# User manual | Chapters1_5

GEFD
Course description:Pictorial introduction
Part III:
Geophysical and Environmental Fluid Dynamics
9:00 in MR 11,
Monday, Wednesday, Friday
Michaelmas 2010
Stuart Dalziel
[email protected]
H.0.11 01223 337911
Department of Applied Mathematics and Theoretical Physics
University of Cambridge
Cambridge CB3 0WA
UK
http://www.damtp.cam.ac.uk/lab/people/sd103/
–1–
GEFD
Course description:Pictorial introduction
Course description
Fluid dynamics is an essential ingredient of the environment in which we live,
and improving our understanding is essential in any attempt to avert both
localised ecological disasters and climate change. At the largest scales, the
Earth’s rotation plays an important role, while density stratification influences
the flow over scales ranging from millimetres to the depth of the atmosphere.
This course focuses on the fundamental processes controlling these geophysical
and environmental flows, whist recognising the immense intellectual
challenges that lie before us.
We begin the course by considering systematic flows driven by or inhibited by
density differences. Internal gravity waves, where the density variations {\it
within} the fluid provide the restoring force, radiate energy vertically as well as
horizontally, and their interaction with boundaries can focus this energy and
cause mixing far from where the energy was input. Imposed horizontal density
Localised sources of density variation can drive convection or create plumes
(steady or time-dependent) that are, in turn, influenced by an ambient density
stratification. We shall also study mixing, the effect of turbulence on density
stratifications, and how the stratification alters the characteristics of the
turbulence.
If the timescale for the motion is not short compared with a day, then the
Earth’s rotation can begin to play a role. The additional timescale this
introduces modifies the dynamics in a profound way for both homogeneous
and density stratified flows. The Coriolis force (a fictitious force arising from
our use of a frame of reference rotating with the planet) causes a moving parcel
of fluid to experience a force directed to its right in the Northern hemisphere
(or its left in the Southern hemisphere). We shall explore how the Coriolis
force introduces new wave mechanisms (e.g. inertial and Rossby waves), how
the spread of a gravity current is arrested to form a front’ that then becomes
unstable, and how turbulence starts to become two-dimensional. While rotation
can complicate the dynamics, at high rotation rates it can also simplify it.
–2–
GEFD
A combination of classical examples and current research will be used to
illustrate these topics. Laboratory demonstrations of some examples will also
be included in the non-examinable course ‘Demonstrations in Fluid Mechanics’
(Thursdays at 2pm throughout Michaelmas term).
Examples Class
• To be arranged…
Surgery
• Best to arrange after lecture or by e-mail.
–3–
GEFD
–4–
GEFD
Introduction:Pictorial introduction
1 Introduction
Aims:
• To develop an understanding of the kinds of motion possible in density
stratified fluids, and to understand how rotation modifies both stratified and
homogeneous flows
• To derive simple yet powerful mathematical models for describing these
flows
1.1 Pictorial introduction
warm air
Air warmed over land
Cooler air over sea
river water
Flow through porous
medium
Sedimentation &
resuspension
Figure 1: Sketch of coastline showing a range of buoyancy-driven flows.
–5–
GEFD
Introduction:Pictorial introduction
(a)
(b)
Figure 2: Internal gravity waves. (a) From an oscillating cylinder, and (b) reflecting from a
rough topography.
(a)
(b)
Figure 3: (a) Gravity current of dense fluid. (b) Particle-laden intrusion along an interface
between two homogeneous layers.
Figure 4: (a) Kelvin-Helmholtz and (b) Rayleigh-Taylor instabilities.
–6–
GEFD
Introduction:Pictorial introduction
(a)
(b)
Figure 5: (a) Particle streaks in ‘classical’ turbulence. (b) Structure formation in rotating
turbulence.
(a)
(b)
Figure 6: (a) Dense thermal and (b) buoyant plume, both in a homogeneous, quiescent
environment.
–7–
GEFD
Introduction:Revision of basic concepts
(a)
(b)
Figure 7: (a) Baroclinic instability and (b) baroclinic vortices on a topographic beta plane.
1.2 Revision of basic concepts
1.2.1 Equations of motion
Revision of basic concepts will help with later material
w + ∂w/∂z δz
v + ∂v/∂y δy
u – ∂u/∂x δx
u + ∂u/∂x δx
v – ∂v/∂y δy
w – ∂w/∂z δz
Figure 8: Fluid element (2δx × 2δy × 2δz) and velocities in and out.
1.2.1.1 Continuity
Conservation of volume for an incompressible fluid
–8–
GEFD
Introduction:Revision of basic concepts
Flux_in = (u − ∂u/∂x δx) 4 δy δz
+ (v − ∂v/∂y δy) 4 δx δz
+ (w − ∂w/∂z δz) 4 δx δy
+ O(δ4)
Flux_out = (u + ∂u/∂x δx) 4 δy δz
+ (v + ∂v/∂y δy) 4 δx δz
+ (w + ∂w/∂z δz) 4 δx δy
+ O(δ4)
Rate_of_accumulation = 0 if incompressible
Equate: Rate_of_accumulation = Flux_in – Flux_out:
⎛ ∂u ∂v ∂w ⎞
0 = −⎜ + +
⎟ = −∇ ⋅ u
∂
∂
∂
x
y
z
⎝
⎠
1.2.1.2 Conservation of mass
Flux_in = (u − ∂u/∂x δx)(ρ − ∂ρ/∂x δx) 4 δy δz
+ (v − ∂v/∂y δy)(ρ − ∂ρ/∂y δy) 4 δx δz
+ (w − ∂w/∂z δz)(ρ − ∂ρ/∂z δz) 4 δx δy
+ O(δ4)
Flux_out = (u + ∂u/∂x δx)(ρ + ∂ρ/∂x δx) 4 δy δz
+ (v + ∂v/∂y δy)(ρ + ∂ρ/∂y δy) 4 δx δz
+ (w + ∂w/∂z δz)(ρ + ∂ρ/∂z δz) 4 δx δy
+ O(δ4)
Rate_of_accumulation = 8 δx δy δz ∂ρ/∂t
Equate: Rate_of_accumulation = Flux_in – Flux_out:
⎛ ∂ρ
⎛ ∂u ∂v ∂w ⎞
∂ρ
∂ρ
∂ρ ⎞
= −⎜⎜ u
+v
+ w ⎟⎟ − ρ ⎜⎜ + +
⎟⎟ = −∇ ⋅ (ρu )
∂t
∂y
∂z ⎠
⎝ ∂x
⎝ ∂x ∂y ∂z ⎠
⇒
∂ρ
+ ∇ ⋅ (ρu ) = 0
∂t
–9–
(1)
GEFD
Introduction:Revision of basic concepts
⎛∂
⎞
⎜ + u ⋅ ∇ ⎟ ρ = − ρ∇ ⋅ u
⎝ ∂t
⎠
or
(2)
In general, the rate of accumulation of some quantity φ is balanced by the
divergence of the fluxes Fφ and possibly a source term.
∂φ
+ ∇ ⋅ Fφ = source .
∂t
(3)
This is often referred to as ‘conservative form’. For conservation of mass, we
identify φ with ρ and note that the flux Fρ = ρu due to advection of density, and
the source term is zero.
1.2.1.3 x−Momentum
The x momentum is ρ u.
∂ρ u
+ ∇ ⋅ ( uρ u ) = source
∂t
Hence
(4)
Pressure provides a source
p - ∂p/∂x δx
so
p + ∂p/∂x δx
∂ρ u
∂p
+ ∇ ⋅ ( uρ u ) = − + source .
∂t
∂x
(5)
For a viscous fluid, we need to take into account viscous stresses: μ∇u
so
∂ρ u
∂p
+ ∇ ⋅ ( uρ u ) = − + ∇.( μ∇u ) .
∂t
∂x
(6)
Expand, for a Newtonian fluid
∂p
⎡ ∂ρ u
⎤
⎡ ∂u
⎤
+ ∇ ⋅ ( uρ ) ⎥ + ρ ⎢ + ( u ⋅ ∇ ) u ⎥ = − + μ∇ 2u
u⎢
∂x
⎣ ∂t
⎦
⎣ ∂t
⎦
– 10 –
(7)
GEFD
Introduction:Revision of basic concepts
Recalling mass conservation shows
1 ∂p
∂u
+ ν∇ 2 u .
+ (u ⋅ ∇ )u = −
∂t
ρ ∂x
(8)
Similarly for y momentum…
1.2.1.4 z−Momentum
Derivation essentially the same as for x-momentum, except have an additional
source term due to gravity. Often write
∂w
1 ⎛ ∂p
⎞
+ ( u ⋅ ∇ ) w = − ⎜ + ρ g ⎟ + ν∇ 2 w ,
ρ ⎝ ∂z
∂t
⎠
(9)
where g acceleration due to gravity.
Tempting to write
∂w
1 ∂
+ (u ⋅ ∇ ) w = −
( p + ρ gz ) + ν∇ 2 w
∂t
ρ ∂z
(10)
for a homogeneous fluid, BUT WATCH OUT if ρ = ρ(x)!
1.2.2 Boussinesq approximation
The momentum equations are combined and written as
∂u
1
+ (u ⋅ ∇ )u = − ∇p − gzˆ + ν∇ 2u .
∂t
ρ
(11)
In many cases of interest in this course we may write ρ = ρ0 + ρ′, where ρ0 is a
the convective accelerations ∂u/∂t + u⋅∇u << g, then we can apply the
Boussinesq approximation. Under this approximation, then density differences
are important only when multiplied by g. We may thus rewrite (11) as
∂u
1
+ (u ⋅ ∇ )u = − ∇( p + ρ 0 gz ) − g ′zˆ + ν∇ 2u ,
ρ0
∂t
– 11 –
(12)
GEFD
Introduction:Revision of basic concepts
where g′ is the ‘reduced gravity’ defined by
g′ =
ρ − ρ0
g.
ρ0
(13)
For flows where the density is constant, it can be more convenient to write
them as
∂u
1
+ (u ⋅ ∇ )u = − ∇( p + ρgz ) + ν∇ 2u .
∂t
ρ
(14)
Caution!
1.2.3 Scalar transport
Suppose the flow contains a solute described by the mass concentration S (i.e.
the mass of solute per unit volume is ρS). In general S is not only advected by
the flow, but also diffuses following Fick’s law where there is a diffusive mass
flux of solute or heat proportional to its gradient:
FS = −ρκS ∇S,
(15)
where κS is the diffusivities of the solute.
Figure 10: Fick’s law. The number of ‘particles’ of solute moving left in high
concentration regions is greater than in low concentration regions, leading to a net flux
towards the low concentration regions.
Combining these fluxes with the normal advected flux ρSu gives
∂ρS
+ ∇ ⋅ (ρSu − ρκ S ∇S ) = 0 .
∂t
Rearranging
– 12 –
GEFD
Introduction:Revision of basic concepts
⎡ ∂S
⎡ ∂ρ
⎤
⎤
+
u
⋅
∇
−
∇
⋅
∇
ρ
ρ
(
)
(
ρκ
)
+
S
S
S
S
⎢⎣ ∂t
⎢⎣ ∂t + ∇ ⋅ (ρu )⎥⎦ = 0 .
⎥⎦
The second term vanishes from mass conservation (2). If ρκS is constant (this is
often a good approximation), then the first term may be simplified to the
∂S
+ (u ⋅ ∇ )S = κ S ∇ 2 S .
∂t
(16)
1.2.4 Heat
Heat (enthalpy) H per unit mass may be treated in a similar way to a solute.
Here the diffusive flux is due to temperature, leading to
∂ρH
+ ∇ ⋅ (ρHu − ρκ H ∇H ) = 0 ,
∂t
where κH is the diffusivity of the heat. The principal contribution to the heat is
from the temperature of the fluid. Introducing the heat capacity per unit mass cp
such that ∂H/∂T = cp (at constant pressure) then we can write κH∇H as cpκT∇T
and note that κH = κT. Simplifying using mass conservation yields
∂T
⎡
⎤
⎡ ∂ρ
⎤
(
)
(
)
(
)
c
c
T
c
T
H
ρ
+
ρ
u
⋅
∇
−
∇
⋅
ρ
κ
∇
+
+
∇
⋅
ρ
u
p
p
p
T
⎢⎣
⎥⎦
⎢⎣ ∂t
⎥⎦ = 0
∂t
which, for constant ρcpκT simplifies to
∂T
+ (u ⋅ ∇ )T = κ T ∇ 2T .
∂t
(17)
1.2.5 Equation of state
Often we express the density ρ in terms of an equation of state that depends on
properties such as the concentration of a solute, temperate, and pressure. For
the present discussion we shall assume the flow is incompressible so that
∂ρ/∂p = 0 and suppose ρ = ρ(S,T).
In many situations the density changes due to the solute and temperature are
small, allowing us to linearise this equation of state to
– 13 –
GEFD
Introduction:Vorticity equation
ρ(S,T) ≈ ρ0(1 − α(T−T0) + β(S − S0))
(18)
where ρ0 = ρ(S0,T0).
If there are variations in only one component (i.e. S or T but not both) then we
may rewrite the advection diffusion equation in terms of density
∂ρ
+ (u ⋅ ∇ )ρ = κ∇ 2 ρ ,
∂t
(19)
where κ is the appropriate diffusivity.
1.3 Vorticity equation
Take the curl of the momentum equation (11):
1
∂ω
+ (u ⋅ ∇ )ω = (ω ⋅ ∇ )u + 2 ∇ρ × ∇p + ν∇ 2ω ,
∂t
ρ
(20)
where ω = ∇×u is the vorticity.
However, under the Boussinesq approximation, where the convective
accelarations are much smaller than g, and so density differences are important
only when multiplied by g, we can take the curl of (12). This eliminates the
pressure (∇×∇ ≡ 0) and yields
∂ω
g
+ (u ⋅ ∇ )ω = (ω ⋅ ∇ )u + zˆ × ∇ρ + ν∇ 2ω .
∂t
ρ0
Vortex
stretching
1.4 Energy equation
Baroclinic
torque
Dot the momentum equation with the velocity
⎡ ∂ρu
⎤
u⋅⎢
+ (u ⋅ ∇ )ρu = −∇p − ρgzˆ + ρν∇ 2u ⎥
⎣ ∂t
⎦
– 14 –
(21)
GEFD
Introduction:Bernoulli equation
⎛∂
⎞
⎛∂
⎞
u ⋅ ρgzˆ = ⎜ + u ⋅ ∇ ⎟(ρgz ) − gz ⎜ + u ⋅ ∇ ⎟ ρ
⎝ ∂t
⎠
⎝ ∂t
⎠
⎛∂
⎞
Noting that
= ⎜ + u ⋅ ∇ ⎟(ρgz ) − gzκ∇ 2 ρ
⎝ ∂t
⎠
∂ρ
⎛∂
⎞
= ⎜ + u ⋅ ∇ ⎟(ρgz ) − ∇ ⋅ ( gzκ∇ρ ) + gκ
∂z
⎝ ∂t
⎠
1
2
2
u ⋅ ∇ 2u = ∇ 2 u − (∇u ) , and that for an incompressible flow u⋅∇p = ∇⋅(up),
2
then
Viscous
ΔPE ← Diffus.
Work
dissipation
density change
(
)
(
)
∂ρ
2
2
⎛∂
⎞
2
− ρν (∇u )
⎜ + u ⋅ ∇ ⎟ 12 ρ u + ρgz + ∇ ⋅ up − 12 ρν∇ u − gzκ∇ρ = − gκ
∂z
⎝ ∂t
⎠
Kinetic
energy
density
Potential
energy
density
Energy
flux
1.5 Bernoulli equation
For inviscid flow of uniform density can write energy equation as
1
⎡ ∂u
⎤
u ⋅ ⎢ + (u ⋅ ∇ )u + ∇( p + ρgz )⎥ = 0 ,
ρ
⎣ ∂t
⎦
Rearranging
2
1 ∂u
p
⎛
⎞
2
+ (u ⋅ ∇ )⎜ 12 u + + gz ⎟ = 0 .
2 ∂t
ρ
⎝
⎠
If u = ∇ϕ, then
⎛ ∂
∂
∂ ⎞
2
⎜⎜ u + v + w ⎟⎟ϕ = u
∂y
∂w ⎠
⎝ ∂x
so
– 15 –
Diffusion
raising
centre of
mass
GEFD
Introduction:Bernoulli equation
(u ⋅ ∇ )⎛⎜ ∂ϕ + 12 u 2 +
⎝ ∂t
⎞
+ gz ⎟ = 0 .
ρ
⎠
p
(22)
Definition Bernoulli potential
B=
∂ϕ 1 2 p
+ u + + gz
∂t 2
ρ
(23)
is conserved along a streamline.
B = ½|u|2 + p/ρ + gz.
– 16 –
(24)
GEFD
Surface and interfacial waves:Linear water waves
2 Surface and interfacial waves
2.1 Linear water waves
Reminder of what you already know
λ
cp
H
Figure 11: Sketch of linear wave.
Assumption: Flow inviscid, irrotational
Boundary conditions:
• At bottom (z = −H), w = 0.
• At free surface (z = η), w = ∂η/∂t + u∂η/∂x, and p = p0.
Interior
Irrotational
u = ∂ϕ/∂x, w = ∂ϕ/dz and ∇2ϕ = 0.
2.1.1 Linear waves
Assumption: Wave amplitude small compared with the depth
Unperturbed state:
u = 0,
– 17 –
GEFD
Surface and interfacial waves:Linear water waves
p = p0 − ρgz
For linear perturbations u′, v′, w′, p′:
∇⋅u = 0 →
∂u′ ∂v′ ∂w′
+
+
=0
∂x ∂y ∂z
(25)
∂u
1
+ (u.∇ )u = − ∇( p + ρgz ) →
∂t
ρ
→
∂u′
1 ∂p′
=−
ρ ∂x
∂t
(26a)
→
∂v′
1 ∂p′
=−
ρ ∂y
∂t
(26b)
→
∂w′
1 ∂p′
=−
ρ ∂z
∂t
(26c)
Consider motion confined to x − z plane (for plane waves, can rotate our
coordinate system so that no motion or variation in y direction)
Eliminating p′:
∂ ⎛ ∂u′ ∂w′ ⎞
−
⎜
⎟=0
∂t ⎝ ∂z
∂x ⎠
Eliminating u′
∂ ⎛ ∂2
∂2 ⎞
⎟ w′ = 0
⎜
+
∂t ⎜⎝ ∂x 2 ∂z 2 ⎟⎠
At surface (z = 0)
At bottom (z = −H)
(27)
w′ = ∂η/∂t
w=0
Suppose there is a wave of the form η = η0 ei(kx − ωt),
[
Look for solution of the form w′(x, z , t ) = ℜ wˆ (z )e i (kx−ωt )
]
⎛ 2
d 2 wˆ ⎞ i (kx−ωt )
∂ 2
=0
∇ w′ = −iω ⎜⎜ − k wˆ + 2 ⎟⎟e
dz
∂t
⎠
⎝
– 18 –
GEFD
Surface and interfacial waves:Linear water waves
wˆ = A cosh (kz ) + B sinh (kz )
⇒
⎛
tanh (kz ) ⎞
⎟⎟
w′ = 0 at z = −H ⇒ wˆ = A cosh (kz )⎜⎜1 +
(
)
tanh
kH
⎠
⎝
Since w′ = dη/dt = −iω η0 ei(kx − ωt), then A = −iωη0 so
⎛
⎛
tanh (kz ) ⎞ i (kx−ωt ) ⎞
⎟
⎟⎟e
w′ = ℜ⎜⎜ − iωη0 cosh (kz )⎜⎜1 +
⎟
⎝ tanh (kH ) ⎠
⎠
⎝
⎛
tanh (kz ) ⎞
⎟⎟ sin (kx − ωt )
= ωη0 cosh (kz )⎜⎜1 +
(
)
tanh
kH
⎝
⎠
(28)
η = ℜ(η0 ei (kx−ωt ) ) = η0 cos(kx − ωt )
(29)
Also have p′ = p0 + ρgη at z = 0.
Using continuity (25) to eliminate u′ from (26a)
∂ 2 w′
∂ 2η
= g 2 at z = 0
∂t∂z
∂x
⇒
⎡
ω 2 kη0 ⎢sinh (kz ) +
⇒
Dispersion relation
⎣
⎡
⎣
(30)
cosh kz ⎤
2
=
k
gη0 at z = 0,
tanh kH ⎥⎦
ω 2 ⎢0 +
1 ⎤
= kg
tanh kH ⎥⎦
ω 2 = gk tanh (kH )
(31)
Exercise: Derive the dispersion relationship by introducing the velocity
potential ϕ and work with this rather than the velocity perturbations u′ and w′.
2.1.2 Phase velocity
The various flow properties in the wave (e.g. surface height or velocity) are all
proportional to eiφ, where φ = k⋅x − ωt is the phase of the wave. Clearly
– 19 –
GEFD
Surface and interfacial waves:Linear water waves
∂φ
∂φ
= ki and
= −ω
∂t
∂xi
Consider the identity
∂φ ∂φ ∂φ ∂φ
−
= 0,
∂xi ∂t ∂t ∂xi
and substitute for φ in each half of this expression
ki
∂φ
∂φ
+ω
= 0.
∂t
∂xi
Multiplying through by ki/|k|2 and rearranging
⎡ ∂ ωk ∂ ⎤
⎡∂
⎤
⎢ + 2i
⎥φ = ⎢ + c p ⋅ ∇ ⎥φ = 0
⎣ ∂t
⎦
⎢⎣ ∂t k ∂xi ⎥⎦
where
cp =
ω
k
2
(32)
k
is the phase velocity.
For the surface waves considered here
Phase velocity
12
ω
⎛g
⎞
c p = = ⎜ tanh (kH )⎟ .
k ⎝k
⎠
(33)
2.1.3 Superposition
Since these are linear waves, we can use linear superposition to construct more
complex wave fields.
Consider the linear superposition of two modes differing in wave number by
2δk and frequency by 2δω:
η = cos[(k + δk)x − (ω + δω)t] + cos[(k − δk)x − (ω − δω)t].
– 20 –
GEFD
Surface and interfacial waves:Linear water waves
This may of course be expressed in terms of the sum and difference,
η = 2 cos[δk x − δω t] cos[kx − ωt].
Figure 12: Sketch showing a group of waves created by linear superposition.
If δk << k, then
2 cos[δk x − δω t] ≈ 2 cos[(x − ∂ω/∂k t)δk]
and we may consider this as a train of approximately sinesoidal waves of
frequency ω and wavenumber k with an amplitude envelope that moves at
speed ∂ω/∂k.
12
⎤⎛ g
∂ω 1 ⎡
kH
⎞
= ⎢1 +
cg =
⎥⎜ tanh (kH )⎟
∂k 2 ⎣ cosh (kH )sinh (kH ) ⎦⎝ k
⎠
⎤
1⎡
kH
Group velocity
= ⎢1 +
⎥c p
2 ⎣ cosh (kH )sinh (kH ) ⎦
1⎡
2kH ⎤
= ⎢1 +
⎥c p
2⎣
sinh (2kH ) ⎦
2.1.4 Group velocity
From our definition of the phase φ we have
Eliminating φ between these
∂φ
∂φ
= ki and
= −ω .
∂xi
∂t
∂ki ∂ω
+
= 0.
∂t ∂xi
Since ω = ω(k) then
– 21 –
(34)
GEFD
Surface and interfacial waves:Linear water waves
∂ki ∂k j ∂ω
+
= 0,
∂t ∂xi ∂k j
∂k j
∂k
∂ 2φ
= i , so
but
=
∂xi ∂xi ∂x j ∂x j
∂ki ∂ki ∂ω ⎛ ∂
⎞
+
= ⎜ + c g ⋅ ∇ ⎟k = 0
∂t ∂x j ∂k j ⎝ ∂t
⎠
where
c gi =
∂ω
∂ki
(35)
is the group velocity. The group velocity can be interpreted as the speed at
which the wavenumber propagates.
For a broad class of waves the group velocity also represents the speed and
direction of energy propagation. Moreover, an observer moving at cg will see a
constant wavenumber.
z
z
(a) (b)
x
t
t
cg
cp
x
(c)
Figure 13: Group of surface waves. (a) Snap shot, (b) time series at left-hand end, (c) time
series at surface.
– 22 –
GEFD
Surface and interfacial waves:Linear water waves
2.1.5 Effects of dispersion
As the dispersion relation means that different wave modes travel at different
speeds, a given spectrum of waves will separate out into its component parts.
z
x
(a)
z
t
(b)
(c)
– 23 –
GEFD
Surface and interfacial waves:Linear water waves
cp
t
x
cp
(d)
Figure 14: Train of surface waves. (a) Snap shot, (b) time series at left-hand end, (c) at
right-hand end, and (d) time series at surface at left.
2.1.6 Limits
g
,
k
Deep water
cp =
Shallow water:
c p = gH ,
– 24 –
1
cg
2
(36)
cg = c p
(37)
cg =
GEFD
Surface and interfacial waves:Linear water waves
Figure 15: Comparison of the speed of linear waves with the deep and shallow water
limits.
Figure 16: Waves on a free surface (from Album of Fluid Motion, M. Van Dyke)
2.1.7 Orbits
From (28) we have
tanh (kz ) ⎞
⎛
w′ = ωη0 cosh (kz )⎜1 +
⎟ sin (kx − ωt ) ,
tanh
(
)
kH
⎝
⎠
– 25 –
GEFD
Surface and interfacial waves:Linear water waves
and noting that ∂u′/∂x + ∂w′/∂z = 0, then
∂u ′
∂w′
cosh (kz ) ⎞
⎛
=−
= −ωη 0 k ⎜ sinh (kz ) +
⎟ sin (kx − ωt ).
tanh
(
)
∂x
∂z
kH
⎝
⎠
Integrating with respect to x
cosh (kz ) ⎞
⎛
u′ = ωη 0 ⎜ sinh (kz ) +
⎟ cos(kx − ωt ) .
tanh
(
)
kH
⎝
⎠
The velocity at a given point rotates with the passage of each wave. At the
surface (z = 0), the magnitudes of both u′ and w′ are not zero (ωη0). However,
at the bottom z = −H the vertical velocity vanishes, but the horizontal velocity
does not.
Suppose the location of a fluid particle is x = X(t), z = Z(t). Clearly
dX
=u
dt
and
dZ
= w.
dt
Assuming η0k is small (necessary for linear wave approximation), then fluid
Figure 17: Sketch of fluid particle orbits beneath a wave.
The above picture is valid only for short times and so long as the excursions of
fluid particles are small compared with their wave length.
– 26 –
GEFD
Surface and interfacial waves:Linear water waves
2.1.8 Stokes drift
Note that the fluid at the crest of a wave is moving in the direction of the wave,
whereas the fluid in the trough of a wave is moving back in the opposite
direction.
cp
Figure 18: Illustration of the Stokes drift mechanism.
We can write the velocity potential for linear waves on deep water as
ϕ =a
ω
k
e kz sin ( kx − ωt ) ,
with ω2 = gk, as before. The Lagrangian position of a fluid parcle is governed
by
dX
= u ( X , Z ,t )
dt
and
dZ
= w( X , Z ,t ) ,
dt
but we can expand u(X,Z,t) aboutu(x,z,t), etc. We, without loss of generality,
begin by setting X(0) = x, Z(0) = z. Here we are interested primarily in the
horizontal drift, so are interested in
u ( X , Z , t ) = u ( x, z , t ) + ( X − x )
∂u
∂u
+ (Z − z) +
∂x
∂z
or more specifically the difference between u(X,Z,t) and u(x,z,t).
To the first order we have
– 27 –
,
GEFD
Surface and interfacial waves:Linear water waves
X −x=∫
dX
dt = ∫ u ( X , Z , t ) dt
dt
≈ ∫ u ( x, z , t ) dt = ∫ − aω e kz cos ( kx − ωt ) dt = − ae kz sin ( kx − ωt )
Z−z=∫
dZ
dt = ∫ w ( X , Z , t ) dt
dt
≈ ∫ w ( x, z , t ) dt = ∫ aω ekz sin ( kx − ωt ) dt = ae kz cos ( kx − ωt )
So
u S = u ( X , Z , t ) − u ( x, z , t )
= ( −ae kz sin ( kx − ωt ) ) ( − akω e kz sin ( kx − ωt ) )
+ ( ae kz cos ( kx − ωt ) ) ( akω e kz cos ( kx − ωt ) ) .
= a 2 kω e 2 kz ( sin 2 ( kx − ωt ) + cos 2 ( kx − ωt ) )
= a 2 kω e 2 kz
This difference in velocity, referred to as Stokes Drift, is independent of time
and is second order in the amplitude of the wave.
Exercise: Compute the Stokes drift when the depth is finite.
2.1.9 Boundary layers
At z = −H, a no-slip boundary layer forms. From dimensional analysis, we can
estimate the thickness of this layer as
δ ~ (ν/ω)1/2.
Detailed analysis of the structure of the boundary layer shows an exponentially
decaying sinusoid:
[
]
u ′′ = u′ℜ 1 − e −(1+i ) z′ 4δ ,
where z′ is the distance from the bottom.
– 28 –
GEFD
Surface and interfacial waves:Linear water waves
At the surface we assume there is a stress-free boundary, and so we expect the
normal gradient in velocity to vanish. However, due to the curvature in the
surface, the normal gradient of the potential flow solution does not vanish, and
a boundary layer forms. This boundary layer has a similar structure to that on
the bottom, with a decaying sinusoidal structure.
2.1.10 Energy
Consider the kinetic energy, perturbation potential energy and energy flux per
unit area:
η
KE =
∫ ρ (u
1
2
2
+ v 2 + w2 )dz ,
−H
PE =
η
0
η
−H
−H
0
∫ ρgz dz − ∫ ρgz dz = ∫ ρgz dz ,
η
F=
∫ u( p − ρgz ) dz .
−H
These may be calculated from the above solution by substitution and
integration. Normally we consider these integrated over a wave period or wave
length to find that
KE = PE ,
(38)
and that the flux of energy in the direction of wave propagation is
F = E cg.
(39)
The first of these, the equipartition of energy between kinematic and potential
energy, is found frequently in a broad class of flows.
Examples sheet question: Prove the equipartition of energy and the energy flux.
– 29 –
GEFD
Surface and interfacial waves:Waves on an interface
2.2 Waves on an interface
Interfaces characterised by sudden changes in density are a common feature of
density stratified fluids. As with the free surface on a quiescent fluid, a
perturbed interface will experience a restoring force due to gravity that acts to
return the surface to horizontal.
λ
H2
ρ2
cp
H1
ρ1
Figure 19: Sketch of wave on an interface between two layers of fluid.
2.2.1 Dispersion relation
As before we consider the undisturbed state
u1 = 0 = u2,
p = p0 − ρ1gz
for −H1 < z < 0,
p = p0 − ρ2gz
for 0 < z < H2.,
across the interface. The dispersion relation
−1
⎡ ρ1
ρ2 ⎤
+
ω 2 = ( ρ1 − ρ 2 ) gk ⎢
⎥ ,
tanh
kH
tanh
kH
1
2⎦
⎣
shows that the wave frequency depends on the depth of both layers.
Examples sheet question: Derive the dispersion relation
– 30 –
(40)
GEFD
Surface and interfacial waves:Waves on an interface
2.2.2 Properties of interfacial waves
The phase velocity and group velocity are determined in the same way as for
the surface waves:
2
10
ω
1
8
0
0.4
cp
0.2
4
-0.2
0
h
8
0
6
-0.4
10
λ
6
-0.4
4
-0.2
0
2
h
0.2
λ
2
0.2
0.4
0.4
Figure 20: Normalised frequency and phase velocity for waves on an interface in the
Boussinesq limit. Here h = ½ (H1 − H2)/(H1 + H2) and the wavelength λ = 2π/k.
For a given total depth, the phase velocity cp = ω/k is maximum when the
layers are of equal depth.
As the depth of one of the layers becomes very large compared with the other,
then the waves begin to behave like those on a free surface with a depth equal
to the shallower layer, but with g replaced by g′.
Waves on an interface also exhibit other properties that are similar to surface
waves. For example, the particle orbits, the existence of boundary layers, and
the separation of wave modes of different frequencies by dispersion.
2.2.3 The short wave limit
In the short wave (deep water) limit, kH1, kH2 >>1 so that tanh kHi → 1.
Examining (40) shows
ω2 → ½ g′k
where g′ = 2g(ρ1 − ρ2)/(ρ1 + ρ2), giving the phase and group velocities as
cp = (g′/2k)1/2 and cg = ½ cp.
This is identical to (36), except with ½ g′ replacing g.
– 31 –
(41)
GEFD
Surface and interfacial waves:Kelvin ship waves
2.2.4 The long wave limit
In the long wave limit, kH1, kH2 <<1 and tanh kHi/k → Hi, so for a Boussinesq
flow,
c p = cg =
H1 H 2
g′ .
H1 + H 2
(42)
2.3 Kelvin ship waves
Consider a ship moving with speed Ship speed: U through deep water.
U
U/2
α
θ
r
Figure 21: Definition sketch for Kelvin ship wave problem.
For waves generated at O and travelling in direction θ to be stationary in frame
of reference of ship:
U cos θ = cp.
The group velocity (speed at which energy propagates) is
cg = cp/2
so distance travelled by wave packet in time x/U for ship to travel a distance
x is
r = cg t = ½ Ut cos θ = ½ x cos θ.
– 32 –
(red vector)
GEFD
Surface and interfacial waves:Kelvin ship waves
and the wave reaches a distance r sin θ from the path of the ship at a distance
x – r cos θ behind the ship.
The angle made to the path of the ship is therefore
1
r sin θ
x cosθ sin θ cosθ sin θ
.
=2 1
=
tan α =
x − r cosθ x − 2 x cos 2 θ 2 − cos 2 θ
The maximum angle α made therefore requires d(tan α)/dθ = 0
6cos ( 2θ ) − 2
d
tan
α
=
=0
(
)
2
dθ
( cos ( 2θ ) − 3)
which gives θ = ½ cos−1(1/3). This, in turn, yields sin α = 1/3 and an angle of
α = 19.47° for the causality envelope.
Note: waves inside the causality envelope will be the result of interference
between the two waves that can reach that point.
Figure 22: Sketch of the envelope of the Kelvin ship waves.
Figure 23: Ship waves.
– 33 –
GEFD
Surface and interfacial waves:Kelvin ship waves
Figure 24: Mach shock waves – non dispersive; angle changes with speed.
Figure 25: Kelvin ship waves due to islands. (a) Amsterdam Island, Southern Indian
Ocean. (b) Prince Edward Islands, Southern Indian Ocean.
– 34 –
GEFD
Internal gravity waves:The minimal mathematics version
3 Internal gravity waves
Continuously varying stratifications exist in many places in our environment.
3.1 The minimal mathematics version
Vertical oscillations
We begin by considering the forces on a parcel of fluid of volume V displaced
vertically by a distance ζ in a linear density gradient described by ∂ρ/∂z.
ζ
ρ(z)
Figure 26: A parcel of fluid displaced vertically in a linear stratification.
If we ignore continuity and any effects this might have on the forces acting on
the fluid parcel, then we may apply Newton’s laws, principally
Force = mass × acceleration.
Buoyancy:
FB = Vg
∂ρ
ζ
∂z
Mass:
m = ρ0V
Acceleration:
d 2ζ
a= 2
dt
– 35 –
GEFD
Internal gravity waves:The minimal mathematics version
⇒
d 2ζ ⎛ g ∂ρ ⎞
⎟ζ = 0
+ ⎜−
dt 2 ⎜⎝ ρ 0 ∂z ⎟⎠
so
ζ = A cos Nt + B sin Nt
Definition
N= −
g ∂ρ
ρ 0 ∂z
(43)
(44)
is the ‘buoyancy frequency’ or ‘Brunt-Väisälä frequency’.
In general we cannot neglect continuity. However, if instead of an isolated
fluid parcel we consider an entire slab of fluid, then we can apply the same
simple analysis and satisfy continuity.
ζ
ρ(z)
Figure 27: A slab of fluid displaced vertically in a linear stratification.
Since we are looking at linear waves, we need not concern ourselves about any
changes in the background density gradient ∇ρ. One consequence of this is that
we can displace neighbouring slabs by differing amounts to build up an
arbitrary profile of displacements.
Oscillating at an angle
What happens if we displace the slab at an angle rather than vertically?
Continuity will ensure it returns along the same trajectory as it was displaced.
– 36 –
GEFD
Internal gravity waves:The minimal mathematics version
θ
ρ(z)
Buoyancy:
so
ζ
Fθ = V ( g cos θ )
∂ρ
(ζ cos θ )
∂z
downward
ζ = A cos ωt + B sin ωt
and the dispersion relationship is
Definition
ω
N
= cos θ .
Forced oscillations
Suppose we force an oscillation at angular frequency ω < N. The fluid will
respond most easily by fluid motions that are in resonance with this.
– 37 –
(45)
GEFD
Internal gravity waves:The minimal mathematics version
ω
ρ(z)
Figure 28: Cartoon of St. Andrew’s cross wave pattern.
Motion is confined to beams at angles given by (45) to the vertical. These
waves combine at the cylinder to satisfy continuity.
If a sphere rather than a cylinder is oscillated, the wave beams form cones with
vertical axes of rotation.
Energy and causality
Clearly in the above example the energy driving the flow originates from the
cylinder, and propagates outwards along the wave beams to cause flow
elsewhere. Hence the group velocity is directed out along these wave beams.
This is a statement of the principle of ‘causality’
Phase propagation
Since the group velocity propagates away from the cylinder, we can envisage
the causality envelope expanding following an impulsively started oscillation.
– 38 –
GEFD
Internal gravity waves:The minimal mathematics version
cg
cp
Figure 29: Sketch showing why phase velocity is normal to group velocity.
This gives us two of the important characteristics of internal gravity waves,
namely
Group and phase velocities are normal
cp⋅cg = 0
(46)
Vertical components of group and phase velocities are in opposite
directions
cpz cgz ≤ 0.
(47)
Since the phase velocity is normal to the group velocity, and the wavenumber
vector is necessarily aligned with the phase velocity, then the angle between
the wavenumber vector and the horizontal is therefore also θ. Thus we may
write
cos θ =
kx ω
= .
k
N
(48)
As we shall see later, for a wave with components in both horizontal directions,
we may rewrite (48) as
– 39 –
GEFD
Internal gravity waves:A more rigorous derivation
2
k x2 + k y2
k z2
⎛ω ⎞
=1− 2 .
⎜ ⎟ =
2
⎝N⎠
k
k
(49)
3.2 A more rigorous derivation
Equations
Consider an, incompressible, inviscid, non-diffusive stratified flow. In this
limit, we may write the continuity and vorticity equations (2), (19), (12) and
(21) as
∇⋅u = 0,
⎛∂
⎞
⎜ + u ⋅ ∇ ⎟ρ = 0 ,
⎝ ∂t
⎠
∂u
1
+ ( u ⋅ ∇ ) u = − ∇ ( p + ρ0 gz ) − g ′zˆ
∂t
ρ0
∂ω
g
+ (u ⋅ ∇ )ω = (ω ⋅ ∇ )u + zˆ × ∇ρ
∂t
ρ0
Suppose we have a quiescent fluid with background density stratification given
the equations to
∇⋅u′ = 0,
ρ
∂ρ ′
= w′ 0 N 2 ,
∂t
g
1
ρ′
∂u′
= − ∇p′ − g zˆ ,
∂t
ρ
ρ0
∂ω′ g
=
zˆ × ∇ρ ′ .
∂t
ρ0
Taking the Boussinesq limit and defining the buoyancy perturbation
– 40 –
GEFD
Internal gravity waves:A more rigorous derivation
σ =−
g
ρ0
ρ′,
(50)
we may write
∇⋅u′ = 0,
(51)
∂σ
= − w′N 2 ,
∂t
(52)
∂u′
1
= − ∇p′ − σzˆ ,
∂t
ρ0
(53)
∂ω′
= − zˆ × ∇σ .
∂t
(54)
Eliminating u, v, σ and ω by (^z×∇)(^z×∇)(52) − (^z×∇)(∂/∂t)(54) gives
⎡ 2 ∂2
2 2 ⎤
N
∇
+
∇ H ⎥ w′ = 0 ,
⎢ ∂t 2
⎣
⎦
(55)
∂2
∂2
where ∇ ≡ 2 + 2 is the horizontal Laplace operator.
∂x
∂y
2
H
Plane wave eigenvalue problem
Recall that in §..2.1.1 we considered horizontally periodic solutions with some
vertical structure. We do a similar thing here and look for plane wave solutions
of the form
[
i (k x + k
w′(x, t ) = ℜ wˆ ( z )e
x
y y −ωt
)
],
(56)
where kx, ky are the horizontal wave numbers. Substitution gives the eigenvalue
problem
2
⎞
d 2 wˆ
2
2 ⎛N
⎟⎟ wˆ = 0 .
⎜
(
)
1
+
k
+
k
−
x
y
⎜ ω2
dz 2
⎠
⎝
– 41 –
(57)
GEFD
Internal gravity waves:Wave velocities
We will concentrate for the moment on the case of constant buoyancy
frequency N. In this case we have the general solution
[
]
wˆ = ℜ Ae −inz + Beinz ,
(58)
for complex coefficients A, B and
⎛ N2 ⎞
n = (k + k )⎜⎜ 2 − 1⎟⎟ .
⎠
⎝ω
2
2
x
2
y
(59)
ω>N
In this case, the right-hand side of (59) is negative, so n is imaginary, and (58)
shows the flow to decay exponentially away from the source. There are no
waves – the disturbances remain trapped or localised – although the flow still
responds in an oscillatory manner.
ω<N
For ω < N, the right-hand side of (59) is positive, so n is real and (58)
represents sinusoidal disturbances. Noting that n is simply the vertical
wavenumber kz, then
⎛ N2 ⎞
k = (k + k )⎜⎜ 2 − 1⎟⎟ ,
⎠
⎝ω
2
z
2
x
2
y
(60)
for real constants A, B.
We may rearrange (60) to obtain the dispersion relation
ω2
N2
=
k x2 + k y2
k x2 + k y2 + k z2
= 1−
k z2
k
2
,
as found previously (see (49)).
3.3 Wave velocities
Recall the definition of the phase velocity from (32). As this definition is
general, we have
– 42 –
(61)
GEFD
Internal gravity waves:Wave velocities
cp =
⇒
ω
k
2
k=N
(k
(k
2
x
2
x
+ k y2 )
12
+k +k
2
y
)
2 32
z
k.
(62)
|cp| = N/|k| cos θ.
Similarly, from (35) we have the group velocity
1 ∂ω 2
∂ω
cg =
=
∂k 2ω ∂k
N 2 ∂ ⎛⎜
k32 ⎞⎟
1−
=
2ω ∂ki ⎜⎝ k j k j ⎟⎠
k3 ⎞⎟
N 2 ⎛⎜ k32 ki
2
2
=
−
2ω ⎜⎝ (k j k j )2
k j k j ⎟⎠
⎞
N 2 k3 ⎛⎜ k3ki
⎟
=ω 2
−
δ
3i ⎟
⎜
ω k jk j ⎝ k jk j
⎠
⎤
ω ⎡N2
= 2 ⎢ 2 (ki − k3δ 3i ) − ki ⎥
k ⎣ω
⎦
⎤
ω ⎡N2
= 2 ⎢ 2 (k − k z zˆ ) − k ⎥
k ⎣ω
⎦
⇒
|cg| = N/|k| sin θ.
Note that
– 43 –
(63)
GEFD
Internal gravity waves:Wave velocities
ω
⎤
ω ⎡N2
c p + c g = 2 k + 2 ⎢ 2 (k − k z zˆ ) − k ⎥
k
k ⎣ω
⎦
=
N2
ωk
2
(k − k z zˆ )
⎛ kx ⎞
N ⎜ ⎟
k
=
2 ⎜ y⎟
ωk ⎜ ⎟
⎝0⎠
2
so
Key result
|cp + cg| = N/|k|
(64)
cpz = −cgz.
(65)
and
Key result
We also see that
ω
⎤
ω ⎡N2
c p ⋅ c g = 2 k ⋅ 2 ⎢ 2 (k − k z zˆ ) − k ⎥
k
k ⎣ω
⎦
⎤
ω ⎡N2
= 2 ⎢ 2 (k ⋅ k − k z k ⋅ zˆ ) − k ⋅ k ⎥
k ⎣ω
⎦
=
Key result
2
ω ⎡ k
(k
2 ⎢ 2
2
k ⎢⎣ k x + k y
=0
2
x
2⎤
+ k y2 ) − k ⎥
⎦⎥
cp⋅cg = 0.
– 44 –
(66)
GEFD
Internal gravity waves:Motion of fluid particles
θ
k
cp
cg
θ
N/|k|
Figure 30: Geometric representation of relationship between phase and group velocities.
3.4 Motion of fluid particles
In §..3.1 we saw that internal gravity waves can be considered in terms of slabs
of fluid sliding at an angle θ to the vertical. We now look in more detail at this
fluid motion.
For simplicity we consider the motion in a rotated coordinate system, x′, y′, z′
such that the wavenumber is k′ = (k′,0,0).
z'
θ
k
x'
cp
cg
θ
N/|k|
Figure 31: Rotated coordinate system for internal waves.
In this coordinate system, flow variables have the form
u′ = (uˆ ′, vˆ′, wˆ ′)ei (k ′x′−ωt ) = (uˆ ′, vˆ′, wˆ ′)eiφ ′ ,
– 45 –
GEFD
Internal gravity waves:Motion of fluid particles
p′ = pˆ ′ei (k ′x′−ωt ) = pˆ ′eiφ ′ ,
σ ′ = σˆ ′ei (k ′x′−ωt ) = σˆ ′eiφ ′ ,
where variables with a hat are functions of z′, and φ′ = k′x′ − ωt is the phase.
Continuity ∇.u′ = 0 ⇒ u^′ = ^v′ = 0.
The linearised momentum equation (53)
∂u′
1
= − ∇p′ − σzˆ
∂t
ρ
becomes
pˆ ′ =
iρ 0 sin θ
σˆ ′ ,
k'
σˆ ′ = −
iω
wˆ ′ .
cos θ
Particle displacements η′ satisfy
∂η ′
= w′ .
∂t
Suppose the motion is driven by
η ′ = ℜ(η 0 eiφ ′ ) = η 0 cos φ ′ ,
then
w′ = ℜ(− iωη 0 eiφ ′ ) = ωη 0 sin φ ′ ,
⎛ ω 2η 0 iφ ′ ⎞
ω 2η 0
σ ′ = ℜ⎜⎜ −
e ⎟⎟ = −
cos φ ′ ,
cos
θ
cos
θ
⎝
⎠
⎛ iρ 0ω 2η 0 tan θ iφ ′ ⎞
ρ 0ω 2η 0 tan θ
sin φ ′ .
p′ = ℜ⎜⎜
e ⎟⎟ = −
′
′
k
k
⎝
⎠
– 46 –
GEFD
Internal gravity waves:Motion of fluid particles
k
p’<0 p’>0
φ’=0
Displacemen
σ’<0
φ’=2
isopycnal
Figure 32: Sketch showing the motions in an internal gravity wave.
Again it can be shown that there is an equipartition between kinetic and
potential energy, and that the energy flux is equal to the group velocity
multiplied by the energy density.
Exercise: Prove the equipartition of energy and that the group velocity is the
speed of energy propagation.
– 47 –
GEFD
Internal gravity waves:Oscillating cylinders
3.5 Oscillating cylinders
– 48 –
GEFD
Internal gravity waves:Oscillating cylinders
Figure 33: Time series of steady internal gravity waves generated by oscillating circular
cylinder. Left column is ∂ρ′/∂x, right column is ∂ρ′/∂z.
– 49 –
GEFD
Internal gravity waves:Oscillating cylinders
Figure 34: Time series of start-up transients generated by oscillating square cylinder.
The impulsive start of the cylinder generates an entire spectrum of wave
modes. Examining (63) tells
– 50 –
GEFD
Internal gravity waves:Oscillating cylinders
ω N2
c g = 2 2 ( k − k z zˆ ) − k
k ω
=
⎞
ω ⎛N
ˆ
k
−
−
k
z
k
(
)
⎟
z
2 ⎜
2
k ⎝ω
⎠
2
2 12
N2 2
ω ⎡N4 2
2⎤
2
= 2 ⎢ 4 ( k x + k y ) − 2 2 ( k x + k y2 ) + k ⎥
ω
k ⎣ω
⎦
12
12
2
2
⎤
⎞ kx + k y
ω ⎡N2 ⎛ N2
= ⎢ 2 ⎜ 2 − 2⎟
+
1
⎥
2
k ⎢⎣ ω ⎝ ω
k
⎥⎦
⎠
ω ⎡N2 ⎤
= ⎢ 2 − 1⎥
k ⎣ω
⎦
12
,
N
= sin θ
k
ωêN
(67)
1
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1
»cg »
Figure 35: Variation in the magnitude of the group velocity with frequency.
The lower frequency modes transport energy away from the source more
rapidly.
– 51 –
GEFD
Internal gravity waves:Leewaves
3.6 Leewaves
3.6.1 Extended range of hills
Consider the passage of a stratified atmosphere over a sinusoidal mountain
range.
U
∂ρ/∂z
λ
It is clear that an observer moving with the mountains is going to experience a
constant phase. The frequency of the forcing must therefore be
ω = kU = 2πU/λ.
Hence the wave crests are aligned at an angle
θ = cos−1 ω/N .
to the vertical. For the wave field to appear stationary to this observer, the
phase velocity (and hence wavenumber vector) must be aligned with the wave
crests.
From our development of the theory, it is generally easier to work in a frame of
reference where the undisturbed fluid is at rest.
– 52 –
GEFD
Internal gravity waves:Leewaves
∂ρ/∂z
λ
-U
Figure 36: Sketch of mountains moving beneath a stratified fluid.
In this frame the angle made by the wave crests remains the same, but now the
group velocity cg′ = cg − U is aligned is aligned with the crests. The vertical
components of the phase and group velocities are equal (and opposite) in both
frames.
U
θ
cp
cp
cp’ = cp − U
−U
−U
cg
cg
cg’ = cg − U
θ
−U
cp’ = cp − U
cg’ = cg − U
Figure 37: Geometrical interpretation of relationship between wave vectors in different
frames of reference.
– 53 –
GEFD
Internal gravity waves:Leewaves
From geometrical arguments, it is clear that
cp + cg = − (cp′ + cg′) = U.
Recalling (64) that |cp + cg| = |cp′ + cg′| = N/|k′|, then
|k′| = N/U.
Since kx′ must be determined by the mountain wavelength λ,
kx′ = 2π/λ,
then
ω = kx′U = 2πU/λ.
From the dispersion relation,
k z′ = k x′
N2
ω2
−1.
Geometrically, we have
k x′
U2
= 2π
c′p = U cos θ = U
k′
λN
k z′
U2
c′g = U sin θ = U
= 2π
k′
λN
N2
ω2
−1 ,
so
c′p = (− U cos 2 θ , − U cos θ sin θ ) ,
c′g = (− U sin 2 θ , U cos θ sin θ )
– 54 –
GEFD
Internal gravity waves:Leewaves
-U
cg′
cp′
λ
Figure 38: Sketch of wave crests and wave velocities for sinusoidal mountains moving
under stratified flow. Frame of reference where the fluid is at rest.
Moving back to the mountain’s frame of reference, the pattern of wave crests
looks geometrically the same, but the crests are stationary.
U
cp
cg
λ
Figure 39: Sketch of wave crests and wave velocities for stratified flow moving over
mountains. Frame of reference where the mountains are at rest.
Figure 40: Experiment showing the distortion of density surfaces due to stratified flow
over a single mountain.
– 55 –
GEFD
Internal gravity waves:Leewaves
(a)
(b)
www.dirauxannex.org/weather.html
Figure 41: Mountain lee waves.
3.6.2 Causality
If we are dealing with linear waves, then we can always use linear
superposition to construct the wave field associated with a particular feature.
The simplest such case is the lee waves generated by a point source located at
x = 0.
In this section we shall concern ourselves more with where the waves can be
located rather than their precise form. The analysis here is very similar to that
of Kelvin ship waves in §2.3 and makes use of the principle of stationary
phase.
Suppose that at t = 0 a small, compact isolated obstacle starts to move at speed
U in the negative x direction so that its position is given by X(t) = −Ut for t ≥ 0.
Fourier decomposition shows that the obstacle has the possibility of exciting
all horizontal wavenumbers k and all frequencies ω. Clearly we will only have
waves produced for ω ≤ N. Moreover, as with the Kelvin ship wave problem,
we are most interested in the waves that are phase-locked with the obstacle,
and so we need consider only the waves moving in the same direction as the
obstacle.
– 56 –
GEFD
Internal gravity waves:Leewaves
-U
θ
r = |cg|(t – ts)
cp
θ
X = −U t
X = Xs
At some time t = ts the obstacle releases waves from x = Xs. Consider the waves
emitted with a frequency ω < N such that they propagate at an angle
θ = cos−1(ω/N). The phase of these waves will not appear to have changed
provided the magnitude of the phase velocity
cp =
N
cosθ
k
is equal to U cosθ (see figure 42a). This relation immediately restricts the
wavenumber to
k =
N
.
U
We can get to this relation via a different route. Recal the phase is defined as
φ = k⋅x − ω t,
which gives the differential expression
dφ =
∂φ
∂φ
∂φ
∂φ
dki +
dxi +
dω +
dt .
∂ki
∂xi
∂ω
∂t
For an observer moving with speed U in the x direction, the wave will appear
steady if the phase and wavenumber do not change (i.e., dφ = 0, dk = 0). The
– 57 –
GEFD
Internal gravity waves:Leewaves
dispersion relation then means that dω also vanishes. Moreover, dx = U dt
while dy = 0 = dz for the point under observation. Thus we have
⎛ ∂φ ∂φ ⎞
+
0 = ⎜U
⎟ dt .
∂
∂
x
t
⎝
⎠
Using the definition of φ, we rewrite this as
Uk x − ω = U k cosθ − ω = 0
where θ is the angle between the wavenumber vector and the x axis (hence
kx = |k|cosθ). Recalling that the dispersion relation gives ω = N cos θ for
internal waves, then this condition becomes
k =
N
U
as before.
By time t the energy in the wave has propagated a distance
r = c g ( t − ts ) =
N
sin θ ( t − ts )
k
,
= U ( t − ts ) sin θ
so that the energy all lies on a circle spanning between the x = Xs where the
wave was released and the current position of the obstacle.
As the obstacle continues to move, it continues to release waves in the same
pattern.
– 58 –
GEFD
Internal gravity waves:Leewaves
τs /T
0.0
1.0
1000.0
800.0
z
600.0
400.0
200.0
0.0
0.0
500.0
1000.0
1500.0
2000.0
x
Figure 43: Causality envelopes and phase shells for internal gravity waves.
Causality surfaces are semicircular hemispherical in 2D, or hemispherical in
3D. Within these the surfaces of constant phase for quarter circles/spheres
around the current position of the obstacle.
This analysis appears to suggest the waves have a sharp front at the causality
envelope emitted when the obstacle started moving at t = 0. This is slightly
misleading, however, as the analysis focuses on the steady motion of the
obstacle and the permanent waves this motion produces.
– 59 –
GEFD
Internal gravity waves:Reflections
The impulsive start at t = 0 emits additional transient waves that are not
described by the principle of stationary phase used here. We shall not analyse
these waves but note that the transient waves merge with the permanent waves.
Causality
envelope
Transient
waves
Permanent
waves
X = −U t
Figure 44: Surfaces of constant phase and the matching between the permanent and
transient waves for an impulsively-started obstacle.
The analysis for different acceleration profiles or for propagation in different
directions can be handled in a similar manner, provided the acceleration is
3.7 Reflections
As the frequency of an internal gravity wave is preserved upon reflection from
a boundary, then the angles of the group and phase velocity are also preserved
(within a sign).
3.7.1 Properties of reflected beams
For reflection from horizontal and vertical boundaries, this means the angle of
incidence equals the angle of reflection, wavenumber and other wave
properties are preserved.
– 60 –
GEFD
Internal gravity waves:Reflections
cp
cg
cg
cp
∇ρ′
u′
cp
cg
∇ρ′
u′
∇ρ′
∇ρ′
u′
u′
cp
cg
Figure 45: Sketches of reflection of wave from (a) horizontal and (b) vertical boundaries.
cp
cg
cg
cp
α′
cg
cp
∇ρ′
u′
∇ρ′
∇ρ′
u′
u′
u′
∇ρ′
α′
cp
cg
Figure 46: Sketches of reflection of wave from sloping boundaries.
Examples sheet question: Determine characteristics of reflected waves
The quantity
– 61 –
GEFD
Internal gravity waves:Reflections
γ=
sin (θ + α )
sin (θ − α )
characterises the ‘focusing’ power of the relection. For a ‘focusing reflection’
(γ > 1), the reflected wave number kr is related to the incident wavenumber ki
by
kr = γ ki,
and the reflected (Er) and incident (Ei) energy densities are related by
Er = γ Ei.
Obviously, for a defocusing reflection, the recirpocal of γ is appropriate.
3.7.2 Subcritical and supercritical reflections
The terms ‘subcritical’ and ‘supercritical’ are applied to wave reflections in an
oceanographic context to distinguish changes in the vertical direction of
propagation following reflection.
In a subcritical reflection, the boundary has a shallower slope than the wave,
and the vertical propagation of the wave is reversed by the reflection. The
horizontal direction of propagation is maintained.
In a supercritical reflection, the boundary has a steeper slope than the wave,
and the vertical direction of propagation is maintained, but the horizontal
direction is reversed.
Wave reflection from a ‘critical slope’ represents the limiting case. If angle
between boundary and vertical, α → ±θ, then linear wave theory predicts
γ → ∞. This inevitably leads to the development of nonlinear effects with the
potential for wave breaking and mixing in addition to disturbances being able
to propagate both up and down the slope..
– 62 –
GEFD
Internal gravity waves:Ray tracing
Figure 47: Near-critical reflections from a slope.
3.8 Ray tracing
Since the frequency is preserved upon reflection, then the angle to vertical is
conserved, and so waves tend to propagate along well-defined rays.
Repeated reflections can lead to trapping, increased amplitude and breaking.
(a)
(b)
Figure 48: Sketch showing trapping of internal waves. (a) Subcritical reflections, and (b)
supercritical reflections.
While rays can be traced in either direction, must take note of causality to
determine which direction is appropriate.
3.9 Wave attractors
Waves can reflect many times in an enclosed basin.
– 63 –
GEFD
Internal gravity waves:Wave attractors
3.9.1 Rectangular basins
(a)
(b)
Figure 49: Eigen modes in a square basin. (a) Mode 1,1; (b) mode 1,2.
If the basin is rectangular, then rays will make closed circuits if the width X
and height Y of the basin are related such that
X n
= tan θ
mY
For some pair of positive integers m and n. This corresponds to eigen modes of
the basin.
If there is no integer relation, then there are no closed circuits and the waves do
not correspond to the eigen modes of the basin.
– 64 –
GEFD
Internal gravity waves:Wave attractors
Figure 50: Ray with θ = 60° in a square basin.
– 65 –
GEFD
Internal gravity waves:Wave attractors
3.9.2 Trapezoidal basin
The basic idea
For downward propagating waves reflecting from the slope, the beams are
focused, not only reducing the wave length and increasing the energy density,
but also focusing them onto an ‘Internal Wave Attractor’. In the case shown
here, we have a mode 1,1 attractor, with the attractor taking a simple (although
somewhat distorted) diamond shape.
Iterative map
x1,y1
x0,y0
x2,y2
x4,y4
x3,y3
Let a = 1/tanα and b = 1/tanθ, then starting on the slope at (x0,y0):
x0
– 66 –
GEFD
Internal gravity waves:Wave attractors
y0 = a x0
x1 = x0 +
Y − y0 Y + ( b − a ) x0
=
b
b
y1 = Y
x2 = X
y2 = y1 − b ( X − x1 ) = 2Y − b ( X − x0 ) − ax0
x3 = X −
( a − b ) x0 − 2Y
y2
= 2X +
b
b
y3 = 0
For x4, need to solve
⇒
y4 = a x4 = b(x3 − x4),
x4 =
2 ( bX − Y ) + ( a − b ) x0
a+b
y4 = a x4.
For a closed loop to exist, x0 = x4, requiring
x4 = X −
Y
.
b
Using the analogy with the General Iteration method for solving an equation,
i.e. the solution of xn+1 = g(xn) will converge iff |g′(x)| < 1, then an attractor will
exist if and only if
– 67 –
GEFD
Internal gravity waves:Wave attractors
a − b sin (θ − α )
=
< 1.
a + b sin (α + θ )
This is satisfied for all focusing reflections.
Not always an attractor
But not all angles will generate a mode 1,1 attractor.
Figure 51: No attractor when two reflections in vertical.
Figure 52: Attractor close to mode 1,1 limit.
Limiting cases
When analysing an attractor, it is useful to start in one of the corners: waves
reflected at a corner will return along the path on which they arrived.
– 68 –
GEFD
Internal gravity waves:Wave attractors
Figure 53: Possible limiting cases for mode 1,1 and mode 1,2.
For higher modes, must end up with a greater number of focusing than
defocusing reflections in a circuit.
If reflection from left-hand wall is subcritical, then waves focused into top-left
corner rather than an attractor.
3.9.3 More complex geometries
Internal wave attractors exist for some range of frequencies in all strictly
convex basins (and many basins that are not strictly convex).
– 69 –
GEFD
Internal gravity waves:Wave attractors
Parabolic
Maas (2005) Int. J. Bif. Chaos 15.
Ray tracing proceeds in the same way as for a trapezoidal basin, but the
curvature of the walls allow two reflections, one supercritical and one
subcritical. The supercritical reflection is focusing, but the subcritical reflection
is defocusing. However, the focusing power is stronger than the defocusing
power, so an attractor exists.
– 70 –
GEFD
Internal gravity waves:Wave attractors
Three-dimensional
Maas (2005) Int. J. Bif. Chaos 15.
3.9.4 Energy spectrum for attractor
We return now to the simple rectangular attractor in a two-dimensional
trapezoidal basin.
Inviscid spectrum
We have seen that when an attractor exists, all rays are focused in to that
attractor. As we saw in §3.7, for each reflection from the slope the
wavenumber is increased from kn-1, say, to
kn = γ kn-1,
– 71 –
GEFD
Internal gravity waves:Wave attractors
γ=
where
sin (θ + α )
sin (θ − α )
.
Similarly, the energy density associated with wavenumber k increases as
En = γ En-1.
Hence, if the disturbance is forced with some wavenumber k0, then we expect
to see an attractor with an energy spectrum that grows linearly with the
wavenumber:
En = γ n E0 = kn E0/k0.
If A is the wave amplitude then
A ~ k1/2
and the wave steepness, kA, (a measure of ratio of amplitude to wavelength)
increases as
kA ~ k3/2.
This suggests that the waves will rapidly become nonlinear and potentially
break. However, energy is also being taken to smaller scales, so even a small
viscosity can start to have an effect.
Viscous spectrum
In the absence of viscosity, the energy flux F is constant and related to the
energy density E by
F = cg E,
where the group velocity cg is independent of wave amplitude.
The energy equation, which may be written as
2
⎛
⎞
1∂u
p
2
+ ( u ⋅ ∇ ) ⎜ 12 u + + gz ⎟ = ν ⎡∇ 2
⎣
ρ
2 ∂t
⎝
⎠
– 72 –
(
1
2
u
2
)
2
− ( ∇u ) ⎤ ,
⎦
GEFD
Internal gravity waves:Wave attractors
also constrains the motion. For a steady wave, averaged over a wave period,
the left-hand side is simply the ∇⋅F.
If we employ a boundary-layer approximation for a wave beam where the wave
properties change only slowly in the along-beam direction ξ compared with the
cross-beam direction ζ, and average across the width of the wave beam, then
we can approximate the energy equation as
2
⎛ du ⎞
dF
∇⋅F ≈
= −νρ 0 ⎜
⎟ ,
dξ
d
ζ
⎝
⎠
where ~
u is the along-beam component of the velocity. Now since
1
1
ρ0 u 2 = KE = E , then if we consider a single Fourier mode with
2
2
wavenumber kn we can rewrite the divergence of the energy flux as
dFn
dE
F
= cn n = −ν kn2 En = −ν kn2 n ,
dξ
dξ
cn
where cn is the magnitude of the group velocity for wavenumber kn.
Now suppose the distance around the attractor (between successive focusing
reflections from the slope) is L, then a ‘packet’ of energy will have
wavenumber kn for a period
Tn = L/cn.
Noting that the magnitude of the group veloctiy is
cn =
N
sin θ ,
kn
so Tn ~ kn. We may integrate the energy equation over Tn to determine the
energy at the end of the loop relative to the start as
3
Eend
= e −ν kn L
Estart
N sin θ
.
(We shall derive this decay more rigourously in a different way a little later.)
– 73 –
GEFD
Internal gravity waves:Wave attractors
Incorporating this into the equation following an energy packet from one
reflection to the next gives at the end of the nth loop
3
En = γ e −ν kn L
= γ e −νγ
3 3
kn −1L
= γ ne
=
k
e
k0
N sin θ
N sin θ
(
En−1
En−1
)E
−Γ γ 3 n −1
0
⎛ ⎛ k ⎞3 ⎞
−Γ⎜ ⎜ ⎟ −1⎟
⎜ ⎝ k0 ⎠ ⎟
⎝
⎠
E0
1 ν k03 L
ν L ( 2π ) 1
Γ= 3
=
,
γ − 1 N sin θ N λ02 λ0 γ 3 − 1 sin θ
3
where
and λ0 is the scale at which energy is supplied (e.g. the scale of the basin). The
spectrum for the wave steepnes (kA) now scales as
kn An = k0 A0γ
3
(
)
3n
1
3 n 2 − 2 Γ γ −1
e
⎛k ⎞
= k0 A0 ⎜ ⎟ e
⎝ k0 ⎠
2
– 74 –
⎛ ⎛ k ⎞3 ⎞
− 12 Γ⎜ ⎜ ⎟ −1⎟
⎜ ⎝ k0 ⎠ ⎟
⎝
⎠
.
GEFD
Internal gravity waves:Decay along a beam
kA
Figure 54: Comparison of the spectrum of steepness for internal wave attractors between
model and experiment.
We may obtain the peak this steepness spectrum by differentiating and setting
equal to zero.
3.10 Decay along a beam
As we have seen with the attractors, a knowledge of how viscosity affects a
beam can be necessary. Here we will derive the behaviour for a beam in a
viscous fluid in a more rigorous manner.
We can write the linearized equations, including viscosity, as
∂u 1 ∂p
+
= ν∇ 2 u ,
∂t ρ0 ∂x
∂w 1 ∂p
+
− b = ν∇ 2 u ,
∂t ρ0 ∂z
∂b
+ N 2w = 0 ,
∂t
∇⋅u = 0,
where b = −g (ρ − ρ0)/ρ0 is the buoyancy.
– 75 –
GEFD
Internal gravity waves:Decay along a beam
Introducing the streamfunction ψ defined such that u = (−∂ψ/∂z, ∂ψ/∂x) allows
us to rewrite the system as
∂b
∂ψ
+ N2
= 0,
∂t
∂x
∂ 2
∂b
∇ ψ − − ν∇ 4ψ = 0 .
∂t
∂x
It is convenient to take N = 1 and introduce a rotated coordinate system with ξ
in the direction of cg and ζ across it (the direction of cp),
ξ = x sin θ + z cos θ,
ζ = −x cos θ + z sin θ,
giving
∂b ∂ψ
∂ψ
sin θ −
cosθ = 0 ,
+
∂t ∂ξ
∂ζ
∂ 2
∂b
∂b
sin θ +
cosθ −ν∇ 4ψ = 0 .
∇ψ −
∂t
∂ξ
∂ζ
We now assume that ν is small and introduce a power series plane-wave
expansion for ψ and b of the form
ψ = (ψ 0 + εψ 1 +
b = ( b0 + ε b1 +
) eiωt ,
) eiωt
where ε = ν/2.
We will also introduced the scaled along-beam variable
χ=
ε
sin θ
ξ.
Substitution into the linearised equations and collection in terms of powers of ε
gives
– 76 –
GEFD
Internal gravity waves:Decay along a beam
∂ψ 0
= ib0 ,
∂ζ
ω
(68)
∂ψ 0
∂ψ 1
− iωb1 =
,
∂ζ
∂χ
(69)
∂ 4ψ 0 ∂b0
∂ 2ψ 1
∂b1
.
iω
+ω
=2
+
2
4
∂ζ
∂ζ
∂ζ
∂ζ
(70)
We can use (68) to eliminate b0 from (70), and differentiate (69) (and multiply
by i) to elimanate b1, thus obtaining
∂ 4ψ 0
∂ 2ψ 0
.
=i
∂ζ 4
∂ζ∂χ
Integrating once gives
∂ 3ψ 0
∂ψ 0
i
=
+ f (χ ),
∂ζ 3
∂χ
(71)
Where f(χ) is an unknown function. (Note that b0 satisfies the same equation
with f(χ) = 0.)
The form of (71) allows solution by separation of variables. Let
ψ0 = F(ζ) G(χ), then the homogeneous problem requires
F ′′′
G′
= i = −ik 3 .
F
G
Our choice of the ‘constant’ −ik3 reflects our wish for wave-like solutions (with
wavenumber k) in cross-beam ζ direction. The solution then gives
3
ψ 0 = Aeikζ −k χ ,
For some constant A. Recalling that χ =
ε
ν
ξ demonstrates the
sin θ
2sin θ
amplitude of the wave decays exponentially in the direction of cg, in agreement
with the energy-based argument used in §3.9.4.
– 77 –
ξ=
GEFD
Internal gravity waves:Decay along a beam
Using the idea of linear superposition, we can write the streamfunction for
some arbitrary disturbance as an integral of Fourier components, with
amplitude A(k), as
∞
⎛
ν k3 ⎞
ψ 0 = ∫ A ( k ) exp ⎜ ikζ −
ξ ⎟ dk .
θ
2sin
⎝
⎠
−∞
If we take A(k < 0) = 0 so that we have the beam propagating in the positive ξ
direction, and select A(k > 0) = a = 1 to represent a point source (deltafunction), we can compute the structure of the beam.
– 78 –
GEFD
Internal gravity waves:Decay along a beam
Figure 55: Stream function (colour) and velocity (vectors), in rotated, scaled coordinates,
for wave from a point source located at the origin. The field is shown at phases of 0, π/4,
π/2 and 3π/4. Note the downward propagation of the peaks in the velocity field.
– 79 –
GEFD
Internal gravity waves:Stokes drift in internal waves
3.11 Stokes drift in internal waves
3.11.1 Analysis of Stokes drift
As we saw in §3.4, fluid parcels in inviscid linear internal gravity waves
execute simple harmonic motion purely in the direction of the group velocity
(energy propagation). As the phase velocity is perpendicular to this, it would
appear that there is no mechanism akin to the Stokes drift discussed in §2.1.8.
However, experiments show that there is a net drift associated with the waves
(at least for an internal wave attractors).
Figure 56: Motion of particles through a Stokes drift mechanism in an internal wave
attractor.
Suppose we allow for the gradual decay of the wave beam due to the action of
viscosity, and that this decay occurs over a length scale that is large compared
with the wavelength. Letting ξ be the along-beam coordinate in the direction of
the group velocity, then we may write the velocity in a monochromatic wave
beam as
U (ξ , t ) = U 0 e −ν k ξ
3
2 N sin θ
cos ωt .
Let Ξ(t) be the Lagrangian position of a particle such that
Ξ = U ( Ξ, t ) ,
then we are interested in the Stokes drift velocity
Us = U(Ξ,t) − U(ξ,t).
– 80 –
GEFD
Internal gravity waves:Stokes drift in internal waves
As for surface waves, we can expand U(Ξ,t) to obtain
2
∂U 1
2 ∂ U
Us = (Ξ − ξ )
+ (Ξ − ξ )
+
∂ξ 2
∂ξ 2
For small Ξ − ξ, we have
Ξ − ξ = ∫ U (ξ , t ) dt =
U0
ω
3
e −ν k ξ
2 N sin θ
sin ωt
Us = −
ν k3
U 02
ω 2 N sin θ
3
e −ν k ξ
N sin θ
sin ωt cos ωt ,
which vanishes when integrated over a wave period.
We must therefore go to higher order to establish the character of the drift.
From the second order term,
Us = −
U 03
ν 2k 6
3
ω 4 N sin θ
2
2
2
e −3ν k ξ
2 N sin θ
sin 2 ωt cos ωt ,
which again integrates to zero over a period.
[Note that we do not need to do the Taylor series expansion but can integrate
the full expression for Us = U(Ξ,t) − U(ξ,t) to obtain
Ξ −ξ = −
− aξ
1
e − aξ sin ωt − e −aU 0e sin ωt ω ,
ω
a
U0
showing that the Stokes drift is zero averaged over a period.]
Superposition of waves with the same frequency but different wavenumbers
does not change things, as we can continue to look at each wavenumber
separately for linear waves. [Different frequencies will not lie in the same
direction.]
– 81 –
GEFD
Internal gravity waves:Stokes drift in internal waves
Why then do we have drift?
Is it the boundary layer approximation that is wrong? Our velocity U(ξ,t) does
not satisfy ∂U/∂x = 0, and so we must have a cross-beam velocity.
U (ξ , ζ , t ) = U 0 e − aξ cos ( kζ − ωt ) ,
Assume
a=νk3/2Nsinθ.
where
Note that for a wave beam in the first quadrant, then k must be negative since
the phase and group velocities have the opposite vertical component.
From continuity, ∂V/∂ζ = −∂U/∂ξ, so
V (ξ , ζ , t ) = − ∫
∂U
a
d ζ = − U 0 e − aξ sin ( kζ − ωt ) .
∂ξ
k
To the leading order the particle displacement is given by
Ξ − ξ = ∫ U (ξ , ζ , t ) dt =
Ζ − ζ = ∫ V (ξ , ζ , t ) dt =
U0
ω
e − aξ sin ( kζ − ωt )
a U 0 − aξ
e cos ( kζ − ωt )
k ω
and the along-beam Stokes drift velocity
Us = (Ξ − ξ )
=a
U 02
ω
∂U
∂U
+ (Ζ − ζ )
∂ξ
∂ζ
e −2 aξ ( cos ( kζ − ωt ) sin ( kζ − ωt ) − cos ( kζ − ωt ) sin ( kζ − ωt ) )
=0
So there is no drift. Although additional terms come in at the second order,
these disappear when averaging over a wave period, so again no drift.
Superposition
The origin of the drift, however, becomes apparent when we consider the
superposition of two modes that satisfy continuity. In particular, assume
– 82 –
GEFD
Internal gravity waves:Stokes drift in internal waves
U (ξ , ζ , t ) = U 0 e − a0ξ cos ( k0ζ − ωt ) + U1e − a1ξ cos ( k1ζ − ωt ) ,
a0 = νk03/2Nsinθ and a1 = νk13/2Nsinθ.
where
Continuity then gives
V (ξ , ζ , t ) = −
a0
a
U 0 e − a0ξ sin ( k0ζ − ωt ) − 1 U1e − a1ξ sin ( k1ζ − ωt ) ,
k0
k1
with the leading order particle displacements
U0
e − a0ξ sin ( k0ζ − ωt ) +
U0
e − a0ξ cos ( k0ζ − ωt ) + a1
Ξ − ξ = ∫ U (ξ , ζ , t ) dt =
Ζ − ζ = ∫ V (ξ , ζ , t ) dt = a0
ω
ω
U1
ω
e − a1ξ sin ( k1ζ − ωt )
U1
ω
e − a1ξ cos ( k1ζ − ωt ) .
The additional terms give new nonlinearities in the expansion of the drift
velocity
Us = (Ξ − ξ )
=
U 0U1
ω
∂U
∂U
+ (Ζ −ζ )
∂ξ
∂ζ
⎡ a0
cos ( k0ζ − ωt ) sin ( k1ζ − ωt )
k
⎣ 0
( k0 − k1 ) e−( a +a )ξ ⎢
0
−
1
⎤
a1
cos ( k1ζ − ωt ) sin ( k0ζ − ωt ) ⎥
k1
⎦
Averaging this over a wave period 2π/ω, gives
Us = −
⎛a a ⎞
U 0U1
( k0 − k1 ) ⎜ 0 + 1 ⎟ e−( a0 +a1 )ξ sin ( ( k0 − k1 ) ζ ) .
2ω
⎝ k0 k1 ⎠
Recalling the definition of ai and that ω = N cosθ,
Us = −
ν U 0U1
( k0 − k1 ) ( k02 + k12 ) e−( a +a )ξ sin ( ( k0 − k1 ) ζ ) .
2
4 N cosθ sin θ
0
– 83 –
1
GEFD
Internal gravity waves:Reflections from rough topography
Hence we have Stokes drift due to the nonlinear interaction of the Lagrangian
paths of two spatially decaying internal wave modes of the same frequency but
different wavenumber.
But is this OK?
Without loss of generality, we shall consider a wavebeam in the first quadrant
such that k0 < k1 < 0. The Stokes drift is therefore positive when sin(|k0 − k1|ζ)
is negative, and vice versa.
The key thing, however, is that this steady drift has a vertical component that
implies a steady vertical flux of fluid, and hence mass (at least in the absence
of mass diffusion).
But the drift does not take into account the buoyancy forces that must also act
on these parcels of fluid. The simplest approach would be to simply ignore the
vertical component of the velocity, U s cosθ . However, even the horizontal
component suggests changes to the density field since it is not divergence free.
This is an active topic of research: there are as yet (October 2009) no papers
published on this Stokes drift phenomenon, although we are able to get a
reasonable agreement between theory and experiments by including a
buoyancy force in the Lagrangian equations.
3.12 Reflections from rough topography
Ray tracing only provides a partial answer if the topography contains
wavelengths comparable to those found in the wavebeam.
2h0
2π/kT
Figure 57: Examples of idealised rough topography.
– 84 –
GEFD
Internal gravity waves:Reflections from rough topography
3.12.1 Subcritical reflection from a sinusoid
Ray tracing suggests that if the slope is everywhere subcritical, then an incident
wavebeam will be reflected forwards. Doing it carefully indicates that there
will be a change in the wavenumber composition of the beam.
Figure 58: Ray tracing for sinusoidal topography.
We can predict the composition of the reflected beam by analysing the
reflection points.
ki
k1
k0
kr
kθ
xi
xr
kT
x0
Figure 59: Definition sketch for reflection from a sinusoid.
Consider a wave beam of wavenumber ki and angle θ reflecting from a
sinusoidal topography with height given by
z = h = h0 sin kTx.
– 85 –
GEFD
Internal gravity waves:Reflections from rough topography
Consider the ray that, in the absence of topography, would reach z = 0 at x = x0.
Clearly, this ray will intersect the topography when
h0 sin kTx = β (xi − x),
where β = 1/tan θ. We define this point as x = x0.
The reflected ray, which would have originated from x = xi in the absence of
the topography, instead appears to originate from x = xr. For properties (such as
phase) that are conserved along the ray, then the topography induces a phase
shift such that the reflected beam has properties associated with xi rather than
xr. This phase shift is equal to 2(xi − x0).
We define δx = xi − x0, so that the incident beam intersects with the topography
when
h0 sin kT x0 = βδ x ,
or
δx=
h0
β
sin kT x0 .
The small-slope approximation means that we need not worry about the effects
of focusing or defocusing on the beams.
Suppose the particle excursion (in the direction of the incident beam) in the
incident beam projected onto z = 0 are given by
ηi(x) = η0 sin (k0x − ωt),
then the excursions in the reflected beam projected onto z = 0 will be given by
ηr(x,t) = −ηi(x + 2δx).
Hence
ηr ( x, t ) = −η0 sin ( k0 ( x + 2δ x ) − ωt )
⎛ ⎛
⎞
h
⎞
= −η0 sin ⎜ k0 ⎜ x + 2 0 sin kT x ⎟ − ωt ⎟
β
⎠
⎝ ⎝
⎠
– 86 –
GEFD
Internal gravity waves:Reflections from rough topography
If we assume also that k0h0/β is small, then we can expand this as
⎛
ηr ( x, t ) = −η0 ⎜ sin ( k0 x − ωt ) + 2
⎝
⎞
cos ( k0 x − ωt ) sin kT x ⎟
β
⎠
k0 h0
Using trig identities,
ηr ( x, t ) = −η0 sin ( k0 x − ωt )
−η0
k0 h0
β
( − sin ( ( k
0
− kT ) x − ωt ) + sin ( ( k0 + kT ) x − ωt )
)
,
so the reflected beam has components at ki, ki − kθ, and ki + kθ.
Causality
Clearly the sign of k0 − kT (or equivalently ki − kθ) depends on the relative
magnitude of k.
Causality requires the reflected waves have their group velocity directed
upwards. This in turn means that the phase velocity must be directed towards
the right for components of the transmitted beam.
However, while the terms containing k0x − ω t and (k0 + kT)x − ω t clearly
satisfy this requirement, the term containing (k0 − kT)x − ω t will only have the
phase velocity towards the right when k0 > kT. (Of course, a flat bottom with
kT = 0 will always satisfy this condition.)
What happens?
If k0 < kT (equivalently, ki < kθ), then the (k0 − kT) wave propagates to the left.
Since the group velocity must still be directed away from the bottom, then the
reflected wave is directed back along the incident wave.
There is therefore back scatter.
3.12.2 Supercritical reflection from a sinusoid
The slope at the steepest point of a sinusoidal topography of the form
z = h = h0 sin kTx.
– 87 –
GEFD
Internal gravity waves:Reflections from rough topography
is kTh0, whereas the slope of the incoming wave beam is β = 1/tan θ. If
kTh0 > β
then the topography can be said to be supercritical. Note that it is never
supercritical everywhere as the crests (and troughs) are horizontal.
With a supercritical topography, the first reflection will be upwards at some
places, and downwards at others.
• Rays that are reflected upwards in the first reflection are transmitted
forwards.
• Rays that are reflected downwards (supercritical) in the first reflection will
undergo further reflections.
Figure 60: Ray tracing showing the first reflection from a supercritical sinusoid.
– 88 –
GEFD
Internal gravity waves:Non-linear stratification
Figure 61: Ray tracing showing multiple reflections from a supercritical sinusoid.
Figure 62: Bounds on the forward and backward reflection coefficients from a sinusoid.
3.13 Non-linear stratification
If the stratification changes over length scales much greater than the
wavelength, then ray-tracing theory can be used.
– 89 –
GEFD
Internal gravity waves:Non-linear stratification
Consider the case
N(z) = N0 e−z/H
where kH >> 1, and H is the characteristic depth of the stratification. In this
case, we have
cosθ(z) = ω/N(z) = (ω/N0)ez/H.
Now, since dX/dz = tan θ, where X(z) describes the location of a wave beam,
then
1 − cos 2 θ
dX
=
cos θ
dz
,
−z / H
e
= 1 − cos 2 θ 0 e 2 z / H
cos θ 0
where θ0 is the angle of the beam at z = 0. Integrating yields
He −2 z H
X (z ) = a +
1 − cos 2 θ 0 e 2 z H
2 cos θ 0
+
[
(
H cos θ 0
ln e −2 z H 2 − e 2 z H cos 2 θ 0 + 2 1 − e 2 z H cos 2 θ 0
4
)]
Note that X(z) exists only so long as 1 − cos2θ0e2z/H ≥ 0. This is simply a
statement that ω ≤ N(z). Solving this for equality gives the maximum height to
which the wave can propagate as
zmax = −½ H ln(cosθ0).
– 90 –
GEFD
Internal gravity waves:Shear flows
z
0.2
X
-6
-5
-4
-3
-2
-1
-0.2
-0.4
-0.6
-0.8
-1
Figure 63: Ray theory prediction for wave beam in exponential stratification.
Figure 64: Experiment showing wave beams in exponential stratification.
The waves reflect from the level where ω = N(z).
3.14 Shear flows
In a shear flow, we cannot find a frame of reference in which the undisturbed
fluid is at rest, and so we must approach the problem slightly differently to
what we have done until now.
In particular, rather than considering a time-dependent problem in a frame
where there is no background motion, we shall consider steady disturbances in
a frame where the background motion is also steady.
– 91 –
GEFD
Internal gravity waves:Shear flows
3.14.1 Sheared base state
Suppose the background motion is U = (U(z),0,0). Recalling the equations of
motion:
∇⋅u = 0,
⎞
⎛∂
⎜ + u ⋅ ∇ ⎟ρ = 0 ,
⎠
⎝ ∂t
∂u
1
+ ( u ⋅ ∇ ) u = − ∇ ( p + ρ0 gz ) − g ′zˆ
∂t
ρ0
∂ω
g
+ (u ⋅ ∇ )ω = (ω ⋅ ∇ )u + zˆ × ∇ρ
∂t
ρ0
and linearising steady motion about u = U for a plane-wave in the (x,z) plane
reveals
∇⋅u′ = 0,
ρ
∂ρ ′
∂ρ ′
+U
= w′ 0 N 2 ,
∂t
∂x
g
∂u ′
∂u′
dU
1 ∂p′
,
+U
+ w′
=−
∂t
∂x
dz
ρ0 ∂x
1 ∂p′
ρ′
∂w′
∂w′
+U
=−
−g ,
∂t
∂x
ρ0 ∂z
ρ0
Defining
⇒
b=−
g
ρ0
ρ′,
∇⋅u′ = 0,
(72)
∂b
∂b
+U
= − w′N 2 ,
∂t
∂x
(73)
– 92 –
GEFD
Internal gravity waves:Shear flows
∂u ′
∂u′ dU
1 ∂p′
,
+U
+
w′ = −
∂t
∂x dz
ρ0 ∂x
(74)
∂w′
∂w′
1 ∂p′
+U
=−
+b,
∂t
∂x
ρ0 ∂z
(75)
Eliminate u′ from (74) using continuity,
∂ ⎞ ∂w′
∂w′ 1 ∂ 2 p′
⎛∂
−U′
=
,
⎜ +U ⎟
∂x ⎠ ∂z
∂x ρ0 ∂x 2
⎝ ∂t
(76)
and eliminate b using D(75)/Dt and (73):
2
1 ⎛∂
∂ ⎞
∂ ⎞ ∂p′
⎛∂
2
′
′
U
w
N
w
U
+
+
=
−
+
.
⎜
⎟
⎜
⎟
∂x ⎠
∂x ⎠ ∂z
ρ0 ⎝ ∂t
⎝ ∂t
Finally, eliminate p′ by ∂2/∂x2(77) and D/Dt ∂/∂z(76):
2
2
2
∂ ⎞ ∂ 2 w′
∂ ⎞ ∂ 2 w′
∂
∂ ⎞ ∂ 2 w′
⎛∂
⎛∂
⎛
2 ∂ w′
+N
+ ⎜ +U ⎟
+ U ′⎜ + U ⎟
⎜ +U ⎟
2
2
2
∂x ⎠ ∂x
∂x
∂x ⎠ ∂z
∂x ⎠ ∂x∂z
⎝ ∂t
⎝ ∂t
⎝ ∂t
∂ ⎞ ⎛ ∂ 2 w′
∂w′ ⎞
⎛∂
− ⎜ + U ⎟ ⎜U ′
+ U ′′
⎟=0
∂x ⎠ ⎝ ∂x∂z
∂x ⎠
⎝ ∂t
2
⇒
2
∂ ⎞ ⎛ ∂2
∂2 ⎞
∂ ⎞ ∂w′
⎛∂
⎛∂
2 ∂ w′
′
′′
U
w
N
U
U
+
+
+
−
+
= 0.
⎟
⎜
⎟ ⎜
⎜
⎟
∂x ⎠ ⎝ ∂x 2 ∂z 2 ⎠
∂x 2
∂x ⎠ ∂x
⎝ ∂t
⎝ ∂t
This relationship can be compared with (55) that we developed in §3.2, viz.
⎡ 2 ∂2
2 2 ⎤
+
∇ H ⎥ w′ = 0 ,
∇
N
⎢ ∂t 2
⎣
⎦
if we consider the limit of U = 0:
2
∂2 ⎛ ∂2
∂2 ⎞
2 ∂ w′
+
= 0.
⎜
⎟ w′ + N
∂t 2 ⎝ ∂x 2 ∂z 2 ⎠
∂x 2
– 93 –
(77)
GEFD
Internal gravity waves:Shear flows
Simplification is also possible if the flow is steady in the selected frame of
reference,
⎛ N 2 U ′′ ⎞ ∂ 2 w′
∂2 ⎛ ∂2
∂2 ⎞
+ 2 ⎟ w′ + ⎜ 2 −
⎟ 2 = 0,
2 ⎜
2
U ⎠ ∂x
∂x ⎝ ∂x
∂z ⎠
⎝U
which may be integrated twice with respect to x to give
⎡ ∂2
∂ 2 N 2 U ′′ ⎤
⎢ ∂x 2 + ∂z 2 + U 2 − U ⎥ w′ = 0 .
⎣
⎦
(78)
When U and N are constant, we can search for solutions of the form
w′ = w0ei(kx + mz), which requires
⎡
N2 ⎤
2
2
⎢ − ( k + m ) + U 2 ⎥ w′ = 0
⎣
⎦
(79)
or |k| = N/U as we found in §3.6. Note that the same form of solution is valid
for other stratifications and shears provided
N 2 U ′′
−
= const .
2
U
U
Exercise: Derive the equivalent expression s for a steady, three-dimensional
disturbance.
3.14.2 Ray tracing in a shear flow
In many cases the length scale over which stratification and velocity change is
large compared with the wavelength of the internal waves. In such cases we
can consider the important wave properties as being instantaneous in the local
state of the background flow. This is the Liouville-Green, WKB or WKBJ
(Wentzel–Kramers–Brillouin–Jeffreys) approximation.
Rays follow the group velocity relative to the fluid. However, in our frame of
reference the waves appear stationary, and so the phase velocity must be
parallel to the wave crests and the group velocity antiparallel with the
wavenumber vector. We can thus take the trajectory of a wave propagating
with the mean flow as
– 94 –
GEFD
Internal gravity waves:Shear flows
cgz
dz
m
=
= ,
dx U − cgx k
where cg is the group velocity relative to the fluid.
[If we write the group velocity relative to the frame of reference as Cg, then
Cgz = cgz and Cgx = U − cgx so that
cgz
Cgz m
dz
=
=
= .]
dx U − cgx C gx k
Recalling from (79) that k2 + m2 = N 2/U 2, and assuming that k is conserved
along the ray and U″/U N2/U2, then
m2 = N 2/U 2 − k2
1
and
2
⎞ 2
dz ⎛ ⎛ N ⎞
= ⎜⎜
⎟ − 1⎟⎟ .
dx ⎜⎝ ⎝ kU ⎠
⎠
Clearly waves cannot propagate through a level where kU/N = 1.
In the atmosphere, U/N tends to increase with height (primarily through U
increasing), and waves that are able to propagate upwards from near the ground
may reach this critical height zc at which kU/N = 1.
Using a Taylor-Series expansion near zc shows
dz ⎛
d ⎛ N 2 ⎞⎞
= ⎜ ( z − zc ) ⎜ 2 2 ⎟ ⎟
dx ⎝
dz ⎝ k U ⎠ ⎠
⇒
1
2
1 d ⎛ N2 ⎞
2
,
−
z = zc +
x
x
(
)
0
⎜
⎟
4 dz ⎝ k 2U 2 ⎠ z = z
c
indicating that rays starting near the ground are bent over and travel back
towards the ground. Note that their vertical wavenumber tends towards zero as
they approach this level, and the beams travel more nearly vertical relative to
the fluid.
– 95 –
GEFD
Internal gravity waves:Shear flows
When sketching waves, remember that in the frame of reference of the forcing
the wave field is steady and so cp is along the wave crests (normal to k) and cg
is antiparallel to k (with opposite signs for the vertical component).
U/N
zc
k
Cp
Cg
Figure 65 Sketch of ray reflecting from a critical layer. Group and phase velocities are
shown in the frame in which the wave field is steady (the frame of the source of the
disturbance). Orientations and not magnitudes are shown.
Note that with m → 0 the vertical scale of the beam is no longer small
compared with the scale over which U and N change, so the approximation
breaks down. Nevertheless, to leading order (and qualitatively), the prediction
remains good.
If the velocity reverses, then where U → 0,
m2 = N 2/U2 − k2 → ∞,
the wave ceases to propagate and is absorbed.
This is referred to as critical layer absorption. The waves cease to propagate
vertically and deposit their energy into the mean flow.
– 96 –
GEFD
Internal gravity waves:Shear flows
U/N
za
k
Cp
Cg
Figure 66: Critical absorption of a ray propagating in a shear flow of the form
U(z) = U0 − β z. Group and phase velocities are shown in the frame in which the wave
field is steady (the frame of the source of the disturbance). Orientations and not
magnitudes are shown.
−20
0
20
U(z) mms-1
300.0
z (mm)
250.0
200.0
150.0
100.0
50.0
0.0
0.0
100.0
200.0
300.0
400.0
x (mm)
(a)
– 97 –
500.0
600.0
700.0
800.0
GEFD
Internal gravity waves:Shear flows
−20
0
20
U(z) mms-1
300.0
z (mm)
250.0
200.0
150.0
100.0
50.0
0.0
0.0
100.0
200.0
300.0
400.0
500.0
600.0
700.0
800.0
400.0
500.0
600.0
700.0
800.0
x (mm)
(b)
−20
0
20
U(z) mms-1
300.0
z (mm)
250.0
200.0
150.0
100.0
50.0
0.0
0.0
100.0
200.0
300.0
x (mm)
(c)
Figure 67: Critical layer absorption. (a) Uniform flow without absorption; (b) critical layer
reflection due to increased velocity, and (c) critical layer absorption due to a reversal in
velocity. In both cases the upward propagating component of ∂ρ′/∂x is shown. There is a
rigid lid at z = 0 that reflects downwards the waves in (a), but there is no downward
propagating signal in (c).
Note that U(z) → 0 causes m → ∞ and so Cgx → 0 (cgx → U).
More generally, for time dependent problems, we can see from the expression
for geometric ray tracing,
cgz
dz
,
=
dx U − cgx
that critical layer absorption occurs when
– 98 –
GEFD
Internal gravity waves:Shear flows
U(z) − cgx → 0.
The above assumed that U″ was negligible. However, this may not be the case.
Recall (78):
⎡ ∂2
∂ 2 N 2 U ′′ ⎤
⎢ ∂x 2 + ∂z 2 + U 2 − U ⎥ w′ = 0 .
⎣
⎦
We may thus generalise the above analysis using
S2 = N2/U2 − U″/U > 0
to indicate where reflection will occur for a beam propagating with U. The
condition for absorption remains U = cgx.
3.14.3 Three-dimensional forcing
In three dimensions, linearising about a mean wind U(z) in the x direction leads
to
⎡
d 2U
N2
2
2
2
2
2 ⎤
ω
−
+
+
+
−
+
kU
k
l
m
k
k
l
(
)
(
)
(
)⎥ w = 0
⎢
2
ω
−
dz
kU
⎣
⎦
giving the equivalent of (79) for a three-dimensional wave as
N 2 k 2 + l 2 1 d 2U
k +l +m = 2
−
U
k2
U dz 2
2
2
2
(80)
One way to continue is to linearise the boundary z = h(x,y) so that
w0(x,y) = U dh/dx.
This requires that ∇h is everywhere small, and h is small compared with the
shortest horizontal length scales. In particular, if we compute H(k,l), a twodimensional Fourier transform of h(x,y), then we require kH, lH 1.
We can use this Fourier transform, H, to compute the wave field.
– 99 –
GEFD
Internal gravity waves:Shear flows
We must note first that causality requires the phase velocity pointing upstream
(i.e. the group velocity must be upstream relative to the fluid, or else the
disturbance is washed downstream). Taking U > 0, we then discard
components of H with k > 0.
As the disturbance propagates vertically, geometric arguments demonstrate that
the horizontal components of the wavenumber are preserved. Hence, for a
given k, l, we can calculate the corresponding vertical wavenumber m as the
negative root (from (80)) of
⎛ N2
⎞
m = ⎜ 2 2 − 1⎟ ( k 2 + l 2 ) .
⎝k U
⎠
2
We use the negative root to satisfy causality. As we are interested in
propagating modes, we require m to be real, and so this places upper bounds on
k and l.
If we take U = const, then we can write the Fourier transform of the vertical
velocity as
N
⎧
imz
⎪Uke H ( k , l ) , − < k < 0,
W ( k , l, z ) = ⎨
.
U
⎪⎩
otherwise,
0,
The eimz term represents the phase change of the waves as different modes
propagate vertically. The velocity of the waves is then recovered by computing
the inverse transform.
3.14.4 Effect of viscosity
As we have seen in §3.9.4 and 3.10, the amplitude of an internal wave decays
along the beam due to viscosity. In two dimensions the decay rate along a ray
is given by ½ ν |k|3/(N sin θ).
For three dimensions we will use the energy-based approach. Viscous
dissipation causes the energy in a given mode to decay as
dE
2
= −ν k E ,
dt
– 100 –
GEFD
Internal gravity waves:Shear flows
where the derivative is following a given parcel of energy. Making use of d/dt
= (dz/dt) d/dz for steady waves, where dz/dt is the vertical velocity of the
parcel of energy, i.e., the vertical component of the group velocity
cgz = (N/|k|) cosθ sinθ = (N/|k|3) m(k2+l2)1/2,
then we can rewrite this decay of energy as
cgz
dE
2
= −ν k E .
dz
Noting that the energy density is proportional to the square of the wave
amplitude, provides a viscous prediction for the Fourier transform of w′:
N
⎧
( im− β ) z
H ( k , l ) , − < k < 0,
⎪Uke
W ( k , l, z ) = ⎨
U
⎪⎩
otherwise,
0,
where the vertical dissipation rate is
β=
ν
k
5
2 N m k 2 + l 2 12
(
)
.
Note that β is maximum (infinite) waves with a purely vertical or horizontal
wavenumber vector, and is smallest (slowest decay) for waves at ~45° (exactly
45° in 2D) to the horizontal. (The absolute value of m is used since m is
negative.)
– 101 –
GEFD
Internal gravity waves:Shear flows
-1.0
−2.0
0.0
2.0
4.0
x/R
6.0
0
8.0
bx /max(bx )
10.0
1.0
5.0
300.0
4.0
200.0
3.0
z/R
z (mm)
250.0
150.0
2.0
100.0
1.0
50.0
0.0
−100.0
0.0
100.0
200.0
300.0
x (mm)
400.0
500.0
600.0
700.0
0.0
Figure 68: Viscous prediction of wave field in a uniform flow obtained by linearising a
hemisphere.
3.14.5 Blocking
If the amplitude of the obstacle generating the lee waves is not negligible, then
linearising the lower boundary is not appropriate.
In two dimensions, the upstream flow may not have sufficient energy to be able
to rise up and over the obstacle, except for heights close to the height of the
obstacle. At lower level the flow is said to be ‘blocked’.
The formation of this blocking region is time-dependent, but will ultimately
influence the upstream conditions. We shall talk about related things when
dealing with hydraulics later in the course.
In three dimensions, the flow may prefer to go around the obstacle rather than
over it. The flow around the obstacle will be approximately two-dimensional
(horizontal), except near the top of the obstacle. Only the flow over the top will
have the vertical excursions necessary to generate internal gravity waves.
The height of the streamline that separates the flow around from the flow over
can be estimated as
zs = (1 – A F)h0,
– 102 –
GEFD
Internal gravity waves:Shear flows
where h0 is the height of the obstacle, F = U/Nh0 is the Froude number (more
on this later), and A is an O(1) constant.
-0.1
−2.0
0.0
2.0
4.0
x/R
6.0
0
8.0
bx /N2
10.0
0.1
5.0
300.0
4.0
200.0
3.0
z/R
z (mm)
250.0
150.0
2.0
100.0
1.0
50.0
0.0
−100.0
0.0
100.0
200.0
300.0
x (mm)
(a)
– 103 –
400.0
500.0
600.0
700.0
0.0
GEFD
Internal gravity waves:Shear flows
-0.1
−2.0
0.0
2.0
4.0
x/R
bx /N2
0
6.0
8.0
0.1
10.0
5.0
300.0
4.0
200.0
3.0
z/R
y (mm)
250.0
150.0
2.0
100.0
1.0
50.0
0.0
−100.0
0.0
100.0
200.0
300.0
400.0
x (mm)
500.0
600.0
0.0
700.0
(b)
20.0
20.0
Mean spectrum
N/U
2π/H
2π/R
15.0
bx (incident)
bx (reflected)
bx (essentially hori
bx (cylinder)
N/U
2π/H
2π/R
10.0
Power
Power
15.0
5.0
0.0
0.0
10.0
5.0
20.0
40.0
60.0
80.0
k (m−1)
100.0
120.0
140.0
0.0
0.0
20.0
40.0
60.0
80.0
k (m−1)
100.0
120.0
140.0
Figure69: (a) Vertical velocity of the internal wave field computed by assuming a
linearized hemispherical cap. (b) The upward propagating component of w from an
experiment of stratified uniform flow over a hemisphere. (c) Comparison between w
spectra for linearized spherical cap and experiment.
– 104 –
GEFD
Internal gravity waves:Columnar modes
Figure 70: Cross-channel structure of w at 50 mm intervals above zs for a spherical cap.
Note the ship-wave-like pattern at low level, but unlike ship waves, the dispersion relation
here allows information to propagate upstream of the hemisphere… Maybe this is an
alternative explanation for the pictures of waves near the Prince Edward Islands.
3.15 Columnar modes
In some cases, reflections from the upper and lower boundaries produce a
superposition of upward and downward propagating waves that combine to
produce an essentially horizontal motion.
Consider the limit ω << N. In this case the wavenumber vector is nearly
vertical, and the group velocity horizontal. Due to the reflections, only standing
waves can persist. Recall the group velocity (63),
cg =
∂ω
∂k
⎤
ω ⎡N2
= 2 ⎢ 2 (k − k z zˆ ) − k ⎥
k ⎣ω
⎦
and consider the zero frequency (vertical wavenumber) limit,
– 105 –
GEFD
Internal gravity waves:Columnar modes
⎛ N 2 k − k zˆ ⎞
z ⎟
lim c g = lim⎜
2
⎜
⎟
ω →0
ω →0 ω
k
⎝
⎠
⎛N⎞
= lim⎜⎜ ⎟⎟xˆ
ω →0 k
⎝ ⎠
NH
=
xˆ
πn
where H is the depth of the flow and n is the mode number. Note that the
lowest mode (n = 1) corresponds to a wavelength λ = 2H. (The phase velocity
vanishes.)
n =1
n =2
n =3
n =4
Figure 71: Sketch of columnar wave modes.
Frans de Rooij (DAMTP)
Figure 72: Columnar wave modes generated by intrusion (from the right) in linear
stratification.
– 106 –
GEFD
Effects of rotation:Coriolis force and rotation
4 Effects of rotation
4.1 Coriolis force and rotation
Motion in a rotating system obeys the same laws as that in a non-rotating
system. Newton’s laws of motion continue to apply.
Ball
Imagine two people having a game of catch. One person is standing on a
roundabout (or merry-go-round) that is rotating anticlockwise, and the other is
standing in the playground. Both are equally accurate in their throws at a
stationary target. The observe that
(i) When the person in the playground throws the ball, he misses the person on
(ii) When the person on the roundabout throws the ball, she hits the person in
the playground.
The cause of (i) is obvious: the girl on the roundabout moves while the ball is
in flight. This is obvious to the boy in the playground, so he asks the girl to
stand at the centre of the roundabout.
However, the girl on the roundabout is confused. To her, it appears like the ball
curves while it is in flight towards her. Moreover, even though she is able to hit
the boy when she throws, it appears that the boy moves and the the ball again
curves sidewise rather than following a straight line.
– 107 –
GEFD
Effects of rotation:Coriolis force and rotation
It appears to the girl that the ball is always accelerated to the right of the path it
is taking. The girl therefore has to imagine that there is a force acting on the
ball to cause this acceleration, even though she knows the ball is really
travelling in a straight line.
The apparent force (also known as a fictitious force) is due to the girl being in
a non-inertial frame of reference. This force is known as the ‘Coriolis force’.
Pendulum
In the absence of air to contaminate things, a pendulum that is free to move in
any direction but is released from rest away from its point of equilibrium will
undergo oscillations in the plane containing both the release point and the
equilibrium point.
However, if we observe the pendulum over the course of a day, we see that this
plane slowly rotates. The motion remains parallel to the original direction
relative to the universe rather than in the same orientation relative to the
building since the building is on a rotating planet.
It is the vertical component of the earth’s rotation that is important here.
– 108 –
GEFD
Effects of rotation:Equations in the rotating frame
At the pole, the earth’s axis of rotation is in the vertical direction and in one
day the plane of the pendulum’s motion appears to rotate through an angle of
2π.
At the equator, the plane of the pendulum’s motion does not appear to rotate.
At a latitude θ, the plane of the pendulum’s motion appears to rotate through an
angle of 2π sinθ.
This apparent rotation can again be described by introducing the Coriolis force
for the rotating (non-inertial) frame of reference.
4.2 Equations in the rotating frame
4.2.1 Simple derivation
Suppose u = U + f × x and substitute into the momentum equation:
∂U
+ ( U ⋅ ∇ ) U + ( U ⋅ ∇ )( f × x ) + ( ( f × x ) ⋅ ∇ ) U + ( ( f × x ) ⋅ ∇ ) ( f × x )
∂t
1
= − ∇ p + ν∇ 2 U
ρ
Now (U⋅∇)(f×x) = Uj ∂/∂xj εikl fk xl = εikl fk Uj δjl = εikl fk Ul = f × U,
and ((f×x)⋅∇)(f×x) = εjkl fk xl ∂/dxj εimn fm xn =εjkl fk xl εimj fm= f×(f×x)
So
1
⎡∂
⎤
+
f
×
x
⋅
∇
U
+
U
⋅
∇
U
+
f
×
U
+
f
×
f
×
x
=
−
∇p + ν∇ 2 U
(
)
(
)
(
)
(
)
⎢⎣ ∂t
⎥⎦
ρ
The term in square brackets in just ∂/∂t in a frame of reference coincident with
x but rotating at angular frequency ½ f. Moving to this rotating frame of
reference,
∂U
1
+ ( U ⋅ ∇ ) U = − ∇p − f × ( f × x ) − f × U + ν∇ 2 U
ρ
∂t
The cyclostrophic term (centrifugal force) is normally absorbed into the
pressure.
– 109 –
GEFD
Effects of rotation:Equations in the rotating frame
For most geophysical flows, the essentially horizontal nature of the flow means
that only the vertical component of the rotation important, so f = 2 Ω sin σ ^z,
where σ is the latitude. Hence,
∂u
1 ∂p
+ (u ⋅ ∇ ) u = −
+ fv + ν∇ 2u
ρ ∂x
∂t
∂v
1 ∂p
+ (u ⋅ ∇ ) v = −
− fu + ν∇ 2 v
ρ ∂y
∂t
∂w
1 ∂p
+ (u ⋅ ∇ ) w = −
− g + ν∇ 2 w .
ρ ∂z
∂t
The Coriolis force, given by −f × u, is not a real force. It is known as a
‘ficticious force’. It is introduced since we want to be able to equate forces with
apparent accelerations.
4.2.2 Vorticity equation
As usual, taking the curl of the momentum equation gives the vorticity
equation:
⎡ ∂u
⎤
1
∇ × ⎢ + ( u ⋅ ∇ ) u = − ∇p − f × u + ν∇ 2u ⎥
ρ
⎣ ∂t
⎦
⇒
∂ω
+ ( u ⋅ ∇ ) ω = ( ω ⋅ ∇ ) u − ∇ × ( f × u ) + ν∇ 2 ω
∂t
⇒
∂ω
+ ( u ⋅ ∇ ) ω = ( ω ⋅ ∇ ) u − ( f ( ∇ ⋅ u ) − ( f ⋅ ∇ ) u ) + ν∇ 2 ω
∂t
⇒
∂ω
+ ( u ⋅ ∇ ) ω = ( ω ⋅ ∇ ) u + ( f ⋅ ∇ ) u + ν∇ 2 ω
∂t
⇒
∂ω
+ ( u ⋅ ∇ ) ω = ( ( ω + f ) ⋅ ∇ ) u + ν∇ 2 ω
∂t
It is important to compare the vortex stretching term here,
– 110 –
GEFD
Effects of rotation:Equations in the rotating frame
((ω + f ) ⋅ ∇ ) u ,
with that of the vortex stretching term in the inertial frame of reference
(indicated by ~),
(ω ⋅ ∇ ) u .
The vorticity ω in the rotating frame is the relative vorticity. If we add on to
this the planetary vorticity f, then we have ω + f which is the absolute vorticity
(i.e., the vorticity relative to the inertial frame).
Vortex stretching acts on the absolute vorticity.
For simplicity, consider the vertical vorticity ω and assume that only the
vertical component of f is important.
• Stretching a cyclone (ω and f the same sign) will make the cyclone stronger.
• Stretching stationary fluid will create a cyclone.
• Stretching an anticyclone will make the anticyclone stronger if ω < −f.
However, if ω > −f then the anticyclone will become weaker.
4.2.3 Rossby number
The Rossby number represents the relative strength of the inertial terms
compared with the Coriolis force
Ro =
O ((u ⋅ ∇ ) u )
O (f × u )
=
U2
L=U ,
fU
fL
where U, L and ρ0 are typical velocity and length scales.
Note that we have two timescales in the problem: L/U and 1/f.
If we choose to nondimensionalise our equations so that u*= u/U, x* = x/L,
t* = Ut/L, p* = ρ0U2, etc., then
– 111 –
GEFD
Effects of rotation:Equations in the rotating frame
ρ0U 2 1 ∂p*
U ∂u * U 2 * * *
U *2 *
*
ν
u
+
⋅
∇
=
−
+
+
∇ u
u
fUv
(
)
ρ0 L ρ * ∂x*
L / U ∂t *
L
L2
⇒
∂u *
1 ∂p* fL * ν *2 *
*
*
*
+ (u ⋅ ∇ ) u = − * * + v +
∇ u
*
∂t
UL
ρ ∂x U
⇒
∂u *
1 ∂p* 1 * 1 *2 *
*
*
*
+ (u ⋅ ∇ ) u = − * * +
v +
∇ u .
∂t *
Re
ρ ∂x Ro
where Re =
UL
ν
is the Reynolds number.
We often choose to rewrite this as
⎡ ∂u *
Ro ∂p*
*
*
*⎤
Ro ⎢ * + ( u ⋅ ∇ ) u ⎥ = − * * + v* + E∇*2u *
ρ ∂x
⎣ ∂t
⎦
where E =
ν
fL2
is the Ekman number.
Note that this choice of non-dimensionalisation is not unique, and indeed may
not be the most useful when Ro is small and the flow is dominated by rotation.
Under such circumstances it may be better to select 1/f as the timescale so that
we use uˆ = u U , xˆ = x L , tˆ = ft , pˆ = p ρ0 f 2 L2 , ending up with
∂uˆ
ˆ uˆ = − 1 ∂pˆ + vˆ + E∇
ˆ 2uˆ
+ Ro uˆ ⋅ ∇
∂tˆ
ρˆ ∂xˆ
(
)
∂vˆ
ˆ vˆ = − 1 ∂pˆ − uˆ + E∇
ˆ 2 vˆ
+ Ro uˆ ⋅ ∇
∂tˆ
ρˆ ∂yˆ
(
)
∂wˆ
ˆ wˆ = − 1 ∂pˆ − Ro + E∇
ˆ 2 wˆ
+ Ro uˆ ⋅ ∇
2
∂tˆ
ρˆ ∂zˆ Fr
(
)
– 112 –
GEFD
Effects of rotation:Taylor-Proudman theorem
4.3 Taylor-Proudman theorem
4.3.1 Derivation
Consider steady motion of an inviscid homogeneous fluid when Ro
rotation dominates over inertia). In this limit we have
1 (i.e.
∂u
1 ∂p
+ (u ⋅ ∇ ) u = −
+ fv + ν∇ 2u
∂t
ρ ∂x
∂v
1 ∂p
+ (u ⋅ ∇ ) v = −
− fu + ν∇ 2 v
∂t
ρ ∂y
and there is a balance between the pressure gradient and the Coriolis force.
This situation is referred to as geostrophic balance.
Equating ∂/∂y of the x momentum equation with ∂/∂x of the y equation,
1 ∂2 p
∂v
1 ∂2 p
∂u
−
+f
=−
−f
∂y
∂y
ρ ∂x∂y
ρ ∂x∂y
⇒
⎛ ∂u ∂v ⎞
f ⎜ + ⎟=0
⎝ ∂x ∂y ⎠
Since ∇⋅u = 0, then we conclude that
∂w
= 0.
∂z
Moreover, the vertical momentum equation in the steady inviscid Ro
becomes
∂w
1 ∂p
+ (u ⋅ ∇ ) w = −
− g + ν∇ 2 w
∂t
ρ ∂z
Thus, taking ∂/∂z of the horizontal momentum equations, shows
– 113 –
1 limit
GEFD
Effects of rotation:Taylor-Proudman theorem
−
∂ ⎛ 1 ∂p ⎞
∂v 1 ∂ ⎛ 1 ∂p ⎞
∂v ∂g
∂v
+f
=
−
+f
=
+f
=0
⎜
⎟
⎜
⎟
∂z ⎝ ρ ∂x ⎠
∂z ρ ∂x ⎝ ρ ∂z ⎠
∂z ∂x
∂z
⇒
∂v
=0
∂z
Similarly
∂u
=0
∂z
Thus, in this limit the vertical gradients in the velocity all vanish: the flow is
two-dimensional, but not necessarily two-component as we have not (yet)
placed any restriction on w = w(x,y).
Obviously, if there are boundaries with a horizontal extent then there will be
some restrictions placed on w.
4.3.2 Taylor columns
Consider an obstacle moving slowly through a rapidly rotating, homogeneous,
inviscid fluid that is otherwise at rest. The Taylor-Proudman theorem has
shown us that the motion is two-dimensional, and independent of z.
If the obstacle is moving horizontally, the flow around the base of the obstacle
will be two-dimensional potential flow.
– 114 –
GEFD
Effects of rotation:Taylor-Proudman theorem
With ∂w/∂z = 0 the fluid cannot move up and over the obstacle. The constraint
∂u/∂z = 0 and ∂v/∂z = 0 means instead the flow must follow the same
streamlines everywhere through the depth.
– 115 –
GEFD
Effects of rotation:Taylor-Proudman theorem
The Taylor-Proudman theorem does not dictate the motion of the fluid above
the obstacle, other than it must be two-dimensional. In real fluids it is likely,
however, that the fluid will be at rest (or close to rest) relative to the obstacle.
– 116 –
GEFD
Effects of rotation:Rotating boundary layers
Figure 73: Taylor column(?) forming a von Karmen vortex street behind the Guadalupe
Islands (to the west of the Baja peninsula). (veimages.gsfc.nasa.gov)
4.3.3 Vertical motion
If we move the obstacle vertically rather than horizontally, Taylor columns still form: the
fluid above and below the obstacle moves at the same speed as the obstacle.
Obviously this statement must be modified in the presence of any horizontal boundaries!
4.4 Rotating boundary layers
4.4.1 Ekman layers
Suppose there is a horizontal no-slip boundary moving with velocity U =
(U,0,0) at z = 0. Near this boundary the flow will be essentially horizontal. We
shall assume that the flow remains laminar and that the velocity u → 0 far from
the boundary.
Assuming rotation about the (vertical) z axis, consider the horizontal flow near
a rigid boundary at z = 0:
If the system is not rotating, then
∂u
d 2u
=ν 2 .
∂t
dz
– 117 –
GEFD
Effects of rotation:Rotating boundary layers
If U is imposed instantaneously, then the boundary layer adopts the well
known error function form that grows in thickness as δ ~ (νt)1/2. However,
there is no steady boundary layer solution.
If the system is rotating, the boundary layer equations are
∂u
d 2u
= fv + ν 2
dz
∂t
∂v
d 2v
= − fu + ν 2
∂t
dz
∂u ∂v ∂w
+ +
=0
∂x ∂y ∂z
For a steady solution we can eliminate v to write
d2 ⎛
d 2u ⎞
d 2v
d4
f2
d4
u +ν 4 u = 0
⎜ fv + ν 2 ⎟ = f 2 + ν 4 u =
dz 2 ⎝
dz ⎠
dz
dz
dz
ν
⇒
d 4u f 2
d 4u 4
u= 4 + 4u=0
+
dz 4 ν 2
dz
δ
1
⎛ ν ⎞ 2
δ = ⎜2 ⎟ .
⎝ f ⎠
where
[Note the similarity with the oscillating boundary layer case where δ = (ν/ω)1/2,
especially since here f = 2Ω and so δ = (ν/Ω)1/2.]
The general solution is therefore
z
z⎞
z
z⎞
⎛
⎛
u = e − z δ ⎜ A cos + B sin ⎟ + e + z δ ⎜ C cos + D sin ⎟ .
δ
δ⎠
δ
δ⎠
⎝
⎝
Imposing u → 0 as z → ∞ gives C = 0 = D while u(z=0) = U gives A = U.
Substituting into the y momentum equation
– 118 –
GEFD
Effects of rotation:Rotating boundary layers
d 2v f
2
=
=
u
u
δ2
dz 2 ν
2
z
z⎞
⎛
= 2 e − z δ ⎜ U cos + B sin ⎟
δ
δ
δ⎠
⎝
Solving
z
z⎞
⎛
v = a + bz + e − z δ ⎜ B cos − U sin ⎟ .
δ
δ⎠
⎝
Again, v → 0 as z → ∞, so a = 0 = b, while v(z=0) = 0, so B = 0. Thus
z
u = Ue − z δ cos ,
δ
z
v = −e − z δ U sin .
δ
The amplitude of the motion in this steady boundary layer decays exponentially
away from the boundary, with the orientation rotating over the same length
scale (δ).
Due to its structure, this Ekman boundary layer is often referred to as an
Ekman spiral.
– 119 –
GEFD
Effects of rotation:Rotating boundary layers
To determine the net flux in the direction of the boundary, we integrate
∞
∞
Fx = ∫ u dz = ∫ Ue − z δ cos
0
=
0
z
δ
dz
,
Uδ
2
∞
∞
Fy = ∫ v dz = ∫ −Ue − z δ sin
0
=−
0
z
δ
dz
Uδ
2
,
and see that the net flux is oriented 45° to the right (in the northern hemisphere)
of the motion of the boundary.
4.4.2 Ekman pumping
Consider a horizontal boundary that is stationary in the rotating frame beneath
a vortex in which there is constant relative vorticity ω (corresponding to a solid
body relative rotation rate Θ = ω/2). For simplicity, we shall assume that ω is
positive, which means that in an inertial frame the fluid is rotating more rapidly
than the boundary.
It is easiest here to move (temporarily) into a frame of reference moving with
the vortex. In this frame the boundary appears to be moving backwards with
rotation rate −Θ (vorticity −ω). Specifically, the velocity of the boundary in
polar coordinates is
U = (U r ,Uθ )
ω ⎞.
⎛
= ( 0, −Θr ) = ⎜ 0, − r ⎟
2 ⎠
⎝
– 120 –
GEFD
Effects of rotation:Rotating boundary layers
f/2 + ω
f/2
ω
F
−ω
U
Figure 74: Schematic showing (a) vortex in rotating frame, (b) motion of boundary relative
to vortex, (c) Velocity at boundary and direction of flux within boundary layer in frame
rotating with vortex.
From our analysis in the last section, the flux within the boundary layer is
directed 45° to the right of this and is thus
F = ( Fr , Fθ )
⎛U δ U δ ⎞
=⎜ θ , θ ⎟
2 ⎠
⎝ 2
Θrδ
ω rδ
=−
(1,1) = −
(1,1)
2
4
Because this flux has a radial component it is not divergence free in the plane
of the boundary. In particular,
1 ∂
1 ∂F
( rFr ) + θ
r ∂r
r ∂θ
Θδ 1 ∂ 2
=−
(r )
2 r ∂r
1
= −Ωδ = − ωδ
2
∇⋅F =
– 121 –
GEFD
Effects of rotation:Rotating boundary layers
This negative two-dimensional divergence implies an accumulation of fluid in
the boundary layer. However, as the boundary layer thickness is set by
δ = (2ν/f), the divergence of the flux must be balanced by a vertical velocity
⎛ ν ⎞
we = Θδ = Θ ⎜ 2 ⎟
⎝ f ⎠
1
2
⎛ν ⎞
= Θ⎜ ⎟
⎝Ω⎠
1
2
out of the boundary layer. Note that this velocity is independent of the position
within the vortex, and of the location of the axis of rotation.
While we have undertaken this calculation in the frame rotating with the
vortex, the divergence of the flux and vertical velocity applies equally to the
frame rotating with the boundary.
The results calculated are the same if consider an anticyclone with relative
vorticity ω < 0, although of course the direction of the vectors is reversed and
fluid is drawn into the boundary layer rather than being expelled from it.
Provided the boundary layer thickness δ is thin compared with the other scales
of motion, and the Eulerian timescale T is long compared with 1/f, then we may
apply the same analysis when ω just above the boundary layer is varying with
position and time.
4.4.3 Spin-up
Consider an open container of liquid of depth H in solid body rotation with
angular velocity Ω. Suppose that at t = 0 the angular velocity is increased to
Ω + ΔΩ. How is the increased speed communicated to the interior of the fluid?
If it were simple molecular diffusion in an inertial frame, then it would take a
time of order H2/ν for the boundary layer to grow through the depth. However,
in the rotating frame the boundary layer is fixed at δ = (ν/Ω)1/2 and does not
grow with time. Does this mean the moving boundary does not affect the flow?
The fluid will not instantaneously change its angular velocity as the increase in
rotation rate of the container is only communicated through viscous effects. At
t = 0+ the flow will look like that analysed in the previous section: in the frame
of reference of the interior (still rotating at Ω = f/2), the lower boundary will
have a relative vorticity ω = 2ΔΩ.
– 122 –
GEFD
Effects of rotation:Rotating boundary layers
Ω
2ΔΩ
The flux outwards through the Ekman layer leads to a drawing down of fluid
into the boundary layer from above. This is referred to as Ekman pumping. The
drawing down of fluid then stretches the columns of fluid in the interior.
Ω
Ω
2ΔΩ
2ΔΩ
The stretching of a column of fluid amplifies the column’s absolute vorticity,
2Ω.
To increase the absolute vorticity of the interior from 2Ω to 2(Ω + ΔΩ) we
must stretch the column from a height H to H(1 + ΔΩ/Ω).
If we take ΔΩ
Ω, then initially the velocity into the boundary layer is
we = ΔΩ(ν/Ω)1/2. The necessary stretching will therefore require a time
– 123 –
GEFD
Effects of rotation:Rotating boundary layers
τ=
ΔH ΔΩ
1
=
H
1/2
we
Ω
ΔΩ (ν / Ω )
1/2
1/2
1/2
1/2
2
⎛ 1 ⎞ ⎛H ⎞
=⎜ ⎟ ⎜
⎟
⎝Ω⎠ ⎝ ν ⎠
⎛ 2 ⎞ ⎛ H2 ⎞
=⎜ ⎟ ⎜
⎟
⎝ f ⎠ ⎝ ν ⎠
21/2 −1/2
E
=
f
This is the Ekman spin-up time.
4.4.4 Stewartson layers
What happens to the flux moving outwards in the Ekman layer? If it meets the
side walls of the container, then it must be deflected upwards into the
Stewartson layers. The flux within these Stewartson layers is convergent,
driving a horizontal return flow into the interior of the container.
As the fluid within the Stewartson layers will be rotating at the new speed of
the container (having been in contact with the container), then there will be an
inward propagation of fluid that has been spun-up by this mechanism, at the
same time as the interior of the flow is spun-up by vortex stretching.
If we increase the rotation rate by 10%, then we must stretch the interior by
10%, so that 10% of the fluid passes through the Ekman boundary layers and
Stewartson layers. This fluid that has passed through the boundary layers will
then occupy the outer 10% of the area, or about 5% of the radius.
– 124 –
GEFD
Effects of rotation:Rotating boundary layers
Ω + 2ΔΩ
Figure 75: The circulation responsible for spinup. The interior fluid (blue) is stretched, by
Ekman pumping drawing interior fluid into the Ekman layer (clear). This fluid is then
returned through the Ek1/4 boundary layer (green) to form a ring of fluid already spun up
(pink).
Examples Sheet: Determine the structure of the Stewartson layers.
The structure of the Stewartson layers is determined to both match the
boundary conditions on the walls of the container, and (more importantly) to
match the fluxes provided by the Ekman layer.
E1/3 layer
The inner Stewartson layer transports vertically the O(E1/2) flux thrown out by
the O(E1/2) thick Ekman layer on the base of the tank. This boundary layer is
able to satisfy a no-slip boundary conditions on the vertical velocity, and a no
normal flux condition on the radial velocity, but is unable to match the
azimuthal velocity of the internal flow with the no-slip boundary condition on
the wall.
The width of the inner Ekman layer scales as E1/3. Introducing a scaled
coordinate normal to the wall of ξ = E−1/3(R−r) leads to the dimensionless
boundary layer equations
∂v
∂w
+ E1/3
= 0,
∂ξ
∂z
– 125 –
GEFD
Effects of rotation:Rotating boundary layers
E
1/3
∂ 2u
− v = 0,
∂ξ 2
E1/3u +
E
1/3
∂p
= 0,
∂ξ
∂ 2 w ∂p
+
= 0,
∂ξ 2 ∂z
where u, v and w are the tangential, normal and vertical velocities.
The flux of fluid supplied by the Ekman layer is carried vertically in the E1/3
Stewartson layer, but is gradually expelled from it in the present case due to the
condition that w = 0 at the free surface. Thus we find ∂w/∂z ~ const and so v
increases linearly with distance from the boundary. The implied cubic change
in u with the stretched coordinate ξ cannot match a no-slip condition at the wall
with the O(1) velocity in the interior. Indeed, the tangential velocity u is only
of O(E1/6) within the E1/3 boundary layer.
Ek1/4 boundary layer
The outer Stewartson layer transports radially back to the interior the O(E1/2)
flux thrown out by the O(E1/2) thick Ekman layer on the base of the tank. Thus,
the radial velocity within this boundary layer must be O(E1/2) since the height
is O(1). The boundary layer must also match onto the O(1) azimuthal flow in
the interior. The O(E1/6) azimuthal flow at the outer edge of the inner E1/3
boundary layer is u ~ 0 at z = 0 for the outer boundary layer.
The width of the outer Ekman layer scales as E1/4. Introducing a scaled
coordinate normal to the wall of χ = E−1/4(R−r) leads to the dimensionless
boundary layer equations
∂v
∂w
+ E1/4
=0
∂χ
∂z
E
1/2
∂ 2u
−v =0
∂χ 2
– 126 –
GEFD
Effects of rotation:Inertial waves
E1/4u +
∂p
=0
∂χ
∂p
= 0.
∂z
The vanishing vertical pressure gradient means that ∂u/∂z also vanishes, and
hence ∂v/∂z = 0, matching the interior under the Taylor-Proudman hypothesis.
(Thus the flux from the Stewartson layers to the interior is uniformly
distributed throughout the depth.)
Note that the vertical velocity is of O(E1/4) so that the vertical flux in the
boundary layer is O(E1/2), again matching that of the Ekman layer. However,
the E1/4 boundary layer alone is not able to match the other boundary
conditions, and so both boundary layers are required.
Occurrence of Stewartson layers
Although there are other possible scalings for other Stewartson layers under
different conditions, the two types we have discussed here are the most
common. They are driven by the thinner, more intense E1/2 Ekman layer on a
horizontal or near-horizontal boundary. They will occur, in general, whenever
there is a sudden change in the boundary condition and a corresponding change
in either the horizontal or vertical velocity of the interior flow.
4.5 Inertial waves
4.5.1 Inertial circles
If we impart a horizontal velocity u to a parcel of fluid, then it will begin to
accelerate in the negative y direction due to the Coriolis force ρfu. For a small
velocity in an inviscid homogeneous fluid we therefore have
∂u
= fv
∂t
∂v
= − fu
∂t
from which we can eliminate v to obtain
– 127 –
GEFD
Effects of rotation:Inertial waves
∂ 2u
+ f 2u = 0 .
2
∂t
Similarly,
∂ 2v
+ f 2v = 0 .
2
∂t
Thus we will have an oscillatory motion of angular frequency f in both
directions. The signs of the first order equations show that v leads u by 90°, so
that the fluid parcel executes anticyclonic circular motions in the horizontal
plane.
There is no coupling between different horizontal planes, so the phase of the
motion can be different at different heights.
If the motion is a perturbation about a state of rest in the rotating frame then it
must satisfy ∂u/∂x + ∂v/∂y = 0 and so if the motion is independent of one
horizontal coordinate, it must also be independent of the other.
4.5.2 Displacements at an angle
We can repeat this analysis by imparting a velocity U at an angle α to the
horizontal above the x axis so that u = U cosα. In this case we can write
∂⎛ U ⎞
⎜
⎟ = fv
∂t ⎝ cos α ⎠
– 128 –
GEFD
Effects of rotation:Inertial waves
∂v
= − fU cos α
∂t
and eliminate v to obtain
∂ 2U
2
+
f
cos
α
U = 0,
(
)
∂t 2
and so the frequency of the motion is ω = f cosα. Note that we now have
vertical velocities (where w = U sin α) in addition to the horizontal velocity
and the motion will still be confined to planes, but these are no longer
horizontal.
k
θ
α
We can describe the motion between the different planes in terms of a
wavenumber vector k, that is normal to the motion. As we did for internal
waves, we can define θ as the angle between the wavenumber vector and the
horizontal, and note that θ = 90° − α so that cos α = sin θ and thus we can
write the frequency relationship as
ω
f
= sin θ .
This is the dispersion relationship for ‘inertial waves’. Note the similarity with
the dispersion relationship for internal waves:
– 129 –
GEFD
Effects of rotation:Inertial waves
ω
N
= cosθ .
4.5.3 Linear inertial waves
Linearise the momentum equations around a state of rest for an inviscid flow:
∇⋅u = 0,
∂u
1 ∂p
=−
+ fv
∂t
ρ ∂x
∂v
1 ∂p
=−
− fu
∂t
ρ ∂y
∂w
1 ∂p
=−
∂t
ρ ∂z
Take time derivatives of x and y momentum to eliminate u or v between them:
⎛ 1 ∂p
⎞
1 ∂2 p
1 ∂2 p
∂ 2u
∂v
f
f
fu
=
−
+
=
−
+
−
−
⎜
⎟
∂t 2
∂t
ρ ∂x∂t
ρ ∂x∂t
⎝ ρ ∂y
⎠
⎛ 1 ∂p
⎞
1 ∂2 p
1 ∂2 p
∂ 2v
∂u
f
f
fv
=
−
−
=
−
−
−
+
⎜ ρ ∂x
⎟
∂t 2
∂t
ρ ∂y∂t
ρ ∂y∂t
⎝
⎠
and add ∂/∂x times the first and ∂/∂y of the second:
1 ∂ ⎛ ∂2
∂ 2 ⎛ ∂u ∂v ⎞
∂2 ⎞
∂v ⎞
2 ⎛ ∂u
p
f
+
=
−
+
−
+
⎜
⎟
⎜
⎟.
x
y
ρ ∂t ⎜⎝ ∂x 2 ∂y 2 ⎟⎠
∂t 2 ⎝ ∂x ∂y ⎠
∂
∂
⎝
⎠
From continuity, ∂u/∂x + ∂v/∂y = −∂w/∂z, so
1 ∂ ⎛ ∂2
∂ ⎛ ∂2
∂2 ⎞
2⎞
+ f ⎟w =
⎜
⎜ 2 + 2⎟p
t
ρ
∂z ⎝ ∂t 2
∂
∂y ⎠
⎠
⎝ ∂x
Recall
∂w
1 ∂p
=−
∂t
ρ ∂z
– 130 –
GEFD
Effects of rotation:Inertial waves
so
∂2 ⎛ ∂2
∂2 ⎛ ∂2
∂2 ⎞
2⎞
+ f ⎟w = − 2 ⎜ 2 + 2 ⎟w
⎜
∂z 2 ⎝ ∂t 2
∂t ⎝ ∂x
∂y ⎠
⎠
⇒
2
⎡ ∂2 ⎛ ∂2
⎤
∂2
∂2 ⎞
2 ∂
+
+
+
f
w = 0.
⎢ 2⎜ 2
2
2 ⎟
2⎥
∂y
∂z ⎠
∂z ⎦
⎣ ∂t ⎝ ∂x
^ ei(kx + ly +mz − ωt), then for non-trivial
Assume wave-like modes with w = w
solution
ω 2 ( k 2 + l 2 + m2 ) − f 2 m2 = 0
so the dispersion relation is
f 2 m2
ω = 2 2
= f 2 sin 2 θ .
2
(k + l + m )
2
Note the similarity with internal gravity waves!
4.5.4 Analogy with internal waves
There is an analogy between stratified flows and rotating flows. This is seen
here by comparing the dispersion relationships between internal and inertial
waves. For simplicity, we shall choose our coordinate system so that the y
component of the wavenumber vector (l) vanishes:
ω
N
= cosθ =
k
k
cf
ω
f
= sin θ =
m
k
In both cases, θ is the angle between the wavenumber vector k and the
horizontal. The expressions have the same form, except that the angles differ
by 90°. For example, if in the case of inertial waves we were to measure the
angle α between the wavenumber vector and the vertical, then we would have
ω
N
= cosθ
cf
ω
f
= cos α
Phase and group velocities are similarly related:
– 131 –
GEFD
Effects of rotation:Inertial waves
ω
cp =
inertial
cp =
internal
cg =
⎛ sin θ ⎞
∂ω N
= sin θ ⎜
⎟
∂ki k
⎝ − cosθ ⎠
inertial
cg =
⎛ − sin θ ⎞
f
∂ω
= cosθ ⎜
⎟
∂ki k
⎝ cosθ ⎠
k
k=
2
⎛ cosθ ⎞
N
cosθ ⎜
⎟
k
⎝ sin θ ⎠
internal
ω
k
k=
2
⎛ cosθ ⎞
f
sin θ ⎜
⎟
k
⎝ sin θ ⎠
In both cases cp⋅cg = 0, but whereas for internal waves cpz = −cgz and cpx + cgx =
N/|k|, for inertial waves cpx = − cgx and cpz + cgz = f/|k|
Thus much of the analysis we did for internal waves can be undertaken for
inertial waves by simply rotating things through 90°.
Motion of fluid particle
One of the things that is not simply a rotation through 90° is the motion of a
fluid particle. Whereas in internal waves the motion of the fluid is confined to
the plane in which the wavenumber vector lies, for inertial waves there is
always a motion perpendicular to this plane.
Internal waves: particle motion is one-dimensional in direction normal to k but
in vertical plane containing k. Particles undergo linear simple harmonic motion
in direction parallel to cg.
Inertial waves: particle motion is two-dimensional in plane normal to k.
Particles undergo circular orbits.
– 132 –
GEFD
Effects of rotation:Inertial-gravity waves
z
cg
z
f/|k|
cp
cg
cp
N/|k|
y
y
k
k
x
x
Figure 76: Comparison of particle motion in internal (left) and inertial (right) waves. In
internal waves, the motion is 1D simple-harmonic; a line parallel to k is distorted to a
sinusoid in the direction of cg. In inertial waves, the particles execute orbits in the plane
normal to k; a line parallel to k is distorted into a helix.
4.6 Inertial-gravity waves
This analysis can be extended readily to include both the effects of rotation and
stratification. These waves are referred to as ‘inertial-gravity waves’, or
sometimes to as ‘inertial-buoyancy waves’; the order of ‘inertial’ and
‘buoyancy’/‘gravity’ can also be swapped.
Example sheet: Determine the dispersion relation, phase and group velocities
for linear inertial-gravity waves.
The dispersion relation can be expressed as the vector sum of the internal wave
ω = N cosθ and the inertial wave ω = f sinθ:
ω 2 = N 2 cos 2 θ + f 2 sin 2 θ .
This converges on the relationship for internal waves when f = 0, and on
inertial waves when N = 0.
In 2D the phase and group velocities can be expressed as
– 133 –
GEFD
Effects of rotation:Inertial-gravity waves
N 2 cos 2 θ + f 2 sin 2 θ ) 2 ⎛ cosθ ⎞
(
cp =
⎜ sin θ ⎟ ,
k
⎝
⎠
1
cg =
(N
2
− f 2 ) cosθ sin θ
⎛ sin θ ⎞
⎟.
1 ⎜
2
2
2
2
2 ⎝ − cos θ ⎠
k ( N cos θ + f sin θ )
Note that cp⋅cg = 0, so the phase and group velocities are again orthogonal. If
N > f then the phase and group velocities have the opposite vertical sign. When
N = f, propagation of the waves ceases.
Figure 77: Variation in wave frequency ω for an inertial-gravity wave propagating at 45°.
Figure 78: Variation in wave phase velocity for an inertial-gravity wave propagating at
45°. Left: cpx. Right: cpz.
– 134 –
GEFD
Effects of rotation:Rossby waves
Figure 80: Variation in wave group velocity for an inertial-gravity wave propagating at
45°. Left: cgx. Right: cgz.
4.7 Rossby waves
In this section we shall denote the vertical component of vorticity as ζ to avoid
confusion with the angular frequency of the wave given by ω.
4.7.1 Conservation of absolute vorticity
Imagine a low Rossby number rotating homogeneous fluid at rest confined
between two nearly horizontal boundaries with a small angle α between them.
– 135 –
GEFD
Effects of rotation:Rossby waves
H
η
α
ζ
Deep
Shallow
η
Suppose a column of fluid of height H were displaced towards shallower fluid
by a distance η. The height of this column will be compressed by an amount
αη. Since it is the absolute vorticity that is squeezed (to conserve the absolute
circulation), then the column develops an anticyclonic relative vorticity ζ given
by
ζ+f
f
=
H − αη H
⇒
ζ =−
αη
H
f
The displacement of a column to shallower fluid must be balanced by the
displacement from shallower to deeper fluid, which will generate cyclonic
(positive) relative vorticity through stretching of the fluid column.
4.7.2 Down-slope displacement
Suppose the depth decreases in the y direction and at t = 0 we have a small
displacement in this direction given by η0 = A cos kx, where k is a
wavenumber.
– 136 –
GEFD
Effects of rotation:Rossby waves
Shallower
η0 = A cos kx
Deeper
The fluid displaced in the positive y direction will develop anticyclonic relative
vorticity, while that displaced in the negative y direction (deeper fluid) will
develop cyclonic relative vorticity.
If we assume that the slope α is sufficiently small then we can take H as being
independent of y so that the initial vorticity distribution is
ζ ( x, y , t = 0 ) = −
αf
H
A cos kx .
Noting that ζ = ∂v/∂x since the symmetry ensures that u = 0, then this vorticity
distribution induces a velocity in the y direction given by
v ( x, y , t = 0 ) = −
αf
A cos kx dx
H ∫
.
αf
A sin kx
=−
kH
This velocity will then act to change η0.
More generally, since ∂η/∂t = v and ∂v/∂x = ζ = −αfη/H, then
– 137 –
GEFD
Effects of rotation:Rossby waves
∂ 2η
αf
=−
η.
∂x∂t
H
We shall look for a travelling wave solution of the form η = A ei(kx − ωt), where
ω is the angular frequency of the wave (not the vorticity). Substitution,
kω Aei( kx−ωt ) = −
αf
Aei( kx−ωt ) ,
H
then leads to the dispersion relation
ω=−
αf
kH
.
From this we can determine immediately the phase and group velocities
cp =
cg =
ω
k
=−
αf
k 2H
,
∂ω α f
=
= −c p .
∂k k 2 H
As we are dealing here with linear waves, if the displacement was not parallel
with the down-slope direction then we can simply decompose the motion into
down-slope and cross-slope flows. Suppose (as before) the depth decreases in
the y direction, but now consider a plane-wave perturbation described by the
wavenumber k = (k,l) and
⎛ l
⎜− k
η ( x, y, t = 0 ) = A cos ( k ⋅ x ) ⎜
⎜ k
⎜
⎝ k
⎞
⎟
⎟ = A cos ( k ⋅ x ) ⎛ − sin θ ⎞ ,
⎜ cosθ ⎟
⎟
⎝
⎠
⎟
⎠
where θ is the angle between k and the x axis. As only the y component of η
leads to a change in vorticity (and ζ = ∂V/∂X), then the differential equation
becomes
∂ 2η
αf
=−
η cosθ
∂X ∂t
H
– 138 –
GEFD
Effects of rotation:Rossby waves
where X is the rotated coordinate parallel to k and η is the displacement in the
rotated Y coordinate direction.
y
Y
X
k
x
Substituting for a travelling wave solution then leads to the dispersion relation
ω=−
αf
kH
cosθ = −
αf k
H k2
,
from which the phase and group velocities may be calculated.
4.7.3 Changes in latitude
The Rossby waves we have been looking at up until now are topographic
Rossby waves. However, Rossby waves do not require a change in depth. The
key component is fluid columns develop a relative vorticity when they are
displaced from their equilibrium position.
Recall that we are generally interested in the vertical component of the earth’s
rotation, hence we take
f = f0 sin ϕ
where f0 is the Coriolis parameter at the pole (i.e. f0 = 2×2π/(24×60×60) ≈
1.45×10−4 rad/s). Close to some latitude ϕ0 we set ϕ = ϕ0 + ϕ′ and perform a
Taylor Series expansion to show
– 139 –
GEFD
Effects of rotation:Rossby waves
1
⎛
f = f 0 ⎜ sin ϕ0 + ϕ ′ cos ϕ0 − ϕ ′2 sin ϕ0 +
2
⎝
⎞
⎟.
⎠
Provided ϕ′ remains small and we are not close to the pole (in which case the
linear term vanishes), then we can approximate this as
f = f1 + β y,
where f1 = f0 sinϕ0 and β = ∂f/∂y.
This approximation is referred to as a β-plane.
Rossby waves on a β plane
Recall that the absolute circulation
Γ=
∫ (ζ + f ) zˆ ⋅ dx
is conserved for a column of fluid. This means that in a layer of uniform depth
the absolute vorticity is conserved by a fluid column if it is displaced in
lattitude. However, the relative vorticity will not be conserved since f is a
function of latitude on the β-plane.
A fluid column from low latitude displaced to higher latitude will have a lower
absolute vorticity than its new surroundings, and so it will appear to have an
anticyclonic vorticity. Similarly, a fluid column displaced towards lower
latitudes will have a higher absolute vorticity than its surrounds and will thus
have cyclonic relative vorticity.
– 140 –
GEFD
Effects of rotation:Rossby waves
Higher latitude
(Shallower)
η0 = A cos kx
)
Lower latitude
(Deeper)
There is an obvious parallel with the varying depth case we looked at before. In
ζ =−
αf
H
η
for a parcel displaced a distance η in the y direction, we now have
f1 = ζ + ( f1 + βη )
⇒
ζ = − βη
for a parcel displaced from latitude ϕ0. Equating the topographic slope with β
gives
β=
αf
H
,
leading directly to the dispersion relation
ω=−
β
k
.
The more general case of a disturbance with arbitrary orientation is simply
– 141 –
GEFD
Effects of rotation:Rossby waves
ω=−
β
k
cosθ = − β
k
k
2
,
4.7.4 Atmospheric Rossby waves
Rossby waves occur at planetary scales and play an important role in the
dynamics of our atmosphere.
www.daukas.com
The related phenomenon of ‘topographic steering’ has a major impact on the
weather in North America. Here the atmosphere is squeezed as it passes over
the Rockies, generating anticyclonic vorticity that steers the jet stream
southward, bringing cold air with it.
– 142 –
GEFD
Effects of rotation:Rossby waves
www.atmos.umd.edu
– 143 –
GEFD
Shallow water:Introduction
5 Shallow water
5.1 Introduction
What is shallow water?
Not very deep?
Flows in layers with a free surface or interfaces
Density differences
Motion predominantly horizontal
Waves play a central rôle
“Compressible”
∂u ∂v ∂w
+ +
∂x ∂y ∂z
∂u ∂v
+
∂x ∂y
Examples of shallow water
• Some aspects of
• Atmosphere
• Oceans
• Lakes
• Rivers
• Avalanches
• Volcanic eruptions
• Industrial accidents
• Chemical warfare
• Flow in buildings
– 144 –
GEFD
Shallow water:Introduction
Atmospheric
www…
www.dropbears.com
Figure 83: The Morning Glory, North Australia
(a)
(b)
www.photolib.noa.gov
www…
Figure 84: (a) Cold front. (b) Fog bank.
– 145 –
GEFD
Shallow water:Introduction
(a)
(b)
Figure 85: (a) Thunderstorm anvil. (b) Dissipated thunderstorm.
(a)
(b)
Figure 86: Dust storms in (a) Arizona and (b) Melbourne.
– 146 –
GEFD
Shallow water:Introduction
www.corbis.com
(a)
www.nasa.com
(b)
(c)
Figure 87: Dust storms seen from space: (a) Sahara; (b) and (c) Mars.
– 147 –
GEFD
Shallow water:Introduction
J.E. Simpson, Gravity Currents, CUP (1997)
pigpen.itd.nps.gov
Figure 88 Avalanches.
quake.wr.usgs.gov
volcanoes.usgs.gov
Figure 89: Pyroclastic flows from (a) Mt. St. Helens and (b) Mt. Augustine, Alaska.
– 148 –
GEFD
Shallow water:Introduction
davidhobbyphoto.com
web.ukonline.co.uk
Figure 90: Flow over river weirs.
www…
www…
Figure 91: Bores and hydraulic jumps. (a) Surfing on the River Severn. (b) Qiantang
River, China.
– 149 –
GEFD
Shallow water:Introduction
J.E. Simpson, Gravity Currents, CUP (1997)
Figure 92: Dense gas releases. (a) Thorny Island trials (Freon). (b) Maplan Sands (LNG).
(a)
(b)
Figure 93: Gravity currents in the laboratory. (a) Shallow current in a deep ambient fluid.
(b) Axisymmetric gravity current on a 5° slope.
(a)
(b)
S. Dalziel, DAMTP
Figure 94: Two-layer rotating exchange flow. (a) Elevation showing intersection with near
(light line) and far (dark line) walls. (b) Cross-section.
– 150 –
GEFD
Shallow water:Introduction
How much do we know?
• Less than you might think
• Excellent understanding of steady flows on a smooth, flat surface or through
rectangular channels
• Good descriptions of transients such as gravity currents in simple (ideal?)
conditions
• Reasonable understanding of drag, two-layer effects, rotation, particle-driven
flows
• Poor understanding of transients with topography, entrainment
• Almost no understanding of collisions, complex ambient stratifications,
complex substrates, resuspension, fully three-dimensional flows, interaction
with obstacles
Approaches to modelling
• Integral (box) models
• For transient problems only
• All dynamics in boundary conditions
• Shallow water equations
• Long wave approximation
• Provides full dynamic treatment of depth-averaged equations
• Analytical solution sometimes possible
• Efficient numerics
• Full Navier Stokes
• Requires numerical solution
• Does not assume much, but difficult and expensive to get adequate resolution
and correct behaviour
• Two-dimensional modelling accessible with workstations/PCs
• Three-dimensional modelling accessible by supercomputers, but only for
relatively low Reynolds numbers
– 151 –
GEFD
Shallow water:Shallow water equations
5.2 Shallow water equations
u
h
L
Figure 95: Sketch of shallow water layer.
Simplifications
Similar to the narrow plume assumption that will be used later.
5.2.1 Mathematical definition of “shallow”
Shallow ⇒ vertical scales much less than horizontal scales
h << L
(81)
∇ ⋅u = 0
(82)
∂u
1
+ (u ⋅ ∇ )u = − ∇( p + ρgz ) + ν∇ 2 u
∂t
ρ
(83)
Equations for a homogeneous layer:
Dimensional analysis (2D)
The same analysis applies in 3D
Let the length scale for variations in the flow be X and Z, and the time scale T.
Let the horizontal and vertical velocity scales be U and W, and the pressure
scale P.
Then
continuity:
U W
+ =0
X Z
– 152 –
(84)
GEFD
Shallow water:Shallow water equations
x momentum:
U
U
U 1P
U ⎞
⎛U
+U +W =
+ ν⎜ 2 + 2 ⎟
T
X
Z ρX
Z ⎠
⎝X
(85)
z momentum
W
W
W 1P
⎛W W ⎞
+U +W
=
+ g + ν⎜ 2 + 2 ⎟
T
X
Z ρZ
Z ⎠
⎝X
(86)
Suppose the scale for horizontal variations is much larger than that for vertical
variations.
Assumption:
X/Z → ∞.
(87)
W/U → 0,
(88)
Continuity gives
Key Result
showing that the flow is predominantly horizontal.
Vertical momentum
W
W
W 1P
⎛W W ⎞
+U +W
=
+ g + ν⎜ 2 + 2 ⎟
T
X
Z ρZ
Z ⎠
⎝X
(89)
gives the pressure field as hydrostatic over the whole layer of depth h:
Key Result
∂p/∂z = −ρg.
(90)
and that vertical accelerations are small
∂w/∂t << g.
(91)
We will relax these conditions later
The horizontal momentum (P ~ ρgZ)
U
U
U ρ gZ
U ⎞
⎛U
+U +W =
+ν ⎜ 2 + 2 ⎟
T
X
Z
Z ⎠
ρX
⎝X
• Viscous effects give horizontal velocity as a function of z.
• For large Reynolds number, the viscous term is negligible.
– 153 –
(92)
GEFD
Shallow water:Shallow water equations
• As the pressure term driving the flow is independent of z, then u will tend
towards z independence and we may neglect the w ∂u/∂z term compared with
u ∂u/∂x.
∂u
∂u
1 ∂
∂ 2u
( p + ρgz ) + ν 2
=−
+u
∂t
∂x
∂z
ρ ∂x
p0 + ρgh
5.2.2 Why use this approximation?
• Mathematically simple
• Two-dimensional (planar)
• Three-dimensional (axisymmetric)
• Channel approximation (a generalisation of axisymmetric flow where flow in
a channel of width b(x) and all flow is parallel with this)
• Three-dimensional (general)
• A much more difficult problem
• Computationally cheap
• Difficult to resolve very strong density gradients with CFD
But is it any good?
• Much better than you might think
• Reasonable approximation even when not slowly varying
• Reasonable approximation even when mixing
5.2.3 Derivation from first principles
Assumptions:
• Slowly varying shallow layer
• Free surface (density “zero” above layer)
• Pressure uniform at surface. Take as p = 0
– 154 –
GEFD
Shallow water:Shallow water equations
• Flat bottom
• Inviscid (high Reynolds number)
• From (83) and (92):
1 ∂
∂u
∂u
∂u
∂ 2u
( p + ρgz ) + ν 2
+u +w = −
∂x
∂z
ρ ∂x
∂t
∂z
→ Velocity uniform with depth (p + ρgz independent of z)
h − ∂h/∂x δx
h + ∂h/∂x δx
u − ∂u/∂x δx
h
uh
u + ∂u/∂x δx
2δx
Figure 96: Definition sketch for derivation of shallow water equations.
Derivation:
Continuity
Flux_in = (u − ∂u/∂x δx)(h − ∂h/∂xδx) + O(δx2)
Flux_out = (u + ∂u/∂x δx)(h + ∂h/∂xδx) + O(δx2)
Rate_of_accumulation = ∂h/∂t 2δx + O(δx2) + O(δx2)
Equate Rate_of_accumulation = Flux_in − Flux_out
∂h/∂t 2δx = (u − ∂u/∂x δx)(h − ∂h/∂xδx) − (u + ∂u/∂x δx)(h + ∂h/∂xδx) + O(δx2)
= −2u ∂h/∂x δx – 2h ∂u/∂x δx + O(δx2)
Let δx → 0 (thus ignore terms O(δx2))
∂h
∂h
∂u ∂h ∂
+u
+h
=
+ (uh ) = 0
∂t
∂x
∂x ∂t ∂x
– 155 –
(93)
GEFD
Shallow water:Shallow water equations
Key Result: The flow is “compressible”, even if the fluid is not.
Momentum
Flux_in = ρ (u − ∂u/∂x δx) [(u − ∂u/∂x δx)(h − ∂h/∂xδx)] + O(δx2)
Flux_out = ρ (u + ∂u/∂x δx) [(u + ∂u/∂x δx)(h + ∂h/∂xδx)] + O(δx2)
h − ∂h / ∂x δx
∫
Pressure_force_to_right =
0
∂h
⎛
⎞
ρg ⎜ h − δx − z ⎟dz
∂x
⎝
⎠
2
( )
1 ⎛
∂h ⎞
= ρg ⎜ h − δx ⎟ + O δx 2
2 ⎝
∂x ⎠
h + ∂h / ∂x δx
Pressure_force_to_left =
∫
0
∂h
⎛
⎞
ρg ⎜ h + δx − z ⎟dz
∂x
⎝
⎠
2
( )
1 ⎛
∂h ⎞
= ρg ⎜ h + δx ⎟ + O δx 2
2 ⎝
∂x ⎠
Rate_of_accumulation = ρ ∂uh/∂t 2δx
Equate
Rate_of_accumulation = (Flux_in − Flux_out)
+ (Pressure_force_to_right − Pressure_force_to_left)
and let δx → 0
∂uh ∂ ⎛ 2
1 2⎞
+
⎜ u h + gh ⎟ = 0
∂t
2
∂x ⎝
⎠
(94)
• Compare with primitive momentum equation (inviscid terms):
(
)
∂ρu ∂
+
ρu 2 + p = 0 .
∂t
∂x
…depth plays a similar role to density.
• Not in the “normal” form
• Simplify by applying continuity equation
– 156 –
(95)
GEFD
Shallow water:Shallow water equations
∂u
∂u
∂h
+u
= −g
∂t
∂x
∂x
(96)
Exercise: Convert between ∂(uh)/∂t form and ∂u/∂t form.
• Axisymmetric - watch out for momentum equation
Examples sheet question: Do derivation for axisymmetric flow
• General geometry (small lateral acceleration)
• Use width of flow b(x).
• Width must vary slowly so that lateral accelerations ~ u2/b ∂2b/∂x2 are much
less than the gravitational acceleration g
5.2.4 Derivation by averaging
This will tell us about some things we have ignored
Split velocity into mean and fluctuations:
u = u + u′,
(97)
where
u (x ) = (u , v , w ) =
1
u dA ,
A ∫A
(98)
and A is the cross-sectional area of the channel.
• For this discussion we shall assume A = b×h where b = b(x) is the width of
the channel filled to a depth h = h(x,t).
This assumption is not necessary
The fluctuations are
u′ = u′(x,y,z) = (u′,v′,w′).
Assumption: Fluid is incompressible
• Fluid incompressible
– 157 –
(99)
GEFD
Shallow water:Shallow water equations
Integrate continuity equation across cross-section
b /2 h
∂u ∂v ∂w
dz dy = 0 .
+ +
∫
∫
x
y
z
∂
∂
∂
− b /2 0
(100)
The velocity normal to the walls is zero, so the kinematic boundary condition
gives v(y=±b/2) = ± ½ u(y=±b/2) ∂b/∂z. Thus
b /2 h
h b /2
∂v
∂v
dz
dy
=
dy dz = ∫ v y = 1 b − v y =− 1 b dz
∫
∫
∫
∫
2
2
∂
y
∂
y
0 − b /2
0
− b /2 0
h
1 ∂b
=
u 1 + u y =− 1 b dz
2
2 ∂x ∫0 y = 2 b
h
.
The kinematic boundary condition at the surface gives
w(z=h) = ∂h/∂t + u(z=h) ∂h/∂x, and so
b /2 h
b /2
b /2
∂w
∂h ∂h
dz
dy
=
w
−
w
dy
=
b
+
u z =h dy .
∫
∫
∫
∫
z =h
z =0
∂
z
∂
t
∂
x
− b /2 0
− b /2
− b /2
Now consider
b /2 h
⎤
∂u ∂ ⎡ 1
u
dz
dy
= ⎢
⎥
∂x ∂x ⎣ bh − b∫/2 ∫0
⎦
(
)
∂u
∂b
1 ⎡
= ⎢ ∫ ∫ dz dy + ∫ 12 u y = 1 b + u y =− 1 b dz
2
2
∂x 0
bh ⎣ − b /2 0 ∂x
b /2 h
h
b /2
b /2 h
⎤
∂h
⎛ 1 ∂b 1 ∂h ⎞
+
−
+
u
dz
u
dz
dy
⎥
⎜
⎟ ∫ ∫
∂x − b∫/2 z = h
⎝ b ∂x h ∂x ⎠ − b /2 0
⎦
which can be rearranged to give
– 158 –
GEFD
Shallow water:Shallow water equations
b /2 h
∂u
∂u ⎛ ∂b
∂h ⎞
dz
dy
=
bh
+
h
+
b
⎜
⎟u
∫
∫
∂
x
∂
x
∂
x
∂
x
⎝
⎠
− b /2 0
)
(
b /2
∂b
∂h
− ∫ 12 u y = 1 b + u y =− 1 b dz −
u z = h dz
∫
2
2
∂x 0
∂x − b /2
h
+
These give continuity as
∂u ∂v ∂w
∂u
∂b ⎞ ∂h
⎛ ∂h
+
+
dz
dy
=
bh
+
u
b
+
h
= 0.
⎜
⎟+b
∫
∫
∂
x
∂
y
∂
z
∂
x
∂
x
∂
x
∂
t
⎝
⎠
− b /2 0
b /2 h
x Momentum equation
b /2 h
∂u
∂u
∂u
∂u
+
u
+
v
+
w
dz dy
∫
∫
∂
t
∂
x
∂
y
∂
z
− b /2 0
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
1 ∂
= ∫ ∫−
( p + ρ gz ) + ν ⎜ 2 + 2 + 2 ⎟ dz dy
ρ
∂
x
∂y
∂z ⎠
⎝ ∂x
− b /2 0
b /2 h
Substituting for u etc. and expanding gives
9g or g′
∂u
∂u
∂h
∂ 2u
+u
+ g −ν 2
∂t
∂x
∂x
∂x
⎡ 1 ⎛ ∂u ′
=ν ⎢ ⎜
⎜
⎢⎣ b ⎝ ∂y y =
1
2
∂u′
−
∂y y =−
b
1
2
⎞ 1 ⎛ ∂u′
∂u′ ⎞⎤
⎟+ ⎜
⎟
−
⎟ h ⎜⎝ ∂z z =h ∂z z =0 ⎟⎠⎥
⎥⎦
b⎠
⎡ ∂u ′
∂u′
∂u′ ⎤
− ⎢u ′
+ v′
+ w′ ⎥
∂y
∂z ⎦
⎣ ∂x
• Have assumed the centre-line of the channel is straight (thus v = 0).
• Have assumed width and depth are slowly varying (ignore w and v )
The terms on the left involve mean quantities.
• Viscous term on left negligible as slowly varying (except near ∂ u /∂x = 0).
– 159 –
GEFD
Shallow water:Shallow water equations
Right-hand side represent momentum losses.
⎡ 1 ⎛ ∂u ′
ν⎢ ⎜
⎢⎣ b ⎜⎝ ∂y y =
1
2b
∂u ′
−
∂y y =−
1
2
⎞ 1 ⎛ ∂u ′
∂u ′ ⎞⎤
⎟+ ⎜
⎟
−
⎟ h ⎜⎝ ∂z z =h ∂z z =0 ⎟⎠⎥⎥
b⎠
⎦
• Viscous stresses
⎡ ∂u ′
∂u ′
∂u ′ ⎤
− ⎢u ′
+ v′
+ w′ ⎥
∂y
∂z ⎦
⎣ ∂x
• Reynolds stresses
Parameterise
Assumption: u′ scales on u and length scales scale on depth h and width b.
Use dimensional analysis
⎡ 1 ⎛ ∂u ′
ν⎢ ⎜
⎢⎣ b ⎜⎝ ∂y y =
1
2b
∂u ′
−
∂y y = −
1
2
⎞ 1 ⎛ ∂u ′
u ⎛ h2 ⎞
∂u ′ ⎞⎤
⎟+ ⎜
⎜ ∂z z = h − ∂z z =0 ⎟⎟⎥ ≈ C L ν h 2 ⎜⎜1 + b 2 ⎟⎟
⎟
h
⎝
⎠
⎝
⎠⎥⎦
b⎠
• CL is laminar drag coefficient.
• For Poiseulle flow, CL = 2/3.
u u ⎛ h⎞
⎡ ∂u ′
∂u ′
∂u ′ ⎤
− ⎢u ′
+ v′
+ w′ ⎥ ≈ CT
⎜1 + ⎟
∂
x
∂
y
∂
z
h
⎝ b⎠
⎣
⎦
• CT is turbulent drag coefficient.
• Depends on bottom roughness.
• Typically CT ≈ 0.03 for smooth bottom.
Often h << b, so only need consider depth.
The equations
Assumption: h << b
– 160 –
GEFD
Shallow water:Shallow water equations
Drop over-bar
∂h
∂
+ b −1 (buh ) = 0 ,
∂t
∂x
Continuity:
Momentum:
uu
u
∂u
∂h
∂u
= −C L ν 2 − C T
.
+g
+u
∂t
∂x
h
∂x
h
9g or g′
• Momentum equation does not depend on b
Bottom drag
• Thinning and accelerating
• Slowing & thickening
• Possibly triggering hydraulic jump
See Examples sheet question.
5.2.5 Boussinesq verses non-Boussinesq flows
What happens if the density above the surface is not zero?
Boussinesq approximation:
Assumption: Density differences are small and enter the momentum
equation only through the buoyancy terms.
Continuity equation as before.
Modification to momentum equation
Assumption: The pressure in the ambient fluid is independent of x.
Assumption: The upper layer is deep and we need not consider the flow
within it.
We will relax this restriction later.
Match pressure across interface
– 161 –
GEFD
Shallow water:Shallow water equations
ρ2
Δh
ρ1
Pressure difference at bottom due to height difference Δh:
ρ1gΔh if ρ2 << ρ1
(ρ1 − ρ2) gΔh for any ρ2 ≤ ρ1.
If Boussinesq, momentum equation has (ρ1 − ρ2)/ ρ g ∂h/∂x in stead of g ∂h/∂x
Reduced gravity
g′ =
Key Result
ρ1 − ρ 2
g
1
(
)
ρ
+
ρ
1
2
2
(101)
plays the role of gravity.
Momentum equation (ignoring drag terms)
∂u
∂u
∂h
+u
=0
+ g′
∂t
∂x
∂x
(102)
Key Result Can do everything upside down!
ρ1
Δh
ρ2
Where is the pressure greater?
• Watch out!
– 162 –
GEFD
Shallow water:Hyperbolic system
Finite depth upper layer
• May need to consider flow within upper layer as well
• Wave speed is function of total depth as well as layer depth
Many of the problems, solutions and techniques discussed from this
point on apply to both the large and small density difference limits
with the only difference being g′ replacing g. Whether or not this can
be done will be indicated by
9g or g′
8 g only
8 g′ only
Many of the results will apply equally to miscible and immiscible
fluids. Any which apply only to miscible fluids will be flagged by
8 miscible only
Examples sheet question: Determine the characteristics in a two-layer system
5.2.6 More than one layer
- later
5.2.7 Entrainment
- later
5.3 Hyperbolic system
The mathematical bit
Whitham, G.B. (1974): Linear and nonlinear waves. John Wiley & Sons, New
York. 636pp.
5.3.1 What does “hyperbolic” mean?
Supports waves
– 163 –
GEFD
Shallow water:Hyperbolic system
Waves carry information at a finite speed
5.3.2 A model for traffic flow
We will use a very similar model later when considering sedimentation.
Density of cars: ρ
Car speed: u = u(ρ)
ρ − ∂ ρ/∂x δx
ρ + ∂ ρ/∂x δx
u − ∂u/∂x δx
ρ
u + ∂u/∂x δx
2δx
Figure 97: Definition sketch for traffic flow problem.
Continuity
∂ρ ∂
+ (uρ ) = 0
∂t ∂x
An example
Let u = umax(1−ρ/ρmax) and take umax = 1 and ρmax = 1.
– 164 –
(103)
GEFD
Shallow water:Hyperbolic system
Traffic flux uρ
Ca
rs
pe
ed
u
Traffic density ρ
Figure 98: Traffic speed and flux.
Require solution in space and time
t
x = λt
x
Figure 99: Trajectory through x−t space.
Consider rays defined by x = x0 + λ(t−t0). As a vector, these rays may be
written parametrically as
⎛ t ⎞ ⎛ t0 ⎞ ⎛ 1 ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ s .
⎝ x ⎠ ⎝ x0 ⎠ ⎝ λ ⎠
The derivative along these rays is therefore
– 165 –
(104)
GEFD
Shallow water:Hyperbolic system
⎛ ∂ ⎞
⎜ ⎟
T ∂t
d ⎛1⎞ ⎜ ⎟ ∂
∂
= ⎜⎜ ⎟⎟ ⎜ ⎟ = + λ .
ds ⎝ λ ⎠ ⎜ ⎟ ∂t
∂x
∂
⎜ ⎟
⎝ ∂x ⎠
(105)
∂ρ
dρ ∂ρ
=
+λ
ds ∂t
∂x
(106)
so the density changes as
Rewriting the traffic equation as (and taking umax = 1)
∂ρ ∂
∂ρ
∂ρ
+ (ρ(1 − ρ)) =
+ (1 − 2ρ) = 0 ,
∂t ∂x
∂t
∂x
(107)
and comparing with (106) shows that dρ/ds vanishes for
λ = (1 − 2ρ).
Key Result
(108)
This trajectory is called a characteristic.
t
x
Figure 100: A traffic flow problem with decreasing traffic density.
• Characteristics travel more slowly than traffic (u = (1−ρ/ρmax))
– 166 –
GEFD
Shallow water:Hyperbolic system
t
x
Figure 101: A traffic flow problem with increasing traffic density.
Shock formation
Still have conservation of traffic across shock.
u shock
u right
u left
ρleft
ρright
Suppose shock moving at speed c. In frame of reference of shock, flux of
traffic approaching from left ρleft(uleft−ushock) is equal to flux receding to right
ρright(uright−ushock). Thus
ρleft(1−ρleft−ushock) = ρright(1−ρright−ushock).
Rearranging
– 167 –
(109)
GEFD
Shallow water:Hyperbolic system
u shock =
ρ right (1 − ρ right ) − ρ left (1 − ρ left )
ρ right − ρ left
2
2
ρ right − ρ left ρ right
− ρ left
=
−
ρ right − ρ left ρ right − ρ left
(
= 1 − ρ right + ρ left
=
[(
,
(110)
)
) (
1
1 − 2 ρ left + 1 − 2 ρ right
2
)]
so
ushock = ½ (λleft + λright).
(111)
5.3.3 Shallow water as a hyperbolic system
What are the characteristics?
Shallow water equations:
∂h
∂u
∂h
+u
+h
=0
∂x
∂x
∂t
(112)
∂u
∂u
∂h
+u
+g
=0
∂t
∂x
∂x
(113)
Want to look for trajectories through x,t space along which something does not
change.
As before, write
⎛ t ⎞ ⎛ t0 ⎞ ⎛ 1 ⎞
⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ s ,
⎝ x ⎠ ⎝ x0 ⎠ ⎝ λ ⎠
and hence
d
∂
∂
= +λ .
ds ∂t
∂x
Eliminate ∂/∂t from (112) and (113) using (114):
– 168 –
(114)
GEFD
Shallow water:Hyperbolic system
dh
∂h
∂h
∂u
−λ
+u
+h
= 0,
ds
∂x
∂x
∂x
(115)
du
∂u
∂u
∂h
−λ
+u
+g
= 0,
ds
∂x
∂x
∂x
(116)
rearranging:
dh
∂h
∂u
+ (u − λ )
+h
= 0,
ds
∂x
∂x
du
∂u
∂h
+ (u − λ )
+g
= 0.
ds
∂x
∂x
Eliminate ∂u/∂x by taking (u−λ) times the continuity equation and subtracting
from it h times the momentum equation:
(u − λ )
dh
du
∂h
2
− h + ⎡( u − λ ) − gh ⎤ = 0
⎦ ∂x
ds
ds ⎣
(117)
Suppose we choose λ such to eliminate ∂h/∂x, requiring the term in square
brackets to vanish. This yields
Key Result
λ = u ± (gh)½.
(118)
• Two solutions (c.f. one for traffic flow).
• These are again called characteristics.
The wave speed relative to the fluid is
c = (gh)½,
(119)
so the characteristics may be written as
λ = u ± c.
(120)
Recall from §..2.1.1 that for linear surface waves
cp =
g
tanh kH ,
k
– 169 –
(121)
GEFD
Shallow water:Hyperbolic system
with the long-wave limit
cp = cg = (gh)1/2.
(122)
Thus, the waves supported by the shallow water system are long waves.
But what is conserved along these lines?
Having eliminated ∂h/∂x from (117), we are left with
(u − λ )
dh
du
−h
= 0.
ds
ds
Substituting for λ and noting that h = c2/g gives
d ⎛ c 2 ⎞ c 2 du
c 2 dc c 2 du
c2 d
∓c ⎜ ⎟ −
= ∓2
−
=−
( u ± 2c ) = 0
ds ⎝ g ⎠ g ds
g ds g ds
g ds
Hence, for non-trivial solution (h ≠ 0),
Key Result
u + 2c is conserved along λ = u + c
Key Result
u − 2c is conserved along λ = u − c.
(123)
5.3.4 Matrix formulation
Stepping stone to a more general solution
Rewrite equations using matrices:
⎡ u h⎤⎛ h ⎞ ⎡1 0⎤⎛ h ⎞ ⎛ 0 ⎞
⎢ g u ⎥⎜⎜ u ⎟⎟ + ⎢0 1⎥⎜⎜ u ⎟⎟ = ⎜⎜ 0 ⎟⎟
⎣
⎦⎝ ⎠ x ⎣
⎦⎝ ⎠ t ⎝ ⎠
(124)
By setting ∂/∂t = d/ds − λ∂/∂x
⎛ h ⎞ ⎡ u h⎤ ⎛ h ⎞
⎡1 0⎤ ⎛ h ⎞ ⎛ 0 ⎞
+
−
λ
⎜u ⎟ ⎢g u⎥⎜ u ⎟
⎢0 1⎥ ⎜ u ⎟ = ⎜ 0 ⎟ ,
⎝ ⎠s ⎣
⎦ ⎝ ⎠x
⎣
⎦ ⎝ ⎠x ⎝ ⎠
(125)
Now suppose we take some linear combination of the equations with a view to
eliminating the partial derivatives ∂/∂x. We can express this as
– 170 –
GEFD
Shallow water:Hyperbolic system
⎛h⎞
qT ⎜ ⎟ + qT
⎝ u ⎠s
⎡⎡ u h⎤
⎡1 0⎤ ⎤ ⎛ h ⎞
−
λ
⎢⎢
⎥⎜ ⎟ = 0
⎥
⎢
⎥
⎣0 1⎦ ⎦ ⎝ u ⎠ x
⎣⎣ g u ⎦
We can eliminate both ∂h/∂x and ∂u/∂x either by taking u and h as constant
(trivial), or by requiring
⎡⎡ u h⎤
⎡1 0 ⎤ ⎤
−
qT ⎢ ⎢
λ
⎥
⎢0 1 ⎥ ⎥ = ( 0 0 )
g
u
⎦
⎣
⎦⎦
⎣⎣
This is a generalised eigenvalue problem. Here it is equivalent to the normal
eigenvalue problem
T
⎡ u h⎤
⎡u g ⎤
=
q
⎢g u⎥
⎢ h u ⎥ q = λq .
⎣
⎦
⎣
⎦
For a non-trivial solution we therefore require
u−λ
h
2
= (u − λ ) − gh = 0 ,
g
u−λ
(126)
and λ = u ± c (where c2 = gh), exactly as before.
• In order to understand how to extract the equations along the characteristics
we shall consider a more general approach to hyperbolic systems.
5.3.5 General approach to hyperbolic systems
How to cope with inhomogeneous equations and other
problems
Our matrix formulation in the previous section is of the form
Avx + Bvt = f,
(127)
where vT = (h u).
Let q be some vector.
We may write a linear combination of the equations as
qTAvx + qTBvt = qTf.
– 171 –
(128)
GEFD
Shallow water:Hyperbolic system
This is a generalisation of the procedure where we earlier took (u−λ) times the
continuity equation and subtracting from it h times the momentum equation.
Let us choose q so that all derivatives are in the same direction, s (say), such
that they may be written in the form
mTvs = mT(λvx + vt),
(c.f. the eariler use of
(129)
d
∂
∂
=λ
+ ).
ds
∂x ∂t
Comparing with qTAvx + qTBvt = qTf suggests that if
λ mT = qTA, mT = qTB,
(130)
mTvs = qTf.
(131)
then
Eliminating m from (130) gives qTA = λqTB, or
qT[A − λB] = 0T.
For non-trivial q we require |A − λB| = 0, giving the slope of the
characteristics, λ, as before.
This is then equivalent to us looking for solutions where ∂/∂t = −λ∂/∂x.
The vectors q, which before were chosen to eliminate ∂u/∂x terms from the
equation, are the left-hand eigenvectors of the system.
The system is hyperbolic if all eigenvalues λ are real and distinct.
Equations along characteristics
Recalling mTvs = qTf and mT = (1/λ)qTA = qTB, then
qT[Avs − λf] = 0 ,
or
qT[Bvs − f] = 0.
– 172 –
GEFD
Shallow water:Hyperbolic system
Shallow water
For the shallow water problem, this means we must determine the left-hand
eigenvectors of [A−λB], i.e. qT:
(1
⎡∓ c h ⎤
q )⎢
⎥ = (0 0 ) ,
g
c
∓
⎣
⎦
so
qT = (1 ±c/g).
(132)
Hence, the equation along the characteristics is
⎛
⎛ 0 ⎞⎫⎪
c ⎞⎧⎪⎡ u h⎤⎛ h ⎞
⎜⎜1 ± ⎟⎟⎨⎢
⎥⎜⎜ u ⎟⎟ − λ ⎜⎜ 0 ⎟⎟⎬ = 0 .
g
u
g
⎪
⎝ ⎠⎪⎭
⎦⎝ ⎠ s
⎠⎩⎣
⎝
Expanding
⎛ u dc 2 c 2 du ⎞
⎛ 2uc dc c 2 du ⎞
⎜
⎟
⎜
⎟
+
+
⎛
⎛
⎞
c
c ⎞⎜ g ds
g
ds
⎟ = ⎜1 ± ⎟⎜ g ds g ds ⎟
⎜⎜1 ± ⎟⎟
2
du ⎟
dc
g ⎟⎠⎜
g ⎠⎜ dc
du ⎟ ⎜⎝
⎝
2
+
u
c
+
u
⎜
⎟
⎜
⎟
ds ⎠
ds
⎝
ds ⎠
⎝ ds
uc dc c 2 du
c 2 dc uc du
=2
+
+ ±2
±
g ds g ds
g ds g ds
=0
so
d
⎧
(
)
(2c + u ) on λ = u + c
+
u
c
⎪
ds
.
0=⎨
d
⎪(u − c ) (2c − u ) on λ = u − c
ds
⎩
⎡ u h ⎤⎛ h ⎞ ⎡1 0⎤⎛ h ⎞ ⎛ 0 ⎞
⎢ g u ⎥⎜⎜ u ⎟⎟ + ⎢0 1⎥⎜⎜ u ⎟⎟ = ⎜⎜ 0 ⎟⎟
⎣
⎦⎝ ⎠ x ⎣
⎦⎝ ⎠ t ⎝ ⎠
– 173 –
(133)
GEFD
Shallow water:Hyperbolic system
• Note: it would have been easier algebraically if we had used the alternate
qT[Bvs − f] form of the equation since B is an identity matrix.
Examples sheet question: Determine the evolution equation along
characteristics for an axisymmetric flow.
Higher order problems
Consider the wave equation
∂ 2η
∂ 2η
−γ 2 =0
∂t 2
∂x
(134)
Rewrite as system of first order equations.
Let v = ∂η/∂x,
Let w = ∂η/∂t
then
∂v ∂w
−
=0
∂t ∂x
∂w
∂v
−γ =0
∂t
∂x
Put into matrix form
⎡ 0 − 1⎤⎛ v ⎞ ⎡1 0⎤⎛ v ⎞ ⎛ 0 ⎞
⎢− γ 0 ⎥⎜⎜ w ⎟⎟ + ⎢0 1⎥⎜⎜ w ⎟⎟ = ⎜⎜ 0 ⎟⎟
⎦⎝ ⎠ t ⎝ ⎠
⎦⎝ ⎠ x ⎣
⎣
gives rise to eigenvalue problem
− λ −1
= λ2 − γ = 0
−γ −λ
Thus system hyperbolic with characteristics λ = ±γ1/2.
5.3.6 Implications of hyperbolic nature
Analogy with compressible gas
– 174 –
(135)
GEFD
Shallow water:Hyperbolic system
• Mach number = velocity/sound speed
Froude number
Fr = velocity/wave speed = u/c = u/(gh)1/2
Definition
• Importance of frame of reference
Information propagation
• If |Fr| < 1 then eigenvalues λ = u+c and λ = u−c are of opposite sign and
waves can propagate in both directions. This state is called subcritical flow
• If |Fr| > 1 then eigenvalues are of the same sign and waves can only
propagate in one direction (the same direction as u). This state is called
supercritical flow
• If |Fr| = 1 then either λ = u+c or λ = u−c vanishes and one of the two waves
does not propagate. This state is called critical flow.
Using characteristics for determining solution
λ=u−c
t
λ=u+c
u1,c1
ua,ca
ub,cb
x
If at time t we know
• the λ+ = u + c characteristic emanated from xa at t = 0,
• the λ− = u − c characteristic emanated from xb at t = 0,
and we know ua,ha and ub,hb, then we may calculate u1,c1 directly:
Along the characteristics
– 175 –
GEFD
Shallow water:Hyperbolic system
λ+
u1 + 2c1 = ua + 2ca
λ−
u1 − 2c1 = ub – 2cb,
so
u1 = ½(ua + ub) + (ca − cb),
c1 = ¼(ua − ub) + ½(ca + cb).
λ=u−c
t
λ=u+c
u1,c1
ua,ca
ub,cb
x
Definition: Zone of dependence.
• The region of space-time the state at a given point depends on.
• Triangular region bounded by λ− and λ+ characteristics leading to the point.
Definition: Zone of influence.
• The region of space-time the state of a given point influences.
• Triangular region bounded by λ− and λ+ characteristics emanating from the
point.
Saint Venant solution; will be dealt with in more detail in §5.6.2.
• For t ≤ 0 there is fluid of depth H for x ≤ 0 and no fluid for x > 0
• At t = 0 the fluid is allowed to collapse.
– 176 –
GEFD
Shallow water:Hyperbolic system
At t = 0:
u=0
h = H for x < 0
x=0
At t > 0:
at front, h = 0 ⇒ uf = 2(gH)1/2
LR = (gH)1/2t as rarefaction
wave propagates back at
velocity c = (gH)1/2.
5.3.7 Entrainment
This section applies only to miscible fluids. While immiscible fluids
can entrain each other, the result is a two-phase flow which, given a
chance, will separate again. Most aspects of such flows are beyond the
scope of this course.
Driven by instabilities on the flow, by turbulence within and ambient
turbulence.
Shallow water flows may be unstable to
• short wave instabilities: Kelvin-Helmholtz & Holmboe
• long wave instabilities
Examples sheet question: Analyse a two-layer Boussinesq flow for long wave
instabilities.
Effect on g′ ?
• Reduced density contrast
Effect on h ?
• Increased volume of layer into which fluid entrained
Effect on g′h ?
• For quiescent ambient, entrainment decreases g′ but increase h by
comparable amount so that g′h conserved
– 177 –
GEFD
Shallow water:Hyperbolic system
Momentum
• Bring in fluid with a different (zero) momentum.
Modification to equations
wd
ρ2
we
ρ1
Figure 102: Schematic of entrainment/detrainment for a shallow layer.
Assumption: The ambient density ρ2 is constant.
Continuity
∂
(ρ1 h1 ) + ∂ (u1ρ1 h1 ) = we ρ 2 − wd ρ1 ,
∂t
∂x
(136)
where we is the entrainment velocity, and wd the detrainment velocity.
Momentum (no drag)
∂
(ρ1h1u1 ) + ∂ ⎜⎛ u1ρ1h1u1 + 1 (ρ1 − ρ 2 )gh12 ⎞⎟ = we ρ 2u 2 − wd ρ1u1 ,
∂t
∂x ⎝
2
⎠
(137)
where u2 = 0 for a stationary ambient. (If Boussinesq then rhs: ρ(weu2 − wdu1)).
Eliminate continuity from (137):
u −u
∂u
∂u ρ1 − ρ 2 ∂h 1
∂ ⎛ ρ − ρ2 ⎞
⎟⎟ = − we 1 2 ,
g
+u
+
+ hg ⎜⎜ 1
h
∂t
∂x
ρ
∂x 2 ∂x ⎝ ρ ⎠
where the reduced gravity g′ = g′(x,t).
– 178 –
(138)
GEFD
Shallow water:Hyperbolic system
Note: Need to be a little careful when determining the effect of
horizontal density gradients. To integrate the pressure term over the
cross-section (where the pressure is p0 at some reference height z0):
∂p
∂
dz
=
∫0 ∂x ∫0 ∂x ( p0 + ρ 2 g (z0 − z ) + (ρ1 − ρ 2 )g (h − z ))dz
h
h
∂
((ρ1 − ρ 2 )h ) − z ∂ (ρ1 − ρ 2 )dz
∂x
∂x
0
h
= g∫
h
1
∂
⎡ ∂
⎤
= g ⎢ z ((ρ1 − ρ 2 )h ) − z 2 (ρ1 − ρ 2 )⎥
2 ∂x
⎣ ∂x
⎦0
∂h 1 ∂
⎡
⎤
= gh ⎢(ρ1 − ρ 2 ) + h (ρ1 − ρ 2 )⎥
∂x 2 ∂x
⎣
⎦
Assumption: Linear mixing → conservation of volume
∂h1 ∂
+ (u1 h1 ) = we − wd .
∂t ∂x
(139)
Combining (136) and (139) gives density
w
∂ρ1
∂ρ
+ u 1 = −(ρ1 − ρ 2 ) e ,
∂t
h1
∂x
(140)
which may be rewritten in terms of the reduced gravity as
w
∂g ′
∂g ′
= −g′ e .
+u
h1
∂x
∂t
Assumption: Stationary ambient: u2 = 0
Is system still hyperbolic?
Put into matrix form,
– 179 –
(141)
GEFD
Shallow water:Hyperbolic system
0 0 ⎤ ⎛ g ′ ⎞ ⎡1 0 0 ⎤ ⎛ g ′ ⎞
⎡ u1
⎢ 0 u h ⎥ ⎜ h ⎟ ⎢0 1 0⎥ ⎜ h ⎟
1
1 ⎥⎜ 1 ⎟ + ⎢
⎢
⎥⎜ 1 ⎟
⎢⎣ 12 h1 g ′ u1 ⎥⎦⎜⎝ u1 ⎟⎠ x ⎢⎣0 0 1⎥⎦⎜⎝ u1 ⎟⎠ t
w ⎞
⎛
⎜ − g′ e ⎟
h1 ⎟
⎜
= ⎜ we − wd ⎟ ,
⎜
we ⎟
u
−
⎟
⎜ 1
h1 ⎠
⎝
(142)
and seek eigenvalues λ satisfying
u1 − λ
0
1
2 h1
0
0
[
]
u1 − λ
h1 = (u1 − λ )(u1 − λ ) − g ′h1 = 0 .
g′
u1 − λ
2
(143)
Three real, distinct eigenvalues:
Interfacial waves
λ = u ± (g′h)½,
λ = u.
Exercise: Determine the equation along each of these characteristics.
Turbulent current, quiescent ambient
• Shear instabilities → mixing between layers
• Turbulence → homogenisation of current & entrainment
Overall Richardson number for single sided mixing
Rio =
g ′l
,
u′2
See §12
(144)
and the entrainment across the interface is
we
= f (Rio ) , for Pe >> 1.
u′
(145)
u′ ~ u1,
(146)
l ~ h.
(147)
For a turbulent current
– 180 –
GEFD
Shallow water:Hyperbolic system
so for current
Definition Richardson number
Ri =
g ′h
u12
(148)
• Note Ri = Fr−2.
The rate of generation of mixed fluid will be
Mixed fluid per unit area ~ u1 f(Ri,roughness).
This mixed fluid will be eroded by turbulence within current and mixed into
current. It will not be mixed into the ambient.
Many categories of flow adjust themselves to a given Fr or Ri, thus
we ≈ α u1,
(149)
wd = 0.
(150)
for some constant α.
Note similarity with entrainment into plume.
Strongly turbulent ambient
• Ambient turbulence → entrainment and mixing into ambient
• Richardson number Ri = g′lT/uT2 where lT and uT characterise the turbulence.
we ≈ 0,
(151)
wd = wTurbf(Ri).
(152)
– 181 –
GEFD
Shallow water:Single-layer hydraulic flow
Δρwe/wtur
Ri
Figure 103: Entrainment velocity as a function of Richardson number
Turbulence everywhere
Both we and wd non-zero
• Typically entrainment/detrainment will lead to the stratification being more
complex than simply two-layer.
• The definition of we and wd must be relative to the current interface, but this
is a poorly defined concept and so in general the above approach is of limited
value.
5.4 Single-layer hydraulic flow
5.4.1 Examples
• Flow is slowly varying
• Sharp-crested weirs
• Flow is not slowly varying, but behaves much in the same way, at least
upstream.
• Mountain ranges
• Submarine sills
• Controlled and uncontrolled flows
– 182 –
GEFD
Shallow water:Single-layer hydraulic flow
z
u
h
x
H
y
b
x
Figure 104: Sketch of typical hydraulically controlled flow.
5.4.2 Physical concepts
5.4.2.1 Conservation of basic quantities
Assumption: Steady, shallow water flow of uniform density.
Conservation of mass
Flux along channel conserved.
uhb = Q
Conservation of Bernoulli potential
Recall the Bernoulli potential (24):
8 g only
B = ½|u|2 + p/ρ + gz,
Look at interface or bottom (or, anywhere between due to assumption of
hydrostatic pressure)
9g or g′
B = ½ u2 + g(H+h) = const.
– 183 –
(153)
GEFD
Shallow water:Single-layer hydraulic flow
5.4.2.2 Specific energy
Eliminate velocity in Bernoulli potential:
u = Q/hb
Specific Energy function:
Key Result
Asymptotes:
1 Q2
+ (H + h ) = E = const
2 gb 2 h 2
(154)
9g or g′
E → h for h → ∞
E → ∞ for h → 0.
Figure 105 Specific Energy function for single-layer flow.
For a given channel profile H = H(x), b = b(x), and given flux Q, then the locus
of E as a function of h describes all possible solutions satisfying the
assumptions.
Channel of constant width
Similar results exist for channels with a varying width.
For b = const, we may rewrite (154) as
– 184 –
GEFD
Shallow water:Single-layer hydraulic flow
9g or g′
1 Q2
E−H =
+ h,
2 gb 2 h 2
(155)
so that everywhere along the channel is described by a single curve.
Given an upstream value of h and a knowledge of Q, can trace along curve as
H = H(x) varies.
E
H
Figure 106: Possible flows described by Specific Energy.
Figure 108: Specific Energy function for a channel of constant width.
5.4.3 Single-layer information propagation
What is the Froude number?
Consider, at constant x,
– 185 –
GEFD
Shallow water:Single-layer hydraulic flow
∂E
Q2
= − 2 3 + 1.
∂h
gb h
Rewrite in terms of velocity
9g or g′
u2
∂E
=−
+ 1 = 1 − Fr 2 .
gh
∂h
Key Result
(156)
(157)
• E tells us about the behaviour of long waves even though derived for a
When does specifying all conditions upstream make
sense?
If details of what is happening downstream cannot affect the upstream
conditions
• Requires the flow is supercritical
If flow subcritical, and upstream conditions do not match onto downstream
conditions
• Transients (waves) will propagate from downstream conditions and alter
upstream conditions.
How do things change along the channel?
As E is constant along the channel,
dE ∂E ∂E dQ ∂E db ∂E dH ∂E dh
=
+
+
+
+
= 0.
dx ∂x ∂Q dx ∂b dx ∂H dx ∂h dx
(158)
• Rate of change of E with bottom elevation H must balance rate of change of
E with fluid depth h.
• At crest of sill, dH/dx = 0
• Either dh/dx = 0
• Or ∂E/∂h = 0.
• More complex if both H and b varying and not stationary at same point.
– 186 –
GEFD
Shallow water:Single-layer hydraulic flow
Key Result A transition from subcritical to supercritical flow can only occur
at a constriction.
5.4.4 Hydraulic control
Concept
At a constriction the flow may switch from a subcritical branch to a
supercritical branch.
• Flow downstream of constriction supercritical and may be specified in terms
of Q and B at the constriction.
• Flow upstream of the constriction will adjust so that ∂E/∂h = 0 at the
constriction.
Figure 109: Information propagation for controlled flows.
When will this occur?
Whenever the upstream and downstream conditions do not match.
– 187 –
GEFD
Shallow water:Single-layer hydraulic flow
Examples sheet question: Determine the relationship between the upstream
depth where u ~ 0 and the depth of water over a weir.
5.4.5 Shocks
How do we match with the downstream conditions?
Figure 110: Shock formation in single layer hydraulics.
5.4.5.1 Hydraulic jumps
Why
• Necessary for matching downstream conditions
• Triggered by obstruction or friction
• Slowing down of waves on supercritical flow
– 188 –
GEFD
Shallow water:Single-layer hydraulic flow
• Can not have steady “hydraulic drop”
Key characteristics
Permitted by shallow water equations
• But violate assumptions (especially p hydrostatic)
• Momentum conservation must still apply on larger scale
Not permitted by Specific Energy function
• Energy reduced, even though momentum conserved: an important distinction
• Dissipation
• Downstream specific energy less than upstream of jump ⇒ lower elevation
than equivalent subcritical branch
Conserves momentum, but not easy if triggered by obstacle.
Differences for large and small density differences
Large density difference
• Behaviour of ambient unimportant
• No momentum transfer from ambient.
Small density difference
• Behaviour of ambient important.
• Pressure on face of jump: is it hydrostatic?
• Possibility of entrainment.
– 189 –
GEFD
Shallow water:Single-layer hydraulic flow
Different forms
Undular
• Looks like sequence of solitary waves trying to propagate against
supercritical flow.
• Solitary waves dissipate energy more efficiently through non-hydrostatic
processes.
• As required energy dissipation increases, waves grow in amplitude and
reduce spacing.
Turbulent
• Too many solitary waves → piling up
• Too large an amplitude → breaking
• More efficient at dissipating energy ⇒ more compact
Position
Intersection of characteristics
• Mass and momentum must still be conserved
Figure 111: Position of hydraulic jump using characteristics.
– 190 –
GEFD
Shallow water:Single-layer hydraulic flow
u L hL = u R hR
Mass
Momentum
u L2 hL +
1
1
g ′hL2 = u R2 hR + g ′hR2
2
2
9g or g′
9g or g′
where uL and uR are the velocities relative to the bore/jump.
5.4.5.2 Bores
Relationship to hydraulic jumps
Equivalent to hydraulic jump in frame of reference moving with bore, provided
ambient density negligible
Different if Boussinesq, or less dense than ambient
• Need to consider how the ambient is moved out of the way
• Form drag
Different forms
Undular
• Small amplitude, low dissipation
– 191 –
GEFD
Shallow water:Single-layer hydraulic flow
J.E. Simpson, Gravity Currents, CUP (1997)
Figure 112: Surfing the undular bore on the River Severn. (See also Figure 91.)
Turbulent
• Large amplitude, high dissipation
– 192 –
GEFD
Shallow water:Single-layer hydraulic flow
J.E. Simpson, Gravity Currents, CUP (1997)
– 193 –
GEFD
Shallow water:Resonance
5.5 Resonance
5.5.1 Generalised Helmholtz equation
Consider a general cross-section described by width B(x,z) for z ≥ H(x),
containing fluid in H(x) ≤ z ≤ H(x) + h(x,t). At a given x, the area containing
fluid is therefore
S ( x, h ) =
H ( x )+ h
∫
B ( x, ζ ) d ζ .
(159)
H ( x)
Assuming there is no flow through the boundaries, then continuity requires
∂S
∂u
∂u
dS ∂S ∂h
⎛ ∂S ∂h ∂S ⎞
+S
+u
=
+S
+ u⎜
+ ⎟
∂t
∂x
∂x
∂
∂
∂x ⎠
dx ∂h ∂t
h
x
⎝
∂S ∂h
∂u
⎛ ∂S ∂h ∂S ∂H ∂S ∂B ⎞
=
+S
+ u⎜
+
+
⎟=0
∂h ∂t
∂x
⎝ ∂h ∂x ∂H ∂x ∂B ∂x ⎠
.
(160)
For a shallow-water flow (hydrostatic, slowly varying, no drag, etc.),
momentum conservation gives
∂u
∂u
∂
+ u + g ( H + h) = 0 ,
∂t
∂x
∂x
(161)
where u is the along-channel velocity.
For looking at resonance problems, we consider the linear wave limit. Let
h = h0 + η and take H + h0 = const (i.e. unperturbed state is motionless with a
flat free surface), and define S0(x) = S(x,h0) with b0 = ∂S/∂h at h = h0. The
linearised continuity and momentum equations become
b0
dS
∂η
∂u
+ S0
+ u 0 = 0,
∂t
∂x
dx
∂u
∂η
+g
= 0.
∂t
∂x
– 194 –
(162)
(163)
GEFD
Shallow water:Resonance
Differentiating (162) with respect to time and eliminating u between these
equations yields
∂ 2η
∂ 2η dS0 ∂η
= 0.
b0 2 − S0 g 2 −
g
∂t
∂x
dx ∂x
(164)
Now for standing wave-like solutions, let η = ηˆ ( x ) eiωt . Substituting,
S0′
b0 ω 2
ηˆ ′′ + ηˆ ′ +
ηˆ = 0 ,
S0
S0 g
where S′0 = dS0/dx and η̂ ′ = ∂η̂ /∂x. This is the “generalised Helmholtz
equation”. Recalling that ω = k c = (gS/b)1/2, where k is the wavenumber,
yields
ηˆ ′′ +
S0′
ηˆ ′ + k 2ηˆ = 0 .
S0
The meaning of this equation is more readily understood if we transform it to
eliminate the first derivative. Suppose η̂ = ψϕ, then η̂ ′ = ψ′ϕ + ψϕ′ and
η̂ ″ = ψ″ϕ + 2ψ′ϕ′ + ψϕ″, so
ψ ′′ϕ + 2ψ ′ϕ ′ + ψϕ ′′ +
S0′
(ψ ′ϕ + ψϕ ′ ) + k 2ψϕ = 0 .
S0
Grouping in terms of ψ gives
⎛ ϕ′
ψ ′′ + ⎜ 2
⎝ ϕ
+
⎛ ϕ ′′ S0′ ϕ ′
S0′ ⎞
2⎞
′
+
+
+
ψ
k
⎟
⎜
⎟ψ = 0 .
S0 ⎠
S
ϕ
ϕ
⎝
⎠
0
To eliminate the ψ′ term, we select 2ϕ′/ϕ = −S0′/S0, solution of which gives
ϕ = S0−½ ,
hence we use the transformation
ψ = η̂ S0½ .
– 195 –
GEFD
Shallow water:Resonance
Noting that ϕ′ = − ½ S03/2S0′ and ϕ″ = ¾ S0−5/2S0′2 – ½ S0−3/2S0″, allows this to
be rewritten as
2
⎡
1 ⎛ S0′′ ⎛ S0′ ⎞ ⎞ ⎤
2
ψ ′′ + ⎢ k − ⎜ 2 − ⎜ ⎟ ⎟ ⎥ψ = 0 ,
4 ⎜ S0 ⎝ S0 ⎠ ⎟ ⎥
⎢⎣
⎝
⎠⎦
ψ ′′ + ⎣⎡ k 2 − Q ⎦⎤ψ = 0
or
1 ⎛ S0′′ ⎛ S0′ ⎞
Q = ⎜2 −⎜ ⎟
4 ⎜ S0 ⎝ S0 ⎠
⎝
with
2
⎞
⎟.
⎟
⎠
Horn equation
The same equation also arises when analysing the (linear) acoustics
(compressible) of a horn of cross-sectional area S(x):
continuity
S
∂ρ ∂
+ ( S ρu ) = 0
∂t ∂x
linearising
S
∂ρ
∂
+ ρ0 ( Su ) = 0
∂t
∂x
∂u
1 ∂p
=−
∂t
ρ ∂x
momentum (linearised)
p=ρ
equation of state
c2 =
speed of sound
⇒
S
∂p RT
=
∂ρ M
R
T
M
⇒ p = c2 ρ
∂ρ
∂
1 ⎛ ∂p
∂
⎞
+ ρ0 ( Su ) = 2 ⎜ S
+ p0 ( Su ) ⎟ = 0
c ⎝ ∂t
∂t
∂x
∂x
⎠
∂u
c 2 ∂p
=−
∂t
p0 ∂x
– 196 –
(165)
GEFD
Shallow water:Resonance
⇒
∂2 p
∂ ⎛ c 2 ∂p ⎞
S 2 − p0 ⎜ S ⎟ = 0
∂t
∂x ⎝ p0 ∂x ⎠
⇒
∂ 2 p 2 ∂ 2 p c 2 dS ∂p
−c
−
= 0.
2
2
∂t
∂x
S dx ∂x
This is Webster’s Horn Equation. Note similarity with shallow water resonance
equation (164):
∂ 2η S0 ∂ 2η S0 S0′ ∂η
− g
− g
=0
∂t 2 b0 ∂x 2 b0 S0 ∂x
⇒
2
∂ 2η
2 ∂ η
2 S 0′ ∂η
−
−
=0
c
c
∂t 2
∂x 2
S0 ∂x
Schrödinger equation
Consider Schrödinger equation
2
∂Ψ
i
+
∇2Ψ − V Ψ = 0 ,
∂t 2m
where Ψ = Ψ(x,t) is the wave function, the reduced Planck constant m the
mass of a particle and V = V(x) is the potential energy of the particle. For a
time independent particle in one dimension,
∂2Ψ
+VΨ ,
EΨ = −
2m ∂x 2
2
where E is the total particle energy (kinetic plus potential). Comparing with our
equation for a resonant wave, we see that the wavenumber k2 plays a role
similar to the particle energy, and the geometric quantity Q a role similar to the
potential energy.
Physical meaning
ψ ′′ + ⎡⎣ k 2 − Q ⎤⎦ψ = 0
If k2 > Q, then ψ is an oscillatory function of space – a standing wave.
– 197 –
GEFD
Shallow water:Resonance
If k2 < Q, then ψ is an exponentially decaying function – energy leaks in/out.
5.5.2 Loudspeakers
For a loudspeaker we want a flat response so that all notes are reproduced
equally faithfully. We therefore want to ensure there are no reflections due to Q
increasing beyond k2. Noting that the speed of sound in air is constant, then k
does not change along the horn for a given frequency, and thus we require
1 ⎛ S0′′ ⎛ S0′ ⎞
Q = ⎜2 −⎜ ⎟
4 ⎜ S0 ⎝ S0 ⎠
⎝
2
⎞
⎟
⎟
⎠
is constant to avoid any reflections and resonance. Hence
2
S ′′ ⎛ S ′ ⎞
2 0 − ⎜ 0 ⎟ = ±β 2 ,
S0 ⎝ S0 ⎠
where β is a constant.
Consider first the solution with +β and try a solution of the form S0 = A eλx,
λ2 = β2.
⇒
So λ = ±β and both S0 = A eβx and S0 = B e−βx are solutions, but since the
equation is not linear, superposition cannot be used.
Consider
2
⎛ S0′ ⎞′ S0′′S0 − ( S0′ )
,
⎜ ⎟ =
2
S
S
0
⎝ 0⎠
so can rewrite equation as
2
⎛ S0′ ⎞′ ⎛ S0′ ⎞
2⎜ ⎟ + ⎜ ⎟ = β 2 .
⎝ S0 ⎠ ⎝ S0 ⎠
Integrating once
β x− x
S0′
e ( 0) +1
β
,
= β β ( x− x )
=
0
β
S0
⎛
⎞
e
− 1 tanh
⎜ ( x − x0 ) ⎟
⎝2
⎠
where x0 is an arbitrary constant. Integrating again,
– 198 –
GEFD
Shallow water:Resonance
(
(
S0 = A exp − β x + 2ln e
(
= Ae − β x e
(
= Ae − β x e
(
=A e
β ( x − x0 )
)
β ( x − x0 )
2 β ( x − x0 )
−1
− 2e
−2+e
β ( x − x0 )
))
−1
2
β ( x − x0 )
)
+1
− β ( x − x0 )
)
⎛β
⎞
= 4 A cosh 2 ⎜ ( x − x0 ) ⎟
⎝2
⎠
for some constants A and x0.
Similarly, for β 2 < 0:
⎡β
⎤
S = A cos 2 ⎢ ( x − x0 ) ⎥ ,
⎣2
⎦
The first of these conditions may be viewed as the condition for an exponential
profile for a loud speaker in the horn equation: it emits uniformly at all
frequencies/wavelengths up to k2 = β2.
5.5.3 Taylor basin
Recall the modal equation (164):
S0′
b0 ω 2
ηˆ ′′ + ηˆ ′ +
ηˆ = 0 ,
S0
S0 g
Consider b0(x) = b1 x/L and S0(x) = S1 (x/L)2. (This was first introduced as a
model for the Bristol channel by GI Taylor, 1922.) The modal equation
therefore becomes
b1 L ω 2
2
ηˆ ′′ + ηˆ′ +
ηˆ = 0 ,
x
S1 x g
⇒
where
xηˆ ′′ + 2ηˆ ′ + κηˆ = 0 ,
b1 L ω 2
κ=
.
S1 g
– 199 –
GEFD
Shallow water:Resonance
This is a form of Bessel equation with general solution
ηˆ = A
J1 ( 2κ 1/2 x1/2 )
κ x
1/2 1/2
+ iB
Y1 ( 2κ 1/2 x1/2 )
κ x
1
1/2 1/2
where J1(..) and Y1(..) are the first order Bessel functions of the first and second
kind.
At x = 0 we must have no flow and finite amplitude. Noting that Y1(x) → −∞
as x → 0+, then B = 0. Note J1(x) → 0 as x → 0, and J1(2x)/x → 1, so
ηˆ ( x = 0 ) = A .
At the seaward end we can take an open boundary condition such that
J1 ( 2κ 1/2 x1/2 )
ηˆ ( x = L ) = A
= 0 so that the eigenmodes of the basin are given by
κ 1/2 x1/2
the roots of J1(2κ1/2L1/2) = 0 (recall that κ is a function of ω which we have not
yet specified). Thus
1
⎛ b1 ⎞ 2
1/2 1/2
2κ L = 2 ⎜
⎟ Lω = 3.83171, 7.01559, 10.17347, ...
S
g
⎝ 1 ⎠
– 200 –
GEFD
Shallow water:Gravity currents
5.6 Gravity currents
5.6.1 Description
What are they?
Time dependent.
Growing.
Driven by horizontal density differences
Needs boundary to support the density differences and convert them into
pressure
• In plume, pressure inside plume equal to that outside
• In gravity current, pressure inside current not equal to that outside
Release or generation of material of a different density
• Plume: Gravity takes material away from any boundaries
• Gravity current: Gravity keeps material in contact with boundary
In the environment
“Simple” gravity currents often difficult to see
• Air too clear
• Water too opaque
Their effects may be more obvious
• Sea breeze
• Cold front
• Cold draft through open door
– 201 –
GEFD
Shallow water:Gravity currents
J.E. Simpson, Gravity Currents, CUP (1997)
Figure 114: Roll-up of head of radial gravity current resulting from a microburst
(thunderstorm outflow).
In the laboratory
(a)
(b)
J. Hacker, DAMTP
Figure 115: Cross-channel average for lock-release gravity current. (a) Raw images. (b)
Density.
In the computer
Number of approaches
– 202 –
GEFD
Shallow water:Gravity currents
• Shallow water – subject of this section
• DNS (Direct Numerical Simulation)
• Difficult due to large density gradients and high Reynolds numbers
Härtel, Mieburg & Necker, JFM
Figure 116: DNS of gravity current with no-slip boundary condition.
– 203 –
GEFD
Shallow water:Gravity currents
Härtel, Mieburg & Necker, JFM
Figure 117: DNS of gravity current with free-slip boundary condition.
Features
• Well defined front
• Smooth leading edge, but mixing behind
– 204 –
GEFD
Shallow water:Gravity currents
Effect of friction
• Differences between currents along bottom and along a free surface
S. Dalziel & L. Thomas, DAMTP
Figure 118: Velocity and vorticity fields in gravity current along a no-slip boundary
(experimental).
S. Dalziel & L. Thomas, DAMTP
Figure 119: Velocity and vorticity fields in gravity current along free surface
(experimental).
• What happens to the overrun fluid?
– 205 –
GEFD
Shallow water:Gravity currents
(a)
(b)
J.E. Simpson, Gravity Currents, CUP (1997)
Figure 121: Lobes and clefts. (a) Laboratory experiment. (b) Cold front.
– 206 –
GEFD
Shallow water:Gravity currents
Types
Differences in initial conditions and/or source conditions
• Lock exchange
• Lock release
• Continuous
Differences in parameters
• Turbulent
• Laminar
• Viscous…
Differences in external conditions
• Quiescent ambient
• Turbulent ambient
• Ambient wind
• Stratified
• Rotating
Other
• Topography
• Collisions
5.6.2 Early models
9g or g′
Dimensional analysis
u2 ~ gH
Characteristics
Saint Venant
• Dam break: at t = 0, we have h = H for x ≤ 0 and h = 0 for x > 0.
– 207 –
GEFD
Shallow water:Gravity currents
x=0
Figure 122: Schematic of dam break problem.
Hypothesise that a self-similar flow exists so that the characteristics form an
expansion fan.
λ=u−c
characteristic
t
The ‘backward’ characteristic
λ = u − c = x/t
Along the forward characteristic λ = u+c
u + 2c = 2C
⇒ 3c = 2C − x/t
⇒ h = (1/9)(2C − x/t)2/g
Back
u=0
c=C
λ = −C
Front
u = 2C
c=0
λ = u±2c = 2C
λ=u+c
characteristic
0
x
Figure123: Characteristics for St. Venant solution.
But not observed in practice
• Bottom drag for ρcurrent >> ρambient
• Form drag for ρcurrent << ρambient
• Form drag for Δρ << ρ.
Boussinesq currents
Prandtl:
– 208 –
GEFD
Shallow water:Gravity currents
• Flow in current exceeds front velocity, leading to raised fluid that is assumed
not to fall.
• Propagation speed by dynamic balance (excluding hydrostatic effects).
• Cannot apply in reality.
von Kármán::
• Flow irrotational
• Bernoulli from far down stream to the front, thus uf2 = 2g′Hdownstream.
• Also ideas similar to sharp-crested water waves gave front-ground angle of
600.
• But: force imbalance; drag requires rotational flow
5.6.3 Cavity flows
Brook Benjamin (1968): Gravity currents and related phenomena (JFM 31,
209-248)
• This paper has, it is believed, more or less exhausted the most useful
applications of inviscid-fluid theory to steady gravity currents...
• ...for it must be acknowledged that gravity currents in practice are highly
complex phenomena, generally featuring a great deal of turbulence and
significant mixing of the two layers, so that the interpretation of them
depends vitally on semi-empirical analyses...
U
h
u=0
H
ρ
Figure 124: Definition sketch for a cavity flow.
5.6.3.1 Cavity flows conserving energy
Assumptions:
This was part of the examination in 1998/99
– 209 –
GEFD
Shallow water:Gravity currents
Continuity
Momentum
Energy conservation ⇒ Bernoulli potential conserved
In frame of reference of nose
Stagnation
Point
h
u1
H
u2
ρ
Far upstream (i.e. to the left), the approaching flow is parallel and has speed
u1 = U.
(166)
Far downstream (i.e. to the right), the receding flow is parallel and continuity
gives the speed
u2 = U H/(H - h)
(167)
There is a stagnation point at the leading edge of the cavity
Apply Bernoulli upstream to get upstream pressure on upper surface
p1 = p0 − ½ ρU2
(168)
Apply Bernoulli downstream to get velocity
p2 = p0 = p0 + ρgh − ½ ρu22 ⇒ u22 = 2gh
Need to conserve flow force: Momentum flux plus pressure force.
Flow force upstream
– 210 –
(169)
GEFD
Shallow water:Gravity currents
Flow_force_upstream = ρU2H + p1H + ½ ρgH2
= ρU2H + (p0 − ½ ρU2)H + ½ ρgH2
= p0H + ½ ρU2H + ½ ρgH2
(170)
Flow force downstream (using u22 = 2gh)
Flow_force_downstream = ρu22(H−h) + p2H + ½ ρg(H−h)2
= 2ρgh(H−h) + p0H + ½ ρg(H−h)2
(171)
Equating flow forces
p0H + ½ ρU2H + ½ ρgH2 = 2ρgh(H−h) + p0H + ½ ρg(H−h)2
(172)
Simplifying
½ ρU2H + ½ ρgH2 = 2ρgh(H−h) + ½ ρg(H−h)2
(173)
Now since u2 = U H/(H - h) then U2H = u22(H−h)2/H = 2gh(H−h)2/H, so
ρgh(H−h)2/H + ½ ρgH2 = 2ρgh(H−h) + ½ ρg(H−h)2
⇒
H2h − 2Hh2 + h3 + ½H3 = 2H2h − 2Hh2 + ½H3 − H2h + ½Hh2
⇒
h3 = ½Hh2
⇒ Key Result
h = ½H.
(174)
U = u2(H−h)/H = ½ u2 = ½ (2gh)1/2 = ½ (gH)1/2
(175)
Speed of propagation:
Key Result
Fr =
U
1
=
gh
2
(176)
Rewriting
g (H 2 − h 2 )H
u =
(2 H − h )h
2
2
Downstream of front, Froude number of √2 relative to the front
• ⇒ dissipative hydraulic jump may occur on downstream flow
– 211 –
(177)
GEFD
Shallow water:Gravity currents
Possible to get approximate shape of energy conserving cavity: conformal
mapping etc.
J.E. Simpson, Gravity Currents, CUP (1997)
Figure 125: Air cavity occupying half the depth as predicted by Benjamin.
But is this the only solution?
What happens if you throttle the air?
Figures of throttled cavity flows
Figure 126: Air cavities occupying less than half the depth.
5.6.3.2 Cavity flows with energy loss
Hypothesise a head loss in the ambient as it is accelerated around the current.
• Follows the ideas of hydraulic jumps
• Supercritical downstream flow could form a hydraulic jump
In §5.6.4 we will see that a hydraulic jump is often required by the
hyperbolic nature of the flow.
h
u2
Figure 127: Cavity flow with a hydraulic jump at the back of the head.
Head loss in ambient fluid due to rotational motions in jump at back of head.
• Follows ideas of hydraulic jumps
• Parameterise as uniform across depth of ambient.
– 212 –
GEFD
Shallow water:Gravity currents
Bernoulli downstream of stagnation point now
u22 = 2g(h−Δ)
(178)
Expressions for flow force unchanged.
Equating u2 from flow force balance with u2 from Bernoulli:
(
H − 2h )h 2
Δ=
2(H − h )(H + h )
(179)
Key Result For h > ½H would require Δ < 0 ⇒ flow cannot exist
U/(gH)1/2
0.2192
05
h/H
0
Figure 128: Energy loss in cavity flow - from Benjamin (JFM 31, 1968)
For 0.2192 < h/H < ½, f = u1/(gH)1/2 > ½
• For given front speed between f = ½ and f = 0.5273 then there are two
possible h/H, with the larger one corresponding to greater energy loss.
• Front condition
– 213 –
GEFD
Shallow water:Gravity currents
⎡⎛
h ⎞⎛
h ⎞⎤
1
2
−
−
⎜
⎟
⎜
⎟
⎢
H ⎠⎝
H ⎠⎥
⎝
⎥
Ff = ⎢
h
⎛
⎞
⎥
⎢
⎜1 + ⎟
⎥⎦
⎢⎣
⎝ H⎠
1
2
Froude number for front as function of relative depth of current
Härtel, Mieburg & Necker, JFM
Figure 129: Froude number variations for a lock-exchange gravity current with Grashoff
number (Gr ~ Re2).
• Is uniform head loss appropriate for an infinite ambient?
5.6.3.3 Filled cavities
What happens if the density of the cavity is not negligible?
Boussinesq
– 214 –
GEFD
Shallow water:Gravity currents
• Can repeat Benjamin’s analysis
• Have g′ instead of g
Water into air
• Completely different
• Air moves out of the way almost effortlessly
• Typical gravity current in a film – dominated by surface tension.
5.6.3.4 Uniformity of velocity
How far downstream of the head must you be to recover a
uniform velocity in the ambient?
Consider a wavenumber k wave in a depth H. The horizontal velocity is
⎛
cosh ( kz ) ⎞
u ′ = ωη0 ⎜ sinh ( kz ) +
⎟ cos ( kx − ωt )
tanh ( kH ) ⎠
⎝
(180)
so the instantaneous flux under the crest is
0
flux =
∫
u ′ dz =
η0ω
−H
k
.
(181)
Near the head the response of the ambient fluid will be similar to that of a wave
with amplitude hf,wavenumber k and frequency ω ~ k uf. The flux carried by
this wave matches the flux carried by the current:
η0ω
k
= hu f .
The velocity in the ambient due to this wave is
ua = ω η0 ≈ khf uf,
whereas the velocity under the shallow water approximation is
– 215 –
(182)
GEFD
Shallow water:Gravity currents
us =
hf
H − hf
uf .
The shallow water approximation is only valid if we are a distance x >
~ H − hf
from the front. Closer to the front we have k ~ 1/x, and the ambient fluid
velocity adjacent to the interface will exceed the shallow water velocity by a
factor of
ua H − h f
≈
,
us
x
with ua ~ uf near the top of the head.
– 216 –
GEFD
Shallow water:Gravity currents
5.6.4 Gravity currents and characteristics
λ = u − c = x/t
u + 2c = 2C
⇒ 3c = 2C − x/t
⇒ h = (1/9)(2C − x/t)2/g
⇒ u = (2/3)(C + x/t)
Back
u=0
c=C
λ = −C
λ = u + c = (4/3)C+(1/3)x/t
Back
u f = F fc
u=0
c=C
= C
Front
u = 2C
c=0
λ = ±2C
0
λ = uf c
= (Ff 1)
λ = u c = x/t
u + 2c = 2C
3c = 2C x/t
h = (1/9)(2C x/t)2/g
u = (2/3)(C + x/t)
t
t
u = 2C
c=0
= 2C
λ = u + c = (4/3)C+(1/3)x/t
x
Saint Venant solution
0
x
Gravity current solution
t
0
Rarefaction wave
5.6.5 Modelling gravity currents
5.6.5.1 Integral models
The simplest dynamical model
Approximate shallow water equations by discarding x derivatives
Retain constant Froude number front condition
Describe current purely in terms of its length L and depth h.
– 217 –
x
GEFD
Shallow water:Gravity currents
8 g′ only
Continuous – planar
dL/dt = uf = Ff (g′h)1/2
Front:
Continuity:
(d/dt)(Lh) = Q/b
Can eliminate L from continuity, but end up with nasty nonlinear, second order
8 g′ only
ode for h.
Useful to do part of this process to obtain
Ff g ′1/2 h3/2 + L
dh Q
= .
dt b
• As L becomes large, dh/dt must become small.
Can search for power law solution of the form L ~ tα, h ~ tβ. Substitution and
gives rise to α = 1, β = 0 when Q is constant.
2
⇒
⎛ Q b ⎞ 3
.
h=⎜
⎜ F g ′1/2 ⎟⎟
⎝ f
⎠
⇒
⎛ Ff2 g ′Q ⎞
uf = ⎜
⎜ b ⎟⎟
⎝
⎠
1
3
dL/dt = uf = Ff (g′h)1/2
Must start at finite radius, L = L0, say.
Continuity:
(d/dt)(π(L2 − L02) h) = Q
Exercise: Derive the solution.
Lock release – planar
Similar procedure, except
•
8 g′ only
(184)
8 g′ only′
Continuous – axisymmetric
Front:
(183)
Initial conditions L = L0 and h = h0
• Continuity: Lh = L0h0.
– 218 –
GEFD
Shallow water:Gravity currents
Eliminate depth between continuity and front position
dL
= Ff g ′h
dt
= Ff g ′
L0 h0
L
and solve
2
8 g′ only
/3L3/2 = Ff(g′L0h0)1/2t
L ~ t2/3
Key Result
(185)
Use simple, self-similar geometric form
u = (x/L) uf
Could use any f(x/L) such that
h(x,t) = hf(t) f(x/L),
the difference being a constant in the relationship between L0h0 and the initial
volume.
Singluar perturbation
• Flows tend towards this solution despite different initial conditions.
• Similar procedure, except Lh = const
Lock release – axisymmetric
For lock release, can start at zero or finite radius; continuity: (L2−L02)h = const
Exercise: Derive the solution.
5.6.5.2 Shallow water models
Analytical solution for special cases
• Can do similarity solution for lock release, valid at late time
• Can do early time solution for lock release
– 219 –
GEFD
Shallow water:Gravity currents
Numerical solution for more interesting problems.
• Finite difference
• Finite volume
• Goudnov - makes explicit use of hyperbolic nature
5.6.6 Real life is more complex
5.6.6.1 Topography
Small slopes
z
x
θ
Add in along-slope component of gravity to momentum equation:
∂u
∂u
∂h
+u
= − g cos θ + g sin θ
∂t
∂x
∂x
∂h
≈ − g + gθ
∂x
Upslope
9g or g′
Will always decelerate flow until it stops
• Eventually all available kinetic energy will be converted into potential
energy.
– 220 –
(186)
GEFD
Shallow water:Gravity currents
M. Brown, DAMTP
Figure 130: Time series of vertical integral of density for a gravity current running up a
slope.
Downslope – water into air
Current will accelerate
• Acceleration will be balanced by drag
For water into air,
uu
∂u
∂u
∂h
+u
= − g + gθ − CT
,
∂t
∂x
∂x
h
(187)
8 g only
u = gh
θ
.
CT
Downslope – form drag
When the density of the ambient fluid is not negligible, form drag will be
important.
– 221 –
(188)
GEFD
Shallow water:Gravity currents
Consider a blob of fluid.
u
h
H
Figure 131: Definition sketch for blob of fluid released on a slope.
If there is no bottom drag, and no dissipation in current, then conservation of
the Bernoulli potential requires
g (cavity) or g′′
½u2 + g(H+h) = g(H0 + h0,)
(189)
where H0 and h0 describe the blob of fluid when it is first released.
If Benjamin’s arguments apply, then current has constant Froude number
u = Ff(gh)1/2, so
[½Ff2 + 1]h = h0 + (H0 − H).
For infinite ambient, Benjamin suggests Ff = √2, giving
h = ½[h0 + (H0 − H)],
(190)
Half the available potential energy is converted to kinetic energy.
Recall the equi-partition of energy for other problems.
• Suggests the blob will get taller and narrower.
• Breaking, mixing and turbulence.
Key Result The Bernoulli potential of the current cannot be conserved.
– 222 –
GEFD
Shallow water:Gravity currents
• Bottom drag
• Internal dissipation
Exercise: Calculate the rate of head loss required for the blob to run down the
slope at a constant velocity.
Is the Benjamin-like front condition appropriate?
Can the flow reach an asymptotic state?
What happens at the trailing edge?
What is the rôle of viscous stresses?
(a)
(b)
(c)
(d)
Figure 132: Evolution of a circular patch of dense fluid released on a slope.
– 223 –
GEFD
Shallow water:Gravity currents
A. Ross, DAMTP
Figure 133: Position of front of a circular patch of dense fluid released on a slope.
5.6.6.2 Other complexities
Wind
Not just a change in the frame of reference
• Drag on boundaries
• Wind shear
• Initial or boundary conditions
Ambient turbulence
Entrainment/detrainment
• Similar to §5.3.7
• Generation of a mixed region
• Reynolds stresses
• Scale on imposed turbulence rather than flow velocity.
… = −CAu|uT|/h instead of CTu|u|/h
– 224 –
GEFD
Shallow water:Gravity currents
P. Harichandran, DAMTP
Figure 134: Gravity current propagating into a turbulent environment
– 225 –
GEFD
Shallow water:Gravity currents
Ambient stratification
Gravity currents can form away from boundary as “intrusions”
J.E. Simpson, Gravity Currents, CUP (1997)
Figure 135: Intrusion into two-layer system. The density of the intrusion is mid-way
between that of the two layers.
Intrusions can generate waves
Waves carry energy
Waves may alter upstream conditions
– 226 –
GEFD
Shallow water:Gravity currents
F. de Rooij, DAMTP
Figure 136: Intrusion into linearly stratified ambient.
Frontogenesis
How does a current form from a weak horizontal density gradient
• Seabreeze
• Turn off turbulence
– 227 –
GEFD
Shallow water:Gravity currents
Vorticity equation
∂ω
∂ω
g ∂ρ
∂ω
+u
+w
+ (ω ⋅ ∇ )u + ν∇ 2 u
=−
∂t
∂x
∂z
ρ ∂x
(191)
to Frontogenesis.
Lossy gravity currents
Not “negative entrainment”
Losses due to
• Seepage into substrate
See §10.2.2 and the 1999 exam question
• Evaporation/boiling
• Rain (negative loss, but adds zero momentum)
L. Thomas & B. Marino, DAMTP
Figure 138: Gravity current over a porous substrate.
– 228 –
GEFD
Shallow water:Gravity currents
Reflection
Shallow water cannot fully model a splash
• Splash ⇒ dissipation
• But conservation of momentum
• Sloping boundary?
Bore propagating back up the current
Collision
Related to reflection, but less precise
• Mixing and dissipation
• How do the densities, heights and velocities of the incoming currents affect
things?
Conservation of momentum
– 229 –
GEFD
Shallow water:Gravity currents
A. Ross, DAMTP
Figure 139: Colliding gravity currents with identical release conditions and times.
– 230 –
GEFD
Shallow water:Gravity currents
A. Ross, DAMTP
Figure 140: Colliding gravity currents of unequal densities.
Non-Boussinesq currents
Density of ambient >> density of current - use above theory
Density of ambient << density of current - no dissipation in ambient, but still
get front
– 231 –
GEFD
Shallow water:Gravity currents
– 232 –
`