Fluid Mechanics, SG2214, HT2013
October 9, 2013
Exercise 11: Vorticity, Bernoulli and Stream Function
Example 1: Solid-Body Rotation
Consider the flow in an uniformly rotating bucket with velocity
u = (ωr, 0, 0)
a) Use Bernoulli equation to determine the free surface: What is wrong?
Bernoulli:
p 1 2
+ |u| + gz = C
ρ 2
along streamlines
This gives the surface of constant pressure
z=
ω2 r2
C − p0
−
ρg
2g
The free surface is highest in the center of the bucket, something is wrong. Bernoulli theorem is valid
along a streamline in a steady ideal fluid. If the flow had been irrotational, it would have been valid
everywhere. But now ∇ × ū = (0, 0, 2ω).
b) Use Euler equation to determine the free surface:
1
Du
= − ∇p + g
Dt
ρ
this gives
er :
−
u2θ
1 ∂p
=−
r
ρ ∂r
eθ :
ez : 0 = −
1 ∂p
−g
ρ ∂z
⇒
∂p
= ρω 2 r
∂r
⇒
0=−
1 ∂p
ρr ∂θ
∂p
=0
∂θ
∂f
= −ρg
∂z
⇒
Thus
p(r, z) =
⇒
⇒
p=
1 2 2
ρω r + f (z)
2
f = ρgz + p0
1 2 2
ρω r − ρgz + p0
2
This gives the free surface where p = p0
z=
ω2 r2
2g
Example 2: Flow over a Hill
A hill with the height h has the shape of a half circular cylinder as shown in Figure 1. Far from the hill the
wind U∞ is blowing parallel to the ground in the x-direction and the atmospheric pressure at the ground is
p0 .
a) Assume potential flow and show that the stream function in cylindrical coordinates is of the form
ψ = f (r) sin θ,
where f (r) is an arbitrary function. Calculate the velocity field above the hill.
1
y
x
Figure 1: Streamlines above a hill with h = 100m and U∞ = 5m/s. A paraglider pilot with a sink of 1m/s
will find lift in the area within the dotted line, while soaring along the hill.
The stream function satisfies continuity:
ur =
1 ∂ψ
,
r ∂θ
uθ = −
∂ψ
∂r
The flow is irrotational:
ω = ∇ × ū = 0
∂ψ
∂2ψ 1 ∂2ψ
+r 2 +
=0
∂r
∂r
r ∂θ2
⇒
Introduce the ansatz ψ = f (r) sin θ into equation (1):
1
f 0 sin θ + rf 00 sin θ − f sin θ = 0
r
⇒
1
f 0 + rf 00 − f = 0
r
Make the ansatz f = rn :
1
nrn−1 + rn(n − 1)rn−2 − rn = 0
r
n + n2 − n − 1 = 0
So we have
ψ=
⇒
n = ±1
B
Ar +
sin θ
r
We need two boundary conditions.
1. Free stream:
Ar sin θ = U∞ r sin θ
⇒
A = U∞
2. Streamline on the hill surface:
U∞ h +
⇒
B
=0
h
So we have:
⇒
ψ = U∞
h2
r−
r
B = −U∞ h2
sin θ
Now we can calculate the velocity field above the hill:
1 ∂ψ
h2
ur =
= U∞ 1 − 2 cos θ
r ∂θ
r
∂ψ
h2
uθ = −
= −U∞ 1 + 2 sin θ
∂r
r
2
(1)
b) Derive an equation for the curve with constant vertical wind velocity V .
Constant vertical wind velocity is described by:
h2
h2
h2
V = ur sin θ+uθ cos θ = U∞ 1 − 2 cos θ sin θ−U∞ 1 + 2 sin θ cos θ = −2U∞ 2 sin θ cos θ
r
r
r
r
U∞
r = h −2
sin θ cos θ
V
c) Assume that the density ρ and the gravitational acceleration g is constant. Calculate the atmospheric
pressure at the top of the hill.
Use the Bernoulli equation (valid everywhere) with free stream pressure p0 at the ground:
1 2
1
po + ρU∞
= p + ρ(2U∞ )2 + ρgh
2
2
3 2
⇒ p = po − ρU∞
− ρgh
2
Stokes stream function
Consider a 2D incompressible flow
∂u ∂v
+
= 0.
∂x ∂y
∇ · ū = 0 or
Define the stream function Ψ such that:
u=
∂Ψ
∂y
This means
∂u ∂v
+
=
∂x ∂y
so continuity is always fulfilled. Now we can write
ēx
∂
ū = ∇ × Ψēz = ∂x
0
v=−
∂Ψ
∂x
∂2Ψ
∂2Ψ
−
= 0,
∂x∂y ∂y∂x
ēz ∂Ψ ∂Ψ
∂ =
,
−
,
0
∂z ∂y
∂x
Ψ
ēy
∂
∂y
0
And
ēx
∂
ω̄ = ∇ × ū = ∂x
∂Ψ
ēy
∂
∂y
− ∂Ψ
∂x
∂y
ēz ∂Ψ
∂2Ψ
∂ =
0,
0,
−
−
= −∇2 Ψēz
∂z 2
2
∂x
∂y
0
For irrotational flow
∆Ψ = ∇2 Ψ = 0
In spherical coordinates for axisymmetrical flow, define Ψ
ur =
r2
1 ∂Ψ
sin θ ∂θ
uθ = −
1 ∂Ψ
r sin θ ∂r
Incompressibility is still valid
∇ · ū = 0
Velocity
ū = ∇ ×
Ψ
ēθ
r sin θ
Vorticity
2
1
∂ Ψ sin θ ∂
1 ∂Ψ
ω̄ = −
+ 2
r sin θ ∂r2
r ∂θ sin θ ∂θ
Irrotational
∂ 2 Ψ sin θ ∂
+ 2
∂r2
r ∂θ
3
1 ∂Ψ
sin θ ∂θ
=0
⇒
Example 3: Flow around a Sphere
Consider a sphere with Ψ = 0 at r = a, compute the irrotational velocity distribution when the velocity of
the freestream at infinity is U :
1
r → ∞ Ψ → U r2 sin2 θ
2
⇒ ur → U cos θ uθ → −U sin θ .
Make the ansatz:
Ψ = f (r) sin2 θ
For an irrotational flow we get
2
f =0
r2
From the boundary conditions at infinity we get
f 00 −
⇒
f = Ar2 +
B
r
1
U
2
A=
On the surface of the sphere (r = a)
1 2 B
Ua +
=0
2
a
⇒
1
B = − U a3
2
This gives
a3
1
2
sin2 θ
Ψ= U r −
2
r
The slip velocity on the sphere is
3
uθ = − U sin θ
2
The radial velocity is of course ur = 0 on the surface.
4