4Inviscid irrotational flow
IB Fluid Dynamics
4.3 Time dependent potential flows
Consider the time-dependent Euler equation
ρ
∂u
∂t
+ ∇
1
2
|u|
2
− u × ω
= −∇p − ∇χ.
We assume that we have a potential flow
u
=
∇φ
. So
ω
= 0. Then we can write
the whole equation as
∇
ρ
∂φ
∂t
+
1
2
ρ|u|
2
+ p + χ
= 0.
Thus we can integrate this to obtain
ρ
∂φ
∂t
+
1
2
ρ|∇φ|
2
+ p + χ = f(t),
where
f
(
t
) is a function independent of space. This equation allows us to correlate
the properties of φ, p etc. at different points in space.
Example (Oscillations in a manometer). A manometer is a U-shaped tube.
H
y = 0
h
We use some magic to set it up such that the water level in the left tube is
h
above the equilibrium position
H
. Then when we release the system, the water
levels on both side would oscillate.
We can get quite far just by doing dimensional analysis. There are only two
parameters
g, H
. Hence the frequency must be proportional to
p
g
H
. To get the
constant of proportionality, we have to do proper calculations.
We are going to assume the reservoir at the bottom is large, so velocities are
negligible. So
φ
is constant in the reservoir, say
φ
= 0. We want to figure out
the velocity on the left. This only moves vertically. So we have
φ = uy =
˙
hy.
So we have
∂φ
∂t
=
¨
hy.
On the right hand side, we just have
φ = −uy = −˙gy,
∂φ
∂t
= −
¨
hy.
We now apply the equation from one tube to the other — we get
ρ
¨
h(H + h) +
1
2
ρ
˙
h
2
+ p
atm
+ gρ(H + h) = f (t)
= −ρ
¨
h(H − h) +
1
2
ρ
˙
h
2
+ p
atm
+ gρ(H − h).
Quite a lot of these terms cancel, and we are left with
2ρH
¨
h + 2gρh = 0.
Simplifying terms, we get
¨
h +
g
H
h = 0.
So this is simple harmonic motion with the frequency
p
g
H
.
Example
(Oscillations of a bubble)
.
Suppose we have a spherical bubble of
radius
a
(
t
) in some fluid. Spherically symmetric oscillations induce a flow in the
fluid. This satisfies
∇
2
φ = 0 r > a
φ → 0 r → ∞
∂φ
∂r
= ˙a r = a.
In spherical polars, we write Laplace’s equation as
1
r
2
∂
∂r
r
2
∂φ
∂r
= 0.
So we have
φ =
A(t)
r
,
and
u
r
=
∂φ
∂r
= −
A(t)
r
2
.
This certainly vanishes as r → ∞. We also know
−
A
a
2
= ˙a.
So we have
φ = −
a
2
˙a
r
.
Now we have
∂φ
∂t
r=a
= −
2a ˙a
2
r
−
a
2
¨a
r
r=a
= −(a¨a + 2 ˙a
2
).
We now consider the pressure on the surface of the bubble.
We will ignore gravity, and apply Euler’s equation at the bubble surface and
at infinity. Then we get
−ρ(a¨a + 2 ˙a
2
) +
1
2
ρ ˙a
2
+ p(a, t) = p
∞
.
Hence we get
ρ
a¨a +
3
2
˙a
2
= p(a, t) − p
∞
.
This is a difficult equation to solve, because it is non-linear. So we assume we
have small oscillations about equilibrium, and write
a = a
0
+ η(t),
where η(t) a
0
. Then we can write
a¨a +
3
2
˙a
2
= (a
0
+ η)¨η +
3
2
˙η
2
= a
0
¨η + O(η
2
).
Ignoring second order terms, we get
ρa
0
¨η = p(a, t) − p
∞
.
We also know that
p
∞
is the pressure when we are in equilibrium. So
p
(
a, t
)
−p
∞
is the small change in pressure δp caused by the change in volume.
To relate the change in pressure with the change in volume, we need to know
some thermodynamics. We suppose the oscillation is adiabatic, i.e. it does not
involve any heat exchange. This is valid if the oscillation is fast, since there isn’t
much time for heat transfer. Then it is a fact from thermodynamics that the
motion obeys
P V
γ
= constant,
where
γ =
specific heat under constant pressure
specific heat under constant volume
.
We can take logs to obtain
log p + γ log V = constant.
Then taking small variations of p and V about p
∞
and V
0
=
4
3
πa
3
0
gives
δ(log p) + γδ(log V ) = 0.
In other words, we have
δp
p
∞
= −γ
δV
V
0
Thus we find
p(a, t) − p
∞
= δp = −p
∞
γ
δV
V
0
= −3p
∞
γ
η
a
0
.
Thus we get
ρa
0
¨η = −
3γη
a
0
p
∞
.
This is again simple harmonic motion with frequency
ω =
3γp
0
ρa
0
1/2
.
We know all these numbers, so we can put them in. For a
1 cm
bubble, we get
ω ≈ 2 × 10
3
s
−1
. For reference, the human audible range is 20
−20 000 s
−1
. This
is why we can hear, say, waves breaking.