4
Two-dimensional ows
If the ow is in the plane, and there are only two independent variables x, y , the ow problem is much simplied. Moreover, we will see that some of the physics is fundamentally
dierent from three dimensions. One fact we know of alread
4.6
4.6.1
The method of images
A vortex next to a wall
4b
1
=0
b
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
4.9
The Joukowski mapping: circles to ellipses
A particularly useful application of the mapping idea concerns the ow around bodies. We
have solved the problem of the ow around a cylinder. Thus if we can nd a conformal
mapping between the unit circle and a
5
Waves and free surface ows
Free surface ows are in some ways very dierent from what we have done before. In
all problems considered so far, the domain D in which to solve the problem is given (for
example some box or the exterior of an aeroplane wing).
6
Nonlinear waves
Note: this material will not be examined
6.1
Quadratic corrections
So far we have restricted ourselves to the case of waves of small amplitude, which means
we only took terms linear in the wave amplitude into account. Now we will conside
Appendix A: Vector calculus
We shall revise some vectors operations that you should have already met before this
course. These may be presented slightly dierently to the way you have previously seen
them.
A.1 Sux notation and summation convention
Suppose
Appendix B: Curvilinear coordinate systems
q1 increasing
e2
e1
r(q1 , q2 )
q2 increasing
Many problems can be approached more simply by choosing a coordinate system that
ts a given geometry. Instead of writing the position vector r as a function of Cartes
Appendix C: Streamfunctions
If u = 0, then it follows that there exists a vector eld A(x, t) s.t.
u = A.
Conversely, it is clear from this representation that the ow is incompressible. The representation is particularly useful if there are two independent
D Some simple ows and their potentials
D.1 Two dimensional ows
We will give the answer either in Cartesians or polars, depending on which is more convenient.
Reminder:
=
, v=
=
where is the
x
y
y
x
velocity potential and is the streamfunction.
(i) in Cart
Fluid Dynamics 3
2011/12
Sheet 1
Homework to be handed in Friday 21st October: 1,3,4,9.
1. Let a closed loop of particles C (t) be dened by
0 s < 2,
x = a(cos s + t sin s, sin s, 0),
where each value of s corresponds to a dierent uid particle, and a, > 0.
Fluid Dynamics 3
2011/12
Sheet 2
Homework to be handed in Friday 28th October: 1,3,5.
1. The Euler equation is
u
+ (u
t
)u =
1
p,
where u(x, t) is the velocity and p(x, t) the pressure. Consider the coordinate transformation x = x Vt, where V is a const
Fluid Dynamics 3
2011/12
Sheet 3
Homework to be handed in on Friday 4th November: Q 1,3,5,7
1. Consider an axisymmetric, incompressible ow in cylindrical polar coordinates (r, , z ). This
means the velocity eld is given by u = ur + uz z. Assume that the v
Fluid Dynamics 3
2011/12
Sheet 4
Questions 1,2,3 to be handed in on 11th November
z
1. Consider a vortex with centre along the -axis, so the ow eld is of the form u = f (r ) in
cylindrical polars.
(i) Show that u = 0.
(ii) Using Eulers equation in cylindr
Fluid Dynamics 3
2011/12
Sheet 5
Homework to be handed in 18th November: questions 2,4,5
1. Consider the axisymmetric ow produced by the superposition of a uniform stream of speed
U and a three-dimensional source of strength m. Choose cylindrical polar co
Fluid Dynamics 3
2011/12
Sheet 6
Homework to be handed in 25th November: questions 2,3,4
1. (i) Show that in spherical polars, the potential of a uniform stream and of a dipole are (a)
cos
= U r cos , and (b) = 2 , respectively.
r
(ii) By using the expr
Fluid Dynamics 3
2011/12
Sheet 7
Homework to be handed in 2nd December: questions 2,4,5.
z
bubble
1.
R
When a large bubble rises in water, it assumes
the shape of a spherical cap as shown in the Figure. At its lower end it forms an irregular
edge. Inside,
Fluid Dynamics 3
2011/12
Sheet 8
Homework to be handed in 9th December: questions 1,2,3.
1. Consider the two-dimensional, unbounded ow outside a cylinder of radius R centred at the
origin.
(i) Let the complex potential of the ow in the absence of the cyli
Fluid Dynamics 3
2011/12
Sheet 9
Homework to be handed in 16th December: questions 1,3,5.
1. Consider a two-dimensional channel with walls located at y = 0 and y = . A point vortex
with circulation > 0 is placed at z0 = ib, with 0 < b < .
(i) Write down t
Fluid Dynamics 3
2011/12
Sheet 10
Homework to be handed in Friday 20th January: questions 2,5.
1. Consider small amplitude two-dimensional oscillations on the free-surface of an unbounded
uid of innite depth. At large depths, the uid is a rest. The elevat
Fluid Dynamics 3 - Solutions to Sheet 1
1.
(i) For t = 0, C is a circle; the points a(1, 0, 0) and
a(1, 0, 0) remain stationary. For 0 < s <
x-values are pushed to the right, for < s <
2 x-values are pushed to the left. Thus one
obtains the following pic
Fluid Dynamics 3 - Solutions to Sheet 3
1. (i) Using the formula for u = A in cylindrical polars, one obtains
ur =
1
,
r z
uz =
1
r r
So if u increases then the gradient of increases and the streamlines thus get closer together.
(iii) Volume ux, Q, acr
Fluid Dynamics 3 - Solutions to Sheet 4
C
D
1. (i) For the given ow eld, u = f (r ) in cylindrical
polars. Thus, applying the formula from chapter 0,
u=
B
1 u
= 0.
r
A
F
(ii) We have to solve the steady Euler equation
E
( u)u = p/.
According to chapter 0
Fluid Dynamics 3 - Solutions to Sheet 5
Hence on the surface
1.
(i) The ow eld of a uniform stream is u = U z,
and thus
ur =
1
= 0,
r z
uz =
u2 = U 2 1 + cos (1 cos ) +
1
= U.
r r
(iv) Very far upstream the pressure is atmospheric,
and u2 = U 2 . Thus
Fluid Dynamics 3 - Solutions to Sheet 6
(ii) The boundary condition is
1.
(i) (a) The potential of a uniform stream is =
U z , and the result follows from z/r = cos .
(b) The potential of a dipole = z is
u n = U n.
Now
1
x
z
cos
= 3 = 3 = 2 .
r
r
r
r
ui
Fluid Dynamics 3 - Solutions to Sheet 7
(ii) The complex potential is dened as
1.
w = + i = U [r(cos + i sin )+
R2
R2
,
(cos i sin ) = U z +
r
z
(i) Since the pressure inside the bubble vanishes, it is
also zero along a streamline that starts on the axis
Fluid Dynamics 3 - Solutions to Sheet 8
Thus the velocity at z0 is
1.
2
(i) If |z | = R (on the boundary) then z z = R and thus
R2 /z = z . But this means than w(z ) = f (z ) + f (z )
for z on the boundary, and thus
i 1
1
1
+
2 2s 2ih 2(s + ih)
s2
h2
=
+i
Fluid Dynamics 3 - Solutions to Sheet 9
(i) The original vortex has images at z1 = i and z1 = i,
and z1 has images at z0 = 0 (upper wall) and z2 = 2i
(lower wall). In general, the images of zn are at zn1
and zn+1 . Evidently, this generates an innity of i
Fluid Dynamics 3 - Solutions to Sheet 10
(ii) Let (x, t) = H sin(kx t) and adopt
a similar separable form for (x, z, t) =
Z (z ) cos(kx t) which ts with the boundary
conditions on z = 0. Stu this into Laplaces
and separate to give
1. This is just like lec
Fluid Dynamics 3 - Solutions to Sheet 2
y
1. If the point x travels at constant speed V, i.e.
x = Vt, the transformation x = x Vt implies
that x = 0. This means the transformation corresponds to the uid as seen by an observer moving
with velocity V. As a
Fluid Dynamics 3 - 2011/2012
Jens Eggers
Preliminaries
Course information
Lecturer: Prof. Jens Eggers, Room SM2.3
Timetable: Weeks 1-12, Tuesday 2.00 (SM2), Thursday 4.10 (SM1) and Friday 11.10
(SM1). Course made up of 32 lectures.
Oce hours: my oce ho