Inclined to exchange: shocking gravity · PDF file · 2007-11-19X n odd 2 πn...

Inclined to exchange:shocking gravity currents

John Hinch (DAMTP)

and at FAST/Paris:

Jean-Pierre Hulin, Dominique Salin & Bernard Perrin

Thomas Seon & Jemil Znaien

Pipemix - Nottingham 2007 – p.1/25

Experimental setup


Regimes

Vertical : turbulent

Inclined : nosed controlled gravity current

Horizontal : viscous counter-current


Speed of front Vf

Vt =√

Atgd (inertial nose) Vν =Atgd2

ν(viscous)

1.0

0.8

0.6

0.4

0.2

0.0

Vf/Vt

1501000

Vν sin α/Vt = Ret sin α

50

horizontal tube

Tilt angles 0 < α < 30o;At = 3.5 × 10−2 (•), 10−2

(�), 4× 10−3 (�), 10−3 (H),4 × 10−4 (N) for a viscosityµ = 10−3 Pa.s;and µ = 10−3 Pa.s (�), 4 ×10−3 Pa.s (⊞) for a densitycontrast At = 10−2

Vf ≈ 0.014Vν sin α


A little theory

Counter current in a circle at 50% fill

∇2u = f(θ) =

{

−12 in 0 ≤ θ < π,

+12 in π ≤ θ < 2π

and u = 0 on r = 1.

Fourier series

f(θ) = −∑

n odd

2

πnsin nθ,

so

u =∑

n odd

2

πn(n2 − 4)(r2 − rn) sin nθ.


a little theory continued

Hence flux

Q =∑

n odd

1

πn2(n + 2)2

=π

16− 1

2π

= 0.037195

Front velocity for area A = π2

Vf =Q

A= 0.012,

experiment 0.014 – eventually will do better.


Horizontal tube

Xf = Vf

∝ R2

µ

dp

dx

∝ R2

µ

∆ρgR

Xf

Hence

Xf =√

Dt

D = 0.0108Vνd in experiment

1.0

0.8

0.6

0.4

0.2

0.0

-20 -10 0 10 20x/t0.5

25 s

75 s

125 s

175 s

225 s

1-1 0.5-0.5 0

distance from gate valve (m)

1.0

0.8

0.6

0.4

0.2

0.0

Norm

alized co

ncen

tration

25 s

75 s

125 s

175 s

225 s

(a)

(b)

Norm

alized co

ncen

tration


2D spreading in a horizontal channel

µ∂2u

∂y2=

dp±dx

in 0 ≤ y ≤ h and h ≤ y ≤ a,

with u = 0 on y = 0 and a.

No net flux

∫

u dy = 0 and

[

dp

dx

]+

−= ∆ρg

∂h

∂x.

Hence∂h

∂t=

∂

∂x

(

∆ρgh3(a − h)3

3µa3

∂h

∂x

)

.


Spreading in a horizontal channel, continued

0

0.2

0.4

0.6

0.8

1

-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 0.2

h

x/t0.5

HenceXf =

√Dt

with D = 0.017Vνa, cf experiment 0.0108.


Spreading in a horizontal circleFlat interface at y = h(x, t). Flow in cross-section

∇2u = G +

{

−12 in h ≤ y ≤ 1,

+12 in −1 ≤ y < h

and u = 0 on r = 1. G so no net flux.

Numerical (finite difference 80 × 80, sor) for flux Q(h) in h ≤ y ≤ 1.

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

h

Q


Thin layer asymptotics, 1 − |h| ≪ 1

No net flow ⇒ all pressure gradient in thin layer.

Thin layer ⇒ no stress by thick on thin.

Wide compared with thin ⇒ 2D, half parabolic profile.

Q(h) ∼ 32√

2

105(1 − |h|)7/2 .

Symmetric and smooth Q(h), so equally

Q(h) ∼ 4

105

(

1 − h2)7/2

.


Lucky break

Thin layer asymptotic’s4

105= 0.038095

50% fill Q(0) = 0.037195

Hence good approximation (within 1%):

Q(h) ≈ Q(0)(1 − h2)7/2

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

h

Q


Nonlinear diffusion equation

∂A(h)

∂t+

∂Q(h)

∂x= 0,

with cross-sectional area A(h) = 2 cos−1 h − h√

1 − h2, sodA/dh = −2

√1 − h2. Hence

∂h

∂t=

1

2√

1 − h2

∂

∂x

(

(1 − h2)7/2 ∂h

∂x

)

.

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4

h

x/√

Dt

Xf =√

Dt

with D = 0.0108Vνd,experiment 0.0108 ± 0.0008.


Comparison with experiment


Spreading in nearly horizontal tubesFlow now driven by difference in pressure gradients in two fluids

∆ρg

(

cos α∂h

∂x+ sin α

)

.

So with suitable rescaling

∂h

∂t=

1

2√

1 − h2

∂

∂x

(

(1 − h2)7/2

(

∂h

∂x+ 1

))

.

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6

h

x

t = 0.5, (0.5), 5.0.


Motion of front Xf(t)

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

Xf

t

Diffusive spreading only at very short times.

Slow to attain long-time velocity.


Experimental motion of front Xf(t)

6

5

4

3

2

1

0

2 X

f/(d c

ot

α)

6543210

t (8 F(0)Vν sin

2α) /(d cos α)

α = 1o ◦,2o △,3o ▽.

Experiments straighter due to different early behaviour – gate release and a short time

limited by inertia


Long time approximation?

Xf (t) ∼ 0.856t + 1.01 4 < t < 5

0.830t + 1.19 7 < t < 11

0.774t + 1.83 20 < t < 25

0.750t + 2.82 90 < t < 100

0.746t + 3.36 145 < t < 150

Linear fit fails.


Logarithms at long timeCrude model: horizontal + inclined spreading

Xf =VfL

Xf+ Vf ,

with solutionVf t = Xf − L log(1 +

Xf

L).

Vf L

0.745 0.798 0 < t < 5

0.735 0.851 0 < t < 20

0.740 0.812 0 < t < 50

0.742 0.791 0 < t < 100

0.742 0.789 0 < t < 200

Slow logarithmic approach also in experimental data.


Test Vf = 0.742

20

15

10

5

0

0.10.080.060.040.020

Vf

(mm/s)

sinα

0 α (°)2 4

∞

Continuous line is new 0.742 (0.014Vν sin α),dashed is old 0.637 (0.012Vν sin α).


But why Vf = 0.742?

∂h

∂t=

1

2√

1 − h2

∂

∂x

(

(1 − h2)7/2

(

∂h

∂x+ 1

))

.

For hx ≪ 1

∂h

∂t+ c(h)

∂h

∂x= 0 with kinematic wave speed c(h) = 7

2h(1 − h2)2.

Test rarefactionwave solution byplotting h vs x/t.

But must shock 0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

t=2550

100200

h

x/tPipemix - Nottingham 2007 – p.21/25

Shock waves

Speed of shock V (h) =Q(h)

A(h)must equal speed of arriving characteristics c(h).

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

t=2550

100200

h

x/t

Equal at h = 0.23817 with c = V = 0.74172.


Form of shock waves

Switch to frame moving with V , then steady form governed by

∂h

∂x= V

A(h)

Q(h)− 1

0

0.2

0.4

0.6

0.8

1

-80 -70 -60 -50 -40 -30 -20 -10 0 10

t=2550

100200

h

x − Xf (t)


Conclusions

Front is a shockwave:

speed selected by matching onto a rarefaction wave.

Logarithmic approach to long-time:

value depends on initial condition.


And next

Vertical : turbulent

Inclined : nosed controlled gravity current

Horizontal : viscous counter-current


Inclined to exchange: shocking gravity · PDF file · 2007-11-19X n odd 2 πn...

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