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[noise]
Welcome back ah in this lecture we will be
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ah ah reviewing the concept of entropy. So,
let me start with the inequality of clausius.
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Let us consider first ah reversible heat engine
case, which we have discussed in our last
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lecture [vocalized-noise], this is nothing
but a Carnot heat engine. For Carnot heat
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engine; we know that ah QH by QL is equal
to TH by TL and as well as we know that ah
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for a cyclic ah process, your delta Q is nothing,
but QH minus QL which is going to be greater
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than 0 [vocalized-noise].
Now, based on this reversible type of concept,
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you can write it in this way where the del
Q by T for a cycle [vocalized-noise] is nothing,
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but QH by TL minus QL by TL and that has to
be 0. So, for a reversible heat engine [noise]
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[vocalized-noise] you have a condition that
this must be equal to 0 ok.
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Del Q by T should be equal to 0 [vocalized-noise],
in general whether it is a reversible or irreversible
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for all cycle [noise] we can write in this
way [vocalized-noise] and this is nothing,
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but your inequality of clausius. So, in order
for a process to occur this particular inequality
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must satisfy [vocalized-noise] ok and this
was proposed by clausius and [vocalized-noise]
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is valid for all cycles, whether is reversible
or irreversible for both the cases it will
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be valid [vocalized-noise].
Now let us consider Process processes which
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changes the state condition on a from 1 to
2 ok and this is drawn on a P v diagram [vocalized-noise].
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So, if you consider reversible cycle A and
B [vocalized-noise] ok which a could be totally
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or internally reversible [vocalized-noise],
we can write down the cyclic ah value of d
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ki d Q by T equal to 0, because of the reversible
cycle and if you are following 1 to 2 or A
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plus 2 to 1 on B that could be one particular
cycle [vocalized-noise]; similarly you can
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take another cycle and you can write the same
expression but now on C and B. Using 2 relation
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you can show clearly that [vocalized-noise]
this [noise] pointer 2, whether you go from
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or whether you take a path of A or you have
taken the path of C, this should be same in
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order to have the same value of 0 [vocalized-noise]
ok; which essentially means that this term
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is basically independent of the path ok [vocalized-noise]
and this term which we we call it as dS.
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Because of the differential Q ok [vocalized-noise]
which essentially means that the delta S from
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1 to 2 is independent of whichever path you
take, if you integrate it you get basically
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delta S [vocalized-noise]. So, this is independent
of the path only depends on the final state
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point [vocalized-noise] and this is nothing
but the principle of the property ok, that
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the property depends only on the state condition.
So, change in the property also depends only
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on the initial and a final condition not on
the path [vocalized-noise]. So, that is what
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we defined as as the considering this is [vocalized-noise]
reversible, we make use of this internal reversible
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in a condition [vocalized-noise].
That this is true [vocalized-noise] assuming
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that you have reversible process ok [vocalized-noise],
now [vocalized-noise] so for the case of ah
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let us say isothermal process ok.
Where ah the heat is being supplied and temperature
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is fixed your delta S uh, in this case of
the system will be nothing but since the temperature
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is fixed will be nothing but simply Q by T
0 [noise] ok [vocalized-noise].
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So, for the case of here 750 divided by 300
kelvin it comes out to be 2.5 kilo joules
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per kelvin ok [vocalized-noise]. Now let us
consider again a cycle now in this case you
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have a reversible and an irreversible process
which constitute the cycle [vocalized-noise].
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So, we will be making use of the [vocalized-noise]
process inequality [vocalized-noise].
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So, let us say 1 to 2 ok could be reversible
or irreversible and 2 to 1 is basically reversible
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or internally reversible [vocalized-noise].
So, considering this you have 1 to 2 plus
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2 to 1 which is internally reversible and
this for this we know this is going to be
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[vocalized-noise] delta S1 minus S2 [vocalized-noise].
So, we can now make use of this expression
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in order to rewrite in this form[noise] ok.
That means, S2 minus S1 is greater than equal
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to 1 to 2 del Q by t. So, in the differential
form 1 can write [noise] d S greater than
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equal to del Q by T ok [vocalized-noise].
So, which essentially means that entropy is
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increasing ok, it is not just equal to this
but there is another additional terms which
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are coming which is we call it as generation
ok [vocalized-noise].
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So, for the case of a [vocalized-noise] closed
system [noise] [vocalized-noise] entropy change
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during irreversible [noise] process is always
greater than the entropy transfer due to heat
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transfer between the system and surrounding
ok. So, this is the heat transfer due to heat
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transfer between the system and surrounding
[vocalized-noise].
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So, for the irreversible reversible process
there is a certain S generation which is due
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to irreversibility's [noise] ok that is S
generation [vocalized-noise]. Now S generation
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or entropy generation is always a positive
quantity or 0 for reversal process ok and
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it depends on the process. Entropy for an
isolated system during a process is always
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greater than equal to 0 that means, ok [vocalized-noise],
because if I isolated you have this would
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be 0. So, S generation is always greater than
or equal to 0 and thus delta S isolated will
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be greater than equal to 0 ok [vocalized-noise].
So, in other word the entropy never decreases
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entropy always increases or remains constant
and this this is the concept of increase of
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entropy principle.
So, as we know based on this the entropy is
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a property and similar to the other properties
you can also make use of ah thermodynamic
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tables, ah the graphs steam table and so forth
to understand or to calculate the changes
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in entropy as in the case of a changes in
internal energy or enthalpy [vocalized-noise]
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ok.
So, this is an example of ah TS diagram temperature
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versus entropy specific entropy diagram, so
this is be the green line is basically nothing,
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but ah vapor liquid equilibrium. So, this
is your compressed liquid this is your superheated
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liquid this is your 2 phase system liquid
and vapor [vocalized-noise] in a similar case
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as ah you have done that for u for ah [vocalized-noise]
h ok.
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If the quality is given for any point in this
2 region 2 you can find out s by simply saying
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u of f plus x u of f g similarly you can also
write s as [noise] s f plus [noise] x [noise]
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s f g ok. [vocalized-noise] And if you do
not have information for the compressed liquid
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we will make use of the similar approximation
as done for enthalpy, that s is nothing but
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s of ah saturated fluid at T1 ok for a given
state point [vocalized-noise].
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The entropic change of a specific mass m during
a process is simply nothing, but [noise] small
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m multiplied by delta small s ok [vocalized-noise].
So, this you can connect ah get it from the
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tables ok, similar to what you have done that
for delta u or delta h [vocalized-noise] ok.
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Let me just summarize here the entropy, as
I said the entropy changes are due to heat
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and if there is a flowing condition due to
the mass also you [vocalized-noise] you can
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have the entropy change.
So, for the case of a fixed ah mass the entropy
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change is ah due to the heat transfer and
due to irreversibility, for the case of a
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ah internally reversible system the ah entropy
ah will remain constant or the entropy remains
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constant for internally reversible and adiabatic
process; that means, when when you do not
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have any heat transfer [vocalized-noise] and
when there is no irreversibility delta S is
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equal to 0 and this is typically used for
turbine case, which we approximate it to isentropic
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process [vocalized-noise] in this case on
a temperature s plot the vertical draw from
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1 to 2 would indicate the expansion at a constant
entropy [vocalized-noise] ok. So, this will
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be the process for [vocalized-noise] the case
where you have adiabatic process and internal
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reversible process. So, this is for fixed
mass ok.
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So, we can also represent entropy on ah or
entropy changes on a TS diagram for a process,
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internal reversible process as depicted here
represented by the green line from this point
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to that point; area under the curve is basically
associated with [noise] the [noise] heat supplied
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to the system ok [vocalized-noise] and the
differential area would be nothing, but d
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a is equal to T ds is nothing, but del Q ok
[vocalized-noise]. For the internet reversible
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process del Q is nothing, but T [noise] T
ds and from here of course, this information
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can be extracted [vocalized-noise] ok.
So, you can find out the heat supplied or
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heat associated with this by just integrating
this T ds ok [vocalized-noise], you can divide
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by the ah mass to get in ah specific ah per
unit mass ah values ok [vocalized-noise],
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for a constant temperature system it will
be simply T 0 delta S ok [vocalized-noise].
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One can also make use of h s diagram which
is a mollier diagram [vocalized-noise] for
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processes involving adiabatic steady flow
in such case ah the drop in delta h [vocalized-noise]
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ah measures the work associated with this
ah steady flow devices and increase in the
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delta S is measures, the irreversibility associated
with the device [vocalized-noise] ok.
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So, this is often also used in order to explain
the irreversibility's of the devices ok. Now
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ah having done some exercise on this we can
take this information particularly for the
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internal reversible expression of del Q and
T ds [vocalized-noise] and plugging in the
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relation at the first law ok [vocalized-noise].
So, what we are considering is ah del qs heat
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supplied minus the work done by the system,
is is basically the change in the internal
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energy of the system [vocalized-noise] now
we considering internally reversible [noise]
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system ok. So, here [noise] we can make use
of a T ds [noise] [vocalized-noise] and here
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we can make use of [noise] only the boundary
work which is associated with the system ok
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[vocalized-noise].
So, if you do that you have this relation
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which you get ok [noise] or in the per k g
you can write it in this form [vocalized-noise].
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Now from here we can find out ah the d s by
[noise] taking dividing by T [noise] [vocalized-noise]
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and you can find delta S [noise] by integrating
it ok, now you can also make use of the expression
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of enthalpy and write this expression instead
of u you can rewrite in instead of u you can
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rewrite [vocalized-noise] this expression
in terms of h [vocalized-noise].
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So, h is equal to u plus P v in other word
d h is equal to d u plus P d v plus v d P
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you can write d u now as d h minus this term
ok and thus you can get [noise] this expression.
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So, from here as I said we can find out ds
in terms of changes in the d u or changes
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in the ah u and v. Here in this case ah ds
is changing can be related to changes in the
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h and ah P [vocalized-noise].
So, if you integrate it and if you know the
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relations of the internal energy or enthalpy
[vocalized-noise] in terms of the pressure
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in the volume, you should be able to find
out the change in the entropy [vocalized-noise]
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for the simple cases it is easy ok.
So, let us take at the some [vocalized-noise]
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simple examples [vocalized-noise].
So, for the case of a liquid and solid [noise]
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for the case of a liquids and solid, we can
consider ah that the volume hardly changes
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and thus your d v is equal to 0; in that case
if you take out this expression in terms of
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u ds is equal to d u by d T [vocalized-noise]
ah d s is equal to d u by T, it can be simply
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written as C because for the case of liquid
and CP you can approximate CP is equal to
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Cv is equal to C and thus d u is nothing,
but C times d T [vocalized-noise] ok and thus
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you can [vocalized-noise] ah get this expression
of the delta ah S in this form where C can
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be [noise] ah taken out as an average [noise]
[vocalized-noise] for the case of idle gas
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[vocalized-noise] you can write du as dv ah
Cv d T and d h as CP dT.
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Now, this you have 2 expression one is here
and the another [noise] 1 is here which is
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based on h [vocalized-noise] you can plug
in this appropriately ok in order to find
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delta [noise] s here ok. So, you can get this
expressions [noise] ah uh in a very simple
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way [vocalized-noise], but in general if you
want ah delta S you should be able to integrate
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[noise] this term [noise] separately [noise]
ok [noise].
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For the case of u and similarly you can do
that for the case of h [vocalized-noise] and
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what we need is basically relation of this
and since this is something which we are going
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to focus more in the later part of the course
[vocalized-noise], making use of the volumetric
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property in order to create this kind of changes
in delta S and other thermodynamic properties
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[vocalized-noise] ok.
So, let us focus now on the entropy balance
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ok [vocalized-noise]. Now entropy balance
for a given system can be represented in this
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schematic form where the total entropy entering
which is S in minus total entropy living out
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plus total generation, should be the change
in the total entropy of the system ok. In
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other word on a simpler way S in minus out
plus S generation is equal to delta S system
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and this is basically nothing but the increase
in entropy principle ok [vocalized-noise].
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Whereas delta S system is nothing, but s final
minus S initial [vocalized-noise] now what
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are the different mechanism of entropy transfer?
Now beside heat transfer you can have also
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the ah due to the mass for the open system
[vocalized-noise].
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So, S heat for a constant temperature is simply
Q by T ok that is due to the heat ah entropy
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transfer by the heat, if the temperatures
are changing across the boundary you can also
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make use of summation of ah heat at different
part of the boundary and this will be your
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T ah Q k divided by T k summation of that
ok [vocalized-noise]. Entropy transfer due
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to the work which is ordered ah form of energy
is is not is it going to be 0 ok anything
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which is associated; while for example, considering
the shaft the entropy is being generated only
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due to the friction associated with this ok
[vocalized-noise].
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So, in addition to the ah heat transfer you
have also due to the entropic transfer due
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to the mass ok. So, of the fluid is flowing
it also bring in entropy into the control
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volume and that is can be written as simply
S mass is equal to [noise] m times the specific
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entropy. [vocalized-noise] And if it if the
the property of the mass changes during time
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you can make use of [vocalized-noise] these
expressions, which is well known it is ah
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S times density times the volumetric ah normal
velocity multiplied by the cross sectional
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area ok or in general S mass is equal to s
times [noise] the the differential mass ok
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which can be written in this form [vocalized-noise].
So let me get back to the again the entropy
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ah balance here as I said is S in minus S
out plus S generation is equal to delta S
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system, S in minus S out is due to the entropy
transfer by heat and mass [vocalized-noise]
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ok.
This can be written in to rate form [noise]
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ok and this is what we have written also earlier.
So, in summary S in is due to mass and heat
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S out [noise] is due to mass again heat [vocalized-noise]
and delta S system the change [noise] is due
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to the changes in the s m S out plus the S
generation [vocalized-noise].
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For a closed system we can we can write down
S in minus S out, simply as due to the heat
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transfer ok there is no mass entropy due to
the mass transfer and thus this can be written
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as summation of Q k by ah T k plus S generation
is equal to delta S system [vocalized-noise]
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ok.
So, ah for adiabatic close ah system when
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the heat is also 0, supply then S generation
is nothing, but [noise] delta S of system
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00:17:16,520 --> 00:17:20,970
which is adiabatic [vocalized-noise]. For
any closed system plus surrounding [noise]
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we can consider that to be an adiabatic in
other word S generation is basically nothing,
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but summation of delta S system plus surrounding.
If we consider system plus surrounding as
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a combined system which can be considered
as adiabatic, then this expression holds [vocalized-noise]
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ok [vocalized-noise] and delta S surrounding
would be nothing, but Q surrounding divided
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by T surrounding [vocalized-noise].
For the case of control volume beside the
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heat you have to consider the mass element
here and the rest of the terms are straight
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00:17:52,220 --> 00:17:56,360
[fa/forward]forward; whereas, of course S2
minus S1 would be of the control systems you
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00:17:56,360 --> 00:18:00,910
can write this in terms of [noise] rate expression
as well [vocalized-noise] ok.
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00:18:00,910 --> 00:18:06,669
So, ah this is something which I expect that
ah you have al al already gone through earlier
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00:18:06,669 --> 00:18:12,000
it is just recap. So, let me just try to finish
this particular lecture by working on this
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00:18:12,000 --> 00:18:15,100
example.
[vocalized-noise]. So, here we have a frictionless
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piston cylinder device [vocalized-noise] which
contains a saturated liquid [vap/vapor] vapor
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00:18:19,710 --> 00:18:24,830
[noise] mixture of water at 100 degree Celsius
[noise] ok [vocalized-noise], during a constant
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00:18:24,830 --> 00:18:29,840
pressure process 600 kilo joules of heat is
transferred to the surrounding at 25 degree
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00:18:29,840 --> 00:18:33,900
Celsius [vocalized-noise] as a result part
of the way water vapor contains in the cylinder
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00:18:33,900 --> 00:18:40,840
condenses [vocalized-noise] ok. So, some part
of the vapor condensed to the liquid ok and
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00:18:40,840 --> 00:18:45,309
what we have to find out the entropy change
of the water and the total entropy generation
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00:18:45,309 --> 00:18:53,030
during this heat transfer process ok [vocalized-noise].
So, ah for the case of a delta S system, we
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00:18:53,030 --> 00:18:58,810
will simply consider the Q which we know and
the T system because it is a saturated liquid
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00:18:58,810 --> 00:19:04,370
vapor mixture the temperature remains constant,
so this is 100 plus 273 kelvin and the Q is
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00:19:04,370 --> 00:19:09,630
minus 600 which turns out to be this [vocalized-noise].
Now, the for the case of ah total entropy
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generation, we have to also find out entropy
generation outside or particularly the surrounding
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as well [vocalized-noise]. So, the way we
are going to do is we are going to consider
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00:19:18,990 --> 00:19:25,540
a combined extended system where system plus
surrounding is extended system ok [vocalized-noise].
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00:19:25,540 --> 00:19:33,669
So, this is extended system [noise] [vocalized-noise],
such that [noise] here T [noise] b or the
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00:19:33,669 --> 00:19:41,770
boundary is 25 degree Celsius ok [vocalized-noise].
So, in that case if you consider the total
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00:19:41,770 --> 00:19:47,700
energy ah total entropy balance, you have
S in minus S out there is no S in only the
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00:19:47,700 --> 00:19:52,600
heat is being transferred. Now at the boundary
of extended system this is Qout divided by
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00:19:52,600 --> 00:19:57,870
T b which is 25 [vocalized-noise] degree Celsius
plus S generation [vocalized-noise] ok is
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00:19:57,870 --> 00:20:02,179
equal to delta S system, S generation is a
total entropy generation [vocalized-noise].
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00:20:02,179 --> 00:20:07,590
So, you can rewrite you can write in this
way S generation is Q out plus T b [vocalized-noise]
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00:20:07,590 --> 00:20:15,750
plus delta S system. Now delta S system we
know and this entropy generation or the entropy
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00:20:15,750 --> 00:20:20,530
due to the heat is can be written as in this
way [vocalized-noise].
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00:20:20,530 --> 00:20:25,419
So, this plus delta S system which we have
already calculated will give us the total
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00:20:25,419 --> 00:20:31,309
entropy generation during this heat transfer
process [vocalized-noise]. So, the idea is
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00:20:31,309 --> 00:20:36,820
basically to make use of extender system in
order to calculate this ok [vocalized-noise].
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00:20:36,820 --> 00:20:45,320
So, ah I hope that ah this quick exercise
and as well as quick recap of second law of
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00:20:45,320 --> 00:20:53,790
thermodynamics ah and beside the the earlier
ah ah contents which we have covered ah the
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00:20:53,790 --> 00:21:00,330
basic definition of the system ah; then the
first law of thermodynamics, open system close
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00:21:00,330 --> 00:21:05,890
systems and how to make use of the balances
energy and ah then the introduction of the
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00:21:05,890 --> 00:21:11,990
entropy and finally a very quick example [vocalized-noise]
ah with this ah I hope that you are now well
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00:21:11,990 --> 00:21:18,890
ah ah warmed up in order to take more ah ah
fundamental aspect of the [thermodyma/ thermodynamics]
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00:21:18,890 --> 00:21:22,190
thermodynamics which we will be starting from
next lecture onward [vocalized-noise] ah.
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00:21:22,190 --> 00:21:28,159
With the basic ah ah thermodynamics or calculus
related to thermodynamics [vocalized-noise].
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00:21:28,159 --> 00:21:33,259
So, ah with this ah I will stop here and we
will see you in the next lecture [noise].