4Inviscid irrotational flow
IB Fluid Dynamics
4.1 Three-dimensional potential flows
We will not consider Laplace’s equation in full generality. Instead, we will
consider some solutions with symmetry. We will thus use spherical coordinates
(r, θ, ϕ). Then
∇
2
φ =
1
r
2
∂
∂r
r
2
∂φ
∂r
+
1
r
2
sin θ
∂
∂θ
sin θ
∂φ
∂θ
+
1
r
2
sin
2
θ
∂
2
φ
∂ϕ
2
.
It is also useful to know what the gradient is. This is given by
u = ∇φ =
∂φ
∂r
,
1
r
∂φ
∂θ
,
1
r sin θ
∂φ
∂ϕ
.
We start off with the simplest case. The simplest thing we can possibly imagine
is if
φ
=
φ
(
r
) depends only on
r
. So the velocity is purely radial. Then Laplace’s
equation implies
∂
∂r
r
2
∂φ
∂r
= 0.
So we know
∂φ
∂r
=
A
r
2
for some constant A. So
φ = −
A
r
+ B,
where
B
is yet another constant. Since we only care about the gradient
∇φ
, we
can wlog B = 0. So the only possible velocity potential is
φ = −
A
r
.
Then the speed, given by
∂φ
∂r
, falls off as
1
r
2
.
What is the physical significance of the factor
A
? Consider the volume flux
q across the surface of the sphere r = a. Then
q =
Z
S
u · n dS =
Z
S
u
r
dS =
Z
S
∂φ
∂r
dS =
Z
S
A
a
2
dS = 4πA.
So we can write
φ = −
q
4πr
.
When
q >
0, this corresponds to a point source of fluid. When
q <
0, this is a
point sink of fluid.
We can also derive this solution directly, using incompressibility. Since
flow is incompressible, the flux through any sphere containing 0 should be
constant. Since the surface area increases as 4
πr
2
, the velocity must drop as
1
r
2
,
in agreement with what we obtained above.
Notice that we have
∇
2
φ = qδ(x).
So φ is actually Green’s function for the Laplacian.
That was not too interesting. We can consider a less general solution, where
φ depends on r and θ but not ϕ. Then Laplace’s equation becomes
∇
2
φ =
1
r
2
∂
∂r
r
2
∂φ
∂r
+
1
r
2
sin θ
∂
∂θ
sin θ
∂φ
∂θ
= 0.
As we know from IB Methods, We can use Legendre polynomials to write the
solution as
φ =
∞
X
n=0
(A
n
r
n
+ B
n
r
−n−1
)P
n
(cos θ).
We then have
u =
∂φ
∂r
,
1
r
∂φ
∂θ
, 0
.
Example. We can look at uniform flow past a sphere of radius a.
U
x
r
θ
We suppose the upstream flow is u = U
ˆ
x. So
φ = Ux = Ur cos θ.
So we need to solve
∇
2
φ = 0 r > a
φ → Ur cos θ r → ∞
∂φ
∂r
= 0 r = a.
The last condition is there to ensure no fluid flows into the sphere, i.e.
u ·n
= 0,
for n the outward normal.
Since
P
1
(
cos θ
) =
cos θ
, and the
P
n
are orthogonal, our boundary conditions
at infinity require φ to be of the form
φ =
Ar +
B
r
2
cos θ.
We now just apply the two boundary conditions. The condition that
φ → Ur cos θ
tells us A = u, and the other condition tells us
A −
2B
a
3
= 0.
So we get
φ = U
r +
a
3
2r
2
cos θ.
We can interpret
Ur cos θ
as the uniform flow, and
U
a
3
2r
2
cos θ
as the dipole
response due to the sphere.
What is the velocity and pressure? We can compute the velocity to be
u
r
=
∂φ
∂r
= U
1 −
a
3
r
3
cos θ
u
θ
=
1
r
∂φ
∂θ
= −U
1 +
a
3
2r
3
sin θ.
We notice that u
r
= u
θ
= 0 when φ = 0, π and r = a.
At the north and south poles, when θ = ±
π
2
, we get
u
r
= 0, u
θ
= ±
3U
2
.
So the velocity is faster at the top than at the infinity boundary. This is why it
is windier at the top of a hill than below.
A
0
A
B
B
0
Note that the streamlines are closer to each other near the top since the velocity
is faster.
To obtain the pressure on the surface of the sphere, we apply Bernoulli’s
equation on a streamline to the surface (a, θ).
Comparing with what happens at infinity, we get
p
∞
+
1
2
ρU
2
= p +
1
2
ρU
2
9
4
sin
2
θ.
Thus, the pressure at the surface is
p = p
∞
+
1
2
ρU
2
1 −
9
4
sin
2
θ
.
At A and A
0
, we have
p = p
∞
+
1
2
ρU
2
.
At B and B
0
, we find
p = p
∞
−
5
8
ρU
2
.
Note that the pressure is a function of
sin
2
θ
. So the pressure at the back of
the sphere is exactly the same as that at the front. Thus, if we integrate the
pressure around the whole surface, we get 0. So the fluid exerts no net force on
the sphere! This is d’Alemberts’ paradox. This is, of course, because we have
ignored viscosity.
In practice, for a solid sphere in a viscous fluid at high Reynolds numbers, the
flow looks very similar upstream, but at some point after passing the sphere, the
flow separates and forms a wake behind the sphere. At the wake, the pressure is
approximately
p
∞
. The separation is caused by the adverse pressure gradient at
the rear.
wake
Empirically, we find the force is
F = C
D
1
2
ρU
2
πα
2
,
where
C
D
is a dimensionless drag coefficient. This is, in general, a function of
the Reynolds number Re.
Example
(Rising bubble)
.
Suppose we have a rising bubble, rising at speed
U
.
We assume our bubble is small and spherical, and has a free surface. In this
case, we do not have to re-calculate the velocity of the fluid. We can change our
frame of reference, and suppose the bubble is stationary and the fluid is moving
past it at
U
. Then this is exactly what we have calculated above. We then steal
the solution and then translate the velocities back by U to get the solution.
Doing all this, we find that the kinetic energy of the fluid is
Z
r>a
1
2
ρ|u|
2
dV =
π
3
a
3
ρU
2
=
1
2
M
A
U
2
,
where
M
A
=
1
2
4
3
πa
3
ρ
=
1
2
M
D
,
is the added mass of the bubble (and
M
D
is the mass of the fluid displaced by
the bubble).
Now suppose we raise the bubble by a distance
h
. The change in potential
energy of the system is
∆PE = −M
D
gh.
So
1
2
M
A
U
2
− M
D
gh = Energy
is constant, since we assume there is no dissipation of energy due to viscosity.
We differentiate this to know
M
A
U
˙
U = M
D
g
˙
h = M
D
gU.
We can cancel out the U’s and get
˙
U = 2g.
So in an inviscid fluid, the bubble rises at twice the acceleration of gravity.