J2006_Termodinamik

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


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, which

are essential for a proper understanding of the subject. We are familiar with most of

these 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 a

quantity 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


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 1.1-1 Fundamental units

Quantity Unit Symbol

Mass kilogram kg

Time 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

3 litre

Velocity meter per second m/s

Acceleration Meter per second

squared

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 = 105 N/m

2 = 10

2 kN/m

2


1.1.1 Force

Newton‟s 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 meter

or 1 J = 1 Nm

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

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


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/m2 and 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 = 105 N/m

2 = 10

5 Pa

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


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/m3 of oil is filled to a tank. Determine the amount of

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

Example 1.1

Solution to Example 1.1

2N/m 05.49

0.01

9.81 x 50

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/m3 and V = 2 m

3

It is obvious that we can eliminate m3 and 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


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


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/s2 or 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 = 3 3

4r

= 3)9.1(3

4

= 28.73 m3

Mass of helium in balloon, m = v

V

= 28.73/5.6

= 5.13 kg


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.


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 can‟t 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 doesn‟t 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 @ 105

cm) = (1 000 000 mm @ 106 mm)

1 hour = 60 minutes = 3600 seconds

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

-3 m

3

1 bar = 1 x 105 N/m

2 = 1 x 10

2 kN/m

2

INPUT


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/s 278.0

s 3600

m 1000

s 3600

j 1 x

km 1

m 1000 x

j

km 1

j

km 1


Example 1.4

Convert 25 g/mm3 to kg/m

3.


1 kg = 1000 g

1 m = 1000 mm

1 m3 = 1000 x 1000 x 1000 mm

3

= 109 m

3

36

3

9

3

39

33

kg/m 10 x 25

m 1000

kg 1x 10 x 25

g 1000

kg 1 x

m 1

mm 10 x

mm

g 25

mm

g 25

How could I convert

g/mm3 to kg/m3?



NEXT INPUT…!

1.5 Convert the following data:

a) 3 N/cm2 to kN/m

2

b) 15 MN/m2 to N/m

2

1.6 Convert 15 milligram per litre to kg/m3.

I hope you’ve learnt something from this unit. Let’s move on to the next topic.

Activity 1B


Feedback To Activity 1B

1.5 a) 1 kN = 1000 N

1 m2 = 100 x 100 = 10

4 cm

2

b) 1 MN = 106

N/m2

1.6 1 kg = 1 000 000 mg

1 m3 = 1000 litre

2

2

4

2

24

22

kN/m 30

m 1000

kN10 x 3

N 1000

kN 1x

m 1

cm10x

cm

N 3

cm

N 3

26

6

22

N/m 10 x 15

MN 1

N10x

m

MN 15

m

MN 15

33-

3

kg/m 10 x 15

m 1

litre 1000 x

mg 000 000 1

kg 1 x

litre

mg 15

litre

mg 15


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 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 the

atmospheric pressure is 95 kPa, determine the pressure inside the cylinder.

2. A force of 8 N is applied continuously at an angle of 30o to 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/m

3. Find also the specific volume of water.

SELF-ASSESSMENT


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



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


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 boundary is 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


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

S

I

S

T

E

M

SYSTEM

Fluid Inlet

Fluid Outlet

SYSTEM

SURROUNDINGS

BOUNDARY


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 intensive or extensive.

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

system such as temperature, pressure and density.

Extensive properties 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 the

same 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 system’s 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 be

dealt 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).



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

S

I

S

T

E

M

ii. _________

i. _____________

iii. _____________


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…..


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 in many forms such as thermal, kinetic, potential, electric,

chemical,…

INPUT


2.4 Work and Heat Transfer

Work transfer 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

p is the pressure exerted on the piston and

A is the area of the piston

If dx is the displacement of the piston and p can be assumed constant

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

W = F x dx

= pA x dx

= p x Adx

= p x dV

= 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 (that

is, 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

F S

I

S

T

E

M

PRESSURE

dx


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

+ ve SYSTEM

SURROUNDINGS BOUNDARY

Figure 2.5 Sign Convention for work transfer

WORK W1

- ve


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 shall assume 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 the gas

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

usually shortened to internal energy. If we are to study thermal effects then

HEAT ENERGY

Q2

-ve

SURROUNDINGS

HEAT

ENERGY

Q1

+ ve

SYSTEM

BOUNDARY


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 increase in 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 the energy 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)


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 kJ Qout = -3 kJ

SYSTEM

Qin is +10 kJ

Qout is –3 kJ

Win is –2 kJ

Wout is +ve


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.



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 from

the 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



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

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

Hence Q - W = 0

W = 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 kJ

W12 = 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.


Problem Solving Methodology There 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 ENTIRE problem carefully and all the way through before starting work on the problem. Make sure that you understand what is being asked.

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

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

Determine the physical principles involved in the particular problem. What are the pertinent equations and how can they be used to determine either the solution or intermediate results that can be further used to determine the solution. Often one equation will be insufficient to solve a particular problem.

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

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

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

Is the result reasonable and are the final units correct?

GOOD LUCK, TRY YOUR BEST.

! Tips


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



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….




OBJECTIVES

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


define the definitions of Boyle’s Law, Charles’ Law and

Universal 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


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, m3 for volume (although liters and

cm3 are often substituted), and the absolute scale of temperature or Kelvin (K).

Two of the laws describing the behavior of a perfect gas are Boyle’s Law and

Charles’ Law.

INPUT


3.1 Boyle’s Law

The Boyle’s Law may be stated as follows:

Provided the temperature T of a perfect gas remains constant, then volume, V of a

given mass of gas is inversely proportional to the pressure P of the gas, i.e. P 1/V

(as shown in Fig. 3.1-1), or P x V = constant if temperature remains constant.

Figure 3.1-1 Graph P 1/V

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

P1 V1 = P2 V2 = constant (3.1)

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

results will be as shown in Fig. 3.1-2. The curve is known as a rectangular

hyperbola, having the mathematical equation xy = constant.

P

P1 1

P2 2

3

P3

V1 V2 V3 V

Figure 3.1-2 P-V graph for constant temperature

P

1/V

P 1/V

PV = constant


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

initial state of 0.22 m3

and 325 kN/m

2 to 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 Charles’s Law may be stated as follows:

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

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

V T, 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 – V diagram as before, the result will be as shown

in Fig. 3.2.

3

2

23

2

112 m 0.421

kN/m 170

kN/m 325m 0.22 x

P

PVV

constant2

2

1

1 T

V

T

V(3.2)

1 2

P

V 0 V1 V2

Figure 3.2 P-V graph for constant pressure process


A quantity of gas at 0.54 m3 and 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 question

V1 = 0.54 m3

T1 = 345 + 273 K = 618 K

V2 = 0.32 m3

3.3 Universal Gases Law

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

pressure remains constant. Boyle’s 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)

K 366

m 0.54

m 0.32 K 618

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


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

imaginary ideal that obeys the law is called a perfect 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 m kg, occupying V m3,

PV = mRT (3.6)

Another form of the characteristic equation can be derived using the kilogram-mole

as a unit. The kilogram-mole is defined as a quantity of a gas equivalent to m kg of

the gas, where M is 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 m kg of a gas we have,

m = nM (3.7)

(where n is the number of moles).

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

mole.

Substituting for m from equation 3.7 in equation 3.6,

PV = nMRT or (3.8)

RT

PV

nT

PVMR


0.046 m3 of gas are contained in a sealed cylinder at a pressure of 300 kN/m

2

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

1.27 MN/m2 and 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

Now Avogadro’s hypothesis states 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/n is 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 symbol Ro.

i.e. (3.9)

or since MR = Ro then,

RR

M

o (3.10)

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

is approximately 22.71 m3. 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

K J/kg 8.25932

4.8314

M

RR o

Example 3.3

TnRPVnT

PVRMR oo or

K J/mole 8314.4273.15 x 1

22.71x 10 x 1 5

0 nT

PVR



From the question

V1 = 0.046 m3

P1 = 300 kN/m2

T1 = 45 + 273 K = 318 K

P2 = 1.27 MN/m2 = 1.27 x 10

3 kN/m

2

T2 = 83 + 273 K = 356 K

R = 0.29 kJ/kg K

From equation 3.6

PV = mRT

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

kg 0.1496318 x 0.29

0.046 x 300

1

11 RT

VPm

K 1346300

10 x 1.27318

3

1

212

2

2

1

1

P

PTT

T

P

T

P



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 volume

of a gas with temperature when the

temperature remains constant.

ii. Boyle’s 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. Ro is the symbol for universal gas constant.

v. The constant R is called the gas constant.

vi. The unit of R is Nm/kg or J/kg.

3.2 0.04 kg of a certain perfect gas occupies a volume of 0.0072 m3 at a pressure

6.76 bar and a temperature of 127 oC. 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



3.1 i. False

ii. False

iii. True

iv. True

v. True

vi. 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 102

kN/m2 P2 = 2.12 bar = 2.12 x 10

2 kN/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

K kJ/kg 3042.0400 x 0.04

0.0072 x 10 x 6.76 2

1

11 mT

VPR

kg/kmol 273042.0

3144.8

R

RM o


From equation 3.6

P2V2 = mRT2

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


PROCEED TO THE NEXT INPUT…..

K 1132.5 0.3042 x 0.04

0.065 x 10 x 2.12 2

222

mR

VPT


3.4 Specific Heat Capacity at Constant Volume (Cv)

The specific heat capacities of any substance is defined as the amount of heat energy

required 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 it

is 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 Cv is 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 Cv are constant for any one gas at all pressures and

temperatures. 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 equation

Q – 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


3.4 kg of gas is heated at a constant volume of 0.92 m3 and temperature 17

oC

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

determine:

a) the heat flow during the process

b) the beginning pressure of gas

c) the final pressure of gas

Given

Cv = 0.72 kJ/kg K

R = 0.287 kJ/kg K

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

Figure 3.4 P-V diagram for reversible constant volume process

Example 3.4


From the question

m = 3.4 kg

V1 = V2 = 0.92 m3

T1 = 17 + 273 K = 290 K

T2 = 147 + 273 K = 420 K

Cv = 0.72 kJ/kg K

R = 0.287 kJ/kg K

P2

P1 1

2

P

V V1 = V2


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/m 6.307

m 92.0

K 290kJ/kgK x 287.0 x kg 4.3

V

mRTP

c) For state 2,

P2V2 = mRT2

2

3

2

22 kN/m 5.445

m 92.0

K 042kJ/kgK x 287.0 x kg 4.3

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 the

temperature 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 Cp are constant for any one gas at all pressures and

temperatures. Equation (3.14) can then be expanded as follows:

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


3.6 Relationship Between The Specific Heats

Let a perfect gas be heated at constant pressure from T1 to 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). Then using equation PV = 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 have

mCp(T2 – T1) = m(Cv + R)(T2 – T1)

Cp = Cv + R

Alternatively, 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 carbon

monoxide (CO), hydrogen (H2), nitrogen (N2), and oxygen (O2). For monatomic


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 by Cv

vv

p

C

R

C

C1

Therefore using equation 3.17, = v

p

C

C, then,

vC

R1

)1(

RCv 3.18

Also from equation 3.17, Cp = Cv hence substituting in equation 3.18,

Cp = Cv = )1(

R

Cp = )1(

R 3.19


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 = 44189

8314

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.



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 100 oC, determine the Cv of the

process.

3.4 A perfect gas is contained in a rigid vessel at 3 bar and 315 oC. 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 15 oC 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




m = 2 kg

Q = 200 kJ

(T2 – T1) = 100 oC = 373 K

Q = mCv(T2 – T1)

kJ/kgK 268.0)373(2

200

12

TTm

QCv


P1 = 3 bar

T1 = 315 oC = 588 K

P2 = 1.5 bar

M = 26 kg/kmol

= 1.26

From equation 3.10,

K J/kg 8.31926

8314

M

RR o

From equation 3.18,

K kJ/kg 1.230K J/kg 1230126.1

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,


Therefore since V1 = V2,

K 2943

5.1 x588

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/kgK 426.0288 x 18.0

17.0 x 130

mT

PVR

ii. From equation 3.10,

kg/kmol 52.19426.0

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




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

Good luck.

1. 1 m3 of air at 8 bar and 120

oC is cooled at constant pressure process until the

temperature drops to 27 oC.

Given R = 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/m2 while the volume changes from

0.20 m3 to 0.006 m

3, determine the work done and the change in internal

energy.

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



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

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


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 the

cylinder 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 to

force 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 fluid

cannot 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


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 never

crosses 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 the

system) 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

STROKE COMPRESSION

STROKE

WORKING

STROKE EXHAUST

STROKE

Figure 4.0 The cycle of an internal combustion engine


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 the

fluid.

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 or

compression 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

v v1

v2

W

1

2


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 above

relation 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 – V diagram in Fig.4.2 it is noted that

the 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 (since pV = C, a constant)

= 2

1V

dVc

= 2

1

11V

dVVp

= 1

211 ln

V

VVp

lumesmaller vo

umelarger vol

= 1

21 ln

V

VmRT (since p1V1 = 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 the

p – V diagram, 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.


Heat transfer:

Energy balance to this case is applied:

U1 + Q = U2 + W

For a perfect gas

U1 = mcvT1 and 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 is

converted 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

p

pmRT = Q


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. For

example, the outer casing of steam engine, steam turbines and gas turbines

are 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

v v1

v2

W

1

2

Thermal insulation


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 adiabatic

process 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 the

volume 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

(since pV = C, a constant)

= 2

1

V

dVc

= 1

2211

VpVp [larger pV- small pV] (4.7)

Note that after expansion, p2 is smaller than p1. In the p – V diagram, 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 the

system. 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).


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 internal

energy. In other words, because of the work output the internal energy of the system

decreases by a corresponding amount.

For a perfect gas, U1 = mcvT1 and 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 = p2V2 and 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, equation

4.6 tells that a decrease in pressure will result in a temperature drop.


In an industrial process, 0.4 kg of oxygen is compressed isothermally from

1.01 bar and 22o C 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 molecular

weight 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( x 260.0 x 4.0

= 52 kJ

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

Therefore, heat rejected, Q = W = 52 kJ


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 contains

0.05 kg of air, calculate the required work input. For air, take = 1.4 and

cv = 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



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


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

= 51020

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


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


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 = 234.5oC

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

8.6015.0

v

i.e. Final volume

v2 = 0.0038 m3

For an adiabatic process,

W = u1 – u2

and for a perfect gas,

W = cv(T1- T2)

= 0.718(295-507.5)

= - 152.8 kJ/kg

i.e. work input per kg = 152.8 kJ

The mass of air can be found using equation pV = mRT

kgx

xx

RT

vpm 018.0

19510287.0

015.01002.13

5

1

11

i.e. Total work done = 0.0181 x 152.8 = 2.76 kJ





1. 0.05 m3 of a perfect gas at 6.3 bar undergoes a reversible isothermal process

to a pressure of 1.05 bar. Calculate the heat flow to or from the gas.

2. Nitrogen (molecular weight 28) expands reversibly in a perfectly thermally

insulated cylinder from 2.5 bar, 200oC to a volume of 0.09 m

3. If the initial

volume occupied was 0.03 m3, calculate the work done during the expansion.

Assume nitrogen to be a perfect gas and take cv = 0.741 kJ/kg K.

3. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is

compressed adiabatically to 5 bar. Determine the following:

a) final temperature

b) final volume

c) work transfer

d) heat transfer

e) change in internal energy

4. A quantity of gas occupies a volume of 0.3 m3 at a pressure of 100 kN/m

2

and a temperature of 20oC. The gas is compressed isothermally to a pressure

of 500 kN/m2 and then expanded adiabatically to its initial volume.

For this quantity of gas determine the following:

a) the heat received or rejected (state which) during the compression,

b) the change of internal energy during the expansion,

c) the mass of gas.

SELF-ASSESSMENT



1. 56.4 kJ

2. 9.31 kJ

3. 222.7oC, 14230 cm

3, 6.56 kJ input, 0 kJ, 6.56 kJ increase.

4. – 48.3 kJ (heat rejected), -35.5 kJ, 0.358 kg


CONGRATULATIONS!!!!

You can now move on to

Unit 5…


NON-FLOW PROCESS

OBJECTIVES

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

thermodynamics


define and calculate the following non-flow processes:

polytropic

constant volume

constant pressure

UNIT 5


5.0 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 never crosses the

boundary. Any process undergone by a close system is referred to as non-flow

process.

The equation for non-flow process is given as follows:

U1 + Q = U2 + W

or, U2 – U1 = Q –W

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 the fluid.

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.

Processes, which are

undergone by a system

when the working fluid

cannot cross the

boundary, are called

non-flow process.

INPUT


5.1 Polytropic process (pVn

= C)

This is the most general type of process, in which both heat energy and work

energy cross the boundary of the system. It is represented by an equation in the

form

pVn

= constant (5.1)

If a compression or expansion is performed slowly, and if the piston cylinder

assembly is cooled perfectly, then the process will be isothermal. In this case the

index n = 1.

If a compression or expansion is performed rapidly, and if the piston cylinder

assembly is perfectly insulated, then the process will be adiabatic. In this case the

index n = .

If a compression or expansion is performed at moderate speed, and if the piston

cylinder assembly is cooled to some degree, then the process is somewhere

between those discussed above. Generally, this is the situation in many

engineering applications. In this case the index n should take some value, which is

between 1 and depending on the degree of cooling.

Some practical examples include:

compression in a stationary air compressor (n = 1.3)

compression in an air compressor cooled by a fan (n = 1.2)

compression in a water cooled air compressor (n = 1.1)

Figure 5.1 Polytropic process

W

Qloss

P

v v1 v2

W

1

2

P1

P2

pVn=C


Equation 5.1 is applied at states 1 and 2 as:

nn VpVp 2211

or

n

V

V

p

p

2

1

1

2 (5.2)

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

given by

2

22

1

11

T

Vp

T

Vp (5.3)

By the manipulation of equations 5.2 and 5.3 the following relationship can be

determined:

1

2

1

1

1

2

1

2

nn

n

V

V

p

p

T

T (5.4)

By examining equations 5.2 and 5.4 the following conclusions for a polytropic

process on a perfect gas can be drawn as:

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.5.1) it is noted that

the volume increases during the process.

In other words the fluid is expands and the expansion work is given by

2

1

pdVW

= 2

1

dVV

cn

(since pVn = C, a constant)

= 2

1

nV

dVc


= 1

2211

n

VpVp [larger pV- small pV] (5.5)

Note that after expansion p2 is smaller than p1. In the p – V diagram, the shaded

area under the process represents the amount of work transfer.

Since this is an expansion process (i.e. increase in 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 as shown in Fig. 5.1.

Heat transfer:

Energy balance is applied to this case (Fig.5.1) as:

U1 – Qloss - W = U2

Qloss = (U1 – U2) – W

or

W = (U1 – U2) - Qloss

Thus, in a polytropic expansion the work output is reduced because of the heat

loses.

Referring to the process represented on the p–V diagram (Fig.5.1) it is noted that

during this process the volume increases and the pressure decreases. For a perfect

gas, equation 5.4 tells us that a decrease in pressure will result in a temperature

drop.

For adiabatic process:

W= 1

2211

VpVp

For polytropic process:

W= 1

2211

n

VpVp


The combustion gases in a petrol engine cylinder are at 30 bar and 800oC

before expansion. The gases expand through a volume ratio (1

2

V

V) of (

1

5.8)

and occupy 510 cm3 after expansion. When the engine is air cooled the

polytropic expansion index n = 1.15. What is the temperature and pressure of

the gas after expansion, and what is the work output?

Example 5.1


State 1 State 2

Data: p1 = 30 bar; T1 = 800 + 273 = 1073 K; n = 1.15

1

2

V

V= 8.5; V2 = 510 cm

3;

t2 = ? p2 = ? W = ?

Considering air as a perfect gas, for the polytropic process, the property relation is

given by equation 5.4 as:

1

2

112

n

V

VTT

=

115.1

5.8

11073

x

= 778.4 K

= 505.4oC

P1= 30 bar

t1 = 800oC Qloss

W

V2 = 510 cm3

p2 = ?

t2 = ?


From equation 5.2

n

V

Vpp

2

112

=

15.1

5.8

1 x 30

= 2.56 bar

Now,

V2 = 510 cm3 = 510 x 10

-6 m

3

and,

1

2

V

V= 8.5

Then,

5.8

10510 6

1

x

V

= 60 x 10-6

m3

Work output during polytropic expansion is given by equation 5.5 as:

W = 1

2211

n

VpVp [larger pV- small pV]

=115.1

)10510()1056.2()1060)(1030( 6565

xxxx

= 330 J

= 0.33 kJ



NEXT INPUT…!

5.1 0.112 m3 of gas has a pressure of 138 kN/m

2. It is compressed to

690 kN/m2 according to the law pV

1.4 = C. Determine the new volume of

the gas.

5.2 0.014 m3 of gas at a pressure of 2070 kN/m

2 expands to a pressure of

207 kN/m2 according to the law pV

1.35 = C. Determine the work done by

the gas during expansion.

5.3 A cylinder containing 0.07 kg of fluid has a pressure of 1 bar, a volume of

0.06 m3 and a specific internal energy of 200 kJ/kg. After polytropic

compression, the pressure and volume of the fluid are 9 bar and 0.011 m3

respectively, and the specific internal energy is 370 kJ/kg.

Determine

a) the amount of work energy required for the compression

b) the quantity and direction of the heat energy that flows during the

compression.

Activity 5A



5.1 Since the gas is compressed according to the law pV1.4

= C, then,

4.1

22

4.1

11 VpVp

4.1

1

2

2

1

V

V

p

p or

4.1/1

2

1

1

2

p

p

V

V

from which,

4.1/1

2

112

p

pVV = 4.1

2

11

p

pV

= 0.012 x 4.1

690

138

= 0.0348 m3

5.2 The work done during a polytropic expansion is given by the expression:

W = 1

2211

n

VpVp [larger pV- small pV]

In this problem V2 is, as yet, unknown and must therefore be calculated.

Now

nn VpVp 2211

n

p

pVV

/1

2

112

or V2 = 0.014 x

35.1/1

207

2070

V2 = 0.077 m3


Work done =135.1

)077.010207014.0102070( 33

xxxx

= 37.3 x 103 Nm

= 37.3 x 103 J

= 37.3 kJ

5.3 a) For a polytropic process,

nn VpVp 2211

In the given case

1 x 0.06n = 9 x 0.011

n

9011.0

06.0

n

n = 1.302

W = 1

2211

n

VpVp

= 1302.1

)0111.0109()06.0101( 55

xxxx

= -13.2 kJ

The negative sign indicates that work energy would flow into the

system during the process.

b) The non-flow energy equation gives

Q – W = U2 – U1

Q – (- 13.2) = ( 370 x 0.07 ) – ( 200 x 0.07 )

Q = - 1.3 kJ

The negative sign indicates that heat energy will flow out of the

fluid during the process.


5.2 Constant volume process

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

approximated as a constant volume process. For example, heating or cooling a

fluid in a rigid walled vessel can be analysed by assuming that the volume

remains constant.

a) Heating b) Cooling

Figure 5.2 Constant volume process (V2=V1)

The general property relation between the initial and final states of a perfect gas is

applied as:

2

22

1

11

T

Vp

T

Vp

If the volume remain constant during the process, V2 = V1 and then the above

relation becomes

2

2

1

1

T

p

T

p

or

1

2

1

2

p

p

T

T (5.6)

From this equation it can be seen that an increase in pressure results from an

increase in temperature. In other words, in constant volume process, the

temperature is proportional to the pressure.

p

v

2

1

Q

p

v

2

1

Q

INPUT


Work transfer:

Work transfer (pdV) must be zero because the change in volume, dV, during the

process is zero. However, work in the form of paddle-wheel work may be

transferred.

Heat transfer:

Applying the non flow energy equation

Q – W = U2 – U1

gives Q – 0 = U2 – U1

i.e. Q = U2 – U1 (5.7)

This result, which is important and should be remembered, shows that the nett

amount of heat energy supplied to or taken from a fluid during a constant volume

process is equal to the change in the internal energy of the fluid.

5.3 Constant pressure process

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

approximated as a constant pressure process. For example, heating or cooling a

liquid at atmospheric pressure may be analysed by assuming that the pressure

remains constant.

Figure 5.3 Constant pressure process

W

Q

P

v v1

v2

W

1 2

v2 – v1

p


Consider the fluid in the piston cylinder as shown in Figure 5.2. If the load on the

piston is kept constant the pressure will also remain constant.

The general property relation between the initial and final states of a perfect gas is

applied as:

2

22

1

11

T

Vp

T

Vp

If the pressure remain constant during the process, p2 = p1 and then the above

relation becomes

2

2

1

1

T

V

T

V

or

1

2

1

2

V

V

T

T (5.8)

From this equation it can be seen that an increase in volume results from an

increase in temperature. In other words, in constant pressure process, the

temperature is proportional to the volume.

Work transfer:

Referring to the process representation on the p-V diagram it is noted that the

volume increases during the process. In other words, the fluid expands. This

expansion work is given by

2

1

pdVW

2

1

dVp (since p is constant)

= p (V2 – V1) (larger volume – smaller volume) (5.9)

Note that on a p-V diagram, the area under the process line represents the amount

of work transfer. From Figure 5.3

W = area of the shaded rectangle

= height x width

= p (V2 – V1) (larger volume – smaller volume)

This expression is identical to equation 5.9


Heat transfer:

Applying the non flow energy equation

Q – W = U2 – U1

or Q = (U2 – U1) + W (5.10)

Thus part of the heat supplied is converted into work and the remainder is utilized

in increasing the internal energy of the system.

Substituting for W in equation 5.10

Q = (U2 – U1) + p(V2 – V1)

= U2 – U1 + p2 V2 – p1 V1 (since p2 = p1 )

= (U2 + p2 V2) – (U1 + p1 V1)

Now, we know that h = u + pv or H = U + pV

Hence

Q = H2 – H1 (larger H – smaller H) (5.11)

Referring to the process representation on the p-v diagram shown in Figure 5.3, it

is noted that heating increases the volume. In other words, the fluid expands. For

a perfect gas, equation 5.8 tells us that an increase in volume will result in

corresponding increase in temperature.

For constant volume process:

W = 0

For constant pressure process:

W = p (V2 – V1)


The specific internal energy of a fluid is increased from 120 kJ/kg to 180 kJ/kg

during a constant volume process. Determine the amount of heat energy

required to bring about this increase for 2 kg of fluid.

Example 5.2


The non flow energy equation is

Q – W = U2 – U1

For a constant volume process

W = 0

and the equation becomes

Q = U2 – U1

Q = 180 – 120

= 60 kJ/kg

Therefore for 2 kg of fluid

Q = 60 x 2 = 120 kJ

i.e. 120 kJ of heat energy would be required.


2.25 kg of fluid having a volume of 0.1 m3 is in a cylinder at a constant

pressure of 7 bar. Heat energy is supplied to the fluid until the volume

becomes 0.2 m3. If the initial and final specific enthalpies of the fluid are

210 kJ/kg and 280 kJ/kg respectively, determine

a) the quantity of heat energy supplied to the fluid

b) the change in internal energy of the fluid

Example 5.3


Data: p = 7.0 bar; V1 = 0.1 m3 ; V2 = 0.2 m

3

a) Heat energy supplied = change in enthalpy of fluid

Q = H2 – H1

= m( h2 - h1 )

= 2.25( 280 – 210 )

= 157.5 kJ

b) For a constant pressure process

W = P(V2 – V1)

= 7 x 105 x ( 0.2 – 0.1)

= 7 x 104 J

= 70 kJ

Applying the non-flow energy equation

Q – W = U2 – U1

gives

U2 – U1 = 157.5 – 70

= 87.5 kJ



NEXT INPUT…!

5.4 The pressure of the gas inside an aerosol can is 1.2 bar at a temperature of

25o C. Will the aerosol explode if it is thrown into a fire and heated to a

temperature of 600o C? Assume that the aerosol can is unable to withstand

pressure in excess of 3 bar.

5.5 0.05 kg of air, initially at 130o C is heated at a constant pressure of 2 bar

until the volume occupied is 0.0658 m3. Calculate the heat supplied and

the work done.

5.6 A spherical research balloon is filled with 420 m3 of atmospheric air at a

temperature of 10o C. If the air inside the balloon is heated to 80

oC at

constant pressure, what will be the final diameter of the balloon?

Activity 5B



5.4 Data: p1 = 1.2 bar; T1= 25 + 273 = 298 K

T2 = 600 + 273 = 873 K; p2 = ?

We can idealize this process at constant volume heating of a perfect gas.

Applying the general property relation between states 1 and 2

2

22

1

11

T

Vp

T

Vp

in this case V2 = V1

Hence, 2

2

1

1

T

p

T

p

or 2

212

T

Tpp

= 1.2 x 298

873

= 3.52 bar

Since the aerosol cannot withstand pressures above 3 bar, it will clearly

explode in the fire.


5.5 Data: m = 0.5 kg; p = 2 bar; V2 = 0.0658 m3

;

T1 = 130 + 273 =403 K

Using the characteristic gas equation at state 2

T2 = mR

Vp 22

= 3

5

10 x 287.0 x 05.0

0658.0 x 10 x 2

= 917 K

For a perfect gas undergoing a constant pressure process, we have

Q = mcp(T2 – T1)

i.e. Heat supplied = 0.05 x 1.005(917 – 403)

= 25.83 kJ

W = p (V2 – V1)

From equation pV = RT

Work done = R (T2 – T1)

= 0.287(917 – 403)

i.e. Work done by the mass of gas present = 0.05 x 0.287 x 514

= 7.38 kJ


5.6 Data: T1 = 10 + 273 = 283 K; T2 = 80 + 273 = 353 K

V1 = 420 m3; V2 = ?

Applying the general property relation between states 1 and 2

2

22

1

11

T

Vp

T

Vp

Since the air is heated at constant pressure p1 = p2

Then,

2

2

1

1

T

V

T

V

or V2 = 1

21

T

TV

= 420 x 283

353

= 523.9 m3

Since the balloon is a sphere, V2 = 3

3

4r

where r = radius of the balloon

Hence,

523.9 = 3

3

4r

Solving gives r = 5 m

Final diameter of balloon, d = 2r = 2 x 5 = 10 m






1. A receiver vessel in a steam plant contains 20 kg of steam at 60 bar and

500oC. When the plant is switched off, the steam in the vessel cools at

constant volume until the pressure is 30 bar. Find the temperature of the

steam after cooling and the heat transfer that has taken place.

2. 0.25 kg of combustion gas in a diesel engine cylinder is at temperature of

727oC. The gas expands at constant pressure until its volume is 1.8 times

its original value. For the combustion gas, R = 0.302 kJ/kgK and

cp = 1.09 kJ/kgK. Find the following:

a) temperature of the gas after expansion

b) heat transferred

c) work transferred

3. A quantity of gas has an initial pressure and volume of 0.1 MN/m2 and

0.1 m3, respectively. It is compressed to a final pressure of 1.4 MN/m

2

according to the law pV1.26

= constant. Determine the final volume of the

gas.

4. A mass of 0.05 kg of air at a temperature of 40oC and a pressure of 1 bar is

compressed polytropicly following the law pV1.25

= C. Determine the

following:

a) final temperature

b) final volume

c) work transfer

d) heat transfer

e) change in internal energy

SELF-ASSESSMENT



1. 233.8oC; 14380 kJ rejected

2. 1527oC; 218 kJ added; 60.4 kJ output

3. 0.01235 m3

4. 158.9oC; 12390 cm

3; 6.82 kJ input; 2.56 kJ rejected; 4.26 kJ increase


CONGRATULATIONS!

!!!…May success be

with you always…

FLOW PROCESS J2006/6/1

FLOW PROCESS

OBJECTIVES

General Objective: To understand the concept of flow process and its application in

steady flow energy equation.


derive the meaning and interpret the steady-flow energy

equation

apply the steady-flow energy equation to :

boiler

condenser

UNIT 6


6.0 STEADY FLOW PROCESSES

In heat engine it is the steady flow processes which are generally of most interest. The

conditions which must be satisfied by all of these processes are :

i. The mass of fluid flowing past any section in the system must be constant with

respect to time.

ii. The properties of the fluid at any particular section in the system must be constant

with respect to time.

iii. All transfer of work energy and heat which takes place must be done at a uniform

rate.

A typical example of a steady flow process is a steam boiler, operating under a constant load

as shown diagrammatically in Fig. 6.0. In order to maintain the water level in the boiler, the

feed pump supplies water at exactly the same rate as that at which steam is drawn off from

the boiler. To maintain the production of steam at this rate at a steady pressure, the furnace

will need to supply heat energy at a steady rate. Under these conditions, the properties of the

working fluid at any section within the system must be constant with respect to time.

Do you know the

conditions for

Steady Flow

Processes ?

INPUT


Figure 6.0 Steam Boiler

6.1 STEADY FLOW ENERGY EQUATION

This equation is a mathematical statement on the principle of Conservation of

Energy as applied to the flow of a fluid through a thermodynamic system.

The various forms of energy which the fluid can have are as follows:

a) Potential energy

If the fluid is at some height Z above a given datum level, then as a result of

its mass it possesses potential energy with respect to that datum. Thus, for

unit mass of fluid, in the close vicinity of the earth,

Potential energy = g Z

9.81 Z

STEAM

OUT

WATER

LEVEL

BOUNDRY

WATER

IN

FURNACE


b) Kinetic energy

A fluid in motion possesses kinetic energy. If the fluid flows with velocity C,

then, for unit mass of fluid,

Kinetic energy =2

2C

c) Internal energy

All fluids store energy. The store of energy within any fluid can be increased

or decreased as a result of various processes carried out on or by the fluid.

The energy stored within a fluid which results from the internal motion of its

atoms and molecules is called its internal energy and it is usually designated

by the letter U. If the internal energy of the unit mass of fluid is discussed

this is then called the specific internal energy and is designated by u.

d) Flow or displacement energy

In order to enter or leave the system, any entering or leaving volume of fluid

must be displaced with an equal volume ahead of itself. The displacing mass

must do work on the mass being displaced, since the movement of any mass

can only be achieved at the expense of work.

Thus, if the volume of unit mass of liquid (its specific volume) at entry is v1

and its pressure is P1, then in order to enter a system it must displace an equal

specific volume v1 inside the system. Thus work to the value P1v1 is done on

the specific volume inside the system by the specific volume outside the

system. This work is called flow or displacement work and at entry it is

energy received by the system.

Similarly, at exit, in order to leave, the flow work must be done by the fluid

inside the system on the fluid outside the system. Thus, if the pressure at exit,

is P2 and the specific volume is v2 the equation is then,

Flow or displacement work rejected = P2v2

e) Heat received or rejected

During its passage through the system the fluid can have direct reception or

rejection of heat energy through the system boundary. This is designated by

Q. This must be taken in its algebraic sense.

Thus,


Q is positive when heat is received.

Q is negative when heat is rejected.

Q = 0 if heat is neither received nor rejected.

f) External work done

During its passage through the system the fluid can do external work or have

external work done on it. This is usually designated by W. This also must be

taken in its algebraic sense.

Thus if,

External work is done by the fluid then W is positive.

External work is done on the fluid then W is negative.

If no external work is done on or by the fluid then W = 0.

Figure 6.1 illustrates some thermodynamic system into which is flowing a fluid with

pressure P1, specific volume v1, specific internal energy u1 and velocity C1. The

entry is at height Z1 above some datum level. In its passage through the system,

external work W may be done on or by the fluid and also heat energy Q may be

received or rejected by the fluid from or to the surroundings.

The fluid then leaves the system with pressure P2, specific volume v2, specific

internal energy u2 and velocity C2. The exit is at height Z2 above some datum level.

Figure 6.1 Thermodynamic system

SYSTEM

(OR CONTROL VOLUME) ENTRY

EXIT Z2

Z1

W

Q

P2 v2

u2 C2

P1 v1

U1C1


The application of the principle of energy conservation to the system is,

Total energy entering the system = Total energy leaving the system

or, for unit mass of substance,

WC

vPugZQC

vPugZ 22

2

22222

2

11111 (6.1)

This is called the steady flow energy equation.

This equation is not applicable to all energy forms. In such cases, the energy forms

concerned are omitted from the energy equation.

In equation 6.1, it was stated that the particular combination of properties of the

form, u + Pv is called specific enthalpy and is designated as h. Thus, the steady flow

energy equation is written as

WC

hgZQC

hgZ 22

2

222

2

111 (6.2)

Steady flow energy equation

Potential energy + Kinetic energy +

Internal energy +Flow or Displacement

energy+ Heat or Work


6.2 APPLICATION OF STEADY FLOW EQUATION

The steady flow energy equation may be applied to any apparatus through which a

fluid is flowing, provided that the conditions stated previously are applicable. Some

of the most common cases found in engineering practise are dealt with in detail as

below.

6.2.1 Boilers

In a boiler operating under steady conditions, water is pumped into the boiler

along the feed line at the same rate as which steam leaves the boiler along the

steam main, and heat energy is supplied from the furnace at a steady rate.

Figure 6.2.1 Steam Boiler

The steady flow energy equation gives

WC

hgZQC

hgZ 22

2

222

2

111

and with the flow rate, m (kg/s) the equation may be written as

12

2

1

2

212

2gZgZ

CChhmWQ (6.3)

STEAM

OUT

BOUNDRY

WATER

IN

FURNACE

SYSTEM

Q

1

1

2

2


In applying this equation to the boiler, the following points should be noted :

i. Q is the amount of heat energy passing into the fluid per second

ii. W is zero since a boiler has no moving parts capable of affecting a

work transfer

iii. The kinetic energy is small as compared to the other terms and may

usually be neglected

iv. The potential energy is generally small enough to be neglected.

v. m (kg/s) is the rate of the flow of fluid.

Hence the equation is reduced to

12 hhmQ (6.4)

FOR BOILER UNDER A STEADY

CONDITION,

WORK = 0

KINETIC ENERGY = NEGLECTED

POTENTIAL ENERGY = NEGLECTED


6.2.2 Condensers

In principle, a condenser is a boiler reverse. In a boiler, heat energy is

supplied to convert the liquid into vapour whereas in a condenser heat energy

is removed in order to condense the vapour into a liquid. If the condenser is

in a steady state then the amount of liquid, usually called condensate, leaving

the condenser must be equal to the amount of vapour entering the condenser.

Figure 6.2.2 Condenser


12

2

1

2

212

2gZgZ

CChhmWQ

Points to note,

i. Q is the amount of heat energy per second transferred from the

system

ii. W is zero in the boiler

iii. The kinetic energy term may be neglected as in the boiler

iv. The potential energy is generally small enough to be neglected

v. m is the rate of the flow of fluid.

Thus the equation is reduced to

12 hhmQ (6.5)

VAPOUR

CONDENSATE

BOUNDARY

WATER

IN

WATER

OUT

SYSTEM


A boiler operates at a constant pressure of 15 bar, and evaporates fluid at the

rate of 1000 kg/h. At entry to the boiler, the fluid has an enthalpy of 165 kJ/kg

and on leaving the boiler the enthalpy of the fluid is 2200 kJ/kg. Determine

the heat energy supplied to the boiler.

Example 6.1



12

2

1

2

212

2gZgZ

CChhmWQ

Q = heat energy per hour entering system

W = work energy per hour leaving system = 0

m = fluid flow rate = 1000 kg/h

h2 = 2200 kJ/kg

h1 = 165 kJ/kg

C1& C2 = neglected

Z1& Z2 = neglected

Thus, the steady flow energy equation becomes

kg

kJ

h

kg

h

kJQ 16522001000

h

kJxQ 610035.2

BOUNDARY

WATER

IN

SYSTEM

Q

1

1

STEAM OUT

2

2


If 65 % of the heat energy supplied to the boiler in example 6.1 is used in

evaporating the fluid, determine the rate of fuel consumption required to

maintain this rate of evaporation, if 1 kg of fuel produces 32000 kJ of heat

energy.

Example 6.2


Heat energy from fuel required per hour = 65.0

10035.2 6x

= 3.13 x 106 kJ/h

Heat energy obtained from the fuel = 32 000 kJ/kg

Fuel required = kJ

kgx

h

kJ

32000

1013.3 6x

= 97.8 kg/h


Fluid enters a condenser at the rate of 35 kg/min with a specific enthalpy of

2200 kJ/kg, and leaves with a specific enthalpy of 255 kJ/kg. Determine the

rate of heat energy loss from the system.

Example 6.3



12

2

1

2

212

2gZgZ

CChhmWQ

For a condenser, W = 0, and the term representing the change in kinetic and potential

energy may be neglected. Therefore the equation is reduced to

12 hhmQ

From the above equation

kg

kJkgQ )2200255(

min35

= - 68 000 kJ/min



NEXT INPUT…!

6.1 In an air conditioning system, air is cooled by passing it over a chilled water

coil condenser. Water enters the coil with an enthalpy of 42 kJ/kg and leaves

the coil with an enthalpy of 80 kJ/kg. The water flow rate is 200 kg/h. Find

the rate of heat absorption by the water in kilowatts.

6.2 In a steady flow system, a substance flows at the rate of 4 kg/s. It enters at a

pressure of 620 kN/m2, a velocity of 300 m/s, internal energy 2100 kJ/kg and

specific volume 0.37 m3/kg. It leaves the system at a pressure of 130 kN/m

2,

a velocity of 150 m/s, internal energy 1500 kJ/kg and specific volume

1.2 m3/kg. During its passage through the system the substance has a loss by

heat transfer of 30 kJ/kg to the surroundings. Determine the power of the

system in kilowatts, stating whether it is from or to the system. Neglect any

change in potential energy.

Activity 6


Feedback to Activity 6

6.1 Data: m = 200 kg/h = kg/s 056.03600

200

h1 = 42 kJ/kg; h2 = 80 kJ/kg

Q = ?

The diameter of the water tube in a cooler is normally constant. Therefore,

there is no change in water velocity and kinetic energy. In general the change

in potential energy is also negligible.

The equation of steady flow is therefore reduced to

12 hhmQ

= 0.056(80 – 42)

= 2.13 kJ/s or kW

The rate of heat absorption by the water is 2.13 kW


6.2 By neglecting the change in potential energy, for a unit mass of

substance, the steady flow energy equation becomes:

WC

vPuQC

vPu 22

2

2222

2

1111 (1)

Q is written negative since 30 kJ/kg are lost to the surroundings.

From equation (1)

Specific W = QCC

vPvPuu

)2

()()(2

2

2

1221111

Working in kilojoules (kJ)

Specific W = (2100-1500)+(620x0.37-130x1.2)+(3

22

102

150300

x

)-30

= 676.75 kJ/kg.

The substance flows at the rate of 4 kg/s

Output (since W is positive) = 676.75 x 4

= 2707 kJ/s or kW





1. A boiler uses coal at the rate of 3000 kg/h in producing steam with a specific

enthalpy of 2700 kJ/kg from feed water with a specific enthalpy of 280 kJ/kg.

The combustion of 1 kg of coal produces 28000 kJ, of which 80% is useful in

producing steam. Calculate the rate at which steam is produced.

2. Fluid with specific enthalpy of 2280 kJ/kg enters a condenser at the rate of

4500 kg/h, and leaves with a specific enthalpy of 163 kJ/kg. If the enthalpy

of the cooling water circulating through the condenser tubes increases at the

rate of 148 000 kJ/min, determine the rate at which heat energy flows from

the condenser to the atmosphere.

SELF-ASSESSMENT



1. 27800 kg/h

2. 646000 kJ/h




always….


FLOW PROCESS

OBJECTIVES

General Objective: To understand the application of steady flow energy equation.


apply the steady-flow energy equation to :

turbine

nozzle

throttle

pump

UNIT 7


7.0 APPLICATION OF STEADY FLOW EQUATION

The steady flow energy equation may be applied to any apparatus through which a

fluid is flowing, provided the conditions stated in Unit 6 are applicable. Some of the

most common cases found in engineering practise will now be dealt with in detail.

Do you know the

application of steady

flow equation for

turbine?

INPUT


7.0.1 Turbine

A turbine is a device which uses a pressure drop to produce work energy

which is used to drive an external load.

Figure 7.0.1 Turbine


12

2

1

2

212

2gZgZ

CChhmWQ

Points to note :

i. The average velocity of flow of fluid through a turbine is normally

high, and the fluid passes quickly through the turbine. It may be

assumed that, because of this, heat energy does not have time to flow

into or out of the fluid during its passage through the turbine, and

hence Q = 0 .

ii. Although velocities are high the difference between them is not large,

and the term representing the change in kinetic energy may be

neglected.

iii. Potential energy is generally small enough to be neglected.

iv. W is the amount of external work energy produced per second.

Q

W

FLUID OUT

FLUID IN

SYSTEM

1

2 BOUNDARY


The steady flow energy equation becomes

- 12 hhmW

or 21 hhmW (7.1)

7.0.2 Nozzle

A nozzle utilises a pressure drop to produce an increase in the kinetic energy

of the fluid.

Figure 7.0.2 Nozzle


12

2

1

2

212

2gZgZ

CChhmWQ

Points to note :

i. The average velocity of flow through a nozzle is high, hence the fluid

spends only a short time in the nozzle. For this reason, it may be

assumed that there is insufficient time for heat energy to flow into or

out of the fluid during its passage through the nozzle, i.e. Q = 0.

ii. Since a nozzle has no moving parts, no work energy will be

transferred to or from the fluid as it passes through the nozzle,

i.e. W = 0.

SYSTEM BOUNDARY

FLUID

OUT

FLUID

IN

1

1

2

2


iii. Potential energy is generally small enough to be neglected.

Hence the equation becomes

20

2

1

2

212

CChhm

Often C1 is negligible compared with C2. In this case the equation becomes

20

2

212

Chhm

or 21

2

2

2hh

C

or 212 2 hhC (7.2)


A fluid flows through a turbine at the rate of 45 kg/min. Across the turbine the

specific enthalpy drop of the fluid is 580 kJ/kg and the turbine loss 2100 kJ/min

in the form of heat energy. Determine the power produced by the turbine,

assuming that changes in kinetic and potential energy may be neglected.

Example 7.1



12

2

1

2

212

2gZgZ

CChhmWQ

Q = heat energy flow into system = -2100 kJ/min

W = work energy flow from system kJ/min

m = fluid flow rate = 45 kg/min

h2 - h1 = -580 kJ/kg

C1& C2 = neglected

Z1& Z2 = neglected

Therefore the steady flow energy equation becomes

-2100 kJ/min – W = 45 kg/min (-580 kJ/kg)

W = (26100 – 2100) kJ/min

= 24000 kJ/min

= 400 kJ/s

= 400 kW


Fluid with a specific enthalpy of 2800 kJ/kg enters a horizontal nozzle with

negligible velocity at the rate of 14 kg/s. At the outlet from the nozzle the

specific enthalpy and specific volume of the fluid are 2250 kJ/kg and

1.25 m3/kg respectively. Assuming an adiabatic flow, determine the required

outlet area of the nozzle.

Example 7.2



12

2

1

2

212

2gZgZ

CChhmWQ

When applied to the nozzle, this becomes

20

2

1

2

212

CChhm

Since the inlet C1 is negligible, this may be written as

21

2

2 2 hhC

= √ [2(2800 – 22500]

= 1050 m/s

Applying the equation of continuity at outlet gives

2

22

v

CAm

14 kg/s = /kgm 25.1

m/s 1050 x 3

2A

A2 = 0.01668 m2



NEXT INPUT…!

7.1 Steam enters a turbine with a velocity of 16 m/s and specific enthalpy

2990 kJ/kg. The steam leaves the turbine with a velocity of 37 m/s and

specific enthalpy 2530 kJ/kg. The heat lost to the surroundings as the steam

passes through the turbine is 25 kJ/kg. The steam flow rate is 324000 kg/h.

Determine the work output from the turbine in kilowatts.

7.2 In a turbo jet engine the momentum of the gases leaving the nozzle produces

the propulsive force. The enthalpy and velocity of the gases at the nozzle

entrance are 1200 kJ/kg and 200 m/s respectively. The enthalpy of the gas at

exit is 900 kJ/kg. If the heat loss from the nozzle is negligible, determine the

velocity of the gas jet at exit from the nozzle.

Activity 7A


Feedback to Activity 7A

7.1 Neglecting the changes in potential energy, the steady flow energy equation

is

2

2

1

2

212

CChhWQ

Q is negative since heat is lost from the steam to the surroundings

specific W =

2

2

2

2

121

CChh - Q

= (2999-2530) + 25102

)3716(3

22

x

= 434.443 kJ/kg

The steam flow rate = 324000/3600 = 90 kg/s

W = 434.443 x 90

= 39099.97 kJ/s or kW

39100 kW

39.1 MW


7.2 The steady energy flow equation for nozzle gives

20

2

1

2

212

CChhm

On simplification,

2

1212 2 ChhC

= 2(1200x 103- 900x

10

3) + 200

2

= 800 m/s


7.0.3 THROTTLE

A throttling process is one in which the fluid is made to flow through a restriction ,

e.g. a partially opened valve or orifice, causing a considerable drop in the pressure

of the fluid.

Figure 7.0.3 Throttling process


12

2

1

2

212

2gZgZ

CChhmWQ

Points to note:

i. Since throttling takes place over a very small distance, the available area

through which heat energy can flow is very small, and it is normally

assumed that no energy is lost by heat transfer, i.e. Q = 0.

ii. Since there are no moving parts, no energy can be transferred in the form of

work energy, i.e. W = 0.

iii. The difference between C1 and C2 will not be great and consequently the

term representing the change in kinetic energy is normally neglected.

iv. The potential energy is generally small enough to be neglected.

INPUT

1

2


The steady flow energy equation becomes

0 = m (h2 - h1)

or h2 = h1 (7.3)

i.e. during a throttling process the enthalpy remains constant.

7.0.4 Pump

The action of a pump is the reverse of that of a turbine, i.e. it uses external

work energy to produce a pressure rise. In applying the steady flow energy

equation to a pump, exactly the same arguments are used as for turbine and

the equation becomes

- 12 hhmW (7.4)

Since h2 > h1, W will be found to be negative.

Figure 7.0.4 Pump

Q

W

SYSTEM

1

2

BOUNDARY

OUTLET

INLET


7.1 Equation of Continuity

This is an equation which is often used in conjunction with the steady flow

energy equation. It is based on the fact that if a system is in a steady state,

then the mass of fluid passing any section during a specified time must be

constant. Consider a mass of m kg/s flowing through a system in which all

conditions are steady as illustrated in Fig.7.5.

Figure 7.1 Mass flowing through a system

Let A1 and A2 represent the flow areas in m2 at the inlet and outlet

respectively.

Let v1 and v2 represent the specific volumes in m3/kg at the inlet and outlet

respectively.

Let C1 and C2 represent the velocities in m/s, at the inlet and outlet

respectively.

Then mass flowing per sec = volume flowing per sec m3/s

volume per kg m3/kg

= s

kg

v

CA

1

11 at inlet

= s

kg

v

CA

2

22 at outlet

i.e. 1

11

v

CAm =

2

22

v

CA (7.5)

1

1

2

2

C1 C2

AREA A2

AREA A1


Example 7.3


For a throttling process, the steady flow energy equation becomes

0 = m (h2 - h1)

or h2 = h1

But h2 = u2 + P2v2

and h1= u1 + P1v1

Therefore the change in specific internal energy

= u2 – u1

= ( h2 - P2v2 ) - ( h1 – P1v1)

= ( h2 - h1 ) – ( P2v2 - P1v1 )

= 0 – ( 1 x 1.8 – 10 x 0.3 ) bar m3/kg

= 120 x 10 Nm/kg

= 120 kJ/kg

Example 7.4


The flow rate of fluid = 45 kg/min

= 0.75 kg/s

The steady flow energy is

12

2

1

2

212

2gZgZ

CChhmWQ

A fluid flowing along a pipeline undergoes a throttling process from 10 bar to 1

bar in passing through a partially open valve. Before throttling, the specific

volume of the fluid is 0.3 m3/kg and after throttling is 1.8 m

3/kg. Determine the

change in specific internal energy during the throttling process.

A pump delivers fluid at the rate of 45 kg/min. At the inlet to the pump the

specific enthalpy of the fluid is 46 kJ/kg, and at the outlet from the pump the

specific enthalpy of the fluid is 175 kJ/kg. If 105 kJ/min of heat energy are lost to

the surroundings by the pump, determine the power required to drive the pump if

the efficiency of the drive is 85 %.


Q = - 105 kJ/min = - 1.75 kJ/s

W = work energy flow (kJ/s)

h1 = 46 kJ/kg

h2 = 1.27 kJ/kg

m = 0.75 kg/s

The kinetic and potential energy may be neglected

Substituting the data above with the steady flow energy equation gives

- 1.75 – W = 0.75 (175 – 46)

W = -1.75 – (0.75 x 129)

= - 98.5 kJ/s

= - 98.5 kW

(N.B. The negative sign indicates work energy required by the pump)

Since the efficiency of the drive is 85 %

Power required by the compressor = 98.5 x 85

100

= 114.8 kW

Turbine 21 hhmW

&

Pump

- 12 hhmW



NEXT INPUT…!

7.3 A rotary air pump is required to deliver 900 kg of air per hour. The enthalpy

at the inlet and exit of the pump are 300 kJ/kg and 500 kJ/kg respectively.

The air velocity at the entrance and exit are 10 m/s and 15 m/s respectively.

The rate of heat loss from the pump is 2500 W. Determine the power

required to drive the pump.

7.4 In activity 7A, for question No. 7.2, if the diameter of the nozzle at exit is

500 mm, find the mass flow rate of gas. The gas density at the nozzle inlet

and exit are 0.81 kg/m3 and 0.39 kg/m

3 respectively. Also determine the

diameter of the nozzle at the inlet.

Activity 7B


Feedback to Activity 7B

7.3 Data : 3600

900m = 0.25 kg/s

h1 = 300 kJ/kg

h2 = 500 kJ/kg

C1= 10 m/s

C2= 15 m/s

Q = 2500 W = 2.5 kW

W = ?


12

2

1

2

212

2gZgZ

CChhmWQ

Neglecting the change in Potential energy since it is negligible

2

2

1

2

212

CChhmWQ

-W = 0.25 [( 500- 300) + (3

22

102

1015

x

)] + 2.5

-W = 52.5 kW


7.4 Data : A2 = 4

5.0 2

= 0.196 m2

1 = 0.81 kg/m3

2 = 0.39 kg/m3

m = ?

d = ?

Mass flow rate at exit,

m = A2 C2 2

= 61.2 kg/s

From the mass balance,

Mass entering the nozzle = mass leaving the nozzle = m

m = A1 C1 1 = A2 C2 2

On substitution

A1 x 200 x 0.81 = 61.2

On simplification

A1 = 0.378 m2

or

d1 = 0.694 m

= 694 mm





1 Steam flows through a turbine stage at the rate of 4500 kJ/h. The steam

velocities at inlet and outlet are 15 m/s and 180 m/s respectively. The rate of

heat energy flow from the turbine casing to the surroundings is 23 kJ/kg of

steam flowing. If the specific enthalpy of the steam decreases by 420 kJ/kg in

passing through the turbine stage, calculate the power developed.

2 A rotary pump draws 600 kg/hour of atmospheric air and delivers it at a

higher pressure. The specific enthalpy of air at the pump inlet is 300 kJ/kg

and that at the exit is 509 kJ/kg. The heat lost from the pump casing is

5000 W. Neglecting the changes in kinetic and potential energy, determine

the power required to drive the pump.

3 A nozzle is supplied with steam having a specific enthalpy of 2780 kJ/kg at

the rate of 9.1 kg/min. At outlet from the nozzle the velocity of the steam is

1070 m/s. Assuming that the inlet velocity of the steam is negligible and that

the process is adiabatic, determine:

a) the specific enthalpy of the steam at the nozzle exit

b) the outlet area required if the final specific volume of the steam is

18.75 m3/kg.

4 Fluid at 10.35 bar having a specific volume of 0.18 m3/kg is throttled to a

pressure of 1 bar. If the specific volume of the fluid after throttling is

0.107 m3/kg, calculate the change in specific internal energy during the

process.

SELF-ASSESSMENT



1. 476 kW

2. 353 kW

3. 2208 kJ/kg; 2660 mm2

4. 175.7 kJ/kg



You can continue with the

next unit…

PROPERTIES OF STEAM J2006/8/1

PROPERTIES OF STEAM

OBJECTIVES

General Objective : To define the properties of wet steam, dry saturated steam and

superheated steam using information from the steam tables.


define the word phase and distinguish the solid, liquid and

steam phases

understand and use the fact that the vaporization process is

carried out at constant pressure

define and explain the following terms: saturation

temperature, saturated liquid, wet steam, saturated steam, dry

saturated steam, , dryness fraction and superheated steam

determine the properties of steam using the P-v diagram

understand and use the nomenclature as in the Steam Tables

apply single and double interpolation using the steam tables

locate the correct steam tables for interpolation, including

interpolation between saturation tables and superheated tables

where necessary

UNIT 8


8.0 Introduction

In thermodynamic systems, the working fluid can be in the liquid, steam or gaseous

phase. In this unit, the properties of liquid and steam are investigated in some details

as the state of a system can be described in terms of its properties. A substance that

has a fixed composition throughout is called a pure substance. Pure chemicals

(H2O, N2, O2, Ar, Ne, Xe) are always pure substances. We all know from experience

that substances exist in different phases. A phase of substance can be defined as that

part of a pure substance that consists of a single, homogenous aggregate of matter.

The three common phases for H2O that are usually used are solid, liquid and steam.

When studying phases or phase changes in thermodynamics, one does not need to be

concerned with the molecular structure and behavior of the different phases.

However, it is very helpful to have some understanding of the molecular phenomena

involved in each phase.

Molecular bonds are strongest in solids and weakest in steams. One reason is that

molecules in solids are closely packed together, whereas in steams they are separated

by great distances.

INPUT


The three phases of pure substances are: -

Solid Phase

In the solid phase, the molecules are;

(a) closely bound, therefore relatively dense; and

(b) arranged in a rigid three-dimensional pattern so that they do not easily

deform. An example of a pure solid state is ice.

Liquid Phase

In the liquid phase, the molecules are;

(a) closely bound, therefore also relatively dense and unable to expand to fill a

space; but

(b) they are no longer rigidly structured so much so that they are free to move

within a fixed volume. An example is a pure liquid state.

Steam Phase

In the steam phase, the molecules;

(a) virtually do not attract each other. The distance between the molecules are

not as close as those in the solid and liquid phases;

(b) are not arranged in a fixed pattern. There is neither a fixed volume nor a

fixed shape for steam.

The three phases described above are illustrated in Fig. 8.0 below. The following are

discovered:

(a) the positions of the molecules are relatively fixed in a solid phase;

(b) chunks of molecules float about each other in the liquid phase; and

(c) the molecules move about at random in the steam phase.

Source: Thermodynamics: An Engineering Approach, 3rd

Ed by Cengel and Boles

Figure 8.0 The arrangement of atoms in different phases

(a) (b) (c)


8.1 Phase-Change Process

The distinction between steam and liquid is usually made (in an elementary manner)

by stating that both will take up the shape of their containers. However liquid will

present a free surface if it does not completely fill its container. Steam on the other

hand will always fill its container.

With these information, let us consider the following system:

A container is filled with water, and a moveable, frictionless piston is placed on the

container at State 1, as shown in Fig. 8.1. As heat is added to the system, the

temperature of the system will increase. Note that the pressure on the system is

being kept constant by the weight of the piston. The continued addition of heat will

cause the temperature of the system to increase until the pressure of the steam

generated exactly balances the pressure of the atmosphere plus the pressure due to

the weight of the piston.

Figure 8.1 Heating water and steam at constant pressure

At this point, the steam and liquid are said to be saturated. As more heat is added,

the liquid that was at saturation will start to vaporize until State 2. The two-phase

mixture of steam and liquid at State 2 has only one degree of freedom, and as long as

liquid is present, vaporization will continue at constant temperature. As long as

liquid is present, the mixture is said to be wet steam, and both the liquid and steam

are saturated. After all the liquid is vaporized, only steam is present at State 3, and

the further addition of heat will cause the temperature of steam to increase at

W

W

W

W

Liquid Steam Superheated

Steam

STATE 1 STATE 2 STATE 3 STATE 4


constant system pressure. This state is called the superheated state, and the steam is

said to be superheated steam as shown in State 4.

8.2 Saturated and Superheated Steam

While tables provide a convenient way of presenting precise numerical presentations

of data, figures provide us with a clearer understanding of trends and patterns.

Consider the following diagram in which the specific volume of H2O is presented as

a function of temperature and pressure1:

Figure 8.2-1 T-v diagram for the heating process of water at constant pressure

Imagine that we are to run an experiment. In this experiment, we start with a mass

of water at 1 atm pressure and room temperature. At this temperature and pressure

we may measure the specific volume (1/ = 1/1000 kg/m3). We plot this state at

point 1 on the diagram.

If we proceed to heat the water, the temperature will rise. In addition, water expands

slightly as it is heated which makes the specific volume increase slightly. We may

plot the locus of such points along the line from State 1 to State 2. We speak of

liquid in one of these conditions as being compressed or subcooled liquid.

1 Figures from Thermodynamics: An Engineering Approach, 3

rd Ed by Cengel and Boles

20

100

300

1

2 3

4

T, oC

v, m3/kg

Compressed

liquid

Saturated

mixture

Superheated

steam


State 2 is selected to correspond to the boiling point (100 oC). We speak of State 2

as being the saturated liquid state, which means that all of the water is in still liquid

form, but ready to boil. As we continue to heat past the boiling point 2, a

fundamental change occurs in the process. The temperature of the water no longer

continues to rise. Instead, as we continue to add energy, liquid progressively

changes to steam phase at a constant temperature but with an increasing specific

volume. In this part of the process, we speak of the water as being a saturated

mixture (liquid + steam). This is also known as the quality region.

At State 3, all liquid will have been vaporised. This is the saturated steam state.

As we continue to heat the steam beyond State 3, the temperature of the steam again

rises as we add energy. States to the right of State 3 are said to be superheated

steam.

Summary of nomenclature:

Compressed or subcooled liquid (Between States 1 & 2)

A liquid state in which the fluid remains entirely within the liquid state,

and below the saturation state.

Saturated liquid (State 2)

All fluid is in the liquid state. However, even the slightest addition of

energy would result in the formation of some vapour.

Saturated Liquid-Steam or Wet Steam Region (Between States 2 & 3)

Liquid and steam exist together in a mixture.

Saturated steam (State 3)

All fluid is in the steam state, but even the slightest loss of energy from the

system would result in the formation of some liquid.

Superheated steam (The right of State 3)

All fluid is in the steam state and above the saturation state. The

superheated steam temperature is greater than the saturation temperature

corresponding to the pressure.


The same experiment can be conducted at several different pressures. We see that as

pressure increases, the temperature at which boiling occurs also increases.2

Figure 8.2-2 T-v diagram of constant pressure phase change processes

of a pure substance at various pressures for water.

It can be seen that as pressure increases, the specific volume increase in the liquid to

steam transition will decrease.

At a pressure of 221.2 bar, the specific volume change which is associated to a phase

increase will disappear. Both liquid and steam will have the same specific volume,

0.00317 m3/kg. This occurs at a temperature of 374.15

oC. This state represents an

important transition in fluids and is termed the critical point.



P = 1.01325 bar

P = 5 bar

P = 10 bar

P = 80 bar

P = 150 bar

P = 221.2 bar

Critical point

374.15

T, oC

v, m3/kg

Saturated

liquid Saturated

steam

0.00317


If we connect the locus of points corresponding to the saturation condition, we will

obtain a diagram which allows easy identification of the distinct regions3:

Figure 8.2-3 T-v diagram of a pure substance

The general shape of the P-v diagram of a pure substance is very much like the T-v

diagram, but the T = constant lines on this diagram have a downward trend, as shown

in Fig. 8.2-4.

Figure 8.2-4 P-v diagram of a pure substance



P

v

Critical

point

Saturated liquid line

Dry saturated steam line

T2 = const.

T1 = const.

COMPRESS LIQUID

REGION

WET STEAM

REGION

SUPERHEATED

STEAM

REGION

T2 > T1

T

v

Critical

point

Saturated liquid line

Dry saturated steam line

P2 = const.

P1 = const. COMPRESS

LIQUID

REGION

WET STEAM

REGION

SUPERHEATED

STEAM

REGION

P2 > P1



NEXT INPUT…!

8.1 Each line in the table below gives information about phases of pure substances. Fill

in the phase column in the table with the correct answer.

Statement Phase

The molecules are closely bound, they are also relatively

dense and unable to expand to fill a space. However they are

no longer rigidly structured so that they are free to move

within a fixed volume.

i._____________

The molecules are closely bound, they are relatively dense

and arranged in a rigid three-dimensional patterns so that they

do not easily deform.

ii.____________

The molecules virtually do not attract each other. The

distance between the molecules are not as close as those in the

solid and liquid phases. They are not arranged in a fixed

pattern. There is neither a fixed volume nor a fixed shape for

steam.

iii.____________

8.2 Write the suitable names of the phases for the H2O in the P-v diagram below.

Activity 8A

P

v

( vi )

( ii )

( iv )

T2 = const.

T1 = const.

( i )

( iii)

( v )

T2 > T1



8.1 i) Liquid Phase

ii) Solid Phase

iii) Steam Phase

8.2 i) Compress liquid region

ii) Saturated liquid line

iii) Wet steam region

iv) Dry saturated steam line

v) Superheated steam region

vi) Critical point


8.3 Properties of a Wet Mixture

Between the saturated liquid and the saturated steam, there exist a mixture of steam

plus liquid (wet steam region). To denote the state of a liquid-steam mixture, it is

necessary to introduce a term describing the relative quantities of liquid and steam in

the mixture. This is called the dryness fraction (symbol x). Thus, in 1 kg of wet

mixture, there must be x kg of saturated steam plus (1 – x) kg of saturated liquid.

Figure 8.3-1 Liquid-steam mixture

The dryness fraction is defined as follows;

where mtotal = mliquid + msteam

(8.1)

(1 - x ) kg of liquid

x kg of steam

total mass = 1 kg

mass total

steam saturateddry of massfraction dryness

total

steam

m

mx

INPUT


8.3.1 Specific volume

For a wet steam, the total volume of the mixture is given by the volume of

liquid present plus the volume of dry steam present.

Therefore, the specific volume is given by,

Now for 1 kg of wet steam, there are (1 – x) kg of liquid and x kg of dry

steam, where x is the dryness fraction as defined earlier. Hence,

v = vf(1 – x) + vgx

The volume of the liquid is usually negligibly small as compared to the

volume of dry saturated steam. Hence, for most practical problems,

v = xvg (8.2)

Where,

vf = specific volume of saturated liquid (m3/kg)

vg = specific volume of saturated steam (m3/kg)

x = dryness fraction

Figure 8.3-2 P-v diagram showing the location point of the dryness

fraction

At point A, x = 0

At point B, x = 1

Between point A and B, 0 x 1.0

Note that for a saturated liquid, x = 0;

and that for dry saturated steam, x = 1.

Sat. liquid

Sat. steam

Sat. liquid

P

v

ts

A B

x = 0.2 x = 0.8

vf vg

Sat. steam

steam wet of mass total

steamdry of volumeliquid a of volume v


8.3.2 Specific enthalpy

In the analysis of certain types of processes, particularly in power generation

and refrigeration, we frequently encounter the combination of properties

U + PV. For the sake of simplicity and convenience, this combination is

defined as a new property, enthalpy, and given the symbol H.

H = U + PV (kJ)

or, per unit mass

h = u + Pv (kJ/kg) (8.3)

The enthalpy of wet steam is given by the sum of the enthalpy of the liquid

plus the enthalpy of the dry steam,

i.e. h = hf(1 – x) + xhg

h = hf + x(hg – hf )

h = hf + xhfg (8.4)

Where,

hf = specific enthalpy of saturated liquid (kJ/kg)

hg = specific enthalpy of saturated steam (kJ/kg)

hfg = difference between hg and hf (that is, hfg = hg - hf )

8.3.3 Specific Internal Energy

Similarly, the specific internal energy of a wet steam is given by the internal

energy of the liquid plus the internal energy of the dry steam,

i.e. u = uf(1 – x) + xug

u = uf + x(ug – uf ) (8.5)

Where,

uf = specific enthalpy of saturated liquid (kJ/kg)

ug = specific enthalpy of saturated steam (kJ/kg)

ug – uf = difference between ug and uf

Equation 8.5 can be expressed in a form similar to equation 8.4. However,

equation 8.5 is more convenient since ug and uf are tabulated. The difference

is that, ufg is not tabulated.


8.3.4 Specific Entropy

A person looking at the steam tables carefully will notice two new properties

i.e. enthalpy h and entropy s. Entropy is a property associated with the

Second Law of Thermodynamics, and actually, we will properly define it in

Unit 9 . However, it is appropriate to introduce entropy at this point.

The entropy of wet steam is given by the sum of the entropy of the liquid

plus the entropy of the dry steam,

i.e. s = sf(1 – x) + xsg

s = sf + x(sg – sf )

s = sf + xsfg (8.4)

Where,

sf = specific enthalpy of saturated liquid (kJ/kg K)

sg = specific enthalpy of saturated steam (kJ/kg K)

sfg = difference between sg and sf (that is, sfg = sg - sf )

REMEMBER! These equations are used very often and

are, therefore, important to remember!

v = xvg

h = hf + xhfg

u = uf + x(ug – uf )

s = sf + xsfg


8.4 The Use of Steam Tables

The steam tables are available for a wide variety of substances which normally exist

in the vapour phase (e.g. steam, ammonia, freon, etc.). The steam tables which will

be used in this unit are those arranged by Mayhew and Rogers, which are suitable for

student use. The steam tables of Mayhew and Rogers are mainly concerned with

steam, but some properties of ammonia and freon-12 are also given.

Below is a list of the properties normally tabulated, with the symbols used being

those recommended by British Standard Specifications.

Table 8.4 The property of steam tables

Symbols Units Description

p bar Absolute pressure of the fluid

ts oC Saturation temperature corresponding to the pressure p bar

vf m3/kg Specific volume of saturated liquid

vg m3/kg Specific volume of saturated steam

uf kJ/kg Specific internal energy of saturated liquid

ug kJ/kg Specific internal energy of saturated steam

hf kJ/kg Specific enthalpy of saturated liquid

hg kJ/kg Specific enthalpy of saturated steam

hfg kJ/kg Change of specific enthalpy during evaporation

sf kJ/kg K Specific entropy of saturated liquid

sg kJ/kg K Specific entropy of saturated steam

sfg kJ/kg K Change of specific entropy during evaporation

These steam tables are divided into two types:

Type 1: Saturated Water and Steam (Page 2 to 5 of steam tables)

Type 2: Superheated Steam (Page 6 to 8 of steam tables)


Complete the following table for Saturated Water and Steam:

t Ps vg hf hfg hg sf sfg sg

oC bar m

3/kg kJ/kg

kJ/kg K

0.01 206.1

0.02337 8.666

100 1.01325

8.4.1 Saturated Water and Steam Tables

The table of the saturation condition is divided into two parts.

Part 1

Part 1 refers to the values of temperature from 0.01oC to 100

oC, followed by

values that are suitable for the temperatures stated in the table. Table 8.4.1-1

is an example showing an extract from the temperature of 10oC.

Table 8.4.1-1 Saturated water and steam at a temperature of 10 oC

t ps vg

hf hfg hg

sf sfg sg

0C bar m

3/kg

kJ/kg

kJ/kg K

10 0.01227 106.4

42.0 2477.2 2519.2

0.151 8.749 8.900

Example 8.1


From page 2 of the steam tables, we can directly read:

t Ps vg hf hfg hg sf sfg sg oC bar m

3/kg kJ/kg

kJ/kg K

1 0.006566 192.6 4.2 2498.3 2502.5 0.015 9.113 9.128

20 0.02337 57.84 83.9 2453.7 2537.6 0.296 8.370 8.666

100 1.01325 1.673 419.1 2256.7 2675.8 1.307 6.048 7.355


Complete the missing properties in the following table for Saturated Water

and Steam:

p ts vg uf ug hf hfg hg sf sfg sg

bar oC m

3/kg kJ/kg kJ/kg kJ/kg K

0.045 31.0 2558

10 0.1944

311.0 5.615


bar oC m


0.045 31.0 31.14 130 2418 130 2428 2558 0.451 7.980 8.431

10 179.9 0.1944 762 2584 763 2015 2778 2.138 4.448 6.586

100 311.0 0.01802 1393 2545 1408 1317 2725 3.360 2.255 5.615

Part 2

Part 2 (Page 3 to 5 of steam tables) is values of pressure from 0.006112 bar to

221.2 bar followed by values that are suitable for the pressures stated in the

table. Table 8.4.1-2 is an example showing an extract from the pressure of

1.0 bar.

Table 8.4.1-2 Saturated water and steam at a pressure of 1.0 bar


bar oC m


1.0 99.6 1.694 417 2506 417 2258 2675 1.303 6.056 7.359

Note the following subscripts:

f = property of the saturated liquid

g = property of the saturated steam

fg = change of the properties during evaporations

Example 8.2


From page 3 to page 5 of the steam tables, we can directly read:


For a steam at 20 bar with a dryness fraction of 0.9, calculate the

a) specific volume

b) specific enthalpy

c) specific internal energy

Example 8.3


An extract from the steam tables


20 212.4 0.09957 907 2600 909 1890 2799 2.447 3.893 6.340

a) Specific volume (v),

v = xvg

= 0.9(0.09957)

= 0.0896 m3/kg

b) Specific enthalpy (h),

h = hf + xhfg

= 909 + 0.9(1890)

= 2610 kJ/kg

c) Specific internal energy (u),

u = uf + x( ug -uf )

= 907 + 0.9(2600 - 907)

= 2430.7 kJ/kg

P

bar

v m3/kg

ts = 212.4 oC

v

u

h

s

vg

ug

hg

sg

x = 0.9

20

uf

hf

sf


Find the dryness fraction, specific volume and specific enthalpy of

steam at 8 bar and specific internal energy 2450 kJ/kg.

Example 8.4


An extract from the steam tables,


8 170.4 0.2403 720 2577 721 2048 2769 2.046 4.617 6.663

At 8 bar, ug = 2577 kJ/kg, since the actual specific internal energy is given as

2450 kJ/kg, the steam must be in the wet steam state ( u < ug).

From equation 8.5,

u = uf + x(ug -uf)

2450 = 720 + x(2577 - 720)

x = 0.932

From equation 8.2,

v = xvg

= 0.932 (0.2403)

= 0.2240 m3/kg

From equation 8.4,

h = hf + xhfg

= 721 + 0.932 (2048)

= 2629.7 kJ/kg

P

bar

v m3/kg

ts = 170.4 oC

v vg

x = 0.932

8


8.3 The internal energy of wet steam is 2000 kJ/kg. If the pressure is 42 bar, what is the

value of dryness fraction?

8.4 Determine the specific volume, specific enthalpy and specific internal energy of wet

steam at 32 bar if the dryness fraction is 0.92.

8.5 Find the dryness fraction, specific volume and specific internal energy of steam at

105 bar and specific enthalpy 2100 kJ/kg.

Activity 8B



8.3 Dryness fraction (x),

u = uf + x(ug -uf)

2000 = 1097 + x(2601 - 1097)

x = 0.6

8.4 Specific volume (v),

v = xvg

= 0.92(0.06246)

= 0.05746 m3/kg

Specific enthalpy (h),

h = hf + xhfg

= 1025 + 0.92(1778)

= 2661 kJ/kg

Specific internal energy (u),


= 1021 + 0.92(2603 - 1021)

= 2476 kJ/kg

8.5 Dryness fraction (x),

h = hf + x hfg

2100 = 1429 + x(1286)

x = 0.52

Specific volume (v),

v = xvg

= 0.52(0.01696)

= 0.00882 m3/kg

Specific internal energy (u),


= 1414 + 0.52(2537 – 1414)

= 1998 kJ/kg


8.4.2 Superheated Steam Tables

The second part of the table is the superheated steam tables. The values of

the specific properties of a superheated steam are normally listed in separate

tables for the selected values of pressure and temperature.

A steam is called superheated when its temperature is greater than the

saturation temperature corresponding to the pressure. When the pressure and

temperature are given for the superheated steam then the state is defined and

all the other properties can be found. For example, steam at 10 bar and 200 oC is superheated since the saturation temperature at 10 bar is 179.9

oC. The

steam at this state has a degree of superheat of 200 oC – 179.9

oC = 20.1

oC.

The equation of degree of superheat is:

Degree of superheat = tsuperheat – tsaturation (8.5)

The tables of properties of superheated steam range in pressure from

0.006112 bar to the critical pressure of 221.2 bar. At each pressure, there is a

range of temperature up to high degrees of superheat, and the values of

specific volume, internal energy, enthalpy and entropy are tabulated.

For the pressure above 70 bar, the specific internal energy is not tabulated.

The specific internal energy is calculated using the equation:

u = h – pv (8.6)

For reference, the saturation temperature is inserted in brackets under each

pressure in the superheat tables and values of vg, ug, hg and sg are also given.

A specimen row of values is shown in Table 8.5.2. For example, from the

superheated table at 10 bar and 200 oC, the specific volume is 0.2061 m

3/kg

and the specific enthalpy is 2829 kJ/kg.

INPUT


Complete the missing properties in the following table for Superheated

Steam:

p

(ts)

t 300 350 400 450

40

(250.3)

vg 0.0498 v 0.0800

ug 2602 u 2921

hg 2801 h 3094

sg 6.070 s 6.364

Table 8.4.2 Superheated steam at a pressure of 10 bar

p

(ts)

t 200 250 300 350 400 450 500 600

10

(179.9)

vg 0.1944 v 0.2061 0.2328 0.2580 0.2825 0.3065 0.3303 0.3540 0.4010

ug 2584 u 2623 2711 2794 2875 2957 3040 3124 3297

hg 2778 h 2829 2944 3052 3158 3264 3370 3478 3698

sg 6.586 s 6.695 6.926 7.124 7.301 7.464 7.617 7.761 8.028

Example 8.5


From page 7 of the steam tables, we can directly read

p

(ts)

t 300 350 400 450

40

(250.3)

vg 0.0498 v 0.0588 0.0664 0.0733 0.0800

ug 2602 u 2728 2828 2921 3010

hg 2801 h 2963 3094 3214 3330

sg 6.070 s 6.364 6.584 6.769 6.935


Steam at 100 bar has a specific volume of 0.02812 m3/kg. Find the

temperature, degree of superheat, specific enthalpy and specific

internal energy.

Example 8.6


First, it is necessary to decide whether the steam is wet, dry saturated or

superheated.

At 100 bar, vg = 0.01802 m3/kg. This is less than the actual specific volume

of 0.02812 m3/kg. Hence, the steam is superheated. The state of the steam is

at point A in the diagram below.

An extract from the superheated table,

p

(ts) t 425

100

(311.0)

vg 0.01802 v x 10-2

2.812

hg 2725 h 3172

sg 5.615 s 6.321

From the superheated table at 100 bar, the specific volume is 0.02812 m3/kg

at a temperature of 425 oC. Hence, this is the isothermal line, which passes

through point A as shown in the P-v diagram above.

P

bar

v m3/kg

ts = 311.0 oC

100 425

oC

vg= 0.01802

v = 0.02812

A


Degree of superheat = 425 oC – 311

oC

= 114 oC

So, at 100 bar and 425 oC, we have

v = 2.812 x 10-2

m3/kg

h = 3172 kJ/kg

From equation 8.6,

u = h – Pv

= 3172 kJ/kg – (100 x 102 kN/m

2)(2.812 x 10

-2 m

3/kg)

= 2890.8 kJ/kg

Note that equation 8.6 must be used to find the specific internal energy for

pressure above 70 bar as the specific internal energy is not tabulated.


8.6 Steam at 120 bar is at 500 oC. Find the degree of superheat, specific volume,

specific enthalpy and specific internal energy.

8.7 Steam at 160 bar has a specific enthalpy of 3139 kJ/kg. Find the temperature, degree

of superheat, specific enthalpy and specific internal energy.

Activity 8C


Feedback to Activity 8C

8.6 From the superheated table at 120 bar, the saturation temperature is 324.6 oC.

Therefore, the steam is superheated.

Degree of superheat = 500 oC – 324.6

oC

= 175.4 oC

So, at 120 bar and 500 oC, we have

v = 2.677 x 10-2

m3/kg

h = 3348 kJ/kg

From equation 8.6,

u = h – Pv

= 3348 kJ/kg – (120 x 102 kN/m

2)(2.677 x 10

-2 m

3/kg)

= 3026.76 kJ/kg

8.7 At 160 bar, hg = 2582 kJ/kg. This is less than the actual specific enthalpy of

3139 kJ/kg. Hence, the steam is superheated.

From the superheated table at 160 bar, the specific enthalpy of 3139 kJ/kg is located

at a temperature of 450 oC.

The degree of superheat = 450 oC – 347.3

oC

= 102.7 oC

At 160 bar and 450 oC, we have v = 1.702 x 10

-2 m

3/kg

From equation 8.6,

u = h – Pv

= 3139 kJ/kg – (160 x 102 kN/m

2)(1.702 x 10

-2 m

3/kg)

= 2866.68 kJ/kg


8.5 Interpolation

The first interpolation problem that an engineer usually meets is that of “reading

between the lines” of a published table, like the Steam Tables. For properties which

are not tabulated exactly in the tables, it is necessary to interpolate between the

values tabulated as shown in Fig. 8.5-1 below. In this process it is customary to use a

straight line that passes through two adjacent table points, denoted by and . If we

use the straight line then it is called “interpolation”.

Figure 8.5-1 Interpolation

The values in the tables are given in regular increments of temperature and pressure.

Often we wish to know the value of thermodynamic properties at intermediate

values. It is common to use linear interpolation as shown in Fig. 8.5-2.

From Fig. 8.5.2, the value of x

can be determined by:

1

12

121

12

12

1

1

xyy

xxyyx

yy

xx

yy

xx

INPUT

f(x)

x

Interpolation

y

x

y2

y

y1

x1 x x2

(x2 , y2)

(x , y)

(x1 , y1)

Figure 8.5-2 Linear interpolation


Determine the saturation temperature at 77 bar.

There are two methods of interpolation:

i. single interpolation

ii. double interpolation

8.5.1 Single interpolation

Single interpolation is used to find the values in the table when one of the

values is not tabulated. For example, to find the saturation temperature,

specific volume, internal energy and enthalpy of dry saturated steam at 77

bar, it is necessary to interpolate between the values given in the table.

Example 8.6


The values of saturation temperature at a pressure of 77 bars are not tabulated

in the Steam Tables. So, we need to interpolate between the two nearest

values that are tabulated in the Steam Tables.

7580

5.290295

7577

5.290

st

5

5.290295

2

5.290

st

5.2905

5.42st

ts = 292.3 oC

P

ts

80

77

75

290.5 ts 295


Determine the specific enthalpy of dry saturated steam at 103 bar.

Determine the specific volume of steam at 8 bar and 220oC.

Example 8.7


hg

2725

103 100

2715 2725

105 100

hg

3 10

52725

2719gh kJ/kg

Example 8.8


From the Steam Tables at 8 bar, the saturated temperature (ts) is 170.4 oC.

The steam is at superheated condition as the temperature of the steam is

220oC > ts.

An extract from the Steam Tables,

p / (bar)

(ts / oC)

t 200 220 250

(oC)

8

(170.4)

v 0.2610 v 0.2933

v

0 2610

220 200

0 2933 0 2610

250 200

. . .

v 027392. m3/kg

P

hg

105

103

100

2725 hg 2715

P

v

250

220

200

0.2610 v 0.2933


Determine the specific enthalpy of superheated steam at 25 bar and

320oC.

8.5.2 Double Interpolation

In some cases a double interpolation is necessary, and it’s usually used in the

Superheated Steam Table. Double interpolation must be used when two of

the properties (eg. temperature and pressure) are not tabulated in the Steam

Tables. For example, to find the enthalpy of superheated steam at 25 bar and

320oC, an interpolation between 20 bar and 30 bar is necessary (as shown in

example 8.9). An interpolation between 300oC and 350

oC is also necessary.

Example 8.8


An extract from the Superheated Steam Tables:

t(oC)

p(bar)

300 320 350

20 3025 h1 3138

25 h

30 2995 h2 3117

Firstly, find the specific enthalpy (h1) at 20 bar and 320 oC;

At 20 bar,

300350

30253138

300320

30251

h

2.30701 h kJ/kg

T

h

350

320

300

3025 h1 3138


0.9 m3 of dry saturated steam at 225 kN/m

2 is contained in a rigid

cylinder. If it is cooled at constant volume process until the pressure

drops to180 kN/m2, determine the following:

a) mass of steam in the cylinder

b) dryness fraction at the final state

Sketch the process in the form of a P-v diagram.

Secondly, find the specific enthalpy (h2) at 30 bar and 320 oC;

At 30 bar,

300350

29953117

300320

29952

h

8.30432 h kJ/kg

Now interpolate between h1 at 20 bar, 320oC, and h2 at 30 bar, 320

oC in order

to find h at 25 bar and 320oC.

At 320oC,

20302025

121

hhhh

h

3070 2

25 20

30438 3070 2

30 20

. . .

h 3057 kJ/kg.

Example 8.9

T

h

350

320

300

2995 h2 3117

P

h

30

25

20

h1 h h2



Data: V1 = 0.9 m3

, P1 = 225 kN/m2 = 2.25 bar,

P2 = 180 kN/m

2 = 1.80 bar

a) Firstly, find the specific volume of dry saturated steam at 2.25 bar.

Note that the pressure 2.25 bar is not tabulated in the steam tables and

it is necessary to use the interpolation method.

From the Steam Tables,

vg at 2.2 bar = 0.8100 m3/kg

vg at 2.3 bar = 0.7770 m3/kg

vg1 at 2.25 bar,

20.230.2

8100.07770.0

20.225.2

8100.01

gv

vg1 0.7935 m3/kg

Mass of steam in cylinder,

1

1

gv

Vm (m

3 x kg/m

3)

0 9

0 7935

.

.

= 1.134 kg

b) At constant volume process,

Initial specific volume = final specific volume

v1 = v2

x1vg1 at 2.25 bar = x2vg2 at 1.8 bar

1(0.7935) = x2 (0.9774)

9774.0

)7935.0(12 x

= 0.81

P

bar

1.80

2.25

v m3/kg

1

2

0.7935 0.9774

v1 = v2


8.8 Determine the specific enthalpy of steam at 15 bar and 275oC.

8.9 Determine the degree of superheat and entropy of steam at 10 bar and 380oC.

8.10 A superheated steam at 12.5 MN/m2 is at 650

oC. Determine its specific volume.

8.11 A superheated steam at 24 bar and 500oC expands at constant volume until the

pressure becomes 6 bar and the dryness fraction is 0.9. Calculate the changes in the

internal energy of steam. Sketch the process in the form of a P-v diagram.

Activity 8D


Feedback to Activity 8D

8.8

250300

29253039

250275

2925

h

2982h kJ/kg

8.9 Degree of superheat = 380oC – 179.9

oC

= 200.1oC

350400

301.7464.7

350380

301.7

s

3988.7s kJ/kg K

8.10 An extract from the superheated steam table:

t(oC)

p(bar) 600 650 700

120 3.159 x 10-2

v1 3.605 x 10-2

125 v

130 2.901 x 10-2

v2 3.318 x 10-2

T

h

300

275

250

2925 h 3039

T

s

400

380

350

7.301 s 7.464


Firstly, find the specific volume (v1) at 120 bar and 650 oC;

At 120 bar,

600700

10 x 915.310 x 605.3

600650

10 x 915.3 222

1

v

2

1 10 x 3.382 v m3/kg

Secondly, find the specific volume (v2) at 130 bar and 650 oC;

At 130 bar,

600700

10 x 901.210 x .3183

600650

10 x 901.2 222

2

v

v2 = 3.1095 x 10-2

m3/kg

Now interpolate between v1 at 120 bar, 650oC, and v2 at 130 bar, 650

oC in

order to find v at 125 bar and 650oC.

At 650oC,

120130120125

121

vvvv

120130

10 x 382.310 x .10953

120125

10 x 382.3 222

v

v = 3.246 x 10-2

m3/kg

T

v

700

650

600

3.159 x 10-2

v1 3.605 x 10-2

T

v

700

650

600

2.901 x 10-2

v2 3.318 x 10

-2

P

v

130

125

120

v v2 v1


8.11 Data: P1 = 24 bar

T1 = 500oC

P2 = 6 bar

x2 = 0.9

Firstly, find the initial internal energy at 24 bar, 500oC. Note that the

pressure 24 bar is not tabulated in the Superheated Steam Tables and it is

necessary to use the interpolation method to find the changes in the internal

energy of steam.

At 500oC,

2030

31163108

2024

31161

u

8.31121 u kJ/kg

Secondly, find the final internal energy at 6 bar where x = 0.9,

u2 = uf2 + x2( ug2 -uf2 )

= 669 + 0.9(2568 - 669)

= 2378.1 kJ/kg

The changes in the internal energy of steam is,

(u2 – u1) = 2378.1 – 3112.8

= - 734.7 kJ/kg

P

u

30

24

20

3116 u1 3108

P

bar

6

24

v m3/kg

1

2

221.8oC

v1 = v2

500oC

158.8oC 500

oC

221.8oC

158.8oC

v1 = v2






1. With reference to the Steam Tables,

i. determine the specific volume, specific enthalpy and specific internal

energy of wet steam at 15 bar with a dryness fraction of 0.9.

ii. determine the degree of superheat, specific volume and specific

internal energy of steam at 80 bar and enthalpy 2990 kJ/kg.

iii. complete the missing properties and a phase description in the

following table for water;

P

bar

t oC

x v

m3/kg

u

kJ/kg

h

kJ/kg

s

kJ/kg K

Phase

description

2.0 120.2 6.4

12.0 1 2784

175 354.6 0.9

200 425

2. With reference to the Steam Tables,

i. find the dryness fraction and specific entropy of steam at 2.9 bar and

specific enthalpy 2020 kJ/kg.

ii. determine the degree of superheat and internal energy of superheated

steam at 33 bar and 313oC.

iii. determine the enthalpy change for a process involving a dry saturated

steam at 3.0 MN/m2 which is superheated to 600

oC and carried out at

constant pressure.

SELF-ASSESSMENT



1. i. v = 0.11853 m3/kg

h = 2600 kJ/kg

u = 2419.8 kJ/kg

ii. degree of superheat = 55 oC

v = 2.994 x 10-2

m3/kg

u = 2750.48 kJ/kg

iii.

P

bar

t oC

x v

m3/kg

u

kJ/kg

h

kJ/kg

s

kJ/kg K

Phase

description

2.0 120.2 0.87 0.7705 2267 2421 6.4 Wet steam

12.0 188 1 0.1632 2588 2784 6.523 Dry sat.

steam

175 354.6 0.9 0.007146 2319.8 2448.1 5.0135 Wet steam

200 425 - 0.001147

2725.6 2955 5.753 Superheated

steam

2. i. x = 0.68

s = 5.2939 kJ/kg K

ii. Degree of superheat = 73.85oC

u = 2769 kJ/kg

iii. h2 – h1 = 879 kJ/kg


THE SECOND LAW OF THERMODYNAMICS J2006/9/1

THE SECOND LAW OF THERMODYNAMICS

OBJECTIVES

General Objective : To define and explain the Second Law of Thermodynamics and

perform calculations involving the expansion and compression of

steam and gases.


state the definition of the Second Law of Thermodynamics

explain the Second Law of Thermodynamics

estimate the efficiency of heat engine

explain entropy and entropy change

sketch processes on a temperature-entropy diagram

understand that Q = h2 – h1 and apply the formula in

calculations

calculate the change of entropy, work and heat transfer of

steam in reversible processes at:

i. constant pressure process

ii. constant volume process

iii. constant temperature (or isothermal) process

iv. adiabatic (or isentropic) process

v. polytropic process

UNIT 9


9.0 Introduction to The Second Law of Thermodynamics

According to the First Law of Thermodynamics as stated in Unit 2, when a system

undergoes a complete cycle, then the net heat supplied is equal to the net work done.

dQ = dW

This is based on the conservation of energy principle, which follows from the

observation on natural events.

The Second Law of Thermodynamics, which is also a natural law, indicates that;

In symbols,

Q1 – Q2 = W (9.1)

To enable the second law to be considered more fully, the heat engine must be

discussed.

Although the net heat supplied in a cycle is equal to the net work done, the

gross heat supplied must be greater than the work done; some heat must

always be rejected by the system.

…The Second Law of Thermodynamics

INPUT


9.1 The heat engine and heat pump

We know from experience that work can be converted to heat directly and

completely, but converting heat to work requires the use of some special devices.

These devices are called heat engines.

A heat engine is a system operating in a complete cycle and developing net work

from a supply of heat. The second law implies that a source of heat supply (or hot

reservoir) and a sink (or cold reservoir) for the rejection of heat are both necessary,

since some heat must always be rejected by the system.

Heat engines differ considerably from one another, but all can be characterised by

the following:

They receive heat from a high-temperature source (for example solar energy, oil

furnace, nuclear reactor, steam boiler, etc.)

They convert part of this heat to work (usually in the form of a rotating shaft, for

example gas turbine, steam turbine, etc.)

They reject the remaining waste heat to a low-temperature sink (for example the

atmosphere, rivers, condenser, etc.)

They operate on a cycle.

A diagrammatic representation of a heat engine is shown in Fig. 9.1-1.

High-temperature

HOT RESERVOIR

Low-temperature

COLD RESERVOIR

HEAT

ENGINE

Q1

Q2

WORK OUTPUT

W = Q1 - Q2

Note:

Q1 = The heat supplied from the

source.

W = The net work done.

Q2 = The heat rejected.

Figure 9.1-1 Part of the heat received by the heat engine is converted to work,

while the rest is rejected to cold reservoir.


Heat engines and other cyclic devices usually involve a fluid that moves to and fro

from which heat is transferred while undergoing a cycle. This fluid is called the

working fluid.

The work-producing device that best fits into the definition of a heat engine are:

The steam power plant

The close cycle gas turbine

By the first law, in a complete cycle,

Net heat supplied = Net work done

Referring to Fig. 9.1-1, from equation dQ = dW, we have,

Q1 – Q2 = W

By the second law, the gross heat supplied must be greater than the net work done,

i.e. Q1 > W

The thermal efficiency of a heat engine is defined as the ratio of the net work done in

the cycle to the gross heat supplied in the cycle. It is usually expressed as a

percentage.

Referring to Fig. 9.1-1,

Thermal efficiency, W

Q1

(9.2)

Substituting equation 9.1,

Q Q

Q

1 2

1

1

2

1

Q

Q (9.3)

It can be seen that the second law implies that the thermal efficiency of a heat engine

must always be less than 100% (Q1 > W ).


Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the

rate of waste heat rejection to a nearby river is 45 MW, determine the net work

done and the thermal efficiency for this heat engine.

Example 9.1


A schematic of the heat engine is given in the diagram above. The furnace serves as

the high-temperature reservoir for this heat engine and the river as the low-

temperature reservoir.

Assumption: Heat lost through the pipes and other components are negligible.

Analysis: The given quantities can be expressed in rate form as;

Q1 = 80 MW

Q2 = 45 MW

From equation 9.1, the net work done for this heat engine is;

W = Q1 – Q2

= (80 – 45) MW

= 35 MW

Then from equation 9.2, the thermal efficiency is easily determined to be

43.75%)(or 0.4375MW 80

MW 35

1

Q

W

That is, the heat engine converts 43.75 percent of the heat it receives to work.

FURNACE

RIVER

HEAT

ENGINE

Q1 = 80 MW

Q2 = 45 MW

W = ?


The first and second laws apply equally well to cycles working in the reverse

direction to those of heat engine. In general, heat only flows from a high-

temperature source to a low-temperature sink. However, a reversed heat engine can

be utilized to pump the heat from a low-temperature region to a high-temperature

region. The reversed heat engine is called heat pump.

In the case of a reversed cycle, the net work done on the system is equal to the net

heat rejected by the system. Such cycles occur in heat pumps and refrigerators. The

equivalent diagram of the heat pump (or refrigerator) is shown in Fig. 9.1-2.

In the heat pump (or refrigerator) cycle, an amount of heat, Q2, is supplied from the

cold reservoir, and an amount of heat, Q1, is rejected to the hot reservoir.

Q1 = W + Q2 (9.4)

In the second law, we can say that work input is essential for heat to be transferred

from the cold to the hot reservoir,

i.e. W > 0

The first law sets no limit on the percentage of heat supplied, which can be

converted into work. Nor does it indicate whether the energy conversion process is

physically possible or impossible. We shall see, though, that a limit is imposed by

the Second Law of Thermodynamics, and that the possibility or otherwise of a

process can be determined through a property of the working fluid called entropy.

High-temperature

HOT RESERVOIR

Low-temperature

COLD RESERVOIR

HEAT

PUMP

Q1

Q2

WORK INPUT

W

Q1 = W + Q2

Figure 9.1-2 Reverse heat engine


9.2 Entropy

The first law applied to a heat engine or energy conversion process is merely an

energy balance. However, the first law does not indicate the possibility or

impossibility of the process, and we know from our everyday experience that some

energy conversions are never observed. In Unit 2, internal energy which is an

important property, arised as a result of the First Law of Thermodynamics. Another

important property, entropy, follows from the second law.

Considering 1 kg of fluid, the units of entropy are given by kJ/kg divided by K. The

the unit of specific entropy, s, is kJ/kg K. The symbol S will be used for the entropy

of mass, m, of a fluid,

i.e. S = ms kJ/K

The change of entropy is more important than its absolute value, and the zero of

entropy can be chosen quite arbitrarily. For example, in the Steam Tables the

specific entropy at saturated liquid is put equal to zero at 0.01oC; in tables of

refrigerants the specific entropy at saturated liquid is put equal to zero at – 40oC.

For all working substances, the change of entropy is given by

dsdQ

T (9.5)

Re-writing equation 9.5 we have

dQ = T ds

or for any reversible process

2

1d sTQ (9.6)

The equation 9.7 below is analogous to equation 9.6 for any reversible process

2

1d vPW (9.7)

Thus, as there is a diagram on areas that represent work done in a reversible process,

there is also a diagram on areas that represent heat flow in a reversible process.

These diagrams are the P-v and the T-s diagrams respectively, as shown in

Figs. 9.2-1 and 9.2-2.


For a reversible process from point 1 to point 2:

in Fig. 9.2-1, the shaded area 2

1d vP , represents work done; and

in Fig. 9.2.2, the shaded area 2

1d sT , represents heat flow.

Therefore, one great use of property entropy is that it enables a diagram to be drawn

showing the area that represents heat flow in a reversible process. In the next input,

the T-s diagram will be considered for a steam.

1

2

P

v dv

P

1

2

T

s ds

T

Figure 9.2.1 Work done Figure 9.2.2 Heat flow


9.3 The T-s diagram for a steam

As mentioned earlier, the zero specific entropy is taken as 0.01oC for steam and

– 40oC for refrigerants only. The T-s diagram for steam is considered here. The

diagram for a refrigerant is exactly similar to the T-s diagram for steam. However,

the zero specific entropy is different.

The T-s diagram for steam in Fig. 9.3-1, shows:

Three lines of constant pressure P1, P2 and P3 ( i.e. ABCD, EFGH and JKLM).

The pressure lines in the liquid region are practically coincident with the

saturated liquid line (i.e. AB, EF and JK), and the difference is usually neglected.

The pressure remains constant with temperature when the latent heat is added,

hence the pressure lines are horizontal in the wet region (i.e. portions BC, FG

and KL).

The pressure lines curve upwards in the superheat region (i.e. portions CD, GH

and LM). Thus the temperature rises as heating continues at constant pressure.

Two constant volume lines (v1 and v2 shown as chain-dotted) are also drawn in

Fig. 9.3-1. The lines of constant volume are concaved down in the wet region

and slope up more steeply than pressure lines in the superheat region.

P3 > P2 > P1

T

s

P3

P2

P1 v1

v2

A B C

D

E

F G

H

J

K L

M

273 K

v1 > v2

Figure 9.3-1 Temperature-entropy diagram for liquid

and steam


In steam tables, the specific entropy of the saturated liquid and the dry saturated

steam are represented by sf and sg respectively. The difference, sg - sf = sfg is also

tabulated. The specific entropy of wet steam is given by the specific entropy of

water in the mixture plus the specific entropy of the dry steam in the mixture.

For wet steam with dryness fraction, x, we have

s = sf(1 – x) + xsg

or s = sf + x(sg – sf )

i.e. s = sf + xsfg (9.8)

Then the dryness fraction is given by

It can be seen from equation 9.9 that the dryness fraction is proportional to the

distance of the state point from the liquid line on a T-s diagram. For example, for

state 1 on Fig. 9.3-2, the dryness fraction,

The area under the line FG on Fig. 9.3-2 represents the latent heat hfg. The area under

line F1 is given by x1hfg.

In unit 8, the enthalpy of wet steam was shown to be given by equation 8.4,

h= hf + xhfg

fg

f

s

ssx

(9.9)

1

11

1FG distance

F1 distance

fg

f

s

ssx

T

(K)

s kJ/kg K s1 sg

s1 = sf1 + x1sfg1

F

sf

G 1 P1

P1

Figure 9.3-2 Temperature-entropy chart for steam


9.4 To show that Q = h2 – h1

The T-s diagram will enable equation Q = h2 – h1 to be expressed graphically, while

areas on the diagram represent heat flow. Assuming that the pressure line in the

liquid region coincides with the saturated liquid line, then the enthalpy can be

represented on the diagram.

Figure 9.4 T-s diagram showing Q = h2 – h1

Referring to Fig. 9.4:

When water at any pressure P, at 0.01oC, is heated at constant pressure, it

follows the line AB approximately; the point B is at the saturation temperature T

at which the water boils at the pressure P. At constant pressure from A to B

Q = hB - hA

= hB (since hA at 0.01oC is approximately zero)

Therefore, we have

area ABFOA = hB

= hf at pressure P

At point B, if heating continues, water will change gradually into steam until

point C and the steam becomes dry saturated. Thus we have

area BCHFB = latent heat

= hfg at pressure P

= hC - hB

At point C, the enthalpy is given by

hC = area ABFOA + area BCHFB

= hg at pressure P

A

B E C

D

P

T

(K)

0 F G H J s kJ/kg K


For wet steam at point E,

hE = hB + xEhfg

i.e hE = area ABEGOA

When dry saturated steam is further heated, it becomes superheated. The heat

added from C to D at constant pressure P, is given by

Q = hD – hC

= area CDJHC

Then the enthalpy at D is

hD = hC + area CDJHC

= area ABCDJOA



NEXT INPUT

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

TRUE (T) or FALSE (F).

STATEMENTS TRUE or FALSE

i. The Second Law of Thermodynamics is

represented by the equation Q1 – Q2 = W.

ii. The heat engine receives heat from a high-

temperature source.

iii. The heat engine convert part of the heat to

internal energy.

iv. The work-producing device of a heat engine

are the steam power plant and a close cycle

gas turbine.

v. A reversed heat engine is called a heat pump.

vi. The work producing device for a heat pump is

the refrigerator.

vii. In heat engines, the net work done must be

greater than the gross heat supplied,

i.e W > Q1 .

9.2 The work done by heat engine is 20 kW. If the rate of heat that enters into

the hot reservoir is 3000 kJ/min, determine the thermal efficiency and the rate

of heat rejection to the cold reservoir.

Activity 9A



9.1 i. True

ii. True

iii. False

iv. True

v. True

vi. True

vii. False

9.2 The given quantities can be expressed as;

W = 20 kW

Q1 = 3000 kJ/min = 50 kJ/s or kW

From equation 9.2

W

Q1

i.e. Thermal efficiency of heat engine is 40 %.

From equation 9.1

W = Q1 – Q2

Q2 = Q1 – W = 50 – 20 = 30 kW

i.e. The rate of heat rejection to the cold reservoir is 30 kW.


PROCEED TO THE NEXT INPUT…

% 40% 100 x 50

20


9.5 Reversible processes on the T-s diagram for steam

In the following sections of this unit, five reversible processes on the T-s diagram for

steam are analysed in detail. These processes include the:






9.5.1 Constant pressure process

The constant pressure process is a good approximation to many of the

common physical processes which we are familiar with. The combustion of

fuel in a boiler, the flow of fluids, the flow of air in ducts and other processes

can be used to illustrate constant pressure.

For example, Fig. 9.5.1 shows a reversible constant pressure process from a

wet steam into the superheat region. It can be seen that, when the boundary

of the system is inflexible, the pressure rises when heat is supplied. For a

constant pressure process, the boundary must move against an external

resistance as heat is supplied.

INPUT

T

s

1

2

s1 s2

Q

P1 = P2

Figure 9.5.1 Constant pressure process

T1

T2


4 kg of steam at 7 bar and entropy 6.5 kJ/kg K, is heated reversibly at

constant pressure until the temperature is 250 oC. Calculate the heat

supplied and show on a T-s diagram the area which represents the heat

flow.

During the constant pressure process, since P is constant,

12

2

1vvPdvPW (9.10)

(Note that this equation was derived and used in Unit 5)

From the non-flow energy equation,

Q = (u2 – u1) + W

Hence for a reversible constant pressure process

Q = (u2 – u1) + P(v2 – v1) = (u2 + Pv2) –( u1 + Pv1)

Now from equation h = u + Pv,

Q = h2 - h1 kJ/kg (9.11)

or for mass, m (kg), of a fluid,

Q = m(h2 - h1)

Q = H2 - H1 kJ (9.12)

Example 9.2


The given quantities can be expressed as;

m = 4 kg

P2 = P1 = 7 bar

s1 = 6.5 kJ/kg K

T2 = 250oC

At state 1

At 7 bar, sg1 = 6.709 kJ/kg K, the steam is wet, since the actual entropy, s1, is

less than sg1 (i.e. s1< sg1).


From equation 9.9


h1 = hf1 + x1hfg1

= 697 + 0.956(2067)

= 2673 kJ/kg

At state 2

The steam is at 250 oC at 7 bar, and therefore superheated. From the

superheated tables,

h2 = 2955 kJ/kg

At constant pressure, from equation 9.11

Q = h2 - h1

= 2955 - 2673

= 282 kJ/kg

Hence for 4 kg of steam,

Q = 4 kg x 282 kJ/kg

= 1128 kJ

i.e. For 4 kg of steam, the heat supplied is 1128 kJ.

The T-s diagram of the process is given below. The shaded area represents

the heat flow.

956.0717.4

992.15.6

1

11

1

fg

f

s

ssx

s (kJ/kg K)

T

(K)

1

2

s1

6.709

s2

Q

P1 = P2 = 7 bar

T2 = 250 oC

= 523 K


9.5.2 Constant volume process

In a constant volume process, the working substance is contained in a rigid

vessel (or closed tank) from which heat is either added or removed. In this

process, the boundaries of the system are immovable and no work can be

done on or by the system. It will be assumed that ‘constant volume’ implies

zero work unless stated otherwise.

From the non-flow energy equation,

Q = (u2 – u1) + W

Since no work is done, we therefore have

Q = u2 – u1 kJ/kg (9.13)

or for mass, m (kg), of the working substance

Q = m(u2 – u1)

Q = (U2 – U1) kJ (9.14)

Note that, all the heat supplied in a constant volume process goes to

increasing the internal energy.

In Fig. 9.5.2, the indicated path is one in which heat is being added in a

constant volume process.

Figure 9.5.2 Constant volume process

T

s

1

2

s1 s2

Q

P1

P2

T2

T1


A wet steam at 10 bar is heated reversibly at constant volume to a pressure

of 20 bar and 250 oC. Calculate the heat supply (in kJ/kg) and show the

process on a T-s diagram, indicating the area that represents the heat flow.

Example 9.3


At state 2

Steam at 20 bar and 250 oC is superheated. From the superheated steam

tables, we have

specific volume, v2 = 0.1115 m3/kg

specific internal energy, u2 = 2681 kJ/kg

At state 1

At 10 bar, we have

v1 = v2 = 0.1115m3/kg

vg1 = 0.1944

Steam at state 1 is wet as v1< vg1, and the dryness fraction is given by

equation 8.2,

v1 = x1 vg1

xv

v1

1 01115

019440 574

g1

.

..

From equation 8.5,

u1 = uf1 + x1(ug1 – uf1)

= 762 + 0.574 (2584 - 762)

= 1807.8 kJ/kg

At constant volume from equation 9.13,

Q = u2 – u1

= 2681 – 1807.8

= 873.2 kJ/kg


The T-s diagram showing of constant volume process is given below. The

shaded area represents the heat flow.

T

s

1

2

s1 s2

Q

P1 = 10 bar

P2 = 20 bar

250 oC


9.5.3 Constant temperature (or isothermal) process

A process that takes place at constant temperature is called an isothermal

process. For example, at the exhaust of a steam turbine, the steam is usually

wet. This steam is subsequently condensed in a unit appropriately known as

a condenser. As the steam is initially wet, this process is carried out

essentially at constant temperature (isothermally).

In an isothermal expansion, heat must be added continuously in order to keep

the temperature at the initial value. Similarly in an isothermal compression,

heat must be removed from the fluid continuously during the process.

A reversible isothermal process will appear as a straight line on a T-s

diagram, and the area under the line must represent the heat flow during the

process. Figure 9.2.3 shows a reversible isothermal expansion of wet steam

into the superheat region.

The shaded area represents the heat supplied during the process.

Q = T(s2 - s1) … T in Kelvin (or K) (9.15)

Note that the absolute temperature must be used. The temperature tabulated in the

steam tables is t oC, and care must be taken to convert this into T Kelvin.

From the non-flow energy equation, work can be expressed by

W = Q - (u2 – u1) (9.16)

Figure 9.5.3 Isothermal process

Q

T

s

1 2

s1 s2

P1

P2

T1 = T2


Steam at 80 bar and enthalpy 2650 kJ/kg expands isothermally and

reversibly to a pressure of 10 bar. Calculate the entropy change, heat

supplied and the work done per kg steam during the process. Show the

process on a T-s diagram, indicating the area that represents the heat flow.

Example 9.4



P1 = 80 bar

h1 = 2650 kJ/kg

P2 = 10 bar

At state 1

At 80 bar, h1 = 2650 kJ/kg, the steam is wet, since the given enthalpy, h1, is

less than hg (i.e. 2758 kJ/kg). From the steam tables, the saturated

temperature of wet steam is 295 oC.

From equation 8.4

h1 = hf1 + x1hfg1

2650 = 1317 + x1(1441)

x1 = 0.925

From equation 9.8

s1 = sf1 + x1sfg1

= 3.207 + 0.925(2.537)

= 5.554 kJ/kg K

From equation 8.5

u1 = uf1 + x1( ug1 - uf1 )

= 1306 + 0.925(2570 - 1306)

= 2475.2 kJ/kg


At state 2

At 10 bar and 295 oC the steam is superheated, hence interpolating

s2 6 926

295 250

7124 6 926

300 250

. . .

s2 = 7.1042 kJ/kg K

u2 2711

295 250

2794 2711

300 250

u2 = 2785.7 kJ/kg

Change of entropy,

(s2 – s1) = 7.1042 - 5.554 kJ/kg K

= 1.5502 kJ/kg K

Then from equation 9.15 we have,

Heat supplied = shaded area

Q = T(s2 - s1)

= 568(1.5502)

= 880.5 kJ/kg

(where T = 295 + 273 = 568 K)

From equation 9.16,

W = Q - (u2 – u1)

= 880.5 - (2785.7 - 2475.2)

= 570 kJ/kg

T

s

300

295

250

6.926 s2 7.124

T

u

300

295

250

2711 u2 2794


The T-s diagram of the isothermal process is given below. The shaded area

represents the heat flow.

Q

T

(K)

s (kJ/kg K)

1 2

s1 s2

P1 = 80 bar

P2 = 10 bar

T1 = T2

295OC

@ 568 K


TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT

INPUT…!

9.3 A rigid cylinder (constant volume) contains steam at 90 bar and 400 oC. The

cylinder is cooled until the pressure is 50 bar. Calculate the amount of heat

rejected per kg of steam. Sketch the process on a T-s diagram indicating the

area, which represents the heat flow.

9.4 Steam at 8 bar, entropy 6.211 kJ/kg K is heated reversibly at constant

pressure until the temperature is 350 oC. Calculate the heat supplied, and

show on a T-s diagram the area which represents the heat flow.

9.5 Dry saturated steam at 100 bar expands isothermally and reversibly to a

pressure of 10 bar. Calculate the heat supplied and the work done per kg of

steam during the process. Show the process on a T-s diagram.

Activity 9B




P1 = 90 bar

T1 = 400 oC

P2 = 50 bar

At state 1

Steam at 90 bar and 400 oC is superheated, and the specific volume from the Steam

Tables is,

v1 = 0.02991 m3/kg.

For superheated steam above 70 bar, the internal energy is not tabulated in the

superheated steam tables and it is found from equation 8.6 that,

u1 = h1 – p1v1

= 3118 – (90 x 102 x 2.991 x 10

-2)

= 2848.8 kJ/kg

At state 2

At P2 = 50 bar and v2 = 0.02991 m3/kg, the steam is wet, and the dryness fraction is

given by equation 8.2 as:

v2 = x2vg2

From equation 8.5

u2 = uf2 + x2(ug2 - uf2)

= 1149 + 0.758(2597 – 1149)

= 2246.6 kJ/kg

758.003944.0

02991.0

2

22

gv

vx


At constant volume from equation 9.13

Q = u2 – u1

= 2246.6 - 2848.8

= - 602.2 kJ/kg

i.e. The amount of heat rejected per kg of steam is 602.2 kJ/kg.


P1 = P2 = 8 bar

s1 = 6.211 kJ/kg K

T2 = 350 oC

At state 1

At 8 bar, sg1 = 6.663 kJ/kg K, the steam is wet, since the actual entropy, s1, is less

than sg1 (i.e. s1< sg1).

From equation 9.9


h1 = hf1 + x1hfg1

= 721 + 0.9(2048)

= 2564.2 kJ/kg

T

s

2

1

s2 s1

Q

P2 = 50 bar

P1 = 90 bar

400 oC

9.0617.4

046.2211.6

1

11

1

fg

f

s

ssx


At state 2

The steam is at 350 oC at 8 bar, and therefore superheated. From the superheated

tables,

h2 = 3162 kJ/kg

At constant pressure, from equation 9.11

Q = h2 - h1

= 3162 - 2564.2

= 597.8 kJ/kg

i.e. The heat supplied is 597.8 kJ/kg

The T-s diagram of the process is given below. The shaded area represents the heat

flow.

s (kJ/kg K)

T

(K)

1

2

s1

6.211

s2

Q

P1 = P2 = 8 bar

T2 = 350 oC = 523 K


9.5 From the Steam Tables at 100 bar, steam is dry saturated,

s1 = sg1 = 5.615 kJ/kg K and t1 = 311oC

At 10 bar and 311oC the steam is superheated, hence interpolating

300350

124.7301.7

3000311

124.72

s

s2 = 7.163 kJ/kg K

Then we have,

Heat supplied = shaded area

i.e. Q = T(s2 – s1)

= 584 (7.163 – 5.615)

= 584 x 1.548

= 904 kJ/kg

(where T = 311 + 273 = 584 K)

T

s

350

311

300

7.124 s2 7.301

Q

T

(K)

s (kJ/kg K)

1 2

s1 s2

P1 = 100 bar

P2 = 10 bar

T1 = T2

(311 + 273) =584 K


To find the work done it is necessary to apply the equation,

W = Q - (u2 – u1)

From the Steam Tables, at 100 bar, steam is dry saturated

u1 = ug = 2545 kJ/kg

At 10 bar and 311oC, interpolating becomes

300350

27942875

300311

27942

u

u2 = 2811.8 kJ/kg

Then,

W = Q - (u2 – u1)

= 904 – (2811.8 – 2545)

= 637.2 kJ/kg

i.e. Work done by the steam = 637.2 kJ/kg


PROCEED TO THE NEXT INPUT…

T

u

350

311

300

2794 u2 2875


9.5.4 Reversible adiabatic (or isentropic) process

For a reversible adiabatic process, the entropy remains constant, and hence

the process is called an isentropic process. During this process, no heat is

transferred to or from the fluid and this process will always appear as a

vertical line on a T-s diagram.

For an isentropic process,

s1 = s2

From the non-flow equation,

Q - W = (u2 – u1)

and for an isentropic process

Q = 0

Therefore, we have

W = u1 -u2 kJ/kg (9.17)

In an isentropic process, all the work done to the fluid can be found in

equation 9.17 by evaluating u1 and u2 from the Steam Tables.

An isentropic process for superheated steam expanding into the wet region is

shown in Fig. 9.5.4.

INPUT

T

s

1

2

s1 = s2

P1

P2

Figure 9.5.4 Adiabatic (or isentropic) process


Steam at 30 bar, 250 oC expands isentropically in a cylinder behind a piston

to a pressure 10 bar. Calculate the work done per kg of steam during the

process. Show the process on a T-s diagram.

Example 9.5


At state 1

From the superheat tables, at 30 bar, 250 oC, we have

s1 = 6.289 kJ/kg K

u1 = 2646 kJ/kg

At state 2

For an isentropic process s1 = s2, therefore we have

s1 = s2 = 6.289 kJ/kg K

At 10 bar and s2 = 6.289 kJ/kg K, the steam is wet, since s2 is less than sg2

(i.e. sg2 = 6.586 kJ/kg K).


From equation 8.5

u2 = uf2 + x2(ug2 – uf2)

= 762 + 0.933 (2584 - 762)

= 2461.9 kJ/kg

For an adiabatic process, from equation 9.17

W = u1 -u2

= 2646 - 2461.9

= 184.1 kJ/kg

i.e. Work done by the steam is 184.1 kJ/kg.

933.0448.4

138.2289.6

2

22

2

fg

f

s

ssx

T

s

1

2

s1 = s2

P1 = 30 bar

P2 = 10 bar

250 oC


9.5.5 Polytropic process

It is found that many processes in practice follows the reversible law in the

form pvn = constant, where n is constant. Both steam and perfect gases obey

this type of law closely in many processes. Such processes are internally

reversible.

For a polytropic process, the equations below are considered

Work done, Wp v p v

n

1 1 2 2

1 (9.18)

Heat transfer, Q = (u2 - u1) + W (9.19)

Equation 9.18 is true for any working substance undergoing a reversible

polytropic process. It follows also that for any polytropic process, we can

write

p1v1n

= p2v2n

= C

p

p

v

v

n

1

2

2

1

;

v

v

p

p

n2

1

1

2

1

T

T

p

p

n

n2

1

2

1

1

; T

T

v

v

n

2

1

1

2

1

A polytropic process for wet steam expanding from a high pressure to a low

pressure is shown in Fig. 9.5.5. To find the change of entropy in a polytropic

process for a steam when the end states have been fixed using p1v1n

= p2v2n,

the entropy values at the end states can be read straight from the tables.

Figure 9.5.5 Polytropic process

T

s

1

2

s2

P1

P2

s1

Q


In a steam engine, the steam at the beginning of the expansion process is at

10 bar and dryness fraction 0.9. The expansion follows the law

pv1.1

= constant, down to a pressure of 0.4 bar. Calculate the change of

entropy and work done per kg of steam during the process. Show the

process on a T-s diagram.

Example 9.6


At 10 bar, from equation 9.8

s1 = sf1 + x1sfg1

= 2.138 + 0.9 (4.448)

= 6.1412 kJ/kg K

At 10 bar, vg1 = 0.1944 m3/kg, then from equation 8.2

v1 = x1 (vg1)

= 0.9 (0.1944)

= 0.175 m3/kg


1

2

1

1

2

p

p

v

v, we have

/kgm 3.2654.0

10175.0 3

1.1

1

1.1

1

2

112

p

pvv

At 0.4 bar, and v2 = 3.265 m3/kg, the steam is wet, since vg2 = 3.992 m

3/kg .

From equation 8.2

v2 = x2 ( vg2 )

xv

vg

2

2

2

3 265

3 9920 82

.

..



s2 = sf2 + x2sfg2

= 1.026 + 0.82 (6.643)

= 6.4733 kJ/kg K

Change of entropy,

(s2 – s1) = 6.4733 - 6.1442

= 0.3321 kJ/kg K

i.e. Increase in entropy, (s2 – s1) is 0.3321 kJ/kg K.

Hence work done by the steam, from equation 9.18

Wp v p v

n

1 1 2 2

1

1.0

6.130175

11.1

3.265)x 10 x (0.40.175)x x10(10 22

= 444 kJ/kg

i.e. Work done by the steam is 444 kJ/kg.

T

s

1

2

s2

P1 = 10 bar

P2 = 0.4 bar

s1

pv1.1

= C


9.6 In a steam engine, steam at 110 bar, 400oC expands isentropically in a

cylinder behind a piston until the pressure is 3 bar. If the work output during

the expansion process is 165.5 kJ/kg, determine the final temperature of the

steam. Show the process on a T-s diagram.

9.7 In the cylinder of a steam engine, wet steam expands from 8 bar, dryness

fraction 0.87 to 0.5 bar according to a law pv1.02

= C. Determine the per kg

of steam for the following:

i. change of entropy

ii. work done

iii. heat flow to or from the cylinder walls

Show the process on a T-s diagram.

Activity 9C


Feedback To Activity 9C


P1 = 110 bar

t1 = 400 oC

P2 = 3 bar

W = 165.5 kJ/kg

Steam at 110 bar and 400 oC is at superheated region. The property tables for this

condition do not list down the specific internal energy (u) and therefore it must be

calculated from

u1 = h1 – p1v1

From the Superheated Steam Tables, at 110 bar and 400 oC

h1 = 3075 kJ/kg

v1 = 2.350 x 10-2

m3/kg

Hence, u1 = 3075 – (110 x 102 - 2.350 x 10

-2)

= 2816.5 kJ/kg

For an adiabatic process, from equation 9.17

W = u1 - u2

u2 = u1 - W

= 2816.5 – 165.5

= 2651 kJ/kg

From the property tables for steam, ug at 3 bar is 2544 kJ/kg and hence the steam at

state 2 must still be at superheat region.

From the property tables for superheated steam, u2 = 2651 kJ/kg when the

temperature is 200 oC.

Hence, t2 = 200 oC


T

s

1

2

s1 = s2

P1 = 110 bar

P2 = 3 bar 400 oC

t2 = 200 oC

Note : In this activity, although there is no heat

transfer in an adiabatic process, the pressure,

volume and temperature of the working fluid are

changed. The work transfer during an adiabatic

process is equal to the change in the internal energy

of the fluid.


9.7 At 8 bar, we have

s1 = sf1 + x1sfg1

= 2.046 + 0.87 (4.617)

= 6.063 kJ/kg K

u1 = uf1 + x1(ug1 – uf1)

= 720 + 0.87 (2577- 720)

= 2335.6 kJ/kg

At 8 bar, vg1 = 0.2403 m3/kg, then from equation 8.2

v1 = x1 (vg1)

= 0.87 (0.2403)

= 0.2091 m3/kg

Then from equation02.1

1

2

1

1

2

p

p

v

v, we have

/kgm 3.1695.0

82091.0 3

02.1

1

02.1

1

2

112

p

pvv

At 0.4 bar, and v2 = 3.169 m3/kg, the steam is wet, since vg2 = 3.239 m

3/kg .

From equation 8.2

v2 = x2 ( vg2 )

978.0239.3

169.3

2

22

gv

vx

Then at 0.5 bar, we have

s2 = sf2 + x2sfg2

= 1.091 + 0.978 (6.502)

= 7.4499 kJ/kg K

u2 = uf2 + x2(ug2 – uf2)

= 340 + 0.978 (2483- 340)

= 2435.9 kJ/kg


i. Change of entropy,

(s2 – s1) = 7.4499 - 6.063

= 1.3869 kJ/kg K

i.e. Increase in entropy, (s2 – s1) is 1.3869 kJ/kg K.

ii. Work done by the steam,

Wp v p v

n

1 1 2 2

1

02.0

45.15828.167

11.02

3.169)x 10 x (0.50.2091)x x10(8 22

= 441.5 kJ/kg

i.e. Work done by the steam is 441.5 kJ/kg.

iii. Heat flow,

Q = (u2 – u1) + W

= (2435.9 – 2335.6) + (441.5)

= 541.8 kJ/kg

i.e. Heat flow from the cylinder walls is 541.8 kJ/kg.

T

s

1

2

s2

P1 = 8 bar

P2 = 0.5 bar

s1

pv1.02

= C





Good luck.

1. Heat is transferred to a heat engine from a hot reservoir at a rate of 120 MW.

If the net work done is 45 MW, determine the rate of waste heat rejection to a

cold reservoir and the thermal efficiency of this heat engine.

2. Steam at 7 bar, entropy 6.5 kJ/kg K is heated reversibly at constant pressure

until the temperature is 250 oC. Calculate the heat supplied per kg of steam

and show on a T-s diagram the area, which represents the heat flow.

3. Steam at 70 bar, 300 oC expands isentropically in a cylinder behind a piston

to a pressure 20 bar. During the process, determine the:

i. dryness fraction at final state

ii. initial and final specific internal energy

iii. work done per kg of steam


4. 0.05 kg of steam at 10 bar with dryness fraction 0.84 is heated reversibly in a

rigid vessel until the pressure is 20 bar. Calculate the:

i. change of entropy and

ii. heat supplied

Show the area, which represents the heat supplied on a T-s diagram.

5. Steam at 20 bar, 250 OC undergoes a reversible isothermal process to a

pressure of 30 bar. Calculate the heat flow per kg of steam and state whether

it is supplied or rejected. Sketch the process on a T-s diagram.

6. A steam engine which receives steam at 4 bar and dryness fraction 0.8 is

expanded according to the law pv1.05

= constant to a condenser pressure of 1

bar. Calculate the change of entropy per kg of steam during the expansion.

Sketch the process on a T-s diagram.

SELF-ASSESSMENT


7. Steam at 30 bar, 300oC expands isothermally and reversibly to a pressure of

0.75 bar. The steam is then compressed according to the law pv1.05

= constant,

until the pressure is 10 bar. Calculate per kg of steam the:

i. total change of entropy

ii. net heat flow

iii. net work done

Sketch the processes on a T-s diagram.


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

1. Q2 = 75 MW

= 37.5 %

2. Q = 274 kJ/kg

3. i. x2 = 0.896

ii. u1 = 2634 kJ/kg and u2 = 2423.9 kJ/kg

iii. W = 210.1 kJ/kg

4. i. 0.704 kJ/kg K

ii. 36.85 kJ

5. Q = -135 kJ/kg

6. (s2 – s1) = 0.381 kJ/kg K

7. ∑ ∆s = (s2 – s1) + (s3 – s2)

= (1.808) + (-0.938)

= 0.87 kJ/kg K

∑ Q = Q12 + Q23

= (1035.98) + (-573.15)

= 462.83 kJ/kg

∑ W = W12 + W23

= (975.98) + (-692.5)

= 283.48 kJ/kg



THE SECOND LAW OF THERMODYNAMICS

OBJECTIVES

General Objective : To define and explain the Second Law of Thermodynamics and

perform calculations involving the expansion and compression of

perfect gases.


sketch the processes on a temperature-entropy diagram

calculate the change of entropy, work and heat transfer of

perfect gases in reversible processes at:






UNIT 10


10.0 The P-V and T-s diagram for a perfect gas

Property diagrams serve as great visual aids in the thermodynamic analysis of

processes. We have used P-V and T-s diagrams extensively in the previous unit

showing steam as a working fluid. In the second law analysis, it is very helpful to

plot the processes on diagrams which coordinate the entropy. The two diagrams

commonly used in the second law analysis are the pressure-volume and temperature-

entropy.

Fig. 10.0-1 shows a series of constant temperature lines on a P-V diagram. The

constant temperature lines, T3 > T2 > T1 are shown.

Figure 10.0-1 The constant temperature lines on a P-V diagram for a perfect gas

Since entropy is a property of a system, it may be used as a coordinate, with

temperature as the other ordinate, in order to represent various cycles graphically. It

is useful to plot lines of constant pressure and constant volume on a T-s diagram for

a perfect gas. Since changes of entropy are of more direct application than the

absolute value, the zero of entropy can be chosen at any arbitrary reference

temperature and pressure.

INPUT

T3 > T2 > T1

P

V

T1

T2

T3

Constant temperature lines


Fig. 10.0-2 shows a series of constant pressure lines on a T-s diagram and Fig.10.0-3

shows a series of constant volume lines on a T-s diagram. It can be seen that the

lines of constant pressure slope more steeply than the lines of constant volume.

Note:

Fig. 10.0-2, shows the constant pressure lines, P3 > P2 > P1;

Fig. 10.0-3, shows the constant volume lines, v1 > v2 > v3.

As pressure rises, temperature also rises but volume decreases; conversely as the

pressure and temperature fall, the volume increases.

T

s

P1

P2

P3

Figure 10.0-2

Constant pressure lines on a T-s diagram

T

s

v2

v1

v3

Figure 10.0-3

Constant volume lines on a T-s diagram


10.1 Reversible processes on the T-s diagram for a perfect gas

The various reversible processes dealt with in Units 4 and 5 will now be considered

in relation to the T-s diagram. In the following sections of this unit, five reversible

processes on the T-s diagram for perfect gases are analysed in detail. These

processes include the:

i. constant pressure process,

ii. constant volume process,

iii. constant temperature (or isothermal) process,

iv. adiabatic (or isentropic) process, and

v. polytropic process.

10.1.1 Reversible constant pressure process

It can be seen from Fig. 10.1.1 that in a constant pressure process, the

boundary must move against an external resistance as heat is supplied; for

instance a fluid in a cylinder behind a piston can be made to undergo a

constant pressure process.

During the reversible constant pressure process for a perfect gas, we have

The work done as

W = P(V2 – V1) kJ (10.1)

or, since PV = mRT , we have

W = mR(T2 - T1) kJ (10.2)

The heat flow is,

Q = mCp(T2 – T1) kJ (10.3)

The change of entropy is, then

S2 – S1 = mCp lnT

T

2

1

kJ/K (10.4)

T

s

P1= P2

v2

v1

1

2

s1 s2

Q

Figure 10.1.1 Constant pressure process on a T-s diagram


Nitrogen (molecular weight 28) expands reversibly in a cylinder behind a

piston at a constant pressure of 1.05 bar. The temperature is initially at

27oC. It then rises to 500

oC; the initial volume is 0.04 m

3. Assuming

nitrogen to be a perfect gas and take Cp = 1.045 kJ/kg K, calculate the:

g) mass of nitrogen

h) work done by nitrogen

i) heat flow to or from the cylinder walls during the expansion

j) change of entropy

Sketch the process on a T-s diagram and shade the area which represents

the heat flow.

or, per kg of gas we have,

s2 – s1 = Cp lnT

T

2

1

kJ/kg K (10.5)

Example 10.1



T1 = 27 + 273 K = 300 K

P1 = P2 = 1.05 bar (constant pressure process)

V1 = 0.04 m3

T2 = 500 + 273 = 773 K

M = 28 kg/kmol

Cp = 1.045 kJ/kg.K

a) From equation 3.10, we have

kg 0.0471300 x 0.297

0.04x 10 x 1.05

have we,= sinceThen

K kJ/kg 0.29728

3144.8

2

1

11

RT

VPm

mRTPV

M

RR o


b) the work done by nitrogen can be calculated by two methods. Hence,

we have

Method I:

From equation 10.2, work done

W = mR(T2 - T1)

= 0.0471 x 0.297 (773 - 300)

= 6.617 kJ

Method II:

For a perfect gas at constant pressure, 2

2

1

1

T

V

T

V

kJ 6.615

0.04)-(0.10310 x 1.05

)(

done work 10.1,equation From

m 0.103300

773 0.04

2

12

3

1

212

VVPW

T

TVV

c) From equation 10.3, heat flow

Q = mCp(T2 - T1)

= 0.0471 x 1.045 (773 - 300)

= 23.28 kJ

d) From equation 10.4, change of entropy

s2 - s1 = mCp lnT

T

2

1

0.0471 x 1.045 ln 773

300

kJ / K0 0466.

The T-s diagram below shows the constant pressure process. The shaded

area represents the heat flow.

T

s

P1 = P2 = 1.05 bar

v2 = 0.103 m3

v1 = 0.04 m3

1

2

s1 s2

Q

T1 = 300 K

T2 = 773 K


10.1.2 Reversible constant volume process

In a constant volume process, the working substance is contained in a rigid

vessel (or closed tank) from which heat is either added or removed. It can be

seen from Fig. 10.1.2 that in a constant volume process, the boundaries of the

system are immovable and no work can be done on or by the system. It will

be assumed that ‘constant volume’ implies zero work unless stated otherwise.

During the reversible constant volume process for a perfect gas, we have

The work done, W = 0 since V2 = V1.

The heat flow

Q = mCv(T2 – T1) kJ (10.6)

The change of entropy is therefore

S2 – S1 = mCv lnT

T

2

1

kJ/K (10.7)

or, per kg of gas we have,

s2 – s1 = Cv lnT

T

2

1

kJ/ kg K (10.8)

T

s

v1 = v2

P1

1

2

s1 s2

Q

P2

Figure 10.1.2 Constant volume process on a T-s diagram


Air at 15oC and 1.05 bar occupies a volume of 0.02 m

3. The air is heated at

constant volume until the pressure is at 4.2 bar, and then it is cooled at

constant pressure back to the original temperature. Assuming air to be a

perfect gas, calculate the:

a) mass of air

b) net heat flow

c) net entropy change

Sketch the processes on a T-s diagram.

Given:

R = 0.287 kJ/kg K, Cv = 0.718 kJ/kg K and Cp = 1.005 kJ/kg K.

Example 10.2



T1 = 15 + 273 K = 288 K

P1 = 1.05 bar

Process 1 - 2 (constant volume process): V1 = V2 = 0.02 m3

Process 2 - 3 (constant pressure process) : P2 = P3 = 1.05 bar T3 = T1 = 288 K

a) From equation 3.6, for a perfect gas,

kg 0.0254288 x 0.287

0.02 x 10 x 1.05 2

1

11 RT

VPm

b) For a perfect gas at constant volume, 2

2

1

1

T

P

T

P , hence

K 115205.1

2.4288

1

212

P

PTT

From equation 10.6, at constant volume

Q12 = mCv(T2 – T1) = 0.0254 x 0.718 (1152 – 288) = 15.75 kJ

From equation 10.3, at constant pressure

Q23 = mCp(T3 – T2) = 0.0254 x 1.005 (288 – 1152) = -22.05 kJ

Net heat flow = Q12 + Q23 = (15.75) + ( -22.05) = -6.3 kJ


c) From equation 10.7, at constant volume

S2 – S1 = mCv lnT

T

2

1

From equation 10.4, at constant pressure

S3 – S2 = mCp

2

3lnT

T

Net entropy change, (S3 – S1) = (S2 – S1) + (S3 – S2)

= (0.0253) + (-0.0354)

= - 0.0101 kJ/K

i.e. decrease in entropy of air is 0.0101 kJ/K.

Note that since entropy is a property, the decrease of entropy in example

10.2, given by (S3 – S1) = (S2 – S1) + (S3 – S2), is independent of the

processes undergone between states 1 and 3. The change (S3-S1) can also

be found by imagining a reversible isothermal process taking place between

1 and 3. The isothermal process on the T-s diagram will be considered in

the next input.

kJ/K 0253.0

288

1152ln 0.718 x 0.0254

kJ/K 0354.0

1152

288ln 1.005 x 0.0254

T

s

P2 = P3 = 4.2 bar

v1 = v2 = 0.02 m3

v3

3

2

s3 s2

T1 = T3 = 288 K

T2 = 1152 K

1

P1 = 1.05 bar

s1


10.1.3 Reversible constant temperature (or isothermal) process

A reversible isothermal process for a perfect gas is shown on a T-s diagram

in Fig. 10.1.3. The shaded area represents the heat supplied during the

process,

i.e. Q = T(s2 - s1) (10.9)

For a perfect gas undergoing an isothermal process, it is possible to evaluate

the entropy changes, i.e. (s2 – s1). From the non-flow equation, for a

reversible process, we have

dQ = du + P dv

Also for a perfect gas from Joule’s Law, du = Cv dT,

dQ = Cv dT + P dv

For an isothermal process, dT = 0, hence

dQ = P dv

Then, since Pv = RT, we have

v

vRTQ

dd

T

s

P2

v2 v1

1 2

s1 s2

Q

Figure 10.1.3 Constant temperature (or isothermal) process on a T-s diagram

P1

T1 = T2


Now from equation 9.5

2

1

2

1

dd d2

112

v

v

v

v v

vR

Tv

vRT

T

Qss

i.e.

2

1

1

212 ln ln

p

pR

v

vRss kJ/kg K (10.10)

or, for mass, m (kg), of a gas

S2 – S1 = m(s2 – s1)

i.e.

2

1

1

212 ln ln

p

pmR

v

vmRSS kJ/K (10.11)

Therefore, the heat supplied is given by,

2

1

1

212 ln ln

p

pRT

v

vRTssTQ

or, for mass, m (kg), of a gas

2

1

1

212 ln ln

p

pmRT

v

vmRTSSTQ

In an isothermal process,

(U2 – U1) = mCv (T2 - T1)

= 0 ( i.e since T1 = T2)

From equation Q - W = (U2 – U1),

W = Q (10.12)

0


0.85 m3 of carbon dioxide (molecular weight 44) contained in a cylinder

behind a piston is initially at 1.05 bar and 17 oC. The gas is compressed

isothermally and reversibly until the pressure is at 4.8 bar. Assuming

carbon dioxide to act as a perfect gas, calculate the:

c) mass of carbon dioxide

d) change of entropy

e) heat flow

f) work done

Sketch the process on a P-V and T-s diagram and shade the area which

represents the heat flow.

Example 10.3



V1 = 0.85 m3

M = 44 kg/kmol

P1 = 1.05 bar

Isothermal process: T1 = T2 = 17 + 273 K = 290 K

P2 = 4.8 bar


kJ/kgK 189.044

3144.8

M

RR o

Then, since PV = mRT, we have

kg 1.628290 x 0.189

0.85x 10 x 1.05 2

RT

PVm

b) From equation 10.11, for m kg,

kJ/K 0.46768.4

05.1ln 0.189 x 1.628ln

2

112

p

pmRSS


c) Heat rejected = shaded area on T-s diagram

= T (S2 – S1)

= 290 K(-0.4676 kJ/K)

= -135.6 kJ (-ve sign shows heat rejected from the system to

the surroundings)

d) For an isothermal process for a perfect gas, from equation 10.12

W = Q

= -135.6 kJ (-ve sign shows work is transferred into the system)

T

s

P1 = 1.05 bar

2 1

s2 s1

Q

P2 = 4.8 bar

T1 = T2 = 290 K



NEXT INPUT…!

10.1 0.1 m3 of air at 1 bar and temperature 15

oC is heated reversibly at constant

pressure to a temperature of 1100oC and volume 0.48 m

3. During the process,

calculate the:

a) mass of air

b) change of entropy

c) heat supplied

d) work done

Show the process on a T-s diagram, indicating the area that represents the

heat flow.

Given, R = 0.287 kJ/kg K and Cp = 1.005 kJ/kg K.

10.2 0.05 kg of nitrogen (M = 28) contained in a cylinder behind a piston is

initially at 3.8 bar and 140 oC. The gas expands isothermally and reversibly

to a pressure of 1.01 bar. Assuming nitrogen to act as a perfect gas,

determine the:

a) change of entropy

b) heat flow

c) work done

Show the process on a T-s diagram, indicating the area which represents the

heat flow.

Activity 10A




P1 = P2 = 1 bar (constant pressure process)

T1 = 15 + 273 K = 288 K

V1 = 0.1 m3

T2 = (1100 + 273) = 1373K

V2 = 0.48 m3

R = 0.287 kJ/kg.K

Cp = 1.005 kJ/kg.K

a) From equation PV =mRT, we have

kg 0.121288 x 0.287

0.1x 10 x 1

2

1

11 RT

VPm

b) From equation 10.4, change of entropy

s2 - s1 = mCp lnT

T

2

1

kJ/K 1899.0

288

1373ln 1.005 x 0.121

c) From equation 10.3, heat flow

Q = mCp(T2 - T1)

= 0.121 x 1.005 (1373 - 288)

= 131.9 kJ


d) The work done by air can be calculated by using two methods which

give the same results.

Method I:

From equation 10.2, the work done

W = mR(T2 - T1)

= 0.121 x 0.287 (1373 - 288)

= 38 kJ

Method II:

kJ 83

0.1)-(0.48101.x

)(

done work the10.1,equation From

2

12

VVPW

The T-s diagram below shows the constant pressure process. The shaded

area represents the heat flow.

T

s

P1 = P2 = 1bar

v2 = 0.48 m3

v1 = 0.1 m3

1

2

s1 s2

Q

T1 = 288 K

T2 = 1373 K



m = 0.05 kg

M = 28 kg/kmol

P1 = 3.8 bar

Isothermal process: T1 = T2 = (140 + 273 K) = 413 K

P2 = 1.01 bar


kJ/kgK 297.028

3144.8

M

RR o

From equation 10.11, for m kg of gas,

kJ/K 01968.0

01.1

8.3ln 0.297 x 05.0

ln 2

112

p

pmRSS

b) Heat flow = shaded area on T-s diagram

= T (S2 – S1)

= 413 (0.01968)

= 8.1278 kJ

c) For an isothermal process for a perfect gas, from equation 10.12

W = Q

= 8.1278 kJ

T

s

P1 = 3.8 bar

1 2

s1 s2

Q

P2 = 1.01 bar

T1 = T2 = 413 K


10.1.4 Reversible adiabatic (or isentropic) process

In the special case of a reversible process where no heat energy is transferred

to or from the gas, the process will be a reversible adiabatic process. These

special processes are also called isentropic process. During a reversible

isentropic process, the entropy remains constant and the process will always

appear as a vertical line on a T-s diagram.

For a perfect gas, an isentropic process on a T-s diagram is shown in Fig.

10.1.4. In Unit 4 it was shown that for a reversible adiabatic process for a

perfect gas, the process follows the law pv = constant.

Since a reversible adiabatic process occurs at constant entropy, and is known

as an isentropic process, the index is known as the isentropic index of the

gas.

INPUT

T

s

P2

v2

v1

1

2

s1 = s2

Figure 10.1.4 Reversible adiabatic (or isentropic) process on a T-s diagram

P1

T1

T2


For an isentropic process,

Change of entropy, s2 - s1 = 0

Heat flow, Q = 0

From the non-flow equation,

dQ - dW = dU

dW = -dU

= -mCv dT

= -mCv(T2 - T1)

W = mCv(T1 -T2) (10.13)

or, since 1

RCv , we have

1

)( 21

TTmRW (10.14)

or, since PV = mRT, we also have

1

2211

VPVPW (10.15)

Note that the equations 10.13, 10.14 and 10.15 can be used to find the work

done depending on the properties of gases given. Each equation used gives

the same result for a work done.

Similarly, equation 10.16 can also be used to determine the temperature,

pressure and volume of the perfect gases.

T

T

P

P

V

V

2

1

2

1

1

1

2

1

(10.16)


In an air turbine unit, the air expands adiabatically and reversibly from 10

bar, 450 oC and 1 m

3 to a pressure of 2 bar. Air is assumed to act as a

perfect gas. Given that Cv = 0.718 kJ/kg K, R = 0.287 kJ/kg K and = 1.4,

calculate the:

a) mass of air

b) final temperature

c) work energy transferred


Example 10.4



P1= 10 bar

V1 = 1 m3

T1 = (450 + 273) = 723K

P2 = 2 bar

Cv = 0.718 kJ/kg K

R = 0.287 kJ/kg K

= 1.4

Isentropic process, s2 = s1

a) From equation PV = mRT, for a perfect gas

kg 82.4723 x 0.287

1x 10 x 10

2

1

11 RT

VPm

b) The final temperature can be found using equation 10.16

K 5.456

10

2 x 723

x

4.1

14.1

1

1

212

P

PTT


c) The work energy transferred can be found using equation 10.13

W = mCv(T1 -T2)

= 4.82 x 0.718 (723 – 456.5)

= 922 kJ

Similarly, the equation 10.14 gives us the same result for the value of

work energy transferred as shown below,

1

)( 21

TTmRW

kJ 922

14.1

)5.456723(287.0 x 82.4

T

s

P2 = 2 bar

v2

v1 = 1 m3

1

2

s1 = s2

P1 = 10 bar

T1 = 723 K

T2 = 456.5 K


10.1.5 Reversible polytropic process

For a perfect gas, a polytropic process on a T-s diagram is shown in Fig.

10.1.5. In Unit 5 it was shown that for a reversible polytropic process for a

perfect gas, the process follows the law pvn

= constant.

For a reversible polytropic process,

Work done by a perfect gas is,

1

2211

n

VPVPW (10.17)

or, since PV = mRT, we have

W

mR T T

n

1 2

1 (10.18)

Change of internal energy is,

U2 -U1 = mCv(T2 -T1) (10.19)

The heat flow is,

Q = W + U2 -U1 (10.20)

T

s

P2

v2 v1

1

2

s1

Figure 10.1.5 Reversible polytropic process on a T-s diagram

P1

T1

T2

A B

s2

sA

sB


It was shown in Unit 5 that the polytropic process is a general case for perfect

gases. To find the entropy change for a perfect gas in the general case,

consider the non-flow energy equation for a reversible process as,

dQ = dU + P dv

Also for unit mass of a perfect gas from Joule’s Law dU = CvdT , and from

equation Pv = RT ,

v

vRTTCQ v

d dd

Then from equation 9.5,

v

vR

T

TC

T

Qs v d d d

d

Hence, between any two states 1 and 2,

1

2

1

212 lnln

dd 2

1

2

1 v

vR

T

TC

v

vR

T

TCss v

v

v

T

Tv (10.21)

This can be illustrated on a T-s diagram as shown in Fig. 10.1.5. Since in the

process in Fig. 10.1.5, T2 < T1, then it is more convenient to write the

equation as

2

1

1

212 lnln

T

TC

v

vRss v (10.22)

There are two ways to find the change of entropy (s2 – s1). They are:

a) According to volume

It can be seen that in calculating the entropy change in a polytropic

process from state 1 to state 2 we have in effect replaced the process

by two simpler processes; i.e. from 1 to A and then from A to 2. It is

clear from Fig. 10.1.5 that

s2 - s1 = (sA - s1) - (sA - s2)

The first part of the expression for s2 -s1 in equation 10.22 is the

change of entropy in an isothermal process from v1 to v2.

From equation 10.10

(sA - s1)

R

v

vln

2

1

(see Fig. 10.1.5)


In addition, the second part of the expression for s2 -s1 in equation

10.22 is the change of entropy in a constant volume process from T1

to T2,

i.e. referring to Fig. 10.1.5,

(sA - s2)

C

T

Tv ln1

2

s2 - s1

R

v

vC

T

Tvln ln2

1

1

2

kJ/kg K (10.23)

or, for mass m, kg of gas we have

S2 - S1

2

1

1

2 lnlnT

TmC

v

vmR v kJ/K (10.24)

b) According to pressure

According to pressure, it can be seen that in calculating the entropy

change in a polytropic process from state 1 to state 2 we have in

effect replaced the process by two simpler processes; i.e. from 1 to B

and then from B to 2 as in Fig. 10.1.5. Hence, we have

s2 - s1 = (sB - s1) - (sB - s2)

At constant temperature (i.e. T1) between P1 and P2, using equation

10.10,

(sB - s1)

R

p

pln

1

2

and at constant pressure (i.e. P2) between T1 and T2 we have

(sB - s2)

C

T

Tp ln1

2

Hence,

s2 - s1

R

p

pC

T

Tpln ln1

2

1

2

kJ/kg K (10.25)


or, for mass m, kg of gas we have

S2 - S1

2

1

2

1 lnlnT

TmC

p

pmR p kJ/K (10.26)

Similarly, the equation 10.27 can also be used to determine the

temperature, pressure and volume of the perfect gases in polytropic

process.

1

2

1

1

1

2

1

2

nn

n

V

V

P

P

T

T (10.27)

Note that, there are obviously a large number of possible equations for the change of entropy in a polytropic process, and it is stressed that no attempt should be made to memorize all such expressions. Each problem can be dealt with by sketching the T-s diagram and replacing the process by two other simpler reversible processes, as in Fig. 10.1.5.


0.03 kg of oxygen (M = 32) expands from 5 bar, 300 oC to the pressure of 2

bar. The index of expansion is 1.12. Oxygen is assumed to act as a perfect

gas. Given that Cv = 0.649 kJ/kg K, calculate the:

a) change of entropy

b) work energy transferred


Example 10.5



m = 0.03 kg

M = 32 kg/kmol

P1= 5 bar

T1 = (300 + 273) = 573K

P2 = 2 bar

Cv = 0.649 kJ/kg K = 649 J/kg K

PV1.12

= C


K J/kg 26032

8314

M

RR o

Then from equation R = Cp - Cv , we have

Cp = R + Cv

= 260 + 649

= 909 J/kg K

From equation 10.27, we have

T

T

p

p

n

n2

1

2

1

1

T TP

P

n

n

2 1

2

1

1 1 12 1

1 12

5732

5519 4

.

.

. K


From equation 10.26, the change of entropy (S2 - S1 ) is,

S2 - S1 = (SB - S1) - (SB - S2)

J/K 4.47

)68.2()15.7(

519.4

573ln x 909 x 0.03

2

5ln x 260 x 0.03

lnln2

1

2

1

T

TmC

p

pmR p

b) From equation 10.18, we have

W

mR T T

n

1 2

1J 3.417

1

519.5)(573 260 x 0.03

n

T

s

P2 = 2 bar

1

2

s1

P1 = 5 bar

T1 = 573 K

T2 = 519.4 K

B

s2

sB


TEST YOUR UNDERSTANDING BEFORE YOU PROCEED TO THE SELF-

ASSESSMENT…!

10.3 0.225 kg of air at 8.3 bar and 538 oC expands adiabatically and reversibly to a

temperature of 149 oC. Determine the

a) final pressure

b) final volume

c) work energy transferred during the process


For air, take Cp = 1.005 kJ/kg K and R = 0.287 kJ/kg K.

10.4 1 kg of air at 1.01 bar and 27 oC, is compressed according to the law

PV1.3

= constant, until the pressure is 5 bar. Given that Cp = 1.005 kJ/kg K

and R = 0.287 kJ/kg K, calculate the final temperature and change of entropy

and then sketch the process on a T-s diagram.

Activity 10B




m = 0.225 kg

P1 = 8.3 bar

T1 = 538 + 273 K = 811 K

T2 = 149 + 273 K = 422 K

Cp = 1.005 kJ/kg K

R = 0.287 kJ/kg K

Adiabatic / isentropic process : s2 = s1


Cv = Cp – R = 1.005 – 0.287 = 0.718 kJ/kg K

Then, from equation 3.17, we have

4.1718.0

005.1

v

p

C

C

For a reversible adiabatic process for a perfect gas, PV = constant.

From equation 10.16

bar 0.844

811

422)3.8(

14.1

4.1

2

1

1

2

1

2

P

T

T

P

P


b) From the characteristic gas equation PV = mRT, hence, we have at

state 2

3

2

2

22 m 0.323

10 x 0.844

422 x 0.287 x 0.225

P

mRTV

c) The work energy transferred can be found from equation 10.13

W = mCv(T1 -T2)

= 0.225 x 0.718 (811 – 422)

= 62.8 kJ

Similarly, the equation 10.14 gives us the same result for the value of

work energy transferred as shown below,

1

)( 21

TTmRW

kJ 8.62

14.1

)422811(287.0 x .2250


T

s

P2 = 0.844 bar

v2

v1

1

2

s1 = s2

P1 = 8.3 bar

T1 = 811 K

T2 = 422 K


m = 1 kg

P1= 1.01 bar

T1 = (27 + 273) = 300 K

P2 = 5 bar

Cp = 1.005 kJ/kg K

R = 0.287 kJ/kg K

PV1.3

= C

From the quantities given, we can temporarily sketch the process as shown in

the diagram below.

From equation 10.27, we have

T

T

p

p

n

n2

1

2

1

1

K 434

01.1

5)300(

3.1

13.1

1

1

212

n

n

P

PTT

From equation 10.25, the change of entropy (S1 – S2) is,

T

s

P2 = 5 bar

2

1

s2

P1 = 1.01 bar

T1 = 300 K

T2 = ? B

s1

sB


S1 – S2 = (SB – S2) - (SB – S1)

kJ/K 088.0 -

)459.0()371.0(

1.01

5ln x 0.287 x 0.1

300

434ln x 1.005 x 0.1

ln ln1

2

1

2

p

pmR

T

TmCp

From the calculation, we have S1 – S2 = - 0.088 kJ/K. This means that S2 is

greater than S1 and the process should appear as in the T-s diagram below.


PROCEED TO THE SELF-ASSESSMENT….

T

s

P2 = 5 bar

2

1

s2

P1 = 1.01 bar

T1 = 300 K

T2 = 434 K B

s1

sB





Good luck.

1. A quantity of air at 2 bar, 25oC and 0.1 m

3 undergoes a reversible constant

pressure process until the temperature and volume increase to 2155 oC and

0.8 m3. If Cp = 1.005 kJ/kg K and R = 0.287 kJ/kg K, determine the:

i. mass of air

ii. change of entropy

iii. heat flow

iv. work done

Sketch the process on a T-s diagram and shade the area which represents the

heat flow.

2. A rigid cylinder containing 0.006 m3 of nitrogen (M = 28) at 1.04 bar and

15oC is heated reversibly until the temperature is 90

oC. Calculate the:


ii. heat supplied

Sketch the process on a T-s diagram. For nitrogen, take = 1.4 and assume it

as a perfect gas.

3. 0.03 kg of nitrogen (M = 28) contained in a cylinder behind a piston is

initially at 1.05 bar and 15 oC. The gas expands isothermally and reversibly

to a pressure of 4.2 bar. Assuming nitrogen to act as a perfect gas, determine

the:


ii. heat flow

iii. work done


heat flow.

SELF-ASSESSMENT


4. 0.05 kg of air at 30 bar and 300oC is allowed to expand reversibly in a

cylinder behind a piston in such a way that the temperature remains constant

to a pressure of 0.75 bar. Based on the law pv1.05

= constant, the air is then

compressed until the pressure is 10 bar. Assuming air to be a perfect gas,

determine the:

i. net entropy change

ii. net heat flow

iii. net work energy transfer

Sketch the processes on a T-s diagram, indicating the area, which represents

the heat flow.

5. a) 0.5 kg of air is compressed in a piston-cylinder device from 100

kN/m2 and 17

oC to 800 kN/m

2 in a reversible, isentropic process.

Assuming air to be a perfect gas, determine the final temperature and

the work energy transfer during the process.

Given: R = 0.287 kJ/kg K and = 1.4.

b) 1 kg of air at 30oC is heated at a constant volume process. If the heat

supplied during a process is 250 kJ, calculate the final temperature

and the change of entropy. Assume air to be a perfect gas and take

Cv = 0.718 kJ/kg K.

6. 0.05 m3 of oxygen (M = 32) at 8 bar and 400

oC expands according to the law

pv1.2

= constant, until the pressure is 3 bar. Assuming oxygen to act as a

perfect gas, determine the:

i. mass of oxygen

ii. final temperature

iii. change of entropy

iv. work done


heat flow.



1. i. m = 0.2338 kg

ii. S2 – S1 = 0.4929 kJ/K

iii. Q = 500.48 kJ

iv. W = 140 kJ

2. i. S2 – S1 = 0.00125 kJ/K

ii. Q = 0.407 kJ

3. i. S2 – S1 = -0.0152 kJ/K

ii. Q = - 4.3776 kJ

iii. W = - 4.3776 kJ

4. i. S = (S2 – S1) + (S3 – S2)

= 0.0529 + 0.0434

= 0.0963 kJ/K

ii. Q = Q12 + Q23

= (30.31) + (-18.89)

= 11.42 kJ

iii. W = W12 + W23

= (30.32) + (-21.59)

= 8.73 kJ

5. a) T2 = 525.3 K, W = - 84.4 kJ

b) T2 = 651 K, (S2 – S1) = 0.5491 kJ/K

6. i. m = 0.23 kg

ii. T2 = 571.5 K

iii. S2 – S1 = 0.0245 kJ/K

iv. W = 30.35 kJ


THE STEAM POWER CYCLE J2006/11/1

THE STEAM POWER CYCLE

OBJECTIVES

General Objective : To understand and apply the concept of steam power cycle in

thermodynamics


define, derive, calculate and differentiate the following heat

engine cycle:

Carnot cycle

Rankine cycle

UNIT 11


11.0 INTRODUCTION

team is the most common working fluid used in heat engine cycles because

of its many desirable characteristic, such as low cost, availability, and high

enthalpy of vaporization. Other working fluids used include sodium, potassium, and

mercury for high-temperature applications and some organic fluids such as benzene

and the freons for low-temperature applications. The majority of this chapter is

devoted to the discussion of steam power plants, which produce most of the electric

power in the world today.

Steam power plants are commonly referred to as coal plants, nuclear plants or

natural gas plants, depending on the type of fuel used to supply heat to the steam. But

the steam goes through the same basic cycle in all of them. Therefore, all can be

analysed in the same manner.

In this chapter it can be shown that there is an ideal theoretical cycle which is the

most efficient conceivable; this cycle is called the Carnot cycle. The highest thermal

efficiency possible for a heat engine in practice is only about half that of the ideal

theoretical Carnot cycle, between the same temperature limits. This is due to the

irreversibilities in the actual cycle, and to the deviations from the ideal cycle, which

are made for various practical reasons. The choice of a power plant in practice is a

compromise between thermal efficiency and various factors such as the size of the

What is the most

common working

fluid used in heat

engine cycle?

INPUT

S


plant for a given power requirement, mechanical complexity, operating cost and

capital cost.

Fig 11.0 Model of a steam plant


11.1 The Carnot cycle

From the Second Law of Thermodynamics it can be derived that no heat engine can

be more efficient than a reversible heat engine working between the same temperature

limits. Carnot, a French engineer, has shown in a paper written in 18241 that the most

efficient possible cycle is one in which all the heat supplied is supplied at one fixed

temperature, and all the heat rejected is rejected at a lower fixed temperature. The

cycle therefore consists of two isothermal processes joined by two adiabatic

processes. Since all processes are reversible, then the adiabatic processes in the cycle

are also isentropic. The cycle is most conveniently represented on a T-s diagram as

shown in Fig. 11.1.

Fig 11.1 The Carnot cycle

1 This paper, called „Reflection on the Motive Power of Heat‟ was written by Carnot before the enunciation of

the First and Second Laws of Thermodynamics. It is a remarkable piece of original thinking, and it laid the

foundations for the work of Kelvin, Clausius and others on the second law and its corollaries.

T

s

1

2 3

4 T1

T2

A B

Boiler

Turbine

Condenser

Compressor W12

Q23

W34

Q41

1

2 3

4


A brief summary of the essential features is as follows:

4 to 1: The heat energy is supplied to the boiler resulting in evaporation of the

water, therefore the temperature remains constant.

1 to 2: Isentropic expansion takes place in the turbine or engine.

2 to 3: In the condenser, condensation takes place, therefore the temperature

remains constant.

3 to 4: Isentropic compression of the wet steam in a compressor returns the

steam to its initial state.

The plant required and the numbers referring to the state points for the Carnot cycle is

shown in Fig. 11.1. The steam at the inlet to the turbine is dry saturated. The steam

flows round the cycle and each process may be analysed using the steady flow energy

equation where changes in kinetic energy and potential energy may be neglected.

i.e. h1 + Q = h2 + W

In this statement of the equation the subscripts 1 and 2 refer to the initial and final

state points of the process; each process in the cycle can be considered in turn as

follows:

Boiler:

h4 + Q41 = h1 + W41

Therefore, since W = 0,

Q451 = h1 – h4 (11.1)

Turbine:

The expansion is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2), and h2 can

be calculated using this latter fact. Then

h1 + Q12 = h2 + W12

W12 = ( h1 – h2) (11.2)


Condenser:

h2 + Q23 = h3 + W23

Therefore, since W = 0

Q23 = h3 – h2

i.e. Q23 = - ( h2 – h3 )

Heat rejected in condenser = h2 – h3 (11.3)

Compressor:

H3 + Q34 = h4 + W34

The compression is isentropic ( i.e. s3 = s4 ), and adiabatic ( i.e. Q = 0 ).

W34 = ( h3 – h4 ) = -( h4 – h3 )

i.e. Work input to pump = ( h4 – h3 ) (11.4)

11.1.1 Thermal efficiency of Carnot cycle

The thermal efficiency of a heat engine, defined in chapter 9, was shown to be given

by the equation,

1

21Q

Q

In the Carnot cycle, with reference to Fig. 11.1, it can be seen that the heat supplied is

given by the area 41BA4,

i.e. Q1 = area 41BA4 = T1(sB- sA)

Similarly the heat rejected, Q2, is given by the area 23AB2,

i.e. Q2 = area 23AB2 = T(sB – sA)

Hence we have

Thermal efficiency of Carnot cycle, )(

)(1

1

2

AB

ABcarnot

ssT

ssT


i.e. 1

21T

Tcarnot (11.5)

or boiler in the suppliedHeat

outputNet work carnot

i.e. 41

3421 )()(

hh

hhhhCarnot

(11.6)

11.1.2 The work ratio for Carnot cycle

The ratio of the net work output to the gross work output of the system is called the

work ratio. The Carnot cycle, despite its high thermal efficiency, has a low work

ratio.

Work ratio workgross

net work (11.7)

i.e. Work ratio = )(

)()(

21

3421

hh

hhhh

(11.8)

The work output of the Carnot cycle also can be found very simply from the T-s

diagram. From the first law,

WQ

therefore, the work output of the cycle is given by

W = Q1 – Q2

Hence for the Carnot cycle, referring to Fig. 11.1,

Wcarnot = area 12341 = (T1 - T2)(sB – sa) (11.9)


This cycle is never used in practice owing to:

1. The difficulty in stopping the condensation at 3, so that subsequent

compression would bring the state point to 4.

2. A very large compressor would be required.

3. Compression of wet steam in a rotary compressor is difficult as the water

tends to separate out.

4. Friction associated with the expansion and compression processes would

cause the net work done to be very small as compared to the work done in the

turbine itself.

The Carnot cycle is modified to overcome the above difficulties and this modified

cycle, known as the Rankine cycle, is widely used in practice.

Thermal efficiency of Carnot cycle,

)(

)(1

1

2

AB

ABcarnot

ssT

ssT

1

21T

Tcarnot


What is the highest possible theoretical efficiency of a Carnot cycle with a hot

reservoir of steam at 200oC when the cooling water available from condenser

is at 10oC?

A steam power plant operates between a boiler pressure of 42 bar and a

condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency

and the work ratio for a Carnot cycle using wet steam.

Example 11.1


From equation 11.5,

1

21T

Tcarnot

= 273200

273101

= 473

2831

i.e. Highest possible efficiency = 1 – 0.598

= 0.402 or 40.2 %

Example 11.2


A Carnot cycle is shown in the figure given in the next page.

T1 saturation temperature at 42 bar

= 253.2 + 273 = 526.2 K

T2 saturation temperature at 0.035 bar

= 26.7 + 273 = 299.7 K



1

21

T

TTcarnot

= 2.526

7.2992.526

= 0.432 or 43.2 %

Also,

Heat supplied = h1-h4 = hfg at 42 bar = 1698 kJ/kg

Then,

432.0Q

Wcarnot

W = 0.432 x 1698

i.e. W = 734 kJ/kg

To find the gross work of the expansion process it is necessary to calculate h2, using

the fact that s1 = s2.

From the Steam Tables,

h1 = 2800 kJ/kg and s1 = s2 = 6.049 kJ/kg K

From the equation

s2 = 6.049 = sf2 + x2sfg2 = 0.391 + x2 8.13

x2 = 696.013.8

391.0049.6

T

s

1

2 3

4 526.2

299.7


Then,

h2 = hf2 + x2hfg2 = 112 + 0.696 x 2438

= 1808 kJ/kg

Hence, from equation 11.2,

W12 = (h1 – h2)

= (2800 – 1808)

= 992 kJ/kg

Therefore, using equation 11.7,


net work

= 992

734

= 0.739



NEXT INPUT…!

11.1 Describe the suitable components of a simple close cycle steam plant as

illustrated in the figure below.

11.2 A steam power plant operates between a boiler pressure of 40 bar and a

condenser pressure of 0.045 bar. Calculate for these limits the cycle efficiency

and the work ratio for a Carnot cycle using wet steam.

Activity 11A

(a)

(b)

(c)

(d) W12

Q23

W34

Q41

1

2 3

4



11.1 The components of a simple close cycle steam plant as illustrated in the figure

are:

a) The boiler, where water is converted into steam at a constant pressure

and temperature by the heat energy is received from the combustion of

the fuel.

b) The engine or turbine, in which the steam expands to a low pressure

causing work energy to be available.

c) The condenser, in which heat energy flows from the low pressure

steam into the condenser cooling water, resulting in the steam being

condensed.

d) The feed pump or the compressor, which returns the water into the

boiler.

11.2 A Carnot cycle is shown in the figure below.

T1 saturation temperature at 40 bar

= 250.3 + 273 = 523.3 K

T2 saturation temperature at 0.045 bar

= 31.0 + 273 = 304.0 K

T

s

1

2 3

4 523.3

304.0



1

21

T

TTcarnot

= 3.523

3043.523

= 0.419 or 41.9 %

Also,

Heat supplied = h1-h4 = hfg at 42 bar = 1714 kJ/kg

Then,

419.0Q

Wcarnot

W = 0.419 x 1714

i.e. W = 718.2 kJ/kg

To find the gross work of the expansion process it is necessary to calculate h2, using

the fact that s1 = s2.

From the tables,

h1 = 2801 kJ/kg and s1 = s2 = 6.070 kJ/kg K

From the equation

s2 = 6.070 = sf2 + x2sfg2 = 0.451 + x2 7.98

x2 = 704.098.7

451.0070.6

Then,

h2 = hf2 + x2hfg2 = 130 + 0.704 x 2428

= 1839.3 kJ/kg

Hence, from equation 11.2,

W12 = (h1 – h2)

= (2801 – 1839.3)

= 961.7 kJ/kg

Therefore, using equation 11.7,



net work

= 7.961

2.718

= 0.747


11.2 Rankine Cycle

Many of the impracticalities associated with the Carnot cycle can be eliminated by

condensing it completely in the condenser, as shown schematically on a T-s diagram

in Fig. 11.3. The cycle that results is the Rankine cycle, is the ideal cycle for vapour

power plants. The ideal Rankine cycle does not involve any internal irreversibility

and consists of the following processes:

4,5 to 1: Constant pressure heat addition in a boiler

1 to 2: Isentropic expansion taking place in the turbine or engine

2 to 3: Constant pressure heat rejection in the condenser

3 to 4: Isentropic compression of water in the feed pump

From a comparison made between Fig. 11.1 and Fig. 11.2, the similarities between

the Carnot and the Rankine cycles can be clearly seen. In the Rankine cycle, the

exhaust steam is completely condensed into water in the condenser. It actually

follows the isentropic expansion in the turbine. This water is then pumped into the

boiler by a boiler feed pump. After the feed pump, since the water is not at the

INPUT

In the Rankine cycle, the exhaust

steam is completely condensed into

water in the condenser.


saturation temperature corresponding to the pressure, some of the heat energy

supplied in the boiler is taken up by the water as sensible heat before evaporation can

begin. This results in the boiler process being no longer completely isothermal; the

process is, therefore, irreversible, causing the Rankine cycle to be an irreversible

cycle and to have a lower efficiency than the Carnot cycle.

Fig 11.2 The Rankine cycle

Boiler

Turbine

Condenser

Pump W12

Q23

W34

1

2 3

4

5

Q451

T

s

1

2 3

4

T1

T2

A B

5

p1

p2


The plant required and the numbers referring to the state points for the Rankine cycle

is shown in Fig. 11.2. The steam at the inlet to the turbine may be wet, dry saturated,

or superheated, but only the dry saturated condition is shown in Fig. 11.2. The steam

flows round the cycle and each process may be analysed using the steady flow energy

equation where changes in kinetic energy and potential energy may be neglected.

i.e. h1 + Q = h2 + W

In this statement of the equation the subscripts 1 and 2 refer to the initial and final

state points of the process; each process in the cycle can be considered in turn as

follows:

Boiler:

h4 + Q451 = h1 + W

Therefore, since W = 0,

Q451 = h1 – h4 (11.10)

Turbine:

The expansion is adiabatic (i.e. Q = 0), and isentropic (i.e. s1 = s2 ), and h2 can

be calculated using this latter fact. Then

h1 + Q12 = h2 + W12

W12 = ( h1 – h2) (11.11)

Condenser:

h2 + Q23 = h3 + W23

Therefore, since W = 0

Q23 = h3 – h2

i.e. Q23 = - ( h2 – h3 )

Heat rejected in condenser = h2 – h3 (11.12)


Pump:

h3 + Q34 = h4 + W34

The compression is isentropic ( i.e. s3 = s4 ), and adiabatic ( i.e. Q = 0 ).

W34 = ( h3 – h4 ) = -( h4 – h3 )

i.e. Work input to pump = ( h4 – h3 ) (11.13)

This is the feed pump term, and as the quantity is small as compared to the

turbine work, W12. The feed pump is usually neglected, especially when the

boiler pressures are low.

The net work done in the cycle, W = W12 + W34

i.e. W = ( h1 – h2) – ( h4 – h3 ) (11.14)

Or, if the feed pump work is neglected,

W = ( h1 – h2 ) (11.15)

11.2.1 Thermal efficiency of Rankine cycle

Rankine efficiency,

boiler in the suppliedHeat

outputNet work R (11.16)

i.e. 41

3421 )()(

hh

hhhhR

or )()(

)()(

3431

3421

hhhh

hhhhR

(11.17)

If the feed pump term, (h4 – h3) is neglected, equation (11.17) becomes

)(

)(

31

21

hh

hhR

(11.18)


When the feed pump term is included, it is necessary to evaluate the quantity, W34.

From equation (11.13)

Pump work = -W34 = f (p4 – p3) (11.19)

11.2.2 The work ratio for Rankine cycle

It has been stated that the efficiency of the Carnot cycle is the maximum possible, but

that the cycle has a low work ratio. Both efficiency and work ratio are criteria of

performance. The work ratio is defined by


net work

i.e. Work ratio = )(

)()(

21

3421

hh

hhhh

or Work ratio =)(

)(v)(

21

34f21

hh

pphh

(11.20)

11.3 Specific steam consumption

Another criterion of performance in steam plant is the specific steam consumption. It

relates the power output to the steam flow necessary to produce steam. The steam

flow indicates the size of plant with its component part, and the specific steam

consumption is a means whereby the relative sizes of different plants can be

compared.

The specific steam consumption is the steam flow in kg/h required to develop 1

kW,

i.e. W x (specific steam consumption, s.s.c.) = 1 x 3600 kJ/h

(where W is in kJ/kg),


A steam power plant operates between a boiler pressure of 42 bar and a

condenser pressure of 0.035 bar. Calculate for these limits the cycle efficiency,

the work ratio, and the specific steam consumption for a Rankine cycle with

dry saturated steam at entry to the turbine.

i.e. s.s.c. = kg/kwh3600

W

or. s.s.c. = kg/kwh)()(

3600

3421 hhhh (11.21)

Example 11.3


The Rankine cycle is shown in the figure below.

As in example 11.2

h1= 2800 kJ/kg and h2 = 1808 kJ/kg

Also, h3 = hf at 0.035 bar = 112 kJ/kg

Using equation 11.18, with v = vf at 0.035 bar

T

s

1

2 3

4

526.2

299.7

A B

5

42bar

0.035 bar


Pump work = -W34 = f (p4 – p3)

= 0.001 x ( 42 – 0.035) x 102

= 4.2 kJ/kg

Using equation 11.11

W12 = h1 – h2 = 2800 – 1808 = 992 kJ/kg

Then using equation 11.17

)()(

)()(

3431

3421

hhhh

hhhhR

= )2.4()1122800(

)2.4()992(

= 0.368 or 36.8 %

Using equation 11.7


net work

= 992

4.2-992

= 0.996


s.s.c. = )()(

3600

3421 hhhh

= )2.4()992(

3600

= 3.64 kg/kW h



NEXT INPUT…!

11.3 Based on the diagram below, describe the four-stage processes that represent a

steam plant operating on an ideal Rankine cycle.

11.4 A steam power plant operates between a boiler pressure of 40 bar and a

condenser pressure of 0.045 bar. Calculate for these limits the cycle

efficiency, the work ratio, and the specific steam consumption for a Rankine

cycle with dry saturated steam at entry to the turbine.

Activity 11B

Boiler

Turbine

Condenser

Pump W12

Q23

W34

1

2 3

4

5

Q451



11.3 The ideal Rankine cycle does not involve any internal irreversibility and

consists of the following four-stage processes:

Stage 1.

4 to 1: Constant pressure heat addition in a boiler.

Stage 2.

1 to 2: Isentropic expansion that takes place in the turbine or engine.

Stage 3.

2 to 3: Constant pressure heat rejection in the condenser.

Stage 4

3 to 4: Isentropic compression of water in the feed pump

11.4 The Rankine cycle is shown in the figure below.

As in activity 11.2

h1= 2801 kJ/kg and h2 = 1839.3 kJ/kg

T

s

1

2 3

4

523.3

3040.

A B

5

40bar

0.045 bar


Also, h3 = hf at 0.045 bar = 130 kJ/kg

Using equation 11.18, with v = vf at 0.045 bar

Pump work = -W34 = f (p4 – p3)

= 0.001 x ( 40 – 0.045) x 102

= 4.0 kJ/kg


W12 = h1 – h2 = 2801 – 1839.3 = 961.7 kJ/kg

Then using equation 11.17

)()(

)()(

3431

3421

hhhh

hhhhR

= )0.4()1302801(

)0.4()7.961(

= 0.359 or 35.9 %

Using equation 11.7


net work

= 961.7

4.0-961.7

= 0.996


s.s.c. = )()(

3600

3421 hhhh

= )0.4()7.961(

3600

= 3.76 kg/kW h





1. Explain why the Rankine cycle and not the Carnot cycle is taken as the ideal

cycle for steam plant. Sketch the T-s diagram for these cycles when using

steam as the working fluid.

2. What is the highest thermal efficiency possible for a Carnot cycle operating

between 210o C and 15

o C.

3. A steam power plant operates between a boiler pressure of 30 bar and a


the work ratio and the specific steam consumption for a Carnot cycle using

wet steam.

4. A steam power plant operates between a boiler pressure of 30 bar and a


the work ratio, and the specific steam consumption for a Rankine cycle with

dry saturated steam at entry to the turbine.

5. In a steam power plant, dry saturated steam enters the turbine at 47 bar and is

expanded isentropically to the condenser pressure of 0.13 bar. Determine the

Rankine cycle efficiency when

a) the feed pump work is neglected

b) the feed pump work is taken into account

SELF-ASSESSMENT



1. Carnot cycle is never used in practice owing to:

1. The difficulty in stopping the condensation at 3, so that subsequent

compression would bring the state point to 4.

2. A very large compressor would be required.

3. Compression of wet steam in a rotary compressor is difficult as the water

tends to separate out.

4. Friction associated with the expansion and compression processes would

cause the net work done to be very small as compared to the work done in the

turbine itself.

The Carnot cycle is modified to overcome the above difficulties and this modified

cycle, known as the Rankine cycle, is widely used in practice.


T

s

1

2 3

4 T1

T2

T

s

1

2 3

4

T1

T2

A B

5

p1

p2

Carnot cycle Rankine cycle


2. 40.37 %

3. 40.4 %, 0.771, 4.97 kg/kW h

4. 34.6 %, 0.997, 3.88 kg/kW h

5. 33.67 %, 33.55 %

CONGRATULATIONS!

!!!…May success be

with you always…

J2006_Termodinamik

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