INDEX

S. No Topic Page No.

Week 1
1 Basic concepts and definitions - Part 1 1
2 Basic concepts and definitions - Part 2 13
3 Basic concepts and definitions - Part 3 21
4 Tutorial problems on exact and inexact differential 30
5 Basic concepts and definitions - Part 4 34
Week 2
6 Work - Part 1 38
7 Work - Part 2 44
8 Work - Part 3 48
9 Work - Part 4 51
10 Work - Part 5 60
11 Tutorial problem on 'Work' - Part 1 64
12 Tutorial problem - Part 2 67
13 Tutorial problem on 'Work' - Part 3 72
14 Tutorial problem on 'Work' - Part 4 75
Week 3
15 Zeroth law of thermodynamics 77
16 Methods of temperature measurement 83
17 Modes of heat transfer 90
18 Tutorial problem on 'Modes of heat transfer 99
19 Tutorial problem on 'Methods of temperature measurement 101
20 First law of thermodynamics 104
21 Tutorial problem 111
22 Tutorial problem 113
Week 4
23 Heat and work interactions for a system 117
24 Tutorial problem - Part 1 119
25 Pure substance 120
26 Tutorial problem - Part 2 126
27 Ideal gas - Part 1 130
28 Ideal gas - Part 2 134
29 Tutorial problem - Part 3 142
30 Tutorial problem - Part 4 146
31 Tutorial problem - Part 5 149
32 Specific heats at constant pressure and constant volume 151
33 Tutorial problem - Part 6 153
34 Tutorial problem - Part 7 156
Week 5
35 Ideal gas - Part 3 159
36 Ideal gas - Part 4 162
37 Ideal gas - Part 5 166
38 Tutorial problem - Part 1 170
39 Tutorial problem - Part 2 172
40 Tutorial problem - Part 3 177
41 Tutorial problem - Part 4 180
42 Beyond ideal gases - Part 1 185
43 Beyond ideal gases - Part 2 190
Week 6
44 Two phase system - Part 1 194
45 Two phase system - Part 2 200
46 Two phase system: water and steam 212
47 Tutorial problems (2 numbers) 222
48 Tutorial problem - Part 1 231
49 Tutorial problem - Part 2 236
50 Tutorial problem - Part 3 239
Week 7
51 Tutorial problems on two-phase systems(2 numbers) 242
52 Tutorial problem (1 number) 249
Rate equation of the first law of thermodynamics for a control mass and
53 a control volume 252
Energy equation for a steady-state, steady-flow process in selected
54 engineering devices 261
55 Tutorial problems (3 numbers) 271
56 Tutorial problem - Part 1 279
57 Tutorial problem - Part 2 282
Week 8
58 Quasi-static process revisited: Work against an external force 285
Second law of thermodynamics: limitations of the first law of
59 thermodynamics 290
60 Second law of thermodynamics: direct and reverse heat engine 299
61 Second law of thermodynamics: Kelvin-Planck and Clausius statements 309
62 Second law of thermodynamics: reversible process 313
Week 9
63 Second law of thermodynamics: Carnot's cycle and theorems 317
64 Second law of thermodynamcis: absolute temperature scale 323
65 Tutorial problems (2 numbers) 328
66 Tutorial problem (1 number) 331
67 Tutorial problem (1 number) 333
68 Tutorial problem (2 numbers) 337
69 Second law of thermodynamics: Clausius's inequality 340
70 Entropy - Part 1 344
71 Tutorial problem (1 number) 351
Week 10
72 Entropy - Part 2 354
73 Entropy - Part 3 358
74 Entropy - Part 4 365
75 Tutorial problem (1 number) 369
76 Tutorial problem (1 number) 373
77 Tutorial problems (2 numbers) 376
Week 11
78 Entropy - Part 5 380
79 Entropy - Part 6 384
80 Entropy - Part 7 391
81 Exergy - Part 1 396
82 Exergy - Part 2 402
83 Exergy - Part 3 408
Week 12
84 Thermodynamics cycles: Rankine cycle 416
85 Tutorial problem (1 number) 426
86 Thermodynamics cycles: Brayton cycle 435
87 Tutorial problem (1 number) 443
88 Thermodynamics cycles: vapor compression refrigeration cycle 446
89 Tutorial problem (1 number) 451
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 01
Basic concepts and definitions – Part 1

Hello, and Welcome to this course on Thermodynamics. We are going to look at several aspects
of thermodynamics over the next 12 weeks or so.

(Refer Slide Time: 0:25)

To look at the course contents, we also have about 11 or 12 topics listed here. So we will roughly
do about one topic a week and that will get us through the course. For those of you who are
watching this video online, what I would like to remind you is that you can always play this
video slower or faster as you want it.

I have a cap to audience here and I am going to do it at some speed, but you may want to either
slow it down or speed it up based on how it is convenient for you. One of the other things I
would like to sort of remind you is that you can always pause the video. I would encourage you
to pause and write down any equations, and I would also encourage you to pause and solve the
tutorial problems yourself before looking at the solutions which we are going to do in this
course.

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This is especially important because thermodynamics is a subject where we would sort of look at
the concepts but it is in theory. But it is much easier to find out our understanding of the concept
by solving out problems where any, if you find any doubts or the finer points sort of get clarified.
So, I sort of recommend that you pause in between, note down equations, solve tutorials with the
problem statements given and then go ahead.

(Refer Slide Time: 1:46)

To look at the text books which are here, there are several text books which I have listed here.
The book by Turns is quite nice. It has a lot of interesting pictures as well as data presented in an
interesting way. I am also going to use a lot of figures from there in later parts of this course
wherever it is not attributed, it is either made by us or from Turns.

This course I have been teaching earlier with several of my colleagues. Professor Sundarajan,
Professor Raghwan, Professor Srinivasan, and Professor Babu who have, some of them have
taught, and whom I have taught with. Professors Babu’s book on Fundamentals of Engineering
Thermodynamics is quite interesting and I think it will sort of be in a similar fashion to which we
go about in this lecture.

So, that may be a book which you may be interested in. As a student I used to follow a book by
Van Wylen and others. It is I think gone into several editions now including I think 7th edition
with Sonntag and Borgnakke and others, and there are also several other books. It does not

2
matter which book you follow as long as you follow one consistently you should be quite alright
in this course.

(Refer Slide Time: 2:54)

At the end of this course, what we expect is that if you go through all of the lectures and solve all
of the tutorial problems, you should be able to analyze thermodynamics processes and cycles
involving heat and work. We will see what heat and work are in a bit. We expect that you should
be able to evaluate the performance of thermodynamic systems, and we also expect that you
should be able to identify hypothetical situations which violate the laws of thermodynamics.

This I think is quite important because very often we see sort of newspaper articles and so on,
which or for example we come across students projects, where students often think that for
example you could run a generator which gives power to run a motor and then you use a motor
to run a generator and do this in a cycle without any other input from anywhere else, and
thermodynamics says that this is not something which you can do or there is often a conception
that if you have, for example, an electric car and your batteries are running out of juice.

So, you can put like a turbine on the car and use wind power to charge the battery and then drive
the car forward. If that is the only thing which you want to do, then thermodynamics says that it
is not possible again. So, there are several scenarios like this which we come to think of has
really interesting things but then we look at in detail we often see that they are not possible

3
because thermodynamics says it cannot be true and if it says so it must be true. So, that is
something.

(Refer Slide Time: 4:32)

In this week, we are going to look at the scope of thermodynamics in general and this course in
particular, what we will also look at is the basics of thermodynamics, we will look at some
nomenclature, definitions, we will look at things like what is called as a system, a control
volume, we will look at properties and in that context we will look at exact and inexact
differentials, we will define a state and a process and so on. So that is essentially what we going
to do this week.

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(Refer Slide Time: 5:00)

But before that let us define thermodynamics. Thermodynamics is defined variously in different
text books and so on but as such it is a science which deals with energy and its conversion and
transformation. So it deals with energy, it deals with the transformation of energy from one form
to another, conversion to one form to another, and in the context of energy we have various ways
in which you can transfer energy, and we roughly categorize them as heat or work.

We will look at the distinction between these in a bit of detail, we will look at each of these over
at least a week for each of these so that we understand the distinction and when we later on go to
the second law, we will see why we have this distinction. When you deal with thermodynamics,
there are various approaches to deal with thermodynamics. So one of them which what is called
as macroscopic approach, and another is what is called is microscopic approach. Macroscopic
approach is also what is called as a classical thermodynamics approach and that is what we are
going to follow in this course. Here we assume that matter exists in a continuum.

So, we will see what this means. So, what essentially this means is that if I have this marker for
example, what we know is that this is composed of molecules and atoms and so on, but in
continuum approach we do not care what is there inside it, what we will say is this is a solid and
that is all we need to know about it. We can find out some properties of the solid but we will not
worry about what it is composed of and the finite molecular details of this.

5
Similarly, if I want to look at for example a gas, the air inside this room, I will not worry about
what it is composed of, I will say air is something which continuous and which is there in this
room and I can study it. So, another way of looking at it is to look at the variation of say density
with respect to the size of some box, the volume of some box. Let us say I have this a big box
which is this entire room around me, it has air inside it. I can find out the density of air in this
room, I have some value for it. I find out the mass of, total mass of air, and the total volume of
this room divide the mass by the volume I get some density and that has some value.

If I now cut take only half of this room and find the density of one half of this room. It may or
may not be equal to the density which I got when taking the entire room. The reason for that is
that for example I have an air conditioner sitting here, which is putting out cold air, the cold air
has a slightly higher density, then say the hot air next to the light because next to the light, the
light bulb is heating the air around it. That air is slightly lighter there.

So, if I cut this room into half, the density will be slightly smaller. If I now cut this room into
another small portion, the density will be slightly more different it may be slightly higher or
smaller. So as I keep reducing the size of this box, finally I have some box which is this size, it
has some density I take a smaller box which has smaller, different density, either smaller or
larger and so on and so forth.

So this density may be varying and finally at some point I will come to a size of a box which is
so small that it has only a few molecules in it, let us say it has 10 molecules in it. Now, if one
molecule, these molecules in the air are all in motion, they are moving around, if one molecule
comes in extra, if there is 11th molecule, then the density inside this box becomes very high, if
like two molecules go out the density becomes very low and so on. And since molecules are
coming in and going out density is beyond some point changing a lot.

And if I come to extremely small boxes, what I see is I may get to a size of a box, where there is
either no molecule or there is a molecule and or if I go to even smaller box I may come to a size
where there is nothing there or for example once in a while a nucleus of a molecule or an atom is
inside it. And the density of a nucleus is extremely high, so I can consider it as very very high or
close infinity for purpose when approximation or it is 0 because there is nothing there.

6
So, what we see is the density will vary like this at very small volumes as we go to reasonably
large volumes where we have like say millions of molecules the density is fairly constant. So in
our assumption of a continuum, what we want to do is, we want to look at volumes which are
large enough that the density is not changing too much.

So we will not look at this region, we will only look at say somewhere over here, and we want a
box which is big enough that is has so many molecules but you do not have to worry if there is
one molecule more or some hundred molecules more or less. At the same time I do not want to
have so large space that the properties, for example, density is changing within that space
because I want to sort of define as an average density.

So what we mean by the continuum approach is that we are looking at somewhere here, where
we can define a finite density for it. So mathematically we say that this the size of this goes
to some, for example dash which is the smallest volume at which the density is not changing
too much and I find out that mass by the volume, mass of that box by the volume of that box and
that will be defined as a density.

So to put it more simply we are concerned with the average effect of a very large number of
molecules, we are not going look at individual molecules. So, for example this marker again
what I am concerned with is that it is there, I do not need to know how many molecules it is
composed of. The net effect of all of these molecules is that it has some shape, I can press it, it
does not deform very much, it has some temperature, it has some mass and it is sitting in some
position and so on. That is what we are interested in.

In contrast to this, in many other courses, for example in courses on Statistical Thermodynamics
or in courses in Chemical Engineering, or Chemistry or Physics and so on, we often use a
microscopic approach, where we look at individual molecules. In one limit of it what we can
think of is that if I knew where every single molecule is in this marker or in this room then I can
predict what happens, I can predict what will be the temperature, what will be the pressure and
so on.

But if I have every molecule I can solve equations for those molecules combine all with and find
out any information I need. But the number of molecules is so high that that is indefinitely not a
feasible approach. So, the next thing which could be to do is I could use a statistical approach,

7
where I club the behavior of say a million molecules or a billion molecules and so on or much
more that and say that in a statistical sense all of these molecules behave in such a fashion, and
then use a kinetic theory and statistical thermodynamics to predict what happens.

That is also an approach which I could use. In this particular course, what we are going to do is
we are going to confined ourselves with the microscopic approach but it is important that you
know that this is not the only way in which we could approach thermodynamics there could be
other ways to do this as well.

(Refer Slide Time: 12:39)

Before we go forward what I would also suggest is that you should sort of think about why this
course or these concepts are important in your field of Engineering. I would encourage you to
may be write some of these in the forum and share your thoughts with others who are taking this
course.

I come from Mechanical Engineering background and anything which moves has usually some
prime mover to it. We are usually interested in converting heat energy or chemical energy into
work into something which is moving and thermodynamics plays a very important role in this.

There are other important aspects which may be useful to your field, for example, if you are
designing a chip you may be worried about how much energy you can dissipate so that it does
not burn out, if you are designing, say a bridge or a bunker something like that may be you are

8
interested in how much energy of an impact it can take before it fails and so on. So there may be
various aspects which maybe you may want to think about which sort of clarifies to you why it is
important to learn about this.

(Refer Slide Time: 13:48)

We will go on to some definitions. So first definition is what we have is what is called as a

control volume, it is also often called as an open system. So we define it as a region in space,
which is separated from its surroundings by either a real or an imaginary boundary and that
boundary is called as control surface.

So here we have some arbitrarily shape region it has some boundary, which we call as a control
surface, whatever is inside it is our control volume or open system, whatever is outside this
boundary is what we call as a surroundings. So what we would typically do is take this region as
something we are interested in. We have another definition of a system which is also often called
a closed system or a control mass. This is specifically identified fixed mass which is separated
from its surroundings by a real or imaginary boundary.

So what we see is in this case, we are talking of a fixed mass. In this case, we are talking of a
region in space. So, in the case of a control volume the mass may not be fixed, it may be varying.
In the case of a control mass or what we more commonly just call as a system, the mass is fixed.
So that is the distinction between these two, here the mass can vary, here the mass is fixed. And

9
both of these kind of systems either control volume or a control mass, both of them can have
interactions in the form of energy transfers.

So the open system or control volume which is essentially this definition here can have mass
transfer that is mass can come in and go out or and it can have heat transfer you can heat it or
you can cool it. It can have work transfer you can do work on it or it can give out some work.
Whereas a closed system or a control mass can have heat transfer and work but it cannot have
mass transfer. So an example of a control volume would be for example, a water bottle, with the
cap either open or closed, you can pour liquid into it or you can take liquid out of it.

So that is essentially mass transfer. You can heat it while putting it on a stove or may have an
induction, immersion rod which you put into it, so that is heat transfer and you can compress the
substance inside it or by pressurizing it or you can have some other kind of a fan which is sitting
inside it and doing some work or something like that. So you can have mass and heat transfer or
you can have work.

In the case of a control mass you have the mass which is fixed, so you cannot put in or take out
mass but you can heat it, for example you have a pressure cooker with the weight ON, you can
heat what is inside it if you had some kind of a fan inside it or something like that you can do
work or if you have a heating coil you can do work, so that is what you see in these two cases. If
you have a system, which is essentially a control mass, where there is no heat transfer, there is no
work transfer and there is no mass transfer, then we call that as an isolated system.

So essentially something which is not interacting with the surroundings because there is no
transfer of heat, work or mass, we call it an isolated system. And by our definition the system is
what we are interested in, everything else outside the system is the surroundings. So, for example
if I am interested in this class room where I am sitting my surroundings include the classrooms
around here, my surroundings include the buildings, in the next compound, it includes rest of the
city, rest of the country, the rest of the Milky Way and any other stars which are outside and so
on. So by definition if I look at the system plus the surroundings that is the universe, it is
whatever there is.

10
(Refer Slide Time: 17:47)

So here I have an example of a control volume, this is this volume. In some books a control
volume is defined as a rigid surface which does not change in volume but more generally a
control volume can be any volume which is also deforming, we have ways of taking into account
the deformation of the control volume but for a first-cut approximation we usually deal with
rigid volumes as control volumes.

So, this is for example, a balloon the surface is the surface of the balloon. The volume is a
volume inside the surface and in this case we are seeing a balloon where the neck has been
released, the air inside is going out. So, there is transfer of mass as in the form of an air jet.

11
(Refer Slide Time: 18:30)

An example of a system or a control mass is shown over here. This is actually representative of
an engine which is seen for example two wheelers, and four wheelers all our cars and trucks and
so on. So this is what we call as a cylinder and this is what we would call as a piston, and we see
that we have some amount of air trapped over here between in the cylinder and the piston which
is essentially this is our system.

So, you have some air which is trapped here, which is our system, we can heat this air, we can
cool it, we can push this piston up and in pushing that up we can make the volume of this air
smaller or we can pull the piston down and make this volume larger but the mass of the air is
fixed as long as the values are closed. So, we are interested in a fixed mass here which we can
heat or cool.

So there is no exchange of mass but we can expand or compress the air and we can have it do
work or exchange heat and so on. Of course, you can think of many other systems this is a
system over here, fixed quantity of mass, I can lift it, I can lower it, I can heat it, I can cool it and
so on. So practically anything you want can be a system or a control volume.

12
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 02 - Basic concepts and definitions – Part 2

(Refer Slide Time: 0:17)

We define a property as a measurable macroscopic characteristic of a system. We are only

looking at macroscopic thermodynamics. More importantly, thermodynamic properties relate to
energy. We said that thermodynamics is a science which deals with conversion and
transformation of energy. So, we are interested in properties which deal with energy. Some
examples of those are pressure, temperature and volume. We will look at several other properties
like enthalpy and entropy and so on as we go in this course. We have a further classification of
these properties as extensive, intensive and specific.

Extensive properties depend on the extent of mass. Intensive properties are independent of the
quantity of mass. And a specific property is an extensive property which, by some means, we
make independent of the quantity of mass.

Let us look at an example of all of these. Let us say I have a box, which contains some mass. If I
cut this box into two, does the mass of each part change? The mass of each part does change. So,
mass itself is an extensive property. When I cut the box into two parts, each part has a volume

13
which is different from the initial volume of the entire full box. Therefore, volume also is an
extensive property. Let us say I was measuring temperature. Initially, the temperature
everywhere in the box was uniform. It had some value. I cut the box into two parts,
hypothetically. Does the temperature of these two parts change? We do not expect the
temperature to change. Therefore, temperature is independent of the size of the box or the mass
of the box. Hence, temperature is an intensive property. Temperature does not change depending
on how much of the quantity of a system I take. If I make the system into half or into three parts
or four parts, it still is the same. Similar conclusions can be drawn in the case of pressure. If I cut
the box into two halves, the pressure of each half will still be the same as it was before. It is
independent of mass. Hence, pressure is an intensive property.

Usually, we do not deal with extensive properties much. There is a way to make an extensive
property independent of mass. For example, a volume can be made independent of the quantity
of mass by dividing it by its mass. It is called a specific volume, which we represent as v.
Similarly, I can find other properties on a specific basis, for example, specific energy, specific
enthalpy, specific entropy and so on. We will discuss these properties later on in the course.

For extensive properties, we usually use capital letters, for example, V denotes extensive
property (volume). There are of course some exceptions. For example, we use m for mass
because we use M for molecular weight. We use T for temperature, which is an intensive
property, because we use t for time. But for the other properties, we use small letters usually for
intensive properties, for example, p is pressure (intensive property) which is independent of the
amount of mass.

For specific properties, which are also independent of the mass, we use small alphabets, for
example, we use v for specific volume. We use V for total volume. So, if I was to remove the
exceptions, we use, for example, v, p and so on for properties which are independent of mass
(intensive properties). We use capital alphabets for properties which are dependent on the
quantity of mass (extensive properties).

(Refer Slide Time: 6:25)

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A state is the set of all properties required to completely describe the system. For example, when
somebody asks you where you are, you need to give them information so that they can find you.
You need to give sufficient information. Similarly, we can think of a state as all the things which
you need to tell so that somebody can reproduce exactly the same system which you have, if they
want to get there.

For example, you have air in this room. If somebody needs to recreate this room, they need to
know what the air is composed of, what its temperature is, what its pressure is, and so on. So, all
of the properties required to describe the system together is called the state. Since we are trying
to define the properties, the property should not be changing. Because, by the time I measure it,
it should not be some other value.

So, we need the properties to be constant. Or, if they are changing, the changes should be so slow
that we can sort of specify their values after some time. For example, after a day, maybe the
value of a property is something else, but at least today or at least at this point of time, it is some
fixed value. After one hour, it may be slightly different. That is alright. In order to measure
properties, we need the system to be in equilibrium, which means properties should not be
changing. When the system is in equilibrium, the properties have definite values. Any two
independent properties can be used at axes and the state can be plotted as a point.

15
We are used to, for example, plotting graphs of say, y versus x. If some distance in y is 2
centimeters and some distance in x is 2 centimeters, you can plot it and say that this represents
that point where y is 2 and x is 2. In a similar fashion, we can plot various properties on graphs
and they help us to visualize processes which happen. We can plot any property versus any other
property. Which set of graphs we use depends on what we can see from the graph. For example,
if the temperature is 300 or 300 K and volume is 5 , I can plot this point on a graph, and
say, this represents my state as far as temperature and volume are concerned. If the state changes
after some time to some other value, I can also plot that. Similarly, I can also plot pressure versus
volume. I can show what happens when the pressure changes or when the volume changes.

One example is that of a bicycle pump. I am filling a tyre of my cycle. Initially, the pressure
inside my bicycle pump is low and the volume is high. Then I press the plunger down. Let us
say, for the moment, I keep the outlet closed. When I press the plunger down, in the bicycle
pump, the pressure increases and the volume decreases. I can plot this on a graph. Initially, let us
say, pressure is 1 bar, volume is some 5 units. After a while, the pressure, when I press the
plunger down, increases and the volume reduces. It means the new point must be higher on the p
axis than the previous point and lower on the V axis.

(Refer Slide Time: 10:46)

A phase is a quantity of matter which is homogenous throughout. We talk of, for example, a
solid phase, a liquid phase, a gaseous phase for the air around. We can also talk of plasmas. In

16
this course, we are just interested in a solid phase, a liquid phase, and a vapor phase, and we look
at some changes in phase between these.

A path is a locus of all the intermediate states which the system passes through when there is a
change of state. For example, we just looked at a p-V diagram for a bicycle pump. Initially, the
pump has some volume and some pressure. I press the plunger in. After sometime, it has some
other volume and some other pressure. We have the initial state and the final state. If I do this
process really slowly, I can also measure, at any time in between, pressure and volume at each
intermediate state. I can connect all of those points. The locus formed by those points is
essentially called as a path. It is a locus of all the intermediate states which the system went
through between the initial state and the final state. Whenever there is a change in properties of a
system, we say a process occurs. So, a process occurs when a system changes from one state to
another. In a process, any property of a system can change, for example, the pressure can change,
the volume can change, the temperature can change, mass can change, and so on. But there are
some special processes where some property remains constant.

If I take a system, then by definition, the mass is constant, because when I talk of systems, I am
generally talking of closed systems. So, the mass is constant. Suppose I have some mass of a gas
in a rigid chamber. The volume is fixed as the walls are rigid. So any process happening inside
this chamber is a constant-volume process. Such a process is also known as an isochoric process.
Similarly, I can have some other process where the pressure is constant, and we would call that
an isobaric process. A process where the temperature is constant is called as an isothermal
process.

A process where there is no heat transfer across the boundary of a system is called an adiabatic
process. In an adiabatic process, properties may change. We also have cyclic processes, which
we will define later.

In order to define a state, we said that you need to have an equilibrium, which means things
should be constant as a function of time. In order to have equilibrium, you should have
mechanical equilibrium i.e. all forces must be balanced. You should have thermal equilibrium
which means there should not be any difference in temperatures. You should have chemical
equilibrium: if there are reactions, rate of forward and reverse reactions should be the same. You

17
should have phase equilibrium: if I have a mixture of two phases, for example, water and ice, the
quantity of ice and the quantity of water should be constant i.e. there should not be net melting of
ice into water or net freezing of water into ice. In order to have all of these equilibria and still
have some change, we need to consider a quasi-static process. This process is such that the states
achieved by the system are always near equilibrium. Anything which changes, changes so slowly
that you can assume that the system is always under equilibrium.

(Refer Slide Time: 14:57)

Let us look at a cyclic process. Let us say you have a piston and a cylinder, which is usually our
most favored configuration for describing a lot of things. We will keep going back to piston-
cylinder arrangements.

The piston-cylinder arrangement has some pressure and some temperature and volume. We call
it as state 1. Very slowly, the temperature inside changes after some time. This process is
happening slowly. It is a quasi-static process. The system is now in a new state (state 2). After
some time, let us say I pull the piston down. The volume has increased. Temperature also might
have changed. The pressure also might have changed.

This is some state 3. After some time, some other property changes and I come to some state 4.
After a while, properties change in such a way that the system comes back to whatever the set of
properties it had initially. It comes back to state 1. This kind of a process, where we reach the

18
initial state at the end of the process, is what we call as a cyclic process. So, the final properties
are the same as initial properties. If initially, the pressure was , finally, the pressure is again .
If I want to find out the change in pressure, it is essentially 0. Similarly, change in mass is 0,
change in volume is 0, change in temperature is 0 and so on. So, there is no change in any
property between the initial and the final state.

(Refer Slide Time: 17:39)

We said that there has to be equilibrium in order to measure the properties of a system.
Consider a box as shown in the figure where one side is hot, the other side is cold. If I ask you to
measure the temperature of the box, you may not know where to measure as different parts of
the box have different temperatures. So, we are not able to find out a single average value, which
is representative of the temperature of this entire box. Whereas, if we leave it for a while, the
temperature inside will be uniform. Now, if I ask you to measure the temperature of this box
again, you can measure anywhere. Anywhere you measure, the value is the same. In this case,
the box has reached thermal equilibrium. Hence, we are looking for systems under equilibrium,
or very close to equilibrium, so that we can define the properties of the system.

19
(Refer Slide Time: 18:43)

In a similar fashion, we also need to have mechanical equilibrium. If I have this piston cylinder-
arrangement (shown in figure) and I was moving this piston really fast, either down or up, then
the pressure near this piston and the pressure far away from this piston would not be the same.
Then, if I ask you to tell me the pressure of the system, you would not be able to give me a single
average value for the pressure. Such kind of a system is not very useful, because we do not know
how to deal with it. Whereas, if the piston is moving slowly or if it is stationary, then everything
inside is essentially the same. The pressures at different points would be the same or
approximately the same if it is moving very slowly. Then, I can say any of the pressure values is
a representative of the system.

So, we are looking for systems which are close to equilibrium even if they are changing or at
equilibrium. Then we can define the properties very clearly. It is these kind of systems which we
are going to deal with in this course. Quasi-static near equilibrium processes can be dealt with
efficiently.

20
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 03
Basic Concepts and definitions -Part 3

Welcome back to the thermodynamics course. In the last class, we looked at the basics of
thermodynamics. We looked at what thermodynamics deals with and definitions such as state
and path and process and so on.

(Refer Slide Time: 00:37)

We will go on to what are called exact and inexact differentials. These are also called as state
and path functions: the exact differential corresponds to a state function and the inexact once
corresponds to a path function. We use different notations to represent them.

We use the alphabet d to represent an exact differential. For example, we write to

represent an exact differential. In contrast, for inexact differentials, we use Greek letter . We
write for an inexact differential. In some books, the inexact differential is also written
with a .

The properties are exact differentials and they are only defined at a state. Of course, a state
requires the system to be in equilibrium. Once you have a system in equilibrium, the
properties only depend on that particular state. In contrast to that, work and heat interactions
are inexact and they depend on the process or the path.

21
(Refer Slide Time: 02:04)

Figure 1: The map of India.

It sounds a little abstract at the moment, but we can look at it with the help of a map. As we
discussed earlier, the state is something which depends on the properties, and you can plot
any properties on a map. For example, Fig. 1 shows the map of India, a physical map, which
shows the latitudes and the longitudes. If I want to specify what in the thermodynamics
would be called a state, I can think of it as a location on a map.

A point on a pressure-temperature diagram has pressure of a certain value and temperature of

a certain value. I can think of it as equivalent to a location/point on a map. Here, pressure and
temperature are analogues to a latitude and a longitude of certain values. If I give the latitude
and longitude values of a particular location on a map, everybody will be able to find that
particular place. Corresponding to this location, there is a certain temperature associated with
it (may vary with time), there is a certain altitude associated with it, and there is a certain
velocity of wind. There are a lot of other properties associated with it. So, as long as I give
the latitude and longitude, everybody knows that this is the particular place we have to go to.

Instead of giving these two properties (latitude and longitude), I could give name of the place,
for example, Hyderabad. Is one attribute (the name of the place in this case) enough to locate
the place on the map? In this case, it may not be enough, because there is a place called
Hyderabad in India which all of us are aware of. There is also a place called Hyderabad in
Pakistan. Hence, I may need to tell you something else. If I go to the place called Hyderabad,

22
which is at a latitude of around 20 degrees, then I would be able to find out that this place is
in India.

So, if I told you the name of the place and one other attribute, for example, latitude, then you
may be able to find out the place. In some cases, it may be sufficient to tell you only one
property/attribute to locate the place, for example, Mount Everest. There is only one Mount
Everest and the name itself is sufficient to find its location.

In a similar analogy, we will find later on, that under certain conditions, where if I just tell
you one property, that is sufficient to identify a state in the thermodynamics sense. The
variables we usually use are pressure, temperature, specific volume, etc. We may need to
specify two of them or one of them or more than two of them in order to specify the state.

We can plot all of the thermodynamics properties on maps, for example, we can have a
pressure versus volume map. If I have pressure which has a certain value a and volume of a
certain value b and if I go to a location which has a pressure of a` and volume of b`, this new
location is a distinct location which can be found.

State functions or properties are represented as exact differentials. A small infinitesimal

change in pressure is represented as dp, or a small infinitesimal change in volume is
represented as dv.

In contrast to these, I could ask other questions. For example, if I go from Chennai to Delhi,
what is the distance I would travel? The distance of course depends on how I go. I could take
a road which goes more or less a straight line. I could go first to Calcutta and then I go to
Delhi. So, the distance I travel to go from one point to another would naturally depend on
how I go there. Different paths imply different distances. So, these kinds of functions which
depend on the path which I take are called as path functions. We will see later on that work
and heat are two examples of path functions.

Path functions are represented by the alphabet . For example, in the case of work,
infinitesimal small work is represented as . We will see later on that we use alphabet q to
represent heat. So, a small quantity of heat which is transferred is represented as .

23
So, in the case of the state functions or the properties, we talk of a certain location on the map
having a certain value, whereas in the case of path functions, we talk of a certain path being
followed, when going from one location to some other location on the map.

(Refer Slide Time: 08:42)

Figure 2

Let’s see how to describe these things mathematically. In Fig. 2, the z value (altitude or
height) is a function of x only as the slope is constant along x direction. However, it is not a
function of y as it is not changing in y direction for a given x.

So, the height is varying as I go along the x direction. As I go along the y direction, the
altitude values are not changing at all. Hence, z is a function of only x.

In this case, the slope is constant in the x direction. Let’s call the altitude of any point on this
surface as dz (with respect to origin). To find dz, I would find the slope of this plane, which is

essentially (which is constant), and multiply it by the distance dx. This would tell me what

is dz. I can find as and multiply that by dx to get the value of dz. This is a special

case where the slope is constant.

24
(Refer Slide Time: 11:42)

Figure 3

The more general case is where the slope, even in one direction, is changing (Fig. 3). As we
go along the x direction, the value of z changes. However, as we go along the y direction (for
a given x), the value of z does not change. Mathematically, we say z is a function of x. For
any given value of x, if I go along the y direction, the value of z does not change.

So, z is not a function of y, but z is a function of x. Here again, it is a function of only one
variable, which is x. In a similar fashion to what we wrote earlier, if I go small distances

along the x-axis, I can say . But in this case, the slope ( ) along the x-axis itself

is changing. Hence, I cannot write for large distances.

In this case, I can use the expression only when dx is a small distance so that is

constant over that small dx. For a large distance, ∫ ∫ . Basically, we find a

slope over a very short distance (so that the slope can be considered constant over that
distance) and multiply it by that short distance to obtain local dz. We repeat the same
procedure till we reach the point whose height/altitude we want to find. Then, we just add all

the local i.e. we integrate. Hence, ∫ ∫ . In Fig. 3, (the height of

point 2 above origin) is the sum of 5 different where, in each case, dx is 1 unit long and

the slope is constant over that 1 unit.

25
The height at a particular point on the graph can be found out using any path available. For
example, the height of the point 2 above the origin in Fig. 3 can be found out by going along
the edge of the surface (y=0) or going along x=0 till y=2, then going along x till x=5 (at y=2),
and then going along y till y=0 (at x=5). There are infinitely many such paths. In all these
cases, the height of the point 2 above the ground is the same.

The height/altitude of any point on such a graph/surface is a state function or a property. It is

independent of the path used to calculate it.

(Refer Slide Time: 19:10)

Figure 4

In the previous graph (Fig. 4), since the slope ( ) is constant, it can be taken out of the

integral in the general expression ∫ ∫ .

So, the height of any point a on such surfaces can be found with respect to any other point b
provided that b’s height is known with respect to the ground or it can be calculated using the
same procedure mentioned before.

However, while calculating the height of a point with respect to the origin for example, the
distance travelled depends on the path taken from the origin to the point, to calculate the
height. This is an example of a path function, because it depends on the path I travel by.

26
Similarly, when we go by a car from one location to another by different routes (assuming car
characteristics are the same all the time), the amount of petrol needed depends on the distance
travelled. Hence, the amount of petrol needed is a path function.

(Refer Slide Time: 25:47)

Figure 5

In Figs. 4 and 5, for a given x, if I move along y, z does not change. So, these are the cases
where z is a function of only one variable.

(Refer Slide Time: 26:04)

Figure 6

27
A more general case would be, where z changes, if I go along x and/or y as shown in Fig. 6.
For y=1, if I go along x from 0 to 5, the z value keeps changing. Similarly, for x=1, if I go
along y from 0 to 5, the z value keeps changing.

So, z is a function of x and y, which is to say z varies if I go along x and/or y. For measuring
a change in z, I need to account for changes in z along both the directions. In such a case, the

change in altitude/height i.e. is expressed as ∫( ) ∫( ) , where ( )

represents a change in z with respect to x when y is kept constant (partial derivative of z with

respect to x) and ( ) represents a change in z when with respect to y when x is kept

constant (partial derivative of z with respect to y).

I want to go to a point (x=2, y=2) from (x=0, y=0) on the surface shown in Fig. 6. There are
many ways to do that. Let’s discuss two of those. I can go from x=0 to x=2 along y=0, and

then from y=0 to y=2 along x=2. In this case, ∫ ( ) as the slope is 0 and

∫ ( ) has some value. Another way that I can go to a point (x=2, y=2) from (x=0,

y=0) is to go from y=0 to y=2 along x=0, and then from x=0 to x=2 along y=2. In this case,

∫ ( ) as the slope is 0, while ∫ ( ) has some value. However, in both

the cases, the height of the point (x=2, y=2) with respect to (x=0, y=0) would be the same.

There are ways where both the terms, ∫ ( ) and ∫ ( ) , are not 0. In all these

cases, these integrals can be calculated as discussed before. It is also possible that we know
the functional form of z = f(x,y). In such a case, finding derivatives and integrations become
easier.

28
(Refer Slide Time: 32:38)

Figure 7

Finding between any two points on the surface in Fig. 7 can be done in multiple ways.
However, no matter what path you take, would be the same between those two points.

Again, the distance travelled between any two points depends on the path taken. It is a path
function. We don’t have the expression for distance travelled as we have for the height
difference between two points.

29
 Thermodynamics
Professor. Anand T. N. C
Department of Mechanical Engineering,
Indian Institute of Technology, Madras.
Lecture 04
Tutorial problems on exact and inexact differential
(Refer Slide Time: 0:19)

Figure 1

If we have z as a function of x and y, we should be able to write it in the form ( )

( ) . ( ) represents derivative of z with respect to x keeping y constant (partial derivative

of z with respect to x) and ( ) represents derivative of z with respect to y keeping x constant

(partial derivative of z with respect to y). How do we know that dz is an exact differential or
inexact differential?

Let’s say we have ( ) and ( ) . Now, differentiate M with respect to y and N with

( ) ( )
respect x. Therefore, and . Now, compare and . If

, dz is an exact differential. However, if they are not equal, dz is an inexact differential.

30
Physically, exact differentials can be used to represent points on a graph. They represent
properties or state functions. Inexact differentials can be used to represent path functions.

We want to know if ∫ is an exact or inexact differential.

Let’s write it as ∫ . Here, z is a function of two variables p and V.

Hence, ( ) ( ) . Let’s write dz as . Here, M=0 as there is

no dp term in the expression and N=p. Apply the same strategy as before.

( ) ( )
and .

Hence, . So, dz is not exact in this case. Hence, it is a path function.

We encourage you to do a similar exercise with the expression ∫ . This is also an inexact
differential.

We will see if the third expression ∫ is an exact or inexact differential. Let’s follow

the same steps. Here again, z is a function of two variable p and V. So, ( )

( ) .

Therefore, ( ) and ( ) . Now, and . Hence,

. Therefore, dz is an exact differential. z represents a property or a state function.

The value of pdV+Vdp does not depend on the path followed by a system. It depends only on the
state of the system. In this case, the integral depends only on the values of pdV+Vdp at states 1
and 3. It does not depend on how you go from 1 to 3.

31
(Refer Slide Time: 8:51)

Figure 2: The map of India

Figure 3

If we were to look at the map of India again (Fig. 2), the difference in pressure and density at two
different locations at a particular time does not depend on the path followed in going from one
location to another because density and pressure are state functions.

I request you to find out if the expression ∫ represent an exact or inexact

differential. This differential is inexact.

32
(Refer Slide Time: 10:43)

Figure 4

In Fig. 4, we have a function ( ) ( ) . Let’s say this expression was written for

an ideal gas. We want to find out whether ɸ is a property or not. If it is not a property, make
changes in the expression such that ɸ becomes a property. Here, R is a constant (gas constant).
Apply the same procedure as we applied in the earlier problems. There are many ways ɸ could
be made a property.

33
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 05
Basic concepts and definitions – Part 4
(Refer Slide Time: 0:18)

We will now see how to measure pressure. We will see rigorous definition of temperature a little
later on in the course, and at that point of time we will see how to measure temperature. Volume
is relatively easy to measure.

34
(Refer Slide Time: 01:00)

Figure 1

Pressure is defined as the normal force exerted by the fluid per unit area. We are mainly
interested in fluids here, liquids and gases. We would look at it in that context.

The SI unit of pressure is Pa (pascal). 1 Pa is a force of 1 newton being exerted on an area of 1

meter squared. We will commonly use kilopascal and megapascal. 1 kilopascal equals 1000 Pa,
while 1 megapascal is 106 Pa. Another commonly used unit is a bar, 1 barometer pressure. 1 bar
equals 105 Pa. The value of 1 bar is close to the value of another unit of pressure which 1 atm
(atmosphere). 1 atm is the pressure on the surface of earth at sea level.

1 atm equals 101325 Pa or 1.01325 bar. Many times 1 bar and 1 atm are taken approximately
equal in calculations as their values are close to each other. However, it is not recommended as it
introduces error in the calculations.

The pressure can be either the absolute pressure or the gauge pressure. The gauge pressure is the
pressure with respect to the atmosphere. It is commonly measured using a manometer or a
bourdon gauge.

A U-tube manometer connected to a cylinder at a pressure higher than the atmospheric pressure
is shown in Fig. 1.

35
The U-tube contains some fluid (a liquid most of the times) which does not evaporate much and
does not stick to walls of the tube. One end of the U-tube is connected to the cylinder, while the
other end is open to atmosphere. If the pressure inside the cylinder is higher than the atmosphere,
the level of the fluid inside the U-tube in the arm on the cylinder-side (right arm) goes down,
while it goes up in the arm which is open to the atmosphere (left arm). Balancing pressures at the
level of fluid in the right arm gives

Pressure of the gas in the cylinder = atmospheric pressure acting on the manometric fluid in the
left arm + pressure due to height of the manometric fluid in the left arm above that in the right
arm = .

If the pressure of the gas is measured with respect to the atmosphere, it is gauge pressure which
in this case is . If you add the atmospheric pressure to it, then you get the absolute
pressure which in this case is .

Another device which is often used to measure the pressure is the bourdon gauge which is shown
in Fig. 1. The blue-colored tube inside this gauge has an oval cross-section. When the gauge is
attached to a system whose pressure is to be measured, the blue-colored tube bends/deforms
which in turn rotates a gear mechanism which in turn rotates a needle. This needle rests on some
value of pressure printed on the dial. A bourdon gauge gives you the gauge pressure.

36
(Refer Slide Time: 7:58)

Figure 2.

We use barometer to measure the atmospheric pressure. Figure 2 shows the schematic of the
barometer. The setup contains a long tube and a bath, both filled with mercury. Initially the tube
is filled with mercury completely. Then it is inverted into the bath. The level of mercury in the
tube falls until it reaches an equilibrium position. If we do a pressure balance at the level of
mercury in the bath, we get

Pressure acting on the mercury in the bath (atmospheric pressure, ) = pressure due to height
of mercury column in the tube + pressure due gas above the mercury column in the tube (which

is usually negligible as mercury has small vapor pressure) =

Since we know the value of patm, we can calculate the value of h. It comes out to be around 760
mm.

It is also useful to get a feel of these numbers.

Suppose you have a mass of 1 kg. Its weight, assuming g=10 m/s2, is 10 N. When this weight is
kept on a surface area of 1 cm by 1 cm, the pressure acting on that surface area is 1 bar.

37
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 06
Work - Part I
(Refer Slide Time: 00:18)

In the last class, we looked at inexact and exact differentials. In this class, we are going to
look at the concept of work. This word is quite familiar with us. We have looked at this in
mechanics, where for example, we define work as the force into the displacement in the
direction of the force. Mathematically, that would be ∫ , where ds is a small displacement
and F is the force in the direction of the displacement.

In thermodynamics, our way of looking at work is slightly different. We know that power is
rate of doing work, which is a definition which we are going to use in both mechanics and
thermodynamics. But, for work itself, our definition in thermodynamics deals with energy.
We look at work as an energy interaction between a system and its surroundings. We defined
the system earlier as the mass which we are interested in and the surroundings as anything
else which is external to the system.

We are going to look at work as an energy interaction. We have not yet defined what energy
is. We will formally define it when we go on to the first law. It may seem like a chicken and
egg thing, where we define energy in terms of work and work in terms of energy.

38
(Refer Slide Time: 01:46)

We say that work is said to be done by a system during a given operation, if the sole effect,
external to the system, could be reduced to the raising of a weight through a distance. We are
not interested in what happens inside the system during the operation. We are only interested
in what happens outside the system.

This emphasizes that work is interaction between a system and its surroundings. There does
not actually have to be a weight which is raised. Whatever is being done by the system on its
surroundings can be thought of as raising a weight.

39
(Refer Slide Time: 03:32)

Figure 1
In Fig. 1, there are at least two systems which are marked by the dotted lines. One is a
battery, and second is gas surrounding a paddle wheel which is confined inside a box. There
is also a third system which is a motor. The battery is connected to the motor. The motor is
connected to a paddle wheel. The paddle wheel is rotating inside a gas.

As the circuit is turned on, there is energy flow from the battery. The battery makes the motor
run. When the motor runs, the shaft spins. The fan rotates and the gas inside the box is
rotated. This definitely does not look like there is any ‘raising of a weight’. However, we can
attach a weight to a string and the string can be wound around the shaft. Instead of a fan, we
now have a weight. When the motor spins the shaft, the weight gets raised. Instead of
rotating/circulating the gas inside the box, the motor is now raising a weight. Hence, the work
is being done by the battery on the gas (we assume that all the components in Fig. 1 are ideal
so that there are no losses (e.g. friction).)

In this case, the battery, which is connected to the shaft, which is connected to the fan, is
essentially doing work on the gas, because it is reducible to the ‘raising of a weight’ external
to the system (battery in this case).

We have a sign convention with respect to work. Not everybody agrees on the sign
convention. Different books use different sign conventions.

40
Work done is positive if it is done by the system on its surroundings. The work done is
negative if it is done on the system by the surroundings. This is the sign convention we are
going to use throughout the course.

(Refer Slide Time: 07:47)

There are various forms of work. One of them is the displacement work. This is the work
involved when the system boundary sweeps a volume while interacting with its surroundings.
It can be visualized using a piston-cylinder arrangement.

Assume that the piston is frictionless and free to move. When the piston moves, the volume
inside the cylinder (control volume) increases or decreases. As the piston moves, it sweeps a
volume, which may be against the surroundings or against a system. If the piston moves
upwards (or outwards), then the volume of the control volume (or a system) increases. Then
we say that the work done is positive. The system increases its volume at the cost of its
surroundings. It is positive for the system and negative for the surroundings. If the piston
moves downwards (or inwards), it reduces the volume of the system. Then, the work would
be negative. The work can be evaluated only for a quasi-static process which will be defined
later.

We usually neglect magnetic, gravitational and electric forces in the analysis of the systems
we are going to look at. However, gravity forces may play important role in a few cases.

41
(Refer Slide Time: 10:05)

Figure 2
Work is done when there is interaction between the system and the surroundings, and the sole
effect external to the system is reducible to lifting of a weight.

Consider a piston-cylinder arrangement shown in Fig. 2. The system consists of gas inside
this arrangement. (A common example of a piston-cylinder would be a plastic medical
syringe.). The arrangement is assumed to be frictionless and, the piston is assumed to be
massless (the mass of the piston is negligible compared to whatever we are going to deal
with).

There is a big block siting on the piston. This block has some mass. So, the pressure inside
must be high because, the block is sitting on the piston, and the system is in equilibrium.

If the block is removed suddenly from the top of the piston, the piston moves up until the
pressure inside becomes equal to atmospheric pressure (outside pressure is atmospheric
pressure). The gas inside expands and the volume of the system increases. The cylinder is tall
enough that the piston does not jump out.

However, during this upward movement of the piston, no weight is raised (neglecting the
mass of air outside or assuming that the experiment was carried out in vacuum).

Work is done only if there is raising of a weight external to the system. In this experiment, no
weight is raised. Hence, the work done by the system must be close to 0 (ignoring the
atmospheric pressure around or assuming that the experiment was carried out in vacuum).

42
Now, cut the block into two. Remove one of the blocks suddenly. The piston moves up
raising the weight of other block and stops when the system reaches new equilibrium. In this
case, work is done by the system because there is raising of a weight by the system.

What would happen if we cut the block into smaller and smaller pieces (or infinitesimally
small pieces) and remove each piece one by one?

(Refer Slide Time: 15:57)

Every time a piece is removed from the top of the piston, it moves up slightly raising the
weight of the remaining pieces and the system attains a new equilibrium. In this case, the
system is not far from the equilibrium. As a piece is removed, there is a slight imbalance
between the forces inside and outside, and the piston moves up to balance the forces again.
The process is slow. There is always a resistance to the movement of the piston. The work is
done by the system as there is raising of weight. This kind of a process is called as a quasi-
static process or a fully-resisted process or a near-equilibrium process. In this process, the
system is never far away from the equilibrium. Since the system is almost always at
equilibrium in such a process, its properties can be measured (the properties of a system can
be measured only when it is in equilibrium). Since the properties of a system can be
measured, work done by the system can be calculated.

43
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 07
Work - Part II
(Refer Slide Time: 00:19)

Figure 1
Consider a piston-cylinder arrangement as shown in Fig. 1. At any spatial point, p represents
the pressure. There is atmospheric pressure on the right side of the piston.

Initially, the piston is at location 1. The volume on the left hand side of the piston is and
the pressure is . Suppose the contents on the left hand side of the piston-cylinder
arrangement (the system) underwent a process during which the piston moved out and
reached point 2. Here, the volume is and the pressure is . The pressure acting on the
piston at any given point of time, as usual, is given as the force acting on the piston at that
point in time divided by the piston cross-sectional area, A.

Initially, since the system is in equilibrium, the pressure values inside and outside are equal.
Let’s say pressure inside is . Assume that there is a slight imbalance in the forces on the
two sides of the piston and the piston moves a small distance in the x-direction (Fig. 1)
(we use d to represent exact differentials). There is an expansion of the system. In this case,
the work done is where F is the force acting on the piston and is the small
displacement. Over the small time interval during which the piston moved by , the
pressure can be taken as . Now, where A represents the cross-

44
sectional area of the piston and represents the small amount of work done by the system
(we use to represent inexact differentials).

The term represents the volume swept by the piston as it moved. It is the volume of the
surroundings that was displaced. So, , where represents the volume swept by
the piston. is infinitesimally small.

The displacement of the piston from location 1 to 2 can be divided into small parts such as
As the volume of the system increases from to , the total work done by the system is

equal to the sum of all such s which can be obtained by integrating . Hence, ∫

equals the total work done by the system as it expands from to .

In reality, the pressure inside the system is changing. The volume of the system is also
changing. By considering small time intervals during each of which the piston moves , and
assuming that the pressure stays constant over that time interval, we can obtain the total work
done by the system by adding the work done over such small time intervals. The cost of
assuming the pressure to be constant over those small time intervals for the accurate
calculation of work can be reduced by choosing even smaller time intervals.

The work done is represented as 1 or . So, displacement work, ∫ . The

system expands quasi-statically from to .

(Refer Slide Time: 06:11)

Figure 2

45
We can represent a process on a graph, e.g., a p-v graph for a system consisting of some gas
where v represents specific volume (specific volume = total volume/mass). Figure 2 shows a
process undergone by a piston-cylinder system as it expands from (corresponding pressure
is and the specific volume is ) to (corresponding pressure is and the specific
volume is ) on a p-v diagram. The process is quasi-static. The piston is moving slowly.

For this process, the total work done by the system is ∫ . is essentially the

area under the curve for a small change dV. Hence, ∫ gives the area under the curve

joining points ( ) and ( ). Also, ∫ ∫ ∫ as the

mass of the system is constant. Hence, the work done per unit mass i.e. specific work is

∫ . Thus, the area under a curve on p-v graph gives specific work. The area under

the curve can be found out if we know the equation of the curve. The curve can be plotted by
experimentally measuring the values of p and v.

(Refer Slide Time: 10:47)

Figure 3
Since the area under the curve depends on the curve, different curves will have different
areas. Hence, the work done also would be different for different curves. In Fig. 3, the system
changes its state from 1 to 2 through three different paths, a, b and c. In all the three
processes, the work done would be different. Hence, work is a path function. It is analogous
to the distance travelled between two points on a map since distance travelled depends on the
path taken to travel. However, the displacement is independent of the path. Since work is a

46
path function, we cannot talk of work on a point. We consider work transfer along a path the
system takes when it goes from state 1 to some other state 2.

47
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 08
Work - Part III
(Refer Slide Time: 00:19)

Work done by a system is represented as 1W2 or W1-2, and ∫ . Specific work is

represented as 1w2 or w1-2, and ∫ (mass of the system is fixed). The integral in

the expression for work can be calculated if the functional form of p in terms of V is known
or p is constant.

A system may undergo various processes. For different processes, the expressions for work
done are different. Let’s look at some common processes.

48
(Refer Slide Time: 01:30)

Figure 1
The horizontal line (yellow colored) in Fig. 1 represents a process on a p-V diagram where
pressure remains constant for any value of volume. The process can be represented as p = c
where c is some constant. All the horizontal lines on such a p-V diagram represent constant
pressure processes.

The vertical line in Fig. 1 represents a process where volume remains constant for any value
of pressure. All the vertical lines represent constant volume processes.

One of the other processes which is commonly encountered in thermodynamics is an

isothermal process. It is represented as for an ideal gas. We will look at the
concept of an ideal gas a little later. The equation on a p-V graph represents
a rectangular hyperbola. In Fig. 1, the green curve is rectangular hyperbola, .
It represents an isothermal process for an ideal gas.

The process expressed as on a p-V diagram is called a polytropic process.

We can obtain expressions for different processes by using different values of n. For
example, if ( is the ratio of specific heats), we get an adiabatic process for an ideal
gas. Isobaric, isochoric and isothermal processes can also be thought of as special cases of a
polytropic process.

49
n = 1 in results in the expression for an isothermal process. n = 0 implies p

= constant which is an isobaric process. Rewriting as and

setting n = , we get isochoric process (V = constant).

Let’s see how to calculate work done in each of these processes.

Isobaric process:
 p = constant
 ∫ ∫

Isochoric process:

 V = constant
 ∫ (dV = 0 as V = constant)

Isothermal process:
 pV = c (c is constant)

 ∫ ∫ ∫
(since )

Polytropic process:
 (c is constant)

 ∫ ∫ [ ]

…….(1)
 Work done in all the other processes discussed in this lecture can be obtained from the
expression of the work done in a polytropic process by setting the value of n
appropriately. For example, for an isobaric process, n=0 and p=constant. Hence,
equation 1 implies .

Adiabatic process:

 Equation 1 implies

50
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Work - Part 4
(Refer Slide Time: 0:17)

We looked at how to calculate the work when you have displacement work. For example,
when you have gas inside a piston-cylinder arrangement, and the gas expands moving piston
outwards against some external resistance (for example, atmospheric pressure), then the work
is given by ∫ . Work is a path function.

We obtained the mathematical expressions for the work done in various common processes.
Please see the notes for the last lecture.

51
Let’s look at some problems on the concepts covered in the last few lectures.
(Refer Slide Time: 2:29)

Figure 1

Question 1. What is the cyclic integral of a property?

Answer: The cyclic integral of a property is zero. In a cyclic process, the final state is the
same as the initial state. Since properties are state functions, their values do not change after
the cyclic process. Hence, the cyclic integral of a property (property value at the final state –
property value at the initial state) is zero.

For a cyclic process 1-2-3-4-5-6-7-1 on a p-V diagram shown in Fig. 1, the cyclic integral of

volume is given as ∮ ∫ ∫ ∫ ∫ ∫ ∫ ∫
( ) ( ) ( ) ( ) ( ) ( ) ( ) . A
similar conclusion can be drawn for other properties.

Question 2. What is the work done in a cycle?

Answer: Work is a path function. Though the states of a system are the same at the end of a
cyclic process, the path followed by the system could be different. Hence, the work done in a
cycle would be non-zero.

Consider the cyclic process shown in Fig. 1, 1-2-3-4-1. The displacement work for the
process is given as,

52
 ∮ ∮ ∫ ∫ ∫ ∫

∫ ∫ ∫ ∫ represent areas under the curves 1-2, 2-3, 3-4 and 4-1
respectively. Since the volume is increasing in the processes 1-2 and 2-3,

∫ ∫ would be positive. However, the volume is decreasing in the processes

3-4 and 4-1. Hence, ∫ ∫ would be negative. Adding all the integrals gives a
non-zero number which is the area of the region enclosed by the curves on a p-V diagram
(colored yellow in Fig. 1). It is the net work done by the system.

What if the system went along 1-2 and 2-3 and retraced the same path to come back to state?
Would the work done in that case be zero?

Question 3: What is a polytropic process?

Answer: A process which can be represented using the equation or

is termed as a polytropic process where p represents pressure, V represents
volume and v represents specific volume of a system (the mass is fixed).

for a system can be obtained by dividing each term in by

. Then, the left hand side becomes ( ) and the right hand side becomes a new

constant ( ) as the mass is fixed.

53
(Refer Slide Time: 12:37)

Work is an interaction between two systems or a system and the surroundings. Hence, we
require a donor (somebody to do the work) and a receiver (somebody to receive the work or
something on which work can be done) and interaction between the donor and the receiver.

The work interaction for the system plus the work interaction for the surroundings equal zero.
The work done by the system on the surroundings is positive, while the work done by the
surroundings on the system is negative.

If we have n systems having work interactions with each other, then

For a system, all other systems form surroundings. Some of the
systems may have positive work output, while others may have negative or zero.

For work to be done by a system, there must be a receiver system which receives work.
Otherwise, work done by a system is zero.

54
(Refer Slide Time: 15:01)

Newton’s second law of motion implies F = ma where m is the mass of the object under
consideration, F is the force acting the object and a is the acceleration of the object.

At equilibrium condition, summation of all the forces equals zero. The forces are balanced. In
thermodynamics, we look at systems which are always in equilibrium or very close to
equilibrium. Only at such conditions, we can calculate work interaction for a system.

Just because the forces acting on the object/system are absent or their summation is zero does
not mean that the velocities are zero. You can have an object moving at a constant velocity
when the forces are absent or their summation is zero. The absence of forces or their
summation being zero implies that the acceleration of the object is zero.

Work interaction for a system does not have to be zero if the summation of the forces acting
on it is approximately zero.

We are looking at quasi-static processes. The system is always near equilibrium. There is
slight imbalance of forces because of which we get work.

Work and heat are modes of energy transfer. We will define heat a little later. Energy transfer
can be work transfer or heat transfer or combination of both depending on the choice a
system. For example, in a typical piston-cylinder arrangement, a system can consist of only
the contents enclosed by the piston and the cylinder or it can consist of those contents as well
as the piston. Though energy transfer (sum of heat and work transfer) for the system in both

55
the cases is the same, its components, energy transfer by heat and work interactions, may be
different.

(Refer Slide Time: 18:12)

Figure 2
Let’s answer the question in Figure 2.

The apple falls under gravity. It covers a distance of 1.5 m before it hits Newton’s head. We
need to calculate the work done by the apple when it falls from a tree to a location just above
Newton’s head (it has not touched his head yet). Assume that the resistance due to air is
negligible.

During its fall, there is nothing that resists the apple’s movement. Hence, the work done by
the apple is zero. As pointed out before, there should be a donor system, a receiver system
and some interaction between the two. In this case, there is a donor system, the apple.
However, there is no receiver system. Hence, the work done is zero. So, as long as we are
considering the motion of the apple till the location just above Newton’s head, the work
interaction is zero. This process, the fall of the apple, is not quasi-static. There are many other
processes which are not quasi-static. Let’s discuss some of those.

56
(Refer Slide Time: 21:33)

Consider a membrane separating a box into two chambers. On one side of the membrane,
there is vacuum, while on the other side, there is gas (at some pressure pg). Initially, the
membrane separates them. If the membrane is punctured, the gas expands into the other
chamber (the volume of the gas increases) against no resistance as the other chamber initially
had vacuum. In this case, as there is no receiver system to receive work, the work done by the
gas is zero. This is known as unrestrained expansion. The example we just considered is a
case of completely unrestrained expansion. We can also have partially restrained expansion
where there is some resistance to the expansion.

In the above example, instead of vacuum in one of the chambers, let’s have some gas at a
pressure less than that of the gas in the other chamber. If the membrane is punctured now, the
gas at higher pressure will expand against some resistance. This is the case of partially
restrained expansion. In this case, the work is not completely zero. However, the work done
is not equal to ∫ where dV is the change in volume of gas at higher pressure.

57
(Refer Slide Time: 24:02)

Figure 3
Let’s look at an example of un-resisted, partially-resisted and fully-resisted processes (Fig.
3).

A man jumps from a plane and falls under gravity. Initially, he doesn’t deploy a parachute.
Assume that the air resistance is negligible. In this case, the work interaction for the person is
zero.

Assume that for the later part of the person’s fall, the air is dense. It offers some resistance to
the fall. Let’s say the person’s mass is 60 kg. Hence, his/her weight is around 600 N (taking g
= ). Assume that the resistance offered by air is 50 N. Hence, the resultant downward
force pulling the person down is 550 N. The balanced force is 50 N. If the person falls by 10
m, then the work interaction would be . The work interaction happens
because of the balanced force only. Un-resisted force does not do any work.

Now, the person deploys parachute and drifts slowly. The person is not accelerating now.
Hence, the person’s fall is fully resisted. Fully-resisted force is 600 N (the weight of the
person). At this condition, if the person covers 10 m, the work interaction would be
.

In un-resisted processes, there is no work interaction because nothing is accepting the work.
In partially-resisted processes, we have some amount of resistance. Hence, the surroundings
is accepting some amount of work. The work is done by the force which is resisted. In fully-

58
resisted processes, there is resistance to the complete force and therefore, all of that force is
taking part in doing some work interaction.

59
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Work - Part 5

(Refer Slide Time: 0:17)

There are forms of work other than displacement work: gravitational work, shaft work,
electrical work, surface tension work and spring work.

Gravitational work

We looked at an example of gravitational work being done when we analyzed the case of a
falling apple or a falling person. The gravitational work is given by the expression mgh,
where m is the mass of the falling object, h is the distance or height through which the object
falls and g is acceleration due to gravity. When the object is lowered in a gravity field, its
energy reduces (the work is done by the object), whereas its energy increases when it is
raised (work is done on the object).

Shaft work

If we apply torque T to a shaft which is free to rotate, it may rotate through some angle .
Then, the shaft work is given as . where N represents revolutions per second of
the shaft and t is the time it rotates for. Hence, shaft work = .

60
Electrical work

When a steady current I flows out of the system at point A and reenters at point B, the system
does a positive electrical work. The system could be a battery sending out current through a
load. The system could be a load with resistance R. Below are given the expressions for the
electrical work: , where I is the current flowing through the system,
R is its resistance, V is the voltage which is causing the current I to flow and R is the
resistance of the system ( according to Ohm’s law). The choice of a formula to use
depends on the system; for a battery, is more suitable, whereas for a
resistor, is a correct choice.

Surface tension work (liquids)

Surface tension tries to minimize the surface area of a liquid if disturbed from equilibrium
condition. It resists the increase in surface area.

Let’s look at an example. Consider a rectangular frame one side of which is movable, i.e. it
can slide. Use soap solution to make a rectangular soap film which anchors on the four sides
of the frame. Now, slide the movable side so that the area of the soap film increases. When
you leave/free the movable side, it comes back to its original position (it slides inwards). This
happens because surface tension opposes the increase in surface area of the liquid film and
acts to minimize it. This work is given as , where is the surface tension of the liquid
and is the increase in surface area (caused by the movement of the movable side) of the
liquid.

Spring work

A spring opposes its contraction or extension from its equilibrium position. When it is
stretched or contracted from its un-stretched equilibrium position, work is done on the spring
(work interaction for the spring is negative).

The force with which the spring opposes its contraction or extension is proportional to its
displacement from the equilibrium position, . We can write , where k is a spring
constant. Work done by the spring can be obtained by integrating the expression over
the displacement of the spring (from its equilibrium position) as ∫ ∫
(k is constant).

61
(Refer Slide Time: 6:18)

Figure 1
Writing correct units for different quantities is of utmost importance. It may lead to wrong
calculations and eventually result in wrong answers of the numerical problems if appropriate
units are not assigned to quantities. The cost is even higher if this happens in actual practice.

Let’s try to figure out what the units in Fig. 1 represent.

1. mPa – millipascal, MPa – megapascal (The symbol or the first letter of the symbol is an
upper-case letter when the name of the unit is derived from the name of a person, e.g., pascal
is named after Louie Pascal.)
2. mN – millinewton, nm – nanometer, – newton meter (Symbols for units formed
from other units by multiplication are indicated by means of either a half-high (that is,
centered) dot, e.g., .)
3. mj – m could be for milli, j is for? (it is not millijoule)
4. kJ – kilojoule, KJ – Kelvin joule
5. Kg – Kelvin gram, kg –kilogram
6. sec – s could be for second, e and c are for?
7. Cms – m and s could be for meter and second, C is for (Unit symbols are unaltered in
plural, e.g., cms does not mean centimeters.)

Students are encouraged to go through the Guide for the Use of the International System
of Units (SI) from NIST for more information. It is freely available.

62
(Refer Slide Time: 10:16)

Few important points/definitions to remember:

1. Definitions of a quasi-static process, a property, exact and inexact differentials

2. Math gives you a number. Physics gives meaning to it. (In the example of the apple falling
on Newton’s head, we can write . Math gives us a number.
However, we know that the work interaction is zero in that case. Physics gives meaning to
numbers.)

3. Do not mess up with units.

63
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Tutorial Problem on ‘Work’ - Part 1

(Refer Slide Time: 0:16)

Solution:
We have a frictionless piston-cylinder mechanism containing gas.

Initial volume of the gas,

64
The piston lifts up only when the pressure inside becomes 1.5 bar. Till then, the process is
constant volume process. Let’s call this state at the end of the constant volume process as
state 2.

After this, the piston moves 10 cm and hits the spring. It moves against a constant
atmospheric pressure of 1 bar and constant weight of piston. Hence, this expansion process is
a constant pressure process. Let’s call the state after this constant pressure process as state 3.

( )

Heating of the gas is still continued till the spring is compressed by 1 cm. This is state 4.

( )

The spring is linear. As it gets compressed, it applies force on the piston which is
proportional to its movement. Hence, the pressure of the gas also varies linearly.

65
The total work interaction for the gas is the hatched area shown in the following figure:

Total work interaction = Area of the rectangle + Area of the trapezium

= ( ) ( ) ( )
( ) ( )

(Refer Slide Time: 4:02)

66
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Tutorial Problem - Part 2
(Refer Slide Time: 0:16)

Problem solving methodology

1. Identify system(s)/control volume(s)

2. Identify work or heat interactions or forces acting on the system or control volume

3. Collect the information regarding the initial state of the system or control volume

4. Collect the information regarding the process the system is undergoing. Obtain the
information regarding property relations.

5. Collect the information regarding the final state of the system or control volume.

6. Draw state diagrams (p-V, p-T, etc.) if it helps.

7. Use conservation laws, e.g., conservation of mass, energy, etc.

8. Is trial and error necessary?

We may have to go through these steps more than once.

67
(Refer Slide Time: 4:25)

Figure 1

Solution of the problem given in Fig. 1:

The battery does work on the motor. The motor does work on the hoist. The hoist does work
on the mass. If we consider the motor and the hoist as separate systems, both are receiving
work as well as doing work. If we consider the battery as a system, it is doing work. If we
consider the mass as a system, work is being done on the system.

Work interaction for the battery:

The motor does work on the hoist and receives work from the battery.

Work interaction for the motor:

( ) ( ) ( )

The work interaction for the motor is negative. It means that work is being done on the
motor. There are friction losses in the motor. Some of the work it receives from the battery
through the work interaction is dissipated as heat energy because of frictional losses.

68
The hoist does work on the mass and receives work from the motor.

Work interaction for the hoist:

The work interaction for the hoist is negative. The hoist has friction losses. Some work it
receives from the motor gets dissipated as heat loss because of friction.

Work is done on the mass.

Work interaction for the mass:

Sum of work interactions for all the components should be zero.

( ) ( )
( ) kJ

(Refer Slide Time: 10:21)

69
(Refer Slide Time: 12:31)

70
(Refer Slide Time: 17:30)

71
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 13
Tutorial Problem on ‘Work’ – Part 3
(Refer Slide Time: 0:24)

Figure 1

Solution of the problem in Fig. 1:

Initial state:

The piston requires 150 kPa to lift.

The system consists of the mass of nitrogen. Initially, it is restricted to tank A. Finally, at the
end of the process, the nitrogen occupies volumes in both tanks. The mass is constant.

The provision of the valve in the setup allows the process to be quasi-static. The valve can be
opened slowly so that the system is never far away from the equilibrium. Without a valve, the
process would not be quasi-static.

Assumptions:

1. The pipe is very small and the mass of nitrogen there is negligibly small. 2. There are no
heat losses. 3. The process is quasi-static. 4. There are no losses because of the valve.

72
We open the valve slowly. The piston in tank B doesn’t lift up until the pressure reaches 150
kPa. Once it reaches 150 kPa, the piston starts moving up, and this process is at constant
pressure of 150 kPa. The upward movement of the piston stops when the pressure below it
equals 150 kPa.

Hence, the final pressure is 150 kPa.

The initial and final states are related through pV = constant.

Hence,

It is a constant pressure process for nitrogen. Hence, the work interaction for nitrogen,

∫ ∫ ( ) ( )

How would you calculate the work interaction for atmosphere in this process if the
atmospheric pressure is 100 kPa?

It is a constant pressure process for the atmosphere.

∫ ( ) (the change in volume is

negative for the atmosphere)

In the above problem, the sum of work interactions for the nitrogen, the piston and the
atmosphere should be zero.

(work is done on the piston)

73
(Refer Slide Time: 6:54)

74
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 14
Tutorial Problem on ‘Work’ – Part 4
(Refer Slide Time: 00:18)

Figure 1.

Solution of the problem in Fig. 1:

A spherical droplet changes shape from a sphere to a disklike structure because of the external
aerodynamics force.

Volume of the spherical droplet, ( )

Volume is conserved (liquids are incompressible).

(h is thickness of the disk)

Hence,

Here, the surface area of the droplet increased against the force of surface tension. We need to
calculate the change in surface area as the expression for work interaction due to surface tension
is , where is surface tension and is the change in surface area.

75
Surface area of sphere,

Surface area of disk-shaped droplet,

Work interaction for the aerodynamic force is positive, whereas it is negative for the droplet.

( )

(Refer Slide Time: 06:32)

76
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 15
Zeroth law of thermodynamics
(Refer Slide Time: 00:13)

Thermodynamics is a science that deals with energy and its conversion and transformation. So
far, we have looked at only one form of energy transformation which is work. Now, we will look
at the other one, heat. So, there are two modes of energy transfer, work and heat.

Based on the choice of a system or a control volume, the energy transfer may be either in the
form of heat or work or both.

77
(Refer Slide Time: 01:02)

Heat is an interaction between a system and its surroundings which takes place only because of a
temperature difference between the two. It is energy transfer across a boundary.

We do not talk of heat transfer within the system. We only talk of heat transfer between a system
and its surroundings. Heat is transferred across a boundary by virtue of temperature difference
between two systems or a system and its surroundings. If there is no temperature difference
between the system and the surroundings, there will be no net heat transfer.

78
(Refer Slide Time: 01:49)

From experience, we know that a hot cup of coffee or tea cools down in a room at a temperature
lower than the coffee or the tea. An ice cream, which is cold, starts melting in a room at higher
temperature, i.e. the temperature of the ice cream increases. What we see is that the spontaneous
heat transfer occurs between a system and the surroundings when the two are at different
temperatures. Spontaneous heat transfer always occurs from a higher temperature to a lower
temperature. Hence, the hot tea cools down. The cold ice cream warms up.

Similar to work transfer, heat transfer is also a path function. It also depends on what path we
take in going from one state to another. Heat transfer is an occurrence. It's not a property. It is
not something which is stored. It is something which happens while a process is going on.

We know that a small amount of work is represented as . To obtain the total work, we
integrate . Similarly, a small amount of heat is represented as . To obtain the total heat
transfer during a process, we integrate from state 1 to state 2 along a particular path. This
total heat transfer during a process is represented as 1Q2 or Q1-2.

Heat transfer per unit mass is represented as q. It is not a property.

The rate of heat transfer is represented as ̇

79
(Refer Slide Time: 05:52)

The SI unit of heat transfer is J (joule). Traditionally, the unit used for heat transfer is calorie. In
this course, we are going to use the SI unit. There is a conversion factor of 4.2 between a calorie
and a joule.

The unit of rate of heat transfer is J/s (joule per second) or W (watt).

In the case of work transfer, work done by the system on the surroundings is positive, whereas
the work is negative if it is done on the system.

Heat transfer is positive if it is into the system, whereas it is negative if it is from the system to
the surroundings.

In general, heat transfer leads to an increase or decrease of the temperature of the system.
However, it is not true always. There are exceptions. During a phase change process, even
though there is heat transfer from the system, the temperature of the system stays constant (e.g.
melting of ice into water at 1 atmosphere).

80
(Refer Slide Time: 08:59)

Figure 1

Answer of the question in Fig. 1:

If the blocks are kept in an air-conditioned room for a long time, we expect that the blocks attain
the temperature of the room. Let’s assume that the temperature of the room is constant.

However, from our experience, we feel the copper block is colder than the wooden block. We
have a similar experience on a cold day when we touch the metal and the rubber/plastic parts of a
cycle. We feel the metal parts are colder compared to the rubber/plastic parts.

Actually, the copper and the wooden blocks attain the same temperature as room when they are
kept in that room (which is at constant temperature) for a sufficiently long time. We feel copper
block colder compared to that of wood because the rate of heat transfer from our hand to copper
block is higher than that for wooden block. This happens because the thermal conductivity of
metals is higher compared to non-metals. More heat is taken away from our hand when we touch
the copper block than when we touch the wooden block in the same period of time. It shows that
we are not good at sensing temperatures of objects.

81
(Refer Slide Time: 11:21)

We need special instruments to measure temperature. We need to calibrate these instruments

using some fixed reference points. For example, the temperature of an ice and water mixture at
atmospheric pressure can be a reference point. The temperature of such mixture measured by
different instruments is given some constant value. If we don’t have such common reference
point, the temperatures measured by different instruments would be different.

We usually call the temperature of ice and water mixture at atmospheric pressure as 0 degree
centigrade. When we insert our instrument (e.g. thermometer) into this ice and water mixture,
and leave it there for a long time, we assume that the instrument attains the same temperature as
the ice and water mixture. Similarly, when we put another instrument into the same ice and water
mixture and leave it there for a long time, it also attains the temperature of the mixture. When
these two instruments are brought in contact with each other, there would be no heat transfer
between the two instruments as there temperatures are the same.

This is what the zeroth law of thermodynamics is all about. If a system A is in thermal
equilibrium with system B and the system B is in thermal equilibrium with a system C, then the
systems A and C are in thermal equilibrium with each other.

This law forms the basis of thermometry.

82
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 16
Methods of Temperature Measurement
(Refer Slide Time: 00:16)

Let’s look at some methods of temperature measurement. The commonly known device is a
thermometer. Think of ways in which you can measure temperature of an object.

83
(Refer Slide Time: 00:46)

The most familiar device for temperature measurement is a liquid-in-glass thermometer. It is a

hollow tube with a bulb-like structure at one of the ends. In that bulb-like structure, a liquid is
kept which expands on receiving heat. If the bulb of the thermometer is brought into contact with
some hot object whose temperature we want to measure, the liquid expands and travels a certain
length of the tube and its length inside the tube gives temperature of the object. Let’s see how it
is done.

Here, we are trying to express the length of the liquid inside the tube as a function of temperature.
Let’s say it is , where l represents the length of the liquid inside the tube, T represents
the temperature of the object, a and b are constants which need to be determined through
calibration.

For finding a and b, we need to use fixed point temperatures such as ice and steam point of water
at atmospheric condition. Ice point is 0 and the steam point is 100 (these are arbitrary
values). Now, we put the thermometer in mixture of ice and water at 0 and let the
thermometer attain thermal equilibrium with the mixture. The liquid inside the thermometer will
rise to some height. We mark that as 0 . Now, we put the thermocouple into boiling water at
100 and let the two attain the thermal equilibrium. The liquid inside the thermometer will rise
to a new height. We mark this as 100 Since we know l and T at two different points, we can

84
calculate a and b. In this way, we can measure temperature using liquid in glass thermometer
assuming a linear relationship between l and T.

Let’s call l corresponding to T= 0 as l0 and l corresponding to T=100 as l100. Then, a=l0 and

b=(l100-l0)/100. Hence, . In this way, the temperature of the object can be

calculated from the length of the liquid column inside the thermometer.

Assumptions made for such measurement are: (1) the cross-section of the thermometer tube is
uniform, (2) thermometer liquid does not boil or freeze in the range of temperature to be
measured, (3) the relationship between the liquid column length and temperature is linear.

For measuring temperature between 0 and 100 , the above thermometer will work
satisfactorily as the liquid used inside the thermometer does not boil or freeze in this range. What
if we want to use this thermometer to measure temperature above 100 ? Then it may not give
us correct reading.

The liquid inside the thermometer should have high boiling point and low melting point
compared to the temperature to be measured. For measuring 100 , we cannot use water inside
the thermometer as it itself will start boiling at 100 giving wrong column length. Hence,
commonly used liquids inside thermometers are gallium, mercury (liquid metals) which stay
preferably in liquid phase at temperature to be measured. Gallium’s melting point is around 30
. Hence it should be used for measurement at higher temperature range.

85
(Refer Slide Time: 06:51)

Another type of thermometer is ideal gas thermometer. We will look at the concept of ideal gas
in a lot more detail in upcoming lectures. An ideal gas follows the following expression:
, where p is the gas pressure, V is the gas volume, m is the gas mass, R is the specific
gas constant and T is the gas temperature.

Fox a fixed mass of the ideal gas, if we somehow keep volume constant, then temperature
depends only on pressure. Here, we find temperature by measuring the pressure of the gas (in
liquid-in-glass thermometer, we find temperature by measuring the length of the liquid column).
Pressure is a function of temperature, (the expression is similar to the one in the case
of liquid-in-glass thermometer). There are many devices to measure pressure (bourdon gauge,
manometer). Hence, by keeping volume of the gas constant, putting the thermometer into
different substances till thermal equilibrium is reached and measuring the corresponding pressure,
and calibrating, we can measure the temperature of the substance.

Similarly, we can have constant pressure ideal gas thermometer. Here, the pressure of the gas is
kept constant while its volume changes as a function of temperature. In this case, the expression
̅
is written as ( ) , where ̅ is universal gas constant and M is

the molecular weight of the gas (we will learn more on this later in the course). As we did
calibration in the case of liquid-in-glass thermometer by putting it into ice and water mixture at
ice point and boiling water at steam point at atmospheric conditions and found the constants a

86
and b, we can follow the similar procedure for constant pressure ideal gas thermometer and find

the constants a and b from the following equation: (we have

100 divisions on the scale here, hence there is 100 in the denominator). For very low
temperatures, pressure also tends to 0. At such conditions, this thermometer is fairly reliable.

The temperature can be associated with kinetic energy of the molecules if we look at the
molecular picture. Molecules move in all the three directions (x, y and z) and their corresponding
kinetic energies can be expressed as ̅ ̅ ̅ which can also be expressed in terms

of Boltzmann constant and temperature as for a monoatomic gas, where k is Boltzmann

constant. Molecules move fast at high temperatures and slow at low temperatures. Their kinetic
energy is associated with temperature. A gas has pressure because the molecules collide (with
each other as well as the walls of the container).

(Refer Slide Time: 10:58)

Figure. 1

There are thermometers based on thermoelectric effect. Consider wires of 2 metals M1 and M2.
Connect them as shown in Fig. 1. It forms 2 junctions. If these two junctions are kept at different
temperatures, voltage (emf) is generated which can be measured. This emf is a function of

87
temperature which can be expressed as , where a, b and c are constants
which can be found from calibration. In this case, as there are three constants, you will need 3
known temperatures to calculate them.

(Refer Slide Time: 12:23)

There are resistance thermometers. The resistance of a piece of wire changes with temperature.
For most substances, it increases with temperature. Here, we can express resistance as a function
of temperature. We keep the piece of wire in contact of the object whose temperature we want to
measure, let the thermal equilibrium reach, and measure its resistance, which can be converted
into temperature using the mathematical relation between resistance and temperature, for
example, , where A and B are constants.

There are liquid crystals which change color as temperature changes. Hence, these crystals can
be painted onto some plate and the plate can brought into contact with the object whose
temperature we want to measure, then the color change would indicate the temperature of the
object.

For measuring high temperatures, we use pyrometers. These contain filaments whose color
changes with change in temperature.

88
(Refer Slide Time: 14:06)

Figure 2.
In all the above methods of temperature measurement, we used some known temperatures for
calibration. These are called fixed points in thermometry. These fixed points are usually phase
change points (we will learn more on phase change later in the course) as these are easy to
reproduce in different labs. Figure 2 mentions some of these fixed points, which are used for
calibrating temperature measurement devices.

89
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 17
Modes of Heat Transfer
(Refer Slide Time: 00:18)

Conduction, convection and radiation are the modes of heat transfer.

(Refer Slide Time: 00:32)

90
Conduction is the energy transfer (heat transfer) between molecules (or through molecules). In
liquids and gases, the molecules or atoms are free to move. Hence, conduction happens because
of their movement. In solids, the molecules or atoms are fixed. However, they can vibrate and
transfer energy from one to another through those vibrations.

Heat transferred through conduction is expressed using Fourier’s law of conduction:

̇ , where ̇ is the rate of heat transfer, k is the thermal conductivity of the material, A

is cross-sectional area of the material perpendicular to the direction of heat transfer, and is the

gradient of temperature. The unit of ̇ is watt.

(Refer Slide Time: 01:14)

Figure 1.

Figure 1 shows typical values of k for different materials. Metals have the highest values of k,
whereas the gases have the least. Hence, for a given temperature difference, conduction is more
in the case of metals compared to that for a gas. The unit of k is W/m K

91
(Refer Slide Time: 02:11)

Figure 2.
Consider a block as shown in Fig. 2. T1 represents the temperature of its left wall and T2
represents the temperature of its right wall. If T1> T2, the slope, which is ⁄ and is also the
gradient of temperature, is negative, and the heat transfer is considered positive (heat transfer
happens in the positive x direction). If T2> T1, the slope is positive, and the heat transfer is
considered negative (heat transfer happens in the negative x direction). Hence, we have the
negative sign in the expression for the Fourier’s law. For solids, conduction is significantly
strong.

92
(Refer Slide Time: 04:40)

Figure 3.
In the case of flowing media like liquids and gases, we also have convective heat transfer
because the substance can flow (conduction is also there). In this case, ̇ , where h is the
convective heat transfer coefficient, A is the area of material which is exposed to the flow (of
liquid or gas), is the temperature difference between the material and the ambient. It is also
known as Newton’s law of cooling. For a surface (of a material) shown in Fig. 3 where a fluid at
temperature is flowing over it, is (if ). The unit of h is .

Natural convection happens if there is a temperature difference between a substance and the
surroundings. This temperature difference creates density gradient in the surrounding gas or
liquid setting up the natural convection currents which take away heat from the substance
(assuming the temperature of the substance is higher compared to the surroundings’). Typical h
values for natural convection in the case of liquid and gas (which could be the surrounding
medium around the considered substance) are shown in Fig. 3.

If there is an external agency (such as a fan or blower) to force liquid or gas over the considered
item, heat is transferred away through forced convection. Typical h values for such cases for
liquids and gases as surrounding medium are shown in Fig. 3. We have even higher h values
during phase changes, for example, water boiling in a pan. It is the most efficient way of cooling
a hot substance.

93
(Refer Slide Time: 08:23)

Figure 4.
Identify if it is a forced or natural (free) convection in the images shown in Fig. 4.

(a) hair dryer – forced convection

There is a fan inside a hair dryer which blows out air.

(b) radiator of the car – forced convection

There is forced flow of air around the radiator because of either a fan behind it or velocity
difference between the car and the ambient.

(c) room heater - natural convection

If the heat transfer from the heater to the room is considered, it is free convection if the room air
is stagnant (there is not much air movement).

(d) Fins on electronic instrument – free convection

Fins are used to increase the surface area in order to facilitate the heat transfer (as ̇ ) . Here,
there is no fan shown for forcing away the hot air near the fins. Hence, it is a case of free
convection.

(e) A person standing in a cold room – natural convection

94
It shows a schlieren image of a person. As there is temperature gradient between the body and
the surroundings, natural convection currents are setup in surroundings taking away the heat
from the body. You can see strong currents near the head and open skin (or exposed surfaces of
the skin).

(Refer Slide Time: 11:55)

Figure 5.

Radiation is another mode of heat transfer. Here, the energy is transmitted as electromagnetic
waves through space. It does not need any medium for heat transfer. The rate of heat transfer
from a surface is ̇ , where is the emissivity of the surface, is Stefan Boltzmann
constant, A is the surface area and Ts is the surface temperature. .

Black body is the best emitter and the best absorber of the energy. Its emissivity is 1. Emissivity
value denotes how good or bad emitter a given surface is compared to the black body. Emissivity
values for different types of surfaces are shown in Fig. 5.

In space, radiation is the only mode of heat transfer. Hence, for space applications, materials are
chosen keeping in mind their emissivity values. However, in our daily lives, the heat transfer
happens in a combination of the 3 modes mentioned above.

95
(Refer Slide Time: 14:11)

Figure 6.

A gray body surface emits radiation at all the wavelengths as the black body. However, the ratio
of the emission from a gray body to black body is fixed at all the wavelengths and it is less than
1. This ratio is called emissivity. Figure 6 shows spectral emission power as a function of
wavelength for a gray and a black body.

(Refer Slide Time: 15:06)

Let’s discuss the concepts of the specific heat and the latent heat.

96
Specific heat is the amount of heat needed to raise the temperature of a unit mass of a substance
by unit change in temperature. Its unit is J/kgK. The heat needed depends on the substance
whose temperature we want to raise by a unit. The amount of heat needed to raise the
temperature of 1 kg of water by 1 K (or 1 ) is larger than the amount of heat needed to raise the
temperature of 1 kg of air by 1 K (1 ). The specific heat may also depend on the temperature.
For example, the amount of heat needed to raise the temperature of 1 kg water from 16 to 17
is different than raising its temperature from 90 to 91 .

The latent heat is the amount of heat we need to give in order to change the phase of a substance.
If I have 1 kg of ice and I want to change it into 1 kg of water at the same temperature, the
amount of heat I need to give is the latent heat of melting. It would be the same as the latent heat
of freezing (amount of heat which need to be removed) if I had 1 kg of water at 0 and I want
to convert it to ice at the same temperature. Similarly, we have the latent heat of boiling or latent
heat of condensation which is the amount of heat needed to convert the given liquid substance
into gaseous phase (or convert given gaseous substance into liquid phase) at the same
temperature. This also depends on the substance under consideration.

(Refer Slide Time: 17:18)

Figure 7.

97
Answers for the questions in Fig. 7:

1. In phase change processes, the temperature of the substance does not increase even when it
receives heat, for example, conversion of ice at 0 to water at 0 . Hence, the temperature of
the object does not always have to increase when heat is given to it.

2. It is not the case always. The temperature of the object can also be increased by doing work on
it. The temperature of gas inside an insulated piston-cylinder arrangement would increase if the
piston is moved inside (work is done on the gas). Think of other such examples.

(Refer Slide Time: 19:35)

We know that displacement work is given as ∫ ∫ . In a similar way, can we write an

expression for ? Such expression exists and it is ∫ ∫ , where S represents entropy,
which we will discuss in more detail later on in the course.

The area under a curve on a p-V diagram gives work for a quasi-static process. In a similar way,
we will see that for a reversible quasi-static process, the area under a curve on a T-S diagram
gives heat transferred.

98
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 18
Tutorial Problem on ‘Modes of heat transfer’
(Refer Slide Time: 00:15)

Figure 1

Solution of the problem shown in Fig. 1:

The grille loses heat to the surroundings by convection as well as radiation.

We assume that the refrigerator has reached steady state operation. Also, the temperature of the
surroundings does not change.

Let’s first calculate the heat lost due to convection.

̇ ( )

Hence, ̇

99
Let’s calculate the heat lost due to radiation.

Assume that the grille is a black body. Hence,

̇ ( ) ( ) (temperatures are
in kelvin)

Hence,

Total heat lost =

Hence, the heat interaction for the grille, taking grille as a system, is -293 kJ while it is +293 kJ
for the surroundings as the grille is losing heat and the surroundings is absorbing heat.

100
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 19
Tutorial problem on 'Methods of temperature measurement'
(Refer Slide Time: 0:15)

Figure 1.

Solution of the problem in Fig. 1:

The thermocouple is calibrated at ice and steam points.
Assuming linear relationship between e and T, we can write,

,
where m is the slope and C is a constant.

Let’s say at T = 0 and at T = 100

Hence, and

Hence, ………(1)

101
We can find the values of and from the correct expression for the emf generated by the
thermocouple because at the points of calibration the emf values shown by the equation 1
(obtained assuming linear relationship between e and T) and the correct equation must match.

Hence, from the correct equation, and

Now, for a temperature of , emf given by the correct equation is,

(emf given by equation 1 would be 3.5 for

T = 50 )

We are asked to find out the temperature given by equation 1 for e = 2.91 mV
Equation 1 implies,

Hence, the thermocouple would read a temperature of 41.5 instead of 50 as we don’t know
the correct relationship between emf and T and we assumed the linear relationship between emf
and T. Hence, it is recommended that the calibration should be carried out over a wide range of
temperatures to get enough data so that the relationship between emf and T is known to a
significant degree of accuracy.

102
(Refer Slide Time: 10:00)

103
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 20
First law of thermodynamics
(Refer Slide Time: 0:20)

Figure 1.
Let’s look at the first law of thermodynamics which is also known as the law of conservation of
energy. Before formally stating the first law, we should look at the experiments Joule did
because these experiments lead to terms involved in the first law of thermodynamics.

Joule had a mechanism as shown in top left corner of Fig. 1. It consisted of a rigid chamber filled
with say water with paddle wheel inside it. The paddle wheel was connected to a mass by a
string which went over a pulley as shown. As the mass falls down through height h, the paddle
wheel rotates and churns water. Here, the rigid chamber is the system. As the mass lowers and
the paddle wheel rotates, work is done on the system. We can calculate this work. We call small
amount of work as .

Let’s do an experiment. Let the mass fall down through some height h. This movement of the
mass rotates the paddle wheel, and the water is churned. Because of this churning, the
temperature of the water increases. Now, let the water dissipate this extra gained heat through the

104
walls of the system and return to its initial state (which is the state which it had before the mass
was lowered). So, after the process, the system returns to the same original state. Hence, it is a
cyclic process. The work done in lowering the mass (which is also the work done on the system)
and heat dissipated by the water through the walls of the rigid chamber to return to its initial state
can be measured. Joule did a similar experiment and found that the work done and heat
dissipated/transferred by the water were proportionate. Joule used different units for work and
heat. However, when you use SI units for heat and work and take the cyclic integral of heat and
work, they come out to be equal. Mathematically, ∮ ∮ . So in a cyclic process, the
work transfer and heat transfer are equal. This is the first law of thermodynamics for a cyclic
process. There is no proof of this law. However, there has not been a case where this law was
proven wrong. What about a non-cyclic process?

Consider a system undergoing a process from some state 1 to 2 through path A and coming back
to state 1 through path B. The system undergoes a cyclic process. According to first law for a
cyclic process,

∮ ∮

We can split the work and heat transfers for a process 1 to 2 through path A and for a process 2
to 1 through path B as,

1W2A + 2W1B = 1Q2A + 2Q1B …..(1)

1Q2A - 1W2A = -(2Q1B - 2W1B)

105
(Refer Slide Time: 8:54)

Figure 2.
Now, let the system go from 1 to 2 through other path C as shown Fig. 2 and come back through
path B. Hence, according to the first law,

1W2C + 2W1B = 1Q2C + 2Q1B …..(2)

Eq. (1) – Eq. (2) implies,

1W2A - 1W2C = 1Q2A - 1Q2C

1Q2A - 1W2A = 1Q2C – 1W2C

[ ] [ ] [ ]

It shows that the difference between the heat transfer and the work done for a process is the same
irrespective of the path taken to carry out the process though work and heat themselves are path
functions. We know that the properties are state functions. Their values depend only on the end
states. Here, the difference between the heat transfer and the work done during the process
depends only on the end states and not on the path taken by the process. Hence, this difference
must be representing some property. We call it energy. For a system undergoing a process,
difference between heat transferred and work done is the difference in energy between the end
states of that process.

106
(Refer Slide Time: 13:43)

Figure 3.
In the differential form, we can write as,

,…..(3)

where and are small amounts of heat transferred and work done and dE represents small
change in energy. Equation 3 is an expression for the first law of thermodynamics for a process
(non-cyclic). From eq. 3, we can obtain the expression for the first law of thermodynamics for a
cyclic process by taking cyclic integral and setting dE = 0 as E is a property and for a cyclic
process change is property is 0, i.e. dE = 0.

Eq. 3 gives us the change in energy between two states when integrated over a process. It does
not give the absolute value of energy. E in dE can be split into kinetic energy (KE), potential
energy (PE) and internal energy (U) as,

…..(4)

The object has kinetic energy by virtue of its motion (the object has velocity). The object has
potential energy when it is lifted with respect to some reference/datum. There are other forms of
energy which are lumped together in the form of U. We will discuss these forms later on. Hence,

107
change in energy (dE) of a system undergoing a process equals change in kinetic energy (dKE),
change in potential energy (dPE) and change in internal energy (dU).

Energy (E) is extensive property. It has the same units as work or heat. Specific energy (e) has
units J/kg.

Consider an object (a system) at rest. Suppose it starts moving horizontally and attains velocity
V. There is change in its kinetic energy. There is no change in its potential and internal energy.
To move the object, work was done on it. Assume that there is no heat transfer involved. Then
Eqs. 3 and 4 give, . Work is force into displacement, i.e. mass into acceleration into

displacement. . Hence, .

Integrating it from V=0 to V=V, ∫ . Here, work done on the object

(system) is equal to the change in its kinetic energy.

(Refer Slide Time: 20:20)

Figure 4.
Consider an object (a system) which is lifted through height h from some datum as shown in Fig.
4. In this process of lifting, there is only a change in potential energy. There is no heat transfer.
There is no change in the temperature of the object. Eqs. 3 and 4 give . So change
in the potential energy of the object is negative of the work done on the object (the system). We
are raising the object against gravity. Hence, the small amount of work done to raise the object

108
through small height dh is , where m is the mass of the object and g is the
acceleration due to gravity. To get the total change in potential energy in the entire process of

lifting the object through height h, we need to integrate as, ∫

∫ . The work done is negative as it is being done on the system.

Let’s discuss about internal energy. It is something which exists because of interaction between
molecules. The molecules of a substance may have vibrational energy, translational energy or
rotational energy. All these constitute internal energy for a substance. We will look at it in a
more detail when we discuss the concept of ideal gas.

(Refer Slide Time: 24:49)

109
 Figure 5.

When changes in kinetic and potential energy of a system undergoing a process are negligible
and there are heat and work transfers, eqns. 3 and 4 give . Consider a piston-
cylinder arrangement as shown in Fig. 5. The system is the gas inside the cylinder as marked by
the dashed line. Assume that there is heat transfer of 100 J to the system and the gas expands
inside moving the piston and doing work of 50 J. According to first law for a process,
. After integration, 1Q2 – 1W2 = . Hence, the internal energy of
the system has increased over the process by 50 J. Here, the piton-cylinder system itself is at rest.
Also, it is not being raised during the process. Hence, the changes in kinetic and potential energy
are ignored. Thus, difference of heat transferred and work done is the difference between internal
energy of the system between the end states. In the above example, signs of heat transferred and
work done are taken according to the convention mentioned in previous lectures: heat entering
the system is positive and work done by the system on the surroundings is positive.

In upcoming lectures, we will come across a combination of internal energy, pressure and
volume a lot of times. This combination is also a property called as enthalpy (H) and H = U+pV.
Specific enthalpy is the enthalpy per unit mass denoted as h=u+pv, where h is specific enthalpy,
u is specific internal energy and v is specific volume.

110
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 21
Tutorial problem – Part 1
(Refer Slide Time: 00:16)

Figure 1.

Solution of the problem in Fig. 1:

F = 180 N, r = 40 mm = 0.04 m, t = 8 min = 480 s, N = 250 rpm = 250/60 rps
Initially, the system is only the metallic workpiece. Finally the system is the machined workpiece
and the chips.
Torque is acting on the workpiece.

Cutting torque,

Work done, ( ) ( )

Here, work is done on the system. Hence, W = -90.4 kJ

Now,

According to the first law for a process, (ignoring changes in potential and
kinetic energy).

Integrating, ( )

111
Heat transfer is negative. The workpiece gets heated during the turning operation and its
temperature increases. Hence, it loses heat to the surroundings.

(Refer Slide Time: 4:31)

112
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 22
Tutorial problem – Part 2
(Refer Slide Time: 0:14)

Figure 1.

Figure 2

113
Solution of the problem in Fig. 1:

(ignoring the potential and kinetic energy

changes in air),

Case a: the system consists of the air only.

The schematic of the system is drawn in top right hand corner of Fig. 2.

The system is expanding against the constant weight of the piston and constant atmospheric
pressure quasi-statically (assume that the process is quasi-static). Hence, it is a constant pressure
process. The piston is frictionless.

Hence, pressure of the air inside the system,

Work done, ∫ ∫

First law for a process, . Here, .

Integrating,

Case b: the system consists of the air and the piston

Now, the piston is internal to the system. Since the entire assembly including the piston is
insulated, there is no heat loss to the surroundings. Hence, in this case also, the heat transfer from
the resistor to the system would be Q = 22.56 kJ.

Let’s check if it comes out to be the same in this case.

In this case, work done would be against the constant atmospheric pressure.

Hence, ∫

114
The first law for a process in the integrated form, . Here, the potential
energy of the system changes as the piston is part of the system and it is raised against gravity.

Hence, .

We don’t know . But we know change in volume of air, .

Now,

Hence,

Based on the choice of the system, heat transferred, work done and change in energy can change.

(Refer Slide Time: 15:51)

115
116
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No 23
Heat and Work Interactions for a System

(Refer Slide Time: 0:23)

Figure 1.
Let us look at distinctions between heat and work.

Consider a stationary rigid insulated chamber as shown in top left hand corner of Fig. 1.
There is a heating coil inside. Consider a system which includes the contents of the rigid
chamber and the heating coil. The coil is connected to a battery of 12 V which sends in a
current of 10 A for 60 s through the coil. Is it a work interaction or heat interaction for the
system?

For answering this question, we need to consider what is crossing the boundary of the system.
In this case, the voltage of the battery is pushing current across the boundary of the system.
The battery is doing work on the system (actually the coil which is the part of the system)
which is W = VIt = . Since this is work done on the system it is
negative, i.e. W = -7.2 kJ. The work interaction is negative here. What about the heat
interaction?

As mentioned before, the chamber is insulated. The heating coil heats the contents of the
chamber. But there is no heat transfer across the boundary of the system as it is insulated. The

117
only heat transfer is between the heating coil and other contents of the system. But all these
are internal to the system. Any heat transfer or work transfer has to happen across the
boundary. Hence in this case, the heat transfer is 0. We can calculate change in internal using
the first law in integrated form, Q-W = (there are no changes in potential and kinetic
energy). Hence, ( ) . The internal energy of the system increased
by 7.2 kJ.

What if we exclude the heating coil from the system?

The current is still flowing through the coil. However, it is not crossing the boundary of the
system. Hence, the work interaction for the system is zero (there is also no shaft work,
gravitational work, electric work or magnetic work). As the current flows through the coil, it
becomes hot as the battery is doing work on the coil. The work done by the battery on the coil
is again W = VIt = 7.2 kJ. Assuming steady state operation of the coil, for the coil would
be 0. Hence, the first law written for the coil implies Q = W = -7.2. It means the coil loses 7.2
kJ of heat to the surroundings which is our system. The heat crosses the boundary of our
system. Hence, there is a heat interaction. The heat transfer across the walls of the rigid
chamber is zero as the walls are insulated. Hence, for such a system (which excludes heating
coil), there is a heat interaction, but no work interaction. The change in internal can be
calculated in the similar fashion. The first law in the integrated form is Here,
as W=0.

Hence, whether a system has work interaction or heat interaction or both the interactions
depends on the choice of the system. The formulae need to be written consistently according
to the definition of the system.

118
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No 24
Tutorial problem - Part 1

(Refer Slide Time: 00:14)

Figure 1.

Solution of the problem in Fig. 1:

The system undergoes a cyclic process.

For the process 2-3, the first law in the integrated form is . Hence, Q2-3 =
670+230=900 kJ.

Similarly, for the process 3-4, .

Similarly, for the process 4-1, .

For a cyclic process 1-2-3-4-1, , i.e. . Hence,

.

Since it is a cyclic process, according to the first law of thermodynamics, ∮ ∮

. Hence, . Thus,

The summation of all W values is +540 kJ. Hence, this cycle gives positive work, and it is a
power producing cycle. The system also takes in 540 kJ of heat during the cyclic process.

119
 Thermodynamics
Professor Anand TNC
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No 25
Pure Substance

In the previous lecture, we introduced the first law of thermodynamics. For a cyclic process,
∮ ∮ in SI units. For a system undergoing a non-cyclic process, the difference
between the heat transferred and the work done is the change in energy of the system. Work
and heat are two different things. We will see how they are different when we look at the
second law of thermodynamics. Before that, let’s look at the properties of a pure substance.

A pure substance is a substance which has uniform composition as well as chemical

aggregation throughout the process under consideration. We are interested in the composition
and chemical aggregation of the substance only for the duration of the process.

We discussed that the state of a system is defined in terms of its properties, e.g., pressure,
temperature, volume, energy, enthalpy, etc. How many properties do we need to define the
state of a system? The number of properties needed to define the state uniquely depends on
the type of substance which constitutes the system. For example, for a chemically reacting
system (e.g. methane burning in oxygen forming carbon dioxide and water), we may need a
certain number of properties such as temperature, pressure or mass. For a falling body, we
may need a different set of properties such as the height through which the body falls, its
temperature, etc.

120
(Refer Slide Time 1:24)

(Refer Slide Time 2:16)

121
(Refer Slide Time 3:40)

Figure 1.
Let’s try to answer the questions shown in Fig. 1.

1. Is air a pure substance?

Air is a mixture of around 76 % nitrogen by mass, 23% of oxygen by mass and some other
gases. At room temperature and atmospheric pressure, air is a mixture of all these gases. For
time being, assume it is just the mixture of nitrogen and oxygen. The composition of air is
uniform. If we start cooling the air, at around -183 degree Celsius, oxygen will become liquid
first. Hence, we have a mixture of liquid oxygen and gaseous nitrogen. The compositions are
different. Hence, at this temperature, the air is not a pure substance. However, for the most
practical applications, we do not have such low temperatures, and air behaves as a pure
substance. Hence, air is or is not a pure substance based on the conditions during the process.

2. Is a combusting H2 - O2 mixture a pure substance?

After combustion, we are left with liquid water as a product of the chemical reaction between
H2 and O2. The composition changes during the process. Hence, a combusting H2 - O2
mixture is not a pure substance.

3. Is mixture of steam and water a pure substance?

122
The chemical composition of steam (water in vapour form) and liquid water are the same. As
far as the chemical composition does not change during the process, a mixture of steam and
water is a pure substance.

We are interested in finding out the properties of such pure substances during a process.

(Refer Slide Time 7:39)

We can use a two-property rule for a system consisting of a pure substance (which in some
sense a reduction of the Gibb’s phase rule). It is stated as following: the state of a pure
substance of given mass can be specified in terms of two independent properties in the
absence of effects due to gravity, motion, electricity, magnetism, elastic deformation, etc.

Our goal while solving the tutorial problems is to find these two independent properties
which define the state of a system (made of a pure substance) under consideration. These two
properties can then be used to find other properties because the state of the system is known.

123
(Refer Slide Time 10:10)

Figure 2.

While implementing the two-property rule, we assume that there is only one mode of energy
storage, and it is the internal energy (because all other forms of energy are not there if we
look at the statement of the two-property rule).

Consider a piston-cylinder assembly consisting of air as shown in Fig. 2. The conditions are
such that the air is a pure substance here. If we know two independent properties such as
pressure and temperature (which are intensive) of this system, we can define the state of the
system and find out other properties such as specific volume, absolute internal energy or
enthalpy (with respect to some datum).

We will often come across a simple compressible system containing a pure substance. For
such a system, electric, magnetic or gravity forces and changes in potential and kinetic energy
are not considered in the analysis.

124
(Refer Slide Time 13:08)

Figure 3.
According to two-property rule, once we know two independent properties of a system, the
state of the system can be defined. Using these two properties, we can find other properties
through property relations.

Based on these property relations, we can classify the substances as shown in Fig. 3: ideal
gas, a mixture of ideal gases, steam/water substance (for the purpose of this course, we are
going to restrict ourselves to these substances).

We will cover the concept of ideal gas and a mixture of ideal gas a little later in the course.
We already saw that steam (water vapor) or water or a mixture of steam and water is a pure
substance. There can be other sort of mixtures such liquid ammonia and gaseous ammonia.

125
 Thermodynamics
Professor Anand TNC
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No 26
Tutorial problem - Part 2
(Refer Slide Time 0:13)

Figure 1.

Solution of the problem in Fig. 1:

(as the separating wall is rigid)

Since the separating wall is perfectly conducting, the temperature of air and nitrogen is equal
all the time, . Also, .

As the air gets compressed, it gets heated and transfers some amount of heat to nitrogen.
There is no work interaction for nitrogen as the separating wall is rigid.

Using pv = 288 T, we can calculate .

126
 (since )

Also, .

Therefore,

For nitrogen, the first law is . As the work interaction is zero for nitrogen,
.

After integration, . . Hence, we need the mass of nitrogen.

Now,

For nitrogen, ( )

Heat interaction for nitrogen is positive as it is receiving heat from air. The heat interaction
for the air is negative of the heat interaction for nitrogen (air is losing heat to nitrogen).

The first law for air in the integrated form is We are asked to find out W.

For air, . We need the mass of air.

Now,

Now, ( )

Therefore, for air, ( )

Work interaction for air is negative as work is being done on the system (air).

Therefore, for air,

127
(Refer Slide Time 5:03)

128
(Refer Slide Time 11:45)

129
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No 27
Ideal gas - Part 1
(Refer Slide Time 0:14)

Figure 1.

While solving problems, we are going to use a particular methodology which is listed in Fig.
1.

1. We are going to find out whether there are one or more systems or control volumes or
whether we can deal with the substance considering it to be a system or a control volume.

2. We will then identify energy transfers across the boundary in the form of either work or in
the form of heat. For this, we may also need to find out all the forces acting on the system so
that we can find out work interactions.

3. We will need to list out what we know about the initial state before the process happens,
e.g., properties of the system such as pressure, temperature, volume, etc.

4. This step asks about the thermodynamic model (TD model) to be implemented, for
example, a property relation for an ideal gas.

In this lecture, we are going to discuss in detail what the 4th step of the problem solving
methodology is all about. After that, we will go through the methodology again.

130
(Refer Slide Time 1:48)

What is an ideal gas which many of us have heard about or studied at some point in school?

(Refer Slide Time 1:57)

Any gas can be treated as an ideal guess, if its molecules exert negligible attractive forces on
each other. So, the molecules of an ideal gas do not affect each other. This assumption is
valid if the molecules are separated and they are sufficiently far apart, which means the
volume as well as specific volume of this gas must be very large. For an ideal gas, specific
volume and its pressure .

131
Ideal gas follows the Boyle’s law which is stated as follows: pressure of the gas into volume
of the gas is constant if its temperature is constant. Mathematically, at
constant temperature (isothermal process). Ideal gas also follows Charles’ law which is stated
as follows: volume of the gas is directly proportional to its temperature at constant pressure.
Mathematically, at constant pressure (specific volume v is used in both the expression

as we are dealing with the system here). Putting these two laws together, we get

Ideal gas follows the above relation.

(Refer Slide Time 3:47)

For an ideal gas, ̅ , where p is the pressure in pascal, V is volume in m3, n is the
number of moles in kmol (kilomole), ̅ ( ) is the universal gas constant in J/kmol K, and T
is the temperature in kelvin. ̅ . The expression ̅ can
be used when the quantity of gas is given in moles or kilomoles. However, we do not use this
expression often. The expression which is used more commonly is , where m is
the mass of the gas in kg and R is the specific gas constant in J/kg K. The specific gas
constant is different from the universal gas constant. The value of the universal gas constant
is the same irrespective of the gas under consideration which is 8314.5 for all the
gases. However, this is not true for the specific gas constant. It changes as the gas changes.

Dividing by n throughout the expression ̅ , we get ̅ ̅ , where ̅ ( ) is the

molar specific volume. Dividing by m throughout the expression , we get

132
 , where v is the specific volume. Since , , where is density of the

gas.

Specific gas constant is obtained from the universal gas constant by dividing the universal gas
constant by the molecular weight of the gas under consideration. For example, the specific

gas constant for nitrogen is .

133
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture No 28
Ideal Gas - Part 2
(Refer Slide Time: 00:12)

Figure 1.

In the previous lecture, we saw that, for an ideal gas, . If we have a system

consisting of ideal gas, then mass of the gas m is constant. Then, ,

because R is a specific gas constant. As long as the composition does not change, R is
constant. A consequence of this is that the internal energy U and the enthalpy H of an
ideal gas are only a function of temperature.

Consider a rigid container consisting of ideal gas as shown in Fig. 1. It is our system. Add
heat Q to this system. There are no changes in kinetic and potential energy of the system
as it is not moving or getting lifted. The first law for this system can be written as
(we are writing dE as dU as there are no changes in kinetic and potential
energy). Since it is a simple compressible system, (we are considering
only the expansion or contraction work). Since the system is rigid, dV = 0. Hence,
, i.e., heat transferred changes the internal energy of the system.

134
Let’s define specific heat for a system. It is defined as the amount of heat needed to raise
the temperature of unit mass of a system by unit value. Specific heat can be represented
as Cp (specific heat at constant pressure) or Cv (specific heat at constant volume). In the

example of the system considered above, . Hence, dU = mCvdT. For a

system, m is fixed. If Cv is also constant, then U is a function only of temperature.

(Refer Slide Time: 05:46)

135
According to the kinetic theory of gases, if the number of translational, rotational and
vibrational degrees of freedom per molecule is equal to ‘D’, then internal energy per
̅
molecule = , where k is the Boltzmann constant ( where ̅ is the

universal gas constant, NAV is the Avogadro’s number which is molecules

per kilomole). The degree of freedom refers to the number of ways in which the molecule
is free to move.

̅
For 1 kilomole of the gas, the molar internal energy is given as ̅

Similarly, molar enthalpy ̅ ̅ ̅ , where ̅ = specific molar volume (n is

number of moles). We know that ̅ . Hence, ̅ ̅ . Thus, ̅ ̅ ̅ .

Substituting the expression for ̅, ̅ ̅ .

(Refer Slide Time: 08:55)

̅
Molar internal energy can also be expressed as follows: ̅ ̅ , where ̅ is
̅
the molar specific heat and ̅ . Similarly, molar enthalpy can be expressed as
̅ ̅
̅ ̅ ( ) , where ̅ ( ) . The unit of ̅ and ̅ is J/kmol.

136
 ̅
The specific internal energy can be represented as , where M is the molecular
̅ ̅
weight of the gas. Substituting for ̅, , where ( ) is the
̅
specific gas constant. Similarly, the specific enthalpy is expressed as
̅
( ) ( )

(Refer Slide Time: 10:53)

̅
The ratio of specific heats is usually termed as gamma and expressed as ̅

A diatomic molecule of the gas (which looks like a dumbbell) has three translational
degrees of freedom (it can move along x, y or z direction), 2 rotational degrees of
freedom (It can rotate around any two of the axes which are perpendicular to the bond
joining two atoms. However, the rotation around the axis along the bond between two
atoms does not much contribute to energy, which results in it having only 2 rotational
degrees of freedom.). At low temperatures, the atoms of a diatomic molecule do not
vibrate much along the bond direction, i.e., the vibration mode is not excited. Hence, D =
5.

137
For a monoatomic gas, D=3, because there are three translational degrees of freedom
(there can be movement in x, y and z direction). The rotational degrees of freedom are
zero because rotation about any axis does not much contribute to the energy of the atom.

(Refer Slide Time: 13:37)

Nitrogen is a diatomic gas. Here, at low temperatures. For a

monoatomic gas, .

138
(Refer Slide Time: 15:11)

Let’s find the relation between Cp, Cv and R.

We know that, h = u + pv = u + RT (pv = RT is the ideal gas equation). Writing h and u

in terms of Cp and Cv, CpT = CvT+RT. Hence, Cp = Cv + R or Cp - Cv = R. If the unit of
Cp and Cv is J/kg K, then R will also have unit J/ kg K. If we are using molar specific
heats, the unit of Cp and Cv is J/ kmol K. In that case, ̅ ̅ ̅ and ̅ is the universal

gas constant. Now, Hence, Thus, . Also,

̅
̅ Since, . Also, ̅ ̅

139
(Refer Slide Time: 18:22)

Let’s calculate for nitrogen using all the relations we developed.

̅
.

140
(Refer Slide Time: 21:08)

The number of degrees of freedom of a gas may change with temperature because the
vibrational mode gets excited at high temperatures, i.e. contribution of vibrational energy
to the total energy becomes significant. Hence, as temperature increases, the fraction of
molecules with vibrational excitation will increase and this results in D (and so and
) varying as function of temperature.

141
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 29
Tutorial Problem - Part 3
(Refer Slide Time: 00:16)

Figure 1.

142
Solution of the problem given in Fig. 1:

Initial state:

Final state:

It is also given that and Cv is 0.718 kJ/kg K.

The system is undergoing a polytropic process. Assuming the system to be a simple

compressible system, the work done by the system is,

∫ ∫ ∫ ∫ [ , m is

constant, i.e., dm=0, hence, ]

We have integrated this before.

( ) ………(1)

We need p1 and p2.

Hence, .

Similarly, .

Substituting the values of p1, p2 and m in (1) and using SI units for all the quantities,

143
(Refer Slide Time: 13:22)

We are also asked to calculate the heat interaction.

The first law of thermodynamics for the given system is,

(There are no changes in kinetic and potential energy. Hence, dE can

be replaced by dU)

Hence,

Integrating,

144
The heat interaction is negative for the system. Hence, the system is losing heat.

The work interaction is positive for the system. Hence, the system is doing work.

Here, the system is losing energy by giving out heat and doing work.

145
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 30
Tutorial Problem - Part 4
(Refer Slide Time: 00:12)

Figure 1.

Figure 2.

146
Solution of the problem given Fig. 1:

Tank A:

Tank B:

After the mixing is over, the gas achieves an equilibrium state. The final temperature is
. Also, .

We are asked to calculate the heat transfer during the process.

Consider both the tanks as our system. During the process, there is no change in kinetic
and potential energy of the system. There is no shaft work, electric work, magnetic work
or surface tension work. The transfer of gas within the system does not qualify as work
interaction. There is no deformation of volume. Hence, for the process, work interaction
W = 0.

The first law for the system in the integrated form is,

Since W = 0,

For an ideal gas, = Internal energy at the final state – internal energy
at the initial state ……(1)

We know the masses at the initial state as well as the final state. At the final state, the
mass is, .

To calculate , we need to know specific gas constant for and the specific heat ratio

as we are using the formula .

147
 ̅
.

Hence, .

We know the temperatures at the initial and final states.

(1) implies,

[( ) ] [ ]
[ ( )]
[ ( ) ( )]

Since is positive, heat is entering the system.

For calculating final equilibrium pressure pf, we use the ideal gas equation,
. We need to know the total volume to calculate pf.

Total volume V is the sum of volumes of tank A and B.

Similarly, .

Therefore,

Now,

148
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 31
Tutorial Problem - Part 5
(Refer Slide Time: 00:12)

Figure 1.

149
Solution of the problem given in Fig. 1:

Assumption: working substance in the cylinder is an ideal gas.

We write the ideal gas equation for two states of the system and take their ratio to
calculate T2, the final temperature.

Hence,

Heat is transferred into the system. The volume of the system is increasing. Hence, work
is done by the system. It is a constant pressure process.

∫ ∫ ( )

The first law in the integrated form is,

(There are no changes in kinetic and potential energy. Hence, )

Hence, the final internal energy of the system is lower than its initial internal energy. The
system is losing energy during the process.

150
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 32
Specific heats at constant pressure and constant volume

Let’s look at the definitions of specific heat at constant pressure (Cp) and specific heat at
constant volume (Cv). Cp is the amount of heat needed for a unit value rise in temperature
of a unit mass of a substance at constant pressure. Similarly, Cv is the amount heat needed
for a unit value rise in temperature of a unit mass of a substance at constant volume.

(Refer Slide Time: 00:48)

Mathematically, and , where m is the mass of the substance, Q is the

amount of heat transferred and is the quantity by which the temperature is raised.

Let’s look at Cv in detail.

Consider a stationary rigid container which contains some gas. Transfer some amount of
heat Q to the container. There is no work interaction here (there is no electrical work,
magnetic work, surface tension work, etc.). The first law in the integrated form is

151
 . The system is not moving or getting lifted. Hence, changes in kinetic and
potential energy are zero. Thus, which can also be written as , where
m is the mass of the system content and u is the specific internal energy. We know that

. Hence, which implies In the differential form,

at V = constant.

Let’s look at Cp in detail.

Consider a frictionless piston-cylinder arrangement (which is the system under

consideration). Heat Q is transferred to the system. The system undergoes a constant
pressure process and the piston moves. Hence, there is work interaction. The first law in
the integrated form for this system is (there are no changes in kinetic and
potential energy). Now, . In the differential form,
(V = mv and m is constant, it can be taken out).
Cancelling out m, (adding dp doesn’t
change anything as dp=0 as it is a constant pressure process). Now,

( ) ( ) , where h is specific enthalpy. Hence, . We had

defined earlier that . For a constant pressure process as in the case of the

frictionless piston-cylinder system above, . Hence, .

152
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 33
Tutorial Problem - Part 6
(Refer Slide Time: 00:13)

Figure 1.

153
Solution of the problem in Fig. 1:

The piston-cylinder assembly and the rigid chamber contain the same ideal gas.

Expansion in the piston-cylinder assembly happens at constant pressure.

The first law for the piston-cylinder arrangement,

(there are no changes in kinetic and potential energy)

Now, ( ) ( ) (adding Vdp does

not change anything as dp=0)

Integrating,

…….(1)

Heat is transferred to the rigid chamber. This is a constant volume process. The first law
for the rigid chamber,

(there is no shaft work, electrical work, magnetic work, surface tension

work, etc.)

Integrating,

…..(2)

We are asked to find the difference between the heat supplied to the two systems. Hence,

̅
( ) = 300 (

, where 25 kg/kmol is the molar mass of the gas)

We see that for the same temperature rise, the amount of heat supplied to the piston-
cylinder assembly is more than the rigid chamber. It is because the heat supplied to the

154
rigid chamber goes into heating the gas alone, whereas in the case of the piston-cylinder
assembly, the supplied heat is used to do work in addition to raise the temperature of the
gas.

155
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture - 34
Tutorial Problem - Part 7
(Refer Slide Time: 00:13)

Figure 1.

156
Solution of the problem given in Fig.1:

The solid aluminium block at is heated. As the temperature reaches 660 , the
block starts melting. During melting, the temperature remains constant. Heating is
continued till the temperature of liquid aluminium reaches 700 . So the heat supplied is
used to increase the temperature of solid aliminimum block from 15 to 660 , melt
aliminium, and then increase the temperature of liquid aliminium from 660 to 700 .

Hence, total amount of heat transferred, ̇ .

For solids or liquids, the volume is small compared to the gas if we consider equal mass
of each, because the density of solids or liquids is significantly larger than gas.

The expression for specific enthalpy is, h = u+pv. Since v is very small for a solid or

liquid, . We also know that, and . Since ,

for a solid or a liquid.

The heat supplied can be expressed mathematically as,

Hence,

( ) ( )

This is the mass of aluminium we can melt if all the heat supplied goes into heating and
melting the aluminium.

157
But, It means 70% of heat goes into melting aluminium. The remaining
is lost. Hence, the actual amount of molten aluminium is
in an hour.

158
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 35
Ideal Gas – Part 3

(Refer Slide Time: 0:15)

We looked at the concepts of ideal gas and pure substance.

(Refer Slide Time: 0:42)

We also saw that the number of degrees of freedom may change with temperature. The
vibrational modes get excited at higher temperatures making the number of degrees of

159
freedom (D) a function of temperature. Hence, Cp, Cv and (ratio of specific heats) for a gas
may change with temperature.

(Refer Slide Time: 1:12)

Figure 1.

Figure 1 shows typical values of for typical gases.

As the temperature increases, the values of Cp and Cv for dry air individually increase,
whereas the value of reduces. For a triatomic gas such as , the value of at 20 is 1.3.
However, this value decreases as the temperature increases. A similar trend is seen in the case
of argon.

160
(Refer Slide Time: 3:10)

Figure 2.
For a perfect gas, Cp and Cv are constant. They do not change with temperature. The plot of
Cp or Cv versus temperature is a line parallel to T axis. Since Cp and Cv are constant, their
ratio is also constant. The difference between Cp and Cv, which is the specific gas constant
R, is also constant. Such a gas is often called calorically perfect gas. For an ideal gas, specific
internal energy u = CvT and specific enthalpy h=CpT, where T is temperature. If Cp and Cv
are constant, the plots of u or h versus T are straight lines passing through origin (see Fig. 2).

For a semi-perfect gas (which is another name for an ideal gas), Cp and Cv are functions of
temperature only. Now, h = Cp(T)T and u = Cv(T)T. Hence, the plots of u or h versus T are
no longer straight lines (see Fig. 2).

161
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 36
Ideal gas - Part 4

(Refer Slide Time: 0:14)

Let’s look at the mixture of ideal gases.

Take the example of air. It is a mixture of nitrogen, oxygen, and other gases. Under certain
conditions, these gases can be treated as ideal gases. Does a mixture of ideal gases behave
like an ideal gas? Can we find its properties?

162
(Refer Slide Time: 1:18)

Consider a system where we have a mixture of N gases. and

represent mass and number of moles of each gas.

Now, total mass of the mixture ∑ , where

mi represents the mass of the ith species. Similarly, total number of moles
∑ , where ni represents the number of moles of the ith
species.

Mole fraction of a particular gas in the mixture is the ratio of the number of moles of that
particular gas to the total number of moles in the mixture. Mathematically, mole fraction of
the ith species, ∑
. Similarly, mass fraction of a particular gas in the mixture is

the ratio of the mass of that particular gas to the total mass of the mixture. Mathematically,
the mass fraction of the ith species, ∑
.

Also, ∑ and ∑ . Check the validity of these two expressions in the

case of a mixture of two gases.

(Refer Slide Time: 7:33)

163
We can convert number of moles of a gas to its mass by multiplying number of moles by
molecular weight (M kg/kmol) of that particular gas. Mathematically,
. Now,
∑ .

The mass fraction of the ith species, ∑

. Dividing the numerator and the

denominator by total number of moles n, ∑ ∑

(as n is constant, it

can be taken inside the summation sign in the denominator).

The molecular weight of the mixture ∑ .

(Refer Slide Time: 12:42)

164
Let’s find a way to convert mole fraction into mass fraction.

Consider a mixture of gases which has a mass of 1 kg. In this case, the mass of each gas in
the mixture equals its mass fraction, i.e., We know that the
number of moles of a particular gas equals its mass divided by its molecular weight, i.e.,
. In the case of the mixture considered above (whose total

mass is 1 kg), . Now, the total number of

moles of the mixture ∑ . Hence,

the mole fraction of the ith species is . Here, the molecular weight of the
∑

mixture would be, as the total mass of the mixture is 1 kg.

∑

165
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 37
Ideal gas - Part 5

(Refer Slide Time: 0:16)

Figure 1.
Let’s look at the Dalton’s law of partial pressures.

Consider a mixture of gases A, B and C occupying a volume Vm at temperature Tm.

According to Dalton’s law of partial pressures, the pressure exerted by this mixture equals the
sum of partial pressures of the gases A, B and C. Partial pressure of the gas A is the pressure
exerted by A when it occupies volume Vm individually at the mixture temperature Tm. In the
similar way, partial pressures of B and C are calculated. Hence, according to Dalton’s law,
the pressure exerted by mixture of A, B and C when it occupies volume Vm is
where, represent partial pressures of A, B and C.

Similarly, we have Amagat’s law of partial volumes.

According to this law, the total volume of the mixture of gases at constant temperature and
pressure is equal to the sum of individual partial volumes of constituent gases. At the mixture
pressure and temperature, the constituent gases occupy different volumes individually.
Hence, for a mixture of gases A, B and C at constant temperature Tm and pressure pm, the

166
total volume , where are the volumes these gases occupy
individually at the mixture temperature Tm and pressure pm.

(Refer Slide Time: 2:30)

Consider a mixture of N ideal gases which is at temperature Tm and occupies volume Vm.
represent the number of moles of those gases and the total number of moles of
the mixture . Since these are ideal gases, they satisfy ideal gas
equation ̅ . So, we can write, for individual gases, ̅ ,
̅ ̅ . According to Dalton’s law of partial pressures, the pressure of
the mixture . The mixture also satisfies the ideal gas
equation. Hence, for the mixture, ̅ . If we take the ratio of the pressure of the
̅
th
i gas to the pressure of the mixture, ̅ =mole fraction of the ith gas.

Consider again a mixture of N ideal gases which is at pressure pm and temperature Tm. So, we
can write, ̅ , ̅ …… ̅ . For the mixture,
̅ , where (according to Amagat’s law). If we take the ratio of the
̅
th
volume of the i gas to the volume of the mixture, ̅ = mole fraction of the

ith gas.

167
(Refer Slide Time: 6:25)

We need to find the properties of a mixture of ideal gases such as energy, enthalpy, etc.

Consider a mixture of N ideal gases. Let and represent the

number of moles and the masses of individual gases. The mass of the mixture
and total number of moles of the mixture Similarly,
represent specific internal energy of individual gases. Now, the internal energy
of the mixture ∑ . The
∑
specific internal energy of the mixture ∑ , where is the

mass fraction of the ith gas. We can write similar expressions for the enthalpy (H) and the
specific enthalpy (h) of the mixture. So,
∑
∑ and ∑ .

The internal energy of the mixture can also be written as ̅ ̅ ̅

̅ ̅ ∑ ̅ , where ̅ represents the molar specific internal energy (its

unit is J/kmol) of the mixture and ̅ ̅ ̅ represent molar specific internal energy of
∑ ̅
individual gases. Now, ̅ ∑ ̅ , where is the mole fraction of

the ith gas. Similarly, molar specific enthalpy of the mixture ̅ ∑ ̅.

168
In the similar fashion, for the mixture, ∑ (the unit of Cp is J/kg K) and
̅ ∑ ̅ (the unit of ̅ is J/kmol K). Similarly, we can write the expressions for Cv
̅
and ̅ . The ratio of specific heats for a mixture ̅
.

169
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 38
Tutorial problem - Part 1

(Refer Slide Time: 00:15)

Figure 1.

Solution of the problem given in Fig. 1:

Partial pressure of gases, and where x and pmix represent the

mole fraction and mixture pressure. Also, and .

We can find mole fraction using molecular weight of the mixture which in turn needs mass
fractions.

Molecular weight of the mixture,

∑ ⁄

Now, and

Hence, 0.026 and .

170
To calculate partial pressures, we need mixture pressure.
Assume that both the gases are ideal gases. Hence, the mixture is also an ideal gas.

̅
Now,

Hence,

Hence,

(Refer Slide Time: 5:05)

171
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 39
Tutorial Problem – Part 2

(Refer Slide Time: 0:13)

(Refer Slide Time: 0:37)

Figure 1.

Solution of the problem in Fig. 1:

for a diatomic gas.

172
Now, and

Molecular weight of the mixture, ∑

Mass fractions, and

̅
We can find using different ways, ∑ or

̅
We use the second option,

(Refer Slide Time: 6:01)

173
We need to find Cp and Cv of the mixture. Let’s also find ̅ and ̅ of the mixture.

( ̅⁄ ) ̅
Now, and ̅

̅
Now, and ̅ ̅ ̅

In the similar fashion, ̅ ,

̅ ( for both the gases)

174
Now, ∑ .

̅ ∑ ̅ ( can also be
̅
calculated as )

Now,

̅ ̅ ̅

̅
̅

There are various ways to calculate and . You get the same values through all the
ways if you consider sufficient digits after decimal point.

(Refer Slide Time: 21:05)

175
We are also asked to calculate and W.

It is a constant pressure process. .

( ) (
)

According to the first law in the integrated form, .

176
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 40
Tutorial Problem - Part 3

(Refer Slide Time: 00:17)

Figure 1.

177
Solution of the problem in Fig. 1:

Both the gases can be treated as ideal gases. Hence, they follow ideal gas relationship. We are
asked to calculate volume of each tank.

( ̅⁄ )
So, = volume of the tank containing Argon

Similarly, = volume of the tank containing Oxygen

After mixing,

The mixture is also an ideal gas. Hence, .

̅ ̅ ̅
Now, g [Mmix = 1/summation(yi/Mi)]
∑

Therefore,

178
The work interaction for this process of mixing is 0, i.e., W=0 (there is no shaft work,
electrical work, magnetic work, surface tension work, etc.).

The first law in the integrated form, (there are no changes in kinetic and
potential energy).

Since W=0, ( ) (

) ( ) ( )

Q is positive. Hence, there is a flow of heat into the system (both the tanks considered
together) during the process of mixing and the internal energy increases.

179
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 41
Tutorial Problem - Part 4
(Refer Slide Time: 00:15)

Figure 1.
Solution of the problem in Fig. 1:

The polytropic process follows, .

The system here is the mixture of gases inside the piston-cylinder assembly. We assume that the
mixture is an ideal gas.

We know the expression for the work done by the system in a polytropic process.

(for an ideal gas, )

180
We need to find .
̅ ̅ ̅
Now,
⁄∑ ⁄ ⁄

Now,

The work interaction is negative. Work is done on the system.

The first law in the integrated form for a process, (there are no changes in kinetic
and potential energy of the system). To calculate Q, we need to calculate .

There are different ways to find out ; or ∑ .

Since the mixture is made of diatomic gases, for the mixture. Hence,

∑ should give the exact value if we consider sufficient digits after the decimal
point in all the calculations.

Now,

The first law implies

The heat interaction is also negative. It means heat is leaving the system. However, the change in
internal energy is positive. It means the internal energy of the system increased after the process.
Since, for an ideal gas, internal energy is only a function of temperature, the temperature has also
increased after the process.

181
(Refer slide Time: 04:25)

182
(Refer Slide Time: 12:38)

183
184
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 42
Beyond Ideal Gases - Part 1

In the previous lecture, we covered concepts of ideal gas and a mixture of ideal gases. Let’s
look at the case where the gas is not ideal.

(Refer Slide Time: 00:59)

Figure 1.
We know, for an ideal gas, or , where p, V, m, R, T and v represent
pressure, volume, mass, specific gas constant, temperature and specific volume of the gas.
For an isothermal process, as R is constant for a gas and m is fixed for a
system. Also, . On a p-V diagram, pV=constant represent a rectangular
hyperbola. Similarly, on a p-v diagram, pv=constant represent a rectangular hyperbola (see
Fig. 1). As the temperature increases, the value of the constant in pv = constant increases, and
we get different curves (isotherms) (see Fig. 1). For the process 1-2 (Fig. 1), which is an
isobaric process, the specific volume (and the volume) increases along with temperature
(temperature increases from 300 K to 600 K). For the process 2-3 (Fig. 1), the specific
volume (and volume) is constant, but the pressure as well as the temperature decreases
(temperature decreases from 600 K to 300 K). For the process 3-1 along the isotherm, the

185
temperature remains constant (T=300 K), but the specific volume decreases and the pressure
increases.

(Refer Slide Time: 05:16)

Figure 2.

The ideal gas equation can also be written as ( ). If, for a process,

p=constant, then T=constant v, which is a straight line with slope passing through the

origin on a T-v diagram (see Fig. 2). For a larger value of p, the slope is also higher (Fig. 2).
Different isobars are shown in Fig. 2. For the process 1-2 (Fig. 2), the temperature is
constant, but the specific volume decreases and the pressure increases (pressure increases
from 500 kPa to 1000 kPa). For the process 2-3, the specific volume is constant, but the
temperature as well as the pressure decreases (pressure decreases from 1000 kPa to 500 kPa).
For the process 3-1, the pressure stays constant as it happens along an isobar p=500 kPa, but
the specific volume and the temperature increase.

186
(Refer Slide Time: 10:30)

Figure 3.

̅
For an ideal gas, where . Here, R is the specific gas constant, ̅ is the

universal gas constant, M is the molecular weight. For a perfect gas, and are constants.
and versus T are lines parallel to the T axis. Hence, is always constant which
is R. For an ideal gas, and are functions of temperature [they can be represented as
and ]. As temperature increases, both, and , tend to increase.
However, the difference, stays the same which is R. Figure 3 shows the
variation of and with temperature (from Tref to T1). For an ideal gas,
∫ change in specific internal energy = area under the curve. For a prefect gas,
is constant. Hence, . Similarly, for an ideal gas, ∫ change in
specific enthalpy. Again, for a perfect gas, .

187
(Refer Slide Time: 15:51)

Figure 4.

For an ideal gas, or . The ratio is called as Z, the compressibility factor,

which has a value of 1 for an ideal gas. For a real gas, Z does not have to be 1. For an ideal
gas, the value of Z is 1 irrespective of the value of pressure (Fig. 4). For nitrogen at T=300 K,
the value of Z is close to 1 over pressures from p=0 atm to p=40 atm. It means that nitrogen
behaves as an ideal gas at these conditions. The Z value for water vapor at T=600 K deviates
from 1 as the pressure increases (Fig. 4). Similar conclusion can be drawn in the case of
carbon dioxide (Fig. 4). Hence, these gases cannot be considered as ideal gases at these
conditions. The value of Z is the measure of deviation of the behavior of a gas from the ideal
gas behavior.

There are some reasons why gases do not behave in ideal fashion. At low temperatures, due
to intermolecular attractive forces, the pressure is lower than for an ideal gas. Also, the
volume occupied would be smaller than for the ideal gas.

188
(Refer Slide Time: 19:31)

How to deal with real gases? One way is use compressibility charts which we will study in
the next class.

189
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering,
Indian Institute of Technology Madras
Lecture 43
Beyond Ideal Gases - Part 2
(Refer Slide Time: 00:12)

Figure 1.
Figure 1 shows a plot of compressibility factor Z versus reduced pressure at different reduced
temperatures. Reduced pressure (pR) equals the pressure divided by the critical pressure (
). Similarly, reduced temperature (TR) equals the temperature divided by the critical

temperature ( ).

We know that when the molecules are far apart from each other (the pressure is low), there is not
much attractive force between the molecules. At such condition, the behaviour of the gas is close
to ideal gas behaviour. In the graph shown in Fig. 1, towards left (pR < 0.5), where the pressure is
low, we see that the value of Z is close to 1 irrespective of temperature. At high temperatures (for
example, TR = 2), the value of Z is close to 1 irrespective what the pressure is. Hence, at very
low pressures (pR < 0.5) and very high temperatures (TR = 2), the gas behaves like an ideal gas.
The graph shown in Fig. 1 has data points for different gases like methane, ethane, propane,
nitrogen, etc. For any particular value of pR and TR, the value of Z for all these gases is the same.

190
We know that . Hence, . If Z=1 (if the gas is ideal), . Hence, Z gives the

ratio of the specific volume occupied by the real gas to the specific volume occupied by the ideal
gas (it applies to volumes also if we consider a system where mass is fixed). The value of Z
corresponding to pR=2 and TR=1.1 is around 0.4. It means that the actual volume equals 0.4
times the volume occupied by the ideal gas at the same pressure and temperature conditions.
Hence, the Z is essentially the correction for the volume occupied by the real gas. If we know the
volume of the real gas in advance, then by taking the ratio of the actual volume and the volume
calculated by ideal gas relation, we can calculate Z, which can be used to calculate pressure and
temperature of the gas. This is one way to deal with real gases.

We also have tables for different gases (air, nitrogen, methane, etc.) which present gas properties
at different conditions. We can use these properties to calculate other properties.

(Refer Slide Time: 08:29)

So, we know how to deal with a perfect gas ( and are constant, ,
etc.), an ideal gas ( ( ) ( ) ( ) ( ) , etc.) and a non-
ideal gas ( tables, compressibility chart). For the purpose of this course, we will
employ perfect gas model because we do not alwasy know the values of and as functions
of temperature. We can use these models for liquid vapors also.

191
How to deal with a mixture of vapour and liquid? Why do we need to deal with mixtures of
vapour and liquid?

(Refer Slide Time: 11:40)

Figure 2.
We see that mixtures of vapour and liquid are fairly common in real life applications. One of
those applications is shown in Fig. 2. It is a steam turbine and it is used for producing electricity.

(Refer Slide Time: 12:04)

Figure 3.

192
In a coal power plant, Rankine cycle is employed for power production. We will look at this
cycle in a lot of detail towards the end of this course.

The Rankine cycle consists of 4 components: a pump, a boiler, a turbine and a condenser (see
Fig. 3). A pump takes in low pressure water and gives out high pressure water. This high
pressure water enters boiler where it boils and we get high pressure steam at the exit of the
boiler. This high pressure steam is expanded in the turbine which drives generator to generate
electricity. At the exit of the turbine, we get low pressure low temperature steam which is sent to
the condenser where the steam condenses into water. This water is sent back to the pump and the
cycle continues. In the boiler and the condenser (also in the turbine and the pump in small
traces), at some point in time, we have a mixture of water and vapour. Hence, we need to know
how to deal with the mixture of liquid and its vapour.

193
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 44
Two Phase System - Part 1
(Refer Slide Time: 00:14)

Figure 1.

Figure 1 shows a frictionless piston-cylinder arrangement containing liquid water. There

is a weight w on the piston. If there is expansion or contraction in this system, it is going
to happen at constant pressure (because the mass of the piston and w are constant).
We have pure water at A and we start heating it. During this process, the temperature of
the water is increasing and the pressure is constant. The volume of the water will increase
slightly because most substances, when heated, slightly expand. So, the volume at B is
slightly more than the volume at A. The temperature at B is higher than the temperature
at A. The water is still in liquid state. If we continue to heat, we reach boiling point of
water at given pressure (for example, at 1 atm, the boiling point of water is 100 ). Let
us say, at B, we have reached the boiling point. The water starts forming vapor as we
continue to heat. The steam (water vapor) occupies some volume and the volume at C is
larger than that at B. As we continue to heat, more and more water becomes steam. The
volume keeps increasing and the water level goes down. The specific volume of steam is
larger than that of water. After some time, all the water becomes steam (state E). If we

194
continue to heat, the temperature of steam as well as its volume increases. This is how the
water boils at constant pressure.
Figure 1 also shows the boiling process on a T-v diagram. At state A, the water is in

liquid state. The temperature and the specific volume ( ) are low. The

temperature increases from state A to state B (state B is at boiling point). During this
process, the volume also increases slightly. The increase in temperature is significantly
larger compared to the increase in volume. The curve between states A and B on the T-v
diagram is very steep. The heat supplied during the process between states A and B is
called sensible heat. We have pure water at all the states between A and B. Hence, it is a
liquid phase (single phase) region. At state B, we reach the boiling point. The heat added
after that is used up to convert all the water into steam. This heat is called as latent heat.
The temperature does not increase until all the water gets converted into steam. During
the sensible heating period, the heat supplied is . However, we cannot use
such equation for the latent heat. Usually, it is obtained from tables. After state B, the
specific volume increases as the temperature remains constant (the pressure is also
constant). From state B to state E, only the specific volume increases at constant
temperature and constant pressure. Hence, all the states, B, C, D, and E, lie on a line
parallel to the specific volume axis. At state B, we have pure water (saturated water) at
boiling point. At state E, we have pure vapor (saturated vapor) at boiling point. At all the
states between B and E, we have a mixture of liquid water and vapor (it is a two-phase
region). As we continue to heat, the temperature and specific volume of the steam
increases (state E to state F is a gas phase (single phase) region). As in the liquid phase
region (state A to state B) the increase in temperature is significantly larger compared to
that of the specific volume. However, for the same temperature difference, the increase in
specific volume for the gas phase region is more compared to that of the liquid phase
region. All the states A, B, C, D, E and F lie on an isobar.

195
(Refer Slide Time: 09:52)

To summarize, water contained in a constant pressure device is heated slowly in the

previous experiment. The volume changes continuously. Temperature increases when
water is either completely in liquid phase or completely in vapor phase. During the phase
change, temperature and pressure remain constant. The heat transfer happens without a
temperature change when you have a mixture of liquid and vapor at constant pressure.

(Refer Slide Time: 10:26)

Figure 2.

196
Figure 2 shows different isobars on a T-v diagram. The boiling point increases with
pressure. As the molecules of the liquid are forced to stay closer at higher pressure, it
requires more energy to convert it into vapor. At 0.1 MPa (1 bar), the boiling point is
around 373 K (100 ). At 10 MPa (100 bar), the boiling point is around 600 K. If we
join all the points where boiling starts (e.g. state B in Fig. 1), we get saturated liquid line
(Fig. 2). This line separates liquid water on the left and the mixture of water and vapor on
the right. Similarly, if we join all the points where boiling ends (e.g. state E in Fig .1), we
get saturated vapor line. This line separates liquid and vapor region on the left and pure
vapor region on the right. The saturated liquid line and saturated vapor line form liquid-
vapor dome. Inside the dome, we have a mixture of liquid and vapor. At low
temperatures (and low pressures), the difference between specific volumes of a pure
liquid and pure vapor is large. As the temperature (and pressure) increases, this difference
reduces. At some temperature and pressure this difference becomes 0. The specific
volume of the liquid and the vapor is the same. The temperature, pressure and volume
corresponding to this point are called critical temperature (Tc), critical pressure (pc) and
critical volume (Vc). The saturated liquid line and the saturated vapor line meet at the
critical point.

The latent heat of vaporization/evaporation reduces as the temperature and pressure

increases. At the critical point, the latent heat becomes 0. At pressures and temperature
above pc and Tc, we can’t distinguish between a liquid and vapor. We call it a
supercritical fluid. The volume and temperature of a supercritical fluid increase with
heating.

The saturated liquid line is much steeper compared to the saturated vapor line. For a
given increase in pressure, the change in the specific volume of the saturated liquid is
much smaller compared to that of the saturated vapor, because liquids are incompressible
whereas the gases are compressible.

197
(Refer Slide Time: 19:27)

(Refer Slide Time: 19:53)

Figure 3.

Figure 3 shows all the important regions and terminologies we learnt till now on a T-v
diagram. On the left of the saturated liquid line, we have compressed or subcooled liquid.
On the right of saturated vapor line, we have superheated vapor. The isobar passing

198
through the critical point is the critical isobar. At high volumes and temperature much
higher than the critical point, the fluid can be treated as ideal gas. Essentially, far away
from the liquid-vapor dome towards right on a T-v diagram, we can treat vapor/fluid as
an ideal gas (even if we have pressure is less than the critical pressure).

199
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 45
Two Phase System - Part 2

We looked at a steam-water system in the previous lecture.

(Refer Slide Time: 00:14)

Figure 1.

During the phase change process (state B to state E), the temperature and the pressure
stay constant (see Fig. 1). The specific volume change is very small from state A to state
B. This change is large from state B to state E, whereas, it is moderate from state E to
state F. The liquid on the left of saturated liquid line is called compressed liquid because
at a given temperature, it is compressed so much that it cannot become vapor.

200
(Refer Slide Time: 03:54)

(Refer Slide Time: 04:17)

Figure 2.

We looked at the graph shown in Fig. 2 when we discussed ideal gases. It shows isobars
for an ideal gas. The equation of those lines is . In Fig. 2, an isobar on a liquid-

vapor dome is also drawn. It is not a straight line in this region. In the two-phase region,
it is straight and parallel to the v axis. Outside the dome and near to it, it is curved. It

201
becomes straight line ( ) as shown in the pink graph far away towards right from

the liquid-vapor dome.

(Refer Slide Time: 05:24)

Figure 3.

Figure 3 shows a p-v diagram for a mixture of liquid and vapor (we will look at solids a
little later). Here, we have a liquid-vapor dome the topmost point of which is the critical
point as shown in Fig. 3. Different isotherms are shown. The temperature increases in the
direction of the red arrow. At low temperature, the specific volume is low. As the
temperature increases, the pressure and the specific volume at which the boiling starts
also increase. Within the liquid-vapor dome (if the boiling happens at constant pressure
as considered in the previous lecture), the temperature also remains constant. The
isotherm and the isobar overlap in the liquid-vapor dome. They are parallel to the v-axis.

If you want to boil water at constant temperature (along an isotherm), you need to change
pressures which fall on that particular isotherm as shown in Fig. 3.

The isotherm which goes through the critical point is a critical isotherm. The critical
isobar is a line parallel to the v-axis on a p-v diagram (Fig. 3).

202
As in the T-v diagram, we have a saturated liquid line and a saturated vapor line on the p-
v diagram too which separate liquid region (single phase region), a mixture of liquid and
vapor (two-phase region), and gas/vapor region (single phase region).

(Refer Slide Time: 08:21)

Figure 4.

Figure 4 shows isotherms for an ideal gas which can be represented by the equation
(which represents rectangular hyperbola on a p-v diagram). We
discussed this when we studied the concept of an ideal gas.

203
(Refer Slide Time: 08:37)

Figure 5.

Figure 5 shows isotherms which are close to a saturated vapor line on a p-v diagram. The
isotherms close to the saturated vapor line are not hyperbolic. However, as you go away
from the saturated vapor line (essentially the liquid-vapor dome) towards right on a p-v
diagram, isotherms look like hyperbola (Fig. 5). In this region, the steam/vapor can be
treated as ideal gas. Hence, in the region close to the liquid-vapor dome, we cannot use
ideal gas relation to find gas/steam/vapor properties. We can get those properties from
tables.

204
(Refer Slide Time: 09:34)

Figure 6.

A saturated state in general is a state at which there is a change of phase which may occur
without a change in pressure or temperature; if the pressure is fixed, the temperature is
fixed, or if the temperature is fixed, the pressure is fixed. The change of state can be
either from solid to liquid or vice versa, liquid to vapor or vice versa, solid to vapor or
vice versa. The saturation temperature is a temperature at which vaporization takes place
at a given pressure. Similarly, the saturation pressure is a pressure at which vaporization
takes place at a given temperature. For example, at 1 bar pressure, the boiling point of
water is 99.6 . Hence, 1 bar is the saturation pressure at 99.6 and 99.6 is the
saturation temperature at 1 bar pressure.

If the temperature for a given pressure is below the boiling point, the fluid is in subcooled
region/compressed liquid region. For a given pressure if the temperature is above the
boiling point, the fluid is in superheated region. Similarly, for a given temperature, if the
pressure is higher than the pressure at which boiling happens (saturation pressure), then
the fluid is in the compressed liquid/subcooled liquid zone. If the pressure is lower than
the saturation pressure at a given temperature, the fluid is in superheated state. If the fluid
is at saturation pressure at given temperature, then the fluid is in the two-phase region.

205
(Refer Slide Time: 12:21)

Figure 7.

Let’s look at a p-T diagram. We know that during phase change, the pressure and the
temperature remains constant. This pressure and the corresponding saturation temperature
form a point on p-t diagram. Similarly, different pressures and the corresponding
saturation temperature can be represented by different points on a p-T diagram. If we join
all these points, we get curves as shown in Fig. 7, which are for a substance which
contract on freezing. On a p-T diagram, you don’t get a liquid-vapor dome (two phase
region). The curves in Fig. 7 show a fusion line, a sublimation line and a vaporization
line. A substance changes phase from solid (S) to liquid (L) across the fusion line. A
substance changes phase from liquid (L) to vapor (V) across the vaporization line. A
substance changes phase from solid (S) to vapor (V) across the sublimation line. These
lines separate regions of three phases of a substance. These lines meet at the triple point
where a mixture of solid, liquid and gaseous phases of a substance exists in equilibrium.
Till now, we were looking at only the conversion from liquid to vapor. For example, in
Fig. 7, the complete conversion of a given amount of liquid substance to vapor at
saturated conditions happen at a point on the vaporization line (L-V line). A critical point
is also shown in Fig .7.

206
Consider a substance at pressure below critical pressure and above triple point pressure
and heat it at constant pressure. Initially, the substance is in solid state. As we keep
heating it, its temperature rises. As we continue to heat, the temperature hits the melting
point (S-L curve or the fusion line). The complete conversion from solid to liquid
happens at constant temperature (a point on the fusion line represents this entire process).
Once the entire solid converts into liquid and we continue to heat, the temperature of the
liquid rises till it reaches the vaporization line (L-V line). Again, the complete conversion
from liquid to vapor happens at a constant temperature. As the heating continues, the
temperature of the vapor increases. If we consider the substance at pressure higher than
the critical pressure and heat it, we see only a conversion from solid to supercritical fluid.
At very low pressures (less than triple point pressure), we see a conversion from solid to
vapor (across S-V line or sublimation line), which is known as sublimation.

Similarly, consider a substance at very low temperature (less than the triple point
temperature) and increase the pressure at that temperature. At very low temperature and
low pressure, the substance is in vapor state. As the pressure increases, the molecules are
forced to come closer and closer. At a pressure corresponding to S-V line, the vapor starts
getting converted into solid. This conversion, again, happens at constant pressure and
temperature (it is a point on S-V line). As the pressure increases, the substance stays in
the solid region.

A substance at a temperature above triple point temperature, below critical point

temperature and low pressure is in a vapor state. As the pressure increases at the same
temperature, the vapor starts becoming liquid at a point on the S-L line. During this
conversion, the pressure and the temperature remain constant. As the pressure increases,
the liquid starts converting into solid at a point on S-L line. Again, the pressure and the
temperature remain constant during this conversion. After that as the pressure increases,
the substance stays in the solid phase.

All the above discussion is for a substance which contracts on freezing.

207
(Refer Slide Time: 21:19)

Figure 8.

Figure 8 shows a p-T diagram for water. Water expands on freezing (anomalous behavior
of water). Freezing point ( ) and the boiling point (100 ) at 1 atm are shown in Fig.
8. All the discussion above in the case of a substance which contracts on freezing is valid
for water also. For example, at a pressure of 1 atm and at low temperature, we have ice.
As we heat it at a constant pressure, it becomes liquid and ultimately vapor. We can have
ice at 100 at 10 GPa. As the pressure reduces at 100 , this ice becomes liquid and
ultimately vapor at 100 . There is a special region on the p-T diagram of the water. At

208
temperature in this region and at very high pressure, we have ice. As the pressure reduces
at constant temperature, the ice becomes liquid, then solid and then vapor. This region is
drawn in the magnified view in Fig. 8. This happens because of the anomalous behavior
of water. For water, the temperature and pressure at the critical point are 374 and
220.6 bar.

The fusion line, the vaporization line and the sublimation line represent states at which
two phases of a substance are in equilibrium. The properties of a substance in each phase
do not change at equilibrium. For example, if we have a mixture of water and its vapor in
equilibrium at certain temperature and pressure, the properties of liquid water and vapor
such as pressure, temperature, mass, specific volume, etc. do not change with time. For
all substances, there is one particular state where we can have a substance in liquid, vapor
and solid form in equilibrium with each other and that state is called a triple point. For
water, the triple point corresponds to temperature of 0.01 and pressure of 611 Pa. At
these conditions, we can find ice, liquid water and steam in equilibrium.

(Refer Slide Time: 33:19)

Figure 9.

We have been looking at p-v, p-T and T-v diagrams. There are three variables, p,v and T.
It means that we can show a state of a substance on a surface in three dimensional p-v-T

209
space. Such a p-v-T surface is shown in Fig .9 for a substance which contracts on
freezing. Figure 10 shows the p-v-T surface for a substance which expands on freezing
such as water along with the p-v-T surface for a substance which contracts on freezing.

(Refer Slide Time: 35:57)

Figure 10.

(Refer Slide Time: 36:33)

Figure 11.

210
Figure 11 shows projections of a p-v-T surface on p-v, p-T and T-v planes which are our
regular p-v, p-T and T-v diagrams. Such a p-v-T surface can be printed using a 3D
printer.

211
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 46
Two Phase System: Water and Steam
(Refer Slide Time: 00:14)

Figure 1.

Let’s look at a mixture of liquid water and steam which is used in a lot of applications.
Whatever we study for the mixture of water and steam will also be applicable for mixtures of
other liquids and their vapour.

Consider a closed container as shown in Fig .1 which contains a mixture of liquid and vapour
at equilibrium (half of the container is invisible because of some display issue). The mass of
the vapour is mvap and the mass of the liquid is mliq. The mass of the mixture is
. Our system is made of this liquid and vapour. For such a system, dryness

fraction or quality x is defined as = fraction of the mixture that is in

vapour phase. We had used the alphabet x to represent mole fraction when we studied ideal
gases. Be careful!

(Refer Slide Time: 03:03)

212
 Figure 2.

In general, we use the alphabet f to represent liquid (fluid) and g to represent gas (these
notations are used in all the books). Now, the mass of the mixture m = and the

quality . The volume of the mixture . Hence,

( ) , where v represents specific volume of the mixture. Dividing by m

throughout, ( ) . Hence, ( ) ( )

, where is . When we have a mixture of a liquid and vapour in

equilibrium, the liquid and vapour are called as saturated liquid and saturated vapour/dry
saturated vapour, respectively. All the properties with subscripts f and g are properties of
saturated liquid and saturated vapour, which are the states on a T-v diagram where an isobar
(with pressure less than critical pressure) cuts the liquid-vapor dome. The intersection of such
an isobar and the saturated liquid line represents the state where we have saturated liquid and
the quality x=0 here. The intersection of that isobar with the saturated vapour line represents
the state where we have saturated vapour/dry saturated vapour and the quality x=1 here. The
quality/dryness fraction is associated with only the two-phase region and those two states
where we have only liquid (x=0) and only vapour (x=1) at saturated conditions. It is not
associated with the subcooled region or the superheated region.

213
(Refer Slide Time: 09:29)

Let’s look at steam tables (these are for water). We can obtain properties of liquid water,
steam and a mixture of water and steam from such tables. Similarly, we can have tables for
other liquids such as refrigerants or for gases such as nitrogen, oxygen, argon, etc.

(Refer Slide Time: 10:09)

Figure 3.

Figure 3 shows a snippet of steam tables for saturated water listed by temperature. For
saturated water (only water, no vapour), . For saturated vapour (only vapour, no water),
x=1. For a mixture of saturated water and saturated vapour, .

214
(Refer Slide Time: 10:51)

Figure 4.

215
Figure 4 shows a snippet of steam tables for saturated water listed by pressure.

In the pressure table, the first column lists pressure (unit is megapascal). Other columns list
properties such as temperature, specific volume, specific internal energy, specific enthalpy,
and specific entropy. Temperature is in . vf represents specific volume of saturated water,
while vg represents specific volume of saturated vapour (unit – m3/kg). uf represents specific
internal energy of saturated water, while ug represents specific internal energy of saturated
vapour (unit – kJ/kg). hf represents specific enthaply of saturated water, while hg represents
specific enthalpy of saturated vapour (unit – kJ/kg), and hfg = hg-hf = latent heat of
vaporization. Similarly, sf represents specific entropy of saturated water, while sg represents
specific entropy of saturated vapour (unit – m3/kg) and sfg = sg - sf (we will look at the
concept of entropy when we look at the second law of thermodynamics). Let’s read a row
shown in the table shown in Fig.4.

For a pressure of 0.1 MPa, the saturation temperature is 99.6 . The specific volume of the
saturated liquid (x=0) is 0.001043 m3/kg and the specific volume of saturated vapour (x=1) is
1.694 m3/kg. Similarly, uf = 417.4 kJ/kg and ug = 2505 kJ/kg, hf = 417 kJ/kg, hg = 2674.9
kJ/kg and hfg = 2257.4 kJ/kg. Similarly, we can find property values at other pressures. For
example, at 0.2 MPa, the saturation temperature is 120.2 . The specific volume of the
saturated liquid (x=0) is 0.001061 m3/kg and the specific volume of saturated vapour (x=1) is
0.8857 m3/kg. Similarly, uf = 504.5 kJ/kg and ug = 2529.1 kJ/kg, hf = 504.7 kJ/kg, hg = 2706.5
kJ/kg and hfg = 2201.5 kJ/kg. The latent heat of vaporization has reduced at higher pressure.
The last value in the pressure column is that of critical pressure; 22.064 MPa and

216
 =373.95 . At the critical point, the specific volume of the liquid and the specific volume
of the gas are the same. Similarly, the specific internal energy and specific enthalpy of the
liquid and the vapor are the same.

(Refer Slide Time: 19:33)

217
(Refer Slide Time: 20:00)

Figure 5.

Figure 5 shows the snippet of the steam tables listed by temperature. Here, the first column is
of temperature. The properties can be read in a similar way we discussed in the case of the
tables listed by pressure. The choice of the tables listed by pressure or temperature depends
on convenience. If the properties at temperature and pressure we are looking for are not
listed, then we can use linear interpolation to find properties at the required temperature and
pressure. The last entry in the tables listed by temperature is the critical point.

Using these tables, we can find properties of liquid water and vapour at saturation, and using
these properties we can also find properties of a mixture of liquid water and vapour.

218
(Refer Slide Time: 24:08)

219
(Refer Slide Time: 25:01)

Figure 6.

There are tables to find properties of superheated vapour. Figure 6 shows tables for
superheated steam. Let’s look at the table in the top left corner of Fig .6. The first row reads
the pressure and the corresponding saturation temperature; p=0.01 MPa and T = 45.8 . This
table is for a pressure of 0.01 MPa and temperatures larger than the saturation temperature
corresponding to p=0.01 MPa which is 45.8 . At T > 45.8 , we will have only the
superheated vapour. Hence, we can’t see properties of liquid in this table. The table mentions
the volume, internal energy, enthalpy and entropy of vapour only. So, this table is for p=0.01
MPa and T > 45.8 (T=45.8 is the saturation temperature for p=0.01 MPa). Similarly, we

220
have tables for different pressures and the properties are mentioned only for temperatures
greater than the corresponding saturation temperatures.

(Refer Slide Time: 28:54)

221
 Thermodynamics
Professor. Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 47
Tutorial Problems (2 numbers)

Let’s look at few tutorial problems on steam and water mixtures. It is recommended to have a
hardcopy of steam tables.

(Refer Slide Time: 00:41)

Figure 1.

Solution of the problem in Fig. 1:

(a) 5 bar, 151.9
Let’s go to the steam tables listed by pressure. We don’t have a value of 5 bar or 0.5 MPa in
the tables. We will find the properties by linear interpolation. The properties at 0.5 MPa are
the average of those at 0.4 MPa and 0.6 MPa. After taking average, saturation temperature at
5 bar is 151.2 . Since the given temperature, 151.9 , is greater than 151.2 , the state at 5
bar and 151.9 lies in the superheated zone. The intersection of the isotherm T=151.9
and the isobar p=5 bar represents the state asked in the problem on a p-v diagram as well as
on the T-v diagram. Figure 2 shows the state (point a) on a p-v and T-v diagram.

222
(b) 5 bar, 200

At 5 bar, the saturation temperature is Tsat = 151.2 . The given temperature is greater than
Tsat. Hence, the state corresponding to 5 bar and 200 lies in the superheated zone. Figure 3
shows the state (point b) on a p-v and T-v diagram. The degree of superheat is more in the
case b than a. Degree of superheat equals the difference between the saturation temperature
and the actual temperature of the superheated vapour.

(c) 2.5 MPa, 200

The pressure 2.5 MPa is not listed in the pressure table. Let’s look at a temperature table.

At 200 , the saturation pressure is psat = 1.55 MPa. The given pressure is higher than psat.
Hence, the state lies in the compressed liquid region. Figure 3 shows the state (point c) on a
p-v and T-v diagram.

(d) 4.8 bar, 160

Let’s look at the table listed by temperature.

At 160 , the saturation pressure is psat = 0.6182 MPa = 6.182 bar. The given pressure, 4.8
bar, is less than psat. Hence, the point lies in the superheated region. Figure 3 shows the state
(point d) on a p-v and T-v diagram.

Within the liquid-vapour dome, the pressure and temperature are not independent. Hence, to
locate the state on a p-v or T-v diagram within the liquid-vapor dome, along with pressure or
temperature some other property such as internal energy, specific volume, specific enthalpy
or specific entropy needs to be given.

223
(Refer Slide Time: 01:15)

(Refer Slide Time: 02:59)

Figure 2.

224
(Refer Slide Time: 10:49)

Figure 3.

225
(Refer Slide Time: 20:53)

Figure 4.

Solution of the problem given in Fig. 4:

=pressure at the critical point
The system is the mixture inside the container.
Since it is a rigid container, the work done W = 0.
The first law in the integrated form, (there are no changes in kinetic and potential
energy).

226
The process is shown on a T-v diagram in Fig. 4. It is a constant volume process. The initial
state is the point of intersection of 2 bar isobar and a vertical line passing through the critical
point on a T-v diagram.

Since the mass is fixed and it is a constant volume process, = specific volume
at the critical point = 0.003106 m3/kg.

Now, [ ( )]

Hence, 0.0023 (the values of the specific volumes are

obtained from the steam tables)

Now, ( ) [ { ( )} ] [

( )] ( =specific internal energy at the critical

point)

Since Q is positive, the heat is transferred to the system.

227
228
(Refer Slide Time: 30:55)

229
(Refer Slide Time: 36:19)

230
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 48
Tutorial Problems – Part 1
(Refer Slide Time: 00:14)

Figure 1.

Solution of the problem in Fig. 1:

231
The piston moves when the pressure becomes 1 MPa. Hence, the pressure outside plus the
pressure due to the weight of the piston equal 1 MPa.

The initial state is

At 0.2 MPa, (from steam tables). Hence, .

Thus, at state 1, we have a mixture of water and vapour (saturated mixture).

Hence, the quality

Until the pressure becomes 1MPa, the process is a constant volume process. Let’s calculate

the heat required for this constant volume process between and

The first law for this constant volume process in the integrated form, (W = 0 and
there are no changes in kinetic and potential energy)

Now, ………(1)

Now, [ ( )]

At . Hence, . Thus, state 2

( ) lies inside the liquid-vapour dome.

Now, (The quality has increased in this process. It means there is more

vapour now compared to that at the initial state.)

So, [ ( )] .

Now, (1) implies .

Out of the total heat supplied, i.e., 2500 kJ, 962.65 kJ is used up in a constant volume process
raising its pressure from 0.2 MPa to 1 MPa (temperature also rises). The remaining heat,
which is 2500-962.65=1537.35 kJ, is going to be used in a constant pressure process.

232
Writing the first law,

For a constant pressure process,

Integrating, .

Now, is the enthalpy at the state where p=1 MPa and v=0.1 m3 (x2=0.512).

Hence,

( )

The state 3 is at 1 MPa and h3 = 3331.32 kJ/kg. At 1 MPa, hg = 2777 kJ/kg. Since, h3>hg(at 1
MPa), the state 3 is in the superheated region.

We are also asked to calculate the work done. The work is done only in the second phase of
the entire process where pressure was constant.

Writing again the first law for the constant pressure process in the integrated form,

…..(2)

We need to calculate u3. h = 3331.32 kJ/kg lies between the enthalpy values at 400 and
450 in the table for superheated steam at 1 MPa. Hence, interpolation gives,

We know h3. Solving the above equation, T3 = 431.3 and u3=3009.85 kJ/kg.

Hence,(2) implies .

The entire process on a p-v diagram is shown in Fig. 3. State 3 is the intersection of the
isotherm T3=431.3 and the isobar p=1 MPa in the superheated zone.

233
(Refer Slide Time: 13:07)

234
(Refer Slide Time: 25:37)

Figure 2.

235
 Thermodynamics
Professor Anand TNC
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 49
Tutorial Problem Part-2

(Refer Slide Time: 0:15)

Figure 1.

236
Solution of the problem in Fig .1:

The insides of the pressure cooker form our system.

Initial state is in the two-phase zone since there is a mixture of water and steam in equilibrium.

237
Now, where vf and vg can be found from steam tables at 100

. At 100 , vf = 0.001044 m3/kg and vg = 1.672 m3/kg. Hence, .

Now,

The mixture is heated until the pressure reaches p2 = 2 MPa at constant volume. Hence, v1 = v2.

Now,

At state 2 (final state), p2 = 2 MPa, . Now, .

Hence, the state 2 is inside the liquid-vapor dome.

Now,

The final temperature T2 = the saturation temperature at 2 MPa = 212.4 .

Now, mass of the vapor at state 1, = . Similarly, . Since the mass is

fixed and , the mass of the vapor at state 2 is larger than that at state 1.

238
 Thermodynamics
Professor Anand TNC
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 50
Tutorial Problem Part-3

(Refer Slide Time: 00:13)

Figure 1.

239
Solution of the problem in Fig. 1:

240
At 0.5 MPa, and (These values are the average of those at 0.4

MPa and 0.6 MPa. In the tables provided, 0.5 MPa is not listed. Hence, we did linear
interpolation.)

Now, . Hence, the state 1 (initial state) is inside the liquid-vapor dome.

Now,

This is a constant pressure process. At p2 = 0.5 MPa, the temperature is 300 . Now, >
saturation temperature at 0.5 MPa which is 151.8 . Hence, the state 2 lies in the superheated
zone. We are asked to calculate Q and W.

For a constant pressure process, we know that ……(1)

Also, the work done ……(2)

Now, at state 2, (from tables for superheated steam)

Now, ( )

Now, (1) implies

(2) implies

241
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 51
Tutorial problems on two-phase systems(2 numbers)

(Refer Slide Time: 0:15)

Figure 1.

Figure 2.

Solution of the problem in Fig. 1:

Since the steam is going to condense, it must be in superheated zone initially.

242
At 0.7 MPa and , (it is average of values at 0.6 MPa and 0.8
MPa)

The initial state (state 1) is shown on a T-v diagram in Fig. 1. The isobar of 0.7 MPa is also
shown.

Now, .

Since the vessel is rigid, the cooling is a constant volume process. The condensation will start
at the state where vertical line through the state 1 intersects the saturated vapor line. It is
shown in Fig. 1. From steam tables, we need to find the state on the saturated vapor line

where the specific volume is .

lies between vg values at 130 and 140 in the tables provided.

Interpolating, we get the temperature as T=139.4 . This is the temperature at which the
condensation starts. The corresponding saturation pressure for T=139.4 is around 0.36
MPa. But, the cooling continues till the pressure reaches p2 = 0.05 MPa. It means the final
state (state 2) is inside the liquid-vapor dome. We need to calculate the quality for p2=0.05
MPa and v2 = 0.5176 m3/kg.

and

The mass of vapor at state 2 is

Hence, the mass of liquid (condensate) at the state 2, mf = 1.93-0.296=1.63 kg.

Hence, the fraction of the total mass that condenses = 1.63/1.93 = 0.84

Volume occupied by the saturated liquid at the final state, (vf

is the specific volume of saturated liquid at 0.05 MPa).

The T-v diagram drawn in Fig. 1 is corrected in Fig. 2 as the quality at the final state is low,
but it is shown quite high in Fig. 1.

A large mass of liquid (1.63 kg) occupies a tiny volume ( ) and a small mass
of vapor (0.296 kg) occupies a large volume (1-0.00167 =0.998 m3).

243
244
245
(Refer Slide Time: 24:27)

Figure 3.

246
Solution of the problem in Fig. 3:

(a) Steam tables

At 20 MPa, the saturation temperature, Tsat is 365 . Since, 400 > Tsat, the state is in the
superheated zone. From the tables for the superheated steam, v = 0.00995 m3/kg.

(b) Ideal gas equation

( ̅⁄ ) ( ⁄ )
.

247
This value quite far from the actual value calculated using the tables.

(c) Generalized compressibility chart

The corresponding Z value for the above pR and TR values is around Z=0.5.

Now, 0.0088 m3/kg

This value is closer to the actual value compared to the one obtained using the ideal gas
equation. If the tables are available, we should always prefer tables to get the property values.

(Refer Slide Time: 30:21)

Figure 4.

248
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 52
Tutorial problems (1 numbers)

(Refer Slide Time: 0:13)

Figure 1.

249
Solution of the problem in Fig. 1:

Compartment A (air):

Compartment B (steam-water mixture):

A schematic of the system is drawn in Fig. 1.

The total volume is conserved. Hence, …..(1)

Air is considered as an ideal gas. Hence, ……(2)

( ̅⁄ )
We can calculate as ∑

) )

We need to find the initial volume for the compartment B, .

For and x1 = 0.5, ( ) . Now,

(1) implies …….(3)

250
(2) implies ( for air assuming it to be a
diatomic gas) )

(3) and (4) are always true.

Heat is added to the steam-water mixture until it reaches saturation. There is no heat transfer
between A and B. The air in A undergoes adiabatic compression.

During the process, the pressure in A and B is always equal, .

We don’t have enough information to arrive at the answer. Hence, we will do it iteratively.

Assume that the final pressure in B, . As there is only vapor at state 2 (final state)
in B, .

(3) implies . It should satisfy (4). Now, . Hence, the

guess is not good. Let’s make another guess.

Assume that the final pressure in B, . Now,

. (3) implies . Now, which is close to the
actual value from (4).

Hence, .

The temperature in B is the saturation temperature at 0.14 MPa which is .

The temperature in A can be calculated using the ideal gas relation.

Students are encouraged to try out more guesses for so that it satisfies the condition
in (4) accurately.

251
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 53
Rate equation of the first law of thermodynamics

(Refer Slide Time: 0:15)

Today we look at the rate equation of the first law of thermodynamics and then go on to the
first law for the control volume.

(Refer Slide Time: 0:24)

We have been writing the first law of thermodynamics as . Dividing by (t

is time) and taking limit as , we get

252
 ̇ ̇ , where is the change in

energy with time, ̇ is the rate of heat transfer, ̇ is the rate of doing work (power), cm
represents control mass. This is the rate equation of the first law for a control mass. We have
been using this equation without specifically mentioning that it is the rate equation of the first
law.

(Refer Slide Time: 2:30)

Figure 1.
(Refer Slide Time: 2:45)

Figure 2.

253
If we take the bulb as shown in Fig. 2 as our system, the work done on the bulb by a battery is
. The power is ̇ . Assuming steady state conditions, the rate of change of

energy of the system (bulb) equals 0, i.e., . Hence, ̇ ̇ , i.e., the rate at which

heat (electromagnetic radiation) is emitted by the bulb equals the rate at which the work is

done on the bulb. can be replaced by in this case as there is no change in kinetic and

potential energy of the system during the process. A similar analysis can be done in the case
of the fan shown in Fig. 1.

(Refer Slide Time: 5:29)

(Refer Slide Time: 5:34)

Figure 3.

254
Let’s look at the battery shown in Fig. 3.
The battery is the system. The battery does work on the surroundings (bulb, fan, etc.) by
pushing current across its boundary. Hence, we have ̇ across the system boundary. We also
have ̇ across the system boundary as the battery loses heat. For the battery, the rate equation

of the first law is ̇ ̇ (there are no changes in kinetic and potential energy of the

system). Since the battery is doing work on the surroundings and losing heat, its internal
energy will go down with time. After some time, the battery will stop working and losing
heat. This is an unsteady process. Total work done or the heat lost or the total change in
internal energy can be found out by integration. If the ̇ and ̇ are constant over a small

interval of time , the work done and heat lost in that time interval is ̇ and
̇ .

(Refer Slide Time: 9:20)

Figure 4.
Let’s do the control volume analysis.

Consider a control volume with a single inlet and outlet as shown in Fig. 4. The rates at
which the mass comes in and goes out of the control volume are ̇ and ̇ . Similarly, we
can have control volumes with multiple inlets and outlets as shown in Fig. 4.

255
(Refer Slide Time: 11:15)

(Refer Slide Time: 11:26)

Figure 5.

The rate equation for the mass conservation in control volume is ∑ ̇ ∑ ̇ ,

where ̇ is the rate at which the mass enters the control volume, ̇ is the rate at which the

mass leaves the control volume (i is for inlet and e is for exit/outlet), is the rate of

change of mass within the control volume. Summation symbol should be used when you have
more than one inlet and outlet. For example, for a control volume with 2 inlets and 2 outlets,

̇ ̇ ̇ ̇ . For a control volume with a single inlet and outlet,

256
 ̇ ̇ . If ̇ and ̇ are constant, then the change in mass of the control volume

over time interval equals ̇ ̇ . If ̇ and ̇ are not constant and we know
their variation with time, then also we can calculate by integrating the rate equation for
mass conservation.

(Refer Slide Time: 17:37)

Figure 6.
Similar to the rate equation for mass conservation, we have the rate equation for energy
conservation for a control volume. Figure 6 shows three control volumes, a bathtub, a balloon
and a compressor across the boundaries of which there is a mass transfer. The system
boundary in the case of balloon changes (the balloon expands or contracts), whereas in the
other two examples it doesn’t change. The volumetric flow rate into and out of the control
volume is ̇ ̅ ∫ ̅̅̅̅̅̅̅ , where ̅ is the flow velocity (m/s), A is the area (m2)
perpendicular to the flow direction. In the case where the velocity is not uniform over a given
cross-sectional area, we have to integrate local velocity over that area to get the volume flow
rate, e.g., to get the volumetric flow rate through a pipe where the fully developed velocity
profile is not uniform, we need to integrate local velocity over the cross-sectional area of the
pipe. If the fluid is incompressible, we can obtain the mass flow rate as ̇ ̇ ̅
̅
, where is the density of the fluid and v is the specific volume of the fluid. In the case the

density is varying, we need to integrate to get the mass flow rate.

257
(Refer Slide Time: 22:10)

Figure 7.
For a control volume, in addition to energy transfer across its boundaries (in the form of heat
and work), there is a transfer of mass also, which does not happen in the case of a control
mass. The incoming and outgoing mass brings in and takes out energy with it from the
control volume.

Figure 7 shows a control volume with a single inlet and outlet. The mass enters the control
volume at the rate ̇ and exits at the rate ̇ . The rate equation for the conservation of mass

is ̇ ̇ . The net heat and work transfer across the control volume boundaries are

represented as ̇ and ̇ . In addition to ̇ there are two more terms

related to work, which are ̇ and ̇ .

How does mass enter and leave the control volume? The fluid/liquid does not enter the
control volume on its own. It has to be pushed. Some work has to be done on the control
volume to make the fluid enter it (surroundings do work on the control volume by pushing
the fluid in). Similarly, this fluid which has entered the control volume has to be pushed out,
otherwise it is just going to sit there. Hence, the control volume has to do some work. Here,
the control volume does work on the surroundings by pushing the fluid out. These work
interactions are related to the flow work, ̇ .

258
(Refer Slide Time: 26:11)

Figure 8.
The fluid entering the control volume brings with it internal energy (u), kinetic energy (k.e.)
and potential energy (p.e.). Similarly, the fluid leaving the volume takes out with it the same
three forms of energy. The specific energy of the fluid e = u+k.e.+p.e = ̅ ,

where Z is the height from some reference.

The rate of flow work is power, which is given as ̇ ̅ , where F is the force with

which the fluid is pushed and ̅ is the fluid velocity. Now, , where p is the pressure
̇
and A is the area. Hence, ̇ ̅ ̅ ̇ ̇ , where ̇ is the

volumetric flow rate, ̇ is the mass flow rate, is the fluid density and v is the specific
volume of the fluid (the fluid is considered incompressible). The rate of flow work at the inlet
is ̇ ̇ and the rate of flow work at the outlet is ̇ ̇ . The flow

work per unit mass is . The rate of change of energy for the control volume ( )

is the rate at which heat is transferred to the control volume ( ̇ ) minus the rate at which
work is done by the control volume in some form ( ̇ ), maybe as shaft work or as electrical
work, etc. plus the mass flow rate of the fluid coming in ( ̇ ) into the energy which is being
brought in by the incoming mass ( ) minus the mass flow rate of the fluid going out ( ̇ )
into the energy which is being taken out by the outgoing mass ( ) plus any flow work which
is coming in and as well as going out ( ̇ ).

259
 ̇ ̇ ̇ ̇ ̇ ̇ ̇ ̇ ̇ ̇

Substituting the expression for e and h=u+pv (h is enthalpy of the fluid) at the inlet and
outlet,

̇ ̇ ̇ ( ̅ ) ̇ ̅

This is the first law for a flow process. For a control volume with multiple inlets and outlets,
the above equation becomes,

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )

260
 Thermodynamics
Professor Divya A
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 54
Energy equation for a steady-state

(Refer Slide Time: 0:28)

The first law for a control volume is written here:

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )….(1)

This expression can be simplified based on the application.

If the mass flow rates are 0, equation (1) reduces to the expression of the rate equation of the

first law for a control mass which is ̇ ̇ . By integrating with respect to time

(assuming we know the variation of ̇ and ̇ with respect to time), we can obtain the
expression for the first law, . So, we can obtain all the previous expressions we
used from the expression of the first law for a control volume (equation (1)).

261
(Refer Slide Time: 2:40)

Let’s look at some other simplifications of the first law for a control volume.

For a steady state steady flow process, (1) state of mass at each point in the control volume
does not change with time, (2) mass flux and the state of mass at the control surface do not
vary with time, (3) rate of heat and work transfer are constant with time. In essence, in a

steady process, nothing changes with time. Hence, in equation (1), . ̇ and ̇ do not

change with time (they may not be 0 individually). The term in the rate equation of mass

conservation ( ∑ ̇ ∑ ̇ ) becomes 0. Hence, ∑ ̇ ∑ ̇ ∑ ̇ . If we have

only one inlet and outlet, which is common in many practical applications, then ̇ ̇

̇ . Dividing by ̇ , equation (1) reduces to ( ̅ ) ( ̅

), where q and w are specific heat and work interactions. It is recommended to remember

only the general form of the first law for a control volume.

262
(Refer Slide Time: 8:36)

(Refer Slide Time: 8:44)

263
(Refer Slide Time: 8:52)

Figure 1.
Let’s apply the first law for a control volume to an adiabatic throttling process.

A capillary tube is shown in Fig. 1 (which is also used in refrigerators). If a fluid is pushed
through such a long capillary tube (it is not shown as long in the Fig. 1, but it is usually quite
long) having a small cross-section, pressure drops significantly. This process is called
throttling. If the fluid is a non-ideal gas, the temperature also drops. This process is
implemented in the refrigerators. Similarly, a valve in a pipe carrying some fluid can block
the fluid flow and there is a pressure drop across the valve. This is also a throttling process.
Let’s apply the first law to a control volume around a valve in a pipe (Fig. 1). It has one inlet
and one outlet (hence, ̇ ̇ ̇ ). This process is assumed to be a steady process

( ). Usually, these pipes and tubes are insulated. Hence, the process is also adiabatic

( ̇ ). In this process, we do not have any work interaction ( ̇ ). The only work
interaction for pushing the fluid in and out of the control volume is included in the enthalpy
term in the equation ( ). Hence, the first law for a control volume (Eq. 1) undergoing an
adiabatic throttling process is,

̇ ( ̅ ) ̇ ( ̅ )

Hence, ( ) ( ). Usually, for such tubes and pipes, the changes in

Z are very small, i.e., is very small. Hence, there is no appreciable change in specific

264
potential energy of the fluid, i.e., g( ) . If we assume that the change in specific

kinetic energy ( ) is also small compared to the enthalpy values, the equation reduces

to . However, if the velocity values at the inlet and the outlet are given and they are
significantly different, the change in specific kinetic energy should not be neglected.

In an ideal adiabatic throttling process, the enthalpy does not change, i.e., .

(Refer Slide Time: 16:33)

Figure 2.
Figure 2 shows turbines. In a turbine, high pressure high temperature gases expand. While
expanding, they rotate blades connected to a shaft, doing work. A turbine has a single inlet
and outlet (hence, ̇ ̇ ̇ ). A turbine is represented by a diverging section as shown
in Fig. 2. 1 represents inlet and 2 represents outlet. It gives out power, ̇ . The turbine

runs at steady state except when it starts or stops ( ). For an ideal turbine, the process

is adiabatic ( ̇ ). The turbine produces power ( ̇ ). The first law (Eq. 1) for a turbine
reduces to,

̇ ̇ ( ̅ ) ̇ ( ̅ )

̇ ̇( ) ̇ ( ) ̇ ( )

265
The terms and individually are not zero. But, their differences are not

appreciable. Hence,

̇ ̇( ) ̇ ( )

The expression for the power in the case of a compressor is exactly the same as that for the
turbine. Hence,

̇ ̇( ) ̇ ( ).

However, in the case of turbine, and ̇ is positive, whereas in the case of a

compressor, and ̇ is negative. In the case of a turbine, the work is done
by the control volume (gases expand rotating the shaft), while in the case of a compressor, the
work is done on the control volume (a fluid is pressurized). A compressor is represented by a
converging section as opposed to a turbine.

(Refer Slide Time: 23:05)

266
(Refer Slide Time: 24:34)

Figure 3.
Figure 3 shows a cross-section of a turbine. A turbine contains a nozzle and a diffuser. A
nozzle has a reducing cross-sectional area, while a diffuser has an increasing cross-sectional
area. In the nozzle, the fluid velocity increases, while in the diffuser, the fluid velocity
decreases. The nozzle and diffusers are not power producing devices ( ̇ ). They have a
single inlet and outlet ( ̇ ̇ ̇ ). These devices operate at steady state conditions

except at the start and the stop ( ). These devices are insulated ( ̇ ). Hence, for

nozzles and diffusers, equation (1) reduces to,

̇ ( ̅ ) ̇ ( ̅ ). Hence,

̇( ) ̇ ( ) ̇ ( ) The change in specific potential energy is not

appreciable in these devices. Hence, . Hence, . If the

specific kinetic energy at the exit is larger than the inlet (which happens in a nozzle),
would be positive, whereas would be negative if the specific kinetic energy at the exit
is smaller than the inlet (which happens in a diffuser).

267
(Refer Slide Time: 25:13)

(Refer Slide Time: 26:04)

268
(Refer Slide Time: 31:27)

Figure 4.
Figure 4 shows a cut section of a heat exchanger. Heat exchangers are used in many
applications such as cars, refrigerators, power plants, etc. A schematic of a heat exchanger is
shown in Fig. 4. It has more than one inlet and outlet. For a steady state operation of the heat

exchanger (shown in the schematic), . Hence, ∑ ̇ ∑ ̇ or ̇ ̇ ̇

̇ . Inside a heat exchanger, heat transfer happens between a hot and cold fluid. Hence, there
are separate inlets and outlets for a cold fluid and a hot fluid. The heat transfer across the
boundaries of the heat exchanger (which appears in the equation of the first law for control
volumes) is 0, ideally (hence, ̇ ). There is no work interaction for a heat exchanger

( ̇ ). As the operation is at steady state conditions, . Hence, equation (1) reduces

to,

∑ ̇ ( ̅ ) ∑ ̇ ( ̅ ) . The changes in kinetic

and potential energy are not usually appreciable. Hence, ∑ ̇ ∑ ̇ . Thus, ̇

̇ ̇ ̇ .

269
(Refer Slide Time: 36:51)

Figure 5.
Figure 5 summarizes the simplified versions of the first law for a control volume in the cases
we discussed above.

(1) For an adiabatic throttling process and high pressure drop,

( ̇ ̇ )
(2) For turbines and rotary compressors,
̇ ( ̇ )
(3) For nozzles and diffusers,

( ̇ ̇ )

(4) For heat exchangers,

∑ ̇ ∑ ̇ ( ̇ ̇ )

270
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 55
Tutorial Problem (3 Numbers).

(Refer Slide Time: 00:21)

Figure 1.

271
Solution of the problem in Fig. 1:

̇ ̇

A schematic of the turbine is shown in Fig. 1.

272
The turbine is our control volume. The first law for a control volume,

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )…(1)

The process is adiabatic ( ̇ ). We assume that the process is steady ( ). Since we are

not given any information regarding the velocities and heights at inlet and outlets, we will
assume that the changes in kinetic and potential energy are negligible. Hence, equation (1)
implies,

̇ ̇ ̇ ̇ ….(2)

Now, ̇ ̇ ̇ ̇ ̇

We will obtain values of from steam tables.

At the inlet, at 350 , the saturation pressure =16 MPa. Since the steam is

in superheated zone. From the tables for superheated steam, .

At the outlet 2, at 250 , . Since , the steam is in superheated zone.

From the tables for superheated steam, .

At the outlet 3, [ ( )]

(2) implies ̇

273
(Refer Slide Time: 13:04)

Figure 2.

Solution of the problem in Fig. 2:

Conditions at the inlet and the outlet of the air compressor and the turbine are shown in Fig. 1.
The control volume includes both, the air compressor and the turbine.

The first law for a control volume

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )…(1)

274
Heat losses from the control volume and the changes in kinetic and potential energy can be
neglected. Assume steady state conditions. Hence, equation (1) reduces to

̇ ∑ ̇ ∑ ̇

For the compressor and the turbine, . The air and steam do not mix. Hence, ̇

̇ ̇ and ̇ ̇ ̇

Hence,

̇ ̇ ( ) ̇ …(2)

Now, at the inlet of turbine, at 10 MPa, . Since the state is in

superheated zone. From the tables for superheated steam, .

At the exit of turbine, [ ( )]

Assuming air an ideal gas and is constant, and .

( assuming air to be a diatomic gas)

(2) implies ̇

The turbine produces 23966 kW of power. The compressor consumes 2570 kW. 21395 kW is
available for the generator.

275
(Refer Slide Time: 22:02)

Figure 3.

276
Solution of the problem in Fig. 3:

Out control volume includes the condenser (heat exchanger). A schematic is drawn in Fig. 3. It
has two inlets (1 and 3) and two outlets (2 and 4).

Assume that the operation is at steady state conditions. The river water and the steam do not mix.

Hence, ̇ ̇ and ̇ ̇ .

The first law for a control volume,

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )…(1)

For a heat exchanger, ̇ ̇ . Since the operation is at steady state conditions, .

Also, the changes in the kinetic and potential energy are negligible. Hence,

∑ ̇ ∑ ̇ ̇ ̇ ̇ ̇

̇
Since, ̇ ̇ and ̇ ̇ , ̇ ̇ ̇ mass

flow rate of the cooling water.

277
Now, [ ( )]

Hence, ̇ = mass flow rate of the cooling water.

278
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 56
Tutorial Problems - Part 1

(Refer Slide Time: 00:13)

Figure 1.

279
Solution of the problem in Fig. 1:

A schematic of the turbine is drawn in Fig. 1. The turbine is a control volume here.

The first law for a control volume,

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )…(1)

Assuming steady state and adiabatic operation,

̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ ),

280
 ̇ ̇ ̇.

̇
Hence, mass flow rate of the steam = ̇ ̅ ̅
….(2)
( ) ( )

At the inlet, at 0.2 MPa, . Since, 400 > , the steam is in superheated zone.

From the tables for superheated steam, .

At the outlet, [ ( )]

(2) implies ̇
( ) ( )

̇
Now, ̇

| |

| |

̅ ̅ | |
| | | | . Hence,

| |
Similarly,

Hence, the major contribution to power is from enthalpy difference. The changes in kinetic and
potential energy are negligibly small.

281
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 57
Tutorial Problems - Part 2

(Refer Slide Time: 00:12)

Figure 1.

282
Solution of the problem in Fig. 1:

The tank and the turbine form a control volume as shown in Fig.1. Here, there is no inlet to the
control volume. There is only one outlet. It is an unsteady process. The control volume loses
mass with time.

The rate equation for the mass conservation is ̇ ̇ . Since there is no inlet,

̇ .

The first law for a control volume,

̇ ̇ ∑ ̇ ( ̅ ) ∑ ̇ ( ̅ )…(1)

Neglecting heat loss, KE and PE changes, setting ̇ ,

̇ ̇ ̇ ……(2)

Since there are no changes in kinetic and potential energy, E = U. Hence,

( )
̇ ̇ ( ) ,………….(3)

where is the temperature at the exit and air is assumed as an ideal gas.

283
We need to know the mass and the temperature of the control volume before and after the
process took place.

Initially, mass in the control volume before the process

took place.

Here, the air (ideal gas) undergoes an adiabatic process. Hence, ( ) . is the pressure

inside the control volume after the process took place and Substituting all the
values, temperature at the exit of the control volume. is lower than
as the air expands adiabatically. Now, we can calculate .

Now, we can integrate (3) with respect to time,

̇ [ ] [ ] ( ) ( ) ( and are constant)

Now, and .

Substituting all the values,

̇ .

284
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 58
Quasi-static Process Revisited: Work Against an External Force

Let’s revisit the concept of a quasi-static process.

(Refer Time Slide: 00:34)

Figure 1.

Consider an insulated piston-cylinder arrangement as shown in Fig. 1 in the top left corner.
The first law for such a system is as . The pressure outside is the
atmospheric pressure, patm, the pressure inside the arrangement is p, the mass of the piston is
mp and the area of the piston facing the gas inside the arrangement is Ap. If the system is in
equilibrium, the force balance gives,

……(1)

In such a case, the piston will not move. Hence, there is no work interaction. If we want to
move this piston, for example, in a downward direction, we need to apply external force on
the piston in the downward direction. This force must be slightly higher than the force ( )
given by the equation (1) so that the piston moves small distance . In this process, the
pressure p also increases. If we want to move the piston again by (compress more), we

285
need to increase the force slightly as the pressure p inside has increased. Hence, this external
force needs to keep increasing slowly if we want to compress more and more. In a similar
fashion, if we want to move the piston upwards, the external force needed keeps on
increasing because the pressure inside decreases as the piston moves up. Note that the process
is adiabatic and quasi-static in both the cases considered above.

Consider a similar piston-cylinder arrangement. Equation (1) is valid for this arrangement.
Now, move the piston down by external force and pin it to the new location. Here, the
pressure inside is greater than p from (1). Here, the extra force to hold the piston in its
position is provided by the pin. As soon as the pin is released, the piston shoots up. However,
this sudden upward movement of the piston is not slow. The process is not quasi-static. For
such a process, we cannot calculate ∫ .

Now assume that the upward movement of the piston is quasi-static or slow (after the release
of the pin). In such a process (movement of the piston from the moment the pin is released till
the achievement of the equilibrium), we can assume that the atmospheric pressure is constant
throughout the process because the piston is moving very slowly. If the piston were moving
very fast, the pressure close to the piston surface wouldn’t be atmospheric (it would be higher
than the atmospheric pressure). The mass of the piston is anyways constant. Here, we don’t
know the pressure inside, but we know the pressure outside (patm) and the force due to the
mass of the piston on the gas inside the system. Hence, work done in this process is

∫ ∫ ∫( ) . Here, dV is the change in volume of the

surroundings which also equals the change in volume of the system. For the system, dV is
positive as it is expanding, while dV is negative for the surroundings.

286
(Refer Time Slide: 11:20)

Figure 2.

We had looked at an example of a paratrooper jumping from a plane while discussing work
interaction. Initially, the paratrooper does not deploy the parachute (he/she falls freely).
Assuming negligible air resistance, the work interaction for the paratrooper is 0. Assume that
the weight of the paratrooper is 600 N. Now, the paratrooper deploys the parachute,
decelerates and falls through 10 m vertically. Suppose the air resistance during this fall is 200
N because of the parachute. The work interaction here is because of the air resistance, which
is . After the parachute is completely deployed and the person is no
more accelerating/decelerating, the air resistance is 600 N. Here, the work interaction is 600
N .

287
(Refer Time Slide: 13:51)

Similarly in the case of the piston-cylinder arrangement considered above, after the release of
the pin, the movement of the piston is resisted by the weight of the piston and the
atmospheric pressure. During the quasi-static process, the movement of the piston is fully-
resisted and it is resisted by the pressure .

(Refer Time Slide: 14:26)

Figure 2.

Consider a piston-cylinder arrangement system as shown in Fig. 2. One half is insulated. The
piston is also insulated. At equilibrium, Heat is now transferred slowly

288
to the section B. pB increases. The piston moves up resulting in the increase of p A. Here, the
piston movement is fully-resisted. We can calculate the work interaction for both A and B.
For A it is negative, whereas for B, it is positive. The work interactions are equal.

So, we can calculate work in quasi-static (near-equilibrium, slow) processes only. During the
process, the work interaction is only due to the resistance by the external force. In non-quasi-
static processes, we can’t measure the properties as the process is very fast and we have
multiple values of a single property inside or outside the system. Hence, we can’t calculate
the work interaction.

289
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 59
Second Law of Thermodynamics: Limitations of the First Law of Thermodynamics
(Refer Time Slide: 00:14)

(Refer Time Slide: 00:37)

Figure 1.

290
For systems shown in Fig. 1, the rate equation of the first law is ̇ ̇ If there are no

changes in kinetic and potential energy during the process, E = U.

(Refer Time Slide: 01:21)

Figure 2.

For a heat exchanger as shown in Fig. 2, after making some assumptions, which were discussed
in the previous lectures, the first law for a control volume reduces to,

∑ ̇ ∑ ̇

291
(Refer Time Slide: 02:56)

Figure 3.

For a nozzle or diffuser, the first law for a control volume reduces to,

̅ ̅̅̅
(see the previous lectures for assumptions made to reach this expression)

(Refer Time Slide: 03:42)

Figure 4.

292
Consider a closed rigid chamber (control volume) where we have air and fuel. There are no inlets
and outlets. The air and fuel burn together to form products (assume the temperatures of the
reactants and the products remain the same throughout the reaction). As the chamber is rigid,
work interaction is 0. If the chamber is insulated, then heat interaction is also 0. In such a case,

the first law for a control volume is , which implies E1=E2. So, the energy content does

not change.

(Refer Time Slide: 05:15)

Figure 5.
If the burning of a air-fuel mixture happens in an insulated piston-cylinder arrangement as shown
in Fig. 5 and the piston is free to move, the work interaction is not 0. The process is a constant
pressure process. Here, E = U. Hence, dU + pdV=0 (dU = and ). Hence, dH =
0, which implies H1=H2. During the process of burning, the temperature inside increases. Figure
5 shows the plot of enthalpy versus temperature for reactants and products. So, reactants and
products can have the same enthalpy provided that the products are at higher temperature.
Reactants can also form products at the same temperature if the heat transfer is allowed across
the boundary of a system or control volume.

293
(Refer Time Slide: 08:26)

Figure 6.

Figure 6 shows a dye mixing in water in a beaker. The mixing process is adiabatic. There is no
work interaction. For such a process, the first law implies E1=E2, i.e., the energy content does not
change during the process. So, the first law allows us to tell something about the final state if we
know something about the initial state.

(Refer Time Slide: 09:19)

294
The first law is a relation between heat and work interaction. Knowing initial state of a system or
a control volume, we can calculate it final state or vice versa. However, the first law does not say
anything about the direction of the process. It does not talk about the feasibility of the process.

(Refer Time Slide: 09:54)

In the example of mixing of the dye in water we considered above, the first law says E 1=E2, i.e.,
the energy of the control volume/system do not change in the process of mixing. The first law is
also valid for a process where the mixed dye in water separates out and agglomerates at one
location. Here also, E1=E2. But, we do not see that happening naturally.

295
(Refer Time Slide: 10:44)

In the example of reactants burning into products (air-fuel mixture burning into products) in a
piston-cylinder arrangement we considered above, the first law says H1=H2, i.e., enthalpy of the
reactants equals enthalpy of the products. The first law is also valid for a reaction where the
products of the combustion in the above case form reactants. However, it doesn’t happen
spontaneously. We need to do something for this to happen. The first law does not say anything
about the direction of a process.

(Refer Time Slide: 11:35)

Figure 7.

296
The combustor in Fig. 7 burns a mixture of air and fuel and forms exhaust gases. If we send in
the exhaust gases through the exit of the combustor, we do not get fuel and air. However, the
first law is still valid for such a process (i.e. sending exhaust gases through the exit of the
combustor and getting back fuel and air).

Hence, the analysis of feasibility and direction of a process is not given by the first law.

(Refer Time Slide: 12:11)

(Refer Time Slide: 12:27)

Figure 8.

297
We know that heat transfer occurs from a hot substance to a cold substance. For example, a cup
of hot coffee kept in an atmosphere whose temperature is lower than that of the coffee, cools
down and achieves the temperature of the atmosphere after some time, and stays at that
temperature. However, we do not see the reverse process happening, i.e., we do not see the
coffee, which is at the atmospheric temperature now, becoming hot and attaining its initial
temperature. However, the first law is valid for both the processes. It does not say anything about
the direction in which the process tends to happen naturally. The first law can be used to
calculate the heat transfer in the process where the hot coffee cools down from to
atmospheric temperature of as well as the process where the coffee at becomes hot
and attains naturally (work interactions are 0 for these processes).

We see spontaneous heat transfer happening from a hot substance to a cold substance, but not
from a cold substance to a hot substance. A spontaneous process happens only in one direction.
The first law is not enough to tell if a process will happen or not.

298
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 60
Second Law of Thermodynamics: Direct and Reverse Heat Engine
(Refer Time Slide: 00:13)

We discussed Joule’s experiment when we introduced the first law of thermodynamics. A falling
mass rotates a shaft having a fan at its one end. The fan churns fluid inside a rigid closed
chamber. The temperature of the fluid increases after the process. Here, the system consists of
the fluid inside the rigid chamber. The work is done on the system. Hence, the work interaction
is negative for the system. The system also loses heat and comes back to its original state. Thus,
heat interaction is also negative for the system. Here, both the interactions, work as well as heat,
are negative.

A system for which both the interactions, heat as well as work, are positive is called a direct heat
engine. This engine takes in heat and gives out work. A piston-cylinder assembly where a piston
is free to move can be a direct heat engine. We can heat the gas inside this system. The gas
expands and moves the piston doing work on the surroundings. Here, heat and work interactions
are positive.

A reverse heat engine transfers heat from a source at low temperature to a source at high
temperature using negative work interaction (work is done on the engine here). Refrigerators and

299
heat pumps are examples of a reverse heat engine. In refrigerators, heat is taken out of the region
which needs to be kept cold (insides of the refrigerator) and it is rejected to the surroundings
which is at higher temperature.

In a direct heat engine, we give in heat and get work out, whereas in a reverse heat engine, we
give in work and extract heat out.

(Refer Time Slide: 05:07)

Figure 1.

A power plant is a direct heat engine. Figure 1 shows components of a steam power plant. It runs
on a Rankine cycle which will be discussed in detail later in the course. There is boiler which
takes in high pressure water coming from the pump. The boiler is a heat exchanger where high
pressure water is heated by hot gases (which are generated by burning air and fuel) and
converted into high pressure superheated steam. This superheated steam enters turbine where it
expands and produces power. This power is used to run an electric generator to produce
electricity. The steam comes out of the turbine at low pressure and low temperature and enters a
condenser (which is also a heat exchanger). In the condenser, the steam is converted back to
liquid water. This water then, is sent to the pump which pressurizes it. The high pressure water
again enters the turbine and the cycle continues.

300
Here, the boiler takes in amount of heat. The turbine generates some amount of work. A
fraction of that work is used to run the pump. The remaining is the net work output of the
turbine, Wnet. In the condenser, as the steam converts back into water, some heat is given out
which is represented as QL. This heat equals latent heat of vaporization. So, we have heat input
as QH, heat output/rejection as QL (QH > QL), and net work output as Wnet. Hence, a steam power
plant is a direct heat engine.

(Refer Time Slide: 08:58)

Figure 2.

Figure 2 shows schematics representing a direct and reverse heat engines. In a direct heat engine
(e.g. a steam power plant), the boiler takes in heat Q1 at high temperature (T1). The turbine
produces work WT. The pump takes in work WP. The net work produced is W = WT – WP. The
condenser gives out heat Q2 at low temperature (T2). So, the direct heat engine takes in some
heat (Q1) from a high temperature source/reservoir at T1, gives out net work W, and rejects some
amount heat (Q2) to the low temperature reservoir at T2. Also, Q2 < Q1. The heat engine shown
here, runs in a cyclic process.

Similarly, we have a reverse heat engine where some amount of heat is absorbed from a low
temperature reservoir, work is done on the heat engine and some amount of heat (which is
greater than what is absorbed) is rejected to a high temperature reservoir. In Fig. 2, components

301
of a refrigerator are shown. The evaporator takes in Q2 amount of heat from a low temperature
reservoir at T2. Because of this, the refrigerant inside the evaporator vaporizes. The compressor
takes in WC amount of work to compress the refrigerant vapor to high pressure and temperature.
The condenser rejects Q1 amount of heat to a high temperature reservoir at T2. Because of this,
the refrigerant converts into liquid. The expansion device cools the refrigerant (no work
interaction is involved for the expansion device). Then, it enters the evaporator and the cycle
repeats. We will discuss this refrigeration cycle in more detail later. So, a reverse heat engine
takes in heat Q2 from a low temperature reservoir at T2, takes in some work W (for a
compressor), and rejects heat Q1 to a high temperature reservoir at T1.

The efficiency of a direct heat engine is defined as what we want or we are interested in divided
by what we need to give. For a direct heat engine, the efficiency (Fig. 2). The purpose

of a heat engine is to produce net work by taking in heat.

For a reverse heat engine, we have coefficient of performance (COP). Similar to the efficiency of
a heat engine, COP is defined as what we want or what we are interested in divided by what we
need to give. However, it depends on a device. For example, for heat pumps which are used in
cold countries to heat insides of homes, we are interested in the amount of heat given by the heat
pump (which is a reverse heat engine). For giving out some amount of heat, we need to give in
some work (Wnet or compressor work). Hence, for a heat pump, (Fig. 2). For a

refrigerator, we are interested in the amount heat it takes in (or removes) from the region to be
kept cold. For removing that heat, we need to give in some work (Wnet or compressor work).
Hence, for a refrigerator, (Fig. 2).

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(Refer Time Slide: 16:52)

Figure 3.
For a direct heat engine, which runs in a cyclic process, as shown in Fig. 3, the first law states
∮ ∮ . Hence, ( ) ( ) considering the sign conventions for heat
and work transfer. We can also write it as | | | | (we have taken out
the negative sign already).

(Refer Time Slide: 21:18)

Figure 4.

303
Similarly for a reverse heat engine which runs in a cyclic process as shown in Fig. 4, the first law
reduces to . Multiplying by -1, we get, . It can also be written
as | | | |.

(Refer Time Slide: 23:36)

Figure 5.

Figure 5 shows the schematics of a direct heat engine and a revere heat engine (running in a
| | | |
cycle). For a direct heat engine, , where and are the heat

rejected to the low temperature reservoir at TL and the heat absorbed from the high temperature
reservoir at TH, respectively. Here, according to the first law for a cyclic process,
| |.

| |
For a heat pump (which is a reverse heat engine), | |
, where QH is the heat

rejected to the high temperature reservoir at TH, QL is the heat absorbed from the low
temperature reservoir at TL, Win is the work input to the heat pump.

For a refrigerator (which is a revere heat engine), | |

.

304
(Refer Time Slide: 29:16)

The heat engine essentially converts heat into work. This work, then, can be converted into, for
example, electricity. The electricity can be used to do work, for example, run household
electrical appliances.

Why do we have to run the heat engine in a cycle?

Consider a piston-cylinder arrangement containing gas where the piston is free to move. If we
give in some amount of heat Q, the gas expands moving the piston and we get work W. If we
need a large quantity of work, we need to give in large amount of heat and the displacement of
the piston has to be large, because we know that the displacement work is given by ∫ . The
change in volume has to be large if we want a large quantity of work (assume that the process is
a constant pressure process). Giving in large amount of heat is possible. However, for the change
in volume to be large, the cylinder length has to be large. Such a long cylinder is difficult to
manufacture and expensive.

The solution for this problem is to run the engine in a cycle. Give in some amount of heat
initially. Let the piston move and extract some work W. Now, bring the piston back to initial
location (by extracting out heat maybe). Give in some heat again, let the piston move, extract
work and bring the piston back to its initial location. This is a cyclic process. This way we do not
need a long cylinder to extract a large quantity of work.

305
Hence, a direct and a reverse heat engine run in a cycle.

(Refer Time Slide: 33:08)

(Refer Time Slide: 33:23)

306
(Refer Time Slide: 33:30)

Knowing this we can on to the Kelvin-Planck and Clausius statements of the second law of
thermodynamics.

The Kelvin-Planck statement says it is impossible for a device operating in a thermodynamic

cycle to produce work while having heat interaction with a single reservoir at any temperature.
(Refer Time Slide: 33:58)

307
The Clausius statement says it is impossible for any device operating in a thermodynamic cycle
to transfer heat from a low temperature region to a high temperature region without the help of
work interaction from the surroundings.

(Refer Time Slide: 34:20)

We will look at these in detail in the next lecture.

Let’s look at the concept of a perpetual motion machine of the first kind. The first law tells us
that energy is conserved. It can neither be created nor destroyed, but you can transform it from
one form to another.

Consider a hypothetical machine which keeps generating energy out of nothing. Such a machine
would violate the first law, because the first law says energy cannot be created or destroyed. The
machine which violates the first law of thermodynamics is called a perpetual motion machine of
the first kind (PMM1). Such a machine can keep on producing energy forever.

308
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 61
Second Law of Thermodynamics: Kelvin-Planck and Clausius Statements

(Refer Time Slide: 00:13)

We looked at the concept of heat engines in the last lecture. We will look at the Kelvin-Planck
and Clausius statements of the second law of thermodynamics in the context of heat engines.

(Refer Time Slide: 01:13)

Figure 1.

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Figure 1 shows a direct heat engine operating between a high temperature reservoir at TH and a
low temperature reservoir at TL in a cycle. It takes in QH amount of heat from the reservoir at TH
and rejects QL amount of heat to the reservoir at TL, while producing net work W.

Kelvin-Planck statement says it is impossible for a device operating in a thermodynamic cycle to

produce work while having heat interaction with a single reservoir at any temperature.

When we looked at a direct heat engine running in a cycle, we saw that the engine interacts with
two heat reservoirs, a low temperature reservoir and a high temperature reservoir. It takes in heat
from a high temperature reservoir and rejects some heat to a low temperature reservoir while
producing some work. According to the Kelvin-Planck statement, it is impossible for such an
engine to produce work while interacting with a single reservoir at any temperature.

(Refer Time Slide: 02:41)

The engine violating the Kelvin-Planck statement is called PMM2, perpetual motion machine of

second kind. The efficiency of a heat engine . As QL is 0 for PMM2, . It

means . According to the Kelvin-Planck statement, QL cannot be 0. Hence, the

efficiency of a heat engine (running in a cycle) will always be less than one and .

The first law says you cannot get energy from nothing, but you can convert energy from one
form to another. The second law says that you can convert heat into work (one form of energy

310
into another) using a cyclic process, but the complete conversion is not possible. So, that is a
consequence of the second law.

(Refer Time Slide: 06:46)

Figure 2.

Another statement of the second law is the Clausius statement which is given in the context of a
heat pump or a refrigerator. A heat pump or a refrigerator transfers heat from a low temperature
reservoir to a high temperature reservoir by taking in some work.

The Clausius statement says it is impossible for any device operating in a thermodynamic cycle
to transfer heat from a low temperature region to a high temperature region without the help of
work interaction from the surroundings.

A heat pump or a refrigerator operating in a cycle cannot transfer heat from a low temperature
reservoir to a high temperature reservoir without any work interaction with the surroundings.
Figure 2 shows a reverse heat engine operating in a cycle. COP for a heat pump is

and COP for a refrigerator is . According to the Clausius statement, W cannot be 0.

Hence, the COP for a heat pump or a refrigerator cannot be infinite.

311
(Refer Time Slide: 10:04)

Figure 3.

We usually see a sticker with stars on an air conditioner or a refrigerator. These stars are linked
to the COP of the air conditioner or a refrigerator in some average sense. Figure 3 shows tables
where the COP values/EER values (EER is linked to COP) corresponding to different star ratings
are mentioned at different times. It seems that for the same star rating, the EER value is
improving with time. In 2011, for a rating of 1 star, EER value lies between 2.3 and 2.5, whereas
for the same rating in 2012 and 2013, EER value is higher. It means that we are improving the
quality of refrigerators.

312
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology Madras
Lecture 62
Second Law of Thermodynamics: Reversible Process
(Refer Time Slide: 00:13)

We looked at direct heat engines. If the processes are reversed, it can run as a reverse heat engine
(a heat pump or a refrigerator).

A reversible process is a process that once having taken place can be reversed and in doing so
leave no change in either system or surroundings.

A reversible process has the maximum efficiency of all the other processes.

Rapidly occurring processes (non-quasi-static processes) cannot be reversible. A reversible

process should be a quasi-static process. However, not all the quasi-static processes are
reversible. We saw that in a non-quasi-static process, a system may be partially resisted. Hence,
we may not get the maximum amount of work possible.

Factors like friction, unrestrained expansion, heat transfer through a finite temperature change,
mixing of two different substances and chemical reaction (combustion) make the process
irreversible.

313
If friction is involved in a process, it will be present during the process reversal also. Friction
generates heat. It is independent of the direction of the process. Hence, the process cannot be
reversed without affecting the system and surroundings.

Unrestrained expansion is a non-quasi-static process. It cannot be reversed without leaving a

trace on the system or the surroundings. For example, if a piston in a frictionless vertical piston-
cylinder assembly moves up suddenly, we need to work on the assembly to bring back the piston
to its original location. We cannot do this without affecting the system or the surroundings.

A process of heat transfer from a hot substance to a cold substance happens naturally. However,
transfer of heat from the cold substance to the hot substance can happen only when the work
interaction with the surroundings is involved according to the Clausius statement of the second
law of thermodynamics. So, the reverse process leaves a trace on the surroundings. Hence, it is
irreversible. However, isothermal heat transfer (as in phase change) is reversible. Heat transfer
through negligible temperature difference is reversible.

A chemical reaction converts reactants into products. To get back reactants from products, we
need to do something which may leave a trace on the system as well as surroundings. Hence, it is
irreversible.

Also, for the process to be reversible, it should be fully resisted (?).

(Refer Time Slide: 08:43)

314
We further classify these irreversibilities as internal and external.

A process is said to be internally reversible if no irreversibilities occur within the boundaries of

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

If a system follows a particular path during a process from state 1 to 2 on, for example, a p-v
diagram and it follows exactly the same path during the reverse process, the process is internally
reversible.

A process is called externally reversible if no irreversibilities occur outside the system

boundaries during the process. Heat transfer between a reservoir and a system is an externally
reversible process if the outer surface of the system is at the temperature of the reservoir. There
should not be any finite temperature gradient between those two.

(Refer Time Slide: 10:51)

Figure 1.

Figure 1 shows some systems undergoing a process. The images of a hot soldering iron and the
bullet being fired from a gun are taken using the technique called shlieren imaging.

315
The temperature of soldering iron is significantly higher than the surroundings. Hence, the
process of heat transfer is highly irreversible. Also, the soldering iron does not become hot just
by keeping it in hot gas.

Figure 1 shows the shockwaves being generated because of the gun firing the bullet. This process
is highly irreversible as (1) there is a presence of friction because of air viscosity, (2) combustion
is involved which is highly irreversible, (3) there is generation of sound and light. Also, there is
no way that the bullet can be put back into the gun, i.e., we cannot reverse the ‘firing a bullet’
process.

(Refer Time Slide: 12:15)

Figure 2.
Figure 2 shows a dye mixing in a liquid. If we stir it, there is no way to reverse the process, i.e.,
unmix the dye and the liquid. The process is highly irreversible.

316
 Thermodynamics
Professor Anand T N C
Department of Mechanical engineering
Indian Institute of Technology, Madras
Lecture 63
Second law of Thermodynamics: Carnot's cycle and theorems

We will now look at the Carnot's cycle. Sadi Carnot was an engineer in the USA. He worked in
the 1800s and came up with a thought experiment of what would be the most efficient engine or
a refrigerator.

(Refer Slide Time: 00:35)

Figure 1.

A system shown in Fig. 1 undergoes a cyclic process. It receives and rejects heat isothermally
(reversibly). Also, it receives work from the surroundings and does work on the surroundings,
reversibly. There are 4 reversible processes. Two of the processes involve heat transfer and the
other two processes involve work transfer. According to Carnot, such a system would have the

highest efficiency ( ) of all.

317
(Refer Slide Time: 01:50)

Figure 2.

Consider a piston-cylinder arrangement shown in Fig. 2. Add heat isothermally. The piston
moves and produces work. The state of the system changes from 1 to 2. This process is
isothermal heat addition or isothermal expansion. From the state 2 to 3, the piston moves (the
system expands) and produces work, adiabatically. The process from the state 2 to 3 is an
adiabatic reversible process (it is an isentropic process which we will study later). From the state
3 to 4, heat is rejected isothermally. The piston moves in the opposite direction (the system gets
compressed, i.e. work is done on the system). The process from 3 to 4 is isothermal heat
rejection or isothermal compression. In the process 4 to 1, the piston moves again compressing
the system. Hence, work is done on the system. The process 4-1 is reversible and adiabatic. All
the processes are reversible. Two of them are isothermal and the other two are adiabatic. These 4
processes form the Carnot’s cycle.

318
(Refer Slide Time: 07:17)

Figure 3.

Figure 3 shows a schematic of the engine running on the Carnot’t cycle. It takes in heat QH from
the high temperature source at TH and rejects heat QL to the low temperature source at TL. It
generates net work W. Figure 3 also shows a T-S (temperature-entropy) and a p-V diagram of
the Carnot’s cycle. We will look at the T-S diagram later on.

The process 1-2 is an isothermal heat addition/isothermal expansion (for an ideal gas, it would
follow pV = constant). For the process 1-2, the heat added is represented as 1Q2 and the work

319
done is represented as 1W2. The process 2-3 is reversible adiabatic expansion (for an ideal gas, it
would follow ). The work done for 2-3 is represented as 2W3. The heat
interaction for 2-3 is 0, i.e., 2Q3 = 0. The process 2-3 is steeper than the process 1-2 on the p-V
diagram. The process 3-4 is isothermal heat rejection/isothermal compression. The heat rejected
and the work done in the process 3-4 are represented as 3Q4 and 3W4. The process 4-1 is
reversible adiabatic compression. Heat and work interactions for the process 4-1 are represented
as 4Q1 and 4W1. For 4-1, 4Q1 = 0.

(Refer Slide Time: 12:11)

320
Carnot also came up with 2 theorems:

(1) No engine can be more efficient than a reversible engine (which is essentially the Carnot
engine) operating between the same high temperature and low temperature reservoirs.

(2) The efficiency of all reversible engines operating between the same constant temperature
reservoirs are the same irrespective of the working fluid. The efficiency of the reversible engine
depends only on the temperature of the heat source (high temperature reservoir) and the
temperature of the heat sink (low temperature reservoir).

According to the Carnot’s theorem, a reversible engine using an ideal gas and another reversible
engine using a non-ideal gas operating between the same high temperature and low temperature
reservoirs have the same efficiency. The efficiency of the reversible engine is independent of the
working fluid. Hence, it depends only on the temperature of a heat source and a heat sink.

(Refer Slide Time: 14:18)

Figure 4.

The violation of the Carnot’s theorems violates the second law of thermodynamics. It can be
proved in many ways. Let’s look at one of those ways.

321
Consider a reversible (R) and irreversible (I) heat engine between the same heat source (at TH)
and the heat sink (at TL) (see the top left corner of Fig. 4). Both the engines, the reversible as
well as the irreversible one, take in QH amount of heat from the heat source. The reversible
engine rejects heat QL,rev to the heat sink, whereas the irreversible engine rejects QL,irrev to the
same heat sink. The work done by the reversible heat engine is Wrev, whereas the work done by
the irreversible heat engine is Wirrev. Now, run the reversible heat engine in a reverse fashion so
that it acts as a heat pump or a refrigerator. It takes in QL,rev from the heat sink at TL and rejects
heat QH to the heat source at TH. It also takes in work Wrev (see the top right corner of Fig. 4).
We cannot do such process reversal with the irreversible engine (I).

Assume that the irreversible engine (I) is more efficient than the reversible engine (R) (this
assumption is in violation of the Carnot’s theorem). As the heat pump or a refrigerator R gives
out QH to the reservoir at TH and the irreversible heat engine I takes in heat QH from the same
reservoir, we can remove that reservoir from the analysis (see bottom left of Fig. 4). Since, W irrev
> Wrev (we assumed that the irreversible engine I is more efficient than the reversible engine R),
QL,irrev < QL,rev (according to the first law). From Wirrev, Wrev can be used to drive the heat pump
R. Now, we combine the heat pump R and the irreversible heat engine I into one system which
takes in heat QL,rev - QL,irrev from the reservoir at TL and does work Wirrev - Wrev. So, this system,
operating in a cycle, gives out work while taking heat from a single heat reservoir. This violates
the Kelvin-Planck statement of the second law which says that it is impossible for a system
running in a thermodynamic cycle to produce work with heat transfer from a single reservoir. It
means our assumption that the irreversible engine (I) is more efficient than the reversible engine
(R) (which violates the Carnot’s theorem) is not true as it violates the second law. Hence, the
violation of the Carnot’s theorem violates the second law.

The Carnot’s heat engine or the Carnot’s heat pump (or a refrigerator) are the most efficient heat
engine or a heat pump between the given heat source and heat sink. All the reversible heat
engines between the same two temperature reservoirs must have the same efficiency. It can also
be proved in the same way as we discussed above, by assuming that one of the engines have
more efficiency than the other.

322
 Thermodynamics
Professor Anand T N C
Department of Mechanical engineering
Indian Institute of Technology, Madras
Lecture 64
Second law of Thermodynamics: Absolute Temperature Scale

(Refer Slide Time: 00:14)

Figure 1.

323
The concept of the Carnot engine and the second law give us a concept of an absolute
temperature scale which is essentially a temperature scale which is independent of the working
substance.

Consider temperature reservoirs at temperatures T1 and T3 as shown in Fig. 1. We have a

Carnot’s engine C working between temperature reservoirs at T1 and T3 and producing work WC.
The engine takes in Q1 amount of heat and gives out Q3 amount of heat. Its efficiency is .

Consider two more reversible heat engines (Carnot’s engines) A and B (Fig. 1). The engine A
operates between the temperature reservoirs T1 and T2, whereas the engine B operates between
the temperature reservoirs at T2 and T3. The engine A takes in amount of heat and rejects
amount of heat. It also gives out amount of work. The engine B takes in amount of heat
and rejects amount of heat while giving out amount of work. The efficiency of engine A

is and that of B is . According to the Carnot’s theorem, the efficiency of a

reversible engine operating between two temperature reservoirs depends only on the temperature
of those reservoirs. Hence, ( ) ( ) ( ). Now,

( ) ( ) ( ) ( ) ( ). The term T2 cancels out in

the multiplication completely. Kelvin assumed the forms of these functions as ( )

( ) ( ) ( )
( ) ( ) so that they satisfy ( ) ( ) ( ).
( ) ( ) ( )

324
(Refer Slide Time: 07:50)

Figure 2.

This gives us the absolute temperature scale (Kelvin scale). We know that for a reversible heat
engine operating between a heat source at T1 and a heat sink at T2, the ratio of the heat received
(Q1) from the heat source and the heat rejected (Q2) to the heat sink equals the ratio of
temperatures, i.e., . For the Kelvin scale, the triple point of water is taken as the standard

reference point. For the Carnot’s engine operating between the temperature reservoirs at T and
the triple point ( ), . The triple point of water is arbitrarily taken as

325
 in the Kelvin scale. Hence, the plot of T (in Kelvin) versus is a straight line

passing through the origin because (see Fig. 2). Thus, any

temperature T can be determined by running the Carnot’s engine between the reservoir at that
temperature (T) and the reservoir at and measuring the heat transfers. The smallest
possible value of is 0, and the corresponding temperature is the absolute 0, i.e., T = 0 K. On
this scale, the melting point of water is 273.15 K and the boiling point of water is 373.15 K. The
conversion between the Celsius scale and the Kelvin scale takes place according to ( )
( ) . The Kelvin scale is independent of peculiar characteristics of any particular
substance. However, measuring temperature in this way is difficult as making Carnot’s engine is
difficult. Hence, practically, ideas gases are used to make a scale which is also fairly close to the
Kelvin scale, i.e., absolute temperature scale.

(Refer Slide Time: 13:48)

Figure 3.

All the processes satisfy the first law irrespective of whether the process is possible or not. The
first law can be used to calculate work or heat transfers involved in a process.

The coffee at , when kept in an atmosphere of , cools down to after some time,
spontaneously (i.e., on its own or naturally). The coffee cannot attain spontaneously. The

326
coffee can attain if some work is done on it according to the Clausius statement of the
second law of thermodynamics. The second law of thermodynamics provides unambiguous
criteria for the direction of spontaneous change, for example, hot coffee cools spontaneously and
spontaneous heating of coffee in cooler surroundings is impossible. The first law, on the other
hand, is applicable to the process of hot coffee getting cooled spontaneously and the spontaneous
heating of the coffee in cooler surroundings. The time, which flows in only one direction, is
linked to the second law.

327
 Thermodynamics
Professor Anand T N C
Department of Mechanical engineering
Indian Institute of Technology, Madras
Lecture 65
Tutorial Problems (2 numbers)

(Refer Slide Time: 00:28)

Figure 1.

Solution of the problem in Fig. 1:

A schematic of the engine is shown in Fig. 1.

Since the engine runs on the Carnot’s cycle,

328
(Refer Slide Time: 05:31)

Figure 2.

Solution of the problem in Fig. 2:

The first law for a heat engine, ∮ ∮ .

Hence, | |

| |

329
Out of all the engines operating between the given temperature sources, the Carnot’s engine or a
reversible engine has the maximum efficiency. For the Carnot’s engine,

330
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 66
Tutorial Problems (1 numbers)

(Refer Slide Time: 00:18)

Figure 1.

331
Solution of the problem in Fig. 1:

For the reversible heat engine,

For the reversible refrigerator,

The schematic of the entire of system is drawn in Fig. 1.

Since the heat engine is reversible,

According to the first law for the heat engine,

| | = net work output of the heat engine alone

(positive for the engine).

Out of this net work output , some is used to run the refrigerator. The net work output of the
combined heat engine-refrigerator system is 350 kJ.

Hence, the work input to the refrigerator is (this is negative for the
refrigerator).

Now, we know the work input to the refrigerator. According to the first law for the reversible
refrigerator,

| | | | ….(1)

Since the refrigerator is reversible,

Now, equation (1) implies and

332
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 67
Tutorial problems (1 numbers)

(Refer Slide Time: 00:13)

Figure 1.
Solution of the problem in Fig. 1:

Constant volume tank A,

Constant pressure device B,

Air is assumed to be an ideal gas.

Here, the temperature of the temperature reservoirs is changing, unlike the problems we saw
previously, because the reservoirs are finite in size.

The heat engine operating between the temperature reservoirs will give maximum work if the
engine runs reversibly.

333
The engine will stop working when the temperature of both the reservoirs become equal, i.e., the
final temperature Tf of both the reservoirs is the same.

We can apply the first law to the constant volume tank A and the constant pressure device B
separately. The tank A and the constant pressure device B are closed systems.

The first law for the tank A, (there are no changes in the kinetic and potential
energy of the system). As the tank is rigid, the process is a constant volume process for tank A.
Hence, . Thus, . is negative for the tank A.

The piston in the constant-pressure device moves as the device receives heat from the engine.
The first law for the constant pressure device B,
. is positive for constant pressure device B.

Now, apply the first law to the heat engine.

∮ ∮ | | | | (considering the sign

convention)

Integrating,

|∫ | |∫ | | ( )| | |…..(1) ( and
are assumed constant. is positive for the engine whereas is negative)

We are asked to find the maximum work. Hence, the heat engine must be reversible.

Hence, (The temperature of the tank A goes down.

Hence, dTA would be negative. To make the numerator positive on the left hand side, minus sign
is added.)

Integrating,

(i – initial, f - final)

334
Substituting the values,

( and assuming

and (diatomic gas))

Hence, Tf = 609 K.

Equation (1) implies,

(Refer Slide Time: 11:02)

335
336
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 68
Tutorial problems (2 numbers)

(Refer Slide Time: 00:13)

Figure 1.
Solution of the problem in Fig. 1:

The schematic showing the engines and the temperature reservoirs is drawn in Fig. 1.

In any cycle, heat rejected by the Carnot’s engine 1 equals the heat received by the Carnot’s
engine 2.

Since these are Carnot’s engines,

( and are the magnitudes of the heat transfers)

Similarly,

337
(Refer Slide Time: 04:54)

Figure 2.
Solution of the problem in Fig. 2:

We are going to assume that the temperature of this refrigerator is the temperature of the
watermelons. We are cooling only the watermelons and not the insides of the refrigerator or there
is nothing else in the refrigerator other than the watermelons.

The schematic of the refrigerator is drawn in Fig. 2.

The magnitude of heat needed to be removed from the watermelons,

( )

Now,

Now, ̇ ̇

338
If the fridge was at the temperature of the watermelons initially, it is going to take longer than
the time calculated above because there is some air around the watermelons which is also getting
cooled along with the watermelons. Also, there are always heat losses as the insulation of the
fridge is not perfect/ideal which also leads to consumption of more time for cooling.

However, if the fridge is already very cold, it might take less time to cool watermelons than what
we calculated.

339
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 69
Second Law of Thermodynamics: Clausius’s Inequality

Let’s look at the Clausius’s Inequality.

(Refer Slide Time: 00:27)

Figure 1.
For a Carnot’s engine running between temperature reservoirs at TH and TL, the ratio of heat
taken in (QH) from the reservoir at TH and the heat rejected (QL) to the reservoir at TL equals the
ratio of the temperatures, i.e., , which can also be written as (Fig. 1). Here, we

are taking absolute values of QH and QL.

According to Clausius, ∮ for a cyclic process. Let’s look at the proof of this inequality.

Let’s expand ∮ for a Carnot’s cycle shown on a p-v diagram in Fig. 1. A Carnot’s cycle

contains 4 reversible processes, two of which are isothermal and the other two are adiabatic. For
adiabatic processes, is 0. Hence, there are heat transfers (QH and QL) in only two of the
processes. The Carnot’s heat engine takes in QH amount of heat and rejects QL amount heat. For

340
the Carnot’s cycle, we can write, ∮ ∑ . For a Carnot’s engine (reversible

engine), according to the first law, work done | |. Hence, | | . Also, for

a Carnot’s engine (reversible engine), and are equal in magnitudes but opposite in sign.

is positive as is positive and is negative as is negative (TH and TL are always positive as

they are taken in Kelvin scale). Hence, ∮ ∑ for a reversible heat engine.

This expression holds true for a reversible heat pump as well as a reversible refrigerator.

For a reversible heat pump or a reversible refrigerator taking in heat from the temperature
reservoir at and rejecting heat to the temperature reservoir at while taking in some

work W, is negative (as is negative) and is positive (as QL is positive). However, their

magnitudes are equal ( ). Hence, for a reversible heat pump or a reversible refrigerator

also, ∮ ∑ . Also, | | | | for a reversible heat pump or a refrigerator

as, according to the first law, | | | | | |. The p-v diagram for a reversible heat pump or
a reversible refrigerator is also shown in Fig. 1.

In essence, ∮ for a cyclic process involving reversible processes is 0, i.e., ∮ .

341
(Refer Slide Time: 12:13)

Figure 2.
What if a cyclic process involves irreversible processes?

We know that, according to the Carnot’s theorem, of all the engines operating between the given
two temperature reservoirs, reversible engines have the highest efficiency.

Consider a reversible heat engine and an irreversible heat engine operating between temperature
reservoirs at TH and TL (see Fig. 2). Both the engines take in QH amount of heat from the
reservoir at TH. As the reversible heat engine is more efficient than the irreversible engine, the
work output of the reversible heat engine is more than that of the irreversible heat engine, i.e.,
. Hence, according to the first law, the heat rejected by the reversible heat engine
to the reservoir at TL is less than that rejected by the irreversible engine, i.e., .

Now, for the reversible engine, ∮ ( is positive and is negative, and

they are equal in magnitude). For the irreversible heat engine, we have ∮ . We

know that (considering magnitudes). We also know that (considering

magnitudes) as . Hence, for the irreversible heat engine, taking magnitudes,

, and is positive and is negative. Hence, for the irreversible heat engine,

342
We get a similar expression for an irreversible heat pump and irreversible refrigerator.

(Refer Slide Time: 17:44)

So, ∮ for a cyclic process involving reversible processes and ∮ for a cyclic process

involving irreversible processes. Combining these two, we get the Clausius’s inequality,

∮ for a process.

343
 Fa
Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 70
Entropy Part 1

(Refer Slide Time: 00:14)

Figure 1.

Figure 2.

344
We know that ∮ for a reversible cyclic process.

Two states, states 1 and 2, of a certain system are shown on a p-v diagram in Fig. 1. The system
can change the state by three reversible paths, A, B or C.

For the reversible cycle 1-A-2-B-1, ∮ ∫ ∫ ….(1)

Similarly, for the reversible cycle 1-A-2-C-1, ∮ ∫ ∫ ….(2)

Subtracting (2) from (1),

∫ ∫ .

This is true for any reversible path between the states 1 and 2, i.e., ∫ is the same for any

reversible path between the same initial and final state. In other words, ∫ is path independent

for a reversible process. We know that for a fixed initial and final state, the difference in a
property values at the initial and final state is independent of the path taken to reach from the
initial state to final state. Also, the cyclic integral of a differential change in a property of a

system undergoing a cyclic process is 0. Hence, must be representing a change in some

property of a system undergoing a reversible process. We assign letter S to this property and

write (‘rev’ is written to indicate that the process is reversible). After integrating,

∫ . Here, δQ is a path function, T is a state function or a property, but represents a

change in a property for a reversible process. We call this property as entropy (S).

In mathematics, inexact differentials are converted to exact differentials by multiplying with

integrating factors. Here, δQ is an inexact differential. By multiplying δQ with the integrating
factor 1/T, we get dS which is an exact differential.

From the first law of thermodynamics, we get the property called energy, while from the second
law, we get entropy. Entropy is defined in a different way in statistical thermodynamics. In
classical thermodynamics, it is the ratio of heat transferred to the temperature at which the heat
was transferred in a reversible process.

345
What about an irreversible process?

An irreversible process cannot be plotted on any property diagram (e.g. p-v, p-T, etc.) because
we don’t know the properties of a system during this process. We cannot measure the heat

transferred and the temperature for the system during such a process. Hence, we cannot find ∫

for an irreversible process.

Figure 2 shows states 1 and 2 of a certain system on a p-v diagram. The path A from the state 1
to 2 is reversible. It is shown with a solid line. The path B joining states 1 and 2 is irreversible. It
is shown with a dotted line. Even though it is shown with a dotted line, it does not mean that
property values (in this case, p and v) of the system during that process lie on that dotted line.
We just don’t know the system properties during the irreversible process. The dotted line could
have been drawn to take any arbitrary path. Also, we cannot calculate the area below the dotted
line.

For the reversible path A, ∫ difference in the entropy values at the state 2

and the state 1 ( may have a positive or a negative value depending on the values at the state 1
and 2, and the direction the path is traversed, i.e., whether the system goes from state 1 to 2 or
from 2 to 1). The magnitude of is the same for the given initial and final states irrespective of
the path taken. Hence, for the irreversible path B, . However, for the

irreversible path B, ∫ , because we cannot measure the heat transfer and the

temperature for an irreversible process. Hence, for calculating for any irreversible

process between the given states, we should calculate ∫ along any reversible path between the

same states to get which also equals . But ∫ .

346
(Refer Slide Time: 12:13)

The unit of entropy (S) is J/K (joule/kelvin). The unit of specific entropy (s) is .

∫ where m is the mass of the system.

In the context of this, it is also useful to look at the third law of thermodynamics. Again, there
are various statements of this law to take into account some very specific things. However, for
the purpose of this course, we have the following statement: for pure substances, if we reach a
temperature of absolute 0, the entropy of the substance would be 0. This gives some reference to
calculate entropy.

347
Let’s see how to calculate changes in entropy for various substances. Consider simple
compressible systems. The first law, by not considering changes in kinetic and potential energy,
for a process is, . There is only pdV type of work. Hence, .

Now, for a reversible process, . Hence, Dividing by mass of the

system throughout, we get, . We also know that h = u + pv. Hence,

. Substituting for du, . and are
known as Gibb’s relations. To calculate change in entropy even for an irreversible process, we
can use Gibb’s relations as these are property relations (?). Properties are independent of path.
Change in a property depends only on the initial and the final state.

Let’s use these relations to calculate entropy change for a solid or a liquid.

For a solid or a liquid, specific volume v is very small and it can be ignored. Changes in specific
volume are even smaller during a process. We have . Setting dv=0, Tds = du.

Hence, (C is specific heat). Therefore, . Integrating,

( ) ( – final temperature, - initial temperature). However, for a process of phase

change, where temperature does not change, T = latent heat of vaporization.

348
(Refer Slide Time: 22:51)

We can use Gibb’s relations to calculate the entropy change for an ideal gas. We have Tds =

du+pdv. Now, . We know, for an ideal gas,

. Hence, . Integrating, ∫ ∫

( ) ( )( is assumed constant and written as . If it is not constant, it cannot

be taken out of the integral).

349
We also have (for an

ideal gas, ). Integrating, ∫ ∫ ( ) ( )( is

assumed constant and written as . If it is not constant, it cannot be taken out of the integral.).

We have been showing processes undergone by a system by drawing curves on a p-v diagram for
quite some time now. We know that small displacement work is represented as (on a
unit mass basis). The term pdv represents the area under the infinitesimally small section of the
curve representing a process on a p-v diagram with respect to v axis. Integrating, we get
∫ ∫ area with respect to v axis under the entire curve representing the process on a
p-v diagram = amount of work done/received by the system. For calculating this area, the
process must be quasi-static.

We know that, for a reversible process, (on a unit mass basis). Hence, (this

function looks similar to ). Hence, in line with discussion above, ∫

∫ area with respect to s axis under the entire curve representing the reversible process on
a T-s diagram = amount of heat transfer to or from the system.

350
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 71
Tutorial Problem (1 number)

(Refer Slide Time: 00:14)

Figure 1.

Solution of the problem in Fig. 1:

Lake is very big. Hence, the lake water’s mass and temperature do not change because of the
Aluminium block. The block attains equilibrium with the lake water and attains lake water’s
temperature.

We have, Tds = du + pdv

For solids and liquids, . Hence, TdsAl = du = CdT .

Integrating, = change in entropy of the Aluminium block.

is negative as the Aluminium block is losing heat. The block’s entropy is decreasing.

351
The process is isothermal for the lake water. Hence, it is reversible.

For the lake water, . Integrating, .

Now, heat gained by the lake water = heat lost by the aluminium block = ( )
.

Heat gained by the water is + 6 MJ. Hence, .

Total entropy change during the process = entropy change of the universe = entropy change for
the aluminium block + entropy change for the lake water = .

The entropy changes for the lake water and the aluminium block are not equal. The entropy is
not conserved. We will soon see the principle of increase of entropy in this context. The only
way to reduce the entropy of a system is to transfer heat out of the system. For an isolated
system, the entropy cannot reduce. Hence, the entropy of the universe cannot decrease as it is the
isolated system. If all the processes are reversible, then the change in entropy of the system is 0.

(Refer Slide Time: 06:36)

352
353
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 72
Entropy Part 2

We will see how to find out entropy values of steam in a steam table.

(Refer Slide Time: 00:41)

Figure 1.

354
We know how to find enthalpy, specific volume and internal energy of steam-water mixture and
superheated steam using steam tables. However, we have not yet read values of entropy from the
steam tables. Figure 1 shows a snippet of steam table. The last three columns list entropy of the
saturated liquid (sf), entropy difference between saturated liquid and saturated vapor (sfg) and
entropy of the saturated vapor (sg). For example, at 0.1 MPa, the entropy value of the saturated

liquid sf = 1.3028 , the entropy value of the saturated vapor sg = 7.3588 and the

difference between sf and sg, i.e., sfg = 6.056 . Students are encouraged to find values of

entropy at different conditions for the sake of practice. Also, the entropy of a steam-water
mixture can be found out in the same usual way as we did for the other properties.
( ) where x is the quality of the steam-water mixture. All the
properties are calculated with respect to some datum values which can be found in a steam table.
Some steam tables consider triple point of water as the datum point. While finding out properties
from the steam tables, students should use only one steam table as the datum points may be
different for different tables. The values of the properties could be different at the same pressure
and temperature in different tables.

(Refer Slide Time: 05:15)

Figure 2.

355
We have come across processes for a system where some property remains constant during the
process, e.g., isobaric process (p=constant), isochoric process (v=constant), isothermal process
(T=constant), isenthalpic process (h=constant). For adiabatic process, heat transfer Q=0.

Let’s look at an isentropic process for a system. In this process, s = constant or . We

know that, for a reversible process, or . For the process to be

isentropic, ds should be 0, which leads to . Hence, the isentropic process is reversible and
adiabatic.

Let’s recap how different processes for a system look on a p-v diagram.

An isobaric process (a constant pressure process, p = c) is represented by a line parallel to the v

axis. Similarly, an isochoric process (a constant volume process, v = c) is represented by a line
parallel to the p axis. An adiabatic process for an ideal gas ( ) is represented by a curve
which is steeper than the curve representing an isothermal process for an ideal gas ( ) as
shown in Fig. 2. If the adiabatic process is also reversible, then the same curve ( )
represents isentropic process (s = c) for an ideal gas. For two-phase mixtures, isobars and
isotherms are lines parallel to the v axis inside liquid-vapor dome. Hence, the nature of the curve
representing a process on a p-v diagram depends on the nature of the substance.

Let’s see how different processes look on a T-s diagram.

A line parallel to the s axis represents an isothermal process on a T-s diagram. Similarly, a line
parallel to the T axis represents an isentropic process on T-s diagram. For two-phase mixtures,
isobars and isotherms are lines parallel to the s axis inside a liquid-vapor dome. For an ideal gas,
we have a relation, . For a constant pressure process, dp=0. Hence,

. Hence, the slope of a curve representing a constant pressure process on

a T-s diagram is . For a perfect gas, as Cp is constant, the slope increases with increase in

temperature. A curve representing a constant pressure process for a perfect gas at low
temperatures is less steeper than the curve representing a constant pressure process for a perfect

gas at higher temperatures as the slope .

356
For an ideal gas, we also have . For a constant volume process, dv=0. Hence,

. Hence, the slope of a curve representing a constant volume

process on a T-s diagram is . For a perfect gas, as Cv is constant, the slope increases with

increase in temperature. A curve representing a constant volume process for a perfect gas at low
temperatures less steeper than the curve representing a constant volume process for a perfect gas

at higher temperatures as the slope .

For an ideal gas (and perfect gas), . Hence, if a constant volume and a

constant pressure process starts at the same point on a T-s diagram, the curve representing the
constant volume process would be steeper compared to that representing the constant pressure
process (see Fig. 2).

357
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 73
Entropy Part 3

(Refer Slide Time: 00:13)

Figure 1.

Figure 2.

358
Let’s define isentropic efficiency for different engineering devices.

For a turbine, after simplifications, the first law for a control volume, as discussed in previous
lectures, reduces to, ̇ ̇( ) . In addition to assuming the gas
expansion process in a turbine adiabatic, if we assume that the process is also reversible, then the
change in entropy (dS) for a gas would be zero. The process would be isentropic. The isentropic
process (1-2s) for a gas expanding from the initial pressure pi to the final pressure pe in a turbine
is shown by a line parallel to the T axis on a T-S diagram in Fig. 2. However, if the process is not
reversible, the entropy would increase during the process. The process takes place between the
same two pressures pi and pe. The temperature at the exit for an irreversible process would be
higher than the temperature at the exit for the reversible process. The irreversible process is
shown by dotted line joining the states 1 and 2 on a T-S diagram in Fig. 2. For the reversible
process, , whereas for the irreversible process, . Real life processes
are irreversible. Hence, represents the actual work output of the turbine.
represents the maximum possible work output from a turbine. Isentropic efficiency for a
( )
turbine is defined as . Though we mentioned that
( )

for an irreversible process, the entropy increases, we have not yet proved it. Let’s look at the
proof.

(Refer Slide Time: 07:54)

Figure 3.

359
According to Clausius, for a reversible cycle, ∮ , and for an irreversible cycle, ∮ .

Consider a system undergoing a cyclic process. States 1 and 2 of the system are shown on some
property diagram (say p-v diagram) in Fig. 3. Paths A and B between states 1 and 2 are

reversible, whereas the path C is irreversible. For the path 1-A-2-B-1, ∮ ∮ ∫

∫ . For the path 1-A-2-C-1, ∮ ∫ ∫ . For a cyclic process, ∮ as

S is a property. However, if the cycle is irreversible as 1-A-2-C-1, ∮ . Hence, ∮ ∮ .

So, for an irreversible cycle, entropy change over a cycle is not 0. It is greater than 0, i.e.,

entropy increases in an irreversible cycle. We can also write if the process is

irreversible. For the reversible process, . So, we can write, in general for a

process. Hence, for an irreversible process, the entropy increases.

It can be proved in another way. For the path 1-A-2-C-1, ∮ ∫ ∫ .∫

as the path A is reversible. Hence, ∫ . Multiplying by -1, ,

∫ . If the path C were reversible, ∫ ∫ ∫ .

Hence, for the same states 1 and 2, the difference in entropy between those two states for a

reverible process equals∫ . However, the difference in entropy between those states for an

irreversible process is greater than ∫ . It means that the entropy difference between the same

two states for an irreversible process is larger than the entropy difference between the same two

states for a reversible process. In genral, for a process, .

360
For an isolated system, there is no heat transfer ( ). Hence, . The universe is an
isolated system. Hence, for the universe, , i.e., the entropy can remain constant or
increase. This is the principle of increase of entropy. If the processes are reversible, then dS=0. If
the processes are irreversible, which is the case in reality, dS > 0. Here, we see that there is a
directionality associated with a process, i.e., a process happens such that the entropy increases or
remains constant, but it doesn’t decrease. This is unlike other physical laws which cannot
determine anything regarding the direction or the feasibility of the process. For example, we can
use the first law for calculating the heat transfer for hot coffee getting cooled in a cold
environment. We can also calculate the heat transfer if the hot coffee gets hotter in the cold
environment spontaneously. However, this process does not happen as it violates the principle of
increase of entropy. Hence, the processes always happen in a direction where the entropy
remains constant or increase. The principle of increase of entropy is a consequence of the second
law.

If we transfer heat into the system reversibly ( ), its entropy increases. If the process
undergone by a system is adiabatic but irreversible, then also, the entropy of the system increases
(as for an irreversible process). For a closed system (fixed mass), the only way of
reducing the entropy is to remove heat. However, for an open system, where there could be a
transfer of mass also across its boundaries, the entropy can also be reduced by taking mass out of
the system. However, the specific entropy does not change by mass transfer.

361
(Refer Slide Time: 21:59)

Figure 4.

Figure 5
Let’s define isentropic efficiencies for engineering devices.

For a compressor, after simplifications, the first law for a control volume gives a similar
expression as that for a turbine, . However, here, work is done on the compressor
( ). The adiabatic reversible (isentropic) compression process 1-2s is shown by a line
parallel to the T axis on a T-S diagram (Fig. 4) between the inlet pressure p1 and the outlet

362
pressure p2. represents the minimum work input that the compressor needs if the
process is isentropic. However, as mentioned before, in reality, the processes are irreversible and
the entropy increases. Irreversible compression process is shown by a dotted line 1-2 between the
same pressures p1 and p2 (Fig. 4) on a T-S diagram. The temperature and the entropy at state 2 is
larger than the temperature and the entropy at 2s. The isentropic efficiency of a compressor is

defined as . Hence, for an irreversible compression, we need to

give in more work than the reversible compression. For an ideal gas, ( )
and ( ).

Similarly, for a nozzle, we have (see a lecture on the first for a control volume

for the derivation of this expression). In a nozzle, the gas expands from some initial pressure pi to
pe. Reversible adiabatic (isentropic) expansion of a gas in a nozzle from pressure pi to pe is
shown by a line 1-2s parallel to T axis on a T-S diagram in Fig. 5. In the irreversible expansion,
the entropy increases. The irreversible expansion in a nozzle between pressure pi and pe is shown
by a dotted line joining 1 and 2 on a T-S diagram in Fig. 5. The isentropic efficiency of a nozzle

is defined as . Sometimes, the inlet kinetic energy is very small

compared to that at the exit. Then . Again, for an ideal gas, (

) and ( ).

For a pump, the expression for the isentropic efficiency is the same as the compressor. For a
pump compressing a liquid . This is the work input. We have . If
the pump works isentropically, ds=0. Hence, . As the work input to a pump is the
difference between enthalpies at the inlet and outlet, ( ) Total work
input to the pump can be obtained by integrating, ∫ ( ), where pe and pi
are the pressure values at the exit and inlet of the pump. As liquids are considered
incompressible, the specific volume v can be taken out of the integral for calculating work input
to the pump. However, we cannot do this for a compressor as the gases are not incompressible. v
changes with pressure. In that case, we need to know the variation of v with pressure to calculate
∫ .

363
We know that the displacement work for a system like a piston-cylinder arrangement is given by
∫ . For an isentropic flow device such as a pump, ∫ .

364
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 74
Entropy Part 4

Let’s look at a few other concepts related to entropy.

(Refer Slide Time: 00:30)

We know that, for a process, . The equality is for a reversible process, whereas the

inequality is for an irreversible process. The inequality can be converted to equality as

, where is the entropy generated in the irreversible process. for a

reversible process. For an irreversible process, . Hence, for a system

going from state 1 to 2, heat transferred in an irreversible process is less than the heat transferred
in the reversible process by , i.e., . Applying the first law to
an irreversible process, . Similarly, . Hence,
or , where
is termed as a lost work. Hence, in an irreversible process between the two states of a system, we
get less work than the reversible process between the same two states by . Hence,
is called as lost work.

365
(Refer Slide Time: 07:28)

Figure 1.

For a Carnot’s engine operating between a temperature reservoir at T (heat source) and T0 (heat
sink) and giving W as work output, the efficiency of the engine is given as ,

where QL and QH are the heat transfers with the temperature reservoirs at T0 and T, respectively.

A Carnot’s cycle is shown on a T-S diagram in Fig. 1. We already know that, on a T-S diagram,
the area under a curve represents the heat transfer for that process. The Carnot’s cycle involves
two heat transfers, QL and QH. For the Carnot’s cycle shown in Fig. 1, and
. (in magnitude). Hence, | | .
According to the first law, ∮ | | ∮ . Hence, .
This is the maximum work output for a reversible heat engine operating between the temperature
reservoirs at T and T0. For a given T, reducing T0 increases the work output. Hence, there is
always an amount of heat equal to which cannot be converted to work. The total energy
(heat) available for conversion to work is , which is termed as available energy. Out of that,
the energy (heat) which cannot be converted into work is , which is termed as unavailable
energy. Usually, T0 represents the temperature of the ambient. In cold countries, ambient
temperature is lower than hotter countries. Hence, the unavailable energy is lower. Here, T and

366
T0 are taken as constants. What if we have a constant pressure heat source with varying
temperature and heat sink at constant temperature T0?

The variation of T with S is shown in Fig. 1 on a T-S diagram (right top corner). If the variation
of T with S is known, we can operate a series of Carnot’s engines between the heat source with
varying temperature and heat sink at T0. For all the engines, heat sink temperature remains at T0.
However, for every subsequent engine, the temperature of the heat source increases as shown on
the T-S diagram in Fig. 1. In this case, the total available energy for the conversion into work is
∫ and the unavailable energy is . Hence, the available energy for conversion into work
is ∫ .

We know that we get the maximum work output in a reversible process. However, in reality, the
processes are irreversible. We have already seen that . The extent of
irreversibility I is defined as .

(Refer Slide Time: 16:58)

Figure 3.
Consider a flow device with a single inlet and outlet. The device is operating at steady state
conditions (dE/dt = 0, dm/dt=0). Hence, the mass flow rates at the inlet and the outlet are equal,
i.e., ̇ ̇ ̇ . The device is adiabatic ( ̇ ) and there is no work transfer ( ̇ ).
Hence, the first law for a control volume in the case the device considered above reduces to,

367
 ….(1)

We know that Tds = dh-vdp. Assuming the process to be isentropic, dh = vdp. Integrating,

∫ ∫ . Assuming the substance incompressible, ( ). For the flow device

considered above, . Hence, equation (1) implies,

. This is also known as Bernoulli’s equation. This equation can be used for devices

such as nozzles, diffusers, pipes, etc. to get property values. However, keep in mind the
assumptions we made to arrive at this equation.

368
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 75
Tutorial Problem (1 number)

(Refer Slide Time: 00:30)

Figure 1.
We know that an isothermal heat addition is a reversible process as the temperature difference
across which the heat is being transferred is very small (assuming other sources of
irreversibilities are absent). Isothermal process is represented by a line parallel to the S axis on a
T-S diagram as shown in Fig. 1 (process 1-2 and process 2-1). Similarly, an isentropic process
(reversible and adiabatic) is represented by a line parallel to the T axis on a T-S diagram as
shown in Fig. 1 (process 1-3 and process 3-1). We have already discussed the sources of
irreversibilities such as friction, a finite temperature difference, unrestrained expansion, etc.

How do we achieve and represent a reversible constant pressure heat addition process on a T-S
diagram?

A curve for a constant pressure process which starts at state 1 is shown in Fig. 1. It is not parallel
to either T axis or S axis. One way to achieve this process is to have a device/temperature
reservoir outside whose temperature is changing with the system’s temperature such that the

369
temperature difference is extremely small leading to isothermal heat transfer, which is also
reversible.

To represent this process on a T-S diagram, draw a large number of isotherms and isentropes and
connect them in a fashion shown by a blue zigzag line as shown on a T-S diagram in Fig. 1.
During the isothermal part of the process, we have reversible heat addition. During the isentropic
part of the process, we do not have any heat addition, but the process reversible. Hence,
combining such isotherms and isentropes, we can obtain a reversible constant pressure heat
addition process approximately. The closer the isotherms and isentropes to each other, the better
is the approximation. In a similar way, we can draw a reversible constant volume process. We
can draw isotherms on a p-V diagram using the similar method.

(Refer Slide Time: 06:02)

Figure 1.

370
Solution of the problem in Fig. 1:

For hydrogen,

H2 is contained in a closed rigid tank. Hence, there is no volume change during the process.

A schematic of the reservoir at 300 and the tank of H2 is shown in Fig. 1.

For hydrogen, (as dv=0)

Hence, . Integrating, ( ) ( )….(1)

We need to calculate Cv and the mass m for hydrogen.

⁄
( and for a diatomic gas like hydrogen is 1.4)

Assuming hydrogen to be an ideal gas, ( ⁄ )

Equation (1) implies ( ) .

371
For the reservoir, the heat transfer is reversible as it happens isothermally. The boundary of the
system through which the heat is entering is also at 300 , i.e., at the reservoir temperature.

For a reversible process, . T = 400 We need to calculate Qrev.

The heat gained by hydrogen equals the heat lost by the reservoir.

Writing the first law for the process undergone by hydrogen,

(There is no change in kinetic and potential energy of the system, hence dE=dU. is 0 as it is
a constant volume process.)

Now, . Integrating, ( )

This heat transfer is negative for the reservoir as heat is leaving it.

Hence,

Now,

The process for the reservoir is reversible. However, the process for hydrogen is irreversible.
Hence, is positive. The entropy of the universe (isolated system) increases because of
the irreversible process.

372
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 76
Tutorial Problem (1 number)

(Refer Slide Time 00:14)

Figure 1.

373
Solution of the problem in Fig. 1:

We are not given the direction of entry and exit of air into the device. Let’s assume that the air
enters at A and leaves at B.

̇ ̇

The first law for a control volume is,

̇ ̇ ̇ ( ) ̇ ( )

The device is operating at steady state conditions (dE/dt=dm/dt=0). Hence, ̇ ̇ ̇.

. Hence,

̇ ̇ ( ) ̇ ( ) ̇ ( ) ̇ ( ) (assuming air as an

ideal gas)

( assuming air to be a diatomic gas and R=287 J/kg K).

Substituting the values,

̇ ( ) ( ) kW

̇ kW if A is the inlet and B is outlet, whereas ̇ if B is the inlet and A is

the outlet. We still don’t know the inlet and outlet.

To determine the inlet and outlet, we will calculate the entropy change for the air as it enters and
leaves the device. Also, the process given in the problem is adiabatic. Hence, if there is going to
be entropy change, it is going to be because of irreversibilities. Hence, entropy at the outlet must
be larger than the entropy at the inlet. If the entropy change for air between the inlet and outlet
comes out to be negative for a choice of inlet and outlet, that choice is then wrong, and the flow
happens in the opposite direction.

374
For an ideal gas, we have,

( ) ( ).

Hence, ( ) ( ) (taking A as the inlet and B as the

outlet).

Now, ̇ .

The entropy is decreasing as air enters at A and leaves at B. Hence, our choice of inlet and outlet
is wrong. The air enters at B and leaves at A. Hence, the work output ̇ .

Work is done on the device.

375
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 77
Tutorial Problem (2 number)

(Refer Slide Time: 00:13)

Figure 1.

Solution of the problem in Fig. 1:

Process 1-2, the process of condensation, is shown on a T-v diagram in Fig. 1. During this
process, the temperature and pressure remain constant.

From the steam tables, at 100 , the saturation pressure is 1.014 bar. Hence, at 100 and 1.014
bar, the difference between entropy of the saturated vapor and the saturated liquid is s fg = 6.0469
kJ/kg K. For the process from 1 to 2 on the T-v diagram (Fig. 1), .

Hence, .

For the surrounding air, the process is isothermal as the surrounding atmosphere is huge. The
heat transfer to the air in the condensation process will not change the air’s temperature. Hence,
the process is reversible for the air.

376
Hence, integrating gives . T = 298 K. We need to find Qrev.

The heat lost by vapor and water in the condensation process equals the heat gained by the air.
The heat transfer is negative for the water and vapor, whereas it is positive for the air.

The heat transfer in the condensation process

( is the enthalpy of vaporization. It is positive for vaporization and negative for
condensation).

Now, ( is positive for air)

= net increase in entropy of water plus surrounding.

The entropy of the universe has increased as the process of condensation is irreversible because
heat transfer occurs because of a finite temperature difference between the system (condensing
water and vapor) and surrounding air.

(Refer Slide Time: 02:32)

377
(Refer Slide Time: 08:28)

Figure 2.

Solution of the problem in Fig. 2:

A schematic of a closed tank containing water receiving heat from a heat pump is shown in Fig.
2.

For the closed tank containing water,

The heat addition process for water is a constant volume process. Hence, work interaction is 0.
The first law in the integrated form is …..(1)

378
From steam tables, at 20 , . Since, , we have a

mixture of water and steam at state 1. Hence, the qaulity at state 1, .

Hence, [ ( )] .

At . Again, . Hence, we have a

mixture of water and steam at state 2. Hence, .

Hence, [ ( )]

Equation (1) implies =heat transfer to the water=heat given

out by the reversible heat pump = QH.

For the reversible heat pump (see the schematic in Fig. 2), .

According to the first law for the pump, | | | | | | kJ.

This is the work input to the pump. Hence, W = -718 kJ.

379
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 78
Entropy Part 5

(Refer Slide Time: 00:16)

We will look at the concept of entropy in a bit more detail.

(Refer Slide Time: 00:36)

Figure 1.

An isolated system does not have any interaction with the surroundings. The entropy of an
isolated system can remain constant or increase, but it cannot decrease.

380
Figure 1 shows a system at temperature T and surroundings at temperature T0. There is a
transfer of heat from surroundings to the system. Hence, . The system does work on
the surroundings. The heat transfer for the surroundings equals heat transfer for the system.
We can write (considering only magnitudes). We know that, for a

process, . For the system, . The heat transfer for the surroundings, which

is huge in size, can be assumed isothermal as its temperature does not change because of the

heat transfer. Hence, the process is reversible for the surroundings. Hence,

(the heat transfer is negative for the surroundings). The net change in entropy is,

( ) ( ) ( ). If T=T0, then

because the heat transfer is isothermal and hence reversible. When T > T 0, there is a finite
temperature difference between the surroundings and the system, and If T < T0,
then would be negative and would be positive. In that case also,
Hence, as long as there is heat transfer through finite temperature difference between the
system and surroundings, net change in the entropy which is the sum of the entropy change
for the system and surroundings is greater than zero. In other words, the entropy change for
the universe is greater than zero. The entropy change for the universe is zero if the process is
reversible for the system as well as the surroundings. Hence, for heat transfer through a finite

temperature difference, entropy generated is ( ). Entropy is not conserved

unlike energy.

(Refer Slide Time: 08:45)

381
According to the Clausius’ inequality, . The equality is for a reversible process,

whereas the inequality is for an irreversible process. For the irreversible process, we can write

. For a reversible process, . For an irreversible process,

. can be positive or negative based on our sign convention.

(Refer Slide Time: 10:21)

Figure 2.
Figure 2 shows heat sources at temperature TH and heat sinks at temperature TL. There can be
heat transfer from the heat source to the heat sink by running a Carnot’s engine between
them. There can also be heat transfer from the heat sink to the heat source if we run the same
Carnot’s engine in reverse (as a heat pump) between them. The directions of heat and work
interactions for the Carnot’s engine and the reverse Carnot’s engine (heat pump) would be
opposite of each other, but the magnitudes would be equal. This is reversible heat transfer. It
does not leave any trace on the system as well as surroundings. The process is internally and
externally reversible.

We can also transfer heat from the heat source to the heat sink by bringing them in contact.
Heat transfer happens spontaneously. There is no work interaction involved. However, if we
want to transfer the same amount of heat back to the heat source from the heat sink, we need
to run a heat pump which needs work input. This reverse process is not spontaneous. The
work input to the pump comes from the surroundings. Here, we cannot just reverse the

382
process and transfer the same amount of heat from the heat sink to the heat source without
affecting the surroundings. Hence, this heat transfer process is irreversible as it leaves a trace
on the surroundings through the work input needed for the pump. Hence, the heat transfer
through a finite temperature difference is an irreversible process.

383
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 79
Entropy Part 6

(Refer Slide Time: 00:16)

Let’s recap some of the concepts.

For a process, the first law is written as . For a cyclic process, the first law is

written as ∮ ∮ Using the second law, we were able to obtain ∮ for a

reversible cycle. A change in entropy for a reversible process is the ratio of heat transfer to

the temperature at which heat transfer takes place. For a process, . In reality, most of

the processes are irreversible, and entropy is generated in the process. Hence, almost all the
processes in practice increase the entropy of the universe.

We use a lot of engineering devices. Our goal is to increase the efficiency of these devices.
We know that the reversible process produces the maximum possible work output (power)
and consumes the least work (power) for the given conditions. The irreversible processes,
because of entropy generation, are less efficient than the reversible processes. However, as
mentioned before, almost all the practical processes taking place in devices are irreversible.
Hence, the goal is to achieve a process with as little entropy generation as possible in the
system as well as surroundings so that the efficiency of the device increases.

384
(Refer Slide Time: 04:15)

Figure 1.
In statistical thermodynamics, we define entropy in terms of probability. The net increase in
entropy that occurs during an irreversible process can be associated with a change of state
from a less probable state to a more probable state.

In statistical thermodynamics, entropy is defined as , where k is the Boltzmann

constant, w is the thermodynamic probability (number of possible configurations). Let’s take
an example.

Figure 1 shows a box with a partition separating gas and vacuum. The box is well insulated
and rigid. There is no heat and work interaction. It means the internal energy of the gas
remains constant. There are also no changes in kinetic and potential energy of the system
(gas). As the partition is removed, the gas molecules move into the vacuum side. Initially,
when the partition was there, suppose that the probability of finding a gas molecule anywhere
in the rigid box is p1. When the partition is removed, the gas molecules occupy the entire
space available in the box. Now, the probability of finding a gas molecule anywhere in the
rigid box is more than p1. Hence, removing the partition increases the entropy of the system.
The number of possible configurations has increased. This process is unrestrained expansion
of the gas, which, we know, is irreversible.

Mixing of two gases is also an irreversible process. Figure 1 shows an insulated and rigid box
with a partition separating nitrogen and oxygen. The partition divides the volume equally. As
the partition is removed, the gases mix (given sufficient time). Now, the probability of

385
finding nitrogen as well as oxygen molecule anywhere in the rigid box has increased as
compared to the case when those molecules were separated by the partition, i.e., it is more
likely that we will find a nitrogen or oxygen molecule anywhere in the rigid box than to find
it in their respective halves when the partition was present. The number of possible
configurations for the molecules of both the gases has increased. Hence, the mixing of
nitrogen and oxygen after removing the partition increases the entropy of the system
(nitrogen and oxygen). Bringing back the nitrogen and oxygen molecules from their mixture
to their initial respective halves when the partition was present needs a lot of work to be done
on the system. Hence, the mixing of gases is irreversible.

(Refer Slide Time: 08:46)

Figure 2.
Figure 2 shows houses (a house-like structure) with 5 floors (0 to 4) and 4 objects in it. This
configuration has only one type of energy, e.g., potential energy. The objects have only
potential energy. The units of potential energy the object has depends on which floor the
object is. For example, the object on the 0th floor has 0 units of potential energy, whereas the
object on the 4th floor has 4 units of potential energy.

In the first configuration, 3 objects are on the 0th floor and 1 object is on the 2nd floor. Hence,
the total energy is 2 units. In the second configuration, 2 objects are on the 0th floor and the
other 2 are on the first floor. Hence the total energy is 2 units. These are the only two possible
and equally likely configurations with four objects having 2 units of energy in total. The
entropy corresponding to these configurations is S. Let’s add 2 units of energy (heat). Now,

386
we have 4 units of energy in total. What are the possible configurations of the objects having
4 units of energy in total? The table in Fig. 2 shows 5 equally likely configurations of the
objects. For example, one of the configurations has all the 4 objects on the 1st floor (each
object has one unit of energy). Each of the configurations has 4 units of energy in total.
Initially, we had 2 possible configurations with 2 units of energy in total. After adding 2 units
of energy (heat), the number of possible configurations increased to 5. Hence, the entropy of
the system has increased. The more the number of the possible configurations, the larger is
the entropy.

(Refer Slide Time: 14:29)

Figure 2.
What if instead of adding 2 units of heat we add 2 units of work (frictionless, quasi-static)?

The upper half of Fig. 2 shows the initial configurations having 2 units of energy each. The
corresponding entropy is S. When we do boundary work on the system (the house), it gets
stretched in the vertical direction (the boundary gets stretched in the vertical direction) as we
are considering only the potential energy. The boundary cannot stretch in the horizontal
direction as it would not increase the potential energy of the system. The stretched houses are
now double the height of the original houses. In this new building, an object on the 0th floor
still has 0 unit of energy. However, the object on the 1st, 2nd, 3rd and 4th floor has 2, 4, 6 and 8
units of potential energy as the houses’ height is doubled now. We see that the number of
possible configurations are still the same after doing boundary work of 2 units. In the first
configuration, the three objects on the 0th floor have 0 unit of energy each, whereas the object

387
on the 2nd floor has 4 units of energy. In the second configuration, the two objects on the 0th
floor have 0 unit of energy each, whereas the two objects on the 1st floor have 2 units of
energy each adding to 4 units in total. As the number of possible and equally likely
configurations has not changed after doing a boundary work of 2 units, the entropy of the
system does not increase. So, adding 2 units of energy through heat transfer increased the
entropy of the system, whereas adding 2 units of energy through work transfer (frictionless,
quasi-static) did not.

(Refer Slide Time: 17:49)

This brings us to grades of energy. According to the first law, energy can neither be created
nor destroyed, but it can be transformed from one form to another.

It is easy to convert 100 J of work to 100 J of heat. For example, 100 J of work from a battery
can be converted to 100 J of heat using a resistance, assuming steady state operation. So, the
complete conversion of work to heat is possible. It is also possible in a cyclic process (Joule’s
experiment).

Can we convert 100 J of heat to 100 J of work in a cyclic process?

It is not possible according to the second law. In a cyclic process, if you want to convert heat
to work, some amount of heat needs to be rejected, i.e., out of the heat supplied, some gets
converted to work, and the remaining heat needs to be rejected. The complete conversion of
heat to work is not possible in a cycle as it violates the second law. The maximum possible
conversion of heat to work can happen in a Carnot’s engine.

388
There is a distinction between work and heat. Heat and work are the forms of energy. The
first law says that the energy can be transformed from one form to another. However, there
are restrictions on the complete conversions from one form to another as we just saw.

So, we can have a cyclic process converting work into heat completely. However, we cannot
have a cyclic process converting heat into work completely. Work is useful because we can
use it either as work or we can convert it into heat. Heat is not as useful because we can use it
as heat, but we cannot convert it back on into work completely. Hence, we have grades of
energy.

Mechanical work, potential energy, kinetic energy and electrical energy are high grades of
energy. They can be converted into each other among themselves. For example, (a)
mechanical work can be converted into electrical energy, potential energy, and kinetic
energy, (b) potential energy of the falling object gets converted into kinetic energy. Some
losses do happen during conversion.

Chemical energy is low grade energy. In most of the applications, we cannot convert
chemical energy directly into work. We usually convert chemical energy into heat and then
convert heat into work. For example, chemical energy of fuels is converted into heat by
burning them, and the heat released can be converted to work (e.g. thermal power plant).
However, the losses here are larger compared to conversions of high grades of energy among
themselves.

Heat is the lowest grade of energy. Its complete conversion to other useful forms of energy is
not possible.

The first law does not put any restriction on the complete conversion of heat into work and
vice versa, but the second law does, as mentioned before.

In India, we get most of the electricity from thermal power plants where chemical energy of
coal is converted into heat, which is then converted into mechanical work, which finally is
converted into electricity. The electricity is then transmitted though wires to our homes. We
run various electrical appliances using the electricity. There are lot conversions from one
form of energy to other at the power plant. Assume efficiency of thermal power plant to be
around 30 % (quite close to the actual one). It means that for running a bulb of 100 W at our
home, we need to burn coal to give around 330 W of heat at the power plant. Hence, when we
forget to turn off a 100 W bulb while leaving our home, we are actually wasting around 330

389
W of power. Hence, it is advised to turn off electrical appliances when they are not in use. In
essence, we are converting low grade energy to high grade energy at the power plant. This
high grade energy is again converted into low grade energy at our homes (almost all the
times) through electrical appliances.

390
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 80: Entropy Part - 7
(Refer Slide Time: 00:25)

Figure 1.
Let’s look at a reversible adiabatic process for an ideal gas.

Figure 1 shows a deformable and insulated chamber containing ideal gas. The volume of the
chamber can change. We have the Gibb’s relations,

. For a reversible adiabatic process, ds=0. Hence,

. For an ideal gas, and . Hence, For an

ideal gas,

391
(Refer Slide Time: 03:36)

Hence, ( ) . Integrating, ( )

( ) ( ) , where c is a constant. Using the properties of logarithms, we get,

. We know that for an ideal gas, . Hence,

, where . Hence, for an ideal gas undergoing a reversible adiabatic

process, , where k is a constant. We have used this relation before. However, we did
not look at the formal proof. For using the expression , the process also has to be a
quasi-static process.

(Refer Slide Time: 06:34)

Figure 2.

392
Let’s look at the rate equation of entropy. We have . Dividing each term by

small time interval , we get, . We saw a similar expression when we

looked at the rate equation of the first law for a control mass, . Taking limit as

we get the familiar form, ̇ ̇

In the rate equation of entropy, the temperature at which the heat transfer occurs needs to be
considered, unlike the first law. The heat transfer is associated with the temperature at which
it happens. A control mass can have heat transfer through its boundary, and the boundary
may have different temperatures in different parts along its length. Hence, we need to sum up

individual terms for each part. Hence, the rate equation of entropy for a control mass as

is ∑ ̇ ̇ (cm represents control mass).

(Refer Slide Time: 08:55)

Figure 3.
Let’s solve the problem given in Fig. 3.

Consider heater as our system. The rate equation of the first law is ̇ ̇.

The heater is operating at steady state conditions. Hence, dE/dt=0. Hence, ̇ ̇ . We know,
̇ (work is being done on the system). Hence, ̇ (the heater is
losing heat). The temperature at which the heater is losing heat is 550 K. This temperature
stays constant.

393
The rate equation of second law (or the rate of equation of entropy) is, ∑ ̇ ̇ .

as steady state conditions have been achieved. Hence, ̇ ( ) =

rate of total entropy generation.

So, energy is conserved, i.e., electric power is getting converted into heat, completely.
However, entropy is generated in this process which is irreversible.

(Refer Slide Time: 12:41)

Figure 4.
Let’s look at the rate equation of entropy for a control volume.

We have already looked at the rate equation of the first law for a control volume (energy
equation). It is basically an equation for energy conservation. However, we know that entropy
is not conserved unlike energy. It is generated in irreversible processes. Hence, we need to
consider entropy generation while writing the rate equation of entropy for a control volume.
The rate equation of entropy for the control volume shown in Fig. 4 is,

̇
∑ ̇ ∑ ̇ ∑ ̇ , where ∑ ̇ representes the rate at which

entropy enters the control volume, ∑ ̇ represents the rate at which entropy leaves the
̇
control volume, ∑ represents the entropy transfer because of heat transfer, and ̇

represents the rate of entropy generation because of irreversibilities in the control volume.
There is no term related to work in the equation as the work is an orderly fashion of adding or
removing energy. Heat is a disorderly fashion of adding or removing energy which can

394
change the entropy of a system or control volume as entropy essentially is a measure of
disorder. The rate equation of entropy for a control volume is quite different compared to the
rate equation of the first law for a control volume. Students are encouraged to compare the
two equations and find the differences.

(Refer Slide Time: 16:02)

For a steady adiabatic process and a control volume with one inlet and one outlet, the rate
equation of entropy for a control volume reduces to,

̇ ̇ ̇

Since the process is steady, dm/dt=0. Hence, ̇ ̇ ̇ . Dividing by ̇ throughout,

. Hence, . The equality holds true for the reversible process, whereas
the inequality holds true for an irreversible process. Hence, the entropy at the exit of a control
volume is either equal to the entropy at the inlet or greater than it.

395
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Exergy – Part 1

(Refer Slide Time 0:16)

Let’s look at the concept of exergy or availability. It is essentially a way of looking at both the
first and second laws together.

(Refer Slide Time 0:36)

Figure 1.

We looked at the concept of available energy before. Let’s recap it.

396
Figure 1 shows a reservoir at temperature T which gives out Q amount of heat. How much of this
heat Q can be converted to work? We know that maximum possible conversion of Q into work
happens if we use a reversible heat engine. Figure 1 shows a reversible heat engine running
between the reservoir at temperature T and environment at temperature T0 giving out work Wrev.
The engine takes in Q amount of heat and rejects Q0 amount of heat. According to the first law,
(taking absolute values). We know that for a reversible heat engine in

Fig. 1. Also, ∮ ∮ . Hence, . Hence,

( ).

(Refer Slide Time 03:11)

Figure 2.

Figure 2 shows a T-S diagram for the reversible heat engine of Fig. 1. We know that the
magnitude of the amount of heat rejected to the environment is | |. This heat is not
converted to work. The temperature reservoir at T supplies heat for conversion to work. The

maximum possible work output is ( ). The exergy ( ) of the constant

temperature reservoir at T, in this case, is the maximum possible work output we can get from it

which is ( ).

397
(Refer Slide Time 04:04)

Exergy ( ) is the maximum possible work obtainable as a system interacts with a reference
environment to achieve equilibrium. Alternatively, we can also define it as the minimum
possible work you need to give in order to take the system which is in equilibrium with the
reference environment to whatever the state it was in.

(Refer Slide Time 04:38)

If the system and surroundings are in equilibrium, then there is no potential for use (i.e. no
potential to get work as output). For example, a weight lying on the ground has no potential for
use. However, a falling weight can be used to get some work output. Similarly, a combusting

398
flame which generates hot gases can be used to get work. Hence, if the system and surroundings
are in equilibrium, we won’t get any work output.

(Refer Slide Time 06:06)

We have been talking about the system and the surroundings. The surroundings is essentially
called as exergy reference environment. The surroundings (exergy reference environment) can be
divided into immediate surroundings and the undisturbed surroundings. The immediate
surroundings gets affected by the system, whereas the undisturbed surroundings which is far
from the system stays undisturbed. In general, when we talk of the reference environment, we
mean undisturbed surroundings.

For doing exergy analysis, the idealized exergy reference environment is assumed to be large and
uniform. We consider only simple compressible systems. The temperature and pressure of the
reference environment are fixed. The common values are 25 and 1 atm. However, these can
be the temperature and pressure at the place where the system is. The reference environment is
assumed to be at rest.

399
(Refer Slide Time 07:19)

Figure 3.

We have come across a heat source at constant temperature quite a lot of times. We briefly
looked at a heat source at constant pressure also. Let’s look at it in more detail.

Constant pressure heat sources are quite common (e.g. a boiler and a condenser). In these heat
sources, pressure remains constant, whereas the temperature changes. Figure 3 shows a heat
exchanger (a constant pressure heat source). To get maximum work from such a heat source, we
need to run a reversible heat engine between the heat source and the surrounding environment at
T0 (T0 is constant) as shown in Fig. 3. This heat engine takes in heat Q from the heat source,
gives out work Wrev and rejects heat Q0 to the environment. According to the first law,
(considering the absolute values). The entropy change for the cycle is given as

∫ . The temperature of the heat source is changing here. Hence, we need to run a

sequence of reversible heat engines.

400
(Refer Slide Time 9:55)

Figure 4.

Figure 4 shows a T-S diagram where the temperature of the heat source is varying (as in the case
of the reversible heat engine of Fig. 3), whereas the temperature of the surrounding environment
is constant. In this case, we can divide the region on the T-S diagram into rectangles as shown.
Each rectangle corresponds to a Carnot’s cycle (reversible cycle). The first rectangle (the
leftmost rectangle) corresponds to a Carnot’s cycle C1 which takes in heat at temperature T and
rejects heat at temperature T0. The next rectangle corresponds to a Carnot’s cycle C2 which takes
in heat at slightly higher temperature than C1 and rejects heat at T0. The next rectangle
corresponds to C3 which takes in heat at slightly higher temperature than C2 and rejects heat at
T0. In this way, the green part of the T-S diagram in Fig. 4 is composed of many Carnot’s cycles
taking in heat at higher and higher temperatures and rejecting heat at T0. Hence, in this case,

∫ (considering absolute values) as the engines are reversible. We can also write

. Hence, exergy of the constant pressure reservoir is

.

401
 Thermodynamics
Exergy – Part 2
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras

(Refer Slide Time 00:15)

Figure 1.
Figure 1 shows an arbitrarily shaped control volume. The process happening inside the control
volume is not ideal (which is also true in practice). Let’s call this control volume as the actual

control volume. The rate equation of mass conservation for the control volume is ∑ ̇

∑ ̇ (cv represents control volume). The rate equation of the first law for a control volume is

∑ ̇ ̇ ∑ ̇ ( ) ∑ ̇ ( ) , where ̇

represents the actual work interaction for the non-ideal process. There could be more than one
heat sources. Also, there could be different amount of heat transfers through different parts of the
boundary at different temperatures. Hence, there is a summation symbol before ̇ . For the rate
equation of entropy, we also need to consider the temperatures of different parts of the boundary.
̇
The rate equation of entropy for the control volume is ∑ ∑ ̇ ∑ ̇

̇
̇ , where ̇ is the entropy generation rate for the non-ideal process. can be

402
positive or negative depending on the direction of heat transfer (i.e., into or out of the control
volume). The mass of the substance entering the control volume brings entropy into the control
volume, whereas the mass leaving the control volume takes entropy out of the control volume.
The surroundings is at temperature T0.

(Refer Slide Time 05:33)

Now, consider a similar control volume as shown in Fig. 1. However, the process in this control
volume is reversible (ideal process). That is the only difference between the control volume of
Fig. 1 (actual control volume) and this ideal control volume. Let’s compare conservation
equations for this control volume or the ideal control volume with that of Fig. 1 (actual control
volume).

The mass conservation equation is the same for both the control volumes. The terms on the left
hand side of the energy equation and the entropy equation, i.e., dE/dt and dS/dt are also the same
for both the control volumes as the states achieved during the processes in both the control
volumes are exactly the same. Time taken for the processes is also the same. The only difference
comes through the term ̇ in the entropy equation. For the actual control volume of Fig. 1,
̇ as the process is irreversible. However, for the ideal control volume ̇ as the
process is reversible.

403
(Refer Slide Time: 07:12)

Hence, though the left hand side of the entropy equation (dS/dt) is the same for the actual control
volume and the ideal control volume, the right hand sides are different for them because of ̇ .
̇ for the ideal control volume, whereas ̇ for the actual control volume. Hence, if
the left hand sides are the same for both the control volume, we need to compensate for ̇
on the right hand side of the entropy equation for the ideal control volume, i.e., we need add
something on the right hand side which will add entropy to the control volume. As the processes
are reversible in the ideal control volume, we cannot have entropy generation because of
irreversibilities. Another way of adding entropy is to add heat. If we are going to add heat, it also
needs to be reflected in the energy equation.

We need to add heat reversibly to increase entropy of the ideal control volume as the process in
the ideal control volume is reversible. This heat addition can be from the ambient which can be
assumed as a constant temperature reservoir. Let’s represent the heat added as . This heat
transfer happens at temperature T0 of the ambient. Because of this reversible heat addition, the
work term in the energy equation for the ideal control volume changes.

404
(Refer Slide Time 10:43)

Figure 2.
So, the rate of entropy transfer into the ideal control volume because of reversible heat transfer
̇
is which equals ̇ of the actual control volume (Fig. 2). Hence, ̇

̇ . After inserting the term of ̇ in the energy equation, we get, for the ideal control
volume,

∑ ̇ ̇ ̇ ∑ ̇ ( ) ∑ ̇ ( ) (icv is ideal

control volume. Also, ).

Comparing this equation with the energy equation for the actual control volume (Fig. 1), we get
̇ ̇ ̇ . Hence, ̇ ̇ ̇ . Hence, the work transfer rate
(power) in the ideal control volume is larger than that in the actual control volume by ̇ . An

interesting thing to note here is that the ideal control volume has heat transfer from ambient even
if the actual control volume undergoes an adiabatic process.

405
(Refer Slide Time 13:11)

Figure 3.

Now, ̇ ̇ ̇ ∑ ̇ ∑ ̇ ( ) ∑ ̇ (

) ̇ ∑ ̇ ∑ ̇ ( ) ∑ ̇ ( )

̇ ∑ ̇ ∑ ̇ ( ) ∑ ̇ ( ) (

̇
∑ ∑ ̇ ∑ ̇ ) ….(taking from equation 1 and and ̇ from equation 2)

We can rewrite it as,

̇ ∑( ) ̇ ∑ ̇ ( ) ∑ ̇ ( )

( ).

For a steady flow device having single inlet and outlet, we can divide all the terms by ̇
( ̇ ̇ ̇ ). Hence, we get,

̇
∑( ) ( ) ( )
̇

406
(Refer Slide Time 15:21)

407
 Thermodynamics
Exergy – Part 3
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras

(Refer Slide Time 0:13)

We studied the extent of irreversibility before. For a control volume, irreversibility is the
difference between the work transfer rates (power) for a reversible process and an irreversible
process. Mathematically, ̇ ̇ ̇ . As we saw in the last lecture, ̇ ̇
̇ ̇ ̇ ̇ is the loss in the work output because of the process being

irreversible.

408
(Refer Slide Time 1:16)

As we saw in the last lecture, for a control volume, we have,

̇ ∑( ) ̇ ∑ ̇ ( ) ∑ ̇ ( )

( )

For a control mass, there is no flow of mass in or out. Hence,

̇ ∑( ) ̇ ( )

For the control mass undergoing a process from state 1 to 2, integrating the above equation from
state 1 to 2 gives,

1 ∑( )1Q2j ( ) ( ).

409
(Refer Slide Time 3:01)

As discussed before, if a system is in equilibrium with the reference surroundings, it has no

potential to give work as output. For extracting the maximum work from a system or a control
volume which are not in equilibrium with the surrondings, the system or the control volume
needs to reach a dead state. When a system or a control volume achieves a dead state, they are in
mechanical, thermal and chemical equilibrium with the surroundings. At a dead state, a system is
at rest. It doesn’t have kinetic energy. For a control volume at a dead state, the substance flowing
through the control volume does not have kinetic energy at the exit of the control volume. It is
not practically possible because, then, the substance will not move out of the control volume.
However, at a dead state, the kinetic energy at the exit of the control volume should be as
minimum as possible because we want to extract the maximum work from the control volume.
At a dead state, a system has 0 potential energy. For a control volume at a dead state, the
potential energy at the exit of the control volume is 0. The exit of the control volume should be
on the ground (local ground level), ideally. We get the maximum work from the system or the
control volume when they reach the dead state at the end of the process.

410
(Refer Slide Time 5:36)

Let’s look at flow exergy.

Consider a control volume with a single inlet and outlet. There is no heat transfer. The exit of the
control volume is at dead state. We have the expression for the maximum possible work from a
control volume,

̇
∑( ) ( ) ( )
̇

Since there is no heat transfer and the exit is at a dead state,

( ) ( ), where is the reference

enthalpy, is the local ground level, is the local reference entropy (entropy of the local
surroundings). This expression represents the flow exergy for a control volume with single inlet

and outlet and having no heat transfer. The expression can be termed as

exergy at the inlet, , of a control volume, whereas the expression can be

termed as the exergy at the outlet, of the control volume. If the exit is at a dead state then
. Hence, we can write if the exit is not at a dead state. Hence,
reversible work from a single inlet-outlet steady state flow device with no heat transfer equals
the decrease in exergy between its inlet and outlet.

411
(Refer Slide Time 09:22)

We know the expression for the maximum available rate of work from a control volume,

̇ ∑( ) ̇ ∑ ̇ ( ) ∑ ̇ ( )

( ).

The last term, , represents storage terms for a control volume. The reversible work

from a storage effect due to a change of state in a control volume can also be used to find an
exergy. The expression we obtain can also be treated as the exergy of a system/control mass.

Here, the volume may change, and work is done against the surroundings. This work is
represented as ̇ ̇ , where p0 is the pressure of the surroundings and ̇ is the rate of
change of volume. The maximum possible rate of work from the storage terms is,

̇ ̇ ̇ ( ) ̇

Integrating between the initial state to a dead state gives exergy (maximum possible work) as,

(( ) ( ) ( )) ( ) ( )

412
Specific exergy is represented as,

( ) ( )

(Refer Slide Time 12:08)

Few important points regarding exergy are mentioned below:

1. Exergy is defined for a system and environment together. The same system in different
environments can have different exergy.

2. Exergy cannot be negative.

3. Exergy is not conserved (as the expression for exergy involves entropy terms, and entropy is
not conserved for irreversible processes)

4. Exergy is the maximum work obtainable from a system or a control volume when they
achieve a dead state from their current state. It is also the minimum work required to take the
system or control volume from a dead state to their original state.

413
(Refer Slide Time 14:01)

Figure 1.
(Refer Slide Time 15:49)

Solution of the problem in Fig. 1:

Dead state: ( )

State 1:

Substance: steam

414
The mass is fixed.

Specific exergy, ( ) ( ) …(1)

Also, .

Hence, we need to find , and at state 1 from steam tables.

[ ( )]

Similarly ( )

Similarly,

Now,

Similarly, we need . At a pressure of 1 bar, the saturation temperature Tsat=99.6 .

T0 < Tsat. Hence, the steam is in liquid state at the reference conditions, and its properties are
given in a table in Fig. 1.

Equation (1) implies,

( )

Students are encouraged to look at contributions from different terms towards exergy.

415
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture – 84
Thermodynamics Cycles: Rankine Cycle

(Refer Slide Time: 00:12)

Let’s study the basic Rankine cycle.

(Refer Slide Time: 00:29)

Figure 1.

416
Figure 1 shows a power plant which employs the Rankine cycle. It has a boiler which takes in
air-fuel mixture (fuel could be coal, natural gas, etc). The air-fuel mixture is burned and hot
gases form inside the boiler. These hot gases give their heat to water and convert water into
steam. The hot gases then, are sent out. In some plants these gases are used to preheat the air
entering the boiler for forming air-fuel mixture. These hot gases contain harmful emissions. They
are sent to pollution control devices for removing those harmful emissions after which they are
sent to stack where they cool. Then, they get out into the environment as clean combustion gases.
The steam, after leaving the boiler, enters 1 or 2 turbines. The steam is expanded inside the
turbine generating shaft work. The shaft is connected to electrical generator for generating
electricity. The steam leaving turbine is at low pressure and low temperature. This low-pressure-
low-temperature steam enters a condenser where it condenses back into water. Inside the
condenser, chilled water is used to cool down the steam. The steam loses heat to this water
which, after leaving the condenser, enters the cooling tower where it is sprayed to cool it down
through evaporative cooling. In this process, some water is lost because of evaporation. Hence,
we need to add make up water. The condensed water in the condenser is sent back to the boiler
through a pump. Sometimes the water entering the boiler is preheated through a feedwater
heater, which uses a small quantity of steam from the turbine to heat water.

(Refer Slide Time: 04:44)

Figure 2.

417
Essentially, the system consists of 4 components, a boiler (a steam generator), a turbine, a
condenser and a pump as shown in Fig. 2. The system here undergoes a cyclic process. The same
fluid flows through all the components. The mass of the fluid is constant (in reality, we keep
supplying fluid at some places in the cycle as we lose fluid because of some problems or
necessities). This system consisting of the four components is a closed system, even though the
components individually are open systems as the fluid enters and leaves them at certain mass
flow rates. The system interacts with the surroundings through heat and work interactions. We
have heat interactions for the boiler and the condenser, and work interactions for the turbine and
the pump.

(Refer Slide Time: 06:20)

Figure 3.
The pump receives the low temperature low pressure liquid from the condenser and gives out
high pressure low temperature liquid. This liquid enters the boiler. The boiler is made of 3 parts,
economizer, boiler and superheater (Fig. 3). We will study them in detail later on. The boiler
gives out high pressure high temperature steam. This team expands inside the turbine and comes
out as low pressure low temperature steam. This steam enters the condenser and comes out as
low pressure low temperature liquid, which is sent to the pump, and the cycle continues.

418
(Refer Slide Time: 09:25)

Figure 4.
We know that the Carnot’s cycle is the most efficient cycle between a given heat source and a
heat sink. The Carnot’s cycle consists of two isothermal processes and two isentropic processes.
The hand-drawn T-S diagram for the Carnot’s cycle is shown in Fig. 4. 1-2 is a reversible
adiabatic process which is compression. 2-3 is an isothermal process where we have heat
addition. 3-4 is a reversible adiabatic process where we have expansion. 4-1 is an isothermal
process where we have heat rejection. So, in the case of the Carnot’s cycle, the process 1-2
happens inside a compressor, and the process 3-4 happens inside a turbine. The processes 2-3
and 4-1 happen inside isothermal heat exchangers. But there are practical issues with the
Carnot’s cycle.

Isothermal heat transfer involving a single phase substance is not achievable. Hence, the
processes 2-3 and 4-1 are not achievable for a single phase substance (i.e. for only gas or only
liquid). However, as we have seen earlier, isothermal heat transfer is possible during a phase
change process inside a liquid-vapor dome. In this process, the temperature and pressure, both
remain constant.

419
(Refer Slide Time: 10:35)

Figure 5.
Figure 5 shows the hand-drawn T-S diagram of the Carnot’s cycle inside the liquid-vapor dome.
Here, we are dealing with a mixture of liquid and vapor. The process 3-4, which is isentropic
expansion, happens in a turbine. It is difficult to design turbine which works with a mixture of
liquid and vapor. The turbines rotate at very high speeds. The liquid droplets in a two phase
mixture can damage the blades of the turbine. Turbines are designed to work with single phase,
i.e., only gases or only liquids.

420
(Refer Slide Time: 15:05)

Figure 6.
The solution is to move the process of expansion outside the liquid-vapor dome, where the
turbine deals with only a gas phase. The process of heat addition 2-3 is extended till state 3` (Fig.
6). The process 2-3-3` is isobaric. The portion 3-3` cannot be isothermal as it is difficult to
achieve isothermal process in a single phase region, as discussed before. Now, the process 3`-4`,
which is in a single phase region, happens in the turbine. So, the states 3 and 4 in the Carnot’s
cycle are changed to 3` and 4`. Hence, the heat rejection process is now from 4` to 1 instead to 4
to 1.

421
(Refer Slide Time: 18:58)

Figure 7.
There is still a problem with the compression process from 1 to 2 (Fig. 7). It is difficult to
compress a mixture of liquid and vapor. The compression devices work with only gases or only
liquids. Hence, it is logical to let the heat rejection process 4`-4-1 end at a state on the saturated
liquid line, which is 1` (Fig. 7). Now, we can have a pump compressing the liquid from 1` to 2`
isentropically. The state 2` is located at the intersection of isentrope starting at 1` and the isobar
going through states 2, 3 and 3`. We also know that the compression work for an isentropic
pump is given as | | ∫ . Since the specific volume of liquid is far smaller than the gas,
the work needed for compressing the liquid is significantly small compared to compressing a
gas.

422
(Refer Slide Time: 20:44)

Figure 8.
After incorporating all the changes suggested above, the cycle becomes 1`-2`-2-3-3`-4`-4-1-
1`(hand-drawn cycle in Fig. 8). Let’s rename state 1` by 1 and erase 1. The Rankine cycle (1-2`-
2-3-3`-4`-1) is shown on the T-S diagram in Fig. 8 (top right corner).

(Refer Slide Time: 24:15)

Figure 9.

423
We have four processes in the Rankine cycle:

1. Process 1-2` – isentropic compression of low pressure low temperature liquid in a pump

2. Process 2`-2-3-3` - constant pressure heat addition in a boiler

The process 2`-2 happens inside the economizer where only liquid is heated at constant pressure
(Fig. 9). The process 2-3 happens inside the boiler where heat addition happens at a constant
pressure and constant temperature as it is a phase change process (Fig. 9). The process 3-3`
happens inside the superheater where only steam is heated at a constant pressure (Fig. 9).

3. Process 3`-4` - isentropic expansion of high pressure high temperature steam in a turbine

4. Process 4`-1 – constant pressure heat rejection in a condenser

(Refer Slide Time: 24:20)

Figure 10.
Figure 10 shows formulae for calculating heat and work interactions for the Rankine cycle on
unit mass basis. We have already derived these formulae in previous lectures.

1. Turbine work output: (isentropic efficiency needs to be considered if given,

which we have discussed in the previous lectures)

424
2. Work input to pump: | | (isentropic efficiency needs to be
considered if given, which we have discussed in the previous lectures)

Work input to a pump is negative as work is being done on the pump. For a non-isentropic
process, | | is not applicable.

3. Heat transfer in the boiler: ( is positive)

4. Heat transfer in the condenser: ( is negative)

| |
5. Efficiency of the cycle,

(Refer Slide Time: 28:21)

425
 Gmail
Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture – 85
Tutorial Problem (1 number)

(Refer Slide Time: 00:13)

(Refer Slide Time: 00:21)

Figure 1.

426
Solution of the problem in Fig. 1:

Turbine inlet:

Turbine exit:

We have saturated water at the condenser exit. It would help in solving the problem if we draw
this cycle on a T-S diagram. Hence, we need to know the states.

At 4 MPa, Tsat = 250 . Since 400 > Tsat, the steam is in superheated zone at the inlet of the
turbine. At the turbine inlet, h = 3214.5 kJ/kg and s = 6.771 kJ/kg K

The condensation process happens at 15 kPa. If the expansion process in the turbine is isentropic
(from 4 MPa to 15 kPa), the entropy at the inlet of the condenser is s = 6.771 kJ/kg K.

At 15 kPa, sg = 8.007 kJ/kg K. As 6.771 < 8.007, the inlet to the condenser lies inside the liquid-
vapor dome. Now, we can qualitatively draw the Rankine cycle on a T-S diagram as shown in
Fig. 2.

The processes are:

Process 1-2: isentropic compression in a pump

Process 2-3-4-5: constant pressure heat addition in a boiler

Process 5-6: isentropic expansion in a turbine

Process 6-1: constant pressure heat rejection in a condenser

Heat supplied to the working fluid in a boiler, (enthalpy at the inlet of the turbine –
enthalpy at the exit of the pump)

427
The state 2 has the same entropy as state 1. We also know the pressure at state 2. Knowing
pressure and entropy values at state 2, we can get h2 from the tables for subcooled liquids.
However, we don’t have those tables here.

We can also get h2 through the expression , where wp is the work input to the
pump. If we consider the isentropic compression, ( )

( )

Hence, ( )

Hence, heat supplied to the working fluid in a boiler,

Net work output on a unit mass basis = | | | |( is the turbine work

output)

For calculating , we need the quality . We know that [

] [ ]

Hence, .

Hence, net work output on a unit mass basis = | | | |

(considering isentropic processes)

Hence,

428
(Refer Slide Time: 03:07)

(Refer Slide Time: 05:48)

429
(Refer Slide Time: 05:57)

(Refer Slide Time: 06:16)

Figure 2.

430
431
(Refer Slide Time: 17:20)

(Refer Slide Time: 21:10)

432
433
434
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture – 86
Thermodynamics Cycles: Brayton Cycle
(Refer Slide Time: 00:17)

Figure 1.
Let’s study the Brayton cycle which is used in gas turbine engines. Figure 1 shows a turbofan
engine used on aeroplanes, which runs on the Brayton cycle.

435
(Refer Slide Time: 01:01)

Figure 2.
Figure 2 shows a gas turbine engine on an aeroplane. It consists of a diffuser, a compressor, a
combustor, a turbine and a nozzle. The diffuser slows down the flow of air and increases its
pressure. Compressor further increases the pressure through multiple stages. In the combustor, a
fuel is added to high pressure air and burned, producing hot gases. These high-enthalpy-high
temperature-high-pressure gases are expanded partially through the turbine generating work.
This work can be used to run the compressor upstream. The gases after coming out of the turbine
expand in the nozzle giving thrust. Here, the gas enters and leaves the system. It does not
undergo a cyclic process.

436
(Refer Slide Time: 03:11)

Figure 3.
We replace the actual process (shown in the top left corner of Fig. 3) in the gas turbine engine by
a hypothetical one (shown in the bottom left corner of Fig. 3) where air undergoes a cyclic
process. This cycle is called air standard cycle or the basic air standard Brayton cycle or the
Joule cycle.

The compressor takes in air at low pressure and temperature, compresses it and gives out high
pressure and high temperature air. At point 2, this air enters a heat exchanger where heat is added
to it at constant pressure (in reality, fuel is added and burned). At point 3, we have high pressure
high temperature air. This air enters a turbine, expands and comes out as low pressure low
temperature air. During expansion in the turbine, work is produced. Some of that work is used to
run the compressor. Depending on the application, the work is used to generate electricity. The
low pressure low temperature air at point 4 enters a heat exchanger where it rejects heat at
constant pressure, and comes out, at even lower temperature, at point 1, where it enters the
compressor and undergoes the cycle again.

To get the maximum work, we would like to have a Carnot’s cycle as shown on a T-S diagram in
Fig. 3. However, as mentioned before, isothermal heat addition and rejection is not possible if we
are dealing with single phase of a substance. Here, the working substance is air. However,
isobaric heat transfer is possible. Hence, the processes 2-3 and 4-1, where heat transfers occur,

437
are converted to isobaric processes (Fig. 3). The processes 1-2 and 3-4, where compression and
expansion occurs, can remain isentropic. Hence, the basic Brayton cycle consist of two isobaric
and two isentropic processes. On a T-S diagram, the isobaric processes, 2-3 and 4-1, diverge
from each other (we have discussed this before).

(Refer Slide Time: 08:03)

Figure 4.
Figure 4 shows the Brayton cycle on a p-v and a T-s diagram. On the T-s diagram, the length 1-2
is smaller than the length 3-4 as the isobars diverge. We know that the processes 1-2 and 3-4
may not be isentropic. In this case, isentropic efficiencies are given. Non-isentropic processes are
shown by dotted lines on the T-s diagram in Fig. 4. Considering non-isentropic processes, the
cycle is 1-2`-3-4`-1.

438
(Refer Slide Time: 11:28)

(Refer Slide Time: 14:00)

Figure 5.
Figure 5 shows expressions/formulae for work and heat interactions, efficiencies, pressure ratio
associated with the Brayton cycle.

The expressions for the work and heat interactions can be derived from the expression of the first
law for a control volume (we have already looked at such derivations in the case of a

439
compressor, turbine, heat exchangers, etc.). Below are mentioned those expressions for the
Brayton cycle shown in Fig. 5:

1. Turbine work output: (Isentropic efficiencies need to be considered if given)

Assumptions made to arrive at this expression are (a) steady state operation, (b) adiabatic
process, (c) kinetic and potential energy changes are negligible, (d) turbine has a single inlet and
outlet

2. Compressor work input: | | (Isentropic efficiencies need to be

considered if given. The assumptions made here are the same as those for the expression of
turbine work output)

3. Heat addition:

4. Heat rejection: | |

| | | |
5. Efficiency: [assuming air as an ideal

gas, | | and ]

6. Pressure ratio: ( ) ( )

1-2 and 3-4 are reversible adiabatic processes. Air is considered as an ideal gas. Using
and , one can derive the above pressure ratio expression.

7. Efficiency of the cycle is also given by,

( ⁄ )

Students are encouraged to derive this expression (Hint: use expressions in 5 and 6). The
expression for the efficiency looks similar to the one for the Carnot’s cycle. The efficiency is the
function of temperatures at the beginning and end of the compression process.

From 6, taking , we can conclude that the pressure ratio is higher than the temperature
ratio, i.e., the pressure changes a lot from 1 to 2, but temperature does not change significantly.

440
(Refer Slide Time: 23:18)

441
(Refer Slide Time: 23:30)

Figure 6.
Figure 6 shows a plot of theoretical efficiency versus pressure ratio for the Brayton cycle. As the
pressure increases, the theoretical efficiency increases. We know that the efficiency of such a
cycle cannot be 100 %. Figure 6 also shows overall pressure ratios employed in the actual
engines on aeroplanes over the years. We can see that with time, attempts are being made to
increase the pressure ratio, and hence, the efficiency of the Brayton cycle.

442
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 87
Tutorial problem (1 number)
(Refer Slide Time 00:15)

Figure 1.

Solution of the problem in Fig. 1:

The Brayton cycle is shown on a T-s diagram in Fig. 1.

We have, ( ) and ( )

Hence, ( ) = temperature at the end of the compression process

( ) = temperature at the end of the expansion process.

We can also find turbine work output , compressor work input , amount of heat
added , amount of heat rejected and efficiency (η).

443
η

(Refer Slide Time 05:00)

(Refer Slide Time: 05:38)

444
(Refer Slide Time: 05:49)

(Refer Slide Time 10:56)

445
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 88
Thermodynamics cycles: Vapor compression refrigeration cycle
(Refer Slide Time: 00:12)

Figure 1.

Let’s study the refrigeration cycle, also known as the vapour compression refrigeration
system cycle (VCRS).

Figure 1 shows the back of a household refrigerator which implements the VCRS cycle. The
system here consists of 4 components: a compressor, a condenser, an expansion valve and
evaporator.

446
(Refer Slide Time: 01:32)

Figure 2.

Figure 2 shows the schematic of the components of the VCRS cycle.

The compressor takes in a refrigerant, compresses it and gives out high pressure high
temperature refrigerant. Ideally, this compression process is isentropic. The refrigerant then
enters the condenser, rejects heat and becomes liquid. The heat rejection process is a constant
pressure process. Then, the refrigerant enters the expansion valve (or a throttle valve). The
pressure as well as the temperature of the liquid refrigerant drop significantly in the
expansion valve. We are left with the mixture of liquid refrigerant and its vapour at the exit of
the expansion valve. This low pressure low temperature mixture, then, enters the evaporator,
where it absorbs heat from the substances inside the refrigerator which we want to cool, and
becomes saturated vapour. This saturated vapour enters the compressor and the cycle
continues.

447
(Refer Slide Time: 03:35)

Figure 3.

Figure 3 shows the processes discussed above on the T-S diagram. The process1-2 happens
inside the compressor. The process 2-3 happens inside the condenser. The process 3-4
happens inside the expansion valve (or a throttle valve). The process 4-1 happens inside the
evaporator. The compressor needs the work input. Heat is rejected in the condenser, whereas
heat is absorbed in the evaporator. The throttling process in the expansion valve is highly
irreversible. There is no work interaction associated with it. Ideally, the throttling process is
adiabatic (we discussed this before).

Ideally, we would like to have a reverse Carnot’s engine (refrigerator) so that we have the
maximum COP possible between the given heat source and heat sink. As discussed before,
we cannot have isothermal heat transfer when we are dealing with a single phase of a
substance. Hence, the processes where heat transfer takes place, processes 2-3 and 4-1,
should be inside the liquid vapour dome of the refrigerant used, where the working substance
is a mixture of liquid and vapour, and isothermal heat transfer is possible. Also, as pointed
out before, a compressor cannot compress a mixture of liquid and vapour without getting
damaged. Hence, the compression process 1-2 lies outside the liquid-vapor dome on the
vapor side (Fig. 3). Ideally, the compression is isentropic. The compression happens till the
state 2. The entire heat rejection process 2-3 cannot happen isothermally as some part of it
lies in the single phase region on the vapor side of the dome (Fig. 3). As said before, it is

448
difficult to have isothermal heat transfer with a single phase of a substance. However,
isobaric heat transfer is possible with a single phase of a substance. Hence, the states 2 and 3
lie on an isobar, and we have isobaric heat rejection in the process 2-3. The temperature at
state 2 is higher than the temperature at state 3. The process 3-4 is an expansion process. We
could have done this expansion in a turbine and get some work output. However, as
mentioned before, a turbine cannot work with a mixture of liquid and vapor without getting
damaged. Even then, if we decide to employ a turbine for this expansion process, the work
output we get will be significantly small. The process 3-4 is quite close the saturated liquid
line. Hence, the specific volume of the liquid-vapor mixture which is going to expand in the
turbine is small. If we consider isentropic expansion in the turbine, the magnitude of the work
output through expansion from the condenser pressure to the evaporator pressure is given by
|∫ |. Since the specific volume is small for the process 3-4, the work output is also small.
Also, the turbine is a costly device. This will increase the cost of a refrigerator significantly
even though the work output is very small. Hence, instead of a turbine, the expansion process
happens through a throttling valve or a throttling tube. This tube is small in diameter and
quite long. There is a large pressure drop as the refrigerant passes through the tube because of
viscosity (friction). If the refrigerant flows through the tube fast, the process can be assumed
to be adiabatic. We had also concluded that the throttling process is an isenthalpic process
based on some assumptions. If the conditions are properly chosen, then along with pressure,
the temperature of the refrigerant also drops significantly. The process 3-4 is an irreversible
process which is represented by a dotted line on the T-S diagram in Fig. 3. The process 4-1
happens inside the evaporator where the refrigerant takes in heat, evaporates and enters the
compressor as saturated vapor. In this way, the VCRS cycle continues. It runs in an
anticlockwise direction on the T-S diagram unlike the Rankine cycle.

The main processes of the VCRS cycle shown on the T-S diagram in Fig. 3 are mentioned
below:

1. Process 1-2: isentropic compression of the refrigerant in the compressor

2. Process 2-3: constant pressure heat rejection by the refrigerant in the condenser

3. Process 3-4: irreversible expansion of the refrigerant in the throttling valve (or tube)

4. Process 4-1: constant pressure heat addition to the refrigerant in the evaporator

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Let’s call the heat rejected QH, heat added QL and the work input to the compressor Wc.
There is no work interaction associated with the expansion valve. Applying the first law to
the cycle, ∮ ∮ | | | | | |. We also know that the
work input to the compressor on a unit mass basis is given by . Hence,
( leaves the system and enters the system [the four components of
the VCRS cycle and the refrigerant inside form the system)]. The expression
can be obtained using the first law for a control volume. We have discussed this before.

If we apply the first law for a control volume to the condenser (heat exchanger) and make
assumptions as discussed in the previous lectures, we get,

| | | |

Similarly, for the evaporator (heat exchanger),

The coefficient of performance, COP, for the refrigerator is , whereas the

COP for the heat pump is .

450
 Thermodynamics
Professor Anand T N C
Department of Mechanical Engineering
Indian Institute of Technology, Madras
Lecture 89
Tutorial problem (1 number)
(Refer Slide Time: 00:12)

Figure 1.

Solution of the problem in Fig. 1.

The VCRS cycle is shown on the T-S diagram in Fig. 1.

Compression work =

Refrigerating effect = heat received by the refrigerant in the evaporator =

Heat rejected in the condenser =

COPVCRS = | |

Cycle efficiency = ( is the COP of the reverse Carnot’s cycle

operating between the same two temperature reservoirs as the given VCRS cycle)

We can obtain enthalpy values in the above formulae from the tables for the refrigerant
R134a. R134a stands for 1, 1, 1, 2- tetrafluoroethane.

451
Different tables have different fixed reference points. Properties at other conditions are
measured with respect to this reference point. At the same condition, the properties may be
different in different tables because the reference points are different for different tables.
However, the difference between a property at two different conditions is the same
irrespective of the table used. It is recommended that the students continue using one table
throughout.

We need enthalpy values at states 1, 2, 3 and 4.

We have saturated vapor at state 1. Hence, ( )

We know that states 2 and 3 lie on an isobar. State 2 is in superheated zone. At 30 , the
saturation pressure psat=771 kPa. The process 1-2 is isentropic. Hence,
( ) . Now, we need enthalpy (h2) at the state where p=771 kPa and

s=1.7308 .

If we have table with better resolution for pressure and temperature values, it would be easy

to get enthalpy value at p=771 kPa and s=1.7308 , which is in the superheated zone.

However, we do not have such table here. In such a case, we can do interpolation and
calculate values at the required condition or use the value at the condition which is closest to
the required condition and listed in the table. In the table we have for superheated refrigerant,

at p=750 kPa and T=35 , the entropy value is very close to s=1.7308 . Hence, we will

use enthalpy value at p=750 kPa and T=35 , which is h2 = 420.8 kJ/kg.

At state 3, we have saturated liquid at 30 . Hence, h3 = ( )

Throttling is an isenthalpic process. Hence,

Now, we have the values of enthalpy at all the states.

Hence, Compression work =

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Refrigerating effect = heat received by the refrigerant in the evaporator =

Heat rejected in the condenser =

COPVCRS = | |

For calculating cycle efficiency, we need .

(This is the maximum COP

achievable between the temperature reservoirs at -5 and 30 .)

Hence, cycle efficiency =

Hence, the VCRS cycle in the problem is around 81 % as efficient as the Carnot’s cycle. In
reality, the COP of the VCRS cycle is even lower as the process in the compressor is also
irreversible.

(Refer Slide Time 05:27)

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