WEBVTT
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In the last few lectures, you have been exposed
to cycle analysis, and the associated aerothermodynamics
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of flow through the jet engines; various kinds
of jet engines, which are used on aircraft
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for flying. These aero thermodynamics cycle
analysis must have been exposed to you all
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kinds of component analysis, engine analysis
as a whole; and this has been covered by professor
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Pradeep.
In today’s lecture, we will take a look
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at one of the components and that is compressors,
and try to look at it from thermodynamic point
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of view; one of the reasons is that compressor
is a working machine, it actually absorbs
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work supplied by their turbine, and then converts
that work to pressurization of air. Now, that
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is one of the intentions of the compressor
inside the gas turbine engine. The other purpose
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is that the compressor as a working medium
is also part of the jet engine, which incidentally
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is also heat engine. So all affairs connected
to any kind of heat engine needs to be analyzed
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in the overall matrix of thermodynamics. So,
any component, which is there inside a jet
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engine must necessarily be subject to a scrutiny
using fundamental thermodynamic laws. And
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this is what to some extent has been done
in great detail through the cycle analysis.
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And today we will look at thermodynamics of
compressors; and in the next lecture we will
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look at thermodynamics of turbines.
So let us take a look at thermodynamics of
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compressors. The compressor thermodynamics
essentially comes from the fact that you have
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a flow which is in our case essentially air
which is performing work, and this work is
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available or made available by running the
turbine, and the depiction of this work the
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expression of this work can be put down in
form of thermodynamic expressions, which means
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the work that is transferred from turbine
to compressor, and the work that is done by
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the compressor in the form of compression
of pressurization can be expressed in terms
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of thermodynamic parameters. And this is what
we intend to look at a closely today through
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various methods of fundamental thermodynamic
relations.
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Let us take a look at fundamental thermodynamics
of compressors, one of the reasons I said
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that we would need to analyze the compressors
and of course, later on turbines through thermodynamic
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means is because it is part of the heat engine
and they need to be subject to the scrutiny
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under thermodynamic matrix, and if we can
show that they are amenable to fundamental
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thermodynamic rules and laws then it is possible
to provide a prediction mechanism of the amount
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of what they are doing the amount of work
that is transferred between turbine and compressor,
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and the efficiency with which this work is
being transferred or this work is being done.
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Now this is the prediction that is required
by the engine designer because, he needs to
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know how each of these components is going
to perform, and then overall how the engine
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is going to perform. So, this prediction mechanism
that thermodynamics can provide is an important
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aspect of the development and creation of
the engine. So this is one of the reasons
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why we need to have a good look at the thermodynamics
of compressors and turbines; this also allows
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another important aspect of creation of the
engine, and that is it allows for optimization
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of the cycle design, you have done all the
cycle analysis over the last few lectures
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much of this cycle cycle analysis would be
subjected to an engine design effort and during
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that process is inevitable.
Now, a days that an optimization would need
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to be arrived at so the final design is invariably
an optimized cycle design which is the fundamental
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basis of the development of the engine.
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So this optimization is another possibility
that thermodynamic analysis throws up let’s
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take a look at some of the fundamental issues,
if we have a simple compressor in which you
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have let us say to begin with a closed system
something which many people are familiar with
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you have a closed system in which work is
expected to be done, either by let us say
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a movement of a piston or by expansion of
gases.
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It could be negative expansion and the work
done can be expressed in terms of the motion
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of the piston or the change of volume of the
enclosed gas. Now, those expressions are written
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down there, and it it shows that for a closed
system it is it is comparatively simple probably
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to write down the work that needs to be performed
to do the process of compression, now in open
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systems the work done by an open system or
a open compression is a little more involved
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to begin with the flow is coming into a system
which is let us say the square red system
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that we have and its coming in with a pressure
p 1 volume v 1 and it has a mass flow rate
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m.1, and it is exiting the system with pressure
p 2 volume v 2 and a mass flow rate m.2.
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It is possible under many systems m1 would
be equal to m 2 but that is a separate issue
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the work done by this kind of an open system
in which it is flowing the fluid is flowing
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its coming in and going out on a continuous
basis. The work done is expressed in terms
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of p 1 v 1 minus p 2 v 2 plus integral of
v 1 to v 2 of p dv. Now, the third term is
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the work done for the compression process
the first term is the work done for the air
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entry; and the second term is the work done
for air exiting the system.
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So in a open system it enters the system with
a certain amount of work, it exits a system
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with a certain amount of work, and during
the process of its resident time within the
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system a certain amount of work is done, and
in our case for the purpose of compression
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which means in the process of the flow going
from 1 to 2 it experiences compression or
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pressurization. So this is a simple way of
writing down how the work can be expressed
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for a compression process.
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If we show the thermodynamic process of compression
in p v diagram something you have already
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done before in your cycle analysis, let’s
take a quick look at the same thing may be
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with a different idea in our mind, the process
from 1 to 2 in the p v diagram is typically
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a polytrophic process a real process the corresponding
ideal process is 1 to 2 prime and that is
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often called the adiabatic process. If it
is a reversible adiabatic we would probably
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call it isentropic process, now these are
the two processes to which with which we could
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start off and the difference between the two
is delta h that is the work extra work that
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is needed to be done.
To accomplish the pressurization from pressure
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1 to pressure 2 from line 1 to line 2; and
this process of pressurization requires that
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much of extra work, if the work done is not
ideal that is adiabatic but polytrophic now
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on the right hand side you would see a number
of other possibilities in a p v diagram you
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could have a process which is isothermal in
between we have the process which is adiabatic
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the so called ideal process, and the third
process which is a theoretical possibility
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is an isochoric process, where the specific
volume remains constant and the pressurization
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goes from 1 to 2 triple point. So 1 to 2 double
prime is the isothermal process, where the
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temperature remains constant and1 to 2 triple
prime is the isochoric process, where the
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specific volume remains constant now as you
can see.
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The index which is normally used for signifying
the process, in case of adiabatic process
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normally this index is gamma, in case of other
processes we often might like to call it some
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k in case of a general polytrophic process
non adiabatic process k is likely to be greater
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than gamma, in case on isothermal process
k would be equal to 1 and in case of isochoric
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process the k would be infinity.
So, these are the theoretical possibilities
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in which the polytrophic process is a realistic
process, which does not confound to any of
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these adiabatic isothermal or isochoric theoretical
possibilities but a general polytrophic process.
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If we say that p 1, v 1, T 1 are the inlet
or the initial conditions with which the mass
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flow of m.1 is coming in, and P 2, v 2 and
T 2 are the exit or final conditions with
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a mass flow of m 2 the work done by the system
is given by integral 1 to 2 dou w that is
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the work done would be equal to minus of integral
1 to of v dp, which we can also write down
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as minus of integral 1 to 2 of dp by row row
of course, is the density of the gas or air.
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And V of course, is the specific volume of
gas or air, now in a real compressor as I
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just mentioned a little while ago the flow
is likely to be quasi static, which means
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m.1 may not be equal to m2. And there is a
possibility that the flow may not be adiabatic
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that means it could have some loss of heat
and some unused energy that is let out at
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the outlet that means certain amount of energy
that was used for the intended purpose of
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compression.
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Is now available as a exhaust energy and goes
out with the outgoing flow, so that is what
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is likely to happen in a real compressor which
is we are generally calling a polytrophic
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process. If you look at the p v diagram and
on the side you have a system in which the
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flow is coming in with P 1 V 1, and let us
say a velocity C 1. In which case it could
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be an actual velocity C a1 coming straight
into let us say a ducting system, and then
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we say that this process now has a possible
heat addition or rejection.
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Now this heat addition is shown as delta Q
R and the heat rejection is shown as delta
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Q Q subscript small q, so the heat net heat
added to the system is then given by delta
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Q is equal to delta Q R minus delta Q subscript
small q, and this Q R that is said to be added
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to the fluid that is flowing in inside the
system may be assumed to have been added by
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the fluid friction, so that is one of the
possibilities and Q subscript small q is the
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heat loss to the surrounding which often happens
due to the normal heat transfer to the surroundings
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another result of this we can write down from
thermodynamics that delta Q the net heat that
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is added to the system is C r into dT, where
C r is the specific of the particular fluid
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gas or air.
For this particular real situation we know
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that specific heat for a constant process
pressure is known as c p constant volume,
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it would be called c v this is a process which
is neither constant pressure nor constant
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volume some real process, and we simply call
it we are simply calling it c r for the time
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being.
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So, the process that we are seeing here is
a little more realistic process subjected
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to a compression, now if we proceed along
this we can say that the work done in any
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real process may be split up in work done
for two ideal processes one is the energy
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added to the fluid that is by delta Q is equal
to C p dT minus V dp. And we we showed the
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V dp in the earlier diagram this is your v
dp and this is of course, delta p dv.
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So, if we add the those work done to c p dt
which is the energy added to the fluid and
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energy taken away from the fluid can be put
down in terms of c v dt plus p dv, then for
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an isentropic process the work done can be
written down simply as delta q is equal to
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c n dt where c n is the specific heat of the
isentropic process.
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Let us say, let us put it that way and then
the net energy transaction in an isentropic
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process is typically 0 because, it is an adiabatic
process, so delta q is actually the heat added
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or subtracted from the system is likely to
be 0 in which case c p dt would be equal to
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v dp and c v dt is would be equal to minus
p dv from the above equations.
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If we do that and then use the concept of
isentropic index, which you have used in your
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cycle analysis this is normally defined as
gamma is equal to C p by C v. And now, we
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can see that it can be written down in terms
of v dp divided by minus p dv, where C p is
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the specific heat at constant pressure process
and C v is the specific heat.
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At the constant volume for the air or gas
it it depends on whether it is air or what
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kind of gas, so the value of C p and C v would
depend on the values of properties of the
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air or gas they also depend on the conditions
of the air or gas notably the temperature
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of the air or gas.
In if we do that the concept of specific heat
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ratio can now be extended, and we can write
down that the polytrophic index for a real
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process can also be defined for a real process
as k just like gamma and the it is also a
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ratio of V dp by P dv, so that ratio remains
as the fundamental definition of the specific
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heat ratio, and now we are saying it is a
real process a polytrophic process and we
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say.
We designate it by k small k, and this can
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now be written down from the earlier equations
as C n minus C p, where C n is the specific
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heat for a isentropic process, C p is that
for a constant pressure process and that is
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divided by C n minus C v and that would be
equal to from the earlier equations 1 can
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derive gamma multiplied by 1 minus d Q by
c p dT and the whole thing divided by 1 minus
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gamma into dQ by c p dT.
So that is in a net shell the fundamental
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definition or concept of polytrophic index
similar to the isentropic index gamma, which
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is used in cycle analysis specially in ideal
cycle analysis thus. We say that if delta
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q is equal to plus or minus delta Q R plus
delta Q q and if that is greater than 0, that
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means that the net that is going into the
system is greater than 0 then the polytrophic
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index k would be greater than gamma.
If the delta q is 0 then it is an isentropic
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process and k would be equal to gamma, if
the process is isothermal then the value of
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k is 1 for the ideal gases, and this is what
we saw in 1 of the earlier p v diagrams, so
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this is what comes out from the definition
of polytrophic index as its used for the gases
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ideal or real used in various cycle analysis
of the engines.
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Now, you we look at the t s diagram or more
appropriately referred to as temperature entropy
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diagram, which can also be referred to as
enthalpy entropy diagram in the sense the
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enthalpy entropy. And the temperature entropy
diagrams looks exactly same except that in
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one case the y axis is enthalpy, and the other
case the y axis is temperature in both the
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cases the x axis is entropy.
Now, if we put the various compression processes
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we were talking about earlier, and we can
see here that the isentropic process, where
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k is equal to gamma goes straight up from
1 to 2 prime and then we have the polytrophic
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process, where k is greater than gamma and
that is from 1 to 2 that is a real process,
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and then we have k equal to 1 which is the
isothermal process that is 1 to 2 double prime
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over here.
And then k equal to infinity where change
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of specific volume that is a is 0 that is
isochoric process; and it goes from 1 to 2
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triple prime, so all the processes we had
talked about earlier are now shown here on
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the t s diagram and those are the typical
lines or the path through which the compression
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process would proceed from 1 to 2 2 being
the end of the compression process.
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Now in case of compressor that we are talking
about in in typical jet engines the working
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of compression is accomplished in a open volume;
and there is a continuous flow of high speed
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air or gas in case of compressor it is air,
and we often adopt the so called isentropic
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process as a reference or a bench mark to
categorize or find out the goodness or badness
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of a real compression process.
This comparison between the ideal compression
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process or isentropic compression process,
and the real compression process is referred
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to as the isentropic efficiency of the process,
so isentropic efficiency of the process is
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essentially a comparison between real process
and isentropic process.
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And hence we need to know a priory what is
the isentropic work done before we know what
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is a real work that needs to be done because,
that then we will know what the isentropic
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efficiency is and we will know how close we
are to an ideal isentropic process and in
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a sense how close we are to the original joule
bration cycle which is the fundamental matrix
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of jet engines.
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If we look at the thermodynamic depiction
of row by row process of a compression through
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router and stator, we know the rotor does
work, and stator does not do any work really
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speaking, we will be talking about those things
in great detail in the chapters to come the
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thermodynamic paths taken by the air through
these compressors through router and stator
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can also be shown in terms of the T S diagram.
The path through the router is actually the
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work done, so it in isentropic work is is
the vertical work along the vertical line,
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and then the stator what is a horizon to line
where no work is being done and the work is
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essentially isothermal. So, this is the isothermal
work that we were talking about and this is
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the isentropic work that we were talking about
earlier.
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So along this k is equal to gamma along this
k is actually equal to 1, so it is a combination
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of these two that make up for the entire compression
process through router and stator, so that
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is how the work through the compression process
actually proceed it is k equal to gamma, and
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then k equal to 1 then again k equal to gamma.
And then again k equal to 1 so it is the step
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by step process by which the compression actually
proceeds through the thermodynamically in
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the typical T S diagram, if one does a rigorous
analysis what of course, happens is that the
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exit kinetic energy of the compressors is
often of the same order as the inlet kinetic
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energy, and as a result of which its expected
that the entire work that is supplied would
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be converted to pressure.
As a result of which what is often done is
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through this steps an average line is drawn,
and that average line is what we see in our
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cycle diagram representing the entire compression
process actually proceeds along the line that
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we have shown, if it is real process it proceeds
along the dotted lines. And then it reaches
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the end of compression what we show in a cycle
diagram is we simply connect the entry point,
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and the exit point through an average curve
and that is depicted as a compression process
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in a T S diagram.
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So rigorous depiction of the compression process
would have to be progressed in a step by step
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manner, let us take a look at how you show
the efficiencies of the compression, the high
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efficiency of the compressor and the turbine
that we would like to achieve actually takes
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the cycle pretty close to the joule brayton
ideal cycle closer, we are to the ideal cycle
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closer we can say that at the option of the
brayton cycle as the basis of a jet engine
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is fundamentally viable preposition.
Further we are away from the brayton cycle
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in actual engine operation adoption of the
brayton cycle is then not a acceptable or
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it become remains of questionable matrix of
analysis of jet engine. So, only way we can
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make sure that the jet engine is pretty close
to a brayton cycle is by ensuring that the
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compressors. And the turbines another components
of the jet engine do have very high efficiency
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operation, and they are pretty close to the
isentropic process; and this efficiency that
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we are talking about is indeed the isentropic
efficiency which I introduced just a few minutes
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back this thermodynamic efficiency or what
we call the isentropic efficiency can be shown
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in terms of the work that is being done ideally
the work that is being done really. And the
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real and the ideal work can be put together
in one relation; and that relation captures
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the essence of the work that is going on through
the compression process.
29:16.169 --> 29:22.929
So, let’s take a look at this thermodynamic
efficiency definition now this definition
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to understand this definition a littlie, let
us take a look at the thermodynamic depiction
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of the compression process and this time I
have put a h s diagram which is the enthalpy
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entropy diagram very similar to the t s diagrams
that we have looked at and what you shown
29:41.009 --> 29:48.850
here of course, are the various work that
is being done. And as you can see here this
29:48.850 --> 29:57.320
is the work ideal work that is being done
ideal adiabatic work in a stage, and then
29:57.320 --> 30:03.860
you have the work split up in rotors and rotor
and stator, and then of course you have the
30:03.860 --> 30:10.179
real work done over here in h that is a real
work done through the stage again that can
30:10.179 --> 30:18.289
be split up in rotor and stator that is H
1 and H 2. And then of course, H subscript
30:18.289 --> 30:24.830
0 is the total work done in term of the stagnation
parameters through the particular stage.
30:24.830 --> 30:30.850
So, all the various kinds of work that this
particular compressor stage is doing is being
30:30.850 --> 30:39.460
shown here, the blue line that is going from
00 to 01 and then to 02 is the total change
30:39.460 --> 30:46.250
parameter that is stagnation change parameter.
And hence can be called the stagnation work
30:46.250 --> 30:54.450
done from 00 to 01, and then from 01 to 02
which is of course, the stator where no work
30:54.450 --> 30:59.760
is being done, so the actual work being done
is form 00 to 01.
30:59.760 --> 31:08.850
The corresponding static work is from 0 to
1, and then from 1 to 2, so these works are
31:08.850 --> 31:15.730
all shown here, and they are written down
over here, the last three are the kinetic
31:15.730 --> 31:25.690
energies at the various stations shown as
H c 0 and H c 2 and H c 1, so these are the
31:25.690 --> 31:32.230
three kinetic energies that the flow carries
through the compressor at stations 01 and
31:32.230 --> 31:36.730
2.
Now, one the basis of this depiction of the
31:36.730 --> 31:42.389
work done through the compressor stage the
definition of the efficiencies can now be
31:42.389 --> 31:47.379
looked at again this is something very similar
to what you must have already done in your
31:47.379 --> 31:53.159
cycle analysis. Let us take a look at it and
understand how we can make use of this in
31:53.159 --> 32:01.559
our thermodynamic analysis of compressors.
Now, what is shown here is the efficiency
32:01.559 --> 32:09.820
of compressor in terms of the work done, this
is the ideal work done, this is the real work
32:09.820 --> 32:21.080
done and work at station 0 this is work at
station 2, and the numerator is the ideal
32:21.080 --> 32:27.929
work and denominator is the real work done,
and the ratio of the 2 is the efficiency which
32:27.929 --> 32:33.759
we have called isentropic efficiency.
The work done as you know can be written down
32:33.759 --> 32:40.110
in terms in terms of C p into temperature
differential, we assume that the c p of real
32:40.110 --> 32:46.830
and ideal processes are same, and that is
a an assumption for the time being the temperature
32:46.830 --> 32:52.470
differential or there ideal temperature differential.
And the real temperature differentials come
32:52.470 --> 32:58.970
into the picture, and the C p gets cancelled
out on what we have is the temperature ratios
32:58.970 --> 33:06.239
the ideal temperature ratio to the real temperature
ratio across the compression process the ideal
33:06.239 --> 33:11.479
temperature ratio can also be shown in terms
of the pressure ratio across the compression
33:11.479 --> 33:17.779
process. Now, we can see that the compression
ratio that we intend to achieve comes into
33:17.779 --> 33:23.919
the picture over here, and if this is the
compression ratio we intend to achieve.
33:23.919 --> 33:30.129
This can be put here and would give us the
ideal work that is necessary the denominator,
33:30.129 --> 33:35.440
of course is the real work that is actually
being consumed and that ratio of the two of
33:35.440 --> 33:43.759
course, is the overall isentropic efficiency
of the compression process. Now, this compression
33:43.759 --> 33:54.229
work can be shown in terms of two mean indices
specially, if they are finite processes as
33:54.229 --> 34:01.669
it is normally in a stage, if we consider
a stage which is a small change in pressure.
34:01.669 --> 34:10.030
We may consider it a finite process, and then
the indices that we have talked about k can
34:10.030 --> 34:17.740
be shown in two different ways, the first
one is often called thermodynamic the first
34:17.740 --> 34:25.200
mean index k 1, and that is something which
we have defined before and that is integral
34:25.200 --> 34:33.310
1 to 2 minus v dp divided by integral 1 to
2 p dv of the finite compression process,
34:33.310 --> 34:40.629
there is an another definition. And that is
the second mean index often referred to as
34:40.629 --> 34:52.250
k 2, and that is defined as ln that is natural
logarithm p 2 by p 1 divided by ln of v 2
34:52.250 --> 34:57.589
by v 1 v 2 of course, v of course, is the
specific volume.
34:57.589 --> 35:05.339
And this can be written down in terms of what
is known as the polytrophic relation that
35:05.339 --> 35:13.750
is p 1 v 1 to the power k 2 is equal to p
2 v 2 to the power k 2, which means we can
35:13.750 --> 35:20.339
write for a compression process p v to the
power k is equal to constant. So that is another
35:20.339 --> 35:27.530
way of showing the polytrophic process in
case of isentropic process it could be p v
35:27.530 --> 35:37.700
to the power gamma equal to constant in our
analysis of jet engines without getting into
35:37.700 --> 35:45.280
very minute thermodynamic analysis. It is
often customary to assume that the average
35:45.280 --> 35:52.280
value of the index as shown above the two
of them are equal to each other, and hence
35:52.280 --> 36:00.250
k 2 is equal to k 1 is equal to k in all designed
computations of a typical jet engine, so the
36:00.250 --> 36:06.109
two mean indices their definitions are slightly
different from each other can be considered
36:06.109 --> 36:07.670
to be one and the same.
36:07.670 --> 36:15.020
So, if we can get a value of the index through
one of the mechanisms, we can use it in the
36:15.020 --> 36:22.089
other relation that is p v to the power k
and carry on our analysis thermodynamic analysis
36:22.089 --> 36:30.910
of compression process. Now, this total head
based compression process that we have talked
36:30.910 --> 36:37.040
about let us have a quick look at that H S
diagram that we had a look at.
36:37.040 --> 36:43.240
You see the blue line shows the compression
process where the changes are occurring in
36:43.240 --> 36:51.130
stagnation parameters form 00 to 01 and from
01 to 02 which means there is a total temperature
36:51.130 --> 37:01.089
change from T 00 to T 01. Similarly, there
is a total pressure change from P 0 to P 01;
37:01.089 --> 37:09.380
and then to P 02 and there is a enthalpy change
form H 0 to H 01 and H 02.
37:09.380 --> 37:14.880
Those are the stagnation properties stagnation
temperature stagnation pressure stagnation
37:14.880 --> 37:21.780
enthalpy, now those stagnations changes are
what is shown here through this blue line,
37:21.780 --> 37:29.770
if we show a thermodynamic process in our
cycle through the stagnation property chain
37:29.770 --> 37:38.890
then it stands to reason that we should be
using a total head based specific heat which
37:38.890 --> 37:45.670
means instead of C r that we have written
down earlier for v l process. We should be
37:45.670 --> 37:56.220
writing C r, now this C or for our total head
bases specific heat definition can now be
37:56.220 --> 38:08.430
written down as C r to the power ln T 2 by
T 1 divided by ln T 02 by T 01, such that
38:08.430 --> 38:19.150
delta S S of course, is the entropy is equal
to C r into ln T 3 by T 1 across the entire
38:19.150 --> 38:28.290
compression process and that would be equal
to C 0r into ln T 03 by T 01.
38:28.290 --> 38:37.000
T 03 being the exit condition, T 01 being
the entry condition, so it is possible to
38:37.000 --> 38:43.349
actually come out with a specific heat based
on total temperature based; and total pressure
38:43.349 --> 38:51.900
based process in which specific heat is now
being shown as C 0r as oppose to C r as we
38:51.900 --> 39:00.089
have done before. Now that brings us to the
point that if we do that then the specific
39:00.089 --> 39:07.420
heat ratio that we have talked about k for
a polytrophic process also should be shown
39:07.420 --> 39:17.630
in terms of the total change of parameters,
and we can say it is depicted as k 0 for a
39:17.630 --> 39:25.520
real polytrophic process this k 0 can be written
down in terms of k 0 by gamma minus divided
39:25.520 --> 39:36.730
by k 0; and that would be equal to k by gamma.
Minus 1 divided by k into ln T 2 by T 1 divided
39:36.730 --> 39:42.960
by ln T 02 by T 01.
Now, these are all derivable definitions form
39:42.960 --> 39:49.650
the earlier thermodynamic definitions that
we has just introduced what happens is if
39:49.650 --> 39:59.760
you do that; if you introduce this specific
heat ratio as a total parameter, then if the
39:59.760 --> 40:05.000
entry and the exit mach numbers through the
compression process are equal.
40:05.000 --> 40:11.650
As we have said it sometimes could be equal
then for all practical purposes k 0 is equal
40:11.650 --> 40:20.059
to k, and we do not have to bother about k
0; however if m 1 is greater than m 3 then
40:20.059 --> 40:25.940
k 0 is likely to be greater than k 1, and
in many practical compressors quiet often
40:25.940 --> 40:36.599
it may well be, so if for example, if k is
equal to 1.5 and the value of k 0 will vary
40:36.599 --> 40:46.740
quiet a lot above 1.5 if m 1 is greater than
m 3. And this variation along with the mach
40:46.740 --> 40:54.490
number or velocity variation across the compressor
then would have to be factored into the rigorous
40:54.490 --> 41:00.960
cycle analysis, that needs to be done to get
a more realistic appraisal of the work done
41:00.960 --> 41:04.819
and the efficiency of the compression process.
41:04.819 --> 41:11.940
Let us quantify the effect of these changes,
if we have a change of k as we have seen k
41:11.940 --> 41:20.470
varies away from the ideal value which is
1. 4 and if for example, we have a compression
41:20.470 --> 41:27.920
process where it is ideally 1.4 that is air
we say that the compression has an efficiency
41:27.920 --> 41:35.540
of one, now if the value of k is 1.5 the effeminacy
starts falling. And if it goes all the way
41:35.540 --> 41:43.559
up to 1.8 the efficiency of the same compression
process would fall all the way to point 644
41:43.559 --> 41:48.599
which is only 64. 4 percent, which is rather
low efficiency.
41:48.599 --> 41:55.520
So, we can see that the compression efficiency
would be dependent quietly a lot on the value
41:55.520 --> 42:03.390
of k, and the value of k of course, depends
on the real polytrophic process that the compression
42:03.390 --> 42:13.039
is executing, if we look at k 0. And its effect
on the compression efficiency we see that
42:13.039 --> 42:26.200
has the value of k 0 changes from k to 1 value
to another, and as the compression ratio changes
42:26.200 --> 42:33.109
the value of efficiency changes substantially.
If we have a compression ratio which is very
42:33.109 --> 42:42.650
let us say one ideal that means no compression
at all we say that the efficiency is one,
42:42.650 --> 42:50.390
and when the value of k 0 changes it remains
1, but once the compression ratio increases
42:50.390 --> 42:59.700
from 1.2, 1.5 which are pressure ratios, and
then multistage pressure ratios of 2 4 6 and
42:59.700 --> 43:06.930
10 for a value of k which is now let us say
1.5.
43:06.930 --> 43:13.910
The or k 0 for example, if the value is 1.5
the efficiencies won’t start falling and
43:13.910 --> 43:21.780
we can see that the efficiency falls all the
way up to 94 percent, if the efficiency ratio
43:21.780 --> 43:28.700
that are looking at if the value of k now
changes to 1.8 the efficiency would start
43:28.700 --> 43:34.299
falling substantially, and this efficiency
ratio that we are looking at falls from 1
43:34.299 --> 43:36.960
to 0.81.
43:36.960 --> 43:45.200
So these are the various effects of the value
of k and k 0 if we use k 0 for total based
43:45.200 --> 43:53.920
cycle analysis, and we see that they have
substantial effect on the efficiency computation
43:53.920 --> 44:03.839
that one may like to do for cycle analysis
of jet engines. Now, we take a look at what
44:03.839 --> 44:10.960
happens? If we have a compression process
with a certain amount of heat transfer what
44:10.960 --> 44:20.300
does happen is that when the process of heat
transfer is going on the compression is affected,
44:20.300 --> 44:27.789
now the blue line one to two adiabatic is
indeed the ideal process or we may call isentropic
44:27.789 --> 44:35.000
process.
On the other hand the red line 1 to 2 is the
44:35.000 --> 44:41.260
polytrophic process, where heat transfer across
the process is still 0, so it is a polytrophic
44:41.260 --> 44:49.530
process where there is no heat transfer isentropic
process of course, has no heat transfer anyway,
44:49.530 --> 44:54.890
and then we consider two processes one in
which the heat transfer is greater than 0
44:54.890 --> 44:59.980
another in which the heat transfer is less
than 0.
44:59.980 --> 45:04.700
So in one case the compression process or
compressor is subjected to to certain amount
45:04.700 --> 45:10.300
of heating, and in the other case the compressor
is subjected to certain amount of cooling.
45:10.300 --> 45:19.289
Now, we can see that the area under the line
1 to 2 is indeed the extra work that is done
45:19.289 --> 45:25.990
to overcome the losses of a real process or
polytrophic process, and that is the extra
45:25.990 --> 45:33.599
work that one has to do to overcome this path
1 to 2 for the time being we are considering
45:33.599 --> 45:41.230
this to be a linear path 1 to 2, we have seen
before that path could be curved for a long
45:41.230 --> 45:45.970
compression process.
Let us consider a small compression process
45:45.970 --> 45:53.069
in which this path is let us say a linear
path, and then we can say each of these can
45:53.069 --> 46:02.089
be considered as rectangles or triangles an
as a result of this. Ee have a heat transfer
46:02.089 --> 46:09.680
now being introduced into the compression
process in which extra work is necessary to
46:09.680 --> 46:17.630
overcome the work done in a polytrophic process
As we can see now, if you introduce heating
46:17.630 --> 46:23.500
you would actually need to do more work because,
the entropy rise or the loss in such a process
46:23.500 --> 46:29.270
be even higher and you would actually need
to do more work, because the path from 1 to
46:29.270 --> 46:37.730
2 prime. Now is longer to accomplish the same
pressurization from P 1 to P 2, see in all
46:37.730 --> 46:43.760
these processes the pressurization is same
from P 1 to P 2; this path is longer which
46:43.760 --> 46:51.079
means you would need to do more work the temperature
rise from T 1 to T two prime is higher, so
46:51.079 --> 46:58.260
you would actually need to do more work.
On the other hand, if you have a cooled compression
46:58.260 --> 47:04.820
process the area under that is smaller than
the earlier polytrophic process, and as a
47:04.820 --> 47:10.520
result of which you have a lesser work to
be done for the compression process, and this
47:10.520 --> 47:16.039
means that if you have a certain amount of
cooling introduced in the compression process
47:16.039 --> 47:23.069
you can actually get away with doing less
of compression work. And this is the thermodynamic
47:23.069 --> 47:30.710
concept which has given rise to various kinds
of cooled compression process or in case of
47:30.710 --> 47:35.039
jet engines or various kinds of gas turbine
engines.
47:35.039 --> 47:42.019
The concept of inter cooling or even simple
cooling has been used for many many years,
47:42.019 --> 47:50.630
wherever there is a lot of the working condition
is very hot as it is in places like India
47:50.630 --> 47:57.720
you often refer to cooling of the flow that
is going into the compressor. And this cooling
47:57.720 --> 48:05.380
as we can see now form pure thermodynamics
introduces a simple thermodynamic basis, which
48:05.380 --> 48:12.519
allows the compressor to do the same amount
of compression with lees work necessary to
48:12.519 --> 48:19.210
be supplied by the turbine; and this is what
thermodynamics tells us that if you have some
48:19.210 --> 48:26.630
method by by which you can accomplish a little
bit of cooling actual work of compression
48:26.630 --> 48:33.140
can be done with less of work supplied.
There is a process here which is which shows
48:33.140 --> 48:39.380
that if you do a lot of cooling which can
one call as super cooling you can get away
48:39.380 --> 48:48.109
with doing a lot less work to accomplish the
same pressurization from 1 to 2 triple prime,
48:48.109 --> 48:54.059
so the pressurization would indeed be achieved,
but its possible that you would be arriving
48:54.059 --> 48:59.930
at a temperature of a lower value at the end
of the compression process even lower than
48:59.930 --> 49:09.670
the ideal in which case for a jet engine purpose
that may not be a very desirable situation.
49:09.670 --> 49:14.880
Because the flow from the compressor is going
to combustion chamber, and a super cooled
49:14.880 --> 49:20.530
flow going into the combustion chamber would
require additional heating in the combustion
49:20.530 --> 49:28.579
chamber for the flow to be taken to high temperature,
so super cooling may not be a desirable phenomenon
49:28.579 --> 49:31.569
in a typical jet engine operation.
49:31.569 --> 49:38.859
But a certain amount of cooling, as we can
see is certainly a desirable phenomenon for
49:38.859 --> 49:49.640
more efficient compression process, we can
summarize all this by simply saying that the
49:49.640 --> 49:55.940
as we have said those diagrams are in terms
of linear diagrams. And hence the final enthalpy
49:55.940 --> 50:06.141
can be written down in terms of the triangle
274 which is shown here in terms of 2 and
50:06.141 --> 50:16.490
7 and 4, and that is the triangle we are approximating
it as a triangle, and that is the final enthalpy
50:16.490 --> 50:29.540
the initial enthalpy is the triangle 103 and
that is shown here in terms of 03 and 1 and
50:29.540 --> 50:33.990
that is the triangle with which the flow is
entering the system.
50:33.990 --> 50:40.700
So, it enters the system through one goes
out through the system from two, so this is
50:40.700 --> 50:48.930
your energy at the entry point, and this is
the energy at the exit point, so these are
50:48.930 --> 50:54.520
the various phenomena that is going on in
a typical compression process as shown in
50:54.520 --> 51:00.440
a little simplified thermodynamic diagram
where the processes are considered to be linear.
51:00.440 --> 51:07.590
So that some of those can be simply referred
to as triangles or rectangles, and as a result
51:07.590 --> 51:15.019
of which those some of those total enthalpy
change can now be shown in terms of the area
51:15.019 --> 51:26.019
2 5 6 4 2 and that is 2 5 6 4 2, and that
is a area which is the total enthalpy change
51:26.019 --> 51:32.289
in terms of total parameter H 0.
So, this is what we learnt from simple thermodynamic
51:32.289 --> 51:38.890
analysis that you can use a thermodynamics
as a matrix try to figure out what is going
51:38.890 --> 51:44.470
on it terms of compression process, and then
see whether you can get your compression process
51:44.470 --> 51:51.329
done a little more efficiently a little more
effectively. So, that it serves the purpose
51:51.329 --> 51:57.700
of the jet engine, and brings the jet engine
closer to the ideal brayton cycle which is
51:57.700 --> 52:04.650
the matrix on which the jet engine is functioning.
So, this is what we have done in today’s
52:04.650 --> 52:12.480
class the thermodynamics of compression process,
and we can see that the cooled thermodynamic
52:12.480 --> 52:18.690
compression can give us some benefits. In
the next class what we will be doing is thermodynamics
52:18.690 --> 52:26.500
of turbines similar thermodynamic analysis.
We will then refer to in thermodynamic diagrams
52:26.500 --> 52:32.119
and see whether we can learn something from
a fundamental thermodynamic understanding,
52:32.119 --> 52:40.130
and make the turbine do work in a more efficient
manner to conform to our ideal cycles that
52:40.130 --> 52:46.869
we have studied over the last few lectures.
So, in the next class we will look at turbines.