
English: 
Hello welcome to my talk, All about Fluids. This talk is on an interesting and
important topic: 'Is a Navier-Stokes equation correct?'
We all know very well that Navier-Stokes equation is correct as the
universal governing equation for fluid motion since numerous examples have
confirmed it. However some recent works have shown
there are some inconsistencies behind Navier-Stokes equation.
for instance, the Cauchy's symmetric stress tensor has been fully adopted to
fluid nearly 200 years ago. But it is required for fluids?

English: 
Hello welcome to my talk All about Fluids.  this talk is on an interesting and
important topic: 'Is a navier-stokes equation correct?'
We all know very well that navier-stokes equation is correct as the
universal governing equation for fluid motion since numerous examples have
confirmed it. However some recent works have shown
there are some inconsistencies behind navier-stokes equation.
for instance the Cauchy's symmetric stress tensor has been fully adopted to
fluid nearly 200 years ago. But it is required for fluids?

English: 
or it is correct for fluid?  In this talk I will show you what are the inconsistencies
behind the Navier-Stokes equation and the examples for illustrating the
inconsistencies including some from the textbooks.
Is Navier-Stokes equation correct?
the answer for this is YES and NO. For YES, that's because the
Navier-Stokes equation as the governing equation for flow motion has been
proven by numerous examples experimental data, analytical solutions and the
numerical simulations. For NO, that's because there are still some
inconsistencies behind Navier-Stokes equation, specifically on the symmetric

English: 
or it is correct for fluid?  In this talk I will show you what are the inconsistencies
behind the navier-stokes equation and the examples for illustrating the
inconsistencies including some from the textbooks.
Is Navier-Stokes equation correct?
the answer for this is YES and NO. For YES, that's because the
navier-stokes equation as the governing equation for flow motion has been
proven by numerous examples experimental data, analytical solutions and the
numerical simulations. For NO, that's because there are still some
inconsistencies behind navier-stokes equation, specifically on the symmetric

English: 
stress tensor for fluid. It is well known the Cauchy symmetrical
stress tensor was originally developed for the elastic material, the solid and
then fully adopted to the fluid by Cauchy himself
but how we can extend exceptions for the asymmetric stress tensors found in real fluids?
Is this Cauchy's law for angular motion also a pre-requisite for fluids?
can we have a different way to derive Navier-Stokes equation?
In this talk I will answer all these questions.
First let us look at a simple classical flow
the Couette flow. The Couette flow is confined between two large parallel

English: 
stress tensor for fluid. It is well known the Cauchy symmetrical
stress tensor was originally developed for the elastic material, the solid and
then fully adopted to the fluid by Cauchy himself
but how we can extend exceptions for the asymmetric stress tensors found in real fluids?
Is this Cauchy's law for angular motion also a pre-requisite for fluids?
can we have a different way to derive navier-stokes equation?
In this talk I will answer all these questions.
First let us look at a simple classical flow
the Couette flow. The Couette flow is confined between two large parallel

English: 
plates. The lower plate is fixed and upper plate moves at a constant speed at a
distance h.  the move speed of the upper plate is assumed small,  so the flow
confined between these two plates is laminar. So based on the analysis of the
problem, we can see the velocity components v and w are zero,
and u is the function of y only. As such the navier-stokes equation can be
simplified to this simple partial differential equation. Apply the boundary
condition we found the solution as this. This is a very simple flow: from the plot
we can see the fluid velocity increases linearly with y.

English: 
plates. The lower plate is fixed and upper plate moves at a constant speed at a
distance h.  the move speed of the upper plate is assumed small, so the flow
confined between these two plates is laminar. So based on the analysis of the
problem, we can see the velocity components v and w are zero,
and u is the function of y only. As such the Navier-Stokes equation can be
simplified to this simple partial differential equation. Apply the boundary
condition we found the solution as this. This is a very simple flow: from the plot
we can see the fluid velocity increases linearly with y.

English: 
In this slide, we are going to look at the shear stress
in the classical Couette flow. From the Cauchy's symmetric stress tensor in the classical Couette flow,
we can easily calculate the horizontal shear stress TAU_12 as this.
And from the requirement of the symmetrical stress tensor,
a vertical shear stress,
TAU_21 is calculated like this. It can be seen both horizontal and vertical
shear stresses are non-zero. If we look at this, based on the symmetric stress tensor,
We have TAU_12= TAU_21, and TAU'_12 = TAU'_21.
we will go back to the definition of fluid.

English: 
In this slide, we are going to look at the shear stress
in the classical Couette flow. From the Cauchy's symmetric stress tensor in the classical Couette flow,
we can easily calculated the horizontal
shear stress TAU_12 as this.
And from the requirement of the symmetrical stress tensor,
a vertical shear stress
TAU_21 is calculated like this. It can be seen both horizontal and vertical
shear stresses are non-zero. If we look at this, based on the symmetric stress tensor,
We have TAU_12= TAU_21, and TAU'_12 = TAU'_21.
we will go back to the definition of fluid.

English: 
From the fluid definition, it states: a fluid cannot bear any shear stress at rest and
under a shear stress the fluid flows in the direction of the applied stress.
From the definition, it is easy to understand the horizontal shear stress TAU_12
because we have the horizontal flow in the Couette flow, but for symmetrical stress
tensor, there must be a nonzero vertical shear stress Tau_21. However,
why this vertical shear stress does not cause a vertical flow?
if from the fluid definition
in the Couette flow, it can be concluded vertical shear stress TAU_21 is 0.  The shear
stress tensor in this case would be asymmetric

English: 
From the fluid definition, it states: a fluid cannot bear any shear stress at rest and
under a shear stress the fluid flows in the direction of the applied stress.
From the definition, it is easy to understand the horizontal shear stress TAU_12
because we have the horizontal flow in the Couette flow, but for symmetrical stress
tensor, there must be a nonzero vertical shear stress TAU_21. However,
why this vertical shear stress does not cause a vertical flow?
if from the fluid definition.
in the Couette flow, it can be concluded vertical shear stress TAU_21 is 0.  The shear
stress tensor in this case would be asymmetric

English: 
i.e., TAU_12 is not equal to TAU_21, since in the flow TAU_12 is non zero, but the
vertical shear stress tensor is zero.
there may be an argument in the Couette flow there is a vertical shear stress, it is
because the upper plate stops the flow moving in the vertical direction.
The answer for this is NO,  because in reality whether there will be a flow flowing out
from the opening totally depends on the pressure of the flow between the plates
if the flow speed is large enough, there might be a suction of the flow in the
opening, instead of the flow flowing out.
Look at this example, in this example we

English: 
i.e., TAU_12 is not equal to TAU_21, since in the flow TAU_12 is non zero, but the
vertical shear stress tensor is zero.
there may be an argument in the Couette flow there is a vertical shear stress, it is
because the upper plate stops the flow moving in the vertical direction.
The answer for this is NO, because in reality whether there will be a flow flowing out
from the opening totally depends on the pressure of the flow between the plates
if the flow speed is large enough, there might be a suction of the flow in the
opening, instead of the flow flowing out. Look at this example, in this example we

English: 
do have two openings the small vertical tubes. If there is a vertical stress TAU_21
in the pipe, then there must be vertical flows in the tubes, regardless of the
height of the tubes. Obviously this is not true
the height of the fluid in the tubes totally depends on the pressure of the
fluid in the pipe.
Now we look at a situation for a shear stress applied on an element as seen in the figure.
and we will see what will happen if the element is a solid and if the element is a fluid.

English: 
do have two openings the small vertical tubes. If there is a vertical stress TAU_21
in the pipe, then there must be vertical flows in the tubes, regardless of the
height of the tubes. Obviously, this is not true:
the height of the fluid in the tubes totally depends on the pressure of the
fluid in the pipe.
Now we look at a situation for a shear stress applied on an element as seen in the figure.
and we will see what will happen if the element is a solid and if the element is a fluid.

English: 
If the element there is an elastic material for instance a solid under the shear stress TAU_23,
there will be a deformation of the element
however due to the elasticity of the element,  there will be a shear stress
TAU_32  produced to balance the applied shear stress and TAU_32 equals
to TAU_23. This is required by the Cauchy's symmetric stress tensor.
and if this element is a fluid,  the applied shear stress will cause a flow of
element. From the definition of the fluid and because of the flow, a fluid

English: 
If the element there is an elastic material for instance a solid under the shear stress TAU_23,
there will be a deformation of the element
however due to the elasticity of the element, there will be a shear stress
TAU_32 produced to balance the applied shear stress and TAU_32 equals
to TAU_23. This is required by the Cauchy's symmetric stress tensor.
and if this element is a fluid, the applied shear stress will cause a flow of
element. From the definition of the fluid and because of the flow, a fluid

English: 
friction is induced to balance the applied shear stress. Hence in this case there
will be no shear stress pair to balance the applied stress in fluid.
here some examples for the inconsistencies of the symmetrical
stress tensor introduced. Kundu 's book 'Fluid Mechanics'
page 126, there is a statement: the stress tensor symmetry is violated
in the electrical field on polarized fluid molecules where anti-symmetric
stresses must be included in the analysis
Another example is the Wilcox's book 'Turbulence Modeling for CFD', on the page 39
there is a statement:  the symmetric stress tensor is the for simple viscous flows,

English: 
friction is induced to balance the applied shear stress. Hence in this case there
will be no shear stress pair to balance the applied stress in fluid.
here some examples for the inconsistencies of the symmetrical
stress tensor introduced. Kundu's book 'Fluid Mechanics'
page 126, there is a statement: the stress tensor symmetry is violated
in the electrical field on polarized fluid molecules where anti-symmetric
stresses must be included in the analysis.
Another example is the Wilcox's book 'Turbulence Modeling for CFD', on the page 39
there is a statement: the symmetric stress tensor is the for simple viscous flows,

English: 
but not for some anisotropic liquids. there's no reason (given) for
the statement. Therefore there will be a question if Cauchy's law of angular
motion is the universally correct for fluid, why exceptions we have as above.
We never see any exceptions for Newton's
laws of motion.
As we know from the Newton's definition of fluid friction, the fluid viscous
force is caused due to the different flow speeds at the two joint spatial
locations. Here we will take two neighboring points in the field: A and B.

English: 
but not for some anisotropic liquids. there's no reason (given) for
the statement. Therefore, there will be a question if Cauchy's law of angular
motion is the universally correct for fluid, why exceptions we have as above.
We never see any exceptions for Newton's laws of motion.
As we know from the Newton's definition of fluid friction, the fluid viscous
force is caused due to the different flow speeds at the two joint spatial
locations. Here we will take two neighbouring points in the field: A and B.

English: 
And the relative velocity at a certain time is calculated here.
For the point A, its position and velocity are given as this, and B the
neighboring point its position and the velocity can be simply expressed as the
position and the velocity of point A plus the small increments, such as
delta x , delta y , delta z; delta u, delta v and delta w. For the relative
increment it is given by this, the velocity vector increment can be
expressed as this and this, it is a dot product of the strain rate tensor
and the relative position vector delta r. Here we can see the strain rate

English: 
And the relative velocity at a certain time is calculated here.
For the point A, its position and velocity are given as this, and B the
neighbouring point its position and the velocity can be simply expressed as the
position and the velocity of point A plus the small increments, such as
delta x, delta y, delta z; delta u, delta v and delta w. For the relative
increment it is given by this, the velocity vector increment can be
expressed as this and this, it is a dot product of the strain rate tensor
and the relative position vector delta r. Here we can see the strain rate

English: 
tensor is not symmetrical. Based on the strain rate tensor, a new flow
friction tensor can be defined as this. this can be also called the fluid
stress tensor, but it is different from the normal symmetric stress tensor.
This tensor is based on the Newton's definition of the fluid friction.
it is from the strict physical analysis for the different velocity between two
neighbouring points in fluids. A symmetry for the stress tensor is not required here
from the stress tensor, the fluid stress force on a unit surface is

English: 
tensor is not symmetrical. Based on the strain rate tensor, a new flow
friction tensor can be defined as this . this can be also called the fluid
stress tensor, but it is different from the normal symmetric stress tensor.
This tensor is based on the Newton's definition of the fluid friction.
it is from the strict physical analysis for the different velocity between two
neighboring points in fluids. A symmetry for the stress tensor is not required here
from the stress tensor , the fluid stress force on a unit surface is

English: 
calculated as this. in a plot,  the stress force on a unit surface is given by this.
From the stress force on a unit surface, it  can be seen stress force in
x-direction Fx is caused due to the increment of the velocity component
u only and the increment can be in all x, y or z direction as seen in the
differentiations with regard to x, y and z. Similarly we have the stress
forces in y and z directions, Fy, Fz, are only related to the corresponding
velocity component increments of v and w.

English: 
calculated as this. in a plot, the stress force on a unit surface is given by this.
From the stress force on a unit surface, it can be seen stress force in
x-direction Fx is caused due to the increment of the velocity component
u only and the increment can be in all x, y or z direction as seen in the
differentiations with regard to x, y and z. Similarly, we have the stress
forces in y and z directions, Fy, Fz, are only related to the corresponding
velocity component increments of v and w.

English: 
Now we look at the special surfaces over cube S1, S2 and S3, on which the normal
vectors all have only one component, that is, on S1, n1 equals to 1 and n2, n3 are 0
so the force on the S1 has 3 components and same on S2 and on S3. so if we draw
this stress force components on the surfaces, we can have the drawing as this,
So from this drawing, it can be seen that the new stress tensor has a better physical
significance than that of the symmetric stress tensor. For example, we

English: 
Now we look at the special surfaces over cube S1, S2 and S3, on which the normal
vectors all have only one component, that is, on S1, n1 equals to 1 and n2, n3 are 0
so the force on the S1 has 3 components and same on S2 and on S3. so if we draw
this stress force components on the surfaces, we can have the drawing as this,
So from this drawing, it can be seen that the new stress tensor has a better physical
significance than that of the symmetric stress tensor. For example we

English: 
can see the stress force acts in x direction on the surface of a constant y
so you can see the x direction because of the velocity increment on the y surface here.
Now we look at the total stress tensor. this is very similar
to the symmetric stress tensor, the total stress tensor contains basically
three terms: the term from the pressure; the term for representing the fluid
the compressibility, LAMBDA is the second viscosity coefficient; and this is
the friction stress tensor, but here the friction stress tensor is not

English: 
can see the stress force acts in x direction on the surface of a constant y
so you can see the x direction because of the velocity increment on the y surface here.
Now we look at the total stress tensor. this is very similar
to the symmetric stress tensor, the total stress tensor contains basically
three terms: the term from the pressure; the term for representing the fluid
the compressibility, LAMBDA is the second viscosity coefficient; and this is
the friction stress tensor, but here the friction stress tensor is not

English: 
symmetric. And the divergence of the total tensor is given by this.
Now we look at the last three terms for each component in the divergence of
the total tensor. And also apply the newly defined stress tensor so we have the
equation like this, and this and this for velocity component u.
Similarly, you can calculate the last three terms in the second component of the
divergence of the total tensor, and the third term as this.
Now we consider the incompressible flows. So put all these together we have the momentum equations

English: 
symmetric. And the divergence of the total tensor is given by this.
Now we look at the last three terms for each component in the divergence of
the total tensor. And also apply the newly defined stress tensor so we have the
equation like this, and this and this for
velocity component u.
Similarly, you can calculate the last three terms in the second component of the
divergence of the total tensor,  and the third term as this.
Now we consider the incompressible flows. So put all these together we have the momentum equations

English: 
for the incompressible flow.  write this in a vector form, it is this and you can
see from here the substantial derivative is used in this way.
For the compressible flow the moment, the equation can be derived from the transport theorem,
with the assumption of the second viscosity coefficient LAMBDA as 1/3
of the fluid viscosity MU.  write this equation into the vector form,
so we have the momentum equation for the compressible flow. this is same equation as

English: 
for the incompressible flow.  write this in a vector form, it is this and you can
see from here the substantial derivative is used in this way.
For the compressible flow the moment, the equation can be derived from the transport theorem,
with the assumption of the second viscosity coefficient LAMBDA as 1/3
of the fluid viscosity MU.  write this equation into the vector form,
so we have the momentum equation for the compressible flow. this is same equation as

English: 
the momentum equation in the Navier-Stokes equation for compressible flow
We will discuss the symmetric and asymmetric shear stresses.
The asymmetric strain rate tensor can be divided into a symmetric and
antisymmetric strain rate tensor as this,
the first part the symmetrical strain rate tensor was actually
employed by Stokes in deriving the fluid dynamic equation: the component of
the strain rate symmetric. the second term 
is the anti- symmetric strain rate
tensor which corresponding to the rotation motion of the fluid, with the
rotating component Omega 1, Omega 2 and Omega 3, the vector Omega is half of the

English: 
the momentum equation in the Navier-Stokes equation for compressible flow
We will discuss the symmetric and asymmetric shear stresses.
The asymmetric strain rate tensor can be divided into a symmetric and
antisymmetric strain rate tensor as this,
the first part the symmetrical strain rate tensor was actually
employed by Stokes in deriving the fluid dynamic equation: the component of
the strain rate symmetric. the second term is the anti- symmetric strain rate
tensor which corresponding to the rotation motion of the fluid, with the
rotating component Omega 1, Omega 2 and Omega 3, the vector Omega is half of the

English: 
curl of velocity vector at this.  it can be understood in the original
Navier-Stokes equation, the rotational motion has been excluded in deriving
the fluid dynamic equation: the rotation motion might not cause the shear stress
in solid, but this is not correct for fluids. We will see this in the next slide
In this slide, a discussion will be made for the difference between the solids and
fluids regarding the pure rotational motion whether they should be included in the stress tensor.
Now imagine a rotating circular cylinder as in the drawing.
if the rotating cylinder is a solid and

English: 
curl of velocity vector at this.  it can be understood in the original
navier-stokes equation,  the rotational motion has been excluded in deriving
the fluid dynamic equation: the rotation motion might not cause the shear stress
in solid, but this is not correct for fluids. We will see this in the next slide
In this slide, a discussion will be made for 
the difference between the solids and
fluids regarding the pure rotational motion whether they should be included in the stress tensor.
Now imagine a rotating circular cylinder as in the drawing.
if the rotating cylinder is a solid and

English: 
the cylinder rotates around its center axis. Obviously, there will be
no shear stress acting on the solid. Therefore, it is reasonable if the rotational
motion is excluded in calculating the shear stress in solid. However, if the
rotating cylinder is full of fluid, and rotates around its central axis,
this would be a special Taylor-Couette flow for which R1 equals to 0 and Omega 1 equals
to Omega 2. The flow in the cylinder will eventually
rotates with the cylinder, since the different velocities exist between
two different radial points as seen here, u1 and u2.

English: 
the cylinder rotates around its center axis.  obviously there will be
no shear stress acting on the solid. Therefore it is reasonable if the rotational
motion is excluded in calculating the shear stress in solid. However if the
rotating cylinder is full of fluid, and rotates around its central axis,
this would be a special Taylor-Couette flow for which R1 equals to 0 and Omega 1 equals
to Omega 2. The flow in the cylinder will eventually
rotates with the cylinder,  since the different velocities exist between
two different radial points as seen here, u1 and u2.

English: 
therefore there must exist a fluid friction in the rotating flow.
In this regard the exclusion of the shear stress due to the pure rotation motion in the
fluid it's not correct.

English: 
Therefore, there must exist a fluid friction in the rotating flow.
In this regard the exclusion of the shear stress due to the pure rotation motion in the
fluid it's not correct.
