university of south carolina erosion rate formulation and modeling of turbidity current ao yi and...
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UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Erosion rate formulation and Erosion rate formulation and modeling of turbidity currentmodeling of turbidity current
Ao Yi and
Jasim Imran
Department of Civil and Environmental Engineering
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Recent Progress in numerical modeling of density currents considering the vertical
structure• Stacey & Bowen (1988): Temporal evolution of vertical structure of density and turbidity current
using zero Equation turbulence model. Convective and diffusive terms in the flow direction neglected.
• Meiburg et al. : (2000) Used DNS to solve the Navier Stokes equations for density current at low Reynolds number (with Boussinesq assumption)
• Imran, Kassem, & Khan AND Kassem & Imran (2005): Used the commercial flow solver FLUENT to simulate density current in confined and unconfined straight and sinuous channel. The model used RANS Equations and k-e turbulence closure
• Felix (2001): Solved the RANS equations for lock exchange particulate currents at large scale using 2 equation Mellor-Yamada level 2.5 model
• Choi and García (2002): Simulated continuous dilute saline density current on a ramp by solving 2-D steady state boundary layer equation using the 2 equation k-epsilon turbulence model
• Huang, Imran & Pirmez (2005): Solved 2 equation k-epsilon model for turbidity current for well-sorted sediment. Considered bed level change by adjusting the bed boundary and the grids during the computation. Successfully reproduced Garcia’s (1990) experiment including the flow structure and bed level changes.
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
For conservative density currents, numerical modeling has certainly reached very high level of
sophistication.
Where are we in terms of modeling turbidity currents and the morphological changes at the
field scale ?
Does not matter how much detail of the current one can churn out from the most sophisticated
solution of the Navier-Stoke Equations, there is no escape from van Rijn or Akiyama-Fukushima, or Garcia-Parker or Smith-McLean or some other ‘empirical relationship’ that defines sediment
entrainment from the bed!
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Suppose we make numerical model runs
keeping the boundary conditions and computational grid same but choose different entrainment relationships in different runs
Can we expect to see the same result?
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
We need to pay serious attention to
• the modeling of turbulent kinetic energy (TKE)
&• the sediment entrainment from the bed
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Peak velocity region poses a strong barrier to
mixing
Barrier to diffusion
If there is a sediment pick up, can The particles diffuse above the ‘fish-trap’?
t (m/s)2
Dep
th,
z(m
)
0 1E-05 2E-05 3E-050
0.02
0.04
0.06
0.08
0.1
0.12
0.14k-eps ( 2 eqn model)
k (m/s)20.0E+00 1.0E-04 2.0E-04 3.0E-040
0.025
0.05
0.075
0.1
0.125
0.15
0.175
0.2
k-eps ( 2 eqn model)
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
How turbulence closure affects the result?
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Consider three different closure models
k – l model ( 1 eqn model) Turbulence kinetic energy (k) is calculated using transport equation where turbulence length scale (l) is calculated using an empirical formula. q2– l model ( 1 eqn model) Turbulence velocity scale (q) is calculated using transport equation where turbulence length scale (l) is calculated using an empirical formula.
k – ε model ( 2 eqn model) Both turbulence kinetic energy (k) and its dissipation rate (ε) are calculated using transport equations.
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
u/U
z/h
0 0.5 1 1.5 20
0.5
1
1.5
2
DAPER6 (Garcia 1993)
k-l ( 1 eqn model)
q2-l ( 1 eqn model)
k-eps ( 2 eqn model)
Supercritical flow condition :vertical section is taken at x = 3.0 meter from inletafter 2400 sec
u/U
z/h
0 0.5 1 1.5 20
0.5
1
1.5
2
DAPER6 (Garcia 1993)
k-l ( 1 eqn model)
q2-l ( 1 eqn model)
k-eps ( 2 eqn model)
Subcritical flow condition :vertical section is taken at x = 8.0 meter from inletafter 2400 sec
c/C
z/h
0 0.5 1 1.5 2 2.5 30
0.5
1
1.5
2
2.5
3
DAPER6 (Garcia 1993)
k-l ( 1 eqn model)
q2-l ( 1 eqn model)
k-eps ( 2 eqn model)
Subcritical flow condition :vertical section is taken at x = 8.0 meter from inletafter 2400 sec
c/C
z/h
0 0.5 1 1.5 2 2.5 30
0.5
1
1.5
2
2.5
3
DAPER6 (Garcia 1993)
k-l (1 eqn model) : with bridge
k-l ( 1 eqn model)
q2-l ( 1 eqn model)
k-eps ( 2 eqn model)
Supercritical flow condition :vertical section is taken at x = 3.0 meter from inletafter 2400 sec
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Let us revisit the Parker et al. (1986) work
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Four equation model
Uex
hU
t
hw
2*
22
2
1uRghCS
x
ChRg
x
hU
t
hU
)( 0CrEvx
hUC
t
hCss
)(2
1
2
12
1
0
032
*
CrERghvRghCUehCRgv
heUUux
hKU
t
hK
ssws
w
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Three erosion rate relationships
1. Akiyama and Fukushima (1985) –E-I
0
1103
3.01012
Z
ZZE c
s
c
mc
m
ZZ
ZZZ
ZZ
sp v
uRZ *5.0
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
2. Garcia and Parker (1993)-E-II
57
57
1033.41
103.1
Z
ZEs
2*1
ps
Rv
uZ
36.2for 23.1
586.0
36.2for 6.0
0.1
2
1
2
1
p
p
R
R
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
3. Smith and McLean (1977) –E-III
11
1
65.0
*
*
0
*
*
0
c
s
c
s
sE
0024.00
;
ss RgD
u**
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Four Equation model under steady-state condition
Ri
RirU
v
Uu
RieRiS
dx
dhes
w
1
121
)21
2( 02
2*
1
es
hU
v
dx
d
h
U
Ri
RirU
v
U
uRieRiS
dx
dUes
w
1
12
1)
2
11( 02
2*
h
URir
U
v
U
vRi
U
h
U
Ke
U
uRie
dx
dK
ess
ww
2
0
30
22
2*
]12
1
)1(2
1[
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
What is ignition condition?
For known values of current thickness, sediment size, and bed slope, there is a combination of velocity, turbulent kinetic energy, and concentration at which the flow will ignite or become erosional.
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Set the left side of these equations equal to zero and find solution for U,and K
Pick the lowest positive value
1
es
hU
v
dx
d
h
U
Ri
RirU
v
U
uRieRiS
dx
dUes
w
1
12
1)
2
11( 02
2*
h
URir
U
v
U
vRi
U
h
U
Ke
U
uRie
dx
dK
ess
ww
2
0
30
22
2*
]12
1
)1(2
1[
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Ignition Condition
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
103
104
105
106
0.2
1
2
UI (
m/s)
h0/ D
s
E-I I I / 0.1E-I I I / 0.05
E-I I I / 0.025
E-I / 0.025
E-I / 0.05
E-I / 0.1
E-I I / 0.025
E-I I / 0.1
E-I I / 0.05
Ds = 0.1 mm
E-II-I
E-III-I
E-III-II
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
103
104
105
106
0.1
0.5
1
UI (
m/s)
h0/ D
s
E-I I I / 0.1
E-I I I / 0.05
E-I I I / 0.025
E-I / 0.025E-I / 0.05
E-I / 0.1
E-I I / 0.025
E-I I / 0.1E-I I / 0.05
Ds = 0.03 mm
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
UI vs. h0/Ds for Ds = 0.03 mm, 0.06 mm, and 0.1 mm. S and Cf* are respectiv
ely set equal to 0.05 and 0.004
103
104
105
106
0.1
0.5
1
1.5
h0/ D
s
UI
E-I / 0.1mm E-I I / 0.1mm
E-I I I / 0.1mm
E-I / 0.06mm
E-I I / 0.06mm
E-I I I / 0.06mm
E-I I / 0.03mm
E-I / 0.03mm E-I I I / 0.03mm
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Phase diagram computed for the case: Ds = 0.1 mm, h0 = 5 m, Cf* = 0.004 an
d S = 0.1. A-U0 = 0.6 m/s, 0 = 2.0×10-4 m 2 /s, and K0 = 8.0×10-3 m 2/s 2
0 0.5 1 1.5 20
0.5
1
1.5
2
0 0.5 1 1.5 20
0.5
1
1.5
2
/ I
U/U I
Subcritical
E-Combine/0.1 mm
Igniting field
subsidingfieldCL
CL
AGL-E-I
AGL-E-II
AGL-E-III
A-III
A-II
A-I
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Phase diagram computed for the case: Ds = 0.03 mm, h0 = 1 m, Cf* = 0.004
and S = 0.05. C-U0 = 0.2 m/s, 0 = 2.0×10-5 m 2/s, and K0 = 1.0×10-3 m 2/s 2
0 0.5 1 1.5 20
0.5
1
1.5
2
U/U I
/ I
Subcritical
E-Combine/0.03 mm
Igniting field
Subsidingfield
CL
CL
AGL-E-III
AGL-E-II
AGL-E-I
C-I
C-II
C-III
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Cross field for case Ds = 0.1 mm, h0 = 5 m, Cf* = 0.004, and S = 0.1
10-6
10-5
10-4
10-3
10-2
0.5
1
U(m
/s)
(m2/ s)
Subcritical
AGL-I
AGL-E-II
AGL-E-III
Ds = 0.1mm
Subsiding field
Igniting field
A
Cross field
B
A-E-I
B-E-I B-E-I I
B-E-I I I
A-E-I I I
A-E-I I
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Cross field for case Ds = 0.03 mm, h0 = 1 m, Cf* = 0.004, and S = 0.05 10
-610
-510
-410
-3
0.2
0.4
0.6
0.8
U(m
/s)
(m2/ s)
Subcritical
AGL-I
AGL-E-II
AGL-E-III
Ds = 0.03mm
Subsidingfield
Igniting field
C
Cross fieldD
D -E-I
C-E-ID -E-I I
D -E-I I I
C-E-I IC-E-I I I
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0.3
0.4
0.5
0.6
0.7
U (
m/s
)
0 100 200 300 400 50010
-4
10-3
10-2
x (m)
(m2 /
s)
5
10
15
h (
m)
E-I
E-II
E-III
A
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0.1
0.2
0.3
0.4
U (
m/s
)
0 100 200 300 400 50010
-6
10-5
10-4
10-3
x (m)
(m2 /
s)
2
4
6
8
10
h (
m/s
)
E-I
E-II
E-III
C
UNIVERSITY OF SOUTH CAROLINAUNIVERSITY OF SOUTH CAROLINA
Conclusions
• 1. The ignition values obtained with different models can vary widely.
• 2. A turbidity current predicted to be subsiding by one entrainment relation could turn out to be igniting when a different entrainment model is used.
• 3. Important implication in the numerical modeling of turbidity current
• 4. Further research on the topic of sediment entrainment is crucial
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ACKNOWLEDGMENT
Funding from the National Science Foundation
(OCE-0134167) is gratefully acknowledged.