# Mod-01 Lec-17 Aero elasticity

Del square phi over delta plus del phi dot
del of del phi dot del phi over 2 is 0. Then a square is equal to a infinity square
plus gamma minus 1 over 2 u infinity square minus del phi dot del phi minus 2. And then the pressure expression p equals
p infinity into 1 plus gamma minus 1 over 2, a
infinity square, u infinity square minus del phi dot del phi minus 2 delta phi over delta
t power gamma over gamma minus 1. And then you have the boundary condition,
which is the boundary condition is delta f
over delta t plus q dot del f 0, where q is velocity which is del phi. Now, this is our exact non-linear potential
equation. The kind of problems which we have to deal in aero elasticity, that makes
simplification to this entire set of equations and
because this is highly non-linear. So, you have to go for the small disturbance potential
equation, which is the linearize equation which will get. So, please note that is the
small disturbance. Why is a small disturbance, because
we are going to treat with wings limiting surface. Limited surfaces we can see aerofoil surfaces,
this is along x, this is you and you can assume this is the wing, so you can go and
this is the wing. Now, the f is the equation to
the surface of the body and which basically create the disturbance in the flow. Now, this
approximation in representing f is actually responsible for reducing our equations to
that small perturbation because how we applied.
Since, it is a relatively thin surface, we will
write this f which is a function of x comma, y comma, g comma t, as z is separate minus
f of because equation of the surface. Now, you can have a upper surface or the top
surface and this is the bottom surface. Now, if you have a top surface that f t is
for top surface, which will be a function of x
comma, y comma t, this is top and the bottom is you can write f b, which is the x comma.
Now, this is how we are saying that I am explaining somehow the result is coming out
and I am writing the set, which is the location on the surface is basically a function of
x and y. Now if I have this type of representation
f t and f b, I will go back to the boundary condition. Now, what is delta f by delta t and this expression
if I write, this is delta f over delta t plus q is del phi, which you can write it
as U i plus V j plus W k. I am going to use the u
a little bigger U because again I will split that into 2 because one is the free stream
velocity, which is along the x axis. This I am going to write as U infinity plus lower
case U i plus V j plus W k, that means my potential
gives me the total velocity gradient of potential and my u infinity is the velocity
per feet. This lower case U V W is essentially bottom
perturbations due to the wing are there aerofoil. Due to this submerge structure which
is present in the flow. Now, if I write this because u infinity is the velocity of this
whole thing, your speed of the aircraft or anything the wing and these are bottom perturbation
in the flow due to the presence of the wing. Now, I substitute this back here, you will
get u infinity plus lower case u del of phi delta
f by delta x because these are dot product then, I will have v delta f over delta y plus
w delta f over delta g, this is 0. Now, I am
going to substitute for f. This expression if I
substitute this expression because f is the equation of the surface at any time. If I
substitute and then delta f by delta g because this delta g by delta g that is 1 g, so that
is 0. You will have w which is actually the flow
velocity. It will become like this w, which is the flow
velocity is going to become this you will take it us delta f by delta g. Here all are
lower case f because f is g f minus x g is the
independent variable. Therefore, that is not a function of time. Similarly, here 0 not
come here also 0 not come they will have a minus
sign. So, you transfer the all term right hand
side you will get delta f by delta t plus U infinity plus lower case u x plus v delta
f by delta y.
Now, please understand on g plus f of x comma y comma t. This is the surface; this
condition is on this surface. W is the flow velocity right and you know that f is this
function. Now, you are going to make some assumptions here that are if the slop of the
surface with respect to x because we have drawn this is g y and the x. Our aerofoil
this slop with respect to x delta f by delta x
are delta f by delta y. They are small because it is
not a large slop. So, these slops are small gradients. And then
you say this is again you are making an assumption that my u and v are also relatively
small in comparison to U infinity. So you say my U, V and w this is an assumption I
making. Now, if I neglect terms which are products of small quantities that means this
is the higher order term. Similarly, this is
also higher order term. I will get here simply neglecting w approximately becomes plus
U infinity delta f by delta x. That is I am eliminating higher orders terms.
This is my approximation later I will say this is equal to that is all. Now, I know
the surface equation f of x y, this function I will
substitute u infinity I know, I can know this. Only thing is that my flow velocity must be
equal to this flow velocity normal whatever we said W, W is along the g direction. Here
again you make another approximation that means W must be what? W must be among
every surface point, but you are going to make an assumption. You are going to expanded that w, which is
the function of f x comma y comma g, whether it is a top surface or a bottom surface
that will depend on g, if you put it top because you want to calculate the flow velocity
on the surface. So, this can be if I put a top g t that means
this is the top surface and you are going to
expanded like a Mac Lorene series because this will write it W, which is x comma y
comma 0 plus. This is an expansion about this g axis plus you will write g top delta W by
delta z hat x comma y comma 0 plus gamma t plus higher order terms again. Similarly, for the bottom surface you will put this is
d. You say that these are thickness is not very
large and the slops delta w by delta g, that is the variation of the flow velocity with
respect to g is not very large. There is no sharp gradient in any of districts,
this is an assumption. Then you say I neglect this term, I neglect y term. Then
I say that w at x comma y comma g top surface t
is 0 plus and I going to write similarly, for the bottom w x comma y comma 0 minus
comma t. I am simply satisfying the condition on this line. Please understand I am just
explain only on x y plane. I am satisfying the condition I am not really going on then
satisfying the condition at the top surface. This is the aerofoil are the entire wing plane,
I will satisfy on this that is why here I will
write it on g equal 0, this I am satisfying the
condition. g is equal to 0 when I put plus that is the
top surface then, I put minus that is the bottom
surface. So, as far as I am concern my w is only on this plane, I am considering. Then
you may ask what happens at the thickness of this. That is why now you say you have
to do one approximation later we will go to that.
The assumption we made is this surfaces are thin and we can make on the slops whatever
the aerofoil shape, which you make the slops are also very small.
Therefore, we are neglected the higher order term that is the non-linear terms because
another assumption that velocity U V W, which is are bottom perturbation quantities are small in comparison to the slop field which
is U infinity. So, this is the basic assumption. Now, we say all my quantities that is the
disturbance due to the presence of my aerofoil, they are small.
So, this is all the small disturbance potential flow. For that I am going to now modify the
entire set of non-linear equations into corresponding linear equations. They will become
a very simplified linear equation and that is what we will use for our solution of the
flow supersonic, flow subsonic flow everything.
So, please understand that the assumptions we make everything is purely based on the
boundary condition because you say my surface is thin. It is not going to disturb the flow
you are not going to put a building over there and then try to solve this problem. In the
sense you have so much of disturbance. We are saying our surfaces are thin surfaces
lifting because that is where aero elasticity problems are important, that is why this
assumption is made on. Thin aerofoil that is could done and the finally,
the proof is if you can get a solution which is good, which is matches with an experiment
that is where the theory gets validated. Now, let us get the formulation
for a small disturbance potential theory that means the reduction of this equations as well
as all the. So, the boundary conditions please understand this has been modified to
this. This is my boundary condition, now w is the
flow velocity please understand, flow velocity. And we are looking w is flow velocity
everywhere, but we are matching the flow velocity through the surface are g is
equal to 0 plus minus only along the up to cart
another wings fan that is all. If it is 3 d problem if it is wings fan, but it is a
2-d problem, its only cart. Now, there will be another
what happens to the off body boundary condition.
There is something like that we do when what happens to the thickness. Does it do
anything, that we will do off body boundary condition later because right now we will
say this is my boundary condition. We will reduce the set of equations and then if
possible we can do right away itself the tough body also because that will also give me
we can do both. Let me finish the linearization first and
then I will take because that is a very interesting condition. Please understand this is the key
we have converted this to this. Now, what is
w? W is delta phi by delta g that is all delta phi by delta g at g equal to 0 plus or g equal
to 0 minus. But please understand on the aerofoil g top surface bottom surface they are
different do not mix this two. That is why this is very essential that when you put the
top surface bottom surface, the s is correspondingly
going to change f for top surface it will be top surface and f d it will be bottom surface.
Now, let us take the reduction of the equation because I will keep this. This I
leave as it is because this is important your boundary condition. We are going to now use the condition that
a speed of sound a infinity plus a hat. Some small these are small disturbance that is
the small perturbation quantity and the pressure s p infinity plus p small and density rho,
rho infinity plus rho hat. And your q that is the
velocity, velocity is u infinity i plus where q is U i plus V j plus W k. Your free is U
infinity x plus p hat. And you know that q prime this is nothing but del phi hat. Please
note that del phi hat is this like what we wrote earlier.
So, from the boundary condition only, I am bringing it here. Now why I am saying from
the boundary condition you may ask the question. See, I am saying that the disturbance
because this boundary condition is only near the surface of the wing. If I can make an
assumption that my U V W, that is the perturbation velocities, they are small near the surface the boundary. Then I can make an assumption
that far away from that they can be small.
Please understand this is an assumption because I say that disturbance to the flow is
created by the presence of right, near the aerofoil the disturbances are small. Therefore,
I say that far away from that the disturbance
will be still smaller. Therefore, I am making all these assumptions that everywhere U V
W are small in comparison with u infinity, but
see this is a very interesting question, you know which comes. The disturbance is small
near the boundary, which is the disturbance casing surface, but how can say that far away
it will be less. This is an assumption you make because this
is there a counter example where it is not because this is very interesting. Suppose
you take the tsunami, the disturbance wherever it is called this small, but then it built
up far away from there and then you have floods only somewhere several miles away from the
point where, the earth quake anything has happened us you follow. Now, the question
is there how can you say that the disturbance is small there, but then it built there catastrophe
in some several hundred miles away from that. It is a very difficult thing to
answer. See, that is the surface where, but here we
are making that assumption that whatever near the aerofoil disturbance small it is not going
too far away. Now, with this assumption what we are going to do is, we are going to
substitute these quantities because phi you know this I am going to put this back there
del square phi. Now, when I put del square phi term by term if we go then, you will find
everything will be ok. The first term del square because I am going to substitute this.
When I substitute this that will become del square phi is going to become del square phi
hat there is no problem. That is the first term because this is little
bit tough algebra and the second term, which is
del square I will write it here del square phi over delta t square plus delta by delta
t of del phi del phi. This is a second term plus del
phi dot del of del phi dot del phi over rho. This
is because when I substitute this, this is independent of time because U infinity x is
not there. So, this will become del square phi
hat over delta t square, this is nothing but q dot
q. So, q dot q is q square. So, you will get when you differentiate that will be 2 q. So,
you can write it as plus 2 q dot delta by delta
t of q. Q is given here, so I can write is as 2 u
infinity i plus, this q prime is what? Q prime is del phi. So, I am going to put plus del phi hat dot
delta by delta t of U infinity i plus again del phi
hat, that is this term same. Then you have to write last term. Last term is plus del phi is nothing but U
infinity i plus del phi hat dot del of this is del
phi dot del phi that is nothing but this dot this. This dot this is nothing but U infinity
square over 2 plus U infinity i dot del phi hat plus half del phi hat dot del phi hat.
And this is what you do is, you neglect all the
non-linear terms. This will essentially reduce, this will again become this term will remind
as it is, you will have what you do is? You substitute everywhere delta by delta t
this is gone because there will go to 0 there is
no place to sent, but this del phi is nothing but delta phi by delta x I delta phi by delta
x j etcetera, etcetera. Then this is also del
phi, this del phi, this del phi they will become
higher order terms. So, you take only U infinity i and the i of this term that will be a
linear term. So, you neglect all the non-linear terms from
here retain only the linear terms. So, if you
want I will write the whole thing. Then, we will del square phi hat over delta t square
plus 2 U infinity i plus del phi hat dot. You will have del square phi hat by delta
t delta x i plus del square phi hat delta t delta y
j plus del square phi hat by delta t delta g k plus
you will have this entire term. That entire term again you can write it as
plus U infinity i plus del phi hat dot. See this is
u infinity; this is del of u infinity square because u infinity is a constant. So, this
will just drop up it will not exist, it has this gradient
of the constant so, that will go up. So, you will have dot u infinity del square phi hat
by delta x square i plus del square phi hat by
delta y delta x j plus del square phi hat by delta g delta x k plus half this whole
thing. Because I am taking dot product of what this
into this because this term as it is remaining here, please understand.
I am taking this del i, I will have only i term because this is U infinity i. So, I will
get only delta phi by delta x phi hat, that is
why I have delta phi hat by delta x. This delta by
delta x I, this is delta phi hat delta x then delta by delta y j delta x delta by delta
g k and U infinity is just U infinity there. Because
this term only i term will survivor or the sub
term will be 0 because i dot j is 0, i dot k is 0. So, this will simply means u infinity
delta phi by delta x delta phi and that you take
a gradient a. And this is half the entire term that
you can take it as half, you allow here i j k.
So, you will write it as delta by delta x of u square plus v square plus w square i
plus delta by delta y of u square plus v square
plus w square j plus delta by delta g of u square
plus v square plus w square k. So, you see this is very complicated expression. You
simply throw away all the non-linear terms that is what is done. Please understand
otherwise if you expand this is going to be very big. So, you neglect the non-linear
expression and write it as I will write the final result here. The final expression will be approximately
del square p hat delta t square plus 2 U infinity that is this term del square phi
hat by delta t delta x plus. You see this term U
infinity i and u infinity square delta phi by delta x plus u infinity square del square
phi hat away that is all, rest of the terms I
am neglecting it. So, you see when I neglect rest of
the terms, this becomes very simple. This is linear operator, so erase is this part
and I will write my equations here.
Now, you will also have to see what I am going to do with this term because here it is a
square, a square is again a function of phi, this is not an independent. Now, I say the
perturbation and other things I am going to assume a square is a infinity square because
the moment I make that it is not. Then it is again an non-linear equation because I
am going to get phi here and this will be denominator. So, it will become again on non-linear that
is why I will write this term on this and a
square as simply a infinity. So, my small perturbation it will be del square phi 1 over
a infinity square del square phi hat by delta
t square plus 2 U infinity delta phi del square phi by delta t delta x plus U infinity square
del square phi hat by delta x square. Please understand, this is my
unsteady small disturbance linearize potential flow equation
because this is the linear. And this equation please understand this is
linear completely absolutely no problem. You have a linear equation. Now, what we are going
to do with the pressure expression. So, let us erase this part and we will write the
pressure expression as well as the disturbance because even though I used here infinity square,
but I can get an expression for local speed of sound. I can get an expression for
a local speed of sound. I am going to write that as a as some a infinity
plus a. So, I substitute that here what happen to a infinity square we will cancel
it out, so you will have a hat square plus 2 a
infinity a hat equals gamma minus 1 over 2. you will have u infinity square minus, this
is a del phi dot del phi which is nothing but
again U infinity plus U whole square plus V
square plus W square minus 2 delta phi over delta t.
Now, what I can do is I can again make a approximation that this a hat is small.
Therefore, I throw this term out and I am just using only this term, which is like my
hat will become gamma minus 1 over 4 a infinity.
This is what U infinity square will go away leaving behind you will have minus 2
U infinity U. So, I am going to use only that term minus 2 U infinity U and rest of the
V square W square I am saying that they are small I neglect that minus 2 delta phi over
delta t. So, this is what I called it as change in
the local speed of sound. Now, if you see u is
delta phi by delta x, this term that is del phi hat by delta x that is what and this is
anyway phi becomes this phi hat because there is
no given prentices independent of time. Now, you see U infinity by a infinity, this is
the Mac number. Schrodinger is the Mac number is not very large. This speed of sound variation is not greatly
affected, that is why Mac number, but very large what is very large is another question,
that is why we will use. Please understand we are going to use this same equation for
supersonic flow. We, are not going to use some other, can you Schrodinger the Mac number
is not very large below Mac number 3 because there are another problems heating
lawful heating everything will come way of isentropic flow.
We made of isentropic flow as an assumption. So, based on that assumption which we
have to less than Mac 3 less than 3, this equation are parallel equal. And that is why
these are used potential flow small disturbance.
Now, let us go to the pressure expression because this is the changing a local speed
of sound this is just for exercise. Now, let us
look at I erase this part. Let me write the pressure expression.
The pressure is this I am again making assumption that all these are small quantities. So,
gamma minus 1, I am going to expand by binomial. When I expand by binomial this will
here inside and I written an only 1 term. I will not write more than any term because
this 1 plus something power x something power.
I said that this quantity is a small quantity. Therefore, I will get my pressure is p infinity
times 1 plus because if I take this, this will
become gamma over you can take it actually after simplifying. You will get a infinity
square because U infinity U infinity will go up square and the factor 2 will cancel
with the factor 2 with another factor 2 here, like
what we did earlier. You will get basically minus u infinity into
lower case u, which is delta phi hat over delta
x minus delta p hat over delta t. This I can write it as p minus p infinity what is this?
This is a infinity square is gamma, so you will
get t infinity gamma over a infinity square is
nothing but rho infinity. So, I will write this is rho infinity times u infinity delta
phi hat over delta x plus delta phi hat over delta
t, this is basically the change in pressure. Change in pressure at any location in the
flow this is my I can get the pressure quite efficient if you want c p. C p is nothing
but p minus p infinity over half rho infinity U
infinity square. You divide by that, so you will get minus 2 over U infinity delta phi
hat over delta x minus 2 over U infinity square
delta phi hat over delta t. Now, you see this is
my equation and my boundary conditions are W s what? Delta f by delta t plus u infinity delta f
over delta x, but this w is delta phi hat over delta z
at z is equal to 0. You can have plus or minus because f top f bottom that is all. Now, this
completes your entire problem. Now, is that all is a question
you follow. Now, you have seen I will not using the hat from now on
because it is not necessary I have to because when I use phi, you assume that it is the
perturbation only. Just for again everywhere I have to put hat,
hat, hat that is just for convenience, I am eliminating the hat from the all these equations.
I will use only this equation, this equation and this equation that is all. My
local speed of sound small disturbance you can
get always what is the local speed of sound that is if you want to get it. So, now this is your you can say your linearize
You say it large in comparison to your perturbation. When we solve our equations
whether it is a supersonic 2 d or subsonic, we
are going to use only this, but then if you see in ((Refer Time: 49:41)) you go that means
you are going to say a infinity is very large. So, this term goes to 0. I am left with just
a plus equation del square phi equal to 0 whether, it is a steady flow or unsteady flow.
on only the boundary condition. If my boundary condition f is independent of t that
is all my boundary condition w is U infinity delta f by delta x that is all. And my pressure
this will go up, this term will go up rho infinity u infinity delta phi by delta x.
So, I need to get only this change in pressure. Now, you will get pressure on the top
surface pressure on the bottom surface and then you subtract that and you will get your
left and you integrate over the cart of the airfoil. So you find finally, all over entire
unsteady aerodynamic equations is reducing to solving this. Only thing is thickness till
we are not protecting because we say this w the flow velocity on the surface, but that
f we have differentiate between comma top surface
and the bottom surface. Now, what we will do is, we with split that into the 2
parts the boundary condition itself taking into
account the thickness of the aerofoil and then argument, we will say that the thickness
is not this is ready to solve. So, at symmetry and anti symmetry see we say
this is our x, this is our y and this is a g
and if we say this is my aerofoil, let me correct out of let me not bother about the
way. This is the top surface, this is the botto1m
surface, but I can write this as some good point I take it. So, I am going to write the
upper surface f top surface, which is our is f is
z top surface which is the function of x comma, y comma t because this is the function of
x. I am going to write it of 2 parts, one is g t.
You take it as t is thickness, this is g t you may say g t by 2, but the 2 is the absurd
in g itself plus g lifting, this is the that is
although I can spilt it into 2 parts. One is the
thickness another one is the lifting component that means I am writing the surface. This
surface although this is this surface plus the small thickness. And similarly, f bottom
this is g bottom, which I will write this minus
because the thickness is negative. So, this is
the mid plane plus some thickness mid plane minus thickness. Thickness is actually g t
by t, but it does not matter. Now, if I am going to use this condition that,
I can spill my top surface and bottom surface into 2 parts rather although I can
add this 2. Then I am going to substitute one is
the lifting case, another one is the thickness case because I am writing in a linear
combination and all my equations are linear please understand. This is a linear, this
is linear, this is also linear, there is no non-linearity
is the problem. Therefore, I am saying I am splitting my problem into thickness case
separately, lifting case separately. Now, if I
look at the boundary condition this is what that delta phi by delta g. If I write thickness
case thickness case, what is the boundary condition? Boundary condition will be upper surface and
then lower surface we will have 2 conditions. In the upper surface I will have
what I have to substitute here in the f upper surface top surface, top surface is Z t. So,
I will put delta Z t over delta t plus U infinity delta Z t over delta x equals delta p over
delta g, g equal to 0 plus. And on the lower surface which is the bottom surface I am going
to put, what is the bottom surface for thickness is a minus Z t.
So, I will have minus that minus delta Z t minus U infinity delta Z t over delta x equal
delta p over delta phi over delta g at phi equal 0 minus. Now you look at this. This
is delta phi by delta g is anti symmetric because
0 plus, 0 minus. Zero plus is positive 0 minus is negative that means delta phi by
delta g is for thickness case is anti symmetric, if this is anti symmetric phi is symmetric.
Phi is symmetric with respect to what, only respect to Z t symmetric, anti symmetric with
respect to Z. If phi is symmetric that means the top surface phi bottom surface phi
should have the same value. Now, if I go to the pressure
p minus p infinity this is if I take in p top surface it will be p top surface minus
p, p bottom surface it will be p bottom surface minus p. But my phi is symmetric with respect
to g, but delta phi by delta x is same that means as 0 plus on 0 minus will have a same
value. Therefore, it will not give me any lift. I cannot get any lift because of the
thickness. So, the thickness effect has no meaning. Only
for all this linear problems I am telling you
because it will not give. Phi is symmetric therefore, the pressure is also phi is symmetric
means what pressure is also symmetric because pressure is delta phi by delta x only. With
respect to Z, it is a symmetric that plus that minus it will have a same value, symmetric
value. Therefore, this will not contribute anything to the pressure.
Now, if you go to the lifting case you will get delta phi by delta g, it will be symmetric.
Therefore, if delta phi by delta g is symmetric that means phi is anti symmetric. If phi is
anti symmetric then this is anti symmetric top surface bottom surface will have different
value. Then you can get a pressure difference and that is what you will get itself the
upper force. So, we will write that part in the lifting
case now. You got it now, this is please understand this is the place where, your wing
is there your surface because we are applying this boundary condition only on that
not on free flow this is on the body. Thickness case and then let me write the other
case I erase this part after that we will go
to this is in the lifting case you will have upper surface as usual. So, we will have delta g del over delta t
plus U infinity g l over delta x, which is 0 plus
and lower minus that is a minus sign sorry 0 plus sign plus U infinity 0 minus. Therefore,
not this out you will have delta phi over delta g symmetric with respect to g. So, phi
is anti symmetric with respect to g. If phi is
anti symmetric with respect to g, I can write my pressure because this is the upper square
lower square I can take. So, upper surface means p u minus p infinity, I will get one
value, lower surface again p l minus p infinity is this. So, if I subtract upper minus lower
which is the delta p change. In the pressure differential pressure p upper,
this is essentially I write this term because p
infinity is anyway come and that will go off. You will have minus rho infinity delta phi
over delta t plus delta x this is g equal 0 plus minus of minus. So, I will get plus rho infinity u infinity
delta phi over delta x plus at g equals 0 so on. If
phi is anti symmetric means then what will happens? This term will become minus on
the top surface p s this will become minus. So, your net term will become minus 2 rho
infinity u infinity delta phi hat over delta x
delta t g equal to 0 plus. So, you have a pressure difference that is because of your
lifting thickness will not contribute. Now, the question
is what of that of half body boundary condition. This is on the body. Now, off body
boundary conditions you have to again write it for the lifting case and thickness
case because this is essential if you want to
solve rueful problems. So, I will write that part I will erase this here, it not necessary.
I will write here off body boundary conditions.
Off body conditions you can say. See off the body means your velocity must
be continuous and the pressure must be continuous that means let us look at it is
only the g equal to 0 plane is important. So, g
equal to 0 planes. You will have pressure upper is equal to pressure lower, it must
be there and velocity must be continuous. So,
you must get delta phi over delta g upper surface must be equal because particular this
is very important for the q dot n because velocity is continuous null fall velocity
have to. Now, if you take condition if you look at
here. From here thickness case, thickness case
what does it say delta phi by delta g is anti symmetric that means, but I want delta phi
by delta g to be same, which means thickness
case what W 0 because there it says anti symmetric delta phi by delta g. Here it says
that means I want w 0 that means there should be no vertical velocity in the backside.
Now, if you go to the lifting case, lifting case this phi is anti symmetric that means
phi is anti symmetric means what? Your pressure upper
and lower are opposite whereas, here it says p u must be equal to p n, which means
the pressure should be 0. Pressure should be
0 means that pressure should be p infinity that is all. So, there is no pressure change.
So, for lifting case
pressure p minus p infinity must be 0, this is off body boundary condition.
This is on g equal to 0, this is off body. Thickness case says that w must be 0 and then
lifting case says that p should be actually p infinity. Now, this is what is used in getting
solutions later off body boundary conditions.

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