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In the last class we had seen the performance
of a turbojet we derive equations for the
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non dimensional thrust as well as we had derived
expressions for ISP of a turbojet. Now let
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us put in some typical numbers and see what
is it that we get out of it and we will see
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some interesting facts effect.
So firstly we will look at optimal, optimal
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expansion of flow in the nozzle here there
are two conditions that we need to look at
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1 is sea level thrust or sea level is been
what happens at cruise conditions so
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00:01:28,979 --> 00:02:03,350
at sea level P0=1 atmosphere 1011.3mPa and
T0 to T if we
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assume an IS, IS a atmosphere model then it
will be 288.15K and m?a0 at sea level before
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take at take off conditions.
And now if we assume the compression ratio
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pc to be very low that is spicy to be of the
order of from 4 then my t c would be something
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like 1.484 and if we take Tt4 to be 1000K
then this makes data b is equal to something
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like 3.4 and we know the Q of the fuel this
is kerosene so 42MJ/kg and let us take the
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Cp of air and flue gases to be the same and
let us take it to be 1kJ/kgK.
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So with this conditions now you have to find
out what is it at the cruise altitude of around
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11 kilometers, so altitude what are the things
that are going to change the things that are
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going to change are these two only atmospheric
conditions and at cruise conditions this Mach
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number will also change so we need to change
the only three parameters P0 would be lower
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and it would be somewhere around 22.6kPa and
T0 would be
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this is a you have taken ISA International
Standard Atmosphere so therefore we get this.
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Now if we use this and calculate F/m?a? and
ISP we have all the parameters here to calculate
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both of them or we will get this kind of a
table.
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M0=0 then I will
also put v7 v0 both these in m/s and lastly
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ISP(kNs/kg ) so if we take the sea level performance
that is at sea level M0 =0 and using this
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we will get F/m?a? as 1.8 and this would be
something like 612m/s and this would be 0and
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ISP would be something like this. Now at 11km
M0 would be 0.8 and this would reduce slightly
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to1.76 okay, now there are two things to notice.
One is that irrespective of the altitude you
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see that f/m?a? is nearly constant right,
this happens because of two things, one notice
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that at sea level this is large the velocity
differential is large okay, the mass flow
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rate is very small velocity differential is
large mass flow rate is very small so you
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end up getting a high thrust because of the
velocity differential.
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At a higher altitude the velocity differential
is reduced but you have air coming in now
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at this velocity therefore what you get as
a larger mass flow rate, so you get a larger
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mass flow rate with a smaller velocity differential
therefore you get nearly the same thrust or
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same non dimensional thrust .
What happens to Isp, Isp is come down which
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means that SFC is gone up if I were to write
SFC also here SFC would be in kg/kg hr this
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would be something like 0.8 this would be
like 1.06 so we are cruising at an altitude
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where an the SFC is low why do we do that,
I think it would be more beneficial if we
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were to stay on ground or close to ground
and carry out our operations in it from the
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looks of it what we see is that the Ips come
down or SFC has increased.
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So why do you think this is happening what
this is desirable not desirable what is the
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opinion you would not say something no. So
what happens, no not really well actually
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the reason why we operate it at a higher altitude
is primarily the drag is less at a higher
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altitude right, you are not looking at drag
of the engine Percy you are looking at the
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drag of the entire aircraft where in when
I was discussing about turbofan engines I
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said G goes in for a bypass ratio of 9 and
says that the engine of the nasal drag is
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very small compared to the overall drag.
And therefore they go in for a large bypass
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ratio right, now what we are looking at is
the drag of the entire aircraft goes down
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as altitude increases because the load go
decreases as you go up in altitude right.
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Now density decreases drag decreases but your
SFC will increase because what we have set
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in here is if you notice I did not change
this part Tt4, Tt4 is the same at both the
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altitudes.
And if you look at ?b in this case ?b would
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be because T0 is lower, so ?b for Tt4=1000K
would be 4.63 that is what we are saying is
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you need more fuel to bring the air from a
lower temperature after compression to 1000K
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because we have set that temperature so therefore
we will find that the SFC will be higher,
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right, okay and that is the reason we see
this now we can look at similar numbers for
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choked flow through the nozzle the only thing
that changes would be I will just put them
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here itself.
Now here we will assume a slightly larger
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compression ratio of round 12 which means
that how sea will also go up it would be as
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much we can also increase the turbine inlet
temperature
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this means ?b will go up to
the rest of the quantities remain the same
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here again sorry, I had written under K here
do not point it out it was to be 1000K earlier
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and that makes it 4.63 okay, and that is why
we saw that change in ISP or SFC.
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Now if we change it to 1200K ?b would be 5.55
and this would mean correspondingly all these
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quantities will change, because now we have
choked flow in the nozzle the thrust consists
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of two parts, one is the convective flux part
and the pressure thrust part.
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So this will again have two parts 1 point
and at an altitude okay, this is what we see
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notice that as you go up in altitude the fraction
of the pressure thrust is increasing it is
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a lower fraction here right or if you look
at this number it is a lower fraction here
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compared to a higher fraction here right,
the overall thrust why does this happen, why
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does pressure thrust part increase as you
go up in altitude.
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Ambient pressure if you see is 1/5th so the
pressure cross part is going to increase if
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you go up in altitude and again you see the
same feature here that SFC has increased and
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that is primarily because you need to heat
the air from a lower temperature to a set
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temperature of 1200k here instead of 1000
earlier so therefore you will find that SFC
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is higher in this case okay right, okay. Then
we have dealt with two things we have looked
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at the optimal fulfill optimally expanded
flow through the nozzle and we looked at choked
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nozzle, okay.
Now let us look at what happens if you have
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the after burner on and the other method of
increasing the thrust that is water-methanol
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injection let us look at these two methods
and how to derive expressions for them. So
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firstly let us look at what happens if we
have the afterburner on.
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I will retain these set of numbers because
I want to use this to look at what happens
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without a afterburner with the afterburner
conditions, so I will retain this numbers
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and we will get back to a similar table at
the end of the discussion on afterburners.
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Now if we look at the TS diagram with the
afterburner on we will again assume ? to be
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1 okay, so you have isentropic compression
through the intake as well as through the
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compressor then you have heat addition in
the main combustor expansion through the turbine
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so this is 0, 2okay, so this is the TS diagram
for efficiencies 1 and with the afterburner
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switched on okay, now what do we have to do
if you have to analyze this okay.
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Firstly our expression for F/m0a0 the non
dimensional thrust would be M0 into in this
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case [1+F+] I will call, I will introduce
a new parameter fab M7/M0(vT7/T0-1) right,
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this is the expression for non dimensional
thrust now if we put fab=0 this is the expression
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that we had derived for a turbojet without
the afterburner switched on right, when fabs
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goes to 0 this is the expression that we get.
Now we have fab where a fab is nothing but
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M0 fuel that is added in the afterburner divided
by m0f okay, so it is the fuel ratio in the
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after 1 so what do you think of the value
of this plus this what is the maximum of F
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and f+ab that it can go to it can maximum
go to stoichiometric which is what is the
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value
it is 0.067which is the stoichiometric value
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you cannot burn more than that so it goes
to something like 0.67.
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Now if you put a 0.67 I can still use the
condition that f+ fab is less than 1 right,
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with an error of around 6.7% so I will use
that i will say f+fab is very much less than
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1 okay, so that we can take this terms of
and I will also introduce another parameter
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here just like we had Tt4/T0 as a control
parameter we have a new control parameter
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that is Tt6/T0 right, what is the maximum
temperature in the cycle divided by T0 the
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minimum temperature.
So I will call Tt6/T0 as equal to ? be remember
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we had Tt4/T0=?b burner similar to that I
am introducing a new control parameter to
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account for the after burn okay. So using
this I need still these ratios T7/T0 and m7/m0.
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So I can get T7/T0 as Tt7.Tt6 by okay, Tt
sorry this has to be Tt6 okay, fine these
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two cancel off and I will get T7/T0 what is
T7/Tt7 it is the ratio of static to stagnation
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so I can express this in terms of Mach number
this would be 1+1/1 ?-1/2 m72 what is T7/Tt7/Tt6
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this is flow through the nozzle right.
So flow through the nozzle we have assumed
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the flow to be isentropic so TT the total
temperature does not change so this is 1 and
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Tt6/Tt0 we have defined it as ?ab, so
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I get T7/T0 as equal to ?ab/m72. Now we need
to do cascading of pressures to get the other
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parameter yes, we always choke yes so while
we are doing for the optimal because in case
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of optimally expanded if you use the conversion
also it is still choke.
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We will do both cases okay, first we will
take the optimal expanded flow then we look
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at what happens if the nozzle is choked the
nozzle being choked does not essentially mean
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that P7 =P0 it can be that P7 is greater than
P0 even if the nozzle is choked right, so
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we are taking a special case of choked flow
wherein the exit pressure is equal to the
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ambient pressure.
Now if we do cascading of pressures
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I get P7/P0 is equal to okay, this is what
we get and
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the first term here is ratio again ratio static
to stagnation pressure this is flow through
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a nozzle flow through jet pipe and then flow
through a turbine, flow through the main combustor
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flow through compressor and this would be
flow through intake and that is ?0?/?-1.
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So we will put that down here and we know
because we are considering optimally expanded
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flow what is P7/P0 this would be equal to
unity so we will use that.
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So we can write 1=1/M72 into flow through
nozzles we are assuming all efficiencies to
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be 1 then flow through afterburner again 1,
then flow through compressor you get pt then
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sorry flow through turbine then you have flow
through combustor which is 1 then you have
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flow through a compressor pc and you have
flow through the intake which is 1 and lastly
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?0?/?-1 and here also p. So I get I can rewrite
these two in terms of tt so I will get 1=1/
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or 1+?-1/2 M72= tt tc eternal okay, and I
know from the definition of ?0 that 1+? – 1/2
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m02=?0. So using these two I can get my expression
for M7/M0.
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So
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I can write M7 as equal 2/?-1 tt tc ?0-1and
M0 similarly would be
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and we required in our expression for T7/T0
this term so I can get that too I can write
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T7/T0 as equal to ?ab/ tt tc ?0-1 okay, so
sorry no -1 sorry, so
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this is the expression that we have now we
can substitute back and get our expression
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for the non-dimensional thrust which is f/moa0=
or I -, I can rewrite this part by taking
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this out as common I can rewrite this as okay.
So this is the expression that we get for
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the non dimensional thrust, now there is still
1 part missing we know that there is a turbine
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compressor power balance so these two are
not independent variables and they are connected,
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okay so let us get that expression also.
We had derived that expression for a case
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without the afterburner on do you think it
will change with the afterburner being switched
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on, because this is something that happens
downstream of the turbine it does not get
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affected and therefore whatever we are derived
earlier holds good.
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So from
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we get tt=1-?0/ ?b tc-1 and if you substitute
this expression back then we will get F by
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okay, this is the overall expression that
we get. Notice here that ?b is now a parameter
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of very small consequence right, the temperature
at the end of the combustor is not such an
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00:39:10,760 --> 00:39:16,730
important parameter whereas for the thrust
the temperature at the end of the afterburner
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00:39:16,730 --> 00:39:28,041
is a more important parameter okay. Now let
us look also at the ISP part if you have to
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calculate ISP we know from our previous expression
for non dimensional specific impulse that
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I need 1/F F/m0a0.
So here this was the expression that we had
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without the afterburner how would this change
with the afterburner I need to include here
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00:39:55,820 --> 00:40:08,920
f+fab for the case with the afterburner okay,
so with the afterburner on you have this additional
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term here okay, now how do we determine this
quantity, how do we determine this quantity
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what do we need to do, what did we do to get
1/F energy balance across the combustor.
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Now what do we need to do, energy balance
across the combustor plus the afterburner
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so let us do that.
By taking energy
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afterburner we get (m?f+m?fab)Q okay, must
be equal to m?a(1+f)Cp(Tt4-Tt3)+m?a(1+f+fab)
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okay, this is the portion that gets added
on in the afterburner into Cp(Tt6-Tt5) okay,
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this is the expression that we get. Now we
know that we can use f is very much less than
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1 and we also know that f+fab is also very
much less than 1, so using this I can eliminate
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these two and therefore I will get f+ fab
that is you take it here and divide this by
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00:43:26,650 --> 00:43:48,750
m?a you will get this into Q/Cp must be equal
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00:43:48,750 --> 00:44:17,210
to Tt6-Tt5+Tt4-Tt3. Now if you look at the
TS diagram.
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If we look at the TS diagram okay, this is
the TS diagram we have and what we have here
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is Tt6-Tt5 that is this minus this plus Tt4-Tt3
right, now I can add this part what you will
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get is you will get as though it is a continuous
rise from Tt3 to Tt6 okay, if I add this part
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but we know that compressor turbine power
balance Cp being the same this portion in
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this portion is same you have to subtract
1of them.
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If you are adding this you to subtract this
okay and I also know that the Tt2 must be
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equal to Tt0 right, because flow through the
what do you have flow through the intake what
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kind of process is that it is an isentropic
process because we are assuming all efficiencies
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to be 1, is an isentropic closest total temperature
does not change okay, so you get Tt2=Tt0 so
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we will use that here and do a little bit
of jugglery as i said I will add –Tt5-Tt4-
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okay.
If you look at this I have not d1anything
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this is what is this part turbine, what in
the turbine work in the compressor these two
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must be equal right. So I can take this out
these two are equal now I know that Tt2/Tt0=1
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so this goes to 0 okay, so by doing this manipulation
have been able to or get this as Tt6-Tt0 which
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00:47:42,890 --> 00:47:54,980
is nothing but you are looking at raising
the temperature from here to Tt6 okay, so
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this was TS diagram so
I can right now (f+fab)Q/Cp=Tt6-Tt0.
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Now if I divided by T0 on both sides I get
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00:49:13,430 --> 00:49:32,450
what is Tt6/T0 this is ?ab and what is this,
this is ?0, so we get a new expression for
186
00:49:32,450 --> 00:50:15,610
we are looking for this expression 1/f+fab=Q/CpT0
?ab-?0 okay, and following this I can now
187
00:50:15,610 --> 00:50:36,690
write ISP/a0=Q/CpT0 (?ab-?0) into okay, so
this is our expression for ISP with the afterburner
188
00:50:36,690 --> 00:50:39,500
switched on, I will stop here and continue.