Dr YZhou Lecture Notes 205
Groundwater Hydraulics
w
0
L
x
h
h
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K
h
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Table of Contents
1. Introduction...............................................................................................................................................1
1.1 Hydrological Cycle...........................................................................................................................1
1.2 Aquifer systems ................................................................................................................................3
1.2.1 Aquifer, aquitard, and aquiclude ................................................................................................3
1.2.2 Aquifer types..............................................................................................................................3
1.3 Vertical distribution of groundwater .............................................................................................4
1.4 Groundwater flow systems ..............................................................................................................6
2. Fundamental Equations of Groundwater Flow.......................................................................................9
2.1 Darcy's Law......................................................................................................................................9
2.2 Generalization of Darcy's Law......................................................................................................13
2.3 Equation of continuity ...................................................................................................................15
2.4 Basic equations for steady incompressible flow...........................................................................17
2.5 Basic equations for non-steady compressible flow ......................................................................18
3. Steady Groundwater Flow in Aquifers...................................................................................................23
3.1 Groundwater flow in a confined aquifer......................................................................................23
3.1.1 Conceptual hydrogeological model..........................................................................................23
3.1.2 Mathematical model.................................................................................................................23
3.1.3 Analytical solution ...................................................................................................................24
3.1.4 Analysis of solution..................................................................................................................25
3.2 Groundwater Flow in an Unconfined Aquifer.............................................................................27
3.2.1 Conceptual hydrogeological model..........................................................................................27
3.2.2 Mathematical model.................................................................................................................27
3.2.3 Analytical solution ...................................................................................................................28
3.2.4 Analysis of solution..................................................................................................................29
3.3 Groundwater flow in a semi-confined aquifer.............................................................................32
3.3.1 Conceptual hydrogeological model..........................................................................................32
3.3.2 Mathematical model.................................................................................................................32
3.3.3 Analytical solution ..................................................................................................................33
3.3.4 Analysis of result......................................................................................................................34
4. Steady Groundwater Flow to Wells ........................................................................................................37
4.1 Groundwater flow to a well in a confined aquifer.......................................................................37
4.1.1 Conceptual hydrogeological model..........................................................................................37
4.1.2 Mathematical model.................................................................................................................37
4.1.3 Analytical solution ...................................................................................................................39
4.1.4 Analysis of results....................................................................................................................40
4.1.5 Application to pumping test .....................................................................................................41
4.2 Groundwater flow to a well in an unconfined aquifer ................................................................43
4.2.1 Conceptual hydrogeological model..........................................................................................43
4.2.2 Mathematical model.................................................................................................................43
4.2.3 Analytical solution ...................................................................................................................44
4.2.4 Analysis of results....................................................................................................................46
4.2.5 Application to pumping test .....................................................................................................46
4.3 Groundwater flow to a well in a semi-confined aquifer..............................................................48
4.3.1 Conceptual hydrogeological model..........................................................................................48
4.3.2 Mathematical model.................................................................................................................48
4.3.3 Analytical solution ...................................................................................................................49
1
4.4.4 Analysis of results ................................................................................................................... 52
4.4 Methods of superposition and images.......................................................................................... 55
4.4.1 Method of superposition.......................................................................................................... 55
4.4.2 Method of images .................................................................................................................... 61
4.5 Flow net .......................................................................................................................................... 65
4.5.1 Potential function and equipotential lines................................................................................ 65
4.5.2 Stream function and stream lines............................................................................................. 66
4.5.3 Flow net................................................................................................................................... 66
5. Non-steady Groundwater Flow in Aquifers........................................................................................... 71
5.1 Basic equations for non-steady flow in aquifers.......................................................................... 71
5.1.1 Confined aquifer...................................................................................................................... 71
5.1.2 Unconfined aquifer.................................................................................................................. 72
5.2 Non-steady Groundwater Flow in a confined aquifer................................................................ 75
5.2.1 Conceptual hydrogeological model ......................................................................................... 75
5.2.2 Mathematical model ................................................................................................................ 75
5.2.3 Analytical solution................................................................................................................... 76
5.2.4 Analysis of the result ............................................................................................................... 77
5.3 Non-steady Groundwater Flow in an unconfined aquifer ......................................................... 78
5.3.1 Conceptual hydrogeological model ......................................................................................... 78
5.3.2 Mathematical model ................................................................................................................ 78
5.3.3 Analytical solutions ................................................................................................................. 79
5.3.4 Analysis of results ................................................................................................................... 81
6. Non-steady Groundwater Flow to Wells................................................................................................ 83
6.1 Groundwater flow to a well in a confined aquifer ...................................................................... 83
6.1.1 Conceptual hydrogeological model ......................................................................................... 83
6.1.2 Mathematical model ................................................................................................................ 84
6.1.3 Analytical Solution: Theis formula (1935).............................................................................. 85
6.1.4 Analysis of the solution ........................................................................................................... 86
6.1.5 Simplification of Theis formula: Jacob's formula.................................................................... 87
6.1.6 Applications............................................................................................................................. 87
6.2 Non-steady groundwater to a well in an unconfined aquifer..................................................... 94
6.2.1 Conceptual hydrogeological model ......................................................................................... 94
6.2.2 Mathematical model ................................................................................................................ 94
6.2.3 Analytical Solution.................................................................................................................. 96
6.2.4 Analysis of results ................................................................................................................... 97
6.2.5 Application of the solution ...................................................................................................... 98
6.3 Non-steady groundwater flow to a well in a semi-confined aquifer .......................................... 99
6.3.1 Conceptual hydrogeological model ......................................................................................... 99
6.3.2 Mathematical model ..............................................................................................................100
6.3.3 Analytical solution................................................................................................................. 101
6.3.4 Analysis of the results............................................................................................................ 102
6.3.5 Application of the solution .................................................................................................... 104
7. References............................................................................................................................................. 110
8. Annexes................................................................................................................................................. 111
T.1 Well function of confined aquifers ............................................................................................. 111
T.2 Well function of semi-confined aquifers .................................................................................... 112
T.3 Function
e
(x)
K0
x
..................................................................................................................... 113
T.4 Function
I
(x)
I
(x),
10
, (x)
K
(x),
K 10
...................................................................................... 114
9. Exercises ............................................................................................................................................... 116
2
9.1 Fundamental Equations of Groundwater Flow.........................................................................116
9.2 Steady Groundwater Flow in Aquifers ......................................................................................117
9.3 Steady Groundwater Flow to Wells............................................................................................120
9.4 Non-steady Groundwater Flow to Wells....................................................................................127
3
4
1. Introduction
1.1 Hydrological Cycle
A schematic view of the global hydrologic cycle is shown in Figure 1.1. This diagram shows
the interactions and mass transfers (water in different states) that occur between the
atmosphere, land surfaces, and oceans.
E
T
=
E
vapot
r
a
n
s
p
i
r
a
t
i
o
n
;
E
=
E
v
a
p
or
a
t
i
o
n
;
I
=
I
n
f
i
l
t
r
a
t
i
o
n
;
S
R
=
S
u
rf
a
c
e
ru
n
o
f
f
;
R
F
=
R
e
t
u
rn
f
l
o
w
f
r
o
m
i
r
ri
g
a
t
i
o
n
;
N
=
N
a
t
u
r
a
l
r
e
p
l
e
n
i
s
h
m
e
n
t
.
Figure 1.1Schematic diagram of hydrologic cycle (from Bear and Verruijt, 1987)
1
Atmospheric water and solar energy provide the main inputs for the generation of
precipitation, which falls over the land and oceans. Rainfall can either infiltrate into the soil
system, percolate to deeper ground water, evaporate or transpire through vegetation back to
the atmosphere, or run into the nearest streams or rivers. Infiltrating water is the main source
of recharge to the root zone and groundwater aquifers below. Rivers can also recharge
aquifers or can act as discharge points for aquifer outflows. The ocean is the ultimate receptor
of surface and groundwater contributions from surrounding land masses and provides the
main source of water for evaporation back to the atmosphere.
Table 1.1 shows the distribution of water throughout the earth. Clearly, the oceans and icecaps
dominate as sources of water.
Table 1.1 Distribution of World's estimated water resources
Location Surface
area (km
2
)
x 10
6
Water
Volume
(km
3
) x 10
6
Percentage
of total (%)
Equivalent
depth (m)
Residence
time
Surface water
Lakes and
reservoirs
1.55 0.13 <0.01 0.25 -10 years
Swamps <0.1 <0.01 <0.01 0.007 1-10 years
River channels <0.1 <0.01 <0.01 0.003 -2 weeks
Subsurface water
Soil moisture 130 0.07 < 0.01 0.13 2 weeks – 1
year
Groundwater 130 60 4 120 2 weeks –
10,000 years
Icecaps and
glaciers
17.8 30 2 60 10 – 1000
years
Atmosphere 504 0.01 <0.01 0.025 -10 days
Oceans and
seas
361 1370 94 2500 -4000 years
Total 1460 100 2680
Source: Nace, 1971
According to these figures, the present amount of free water on earth equals 1460 million km
3
corresponding to a layer 2680 m thick over the whole earth surface. It seems that the supply
of water on earth is nearly inexhaustible. However, a completely different picture obtained
when water quality is taken into account. Of the total amount of free water in earth, more than
94% is salt water, which cannot be used for agriculture or for domestic and industrial
consumption. The water present as snow or ice (2%), and as water vapour in the atmosphere
(.001%), cannot be used for these purposes either, and so, for the use of our society only the
2
remaining 4% fresh liquid water is available. It will be clear from this figure that the
development of world's economy must rely on fresh water resources.
The storage of fresh water on earth consists mainly of groundwater. Even if we consider only
the most active groundwater storage, 4 million km
3
instead of 60 million km
3
, groundwater
still accounts for 95%, the surface water in rivers and lakes accounts only 3.5 % of the total
quantity and is most vulnerable to pollution. On the other hands, groundwater is of more
homogeneous and stable spatial and temporal distribution and good quality. In the future,
people will rely more and more on groundwater for water supply.
1.2 Aquifer systems
1.2.1 Aquifer, aquitard, and aquiclude
Groundwater is stored in geological formations. According to the volume of storage and
capacity of transmitting water, geological formations are distinguished relatively as aquifer,
aquitard and aquiclude.
An aquifer is defined as a saturated permeable geological unit that stores a big quantity of
water and is permeable enough to yield economic quantities of water to wells. The most
common aquifers are unconsolidated sand and gravel, but permeable sedimentary rocks such
as sandstone and limestone, and heavily fractured or weathered volcanic and crystalline rocks
can also be classified as aquifers.
An aquitard is a geological unit that is permeable enough to transmit water in significant
quantities when viewed over large areas and long periods, but its permeability is not sufficient
to justify production wells being placed in it. Clays, loams and shale are typical aquitards.
An aquiclude is an impermeable geological unit that dose not transmit water at all. Dense
unfractured igneous or metamorphic rocks are typical aquicludes. In nature, truly
impermeable geological units seldom occur; all of them leak to some extent, and must
therefore be classified as aquitards. In practice, however, geological units can be classified as
aquicludes when their permeability is several orders of magnitude lower than that of an
overlying or underlying aquifer.
1.2.2 Aquifer types
There are three main types of aquifer: confined, unconfined, and semi-confined aquifers
(Figure 1.2).
3
Confined aquifer
A confined aquifer (Figure 1.2A) is bounded above and below by an aquiclude. In aquifer, the
pressure of the water is usually higher than that of the atmosphere, so that if a well taps the
aquifer, the water in the well rises above the top of the aquifer, or even above the ground
surface. The aquifer then is called artesian aquifer.
Unconfined aquifer
An unconfined aquifer (Figure 1.2B), also known as a phreatic aquifer, is bounded below by
an aquiclude, but is not restricted by any confining layer above it. Its upper boundary is the
water table, which is free to rise and fall. Water in a well penetrating an unconfined aquifer is
at atmospheric pressure and does not rise above the water table.
Semi-confined aquifer
A semi-confined aquifer (Figure 1.2C and D), also known as a leaky aquifer is an aquifer
whose upper and lower boundaries are aquitards, or one boundary is an aquitard and the other
is an aquiclude. Water is free to move through the aquitards, either upward or downward. If a
leaky aquifer is in hydrological equilibrium, the water level in a well tapping it may coincide
with the water table. The water level may also stand above or below the water table,
depending on the recharge and discharge conditions.
In deep sedimentary basins, an interbedded system of permeable and less permeable layers,
that form a multi-layered aquifer system (Figure 1.2E), is very common. But such an aquifer
system is more a succession of leaky aquifers, separated by aquitards, rather than a main
aquifer type.
1.3 Vertical distribution of groundwater
The vertical distribution of groundwater may generally be categorized zones of aeration and
saturation (Figure 1.3). The saturated zone is one in which all voids are filled with water
under hydrostatic pressure. In the zone of aeration, the interstices are filled partly with air,
partly with water. The saturated zone is commonly called the groundwater zone. The zone of
aeration may ideally be subdivided into several sub zones. Todd classifies these as follows.
Soil water zone
A soil water zone begins at the ground surface and extends downward through the major root
band. Its total depths are variable and dependent on soil type and vegetation. The zone is
unsaturated except during periods of heavy infiltration. Three categories of water
classification may be encountered in this region: hygroscopic water, which is adsorbed from
the air; capillary water, held by surface tension; and gravitational water, which is excess soil
water draining through the soil.
4
Figure 1.2 Different types of aquifers (from Kruseman and de Ridder, 1990)
Intermediate zone
This belt extends from the bottom of the soil-water zone to the top of the capillary fringe and
may change from non-existence to several hundred meters in thickness. The zone is
essentially a connecting link between a near-ground surface region and the near-water-table
region through which infiltrating fluids must pass.
Capillary zone
A capillary zone extends from the water table to a height determined by the capillary rise that
can be generated in the soil. The capillary band thickness is a function of soil texture and may
fluctuate not only from region to region but also within a local area.
Saturated zone
5
In the saturated zone, groundwater fills the pore spaces completely and porosity is therefore a
direct measure of storage volume. Part of this water (specific retention) cannot be removed by
pumping or drainage because of molecular and surface tension forces. Specific retention is the
ratio of volume of water retained against gravity drainage to gross volume of the soil.
Figure 1.3 Subsurface moisture zones (from Bear and Verruijt, 1987)
1.4 Groundwater flow systems
Groundwater flows are usually three-dimensional. Unfortunately, the solution of such
problem by analytic methods is complex unless the system is symmetric. In other cases, space
dependency in one of the coordinate directions may be so slight that assumption of
two-dimensional flow is satisfactory. Many problems of practical importance fall into this
class. Sometime one-dimensional flow can be assumed, thus further simplifying the solution.
Fluid properties such as velocity, pressure, temperature, density, and viscosity often vary in
time and space. When time dependency occurs, the issue is termed an unsteady flow problem
and solutions are usually difficult. On the other hand, situations where space dependency
alone exists are steady flow problems. Only homogeneous (single-phase) fluids are
considered here.
Boundaries to groundwater flow systems may be fixed geologic structures or free water
surface that are dependent for their position on the state of the flow. A hydrologist must be
able to define these boundaries mathematically if the groundwater flow problems are to be
solved.
Porous media through which groundwater flow may be classified as isotropic, anisotropic,
heterogeneous, homogeneous, or several possible combinations of these. An isotropic
medium has uniform properties in all directions from a given point. Anisotropic media have
one or more properties that depend on a given direction. For example, permeability of the
6
medium might be greater along a horizontal plan than along a vertical one. Heterogeneous
media have non-uniform properties of anisotropy or isotropy, while homogeneous media are
uniform in their characteristics.
Figure 1.4 Definition sketches of groundwater systems (from Toth, 1963)
7
8
2. Fundamental Equations of Groundwater Flow
2.1 Darcy's Law
Darcy's experiment
In 1856, A French engineer Darcy did a laboratory experiment as sketched in Figure 2.1. A
one-dimensional steady flow in the sand column is created by keeping water levels in the left
and right reservoirs constant.
Reference level
Flow
Area A
L
n
2
n
1
P
1
/Dg
P
2
/Dg
z
2
z
1
Q
Figure 2.1 Darcy's experiment
Darcy measured the total discharge through a sand column when changing the difference
between water levels in two reservoirs. He found that the total discharge was proportional to
the difference of water levels:
L
-
A K - = Q
12
??
(2.1)
where:
Q : total discharge, [L
3
T
-1
];
A : cross-sectional area, [L
2
];
K : coefficient of permeability, [LT
-1
];
9
φ
1
, φ
2
: water levels in the left and right reservoirs, [L];
L: length of the sand column (distance), [L].
Equation (2.1) is the famous Darcy's law.
From basic hydraulics it is known that φ
1
is also the groundwater head at the left end of the
sand column, and φ
2
the groundwater head at the right end of the sand column.
Groundwater head
In general, groundwater head is defined as the elevation head plus the pressure head, i.e.
g
p
+ z =
ρ
?
(2.2)
where
z : elevation of the point concerned above the reference level, [L];
p : pressure in the fluid at that point, [ML
-1
T
-2
];
ρ : density of fluid (mass per unit volume), [M L
-3
];
g : acceleration of gravity, [LT
-1
]. g = 9.81 m/sec
2
.
The quantities z and p/ρg are usually called the elevation head and pressure head,
respectively.
Specific discharge
The specific discharge is defined as the discharge per unit of cross-sectional area and denoted
by q. It follows from Equation (2.2) that,
l
K - = q
?
??
(2.3)
By taking the limit ?l → 0, the equation (2.3) becomes,
dl
d
K - = q
?
(2.4)
where
q : specific discharge, [LT
-1
];
dφ /dl: gradient of groundwater head (hydraulic gradient), [-].
10
Equation (2.4) is the differential formulation of the Darcy's law. It expresses a linear
relationship between the specific discharge and hydraulic gradient.
Velocity
Although the specific discharge has the dimension of the velocity, it is not the actual velocity
of the groundwater flow. The total cross-sectional area is A, but the area through which the
groundwater can flow is only n*A. n is the porosity of the sand. Hence, in Darcy's
experiment, the mean velocity, v, of the flow can be computed as
n
q
=
nA
Q
= v
(2.5)
The actual velocity is always greater than the specific discharge since the porosity is smaller
than one.
Validity of Darcy's law
The linear relationship between the specific discharge and hydraulic gradient suggests that the
Darcy's law can be applied to the laminar fluid flow. Therefore, Renolds number can be used
as an indicator of validity of the Darcy's law. The Renolds number for porous medium is
defined as
ν
d q
=
Re
(2.6)
where
d: average grain diameter, [L],
ν: kinematic viscosity, [L
2
T
-1
]; it is related to dynamic viscosity with ν = η/ρ;
η : dynamic viscosity, [ML
-1
T
-1
].
Experiments show that when R
e
< 10, flow is laminar fluid flow and Darcy's law can be
applied. Practical experiences show that Darcy's law can be applied to most cases of
groundwater flow in porous medium.
Intrinsic permeability
In Darcy's experiments, if using different fluids flowing through the same sand column,
different values of the coefficient of permeability will be measured. It means that the
coefficient of permeability does not only depend on the characteristics of the medium, but
also on the properties of fluid. Experiments show that the property of fluids influencing the
coefficient of permeability is the kinematic viscosity. The following relation is usually used:
11
ν
κg
=K (2.7)
where κ is called the intrinsic permeability with a dimension of [L
2
], depending only on the
characteristics of the medium.
It is known that the kinematic viscosity varies with the temperature and density of fluids. In
the cases of flow of hot groundwater or salt groundwater, coefficient of permeability will be a
function depending on the variation of the kinematic viscosity. However, the intrinsic
permeability is a constant. Therefore, in these cases, the following alternative formulation of
Darcy's law is more convenient:
dl
d
g
- =
?
ν
κq (2.8)
For the case of fresh groundwater flow, the introduction of the intrinsic permeability has no
advantages since the kinematic viscosity is a constant. Therefore, in the case of fresh
groundwater flow, the coefficient of permeability is used.
The analogy of laminar flow in tubes with groundwater flow in porous medium indicates that
the intrinsic permeability is proportional to squared diameter of grain size (d
2
) and porosity
(n). In practice, empirical formula may be used to calculate the value of intrinsic permeability.
One of such formula is Kozeny-Carman's formula:
)n-(1
n
d
c =
2
3
2
κ
(2.9)
where
d : average diameter of particle size, [L];
n : porosity, [-];
c : constant to be determined with the experiment.
Equation (2.9) explains why the permeability of clay is small and gravel is large. Although the
porosity of clay is very large, the pore space is very small resulting of low permeability. On
contrary, the pore space of gravel is very large and therefore, the permeability of gravel is
high. Table 2.1 gives range of permeability for common porous medium.
In reality, the structure of porous medium is so complicated that no direct formula can be used
to calculate the permeability. Its value is usually determined with pumping tests which are
introduced in chapters 4 and 6.
12
Table 2.1 Range of permeability for common porous medium
Medium κ (m
2
) K (m/day)
Clay 10
-17
-10
-15
10
-5
-10
-3
Silt 10
-15
-10
-13
10
-3
-10
-1
Sand 10
-12
-10
-10
10
-1
-10
2
Gravel 10
-9
-10
-8
10
4
-10
2
2.2 Generalization of Darcy's Law
Darcy's law in the differential form of Equation (2.4) expresses the relationship between the
specific discharge and hydraulic gradient for the uniform flow in one dimension. Darcy's law
can be extended to more general cases of groundwater flow.
For three dimensional flow in isotropic porous medium, Darcy's law will be written as:
x
K - =
x
?
??
q
z
K - = q
z
?
??
(2.10)
y
K - = q
y
?
??
where q
x
, q
y
, and q
z
are three flow components in x, y, and z directions, respectively. The
groundwater head will be a function of x, y, and z coordinates and is defined as:
g
z)y,p(x,
+ z =z)y,(x,
ρ
? (2.11)
where ρ is assumed a constant.
In cases of three dimensional groundwater flow in anisotropic medium with principle
directions of the permeability in x, y, and z directions, Darcy's law can be generalized as:
x
K
- = q
xx
x
?
??
13
y
K
- = q
yy
y
?
??
(2.12)
z
K
- = q
zz
z
?
??
where K
xx
, K
yy
, and K
zz
are anisotropic permeability in three principle directions, respectively.
The total specific discharge q can be considered as a vector consisting of three components
q
x
, q
y
, and q
z
. Therefore, q can be expressed as:
??K - = q
(2.13)
where:
?
?
?
?
?
?
?
?
?
?
?
?
K
00
0
K
0
00
K
=K
zz
yy
xx
is so-called permeability tensor and
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
??
?
??
??
z
y
x
=
is the hydraulic gradient.
The geometric illustration of Equation (2.13) is shown in Figure 2.2.
14
x
qx
qy y
z
q
qz
Figure 2.2 Geometric illustration of composition of specific discharge
2.3 Equation of continuity
Groundwater flow also satisfies the principle of conservation of mass, or mass balance.
Taking a small parallelepiped in porous medium, groundwater flow through the
parallelepiped must obey the mass balance stated as:
total mass in - total mass out = change of mass storage
If the density of groundwater is a constant, mass balance will be identical to water balance.
Water balance states that:
total flow in - total flow out = change of water storage
From the principle of mass balance, the so-called equation of continuity can be derived.
Figure 2.3 shows the mass fluxes through six sides of a parallelepiped in porous medium.
15
)x
)y
)z
Dq
y
Dq
x
Dq
z
Dq
z
+)(Dq
z
)
Dq
y
+)(Dq
y
)
Dq
x
+)(Dq
x
)
Figure 2.3 Mass fluxes through a parallelepiped in porous medium
First, consider flow in y-direction only. The mass carried into the left face of the element by
the specific discharge q
y
is:
zxq
y
??ρ
The mass carried out from the right face of the element is:
zxy
y
)q(
+ q
y
y
??
?
?
?
?
?
?
?
?
?
?
ρ?
ρ
Hence the increase of mass (per unit time) due to the flow in y-direction will be:
zyx
y
)q(
- = zxy
y
)q(
+ q - zxq
yy
yy
???
?
ρ?
??
?
?
?
?
?
?
?
?
?
?
ρ?
ρ??ρ
(2.14)
Similarly, the increase of mass due to the flow in x and z directions can be derived as:
(2.15)
16
zyx
x
)q(
- = zyx
x
)q(
+ q - zyq
xx
xx
???
?
ρ?
??
?
?
?
?
?
?
?
?
ρ?
ρ??ρ
and
zyx
z
)q(
- =y xz
z
)q(
+ q -y xq
zz
zz
???
?
ρ?
??
?
?
?
?
?
?
?
?
ρ?
ρ??ρ
(2.16)
The total mass in the parallelepiped ?x ?y ?z is:
z)nyx( = M ???ρ?
The change (increase) of mass storage per unit time in the parallelepiped is:
z)n]yx([
t
=
t
M)(
???ρ
?
?
?
??
(2.17)
According to the mass balance, the change of mass storage must be equal to total increase of
mass due to flow in x, y and z directions. Thus, the summation of equations (2.14), (2.15) and
(2.16) equals (2.17) as:
zy x ]
z
)q(
+
y
)q(
+
x
)q(
[- = n])zyx([
t
z
y
x
???
?
ρ?
?
ρ?
?
ρ?
???ρ
?
?
(2.18)
Equation (2.18) is the general equation of continuity of groundwater flow. The equation of
continuity is second fundamental equation of groundwater flow. The equation of continuity
combined with Darcy's law will result in basic equations describing groundwater flow in
porous medium.
2.4 Basic equations for steady incompressible flow
The groundwater flow will be in steady state when there is no change of mass storage. In this
case, the mass balance states that total mass in = total mass out.
Since ?(?M)/?t = 0 in the case of steady flow, the equation of continuity become:
(2.19)
17
0 =
z
)q(
+
y
)q(
+
x
)q(
z
y
x
?
ρ?
?
ρ?
?
ρ?
Furthermore, if the fluid is incompressible, the density ρ is a constant and equation (2.19)
reduces to:
0 =
z
q
+
y
q
+
x
q
z
y
x
?
?
?
?
?
?
(2.20)
Equation (2.20) is the equation of continuity for steady incompressible groundwater flow.
Substitution of Darcy's law, equation (2.10) into (2.20) gives:
0 =
z
+
y
+
x
2
2
2
2
2
2
?
?
?
?
?
?
?
?
?
(2.21)
which is the basic differential equation of steady incompressible flow in homogeneous
isotropic porous medium. It is noted that Equation (2.21) is the standard Laplace equation and
is often written in abbreviation form:
?
0 =
2
?
(2.22)
Substitution of Darcy's law, equation (2.11) into (2.20) will give the differential equation of
steady incompressible flow in anisotropic porous medium, which is:
0 = )
z
K
(
z
+ )
y
K
(
y
+ )
x
K
(
x
zzyyxx
?
??
?
?
?
??
?
?
?
??
?
?
(2.23)
When the porous medium is anisotropic but homogeneous, equation (2.23) reduces to:
0 =
z
K
+
y
K
+
x
2
2
zz
2
2
yy
2
2
xx
?
?
?
?
?
?
?
?
?
K
(2.24)
2.5 Basic equations for non-steady compressible flow
In the case of non-steady flow, the storage of mass is changed since the total mass in is not
equal to the total mass out.
18
From the principles of Soil Mechanics, it is known that the total pressure, σ
z
, from overlying
geological strata is balanced by the effective pressure, σ
z
', of grains and pore pressure, p, of
water. i.e.
p + =
zz
σσ ′ (2.25)
Since the total pressure can be assumed to be independent of time, any increase of pore
pressure will result in a decrease of effective pressure. The consequences of such a change
are:
(1) water is compressed due to the increase of pore pressure. Therefore, the density of
water is a function of time;
(2) grain skeleton is expanded due to the decrease of effective pressure. Hence, the
porosity and size of grain skeleton are functions of time.
Taking the above effects into consideration, the change of mass storage in equation (2.17) can
be expressed as:
z)yx(
t
n +
t
n
z)yx( +
t
z)nyx( = z)n]yx([
t
???
?
?
ρ
?
?
???ρ
?
ρ?
??????ρ
?
?
(2.26)
Compressibility of water
From the elastic theory, the relative change of density of water is proportional to the change
of pore pressure, i.e,
dp =
d
w
β
ρ
ρ
(2.27)
where β
w
is the coefficient of compressibility of water. Its value is about 0.5x10
-9
m
2
/N.
From (2.27) it follows that:
t
p
=
t
p
dp
d
=
t
w
?
?
ρβ
?
?ρ
?
ρ?
(2.28)
Compressibility of soil
From the theory of elasticity, soil will be compressed under the pressure. When sample is
confined horizontally, the vertical relative compression is proportional to the increase of
pressure, i.e
19
'
z
d - =
z
zd
σα
?
?
(2.29)
where:
σ
z
': vertical pressure on soil with area (?x ?y) and height ?z;
α : coefficient of compressibility of soil. Common values are 10
-8
-10
-7
m
2
/N for sand, and
10
-7
-10
-6
m
2
/N for clay.
In many practical cases usually the horizontal deformations of the soil skeleton is much
smaller than the vertical deformation. Therefore, it can be assumed that ?x, ?y are constant
and only ?z changes. Under this assumption,
t
zyx - =y x
t
z)(
=yx
t
z)(
= z]yx[
t
zz
z
?
σ
?
???α??
?
σ
?
σ
?
??
??
?
??
???
?
?
′′
′
(2.30)
Since σ
z
= σ
z
' + p is assumed to be independent of time, i.e.
0 =
t
z
?
σ
?
so that
t
p
- =
t
z
?
?
?
σ
?
′
(2.31)
Substitution of (2.31) into (2.30) yields:
t
p
zyx = z]yx[
t ?
?
???α???
?
?
(2.32)
Change of porosity
Although the size ?x ?y ?z changes with compression, it is assumed that only pore space is
changed, but the total value of grains, V
s
, is kept constant since grains can be considered
completely incompressible. The assumption of V
s
= (1-n) ?x ?y ?z constant gives:
20
0 =
t
z)(
yxn)-(1 + zyx
t
n)-(1
=
t
V
s
?
??
?????
?
?
?
?
(2.33)
From equation (2.33) with (2.32), the relation between the changes of the porosity with the
change of the pore pressure is found as:
t
p
n)-(1 =
t
z)(
z
n-1
=
t
n
?
?
α
?
??
??
?
(2.34)
Equation of continuity of non-steady compressible flow
Substitutions of equations (2.28), (2.32) and (2.34) into (2.26) gives:
t
p
]n + [ zy x = z)n]yx([
t
w
?
?
βαρ??????ρ
?
?
(2.35)
Substitution of (2.35) into (2.18) gives the general equation of continuity for non-steady
compressible flow as:
]
z
)q(
+
y
)q(
+
x
)q(
[- =
t
p
)n + (
z
y
x
w
?
ρ?
?
ρ?
?
ρ?
?
?
βαρ
(2.36)
Basic equation of non-steady compressible flow
In case of compressible flow in homogeneous isotropic porous medium, Darcy's law of
equation (2.10) will be replaced by:
x
p
g
K
- = q
x
?
?
ρ
y
p
g
K
- = q
y
?
?
ρ
(2.37)
z
p
g
K
-K - = q
z
?
?
ρ
Substitution of (2.37) into (2.36) results in:
z
g + ]
z
p
+
y
p
+
x
p
[K =
t
p
)n + g(
2
2
2
2
2
w
?
ρ?
?
?
?
?
?
?
?
?
βαρ
(2.38)
21
If the vertical gradient of fluid density can be neglected (for example in one fluid flow),
Equation (2.38) reduces to:
t
p
K
)n + g(
=
z
p
+
y
p
+
x
p
w
2
2
2
2
2
?
?β
αρ
?
?
?
?
?
?
(2.39)
which is a basic equation of non-steady compressible flow in homogeneous isotropic
medium.
Using the relation of
g
p
+ z =
ρ
?
an equation in terms of head can be found as:
tK
S
=
z
+
y
+
x
s
2
2
2
2
2
?
??
?
?
?
?
?
?
?
?
?
(2.40)
where is usually called the storativity or specific storage coefficient. It is
defined as the volume of water stored in a unit volume of soil by a unit increase of the head. It
has a dimension of [L
)n + g( =
S
w
s
βαρ
-1
]. The physical explanation of the storativity is that with the increase of
head (1) more water is stored in pore space since water is compressed; and (2) pore space is
enlarged since the grain skeleton is expanded. Since the expansion and compression of water
and soil skeleton is elastic, S
s
is called elastic storage.
22
3. Steady Groundwater Flow in Aquifers
3.1 Groundwater flow in a confined aquifer
3.1.1 Conceptual hydrogeological model
- The aquifer is confined by two impermeable layers on the top and bottom;
- The aquifer consists of homogeneous porous medium ( K = constant);
- The aquifer is of uniform thickness (H);
- The aquifer is bounded with two parallel rivers on the left and right with constant river
stage.
n H
H
0
H
L
x
L
0
x
K
Figure 3.1 A confined aquifer
Under these assumptions, groundwater flow in the aquifer is one-dimensional steady flow
from the left river through the aquifer to the right river.
3.1.2 Mathematical model
A mathematical model describing steady groundwater flow in the aquifer consists of:
- a partial differential equation which is a governing equation of groundwater flow; and
- boundary conditions which are external influences on groundwater flow in aquifers.
The governing equation of groundwater flow can be derived with equation of continuity and
Darcy's law. In section 2.4 the basic equation of steady groundwater flow in homogeneous
23
medium is derived as in (2.21), which is applicable to the steady flow in this confined aquifer.
However, since flow in this case is only in x-direction, equation (2.21) is reduced to
0 =
x
2
2
?
?
?
(3.1)
which is the governing equation of one-dimensional steady flow in the confined aquifer.
Furthermore, groundwater head in the aquifer has to satisfy the following two boundary
conditions:
H
= |
H
= |
L
L=x
0
0=x
?
?
(3.2)
Equations (3.1) and (3.2) form the mathematical model of one-dimensional steady flow in the
confined aquifer.
3.1.3 Analytical solution
The solution of the mathematical model gives the distribution of groundwater head in the
aquifer.
The integration of equation (3.1) results in:
c
+x
c
=
21
?
c
1
and c
2
are constants of the integration and must be determined by the boundary conditions.
Application of the boundary conditions of (3.2) gives:
L
H
-
H
- =
c
H
=
c
L0
2
01
The final solution becomes:
x
L
H
-
H
-
H
=
L0
0
? (3.3)
which shows that the distribution of groundwater head in the cross-section is a straight line
from left river stage to the right river stage.
24
3.1.4 Analysis of solution
Specific discharge
The specific discharge can be calculated with Darcy's law and equation (3.3) as:
L
H
-
H
K = q
L0
(3.4)
It shows that the specific discharge is a constant through the cross section of the aquifer.
Unit width discharge
The total discharge through whole thickness of the aquifer with unit width is:
L
H
-
H
T =
L
H
-
H
HK = q . H = Q
L0L0
(3.5)
where T is called the transmissivity with a dimension of [L
2
/T]. The transmissivity is a aquifer
parameter describing the capacity of aquifer to transport water. Its value is unit width aquifer
discharge when the hydraulic gradient is one.
It is noted that in this case the unit width discharge is the same as the recharge of the left river
to the aquifer and the discharge of the aquifer to the right river.
Average velocity
The average velocity of groundwater flow in the aquifer is
L
H
-
H
n
K
=
n
q
= v
L0
ee
(3.6)
where n
e
is the effective porosity. In this case, the groundwater flow velocity is also a
constant.
Travel time
From the definition of the velocity of flow:
dt
ds
= v (3.7)
The travel time can be calculated as the following integration:
25
v
ds
=t ∫
(3.8)
where ds is the distance of a water particle moving along the flow line.
In this case, the travel time of a water particle from the
left river to the right river is:
)
H
-
H
K(
Ln
=
q
L
n
=
v
dx
=t
L0
2
ee
L
0
∫
(3.9)
It shows that the travel time depends on the distance between two rivers, the effective
porosity, the permeability, and the head difference of the two river stages.
Example
In order to have an idea of range of values of groundwater discharge, velocity and travel time,
some hypothetical values of aquifer parameters are assumed. Table 3.1 gives the results.
Table 3.1 Groundwater discharge, velocity and travel time for a hypothetical example
H
o
H
L
H K T L n
e
q Q v t
m m m m/d m
2
/d m _ m/d m
3
/d m/d d
-----------------------------------------------------------------------------------------------------------
20 19 10 10 100 1000 0.2 0.01 0.1 0.05 20000
20 19 10 100 1000 1000 0.2 0.10 1.0 0.50 2000
20 19 100 10 1000 1000 0.2 0.01 1.0 0.05 20000
The value of permeability falls in the range of sand. It shows that groundwater flow is very
slow. It may take many years for water to flow from the left river to the right river with a
distance only 1000 meters!.
26
3.2 Groundwater Flow in an Unconfined Aquifer
3.2.1 Conceptual hydrogeological model
- An unconfined aquifer is above a horizontal impermeable base;
- The porous medium is homogeneous (K = constant);
- The aquifer receives uniform recharge (w = constant) on the top; w is defined as amount
of water entering to aquifer per unit length and width per unit time.
- The aquifer is bounded by two rivers of constant stages h
0
and h
L
.
- Although flow is two-dimensional in the cross-section, vertical flow velocity is much
smaller than the horizontal flow so that the flow is assumed to be one-dimensional
horizontal flow (Dupuit's assumption).
w
0
L
x
h
h
0
K
h
L
x
w)x
x+)x
Figure 3.2 An unconfined aquifer
3.2.2 Mathematical model
The governing equation of steady flow in an unconfined aquifer can be derived with the
principle of continuity and Darcy's law.
If we choose the reference level at the bottom of the aquifer, the height of water table equals
the saturated thickness of the aquifer. Taking a parallelepiped with length of ?x and height h
(Figure 3.2), water balance of the parallelepiped gives:
27
x
x
)q(h
+ hq =x w+ hq
x
xx
?
?
?
?
(3.10)
Substitution of Darcy's law into (3.10) results in:
0 =
K
w
+ )
x
h
(h
x ?
?
?
?
(3.11)
which is the governing equation of groundwater flow in the unconfined aquifer. It is noted
that equation (3.11) is not linear.
The boundary conditions are:
h
= |h
h
= |h
L
L=x
0
0=x
(3.12)
Equations (3.11) and (3.12) form the mathematical model for one-dimensional steady
groundwater flow in the unconfined aquifer.
3.2.3 Analytical solution
Equation (3.11) can be rearranged as:
0 =
K
2w
+
x
h
2
22
?
?
(3.13)
Integration of equation (3.13) will give:
c
+ x
c
=
x
K
w
+
h
21
22
(3.14)
where c
1
and c
2
are constants of the integration and must be determined by the boundary
conditions. Application of the boundary conditions of (3.12) gives:
h
=
c
0 =x
0
2
2
(3.15a)
28
and
L
K
w
+
L
h
-
h
- =
c
L =x
L
2
0
2
1
(3.15b)
Substitution of c
1
and c
2
into (3.14) gives:
x
K
w
-)x
K
wL
-
L
h
-
h
( -
h
=
h
2L
2
0
2
0
2
2
(3.16)
which is not a straight line.
The unit width discharge in the aquifer will be:
wx+ )
K
wL
-
L
h
-
h
(
2
K
=
x
h
Kh- = Q
L
2
0
2
x
?
?
(3.17)
which is not a constant through the aquifer.
3.2.4 Analysis of solution
Water divide
At water divide x = d, 0 =
x
h
?
?
, resulting the unit width discharge Q
d
= 0. From (3.17) it gives:
L
h
-
h
2
K
-
2
wL
= wd
L
2
o
2
(3.18)
The location of water divide will be
wL
h
-
h
2
K
-
2
L
= d
L
2
0
2
(3.19)
Equation (3.19) indicates that the location of the water divide is close to the river with higher
river stage since d < L/2 when h
0
> h
L
. The water divide will be in the middle of the aquifer
only when two river stages are the same (h
0
= h
L
).
29
At water divide, water table is highest h = h
max
:
d
K
W
- )d
K
L w
-
L
h
-
h
( -
h
=
h
2L
2
0
2
0
2
max
2
(3.20)
When h
0
= h
L
, the highest water table will be:
4K
L
w
+
h
=
h
2
0
2
2
max
(3.21)
Discharge to rivers
Discharge to left river can be found from (3.17) when x = 0:
]
L
h
-
h
2
K
-
2
wL
[ - = Q
L
2
0
2
L
(3.22)
The sign "-" indicates that the flow is in the opposite of x-direction.
Discharge to right river is found when x = L in (3.17):
L
h
-
h
2
K
+
2
wL
= Q
L
2
0
2
R
(3.23)
Equations (3.22) and (3.23) indicate that discharge to both rivers consists of two flow
components: first component induced by the recharge and second component induced by head
differences between two rivers. When h
0
> h
L
, discharge to the left river is smaller than to the
right river since flow induced by head difference is from left river to right river. Only when h
0
= h
L
, discharge to two rivers will be the same as:
wL
2
1
= |Q| = |Q|
RL
(3.24)
and the total discharge to two rivers is:
wL= |Q| + |Q| = Q
RL
(3.25)
which equals the total recharge to the aquifer!.
30
No recharge
In the case where there is no recharge to the aquifer, w = 0, water table (3.16) reduces to:
x
L
h
-
h
-
h
=
h
L
2
0
2
0
2
2
(3.26)
which has a parabolic shape.
In this case, the flow occurs only from the left river to the right river with unit width discharge
as:
L
h
-
h
.
2
h
+
h
.K =
L
h
-
h
2
K
= Q
L0L0L
2
0
2
(3.27)
which is the well-known Dupuit formula derived in 1863. It indicates that the unit width
discharge is a constant and can be obtained using Darcy's law with average aquifer thickness,
(h
0
+ h
L
)/2, and average hydraulic gradient, (h
0
- h
L
)/L.
Although the unit width discharge is constant, flow velocity increases at the downstream since
the saturated thickness at the downstream is much smaller.
31
3.3 Groundwater flow in a semi-confined aquifer
3.3.1 Conceptual hydrogeological model
- A semi-confined aquifer separated with an overlying phreatic aquifer by a semi-
permeable layer;
- The semi-confined aquifer is homogeneous (K) with uniform thickness (H);
- The semi-permeable layer is homogeneous with constant vertical hydraulic conductivity
(K') and thickness (d);
- The aquifer system is bounded by a lake on the left with constant level (φ
2
) and extends
infinite on the right;
- Groundwater in the phreatic aquifer leaks to semi-confined aquifer. Flow in semi-
permeable layer is assumed only vertical and flow in semi-confined aquifer is assumed
only horizontal to the lake;
- Although groundwater in the phreatic aquifer leaks to semi-confined aquifer, water table
in phreatic aquifer is assumed to keep constant (φ
1
).
Dike Polder
Lake
0 x
Semi-confined aquifer
phreatic aquifer
K'
d
z
n
2 n
n
1
Figure 3.3 A semi-confined aquifer
3.3.2 Mathematical model
The governing equation of groundwater flow in the semi-confined aquifer can be derived
from the water balance equation as:
q = )q(H
x
Lx
?
?
(3.28)
32
where q
L
is the leakage from the phreatic aquifer to the semi-confined aquifer. It can be
defined from Darcy's law as:
]
T
[L
d
-
K = q
1-1
L
??
′
(3.29)
If we define a new parameter:
c = d/K'
as the resistance of the semi-permeable layer, (3.29) becomes:
c
-
= q
1
L
??
(3.30)
It is clear that the amount of leakage depends on the head difference in two aquifers and the
resistance of the semi-permeable layer.
Substituting (3.30) to (3.28) and replacing q
x
with Darcy's law gives:
∞≤
λ
??
?
?
?
x 0 0 =
-
-
x
2
1
2
2
p
(3.31)
where
λ
. The parameter λ with a dimension of length is called leakage factor. Tc =
2
The boundary conditions to be satisfied are:
??
20=x
= |
(3.32a)
??
∞→ 1x
= |
(3.32b)
Equations (3.31) and (3.32) form the mathematical model of groundwater flow in a semi-
confined aquifer.
3.3.3 Analytical solution
The general solution for Equation (3.31) is:
33
)
x
(-exp
c
+ )
x
(exp
c
= -
21
1
λλ
??
(3.33)
The constants c
1
and c
2
can be determined with boundary conditions (3.32):
0 =c
1
) - ( - =c
212
??
The solution becomes:
)
x
(- exp ) - ( - =
211
λ
????
(3.34)
3.3.4 Analysis of result
Specific discharge
The specific discharge can be calculated with Darcy's law as:
)
x
(- exp
-
K- =
x
K - = q
21
x
λλ
??
?
??
(3.35)
The minus sign indicates that flow is in the opposite of x direction.
Unit width discharge
The unit width discharge can be calculated from (3.35) as:
)
x
(- exp
) - ( T
- = qH = Q
21
xx
λλ
??
(3.36)
The discharge increases along flow direction since more leakage comes in.
Unit width discharge to the lake
The maximum discharge will be at x = 0, which is discharge to the lake:
λ
??
21
o
-
. T - = Q
(3.37)
It is interesting to note that the leakage factor λ functions as a
34
distance determining the hydraulic gradient with a head difference of φ
1
-φ
2
. From (3.34) it
can be also demonstrated that when x = 4λ, φ = φ
1
-(φ
1
-φ
2
)*0.0183. It follows that
groundwater head in the semi-confined aquifer is practically equal to the water table in the
phreatic aquifer at a distance of 4 times of leakage factor. Therefore, the value of the leakage
factor appears to determine the zone of influence of the lake on the aquifer system.
Velocity
The velocity of groundwater flow can be calculated as
)
x
(- exp
-
n
K
- =
n
q
=
v
21
ee
x
x
λλ
??
(3.38)
The velocity increases towards the lake.
Travel time
The travel time of a groundwater particle at a distance of x to the lake can be calculated as:
e
dx
]
-
n
K
[ - =
v
dx
=t
x/-
0
x
1-21
ex
0
x
λ
∫
λ
??
∫
which will be:
1] -
e
[
-
K
n
=t
x/
21
2
e λ
??
λ
(3.39)
The maximum travel time of a water particle from the zone of influence to the lake will be:
??
λ
?
21
2
e
max
-
K
n
54
t
(3.40)
35
36
4. Steady Groundwater Flow to Wells
4.1 Groundwater flow to a well in a confined aquifer
4.1.1 Conceptual hydrogeological model
- A confined aquifer is located in a circular island with a radius of R;
- The aquifer is homogeneous and isotropic (k = constant) with uniform thickness (H =
constant);
- In the centre of the island a pumping well fully penetrates the whole thickness of
aquifer operating with a constant pumping rate, Q
0
[L
3
T
-1
];
- The aquifer is surrounded by a lake with constant level φ
0
.
- Before pumping, groundwater head is in the same level in the lake. After pumping
with a constant rate, groundwater flows in radial direction from the lake towards the
well. Steady state flow condition is reached after some time of pumping.
n
w
H
r
Q
0
2R
n
0
n
0
2r
w
Figure 4.1 A well in a confined aquifer
4.1.2 Mathematical model
It is convenient to introduce polar coordinates r and θ to describe radial flow to wells. The
basic equation can be derived from Equation (3.21) using the chain rule of differentiation with
the relationship:
37
x = r cos(θ)
y = r sin(θ)
However it is instructive to derive the equation directly in polar coordinates. For radial flow,
Darcy's law states:
dr
d
K - = q
r
?
(4.1)
The total discharge (Q
r
) through a circular ring of radius r and height H equals:
q Hr 2 = Q
rr
π (4.2)
Substituting (4.1) to (4.2) results:
dr
d
r T 2 - = Q
r
?
π
(4.3)
From the principle of continuity (water balance), Q
r
is constant, i.e.
0 =
dr
dQ
r
resulting in:
0 = )
dr
d
r (
dr
d ?
(4.4)
which is the basic equation of steady radial flow in a confined aquifer.
Two boundary conditions will have to impose on this special case. The lake forms a outer
boundary and it is assumed that groundwater head at outer boundary is constant:
??
0R =r
= | (4.5a)
The pumping well consists of an inner boundary. The total discharge across well screen must
be equal to pumping rate:
Q - = |Q
0
r
w
=rr
(4.5b)
38
Equations (4.4) and (4.5) form mathematical model of radial groundwater flow to a well in a
confined aquifer.
4.1.3 Analytical solution
Successive integration of Equation (4.4) will give:
c
=
dr
d
r
1
?
(4.6a)
and
c
+r ln
c
=
21
?
(4.6b)
The constants c
1
and c
2
can be determined with two boundary conditions. The inner boundary
condition (4.5b) gives,
Q =
dr
d
r T 2
0
?
π
Comparing with Equation (4.6a), we can find:
T2
Q
=
dr
d
r =
c
o
1
π
?
(4.7a)
The outer boundary condition gives:
)Rln(
T2
Q
- =
c
c
+ )Rln(
c
=
0
0
221
0
π
??
(4.7b)
The final solution is obtained by substituting (4.7) into (4.6b):
??
π
??
0
0
0
)
r
R
ln(
T 2
Q
- = p
(4.8)
Groundwater head in the well is:
)
r
R
ln(
T 2
Q
- =
w
0
0w
π
??
(4.9)
39
Usually, the difference of groundwater heads before and after pumping is defined as
drawdown:
?? - = s
0
(4.10)
The shape of drawdown around the well is a cone of depression and it is described by:
0 s )
r
R
ln(
T 2
Q
= s
0
f
π
(4.11)
The maximum drawdown will occur at the pumping well and is:
)ln(
T 2
Q
=
s
w
r
R0
w
π
(4.12)
4.1.4 Analysis of results
Total groundwater flow
The specific discharge can be found with Darcy's law as
r
1
.
H 2
Q
- =
dr
d
K - = q
0
r
π
?
(4.13)
The total groundwater flow through any circular ring with height of the thickness of the
aquifer Q
r
is:
Q - = )
r
1
H 2
Q
- ( Hr 2= q Hr 2 = Q
0
0
rr
π
ππ
(4.14)
which is exactly the pumping rate.
Velocity
The groundwater flows in radial direction with velocity of
r
1
H
n
2
Q
- =
n
q
=
v
e
0
e
r
r
π
(4.15)
The velocity increases towards the well since the area of the circular ring decreases.
40
Travel time
The travel time of a water particle at the outer boundary to reach the well can be calculated as
)
r
-
R
(
Q
H
n
=dr r
Q
H
n
2
- =
v
dr
=t
w
2
2
0
e
r
w
R
0
e
r
r
w
R
π
∫
π
∫
(4.16)
4.1.5 Application to pumping test
The equation (4.11) can be applied to determine the transmissivity (T) from a pumping test.
Figure 4.2 shows a design of pumping test with a pumping well and two observation wells.
If a constant rate (Q
0
) is applied to the pumping well, groundwater heads in pumping well and
observation wells will reach steady state after some time of pumping. The observed steady
groundwater heads in two observation wells will satisfy the equation (4.11):
)
r
R
ln(
T 2
Q
- =
1
0
01
π
??
(4.17)
and
)
r
R
ln(
T 2
Q
- =
2
0
02
π
??
(4.18)
0
OW1 OW2
Q
0
T
n
1
n
2
n
w
r
w
r
1
r
2
s
2
s
1
s
w
Figure 4.2 A system of pumping test
41
From equations (4.17) and (4.18) we can find:
)
r
r
ln(
)-(2
Q
= T
1
2
12
0
??π
(4.19a)
or
)
r
r
ln(
)
s
-
s
(2
Q
= T
1
2
21
0
π
(4.19b)
In case of one observation well, the observed head in pumping well itself can be used, which
will result in:
)
r
r
ln(
)-(2
Q
= T
w
1
w1
0
??π
(4.20a)
or
)
r
r
ln(
)
s
-
s
(2
Q
= T
w
1
1w
0
π
(4.20b)
However, it should bear in mind that the transmissivity will be underestimated with equation
(4.20) since the drawdown in the well does not represent the drawdown in the aquifer because
of the so-called well loss. The well loss is mainly caused by gravel pack and well screen.
Once the transmissivity is determined, the hydraulic conductivity can be calculated with the
known aquifer thickness as:
K = T / H
In reality, a confined aquifer with a shape of the circular island rarely exists. However, for
engineering purpose, the drawdown caused by the pumping in a certain distant is negligible
and can be considered practically as zero. This distant is called the radius of influence and
denoted as the same R. The radius of influence can be determined from Equations (4.17) or
(4.18) with pumping test data.
42
4.2 Groundwater flow to a well in an unconfined aquifer
4.2.1 Conceptual hydrogeological model
- An unconfined aquifer is located in a circular island;
- The aquifer is homogeneous and isotropic (K = constant);
- In the centre of the island, a pumping well fully penetrates the aquifer with a constant
pumping rate: Q
0
;
- The aquifer is surrounded by a lake with constant level (H
0
);
- The aquifer is bounded with an impermeable base and there is no recharge on the top.
- Before the pumping, water table is the same as the lake water level; in radial direction
towards the well and a steady state condition is reached after some time of pumping.
H
0
H
0
R
K
2r
w
Qo
r
h
w
0
Figure 4.3 A well in an unconfined aquifer
4.2.2 Mathematical model
The total discharge through a circular ring of radius r is:
q .
A
= Q
r
r
r
(4.21)
where the specific discharge is:
43
dr
dh
K - = q
r
(4.22)
and the area of the circular ring is:
hr 2 =
Ar
π
(4.23)
Substituting (4.22) and (4.23) into (4.21) gives
(4.24)
Q
r
is a constant since there is no water entering into the aquifer
from top or bottom. Thus,
dr
dh
h K r 2 - = Q
r
π
0 =
dr
Qd
r
will result in:
0 = )
dr
dh
h (r
dr
d
(4.25)
Equation (4.25) is the governing equation of groundwater flow towards a well in an
unconfined aquifer.
The boundary conditions are:
H
= |h
0
R =r
(4.26a)
Q- = |Q
0
r
w
=r r
(4.26b)
4.2.3 Analytical solution
Successive integration of Equation (4.25) gives:
c
=
dr
h
d
r
1
2
(4.27a)
44
and
c
+ )rln(
c
=
h 21
2
The constants c
1
and c
2
can be determined with two boundary conditions. The inner boundary
condition (4.26b) gives:
Q =
dr
dh
rKh 2
0
π
Comparing with Equation (4.27a), we can find:
K
Q
=
dr
h
d
r =
c
0
2
1
π
(4.28a)
The outer boundary condition gives:
)Rln(
K
Q
-
H
=
c
c
+ )Rln(
c
=
H
02
0221
2
0
π
The final solution is obtained by Substituting (4.28) into (4.27b):
)
r
R
ln(
K
Q
-
H
=
h
0
0
2
2
π
(4.29)
Since:
s)-
H
s(2 = h)-
H
h)(+
H
( =
h
-
H 000
2
0
2
from (4.29) we can find:
)
r
R
ln(
K
H
2
Q
= )
H
2
s
- s(1
0
0
0
π
(4.30)
where s is drawdown defined as s = H
0
- h. It shows that the drawdown is not a simple
logarithmic function of radial distance.
In the case where s << 2H
0
, (4.30) reduces to:
)
r
R
ln(
T2
Q
= s
0
π
(4.31)
45
where T = H
0
K. Equation (4.31) is identical to (4.11). It means that if the drawdown caused
by the pumping is much smaller than the initial saturated thickness of the aquifer, the
drawdown in the unconfined aquifer is identical to the drawdown in the confined aquifer.
4.2.4 Analysis of results
Total discharge
It can be shown that the total discharge through any circular ring of radius r equals the
pumping rate. Substituting (4.28a) into (4.24) gives Q . Q - =
0r
Specific discharge
The specific discharge can be calculated as:
rh2
Q
- =
A
Q
= q
0
r
r
r
π
(4.32)
Velocity
The velocity is:
rh
n
2
Q
- =
n
q
=
v
e
0
e
r
r
π
(4.33)
Comparing with velocity in the confined aquifer indicates that the velocity in the unconfined
aquifer increases faster towards the pumping well since cross-sectional area of the circular
ring decreases faster with the decrease of both of radius and saturated thickness.
4.2.5 Application to pumping test
A similar design of a pumping test as in the confined aquifer can be conducted in the
unconfined aquifer (Figure 4.4).
Application of equation (4.24) into two observation wells gives:
(4.34a)
(4.34b)
)
r
R
ln(
k
Q
-
H
=
h
1
0
0
2
1
2
π
)
r
R
ln(
k
Q
-
H
=
h
2
0
0
2
2
2
π
46
From (4.34a) and (4.34b) it can be found that:
)
r
r
( ln
)
h
-
h
(
Q
=K
1
2
1
2
2
2
0
π
(4.35)
In terms of drawdown, (4.35) becomes:
)
r
r
ln(
)
s
-
s
-
H
)(2
s
-
s
(
Q
=K
1
2
21021
0
π
(4.36)
In case of one observation well, measurements in the pumping well can be used. Equations
(4.35) and (4.36) become:
)
r
r
ln(
)
h
-
h
(
Q
=K
w
1
w
2
1
2
0
π
(4.37)
and
)
r
r
ln(
)
s
-
s
-
H
)(2
s
-
s
(
Q
=
w
1
1w01w
0
π
K
(4.38)
The radius of influence can be determined also from (4.34) with the pumping test data.
OW1
OW2Q0
s
1
s
2
h
2
h
1
r
1
r
2
Figure 4.4 A pumping test in an unconfined aquifer
47
4.3 Groundwater flow to a well in a semi-confined aquifer
4.3.1 Conceptual hydrogeological model
- The semi-confined aquifer is homogeneous and isotropic with uniform thickness;
- The semi-permeable layer is homogeneous with constant vertical hydraulic
conductivity (K') and uniform thickness (d);
- A pumping well fully penetrates the semi-confined aquifer with constant pumping rate
Q
0
;
- Before the pumping, groundwater head in the semi-confined aquifer is the same as the
water table in the phreatic aquifer. After pumping, groundwater in the phreatic aquifer
leaks into semi-confined aquifer. Flow is assumed to be vertical in the semi-permeable
layer and is radial towards pumping well in the semi-confined aquifer;
- The aquifer system extends to infinite laterally.
n
0
H
R
2r
w
Q
0
T n
w
Figure 4.5 A pumping well in a semi-confined aquifer
4.3.2 Mathematical model
Taking a circular ring of radius r and thickness ?r and height H from the semi-confined
aquifer, the water balance states:
48
r
dr
dQ
+ Q = Q + Q
r
rLr
?
(4.39)
where Q
r
is lateral flow and Q
L
is the leakage. They can be determined with Darcy's law as:
dr
d
rHK 2 - = Q
r
?
π (4.40)
(4.41)
c
-
r r2 = Q
0
L
??
?π
Substituting (4.40) and (4.41) into (4.39) gives:
0 =
c
-
r r2 -r ]
dr
d
+
dr
d
[r T2 -
0
2
2
??
?π?
??
π
(4.42)
The governing equation of radial flow towards a pumping well in the semi-confined aquifer
can be found from (4.42) as:
0 =
-
-
dr
d
r
1
+
dr
d
2
0
2
2
λ
????
(4.43a)
where λ is the leakage factor.
Two boundary conditions are:
T2
Q
= |
dr
d
r
0
r
w
=r
π
?
(4.43b)
and
??
∞→ 0 r
= |
(4.43c)
4.3.3 Analytical solution
With the following transformations:
?? - = s
0
and
49
λ x =r
Equation (4.43) is simplified as:
0 = s -
dx
ds
x
1
x
d
s
d
2
2
+
(4.44a)
T2
Q
= |
dr
ds
r
0
r
w
=r
π
?
(4.44b)
(4.44c)
0 = | s
r ∞→
The differential equation (4.44a) is a standard mathematical physical problem and its general
solution is:
(x)
K
c
+ (x)
I
c
= s
0201
(4.45)
where:
I
0
(x) is called the first kind of Bessel function of order zero;
K
0
(x) is called the second kind of Bessel function of order zero;
c
1
and c
2
are constants to be determined by boundary conditions.
The Bessel functions of order one I
1
(x) and K
1
(x) are related to I
0
(x) and K
0
(x) by
(4.46a)
(x)
I
= (x)
I
dx
d
10
(x)
K
- = (x)
K
dx
d
10
(4.46b)
Figure 4.6 and 4.7 show the behaviour of functions I
0
(x), I
1
(x), K
0
(x) and K
1
(x). It can be
observed that when x approaches to infinite, I
0
(x) and I
1
(x) approach to infinite, and K
0
(x) and
K
1
(x) approach to zero.
50
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
45.0
50.0
55.0
60.0
65.0
70.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
x
B
essel
f
u
n
c
t
i
o
n
I
(
x)
I0(x) I1(x)
Figure 4.6 Bessel function I
0
(x) and I
1
(x).
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
x
B
e
s
s
e
l
f
unc
t
i
on K
(
x
)
K0(x) K1(x)
Figure 4.7 Bessel function K
0
(x) and K
1
(x).
51
Replacing x with r/λ in equation (4.45) gives:
)
r
(
K
c
+ )
r
(
I
c
= s
0201
λλ
(4.47)
To satisfy the boundary condition (4.44c), c
1
must be zero since when r approaches to infinite,
I
0
(x) approaches to infinite while K
0
(x) approaches to zero. Hence the solution reduces to:
)
r
(
K
c
= s
02
λ
(4.48)
c
2
can be determined with the boundary condition (4.44b) as:
)/
r
(K)/r(
1
T2
Q
=
c
ww
1
0
2
λλπ
(4.49)
Substituting (4.49) into (4.48) gives:
)/
r
(K)/r(
)(r/
K
T2
Q
= s
ww
1
00
λλ
λ
π
(4.50)
which is the solution of drawdown caused by pumping in a semi-confined aquifer. It indicates
that the drawdown increases towards the well following the shape of the second kind of
Bessel function of order zero.
Putting r = r
w
in (4.50) gives the drawdown at the pumping well:
)/
r
(K)/r(
)/
r
(
K
T2
Q
=
s
ww
1
w00
w
λλ
λ
π
(4.51)
4.4.4 Analysis of results
Simplification of the solution
In the practical problem, the radius of the well is usually much smaller than the leakage factor.
The following approximation holds:
52
)
r
(1.123 ln )
r
(
K
1 )
r
(
K
r
: 1
r
w
w
0
w
1
w
w
λ
≈
λ
≈
λλ
λ
pp
The solutions are simplified as:
)
r
(
K
T2
Q
= s
0
0
λπ
(4.52)
and
)
r
(1.123 ln
T2
Q
s
w
0
w
λ
π
≈
(4.53)
In the vicinity of the well where r/λ << 1 holds,
)
r
(1.123 ln = )
r
(
K0
λ
λ
Equation (4.52) becomes:
λ
λ
π
r )
r
1.123
ln(
T2
Q
= s
0
pp
(4.54)
Equivalent radius of influence: R
eq
Let , Equation (4.54) becomes: λ 1.123 =
Req
)
r
R
ln(
T2
Q
= s
eq
0
π
(4.55)
Comparing Equation (4.55) with (4.11) for a well in a confined aquifer shows that two
formulas are of the same form. It indicates that the drawdown in the vicinity of a semi-
confined aquifer with a leakage factor λ corresponds to the drawdown near a well in a circular
53
confined aquifer of the same transmissivity with a radius R = 1.123 λ. Therefore, R
eq
= 1.123
λ is called the equivalent radius of influence.
Specific discharge
The specific discharge can be determined with Darcy's law as:
)/
r
(
K
)(r/
K
H
r
2
Q
- =
dr
ds
K = q
w1
1
w
0
r
λ
λ
π
(4.57)
and the total discharge is:
)/
r
(
K
)(r/
K
r
r
Q - = qrH2 = Q
w1
1
w
0rr
λ
λ
π
(4.58)
It shows that the total discharge increases towards the well since more leakage from the top
aquifer joins the flow. The maximum total discharge is found from equation (4.58) when r
equals r
w
, which is exactly the pumping rate.
Velocity
Groundwater flow velocity can be calculated with:
)/
r
(
K
)(r/
K
n
H
r
2
Q
- =
n
q
=
v
w1
1
ew
0
e
r
r
λ
λ
π
(4.59)
The maximum velocity is found when r equals r
w
,
n
H
r
2
Q
- =
v
ew
0
w
π
54
4.4 Methods of superposition and images
4.4.1 Method of superposition
1. Principle
For a linear system governing by a linear partial differential equation as
0 =
y
+
x
2
2
2
2
?
?
?
?
?
?
or
0 =
y
s
+
x
s
2
2
2
2
?
?
?
?
(4.60)
if s
1
and s
2
are solutions of (4.60), the linear combination of
s
+
s
= s
21
βα
is also the solution of (4.60).
2. Application to groundwater flow to wells
The principle of the superposition is applied to two cases of wells operating in a confined
aquifer to find the distribution of the drawdown. The same procedure can be also applied to
unconfined and semi-confined aquifers.
Case 1: Two pumping wells in a confined aquifer
Conceptual hydrogeological model
Figure 4.8 shows two pumping wells operating in a confined aquifer. The assumptions are:
- The aquifer is homogeneous and isotropic;
- The aquifer is bounded with circular constant head boundary;
- A system of two wells is operating with the same rate at a distance of 2p. p << R.
Drawdown distribution
55
Since the distance of wells to the centre of the aquifer (p) is very small, pumping wells can be
considered to be located in the centre of the aquifer. Therefore, the solution of a single well in
a confined aquifer can be applied to each of two pumping wells:
)
r
R
ln(
T2
Q
= s
0
π
y +
x
=r
2
2
y
A
r
2
Q
0
p
well 2
well 1
x
x
r
1
Q
0
Q
0
Q
0
p
Figure 4.8 Two pumping wells in a confined aquifer
Taking any point A in the aquifer, the pumping well 1 ( x = + p, y = 0) will cause drawdown
s
1
at point A as:
y + p) -(x =
r
)
r
R
ln(
T2
Q
=
s
22
1
1
0
1
π
From pumping well 2 ( x = - p, y = 0),
y + p) +(x =
r
)
r
R
ln(
T2
Q
=
s
22
2
2
0
2
π
56
According to the principle of superposition, the total drawdown caused by two pumping wells
at point A will be:
)
rr
R
ln(
T2
Q
=
s
+
s
= s
21
2
0
21
π
or
)
]y + p)+(x[ ]y + p)-(x[
R
( ln
T2
Q
= s
2222
2
0
π
Analysis of solution
1) Groundwater divide
From
]
y + p)+(x
p+x
+
y + p)-(x
p-x
[
T2
Q
- =
x
s
2222
0
π?
?
it can be shown that when x = 0, 0 =
x
s
?
?
. It indicates that a groundwater divide exists at the
middle of two pumping wells along with the y axis.
2) Drawdown near a well
At a point near a well of radius r << p, which will have coordinates of
p r ,sinr =y ,cosr + p =x ppθθ
substituting (4.64) into (4.63) gives the drawdown as:
)
r
2p/
R
ln(
T2
Q
= s
2
0
π
Let
2p
R
=
R
2
eq
, equation (6.65) becomes:
57
)
r
R
ln(
T2
Q
= s
eq
0
π
which is of the same form of the drawdown in a confined aquifer with a single pumping well.
R
eq
is the equivalent radius of influence for a system of two wells operating in a confined
aquifer.
Case 2: A pumping and an injecting wells in a confined aquifer
Conceptual hydrogeological model
Figure 4.9 shows a pumping well and an injecting well operating in a confined aquifer. The
assumptions are:
- aquifer is homogeneous and isotropic;
- confined aquifer is bounded with circular constant head boundary;
- one pumping well and one injecting well are operating with a distance of 2p, p << R.
Drawdown distribution
The drawdown at any point A, caused from the injecting well is:
y + p)-(x =
r
)
r
R
ln(
T2
Q
- =
s
22
1
1
0
1
π
and from pumping well is:
y + p)+(x =
r
)
r
R
ln(
T2
Q
=
s
22
2
2
0
2
π
The total drawdown will be:
)
y + p)+(x
y + p)-(x
ln(
T2
Q
=
s
+
s
= s
22
22
0
21
π
58
y
A
-Q
0
Q
0
p
well 2
well 1
x
r
1
x
0
r
2
p
Q
0
-Q
0
Figure 4.9 One pumping and one injecting wells in a confined aquifer
Analysis of solution
1) the drawdown s is independent of R
Since the pumping rate equals injecting rate, groundwater flows from injecting well to
pumping well. Therefore, it doesn't need recharge from the boundary. Hence, the solution will
be also valid for an infinite confined aquifer.
2) At x = 0, s = 0
It shows that at the middle of two wells, groundwater head keeps constant along y-axis.
Groundwater head decreases in the left half of the aquifer due to the pumping while
groundwater head increases in the right half of the aquifer due to the injection.
3) Drawdown near the pumping well
At a point near the pumping well of radius r << p, which will have coordinates of
x = p + r cos(0), y = r sin(0), r << p, the drawdown becomes:
)
r
2p
ln(
T2
Q
= s
0
π
59
Let R
eq
= 2p, equation (4.70) becomes:
)
r
R
ln(
T2
Q
= s
eq
0
π
which is of the same form of the drawdown in a confined aquifer with a single pumping well.
R
eq
is the equivalent radius of influence for a system of one pumping well and one injecting
well operating in a confined aquifer.
4) Minimum travel time from injecting well to pumping well
It can be understood that a water particle travelling from injecting well to pumping well along
the x-axis will take the minimum time since the distance is shortest. Since y = 0 along the x-
axis, Equation (4.69) reduces to:
p < |x| )
x+p
x-p
ln(
T2
Q
= s
0
π
The specific discharge along the x-axis can be found with Darcy's law as:
x
-p
2p
T2
Q
- =
x
s
K =
x
K - = q
2
2
0
x
π?
?
?
??
The velocity is:
x
-p
2p
H
n
2
Q
- =
v
2
2
e
0
x
π
It indicates that groundwater flow velocity is smallest at the middle of the wells and increases
symmetrically towards the injecting well and pumping well.
The travel time can be found with the following integration:
dx )
x
-p(
Qp
H
n
=
v
dx
=t
2
2
p
p-
0
e
x
-p
p
∫
π
∫
Resulting in:
Q3
pH
n
4
=
0
2
e
π
t
(4.75)
60
4.4.2 Method of images
The method of images is developed to find the distribution of drawdown caused by pumping
in aquifers bounded by constant head or no-flow boundaries. The method is explained with
two cases of confined aquifers. However, the same procedure can be applied to unconfined
and semi-confined aquifers.
Case 1: A pumping well near a constant head boundary
Conceptual hydrogeological model
Figure 4.10 shows a pumping well near a constant head boundary in a confined aquifer. The
assumptions are:
- The confined aquifer is homogeneous and isotropic;
- The aquifer is bounded by a straight constant head boundary
(river, canal, ...) on the left;
- The aquifer extends towards infinity in right half;
- A pumping well is located near the boundary with a constant pumping rate.
Q
0
0
0
+p
x
x
p
s=0
y
Figure 4.10 A pumping well near a constant head boundary in a confined aquifer
61
Conditions for the solution of drawdown
From the physical understanding of the problem, the solution of the drawdown must have the
following properties:
1) it must satisfy the Laplace's equation for x > 0, except in the point (x=p, y=0);
2) s = 0 when x approaches the infinity;
3) s = 0 when x = 0 along the y-axis;
4) the amount of water leaves the aquifer at point (x=p, y=0) must equal pumping rate.
It can be proven that:
0_ x
y + )p-(x
y + )p+(x
ln
T2
Q
= s
22
22
0
π
(4.76)
satisfies the above four conditions.
Method of image
The solution of drawdown (4.76) can be obtained with the following procedure:
- replacing the semi-infinite aquifer with an fictitious infinite aquifer with the same
hydrogeological parameters;
- considering the constant head boundary as a mirror;
- putting an imaginary injecting well at the image location of the pumping well, i.e. at
point (x=-p, y=0);
- giving the injecting rate equal to the pumping rate;
- using the principle of superposition to find the solution.
The real aquifer system is now transformed into a fictitious system of one injecting well and
one pumping well operating in an infinite confined aquifer. According to the principle of
superposition, the drawdown caused by pumping well is
)
y + )p-(x
R
ln(
T2
Q
=
s
22
0
1
π
and by injecting well
)
y + )p+(x
R
ln(
T2
Q
=
s
22
0
2
π
Using the principle of superposition to the region occupied by the original aquifer, the total
drawdown will be
62
y + )p-(x
y + )p+(x
ln
T2
Q
=
s
+
s
= s
22
22
0
21
π
which is exactly the solution of (4.76).
Since the line of x = 0 (y- axis) is considered as a mirror and the location of an imaginary
injection well is the image of the location of the pumping well, the method is, therefore, called
the method of image.
Case 2: A pumping well near an impermeable boundary
Conceptual hydrogeological model
Figure 4.11 shows a pumping well near an impermeable boundary in a confined aquifer. The
assumptions are:
- confined aquifer is homogeneous and isotropic;
- the half circular aquifer is bounded by a straight impermeable boundary on the left;
- a pumping well is located at the distance of p from the boundary with constant
pumping rate, R >> p.
T
0
p
n
0
R
Q
0
0
q=0
x
y
Figure 4.11 A pumping well near an impermeable boundary in a confined aquifer
63
Conditions for the solution of drawdown
From the physical understanding of the problem, the solution of the drawdown must have the
following properties:
1) it must satisfy the laplace's equation for x > 0, except in the point (x=p, y=0);
2) s = 0 when r = R at the circular head boundary;
3) q
x
or ds/dx = 0 at x = 0 along the y-axis;
4) the amount of water leaves the aquifer at point (x=p, y=0) must equal pumping rate.
Method of image
The solution of the drawdown can be obtained with the following procedure:
- replacing the half-circular aquifer with an fictitious circular aquifer with the same
hydrogeological parameters;
- considering the impermeable boundary as a mirror;
- putting an imaginary pumping well at the image location of the real pumping well, i.e.
at point (x=-p, y=0);
- giving the imaginary pumping rate equal to the real pumping rate;
- using the principle of superposition to find the solution.
The real aquifer system is now transformed into a fictitious system of two pumping wells
operating in a circular confined aquifer. According to the principle of superposition, the
drawdown caused by the real pumping well is:
)
y + )p-(x
R
ln(
T2
Q
=
s
22
0
1
π
and by the imaginary pumping well:
)
y + )p+(x
R
ln(
T2
Q
=
s
22
0
2
π
According to the principle of the superposition, the total drawdown will be:
)
]y + )p-[(x ]y + )p+[(x
R
ln(
T2
Q
=
s
+
s
= s
2222
2
0
21
π
(4.77)
The solution (4.77) is valid to the half of region occupied by the real aquifer.
64
4.5 Flow net
4.5.1 Potential function and equipotential lines
1. Potential function
In two-dimensional flow, Darcy's law states:
y
K- = q ,
x
K - = q
yx
?
??
?
??
(4.78)
If the media is homogeneous (K=constant), we define a new function:
K = ?Φ
(4.79)
Substituting (4.79) into (4.78) gives:
y
- = q ,
x
- = q
yx
?
Φ?
?
Φ?
(4.80)
Where Φ is called the groundwater potential.
The potential function Φ satisfies the Laplace's equation in steady flow:
0 =
y
+
x
2
2
2
2
?
Φ
?
?
Φ
?
(4.81)
2. Equipotential lines
It is possible to represent the potential function Φ(x,y) by drawing lines of constant Φ in the
x,y plane. These lines of constant Φ are called the equipotential lines. It is natural to choose a
constant interval between the values of the potential, i.e.
?Φ
ΦΦΦΦ
= ... = - = -
3221
65
4.5.2 Stream function and stream lines
1. Stream function
The stream function Ψ(x,y) is defined as
x
- = q
y
- = q
yx
?
ψ?
?
ψ?
(4.82)
From (4.80), it can be found the relation
x
- =
y
y
=
x ?
Ψ?
?
Φ?
?
Ψ?
?
Φ?
(4.83)
The stream function Ψ(x,y) also satisfies the Laplace equation:
0 =
y
+
x
2
2
2
2
?
Ψ
?
?
Ψ
?
(4.84)
2. Streamlines
It is also possible to represent the stream function Ψ(x,y) by drawing lines of constant Ψ(x,y).
The lines of constant Ψ are called streamlines. When drawing the streamlines, it is also
chosen the constant interval, i.e.
?Ψ
ΨΨΨΨ
= ... = - = -
3221
(4.85)
4.5.3 Flow net
In a two-dimensional flow field, we can now draw a set of equipotential lines and stream
lines. The equipotential line and stream lines will form a set of orthogonal curve lines. These
orthogonal lines are called flow net.
66
n
s
"
y
x
M1
M2
M3
Figure 4.12 Definition of flow net
In any interception point of flow net, two tangent directions can be introduced:
- tangent of potential line: n
- tangent of stream line: s
It can be proven that:
q - =
s
0 =
n ?
Φ?
?
Φ?
(4.86)
0 =
n?
Φ?
means that there are no change of Φ in the direction of n (equi-potential). Therefore,
the specific discharge will be zero.
q- =
s?
Φ?
means that flow is in the direction of s which is perpendicular to the potential line,
and the specific discharge follows the Darcy's law.
It can also be proven that:
q - =
n
0, =
s ?
Ψ?
?
Ψ?
(4.87)
67
0 =
s?
Ψ?
means that Ψ does not vary in s direction which is the direction of flow. Hence, the
direction of flow is everywhere tangent to lines of constant Ψ. That is why the lines of
constant Ψ are called scream lines.
Comparing (4.86), we can find:
n
=
s ?
Ψ?
?
Φ?
This means that in the x,y plane, if lines of constant Ψ and Φ are drawn at intervals of ?Ψ and
?Φ, Then:
n
=
s ?
?Ψ
?
?Φ
(4.88)
Where ?s: distance between two potential lines;
?n: distance between two streamlines.
If we chose ?Ψ?Φ = , then n = s ?? . This means that in a two-dimensional flow field, the
streamlines and equipotential lines will constitute a network of elementary squares. An
example can be seen in Figure 4.13.
The constant flow q between two-adjacent flow lines (a flow channel) is
???
?
??
K =n .
s
K = q (4.89)
If there are m flow channels, total flow will be
??mK = mq = Q
(4.90)
If the total drop of groundwater head in the entire region is φ
1
- φ
2
and there are n squares
between φ
1
and φ
2
, then
n
-
=
21
??
??
(4.91)
Equation(4.90) becomes:
68
)-(K
n
m
=
n
-
mK = Q
21
21
??
??
(4.92)
Q is the total discharge per unit thickness of the aquifer.
Figure 4.13 Examples of flow net
Flow nets are useful graphic methods to display streamlines and equipotential lines. It is also
the approximate method to calculate the flow in a complicated flow field. To sketch the flow
net, boundary conditions must be taken into account:
1) impermeable boundary: flow lines should be parallel to the impermeable boundary or
equipotential lines should be perpendicular to the impermeable boundary.
69
2) constant head boundary is an equipotential line. Therefore, stream line should be
perpendicular to the constant head boundary
3) Groundwater table in an unconfined aquifer is a stream line.
70
5. Non-steady Groundwater Flow in Aquifers
5.1 Basic equations for non-steady flow in aquifers
5.1.1 Confined aquifer
1) General equation for non-steady flow in a confined aquifer
In Chapter 2, general equation for no-steady flow in a confined aquifer is derived as:
t
S
= )
z
(K
z
+ )
y
(K
y
+ )
x
(K
x
s
?
??
?
??
?
?
?
??
?
?
?
??
?
?
(5.1)
where
φ(x,y,z,t): groundwater head, a function of space (x,y,z) and time;
K: hydraulic conductivity;
S
s
: specific storage: the volume of water stored (or released) in an unit volume of
soil by an unit increase (or decrease) of the head;
2) Two-dimensional non-steady flow in a confined aquifer
If vertical flow velocity,
z
K - =
v
z
?
??
, can be neglected, and the thickness of the confined
aquifer is uniform, Equation (5.1) can be reduced to:
t
S = )
y
(T
y
+ )
x
(T
x ?
??
?
??
?
?
?
??
?
?
(5.2)
where:
T=KH: transmissivity;
S=S
s
H : storage coefficient: the volume of water stored (or released) in a
column of soil with unit area and height of the thickness of aquifer due
to the unit increase (or decrease) of the head.
If the aquifer is homogeneous and isotropic (T = constant), Equation (5.2) can be simplified
as:
t
T
S
=
y
+
x
2
2
2
2
?
??
?
?
?
?
?
?
(5.3)
3) One-dimensional flow
71
For one dimensional flow, equation (5.3) will be:
t
T
S
=
x
2
2
?
??
?
?
?
(5.4)
5.1.2 Unconfined aquifer
1) Hydrogeological characteristics
When water table decreases in an unconfined aquifer, groundwater will be drained under
gravity. The drainage of groundwater occurs only in the zone of water table changes. The
amount of groundwater released by gravity drainage will depend on the specific yield, μ.
Specific yield: the volume of water released from the unit volume of soil by gravity drainage,
defined as:
V
V
=
w
μ
where V is the volume of soil and V
w
is the volume of water released from the soil.
Porosity: the percentage of pore space defined as
V
V
=n
p
where V
p
is the volume of pore space.
Effective porosity: the percentage of pore space with full connections defined as
V
V
=
n
e
e
where V
e
is the volume of pore space with full connections.
The following relations hold:
n
n <
n
>n
ee
≈μμ
2) Two-dimensional non-steady flow in an unconfined aquifer
Figure 5.1 shows a soil column taking from an unconfined aquifer, water balance components
are:
Excess flow in x direction: dtdx
x
Q
- =dx)dt
x
Q
+ Q( -dt Q
xx
xx
?
?
?
?
(5.6)
72
Excess flow in y direction: dtdy
y
Q
- =dy)dt
y
Q
+ Q( -dt Q
yy
yy
?
?
?
?
(5.7)
Changes of the storage: dydx dt
t
h
?
?
μ (5.8)
h
y
x
dy
h
Qx
dh
dx
Figure 5.1 A soil column from an unconfined aquifer
Water balance states:
dydx dt
t
h
=dt dy]
y
Q
+dx
x
Q
[ -
y
x
?
?
μ
?
?
?
?
(5.9)
From Darcy's law, we can find:
dy )
x
h
h (K
x
- =
x
Q
dy
x
h
K h - = Q
x
x
?
?
?
?
?
?
?
?
(5.10)
dx )
y
h
h (K
y
- =
y
Q
dx
y
h
K h - = Q
y
y
?
?
?
?
?
?
?
?
(5.11)
Substituting (5.10) and (5.11) into (5.9) gives:
73
t
h
= )
y
h
h (K
y
+ )
x
h
h (K
x ?
?
μ
?
?
?
?
?
?
?
?
(5.12)
which is the general equation for two dimensional flow in an unconfined aquifer without
vertical infiltration. It is clear that the equation for unconfined aquifers is not linear.
3) One-dimensional flow
In one dimensional flow case, Equation (5.12) reduces to
t
h
= )
x
h
h (K
x ?
?
μ
?
?
?
?
(5.13)
for nonhomogeneous aquifer, and
t
h
= )
x
h
(h
x
K
?
?
μ
?
?
?
?
(5.14)
for homogeneous aquifer.
74
5.2 Non-steady Groundwater Flow in a confined aquifer
5.2.1 Conceptual hydrogeological model
- The confined aquifer is homogeneous and extends to infinitely;
- A river forms the head boundary of the aquifer on the left;
- When t=0, groundwater head is the same as the river level;
- When t>0, river level suddenly increases by ?φ;
- Groundwater head in the aquifer will increase in response of the sudden change of
water level in the river.
n
0
)n
T=H K
x
Figure 5.2 A confined aquifer with change of river stage
5.2.2 Mathematical model
The mathematical model for this case is:
0 > x , 0 > t
t
T
S
=
x
2
2
?
??
?
?
?
(5.15a)
(5.15b)
(5.15c)
(5.15d)
0 > t (t) + = |
00 =x
δ????
0 x
= | ??
∞→
00 =t
= | ??
where
φ(x,t): groundwater head as a function of x and t;
T: transmissivity;
75
S: storage coefficient;
?φ: sudden increase of river level;
δ(t): unit-step function defined as
δ |{ =(t)
0 < t 0
0 > t 1
5.2.3 Analytical solution
Partial differential equation (5.15) is a standard mathematical physical equation and the
solution is found with Laplace transform as
]
S/4Tt
x
[ erfc + =
0
φ?φφ (5.16)
erfc{μ} is called the complementary error function defined as
dy )y(-exp
2
= ][ erfc
2
∫
π
μ
∞
μ
(5.17)
μ is defined for this case as
4Tt/S
x
= μ (5.18)
The characteristics of the complementary error function is shown in Figure 5.3. erfc(μ) = 1
when μ = 0. erfc(μ) = 0 when μ approaches infinity.
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
u
C
o
m
p
l
e
m
e
n
t
ar
y er
r
o
r
f
u
n
c
t
i
o
n
er
f
(
u
)
Figure 5.3 Complementary error function
76
5.2.4 Analysis of the result
1) Changes of groundwater head with time
Equation (5.16) shows that groundwater head in the aquifer increases with time. For any point
in the cross-section (x > 0), when t increases, μ decreases and, therefore, erfc(μ) increases,
resulting the increase of φ.
In special case when t approaches infinity, ???? + =
0
since erfc(μ) = 1 when μ = 0. It
indicates that groundwater flow in the aquifer will reach new steady state flow if the river
level stays at the new level.
2) Distribution of groundwater head in the cross-section
For any time, t > 0, with increase of x, μ increases and erfc[μ] decreases, and therefore
groundwater head decreases. When x approaches to infinity, groundwater head approaches φ
0
.
3) Unit-width discharge
From
e
Tt/s
- =
x
S/4Tt
x
2
-
π
??
?
??
(5.19)
we can find the specific discharge as:
e
Tt/s
K
= q
S/4Tt
x
2
-
x
π
??
The unit-width discharge will be:
e
)
t
ST
( =
e
Tt/S
T
= Hq = Q
S/4tT
x
2
-
1/2
s/4Tt
x
2
-
xx
π
??
π
??
(5.20)
It shows that Q
x
decreases with increase of x. When x = 0, the maximum discharge is found
as
)
t
ST
( = Q
1/2
0
π
??
(5.21)
Q
0
decreases with increase of time.
77
5.3 Non-steady Groundwater Flow in an unconfined aquifer
5.3.1 Conceptual hydrogeological model
- An unconfined aquifer between two canals is homogeneous;
- Water levels in canals are constant;
- Starting from t=0, aquifer receives the uniform infiltration (w);
- Water table in the aquifer increases with the infiltration;
- The distribution of water table will be symmetrical with a water divide at the middle
of the aquifer;
L
H
H
K μ
h
z
w
0
x
L
Figure 5.4 An unconfined aquifer between two canals
5.3.2 Mathematical model
From steady state solution (3.16) it is known that the distribution of water table is
symmetrical with the middle of aquifer as the water divide. Therefore, only the solution for
the right half of the aquifer is derived. If the middle of the aquifer is taken as the origin of the
x-coordinate, the mathematical model for this case is:
78
o > t 0, > x
t
h
= w+ )
x
h
(Kh
x ?
?
μ
?
?
?
?
(5.22a)
0 > t 0 = |
x
h
0 =x
?
?
(5.22b)
0 > t H = |h
L =x
(5.22c)
0 > x H = |h
0 =t
(5.22d)
where
h(x,t) : water table as a function of x and t;
K : hydraulic conductivity;
μ : specific yield.
Since partial differential equation (5.22a) is not linear, some linearization has to be made to
find the solution.
In the case where h - H << h, T = Kh can be considered as a constant. Therefore, equation
(5.22a) is simplified as
t
h
= w+
x
h
T
2
2
?
?
μ
?
?
(5.23)
5.3.3 Analytical solutions
1) Solution with Laplace transform
With Laplace transformation, the solution of equation (5.23) is found as:
]
4SL
Tt
)1+[-(2k exp ]
2L
x
1)+[(2kcos
)1+(2k
)(-1
T
wL
16
-
2T
)
x
-
L
w(
+ H =h
2
2
2
3
k
0=k
3
22
π
π
∑
π
∞
(5.24)
2) Schaperg's approximate solution
The solution is
79
}
])[L(S/2Ttcosh
])h[x(S/2Ttcos
- 1 {2t
S
w
+ H =h
1/2
1/2
(5.25)
The steady state solution is found by forcing t to approach infinity,
2T
)
x
-
L
w(
+ H =h : t
22
∞→
(5.26)
3) Barakel's solution
It assumes that the solution will have the form of
f(t)
2T
)
x
-w(L
+ H =h
2
2
(5.27)
It is clear that f(t) must satisfy the following two conditions:
- initial condition: f(t) = 0 when t = 0; and
- steady state condition: f(t) = 1 when t→ 4.
If only the total water balance is required in the whole section of aquifer, equation (5.23) is
relaxed to:
0 =dx ) w-
x
h
T -
t
h
(S
2
2
L
0
?
?
?
?
∫
(5.28)
Substituting (5.27) into (5.28) gives:
1) - (f
SL
3T
- =
dt
df
2
(5.29)
which is an ordinary differential equation of function f(t). The solution of (5.29) gives:
)
L
3Tt/S - ( exp - 1 = f(t)
2
The Braket's approximate solution will be:
)]
SL
(-3Tt/ exp -[1
2T
)
x
-
L
w(
+ H =h
2
22
(5.31)
80
5.3.4 Analysis of results
The flow characteristics are analyzed based on Brakel's approximate solution.
1) Fox a fixed x > 0, with increase of time t, h increases due to the infiltration. Especially
when t approaches infinity, a steady state solution is found as:
2T
)
x
-
L
w(
+ H =h
22
2) Fox any time t > 0, with increase of x, h decreases. h = H when x = L.
3) Unit-width discharge can be found with Darcy's law.
)]
SL
(-3Tt/ exp - 1 [
T
wx
- =
x
h
2
?
?
)]
SL
(-3Tt/ exp-[1
h
wx
= q
2
x
(5.32) )]
SL
(-3Tt/ exp-[1 wx = qh = Q
2
xx
It shows that Q
x
increases with increase of x.
When x = L,
)]
SL
(-3Tt/ exp-[1 wL= Q
2
L
The maximum discharge is found when t approaches infinity: Q
L
= w L.
81
82
6. Non-steady Groundwater Flow to Wells
6.1 Groundwater flow to a well in a confined aquifer
6.1.1 Conceptual hydrogeological model
- The confined aquifer is homogeneous and isotropic;
- The aquifer extends to infinity;
- A pumping well fully penetrates the aquifer with constant pumping rate;
- Before the pumping, groundwater head in the aquifer is horizontal;
- After pumping with constant rate, a cone of depression of groundwater head is
developed in space with time;
- The aquifer extends to the infinity where groundwater head keeps constant.
dr
r
Q
0
n
0
S
H K
r
dr
r
Figure 6.1 Non-steady flow to a well in a confined aquifer
83
6.1.2 Mathematical model
The mathematical model can be developed with transformation of Equation (5.2) into polar
coordinates. However, it is developed here with the principle of water balance in order to
understand the physics of the flow.
The water balance for non-steady groundwater flow states
inflow - outflow = changes of storage
Taking a circular ring from the aquifer, the water balance components are:
inflow to a circular ring: Q
r
dt
outflow from a circular ring: dt dr)
r
Q
+ Q(
r
r
?
?
changes of storage: dt
t
.rdr 2 . S = dV
?
??
π
The water balance equation will be:
dt
t
.rdr 2 S. =dt dr)
r
Q
+ Q( -dt Q
r
rr
?
??
π
?
?
(6.1)
Since
r
rHK 2 - = Q
r
?
??
π (6.2)
)
r
(r
r
T2 - =
r
Q
r
?
??
?
?
π
?
?
(6.3)
Substituting (6.2) and (6.3) into (6.1) gives:
t
S = )
rr
1
+
r
( T
2
2
?
??
?
??
?
?
?
(6.4a)
which is the governing equation for non-steady radial flow to the well in a confined aquifer.
The boundary conditions are:
84
??
∞→ 0r
= | t)(r,
(6.4b)
T2
Q
= )
r
(r
lim
0
0r π?
??
→
(6.4c)
The initial condition is:
(6.4d)
??
00=t
= | t)(r,
Equation (6.4a,b,c,d) consist of mathematical model for non-steady radial flow to a well in a
confined aquifer.
6.1.3 Analytical Solution: Theis formula (1935)
Theis found the solution of Equation (6.4) as:
du
u
e
T4
Q
- =
-u
u
0
0
∫
π
??
∞
(6.5)
where
4Tt
S
r
=u
2
(6.6)
is an intermediate variable.
Let:
4Tt
S
r
=u du
u
e
= W(u)
2-u
u
∫
∞
(6.7)
W(u) is called the Theis well function. Equation (6.5) becomes
W(u)
T4
Q
- =
0
0
π
??
(6.8)
or in terms of drawdown as:
85
W(u)
T4
Q
= s
0
π
(6.9)
Equation (6.8) or (6.9) is well-known Theis formula. The series expression of the Theis well
function is:
n! .n
u
)(-1
+ (u) ln - 0.577216 - = (u)W
n
1+n
1=n
∑
∞
(6.10)
From the definition (6.7) it is clear that W(u) decreases with the increase of u. Especially
W(u) approaches zero when u approaches infinity. The values of Theis function is usually
tabulated in a table as in Annex T.1.
6.1.4 Analysis of the solution
1) Changes of drawdown with r for any time t = t
1
For a given time t = t
1
, u will be a function of only r:
t
4T
S
r
=u
1
2
. Equation (6.9) shows that the
drawdown decreases with the increase of radial distance r. The drawdown becomes zero when
r approaches infinity. The maximum drawdown occurs at the well (r = r
w
). It is clear that a
cone of depression is created with pumping.
2) Changes of drawdown with time t for r = r
1
For a given radial distance r = r
1
, u will be a function of only t:
4Tt
S
r
=u
2
1
. Equation (6.9)
shows that the drawdown increases with the increase of time. It is clear that the cone of
depression is developing deeper and larger with continuous pumping.
3) Groundwater flow through a circular ring
Since
e
T2
Q
- =
r
r
4Tt
S
r
2
-0
π?
??
the total discharge will be:
86
e
Q =
r
Tr 2 - = Q
4Tt
S
r
2
-
0r
?
??
π
(6.11)
It is clear that Q
r
< Q
0
for any time t. That is the difference from the steady state flow. In the
non-steady flow, groundwater pumped from the aquifer is from the release of the aquifer
storage due to the decrease of groundwater head.
6.1.5 Simplification of Theis formula: Jacob's formula
When 1 ?
4Tt
S
r
=u
2
, )
S
r
2.25Tt
ln( W(u)
2
≈ , Equation (6.9) reduces to:
)
S
r
2.25Tt
ln(
4Tt
Q
= s
2
0
(6.12)
or
)
S
r
2.25Tt
log(
4Tt
Q 2.3
= s
2
0
(6.13)
Equation (6.12) is called Jacob's formula. It is only a good approximation of Theis formula
near pumping well and after long time of pumping. In practice, Jacob's formula is often used
since it is a simple logarithm function.
6.1.6 Applications
1) Prediction of s or Q
0
Once the hydrogeological parameters (T and S) are known, the drawdown cause by given
pumping rate can be predicted with:
W(u)
T4
Q
= s
0
π
In some cases, for example, dewatering a building site, the drawdown is needed to keep
below a certain limit. The required pumping rate can be calculated with
87
W(u)
Ts4
= Q
0
π
2) Pumping test analysis with Jacob method
A: Drawdown-time approach
Jacob's formula (6.13) can be rearranged as:
)tlog(
T4
Q2.3
+ )
S
r
2.25T
( log
T4
Q2.3
= s
0
2
0
ππ
(6.14)
Giving the radial distance, Equation (6.14) is a straight line on a semi-logarithm paper with
the slope as
T4
Q2.3
0
π
.
It is clear that the plot of the observed drawdown (linear scale) against time (logarithmic
scale) from an observation well (r is known) on a semi-logarithm paper will show a straight
line (Figure 6.2).
10
1
)logt=1
i=)s/)logt
t
logarithm scale
)s
t
0
s
10
2
10
3
10
0
Figure 6.2 Drawdown-time plot on a semi-logarithmic paper
Procedures for the determination of aquifer parameters are as following.
88
a: Draw a straight line fit to drawdown-time plot and determine the slope of the straight
line as:
)tlog(
s
=
)t(log - )t(log
s
-
s
= i
12
12
?
?
Usually one logarithmic cycle is taken so that ?log(t) = 1 and i = ?s.
Since
T4
Q 2.3
= i
0
π
the transmissivity is determined by:
i
Q
0.183 =
i4
Q 2.3
= T
00
π
(6.15)
b: Extend the straight line to intersect the t-axis and read t
0
. The point of intersection
means that s = 0 when t = t
0
, i.e.
0 = )
S
r
t
2.25T
log(
T4
Q 2.3
= s
2
00
π
(6.16)
which is only possible when
1 =
S
r
t
2.25T
2
0
resulting
r
t
2.25T
= S
2
0
(6.17)
B: Drawdown-distance approach
Jacob's formula (6.13) can be rearranged as:
)rlog(
T4
Q2.3
- )
S
t2.25T
log(
T4
Q2.3
= s
00
ππ
(6.18)
89
Given the specific time t, Equation (6.18) is a straight line on a semi-logarithm paper with the
slope as
T2
Q2.3
0
π
.
It is clear that the plot of the observed drawdown (linear scale) against radial distance
(logarithmic scale) from several observation wells on a semi-logarithm paper will show a
straight line (Figure 6.3).
10
1
0
r
0
logarithic scale
)s
s
r
)log r=1
i= )s / )log r
10
2
10
3
10
0
Figure 6.3 Drawdown-distance plot on a semi-logarithmic paper
Procedures for the determination of aquifer parameters are as following.
a: Draw a straight line fit to drawdown-distance plot and determine the slope of the
straight line as:
)rlog(
s
=
)r(log - )r(log
s
-
s
= i
12
12
?
?
Usually one logarithmic cycle is taken so that ?log(r) = 1 and i = ?s.
Since:
90
T2
Q 2.3
= i
0
π
the transmissivity is determined by
i
Q
0.366 =
i2
Q 2.3
= T
00
π
(6.19)
b: Extend the straight line to intersect the r-axis and read r
0
. The point of intersection
means that s = 0 when r = r
0
, i.e.
0 = )
S
r
t2.25T
log(
T4
Q 2.3
= s
0
2
0
π
This indicates:
1 =
S
r
t2.25T
0
2
Resulting in:
r
t2.25T
= S
0
2
(6.20)
3) Pumping test analysis with Theis method
Method of Type Curve Matching
Taking logarithm transformation of Theis drawdown formula:
W(u)
T4
Q
=
0
π
s (6.21a)
4Tt
S
r
=u
2
(6.21b)
gives:
)
T4
Q
log( + W(u)log = )slog(
0
π
(6.22a)
91
)
4T
S
r
log( + )
u
1
log( = )tlog(
2
(6.22b)
On double logarithm paper, curve log(s) – log(t) is of the same shape of curve log W(u) -
log(1/u), only with some vertical and horizontal shift. The curve log W(u) - log (1/u) is called
type curve and can be prepared with the table of Theis Well Function. The curve log(s) –
log(t) can be plotted on transparent double logarithm paper with measurements from an
observation well in a pumping test. After two curves are matched by moving log(s) – log(t)
curve while keeping two coordinate systems parallel (Figure 6.4), any common point on the
matched curves can be found and the following data can be read:
s', t' on log s - log t curve
W'(u), (1/u)'on log W(u) - log (1/u) curve
These four values satisfy the Equations (6.21a) and (6.21b). From Equation (6.21a), the
transmissivity can be determined:
(u)W
s4
Q
= T
0
′
′π
(6.23)
From equation (6.21b), the storage coefficient is found:
)(1/u
r
t 4T
= S
2
′
′
(6.24)
Figure 6.4 Theis type curve match (from Kruseman and de Ridder, 1990)
92
Procedure:
The procedures of application of Theis type curve match method into analysis of pumping test
data are:
a: Prepare Theis type curve W(u) - 1/u on a double logarithm paper;
b: Plot drawdown time measurements from a pumping test on the double logarithm
paper with the same scale as type curve;
c: Match curve log(s) – log(t) with type curve log W(u) – log(1/u)
while keeping two coordinate systems parallel;
d: Find a common point on the matched curves and read the following data
s', t', W'(u), (1/u)';
e: Calculate transmissivity and storage coefficient with equations (6.23) and (6.24),
respectively.
93
6.2 Non-steady groundwater to a well in an unconfined aquifer
6.2.1 Conceptual hydrogeological model
- Unconfined aquifer is homogeneous and isotropic, extends to infinite laterally;
- There is vertical recharge or leakage;
- Water table is horizontal before pumping;
- When pumping with a constant rate, water table decreases with time;
- It is assumed that the drawdown is very small comparing to the thickness of aquifer,
vertical flow can be neglected.
dr
r
Q
0
r
r+dr
r
s(r,t)
h(r,t)
h
0
K . S
Figure 6.5 Non-steady radial flow to a well in an unconfined aquifer
6.2.2 Mathematical model
Water balance components for a circular ring taking from the aquifer are:
flow in: Q
r
dt
flow out: dr)dt
r
Q
+ Q(
r
r
?
?
94
Changes of storage: dV
Water balance gives
dV =dt dr]
r
Q
+ Q[ -dt Q
r
rr
?
?
(6.25)
Since:
(6.26)
r
h
rkh 2 - = Q
r
?
?
π
]
r
h
h + )
r
h
(h
r
[rk 2 - =
r
Q
r
?
?
?
?
?
?
π
?
?
(6.27)
dt
t
h
rdr 2 = dV
?
?
μπ (6.28)
substituting (6.26), (6.27) and (6.28) into (6.25) gives
t
h
= ]
r
h
r
1
+
r
h
[Kh
22
2
22
?
?
μ
?
?
?
?
(6.29a)
which is the governing equation of non-steady radial flow to a well in an unconfined aquifer.
It is noticed that the equation is not linear.
Initial condition:
h
= |t)(r,h
0
0=t
(6.29b)
Boundary conditions:
h
= |t)(r,h
0
t ∞→
K
Q
= )
r
h
(r
lim
0
2
0r π?
?
→
(6.29c)
(6.29d)
Equations (6.29a,b,c,d) consist of mathematical model for non-steady groundwater flow to a
well in an unconfined aquifer.
95
6.2.3 Analytical Solution
Analytical solution of the model (6.29) can only be found with the following linearizations:
1) Let T = K h. T is a function of time t and radial r in general. However, if s(r,t) << h
0
, T
can be approximated as a constant of T = K h
m
. h
m
is average saturated thickness of
the aquifer.
2) let
h
2
1
= ,
h
2
1
=
2
0
0
2
??
The original model will be transformed into:
t
= )
r
r
1
+
r
( T
2
2
?
??
μ
?
??
?
?
?
(6.30a)
??
00=t
= |
(6.30b)
??
∞→ 0t
= |
(6.30c)
K2
Q
= )
r
(r
lim
0
0r π?
??
→
(6.30d)
The transformed model (6.30) is a linear model and is similar with the mathematical model
for confined aquifer. By comparing two models, the solution can be found as:
W(u)
K4
Q
- =
0
0
π
??
(6.31)
Replacing φ with h in equation (6.31) gives:
W(u)
K2
Q
-
h
=
h
0
0
2
2
π
(6.32)
where:
4Tt
r
=u
2
μ
(6.33)
Equation (6.32) is the approximate solution of groundwater table. The drawdown will be:
96
W(u)
K2
Q
= s)-
h
s(2
0
0
π
(6.34)
or
W(u)
T4
Q
= )
h
2
s
-s(1
0
0
π
(6.35)
where T = k h
0
. Equation (6.35) is similar as Theis formula.
6.2.4 Analysis of results
1) Jacob formula for unconfined aquifer
when )
r
2.25Tt
ln( W(u)1, ?
4Tt
r
=
2
2
μ
≈
μ
u , equation (6.35) becomes:
)
r
2.25Tt
ln(
T4
Q
= )
h
2
s
-s(1
2
0
0
μπ
(6.36)
or
)
r
2.25Tt
log(
T4
Q2.3
= )
h
2
s
-s(1
2
0
0
μπ
(6.37)
2) When s
w
/ h
0
< 0.1, s
w
is the drawdown in the well, 0
h
2
s
0
≈ . Equations (6.35) and
(6.37) become:
(6.37)
W(u)
T4
Q
= s
0
π
)
r
2.25Tt
log(
T4
Q2.3
= s
2
0
μπ
(6.38)
Equations (6.37) and (6.38) are the same as Theis and Jacob formula for the confined aquifer.
However, since μ >> S, the same pumping rate will cause smaller drawdown in the
unconfined aquifer than in the confined aquifer.
97
3) When 0.1 < s
w
/ h
0
< 0.3, Equations (6.35) and (6.37) have to be applied.
Let
)
h
2
s
-(1 s =
s
0
c
(6.39)
W(u)
T4
Q
=
s
0
c
π
(6.40)
or
)
r
2.25Tt
log(
T4
Q2.3
=
s
2
0
c
μπ
(6.41)
Therefore, the measurements of drawdown (s) from the observation wells can be modified
with formula (3.39) to get s
c
. s
c
will satisfy Theis and Jacob type of formula for the confined
aquifer.
4) When s
w
/h
0
< 0.3, the vertical flow component near the well can't be neglected, Theis
type formula can't be applied.
6.2.5 Application of the solution
1) Predication of drawdown with proposed pumping rates
When parameters T and μ are known, given Q
0
, s(r,t) can be predicted with Equations (6.35)
or (6.36).
2) Determination of pumping rate with the require s(r,t)
When parameters T and μ are known, given s(r,t), Q
0
can be determined with Equations
(6.35) or (6.36).
3) Analysis of pumping test
In cases where the drawdown is much smaller than the saturated aquifer thickness, Theis type
curve matching and Jacob straight line methods can be applied to the analysis of pumping test
data from an unconfined aquifer to determine parameters T and μ. However, the drawdown
measurements from the observation wells should be modified with formula (6.39).
98
6.3 Non-steady groundwater flow to a well in a semi-confined
aquifer
6.3.1 Conceptual hydrogeological model
- Semi-confined aquifer is homogeneous and isotropic with uniform thickness;
- Semi-permeable layer is homogeneous with uniform thickness (d);
- Before pumping, groundwater head in the semi-confined aquifer is the same as the
water table in the above phreatic aquifer;
- When pumping in the semi-confined aquifer with constant rate Q
0
, water in the
unconfined aquifer leaks to the semi-confined aquifer while water table keeps
constant;
- It is assumed that groundwater flow in the semi-confined aquifer is horizontal radial
flow while the flow in the semi-permeable layer is only vertical.
- The changes of storage in semi-permeable layer are negligible.
Q
0
n
0
K',d
K,H
n(r,t)
rdr0
Semi-confined
Semi-permeable
Unconfined
Figure 6.6 Non-steady radial flow to a well in a semi-confined aquifer
99
6.3.2 Mathematical model
Taking a circular ring in the semi-confined aquifer, water balance components are:
flow in: dt Q +dt Q
lr
flow out: dt dr]
r
Q
+ Q[
r
r
?
?
Changes of storage: dV
Water balance gives
dV =dt dr
r
Q
-dt Q
r
l
?
?
(6.42)
Since
(6.43)
r
rkH2 - = Q
r
?
??
π
]
r
+
r
[r T2 - =
r
Q
2
2
r
?
??
?
?
?
π
?
?
(6.44)
c
-
rdr 2
d
-
k ]
r
- )dr+(r[ = Q
0022
l
??
π≈
??
′ππ
(6.45)
dt
t
rdrS2]
r
-)dr+(r[dt
r
S = dV
2
2
?
??
π≈ππ
?
??
(6.46)
Substituting (6.44), (6.45), and (6.46) into (6.42) gives:
t
S =
c
-
+ )
r
r
1
+
r
( T
0
2
2
?
??
??
?
??
?
?
?
(6.47a)
which is the governing equation of non-steady radial flow to a well in a semi-confined
aquifer.
100
Initial condition:
??
00=t
= |t)(r, (6.47b)
Boundary conditions:
??
∞→ 0r
= |t)(r, (6.47c)
T2
Q
= )
r
(r
Lim
0
0r π?
??
→
(6.47d)
Equations (6.47a,b,c,d) consist of the mathematical model of non-steady radial flow to a well
in the semi-confined aquifer.
6.3.3 Analytical solution
Hantush-Jacob solution
The solution is:
]dz
z4
r
- [-z exp
z
1
T4
Q
- =
2
2
u
0
0
λ
∫
π
??
∞
(6.48)
where is leakage factor; and c T =
2
λ
4Tt
S
r
=u
2
Let:
]dz
z4
r
- [-z exp
z
1
= )
r
F(u,
2
2
u λ
∫
λ
∞
(6.49)
Equation (6.48) becomes:
)
r
F(u,
T4
Q
- =
0
0
λπ
??
(6.50)
or
)
r
F(u,
T4
Q
= s
0
λπ
(6.51)
101
Where )
r
(u, F
λ
is the well function of the semi-confined aquifer. It is a function of r and t,
depending on aquifer parameters T, S, and λ. Its value is usually tabulated as in annex T.2.
6.3.4 Analysis of the results
Type curves of the well function F(u,r/λ) is shown in Figure 6.7.
Figure 6.7 Type curves of the well function F(u,r/λ) ((from Kruseman and de Ridder, 1990)
1) Change of drawdown with time
For a fixed r, ↑↑
λ
↑↓↑ s _ )
r
F(u, _
u
1
_
t4T
S
r
=u _ t
2
. Drawdown increases with the increase
of time.
2) Change of drawdown with radial distance
102
For a give time, ↓↓
λ
↓↑↑ s _ )
r
F(u, _
u
1
_ u _ r .
Drawdown decreases with the increase of distance.
3) Influence of leakage factor on drawdown
For given r and t, ↑↑
λ
↓
λ
↑ s _ )
r
F(u, _
r
_ . λ
Drawdown is larger when leakage factor is larger.
4) Comparison with the confined aquifer
For the same u, )
r
F(u, > W(u)
λ
. It indicates that with same pumping rate Q
0
, drawdown in
confined aquifer is lager than that in the semi-confined aquifer. When
W(u) )
r
F(u, , 0, k →
λ
∞→λ→′ . Equation (6.51) becomes Theis formula. Therefore,
confined aquifer is a special case of semi-confined aquifers.
5) Steady state drawdown
When )
r
(
K
2 )
r
F(u, 0,u ,t
0
λ
≈
λ
→∞→ . Groundwater flow becomes steady when leakage
induced by the development of cone of depression equals the pumping rate. The steady state
drawdown is:
)
r
(
K
T2
Q
=
s
0
0
max
λπ
(6.52)
where K
0
is the second type of Bassel function of order zero.
103
6.3.5 Application of the solution
1) Predication of drawdown with given pumping rate
When parameters T, S and c are known, drawdown caused by given Q
0
can be predicted with
equation (6.51). The value of F(u, r/λ) can be found from table T.2.
2) Determination of pumping rate with given drawdown
When parameters T, S and c are known, given s, the required pumping rate Q
0
can be
determined with
)
r
F(u,
ts4
= Q
0
λ
π
(6.53)
3) Analysis of pumping test data
A: Walton's method - type curve matching method
The method is similar as Theis type curve matching method. Figure 6.8 shows the match of
the observation data to one of type curves.
Procedures are:
- Plot on log-log paper F(u,r/λ) versus 1/u for different values of {r/λ}: this gives a
family of type curves;
- Plot for one of the observation wells the drawdown versus the corresponding time t on
another sheet of log-log paper of the same scale. this gives the observed data time-
drawdown data curve;
- Match the observed data curve with one of the type carves, read the value (r/λ);
- Select a match point A and read for A the values ;t , s , )
u
1
( , )
r
(u,F ′′′
λ
′
- Calculate )
r
(u,F
s4
=
0
λ
′
′π
Q
T ;
- Calculate
)
u
1
(
r
t
=
2
′
′4T
S ;
- Calculate
)(r/
r
=
′λ
λ ;
- Calculate
T
=
2
λ
c ;
104
- Repeat the procedure for all observation wells. The calculated values of T, S, c should
show reasonable agreement.
Figure 6.8 Walton's type curve matching method (from Kruseman and de Ridder, 1990)
B: Hantush's inflection-point method
Hantush's inflection point method is based on the characteristic of s-log(t) cure. It can be
proved that, an inflection-point exists on the s-log(t) curve (Figure 6.9). At the inflection-point
p, the flowing relations hold:
)
r
(
K
T4
Q
=
s
0.5 =
s
0
0
p
λπ
max
(6.54)
105
λ2
r
=
u
p
(6.55)
The slope at the inflection point is:
e
T4
Q2.3
=
s
r/-0
p
λ
π
?
(6.56)
and
)
r
(
K
e
=
s
s
2.3
0
r/-
p
p
λ?
λ
(6.57)
With equation (6.54), (6.57), (6.58) and (6.59), a procedure is developed to calculate T, S and
c from pumping test data.
Procedures are:
- Plot the observed s versus t on a semi-log paper (t on logarithmic scale);
- Estimate the maximum drawdown s
max
by extrapolating the curve;
- Calculate s
p
= 0.5 s
max
. From s
p
, locate the inflection point p on the s-log(t) curve;
- Read the value of t
p
at the inflection point from the time-axis;
- Draw a tangent line of the curve through inflection point and Determine the slope ?s
p
of the tangent line at the inflection point;
- With values of s
p
and ?s
p
, determine (r/λ) from equation
)
r
(
K
e
=
s
s
2.3
0
r/
p
p
λ?
λ
using the table e
x
K
0
(x) (annex T.3) by interpolation.
- Calculate
)(r/
=
λ
r
λ ;
- With the values ?s
p
, Q
0
, and (r/λ), calculate T from
e
s
4
Q2.3
= T
r/-
p
0 λ
?π
- With the values T, t
p
, r and λ, calculate S from
λr
t
2T
= S
p
- With values λ and T, calculate c from
106
T
= c
2
λ
Figure 6.9 Hantush's inflection-point method (from Fetter, 1994)
Example: Pumping test of Dalem with Q
0
=761 m
2
/d, r=90 m
1) Plot s-logt curve on semi-log paper (Figure 6.10);
2) Find s
max
by extrapolation: s
max
=0.147 m;
3) Calculate the drawdown at inflection point:
s
p
= 0.5 s
max
= 0.0735 m;
4) Locate the inflection point on s-logt cure and read:
t
p
= 2.8 * 10
-2
d;
5) Draw a tangent line of the curve through inflection point and find the slop of the
tangent line m 0.072 =
s
p
?
Using Equation
)
r
(
K
e
= 2.34 =
0.072
0.0735
2.3 =
s
s
2.3
0
r/
p
p
λ?
λ
from Table T.3, find (r/λ) = 0.15
107
then m 600 =
0.15
90
=
0.15
=
r
λ
6) Calculate T from equation
/d
m
1665 =
e
*
0.072*3.14*4
761 * 2.3
=
e
s
4
Q2.3
= T
20.15-
r
-
p
0
λ
?π
7) Calculate
d 216 =
1665
)(600
=
T
= C
2
2
λ
8) From
t
4T
S
r
=
2
r
=
u
p
2
p
λ
Calculate:
10
*1.7 =
90*600*2
10
*2.8*1665*4
=
r2
t
4T
= S
3-
-2
p
λ
9) Verify the extrapolated s
max
:
Chose a time t=0.1 d, calculate s (t = 0.1) with
0.02 =
10
*1665*4
10
*1.7*
90
=
t4T
S
r
=u
1-
-322
0.15 =
r
λ
m 0.113 = 3.11 *
1665*3.14*4
761
= )
r
F(u,
T4
Q
= 0.1)=s(t
0
λπ
The point (0.1,0.113) falls on the s-log t cure. It justified that the extrapolated S
max
is
ok.
108
Figure 6.10 Analysis of pumping test data from Dalem (from Kruseman and de Ridder, 1990)
109
7. References
Bear, J. 1972, Dynamics of Fluids in Porous Media, American Elservier, New York.
Bear, J. 1979, Hydraulics of Groundwater, McGraw-Hill, New York.
Bear, J. and A. Verruijt, 1987, Modelling Groundwater Flow and Pollution, D. Reidel
Publishing Company.
Fetter, C.W., 1994, Applied Hydrogeology, 3
rd
Edition, Macmillian College Publishing
Company, New York.
Kruseman, G.P. and N.A. de Ridder, 1989, Analysis and Evaluation of Pumping Test Data,
Publication 47, International Institute for Land Reclamation and Improvement/ILRI,
Wageningen.
Verruijt, A. Theory of Groundwater Flow, The MacMillan Press Ltd, London.
110
8. Annexes
T.1 Well function of confined aquifers
111
T.2 Well function of semi-confined aquifers
112
T.3 Function
e
(x)
K0
x
113
T.4 Function (x),
I
, (x)
I
10
(x)
K
(x),
K 10
x I0(x) I1(x) K0(x) K1(x)
0.00 1.0000 0.0000
0.10 1.0025 0.0501 2.4271 9.8538
0.20 1.0100 0.1005 1.7527 4.7760
0.30 1.0226 0.1517 1.3725 3.0560
0.40 1.0404 0.2040 1.1145 2.1844
0.50 1.0635 0.2579 0.9244 1.6564
0.60 1.0920 0.3137 0.7775 1.3028
0.70 1.1263 0.3719 0.6605 1.0503
0.80 1.1665 0.4329 0.5653 0.8618
0.90 1.2130 0.4971 0.4867 0.7165
1.00 1.2661 0.5652 0.4210 0.6019
1.10 1.3262 0.6375 0.3656 0.5098
1.20 1.3937 0.7147 0.3185 0.4346
1.30 1.4693 0.7973 0.2782 0.3725
1.40 1.5534 0.8861 0.2437 0.3208
1.50 1.6467 0.9817 0.2138 0.2774
1.60 1.7500 1.0848 0.1880 0.2406
1.70 1.8640 1.1963 0.1655 0.2094
1.80 1.9896 1.3172 0.1459 0.1826
1.90 2.1277 1.4482 0.1288 0.1597
2.00 2.2796 1.5906 0.1139 0.1399
2.10 2.4463 1.7455 0.1008 0.1227
2.20 2.6291 1.9141 0.0893 0.1079
2.30 2.8296 2.0978 0.0791 0.0950
2.40 3.0493 2.2981 0.0702 0.0837
2.50 3.2898 2.5167 0.0623 0.0739
2.60 3.5533 2.7554 0.0554 0.0653
2.70 3.8417 3.0161 0.0493 0.0577
2.80 4.1573 3.3011 0.0438 0.0511
2.90 4.5027 3.6126 0.0390 0.0453
3.00 4.8808 3.9534 0.0347 0.0402
3.10 5.2945 4.3262 0.0310 0.0356
3.20 5.7472 4.7343 0.0276 0.0316
3.30 6.2426 5.1810 0.0246 0.0281
3.40 6.7848 5.6701 0.0220 0.0250
3.50 7.3782 6.2058 0.0196 0.0222
3.60 8.0277 6.7927 0.0175 0.0198
3.70 8.7386 7.4357 0.0156 0.0176
3.80 9.5169 8.1404 0.0140 0.0157
3.90 10.3690 8.9128 0.0125 0.0140
4.00 11.3019 9.7595 0.0112 0.0125
4.10 12.3236 10.6877 0.0100 0.0111
4.20 13.4425 11.7056 0.0089 0.0099
4.30 14.6680 12.8219 0.0080 0.0089
4.40 16.0104 14.0462 0.0071 0.0079
4.50 17.4812 15.3892 0.0064 0.0071
4.60 19.0926 16.8626 0.0057 0.0063
4.70 20.8585 18.4791 0.0051 0.0057
4.80 22.7937 20.2528 0.0046 0.0051
4.90 24.9148 22.1993 0.0041 0.0045
5.00 27.2399 24.3356 0.0037 0.0040
5.10 29.7889 26.6804 0.0033 0.0036
5.20 32.5836 29.2543 0.0030 0.0032
5.30 35.6481 32.0799 0.0027 0.0029
5.40 39.0088 35.1821 0.0024 0.0026
5.50 42.6946 38.5882 0.0021 0.0023
5.60 46.7376 42.3283 0.0019 0.0021
5.70 51.1725 46.4355 0.0017 0.0019
5.80 56.0381 50.9462 0.0015 0.0017
5.90 61.3766 55.9003 0.0014 0.0015
6.00 67.2344 61.3419 0.0012 0.0013
114
115
9. Exercises
9.1 Fundamental Equations of Groundwater Flow
Problem 1
In a laboratory Darcy's experiment is used to determine the permeability of a sand sample.
The length of the sample is 20 cm and its cross-sectional area is 10 cm
2
. The difference in
head between the two ends is 25 cm, and the amount of water flowing through the sample in
5 min is measured to be 75 cm
3
. Please calculate the value of the hydraulic conductivity.
Problem 2
For a certain type of sand the hydraulic conductivity has to be measured to be K =2x10
-4
m/s,
with water at a tempreture of 20
o
C (kinematic viscosity v = 10
-6
m
2
/s). What would be the
value at a tempreture of 5
o
C (kinematic viscosity v = 1.5x10
-6
m
2
/s).
Problem 3
For a certain type of sand the hydraulic conductivity has been measured to be k=2 * 10
-4
m/s.
The porosity of the sand was n = 0.40. What would be the hydraulic conductivity if the sand
were compacted (for instance by vibration) so that the porosity is reduced to 0.35, without
crushing the particles?
116
9.2 Steady Groundwater Flow in Aquifers
Problem 1
Two rivers located 1000 m apart fully penetrate a confined aquifer (Figure 1). The
confined aquifer consists of gravels in left part (500 m) and sand in right part (500 m). The
parameters of the aquifer are:
Hydraulic conductivity: K
1
= 200 m/day; K
2
= 50 m/day;
Effective porosity: n
1
= 0.25; n
2
= 0.30;
Thickness of the aquifer: H = 10 m;
Water level in left river H
0
= 20 m;
Water level in right river H
L
= 18 m;
H0
HL
500 m 500 m
L
k1
n1
k2
n2
Figure 1 Groundwater flow in a confined aquifer
a) determine the distribution of groundwater head;
b) calculate the velocity of groundwater in the aquifer. Is the velocity same in
the two parts of aquifer ?
c) calculate the unit width discharge to the right river;
d) calculate the travel time of a water particle travelling from left river to right river.
117
Problem 2
Two rivers located 1000 m apart fully penetrate a phreatic aquifer (Figure 2). The
parameters of the aquifer are:
Hydraulic conductivity K = 0.5 m/day;
Uniform recharge w = 1.369 * 10
-4
m/day;
Water level in left river h
0
= 20 m;
Water level in right river h
L
= 18 m;
d
L
h0
hL
w
w
ww
Recharge
hmax
x
0
Figure 3.2 Groundwater flow in an unconfined aquifer
(a) derive the formula for the calculation of unit width discharge Q.
(b) determine the location (d) and height (h
max
) of the water divide.
(c) what is the unit width discharge of the aquifer into left and right rivers ?
118
Problem 3
A dike (Figure 3) was built to separate a lake from low-lying land so that the land can be
reclaimed for agricultural use. This low-lying land is called a polder in the Netherlands. The
drainage has to be implemented to keep the required water level in the polder. The amount of
drainage will depend on the seepage from the lake. It is assumed that the dike itself is
completely impermeable, and that the soil consists of a permeable aquifer (semi-confined)
with an overlying layer of low permeability (semi-permeable layer).
Data for the semi-confined aquifer are: hydraulic conductivity K = 10 m/d; thickness H = 14
m. Data for the semi-permeable layer are: hydraulic conductivity k' = 0.001 m/d; thickness d
= 2m. Other data are: water level in the lake φ
1
= 20 m; water level in the polder φ
4
= 16 m;
the width of the dike 2L = 4 m.
Figure 3.3 Groundwater flow under the dike
a) derive the formula of groundwater head and unit width discharge for the left part of
the aquifer from negative infinite up to the left side of the dike ( - ∞ < x < - L).
b) derive the formula of groundwater head and unit width discharge for the middle part
of the aquifer under the dike ( - L < x < + L).
c) derive the formula of groundwater head and unit width discharge for the right part of
the aquifer from the right side of the dike up to positive infinite ( + L < x < + ∞ ).
d) derive the formula and calculate φ
2
, φ
3
and the unit width discharge at x = + L (Q
0
).
Q
0
will be the seepage to the polder.
119
9.3 Steady Groundwater Flow to Wells
Problem 1
In the centre of a circular confined aquifer of thickness H = 20 m, a well is constructed to
pump water with a constant rate Q
0
= 250 m
3
/day. The aquifer consists of two regions with
different hydraulic conductivity and effective porosity (see Figure). The values of parameters
are:
K
1
= 10 m/day; n
1
= 0.25; K
2
= 5 m/day; n
2
= 0.3; φ
0
= 25 m; R = 5,000 m; r
w
= 0.5m; r
a
= 50
m.
Figure 4.1 Groundwater flow to a well in a non-homogeneous confined aquifer
a) calculate the groundwater head and drawdown at r = r
a
;
b) calculate the head and drawdown at the pumping well (r = r
w
);
c) calculate groundwater velocity at r = r
a
and r = r
w
;
d) calculate the total time of a water particle travelling from the boundary (lake) to the
pumping well.
120
Problem 2
A well is constructed to pump water from a confined aquifer with a thickness of 20 m. Two
observation wells, OW1 and OW2, are constructed at distance of 100 m and 1000 m,
respectively (Figure 2). With a pumping rate of 300 m
3
/day, the steady drawdown is observed
as 2 m in OW2 and 8 m in OW1.
r1
r2
Q0
T ?
H=20m
Impermeable base
OW1 OW2
s1
s2
2rw
Pumping Well
Original
groundwater
head
Figure 4.2 Pumping test in a confined aquifer
(a) determine the transmissivity and hydraulic conductivity of the aquifer from the
measurements.
(b) determine the radius of influence (R) from the measurements.
(c) predict the drawdown at the pumping well (r
w
= 0.1 m) when the pumping rate
increases to 500 m
3
/day.
121
Problem 3
A pumping test is conducted in an unconfined aquifer to determine the hydraulic
conductivity and the proper pumping rate. The groundwater table before the pumping is 20
m above the impermeable bottom. The radius of the pumping well is 0.1 m and the
distance between the pumping and observation well is 1000 m. With a pumping rate of 300
m
3
/day, the steady drawdown is observed as 1 m in the observation well OW1 and 3 m in
the pumping well.
3m
H0
1000m
Q0
OW1
2rw
1m
Drawdown curve
of pumping test
Initial water
table.
Figure 4.3 Pumping test in an unconfined aquifer
(a) determine the hydraulic conductivity of the aquifer from the pumping test data.
(b) determine the pumping rate if the maximum drawdown at the pumping well is
allowed to be 6 m.
(c) predict the possible drawdown at the observation well with the new pumping rate.
122
Problem 4
A well with a radius of 0.1 m is constructed in a semi-confined aquifer to pump
groundwater with a pumping rate of 300 m
3
/day. The thickness of the semi-confined
aquifer is 20 m and hydraulic conductivity is 0.98 m/day. The thickness of the semi-
permeable layer is 1 m and vertical hydraulic conductivity is 0.0001 m/day.
Q0
2rw
n0
100m
K
d,k'
Figure 4.4 Groundwater flow to a well in a semi-confined aquifer
a) calculate the drawdown at the pumping well and at a distance of 100 m away from
the pumping well with the exact solution;
b) check whether the approximate solution can be applied in this case. if so, calculate
the drawdown at the pumping well and at a distance of 100 m away from the
pumping well with the approximate solution and compare the results with the exact
solution;
c) calculate the equivalent radius of the influences.
d) assume that a confined aquifer of the same transmissivity has a circular recharge
boundary with a radius which is equal to the equivalent radius of this semi-confined
aquifer, calculate the drawdown at the distance of 100 m away from the pumping
well with the same pumping rate of 300 m
3
/day. Compare the drawdown calculated
in a) and b).
123
Problem 5
Four wells of the same discharge Q
o
are operating in an unconfined aquifer. In the corner
points of a square with sides 2a. The aquifer is bounded externally by a circle of radius R,
the centre of which coincides with the centre of the square. The radius R is so large
compared to the dimension so that the individual wells can be central. Establish, by means
of superposition, formula for the height of the water table in (i) the centre of the square,(ii)
the mid-point of a side of the square.
This system is used to lower the table to permit the excavation of a building pit in the form
of a square with sides 40m, the a depth of 4m below the original water table, which was at
10m above the impermeable base. What should be the discharge of each well in order to
keep the bottom of the building pit dry, if k = 10
-7
m/s and R = 2000m ?
Well 3
Well 1 Well 2
Well 4
2a
a
a
%2a
%5a
Figure 4.5 A system of 4 wells to lower water table for excavation
124
Problem 6
In a semi-confined aquifer, bounded by a straight impermeable boundary, a well is
operating at a distance of 500m from the impermeable boundary. The radius of the well is
r
w
=0.50m, the transmissivity of the aquifer is T=2x10
-3
m
2
/s, the resistance of the confining
layer is c=2x10
9
s, and the discharge of the well is Q
o
=10
-2
m
3
/s Calculate the drawdown in
the well.
Q0
Q0
y
p
p
r2
r1
x
Figure 4.6 A pumping well near a straight impermeable boundary
125
Problem 7
A company wishes to extract water at a discharge of Q
o
= 3.14x10
-3
m
3
/s from a
completely confined aquifer, which is bounded by a straight canal. The transmissivity of
the aquifer is 2x10
-3
m
2
/s. In order to obtain water of optimal quality it is desirable to
construct the well as far from the canal as possible. On the other hand, however, local
authorities require that at a distance of 400 m from the canal the ground water head in the
aquifer may not be lowered by more than 0.10m. At what distance from the canal should
the well be located.
- Q0
Q0
p
p
r2
r1
x
400m
s < 0.1
Figure 4.7 A pumping well near a straight canal
126
9.4 Non-steady Groundwater Flow to Wells
Problem 1
A well is pumping groundwater from an infinite confined aquifer with an abstraction rate
of 3140 m
3
/day. The transmissivity and storage coefficient of the aquifer are 2000 m
2
/day
and 2.0*10
-4
, respectively. Find the drawdown at the distance of 300 m from the well when
the time after pumping is 10 days, 20 days and 30 days.
Problem 2
A coal mine is underlain by a confined sandstone aquifer. In order to mine the coal,
groundwater head in the sandstone aquifer has to be lowered 300 m. The transmissivity
and storage coefficient of the aquifer are found to be 400 m
2
/day and 1.0*10
-3
, respectively
from a pumping test. A circular battery of wells with a radius of 100 m is designed to
dewater the aquifer in one year.
Figure 6.1 Lowering of groundwater head by a circular battery of wells
(1) Find the total daily pumping rates neglecting the point abstraction;
(2) How many wells are needed with a capacity of pumping rate of 12071 m
3
/day.
127
Problem 3
A pumping test is conducted in a confined aquifer with a pumping well and two
observation wells. While it pumps groundwater from the pumping well with a constant rate
of 60 m
3
/hour, drawdown in two observation wells are measured with time. The data is
tabulated in the following table:
Table 6.1 Pumping test data
Measurement Time past after Drawdown in observation wells
Date Time pumping: min. OW1 (43m) m OW2 (125m) m
8/6 13:30 0.0 0.0 0.0
13:40 10.0 0.73 0.16
13:50 20.0 1.28 0.48
14:00 30.0 1.53 0.54
14:10 40.0 1.72 0.65
14:30 60.0 1.96 0.75
14:50 80.0 2.14 1.00
15:10 100.0 2.28 1.12
15:30 120.0 2.39 1.22
16:00 150.0 2.54 1.36
17:00 210.0 2.77 1.55
18:00 270.0 2.99 1.70
19:00 330.0 3.10 1.83
20:10 400.0 3.20 1.89
21:00 450.0 3.26 1.98
9/6 00:15 645.0 3.47 2.17
04:00 870.0 3.68 2.38
06:00 990.0 3.77 2.46
Please find the aquifer parameters ( T and S) with:
(1) Jacob straight-line graph method.
(2) Theis type curve matching method.
128
Problem 4
A square excavation (375 m on a side) is to be dewatered by the installation of four wells
at the corners. Point A is in the middle and point B is on side equidistant from two of
wells. For an allowable pumping period of 24 hours determine the pumping rate required
to produce a minimum drawdown of 4 m everywhere within the limits of the excavation.
The semi-confined aquifer has a transmissivity of 18 m
2
/day, a storage coefficient of
7.0*10
-5
, and a resistance 1800 days.
Well 3
Well 1 Well 2
Well 4
2a
a
a
%2a
%5a
Figure 6.3 A square excavation
129
Problem 5
A pumping test is conducted in an infinite semi-confined aquifer with a pumping well and
an observation well. The distance between the pumping well and the observation well is
17.34 m. While it pumps groundwater from the pumping well with a constant rate of 5530
m
3
/day, drawdown in the observation well are measured with time. The data is tabulated in
the following table:
Table 3 Pumping test data in a semi-confined aquifer
t (min) s (m) t (min) s (m) t (min) s (m)
1 1.30 10 2.41 100 3.37
2 1.63 20 2.72 200 3.55
3 1.80 30 2.95 300 3.63
4 1.93 40 3.05 400 3.69
5 2.10 50 3.15 500 3.70
6 2.18 60 3.20 600 3.72
7 2.25 70 3.25 700 3.73
8 2.30 80 3.30 800 3.73
9 2.36 90 3.34 900 3.73
1000 3.73
(1) Determine T, S and c with Walton's Type Curve Matching method;
(2) Determine T, S and c with Hantush's inflection point method.
130
131
| 课件名称： | 地下水力学（英文版） |
| 课件分类： | 水利 |
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