WEBVTT
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Let us start this lecture a thought process
from Rudolf Clausius he says heat can never
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pass from a colder to a warmer body without
some other change connected here with occurring
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at the same time. And of course that we have
seen as a part of what you call Clausius statement
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it is impossible to transfer the heat from
the cold region to the hot region without
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any external aid. And we have basically discuss
about various aspects of second law of thermodynamics
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including Kelvin Planck statement, Clausius
statement and then we moved into the Carnot
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cycle analysis.
And we also found out that the thermal efficiency
01:08.750 --> 01:19.899
of a Carnot engine is equal to 1-TL / TH and
which will give the maximum efficiency whenever
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an engine is operated between a thermal reservoir
and rather a source and a sink. So we looked
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at basically two corollaries that is a Carnot
corollary 1 and Carnot corollary 2. And we
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will be now taking up an example to see how
you can apply the Carnot corollary and solve
01:53.790 --> 01:54.790
a problem.
01:54.790 --> 02:03.369
So let us take an example, so this example
generally you know it is related to again
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you know claim being made by an inventor.
And which is things one has to clarify and
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one has to invoke the second of thermodynamics,
the example is like this then in winter claims
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to have design and develop a new heat engine
that can produce 45 kilowatt of power by observing
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52 kilowatt of heat from a thermal reservoir
at 700 Kelvin while rejecting heat to the
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sink at 300 Kelvin is it true or false that
you need to find it out.
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If you look at this problem the question arises
whether you know first law of thermodynamic
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being, you know satisfied or not right and
then we need to look at also whether the second
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law of thermodynamics is being satisfied or
not. If you look at like it is basically what
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you call absorbing 52 kilowatt of power from
a source from a thermal reservoir at 700 Kelvin
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and it produces 45 kilowatt of power that
means it is whatever heat being observed is,
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you know some portion is going to seek and
45 kilowatt of the power is being produced
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and sink is that 300 K.
That means it looks to be, it may be true.
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Let us look at whether it is true or not,
so for that what we will have to we will have
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to find out the thermal efficiency, thermal
efficiency is nothing but W./Q. and the work
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produced by this heat engine is 45 and heat
observed from the source is 52 the efficiency
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is 86.5% quite a bit. And now what you will
have to do, you will have to basically invoke
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the Carnot corollary 2 which says no heat
engine can be more efficient than a reversible
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engine operating between two thermal reservoir
right.
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Suppose, you know this engine and then this
is the actual engine or its efficiency is
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86.5 and if you say that Carnot heat engine
is being operated between the temperature
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700 Kelvin that is the source and the sink
is 300 Kelvin then what would be the efficiency
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would be efficient should be equal to 1-TL/TH
and the TL is 300 Kelvin right and the TH
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the temperature of the source is 7oo Kelvin
and you are getting the efficiency of 57.1%
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that means a reversible engine is giving 57.1%
and the inventor is claiming is efficiency
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to be 86.5 therefore it is not really possible
right.
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And hence the claim of this inventor is false.
So by, you know using a very simple way one
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can really find out whether it is true or
not.
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So let us now see how we can apply you know
thermodynamic apply the second law of thermodynamics
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for a temperature scale which we have already
discussed earlier let us devise a temperature
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scale which would not be dependent on the
properties of the substance that we have already
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defined as a thermodynamic temperature scale,
if you recall that is the thermodynamic temperature
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scale which would not be dependent on the
properties of the substance like you know
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what we use.
Let us say in a mercury and glass thermometer,
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so if you look at the temperature what it
will measure it will be dependent on the expansion
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of the, what you call mercury. But if I use
alcohol then you know that expansion will
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be different, so it will be dependent on the,
what you call substance the thermometric fluid
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rather, but however we need to devise a temperature
scale which will be not dependent on the properties
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of substance.
So why is it required we know because we have
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already know that whenever this mercury glass
and alcohol glass thermometer will be used
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even though, you know it will be calibrated
at the ice point and boiling point, but however
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it will give you different, you know what
you call temperature at the at other points,
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you make elevator the same ice point and boiling
point and assign those values, but however
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if you go deviate from this ice point and
boiling point it may give you different what
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you call values.
For example like mercury in glass thermometer
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will give you 50.12°C whereas the alcohol
in glass thermometer will give 50.45°C, of
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course you may say look it is not much, but
in accuracy wise it is not good. So therefore
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it indicates that what you call this temperature
scale, you know is basically dependent on
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the properties of the thermometric fluid right,
because mercury and alcohol are two different
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and their properties are different therefore.
So therefore we need to have a scale which
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is independent of the what you call properties
of the thermometric fluid or the substance.
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So for that of course we know that alcohol
have different T-V relations it is very much
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required to devise temperature scale. And
how can we devise such a temperature scale
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for that what we will do by using the reversible
engine concept from the Carnot theorem 2 that
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we have seen that the efficiency of a reversible
engine or what we call Carnot engine is basically
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1 - QL / QH which is equal to 1minus TL/TH.
And we say that QH/ TL is basically a function
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of TH and TL temperature both the source and
the sink temperature it is function.
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So let us consider the three reversible engine
that is reversible engine 1 2 & 3 which is,
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you know let us say the reversible engine
is observed what you call Q1 amount of heat
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and rejects Q2 amount of weight to the sink
2 right, and T1 is greater than T2. And it
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does the W1 amount of work right, and let
us say that whatever heat is coming from what
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we call the sink 2 the sink 2 then that will
be observed by another reversible engine and
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it rejects the Q3 amount of heat to the sink
T3 that if you look at so far the reversible
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engine two is concerned this is this sink
is not a sink this is basically a source for
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the reversible engine right.
And it is a thermal reservoir I mean you call
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it scenes or I should call rather it is a
thermal reservoir right. So it is basically,
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you know thermal reservoir right, thermal
which can act as a source as can a sink depending
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on the word let us say that another engine
that is a third engine which observed the
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same amount of Q1 from the source that is
T1 which is at the T1 and it rejects Q3 amount
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of heat to the sink 3 and it produces W3 amount
of work.
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Now if we look at for the engine 1 we know
that Q1 Q2 is a function of T1 and T2 if it
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is reversible engine that we have already
seen. And similarly for the engine 2 the Q2
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and Q3 ratio is a function of T2 and T3, T2
is what you call the thermal reservoir 2 temperature
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which will be acting as a source for the reversible
engine 2 while it will be acting as a sink
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for the first reversible engine. So similarly
for the engine 3 we can say that the key one
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Q1/Q3 is basically the function of T1 and
T3 right.
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Now if you say that like Q1/Q3=Q2 = Q1/Q2
x Q2/Q3 it is possible only if you know that
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we can say from this that it is basically
Q1 by Q3 is a function of T1 T3 and Q1 by
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Q2 is a function of T1 by T2 and Q2 by Q3
is a function of T2 by T3. So if you look
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at the this portion that is if you look at
this the if you look at LHS you know this
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is your LHS is the independent of the temperature
T1 and you know it is a LHS is a function
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of T1 T3 and the LHS is for basically independent
of temperature T2.
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Only if function as the following form I mean
that is only possible right that if I say
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that this function that is F1 T the f this
function is basically of is equal to g/g is
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another function of temperature T1 and is
also a ratio of the temperature T1 and T2
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and f2 the function f which is a function
of T to these basically gT2 2 / gT3 now if
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I say then only we can say that because this
if it is there then I can it will cancel it
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out so that we can say that Q1 by Q3 is function
of T1 by T3 right two functions are there.
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And so if you look at if you take a you know
of course it can be any function you can say
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in the more general term we can write down
that equation for this reversible engine is
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basically QH by Ql is equal to the g which
is a function of the and the function of PL
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right.
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And where gT can have several forms right
you can think of any forms but by choosing
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an appropriate function one can define a temperature
and which is can be considered as independent
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of the physical thermometric properties right
and the if you look at Lord Kelvin he had
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proposed a simple function he says that gTH
that is the function is equal to the temperature
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then he says I am then the selected what you
call the relations that QH by Ql will be equal
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to PH by TL you may see that you know people
have used a complex function and they have
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arrived at similar relationship.
So therefore this is the scale which we call
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it as basically Kelvin scale which varies
from 0 to the ? right and then expression
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QH by Ql is does not really on protocol depend
on the properties of the thermal when deployed
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and it can be you know even though it is a
relationship even though this relationship
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it does not really define the thermodynamic
scale completely but it gives a relationship
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so in 1954 the international conference on
weights and the major assigned value 270 3.16
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to the triple point of water right.
As a result like you can get one point here
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then after that you can find out that magnitude
of Kelvin is basically defined as the one
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by 270 3.16 Kelvin between absolute zero and
triple point of water triple point of water
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you know like 0.010 C right so therefore he
has defined that you know one over 70 and
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that 0 is considered to be there Kelvin a
absolute zero so by this way one can use the
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basically the second law of thermodynamics
to define a thermodynamic temperature scale
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which will be independent of the thermometric
flow so let us take an example.
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Like whether this relationship can be used
for solving some other problem we have talked
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about a temperature scale that is of course
to you know to define a scale Kelvin scale
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basically defined from that and then you can
use it but here we will be applying that for
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solving a problem let us say a heat engine
operated between thermal reserve at 500 Kelvin
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and ambient air at 300 Kelvin is used to drive
a refrigerator to produce a temperature of
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250 Kelvin by extracting heat of 100 kilo
watt from the cold region right.
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See in this heat engine is basically produced
giving some walk it is producing some work
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and that work is being utilized to run the
refrigerator and assuming the all the processes
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to be reversible and we need to determine
the power produced by a heat engine right
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and the ratio of heat energy observed by the
refrigerant system to the heat energy observed
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by the heat engine and we will have to also
find out the amount of heat rejected to ambient
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air so let us say that that is a source
that is T is equal to 500 Kelvin and it is
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observing some amount heat.
Let us say key one and there is a reversible
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engine which is producing amount of work w
dot and it is rejecting certain amount of
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it to the sink that is your sink, sink is
which at what temperature that is a 300 Kelvin
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and of course this sink you can say the same
thing right and there is a refrigerator right
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I say that RE, RE or I can say this is basically
he reverse heat engine or the refrigerator
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and this region is the cold region which is
at 250 Kelvin and what it does it observe
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certain amount of it.
Let us say this is Q3 and it is rejecting
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certain amount of heat that is Q dot for right
what we will have to find out to find and
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this work is being supplied right if you look
at this work is being supplied what is that
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that is w that means you can say that this
amount of work is being I think this amount
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of it being supply now what is it we found
out to find basically we need to find the
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power produced w right and that is WHE is
equal to w RE, RE is the RHE let me put it
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that means reverse heat engine which is nothing
but a refrigerator right.
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And how we will find out that how we will
find it out that we know that coefficient
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of performance of the refrigerator what is
that SI of P of refrigerator if I say simply
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are HE or I can say R is equal to what is
that what is that if I say this is TL this
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is TL and this is TH right this I can say
TR right can I say that what will be coefficient
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of performance of the refrigerator that is
nothing but your TRTL - TR yes or no we have
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already seen the coefficient of what you call
reversible refrigerator right.
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Are reverse heat engine whatever you call
so that is equal to what you call TR is 250
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this is 300 this is 250 is equal to what is
equal to what that is Q3 / W yes or no what
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is Q3 here what is Q3 Q3 is 100 kilo watt
is not this is your Q3 so that is 100 / w
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dot so W that would be what and what is these
values this one 250 /350 it will be nothing
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but your 5 this is equal to 5.
And what is these values this one 250/350
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it will be nothing but your 5 this is equal
to 5 so ? that would be what 20 how much it
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will be w that will be 20 kilo watt, so I
mean we got now the a so if you look at this
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is your a we got now we need to find out b,
what is b, b is the ratio of the heat absorbed
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by the refrigeration system are much heat
being observed it is 3 divided by the heat
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observed by the heat engine what it would
be what it is it is it is Q?1 yes or no right.
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So how we will find out that I can write down
this Q3/? yes or no similarly I can write
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down Q?1/W1? so if you look at Q?1 is it we
know this one Q?1W we know or not we really
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do not know but we know this Q?3 we know and
? that we know that is nothing but your word
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COP of reverse engine reverse heat engine
or the refrigerator right, and how I can get
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this Q?1W how, basically that is nothing but
your Carnot cycle efficiency.
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What is that Carnot cycle efficiency cannot
I am just writing c is equal to by definition
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is yes or no that is equal to 1 minus PL/PH
that will be equal to what 1-3 under / 500
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will be equal to what 0.4 can I say so therefore
the Q1 will be equal to leave at 20 right
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Q1 will be equal to 20/0.4 is equal to 50
kilowatt yes or no, so now we will have to
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find out basically the what you call if we
know this Q1 then I can substitute over here
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that is Q?/Q?3/W is 5 and this is 50 / 20
that will be getting what will be this will
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be 2 yes or no.
This will be 2 so we need to now find out
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that is C, Q?2 that is the amount of heat
rejected amount of heat rejected will be Q?2+Q?4
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that is the amount of heat rejected to the
ambient air is it not Q?2 and Q?4 so how we
28:37.600 --> 29:04.769
will get this Q?2 we know that Q?2/Q?1 is
equal to TL/TH yes or no right. Similarly
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we can also find out Q?4/Q?3 is equal to T4
is Tl divided by Tr from this you can substitute
29:27.419 --> 29:42.520
these values you can get Q? is basically Q?1TL/TH+Q?3
TL/Tr.
29:42.520 --> 29:54.480
If you substitute these values you will get
basically because you know that Q?1 is known
29:54.480 --> 30:03.440
so TL and TH is known right Q?3 is known or
not Q?3 is given to you 100 kilowatt so TL
30:03.440 --> 30:07.870
you know Tr you know all those things you
know he will substitute this value you will
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get something 150 kilo watt, right. So now
you know how you can apply what you call this
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whatever we have derived the relationship.
That is basically Q from the what we call
30:29.739 --> 30:35.970
thermodynamic temperature scale we have learned
that Q? or the heat ratio is equal to the
30:35.970 --> 30:43.710
temperature ratio whenever the engine is operated
between the source temperature and the sink
30:43.710 --> 30:49.210
temperature right, reversible it should be
reversible that you should keep in mind. So
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now what we will see how we can up you know
take this use this second law of thermodynamics
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for other purposes.
31:02.289 --> 31:07.539
And we have you know we will have to apply
the second law of thermodynamics for analysis
31:07.539 --> 31:12.669
of engineering problems and this whatever
we have done is basically cyclically operating
31:12.669 --> 31:20.149
device not that all the cycling operatic device
will be handling but we will be handling also
31:20.149 --> 31:30.970
the what you call various cases for that let
us consider the corner cycle right, and which
31:30.970 --> 31:43.259
will be in this case the heating in observe
the Q1 amount of heat, right and it rejects
31:43.259 --> 31:50.610
Q2 amount of heat to the sink which is at
temperature T2 and it does the W amount of
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work because is the reversible engine right.
So for Carnot cycle we know that from this
32:01.900 --> 32:12.019
the Q1/Q2=T1/T2 and from this what will we
can write down we can write down basically
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here we can say Q1/T1= Q2/T2 and I can take
this out that is nothing but Q1/T1-Q2/T2=0
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this is one relationship which we can clear.
So and of course this is mean for the reversible
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engine only right.
So in this case if you look at if I take this
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heat is entering into the heat engine right,
if I take this my system right, it is entering
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what it would be it will be positive and if
it is going out it will be negative right
33:03.590 --> 33:13.019
from the system if I take this is my system
right, so then we can write down this expression
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as you know Q1/T2+Q2/T2=0 and this is of course
between the two what you come to thermal reserve
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one is of course the source other is a sink.
But there might be several deserve worse right
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one can think of so we can write down this
expression that what we have derived for the
33:37.529 --> 33:44.940
reversible engine which is operated between
two thermal reservoirs it can be n number
33:44.940 --> 33:50.350
of you know thermal reservoir we can say that
engine can operate so therefore I can write
33:50.350 --> 33:57.160
down i is equal to 1 to n right, and then
this I can write down I mean from here I can
33:57.160 --> 34:07.149
write down that Qi/Ti some of the heat and
then you know heat and temperature ratio is
34:07.149 --> 34:10.970
equal to 0.
34:10.970 --> 34:17.919
So let us look at because all the time we
need not to what you call use the heat engine
34:17.919 --> 34:25.020
we will have now considering irreversible
heating irreversibility. In earlier we have
34:25.020 --> 34:32.910
considered the reversible heat engine and
again it is taking the this irreversible heat
34:32.910 --> 34:44.350
engine is taking the heat Q1 from the source
which is at temperature T1and it is rejecting
34:44.350 --> 34:54.179
the Q?2 amount of heat to the sink and it
produces double amount of work.
34:54.179 --> 35:04.900
So for the an irreversible engine we know
that the according to the cognate corollary
35:04.900 --> 35:13.850
right, that is the thermal efficiency of the
irreversible engine always will be less than
35:13.850 --> 35:20.770
the reversible heat engine, so that means
if you look at the irreversible engine efficiency
35:20.770 --> 35:38.430
will be what 1-Q?2/Q1 and less than the 1-Q2/Q1
right. And we know for the reversible heat
35:38.430 --> 35:48.050
engine the what will be the efficiency, efficiency
will be 1-T2/T1 right, yes or no from the
35:48.050 --> 35:53.700
current because the reversible engine is your
Colonel this is corresponding to what you
35:53.700 --> 36:00.690
call Carnot efficiency right.
Carnot efficiency is not equal to, is equal
36:00.690 --> 36:09.660
to 1-T2/T1 so therefore I can write down can
I substitute here in this case no I cannot
36:09.660 --> 36:16.950
because that is it this one is mean for the
irreversible heat engines therefore I cannot
36:16.950 --> 36:25.390
really apply this thing. So then I can write
down here you know if you look at so this
36:25.390 --> 36:34.240
is less than that so therefore I can write
down here that is Q2’/Q1 is greater than
36:34.240 --> 36:56.090
T2 by T1 that is being written here okay.
So I can write down that you know like Q2’/T2
36:56.090 --> 37:06.310
is greater than Q1/T1.
37:06.310 --> 37:13.130
So using the sign convention what we have
done we have already discussed right in this
37:13.130 --> 37:23.490
case Q1 is basically being observed and Q2
dash is given to the sink that means it is
37:23.490 --> 37:31.240
going out of the out of the engine irreversible
engine. So therefore I can you know take this
37:31.240 --> 37:43.150
negative and therefore I can write down that
that Q1/T1 + Q2’/T2 is less than zero, so
37:43.150 --> 38:02.040
this is mean for what for irreversible engine
and if it is I at number of engines irreversible
38:02.040 --> 38:10.750
engine I can write down summation of QI’
/ TI less than zero right.
38:10.750 --> 38:17.210
So if you combine this equation that is the
reversible engine which is equal to basically
38:17.210 --> 38:23.510
summation of ki y x TI is equal to 0 that
is reversible and irreversible engine will
38:23.510 --> 38:31.320
be less than equal to 0, if I just combine
this thing I can write down this expression
38:31.320 --> 38:41.000
that is summation of this heat interaction
basically a queue I it is the you know engines
38:41.000 --> 38:53.390
t r is less than equal to 0 and that is nothing
but your Clausius inequality this is your
38:53.390 --> 39:00.820
Clausius inequality what if state Clausius
inequality?
39:00.820 --> 39:09.590
States that whenever a system undergoes a
cyclic change however the complex cycle it
39:09.590 --> 39:19.000
maybe the algebraic sum of all the heat interaction
/ the respective absolute temperature at which
39:19.000 --> 39:28.020
heat interaction take place consider for the
entire cycle is less than equal to 0 that
39:28.020 --> 39:35.340
is the Clausius inequality I mean which just
now we have derived from taking very simple
39:35.340 --> 39:42.190
concept of the vertical Carnot engine that
is a reversible engine and we have also considered
39:42.190 --> 39:47.550
the irreversible engine and done that and
that is nothing but you're mathematically
39:47.550 --> 39:54.890
one can write down the queue I divided by
T are some of over all the this thing is less
39:54.890 --> 40:07.570
than equal to 0 if this is basically mathematical
expression of the classiest inequality.
40:07.570 --> 40:14.270
And keep in mind that this is the basically
the second law of thermodynamics which can
40:14.270 --> 40:24.110
be applied not only for the heat engine for
the also for the refrigeration system and
40:24.110 --> 40:34.060
which must be cyclically operated that means
it is applied for the cyclically right and
40:34.060 --> 40:41.910
let us consider an example a reversible heat
engine operates between3d servers that is
40:41.910 --> 40:50.720
at 600 Kelvin400 Kelvin and 300 Kelvin it
receives two thousand five hundred kilo Joule
40:50.720 --> 40:59.860
of heat energy from the thermal reserve at
six hundred and does the thousand kilojoules
40:59.860 --> 41:05.470
of walk and we need to determine heat interaction
with other two reserve that is 400 Kelvin
41:05.470 --> 41:13.960
and 300 Kelvin that is the other two receivers
we need to you know. So question arises how
41:13.960 --> 41:16.280
to go about it.
41:16.280 --> 41:23.631
So let us say that is the heat engine this
is your heat engine basically you know I can
41:23.631 --> 41:35.160
take this is my best system I can say right
heat be observed q1 to 2500 you will and some
41:35.160 --> 41:43.730
heat being rejected to 400 Kelvin and this
is again rejects q3 to the 300kelvin thermal
41:43.730 --> 41:50.780
reserve right and it does the thousand kilo
Joule of cortical walk and keep in mind that
41:50.780 --> 42:01.120
this is a reversible heat engine right.
So how will go about it will basically first
42:01.120 --> 42:08.520
we will have to look at the first law of thermodynamics
right, so according to fast law what is says
42:08.520 --> 42:21.350
W is equal to Q1 - Q2 - Q3 right. So if you
look at the this is given w is given and Q1
42:21.350 --> 42:29.970
is given and Q2 and Q3 are not given so you
cannot really proceed now right you can say
42:29.970 --> 42:34.170
look this is just an energy balance it is
not like then what we'll have to do you will
42:34.170 --> 42:45.760
have to apply now second law of thermodynamics
right and of course for this we will have
42:45.760 --> 42:55.740
to use the classiest inequality that is dQ
by T over the cycle is equal to the Q1 divided
42:55.740 --> 43:06.780
by T1 right minus Q2 divided by T2 - Q3/3
right yes or no is equal to 0 because this
43:06.780 --> 43:14.890
is a reversible eating it.
And in this case this Q1 is given and T1 of
43:14.890 --> 43:23.270
course P2 is given and T3 is given but Q2
and Q3 is not are not known but however you
43:23.270 --> 43:28.530
know you can get an expression from here and
we have already derived an expression from
43:28.530 --> 43:36.600
first law of thermodynamics that is Q2 plus
Q3 is equal to 15 hundred kilo Joule. So if
43:36.600 --> 43:39.760
we just solve this two equation we can get
that.
43:39.760 --> 43:51.950
Let us see that is we are getting 2500 that
is q1 divided by 6 and 8 - Q2 / 400 - Q 300
43:51.950 --> 44:01.300
and they equal to 0 why here it is minus because
heat is going from the engine to the sink,
44:01.300 --> 44:09.450
so therefore Q 2 we have taken minus and so
also the Q3 the sign convention and that is
44:09.450 --> 44:19.080
your what you call equation 2 we are getting
3 Q2 plus for Q3 is equal to five thousand
44:19.080 --> 44:25.570
and if you already we have you know derived
this thing that is from the first law of thermodynamic
44:25.570 --> 44:32.890
if we just minus here you know this one so
what you will cancel this will cancel it out.
44:32.890 --> 44:44.830
So we will get 7 q 3 is equal to 500 right
and from that Q3 is equal to from this what
44:44.830 --> 44:56.060
will say you will get Q3 is equal to 500 kilo
Joule and from the other expression you can
44:56.060 --> 45:04.320
get from the equation am like a from here
any one of them that Q2 is equal to thousand
45:04.320 --> 45:13.460
Kelvin. So that till now what we have seen
we have seen that how you can apply the second
45:13.460 --> 45:19.990
law of thermodynamics for the cyclic process
right but not all the processes should be
45:19.990 --> 45:27.790
cyclic in nature there will some processes
which will be non-cyclic in nature right.
45:27.790 --> 45:36.100
Now question arises how to go about it, if
you remember that we are derived the what
45:36.100 --> 45:46.660
you call the processes first law of thermodynamics
for the cyclic process for first and then
45:46.660 --> 45:57.590
we apply it for the non cyclic process what
do we do there at that time could you recall
45:57.590 --> 46:05.770
how do you do we have done that already for
the first law of thermodynamics a cyclic integral
46:05.770 --> 46:12.420
of the heat will be proportional to cycle
integral of the work and then we say that
46:12.420 --> 46:20.000
it is equal to the JJ is there your joules
constant right and then from that we have
46:20.000 --> 46:26.270
derived expression that is de is equal to
dQ minus DW that is the first law of thermodynamics
46:26.270 --> 46:34.060
for non cyclic process yes or no right.
So how did you we arrived when that will be
46:34.060 --> 46:43.220
following the similar procedure right we will
be following the similar procedure to do that,
46:43.220 --> 46:51.220
so I think we will stop over here but let
us recall that how what we did we basically
46:51.220 --> 47:00.670
looked at the how to apply the first law of
thermodynamics for acyclic process right and
47:00.670 --> 47:11.480
we derived the classiest inequality relationship
right, we say that whenever a what you call
47:11.480 --> 47:20.190
reversible engine whenever the heat engine
will be interacting with the particle system
47:20.190 --> 47:26.460
and is whenever it will be interacting with
the thermal reservoirs right source and see
47:26.460 --> 47:35.820
however the complex cycle it may be the sum
of the heat interaction and the respective
47:35.820 --> 47:42.770
thermal reservoir ratio will be equal to less
than equal to 0 right.
47:42.770 --> 47:53.890
So that is the thing what we have seen so
if you look at we have seen whenever system
47:53.890 --> 48:00.650
undergoes a cyclic change however the complex
cycle may be the algebraic sum of all heat
48:00.650 --> 48:08.390
interaction / the respective absolute temperature
at which heat interaction take place consider
48:08.390 --> 48:14.380
for the entire cycle is less than equal to
0that is a Clausius inequality and before
48:14.380 --> 48:20.660
that of course we have seen that how one can
devise the thermodynamic temperature scale
48:20.660 --> 48:26.640
by using the second law of thermodynamics.
So in the next lecture what we will be discussing
48:26.640 --> 48:34.140
basically how we will apply the second law
of thermodynamics for the non cyclic process
48:34.140 --> 48:40.340
thank you very much.