Rocket Propulsion

Prof. K . Ramamurthi
Department of Mechanical Engineering
Indian Institute of Technology, Madras

Lecture 13
Divergence Loss in Conical Nozzles and the Bell Nozzles

Good morning. In today’s class, we continue with nozzles. First, I will review what we
have done so far and then, we will see whether there are any gaps, anything which has
happened in nozzle development, which I have not covered. We said that nozzle is
something like a vent, and we first learnt how to calculate the jet velocity. How did we
calculate? We said that we have a chamber which is at a pressure pc. Suppose, at the exit
I have pressure pe; I can calculate the jet velocity VJ. We wanted VJ to be as high as
possible and to be able to get a high value, I need this vent or opening to be in the form
of a convergent divergent shape and we also put a condition that the minimum area
which we called as a throat should have a Mach number equal to 1.

(Refer Slide Time: 00:28)

We called this nozzle as a de Laval nozzle or a convergent divergent nozzle. In the

convergent part, the Mach number was less than 1; in the divergent the Mach number
was greater than 1. After having done this, we said at the exit I have exit pressure which
is pe; the ambient pressure is pa and if the exit pressure does not match to the ambient
pressure i.e., is not equal to the ambient pressure, I could have some shortcomings in the
nozzle performance either due to under expansion or due to over expansion. After this,
we also took a look at what is the flow through the nozzle. In fact, we got an equation for
m dot is equal to 1 over C star into p c into At was the mass flow rate, and we got an
expression say m dot by At which is mass flux as equal to chamber pressure pc into 1
over C star here. C star had units of meter per second and we were able to correlate it
with the pressure built in the chamber.

The pressure built in the chamber is related to the mass generated or rather to the mass
flow rate through the nozzle. We developed a relation, which we called as a transfer
function C star as the characteristic velocity of a rocket. Let us take one example just to
clarify things. Suppose, I have a rocket in which the mass of propellant is let us say Mp.

(Refer Slide Time: 02:17)

Let the rocket fire steadily for a period of t seconds. The mass flow rate through the
nozzle is therefore equal to Mp divided by t; let us say Mp is in kilograms; this flow rate
is in kilogram per second. If the pressure built in the rocket chamber is pc and if the
nozzle throat area is At, I can directly say that Mp by t is equal to Mp dot is equal to 1
over C star into the chamber pressure pc into At. I can therefore determine this value of
C star. If I go and do an experiment and determine the value of C star based on the
measured value of mass of propellant, time and pressure and then, I start comparing the
value of C star which I actually measure to the C star which we derived ideally, both
would not be equal. We determined C star is equal to under root RTc by capital gamma;
we will find that the ideal value may be a little more than the actual experimental value
because, some flow losses are taking place. The measured value to the ideal theoretical
value was called it as C star efficiency of a rocket or rocket chamber. We also said that
the nozzle has a divergent portion and we can also write the thrust of a nozzle as equal to
in terms of the chamber pressure in terms of At and we put it in terms of a coefficient.
We derived the equation for this coefficient. We called this as CF ideal. Further, we
found that the thrust also depends on the exit pressure and the ambient pressure. I show
exit pressure pe and the ambient pressure pa here. Always pe may not be equal to pa, but
the thrust is a maximum in pe is equal to pa; we called this ideal thrust coefficient as CF0
when the exit pressure was equal to pa.

(Refer Slide Time: 03:56)

In a similar manner we have a rocket, which is firing. I make some measurement using
transducers; I measure what is the thrust what is is generated by rocket then, I determine
the measured value of CF. I also get the value of CF, which I calculated using the ideal
theory. The ratio of the experimental to the theoretical value was called as the thrust
correction factor eta F. This is a summary of what we did. We also did something which
was important. We said that instead of specifying the exit pressure and the chamber
pressure, I can also specify a rocket in terms of a nozzle exit area Ae divided by the
throat area At.
(Refer Slide Time: 05:54)

Are there any questions on what we have done so far? Mind you all that we have done is
for an ideal case, an adiabatic nozzle. And one-dimensional flow you always said that
flow is going straight like this right and all the mass flow rate is contributing to the thrust
and expansion. Are there any questions so far?

Your question is why we need to couple the C star with CF to determine the value of
specific impulse when the specific impulse can be readily evaluated based on nozzle
flow.

Let us first clarify that C star is something which tells how much chamber pressure is
developed when we provide a certain mass flow rate through the nozzle. The nozzle is
identified by the throat area. In other words, the transfer function between mass per unit
flow rate through the nozzle or mass flux through the nozzle at the throat to the chamber
pressure gives me the value of the C star. We could get the chamber pressure for a given
mass flux at the throat and this is the transfer function. What does it tell us? When we
looked at the expression for C star, it was under root RTc by the value of capital gamma
which is a function of gamma being under root gamma 2 over gamma plus 1 to the
power gamma minus 1 gamma plus 1 divide by gamma minus 1. What does it really tell?
It tells, supposing I have a mass flow rate through the nozzle - mass flux through the
nozzle throat,

(Refer Slide Time: 06:54)

what is the value of chamber pressure that we get? To get a high value of VJ, I need a
high value of chamber pressure. Therefore, this C star tells you the capacity of whatever
you have in the chamber to generate a high pressure. Therefore, C star is not a function
of nozzle performance, but more like what how a chamber can build up high pressure.
All what it tells you is let us take one or two small examples.

Let us take an example of a rocket, which burns a liquid fuel. We will study about
propellants in the next series of classes.

Suppose, I have a tank containing let us say liquid kerosene I call it as kerosene. I have
another tank containing oxygen. I introduce them or rather force or push them in a
chamber and allow it to burn to generate a high value of chamber pressure pc. The value
of C star for the kerosene and oxygen introduced in the chamber will tell us the capacity
of propellants to generate a given pressure pc.
(Refer Slide Time: 09:07)

A higher value of chamber pressure pc will be obtained with a higher value of C star. But
then it is for the chamber. However, we did specify the mass flow rate and the throat
diameter. We did not look at the divergent part of the nozzle even though we did solve
for the nozzle flow. We has written that the mass flow rate M dot is equal to rho t At into
Vt. Therefore, we did not never really looked at the divergent part of the nozzle; we
looked only up to the throat. Therefore, C star is representative of the chamber to be able
to generate high pressure gases when some mass is flowing. If instead of having
kerosene and oxygen suppose, I have say liquid hydrogen and liquid oxygen may be the
C star could be higher. Therefore, I would prefer this to this and C star therefore
becomes a capacity of the chamber for a given propellant to generate high pressure
gases.

Normally, the value of C star is around 2000 to 3500 m/s with a lower performing rocket
or propellant, which is not that good, will give a lower value of C star. Propellants,
which are extremely energetic, will give higher values of C star.

Now, what is CF? We defined CF as equal to thrust divided by pc into At. In other words,
if we had terminated the rocket at the nozzle throat itself we would perhaps have got a
lower thrust than in a convergent divergent nozzle. Let us qualify this further. If I have a
rocket nozzle and I terminate it at the throat where the Mach number is equal to 1 and
now I want to find the thrust. We have chamber pressure pc and now for all practical
purposes pc is acting on all this area. The pressure is also acting normally over the head
end over here and over the convergent part of the nozzle. The force generated from the
pressure gets cancelled as shown, except over the throat area.

The thrust from the unbalanced pressure is over the throat area and is equal to pc into At.
This gives the order of magnitude since we did not consider the variations in pressure
along the length of the nozzle. But we have the divergent part like this because of which
the thrust would have increased. The increase comes from the pressure acting on the
walls of the divergent.

(Refer Slide Time: 11:38)

We can write F is equal to CF into pc into At. The value of pc into At was the thrust if the
nozzle was truncated at the throat. Therefore, CF is something like thrust magnification
due to the divergent part of the nozzle and therefore CF is a quality factor for nozzle. In
fact the values of CF for most nozzles are between 1.2 to something like 3 or 4. We will
work through some examples in the later part of this class. Therefore, we conclude that
CF is a quality factor for a nozzle while C star is a quality factor on the capacity to
generate the pressure in the combustion chamber of a rocket. The product of CF and C
star is the net specific impulse Isp. What is Isp? It is the total thrust divided by the mass
flow rate.
(Refer Slide Time: 13:34)

Let us put the expressions down again. Force is equal to thrust is equal to CF into pc into
At. Well, pc can be written as in terms of mass flow rate M dot is equal to 1 over C star
into pc into At. Therefore pc can be written as equal to M dot into C star divided by At.
Substituting this value of pressure in the expression for thrust F, we observe that At and
At get cancelled and we have thrust or force divided by M dot is equal to CF into C star.
The value of mass of propellant is M dot into time while force is impulse per unit time.
Either way the specific impulse is impulse per unit mass of propellant or thrust per unit
mass flow rate and works out to be the product of CF and C star.

Therefore, the specific impulse of a rocket has two things in it; the capacity of chamber
to generate high pressure and high temperature gases and how you expand the gases to
get high velocity. Therefore, let us keep this terminology very clear. I have nozzle factor
and a chamber factor, which gives me the net Isp. I will dwell on this further after a
couple of minutes. But does this answer your specific question, why C star? Why CF and
what is the relation? How I got the specific impulse to depend on C star and CF?

Let us take one example: let me take the example of a particular nozzle which operates
let us say in vacuum, and let me take another nozzle which operates on the ground; let us
say at Chennai which is at sea level. I have a chamber, which generates high pressure
and high temperature gases. The exit pressure is equal to say pe. At sea level the ambient
pressure is equal to pa which is equal to 100 kPa or 0.1 MPa. Now, I want you to tell me
what is the relation between let us say Isp when the rocket operates at sea level and in
vacuum. I want Isp at sea level condition and at very high altitude conditions where the
pressure is almost zero.

We can the thrust developed as equal to m dot VJ plus I have pe minus pa into what? Ae.
The thrust comes from the momentum thrust plus the exit pressure minus pa into the exit
area where Ae is the exit area of the nozzle. Mind you we derived this, and we said we
had control volume and therefore, we found a pressure thrust in addition to momentum
thrust.

(Refer Slide Time: 15:33)

Now, we would likw to write the equation for thrust of the rocket using the same nozzle,
when it operates in vacuum instead of operating on the ground at sea level conditions.
Let us say the nozzle now operates in the vacuum, the same nozzle, the same area ratio
this is the value of the exit pressure is the same and it is pe. The chamber pressure
remains the same at pc. The thrust now becomes m dot into VJ plus the pressure thrust pe
into Ae.

Now, I want to find out what is the specific impulse at sea level. Specific impulse at sea
level is therefore equal to VJ plus I have pe minus pa divided by m dot into the value of
Ae at sea level.
(Refer Slide Time: 17:24)

Now, what would be the value of specific impulse for the same nozzle functions in
vacuum; I call it as vacuum specific impulse. Vacuum is equal to what would be the
value? pa goes to 0 it is vacuum therefore, I have VJ plus pe divided by m dot into Ae. In
other words, the same nozzle when it is fired in vacuum gives me a higher thrust
because, minus pa is missing over here. Can I relate these two? Let us say Isp at vacuum
with the Isp at sea level. I find therefore, the specific impulse of a given rocket operating
in vacuum is greater than when it operates on the ground. In other words, the specific
impulse corresponding to operation vacuum is greater than when the same rocket
operates at sea level conditions. Now, I want to derive a slightly modified relationship
relating the two.

Therefore, we write Isp at vacuum is equal to VJ plus pe, what is the value of m dot? m
dot is equal to 1 over C star into pc At. If in terms of the C star, the Isp becomes the
following: I have the value of Ae here; Ae by At is the nozzle ratio epselon. The vacuum
specific impulse becomes VJ plus I have C star into pe by pc into epselon. This is the
nozzle area ratio over here. In other words, compared to a nozzle which gave me VJ plus
this value here, I get a much higher value and if this particular nozzle at sea level was
such that I have optimum expansion namely pe was equal to pa, the Isp at sea level
would have been just VJ alone.
All what I am telling is, if the nozzle was such that that the exit pressure was same as the
ambient pressure for which we told ourselves the CF is a maximum, we would have got
the value I sp is equal to VJ alone whereas, the same nozzle when I fly in vacuum, I get
an additional contribution coming over here. Therefore, now the question comes how do
I specify the specific impulse? If I tell if I say my rocket is flying in vacuum, I get a
higher value of specific impulse. If it is tested on ground, I get a different value.
Therefore, I must be clear in my terminology and therefore, two types of specific
impulses are given; one is Isp corresponding to sea level operation and the second is Isp
corresponding to vacuum. Therefore, whenever the performance of a rocket is specified
we must be careful to know whether sea level or vacuum operation is being specified.

(Refer Slide Time: 20:00)

Therefore, there are two ways of specifying the specific impulse whether a vacuum
specific impulse or sea level specific impulse but then, there is another problem. If I have
a higher value of chamber pressure, I get a higher value of expansion ratio and I can get
a higher value of the specific impulse. Therefore, I also need some terminology which
says a standard chamber pressure and the standard chosen is we specify specific impulse
for pc equal to 7 MPa or 70 bar pressure; that means, specific impulse is normally
specified when the chamber pressure is equal to 70 bar. The choice of 70 bar comes as it
is about 1000 psi in the FPS system of units.
(Refer Slide Time: 22:40)

When the ambient pressure is equal to 1 bar or 100 kPa pressure, we imply sea level
conditions whereas, when I talk in terms of vacuum, we imply very rarified atmosphere.
However, for each of these two conditions, we specify chamber pressure as 7 MPa. A
rocket can fire at different pressures, but if we are to compare something, we need some
standard and the standard is a chamber pressure of 70 bar and an ambient pressure of 1
bar for sea level Isp and 0 bar for vacuum specific impulse. The vacuum specific impulse
is higher than the sea level specific impulse, which is essentially VJ when the exit
pressure is equal to ambient pressure. Are there any other questions on what we have
done?

See so far, we have been talking of only one - dimensional flow in the nozzle. We
discussed the divergent part and sketched it as a diverging cone.

I have the convergent part, but I am really not bothered about convergent because,
anyway at throat I had a Mach number of one and the flow lines stream out almost
axially. We had a chamber of pc. Now, my divergent part you know something like it is
diverging out, and if I look at the flow which is taking place, the gas which is flowing
near to the wall will have a direction along the wall while for the gas along the center
line the flow would be along the axis as per symmetry. Therefore, we may not be
justified in assuming one - dimensional flow. It is really not correct and we have to make
some corrections for may be the radial flow or for the divergence in the flow.
There is a simple way of doing this. The thrust is not going to be in the axial direction
according to the figure. A component of thrust is going in this direction along the wall; it
gets balanced out and only my effective thrust in the axial direction. How do I get that
value?

Well, there is an actual flow over here; let us assume that the half divergent angle of the
nozzle divergent is alpha. Now, the flow near the wall will be alpha. On an average, the
mean flow direction could be alpha by 2 because here, it is alpha along the wall and zero
along the center line of symmetry. On an average the mean flow we can assume makes
an angle of alpha by 2. I can also define a small element and do the problem by
integrating it out, but the approximation is sufficient for me to give an answer. In other
words, on an average the flow leaves at at an angle equal to alpha by 2. Is it ok?

We would like to determine the thrust. Let us assume that the nozzle is adapted; that
means, ambient pressure is equal to pe here; F is equal to m dot into VJ over here. What
is the mass which flows along the axis now? It is equal to m dot into Cos alpha by 2, that
is the actual mass flow rate because, on an average some goes at alpha, some goes at 0;
the mean direction is alpha by 2.

(Refer Slide Time: 24:00)

The average mass flowing along the axis is m dot into Cos alpha by 2; VJ is again
corresponding to this over here is equal to VJ Cos alpha by 2. Therefore, the thrust due to
the divergence being at an angle alpha will therefore be F is equal to m dot VJ Cos
squared alpha by 2. Is it all right? All what did we told was flow is not all along the axis.
Flow along the wall is at an angle alpha; on an average the flow is alpha by 2 and
therefore, the mass component along the axis is equal to m dot into Cos alpha by 2. The
axial velocity on an average is equal to VJ Cos alpha by 2. Therefore, the product of m
dot Cos alpha by 2 into VJ Cos alpha by 2, this make Cos square alpha by 2.

We would like to simplify this expression. We use the term Cos 2 theta is equal to 2 Cos
square theta minus 1 or I have Cos square theta is equal to 1 plus Cos 2 theta divided by
2. And therefore, I can write Cos square alpha by 2 is equal to 1 plus Cos alpha divided
by 2. This trigonometric manipulation is done because, I can express it in terms of the
half divergence angle of the nozzle. Mind you the total divergence is 2 alpha; I said
alpha is equal to half divergence angle over.

(Refer Slide Time: 27:35)

Therefore now, I get thrust of a rocket F is equal to m dot VJ into 1 plus Cos alpha by 2.
The term 1 plus cos alpha by 2 shown within the circle is the loss factor due to the
divergence. It is denoted by the Greek symbol lambda, and we say lambda corresponds
to divergence loss or F is equal to m dot VJ into lambda. Well, lambda is something we
say loss due to divergence, but I am not really looking at a loss; see actually, I am just
multiplying it by a factor and it is the diverging loss factor. Therefore, if I have to have a
loss, the loss should be something different.
W e say lambda is due to availability of the axial thrust out of the total due flow
vectoring. What is the non-available part? The thrust not available is equal to 1 minus
lambda. In other words, lambda tells us the fraction of the available thrust, which we call
as divergence loss factor. The loss of thrust expressed explicitly is equal to 1 minus
lambda. Therefore, we have defined two terms and what are the two terms for the actual
divergence effects? We defined lambda as the divergence loss factor or rather we have to
multiply the value of m dot VJ by lambda to determine be the thrust in the convergent
divergent nozzle.

(Refer Slide Time: 30:40)

What is not available? The thrust, if the entire flow was axial everything would have
been m dot VJ therefore, 1 minus lambda is not available. Therefore, we define capital
delta as 1 minus lambda for losses. Now, why are we doing all this? We would like to
have a nozzle in which we do not have too much of loss due to the divergent. Therefore,
let us put some numbers for the losses.

If the half divergence angle alpha of a nozzle is 0 and the loss would be zero. This is not
possible because, I have a parallel walls for the nozzle. Alpha could be 5 degrees; it
could be 10; it could be 15; it could be 20; it could be 25 let us say 30. In other words, I
am looking at different nozzles for which let us say half divergence angle varies from 0
degree in which case I have no divergence to other angles.
When alpha is 0, Cos alpha is 1; therefore, the value of 1 plus Cos alpha by 2 is 1; that
means, the entire thrust is available and the loss coefficient is 0 or in terms of percentage
loss it is 0 percent.

When I have the value of alpha equal to 5, the value of lambda is 1 plus Cos alpha by 2;
Cos alpha is around 0.99 and the value of lambda comes out to be 0.9988, and if I have 1
minus lambda, it is equal to 0.12 percentage; that means, 0.0012. If the angle alpha is 10
degrees, lamda is equal to 0.9924 and the loss which comes out to be 0.76 percent. Let us
put few more values: for 15 degrees, the value is 0.9830 1 plus Cos alpha by 2 and the
loss is 1.7 percent. If alpha is 20 degrees lambda is 0.9699 1 minus lambda gives me
value of around 3 percent. If it is 25 degrees, lambda is 0.9537; I will qualify these
numbers and the loss is 4.63 percent or 0.0463. Well, the last value of alpha, we take is
30 degrees for which lambda is 0.933 and the loss 1 minus lambda comes out to be 6.7
percent. If I were to put one more angle let us say 35 degrees, the value is 0.9066 and the
value is something like a 9.04 percent. What is it that we are tabulating here? We are
considering the divergent angle of the nozzles to vary from 5 degrees to 35 degrees and
for each of the values, I get the divergence coefficient and also I am putting the loss in
percentage. You find when I go for 5 degree divergent, I am losing just 0.12 percent
thrust; for 10 degree I am losing 0.76 percent thrust; when I come to 15 degree, I have
lost already 1.7 percent thrust; when I go to 20 the loss is quite high. We have lost all
those 3 percent of the thrust; when it goes to 25 it becomes 4, 6 and so on.

In other words, it does not appear meaningful to have any divergence angle greater than
20 degrees since the losses become substantial. In fact, I were to compare 10 and 15
degree nozzles, the loss for the 15 degree nozzle is 1.75 times more.
(Refer Slide Time: 32:16)

If the half divergent value was 20 degrees, the loss is quite heavy here. Therefore, you
know the general practice therefore is to adopt some value around 15 degrees such that
the loss is somewhat small. What loss? The divergence loss, but that is not the only
reason. Let us try to put one more reason on to it. Let us consider the divergent part that I
show here.

(Refer Slide Time: 37:27)

We have seen that if the divergence angle is very small say of 5 degrees or even 1
degree, the loss factor is almost going to be negligibly small. Therefore, why not have
such a small angle. There is another implication as shown in the figure. We show the
center line alone along the axis of the nozzle; we have the throat; the throat radius is rt;
the exit value of radius is re and what is alpha? alpha is a value of this angle.

In other words, the value of re minus rt which is this value divided by the length of the
divergent Ld is equal to tangent of alpha. The divergent length Ld is equal to re minus rt
divided by the tangent of the value of alpha. Is it all right? All what we are saying is, if I
have a small angle for alpha, a nozzle for the same exit diameter and throat diameter will
have a much longer one. If my angle is 0 degree, the length will be infinity. Therefore,
let us put the same thing down over here into this plot. What do we find? Well, we just
put the length of the divergent for a particular case of throat and exit diameter of the
nozzle i.e., Ld divided by re at exit minus radius rt at the throat. This for the zero degree
nozzle 0 is infinity.

If I have something like 5 degrees, the value becomes 11.43 the. If it 10 degrees, it is
5.67. If it is 15 degrees, it is 3.73. If it is 20 degrees, it is 2.75; 25 it is 2.14; 30 degrees it
is 1.73 and if it is 35, it is 1.43. What does this mean? The length of the nozzle is very
large if the angle is very small and the nozzle length reduces, but if compare these things
for a 15 degree nozzle. If I compare 3.73 with 2.75 well then, change is not as rapid as it
is between 11.53 and 5.67 for smaller values of alpha. As the nozzle length becomes
longer and longer, the mass of the nozzle becomes larger. We had also found earlier that
delta v, the ideal velocity provided by a rocket is equal to you Isp or VJ into logarithm of
initial mass to final mass of the rocket. The mass of the rocket will go up as the length of
the nozzle increases. The mass of a nozzle and therefore of a rocket of small angle of
nozzle divergence will be more. And as the inert mass of the rocket increases, the ideal
velocity provided by the rocket will decrease. The general practice is therefore to choose
a divergence angle around 15 degrees. Mind you it is just based on the premise that, I do
not lose any further: I do not lose too much of thrust because of enhanced angle, but at
the same time, I do not enhance too much of mass of the nozzle. I have lost only 1.7
percent of thrust and and the nozzle weight does not go up drastically as it is if we go for
smaller angles. Therefore, based on this divergence analysis, we can summarize that a
conical nozzle will normally have a semi divergence angle of 15 degrees. We will not go
for smaller angles because in that case, the nozzle becomes long and mass would go up;
we will not go for larger values of angle because, if we go for larger angles, we will lose
more by the thrust. Therefore, the optimum for a conical nozzle is generally kept at 15
degrees semi divergent angle.

(Refer Slide Time: 41:20)

Does it make sense? Now, if this part is clear I just have a few more things to tell in a
nozzle. The question is, why did we address this divergence problem in such a major
way like for instance. We said that the divergent part of nozzle is something like this
with this angle being alpha; you do not want to lose thrust and therefore require a smaller
value of alpha. What prevents us from having a nozzle in which I can bring it back like
this; I can initially expand it out with larger angles and reduce the angle at the exit. We
have the throat here; I have a conical nozzle. If we could have a small value of
divergence angle at the exit, we would not loose out by the divergence loss.
(Refer Slide Time: 42:28)

What is being said is that in the initial stages of the divergent we provide a larger
divergence angle and reduce it later on like this; that means, we have initially a higher
rate of expansion followed by a smaller rate of expansion. The shape of the divergent
then looks something like a bell: the shape of the nozzle looks like a bell here. This is
the cente line. Initially, we have larger expansion angle. We decrease the divergence
angle such that the flow goes out more axially.

In other words, I have a contour for the shape of the nozzle and such nozzles are known
as contour nozzles or simply as bell nozzles. I think we could discuss further for a couple
of minutes on the contour nozzles. We initially expand out the gases using larger values
of divergence angles. Let us plot the pressure distribution along the length of the nozzle.
We know how to do it. The pressure at the throat is equal to 2 over gamma plus 1
divided by we had an expression gamma by gamma minus 1. Let us first consider a
conical nozzle divergent with a divergence angle of 15 degrees. We are only following
up with the one - dimensional analysis. We plot the value of pressure in the nozzle as a
function of distance over here. This is from the chamber; this is at the throat t and you
have the divergence over here.

Now, the pressure keeps falling as we progress towards the nozzle exit. Let us say this is
the chamber pressure value at the throat; we know how to calculate it. It is 2 over gamma
plus 1 into gamma by gamma minus 1. The pressure keeps falling further and this is the
exit value of the pressure. This is for a conical nozzle. At the entrance to the divergent,
wherein the pressure is still quite high, we have a more rapid expansion in the case of a
contour nozzle. Since the pressure is high the flow cannot readily separate from the walls
of the divergent. The divergent wall continues to guide the flow. Afterwards, the angle is
reduced and the flow is guided to be more axial.

(Refer Slide Time: 44:34)

In this case it is seen possible for us to even reduce the length of the nozzle further and
therefore, these bell nozzles are specified in terms of let say 80 percent bell or they say
70 percent bell. What is meant is the following. A bell nozzle whose length is 80 percent
or 70 percent of a conical nozzle is all what is required. We can terminate the length here
itself because we are able to get the exit pressure or equivalently the exit area ratio. We
can more effectively employ a bell nozzle than a conical nozzle.
(Refer Slide Time: 46:51)

The exit divergence angle of a bell nozzle could be between 2 degrees to something like
5 degrees, thus giving a very low divergence loss. Initially, we expand it rapidly; here we
provide a large value may be 20 degrees to 50 degrees. We can make the nozzle a little
more stubby or shorter with and the length of the bell nozzle being a fraction of the
conical nozzle. To repeat, 80 percent bell nozzle means, the length of the bell is 0.8 times
that of an equivalent conical nozzle. What do we do in a bell nozzle? We immediately
expand downstream of the throat where the pressure is higher and allow the divergence
to be smaller in the region of the exit such that we have less divergence loss. In fact, one
paper which we could read on this subject is by G V R Rao. He worked on it at
Rocketdyne a long time ago. The paper is on exhaust nozzle contour for optimum thrust
and is refereed to as Rao nozzle.
(Refer Slide Time: 48:16)

This was published in a journal known as jet propulsion which preceded the a AIAA
journal. The volume number is 38 and the year of publication is 1958. The page number
is 377 to 381.

In a bell nozzle we initially expand the gas rapidly; you have something like raid
expansion and then, you have something like less expansion or compression which does
not lead to adverse pressure gradients near the walls. We can afford to have a shorter
nozzle compared to a conical nozzle, which is the core of bell nozzle. Most rockets
make use of the bell nozzles. If we have a bell nozzle, none of the earlier criterion like
flow separation are relevant because we have higher pressure gradient along the wall.
Therefore, whenever we talk of Somerfield criterion saying exit pressure is 0.4 times the
ambient; it is more applicable for conical nozzle and not exactly for a bell nozzle. I think
this is all about nozzles; conical nozzle and contour nozzle. I want to spend another few
minutes on different types of nozzles.
(Refer Slide Time: 50:13)

The ambient pressure decreases as the rocket moves up to higher altitudes. We had said
that for maximum thrust coefficient, the exit pressure at the nozzle should be the same as
the ambient pressure. Is it possible to have a different type of nozzle all together which
can adapt to the altitude of operation. Can we make a nozzle to adapt to different
altitudes starting from 0 kilometers and keep going up to 10 kilometers or 100 kilometers
height? We shall deal with this in the next class and work out one or two small problems.