So welcome to this lecture number 26 of our
course on fundamentals of transport processes.
We were discussing high Reynolds number flows,
potential flows, if you recall the previous
section was on low Reynolds number flows where,
inertial effects are neglected. This section
is on high Reynolds flows where viscous effects
are neglected.
So these are potential flows, they have two
characteristics; one is they are inviscid
that is the viscosity is equal to 0. So we
completely neglect the viscous terms in the
conservation equation and there also irrotational
del cross u which is the vertisity is equal
to 0 everywhere in the flow. So therefore,
since the vertisity is equal to 0, the anti-symmetric
part of the rate of deformation tensor is
equal to 0 everywhere in the flow. We derived
the conservation equations for this case,
since the flow is a rotational the velocity
can be written as the gradient of a potential
because the curl of a gradient of a scalar
is always equal to 0.
Once we do that we get the mass conservation
equation, the divergence of velocity is equal
to 0 reduces to del square phi is equal to
0, phi is called the velocity potential phi
is called the velocity potential. The momentum
conservation equation in the absence of inertia
reduces to rho times partial u by partial
t plus u dot grad u is equal to minus grad
p neglect the viscous terms and therefore,
you just get the body force. And we had we
had simplified this non-linear term for the
case where the vertisity is 0 or the anti-symmetric
part of the rate of deformation tensor is
equal to 0. And once we do that we get an
equation of the form the gradient of p plus
half rho u square plus rho partial phi by
partial t plus the potential, here we are
assuming that the force is given by the gradient
of a potential. So it is a conservative force
f is equal to minus grad v. So the force is
a conservative force.
So this whole thing is equal to 0 or p plus
half rho u square plus rho partial phi by
partial t plus v is equal to some constant
p naught. So the gradient of some function
is equal to 0 that function itself is equal
to a constant everywhere within the flow field
of course, there pressure itself is known
only to within an unknown constant. So we
had to specify the pressure at some boundary
and then you know the pressure everywhere
within the flow. As I said we have neglected
the viscous terms of the conservation equation
and therefore, it is not possible to satisfy
the zero normal velocity and zero tangential
velocity boundary conditions at the surface.
The original Navies Stokes equations that
we had was second order differential equations
in the velocity. When we neglected the viscous
terms we basically get a first order equation
in the velocity or a second order equation
in the potential, in the original equations
we could satisfy both tangential and normal
velocity or stress conditions. In the modified
equations we can only satisfy the normal velocity
and normal stress conditions.
The stress itself in this case is given by
the stress tensor is equal to just the pressure
part, since we have neglected the viscosity
there is no viscous stress in this particular
case, the stress is determined purely by the
pressure and it is isotropic it is an isotropic
tensor and therefore, at solid surfaces we
can only satisfy the boundary condition that
the normal velocity is equal to the velocity
of surface along the unit normal itself where
capital and is the unit normal to the surface.
This is the equivalent of the zero normal
velocity boundary condition where n is the
normal and we cannot satisfy the tangential
boundary condition because, we reduced the
equation from a second order to a first order
physically, the reason is because when we
neglect viscosity we neglect the diffusion
of momentum, in the absence of momentum diffusion
there cannot be any transfer of momentum perpendicular
to the flow.
There is of course, convective transport of
momentum along the flow direction, however
there is no momentum transport perpendicular
to the flow and the no-slip condition of the
zero tangential velocity boundary condition
at the surface requires a transfer of momentum
perpendicular to the flow. So that the flows
gets stopped at the surface, since we do not
have that mechanism, we do not have a zero
tangential velocity boundary condition for
these potential flows.
For these potential flow solution equations
we had we had proved various theorems for
example, you can show that the solution is
unique, you cannot have two different solutions
of the potential flow equation that satisfy
the same boundary conditions. You can show
that for the velocity in the fluid to be non-zero,
you have to have a non-zero normal velocity
at the bounding surfaces because the kinetic
energy as you recall the kinetic energy for
the flow can be written just as a surface
integral half rho integral over the volume
of u square.
Using the fact that the velocity is the gradient
of a potential I can write this as half rho
integral over the bounding surface as of this
volume of u i times the unit normal times
the potential u dot n times phi therefore,
the if the normal velocity at all bounding
surfaces is equal to 0, the kinetic energy
is 0 and therefore, the velocity is equal
to 0 at each point in the flow. We also showed
that for a potential flow solution the kinetic
energy of the flow is smaller than the kinetic
energy of any other flow which does not necessarily
satisfy the potential flow conditions.
So therefore, the potential flow has the minimum
kinetic energy of all possible flows that
satisfies the conservation equations call
the minimum energy theorem and finally, we
were solving the potential flow solutions
for the flow around a solid object, as usual
we take the sphere as the simplest solid object
because we can use tensor symmetries in order
to determine the velocity profile of the sphere.
So let us go back to that solution for the
flow around a sphere. So I have a sphere that
is translating with some velocity u vector.
In potential flow and I need to find the solution
for the potential flow equations. The boundary
conditions as I mentioned are u i n i is equal
to capital U i n i or u dot n. For the fluid
is equal to the sphere velocity u dotted with
the unit normal at that particular location
on the surface and of course, we require that
u goes to 0 as the distance goes to infinity
as the distance from the surface of the sphere
becomes larger and larger, since this sphere
is moving in a quiescent fluid. We require
that the velocity goes to 0 as r goes to infinity
and this is the local unit normal at each
point in the surface.
Now without loss of generality of course,
you can place your the origin of the coordinate
system at the center of the sphere. So that
the surface of the sphere is basically given
by r is equal to capital R. So I use a spherical
coordinate system in which the center of the
coordinate system is at the center of the
sphere.
And then I have my radius vector r and angle
theta the configuration is axis symmetric
as you go around this axis which is parallel
to u that I use the velocity direction itself
as the axis from my coordinate system that
is the simplest thing to do. Once you have
chosen that as the axis there should be no
variation as you go around that axis. So therefore,
there is no dependence upon the phi coordinate.
So it becomes an axis symmetric problem. We
solved the potential flow equations in the
last lecture del square phi is equal to 0
with this boundary condition, solutions are
of two types the growing and the decaying
harmonics of course, we cannot have the growing
harmonics because we require the velocity
to go to 0 far away therefore, the solution
has to be linear in the decaying harmonics.
In addition the velocity field is linear in
the velocity of the sphere and this is the
gradient of potential is equal to the velocity,
the potential is also linear in the velocity
of the sphere. So therefore, my potential
function has to be something that is linear
in the velocity of the sphere as well as one
of the harmonics.
It has to be linear in the velocity of the
sphere as well as in one of the harmonics.
The only way to get a potential that is linear
in the velocity of the sphere as well as one
of the harmonics is to multiply it by dot
the velocity vector with the first vector
sorry call harmonic, where A is some constant
which will be determined from the boundary
conditions. So since this vector spherical
harmonic dotted with capital U gives you a
scalar. This is the only possible solution,
you cannot get solutions of any other form,
this satisfies the Laplace equation because
the vectors spherical harmonics is a solution
of the Laplace equation.
So this is the final solution for the velocity
field. Now of course, we determine this constant
A by finding out what is the normal velocity
boundary condition on the surface of the sphere
the unit normal is the outward radial direction,
it is along the radial direction therefore,
the unit normal around the surface of the
sphere can be written as the displacement
vector, at the displacement vector to that
point on the surface divided by the radius,
since the unit normal is along the displacement
vector and it has unit modules it is the unit
normal is equal to the displacement vector
divided by its magnitude which is the radius
and the surface of the sphere.
Of course at the surface of the sphere this
is just going to be equal to x i by capital
R, capital R is the radius of the sphere.
So we can impose this boundary condition u
i n i is equal to capital U i n i as you recall
u i is equal to partial phi by partial x i
is equal to A U j into delta ij by r cubed
minus 3 x i x j by r power 5 therefore, the
boundary condition requires that A U j 
this is into the unit normal, unit normal
in this case is equal to x i by r is equal
to capital U i x i by r at r is equal to capital
R. So that is the boundary condition from
that boundary condition you find out the value
of A.
We solved this in the previous lecture and
we got that A is equal to minus R cubed by
2. You can easily verify that for A is equal
to minus R cubed by 2 this boundary condition
is identically satisfied therefore, this gives
the solution for the potential minus R cubed
by 2 by r cubed and the velocity 
that gives the solution for the potential
and for the velocity. Next we had determined
the total kinetic energy of the flow in the
last lecture.
So as I said the total kinetic energy half
rho integral over the volume of d v times
u i square, this is the integral over the
entire fluid volume, that is this the integral
over all of the fluid that is located outside
this sphere all of the fluid that is located
outside this sphere. I can also write this
as half rho integral ds of u i n i phi and
let me just put this n i in blue here for
a reason, this n i have put it in blue for
a reason because that is the outward unit
normal to the fluid volume. The fluid is outside
this sphere extending up to infinity, the
fluid is outside the sphere extending up to
infinity obtaining the surface integral over
this fluid volume, one has to take this carefully.
I am going through this in detail because
we will see such volume integrals again and
again in our analysis of potential flow. The
fluid volume is the volume that is in between
the surface of the sphere in between the surface
of the sphere and the surface at infinity
very far away from the sphere. I will call
this surface of the sphere as s and this as
s infinity. So the fluid is in between the
sphere and far away the surface far away.
So I have to do this integral over these two
surfaces and these two surfaces there is a
fluid in between these two surfaces, recall
that when we when we did the divergence theorem
and we did the divergence theorem just to
make this concept clear, we said that integral
over the volume of d v of del dot A is equal
to integral over the surface ds of n dot A
that was the divergence theorem. How did we
prove it? If you recall we took some volume
with an outward unit normal n we took this
volume with an outward unit normal n, then
we divide this volume into small little bits.
And we looked at two adjacent volumes we looked
at two adjacent volumes and we found out contributions
to integral of divergence of A over these
two adjacent volumes reduce to surface integral
over the surfaces of these two volumes. Now
for these two volumes they have this common
surface they have this common surface, the
flux through the surface is the same because
what leaves one volume comes into the other
volume right. So therefore, the value of A
at these two surfaces is exactly the same.
Whereas the unit normal’s the outward unit
normal’s are in opposite directions. The
outward unit normal’s are in opposite directions
whereas, the value of the vector A at that
particular location was the same therefore,
the integral over these internal surfaces
which are between two adjacent volumes exactly
cancel out and all I am left with is the integral
over the outside surface all I am left with
is the integral over this outside surface,
that give me the divergence theorem. This
was done for an object which is what is called
singly connected that is you have one object
with just one surface surrounding it, this
could be extended to multiple surfaces as
well as follows 
I could have for example, one outer surface
and one inner surface and there is a volume
between these two there this a volume between
these two there is an outer surface and an
inner surface and I could also what is the
value of integral d v divergence of A over
this surface I am sorry over this volume which
has two surfaces, one surface s inside, the
other surface s infinity, outward unit normal
to the volume for the surface outside it is
directed outward to the volume, for the surface
inside this is directed into the sphere that
is at the center.
The outward unit normal to the volume that
is it goes from the surface, it does not it
is in opposite direction to the direction
in which this volume is located therefore,
it is inverted this point I could do my exact
same calculation, divide the surface into
large number of volumes 
similar to what I had for the divergence theorem
I could divide into large number of volumes,
then calculated integral of d v times del
dot A over each of these volumes.
What you will find is that for two volumes
that are adjacent to each other, once again
those that have a common surface this integral
will cancel out those that have a common surface
this integral will cancel out, it will not
cancel out when there is no common surface
when the surface is actually bounding the
volume. In this case there are two surfaces
bounding the volume, one outside one inside.
So therefore, one has to take the integral
over these two surfaces what I have called
s infinity here and what I call s inside over
here, for s infinity the unit normal is facing
radically outward, where as for the surface
s itself it is directed into the sphere.
What is outward to the fluid volume is inward
to the sphere. So therefore, the unit normal
is inward to the sphere for this sphere itself.
So for these two integral over these two surfaces,
for the integral over these two surfaces this
divergence theorem states that, so this has
to be is equal to half rho integral over the
surface at infinity ds u i n i phi plus half
rho integral over the surface S ds u i n i
phi, where this n i for the surface s is directed
inwards therefore, the the the the value of
the unit normal at the surface is directed
inwards which means that this unit normal
is equal to 
minus x i by r into phi, because it is directed
opposite to the displacement vector in the
coordinate system whose origin is at the center
of the sphere.
So therefore, we can write this as a sum of
two integrals, the first integral over the
surface at infinity. If you recall the velocity
field that we have just derived u i is equal
to minus R cubed by 2 u j delta ij by r cubed
minus 3 x i x j by r power 5 as you can see
this velocity decreases proportional to 1
over r cubed as r goes to infinity.
This velocity decays as 1 over r cubed as
r goes to infinity, my potential was a dipole
and therefore, this is the next higher term
the quadruple term. So this velocity decreases
as 1 over r cubed. Over this surface at infinity
the surface at infinity that I have here as
r goes to infinity the surface area increases
as r square because the surface of a sphere
the surface area is proportional to r square
whereas, the velocity itself decreases as
1 over r cubed.
So in the limit as r goes to infinity this
gives me a contribution that goes to 0, because
the surface is increasing as r square velocity
is decreasing as 1 over r cubed the product
of the two goes as 1 over r and therefore,
it goes to zero in the limit as r goes to
infinity. So for that reason this contribution
over the surface far away actually goes to
0 as r goes to infinity and I am left with
this second contribution minus half rho integral
over surface ds u i phi into minus x i by
r and I know that on the surface u i n i.
So this u i n i is equal to capital U i times
n i I should this sign must take here into
phi and therefore, I can write this as minus
half rho integral over the surface s of capital
U i x i by r times phi because I know that
small u i times x i is the same as capital
U i times x i and if I substitute my value
for phi I will get this as phi goes as minus
r cubed by 2.
So I will get minus half rho integral S ds
u i x i by r into minus R cubed by 2 u j x
j by r cubed and this is taken at the value
where capital R is equal to small r because
the surface of the sphere is at the location
where capital R is equal to small r. So this
just becomes equal to rho u i u j by 4 R integral
ds x i x j integral over the surface of x
i times x j we know how to do is equal to
a times delta ij and if you multiply both
sides by delta ij finally, you will find that
a is equal to 4 phi by 3. So you get 4 by
3 r power 4 times delta i j.
And finally, the result we got in the last
lecture is equal to half times the added mass
times u i square, where the added mass is
equal to 2 by 3 pi R cubed rho, is equal to
2 by 3 pi R cubed times that density of the
fluid. So 2 by 3 pi R cubed is half the volume
of the sphere. So 2 by 3 pi R cubed times
rho is half the volume of fluid displaced
by the sphere. So therefore, the additional
kinetic energy due to the fluid flow because
when the sphere is moving.
There is a kinetic energy associated with
the sphere itself but, there is also an additional
kinetic energy associated with the flow of
the fluid and this additional kinetic energy
is equal to half the mass of fluid displayed
by the sphere or it is equal to the density
of the fluid times half the volume of the
sphere. So this is a general result because
even if you had a more complicated object
you would have higher order terms in this
expansion.
For the potential, however the leading term
would still go as 1 over R cubed and if you
do the calculation you will find that the
kinetic energy is equal to added mass times
half u square, where the added mass is some
other fraction not half could be some other
fraction of the volume of mass of that fluid
displaced by the object. So that gives us
the total kinetic energy of the flow.
What about the force exerted by the fluid
on the sphere or the sphere on the fluid the
equivalent of the drag force, that we had
calculated previously for viscous flows. In
other words to move this sphere within the
fluid what is the force that needs to be exerted
on the sphere? So let us look at that calculation
the force calculation. So I have a sphere
which is moving with a velocity U and I want
to calculate the net force exerted by the
fluid on the sphere and I have of course,
a solution for the velocity field due to the
for the potential as well as the velocity
field.
So that potential solution in the velocity
solution both satisfy the zero the normal
velocity boundary conditions at the surface
and I will just use that to calculate the
total force. So how do I calculate the total
force? I have to use the equation for the
stress f is f i acting at a surface is equal
to T ij n j the stress is purely due to pressure.
So it is equal to minus p delta ij n j is
equal to minus p n i. So this is the force
acting per unit area on the surface the total
force is calculated as integral over the surface
minus p times the unit normal, the pressure
of course, is given by the Bernoulli equation.
In this particular case we will neglect the
body force since because we will assume there
is no gravitational acceleration for the present
as I showed you in my Bernoulli equation p
plus half rho u i square plus rho partial
phi by partial t plus V is equal to p naught.
So provided the force is conservative I can
combine these two terms to define a new pressure
which I will call pi.
So these two terms can in general be combined,
the potential plus the fluid pressure can
be combined to get pressure which incorporates
the body force terms, for this particular
calculation we will assume for the present
that the potential is identically equal to
0. So with that p is equal to minus p naught
minus half rho u i square minus rho partial
phi by partial t, a constant integrated so
if, so there are three parts here, one is
the constant this is the kinetic energy and
this is the acceleration this is the acceleration
term.
If I take this constant pressure and integrated
over the surface I will get 0 because if I
take p naught times n i and p not is a constant
there is equal force is acting in all directions
on the surface and therefore, the net force
due to the pressure will end up being equal
to 0 because there is a constant acting on
all sides of the surface therefore, the net
force exerted will be 0. We only non-zero
contributions you will get are due to the
velocity and the time derivative of the potential
due to minus half u rho u square and rho times
d phi by d t.
So the velocity of course, we have we know
what the velocity is and we can integrate
this out over the surface minus half rho u
i square integrate over the surface which
says that kinetic energy density integrate
over that surface. What about the time derivative
of the potential? Here is where one has to
be careful in going through the calculation,
the reason is as follows my sphere is moving
with a constant velocity u. So I had got my
solution for the potential as phi is equal
to minus R cubed by 2 u j x j by r cubed of
course, the velocity u could be depend upon
time and therefore, you could get a time dependence
in the equation for the pressure because d
phi by d t there could be a contribution due
to the velocity itself.
So I could have a contribution of the form
d phi by d t is equal to minus R cubed x j
by 2 r cubed times d u by d t. However if
we had a steady velocity is the time derivative
of the potential equal to 0 we had a steady
velocity is the time derivative of the potential
equal to 0 or if d u d t is equal to 0 does
it mean that partial phi over with respect
to t is equal to 0 turns out it is not.
And the reason is as follows, the reason it
is not is because I have solved the problem
in a coordinate system with origin fixed at
the center of the sphere, my origin of the
coordinate system was fixed at the center
of the sphere and the sphere itself is moving
with a constant velocity, this sphere itself
is moving with a constant velocity. So therefore,
if I look after sometime delta t if I look
after sometime delta t this sphere would be
at a new position after sometime delta t this
sphere would be at a new position and the
origin of the coordinate system would also
be at a new position, it would at x is equal
to x naught plus u delta t.
So the origin of the coordinate system is
also moving when this sphere moves and I have
calculated the potential in this moving coordinate
system. What is partial phi by partial t?
Partial phi by partial t is the potential
the variation potential at a fixed observation
if I look at one particular location x.
I call this is x naught right? I look at one
particular location x and find out what is
the difference in potential between the time
t and the time t plus delta t divided by delta
t, that is my partial phi by partial t at
that location. If the sphere velocity is a
constant is that non-zero? Of course,it is
not because even though I am sitting at a
particular location the sphere has moved therefore,
the vector distance between the center of
this sphere and the observation location has
moved. Even though I am sitting at one particular
location because my solution is in a coordinate
system whose origin at the center of this
sphere, the origin of the coordinate system
has moved the observation point has not moved
but, because the origin of the coordinate
system has moved there is going to be a variation
in the potential at their that fixed location.
So that second term has to be taken into account
when we calculate the time derivative of the
potential, when we are doing the calculation
in a reference frame where there is a moving
particle. How much is that difference in potential?
That can be shown by a simple construction.
Let us say that my original location was x
naught and this sphere moved to a new location
x naught prime, where this displacement was
delta x naught. My observation point is at
the location x, so vector at initial time
was is equal to this one, the radius vector
after the time delta t when this origin of
the coordinate system has moved a distance
delta x naught is this one. So the change
in radius vector is basically equal to the
final minus the initial, the change in radius
vector is equal to the final minus the initial,
that is this is a change in the radius vector
the final one minus the I should I should
be careful here that is actually this one.
So this is the final and by initial had a
larger angle of inclination. So my initial
one had and something like this and this was
the change in the radius vector, that same
change in radius vector can be obtained. So
I am getting a change in radius vector, if
I move the point x naught by delta x naught
keeping x stationary, that same change in
radius vector can be obtained if I keep x
naught stationary and move x by distance minus
delta x naught.
That same change can be obtained if I keep
x stationary and move the observation point
by a distance minus delta x naught. You can
see that this modified radius vector is identical
to this one that I had. So what I get by moving
the origin by a distance plus delta x naught
is identical to what I get by moving the observation
point by a distance minus delta x naught.
So therefore, the change in potential is equal
to the change in displacement times the gradient
of the potential.
So what is the change in potential between
this observation point and the observation
point after I get after moving that distance,
that change in potential is going to be equal
to delta phi is equal to minus delta x naught
the distance moved times grad phi, that is
delta phi change in potential between these
two locations is equal to the displacement
which is minus delta x naught times grad phi
that is taking place in a time delta t.
Therefore, the rate of change of potential
which is delta phi by delta t is equal to
minus delta x naught by delta t times grad
phi, delta x naught by delta t is just the
velocity of the sphere because it moves a
distance this velocity this sphere moves a
distance delta x naught. So delta x naught
by delta t is just the velocity of the sphere.
You note that this grad x so therefore, this
becomes minus the velocity of the sphere and
grad phi grad phi is just the fluid velocity
because I said that the velocity can be written
as the gradient of a potential.
So this becomes minus the sphere velocity
dotted with the fluid velocity, this will
be the additional contribution whenever you
are working in a reference frame in which
the origin of the reference frame is moving
in time. So I get this additional contribution
to partial phi by partial t minus U dot U
or minus capital U i dotted with small u i.
If you recall even when we did viscous flows
we worked in a coordinate system which was
fixed at the origin of the sphere in that
case of course, the flow was quasi steady,
you had neglected the time derivatives and
therefore, we could always calculate the stresses
because we are neglected the time derivative
terms anywhere.
In this particular case there is a time derivative
of the potential that enters into the calculation
and when you have a time derivative in a reference
frame that is moving, you have this additional
term which comes in to the time derivative
of the potential that has to be included.
So therefore, if I calculate the pressure
in this reference frame, calculate the pressure
using my potential which was calculated in
a moving reference frame I get minus half
rho u i square minus R cubed x j by 2 I should
have a density there plus rho U j times u
j.
So this first term here is due to the variation
of the velocity with respect to time, the
second term was because the coordinate system
the origin of the coordinate system is moving.
So for a steady flow for a steady flow the
pressure is just equal to p is equal to p
naught minus half rho u i square plus rho
U i times u i and of course, the net force
is obtained as the F i is equal to integral
ds of the unit normal times p naught minus
half rho u j square plus rho.
So this is the net force that is exerted on
the sphere. Now with this expression of the
net force of course, one can calculate of
the net force by actually doing the integral
by actually taking the solutions for the velocity
of the sphere and doing this integral in order
to find out what is a net force exerted on
the sphere? However it is easy to show just
using symmetries that this net force has to
be equal to 0 the solution is as follows.
If I have a sphere if I have a sphere which
is moving in the direction capital U the velocity
the velocity of the flow around this sphere
if I look at the velocity of the flow around
the sphere, the velocity of this sphere it
has to satisfy the normal velocity boundary
condition. So; that means, this is the normal
velocity of the sphere has to be equal to
the normal velocity of the fluid at the surface.
And I will get a normal velocity which looks
something like this, because the sphere is
moving in this direction. The normal component
of the velocity will be along the plus along
the along the axis in this direction and along
the axis in this direction. You do have a
tangential velocity of course, but, you would
expect based up on symmetry that tangential
velocity is symmetric about this axis. So
if I had a tangential velocity that looks
something like this over here just based up
on symmetry you would expect that the same
thing is there on the other side as well.
So would expect a tangential velocity to come
down along this axis and go up along that
axis, the point is that the magnitudes of
these two velocities have to be the same;
that means, that if I take two positions which
are at an angle theta here and the same angle
theta here can I take two positions which
are angle theta with respect to the plus velocity
axis same angle theta with respect to the
minus velocity axis u square on these two
sides is exactly the same. So u square here
is the same as u square here.
In addition u dot n is also the same because
the normal velocity boundary condition, I
required that small u dot n is equal to capital
U dot n n i am sorry capital U dot n along
the surface along, the front surface is the
component this component along the front surface.
So when we are on the rear surface the component
is in the same direction. So small u dot capital
U small u dot capital U is exactly the same
on these two points on the surface.
So the velocity square is the same on these
two points the velocity square is the same
on these two points just drawn symmetry u
dot n is also the same on these two points
just drawn symmetry. If you look at the net
force there are two components, one is along
capital U here there is perpendicular to capital
U. So let us look at the net force along the
direction of capital U. So it is equal to
half u square plus u dot u times the unique
normal, this unit normal is the outward unit
normal. So that outward unit normal at the
point on the top stream side it is along this
direction the downstream side it is along
this direction.
The component of the unit normal along the
u direction on the upstream side it is here,
that is on the down streamside it is pointing
backwards, the velocity square and u dot n
are both the same; that means, that the pressure
on those two sides is the same, unit normal
is pointing in or the component of unit normal
along u is pointing in opposite directions;
that means, that at these two points unit
normal times pressure is exactly equivalent
magnitude in opposite in direction.
Therefore, the net force due to these two
points is equal to 0, the same holds for any
point on the circle at an angle theta on the
sphere, there is an equivalent point which
makes an angle with the downstream side that
same angle theta, u square as well as u dot
capital U are exactly the same on those two
points. The component of the unit normal along
the flow direction is opposite on those two
therefore, the net force just one symmetry
has to be equal to 0. So this basically tells
us that for a sphere that is translating at
a constant velocity in a fluid the net force
is identically equal to 0.
The net force along the direction of u is
equal to 0 and because the of the axis symmetry
you can see that the net force perpendicular
to u also has to be equal to 0 because the
flows perfectly axis symmetric around this
axis. So for a sphere moving at a constant
velocity we find the result that the net force
is equal to 0, it goes by the name of the
d Alembert’s paradox. The net force on a
sphere which is moving at constant velocity
has to be identically equal to 0, this is
true not just for a sphere.
This is true for any object. I will show you
that in the next lecture that for any object
that you take in three dimensions the net
force exerted by the object on the fluid at
constant velocity in potential flow is equal
to 0, the reason is because in the potential
flow equations we have neglected the viscous
terms. The viscous terms are responsible for
energy dissipation and it is because you need
to compensate for the energy dissipation that
we need to do work in order to in order to
move the object.
Because the work done due to the motion of
the object and due to the force exerted on
the object has to balance the dissipation
of energy within the fluid. In potential flow
we have neglected energy dissipation therefore,
there should be no net work done for moving
an object at constant velocity because we
are moving at a constant velocity the kinetic
energy remain of the fluid remains the same.
We have neglected viscous dissipation and
if the kinetic energy remains the same; that
means, that you do not need do work to move
it at constant velocity of course, if the
particle is accelerating you do need to work
in order to accelerate the particle in the
fluid and the work required for acceleration
basically comes about by incorporating the
time derivative term. So the time derivative
term F i for doing for doing work due to acceleration
is an integral over the surface of if you
recall this is equal to the unit normal times
the pressure which is minus rho d phi by d
t for an accelerating object where d u d t
is not is equal to 0.
So if you recall for a steady flow we had
neglected this term in the equation for the
pressure and we showed that the net force
due to all the other terms is identically
equal to 0. I should remove the p naught here,
there should be p is equal to p. We showed
that the network due to all of these other
terms is identically equal to 0 therefore,
the net contribution that comes can come in
only due to this particular term.
So this equal to minus p is equal to integral
ds n i into R cubed by 2 d u j by d t x j
by r cube times rho and once again you can
do this integral quite easily. This is calculated
on the surface of the sphere and therefore,
this R cubed and this is calculated at r is
equal to capital R. So this will become integral
ds rho by 2 d u j by d t integral ds the unit
normal on the surface is equal to x i by r
times x j and once again we have integral
over the surface of x i x j 4 by 3 pi r power
4 delta ij and I had put that n it is quite
easy to see that this is just equal to the
added mass times d U j by d t.
Where the added mass is exactly what we had
got from our calculation of the kinetic energy
2 by 3 pi R cubed rho by just half the mass
of the fluid that is displaced by this sphere.
So therefore, the force for an accelerating
sphere is equal to the added mass times the
acceleration. The kinetic energy is the half
added mass times velocity square. So it is
all consistent with each other. So this added
mass for a sphere is 2 by 3 pi R cubed rho
one half of the mass of the fluid displaced
by this sphere of course, for objects of other
shapes will have a different added mass but,
one can be guaranteed that the added mass
due to obtained from the force exerted on
the sphere as well as from the kinetic energy
will end up being the same.
So for a sphere for a sphere moving in three
dimensions the net force exerted by the sphere
on the fluid is equal to 0. If it is accelerating
then there is a net force which is equal to
the added mass times the acceleration. What
about for objects of other shapes? Is the
added is the force required to be exerted
by that object also 0 under conditions of
steady flow. So that is the question that
you will first take up in the next lecture.
We will try to show that this is a general
result for an object of any shape moving in
any direction under potential flow conditions
the net force exerted by the object on the
fluid. If the velocity is steady that net
force has to be is equal to 0, both along
the velocity direction as well as perpendicular
to the velocity direction. Once you completed
that we will go on to analyzing two dimensional
potential flows, where we look only at variations
in two directions, there are certain simplifications
you can use there, you can use complex variables
to get simple solutions for these equations.
So that we will look at in the next lecture.
So we will continue our discussion of potential
flow in the next lecture. We will see you
then.
