Lectures for MA3D1, FLUID DYNAMICS
Term 1, Weeks 1-10, 2007.
R.M. Kerr, University of Warwick Robert.Kerr@warwick.ac.uk
Tel: 74718, Messages: 22014
April 15, 2008
Preliminaries
0.1
Class times and places
• Tuesdays 10:05-10:55 MS.04
• Support classes: starting in week 3. Time to
be arranged.
• Tuesdays 16:05-16:55 MS.03
0.2
• Thursdays 16:05-16:05 L4
0.3
Office hours
• Tuesdays 14:00-15:00 C2.29
Web sites
• Reading notes: http://www.eng.warwick.ac.uk/staff/rmk/teaching/MA/MA3D1 0708.pdf
0.4
Main TEXTS
The book that probably covers the greatest portion of this course is:
• D.J. Acheson, Elementary Fluid Dynamics, OUP. (The main text. Excellent and affordable.)
Other sources
• G.B. Batchelor, Fluid Dynamics, CUP. (A classic. used by most Applied Mathematics Depts in
UK. Used by Engineering at Cornell University, Ithaca, New York in 1978.)
• L.D. Landau and E.M. Livshitz, Fluid Dynamics, OUP. (Used by Physics Depts throughout the
world, including Cornell.)
• D.J. Tritton, Physical Fluid Dynamics (Second Edition), Oxford Science Publs. (The emphasis is
on the physical phenomena and less on the mathematics. All the geophysical fluid dynamics will be
from here.)
• A.R. Paterson, A First Course in Fluid Dynamics, CUP. (Affordable and easier than Acheson.)
0.5
Aims:
An important aim of the module is to provide an appreciation of the complexities and beauty of fluid
motion. This will be brought out in lectures, computer demonstrations and visualisations, web pages.
0.6
Objectives:
It is expected that by the end of this module students will be able to:
1.) understand the underlying mathematics of the physical processes in a number of different fluid flows.
2.) come up with qualitative and quantitative solutions for particular fluid dynamics problems ranging
from simple laminar flows to fully developed turbulence.
3.) use concepts and mathematical techniques learned from this course for analysis of other partial
differential equations arising for example in plasma physics or nonlinear optics.
0.7
Syllabus:
• Kinematics of Fluid Motion. Specification of the flow by field variables; vorticity; stream function;
strain tensor; stress tensor.
• Conservation Laws. Conservation of mass, momentum and energy and equations of motion deduced
from these laws; Bernoulli’s equation.
• Vorticity. Vortex lines and vortex tubes, Kelvin’s circulation theorem, vorticity equation, interaction
of vortices, vortex sheet.
• Dimensional analysis. Reynolds number, Rayleigh number, Ekman number, Rossby number, etc.
• Laminar flow. Flow in a pipe; shear flows; flow due to an oscillating plate; Stokes flows of very
viscous fluids.
• Boundary layers. Prandtl’s boundary layer theory; flow separation. Taylor-Proudman theorem and
Eckman boundary layer in rotating fluids.
• Instability and waves. Raleigh-Taylor and Kelvin-Helmholtz instabilities; stability of parallel flows.
Inertia-gravity and internal waves. Waves on a deep water. Sound.
• Geophyiscal Fluid Dynamics. Stratified, rotating flows, balanced flows.
0.8
Assessment
• Three-hour written examination.
• Preparation:
– Mathstuff currently has past example problems, answers and exam solutions.
– I will add past exams.
check.
− In the past hard copies were available in the general office. I will
Where are there fluids?
Everywhere
The Earth: the oceans, lower atmosphere and upper atmosphere
are all fluids, but with very
different properties.
The sun in extreme ultraviolet
Th
mos: Crab nebula
Jupiter
Laboratory: Geophysics
Rayleigh-Taylor oil slick
Double diffusion: salt
fingers
Rotating flows: Merging Taylor columns
Laboratory: Fundamental Engineering Flows
Wakes: behind a sphere
Shear flow: Karman wake
Venturi pipe
Turbulent jet
Military and aviation
Wingtip vortex roll-up
NASA X-15 experimental jet
NASA experimental hypersonic
aircraft
Atomic bomb over Nagasaki
This is all physics.
Why are pure mathematicians interested in this problem?
Navier-Stokes is worth $1M
Central mathematical object:
compressible Navier-Stokes equation
∂ρ
+ ∇ · (ρv) = 0
∂t
∂v
ρ
+ v·(∇v) = µ4v + (λ + 13 µ)∇(∇ · v) − ∇p + f
∂t
∂θ
+ ∇θ · v = −p∇ · v + κ4θ + Φ
ρcv
∂t
Euler, Stokes and Navier
(0.1)
Central mathematical object:
compressible Navier-Stokes equation
∂ρ
+ ∇ · (ρv) = 0
∂t
∂v
ρ
+ v·(∇v) = µ4v + (λ + 13 µ)∇(∇ · v) − ∇p + f
∂t
∂θ
ρcv
+ ∇θ · v = −p∇ · v + κ4θ + Φ
∂t
The fields for a compressible gas are
Fields :
ρ(x, t)
p(x, t)
mass density
pressure
vi(x, t) 3 components of velocity
θ(x, t)
temperature
where
p = (γ − 1)ρθ
internal energy/unit mass
f
forcing
Φ(x, t)
Parameters :
λ
bulk viscosity
κ
thermal diffusivity
µ
cv
coeff of viscosity
specific heat
Simplifications:
• Incompressible Navier-Stokes equations
∂v
1
+ v·(∇v) = − ∇p + ν4v + +f
∂t
ρ
∇·v = 0
ρ = constant and ν = µ/ρ is the Newtonian viscosity
(0.2)
This is what the Clay Prize is for
• Euler’s equations or inviscid (ν = 0) incompressible Navier-Stokes
equations
∂v
1
+ v·(∇v) = − ∇p + f
∂t
ρ
(0.3)
∇·v = 0
ρ = constant and ν = 0
1
Origin of equations
The physical laws that govern fluid flow are deceptively simple.
a) Start with Newton’s second law of motion which states that:
MASS × ACCELERATION = APPLIED FORCE
In fluid mechanics we prefer to use the equivalent form of
RATE OF CHANGE OF MOMENTUM = APPLIED FORCE
b) Add: Principles of conservation of:
mass, momentum and energy
c) Apply: Concept of the continuum.
d) Physical laws that determine forces:
body forces due to the action of gravity
pressure forces: internal forces
viscous forces.
• Generalise viscous force: Replace f ν = ν4v by f ν = ∇S where S is the
viscous stress tensor.
• To determine the viscous forces we need to supplement Newton’s laws of motion
with a constitutive law.
• In simple terms the constitutive law for an incompressible Newtonian fluid states
that:
1 ∂ui ∂uj
• VISCOUS STRESS ∝ RATE OF STRAIN Sij = [S] = ν
+
2 ∂xj ∂xi
ν ∂ui ∂uj
or∇S = ∇
+
= ν4v
2 ∂xj ∂xi
• This applies to air (lower atmosphere) and water.
• Which brings us to the Reynolds number:
– Given a velocity scale U
– A length scale L
– And the viscosity ν
UL
– The Reynolds number Re =
ν
determines how turbulent or laminar a flow is.
REYNOLDS NUMBER
∂v
1
+ v·(∇v) = − ∇p + ν4v
∂t
ρ
• What dimensionless number describes this equation.
Start with:
• Estimate: |v| ∼ U , a velocity scale and ∇ ∼ L1 , an inverse length scale.
• We want to balance the convective term: (v·∇)v ∼
U2
L
• Against the viscous term: ν4v ∼ ν LU2
convective
U 2/L U L
• The Reynolds number is Re ∼
∼
=
viscous
νU/L
ν
• Originated by Osborne Reynolds, University of Manchester, 1883.
• If Re >> 1 can viscosity be ignored?
• NO: viscosity underlies all large Reynolds number, turbulent flows.
• Low Reynolds number: plane channel: Pouseuille flow
– Assume steady fluid with viscosity ν forced by a constant pressure gradient
dP/dx < 0 between two fixed plates at y = −L and y = +L then
2
2
1 dP
dU
y
2 −dP/dx
0=−
+ν 2
U =L
1− 2
(1.1)
ρ dx
dy
2ρν
L
$:
$"+2, 4
"+252,5 04
3
0
6+2 4
1+25 .+2,5772
869/.
4:
1+7
Why do such simple physical laws lead to such complex flows? There
is no simple answer. However, part of the reason is the
• effect of viscosity at high Reynolds number.
– Since the Reynolds number is a measure of the ratio of inertial forces to viscous
forces, we might expect viscous effects to become progressively weaker as the
Reynolds number is increased.
– What actually happens, instead, is that the viscous effects become concentrated
in a thin layers or vortices.
• Near boundaries, the layer which forms at the surface becomes progressively thinner as the Reynolds number is increased.
– Viscosity + shear at the surface generates vorticity in the boundary layer.
– Then, owing to the effects of the combination of pressure and viscous forces:
the boundary layer inevitably separates from the surface and
the thin vortical layers move into the flow field.
– These form complex structures.
• The most extreme manifestation of this is turbulent flow which is a
highly complex, unsteady, three-dimensional flow.
• Turbulent flow is the subject of a 4th Year MEng Module.
1.1
Continuum model
• These simple physical laws are, of course, merely theoretical models. The principal
theoretical assumption is that the fluid consists of continuous matter – the so-called
continuum model.
• At the microscopic level, a fluid is not a continuum, but consists of myriads of
individual molecules.
In most engineering applications even a tiny volume of fluid (measuring, say, 1
µm3) contains a large number of molecules.
• However, a typical molecule travels on average a very short distance (known as the
mean free path) before colliding with another.
In typical engineering aerodynamics application the m.f.p. is less than 100 nm,
which is very much smaller than any relevant scale characterising quantities of engineering significance.
Owing to this disparity between the m.f.p. and relevant length scales, we may
expect the equations of fluid motion, we can apply averaging.
This is the continuum model and is obeyed to great precision by the
fluid flows found in almost all engineering applications.
• This expectation is supported by experience.
When pure homogeneous fluids are involved the only exceptions are very small-scale
motions (e.g., nanotechnology),
where the relevant scales are comparable to the m.f.p.; and rarefied gas dynamics
(e.g., re-entry vehicles),
where there are so few molecules present that the m.f.p. becomes comparable to
the dimensions of the vehicle.
• In engineering and for some other scientific applications we often wish to understand
flows of mixtures of pure fluids and particulate matter, e.g.,
blood, toothpaste, slurries.
These are usually modelled as fluids that obey more complex constitutive
laws termed collectively
non-Newtonian fluids.
bredWe will not consider non-Newtonian fluids in any detail on this course.
The physical laws governing fluid flows may be simple but, even for the
simplest geometries, such as the flow past a flat plate or around a sphere, the
resulting fluid flows can be exceeding complex especially at high Reynolds
numbers. This complexity still presents a major challenge to the engineer.
1.2
1.2.1
Vector Algebra
Scalar product
~ If
Or dot product. Noteboldface represents vectors: ()
~ = A = (A1, A2, A3) and B = (B1, B2, B3)
A
X
then A · B =
AiBi
i
2
From this it follows that A2 = |A| = A21 + A22 + A23 = A · A
and A · B = AB cos θAB
where θAB is the angle between the vectors A and B.
A · B = AB cos θAB
The following diagram shows the
angle that gives cos θAB that comes
from the scalar product between
two vectors A and B in an arbitrary plane through 3D space.
1.2.2
Vector product
Or cross product of two vectors A and B yields a third vector C and is written two
ways
C = A × B or C = A ∧ B = (A2B3 − A3B2, A3B1 − A1B3, A1B2 − A2B3)
The second is preferred by some mathematicians because it refers to a generalisation
of this concept to higher dimensions. Physicists and engineers use the first notation.
• In general C is perpendicular to both A and B.
• That is: C · A = 0 and C · B = 0
• Assume A = (A1, A2, 0) and B = (B1, B2, 0) then C = (0, 0, C3)
This
is
represented
by
the following figure where
|C| = |A||B|| sin θ|.
Note
that for a small angle between A
and B, where sin θ is small, that
the magnitude of C is small.
1.2.3
Vector Calculus and Gradient operator
The gradient operator is a vector consisting of derivatives. Defining what is sometimes
called grad or del=
X
∂
∂
∂
∂
~ =
∇
,
,
=
ei
∂x1 ∂x2 ∂x3
∂xi
i
∂
(grad)i = ∇i =
∂xi
∇ is called nabla in some typesetting languages, including the one that created this
document called LaTeX.
The GRADIENT is this operator applied to a scalar φ to create a vector
~ = ∇φ
~ = ∂φ , ∂φ , ∂φ
A
∂x1 ∂x2 ∂x3
~
The DIVERGENCE is the equivalent of a scalar product using ∇
∂A1 ∂A2 ∂A3
~
~
∇·A=
+
+
∂x1 ∂x2 ∂x3
It can be shown that if the variables (x1, x2, x3) are transformed to a new orthogonal
coordinate system (x01, x02, x03) that the divergence is invariant.
~
The CURL is the equivalent of a vector product using ∇
∂
∂
∂
~
,
,
× (u1, u2, u3) =
curl ~u = ∇ × ~u =
∂x1 ∂x2 ∂x3
∂u1 ∂u3
∂u2 ∂u1
∂u3 ∂u2
,
,
−
−
−
∂x2 ∂x3
∂x3 ∂x1
∂x1 ∂x2
1.3
Streamlines
• Streamline: A curve which is tangential to v(x, t) at every point. Use v =
(v1, v2, v3) = (u, v, w)
• then on a streamline parameterised by s: x(s) = (x(s), y(s), z(s)) solve:
dx/ds dy/ds dz/ds
=
=
u
v
w
(1.2)
1.4
Frames of reference
• Given Eulerian coordinates x(t) = ϕ(X, t) then X is the Lagrangian coordinate.
• There is the inverse transform: X = ψ(x, t): Eulerian→ Lagrangian
• Lagrangian coordinates follow the flow.
• Material/Convective Derivative Given f (x(t), t)) = f (X, t) then
d
Dtf = f (x(t), t)) = ∂tf + (v·∇)f
dt
Applied to the velocity
(1.3)
d
v(x(t), t)) = ∂tv + (v·∇)v
dt
then Navier-Stokes becomes Dtv = − ρ1 ∇p + ν4v
Dtv =
• The pressure gradient and the viscous Laplacian are taken in the Eulerian frame.
∂f
• I will often use:
= ∂tf = ft
∂t
2
Plane Shear Flow
Consider the following velocity field: u = (u(y, t, 0, 0)
• Only one velocity in the streamwise direction x: u
• Only one directional dependence: spanwise y
• Incompressibility automatically satified:
∇·u = ∂y u = 0
∂
u=0
• Advection goes to zero: (u·∇)u = u ∂x
• Pressure gradient simplifies to: ∇p = (∂xp, 0, 0)
∂2
• Viscosity only depends on y: ν4u = 2 u
∂y
• Therefore the momentum equation reduces to:
1
2
∂tu = − ∂xp + ν∂yy
u
ρ
P
Steady shear flow Assume: ∂xp = P,x =const, then solve ∂yy u =
ρν
P,x 2
y + Ay + B
Solution :u =
2ρν
2.1
Pouseuille Flow
• Channel flow
• Similar to 3D pipe flow.
p2 − p1
• P,x = ∂xp =
L
• Boundary conditions: u(0) = u(h) = 0
• These fix B and A:
B=0
• Solution is:
P,x
h
A=−
2ρν
P,x
u=
y(y − h)
2ρν
• Note: u > 0 when P,x < 0 or p2 < p1
2.2
Couette Flow
• Flow between moving plates
• NOT pressure driven:
P,x = ∂xp = 0
• Boundary conditions: u(0) = 0 and u(h) = U
• These fix B and A:
B=0
• Solution is:
U
A=−
h
U
u= y
h
2.3
Turbulent Shear Flow
2.3 Turbulent Shear Flow
• By increasing the Reynolds, then previous assumptions break down.
• All three velocity components are significant.
• All directions are significant.
• Sharp gradients at the walls.
• Plane Poiseuille flow is linearly unstable to infinitesimal 2D disturbances for
Reynolds based on channel half-width
of about Re ≈ 5000
• Experimentally: nonlinearly unstable
to finite amplitude 3D disturbances for
Re ∼ 300 − 600
• Couette flow is not linearly unstable.
• Experimentally:
Pouseuille.
about the same as
2.3 Turbulent Shear Flow
• By increasing the Reynolds, then previous assumptions break down.
• All three velocity components are significant.
• All directions are significant.
• Sharp gradients at the walls.
• Plane Pouseuille flow is linearly unstable to infinitesimal 2D disturbances for
Reynolds based on channel half-width
of about Re ≈ 5000
• Experimentally: nonlinearly unstable
to finite amplitude 3D disturbances for
Re ∼ 300 − 600
• Couette flow is not linearly unstable.
• Experimentally:
Pouseuille.
about the same as
• In engineering the viscosity ν is replace
by an eddy viscosity νe so that the laminar solutions can be used.
• But then one needs extra equations to
find νe.
3
Steady vortex flows
• Vorticity is the curl of the velocity ω = ∇ × u
• An alternate (rotation) form of the Euler equations comes from the vector identify:
• One obtains:
(u·∇)u = (∇ × u) × u + 21 ∇u2 = ω × u + 12 ∇u2
(3.1)
1
∂tu + ω × u = − ∇(p + 12 ρu2 + λ)
ρ
(3.2)
The pressure head ph = p + 21 ρu2 and λ is an arbitrary scalar.
• By taking the curl of the Navier-Stokes equation, the vorticity equation can be
derived.
h ∂u
i
+ ν4u
∇·u = 0
∇×
+ (u·∇)u =
− ρ1 ∇p
∂t
∂ω
+ (u·∇)ω =
∂t
T ime
Advection
(ω·∇)u
V ortex
Stretching
+
ν4ω
∇·ω = 0
V iscous
Damping
Incompressibility
(3.3)
Pressure is eliminated by taking the curl of the divergence ∇ × ∇p = 0.
Leonardo sketch of 3D vortices. Tornado.
Mount Etna, Sicily ejects vortex rings.
3.1
Bernoulli Theorem
• For steady, ideal flow the pressure head is constant.
Steady :
ω × u = −∇H where
∂tu = 0
Ideal :ν = 0
1
H = p + 21 u2 + λ
ρ
(3.4)
• Take scalar product with u and use the vector identity: u·(ω × u) = 0
h
i
u· ω × u = −∇H ⇒ (u·∇)H = 0
• Recall that ∂t = 0 (steady), therefore (u·∇)H = 0
• or: If an ideal fluid is a steady flow, then H is constant along streamlines
• H can be different constants for different streamlines.
• For steady, irrotational, ideal flow then ∇H = 0 everywhere.
• Therefore: If an ideal fluid is a steady, irrotational flow, then H is constant
throughout the fluid.
3.1.1
Applications of Bernoulli: Nozzle
• Assume the pipe is horizontal to gravity does not play a role: z 1 = z 2
• Assume different areas: A2 < A1, but the same mass flux: u1A1 = u2A2
This implies: u2 > u1
• Apply Bernoulli: H =
p 1 2
+ 2 u =const ⇒ p1 > p2
ρ
3.1.2
Flow down a slope
• This time gravity matters: H =
p 1 2
+ 2 u + z =const
ρ
And: z 1 > z 2
• On the free surface the pressure is everywhere atmospheric:
p1 = p2 = patm
• Therefore as z decreases: u increases: u2 > u1
3.1.3
Lift on an aerofoil
• Incoming flow is irrotational: ω ≡ 0 for x < A: H=const
• Mass flux is uniform in height.
• In the thin aerofoil approximation this implies: u1x=const between A and B
• Aerofoil shape implies: u1y increases.
q
• Therefore: |u| = u21x + u21y increases.
p 1 2
• Apply Bernoulli: H = + 2 u =const: p1 decreases.
ρ
• Therefore p2 > p1 ⇒ Net force is upwards: L=Lift.
3.2
Bernoulli for time dependent, irrotational (ω = 0)
• For an irrotational (ω = 0) flow the velocity field can be written in terms of a
potential φ
u = ∇φ
• Substituting this into the Euler equation and removing ∇ from both sides, one gets
p 1
∂tφ + + 2 (∇φ)2 + λ = f (t)
(3.5)
ρ
• One can assume f (t) =const without loss of generality. This is because φ is defined
only up to an arbitrary function of time t, which can always be chosen so that
f (t) =const.
• Therefore
p 1
+ 2 (∇φ)2 + λ = f (t)
ρ
• For an incompressible fluid this can be used to find p given u
∂tφ +
• When p is given by an equation of state, this can be used to find u.
(3.6)
4
Conservation of mass, energy and production of enstrophy
• Conservation of mass requires that the density ρ(x, t) obey
ρt + ∇·(ρu) = 0 or ρt + (u·∇)ρ + ρ∇·u = Dtρ + ρ∇·u = 0
(4.1)
If the flow is incompressible ∇·u = 0 then Dtρ = 0 or density does not change in
time.
• Energy conservation. Define the kinetic energy density as E(x, t) = 12 ρu2. Its
equation can be gotten as follows: (Use: −∇·(up) = −u·∇p − p(∇·u))
i
h ∂u
1
+ (u·∇)u = − ρ ∇p +
ν4u
u·
∂t
∂E(x)
+ (u·∇)E(x) = − ρ1 ∇·(up) −
ν(∇u)2
∂t
T ime
Advection
P ressure
Dissipation
T ransport
+ν∂j (u∂j u)
(4.2)
Dif f usive
T ransport
• The two transport terms are pure divergences:
Therefore they are conservative except possibly at boundaries.
• There is only one dissipative term:
Converts fluid kinetic energy into heat.
• I will discuss gravity and conservative forces when I get to the Boussinesq approximation.
4.1
Effect of boundaries on Euler and Navier-Stokes
• There are two boundary terms:
• First rewrite u·∇E + u·∇ ρ1 p as
1
∇· u( p + 12 |u|2) = ∇·(uH)
ρ
(4.3)
• Then integrate over a FIXED volume Ω with surface ∂Ω
Z
Z
Z
Z
∂t E(x, t)dV = − ∇·(uH)dV − ν(∇u)2dV + ν∇(u∇u)dV (4.4)
Ω
Ω
Ω
Ω
R
• Apply divergence theorem to the integrals of divergences: Ω ∇·f dV =
Z
Z
Z
∂t E(x, t)dV = −
uH·dS − + ν
(u∇u)·dS
Ω
∂Ω
∂Ω
Z
where = − ν(∇u)2dV is the rate of energy dissipation
R
∂Ω f ·dS
(4.5)
Ω
• In Euler there are free-slip boundary conditions where the normal velocity at a
surface is zero: u·dS = 0
R
– Implies ∂t Ω E(x, t)dV = 0, energy conservation.
• In Navier-Stokes there are no-slip or rigid boundary conditions where u = 0 on
∂Ω.
R
– Implies ∂t Ω E(x, t)dV = −, internal energy dissipation.
4.2
Enstrophy production by vorticity equation
Just as for the Navier-Stokes equation for velocity we can create a budget equation by
taking the dot product of ω
~ with the entire equation.
∂ω
ω
~·
+ (u·∇)ω = (ω·∇)u +
ν4ω]
∂t
(4.6)
∂Ω(x)
+∂j (ω∂j ω)
+ (u·∇)Ω(x) =
ωieij ωj
−
ν(∇~ω )2
∂t
T ime
Advection
Enstrophy
Dissipation Dif f usive
P roduction
T ransport
∂ui ∂uj
eij = 12
+
is the strain and Ω(x) = 21 |ω(x)|2 is the enstrophy density
∂xj ∂xi
• Equation is similar to, but not identical to energy equation.
• Enstrophy equation has a production term instead for pressure transport.
• Analogy in line: vortex lines instead of streamlines.
dx/ds dy/ds dz/ds
=
=
ωx
ωy
ωz
• A vortex ring or a tornado marks the path of the vortex line.
(4.7)
4.3
Conservation in Two Dimensions
u = (u, v, 0)
ω = (0, 0, ω)
Consider the 2D flow:
• Strong differences with three-dimensional vorticity and enstrophy equations:
• There is also only one component of vorticity.
• The two-dimensional vorticity equation is a scalar equation.
∂ω
+ (u·∇)ω = ν4ω
∂t
(4.8)
• It is moved around exactly as a scalar such as temperature.
• For ν = 0: Vorticity is a Lagrangian invariant:
It is conserved along fluid paths: Dtω = 0
• Therefore: any function of vorticity is a Lagrangian invariant:
Dtf (ω) ≡ (∂t + (u·∇))f (ω) = 0
Z
Integrating: ∂t
f (ω)dA = 0
(4.9)
• In particular:
The inviscid vorticity equation conserves ENSTROPHY:
Z
2
1
|ω|
dA
Ω=
2
Z
• In addition to ENERGY:E =
(4.10)
2
1
|u|
dA
2
• Vortex stretching and enstrophy production DO NOT EXIST in
2D.
Z
(ω·∇)u = 0
dAωieij ωj ≡ 0
• Thus, the inviscid two-dimensional fluid equations conserve
Two quadratic invariants ENERGY E and ENSTROPHY Ω.
• This strongly influences the type of flow.
• This is one reason that in 2D:
Vorticity tends to form isolated point-like structures.
• And strongly influences nearly 2D systems like geophysical flows.
4.4
2D streamfunction
• In general a two-dimensional vector field can be written as a combination of solenoidal
and potential components
u=∇
∇φ
| ×
{zψ ẑ} + |{z}
solenoidal
(4.11)
potential
• Apply incompressibility: ∇·u = ∇·(∇ × ψ ẑ + ∇2φ = 4φ = 0
• Therefore: φ can be at most a linear function of position and we will take it to be
zero.
• Furthermore vorticity is:
ω = ∇ × u = ∇ × ∇ × ψ ẑ = −4ψ
By a vector identity.
• One can write
u = ∂y ψ
or u = ∇⊥ψ
v = −∂xψ
Example: Shear flow
• One component of velocity
u = (f (y), 0)
• Vorticity: ω = −f 0(y)
• Stream
Z function: u = ∂y ψ ⇒
ψ=
f (y)dy
4.5
Kelvin circulation
Define C(s, t) as a closed curve in 3-space that is parameterised by s and evolves in
time.
Consider an ideal flow in the presence of a conservative force (e.g. g = −∇λ) and let C(s, t) be a closed material contour. Then
the circulation
I
Γ=
u·dx
(4.12)
C (s,t)
is independent of time.
Kelvin’s circulation theorem
Many consequences of Kelvin’s theorem were preceded by work of Helmholtz and Cauchy.
In particular (Helmholtz, 1858) that
• Fluid elements initially free of vorticity remain free of vorticity.
• Vortex lines move with the fluid.
Proof:
• Consider
dΓ
d
=
dt
dt
I
I
u·dx =
Dtu·dx
C (s,t)
C (s,t)
The time derivative of dx and DtC(s, t) do not need to be considered due
to properties of the time derivatives of materials volumes to be considered in
Continuum Mechanics.
p
+ λ . Then
• Integrate using Euler’s equation: Dtu = −∇
ρ
I
dΓ
p
p
=−
∇
+ λ ·dx =
+λ
dt
ρ
ρ
C (s,t)
C
where [. . . ]C denotes integration over one loop of C.
• Because p and λ are singled-valued functions [. . . ]C = 0
Remarks:
• Constant density is not essential:
Kelvin’s established this results for compressible fluids as well.
• It does not require that C be simply connected.
This will be important for vortex shedding.
• It applies only to inviscid fluids or where viscosity is negligible.
I
• Now apply Stokes theorem:
Z
u·dx =
Z
(∇ × u)·ndS =
ω·ndS
C
S
S
where S is a surface bounded by the curve C. The n are normals to that surface.
• Kelvin’s circulation theorem is true in three dimensions,
but the consequences are more obvious in two dimensions.
• In 2D, Euler advects circulation exactly as any density.
5
2D vortices
Circular vortex
• Consider a vortex in cylindrical coordinates.
• Because ur = 0 and ∂θ f (r) = 0:
it is stationary:
Dtω = ∂tω + ur ∂tω +
uθ
∂θ ω = 0
r
• Any f (r) is a solution.
• Find uθ = g(r) using Stokes theorem:
I
Z
Z
u·dx =
ω·ndS =
ωdS
C(r)
S(r)
ZS(r)
r
⇒ uθ ·2πr =
= 2π
f (s)sds
0
Divide
by
1
⇒ uθ = g(r) =
r
2π
Z
r
f (s)sds
0
Z
Azimuthal velocity: uθ = −∂r ψ ⇒ Stream function: ψ(r) = −
g(r)dr
Rankine vortex

 Cr, r < a
uθ = Ca2

, r ≥ a.
r
(5.1)
This corresponds to a vorticity distribution of ω =
2C, r < a
0, r ≥ a.
(
Stream function : ψ =
−Cr2/2
r<a
−Ca2 log(r/a) r ≥ a
C is chosen so that ψ is continuous at r = a.
(5.2)
Point vortex
Delta-function distribution of vorticity:
ω = Γδ(x)δ(y)
(5.3)
Γ
uθ =
(5.4)
2πr
Irrotational flow everywhere except r = 0
Stream function : ψ = −
Γ
log(r)
2π
(5.5)
• The vorticity field can be replaced by point vortices with discrete
circulation.
• Consider the motion of two point vortices of the same sign and equal strength.
Their respective motion revolves about a common centre of motion between them.
• Next: two vortices of opposite sign and the same strength. These will
propagate together in the direction perpendicular to the cord that separates them.
Vortex dipole
Two opposite point vortices propagating in
time.
ω = Γδ(x−U t)δ(y−d/2)−Γδ(x−U t)δ(y+d/2)
Γ
• Centerline velocity: U =
2πd
• Vorticity moves with the fluid.
• Velocity at each vortex is induced by the
other.
Stream function:
Γ
Γ
2
2 1/2
2
2 1/2
ψ = − log (x − U t) + (y − d/2)
log (x − U t) + (y + d/2)
+
(5.6)
2π
2π
“Gas” of point vortices
ω(x) =
N
X
Γiδ x − xi(t)
(5.7)
i=1
X ẑ × (xi − xj )
d
x i = ui =
Γj
dt
|xi − xj |2
(5.8)
i6=j
Stream function :
• Each vortex moves under the collective action of all the other vortices.
X Γi
log |x − xi|
(5.9)
ψ=−
2π
i
• 2D Euler can be replaced by Vortex-particle codes.
• The major difference between a ”vortex” particle code and a grid-point calculation
is that with vortices the velocity field is generated by the point vortices.
• Numerical codes using this method are excellent for showing the behaviour of
coherent structures that are observed in experiments in shear flows.
5.1
Complex potential
• Let us generalise the stream function using a
Complex potential: χ = φ + iψ
Complex velocity:
u − iv
Complex position: z = x + iy
∂y ψ
u
∂xφ
=
=
Then ∂z χ =
−i∂y φ
i∂xψ
−iv
• These are Cauchy-Riemann equations.
• If the flow is incompressible: ∇·u = 0 ⇒ 4φ = 0
• If the flow is irrotational: ∇ × u = 0 ⇒ 4ψ = 0
Cauchy, Riemann, Laplace
(5.10)
5.1.1
Examples of potential flow
• Uniform flow at an angle α:
(u, v) = U0(cos α, sin α) then
∂z χ = U0(cos α − i sin α) = U0e−iα ⇒
χ = U0ze−iα
From this we can determine the potential
and stream functions:
φ = Re(χ) = U0(x cos α+y sin α)
ψ = Im(χ) = U0(y cos α−x sin α)
Pure strain near a stagnation point:
u=
αx
v=
−αy
w=
0
φ=
0.5α(x2 − y 2)
ψ=
0.5α(xy)
Show :
1 2
χ = αz
2
(5.11)
• Point vortex can also be a potential flow.
– Flow is irrotational everywhere except r = 0.
– In this case the potential is not single-valued.
– Therefore, exclude r = 0 and allow a multiply-connected domain.
Γ
φ= θ:
or θ + 2nπ
2π
Γ for closed loops that encompass r = 0
Γ=
0
otherwise
Γ
Recall: ψ = − log r. Therefore:
2π
Γ
χ = φ + iψ = (θ − i log r)
2π
−iΓ
=
(log r + iθ)
2π
−iΓ
=
log z
2π
I
I
Z 2π
dχ
Γ
Note :
dz = ∂z φdz =
dθ = Γ
dz
θ=0 2π
(5.12)
• Irrotational flow past a cylinder
2
a
– Define: χ = U0 z +
z
– Consider the circle zz = a2,
(z =complex conjugate of z)
a2
– then z =
and
z
– If ψ ≡ 0 or constant on |z| = a then
– ⇒ |z| = a is a streamline.
Proof:
χ = U0
a2
z+ 2
a /z
= U0(z + z)
= 2U0x = φ + iψ
φ = 2U0x and ψ = 0
(5.13)
U0a2
• The secondary
flow is a
z
”vortex dipole”
• Two vortices of opposite sign
placed infinitesimally close to
one another.
5.1.2
Examples of analytic functions
• What types of functions can be used in this way?
• Only analytic functions:
–z
– 1/z
– zn
– log z
– Any combination of the above.
– No other choices.
• This can directly treat flow and lift around cylinders.
• What about other bodies?
• Conformal mapping
6
Classical aerofoil theory
(out of Acheson)
1894 F.W. Lanchester: ‘The soaring of birds and the possibilities of mechanical flight’
– Meeting of the Birmingham Natural History and Philosophical Society.
– Introduces circulation theory of lift.
1987 Written version is rejected by the Physical Society.
1901 Wright brothers fail in glider design.
1902 Kutta: ‘Lifting forces in flowing fluids’.
– Solution of 2D irrotational flow past a circular arc.
– Circulation around a surface.
– Question of finite velocity at the trailing edge.
– Connection between circulation and lift is recognized.
1903 17 December: Wright brothers fly.
1904 Prandtl: Boundary layer paper to the: 3rd International Congress of Mathematicians, Heidelberg.
ukovski)
1905 Zhukovski lift theorem. (Zhukovski=Joukowski=
1907 Lanchester publishes Aerodynamics
6.1
Irrotational flow past a cylindrical cylinder
2
a
z
iΓ
log z
2π
(6.1)
For arbitrary Γ:(see Acheson
(4.38))
a2
(6.2)
Show : ur = U0 1 − 2 cos θ
r
Consider : χ = U0 z +
−
• Note: ur = 0 at r = a
2
a
Γ
and uθ = −U0 1 + 2 sin θ+
(6.3)
r
2πr
• One can add arbitrary axially symmetric irrotational flow that decays to 0 at
infinity.
• This is formally equivalent to placing a
point vortex place at r = 0.
• At (x, y) on the surface, or r = a in this
case, we want:
– un = u·n̂ = 0 and
– ut = u − un to be free-slip and
• Stagnation points move from sides,
change.
to bottom, to beneath.
• Note the positions of the stagnation
Summary of flow past cylinder
• Consider: (Acheson (4.32))
2
a
iΓ
χ = U0 z +
log z
−
z
2π
• For arbitrary Γ: (Acheson (4.38))
2
a
• Show: ur = U0 1 − 2 cos θ
r
– Note: ur = 0 at r = a
2
a
Γ
• and uθ = −U0 1 + 2 sin θ +
r
2πr
• One can add arbitrary axially symmetric irrotational
flow that decays to 0 at infinity.
• This is formally equivalent to placing a point vortex
place at r = 0.
• At (x, y) on the surface, or r = a in this case, we
want:
– un = u·n̂ = 0 and
– ut = u − un to be free-slip and change.
• Note the positions of the stagnation points.
6.2
Force on the cylinder and L=lift
• The surface is a streamline, therefore: p + 21 ρu2 =C
• At r = a: ur = 0, so only uθ matters:
Γ
uθ (r = a) = −2U0 sin θ +
2πa
(6.4)
• pressure is
p(r = a) = C − 21 u2θ = C − 2U02 sin2 θ +
U0 Γ
sin θ
πa
(6.5)
Consider a steady flow (uniform at ∞ with speed U in
the x-direction) past a 2D body of arbitrary cross-section. Let the circulation
around the body be Γ. Then the force F acting on the body will have the following
components: (Acheson (4.41))
(6.6)
F = (0, −ρU0Γ)
Z 2π
Proof: (simple)
• Net force in x: Fx = −
p cos θ(adθ) = 0.
Zhukovski lift theorem
0
Because this is symmetric about θ = π/2 (vertical). But:
Z 2π
• Net force in y: Fy = −
p sin θ(adθ) = −ρU0Γ.
0
6.3
Conformal mapping: general shapes
• Choose Z = f (z) and an inverse z = F (Z), then
• X(Z) = χ{F (Z)} is an analytic function of Z.
• We can write: Z = X + iY , X(Z) = Φ(X, Y ) + iΨ(X, Y ) and (Acheson (4.48))

 

∂Φ
∂Ψ
 ∂X   ∂Y 

 

u∗(X, Y ) = 
(6.7)
=

 ∂Φ   ∂Ψ 
−
∂Y
∂X
• Definition: A conformal map is a transformation Z = f (z) such that preserves
local shapes and angles, if not their magnitude or orientation.
(δz)n (n)
f (z0) + O(δz)n+1
• Consider the Taylor expansion: δZ =
n!
and two small elements: δz1 and δz2.
Then the angles obey: arg(δZ2) − arg(δZ1) = n[arg(δz2) − arg(δz1)]
• If linear (n = 1), then the magnitude of the angle is preserved.
1
c2
1
1 2
2 2
• Zhukovsky transformation: Z = z + (4.52)
z = 2 Z + ( 4 Z − c ) (4.53)
z
2
Z − 2c
z−c
=
(exercise 4.8)
Note that as (z, Z) → ∞, Z → z
Z + 2c
z+c
c2
6.4 Zhukovsky transformation: Z = z +
z
• I will now consider how to transform from:
– Flow at an angle around a cylinder with lift into:
– Flow around an aerofoil at an angle of attack.
Symmetric aerofoil: c = a where a= radius of the pre-mapped circle.
• A cylinder with
– Displaced by λ
– At an angle α
– With a virtual point vortex:
circulation Γ
• Following Nazarenko I will use (χ, X)
instead of (w, W ) for the mapped and
true complex potentials.
• Resulting complex potential (after 4.59
in Acheson) is:
2
(a + λ) iα
iΓ
χ = U0 (z + λ)e−iα +
e −
log(z + λ)
(6.8)
z+λ
2π
• Further special case: λ = 0, a flat plate.
GOAL: Calculate lift with these representations.
• Apply Zhukovski transformation, but do not calculate new X
• Instead calculate new ∂Z X = U + iV
From Acheson (4.60)
)
−1 ( "
2 #
2 −1
∂χ dZ
iΓ
a
a
+
λ
∂Z X =
= U0 e−iα −
eiα −
1− 2
∂z dz
z+λ
2π(z + λ)
z
1
(6.9)
1
1 2
2 2
• Substitute: z = 2 Z + ( 4 Z − a ) to get U and V .
a2
• Problem: Solution is singular where 1 − 2 = 0: z = ±(a, 0)
z
• Because we have displaced by λ, z = −a is inside the original circle.
– So we can ignore z = −a
• What about z = a?
– Resolution of the singularity problem will give us lift
• Steps:
1.) Kutta-Zhukovski hypothesis or condition:
Find critical circulation: Γk that removes singularity.
2.) Blasius theorem: How to find force if given aerofoil and χ.
Applications.
3.) Kutta-Zhukovski lift theorem: Lift/span=L = −ρU Γ
General case: c 6= a, λ 6= 0
6.5
Step 1: Kutta-Zhukovski hypothesis:
The critical value of circulation Γk for a wing in a flow of speed U0, at an angle
of attack α and length L is Γk = −πLU0 sin α
• The determining factor: choose Γ such that there is no singular at the trailing edge.
Example: symmetric aerofoil.
– Set z = (a, 0), this is true if the numerator=0: (Acheson (4.61))
2iU0 sin α +
iΓ
=0:
2π(a + λ)
Γ = −4πU0(a + λ) sin α
(6.10)
• Find L:
– In the original circle the stagnation points go towards z0 = z + λ = ±a.
– In the rotated and displaced frame this corresponds to z = (±a − λ)e−iα
– In the physical coordinates Z, the separation between these points is:
2
2
a
a
− −a − λ +
L = f (a − λ) − f (−a − λ) = a − λ +
a−λ
−a − λ
2a3
= 2a + 2
→ 4a as λ → 0
2
a −λ
– So 4(a + λ) → 4a → L as λ → 0 and Γk ≈ −πU0 sin α
6.6
Step 2: Blasius theorem:
Let there be a steady flow with a complex potential χ(z) about some fixed body which has
a closed contour C at its boundary. Then
I
(4.62)
Fx − iFy = 12 iρ (∂z χ)2dz
C
Proof:
• Define the following quantities using θ:
the angle of the tangent to C with the xaxis.
- A small segment dz = |dz|eiθ = dx + idy of the contour C.
- The force on dz: δF = δFx − iδFy = −p(sin θ + i cos θ)|dz| = −pie−iθ |dz|
- C is a streamline so u − iv = qe−iθ = dχ/dz where q 2 = (dχ/dz)2 e2iθ .
• Use Bernoulli: p + 21 ρq 2 = k or −p = 12 ρq 2 − k to get:
iδFx − iδFy
= ( 21 ρq 2 − k)ie−iθ |dz|
2
2
dχ
dχ
1
1
2
iθ
−iθ
substituting in q
e |dz| − kie |dz| = 2 iρ
dz − ikdz
2 iρ dz
dz
I
I
• The integral of k is 0:
kdz = k (dx − idy) = 0, leaving the result.
C
C
Applications of Blasius
Uniform flow past a cylinder
• Already done by explicit integration in (4.41) (my sec 6.2): Fy = −ρU0Γ
• To use Blasius start with the complex potential (4.32) (my sec 6.1):
2
a
iΓ
χ = U0 z +
−
log z
z
2π
2
iΓ
dχ
a
= U0 1 − 2 −
giving
• Use
dz
z
2πz
2
2
4
2
2a
a
iU0Γ
Γ
dχ
a
= U02 1 − 2 + 4 −
1− 2 − 2 2
dz
z
z
πz
z
4π z
(A4.32)
(6.11)
iU0Γ
• Substituting into (4.62), the ONLY term with 1/z is −
:
πz
I
iU0Γ
1
Fx − iFy = 2 iρ −
dz
πz
C
• By simple residue calculus:
I
iU0Γ
iU0Γ
1
−
= iρU0Γ
iρ
dz
=
2πi
−
2
πz
π
C
• Therefore −iFy = iρU0Γ or Fy = −ρU0Γ
(6.12)
6.7
Step 3: Kutta-Zhukovski theorem:
Consider a steady flow (uniform at ∞ with speed U0 in the x-direction) past a 2D
body of arbitrary cross-section. Let the circulation around the body be Γ. Then
the force F acting on this body will have the following components,
Fx = 0 ← (D’Alembert’s Paradox)
Fy = −ρU0Γ
Proof: The idea now is to show this for any complex potential.
• Therefore, generalise the gradient of the complex potential applications of Blasius
a1 a2
to a Laurent series where ∂z χ = U0 + + 2 + . . .
z
z
I
(∂z χ)2dz is: 4πiU a1 so that −iFy = −2πρU a1
• Again, the only surviving term in
C
• But we want to give a1 in terms of Γ
• Recall the point vortex potential and how the integral of the potential around a
I
contour is the circulation, then I
dχ
a1
Γ=
dz =
dz = 2πia1
dz
z
C
c
• Substitute above and get: Fx − iFy = iρU0Γ
• Total force: combine Kutta-Zhukovski hypothesis and theorem:
Fy = −ρU0Γ = −ρU0(−πLU0 sin α) = πU02ρL sin α
(6.13)
6.8
D’Alembert’s Paradox: about 1748
• In our ideal solutions (no drag) there is:
• Net circulation and lift.
Why is this a paradox?
• Start with no flow:
pressures upstream p1 and downstream p2 at large distances from the object are
the same.
• Net circulation is zero Γ = 0.
• Suddenly start moving an object in a fluid.
• Experimental:
– Object suddenly has net circulation Γ 6= 0.
− And there is drag: Fx 6= 0
• Where are the missing −Γ and drag? This is the paradox.
• Example: for flow around a cylinder (Sec. 6.2),
U0 Γ
sin θ
πa
At (x±, y) = (±a, 0), sin θ = 0, so p+ = p− = C and u+ = u− = U0: drag=
Z
Z
(Acheson (4.75))
D=
(p+ + ρu2+)dS −
(p− + ρu2−)dS = 0
(6.14)
p = C − 2U02 sin2 θ +
S+
S−
• The source of the drag is the boundary layer.
• This forms separated vortices.
• The momentum loss (drag) and −Γ are in the separated vortices.
• ΓA = −ΓB , net circulation inside C: ΓC = ΓA + ΓB = 0
• Extend C to ∞, then outside C in (4.75):
p2 < p1 and there is drag.
• How much drag does the boundary layer create?
– It is small if the boundary layer is laminar,
– Large if turbulent, or:
– Can change counterintuitively if there is separation or reattachment.
– Left: Turbulent boundary layer. Right: Reattachment at higher velocities.
– Reattachment: Vortices at top and bottom on left do not appear in right.
– Drag crisis: Counterintuitively, even
when velocity increases, due to reattachment the drag decreases.
• Large parts of modern aerodynamics is concerned with separation and turbulent
boundary layers: we don’t know how to properly calculate these.
Trailing vortices
• Circulation is actually shed from wing
tips.
• This forms trailing vortices, that we see
as contrails.
• These can be hazardous to following aircraft.
Viscous solutions
7
7.1
Exact vortex solutions
• Viscous decay of a 2D point vortex.
Γ0 −r2/4νt
Let ω =
e
4πνt
Solves: ∂tω = ν4ω = r−1∂r (r∂r ω)
t=0:
Γ0 −r2/4νt
ω(t = 0) = lim
e
= Γ0δ(x)
t→0 4πνt
Preserves circulation:
Z
Z
Γ = ωdA = 2π
∞
Z
∞
ωrdr = 2π
0
Let z = r2/4νt then dz = r/2νt so Γ = Γ0
0
Z
Γ0 −r2/4νt
e
rdr
4πνt
∞
e−z dz = Γ0
0
• (Note propagating image vortex example problem.)
(7.1)
(7.2)
• 3D example: Burgers vortex.
Let f (r) = 1 − e−αr
2 /4ν
uz = αz
uz = αz
ux = − 12 αx −
uy =
− 21 αy
Γy
f (r)
2
2πr
Γx
f (r)
+
2πr2
ur = − 12 αr
or
(7.3)
Γ
uθ =
f (r)
2πr
– Flow is incompressible: ∇·u = 0
– ur and uz with Γ = 0 is a 3D potential flow: φ = 12 αz 2 − 14 r2
αΓ0 −αr2/4ν
– Vorticity is: ω =
e
ez
4πν
7.2
Stokes (creeping) flow.
• Re = U L/ν << 1
• High viscosity, small length scales or
small velocites.
• No separation, no turbulence.
• Important for lubrication.
• Stokes flow: 4ω = 0
• Add ideal assumption 4ψ = −ω
• Implies you must solve 42ψ = ∇4ψ = 0
• 4th-order equation. This generally true for all analytic solutions of the linearised
Navier-Stokes equations.
7.3
Hele-Shaw flow/Darcy’s Law
• Consider Couette flow again. This time a general 2D ∇p.
1
u2D (z) = − z(h − z)∇p
2µ
Z h
1
h2
• Integrate across the channel to get: u =
dhu(z) = −
∇p
h 0
12µ
• For porous media, assume h is very small and in general u = −C∇p
• Laboratory example
– Two glass plates with a narrow gap h
– h << 1 ⇒ Re = U h/ν << 1
– Ink added to glycerol: interface between
two liquids of different viscosity.
– Viscous fingers.
(7.4)
7.3.1
Spilt beer:
• Axi-symmetric and surface h(t) decreases.
– There is some vertical velocity at the
outer edge of the drop.
If u = ur (r, z, t)er + uz (r, z, t)ez by
continuity:
1
∂r (rur ) + ∂z uz = 0
r
∂r ur ∼ ur /a
∂z uz ∼ uz /h
∂r p = νρ∂zz ur
∂z p = νρ∂zz uz
a νρ
νρ
∂r p ∼ 2 ur ∼
uz
h
h h2
⇒ ∂r p >>
νρ
∂z p ∼ 2 uz
h
∂z p
• This is similar to the scaling of the aerodynamic boundary layer.
(7.5)
7.3.2
Flow around a flat plate:
• Potential, so use Zhukovsky transform equation
2
2
cos α 1 − az 2 − i sin α 1 + az 2
U − iV = U0
2
1 − az 2
(7.6)
• There is a singularity at the ends: Z = X + iY = ±2a
– and there is no Γ to resolve it.
– Nonetheless, photographs reveal that the flow resolves this somehow.
– Use Acheson 4.6a/4.6b to illustrate flow.
• Velocity jump and vortex sheet:
a
– Recall: true coordinates Z = X + iY = z +
z
– Parameterise position on the flat plate by: z = aeiθ so Z = X + i0 = 2a cos θ
cos θ
– Show: U − iV = cos α − sin α
+ i0
sin θ
– There is a jump in U -velocity across the plate:
cos θ
∆U = U+ − U− = −2 sin α
sin θ
– While there is no net circulation within the flat plate, satisfying Darcy’s law,
there is a vortex sheet.
7.3.3
Porous media:
• C is a constant dependent upon the specific medium and fluid:
3
Vs2
C=
(1 − )2 µS 2k
(7.7)
• = Vv /(Vs + Vv ), Vs, Vv are the volumes of the solids and voids respectively, S is
the surface area between the two and k is the Kozeny constant.
• Assuming the voids behave like capillary tubes, Vv /S = Re/2 where Re is the grain
size.
• Given ≈ 0.3 − 0.5, k = 4 − 6, Re = 0.25 − 2mm for different types of sand
and ν = 10−6m2/s or µ = 10−9kg/m-s for water (for air µ =) then C can be
determined. (≈ .1Re2/µ = 10−4)
• Flow will always be irrotational: ∇ × u2D = ω = 0.
– u always obeys potential flow.
– Flow around an object will never have internal vorticity.
– There will never be separation at the stagnation points.
• Application to seepage (ground flow) underneath a resevoir.
Consider a reservoir whose bottom depth approaching a dam is given by D(x) =
D0 − ax or the height above some reference is z = z0 + ax.
– Assume that the hydrostatic pressure at the bottom of the reservoir goes as
pb = ρgD(x) = ρg(D0 − ax) = p0 − ρgz = p0 − ρgz0 − ρgax
where ρ = ρw
is the density of water.
– Assume that the pressure at a height z in the porous material underneath the
reservoir goes as p = p0 − ρg(z0 + ax) + ρg(1 − )(z0 + ax − z) + ρa2z where the
density of the porous material is ρg = ρw Vs/(Vs + Vv0) = (1 − )ρ and << 1.
The terms are:
a) Pressure from the reservoir: p0 − ρg(z0 + ax)
b) Pressure from the saturated soil underneath the reservoir: ρg(1−)(z0 +ax−
z)
c) An additional term is to ensure conservation of mass. This is negligible. ρa2z
– This simplifies to: p(x, z) = p0 − ρgz0 − ρgax − ρg(1 − )z + ρga2z
– What is the velocity as a function of x and z?
7.4
Impulsively moved plate
• Another example of a solution in similarity variables:
• Another example of a solution in similarity variables: η = y/(νt)1/2
2
u: heat equaNS reduces to: ∂tu = ν∂yy
tion
• Advection u∂xu = 0 because u is small.
• No pressure because equation is invariant with respect to translation in x.
• If u = f (η) then
y
2t(νt)1/2
0
0
1 1
00
∂yy u = ∂y f (η)∂y η = f (η)
= f (η)
νt
(νt)1/2
• Reduces to:
∂tu = f 0(η)∂tη = −f 0(η)
f 00 + 12 ηf 0 = 0 implies
0
• f = Be
−η 2 /4
Z
η
−s2 /4
e
. Integrating again gives: f = A + B
0
ds
• Now apply initial condition: u(y, 0) = 0, y > 0
u(0, t) = U,
t>0
• Boundary conditions:
u(x = ∞, t) = 0, t > 0
Z η
2
e−s /4ds then
• If f = A + B
0
Z η
1
2
e−s /4ds
u(η) = U 1 − 1/2
π
0
(7.8)
• This solution is self-similar:
1/2
y2
y1
t2
– Solution at t2 > t1 is u
=u
where y2 = y1
t1
(νt2)1/2
(νt1)1/2
– New y2 is stretched y1.
U
−y 2 /4νt
• Be able to find the vorticity: ω = −∂y u =
e
(πνt)1/2
• Note: The characteristic penetration depth is: (νt)1/2.
8
Aerodynamic boundary layers
Prandtl, August 1904, Third International Congress of Mathematicians, Heidelberg,
‘On the motion of fluids of very small viscosity’.
Published 1905 in 7.5 pages with 2 pages of photographs.
... if the viscosity is very small, and the path of the fluid along the wall is
not too long, then the velocity will assume its usual value at a very short
distance from the wall. In the thin transition layer, the sharp changes of
velocity produce notable effects, despite the small coefficient of viscosity.
These problems are best tackled by making an approximation in the governing
differential equation. If µ is taken to be of second order in smallness, then the
thickness of the transition layer becomes of the first order in smallness, and
so too do the normal components of velocity. The pressure different across
the layer may be neglected, as may be any bending of the streamlines. The
pressure distribution of the free fluid will be impressed on the transition layer.
For the two-dimensional problems, with which I have hitherto been solely
concerned, we get, at any particular position the differential equation
∂u
∂u
∂ 2u
dp
=− +µ 2 ,
ρ u +v
∂x
∂y
dx
∂y
to which must be added ∇·u = 0
• Conditions:
– The viscosity is very small.
– The path of the fluid along the wall is not too long.
– The velocity will assume its usual value very close to the wall.
Usual value: free-stream velocity. (very short distance from the wall not very
close to the wall)
• Effect:
– In the thin transition layer,
– sharp changes of velocity with notable effects,
– despite the small coefficient of viscosity.
• Approximation:
– µ viscosity order 2
– δ transition layer thickness (BL thickness) order . (δ << L, x-distance)
– v normal order – dp/dy neglect, as well as any bending of streamlines
– dp/dx of the free-fluid is impressed upon the BL and depends only upon x.
– Then steady-NS reduces to
∂u
∂u
dp
∂ 2u
ρ u +v
=− +µ 2 ,
(8.1)
∂x
∂y
dx
∂y
Let us assume that, as is usual, dp/dx is given as a function of x, and further,
that the velocity u is given as a function of y at some initial value of x. Then
we can determine numerically, for each u, the associated ∂u/∂x, and with one
of the known algorithm we can then proceed step-by-step in the x-direction.
• Approximation:
– U ∼ u ∼ O(1)
L ∼ O(1)
– µ viscosity order 2
− δ (BL thickness) order .
– v normal order – dp/dx: Assume determined by Bernoulli and free-stream velocity U (x) on a
streamline at the edge of the boundary layer:
1 dp
dU
1 2
p + 2 U =constant:
=U
ρ dx
dx
∂u
∂u 1 dp
∂ 2u
∂ 2u
u
+
v
=
−
+ ν 2 + ν 2
∂x
∂x
ρ
dx
∂x
∂y
1
u2
12
uv
1
∼O
∼1
∼O
∼1
U 2 ∼ O(12)
2 2 ∼ 2
2 2 ∼ 1
L
1
δ
1
∂u
∂v
+
=
0
∂x
∂y
u
v
1
∼O
∼1
∼O
∼1
L
1
δ
However, a difficulty ... singularities that appear on the fixed boundary.
The simplest case of the flows under discusssion is that of water streaming
along a thin flat plate. Here, variable reduction is possible and one can write
u = U h(y/x1/2).
dp
dU
∼U
= 0 giving
dx dx
∂u
∂u
∂ 2u
u +v
=ν 2
∂x
∂y
∂y
• Scaling variable will be η = y/g(x)
• Now do a flat plate where:
– It will be found that this is g(x) = (2νx/U )1/2.
– Replace x/U by t and this looks like the scaling variable for diffusion:
η = y/(2νt)1/2
• Use stream function: u = ∂ψ/∂y, v = −∂ψ/∂x, where ψ = U g(x)f (η)
00 0
00
000
f
f
g
f
u = U f 0(η)
∂xu = −U 2
∂y u = U
∂y2u = U 2
g
g
g
∂η
y
v = −ψ,x = −U g 0f + gf 0
= −U g 0g − f 0g 0 = U (ηf 0 − f )g 0
∂x
g
000
00
0
y
f
f
U
gg
Combine: −U 2f 0f 00 2 g 0 + U 2(ηf 0 − f )g 0 = νU 2 or f 000 +
f f 00 = 0
g
g
g
ν
U gg 0 00
f +
ff = 0
(8.2)
ν
• Need to choose U gg 0/ν and solve.
νx
1 2
0
– Choose U gg /ν = 1 or 2 g =
+d
U
– Choose d = 0 by matching to the outer solution (I will explain later). Get
000
f 000 + f f 00 = 0
– Solve this NUMERICALLY. Even in 1904 this 1D ODE could be done tediously
with a slide rule.
– Integrate along the plate and get the drag
p
√ 00
1/2 1/2 3/2
(8.3)
D/b = 2 2f (0)ρν L U = 1.1 µρLU 3
√ 00
where 1.1 is a constant 2 2f (0) found numerically.
By numerical integration of the resulting differential equation we obtain an
expression for the drag
q
D = 1.1 × b µρ`u30
(8.4)
where b is the breadth, ` the length of the plate, and u0 is the velocity of the
undisturbed flow relative to the plate. The velocity profile is shown.
For practical purposes, the most important result of these investigations is that in certain cases, at a point wholly determined by the external conditions, the flow separates from the
wall. (second figure) A fluid layer which is set into rotation
by friction at the wall thus pushes itself out into the free fluid
where, in causing a complete transformation of the motion, it
plays the same role as a Helmholtz surface of discontinuity. A
change in the coefficient of the viscosity µ produces a change n
the thickness
p of the vortex layer (this thickness being proportional to µ`/ρu), but everything else remains unchanged, so
that one may, if one so wishes, go to the limit µ → 0 and still
obtain the same flow picture.
Separation can only occur if there is an increase in pressure
along the wall in the direction of the stream.
Both the Navier-Stokes equations and the boundary layer equations are coupled,
nonlinear partial differential equations. But:
Due to Anderson: The Navier-Stokes equations have what mathematicians
call elliptic behavior. That is to say, the complete flow field must be solved
simulataneously, in accord with specific boundary conditions defined along
the entire boundary of the flow. (This is done through the pressure where
distant boundary conditions affect the local flow) In contrast, the boundarylayer equations have parabolic behavior, which affords tremendous analytical
and computational simplification. They can be solved step-by-step by marching downstream from where the flow encounters a body, subject to specified
inflow conditions at the encounter and specified boundary conditions at the
outer edge of the boundary layer. The systematic calculation yields the flow
variables in the boundary layer, including the velocity gradient at the wall surface. The shear stress at the wall, hence the skin-friction drag on the surface,
is obtained directly from those velocity gradients.
Steps in drag calculation:
• Calculate skin friction drag.
• Calculate point of separation.
• Calculate the form drag.
• Separation is related to:
– Curvature of the aerofoil.
– Angle of attack: too much angle, then separation, then stall.
– Adverse pressure gradient.
• History post-1905 (ignoring World Wars)
3/2
– Lanchaster (1907) independently found that the drag goes as µ1/2u0 and discussed separation.
– Blasius (1908) solves BL equations for a flat plate and a circular cylinder with
more accurate skin-friction drag.
– von Karman (1921) find momentum integral formulation.
– Lamb (1932) finally puts a 7 page section on boundary layer theory into his
book.
• Influence of Prandtl’s work on mathematics.
– First treatment of nonuniform asymptotic expansions of differential equations
that have a small parameter.
– Perturbative expansions that are well-behaved “uniformly valid” are relatively
straightforward.
– The boundary layer equations are the proto-typical example of expansions that
are not uniformly valid.
– This has led to “singular perturbation theory”.
Figure 1 from Physics Today. Pressure and shear-stress distributions
are responsible for the force exerted on a body in a fluid flow. The pressure (top) acts
normal to the surface; the shear stress (bottom) acts tangentially.
Figure 2 from Physics Today. A fluid flow may be viewed as comprising
two parts. In a thin boundary layer (blue) adjacent to the surface, the effects of
friction are dominant. Outside the boundary layer, the flow is inviscid. The blowup of
the boundary layer shows how the flow velocity v changes as a function of the normal
distance n. From zero at the surface to the full inviscid-flow value at the outer edge.
Figure 3 from Physics Today.
The boundary layer can separate from the top surface of an airfoil if
the angle of attack is greater than the
so-called stall angle. The upper dark
region that trails downstream from the
separation point is the remnant of the
boundary layer that originally formed
on the top surface of the airfoil.
The lower dark region that trails downstream from the trailing edge of the airfoil is
the remnant of the boundary layer over the bottom surface. When separated, these
two dark regions are called shear layers, and they form the upper and lower boundaries
of the separated flow region. Between the shear layers is a dead-air region. Due to
the considerable flow separation illustrated here, the lift of the airfoil is dramatically
reduced: the airfoil is stalled. The blowup shows the flow’s velocity profile above the
separation point.
9
Compressible equations, sound, gas dynamics
Recall the full compressible Navier-Stokes equations are (0.1):
∂ρ
+ ∇ · (ρv) = 0
∂t
∂v
ρ
+ v·(∇v) = µ4v + (λ + 13 µ)∇(∇ · v) − ∇p + f
∂t
∂θ
+ ∇θ · v = −p∇ · v + κ4θ + Φ
ρcv
∂t
• Assume a polytropic gas: p = (γ − 1)ρθ.
γ = cp/cv is the ratio of specific heats.
∂p
+ v · ∇p + γp∇·v + . . .
∂t
p
• Entropy is η = cv log γ
ρ
• Then
ρDtη = θ−1∇·(κ∇θ)
• If there is no thermoconductivity (κ = 0):
• In an isentropic fluid η =constant for fluid parcels.
(9.1)
• Sound using linearised equations.
– Assume isentropic flow: pρ−γ =constant.
– Divide all quantities into their mean values and first-order perturbations
u = u1,
p = p0 + p1,
−γ
– This implies p0ρ−γ
0 = (p0 + p1 )(ρ0 + ρ1 )
ρ1
p1
=γ
p0
ρ0
ρ = ρ0 + ρ + 1
p1
ρ1
−γ
≈ p 0 ρ0 1 + −γ
p0
ρ0
– Order terms must be zero so
s r
p0
∂p
– Define cs = γ =
then
p1 = c2s ρ1
and
ρ0
∂ρ S
– Write the order equations for u1 and ρ1:
∂ρ1
∂u1
ρ0
= −∇p1,
+ ρ0∇·u1 = 0
∂t
∂t
∂tp1 = c2s ∂tρ1
– Take the divergence of the u1 equation and use the ρ1 equation to replace ∇·u1:
∂
∂ρ1
ρ0 ∇·u1 = −4p1
∇·u1 = −
∂t
∂t
∂ 2 ρ1
– to get ρ0 2 = 4p1 and finally replace ρ1 with p1 to get
∂t
∂ 2 p1
2
=
c
(9.2)
s 4p1
∂t2
– In 1D this reduces to:
2
∂ 2 p1
2 ∂ p1
= cs 2
(9.3)
2
∂t
∂x
– and has the general solution of a supposition of waves traveling in two directions
p1 = f (x − cst) + g(x + cst)
| {z } | {z }
right travel
left travel
– cs is now identified as the speed of sound:
For 20◦C and standard atmospheric pressure (about 1000 millibars):
cs = 340ms−1=1224 km/s.
– Waves travel without change of shape: No dispersion.
– Particles themselves just oscillate back-and-forth.
(9.4)
• Dispersion
– In general f (x, t) and g(x, t) are wave packets.
Z
f = dk f˜(k) cos(kx − ω(k)t + δ(k))
Z
g = dkg̃(k) cos(kx + ω(k)t + δ(k))
2π
– The wavenumber k is related to the wavelength λ by λ =
k
– The frequency ω(k) and the phase δ(k) are wavenumber dependent.
For sound waves the frequency is linear: ω(k) = csk.
ω
– More generally the phase speed=cph = is k dependent.
k
The phase speed describes motion within the wave packet.
∂ω
– The group velocity cg =
∂k
This describes how the whole wave packet moves.
cg describes energy transport.
– When cph 6= cg , usually when ω is strongly k dependent:
Waves are dispersive.
It looks like waves are created at one end and go out the other.
The entire wave packet becomes broader.
This describes surface waves, which I might come to later.
(9.5)
• The compressible Navier-Stokes equations are “solved”:
– In the sense that we know that the smallest scale structures are shocks.
– Practical solutions are again an example of singular perturbation problems:
Local shock solutions must be matched to outer solutions.
• 1D and homoentropic: pρ−γ = a20
∂tu + u∂xu = −ρ−1∂xp
∂tρ + u∂xρ = −ρ∂xu
= (γa20ργ−1)1/2
– Define the variable: a = ±(γp/ρ)1/2
p ∂xρ 2
2 p ∂xa
2
p
−ρ−1∂xp = ∂x
+
= a∂xa +
=
a∂xa
ρ
ρ ρ
γ
γ − 1ρ a
γ−1
– Substitute in
2
∂tu + u∂xu = −
a∂xa
γ−1
2
{∂ta + u∂xa} = −a∂xu
γ−1
– Add with ±a to get
h
i
2
∂tu + (u + a)∂x u +
a = 0
Started with 2 equations
γ
−
1
h
i
2
∂tu + (u − a)∂x u −
a = 0
Ended with 2 equations
γ−1
• Characteristics
– Lines defined by u ± a are characteristics along which disturbances move.
2
–u±
a is conserved along characteristics.
γ−1
– Burgers’ equation: paradigm for shock formation:
∂tu + u∂xu = ν∂xxu
– If there is inflow from two directions the solution goes to
u = tanh(x) =
sinh(x)
cosh(x)
– Velocities profiles steepen until a discontinuity: a shock forms.
– With a narrow singular perturbation where viscosity dissipates energy: the
shock.
– a > 0 (a same sign as u) Shock: compression.
– a < 0 (a opposite sign as u) Rarefaction: expansion..
–
10
Buoyancy induced waves
• Now I will gradually add geophysical terms:
– Surface effects.
Shallow water equations.
– Rotation: 2Ωẑ × u: Coriolis acceleration.
−Ω2ẑ × ẑ × r: Centrifugal acceleration.
Rossby waves. Ekman boundary layers.
– Buoyancy: A linearised density or temperature.
Interaction of the vertical velocity with a mean gradient of this field.
A gravitational forcing on Navier-Stokes by this field.
Internal or gravity waves.
Either thermal convection or stratification.
– Shears.
Kelvin-Helmholtz instability.
– Rotation on a sphere
Geostrophic flow.
10.1
Deep water equations
• Surface waves on deep water
– Define the vertical position of a free surface as: z = η(x, t) and
F (x, z, t) = z − η(x, t) = 0. For a given particle on the surface:
∂F
+ (u·∇)F = 0
on z = η(x, t)
∂t
– F = 0 is on a sloping surface. Therefore: ∂xF = −∂xη and ∂z F = 1.
– Add ∂tF = −∂tη to transform DtF = 0 into (Acheson (3.18))
∂η
∂η
+u
=w
∂t
∂x
on z = η(x, t)
(10.1)
• Bernoulli equation
– Assume flow is irrotational: ω = 0 and u = ∇φ so: (u·∇)u = 12 ∇u2
– Assume a gravitational potential ρχ = ρgz
∂
p 1 2
– Then Euler’s equation becomes: ∇φ = −∇
+ 2u + χ
∂t
ρ
leading to (Acheson (3.19))
∂φ p 1 2
+ + u + χ = G(t)
∂t ρ 2
– G(t) is an arbitrary function of time alone.
(10.2)
∂φ p 1 2
Bernoulli equation
+ + u + χ = G(t)
∂t ρ 2
– This equation has many uses. For a barotropic fluid p = f (ρ):
Combine with density equation: exact solutions can be found.
– For the free surface we will assume p = p0 + p1, p0 = patmos=constant.
p0
Choose G(t) − + gz0 = 0, use χ = gz = g(η + z0) to get (Acheson (3.20))
ρ
∂φ 1 2
on y = η(x, t)
(10.3)
+ 2 (u + w2) + gη = 0
∂t
– Note: φ is a full 2D function of (x, z) satisfying the Laplace equation: (Acheson
(3.24))
∂ 2φ ∂ 2φ
∇·u = 0 = ∇·(∇φ) = 2 + 2 = 0
(10.4)
∂x
∂z
– p1 will be absorbed into the potential φ.
• Recall sound waves: motion u in (x, y) restored by pressure gradient ∇p.
• What will restore motion for surface waves?
Gravitational pressure ρgη
(where pressure effectively enters through the potential φ)
• Linearise. Use Laplace not wave equation: Dispersive waves. (Acheson (3.21-3.22))
– All terms are order : u = u1, v = v1, η = η1, φ = φ1. So drop and 1.
∂η
∂φ
∂η ∂φ
Use
=w=
:
=
(10.5)
∂t
∂z
∂t
∂z
∂φ
Quadratic terms: 21 (u2 + v 2) ∼ 2 → 0 :
+ gη = 0
(10.6)
∂t
∂ 2φ
∂φ
– These combine to form 2 + g
= 0. Not a wave equation:
∂t
∂z
– We will be looking for surface waves of the form η = A cos(kx − ωt).
– Or φ(x, z) = f (z) sin(kx − ωt)
(Don’t expand f (z) = f (z0) + ηf 0 because φ is already order )
– Apply Laplace: 4φ = 0 to get f 00 − k 2f = 0.
– General solution: f = Cekz + De−kz
Solutions like this come up as particular solutions that satisfy boundary conditions to solve many problems with Fourier decompositions.
– Need f (z) → 0 as z → −∞, therefore D = 0 and
φ(x, z) = Cekz sin(kx − ωt) and substitute in (3.21), (3.22) at z = 0 to get
– Ck = Aω and −Cω + gA = 0 giving (Acheson (3.25-3.26))
Aω kz
φ=
e sin(kx − ωt)
and
ω 2 = gk
k
(10.7)
• Particle paths and restoring forces.
.
.
– u = x = ∂xφ = Aωekz cos(kx − ωt) and w = z = ∂z φ = Aωekz sin(kx − ωt)
(u, w) are restored by pressure gradients through φ
– Locally define (x, z) = (x, z) + (x0, z 0) then (Acheson (3.27))
x0 = −Aekz sin(kx − ωt)
z 0 = −Aekz sin(kx − ωt)
– Particle paths are circular.
g 1/2
∂ω 1
ω
2
and cg =
• Dispersion: ω = gk and cph = =
= cph
k
k
∂k 2
– Long waves (λ large, k small) move faster than short waves.
– Wave crests move at twice the group velocity.
– Can be seen by throwing a rock in water.
(10.8)
ρ1 − ρ2
– Modify gravity by looking at the interface between oil and water: g̃ = g
ρ1 + ρ2
– In film clip note how particles ride over the wave crests.
10.2
Boussinesq approximation and internal gravity waves
• Take the Euler equations in a:
– Background gravitational field.
– Density change smooth, not sharp.
– Assume incompressible, small perturbations and
– NO pressure equation.
∇p
Du
=−
+g
Dt
ρ
• Small perturbations
∇·u = 0
Dρ
=0
Dt
(10.9)
– As before: u = u1 p = p0 + p1, ρ = ρ0 + ρ0. Keep terms of order (Acheson
(3.81))
∂u1
∂p1
∂w1
∂p1
ρ0
=−
,
ρ0
=−
− ρ1g,
∂t
∂x
∂t
∂z
(10.10)
∂u1 ∂w1
∂ρ1
dρ0
+
= 0,
+ w1
=0
∂x
∂z
∂t
dz
• Dispersion relation for internal waves: Boussinesq (from Tritton’s book)
∂u1
∂p1
ρ0
=−
,
∂t
∂x
∂u1 ∂w1
+
= 0,
∂x
∂z
– Assume
ρ00
ρ0
=
1
H
∂w1
∂p1
ρ0
=−
− ρ1g,
∂t
∂z
∂ρ1
dρ0
+ w1
=0
∂t
dz
(10.11)
<< m the vertical wavenumber.
– Then w1 = w̃1(k, l)ei(kx+mz−ωt)
ρ0ω ũ1 = k p̃1
ik ũ1 + imw̃1 = 0,
(Acheson (3.83))
ρ0iω w̃1 = imp̃1 + ρ̃1g
−iω ρ̃1 + w̃1ρ00 = 0
(10.12)
– Solve
k 2 ρ0 ω
g k2
N2
k2
1 − 2 w1 = 0
p1 = mw1+
w1 −
ρ1 = mw1+
0 = mw1+ku1 = mw1+
ρ0 ω
ρ0 ω m
im
m
ω
N 2k 2
2
– Giving
(10.13)
ω = 2
2
k +m
– where N is the buoyancy frequency or Brunt-Väisala frequency
N2 = −
g dρ0
ρ0 dz
(10.14)
– Further info: http://www.physics.uwo.ca/~whocking/p103/grav_wav.html
• Anelastic approximation (what Acheson is doing)
– Alternatively assume w1 = w̃1(z)ei(kx−ωt) then
ρ0ω ũ1 = k p̃1
ik ũ1 +
w̃10
= 0,
– Solve (using ρ̃1 =
ρ0ω ũ01 + ρ00ω ũ1 = k p̃01
ik ũ01
+
w̃100
= 0,
ρ0iω w̃1 = p̃01 + ρ̃1g
−iω ρ̃1 +
0
w̃1 dρ
dz
=0
(10.15)
ρ00
−i ω w̃1)
0
2
0
2
ρ
k
ρ
k
p̃01 = w̃100 + 0 w̃10 + i
p̃01
0 = ik ũ01 + w̃100 = w̃100 − ik 0 ũ1 + i
ρ0
ρ0 ω
ρ0
ρ0 ω
ρ00 0
ρ00 0
k 2g ρ̃1
k 2 gρ00
00
00
2
2
= w̃1 + w̃1 − k w̃1 − i
= w̃1 + w̃1 − k w̃1 − 2
w̃1 (10.16)
ρ0
ω ρ0
ρ0
ω ρ0
– Gives (Acheson (3.84))
2
0
ρ
N
w̃100 + 0 w̃10 + k 2
− 1 w̃1 = 0
(10.17)
2
ρ0
ω
∗ Ideal gas: P = ρRdT
∗ Assume constant temperature=300◦K.
Very bad assumption.
dP
∗ Hydrostatic balance:
= −ρg ⇒
dz
dP
g
∗
= −P , c = RdT and P = P0e−zg/c
dz
c
∗ Assume ρ0 ∝ e−z/H and
∗ w1 ∝ ez/2H ei(kx+lz−ωt).
(No imaginary in dispersion (10.18))
∗ That is w̃1(z) ∝ ez/2H+ilz .
∗ Note: w1 increases with height:
∗ eventually there is wave overturning.
∗ And p̃1 ∝ ez/H and ρ̃1 ∝ ez/H
(Not e−z/2H as in Acheson (3.88))
– Take (10.17)
w̃100 +
ρ00
ρ0
w̃10 + k 2
2
N
− 1 w̃1 = 0
2
ω
– then becomes
2
2il
1
N
1
1
2
2
+
−
l
−
+
il
+
k
−1
w̃1 = 0
4H 2 2H
H 2H
ω2
– or (Acheson (3.87))
N 2k 2
ω = 2
(10.18)
2
2
(k + l + (1/4H ))
– In the laboratory, H >> λ = 2π/(k 2 + l2) therefore (Acheson (3.89))
N 2k 2
2
ω→ω = 2
(10.19)
2
(k + l )
2
• Linearised pressure
– How were we able to eliminate pressure and density together without an equation
of state or adiabatic assumption?
– Apply incompressibility ik ũ1 +w̃10 = 0 to the two velocity perturbation equations
and get:
iρ0ωk ũ1 + iρ0ω w̃10 = (∂z2 − k 2)p̃1 + ρ̃01g = 0
– The derivation of the internal wave equation:
avoided solving for ρ̃1 and p̃1 directly,
– But once the solutions of all the components are found, they could be substituted
into this equation and shown to satisfy it.
– This is (in the end) how we usually solve for the effects of density and thermal
variations in ordinary fluids.
• Pressure and Poisson equation
– In general the pressure for an incompressible fluid is found by:
1
ρ1
∂u
∇·
+ (u·∇)u = − ∇p + ν4u − ∇χ
here ∇χ = g
∂t
ρ0
ρ0
∂
– ∇·u = 0
4∇·u = 0
∇·(∇p) = 4p
ui,j = eij − 12 ijk ωk
∂t – ∇· u·∇)u = ui,j uj,i + uj ui,ij = ui,j uj,i = ui,j uj,i = e2 − 12 ω 2 therefore
−4p/ρ0 − 4χ = ui,j uj,i = e2 − 21 ω 2
(10.20)
• Group and phase velocities.
– Compare cph and cg .
cph
Nk
ω
= k̂ = 2
|k|
(k + l2)
ωl
cg = ∇k ω = 2
(l, −k)
(k + l2)k
– Note cg ∝ (l, −k) is perpendicular to k = (k, l)
– Wave crests move in one direction, but
wave packet moves perpendicular to that.
(10.21)
• Gravity waves emitted by a moving ball.
• Density stratification of water obtained
using salt.
• Internal waves propagate away from
ball.
• Crests are perpendicular to direction of
propagation, but appear radially.
• Upward propagating internal waves,
eventually overturning and through turbulence dissipating.
• Are the primary source of drag in atmospheric models.
• Internal waves in the sky marked by clouds.
• Have you ever seen this whilst flying?
• If not, take a look out the window.
• Clouds are where air is going up.
• Often, later in the day, this is where thunderstorms will form.
• Surface patterns mark internal wave crests
• Thermocline is the warmer upper 100m of ocean.
• Below the thermocline the ocean is strongly stratified by temperature.
• Stratification ⇒ internal waves.
• Leaves a large-scale imprint on ocean surface.
10.3
Shallow water equations.
• Objective: Dam break (Hydraulic jump or bore is too difficult)
• Return to surface equations but do NOT assume:
– Small-amplitude waves
• DO assume: height h0 << λ wavelength.
• Obtain pressure via:
− Taken with respect to surface. z = h
∂
u=0
– Fluid moves horizontally as a unit:
∂z
∂p
= −ρg ⇒ p = p0 − ρg[z − h(x, t)]
(10.22)
∂z
– then
∂u
∂u
∂u
1 ∂p
∂u
∂u
∂h
+u +w
=−
⇒
+u
= −g
(10.23)
∂t
∂x
∂z
ρ ∂x
∂t
∂x
∂x
∂u ∂w
∂u
– Integrate incompressibility:
+
= 0 to get w = − z + f (x, t) with
∂x
∂z
∂x
f (x, t) = 0
– Apply to the previous kinematic condition at the free-surface: z = h
∂h
∂h
∂u
∂h
∂
+u
= w = −h
implies
+ (uh) = 0
(10.24)
∂t
∂x
∂x
∂t ∂x
– Hydrostatic balance.
10.3.1
Characteristics
∂u
∂u
∂h
∂h
∂h
∂u
+u +g
=0
+u +h
=0
∂t
∂x
∂x
∂t
∂x
∂x
• Introduce c = (gh)1/2
∂u
∂u
∂c
∂2c
∂2c
∂u
+ u + 2c
=0
+u
+c
=0
∂t
∂x
∂x
∂t
∂x
∂x
• Add and subtract to get ∂
∂
+ (u + c)
(u + 2c) = 0
∂x ∂t
∂
∂
(u − 2c) = 0
+ (u − c)
∂t
∂x
(10.25)
• The method of characteristics can be applied again.
• There will be something like shocks.
• Define: x(s) = φ(X, s) = X + (u + c)s where φ(X, 0) = X. Also: t = t0 + s.
dt
dx
• Then
= 1 and
= u + c and
ds
ds
d
dt ∂
dx ∂
+
(u + 2c) = 0 or
(u + 2c) = 0
(10.26)
ds ∂t ds ∂x
ds
u ± 2c is constant along positive/negative characteristic curves defined by
dx/dt = u ± c
10.3.2
Dam break
• Start with a block of stationary fluid at t = t0 = 0.
• Velocity at each point is a supposition of the two characteristics
crossing at that point. DRAW
• Assume all characteristics before this time are straight lines before t = t0 = 0.
√
.
* x = ±c0 = gh0
* t < 0: at each point: u+ + 2c = 2c0, u− − 2c = −2c0 such that u = u+ + u− = 0
* This defines envelope of undisturbed fluid.
• t > 0: two outer boundaries:
.
−) Undisturbed fluid where h = h0 and c = −c0: x− = −c0 and x− = −c0t
√
.
+) Tip of outflow: h = 0 and c = gh = 0: x+ = 2c0 and x+ = 2c0t.
* This defines envelope of disturbed fluid. DRAW
• Disturbed fluid: at each point there are two characteristics
+) Positive characteristic emanating from the undisturbed fluid,
Therefore u + 2c =constant= 2c0.
−) Negative characteristic: u − 2c =constant= k, k is variable.
• Add and subtract
k
u + 2c = 2c0
2
c0 k
u − 2c = k
−
c=
2
4
– Both u− and c− are constant for a given k.
c0 3k
.
– Find the velocity of the negative characteristic: x−(k, t) = u− − c− = +
2
4
• k at limiting negative characteristics at the outer boundaries:
c0 3k
:
k = −2c0
Inner: u − c = −c0 = +
2
4
c0 3k
Outer: u − c = 2c0 = +
:
k = 2c0
2
4
* Recall: Both characteristics emanate from x0 = t0 = 0.
* Therefore, to avoid crossing, all negative characteristics must
emanate out of x0 = t0 = 0.
c0 3k
x−(k, t) =
t
DRAW
+
2
4
)
u = c0 +
* Given (x, t) within the disturbed fluid one can find a unique negative characteristic defined by k.
3k x c0
4 x 2c0
= −
and u − 2c = k =
−
4
t
2
3t
3
• Finding (u, c) as a function of (x, t)
4 x 2c0
−) Last page: u − 2c = k =
−
3t
3
+) Recall: u + 2c = 2c0
* Add and substract:
2 x c0
2 x
u=
− + c0 =
+ c0
3t
3
3 t
1
1
x
c
x
0
+ 12 =
c = 12 c0 −
2c0 −
3t
3
3
t
c2
x 2
1 p
h= =
2 gh0 −
g
9g
t
dx+(t, t1)
• Speed of positive characteristics: u + c = 2c0 − c =
dt
* where t1 is the time the characteristic leaves the undisturbed fluid.
* (line x = −c0t) or x+(t1, tt) = −c0t1
then
* Define u+(t, t1) = u(t, t1) + c
1 x+ 4c0
.
x + = u+ = u + c =
+
3 t
3
. 1y
* Solve using integrating factors: Assume x+(t) = yf (t) where y = .
3t
.
4 −1/3
1/3
1/3
* y = t so x+ = t f (t). Show that f (t) = c0t
and f = 2c0t2/3 + a0.
3
• Put pieces of positive characteristics together:
x+ = yf (t) = t1/3 a0 + 2c0t2/3 = a0t1/3 + 2c0t
1 −2/3
1 x+ 4c0 1 −2/3 2c0 4c0 1 −2/3
.
+ 2c0
+
= a0t
+
+
= a0t
+ 2c0
x+ = a0t
3
3 t
3
3
3
3
3
Solution checks. What is a0?
1/3
x+(t1, t1) = −c0t1 = a0t1 + 2c0t1
2/3
x+(t1, t1) = −3c0t1 t1/3 + 2c0t
2/3
⇒ a0 = −3c0t1
DRAW
As an additional check, the velocity of the positive characteristic should be contin.
uous with its value emanating from the undisturbed fluid. i.e. x+(t1) = c0:
1
.
2/3
x+(t) = a0t−2/3 + 2c0 = −c0t1 t−2/3 + 2c0 = c0 at t = t1
3
And for t1 = 0 (the leading positive characteristic): a0 = 0 and x+(t, 0) = 2c0t
11
Rotation
These problems are best Include Ekman layers, Ekman number, Taylor columns,
Taylor-Proudman (p 273), Rossby number, Rossby radius. Vortex stretching.
• Start with the Euler equations in the inertial frame I
(what we have used up to now)
• and rotating frame R.
D
1
D
uI = − ∇p − gr̂ =
uR + Ω × (Ω × r) + 2Ω × uR
Dt
ρ
Dt
where Ω is the angular velocity
• Dropping the subscript R then for this chapter:
D
1
u = − ∇p − Ω × (Ω × r) − 2Ω
× uR} −gr̂
|
{z
{z
}
|
Dt
ρ
centrigular force
(11.1)
(11.2)
Coriolis force
• Recall hydrostatic balance:
– Once the reference pressure and density are in balance,
– then the only effect of density is a linear term.
• We want to do the same with the centrifugal term.
Ω × (Ω × r) = −∇( 21 Ω2R2)
(11.3)
• Centrifugal term: Ω × (Ω × r) = −∇( 12 Ω2R2)
– Where R = r cos φ is the distance from the axis of rotation.
– θ=latitude.
– Because this is now a gradient: it is absorbed into the pressure.
– We don’t feel this: the earth (which is mostly a fluid) is thicker at the equator.
– We will return to centrifugal for deviations from geostrophy.
• Therefore the difference between non-rotating and rotating fluids is the Coriolis
term.
• Derivation of rotating frame equations
Dr
Dr
=
+Ω×r
u I = uR + Ω × r
Dt I
Dt R
DuI
DuI
=
+ Ω × uI
ω I = ω R + ∇ × (Ω × r) = ω R + 2Ω
Dt I
Dt R
DuI
DuR
DΩ
Dr
=
+
×r+Ω×
Dt I
Dt R Dt
Dt R
+Ω × (uR + Ω × r)
DuR
=
+ 2Ω × uR + Ω × (Ω × r)
Dt R
DΩ
+
×r
Dt
• Choose characteristic velocity U and length L. Important numbers:
U
=Rossby number
Ratio of inertial |u·∇u| to Coriolis forces.
– Ro =
ΩL
ν
=Ekman number
Ratio of viscous |ν4u| to Coriolis forces.
– Ek =
2
ΩL
11.1
Geostrophic flow:
− Consider both Ro and Ek small: usually ΩL large.
Euler becomes balance between Coriolis and pressure forces. 2Ω × u = − ρ1 ∇p
− Coriolis force 2Ω × u is always perpendicular to flow u
− When in balance atmospheric flow is perpendicular to velocity.
− If you know the pressure map isobars, then velocity follows the pressure contours.
• Geostrophic wind: Take orders of magnitude:
– L ∼ 1000km, H = 10km, t ∼ 105s=(a day), u⊥ = 10ms−1, w ∼ 10−2ms−1
Du⊥
1
= f| k̂ {z
× u⊥} + ∇p
ρ
|Dt
{z }
−3
|{z}
10−4
10
10−3
Dw
1
− |{z}
g + ∂z p = 0
Dt
ρ
|{z}
|{z}
10−7
(11.4)
10
10
∂p ∂p
1
− ,
(fc = 2Ω sin φ)
(11.5)
ρfc
∂y ∂x
– The geostrophic wind component blows parallel to isobars.
– Around a low pressure in the N. Hemisphere to the left (anti-clockwise).
– Around a high pressure in the N. Hemisphere to the right (clockwise).
– Opposite sense if downunder.
q
– G = Ug2 + Vg2
G = (Ug , Vg ) =
– Often pressure is given in terms of the height Z above the ground of a given
pressure level.
– Then
g
∂Z ∂Z
G = (Ug , Vg ) =
− ,
(11.6)
fc
∂y ∂x
– Note: The strength of the geostrophic wind depends upon the gradient of the
pressure.
11.2
Taylor-Proudman theorem
− In geostrophic flow: ∇ × ∇p = 0 implies ∇ × (Ω × u) = 0
− Expand: ∇ × (Ω × u) = Ω·∇u − u·∇Ω + u(∇·Ω) − Ω(∇·u)
− Use: ∇Ω = 0 and ∇·u = 0 implies Ω·∇u = 0
− Ω is parallel to ẑ not radial: implies ∂u/∂z = 0: Taylor-Proudman theorem
− This implies that away from boundaries, no velocity variations in ẑ.
∂z u = ∂z v = ∂z w = 0
− If there are boundaries in z such that w = 0 then ∂z u = ∂z v = w = 0
− Diagram of flow between two rotating plates.
Joseph Proudman: Based
at University of Liverpool
The interior flow obeys
the
Taylor-Proudman
theorem.
11.3
Taylor columns
• Transverse Taylor column.
– Motion perpendicular to the axis of rotation.
– Flow past an obstacle embedded in a rotating
fluid,
but not extending from top to bottom.
– Taylor column forms over the obstacle.
– Flow inside it behaves as a single rotating
body.
– Flow around it behaves as if the Taylor column is an aerofoil. Zhukovsky.
Red
spot
real colour
• Astrophysical applications?
– It was originally thought the bands on
Jupiter, Saturn, and the Sun
where a signature of underlying Taylor
columns.
– This is definitely not the case with the Sun.
– I believe that the current theories for:
Jupiter and Saturn require:
something more complicated.
– It is still believed to be responsible for:
the Red Spot on Jupiter.
• Longitudinal Taylor column.
– Obstacle moves along the axis of rotation.
– Columns above and below have
zero rotation.
Red
spot
false colour
Red
spot
simulated
11.4
Outline of Geoffrey Ingram Taylor’s (1886-1975) contributions
• Introductory remarks
– Overview of G.I. Taylor’s research
– State of fluid mechanics in 1900
• Taylor’s first two papers
– Diffraction at low light levels
– Regularisation of shocks
• Instabilities
– Taylor-Couette flow
– Saffman-Taylor problem
– Rayleigh-Taylor instability
• Turbulence
– Eddy diffusivity in the atmosphere
– Diffusion by continuous movements
– Statistical theory of turbulence
– Vortex breakdown
• Rotating flows
– Taylor-Proudman theorem
– Particle motion and Taylor columns
• Dispersion in laminar flows
– Taylor-Aris dispersion
– Measurement of molecular diffusivities
• Solid mechanics
− Dislocations and the strength of solids
• Swimming at low Reynolds numbers
• Drops and bubbles
– Drop deformation and breakup
– Viscosity of mixtures; emulsions
• Electrohydrodynamics
• Surface tension
− Leaky dielectric model
− Thin films, peeling, water bells.
• Explosions and Shocks
11.5
Gradient wind
• In the atmosphere around a low pressure cell, the wind is slower than geostrophic
and call subgeostrophic.
• Around highs, the wind is faster and called supergeostrophic.
• The steady-state wind (geostrophic+correction) is called the
gradient wind=(Ur , Vr ).
• It is found by
1 ∂p
Vr |Vr |
+fcVr +s
ρ ∂x
R
1 ∂p
Ur |Ur |
0= −
−fcUr −s
ρ ∂y
R
0= −
• R is distance from high or low
• s is a sign given by
Flow around a
(11.7) Hemisphere Low
High
Northern
1
-1
Southern
-1
1
2
M
r
• Define Mr = Ur2 + Vr2.
Equation for Mr is: Mr = G ±
fcR
• Solution for cyclonic flow around a low • Solution for anti-cyclonic flow around high
• (anti-clock: NH, clock:SH)
• (clock: NH, anti-clock:SH)




s
s
4G 
4G 


Mr = 0.5fcR −1 + 1 +
(11.8) Mr = 0.5fcR 1 − 1 −
(11.9)
fcR
fcR
p
• Solution for cyclonic flow around a low • Solution for anti-cyclonic flow around high
• (anti-clock: NH, clock:SH)

s
Mr = 0.5fcR −1 +
1+

4G 
fcR
•R→∞
2
2G
G
Mr → 0.5fcR −1 + 1 +
− 2 2
fcR f c R
• (clock: NH, anti-clock:SH)

s
Mr = 0.5fcR 1 −

1−
4G 
fcR
•R→∞
2
2G
G
Mr → 0.5fcR 1 − 1 +
+ 2 2
fcR f c R
G2
=G−
<G
fcR
• Mr → G as R → ∞
G2
=G+
>G
fcR
• Mr → G as R → ∞
• Mr → 0 as R → 0
• There are NOT solutions for all R
• There are solutions for all R
• Limiting R = 4G/fc.
• There can be strong gradients of pres- • How does Nature resolve this?
sure and
• Pressure gradients and G → 0 as R → 0
• Strong geostrophic winds in low pres- • Low geostrophic winds in high pressure cells.
sure cells.
• Consistent with our observations of
storms.
11.6
Axisymmetric gradient wind
• In cylindrical coordinates the horizontal components of the equation of motion are:
∂ur
∂ur uθ ∂ur u2θ
1 ∂p
+ ur
+
− − fcuθ = −
∂t
∂r
r ∂θ
r
ρ ∂r
(11.10)
∂uθ
∂uθ uθ ∂uθ ur uθ
1 1 ∂p
+ ur
+
+
+ fcur = −
∂t
∂r
r ∂θ
r
ρ r ∂θ
• Derive the gradient wind:
– Assume steady (∂/∂t = 0) and axisymmetric (∂/∂θ = 0, ur = 0) flow (i.e.
circular motion).
Then these equations reduce to:
u2θ
1 ∂p
+ fcuθ −
=0
r
ρ ∂r
which is known as the gradient wind equation.
1 ∂p
– The geostrophic wind G =
fc ∂r
– Physically it represents a balance between the centrifugal force, the coriolis force
and the pressure gradient forces.
r ∂p
2
– Rewrite as uθ + fcruθ −
=0
ρ ∂r
– This quadratic equation is easily solved to yield
s
"
#
fcr
4 ∂p
uθ =
−1 ± 1 + 2
2
fc rρ ∂r
• Which root to take?
– As r → ∞ we want the solution to converge to the geostrophic wind:
1 ∂p
v = lim uθ =
r→∞
fcρ ∂r
– The positive root yields v =
1 ∂p
fcρ ∂r
– The negative root yields v = −fcr 6=
1 ∂p
fcρ ∂r
– So take the positive root:
"
fcr
−1 +
uθ =
2
s
4 ∂p
1+ 2
fc rρ ∂r
– For low pressure ∂p/∂r > 0 and no restrictions on v.
1 ∂p fc2r
≤
– For high pressure it is required that: −
ρ ∂r
4
#
11.7
Ekman boundary layers
• The Ekman spiral refers to currents or winds near a horizontal boundary in which
the flow rotates towards the interior flow as one moves away from the boundary.
• During the expedition of the Fram, Fridtjof Nansen had observed that:
– Icebergs tend to drift not in the direction of the prevailing wind
– but at an angle of 20-40 to the right.
– Bjerknes invited Ekman, still a student, to investigate the problem and, in 1905,
– Ekman explains the phenomenon in terms of:
– The balance between frictional effects in the ocean and the Coriolis force, which
arises from planetary rotation.
• Assume:
– An interior flow in geostrophic equilibrium, varying slowly in x and y:
– An ageostrophic component varying slowly in
(x, y),
but possibly rapidly in z, and no vertical velocity.
– ∇⊥(G + u1) ≈ 0 and w1 ≈ 0
• Then −2Ωv1 = ν∂zz u1
2Ωu1 = ν∂zz v1
• This could lead to a 4th-order equation.
Vilhelm Bjerknes, Vagn Ekman
• Use complex variables: Z = u1 + iv1
ν∂zzZ = 2ΩiZ
−(1+i)z∗
• Solution is: Z = Ae
+ Be
(1+i)z∗
where z∗ =
(11.11)
Ω 1/2
z
ν
• Ae−(1+i)z∗ is the solution from the
bottom boundary.
* u + iv = Ug + iVg + Z. If Vg = 0,
* (u, v) = (0, 0) at z∗ = 0 and
* (u, v) → Ug as z∗ → ∞ then A = −1
* u = Ug (1 − e−z∗ cos z∗)
*
→ Ug z∗ for z∗ ≈ 0
* v = Ug e−z∗ sin z∗
*
→ Ug z∗ for z∗ ≈ 0
* u/v → 1 for z∗ ≈ 0
* Flow at surface is at 45◦ to interior flow
• Be(1+i)z∗ is the solution from the
top boundary.
• Or allow A and B to be slowly varying functions of x and y.
• Further reference: www.met.reading.ac.uk/courses/MTMG49/2.2_ekman.doc
Visualization of the Ekman spiral for:
Geostrophic wind = 10 m s−1 in the xdirection. In the northern hemisphere this
would correspond to low pressure to the
north and high pressure to the south. Eddy
diffusivity for momentum Km = 10 m2 s−1.
Coriolis parameter p
fc = 104 s−1. Ekman
layer depth hv = π 2Km/fc=1400m.
a) The Ekman spiral in laminar flow, exact solution;
b) Typical example of the Ekman spiral in turbulent flow (from Prandtl 1965).
• Note that initially (in the surface layer) the wind speed increases markedly
with altitude, but the direction of the wind doesnt change much.
• Thus the assumption in Monin-Obukhov theory that rotation is unimportant in
the surface layer appears to be reasonable.
• In the mixed layer, the wind speed increases only slowly with height, but most of
the change in direction occurs here.
11.8
Nansen, Jackson, Bjerknes, Ekman
The Fram on
Amundsen’s
Antarctic expedition.
Nansen.
Jackson meeting
Nansen.
International Herald Tribune, August 14 1896
CHRISTIANIA Dr. Nansen has arrived at Vardoe after a successful expedition. Dr.
Nansen reports that he and Lieutenant Johansen left their ship, the Fram, in March
1895 to explore north of the Fram’s route. They journeyed through the unknown Polar
Sea and explored the north of Franz-Josef Land. No land was seen north of 82deg.
of latitude, only ice. They wintered in Franz-Josef Land, living on whale meat and
bear meat. Here they met the Jackson-Windward expedition, which brought them to
Vardoe. The Fram is reported to be a good ice ship.
My notes: he actually reached 86◦140N while the Fram drifted to 84◦40N. Fram means
forward. Used by Fridtjof Nansen, Otto Sverdrup, Roald Amundsen.
11.9
Ekman pumping
• Call the wind at the surface due to the Ekman spiral the
• Boundary-Layer Gradient Wind: M BLG
• The solid arrows represent the gradient wind M r determined by the balance between rotation and the pressure gradient.
• Around a low there is a local circulation δΩ
• The lighter arrows are M BLG, with the 45◦ correction to M r determined by the
rotation of the surface Ω.
• There is a component pointed inwards: Surface flow around a low is convergent.
• Around a high, M BLG points outwards: Surface flow around a high is divergent.
• Flow between two plates rotating separately.
– General solution for (u, v) given a relative internal flow:
Gr = (Ur , Vr ) = 12 ΩR(− sin θ, cos θ) =
= 21 Ω(−y, x)
(u, v) = Gr (1 − e−z∗ cos z∗) + ẑ × Gr e−z∗ sin z∗
– Consider the divergence of velocity when Gr is slowly varying.
∂z w = −(∂xu+∂y v) = − (∂xUr + ∂y Vr )(1−e−z∗ cos z∗)+(∂xVr −∂y Ur )e−z∗ sin z∗
{z
}
|
=0
– Let (∂xVr − ∂y Ur ) = ωr . Integrating towards the interior vertical velocity wI :
ν 1/2 Z z∗
ν 1/2 ω
r
−z∗
−z∗
w(z) =
dz∗ωr e sin z∗ =
1 − e (cos z∗ + sin z∗)
(11.12)
Ω
Ω
2
0
ν 1/2
ν 1/2
(∂xVr − ∂y Ur ) = 21
ωr
– Then as z → ∞, wI → w|z=∞ → 21
Ω
Ω
– ωr is relative vorticity for interior flow compared with the boundary.
– Recall ωr = 2δΩI . If ΩI = 12 (ΩT + ΩB )
then δΩI = ΩI − ΩB = 12 (ΩT − ΩB ) = 21 Ω
– Therefore ωr = 2δΩI = Ω and
ν 1/2
Ω = 21 (νΩ)1/2 (11.13)
wI = 21
Ω
• Rates of convergence and divergence: Total profiles between rotating plates.
– From the bottom:
(uT , vT ) = ΩB ×R+(u, v) = ΩB −y, x
+ 12 Ω(1−e−z∗
cos z∗) −y, x
− 12 Ωe−z∗
sin z∗ x, y
– Define UBL as the component of uT along an isobar.
– Define VBL as the perpendicular component of uT in or out.
1
−z∗
cos z∗, − sin z∗
(U, V )BL = −ΩI R 1, 0 + 2 ΩRe
Ω 1/2
– From the top at L with L∗ = ν
L:
1
(z∗ −L∗ )
(U, V )BL = −ΩI R 1, 0 − 2 ΩRe
cos(z∗ − L∗), sin(z∗ − L∗)
i
h
1/2
−z∗
z∗ −L∗
1
wBL = 2 (νΩ) 1 − e (cos z∗ + sin z∗)
or e
(cos(z∗ − L∗) + sin(z∗ − L∗))
Profiles of the velocity along isobars U ,
perpendicular to isobars V and the vertical
velocity W for an imaginary flow between
two plates.
• To those of us in the N.Hem the
Earth rotates anti-clockwise.
• A low pressure cell also rotates anti-clockwise
– So the net effect is the atmosphere above the ground in a low rotates faster
than the ground below:
– Ageostrophic flow V points in at the bottom:
dragging pressure lines in, steepening pressure gradients.
• A high pressure cell also rotates clockwise
– So the net effect is the atmosphere above the ground in a high rotates slower
than the ground below:
– Ageostrophic flow V points out at the top:
pushing pressure lines out, decreasing pressure gradients.
Streamlines of the secondary circulation forced by frictional convergence in the
planetary boundary layer for a cyclonic vortex in a barotropic atmosphere. This is
known as Ekman pumping and is responsible for the spin down of weather
systems.
12
Rossby waves
Rossby at a rotating tank.
Carl-Gustav Rossby
• Rotating fluids have an intrinsic stability and inertial waves.
• Displaced fluid will return, but not directly
because the Coriolis force acts perpendicular to the motion.
• A particle returns to its original position twice during every revolution of the fluid:
ωmax = 2Ω.
• Rossby waves are also called planetary waves: example: large-scale meanders
in the jet stream seen in 500hPa pressure height (10km).
• Coastal Rossby waves in the Gulf Stream off of Labrador visualised by 18◦ water.
12.0.1
Background
• 2D shallow water equations derivation (followed by full 3D laboratory)
∂v ∂u
–ζ=
− = the relative vertical component of vorticity
∂x ∂y
– In the define the potential vorticity as (ζ + Ω)/h
– Just as 2D Euler conserves circulation:
in shallow water the potential vorticity is conserved.
– From this one gets by several means the Charney-Hasegawa-Miwa equation
∂
∂
∂
+u +v
ζ + βv = 0
(12.1)
∂t
∂x
∂y
• The CHM (Holton (7.89)) equation appears in three contexts:
– Global, geophysical: Assume the fluid is barotropic (ρ = ρ(p)) and in a β-plane.
∗ Waves are Rossby or planetary waves.
∗ The equations are the Charney equations.
– Laboratory: A rotating fluid with a sloping bottom.
∗ Waves are “ topographic Rossby waves”
∗ Trapped near a coastline they are “continental shelf waves”
– Plasma physics. B × u replaces Ω × u
∗ Hasegawa-Miwa equations and drift waves, for tokamaks.
12.1
Difference between barotropic and baroclinic instability
• A barotropic atmosphere is:
– One in which the density depends only upon the pressure ρ = ρ(p) so that
isobaric surfaces are also surfaces of constant density.
– Barotropy provides a very strong constraint on the motions in a rotating fluid;
Large-scale motion depends only upon horizontal position and time, not height.
– In an inviscid barotropic fluid of constant depth, the Rossby wave is:
An absolute vorticity (ζ + fc)/h conserving motion.
– Rossby waves owe their existence to a restoring force from either:
∗ N-S variation of the Coriolis parameter, the β-effect: planetary waves.
∗ Or variations of h: coastal waves.
• In a baroclinic atmosphere:
– The density depends upon both the temperature and the pressure ρ = ρ(p, T ).
The geostrophic wind generally has a vertical shear. This shear is from:
Horizontal temperature gradients and the thermal wind equation.
– In practice, the equations are written on isentropic surfaces (pressure adjusted
by hydrostatic equilibrium) and look similar to the shallow water equations.
(ζ + fc) θ
– Then a Rossby wave is a isentropic P V : ζIP V =
conserving motion
ρ dz
owing its existence to the isentropic gradients of potential vorticity.
12.2
Rossby waves from the 2D shallow water equations
• Start with the 2D shallow water equations:
∂u⊥
Dh
+ u⊥·∇u⊥ = −g∇h
+ h∇·u⊥ = 0
∂t
Dt
– take curl
D(ζ + fc)
∂u⊥
∇×
+ u⊥·∇u⊥ = −g∇h
or
+ (ζ + fc)∇·u⊥ = 0
{z
}
|
∂t
Dt
vortex stretching
– Combine to get P V = (ζ + fc)/h=potential vorticity equation
(ζ + fc)
D(ζ + fc)/h
D(ζ + fc)/h (ζ + fc)
+
∇·u⊥−
h∇·u
=
=0
⊥
2
Dt
h
h
Dt
• More generally (see Holton equations 4.14 to 4.23):
conserves P V
– Assume fluid is barotropic (ρ = ρ(p)) and get:
– Barotropic (Rossby) Potential Vorticity Equation
Dh(ζg + fc)
∂w
= (ζ +fc)
Dt
∂z
• To get planetary Rossby waves: work in a β-plane,
– assume h = h0 and assume vertical gradients are small: ∂z w = −∇·u⊥ ≈ 0.
Dfc
∂fc
=v
= vβ or fc = βy.
– β-plane is when
Dt
∂y
– Then Holton (4.23) becomes Holton (7.89), the Charney-Hasegawa-Miwa eqn.
∂
∂
∂
+u +v
ζ + βv = 0
∂t
∂x
∂y
• Now rewrite in terms of a stream function:
– Holton assumes: u = u + u0, v = v 0, u0 = ∂y ψ, v 0 = −∂xψ where ζ = −4ψ
and ψ is the stream function.
∂
∂
∂
∂
∂
∂ψ
+u +v
ζ + βv = 0 becomes
+u
4ψ + β
=0
∂t
∂x
∂y
∂t
∂x
∂x
iφ
– Assume ψ = < ψe and φ = kx + ly − ωt then (Holton (7.91))
(−ω + ku)(−k 2 − l2) + kβ = 0 or ω = uk −
βk
k 2 + l2
(12.2)
– Relative (to u from geostrophic balance) frequency: ωr = ω − uk is
βk
K2
– If l = 0: phase and group velocities:
ωr = −
K 2 = k 2 + l2
ωr
β
ωr β
cph =
=−
cg = ∇k ωr = − = = −cph
k
k
k
k
– Pattern goes west while disturbance goes east.
• Alternatively assume Ω=constant and h = h0 + H(y)
where H(y) is the depth of the layer for laboratory and coastal waves.
(12.3)
(12.4)
• Laboratory waves and coastal waves.
– Alternatively assume fc=constant and
h = h0 + H(y)
– where H(y) is the depth of the layer.
– Flow between not quite parallel plates.
Or basin with a slope.
wh
– ∂z w =constant=
= vh tan α = vγ
h
is used in the Barotropic Potential Vorticity Equation.
– Assume h0 is small.
– Then Dt((ζ + fc))/h) becomes H0Dtζ − fcv∂y H = 0. Follow (Tritton (16.57))
and
– use γ = ∂y H, assume ζ = ∂xv if no y dependence for (u, v) and fc = 2Ω then
2Ωγ
∂ 2v
2Ωγ
∂tζ −
v = 0 or
−
v=0
(12.5)
h0
∂x∂t
h0
2Ωγ
– Assume v = v0 exp(i(ωt − kx)), then ω =
kh
– Holton (7.80) reduces to this if u = 0, l = 0 and we reverse the sign of k.
i.e. only negative k is allowed.
• Shallow water Rossby waves have the following unique properties:
1.) They are linear. There is no dependence upon the amplitude of the waves.
2.) They are dispersive.
3.) Wave crests can move in only one direction with respect to mean velocity.
cph − u =
k
ωr
k̂ = −β 2
k̂
2
3/2
K
(k + l )
(12.6)
4.) u and v are π/2 out of phase.
• Why only in one direction?
– We want to conserve P V = (f +ζ)/H. If the fluid moves into a region of higher
f /H, then the relative vorticity ζ must decrease, resulting in either clockwise
or anti-clockwise motion.
ω
β
– If l = 0, the phase speed cph = = − 2 [Holton (7.88)] points west.
k
k
– Rossby waves travel in only one direction: west on the Earth. Why?
– Consider a wave conserving ζ + f0 with ζ|t=0 = 0 displaced by δy at δt.
– Then (ζ + f )|δt = f0 so ζ|δt = f0 − fδt = −βδy.
– Positive displacement creates negative ζ that pushes crest to west.
– Negative displacement creates positive ζ that pushes crest to west.
• See below for full group velocity cg .
12.3
Inertial waves in a 3D tank
• Consider the linearised rotating equations
∂u1
1
+ 2Ω × u1 = − ∇p1,
∂t
ρ
∇·u1 = 0
(12.7)
• Use the following three equations to get a single equation for p1 which will give us
a dispersion relation.
∂u1
1
∂u1 ∂v1
1
+ 2Ω × u1 = − ∇p1
−
= − 4p1
⇒ 2Ω
∇·
∂t
ρ
∂y
∂x
ρ
∂u1
∂u1 ∂v1
1
∂ ∂u1 ∂v1
∇×
+ 2Ω × u1 = − ∇p1
−
= 2Ω
+
⇒
∂t
ρ
∂t ∂y
∂x
∂x
∂y
∂
∂2
∂ ∂w1
∂
Ω × u1 = 0, ⇒
p
=
−ρ
=
ρ
(∇·u1)
1
2
∂z
∂z
∂z ∂t
∂t
• Then
2
2
∂u1 ∂v1
∂
∂
∂2
2 ∂
4Ω 2 p1 = 4Ωρ ∇·u1 = 2Ωρ 2
−
= − 2 4p1
∂z
∂t
∂t
∂y
∂x
∂t
• Or
ikx+ly+mz−ωt
• If p1 ∝ e
2
2
∂
2 ∂
4p1 + 4Ω 2 p1 = 0
∂t2
∂z
(12.8)
4Ω2m2
then the dispersion relation is ω = 2
k + l 2 + m2
2
(12.9)
4Ω2m2
•ω = 2
is very similar to the internal-gravity wave dispersion relation:
2
k +m
2 2
N
k
ω2 = 2
k + m2
• Except that one cannot have internal-gravity waves without a horizontal wavenumber k
2
• And you cannot have inertial waves without a vertical wavenumber m.
• In the environmental one gets rapidly rotating 3D fluids only in a few Arctic seas.
• My recollection is that the main problem is how to get rid of the fast inertial waves
in the dynamics.
• The only mathematical theory that gives the most common approximation associated with geostrophic flow: quasi-geostrophy, can be proven rigourously only
in the infinite rotation (Ro → 0) limit.
13
Instabilities: Shear, convection and stratification
a) Instability of inviscid continuous parallel shear.
Kelvin-Helmholtz instability. Rayleigh criterion.
b) Stratified shear instability:
Richardson criterion. Richardson number.
c) Viscous shear instability. Orr-Sommerfeld equation.
Tollmein-Schlichting waves. Finite-amplitude disturbances.
d) Thermal convection. Rayleigh-Bénard instability. Bénard cells.
e) Landau equation. Bifurcations, Hopf. Period doubling. Chaos.
f) Real 3D turbulence.
13.1
Inviscid Kelvin-Helmholtz
A smooth shear with an inflection point
(U 00 = 0) This is covered for unstratified
shears in Acheson 9.5 and I will lecture on.
• Mixing layer behind a splitter plate.
• Use
linearised
interface
conditions:∂y φ1 = ∂tη, ∂y φ2 =
∂tη + U ∂xη (Acheson problem 3.6)
ρ1(∂tφ1+gη)−ρ2(∂tφ2+U ∂xφ2+gη) = T η,xx
• T is the surface tension.
1
∂tu1 + U ∂xu1 + w1∂z U = − ∂xp
ρ
1
∂tw1 + U ∂xw1 = − ∂z p
ρ
∂xu1 + ∂z w1 = 0
• Assume monochromatic solutions
i(kx−ωt)
u1 = Re ũ(z)e
i(kx−ωt)
w1 = Re w̃(z)e
i(kx−ωt)
p1 = Re p̃(z)e
• Linearise Euler:
• Gives
1
−i(ω − U k)ũ + w̃U = − ik p̃
ρ
1
−i(ω − U k)w̃ = − p̃0
ρ
0
ik ũ + w̃ = 0
(13.1)
(13.2)
0
• Take z-derivative of (KH3a) and (KH3c), grouping all U 0 terms together:
1 0
0
0
0
00
−i(ω − U k)ũ + U (ik ũ + w̃ ) + w̃U = − ik p̃
ρ
ik ũ0 + w̃00 = 0
(13.3)
(13.4)
In (KH4a) use (KH3b) to replace p̃0, use (KH3c) to eliminate the U 0 term and
replace ũ0 using (KH4b)
(ω − U k)w̃00 + (kU 00 − k 2(ω − U k))w̃ = 0
• to get the master equation: Acheson (9.48)
00
kU
w̃00 +
− k 2 w̃ = 0
ω − Uk
(13.5)
• Choose no flow through boundary conditions w̃(L) = w̃(−L) = 0
• In general this requires a numerical solution.
• All I am interested in now is a general condition upon U 00 that is necessary
condition
• Integrate the master equation across the domain
Z L
Z L
00
kU
∗ 00
+
− k 2 |w̃|2dz = 0
w̃ w̃ dz
ω − Uk
| −L
| −L {z
}
{z
}
by parts
split ω=ωR +ωI
(13.6)
h
iL
w̃0w̃∗
−
−L
| {z }
=0 due to BC
Z
L
−L
|w̃|2dz +
Z L
−L
ωR − U k − iωi)kU 00
2
2
−
k
|
w̃|
dz
|ω − U k|2
• For determining stability, only the imaginary part matters.
0
Z
L
• For instability we need ωI > 0, which means
[] = 0.
−L
L
U 00|ṽ|2
ωI k
dz =
2
|ω
−
U
k|
−L
Z
• If k > 0, instability is only possible if the integral=0. That is
Z L
U 00|ṽ|2
k
dz = 0
2
−L |ω − U k|
(13.7)
• This is only possible if U 00 = 0 changes sign.
• This is for instability U (z) must have an inflection point somewhere
• Rayleigh’s Inflection Point Theorem (1880) A necessary condition for
the linear instability of an inviscid shear flow U (z) is that U 00(z) should change
sign somewhere in the flow.
13.2
Inviscid Richardson criteria
• Now add the density equation
∂tu1 + U ∂xu1 + w1∂z U = −ρ−1
0 ∂x p
−1
∂tw1 + U ∂xw1 = −ρ−1
∂
p
−
ρ
z
1
0
0 ρ1 g
∂xu1 + ∂z w1 = 0
h
i
∂tρ1 + U ∂xρ1 + w1ρ00 = 0
ρ1 = Re ρ̃(z)ei(kx−ωt)
• Gives
(ω − U k)ũ + iw̃U 0
ρ0i(ω − U k)w̃
ik ũ + w̃0
−i(ω − U k)ρ̃ + w̃ρ00
=
=
=
=
ρ−1
0 k p̃
p̃0 + ρ̃g
0
0
(13.8)
(13.9)
• Following my earlier derivation of Acheson’s internal-gravity waves in an anelastic
atmosphere
(ω − U k)ũ + iω w̃U 0 = ρ−1
0 k p̃1
−2 0
0
(ω − U k)ũ0 + iw̃U 00 = ρ−1
0 k p̃ −ρ0 ρ0 k p̃
ρ0i(ω − U k)w̃ = p̃0 + ρ̃g
ik ũ1 + w̃10 = 0,
ik ũ01 + w̃100 = 0,
−i(ω − U k)ρ̃ + w̃ρ00 = 0
(13.10)
0
−2 0
0 = w̃00 + ik ũ0 = w̃00 + ik(ω − U k)−1 − iw̃U 00 + ρ−1
k
p̃
−
ρ
0
0 ρ0 k p̃ =
00
w̃ +ik(ω−U k)
−1
00
−iw̃U +k
i(ω−U k)w̃−ρ−1
0 ρ̃g
0
−ρ−1
ρ
0
0
(ω−U k)ũ+iω w̃U
• Define c = ω/k then the master equation is (Acheson problem 9.2)
0
2
00
0
0
ρ
U
ρ
U
N
0
2
w̃00 + 0 w̃0 +
+
+
−
k
w̃ = 0
ρ0
(c − U )2 c − U ρ0 (c − U )
0
=
(13.11)
• Problems: Derive Acheson (3.84) and (9.48) from this.
• We now integrate this in a similar manner, but first:
– Neglect ρ00/ρ0 = −N 2/g terms (Boussinesq). The one gets the Taylor-Goldstein
equation (Acheson problem 9.2):
00
2
U
N
2
+
+
−k
w̃ = 0
(13.12)
w̃00 +
2
(c − U )
c−U
– Make the variable change: w̃ = (U − c)nq.
2
Use w00 = n(n−1)(U −c)n−2U 0 q+nU 00(U −c)n−1q+2nU 0(U −c)n−1q 0 +(U −c)nq 00
– [(U − c)2nq 0]0 = 2nU 0(U − c)2n−1q 0 + (U − c)2nq 00
= (U − c)nw̃00 − n(n − 1)(U − c)2n−2U 02q − nU 00(U − c)2n−1q
2
2
∗
2n 0 0
2
2
02
2n−2
00
2n−1
−q [(U −c) q ] +k |q| = N +n(n−1)U (U −c)
+(n−1)U (U −c)
|q|
Z L
02
2n
2
2
• Integrate and use parts on first term:
(U − c) |q | + k |q| dz
−L
Z
L
=
2
N + n(n − 1)U
02
(U − c)
2n−2
00
+ (n − 1)U (U − c)
2n−1
2
|q| dz (13.13)
−L
Choose n = 1/2 and c = cR + icI find the following then
U − cR + icI
2n−2
−1
2n−1
2n
(U −c)
= (U −c) =
(U
−c)
=
1
(U
−c)
= U −cR −icI
(U − cR )2 + c2I
(13.14)
#
Z L
Z L"
1 02
2
02
N
−
2
2
2
4U
|q|
dz
−icI
|q | + k |q| dz = icI
2 + c2
(U
−
c
)
R
−L
−L
I
• If N 2 > 41 U 02 then both terms in the brackets are positive, so cI = −cI ⇒ cI = 0
• This is usually written in terms of the Richardson number
2
g∆θ
∆u
/θ
Ri =
∆z
∆z
• with Ri >
1
4
(13.15)
required for stability.
• The reality of the atmosphere is much more complicated with many definitions of
Richardson number depending on different stability mechanisms.
• This is the mean-flow Richardson number. There is also a gradient Richardson
number and a flux Richardson number
• There is also the Bulk Richardson number which tells how unstable an entire atmospheric boundary layer probably is.
13.3
Viscous Kelvin-Helmholtz
• Linearise Navier-Stokes:
1
∂tu1 + U ∂xu1 + w1∂z U = − ∂xp + ν(∂x2 + ∂z2)u1
ρ
1
∂tw1 + U ∂xw1 = − ∂z p + ν(∂x2 + ∂z2)w1
ρ
∂xu1 + ∂z w1 = 0
(13.16)
• Assuming monochromatic solutions gives
1
−i(ω − U k)ũ + w̃U 0 = − ik p̃ + ν(ũ00 − k 2ũ)
ρ
1 0
−i(ω − U k)w̃ = − p̃ + ν(w̃00 − k 2w̃)
ρ
ik ũ + w̃0 = 0
(13.17)
• Note: We now have some second-derivatives from the viscosity.
• This will be dealt with by adding yet more derivatives to a new master equation:
The Orr-Sommerfeld equation
• The end result will be a 4th-order equation in the stream function whose solutions
are Tollmein-Schlichting waves: both stable and unstable.
• Use ũ = ∂y ψ̃ = ψ̃ 0, w̃ = −∂xψ̃ = −ik ψ̃ and take the curl (−∂y , ∂x)· = (−∂y , ik)·
of (13.16).
h
i
1
−∂y − i(ω − U k)ψ̃ 0 −ik ψ̃U 0 + ik p̃ − ν(ψ̃ 000 − k 2ψ̃ 0)
ρ
h
i
(13.18)
1 0
00
2
+ik i(ω − U k)ik ψ̃
+ p̃ + ikν(ψ̃ − k ψ̃) = 0
ρ
• Yielding
νk 4ψ = 0
νψ 0000 − νk 2ψ 00 + i(ω − U k)ψ 00 + ikψU 00
− ik 2(ω − U k)ψ + νk 2ψ 00 −
• Finally
iν(ψ̃ 0000 − 2k 2ψ̃ 00 + k 4ψ̃) + (U k − ω)(ψ̃ 00 − k 2ψ̃) − U 00k ψ̃ = 0
(13.19)
• This is commonly written
(U − c)(ψ̃ 00 − k 2ψ̃) − U 00ψ̃ = −
i
(ψ̃ 0000 − 2k 2ψ̃ 00 + k 4ψ̃)
kRe
(13.20)
i
(U − c)(ψ̃ − k ψ̃) − U ψ̃ = −
(ψ̃ 0000 − 2k 2ψ̃ 00 + k 4ψ̃)
kRe
• Consider a disturbance of wavenumber k = α within a shear layer of thickness δ
with a characteristic velocity Um and with viscosity ν.
00
2
00
• Then simply by the bending of the sinusoidal disturbance by the shear
• One can determine if the disturbance has an inflection point and use the inviscid
Rayleigh criteria to determine if the shear layer is unstable.
• This gives an Rcrit, a Reynolds number below which the flow will always be stable.
• And an envelope for possible instability for higher R.
• Viscous terms will tend to suppress this instability, but there will always be some
regime of wavenumber α and Reynolds number R for instability.
• This must be found numerically.
• I did this once for plane Poiseuille flow.
Channel flow with a pressure gradient.
• The reason was to compare the growth rates
from my newly written channel code with
the prediction of linear theory.
• Poiseuille instabilities:
– Squires theorem: 2D flow is sufficient for finding the minimum Reynolds
number for absolute instability to infinitesimal disturbances for viscous
flows.
– Numerical found that unstable for some wavenumber for R > 5772
– But will be unstable to finite 3D disturbances for R ∼ 500
– Inviscid Rayleigh criteria seems to be more useful even if it is only necessary,
not sufficient.
13.4
Thermal convection
This time in the linearised equations we will use the Boussinesq approximation from
the start. That is:
• ρ1 = −αρ0T (density decreases as T increases) and
• The only spatial derivatives for density will through T1 where
dT0
=constant.
• T = T0(z) + T1 and
dz
• There will be no mean velocity U = 0 and therefore no convective derivative (u·∇)
for the velocity equations.
• There will be vertical advection associated with dT0/dz
where αT00 , αT1 << 1.
• And I will now add viscosity ν and thermal diffusion κ
ρ0∂tu⊥1 = −∇⊥p1 + ν(∂x2 + ∂z2)u⊥1
∂xu1 + ∂z w1 = 0
ρ0∂tw1 = −∂z p1 + ν(∂x2 + ∂z2)w1 − αρ0T1g
(13.21)
∂tT1 + w1T00 = κ(∂x2 + ∂z2)T1
(13.22)
Taking the curl twice of the velocity gives equations for the Laplacian of the velocity
components while eliminating pressure:
i
h ∂
∇ × ∇ × ( − ν4)u = −αT1g ẑ
∂t
∂xu1 + ∂z w1
ρ0∂tu1
ρ0∂tv1
ρ0∂tw1
∂tT1 + w1T00
= 0
= −∂xp1 + ν4u1
= −∂y p1 + ν4v1
= −∂z p1 + ν4w1 − αρ0T1g
= κ4T1
i
h ∂
∇ × ∇ × ( − ν4)u = −αT1g ẑ
∂t
• Use ∇ × ∇ × T1ẑ = ∇ × (∇T1 × ẑ) = −4T1ẑ)
• And take the vertical component
2
2
∂
∂
∂
− ν4 4w1 = αg
+
T1
2
2
∂t
∂x
∂y
∂
− κ4 T1 = −w1T00
Rewrite (13.22) ∂tT1 + w1T00 = κ4T1 as
∂t
∂
• And finally operate on (13.23) by ( ∂t − κ4)
i
h
2
2
∂
∂
∂
∂
T1
− κ4
− ν4 4w1 = αg
+
2
2
∂t
∂t
∂x
∂y
• to get Acheson (9.22)
∂
∂
dT0 ∂ 2
∂2
− ν4
− κ4 4w1 = −αg
+
w1
∂t
∂t
dz ∂x2 ∂y 2
(13.23)
(13.24)
2
2
∂
∂
dT0 ∂
∂
− ν4
− κ4 4w1 = −αg
+
w1
2
2
∂t
∂t
dz ∂x
∂y
is second order in time. If the eigenfunctions were simple we could perhaps solve
it.
• However, the equation is 6th-order. Very difficult.
– The usual physical condition is no-slip boundary conditions.
– This can be solved numerically by the same methods as used for finding TollmeinSchlichting waves.
– However:
• Free-slip or stress-free boundary conditions:
– I will now outline that there is an easy solution for this case.
– This will help demonstrate the idea of a critical wavenumber kc
– The relevant dimensionless parameter: the Rayleigh number.
– I will start mid-page 310 of Acheson. Originally due to Rayleigh (1916).
– Free-slip boundaries: w = ∂z u = ∂z v = ∂z2w = ∂z4w = 0 at z = 0, d.
Nπ
– Assume w1 = sin(kz z)eik⊥·x⊥ est with kz =
d
2
– Using k 2 = a2∗ and k⊥
= a2
dT0 2
2
2
then (13.24) becomes
(s + νk )(s + κk ) + αg
k =0
dz ⊥
2
– Using k 2 = a2∗ and k⊥
= a2 then (13.24) becomes
dT0 2
(s + νk )(s + κk ) + αg
k =0
dz ⊥
dT0 ∆T
– This is quadratic in s, therefore solve for s using
=
dz
d
1/2
2
1
αg∆T k⊥
1
4
s = − (ν + κ)k 2 ± (ν + κ)2k 4 +
−
νκk
2
4
d k2
2
2
(13.25)
(13.26)
2 1/2
1
αg∆T k⊥
1
= − (ν + κ)k 2 ± (ν − κ)2k 4 +
2
4
d k2
– If ∆T > 0 (convection), then s is real.
– If ∆T < 0 (stratification), there is possibly
2
– Growth s > 0 if: 1
k
αg∆T
⊥
(ν − κ)2k 4 +
4
d k2
2
αg∆T k⊥
d k2
αg∆T d3
Ra =
κν
waves: internal-gravity waves.
1
> (ν + κ)2k 4
4
>
>
αg∆T d3
– Where Ra =
is the Rayleigh number.
κν
νκk 4
6
4k
d 2
k⊥
– Acheson explains how to show that the critical horizontal wavenumber
k⊥c
27π 4
π
and Rac =
= 657
=√
4
2d
• No-slip boundary conditions
– u = v = w = ∂z w = 0 at z = 0, d
– ∂z w = 0 because of incompressibility: ∂z w = −∂xu − ∂y v = −0 − 0 at z = 0, d
– In this case Rac = 1708
– Unlike plane shear instabilities where Rayleigh’s first guess is closer to the real
case,
for convection, this is the physically relevant case.
• Bénard cells in the laboratory (left) and in the atmosphere (right).
L Take a stainless steel frying pan. Fill with a thin layer of oil and use low heat.
Sprinkle chocolate powder on the surface and see the patterns.
R Start with an atmosphere over a warm ocean. The moist air will rise until a cloud
layer forms. Marine stratocumulus. There will be radiative cooling in the cloud
layer, creating Bénard cells.
• Dislocations in low Rayleigh number Rayleigh-Bénard flow.
Assume you start with a uniform roll pattern which cannot perfectly match the
boundary conditions.
Then dislocations will form:
imperfections in the cellular pattern.
• These can lead to intricate patterns.
• These can lead to intricate patterns.
• Show Ecke: Thermochromic liquid crystal visualization of the temperature field of
convection.
• My pictures
Cross-sections of temperature
fluctuations
through the ends and
centre from two directions
for the Ra = 8 × 107,
AR = 1, P r = 0.7
calculation.
Arrows
represent velocities in the
planes.
The geometry
is cubic, the top and
bottom boundary conditions are no-slip, constant
temperature and the
sidewalls are free-slip and
insulating.
Volume rendering of temperature for one time from
the Ra = 2 × 107 calculation. Note the dominant large-scale diagonal
pattern, which persists for
all Rayleigh numbers simulated and all times.
Colour contour plots in horizontal planes of the temperature and vertical velocity for
Ra = 2 × 107 at twc/8 = 1.47.
Each plot is scaled to its own
minimum and maximum. In
this plate levels are near the
lower surface; BOT1: z=-.97,
BOT2: z=-.83, BOT3: z=-.61.
Note the dominant convective
structure, buoyant sheets leaving a surfaces and plumes colliding with the surfaces, producing a cellular pattern, with
fine networking near the surfaces.
Vertical
x, z
crosssections of T at 12 times.
The vertical direction z
has been stretched by a
factor of 2. Times run
up the column on the left
from bottom to top, then
up the column on right.
14
Turbulence statistics: 3D versus 2D
• 3D turbulence is a statistical system
– But does not obey classical thermodynamics.
– Nor is it chaos (A specific mathematical invention)
– Often we idealise by assuming it is in a statistically steady state. But:
– The dynamics more closely corresponds to non-equilibrium statistical mechanics.
– Or maybe critical phenomena (related to phase transitions).
– What characterises all types of turbulence is: Cascades
– The scale of energy or some other property changes with time.
“Big whirls have little whirls,
which feed on their velocity,
and little whirls have lesser whirls,
and so on to viscosity”.
Compare this with Jonathan Swift’s flea sonnet:
“A flea hath smaller fleas that on him prey;
And these have smaller yet to bite ’em,
And so proceed ad infinitum.”
However, this is not what 3D turbulence looks like.
Vortex tubes dominate
turbulence, up to the
largest Reynolds number simulated. Where
do they come from?
Decaying 20483 by M.
Taylor, Los Alamos
Zooming in on the structures. The statistical description could be independent of
the scale you are looking at: scale-invariance.
From 1985 paper with first 1283 calculations.
Three images of 3D forced 643 turbulence in a box.
Lines show vector directions for quantities above a
threshold. In all three frames the primary field, vorticity, is shown using blue lines .The frames at the
top are at t = 1.50, but different perspectives and
different secondary fields. Yellow Yellow for scalar
gradients and red for compressive strain, White is
the overlap between the scalar or compressive fields
and two interacting regions of vorticity.
Top frames: two vortex regions are strongly interacting.
Bottom: reconnection leaves one vortex tube and a
straining region.
14.1
3D Energy Cascade
• Kolmogorov’s proposal was that between the large energy-containing spatial scales
and the small dissipation scales:
– There are many steps.
– Image a stream of water “cascading” down a water fall.
– In each cascade step the energy in the eddies becomes more and more disordered, isotropic and homogeneous.
– Therefore, for sufficiently high Reynolds numbers:
∗ No matter how anisotropic and inhomogeneous the energy containing scales
are.
∗ Small length-scale (high wave-number), turbulence is assumed to be isotropic,
homogeneous and statistically independent of the large-scale turbulence.
∗ With constant energy input one hopefully achieves universal equilibrium.
∗ Equilibrium statistical systems are something statistical physicists know about.
– This is why so much effort has been expended on understanding isotropic turbulence by a segment of the physics community (including me).
14.2
3D Energy Spectrum
• Define
Ẽ(k) =
• Define dẼ(|k|)d|k| =
R
1
2
Z
< u(x)·u(x + r) > e−ik·r d3r
dΩẼ(k) where dΩ is the solid angle.
• The energy cascade in Fourier space is visualised as:
• Universality hypothesis: Assume that the statistics in an inertial subrange
depend only upon the length scale and the rate of energy cascade through the
spectrum.
• This energy cascade rate equals both:
The energy input at wavenumber kf and energy dissipation= at wavenumber kη .
• The length scale is 1/k where kf << k << kη
2 2
L
U
=
Dimensionally : =
T
T3
dẼ
∼ a k b
d|k|
L3
T2
L : 3 = 2a − b
∼
L2
T3
a b
1
L
T : 2 = 3a
• Solution: b = −5/3, a = 2/3
• Then the only dimensionally consistent energy spectrum is:
dẼ/d|k| = C2/3k −5/3
(14.1)
where C is hopefully a universal constant
of order unity.
• This is the Kolmogorov spectrum.
"
#
2
dẼ
U
=
d|k|
k
=
3
L
T2
• In 3D turbulence there is a forwards energy cascade:
– Small scale structures form.
– Energy moves from large scales to small scales.
– But not by the mechanism that Richardson imagined.
• In 2D, there is a Richardson cascade in reverse.
– Large scale structures form.
– Energy moves from small scales to large scales.
– The butterfly effect. A butterfly flaps its wings in Boulder. and the weather
changes in Brazil.
– Not quite true: we now believe the scale for this to happen is the synoptic scale
of about 600km.
• The theory of 2D turbulence predicts two spectral regimes:
• For k < kf , large length scales, energy should cascade to large scales.
• Because it is an energy cascade: Ẽ(k) ∼ k −5/3 predicted.
• for k > kf there is an enstrophy cascade. Mean square vorticity cascades to small
scales.
• Dimensional analysis predicts: Ẽ(k) ∼ k −3
• Both regimes can be produced in simulations with:
an energy sink at large scales
and viscous dissipation at small scales.
• 2D inertial subranges
– Backwards energy cascade gives the same scaling as forwards cascade:
dẼ(k)
= C2d2/3k −5/3
dk
with a different constant C2d.
– Recall that in 2D Euler: conservation of the circulation and all higher moments
of vorticity.
– Forwards cascade is postulated to be of enstrophy, the mean square vorticity.
Z
dΩ̃(|k|)
Ω
1 1
∼
= dθω(|k|)2 = 2|k|2Ẽ(|k|)
d|k|
k
T 2 1/L
1
d
[Ω] = T 3
– Let Ω = Ω:
dt
a b
dΩ̃(|k|)
L
1
1
– If
∼ aΩk b then
∼
d|k|
T2
T3
L
– Solution: b = −1, a = 2/3
– Then the only dimensionally consistent energy spectrum is:
2/3
2/3
dΩ̃/d|k| = CΩ k −1 or dẼ/d|k| = CΩ k −3
(14.2)
• In numerical simulations:
– Theses regimes are seen in the right order in if
– There is large-scale forcing.
– Or as a transient until lowest wavenumbers saturate.
– High wavenumber regime is steeper
than k −3.
• If there is no large-scale energy sink:
– Energy builds up at the largest scales.
– Saturation spectrum is k −3.
– Two backwards cascade regimes.
– Figures from Tran and Bowman, Physical Rev E 69 036303 (2004)
• Order of the regimes in the environment.
• Originally it was thought that k −3 for k > kf and k −5/3 for k < kf . But
• Nastrom, Gage, Jasperson
• Is this therefore 2 backwards cascades regimes?
• What direction does the energy flow in the environmental -5/3 regime?