Click Here Full Article Seepage from a special class of...
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Seepage from a special class of a curved channel
with drainage layer at shallow depth
Bhagu R. Chahar1
Received 23 February 2009; accepted 16 June 2009; published 29 September 2009.
[1] In the present study an inverse method has been used to obtain an exact solution forseepage from a curved channel passing through a homogeneous isotropic porousmedium underlain by a drainage layer at shallow depth whose boundary maps along acircle onto the hodograph plane. The solution involves inverse hodograph andSchwarz-Christoffel transformation. The solution also includes a set of parametricequations for the shape of the channel contour and loci of phreatic line. Variation in theseepage velocity along the channel contour is presented as well. All these expressionsinvolve improper integrals along with accessory parameters. A particular solutioncorresponding to the water table below the top of the drainage layer has also been deducedfrom the general solution.
Citation: Chahar, B. R. (2009), Seepage from a special class of a curved channel with drainage layer at shallow depth, Water Resour.
Res., 45, W09423, doi:10.1029/2009WR007899.
1. Introduction
[2] Study of seepage from curved channels is importantbecause of its applications in areas of irrigation engineering,hydrology, reservoir management, and groundwater re-charge. Seepage from curved channels has been estimatedby several investigators for different boundary conditionsusing analytical methods [Aravin and Numerov, 1965;Ilyinskii and Kacimov, 1984; Polubarinova Kochina,1962]. Using an inverse method, Chahar [2006] presentedan exact solution for seepage from a curved channel whoseboundary maps along a circle onto the hodograph plane.The resulting channel possesses many interesting and usefulproperties. The channel shape is an approximate semiellipsewith top width as major axis and twice of depth of flow asminor axis and vice versa. Actually, it always lies betweenan ellipse and a parabola and its any coordinate is nearlyexact to the average of coordinates of corresponding ellipseand parabola. Also, this shape is non self intersecting and isfeasible from a slit (very narrow and deep channel) to a strip(very wide and shallow channel) unlike the Kozeny’strochoid shape, which is self intersecting for top width todepth ratio greater than 1.14 [Kacimov, 2003]. The seepagefunction of this channel is a linear combination of seepagefunctions for a slit and a strip [Chahar, 2001, 2007a]. Asemicircle is a special case of a semiellipse, consequently attop width to depth ratio equal to 2 the curved channel can beapproximated into a semicircular channel. Moreover, thequantity of seepage from this class of curvilinear channel issimply top width times the maximum velocity at the centerof the channel. Additionally, Chahar and Vadodaria [2008]observed that the variation in the drainage quantity from aponded surface to an array of empty ditches is like the shape
of this curved channel. The hydraulic properties of this typeof channel and sections corresponding to minimum area,minimum seepage and maximum hydraulic efficiency havebeen investigated by Chahar [2007b]. The solutions givenby Chahar [2006, 2007b] are applicable for the curvilinearbottomed channel passing through a homogeneous andisotropic porous medium of infinite extent. However, inmost alluvial plains the soil is stratified. In many cases,highly permeable layers of sand and gravel underlie the toplow permeable layer of finite depth. In that case the lowerlayer of sand and gravel acts as a free drainage layer for thetop seepage layer. The seepage from a canal runningthrough such stratified strata depends on the position ofwater table with respect to the drainage layer. If the watertable is below the top of the drainage layer then the seepageis much more than that in homogeneous medium of verylarge depth. The difference in quantity of seepage becomesappreciable when the drainage layer lies at a depth less thantwice the depth of water in the canal [Chahar, 2007a]. Onthe other hand, if the water table is above the top of thedrainage layer then the seepage is less depending on thehead difference between the water table and the watersurface in the channel. For that reason it is pertinent toinvestigate the seepage from a curved section whose bound-ary maps along a circle onto the hodograph plane andpassing through a porous medium of finite depth underlainby a drainage layer. Using the inverse method, presentedherein is an exact solution for seepage from the abovementioned curvilinear channel.
2. Analytical Solution
[3] Consider seepage from a curvilinear bottomed sym-metrical channel of top width T (m) and water depth y (m)passing through a homogeneous isotropic porous mediumof hydraulic conductivity k (m/s) underlain by a drainagelayer at a depth d (m) below the water surface is shown inFigure 1a. The position of the water table far from thechannel is dw (m) below the water surface and dw < d. Thus
1Department of Civil Engineering, Indian Institute of Technology Delhi,New Delhi, India.
Copyright 2009 by the American Geophysical Union.0043-1397/09/2009WR007899$09.00
W09423
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the boundary condition corresponds to condition A asdefined by Bouwer [1978] since the water table undisturbedby seepage from the channel is above the top of drainagelayer or pressure head at the drainage layer is d � dw (m)above atmospheric pressure. The pattern of seepage from
the channel is shown in Figure 1a, where a0 is the point ofinflection or maximum velocity point on the phreatic lineb0a0a00 at a depth di (m) below the water surface. The effectsof capillarity, accretion due to infiltration, and evaporationare ignored. In view of the significant length of the channel,
Figure 1. Seepage from a curvilinear bottomed channel with drainage layer at shallow depth and thewater table above the top of the drainage layer.
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the seepage flow can be considered 2-D in the verticalplane. Because of vertical symmetry, the solution for thehalf domain (c0f 0a00a0b0c0) is sought.[4] Let the complex potential be defined as W = 8 + iy
where 8 = velocity potential (m2/s) which is equal tonegative of hydraulic conductivity times the head h (m)plus C0 (a constant) (i.e., 8 = �kh + C0) and y = streamfunction (m2/s) which is constant along streamlines. If thephysical plane is defined as Z = X + iY then Darcy’s lawyields u = @f/@X = �k@h/@X and v = @f/@Y = �k@h/@Y;where u and v are velocity or specific discharge vectors inX and Y directions, respectively. In the velocity hodographplane (dW/dZ = u � iv), the phreatic line b0a0a00 will mapalong a double arc of circle of diameter k with a0 as reversalpoint. The channel boundary b0c0d 0 is an equipotential line,so the seepage velocity is normal to the boundary and inhodograph plane it will map a curvilinear path. However,the exact shape of the curve is not known. It is assumed thatthe channel boundary maps along a circle of diameter cin the hodograph plane and c > k. The top of the drainagelayer a00f 0e00 will represent a cut of length k0 in the hodographplane as shown in Figure 1b. It should be noted that theordinate of f 0 and the position of point a0 or e0 on the circle ofdiameter k are unknown in the hodograph plane. Thus thereare three unknown parameters c, k0 and k00 in the hodographplane, which have to be established. The inverse hodographdZ/dW (Figure 1c) for half of the physical flow domain hasbeen drawn following the standard steps [Harr, 1962;Strack, 1989]. The complex potential W (Figure 1d) forhalf of the physical flow domain has been drawn assumingC0 = 0, so that 8 = 0 along the channel contour c0b0 and kdwalong the water table. The dZ/dW plane and W plane havebeen mapped onto the upper half of an auxiliary (Im z > 0)plane (Figure 1e) using the Schwarz-Christoffel conformaltransformation [Harr, 1962; Polubarinova-Kochina, 1962].
2.1. Mapping in Various Planes
[5] Mapping of dZ/dW plane on the z plane (refer toAppendix A for details) results in
dZ
dW¼ c� k 0
ck 0F1 a;bð Þ
� �Zz0
t � 1ð Þdtffiffit
pt � bð Þ1:5
� i
cð1Þ
where t = dummy variable; and a, b = accessoryparameters. The W plane mapping on the z plane is
W ¼ kdwffiffiffib
p
2Kffiffiffiffiffiffiffiffiffia=b
p� � Zz
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ð2Þ
where K(ffiffiffiffiffiffiffiffiffia=b
p) = complete elliptical integral of the first
kind with modulus (ffiffiffiffiffiffiffiffiffia=b
p) [Byrd and Friedman, 1971].
Consequently the physical plane mapping becomes
Z ¼ kdwffiffiffib
p
2cKffiffiffiffiffiffiffiffiffia=b
p� � i c� k 0ð Þk 0F1 a; bð Þ
Z�z
0
F4 t;a;bð Þdt
0@
þZ�z
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p!
� i y ð3Þ
where t = dummy variable; and F1(a, b) and F4(t, a, b) aredefined by equation (A9) and equation (A17), respectively.
2.2. Shape of Channel
[6] Separating real and imaginary parts in equation (3)gives
X ¼ kdw
cKffiffiffiffiffiffiffiffiffia=b
p� �F sin�1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z
�z þ a
s;
ffiffiffiffiffiffiffiffiffiffiffiffib � ab
s !ð4aÞ
Y ¼ k c� k 0ð Þdwffiffiffib
p
2ck 0F1 a;bð ÞKffiffiffiffiffiffiffiffiffia=b
p� � Z�z
0
F4 t;a;bð Þ dt � iy ð4bÞ
where F (sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z= �z þ að Þ
p,
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p) = incom-
plete elliptical integral of the first kind with modulusffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
pand amplitude sin�1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z= �z þ að Þ
p[Byrd
and Friedman, 1971]. At the point b0 (z = �1; Z = T/2),equations (4a) and (4b) yield
c ¼2kdwK
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p �T K
ffiffiffiffiffiffiffiffiffia=b
p� � ð5Þ
c� k 0ð Þck 0F1 a;bð Þ ¼
2yKffiffiffiffiffiffiffiffiffia=b
p� �kdw
ffiffiffib
p�Z1
0
F4 t;a; bð Þ dt ð6Þ
and
1
k 0¼
Kffiffiffiffiffiffiffiffiffia=b
p� �kdw
T
2Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p �þ 2yF1 a; bð Þffiffiffib
p�Z1
0
F4 t;a;bð Þ dt
0@
!
ð7Þ
Substitution of c and k0 in equations (4a) and (4b) results in
X
T¼ 1
2Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ
p=b
�F sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z= �z þ að Þ
p;ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p� �ð8aÞ
and
Y
y¼ �1þ
Z�z
0
F4 t;a; bð Þ dt�Z1
0
F4 t;a;bð Þ dt ð8bÞ
Equations (8a) and (8b) in parametric form can be used toplot the channel shape by varying parameter z in the range�1 < z 0.
2.3. Position of Phreatic Line and Inflection Point
[7] The physical plane equation yields the following pa-rametric equations for the phreatic line b0a0a00 (b z 1):
X ¼ T
2þ y
Z1z
F5 t;a;bð Þ dt�Z1
0
F4 t;a; bð Þ dt ð9aÞ
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and
Y ¼ � dwffiffiffib
p
2Kffiffiffiffiffiffiffiffiffia=b
p� � Z1z
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p¼ � dw
Kffiffiffiffiffiffiffiffiffia=b
p� �F sin�1ffiffiffiffiffiffiffiffib=z
p;ffiffiffiffiffiffiffiffiffia=b
p� �ð9bÞ
where F5(t, a, b) is defined by equation (A26). At theinflection point a0 (z = 1; Z = X � idi), equation (9b) resultsin
di ¼dw
Kffiffiffiffiffiffiffiffiffia=b
p� �F sin�1ffiffiffib
p;ffiffiffiffiffiffiffiffiffia=b
p� �ð10aÞ
Utilizing equation (2) at the point a0 (z = 1; W = kdi + iqs/2)to get
di ¼dw
Kffiffiffiffiffiffiffiffiffia=b
p� � Kffiffiffiffiffiffiffiffiffia=b
p� ��
� F sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� bð Þ= 1� að Þ
p;ffiffiffiffiffiffiffiffiffia=b
p� ��ð10bÞ
Equating equations (10a) and (10b) gives an identity
F sin�1ffiffiffib
p;ffiffiffiffiffiffiffiffiffia=b
p� �þ F sin�1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� bð Þ= 1� að Þ
p;ffiffiffiffiffiffiffiffiffia=b
p� �¼ K
ffiffiffiffiffiffiffiffiffia=b
p� �ð11Þ
Equation (1) of the inverse hodograph plane at the inflectionpoint a0 (z = 1; dZ/dW = k00 � i/k) after substituting c and k0
yields
kk 00 ¼4yK
ffiffiffiffiffiffiffiffiffia=b
p� �dw
ffiffiffib
p
0@
1A
ln1þ
ffiffiffiffiffiffiffiffiffiffiffiffi1� b
pffiffiffib
p� �
þffiffiffiffiffiffiffiffiffiffiffiffi1� b
p
b
� ��Z10
F4 t;a; bð Þdt
ð12Þ
2.4. Variation in Seepage Velocity
[8] The distribution of the velocity of seeping waternormal to the channel contour can be found by usingequation (1) along c0b0 (�1 < z 0) as
V
k¼ 1
Kffiffiffiffiffiffiffiffiffia=b
p� � dwy
�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2F2 z;bð Þffiffiffib
p�Z1
0
F4 t;a;bð Þ dt
0@
1A2
þ T
2y
�K
ffiffiffiffiffiffiffiffiffiffiffiffib � ab
s ! !2
vuuutð13Þ
This equation reduces to equation (4) for z = 0 (at the centerof the channel c0) yielding maximum velocity (Vmax =assumed c).
2.5. Quantity of Seepage
[9] The steady seepage discharge per unit length ofchannel qs (m
2/s) can be expressed in the following simplestform:
qs ¼ k T þ Ayð Þ ¼ kyFs ð14Þ
where A = Vedernikov’s parameter [Harr, 1962] and Fs =seepage function [Chahar, 2001, 2006], which is adimensionless function of channel geometry and boundaryconditions [Chahar, 2007a]. Applying equation (2) at thepoint a00 (z = b; W = kdw + iqs/2) gives
qs ¼ 2kdwK
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ
p=b
�K
ffiffiffiffiffiffiffiffiffia=b
p� � ð15Þ
Therefore the seepage function is
Fs ¼qs
ky¼ 2
dw
y
Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p �K
ffiffiffiffiffiffiffiffiffia=b
p� � ð16Þ
Operating equations (5) and (15) results in
qs ¼ Tc ð17Þ
This equation reveals that the seepage from this class ofcurvilinear channels is simply top width times the seepagevelocity at the center of the channel. The same was true forhomogeneous medium of infinite depth (no drainage layer)case [Chahar, 2006].
2.6. Accessory Parameters
[10] Overall there are two unknowns i.e., accessoryparameters a and b. Other parameters c, k0, and k00 are partof solution as established by equations (5), (7), and (12),respectively. Two equations are required to determine theseaccessory parameters. One equation can be obtained fromthe physical plane. The physical plane equation at the pointf 0(z = a; Z = �id) gives
d
y¼
Ra0
F6 t;a;bð Þdt
R10
F4 t;a;bð Þ dtþ T
y
Kffiffiffiffiffiffiffiffiffia=b
p� �2K
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p �þ 1 ð18Þ
See equation (A24) for the definition of F6(t, a, b). For thesecond equation, the property of the slope of phreatic line atthe inflection point can be used (see Appendix A). Theslope at the point of inflection is maximum, hence kk00 ismaximum. Therefore the determination of the accessoryparameters a and b involves maximization of right-handside of equation (12) subject to equation (18). Thus it is anonlinear optimization problem subjected to a nonlinearequality constraint. Once a and b are known for specifiedd/y, dw/y and T/y, the relevant relations can be used to findc/k, k0/k, kk00, the variation in the seepage velocity, thequantity of seepage, the loci of phreatic lines, the positionof inflection point and the shape of the channel contour.
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2.7. Special Case: Condition A0
[11] When the water table is below the top of the drainagelayer (dw � d) then atmospheric pressure prevails thereleading to condition A0 [Bouwer, 1978]. Figure 2 shows thealtered physical, hodograph, and inverse hodograph planes.For this condition b = 1 and hence, the mappings in variousplanes become
dZ
dW¼ 1
p1
k� 1
c
� �Zz
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � 1ð Þ
p � i
cð19Þ
W ¼ kd
2Kffiffiffiffia
pð Þ
Zz
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � 1ð Þ
p ð20Þ
Z ¼ kd
2Kffiffiffiffia
pð Þ
�Zz0
1
p1
k� 1
c
� �Z t
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � 1ð Þ
p � i
c
0@
1A dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t t � að Þ t � 1ð Þp
� i y
ð21Þ
The Z plane mapping at the point b0 gives
c ¼2kd K
ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p �T K
ffiffiffiffia
pð Þ ð22aÞ
1
p1
k� 1
c
� �¼ yK
ffiffiffiffia
p ��2kd
Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð22bÞ
and
2
Kffiffiffiffia
pð Þ
d
y� 1
Kffiffiffiffiffiffiffiffiffiffiffiffi1� a
p � T
y¼ p
�Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð23Þ
Equation (23) gives accessory parameter a for specified d/yand T/y.
Figure 2. Seepage from a curvilinear bottomed channel with drainage layer at shallow depth and thewater table below the top of the drainage layer.
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[12] From equation (21) in the range �1 < z 0, theparametric equations for the shape of the channel contourare
X
T¼ 1
2Kffiffiffiffiffiffiffiffiffiffiffiffi1� a
p �F sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z= �z þ að Þ
p;ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p� �ð24aÞ
and
Y
y¼ �1þ
Zsinh�1ffiffiffiffiffi�z
p
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p �Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð24bÞ
The real and imaginary parts of equation (21) in the range1 z < 1, give the following parametric equations forthe phreatic line:
X ¼ T
2þ y
Z1cosh�1
ffiffiz
p
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficosh2 t � a
p�Z1
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð25aÞ
and
Y ¼ �d
Kffiffiffiffia
pð ÞF sin�1
ffiffiffiffiffiffiffiffi1=z
p;ffiffiffiffia
p� �ð25bÞ
At the drainage layer a0 (z = 1; X = B/2), equation (25a)yields
B
y¼ T
yþ 2
Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficosh2 t � a
p�Z1
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð26Þ
The seepage velocity for this condition reduces to
V
k¼
pKffiffiffiffiffiffiffiffiffiffiffiffi1� a
p �þ T
y
Z10
t=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p� �dt
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2K
ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p �sinh�1
ffiffiffiffiffiffiffi�z
pÞ
�2þ T
y
Z10
t=ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p� �dt
0@
1A
2vuuut
ð27Þ
Therefore the maximum velocity (Vmax = assumed c) at thecenter point c0 (z = 0) is
Vmax ¼ k þ p ky
TK
ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p� ��Z10
t. ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
aþ sinh2 tp� �
dt
ð28Þ
On substituting y from equation (23) in equation (28), itbecomes identical to equation (22a).
[13] Adopting z = a and dZ/dW = �i/k0 in equation (19)to get k0 or seepage velocity at the point f0 as
k
k 0¼ TK
ffiffiffiffia
pð Þ
2dKffiffiffiffiffiffiffiffiffiffiffiffi1� a
p � 1þ 2 sin�1 ffiffiffiffia
p
p
� �� 2 sin�1 ffiffiffiffi
ap
pð29Þ
Using b = 1 in equation (15) gives the quantity of seepage:
qs ¼ 2kdK
ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p �K
ffiffiffiffia
pð Þ ð30Þ
Therefore the seepage function from equations (23) and (30)is
Fs ¼T
yþ pK
ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p� ��Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð31Þ
and Vedernikov’s parameter is
A ¼ Fs �T
y¼ pK
ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p� ��Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ð32Þ
2.8. Special Case: Condition A0 With Infinite Depthof Drainage Layer and Water Table
[14] When both the depth of the drainage layer and theposition of water table lie at very large depth (d ! 1 anddw ! 1), then a = b = 1 and the above expressions reduceto the solution obtained by Chahar [2006] for homogeneousand isotropic porous medium of infinite extent.
2.9. Example
[15] Consider a curved channel having T = 2y; d = 2y anddw = 1.5y. The determination of the accessory parameters aand b consists of maximization of kk00 subject to equation(18). This constrained optimization problem has been con-verted into unconstrained minimization problem by penaltyfunction to have augmented function as
F a; bð Þ ¼ �4yK
ffiffiffiffiffiffiffiffiffia=b
p� �dw
ffiffiffib
p
0@
1A ln
1þffiffiffiffiffiffiffiffiffiffiffiffi1� b
pffiffiffib
p� �
þffiffiffiffiffiffiffiffiffiffiffiffi1� b
p
b
� �
�Z10
F4 t;a;bð Þ dt þ l abs
1� d
yþ
Za0
F6 t;a; bð Þdt
Z10
F4 t;a;bð Þ dt
þ T
y
Kffiffiffiffiffiffiffiffiffia=b
p� �2K
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þb
p �þ 1
!ð33Þ
where l = penalty function. Minimization functionfminsearch of MATLAB [The MathWorks, 2007] has beenused to minimize equation (33) with respect to a and b forT/y = 2; d/y = 2; dw/y = 1.5; and l = 10. The resultant valueshave been a = 0.0179 and b = 0.1388. With these values ofaccessory parameters in equations (5), (7), (12), (10a), and(15) give c = 2.2689k; k0 = 1.3618k; k00 = 0.8874/k; depth
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W09423 CHAHAR: SEEPAGE FROM CURVED CHANNELS W09423
of inflection point = 0.3534y; and seepage discharge =4.5378ky, respectively.[16] If d < dw then solving equation (23) with fzero
function of MATLAB [The MathWorks, 2007], the acces-sory parameter = 0.2182, therefore equations (26), (28), and(30) yield: B = 3.7911y; Vmax = 2.6576 k; and qs = 5.3152ky,respectively. Thus the seepage is 17.13% higher than theprevious case.[17] If both the drainage layer and water table were at
infinite depth then B = 4.6938y; Vmax = 2.3469k; and qs =4.6938ky. Figure 3 shows a comparison between the lasttwo cases; that is, the water table is below the top of thedrainage layer (d < dw) and both the drainage layer andwater table are at infinite depth (d ! 1 and dw ! 1). Itcan be observed that the channel without drainage layer isinscribed in the corresponding channel with drainage layer,this is true for all T/y and d/y ratios. The same thing isapplicable for the phreatic lines, and the velocity distribu-tion curves. Though the perimeters do not differ very muchbut the phreatic lines and the velocity curves do. Thedifferences increase with increase in T/y or decrease ind/y. For same T/y and d/y ratios, qs = 6.9702ky and B =4.4410y in a rectangular channel, qs = 5.5028ky and B =3.7884y in a trapezoidal channel and qs = 4.2593ky andB = 3.3126y in a triangular channel [Chahar, 2007a]. Thecomparison shows that the seepage from the present curvedchannel is higher (19.87%) than from the inscribing trian-gular channel and less (31.14%) than from the comprisingrectangular channel; thus it complies with the comparisontheorem [Kacimov, 2003].
3. Conclusions
[18] An exact analytical solution for the quantity ofseepage from a curved channel passing through a porousmedium underlain by a drainage layer at shallow depth
whose boundary maps along a circle onto the hodographplane can be obtained using an inverse method along withinverse hodograph and Schwarz-Christoffel transformation.The exact analytical expressions for the quantity of seepage,the shape of the channel contour, the variation in theseepage velocity normal to the channel contour, and lociof the phreatic lines involve improper integrals along withthe accessory parameters. The quantity of seepage from achannel whose boundary maps along a circle onto thehodograph plane is simply top width times the seepagevelocity at the center of the channel irrespective of it passesthrough a homogeneous medium of infinite depth (nodrainage layer case) or it underlain by a drainage layer withwater table above or below the top of the drainage layer.
Appendix A: Mapping Details
A1. Mapping of Inverse Hodograph Plane
[19] Mapping of dZ/dW plane on the z plane results in
dZ
dW¼ C1
Zz0
t � 1ð Þdtffiffit
pt � bð Þ1:5
þ C2 ðA1Þ
where C1 and C2 = constants. Using the values at pointc0 (z = 0; dZ/dW = �i/c) gives C2 = �i/c and then atthe point f 0 (z = a; dZ/dW = �i/k0) yields
C1 ¼ i1
c� 1
k 0
� ��Za0
t � 1ð Þdtffiffit
pt � bð Þ1:5
¼ c� k
ck 0
� ��F1 a;bð Þ ðA2Þ
Substitution of C1 and C2 in equation (A1) givesequation (1), where for the range 0 < z b,
Zz0
t � 1ð Þdtffiffit
pt � bð Þ1:5
¼ �i
Zz0
1� tð Þdtffiffit
pb � tð Þ1:5
¼ �iF1 z;bð Þ
¼ �i2 sin�1
ffiffiffizb
sþ 1� b
b
ffiffiffiffiffiffiffiffiffiffiffiffiz
b � z
s !ðA3Þ
dZ
dW¼ �i
c� k
ck 0
� �F1 z; bð ÞF1 a;bð Þ �
i
cðA4Þ
for the range �1 < z 0,
Zz0
t � 1ð Þdtffiffit
pt � bð Þ1:5
¼Z�z
0
1þ tð Þdtffiffit
pt þ bð Þ1:5
¼ F2 z;bð Þ
¼ 2 ln
ffiffiffiffiffiffiffi�zb
sþ
ffiffiffiffiffiffiffiffiffiffiffiffib � zb
s !þ 2
1� bb
ffiffiffiffiffiffiffiffiffiffiffiffi�z
b � z
s
ðA5Þ
dZ
dW¼ c� k 0
ck 0
� �F2 z; bð ÞF1 a;bð Þ �
i
cðA6Þ
Figure 3. Comparison of channels with and without thedrainage layer.
W09423 CHAHAR: SEEPAGE FROM CURVED CHANNELS
7 of 12
W09423
and otherwise,Zzb
t � 1ð Þdtffiffit
pt � bð Þ1:5
¼ F3 z;bð Þ ¼ 2 ln
ffiffiffizb
sþ
ffiffiffiffiffiffiffiffiffiffiffiffiz � bb
s !
þ 21� bb
ffiffiffiffiffiffiffiffiffiffiffiffiz
z � b
sðA7Þ
dZ
dW¼ c� k 0
ck 0
� �F3 z;bð ÞF1 a;bð Þ �
i
kðA8Þ
Therefore
F1 a; bð Þ ¼Za0
1� tð Þdtffiffit
pb � tð Þ1:5
¼ 2 sin�1
ffiffiffiffiab
rþ 1� b
b
ffiffiffiffiffiffiffiffiffiffiffiffia
b � a
r� �
ðA9Þ
The branch cuts in equations (A2)–(A9) are selectedsuch that both
ffiffit
pand
ffiffiffiffiffiffiffiffiffiffiffit � b
pare negative.
A2. Mapping of Complex Potential Plane
[20] The W plane mapping on the z plane is
W ¼ C3
Zz0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p þ C4 ðA10Þ
Using the values of W and z at points c0 (z = 0; W = 0) andf 0 (z = a; W = kdw), the constants C4 = 0 and
C3 ¼ kdw
�Za0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
p ¼ 0:5kdwffiffiffib
p=K
ffiffiffiffiffiffiffiffiffia=b
p� �
ðA11Þ
where
Za0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
p ¼ 2ffiffiffib
p Kffiffiffiffiffiffiffiffiffia=b
p� �ðA12Þ
The branch cuts are selected such thatffiffiffiffiffiffiffiffiffiffiffit � a
pis posi-
tive while bothffiffit
p, and
ffiffiffiffiffiffiffiffiffiffiffit � b
pare negative as before.
After substituting C3 and C4, equation (A10) results inequation (2). Differentiating equation (2) with respect to zgives
dW
dz¼ kdw
ffiffiffib
p
2Kffiffiffiffiffiffiffiffiffia=b
p� � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz z � að Þ z � bð Þ
p ðA13Þ
A3. Mapping of Physical Plane
[21] Since
dZ
dz¼ dZ
dW
dW
dzðA14Þ
Substitution of dZ/dW from equation (1) and dW/dz fromequation (A13) results in
dZ
dz¼ kdw
ffiffiffib
p
2Kffiffiffiffiffiffiffiffiffia=b
p� � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz z � að Þ z � bð Þ
p
c� k 0
ck 0F1 a;bð Þ
Zz0
t � 1ð Þdtffiffit
pt � bð Þ1:5
� i
c
0@
1A ðA15Þ
Integrating equation (A15) along the channel contour c0b0
(�1 < z 0), gives equation (3) wherein
Zz0
Zt0
t � 1ð Þdtffiffit
pt � bð Þ1:5
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p0@
1Adt
¼Z�z
0
Z�t
0
1þ tð Þdtffiffit
pt þ bð Þ1:5
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p0@
1Adt ðA16Þ
F4 t;a;bð Þ ¼Z�t
0
1þ tð Þdtffiffit
pt þ bð Þ1:5
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p¼ F2 t; bð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t t þ að Þ t þ bð Þp ðA17Þ
Equation (3) is the mapping along the channel contour c0b0.The real part of it is
X ¼ kdwffiffiffib
p
2cKffiffiffiffiffiffiffiffiffia=b
p� � Z�z
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p ðA18Þ
where
Z�z
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p ¼ 2ffiffiffib
p F sin�1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z
�z þ a
s;
ffiffiffiffiffiffiffiffiffiffiffiffib � ab
s !
ðA19Þ
Plugging equation (A19) in equation (A18) results inequation (4a). At the point b0 (z = �1; Z = T/2)equations (4a) and (4b) give
T
2¼ kdw
ffiffiffib
p
2cKffiffiffiffiffiffiffiffiffia=b
p� � Z10
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p¼
kdwKffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p �cK
ffiffiffiffiffiffiffiffiffia=b
p� � ðA20Þ
y ¼ k c� k 0ð Þdwffiffiffib
p
2ck 0F1 a;bð ÞKffiffiffiffiffiffiffiffiffia=b
p� � Z10
F4 t;a; bð Þ dt ðA21Þ
respectively, where
Z10
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ
p ¼ 2ffiffiffib
p Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p� �ðA22Þ
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W09423 CHAHAR: SEEPAGE FROM CURVED CHANNELS W09423
Manipulation of equations (A20) and (A21) yieldsequations (5), (6), and (7).[22] Physical plane equation along the centerline c0f 0 (0
z a) converts to
Z ¼ �i kdwffiffiffib
p
2cKðffiffiffiffiffiffiffiffiffia=b
pÞ
c� k 0
k 0F1 a;bð Þ
Zz0
F6 t;a;bð Þdt
þZz0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
p!
� iy ðA23Þ
where
F6 t;a;bð Þ ¼Zt0
1� tð Þdtffiffit
pb � tð Þ1:5
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
p
¼ F1 t;bð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
pðA24Þ
Equation (A23) at the point f0 (z = a; Z = �id) givesequation (18).
A4. Position of Phreatic Line and Inflection Point
[23] For the phreatic line segment b0a0a00 (b z 1),the Z plane mapping equation (3) changes to
Z ¼ T
2þ kdw
ffiffiffib
p
2Kffiffiffiffiffiffiffiffiffia=b
p� �
c� k 0ð Þck 0F1 a; bð Þ
Z1z
F5 t;a; bð Þdt
� i
k
Z1z
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p!
ðA25Þ
where
F5 t;a;bð Þ ¼Ztb
t � 1ð Þdtffiffit
pt � bð Þ1:5
1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p
¼ F3 t; bð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
pðA26Þ
Separating real and imaginary parts of equation (A25) alongwith values of c and k0 results in equations (9a) and (9b)where
Z1z
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ¼ 2ffiffiffib
p F sin�1ffiffiffiffiffiffiffiffib=z
p;ffiffiffiffiffiffiffiffiffia=b
p� �ðA27Þ
A5. Inflection Point
[24] The phreatic line equation (9b) at the inflection point(z = 1) yields equation (10a) where
Z11
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ¼ 2ffiffiffib
p F sin�1ffiffiffib
p;ffiffiffiffiffiffiffiffiffia=b
p� �ðA28Þ
In the hodograph plane, the slit a00a0b0 at point a0 on thecircumference of the circle of diameter k is a consequence ofthe velocity at the point of inflection representing themaximum velocity along the phreatic line, thus the slope atthe inflection point is a maximum. The slope at any point onthe phreatic line can be obtained as
dY
dX¼ df=dX
df=dY¼ u
v¼ u= u2 þ v2ð Þ
v= u2 þ v2ð Þ ¼ ku
u2 þ v2ðA29Þ
wherein u2 + v2 � kv = 0 along the phreatic line [Harr,1962]. In the inverse hodograph plane the abscissa axis isu/u2 + v2) and the ordinate axis is v/(u2 + v2). It is evidentfrom equation (A29) that the slope at any point on thephreatic line is simply k times the abscissa value at thatpoint in the inverse hodograph plane, therefore the slope atthe inflection point is
dY
dX
� �inflection Point
¼ kk 00 ðA30Þ
where k00 = location of the inflection point a0 along theabscissa in the inverse hodograph plane (Figure 1c). Sincethe slope at the inflection point is a maximum, kk00 should bemaximum.
A6. Variation in Seepage Velocity
[25] The distribution of the velocity of seeping waternormal to the channel contour can be found by usingequation (1) along c0b0 (�1 < z 0) as
dZ
dW¼ 1
u� iv¼ c� k 0
ck 0
� �F2 z;bð ÞF1 a;bð Þ �
i
cðA31Þ
Substituting value of c and k0 and then manipulating yields
uþ i v
u2 þ v2¼
2yKffiffiffiffiffiffiffiffiffia=b
p� �kdw
ffiffiffib
p F2 t;bð ÞZ10
F4 t;a; bð Þ dt
� iT
2kdw
Kffiffiffiffiffiffiffiffiffia=b
p� �K
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p � ðA32Þ
Separating real and imaginary parts and then squaring andadding give equation (13).
A7. Quantity of Seepage
[26] Equation (2) at the point a00 (z = b ; W = kdw + iqs/2)generates
kdw þ qs
2i ¼ kdw
ffiffiffib
p
2Kffiffiffiffiffiffiffiffiffia=b
p� � Zb
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p¼ kdw þ
ikdwKffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p �K
ffiffiffiffiffiffiffiffiffia=b
p� � ðA33Þ
W09423 CHAHAR: SEEPAGE FROM CURVED CHANNELS
9 of 12
W09423
because
Zb0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ¼Za0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
p þ i
Zba
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ b � tð Þ
pðA34Þ
Zba
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ b � tð Þ
p ¼ 2ffiffiffib
p Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b
p� �ðA35Þ
Equating imaginary parts in equation (A33) leads toequation (15). The real part of equation (2) at the point a0
(z = 1; W = kdi + iqs/2) yields equation (10b) wherein
Z10
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ¼Za0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ b � tð Þ
p
þ i
Zba
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ b � tð Þ
p
þZ1
b
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ðA36Þ
Z1b
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ
p ¼ 2ffiffiffib
p F sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� bð Þ= 1� að Þ
p;ffiffiffiffiffiffiffiffiffia=b
p� �
ðA37Þ
A8. Special Case: Condition A0
[27] When the water table is below the top of the drainagelayer (dw � d) then there will be no inflection point on thephreatic line (Figure 2). The point a00 vanishes afterapproaching to a0 in various planes, which result to b = 1.With b = 1, the mappings in various planes onto z planeconvert into equations (19)–(21). Along the channel con-tour c0b0 (�1 < z 0), equation (21) becomes
Z ¼ kd
2Kffiffiffiffia
pð Þ
1
p1
k� 1
c
� �Z�z
0
Z t
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ 1ð Þ
p �i dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p
� i
c
Z�z
0
i dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p!� i y ðA38Þ
Sorting out real and imaginary parts gives
X ¼ kd
2cKffiffiffiffia
pð Þ
Z�z
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p¼ kd
cKffiffiffiffia
pð ÞF sin�1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z
�z þ a
s;ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p !
ðA39Þ
Y ¼ kd
2pKffiffiffiffia
pð Þ
1
k� 1
c
� �Z�z
0
Z t
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ 1ð Þ
p dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p � y
ðA40Þ
where
Z�z
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p ¼ 2F sin�1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z
�z þ a
s;ffiffiffiffiffiffiffiffiffiffiffiffi1� a
p !
ðA41Þ
Equations (A39) and (A40) at the point b0 (z =�1; Z = T/2)give equations (22a), (22b), and (23), where
Z10
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p ¼ 2Kffiffiffiffiffiffiffiffiffiffiffiffi1� a
p� �ðA42Þ
Z10
Z t
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ 1ð Þ
p dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p ¼ 4
Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
pðA43Þ
Equations (22a) and (A38) produce
Z ¼ T
4Kffiffiffiffiffiffiffiffiffiffiffiffi1� a
p � Z�z
0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ
p
þ iy
2
Z�z
0
sinh�1ffiffit
pdtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t t þ að Þ t þ 1ð Þp �Z1
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p � i y ðA44Þ
Separating real and imaginary parts and simplifying giveequations (24a) and (24b).[28] An alternate relation for the accessory parameter can
be obtained by using physical plane equation along thecenterline c0f 0 (0 z a) as
Z ¼ kd
2Kffiffiffiffia
pð Þ
�2 i
p1
k� 1
c
� �Zz0
sin�1ffiffit
pdtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t a� tð Þ 1� tð Þp
� i
c
Zz0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit a� tð Þ 1� tð Þ
p!
� i y ðA45Þ
Combining equations (22a) and (22b) in equation (A45) atthe point f 0 (z = a; Z = �id) yields
2
Kffiffiffiffia
pð Þ
d
y� 1
Kffiffiffiffiffiffiffiffiffiffiffiffi1� a
p � T
y
¼ 2
Kffiffiffiffia
pð Þ 1þ
Zsin�1 ffiffiffiap
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia� sin2 t
p�Z1
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p0B@
1CAðA46Þ
In lieu of equation (23), equation (A46) can be used tofind the accessory parameter a for specified d/y and T/y.
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W09423 CHAHAR: SEEPAGE FROM CURVED CHANNELS W09423
Comparing equations (23) and (A46) leads to the followingidentity:
Z10
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p þZsin�1 ffiffiffiap
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffia� sin2 t
p ¼ Kffiffiffiffia
p �p2
ðA47Þ
For the phreatic line segment a0b0 (1 z < 1), the Z planemapping equation (21) changes to
Z ¼ kd
2Kffiffiffiffia
pð Þ
Zz1
2
p1
k� 1
c
� �cosh�1
ffiffit
p� i
k
� �dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t t � að Þ t � 1ð Þp þ T
2
ðA48Þ
Using equation (22a) in equation (A48) and manipulatingcapitulate
Z ¼ y
Z1cosh�1
ffiffiz
p
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffifficosh2 t � a
p�Z1
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p
� i d
2Kffiffiffiffia
pð Þ
Z1z
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � 1ð Þ
p þ T
2ðA49Þ
Equating real and imaginary parts gives equations (25a) and(25b) where
Z1z
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � 1ð Þ
p ¼ 2F sin�1ffiffiffiffiffiffiffiffi1=z
p;ffiffiffiffia
p� �ðA50Þ
Another relation for the seepage width B (m) at the drainagelayer can be obtained using physical plane mapping alongthe drainage layer f 0a0 (a z 1) as
Z ¼ kd
2Kffiffiffiffia
pð Þ
2
p1
k� 1
c
� �Zza
sin�1ffiffit
pdtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
t t � að Þ 1� tð Þp
þ 1
c
Zza
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ 1� tð Þ
p!
� i d ðA51Þ
and then at point a0 (z = 1; Z = �id + B/2) as
B
y¼ T
yþ 2
Zp=2sin�1 ffiffiffiap
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffisin2 t � a
p�Z1
0
t dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiaþ sinh2 t
p ðA52Þ
Rewriting equation (19) for the velocity of seeping waternormal to the channel contour c0b0 (�1 < z 0) as
1
u� iv¼ 2
p1
k� 1
c
� �sinh�1
ffiffiffiffiffiffiffi�z
p� i
cðA53Þ
Substituting value of c and manipulating gives equation (27).
A9. Special Case: Condition A0 With Infinite Depthof Drainage Layer and Water Table
[29] If the porous medium is homogeneous and isotropicof infinite extent then both the depth of the drainage layerand the position of water table lie at very large depth (d!1and dw ! 1). For d ! 1 and dw ! 1; a = b = 1 and themappings onto z plane result in [Chahar, 2006]
dZ
dW¼ 1
p1
k� 1
c
� �Zz0
dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � 1ð Þ
p � i
cðA54Þ
W ¼ qs
2p
Zz0
dt
t � 1ð Þffiffit
p ¼ qs
2pln
ffiffiffiz
p� 1ffiffiffi
zp
þ 1
�������� ðA55Þ
Z ¼ qs
p ctan�1
ffiffiffiffiffiffiffi�z
p� i y� 2 qs
p2
1
k� 1
c
� � Zsinh�1ffiffiffiffiffi�z
p
0
t dtcosh t
0BB@
1CCA
ðA56Þ
Using equation (A54) at the point b0 (z = �1; Z = T/2) andsimplifying generate
c ¼ k
TT þ p2
4Gy
� �ðA57Þ
qs ¼ kyp2
4Gþ T
y
� �ðA58Þ
where G = 0.915965594. . . = Catalan’s constant. Theequation for the shape of the channel contour is
Y
y¼ 1
2G
Zsinh�1 tan pX=Tð Þ
0
t dtcosh t
� 2G
0B@
1CA ðA59Þ
The parametric equations for the phreatic line are
X ¼ T
2þ y
2G
Z1cosh�1
ffiffiz
p
t dtsinh t
ðA60Þ
Y ¼ T þ p2
4Gy
� �ln tanh
cosh�1ffiffiffiz
p
2
� �ðA61Þ
The width of seepage flow at infinity B (m) comes out
B ¼ T þ p2
4Gy ðA62Þ
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Finally the seepage velocity in this condition is given by
V ¼ k T=yþ p2=4Gð Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip sinh�1 tan pX=Tð Þ �
=2G �2þ T=yð Þ2
q ðA63Þ
[30] Acknowledgments. The author is grateful to A. R. Kacimov andother anonymous reviewers of an earlier version of the manuscript for theirconstructive suggestions, which resulted in a significant improvement of thepresent manuscript. The help rendered by G. C. Mishra in getting animportant condition at the inflection point is heartily acknowledged.
ReferencesAravin, V. I., and S. N. Numerov (1965), Theory of Flow in UndeformablePorous Media, Isr. Program for Sci. Transl., Jerusalem.
Bouwer, H. (1978), Groundwater Hydrology, McGraw Hill, New York.Byrd, P. F., and M. D. Friedman (1971), Handbook of Elliptic Integrals forEngineers and Scientists, Springer, Berlin.
Chahar, B. R. (2001), Extension of Vederikov’s graph for seepage fromcanals, Ground Water, 39(2), 272 – 275, doi:10.1111/j.1745-6584.2001.tb02308.x.
Chahar, B. R. (2006), Analytical solution to seepage problem from a soilchannel with a curvilinear bottom, Water Resour. Res., 42, W01403,doi:10.1029/2005WR004140.
Chahar, B. R. (2007a), Analysis of seepage from polygon channels,J. Hydraul. Eng., 133(4), 451–460, doi:10.1061/(ASCE)0733-9429(2007)133:4(451).
Chahar, B. R. (2007b), Optimal design of special class of curvilinear bot-tomed channel section, J. Hydraul. Eng., 133(5), 571–576, doi:10.1061/(ASCE)0733-9429(2007)133:5(571).
Chahar, B. R., and G. P. Vadodaria (2008), Drainage of ponded surface byan array of ditches, J. Irrig. Drain. Eng., 134(6), 815–823, doi:10.1061/(ASCE)0733-9437(2008)134:6(815).
Harr, M. E. (1962), Groundwater and Seepage, McGraw-Hill, New York.Ilyinskii, N. B., and A. R. Kacimov (1984), Seepage limitation optimizationof the shape of an irrigation channel by the inverse boundary valueproblem method (in Russian), Izv. Ross. Akad. Nauk, 3, 74–80, (J. FluidDyn., Engl. Transl., 19(4), 404–410.)
Kacimov, A. R. (2003), Discussion on ‘‘Design of minimum seepage lossnonpolygon canal sections,’’ by P. K. Swamee and D. Kashyap, J. Irrig.Drain. Eng., 129(1), 68 – 70, doi:10.1061/(ASCE)0733-9437(2003)129:1(68.2).
Polubarinova-Kochina, P. Y. (1962), Theory of Ground Water Movement,Princeton Univ. Press, Princeton, N. J.
Strack, O. D. L. (1989), Groundwater Mechanics, Prentice Hall, Engle-wood Cliffs, N. J.
The MathWorks (2007), The language of technical computing, version7.4.0.287(R2007a), Natick, Mass.
����������������������������B. R. Chahar, Department of Civil Engineering, Indian Institute of
Technology Delhi, Hauz Khas, New Delhi, 110 016, India. ([email protected])
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