Flow. Soil Mechanics. Chapter 2. Universitat Politècnica de Catalunya BARCELONATECH. Escola Tècnica Superior d Enginyers de Camins, Canals i Ports

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1 Universitat Politècnica de Catalunya BARCELONATECH Escola Tècnica Superior d Enginyers de Camins, Canals i Ports Soil Mechanics Chapter 2 Flow

2 Chapter 2 1. Basic concepts 2. Differential equation of flow 3. Seepage forces 4. Solving flow problems 5. Piezometers 6. Filters and earth dams 7. Flow in unsaturated soils Exercises Laboratory: practice 2 (permeability)

3 2.1 Basic concepts

4 Total and effective stress Relative to saturated soils In saturated soils there are two phases (solid, water) that may be seen as two superposed continuum media. hypothesis of the mechanics of a continuum medium Stresses: total, σ, acting externally as if there were one phase only (e.g. σ vert ) pore water pressure, p w

5 Total and effective stress = + saturated soil solid skeleton water continuum medium??

6 Total and effective stress p w pore water pressure σ total stress stress acting as if there were only one phase

7 Total and effective stress Define effective stress as: σ' = σ p w Using tensorial index notation: σ' ij = σ ij p w δ ij Note: water has no shear strength

8 Sign criterion In Soil Mechanics the sign criterion is opposite to the one used in the theory of the mechanics of a continuum medium: + compressive stresses are positive tensile stresses are negative

9 Effective stress principle Karl von Terzaghi formulated in the 1930 s the following principle, based on experimental observations: deformation of a soil depends exclusively on changes of the efective stresses assuming that solid particles do not deform and that water is incompressible

10 Physical interpretation of effective stresses σ p w σ : total stress applied externally σ : stresses acting on particle contacts, on a total area A m (in yellow) p w : pore water pressure, acting on a total area A w (in blue) A t : total area σ A t = A m + A w A t

11 Physical interpretation of effective stresses Force equilibrium: A A p A ( R A) t m w w Am Aw ( R A) pw A A A t t t Two possible cases: a) Soil under formation no contact between particles, A m = 0, A w = A t ( R A) ( R A) pw A A t t

12 Physical interpretation of effective stresses b) Contact between particles, electrostatic forces are small (R A 0): A m pw At A A A m pw 1 At w t A A m t ( 10 MPa) p w ( 1MPa) A A A pw pw A A A m m m t t t Effective stresses may be interpreted as the contact stresses between solid particles.

13 Comments In the soil there are three stress fields: total stresses σ ij effective stresses σ' ij pore water pressures p w δ ij In tensorial notation, σ' ij = σ ij p w δ ij decomposing σ ij = pδ ij + s ij p = mean stress = (σ 11 + σ 22 + σ 33 )/3 s ij = stress deviator tensor σ' ij = (p p w )δ ij + s ij water does not affect the deviatoric stresses

14 Example 1 raising water level h H 0 A sat 0 w 0 0 A A w 0 A H H 1 A h H sat 0 w 0 w sat p h H h H h h w p w h H 1 A sat 1 w 1 1 A A w 0 A A sat 1 w 1 w sat p h H h H h h w p w 0 no deformation 1 0 A A A

15 Example 2 water table drop h 0 h 1 h 0 A 0 0 A A w sat sat 0 A h p h h h p w A A 0 h h h h w sat w 0 w 0 w 1 A 1 1 A A w sat sat 0 A h p h h h p w 1 h h h h w sat w 1 w 1 w 0 there is deformation 1 0 A A A w h

16 Example 3 fast increase ( undrained ) of the external pressure p allow water to escape ( drainage ) p p 0 p 0 p 0 p p 0 0 p w0 0 p 1 0 w1 0 p p p p p 2 0 p w2 p σ' = 0 σ' p 1 0 p

17 Piezometric level, water table and stresses in the soil

18 Piezometric level The piezometric level (or total or hydraulic head) is a measure of the energy of water per unit weight Represented by the symbols h or φ Given in length units To evaluate the piezometric level, the vertical axis (z) must point upwards: We use Bernouilli s formula, from Hydraulics:!!

19 Piezometric level 2 pw v h z 2g potential energy w pressure term kinetic energy h = z + negligible in soils, not considered Origin of the vertical axis z is arbitrary (but it must always point upwards!! ) Do not confuse phreatic level (water table) with piezometric level: phreatic level: free surface of ground water at atmospheric pressure (p w = 0) piezometric level: measure of ground water energy p w w

20 Piezometric level piezometric level water table unconfined aquifer confined aquifer

21 Stress distribution on horizontal soil Equilibrium of vertical forces: dz σh nat dz 1 σ V σ + V σh z σ z V dz V V 1 natdz 1 V dz 1 z V dz natdz z Integrate with respect to depth: V 0 z nat ( z) dz Unit weight changes with depth, but it is constant piecewise the previous integral reduces to a summation, with unit weights constant within each stratum of the same material σ' = σ p w p w we ll see how to calculate it in this chapter; σ we ll see it in later chapters, at this point we ll see how it is done in one particular case (horizontal, very extensive soil)

22 Stress distribution on horizontal soil z 0 WT p ( z z ) w and about the horizontal stresses, σ H, σ' H? w 0 z With water present, one can calculate the effective vertical stresses: ( z) ( z) p ( z) 0 z nat ( z) dz If the stratum is homogeneous ( nat is constant) and the water table is at the surface (z 0 = 0): ( z) ( z) p ( z) nat w w w z z 0 dz z z z w weight of submerged soil

23 Horizontal stresses Most soils have formed by sedimentation into layered sequences, resulting in very extensive horizontal surfaces. During material deposition, the total stresses (both vertical and horizontal) increase due to the added weight. Vertical deformations appear as a result but because of symmetry, the horizontal deformations are zero. If there is water in the soil, we also know that σ' H = σ H p w and σ' V = σ V p w

24 Horizontal stresses With these conditions horizontal soil surface water at rest zero lateral deformation there is a linear relation between the increments of vertical and horizontal stresses Define coefficient of earth pressure at rest, K 0 : σ K 0 = σ under zero lateral deformation conditions H V σ V σ H σ' H σ' V linear

25 Horizontal stresses σ' H σ' V normally consolidated soil: K 0 ~ A soil is called normally consolidated (NC) when the current stresses are the historical highest A soil is called overconsolidated (OC) when the current stresses are less than the historical highest overconsolidated soil: K 0 > 1

26 Horizontal stresses During one unloading-reloading cycle, the vertical stresses decrease during the unloading (e.g. after a glacier that had induced large vertical stresses melts or recedes) However, the horizontal stresses do not decrease as much: they remain locked. That s why for OC soils K 0 may have much larger values.

27 Stress evaluation Evaluate the total vertical stress and the pore water pressure: σ V, p w Evaluate the effective vertical stress: σ' V = σ V p w Evaluate the effective horizontal stress: σ' H = K 0 σ' V Evaluate the total horizontal stress: σ H = σ' H + p w

28 Example A 3 m 5 m K K nat 0 nat 0 20 kn/m kn/m B C D w WT 2 m 3 10 kn/m 4 m K nat 0 19 kn/m E

29 Example Point z σ V p w σ' V K 0 σ' H σ H A B C C D D E operations (1) (2) = Σ nat z (3)= w (zz 0 ) (4)=(2)-(3) (5) (6)=(5)x(4) (7)=(6)+(3)

30 Example WT z σ V p w σ' V σ' H σ H (stresses in kpa)

31 2.2 Differential equation of flow

32 Water flow flow rate A p = void area A t = total area A p * vw A t vw v * w = real water velocity v w = equivalent water velocity in a continuum medium Q = total real discharge Q = v w A * p

33 Water flow flow rate A p * vw A t vw Define unit flow rate: q Q vw A Define surface porosity: n sup A A p t t n v w A t A p q n v * w

34 Water flow flow rate q has dimensions of velocity (e.g. m/s) q is sometimes called Darcy s velocity or seepage rate or discharge velocity q is actually a vector quantity: q Usually this fictitious water velocity is used instead of the real velocity.

35 Water mass conservation equation

36 Equation of conservation of water mass in a saturated soil Considering a fixed volume of soil, the change of the mass of water within that volume in a certain time must be equal to the difference between the mass of water leaving the volume and the mass of water entering the volume. Mass of water in a differential element: dm ndv w w density volume of water q ds Total mass of water: M w V ndv w V S

37 Equation of conservation of water mass in a saturated soil Balance of the total mass of water crossing the surface S of the reference volume: S w q ds This balance must equal the change in time of the total mass of water within the volume: V wndv S w t q ds V q S ds

38 Equation of conservation of water mass in a saturated soil Why there is a minus sign in that equation? since q is positive toward the outside of the element (i.e. if water flows out of the element), a positive balance means that more water flows out from than into the element, and therefore there is a net reduction of water mass within the element; thus the change of the water mass per unit of time is then negative S wqds 0 wndv 0 t V V q S ds

39 Equation of conservation of water mass in a saturated soil t V ndv q ds q V w S w V w Gauss' theorem t n w dv 0 V q w note :,, x y z n q w w t continuity equation of the mass of water in a saturated soil 0 dv V q S ds

40 Equation of conservation of water mass in a saturated soil Particular cases of the continuity equation t n q 0 if water is incompressible, ρ w = ct.: w n q 0 t if soil does not deform because of water flow, n = ct.: q 0 if flow is one-dimensional, q = (q x,0,0): dqx q 0 0 qx ct. dx w

41 Equation of motion: Darcy s law

42 Darcy s law Henri Darcy ( ), French engineer from Dijon, worked on the improvement of potable water supply in that city. During his work, he came across a relationship that we commonly know today as Darcy s law, which he published in a research paper: Henry Darcy, Détermination des lois d'écoulement de l'eau à travers le sable. Les Fontaines Publiques de la Ville de Dijon, Paris, Victor Dalmont, pp

43 Darcy s law H 1 S A B reference level sample L Q H 2 Loss of water energy is due mainly to friction between water and solid particles energy at point A: h A H 1 energy at point B: h B H 2 Steady state, water flows continuously, H 1 and H 2 are kept constant during the test Measure the discharge Q [m 3 /s] at the exit point

44 Darcy s law H 1 S A B sample L Q H 2 Darcy carried out his experiments with different types of soil, changing the parameters S, L, H 1 (=h A ) and H 2 (=h B ) He realized that the measured discharge Q was directly proportional to the difference h A h B the section of the sample, S and inversely proportional to reference level the length of the sample, L

45 Darcy s law H 1 S A B sample L Q H 2 Therefore: Q h A h L The proportionality constant (K) is called permeability: Q h B L S h A B K S Darcy s law reference level K has dimensions of a velocity [m/s]

46 Darcy s law q Q h h h S L L A B K K H 1 S A B reference level sample L Q H 2 Kh Kgradh the flow vector (indicating the direction of water movement) has the same direction as, but opposite sense than, the hydraulic head gradient vector: if h A > h B, flow is from A B grad h is from B A

47 Darcy s law Darcy s law written as soils that are q= -K h is valid only for homogeneous (properties are independent of the point where they are measured) isotropic (properties are independent of the direction along which they are measure) In that case, K is a scalar constant But it the soil is heterogeneous and anisotropic, the permeability must be generalized in the form of a tensor of permeability

48 Darcy s law q K qx K11 K12 K13 h x q K K K h y K ij h y q z K31 K32 K 33 h z K ji tensor of permeability K K T heterogeneous soil: homogeneous soil : K K ( x, y, z) K ij ij ij constant

49 Darcy s law In the space of principal directions of permeability: K Kx K y K z In stratified sedimentary soils K x = K y > K z K x = K y K z

50 Factors affecting K Void ratio (e): K ~ log (e) Grain-size distribution, particle size (related to e) Roughness of the particles Soil structure (in clays): flocculated, disperse Fluid characteristics: viscosity, unit weight Degree of saturation: S r K

51 Factors affecting K Kozemy-Karman theoretical model, based on a pipe network with laminar flow: q 3 Q 2 e fluid C ds h A 1 e average size of particles fluid viscosity K C d 2 s 3 e 1 e fluid depends on the specific permeability of the soil depends on the fluid

52 Limitations to Darcy s law It has been assumed: slow, laminar motion, no turbulences kinetic energy is negligible If motion is fast (e.g. in breakwaters): the relation q grad h is no longer linear in this case: aq 2 + bq = Kgrad h q linear non-linear -grad h

53 Limitations to Darcy s law In very fine soils (clays), water movement becomes influence by the surface of clay particles ( double layer) with small gradients, water does not flow there exists a threshold gradient I 0 q = -K(grad h I 0 ) if grad h > I 0 q = 0 if grad h < I 0 q -grad h I 0

54 Typical values of K Type of soil Clean gravel Sand Silt Clay Permeability K 1 10 cm/s cm/s cm/s cm/s

55 Typical values of K Clays are impermeable but the concept of impermeability is relative silts are impermeable if compared to gravels... K is difficult to measure, large variability indirect measurements: K = f(e) or correlations: K (cm/s) 100 (D 10 in cm) 2 D 10 : 10% in weight with a size less than D 10

56 Permeability tests Field tests: pumping, injection, etc. for coarse sand and gravel A Q L H H Q K A L Constant-head permeameters Falling-head permeameters

57 Constant-head permeameter In essence, it is Darcy s apparatus H 1 reference level S L Q H 2 Q S K H K L H H H 1 2 Q L S h constant Slow test: acceptable only for large K: gravel, coarse sand, with K = cm/s

58 Falling-head permeameter a a dh q Adt; q h A L dh h h A Q q A K h L h a dh K A dt L ln h dh A K dt h a L h h 2 1 t 1 t ln t 2 t2 t1 L 2 A K h A K 1 K L a L h a faster test acceptable up to silts with K = 10-1 to 10-4 cm/s K a L ln A t h h t

59 Differential equation of flow

60 Flow equation The following equations are available: continuity (conservation of mass) t n q w 0 Darcy s (equation of motion) q Kh w Combining both: t wn wk h 0 most general form of the flow equation

61 Flow equation Assuming that the soil does not deform (because of groundwater flow) and that water is incompressible, n and ρ w are constants: K h 0 If the reference axis are the principal directions of permeability, Kx 0 0 K 0 K y K z h h h Kx K y Kz x x y y z z flow equation for an heterogeneous and anisotropic soil 0

62 Flow equation If the soil is homogeneous but anisotropic, K x, K y and K z are constant but different: h h h K K K x y z x 0 2 y 2 z 2 flow equation for an homogeneous and anisotropic soil

63 Flow equation If the soil is homogeneous but anisotropic, K x, K y and K z are constant but different: h h h K K K x y z x 0 2 y 2 z 2 flow equation for an homogeneous and anisotropic soil Finally, if the soil is isotropic, K x = K y = K z = K, constant: h h h x y z h flow equation for an homogeneous and isotropic soil (Laplace equation)

64 Boundary conditions On the boundary of the domain where the flow equation needs to be solved, we may prescribe (piecewise): either the hydraulic head h or the flow rate q or a combination of both Two examples: one-dimensional two-dimensional

65 Boundary conditions One-dimensional large excavation: z h h A B z A B A p p wa w p wb zb zb w wa w 2 dh 1D 0 h Az B 2 dz h Az B A0 B B A A h Az B Az h A B B B A h h B A h z ha zb h B h z B A

66 Boundary conditions Two-dimensional earth dam: H 1 z A B domain impervious surface p water table, p w = 0 w 1 w ha B z z H1 w w p w 2 w hde z z H2 w w H H z z running surface p w = 0 C D E H 2 Specified hydraulic head: AB: h = H 1 BC: h = z CD: h = z DE: h = H 2 Specified flow rate: AE: q z = 0

67 Boundary conditions Specified flow rate: nx qn f ( x, y, z) qn h h h q K x, K y, K z n n y q Kh x y z n z in the earth dam example: h h on the AE line, n (0, 0,1) qz 0 K z 0 0 z z conditions upon specified flow rate imply conditions on the derivatives of the hydraulic head h

68 Boundary conditions Mixed conditions: affect both the hydraulic head h as well as its derivatives: h A( x, y, x) h( x, y, z) B( x, y, z) ( x, y, z) n where A(x,y,z), B(x,y,z) and λ(x,y,z) are known functions

69 Example with heterogeneous soil H 1 H 2 H 1 H 2 Sand Silt Gravel excavate: s C B z A K 1 K 2 WT WT K 1 K 2 p h A > h C flow from A to C but h B is not known since, because of water flow, pressure in B is not the hydrostatic and therefore wa 1 2 w ha za 0 H1 H2 w w pwc hc zc H1 H2 s 0 H1 H2 s h B w z B H H p wb hydrostatic w

70 Example with heterogeneous soil H 1 H 2 H 1 H 2 Sand Silt Gravel excavate: s C B z A K 1 K 2 WT WT K 1 K 2 Work with two domains: 1 Sand 2 Silt There are four boundary conditions: h h 1 2 z z h h z z h h z z h h h B az b C A cz d z zb h B C A B There are five unknowns: a, b, c, d and h B

71 Example with heterogeneous soil H 1 H 2 H 1 H 2 Sand Silt Gravel excavate: s C B z A K 1 K 2 WT WT K 1 K 2 One extra equation is needed: flow continuity at point B: hc hb q1 K1h1 K1 H s q K h K q q K C B B A H1 s H 2 h 1 B h H h h h h K 2 A

72 Example with heterogeneous soil H 1 H 2 H 1 H 2 Sand Silt Gravel excavate: s C B z A K 1 K 2 WT WT K 1 K 2 If the silt layer is much more impervious than the sand one: K2 K1 hc hb the main hydraulic head loss occurs between A and B

73 2.3 Seepage forces

74 Introduction Water in motion applies forces on the soil particles. These forces are per unit of volume, same as the self-weight. The total stresses do not change due to water flow (the weight of the soil, solid + water, remains constant). Equilibrium imposed in terms of total stresses.

75 Equilibrium equations z x xz z z z x nat z dz Same in the other directions xz xz dz z x x dx x 3 ij ij, j bi bi j1 x j b (0,0, ) nat x xy xz 0 x y z xy y yz 0 x y z xz yz z nat x y z 0 Equilibrium in terms of total stresses

76 Equilibrium equations z x xz z z z x nat z dz Same in the other directions xz xz dz z x x dx x p ij ij w ij x xy xz pw 0 x y z x xy y yz pw 0 x y z y xz yz z pw nat 0 x y z z Equilibrium in terms of efective stresses

77 Equilibrium equations p w h z pw h z w w pw h pw h pw h w w w 1 x x y y z z x xy xz h w 0 x y z x xy y yz h w 0 x y z y xz yz z h w 1 nat 0 x y z z

78 Seepage forces 1 ij i j x w y z na w w t w w b x b b h b h x h z h z h y z b i forces due to seepage the submerged forces weight In terms of effective stresses : In that case : Powered by TCPDF (

79 Seepage forces b wh iz forces due to seepage the submerged forces weight The sense of the seepage forces ( w h ) is opposite to that of the hydraulic gradient vector. In an isotropic soil, q has the same direction but opposite sense to the gradient. Therefore, the seepage forces have the same direction and sense as the flow. The forces due to the submerged weight are always vertical and pointing downwards.

80 Critical hydraulic gradient area where the seepage forces may be significant z H We assume that in the shaded area, because of symmetry, there are no shear stresses and that the flow has only a vertical component (1D) h = h(z)

81 Critical hydraulic gradient z H Equilibrium (1-D): d z dz w Define dh I 0 dz dh dz negative Therefore d z dz w I

82 Critical hydraulic gradient Integrate the diff. eq.: I z C z w z H Boundary conditions: z H 0 C I H Therefore z w I H z z w

83 Critical hydraulic gradient H z I = 0 Water in equilibrium, no flow: I = I 2 > I 1 > 0 I = I 1 > 0 z z H z I = I crit If I is big, it may happen σ I w 1 z = 0 I w 2 that: σ z = 0 I I I w 0 crit w critical hydraulic gradient Warning: this is valid only where 1D-flow conditions can be assumed!!!

84 Critical hydraulic gradient Definition of critical hydraulic gradient: It is the hydraulic gradient for which the effective vertical stresses become zero. In the previous example: I z 0 w Usually, the submerged unit weight is of the order of 10 kn/m 3. Therefore: I crit w I 1 crit w

85 Liquefaction When σ' z = 0 all contact between solid particles is lost: this phenomenon is called liquefaction (also boiling or quick condition) Actually, when I 0.5 conditions become already dangerous with a violent and visible agitation of the sand particles quicksands This is common with fine sand or silt, but no so much with gravel or clay

86 Liquefaction Gravel: The permeability K is very high and very high flow rates (that are uncommon) would be necessary to reach critical conditions Clay: Material cohesion (not taken into account so far) helps in preventing critical conditions The permeability K is small. It takes a long time to reach a steady-state flow the time factor has not been taken into consideration either

87 Liquefaction Prevention: construct a weighted filter above the area in which seepage emerges from the ground σ 0 Same differential eq.: I z C z w Boundary conditions: z H 0 Therefore z C I H w 0 z H 0 I H z z w

88 Liquefaction σ 0 z I > I crit, but it is OK!, since σ' z > 0 I w 2 I = 0 Water in equilibrium, no flow: I w 1 0 z H z I = I 1 > 0 H z σ I = I crit σ z = 0 I = I 2 > I 1 > 0 Warning: this is valid only where 1D-flow conditions can be assumed!!!

89 2.4 Solving flow problems

90 Methods For a homogeneous and isotropic soil, the differential flow equation is the Laplace equation: To solve the Laplace equation, several methods are available: analytical numerical analogical graphical 2 0 boundary conditions small-scale model tests

91 Methods Analytical methods 1D: simple if K = constant 2D: separation of variables complex variable variation calculus fragment method (aproximate) Numerical methods finite differences finite elements

92 Methods Analogical methods the Laplace equation appears also in the field of electricity: permeability conductivity hydraulic head potential (voltage) Graphical methods Small-scale model tests problems to reduce the size of the particles

93 Analytical method (1D) Useful simplification: Dupuit s hypothesis Consider the case in the figure: unconfined aquifer, h = h(x,z) L >> H 1 H 2 approximate h = h(x,z) h(x) Dupuit Q H 1 z h h(x) x Q H 2 K L

94 Q H 1 z h h(x) x Q H 2 L K Although h h(x), the Laplace equation cannot be directly applied, since Dupuit s hypothesis is only an approximation Procedure: dh Q ct. q h K h dx Q K 1 2 2Q K d h dx 2 2 h x C boundary conditions : x 0 h H x L h H 1 2 2Q H 0 C C H K Q 2 H 2 L H1 Q K h H1 H1 H 2 L x K H L H

95 Analytical method (1D) In a confined aquifer, the flow is strictly one-dimensional and therefore the Laplace equation can be directly applied: 2 2 dh ( ) 0 2 h h x h h Ax B dx H H H A0 B B H H A L H A h H H H L x L H 1 z x Q h = h(x) K Q H2 L

96 Graphical method preliminaries Assume: Two-dimensional flow Homogeneous and isotropic soil K constant 1. Equipotential lines: qx K x q K define K velocity potential qy K y q is perpendicular to the lines x, y constant (equipotentials)

97 Graphical method preliminaries 2. Flow lines q 0 w q w q w ( x, y) w (0,0, w ) i j k w3 w3 w,,0 x y z y x 0 0 w define x, y xy, constant q,,0 y x w flow lines

98 Graphical method preliminaries 3. The flow vector is tangent to the flow lines q i j y x ds dxi dyj on a flow line Compute the cross product : q ds i j k dsq dx dy 0 dx dy k dk x y 0 y x On a flow line constant d 0 ds q constant

99 Graphical method preliminaries 4. Flow lines (Ψ=ct.) are perpendicular to the equipotential lines (Φ=ct.) If they are, so will be thegradients of and :,, x y x y x x y y q q q q x y x y 0 q y q x q x q y

100 Graphical method preliminaries 5. Given two flow lines, the discharge between them is constant y x C Q AC Q CB B Ψ=Ψ B Q A Ψ=Ψ A Within triangle ABC : Q Q Q C C C Q q dy dy d y AC A x A A A C B B B Q q dx dx d x CB C y C C C B AC CB Ψ A and Ψ B are constants Q A C C B A B

101 Graphical method Consider a flow channel such the one in the figure, plane state (2D) The discharge within the channel, per unit of length perpendicular to the drawing is: Qch A K I a 1 K b total K K t total t ch total s s t a a b n b n a Q n Q K n b n n area permeability gradient number of jumps ( equipotentials) number of flow channels s If Q total a b a Φ=φ 1 Φ=φ 2 b Ψ=Ψ A Ψ=Ψ B flow channel total potential drop between the channel exit and entry points s 1 Q n K total t 0 n s n

102 Graphical method The previous formula is valid only when all channels are limited by the same equipotential line both at the entry and exit points Otherwise it is necessary to compute separately the discharge at each channel, and add them to obtain the total discharge of the flow net: Q total n t 0 ns K n i1 s i

103 Graphical method example z H h 1 h 2 Objective: Calculate the total discharge through the flow net K Calculate the pore water pressure at each point within the domain

104 Graphical method example z H h 1 h 2 Procedure: Draw an orthogonal flow net: flow lines are perpendicular to equipotential lines, with cells as much square as possible K

105 Graphical method example 1: Flow lines 2: Equipotential lines h 1 h 2 φ 0 K φ 8 H z φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7

106 Graphical method example Orthogonal flow net draw flow lines (only a few!) first Then draw the equipotential lines perpendicular to the already drawn flow lines, such that the curved cells are as much square as possible adjust the flow net as necessary The flow net is the same independently of the headwater (h 1 ) and tailwater (h 2 ) elevations.

107 Graphical method example 1. Seepage 2. Porewater pressure h 1 h 2 φ 0 K φ 8 z H A z A φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7

108 Graphical method example h 1 h 2 SEEPAGE z φ 0 H K φ 8 Within the flow channel: H h H h h h n n s t total φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7 total 5 Q nt K Kh h n 8 s 1 2

109 Graphical method example z φ 0 H h 1 h 2 K φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7 A z A φ 8 POREWATER PRESSURE At point A: p wa w A A n A 0 H h n 0 1 h h 11 h 1 h2 A H h h1 h2 5h1 11h2 pw A w H h1 za w H za A z A

110 Graphical method example Exercise: find the condition between h 1 and h 2 that prevent quicksand conditions z φ 0 h 1 h 2 A H K φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7 z A φ 8 The zones where quicksand conditions due to critical hydraulic gradient may appear are the ones close to the sheet piles wall and to the ground surface, at the exit of the flow net. Choose a point in that zone were there are data, e.g. point A in the figure.

111 Graphical method example z φ 0 h 1 h 2 A H K φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7 z A φ 8 Let A A A wa H h h h 8 h1 7h2 H 8 A A nat A w wa w A A A z H z p z h p z h 7h h 7h p wa w H H za w za 8 8

112 Graphical method example z φ 0 H h 1 h 2 A z A K φ 8 z h A nat A w h 7h w A A p z wa w h 2 z h 8 A 1 2 φ 1 φ 2 φ 3 φ 4 φ 5 φ 6 φ 7 A 0 h h 8z 1 2 A w

113 Graphical method The total discharge Q calculated with this method is mostly independent on the quality of the flow net, since errors tend to compensate. However, porewater pressures, and therefore the critical hydraulic gradient, are more sensitive to the quality of the flow net.

114 Soils with transverse anisotropy The differential equation becomes: K x h x h y 2 2 K 0 2 y 2 Therefore the graphical method cannot be applied However, with a change of variables: one gets: K y xt x ; yt y K x h h x K t h x x x x K 2 2 h K y h x K x t t x 2 2 x t y

115 Soils with transverse isotropy Substitution into the differential equation leads to: K x K y h h h h K 2 y Kx xt yt xt yt 0 0 which is the Laplace equation in the transformed space (x t,y t ) In this transformed space, the graphical method can be used: draw the complete original geometry (domain, geologic elements, discontinuities, etc.) in the transformed space apply the graphical method in this space undo the transformation

116 Soils with transverse isotropy y Assume K x 9K y y t x t K y x x 9K 3 y K y y t y K x x x t Do the change of variables Draw the flow net with the transformed geometry Undo the transformation orthogonality is lost

117 Equivalent permeability To use the previous formulas to evaluate the total discharge Q, an equivalent permeability K e is needed Consider two cases: horizontal flow vertical flow

118 Equivalent permeability y b y t b t a Q x K x a t Q xt K e? h 1 h 2 h x 1 h 2 h h h1 h2 Q K a b b 1 2 Qx Kx a y Qx Qxt at a bt b K x t x e e t xt e t t a a a K K K b b K Ke K x K y y b K x K x t

119 Equivalent permeability y a y t a t h 2 h 2 b Q y b t Q yt h 1 K y 1 2 Qy K y a x K e? h h h1 h2 Q K a b b K Q Q a a b b y y yt t t Kx h 1 yt e t t K y a a a K K K K K K K t x y e e e x y b bt b x t

120 Equivalent permeability Therefore, to calculate the total discharge Q through a flow net in the transformed space, one needs to use this equivalent permeability: K Q e ch K x K a t Ke bt y n total s

121 Heterogeneous (stratified) soil K 1 K 2 α 1 A α 2 Q 1 C B D Q 2 Continuity : 1 2 hb hd Q1 AB1 q1 AB K1 BD ha h Q2 CD1q2 CD K2 AC Also : h h A C Q Q hb ha hc hd h B h D C AB CD AB 1 CD 1 Q1 Q2 K1 K2; ; BD AC BD tan AC tan tan 2 K K 2 1 tan 1 1 2

122 Heterogeneous (stratified) soil K 1 K 2 K K sand clay K 1 K 2 K K clay sand

123 Heterogeneous (stratified) soil K 1 K 2 a 1 Q a 2 b 2 b 1 Q Q a1 K1 b 1 Q a2 K2 b 2 Results in : a b h h a K K b2 continuity within a channel flow If a b 1 1 a 1 b K K Therefore the size of the cells needs to be adjusted in order to use the graphical method

124 2.5 Piezometers

125 Introduction A piezometer is a device (standpipe, tube, vibrating wire piezometer, manometer) used to measure the porewater pressure in the ground. The ideal piezometer: chamber (tube) to collect water allows measuring water level porous element (water continuity) can measure positive (saturated soil) as well as negative pressures (unsaturated soil) with an acceptable margin of error fast response to changes in environmental conditions stable for a long time must cause a minimum possible interference with the soil

126 Introduction Water pressure inside the device is different from the pressure in the ground therefore a water flow from the ground toward the device will be established in order to reach hydraulic equilibrium The equilibrium is reached after a certain time interval time-lag, response time depending on: ground permeability shape and material of the piezometer It works the same way as a falling-head permeameter

127 Falling-head permeameter a a dh q Adt; q A h K L dh h Q q A K h L h a dh K A dt L dh A K dt h a L A L 2 A K h A K h2 1 h ln h t ln t2 t1 h 1 t a L h a t 1 2 L K a L ln A t h h t

128 Piezometers a For a piezometer: H = H 0 -y H 0 dy y t = t+dt t = t t = t 0 Q F K H 3 m m m m s s shape factor obtained from tests + theory Q a dy Q dt F K H y dt dy F K dt H y a 0 0 y F K ln H0 y t y0 a t t 0

129 Piezometers H = H 0 -y H 0 a dy y t = t+dt t = t t = t 0 y F K ln H0 y t y0 a Then : Base time : ln T t t 0 H0 y F K t 0 t H0 a H0 a a F K H F K time needed to fill the total volume with the initial flow rate 0 Q If t 0 H t H 0 ln H T H 0 0 e tt y H 0 1 e tt t t y T 63% H 0 y 2.3T 90% H 0

130 Shape factors (1) K D impervious K impervious D spherical chamber F = 2πD semi-spherical chamber F = πd

131 Shape factors (2) imp. D imp. K D K flat bottom above interface F = 2D flat bottom on homogeneous ground F = 2.75D

132 Shape factors (3) imp. K v D imp. K L L K v D K soil (K v ) at bottom of pipe flat bottom above interface 2D F= 8 L K soil (K v ) at bottom of pipe flat bottom on homogeneous ground 2.75D F= 11 L K 1+ π D K 1+ π D K v v

133 Shape factors (4) imp. L D imp. K L D K perforation continues from interface 2πL F= 2 2L 2L ln + 1+ D D perforation continues from interface homogeneous ground 2πL F= 2 L L ln + 1+ D D

134 Shape factors (5) impervious L 2R impervious K total penetration into the pervious layer (R 2 = influence radius) 2πL F= ln R R 2

135 Piezometer types Observation well, standpipe Casagrande piezometer Closed circuit hydraulic piezometer Piezometric cells pneumatic vibrating wire with extensometers

136 Observation well (1) tube screened section protective cap mortar granular material cap A borehole with a simple pipe ( = 25 to 50 mm) is useful to locate the water table in high permeability ground (K > 10-6 m/s) It is a very common type because of its simplicity and low cost Big time-lag, not useful to measure rapid changes of water level

137 Observation well (2) plastic tube porous element protection cap granular material bentonite sealing sand filter Care must be taken in stratified soils with layerdependent piezometric levels, so that readings do not lose meaning (connected aquifers) The porous element where readings are taken must be isolated from the other aquifers, e.g., by a bentonite seal.

138 Casagrande Piezometer double PVC pipe (15 mm) earth filling sealing clean sand sealing drain (clean sand 1-4 mm) Casagrande Piezometer (measures in cm) Measures the porewater pressure of the ground. Made of a porous element (L 300 mm, 38 mm) connected to the surface with a twin pipe system. The porous elements is surrounded by clean sand, and it is confined above and below by a bentonite seal. The existing air in the circuit is eliminated through the twin pipe system.

139 Piezometers Open pipe & Casagrande

140 Hydraulic piezometer It is made of a porous element connected to a remote reading station with two flexible pipes ( 3 mm) along which airless water circulates. The water pressure at the tip of the piezometer is measured by a transducer. Advantages: reliable for long time-periods possibility of remote readings can be used in partially saturated soils Careful installation is necessary to avoid cavitation problems in the circuit

141 Piezometric cells In low-permeability grounds, or where an acceptable response time (few hours) is required, piezometric cells may be necessary in order to have pressure transducers directly at the measuring point. Types: pneumatic vibrating wire with extensometers

142 Pneumatic piezometer (1) plastic tubes compressed air water compressor air-flow indicator diaphragm filter A diaphragm inside the intake element deforms because of the surrounding groundwater pressure. The pressure can be measured by applying on the opposite side of the diaphragm an equal pressure by means of a gas. Direct reading of the pressure that restores equilibrium. Simple and stable Small time-lag

143 Pneumatic piezometer (2) Typical installation and operating scheme of a pneumatic piezometer

144 Vibrating wire piezometer (1) Consistis of a metallic diaphragm that deforms because of the external pressure. The deformation is measured with a vibrating wire extensometer; the measure is sent to a frequency indicator on the surface through an electric connexion. Long lifespan and high sensibility to environmental conditions. Must take into account: possible corrosion of metallic elements, and damage to electric wiring due to large ground movements.

145 Vibrating wire piezometer (2)

146 Piezometer with extensometers The diaphragm deformation is measured with an electric extensometer by changes of electric resistance.

147 Multi-level piezometer

148 Multi-level piezometer

149 Applications concrete dams retaining walls pumping earth dams typical installation

150 Time-lag Basic hypotheses: K is constant, isotropic soil No hydraulic head loss within the piezometer Soil does not deform The time factor (base time, T = a/fk) is the fundamental parameter: If K is small (clays), T is high In practice, the time-lag can vary from just minutes, to days or months

151 Time-lag (for 90%) Type of soil Sand Silt Clay Permeability (cm/s) Standpipe 5 cm Soil within standpipe: L = 3 = 15 cm Standpipe 5 cm Flat bottom Standpipe 5 cm Prolonged borehole: L = 3 = 15 cm Standpipe 5 cm Prolonged borehole: L = 12 = 60 cm Piezometer: 10 mm with porous tip, L = 45 cm = 38 mm Piezometer: 10 mm with porous tip and sand filter, L = 108 cm = 15 mm Mercury manometer mm single pipe, porous tip, L = 62 mm = 32 mm Mercury manometer mm single pipe, porous tip, L = 45 cm = 58 mm 6 m 1 h 10 h 4.2 d 0.6 m 6 m 1 h 10 h 4.2 d 1.5 m 15 m 2.5 h 25 h 10 d 6 m 1 h 10 h 4.2 d 42 d 3 m 30 m 5 h 50 h 21 d 12 m 2 h 20 h 8.3 d 83 d 2 m 20 m 3.3 h 33 h 14 d 6 m 1 h 10 h 42 d 1 Reduce to half for mercury manometers, U-tube (1.5 mm) or manometers Bourdon 12 cm m: minutes, h: hours, d: days

152 homogeneous clay fine silty sand, silts stratified clay sand Time-lag 1 hour 10 hours 1 day 10 days 1 month (hores)

153 2.6 Filters and earth dams

154 Filters Introduction Earth and rockfill dam at Schofield, Utah, USA (1926) The dam failed because of erosion of the finer material through the dumped rock on the downstream side. It helped to understand the role of filters between finer and coarser materials.

155 Earth and rockfill dam at Nantahala, North Carolina, USA (1942)

156 Earth and rockfill dam at Bear Creek, USA (1953)

157 Filters Main requirements: avoid erosion enough permeability Spherical particles: D 6.5D If a b, the small particle will pass through

158 Filters

159 % passing Powered by TCPDF ( Criteria based on grain-size curves 85 soil filter 15 D 85f D 85s D 15f D 15s

160 Criteria based on grain-size curves Totally empirical (Terzaghi) Particles with size > 2 cm must be previously removed Relative to erosion: Relative to permeability: D D 15,filter 85,soil 2 [Since K D 15, makes sure that K filter > K soil ] 4 to5 D 15,filter 4 to5 20 D 15,soil Additional rule (USCE): D D 50,filter 50,soil 25

161 Filters to protect earth dams cores Sherard and Dunnigan, 1989 Concentrated leak flowing in dam impervious core toward filter Very high gradient develops here after seal forms on surface ab (Path Y) a b "Critical" downstream filter Impervious dam core Seal forms when surface (ab) plugs Filter face Concentrated leak through a crack open in the core

162 Non-erosion filter test

163 Test for group 1 soils: clay and fine silt Conditions existing immediately after beginning of test

164 Conditions existing before reaching equilibrium

165 Final conditions in a positive test

166 Final conditions in a negative test

167 Clay and fine silt (group 1) d 85 : fine-grain soil to be protected D 15b : filter s critical D 15 If: D 15,filter D 15b correct D 15,filter > D 15b erosion

168 Silt and sandy clay (group 2) D 15b = mm

169 Silty and clayey sands (group 3)

170 D 15b for the tested soils TYPE OF SOIL FINE CONTENT (% < #200) D 15b FROM TESTS GROUP 1: CLAY AND FINE SILT D 15b = 7d 85 to 12d 85 ( 9d 85 ) GROUP 2: CLAY AND SILTY SAND D 15b = 0.7 to 1.5 mm GROUP 3: SAND WITH SILT AND CLAY, LOW FINE MATERIAL CONTENT 0 15 D 15b = 7d 85 to 10d 85 GROUP 4: SAND WITH SILT AND CLAY Intermediate between group 2 and group 3 depending on fine content

171 Dispersive soils Some fine grained natural soils are highly subject to erosion (dispersive). They usually have a high percentage of sodium salts dissolved in the pore water. The clay particles of these soils can pass easily to the water without much need for this water to be in motion. Fine soils of this type can be usually found in dry climates.

172 Dispersive soils identification Pinhole test : distilled water 1 mm If the water comes out dirty and there is erosion in the hole, then the soil is dispersive. If the water comes out clear, then the soil is not dispersive

173 Dispersive soils identification Salts content: subject to erosion non-subject to erosion [remember double layer : d 1 K vn0 0.5 val Na + = 1 val Ca ++ = 2 ]

174 Dispersive soils identification Adding lime Ca(OH) 2 (between 2 and 4% in weight) inhibits dispersibility Contact of dispersive clay masses with zones where water may flow (such as in earth dam s cores) must be protected by thick layers of nondispersive treated clay.

175 Homogeneous earth dams toe drain chimney and toe drain Can be completely homogeneous only if the height is H < 6-7 m Low or medium height Convenient to have an inner drainage system to: reduce water pressure inside the downstream slope (stability) avoid uncontrolled downstream surface water flow avoid backward-erosion of the dam (piping)

176 Homogeneous earth dams K h = K v K h = 4K v K h = 9K v Anisotropic permeability reduces efficiency of a toe drain

177 Homogeneous earth dams Chimney drains are always efficient In any case, drainage must be designed to evacuate all water from seepage

178 Homogeneous earth dams Anisotropy: construction of the flow net a) K foundation = K dam b) K foundation = 30K dam

179 Homogeneous earth dams Vega Dam Sherburne Lakes Dam, Montana, USA, 1916

180 Thin clay core dams Convenient when granular materials are more abundant than finer ones. Coarser materials are more stable and easier to place than finer ones, that need to be compacted under controlled water content conditions.

181 Thin clay core dams Natural slopes with gravel and rockfill: Smooth gravel Rough gravel Rough gravel or rockfill compacted in relatively thin layers The thickness of the impervious core is determined by: risk of erosion originating from fissures (differential settlement) maximum admissible seepage construction problems 1 1

182 Thin clay core dams Empirically accepted core thickness: e 0.3 to 0.5 H satisfactory H e Two designs: e 0.15 to 0.2 H satisfactory for finer materials e < 0.1 H seldom used core sloping upstream central core

183 Thin clay core dams Yale Dam, Washington, USA (1952) Success Dam, California, USA (1960)

184 Thin clay core dams Gepatsch Dam, Austria (1963)

185 Thick clay core dams core core rockfill rockfill Improved safety against internal erosion Extended contact between core and foundation (smaller gradients in general) drain blanket Larger volume of finer materials needed

186 Thick clay core dams Ice House Dam, California, USA Presa de San Lorenzo, Perú

187 Thick clay core dams Presidente Alemán Dam, México

188 Shoulder saturation Through the supporting foundation stratum

189 Shoulder saturation Effect of permeability contrast between core and downstream shoulder

190 Shoulder saturation Effect of permeability contrast between core and downstream shoulder

191 Shoulder saturation Effect of permeability contrast between core and downstream shoulder, and of anisotropic permeability

192 Shoulder saturation Effect of permeability contrast between core and downstream shoulder, and of anisotropic permeability

193 Rapid drawdown It is the worst stability condition affecting the upstream slope Theoretical porewater pressure in the upstream slope of an homogeneous earth dam after rapid drawdown (Terzaghi): (a) full reservoir; (b) after drawdown p w >0 p w >0

194 Rapid drawdown Influence of the lower stratum permeability in the theoretical porewater pressures (Cedergren): (a) stratum and dam have the same permeability; (b) permeability of stratum is 20 times that of the dam

195 Rapid drawdown Solution: place chimney drains near the upstream slope surface, to force near vertical flow after drawdown

196 Earth dams: project issues Steps of design exploration (situation, available materials) compare different initial designs estimate cost and safety of those initial designs chose a reasonable combination

197 Earth dams: project issues Availability of materials large amounts of clay homogeneous dam or thick core dam large amounts of sand/gravel thin core abundant rockfill large shoulders, rockfill dam with impervious screen several materials zoned dam mixed materials use in protected zones (with drains and/or filters)

198 Earth dams: project issues Foundations: can be build on any type of soil but the following must be considered: differential settlements seepage control in permeable grounds quicksand conditions interaction between dam and rock foundation, especially in fractured rock sites

199 Earth dams: project issues Weather in rainy climates, correct compaction is more difficult to achieve if freezing occurs, clayey materials cannot be compacted (but gravel or rockfill materials can still be properly placed) in dry areas optimum water content for compaction can be better controlled

200 Earth dams: project issues Shape of the valley in narrow valleys it is more difficult to construct a dam; in that case rockfill dams perform better care must be taken of differential settlements critical sections

201 2.7 Flow in unsaturated soils

202 Introduction meniscus surface tension Origin: natural soils, above water table, dry climates,... artificial soils compacted Air in the larger pores Water in the smaller pores

203 Surface tension Surface tension develops when there is an interface (contact surface) between two different media. This surface (meniscus) acts as a membrane: σ σ σ: surface tension [σ] = F/L

204 Surface tension For water/air contact: σ N/m at 20ºC σ N/m at 60ºC σ depends on temperature, air pressure, etc. The surface tension σ is a measure of the needed work supply for the system to produce a unit increase of the contact surface.

205 Surface tension The shape of the meniscus depends on the materials that form the interface: θ>90º θ<90º wetting liquid non-wetting liquid θ θ 1 2 on an incline and on the history as well (flow direction): water drops water rises

206 Surface tension Because of the membrane, there is a discontinuity on the pressure field: σ σ p air p water Define capillary pressure (also known as suction) as: p c = p air p water (> 0 in general)

207 Surface tension Forcing equilibrium (or minimizing energy) leads to: pc pa pw where R R R2 R R R 1 2 where R 1 and R 2 are the curvature radii of the meniscus

208 Surface tension In a capillary tube: R r cos 2 cos r The capillary pressure depends on: p c θ : depends on wheter the water rises or drops history, flow σ : characteristic of the water/air interface r : size of the capillary tube σ θ R meniscus circular r

209 Capillary rise

210 Capillary rise h c p w? p w = 0 cos 2 2 r r hc w hc resulting force weight of the water of surface tension column within the capillary tube When the capillary tube is submerged in a water tank, the water level within the capillary tube rises (h c ) Assuming that the weight of water within the capillary tube is fully sustained by the surface tension leads to: 2 cs o r w capillary rise height

211 Capillary rise h c p w? p w = 0 Once level h c has been reached, water is in equilibrium: the piezometric level remains constant! Actually, water pressure within the capillary tube is negative

212 Water above phreatic level z WT p w = 0 soil acts as if there were capillary tubes, although irregular Above phreatic level (water table) a pressure p c develops (capillary pressure suction) and water rises up to the capillary rise height p c capillary pressure s suction

213 Water above phreatic level Remember that for a capillary tube In the soil, p c 2 cos r σ does not change much θ depends on water motion (rise/drop) r depends on the type of soil, on amount of water in the soil, and on history

214 Water above phreatic level In general, a decrease in the degree of saturation (S r ) results in meniscus with smaller radii : r r

215 Water above phreatic level For a given soil, p c = f(s r ); also, p c = f(w) These relationships are obtained experimentally: hysteresis p c = p a p w drying the soil e.g., when the water level drops in an initially saturated soil wetting the soil e.g., if water is added to an initially dry soil S r WATER RETENTION CURVE 100%

216 Water above phreatic level Hysteresis is due to the term cos θ and to geometry: for the same suction s, the degree of saturation, S r, increases when the soil gets dryer: direction of flow water reaches this level if the soil is initially dry and water is added water is retained here when drying an initially saturated soil direction of flow

217 Water above phreatic level In soils the shape and size of capillary tubes is irregular grain-size dependence If the pores are small and the meniscus radii are also small, then large suctions result If suction is big, there are additional forces between soil particles that provide an apparently larger strength to the soil

218 Water above phreatic level forces due to suction in yellow, forces of mechanical nature saturated soil: no suction dry soil: no suction resultant of the suction forces: increase of contact forces between particles and apparent increase of strength

219 Water above phreatic level In unsaturated soils, the meniscus produce an increase of effective stresses and therefore an apparent increase of strength. pw 2 pc pa pw pc pc R if pa 0 pc p w if S r decreases, the radius R* also decreases and therefore the suction p c and the effective stress σ' also increases

220 Water above phreatic level Examples: beach sand: near the water line, the sand is harder than far from the water (dry sand) or than submerged sand (saturated sand) it is with this partially saturated sand that one can build a sand castle!! with partially saturated sands it is possible to have almost vertical slopes but it the slopes get wet or dry, they immediately collapse

221 Water above phreatic level z B A p w WT water moves from points with more energy (φ) to points with less energy if equilibrium is reached, then φ A = φ B : in this zone the porewater pressure is negative: p w = -p c < 0 p pc z w z pc pa pw w w Point A: pc pa pw 0 ; z 0 A 0 p cb A B zb Point B: z w B B w p cb thus in equilibrium to each height corresponds one suction: z = f(p c )

222 Water above phreatic level Since p c = p c (S r ) z = f(s r ) z z B S r S rb 100%

223 Water above phreatic level In practice, this curves are of this type: saturated zone above phreatic level z h c : capillary rise height the capillary rise height is very well defined when the pore size is uniform S r 100%

224 Water above phreatic level h c depends on the type of soil: Type of soil coarse Sand medium fine Silt Clay h c 2 to 5 cm 15 to 30 cm 40 to 70 cm 70 to 150 cm > 4 m correlations: C hc C 0.1cm to 0.5cm D e 10 poorly graduated 2 2 well graduated

225 Water above phreatic level z The curves p c S r and z S r are equivalent (assuming water continuity not valid for isolated water drops ) Similar to water retention curves (p c S r ), the curve z S r depends on history: remaining dry soil z some water remains S r S r initially dry soil, with rising phreatic level ( wetting ) 100% 100% initially saturated soil, with dropping phreatic level ( drying )

226 Suction Define capillary potential as p c w given in units of length; in fine-grained soils, it may be very high. Often the measuring unit is pf: pf pf log p c w in cm p cm m MPa c pc 10 w common for mediterranean climates