Environmental Engineering Reference
In-Depth Information
0.0
-0.4
-0.8
E = .9631E+09
E = .9631E+06
E = .9631E+05
0
200
400
600
800
Production time (days)
Figure 14.11.
Effects of reservoir compaction on reservoir pressure, for three different
reservoir rocks (redrawn with permission from [LEW 98b])
14.3.5.2.
Immiscible pollutant
DNAPL exist as separated fluid phases in ground water and are classified into a
single class of subsurface hazardous pollutants. With the numerical model of the
previous example it is possible to simulate a situation where such a fluid, e.g. PCB
oil, which is heavier than water, displaces water in a phreatic aquifer.
The test example involves a vertical sand column of 1 m in height, initially fully
saturated with water, with a porosity of 0.4, a bulk modulus
K
s
= 1 x 10
12
N/m
2
,
Young's modulus
E
= 1.3 x 10
6
N/m
2
and a Poisson ratio ν = 0. At time instant
t
= 0, the pollutant is injected continuously at the top of the column where a fixed
saturation degree of 0.5 is maintained for the pollutant. The result is a displacement
of water from the column by the encroaching NAPL. At the bottom of the column
the water flows out with a capillary pressure equal to zero, i.e. water and
pollutant pressures are constant and equal to the reference pressure. The viscosity
of the pollutant is 0.57 x 10
-3
Ns/m
2
and its density 1,461 kg/m
3
. Water density
is 1,000 kg/m
3
and its viscosity 1.0 x 10
-3
Ns/m
2
. The following two values of
the intrinsic soil permeability have been taken into account: 1.0 x 10
-11
m
2
and
1.5 x 10
-12
m
2
. Brooks and Corey [BRO 66] functions are used to relate the capillary
and pore pressure distribution index to saturation and relative permeability. In
particular, the irreducible saturation is set to
S
wc
= 0.086, pore size distribution index
is λ = 2.0 and bubbling pressure
p
b
= 0.1784. The lateral surfaces are impermeable
to fluids and have fixed horizontal displacements.
The vertical and horizontal displacements are also fixed at the bottom surface.
On the top surface
p
w
= p
ref
−
p
sw
=0.5
, the air pressure
p
a
= p
ref
. p
sw
=0.5
means the