click here full article seepage from a special class of...

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

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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


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

W09423 CHAHAR: SEEPAGE FROM CURVED CHANNELS

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and

Y ¼ � dwffiffiffib

p


p� � Z1z

dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ t � bð Þ

p¼ � dw


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


p� �F sin�1ffiffiffib


p� �ð10aÞ

Utilizing equation (2) at the point a0 (z = 1; W = kdi + iqs/2)to get

di ¼dw


p� � Kffiffiffiffiffiffiffiffiffia=b

p� ��

� F sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� bð Þ= 1� að Þ


p� ��ð10bÞ

Equating equations (10a) and (10b) gives an identity

F sin�1ffiffiffib


p� �þ F sin�1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� bð Þ= 1� að Þ


p� �¼ K


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


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


p� � dwy

�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2F2 z;bð Þffiffiffib

p�Z1

0

F4 t;a;bð Þ dt

0@

1A2

þ T

2y

�K


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


p� � ð15Þ

Therefore the seepage function is

Fs ¼qs

ky¼ 2

dw

y

Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b

p �K


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


p� �2K


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


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


p �Z10


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


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


p�Z1

0


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


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


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


p �K

ffiffiffiffia

pð Þ ð30Þ

Therefore the seepage function from equations (23) and (30)is

Fs ¼T

yþ pK


p� ��Z10


p ð31Þ

and Vedernikov’s parameter is

A ¼ Fs �T

y¼ pK


p� ��Z10


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


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


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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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


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


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


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


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.


7 of 12

W09423

and otherwise,Zzb


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


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


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


p� �

ðA11Þ

where

Za0


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


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


p� � 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiz z � að Þ z � bð Þ

p

c� k 0

ck 0F1 a;bð Þ

Zz0


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


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


p� � Z�z

0


p ðA18Þ

where

Z�z

0

dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ bð Þ

p ¼ 2ffiffiffib

p F sin�1


�z þ a

s;


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


p� � Z10


p¼

kdwKffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b

p �cK


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


p ¼ 2ffiffiffib

p Kffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b

p� �ðA22Þ

8 of 12


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


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


p� �

c� k 0ð Þck 0F1 a; bð Þ

Z1z

F5 t;a; bð Þdt

� i

k

Z1z


p!

ðA25Þ

where

F5 t;a;bð Þ ¼Ztb


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


p ¼ 2ffiffiffib

p F sin�1ffiffiffiffiffiffiffiffib=z


p� �ðA27Þ

A5. Inflection Point

[24] The phreatic line equation (9b) at the inflection point(z = 1) yields equation (10a) where

Z11


p ¼ 2ffiffiffib

p F sin�1ffiffiffib


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


p� �K


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


p� � Zb

0


p¼ kdw þ

ikdwKffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffib � að Þ=b

p �K


p� � ðA33Þ


9 of 12

W09423

because

Zb0


p ¼Za0


p þ i

Zba

dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t � að Þ b � tð Þ

pðA34Þ

Zba


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


p ¼Za0


p

þ i

Zba


p

þZ1

b


p ðA36Þ

Z1b


p ¼ 2ffiffiffib

p F sin�1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1� bð Þ= 1� að Þ


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


�z þ a

s;ffiffiffiffiffiffiffiffiffiffiffiffi1� a

p !

ðA39Þ

Y ¼ kd

2pKffiffiffiffia

pð Þ

1

k� 1

c

� �Z�z

0

Z t

0


p dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ

p � y

ðA40Þ

where

Z�z

0


p ¼ 2F sin�1


�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


p ¼ 2Kffiffiffiffiffiffiffiffiffiffiffiffi1� a

p� �ðA42Þ

Z10

Z t

0


p dtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit t þ að Þ t þ 1ð Þ

p ¼ 4

Z10


pðA43Þ

Equations (22a) and (A38) produce

Z ¼ T

4Kffiffiffiffiffiffiffiffiffiffiffiffi1� a

p � Z�z

0


p

þ iy

2

Z�z

0

sinh�1ffiffit

pdtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

t t þ að Þ t þ 1ð Þp �Z1

0


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


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


p0B@

1CAðA46Þ

In lieu of equation (23), equation (A46) can be used tofind the accessory parameter a for specified d/y and T/y.

10 of 12


Comparing equations (23) and (A46) leads to the followingidentity:

Z10


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


p�Z1

0


p

� i d

2Kffiffiffiffia

pð Þ

Z1z


p þ T

2ðA49Þ

Equating real and imaginary parts gives equations (25a) and(25b) where

Z1z


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


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


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


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


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


11 of 12

W09423

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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