), is about 1.3. For some hydro-

carbons the value of is quite low (e.g. for ethane (C2H6), = 1.22, and for iso-

butane (C4H10), = 1.11.

Some useful relationships between Cp, Cv, R, and can be derived.From equation 3.17

Cp- Cv=R

Dividing through byCv

vv

p

C

R

C

C=1

Therefore using equation 3.17, =v

p

C

C, then,

vC

R=1

)1( =

RCv 3.18

Also from equation 3.17, Cp= Cvhence substituting in equation 3.18,

Cp= Cv =)1(

R

Cp=)1(

R 3.19



63/132


Example 3.5


From equation 3.16

R = Cp- Cv

i.e. R = 0.846 0.657 = 0.189 kJ/kg K

or R = 189 Nm/kg K

From equation 3.10

M=R

R0

i.e. M= 441898314 =

A certain perfect gas has specific heat as follows

Cp= 0.846 kJ/kg K and Cv= 0.657 kJ/kg K

Find the gas constant and the molecular weight of the gas.



64/132



NEXT INPUT!

3.3 Two kilograms of a gas receive 200 kJ as heat at constant volume process. If

the temperature of the gas increases by 100oC, determine the Cv of the

process.

3.4 A perfect gas is contained in a rigid vessel at 3 bar and 315oC. The gas is

then cooled until the pressure falls to 1.5 bar. Calculate the heat rejected per

kg of gas.

Given:

M= 26 kg/kmol and = 1.26.

3.5 A mass of 0.18 kg gas is at a temperature of 15oC and pressure 130 kN/m

2.

If the gas has a value of Cv= 720 J/kg K, calculate the:

i. gas constant

ii. molecular weight

iii. specific heat at constant pressure

iv. specific heat ratio

Activity 3B



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m= 2 kg

Q= 200 kJ

(T2 T1) = 100 oC = 373 K

Q= mCv(T2 T1)

( )kJ/kgK268.0

)373(2

200

12

==

=TTm

QCv


P1= 3 bar

T1= 315 oC = 588 KP2= 1.5 bar

M = 26kg/kmol

= 1.26

From equation 3.10,

KJ/kg8.31926

8314===

M

RR o

From equation 3.18,

KkJ/kg1.230KJ/kg1230126.1

8.319

)1(==

=

=

RCv

During the process, the volume remains constant (i.e. rigid vessel) for the

mass of gas present, and from equation 3.4,



66/132


Therefore since V1= V2,

K2943

5.1x588

1

212 ===

P

PTT

Then from equation 3.12,

Heat rejected per kg gas, Q= Cv(T2 T1)

= 1.230(588 294)

= 361.6 kJ/kg

3.5 From the question

m= 0.18 kg

T= 15 oC = 288 K

V= 0.17 m3

Cv= 720 J/kg K = 0.720 kJ/kg K

i. From equation 3.6,

PV = mRT

kJ/kgK426.0288x18.0

17.0x130===

mT

PVR

ii. From equation 3.10,

kg/kmol52.19426.0

3144.8===

=

R

RM

M

RR

o

o

iii. From equation 3.16,

R = Cp- Cv

Cp= R + Cv= 0.426 + 0.720 = 1.146 kJ/kg K

iv. From equation 3.17,

59.1720.0

146.1===

v

p

C

C

2

22

1

11

T

VP

T

VP=



67/132


You are approaching success. Try all the questions in this self-assessment

section and check your answers with those given in the Feedback to Self-

Assessment on the next page. If you face any problem, discuss it with your lecturer.

Good luck.

1. 1 m3of air at 8 bar and 120 oC is cooled at constant pressure process until the

temperature drops to 27o

C.GivenR= 0.287 kJ/kg K and Cp= 1.005 kJ/kg K, calculate the:

i. mass of air

ii. heat rejected in the process

iii. volume of the air after cooling.

2. A system undergoes a process in which 42 kJ of heat is rejected. If the

pressure is kept constant at 125 kN/m2while the volume changes from

0.20 m

3

to 0.006 m

3

, determine the work done and the change in internalenergy.

3. Heat is supplied to a gas in a rigid container.The mass of the container is 1 kg

and the volume of gas is 0.6 m3. 100 kJ is added as heat. If gas has

Cv= 0.7186 kJ/kg K during a process, determine the:

i. change in temperature

ii. change in internal energy

SELF-ASSESSMENT



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1. i. m= 7.093 kg

ii. Q= 663 kJ

iii. V2= 0.763 m3

2. W= -24.25 kJ

(U2 U1) = -17.75 kJ

3. i. (T2 T1) = 139.2 K

ii. (U2 U1) = 100 kJ



always.

Feedback To Self-Assessment



69/132

NON-FLOW PROCESS J2006/4/1

NON - FLOW PROCESS

OBJECTIVES

General Objective : To understand and apply the concept of non-flow process in

thermodynamics


define and describe the differences between the flow and the

non-flow processes

identify heat and work in reversible process

define and calculate the following non-flow processes :

constant temperature (Isothermal) adiabatic

UNIT 4



70/132


4.0 INTRODUCTION

nce a fluid has entered a system, it may be possible for it to undergo a series

of processes in which the fluid does not flow. An example of this is thecylinder of an internal combustion engine. In the suction stroke, the working

fluid flows into the cylinder in which it is then temporarily sealed. Whilst the

cylinder is sealed, the fluid is compressed by the piston moving into the cylinder,

after which heat energy is supplied so that the fluid possesses sufficient energy toforce the piston back down the cylinder, causing the engine to do external work. The

exhaust valve is then opened and the fluid is made to flow out of the cylinder into the

surroundings. Processes which are undergone by a system when the working fluidcannot cross the boundary are called non-flow process. This process occurs during

the compression and the working stroke as mentioned in the above example (refer to

Fig. 4.0).

O

What is a non

flow process?

INPUT



71/132


4.1 Differences Between The Flow and Non-flow processes

4.1.1 Flow Process

In an open system, not only the energy transfers take place across the

boundary, the fluid may also cross the boundary. Any process undergone by

an open system is called a flow process. This process may be sub-divided

into an unsteady flow process and steady flow process. The general equation

is shown below,

WC

vPugZQC

vPugZ ++++=++++22

2

22222

2

11111

4.1.2 Non-flow process

In a close system, although energy may be transferred across the boundary in

the form of work energy and heat energy, the working fluid itself nevercrosses the boundary. Any process undergone by a close system is referred to

as the non-flow process.

If the fluid is undergoing a non-flow process from state (1) to state (2) then

the terms from the general equation for p1V1 and p2V2 (which represent the

amount of work energy required to introduce and expel the fluid from thesystem) will be zero, since the fluid is already in the system, and will still be

in the system at the end of the process. For the same reason, the changes in

SUCTION

STROKECOMPRESSION

STROKE

WORKING

STROKE EXHAUST

STROKE

Figure 4.0The cycle of an internal combustion engine



72/132


kinetic and potential energies of the fluid will also be zero. Thus the equation

becomes

U1+ Q = U2+ W

or, U2 U1 = Q W (4.1)

In words, this equation states that in a non-flow process, the change in the

internal energy of the fluid is equal to the nett amount of heat energy

supplied to the fluid minus the nett amount of work energy flowing from thefluid.

This equation is known as the non flow energy equation, and it will now be

shown how this may apply to the various non-flow processes.

4.2 Constant temperature (Isothermal) process (pV = C)

If the change in temperature during a process is very small then that process may be

approximated as an isothermal process. For example, the slow expansion orcompression of fluid in a cylinder, which is perfectly cooled by water may be

analysed, assuming that the temperature remains constant.

Figure 4.2 Constant temperature (Isothermal) process

W

Q

P

vv1

v2

W

1

2



73/132


The general relation properties between the initial and final states of a perfect gas are

applied as follows:

2

22

1

11

T

Vp

T

Vp=

If the temperature remains constant during the process, T1 = T2 and the aboverelation becomes

2211 VpVp =

From the equation we can know that an increase in the volume results in a decrease

in the pressure. In other words, in an isothermal process, the pressure is inversely

proportional to the volume.

Work transfer:

Referring to the process represented on the p Vdiagram in Fig.4.2 it is noted thatthe volume increases during the process. In other words the fluid is expanding. The

expansion work is given by

=2

1

pdVW

= 2

1

dVV

c (sincepV = C, a constant)

= 2

1V

dVc

=

2

1

11V

dV

Vp

=1

211 ln

V

VVp

lumesmaller vo

umelarger vol

=1

21 ln

V

VmRT (sincep1V1= mRT1)

=2

11 ln

p

pmRT (since

2

1

1

2

p

p

V

V= ) (4.2)

Note that during expansion, the volume increases and the pressure decreases. On thep Vdiagram, the shaded area under the process line represents the amount of work

transfer.

Since this is an expansion process (i.e. increasing volume), the work is done by the

system. In other words the system produces work output and this is shown by the

direction of the arrow representing W.



74/132


Heat transfer:

Energy balance to this case is applied:

U1+ Q = U2+ W

For a perfect gas

U1= mcvT1and U2= mcvT2

As the temperature is constant

U1= U2

Substituting in the energy balance equation,

Q = W (4.3)

Thus, for a perfect gas, all the heat added during a constant temperature process isconverted into work and the internal energy of the system remains constant.

For a constant temperature process

W=1

21 ln

V

VmRT = Q

or

W=2

11 ln

ppmRT = Q



75/132


4.3 Adiabatic process (Q= 0)

If a system is thermally well insulated then there will be negligible heat

transfer into or out of the system. Such a system is thermally isolated and a

process within that system may be idealised as an adiabatic process. Forexample, the outer casing of steam engine, steam turbines and gas turbinesare well insulated to minimise heat loss. The fluid expansion process in such

machines may be assumed to be adiabatic.

Figure 4.3 Adiabatic (zero heat transfer) process

For a perfect gas the equation for an adiabatic process is

pV= C

where = ratio of specific heat =v

p

C

C

The above equation is applied to states 1 and 2 as:

2211 VpVp =

=

2

1

1

2

V

V

p

p (4.4)

W

P

vv1

v2

W

1

2

Thermal insulation



76/132


Also, for a perfect gas, the general property relation between the two states is given

by the equation below

2

22

1

11

T

Vp

T

Vp= (4.5)

By manipulating equations 4.4 and 4.5 the following relationship can be determined:

1

2

1

1

1

2

1

2

=

=

V

V

p

p

T

T (4.6)

By examining equations 4.4 and 4.6 the following conclusion for an adiabaticprocess on a perfect gas can be drawn:

An increase in volume results in a decrease in pressure.An increase in volume results in a decrease in temperature.An increase in pressure results in an increase in temperature.

Work transfer:

Referring to the process represented on the p-V diagram (Fig.4.3) it is noted that thevolume increases during the process.

In other words, the fluid expanding and the expansion work is given by the formula:

=2

1

pdVW

= 2

1

dVV

c

(sincepV= C, a constant)

= 2

1

V

dVc

=1

2211

VpVp [largerpV- smallpV] (4.7)

Note that after expansion, p2is smaller thanp1. In thep Vdiagram, the shaded

area under the process represents the amount of work transfer.

As this is an expansion process (i.e. increase in volume) the work is done by thesystem. In other words, the system produces work output and this is shown by the

direction of the arrow representing W(as shown in Fig 4.3).



77/132


Heat transfer:

In an adiabatic process, Q = 0.

Applying an energy balance to this case (Fig.4.3):

U1- W = U2

W = U1 U2

Thus, in an adiabatic expansion the work output is equal to the decrease in internalenergy. In other words, because of the work output the internal energy of the system

decreases by a corresponding amount.

For a perfect gas, U1= mcvT1and U1= mcvT1

On substitution

W = mcv(T1-T2) [larger T- smaller T] (4.8)

We know

cp- cv= R

or

cv=1

R

Substituting in equation 4.8

1

( )21

=

TTmRW (4.9)

But, mRT2= p2V2and mRT1= p1V1

Then the expression for the expansion becomes

1

2211

=

VpVpW (4.10)

Referring to the process represented on the p-V diagram it is noted that during this

process the volume increases and the pressure decreases. For a perfect gas, equation4.6 tells that a decrease in pressure will result in a temperature drop.



78/132


In an industrial process, 0.4 kg of oxygen is compressed isothermally from1.01 bar and 22

oC to 5.5 bar. Determine the work done and the heat transfer

during the process. Assume that oxygen is a perfect gas and take the molecularweight of oxygen to be M = 32 kg/kmole.

Example 4.1


Data: m = 0.4 kg; p1= 1.01 bar; t1= 22oC

p2= 5.5 bar; W= ? Q= ?

From the equation

R=M

R0

=32

8314

= 260 J/kgK

= 0.260 kJ/kgK

For an isothermal process

Work input,

W = mRTln1

2

p

p

=01.1

5.5ln)27322(x260.0x4.0 +

= 52 kJ

In an isothermal process all the work input is rejected as heat.

Therefore, heat rejected, Q = W= 52 kJ



79/132


In a thermally insulated reciprocating air compressor, air at 0.98 bar and 20oC

is compressed into one sixth of its original volume. Determine the pressure and

temperature of the air after compression. If the compressor cylinder contains0.05 kg of air, calculate the required work input. For air, take = 1.4 andcv = 0.718 kJ/kgK.

Example 4.2


Data : p1= 0.98 bar; T1= 20 + 273 = 293 K

;6

1

1

2 =V

V m= 0.05 kg; W= ?

As the cylinder is well insulated the heat transfer is negligible and the process may

be treated as adiabatic.

Considering air as a perfect gas

From equation 4.4,

=

2

1

1

2

V

V

p

p

p2= 0.98 x 61.4

= 12 bar

From equation 4.6,

1

2

1

1

2

=

V

V

T

T

T2= 293 x 60.4

= 600 K

= 327oC

Re-writing equation 4.8 for an adiabatic compression process

W = mcv(T2-T1) [larger T- smaller T]= 0.05 x 0.718 (600-293)

= 11 kJ



80/132



NEXT INPUT!

4.1 In the cylinder of a large engine, 1.0 kg of carbon dioxide at 527o C and

20 bar expands isothermally to a pressure of 1.4 bar. What is the final volume

of the gas?

Take R = 189 Nm/kgK for carbon dioxide.

4.2 1 kg of nitrogen (molecular weight 28) is compressed reversibly and

isothermally from 1.01 bar, 20oC to 4.2 bar. Calculate the work done and the

heat flow during the process. Assume nitrogen to be a perfect gas.

4.3 Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015 m

3, is

compressed reversibly and adiabatically by a piston to a pressure of 6.8 bar.

Calculate the final temperature, the final volume, and the work done on the

mass of air in the cylinder.

Activity 4



81/132


Feedback to Activity 4

4.1 Data: m = 1.0 kg; T1= 527 + 273 = 800 K

p1= 20 bar; p2= 1.4 bar; V2= ?

Carbon dioxide is a perfect gas and we can apply the following characteristic

gas equation at state 1.

p1V1= mRT1

V1=1

1

p

mRT

=5

1020

8001891

x

xx

= 0.0756 m3

Applying the general property relation between state 1 and 2

2

22

1

11

T

Vp

T

Vp=

For an isothermal process T1= T2

Hence,

2211 VpVp =

V2= 0756.0x4.1

20

V2= 1.08 m3



82/132


4.2 Data: m=1kg; M= 28 kg/kmole p1= 1.01 bar;

T1= 20 + 272 = 293 K; p2= 4.2 bar

From equation

R=M

R0

=28

314.8

= 0.297 kJ/kgK

The process is shown on a p-v diagram below. When a process takes place

from right to left on a p-v diagram the work done by the fluid is negative.

That is, work is done on the fluid.

From equation 4.2

W=2

11 ln

p

pmRT

= 1 x 0.297x293x ln2.4

01.1

= -124 kJ/kg

For an isothermal process for a perfect gas,

Q = W= -124 kJ/kg

p

v

4.2

1.01

pV=C



83/132


4.3 Data: p1=1.02 bar; T1=22 + 273 = 295 K;

v1= 0.015 m3; p2= 6.8 bar

From equation 4.6

1

1

2

1

2

=

p

p

T

T

T2 = 295 x

4.1/)14.1(

02.1

8.6

= 507.5 K

(where for air = 1.4)i.e. Final temperature = 507.5 273 = 234.5

oC

From equation 4.4

=

2

1

1

2

V

V

p

p or

/1

1

2

2

1

=

p

p

v

v

4.1/1

2 02.1

8

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