environmental fluid dynamics - university of …€¦ · environmental fluid dynamics ......

Stuart Dalziel 1

Environmental Fluid Dynamics

Stuart B Dalziel, Joanne M Holford & Gary R HuntDepartment of Applied Mathematics and Theoretical PhysicsThe University of CambridgeSilver StreetCambridge CB3 9EWENGLAND

Phone: (44)(1223) 337911; Fax: 337918

Lent 1999

Tuesdays and Thursdays at 12:00Syndics Room

Outline

Part 1: Horizontal flows, porous media & particles

• SBD

• 8 lectures

Part 2: Plumes & jets

• GRH

• 4 lectures

Part 3: Stably stratified flows

• JMH

• 4 lectures

Environmental Fluid Dynamics Lent 1999

Stuart Dalziel 2

Contact

Stuart Dalziel:

• [email protected]

• Ph. 337911

• F46(s)

Gary Hunt:


• Ph. 337858

• F49(s)

Joanne Holford:


• Ph. 337858

• F49(s)

Course web site

• http://www.damtp.cam.ac.uk/user/fdl/people/sd103/lectures/maths/

Lab demonstration course

• Mon & Wed @ 11:00 for first half term

• Not necessary for EFD course, but it may help

• Includes visits to Aerodynamics and Chemical Engineering

Examples sheets and classes

• Two examples sheets, one for each half of course

• Not yet ready

• First examples class: 2:00 to 4:00, Thursday 11 February, Syndics Room


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Revision class

• Sometime next term

Surgery

• 2:00 to 3:00, Wednesday afternoons

• Best to check by e-mail.

Examination

• Not “Open Book”

• Questions covering entire course

• Final format (e.g. number of questions) not yet determined.

PhD’s

• Geophysical & Environmental Fluid Dynamics

• Combination of laboratory experiments, theoretical and/or numerical modelling

• NERC/EPSRC/EU/Industry


Stuart Dalziel 4

Course summaryWhereas the large-scale dynamics of the atmosphere and oceans affect the climate and weather

we experience, smaller scale processes which do not feel the earth’s rotation can have a significantimpact on the local environment. This course concentrates on the fluid dynamics of these processeswhich may range in scale from the flow through a doorway up to the accidental release of apollutant which may spread to contaminate a region many kilometres in size.

The buoyancy effects associated with density differences are central to most of the flows whichwill be discussed. These density differences may be due to differences in temperature, concentrationof a solute, composition or the presence of a second phase. For some flows the density differencesare large and the differences in inertia important. For others the density differences are small and theBoussinesq approximation may be applied.

Many of the flows are high Reynolds number and turbulent and a brief discussion on the mannerin which the density differences modify the turbulence will be presented. However, for most of theflows parameterisations are introduced which model the physical processes associated withturbulence in a more straight straightforward manner.


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Part 1 (8 lectures):

Shallow water equations

• Derivation

• Implications

Single-layer hydraulics

• Classical flow over a weir

Gravity currents

• Front conditions

• Boussinesq verses non-Boussinesq

Two-layer flows

• Derivation

• Straits and doorways

Porous media

• Darcy’s Law

• Ground water

Sedimentation & resuspension

• Mechanisms

[Complex gravity currents]

• Lossy currents

• Thermodynamic effects

• Particle-driven currents


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1 Shallow water flows

1.1 Introduction

1.1.1 Description

• Examples

◊ Flow over weir

◊ Gravity current

◊ Some aspects of the atmosphere and oceans

∗ Sea breeze

∗ Flow through straits

∗ Rotation may be importat

• Key features

◊ Length scales

◊ Accelerations

1.1.2 How much do we know?

• Less than you might think

• Excellent understanding of steady flows on a smooth, flat surface or through rectangular channels

• Good descriptions of transients such as gravity currents in simple (ideal?) conditions

• Reasonable understanding of drag, two-layer effects, rotation, particle-driven flows

• Poor understanding of transients with topography, entrainment

• Almost no understanding of collisions, complex ambient stratifications, complex substrates,resuspension, fully three-dimensional flows, interaction with obstacles

1.1.3 Approaches to modelling

• Integral (box) models

◊ For transient problems only

◊ All dynamics in boundary conditions

• Shallow water equations

◊ Long wave approximation

◊ Provides full dynamic treatment of depth-averaged equations

◊ Analytical solution sometimes possible

◊ Efficient numerics

• Full Navier Stokes

◊ Requires numerical solution


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◊ Does not assume much, but difficult and expensive to get adequate resolution and correctbehaviour

◊ Two-dimensional modelling accessible with workstations/powerful PCs

◊ Three-dimensional modelling accessible by supercomputers, but only for relatively lowReynolds numbers

1.2 Shallow water equations

• Mathematically simple

• Two-dimensional (planar)

• Three-dimensional (axisymmetric)

◊ Channel approximation (a generalisation of axisymmetric flow where flow in a channel ofwidth b(x) and all flow is parallel with this.

• Three-dimensional (general)

◊ A much more difficult problem

1.2.1 Derivation - simple

• First principles

u + ∂u/∂x δx

2δx

h + ∂h/∂x δx

huh

h − ∂h/∂x δx

u − ∂u/∂x δx

• Assume

◊ Inviscid

◊ Unidirectional (parallel to x axis)

• Derivation

◊ Continuity

Flu_ in = (u − ∂u/∂x δx)(h − ∂h/∂xδx)

Flux_out = (u − ∂u/∂x δx)(h − ∂h/∂xδx)

Rate_of_accumulation = ∂h/∂t 2δx

Equate Rate_of_accumulation = Flux_in − Flux_out and let δx → 0 (thus ignore terms O(δx2))

( )∂∂

∂∂

∂∂

∂∂

∂∂

h

tu

h

xh

u

x

h

t xuh+ + = + = 0


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◊ Momentum

Flux_in = (u − ∂u/∂x δx) [(u − ∂u/∂x δx)(h − ∂h/∂xδx)]

Flux_out = (u + ∂u/∂x δx) [(u + ∂u/∂x δx)(h + ∂h/∂xδx)]

( ) ( )Pressure force to right g h h x x z dz g h h x xh h x x

_ _ _ / //

= − − = −−

∫ ρ ∂ ∂ δ ρ ∂ ∂ δ∂ ∂ δ

0

21

2

( ) ( )Pressure force to left g h h x x z dz g h h x xh h x x

_ _ _ / //

= + − = ++

∫ ρ ∂ ∂ δ ρ ∂ ∂ δ∂ ∂ δ

0

21

2

Rate_of_accumulation = ∂uh/∂t 2δx

Equate Rate_of_accumulation = (Flux_in − Flux_out) + (Pressure_force_to_right − Pressure_force_to_left) and let δx → 0

∂∂

∂∂

uh

t xu h gh+ +

=2 21

20

• Alterative expressions for momentum equation

∂∂

∂∂

∂∂

u

tu

u

xg

h

x+ + = 0

Exercise: Convert between ∂(uh)/∂t form and ∂u/∂t form.

• Axisymmetric - watch out for momentum equation

Examples sheet question: Do derivation for axisymmetric flow

• General geometry (small lateral acceleration)

◊ Use width of flow b(x).

◊ Width must vary slowly so that lateral accelerations ~ u2/b ∂2b/∂x2 are much less than thegravitational acceleration g

1.2.2 Hyperbolic system

• Want to look for trajectories through x,t space along which something doesn’t change.

∂∂

∂∂

∂∂

h

tu

h

xh

u

x+ + = 0

∂∂

∂∂

∂∂

u

tu

u

xg

h

x+ + = 0

◊ Consider rays defined by x = x0 + λ(t−t0). As a vector, these rays may be written as

( )x

t

x

tt t

=

+ −

0

00 1

λ.

The derivative along these rays is therefore


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d

dsx

tx t

T

=

= +λ

∂∂∂∂

λ∂∂

∂∂1

.

◊ We look for solutions where the along-ray derivative d/ds vanishes. Thus we require ∂/∂t = −λ∂/∂x. Substitution in reveals

− + + =λ∂∂

∂∂

∂∂

h

xu

h

xh

u

x0

− + + =λ∂∂

∂∂

∂∂

u

xu

u

xg

h

x0

⇒

( )uh

xh

u

x− + =λ

∂∂

∂∂

0

( )uu

xg

h

x− + =λ

∂∂

∂∂

0

Eliminate ∂u/∂x by taking (u−λ) times the continuity equation and subtracting from it h timesthe momentum equation:

[(u−λ)2 − gh](∂h/∂x) = 0

◊ Two solutions:

∂h/∂x = 0

λ = u ± (gh)½

The first of these is of no interest.

Let c = (gh)½,

λ = u ± c

• Trajectories described by x = λt, where λ = u ± c are called characteristics.

• Now need to determine what is conserved along these lines. To achieve this we need to considerthe same manipulations of the original continuity and momentum equations as we undertook toobtain the characteristics, i.e. taking (u−λ) times the continuity equation and subtracting from it htimes the momentum equation. Since u − λ = –+c and h = c2/g, we may simplify this operation byjust adding ±c/g times the momentum equation to the continuity equation:

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

h

tu

h

xh

u

x

c

g

u

t

uc

g

u

x

c

g

h

x+ + ± ± ± = 0

Reducing h to c2/g gives

( ) ( ) ( )

2 2 2

2 2

0

22c

g

c

t

uc

g

c

xc

u

x

c

g

u

t

uc

g

u

x

c

g

c

x

c

g tc u u c

xc u

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+ + ± ± ±

= ± + ± ±

=


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For a non-trivial solution this requires u ± 2c to be conserved along the λ = u ± c characteristics.

◊ Along λ = u+c, the quantity u + 2c is conserved.

◊ Along λ = u−c, the quantity u − 2c is conserved.

1.2.2a Matrix formulation

• Start by writing our equations in a matrix formulation

u h

g u

h

u

h

ux t

+

=

1 0

0 1

0

0

and note that our earlier ∂/∂t = −λ∂/∂x leads to the generalised eigenvalue problem

u h

g u

h

u

h

ux x

−

=

λ

1 0

0 1

0

0,

which, for a non-trivial solution, requires

( )u h

g uu gh

−−

= − − =λ

λλ 2 0 ,

and λ = u ± c (where c2 = gh), exactly as before.

• In order to understand how to extract the equations along the characteristics we shall consider amore general approach to hyperbolic systems.

1.2.2b General approach to hyperbolic systems

• Our matrix formulation in the previous section is of the form

Avx + Bvt = f,

where vT = (h u).

• Let l be some vector, then we may write a linear combination of the equations as

lTAvx + lTBvt = lTf.

This is a generalisation of the procedure where we earlier took (u−λ) times the continuityequation and subtracting from it h times the momentum equation.

• Let us choose l so that all derivatives are in the same direction, s (say), such that they may bewritten in the form

mTvs = mT(cosθ vx + sinθ vt),

◊ Comparing with lTAvx + lTBvt = lTf suggests that if

cosθ mT = lTA, sinθ mT = lTB,

then

mTvs = lTf.

◊ Eliminating m gives lTA = λlTB, where λ = cotθ, or

lT[A − λB] = 0T.


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◊ For non-trivial l we require |A − λB| = 0, giving the slope of the characteristics, λ, as before.This is then equivalent to us looking for solutions where ∂/∂t = −λ∂/∂x. The vectors l, whichbefore were chosen to eliminate ∂u/∂x terms from the equation, are the left-hand eigen vectorsof the system. The system is hyperbolic if all eigenvalues λ are real.

• Recalling mTvs = lTf and mT = lTA/cosθ = lTB/sinθ, then the equations along these characteristicsare

lT[Avs − f dx/ds] = 0 = lT[Avs − cosθ f] ,

or

lT[Bvs − f dt/ds] = 0 = lT[Bvs − sinθ f].

• For our present problem this means we must determine the left-hand eigen vectors of [A−λB],i.e. lT:

( ) ( )1 0 0lc h

g c

m

m

= ,

so

lT = (1 ±c/g).

• Hence, the equation along the characteristics is

10

00±

+

=

c

g

u h

g u

h

us

cosθ .

Expanding

1 1

2

2

2 2

0

2 2

2

2

2 2

±

+

+

= ±

+

+

= + + ± ±

=

c

g

u

g

dc

ds

c

g

du

dsdc

dsu

du

ds

c

g

uc

g

dc

ds

c

g

du

ds

cdc

dsu

du

ds

uc

g

dc

ds

c

g

du

ds

c

g

dc

ds

uc

g

du

ds

so

( ) ( )

( ) ( )0

2

2=

+ + = +

− − = −

u cd

dsc u u c

u cd

dsc u u c

on

on

λ

λ.

u h

g u

h

u

h

ux t

+

=

1 0

0 1

0

0

◊ Note: it would have been easier algebraically if we had used the alternate lT[Bvs − sinθ f] formof the equation since B is an indentity matrix.


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1.2.2c Implications of hyperbolic nature

• Analogy with compressible gas

◊ Mach number = velocity/sound speed

• Froude number

Fr = velocity/wave speed = u/c = u/(gh)1/2

• Information propagation

◊ If |Fr| < 1 then eigenvalues λ = u+c and λ = u−c are of opposite sign and waves can propagatein both directions. This state is called subcritical flow

◊ If |Fr| > 1 then eigen values are of the same sign and waves can only propagate in onedirection (the same direction as u). This state is called supercritical flow

◊ If |Fr| = 1 then either λ = u+c or λ = u−c vanishes and one of the two waves does notpropagate. This state is called critical flow.

x

t

u1,c1

ua,ca ub,cb

λ = u + c

λ = u − c

◊ We may calculate u1, c1 = (gh1)1/2 at x = x1, t = t1 directly from quantities at t = 0 if we know

where the λ = u + c and λ = u − c characteristics which pass through x1,t1 were at t = 0 and thevalues of u and c at thos points. In the above figure the λ = u+c characteristic came fromx = xa where u = ua and c = ca, whereas the λ = u−c characteristic came from x = xb whereu = ub and c = cb. Noting that u + 2c is conserved along the λ = u + c characteristic, andsimilarly for the other characteristic, we obtain a pair of equations giving u1 and c1:

u1 + 2c1 = ua + 2ca,

u1 − 2c1 = ub − 2cb.

◊ Zone of dependence. The conditions at x1,t1 are influence only by conditions within thetriangle described by the two characteristics approaching this point. To illustrate this, considerthe solution along the λ = u + c characteristic emanating from xa at t = 0. The initial slope ofthis characteristic is just λ = ua + ca. A short time later the values of u and c will have changedbecause of the need to conserve u − 2c along the λ = u − c characteristic which will haveoriginated from a point to the right of xa. Thus the slope of the λ = u + c characteristic at this


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later time depends also on the conditions immediate to the right of xa. As we move closer to t1

the values of u and c are influenced more and more by values originating to the right of xa andpropagating along the corresponding λ = u − c characteristic. Thus the information whichultimately gives us the values u1, c1 has come from the region between the two characteristics.This is called the zone of dependence for the point x1,t1. The conditions outside this region donot affect the solution.

◊ Zone of influence. In a similar way to the zone of dependence, the conditions at xa, say, areonly capable of influencing the future solution for points which lie between the λ = u + c andλ = u − c characteristics emanating from xa.

• Saint Venant solution; will be dealt with in more detail in §1.4.2.

◊ For t ≤ 0 there is fluid of depth H for x ≤ 0 and no fluid for x > 0

◊ At t = 0 the fluid is allowed to collapse.

At t = 0:u = 0h = H for x < 0

At t > 0:at front, h = 0 ⇒ uf = 2(gH)1/2

LR = (gH)1/2t as rarefactionwave propagates back atvelocity c = (gH)1/2.

x = 0

1.2.3 Derivation by averaging

• Start with Navier-Stokes

The shallow water equations may be derived in a number of ways. Here we express the velocity

field as u = u + u’. Here u = u (x) = ( u , v , w ) is the mean velocity, calculated over the width

b(x) and depth h(x) of the current, at some location x, and u’ = u’(x,y,z) = (u’,v’,w’) are thefluctuations about this mean. The continuity equation integrated over the cross-section of theflow gives

bhu

xw w dy v v dzz h zb

b

y b y b

h∂∂

+ − + − == =−

= =−∫ ∫001

2

12

12

12

0 ,

where we have assumed the depth h is constant across the width of the current and used the

identity u’ = 0 . The span-wise horizontal and vertical velocities at the boundaries are related to

the stream-wise horizontal velocity at the boundaries through wz=h = ∂h/∂t + uz=h∂h/∂x, wz=0 = 0and vy=½b = -vy=-½b = ½uy=½b∂b/∂x so that the continuity equation becomes

bhu

xu b

h

xh

b

xb

h

t

∂∂

∂∂

∂∂

∂∂

+ +

+ = 0 .


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Similar substitution into the x momentum equation leads to

∂∂

∂∂

∂∂

ν∂∂

ν∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

u

tu

u

xg

h

x

u

x

b

u

y

u

y h

u

z

u

zu

u

xv

u

yw

u

zy b y b z h z

+ + −

=′

−′

+

′−

′

− ′′

+ ′′

+ ′′

= =− = =

2

2

0

1 1

12

12

.

This equation has been written so as to separate the behaviour of the mean from the fluctuations.The last term on the left-hand side will be negligible, due to the slowly varying assumption,except in the neighbourhood of turning points in the velocity field. In most treatments of theshallow water equations, it is simply discarded. Note that we have assumed that the centre-line of

the channel is straight (thus v = 0) and that the width and depth are slowly varying (allowing us

to ignore w and replace the integrals of a derivative with the derivative of an integral in the

averaging procedure).

The right-hand side of represents momentum losses. The first term in square brackets gives theshear stresses on the surroundings, while the second term in square brackets are the so-calledReynolds stresses. It is not practical to retain either of these terms in this form. Instead we note

that we expect the fluctuations u’ to scale on u and the length scale of these fluctuations to vary

as the thickness of the pool (typical vortex structures like being approximately circular and adoptthe largest scale available). Typically h << b so that the shear stress on the lateral boundaries may

be ignored and we rewrite these terms as −CLν u /h2 and −CT u | u |/h, where ν is the kinematic

viscosity, CL is a laminar drag coefficient (which may be calculated as CL = 2/3 from Poiseulleflow with appropriate boundary conditions) and CT a turbulent drag coefficient. The value of CT

will depend on the bottom roughness, but a typical value for a smooth bottom is CT = 0.03.Except for very thin fluid layers, or fluids with a large viscosity, the turbulent drag will generallydominate over the laminar drag term.

For clarity we now drop the overbars on the mean velocities and rearrange the continuity andmomentum equations as

( )∂∂

∂∂

h

tb

xbuh+ =−1 0 ,

and

∂∂

∂∂

∂∂

νu

tu

u

xg

h

xC

u

hC

u u

hL T+ + = − −2 ,

where we have discarded the viscous term involving stream-wise derivatives of u as this willgenerally be negligible compared with the other terms due to the slowly varying assumption. Wehave also discarded the lateral drag terms as these will, in general, be small compared with thebottom drag terms, except in a narrow channel where b <~ h. One important feature of themomentum equation is that it does not depend on the domain width b(x).

1.2.4 Drag

• Effects

◊ Slowing & thickening


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∗ Possibly triggering hydraulic jump

◊ Lobes & clefts in gravity currents

◊ Turbulence → entrainment?

1.2.5 Boussinesq flows

• Match pressure across interface

∆h

ρ2

ρ1

◊ Pressure difference at bottom due to height difference ∆h:

∗ ρ1g∆h if ρ2 << ρ1

∗ (ρ1 − ρ2) g∆h if ρ1 − ρ2 << ρ1,ρ2

• Momentum equation has (ρ1 − ρ2)/ ρ g ∂h/∂x in stead of g ∂h/∂x

• Reduced gravity

( )′ =−+

g gρ ρρ ρ1 2

12 1 2

plays the role of gravity.

• Momentum equation (ignoring drag terms)

∂∂

∂∂

∂∂

u

tu

u

xg

h

x+ + ′ = 0

• Finite depth upper layer

◊ May need to consider flow within upper layer as well

◊ Wave speed is function of total depth as well as layer depth

From here on we unify the treatment of flows with small and large density differences.Many aspects of these flows are identical except that the relevant gravitationalacceleration is g for flows with large density differences and g’ for flows with smalldensity differences. Aspects of the flow which apply to only one or the other, and anydifferences which arise will be flagged explicitly. We will not consider flows whereboth the layers are playing an active role until §1.5.

Examples sheet question: Determine the characteristics in a two-layer system


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1.2.6 Entrainment

This section applies only to miscible fluids. While immiscible fluids can entraineachother, the result is a two-phase flow which, given a chance, will separate again.Such flows are beyond the scope of this course.

• Driven by instabilities on the flow and turbulence within

◊ Unstable to short wave instabilities: Kelvin-Helmholtz & Holmboe

◊ Long wave instabilities also possible

Examples sheet question: Analyse a two-layer Boussinesq flow for long wave instabilities.

• Effect on g’

◊ Reduced density contrast

• Effect on h

◊ Increased volume of layer into which fluid entrained

• Effect on g’h

◊ For quiescent ambient, entrainment decreases g’ but increase h by comparable amount so thatg’h conserved

• Momentum

( ) ( )∂∂

ρ∂∂

ρ ρ ρt

hx

u h w we d1 1 1 1 1 2 1+ = −

where we is the entrainment velocity of ambient fluid into the current and wd is the detrainmentvelocity if current fluid into the ambient.

( ) ( )∂∂

ρ∂∂

ρ ρ ρ ρ ρt

h ux

u h u gh w u w ue d1 1 1 1 1 1 1 1 2 12

2 2 1 1

1

2+ + −

= −

∗ Boussinesq ⇒ rhs is ρ(weu2 − wdu1)

∗ Need a third equation! If linear mixing then can use conservation of volume

( )∂∂

∂∂

h

t xu h w we d

11 1+ = −

∗ Combination with mass equation leads to

∂ρ∂

∂ρ∂

1 1

1tu

x

w w

he d+ = −

−

Exercise: Eliminate temporal derivatives of ρ1 and h1 from momentum equation.

• Typically entrainment/detrainment will lead to the stratification being more complex than simplytwo-layer.

• The definition of we and wd must be relative to the current interface, but this is a poorly definedconcept and so in general the above approach is of limited value.


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[1.2.7 Rotation]

This section is not examinable.

• Balance between cross-channel pressure gradient g’∂h/∂y and Coriolis force fu gives cross-channel slope.

• Form of cross-channel profile depends on potential vorticity distribution

1.3 Single-layer hydraulic flow

1.3.1 Examples

• Broad-crested weirs

◊ Flow is slowly varying

• Sharp-crested weirs

◊ Flow is not slowly varying, but behaves much in the same way, at least upstream.

• Mountain ranges

• Submarine sills

• Controlled and uncontrolled flows

H

u

h

b

x

z

x

y

1.3.2 Conservation of mass

uhb = Q

1.3.3 Conservation of Bernoulli potential

• Inviscid Navier Stokes momentum equation (Euler equation), using suffix notation:

( )∂∂

∂∂ ρ

∂∂

ρu

tu

u

x xp gzi

ji

j i

+ + + =1

0


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Multiply by ui to obtain energy equation (suffix notation)

( )uu

tu u

u

xu

xp gzi

ii j

i

ji

i

∂∂

∂∂ ρ

∂∂

ρ+ + + =1

0,

and rearrange

∂∂

∂∂ ρ

∂∂ ρt

ux

u up

gzt

pgzj

ji i+

+ +

− +

=

1

20 ,

For steady flow the material derivative is conserved, thus

1

22u + + =

pgz const

ρ

◊ For channel flow (|u|2 = u2):

½ u2 + p/ρ + gz = const.

• Look at interface or bottom (or, anywhere between due to assumption of hydrostatic pressure)

½ u2 + g(H+h) = const.

1.3.4 Specific energy

• Eliminate velocity:

u = Q/hb

( )1

2

2

2 2

Q

gb hH h E const+ + = =

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

Depth (h)

Spec

ific

Ene

rgy

(E)

Specific Energy function.

1.3.5 Hydraulic control

• Mechanism

◊ For example, constant width, varying bottom elevation H:


Stuart Dalziel 19

½Q2/(gh2b2) + h = E − H

Moving along channel moves the specific energy curve up and down without changing shape

◊ Subcritical and supercritical branches

∗ Subcritical to right of turning point

∗ Subcritical to left of turning point

∗ Critical at turning point:

∂E/∂y = −Q2/(gb2h3) + 1 = 0

⇒ u2 = (Q/bh)2 = gh

◊ Can only swap branches by passing through turning point

• Implications

• Froude number

⇒ Fr2 = 1 − ∂E/∂h

• How do we match with the downstream conditions?

Examples sheet question: Determine the relationship between the upstream depth where u ~ 0 andthe depth of water over a weir.

1.3.6 Hydraulic jumps

• Why

◊ Necessary for matching

◊ Triggered by obstruction or friction

◊ Slowing down of waves on supercritical flow

◊ Can not have “hydraulic drop”

• Key characteristics

◊ Energy reduced, but momentum conserved: an important distinction

◊ Dissipation

◊ Downstream specific energy less than upstream of jump ⇒ lower elevation than equivalentsubcritical branch

◊ Conserves momentum, unless triggered by obstacle.

• Different forms

◊ Undular

◊ Turbulent

1.3.7 Bores

• Equivalent to hydraulic jump in frame of reference moving with bore, provided ambient densitynegligible

• Different if Boussinesq, or less dense than ambient


Stuart Dalziel 20

• Hyperbolic system

x

t

u1,c1

ua,ca ub,cb

λ = u + c

λ = u − c

uc,cc

◊ A shock (hydraulic jump or bore) forms when either two sets of λ = u+c or two sets of λ = u−c characteristics intersect, conveying to the solution two conflicting sets of information. In thecase illustrated above, the λ = u+c characteristics from both xa and xb carry u + 2c informationto the point x1,t1. The λ = u−c characteristic from xc also carries information. The net result isthat there are two values for u1,c1 predicted, and a discontinuity forms in the solution. To theleft this is

uL1 + 2cL1 = ua + 2ca,

uL1 − 2cL1 = uc − 2cc,

while to the right it is

uR1 + 2cR1 = ub + 2cb,

uR1 − 2cR1 = uc − 2cc.

The position of this shock then propagates as λ = ½ (uL1+cL1 + uR1+cR1).

1.4 Gravity currents

1.4.1 Description

• Examples

Need some pictures!

• Features:

◊ Head

◊ Effect of friction

◊ Lobes & clefts

• Types

◊ Lock exchange


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◊ Lock release

◊ Continuous

◊ Turbulent

◊ Laminar

◊ Viscous…

1.4.2 Early models

• Dimensional analysis: u2 ~ g’H

• Saint Venant

◊ Dam break: at t = 0, we have h = H for x ≤ 0 and h = 0 for x > 0.

∗ Hypothesise that a self-similar flow exists so that the characteristics form an expansion fan.

x

t

0

u = 0c = Cλ = −C

u = 2Cc = 0λ = u±2c = 2C

Front

Backλ = u − c = x/tu + 2c = 2C (along the λ = u + c characteristic)⇒ 3c = 2C − x/t⇒ h = (1/9)(2C − x/t)2/g

λ = u + ccharacteristic

λ = u − ccharacteristic

• Prandtl: Flow in current exceeds front velocity, leading to raised fluid which is assumed not tofall. Propagation speed by dynamic balance (excluding hydrostatic effects) - can not apply inreality.

• von Kármán:

◊ Flow irrotational

◊ Bernoulli from far down stream to the front, thus uf2 = 2g’Hdownstream.

◊ Also ideas similar to sharp-crested water waves gave front-ground angle of 600.

◊ But: force imbalance; drag requires rotational flow

1.4.3 Energy conserving cavity flows

• Brook Benjamin (1968): Gravity currents and related phenomena (JFM 31, 209-248)

◊ The answer to everything?


Stuart Dalziel 22

This paper has, it is believed, more or less exhausted the most useful applications of inviscid-fluid theory to steady gravity currents...

...for it must be acknowledged that gravity currents in practice are highly complexphenomena, generally featuring a great deal of turbulence and significant mixing of the twolayers, so that the interpretation of them depends vitally on semi-empirical analyses...

• Energy conserving

◊ Bernoulli to left (u1) and right (u2) of stagnation point

p1 = −½ ρu12

u22 = 2g(H−h)

◊ Pressure force upstream (hydrostatic less Bernoulli dynamic):

p1H + ½ρgH2 = ½ρ(−u12H + gH2)

◊ Flow force upstream

S1 = ½ρ(u12H + gH2)

◊ Flow force downstream

S2 = ρ(c22h + ½gh2)

• Continuity

u1H = u2h

• gives

( )( )u

g H h H

H h h22

2 2

2=

−−

• Equating with downstream Bernoulli gives h = H and h = ½H and thus a Froude number of √2relative to the front

⇒ dissipative hydraulic jump may occur

• Possible to get approximate shape of energy conserving cavity: conformal mapping etc.

• What happes if you throttle the air?

• Boussinesq

• Water into air

1.4.4 Cavity flows with energy loss

• Head loss in downstream (accelerated) flow; follows ideas of hydraulic jumps

◊ Head loss due to irrotational motions in jump at back of head

u22 = 2g(H−h−∆)

• Flow force unchanged; equating u2:

( )( )( )∆ =

− −−

2

2 2

2h H H h

h H h


Stuart Dalziel 23

• For h < ½H, ⇒ ∆ < 0 ⇒ flow can not exist

Energy loss in cavity flow - from Benjamin (JFM 31, 1968)

For ½ < h/H < 0.7808, f = u1/(gH)1/2 > ½

◊ For given front speed between f = ½ and f = 0.5273 then there are two possible h/H, with thelarger one corresponding to greater energy loss.

• Front condition (hf is height of front; hf = H−h in Benjamin’s model)

( )( )( )Fh h

hf

f f

f

=− −

+

1 2

1

1

2


Stuart Dalziel 24

0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.5000.00

0.20

0.40

0.60

0.80

1.00

1.20

1.40

Fractional depth

Fron

t Fro

ude

num

ber

(Ff)

Froude number for front as function of relative depth of current

• Is uniform head loss appropriate for infinite ambients?

1.4.5 Gravity currents and characteristics

x

t

0

u = 0c = Cλ = −C u = 2C

c = 0λ = ±2C

Front

Backλ = u − c = x/tu + 2c = 2C⇒ 3c = 2C − x/t⇒ h = (1/9)(2C − x/t)2/g⇒ u = (2/3)(C + x/t)

λ = u + c = (4/3)C+(1/3)x/tx

t

0

u = 0c = Cλ = −C u = 2C

c = 0λ = ±2C

uf = Ffc

Back

λ = u − c = x/tu + 2c = 2C⇒ 3c = 2C − x/t⇒ h = (1/9)(2C − x/t)2/g⇒ u = (2/3)(C + x/t)

λ = uf−c = (Ff−1)

Head

λ = u + c = (4/3)C+(1/3)x/t

Saint Venant solution Gravity current solution


Stuart Dalziel 25

x

t

0

Rarefaction wave

• Head controlling or controlled?

1.4.6 Integral models

• Continuous

◊ Front: dL/dt = uf = Ff (gh)1/2

◊ Continuity: (d/dt)(Lh) = Q

◊ Can eliminate L from continuity, but end up with nasty nonlinear, second order ode for h.

◊ Useful to do part of this process to obtain

F g h Ldh

dtQf

1 2 3 2/ / + =

and note that as L becomes large, dh/dt must become small.

◊ Can search for power law solution of the form L ~ tα, h ~ tβ. Substitution and gives rise toα = 1, β = 0 when Q is constant.

• Lock release

◊ Similar procedure, except Lh = const

• Axisymmetric flows

◊ For continuous, must start at finite radius L0; continuity then gives (d/dt)[p(L2−L02) h] = Q

◊ For lock release, can start at zero or finite radius; continuity: (L2−L02)h = const

1.4.7 Shallow water models

• Analytical solution for special cases


Stuart Dalziel 26

• Numerical solution for more interesting problems.

◊ Finite difference

◊ Finite volume

◊ Goudnov - makes explicit use of hyperbolic nature

1.4.8 Topography

• Upslope/downslope

◊ How small does the slope have to be before it is “easy”?

◊ Downslope acceleration can be balanced by drag

◊ Upslope deceleration can not leady to steady flow

• Front condition

◊ For Benjamin it matters how far before parallel flow is established

• Asymptotic states

1.4.9 Reflection

• Shallow water can not model a splash

• Splash ⇒ dissipation, but conservation of momentum

• Sloping boundary?

1.4.10 Wind

• Frame of reference

◊ Initial value problem, or boundary value problem?

◊ Drag to boundaries?

• Wind shear

1.4.11 Ambient turbulence

• Entrainment/detrainment

◊ Similar to §1.2.6, but with we and wd given by imposed turbulence and stability of interface.

• Reynolds stresses

◊ Scale on imposed turbulence rather than flow velocity.

… = −CTu|uT|/h

◊ Can balance pressure gradient

1.4.12 Frontogenesis

• How does the seabreeze become established?

1.4.13 Lossy gravity currents

• Not “negative entrainment”


Stuart Dalziel 27

• Steady states possible

• Losses due to

◊ Seepage into substrate

◊ Evaporation/boiling

◊ Rain (negative loss, but adds zero momentum)

1.4.14 Non-Boussinesq currents

• Density of ambient >> density of current - use above theory

• Density of ambient << density of current - no dissipation in ambient, but still get front

1.5 Two-layer flows

1.5.1 Yih and energy conservation

• Lock exchange: dEp/dt = dEk/dt

• Figure 14 from JFM paper

1.5.2 Two-layer hydraulics

• One interface, two waves

b(x)

H(x)

h1(x)

h2(x)

u1(x)

u2(x)

Datum

ρ2

ρ1

p = 0

Examples sheet problem: Derive equations for two-layer fluid and look at hyperbolic nature

• Co-flowing layers and exchange flows

• Hydraulic functional

◊ Similar to specific energy of single-layer hydraulics


Stuart Dalziel 28

Examples sheet problem: Derive functional and look at its properties

Along-channel profile and hydraulic functional in two-layer flow over a sill

◊ Composite Froude number G2 = F12 + F2

2

• Width

• Depth

• Barotropic flow

◊ If strong enough, can bring one layer to rest

1.5.3 Non-rectangular cross-section

• Hydraulic radius

he = Awet/bwet

Examples sheet problem: Show how the Froude number scales on the hydraulic radius for single-layer flow


Stuart Dalziel 29

Along-channel profile of two-layer exchange flow in parabolic cross-section

1.5.4 Hydraulic jumps and bores

Range of possible subcritical states which may be matched by a hydraulic jump from exchange flow in aparabolic cross-section

• Long wave instability

Examples sheet problem: Behaviour when F∆ ≡ (u1−u2)2/[g’(h1+h2)] > 2


Stuart Dalziel 30

1.5.5 Rotation

1.5.6 Gravity currents

2 Porous media

2.1 Darcy’s law

• Derivation

◊ Navier-Stokes

( )∂∂

∂∂ ρ

∂∂

ρ ν∇u

tu

u

x xp gz ui

ji

j ii+ = − + +

1 2

◊ Low Reynolds number ⇒ neglect inertia

◊ Quasi-steady: changes in flow slow compared with flow around a typical element of themedium of size a (∂/∂t << u/a) ⇒ neglect ∂/∂t.

( )01 2= − + +ρ

∂∂

ρ ν∇x

p gz ui

i

◊ Viscous drag feels length scale a of elements in medium, so |∇2u| ~ |u/a2|. Substitute andintroduce constant of proportionality −α:

( )ua

xp gz

i

= − +α

µ∂

∂ρ

2

.

• Darcy velocity is defined as

u ≡ Q/A,

where Q is the volume flux and A is the gross cross-sectional area carrying this flux. As onlysome fraction φ, the porosity, is capable of containing fluid (the balance 1−φ is solid), the meanvelocity of a fluid element will be greater than u. The relationship between the mean orinterstitial fluid velocity uint is

u = φuint.

• The permeability k is normally defined in terms of the Darcy velocity so that

( )uk

xp gz

i

= − +µ

∂∂

ρ .

From the derivation above we see that k ~ a2.

2.2 Free surface problems

2.2.1 Steady flow

• Summer School experiment


Stuart Dalziel 31

h0

h1

◊ Shallow water model gives p + ρgz = ρgh and the velocity field purely horizontal, so

u gk h

x= −ρ

µ∂∂

and for steady flow

Q = uh

2.2.2 Integral model for a porous gravity current

• Integral model

φdL

dtu M

h

L= = 0

where M = ρgk/µ. Thus

⇒ φL2 = 2Mh0(t−t0)

2.2.3 Full time dependent model for a porous gravity current

• Continuity

( )φ∂∂

∂∂

h

t xuh+ = 0

• Darcy’s law

( )uk

xp gz

kg

h

xM

h

x= − + = − = −

µ∂∂

ρµ

ρ∂∂

∂∂

◊ Look to see if hyperbolic:

u h

M

h

u

h

u ux t

−

+

=

0

1 0

0 0

0

For non-trivial solution, require |A − λB| = 0, but

u h

MMh

−−

= ≠λ

00 ,

so system not hyperbolic. (This is no surprise.)

• Eliminate u:


Stuart Dalziel 32

φ∂∂

∂∂

∂∂

φ∂∂

∂∂

∂∂

φ∂∂

∂∂

h

tM

xh

h

x

h

tMh

h

xM

h

x

h

tM

h

x

−

= − −

= − =

2

2

2

2 2

2

1

20

• Assume separable h(x,t) = X(x)T(t) then

( )φXT M X T′ −″

=1

202 2

Thus

( )φ λ

ξξ

′= − =

″=

′′T

TM

X

XM2

2

1 2

1

2

1

2 /

for any constant λ and ξ ≡ X2

• The equation for time has the solution

( )φλ

Tt t= − 0 ,

where t0 is an arbitrary constant. Rearranging

( )Tt t

=−φ

λ 0

.

• The equation for space ½Mξ" + λξ1/2 = 0 has no general solution for arbitrary λ. However,ξ = a2(x−x0)

4 is a solution when λ = −6Ma, giving rise to X = ±a(x−x0)2

( )( )h XT

x x

M t t= =

−

−m

φ 0

2

06

• Similarity solution with η = x2/t, but no general solutionto second-order nonlinear ode whichresults. Can not use superposition.

• The form x2/t comes as no surprise in light of integral model results.

2.2.4 3D porous medium

• Continuity: ∇.u = 0, thus

∇2p = 0

• Free surface ⇒ free boundary problem with p = 0 on surface.


Stuart Dalziel 33

2.3 Seepage

h

ζ

p = 0

p = 0

Air

Air in porousmatrix

Water

Water inporous matrix

Example sheet question: Analyse the penetration into the substrate

2.4 Heterogeneous media

• Parallel

Examples sheet question: Free surface flow in two-layer medium

◊ Typical aquifers

• Series

• More complex

• Fractured

◊ Geothermal fields

• Anisotropic

2.5 Two phase flows

• Relative permeability

◊ Effective permeability for phase i ki = krik is a function of the volume fraction ξ. Darcy’s lawwritten as

( )uk k

p gziri= − ∇ +µ

ρ


Stuart Dalziel 34

ξ

kri

1

0

10

◊ Phases may be immobile if at sufficiently low concentrations.

∗ Held as droplets/buttles in voids by surface tension if non-wetting and discontinuous

∗ Held in film if wetting

• Fountain

◊ If pressure gradient less than hydrostatic but greater than vapourstatic

Immobile phase

2.6 Surface tension

• Pressure jump across meniscus proportional to curvature

• Contact angle θ fixed by three phases involved


Stuart Dalziel 35

Liquid

Air

Contactangle

p = 0

p = −∆pη(x)

x = −½βa x = ½βa

◊ For 2D gap βa in width

∆p = −γ d2η/dx2

dη/dx = ±(π/2 − θ) at x = ±½ βa

From boundary conditions

∆pa

= −

γβ

πθ

2

• Wetting verses non-wetting

Liquid: p = ∆p < 0

Air: p = 0

Increasing |∆p|

Wetting

Decreasing |∆p|

Air: p = 0

Liquid: p = ∆p > 0

Non-wetting

• If imposed pressure high enough, then meniscus can move across grains and from one grain toanother.

• Viscous fingering:

◊ Water into air → stable interface

◊ Air into water → unstable interface

◊ The difficulty geting oil out!


Stuart Dalziel 36

3 Sedimentation & Resuspension

3.1 Flux relationship

3.1.1 Stokes settling

• Low Reynolds number limit

• Drag bounded by in-scribed and circumscribed sphere

• Settling flux Flux = Vsφ

3.1.2 Boycott effect

• Organised flow established

Cross-tube densitygradient → baroclinicvorticity generation

Particles settledownwardacross tube

Organised flowestablished

3.1.3 Hindered settling

• Sedimentation balanced by corresponding flux of fluid in opposite direction

• Empirical relationship: Flux = Vsφ(1−φ/φMax)α


Stuart Dalziel 37

Key φ φ(1-φ)1

φ(1-φ)2

φ(1-φ)3

φ(1-φ)4

Concentration

Flu

x

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.900.000

0.020

0.040

0.060

0.080

0.100

0.120

0.140

0.160

0.180

0.200

0.220

0.240

• Low concentration model

• High concentration porous model

3.2 Shock formation

• Hyperbolic system

( )[ ]∂φ∂

∂∂

φ φα

t xVs+ − =1 0

◊ Let x + λt = 0, then

( )[ ]∂∂

φ φ λφα

xVs 1 0− − =

◊ Single characteristic λ = Vs(1−φ)α

◊ Two possible states for given flux → shock formation

3.3 Turbulence

• Well mixed if u’ >> Vs

• Velocities normal to wall suppressed by wall

◊ Occurs over integral length scale

◊ Velocities parallel enhanced by 50% if inviscid

• Viscous sublayer of thickness δ ~ ν/u’ adjacent to wall.

• Entrainment/detrainment

3.4 Sedimenting gravity currents

• Particle continuity equation

( )∂ φ∂

∂∂

φ φh

t xuh Vs+ = −


Stuart Dalziel 38

⇒∂φ∂

∂φ∂

φt

ux

V

hs+ = −

• Bulk density

ρ = ρ0 + φρp

• Reduced gravity, for φ << 1

g’ = φg ρp/ρ0

• Integral model

Lh = M =const

( )d

dtLh M

d

dtV Lsφ

φφ= = −

dL/dt = Ff(φg ρp/ρ0)1/2h1/2 = Ff(φg’0M)1/2L−1/2,

where g’0 = g ρp/ρ0.

◊ Can eliminate either φ or L between these two ode’s, e.g.

d

dtL

dL

dt

V

ML

dL

dts

+

=

22

2

0

but there is no closed form solution to this.

◊ Noting that dt = −M/(VsφL) dφ, then

dL F g M LM

V Ldf

s

= − ′ −0

1 2 1 2 1 2 1 2/ / / /φφ

φ

⇒ L dL d3 2 1 2/ /= − −βφ φ

where β = Ff g’01/2M3/2/Vs.

◊ Integrating

( ) ( )2

525 2

05 2 1 2

01 2L L/ / / /− = − −β φ φ

where L = L0 when φ = φ0.

◊ Run out length: φ → 0, gives

L

L L0

5 2

001 21

5

→ +

/

/βφ

⇒L

L

F g M

L V

F g M

L Vf

s

f

s0

01 2 3 2

001 2

2 5 20

3

02 2 0

1 5

15 25

→ +′

≈

′

/ /

/

/ /

φ φ

◊ Time depedence - must do numerically. Can do with shallow water formulation or integralformulation.


Stuart Dalziel 39

3.5 ResuspensionLift

u

Buoyancy

Drag

• Forces

Lift ~ ρf u2 a2

Drag ~ ρf u2 a2

Buoyancy ~ (ρp−ρf)g a3

Cohesion

Rolling resistance

• No motion until

◊ Large “particles”: Drag ~ Rolling

◊ Medium particles: Lift ~ Buoyancy

◊ Small particles: Lift ~ Cohesion

dust sand pebbles

Critica

l vel

ocity

• In atmosphere, sand particles are more easily resuspended than fine dust or large pebbles.

• Once a particle is suspended, then will remain suspended until it gets in boundary layer wherethere is insufficient lift force.

• Shields parameter

θ = ρf u2 /(ρp−ρf)ga

◊ Velocity scale normally taken as the friction velocity for logarithmic profile

• Particles enhance momentum transport → change in boundary layer structure.


Stuart Dalziel 40

3.6 Saltation

3.7 Impacting bodies

3.8 Resuspension by gravity currents

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