4Inviscid irrotational flow
IB Fluid Dynamics
4.2 Potential flow in two dimensions
We now consider the case where we have two-dimensional world. We use polar
coordinates. Then we have
∇
2
φ =
1
r
∂
∂r
r
∂φ
∂r
+
1
r
2
∂
2
φ
∂θ
2
.
The gradient is given by
u = ∇φ =
∂φ
∂r
,
1
r
∂φ
∂θ
.
The general solution to Laplace’s equation is given by
φ = A log r + Bθ +
∞
X
n=1
(A
n
r
n
+ B
n
r
−n
)
(
cos nθ
sin nθ
.
Example
(Point source)
.
Suppose we have a point source. We can either solve
∇
2
φ = qδ(r),
where
q
is the volume flux per unit distance, normal to the plane, or use the
conservation of mass to obtain
2πru
r
= q.
So
u
r
=
q
2πr
.
Hence we get
φ =
q
2π
log r.
Example
(Point vertex)
.
In this case, we have flow that goes around in circles:
Since there is no radial velocity, we must have
∂φ
∂r
= 0. So
φ
only depends on
θ
.
This corresponds to the solution φ = Bθ.
To find out the value of B, consider the circulation around a loop
K =
I
r=a
u · d` =
Z
2π
0
B
a
· a dθ = 2πB.
So we get
φ =
K
2π
θ, u
θ
=
K
2πr
.
It is an exercise for the reader to show that this flow is indeed irrotational, i.e.
∇ × u
= 0 for
r 6
= 0, despite it looking rotational. Moreover, for any (simple)
loop c, we get
I
c
u · d` =
(
K the origin is inside C
0 otherwise
.
We can interpret this as saying there is no vorticity everywhere, except at the
singular point at the origin, where we have infinite vorticity. Indeed, we have
I
c
u · d` =
Z
S
ω · n dS,
where S is the surface bounded by c.
Example (Uniform flow past a cylinder). Consider the flow past a cylinder.
U
x
r
θ
We need to solve
∇
2
φ = 0 r > a
φ → Ur cos θ r → ∞
∂φ
∂r
= 0 r = a.
We already have the general solution above. So we just write it down. We find
φ = U
r +
a
2
r
cos θ +
K
2π
θ.
The last term allows for a net circulation
K
around the cylinder, to account for
vorticity in the viscous boundary layer on the surface of the cylinder. We have
u
r
= U
1 −
a
2
r
2
cos θ
u
θ
= −U
1 +
a
2
r
2
sin θ +
K
2πr
.
We can find the streamfunction for this as
ψ = Ur sin θ
1 −
a
2
r
2
−
K
2π
log r.
If there is no circulation, i.e.
K
= 0, then we get a flow similar to the flow around
a sphere. Again, there is no net force on the cylinder, since flow is symmetric
fore and aft, above and below. Again, we get two stagnation points at A, A
0
.
A
0
A
What happens when
K 6
= 0 is more interesting. We first look at the stagnation
points. We get
u
r
= 0 if and only if
r
=
a
or
cos θ
= 0. For
u
θ
= 0, when
r
=
a
,
we require
K = 4πaU sin θ.
So provided
|K| ≤
4
πaU
, there is a solution to this problem, and we get
stagnation points on the boundary.
For
|K| >
4
πaU
, we do not get a stagnation point on the boundary. However,
we still have the stagnation point where
cos θ
= 0, i.e.
θ
=
±
π
2
. Looking at the
equation for u
θ
= 0, only θ =
π
2
works.
Let’s now look at the effect on the sphere. For steady potential flow, Bernoulli
works (i.e.
H
is constant) everywhere, not just along each streamline (see later).
So we can calculate the pressure on the surface. Let
p
be the pressure on the
surface. Then we get
p
∞
+
1
2
ρU
2
= p +
1
2
ρ
K
2πa
− 2U sin θ
2
.
So we find
p = p
∞
+
1
2
ρU
2
−
ρK
2
8π
2
a
2
+
ρKU sin θ
πa
− 2ρU
2
sin
2
θ.
We see the pressure is symmetrical fore and aft. So there is no force in the
x
direction.
However, we get a transverse force (per unit length) in the
y
-direction. We
have
F
y
= −
Z
2π
0
p sin θ(a dθ) = −
Z
2π
0
ρKU
πa
sin
2
θa dθ = −ρUK.
where we have dropped all the odd terms. So there is a sideways force in the
direction perpendicular to the flow, and is directly proportional to the circulation
of the system.
In general, the magnus force (lift force) resulting from interaction between
the flow U and the vortex K is
F = ρU × K.