Mass-Conservative Finite Volume Methods for the Richards equation - - PowerPoint PPT Presentation

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Mass-Conservative Finite Volume Methods for the Richards equation - - PowerPoint PPT Presentation

Mass-Conservative Finite Volume Methods for the Richards equation Lecturer: G. Manzini 1 1 IMATI - CNR, Pavia Siena 2006 (Joint collaboration with Stefano Ferraris, Univ. of Turin, Italy) R + , Richards equation: mixed formulation


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SLIDE 1

Mass-Conservative Finite Volume Methods for the Richards’ equation

Lecturer: G. Manzini1

1IMATI - CNR, Pavia

Siena 2006

(Joint collaboration with Stefano Ferraris, Univ. of Turin, Italy)

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SLIDE 2

Richards’ equation: mixed θ−ψ formulation

  • The solution ψ(x, t) satisfies

∂θ(ψ) ∂t − div

  • K(ψ)∇(ψ + z)
  • = s,

in Ω ×

R+,
  • Ω ⊂
Rd, computational domain
  • ψ, hydraulic pressure head;
  • θ(ψ), volumetric water content;
  • K(ψ), hydraulic conductivity tensor;
  • z, vertical coordinate (positive upward)
  • s, production/consumption source term;
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SLIDE 3

Richards’ equation: head formulation

  • The solution ψ(x, t) satisfies

η(ψ)∂ψ ∂t − div

  • K(ψ)∇(ψ + z)
  • = s,

in Ω ×

R+,

where

  • η(ψ) = ∂θ(ψ)/∂ψ is the general storage term.
  • Remark: the two formulations are theoretically equivalent,

but leads to different numerical discretizations.

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SLIDE 4

To complete the model. . .

. . . we must assign:

  • a proper set of boundary conditions:

Dirichlet BCs: ψ = gD,

  • n

ΓD ×

R+,

Neumann BCs: −n · K(ψ)∇(ψ + z) = gN,

  • n

ΓN ×

R+,

where ∂Ω = ΓD ∪ ΓN, and gD and gN are, respectively, Dirichlet and Neumann boundary data;

  • an initial solution:

ψ(x, t = 0) = ψ0(x) for x ∈ Ω,

  • the characteristic relations for θ(ψ), K(ψ), and η(ψ) depending
  • n soil nature: Van Genuchten, e.g. HYDRUS-2D, 1999,

Brooks-Corey, Haverkamp (1977), and many other models. . .

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SLIDE 5

Extension to multi-layered soils

Multi-layered soils are taken into consideration by assuming that Ω be partitioned into a set of soil zones (sub-surface layers) Ωl, l = 1, . . . , L such that Ω =

  • l=1,...,L

Ωl, showing different constitutive relations: θ(ψ)|Ωl = θl(ψ), K(ψ)|Ωl = Kl(ψ), η(ψ)|Ωl = ηl(ψ).

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SLIDE 6

The finite volume framework

Let Th =

  • T
  • be a suitable triangulation of Ω;

Let us integrate over any control volume T ∈ Th to obtain

  • T

∂θ(ψ) ∂t dV −

  • T

div

  • K(ψ)∇ (ψ + z)
  • dV =
  • T

s dV; then, apply the Gauss-Green divergence theorem

  • T

∂θ(ψ) ∂t dV −

  • ∂T

n ·

  • K(ψ)∇ (ψ + z)
  • dS =
  • T

s dV, introduce cell averages and split face contributions to flux balance d dt

  • T

θ(ψ) dV −

  • e∈∂T

|e|

  • ∂T

n ·

  • K(ψ)∇ (ψ + z)
  • dS =
  • T

s dV.

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SLIDE 7

Finally, let us introduce the cell-average approximation of the pressure field, e.g. ψh ≡ {ψT}, and the numerical flux, Gij(ψh) = |eij| nij · Kij

  • G⋄

ij(ψh) + ζ

  • ,

where ζ = ∇z is the versor along Z, 1 Kij(ψh) = 1 2

  • 1

K(ψj) + 1 K(ψi)

  • ,

and G⋄

ij(ψh) is the face-based gradient on the diamond Dij that is

given by G⋄

ij(ψh) = G(n) ij (ψh)nij + ψβ − ψα

|eij| tij.

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SLIDE 8

The normal component G(n)

ij (ψh) of the face-based gradient G⋄ ij(ψh) is

G(n)

ij (ψh) =

1 |Dij|

  • k

w(n)

ij,k ψk + g(n) ij

where the weights

  • w(n)

ij,k

  • are obtained by the Least Squares

reconstruction algorithm. Rewriting

  • G(ψ)ψ + g(ψ)
  • i =
  • eij∈∂Ti

|eij| nij · KijG⋄

ij(ψh),

  • h(ψ)
  • i =
  • eij∈∂Ti

|eij| nij · Kijζ, we finally obtain the vector formulation dθ(ψ) dt

  • i −
  • G(ψ)ψ + g(ψ) + h(ψ)
  • i = si(ψ).
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SLIDE 9

Time discretization

  • Crucial issue: we are advancing in time θ(ψ), but we need ψ

for flux balancing: dθ(ψ) dt

  • i
  • non-linear

function of ψ +

  • G(ψ)ψ + g(ψ)
  • non-linear

flux term + h(ψ) gravity term

  • i = si(ψ).
  • the simplest approach is to switch to a discrete form of the

head-based formulation by using dθi(ψi) dt ≈ η(ψi)dψi dt .

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SLIDE 10
  • Using the implicit Euler F.D. scheme for the time derivative give

us η(ψn+1

i

)ψn+1

i

− ψn

i

∆t −

  • G(ψn+1)ψn+1 + g(ψn+1) + h(ψn+1)
  • i = s(ψn+1).
  • This non-linear algebraic problem can be solved by using a

fixed-point iterative scheme (Picard):

  • diag
  • η(ψn+1,m

i

)

  • − ∆tG(ψn+1,m)
  • ψn+1,m+1 =

η(ψn+1,m)ψn + ∆t

  • si(ψn+1,m) + g(ψn+1,m) + h(ψn+1,m)
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SLIDE 11
  • The (so-called) delta-form is obtained by solving for δm+1,m
  • diag
  • η(ψn+1,m)
  • − ∆tG(ψn+1,m)

ψn+1,m+1 − ψn+1,m

  • =δm+1,m

= +∆t

  • s(ψn+1,m) + g(ψn+1,m) + h(ψn+1,m)
  • −diag
  • η(ψn+1,m)
  • ψn+1,m − ψn

+ ∆tG(ψn+1,m)ψn+1,m

  • Iterate on m to get ψn+1,m+1 from ψn+1,m starting from

ψn+1,0 = ψn until

  • δm+1,m

≤ TOL and, finally, set ψn+1,m+1 = ψn+1,m + δm+1,m This approach is easy, but shows very poor mass conservation!

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SLIDE 12

What is a mass-conservative discretization?

Following M. Celia et al., W.R.R. 1990, let us define:   total additional mass in the domain   =   total mass in the domain at any time t > 0   −   total mass in the domain at initial time t = 0   ,

  • total net flux

into the domain

  • =

flux balance integrated from initial time t = 0 to the current time t

  • MASS BALANCE RATIO

= total additional mass in the domain total net flux into the domain Clearly, it must be MASS BALANCE RATIO=1.

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SLIDE 13

How does the head-based approach perform?

Let us consider, for example, the vertical column infiltration problems proposed by M. Celia et al. W.R.R. 1990: Case 1 Data: column range: [bottom] 0 ≤ ζ ≤ 40 [top] bottom pressure: −61.5 cm top pressure: −20.7 cm initial pressure: −61.5 cm for 0 ≤ ζ < 40 running time: from T = 0 to T = 360 s. Constitutive relationships (Haverkamp et al.): θ(ψ) = α(θs − θr) α + |ψ|β + θr K(ψ) = Ks A A + |ψ|γ (α, β, γ, θs, θr, Ks and A are known parameters.)

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SLIDE 14

Column infiltration problem: Case 1

100 200 300 400 Time-step size (sec) 0.75 0.8 0.85 0.9 0.95 1 Mass Balance Ratio Head-based 2D FV scheme (M. & Ferraris 2004) Head-based 1D FD scheme (Celia et.al., 1990)

Mass balance ratio for different time steps ∆t

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SLIDE 15

How does the head-based approach perform?

Case 2 Data: column range: [bottom] 0 ≤ ζ ≤ 100 [top] bottom pressure: −1000 cm top pressure: −120 cm initial pressure: −1000 cm for 0 ≤ ζ < 100 running time: from T = 0 to T = 1 day. Constitutive relationships (Van Genuchten et al.): θ(ψ) = θs − θr 1 + (ε |ψ|n)m + θr K(ψ) = Ks

  • 1 − (ε |ψ|)n−1

1 + (ε |ψ|n) −m2

  • 1 + (ε |ψ|n)

m

2

(m, n, ε, θs, θr, and Ks are known parameters.)

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SLIDE 16

Column infiltration problem: Case 2

1000 2000 3000 4000 Time-step size (sec) 0.4 0.5 0.6 0.7 0.8 0.9 1 Mass Balance Ratio Head-based 2D FV scheme (M. & Ferraris, 2004) Head-based 1D FD scheme (Celia et.al. 1990)

Mass balance ratio for different time steps ∆t

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SLIDE 17

How can we improve this behavior?

  • Main Idea: M. Celia et al. observed that mass conservation may

be lost in the head-based formulation because of a poor approximation of the time derivative of θ(ψ): ∂θ(ψ) ∂t

  • n+1

i

≈ 1 |Ti|

  • Ti

∂θ(ψ) ∂t

  • n+1

i

dV ≈ d dt θ(ψi(t))

  • n+1

≈ η(ψn

i )ψn+1 i

− ψn

i

∆t + O(|δψ|)

  • Better strategy: develop θ(ψ) to second-order in ψ-variation in

the cell-average integral θn+1,m+1

i

= 1 |Ti|

  • Ti

θ(ψn+1,m+1

i

) dV

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SLIDE 18

How can we improve this behavior?

A straightforward calculation gives θn+1,m+1

i

= 1 |Ti|

  • Ti

θ(ψn+1,m+1

i

) dV = 1 |Ti|

  • Ti

θ

  • ψn+1,m

i

+ (ψn+1,m+1

i

− ψn+1,m

i

)

  • = δm+1,m

i

  • dV

= 1 |Ti|

  • Ti
  • θ(ψn+1,m

i

) +

  • ∂θ

∂ψ

  • n+1,m

i

  • ψn+1,m+1

i

− ψn+1,m

i

  • + O(|ψ|2)
  • dV

≈ θn+1,m

i

+ η(ψn+1,m)δm+1,m

i

+ O(|δψ|2).

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SLIDE 19
  • Let us start again from

dθ dt

  • i −
  • G(ψ)ψ + g(ψ) + h(ψ)
  • i = si(ψ),
  • and discretize directly the time derivative of θ(ψ) to get

θn+1

i

− θn

i

∆t −

  • G(ψn+1)ψn+1 + g(ψn+1) + h(ψn+1)
  • i = si(ψn+1).
  • Use Picard iterations for solving the non-linear problem

θn+1,m+1

i

− θn

i

∆t −

  • G(ψn+1,m)ψn+1,m+1 + g(ψn+1,m) + h(ψn+1,m)
  • i = si(ψn+1,m).
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SLIDE 20
  • Re-write Picard iterations in δ-form,

θn+1,m+1

i

− θn+1,m

i

∆t −

  • G(ψn+1,m)
  • ψn+1,m+1 − ψn+1,m

+ g(ψn+1,m) + h(ψn+1,m)

  • i =

si(ψn+1,m) + G(ψn+1,m)ψn+1,m

  • i − θn+1,m

i

− θn

i

∆t , (recall δm+1,m = ψn+1,m+1 − ψn+1,m).

  • We will use the previously derived second-order development

θn+1,m+1

i

≈ θn+1,m

i

+ η(ψn+1,m)δm+1,m

i

+ O(|δψ|2).

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SLIDE 21
  • Finally, we obtain the vector expression:
  • diag
  • η(ψn+1,m)
  • − ∆tG(ψn+1,m)
  • δm+1,m =

∆t

  • s(ψn+1,m) + g(ψn+1,m) + h(ψn+1,m)
  • θn+1,m − θn

+ ∆tG(ψn+1,m)ψn+1,m,

  • to be compared with the head-based discretization:
  • diag
  • η(ψn+1,m)
  • − ∆tG(ψn+1,m)
  • δm+1,m =

∆t

  • s(ψn+1,m) + g(ψn+1,m) + h(ψn+1,m)
  • −diag
  • η(ψn+1,m)
  • ψn+1,m − ψn

+ ∆tG(ψn+1,m)ψn+1,m.

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SLIDE 22

Evaluation of mass conservation, Case 1

100 200 300 400

Time-step size (sec)

0.75 0.8 0.85 0.9 0.95 1

Mass Balance Ratio

Mixed-based (Celia et.al. 1990) 2D FV Standard Head-based 2D FV Standard Head-based 1D FD

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SLIDE 23

Evaluation of mass conservation, Case 2

1000 2000 3000 4000

Time-step size (sec)

0.4 0.5 0.6 0.7 0.8 0.9 1

Mass Balance Ratio

Mixed-based (Celia et.al. 1990) 2D FV Standard Head-based 2D FV Standard Head-based 1D FD

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SLIDE 24

Evaluation of mass conservation, Case 1

10 20 30 40

Z (cm)

  • 60
  • 50
  • 40
  • 30
  • 20

Pressure Head (cm)

t=120s t=240s t=360s t=480s t=600s

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SLIDE 25

Evaluation of mass conservation, Case 1

10 20 30 40

Z (cm)

0.1 0.15 0.2 0.25

Water Content

t=120s t=240s t=360s t=480s t=600s

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SLIDE 26

Evaluation of mass conservation, Case 2

20 40 60 80 100

Z (cm)

  • 1000
  • 800
  • 600
  • 400
  • 200

Pressure Head (cm)

t=12h t=24h t=36h t=48h

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SLIDE 27

Evaluation of mass conservation, Case 2

20 40 60 80 100

Z (cm)

0.1 0.15 0.2

Water Content

t=12h t=24h t=36h t=48h

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SLIDE 28

Multi-layered Calculations

The performance of the Finite Volume method for predicting unsaturated flow in multi-layered calculation is measured in two different test cases:

  • horizontally layered soil (three layers), e.g. Baca, Chung,

Mulla, IJNMF, 1997;

  • curvilinearly layered soil (two layers), M. & Ferraris, AdWR

2004.

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SLIDE 29

Horizontally layered soil

Soil Data: range: 0cm ≤ ζ ≤ 25.5cm [top] surface crust: 25cm ≤ ζ ≤ 25.5cm Ks = 0.0616cm h−1 tilled layer: 15cm ≤ ζ ≤ 25cm Ks = 1.396cm h−1 sub-soil: 0cm ≤ ζ ≤ 15cm Ks = 0.312cm h−1 Constitutive relationships (Brooks-Corey et al.): θ(ψ) =

  • θs

ψ ≥ ψa, θs(ψ/ψa)−1/b ψ < ψa K(ψ) =

  • Ks

ψ ≥ ψa, Ks(ψ/ψa)−1/b ψ < ψa (θs, ψa, and b are known parameters.)

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SLIDE 30

Simulation Data: Case 1 bottom pressure: −80 cm top pressure: 0 cm initial pressure: −80 cm for 0 ≤ ζ < 25.5 cm running time: from T = 0 to T = 4 hours. Case 2 bottom pressure: −1000 cm top pressure: 0 cm initial pressure: −1000 cm for 0 ≤ ζ < 25.5 cm running time: from T = 0 to T = 4 hours.

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SLIDE 31

Horizontally layered soil, Case 1

5 10 15 20 25

Z (cm)

  • 100
  • 80
  • 60
  • 40
  • 20

Pressure Head (cm)

t=30m t=1h t=1h30m t=2h

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SLIDE 32

Horizontally layered soil, Case 1

5 10 15 20 25

Z (cm)

0.35 0.4 0.45 0.5 0.55

Water Content

t=0 t=1h t=2h t=3h t=4h t=1h t=2h t=3h t=4h

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SLIDE 33

Horizontally layered soil, Case 2

5 10 15 20 25

Z (cm)

  • 1000
  • 800
  • 600
  • 400
  • 200

Pressure Head (cm)

t=1h t=2h t=3h t=4h

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SLIDE 34

Horizontally layered soil, Case 2

5 10 15 20 25

Z (cm)

  • 50
  • 40
  • 30
  • 20
  • 10

Pressure Head (cm)

t=1h t=2h t=3h t=4h

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SLIDE 35

Horizontally layered soil, Case 2

5 10 15 20 25

Z (cm)

0.2 0.3 0.4 0.5 0.6

Water Content

t=1h t=2h t=3h t=4h t=0 t=1h t=2h t=3h t=4h

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SLIDE 36

Curvilinearly layered soil

Simulation Data domain: [0, Lx] × [0, Lz], with Lx = Lz100 cm bottom pressure: 0 cm top pressure: 0 cm initial pressure: (100 − z) cm for 0 ≤ z < 25.5 cm vertical BCs: homogeneous Neumann (no flux) running time: from T = 0 to T = 2 days. The curvilinear layer is the curve (x, ζ(x)) such that ζ(x) = 1 10

  • 1 − cos(πx/Lx)
  • + 0.45

Van Genuchten constitutive curves are used, in particular with layer 1: z ≥ ζ(x) Ks = 0.25, layer 2: z ≤ ζ(x) Ks = 2.00.

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SLIDE 37

Curvilinearly layered soil

100 100

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SLIDE 38

Curvilinearly layered soil

100 100 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 0

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 0.25 day

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SLIDE 39

Curvilinearly layered soil

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 0.5 day

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 0.75 day

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SLIDE 40

Curvilinearly layered soil

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 1 day

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 1.25 day

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SLIDE 41

Curvilinearly layered soil

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 1.5 day

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 1.75 day

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SLIDE 42

Curvilinearly layered soil

100

100

0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

T = 2 day

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SLIDE 43

Final Remarks

  • The second-order Finite Volume method based on

diamond discretization of fluxes is valid alternative to more “traditional” numerical methods like FDM, FEM, for solving flow models in partially saturated soils;

  • the mixed θ−ψ formulation of the Richards equation and

the mass-conservative technique by M. Celia et al. W.W.R 1990 makes it possible the development of a very accurate time-marching method;

  • the scheme is naturally adapted to multi-layered soil

calculations.