Seepage
In the precedingchapter.we considcredsome simple casesfor which direct application of Darcy's law was required to calculatethe flow of water through soil. ln many
instances,the flow of water through soil is not in one direction only, nor is it uniform
over the entire area perpendicularto the flow. In such cases,the groundwater flow is
generallycalculatedby the use of graphsreferred to ast'low nets.The concept of the
flow net is based on Laplace's equulion of continuity, which governs the steady flow
condition for a given point in the soil mass.In the following sectionsof this chapter,
cquation of continuity will be presentcdalong with its apthe derivation of Laplace'.s
plication to seepagcproblcms.
7.1
Laplace's Equation of ContinuitY
To derive the Laplace diflercntial cquation of continuity, let us considera singlerow
of shcetpilesthat havebeen driven into a permeablesoil layer,asshownin Figure 7.1a.
The row of sheet piles is assumcdto be impervious. The steady state flow of water
from the upstream to the downstream side through the permeable layer is a twodimensionalflow.For flow at a point,4, we consideran elementalsoil block. The block
has dimensionstlx, tly, an<ldz (length dy is perpendicularto the plane of the paper);
it is shown in an enlargedscalein Figure 7.1b.Let o-,and u. bc the componentsof the
dischargevelocity in the horizontal and vertical directions,respectively.The rate of
flow of water into the elementalblock in the horizontal direction is equal to u, dz dy,
and in the vertical direction it is u. dx dy. The rates of outflow from the block in the
horizontal and vertical directions are, respectively,
/
Ir,*
0u. \
d r d zd t
*
)
and
dD-
( , , + - dt l zz l
178
\
/
dx dy
7.1 Laplaceb Equation of Continuity
179
S h e e tp i l e
'-'*.--"'r',
I
Hy
|
I
Y
- * l -Tt.
.
:
;;
-,
l,r
i
H2
:tr ,
fi
- t
. . t , $
it
[ffif.#
". J*1,'1.'-
Impermeable
layer
$ (,',* \ u,!u,,,-
,',,l:,1r'
I
,ra;(,,,+!*1,ra1,
I
I
I
I
tl:.
l-,/.r;;;l
l.-,/'
(DJ
Figure 7.1 (a) Singlc-row shect piles driven into permeable layer; (b) flow at.4
Assumingthat watcr is incompressibleand that no volume changein the soil mass
occurs,we know that the total rate of inflow should equal the total rate of outflow.
Thus.
l/
iL l\ ? ] ' +#
r-) ttz rty+ (,, .
ar) a, or)
#
lu,ttzrty* u,dx,tyl: 0
or
+ A a". - 0
6x Az
d a^,
(7.r)
180
Chapter 7 Seepage
With Darcy's law, the discharge velocities can be expressedas
. ctn
D-: k,t-: K"dx
(7.2)
and
u":
, a h
K-t-
"Az
(1.3)
where k, and k. are the hydraulic conductivities in the vertical and horizontal directions, respectively.
From Eqs. (7.1), (7.2), and (7.3), we can write
A?h
. a2h
k- ^ *k--"^:U
d?.dx-
k,:
If the soil is isotropic with respect to the hydraulic conductivity- that is,
k,- the preceding continuity equation for two-dimensionalflow simplifiesto
a2h
_a2h
+ - - l l
0x'
7.2
(1.4)
3z'
(7.s)
Continuity Equation for Solution
of Simple Flow Problems
The continuity equation given in Eq. (7.5) can be used in solving some simple flow
problems. To illustrate this, let us consider a one-dimensionalflow problem, as
shown in Figure 7.2,in which a constanthead is maintained acrossa two-layeredsoil
for the flow of water. The head differencebetween the top of soil layer no. 1 and the
bottom of soil layer no. 2 is h1. Becausethe flow is in only the e direction, the continuity equation [Eq. (7.-5)]is simplified to the form
a"h :
^ )
h:
tl
A(.-r A2
(7.6)
Q.7)
where At and A, are constants.
To obtain A1 and A2for flow through soil layer no. I , we must know the boundwhich are as follows:
conditions,
ary
. Condition l: At z : 0, h -- h.
. Condition 2: At z : H,, h : hr.
7.2 Continuity Equation for Solution of Simple Flow problems
:;*1,*.
! l
Water supply
&
v
V
H---
t;.1
,
$
&
#
-u,^
Y
,
l"-
:
,
#
$
$
i
v
-
,$.
= ff.
s.
+
l l
'
I
. . , ,: hl
_ _ _r
:
H'
$
$
Figure 7.2 Flow througha two-layeredsoil
Combining Eq. (1.7) and condition I gives
Az: hr
(7.8)
Similarly,combining Eq. (7.7) and condition 2 with Eq. (7.8) gives
hz: AtHt + hl
or
At:
/h'-h'\
(
.l
/
(7.e)
Combining Eqs. (7.7), (7.8), and (7.9),we obtain
h:
(*\rno,
\
F1, /-
",
(toro<7<H,)
(7.10)
For flow through soil layer no. 2, the boundary conditions are
o
t
Condition l:
Condition 2:
At z : H1, h : h2.
At ; : H, + H2, h : 0.
From condition 1 and F,q. (7.7),
Az:
hz - ArH,
(7.rr)
182
Chapter 7 Seepage
Also, from condition 2 and Eqs. (1.7) and (7.11),
0:A(Hl +H)+(h2-Afit)
A t H t + A t H 2 + h 2 - A 1 H 1: 0
A r -
h.
I
H2
(7.12)
So,from Eqs.(7.7),(7.1i), and(7.12),
h:-1
H,\
/ h,\
/
. ' i l z - h 'r\l l t ; l
( l i r r H<, z < H , - H , )
Ht/
\H)/
(7.1.1)
At any given time, flow through soil layer no. I equalsflow through soil layer no. 2, so
/ h , - h .' \
17,-{t\
- k,\-'s,
tt- kt\-r,
)A
)A
where A - areaof crosssection of the soil
ft 1 - hydraulic conductivity of soil layer no. I
k2 : hydraulic conductivity of soil layer no. 2
or
hz=
hrk,
''(*,.
h)
(1.14\
Eq. (7.1a)into Eq. (7.10),we obtain
Substituting
h: h,(, -
^#r^)
, * ' o ' <z * H 1 )
( 7 l.s )
Similarly,combining Eqs. (7.13) and (7.14)gtves
, - , , 1 6 d - F n ) r o , * H z- z ) ] , " '
H t< z < H l+ H z )
(1.16)
7.3 Flow Nets
183
Example7.1
Refer to Figure 7.2. Given that 111= 300 mm, FIz : 500 mm, and ftr * 600 mm"
and that al z :200 mm, ft : 500mm, determineh at z : 600mm.
Solution
we know that z : 200mm is locatedin soil layerno. 1, soEq. (7.15)is valid.Thus,
k"
h : h , ' (\ t \
k r H , + k r H t/
soo:6oolr
L
k'?(2oo)
I
k r ( 5 0 0+) k 2 ( 3 0 0 ) l
or
k,
o,
: t'*
Becausez - 600mm is located in soil layerno.2,
r
Eq. (7.16)is valid,yielding
h *
,rl*
1
/k,\
La,+ l ;
(
h:
600
r
\
[,roon #/,'on
I
(ry+H2*z)l
J
lH,
I
+ soo* *n).1: 17e'e
mm
Flow /Vefs
The continuity equation IEq. (7.5)] in an isotropic medium representstwo orthogontrl families of curves- that is, the flow lines and the cquipotential lines.A
ltow ine
is a line along which a water particle will travel from upstream to the downstream
side in the permeablesoil medium. An equipotentialline isa line along which the potential head at all points is equal. Thus, if piezometersare placed at different points
along an equipotential line, the water level will rise to the same elevation irrall of
them. Figure 7.3ademonstratesthc definition of flow and equipotential lines for flow
in the permeable soil layer around the row of sheet piles shown in Figure 7.1 (for
n,. -
n_ -
Kr.
A combination of a number of flow lines and equipotential lines is calleda
flow
net. As mentioned in the introduction, flow nets are constructedfor the calculation
of groundwater flow and the evaluation of heads in the media. To complete the
graphic construction of a flow net, one must draw the flow and equipotenti;l lines
in
such a way that
1. The equipotential lines intersectthe flow lines at right angles.
2. The flow elements formed are approximate squares.
Chapter 7 Seepage
t
I
I
H1
H2
f
Flowline
Equipotential
S h e c tp i l c
,
Water lcvcl
W i r t c rl ev c l
d
{
v
H.
t
\
t
, ,lt
Impervious
layer
(D)
Figure 7.3 (a) Definitionof llow linesandcquipotentiallines;(b) completedflow net
Figurc 7.3b showsan cxample of ercompleted flow net. Two more examplesof
flow net in isotropic permeablelayer are given in Figures7.4 and 7.5. In thesefigures,
l/1 is the number of flow channelsin the flow net, and N4 is the number of potential
drops (defined later in this chapter).
Drawing a flow net takes severaltrials. While constructingthe flow net, keep
the boundary conditions in mind. For the flow net shown in Figure 7.3b, the following four boundary conditions apply:
7.4 Seepage Calculation from a Ftow Net
,l,
H2
'ti
Figure 7.4 Flow net under a dam
Figure 7.5 Fkrwnct undcra clanrwith toc filtcr
1. Thc upstream and downstrcam surfaccsof the pcrmeable layer (lines ah and,
d e ) a r ee q u i p o t c n t i a l i n c s .
2. Becauseub and de are equipolential lincs. all the flow lines intersectthem at
right angles.
3 . T h e b o u n d a r yo f t h c i m p c r v i o u sl a y e r- t h a t i s , l i n e
lb- is a flow line,and so is
the surfzrceof the impervious sheet pile, Iine ncr1.
4. The cquipotential lines intersccl ucdand l! at right angles.
7.4
Seepage Calculation from a Flow Net
In any flow net, the strip bctween any two adjacentflow lines is called a
fltw channeL Figure 7.6 showsa flow channel with the equipotential lines forming squareelements. Let h1, hz, ht. hr,.
. , h , , b e t h e p i e z o m e t r i cl e v e r s . o r . " r p o n . r i n g t o t h e
equipotential lines. The rate of seepagethrough the flow channel per unit length
(perpendicular to the vertical section through the permeable layer) can be callulated as follows: Becausethere is no flow acrossthe flow lines.
Lq, : Lq, : Lqr: ... : L,q
(1.17)
186
Chapter 7 Seepage
,..! i ,.t
Lq
. r '
t
h1
h1
t.
T '
l
l .
, . , .
1
2
t
i
I
,:i*
Lq
Figure 7.6
througha flow channel
Seepage
with squareelemcnts
From Darcy'.slaw the flow rate is equal to kiA. Thus, Eq. (7.17) can be written as
-/r,\.
, ( h -. -- ) th^ t-\", '
'),,- .lh,-/r,\,
- k (/ "h ,
(7.r8)
Lt1
o(-',.=),,-- o(\
I
Eq. (7.18) shows that if the flow clements are drawn as approximate squares,the
drop in the piczometric level between any two adiacent equipotcntial lines is the
same. This is called the pcttentioldrop. Thus,
ht
h2: lt2-
(7.1e)
:";
ho:
ht- h.-
and
trt:
. H
k,V,
(1.20)
where H : head difference between the upstreamand downstrcam sides
ly',i: number of potential droPs
In Figure 7.3a,lor any flow channel, H : H r ,f1, and N,r : 6'
If the number of flow channelsin a flow net is equal to N1 the total rate of flow
through all the channelsper unit length can be given by
q:K
, HNI
lv,
(7.21)
Although drawing squareelementsfor a flow net is convenient,it is not always
necessary.Altcrnatively, onc can draw a rectangular mesh for a flow channel, as
shown in Figure 7.7, provided that the width-to-length ratios for all the rectangular
elementsin the flow net are the same.In this case,Eq. (7.18)for rate of flow through
the channel can be modified to
-hr\,
ftr\,
.
.(h,
,(h,
,(h.-ho\,
rq: *(T
)r, o(?
lf bJh : b2l12:full
(1
and .27\ can be modifiedto
)0,:
o(,
z
"
)b,:
(1.22)
n (i.e.,the elementsare not square),Eqs.(7.20)
L,q :
r,(ft,)
(7.23)
7.4 Seepage Calculation from a Ftow Net
aq \*
1g7
: . . .)
0.1:'
,\"
/
L
l
Figure 7.7
Seepage
througha flow channel
with rectangular
elements
h t b b l
I ' =r = r - " ' = n
/N,r
n: kH\*,)"
(7.24)
Figure 7.8 showsa llow net for seepagcaround a singlerow of sheetpiles.Note
that flow channels 1 and 2 have square elements. Hence, the rate of flow throush
thcse two channelscan be obtained from E,q.(7.20):
Lt11
i- 1,t12
,, . !. "n - 4!!
!,
tv,t
N,r
N,t
However' flow channcl 3 has rectangularelements.These elementshave a width-tolength ratio of about 0.311;
hence,from Eq. (7.23)
A,/,:
#H(o.3rt)
Water levcl
TH
I
*
Water table
I
l
Ground surface
I
..,,ii
. . " . jt
. - - . i;
Flow channel I
I
i - l
I
i
Flow channel2
I
b = l
I
Flow channel3
I
I
b
0.38
Impervious
layer
Figure 7.8
Flow net for seepage
around a single row of
sheetpiles
188
Chapter 7 Seepage
So, the total rate of seepagecan be given as
e - Lqrt
Lqz- A4r - -J8+
TV
A flnw net for flow around a singlerow of sheetpiles in a permeablesoil layer is
shownin Figure7.9' Given lhat k, k,. = k = 4'2 x 10-r cm/sec'determine
a. How high (abovethe groundsurface)the water will rise if piezometers
are placed at Pointsa, b, c, and d
b. The rate of seepagethroughflow channelII per unit length (perpendicular to the sectionshown)
Solution
a. From Figure7 '9,weseethat N/ : 3 and No - 6' The head differencebetweenthe upstreamand downstreamsidesis 3.5m, so the headlossfor
eachdrop is 3.5/6: 0.583m.
Point a is locatedon equipotentialline 1, which meansthat the poten'
tial drop at a is 1 X 0.583m. The water in the piezometerat a will rise to
an elevationof (5 - 0.583): 4'417m abovethe groundsurface'
Similarly,the piezometriclevelsfor
b:
( 5 - 2 x 0 . 5 8 3 :) 3 . 3 3 4 m a b o v e t h e g r o u n d s u r f a c e
c = (5 - 5 x 0.583): 2.085m abovethe groundsurface
d : (5 - 5 x 0.583): 2.085m abovethe groundsurface
S h r ' e tp i l c
Impermeable
layer
Figure7.9
5
7.5 Flow Nets in A,nisotropic Soil
189
b. From Eq. (7.20).
H
L ,' q : Nk . =
d
k : 4 . 2 x l 0 - 3 c m / s e=c 4 . 2 x [ 0 - s m / s e c
Aq * (4.2x t0-sx0.5g3): 2.45x 10-5m3/sec/m
7.5
FIow Nets in Anisotropic Soil
The flow net construction describedthus far and the derived Eqs. (7.21) and (7.24)
for seepagecalculation have been basedon the assumptionthat the soil is isotropic.
However, in nature, most soils exhibit some degree of anisotropy.To account for
soil anisotropy with respectto hydraulic conductivity,we must modify the flow net
construction.
The differential equation of continuity for a two-dimensionalflow
[Eq. (7.a)] is
a2h
, ih
^,i,"..,k- ,-ll
For anisotropic soils, k, + kr. In this case,the equation representstwo families of curves that do not meet at 90'. However, we can rewrite the precedingequation as
a2h
*urh:,.,
l
k
\k ,) dx2 dz2
(1.2s)
Substitutingx' : t/k,J4x. we canexpressEq. (7.25)as
d2h
a2h
_ _ f _ :' r l
^
dx
-
t)
-.
)
d7.'
(1.26)
Now Eq. (7.26) is in a form similar to that of Eq. (7.5), with x replaced by x', which
is the new transformed coordinate. To construct the flow net, use the followine
procedure:
1. Adopt a vertical scale (that is, e axis) for drawing the cross section.
2. Adopt a horizontal scale (that is, x axis) such that horizontal scare : {Efh
x
vertical scale.
3. with scalesadopted as in steps 1 and2, plot the vertical section through the
permeable layer parallel to the direction of flow.
4. Draw the flow net for the permeable layer on the section obtained from step 3,
with flow lines intersecting equipotential lines at right angles and the elements
as approximate squares.
190
Chapter 7 Seepage
The rate of seepageper unit length can be calculated by modifyin gEq. (7 .21) to
-HNr
s:vk-kz
where
{
(t.21)
N:
H : total head loss
: number of flow channelsand potential drops, respectively
and ,Ay',i
(from flow net drawn in step 4)
Note that when flow nets are drawn in transformed sections (in anisotropic
soils),the flow lines and the equipotential lines are orthogonal. However,when they
are redrawn in a true section,these lines are not at right anglesto each other. This
f a c t i s s h o w n i n F i g u r e 7 . 1 0 .I n t h i s f i g u r c ,i t i s a s s u m e dt h a t k , : 6 k , . F i g u r e 7 . 1 0 a
shows a flow element in a transformed section.Thc flow element has been redrawn
i n a t r u e s e c t i o ni n F i s u r c 7 . I 0 b .
k_
(,
|
o
Vertical scale= 20 li
H o r i z o n t a sl c a l e=
20(G) = 49 li
Scale20 ft
(b)
Figure 7.10 Aflow element in anisotropic soil: (a) in transformed section; (b) in true section
n
7.7 Uplift Pressure under Hydraulic Structures
191
KH
s/7'
Figure 7'11 Pl<ttof q/kH againstS/'l' I'orflow around a single row of sheetpiles (after
Harr. 1962)
7,6
Mathematical Solution for Seepage
The seepageundcr severalsimple hydraulic structurescan be solvedmathematically.
Harr ( 1962)hasanalyzedmany suchconditions.Figure 7. I 1 showsa nondimensional
plot for the rate o1 seepagearound a single row of sheet piles. In a similar manner,
Figure 7.12is a nondimensional plot for the rate of seepageunder a dam. In Figure 7.11, the depth of penctration of the sheetpile is S, and the thicknessof the peim e a b l es o i l l a y e ri s f ' .
7.7
Uplift Pressureunder Hydraulic Structures
Flow nets can be used to determine the uplift pressure at the base of a hydraulic structure. This general concept can be demonstrated by a simple example. Figrrr"
7.13a shows a weir, the base of which is 2 m below the ground surface.The necesiary
flow net has also been drawn (assuming Ihat k, - k, : k). The pressure distribu_
tion diagram at the base of the weir can be obtained from the equipotential lines as
follows.
192
Chapter 7 SeePage
F-B+l
,]
kil
""'+I
+0.75
0
t0 25
+0 5
;
undera dam (afterHarr, I 962)
Figure 7.72 Scepage
t+ rn ------+
I
l*4:1t..
l,*:;*
:
:
r+m-----l
i
.
b
.
:
1
-
.
b
c
d
e
.
l
3y,. kN/m2
Impermeable
liryer
81, kN/m2
(a)
Figure 7.13 (a) A weir; (b) uplift force under a hydraulicstructure
\
7.8 Seepage through an Earth Dam on an lmpervious Base
193
There are sevencquipotential drops (1vr)in the flow net, and the differencein
the water levelsbetween the upstream and downstream sidesis H : 7 m. The head
lossfor each potential drop is Hl7 :7ll : I m. The uplift pressureat
a (left corner of the base) : (Pressurehead at a) x (y,,.)
:
- l]y,n: 8yu,
[(7 + 2)
Similarly, the uplift pressureat
b : l e - ( 2 ) ( 1 ) ) y , ,7: y , ,
and at
f : l e - ( 6 ) ( 1 ) 1 7: " ,3 y , , ,
The uplift pressureshave been plott"edin Figure 7.13b. The uplift force per unit
length measured along the axis of thc weir can bc calculatedby finding the area of
the pressurediagram.
7.8
Seepage through an Earth Dam
on an Impervious Base
Figure 7.14 showsa homogeneousearth dam resting on an impervious base.Let the
hydraulic conductivity of the compactedmatcrial of which the earth dam is made be
equal to k. The free surfaccof the water passingthrough the dam is given by abcd. lt
is assumedthat a'bc is parabolic.Thc slope of the free surfacecan be assumedto be
equal to the hydraulic gradient. It is also assumedthat, becausethis hydraulic gradie n t i s c o n s t a n tw i t h d e p t h ( D u p u i t . 1 8 6 3 ) ,
. d z
dx
(7.28)
i
Water
A
fevel ,
l
s
i
I
H
I m p e r v i o u sl a y e r
Figure
7.14 Flow through an earth dam constructed over an impervious
base
194
Chapter 7 Seepage
Considering the triangle cde, we can give the rate of seepageper unit length of the
dam (at right anglesto the crosssection shown in Figure 7.14) as
q: kiA
. d z:tand
,
r:
ax
( c e ) ( t ) : L s i na
A:
So
:
Q : k(tana)(L sina)
kLtan a sina
(7.2e)
Again, the rate of seepage(per unit length of the dam) through the section b/is
q - k i A -r ( # ) a x
t): kzddz
(7.30)
For continuous flow,
:
QEtl Q.2e)
4Eq. iz.:o;
or
o,
t
I
-,:
t(Ht
'
1';
k zd z : |
- L 2 s i n 2 a ) - L Ian a sin a(d - L cos a)
L2 sin2ct
2
2
zsln-a
(kL tan a sina) dx
.1-srnd
LI2
H2 cosa
^
)
- k L ranasina
L2 cosa
/
/ . . : - 2 ^ .\
L 4 ( " , " " ) _ L 2 s i n 2o
\cosa/
Ld - L2cosa
L 2 c oos t - 2 L r t *
cos a
l1tryu:n
cos2*
sin2a ,
(1.31)
Following is a step-by-step procedure to obtain the seepage rate q (per unit
length of the dam):
1. Obtain a.
2. Calculate A (see Figure 7.74) and then 0.3A.
7.9 L. Casagrande'sSolution for Seepage through an EarthDam
195
3. Calculate r/.
4. With known valuesof a and d, calculateI from Eq. (7.31).
With known valuesof l-, calculateq from Eq. (j.29).
Example7.3
Refer to the earth dam shown in Figure 7.14.Given that : 45o,a :30",
B : 10 ft,
B
H : 20 ft, lleight of dam : 25 ft, and k : 2 x l0-4 ftlmin, calculate
the seepage
rate q in ft3ldaylft length.
Solution
WeknowthatB :45" anda : 30".Thus.
H
=
tan B
tan 45'
L -
(2s- 20)
'
1 1: 0 . 3 A + '
tan p
: 6 * (2s 20\
t#+
* 6I
0.3a : (0.3)(20)
20ft
+B+*25
tan cY
ro
25
: 64.3ft
tan 30
From Eq. (7.31),
,
r
L
-
_
f
---;-
_- d -l t - : -l ' H 2
rl
cos cr
V cos, cr sinr a
64.3
cos30
From Eq. (7.29),
S : kL tan a sina : (2 x 10 4)(t1.7)(tan
30)(sin30)
- 6.754x 10-4tr3lminlft = 0.973ft3ldaylft
7.9
L. Casagrande's Solution for Seepage
through an Earth Dam
Equation(7.31)isdcrivedonthebasisofDupuit'sassumption(i.e.,r:
rlz/tlx).lLwas
shown by Casagrande( 1932)that, when the downstreamslopeangle
a in Figure 7. l4
becomesgreater than 30', deviations from Dupuit's assumption-b".,r,n"
more no_
t i c e a b l e .T h u s ( s e eF i g . 7 . l a ) , L . C a s a g r a n d c( 1 9 3 2 )s u g g e s t l d
that
. dz.
t-d:srna
whereat : f n\
al
(1.32)
196
Chapter 7 Seepage
So Eq. (7.29) can now be modified as
q : kiA:
k sin a(L sin a) : kL sin2a
(7.33)
Again,
/ s-\
q : k i A- - ( ; :
)rr
, ,t
(7.34)
CombiningEqs.(7.33)and (7.34)yields
a tts
o' : L sin2
1,.'.,,",,' I,
(7.3s)
where s : length of curve a'bc
l
-
s i n 2a ) - L s i n ra ( s - L )
^(U' - 1-2
2'
or
H'
l.
L=.s_
Vr,_r,nro
(7.36)
Wilh about 4-5"/" error,wc can wrilc
t-fi'+u'
(1.37)
CombiningEqs.(7.36)and (7.37)yields
2 - {"Q a 112 lT:E'rco(
a
(7.38)
Once the magnitudeof L is known, the rate 'of seepagecan be calculatedfrom
Eq. (7.33)as
q :
kLsin2 a
Exa mp l e7 .4
method.
SolveExample7.3usingL. Casagrande's
Solution
,:i
From Example7.3,d * 64.3ft, H = 20 ft, and a : 30'. Sol
r = t/8 + g, - {F:p7ri"};
:l6nf
,i
+@-@:r3.17rt
= 6.5*t x 10*4ft3/min/ft= 0.948ft3lday/ft r
q * (2 x 10*4x13.1?){sin230)
Problems
7.10
197
Summary
In this chapter, we studied Laplace's equation of continuity and its application in
solving problems related to seepagecalculation.The continuity equation is the fundamental basis on which the concept of drawing flow nets is derived. Flow nets are
very powerful tools for calculation of seepageas well as uplift pressure under various hydraulic structures.
Also discussedin this chapter (Sections7.8 and 7.9) is the procedure to calculate seepagethrough an earth dam constructedover an impervious base.Section7.8
derivesthe relationshipfor seepagebasedon Dupuit's assumptionthat the hydraulic
gradient is constant with depth. An improved procedure (L. Casagrande'ssolution)
for seepagecalculation is provided in Section7.9.
Prohlems
7.1
7.2
Refer to Figure 7.1-5.Given that
. H t - 6 m
D : 3 m
. H:-1.-5m
Dr:6m
draw a flow net. calculate the seepagelossper meter lcngth of the sheet pile
(at a right angle to the crosssection shown).
Draw a flow net for the single row of sheetpiles driven into a permeable
l a y e r a s s h o w n i n F i g u r e7 . l 5 . G i v e n t h a t
' Hr- 6m
D:2m
. Hz-7m
Dr:5m
calculatethe seepageloss per meter length of the sheet pile (at right angles
to the crosssection shown).
Figure 7.15
198
Chapter 7 Seepage
1 . 5m
t
i- :t:l;
,'n 1J '
:,;,
"',
Figure 7.16
Draw a flow net for the weir shown in Figure 7.16.Calculatc the rate of seepage under thc weir.
For the flow net drerwnin Problcm 7.3. calculatethc uplift forcc at the base
o f t h e w e i r p e r m e t c r l e n g t h ( m e a s u r e da l o n g t h e a x i s )o f t h e s t r u c t u r c .
7.3
7.4
t * ,.:'
Water level y
|
H
. i "' ,'.'i'':
.
!.'.' ..'
i'
.,
Figure 7.17
7.5
7.6
7.7
7.8
A n e a r t h d a m i s s h o w n i n F i g u r e7 . 1 7 .D e t e r m i n c t h e s e e p a g er a t e ,q , l n
m r / d a y i ml e n g t h .G i v e n :e t : u 2 : 4 5 " ,L t - - 5m , H - l 0 m , H r : 1 3 m , a n d
l0 acmiscc.
k-2x
Repeat Problem 7.-5with thc followingi u, - 23o,sz - 35",Lr :'7 m, H : 8 m,
rcmisec.
H , - 1 2m . a n d k - 1 . 5\ l 0
Repeat Problem 7.-5using L. Casagrande'smethod (Section 7.9).
Repeat Problem 7.6 using L. Casagrande'smethod (Section 7.9).
References
Case.cRaNoe, L. (1932). "Naeherungsmcthodenzur Bestimmurg von Art und Menge der
Sickerungdurch geschuetteteDaemme," Thesis,TechnischeHochschule,Vienna.
DupLrrr, J. (1863). Etudes Theoriques et Practiques sur le Mouvement des Eaux dans les
Canaux Decouverts et a Travers les Terrains Permeables,Dunod, Paris.
HaRn, M. E. ( 1962). Ground Water and Seepage,McGraw-Hill, New York.