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In the last class we were discussing about
hybrid rockets or in this class let us look
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at how the burn rate relationship AGOX to
the power of n is obtained so firstly to do
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that we need to understand how the burning
takes place inside the hybrid rocket motor.
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Now if this is the hybrid rocket motor then
oxidizer is injected here right so a boundary
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layer develops and the combustion takes place
inside the boundary layer
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okay
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so this is how the combustion takes place
inside a hybrid rocket motor now if we look
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closely into this combustion zone then we
will be able to look at what are all the processes
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that are taking place in that zone.
So let us say this is the solid fuel that
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we have then let us say we have taken somewhere
in this region right where the boundary layer
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is already established so this is the
boundary layers and let us say this is the
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flames own right now what is happening because
of this boundary layer there is a heat transfer
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to the propellant which is convective in nature
right and that heat transfer is being taken
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up partly for maintaining the temperature
profile in the condensed phase and then partly
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to evaporate the fear.
So if we were to draw the temperature profile
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inside the boundary layer it would go something
like
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this is Ts this is Tc and this is Te the combustion
that is achieved here I mean the temperature
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that is a trade here is the highest simply
because there is combustion and after which
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it is diluted by oxidizer right now the propellant
burns in this direction and there is because
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of that there is a mass addition that is given
by ? Pr. okay mass is getting in and there
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is something known as blowing effect that
we had discussed when we were discussing about
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erosive burning in solid Rockets.
Now so there is heat that is being transferred
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let me call it as Qc okay that is heat that
is being converted into the propellant surface
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now this Qc I can write it as h into Tc - Ts
okay this is the convective heat transfer
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coefficient
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and this is the adiabatic flame temperature
or the combustion temperature and this is
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the surface temperature now this heat as I
said is a part of it goes to maintaining a
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temperature profile and part of it is used
for evaporation so Qc must be equal to
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let me call this direction as X and this direction
is y.
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So
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okay this is the temperature there is the
part of the heat that is taken up to maintain
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the temperature profile and this is the Qs
get seen earlier is the heat of evaporation
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at the surface so this we are seeing with
regards to solid propellant that we can rewrite
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this part as ?P r . CPTS - T? let me call
the temperature
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here as T8.
So I can rewrite this as Qc must be equal
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to ?pr. okay now for ease of our terminology
and other things to follow let me call this
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part as hv hv is nothing but Cp TS - T8 +
Qs now we know that this problem is a boundary
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layer driven problem okay so what we are going
to use is something known as Reynolds analogy
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what it says is there is a similarity between
the momentum boundary layer and the thermal
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boundary layer right and we use that it is
easier to calculate the momentum boundary
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layer there are correlations for it and we
use that to calculate the thermal boundary
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layer.
This can happen when only what Prandtl number
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is close to 1 which in carry gases it is always
close to 1 so we can make this assumption
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so
using approximate
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Reynolds analogy.
That is there is similarity between
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momentum and
heat or momentum and energy
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so we can define something known as a stanton
number
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ue is nothing but the velocity here okay so
h is the heat transfer coefficient ?e is the
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density at the beyond the boundary layer and
Ue is the velocity beyond the boundary layer
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CP is the specific heat this is a non-dimensional
number it is also known as diamond dimensionless
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heat transfer coefficient okay now if we use
the Reynolds analogy what it says is that
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this stanton number indicated by Ch = Cf/2
where Cf is
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nothing but skin friction coefficient okay.
And we know that rundle number is close to
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1 for gases so this term becomes merely Ch
– Cf/2 now we have defined what a stanton
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numbers and we have found out that by Reynolds
analogy we can equate it to this skin friction
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coefficient if we multiply the numerator and
denominator of the stanton number by a ? T
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that is given by Ts - Tc - Ts okay we will
do that and see what we will get.
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So we had stanton number Cf = h/?e ue Cp so
we will multiply both numerator and denominator
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by
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Tc - Ts so it does not change anything to
the left hand side but what is H into TC - TS
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that is the convective heat transfer right
so Qs = h into Tc - Ts so what we will end
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up getting is sorry this is not Cf Ch Ch is
equal to this which one oh okay QC fine now
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what we need to this part is the QC so what
is this part we can define it as some change
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in enthalpy so we will call the ?h = CP TC
- TS so what we will get this we also had
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the other term that is the heat.
That was taken up by the solid let me call
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that as q s that was is equal to ?p r. hv
right but we know that these two must be equal
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right that was the heat balance at the surface
for the heat balance at the surface these
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two must be equal.
So I can now replace in my stanton number
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definition this here right so and stanton
number this also I know is nothing but Cf
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/ 2 by two right so I will use that so I will
get ?p or Cf / = ?p r . hv right or in other
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words I can rewrite a expression for r . as
r . = Cf / 2 x ?e ue ?h/ hv / ?P now we have
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been able to get this expression for r. in
terms of ?V x ue is nothing but in some sense
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mass flux and Cf /2 what we need to further
determine is what is this Cf for the particular
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geometry that we have considered.
So skin friction coefficient
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are without blowing
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we can write it as Cf0/ 2 is equal to in terms
of Reynolds number
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this x indicates the location along the axis
this is for a turbulent flow
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now what happens because of blowing is the
boundary layer in essence tends to get thicken
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okay so the skin friction with blowing will
be different
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and it can be given as Cf /Cf0 where ß varies
from p less than 1ß is nothing but the blowing
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coefficient.
And we can define ß as follows ß is defined
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as ?p r. that is the mass flux that is coming
in from the surface to the mass that is going
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through the port okay. It is also known as
non-dimensional
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so
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we have defined Cf in terms of Cf0 x ß and
we have ß so we can now get the expression
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that we were looking for that is we had said
that
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Ch is equal to this is nothing but ?p r . hv
/ ue that is this expression right.
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So we will treat in use this expression if
you look at this expression and the expression
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for ß right there is some similarity between
the two and we can write ?h \hv I it is nothing
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but ?pr .that is equal to ß right so now
I have this expression for r .here wherein
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I have defined Cf I have defined all other
terms so I can get plug them in and get my
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expression for r .which we will do r. = Cf
/ 2?e ue / ?p right now we know that Cf is
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nothing but Cf 0 x ß so we will use that
here and get our expression for r.
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So r. =1.27 /2 xCf0 x ß0.77 x ?e ue / ?a
x ß right because we know that ? h/h y is
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nothing but ß so Cf0 also we know in terms
of Reynolds number as this expression here
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so if we substitute that we will get r . = ß
is – 0.77 and one so I can write it as ß0.23
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Reynolds number is nothing but ?e ue x x/
µ so I get r.= so we have ?e ue here and
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here so we can Club those two.
And rewrite our expression as r. = and I can
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take the ?p that is here this ? p I can multiply
it back with the r. so I will get ?p r . is
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equal to
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what is ? ue is nothing but the mass flux
in the put right so mass flux let us define
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G as m . /Ap so that will be nothing but ?e
ue
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so you have we can replace this with G right
and we will get -
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s thank you
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okay so we come to this expression there is
0 here.
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So we get this expression where in the burn
rate is related to the mass flux through the
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port right and if you look at when we do experiments
it is easier to characterize this in terms
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of the oxidizer mass flux itself because that
is something that we have control of and we
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know what it is so which is why you will find
that if you take a look at any hybrid related
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literature you will find that the burn rate
is expressed as a function of G ox okay.
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So
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we can recast this in terms of zero X and
that is what you will find in most literature
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as a dot is a function of 0 X to the power
of n and if you look at this expression you
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will find that it depends very weakly on X
the axial location and beta but it is a very
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strong function of the mass flux right now
let us look at the problem that we had said
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yesterday with regards to what happens to
burn rate as we proceed in from the head end
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to the nozzle length.
If you look at this expression here it looks
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like what should happen to the burn rate as
you increase X it should decrease right set
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the opposite yesterday right is that correct
order is that wrong
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if you look at this yes you are saying but
you are forgetting that there is a greater
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dependence on G if you look at what is happening
as you move from the head end of the port
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to the nozzle end there is continuous mass
addition and this is increasing as you go
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from heading to the nozzle end and this is
raised to a power of 0. 8.
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So therefore you will find that the burn rate
as you go from heading to the nozzle length
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will always increase
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okay now people have done people have conducted
experiments and found out that if you take
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the oxidizer mass flux the dependence of burn
rate varies this n varies from 0.42 to 0.7
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one of the things that we have kind of made
an assumption right at the beginning was we
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had assumed something like a flat plate if
we assume a pipe flow kind of a situation
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and redo the calculations.
We will probably get closer to what we what
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is observed experimentally, experimentally
what we can get is in terms of the oxidizer
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mass flux and not the overall flux k overall
flux involves that you take across-section
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and find out what is the fuel plus oxidizer
that is coming which is a lot more difficult
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to do if you are doing an experiment, experiment
you are better off knowing what is your mass
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flow rate of oxidizer and then calculating
the oxidizer mass flux there was another problem
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that I said hybrid Rockets suffer from because
of this n.
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That is if you look at holding even in even
when you hold the oxidizer mass flux constant
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we find that the fuel mass flow rate was changing
now let us look at is there any condition
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at which the oxidizer mass flux if you hold
constant or if you fold the oxidizer mass
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flow rate constant when the fluid flow rate
also become constant.
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We know that the mass flow rate of fuel is
given ?p Ab*I .so if you are taking a port
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configuration ?p into this is the diameter
of the port
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and this is the length of the port
this is the burning surface area into our
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dot is nothing but A into okay now let us
determine what value of n will give us something
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that when you hold the oxidizer mass flow
rate constant you can also get a few mass
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flow rate constant if you have to engineer
such a hybrid then that is a very good situation.
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Because you can operate it at a set specific
impulse even though the diameters are thinking
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so Gio X we know is nothing but fine this
is geo X so if we plug it into this equation
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we will get m dot Fo know this is you have
2 into n right in the denominator and if you
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look at this equation you have D to to the
part of one end d to the power of 2n as very
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obvious that if you have a value of n that
is equal to 0.5 for n equal to 0.5 m dot F
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becomes
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independent of the pore diameter so therefore
you can operate it at a set o by F ratio so
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that you can choose that over F ratio such
that it gives the maximum is B.
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Now what happens if n is greater than 0.5
so it is desirable to have
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in equal 2.5 now for n greater than .5 what
will happen to the mass of here as diameter
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increases it will reduce because you will
have 2 N and you will have DP here so the
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denominator will be greater than the numerator
so therefore as diameters increased mass flow
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rate of here will decrease
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and similarly or for the other case that is
for n less than .5 okay this finishes our
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discussions on hybrid rockets so in a sense
we have learnt three, three kinds of rocket
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engines in this course that is one is solid
propellant the other one is liquid propellant
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and then lastly the hybrid propellant okay
thank you.