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In
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the last class we had looked at cycle analysis
of everything engines now let us look at rocket
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engines in the next few classes here we will
be trying to derive expressions for specific
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impulse and how mass flow rate varies through
a choke nozzle how do we get thrust how do
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we get exit velocity what are the assumptions
that we make and as a consequence of these
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assumptions what do we have to deal with or
what is the error of these assumptions we
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will look at it a little later in the course
okay.
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So now let us look at rocket engines now rocket
engines
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all the three kinds of rocket engines that
we discover discussed in the earlier classes
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that is solid liquid and hybrid all three
of them have the following that is they have
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propellant storage then they have a thrust
chamber
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and lastly a convergent divergent nozzle
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in the solid rocket both these two are in
the same physical location whereas in liquids
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and hybrids all three of them are separate
but one thing common to all of them is this
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part.
That has the nozzle part in the nozzle what
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you have as the terminal energy because of
release because of chemical reactions is now
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converted into kinetic energy so in the nozzle
you have thermal to kinetic energy conversion
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and as this is common to all three kinds of
rockets we can study them exclusively okay
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and without having to pay any need to pay
any attention to what kind of rocket to test
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whether it is a solid rocket or a liquid rocket
or a hybrid rocket. So we will do that now
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we look at the converging diverging nozzle
now before we go there what we are going to
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look at is quasi steady one dimensional analysis.
That is we are going to assume that all changes
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happen along the axis only there are no changes
that are happening in the radial or the azimuth
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direction R and ? directions we do not assume
any changes to be happening so all changes
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are happening along the axis so it is one
done which is not strictly true for rocket
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engine nozzle okay we will see what error
this bring about a little later first as we
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are going to make this assumption then we
are going to assume that the chamber pressure
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and temperature are given to us.
That is let us say if this is the rocket motor
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we will assume that we know chamber pressure
which is called PC and chamber temperature
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TC has known will also take these to be stagnation
quantities
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that is if you look at velocities elsewhere
in the nozzle and if you look at the velocity
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at the entry of the nozzle these velocities
are very small compared to the velocities
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as you go along the nozzle so you can take
them to be stagnation conditions at the entry
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of the nozzle okay.
And we are also going to assume that you are
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also going to assume that as I said these
are known to us firstly and we are also going
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to assume that fluid is an ideal gas
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with constant
thermodynamic properties
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that is the CP ? thermal conductivity all
these do not change as we go from this portion
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of the nozzle to the exit of the nozzle then
we will also assume that the flow is in viscid
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that is viscosities are very small and will
also assume fluid flow is isentropic.
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00:07:30,749 --> 00:07:35,520
So with these assumptions let us see what
we can derive and based on this we will also
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evaluate what are the short comings of some
of these assumptions are these really valid
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and then we will try to estimate what is the
error now from your basic gas dynamics you
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know that for a nozzle for a convergent.
That was in nozzle I can write you buy you
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that is change in velocity as you move along
the nozzle =
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now we know that if M is less than 1 what
should be the dA / A if this has to flow it
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00:08:43,080 --> 00:08:51,500
has to accelerate if M is less than 1 dA/
A has to be negative if fluid has to accelerate
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00:08:51,500 --> 00:09:02,320
and if M is greater than 1 dA/ A has to be
positive if fluid has to accelerate and we
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know that if at the throat at the throat of
the C-D nozzle Mach number is 1 then we call
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it choke nozzle let us now find out what are
the pressure ratio.
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If you have PC here what should be the PA
or what should be this ratio of PC / PA for
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the nozzle to be choked okay
so let us find the critical ratio of this
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pressure for doping
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in order to do this we will solve the energy
equation from here to here okay from the entry
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to the root section and I can write one dimensional
energy equation as one dimensional
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steady state and for in viscid flow with constant
thermodynamic properties I can write the energy
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equation as HT + = UT.
Where C were c there indicates our chamber
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conditions and T indicates throat conditions
C indicates and T indicates throat that is
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we are looking at this to be see this to be
okay now there is no heat addition that is
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taking place from here to here okay and the
composition is the same so if the composition
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is the same then CP has to be the same so
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C to T that is from chamber to throat it means
that CT is same and there is only change in
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sensible enthalpy and it also means that only
there is change
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in enthalpy.
Because there are no reactions that are taking
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place so if all the change from C to T is
only because of sensible enthalpy change then
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I can rewrite this equation as right that
is I have replaced HT / CpTt and HC / CpTc
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we also know that if the throat is if the
flow is choked at the throat then add throat
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M = 1and therefore Ut = 80 that is the speed
of sound which is given by here Rs nothing
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but R = Ru universal gas constant /molecular
weight okay now using all this we can simplify
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this equation further and write it as.
Because we know Ut = 80 and 80 is this I can
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write it bit like this =
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we know that CP / CV = ? and CP - CV = R so
dividing the entire equation / CpTp that is
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dividing my CpTp we will get 1 + ?RTT / 2
okay now we know that R and CP we can get
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through this so let me write that down this
will then become 1 + ? x 1 / 1 - ? this and
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this cancels out so you get 2 = TC/ DT or
simplifying.
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I can write Tc / Tt = ? + 1 / 2 from this
and knowing the connection between CP 2 temperatures
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to pressure for an isentropic flow I can write
Pc / Pt as = so now we can get the critical
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ratio that is required for choking okay and
if you substitute for ? =1.2 which is usually
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the value of ? for burnt gases you get Pc
/ Pt to be 1.7 that is if the chamber ratio
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to the exit pressure or the pressure outside
is greater than this then the nozzle will
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be choked okay
for any pressure ratio PC / or for any pressure
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ratio that is from Pc to ambient wear in it
is greater than or = 1.7.
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Then the fluid flow is all choked flow as
I had said in the previous class the advantage
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of chalk flow is now it becomes independent
of downstream conditions and it is only a
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function of upstream conditions okay so let
us see how we can use this to calculate what
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is the mass flow rate through a choke nozzle
we have calculated what is the ratio that
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is required for the flow to be choked let
us now calculate what is the mass flow rate
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through a choke nozzle.
Now we know that m. which is the mass flow
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rate is given / ?tAt Vt where T as I said
earlier indicates throat conditions. So this
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is the density into area x velocity now let
me normalize ?T and VT with respect to chamber
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conditions why because I know for a choked
flow that it is only dependent on the upstream
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conditions upstream of the throat of the chamber
so if I normalize it with chamber conditions
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I will be able to derive some useful equations
?T and VT with ?C which is the density at
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in the chamber and At which is the acoustic
speed or the speed of sound at the throat.
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So we will get m.= ?T / ?C x At xVtL / At
x what is At I know that At = right so I will
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normalize this also with the chamber conditions
and multiplied /under v TC what is this ratio
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is the ratio of local speed fluid speak to
acoustic speed which is nothing but Mach number
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Mach number at throat for choked condition
is 1 so this is 1 so we get m. = ?T / ?C x
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At okay so now we need to find this ratio
and this ratio we already know for choked
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flow what is this ratio we also know pressure
ratio.
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So density ratio is nothing but ?T / ?C is
nothing but P t / Rt /tc so R cancels out
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you get Pt / Pc x Tc / Tt we have already
derived them that is nothing but x ? + 1 / 2
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this is the temperature ratio this is the
pressure ratio so now if we because these
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two are the same we can add the powers and
we can simplify it as ?T / ?C I will get it
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as
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I also note Tt / Tc which is nothing but.
I need under v so this would be- ½ now I
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know all the ratios that I wanted these two
I can substitute them and get the relation
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for mass flow rate as M0 = I am sorry you
do not remind me I needed to multiply/ if
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I divide this / ?c I need to multiply / ?c
again so this is ?c x v y / ?c x this so I
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will get here ?c At okay now I can deal with
them because their same base but different
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powers I can add the powers.
So I will get
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and further this - I can replace it as + and
this ?c I can write it in terms of pressure
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so I will get PC /RTc x At ? RTc because I
have a - sign here I can write this as 2 / ? +
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1 x ? + 1 now I have RTc here and RTc here
so I can find one of this and write it as
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okay this is very similar to something that
you have already derived in gas dynamics where
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and you will not use Pc you will instead use
P0 A* / T 0right this whole function this
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whole thing is a function of only ? so normally
in rocket literature this is denoted as a
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? of ? function.
So you get m. = PcAC/ c* where C* is called
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the
characteristic
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velocity and Z* is given / 1 /? of ? x Ru/
is nothing but R so C* is given / this where
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? of ? is only a function of
function of ratio specific heats that is okay
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so this is the typical expression that we
use in Rockets when we are discussing Rockets
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for mass flow rate through a tube nozzle this
? of ? is also called as a random capture
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function and varies from 0.642, 0.67 for ? variation
of 1.3.
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Now we know how to get the mass flow rate
through a joke nozzle now if you are collect
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back we had derived expression for the thrust
of the rocket motor when we were discussing
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about the thrust of the turbojet engine so
or an air breathing engine.
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So
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we know that
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rocket engine f = m. Ue+ Ae Pe - Pa where
e indicates exit conditions a indicates ambient
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condition so you have here Ue which is the
exit velocity exit area exit pressure and
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ambient pressure now we in the thrust equation
we already know how to calculate m. for a
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choked nozzles o this part we know we need
to now find out how to get the exit velocity
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and then we need to get something about area
ratios and other things we will look at it
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a little later firstly let us get an expression
for the exit velocity.
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How do we get this expression we know that
the energy equation is valid from the entry
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of the nozzle to the exit of the nozzle so
from that we know we can write from energy
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equation he + Ue2 / 2 =hc we had seen a little
earlier with respect to the throat this equation
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now we apply it to the exit of the nozzle
okay so I can rewrite this as Ue = under v2
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hc – he since we know that there are no
reactions that are taking place inside the
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nozzle and the thermodynamic properties are
constant from the entry of the nozzle to the
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exit of the nozzle all this must be sensible
enthalpy.
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So therefore we can write this in terms of
Ue = 2CpTc x 1 - T/ tc + e / tc we can express
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it in terms of pressure as = under v ? – 1
this part CpTc part I can use e Ue CpTc I
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know = Cp / r x r Pc now this RTc what is
C * C * is nothing but RuTc / molecular weight
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00:39:18,620 --> 00:39:36,710
or RTC right this is nothing but = 1 / ? of
? x underv RTc so this part here RTc must
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be = C *2 ?2 okay so I can put this what is
Cp / up this I can write it as ? / ? – 1
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x so if I substitute back this into the expression
for u I will get V = because this has square
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here and this is in this v I can take it out
and write it like this.
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Now what this tells us is that if we know
the characteristic velocity and the pressure
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ratio we can get the exit velocity right so
if Pe goes to 0 what happens if this will
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Ue will be a maximum when Pe goes to 0 so
when e = 0 U is maximum and is called
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limiting exhaust velocity that is it will
only become a function of the chamber temperature
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the becoming 0 here means in this case if
P goes to 0T U also has to go to 0 so the
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Ue would this then become under v that is
it becomes only a function of the chamber
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00:42:39,540 --> 00:42:45,650
temperature but in reality.
We cannot get a condition where in the exit
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velocity this is not the ambient pressure
this is the exit of the nozzle being at zero
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00:42:53,040 --> 00:42:58,210
pressure this is not possible this will be
some finite number so therefore you cannot
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00:42:58,210 --> 00:43:02,840
get this condition but this is the maximum
that one can get with a convergent divergent
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00:43:02,840 --> 00:43:28,580
nozzle.
Now
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we need to also get one more parameter that
is how does we still do not know how a and
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00:43:38,540 --> 00:43:46,310
PE are connected that is if we know the area
ratio there from the throat to the exit how
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does how does one obtain PE from if knowing
PC how can we obtain PT.
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So let us do that so let us relate Pe / Pc
to Ac/ At to start this what we know is we
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know that mass flow-rate once the nozzle is
choked and the flow is steady the mass flow
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rate at any cross-section is the same and
cannot vary so we know m. = Pc At / C* and
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00:44:42,210 --> 00:44:57,920
is also = ?e Ue Ae where e indicates exit
conditions so from this I can get an expression
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00:44:57,920 --> 00:45:40,210
for Ae / At as Pc okay now I know ?e is nothing
but ?e Pc/RTE and if I substitute for that
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here I will I also know the expression for
Ue which we just derived okay.
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00:45:47,440 --> 00:46:15,010
So if we substitute for both of them here
?e and Ue we will get an expression for A/
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00:46:15,010 --> 00:47:07,170
At eA this is the expression that we get once
we substitute for Ue and ?e now here what
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is again C* x if we take a ? of ? here so
you get C* x ? of ? 2 which is nothing but
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RTC right so we get Pc /Pc x RTe divided / RTc
that is we need to have a ? of ? in the numerator
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to account for this x 2 ? / ? – 1 okay I
can cancel the R’s and Te / Tc I can again
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00:48:16,670 --> 00:48:23,180
express in terms of pressures I already have
a Pc / Pe.
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So I know that Te / Tc is nothing but Pe / Pc
to the power of ? - 1 / ? so if I substitute
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for all this for this here and simplify or
I can.
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Because I can put them together and rewrite
this as now if you notice this is a relation
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connecting area ratios to pressure ratios
mostly we will know what is the geometric
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area ratio of the nozzle that we are taking
and to get the pressure ratio from knowing
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the nozzle nice nozzle area ratio using this
equation is very difficult so therefore we
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have gas dynamic tables which will give you
this or there are plots that will tell you
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how T / Pc varies if you vary Ae / At okay
we look at it and then next class thank you.