Today we will start with solid propellant
rockets and this will be an extremely simple
topic. We will take a look at what are the
elements of a solid propellant rocket, but
primarily, we start from the propellant point
of view, because that is what we have
covered so far. We told that solid propellant
propellants consists of may be double base,
may be composite, may be composite modified
double base, and also nitramine
propellants. Whenever these propellants are
made, you cure the substances and make it
into something like a solid form, and this
solid is what constitutes let us say a solid
propellant. We would like to use it to generate
thrust.
Therefore, what is it we do? We take let us
say a case which could be cylindrical. We
put
the solid propellant inside the case. How
do you put it inside the cylinder or a case?
We
keep the case vertically, We put all the ingredients
into it; we make the propellant
mixture into a sort of a liquid or a slurry,
then I cure it and I get something like a
solid
which is formed within the cylindrical case.
We then attach a nozzle to it and this
becomes a solid propellant rocket.
Therefore, in a solid propellant rocket we
have something like a structural member
which is a case and supports the propellant.
Either we pour the slurry of the mixture may
be a composite slurry consisting of the binder
may be the ammonium per chlorate and
aluminum in into in this. Then we take it
into a furnace and cure it at a temperature
of
between 70 and 200 Celsius depending on the
composition. We form a block may be a
composite solid propellant block which we
call as a propellant grain. And when we cure
the slurry or the mixture in it and make it
as a solid inside a case the rocket or solid
rocket is known as a case bonded rocket.
It is also possible to make the propellant
in the form of a paste and extrude it out
just like
we squish out the tooth paste from its container.
We could thereafter cure this squished
out paste using in a furnace and enclose it
in within a covering. Thereafter we could
introduce it viz., charge it into the case.
In this case, it is known as a free standing
propellant grain. And what is free standing?
The grain propellant inside it is freely
standing it is known as a free standing propellant
grain. It is known as a case bonded
propellant grain when the propellant is cast
in the case. Therefore, without any
qualification, we have just introduced the
word propellant grain. What does grain mean?
The solid propellant block of a particular
configuration in a case is what we call as
a
grain. This grain or propellant grain could
either be directly cured in the case itself
in
which case it is known as case bonded.
And when it is cured outside and we machine
it and then push it into the case the
propellant grain is known as free standing.
We have both free standing and case bonded
grains being used, but when we talk of large
solid propellant rockets invariably we cast
it
in the case itself. But there are other certain
things required. As the propellant burns,
the
hardware or the metal case or whatever be
the case will get heated and therefore it
is
necessary for us to protect the case with
a liner. This liner could be an insulator
on the
inner walls of the case and then we put the
propellant into it 
and then I attach a nozzle
onto the case.
Therefore, I have a case, may be an insulator
or something which will which will make
sure that when something burns the hot gases
generated by the solid propellant do not
come and melt the motor case or the case as
it where and then I have the nozzle and this
what constitutes the solid propellant rocket.
But what is it we want a rocket to do? We
want the rocket essentially to generate some
thrust. The thrust of the rocket is equal
to m
dot into VJ or m dot into the specific impulse.
We learnt how to calculate the specific
impulse and supposing we want to generate
a large force, we need to generate more mass
flow from the burning solid propellant.
Therefore, in today’s class we will first
examine how do you make the solid propellant
grain, which could either be a case bonded
or free standing. If we ignite it and start
burning it, how do we ensure that we have
sufficient amount of mass getting generated
from the propellant. We have just studied
the chemistry of the propellants, but we have
not studied anything about how it will burn
we want to get some feel for how a
propellant surface will burn. How do we calculate
the mass generation rate and what
should be the surface area such that we get
the concerned thrust and this is what we will
be discussing today.
Let us again look at the problem in a slightly
different scenario. However, before that to
repeat if we have a composite propellant like
we have a polybutadiene which is a resin
and the resin which is in liquid form is added
with the solid loading crystals of AP and
aluminum powder. The slurry so formed is a
mixture of all these three constituents and
we heat it to a high temperature to form a
solid block and this solid block. This solid
block is covered with a liner of insulation
and kept inside the case. And what is this
resin? It is something, which when heated
set, and this resin is known as thermosetting
resin. This is because when heated it becomes
hard. Resin is nothing but a polymer
polybutadiene like a glue.
I will show you some examples. But if you
have plastic which is also a resin and if
we
heat it, then it softens therefore we cannot
use such resins for composite propellants
and
the same thing is true when I use nitrocellulose
and nitroglycerine or the insulation. We
make it as a liquid, cure it into a solid
and make it as a block and this block is known
as a
propellant grain. Therefore, we are interested
in finding the burn rate; the rate at which
it
will burn or the rate at which mass gets generated
from a grain. Is it clear? We must
differentiate between a propellant block that
is the solid propellant which is obtained
after curing.
A propellant grain gives us the necessary
m dot of gases generated for providing the
thrust. And this propellant grain could either
be directly bonded in a case or the hardware
of a rocket or it could be prepared elsewhere
in which case we just introduce the grain
inside the case. The later is known as free
standing while the former is known as case
bonded grain.
This is all about the terminologies. Let us
get back and ask ourselves if we have a case
over here in which we have the propellant
block inside it and let us say I have no nozzle.
We want to generate hot gases from propellant
burning. May be this is the exposed
surface area of the propellant. It is this
surface, which burns and we call it as burning
surface area and denote it as Sb. Supposing
this surface is ignited. We would like to
know the rate at which we are generating the
hot gases from it. And if the surface burns
we would like to you have something like r
so many meters per second is the rate at
which a propellant burns. What is this rate?
Supposing, I have a surface, which is like
this, let us say a convex surface and this
is the
propellant. We can take a flat surface of
a propellant, which is the propellant surface
here. Or the other extreme is we take a concave
surface something like this and this is
my propellant here. We could have different
configurations of the surface. Or we could
have hole right through. Let me consider these
cases when I start burning at the surface
of he propellant. At the next instant after
burning commences, the flame comes over here
and then it comes over here.
In other words the burning progresses into
the propellant as it burns. The rate at which
the surface regresses is 
what we call as the burning rate. Rate at
which surface regresses,
but it must be normal to the surface and that
is why we made these different
configurations over here. The surface regresses
in this direction over here; the surface
regresses in this direction normal to the
surface in this direction. In all cases the
surface
evolves normal to it and this constitutes
the burning. Therefore, we call burning rate
as
normal to the surface.
When I say a propellant burns at the rate
r meters per second, we mean that the burning
is normal to the surface. Let us spend some
more time on this. Supposing I have a
propellant let us say a block something like
this, this is my propellant. Let us say now,
I
have the propellant in this configuration.
May be this is my propellant here 
and now I
ignite or I start burning over the exposed
surface. How will the burning take place?
The
burning will take place normal to this surface
along X axis. The burning will take place
normal to this surface along the Y axis. And
what happens to this contact point? Well it
should go like this right. In other words
over here since it is normal to this point
it will
go as a circle (quadrant of a circle) over
here. This is how normal to the surfaces that
the
regression proceeds. May be we will have to
look at this in some detail when we come to
different configurations of the propellant
grain.
Therefore, we define something like a burning
rate in terms of meter per second, which
is normal to the surface. The gas generated
viz., rate of mass of gas which is generated
by the burning is equal to r is the burn rate
so, many meters per second into the burning
surface area so, many meter square. This gives
us meter per second into meter square
which is meter cube per second that is the
volume rate at which the burning or the solid
propellant gets consumed.
Knowing the volume rate at which the regression
takes place we multiply it by the
density of the propellant, which is so much
kilogram per meter cube. I get meter cube
meter cube gets cancelled and we get m dot
g is equal to r into the burning surface area
into rhop so, much kilograms per second. Therefore,
the rate at which mass gets
generated from a surface can be written in
terms of the linear regression rate known
as
burn rate. In terms of a linear dimension
into the burning surface area in this case
the
whole thing is the burning surface area in
this case.
The burning surface area is this and at the
next instant of time may be it is over here
this
is the burning surface area and you have burning
surface area into the burn rate into the
propellant density as the rate at which mass
of gas is generated. And burn rate is but
the
rate of regression of the surface. Therefore,
we say mass generated by burning is equal
to
the burn rate in meters per second surface
area in meter square into propellant density
in
kilogram per meter cube This part must be
clear. Then let us illustrate it through an
example, because we also you will recall we
have been talking of black powder in the
course on explosion also. We said that black
powder consists of an oxidizer like KNO3
and it also consisted of something like charcoal
you say carbon and some sulfur.
We had said that it consists of something
like 75 percent potassium nitrate maybe 15
percent of carbon and sulfur is the remaining
10 percent. And how do you form these
crackers or sparklers which use this black
powder. You have these solids this is an
oxidizer and fuel you grind them together
may be make a good mixture of it may be add
some glue may be in that case it becomes wet.
You sort of cure it make it as a solid and
that is what constitutes a composition. Well
this is also a sort of composite propellant.
Therefore, let us take the example just to
understand a little bit further about burn
rate I
brought a sparkler with me. We will see the
burn rate of this composition. I will just
ignite it I brought two sparklers, because
I thought the effect of aluminum must also
be
clear to us. Let us just ignite it and see
what happens? Well! This is ignition right
you
light a candle. Well, this is a sparkler this
consists of as I said the composition, and
the
composition is again carbon you have sulfur
and you have KNO3 and this it will burn by
itself without the support of air. It is a
propellant and instead of using polybutadiene
it
has resin that is the glue which is used.
Well, you ignite it you ignite it at the surface
tip. How does burning take place? This
surface, which gets ignited and this ignited
surface burns the next surface next adjacent
to it and so on. We find that this surface
is regressing at a particular rate. Well,
in this
case the whole surface is exposed, but because
there is air over here the flame expanded
slightly here. The burning surface area is
perpendicular to this plane and the burning
progresses axially. This is the case of this
composition without metal in it. Let us see
what is the effect when we have when we have
something like a metal in this sparkler,
the same composition is the same.
You see the 
distinct effect of the metal. The intensity
is very much higher. See the metal
having so much energy you find the heat, which
is much higher. Metal has a higher heat
of combustion and that is why metals are included
in solid propellant. Again we are
talking of the same thing the surface regressing
as it is going on and we find that the
propellant regresses it burns. And what is
happening is gas is getting generated the
rate at
which the surface is regressing is the rate
at which the composition is getting consumed.
The energetics is such that you know the sparks
fall on me I feel little warm and is
harmless but with metals in it I get burnt.
Therefore, you find that metals have a definite
role to play and I think this is all what
I wanted to illustrate. We are talking of
a burning
rate, which is linear regression rate of the
propellant. We will come back to our sparklers
when we talk of combustion instability. We
will see how even this can make a cavity
unstable. But, when burnt a propellant regresses
at a given rate and as it regresses mass
gets generated. This is all what we say about
burning rate.
What will be the typical burning rate of the
sparkler? We had a length something like
almost twenty centimeters, it burnt over a
period of one minute. Therefore, the mean
burn rate r is equal to 200 millimeters divided
by 60 seconds, which is about 3 mm per
second or 3.3 mm per second. And this is how
burning rate is calculated; you have a
solid propellant strand; you ignite it and
measure the burning rate. Therefore, with
this
background I think it is time to go ahead
and examine the burn rate of let us say the
double base propellants and may be the composite
propellants. We will try to get into
some details and we will try to make some
sense of how the composition influences the
burn rate.
Let us say that we have a double based propellant
here, and just the same way we ignited
this sparkler we place a fire over here or
a heat source over here start the burning.
What
was the composition of double base propellant?
We told that it consist of nitroglycerine
which is C2H5(ONO2)3 and nitrocellulose C6H5(OH)
and ONO2. We 
had something like
five of this OH in cellulose we put x of ONO2
we removed x of OH and put ONO2 there
we had 5 minus x over here giving C6H5(OH)5-x(ONO2)x.
Now, we ignite the double base propellant
and what happens. The region very near to
the
surface gets heated, because all these propellants
are insulators. Heat does not conduct
deeply into it. And therefore, at the other
end, if we consider this as the cordinate
x, the
temperature will be almost the ambient value.
Only a small thickness of the propellant
near the surface gets heated. And therefore,
a plot of the temperature in the propellant
with the temperature shown on the y axis and
this is my distance along will be something
like this. Let us show it in red; the depth
of the propellant is really not heated. We
have
the the temperature at the surface as Ts and
this is the original ambient temperature.
Let us say that the initial temperature of
the propellant Ti, all what we have done is
just
like we ignited the sparkler we ignite this
particular surface of the propellant. And
therefore, the surface temperature increases
heat gets conducted over a certain thickness
while at the end the temperature has the original
value. These are all insulators. We
could have held the sparkler along the charge
because of its poor thermal conductivity.
Now, what happens when we heat this particular
zone to a higher temperature say Ts.
This particular zone begins to foam as it
were begins and begins to evolve some hot
gases. And the gases, which are evolved are
essentially aldehydes which are COH; could
be formaldehyde could be acetaldehyde CH3COH.
We could get some CO, we could also
get some NO. These are the gases, which are
evolved. Because something is getting
heated these gases tend to come out over here,
and when the gases come out they react
further; that means, I have a zone adjacent
to the surface which is now at temperature
Ts
which forms hot gases and the temperature
continues to increase in the gas region, in
this
particular zone.
Adjacent to this particular zone of propellant,
which is getting heated which foams and
forms hot gases; this zone is known as foam
zone and is the solid zone, which is heated.
And while it is getting heated it decomposes
and generates these gases like aldehydes
may be NO2 may be NO may be CO. These gases
are coming out and they react and
increase the temperature from, the surface
temperature Ts to a higher value of
temperature which we call as T1, the temperature
T1. These gases do not completely
burn and when the pressure over here is less
than, let us say something like 10 MPa that
is about 100 bar.
What happens is the gases are not at very
high pressure and the process of intermediate
products being formed continues without final
products being formed. The gases relax
for some time they begin to mix you know it
takes some time for the gases to burn and
therefore the temperatures remain at T1 level
for quite some time. That means, hot gases
are generated, it reacts increases the temperature
from the surface temperature Ts to T1.
The gases are still reacting and what are
the reactions which could take place? The
reactions, which could take place are CO which
is formed reacting with aldehydes; even
NH2 is formed reacting with CO plus NH2, but
these are not very exothermic reactions.
And therefore, the temperature remains fairly
constant till that time final recombination
reactions take place. This constant temperature
zone is a precursor to the final reaction.
After the reaction takes place, the final
flame temperature Tf is reached.
We have seen in the last class how to calculate
the flame temperature using dissociative
equilibrium using complete combustion. I can
calculate the flame temperature Tf. But
this flame temperature when the ambient pressure
is less than about ten MPa goes in
steps initially it goes to T1, then it reacts,
but the reactions are not very exothermic
and
there after it increases this in this zone.
Actually, CO2, H2O which are complete
combustion products are formed and you get
the maximum temperature Tf. Of course,
you have CO, because these are all fuel rich
substances and therefore, you have
something like a two step reaction process
in the gas medium. This is the gas phase.
This was the within the solid region; this
was outside. We should extend the gas phase
right up to the boundary over here, this is
the gas phase and this is the solid. I will
make
it clear in the next diagram.
What is it we are talking? We consider a propellant
solid block something like this;
Initially the surface gets heated. And there
after some chemical reactions are occurring
in
this zone and the surface temperature increases
to Ts. In the gas phase, the temperature
increases from the surface temperature Ts
to T1 over here. Then I have another zone
over here in which no increase in temperature
we said T1 remains constant over a certain
zone and there after the final chemical reactions
take place it leads to temperature Tf. Let
us plot the temperature distribution again
over here as a function of x.
We said at the surface this is the temperature
starts from the initial value in this zone
it
increases, then it increases still further
then, I have a zone in which the temperature
is
more or less constant and then the temperature
shoots to the Tf over here. This is the way
a reaction takes place. This is all the gas
part of the combustion. This is the solid
part of
it. This solid reaction zone; we already called
as foam zone. The zone in which the initial
temperature increases in the gas phase is
known as the fizz zone, because you are having
some reactions zone it is known as fizz zone.
This is a luminescent zone, because
chemical reactions are happening just like
we saw in this sparkler, some luminosity.
This
luminous zone is known as the fizz zone.
The fizz zone is followed by a dark zone.
Why it is dark? There is no further increase
in
temperature. It looks dark at the low temperature.
There after you have this zone which is
the second luminous zone, because again high
temperature are achieved giving a second
luminous zone. This is how a double base propellant
burns to give you the high
temperature Tf. If we have very high pressures
like exceeding 100 atmospheres that is 10
MPa, the chemical reaction continues in the
dark zone or heat release continues here.
And I could have the temperature going straight
away to Tf at high values of pressures.
That means, the dark zone will be absent when
you have high pressure. We will have this
luminous zone viz., the first luminous zone
followed by the second luminous zone going
to the temperature final temperature which
will be Tf. That means, the temperature
increases like this goes to the surface and
then you have this zone and the temperature
increases over here. This is at very high
pressures whereas at moderate pressures up
to
100 atmospheres well you have this dark zone
and after which the temperature increases
to the final value of Tf.
Well, this picture should be sufficient for
us to be able to calculate the burn rate.
I have a
high temperature T1 over here. The surface
temperature is lower at Ts and heat gets
conducted from the surface. It is this heat,
which further heats this propellant surface
and
causes it to gasify and foam. Therefore, I
am interested in events in the fizz zone wherein
hot gases or heat from this zone gets conducted
over here.
What is the model or what is my mental picture
with which I can solve for the burn rate?
It is necessary to first consider the physical
problem. Like we say in the case of the
sparkler. We looked at how regression took
place.
We tell that heat is generated here. The temperature
in the fizz zone is T1 and the heat
gets conducted from T1 to the surface, which
is at temperature Ts. and if we can
calculate this heat which is conducted then,
we can write the equation at the surface and
then get the burn rate. This is my mental
picture of the problem. The mental picture
might be wrong. If my mental picture is wrong,
the burn rate what I get might be wrong,
But the picture seems to be reasonable this
has been experimentally verified. But this
thickness is what we are talking of. Even
though I show it like this, the total thickness
of
the fizz zone followed by dark zone followed
by the second luminous zone is something
like fraction of an mm. That means, it is
all over here just like what happened in the
sparkler.
You had a rod. You coated the propellant over
it. And what was this propellant? KNO3
plus carbon plus sulfur all bonded together.
We found the reaction distances are all
within a fraction of a mm. We are talking
of distances which are fraction of a millimeter
thick.
Therefore, let us write an equation for it.
It is not difficult at all. We are going to
be
interested in the fizz zone, which is the
zone wherein temperature increases form Ts
to
T1. Why are we not considering the dark zone?
See the temperature is constant here. If
temperature is constant here nothing is going
to come over here into the fizz zone from
the hot zone. If we were say that the temperature
at the edge of the fizz zone is T1 and at
the surface the temperature is Ts, we have
heat conduction from T1 to Ts coming here
and still getting conducted in the solid.
We know that heat is getting conducted along
in the fizz zone to the surface of the
propellant; but in the dark zone since the
temperature is constant no heat is getting
conducted. Therefore, my condition here is
T1 over here, and the heat gets conducted
from T1 as the temperature falls to Ts at
the surface. We would like to write the equation
for the heat conduction.
This is the mental picture or the model that
we have constructed. Well, we are interested
only in the fizz zone; let us forget about
the dark zone and the second luminous zone.
We
now say that let the length of the fizz zone
be capital L. We recognize that we are talking
of things, which are fraction of a mm thick.
Let us say we have the surface at x is equal
to 0; this is the surface over here. We are
interested in let us say a small element over
here within the fizz zone at a distance x
from the surface and of width d x.
We would like to write an equation for this
element and then integrate it to find out
how
the burning progresses. And this is how we
do any heat transfer problem. Therefore, we
are interested in finding out for the surface
area over here equal to unity - 1 meter square
- such that I do not need to carry the value
of surface area as I am doing the problem.
Let
us say there is some heat conduction q dot
here. What is the value of heat conduction?
Rate of heat conduction at x; q dot is equal
to minus it is a gas phase and therefore
thermal conductivity of the gas into dT by
dx by Fourier law. At x plus dx, the rate
q dot
is equal to 
the small increment over dx which is 
d by dx of q dot into dx plus q dot. The
net heat which is leaving this is equal to
minus k g into d two T by dx square into dx
is
the net difference between the heat which
is entering and leaving. Heat which is entering
is equal to k g dT by dx heat which is leaving
is k g dT by dx plus a small increment,
which is d by dx into kg into dT by dx into
d x. Therefore, the net accumulation of heat
in a control volume of width dx is equal to
minus kg d 2 T by dx square into d x.
We should learn to write these equations at
any time for any problem. We will again be
coming to such formulations when wish to calculate
how much ignition energy is there
required to ignite a propellant. When burning
takes place at the surface, heat from the
burning of gases supplies energy to the propellant.
We will erase this part in the fizz zone
and assume the rate of chemical reaction heat
release in the reactions to be equal to q
dot
chemical. So much let us say joules per let
us say per volume. This is joules per meter
cube and is the rate at which heat is getting
generated. And therefore, the amount of heat
which is getting generated in this fizz volume
is going to be q dot chemical into the
volume of this element. And the volume of
this element is unit area into dx which is
equal to q dot chemical into dx is the rate
at which heat is getting generated in this
volume.
Is there anything else I have to consider
in the heat balance relation? When the propellant
burns, we considered the surface to be the
frame of reference. With respect to this
surface propellant, which is regressing or
burning at a value r, hot gases are getting
generated. The gases generated at the surface
are moving out at a velocity ug and this ug
will be different from r. This is because
r is for a solid while ug is for a gas. Let
the mean
density of the gases be rho g. Therefore,
the mass flow rate at which the gases are
entering into this control volume is equal
to m dot g per unit area. This is equal to
rho g
into ug that means, I have so much kilogram
per meter cube into meter per second
therefore, I have so much kilogram per meter
square second. This is the mass flux.
Since we are considering unit area, the mass
rate at which gas is moving or the mass flux
is equal to rho g into ug. And what is the
enthalpy that is gained as it moves through
the
elemental volume.
The gas is moving let us say this particular
temperature here at x is T the temperature
here at the next x plus dx is equal to T plus
d T, because there is a small increase in
temperature. And therefore, the increase in
enthalpy of the gases is going to be rho g
into
ug into specific heat let us say constant
pressure into the temperature at exit which
is T
plus dT minus temperature T at x. Because
for this element at the left hand side the
temperature is T it has increased in temperature
over a small distance dx by dT to T plus
dT. Therefore, the increase in energy or increase
in enthalpy of the gases is equal to rho g
ug into Cp into d T.
Where does this energy come from? The energy
comes from the chemical reactions
which generate so much of heat so much joules
per meter cube and and also some heat
which is getting conducted.
This means, that the enthalpy gained plus
the heat transferred from the element must
be
the net heat that is generated. I now put
the equation down as minus kg into d two T
by
dx square into d x, which is the heat transfer
plus I am getting some heat which is getting
generated q chemical rate into d x, which
is the net heat accumulation. And where is
this
heat supplies this enthalpy which is equal
to rho g into ug into Cp into d T. This becomes
the energy balance equation and we must be
able to write such energy balance for any
problem. We can write this equation as kg
into d two T by dx square. I take it on the
right hand side plus I have rho g into ug
into Cp into dT by dx. dx has come out from
the
left hand side. It is equal to q dot chemical,
where q dot chemical is the energy release
per unit volume in the fizz zone so much joules
per meter square.
If this is the case how do I solve this equation?
See solving this equation is a little
difficult in the sense all what I can solve
this I can solve for temperature distribution,
in
the fizz zone. If I know the value of rho
g into ug and q dot chemical, then only I
can
solve for temperature distribution. Once temperature
distribution is known, I can find out
the temperature at the surface Ts. We can
also get the value of dT by dx at x is equal
to 0
viz., the surface. And then find out the rate
of heat transfer. Therefore solution will
give
us the temperature distribution, but this
temperature distribution is going to be a
function
of the value of the thermal conductivity of
the gas in the fizz zone may be the density
in
the fizz zone may be ug in the fizz zone also,
the value of q dot chemical.
We examine these quantities. Well! The thermal
conductivity of the gas kg should be a
function of pressure of the gas. As the pressure
increases may be the conductivity of the
gas will go up, because we have more number
of molecules in it. The density of the gas
rho g is again a function of pressure. ug
has to be related it to the burn rate at the
surface
and if we know it we will get the value of
the burn rate r. For a given ug we can solve
this
problem for a given dot q chemical. But what
is the rate of heat release? The rate of heat
release q dot chemical in the gas depends
on the pressure p to the power let us say
m, the
exponential of the activation energy of minus
the activation energy E by the universal
gas constant R naught into temperature T.
E is the activation energy. We have a steric
factor depending on the molecules here.
Therefore, q dot chemical will be a pressure
exponential minus E by R naught T. We
therefore find that q dot chemical is a strong
function of pressure or a function of
pressure, rho g is a function of pressure.
The thermal conductivity is also a function
of
pressure. Therefore, we can say that ug, that
we wanted to find out should also be a
function of pressure.
In addition to pressure, we find that ug should
be a function of activation energy and this
activation energy should be a function of
the composition. May be a few more things
may be the density may be other parameters
which are there. But what we get is that the
temperature distribution and the velocity
of the gases ug would be a function of pressure.
What is the value of ug? The gas velocity
ug in the fizz zone of density rho g equals
the
mass, which is moving out and must be equal
to the density of propellant into the
burning rate of the propellant.
We have a solid, which regresses at a rate
r and which the propellant burn rate r. The
mass which is getting generated is equal to
rho p into r, because I have unit surface
area
and the rate at which gas is getting generated
is equal to rho g into ug per unit surface
area. Therefore, the value of the burn rate
r goes as rho g by rho p into ug.
But ug is seen to be a strong function of
pressure and therefore the burn rate is going
to
be again a strong function of pressure and
the effect of the other constituents like
the
composition, activation energy etc.. And therefore,
it is possible for us to write the
burning rate r in let us say meters per second
as equal to a constant into pressure to the
power n, where this constant will take care
of activation energy. May be the other
composition and other features are included
in the constant. And pressure comes over
here. It is possible for us to write the burn
rate of a propellant for the double base
propellant r as equal to a which is now a
constant into pressure to the power n.
And this form of burn rate is what is known
as Vielle’s law. It is also known as Saint
Robert’s law.
Therefore, what is it we have done so far?
We said that the surface regression rate of
a
double base propellant could be written in
terms of the ambient pressure p at which it
burns in the form r is equal to a p to the
power n. a will have the compositional features
may be may be the initial temperature may
be the final temperature. Maybe I should
have written here ug is going to be a function
of the initial temperature, may be the
temperature T1 and so on.
Therefore, I put all the effects other than
pressure into this a factor ‘a’ here that
is the pre
exponent and I have the burn rate law during
r is equal to a pn. It becomes much easier
to
use a law like this rather than keep trying
to solve this equation which is a little bit
more
difficult. And in practice we use this law.
How do we determine the value of n? We do
an experiment at pressure p1 and measure the
burn rate r1 which can be written as a p1
to the power n. We then measure the burn rate
r2 at a pressure which is little bit away
at
p2. We get r2 as equal to a p2 to the power
n. We use these two equations to find out
the
value of n. We get n is equal ln r2 minus
ln r1 divided by ln p2 minus ln p1.
Once we get the value of n, we determine the
value of a and this is how we determine the
burn rate law as r is equal to a pn. Just
like the sparkler experiment, may be at an
ambient
pressure, we put a strand of propellant in
a vessel and do the experiment. We measure
the time taken to regress a particular distance,
measure the burn rate, then determine the
value of n and a.
If we have the ambient pressure in excess
of let us say 10 MPa, then we had said that
there is no such thing as a intermediate T1
zone. The temperature directly goes to
something like Tf over here and more heat
is conducted. And therefore, if we have to
plot the burn rate r let us say meters per
second, as a function of pressure at low
pressures I have n which is something like
this: when pressure goes up it also increases
like this exponentially.
This is how the burn rate varies with pressure
when the pressure corresponds to within
where the dark zone prevails. However, when
the dark zone is absent at higher values of
pressure, something greater than about 10
MPa, the heat transfer increase rapidly and
the
burn rate increases much faster. A plot of
burn rate as a function of pressure would
steepen as shown. But when we add additives
to the propellant, then the dependence
changes.
Let me again repeat this point. We found that
the burn rate of a double base propellant
can be written as r is equal to a pn, where
a is a factor which considers all factors
other
than pressure like temperature of the gases
may be T1 may be the surface temperature
Ts, may be the activation energy and all that.
And pressure is directly taken to a power
n.
Then we said that at low pressure the value
of n is smaller while at high pressures for
which there is no dark zone and you have higher
temperatures driving the heat
conduction the value of n becomes greater.
n is larger when p is greater than about 10
MPa.
But we could also add some salts, like lead
salts to the double base propellant. Then
what
happens is instead of getting a curve like
the one earlier, we get a different pattern.
As a
function of pressure, the burn rate initially
picks up and there after you know it does
not
change with pressure that means, you have
a plateau over here. In other words, I have
burn rate increasing up to a pressure and
it becomes a plateau. Such type of propellants
which exhibit this feature of n is equal to
0 over a certain range of pressures are known
as plateau burning propellants.
Well, some propellants have even a slightly
different behavior and some may have their
burn rates going down. We have r versus p.
In fact, we should have written ln r versus
ln
p over here and that is when I would have
a straight line so, this should have been
ln r
and ln p. Sometimes the burn rate begins to
fall at higher pressures. It all depends on
the
chemical constituent in the propellant and
such propellants which exhibit a drop in
burning rate with increase in pressure are
known as mesa burning propellants.
Therefore, what is it we did in this class
today? We looked at the rate at which mass
gets
generated from a surface. We found out that
the surface area is important - Sb is
important. Then we found for the double base
propellant we have two or three zones
during their burning: like a foam zone followed
by a fizz zone in which temperature
increases to some intermediate T1, then a
dark zone in which the temperature is more
or
less constant followed by a zone in which
temperature further increases.
We looked at the events happening at the at
the fizz zone and not in the foam zone. We
wrote an equation and we were able to say
that the velocity of the gases in this zone
are
dependent on pressure and the initial temperature
may be the T1, may be the surface
temperature Ts, activation energy E and all
that. And since the velocity of gases is
directly proportional to the burn rate r of
the propellant, we said that the law is r
is equal
to a pn which is known as Saint Robert’s
law or Vielle law.
Thereafter we found that since we have the
reactions happening directly at high
pressures; that means, instead of having a
dark zone is not present at high pressures.
That
means, at high pressure the temperature increases
from surface temperature directly to
the flame temperature and more heat gets conducted.
Therefore, n is higher at the higher
pressures. Therefore you have small n followed
by a large n for the double base
propellants. If we add some salts and alter
the mode of burning we could have something
like a plateau burning propellant or a mesa
burning propellant.
In the next class, we will take a look at
composite propellants, and then put the whole
thing of burn rate together and then address
the design of a solid propellant rocket.
