Hello friends. Welcome back for the third lecture of this
particular week where we are reviewing our principles of basic thermodynamics with the
focus of setting up for the concept which we may have to use in the subsequent weeks. Now, in the first and second lecture, we have
discussed about some very fundamental aspects like the concept of thermodynamic system,
property, state, equilibrium. Then, we discussed about the different laws
of thermodynamics like in the first week or I should say in the first lecture, I mentioned
about the zeroth law of thermodynamics which gave us the concept of the property named
temperature and in the previous lecture, I mentioned about both first and second law
of thermodynamics. Now, from the first law of thermodynamics,
we got the concept of internal energy or I should say total energy of a system is a property. And we have also seen how to apply the first
law principle for both closed and open system analysis. The second law was also introduced where generally
we talk about 2 different statements, the Kelvin-Planck statement, which is the suitable
one for heat engines whereas the Clausius statement which is a suitable one for heat
pumps and refrigerators or in one word the reverse heat engines. Now, while the analysis of a system following
the first law of thermodynamics is quite straightforward that is we just have to go for the balance
of energies which are either coming to the system or going out of the system or in we
can say that whatever energy interactions are taking place within the system and surrounding,
we just have to take a balance of that and that should give you an idea about the change
in the total energy content of the system. However, for second law of thermodynamics
which we have to, we do not have such straightforward approach; we generally have 2 kinds of approach
to choose from, one based upon the property called entropy, other based on the concept
called exergy. So, in today's lecture, I am going to talk
about entropy, introduce the property entropy to you and see how we can calculate entropy
for a system. And in the next lecture, which is going to
fourth and last one for this week, I shall be introducing the property exergy and then
we shall be seeing how to analyze a thermodynamic system following the second law of thermodynamics
based upon the exergy approach. So, let us start with our discussion on entropy. Now, in the previous lecture, we have seen
that there are 2 different statements of thermodynamics. And depending on what kind of system we are
talking about, we can choose any one of them. And we know that though the statements look
quite different, they are actually equivalent to each other that is any violation of the
Kelvin-Planck statement leads to violation of the Clausius statement or vice-versa. If you are not sure about how to prove that,
you can refer to any standard thermodynamics book. Now, the first term that I would like to introduce
today which comes straight as corollaries or consequence of the second law of thermodynamics
is the reversible process. Now, the reversible process refers to a process
that can completely be reversed without leaving any trace of proof on the surrounding. It means we are talking about changing the
direction of a process without keeping any mark on the surrounding. If we draw a very basic property diagram,
let us say I draw a property diagram where we are using some properties a1 and a2 based
upon which we are plotting. This a1 and a2 can be any properties of our
choice like pressure and specific volume, pressure and temperature or maybe something
else and let us say this is our initial point 1 and this is state point 2. So, this is our state point 1 and state point
2. So, we are having a process somewhat like
this which has taken place. So, during the process, all properties are
changing if we talk about any standard extensive property say X, then the change in this extensive
property during the process will be equal to X2-X1 and also certain interactions are
taking place between system and surrounding during this. Let us say this is our system, so during this
forward process, the system receives this amount of heat from the surrounding and gives
this much of work output. So, this is the scenario during this process
1 to 2. Now if this process 1 to 2 is a reversible
one, then we can reverse its direction thereby moving from this so-called final state back
to the initial state without leaving any mark on the surrounding. That we say now we are talking about while
the points remain the same 1 to 2, we shall be following exactly the same path but now
the direction will be opposite going back from 2 to 1. Now, what about the change in the properties? Properties are point function that is whatever
maybe the nature of the process, the final change in the property will always remain
the same. That is in the second process, the change
in the property is here we are changing the same property X will be from, will be again
equal to X1-X2, properties being point functions we do not need to bother about them. But we are talking about without leaving any
trace of proof on the surrounding or mark on the surrounding that will be coming in
terms of whatever interactions that are taking place. That is if the process has to be completely
reversed without giving any mark on the surrounding, then it would rather same process. Then, in the first case we have moved from
1 to 2. Now, we are going back from 2 to 1 and to
make this process perfectly reversible, then during the second process the system should
reject the exactly same amount of heat del Q to the surrounding and receives exactly
same amount of work del W from the surrounding. Then, only whatever energy interactions that
has taken place during the forward process 1 to 2, we can completely reverse all those
energy interactions during the backward process. And thereby getting both system and surrounding
back to its initial condition. If there is any mass transfer involved, let
us say del m amount of mass also moved in during the forward process. Then, during the reverse process, exactly
the same amount that is del m amount of mass must move back from the system to the surrounding. So, when we are performing the reverse process
that is going back from 2 to 1, the system is always restored. Because we are back to the initial point,
so all the system properties will go back to its initial value. However, the surrounding may not be restored
if the process is not a reversible one. Because just think about during the forward
process; if we neglect that mass transfer for the moment, then how much is the net energy
that system has received? So, delta E for the system is, it has received
del Q amount of heat and given a del W amount of work. Now, delta E for the surrounding how much
is that? If we consider to have just two systems, one
is that so-called system, other is the surrounding. Then, for the surrounding how much is energy
interaction? It has received del W amount of work and given
a del Q amount of heat. So, once we are performing this reverse process,
the system is coming back to its initial state. So, all system properties are restored. However, to restore the surrounding also back
to its initial condition, we must take away this amount of heat from the surrounding so
that during the reverse process, the energy interaction corresponding to the surrounding
is exactly=del Q-del W thereby restoring the surrounding as well. So, we can identify system to be reversible
only when during the reverse process both the system and the surrounding has been restored
back to its initial condition. Otherwise, it is not a reversible one. The most common situation is the system gets
restored to the initial situation but the surrounding does not and thereby the process
is not a reversible one what we call is an irreversible process. Truly speaking, all practical processes are
irreversible in nature because this particular condition that is without leaving any mark
or trace or proof on the surrounding that is extremely difficult to achieve. Let us take an example. Let us take a cup of tea. So, we have taken a cup of tea, let us say
we have a paper cup which is filled with some tea and now I have dropped that cup to the
floor. So, as I drop the cup to the floor then what
will happen? The tea the liquid that I had in the cup that
gets splashed everywhere that is poured on the floor and there may be scattered drops
also all around. So, if we want to now reverse that process,
then what should happen, that tea which is now scattered everywhere that should come
back to the cup and that cup filled with tea should come back to my hand. Now, that is practically looking quite impossible
and that is why irreversible or reversible processes, the term is only an idealization;
all real processes are irreversible in nature. Now, let me just erase a bit but despite mentioning
that no real processes are reversible, then why you are so much interested with this term? Because it can we proved that and you must
have learnt that in your basic thermodynamics that whenever a heat engine is undergoing
a reversible process, then it is expected to produce the highest possible work output
among all the engines operating between the same two reservoirs. Similarly, when we are talking about energy
absorbing device like a compressor or a pump. If a reversible reversed heat engine consumes
work, we are working on a reversible process then it will consume the least possible work
input among all the reversed engines operating between again the same two reservoirs. So, once we fix up the two reservoirs that
is the upper and lower temperature limits of the operation. Then, for a reversible heat engine you are
going to get the maximum possible work output and from a reversible heat pump or refrigerator,
we need to provide the minimum possible work input. So, that is the most ideal process that we
would always like to achieve. That is why we are very much interested in
a reversible process because reversible process sets up the upper limit of operation both
in case of heat engines and reversed heat engines. But practically there are several hindrances
because of which a process can move away from the reversible nature and the most common
causes are friction because friction is a kind of irreversible effect. Like let us suppose if I have, I move my right
hand over my left one, then what will happen? As I am moving forward there is some kind
of rubbing of friction action that is taking place and because of which some energy is
being lost. So, if I want to reverse this action completely,
then what should happen? Whatever amount of energy that I have lost
because of friction which eventually gets converted to heat that should come back to
my hand thereby enabling my hand to move back into the original position but that never
possible because you know that whenever hand is moving in this direction, friction is acting
opposite to motion. Similarly, when I am moving the hand back
into the original position, again friction is opposing the motion. It is always acting opposite to the motion
and immediately converting that corresponding mechanical energy or lost mechanical energy
to thermal energy thereby making it possible to recover that lost energy. So, friction is the biggest source of irreversibility. Under certain situation, we may have unrestrained
or uncompressed or I should say unrestricted expansion of certain gases, in very specific
situation you may get that. But the third one is a very common source
again, heat transfer through a finite temperature difference. From the Clausius statement of the second
law of thermodynamics, we know that heat can move only from a high temperature body to
a low temperature body. And to have any kind of heat transfer, we
need to have a certain amount of temperature difference between the two bodies. If both the bodies are at the same temperature,
then we can never have any kind of heat transfer because heat transfer requires the temperature
difference as its driving potential. If the temperatures are same, there is no
driving potential so there will be no heat transfer. However, as heat can flow only from high temperature
body to a low temperature body and the reverse is impossible therefore that is also a kind
of irreversibility. Now, the first two friction and unrestrained
expansion or primarily friction are generally referred as the reasons for internal irreversibility. That is if the system is frictionless and
there is no unrestrained expansion, then we can call the system to be internally reversible. Similarly, when there is no heat transfer
through a finite temperature difference, then we call the system to be externally reversible. It is possible for a system to be internally
reversible but externally irreversible and similarly a system can be internally irreversible
because of the presence of friction but maybe externally reversible. Let us take this example here. So, here I have got a system which is kept
at 20 degree Celsius. Now, if I want to transfer some heat from
the surrounding to the system following a reversible manner, then the surrounding has
to be at the same temperature, but if it is at the same temperature then there will be
no heat transfer. So, we can assume that surrounding is at an
infinitesimally small temperature difference higher than the system thereby facilitating
this heat transfer. So, here the heat transfer is taking place
because of an infinitesimally small temperature difference between system and surrounding. So, it is a totally reversible process, of
course assuming there is no friction, it is externally reversible. Now, look at the second example where the
surroundings at a much higher temperature, there is a distinct temperature difference
existing between system and surrounding. However, inside the system temperature is
same everywhere that is system itself is in thermal equilibrium. This is the situation of internally reversible
but externally irreversible process because there is a heat transfer taking place with
finite temperature difference 30 degree on one side, 20 degree on the other side, there
is a finite temperature difference of 10 degree Celsius and this heat transfer is taking place
because of the finite temperature difference, so the system is externally irreversible but
internally reversible. We can also have the other example when a
system is internally irreversible because of the presence of friction, but externally
reversible. Externally reversible is possible when this
heat transfer is taking place because of extremely small temperature difference or if there is
no heat transfer at all. If there is no heat transfer at all, then
this point never comes into picture. So, there also the system can be internally
sorry externally reversible. Therefore, for a system to be externally reversible,
we need to have either an adiabatic process that is no heat transfer or an isothermal
heat transfer process during which system and surrounding are at the same temperature,
that takes us to the concept of a reversible cycle. Practically, all power producing or power
consuming systems has to work on certain cycle, so that they can keep on repeating the same
action over a long period of time. And as we are going to get the largest possible
work output from a reversible process, so a cycle which works only using reversible
cycles sorry reversible processes is expected to give the largest possible work output and
that is the interest of talking about a reversible cycle. So, a cycle is called a reversible if all
the constituent processes are reversible in nature that is, they are frictionless and
none of the processes involved heat transfer with finite temperature difference. The most common cycle that we talk about as
a reversible one is a Carnot cycle, which is again an idealization but it is possible
to have any other kind of reversible cycles also. In a Carnot cycle, we generally consider four
processes to constitute the cycle. Process number 1, which is from this process
or point 4 to 1, this is an adiabatic process and frictionless adiabatic process. So, during this process, it is frictionless
so internally reversible and also adiabatic so externally reversible. Therefore, this 4 to 1 is a perfectly reversible
process or totally reversible process. Next process 1 to 2, 1 to 2 refers to an isothermal
heat addition process, so QH amount of heat gets added to the system during which its
temperature remains constant at TH, 2 to 3 is again an adiabatic process during which
the temperature reduces from this high temperature TH to low temperature TL and 3 to 4 another
isothermal heat rejection process. So, the first process is an adiabatic one,
it is a frictionless adiabatic. Similarly, third one, this is also a frictionless
adiabatic process. This second process 1 to 2 is an isothermal
process performed at the temperature TH. Similarly, this fourth process 3 to 4 is another
isothermal process performed at the temperature TL during which QL amount of heat is rejected
to the surrounding. So, this is how we talk about the Carnot cycle,
now in a Carnot cycle then how we can calculate the efficiency of the cycle. So, you know that efficiency for any heat
engine can be written as network output/heat input or I should say cyclic integral of del
Q, now cyclic integral of del W can be written as cyclic integral of del Q/QH. So, QH is the amount of heat added to the
system and QL is the amount of heat rejected that is 1-QL/QH which is true for any heat
engine. Now, somehow you have to establish a relation
between this QL/QH to this TL/TH and then only we can represent the efficiency of this
Carnot cycle in terms of temperatures only and in this context, couple of Carnot principles
comes into picture. The first Carnot principle says that the efficiency
of an irreversible heat engine is always less than the same for a reversible heat engine
operating between the same 2 reservoirs. It means suppose if we decide 2 reservoirs
one at temperature TH and another at temperature TL and heat engine operating between the two
getting suppose QH amount of heat from the high temperature body rejecting QL to the
low temperature body and producing W amount of work. Then, it is saying that once we have fixed
up these 2 reservoirs, then whatever may be the nature of this heat engine, whatever may
be the magnitude of QH, W and QL, the efficiency of any reversible heat engine working between
these 2 reservoirs will always be greater than any irreversible heat engine that is
a reversible heat engine is going to give the maximum possible efficiency. Second Carnot principle talks about the efficiency
of all reversible heat engines operating between the same 2 reservoirs are the same, means
again we are fixing up 2 reservoirs, one at temperature TH, other at temperature TL. Now, you are talking about two engines, both
are reversible in nature. So, engine 1 is giving you W amount of work
output, engine 2 is giving W2 amount of work output. Both 1 and 2 are reversible nature but they
are having different kinds of heat interaction producing different amount of work. Now, the second Carnot principle says that
efficiency of 1 and efficiency of 2 has to be equal and this whatever may be the nature
of the reversible cycle as long as that is reversible, that efficiency will be always
the same. What does it indicate? So, once we specify the two end temperatures,
the efficiency does not change, whatever may be the magnitude of heat drawn by the system
or whatever may be the working fluid, you will always get the same efficiency. That means this has to be a function of these
2 temperatures TH and TL. Once you fix up these two temperatures, the
efficiency of any reversible heat engine is fixed. From there, we get the concept of a thermodynamic
temperature scale and using the Joule postulate, it can be shown that QL/QH comes to be TL/TH
that is the ratio of heat transfer can be directly equal to the ratio of corresponding
temperatures and accordingly we get efficiency for any reversible heat engine, Carnot cycle,
or any other kind of reversible heat engine, it is 1-T low/T high. So, once you know these two temperatures,
you can calculate the efficiency of any reversible heat engine. The same principle is also applicable to reversed
heat engines. So, like COP for a reversible heat pump, we
know COP for any heat pump is QH/QH-QL, using this equality you can easily relate this one
as TH/TH-TL and similarly COP for a reversible refrigerator, which is generally QL/QH-QL
that will come as TL/TH-TL. Say for example, if we consider a situation
of a heat engine which is operating between the temperatures where TH=1000 Kelvin and
TL=300 Kelvin. Then, without knowing anything else we can
say that the efficiency of this reversible heat engine or the maximum possible efficiency
following this principle once we have fix up these 2 temperatures, the maximum possible
efficiency any heat engine can give is 1-TL/TH that is 1-300/1000 that is 0.7. So, that is the maximum possible efficiency
a heat engine operating between these two temperatures can give you. If the engine is a reversible one, you are
going to get the 70% efficiency. If the engine is an irreversible one, your
efficiency will be much lesser and how much less it will be that depends upon the level
of irreversibility is present in the system. Similarly, if we talk about the example of
a refrigerator, so I give you refrigerator which is operating between the temperatures. On the lower side, the temperature is say
280 Kelvin; on the higher side, the temperature is 300 Kelvin. Then, the maximum possible COP this refrigerator
can have is the one corresponding to its reversible version which is 280/300-280, so 280/20 that
is=14. So, this way we can calculate the upper limit
of efficiency or COP for a heat engine or reversed heat engine respectively once the
two temperature limits are given. Now, let us use this concept to derive another
important principle known as the Clausius inequality for which we consider a heat engine. This TH and TL are the two temperatures, we
have the heat engine which is drawing QH amount of heat rejecting QL amount of heat and producing
W amount of work. This heat engine can be reversible, can be
irreversible, we do not care. Now, if we apply the first law of thermodynamics
on this over a cycle, we know that cyclic integral of del Q=cyclic integral of del W.
Now, amount of heat it is taking is QH-QL=W. So, cyclic integral of del Q, how much is
that? That is QH is amount of heat received by this
system and QL is amount of heat rejected by the system. Now, which one is higher? QH definitely is higher than QL; otherwise
there will be no W. So, this cyclic integral of del Q is greater
than 0. Now, we take the quantity, cyclic integral
of del Q/T, so how much will be used? The T is the absolute temperature so that
is always a positive quantity. So, the symbol of this one will depend upon
the symbol of Q, so you have QH/TH-QL/TL because QL is negative. Now, we have just seen that following the
Carnot principle, for a reversible one, we can say that QL reversible/QH=TL/TH. So, if we use that for first, we consider
the reversible heat engine. So, reversible means we are indicating QL
as QL reversible. Then, using this particular principle, we
can write that by rearranging them, we can write QH/TH will be equal to QL reversible/TL,
which gives us the cyclic integral of del Q/T for a reversible situation to be equal
to 0 okay. Now, we take an irreversible heat engine where
QL=QL irreversible. Now, this we know that work output from the
Carnot principle, the work output in this case will be lesser than the reversible one
that is W irreversible will be less than W reversible. If QH remains the same that means that QL
irreversible has to be greater than QL reversible that is larger fraction of heat will be rejected,
only lesser fraction will be converted to work. So, if we take now QH/TH-QL irreversible/TL,
we can write this one as QH can be converted to using this principle. It can be converted to QL reversible/TL-QL
irreversible/TL sorry I should have put an equal to sign here. So, I am taking 1/TL out of this QL reversible-QL
irreversible. Now, here we have seen that QL irreversible
is greater than QL reversible, so this quantity is less than 0. That means the cyclic integral of de l Q/T
for an irreversible situation is less than 0. So, if we combine this particular one for
reversible one and this particular one for an irreversible one, then together we can
write cyclic integral of del Q/T for any heat engine is less equal to 0 where the equality
holds for a reversible one which is the limiting case. It is possible to prove the same thing for
a heat pump cycle also. Only for a heat pump, you are going to get
for, if this particular situation we have derived for heat engine, if we do the same
thing for heat pump, you are going to get cyclic integral of del-Q less than 0. However, cyclic integral of del Q/T will continue
to be less equal to 0. Therefore, we can conclude that for any kind
of system, heat engines or reversed heat engines, reversible or irreversible; we can always
write cyclic integral of del Q/T is less equal to 0. This is called the Clausius inequality which
sets up with us with the concept of entropy. So, to discuss the concept of entropy, let
us briefly consider again a property diagram where we move from a point 1 to 2 following
this particular process a and then we go back from 2 to 1 following the process b. Let us say this is a reversible cycle that
is both a and b are reversible processes. Then, following Clausius inequality we have
just derived that cyclic integral of del Q/T for a reversible cycle will be equal to 0. That means integral 1 to 2 following a del
Q/T+integral 2 to 1 following b del Q/T will be=0. Now, if we consider another cycle which is
quite similar, again we have the same two endpoints 1 and 2 and the same process ‘a’
that is from 1 to 2 we are going following the same process ‘a’. However, during return we are taking a different
route something like this a ‘c’ which is again a reversible process. So, in the first case, we had reversible cycle
comprising of ‘a’ and ‘b’, in the second case, we had reversible cycle comprising
of ‘a’ and ‘c’. So, in this case also we can use the Clausius
inequality and accordingly we can write 1 to 2 following a del Q/T+2 to 1 now following
c del Q/T=0. So, if we compare these two, then we can say
that for both the situations integral 2 to 1 del Q/T is the same. And this way, we can choose any process to
go back from 2 to 1 as long as that is reversible, the magnitude of this integral 2 to 1 del
Q/T will remain the same and therefore what does that indicate? That definitely indicates that the change
or rather the magnitude of this quantity is independent of the process path that has been
followed that is whatever may be the process path we are following between 2 to 1 as long
as that is reversible, this quantity remain the same and therefore that has to be some
property. Also, cyclic integral of some quantity to
be 0 does mean that that has to be a property and that property is called entropy, which
is generally denoted by the symbol S. So it has to denote change in property S1-S2. So, the change in any property dS is defined
as the quantity del Q/T following a reversible path. As long as you are following a reversible
path, just this del Q/T this quantity is going to give you the change in property. However, one thing you have to be careful,
here we are going from one state point to another state point that is from 1 to 2 or
2 to 1. So, regardless of you are following a reversible
path or irreversible path, the change in entropy will always remain the same because you are
talking about a property. However, the magnitude of that change in entropy,
magnitude of that property can be calculated using the heat transfer value only if you
are following a reversible path. Like take this example where we have plotted
two processes between point 1 and 2. Now, during 1 to 2, the change in entropy
is from point 3 to point 7. So, the delta S, the change in entropy following
this is S2-S1 that is equal to 0.4. Now, if we follow an irreversible process,
even in that case also, the change in entropy will remain the same because again you are
going away from the same initial state back to the same final state. So, the change in entropy which is there also
will remain to be 0.4 kilo joule per Kelvin. However, in the first case, as we are following
a reversible path, so the change in entropy in this case will be equal to this del Q/T. However, in this case, the change in entropy
will not be equal to del Q/T. So, if we are following a reversible path and we are plotting
that on a TS diagram using temperature and entropy at the two axis, then the area under
this particular curve that is this shaded portion will be giving you the amount of heat
transfer involved with this. However, that is not true for the irreversible
path. So, entropy can be defined as the property
which satisfies this particular relation that is the magnitude of this del Q/T following
a reversible path. Now, we consider another cycle which is actually
an irreversible cycle. It comprises of one irreversible and a reversible
process. So, process 1 to 2 is irreversible, in fact
that can be reversible as well but process 2 to 1 is internally reversible. Now, we are applying the Clausius inequality
on this. So, cyclic integral of del Q/T will be less
equal to 0. If 1 to 2 is reversible, then the equality
will hold, otherwise it will be that less than 0 thing. So, if we write this now, break this from
1 to 2 del Q/T+2 to 1 del Q/T the summation has to be less equal to 0. Now, this 2 to 1 is internally reversible. Therefore, del Q/T which let us is consider
to be an irreversible one and this is a reversible one, so during which as it is a reversible
process we can directly relate this one to the change in entropy and we can directly
write this one to be S1-S2 that has to be less equal to 0. That is now integral 1 to 2 del Q/T for this
irreversible path, so this has to be equal to S2-S1 or if we write formally then S2-S1
has to be greater equal to this del Q/T integration from 1 to 2. As long as we are following a reversible path
then the equality will hold and the change in entropy can directly be related to this
del Q/T quantity. However, if we are following an irreversible
path, then del Q/T alone is not sufficient rather, we need some other contribution. That is in differential if we write dS greater
equal to del Q/T. So, informally if we want to avoid this inequality sign, we can write
this onto be dS=del Q/T+del S generation, which relates to the entropy generation. So, during the irreversible process, we can
see that entropy change cannot be related fully to this term but there will be some
additional contribution coming from the entropy generation which actually is a result of all
the irreversibilities that are present inside the system. For a reversible process, we do not need to
bother about this particular part; however, when we are following an irreversible process,
then this entropy generation will always be there and that itself a very important principle
known as the increase in entropy principle. So, even if we are talking about a reversible
process, then there is no entropy generation; however, entropy generation is always a positive
term. At the limiting case of a reversible process,
it can be 0 and therefore during any real process the entropy of the system or entropy
of the system surrounding combination will always keep on increasing. Like let us take the example. So, this is our system, outside we have the
surrounding and during this del Q amount of heat moves from system to the surrounding. So, if suppose T1 is a system temperature,
T2 is a surrounding temperature. Then, entropy change for the system will be
how much? It will be equal to del Q/T1+entropy generation
for the system and entropy change for the surrounding will be del Q/T2+del S generation
for the surrounding. Now, if the limiting case of T1+T2 or rather
T1=T2, how much is the net entropy generation? If we combine these two, dS net will be equal
to, now there is one thing I have missed. Here del Q is positive for the system because
it is receiving heat but it is negative for the surrounding. So, the net entropy generation or net entropy
exchange here is del Q*1/T1-1/T2+del S generation for the system+del S generation for the surrounding. Now, as heat is being transferred from surrounding
to system, T2 has to be greater than T1 thereby getting this quantity to be positive. In the limiting case, when T1 and T2 are equal
that is we are talking about an isothermal heat transfer, this quantity goes to 0. However, this entire quantity is always positive
and thereby this net change in entropy is greater than 0. This is what we call the increase in entropy
principle that is increase sorry the entropy of the system surrounding combination or entropy
of the universe is what we tell in formal thermodynamics term, this always increasing. However, for the system entropy can always
decrease but we are here talking not only about the system, not only about the surrounding,
we are talking about the system surrounding combination which we refer as a universe. So, entropy of the universe is always increasing
because of this entropy generation. So, with this we can have a few important
conclusions. One, real processes can occur only in a certain
direction, which gives to a positive entropy generation or in the limiting case zero entropy
generation for a reversible process. Negative entropy generation is never possible
and therefore any hypothetical thermodynamic process if that indicates negative entropy
generation that is impossible. Secondly, entropy is a non-conserved property
because of the presence of these inequalities and also the presence of this inequality here. So, entropy is a non-conserved property, so
there is nothing like conservation of entropy. We talk about conservation of mass, momentum,
energy but not conservation of entropy. We can only write an entropy balance equation
where we incorporate this entropy generation term but conservation of entropy is never
possible and third, presence of irreversibilities degrade the performance of engineering systems
and this entropy generation gives us an idea to measure such irreversibilities. More about this third point means how to calculate
the irreversibilities using the concept of entropy generation we shall be discussing
in the next lecture where we talk about exergy. Let us quickly do one calculation. Here, I am presenting you with two situations;
in situation ‘a’ you have a source at temperature 800 Kelvin and sink at temperature
500 Kelvin and exchanging 2000 kilo joule of heat. Then, how much is the entropy transfer in
case number ‘a’? So, in case number ‘a’ entropy transfer
associated with the heat transfer, so heat is going away from the source. So, it will be –del Q/T source+del Q/T sink
because it is receiving heat. So, if you take del Q common which is the
same which is 2000 kilo joule*-1/800+1/500. So, I have not calculated the numbers but
you can try to simplify this if I neglect the 0s. So, I have 20, actually I should not have
written kilo joule here. I should put the unit at the end, which is
kilo joule per Kelvin. So, it is 20*1/5-1/8 that is 20*3/40 that
is 3/2, so 1.5 kilo joule per Kelvin. Now, in case b, delta Sb will be following
the same principle 2000 kilo joule of energy is being exchanged, source temperature is
the same -1/800 but sink temperature is now 750 Kelvin. So, kilo joule per Kelvin, so taking it common
if I write this way, so it is 1/7.5-1/8, so you please calculate this, I do not have a
calculator with me now but you please calculate this one and surely we say that in this situation
the value will be coming lesser than the previous case that is you are going to get del Sb less
than delta Sa. So, definitely the second situation is more
irreversible because here the, this is more irreversible because the temperature difference
involved here is of 300 K whereas here it is only 50 K, so much lesser temperature difference
and hence much lesser level of external irreversibility. In the first case, we are having heat transfer
over 300 Kelvin temperature difference; there is only 50 Kelvin in the second case. Therefore, despite this amount of heat transfer
remaining the same, we are having lesser amount of irreversibility in the second case. So, quickly we shall be checking out if your
property relations involving entropy which we have to make use of in the following lectures
in the next week itself. So, here we are talking about one closed system,
then we consider it to be stationary, then we consider a simple compressible system so
that there is no additional effect like gravity, magnetism, electricity, etc and also, we consider
an internally irreversible sorry internally reversible process. So, now if you write the first law of thermodynamics
for this, we can write dU because it is a stationary system. What I have written here? Okay let me correct this, stationary. So dU=del Q-del W. Now, from the definition
of entropy as we are dealing with an internally reversible process, so del Q can be written
as TdS and as we are talking about the simple compressible substance so del W can be written
as PdV that is we can write sorry let me write this way. We can write TdS=dU+PdV or dividing everything
by mass, we can write Td small s=d small u+Pd small v. This is called the first T-dS equation or
the Gibbs equation. Again, we know that h=u+Pv or dh=du+Pdv+vdP. So, if we make use of that in the previous
T-dS equation, we can write TdS=du+Pdv is replaced by dh-vdP. So, this is the second T-dS equation or second
Gibbs equation. These two are very important equations; remember
the conditions for which you have done. We have considered closed stationary system,
a simple compressible one and internally reversible. In that case, we can relate the properties,
internal energy or enthalpy with entropy following this way thereby providing as a way of measuring
the entropy because like if we take the first equation, we can write ds=du/T+P/T dv. And from the second one, we can write dh/T-v/T
dP. So, if we know the PVT relationship between
the substances that we are talking about and also somehow if we can relate the u or h which
temperature, we can calculate the change in entropy. Let us take the example of an ideal gas. So, this is the equation that we have just
written. So, for ideal gas what you know? We know Pv=RT and also for ideal gas, we can
write du=Cv dT and dh=Cp dT. You may have already used this relation but
in next week when we talk about the thermodynamic property relations, this one I shall be developing
again. For the moment, you just take this one for
granted. So, if you put this in the first equation,
then we have ds=Cv dT/T+P/T can be replaced with R/v so R dv/v. That means S2-S1=integral 1 to 2 Cv dT/T+R*integral
R being a constant because there is a gas constant integral 1 to 2 dv/v. Similarly,
from the second equation, we can write ds=Cp dT/T-R v/T can be replaced by dp/P which gives
us the second equation S2-S1=integral 1 to 2 Cp dT/T-R 1 to 2 dP/P. So, this is our equation
number 1, this is equation number 2 and depending upon what kind of information are available
to you, if you know the change in pressure and information about Cp then we use second
one. If we know the change in volume and information
about the Cv then we use the first one. There are generally two kinds of approaches
because Cp and Cv generally functions of temperature. So, there are two approaches, the first one
is called approximate approach. In approximate approach, we consider an average
value of Cp and Cv and proceed accordingly. So, if we take the second one, there we shall
be having according to the approximate approach, S2-S1 will be equal to some Cp average*ln
T2/T1-R ln P2/P1 following this particular equation and the second approach is exact
approach where we consider the temperature dependence of Cp while performing the integration
and in that case we represent this S2-S1 as S20-S10-R ln P2/P1 where this S0 at any particular
temperature is defined as integral absolute zero temperature to T Cp dT/T. So, once you know Cp as a function of T, maybe
a polynomial equation, maybe some other form you can perform this integration and integrated
from 0 to capital T to get the value of S10 and S20. Thankfully, standard values of this S0 are
available for different temperature levels in any standard textbooks, you can refer to
the appendix of your textbooks, the book of Cengel and Boles or the book of Sonntag you
will have the table of this S0. So, to round up the day, we shall be checking
one case where we make use of this entropy calculation following both exact and approximate
approach. So, our situation is air is compressed from
an initial state to a final state. So, we are given with P1=100 kilo Pascal,
T1=290 Kelvin. It is compressed to a final pressure P2 of
600 kilo Pascal and T2 of 330 Kelvin. Actually, it is a quite small temperature
range that we are talking about. So, we have to calculate the entropy change
of air assuming to ideal gas. So, if you follow the approximate approach,
then we have to choose an average value of Cp. Let us take Cp average for air to be=1.006
kilo joule per kg Kelvin. In that case, S2-S1 will be=Cp average ln
T2/T1-R ln P2/P1, you can put the numbers here and according the value that you are
going to get is equal to -0.3842 joule per Kelvin sorry kilo joule per the specific entropy
so kg Kelvin. You can see, here it is negative means entropy
of the system is decreasing. Now, if we follow the exact approach, then
we need the value of S0. So, S0 here S10 means S0 corresponding to
that 290 Kelvin. Let me write it in a better way, so S10 that
is S0 corresponding to 290 Kelvin. I have taken the value from that table; I
will provide you the table later on. So, here the value is 1.66802 kilo joule per
kg Kelvin and S20=S0 corresponding to 330 Kelvin which is 1.79783 kilo joule per kg
Kelvin. So, if you take both these two into account,
then your S2-S1=S20-S10-the second part remains the same P2/P1 which will be coming as -0.3844. So, in this particular case, you can see that
the derivation is extremely small because actually we are talking about only 40 Kelvin
temperature difference. However, you can try to solve the same problem
assuming the final temperature T2 to be something like say 600 Kelvin. Try to solve the same problem using this and
you will find that approximate approach and exact approach are giving widely different
result and that shows when to apply which approach that is when you are talking about
a quite small change in temperature we generally go for the approximate approach assuming an
average value of the specific heat. However, like in several kinds of gas turbine
or IC engine applications, the temperature variation can be very significant. Then, we have to consider the exact variation
is specific heat and so we have to go for the exact approach but as long as the temperature
difference involved is limited to something less than 100 Kelvin, approximate approach
you can easily go for. So, that takes us to the end of today's lecture
where we have talked about the reversible process and irreversibilities, then the reversible
cycle with Carnot cycle as the example of a reversible cycle, then we derived the Clausius
inequality which is formally given as cyclic integral of del Q/T to be less equal to 0
for any kind of system. The equality holds for the limiting case of
a reversible process. Then, we introduced the concept of entropy
and entropy generation. So, we have seen that during any process,
the change in entropy can be written as integral del Q/T+or just del Q/T+entropy generation,
I should write capital S in this case. For a reversible situation, the entropy generation
is not there, then the change in entropy can directly be related to del Q/T and finally
we talked about the T-dS relations, how to calculate entropy using the T-dS relations
and discussed the specific case of ideal gases. So, that is it for the day. In the final lecture of this week, I shall
be talking about the concept of exergy and available work where we shall be using this
concept of entropy calculation, T-dS relations probably and also the irreversibility calculations
to get a complete second law analysis of both closed and open systems. Till then, you revise this lecture and if
you have any query please write back to me. Thank you.
