Welcome back to the video course on fluid
mechanics. So in the last lecture, we were
discussing about the dynamics of fluid flow,
we were discussing about the energy equation
and we were discussing about the corresponding
Bernoulliís equations and its applications.
And we have also seen the various aspects
of the dynamics of fluid flow which we will
be discussing further like analysis of fluids
in motion, though we have already seen in
fluid kinematics without considering the various
forces.
But in dynamics of fluid flow as we have seen,
we will be discussing the various fluid motions
including whenever forces are applied. Some
introductory aspects we have seen various
cases which we will be discussing and then
we have also seen various forces influencing
the fluid motions like gravity force, fluid
pressure force due to molecular viscosity,
surface tension, compressibility Reynolds
stresses etc.
Also as we have seen that fluid motion can
be either 1 D1 dimensions two dimensions,
three dimensions or steady state and transient
or the uniform flow or non uniform flow with
respect to dynamics of fluid flow also.
And then we have also seen the various aspects.
Like how to analyze when we are analyzing
the fluid flow. We will be either using endless
stream of fluid, so that what part of this
stream shall constitute the system which we
will be analyzing. So we have seen two alternatives,
one is behavior of a specific element of fluid
of fixed mass so in a crossed system.
There are two alternatives the way of analyzing
fluid flow, first one is behavior of a specific
element of fluid of fixed mass in a closed
system, and then we can also do the analysis
by defining a system as a fixed region in
space known as control volume through which
fluid flows so that it is an open system.
So this we have seen and now in our analysis
we will be using this finite control volume
analysis, as we have seen and discussed in
the last lecture.
So this finite control volume analysis used
in many of the theoretical development in
fluid mechanics including this dynamics of
fluid flow. The advantages are it is very
easy to interpret physically and use and the
formulas derived from basic laws, we can apply
to the system and as we have seen the description,
how dealing the fluid flow is concerned.
It can be either say Lagrangian description
or Eurasian description but in this finite
control volume analysis, we will be using
the Eurasian description and then the basic
development is based upon the Reynolds transport
theorem, which we discussed earlier in the
fundamental aspects of fluid mechanics in
the earlier lectures. So now in todayís lecture
we will derive some of the fundamental equations
starting from the conservation of linear momentum.
First we will derive the Eulerís equation
for in viscid flow or non viscous flow and
then or shall we discuss the Bernoulliís
equation and its applications. So as far as
the three fundamental principles, the principles
based upon the most of the theories derived
we have already seen the conservation of mass
based upon which the continuity equation is
derived that we have already seen in the previous
lectures, how to derive the continuity equation
based upon the conservation of mass.
So this is one of the other most important
theory based upon which most of the fundamental
laws are derived, conservation of linear momentum.
Today we would discuss, we would derive the
Eulerís equation based upon this conservation
of linear momentum and then further we will
derive the Bernoulliís equation based upon
the Eulerís equation.
So say us? The word itself indicates the conservation
of linear momentum means the title itself
indicates the linear momentum is conserved.
What is momentum? So we all know that momentum
is the quantity of motion an object has, if
the object is in motion then say it has momentum.
So a momentum can be defined as it is a quantity
of motion, an object has when the object is
in motion, then we say that it has momentum.
The momentum of course most of the bodies
the moving bodies have got mass and then it
has got velocities the momentum basically
depends upon the mass and velocity so that
we can express the momentum as mass multiplied
by the velocity.
So here you can see that the momentum depends
upon mass and velocity and then we can express
the momentum is equal to mass into velocity,
and then say a body under motion say a fluid
in motion, then we can say that an impulse
is exerted on a object which is in motion,
when the force acting on the object is a force
acted upon the object so that we can say that
the impulse which gets to the object is proportional
to the force. So impulse is also very important
as far as the momentum of the system is concerned,
linear momentum of the system is concerned.
So the impulse depends upon the time that
the force acts upon the body or upon fluid
which we are considering so that impulse is
directly proportional to time. So that here
as written in the slide, impulse is equal
to proportional to force is equal to force
into time so that impulse is directly proportional
to time. So the momentum is depends upon the
impulse, that when we discuss the conservation
of linear momentum we have to consider an
impulse that means a force applied with respect
to time and then with respect to motion already
it has the body or the fluid has a momentum,
so both this will be playing as far as total
momentum is concerned. So the impulse exerted
on an object is equal to the objects change
in momentum.
So finally we can define the impulse as the
impulse is what exerted impulse is exerted
on an object equals to the objectís change
in momentum. Whenever a fluid is moving so
that we can say that with respect to the motion
say there is already an initial momentum and
then we are applying an extra force on the
fluid or on the particle which you are considering
so that there is a change in momentum. So
this is so called impulse with respect to
the force which is acting upon the fluid or
the body which you are considering. So if
no impulse is exerted on an object then momentum
of the object will not change. These are some
fundamentals of this mechanics, so based upon
which also the fluid mechanics theories are
derived. We can say that whenever there is
no impulse then we can say that the momentum
is not changing, momentum is constant so that
means this leads to the conservation of momentum.
So according to the conservation of momentum
we can say that final momentum is equal to
the impulse which is acted upon the body or
the fluid which we are considering then plus
the initial momentum.
So the bodies or the fluid which is under
motion it has got an initial momentum, so
then the final momentum will be if you exert
an extra impulse so that, there is a change
in momentum. The final momentum will be impulse
plus the initial momentum so while we are
considering the conservation of linear momentum
we have to see that what is the initial momentum
for the fluid for the body which we are considering
or what is the change? What is the impulse
acted upon the body? The total or the final
momentum is equal to impulse plus the initial
momentum. So now as far as any fluid system
which we discuss is concerned if there is
no external force acted on the system then
we can say that the total momentum of the
system will not change.
So we have already seen that the total momentum
is equal to initial momentum plus the momentum
change due to the impulse, we can say that
if there is no external force act on a system
then the total momentum of the system will
not change. So such a system is called an
isolated system.
So the system can be of two types wherever
that is the total momentum of the system is
not changing, there is no external force whatever
the fluid is moving or the body is moving
its all initial moment, we can say that the
total momentum is not changing so such a system
is called a isolated system and the other
type of system is whenever we are exerting
a force, there is a impulse and then there
is a change in momentum so that the normal
whenever we consider the dynamics of fluid
flow most of the time we will be discussing
due to external force is acting with respect
to system but sometimes, also depending on
the problem isolated system will also be considered.
Finally we can say that the Eulerís equation
which we are going to derive is based upon
conservation of linear momentum so the momentum
is we can say that, when we consider a particular
control volume, the momentum is conserved
in a very isolated system. If you isolate
a system and there is no external force then
we can say that the momentum is conserved
in that particular system but as far as the
internal forces are concerned it can never
change.
The total momentum of the system even though
there can be internal forces within the fluid
system which we are considering but it cannot
change the total momentum of the system. So
these are some of the fundamentals when we
consider the linear momentum and the impulse
or the effect of force upon a particular system.
So system can be either isolated system or
system can be wherever an impulse is acted
then there is a change in momentum. We have
seen that internal forces are concerned it
cannot change the total moment of a system.
So now based upon these say the linear momentum
equation we can write as shown in this slide
here.
Say, if F is the resultant force acting on
a fluid mass F is equal to d P by d t where
P is the linear momentum. So the force is
equal to the total derivative of linear momentum
with respect to time. So F is equal to d P
by d t for the particular system. So we can
write the momentum is equal.
So the integral of this V d m, so this is
the basic definition when we consider the
linear momentum equation, and for differential
set of mass say mass of delta m, we can write
delta F is equal to V the total derivative
V into delta m so that this is the velocity
multiplied by the small mass delta m. So here
in this equation p is equal to integral for
the system V d m where V now delta F say for
a differential system of mass which we will
be generally considering while deriving the
equations. So delta F is equal to the D by
D t of V delta m as written here.
So that we can write delta F is equal to delta
m is constant. So delta can be taken out so
delta m D V by D t but here you can see that,
here the V is the velocity and then D V by
D t is the acceleration a of b element which
we are considered considering here. So finally
with respect to this we know that from Newtonís
second law this force is equal to mass into
acceleration, so this delta F also we can
write as delta m into a that means effectively
the change in force or delta F is equal to
the mass into acceleration that is what we
are getting with respect to the equations
here.
So now if you consider the resultant force
acting on a fluid mass, here we consider as
in this slag we consider a fluid element here
of small area delta a here with respect to
x y z axes say the force is acting can be
put as delta F s is written delta F1 and delta
F 2 and then delta F a with respect to normal.
So the arbitrary surface which you are considering,
the resultant force acting on fluid mass is
equal to the time rate of change of linear
momentum of mass. The conservation of linear
momentum we can say now in the previous slide
we have already seen here. So the delta F
which we derived is equal to D into D by D
t of V D m delta F is equal to delta m into
D V by D t again that comes as the mass into
acceleration Newtonís second law delta m
into a from this we consider here the resultant
force acting on a fluid mass shown in this
figure is equal to the time rate of change
of linear momentum of mass.
So finally with respect to the consideration
of linear momentum say now also considering
the Newtonís second law we can write the
resultant force acting on fluid mass is equal
to the time rate of change of linear momentum
mass.
So now as in this figure we consider a mass
of delta m and as far as this figure is concerned
here various kinds of forces will be acting
on this particular fluid element of mass which
we are considering, So this is the fluid element
which we are considering, so the forces generally
are the surfaces forces and the body forces.
The body forces concerned here you can see
with respect to this figure, delta F d with
respect to this particular fluid element which
we are considering, the body forces mainly
due to acceleration, due to the gravity so
that we can write that delta F d is equal
to delta m into g where g is the acceleration
due to gravity and delta m is the small mass
which we are considering.
So this is the body force so for this particular
fluid element so now based upon this, we are
now deriving this Euler equation. Before that
we are now seeing for a particular fluid element
which we are considering when we derive these
kinds of fundamental equations we are considering
a fluid element of mass delta m and then we
are now checking what are the various forces
acting upon that particular fluid element,
So now the surface forces they can be either
normal to the surface which we have considered
or it can be also with respect to the shearing
stresses. Here if you consider the normal
stress with respect to the previous figure
here if you consider this particular fluid
element the normal stresses ,we can write
sigma s is equal to when the limit delta A
approaches to 0, this can be written as delta
F n by delta A.
So the normal stress is equal to delta F n
by delta A when the limit delta A approaches
to 0 with respect to this previous figure
here. The other surface force is called shearing
stresses. So the shearing stresses we can
write as shown in this figure it can be spitted
into these forces F1 in this direction and
F 2 the other direction.
F 1 in this x direction and F 2 in the y direction.
So if you consider that we will be having
two components for the shearing stresses,
these components have first one is tau1, tau1
is limit of delta F1 by delta A as delta A
approaches to zero and tau 2 can be written
as limit delta F2 by delta A as delta A approaches
zero.
So these are the main surface forces, so we
have already seen as far as a fluid element
is concerned the important forces are the
body forces so that body forces mainly due
to acceleration, due to gravity as we have
seen here the body forces delta F is equal
to delta m into g and then the surfaces forces
are concerned there can be normal and shearing
stresses.
Normal stresses we can write with respect
to normal force. So here sigma n is equal
to delta F n by delta A as limit delta A approaches
to 0. Similarly shearing stress is tau1 and
tau 2 we have already here the expression
for tau1 and tau 2. By using all this the
conservation is based upon the conservation
of linear momentum we are going to derive
the one of the fundamental equation of say
fluid mechanics so called Eulerís equation.
So this is mainly for in viscid flow that
means the flow is considered without the viscosity
so that is we derived based upon the conservation
of linear moment. Now to derive this Eulerís
equation let us first consider a fluid element
like this of size delta x delta y delta z
so as shown in this figure the fluid element
is considered.
For the fluid element the forces acting let
us consider now the x direction. So, x is
in this direction, y is this direction and
z is the other direction. The various forces
acting here you can see that as we have seen
earlier the two types of forces we have seen
surfaces forces and body forces. So the surface
forces we have already seen it can be the
normal stress with respect to normal stress
or shear stress here with respect to the normal
stress say for this particular fluid element.
We can see that here for this in this direction
not this phase of the fluid element it can
be written as since x direction is like this,
so x on this phase it can be written as sigma
x minus del sigma x by del x into delta x
by 2 multiplied by this area of this phase
of this fluid element delta by delta z and
then similarly the opposite side of this other
side of this fluid element we can write here
the normal with respect to normal stress the
force can be written as sigma x plus del sigma
x by delta x into delta x by 2 into delta
y delta z.
So this is as far as the normal stress is
concerned. Now also see there will be surface
forces with respect to the shear stress so
here the shearing with respect to the shearing
is concerned say shearing stress is concerned
we can write on this phase as tau y x minus
delta tau y x by del y into delta say delta
y by 2 into delta x into delta z. Similarly
the other direction to be tau y x minus del
tau y x del z by del z by 2 this direction
and here the other direction is tau y x plus
del tau y x by del y into delta y by delta
x into delta z so similarly other phase also
we can.
So all the phases we can write body forces
as far as this particular fluid element is
we can write the surface forces for this particular
fluid element is concerned one is the with
respect to the surface forces with respect
to the normal stress on this phase and other
phase and similarly the shearing stress is
concerned we can write all the components.
So now with respect to the surface forces
say if you take the i j k that means the x
y z direction we can write this total say
surface force delta F s is equal to delta
F x F s F s F x i with respect to i j k the
unit vector i plus delta F s y j plus delta
F z k.
So now we have seen that this particular fluid
element is considered there are three components
x y z. So for the surface force we have seen
so now if you consider the x direction with
respect to the previous figure.
This figure we can write delta F x s x is
equal to delta F s is equal to delta sigma
F x by sigma delta x plus del tau y x by delta
del y plus del tau z by del z into delta x
del y del z so this is the body of surface
forces with respect to the normal stress and
shearing stress.
So similarly delta F s y can be written as
del tau x y by del x plus del sigma y by del
y plus del tau z y by del z into delta x delta
y delta z and similarly in the z component
we can write del F s z is equal to del tau
x z by del x plus del tau y z by del y by
plus del sigma z by del z into delta x del
y del z.
So now with respect to the fluid element we
have seen three components with respect to
x y z direction of the surface forces. We
have seen that as far as the total forces
are concerned there is a surface forces and
body force. So body forces as I mentioned
will be with respect to the acceleration due
to gravity so this also we can put in x y
z direction. So here the body forces are concerned
del F b x in the x direction will be mass
into acceleration due to gravity in x direction
so delta m into g x and in y direction delta
F b y is equal to delta m into g y and delta
F b z is equal to delta m into g z.
So these are the body force component in x
y z direction. Now with respect to the Newtonís
second law also we have seen earlier with
respect to the conservation of linear momentum.
So finally we have come to that if you use
the Newtonís second law we will be getting
effectively this conservation of linear momentum.
So now the various forces we have seen for
the particular fluid element the surface force
and body forces we have already seen various
components of the surface forces and also
we have seen the various components of the
body forces. As per the Newtonís second law
in x y z direction we can write delta F s
F x equal to that means the force in x direction
is equal to delta m into acceleration in x
direction and delta F y is equal to delta
m into a y and delta F z is equal to delta
m into a z, where delta m is the mass of the
fluid element, so if rho is the density of
the fluid element, we can write delta m is
equal to rho into delta x into delta y into
delta z .Now all these parameters are known.
Now if you use the Newtonís second law finally,
we can write the general differential equation
of motion for the fluid in x y z directions.
So by using Newtonís second law we have equated
the forces into the mass into acceleration,
the forces are concerned we have seen the
body forces and surface forces. By equating
by using Newtonís second law if you get then
we can see that this is in x direction the
equation will be rho g x which is the body
force in x direction plus due to the stress
force shearing and normal stress we can write
del sigma x by delta x plus del y del tau
y x by del y plus del tau z x by del z is
equal to mass into acceleration.
So mass is we have already seen say rho into
delta x into delta y into delta z. These deltas
x delta y delta z are there in all this components
which we have derived. So delta m as delta
x delta y delta z and also we have seen delta
F s x delta F s y delta F s z all these are
delta x delta y delta z.
So while using the Newtonís second law and
equating force into mass into acceleration
delta x delta y delta z will be cancelled
on both sides. Finally, the total body forces
and surface forces can be equated mass into
acceleration.
Mass is effectively here now rho since delta
x delta y delta z cancel and acceleration
so we have already seen when we consider the
concept of the Eulerian concept which we have
seen here we are using Eulerian concept in
the derivation of this equation.
So while considering this, Eulerian description
so the acceleration can be written as, the
local acceleration del u by del t plus the
convective acceleration. So, del u by del
t, plus u into del u by del x, plus v into
del u by del y, plus w into del u by del z,
this is the total acceleration in x direction
so that mass into acceleration is rho into
del u by del t plus u into del u by del x
plus v into del u by del y plus del y plus
w into del u by del z.
So this gives the general differential equation
of motion in x direction. Final equation is
rho g x plus del sigma x by del x plus del
tau y x by del y plus del tau z x by del z
equal to rho into del u by del t plus u into
del u by del x plus v into del u by del y
plus w into del u by del z.
So this is the general differential equation
of motion for fluid in x direction. So from
this only we will be deriving other equations
like Eulerís equation or other kinds of equation.
Now this is the x equation in x direction
and the general equation of motion by direction
can be written as rho g y plus del tau x y
by del x plus del sigma y by del y plus del
tau z y by del z equal to rho into del v by
del t plus u into del v by del x plus v into
del v by del y plus w into del v by del z
and the general differential equation of motion
in z direction can be written as rho into
g z plus del tau x z by del x plus del tau
y z by del y plus del tau del sigma z by del
z is equal to rho into del w by del t plus
u into del w by del x plus v into del w plus
del y plus w into del w by del z.
So where u, v and w are the velocity components
in x y z direction when we consider the fluid
motion in three dimensions. Now we got the
general differential equation for fluid in
x y z direction so now from the general differential
equation of motion we will be deriving various
equations. So the first case which we will
be considering here is the case for in viscid
flow so the equation is basic equation is
Eulerís equation.
Here the Eulerís equation for in viscid flow
based upon the general differential equation
of motion will be deriving. So Eulerís equation
is concerned, the equation is valid for in
viscid or non viscous flow that means shearing
stresses is 0. We have already seen the general
equation here.
The shear shearing components are there, so
when we consider the Eulerís equation for
in viscid flow that means there is in viscid
means no viscous non viscous flow. We donít
have to consider the shearing stresses, so
that shearing stresses become 0. So this actually
is not an exact way but we can approximate
many of our fluid flow like wind flow or air
flow or water, we can approximate the viscosity
is small. Sometimes we can neglect the viscosity
and then this Eulerís equation can be utilized
many of the problem but it is an approximation
we are considering say the viscosity of air
or water is small.
So that it can be neglected and then as far
as the general equation which we are considering
here generally differential equation of motion
concerned now for Eulerís equation or in
viscid flow is concerned the shearing stresses
are 0. So that the shearing terms are gone
on the left hand side of this equation and
then we have this normal stress component
so as far as normal stress components are
concerned.
So if you consider normal stress component,
it will be generally the pressure which will
be acting. That we can write for the fluid
element is concerned with respect to x y z
direction we can approximate as minus p is
equal to which is the pressure force minus
p is equal to sigma x is equal to sigma y
is equal to sigma z. Here we use the negative;
we are putting this negative so that component
normal stress will be positive.
So now like this we are approximating the
normal stress component the pressure term
which is coming upon the fluid, so minus p
is equal to sigma x is equal to sigma y is
equal to sigma z. Now after this approximation
based upon the general equation which we have
already derived here, now the shear terms
have gone and the normal stress components
its derivative we are now approximating with
respect to pressure components. Finally this
equation can be put as general equation of
motion for Eulerís for in viscid flows now
reduces to like this so that general equation
of motion in x direction can be written as
rho g x that is body force minus del p by
del x.
So the pressure gradient in the x direction
is equal to rho into del u by del t plus u
into del u by del x plus v into del u by del
y plus w into del u by del z and in y direction
we can write rho g y minus del p by del y
is equal to rho into del v by del t plus u
into del v by del x plus v into del v by del
y plus w into del v by del z and in z direction
rho g z minus del p by del z is equal to rho
into del w by del t plus u into del w by del
x plus v into del w by del y plus w into del
w by del z.
So these are the Eulerís equations for in
viscid flow as I mentioned say this air flow
or the water flow is concerned we can sometimes
neglect the viscosity and then we can use
this Eulerís equation. So that the shearing
terms are not considered so for finding the
Eulerís equation we can utilize but it is
an approximation. So this Eulerís equation,
this name is given to honor Leonhard Euler,
who lived from seventeen naught seven to seventeen
eighty three that is the eighteenth century.
In his honor only these equations are named
as Eulerís equation and this same equation
in x y z direction which we have derived now
we can express as in vector notation as rho
g minus del p is equal to rho del V by del
t plus V dot product del V. Here this is the
vectorial vector notation form of the Eulerís
equation, so where V is the velocity vectors
and g is also with respect to x y z acceleration
the acceleration vector so the vector notation
can be written like this.
So now as I mentioned this Eulerís equation
is 1 of the fundamental equation of fluid
mechanics even though here we consider the
flow as in viscid but many of the practical
cases like water flow or air flow we can approximate
as in viscid and then try to get a solution
the advantage is that, here now we can see
the equations are simple and then we can easily
try to get the solution very much easier.
If we do not consider the shearing terms here
but it is an approximation but still this
Eulerís equation have have lot of applications
in practical areas by wherever the fluid viscosity
can be neglected or where small value of viscosity
is there.
Based upon this Eulerís equation, now we
will be deriving another fundamental equation
called Bernoulliís equation. This Bernoulliís
equation is actually one of the most important
or most useful equations in fluid mechanics.
It is one of the fundamental equations.
So this equation Bernoulliís equation can
be derived by direct application of the Newtonís
second law into a fluid particle moving along
a stream line. We can use the Eulerís equation
and then derive or we can also derive the
conservation of energy. Now we will discuss
we have already seen the Eulerís equation
for in viscid flow. Now based upon the Eulerís
equation here we will derive the Bernoulliís
equation which is one of the most important
equations.
So as I mentioned here, it can be either derived
by direct application of the Newtonís second
law or we can use the Eulerís equation or
we can derive from the conservation of energy.
Bernoulliís equation is concerned, it has
got some restrictions. This Bernoulliís equation
is derived based upon the assumption that
flow is steady state, so time component is
not there rho is steady state and density
is constant so that fluid can be considered
as incompressible.
And then of course we also put the assumption
that fluid is in viscid, so that friction
losses are negligible so this assumption is
actually many of the practical problems it
will be difficult to apply but certain times,
certain places we can approximate the flow
is even though, say the viscosity is to be
considered. Whenever water is concerned since
its viscosity is small or air flow is concerned
viscosity is small.
We can use this Bernoulliís equation so the
assumptions are flow is steady density is
constant and in viscid flow. So actually this
Bernoulliís equation it relates the states
at two points along a single stream line.
Whenever there is a fluid flow is there, if
you consider a fluid flow here in a channel
like this, Here we consider two sections here,
1 and two and fluid is flowing in this direction.
So this Bernoulliís equation is it is relation
between two points if you consider this point
here and this point 1 and two, it is along
a stream line if you consider a streamline
like this, the Bernoulliís equation is an
expression for the states. What is the state
between this position and first position and
second position? So that is what the Bernoulliís
equation states.
So now we will derive here the Bernoulliís
equation based upon the Eulerís equation
which we have already seen earlier. From Eulerís
equation at steady state the equation, we
have already seen earlier here, the Eulerís
equation with respect to vector notation,
if you consider the flow at steady state condition.
Then we can write rho g bar minus del p is
equal to rho V bar dot del V. This is the
Eulerís equation at steady state.
So now to get the Bernoulliís equation we
will be integrating this equation along some
arbitrary streamline so before the integration.
The acceleration due to gravity we can write
as g bar can be with respect to the normal
direction minus g del z so and then also this
other term del dot V so or V dot del V bar
this right hand side of the equation the steady
state equation Eulerís equation can be written
as can be approximated using the vector notation
mathematics as half del V dot V minus V cross
del cross V.
So this is from the vector algebra the mathematics,
vector product, and dot product we can get
this equation, and now Eulerís equation finally,
becomes rho g del z minus del p is equal to
rho by 2 del V dot V minus rho V cross del
cross V. This we can rearrange as deep p by
rho plus 1 by 2 del V square, plus g del z
and that is equal to V cross del cross V.
So now we are approximating the Eulerís equation
into derive the Bernoulliís equation. Now
say let us consider a stream line like this
as shown in this figure here, so if you take
the dot product of each term with a differential
length ds along a streamline as shown here
this is streamline so that equation will become
del p by rho dot d s plus half into del V
square d s plus rho del z d s is equal to
V cross del cross V d s.
Now this d s has a direction along the streamline,
so here the direction is shown here and d
s and V are parallel. So since this streamline
is derived like that, so that we can write
also V cross del cross V is perpendicular
to V bar so that we can write V cross del
cross V d s is equal to zero and then this
d s can be this d s in this previous figure
ds this vector can be spitted into, here it
can be written as, d x i plus d y j plus d
z k. So that finally we can write del p dot
d s is equal to and also d p del p d s can
be written as del p by del x d x plus del
p by del y d y plus del p by del z d z that
is d p.
So that finally after using all these approximations
the Eulerís equation in steady state become
d p by rho plus half d V square g d z is equal
to zero. So after using all these approximations
put forward here in the slides, finally we
get this dp by rho plus half d V square plus
g d z is equal to zero. So now along the streamline
as we discussed here we integrate so that
integral d p by rho plus V square by 2 plus
g z is equal to constant.
Finally, this gives the Bernoulliís equation,
so Bernoulliís equation is actually here
are three terms: one is with respect to pressure,
other one is the velocity, with respect to
velocity square, and the position head gz
with respect to z. So this is the integral
form this is the general Bernoulliís equation.
As we have seen this is valid for incompressible
and inviscid fluid.
Actually, we can write this general equation,
since rho term is there, This is valid for
compressible and incompressible viscous fluid
so that, we can write p by rho g plus V square
by 2 g plus z is equal to constant and now
between two sections if you consider two sections
as we have seen here in the previous here.
If you consider the section 2, section 1 and
section 2 then finally, we can write p1 by
gamma plus V1 square by 2 g is equal to z
plus z 1 is equal to p2 by gamma plus V2 square
by 2 g plus z1 z2. So here this if we consider
in this figure here this is the datum and
here z1 the position head and this is z2 position
head. So that finally with respect to Bernoulliís
equation and p1 is the pressure here V1 is
the pressure velocity here, and p2 is the
pressure here and V2 is the velocity on the
section 2. Finally the equation become p1
by gamma plus V1 square by 2 g plus z1 is
equal to p2 by gamma plus V2 square by 2 g
plus z2 where gamma is the specific weight
of the liquid. So this is the general equation
when we consider between the sections.
So the Bernoulliís equation, actually when
we derive the equation earlier for general
one but it is restricted to in viscid flow
or non viscous flow, and it is in steady state
and the general application is for incompressible
flow and flow along a streamline. These are
some of the restrictions as far the Bernoulliís
equation is concerned. The equation is applicable
for in viscid flow since from Eulerís equation
we have derived and steady state only.
We have considered so it is generally applicable
to incompressible flow and the generally take
a streamline and then flow along a streamline
is the equation is derived. Here in the previous
slide you can see here, there are three terms
one is say p1 by gamma, that is actually so
called the pressure head, and then second
1 is V1 square by2 g, which is so called velocity
head, and then z1 and z2 which is considered
here that is the datum head.
So the Bernoulliís equation holds three terms
the pressure head the velocity head and datum
head. Now with respect to this we can see
that the work done on a particular by all
forces acting on the particle is equal to
the change of kinetic energy of the particle.
So if you consider particular particles which
were on the streamline which we are considering,
with respect to the equation, we can say that
the work done on a particle by all forces
acting on the particle is equal to the change
of kinetic energy of the particle.
So now also the Bernoulliís equation can
be derived as we have seen we have derived
now the Bernoulliís equation based upon the
Eulerís equation for the in viscid flow but
Bernoulliís equation can also be derived
from the basic Newtonís second law or also
we can derive from the conservation of energy.
So from the conservation of energy for the
system which we are considering as we have
seen in this figure, here we can see that
the energy like pressure energy per unit weight
plus kinetic energy per unit weight plus the
potential energy per unit weight is equal
to total energy per unit weight.
So as per the conservation of energy, this
total energy is conserved so that the pressure
the energy per unit weight due to pressure
kinetic or potential that should be considered
so this is the total energy per unit weight.
Total energy in the system does not change
or total head loss does not change. So since
the total energy is not changing we can say
that the total head also does not change.
That we can write from this energy conservation
we can write p by rho g plus u square by2
g plus z is equal to constant. So, this is
coming from this basic energy equation. The
same equation we are getting from the conservation
of energy principles.
So if you consider say a flow in a channel
or flow in a pipe like this, if you consider
two points are shown in this slide here, two
points joined by a streamline, the total energy
per unit weight at 1 is equal to total energy
per unit weight at 2. Total head at one is
equal to total head at 2. That is why we are
getting p1 by rho g plus v1 square by 2 g
plus z1 is equal to p2 by rho g plus v2 square
by 2 g plus z1 for this section 1 and section
2 which we consider here.
So finally the total energy per unit weight
is equal to total energy per unit weight at
2 plus. If there is any loss in case of previous
figure, if there is loss per unit weight that
also we have to consider. It is considered
here as h and in case between section 1 to
section 2 if any work is done that also to
be consider that is w.
So work done per unit weight minus energy
supplied per unit weight so we consider all
the aspects then the equation becomes like
this p1 so with respect to this previous figure
if we consider the work inside between section
1 and 2 also any loss of a energy between
section 1 and 2 and any energy supplied also
considered then we can write the general equation
as p1 by rho g plus u1 square by 2 g plus
z1 is equal to p2 by rho g plus u2 square
by 2 g plus z2 plus h plus w minus q.
So this is the generalized equation when we
consider a real system. So the earlier equation
which we have considered is pressure head
velocity head datum head are considered as
derived here but when you consider a real
system there can be loss of energy or there
can be energy supplied or there can be work
done. So this will be considered as shown
in this general equation.
So now as we have already seen the equation
is applicable across the streamline also if
the flow is irrigational, so we have already
seen we have derived the equation for flow
along the equation is derived for streamlines
along a streamline but it can be also be applied
across streamline if the flow is irrigational.
We have already seen, what is rotational flow?
What is irrigational flow, in that case irrigational
flow, we can say that the Bernoulliís equation
is also applicable for across the streamline
and some of the practical applications, the
restriction frictionless can be considered
then it can accommodate by introducing a loss
of energy term and restriction of irrigational
flow as we have seen here.
So and we can apply many of the cases which
we will be discussing later. Some of the important
applications of Bernoulliís equation will
be discussed in later. So Bernoulliís equation
as we have seen it is a special case of general
energy equation, so general energy equation
which we have seen here.
It is a special case of the general energy
equation. So now with respect to this we will
be discussing some of the applications of
Bernoulliís equation so some of the things
which we should always remember before using
this Bernoulliís equation, are it may be
applied without much error for unsteady flow
also with gradually changing conditions.
We have already derived the equation for steady
state condition but the variation is very
gradual then still sometimes we can use this
Bernoulliís equation for unsteady flow with
gradual change in condition and also for flow
of gases the change in pressure is small fraction
of absolute pressure we can treat as incompressible
and then can be considered the Bernoulliís
equation.
And when all streamlines originate from a
reservoir where the energy constant is everywhere,
the same constant of integration does not
change from one streamline to another. So
that this case also we can consider, so this
has some of the cases which we can consider
the. Bernoulliís equation further equation
will be discussing later. So before closing
todayís say just we will also see a simple
example here. So Bernoulliís equation a small
example will be discussed here.
A fluid of density rho is equal to 980 kilogram
per meter cubic is flowing steadily through
a above tube, here you can see a tube flow
a liquid is flowing through this tube and
diameter at section 1 is ten centimeter and
section 2 it is 12 centimeter and gauge pressure
at section 1 is 260 Kilonewton per meter square
and velocity here at section 1 is u1 is equal
to five meter per second we want to find P2.
So here this simple example shows how the
Bernoulliís equation can be applied. Here
we can see that there is a tube and then varying
the diameter is varying and the liquid is
flowing through the two sections which we
are considered and at one section the velocity
is known, the pressure is known and other
section both section diameters are known we
want to find the pressure at section 2.
So that is what we want to find in particular
problem and the pressure at section 1 is given
as two sixty kilo Newton per meter square.
So here we will be using the Bernoulliís
equation and also the continuity equation.
We apply the Bernoulliís equation along a
streamline joining one and two. Here, as shown
in this figure the tube is assumed to be horizontal,
so that is z1 is equal to z2. The general
equation we can write with respect to Bernoulliís
equation P2 plus rho by 2 u2 square is equal
to P1 plus rho by 2 u1 square so you can see
here.
Section 1 and 2 we are equating the pressure
head and the velocity head z1 is equal to
z2 so datum head is not to be considered.
So section 1 section 2 so that P2 plus rho
by 2 u2 square is equal to p1 plus rho by
2 u1 square u is the velocity P is the pressure
and now the velocity at section 1 is already
unknown u1 is equal to five meter per second
but u2 is not known.
But we can use the continuity equation so
from the continuity equation we can write
A1 u1 is equal to A2 u2 so that we can get
u2 so u2 is equal to A1 u1 A2 so the diameters
d1 is ten centimeter and diameter d2 is twelve
centimeter. So that we can write u2 is equal
to d1 square by d2 square into u1 and if you
substitute all the values here. We will get
the velocity u2 is equal three point four
seven meter per second and then once we the
velocity u2 is known we will substitute back
to this equation here the Bernoulliís equation
so that P2 is equal to P1 plus rho by 2 u1
square minus u2 square.
So in this equation all the values u1 u2 and
p1 is known. So that we can find P2 so P2
is equal to two hundred and sixty six point
three five kilo Newton per meter square and
we can see that when the velocity is decreased
from five meter per second to three point
four seven meter per second per pressure is
increased from two, two sixty kilo Newton
per meter square to two sixty six point three
five kilo Newton per meter square. So this
problem shows a simple case of the application
of the Bernoulliís equations. So very similarly
way numbers of problems practical problems
can be solved using this Bernoulliís equation
further applications of the Bernoulliís equation
will be discussed later.
