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8.3 application to saturated-unsaturated water flow problems 397
396 8 solving saturated/unsaturated water flow problems
4020
Silt
3015
FrequencyFrequency2010
105
00.71.42.12.83.54.24.95.66.3more00
0.20.61.01.41.82.22.63.03.43.84.24.65.05.45.8moreq
Figure 8.27 Frequency distribution for q permeability parameter for all soils analyzed (from Fredlund et al., 2001a).
20
Sand 15
q
Figure 8.29 Frequency distribution for q permeability parameter for silt soils analyzed (from Fredlund et al., 2001a).
Table 8.4 Statistical Analysis of q Soil Parameter for All Soils
Statistical Variableq Soil Parameter
Mean3.29Standard error0.08Median2.96Mode5.61Standard deviation1.40Sample variance1.96Kurtosis14.80Skewness2.35Frequency10
5
00.81.62.43.24.04.85.66.47.2more0
q
Figure 8.28 Frequency distribution for q permeability parameter for sand soils analyzed (from Fredlund et al., 2001a).
sandy silt and 1.81 for the clayey silt. The results indicate that there is less scatter in the tting parameter q as the soil becomes closer to a sand material. The tting parameter q tends toward 1.0 for sand soils.The sand showed a mean q value of 2.37 while the clay soils showed a mean q value of 4.34. The sandy silt showed a mean q value of 2.86 and the clay silt showed a mean q value of 3.58. The results indicate a trend toward a larger q value for soils with higher plasticity and a wider distri- bution of particle sizes. The mean tting parameters for all soil types ranged from 2.4 to 5.6. The statistical results pro- vide a general indication of the range and scatter that can be anticipated when using the Leong and Rahardjo (1997a) equation (i.e., Eq. 8.29) for estimation of the permeability function for an unsaturated soil.The analysis of an even larger database of measured
Range14.39Minimum0.64Maximum15.03Number of soils323
associated with the q parameter. However, it must also be recognized that there will always be considerable scatter in the q parameter. The estimation procedure proposed by Leong and Rahardjo (1997a) is of value, but it does not appear to provide a singular value for the q parameter.
8.2.9 Regression Techniques for Best-Fit Permeability FunctionGardner (1958a) published two empirical forms for a per- meability equation. The rst equation was presented earlier and takes the following form:
ks
SWCC and permeability function measurements wouldbe useful to better understand the mean and variation
kw (ua uw =
1 + ag
(ua
uw
/(w
g nng(8.30)
Table 8.5 Statistics of Permeability Fitting Parameter q for Various Soil Textures
Statistics
ClayClay loam
Loam
SandSandy clay loam
Silty claySilty clay loamSilt loamSandy loamLoam sand
Mean4.343.583.782.372.805.593.223.522.862.67
Median4.712.623.562.362.624.773.183.462.852.59
Mode3.002.803.252.202.754.604.053.152.754.63
Standard deviation1.501.811.160.491.001.311.361.090.840.68
Sample variance2.253.291.340.240.991.731.861.190.710.46
Range5.425.924.142.254.716.334.975.994.204.01
Minimum1.521.841.631.250.641.111.280.831.021.41
Maximum6.947.765.763.495.357.446.256.825.225.42
Number of sets21181249173418743029
where:kw (ua uw = coefcient of permeability as a functionof soil suction,ks = saturated coefcient of permeability,ag = tting parameter related to the inverse of the air-entry value,ng = tting parameter related to the rate of desaturation of the soil,ua uw = matric suction,w = density of water, andg = acceleration due to gravity.
The second Gardner (1958a) permeability equation has the following form:kw (ua uw = ks exp cg (ua uw(8.31) where:
cg = soil parameter related to the exponential decrease in permeability with respect to soil suction.
Inherent in the form of the latter equation is the assump- tion that the air-entry value for the soil is near zero. Equation8.31 should only be used when it is known that the air- entry value of the soil under consideration is approaching zero. Neither of the above permeability functions should be used unless it has been possible to determine the ag andng tting parameters through use of an independent anal-ysis or through direct measurements of the coefcient ofpermeability at various applied soil suction values.
8.2.10 Estimation of Minimum Coefcient of PermeabilityLimited research has been undertaken on the form that the permeability function should take once residual water content conditions are reached and exceeded. A recent
study (Ebrahimi-Birang et al., 2004) has suggested that there should be a lower limit for the water coefcient of permeability and that the magnitude of potential vapor diffusion be taken into consideration.Figure 8.30 shows the results of the method proposed by Ebrahimi-Birang et al. (2004) applied to silty sand. The lower limit that was suggested for liquid water ow was 1 1014 m/s. It was suggested that the same lower limit for the coefcient of permeability might be applicable for all soils. The lower limit for the water coefcient of perme- ability is of importance with respect to numerical modeling of water ow. The lack of a lower limit for the water coef- cient of permeability can give rise to numerical convergence problems during seepage modeling.
8.2.11 Water Storage Modulus for Transient Modeling
2The water storage property of a soil, mw, is dened as the relationship of the slope of the (volumetric) water con- tent and soil suction. The water storage variable is required whenever a transient seepage analysis is performed. The water storage modulus can be obtained through the (arith- metic) differentiation of any of the equations designated for the SWCC (Fig. 8.31).
8.3 APPLICATION TOSATURATED-UNSATURATED WATER FLOW PROBLEMS
Steady- and unsteady-state water seepage through a saturated-unsaturated soil system can be analyzed by solv- ing the governing partial differential equation of seepage. The solution is obtained through use of the numerical mod- eling methods such as nite difference and nite element methods. The approach is similar for steady-state seepage and unsteady-state seepage with the exception that unsteady- state seepage is time dependent, requiring discretization of time (i.e., elapsed time is handled in an incremental elapsed time manner for unsteady-state water seepage).
8.3 application to saturated-unsaturated water flow problems 399
398 8 solving saturated/unsaturated water flow problems
1E-041E-05
Coefficient of permeability, m/s1E-061E-071E-081E-091E-101E-111E-121E-131E-141E-151E-161E-17
Experimental data Fredlund and Xing (1997) Vapor permeability
0.1110100100010,000 100,000106Soil suction, kPa
Figure 8.30 Variations of liquid water and vapor permeability function for a silty sand soil.
Volumetric water content, %40
30
20
10Hysteresis
00.1110100100010,000 100,000106Soil suction, kPa
Water storage modulus (m2w), kPa18
4
2Hysteresis
00.1110100100010,000 100,000106Soil suction, kPa
Figure 8.31 Relationship of water storage functions to drying and wetting SWCCs.
The solution of saturated-unsaturated seepage problems is presented through a series of problems that are progres- sively more complex. The rst seepage problems are of a steady-state nature and these are followed by several unsteady-state (or transient) analysis problems. One- dimensional steady-state problems are rst solved and these are followed by two-dimensional examples. The one- dimensional seepage examples illustrate the application of hydraulic head and moisture ux boundary conditions. The one-dimensional problems are solved using a nite difference technique. Two-dimensional steady-state example problems are then solved using the nite element method.
These example problems are followed by a series of two-dimensional, unsteady-state examples that are solved using the nite element methodology.
8.3.1 Solution of One-Dimensional Flow ProblemsThe differential equation for one-dimensional steady-state ow through a homogeneous, saturated soil can be solved by integrating the differential equation two times. The equation for one-dimensional steady-state ow through an unsaturated soil requires a more complex solution than that for a saturated soil. A numerical solution must be used rather than a closed-form solution. The nite difference method
Figure 8.32 Hydraulic head function h(y) written in nite differ- ence form.
can be used to illustrate the solution to the steady-state ow equation for an unsaturated soil.The seepage differential equation can be written in a nite difference form. Consider the situation where a function h(y) varies in space, as shown in Fig. 8.32. Values of the function at points along the curve describing hydraulic head can be computed using a Taylor series to write forward- difference and backward-difference equations. The forward difference-equation can be written as
Equations 8.34 and 8.35 are called the central-difference approximations for the rst and second derivatives of the function h(y) at point i . These approximations can be used to solve the differential seepage equation. Similar approxi- mations can be derived for a function h(x) in the x -direction when there is ow in the x-direction. The use of an iterative nite difference technique in solving seepage problems is illustrated in the following sections.
8.3.2 Head Boundary Conditions AppliedSteady-state evaporation under a xed head boundary condition from a column of unsaturated soil is illus- trated in Fig. 8.33. A tensiometer is installed near the ground surface to measure the negative pore-water pressure. One-dimensional, steady-state ow is assumed when the ten- siometer reading remains constant with respect to time. The pore-water pressure at the base of the column (i.e., the water table) is equal to zero.The hydraulic head distribution along the length of the column is given by Eq. 7.30 [Chapter 7; i.e.,kwy (d2hw/dy2 + (dkwy/dy (dhw/dy = 0]. This equationcan be solved using the nite difference approximationsshown in Eqs. 8.34 and 8.35. The column length is rst subdivided into n equally spaced nodal points at a distance Ay apart (Fig. 8.33). A central-difference approximation is then applied to the hydraulic head and coefcient- of-permeability derivatives. The nite difference form can
1 dh
Ay2 1 d2h
be written as follows for point i:
hi+1 = hi + Ay
+dy i2!
dy2 i
hw(i+1)
+ hw(i1)
2h
w(i)
Ay3 1 d3h +
+ (8.32)
kwy(i)
(Ay2
3!dy3 i
1 kwy(i+1) kwy(i1) 11 hw(i+1) hw(i1) 1 = 0 (8.36)
The backward-difference equation can be written as
+2 Ay
2 Ay
1 dh
Ay2 1 d2h
hi1 = hi Ay
+dy i2!
dy2 i
where:
where:
Ay3 1 d3h 3!dy3 i
+ (8.33)
kwy(i), kwy(i1), kwy(i+1) = water coefcients of perme-ability in the y -direction atpoints i, i 1, and i + 1, re- spectively, and
i 1, i , i + 1 = three consecutive points spaced at incre-ments Ay.Subtracting the forward- and backward-difference equations and neglecting the higher order derivatives result in the rst derivative of the function at point i:
hw(i), hw(i1), hw(i+1) = hydraulic heads at points i, i 1, and i + 1, respectively.
Equation 8.36 can be rearranged after assuming equal Ayincrements:
1 dh
hi+1 hi1=
(8.34)
8k
wy(i)n h
w(i)
+ 4k
wy(i)
+ kwy(i+1)
kwy(i1)n h
w(i+1)
dy i
2 Ay
+ 4kwy(i) + kwy(i1) kwy(i+1)n hw(i1) = 0(8.37)
Summing the forward- and backward-difference equationsand again neglecting the higher order derivatives give the second derivative of the function at point i:
The hydraulic heads at the external points (i.e., points 1 and n) become the boundary conditions. The hydraulic head at point 1 is zero. The elevation of point n relative
1 d2h
= hi+1 + hi1 2hi
(8.35)
to the datum, hgn, gives the gravitational head at point n.
dy2 i
Ay2
The tensiometer reading near the ground surface indicates
Figure 8.33 One-dimensional, steady-state water ow through an unsaturated soil with a constant-head boundary condition.
8.3 application to saturated-unsaturated water flow problems 401
400 8 solving saturated/unsaturated water flow problems
hpn
Pore-water pressure head, hp
hwnhgn
Hydraulic head, hw
Gravitational
hgn
hpn
vwy
hwn0.9 hwn0.8 hwn0.7 hwn0.6 hwn0.5 hwn
0.4 hwn
0.3 hwn
0.2 hwn
yhead, hg
0.1 hwn
()
Datum 0
(+)
Water table 0
Head, h
vwy
Figure 8.34 Steady-state evaporation from top of unsaturated soil column.
the negative pore-water pressure head at point n (i.e., hpn). The hydraulic head boundary condition at the top and the base of the soil column can be expressed mathematically as follows:
hw(l) = 0.0at y = 0.0 (base)hw(n) = hgn + hpnat y = hgn (top)
The nite difference seepage equation can be written for the n 2 internal points (i.e., points 2, 3, ... , n 1). As a result, there are n 2 equations that must be solved simul- taneously for n 2 hydraulic heads at intermediate points. The illustrated nite difference scheme is called an implicit form. The equation is also nonlinear since the coefcients of permeability kwy are a function of matric suction, which
in turn is a function of pore-water pressure, which is a component of hydraulic head hw. The nonlinear equations require several iterations to produce convergence. The water coefcient of permeability at each node is revised to a new constant value following each solution for the hydraulic heads.For the rst iteration, the kwy values at all points can be set equal to the saturated coefcient of permeability ks . The n 2 linear equations can then be solved simultaneously using a procedure such as the Gaussian elimination technique. The computed hydraulic heads are used to calculate new values for the water coefcient of permeability. The coefcient- of-permeability values at each point must be in agreement with the coefcient of permeability versus soil suction func- tion. The revised coefcient-of-permeability values kwy are then used for the second iteration. New hydraulic heads are then computed for all depths. The iterative procedure is repeated until there is no longer a signicant change in the computed hydraulic heads and the computed coefcients of permeability.Figure 8.34 illustrates typical distributions for the pore- water pressure and the hydraulic head along the unsaturated soil column. Flow is occurring under steady-state evap- oration conditions. The nonlinearity of the ow equation results in a nonlinear distribution of the hydraulic head and the pore-water pressure head. The equipotential lines are not equally spaced along the column. This is different from the
varying coefcient-of-permeability values throughout the unsaturated soil column. The above analysis can also be applied to the steady-state downward ow of water through an unsaturated soil. Once again, the hydraulic head bound- ary conditions at two points along the soil column must be known.
8.3.3 Flux Boundary Conditions AppliedInltration into an unsaturated soil column is another example which can be used to illustrate the solution of the nonlinear differential seepage equation (Fig. 8.35). Steady- state inltration can be established as a result of sprinkling irrigation. Let us assume a constant downward water ux qwy which is less than the saturated coefcient of perme- ability of the soil. Steady-state ow can be described using Eq. 7.30. The hydraulic head distribution can be determined by solving the nite difference form of the steady-state ow equation (i.e., Eq. 8.37). The hydraulic head boundary condition at the ground surface is a computed value. However, the water ux qwy is known and is constant throughout the soil column for steady-state conditions.The soil column can be discretized into n nodal points with an equal spacing Ay (Fig. 8.35). The water ux at point i can be expressed in terms of the hydraulic heads at points i + 1 and i 1 using Darcys law:
hw(i+1) hw(i1)
uniformly spaced equipotential lines for a homogeneous,saturated soil column. The difference is the result of the
qwy = kwy(i)
2 Ay
A(8.38)
Figure 8.35 One-dimensional steady-state water ow through an unsaturated soil with a ux boundary condition.
where:
qwy = water ux through the soil column during the steady-state ow, where ux is assumed to be positive in an upward direction and negative in a downward direction, andA = cross-sectional area of the soil column.
Equation 8.38 can be rearranged as follows:
2 Ay
solved directly starting from a known boundary condition such as point 1 (Fig. 8.35), which has a zero hydraulic head. The base of the soil column is a suitable point to commence solving for the remaining hydraulic heads. Hydraulic heads can subsequently be solved point by point up to the ground surface. Equation 8.41 is nonlinear since the coefcient of permeability kwy varies with the pore-water pressure component of hydraulic head hw. The equation must be solved iteratively with the coefcient of permeability set to constant values that are consistent with the computed
8.3 application to saturated-unsaturated water flow problems 403
402 8 solving saturated/unsaturated water flow problems
hw(i+1) = hw(i1) Ak
wy(i)
qwy(8.39)
pore-water pressures.The coefcient of permeability at each node, kwy, is assumed to be equal to the saturated coefcient of perme-
Substituting hw(i+1) into the ow equation for point iyields the following form: 8kwy(i)n hw(i) + 4kwy(i) + kwy(i+1)
ability, ks , for the rst iteration. The computed hydraulic heads, and subsequently the negative pore-water pressures, are used to revise the coefcients of permeability for the second iteration. The iterative procedure can be repeated
kwy(i1)n
2 Ay hw(i1) Ak
qwy
until there is convergence with respect to the hydraulic heads and the coefcients of permeability. When computing
wy(i)
the hydraulic head at the ground surface, the k
wy(n+1)
value
+ 4kwy(i) + kwy(i1) kwy(i+1)n hw(i1) = 0(8.40)
The hydraulic head can now be solved for point i:
4kwy(i) + kwy(i+1) kwy(i1) 2 Ay
can be assumed to be equal to the kwy(n) value.Typical distributions of pore-water pressure and hydraulichead during steady-state inltration are illustrated in Fig. 8.36. The nonlinear distribution of the pore-water pressure and hydraulic head is produced by the nonlinearity
hw(i) = hw(i1)
2
8kwy(i)
qwyA(8.41)
of the differential seepage equation. The equipotential lines are not uniformly distributed along the soil column. The above analysis is also applicable to steady-state, upward
The nite difference hydraulic head equation (i.e., Eq. 8.41) is in an explicit form. The hydraulic heads can be
ow (e.g., evaporation from ground surface), where the uxqwy is known.
hgn hpnhwn Pore-water Pressure head, hy DatumHydraulic head, hw Gravitational head, hgWater tableqwy
p
hgn
0.9 hwn
0.8 hwn0.7 hwn0.6 hwn0.5 hwn0.4 hwn0.3 hwn
0.2 h
wn
0.1 hwn
(-)
0Head, h
(+)
0
qwy
Figure 8.36 Steady-state inltration through an unsaturated soil with a designated head at the ground surface.
In the case of a heterogeneous, saturated soil, the coef- cients of permeability can be replaced by ks in Eq. 8.41:
4ks(i) + ks(i+1) ks(i1) 2 Ay
a soil mass. The solution can be obtained using closed-form analytical methods, analog methods, or numerical methods. Historically, a graphical method referred to as drawing a flownet has been used to solve the Laplacian equation
hw(i) = hw(i1)
2
8ks(i)
qwy A(8.42)
(Casagrande, 1937).The ownet solution results in two families of curves referred to as ow lines and equipotential lines. The ownet
Equation 8.42 becomes linear when the soil is homoge-
solution has been used extensively to analyze problems
neous:
Ay
shw(i) = hw(i1) Ak qwy(8.43)
involving seepage through saturated soils and is explainedin most soil mechanics textbooks. Boundary conditions for the soil domain must be known prior to the construction of the ownet. Either a head or zero ux is prescribed
Equation 8.43 specializes to a linear distribution of thehydraulic head for a homogeneous, saturated soil column subjected to one-dimensional steady-state ow.
8.3.4 Solution of Two-Dimensional Water Flow ProblemsTwo-dimensional steady-state ow through a homogeneous, isotropic saturated soil must have the x- and y-direction components added together. The result is a partial differ- ential equation referred to as the Laplacian equation. The solution of this equation describes the head at all points in
along the boundary. A boundary condition exception is the case of a free surface. A network of ow lines and equipotential lines is sketched by trial and error in order to satisfy the boundary conditions and the requirement of right-angled, equal-dimensional elements.A hydraulic head boundary condition or an imperme- able boundary condition can readily be imposed for most saturated soil problems. For example, steady-state seepage beneath a sheet pile wall has the boundary conditions shown in Fig. 8.37a. However, the boundary conditions can be more difcult to assign when dealing with unsaturated soils.
Sheet pile wallH1H2
ADatum
BFE
SaturatedCDsoil
HImperviousG(a)
Reservoir levelB
H1Flow lines
Assumed impervious
Free surface and uppermost flowline
ASaturated soil Impervious
Equipotential lines
CDatumDHorizontal drain
Boundary conditions:AB: hw = H1BC: free surface, its location is unkownCD: hw = 0DA: qw = 0
(b)
Figure 8.37 Flownet constructions to solve the Laplacian partial differential equation:(a) steady-state seepage throughout homogeneous, isotropic saturated soil, (b) steady-state seepage throughout homogeneous, isotropic earth dam.
Let us consider steady-state seepage through an earth dam (Fig. 8.37b). In early soil mechanics, the assumption was generally made that there was negligible water ow through the unsaturated soil zone due to its low coefcient of permeability. The phreatic line was assumed to behave as an impervious, uppermost boundary when constructing a ownet. This uppermost boundary (i.e., line BC in (Fig. 8.37b) is considered to be not only a phreatic line but also an uppermost ow line. The uppermost boundary was referred to as a free surface under these special conditions (Freeze and Cherry, 1979). However, the position of the free surface was unknown and its location needed to be approximated prior to constructing the ownet.The position of the free surface was usually determined using an empirical procedure (Casagrande, 1937). The assumption that the free surface is a phreatic line requires that the pore-water pressures be zero along this line. Equipo- tential lines must intersect the free surface (which is also an uppermost ow line) at right angles. In other words, it was assumed that there was no ow across the free surface. The flownet could then be drawn.The ownet technique was mainly suitable for analyz- ing steady-state seepage through isotropic, homogeneous, saturated soils. The ownet technique becomes complex and difcult to use when analyzing anisotropic, heteroge- neous soil systems. There is an inherent problem associated with applying the ownet technique to saturated-unsaturated ow. Freeze (1971) stated that the boundary conditions that are satised on the free surface specify that the pressure head must be atmospheric and the surface must be a streamline. Whereas the rst of these conditions is true, the second is not. The incorrect assumption regarding the uppermost boundary condition can be avoided by recognizing that there can be water ow between the saturated and unsaturated zones (Freeze, 1971; Papagiannakis and Fredlund, 1984).Steady-state ow in the saturated and unsaturated zones can be solved simultaneously using the general partial dif- ferential equation governing water ow through saturated- unsaturated soils. Both zones are treated as a single domain. The water coefcient of permeability in the saturated zone is equal to ks . The water coefcient of permeability kwvaries with respect to the matric suction in the unsaturatedzone. The ownet technique is not applicable to saturated- unsaturated ow modeling since the governing ow equation is not of the Laplacian form. The general ow equation can be solved using a numerical technique such as the nite dif-
Figure 8.38 Typical solutions for saturated-unsaturated ow modeling of various dam sections (after Freeze, 1971).
two-dimensional problems. Figure 8.39 shows the cross section of a dam that has been subdivided into triangular ele- ments. The lines separating the elements intersect at nodal points. The hydraulic head at each nodal point is obtained by solving the governing ow equation for the applied boundary conditions.The nite element formulation for steady-state seepage in two dimensions has been derived using the Galerkin prin- ciple of weighted residuals (Papagiannakis and Fredlund, 1984):
8.3 application to saturated-unsaturated water flow problems 405
404 8 solving saturated/unsaturated water flow problems
ference method or the nite element method. Figure 8.38 shows several typical solutions by Freeze (1971) involvingsaturated-unsaturated ow modeling.
{L} x
T kwx0
8.3.5 Steady-State Seepage Analysis Using Finite
A {L} y
0kwy
Element Method
{L}
The application of the nite element method requires the discretization of the soil mass into elements. Triangular and quadrilateral shapes of elements are commonly used for
x {L}y
dA hwnn
{L}T vw dSp = 0(8.44)S
Reservoir level
10 m
10 m
Datum
Datum
Phreatic line from finite element model Reservoir level9 m10 m
(a)
7 m8 m
(b)
Horizontal drain
6 m5 m4 m3 m2 m1 m
Horizontal drain
Figure 8.39 Discretization of cross section of earthll dam for nite element analysis. (a) Opti- mized nite element mesh. (b) Finite element mesh with superimposed hydraulic head solution.
where:
{L} = matrix of the element area coordinates (i.e.,
L1 = 1/2A (x2y3 x3y2 + (y2 y3 x + (x3 x2 yn L2 = 1/2A (x3y1 x1y3 + (y3 y1 x + (x1 x3 yn
L1L2L3n) and
( + (y y x + (x x yn
L1, L2, L3 = area coordinates of points in the element.
The area coordinates of points in the element are related
L3 = 1/2A
where:
x1y2 x2y11221
to the Cartesian coordinates of nodal points as follows (Fig. 8.40):
xi , yi (i = 1, 2, 3) = Cartesian coordinates of the threenodal points of an element,x , y = Cartesian coordinates of a point within the element,A = area of the element,
kwx00kwy
= matrix of the water coefcients of permeability (i.e., kw ),
hwnn = matrix of hydraulic heads at the hw1 nodal points, that is, hw2 , hw3 vw = external water ow rate in a direc-tion perpendicular to the boundaryof the element, andSp = perimeter of the element.
A simplied form of Eq. 8.44 can be written as follows:
Figure 8.40 Area coordinates in relation to the Cartesian coordi- nates for a triangular element.
[B]T kw A
[B] dA
hwnn S
[L]T vw
dSp
= 0(8.45)
where, [B] , the matrix of the derivatives of the area coor- dinates can be written as
equations. The global equations are solved while satisfying compatibility at each node (Desai and Abel, 1972). Nodal compatibility requires that a particular node shared by the
8.3 application to saturated-unsaturated water flow problems 407
406 8 solving saturated/unsaturated water flow problems
1 1 y2 y3y3 y1y1 y22Ax3 x2x1 x3x2 x1
1(8.46)
surrounding elements must have the same hydraulic head in all of the elements (Zienkiewicz, 1971).Equation 8.45 is nonlinear because the coefcients of per-
Either a hydraulic head or a ow rate must be specied
wnat boundary nodal points. Specied hydraulic heads at the boundary nodes are called Dirichlet boundary conditions. A specied ow rate across the boundary is referred to as a Neumann boundary condition. The second term in Eq. 8.45 accounts for the specied ow rate measured in a direc- tion normal to the boundary. For example, a specied ow rate vw in the vertical direction must be converted to a nor- mal ow rate vw, as illustrated in Fig. 8.41. The normal ow rate is converted to a nodal ow Qw (Segerlind, 1984). Figure 8.41 shows the computation of the nodal ows Qwi and Qwj at the boundary nodes i and j , respectively. A pos- itive nodal ow signies that there is inltration at the node or that the node acts as a source. A negative nodal ow indicates evaporation or evapotranspiration at the node and that the node acts as a sink. When the ow rate across a boundary is zero (e.g., impervious boundary), the second term in Eq. 8.45 disappears.
meability are a function of matric suction, which is related tothe hydraulic head at each nodal point, hwnn. The hydraulicheads are unknown variables in Eq. 8.45 which are solvedby using an iterative procedure. The coefcient of perme- ability within an element is set to a value depending upon the average matric suction at the nodal points. In this way, the global ow equations are linearized and can be solved simultaneously using a Gaussian elimination technique. The computed hydraulic head at each nodal point is again aver- aged to determine a new coefcient of permeability fromthe permeability function kw (ua uw . The above steps arerepeated until the hydraulic heads and the coefcients ofpermeability no longer change by a signicant amount.The hydraulic head gradients in the x - and y -directions can be computed for an element by taking the derivative of the element hydraulic heads with respect to x and y , respectively:
The nite element equation 8.45 can be written for each element and assembled to form a set of global ow
1 ixiy
1 = [B] h
n(8.47)
Figure 8.41 Vertical rainfall converted to applied moisture ow rate across sloping boundary expressed as nodal ows.
where:
ix , iy = hydraulic head gradient within an element in thex - and y -directions, respectively.
The element ow rates vw can be calculated from the hydraulic head gradients and the coefcients of permeability in accordance with Darcys law:
a dam using a saturated-unsaturated nite element seepage analysis. The cross section and discretization of the problem are illustrated in Fig. 8.39. The discretization of the soil con- tinuum is designed in an optimum manner. In other words, there are areas where it is advantageous to have smaller elements and there are other areas where larger elements are adequate. The nonlinear soil properties also inuence the design of the nite element mesh. An automatic nite element mesh design algorithm has been used for all nite
wwhere:
1 vwxvwy
1 = k
[B] hwn
(8.48)
element example problems shown in this book (M.D. Fred- lund, 2005). Consequently, each example problem has its own nite element mesh.Water 10 m high is applied to the upstream of the dam.
vwx,vwy = water ow rates within an element in the x -and y -directions, respectively.
The hydraulic head gradient and the ow rate at nodal points are computed by averaging the corresponding quan- tities from all elements surrounding the node. The weighted average is computed in proportion to the element areas.
8.3.6 Examples of Two-Dimensional, Steady-State Water Flow ProblemsThe following examples are presented to demonstrate the application of the nite element method to steady-state seep- age through saturated-unsaturated soils. Papagiannakis and Fredlund (1984) solved several examples of seepage through
The permeability function used in the analysis is shown as function A in Fig. 8.42. The saturated coefcient of per- meability ks is 1.0 107 m/s. The pore-air pressure is assumed to be atmospheric. Therefore, the matric suction values in Fig. 8.42 are numerically equal to the pore-water pressures. The base of the dam is chosen as the datum.The rst example is an isotropic earth dam with a hori- zontal drainage layer near the downstream toe, as shown in Fig. 8.43. The 10-m-high water on the upstream of the dam gives a 10-m hydraulic head at each node along the upstream face. A zero hydraulic head is specied at nodes along the horizontal drain. Zero nodal ow is specied at nodes along the remaining boundaries. The results of the nite element analysis are presented in Figs. 8.43a and 8.43b.
106
107
Water coefficient of permeability (kw), m/s108
109
1010
1011
Funtion A
Funtion B
1012
1013
0110100Soil suction, kPa
1000
Figure 8.42 Specied permeability functions for analyzing steady-state seepage through earthll dam.
Phreatic line from finite element model
10 m
Datum
Reservoir level
10
910
109
9
876876
876
545342
513214321000
Horizontal drain
Phreatic line from finite element model
(a)
10 m
Datum
Reservoir level
10
109
109
9
87687565
7865
43
1432
124321000
408Horizontal drain
(b)
Figure 8.43 Saturated-unsaturated, steady-state seepage through isotropic earthll dam with horizontal drain: (a) equipotential lines, phreatic line, and nite element mesh in background; (b) equipotential lines, phreatic line, and ow vectors.
The phreatic line resulting from the saturated-unsaturated seepage analysis is in close agreement with the empiri- cal free surface from the conventional ownet construction (Casagrande, 1937). This observation supports the assump- tion that the empirical free surface is approximately equal to a phreatic line. However, water can ow across the phreatic line, as indicated by the nodal ow rate vectors. Water ow across the phreatic line into the unsaturated zone indicates that the phreatic line is not the uppermost ow line as assumed in the ownet solution technique.The difference between the phreatic line (from the nite element analysis) and the free surface (from the ownet technique) decreases as the permeability function for the unsaturated zone becomes steeper. A steep permeability function indicates a rapid reduction in the water coefcient of permeability for a small increase in matric suction. In this case, the quantity of water ow into the unsaturated zone is considerably reduced. This condition approaches the assumption associated with the conventional ownet technique.Equipotential lines extend from the saturated zone through the unsaturated zone, as shown in Fig. 8.43. Changes in hydraulic head between equipotential lines demonstrate that water ows in both the saturated and unsaturated zones. The amount of water owing in the unsaturated zone depends upon the rate at which the coefcient of permeability changes with respect to matric suction.The pore-water pressure heads are computed at all nodes throughout the dam. Pore-water pressure heads are com- puted by subtracting the elevation head from the hydraulic head. Contour lines of equal-pressure heads or isobars are of little value for the interpretation of the computer results for this problem (apart from the phreatic line or zero-pressure line).The ow of water in the saturated and unsaturated zones is approximately parallel to the phreatic line as observed from the ow rate vectors in the central section of the dam. This is not the situation for sections close to the upstream face and the toe of the dam. Near the upstream face of the dam, water ows across the phreatic line from the saturated to the unsaturated zone and continues to ow in the unsaturated zone. The water in the saturated and unsaturated zones then ows essentially parallel to the phreatic line in the central section of the dam. The water in the saturated zone then ows across the phreatic line into the unsaturated zone at the toe of the dam.Figure 8.44 shows steady-state seepage through the above dam cross section when the soil is assumed to be anisotropic. The water coefcient of permeability in the horizontal direction is assumed to be six times larger than in the vertical direction (i.e., kwx = 6kwy). The anisotropyratio is assumed to be constant throughout the dam. Onepermeability function (i.e., function A in Fig. 8.42) is used for the x - and y -directions. The phreatic line is elongated in the direction of the major coefcient of permeability
for the anisotropic case (Fig. 8.44). The saturated zone could exit on the downstream face of the dam for higher anisotropy ratios for the horizontal-to-vertical coefcients of permeability.The third example shows an isotropic earth dam having a core with a lower coefcient of permeability and a hor- izontal drain, as illustrated in Fig. 8.45. The soil forming the shell of the dam has a saturated coefcient of perme- ability ks of 1.0 107 m/s, and the permeability function is in accordance with function A in Fig. 8.42. The core has a saturated coefcient of permeability ks of 1.0 109 m/s and a permeability function in accordance with function B in Fig. 8.42. The boundary conditions used in the analysis are the same as those applied to the previous problems.The results show that most of the hydraulic head change occurs in the region around the core as depicted by the concentrated distribution of equipotential lines in the core zone. As the difference in the coefcients of permeability between the soil and the core increases, greater hydraulic head changes take place in the core. The nodal ow rate vectors also indicate that a signicant amount of water ows upward into the unsaturated zone and over the top of the rel- atively impermeable core (i.e., the siphon effect), as shown in Fig. 8.45b.The fourth example demonstrates the effect of a mois- ture ux boundary condition (i.e., inltration) placed on the isotropic earth dam (Fig. 8.43). The seepage analysis results are presented in Fig. 8.46. Steady-state inltration is simulated by applying a positive nodal ow Qw of 1.0 108 m2/s to each of the nodes along the upper boundaryof the dam. The results can be compared to the case of zero ux across the upper boundary by comparing Figs. 8.43 and 8.46. Inltration results in a rise in the phreatic line. Consequently, the pore-water pressures in the unsaturated zone increase relative to the case where there is zero inl- tration at the ground surface.The fth example demonstrates the development of a seepage face on the downstream of the dam. In this case, there is no horizontal drain and zero nodal ows are specied along the entire lower boundary. There is close agreement between the phreatic line obtained from the nite element analysis and the free surface obtained using the ownet technique (Fig. 8.47). The phreatic line extends to the downstream face of the dam. The phreatic line exits on the downstream face and the portion below the exit point is called the seepage face. The seepage face has a zero pore-water pressure (i.e., atmospheric pressure), boundary condition. In other words, the hydraulic head is equal to the gravitational or elevation head along the seepage face.The location of the exit point is not known prior to per- forming the analysis. Therefore, the location of the exit point must be assumed in order to commence the analy- sis. The exit point is revised following each solution of the nite element solver. The seepage face boundary condition is reevaluated and the solution is repeated.
8.3 application to saturated-unsaturated water flow problems 409
Phreatic line
kwx = 6kwy
10 m
Datum
Reservoir level
10
981098107
987
766554
43
46532
132211000
Horizontal drain
(a)
Phreatic line
kwx = 6kwy
10 m
Datum
Reservoir level
10
981098107
987
766554
43
46532
132211000
410Horizontal drain
(b)
Figure 8.44 Saturated-unsaturated, steady-state seepage through an anisotropic earthll dam with a horizontal drain: (a) equipotential lines, phreatic line, and nite element mesh in the background; (b) equipotential lines, phreatic line, and ow vectors.
Phreatic line
Reservoir level
10
98 7 543
10 m
9 8 7 6 32102
Permeability function A1
10
Datum
8 6 4 2
1
1
000
Core with permeability function B (a)
Horizontal drain
Phreatic line
Reservoir level
10
98 7 543
10 m
9 8 7 6 32102
Permeability function A1
10
Datum
8 6 4 2
1
1
000
411Core with permeability function B(b)
Figure 8.45 Saturated-unsaturated, steady-state seepage through earthll dam with core and horizontal drain: (a) equipotential lines, phreatic line and nite element mesh in background; (b) equipotential lines, phreatic line, and ow vectors.
Horizontal drain
Qw = 108 m2/s
Phreatic line
10 m
Reservoir level
10
10
897
987
6
4565342
10
Datum
987
3
6543
21121000
(a)
Horizontal drain
Qw = 108 m2/s
Phreatic line
10 m
Reservoir level
10
10
897
987
8
6
456534231
10
Datum
976
2
2524311
000
412(b)
Figure 8.46 Saturated-unsaturated steady- state seepage through isotropic earth dam with hor- izontal drain under steady-state rainfall at ground surface: (a) equipotential lines, phreatic line, and nite element mesh in background; (b) equipotential lines, phreatic line, and ow vectors.
Horizontal drain
Phreatic line
10 m
Reservoir level
10
10
98Exit point
798675964
Seepage face
57108
543
Datum
(a)
654
2
1321
Phreatic line
10 m
Reservoir level
10
10
98Exit point
7986
575
4346
Seepage face
1098
7543
1652
Datum
4321
413(b)
Figure 8.47 Saturated-unsaturated steady-state seepage through isotropic earthll dam with impervious lower boundary: (a) equipotential lines, phreatic line, and nite element mesh in background; (b) equipotential lines, phreatic line, and ow vectors.
The nodal ows above the assumed exit point are set at zero while iterating toward a solution. The hydraulic heads at or below the assumed exit point are specied as being equal to the gravitational heads. After convergence, the pore- water pressure head at the node directly above the assumed exit point is examined. A negative pore-water pressure head at this point indicates that the assumed exit point is correct. Otherwise, the seepage face boundary needs to be revised by assuming a higher exit point for the phreatic line. The above procedure is repeated until the correct exit point is obtained. The convergence of the solution to the correct exit point on the downstream face adds one more degree of nonlinearity that needs to be satised when using a numerical solution for saturated-unsaturated seepage problems.The above examples deal with seepage through earth dams. However, steady-state analyses can be applied to many other problems involving saturated-unsaturated ow.
8.3.7 Example of Innite SlopeA slope of innite length is illustrated in Fig. 8.48. Let us consider the case where steady-state water ow is estab- lished within the slope and the phreatic line is parallel to the ground surface. Water ows through both the saturated and unsaturated zones and is parallel to the phreatic line. The direction of the water ow indicates that there is no ow perpendicular to the phreatic line. In other words, the hydraulic head gradient is equal to zero in a direction
perpendicular to the phreatic line. In this case, the lines drawn normal to the phreatic line are equipotential lines.Isobars are parallel to the phreatic line and this condition is similar to that of the central section of a homogenous dam shown earlier. The coefcient of permeability is essentially independent of the pore-water pressure in the saturated zone. Therefore, the saturated zone can be subdivided into several ow channels of equal size. An equal amount of water (i.e., water ux qw) ows through each channel. Lines separatingthe ow channels are referred to as ow lines.The water coefcient of permeability depends on the neg- ative pore-water pressure or the matric suction in the unsat- urated zone. The pore-water pressure decreases from zero at the phreatic line to some negative value at ground sur- face. Similarly, the permeability decreases from the phreatic line to ground surface. Increasingly larger ow channels are therefore required in order to maintain the same quantity of water ow, qw, as ground surface is approached.The water ow in each channel is one dimensional, ina direction parallel to the phreatic line. The coefcient of permeability varies in the direction perpendicular to ow. This condition can be compared to the previous case of water ow through a vertical column. In the case of the vertical column, the coefcient of permeability varied in the ow direction and the equipotential lines were not equally distributed throughout the soil column.The above examples illustrate that equipotential lines and ow lines intersect at right angles for unsaturated ow
8.3 application to saturated-unsaturated water flow problems 415
414 8 solving saturated/unsaturated water flow problems
Figure 8.48 Saturated-unsaturated, steady-state seepage through innite slope with impervious lower boundary.
problems as long as the soil is isotropic. Heterogeneity with respect to the coefcient of permeability results in varying distances between either ow lines or equipotential lines; however, the respective lines must cross at 90.The pore-water pressure distribution in the unsaturated zone can be analyzed by considering a horizontal datum through an arbitrary point (e.g., point A in Fig. 8.49) on the phreatic line. The pore-water pressure distribution in a direction perpendicular to the phreatic line (i.e., in the a- direction) is rst examined. The results are then used to ana- lyze the pore-water pressure distribution in the y -direction (i.e., vertically). The gravitational head distribution in the a-direction is zero at point A( i.e., datum) and increases lin- early to a gravitational head of H cos2 at ground surface. The pore-water pressure head at a point in the a-direction must be negative and equal in magnitude to its gravitational head because the hydraulic heads are zero in the a-direction. Therefore, the pore-water pressure head distribution in the a-direction must start at zero at the datum (i.e., point A) and decrease linearly to H cos2 at ground surface. A pore- water pressure head of H cos2 applies to any point along the ground surface since every line parallel to the phreatic line is also an isobar.The pore-water pressure head distribution in a vertical direction also commences with a zero value at point A and decreases linearly to a head of H cos2 at ground sur- face. However, the pore-water pressure head is distributed along a length H cos in the a-direction, while the head is distributed along a length H in the vertical direction. The
negative pore-water pressure head at a point on a vertical plane can therefore be expressed as follows:
hpi = y cos2 (8.49) where:hpi = negative pore-water pressure head on a vertical plane (i.e., the y -direction) for an innite slope,y = vertical distance from the point under consideration to the datum (i.e., point A), and = inclination angle of the slope and the phreatic line.
When the ground surface and the phreatic line are hori- zontal (i.e., = 0 or cos = 1), the negative pore-water pressure head at a point along a vertical plane, hps, is equal to y. This is the condition of static equilibrium above and below a horizontal water table. The ratio between the pore-water pressure heads on a vertical plane through an innite slope (i.e., hpi = y cos2 ) and the pore-water pres- sure heads associated with a horizontal ground surface (i.e., hps = y) is plotted in Fig. 8.50. The ratio of pressure heads indicates the reduction in the pore-water pressures on a ver- tical plane as the slope becomes steeper.The gravitational head at a point along a vertical plane is equal to its elevation from the datum, y ( Fig. 8.49). The hydraulic head is computed as the sum of the gravitational and pore-water pressure heads:hwi = (1 cos2 y(8.50)
y
-H cos2
Pressure head
aHydraulic head is equal tozero along the a-direction
Datum
HPressure heady
A
H cos2 Gravitationalhead
Saturated zone
Figure 8.49 Pore-water pressure distribution above phreatic line in unsaturated zone during steady-state seepage through innite slope.
8.3 application to saturated-unsaturated water flow problems 417
416 8 solving saturated/unsaturated water flow problems
w B T kA
B dA hwn
n + A
L T L dA hwnnt
w [L]T vS
dS = 0(8.51)
Figure 8.50 Effect of slope inclination on the pore-water pressure distribution along a vertical plane.
The pore-water pressure head equation indicates that there is a decrease in the hydraulic head as the datum is approached. In other words, there is a vertical downward component of water ow. The above analysis also applies to the pore-water pressure conditions below the phreatic line. Using the same horizontal line through point A, positive pore-water pressure heads along a vertical plane can be computed in accordance with Eq. 8.49. The hydraulic head (Eq. 8.50) is zero at the phreatic line and decreases linearly with depth along a vertical plane.
8.3.8 Examples of Two-Dimensional, Unsteady-State Water Flow ProblemsExample problems are presented to illustrate unsteady-state seepage solutions when using the nite element method. The nite element formulation for two-dimensional, saturated- unsaturated seepage analyses is rst presented followed
where:
[B] = matrix of the derivatives of the area coordinates, as shown for the steady-state formulation,[L] = matrix of the element area coordinates (i.e., L1L2L3n), and
2 = wgmw.
Boundary conditions must also be dened for an unsteady-state nite element analysis. Either the hydraulic head or the ow rate must be specied at the boundary nodal points. Specied hydraulic heads at the boundary nodes are called Dirichlet boundary conditions. A specied ow rate across the boundary is referred to as a Neuman boundary condition. The third term in Eq. 8.51 (i.e., the terms just before the equal sign) accounts for the specied ow rates across the boundary. The specied ow rates at the boundary must be projected to a direction normal to the boundary. As an example, a specied ow rate vw in the vertical direction must be converted to a normal ow rate vw. The normal ow rate must in turn be converted to a nodal ow Qw (Segerlind, 1984). A positive nodal ow signies that there is inltration at the node or that the node acts as a source. A negative nodal ow indicates evaporation or evapotranspiration at the node and shows that the node acts as a sink. When the ow rate across a boundary is zero (e.g., impervious boundary), the third term in Eq. 8.51 disappears. The numerical integration of Eq. 8.51 results in the following expression for the saturated-unsaturated seepageequation:[D] hwnn + [E] hwnn = [F ](8.52)
where:
T
by the solution of several example problems. Unsteady-
[D] = stiffness matrix, that is, B
kw B A,
2state problems require that the permeability function and the water storage function (i.e., kw and mw) be known for each soil strata that might become desaturated at any
[E] = capacitance matrix, that is,
A 2 1 1
12 1 2 1 ,1 1 2
elapsed time. Both the permeability function and the water storage function take the form of a nonlinear mathematical function for unsaturated soils.
hwnn = matrix of the time derivatives of the hydraulicheads at the nodal points (i.e., hwnn /t ), and[F ] = ux vector reecting the boundary conditionsT
8.3.9 Unsteady-State Seepage Analysis Using Finite
(i.e., S [L]
vw dS).
Element MethodThe nite element formulation for unsteady-state seepage in two dimensions can be derived using the Galerkin princi- ple of weighted residuals (Lam et al., 1987). The Galerkin
The time derivative in Eq. 8.52 can be approximated usinga nite difference form. The relationship between the nodal heads of an element at two successive time steps At can be expressed using the central-difference approximation
solution to the governing seepage equation is given by the 1
2 [E]
1 2 [E]n
following integrals over the area and the boundary surface of a triangular element:
[D] +At
hwnnt +At =
[D]At
hwn
t + 2 [F ](8.53)
or the backward-difference approximation
nsolving highly nonlinear PDEs. The design of the nite element mesh becomes dynamic, changing as necessary to meet the requirements for convergence to a reliable solution.
1[D] +
[E] hAtwn
t +At =
[E] At
hwnnt
+ [F ](8.54)
The unsteady-state seepage equation is assumed to be solved for one time step once the converged nodal hydraulic heads of the system have been obtained. Having reached
The above time derivative approximations are considered to be unconditionally stable. The central-difference approx- imation generally gives a more accurate solution than that obtained from the backward-difference approximation. However, the backward-difference approximation is found to be more effective in reducing numerical oscillations commonly encountered in highly nonlinear systems of ow equations (Neuman and Witherspoon, 1971; Neuman, 1973). The nite element seepage equation can be written for each element and assembled to form a set of global ow equations. The nite element formulation is solved while
convergence at a particular time step, other secondary quan- tities, such as pore-water pressures, hydraulic head gradi- ents, and water ow rates, can then be calculated using the converged nodal hydraulic heads. The equation for nodal pore-water pressures is uwnn = ( hwnn ynn wg(8.55) where: uwnn = matrix of pore-water pressures at the nodal
satisfying nodal compatibility (Desai and Abel, 1972). Nodalcompatibility requires that a particular node shared by the surrounding elements has the same hydraulic head in all
i.e.points ,
uw1 uw2 , uw3
shared elements (Zienkiewicz, 1971; Desai, 1975b).The global ow equations for the whole system are solvedfor the hydraulic heads at the nodal points, hwnn. However,Eq. 8.52 is nonlinear because the coefcients of permeabilityare a function of matric suction, which is related to the hydraulic head at the nodal points.The hydraulic heads are unknown variables in Eq. 8.52. The equation must be solved using an iterative procedure that involves a series of successive approximations. For the rst approximation, the coefcients of permeability are esti-
ynn = matrix of elevation heads at the nodal points y1 i.e., y .2 y3
The hydraulic head gradients in the x - and y -directions can be computed for an element by taking the derivative of the element hydraulic heads with respect to x and y , respectively:
mated in order to calculate the rst set of hydraulic heads at the nodal points. The computed hydraulic heads are used to calculate the average matric suction within an element. In the subsequent approximations, the coefcient of perme- ability is adjusted to a value corresponding to the computed
wnwhere:
1 ixiy
1 = [B] h n
(8.56)
average matric suction in the element. The adjusted per- meability value is then used to calculate a new set of nodal point hydraulic heads. The above procedure is repeated until both the hydraulic head and the permeability differences for each element are smaller than a specied tolerance between two successive iterations.The above iterative procedure allows the global ow
ix , iy = hydraulic head gradient within an element in thex - and y -directions, respectively.
The element ow rates vw can be calculated from the hydraulic head gradients and the coefcients of permeability in accordance with Darcys law:
equations to be solved using a Gaussian elimination tech- nique. The convergence on hydraulic head and coefcient of permeability is dependent on the degree of nonlinearity of the permeability function and the spatial discretization of the continuum. A steep permeability function generally
where:
1 vwxvwy
1 = k
w B hwn
n(8.57)
requires an increased number of iterations to achieve convergence. A ner discretization in both element size and time step assists in obtaining more rapid convergence with a smaller tolerance. The solution often converges to within a tolerance of less than 1% in a few iterations provided the permeability and water storage functions are not too highly nonlinear. The use of PDE solvers with optimized mesh design and automatic mesh renement is of great value in
vwx, vwy = water ow rates within an element in the x - and y -directions, respectively.
The hydraulic head gradient and the ow rate at nodal points are computed by averaging the corresponding quan- tities from all elements surrounding the node. The weighted average is computed in proportion to the respective elemen- tal areas.
8.3.10 Example of Water Flow through Earth DamThe example problem involving water ow through an earth- ll dam is rst discussed. The soil is assumed to be isotropic
coefcients of permeability are equal to 5.0 106 and
21.0 105 m/s for the liner and the surrounding soil, respectively. A water storage modulus mw of for both soils
8.3 application to saturated-unsaturated water flow problems 419
418 8 solving saturated/unsaturated water flow problems
with respect to its coefcient of permeability and the perme- ability function along with the SWCC and the water storage modulus function are shown in Fig. 8.51. The saturated coef- cient of permeability ks is 1.0 107 m/s. The pore-air pressure is assumed to remain at atmospheric pressure. The base of the dam is selected as the datum for elevation head.The reservoir is initially assumed to be empty so the water level is at 0 m (i.e., the datum). The water level in the reservoir is then instantaneously raised to a level of 10 m above datum. The water level remains constant at 10 m as transient or unsteady-state seepage takes place in the dam. The rise of the phreatic line from the initial steady-state condition to an elapsed time of 1500 days is computed.The development of equipotential lines, the phreatic sur- face, and water ow vectors through the dam are illus- trated for four elapsed times during the transient process. Figure 8.52 shows the development of the phreatic lines, the ow vectors, and the equipotential lines after an elapsed time of 25 days. The dynamically generated nite element mesh corresponding to the converged solution is also shown. Similar plots of the phreatic lines, the ow vectors, and the equipotential lines after an elapsed time of 60 days are shown in Fig. 8.53. Likewise, the development of seepage toward steady-state conditions is shown for elapsed times of 120 and 1500 days in Figs. 8.54 and 8.55, respectively.The increase in the reservoir level results in an increase in pore-water pressures with time. This is demonstrated by the advancement of equipotential lines from upstream to down- stream of the dam with elapsed time. It should also be noted that the equipotential lines extend from the saturated through the unsaturated zones. In other words, water ows in both the saturated and the unsaturated zones as a result of the hydraulic head differences between the equipotential lines. The ow of water in both zones can be observed directly from the ow rate vectors that exist in both the saturated and unsaturated zones. The amount of water owing in the unsat- urated zone depends on the rate of change in the coefcient of permeability with respect to matric suction changes.
8.3.11 Example of Groundwater Seepage below LagoonThe second example problem illustrates unsteady-state groundwater seepage below a lagoon. The lagoon is placed on top of a 1-m-thick soil liner. The geometry of the problem is symmetrical, and the liner and the surrounding soil are assumed to be isotropic with respect to coefcient of permeability. The problem is analyzed by considering half of the geometry. The SWCC is assumed to be the same for the soil liner and the underlying soil and is shown in Fig. 8.56a. The permeability functions for the soil liner and the surrounding soil are shown in Fig. 8.56b. The saturated
is shown in Fig. 8.56c.The discretized cross section of the soil liner and its sur- rounding soil are depicted in Fig. 8.57. Initially the ground- water table was located 5 m below the ground surface. The lagoon was assumed to be instantaneously lled with water to a depth of 1 m. No-ow boundary conditions were assumed along the ground surface outside the lagoon, the bottom boundary, and the vertical axis of symmetry. On the right-hand boundary, a hydrostatic condition was assumed to exist below the groundwater table. A no-ow condition was assumed to exist above the groundwater table to the ground surface. The exit point of the groundwater table was assumed to be xed with elapsed time.The transient modeling process commenced as the lagoon was lled with water to the 1-m height which gave rise to a 1-m pore-water pressure head. The water commenced seeping from the lagoon, causing the groundwater to grad- ually mound to its nal steady-state condition. Figure 8.57 illustrates the transient positions of the water table at an elapsed time of 24 h. Figure 8.57a shows the pressure head contours, including the phreatic line, along with the automat- ically generated nite element mesh required for solving the problem at 24 h of elapsed time. It can be seen that there is a slight rise in the groundwater table 5 m below the ground surface. At the same time there is a phreatic line that has developed at the 0.5-m depth below ground surface as water moves downward from the lagoon. Figure 8.57b shows the equipotential heads along with the ow vectors correspond- ing to an elapsed time of 24 h. The water level in the lagoon remains constant at a 1-m height throughout the transient process.The seepage ow pattern and the development of pore- water pressures in the soil below the lagoon after an elapsed time of 40 h are shown in Figs. 8.58a and 8.58b. At the beginning of the transient process, water moved downward from the lagoon, while the position of the groundwater table showed little effect. With time, the wetting front moves deeper into the soil mass while the groundwater begins to mound upward. After 52 h, the downward wetting front from the lagoon joins the rising groundwater table, as shown in Fig. 8.59. Steady-state conditions are approached after an elapsed time of about 200 h, as shown in Fig. 8.60. The length of time required to approach steady-state seepage conditions is largely controlled by the saturated coefcients of permeability for the soils.
8.3.12 Example of Seepage within Layered Hill SlopeThe third example problem illustrates unsteady-state seepage within a layered hill slope under constant-inltration condi- tions. Rulon and Freeze (1985) studied this problem using a sandbox model of the layered hill slope. Details of the geom- etry of the slope and the applied surface moisture ux are
0.5
Volumetric water content, 0.4
0.3
s = 0.495af = 19.37 kPanf = 3.29mf = 0.828r = 67.6 kPa
0.2
0.1
0.00
106107
Coefficient of permeability, m/s1081091010101110121013101410151016
110100100010,000100,000106Soil suction, kPa (a)
0.012
0110100100010,000100,000106Soil suction, kPa (b)
0.010
Water storage modulus0.008
0.006
0.004
0.002
0.0000.1110100100010,000100,000106Soil suction, kPa (c)Figure 8.51 Permeability and water storage function for analyzing unsteady-state seepage through an isotropic earthll dam. (a) SWCC. (b) Permeability function. (c) Water storage function.
10 m
1Reservoir level22
10 m
9 83
71059 8 7 6531042
98 7 6 5 4103
100
Time = 25 days (a)
10 m
1Reservoir level22
10 m
9 83
71059 8 7 6531042
42098 7 6 5 4103
100
Time = 25 days (b)Figure 8.52 Hydraulic heads after an elapsed time of 25 days. (a) Finite element mesh and hydraulic heads. (b) Velocity vectors and hydraulic heads.
Figure 8.53 Hydraulic heads after an elapsed time of 60 days. (a) Finite element mesh andhydraulic heads. (b) Velocity vectors and hydraulic heads.
10 m
10 m
0
Reservoir level
10
3
2987 6 41098 7 65410239876 542110
Time = 60 days (a)
10 m
10 m
0
Reservoir level
10
3
2987 6 41098 7 65410239876 542110
421Time = 60 days (b)
10 m
10 m
0
Reservoir level
10
10
4 3
87 62109
987 6 5432
98765432110
Time = 120 days
(a)
10 m
10 m
0
Reservoir level
10
10
4 3
87 62109
987 6 5432
98765432110
422Time = 120 days
(b)
Figure 8.54 Hydraulic heads after an elapsed time of 120 days. (a) Finite element mesh and hydraulic heads. (b) Velocity vectors and hydraulic heads.
Figure 8.55 Hydraulic heads after an elapsed time of 1500 days. (a) Finite element mesh andhydraulic heads. (b) Velocity vectors and hydraulic heads.10 m
10 m
0
Reservoir level
10
10
10
987
987
987
Time = 1500 days (a)
6
6554
5643
34221
42310 m
10 m
0
Reservoir level
10
10
10
987
987
987
Time = 1500 days (b)
6
6554
5643
34221
s = 0.495af = 17.05 kPanf = 1.17mf = 0.729r = 367 kPa0.5
Volumetric watercontent, 0.4
0.3
0.2
0.1
8.3 application to saturated-unsaturated water flow problems 425
424 8 solving saturated/unsaturated water flow problems
0.00110100
1000
10,000 100,000
106
103104
Coefficient of permeability, m/s105 106107108109101010111012101310141015
Soil suction, kPa (a)
Soil liner
Soil
1016 0110
100
1000
10,000 100,000
106
0.012
Soil suction, kPa (b)
0.010
Water storage modulus0.008
0.006
0.004
0.002
0.000
0110100
1000
10,000 100,000
106
Soil suction, kPa (c)Figure 8.56 Permeability and water storage function for analyzing unsteady-state seepage below a lagoon. (a) SWCC. (b) Permeability function. (c) Water storage function.
Water level in lagoon
1 m
109998887 67766
Water table
Time = 24 h (a)
Water level in lagoon
1 m
10 10 10 9998887 67766
Water table at 5 m
426 8 solving saturated/unsaturated water flow problems
8.3 application to saturated-unsaturated water flow problems 425
Figure 8.58 Hydraulic heads after an elapsed time of 40 hours. (a) Finite element mesh andhydraulic heads. (b) Velocity vectors and hydraulic heads.
Figure 8.57 Hydraulic heads after an elapsed time of 24 hours. (a) Finite element mesh andhydraulic heads. (b) Velocity vectors and hydraulic heads.
Time = 24 h (b)
Datum
Water level in lagoon 1 m10 10 10 9998878
767
6Water table
6
Time = 40 h (a)
Water level in lagoon
1 m
10 10 10 9998878
767
6Water table at 5 m
6
Time = 40 h (b)
Datum
Water level in lagoon 1 m10 10 10 9998
8786
7
76Water table
6
Time = 52 h (a)
Water level in lagoon
1 m
10 10 10 9998
8786
7
76Water table
6
Figure 8.60 Hydraulic heads after an elapsed time of 200 hours. (a) Finite element mesh andhydraulic heads. (b) Velocity vectors and hydraulic heads.
Figure 8.59 Hydraulic heads after an elapsed time of 52 hours. (a) Finite element mesh andhydraulic heads. (b) Velocity vectors and hydraulic heads.
Time = 52 h (b)
Datum
Water level in lagoon
1 m
1010 109
99876
8.3 application to saturated-unsaturated water flow problems 429
428 8 solving saturated/unsaturated water flow problems
876
Water table
8
76
Time = 200 h (a)
Water level in lagoon
1 m
1010 109
99876
876
Water table
8
76
Time = 200 h (b)
Datum
Figure 8.61 Geometry and boundary conditions for an example involving layered soils in a hillslope.
shown in Fig. 8.61. The physical model was instrumented with pore-water pressure measuring devices. The slope con- sisted of medium sand with a horizontal layer of ne-grained sand. The ne sand had a lower coefcient of permeabil- ity than the medium-grained sand. The intent was to study whether the ne sand layer would impede water ow and create a seepage face on the slope. The steady-state results from the experimental measurements are shown in Fig. 8.62. The steady-state equipotential lines and the phreatic water surface are presented. The experimental study forms a valu- able benchmark problem against which numerical models can be tested.The soil properties of the materials involved are shown in Fig. 8.63. The two sand materials are assumed to have the same SWCCs (Fig. 8.63a); however, the saturated coef- cients of permeability for the two sand materials were different (Fig. 8.63b). The saturated coefcient of perme- ability for the medium sand was 1.4 103 m/s and the ne sands had a saturated coefcient of permeability of5.5 105 m/s (Rulon and Freeze, 1985). The water stor-age function for both sands was obtained by differentiating the volumetric SWCC and the result is shown in Fig. 8.63c. The constant rate of inltration of 2.1 104 m/s was applied at the top of the hill slope in the experimental model
Figure 8.62 Comparison of observed water table and pore-water pressures in experimental tank with that predicted using Neu- mann model under steady-state conditions (from Rulon and Freeze, 1985).
and in the numerical simulation. The results of pore-water pressure measurements can be compared with the computed results from a numerical model. Close agreement has been observed between pore-water pressure measurements and the results of an unsteady-state nite element analysis.An initial steady-state condition was assumed with the water table located at a height of 0.3 m from the toe of the slope. At an elapsed time taken as zero, a constant inltration rate of 2.1 104 m/s was applied to the top portion of the hill slope. The development of the groundwater table from its initial steady-state condition (i.e., at time equal to zero) to an elapsed time of 100 s is shown in Fig. 8.64. The water level at the toe of the slope and the inltration rate at the top of the slope were assumed to remain constant throughout the transient process. No ow boundary conditions were assumed along the remainder of the slope boundaries.The seepage ow pattern and the development of the equipotential lines within the slope continued to change with elapsed time. The nite element mesh was also auto- matically rened as necessary, internal to the analysis. The equipotential lines and the phreatic line corresponding to an elapsed time of 200 s are shown in Fig. 8.65. The results for an elapsed time of 220 s are shown in shown in Fig. 8.66. The results show a gradual rise in the lower phreatic water surface and a gradual change in the equipotential lines. At the start of the inltration process, water inltrates vertically toward the impeding layer. As the inltration of water continues, water moves through the impeding layer, causing the groundwater table to rise. After 240 s a perched water table develops on the impeding layer and moves toward the face of the slope (see Fig. 8.67). After 260 s a wedge-shaped unsaturated zone is formed, and the perched water table moves toward the edge of the slope, as shown in Fig. 8.68. Now, two seepage faces began to develop. One seepage face is near the toe of the slope and the other seepage face develops at the top of the impeding layer. In other words, the presence of the impeding layer results in a complex congura- tion for the groundwater table and the position of the equipo- tential lines. A steady-state condition is established after about 280 s. The steady-state results are shown in Fig. 8.69. There is close agreement between the results of the physical model measurements (Rulon and Freeze, 1985) and the numerical model results. The positions of the developed water table, theseepage faces, and pore-water pressures are similar.The nite element numerical analyses presented for the three example problems have illustrated the way in which unsteady-state modeling can be performed for saturated- unsaturated ow problems. The application of a saturated ow model to each of these problems would be virtually impossible to undertake. However, the use of a combined saturated-unsaturated ow is quite straightforward. The ow systems that developed throughout the unsaturated soil sys- tem can be complex, depending upon the contrast in the coefcients of permeability and the water storage modulus for the different soils.
430 8 solving saturated/unsaturated water flow problems
8.3 application to saturated-unsaturated water flow problems 429
0.5
430 8 solving saturated/unsaturated water flow problems
Volumetric water content, 0.4
0.3
0.2
s = 0.495af = 5.19 kPanf = 20mf = 0.306r = 9.42 kPa
0.1
0.0
0110100
1000
10,000100,000
106
102103
104
Coefficient of permeability, m/s10510610710810910101011101210131014
10151016
Soil suction, kPa (a)
Soil 2
Soil 1
0110100
1000
10,000100,000
106
Soil suction, kPa (b)
0.14
0.12
Water storage modulus0.10
0.08
0.06
0.04
0.02
0.00 0110100100010,000100,000106Soil suction, kPa (c)
Figure 8.63 Permeability and water storage functions for analyzing unsteady-state seepage below a lagoon. (a) SWCC. (b) Permeability functions. (b) Water storage function.
I_nfiltrationVw = 2.1 x 104m/s
0.4
0.4
0.4
Groundwater table
0.3
0.35
0.3
0.35
0.35
0.3 m
0.25
0.25
0.3
4310.25
Time = 100 s
Figure 8.64 Finite element mesh and hydraulic heads after an elapsed time of 100 s.
In_filtrationVw = 2.1 x 104 m/s
0.65
0.65
0.6
Groundwater table
0.35
0.4
0.45
0.4
0.55
0.45
0.55 0.6
0.550.45
0.3 m
0.250.250.25
0.3
0.3
0.3
0.35
0.35
0.4
432Time = 200 s
Figure 8.65 Finite element mesh and hydraulic heads after an elapsed time of 200 s.
I_nfiltration
wV = 2.1 x 104 m/s
0.7
0.4
0.450.5
0.45
0.6
0.65
0.5
0.6 0.65
0.7
0.60.5
0.7
0.3 m
0.25
0.25
0.3
0.3
0.3
0.35
0.35
0.35
0.4
0.4
0.45
433
Time = 220 s
Figure 8.66 Finite element mesh and hydraulic heads after an elapsed time of 220 s.
I_nfiltrationVw = 2.1 x 104 m/s
0.75
0.8
0.8
0.8
0.45
0.50.55
0.5
0.650.7
0.6 0.650.55
0.70.75
0.6
0.7
0.3 m
0.250.25
0.3
0.3
0.35
0.35
0.4
0.4
0.4
0.45
0.45
0.5
0.55
4340.30.35
Time = 240 s
Figure 8.67 Finite element mesh and hydraulic heads after an elapsed time of 240 s.
I_nfiltrationVw = 2.1 x 104m/s
0.85
0.5
0.550.6
0.70.75
0.8
0.6
0.7
0.75 0.8
0.85
0.750.65
0.85
0.3 m
0.250.25
0.3
0.3
0.35
0.35
0.4
0.4
0.4
0.45
0.45
0.45
0.5
0.5
0.55
0.55
0.6
4350.30.35
Time = 260 s
Figure 8.68 Finite element mesh and hydraulic heads after an elapsed time of 260 s.
I_nfiltrationVw = 2.1 x 104m/s
0.95
0.55
0.60.65
0.75
0.85
0.9
0.750.65
0.85
0.95
0.850.7
0.95
0.3 m
0.25 0.3
0.35
0.4
0.4
0.45
0.45
0.5
0.5
0.55
0.55
0.6
0.6
0.65
0.25
0.350.3
0.45
0.5
0.3
0.350.4
436
Time = 280 s
Figure 8.69 Finite element mesh and hydraulic heads after an elapsed time of 280 s.
10.4 formulation of partial differential equations for conductive heat flow 499
10.4.4 Heat Flow Equations Including Vapor FlowThere can also be vapor ow associated with a thermal gra- dient; however, vapor ow is closely associated with ow in the pore-air and pore-water phases. The components related to vapor ow are mentioned below, but further details related to the theory and solution to practical engineering problems are mentioned under the topic of coupling liquid and vapor ow through unsaturated soils.Fredlund and Gitirana (2005) presented the vapor ow in terms of two components: one due to vapor diffusion within the air phase and a second due to advection within the free pore-air. The velocity of vapor ow can be written as follows:
the above equation: the thermal conductivity function, the vapor conductivity function (diffusion), and the vapor advec- tion function. These soil property functions can vary with the stress state (e.g., soil suction), and therefore, the partial differential equations are nonlinear.
10.4.5 Temperature and Heat Flux Boundary ConditionsThere are two common thermal boundary conditions asso- ciated with the solution of heat ow problems: Dirichlet- type boundary conditions (i.e., the primary variable specied is temperature T ) and Neumann-type boundary conditions (i.e., the derivative of the primary variable or the heat ux
vvvdva
k vd uw
k vduwT
qh). The Neumann heat ux can be specied as zero when
y = vy + vy = w
akva u
+yw T + 273.15 y
(10.29)
the boundary is perfectly insulated. The boundary condi- tion can also be considered as a zero heat ux bound- ary under conditions of symmetry for the material being
where:
vv
a y
modeled (e.g., a two-dimensional section out of a laterally extending continuum).Most geotechnical engineering problems involve exposure
y = total velocity of vapor ow, m/s,
vvdy = velocity of vapor ow by diffusion through the airphase, m/s,
vvay = velocity of vapor ow by advection in the air phase,m/s,kvd = pore-water vapor conductivity by vapor diffusionwithin the air phase; which can be written as
to the ground surface where weather conditions are rapidlyand continuously changing. The nighttime and morning peri- ods are generally cool relative to the midday temperatures. Also, the temperatures from one day to the next can change signicantly, as can the conditions from one month to the next. In other words, the temperature boundary condition at ground surface is in an unsteady or transient state.
vvuair w
Dv
, m/s
Some geotechnical engineering problems can be mod-eled using average daily temperature values. However, other
wR(T + 273.15) wkva = pore-water vapor conductivity by advection withinthe free pore-air, which can be written as va a
engineering problems may require a more rigorous represen- tation of the thermal boundary conditions (e.g., a mathemati- cal function approximating changes in temperature through- out the day). Weather stations commonly collect data and
Daw
, m/s
report the information on a daily basis (Fig. 10.13). Quite often the average daily temperature is recorded along with
w = unit weight of water, kN/m3,v = density of the vapor, kg/m3,w = density of water, kg/m3,
uaira = density of air, kg/m3, v = vapor pressure, kPa, ua = pore-air pressure, kPa,uw = pore-water pressure, kPa,T = temperature, C,v = molecular weight of water vapor, 18.016 kg/kmol,R = universal gas constant, 8.314 J/(mol.K),Dv = molecular diffusivity of vapor through soil, m2/s, andDa = (1S)nDvv/RT, kgm/kNs.
The above equation shows that the ow and storage of heat within a saturated-unsaturated soil can be a function of three primary variables: uw, ua , and T . Volume changes associated with the relative phases of the soil are not explicitly pre- sented as primary variables; however, volume-mass changes affect thermal conductivity and volumetric heat capacity. Sev- eral unsaturated soil property functions can be identied in
values for the maximum and minimum temperatures for the day. Temperature data can also be collected on an hourly basis but the amount of data becomes excessive.Temperature as a Dirichlet-type boundary condition can readily be applied at the ground surface. The temperature may be assumed to vary according to a particular pattern during the day. For example, temperature may take the form of a sine function with the peak temperature near noontime. It is also possible to use measured hourly temperatures as input for numerical modeling purposes.Temperature at a weather station is generally recorded at 1 or 2 m above ground surface. It is the ground surface tem- perature, however, that forms the more accurate boundary condition for modeling geotechnical engineering problems. The soil temperature at the ground surface may not be the same as the air temperature above the ground surface. The difference in temperature between the air and the soil can be taken into consideration when performing coupled heat and moisture ow analyses for the calculation of actual evaporation at ground surface. Wilson (1990) suggested a
10.11 aldrich (1956) example of vertical column 519
0.2
0
0.2
Frost depth (m)0.4
0.6
0.8
1
1.2
1.4
Analytical solution (Aldrich, 1956)
Thermal conductivity , = 2.873 W/m/K Volumetric water content = 47% Latent heat = 3.34 108 J/m3 C
SV Heat Solution
1.6
050100
150
200
250
300
350
Time (days)
Figure 10.40 Depth of freezing during soil freezing and subsequent thawing.