in this problem we're taking a first
look at boundary layer flow flow very
close to surfaces in this problem we
were in particular looking at air coming
in and flowing over a flat plate and
it's coming in here from the left with a
very nice smoothly uniformly distributed
velocity 25 meters per second and it's
leaving towards the right and by the end
of the plate it has now a very strange
and complex velocity distribution that
is not uniform anymore the question
we're trying to answer is what is the
thickness of a control volume at the
inlet that has the same mass flow as the
thickness here has at the outlet it's a
bit of a strange question and so I'll
spend some time explaining what the
question really means before I start
answering it let me show you what the
control volume could be to study this
problem we could simply draw a rectangle
on the bottom would be the plate and on
top would be a straight line at the
outlet you would have a velocity
distribution that looks like this like
so with progressively smaller arrows as
you go down and then you would take this
length here and we reproduce it here and
this would be here the inlet velocity
distribution there's a problem with this
control volume which is it doesn't have
a mass balance if you look at the mass
flow exiting of the control volume here
there's deficit here this whole this
whole velocity here is lacking if you
went to integrate velocity with respect to
area at the outlet you will not get the
same number as in the inlet and so the
question becomes where is this mass flow
gone? you cannot create or destroy a mass
in fluid mechanics and so the answer is
the mass flow exits towards the top like
this and with it it carries momentum
it carries mass flow but we have no
information about this mass flow on top
so even though this is a correct control
volume to study the problem it's not a
very useful one because we're lacking
information about what happens on top of
the control volume.
instead to study this problem we would
draw a control volume that would look
like this we will have a flat plate on
the bottom and then the outlet here with
the height Delta and then the inlet we
would have less of a height so then when
we look at this velocity distribution
here at the outlet that looks like this
like so and we take this maximum
velocity here and we reproduce it here
then we have here the same mass flow at
inlet as we have at the outlet and for
this to be correct then we have the top
surface top surface must have kind of
an unknown shape here that joins one
side to the other
woop should not auto correct this shape
here like so in this control volume mass
is conserved we have the same mass flow
incoming as we have outgoing and there
is no flow across the top surface of the control volume so the
question we try to answer today is if
this is known this is the height here
this height is called Delta is the
thickness of the boundary layer then
what is the thickness at the inlet here?
what is this here
this height here which we call h1?
what is this height yes? this is what if
we try to find once we have this height
here it will be a lot easier to
calculate force inside this control
volume which is the question in the next
problem
ok so let's take a look at this let me
push this up here and let's now figure
out what this height h1 is since we
have a mass flow balance problem I can
start by writing a mass balanced
equation the mass of a balance equation
in fluid mechanics for control volumes looks
like this it says that 0 is the total
amount of mass that is created or
destroyed inside the control volume well
it is the sum of two terms it is the
change in time of mass inside the
control volume and so this is here an
integral over the whole volume
with the whole control volume and to
this I had a double integral over the
control surface here and this is then an
integral over an area I don't like to
remember this by heart because I
miss it very often so I just fill in the
terms by looking at the formula sheet
every time and so for mas flow this is
rho simply density in the middle
and in the area it is Rho with a very
annoying term which is V rel here dot
in the relative velocity as a vector dot
an N vector which is unit vector which
is always pointing outwards by
convention ok what does this equation
become in our particular case well the
whole first term here this whole term
there here this is zero because the flow
is steady and so even though this
integral here of the total amount of
mass inside the control volume this
integral is not zero its change in time
is 0 so this whole first term goes away
what about the second term we're going
to split into two I'm going to say here
it is minus for incoming so the
integral over the incoming flow here of
Rho V orthogonal here and I'm going to
put an absolute value and DA this is for
the incoming flow and then I have for
the outgoing flow plus the integral whop
this both of those should not be
straight lines the integral out Rho V
orthogonal da yeah so this is for the
outgoing flow now let's simplify it
further and let's write it out in a way
that's understandable the velocity
incoming
here is anyway orthogonal to the inlet
of the flow so this V orthogonal here
it's going to be V in one so I can just
write it away so the integral over the
inlet flow of Rho 1 V 1
hey yeah plus the integral over the
outgoing flow of Rho 2 V 2 dA like so and
this is still equal to zero
now let's work out step by step what we
have in here we have on the incoming
flow v1 is perfectly uniform because if
you look down here at the bottom left
the v1 is just capital u it is 25 meters
per second everywhere
so the integral of constant Rho
constant V vs area
it's just Rho V area yes here it is
Rho 1 V 1 a 1 like so Plus at the outlet
at the other the velocity is not uniform
it has this very weird distribution it
is the integral of here Rho 2 V 2. V 2 is
the strange velocity distribution which
if I gather up my equation over there is
expressed it has u Y over Delta to the
power of 1 over 7 to the power 1 over 6
dA let's keep this going and now let's
split up
a the Inlet what this term a 1 is
because this is very interesting now
we have minus Rho 1 V 1, V 1 is just u so
and a 1 is 2 components it has on the
width side on the Z side it has length L
2 and on the height side
this vertical side it has the length
we're looking for which is length H 1
and on the right side on the
outlets let's work it out - Rho is
constant and gets out of the integral Rho 2, u is also constant and
can also
be removed from this and we're left with the
integral here over the whole area of Y
over Delta to the power 1 over 6 okay so
now let's take it one step further and
leave this here as such for the inlet
minus Rho 1 u 1 ya to H 1
perhaps let me just stop for a second
and explain again what we're trying to
do here the unknown in this equation is
H 1 we've got all of the other terms and
this is equal to 0 so we know it's down
there on the other side we're only
looking for h1 and we're
formulating this mass balance equation
as such we're saying basically the inlet
has the same mass flow as the outlet what
is the height h1 so that the inlet
has exactly the same mass flow as the outlet so h1 is the unknown in this
whole equation Let me continue now at the
bottom with this I say this
is Rho 2 u2 the integral over the
whole area is done again in two
dimensions one is going to be the
dimension across the plate this is going
to be the dimension L2
nothing changes according to Z so
everything is just multiplied by L 2 and
then I have the integral here of let's
split this into two components so that it's useful Delta to the
power minus 1 over 6 this is just a
number and Y to the power 1 over 6 this
is the coordinate and then we're left
here with the coordinates that is the
vertical direction that is away from the
plate and this is dy like so from where
to where do we do this integral well
where are the limits well we start at
zero and Y is equal to 0 the height of
the plate here and then we move up along
the plate until we reach the height
Delta like so now there are a few
things that we can simplify and remove
from
equation we try to move this up again we
have zero on the other side so a few
terms that are the same one in both
parts of the equation can be removed one
is the density Rho Rho 1 Rho 2 are the
same U 1 and u are the same thing and I
also have here L 2 which is the same on
both sides
so if I divide everything by those three
terms they just get away just go away so
I still here have 0, 0 is equal to minus
I'm just left with h1, our unknown
plus here the integral from 0 to Delta
of Y to the power of minus 1 over
6 Delta to the power minus 1 over 6 y
to the power of 1 over 6 dy it seems a
bit scary it's actually not that
complicated Delta to the power minus 1
over 6 this is just one number it's not
affected by Y and so we leave it alone
and then Y to the power 1 over 6 is
integrated exactly the same way as if we
had Y to the power 2 and so Y to the
power 2 would become 1 over 3 (so 1 over
2 plus 1) of Y to the power 2 plus 1
that's the same thing applies here we
have here minus of h1 at the
inlet plus here Delta to the power
minus 1 over 6 not affected by Y and
then we have 1 over one 6th plus one here of
Y to the power 1/6 plus 1 over here
this is integrated between 0 and Delta
like so all right so just time now
to just group up the numbers because we
have the solution already 0 is equal to
minus H 1 plus let's see what we have
here we have still Delta to the power
minus 1 over 6…
…
…and this is evaluated from 0
to delta and so now we can get finally
what we want to express h1 is
equal to …
…
…
here we have it this is the
result that we're looking for and so now
we can just put in numbers into this we
have h1 is equal to Delta, Delta is 2
centimeters so it is 0.02 meters
multiplied by 6 over 7 and this is equal
to (I put this into the calculator before)
1.71 4 times 10 to the power minus 2
meters and we can just write it like
so h1 is equal to one point
seven one four centimeters okay so this
is the final result again what we're
calculating here is the height if I go
back here the height of the inlet
control volume that is so that it has
exactly the same mass flow as the outlet
over here and so this is how you use the
mass balance equation to calculate
and adjust your control volume
so that you have exactly the same mass
flow outgoing in as the mass flow incoming.
