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In this lecture, I will take up solution of the Integral momentum equations or the Integral
equations of the velocity boundary layer.
My task would be to introduce you to the general solution procedure and then using this procedure,
study the effects of the pressure gradient, suction and blowing. I recall again that Integral
solutions can be obtained for arbitrary variations of the free stream velocity U infinity at
the edge of the boundary layer and the wall velocity at the solid surface; we call it
as suction or blowing.
To refresh our memory, the Integral momentum equation reads as d delta 2 by d x plus 1
over U infinity d U infinity by d x plus 2 delta 2 plus delta 1 equal to C f x by 2 plus
V w by 2. How do we solve this equation? Just recall,
the procedure here is quite the reverse of what we adopted for similarity method. In
the similarity method, a third order similarity equation was solved with appropriate boundary
conditions to obtain the velocity profile.
The velocity profile came out of the solution of the ordinary differential equation and
from which the Integral parameters such as delta 1 delta 2 and C f x were recovered by
integrating and differentiating the profiles. In contrast, in Integral method the velocity
profile is assumed usually a polynomial in y by delta. Such that, it satisfies the boundary
conditions having assumed this profile. We evaluate the Integral parameters delta 1 delta
2 and C f x and substitute them in the Integral momentum equation. The IME is then solved
to obtain delta 2 as a function of x and hence all other parameters as functions of x. So,
the procedure is quite the reverse of the similarity method and to the extent that we
have assumed the velocity profile that which satisfies boundary conditions but none the
less is an approximation to what could be the real velocity profile at a given x.
We say, let u over U infinity, which is the dimensionless velocity and it would be a function
of x and y be a polynomial in eta variable, where eta is y by delta a plus b eta plus
c eta square plus d eta cube and e eta 4. So, there are 5 constants to be determined
a, b, c, d and e and this we do by invoking 5 boundary conditions. The boundary conditions
are as follows - At the wall y equal to 0, u is equal to 0, which will render u equal
to 0
The second condition is that, if you look at the momentum equation, if I write this
equation at y equal to 0, then clearly that term will be 0 because u itself is 0. However,
v will be v w into d u by d y at y equal to 0. This term will survive, U infinity d U
infinity by d x plus nu d 2 u by d y square at y equal to 0. I used this as a boundary
condition, to say that nu d 2 u by d y square at y equal to 0 is equal to v w d u by d y
at y equal to 0 minus U infinity d U infinity by d x.
That is what I have written here as the equation 4, which is the second boundary condition
at the wall. At the edge of the boundary layer, y equal to delta u will equal U infinity and
therefore the left hand side will be equal to 1 and so will each of these etas will be
1. You will have a condition a plus b plus c plus d plus e equal to 1. Also d u by d
y is equal to 0 at the edge of the boundary layer d u d y is 0. To the extent that the
velocity approaches U infinity, asymptotically the continuity of this first derivative survives
d u by d y equal to 0 survives as we increase y. Therefore, the last boundary condition
is that d 2 u by d y square will also be equal to 0.
I have 1 2 3 4 and 5; five boundary conditions which will enable me to determine a, b, c,
d and e. Let us see, what those coefficients look like? Each of these coefficients can
be represented in terms of the coefficient e, a equal to 0 because of u is equal to 0
at y equal to 0, b will be equal to 3 minus e, c will be equal to 3 into e minus 1, d
will be 1 minus 3 e and e will equal 3 times V w star minus lambda plus 6 divided by 6
plus V w star. With these five boundary conditions, if I
were to write out the equation, the velocity profile will look like u over U infinity 6
by 6 plus V w star a function F 1 a second function F 2 multiplied by V w star and a
third function F 3 multiplied by lambda. F 1 is simply 2 eta minus 2 eta cubes plus
eta raised to 4. F 2 is this function F 3 is this function V w star is a dimensionless
V w delta by nu. It is suction and blowing parameter and lambda is delta square by nu
d U infinity by d x is a pressure gradient parameter associated with variation of free
stream velocity U infinity with respect to x.
I did say that we have to choose the values of V w star and lambda very carefully because
as much as we can allow for any variations of V w and U infinity, we must ensure that
the assumed velocity profile is such that u over U infinity is always less than 1 inside
the boundary layer for eta less than 1. So, values of V w star and lambda minimum or maximum
must be such that this condition holds. Remember, V w star can be both positive or negative
depends on suction or blowing and lambda depending on whether it is an adverse pressure gradient
and a favorable pressure gradient can also be positive or negative. We have to choose
these parameters within a certain restricted range only; so that the boundary layer approximations
are well held.
I have plotted 3 graphs. The middle graph is for no suction or blowing that is V w star
equal to 0. When lambda is equal to 0, you will have the flat plate because d U infinity
by d x is equal to 0. As you can see here, this will be the profile, lambda equal to
positive values means accelerating boundary layer because d U infinity by d x then is
positive, lambda negative would be d U infinity by d x because d U infinity by d x is negative
and therefore, a decelerating boundary layer.
If I choose arbitrarily values of lambda, you will see up to lambda equal to 12 the
u over U infinity values are well within 1 and up to minus 12 they are very much within
0 to 1. If I take lambda equal to 30, which is a very highly accelerating flow then u
over U infinity exceeds 1. In fact that is what would happen for all values of lambda
greater than 12. Therefore, such values of lambda are not admissible; likewise, values
less than lambda equal to minus 12 are not admissible because velocity itself will turn
negative this is the importance of this comment which I made here.
If u over U infinity exceeds 1 then delta 1 or delta 2 can become negative. Likewise,
if u over U infinity is negative then delta 2 can be negative and that is not admissible.
This is similar profiles for V w star equal to minus 2, which means the suction case and
again I have plotted several values. Before I do, look at that let me go back again to
V w star equal to 0 value, when lambda equal to 0, I said this is a flat plate solution
but lambda equal to minus 12 gives us 0 gradient at the wall, which means separation must occur.
It is a peculiar value, where lambda equal to 7.052 and the importance of that you will
recognize a short while from now. Let me go back again to the suction profile,
here you will see for lambda equal to 12 plus 12 u over U infinity exceeds 1 and for 20
it exceeds very much. Likewise, here for lambda equal to minus 12 there is a velocity which
goes less than 0. On the blowing side, however up to lambda
equal to 20 or little more, there is a u over U infinity well within 1. So, blowing permits
much higher pressure gradients favorable pressure gradients. On the adverse pressure gradient
side, again lambda equal to minus 2 is the limit.
Integral method, we evaluate three thicknesses - the first one is very well known to you
delta 1, which can be where the integration limit from 0 to infinity is replaced by 0
to delta because from delta to infinity u will remain equal to U infinity. Therefore,
both these Integrals will be equal to 0 between delta and infinity and the limit infinity
can be replaced by delta in both of them. In addition, we define a shear thickness mu
U infinity divided by shear stress. You will notice that it has a length dimension, we
call delta 4 and it is called the shear thickness mu U infinity by tau wall x. The reason for
this is, you will understand very shortly.
We have three thicknesses delta 1, delta 2 and delta 4. We could have recovered value
of delta 4, even in the similarity method because we know how the tau wall x varies
with x. If I substitute our velocity profile u over U infinity equal to all these as a
function of eta into our definitions, then, you will see that delta 1 by delta would become
1 over 4 1 plus e by 5. If I represent e as 3 V w star minus lambda plus 6 over 6 plus
V w star, then it will simply read like that delta 2 by delta. This integration involves
product of u over U infinity square and that would result into 3 by 28 plus e by 70 minus
e square by 252 or what I have given here by simply replacing e.
Most importantly, delta 4 by delta would simply result in 1 minus 3 e 6 plus V w star divided
by lambda plus 12. This is quite easy to see, if we see our equation then remember delta
4 is mu over U infinity divided by tau wall x and this is equal to mu times U infinity
divided by mu times d u by d y at y equal to 0. If we look at our velocity equation
then you will see d u by d y at y equal to 0 will become equal to U infinity into 6 over
6 plus V w star into 2 by delta plus V w star into 0 plus lambda by 6 into 1 over delta.
You will see, if I take delta common I will get this as U infinity by delta into 6 over
6 plus V w star into 2 plus lambda by 6 and delta 4 will become equal to mu mu gets cancelled
here and U infinity divided by U infinity by delta 6 over 6 plus V w star into 2 plus
lambda by 6. Therefore, this and this gets cancelled and delta 4 by delta will read as
what I have shown there 6 plus V w star over lambda plus 12. This shows very clearly that
if lambda is equal to minus 12, delta 4 will become infinity or if delta 4 becomes infinity
then tau wall x must be 0. Therefore, lambda equal to minus 12 represents separation, values
of u over infinity were not applicable in our profiles for lambda less than minus 12.
We reorganize the Integral momentum equation by multiplying each term by u delta 2 by nu.
You will see that our equation would read as U infinity delta 2 by nu d delta 2 by d
x plus U infinity delta 2 by nu divided by 1 over U infinity d U infinity by d x into
delta 2 into 2 plus delta 1 by delta 2 equal to C f x which is by 2, which is tau wall
over rho U infinity square into U infinity delta 2 by nu plus V w by U infinity into
U infinity delta 2 by nu. Then, you will see this term simply becomes
U infinity by nu into d delta 2 square by d x divided by 2. In this case, U infinity
gets cancelled with this and I get delta 2 square by nu d U infinity by d x equal to
rho into nu becomes mu and nu infinity becomes U infinity. So, mu U infinity by tau wall
is delta 2 by delta 4 plus U infinity gets cancelled with that I will be getting V w
delta 2 by nu. Therefore, multiplying throughout by 2 I get U infinity by nu d delta 2 square
by d x equal to 2 times V w delta 2 by nu plus delta 2 by delta 4 minus delta 2 square
by nu oh sorry this should be multiplied by 2 plus delta 1 by delta 2 I forgot d U infinity
by d x into 2 plus delta 1 by delta 2.
Again, the equation maintains its dimensionless form, Now, all I have done is in our derivation
is delta 2 by delta 4 is replaced by s and we call it shear factor; delta 1 by delta
2 is replaced by h, which we call shape factor; delta 2 square divided by nu d u infinity
by d x and that would simply be equal to lambda times delta 2 by delta whole square as a pressure
gradient parameter. Remember, what was lambda? Lambda will be delta square by nu d U infinity
by d x and therefore this parameter which is kappa. kappa will be simply lambda times
delta 2 by delta whole square and that is the pressure gradient parameter and V w plus
will be V w delta 2 by nu is equal to V w star delta 2 by delta because V w star was
simply V w delta by nu. Each term is dimensionless, you can see this is velocity a length dimension
because this is square and so this has a length dimension and this is nu.
So, it is a kind of a Reynolds number, which represents rate of growth of momentum thickness
delta 2; kappa is a pressure gradient parameter. This is really a universal relationship what
we mean that the relationship applies no matter what is the variation of U infinity or the
wall velocity. It is a universal relationship between Integral parameters delta 2 delta
4 and delta 1.
Let me go to the next slide. By universal, we mean that it is typically applicable to
all types of variation on U infinity, including the ones we used in similarity method, U infinity
equal to C x raise to m. kappa which is proportional to d U infinity by d x, when it is 0 implies
a flat plate solution because U infinity equals constant.
On the other hand, the right hand side F k or U infinity over nu d delta 2 square by
d x equal to 0 implies that delta 2 is constant with x. As you will recall from our similarity
method, it represents stagnation point solution because for m equal to 1 all thicknesses delta
1, delta 2, enthalpy thickness, heat transfer coefficient and all are constants with x that
is the characteristic of a stagnation point solution.
Shear parameter equal to 0 delta 2 over delta 4 equal to 0 implies that delta 4 is infinite,
which in turn implies that tau wall x is equal to 0 and it represent separation. All values
of V w plus and lambda for which delta 1, delta 2 and delta 4 are less than 0 must be
discarded because for these values, u over U infinity is either greater than 1 or u over
U infinity is less than 0, which means those are inadmissible values of V w plus and delta
lambda.
Now, for U infinity equal to C x m it is easy to derive that kappa will be equal to delta
2 star square divided by 2, I will show you how this is the case. If U infinity is equal
C x raise to m and kappa is equal to delta 2 star by nu d U infinity by d x. Then, you
will see that this becomes delta 2 square by nu C m x raise to m minus 1 or I can write
this as C x raise to m into delta 2 square by x into m, I can write like that. This is
nothing but C x to m is nothing but U infinity into delta 2 square by x into m, which I can
write as U infinity oh sorry that should be a nu here. So, divide by nu I can write this
as U infinity x by nu into delta 2 by x whole square into m that is equal to delta 2 by
x square into R e x into m.
If you recall, in our similarity method, we had defined delta 2 star as delta 2 by x R
e x to the half. So, essentially becomes delta 2 star square into m and that is what I shown
here. Similarly, you can show that F kappa which is equal to U infinity by nu d delta
2 square by d x can be shown to be equal to 1 minus m delta 2 star square for that value
of m.
Similarly, here delta 2 square is corresponding to the value of m you are concerned with.
You can also show S; for example, what is S? S is equal to delta 2 by delta 4 and that
is equal to delta 2 divided by mu U infinity by tau wall x or that is equal to delta 2
tau wall x divided by mu U infinity. This term can be shown to be equal to f double
prime 0 into delta 2 star, which simply a question of manipulation here and you will
see the tau wall x.
For example, this will be delta 2 into mu times d u by d y at 0 divided by U infinity
into mu. So, mu and mu gets cancel and I can construct here delta 2 by x R e by x by half
and you will get that f double prime. V w plus can also be shown to be B f; the similarity
parameter B f delta 2 star and delta 2 star m are available from similarity methods. These
deductions are very important as we shall see shortly.
Let me plot the results. Remember, I have calculated S V w plus kappa, I have assumed
values of kappa and calculated H also from the velocity profile that we assumed; that
is delta 1 by delta, delta 2 by delta and delta 4 by delta.
For a boundary layer without suction or blowing, the acceleration and deceleration parameters;
acceleration parameter I have gone up to 12 and deceleration I have gone up to minus 12
because I cannot go less than minus 12. Then, the kappa values take these values, this is
0.095 and this goes on to minus 0.157 delta 1 by delta 2. As you can see compared to 0,
where it was about 0.3 with acceleration the displacement thickness reduces and with deceleration
it increases. Same thing holds for a momentum thickness divided by delta. It reduces and
increases, inverse is true for shear thickness. The shear thickness goes on increasing whereas
on the negative side, when there is suction it goes on reducing so much. So that at minus
12, it reduces exactly to 0.
The shape factor H delta 1 by delta 2 is remarkably a constant for highly on acceleration side
but on decelerating side, it goes on increasing quite significantly. This is the right hand
side with some values negative and then positive. Now, what is F k? To remember again F k is
nothing but that value and when that is equal to 0, it means delta 2 is constant. Therefore,
it represents stagnation point solution if you look at here the stagnation point solution
will be somewhere here and its value will be lambda equal to 7.052 and kappa equal to
minus 7824. The similarity solution for delta 2 star m
equal to 0 was 0.663 and delta 2 star m equal to 1 was 0.292. Therefore, kappa for m equal
to 1 will be 0.0841 and F for k m equal to 0, kappa m equal to 0 would be 0.44 will make
use of these numbers very shortly but remember this is the flat plate solution, 7.052 lambda
is a stagnation point solution and minus 12 is a separation solution.
This is what I have plotted here. On the x axis, you have kappa the pressure gradient
parameter and F kappa, which is the rate of growth of momentum thickness parameter on
the y axis. When kappa is positive, we have accelerating
flow or favorable pressure gradient. When kappa is negative, we have the decelerating
flow or adverse pressure gradient. The results are plotted, when there is no suction and
blowing. Therefore, V w star is equal to 0.
You will notice that when kappa is equal to 0, the value of F k must represent the flat
plate solution. When F k is equal to 0, as we just said it must represent the stagnation
point solution. So, the intercept on the y axis represents the flat plate solution whereas
the intercept on the x axis represents the stagnation point solution and F k values turn
negative, when you have very highly accelerating flow, whereas when the flow is decelerating
F k values are positive. Now, using the relationship between kappa
and F k with Integral parameters, I have also plotted the parameters here the similarity
solutions here. You will notice that this is m equal to 1 solution, this is m equal
to 0.33 solution, minus 4 m equal to minus 4 minus 0.04 minus 0.065 minus 0.085 and minus
0.09 is the separation. So, the separation is seen at about kappa
equal to minus. Let us say about minus 0.07 in similarity solutions, whereas in Integral
solution the separation occurs at minus 0.1567. On the Integral solution deviates from similarity
solution for kappa less than 0 because we have allowed for arbitrary variation of U
infinity but for specific variation U infinity equal to C x m the results go along there.
Now, in order to develop close form solution, we can see that at least for very moderate
decelerating flows and accelerating flows, a near straight line approximation can be
made. This was done by a scientist called Thwaite's. It is simply F kappa equal to a
minus b kappa and the a will be the value of kappa equal to 0 and therefore represents
flat plate solution a delta 2 star square m equal to 0, whereas b will be simply when
F k is equal to 0 or the stagnation point solution then b will equal a divided by delta
2 star square of m equal to 1.
If you look at our previous slide, I have said delta 2 square star m equal to 0.663.So,
delta 2 star square would be square of 0.663, which is 0.44 and delta 2 star m equal to
1 is 0.292. So, 0.44 divided by 0.292 will give me this value of 5.17. So, a will become
equal to 0.44 whereas b will equal minus 0.4. F k being equal to U infinity, rate of growth
of momentum thickness is equal to a constant minus another constant times delta 2 square
by nu d U infinity by d x. This is the Thwaite's curve-fit, a universal curve fit for the case
in which suction and blowing are absent. We will make use of this relationship as you
will see on the next slide.
Just see, this was the relationship. I can manipulate these 2 terms; this term and this
term as d by d x of d u delta 2 square U infinity 5.7 equal to 0.44 nu 4 u. To convince you,
let me open up again the equation, then you will see d delta 2 square U infinity equal
to 5.17 by d x will equal delta 2 square into 5.17 into U infinity raised to 4.17 d U infinity
by d x plus U infinity raise to 5.17 into d delta 2 square by d x.
This is what it would mean and if I divide this 1 over nu, U infinity raise to 4.17,
I will get 5.17 into delta 2 square by nu u d U infinity by d x plus U infinity by nu
d delta 2 square by d x that is equal to 0.44. Therefore, you will see that I can write this
equation in this form. If I were to integrate this equation from 0 to x then delta 2 square
U infinity raise to 5.17 del x. We will equal delta 2 square U infinity raise 5 minus x
equal to 0 equal to 0.44 nu 0 to x U infinity raise to 4.17 d x.
So, the solution is applicable to any arbitrary variation of U infinity and restriction imposed
by similarity method is now removed. We use this relationship to calculate delta 2 at
any x because U infinity at that x will be known. You must know delta 2 square at x equal
to 0. If you start from x equal to 0 itself where delta 2 is 0 then of course that term
will be 0. Evaluate kappa from d U infinity by d x. Now, that you know delta 2 square,
you can evaluate our kappa which is.
Once you have evaluated delta 2, you can now evaluate delta 2 square nu d U infinity by
d x because you already know what U infinity x is. For this value of kappa, you can evaluate
S value for knowing the kappa value; you can interpolate to get S value. You get a shear
stress value from which you can evaluate S and delta 4 and from delta 4 you can evaluate
the skin friction coefficient 2 nu over delta 4 U infinity, which is what we wish to evaluate
anyway. That is the purpose. In other words, knowing U infinity as a function
of x, we get delta 2 as a function of x from which we get kappa from which we get S as
a function of x. In fact, we get all other parameters delta 1 as a function of x and
so on and so forth and because we know the shear factor variation with x. We can get
C f x also as a function of x.
In the previous table, I had held V w equal to the no suction and blowing. Now, I am saying,
I am going to set lambda equal to 0, which is the case of a flat plate but allow for
suction and blowing and that is what I have done here.
I allow for blowing parameter to go up to 5 on the suction side; I go up to minus 4.2
delta 1 by delta with blowing compared to V w star equal to 0 the delta 1 by delta increases
as we expect delta 1 by delta decrease. As we increase suction, these also increase where
it is not seen here because I have plotted results only up to second decimal place. On
this side, it reduces but notice that at minus 4.4 delta 2 has already turned negative and
that is not permitted. I cannot go below lambda less than minus 4.2.
So, feasible solutions are possible only for lambda greater than minus 4.2 as you can see
here. The shear stress also has almost vanished here, which means this is where separation
is about to take place. The shape factor is increased enormously to
47.25 from the H average values around 2.7 on the positive side on the blowing side around
this at moderate suction rates it is about 2.4 but it increases very rapidly to 47.5.
This is almost the separation power profile and these are the values of F k.
Likewise, I have now included effects x of both lambda n for a certain V w star, which
is minus 0.2 that means it is a suction case with V w star equal to minus 0.2 and here
I have gone up to 15 but notice that beyond 14.8 or S has become negative therefore this
is not acceptable solution lambda is equal to 15 is not acceptable. On the adverse pressure
gradient side, you will see I have gone up to minus 13 but at minus 12 it is 0 already.
So, this is the separation occurs and minus 13 is minus 0.03 so this is not admissible.
Effect of pressure gradient on at a certain suction state is valid between minus 12 and
14.5 only.
The remarkable feature of this solution is that for very mild acceleration to all the
adverse pressure gradients, the value of V w plus is almost constant. In this, V w plus
is almost constant on the suction side for lambda equal to 0 F k is about 21 you can
see that 0.21 and that must equal delta 2 star square and V w plus equal to minus 0.21.
This amounts to B f equal to minus 0.21 divided by root 0.21 0.458.
This is the blowing side and again you will see for adverse pressure gradients less than
minus 12, you have negative. So, you cannot go below then again on V w plus is remarkably
constant and this will correspond to B f equal to about 0.2619.
For simultaneous variations of V w star and lambda, close form solutions can again be
developed in the regions in which V w plus is constant. You can curve fit F kappa equal
to a minus b kappa or a relationship of this type can be established, where a and b are
functions of V w plus. So, V w star equal to minus 0.2 you get a equal to 0.21, b equal
to 4.2, V w star equal to plus 2 and you will get 0.84 and 7.4. Manipulation would give
again d delta 2 d by d x of this equal to that and therefore you will get a solution.
So, the procedure remains exactly the same as before only the values of a and b change
with value of V w star. I am taking now, a case of flow over a cylinder; it is an impervious
cylinder. There is no suction or blowing; there is an approach velocity V, a potential
theory will show that the free stream velocity U infinity would vary as U infinity by V a
equal to 2 times sine 2 x star where x star is x divided by diameter.
So, 2 x star is nothing but x divided by radius and I will call this F x star. Then, for this
variation of free stream velocity, I can show that delta 2 by d R e D will be simply from
equation here using this relationship, I can determine delta 2. Remember, delta 2 at x
equal to 0 at the stagnation point will be 0 and from there I integrate. So, I can show
that delta 2 by d R e d will be done and kappa will be this.
Our objective is to determine the location of the separation point corresponding to kappa
equal to minus 1567 R e D is equal to V a D by nu, which is the Reynolds number defined
for flow over a cylinder.
Here are the results - U infinity by V a is a sine function which goes and beyond 90 deceleration
sets in whereas on the below 90 degrees there is a flow acceleration. This is the variation
of kappa and you can see it has reached minus 0.1567 at about 108.3 degrees and therefore
it is associated with separation. So, we have located the separation point from stagnation
point from the unknown arbitrary velocity distribution.
Similarly, now consider a cylinder. In which, there is blowing taking place from the cylinder
surface. Then, I can curve fit as I said F kappa in this manner for different values
of V w star and theta separation for V w star equal to 0 was 108 and that goes on reducing.
You will expect where that a flow over cylinder with blowing would the separation would occur
at an earlier location and that is what you see up to V w star at 0.2 the separation point
has advanced. Average skin friction up to the separation point defined in this manner;
it is highest at V w star equal to 0 but with blowing skin friction goes on reducing. As
expected, separation angle is advance with increase in blowing rate with reduction in
average skin friction due to thickening of the boundary layer.
This shows you the power of the Integral method. What it cannot do is to go beyond the point
of separation and complete the analysis of flow over a cylinder. It is a very useful
tool to determine flow through conversion or diversion nozzles. For example, where the
boundary layer, where the free stream velocity would either accelerate or decelerate with
x and you want to determine the thickness of the boundary layers developing on the wall
because it is this thickness, which determines the discharge coefficients of such nozzles.