last time we were discussing about ah the
integral forms of the conservation equations
and as an example we looked into the integral
form of the mass conservation and ah its corresponding
differential form also we revisited and ah
ah we found out that it is possible to convert
one form to the other now ah let us look into
some more examples of the use of the integral
form of the mass conservation equation let
us say that you have ah weight shaped element
like this with the axis oriented along x and
y and the velocity field is a two dimensional
velocity field is given by say u is given
by this v is given by this one and two dimensional
w is zero let us give some names to the faces
of 
the elements here the objective is to find
out that what is the volume flow rate through
a c the dimensions are given let us say this
is 
one meter or just one unit all are given in
some consistent units say this is two these
are given ok
so ah what we also ah are assuming that v
zero is a constant it is not a function of
any other variable that we are looking for
so how should we proceed with this problem
let us say that we are interested to use the
integral form of the mass conservation and
that is one of the natural things that we
should use here because there are the three
faces across which fluid will enter and leave
so the net rate of transport should be given
ah by the integral form of the mass conservation
that is you have d m d t for the system plus
this for the control volume plus integral
of rho 
the left hand side is zero because no matter
whatever system you are considering it is
by definition of fixed mass the right hand
side because we are assuming that ah rho is
not changing with time neither the control
volume is changing with time what is our control
volume let us say that this triangular shaped
element is our control volume so whenever
we are making a control volume analysis it
is important to identify what is the control
volume that we are taking because you may
take different control volumes of course for
this problem this is an obvious choice of
the control volume but there could be problems
where there could be many different choices
of the control volume
so the equation that you are writing should
be pertinent to a particular control volume
that you are chosen and ah that should be
clearly mentioned so this triangular shaped
ah thing ah is a fixed control volume the
volume of that is not changing with time so
its not a deformable control volume neither
the density is changing with time so this
term goes to zero so what remains is ah this
net term which is which is nothing but basically
the net rate of out flow minus inflow of mass
equal to zero so ah if rho is a constant let
us ah take a further make a further simplification
that rho is a constant then this will boil
down to basically integral of v dot n d a
over the control surface equal to zero and
when we say v hear the relative velocity and
absolute velocity are the same because it
is a stationary control volume it is not a
moving control volume so this now we may break
it up into three parts because the control
volume has three different distinct oriented
surfaces
so you can write this as ah the sum of the
effects of a b b c and a c so for a b when
you write what should be this corresponding
expression so let us identify that where is
that a b so you are looking for this face
you are interested to find out what is v dot
n integral of that over the entire area so
one important thing is here n is not a variable
it is like a line oriented along y axis
so n is a constant so for a b what is n so
let us try to identify the what are the normal
directions for the different edges here the
edges are all straight lines so they have
unique normal directions so let us say a b
what is the n for that minus i then b c and
a c 
so for a c ah let us let us say that this
is the unit vector normal so this will have
its components let us say this angle is theta
so if the angle between the normal and the
horizontal is theta then the angle between
the edge and the vertical is also theta right
that means you can say that n is definitely
cos theta i plus sin theta j where you have
tan theta equal to one by two
ok 
so what will this imply this implies ah that
like you have identified the direction normals
for all the faces only thing you require is
the velocity so the velocities for the faces
for a b what is v let us write the in the
same table let us try to write what is the
velocity v cap so for a b what is v cap 
for a b x is zero so u is not there there
is some v ok so it is minus v zero y by l
j for b c similarly you have y equal to zero
so it is v zero x by l i and for a c it is
the real ah the sum of the two components
because ah here x and y are both non zero
so we are not writing it because it is like
ah let us just write it as in general u i
plus v j now to get the integral you have
to keep in mind that if the velocity varies
along that length then you have to ah integrate
it over the length to get the total flow rate
when you come to the so let us start with
a b you can of course evaluate this try to
evaluate this but let us not do that bull
work you see that the velocity along a b is
like a it it is oriented along a b so there
is no normal component of that so there is
no net flux or influx or out flux of flow
across a b because there is no normal component
so if you make a dot product of this ah that
you can clearly make out and ah so that will
not give rise to any net flow same might be
true for b c right so the choice of the axis
here has been such that ah if this the velocity
field then ah that is the case but the same
is not the case for a c right so for a c how
will you find out what is the total or the
net rate of flow yes
so if something in there is there is nothing
that enters here there is nothing that enters
or leaves here you are expecting that there
is nothing enters or leaves here right how
could you verify by doing the integral that
it should be zero no that is what like this
is q a b this is q b c this is q a c like
if you evaluate by doing the dot products
ah will the dot product automatically give
it 
it is expected that the dot product to integral
over that should automatically give it so
that you should check so that it gives you
a confidence of how to calculate that because
here i have given a special type of velocity
field so that actually this problem solution
is not necessary i mean i am trying to go
through this through ah through a formal route
to give an idea of what should be done in
in a case when when ah the the problem solution
deserves that but here actually doesnt deserve
so these are ah this is ah more intuitive
case where nothing is entering and nothing
ah therefore is expected to leave now
if by chance you get something which is leaving
through this and nothing is [enter/entering]
entering so [the/that] that will really violate
the law of mass conservation so that ah one
has to be careful of now let us look into
ah some other problem which is not as trivial
as this one so another example let 
us say there is a tank like this 
the velocity profile at the exit of the tank
is through this pipe is given by this one
and in terms of a local coordinate system
this ah
let us say that ah the local coordinate system
is x one y one it is given as u one is equal
to some u zero in to one minus y one by y
one square by h square where y one ah is the
transverse coordinate and h is this height
and the fluid entering here here the again
in terms of the local coordinates you can
specify it but ah that specification may not
be necessary because it is given that it is
a uniform velocity profile here with a velocity
u infinity which is uniform and let us say
this is h by two 
ok 
the directions of the axis are not given that
means it is not given that what is this angle
theta so theta one it is not given that what
is this angle say theta zero these are not
given
what is given it is given that the density
is a constant we have to find out what is
u not given the width of this figure ah perpendicular
to the plane where it is drawn is uniform
ok so u infinity is given h is given these
two things are given ok 
so how will you go about it again looks like
a situation where mass conservation should
be applied and if you want to apply the integral
form of the mass conservation let us say that
we consider a control volume so what should
be a good choice of the control volume so
a good choice of the control volume is something
where across the surface we are totally confident
about the velocity field so let us say that
we make a choice of the control volume something
like this
so with this choice of the control volume
you can write the law of mass conservation
if you want to write that then ah like it
is because rho is a constant again it will
boil down to a case very similar to the previous
one where ah eventually it will be integral
of v dot n d a over the control surface equal
to zero the remaining terms will not be relevant
so this is the only term that is relevant
now out of the surfaces that you have ah ah
only you have one inflow and one outflow surface
across the other surfaces fluid is not flowing
so those surfaces are not relevant so you
may break it up into integrals one for this
inflow and another for the outflow ok so the
first one is ah a very straight forward let
us do that so when you have ah the ah v dot
n d a see this v is uniform over the area
over which it is flowing for the inflow
so you can take away this v out of the integral
not only that the dot product also because
n is also a constant here so v dot n the entire
thing you can take out of the integral so
what it will become it will become v dot n
into area of the face over which it is coming
so ah let us call that this area of the face
is a here a naught and here the area of the
face is say a one 
so when you write v dot n you have to keep
in mind that what is the direction of n in
along this surface so the direction of n is
opposite to v right so v dot n will give the
minus of the magnitude of v so for the area
zero or area o it will become this term will
become what 
minus u infinity into h by two into the width
let us say b is the width perpendicular to
the plane of the figure then the other area
for a one we cannot have the same consideration
because the velocity varies over a one
so to see how the velocity varies with a one
we are ah ah ah along a one we are already
given that with respect to the local transfers
coordinate how it varies but n is a but you
have to keep in mind that forget about that
functional dependence on y one u one is going
out of the area and what is n n is also oriented
out of the area this means that if you just
take the vector sense v dot n that will give
you the magnitude of v into one because the
dot products are in the same ah dot products
of two vectors in the same ah sense ah that
will be ah leading to that conclusion what
because it is varying with y one now you have
to really do the integration so when you first
have v dot n that will become
u zero into one minus y one square by h square
that will be v dot n and d a d a is what so
you take a small element on the axis y one
so this is the small element of the area so
what is that small element of the area say
at a height y one from the centre line see
the coordinate system is from the central
line so at a height y one from the centre
line say you have taken a small area of width
d y one so the elemental area which is like
the d a given symbolically it is d y one into
b if you integrate that from minus h by two
to two h by two then that will represent that
what happens for the area one
so sum of these two should be equal to zero
clearly from the minus sign of the first term
you can make that it is inflow and the plus
sign of the second term means that it is it
outflow and it is possible to ah complete
disintegration in a very simple way we are
not going into that just ah to save some time
but important thing is that see this is the
variation of u over the section so you can
write it equivalently as integral of u d a
in the fundamental form in a scalar form so
this is this is not like a vector form so
its like just like because eventually with
the dot product it has become a scalar you
have to keep in mind this is this u is nothing
but the component of velocity which is normal
to the area here fortunately all component
is normal to the area there is no component
which is cross that
so this you may express through some quantity
which is called as average velocity so average
velocity is this divided by the area that
means if this velocity was uniform but the
same flow rate was there see if it was uniform
then that uniform velocity times the area
will give you the flow rate that is what the
first term has told us and if it is not uniform
obviously we have to integrated it to get
the flow so if the flow rate where a uniform
in a hypothetical case sorry if the velocity
profile was uniform in a hypothetical case
but the flow rate becoming still the same
as it is in the real case then if you equate
those two flow rates then that equivalent
hypothetical uniform velocity this is called
as average velocity so it is like a equivalent
uniform velocity that would have prevailed
across the section satisfying the same volume
flow rate as it is there in the real case
so ah therefore this is like as good as some
u one average into a one and this is just
like u not into a naught so it is it is just
like a one v one equal to a two v two we have
to keep in mind again that what is v one and
what is v two these are very important again
i am repeating these are fundamentally the
average velocities over the sections one and
two see when we are writing here one and two
or maybe zero and one or whatever subscripts
there is a fundamental difference from what
we wrote for using the bernoullis equation
those we wrote for points one and two now
we are writing for sections one and two so
ah maybe subscript wise they look very similar
but ah meaning is entirely different when
it is uniform it doesnt matter it is as good
as writing for a point because velocity doesnt
vary from one point to the other but you dont
have anything called as area at a point
so it is basically when you write a one no
matter in the context of ah what we write
what we have written earlier ah for a for
two points when you use the bernoullis equation
we should keep in mind that there also a one
was for the section that contained the point
one so it is not that area of a point one
or something like that there we use velocity
at the point one the reason is that in the
bernoullis equation we use velocity at a point
so we had to link it with velocity at a point
now it is like we are linking things through
velocity over an area so if you complete this
problem you will get what is u naught because
all other things are known now let us look
into some other examples where may be we look
into ah different case may be unsteady case
