in this problem we are playing with a
very important equation in fluid
mechanics the navier-stokes equation and
the game that we're playing is that we
are given a fluid flow with a two-component steady flow, a two-component
velocity flow and we are looking for
pressure and this is a pretext to
articulate a little bit the mathematical
mechanics of this physical law there's
no particular meaning in this fluid flow
here that we have if a flow looked like
this we wouldn't even know certainly we
would not calculate pressure based on a
velocity field like this but it's just a
good opportunity to play with with those so
do not focus on the on a meaning of this
field on the top left here but instead
try to focus on what the navier-stokes
equation tells us and does not tell us
about the nature of the fluid flow so
what do we have the navier-stokes equation
is a mass sorry a momentum balance
equation and so what it tells us is that
mass times acceleration is the sum of
forces mass times acceleration in a
fluid flow we write it like this we
write it as Rho times the total time
derivative of velocity like this perhaps
let me write it on the left so we have
more space so like this Rho times the
total time derivative of velocity this
is the acceleration field this is equal
to the sum of three contributors the
contributors are gravity rho g
like this the negative of the gradient of
pressure which is - grad P like
this and then the contribution of shear
and shear is the viscosity multiplied by
the Laplacian of velocity so the second
derivative with respect to space of
velocity this is a three-dimensional
equation in this case we already have
two dimensions on the top so we're going
to split this in only two components yes
so let's write out those two components
for u and V according to that equation
this is for U something that
looks like this density multiplied by
the total time derivative of the X
component of velocity and so this is the
partial derivative with respect to time
of U Plus u multiplied by the partial
derivative with respect to X of U plus V
multiplied by the partial derivative
with respect to Y of U and this is equal
to the sum of three contributions one is
the component of gravity in the X
direction second is minus the partial
derivative of P with respect to X and
third is going to be the laplacian of
the x component of velocity multiplied by
viscosity and the laplacian is the sum
of the second derivative of U with
respect to X plus the second derivative
of U with respect to Y like this there
would be a third component here and also
a third component there if we had three
dimensions but in this case there are
only two and this equation has an
equivalent in the Y direction which is
like this let's write it out
whoop should look like this we have
partial V with respect to T Plus u
partial V over partial X plus V partial
V over partial Y like this yeah this is
equal to Rho G Y minus partial P over
partial Y Plus mu multiplied by
the second derivative with respect to X
plus the second derivative with respect
to Y of V all right so what does that tell
us about pressure where pressure here is
hidden inside two terms one is here and
the other is there how are we going to
try to do is to isolate those two terms
as a function of all the rest in which
we're going to plug in U and V which are
on top here and based on this we are
going to try to find out what P could be
so let's do that let's let's write out
what this could be
I'm gonna number those equations and I'm
gonna call this first equation here 1
and the second equation here 2 and I'm
also going to write on the side if I
have a little bit of space perhaps let's
do it here I'm going to write on the
side the conditions that we have on the
top so if I move this up a little bit
hmm not so much space left we have here
we have U is equal to 8x plus B and V is
equal to minus a y plus CX okay let's
take equation number one top equation
here and let's try to isolate P bar and
start to fill it in with what the data
that we have let me bring it down a
little bit again so we can see the whole
equation like so okay so I think this
first equation equation number one and
let's apply the special condition of
having U which is there so equation one
becomes this let's go let's go
Rho times the derivative of U with
respect to time here is going to be zero
you multiply by the derivative of U with
respect to X it's going to be ax
plus B multiplied by the derivative of U
with respect to X this is going to be a
yes then we have V multiplied by the
derivative of U with respect to Y but if
I derivate this with respect to Y I get
zero so this whole expression is going
to be zero
this is equal to Rho GX in our case in
this particular problem we have no
gravity so we just ignore G and we have
just right away
- grad P of X component of grad P so
partial P partial X with P here being
our unknown and then we have mu here
with second derivative of U with respect
to X I derivated once here with respect
to X to get a and a derivate a with
respect to X to get just zero so this is
going to be zero and the second second
time I'm going to derivate this with
respect to Y two times to also get zero
like so so this simplifies a whole a
whole lot and if I bring it up like so
we should get something like this I'm
going to take this partial P partial X
and move it to the other side so I have
something positive partial P partial X
like this is equal and I'm putting this
to the other side is equal to Rho times
a squared X plus a B like this and all
the rest goes to zero and you check that
I'm not tripping over my own feet yes
this should be fine
like this okay so this is equation one
and then we're going to take equation
two and do the same thing
so right now concerning ourselves with y
component here and then this is y
component we also have the X component
of velocity appearing a couple of times
so let's let's take a good look
equation let me perhaps try to play a
trick well with this cool software and try to
move this down yes so we have a bit more
space to play with equation two like
this yes this is gonna cool so let's
take V here and apply equation two let's
go we have Rho times the change in time
of this velocity here component this is
zero plus u multiplied by the derivative
of this with respect to X so that's going
to be e X plus B this is U and I
multiply this by the partial derivative
of this with respect to X is going to be
just C plus V here
multiplied by the derivative of this
with respect to Y so that's going to be
V yes
- a y plus C X here and I multiply this
by the derivative of V with respect to Y
which is going to be minus A yes okay
and this is equal to gravity is zero in
our case minus the partial derivative of
P with respect to y plus mu x I'm going
to derivate this the expression with
respect to X 2 times the first time I
get just C and the second time I get
zero and then when I derivate with
respect to Y two times the first time I
get minus a and the second time I also
get zero
very convenient and so when we isolate
now this partial P partial Y now we'll
get something like this we get partial P
partial y is equal to minus this whole
term here so I have minus rho x let's
write it out let's write it out let's
see we have AC X plus BC - it's going to
be a plus a squared Y and then minus AC
X - X get in here it's a funny
coincidence almost like this exercise
was engineered to look like this and we
have the ACX going away so that we get
partial T over partial Y is equal to a
minus Rho times BC plus A squared Y okay
so let's take a look now that's the
result that we have we have one
equation which tells us how P changes
with Y like so and we have one equation
that tells us how P changes with X our
mission now is to take both of those
hints as to how P could be and put them
up to one complete model equation now
what we're gonna do is we're going to
integrate we're going to integrate both
of those equations in a very special way
there's a trap
when we do this and the trap is to not
check that those two equations which I'm
gonna number here it's going to be
equation 3 and it's going to be equation 4
the trap is to not check that both
of those equations are compatible with
one another and so the step that we need
do is to make sure that there exists a
pressure
that matches both of those conditions it
could well be that the derivative of
P with respect to Y
specifies conditions that make it so that
it's impossible that the derivative of P
with respect to X matches equation 4 and
a way to check that this is true the way
to check that the continuous function
for P exists is to go like this so I
like to have a warning sign
over here like so and we check that a function P exists
the way to check is to take the
the derivative with
respect to Y of the derivative with
respect to X of P and this must must be
equal to the derivative with respect to
X of the derivative with respect to Y of
P like this and let's do this it's very
simple i derivate this function here
with respect to x there's no X present
in this equation so this is going to be
0 yeah derivate this with respect to Y
there is no function of Y in here so
it's all going to be 0 so I can just
make a big satisfying check like this
yes right verified because both are
equal to 0 this is a necessary
step if we don't do that we could end up
with the very uncomfortable result that
based on which equation we begin the
rest of the problem with we get
different fields for pressure that are
completely incompatible one with the
other
so now now that we make sure that those
who are compatible with one another
we're going to try to draw to find
out what the equation for P is so let's
take equation 3 yeah let's try to
integrate this equation with respect to
Y to obtain a function
for P actually I'm going to change my
mind I'm going to take equation 4
because I can keep it on the screen and
I think like this no no let's go to for
equation 3 okay I have a few issues with
space management's on the screen let's
do it like this
so let me copy equation 3
3 over here
equation 3 tells us that partial P over
partial Y is equal to minus Rho times BC
plus a squared Y I think I mean let me
just check BC plus a squared y yes so
this is equation 3 and based on this
equation 3 we're going to find out what
P is so what do we do we integrate so
let's integrate P with respect to Y to
get a function so we get here minus rho
x b c y plus a squared times one-half of
Y squared and you know what comes after
that it's a big plus very important here
if we had an integration where we have a
complete derivative we just have here a
starting point arbitrary constant behind
us but here is a partial derivative this
is not a complete derivative it's only
the derivative with respect to Y and we
know in the back of our mind that P is a
function of X Y and T so we need instead
of this plus that
here I'm going to erase we need to have
something that is a function of all the
rest that we did not take into account
and this function here will be a
function whoop
this is too thick will be a function of
all the rest so we'll then call this a
function f and this function f is a
function of what we did not have in
there which is X and T some people like
to say that this pure function of X and
T that does not have any constant and
like to add over here a p0 a starting
point some people say that just having a
function like this is enough and it's it
already is this p0 the starting
constant is integrated it's included
already inside this function I'm of that
opinion too so I would not put a p0 over
here but keep in mind that inside this
function there may be it's just a
starting point a basic starting pressure
that is the same everywhere ok this is
highly unsatisfactory we don't want to
have a function of X left over over here
because we know something about the
relation between pressure and X and we
know that because of equation number 4
that we had over here we know that the
derivative of P with respect to X is
equal to that and so what are we going
to do with this thing that we integrated
here we had an integration what do we do
with this to find out what this function
is we're gonna derivate it this is
really funny and so we're gonna take the
derivative of P with respect to X so
that we can find out what is missing bit
here this function of X is because we
know what the derivative of P with
respect to X is according to equation 4
yes so how we do here yes now we let me
change again to this turquoise color
here
and then we equate it with equation
equation number four which is over here
well the equation four and so let's
derivate this again so let's have
partial T with partial X like this we
derivate this with respect to X minus
Rho times the duration of this with
respect to X is going to be zero and the
derivation of this with respect to X it's
also going to be zero plus then here is
going to be the function f prime of X
and T that's the derivative of f with
respect to X and so and this I now know
it's equal to equation four and so I can
come back up here and copy the content
of equation for Rho times a squared X
plus AB which I believe is missing a
minus because this was a minus over here
and I should have carried out the minus over
there so there should be a minus in here
this equation here comes right there and
so I'm gonna copy it over this is equal
to minus Rho times a squared X plus a B
we're almost there
we're almost there we have not F but F
prime the derivative of F with respect
to X this is equal to that so what is f
and how do we get F this is my favorite
part because when I do this in class I
can read the despair in the eyes of the
students who are realizing that after we
integrated our equation and then
derivated it again we're going to
integrate it again and then we're gonna
find what F is by integrating this with
respect
yes so I need to rewrite this equation
here let me let me specify precisely
what we have we have partial F over
partial X is equal to minus Rho times a
squared X plus A B what is F well F is
going to be the integration of this with
respect to X so it's a minus Rho times a
squared times 1/2 of x squared plus a B
X like this and this is plus then a
function G of all the rest but F is only
a function of X and T and so since we
integrated this with respect to X I'm
only left with T and so I have a
function G of T here this function again
includes some starting point for for
pressure so now we have this let me
again describe what we did here we
have integrated with respect to X now
that we have F we can put this term here
this whole equation there at the bottom
back into this equation here that we
have there and this equation
in which I'm going to
insert I'm gonna number it and I think
we were at number 4 so I'm gonna call
this number 5 this equation here is
number 5 right there so we go back to
equation 5
like so and now we rewrite it so we have
P as before so equation 5 on top of the
screen has minus Rho times B C y plus a
squared times 1/2 of Y squared plus this
function f and this function f we
described as so which is minus Rho times
a squared 1/2 of x squared plus a B X
here plus a function G which is a
function of time and so we can now just
rewrite just for aesthetics the final
equation like this this is minus Rho
times B C y plus one-half of a squared Y
squared minus no plus one half of a
squared x squared plus a B X like this
plus a function G of T of time and this
function G again I'm going to add a
little comment on here to say that this
includes oops
this is software bug here we go
this includes starting pressure P0
initial pressure
and that, this expression here is
everything we can say about pressure in
the initial field so I mean let me
square this up
hopefully the software will come after
me to fix it no it won't let me
configure this let me square this up
like so
ah didn't work it's fine
this is everything we know about
pressure given the original velocity
field that's given on the top over here
so again let's let's try to recap
what the navier-stokes equation tells us
and does not tell us about pressure if
you have the complete velocity field
then you can find out what the
derivative of the derivative of P with
respect to x and y are with both of
those derivatives providing you've done
your homework and checked that they are
compatible with one another you pick one
you integrate it with respect to the
parameter that you have on the bottom
here and you get one first expression
this is the expression 5 that we have in
there you're left with a function of the
other variable and this function of the
other variable you can evaluate it
because you've got the other
specification for the derivative of
pressure and playing with those two
informations you combine them together
through an integration and a derivation
to get one complete function of P
with respect to space what the
navier-stokes equation does not tell us
is two things one is how p changes
with time because if you look back on
this equation here or even that one
there you have the gradient of pressure
which specifies how pressure changes in
space but tells us nothing about how
pressure changes in time and if you run a
computational fluid dynamics simulation
you will have a great
deal of trouble predicting how
pressure has changed since the last time
you had values for pressure in order to
solve your equations and the second
thing it doesn't tell us is what the
basis value for pressure yes in other
words in the Navier-Stokes equation only the
change in space of pressure counts and
not the value of pressure itself and so
it turns out in a fluid flow simulation
in incompressible fluid flow you can
increase or decrease the value of the
pressure everywhere by the same number
and it doesn't change at all the
dynamics of the flow this is because
pressure itself does not influence the
dynamics of the flow it is the change in
space of pressure which does so
this is what you can say using just a
basic academic example about the
mechanics of the momentum balance
equation the navier-stokes equation so
here you are
