Welcome back to the video course on fluid
mechanics. In the last lecture we were discussing
about the kinematics of fluid flow. So in
the kinematics of fluid flow as we discussed,
we are studying fluid mechanics without much
consideration to the forces up on which the
fluid flow is governed. So, without much concern
to the forces, we are trying to derive various
equations and various theories as far as the
fluid flow is concerned.
We have already seen the various velocity
field then the acceleration field and also
we have seen how the flow can be expressed
in one dimension, two dimension and three
dimension flow that can be steady state or
unsteady state; all these aspects we have
seen in the previous lecture, also we have
seen the fluid flow can be described in terms
of either lagrangian description or eulerian
description.
In the lagrangian description, we are just
tracking certain fluid particles and seeing
with respect to time what happens to those
fluid particles. But in eulerian approach
what we are doing is we are taking a particular
point or particular section of the fluid flow
and then with respect to space and time we
are just taking what is happening to the fluid
particles passing through the particular point
or section.
Generally, in fluid flow problems, we are
generally using eulerian approach. Since it
is much easier and we are much interested
what happens to the fluid flow at the particular
section or particular point. So, we have also
seen that as far as the fluid flow is concerned,
we can describe with respect to the acceleration
is concerned we can have a local acceleration
and also we have conductive acceleration.
So total acceleration with respect to the
local acceleration, conductive acceleration
and with respect to this we have also discussed
about some numerical examples.
And also we have seen when we discuss the
fluid flow parameters we will be describing
the time rate of change for given particle
with respect to Local and with respect to
conductive cases as shown in this slide and
this is called as the material derivative.
These things we have discussed in the last
lecture, now we have seen that in the fluid
flow can be described in three dimensions
with respect to x, y and z coordinates and
with respect to time, but many times it will
be very convenient if we can describe fluid
flow with respect to the stream lines, since
stream line as we have seen earlier, stream
line is a continuous line drawn tangential
to the velocity vector at every point that
is known as stream line.
Since the flow is with respect to the direction
we are describing the stream lines, it is
always advantages if we can describe the fluid
flow with respect to the stream line coordinate
system. If the velocity vector is defined
as v is equal to the unit vector I into u
plus unit vector j into v plus unit vector
k into w in three dimensions, then the differential
equation for stream line we can defined as
dx by u is equal to dy by v is equal to dz
by w. So this is as far as the three dimensional
flow in x y z directions the velocity component
u v w are concerned.
The differential equation for stream line
is generally described as dx by u is equal
to dy by v is equal to dz by w, so the stream
line is the continuous line drawn that is
tangential to the velocity vector at every
point. So this will be very usefully if we
can say most of the fluid flow if you can
describe in terms of the with respect to stream
line coordinates it will be very advantages
in many of the fluid flow problem.
Here in this slide the stream line coordinates
with respect to stream line coordinates the
flow is described so stream line is a coordinate
system defined in terms of the stream lines
of the flow, here in this slide you can see
this stream lines are drawn with x and y axis
in two dimensions. Here, the stream lines
shows with respect to flow stream lines are
shown here with respect to n is equal to n1
n2 like that, so its normal direction with
respect to that is shown as s1 s2 like that
so you can see just like a flow net where
the stream lines and then its normal lines
which is the potential line generally.
Instead of describing the fluid flow with
respect to this xy coordinate system, here
we can describe the fluid flow with respect
to this s and n, so the stream line and then
its normal into direction of the fluid flow,
so we have much advantages as shown in here,
so the flow is described with respect to the
stream line coordinates.
This has got the advantages as for as the
fluid flow is concerned, the stream lines
give the natural coordinate system. That is
one of the important advantages of the stream
line coordinate system since in natural coordinate
system itself is flow, so the direction itself
taken as the coordinate system so we can describe
many of the fluid flow properties with respect
to this coordinate system and flow is described
as one coordinate along the stream line s
and then second coordinate normal to the stream
lines n.
As described in normal lines and then stream
lines s, here in the previous figure, so s
indicates the stream line and then this normal
line are also drawn. So with respect to this
the flow is described with respect to along
the stream line s and second coordinated normal
to the streamline n. So flow plane is orthogonal
curved net of lines, instead of the Cartesian
coordinates system which we generally described
here, the flow plane is orthogonal in orthogonal
curved net of lines and the advantages that
velocity is always tangent to the s direction
or the stream line directions.
So that we can describe the velocity vector
as v bar is equal to v into s bar which is
the stream line coordinate system. So many
times we will be describing the flow system
with respect to this stream line coordinates
since it has got advantages compare to the
other coordinate system which we generally
use in mechanics.
The acceleration is described as DV bar by
Dt. So that is here as into x bar plus a n
into n barso where as and an are the acceleration
components in the direction of stream line
and its normal direction so the acceleration
is described as x bar plus an n bar, so this
stream line coordinate system is most of the
time used since it has got it soon, advantage
in fluid flow description. Now, before going
to the some of the fundamental theories and
principles with respect to the fluid kinematics,
we will discuss the control volume and system
representations. This we have seen earlier,
but since we are going to derive these equations,
before that we will further discuss, what
the control volume is and what is the system
representation? So, a system is a correction
of matter of fixed identity.
If we consider, for example, here in this
board what is inside whether it is fluid either
it is water or gas what ever inside in this
ball, this is a fixed it has its own fixed
identity and it is a collection of matter
inside fixed boundaries so for example as
the mass of the air drawn into an air compressor
or the air inside this board. So all this
system approach is very much used in fluid
mechanics, in most of the time we will be
interested to see what happens to the fluid
inside a system and then a control volume
as we have discussed earlier control volume
is approach is used in fluid mechanics. So
control volume is the volume in space through
which the fluid flows, for example, if we
consider a small pipe like this here, when
the fluid is for from one direction to another.
So the control volume if we consider between
one section to another section so in between
that space through the fluid flow it is the
control volume here considering as for as
the derivation of various fluid flow theories
are concerned and also some times we can have
deforming of control volume. Here, you can
see this is just like a balloon so it can
deform so that one time there is a very less
fluid or for gas is inside and then if you
blow it up then you can see that it is full.
So this is deform so this is called a just
like in a balloon it is called a deforming
control volume so it either a deflating balloon
or just like a deflating plastic material
as shown in here so this shows the deforming
control volume. So many times instead of a
fixed control volume we can also use deforming
control volume as far as the fluid flow theories
of development of the equations depending
up on the problem. Now before going to the
derivation of the continuity equation based
up on the consideration of mass and other
equations we will discuss one theorem called
Reynolds transport theorem.
This Reynolds transport theorem is one of
the fundamental principles used in a fluid
mechanics based up on which many of the fundamental
theories or fundamental principles are derived.
According to the Reynolds transport theorem,
according to this laws governing fluid motion
using both system concepts and control volume
concepts so we have seen in the case of system
concept and control volume concept, so most
of the time this Reynolds transport theorem
is the law which governs the fluid motion
either in system concept or control volume
concept.
To do this we need an analytical tool to shift
from one representation to the other, that
means when we are describing the fluid flow
from a system approach to the control volume
approach or the from the control volume to
the system so we need a analytical tool which
can easily used so that we can shift from
one representation to the other. So, this
Reynolds transport theorem provides this tool.
Reynolds transport theorem is generally used,
as we have seen most of the time we will be
using either a control volume approach or
a system approach so when we want to shift
from one approach to one representation to
another representation we can use this Reynolds
transport theorem which is described here.
According to the Reynolds transport theorem
it gives the relationship between the time
rate of change of an extensive property for
a system and for a control volume. We can
describe a fluid flow with respect a system
or a control volume. For example, take any
properties like velocity pressure or any of
the other properties of fluid flow then Reynolds
transport theorem use a relationship between
the time rate of as returning this slide this
gives the time rate of change of an extensive
property for a system and for the control
volume. If there is a system inside on which
we are dealing, then we are considering control
volume between that how with respect to time
the property is changing the Reynolds transport
theorem describes. Here, we can see a control
volume.
On this section, the velocities V1 and at
the second section here velocities V2 and
here delta1 is V1 into delta t, that means
the fluid movement with respect to time delta1
is V1 into delta t and here delta2 is equal
to V2 into delta t, then with respect to the
changes taking place here is the control volume
of the fluid, initially it is like this and
after delta t so the control volume is shifting
like this in this slide, delta changes after
t plus delta t, so delta is a control volume
which we are dealing.
The control volume is changing after delta
t time, so after t plus delta t this is the
position. So, this is the initial position
number 1 and this is after sometime delta
t the position is number 2. Finally we can
see that transfers taking place between the
control volumes minus 1, this is the change
taking place with respect to this.
According to the Reynolds transport theorem,
if we consider the system property B is the
extensive parameter which we are dealing.
Here, B system t is equal to B control volume
t, so in the previous slide if B is the extensive
property we are dealing with so B system at
time t is equal to B control volume at time
t. so the system is equal to control volume
plus 2 minus 1, so here in the slide the system
is control volume plus 2 minus 1 gives the
system as far as between this, as mentioned
here, the system is control volume plus 2
which is described in this figure and then
minus 1 as shown in that figure, so B is the
extensive parameter.
According to the Reynolds transport theorem
we can describe with respect to the control
volume and system like this, so DB system
by Dt the total derivative of the extensive
property DB divided by Dt or DB by Dt is equal
to a partial derivative of the extensive property
B with respect to control volume B, del Bcv
by del t plus the property what is the something
adding to the system from the buoying to the
out of the system B out minus B in that is
equal to del B by del t plus rho A2 V2 b2
minus rho A1 V1 b1.
So, we have seen here V1 is the velocity at
this section and V2 is the velocity at this
section and so the control volume is considered
here. Now, for the extensive parameter which
we are described here according to the Reynolds
transport theorem we can write DBsys divided
by Dt with respect to system is given as with
respect to the control volume del Bcv by del
t plus changes taking place with respect to
the inflow and out flow so B out minus B in
so that is with respect to time what happens
whether how much is coming to the system and
how much is going out of the system, so that
is the Reynolds transport theorem.
Finally DB system by Dt can be represent as
del Bcv by del t plus roh A2 v2 b2 minus rho
A1 V1 b1, where rho is the density of fluid
which we are considering here and A2 is the
cross sectional area at section 2 and A1 is
the cross sectional area section 1 and V2
is the velocity at section 2, V1 is the velocity
at section 1.
Finally, we can write this in an integral
form this Reynolds transport theorem can be
written like this DB system DB by Dt is equal
to del by del t of we can integrate with in
the control volume del by delt of integral
roh bdv plus on the surface what happens in
the control surface integral rho b v n hat
which is the union vector dA, that means finally
what happens to the system is given such a
way that here you can see that with respect
to system if you are considering just like
a pipe like this so what happens to the extensive
property which we are dealing is inside the
control volume what happens and then on the
surface whether with respect to surface something
is going inside something is going outside
from the system through the surfaces so what
happens to that surface so that gives Reynolds
transport theorem. The total changes with
respect to time for the extensive property
which we are considering here that means DB
by Dt so B is the extensive property.
So the total change with respect to time what
happens to the extensive property is expressed
as we are considering within the control volume
what happens with respect to time as far as
the fluid inside is concerned that is given
by del by del t of integral with respect to
control volume rho DB and then on the surface
what happens that means what is something
is come adding to the system or something
is going out of the system, so that gives
the integral over the control surface rho
b V n had d, so this is the same as the above
expression so Reynolds transport theorem like
as described in this slide so it gives what
happens with respect to system and then the
control volume inside the system. This way
we can easily represent same you can easily
transform or we can change one system that
means the approach is based up on the system
then we can get equations or we can transform
that into a control volume based up on this
Reynolds transport theorem the parameters
are given with respect to the control volume
we can also transform to the system.
Finally the Reynolds transport theorem what
happens with respect to total system is for
an extensive property what happens with respect
to time is described in terms of the control
volume, inside the control volume and then
with respect to surface what is happening
what is something is coming through or going
out of the surface, so that gives the Reynolds
transport theorem. This theorem is very useful
in derivation of most of the fluid flow theories
and fundamental principles which we will be
discussing later so finally as we can seen
in this slide.
So, DB system that is gives the time rate
of change of an arbitrary extensive parameter
of a system. For example, rate of change of
mass momentum energy extra so this is the
DB by Dt and that is equal to say in del by
del t of integral rho bdv so that is equal
to time rate of change of the amount of B
within the control volume as the fluid passes
through it plus the net flow rate of the parameter
B across the entire control surface. This
slide gives the Reynolds transport theorem.
The time rate of change of an arbitrary extensive
parameter of a system is equal to time rate
of change of the amount of B within the control
volume as the fluid passes through it plus
the net flow rate of the parameter B across
the entire control surface so this theorem
is very useful many of the fundamental development
of the fluid flow theories and principles,
so with respect to the Reynolds transport
theorem the integral counter part of material
derivatives that we have seen.
Reynolds transport theorem with respect to
the counter part of the material derivative,
we can write as D of Dt of the particular
parameter which we are dealing equal to del
by del t of the parameter plus V dot product
del of the parameter, so this include both
the steady and unsteady effects we have seen,
so this gives the Reynolds transport theorem
with respect to the material.
With respect to the material derivative, you
can see that here steady and unsteady conditions
are described. Reynolds transport theorem
will be using many times as far as the fundamental
theories development and demonstration is
concerned. Before going to the continuity
equation based up on the consideration of
mass, we will initially discuss different
approaches fluid flow behavior, so how we
can analyze fluid flow behavior with respect
to different approaches.
In fluid flow analysis, generally we can have
two types of approaches: first is called integral
approach and second one is called differential
approach. Here, the fluid flow behavior, we
are analyzing with respect to either a system
or with respect to a control volume so with
in this we can analyze either we can go for
a integral approach or we can go for a differential
approach, so what is an integral approach
so in an integral approach what we are doing
is we are trying to evaluate the quantities
within a volume of fluid? For example, if
you consider the pipe flow here between this
section 1 and 2 what happens so what we will
be discussing? We are trying to evaluate or
we are trying to identify within the volume
of fluid.
So through this we get integrated equations
expressing behavior of fluid flow for control
volume in a flow field so that is what we
are going in the integral approach. So in
the integral approach there are basically
two steps first one is we have to check the
problem or we have to verify whether the system
is appropriate for a control volume analysis,
so depending up on the problem sometimes we
can go for a control volume analysis sometimes
it may not be possible.
We have to see the problem in detail whether
we can go for a control volume analysis or
we have to go for other types of analysis
so we have to scrutiny the problem whether
the system as appropriate for the control
volume analysis and second step in the integral
approach is we have to examine the behavior
of the control volume. So we have to see whether
the control volume is just like a fixed control
volume like shown in this ball inside it is
a fixed control volume or we have to see just
like a in an aeroplane with respect to when
we doing the analysis the aeroplane also moving,
so whether we are trying to analyze a moving
control volume or we have to see whether just
like we have seen deforming or an elastic
control volume. So these aspects also we have
to analyze or we have to evaluate before we
choose either an integral approach or whether
we have going for a differential approach
The two essential steps before going for integral
approach is we are to scrutiny the problem
we have to verify the problem whether the
system is appropriate for control volume analysis
and secondly we have to see the behavior of
the control volume whether it is fixed one
or whether it is moving one or whether it
is an elastic type of control volume. So first
we have to do these two steps of analysis
and then only we will be going for the integral
approach or whether we will be going for other
types of approach.
So integral form of quantities we can see
that most of the quantities we can described
in terms of integral form just like for example
the volume flow rate it is the integral of
velocity over an areas so if we consider the
volume of the discharge passing through integral
of velocity over an area that is the discharge
for volume of flow rate and second for example
the mass if we consider the mass it is the
integral of density over volume.
So the mass, when we are dealing with the
consideration of mass we can go for integral
form of approach and then if you consider
the force, so it is the integral of stress
over an area and kinetic energy it is the
integral of v square, that means the velocity
square by 2 over each mass element in a volume.
Like this we can analyze the problem and see
whether it is this kind of quantities whether
it is a discharge of volume flow rate or it
is mass or force or kinetic energy what we
are going to deal and then we can choose the
integral approach or differential approach
the other kinds of approach which will be
discussing later so this integral approach
is very useful in many problems where this
kinds of say we can use the appropriate depending
up on the problem.
Now, the second approach commonly used in
fluid flow behavior analysis is called differential
analysis. In the differential analysis, what
we are trying to do is we are trying to get
detailed knowledge of a flow field apply at
a point or very small region so we have already
seen that most of the time we will be going
for an eulerian approach so we will be discussing
what happens in the fluid flow what happens
at a particular point or particular section
what happens.
So, that is our major concern this differential
analysis the second kind of analysis very
important. So what we are doing here is as
far as the flow field concerned which we will
be choosing at particular point or a very
small regions trying to see the variations
of the differential of various fluid flow
properties at that particular point. In this
differential analysis we will be deriving
differential equation which expresses the
behavior or fluid flow behavior.
Differential equations are derived with respect
to theories in differential analysis and then
we are trying to evaluate the unknown dependent
variable at any space point in fluid flow
for all the times, that is what we are doing
the differential analysis so we are trying
to evaluate the unkown depending parameter
dependent variable with respect to space for
all the time.
These two approaches the integral approach
or differential approach, for both of the
approaches are very commonly used in fluid
flow problem so fluid mechanics problem we
will be discussing both the integral approach
and the differential approach for our problems
for the development of fundamental theories
in fluid flow and fluid mechanics which will
be discussing later. Initially, we will be
discussing the integral approach for the consideration
of mass or the derivation of the continuity
equation.
Now, we discuss the conservation of mass,
the continuity equation with respect to the
integral form of approach. We have seen that
just like in fluid mechanics also the fundamental
theories are based up on the consideration
of mass, consideration of momentum and consideration
of energy, so most of the basic principles
of basic theories are developed based up on
conservation of mass momentum and consideration
of energy.
First, we will be discussing here is conservation
of mass and the continuity equation is derived
based up on the conservation of mass. First
we will see the continuity equation with respect
to the integral approach which we have seen
earlier so as I mentioned earlier here say
the mass has the integral of density over
volume. so we can see that here we can easily
use when we scrutinize the problem we can
see that we can easily use the integral approach
so but any way as for as conservation of mass
is or the continuity equation is concerned
we will be discussing also in terms of the
differential approach.
First we will see the integral approach. Amount
of mass in system is according to the conservation
of mass, amount of mass in a system is constant
or the time rate of change of mass is equal
to 0 that is the basic principles of the conservation
of mass. Conservation of mass is the amount
of mass in a system is constant or the time
rate of change of system mass is equal to
0 or we can describe with respect to the total
derivative DM with in the system by Dt is
equal to 0, where t is the time M is the mass
D indicates the total derivative.
As far as the system is concerned, if the
fluid inside this ball or inside in a container
here we can see the fluid inside this container
so with respect to the conservation of mass
we are saying that the amount of mass inside
the system inside this the gas inside this
ball or inside this container it is constant,
so that is the basic principle, so that we
can say that if there is any change of gas
inside this ball or with respect to this container
here we can say the time rate of change of
mass is equal to 0, so that is the conservation
of mass or the continuity equations says so
that we can write this mass of the system
is concerned with respect to whether the air
inside or water inside in a container.
The mass can be defined with in the system
we can defied as we can integrate with respect
to the density and then the volume so the
mass of the system can be written as integral
of rho dv so now have seen the Reynolds transport
theorem earlier So within the system what
happens we have to describing here if we consider
the pipe flow which we have seen.
Within the system we will be describing in
terms of the control volume what happens and
then with respect to the control surface or
the boundaries say whether something is entering
to this system or something is dealing from
the system so that gives the Reynolds transport
theorem so here as far as conservation of
mass is concerned say if we apply the Reynolds
transport theorem we can write D by Dt the
total derivative of the system with respect
to the mass rho dV is equal to del t with
in the control volume which we are considering
here of rho dV plus with respect to the control
surface or the series that means exceed time
inlet of the system plus the control integral
of the control surface of rho with V n hat
dA.
Here, this V indicates the velocity and this
v indicates the volume. Now, here to derive
the basic equation of the continuity based
up on the conservation of mass we are using
the Reynolds transport theorem, so D by Dt
of integrals with respect to system rho dV
can be written as del by del t of the control
volume rho dv plus with respect to control
surface integral of rho V n hat dA, so this
gives the Reynolds transport theorem as far
as the mass which we are considering here.
Now, as we can see in this slide the control
volume it can here, this is the system which
we are considering and the control volume
is inside what is there and then what changes
taking place with respect to whether with
in the control surface that is what we are
describing so control volume expression for
conservation of mass called the continuity
equation.
Now from the Reynolds transport theorem and
since according to the conservation of mass
we can write DM by Dt is equal to 0 so now
if you use the Reynolds transport theorem
here we can write del by del t of integral
of control volume rho dV plus integral over
the control surface rho V n hat dA is equal
to 0. So, that means, to conserve the mass
the time rate of change of mass of the contents
of the control volume plus net rate of mass
flow through the control surface must be equal
to 0 so that gives the Reynolds transport
theorem with respect to the conservation of
mass. Time rate of change of the mass of the
contents of control volume plus net rate of
mass flow through the control surface must
be equal to 0.
From this we can write the mass flow rate
is equal to rho into Q so that we can express
as if V is the velocity and A is the area
cross section so this s equal to rho into
A into V, where rho is the density of the
fluid, so the integral expression from the
integral approach we have seen the continuity
equation can be described by the Reynolds
transport theorem as given this equation del
by del t of integral of control volume of
rho dV plus integral over the control surface
rho Vn dA. So this gives the integral form
of the conservation of mass for the continuity
equation and finally it is steady state condition
you can say that mass change can be written
as rho into AV where V is the velocity here.
Finally, the continuity equation integral
form if the system is at steady state or if
there is no change with respect to the density
that means there is del rho by del t that
means no change of density with respect to
time this can be written as 0.
In most of the fluid flow which will be discussing
will not be any change as far as rho density
concerned, so we generally dealing with incompressible
fluid flow so that del rho by del t is equal
to 0 so finally if the density is constant
we can write A1 V1 is equal to A2 V2 for the
system which we are considered and the system
if it is non uniform or it is not uniform
then we can write rho1 integral over area
A1 V1 d A is equal to rho2 integral A2 V2
d A or the non uniform flow we can write rho1
V1 bar A1 is equal to rho2 V2 bar A2 so this
gives the integral form of the continuity
equation.
As we have seen here we can the general expression
if you deal with compressible flow also and
in the case of incompressible flow you can
write the equation since here del rho by del
t is equal to 0 that we can write the integral
form of the continuity equation as rho V1
bar A1 is equal to rho2 V2 bar A2 so here
what it says is with respect to this is the
system is just like in a pipe system which
we are dealing so the fluid entering with
respect to system that should be equal to
the fluid leaving from the system so that
we have to use this equation here.
The continuity equation as we have seen here
the mass is concerned and we have also seen
that there can be or deforming a control volume
so in that case the with respect to the continuity
equation of the conservation of the mass we
can write DM system Dt is equal to del by
del t of control volume rho dv plus all the
control surface rho w instead of V here we
are using rho w n hat dA is equal to 0.
For the case of the moving or deforming the
control volume this w is the relative velocity
so the system itself in the control volume
is moving or deforming with respect to that
we have to consider changes taking place with
respect to the relative velocity. So, for
moving or deforming control volume we can
write the continuity equation as DM by Dt
is equal to del by delt of integral of rho
dV plus all the integral over the control
surface rho w, where w is the relative velocity
n hat dA where n hat dv is the unit vector
so this gives the continuity equation in the
integral form.
For example, if you consider a pipe with branch
like in this slide so here we can express
the continuity equation like here. At the
section 1 if V1 is the velocity and area of
cross section is A1 and the density here is
rho1 and at section here at three if the velocity
is V3 and area of cross section is A3 and
density is rho3 and then here this location
the density is rho2 and A2 is the area cross
section V2 is the velocity.
Here, we can see as far as system is concerned
there are 2 inflow at 1 and 3 and then there
is an outflow at 2 so finally you can write
rho1 A1 V1 plus rho3 A3 V3 is equal to rho2
A2 V2 so if the density say we can write this
equation as A1 V1 plus A3 V3 is equal to A2
V2 so this way we can write with respect to
the integral form of the continuity equation
based up on the conservation of mass we can
derive the continuity equation for various
system. Now before discussing the differential
approach for continuity equation we will see
the example problems as for as the continuity
equation is concerned.
Now, simple example initially which we are
discussing here is say example 1 is as shown
in figure, water flows steadily from a fire
hose. So, there is a pipe through which a
fire hose is connected there is a constant
supply to the fire hose here and from which
we are talking through a pipe system connected
with the fire hose we are taking water through
this for various purpose so if the nozzle
exit velocity must be at least 10 meter per
second. Here, the exit velocity is 10 meter
per second we want to determine the minimum
pumping capacity for which here is the flow,
so here this is the direction of the flow
we want to determine the minimum pumping capacity
for this simple system so here this is the
fire hose and this is through which the flow
type displace and if we consider the control
volume now the control volume here you can
see that this is the control volume with respect
to the fire hose which we are considering.
So, here the exit velocity is given as 10
meter per second and here the diameter of
the pipe the hose is given as 20 millimeter
so V1 to find minimum pumping capacity such
that the system is supplying a minimum of
10 meter per second with respect to the continuity
equation which we have seen earlier.
Here you have a control volume and then control
system that means the control volume you can
see here in this figure so this gives the
control volume is here and then with respect
to the inlet this is the control surface what
is going inside and what is going outside
one inlet and exit for the system is concerned,
so the continuity equation which we have seen
earlier we can write with respect to integral
form which we are discussed so as del by del
t of integral over the control volume rho
dv plus integral over the control surface
through V n hat dA is equal to 0 according
to the continuity equation, so here the time
rate we are not considering.
So this term goes as 0, since a flow is considered
as study, so that we can write the final system
as integral over the control surface rho Vn
dA that means with respect to the system is
concerned with respect to what is going inside
to this control volume inside the fire hose
there should be going out through the exit
so what is entering go through the exit.
If you consider the second part here, the
continuity equation is integral c over the
control surface rho Vn dA that can be written
as m2 dash minus m1 dash that means what is
entering that is equal to m1. Finally, we
can write rho2 Q2 is equal to rho1 Q1 that
means through the system what changes that
should be through the exit from the system.
So rho2 Q2 is equal to rho1 Q1 so here if
you assume that since here we are considering
water so rho density so that we can write
Q1 is equal to Q2 that is equal to V2 into
A2 so here before the fire force which we
are consider the radius is the diameter of
the hose is given and the velocity is givenso
that we can write finally the discharge that
means the minimum pumping capacity should
be Q1 is equal to Q2 is equal to V2 A2 that
is equal to the velocity is 10 meter per second
10 multiplied by the area cross section of
the pipe that is phi by 4 d square so d is
0.02 or 20 millimeter.
Finally, this is the a pumping capacity for
the fire hose system which we have seen in
this particular figure so here Q is equal
to 0.00314 meter cube per second so this is
the simple problem where we can use this integral
form of the continuity equation. We will also
discuss another simple example which is related
to the conservation of mass based up on the
continuity equation.
So a problem here is a centrifuged pump has
got a discharge capacity of 50 liters per
second as shown in figure it has an axial
inlet diameter of 9 centimeter and then impeller
of 20 centimeter diameter so it can be assume
the 10 percent of outlet area is occupied
by blades so we have to determine the axial
velocity in the inlet pipe and the radio component
of velocity like the outer impeller.
So, this is the problem here. There is a flow
takes place to the centrifuged pump in this
direction and here the inlet is given as 9
centimeter and here you can see that there
will be outlet area blades will be there with
respect to the centrifuged pump so this way
once it enters here and then it should go
through this direction and this direction.
Here, the diameter is given as 20 millimeter
and here also 20 millimeter and here this
impeller total diameter is given as 20 centimeter,
so we want to find the axial velocity entering
through the inlet pipe also the radio component
of the velocity at the outlet of the impeller,
so this is the problem which we are discussing
here. Here the discharge for the problem is
given as 50 meters per second.
So, the discharge can be written as Q is equal
to 0.05 meter cube per second Q is equal to
that is discharge is equal to 50 liter per
second or it can be written as 0.05 meter
cube per second at inlet. So, the inlet is
here. At inlet we can write, the discharge
is equal to area of cross section multiplied
by the axial velocity so that what is entering
that should go through impeller.
We can write Q is equal to Aa into Va is equal
to Vr into Ar, Aa is the area cross section
of the inlet and Va is the axial velocity
and Vr is the radio velocity and Ar is the
area which we will be considering, so the
axial velocity can be obtained as the discharge
Q is known, so that is given as 0.05 meter
cube per second. At inlet the area of cross
section can be written as phi by 4, the diameter
is phi by 4 into 0.09 square, so that gives
the inlet area cross section and then axial
velocity Va is equal to Q divided by Aa so
this is equal to 0.05 divided by this Aa that
is 0.00636 so that gives the axial velocity
as 7.86 meter per second. To find the radial
velocity we can calculate the effective area
with respect to the 10 percent of the outlet
area keeper by blades. We can find the effective
area by taking the diameter of the impellers
is equal to phi into 0.025, since 20 centimeter
is given here and then 0.025 is this depth.
So phi into 0.2 into 0.025, so this is actually
see 0.025 and this is 25 millimeter so phi
into 0.2 into 0.025 into 90% is the effective
area since the 10% is occupied by the blades
into 0.9 so this gives the Ar so from that
we can get the radial velocity is equal to
Q by Ar so that is equal to 3.5367 meter per
second. Like this we can use this integral
approach of the continuity equation of the
consideration of mass to show various examples.
This is the conservation of mass of the continuity
equation which is one of the basic equations
which is used in all the fluid flow analysis,
so this is the integral form. Now we will
be discussing about the differential form
later. Before going to the differential form
of the continuity equation here you can we
will see in this slide the linear motion and
deformation before going to the differential
on the continuity equation will see with respect
to the fluid moment how the linear motion
and deformation takes place.
Here, we can see that a container here delta
y is the size, delta x is the direction and
velocity in the x direction u, velocity in
y direction is v, so with respect to this
say here from one section to another, the
velocity change is u plus del u plus del x
into delta x and velocity change in the from
one section to another, the y direction is
v plus del v by del y into delta y. So the
volumetric dilation rate means with respect
to the volume inside the system dilation,
that means the change or the rate of change
of volume per unit volume can be expressed
by this equation.
Since we are dealing with the velocity in
x y and z direction, del v is equal to del
u by del x plus del v by del y plus del w
by del z, where u v w are the velocity components
in xyz direction. So, the total the rate of
change of volume per unit volume is expressed
as del u by del x plus del v by del y plus
del w by del z.
Now, we can see the volumetric dilation rate
is 0 for an incompressible fluid, so when
we are dealing with incompressible fluid volumetric
dilation rate is 0, so we can write the variation
of velocity in direction of velocity is del
u by del x, del v by del y and del w by del
z, since we are dealing the velocity components
in xyz direction.
So del u by del x gives the velocity x direction,
its radiation, next del v by del y gives the
velocity change in the direction of y, del
w by del z gives the velocity change in the
direction of z for w. So, variation of velocity
in direction of velocity simply cause linear
deformation for fluid element or shape does
not change for the linear variation, so cross
derivative such as del u by del y only cause
the angular deformations, but otherwise the
linear motion is concerned, the variation
is the velocity in direction of velocity simply
cause the linear deformation. So, shape does
not change, now the next lecture we will be
discussing the deformation and then we will
be discussing the and then further we will
be going to the differential approach of the
continuity equation or the conservation of
mass.
