Blade element theory, because earlier we related
the thrust, the inflow, the power, rotor disk
area only induces using momentum theory, but
that does not consider any detail. But now,
we are going to look at the details of the
rotor, rotor operating conditions everything,
but that is the blade element theory, please
understand this is the fundamental. We have
to have this theory for all analysis of helicopters.
And whether it is a fluid dynamics, whether
it is aero elasticity, loads, stability everything,
blade element theory is the fundamental. So,
you have to know the blade element theory,
but to give you a brief history, see it was
proposed by basically started William Froude
in 1878.
Because blade element theory essentially says
that I am going to treat every cross section
of the blade, because you know that the rotor
blade 
is essentially an if you look at that this
is a aerofoil shape. He said, I am going to
look at the cross section of the blade, which
is an aerofoil, because these are all please
understand you know before the flight started,
even the fixed wing, but then the real development
was done by STEFAN DRZEWIECKI or something
it is 1892 to 1920. What he did was the propeller,
these are all for propellers, all the propeller
is going forward forward. It is not hovering;
there is a velocity which is coming. And there
is a omega R which is due to rotation velocity
at any cross section. But he did not take
into account the velocity due to the inflow
or the induced velocity; he just took the
propeller forward velocity and then the omega
R.
And when he calculated the thrust the power
etcetera it was not matching with the experiment,
because please understand propeller experiments
were lot of experiments were conducted earlier
where power and thrust and he found that they
were not matching. Because then he felt there
is something wrong between theory and experiment,
and the difference he thought it is due to
the aspect ratio of the basically the propeller,
because it is not a long propeller it is a
finite propeller.
But he was not sure what is the aerodynamic
characteristic, I must take into account at
every section. So, when he makes some corrections
for aspect ratio, because you know that there
is a correction for aspect ratio in the finite
wing theory. Similarly when the trends were,
but still the results are not good, they are
not matching with the experiment. Then several
people attempted to see how to take the proper
sectional characteristic, because you have
two: one is the aspect ratio effect, another
one is airfoil characteristic. These are the
two things there is a something I have to
make adjustment.
So, in 1915 and 1918 and also 17 that is Betz
and Bothezat and then Fage and Collins these
are the three groups of people, they said
I have to take the induced velocity. But what
induced velocity, I should take at the if
I want to calculate the lift here, the propeller
this is going this is the cross section omega
R is there. So, you may take this is my omega
R and there is a propeller is going axial
velocity of the propeller which you may call
it V plus some induced flow. This at we will
take the induced flow from momentum theory
or something like that. But still, but it
is not aspect ratio effect there started using
it. So, you have an aspect ratio effect was
also brought, but you took the inflow also
again there was a miss match.
So, they found that there is something wrong,
in the sense wrong in the sense something
inflow if I take it from momentum theory;
if I take the aspect ratio they do not match
again they give a trend. So, they try to adjust
some empirical numbers that is why Fage and
Collins, he took the aspect ratio as 6 fix
the number and then he started correcting
the inflow by some empirical factor. This
is how it was going on to match the theory
experiment purely thrust and power that is
all nothing else, but later because Prandl
finite wing theory that is a Prandl vertex.
Now, lifting line theory that was in 1918
when he proposed then they said yes we have
to this is a vortex model, because I should
take lift is O U gamma and the gamma is the
vortices. And you find out the strength of
the gamma from actually the trailing edge,
because you have to adjust lift you say O
U gamma. And then you go back and then take
a vortex finite vortex and then get this strength
of the vortex. Then you use it then they said
yes, we have to use vortex theory to get the
proper value of inflow at the place where
it is going at the rotor disk. That is why
in the beginning the correct treatment of
2 D airfoil theory for rotors came from vortex
theory. Because two dimensional there is a
correction, because the vortex is given induced
velocity in the that is what you get induced
dragger in your fixed wing theory.
So, the vortex theory was dominating in the
beginning of the rotors, because they said
that is the correct treatment. So, everything
went in vortex theory approach momentum theory
was not really looked at. But subsequently
then people realized yes, you can take momentum
theory and then make, but even today if you
want more précised treatment of induced flow
at the rotor disk, induced flow I mean you
all know that the flow normal to the rotor
disk, if you want they get the correct value.
Because correct is something exact is different
there is a vortex model, but that is more
complex, because if you start looking at the
vortex theory as such if I draw this as the
rotor, it is a short, it is a rotor, it is
rotating. The vortex is will be because as
it goes you will find a vortex coming. And
if the lift is varying along you will have
a vortex sheet actually a sheet will come,
because there is a lifting line theory. I
hope you understand some of you may not know
lifting line theory, and then the strength
of the vortex that varies, how the lift variation
and then what is the shape of because this
is going to go in a in the rotor it will go
in a helical shape.
Because in a finite wing case, you will say
my vortex sheet is going like the horseshoe
vortex, because this what do you say this
is a horseshoe vortex in finite wing. The
vortex swept back, but in the case of a rotor
hovering rotor the vortex is piling below
it, it will go in a helical path, because
as it goes around it leaves and this one push
down by the induced velocity by its own interaction.
Now, the structure of the vortex you have
to make approximation, because if you experiment
laws of observations were made on how the
vertices look. And they found it goes in a
some kind of an I will show later some pictures,
it will like this. So, like that this will
go vortex sheet and this what will happen
initially, they will start sheet suddenly,
they will all bunch up become a tip strong
tip vortex. Because you know the strong tip
vortex and how that is convicted down; that
means, you have to take a structure of that,
that is the wake structure. Now, that you
have to know or you say if you have vortex
to vortex interaction, if you consider then
it is called the free wake theory.
But if you say I fix my structure of the vortex,
it is going to be in a cylindrical vertical
cylinder that is all no deformation. Then
I call it prescribed wake either I prescribe
the wake structure that is the relatively
easy, but if I do not prescribe I say let
it evolve by itself that is a free wake theory.
But it is more complex even prescribed wake,
pre wake both are complex. Now of course,
people have developed their own computational
course using vortex theory, today I am telling
you vortex theory is there it is used.
But the beauty is, if I consider the hovering
rotor and then you assume that the vortex
is in a cylinder and it is an infinite number
of blades actuator disk; that means, it is
just a lot of circular continuous vertices,
it is easy for integration that was what was
done in the earlier. Cylindrical wake structure
and infinite number of blades means, there
is no gap between one and the other please
understand one blade gives a wake it will
come down, then there is another blade that
will also give; that means, there will be
several helical wake. And there will be gap
between those wakes, because you get one coming
down then if you say another blade end, that
will come and they will go intertwined.
But if you have infinite number it is almost
like a continuous cylinder when you use a
cylindrical wake like this. The induced flow
calculated by this approach at the rotor disk
as same as momentum theory uniform loading,
uniform thing you take then you get the same
result. So, what happens in momentum theory
is not all that bad. So, we use in this course
momentum theory please understand, momentum
theory is even today in the research, it is
used I will show you later. Now we use it,
because if you want to have vortex theory
then you need to sit under, it is like a pretty
much computational more internship.
Whereas you will find even with a momentum
theory, the result which you predict are reasonable
good, pretty good I would say, but you have
to make some corrections here etcetera with
that your results will be good. And that is
why momentum theory is still used in rotor
analysis, this some people may use it that
is a different point. It is a then school
of school of thought and if you want to complicities,
but it does not mean that everything you were
able to predict. Because the wake structure
unless it is correct and you do not know,
because the viscosity will diffuse the wake
how far you must go down.
So, there is on group of researchers working
on wake, that type of prescribed wake free
wake analysis, other group they said that
I need an inflow. Because now, you see the
research itself was directing towards whether
modeling the inflow accurately, that is the
research I do not do anything else except
my focus is inflow modeling, how do I know
what is the velocity at every point on the
rotor disk. Even today it is a research later
you will find how that itself gets complicated,
another one is an I have a good theory pretty
descent which is easy to incorporate in my
aero-elastic loads responds analysis, because
I need to do that also.
So, you see the research if I somebody focuses
on only aerodynamic related problems, he will
be focusing on only that inflow calculated,
he may not get into rest of the things. But
if you are working on aero-elasticity, loads,
stability etcetera of the vehicle you need
to know the inflow, you say I will calculate
by momentum theory which is easy to incorporate.
But of course, that vary somewhere, but it
is not too bad that is what I am saying, it
still gives a result which are quite close,
but of course, there are always differences.
Now, one can debate on whether one should
use always this theory or the other theory,
then you will find slowly as we proceed even
that has certain draw backs.
Now, I just gave a brief introduction with
the field how the research in inflow calculation
is important particularly, but we will use
not vortex theory, we will use simple momentum
theory. But then we will complicate it not
the uniform inflow that is what we did in
the momentum theory, the flow over the disk
is constant at every location that is uniform
inflow. Later you will see hey that is too
gross on approximation, let me have non-uniform
inflow; that means, it is varying along the
radial direction of course, if you want more
complex in general various other situation
you can say it varies everywhere. So, that
is how the complexity in the momentum theory
also went on.
At today, that is also there is one theory
which is dynamic wake theory which is somewhat
similar to this somewhat similar. So, please
understand that model is what we use dynamic
wake, we do not model this, we use the theory
to get the time varying inflow at the rotor
disk, please understand we use till now constant,
it does not vary with time constant and uniform
everywhere.
Now, you say initial development we learnt
that there is a constant; it is uniform first
theory that is what I am going to discuss.
Now, do not think that is the end of it, because
there is a lot more in the theory in the inflow
modeling alone. So, when we go to the research
level, we do a little bit more complex theory.
Now what is the blade element theory, this
is a rotor blade which is rotating. So, I
said the cross section is this I know what
is the oncoming velocity, it is a 2D dimensional
model that is all 2D airfoil model there is
omega R. Let us first consider, hover and
vertical flight in the sense the disk can
go straight up down is a little, we will do
that later vertical flight and up and down
that has it is one complexities, we will do
one by one slowly. Here I said that this is
the rotor blade which is rotating with an
angular velocity omega, and this is my frame
X and Z and this is Y which is attached to
the blade it keeps going.
Now, I look at a cross section which is at
a distance r from the center of the disk or
you can say hub that is r. So, your velocity
is omega r simple, because please note this
is only an introduction, and you go to actual
blade it is not omega R you have so many other
terms will start coming it. So, that is why
when we start the course you will learn basics
and then at this this is a small elemental
cross section D R.
Because please note my omega R is changing
depending on the radius, that is why that
lower case r is used for running variable
along the span of the blade. Now, I have omega
R and I assume that this is the climb; you
may put a subscript V C to denote it is a
climb velocity. And nu is induced velocity
you do not know that, but you say I know it
somehow, you may get it from any theory that
is where you can use momentum theory, you
can use vortex theory etcetera.
And that is why I mark up which is perpendicular
to the disk. So, now I have a disk and this
blade is going in the disk please understand
it is not coming out or anything like that,
we will do later one by one, this is just
a very, very simple theory. So, this is rotating
and there is a normal flow. So, I have V C
and nu now if this is kept at a pitch angle
theta with respect to the disk on coming flow.
Now, you can get the relative velocity as
well as the angle of effective angle of attack
you can say. So, this is the phi, phi is the
change because of the normal flow.
So, you will have your resultant velocity
U is this is my resultant velocity, but for
simplicity you know, because of every time
we do not want to carry this, you use the
symbol U P that is the velocity perpendicular
to the disk this you call it as V C plus nu
and U T which is omega r this is tangential
velocity, this is the normal velocity instead
of normal N you call it U P. So, this is U
T this is U P this is a general symbol this
is a standard notation that is used. So, that
so, that it is ease of understanding U T is
tangential U P is normal only everything with
reference to the airfoil.
Now, immediately you go back what is my effective,
because you know the tan phi is 
U P over U T tan phi is U P over U T. And
now, I have to define sectional lift sectional
drag that is all, and lift is normal to the
I use simple 2 D airfoil theory lift is normal
to the oncoming flow and drag is along the
oncoming flow. So, now, you see and I define
some psi just to denote, if the azimuth location
this is for one blade, if I have N number
of blades, I will use all the blades everything.
Now, I have got sectional lift and dragger
this is for the please note very very simple
model and this is what I have shown here.
So, U P U T and this is the resultant U and
theta is the pitch angle and alpha is the
effective angle of attack. So, you see effective
angle of attack is what I now you remember,
why I used pilot input gives pitch change
it does not change the angle of attack it
will change indirectly. But he gives only
pitch change in the sense he changes this
angle, but the angle of attack changes depending
on what is the U P, what is the U T etcetera.
I may assume now this induced velocity nu
to be constant over the rotor disk that is
it is not a function of r. It is a hovering
theory, it is hovering or cline, it is not
a function of radial, it is a constant then
you see the phi changes, because of omega
r. So, every cross section because the tangential
velocity is changing. So, automatically my
effective angle of attack which I call it
alpha alpha is theta minus phi which you may
call it theta minus tan inverse U P over U
T and then I am going to make approximation
straight away right here.
So, please understand write at the definition
of angle of attack, you have tan coming in
you make now an approximation, because in
a text book form if I have to give you a closed
form solution. Computationally you can always
take tan inverse no problem, but you know
that has R this is nothing, but what V client
plus nu over omega r as r decreases this is
very large; that means, phi is going to become
more and more as you come near the root, and
that means, the angle of attack whatever because
this value will be classically increasing
theta may be fixed, but this will keep on
increasing.
But then you were going to have problems,
but usually what happens is your aerodynamic
section starts not from the center of the
hub. And another thing is the omega r; the
dynamic pressure due to velocity omega r is
very small near the root. Therefore, you make
an error, but you say it is all right I accept
that error. So, I represent this tan inverse,
because I assume this angle is small.
So, please understand I assume phi is small,
I write it U P over U T tan phi is phi, tan
phi is phi. When phi is small U P over U T,
but I am violating this rule as I come closer
to the hub, but I still say it is all right,
I accept it. Because I know that dynamic pressure
is very low. So, the lift is not very large.
So, I may usually you will find the aerodynamics
section of the rotor blade starts about 20,
25 percent away from the center, because of
the geometric, because you have to have you
have a attachment everything. So, it will
be around 20 percent 20 25 percent and the
error you make in that is not allowed that
is why you make this approximation first theta
minus U P.
Now, you know angle of attack and I can write
the lift per unit span straight from your
aerodynamics, lift per unit span is lift per
unit span please understand, otherwise over
a small element you put a D R, you will have
half rho U square card and C L. This is lift
and the drag will be if you want to put you
can put A D R half this is per unit span this
is the lift C D, this is my 
at any small element.
Now, if I want to get the total lifts I need
to want integrate that is all. Now, I make
more approximation the more approximation
is I am going to call now, U I say omega R
is because when phi is small basically omega
R is large. So, when omega R is large this
is a small quantity. So, I which is basically
UT so, I represent my U as UT resultant velocity
as U T, but that does not been I change the
angle of attack also I keep alpha as this.
Now I know directly I go back, I write my
lift as half rho UT square this is an approximation
I am making. And see now what is my C L, C
L this is from your elementary, C L alpha
lift curve slope tan phi angle of attack.
Now, if I substitute here this values C L
I will get C L alpha into theta is the which
input minus phi is U P over U T into d r.
This is the approximate expression now you
see this is much easier if I write it I take
the half rho CUT square I take it inside.
So, you will have C L alpha U T square is
omega r theta minus U P U T all right. And
UP you know v plus nu U T is o mega r now
it is easy for me to integrate along the radius
on the other hand if I put divided by omega
R and every other factor I do not take it
you know it is going to become a messy stuff
that is what is done first. Now this is my
lift for an elemental length if I want the
total I go back. So, I erase this part.
Total thrust total thrust is total lift that
acts on all the blades in the rotor system;
that means, I assume every blade in hover
behaves the same way; that means, I will simple
put number of blades N all right. And then
I have to integrate from this expression I
will put 0 to capital R the entire expression
you write this you will have the full expression
which is maybe I will write this then I will
do the non-dimensionalization because that
is important.
So, you will have this is N 0 to R half rho
C C L alpha omega r whole square theta minus
U P S V climb plus nu and U T is omega r into
d r this is my thrust rotor thrust. Now, you
surely we non-dimensionalize the quantity,
we do not carry this whole thing when we non-dimensionalize
we use I erase this part here velocity quantities
are normalized with respect to tips speed
and you use omega r.
So, you see CT is thrust coefficient is thrust
divided by rho pi r square, this is the area
of the rotor disk and omega this is my C T.
Now I divide this entire quantity by this,
so now, please understand one thing density
card card can vary along the blade, but if
my today most of the rotor blades actually
they have a constant card except near the
root.
Otherwise your card can vary that is what
your small model card is varying. So, you
cannot use the same expression you have to
put that integration, but normally what is
done is we take some constant card etcetera
and then we calculate. Now when I do this
I get a quantity I will just briefly describe
I will have N 0 to R. So, I am going to divide
divided by rho pi R square omega capital R
whole square this is what I am dividing when
I divide let me write the expression here
C T I will have N there is a half factor outside
any way rho will cancel out.
Now omega R whole square you take it here,
when you take it here this will be omega square
will go up R bar square we write it as I will
say integral leave it is essentially 0 to
1 it will become. Because pi R square is there,
there is a C, there is a d r. So, one of the
R you take it to this that will come R bar.
So, I will make this bracket here this will
be d r bar r bar means r over r. So, integral
is 0 to L right.
Now, inside quantity let me first write it
that is r bar square theta minus I am going
to call this symbol this V C plus nu over
omega capital R by a new symbol which is lambda,
lambda is inflow. Please understand this includes
climb and induced velocity later I may split
this lambda C and then nu some lambda I, but
right now I am calling it as lambda because
this itself we can write it as lambda C plus
lambda I; that means, sorry C is climb I is
induced.
You can split that is later right, now we
will take it as lambda and one omega R one
another omega lambda r bar. So, this is gone
then you are left with pi R. So, I can have
here there is a C and because there is a one
of the R has gone to r bar. Now, if you assume
it is a constant card then you can take out
the constant card means C also outside. So,
I will take the C outside.
Now, this quantity I am going to call it as
sigma which is essentially blade area over
disk, because if I multiply R and square this
is C R is card into span that is a blade area
N is number of blades. So, because please
note all blades are identical. So, you will
have sigma blade area over disk area this
is called solidity solidity ratio.
Now, I can write this entire quantity as sigma
over 2 and then there is a C L alpha I have
to add up. So, C L alpha is there sorry I
forgot to put that C L alpha, because this
is there. And C L alpha, you may call it C
L alpha or some time we call it a by the symbol
C L alpha or a both mean the same. Now, you
had left with completely the solidity ratio
is important, usually blade area divided by
disk area. And this is an important parameter
like your thrust co-efficient, now we have
introduced one more new quantity which is
the solidity ratio of the rotor, because that
is that plays a key role.
As we go along you will see now, this is what
my expression for C T. Now, this is the lift
it adds please understand now I go back here
a little, we make a we made some approximation
what is by lift, lift is normal to the resultant
flow right and drag is along the resultant
flow, and I said in my hub co-ordinate system
X is in the plane of the hub. So, that is
normal to that I should take if I want actually
the force normal to the rotor disk or the
hub I must resolve these two components L
and D along vertical and this X direction
that is the precise. Now, I go and do that
I have written here normal force f g is L
cosine phi minus D sin phi that is a subtract,
because this is the resultant is coming here
this is omega R. So, you have your L your
right I think I should this is L this is D
and this angle is phi, this angle is phi,
this is we said X, this is we said Z. So,
L cos phi minus D sin phi so, I miss drag
drag also into account. And then in-plane
force is L sin phi plus D cos phi now I make
another approximation that because my phi
is small and the drag is usually for aerofoil
is very small it is you can take it at the
most .01.
Whereas lift C L is C L alpha that is 2 phi,
so he says my drag is very small therefore,
F Z please understands F Z I say is L directly.
now that is what I have used only L in defining
the thrush coefficient; that means, basically
this is normal to the rotor disk that is all,
this is approximation I make this. But in
the case of a pin-plane force I say F X, I
cannot neglect lift lift large phi is small.
So, I am gone to write it as L phi plus d,
because phi is small. So, cosine phi I take
it as 1, sin phi I call at as phi whereas,
when I go here D is also small phi is also
small. So, product is much smaller. So, I
neglect that whereas, when I come here I do
not do that. So, this is the now you understand
right at the beginning I make all sorts of
approximations.
That is what I given here I put it and elemental
thrust, because this just to indicate to you
what all we have done L d r, L is lift for
units pan, this is d r here I modifies slightly
and then N is number of blades D T is actually
thrust which is F Z this is per unit area.
So, I multiply by d r is it clear. And then
when I go for torque, because I am not interested
in the drag force I want to see what is the
torque because drag force I want just integrated
I will integrate and then get what is the
torque required to rotate, but later if I
want hub hub loads I need to get the drag
also I will do that later. But right now I
am calculating only the torque is L phi, because
I approximated affects L phi plus D and it
is acting at a distance or from the center.
So, the drag force into distance is elemental
torque and the power into omega, because torque
into omega is power. Now I have all the three
quantities elemental thrust, elemental torque,
elemental power I only shown you this particular
thing.
If I integrate this, before I let me write
the expression C T is here I put what sigma,
now C L alpha what type of aerofoil you are
having at every section where it is a constant
it can vary we are we said card is a constant,
but my aerofoil can change, but we make again
no I am going to have the same aerofoil through
out. Now, C L alpha also comes out which I
call it as using the symbol a because most
of the helicopter literature they use the
symbol a lift curve slope C L alpha is a sigma
a over to 0 to 1 r bar square theta minus
lambda r bar into d r bar. This is my C T
which is the non-dimensional actually the
first expression d T is there right if I integrate
the d T over r I will this is the thrust that
is normalized I get this expression.
Now how my theta is varying along the blade
I said pilot gives an input that is the pilot
input he gives at the root, but the blade
itself can be twisted; that means, there can
have a pre-twisting the blade geometric twist;
that means, what initially pilot gives and
then what is the angle and as it goes what
is the pitch angle at every cross section
of the blade that you have to take it.
Now, here only there are if it is constant,
constant means there is not twist zero twist,
zero twist means you can directly take it
if you integrate which very simple. Assume
zero twist theta is constant everywhere this
is very simple this is what C T becomes sigma
a over 2, theta over 3 minus lambda over two
that is all very simple expression. But you
do not know lambda that is the different thing,
but if it is hover you will see how we get
it from if you have at twist built in the
blade then you have to take that into account.
Usually rotor blades they do with a linear
twist, linear twist means the angle theta
varies linearly from root to tip with a reduction.
So, you can write it in a form where you have
a twist then you can integrate fully now why
do you give twist that is another question.
So, you can have, but then the proof for twist
if you give you vary your inflow also, there
is a little different because we still have
not obtained this value please understand
if you use momentum theory assuming hover
I am just going a little only with this I
am discussing I am not gone to be other two
terms only this term zero twist I am considering
hover, then lambda becomes lambda I, that
is all induced. Because there is no climb
then I will have my C T is sigma a over 2
theta over 3 minus lambda I over 2. But momentum
theory hover gave me what lambda is root of
right sorry lambda I, I should use the symbol
lambda I because this is for hover. Now blade
element theory is relating pitch angle, induced
flow, solidity and aerofoil characteristic.
Because all these are there number of blades
actually this is are non dimensional quantity
if you want then operating condition is there
angle of attack everything is there. Now,
I can substitute for lambda I here.
And then I can get theta directly what should
be the pitch angle I must give if I want to
have this much C T then once I get the once
I know the C T, I can get the inflow. On the
other hand in a experiment if you give a pitch
angle I must know what should be the C T;
that means, I have to do iteration, because
first I assume the value of theta then I do
not know this value I am take it as 0 you
follow.
Then I put the value of theta I get a C T
when I get the C T, I go back and substitute
here, then I get the lambda I put it back
here, then I get the new value of C T go back
do the iteration, that is in a wind tunnel
test if you want correlate hey I have given
this much pitch angle what is my thrust because
these test are done lot of.
So, you see only these two equations you are
using only these two now in helicopters you
understand that you cannot live with only
momentum theory or you cannot leave only with
blade element theory you can replace the momentum
theory by some other theory vortex theory,
but this is essential blade element theory
is essential. So, you always have two theories
either you say I take prescribed wake or free
wake any analysis and then I get the inflow
then use that inflow put it here get the thrust
then go back again to see whether that thrust
gives me that inflow.
Now you see the loop of the calculation in
just a very simple problem, if I give you
a weight of the helicopter and the rotor radius
some C L alpha aerofoil shape etcetera calculates
what should be the pitch angle, you should
be able to do it.
Now I will just compile this staff and then
write it here later we will see the twist
part right. Now untwisted zero twist so, I
have C T and zero twist and hover please understand
because I am doing both C T becomes sigma
a over 2 theta over 3 minus lambda I over
2 and lambda i from momentum theory is root
of C T over 2. Now, I just want to this is
for practical purposes you substitute this
here and then collect the terms of C T on
one side just in this expression if you do
that you will get what, 2 C T over sigma a
this is theta over 3 equals this will go here
plus one over 2.
So, I take the 3 please understand I am taking
the 3 here at 6, Zero twist hover if I know
the thrust coefficient please understand if
I know the thrust coefficient; that means,
I know C T thrust coefficient I know, because
that is the weight of the helicopter I now
you see I want really thrust coefficient I
equate to sometimes people say weight coefficient.
Basically weight is supported by the thrust
you say thrust coefficient is equal to C T
equal to C W please understand. So, I use
this approximation again here if you are given
the weight of the helicopter if it has to
hover W you say that is t that is all weight
is equal to the thrust. So, you will have
C W is C T weight coefficient thrust coefficient
weight coefficient is nothing but W by rho
again pi R square omega R whole square which
is same as if you are given a helicopter with
the radius and with the rotor angular velocity
you can get the C T. Now you see what should
be my pitch angle I must provide for hover
you substitute the value.
Here this quantity you know that a is the
lift curve slope is theoretical value is 2
pi theoretical value is 2 pi this is almost
close to 6. So, you not deserved now you see
C T by sigma this is the mean angle this the
term due to variation in the because of the
inflow. Because of the inflow there is a change
C T over sigma now please understand why this
parameter becomes important thrust coefficient
over sigma this gives you mean angle approximately
straight short.
Because sigma takes N C over pi R C T is thrust
coefficient, now you will immediately say
hey what should be the angle I must provide
for hover just mean value just you you take
its if you negated it is all right that is
why in practical situations helicopters they
always want C T over sigma that is the value
Because that tells you what is angle you must
provide suppose if this value become for the
aerofoil 15 degrees then it is your nearing
stall of the blade. So, it gives you they
do not go you immediately see if I increase
by solidity ratio I reduce the angle for hover.
That means my aerofoil is good, but if I reduce
my solidity I am actually increasing my angle
of operation, but you may stall. So, these
are very important you have to operate and
if you are near the stall angle if you want
to climb up, because this is hover if you
want to I have increased my thrust increase
my thrust I increase my collective, but what
will happened is you will not be able to lift
the blade will stall. So, this is very, very
important C T over sigma has a parameter.
This is also called blade loading why it is
called blade loading I will briefly describe,
because C T is let me erase this why it is
called blade loading thrust divided by rho
pi R square omega R whole square this is C
T sigma is N this is sigma.
Now, this factor you can reduce it in a slightly
different form and which will give you approximately
in a different form of expression that is
I will writing it in this fashion because
if you take it as N C R pi R square what happens
this will become pi R square will go this
will become thrust divided by N number of
blades C R of course, omega R whole square
and rho this is basically for non dimensional
staff, but N C R tells you the blade area.
So this is also called C T over sigma is called
blade loading see you saw disk loading disk
loading was T over A then you saw power loading
now this is blade loading it also tells you
how much each blade lifts how it is shared
by how the load is shared by the blades. So,
you see two non-dimensional parameters giving
you the pitch angle of operation of a blade
in hover that is the you know critical thing.
Now, I want you to note down because this
particular quantity you may call it blade
loading you may call it just mean angle of
attack whatever it may be mean angle of attack
actually mean pitch angle you may call it
I thing I would better mean pitch angle would
be a better mean pitch angle required for
operation. How this value please understand
this is just for you to know we are talking
about hover, how this value vary with forward
speed will limits because we know it is related
to pitch angle pitch angle means if you keep
on increasing your C T I am giving you one
rotor everything dimension everything is fixed
operating condition is fixed omega R is also
known.
But you keep on increasing the weight of the
helicopter what you will do you have to keep
increasing your blade angle up to some blade
angle you will lift, after that blade will
stall; that means, you cannot lift more weight;
that means, that set the stall limit in the
sense the limit the rotor can lift beyond
which that rotor cannot lift because you have
exceed that. Now this is true for even forward
flight, but in forward flight also you have
to lift the weight in addition to overcoming
drag etcetera. Now industry requires this
C T over sigma it is not a constant with forward
speed please understand this varies with forward
speed, this is the what I gave you as a project,
because this is a very tricky stuff. Limit
of the rotor I design a rotor how much you
can take it, because this very, very important
if you are making you cannot just like that
take a rotor put any weight they have to make
sure what is my operating condition and then
I have a envelop well set. So, C T over sigma
plays a very very important role in the helicopters
that is why I am saying and this quantity
is not when as a blade loading.
You tend to get an idea that you know the
weight of the helicopter everything is known;
that means, its constant you will say please
understand tends to give you a feeling that
is the constant; that means, irrespective
of the flight condition it is a constant.
Normally using this quantity you define a
limit with respect to I will just draw the
diagram, but I will not C T over sigma this
is speed forward speed, hover yeah I know
that mean how much this quantity in hover
you can go for a given rotor please understand
this is drawn for a given rotor, given rotor
means sigma is fixed right. These quantities
are fixed even rotor omega R I is also fixed,
but this you keep on increasing up to a point
where blade will the angle you will reach
that the blade will stall and the same think
can happen at every forward speed, because
every forward speed you will find I can keep
on increasing the weight till the point where
my blade will stall; that means, that becomes
your stall limit.
It will something like this it may actually
come down down down; that means, the limit
you can go it is drawn in C T over sigma curve,
but these are actually I had a lot discussion
with about this particular thing finally,
we figure out this is that key this becomes
important, but how do you generate this curve
theoretically that will now we are going to
do it.
But we will give because this requires very
detail if you want actual helicopter thing,
but in a preliminary thing we can do simplistically.
So, this is just for you to know C T by sigma
is also used for drawing a limit of a rotor
stall limit you may say, because beyond which
if you go blade will stall you cannot rotor
will stall you may call it rotor may stall
or blade may stall.
So, it may come and may hit like this if you
say you see if I go I am stalling, because
I crease my C T by sigma gone up, pitch angle
is gone up, blade stalling same thing will
happened at every forward speed, but how do
I get this curve in forward speed with all
complexities that is the key. And they make
sure that when they design the helicopter
your has below here you do not go anywhere
near because you know because you never know
at some operation you may stall and if you
stall loads will go up and actually I do not
know whether you people know there are couple
actions that which came in the video and other
things.
The pilot was doing a turn and usually you
go and climb, but when he started for climbing
he was actually going down climbing mean what
you increase your collective you increase
the collective, but he was going down which
means what blade has gone into stall. So,
these are very critical things in Manuel suddenly
you may stall and then you thing that because
the pilot is that is why it is very, very
tricky of course, he walked out from the accident
in one case in one case it was fatale. So,
these are very important parameters in the
helicopter field. Now let us go to the next
factor which is the basically the torque part
and I will I will also give you a little bit
of this twist before I go to the torque part
I will give the twist just directly from here.
I am now going to write it just for if I have
a linear twist, linear twist please understand
here I use 0 twist if I have a linear twist
theta naught that is what is the value theta
twist into r over R this can be written in
terms of the angle at .75 R. 75 R if you take
it in this fashion please understand if you
take it in this form and put it in the integral
and then integrate you will get the expression
like this which is I use lambda it exactly
same you follow please understand.
If I take .75 R as my reference angle I am
representing theta equal to theta .75 into
R over R minus .75 theta twist this fine this
expression is this is basically same as this
I am putting it in a different form. If I
use this expression in my thrust and integrate
where this you can do yourself because you
do it as a exercise if you take this expression
integrate you will get this expression which
is essentially sigma a by 2 theta by 3 minus
lambda by 2 now you see theta represents the
angle at 75 percent R.
So, why 75 percent is so, critical you know
most of the time we say because that is where
the lift also becomes high around that region
because more near the root lift is small then
as you go towards end the lift will rise up
and then of course, the tip it will drop.
We will come to that tip etcetera later right
now or if you directly use this expression
you will have this because you see if you
take the 3 out theta naught plus 3 by 4 theta
twists that is nothing but .75 angle at .75.
So, the 75 percent of the blade is always
they will say what the angle theta is?
Now, there is something called a ideal twist
I will just mention this and then we will
close that is if my pitch please understand
r over R is in the denominator tip angle I
said as I come in board it will go like this
ideal twist, linear twist is straight line
say it will be that if theta and linear twist
may be like this R over R this may be one.
But ideal twist this is ideal, this is linear
you cannot make it because you know I will
take infinity twist infinity has no meaning,
but up to some point to radius because near
the root it is all not an aerofoil section.
So, you can cut it out here or somewhere place
then you says I can do something like that
I can manufacture. So, you see this I will
give you an exercise which you can try I can
have an ideal twist.
And if I use this my C T will be this expression;
sigma a over 4 theta tip minus 4, expression
is different. Now you see depending on the
twist, the expression for now why ideal twist
is in this form, but why do I call idea, is
another question. The reason is I call it,
if I use this twist my inflow will be uniform
I can prove that using the another again that
theory we have to come, whereas in all these
things I assume that inflow is constant everywhere,
please understand I assumed inflow is constant,
but I do not know whether it is constant or
not.
That means I said lambda I equal to root of
C T over 2 I assumed it, but it not true;
that means, I am making even this writer very
simple expression, I am making a error, I
assume this is constant, but I do not know
whether it is really constant or not; this
is a problem. You can use that, but you make
errors. Now you say how do I later we will
do, now the torque part we will do it in the
next class.
