1
Lecture Notes on Fluid Dynmics
(1.63J/2.21J)
by Chiang C. Mei, MIT
3-7-oscill-BL.tex
3.7
3.7.1
Oscillatory Boundary Layers
Stokes problem
Near the solid bottom under a wave there is a boundary layer. Let the outside flow have the
tangential velocity u = <U (x)e−iωt . Consider the ratio
Uo
uux , vuy
=O
ut
ωL
If
1
U
ν
ωL
ωL2
we let
u = u1 + u2 + . . .
(3.7.1)
and get from the Navier-Stokes equations the leading order approximation
∂
∂ 2 u1
∂u1
= Re U e−iωt + ν 2
∂t
∂t
∂y
y>0
(3.7.2)
subject to the boundary conditions that
u1 → Re U e−iωt
y→∞
(3.7.3)
and
u1 = 0
Let
y=0
(3.7.4)
h
u1 = Re û1 (x, y)e−iωt + U e−iωt
then
−iωU − iω û1 = −iωU + ν
Therefore,
û1 = −U1 (x),
(3.7.5)
d2 û1
dy 2
d2 û1 iω
+ û1 = 0
dy 2
ν
û1 → 0,
i
y→∞
y=0
(3.7.6)
(3.7.7)
(3.7.8)
2
The solution is
or,
The sign of
√
ω
û1 = −U (x) exp −(1 − i)y
2ν
ω
u1 = < U (x) 1 − exp −(1 − i)y
2ν
r
(3.7.9)
e
−iωt
(3.7.10)
−i is chosen so that (3.7.7) is satisfied. The boundary layer thickness is
δ=
3.7.2
r
s
2ν
ω
(3.7.11)
Induced Streaming
If the inviscid outer flow has tangential variation
the boundary layer. By continuity:
dU
dx
6= 0, then there is transverse flow v1 in
i
dU y h
∂u1
dy = ie−iωt
1 − e−(1−i)y/δ dy
dx 0
0 ∂x
(
)
h
i
dU
δ
y−
1 − e−(1−i)y/δ
= −e−iωt
dx
1−i
v1 = −
Z
y
Z
(3.7.12)
which is valid in y ≤ O(δ) only.
Let us examine the second order:
∂u2
∂ 2 u2
dU
∂u1
∂u1
−ν 2 = U
− u1
+ v1
∂t
∂y
dx
∂x
∂y
"
#
dU
∂ (u1 u1 ) ∂ (u1 v1 )
= U
−
+
dx
∂x
∂y
!
Since U, u1 and v1 are simple harmonic in time, the solution for u2 must have zeroth and
second harmonics. Focussing on the zeroth harmonic by taking the average over a period
∂ 2 ū2
dU
∂u1 u1 ∂u1 v1
−
+
−ν 2 = U
∂y
dx
∂x
∂y
!
On the right-hand-side the last two terms u1 u1 , u1 v1 are wave-induced Reynolds stresses. In
particular ρu1 u1 is the rate of transporting x-momentum in the x-directionm, and ρu1 v1 is
the rate of transporting x-momentum in y-direction.
Alternatively:
∂ 2 ū2
1 ∂ 2 1 ∂ 2
∂u1
−ν 2 =
U −
u1 − v 1
∂y
2 ∂x
2 ∂x
∂y
Let
α = (1 − i)/δ
(3.7.13)
3
Since
v1 = ie−iωt
1 dU αy − 1 + e−αy
α dx
∂u1
= αU (x)e−iωt e−αy
∂y
"
#
1
dU α∗ −α∗ y ∂u1
−v1
= Re U ∗
e
αy − 1 + e−αy
∂y
2
dx α
Thus
i
1 d|U |2 h
∂ 2 ū2
−αy
−α∗ y
=
G(y)
≡
1
−
1
−
e
1
−
e
∂y 2
2 dx
dU α∗ −α∗ y +Re U ∗
e
αy − 1 + e−αy
dx α
Z ∞
∂ ū2
ν
=
G(y 0 )dy 0
∂y
y
−ν
ν ū2 =
Z
y
0
= −y
dy
Z
00
∞
Z
∞
G(y 0 )dy 0
y 00
G(y 0 )dy 0 +
y
Z
y
0
y 00 G(y 00 )dy 00
One more integration gives
−ω ū2 = Re F U
dU ∗
dx
1
where F = − (1 − 3i)e−(1−i)η −
2
1
+ (1 + i)ηe−(1−i)η +
2
i −(1+i)η 1
e
− (1 + i)e−2η
2
4
3
(1 − i)
4
Note that as y → ∞, just outside the boundary layer,
"
3
dU ∗
ū2 = − Re (1 − i) U
4ω
dx
#
(3.7.14)
By Taylor expansion we can show that for η 1,
"
η dU ∗
ū2 ≈ Re (1 + i) U
2 dx
#
(3.7.15)
Example : Surface gravity waves
On the free surface of water of contant depth h, let the vertical displacement be
h
ζ = Re A eikx + Re−ikx e−iωt
i
(3.7.16)
4
where R denotes the reflection coefficient. The frequency ω frequency and and the wavenumber k are related by
ω 2 = gk tanh kh
(3.7.17)
The corresponding velocity potential is
"
igA cosh k(z + h) ikx
e + Re−ikx e−iωt
Φ = Re −
ω
cosh kh
#
(3.7.18)
The inviscid horizontal velocty just above the bed boundary layer is
h
i
∂
gkA
Φ(x, −h, t) =
Re eikx − Re−ikx e−iωt
∂x
ω cosh kh
(3.7.19)
We can then identify
gkA ikx
e − Re−ikx
ω cosh kh
For purely progressive waves, R = 0
U=
U=
gkA
eikx
ω cosh kh
(3.7.20)
(3.7.21)
hence
dU ∗
gkA
= −ik
e−ikx
dx
ω cosh kh
The induced streaming velocity is,
3
gkA
ū2 (∞) =
k
4ω
ω cosh kh
(3.7.22)
!2
(3.7.23)
η 1.
(3.7.24)
at the upper edge of the boundary layer, and
η
gkA
ū2 (η) ≈
k
2ω
ω cosh kh
!2
,
near the bottom of the boundary layer. The velocity profile is monotonic in height.
For purely standing waves R = 1, we have
U=
and
gkA
2i sin kx
ω cosh kh
dU ∗
gkA
= −2ik
cos kx
dx
ω cosh kh
Hence
dU ∗
U
=
dx
gkA
ω cosh kh
!2
2k sin 2kx
(3.7.25)
(3.7.26)
(3.7.27)
5
It follows that
gkA
3
ū2 (∞) = − k
4ω
ω cosh kh
and
gkA
η
k
ū2 (η) ≈
2ω
ω cosh kh
!2
!2
sin 2kx
sin 2kx,
(3.7.28)
η 1.
(3.7.29)
Thus near the bottom of the boundary layer, the streaming velocty converges toward points
beneath the amplitude minima. Near the top, the opposite is true. See Figure (3.7.2).
3.7.3
Physics of the Induced Streaming
Take progressive water waves as an example: We have outside the boundary layer,
u∞ = Uo cos(ωt − kx)
(3.7.30)
and inside the boundary layer,
h
u = Uo cos(ωt − kx) − e−y/δ cos(ωt − kx − y/δ)
i
(3.7.31)
6
where the velocity amplitude Uo is related to the surface amplitude A by
Uo =
Aω
gkA
=
ω cosh kh
sinh kh
(3.7.32)
Let us find the induced transverse velocity v
∂u
= Uo sin(ωt − kx) − Uo e−y/δ sin(ωt − kx − y/δ)
∂x
v∞ = −
Now
Z
yδ
0
∂u
1
1
dy = −y Uo sin(ωt − kx) − Uo kδ cos(ωt − kx) + Uo kδ sin(ωt − kx)
∂x
2
2
1
u∞ v∞ = − Uo2 kδ < 0
4
where the sin(ωt−kx) terms in v∞ are out of phase with u∞ by π/2, hence does not contribute
to the mean.
Now consider a slice of boundary layer one wavelength long. Because of periodicity, there
is no net transfer of momentum or forces at two ends x0 and x0 + 2π/k. But the momentum
2
transfer downwards is U4o kδ, causing a positive shear stress. To balance it there must be a
non-zero µ ∂∂yū at all levels y below the top. Hence, the induced streaming velocity is created
and ū 6= 0.
Figure 3.7.1: Reynolds stress and Induced streaming in Stokes layer
.