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Riccardo Rigon, Stefano Endrizzi, Matteo Dall’Amico, Stephan Gruber, Cristiano Lanni
GEOtop: Richards
G. O
Kee
fe, S
ky
wit
h f
lat
wh
ite
clou
d, 1
96
2
Wednesday, June 29, 2011
“The medium is the message”Marshall MacLuham
Wednesday, June 29, 2011
3
Objectives
•Make a short discussion about Richards’ equation (full derivation is
left to textbooks)
•Describe a simple (simplified solution of the equation)
•Analyze a numerical simulation for a linear hillslope
•Drawing some (hopefully) non trivial conclusions
•Doing a brief discussion of what happens when the system becomes
saturated from saturated
Rigon et al.
Richards
Wednesday, June 29, 2011
4
What I mean with Richards ++
First, I would say, it means that it would be better to call it, for
instance: Richards-Mualem-vanGenuchten equation, since it is:
Se = [1 + (−αψ)m)]−n
Se :=θw − θr
φs − θr
C(ψ)∂ψ
∂t= ∇ ·
�K(θw) �∇ (z + ψ)
�
K(θw) = Ks
�Se
��1− (1− Se)1/m
�m�2
C(ψ) :=∂θw()∂ψ
Rigon et al.
Richards
Wednesday, June 29, 2011
4
What I mean with Richards ++
First, I would say, it means that it would be better to call it, for
instance: Richards-Mualem-vanGenuchten equation, since it is:
Se = [1 + (−αψ)m)]−n
Se :=θw − θr
φs − θr
C(ψ)∂ψ
∂t= ∇ ·
�K(θw) �∇ (z + ψ)
�
K(θw) = Ks
�Se
��1− (1− Se)1/m
�m�2
Water balance
C(ψ) :=∂θw()∂ψ
Rigon et al.
Richards
Wednesday, June 29, 2011
4
What I mean with Richards ++
First, I would say, it means that it would be better to call it, for
instance: Richards-Mualem-vanGenuchten equation, since it is:
Se = [1 + (−αψ)m)]−n
Se :=θw − θr
φs − θr
C(ψ)∂ψ
∂t= ∇ ·
�K(θw) �∇ (z + ψ)
�
K(θw) = Ks
�Se
��1− (1− Se)1/m
�m�2
Water balance
ParametricMualem
C(ψ) :=∂θw()∂ψ
Rigon et al.
Richards
Wednesday, June 29, 2011
4
What I mean with Richards ++
First, I would say, it means that it would be better to call it, for
instance: Richards-Mualem-vanGenuchten equation, since it is:
Se = [1 + (−αψ)m)]−n
Se :=θw − θr
φs − θr
C(ψ)∂ψ
∂t= ∇ ·
�K(θw) �∇ (z + ψ)
�
K(θw) = Ks
�Se
��1− (1− Se)1/m
�m�2
Water balance
ParametricMualem
Parametricvan Genuchten
C(ψ) :=∂θw()∂ψ
Rigon et al.
Richards
Wednesday, June 29, 2011
5
Parameters !
Se = [1 + (−αψ)m)]−n
C(ψ)∂ψ
∂t= ∇ ·
�K(θw) �∇ (z + ψ)
�
K(θw) = Ks
�Se
��1− (1− Se)1/m
�m�2
Se :=θw − θr
φs − θrC(ψ) :=
∂θw()∂ψ
Rigon et al.
Richards
Wednesday, June 29, 2011
6
dθw
dψ= −φs
α m n(α ψ)n−1
[1 + (α ψ)n]m+1(θr + φs)
The hydraulic capacity of soil is proportional to the pore-size distribution
SWRC
Derivative
Water content
Rigon et al.
Richards
Wednesday, June 29, 2011
7
Rigon et al.
Richards
Wednesday, June 29, 2011
7
Rigon et al.
Richards
Wednesday, June 29, 2011
7
Rigon et al.
Richards
Wednesday, June 29, 2011
7
Rigon et al.
Richards
Wednesday, June 29, 2011
8
Pedotransfer Functions
Nem
es (
20
06
)
Rigon et al.
Richards
Wednesday, June 29, 2011
9
X - 52 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
Figure 2: Experimental set-up. (a) The infinite hillslope schematization. (b) The initial suction head profile.
Figure 3: The soil-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900 (the total number of
soil-pixels), corresponds to moving from the crest to the toe of the hillslope
Table 1: Physical, hydrological and geotechnical parameters used to characterize the silty-sand soil
Parameter group Parameter name Symbol Unit Value
Physical Bulk density ρb (g/cm3) 2.0
% sand - - 60
% silt - - 40
Hydrological Saturated hydraulic conductivity Ksat (m/s) 10−4
Saturated water content Θsat (cm3/cm−3) 0.39Residual water content Θr (cm3/cm−3) 0.155
water retention curve parameter n [−] 1.881water retention curve parameter β (cm−1) 0.0688
Geotechnical Effective angle of shearing resistance φ� ◦ 38Effective cohesion c� kN/m2 0
D R A F T September 24, 2010, 9:13am D R A F T
Lanni and Rigon
Richards
Wednesday, June 29, 2011
C(ψ)∂ψ∂t = ∂
∂z
�Kz
�∂ψ)∂z − cosθ
��+ ∂
∂y
�Ky
∂ψ∂y
�+ ∂
∂x
�Kx
�∂ψ)∂x − sinθ
��
ψ ≈ (z − d cos θ)(q/Kz) + ψs
Bearing in mind the previous positions, the Richards equation, at hillslope
scale, can be separated into two components. One, boxed in red, relative
to vertical infiltration. The other, boxed in green, relative to lateral flows.
10
The Richards equation on a plane hillslope
Iver
son
, 20
00
; Cord
ano a
nd
Rig
on
, 20
08
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
C(ψ)∂ψ∂t = ∂
∂z
�Kz
�∂ψ)∂z − cosθ
��+ ∂
∂y
�Ky
∂ψ∂y
�+ ∂
∂x
�Kx
�∂ψ)∂x − sinθ
��
ψ ≈ (z − d cos θ)(q/Kz) + ψs
Bearing in mind the previous positions, the Richards equation, at hillslope
scale, can be separated into two components. One, boxed in red, relative
to vertical infiltration. The other, boxed in green, relative to lateral flows.
10
The Richards equation on a plane hillslope
Iver
son
, 20
00
; Cord
ano a
nd
Rig
on
, 20
08
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
C(ψ)∂ψ∂t = ∂
∂z
�Kz
�∂ψ)∂z − cosθ
��+ ∂
∂y
�Ky
∂ψ∂y
�+ ∂
∂x
�Kx
�∂ψ)∂x − sinθ
��
ψ ≈ (z − d cos θ)(q/Kz) + ψs
Bearing in mind the previous positions, the Richards equation, at hillslope
scale, can be separated into two components. One, boxed in red, relative
to vertical infiltration. The other, boxed in green, relative to lateral flows.
10
The Richards equation on a plane hillslope
Iver
son
, 20
00
; Cord
ano a
nd
Rig
on
, 20
08
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
11
The Richards Equation!
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
C(ψ)∂ψ
∂t=
∂
∂z
�Kz
�∂ψ
∂z− cos θ
��+ Sr
Vertical infiltration: acts in a
relatively fast time scale because
it propagates a signal over a
thickness of only a few metres
11
The Richards Equation!
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
Sr =∂
∂y
�Ky
∂ψ
∂y
�+
∂
∂x
�Kx
�∂ψ
∂x− sin θ
��
12
The Richards Equation!
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
Sr =∂
∂y
�Ky
∂ψ
∂y
�+
∂
∂x
�Kx
�∂ψ
∂x− sin θ
��
Properly treated, this is reduced to
groundwater lateral flow, specifically to the
Boussinesq equation, which, in turn, have
been integrated from SHALSTAB equations
12
The Richards Equation!
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
C(ψ)∂ψ
∂t=
∂
∂z
�Kz
�∂ψ
∂z− cos θ
��+ Sr
In literature related to the determination of slope stability this equation
assumes a very important role because fieldwork, as well as theory, teaches
that the most intense variations in pressure are caused by vertical infiltrations.
This subject has been studied by, among others, Iverson, 2000, and D’Odorico
et al., 2003, who linearised the equations.
13
The Richards Equation!
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
14
Decomposition of the Richards equation
In vertical infiltration plus lateral flow is possible under the assumption
that:
Time scale of infiltration
soil depth
constant diffusivity
time scale of lateral flow
hillslope length
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
ψ ≈ (z − d cos θ)(q/Kz) + ψs
Iver
son
, 20
00
; D’O
dori
co e
t al
., 2
00
3,
Cord
ano a
nd
Rig
on
, 20
08
15
s
The Richards equation on a plane hillslope
Rigon et al.
Richards simplified
Wednesday, June 29, 2011
Assuming K ~ constant and neglecting the source terms
16
The Richards Equation 1-D
C(ψ)∂ψ
∂t= Kz 0
∂2ψ
∂z2
D0 :=Kz 0
C(ψ)
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
Assuming K ~ constant and neglecting the source terms
∂ψ
∂t= D0 cos2 θ
∂2ψ
∂t2
16
The Richards Equation 1-D
C(ψ)∂ψ
∂t= Kz 0
∂2ψ
∂z2
D0 :=Kz 0
C(ψ)
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
∂ψ
∂t= D0 cos2 θ
∂2ψ
∂t2
17
The Richards Equation 1-D
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
The equation becomes LINEAR and, having found a solution
with an instantaneous unit impulse at the boundary, the
solution for a variable precipitation depends on the
convolution of this solution and the precipitation.
∂ψ
∂t= D0 cos2 θ
∂2ψ
∂t2
17
The Richards Equation 1-D
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
18
The Richards Equation 1-D
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
For a precipitation impulse of constant intensity, the solution can be
written:
ψ0 = (z − d) cos2 θ
D’O
dori
co e
t al
., 2
00
3
19
ψ = ψ0 + ψs
ψs =
qKz
[R(t/TD)] 0 ≤ t ≤ T
qKz
[R(t/TD)−R(t/TD − T/TD)] t > T
The Richards Equation 1-D
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
In this case the equation admits an analytical solution
D’O
dori
co e
t al
., 2
00
3
20
R(t/TD) :=�
t/(π TD)e−TD/t − erfc��
TD/t�
ψs =
qKz
[R(t/TD)] 0 ≤ t ≤ T
qKz
[R(t/TD)−R(t/TD − T/TD)] t > T
TD :=z2
D0
The Richards Equation 1-D
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
D’O
dori
co e
t al
., 2
00
3
21
TD
TD
TD
TD
Th
e R
ich
ard
s Eq
uat
ion
1
-D
Rigon et al.
Richards super-simplified
Wednesday, June 29, 2011
The analytical solution methods for the advection-dispersion equation
(even non-linear), that results from the Richards equation, can be found
in literature relating to heat diffusion (the linearized equation is the
same), for example Carslaw and Jager, 1959, pg 357.
Usually, the solution strategies are 4 and they are based on:
- variable separation methods
- use of the Fourier transform
- use of the Laplace transform
- geometric methods based on the symmetry of the equation (e.g.
Kevorkian, 1993)
All methods aim to reduce the partial differential equation to a system
of ordinary differential equations
22
Th
e R
ich
ard
s Eq
uat
ion
1
-D
Rigon et al.
Richards 1D
Wednesday, June 29, 2011
23
Th
e R
ich
ard
s Eq
uat
ion
1
-D
Rigon et al.
Richards 1
D
Wednesday, June 29, 2011
Sim
on
i, 2
00
7
24
Th
e R
ich
ard
s Eq
uat
ion
1
-D
Rigon et al.
Richards 1D
Wednesday, June 29, 2011
25
Sim
on
i, 2
00
7
Th
e R
ich
ard
s Eq
uat
ion
1
-D
Rigon et al.
Richards 1D
Wednesday, June 29, 2011
26
A simple application ?
X - 52 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
Figure 2: Experimental set-up. (a) The infinite hillslope schematization. (b) The initial suction head profile.
Figure 3: The soil-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900 (the total number of
soil-pixels), corresponds to moving from the crest to the toe of the hillslope
Table 1: Physical, hydrological and geotechnical parameters used to characterize the silty-sand soil
Parameter group Parameter name Symbol Unit Value
Physical Bulk density ρb (g/cm3) 2.0
% sand - - 60
% silt - - 40
Hydrological Saturated hydraulic conductivity Ksat (m/s) 10−4
Saturated water content Θsat (cm3/cm−3) 0.39Residual water content Θr (cm3/cm−3) 0.155
water retention curve parameter n [−] 1.881water retention curve parameter β (cm−1) 0.0688
Geotechnical Effective angle of shearing resistance φ� ◦ 38Effective cohesion c� kN/m2 0
D R A F T September 24, 2010, 9:13am D R A F T
Rigon et al.
Richards 3D
Wednesday, June 29, 2011
27
X - 52 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
Figure 2: Experimental set-up. (a) The infinite hillslope schematization. (b) The initial suction head profile.
Figure 3: The soil-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900 (the total number of
soil-pixels), corresponds to moving from the crest to the toe of the hillslope
Table 1: Physical, hydrological and geotechnical parameters used to characterize the silty-sand soil
Parameter group Parameter name Symbol Unit Value
Physical Bulk density ρb (g/cm3) 2.0
% sand - - 60
% silt - - 40
Hydrological Saturated hydraulic conductivity Ksat (m/s) 10−4
Saturated water content Θsat (cm3/cm−3) 0.39Residual water content Θr (cm3/cm−3) 0.155
water retention curve parameter n [−] 1.881water retention curve parameter β (cm−1) 0.0688
Geotechnical Effective angle of shearing resistance φ� ◦ 38Effective cohesion c� kN/m2 0
D R A F T September 24, 2010, 9:13am D R A F T
Going back to the simple geometry case
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
28
Conditions of simulation
Wet Initial Conditions
Dry Initial Conditions
Intense Rainfall
Low Rainfall
Moderate Rainfall
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
29
X - 54 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
(a) DRY-Low (b) DRY-Med
(c) DRY-High (d) WET-Low
(e) WET-Med (f) WET-High
Figure 5: Values of pressure head developed at the soil-bedrock interface at each point of the subcritical parallel hillslope. The slope of
the pressure head lines represents the mean lateral gradient of pressure
D R A F T September 24, 2010, 9:13am D R A F T
Simulations result
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
30
Is the flow ever steady state ? X - 54 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
(a) DRY-Low (b) DRY-Med
(c) DRY-High (d) WET-Low
(e) WET-Med (f) WET-High
Figure 5: Values of pressure head developed at the soil-bedrock interface at each point of the subcritical parallel hillslope. The slope of
the pressure head lines represents the mean lateral gradient of pressure
D R A F T September 24, 2010, 9:13am D R A F T
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
31
X - 54 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
(a) DRY-Low (b) DRY-Med
(c) DRY-High (d) WET-Low
(e) WET-Med (f) WET-High
Figure 5: Values of pressure head developed at the soil-bedrock interface at each point of the subcritical parallel hillslope. The slope of
the pressure head lines represents the mean lateral gradient of pressure
D R A F T September 24, 2010, 9:13am D R A F T
Simulations result
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
32
X - 54 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
(a) DRY-Low (b) DRY-Med
(c) DRY-High (d) WET-Low
(e) WET-Med (f) WET-High
Figure 5: Values of pressure head developed at the soil-bedrock interface at each point of the subcritical parallel hillslope. The slope of
the pressure head lines represents the mean lateral gradient of pressure
D R A F T September 24, 2010, 9:13am D R A F T
Simulations result
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
33
LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES X - 55
(a) (b)
Figure 6: Temporal evolution of the vertical profile of hydraulic conductivity (a) and hydraulic conductivity at the soil-bedrock interface
(b) of a soil-pixel located in the mid-slope zone. Results are shown for the case representing DRY antecedent soil moisture conditions, Low
rainfall intensity and parallel hillslope shape of the subcritical (gentle) case
D R A F T September 24, 2010, 9:13am D R A F T
Three order of magnitude faster !
The key for understanding
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
34
When simulating is understanding
•Flow is never stationary
•For the first hours, the flow is purely slope normal with no lateral
movements
•After water gains the bedrock and a thin capillary fringe grows,
lateral flow starts
•This is due to the gap between the growth of suction with respect to
the increase of hydraulic conductivity
•The condition:
is not verified, since diffusivity in the slope normal direction is much lower
than in the lateral direction (after saturation is created)
Lanni and Rigon
Richards 3D for a hillslope
Wednesday, June 29, 2011
35
Another issue
Extending Richards to treat the transition saturated to unsaturated zone.
Since :
At saturation: what does change in time ?
Rigon et al.
Saturation vs Vadose
Wednesday, June 29, 2011
36
Another issue
Extending Richards to treat the transition saturated to unsaturated zone. Which means:
Rigon et al.
Saturation vs Vadose
Wednesday, June 29, 2011
37
If you do not have this extension you cannot deal properly with from
unsaturated volumes to saturated ones.
Or
Rigon et al.
Saturation vs Vadose
Wednesday, June 29, 2011
GEOtop: Richards++
Law
ren
Har
ris,
Mou
nt
Rob
son
Riccardo Rigon, Stefano Endrizzi, Matteo Dall’Amico, Stephan Gruber
Wednesday, June 29, 2011
“I would like to have a smart phrase for any situation.
But I don’t . Actually, I think is not even necessary.
I learned that this save time to listening to what others
have say, and by be silent you learn ”
Riccardo Rigon
Wednesday, June 29, 2011
40
Objectives
•Make a short discussion about what happens when soil freezes
•Introduce some thermodynamics of the problem
•Discussing how Richards equation has to be modified to include soil
freezing.
•Treating some little concept behind the numerics
•Seeing a validation of the model
Rigon et al.
Richards ++
Wednesday, June 29, 2011
41
What I mean with Richards ++
Extending Richards to treat the phase transition. Which means essentially to extend the soil water retention curves to become dependent on temperature.
Unsaturatedunfrozen
Freezingstarts
Freezingproceeds
UnsaturatedFrozen
Rigon et al.
Richards ++
Wednesday, June 29, 2011
42
Rigon et al.
The variable there !
Ice, soil, water and pores
Wednesday, June 29, 2011
43
Uc( ) := Uc(S, V, A, M)
Expression Symbol Name of the dependent variable∂SUc T temperature
- ∂V Uc p pressure∂AUc γ surface energy∂MUc µ chemical potential
dUc(S, V, A, M)dt
=∂Uc( )
∂S
∂S
∂t+
∂Uc( )∂V
∂V
∂t+
∂Uc( )∂A
∂A
∂t+
∂Uc( )∂M
∂M
∂t
Internal Energy
entropy
interfacial area
volume mass Independent variables
dUc(S, V, A, M) = T ( )dS − p( )dV + γ( ) dA + µ( ) dM
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
44
d[U( )+K( )+P ( )]dt = Φ( )
dS( )dt ≥ 0
dS(U, V, M) = 0
potential energy
kineticenergy
internalenergy
energy fluxes at the boundaries
the equilibrium relation becomes:
Assuming:
K( ) = 0 ; P ( ) = 0 ; Φ( ) = 0
Total Energy
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
45
At equilibrium. Gravity. One phase. No fluxes
the equilibrium relation becomes:
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
46
Ti = Tw
pi = pw
µi = µw
The equilibrium relation becomes:
At equilibrium. No gravity. No fluxes. Two phases
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
47
the equilibrium relation becomes:
At equilibrium. Gravity. Two phases. No fluxes
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
48
At equilibrium. Gravity. Two phases. No fluxesSeen in the phases diagram.
Going deeper in the pool we move according to the arrows going from higher to lower positions
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
49
SdT ( )− V dp( ) + Mdµ( ) ≡ 0
dµw(T, p) = dµi(T, p)
Gibbs-Duhem identity
From the equilibrium condition
At equilibrium. Two phases. No gravity. No fluxes. And ... no interfaces.
The equilibrium equation between the phases allows to derive the equations for the curves separating the phases, i.e. to obtain the Clausius-Clapeyron equation:
Internal Energy
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
50
−hw( )T
dT + vw( )dp = −hi( )T
dT + vi( )dp
⇒ dp
dT=
sw( )− si( )vw( )− vi( )
=hw( )− hi( )
T [vw( )− vi( )]≡ Lf ( )
T [vw( )− vi( )]
At equilibrium. Gravity. Two phases. No gravity. No fluxes. And ... no interfaces.
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
51
At equilibrium. Two phases. No gravity. No fluxes. And interfaces.
U( ) := T ( )S − p( )V + γ( )A + µ( )M
If we assume existing a relation between the interfacial area A and the volume, the effect of the surface can be seen as a pressure
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
52
pw = pa − γwa∂Awa(r)∂Vw(r)
= pa − γwa∂Awa/∂r
∂Vw/∂r= pa − γwa
2r
:= pa − pwa(r)
That is, what is seen in the Young-Laplace equation
pa
pw
At equilibrium. Two phases. No gravity. No fluxes. And interfaces.
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
53
dS =�
1Tw
− 1Ti
�dUw +
�pw + γiw
∂Aiw∂Vw
Tw− pi
Ti
�dVw −
�µw
Tw− µi
Ti
�dMw = 0
Ti = Tw
pi = pw + γiw∂Aiw∂Vw
µi = µw
The equilibrium condition becomes:
and, finally:
Putting all together
Rigon et al.
Ice, soil, water and pores
Wednesday, June 29, 2011
54
pw0 = pa − γwa∂Awa(r0)
∂Vw= pa − pwa(r0) pi = pa − γia
∂Aia(r0)∂Vw
:= pa − pia(r0)
pw1 = pa − γia∂Aiar(0)
∂Vw− γiw
∂Aiw(r1)∂Vw
Two interfaces (air-ice and water-ice) should be considered!!!
A closer look
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pi = pa − γia∂Aia(r0)
∂Vw≡ pa
pia(r) = −γia∂Aia(r0)
∂Vw← 0
pw1 = pw0 − γwa∂
∂Vw(Awa(r1)−Awa(r0)) = pw0 + pwa(r0)− pwa(r1)
pw1 = pw0 + ∆pfreez∆pfreez := −γwa∂ ∆Awa
∂Vw= pwa(r0)− pwa(r1)
Considering the assumption “freezing=drying” (Miller, 1963) the ice “behaves like air”:
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−hw( )T
dT + vw( )dpw = −hi( )T
dT + vi( )dpi
�dpw = dpfreez
dpi = 0
LfdT
T=
1ρw
dpfreez
pw1 ≈ pw0 + ρwLf
T0(T − T0)
From the equilibrium condition and the Gibbs-Duhem identity:
From the “freezing=drying” assumption:
freezing conditionunsaturated condition
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Unsaturatedunfrozen
Freezingstarts
Freezingprocedes
UnsaturatedFrozen
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Unsaturatedunfrozen
UnsaturatedFrozen
Freezingstarts
Freezingprocedes
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θw =θs
Aw |ψ|α + 1
θw = θr + (θs − θr) · {1 + [−α (ψ)]n}−m
θmaxw = θs ·
�Lf (T − Tm)
g T ψsat
�-1/b
C l a p p a n d H o r n b e r g e r (1978)
Unfrozen water content
soil water retention curve
thermodynamicequilibrium (Clausius Clapeyron)
+
Luo et al. (2009), Niu a n d Y a n g ( 2 0 0 6 ) , Zhang et al. (2007)
Gardner (1958) Shoop and Bigl (1997)
Van Genuchten (1980) Hansson et al (2004)
ψw =pw
ρw gpressure head:
θw(T ) = θw [ψw(T )]
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T ∗ := T0 +g T0
Lfψw0
Θ = θr + (θs − θr) · {1 + [−α · ψw0]n}−m
ice content: θi =ρw
ρi
�Θ− θw
�
θw = θr + (θs − θr) ·�
1 +�−αψw0 − α
Lf
g T0(T − T
∗) · H(T − T∗)
�n�−m
liquid water content:
Total water content:
depressed m e l t i n g point
A summary of the equations
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−0.05 −0.04 −0.03 −0.02 −0.01 0.00
0.1
0.2
0.3
0.4
Unfrozen water content
temperature [C]
Thet
a_u
[−]
psi_m −5000
psi_m −1000
psi_m −100
psi_m 0
ice
air
water
...
Assume you have an initial condition of little more that 0.1 water content
In the graphics
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−0.05 −0.04 −0.03 −0.02 −0.01 0.00
0.1
0.2
0.3
0.4
Unfrozen water content
temperature [C]
Thet
a_u
[−]
psi_m −5000
psi_m −1000
psi_m −100
psi_m 0
ice
air
water
...
There is a freezing point depression of less than 0.01 centigrades
In the graphics
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−0.05 −0.04 −0.03 −0.02 −0.01 0.00
0.1
0.2
0.3
0.4
Unfrozen water content
temperature [C]
Thet
a_u
[−]
psi_m −5000
psi_m −1000
psi_m −100
psi_m 0
ice
air
water
...
Temperature goes down to -0.015. Then, the water unfrozen remains 0.1
In the graphics
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Unsaturatedunfrozen
UnsaturatedFrozen
Freezingstarts
Freezingprocedes
An over and over again
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The overall relation between Soil water content, Temperature, and suction
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The overall relation between Soil water content, Temperature, and suction
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-3 -2 -1 0 1
0.0
0.2
0.4
0.6
0.8
1.0
n=1.5
temperature [C]
the
ta_
w/t
he
ta_
s [
-]
psi_w0=0 psi_w0=-1000
alpha=0.001 [1/mm]
alpha=0.01 [1/mm]
alpha=0.1 [1/mm]
alpha=0.4 [1/mm]
-10000 -8000 -6000 -4000 -2000 0
0.0
0.2
0.4
0.6
0.8
1.0
n=1.5
psi_w0 [mm]
the
ta_
w/t
he
ta_
s [
-]
T=2 T=-2
alpha=0.001 [1/mm]
alpha=0.01 [1/mm]
alpha=0.1 [1/mm]
alpha=0.4 [1/mm]
T > 0α [mm−1]
n 0.001 0.01 0.1 0.41.1 0.939 0.789 0.631 0.5491.5 0.794 0.313 0.099 0.0492.0 0.707 0.099 0.009 0.0022.5 0.659 0.032 0.001 1.2E-4
T = −2 ◦C
α [mm−1]n 0.001 0.01 0.1 0.41.1 0.576 0.457 0.363 0.3161.5 0.063 0.020 0.006 0.0032.0 4E-3 4E-4 4E-5 1E-52.5 2.5E-4 8E-6 2.5E-7 3.2E-8
24
θw/θs at ψw0=−1000 [mm]
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M. Dall’Amico et al.: Freezing unsaturated soil model 473
M. Dall’Amico et al.: Freezing unsaturated soil model 5
and consequently:
ψ =ψ(T ) =ψw0 +Lf
g T ∗(T −T
∗) (19)
The above equation is valid for T < T∗: in fact, when T ≥
T∗, the freezing process is not activated and the liquid water
pressure head is equal to the ψw0. Equations (17) and (19)
collapse in Eq. (15) for a saturated soil (i.e. ψw0 = 0). Thus
the formulation of the liquid water pressure head ψ(T ) under
freezing conditions, valid both for saturated and unsaturated
soils, becomes:
ψ(T ) =ψw0 + Lf
g T∗ (T −T∗) if T <T
∗
ψ(T ) =ψw0 if T ≥T∗
(20)
which can be summarized using the Heaviside function H( )as:
ψ(T ) =ψw0 +Lf
g T ∗(T −T
∗) ·H(T ∗−T ) (21)
If the soil water retention curve is modeled according to the
Van Genuchten (1980) model, the total water content be-
comes:
Θv = θr +(θs−θr) ·{1+[−α ψw0]n}−m
(22)
where θr (-) is the residual water content. The liquid water
content θw becomes:
θw = θr +(θs−θr) ·{1+[−α ψ(T )]n}−m(23)
Equation (23) gives the liquid water content at sub-zero
temperature and is usually called ”freezing-point depression
equation” (e.g. Zhang et al., 2007 and Zhao et al., 1997).
Differently from Zhao et al. (1997), it takes into account not
only the temperature under freezing conditions but also the
depressed melting temperature T∗, which depends on ψw0. It
comes as a consequence that the ice fraction is the difference
between Θv and θw:
θi = Θv(ψw0)−θw [ψ(T )] (24)
It results that, under freezing conditions (T < T∗), θw and
θi are function of ψw0, which dictates the saturation degree,
and T, that dictates the freezing degree. Equation (24), usu-
ally called “freezing-point depression equation”, relates the
maximum unfrozen water content allowed at a given temper-
ature in a soil. Figure 2 reports the freezing-point depression
equation for pure water and the different soil types according
to the Van Genuchten parameters given in Table 1.
Equations (21) and (17) represent the closure relations sought
for the differential equations of mass conservation (Eq. 6)
and energy conservation (Eq. 8).
Table 1. Porosity and Van Genuchten parameters for water and
different soil types as visualized in Fig. 2
θs θr α n source
(-) (-) (mm−1
) (-)
water 1.0 0.0 4E-1 2.50
sand 0.3 0.0 4.06E-3 2.03
silt 0.49 0.05 6.5E-4 1.67 (Schaap et al., 2001)
clay 0.46 0.1 1.49E-3 1.25 (Schaap et al., 2001)
−5 −4 −3 −2 −1 0 1
0.0
0.2
0.4
0.6
0.8
1.0
Temperature [ C]
wate
r con
tent
[−]
pure water
clay
silt
sand
Fig. 2. Freezing curve for pure water and various soil textures, ac-
cording to the Van Genuchten parameters given in Table 1.
5 The decoupled solution: splitting method
The final system of equations is given by the equations of
mass conservation (Eq. 6) and energy conservation (Eq. 8):
∂Θm(ψw0,T )∂t +∇•Jw(ψw0,T )+Sw = 0
∂U(ψw0,T )∂t +∇• [G(T )+J(ψw0)]+Sen = 0
(25)
The previous system is a function of T and ψw0 and can be
solved by the splitting method, as explained in Appendix B.
In the first half time step, the Richards’ equation is solved
and the internal energy is updated with only the advection
contribution. In the second half, no water flux is allowed,
which makes the volume a closed system, and the internal
energy is updated with the conduction flux in order to find
the new temperature and the new combination of water and
ice contents.
Fig. 2. Freezing curve for pure water and various soil textures, ac-cording to the Van Genuchten parameters given in Table 1.
The above equation is valid for T < T∗: in fact, when T ≥
T∗, the freezing process is not activated and the liquid water
pressure head is equal to the ψw0. Equations (17) and (19)collapse in Eq. (15) for a saturated soil (i.e. ψw0 = 0). Thusthe formulation of the liquid water pressure headψ(T ) underfreezing conditions, valid both for saturated and unsaturatedsoils, becomes:
ψ(T ) = ψw0+ Lfg T ∗ (T −T
∗) if T <T
∗
ψ(T ) = ψw0 if T ≥ T∗
(20)
which can be summarized using the Heaviside function H( )
as:
ψ(T ) = ψw0+Lf
g T ∗ (T −T∗) ·H(T
∗ −T ) (21)
If the soil water retention curve is modeled according to theVan Genuchten (1980) model, the total water content be-comes:
�v= θr+(θs−θr) ·�1+ [−α ψw0]n
�−m (22)
where θr (−) is the residual water content. The liquid watercontent θw becomes:
θw= θr+(θs−θr) ·�1+ [−α ψ(T )]n
�−m (23)
Equation (23) gives the liquid water content at sub-zerotemperature and is usually called “freezing-point depressionequation” (e.g. Zhang et al., 2007 and Zhao et al., 1997).Differently from Zhao et al. (1997), it takes into account notonly the temperature under freezing conditions but also the
Table 1. Porosity and Van Genuchten parameters for water anddifferent soil types as visualized in Fig. 2.
θs θr α n source(−) (−) (mm−1) (−)
water 1.0 0.0 4×10−1 2.50sand 0.3 0.0 4.06×10−3 2.03silt 0.49 0.05 6.5×10−4 1.67 (Schaap et al., 2001)clay 0.46 0.1 1.49×10−3 1.25 (Schaap et al., 2001)
depressed melting temperature T∗, which depends onψw0. It
comes as a consequence that the ice fraction is the differencebetween �v and θw:
θi= �v(ψw0)−θw[ψ(T )] (24)
It results that, under freezing conditions (T < T∗), θw and
θi are function of ψw0, which dictates the saturation degree,and T , that dictates the freezing degree. Equation (24), usu-ally called “freezing-point depression equation”, relates themaximum unfrozen water content allowed at a given temper-ature in a soil. Figure 2 reports the freezing-point depressionequation for pure water and the different soil types accordingto the Van Genuchten parameters given in Table 1.Equations (21) and (17) represent the closure relations
sought for the differential equations of mass conservation(Eq. 6) and energy conservation (Eq. 8).
5 The decoupled solution: splitting method
The final system of equations is given by the equations ofmass conservation (Eq. 6) and energy conservation (Eq. 8):
∂�m(ψw0,T )
∂t+∇•Jw(ψw0,T )+Sw= 0
∂U(ψw0,T )
∂t+∇• [G(T )+J (ψw0)]+Sen= 0
(25)
The previous system is a function of T and ψw0 and can besolved by the splitting method, as explained in Appendix B.In the first half time step, the Richards’ equation is solvedand the internal energy is updated with only the advectioncontribution. In the second half, no water flux is allowed,which makes the volume a closed system, and the internalenergy is updated with the conduction flux in order to findthe new temperature and the new combination of water andice contents.
5.1 Step 1: water and advection flux
Let us indicate with the superscript “n” the quantities at thetime step n, with “n+1” the quantities at the time step n+1:then t
n+1 = tn +�t (�t being the integration interval), and
with “n+1/2” the quantities at the end of the first step (tem-porary quantities). In the first step of the splitting method,
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∂
∂t
�θfl
w(ψw1)
�− �∇ •
�KH
�∇ ψw1 + KH�∇ zf
�+ Sw = 0
Liquid water may derive from
ice me l t ing : ∆θph
water flux: ∆θfl
Volume conservation:
0 ≤ θr ≤ Θ ≤ θs ≤ 1
θr −�θw0 + θi0 +
�1− ρi
ρw
�∆θph
i
�≤ ∆θfl
w ≤ θs −�θw0 + θi0 +
�1− ρi
ρw
�∆θph
i
�
Mass conservation (Richards, 1931) equation:
The Equations: the mass budget
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U = Cg(1− θs) T + ρwcwθw T + ρiciθi T + ρwLfθw
∂U
∂t+ �∇ • (�G + �J) + Sen = 0
�G = −λT (ψw0, T ) · �∇T
�J = ρw · �Jw(ψw0, T ) · [Lf + cw T ]
0 assuming freezing=drying
U = hgMg + hwMw + hiMi − (pwVw + piVi) + µwMphw + µiM
phi
no expansion: ρw=ρi
assuming:0 no flux during phase change
Eventually:
0 assuming equilibrium thermodynamics: µw=µi and Mw
ph = -Miph
conduction
advection
The Equations: the energy budget
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dU
dt= CT
dT
dt+ ρw
�(cw − ci) · T + Lf
�∂θw
∂t
∂θw [ψw1(T )]∂t
=∂θw
∂ψw1· ∂ψw1
∂T· ∂T
∂t= CH(ψw1) · ∂ψfreez
∂T· dT
dt
dU
dt=
�CT + ρw
�Lf + (cw − ci) · T
�· CH(T ) · ∂ψfreez(T )
∂T
�· dT
dt
-3 -2 -1 0 1
020
40
60
80
100
140
alpha= 0.01 [1/mm] n= 1.5 theta_s= 0.4
Temp. [ C]
U [M
J/m
3]
psi_w0=0
psi_w0=-100
psi_w0=-1000
psi_w0=-10000
-3 -2 -1 0 1
alpha= 0.01 [1/mm] n= 1.5 C_g= 2300000 [J/m3 K]
Temp. [ C]
C_
a [
MJ/m
3 K
]
1e+01
1e+02
1e+03
psi_w0=0 psi_w0=-1000
theta_s= 0.02
theta_s= 0.4
theta_s= 0.8 {Capp
The Equations: the energy budget
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∂U(ψw0,T )∂t
− ∂
∂z
�λT (ψw0, T ) · ∂T
∂z− J(ψw0, T )
�+ Sen = 0
∂Θ(ψw0)∂t
− ∂
∂z
�KH(ψw0, T ) · ∂ψw1(ψw0,T )
∂z−KH cos β
�+ Sw = 0
What we do in reality (GEOtop) is 1D
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73
turbulent fluxesInput
boundaryconditions
energybudget
snow/glaciers
precipitation water budget
Outputnew time step
GEOtopworkflow
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Equations and Numerics
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• Finite difference discretization, semi-implicit Crank-Nicholson method;
• Conservative linearization of the conserved quantity (Celia et al, 1990);
• Linearization of the system through Newton-Raphson method;
• when passing from positive to negative temperature, Newton-Raphson method is subject to big oscillations (Hansson et al, 2004)
Numerics
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if ||�Γ(η)m+1|| > ||�Γ(η)m|| ⇒ �ηm+1 � �ηm − �∆η · δ
globally convergent Newton-Raphson
reduction factor δ with 0 ≤ δ ≤ 1. If δ = 1 the scheme is the normal Newton-Raphson scheme
∆η
Numerics
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• Moving boundary condition between the two phases, where heat is liberated or absorbed
• Thermal properties of the two phases may be different
v1 = v2 = Tref (t > 0, z = Z(t))
v2 → Ti (t > 0, z →∞)
v1 = Ts (t > 0, z = 0)
λ1∂v1∂z − λ2
∂v2∂z = Lf ρw θs
dZ(t)dt (t > 0, z = Z(t))
∂v1∂t = k1
∂2v1∂z2 (t > 0, z < Z(t))
∂v2∂t = k2
∂2v2∂z2 (t > 0, z > Z(t))
v1 = v2 = Ti (t = 0, z)( Carlslaw and Jaeger, 1959, Nakano and Brown, 1971 )
The Stefan problem
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M. Dall’Amico et al.: Freezing unsaturated soil model 477M. Dall’Amico et al.: Freezing unsaturated soil model 9
!"#$%&'()*+
,%-./
0
0 10
23 4"#
!"#$%&'()*+
,%-./
0
0 10
23 4"#! "
Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) at various depths (m). The
numerical model in panel (A) uses Newton C-max the one in panel (B) uses Newton Global. Both have a grid spacing of 10 mm and 500
cells. Oscillations are present in panel (B) but not in panel (A) where no convergence is reached.
Fig. 5. Comparison between the simulated numerical and the an-
alytical solution. Soil profile temperature at different days. Grid
size=10 mm, N=500 cells
semi-infinite region given by Neumann. The features of this
problem are the existence of a moving interface between
Fig. 6. Cumulative error associated with the the globally convergent
Newton method. Solid line: cumulative error (J), dotted line: cumu-
lative error (%) as the ratio between the error and the total energy
of the soil in the time step. � was set to 1E-8.
the two phases, in correspondence of which heat is liberated
Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) at various depths (m). Thenumerical model in panel (A) uses Newton C-max the one in panel (B) uses Newton Global. Both have a grid spacing of 10mm and 500cells. Oscillations are present in panel (B) but not in panel (A) where no convergence is reached.
conductive heat flow in both the frozen and thawed regions,(3) change of volume negligible, i.e. ρw= ρi and (4) isother-mal phase change at T = Tm, i.e. no unfrozen water existsat temperatures less then the melting temperature Tm. Theisothermal phase change and uniform thermal characteristicsin the frozen and unfrozen state, may be assumed by im-posing a discontinuity on the freezing front line z = Z(t):θw(z) = 0, θi(z) = 1 λ(z) = λi, CT(z) = ρici for (t > 0, z <
Z(t)) and θw(z) = 1, θi(z) = 0 λ(z) = λw, CT(z) = ρwcwfor (t > 0, z ≥ Z(t)), respectively. The initial conditionsare: Ti(t = 0,z > 0) = +2 ◦C and a substance completely un-frozen: θw(t = 0,z) = 1 and θi(t = 0,z) = 0. The boundaryconditions of Dirichlet type: Ts(t > 0,z = 0) = −5 ◦C andTbot(t > 0,z → ∞) = +2 ◦C for the top and bottom bound-ary, respectively. As Nakano and Brown (1971) did for thecase of an initially frozen soil, in Appendix C we reportedthe complete derivation of the solution both for freezing andthawing processes.As the analytical solution considers the freezing of pure
water, in the numerical scheme we have considered a soilwith porosity θs = 1 characterized by a very steep soil waterretention curve with no residual water content, approachinga step function (Table 1).The domain is composed of 500 cells characterized by a
uniform depth �z = 10mm; the integration time �t = 10 s.We have already shown in Fig. 4a the results of the test with-out the globally convergent method. Figure 5 shows the com-parison between the numerical and the analytical solutionsof the soil temperature profile using the globally convergent
Newton’s method. The analytical solution is represented bythe dotted line and the simulation according the numericalmodel by the solid line. The results are much improved withthe globally convergent method, as the simulated tempera-ture follows the analytical solution very well. The temper-ature evolution shows a change in the slope that coincideswith the separation point between the upper frozen and thelower thawed part. Figure 4b reports the comparison on thetime axis (days) at different depths (m). The numerical sim-ulation result shows oscillations, which begin at the time ofphase change and then dampen with time. In the numericalsolution the temperature starts decreasing only when all thewater in the grid cell has been frozen. Furthermore, Tl isinfluenced by the phase change of Tl+1 by the release of la-tent heat and thus the temperature oscillation continues alsoin the frozen state. Therefore, oscillation amplitude is bothlinked to the grid size and to the time: increasing the gridsize, the oscillation amplitude increases, as the mass of wa-ter to freeze increases before the temperature may decrease.The oscillation amplitude dampens with time as the freezingfront moves away from zl ; it may be reduced but not elimi-nated, as it is embedded with the fixed-grid Eulerian method,where the freezing front may move in a discrete way and notin a continuum as in the reality. In order to test the energyconservation capabilities of the algorithm, the error, definedas the difference between the analytical and the simulated so-lution, was calculated at each time step as the p-norm (p = 1)of all the components and was cumulatively summed for theduration of the simulation. The tolerance � on the energy
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478 M. Dall’Amico et al.: Freezing unsaturated soil model
M. Dall’Amico et al.: Freezing unsaturated soil model 9
Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) at various depths (m). The
numerical model in panel (A) uses Newton C-max the one in panel (B) uses Newton Global. Both have a grid spacing of 10 mm and 500
cells. Oscillations are present in panel (B) but not in panel (A) where no convergence is reached.
−5 −4 −3 −2 −1 0 1 2
−2.5
−2.0
−1.5
−1.0
−0.5
0.0
Temp [ C]
soil
dept
h [m
]
03
15
40
75
SimAn
Fig. 5. Comparison between the simulated numerical and the an-
alytical solution. Soil profile temperature at different days. Grid
size=10 mm, N=500 cells
semi-infinite region given by Neumann. The features of this
problem are the existence of a moving interface between
Fig. 6. Cumulative error associated with the the globally convergent
Newton method. Solid line: cumulative error (J), dotted line: cumu-
lative error (%) as the ratio between the error and the total energy
of the soil in the time step. � was set to 1E-8.
the two phases, in correspondence of which heat is liberated
Fig. 5. Comparison between the simulated numerical and the an-alytical solution. Soil profile temperature at different days. Gridsize= 10mm, N = 500 cells.
balance was set to 1×10−8. Figure 6 shows the cumulatederror in J (solid line) and in percentage as the ratio betweenthe error and the total energy of the soil in the time step. With� set to 1×10−8, after 75 days of simulation, the error in per-centage remains very low (< 1×10−10), suggesting a goodenergy conservation capability of the algorithm.
7.2 Coupled water and energy flux: experimental data
In order to test the splitting time method for solving the cou-pled water and energy conservation, as done by Daanen et al.(2007), the model was tested against the experiment of Hans-son et al. (2004). The soil considered represents a Kana-gawa sandy loam, with the following parameters: θs= 0.535,θr= 0.05, α = 1.11×10−3 mm−1, n = 1.48, Cgs= 2.3×106,Jm−3 K−1, λgs = 2.5Wm−1 K−1 and saturated hydraulicconductivity KH(sat)= 0.0032mm s−1. The column wasconsidered initially unfrozen, with a uniform total water con-tent �v= 0.33 which, given the parameters of the soil reten-tion curve, corresponds to ψw0 = −2466.75mm, and initialtemperature T = 6.7 ◦C uniform. The boundary conditionsare of Neumann type: for the energy balance, at the top aflux F = −28 ·(T1+6) was considered, where T1 is the tem-perature of the first layer, and a zero flux condition at thebottom. For the mass balance equation, a zero flux at bothtop and bottom boundaries were used.Figure 7 shows the comparison of the profile of the total
water content �v. Starting from a thawed condition and auniform water content �v = 0.33, the liquid water content
M. Dall’Amico et al.: Freezing unsaturated soil model 9
Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) at various depths (m). The
numerical model in panel (A) uses Newton C-max the one in panel (B) uses Newton Global. Both have a grid spacing of 10 mm and 500
cells. Oscillations are present in panel (B) but not in panel (A) where no convergence is reached.
Fig. 5. Comparison between the simulated numerical and the an-
alytical solution. Soil profile temperature at different days. Grid
size=10 mm, N=500 cells
semi-infinite region given by Neumann. The features of this
problem are the existence of a moving interface between
0.00
00.
005
0.01
00.
015
0.02
0
time (days)
Cum
ulat
ive e
rror (
J)
0 15 30 45 60 75
Error (%)Error (J)
5e−1
35e−1
11e−1
0C
umul
ative
erro
r (%
)
Fig. 6. Cumulative error associated with the the globally convergent
Newton method. Solid line: cumulative error (J), dotted line: cumu-
lative error (%) as the ratio between the error and the total energy
of the soil in the time step. � was set to 1E-8.
the two phases, in correspondence of which heat is liberated
Fig. 6. Cumulative error associated with the the globally convergentNewton method. Solid line: cumulative error (J), dotted line: cumu-lative error (%) as the ratio between the error and the total energyof the soil in the time step. � was set to 1×10−8.
decreases from above due to the increase of ice content. It isvisible that the freezing of the soil sucks water from below.The increase in total water content reveals the position ofthe freezing front: after 12 h it is located about 40mm fromthe soil surface, after 24 h at 80mm and finally after 50 h at140mm. Similar to Hansson et al. (2004), the results wereimproved by multiplying the hydraulic conductivity by animpedance factor, as described is Sect. 3.1. It was found thatthe value of ω that best resembles the results is 7.Figure 8 shows the cumulative number of iterations re-
quired by the Newton C-max and the Newton global schemesto converge. It is clear that the number of iterations of thenew method is much lower than the previous, indicating thatthis method provides improved performance on the total sim-ulation time.
7.3 Infiltration into frozen soil
The coupled mass and energy conservation algorithm wasfinally tested with simulated rain (infiltration) during thethawing of a frozen soil. The soil geometry is a 20 cmdepth column discretized in 800 layers of 0.25mm depth;the top boundary condition for the energy equation is ofNeumann type, with 10Wm−2 constant incoming flux (nodaily cycle) and the bottom boundary condition is a zeroenergy flux. The initial conditions on the temperature are:Ti(t = 0,z > 0) = −10 ◦C; the soil is considered initially un-saturated, with water pressure head ψw0 given by a hydro-static profile based on a water table at 5m depth. This profile
The Cryosphere, 5, 469–484, 2011 www.the-cryosphere.net/5/469/2011/
Rigon et al.
Equations and Numerics
Wednesday, June 29, 2011
79
M. Dall’Amico et al.: Freezing unsaturated soil model 479M. Dall’Amico et al.: Freezing unsaturated soil model 11
water content [−]
soil
dept
h [m
m]
−200
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
0.25 0.30 0.35 0.40 0.45 0.50 0.55
after 12 hours!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
Sim! Meas
water content [−]
soil
dept
h [m
m]
−200
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
0.25 0.30 0.35 0.40 0.45 0.50 0.55
after 24 hours!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
Sim! Meas
water content [−]
soil
dept
h [m
m]
−200
−180
−160
−140
−120
−100
−80
−60
−40
−20
0
0.25 0.30 0.35 0.40 0.45 0.50 0.55
after 50 hours!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
!
Sim! Meas
Fig. 7. Comparison between the numerical (solid line) and the experimental results (points) obtained by Hansson et al. (2004) of the totalwater (liquid plus ice) after 12 (left), 24 (center) and 50 hours (right).
Fig. 8. Cumulative number of iterations of the Newton C-max andthe Newton global methods on the simulation test based on the ex-perimental results obtained by Hansson et al. (2004)
7.3 Infiltration into frozen soil
The coupled mass and energy conservation algorithm was fi-nally tested with simulated rain (infiltration) during the thaw-ing of a frozen soil. The soil geometry is a 20 cm depthcolumn discretized in 800 layers of 0.25 mm depth; the topboundary condition for the energy equation is of Neumanntype, with 10 W m−2 constant incoming flux (no daily cycle)and the bottom boundary condition is a zero energy flux. Theinitial conditions on the temperature are: Ti(t = 0,z > 0) =−10◦C; the soil is considered initially unsaturated, with wa-ter pressure head ψw0 given by a hydrostatic profile based ona water table at 5 m depth. This profile corresponds, accord-ing to Eq. (23) and (24), to the water and ice contents at eachlevel. The soil texture and thermal parameters are as in theexperiment described in Section 7.2, whereas the saturatedhydraulic conductivity is 0.3 mm s−1. As far as the bound-ary condition on the mass is concerned, the bottom bound-ary is characterized by a no flux condition, whereas the topboundary condition varies along two simulations: zero flux(without rain) and constant 10 mm h−1 flux, resembling aconstant precipitation (with rain). As can be seen in Fig. 9,in the first 6-7 days the behaviors with rain (solid line) andwithout precipitation (dotted line) are almost equal, becausehydraulic conductivity is so low due to the fast ice-saturationof the first layer during cold conditions neat the beginning ofthe experiment. Then, when some ice is melted, hydraulic
Fig. 7. Comparison between the numerical (solid line) and the experimental results (points) obtained by Hansson et al. (2004) of the totalwater (liquid plus ice) after 12 (left), 24 (center) and 50 h (right).
M. Dall’Amico et al.: Freezing unsaturated soil model 11
Fig. 7. Comparison between the numerical (solid line) and the experimental results (points) obtained by Hansson et al. (2004) of the totalwater (liquid plus ice) after 12 (left), 24 (center) and 50 hours (right).
010
000
2000
030
000
4000
0
time (days)
cum
ulat
ive n
umbe
r of i
tera
tion
0 1 2 3 4 5
Newton C=Cmax
Newton global
Fig. 8. Cumulative number of iterations of the Newton C-max andthe Newton global methods on the simulation test based on the ex-perimental results obtained by Hansson et al. (2004)
7.3 Infiltration into frozen soil
The coupled mass and energy conservation algorithm was fi-nally tested with simulated rain (infiltration) during the thaw-ing of a frozen soil. The soil geometry is a 20 cm depthcolumn discretized in 800 layers of 0.25 mm depth; the topboundary condition for the energy equation is of Neumanntype, with 10 W m−2 constant incoming flux (no daily cycle)and the bottom boundary condition is a zero energy flux. Theinitial conditions on the temperature are: Ti(t = 0,z > 0) =−10◦C; the soil is considered initially unsaturated, with wa-ter pressure head ψw0 given by a hydrostatic profile based ona water table at 5 m depth. This profile corresponds, accord-ing to Eq. (23) and (24), to the water and ice contents at eachlevel. The soil texture and thermal parameters are as in theexperiment described in Section 7.2, whereas the saturatedhydraulic conductivity is 0.3 mm s−1. As far as the bound-ary condition on the mass is concerned, the bottom bound-ary is characterized by a no flux condition, whereas the topboundary condition varies along two simulations: zero flux(without rain) and constant 10 mm h−1 flux, resembling aconstant precipitation (with rain). As can be seen in Fig. 9,in the first 6-7 days the behaviors with rain (solid line) andwithout precipitation (dotted line) are almost equal, becausehydraulic conductivity is so low due to the fast ice-saturationof the first layer during cold conditions neat the beginning ofthe experiment. Then, when some ice is melted, hydraulic
Fig. 8. Cumulative number of iterations of the Newton C-max andthe Newton global methods on the simulation test based on the ex-perimental results obtained by Hansson et al. (2004).
corresponds, according to Eqs. (23) and (24), to the waterand ice contents at each level. The soil texture and thermalparameters are as in the experiment described in Sect. 7.2,whereas the saturated hydraulic conductivity is 0.3mm s−1.As far as the boundary condition on the mass is concerned,the bottom boundary is characterized by a no flux condition,whereas the top boundary condition varies along two simula-tions: zero flux (without rain) and constant 10mmh−1 flux,resembling a constant precipitation (with rain). As can beseen in Fig. 9, in the first 6–7 days the behaviors with rain(solid line) and without precipitation (dotted line) are almostequal, because hydraulic conductivity is so low due to the fastice-saturation of the first layer during cold conditions neatthe beginning of the experiment. Then, when some ice ismelted, hydraulic conductivity increases so that some watercan infiltrate. At this point some incoming water may freezebecause the soil is still cold, the ice content is increased (notshown here) and the zero-curtain effect is prolonged. As aresult, infiltrating water provides energy (latent), so that tem-perature rises above 0 ◦C earlier than in the case of withoutrain. This earlier complete thaw is more evident at greaterdepths: in the case without rain, one has complete thawing at≈ −0.1 ◦C (change of slope of dashed curve) because the soilremains unsaturated and is characterized by T ∗ < Tf, while inthe case with rain complete thawing occurs at 0 ◦C becausethere is saturation. When soil is thawed, in the with rain casesoil temperature rises more slowly due to the lower thermaldiffusivity of the soil. It is interesting to notice that wateronly partially infiltrates into the frozen soil (see that dotted
www.the-cryosphere.net/5/469/2011/ The Cryosphere, 5, 469–484, 2011
Rigon et al.
Equations and Numerics
Wednesday, June 29, 2011
80
Endrizzi, Quinton, Marsh, Dall’Amico
Siksik creek (NWT, Canada) photo Endrizzi
Siksik
Applications
Wednesday, June 29, 2011
81
What are the key factors controlling the ground thaw in organic-covered permafrost terrain?
• Peat thickness? thermal and hydraulic properties of the ground (Quinton et al, 2008)
snow evolution, soil moisture, evapotranspiration...
position of the water- and frost-table
• Water movement?
•Topographic complexity (slope, aspect)?
Endrizzi, Quinton, Marsh, Dall’Amico
Siksik
Applications
Wednesday, June 29, 2011
82
0 10 20 30 40 50 60
01
02
03
04
0
thaw depth [cm]
rela
tive
fre
qu
en
cy [
%]
!
!
!
! !
!
! week1
week2
week3
week4
week5
week6
week7
week8
week9
Thaw depth relative frequencies: weekly average
Endrizzi, Quinton, Marsh, Dall’Amico
Siksik
Applications
Wednesday, June 29, 2011
83
• lateral water flow• variable peat thickness
high spatial variability of thaw depth
Endrizzi, Quinton, Marsh, Dall’Amico
Siksik thaw depth
Applications
Wednesday, June 29, 2011
84
week 1
___ model- - - meas
week 3
___ model- - - meas
Endrizzi, Quinton, Marsh, Dall’Amico
Siksik thaw depth
Applications
Wednesday, June 29, 2011
85
week 7
___ model- - - meas
week 9
___ model- - - meas
Siksik thaw depth
Applications
Wednesday, June 29, 2011
86
Stuff learned
•Soil freezing affects infiltration dramatically
•For describing soil freezing you need to deep in thermodynamics
and make several assumptions about the reference state. In this case
the freezing equal drying assumption has been made, and the
Clausius-Clapeyron equation has been generalized (this can be useful
for ET too).
•Phase transitions introduce discontinuities that must be treated
properly from a numerical point of view
Rigon et al.
Applications
Wednesday, June 29, 2011
Riccardo Rigon, Cristiano Lanni, Silvia Simoni
GEOtop: stability of slopes
Blu
rred
Joh
n W
all, r
ed r
oom
Wednesday, June 29, 2011
“Often meteo hydrologic forecasting looks
perturbed by imponderable pangs*, as they were
influenced by horoscopes and astrology”
Valentino Zeichen
* Correction by the authors. The aphorism in its triviality, can in fact be adopted for many sciences, or even for scientists. It is certainly true that “scientific” forecasting are not better than hydrology, above all if they are applied blindly to real system, and the modeler does not visit everyday her problem with its difficulties and irreducible issues. Forecasting is difficult and full of uncertainties. However, o not believe to the contrary: reading horoscopes does not helps you in understanding hydrology. Landslide occurence either.
Wednesday, June 29, 2011
89
•Introducing the problem of physically modelling ladslides with a
physically based model
•Discussing some conceptual result
•Discussing some effects of soil depth variability
•Seeing some application
Objectives
Rigon et al.
Landsliding
Wednesday, June 29, 2011
90
When simulating is understanding
•Flow is never stationary
•For the first hours, the flow is purely slope normal with no lateral
movements
•After water gains the bedrock and a thin capillary fringe grows,
lateral flow starts
•This is due to the gap between the growth of suction with respect to
the increase of hydraulic conductivity
Rigon et al.
Landsliding
Wednesday, June 29, 2011
91
The three dimensional structure of the problem
Landsliding
Wednesday, June 29, 2011
R
A stability criterion
92
Landsliding
Lanni and Rigon
Wednesday, June 29, 2011
93
Symbol Name nickname UnitFoS Factor of Safety fos [/]c� cohesion chsn [M L2 T−2]φc columbian friction angle cfa [/]ηw position of the water table surface pwts [L]z depth of soil ds [L]γs soil/terrain density std [M L−1 T−2 ]γw density of liquid water dlw [M L−1 T−2 ]ζs slope of terrain surface sts [/]
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
Coesione
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
p e n d e n z a
topografica
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
p e s o
specifico
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
a n g o l o d i
attrito
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
spessore del
SSD
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
94
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Infinite SlopeThe equation
Landsliding
Wednesday, June 29, 2011
95
Infinite SlopeThe equation
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Landsliding
Wednesday, June 29, 2011
95
Infinite SlopeThe equation
Cohesion
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Landsliding
Wednesday, June 29, 2011
95
Infinite SlopeThe equation
Cohesion Mohr-Coulomb
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Landsliding
Wednesday, June 29, 2011
95
Infinite SlopeThe equation
Cohesion Mohr-Coulomb Hydrology
FoS =c�
γs z cos ζs sin ζs+
tanφc
tan ζs− γwηw tanφc
γs z tan ζs
Rigon
Landsliding
Wednesday, June 29, 2011
96
Infinite Slope Introducing unsaturated conditionsThe equation
e.g. Lu and Godt, 2008
Landsliding
Wednesday, June 29, 2011
97
LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES X - 59
Table 3: A matrix of the times needed to achieve specific percentages of destabilized hillslope area for a continuous rainfall simulation for
a 5-day period.
A.C. RAIN SHAPE TF5% TF10% TF15% TF30% TF50%
DR
Y
Low
Divergent 41h ◦ ◦ ◦ ◦Parallel 41h ◦ ◦ ◦ ◦
Convergent 41h 60h ◦ ◦ ◦
Med Divergent 14-15h 15-16h 17-18h ◦ ◦
Parallel 14-15h 15-16h 16-17h 18h ◦Convergent 14-15h 14-15h 14-15h 15h ◦
Hig
h Divergent 7-8h 8-9h 9-10h 10-11h 12h
Parallel 7-8h 8h 8-9h 8-9h 8-9h
Convergent 7-8h 7-8h 7-8h 7-8h 8-9h
WE
T
Low
Divergent 3-4h ◦ ◦ ◦ ◦Parallel 3-4h ◦ ◦ ◦ ◦
Convergent 3-4h 4-5h ◦ ◦ ◦
Med Divergent 2-3h 3-4h 4-5h ◦ ◦
Parallel 2-3h 3h 3-4h 4-5h ◦Convergent 2-3h 2-3h 2-3h 2-3h ◦
Hig
h Divergent 1-2h 1-2h 1-2h 3h 5h
Parallel 1-2h 1-2h 1-2h 2-3h 2-3h
Convergent 1-2h 1-2h 1-2h 1-2h 1-2h
◦60h - - 20 h - - - 10 h - - - 5 h - 0h not achieved
D R A F T September 24, 2010, 9:13am D R A F T
Lanni and Rigon
Back to the simple case of planar hilllslope
Wednesday, June 29, 2011
98
X - 60 LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES
Table 4: A matrix of the rain volumes RFi and total water volume VFi (Rain volume + Pre-rain soil-water volume) needed to achieve
specific percentages of hillslope area for a continuous rainfall simulation for a 5-day period.
RAIN SHAPEF5% F10% F15% F30% F50%
DRY WET DRY WET DRY WET DRY WET DRY WET
RF
i(m
3)
Low
Divergent ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦Parallel ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦
Convergent ◦ ◦ ◦ ◦ ◦ ◦
Med Divergent ◦ ◦ ◦ ◦
Parallel ◦ ◦Convergent ◦ ◦
Hig
h Divergent
Parallel
Convergent
VF
i(m
3)
Low
Divergent ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦Parallel ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦
Convergent ◦ ◦ ◦ ◦ ◦ ◦
Med Divergent ◦ ◦ ◦ ◦
Parallel ◦ ◦Convergent ◦ ◦
Hig
h Divergent
Parallel
Convergent
◦15m3 - - 125m3 - - 230m3 - - 350m3 - - > 520m3 not achieved
Table 5: A matrix of the times needed to achieve specific percentages of destabilized hillslope area for a continuous rainfall simulation for
a 5-day period. The case of steep hillslopes
A.C. RAIN SHAPE FT( 5%) FT(10%) FT(15%) FT(30%) FT(50%)
DRYLow Parallel 32h 35.5h 38h 39h ◦High Parallel 7h 7h 7h 7h 7h
WETLow Parallel 0.25h 0.25h 0.25h 0.25h 0.25h
High Parallel 0.25h 0.25h 0.25h 0.25h 0.25h
◦60h - - - - - - - - - - - > 0h not achieved
D R A F T September 24, 2010, 9:13am D R A F T
Lanni and Rigon
Back to the simple case of planar hilllslope
Wednesday, June 29, 2011
A Flash back ...
Two main contenders debated in the last decade about modeling shallow landslides. Just to personalize this, as it is common nowadays
99
into the geological/gemorphological community
The West Coast guys The USGSes
Rigon
Just for fun
Wednesday, June 29, 2011
100
A Flash back ... into the geological/gemorphological community
Bill and Dave assert that shallow landslides can be approximately explained by saturation excess hydrology, lateral flow, and infinity slope stability. They produced the widely used and cited SHALSTAB model.
Montgomery and Dietrich, 1994
Rigon
Just for fun
Wednesday, June 29, 2011
101
A Flash back ... into the geological/gemorphological community
Dick Iverson (S. Baum, J. Godt) insists that the transient vertical effects counts in building a sufficient pore pressure to destabilize hillslope. This broughto TRIGRS
Iverson, 2000; Baum et al., 2002
Rigon
Just for fun
Wednesday, June 29, 2011
102
A Flash back ... with intrusion of hydrologists
One Italian gang tried to reconcile the fighters observing that all
of that above derives from the Richards equation with various
degree of simplifications, and proposed a linear simplified
theory with analytical solutions available that superimpose the
SHALSTAB theory with Iverson’s one, to be convoluted to
precipitations. D’Odorico et al., WRR, 2005; Cordano and Rigon, WRR, 2008
Rigon
Just for fun
Wednesday, June 29, 2011
103
A Flash back ... with intrusion of hydrologists
I.M.H.O., the linear theory works close to saturation but has some
problems of parameter characterizations in more unsaturated
cases, which caused some headaches to us.
To make a long story short, we did our best with simplified
theories*, but finally we had to observe that going directly to a
numerical non linear model was the right and conceptually
simpler choice.
Rigon
Just for fun
Wednesday, June 29, 2011
104
Ground surfaceBedrock surface
Soil-depth variability
Bedrock depression
Lanni et al.
Panola and the soil depth question
Panola
Wednesday, June 29, 2011
105
α = 13° α = 20° α = 30°
Soil (sandy-silt) Ksat = 10-4 m/sBedrock Ksat = 10-7 m/s
Rain Intensity = 6.5 mm/h Duration = 9 hours
Slope
Lanni et al.
Soil properties
Panola
Wednesday, June 29, 2011
106Downslope Drainage efficiency
Lanni et al.
Pressure growing
Panola
Wednesday, June 29, 2011
107
time
t=6h
t=9h
t=7h
Saturated area at the soil-bedrock interface increases very rapidly…..
α = 13°
Lanni et al.
Panola
Wednesday, June 29, 2011
108
time
t=6h
t=9h
t=7h
α = 13°
…..and than the average value of positive pore-water pressure continues to grow
Lanni et al.
Pressure growing
Panola
Wednesday, June 29, 2011
109
Q (m
3 /h)
t=9h
t=18h
t=22h
Hillslope water dischargeo 2 peaks α = 13°
t=6h t=9ht=7h t=14h
Lanni et al.
Wednesday, June 29, 2011
110
1D
3D
No role played by hillslope gradient
1° STEP: Vertical rain-infiltration
2° STEP: Lateral-flow
Infiltration-front propagation
Downslope drainagelimited by bedrock topography
Lanni et al.
Same as in the ideal planar case
Panola
Wednesday, June 29, 2011
111
(FS=1)
(1<FS<1.05)
t=10h
α = 30°
c’ = 0 kPaφ’ = 30°
Lanni, McDonnel, Hoop, Rigon
If you tilt you slide
Panola
Wednesday, June 29, 2011
112
Another case: Duron
Simoni, Dall’Amico, Zanotti and Rigon
Duron
Wednesday, June 29, 2011
113
Land cover
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Wednesday, June 29, 2011
114
Duron
Stratigraphy
Simoni, Dall’Amico, Zanotti and Rigon
Wednesday, June 29, 2011
115
Duron
And a tentative association of those maps withhydrological characters
Simoni, Dall’Amico, Zanotti and Rigon
Wednesday, June 29, 2011
116
Duron
Forecasting of temperaturein a point
In time
Simoni, Dall’Amico, Zanotti and Rigon
Wednesday, June 29, 2011
117
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Soil water content at different depthin a point
Wednesday, June 29, 2011
118
Duron
Pro
bab
ilit
y of
lan
dsl
idin
gSi
mon
i et
al, 2
00
8
Simoni, Dall’Amico, Zanotti and Rigon
Wednesday, June 29, 2011
119
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Pro
bab
ilit
y of
lan
dsl
idin
gSi
mon
i et
al, 2
00
8
Wednesday, June 29, 2011
120
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Pro
bab
ilit
y of
lan
dsl
idin
gSi
mon
i et
al, 2
00
8
Wednesday, June 29, 2011
121
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Pro
bab
ilit
y of
lan
dsl
idin
gSi
mon
i et
al, 2
00
8
Wednesday, June 29, 2011
122
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Pro
bab
ilit
y of
lan
dsl
idin
gSi
mon
i et
al, 2
00
8
Wednesday, June 29, 2011
123
And the snow again !
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Wednesday, June 29, 2011
124
Duron
Simoni, Dall’Amico, Zanotti and Rigon
Temperature of snow !
Wednesday, June 29, 2011
125
Lection Learned
• Simple stability analysis can be successful. Probably not for the right
reasons
• Simple settings give simple results (the total weight of water commands
the creation of large instabilities)
•This is due in the model to the compound of the vanGenuchten and
Mualem theory (which could not be always true)
•Soil depths counts
•On small scales instabilities could be controlled by constraints of local
topography
•Boundary conditions matter (trivial kinematic approaches could not work)
The GEOtop on Landliding
Rigon et al.
Wednesday, June 29, 2011
Thank you for your attention.
G.U
lric
i -
20
00
?
126
Wednesday, June 29, 2011
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