a mathematical and numerical study of roll wavessdalessi/afm08talk.pdfnumerical solution procedure...

IntroductionMathematical Formulation

Stability AnalysisNumerical Solution Procedure

SimulationsSummary

A Mathematical and NumericalStudy of Roll Waves

Serge D’Alessio1 and J.P. Pascal2

1Department of Applied MathematicsUniversity of Waterloo

2Department of MathematicsRyerson University

AFM 2008

Serge D’Alessio and J.P. Pascal Study of Roll Waves



SimulationsSummary

Outline

1 Introduction

2 Mathematical Formulation

3 Stability Analysis

4 Numerical Solution Procedure

5 Simulations

6 Summary




SimulationsSummary

Unstable flow down an inclineUnstable Flow Down an Incline

θ

Unstableuniform flow

θ

Roll waves

Critical conditions for the onset of Instability.Structure of Roll WavesInvestigate the effect of bottom topography

J.P. Pascal and S.J.D. D’Alessio Instability of Laminar Flow Down an Uneven Inclined Plane




SimulationsSummary

Chapter 1. Introduction 5

Figure 1.1: Spillway from Llyn Brianne Dam, Wales [115]. For an idea of the scale, the

width of the spillway is about 75 feet.

The spillway from the LlynBrianne Dam in Wales




SimulationsSummary

Chapter 2. Dynamics of roll waves 32

PSfrag replacements

t (sec)

h (mm)

0 2 4 6 8 10−12

−10

−8

−6

−4

−2

0

2

0 2 4 6 8 10

−20

−15

−10

−5

0

5

PSfrag replacements

t (sec)

t (sec)

h(m

m)

h(m

m)

Figure 2.1: The picture on the left shows a laboratory experiment in which roll waves

appear on water flowing down an inclined channel. The fluid is about 7 mm deep and the

channel is 10 cm wide and 18 m long; the flow speed is roughly 65 cm/sec. Time series

of the free-surface displacements at four locations are plotted in the pictures on the

right. In the upper, right-hand panel, small random perturbations at the inlet seed the

growth of roll waves whose profiles develop downstream (the observing stations are 3 m,

6 m, 9 m and 12 m from the inlet and the signals are not contemporaneous). The lower

right-hand picture shows a similar plot for an experiment in which a periodic train was

generated by moving a paddle at the inlet; as that wavetrain develops downstream, the

wave profiles become less periodic and there is a suggestion of subharmonic instability.

Experiment taken from Balmforth & Mandre (JFM, 2004)




SimulationsSummary

Coordinate system

θ

g

x, u

z, w

h(x, t)

ζ(x)




SimulationsSummary

Equations of motion

∂u∂x

+∂w∂z

= 0

ρ

(∂u∂t

+ u∂u∂x

+ w∂u∂z

)= −∂p

∂x+ gρ sin θ + µ

(∂2u∂x2 +

∂2u∂z2

)1ρ

∂p∂z

+ g cos θ − µ

ρ

∂2w∂z2 = 0

Assumed Re ∼ O(1) and neglected terms O(δ2) and higherwhere δ = H/L is the aspect ratio




SimulationsSummary

Interface conditions

Free surface conditions:

p − 2µ∂w∂z

= 0

µ∂u∂z

= 0

w =∂h∂t

+ u∂h∂x

+ uζ ′(x)

at z = ζ(x) + h(x , t)

Bottom boundary conditions:

u + ζ ′(x)w = 0 and ζ ′(x)u − w = 0 at z = ζ(x)

⇒ u = w = 0 at z = ζ(x)




SimulationsSummary

Integral boundary layer (IBL) method

Depth-integrate equations and introduce flow variables

h(x , t) and q(x , t) =

∫ ζ+h

ζudz

To convert terms∫ ζ+h

ζu2dz ,

µ

ρ

∂u∂z

∣∣∣∣z=ζ

assume the parabolic velocity profile:

u(x , z, t) =3q2h3

[2(h + ζ)z − z2 − (ζ + 2h)ζ

]




SimulationsSummary

Dimensionless equations

In terms of h, q the dimensionless equations become∂h∂t

+∂q∂x

= 0

∂q∂t

+65

∂

∂x

(q2

h

)=

1Fr2

(h − h

∂h∂x

− ζ ′(x)h − qh2

)+

3Fr2

Re2

[72

∂2q∂x2 −

9h

∂q∂x

∂h∂x

+9qh2

(∂h∂x

)2

− 9q2h

∂2h∂x2

−6ζ ′(x)

h∂q∂x

+6ζ ′(x)q

h2∂h∂x

− 3ζ ′′(x)qh

− 6 (ζ ′(x))2 qh2

]

where Fr2 =Re

3 cot θ, Re =

ρQµ

and ζ(x) = ab cos(kbx)




SimulationsSummary

Linear stability: ab = 0 case

The steady-state flow is: qs = hs = 1Imposing disturbances on this steady flow and linearizing yieldsthe dispersion equation

Fr2σ2 +

(21Fr4

2Re2 k2 + 1 + i125

Fr2k)

σ +

(1− 6

5Fr2)

k2

+i(

3k +27Fr4

2Re2 k3)

= 0

where σ is the growth rate and k is the wavenumber of thedisturbance




SimulationsSummary

Linear stability results for ab = 0

The flow is stable if Fr <1√3

while for Fr >1√3

instability

occurs for wavenumbers k < kmax where

kmax =10Re√30Fr2

√3Fr2 − 1

3Fr2 + 35 + 12Fr√

6Fr2 + 25

For large Fr the asymptotic behaviour is

kmax ∼10Re

Fr2√

30(1 + 4√

6)




SimulationsSummary

Neutral stability curves for ab = 0

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

k

Fr

STABLE

Re=20

Re=10

UNSTABLE

Re=5

Neutral stability curveshave a maximum atFr ≈ 0.76286(independent of Re)




SimulationsSummary

Linear stability: ab 6= 0 case

The steady state solution is qs = 1 and hs(x) satisfies

3β[hsh′′s − 2(h′s)2] + (2αh3

s − 4βζ ′ − 125

)h′s

+2βζ ′′hs − 2α(1− ζ ′)h3s = −2α− 4β(ζ ′)2

where α =1

Fr2 and β = 9(

FrRe

)2

An approximate solution can be constructed in the form

hs(x) = 1 + (abkb)h(1)s (x) + (abkb)2h(2)

s (x) + · · ·




SimulationsSummary

Periodic steady state solution

0 0.2 0.4 0.6 0.8 10.9

0.95

1

1.05

1.1

1.15

x

h

Analytical (1 term)Analytical (2 terms)Numerical

Fr = 1, Re = 10,ab = 0.1, kb = 2π




SimulationsSummary

Linear stability: ab 6= 0 case

To study how small disturbances will evolve, introduceperturbations h, q superimposed on the steady-state solutionand linearize equations using

h = hs(x) + h , q = 1 + q

For an uneven bottom, the coefficients in the linearizedequations are periodic functions; hence apply Floquet-Blochtheory to conduct the stability analysis and represent theperturbations as Bloch-type functions having the form

h = eσteiKx∞∑

n=−∞hneinkbx , q = eσteiKx

∞∑n=−∞

qneinkbx




SimulationsSummary

Numerical linear stability results for ab 6= 0

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

0.58

0.6

0.62

0.64

0.66

0.68

ab

Frcrit

kb=π

kb=2π

kb=2.5π

Critical Froude numberas a function of bottomamplitude with Re = 10




SimulationsSummary


0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.40.5

0.55

0.6

0.65

0.7

0.75

ab

Frcrit

Re=3

Re=8

Re=5

Re=10

Re=20

Re=7

Critical Froude numberas a function of bottomamplitude with kb = 2π




SimulationsSummary


0

0.5

1

1.5

2

2.5

3

3.5

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

K

Fr

0

0.5

1

1.5

2

2.5

3

0.5 0.6 0.7 0.8 0.9 1

K

Fr

UNSTABLE

STABLE

ab=0.1

ab=0.2

ab=0

STABLE

ab=0.2

UNSTABLE

ab=0.1ab=0

Neutral stability curvesfor the case withRe = 10 and kb = 2π




SimulationsSummary


0

0.5

1

1.5

2

2.5

3

3.5

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

K

Fr

0

0.5

1

1.5

2

2.5

3

0.5 0.6 0.7 0.8 0.9 1

K

Fr

STABLE

kb=0

UNSTABLE

kb=2.5π

kb=2π

kb=π

STABLE

UNSTABLEkb=0kb=π

kb=2.5π

kb=2π

Neutral stability curvesfor the case withRe = 10 and ab = 0.2




SimulationsSummary

Begin by expressing equations in the form

∂h∂t

+∂q∂x

= 0

∂q∂t

+∂

∂x

(65

q2

h+

α

2h2)

= Ψ(h, q)+χ

(x , h, q,

∂h∂x

,∂q∂x

,∂2h∂x2 ,

∂2q∂x2

)where Ψ = α

(h − q

h2

)and χ = −αζ ′h − 2βζ ′

(ζ ′ − ∂h

∂x

)qh2 − βζ ′′

qh− 2β

ζ ′

h∂q∂x

+β

(76

∂2q∂x2 −

32

qh

∂2h∂x2 −

3h

∂q∂x

∂h∂x

+ 3qh2

(∂h∂x

)2)




SimulationsSummary

Fractional-step method (LeVeque, 2002)

Decouple the advective and diffusive components, first solve

∂h∂t

+∂q∂x

= 0

∂q∂t

+∂

∂x

(65

q2

h+

α

2h2)

= Ψ(h, q)

over a time step ∆t , and then solve

∂q∂t

= χ

(x , h, q,

∂h∂x

,∂q∂x

,∂2h∂x2 ,

∂2q∂x2

)using the solution obtained from the first step as an initialcondition for the second step; the second step returns thesolution for q at the new time t + ∆t




SimulationsSummary

First step

This involves solving a nonlinear system of hyperbolicconservation laws; express in vector form

∂U∂t

+∂F(U)

∂x= b(U)

where U =

[hq

], F(U) =

[q

65

q2

h + α2 h2

], b(U) =

[0Ψ

]Utilize MacCormack’s method to solve this system; this is aconservative second-order accurate finite difference schemewhich correctly captures discontinuities and converges to thephysical weak solution of the problem




SimulationsSummary

First step

LeVeque & Yee (JCP, 1990) extended MacCormack’s method toinclude source terms; this explicit predictor-corrector schemetakes the form

U∗j = Un

j −∆t∆x

[F(Un

j+1)− F(Unj )]

+ ∆t b(Unj )

Un+1j =

12

(Un

j + U∗j

)− ∆t

2∆x

[F(U∗

j )− F(U∗j−1)

]+

∆t2

b(U∗j )

where the notation Unj ≡ U(xj , tn) was adopted, ∆x is the grid

spacing and ∆t is the time step; second-order accuracy isachieved by first forward differencing and then backwarddifferencing




SimulationsSummary

Second step

This reduces to solving the generalized one-dimensional lineardiffusion equation given by:

∂q∂t

=7β

6∂2q∂x2 + S1

∂q∂x

+ S0q + S

where S = −αζ ′h and

S0 = −βζ ′′

h− 2β

ζ ′

h2

(ζ ′ − ∂h

∂x

)− 3

2β

h∂2h∂x2 + 3

β

h2

(∂h∂x

)2

and S1 = −2βζ ′

h− 3

β

h∂h∂x




SimulationsSummary

Computational parameters

The problem is completely specified by Fr , Re, ab and kb;typical computational parameters used were:Computational Domain: 0 ≤ x ≤ L

with λb ≤ L ≤ 300λb , λb =2π

kbGrid Spacing: ∆x = .01Time Step: ∆t = .002




SimulationsSummary

Evolution of flow rate

0 5 10 15 20 25 30 35 40 450.5

1

1.5

2

2.5

3

3.5

4

x

q

t=400

t=495

t=264

t=450

Parameters:ab = 0.1, kb = 2π,Re = 10, Fr = 0.7A subharmonicinstability known aswave coarseningoccurs for L = 45λb




SimulationsSummary

Evolution of flow rate

0 5 10 15 20 25 300.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

x

q

t=450

t=260

Parameters:ab = 0.1, kb = 2π,Re = 10, Fr = 0.7Interruption in wavecoarsening occursfor L = 30λb




SimulationsSummary

Wave spawning

0 50 100 150 200 250 3000

1

2

3

4

5

6

7

8

9

10

x

q

t=700

t=400 Parameters:ab = 0.1, kb = 2π,Re = 10, Fr = 0.7A wave-spawninginstability occurs forL = 300λb




SimulationsSummary

Concluding remarks

A mathematical model of roll waves along with a numericalmethod to solve the model were presentedInvestigated the effect of sinusoidal bottom topography onthe formation of roll wavesBottom topography has a stabilizing effect on the flow forsmall to moderate waviness parametersFuture work includes repeating the analysis for a porouswavy bottom and to include surface tension


a mathematical and numerical study of roll wavessdalessi/afm08talk.pdfnumerical solution procedure...

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