Derivation of the Navier Stokes Equations Video Lecture | Fluid Mechanics for Mechanical Engineering

these are the navier-stokes equations as
they're commonly written in this
screencast we examine their physical
meaning and perform a simple
mathematical derivation based on
Newton's second law while these
equations may look intimidating and
complicated to a lot of people all they
really are is a statement that the sum
of forces is equal to the mass times the
acceleration to make it a little bit
more apparent let's flip the equations
about the equal sign so what we have in
the first equation is the sum of forces
in the x-direction is equal to the mass
times the acceleration in the
x-direction in the second and third
equations of the some forces in the Y
and Z Direction equal to their
respective accelerations these equations
are written for a differential element
of fluid which is infinitesimally small
so as small as we can possibly imagine
the three forces we're concerned with of
forces due to gravity forces due to
differences in pressure and forces due
to the viscosity of the fluid keep in
mind that each of these terms is on a
per unit volume basis so typically we
would say the force due to gravity might
be the weight of something if I took the
weight and divided by the volume what
I'm left with is the density M over V
times gravity we see this occurring for
gravity most explicitly on the
right-hand side of the equation if we
have mass times acceleration if we were
to divide that by the volume of the
fluid we would be left with the density
times the acceleration so we're looking
at the sum of forces is equal to MA on a
per unit volume basis for this
screencast let's deal only with the
x-direction the same mathematics would
apply for the Y in the Z directions but
we'll leave it with two the X direction
to save time let's examine the
right-hand side of the X component the X
component of velocity for a fluid will
call lowercase U and strictly speaking
you could be a function of X Y Z and
time the X component of acceleration for
the fluid is equal to the time
derivative of U is a derivative of U
with respect to time but because U is
not simply a function of time it's also
a function of XY and Z we need to use
the chain rule to perform this
differentiation
so I've got do the partial you with
respect to time plus D u DX times DX DT
Plus D u dy dy DT Plus D u DZ times DZ
DT if I use the definition that DX DT
simply equal to u dy DT is equal to
lowercase v + DZ DT is equal to
lowercase W we're left with something
that looks an awful lot like the right
hand side of the equation above the
first term on the right hand side is
known as a local acceleration and the
remaining three terms are known as the
convective acceleration to think about
what this means physically let's
consider some fluid that's flowing
steadily from left to right through this
two-dimensional constriction and let's
consider a differential element of fluid
which will be infinitesimally small so
I'll make it a real small cube and let's
place it right here to begin with if you
think about the motion of this element
to fluid as it flows through it's
flowing steadily from left to right it's
slow in this region but then it begins
to accelerate because of the
constriction where the fluid is moving
very rapidly here and now when it
reaches the right hand side it slows
down again and it recovers to a steady
velocity as it moves towards the exit if
the flow of fluid is at steady state
then the velocity of the differential
element to fluid at points one two and
three we would see no change over time
but let's examine the constricting
region we've got U is a positive
quantity it's moving from left to right
and D U DX is also a positive quantity
so this term of the convective
acceleration is greater than zero so we
see in the highlighted region that the
fluid element is accelerating from left
to right conversely in the expanding
region although U is positive D u DX is
less than zero it's a negative quantity
the fluid is slowing down within that
region so the convective acceleration
term is less than zero or it would be
the acceleration would be to the left in
the highlighted read
let's examine the forces acting on the
differential element to flu it a little
bit more carefully call this point XY Z
in our differential element to flu it
has length DX a height dy and a depth DZ
the first force we'll consider as
gravity and typically when you draw a
free body diagram gravity will be acting
downward in the Y direction but let's do
an arbitrary case where a component of
gravity for example could act in the X
direction so the force due to gravity is
the mass of our differential element two
fluid times the X component of gravity
let's rewrite the mass is equal to the
density times the volume of our
differential element two fluid DX dy DZ
again have the mass times gravity in the
X direction
let's examine forces acting on the left
and the right sides of our differential
element we could have a normal stress
acting directly to the right on the
right face and a stress acting to the
left outward from the left face the
notation we'll use for these stresses is
Sigma X X and we're going to evaluate
Sigma xx at X plus a distance DX and on
the left face we have Sigma X X
evaluated at X on the top and the bottom
faces we could have a shear stress
acting to the right on the top face and
a shear stress acting to the left on the
bottom face the notation we'll use for
these stresses is tau YX evaluated at y
plus dy for the top of the cube and tau
YX evaluated at Y for the bottom of the
cube and additionally we could have
stresses acting to the right on the
front of the cube and the stress acting
to the left at the back of the cube for
the back of the cube we'll use the
notation tau ZX evaluated at Z and at
the front of the queue we'll use the
notation tau ZX evaluated at Z plus DZ
so continue writing the some forces in
x-direction I've got any X component of
gravity plus a normal stress Sigma X X
evaluated at X plus DX multiplied by the
surface area of the right side of the
cube which is equal to dy DZ because it
acts to the left I'll subtract Sigma X X
evaluated at X multiplied by the same
area dy DZ
then I'll add the force due to the shear
stress at the top of the cube tau YX
evaluated at y plus dy multiplied by its
area which is DX times DC I'll subtract
the shear stress acting on the bottom
face multiplied by its area then I'll
add the force due to the shear stress at
the front of the face and subtract off
the shear stress acting at the rear face
the sum of these forces will equal the
mass times the acceleration in the X
direction or you could write mass is
equal to Rho times DX dy DZ times the
acceleration in the X direction you have
cleaned up that equation divided by the
volume of the differential element and
what I immediately see is that DX dy
disease will cancel out and in some
terms if I simplify a dy and DZ will
cancel out DX cancels out as well as DZ
in this term and so forth as I can't
cancel out terms as I continue to
simplify am left in the neck with this
expression and in the limit of DX dy and
DZ are approaching zero this turns into
a differential form and the resulting
equation represents the sum of forces in
the x-direction due to gravity due to
the normal forces acting on the left and
the right side of the differential
element the shear stresses acting on the
top and the bottom of the element and
the shear stresses acting in the front
and the rear faces of the element so
these are equal to the density times the
acceleration of the differential element
in the x-direction and if I expand the
raishin into its local and convective
components were left with the an
expression on the right hand side if I
did the same analysis for the Y and the
Z directions I would come up with these
three equations which are known as the
equations of motion for a fluid but to
get from these equations to the
navier-stokes equations we need a way to
relate the normal and the shear stresses
to the viscosity of the fluid and the
velocity profiles and this is done using
these equations which are the
constitutive relations for a Newtonian
fluid which we won't get into here but
if we accept them as being true for this
screencast it becomes a series of
algebraic manipulations to arrive at the
navier-stokes equation the first
substitution into the left hand side
gives me the expression at the bottom if
I differentiate the three terms I'm left
with this expression and some additional
manipulations leaves me with this
expression in which I've split this term
into two parts and I've switched the
order in which I differentiate these two
terms so I'm switching dy/dx and DD and
this expression as I continue to
simplify and rearrange terms I'm left
with this expression but what's
interesting is the sum of these three
terms D u DX plus DV dy plus DW DZ is
equal to zero by way of the continuity
equation so that whole term on the right
is identically zero for an
incompressible fluid so what I'm left
with the sum has three forces the force
due to gravity force due to any pressure
differences in the fluid plus all the
forces due to viscosity the sum of these
three is equal to the density of the
fluid multiplied by the X component of
its acceleration and expanding a X out
into its local and convective components
of acceleration I just arrived at the
first of the X component for the
navier-stokes equation so we could do
the exact same thing for the Y and the Z
directions it will arrive at the
navier-stokes equations for those
directions as well if I flip the order
of the equations you're left with the
navier-stokes equations as you'll
commonly see the
and although they may look intimidating
and complicated to begin with if someone
asks you to sum up what the
navier-stokes equations are in words
just simply tell them they're an
expression of the sum of forces is equal
to the mass times the acceleration