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# 3 geotop-summer-school2011

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• A minor observation: Mualem Hydraulic Conductivity Formula is not correct!! Anyway the presentation is good! Emanuele

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### 3 geotop-summer-school2011

1. 1. GEOtop: Richards G. OKeefe, Sky with flat white cloud, 1962 Riccardo Rigon, Stefano Endrizzi, Matteo Dall’Amico, Stephan Gruber, Cristiano LanniWednesday, June 29, 2011
2. 2. “The medium is the message” Marshall MacLuhamWednesday, June 29, 2011
3. 3. 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 3 Rigon et al.Wednesday, June 29, 2011
4. 4. 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 4 Rigon et al.Wednesday, June 29, 2011
5. 5. 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) Water balance ∂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 4 Rigon et al.Wednesday, June 29, 2011
6. 6. 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) Water balance ∂t m 2 Parametric K(θw ) = Ks Se 1 − (1 − Se ) 1/m Mualem −n Se = [1 + (−αψ) )] m ∂θw () θw − θr C(ψ) := Se := ∂ψ φs − θr 4 Rigon et al.Wednesday, June 29, 2011
7. 7. 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) Water balance ∂t m 2 Parametric K(θw ) = Ks Se 1 − (1 − Se ) 1/m Mualem −n Parametric Se = [1 + (−αψ) )] m van Genuchten ∂θw () θw − θr C(ψ) := Se := ∂ψ φs − θr 4 Rigon et al.Wednesday, June 29, 2011
8. 8. 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 ∂θw () θw − θr C(ψ) := Se := ∂ψ φs − θr 5 Rigon et al.Wednesday, June 29, 2011
9. 9. 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 Rigon et al.Wednesday, June 29, 2011
10. 10. Richards 7 Rigon et al.Wednesday, June 29, 2011
11. 11. Richards 7 Rigon et al.Wednesday, June 29, 2011
12. 12. Richards 7 Rigon et al.Wednesday, June 29, 2011
13. 13. Richards 7 Rigon et al.Wednesday, June 29, 2011
14. 14. Richards Pedotransfer Functions Nemes (2006) 8 Rigon et al.Wednesday, June 29, 2011
15. 15. Richardsigure 2: Experimental set-up. (a) The inﬁnite hillslope schematization. (b) The initial suction head pril-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900 9sponds to moving from the crest to the toe of the hillslope Lanni and Rigon Wednesday, June 29, 2011
16. 16. 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. 10 Rigon et al.Wednesday, June 29, 2011
17. 17. 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. 10 Rigon et al.Wednesday, June 29, 2011
18. 18. 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. 10 Rigon et al.Wednesday, June 29, 2011
19. 19. Richards simplified The Richards Equation! 11 Rigon et al.Wednesday, June 29, 2011
20. 20. Richards simplified The Richards Equation! ∂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 11 Rigon et al.Wednesday, June 29, 2011
21. 21. Richards simplified The Richards Equation! ∂ ∂ψ ∂ ∂ψ Sr = Ky + Kx − sin θ ∂y ∂y ∂x ∂x 12 Rigon et al.Wednesday, June 29, 2011
22. 22. Richards simplified The Richards Equation! ∂ ∂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 12 Rigon et al.Wednesday, June 29, 2011
23. 23. Richards simplified The Richards Equation! ∂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. 13 Rigon et al.Wednesday, June 29, 2011
24. 24. 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 14 Rigon et al.Wednesday, June 29, 2011
25. 25. 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 15 Rigon et al.Wednesday, June 29, 2011
26. 26. 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 Rigon et al.Wednesday, June 29, 2011
27. 27. 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 ∂ψ ∂2ψ = D0 cos θ 2 ∂t ∂t2 Kz 0 D0 := C(ψ) 16 Rigon et al.Wednesday, June 29, 2011
28. 28. Richards super-simplified The Richards Equation 1-D ∂ψ ∂2ψ = D0 cos2 θ ∂t ∂t2 17 Rigon et al.Wednesday, June 29, 2011
29. 29. Richards super-simplified The Richards Equation 1-D ∂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. 17 Rigon et al.Wednesday, June 29, 2011
30. 30. Richards super-simplified The Richards Equation 1-D 18 Rigon et al.Wednesday, June 29, 2011
31. 31. Richards super-simplified The Richards Equation 1-D 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 19 Rigon et al.Wednesday, June 29, 2011
32. 32. Richards super-simplified The Richards Equation 1-D In this case the equation admits an analytical solution  q  Kz [R(t/TD )] 0≤t≤T D’Odorico et al., 2003 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 20 Rigon et al.Wednesday, June 29, 2011
33. 33. Richards super-simplified TD The Richards Equation 1-D TD D’Odorico et al., 2003 TD TD 21 Rigon et al.Wednesday, June 29, 2011
34. 34. Richards 1D The analytical solution methods for the advection-dispersion equation (even non-linear), that results from the Richards equation, can be found The Richards Equation 1-D 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 Rigon et al.Wednesday, June 29, 2011
35. 35. D s1 ard Rich The Richards Equation 1-D 23 Rigon et al.Wednesday, June 29, 2011
36. 36. Richards 1D The Richards Equation 1-D Simoni, 2007 24 Rigon et al.Wednesday, June 29, 2011
37. 37. Richards 1D The Richards Equation 1-D Simoni, 2007 25 Rigon et al.Wednesday, June 29, 2011
38. 38. 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. 26 Rigon et al.Wednesday, June 29, 2011
39. 39. Richards 3D for a hillslopeigure 2: Experimental set-up. (a) The inﬁnite hillslope schematization. (b) The initial suction head pr Going back to the simple geometry caseil-pixel hillslope numeration system (the case of parallel shape is shown here). Moving from 0 to 900 27sponds to moving from the crest to the toe of the hillslope Lanni and Rigon Wednesday, June 29, 2011
40. 40. Richards 3D for a hillslope Conditions of simulation Wet Initial Conditions Intense Rainfall Moderate Rainfall Dry Initial Conditions Low Rainfall 28 Lanni and RigonWednesday, June 29, 2011
41. 41. Richards 3D for a hillslope- 54 Simulations result LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES (a) DRY-Low (b) DRY-Med 29 Lanni and Rigon Wednesday, June 29, 2011
42. 42. Richards 3D for a hillslope- 54 Is the flow ever steady state ? LANNI ET AL.: HYDROLOGICAL ASPECTS IN THE TRIGGERING OF SHALLOW LANDSLIDES (a) DRY-Low (b) DRY-Med 30 Lanni and RigonWednesday, June 29, 2011
43. 43. Richards 3D for a hillslope (a) DRY-Low (b) DRY-Med Simulations result (c) DRY-High (d) WET-Low 31 Lanni and RigonWednesday, June 29, 2011
44. 44. Richards 3D for a hillslope (c) DRY-High (d) WET-Low Simulations result (e) WET-Med (f) WET-High F T September 24, 2010, 9:13am D 32 A RValues of pressure head developed at the soil-bedrock interface at each point of the subcritical parallel hillslope. The Lanni and Rigone Wednesday, Junerepresents the mean lateral gradient of pressure head lines 29, 2011
45. 45. Richards 3D for a hillslope 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: RigonWednesday, June 29, 2011
46. 46. Richards 3D for a hillslope 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) 34 Lanni and RigonWednesday, June 29, 2011
47. 47. Saturation vs Vadose Another issue Extending Richards to treat the transition saturated to unsaturated zone. Since : At saturation: what does change in time ? 35 Rigon et al.Wednesday, June 29, 2011
48. 48. Saturation vs Vadose Another issue Extending Richards to treat the transition saturated to unsaturated zone. Which means: 36 Rigon et al.Wednesday, June 29, 2011
49. 49. Saturation vs Vadose Or If you do not have this extension you cannot deal properly with from unsaturated volumes to saturated ones. 37 Rigon et al.Wednesday, June 29, 2011
50. 50. GEOtop: Richards++ Lawren Harris, Mount Robson Riccardo Rigon, Stefano Endrizzi, Matteo Dall’Amico, Stephan GruberWednesday, June 29, 2011
51. 51. “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 RigonWednesday, June 29, 2011
52. 52. 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 40 Rigon et al.Wednesday, June 29, 2011
53. 53. Richards ++ 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. Freezing Unsaturated starts unfrozen Unsaturated Freezing Frozen proceeds 41 Rigon et al.Wednesday, June 29, 2011
54. 54. Ice, soil, water and pores The variable there ! 42 Rigon et al.Wednesday, June 29, 2011
55. 55. Ice, soil, water and pores 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 43 Rigon et al.Wednesday, June 29, 2011
56. 56. Ice, soil, water and pores 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 44 Rigon et al.Wednesday, June 29, 2011
57. 57. Ice, soil, water and pores At equilibrium. Gravity. One phase. No fluxes the equilibrium relation becomes: 45 Rigon et al.Wednesday, June 29, 2011
58. 58. Ice, soil, water and pores At equilibrium. No gravity. No fluxes. Two phases The equilibrium relation becomes:   Ti = Tw pi = pw  µi = µw 46 Rigon et al.Wednesday, June 29, 2011
59. 59. Ice, soil, water and pores At equilibrium. Gravity. Two phases. No fluxes the equilibrium relation becomes: 47 Rigon et al.Wednesday, June 29, 2011
60. 60. Ice, soil, water and pores 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 48 Rigon et al.Wednesday, June 29, 2011
61. 61. Ice, soil, water and pores 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 49 Rigon et al.Wednesday, June 29, 2011
62. 62. Ice, soil, water and pores 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 ( )] 50 Rigon et al.Wednesday, June 29, 2011
63. 63. Ice, soil, water and pores 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 51 Rigon et al.Wednesday, June 29, 2011
64. 64. Ice, soil, water and pores At equilibrium. Two phases. No gravity. No 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 52 Rigon et al.Wednesday, June 29, 2011
65. 65. Ice, soil, water and pores 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 53 Rigon et al.Wednesday, June 29, 2011
66. 66. Ice, soil, water and pores 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!!! 54 Rigon et al.Wednesday, June 29, 2011
67. 67. 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 55 Rigon et al.Wednesday, June 29, 2011
68. 68. 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 56 Rigon et al.Wednesday, June 29, 2011
69. 69. freezing = drying Unsaturated Freezing unfrozen starts Unsaturated Freezing Frozen procedes 57 Rigon et al.Wednesday, June 29, 2011
70. 70. freezing = drying Unsaturated Freezing unfrozen starts Unsaturated Freezing Frozen procedes 58 Rigon et al.Wednesday, June 29, 2011
71. 71. 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/bClapp and Lf (T − Tm ) max θw = θs · Luo et al. (2009), NiuHornberger 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) 59 Rigon et al.Wednesday, June 29, 2011
72. 72. 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 60 Rigon et al.Wednesday, June 29, 2011
73. 73. 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 61 Rigon et al.Wednesday, June 29, 2011
74. 74. 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] There is a freezing point depression of less than 0.01 centigrades 62 Rigon et al.Wednesday, June 29, 2011
75. 75. 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] Temperature goes down to -0.015. Then, the water unfrozen remains 0.1 63 Rigon et al.Wednesday, June 29, 2011
76. 76. freezing = drying An over and over again Unsaturated Freezing unfrozen starts Unsaturated Freezing Frozen procedes 64 Rigon et al.Wednesday, June 29, 2011
77. 77. freezing = drying The overall relation between Soil water content, Temperature, and suction 65 Rigon et al.Wednesday, June 29, 2011
78. 78. freezing = drying The overall relation between Soil water content, Temperature, and suction 66 Rigon et al.Wednesday, June 29, 2011
79. 79. 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 67 Rigon et al.Wednesday, June 29, 2011
80. 80. 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 2.03or 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) clayT ≥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 (ST − T ∗ ) · H(T ∗ − T ) (21) clay 0.46 0.1 1.49 × 10−3 1.25 (S 0.4ion 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 −m1 + [−α ψw0 ] } (22) θi = v (ψw0 ) − θw [ψ(T )] 0.0 It results that, under freezing conditions (Tidual 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 equa1 + [−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 soiually called ”freezing-point depression cording to the Van Genuchten parameters given in Table 1. 68 to the Van Genuchten parameters given in Tg 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 munder 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,
81. 81. 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 69 Rigon et al.Wednesday, June 29, 2011
82. 82. 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 70 Rigon et al.Wednesday, June 29, 2011
83. 83. 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= 0.8C_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. Wednesday, June 29, 2011 -3 -2 {-1 Temp. [ C] 0 1 alpha= 0.01 [1/mm] n= 1.5 C_g= 2300000 [J/m3 K] Capp 71
84. 84. 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.Wednesday, June 29, 2011
85. 85. Equations and Numerics GEOtop workflow turbulent boundary snow/glaciers Input fluxes conditions energy precipitation water budget budget new time step Output 73 Rigon et al.Wednesday, June 29, 2011
86. 86. Equations and Numerics 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.Wednesday, June 29, 2011
87. 87. Equations and Numerics 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.Wednesday, June 29, 2011
88. 88. Equations and Numerics 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.Wednesday, June 29, 2011
89. 89. Equations and Numerics 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 withWednesday, June 29, 2011
90. 90. Equations and Numerics 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 wereWednesday, June 29, 2011