Environmental Engineering Reference
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where:
where:
k
w
=
water coefficient of permeability in the
x
- and
y
-
directions.
v
wx
=
water flow rate across a unit area of the soil in the
x
-direction.
Table 7.5 summarizes the relevant equations for
two-dimensional steady-state flow through unsaturated soils.
Seepage through a dam involves flow through the unsatu-
rated and saturated zones. For the
saturated
portion, the water
coefficient of permeability becomes equal to the saturated
coefficient of permeability
k
s
. The saturated coefficients of
permeability in the
x
- and
y
-directions,
k
sx
and
k
sy
, respec-
tively, may not be equal due to anisotropy. The saturated
coefficients of permeability may vary with respect to location
and give rise to heterogeneity (i.e., different soil zones).
Therefore, the net flux in the
x
- and
y
-directions is
∂
v
wx
∂x
dx dy dz
∂
v
wy
∂y
+
=
0
(7.34)
Substituting Darcy's laws into the net flux equation results
in the following nonlinear partial differential equation:
k
wx
(u
a
−
k
wy
(u
a
−
∂
∂x
u
w
)
∂h
w
∂x
∂
∂y
u
w
)
∂h
w
∂y
+
=
0
(7.35)
where:
7.4.7 Three-Dimensional Flow through
Unsaturated Soil
The real world is defined in terms of three spatial dimen-
sions. It is increasingly becoming appropriate to simulate
real-world flow systems using a three-dimensional flow anal-
ysis. Three-dimensional flow can be formulated by expand-
ing the two-dimensional flow equation to include the third
dimension. The three-dimensional equation is once again
derived based on continuity, and the equation is referred to
as an uncoupled equation of saturated-unsaturated flow.
Let us consider an unsaturated soil having heterogeneous,
anisotropic conditions (Fig. 7.21b). The coefficient of per-
meability at a point varies in the
x
-,
y
-, and
z
-directions.
However, the permeability variations in the three directions
will be assumed to be governed by the same permeability
function. Figure 7.25 shows a cubical soil element with water
flow in the
x
-,
y
-, and
z
-directions. The soil element has
infinitesimal dimensions of
dx
,
dy
, and
dz.
The flow rates
v
wx
,
v
wy
, and
v
wz
are assumed to be positive when water flows
in the positive
x
-,
y
-, and
z
-directions. Continuity for three-
dimensional, steady-state flow can be satisfied as follows:
v
wx
+
k
wx
(u
a
−
u
w
)
=
water coefficients of permeability as a
function of matric suction, the perme-
ability is able to vary with location in
the
x
- direction, and
∂h
w
/∂x
=
hydraulic head gradient
in the
x
-
direction.
For the remainder of the formulations, the coefficients
of permeability
k
wx
u
a
−
u
w
are written
as
k
wx
and
k
wy
, respectively, for simplicity. Equation 7.35
describes the hydraulic head distribution in the
x
-
y
plane for
steady-state water flow. The nonlinearity of the partial dif-
ferential seepage equation becomes clearer when it is written
in an expanded form:
u
w
and
k
wy
u
a
−
∂
2
h
w
∂x
2
∂
2
h
w
∂y
2
∂k
wy
∂y
∂k
wx
∂x
∂h
w
∂x
∂h
w
∂y
k
wx
+
k
wy
+
+
=
0 (7.36)
where:
∂k
wx
/∂x
=
change in water coefficient of permeability in
the
x
-direction.
v
wx
dy dz
v
wy
+
v
wy
∂
v
wy
∂y
∂
v
wx
∂x
dx
−
+
dy
−
Equation 7.36 contains two unknowns, namely, hydraulic
head and the coefficient of permeability. These two variables
are related by another equation referred to as the permeabil-
ity function. The two equations can be solved through use
of an iterative process where the coefficient of permeabil-
ity is first assumed and the hydraulic heads are solved. The
process is repeated with continuously improved assumptions
regarding the coefficients of permeability (through use of the
permeability function). The process is repeated until conver-
gence is reached with respect to the assumed and calculated
coefficient of permeability and hydraulic heads.
For the
heterogeneous
,
isotropic
case, the coefficients of
permeability in the
x
- and
y
-directions are equal (i.e.,
k
wx
=
k
wy
=
v
wz
+
v
wz
dx dy
∂
v
wz
∂z
×
dx dz
+
dz
−
=
0
(7.38)
Table 7.5 Two-Dimensional Steady-State Equations
for Unsaturated Soils
Heterogeneous, anisotropic:
∂
2
h
w
∂x
2
∂
2
h
w
∂y
2
∂k
wy
∂y
∂k
wx
∂x
∂h
w
∂x
+
∂h
w
∂y
=
k
wx
+
k
wy
+
0
Heterogeneous, isotropic:
k
w
∂
2
h
w
∂x
2
k
w
). Therefore, Eq. 7.36 can be written as follows:
k
w
∂
2
h
w
∂x
2
∂
2
h
w
∂y
2
∂k
w
∂x
∂h
w
∂x
∂k
w
∂y
∂h
w
∂y
∂y
2
∂
2
h
w
+
+
+
=
0
∂k
w
∂x
∂h
w
∂x
∂k
w
∂y
∂h
w
∂y
+
+
+
=
0
(7.37)
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