91

Example 2.1: Piezometric Head

Figure 2.17 shows a dam, using equation 2.2: zg

Ph +=ρ

, show that the Piezometric head on

surface AB = H1 at any point on surface AB, ( i.e. show that h1 = h2 = h3 = H1).

Figure 2.17

Answer 2.1

From equation 2.2: zg

Ph +=ρ

33

322

211

1 ,, zg

Phz

gPhz

gPh +=+=+=

ρρρ

313332122211111 ,,,,, dHzgdPdHzgdPdHzgdP −==−==−== ρρρ Then,

1313313

33

3

1212212

22

2

1111111

11

1

HdHddHgdg

zg

Ph

HdHddHgdg

zg

Ph

HdHddHgdgz

gPh

=−+=−+=+=

=−+=−+=+=

=−+=−+=+=

ρρ

ρ

ρρ

ρ

ρρ

ρ

So, on the surface AB h= H1, but the pressure is not the same.

92

Example 2.2: Darcy’s Law and Groundwater velocity A confined aquifer is 50 m thick and 0.5 km wide. Two observation wells are located 1.4 km apart in the direction of flow. Head in well No. 1 is 50m and in well No. 2 is 42m. Hydraulic conductivity is 0.7 m/day. Effective porosity is 0.2.

(i) What is the total flow of water through the aquifer? (ii) What is the actual groundwater velocity between the two wells?

Answer 2.2

(i) From equation 2.4, L

hhKAQ 12 −−=

Where, K = 0.7 m/day, A= 50x500=25,000 m2, h1= 50m, h2= 42m, L= 1,400m.

So, the total flow of water through the aquifer = daymQ /100400,1

8000,257.0 3=−

××−= .

(ii) From equation 2.6, lh

nKu

e ΔΔ

×=

So, the actual groundwater velocity between the two wells = daymu /02.01400

82.07.0

=×= .

Example 2.3: Specific Discharge Figure 2.18 shows two piezometers in a confined aquifer:

(i) What is the direction of the groundwater flow in the Figure? (ii) What is the specific discharge using Darcy’s law, when the hydraulic conductivity is

2x10-3 m/sec?

Figure 2.18

Answer 2.3

(i) The direction of the flow is

(ii) From equation 2.5, dxdhKq −= ,

So the specific discharge = .sec/108.4500

12102 53 mq −− ×=−

×−=

93

Example 2.4: Groundwater Flow into a river A river penetrates a confined aquifer of 10 m thick Figure 2.19. A long drought decreases the flow in the stream by 0.5 m3/sec between gauging stations. A and B some 6000 m apart. On the west side of the river, the piezometer contours parallel the bank and slope towards the river at 0.0004 m/m. The piezometric contours on the east side of the river slope away from the river toward a well field at a slope of 0.0006 m/m.

Figure 2.19

(i) Explain the flow system along this section. (ii) Using Darcy’s law and the continuity equation, compute the transmissivity of the

aquifer through the river section. Answer 2.4 (i) Flow into the river from the aquifer occurs on the west and flow from the river into the aquifer occurs on the east. (ii) From continuity equation, we have: QQQ outin Δ=− Due to the slope of the piezometric surface, we assume the flow into the stream from the aquifer occurs on the west and flow from the stream into the aquifer occurs at the east, or:

eastoutwestin dldhAKQanddl

dhAKQ )()( ==

Thus,

sec/1017.410sec/1017.4sec/1017.4

sec/5.0]/0006.0/0004.0[)106000(

sec/5.0))()((

)(sec/5.0)()(

212

2

3

3

3

mmmKbTbutmK

mmmmmmmK

mdldh

dldhKA

DecreasemdldhAKdl

dhAK

eastwest

eastwest

−−

−

×=××==

×=⇒

−=−×⇒

−=−⇒

−=−

94

Example 2.5: Vertical Hydraulic Conductivity Three geological formations overlie one another with the characteristics shown in Figure 2.20 below. A constant-velocity vertical flow field exists across the three formations. The hydraulic head is 11 m at the top of the formations and 6.5 m at the bottom.

Figure 2.20

(i) Calculate the average vertical hydraulic conductivity? (ii) Calculate the velocity through the entire system?

(iii) Calculate the hydraulic head at the two internal boundaries?

Answer 2.5

(i) According to equation 2.74,

∑=

= n

iii

v

Kb

bK

1/

So, daymKz /2.6

85210

5.020

1850

280=

++=

95

(ii) Now, we find the velocity through the entire system.

daymlhKqz /1096.9

280115.62.6. 2−×=

−×−=

ΔΔ

−=

(iii) Now, we use the calculated velocity in (ii) to find the change in head across each layer.

( ) ( )

Kbqhhorh

Kbqh

hhKbqb

hhKlhKq

zz

zz

−=+−=⇒

−×−=⇒−

−=ΔΔ

−=

1212

1212.

Thus, at the bottom of the first aquifer we get:

.2.102.6

501096.9112

12

mhK

bqhh

layerfirsttheofbottom

z

=

××−=−=

−

At the bottom of the second aquifer we get:

.88.92.6

201096.92.10

sec

2

12

mhK

bqhh

layerondtheofbottom

z

=

××−=−=

−

To check our calculations, we continue:

.50.62.6

2101096.988.92

12

mhK

bqhh

layerthirdtheofbottom

z

=

××−=−=

−

So, our calculations are right since the head at the bottom of the formations is 6.5 m.

96

Example 2.6: Compressibility and Effective Stress A confined aquifer with an initial thickness of 45 m consolidates (compacts) 0.20 m when the head is lowered by 25 m. The porosity of the aquifer is 12 % after compaction.

(i) What is the vertical compressibility of the aquifer? (ii) Calculate the storativity of the aquifer?

(iii) How much water was released from storage for a head drop of 25 m averaged over

the aquifer, assume that the aquifer has an area of 106 m2? Answer 2.6 (i) The given parameter values are mdbandmbmdP 20.0,45,25 === . A pressure head of 25 m of water can be converted to a fluid pressure by multiplying the pressure head by the density of water times the gravitational constant. 223 /000,245/8.9/100025 mNsmmkgmdP =××= From equation 2.19,

NmmN

mm

dPb

db

/108.1/000,245

452.0

282

−×==

=

α

α

(ii) Aquifer Storativity is found from equation 1.26

][ βαρ ngbS +=

The given parameter values are 23 /8.9,/1000,12.0,8.44 smgmkgnmb w ==== ρ

and NmandNm /106.4,/108.1 21028 −− ×=×= βα .

3

283

2102823

109.7)/10806.1)(/9800()8.44(

]/106.412.0/108.1[)/8.9()/1000)(8.44(

−

−

−−

×=

×=

××+×=

NmmNmNmNmsmmkgmS

(iii) From equation 1.30, hASVw Δ= .. The volume of water released from storage of aquifer = 7.9 x 10-3 x 106 m2 x 25 m = 197.5 x 103 m3.

97

Example 2.7: Groundwater Flow Equations Define the model assumptions implicit in the following equations of groundwater motion (i.e. state whether the flow is steady or transient; 1-dimenional or 2-dimensional or 3-dimensional; homogeneous or heterogeneous; isotropic or anisotropic).

(i) 0=⎥⎦⎤

⎢⎣⎡

dxdhh

dxdK (ii) 02

2

2

2

=∂∂

+∂∂

yxφφ

(iii) 2,1,)( =∂∂

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

∂∂

∂∂ i

tSxa

xT

x oij

iji

φφ

Answer 2.7

(i) Homogenous, isotropic, 1-d, steady. (ii) Homogenous, isotropic, can be 3-d, steady. (iii) Heterogeneous, anisotropic, 2-d, transient

Example 2.8: Groundwater Flow Equations Show that for constant density and viscosity the following two equations are related:

( ) *2. QKt

SandQgpktPS opop =∇−

∂∂

=⎥⎦

⎤⎢⎣

⎡+∇⎟⎟

⎠

⎞⎜⎜⎝

⎛∇−

∂∂ φφρ

μρρ

Hints:

*0 )5,)4,1)3,)2,)1 Q

Qz

gPzgkKgSS p

op =+==∇==ρρ

φμρρ

Answer 2.8

( )

[ ]

*20

*2

*

)(,

)(

)(tan(

),,1(,.

.1.

..

QKt

S

gSSthatNoteQKt

gS

tg

tPzgPbut

QQ

thatNoteQ

KtPStconsbytermsalldivide

KgkandzgpzthatNoteQgk

tPS

Qzgpkg

tPSQ

gpkg

tPS

Qggpgk

tPSQgpk

tPS

opoop

ppop

pop

poppop

poppop

=∇−∂∂

⇒

==∇−∂∂

⇒

∂∂

=∂∂

⇒−=

==∇∇−∂∂

⇒

=+=∇==⎥⎦

⎤⎢⎣

⎡∇⎟⎟⎠

⎞⎜⎜⎝

⎛∇−

∂∂

⇒

=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+∇⎟⎟

⎠

⎞⎜⎜⎝

⎛∇−

∂∂

⇒=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

∇⎟⎟⎠

⎞⎜⎜⎝

⎛∇−

∂∂

⇒

=⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

∇⎟⎟⎠

⎞⎜⎜⎝

⎛∇−

∂∂

⇒=⎥⎦

⎤⎢⎣

⎡+∇⎟⎟

⎠

⎞⎜⎜⎝

⎛∇−

∂∂

φφ

ρφφρ

φρφρ

ρρφρ

μρ

ρφφ

μρρρ

ρμρρρ

ρμρρρ

ρρ

ρμρρρ

μρρ

98

Example 2.9: Groundwater Flow Equations Given the piezometric heads in three observation wells located in a homogeneous confined aquifer of constant transmissivity, T= 5000 m2/day.

Well X(m) Y(m) Ф (m) A B C

0 0

200

0 300

0

15 10.4 12.1

(i) Draw the contours of the piezometric surface (ii) Determine the discharge through the aquifer per unit width (magnitude and

direction).

(iii) For the same data above except for T which is replaced by ⎥⎦

⎤⎢⎣

⎡=

108830

K for

aquifer thickness B=50m, what will be the discharge vector? Draw the discharge vector and grad ( Ф ) on your figure for (i).

Hint,

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡+−=

dydhdxdh

KK

KKB

qq

qjdydhi

dxdhTq

yyyx

xyxx

y

x,

Answer 2.9 (i)

99

(ii) ⎥⎦

⎤⎢⎣

⎡+−= j

dydhi

dxdhTq

( ) ( )

eastfrom

daymq

daymyhTq

daymxhTq

y

x

o6.465.7265.76tan

/5.10565.765.72

/65.760300154.105000

/5.720200151.125000

1

222

2

2

=⇒⎟⎠⎞

⎜⎝⎛=

=+=

=−−

×−=∂∂

−=

=−−

×−=∂∂

−=

− θθ

(iii)

( ) ( )

eastfrom

daymq

daymyhk

xhkBq

daymyhk

xhkBq

yyyxy

xyxxx

o78.25885.27467.13tan

/97.30467.13885.27

/467.13300

6.410200

9.2850

/885.27300

6.48200

9.23050

1

222

2

2

=⇒⎟⎠⎞

⎜⎝⎛=

=+=

=⎥⎦⎤

⎢⎣⎡ −

×+−

××−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−=

=⎥⎦⎤

⎢⎣⎡ −

×+−

××−=⎥⎦

⎤⎢⎣

⎡∂∂

+∂∂

−=

− θθ

100

Example 2.10: Groundwater Flow Equations

Let Kx = 26 m/day and Ky = 16 m/day be the principal values of K in an anisotropic aquifer, in the x and y directions respectively for two dimensional flow. The hydraulic gradient is 0.004 in a direction making an angle 30° with the +x axis.

(i) Determine q (ii) What is the angle between grad (Ф) and q.

Hint:

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

Φ

Φ

⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡=

dyddxd

KK

qq

qy

x

y

x

00

Answer 2.10

3

3

10000.2)30(sin004.0)90(cos.

10464.3)30(cos004.0cos.

−

−

×=⋅=−Φ∇=Φ∇=Φ

×=⋅=Φ∇=Φ∇=Φ

o

o

θ

θ

jdyd

idxd

⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

×⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡=

−

−

3

3

10000.210464.3

160036

y

x

qq

q

⎥⎦

⎤⎢⎣

⎡−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

××

××−=⎥

⎦

⎤⎢⎣

⎡=

−

−

032.0125.0

10000.21610464.336

3

3

y

x

qq

q

( ) ( )

eastfrom

daymq

o36.14125.0032.0tan

/129.0032.0125.0

1

22

=⇒⎟⎠⎞

⎜⎝⎛=

=+=

− θθ

101

Example 2.11: Water Table Map Figure 2.21 is a map showing the groundwater elevation in wells screened in an unconfined aquifer at Milwaukee, Wisconsin. The aquifer is in good hydraulic connection with Lake Michigan, which has a surface elevation of 580 ft above sea level. Lakes and streams are also shown on the map.

(i) Make a water-table map with a contour interval of 50 ft, starting at 550 ft. (iv) Why do you suppose that groundwater levels are below the Lake Michigan surface

elevation in part of the area?

Figure 2.21 Base map for example 2.11

102

Answer 2.11

(i) Here is the Water table map,

(ii) Water levels are below Lake Michigan because of pumping from the aquifer.

103

Example 2.12: Steady Flow in a Confined Aquifer

A river and a drainage channel are shown in Figure 2.22. The average elevation of the water surface in the river is 144.6 meters, and in the channel 142.20 meters. The hydraulic conductivity of the confined inter-granular aquifer developed in medium alluvial sand is 3.5x10-4 m/s. The hydrologic cross section in Figure 2.23 shows the relationship between the aquifer, the overlying silty clay (aquitard) and the underlying dense (impermeable) clay.

Figure 2.22 Plan view of the river and the drainage channel

Figure 2.23 Hydrogeological cross-section between the river and the drainage channel shown in Figure 11 as determined by filed investigations.

(i) Calculate the groundwater flow per unit width between the river and the drainage channel?

(ii) What is the height z of the piezometric surface at a midpoint between the river and

the channel?

104

Answer 2.12

(i) According to Darcy law xhwbK

xhAKQ

ΔΔ

−=ΔΔ

−= .....

So, xhbKq

ΔΔ

−= ..

smxq

xq

mbbut

LhhbKq

/1014.4

720)5.14420.142()55.3()105.3(

55.32

6.35.32

)40.13500.139()80.13330.137(,

..

26

4

12

−

−

=⇒

−−=⇒

=+

=−+−

=

−−=

(ii) The position of the potentiometric surface at a midpoint between the river and the channel. Since the aquifer is confined, the piezometric surface is linear.

aslmh

xxx

xh

xKbqhh

xhh

bKq

xhh

bKq

x

x

x

x

4.143

)55.355.3105.3

1014.4(6.144

.

..

4

6

1

1

12

=⇒

−=⇒

−=⇒

−−=⇒

−−=

−

−

105

Example 2.13: Steady Flow in an Unconfined Aquifer Refer to Figure 2.24; the hydraulic conductivity of the aquifer is 1.2 m/day. The value of h1 is 17 m and the value of h2 is 12 m. The distance from h1 to h2 is 4525 m. There is an average rate of recharge of 0.0002 m/day.

(i) What is the average discharge per unit width at x=0 m? (ii) What is the average discharge per unit width at x=4252 m? (iii) Where is the water divide located? (iv) What is the maximum height of the water table?

Figure 2.24

Answer 2.13 (i) at x=0, Use equation 2.111

( )

)(/43.0

)02

4525(/0002.0)4525(2

)1217(/2.122

20

2222

0

22

21

directionxnegativeinmovementimpliessignnegativedaymq

mdmm

mmdmq

xLwL

hhKqx

−=

−−−

=

⎟⎠⎞

⎜⎝⎛ −−

−=

106

(ii) at x= 4525, Use equation 2.111

( )

daymq

mdmm

mmdmq

xLwL

hhKqx

/47.0

)45252

4525(/0002.0)4525(2

)1217(/2.122

24525

2222

4525

22

21

=

−−−

=

⎟⎠⎞

⎜⎝⎛ −−

−=

(iii) From equation 2.112,

( )

.2166)4525(2

1217(/0002.0

/2.12

452522

2222

22

21

metersd

mmdm

dmmd

Lhh

wKLd

=

−−=

−−=

(iv) From equation 2.113,

( )

( )

mh

mmmdm

dmmmmmh

ddLKw

Ldhhhh

7.32

2166)21664525(/2.1

/0002.04525

2166121717

)(

max

222222

max

22

212

1max

=

×−+×−

−=

−+−

−=

107

Example 2.14: The Potentiometric Surfaces for the Steady-State One-Dimensional Groundwater Flow System

(i) Sketch in, as accurately as possible, the potentiometric surfaces for the steady-state one-dimensional groundwater flow systems given below.

Figure 2.25

(ii) For the following hydraulic parameter values determine the distribution of h(x) over

the length of the aquifer in Figure 2.25 (b).

K = 10 m/day h1= 10 m h2 = 5 m Determine the largest abstraction that can be obtained theoretically from the aquifer assuming the equation you derive is valid.

108

Answer 2.14 (i) (a)

(b) There are two possibilities: (1) All Q goes to the left hand side as this side has less head. (2) Q is divided according to length.

(c) There are three possibilities: (1) Q1 as well as Q2 is relatively small. (2) Q1 and Q2 are big, they seem equal. (3) Q2 > Q1

109

(ii)

Apply Dupuit Assumption, see Equation 2.98

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

Lhh

Kq22

21

21

You can follow the derivation of this equation from Darcy’s Law, as shown below.

⇒−=dxdhKhq ⇒−=∫ ∫

L h

h

hdhKdxq0

2

1

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

22

21

22 hh

KqL

Then, ⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

Lhh

Kq22

21

21

Now, we can calculate the total flow in the aquifer, Now, 21 qqQ +=

[ ]

[ ]

007024

0102

7010

101

25

705

750

50525

505

705100

705

25505

505

210

100705

7010

210

22

22

222

2

222

1

=⇒=⇒

=−−=

−+−=

−×+−×=

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

oo

ooo

oo

oo

oo

oo

hh

hhdhdQ

hhQ

hhQ

hh

q

hh

q

widthmdaymQ

qqQ

//64.9

5.214.7)]025[505]0100[

705(

3

21

=⇒

+=−+−=+=

mdaymqTotal //125.3120

251001021 3=⎟

⎠⎞

⎜⎝⎛ −

×=

110

Example 2.15: The Potentiometric Surfaces for the Steady-State One-Dimensional Groundwater Flow System

Sketch in, as accurately as possible, the potentiometric surfaces for the steady-state one-dimensional groundwater flow systems given below.

Figure 2.26

111

Answer 2.15 (a) Recharge = qL and Discharge = qL/2

(b)

(c)

112

Example 2.16: The Potentiometric Surfaces for the Steady-State One-Dimensional Groundwater Floe System Sketch in, as accurately as possible, the potentiometric surfaces for the steady-state one-dimensional groundwater flow systems given below.

Figure 2.27

For Figure 2.27 b, assume that ( )

( )L

hhKBQ

LhhKB

Q

122

121

−+=

−−=

113

Answer 2.16 (a) You need to locate ho at the abstraction line as 21 qqQ +=

01

01

02112

02112

21

22

)(2)(2)(2

hhhh

hhhhhhL

hhKBL

hhKBL

hhKBqqQ

o

o

=⇒−=−⇒

−+−=−⇒

−+

−=

−+=

(b) 322211 qqQandqqQ +−=+=

114

First

)1(3263333

)(3)(3)(

3210

03112

03112

211

hhhhhhhhhh

LhhKB

LhhKB

LhhKB

qqQ

o

o

++=⇒−+−=+−⇒

−+

−=

−−+=

Second

)2(6233333

)(3)(3)(

3210

323012

32312

322

hhhhhhhhhh

LhhKB

LhhKB

LhhKB

qqQ

o

−+=−⇒−+−=−⇒

−+

−=

−+−=

Multiply [(1)x 2] and add (1) to (2)

210

210

3210

3210

94

95

459____________________________________

623

62412

hhh

hhh

hhhh

hhhh

+=⇒

+=

−+=−+

++=

Solving for h3 , substitute ho in (2)

213

2122113

32121

3210

5414

544

914

94

942

956

6294

95

623

hhh

hhhhhhh

hhhhh

hhhh

+=⇒

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ −=⇒

−+=+

−+=−

115

Example 2.17: One-Dimensional Flow Regime For the one-dimensional flow regime shown in Figure 2.28,

(i) Calculate how long it will take a particle to travel from point A to the river B assuming that it travels with the averaged advective velocity of the groundwater flow.

(ii) Sketch in, as accurately as possible, the potentiometric surface for the steady-state one-dimensional groundwater flow systems given in the Figure.

Figure 2.28 (Not to scale)

Answer 2.17 (i) From the continuity equation 321 qqq == ,

)1(3501323131335010

)(13)35(10

)()()(

21

121

121

121

21

+=⇒−=−⇒−=−⇒

−−=

−−=

hhhhh

hhh

sametheisLNoteL

hhKL

hhKqq

A

)2(10032)5(5001510

1515035010)10(15)35(10

)()()(

21

21

21

21

21

31

+−=⇒+−=⇒−=−⇒−=−⇒

−−=

−−=

hhbydividedhh

hhhh

sametheisLNoteL

hhKL

hhKqq

BA

116

Multiply [(1)x 3] and [(2)x 13] and add them

mhh

hh

hh

74.24350,295

____________________________________13003926

050,13969

1

1

21

21

=⇒=

+−=+

+=

Solving for h2 , substitute h1 in (1)

mhhhh

85.1635013)74.2423(3501323

2

221

=⇒+=×⇒+=

Now,

( ) ( )

( ) ( )

( ) ( )

daysdaym

muL

t

daymL

hhnK

u

daysdaym

muLt

daymL

hhnKu

daysdaym

muLt

daymL

hhnKu

B

A

22.236/27.1

300

/27.1300

85.161027.0

15

98.218/37.1

300

/37.1300

74.2485.1625.0

13

44.175/17.1

300

/71.1300

3574.242.0

10

3

33

3

2

3

33

2

22

2

12

2

22

1

11

1

1

1

11

===⇒

=−

×−=−

×−=

===⇒

=−

×−=−

×−=

===⇒

=−

×−=−

×−=

It will take a particle to travel to travel from point A to the river B about 631 days (ii) Note that the sketch is not to scale.

117

Example 2.18: Flow Net Figure 2.29 is a piezometric map of a shallow unconfined groundwater catchment running down to the sea.

(i) Sketch streamlines on this map in order to form a flow net. (ii) Estimate the flow rate of groundwater across the line on the map indicated by the

20 m contour, assuming an aquifer transmissivity of 250 m2/day.

Figure 2.29

118

Answer 2.18

(i)

(ii) widthdldhTQ ..=

* From Figure 2.29, it is approximated (measured) that distance between contour 80 and contour 20 is about 5 cm, i.e. 0.05 m. It is also approximated (measured) that the width of contour 20 is 10 cm (0.1m). Note that the scale is 1:25,000.

[ ]

daymQ

Q

widthdldhTQ

/000,30

000,251.0000,2505.0

2080250

..

3=

××⎥⎦

⎤⎢⎣

⎡×−

×=⇒

=

119

Example 2.19: Flow Nets and Well Operation

(i) The attached map (Figure 2.30) gives the positions of seven observation boreholes in a limestone aquifer, with piezometric levels for each borehole in meters above sea level. The mean stage of the river is 65 m above sea level. Construct a flow net representing the equipotential lines and selected flow lines in the aquifer.

(ii) Pumping tests in wells 2 and 6 have yielded transmissivity values of 110 m2/day and

95 m2/day respectively. The effective saturated thickness of the aquifer in the area is 45 m. Use Darcy’s Law to calculate the flow of groundwater into the river when the potentiometric surface is in the configuration you have sketched on Figure 2.29.

Figure 2.30 Piezometric levels (meters above sea level [masl]) in the Sherburn Limestone Aquifer, 1-7-1992

120

Answer 2.19 (i)

(ii) From the given scale: 1 km = 1.7 cm. So, the distance between well No.2 and well No.6 = 10.5 cm = 6.176 km And the length of the segment of the river that faces the groundwater flow = 11.2 cm = 6.588 km.

[ ]

daymQ

Q

widthdldhTQ

/19.105,3

65886176

654.932

95110

..

3=

×⎥⎦⎤

⎢⎣⎡ −

×⎟⎠⎞

⎜⎝⎛ +

=⇒

=

121

Example 2.20: Analysis of Groundwater Flow Systems INTRODUCTION Analysis of groundwater flow system involves identification of groundwater flow directions and qualification of fluxes. The main method for achieving these objectives is the construction of flow nets. A flow net is an assemblage of contours of groundwater head, with flow-lines drawn at right angles to the contours. It is usual to construct flow nets in two orientations, i.e. a plan view [water table map] and one or more vertical profiles [usually cross sections along the trace of a flow line identified on the water table map]. MATERIALS You are provided with a map showing groundwater levels (in meters above mean sea level) on a single day in July 1990 in an extensive limestone aquifer (lithologically similar to the deep West Bank aquifer). You should bring drawing equipment (rulers, protractors, set-squares, coloured pencils etc.). (Should you complete the manual steps of this exercise, software is available for step 8). WHAT YOU HAVE TO DO 1 Draw water table contours on the map provided. Pay particular attention to the need to

represent cones of depression around abstraction boreholes (i.e. pumping wells). [Hint: the coast-line is fixed head boundary of 0 meter water table elevation].

2 Compute the flow net in plan by adding ten flow lines, perpendicular to the contours. 3 Examine the shape of your flow net:

a. Does the pattern of contours (spacing, orientation) tell you anything about local variations in aquifer permeability?

b. How many distinct discharge areas can you recognize? c. What is the hydrogeological influence of the fault at the southern edge of the

mapped area?

4 Given a mean transmissivity of 250 m2/day, what volume of water flows daily from the aquifer into the ocean? Is the flow rate constant along the coastline, or does it vary from one stretch to another?

5 Given that the abstraction boreholes shown on the map are long-established (with steady drawdowns), and given their combined abstraction rate is around 100,000 m3/day, estimate the amount of recharge this aquifer is receiving.

6 Choose one of the flow lines you have drawn on the map, and for each construct a vertical profile flow net, showing the inferred distribution of groundwater head in the subsurface. You may assume that, from its outcrop position, the base of the aquifer dips eastwards at a gradient of 25 m per kilometer.

7 Assuming a hydraulic conductivity of 25 m/day is valid for both profiles, what would be the average transit time to the coast for a molecule of water entering the aquifer near the outcrop of the base of the limestone? Does the result surprise you? If so, why?

122

Figure 2.31

123

Answer 2.20 (1) and (2)

124

3 a. Closer contour lines indicate lower hydraulic gradient, hence lower permeability. b. Nearly 2. c. It acts as a no-flow boundary. 4

./167,3618667500,17

/667,18000,16000,15

)070(250

/500,17000,106000

)042(250

)2arg()1arg(

..

3

32

31

21

daymQ

daymQ

daymQ

areaedischQareaedischQQ

widthdldhTQ

=+=

=×−

×=⇒

=×−

×=⇒

+=

=

The flow rate along the coastline varies from one stretch to another.

5 The Abstraction = 100,000 m3/day Flow to the ocean = 36,167 m3/day So, theoretically, the amount of recharge that this aquifer is receiving = 136,167 m3/day. 6 The vertical profile flow net for the flow line marked by (1).

7

.219000,80125.0000,10

./125.0000,10

50/25

yearsdayst

daymu

daymiKnQu

===

=

×=×==

125

Example 2.21 Application of Darcy’s Law in Landfills A landfill liner (very low permeable material) is laid on a top of permeable sand unit while the sides of the liner have very contacts with the sand unit and a good clay unit as shown in Figure 2.32. The landfill can be represented by a square with length of 150 m on a side and a vertical depth of 15 m from the surface. Thickness of landfill clay liner = 0.45m. Hydraulic conductivity of landfill clay liner = 8.64x10-4 m/day. The regional groundwater level for the confined sand is located 3 m below the surface. How much water will have to be continuously pumped from the landfill to keep the potentiometric surface at 18 m above sea level within the landfill? (Hint: Assume horizontal flow and use Darcy’s Law only)

Figure 2.32

126

Answer 2.21

Flow into landfill = flow to be pumped

Since flow is horizontal, Darcy’s law can be applied

lhAKQΔΔ

= ..

Flow to landfill will come form the sides of the landfill that have contacts with sand + from the bottom of the landfill because it also has contacts with landfill.

To apply Darcy’s law for horizontal flow, means that head is constant vertically, then

hydraulic gradient for the given situation is

2045.01827

=−

=ΔΔ

=lhi

Note that the thickness of the clay liner = 0.45 m

Now, we have four sides that the landfill has contact with them. These four sides are identical;

daymQ

xmxdaymx

lhAKQsidesfromlandfilloFlow

mxAsidesfourtheofareaTotalmxsideeachofArea

s

s

s

/78

2045001064.8

..int

450011254)(1125150)155.22(

3

24

2

2

=

=

ΔΔ

=

==

=−=

−

Now, we have one bottom that has contacts with the landfill;

daymQ

xmxdaymx

lhAKQbottomfromlandfilloFlow

mxAbottomtheofArea

b

b

b

/389

20500,221064.8

..int

500,22150150)(

3

24

2

=

=

ΔΔ

=

==

−

The flow into landfill = Q s + Q b = 78 + 389 = 467 m3/day, which is the same as

volume of water that should be pumped to keep the landfill hydraulics as shown in the figure.

127

2.22

128

2.23

129

130

2.24

131

Example 2.25

132

133

134

Example 2.26

Example 2.26

135

136

Example 2.27

137

138

139

Example 2.28

140

Example 2.29

141

Example 2.30

142

143

Example 2.31