3 geotop-summer-school2011

GEOtop: Richards
G. OKeefe, Sky with flat white cloud, 1962

Riccardo Rigon, Stefano Endrizzi, Matteo Dall’Amico, Stephan Gruber,
Cristiano Lanni
Wednesday, June 29, 2011

“The medium is the message”
Marshall MacLuham


Richards

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

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Richards

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:

∂ψ
C(ψ)
= ∇ · K(θw ) ∇ (z + ψ)
∂t
m 2
K(θw ) = Ks Se 1 − (1 − Se ) 1/m

−n
Se = [1 + (−αψ) )] m

∂θw () θw − θr
C(ψ) := Se :=
∂ψ φs − θr
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Richards


∂ψ
C(ψ)
= ∇ · K(θw ) ∇ (z + ψ) Water balance
∂t
m 2
K(θw ) = Ks Se 1 − (1 − Se ) 1/m

−n
Se = [1 + (−αψ) )] m

C(ψ) := Se :=
∂ψ φs − θr
4

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Richards


∂ψ
C(ψ)
∂t
m 2
Parametric
K(θw ) = Ks Se 1 − (1 − Se ) 1/m
Mualem

−n
Se = [1 + (−αψ) )] m

C(ψ) := Se :=
∂ψ φs − θr
4

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Richards


∂ψ
C(ψ)
∂t
m 2
Parametric
K(θw ) = Ks Se 1 − (1 − Se ) 1/m
Mualem

−n Parametric
Se = [1 + (−αψ) )] m
van Genuchten

C(ψ) := Se :=
∂ψ φs − θr
4

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Richards

Parameters !
∂ψ
C(ψ)
= ∇ · K(θw ) ∇ (z + ψ)
∂t
m 2
K(θw ) = Ks Se 1 − (1 − Se )1/m

−n
Se = [1 + (−αψ) )]
m

C(ψ) := Se :=
∂ψ φs − θr

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Richards

The hydraulic capacity of soil is proportional
to the pore-size distribution
dθw α m n(α ψ)n−1
= −φs (θr + φs )
dψ [1 + (α ψ)n ]m+1

SWRC

Derivative

Water content 6

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Richards

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Richards

Pedotransfer Functions
Nemes (2006)

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Richards

igure 2: Experimental set-up. (a) The inﬁnite hillslope schematization. (b) The initial suction head pr

il-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900
9

sponds to moving from the crest to the toe of the hillslope
Lanni and Rigon

Richards simplified

The Richards equation on a plane hillslope

∂ψ) ∂ψ)
C(ψ) ∂ψ = ∂
Kz − cosθ + ∂
Ky ∂ψ + ∂
Kx − sinθ
Iverson, 2000; Cordano and Rigon, 2008

∂t ∂z ∂z ∂y ∂y ∂x ∂x

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.

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Richards simplified

The Richards Equation!

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Richards simplified


∂ψ ∂ ∂ψ
C(ψ) = Kz − cos θ + Sr
∂t ∂z ∂z

Vertical infiltration: acts in a
relatively fast time scale because
it propagates a signal over a
thickness of only a few metres

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Richards simplified


∂ ∂ψ ∂ ∂ψ
Sr = Ky + Kx − sin θ
∂y ∂y ∂x ∂x

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Richards simplified


∂ ∂ψ ∂ ∂ψ
Sr = Ky + Kx − sin θ
∂y ∂y ∂x ∂x

Properly treated, this is reduced to
groundwater lateral flow, specifically to the
Boussinesq equation, which, in turn, have
been integrated from SHALSTAB equations

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Richards simplified


∂ψ ∂ ∂ψ
C(ψ) = Kz − cos θ + Sr
∂t ∂z ∂z

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.
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Richards simplified

Decomposition of the Richards equation

In vertical infiltration plus lateral flow is possible under the assumption
that:
soil depth hillslope length

time scale of lateral flow

constant diffusivity

Time scale of infiltration

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Richards simplified

The Richards equation on a plane hillslope

ψ ≈ (z − d cos θ)(q/Kz ) + ψs
Iverson, 2000; D’Odorico et al., 2003,
Cordano and Rigon, 2008

s

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Richards super-simplified

The Richards Equation 1-D
Assuming K ~ constant and neglecting the source terms

∂ψ ∂2ψ
C(ψ) = Kz 0
∂t ∂z 2

Kz 0
D0 :=
C(ψ) 16

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Assuming K ~ constant and neglecting the source terms

∂ψ ∂2ψ
C(ψ) = Kz 0
∂t ∂z 2

∂ψ ∂2ψ
= D0 cos θ
2
∂t ∂t2
Kz 0
D0 :=
C(ψ) 16

Rigon et al.



∂ψ ∂2ψ
= D0 cos2 θ
∂t ∂t2

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∂ψ ∂2ψ
= D0 cos2 θ
∂t ∂t2

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.

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For a precipitation impulse of constant intensity, the solution can be
written:

ψ = ψ0 + ψs
D’Odorico et al., 2003

ψ0 = (z − d) cos θ 2

 q
 Kz [R(t/TD )] 0≤t≤T
ψs =
 q
Kz [R(t/TD ) − R(t/TD − T /TD )] t T

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In this case the equation admits an analytical solution

 q
 Kz [R(t/TD )] 0≤t≤T

ψs =
 q
Kz [R(t/TD ) − R(t/TD − T /TD )] t T

R(t/TD ) := t/(π TD )e −TD /t
− erfc TD /t

z2
TD :=
D0
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TD

TD

TD

TD
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Richards 1D

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
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D
s1
ard
Rich

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Richards 1D
Simoni, 2007

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Richards 1D
Simoni, 2007

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Richards 3D

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.

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Richards 3D for a hillslope

igure 2: Experimental set-up. (a) The inﬁnite hillslope schematization. (b) The initial suction head pr
Going back to the simple geometry case

il-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900
27

sponds to moving from the crest to the toe of the hillslope
Lanni and Rigon


Conditions of simulation

Wet Initial Conditions Intense Rainfall

Moderate Rainfall

Dry Initial Conditions Low Rainfall

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- 54
Simulations result
LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES

(a) DRY-Low (b) DRY-Med

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- 54 Is the flow ever steady state ?
LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES


30

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Simulations result

(c) DRY-High (d) WET-Low

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(c) DRY-High (d) WET-Low
Simulations result

(e) WET-Med (f) WET-High
F T September 24, 2010, 9:13am D 32 A
R
Values of pressure head developed at the soil-bedrock interface at each point of the subcritical parallel hillslope. The
Lanni and Rigon
e Wednesday, Junerepresents the mean lateral gradient of pressure
head lines 29, 2011


The key for understanding

Three order of magnitude faster !
(a) (b)
33

Lanni andTemporal evolution of the vertical proﬁle of hydraulic conductivity (a) and hydraulic conductivity at the soil-bedrock interface
Figure 6: Rigon


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)
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Saturation vs Vadose

Another issue
Extending Richards to treat the transition saturated to unsaturated zone.
Since :

At saturation: what does change in time ?

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Another issue
Extending Richards to treat the transition saturated to unsaturated zone.
Which means:

36

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Or

If you do not have this extension you cannot deal properly with from
unsaturated volumes to saturated ones.

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GEOtop: Richards++
Lawren Harris, Mount Robson

Riccardo Rigon, Stefano Endrizzi, Matteo Dall’Amico, Stephan Gruber


“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


Richards ++

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

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Richards ++


Extending Richards to treat the phase transition. Which means essentially to
extend the soil water retention curves to become dependent on temperature.

Freezing
Unsaturated starts
unfrozen

Unsaturated Freezing
Frozen proceeds

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Ice, soil, water and pores

The variable there !

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Internal Energy
Uc ( ) := Uc (S, V, A, M )
entropy
volume mass Independent variables

interfacial area

dUc (S, V, A, M ) ∂Uc ( ) ∂S ∂Uc ( ) ∂V ∂Uc ( ) ∂A ∂Uc ( ) ∂M
= + + +
dt ∂S ∂t ∂V ∂t ∂A ∂t ∂M ∂t
Expression Symbol Name of the dependent variable
∂S Uc T temperature
- ∂V Uc p pressure
∂A Uc γ surface energy
∂M Uc µ chemical potential

dUc (S, V, A, M ) = T ( )dS − p( )dV + γ( ) dA + µ( ) dM
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Total Energy
internal kinetic potential energy fluxes at
energy energy energy the boundaries

 d[U ( )+K( )+P ( )]
 dt = Φ( )
 dS( )
dt ≥ 0
Assuming:

K( ) = 0 ; P ( ) = 0 ; Φ( ) = 0
the equilibrium relation becomes:

dS(U, V, M ) = 0
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At equilibrium. Gravity. One phase. No fluxes


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At equilibrium. No gravity. No
fluxes. Two phases

The equilibrium relation becomes:


 Ti = Tw
pi = pw

µi = µw

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At equilibrium. Gravity. Two phases. No fluxes


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At equilibrium. Gravity. Two phases. No fluxes
Seen in the phases diagram.

Going deeper in the pool we move according to the arrows
going from higher to lower positions
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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

SdT ( ) − V dp( ) + M dµ( ) ≡ 0 Gibbs-Duhem identity

dµw (T, p) = dµi (T, p) From the equilibrium condition

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At equilibrium. Gravity. Two phases. No gravity.
No fluxes. And ... no interfaces.

hw ( ) hi ( )
− dT + vw ( )dp = − dT + vi ( )dp
T T

dp sw ( ) − si ( ) hw ( ) − hi ( ) Lf ( )
⇒ = = ≡
dT vw ( ) − vi ( ) T [vw ( ) − vi ( )] T [vw ( ) − vi ( )]
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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

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fluxes. And interfaces.
That is, what is seen in the Young-Laplace equation

∂Awa (r) ∂Awa /∂r 2
pw = pa − γwa = pa − γwa = pa − γwa := pa − pwa (r)
∂Vw (r) ∂Vw /∂r r

pa

pw

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Putting all together

The equilibrium condition becomes:

1 1 pw + γiw ∂Aiw
∂Vw pi µw µi
dS = − dUw + − dVw − − dMw = 0
Tw Ti Tw Ti Tw Ti

and, finally:


 Ti = Tw
pi = pw + γiw ∂Aiw
∂Vw

µi = µw

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A closer look
∂Awa (r0 ) ∂Aia (r0 )
pw0 = pa − γwa = pa − pwa (r0 ) pi = pa − γia := pa − pia (r0 )
∂Vw ∂Vw

∂Aia r(0) ∂Aiw (r1 )
pw1 = pa − γia − γiw
∂Vw ∂Vw

Two interfaces (air-ice and water-ice)
should be considered!!!

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freezing = drying

Considering the assumption
“freezing=drying” (Miller, 1963)
the ice “behaves like air”:

∂Aia (r0 )
pi = pa − γia ≡ pa
∂Vw

∂Aia (r0 )
pia (r) = −γia ←0
∂Vw

∂
pw1 = pw0 − γwa (Awa (r1 ) − Awa (r0 )) = pw0 + pwa (r0 ) − pwa (r1 )
∂Vw

∂ ∆Awa
∆pf reez := −γwa = pwa (r0 ) − pwa (r1 ) pw1 = pw0 + ∆pf reez
∂Vw
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freezing = drying

From the equilibrium condition and
the Gibbs-Duhem identity:
dT 1
hw ( ) hi ( ) Lf = dpf reez
− dT + vw ( )dpw = − dT + vi ( )dpi T ρw
T T

From the “freezing=drying” assumption:
Lf
dpw = dpf reez pw1 ≈ pw0 + ρw (T − T0 )
dpi = 0 T0

unsaturated condition freezing condition

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freezing = drying

unfrozen starts

Frozen procedes

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freezing = drying

unfrozen starts

Frozen procedes

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freezing = drying

pw
pressure head: ψw =
ρw g

Unfrozen water content

θw (T ) = θw [ψw (T )]

soil water thermodynamic
+
retention curve equilibrium (Clausius Clapeyron)
-1/b
Clapp and Lf (T − Tm )
max
θw = θs · Luo et al. (2009), Niu
Hornberger g T ψsat
and Yang (2006),
(1978) θs Zhang et al. (2007)
θw =
Gardner (1958) Aw |ψ|α + 1 Shoop and Bigl (1997)

Van Genuchten θw = θr + (θs − θr ) · {1 + [−α (ψ)] }
n −m
(1980) Hansson et al (2004)

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freezing = drying

A summary of the equations

n −m
Total water content: Θ = θr + (θs − θr ) · {1 + [−α · ψw0 ] }
n −m
Lf
liquid water content: θw = θr + (θs − θr ) · 1 + −αψw0 − α (T − T ∗ ) · H(T − T ∗ )
g T0
ρw
ice content: θi = Θ − θw
ρi

depressed g T0
T ∗ := T0 + ψ w0
melting Lf
point

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freezing = drying

In the graphics
...Unfrozen water content

0.4
psi_m −5000

psi_m −1000

0.3
psi_m −100

psi_m 0 air
Theta_u [−]

0.2

ice
0.1

water

−0.05 −0.04 −0.03 −0.02 −0.01 0.00

temperature [C]

Assume you have an initial condition of little more that 0.1 water content
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freezing = drying

In the graphics

0.4
psi_m −5000

psi_m −1000

0.3
psi_m −100

psi_m 0 air
Theta_u [−]

0.2

ice
0.1

water

−0.05 −0.04 −0.03 −0.02 −0.01 0.00

temperature [C]

There is a freezing point depression of less than 0.01 centigrades
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freezing = drying

In the graphics

0.4
psi_m −5000

psi_m −1000

0.3
psi_m −100

psi_m 0 air
Theta_u [−]

0.2

ice
0.1

water

−0.05 −0.04 −0.03 −0.02 −0.01 0.00

temperature [C]

Temperature goes down to -0.015. Then, the water unfrozen remains 0.1
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freezing = drying

An over and over again

unfrozen starts

Frozen procedes

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freezing = drying

The overall relation between Soil water content,
Temperature, and suction

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freezing = drying

The overall relation between Soil water content,
Temperature, and suction

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freezing = drying

T=2 T=-2 T = −2 ◦ C θw/θs at ψw0=−1000 [mm]
1.0

alpha=0.001 [1/mm]
α [mm−1 ]
alpha=0.01 [1/mm]
0.8

n 0.001 0.01 0.1 0.4
theta_w/theta_s [-]

alpha=0.1 [1/mm]
1.1 0.576 0.457 0.363 0.316
0.6

alpha=0.4 [1/mm]
1.5 0.063 0.020 0.006 0.003
2.0 4E-3 4E-4 4E-5 1E-5
0.4

2.5 2.5E-4 8E-6 2.5E-7 3.2E-8
0.2

psi_w0=0 psi_w0=-1000

1.0
0.0

alpha=0.001 [1/mm]

-10000 -8000 -6000 -4000 -2000 0 alpha=0.01 [1/mm]

0.8
theta_w/theta_s [-]
psi_w0 [mm] alpha=0.1 [1/mm]
n=1.5

0.6
alpha=0.4 [1/mm]

T 0

0.4
α [mm−1 ]
n 0.001 0.01 0.1 0.4

0.2
1.1 0.939 0.789 0.631 0.549
1.5 0.794 0.313 0.099 0.049 0.0
2.0 0.707 0.099 0.009 0.002 -3 -2 -1 0 1

2.5 0.659 0.032 0.001 1.2E-4 temperature [C]
n=1.5 24
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different soil types as visualized in Fig. 2
f freezing = drying
(T − T ∗ ) (19)
T∗
θs θr α n source
valid for T T ∗ : in fact, when T ≥
ss is not activated and the liquid water
Dependence 2.50 texture
water 1.0 0.0 4E-1
on (-) (-) (mm−1 ) (-)

to the ψw0 . Equations (17) and (19) sand 0.3 0.0 4.06E-3 2.03
or a saturated soil (i.e. ψw0 = 0). Thus silt 0.49 0.05 6.5E-4 1.67 (Schaap et al., 2001)
liquid water pressure head ψ(T ) under clay 0.46 0.1 1.49E-3 1.25 (Schaap et al., 2001)
alid both for saturated and unsaturated M. Dall’Amico et al.: Freezing unsaturated soil model

1.0
Table 1. Porosity and Van Genuchten parame
(T − T ∗ ) if T T ∗ pure water different soil types as visualized in Fig. 2.
(20)
clay
T ≥T∗

0.8
θs θr α n
silt (−) (−) (mm−1 ) (−)
zed using the Heaviside function H( ) 4 × 10−1
water content [−]
sand water 1.0 0.0 2.50
0.6 sand 0.3 0.0 4.06 × 10−3 2.03
silt 0.49 0.05 6.5 × 10−4 1.67 (S
T − T ∗ ) · H(T ∗ − T ) (21) clay 0.46 0.1 1.49 × 10−3 1.25 (S
0.4

ion curve is modeled according to the
depressed melting temperature T ∗ , which de
) model, the total water content be-
0.2

comes as a consequence that the ice fraction
between v and θw :
n −m
1 + [−α ψw0 ] } (22) θi = v (ψw0 ) − θw [ψ(T )]
0.0

It results that, under freezing conditions (T
idual water content. The liquid water −5 −4 −3 −2 −1 0 1
θi are function of ψw0 , which dictates the s
Temperature [ C] and T , that dictates the freezing degree. Eq
n −m ally called “freezing-point depression equa
1 + [−α ψ(T )] } (23) maximum unfrozen water content allowed a
Fig. 2. Freezing curve for pure water and various soil textures, ac-
Fig. 2. Freezing curve for pureparameters given in Table 1.
cording to the Van Genuchten water and various soil textures, ac- ature in a soil. Figure 2 reports the freezing-
the liquid water content at sub-zero equation for pure water and the different soi
ually called ”freezing-point depression cording to the Van Genuchten parameters given in Table 1.
68
to the Van Genuchten parameters given in T
g et al., 2007 and Zhao et al., 1997). The above equation is valid for T T ∗ : in fact, when T ≥ Equations (21) and (17) represent the
et al. (1997), it takes into account not T ∗ , the freezing process is not activated and the liquid water
Rigon et al. 5 The decoupled solution: splitting method sought for the differential equations of m
under freezing conditions but also the pressure head is equal to the ψw0 . Equations (17) and (19) (Eq. 6) and energy conservation (Eq. 8).
perature T ∗ ,June 29, 2011
which depends on ψw0 . It collapse system of for a saturated soil by ψw0 = 0). Thus
The ﬁnal in Eq. (15) equations is given (i.e. the equations of
Wednesday,

freezing = drying

The Equations: the mass budget

ice melting:
Liquid water may derive ∆θph
from
water flux: ∆θfl

Volume conservation:

 0 ≤ θr ≤ Θ ≤ θs ≤ 1



 θr − θw0 + θi0 + 1 − ρi ph ρi ph
ρw ∆θi ≤ ∆θwl ≤ θs − θw0 + θi0 + 1 −
f
ρw ∆θi

Mass conservation (Richards, 1931) equation:

∂ fl

θw (ψw1 ) − ∇ • KH ∇ ψw1 + KH ∇ zf + Sw = 0
∂t

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freezing = drying consequences

The Equations: the energy budget

U = hg Mg + hw Mw + hi Mi − (pw Vw + pi Vi ) + µw Mw + µi Miph
ph

0 assuming equilibrium thermodynamics:
µw=µi and Mwph = -Miph

0 assuming freezing=drying

no flux during phase change
0 assuming:
no expansion: ρw=ρi
Eventually:
U = Cg (1 − θs ) T + ρw cw θw T + ρi ci θi T + ρw Lf θw

G = −λT (ψw0 , T ) · ∇T conduction
∂U
+ ∇ • (G + J) + Sen = 0
∂t

J = ρw · Jw (ψw0 , T ) · [Lf + cw T ] advection

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freezing = drying consequences

The Equations: the energy budget

140
dU dT ∂θ
w
= CT + ρw (cw − ci ) · T + Lf psi_w0=0
dt dt ∂t psi_w0=-100

100
psi_w0=-1000

psi_w0=-10000

U [MJ/m3]

80
∂θw [ψw1 (T )] ∂θw ∂ψw1 ∂T ∂ψf reez dT
= · · = CH (ψw1 ) · ·

60
∂t ∂ψw1 ∂T ∂t ∂T dt

40
20
psi_w0=0 psi_w0=-1000
1e+03

theta_s= 0.02

0
theta_s= 0.4
theta_s= 0.8
C_a [MJ/m3 K]

-3 -2 -1 0 1
1e+02

Temp. [ C]
alpha= 0.01 [1/mm] n= 1.5 theta_s= 0.4
1e+01

dU ∂ψf reez (T ) dT
= CT + ρw Lf + (cw − ci ) · T · CH (T ) · ·
dt ∂T dt

Rigon et al.
-3 -2 {-1

Temp. [ C]
0 1

alpha= 0.01 [1/mm] n= 1.5 C_g= 2300000 [J/m3 K]
Capp
71

Equations and Numerics

What we do in reality (GEOtop) is 1D

 ∂U (ψw0 ,T )

 ∂t − ∂
∂z λT (ψw0 , T ) · ∂T
∂z − J(ψw0 , T ) + Sen = 0


 ∂Θ(ψw0 )
− ∂
KH (ψw0 , T ) · ∂ψw1 (ψw0 ,T )
− KH cos β + Sw = 0
∂t ∂z ∂z

72

Rigon et al.


GEOtop
workflow

turbulent boundary snow/glaciers
Input
fluxes conditions

energy
precipitation water budget budget

new time step Output

73

Rigon et al.


Numerics

• 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)

74

Rigon et al.


Numerics

∆η

globally convergent Newton-Raphson
if || m+1 || || m ||
Γ(η) Γ(η)
⇒ m+1 m − ∆η · δ
η η

reduction factor δ with 0 ≤ δ ≤ 1.
If δ = 1 the scheme is the normal Newton-
Raphson scheme
75

Rigon et al.


The Stefan problem

 v1 = v2 = Tref
 (t 0, z = Z(t))





 v2 → Ti

 (t 0, z → ∞)





 v1 = Ts

 (t 0, z = 0)




λ1 ∂v1 − λ2 ∂v2 = Lf ρw θs dZ(t)
∂z ∂z dt (t 0, z = Z(t))




 ∂v1
 2
 ∂t = k1 ∂ v21
 (t 0, z Z(t))


∂z


 ∂v
 2 ∂ 2 v2

 ∂t = k2 ∂z2 (t 0, z Z(t))




( Carlslaw and Jaeger, 1959, Nakano and Brown, 1971 ) 

v1 = v2 = Ti (t = 0, z)

• Moving boundary condition between the two phases,
where heat is liberated or absorbed

• Thermal properties of the two phases may be
different
76

Rigon et al.


M. Dall’Amico et al.: Freezing unsaturated soil model
M. Dall’Amico et al.: Freezing unsaturated soil model 477 9

! 23 4# 23 4#
0

0
,%-./

0 10 ,%-./ 0 10

!#$%'()*+ !#$%'()*+

Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) at various depths (m). The
Fig. 4. Comparison panel (A) the analytical solution (dottedpanel (B) uses Newton Global. Both have a grid spacing of 10 mm and 500The
numerical model in between uses Newton C-max the one in line) and the simulated numerical (solid line) at various depths (m).
numerical model in are present in panel (B) but not inthe one in where (B)convergence is Global. Both have a grid spacing of 10 mm and 500
cells. Oscillations panel (A) uses Newton C-max panel (A) panel no uses Newton reached.
cells. Oscillations are present in panel (B) but not in panel (A) where no convergence is reached.
77
conductive heat ﬂow in both the frozen and thawed regions, Newton’s method. The analytical solution is represented by
Rigonchange of volume negligible, i.e. ρw = ρi and (4) isother-
(3) et al. the dotted line and the simulation according the numerical
mal phase change at T = Tm , i.e. no unfrozen water exists model by the solid line. The results are much improved with

Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) various depths (m).
Fig. 4. Comparison between the analytical solution (dotted line) and the simulated numerical (solid line) atat various depths (m).The
The
numerical model in panel (A) uses Newton C-max the one in panel (B) uses Newton Global. Both have a a grid spacing of 10 mm and 500
numerical model in panel (A) uses Newton C-max the one in panel (B) uses Newton Global. Both have grid spacing of 10 mm and 500
cells. Oscillations are present in panel (B) but not in panel (A) where no convergence isis reached.
cells. Oscillations are present in panel (B) but not in panel (A) where no convergence reached.
478 M. Dall’Amico et al.: Freezing unsaturated soil model

1e−10
0.0
0
3

0.020
−0.5

15

Cumulative error (%)
Cumulative error (J)

0.015
40
−1.0
soil depth [m]

5e−11
Sim
An

0.010
75
−1.5

0.005
−2.0

Error (%)

5e−13
0.000
−2.5

Error (J)

−5 −4 −3 −2 −1 0 1 2 0 15 30 45 60 75

Temp [ C] time (days)

Fig. 5. Comparison between the simulated numerical and the an- Fig. 6. Cumulative error associated with the the globally convergent
Fig. 5. Comparison Soil profilethesimulated numerical and the Grid
Fig. 5. Comparisonbetween the simulatedatnumerical days.the an-
alytical solution. between temperature different and an- Fig. 6.6. method. Solid line: cumulative error the globally convergent
Newton (J), dotted line: cumu-
Fig. Cumulative error associated with the the globally convergent
Cumulative error associated with the
size = 10 mm, = 500 cells.
alytical solution.NSoil profile temperature atatdifferent days. Grid Newton method.as Solid line: cumulative errorand the total energy
lative error (%) Solid ratio between theerror (J), dotted line: cumu-
the line: cumulative error
alytical solution. Soil profile temperature different days. Grid Newton method. (J), dotted line: cumu-
of the soil in the time step. was set to 1 × 10−8 .
size=10 mm, N=500 cells
size=10 mm, N=500 cells lative error (%) asas the ratio between the error and the total energy
lative error (%) the ratio between the error and the total energy
of the soil inin the time step. was set toto 1E-8.
of the soil the time step. was set 1E-8.
balance was set to 1 × 10−8 . Figure 6 shows the cumulated
error in J (solid line) and in percentage as the ratio between decreases from above due to the increase of ice content. It is
semi-infinite region given by Neumann. inThetime step.of this
semi-infinite region given by of the soil The features ofWith
the error and the total energy Neumann. the features this visible that the freezing of the soil sucks water from below. 78
problem 1×10−8 , existenceof a simulation, the error in per-
problem are the existencedaysaofmoving interface between
set to are the after 75 of moving interface between the two phases, inin correspondence reveals the position of
The increase in total water content which heat is liberated
the two phases, correspondence ofof which heat is liberated
−10 the freezing front: after 12 h it is located about 40 mm from
Rigon et centage remains very low ( 1 × 10 ), suggesting a good
al. the soil surface, after 24 h at 80 mm and finally after 50 h at
energy conservation capability of the algorithm.
140 mm. Similar to Hansson et al. (2004), the results were

3 geotop-summer-school2011

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