WEBVTT
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and welcome to lecture number eleven of this
lecture series on jet propulsion. We have
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been discussing about the various aspects
of jet engine cycles; we have discussed about
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the ideal cycles, the real cycles, and also
the component performance parameters based
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on which we were able to derive the real cycles.
So, based on our discussion during the last
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several lectures, you must have had some idea
of what is involved in carrying out a cycle
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analysis. And as I have been discussing earlier,
it is also being assumed that you have undergone
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some sort of training in the form of the ideal
cycle analysis; specially, if you have undergone
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the previous course on introduction to aerospace
propulsion.
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So, in today’s class as promised in the
last class we shall be discussing a few problems
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we will basically be having a tutorial session,
we will solve a few problems in today’s
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class. And we will see how we can carry out
a real cycle analysis for a few configurations
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of the jet engine. So, we will basically be
having a problem on a simple turbo jet. And
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then I have configured a problem on a turbo
jet the same problem with after burning.
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We will subsequently solve a problem on turbo
fan, one of the configurations of a turbo
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fan. And then I also have a problem for you
on turbo prop engines. And towards the end
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of the class, I will also give you a few exercise
problems, which you can try and solve based
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on our discussion in the previous classes
as well as based on the tutorial session that
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we have today.
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So, today’s class is basically a tutorial
session, we will solve a few problems from
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the jet engine cycle we will basically be
solving real cycle problems; and not the ideal
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cycles. So, the first problem that we have
for today’s discussion is a simple turbo
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jet. So, let us take a look at the problem
statement.
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So, problem number one states that an aircraft
using a simple turbo jet engine flies at mach
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0.8, where the ambient temperature, and pressure
are 223.3 Kelvin, and 0.265 bar respectively
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the compressor pressure ratio is 8 and the
turbine inlet temperature is 1200 Kelvin the
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isentropic efficiencies of the compressor
are 0.87 the turbine is 0.90 the intake is
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0.93.
The nozzle is 0.95 the mechanical efficiency
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is 0.99 the combustor efficiency is 0.98.
The pressure loss in the combustor is 4 percent
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of the compressor delivery pressure determines
the thrust; and specific fuel consumption
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so, this is the first problem that we shall
be attempting to solve today.
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It is basically on a simple turbo jet engine
a simple turbo jet as you know is a turbo
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engine, which does not have any form of after
burning. So, which means that there is no
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reheating in this particular cycle so, we
have been given a lot of data corresponding
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to this particular turbo jet engine. We have
the compressor pressure ratio the turbine
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inlet temperature. And all the efficiencies
of the components like the compressor the
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turbine the fuser nozzle. And so, one we also
have been given the ambient conditions; and
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the flight mach number so.
Based on this data’s we should be able to
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carry out a cycle analysis; and determine
what is the thrust; and specific fuel consumption
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of such an engine so, given this particular
problem statement, where do we first begin
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to solve this problem so, as you must have
already realized by now the first step towards
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solving such a problem is to get the cycle
diagram first.
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That is on either on a temperature entropy
plot or on a pressure volume plot so, once
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we get the cycle diagram right. And also we
mark the points, where we have the data with
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us. And we would therefore, be able to determine,
which are those points, where we need to find
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the properties of temperature, and pressure,
and therefore will help us in determining
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the exhaust velocity, and hence the thrust,
and fuel consumption.
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So, the first step towards solving this problem
is to draw the cycle diagram. So, this is
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a turbo jet cycle without after burning. And
this problem does not have this particular
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jet engine that we have does not have any
after burning. So, this is the real turbo
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jet cycle that we had discussed in the last
class on a temperature entropy plot. So, I
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will just quickly review the different processes
involved in this particular turbo jet cycle,
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the ambient conditions are denoted by a subscript
a station two corresponds to the intake exit
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or the diffuser exit, which is also the compressor
inlet.
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Station 3 is the compressor outlet station
4 is the turbine outlet 5 is turbine outlet;
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and 7 is the nozzle exit so, process between
a; and 2 is the compression in the intake
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of the diffuser; and the line shown by this
dotted line is the real process or the actual
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process; and because it is an actual process.
And because they have been given diffuser
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efficiency it means that that process is not
isentropic.
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So, process a to two is non isentropic the
compression begins at from station between
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2; and 3 is the compressor again this is also
a non isentropic process, which is why we
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have this dotted line which indicates the
compression process between process states
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3, and 4 we have the combustion chamber or
the heat addition.
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And we have been given the combustion efficiency
as well as the pressure loss in the combustion
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chamber so, this process is no longer a constant
pressure process so, there is a pressure loss
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occurring in the heat addition process or
the combustion process between stations 4,
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and 5 is the turbine. And the turbine process
expansion process is non isentropic; and that
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is why the expansion process is indicated
by this dotted line.
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Between stations 5, and 7 is the nozzle, and
this again is a non isentropic process, and
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basically we get the exhaust dust as a result
of expansion through the nozzle so, these
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are the various processes that are that basically
constitute the turbo jet cycle, and what we
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shall try to do now is that we will identify
those points where we have been given the
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data. And then we will get some idea of which
are those points where we need to find out.
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Temperature, and pressure, and that is of
course, that is what will be required for
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calculating the exhaust velocity so, we have
been given the ambient velocity ambient temperature,
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and pressure, and the mach number, and then
the diffuser efficiency is given the compressor
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pressure ratio is given then we have the turbine
inlet temperature.
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So, these are the different property or parameters
which have been specified in the problem besides
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of course, the efficiencies of all the components
so, based on the ambient temperature, and
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pressure, and the mach number we can calculate
the ambient stagnation conditions, and also
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because intake efficiency is given we can
calculate the intake exit conditions like
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its stagnation temperature and pressure.
And then we have the compressor pressure ratio,
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therefore compressor inlet conditions being
known we can find out the compressor exit
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conditions from the pressure ratio. And the
efficiency then we arrive at the combustion
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chamber where again we have been given the
efficiency the pressure loss; and the exit
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conditions of the combustor that is the turbine
inlet temperature.
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So, from the combustor analysis we can find
out the fuel to air ratio; and then we have
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the turbine inlet condition is known. And
so, we can find out by equating the turbine
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work to the compressor work the turbine exit
conditions. And also the exhaust nozzle which
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will be solved so, these are the steps that
will be that are involved in solving this
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particular problem.
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So, let us begin with the intake, where we
have the intake inlet conditions that is we
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have the ambient temperature pressure Mach
number; and the intake efficiency so, for
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the given ambient conditions that is the temperature,
and pressure, and the mach number we can calculate
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the flight speed, because flight speed will
be equal to Mach number into the speed of
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sound that is square root of gamma RT.
So, from that we can calculate flight speed
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which comes out to be 239.6 meters per second
so, this is calculated, because the Mach number
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is known; and the ambient temperate is known;
so we can calculate the speed of sound, which
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is square root of gamma RT that multiplied
by the Mach number gives us this flight speed.
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And so, once we know the flight speed or the
Mach number we can calculate the intake exit
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stagnation conditions - that is the stagnant
temperature, which is the ambient temperature
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plus V square by 2C p. And so, we the ambient
temperature is known the velocity is known;
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and C p for air we will be assuming as 1.005
kilo joules per kilo grams per Kelvin; and
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for the exhaust products or the combustion
products we will assume it to be 1.47 kilo
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joules per kilo grams Kelvin.
So, based on these when we substitute these
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values; and simplify we get the stagnation
condition as 251.9 Kelvin; and now similarly,
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the stagnation pressure is given by this stagnation
is related to the stagnation temperature to
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be isentropic efficiency of the diffuser so,
P 0 2 by P a is equal to one plus diffuser
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efficiency multiplied by T 0 2 by T a minus
one raised to gamma by gamma minus 1.
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So, diffuser efficiency is given to us it
is 0.93 stagnation temperature we have just
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now calculated ambient temperature is known;
and gamma for air is 1.4, so if we substitute
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these values we get P 0 2 by P a is equal
to 1.482. And since ambient pressure static
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pressure is known stagnation pressure can
be calculated by P a multiplied by 1.482 that
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is 0.265 into 1.482; that is 0. 393 bars so,
what we have calculated now are the exit conditions
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of the intake which will basically be the
inlet conditions of the compressor so, based
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on these conditions that are correspond to
the compressor inlet we can now proceed towards
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calculating the compressor exit conditions.
And how do we calculate that compressor exit
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pressure is straight forward.
Because the pressure ratio is specified so,
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given the pressure ratio we can calculate
the compressor exit stagnation pressure which
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is basically the stagnation pressure multiplied
at the inlet of the compressor multiplied
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by the pressure ratio so, P 0 3 will be equal
to pi c which is the compressor pressure ratio
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multiplied by P 0 2. And how do you calculate
the temperature is basically calculated using
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the compressor isentropic efficiency definition,
which is eta d is equal to T 0 3 minus T 0
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2 T 0 3 s by T 0 2 divided by T 0 3 minus
T 0 2. And if we simplify this we get an expression
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in terms of the pressure ratio; and therefore,
we can calculate T 0 3, we have discussed
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that during the last lecture, when we took
up the cycle analysis of the turbo jet.
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So, compressor exit pressure is basically
just the product of the pressure ratio; and
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inlet stagnation pressure, which is already
calculated as 0. 393 bar. And so, that multiplied
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by the exit by the pressure ratio that is
8.0 gives us the exit stagnation pressure.
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So, compressor exit stagnation pressure is
3.144 bar, and exit stagnation temperature
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is from the efficiency definition.
T 0 3 is equal to T 0 2 multiplied by one
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by eta c which is the isentropic efficiency
of the compressor multiplied by pi c raised
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to gamma minus 1 by gamma minus 1 plus 1 so,
this basically gives us the stagnation temperature
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at the compressor exit so, all these parameters
on the right hand side are known to us, now
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T 0 2 we have calculated in the previous expression
in the previous slide. Eta c is specified
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c is the pressure ratio gamma is 1.4 so, we
substitute for all these values in this expression
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on the right hand side we should be able to
get an expression for we should be able to
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find out the compressor exit stagnation temperature
so, substituting for all these values we get
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T 0 3 is equal to 486.8 Kelvin
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So, this is the stagnation temperature at
the compressor exit so, compressor exit stagnation
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temperature we have related that to the pressure
ratio; and the efficiency based on which we
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can calculate the compressor exit conditions
so, the next component is we have now called
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or carried out the cycle analysis up to the
compressor exit. And so, we now have data
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up to the compressor exit in terms of the
stagnation pressure; and the stagnation temperature
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we also have which is basically the combustion
chamber inlet conditions.
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And for the combustion chamber we have the
combustion efficiency, which has been specified;
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and the pressure loss taking place in the
combustion chamber. And what else we have
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we also have the combustor exit temperature
stagnation temperature, which is the turbine
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in the temperature so, based on this much
data that we have we should be able to calculate
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the fuel to air ratio. And how do we calculate
that if you recall in the last lecture we
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had discussed to calculate the fuel to air
ratio, which is carry out energy balance from
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between the inlet; and exit. So, energy balance
will be able to help us in calculating the
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fuel to air ratio so, enthalpy at inlet which
is the enthalpy of the air coming in from
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the compressor plus the enthalpy or heat of
reaction of the fuel.
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Is equal to enthalpy at the exit of the combustion
chamber so, we solve this; and we should be
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able to get the fuel to air ratio. So, h 0
4 which is the enthalpy at the exit of the
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combustion chamber, which is the turbine inlet
is equal to h 0 3, which is the compressor
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exit plus the burner efficiency or the combustion
chamber efficiency.
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Multiplied by the fuel to air ratio; and the
heat of reaction of the fuel we should be
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able to solve this; and get the fuel to air
ratio so, C p times T 0 4 is equal to C p
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a times T 0 3 eta b into f times Q dot f so,
C p g here corresponds to the specific heat
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of the combustion product C p a corresponds
to the specific heat of air. So, all these
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numbers are known to us now so, f we can simplify.
And that would be equal to C p g by into T
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0 4 divided by C p a into T 0 3 minus one
divided by burner efficiency or combustion
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efficiency into Q dot f into C p a T 0 3 minus
C p g T 0 4 by C p a T 0 3. Now we will assume
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a value of q dot f, because it is not explicitly
stated in the problem usually for jet engines
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the fuel that is used is known as aviation
turbine fuel. And that is a fuel which is
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similar to kerosene.
And for that particular fuel the Q dot f or
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heat of reaction is about forty four mega
joules per kilo gram so, we will assume that
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value here; and rest of the values are already
known to us so, we substitute for all these
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values; and if we substitute for all of them
T 0 4 is basically the turbine inlet temperature
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which is given T 0 3 is compressor exit stagnation
temperature, which we have calculated, and
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C p a, and C p g we have known eta b burner
efficiency is also given so, f is equal to
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if we simplify substitute for all these values,
and simplify we get 0.0198.
17:47.629 --> 17:55.769
Now so, this is the fuel to air ratio what
else is required to be calculated we need
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to know to calculate the combustion chamber
exit stagnation pressure exit stagnation temperature
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is known. That is the turbine inlet temperature
exit stagnation pressure is equal to the pressure
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loss in the combustion chamber multiplied
by the inlet time machine pressure it is given
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in the problem that 4 percent loss occurring
in the combustion chamber so, pi b is basically
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equal to 0.96 so, P 0 4 will be pi b multiplied
by P 0 3 that is 0.96 into 3.144 so, that
18:32.830 --> 18:38.320
is 3.018.
So, this is how we have estimated; and calculated
18:38.320 --> 18:45.600
the properties all that way from the inlet
then proceeding towards the compressor. And
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the combustion chamber the next component
is the turbine, we have the turbine inlet
18:52.369 --> 18:58.269
stagnation temperature, and pressure, and
we also have the mass flow rate in the sense
18:58.269 --> 19:02.570
that we have mass flow rate of air plus mass
flow rate of fuel we have just calculated
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the fuel to air ratio.
So, we should now be able to calculate the
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fractional mass flow rate that is actually
going into the turbine so, all these conditions
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are known for the turbine, now what we need
to calculate are the exit conditions of the
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turbine. And how do we calculate that remember
in a simple turbo jet engine the basic function
19:24.230 --> 19:29.919
of a turbine is only to drive a compressor.
It means the turbine is generating just enough
19:29.919 --> 19:36.210
work to drive the compressor, so turbine work
the work output of the turbine will be equal
19:36.210 --> 19:42.019
to the work input of the compressor. And of
course, there is there is a mechanical efficiency
19:42.019 --> 19:47.049
involved that means the work output of the
turbine is getting diminished by a certain
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fraction which is equal to the mechanical
efficiency.
19:50.730 --> 20:00.200
So, multiplied by the turbine work will be
equal to the compressor work this if we simplify
20:00.200 --> 20:06.840
which should be able to calculate the compressor
turbine exit stagnation temperature. And then
20:06.840 --> 20:11.650
from the efficiency definition we can calculate
the stagnation pressure. And so that is how
20:11.650 --> 20:15.269
we will be proceeding towards calculating
turbine exit conditions.
20:15.269 --> 20:20.390
So, let us do that; and see what the turbine
exit conditions come out to be now since we
20:20.390 --> 20:28.289
know that turbine exit is work done by the
turbine is equal to that of the compressor,
20:28.289 --> 20:33.119
we know that the mechanical efficiency multiplied
by m dot, which m dot mass flow rate of air
20:33.119 --> 20:40.149
plus mass flow rate of fuel this sum of this
multiplied by C p of gases into the temperature
20:40.149 --> 20:47.279
difference across the turbine that is T 0
4 minus T 0 5 is equal to m dot, which is
20:47.279 --> 20:54.009
mass flow rate of air into C p of air into
T 0 3 into T 0 2 which is the differential
20:54.009 --> 20:58.519
temperature across the compressor.
So, left hand side is the work done by the
20:58.519 --> 21:04.149
turbine right hand side is the word done by
the required by the compressor. So, let us
21:04.149 --> 21:10.980
simplify this once you simplify this in this
equation the only unknown is T 0 5. So, if
21:10.980 --> 21:18.269
you simplify this; and we get an expression
in terms of T 0 5; so, T 0 5 on simplification
21:18.269 --> 21:26.749
is equal to C p into well C p g into T 0 4
minus C p a into T 0 3 minus T 0 2 divided
21:26.749 --> 21:35.710
by mechanical efficiency into one plus f,
because we cancel out m dot, so we get 1 plus
21:35.710 --> 21:41.659
f, so the entire all the terms which are involved
on the right hand side are known from our
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cycle analysis so far. So we have C p g, which
is 1147 into T 0 4 which is 1200 minus C p
21:51.340 --> 22:01.299
a, which is 1005 multiplied by T 0 3 that
is 486.8 minus 251.9 divided by mechanical
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efficiency that is 0.99 into 1 plus f that
is 1 plus 0.189, if we calculate this we will
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get the exit temperature as 998.3 Kelvin,
so this is the stagnation temperature at the
22:19.090 --> 22:26.470
turbine exit now, how do we calculate the
stagnation pressure at the exit? We will now
22:26.470 --> 22:32.619
make use of the efficiency definition of the
turbine well efficiency of the turbine isentropic
22:32.619 --> 22:38.899
efficiency of the turbine is defined as eta
t is equal to the inlet temperature stagnation
22:38.899 --> 22:47.360
temperature T 0 4 minus T 0 5, which is the
actual temperature divided by T 0 4 minus
22:47.360 --> 22:54.049
T 0 5 isentropic. Now if we divide this and
convert them into temperature ratios in the
22:54.049 --> 23:01.899
denominator, we have the isentropic temperature
ratio T 0 5 s divided by T 0 4, which is equal
23:01.899 --> 23:10.350
to P 0 5 by P 0 4 raised to the gamma minus
one by gamma from this implication, we should
23:10.350 --> 23:16.500
get an expression for T 0 5, which is the
turbine exit stagnation pressure.
23:16.500 --> 23:22.739
We already derived that in the last lecture;
and so T 0 5 can be expressed in terms of
23:22.739 --> 23:29.239
the efficiency; and the temperature ratio,
so if we do that we get P 0 5 is equal to
23:29.239 --> 23:36.539
P 0 4 by four multiplied by one minus one
by eta t the turbine efficiency into one minus
23:36.539 --> 23:44.750
T 0 4 this raised to gamma by gamma minus
1. So, all these values are known to us T
23:44.750 --> 23:51.029
0 4 is known turbine efficiency is known.
And these two temperatures are also known;
23:51.029 --> 23:58.269
and remember we are going to use gamma here
as one 0.33; and not 1.4, because we are not
23:58.269 --> 24:03.799
dealing with the combustion products; and
so gamma is we are going to assume an average
24:03.799 --> 24:10.739
gamma for all the combustion products; and
that is going to be assumed as 1.33; and for
24:10.739 --> 24:17.789
the average gamma for air is 1.4.
So, if we substitute for all these values
24:17.789 --> 24:23.590
here; and simplify we get 305, which is the
turbine exit stagnation pressure which his
24:23.590 --> 24:34.479
equal to 1.284 bar, so this is the pressure
of the of the exit of the turbine. So we have
24:34.479 --> 24:39.269
data now all the way up to turbine exit, we
also have now the turbine exit stagnation
24:39.269 --> 24:43.259
pressure.
And the temperature; and in this particular
24:43.259 --> 24:48.649
engine configuration there is no after burner;
and since there is no mention of any other
24:48.649 --> 24:55.249
losses after the turbine except the nozzle
efficiency we can assume that all these parameters,
24:55.249 --> 25:02.919
which we have calculated for the turbine exit
is valid for the nozzle inlet also. That means
25:02.919 --> 25:10.379
T 0 5; and P 0 5 will be the same as the nozzle
entry, so we will use these same values at
25:10.379 --> 25:15.960
the nozzle entry to calculate the nozzle exit
condition. And therefore, the velocity of
25:15.960 --> 25:23.340
the jet, but before that we have to make one
check, which is basically to see this nozzle
25:23.340 --> 25:30.629
is operating under choked conditions or not.
Because if it is choking then the exit conditions
25:30.629 --> 25:35.510
get fixed by the critical parameters that
is the exit temperature will be the critical
25:35.510 --> 25:40.919
temperature exit pressure will be the critical
pressure, and density will be critical density
25:40.919 --> 25:43.229
and so on.
So based on which we need to know calculate
25:43.229 --> 25:49.119
the other conditions; if it is un chocked,
then the nozzle exit conditions are to be
25:49.119 --> 25:55.640
calculated using the enthalpy drop across
the nozzle something, you have done in the
25:55.640 --> 25:57.989
last class during the cycle analysis discussion.
25:57.989 --> 26:03.399
So, let us check for the chocking of the nozzle,
so we first check for nozzle chocking. And
26:03.399 --> 26:10.779
how do we do that if you are familiar with
the ideal cycle analysis, you must have had
26:10.779 --> 26:17.739
some experience of calculating the nozzle
pressure ratio. And to check whether its indeed
26:17.739 --> 26:27.110
chocking the nozzle pressure ratio is basically,
P 0 5 by P a P 0 5 is known from our previous
26:27.110 --> 26:32.249
calculation.
That is 1.284; and P a is given as 0.265 bar
26:32.249 --> 26:41.200
so, P 0 5 by P a is equal to 4.845; and that
is the nozzle pressure ratio; and what is
26:41.200 --> 26:47.659
the critical pressure ratio that is P 0 5
by P c how do we calculate this critical pressure
26:47.659 --> 26:54.529
ratio? This is basically calculated by equating
mach number is equal to one. And so, we can
26:54.529 --> 26:58.759
from isentropic relations we should be able
to calculate the critical pressure ratio,
26:58.759 --> 27:04.779
but in that case we have been given a nozzle
efficiency as well as, so there is a nozzle
27:04.779 --> 27:11.070
efficiency that also comes into picture here,
so this basically can be derived when we either
27:11.070 --> 27:17.070
from the nozzle efficiency definition or from
the pressure ratio definition, so critical
27:17.070 --> 27:23.919
pressure ratio is 1 by 1 minus 1 by eta n
into gamma minus 1 by gamma plus 1 this raised
27:23.919 --> 27:28.659
to gamma by gamma minus 1.
So, if you substitute for all these values,
27:28.659 --> 27:34.899
we have the efficiency; and gamma which is
1. 33, then we get a critical pressure ratio
27:34.899 --> 27:43.570
1.914 so, here we have now two pressure ratios
1 is the actual nozzle pressure ratio which
27:43.570 --> 27:51.070
is 4.845; and the critical pressure ratio
which we have calculated as 1.914. So what
27:51.070 --> 27:57.460
we see here is that the nozzle pressure ratio
is greater than the critical pressure ratio;
27:57.460 --> 28:01.970
and what that means that means that is the
nozzle pressure ratio being greater than nozzle
28:01.970 --> 28:09.110
the critical pressure means that the nozzle
is chocking. And why should that be, because
28:09.110 --> 28:16.190
from the pressure ratio that we have it means
that the nozzle entry pressure is much higher
28:16.190 --> 28:21.099
than the critical pressure that is it will
expand only up to the critical pressure ratio,
28:21.099 --> 28:26.899
if we have a pressure ratio higher than that
it will still mean that the exit conditions
28:26.899 --> 28:32.669
of the nozzle are choked or fixed.
Because it is a convergent model, so because
28:32.669 --> 28:38.390
it happens to be a convergent nozzle then
the exit conditions are fixed that is Mach
28:38.390 --> 28:43.049
number will become one mass flow rate will
be maximum mass flow rate. And therefore,
28:43.049 --> 28:49.740
it is choking; and so, if it is chocking then
the nozzle exit conditions are fixed; and
28:49.740 --> 28:54.600
they fixed, because those conditions will
be equal to the critical parameters that is
28:54.600 --> 28:58.570
the critical temperature; and critical pressure;
and the density.
28:58.570 --> 29:03.729
That will also fix the exhaust velocity, because
since Mach number is equal to one the exhaust
29:03.729 --> 29:08.370
velocity will now be equal to square root
of gamma r t which is speed of sound based
29:08.370 --> 29:15.900
on the critical temperature so, in this problem
we now have a scenario where the nozzle is
29:15.900 --> 29:21.730
chocking. And therefore, we now will calculate
the nozzle exit conditions based on the critical
29:21.730 --> 29:26.960
parameters. So, let us see how we can calculate
the critical parameters so, we will calculate
29:26.960 --> 29:32.039
the critical temperature pressure. And density
so, nozzle exit conditions will get fixed
29:32.039 --> 29:33.549
by critical parameters.
29:33.549 --> 29:38.739
So, the critical temperature that is static
temperature T 7 will be equal to critical
29:38.739 --> 29:46.809
temperature T c which is 2 by gamma plus 1
into T 0 5. And it becomes 2 by gamma plus
29:46.809 --> 29:53.499
1, because of the temperature ratios that
is T 0 5 by T 0 c is 1 plus gamma minus 1
29:53.499 --> 29:59.149
by 2 m square; and if we put m is equal to
1 we get this expression so, T c is 2 plus
29:59.149 --> 30:05.580
gamma into T 0 5 so, if we substitute for
all these values 850.7 Kelvin. Similarly,
30:05.580 --> 30:14.179
P 7 or P c is equal to T 0 5 multiplied by
1 by P 0 5 by P c, therefore this comes out
30:14.179 --> 30:21.250
to be 0.671 bar.
The density rho seven is equal to P 7 by RT
30:21.250 --> 30:30.999
7 this on simplification we get 0.275 kilograms
per meter cube, so we have temperature pressure;
30:30.999 --> 30:36.929
and density at the exit, so exhaust velocity
v exit is equal to square root of gamma RT
30:36.929 --> 30:44.249
7. And we substitute, and we find out that
the exhaust velocity is 570.5 meters per second,
30:44.249 --> 30:47.799
so this is the exit velocity.
30:47.799 --> 30:52.909
We can also calculate from what the data that
is known to us this particular parameter that
30:52.909 --> 30:58.450
that is area to mass flow rate ratio, because
this will be required for calculating thrust
30:58.450 --> 31:03.929
so, exit area by mass flow rate is equal to
one by rho seven into V exit, because mass
31:03.929 --> 31:11.409
flow rate is rho into area into velocity.
So, this parameter that is area divided by
31:11.409 --> 31:19.289
mass flow rate is 0.006374 zero meters square
second per kilogram therefore, we now have
31:19.289 --> 31:26.379
the exit velocity the flight mach number;
and the pressures that is P c minus P f critical
31:26.379 --> 31:29.779
parameter pressure is known that is exit pressure
is known that is the ambient pressure is also
31:29.779 --> 31:34.840
known; and they are not equal.
And so, if we substitute for all these values
31:34.840 --> 31:41.479
the specific thrust we can calculate as 1
plus f into V m exit minus V plus A e by m
31:41.479 --> 31:49.960
dot into p c minus p a so, if we substitute
all the values we get 596.25 Newton second
31:49.960 --> 31:56.480
per kilogram so, this will be the thrust specific
thrust which is Newton per that is thrust
31:56.480 --> 32:04.879
per unit mass flow that is 596.25 Newton second
per kilogram so, we have calculated the thrust
32:04.879 --> 32:11.600
what is the next parameter to be calculated
that is fuel consumption or specific fuel
32:11.600 --> 32:16.000
consumption.
So, specific thrust we have calculated. And
32:16.000 --> 32:22.590
that give that is basically 596.25 Newton
second per second; and next parameter to be
32:22.590 --> 32:27.320
calculated is the fuel consumption the specific
fuel consumption we have already calculated
32:27.320 --> 32:32.639
he fuel to air ratio. So that divided by the
specific thrust will give us the specific
32:32.639 --> 32:40.539
fuel consumption, so 0.0198 minus the fuel
to air ratio divided by 596.25. So, specific
32:40.539 --> 32:45.789
fuel consumption comes out to be 3.32 into
ten raised to minus 5 kilo grams per Newton
32:45.789 --> 32:52.460
second. And sometimes this is also converted
to kilograms per Newton hour that is multiplied
32:52.460 --> 32:59.649
by 3600. And but basically I have I have expressed
it the SI units here.
32:59.649 --> 33:05.850
So, we have now completed cycle analysis for
this first problem which was basically a simple
33:05.850 --> 33:11.509
turbo jet engine without any after burning.
And given a certain amount of data, we were
33:11.509 --> 33:17.419
able to calculate the different temperatures;
and pressure across various components finally,
33:17.419 --> 33:23.799
leading to the nozzle exit conditions; and
the nozzle exhaust velocity; and so using
33:23.799 --> 33:28.000
that we were able to calculate the specific
thrust.
33:28.000 --> 33:35.299
And also the specific fuel consumption so,
this is how you would be carrying out real
33:35.299 --> 33:40.759
cycle analysis based on the data that is provided
to us of course, with certain assumptions
33:40.759 --> 33:47.759
like we assumed a constant specific heat for
air all the way up to compressor exit. And
33:47.759 --> 33:53.269
we have also assumed an average specific heat
for the combustion products; and similarly,
33:53.269 --> 33:59.330
the pressure ratio is heat and so on.
And so, the next problem that we will be take
33:59.330 --> 34:06.989
up that is basically on the same turbo jet,
but if the turbo jet were to be operating
34:06.989 --> 34:13.610
in an after burning mode that is if this particular
turbo jet is to operate with an after burner,
34:13.610 --> 34:17.070
then what would be the corresponding thrust,
and fuel consumption.
34:17.070 --> 34:24.649
So, the problem statement for the second problem
statement is determined the thrust and specific
34:24.649 --> 34:29.470
fuel consumption in the above problem if the
engine operates with an after burner. The
34:29.470 --> 34:34.990
nozzle inlet temperature in this case is limited
to 1800 Kelvin all other parameters; and operating
34:34.990 --> 34:41.690
conditions remain unchanged that is if we
assume that other parameters can be fixed
34:41.690 --> 34:48.130
as it is in the previous case, the only change
is that the after the turbine exit we have
34:48.130 --> 34:53.770
now the after burner that is additional fuel
is added in the after burner taking the temperature
34:53.770 --> 35:01.660
to 1800 Kelvin in which case what will be
the new value of the thrust as well as the
35:01.660 --> 35:07.180
fuel consumption. So, before we start solving
this problem, we will first take a look at
35:07.180 --> 35:12.200
the cycle diagram like we did in the previous
case so, we look at a turbo jet with after
35:12.200 --> 35:19.050
burning. And then we shall attempt to solve
this problem. And find out the thrust or the
35:19.050 --> 35:21.700
increase in the thrust, and fuel consumption.
35:21.700 --> 35:30.589
So, for an after burning turbo jet real turbo
jet cycle with after burning the cycle diagram
35:30.589 --> 35:35.540
up to five is the same as what we discussed
in the previous problem, because it is the
35:35.540 --> 35:41.849
same turbo jet engine now at the exit of the
turbine instead of a nozzle now we have an
35:41.849 --> 35:49.410
after burner after burner will increase the
temperature. And so, the temperature which
35:49.410 --> 35:57.299
was at T 0 5 at the exit of the turbine will
now increase; and reach T 0 6.
35:57.299 --> 36:03.779
That is temperature after the exit of the
after burner or the nozzle entry; and after
36:03.779 --> 36:09.510
the after burner we have the nozzle which
expands the combustion product; and generated
36:09.510 --> 36:14.869
the thrust so, we can see that there is a
substantial increase in the temperature that
36:14.869 --> 36:23.030
is expected in the case of after burning turbo
jet. The turbine in that temperature was 1200
36:23.030 --> 36:30.140
Kelvin where as the temperature at the exit
of the after burner is specified as 1800 Kelvin
36:30.140 --> 36:37.500
so, we can afford to have higher temperatures
in the after burner. And therefore, we should
36:37.500 --> 36:40.390
be able to get substantial increase in the
thrust.
36:40.390 --> 36:47.039
So, let us see how much increase in thrust
we should be able to achieve. So, all other
36:47.039 --> 36:53.130
it is also mentioned that the operating conditions
or operating parameters remain unchanged up
36:53.130 --> 36:57.520
to the turbine exit, which means that the
cycle analysis. Of up to the turbine exit
36:57.520 --> 37:04.740
is exactly the same. And so, we will not repeat
the same calculations here. And so, we will
37:04.740 --> 37:12.390
now need to see if the increase in temperature
what kind of how would it affect s the performance,
37:12.390 --> 37:17.599
so the nozzle will continue to be chocked,
because we have calculated the nozzle pressure
37:17.599 --> 37:23.740
ratio. And so, we have seen in the previous
case that the nozzle was chocking, which means
37:23.740 --> 37:28.430
that even if we use an after burner the nozzle
will continue to be chocked. But we will have
37:28.430 --> 37:33.930
an increase in the stagnation temperature,
but we will see if it makes an difference
37:33.930 --> 37:39.120
in the thrust, so what else we will change,
we will have an additional fuel flow rate
37:39.120 --> 37:43.289
in the after burner, because we have to add
additional fuel in the after burner.
37:43.289 --> 37:47.130
Which is what will lead to the increase in
the temperature, therefore the fuel consumption
37:47.130 --> 37:53.340
is going to change, and the total fuel flow
rate will now be equal to the fuel you added
37:53.340 --> 37:59.819
in the combustion chamber that is main combustor
plus the fuel added in the after burner so,
37:59.819 --> 38:04.069
this will directly impact the fuel consumption,
but besides this there will also be a change
38:04.069 --> 38:05.069
in the thrust.
38:05.069 --> 38:11.630
So, to calculate the fuel flow rate we calculated
first in the after burner, because the rest
38:11.630 --> 38:18.980
of the parameters up to the turbine exit remains
unchanged. So if we will use the same principle
38:18.980 --> 38:26.690
if we did for the main combustor that is basically
an energy balance across the after burner,
38:26.690 --> 38:35.140
so enthalpy of the after burner exit is h
0 6 which is equal to h 0 5 plus the efficiency
38:35.140 --> 38:41.589
or burner efficiency into fuel flow rate multiplied
by the heat of reaction or the calorific value
38:41.589 --> 38:46.480
of the fuel.
So, this on simplification we get f 2, which
38:46.480 --> 38:54.220
is the fuel added in the after burner is equal
to C p into T 0 6 by C p g into T 0 5 minus
38:54.220 --> 39:01.990
1 divided by the burner efficiency into calorific
value divided by C p into T 0 5 minus C p
39:01.990 --> 39:09.839
T 0 6 by T 0 5. So, all these numbers are
known to us T 0 6 is specified as 1800 Kelvin
39:09.839 --> 39:15.320
other parameters are known from our previous
cycle analysis. So if we substitute all these
39:15.320 --> 39:24.650
values we get f 2, which is fuel flow rate
in the after burner is equal to 0.02256. And
39:24.650 --> 39:30.190
therefore, the total fuel flow rate is f is
equal f 1 plus f 2, where f 1 is the fuel
39:30.190 --> 39:37.560
plus air ratio in the main combustor that
plus f 2 in the after burner, so the total
39:37.560 --> 39:46.359
fuel flow rate is 0.04236, so that is as you
can see more than 50 percent there is more
39:46.359 --> 39:57.039
than 100 percent increase in the fuel to air
ratio, in this the turbo jet with after burning.
39:57.039 --> 40:03.619
The nozzle exit conditions, we will now calculate
we have already calculated. And seen that
40:03.619 --> 40:10.049
he nozzle is operating under chocked condition.
So, the nozzle exit temperature critical temperature
40:10.049 --> 40:18.049
will now be equal to 2 by gamma plus 1 into
T 0 6 that T 0 6 is given as 1800 Kelvin,
40:18.049 --> 40:24.710
so this is 1545.06 Kelvin. Similarly, the
exit pressure which is the critical pressure
40:24.710 --> 40:32.050
is equal to P 0 5 by 1 by P 0 5 by P c which
is 0.671 bar which is same as what we calculated
40:32.050 --> 40:36.800
in the last problem.
The density will now change, because the temperature
40:36.800 --> 40:44.609
has changed that is P 7 by RT 7 that is 0.151
kilograms per meter cube therefore, the exhaust
40:44.609 --> 40:50.822
velocity will be equal to square root of gamma
RT 7, where T 7 is something which we have
40:50.822 --> 40:58.340
already calculated so, this is equal to 787.9
meters per second. So, at the nozzle exit
40:58.340 --> 41:01.819
we now have the exit velocity; and also the
critical parameters.
41:01.819 --> 41:07.940
So, we now calculate the specific thrust in
the same way as we calculated in the previous
41:07.940 --> 41:17.040
case, we have the specific thrust is equal
to one plus f into V exit minus b plus a m
41:17.040 --> 41:22.869
dot into P c minus P a here, f is different
from what we calculated in the previous problem;
41:22.869 --> 41:28.970
and so, is the exhaust velocity. So, if we
substitute these values we get the specific
41:28.970 --> 41:37.760
thrust as 912.56 Newton second per kilogram.
Similarly, the specific fuel consumption is
41:37.760 --> 41:45.130
f divided by specific thrust this comes out
to be 4.64 into 10 raised to minus 5 kilograms
41:45.130 --> 41:50.569
per second Newton second.
So, if we compare these values with what we
41:50.569 --> 41:56.789
had calculated in the first problem we will
see that after burning it leads to a very
41:56.789 --> 42:02.670
substantial increase in thrust we get about
35 percent increase in the thrust. And this
42:02.670 --> 42:08.160
is also accompanied by a corresponding increase
in the fuel consumption it is with a cost
42:08.160 --> 42:15.960
of 28 percent in the fuel consumption. So,
this is just to highlight that after burning
42:15.960 --> 42:22.579
is something that is used to increase the
thrust or to achieve momentary increase in
42:22.579 --> 42:31.040
thrust or so, to sustain high mach numbers
in supersonic flights. And so, after burning
42:31.040 --> 42:39.400
turbo jets are usually used in these applications
where there is need to increase thrust momentarily.
42:39.400 --> 42:43.069
And so, since we have seen that after burning
also leads to an increase in fuel consumption.
42:43.069 --> 42:50.039
It is not something used in civil aviation
turbo fan normally do not operate in after
42:50.039 --> 42:57.609
burning so, after burning is usually used
or limited to military engines. So, we have
42:57.609 --> 43:04.140
now solved two problems related to turbo turbo
jet engines one was a simple turbo jet engine
43:04.140 --> 43:09.279
without any after burning; and the second
problem was an extension of the first problem
43:09.279 --> 43:14.900
with after burning; and we have seen how we
can calculate; and carry out cycle analysis.
43:14.900 --> 43:21.130
Realistic cycle analysis, because we have
efficiencies of all the components; and how
43:21.130 --> 43:27.690
the cycle analysis can be carried out in a
systematic manner so, we will now take up
43:27.690 --> 43:33.500
an another version of jet engine we will now
discuss about the turbo fan engine we will
43:33.500 --> 43:38.329
solve one problem on turbo fan engine followed
by another problem on turbo prop engine.
43:38.329 --> 43:46.500
So, let us take a look at the third problem,
problem number 3 states that a twin spool
43:46.500 --> 43:52.599
unmixed turbo fan engine has the fan driven
by the LP turbine; and the compressor driven
43:52.599 --> 43:59.170
by the HP turbine the overall pressure ratio
is 25; and the fan pressure ratio is given
43:59.170 --> 44:06.740
as 1.65 the engine has a bypass ratio of 5;
and the turbine in the temperature of 1550
44:06.740 --> 44:14.250
Kelvin. The fan turbine and the temperature
have polytrophic efficiencies of 0.9 the nozzle
44:14.250 --> 44:22.380
efficiency is 0.95; and the mechanical efficiency
of each spool is 0.99 the combustor pressure
44:22.380 --> 44:31.260
loss is 1.5 bar; and the total air mass flow
rate is 215 kilograms per second.
44:31.260 --> 44:34.849
Find the thrust under the sea level static
conditions, where ambient temperature; and
44:34.849 --> 44:44.359
pressure are 1 bar; and 288 Kelvin, so in
this problem to do with an unmixed turbo fan,
44:44.359 --> 44:49.579
which means that there are two separate nozzles
that are used as cold nozzle or the bypass
44:49.579 --> 44:56.040
nozzle or the secondary nozzle. And the primary
nozzle which is like that used in a turbo
44:56.040 --> 45:00.420
jet engine.
So, we will solve these two streams separately.
45:00.420 --> 45:05.279
And then the total thrust will be equal to
sum of the thrust generated by the primary
45:05.279 --> 45:12.890
nozzle; and the thrust developed by the secondary
nozzle so, the steps that are going to be
45:12.890 --> 45:18.869
followed are very similar to what we had done
in the case of a turbo jet engine. So, I will
45:18.869 --> 45:23.310
I will go through this problem little more
quickly a little faster than what we had discussed
45:23.310 --> 45:25.809
for the turbo jet, because the steps are identical.
45:25.809 --> 45:32.170
So, we will begin with the fan; and it is
mentioned in the problem that is we are required
45:32.170 --> 45:39.269
to calculate the thrust under static conditions,
which means that the engine is mach number
45:39.269 --> 45:45.089
or flight speed is 0 therefore, T 0 1 will
be equal to ambient temperature P 0 1 will
45:45.089 --> 45:54.130
be equal to ambient pressure. And therefore,
the fan exit temperature T 0 2 prime is equal
45:54.130 --> 45:59.730
to the inlet temperature.
Multiplied by the fan pressure ratio into
45:59.730 --> 46:05.131
gamma minus 1 divided by in this problem we
have been given the polytrophic efficiency
46:05.131 --> 46:11.430
of the fan. And if you recall during our discussion
little earlier when we were talking about
46:11.430 --> 46:17.480
the component performance, we had discussed
about polytrophic efficiency of the compressor.
46:17.480 --> 46:23.319
And we have seen that we can relate the polytrophic
efficiency to the isentropic efficiency; and
46:23.319 --> 46:28.541
through the ratio of specific heat. So we
are going to use the polytrophic efficiency
46:28.541 --> 46:34.690
here so, we have pi f which is fan pressure
ratio raised to gamma minus 1 divided by the
46:34.690 --> 46:37.579
polytrophic efficiency of the fan multiplied
by gamma.
46:37.579 --> 46:45.670
So, this comes out to be 330.6 Kelvin now
it is given that the overall pressure ratio
46:45.670 --> 46:52.650
is 25; and the fan pressure ratio is known
which is 1.65, therefore the compressor pressure
46:52.650 --> 46:58.400
ratio will be equal to overall pressure ratio
divided by the fan pressure ratio, so we have
46:58.400 --> 47:08.500
25 divided by 1.65. That is 15.15 so, the
compressor exit stagnation temperature T 0
47:08.500 --> 47:14.380
3 will be equal to T 0 2 into the fan the
compressor pressure ratio that is pi c raised
47:14.380 --> 47:20.819
to gamma minus 1 by polytrophic efficiency
of the compressor into gamma so, this temperature
47:20.819 --> 47:31.590
comes out to be 800.1 Kelvin.
So, we will first calculate the secondary
47:31.590 --> 47:36.339
nozzle properties we will calculate the thrust
developed by the secondary nozzle. And then
47:36.339 --> 47:39.770
we will proceed to the primary nozzle; and
calculate the thrust developed by that; and
47:39.770 --> 47:45.880
then add up the two to get the final thrust
generated by the turbo fan.
47:45.880 --> 47:49.500
Now in the case of nozzle as we have seen,
we will first need to determine whether the
47:49.500 --> 47:55.110
nozzle is chocking or not. So, we will calculate
the nozzle pressure ratio; and the critical
47:55.110 --> 48:00.200
pressure ratio; and see if check, if the nozzle
is indeed chocking.
48:00.200 --> 48:04.990
The cold nozzle pressure ratio is the fan
pressure ratio, because that is what is available
48:04.990 --> 48:11.440
at the inlet so, it is 1.65. And critical
pressure ratio is P 0 2 prime by P c which
48:11.440 --> 48:19.380
is one by 1 minus 1 eta n into gamma minus
1 raised to gamma by gamma plus 1 so, for
48:19.380 --> 48:24.589
the cold nozzle we will take gamma as 1.4
where as for the second nozzle we will take
48:24.589 --> 48:31.990
gamma as 1.33. So, if we substitute the nozzle
efficiency; and the value of gamma we get
48:31.990 --> 48:37.920
the critical pressure ratio as one point nine
six five now since the pressure ratio of the
48:37.920 --> 48:42.920
nozzle is less than the critical pressure
ratio it means that the nozzle is not chocking.
48:42.920 --> 48:50.309
So, if the nozzle is not chocking then we
have the nozzle exhaust velocity which will
48:50.309 --> 48:57.369
be equal to square root of 2C p into nozzle
efficiency into the temperature T 0 2 prime
48:57.369 --> 49:03.750
into 1 minus P a by P 0 2 prime raised to
gamma minus 1 by gamma. So, how we have derived
49:03.750 --> 49:11.759
his expression we have discussed in earlier
lecture on cycle analysis this basically comes
49:11.759 --> 49:16.180
from the enthalpy drop across the nozzle;
and if we simplify we get the exhaust velocity.
49:16.180 --> 49:23.819
So, this is equal to two into the C p which
is 1005 nozzle efficiency temperature; and
49:23.819 --> 49:28.260
the pressure ratio so, we get the exhaust
velocity at the fan at the secondary nozzle
49:28.260 --> 49:35.620
at 293.2 meters per second now it is given
that the bypass ratio is 5. Therefore the
49:35.620 --> 49:41.660
cold mass flow m dot c is equal to mass total
mass flow into bypass ratio divided by b plus
49:41.660 --> 49:48.960
1 that is bypass ratio plus 1, which we get
as 179.2 kgs per second therefore, the thrust
49:48.960 --> 49:54.250
developed by the secondary nozzle mass flow
rate into exhaust velocity that is 52.532
49:54.250 --> 49:55.250
kilo newtons.
49:55.250 --> 50:09.990
Now we now move on to the core nozzle or the
core engine so, for the HP turbine we have
50:09.990 --> 50:16.210
the work done work done by turbine to be equated
by the work required by the compressor. So,
50:16.210 --> 50:22.820
HP turbine drives the compressor so, the temperature
across the HP turbine T 0 4 minus T 0 5 prime
50:22.820 --> 50:29.450
is C p a divided by mechanical efficiency
into C p g into T 0 3 minus T 0 2. So, we
50:29.450 --> 50:38.579
simplify this, and we get the HP turbine exit
stagnation temperature which is 1141 Kelvin.
50:38.579 --> 50:44.050
Similarly we can calculate the LP turbine
exit conditions, because LP turbine drives
50:44.050 --> 50:49.329
the fan; and the fan mass flow rate as we
can see here is different from the core mass
50:49.329 --> 50:55.589
flow, because the fan drives a larger amount
of mass flow, so T 0 5 that is LP turbine
50:55.589 --> 50:59.319
exit temperature is 877.8 Kelvin.
50:59.319 --> 51:05.840
Similarly, we proceed towards finding the
exit pressures, we have these pressure ratios
51:05.840 --> 51:12.339
P 0 4 by P 0 5 prime, which is temperature
ratio raised to gamma minus 1 by gamma, which
51:12.339 --> 51:21.769
is 3.902. Similarly, to the LP turbine; and
the turbine inlet pressure is equal to compressor
51:21.769 --> 51:30.850
exit pressure minus delta p pressure loss
in the burner. So, that is 25 minus 1.0 minus
51:30.850 --> 51:37.890
23.5, so P 0 5 is basically P 0 4 divided
by these pressure ratios. And so, we get 1.878
51:37.890 --> 51:43.720
bar, and that the hot. Therefore the hot pressure
ratio of pressure ratio of the hot nozzle
51:43.720 --> 51:44.800
is 1.878.
51:44.800 --> 51:51.440
The critical ratio pressure ratio for this
nozzle is different nozzle is primary nozzle
51:51.440 --> 51:57.259
from that of the secondary nozzle; because
gamma is different here gamma is 1.33. Critical
51:57.259 --> 52:02.619
pressure ratio is actually 1.914 one, so we
find that since the pressure ratio is less
52:02.619 --> 52:08.240
than the critical pressure ratio this nozzle
is also not chocking. So we proceed to find
52:08.240 --> 52:13.390
out the exit velocity exhaust velocity here
in the same manner as we calculated for the
52:13.390 --> 52:18.220
secondary nozzle; and that we get 528.3 five
meters per second.
52:18.220 --> 52:25.000
And the mass flow rate through the hot nozzle
is 35.83. Therefore, thrust developed is 35.83
52:25.000 --> 52:33.180
times the exhaust velocity which is 18.931.
Therefore the total thrust is equal to primary
52:33.180 --> 52:37.420
thrust, which is developed by the primary
nozzle plus that of the secondary nozzle that
52:37.420 --> 52:46.039
is 71.5 kilo Newton’s. So, this is a problem,
which involved a turbo engine; and unmixed
52:46.039 --> 52:50.970
turbo fan engine which was consisting of two
separate streams each of them generating a
52:50.970 --> 52:55.240
thrust; and contributing towards he total
thrust.
52:55.240 --> 52:59.559
The last problem that we will solve today
corresponds to that of a turbo prop engine,
52:59.559 --> 53:04.680
where in an aircraft operating on a turbo
prop engine flies at 200 meters per second.
53:04.680 --> 53:10.539
While ingesting a primary mass flow of 20
kgs per second the propeller of the engine
53:10.539 --> 53:16.630
having an efficiency of 0.8 generates a thrust
of 10000 Newton, while the jet thrust is 2000
53:16.630 --> 53:23.410
Newton the power turbine. And the nozzle have
efficiencies of 0.88, and 0.92 if we remove
53:23.410 --> 53:27.809
the power turbine; and nozzle what would be
the thrust developed by the engine while operating.
53:27.809 --> 53:33.711
Under the same conditions so, here we have
a turbo prop; and we have been given all the
53:33.711 --> 53:38.660
efficiencies of the propeller the turbine;
and the nozzle so, we have required to find
53:38.660 --> 53:43.480
out that if we remove the power turbine; and
the nozzle what will be the thrust developed
53:43.480 --> 53:46.910
by the same engine operating under the same
conditions.
53:46.910 --> 53:54.240
So, in this case we will basically be calculating
the enthalpy drop across the power turbine
53:54.240 --> 53:59.059
plus the nozzle combination. And then we see
that if we remove the power turbine then the
53:59.059 --> 54:03.269
entire enthalpy drop is available for exhaust
through the nozzle.
54:03.269 --> 54:08.460
And that is how we will be able to calculate
the thrust developed by this particular engine.
54:08.460 --> 54:14.420
So, thrust power that is developed by the
propeller we have already discussed this is
54:14.420 --> 54:20.220
our last class thrust power by the propeller
is basically equal to the thrust developed
54:20.220 --> 54:25.789
by the propeller into the velocity this is
equal to the propeller efficiency into the
54:25.789 --> 54:31.970
power turbine efficiency into alpha delta
h into m dot. Therefore, if we simplify this
54:31.970 --> 54:38.119
we get alpha delta h as the thrust by the
propeller, which is given as 10000 Newton
54:38.119 --> 54:45.730
into velocity 200 divided by propeller efficiency
0.8 or turbine efficiency 0.8 into m dot which
54:45.730 --> 54:54.559
is 20. So, alpha delta h is 142045 joules
per kilo gram; and the nozzle thrust is also
54:54.559 --> 54:59.760
given it is 2000 Newton. So, nozzle plus 2000
is equal to m dot into exit velocity minus
54:59.760 --> 55:05.750
V, therefore we get the exit velocity as 300
meters square per second.
55:05.750 --> 55:10.130
We also know that the exit velocity is equal
to square root of two into nozzle efficiency
55:10.130 --> 55:15.360
into 1 minus alpha into delta h so, from this
we can calculate what is delta h, because
55:15.360 --> 55:20.920
alpha delta h we have already calculated so,
delta h comes out to be 190508.49 joules per
55:20.920 --> 55:29.059
kilogram. Now it is mentioned that, now if
you remove the power turbine; and propeller
55:29.059 --> 55:35.680
what will be the thrust which means that this
entire delta h will now be available for expansion
55:35.680 --> 55:40.660
through the nozzle; therefore, the exit velocity
now will change it was earlier three hundred
55:40.660 --> 55:44.660
now the exit velocity is not equal to one
minus alpha delta h, but it is just equal
55:44.660 --> 55:49.990
to delta h so, we get exit velocity is equal
to square root of 2 into efficiency of the
55:49.990 --> 55:55.650
nozzle into delta h which is equal to 592.76
meters per second.
55:55.650 --> 56:01.599
Therefore, the thrust the new thrust will
now be equal to 20 that is mass flow rate
56:01.599 --> 56:11.619
into V exit minus V that is 592.76 minus 200
that is 7855.19 Newton, so this is the thrust
56:11.619 --> 56:16.950
that will be developed by the nozzle if we
were to remove the power turbine; and the
56:16.950 --> 56:18.430
propeller.
56:18.430 --> 56:25.309
So, we have today discussed four problems
we have solved four problems two of them related
56:25.309 --> 56:31.809
to turbo jet one of them to turbo fan; and
one with turbo prop so, I have now four exercise
56:31.809 --> 56:36.480
problems for you. And the first one is to
do with the turbo jet engine which is operating
56:36.480 --> 56:40.599
with a compressor pressure ratio of eight
turbine inlet temperature as twelve hundred;
56:40.599 --> 56:45.559
and mass flow rate of 15 kgs per second.
The altitude of the aircraft; and the speed
56:45.559 --> 56:51.490
is given so, assuming these efficiencies that
is polytrophic efficiencies of turbine; and
56:51.490 --> 56:58.869
compressor as 0.87 intake; and nozzle efficiency
as 0.95 mechanical efficiency as 0.99 combustion
56:58.869 --> 57:04.450
efficiency 0.97 pressure loss as six percent
of compressor delivery. We need to calculate
57:04.450 --> 57:09.380
the net thrust area the nozzle area required
the thrust; and specific fuel consumption
57:09.380 --> 57:17.400
so, the answer to the 0.0713 meter cube meters
square 0.7896 Newton which is the thrust;
57:17.400 --> 57:21.309
and fuel consumption is 0.26 kilograms per
Newton hour.
57:21.309 --> 57:23.829
(Refer Slide Time: 57: 34)
57:23.829 --> 57:32.680
The second problem is the same problem with
turbo with after burning so, at the turbine
57:32.680 --> 57:35.973
exit the gases are reheated to two thousand
Kelvin; and pressure loss is 3 percent calculate
57:35.973 --> 57:43.029
the percentage increase in nozzle area required
if the mass flow rate is to remain unchanged.
57:43.029 --> 57:47.200
And also the percentage increase in net thrust
that means we need to calculate how much area
57:47.200 --> 57:51.470
increase of the nozzle is required if we have
to operate the engine keeping the same mass
57:51.470 --> 57:52.470
flow.
57:52.470 --> 57:59.430
So, the answer to that is 48.3 percent increase
in area; and 64.5 percent increase in net
57:59.430 --> 58:06.859
thrust third problem is to do with a turbo
fan engine, a twin spool turbo fan engine,
58:06.859 --> 58:13.770
its again unmixed above fan the overall pressure
ratio. The fan pressure ratio bypass ratio
58:13.770 --> 58:20.319
is given as three turbine in the temperature
is 1300 power loss is given as 0.25 in the
58:20.319 --> 58:26.299
combustor. Total air mass flow is 115 kgs
per second we need to find out the thrust,
58:26.299 --> 58:32.460
when the ambient pressure and temperature
are 1 bar; and 288 Kelvin the fan; and compressor
58:32.460 --> 58:37.749
turbine efficiencies are 0.9 nozzle has an
efficiency both the nozzles have an efficiency
58:37.749 --> 58:44.140
is 0.95 so, thrust in this case is 47.6 kilo
Newton.
58:44.140 --> 58:51.269
And the last problem is to do with the turbo
prop turbo prop operating under sea level
58:51.269 --> 58:56.049
conditions with flight speed as 0 air flow
entering the compressor is 1 kgs per second
58:56.049 --> 59:02.970
compressor pressure ratio is 12, all the efficiencies
are given turbine in the temperature is 1400
59:02.970 --> 59:08.560
Kelvin stagnation pressure leaving the second
turbine is 4.6 bar.
59:08.560 --> 59:15.089
And we need to calculate the horse power delivered
to the propeller that is by the power turbine;
59:15.089 --> 59:19.380
and the thrust developed by the gases passing
through the engine so, the power developed
59:19.380 --> 59:27.710
by the propeller is 632 kilowatts; and the
thrust developed by the nozzle is 875 Newton.
59:27.710 --> 59:34.970
So, these are four different problems that
I have sorted out for you basically to do
59:34.970 --> 59:39.940
with turbo two of them to do with turbo jet
one to do with the turbo fan. And another
59:39.940 --> 59:45.099
one with turbo prop engine all these problems
are exactly in line with what we have solved
59:45.099 --> 59:49.470
with slight differences here and there. So,
I am sure you would be able to solve these
59:49.470 --> 59:55.640
problems based on what we discussed during
today’s tutorial session, so that brings
59:55.640 --> 1:00:02.270
us to the end of today’s tutorial session.
In the next class, we will be taking up analysis
1:00:02.270 --> 1:00:08.190
of compressors axial flow compressors; we
will begin a simplistic analysis of axial
1:00:08.190 --> 1:00:09.089
flow compressors.