Good morning and welcome you all to this first
session of the course on fluid machines. Now
as I have already mentioned in the brief introductory
session of this course that fluid machine
is a device that converts the stored energy
of fluid into mechanical energy and vice versa.
The stored energy of fluid usually appears
in the form of pressure, velocity or temperature
or inter molecular thermal energy by virtue
of its temperature, while the mechanical energy
is obtained by a rotating shaft.
Now the use of fluid machines, already I mentioned
is very wide in the industrial applications.
The major applications pertain to electric
power generations, aircraft and rocket propulsions
and in varieties of medium and small scale
industries. In electric power generations
as you know turbines are used which are the
main important components or power producing
component of the unit. For aircraft propulsion,
compressors are used, centrifugal compressors
and axial flow compressors.
Again for industries where high pressure air
is required and this is used in almost all
process industries, the centrifugal compressors
are used. Similarly for transmission and distribution
of water or liquid, circulation of liquid,
pumps are incorporated in industrial applications
and almost all industrial applications are
involved in such applications or such operations
of transmission and distribution and circulation
of liquid. So therefore we have already appreciated
starting from a very routine industrial application
to a very high-tech industrial application,
the fluid machines, turbines, pumps and compressors,
they are used.
Now, at the outset of this course, with this
short introductory idea as the preface, I
will start with the classification of fluid
machines. Let us start with the classification
of fluid machines. Now fluid machines are
classified in various ways depending upon
the different aspects. The first classification
is based on the principle of operations and
based on these principle of operation, we
classify the machines into two groups, one
is Rotodynamic machines and other is positive
displacement machines.
Now Rotodynamic machines are those machines
where the energy conversion between the or
energy transfer between the fluid and the
machine takes place due to a rate of change
of angular momentum of the fluid in course
of its continuous flow through a number of
blade passages comprising one or more moving
rows of blades. That means in this category
Rotodynamic machines, there should be or there
has to be a continuous flow of fluid through
blade passages and there has to be a relative
motion between the fluid and the blade.
And this way we can tell that the principle
of Rotodynamic is based on the principle of
fluid dynamics. Whereas in a positive displacement
machine, what is done is that some mass of
fluid is taken as a closed system or is entrapped
under some some locations of the machines
and then by the physical displacement of one
of the boundaries, its volume is changed and
accordingly the pressure change is manifested.
So, or the vice versa, the pressure, high
pressure is being converted to mechanical
energy.
So in this case this energy interaction takes
place by changing the volume of a given amount
of or given mass of fluid as a closed system
by a physical movement of the boundary, that
is
why the name positive displacement machine
has come or is given. This type of machines
based on this type of principles are seen
in the form of a reciprocating piston cylinder.
That means this principle is utilised by executing
a reciprocating motion of the piston in a
cylinder while entrapping certain amount of
air or certain amount of water within it.
And those machines are known as reciprocating
engines or reciprocating pumps or compressors
depending upon the direction of the energy
transfer, that I will tell afterwards.
So based on these two principles of operations,
two different principle of operations we define
one as Rotodynamic machines and another class
as positive displacement machines. Now the
Rotodynamic machines, we will discuss mostly
the Rotodynamic machines in this course which
is based on the fluid dynamical principle
of rate of change of angular momentum of a
continuous flow of fluid through rotating
blade passages can be classified like this
based on the direction of energy transfer.
If you see the classification is like this,
where the input energy is stored energy of
the fluid and the output energy is the mechanical
energy, those machines are known as turbines.
And depending upon the type of fluids used,
they are classified as hydraulic turbine which
is water, gas turbines which is gas, which
are generally the mixture of air and the products
of combustion by burning the fossil fuel for
this purpose and the steam turbines which
use steam, that is used in thermal power stations.
So these are the turbines, hydraulic turbines,
gas turbines and steam turbines operate mostly
on the same principle but use different types
of fluids. Now the fluid machines where input
energy is mechanical energy, that means the
machines which absorb mechanical energy or
take mechanical energy as input and deliver
stored energy of fluid as output is are known
as pumps, compressors, fans and blowers. Actually
there is no generic name like turbines, so
they are termed by different names and different
names have meaning otherwise the depending
upon the different fluids.
For example, pumps which use liquid, mostly
water. Compressors that use air, and in both
the machines deliver stored energy in the
form of fluid pressure. So high-pressure liquid
and high-pressure air we get as the output
from pumps and compressors, where fans and
blowers, they use air but there the stored
energy of fluid is in the form of velocity,
so they deliver high velocity air, that means
high flow of air, which comes from fans and
blowers. So this is the preliminary classifications
of fluid machines.
Now we will develop or deduce a very simple
expression of the energy transfer between
the fluid rotor, and the machine rotor, fluid
machine’s rotor and the fluid. So this thing
we will do now, let us see this picture, now
where we have shown the components of flow
velocity in a generalised fluid machine. Now
this is a generalised rotor or a rotor of
a generalised fluid machines, representative
of a typical rotor of a fluid machine which
is shown as a circular disk rotating with
a constant angular velocity omega and this
is mounted on this shaft.
Now here we consider a radial location R1
where the fluid enters to the machine, that
is the inlet to the fluid. Now fluid inlets
at all points on the radial location R1, so
R1 are the radial locations where the fluid
takes, comes into the machine, that is the
inlet of the fluid. Now
there are certain assumptions in this deductions.
The first assumption is that we will consider
the fluid which is entering into this machine
at different radial locations, that means
over the azimuthal locations at the radial
location R1 is uniform, that is uniform over
the azimuthal location, over the periphery,
that means the fluid inlet is uniform with
azimuthal location. That means that any point
here one, the fluid velocity ‘V’ one is
the representative fluid velocity of inlet.
Similar is the outlet where we define a outlet
at a radial location two where the fluid velocity
‘V’ two represents a, representative velocity
of the outflow velocities since the outflow
velocity around this periphery or around the,,
about the azimuthal location is uniform throughout.
So this uniform flow velocity is one assumption.
Next assumption is that flow is steady.
That means the mass flow rate across any cross-section
at any radial location is same, there is no
mass accumulation, no mass depletion within
the rotor, the flow is in the at the steady
state, no mass, no energy accumulation or
depletion within the rotor. And another very
important thing you have to know that here
we will deduce the expression for the energy
transfer under these assumptions with another
basic consideration is that we are not going
to solve any fluid mechanics or the fluid
dynamics or the flow field for the fluid flow
here.
That means given a prescribed inlet flow field,
we are not finding out the outlet flow field.
Here we will assume both the inlet and outlet,
our flow field is given. So the solution of
the flow of the fluid within the rotor is
not of our concerns, that is very complex
and that requires a CFD tool, so therefore
that is beyond the scope of this course, this
is usually done in an advanced course on fluid
machines or Turbo machines.
So therefore we will accept the solution,
the flow velocity at the outlet at a radial
location of the rotor and this flow is uniform
over the entire azimuthal direction, similarly
at another radial location where the fluid
enters into the machine is uniform over the
entire azimuthal location. And at any point
at the radial location, the inlet velocity
is the representative of the velocity, all
the velocities at all the points are representative
of the mass flow at the inlet, similarly that
at the outlet for outlet velocity, and we
consider the flow to be steady.
So with these assumptions now we will derive
the equations for energy transfer. And that
is based on the principles of rate of change
of angular momentum, that is angular momentum
theorem, that is known as angular momentum
theorem. Angular momentum theorem. Now look
what is the conservation of angular momentum?
If you consider the conservation of angular
momentum, therefore we can see that if we
consider a control mass system, we can tell
that the rate of change of angular momentum
of a control mass system if the angular momentum
is referenced with respect to an inertial
frame of reference equals to the moment or
torque exerted on that controlled mass system.
Now if we apply this same conservation principle
to a control volume, because here in fluid
machines we will apply this principle for
a control volume considering rotor itself
as a control volume. So now we know that when
these conservation principles for any extensive
property of a control mass system is applied
to a control volume, there is a theorem, we
invoke a theorem known as Reynolds transport
theorem.
Which makes the link between the statement
of conservation of such an extensive property
for this change, for a control mass system
to that its change for a control volume system,
this property may be mass, maybe linear momentum,
maybe angular momentum. Let us consider, at
present we are dealing with the angular momentum.
For an angular of momentum, if we invoke the
Reynolds transport theorem, it states like
that if you recall that the Reynolds, according
to this Reynolds transport theorem, the rate
of change of angular momentum of a control
mass system equals to the rate of increase
of angular momentum within a control volume
plus net rate of a flux of angular momentum
from the control volume, at is control at
its surface, that is control surface.
So this is the Reynolds transport theorem
and this is valid for linear momentum, this
is valid for mass also. So therefore we can
write this, therefore this theorem like this
and before that if we write the angular momentum
as M, angular momentum we write angular momentum,
angular momentum nomenclature as M and M per
unit mass as Eta.
Then we can write the theorem, the Reynolds,
according to Reynolds transport theorem that
the rate of change of angular momentum for
a control mass system equals to the rate of
change of angular momentum, if this is Eta,
angular momentum per unit mass Eta rho dV
for a control volume plus this is the rate
of increase, time rate of change of angular
momentum within the control volume dV is the
elemental volume of the control volume, plus
the rate of change of angular, rate of, net
rate of efflux of angular momentum across
the control surface.
What is this, Eta is the angular momentum
per unit mass and this quantity represents
the mass flux over a elemental area dA where,
VR is the velocity vector relative to the
control volume. Here definitely I have forgotten
to tell you, we define the control volume,
the fluid inscribing the rotor, within the
rotor we take as control volume, okay. And
VR is the velocity of the fluid is relative
to the control volume, N is the vector, unit
vector directed radial, directed outward of
the area A. So this is used as the vector,
the area is used as a vector by using this
unit vector whose direction is radial, is
outward, outward of the area DA, radially
outward of the area DA and this is the velocity
relative to the control volume. Okay. Now
here Eta, this angular momentum per unit mass
can be written as R cross V where R is the
radial location or the positional vector of
the point where we are defining the angular
momentum per unit mass and V is the velocity
vector.
If this is referenced with respect to an inertial
frame a static frame of reference, then we
describe the angular momentum M with respect
to a Cartesian, with respect to an inertial
frame of reference. Now here you see V one
is the velocity at the inlet which has got
three components, one is the axial direction
VA one, another is the radial direction VF
one, another is the tangential direction V
W one. Similarly this outlet velocity V two
can be resolved into three components, one
is axial direction VA two, another race radial
direction VF two, another is the tangential
direction VW two.
These are the typical nomenclature for the
components at inlet and outlet, both the velocities
at inlet and outlet can be resolved into three
respective components axial directions, radial
directions and the tangential direction. The
radial direction of component is sometimes
known as flow velocity. Now with this we see
now next, if the flow is steady, this part
is 0, that means rate of change of angular
momentum within the control volume, the time
rate of change, Del Del T is zero because
there is no change within the control volume,
that is the definition of a steady flow.
Under steady state, this becomes zero. Now
we write the left-hand side, that is the DM
DT, that is the left-hand side DM DT control
mass system with this can be written as, this
is the DM DT control mass system can be written
as D DT of R cross V, that is the per-unit
mass DM over the control mass system. This
can be written in this way over the control
mass system.
And this can be written equal to the torque
or moment exerted by the control mass system
on the control mass system, exerted on the
control mass system provided this is defined
or this is referenced with respect to an inertial
frame of reference. Okay so if we define the
velocity with respect to an inertial frame
or static frame of reference and accordingly
the angular momentum, we can tell the rate
of change of angular momentum for a control
mass system. This can be expressed in this
fashion, is equal to the torque or moment
applied to the control mass system.
And the Reynolds transport theorem, therefore
we can write this torque or moment applied,
exerted on the control mass system which coincides
the control volume at that instant. That
means this is the torque or moment exerted
on the control volume itself, so then it becomes
the control surface, that means the this quantity
VR dot N DA. That means this becomes that
a torque or moment exerted on the control
volume, control volume is the fluid containing
in the rotor, this is the control volume,
this is the CV, control volume.
Now we can write the torque or moment exerted
on this CV is equal to the net rate of angular
momentum, a flux from this CD, this is zero
under steady state, net rate of angular momentum,
a flux from the control volume. Now if we
designate outflow and inflow area like R one
radial location inflow R two radial location
outflow, if we can, then the RHS can be written,
sorry I will take another …
Now you can see here that, can you see, yes.
That RHS can be written, let me write this
thing again. Now T is equal to, T is equal
to Eta rho V R dot N DA. Now if we can identify
the outflow and inflow area is separately,
then we can write his as Eta rho VR dot N
DA, A outflow plus Eta rho VR, all nomenclature
are known to you as I have told inflow, A
inflow. Now here you see since the outflow
and inflow, both, let us consider outflow,
the radial location is fixed at R2, at all
outflow points and we consider that the tangential
component of velocity which is perpendicular
to the radial location, that is also constant
at all point.
This is a very simple assumption based on
which we are deducing this expression. So
if that is the case, the first term becomes
equal to, in that case before telling that
we write Eta for example at outflow can be
written as R two into VW two, similarly Eta
at inflow can be written as R one into V W
one. What is Eta, Eta is R cross V, that is
the angular momentum per unit mass, under
this simple condition, simple case, this can
be… So without going for any complex vector
calculations, we can simply and this is the
mass flow rate, this is the mass flow rate
at outflow and the mass flow rate at inflow.
If Eta is constant, comes out of these, so
integration of this will be the mass flow
rate, so therefore torque can be written,
the first one can be written as R two VW two
into M dot outflow because this quantity comes
out, so this is the mass flow rate, similarly
this is M dot outflow and this is M dot inflow,
this quantity, only this quantity. So therefore
this will be with a minus sign, why minus
sign because the outflow and inflow finally
comes with a minus sign, why, because this
is dot product.
If you consider a control volume, let us have
a area DA and let us has a velocity here V
two, the outlet velocity, this is the normal
direction, they are in the same direction.
But whereas this DA, where the normal outward
is like that, whether the velocity is in this
direction. So therefore if you make this dot
product, we will see automatically this term
will come with a negative sign, so therefore
this is the angular momentum outflow, rate
of angular momentum outflow, this is rate
of angular momentum inflow and with a minus
sign which will automatically come by this
sign Convention if you make the vector operation
in case of a very generalised three-dimensional
analysis…
But here it is so simple that Eta outflow
is R two VW two and Eta inflow is R one VW
one, we take it out and simply we just write
it as M outflow and M inflow with a minus
sign. And since it is at steady-state, M outflow
equals to M dot, M dot outflow is equal to
M dot inflow is equal to M dot. So therefore
we can write T is equal to M dot into R two
VW two minus R one VW one. So therefore we
get that the torque, that is or moment applied
on this control volume is equal to M dot R
two VW two minus R one.
The nomenclature is that VW two and VW one
are the tangential velocity component of the
fluid, tangential velocity or the tangential
component of the fluid velocity at the outlet,
tangential component of the fluid velocity
at the inlet, R two and R one are the radial
locations at the outlet and inlet.
Now we know that the energy, rate of energy,
rate of energy, rate of energy, okay, rate
of energy exerted, rate of energy given rather,
givev to the fluid, to the fluid E dot if
I write is nothing but the torque or moment
exerted on the fluid into the angular velocity.
So the rate of energy given to the fluid will
be equal to MW into R two VW two minus R one
VW one into the angular constant angular velocity
of the rotor. This can be written as M dot
VW two U two minus VW one U one.
Where U two is R two omega and U one is R
one omega and they represent the rotor velocity,
the linear velocity of the rotor which is
the radius time the angular speed, this is
the at inlet, this is the velocity of the
rotor, linear velocity of the rotor, at the
outlet this is the linear velocity of the
rotor. So therefore, they represent the linear
velocities of the rotor. So VW two U two minus
V W one U one. So the rate of energy given
to the fluid is can be written like this.
Now usually we express this in terms of the
energy per unit weight.
Now if we write the energy per unit weight
given to the fluid E, energy per unit weight
given to the fluid E is equal to E dot by
M dot G, then we can write it as simply an
algebraic manipulation, energy per unit weight.
Now in fluid mechanics, always people use
these energy per unit weight E which is given
as head H, energy per unit weight is known
as head. The energy per unit weight has a
beauty that its dimension is linear dimension,
that means in meter in SI units.
So therefore in hydraulics and in fluid mechanics
we use the energy per unit weight as head,
usually it is the terminology used in hydraulics,
so here I will use in terms of that in fluid
machine that head, that is energy per unit
weight given to the fluid is expressed as
this. Now
there is a convention, similar to that of
thermodynamics that when the energy, mechanical
energy or work being delivered by a system
or a control volume, it is taken as positive
whether it is added to the system is taken
as negative.
So here also in this similar convention that
when the head, that is energy per unit weight
is delivered by the fluid of the machine,
that means delivered by the fluid to the machine,
then it is considered as positive, while it
is given to the fluid by the machines is considered
to be negative. With that consideration H
is written as VW one Uone minus VW two U two
by its sign convention, not from this formula,
otherwise this would be minus H, H is this
but by sign convention we write H, that means
the positive value of this means the head,
that is energy per unit weight has been delivered
by the fluid.
That means in case of turbines which deliver
mechanical energy, which delivers work, mechanical
work VW two that means for turbines I write,
for turbines therefore VW one U one is greater
than VW two U two, okay. And for pumps or
compressors, for pumps or compressors, just
the reverse, V W one U one is less than VW
two U two, there are negative. So this is
the usual sign convention we write that H,
that is the head, energy per unit weight delivered
by the fluid to the machine is given by this.
And this equation is known as Euler’s equation,
Euler’s LERS, Euler’s equation of fluid
machines.
Euler’s, I think you have understood this
thing, so with a very simple, in a very simple
case with all these assumptions we have derived.
So this is a very simple case, actual cases
are different but this gives a guideline.
Now again I tell that the assumptions are
like that the flow at the inlet is uniform
over the entire azimuthal direction, similarly
at the outlet flow is steady and we have the
solutions of the flow field at the outlet
for a given flow field at the inlet.
But this is at the same time is very generalised
expression in a sense that it does not take
care of the path taken by the fluid in the
rotor and the way the density is varying,
so this does not take care of that. So whether
density varies or not, whether the fluid path
is different or not, whatever path may be
taken, this is the expression, under these
assumptions as just now I have told, this
is the for the head, that is energy per unit
weight delivered by the fluid to the machine.
VW one U one minus VW two U two by G.
The nomenclatures are VW one, VW two at the
tangential component of fluid velocities at
inlet and outlet, U one and U two are respectively
the rotor velocities at inlet and outlet and
this equation is known as Euler’s equation
of fluid machines and this is the basic equation
of energy transfer between the fluid and the
rotor and we will deduce many other equations
from these equations. These equations will
be expressed in terms of different components
of the fluid velocities which I will discuss
in the next class, thank you.
