Contents preface (VII) introduction 1—37

9.3.1. Theory of Seepage
According to Darcy’s law [Eq. (4.4)], the apparent velocities of flow of liquid through porous media are

u = − k	∂h	,	v = − k	∂h	, and	w = − k	∂h	(9.42)
∂x	∂y	∂z

338 IRRIGATION AND WATER RESOURCES ENGINEERING
where, k is the coefficient of permeability, h is the hydraulic head causing the flow, and u, v, and w, respectively, are the x-, y-, and z-components of velocity.
Substituting these in the continuity equation,

∂u	+	∂v	+	∂w	= 0	(9.43)
∂x	∂y	∂z

and assuming that k is a constant, one obtains

∂	^2 h	+	∂	^2h	+	∂	^2h	= 0	(9.44)
∂x^2	∂y^2	∂z^2

This is the well-known Laplace equation which governs the flow of liquid through porous medium. This equation implicily assumes that,
(i) The soil is homogeneous and isotropic,
(ii) The voids are completely filled with water,
(iii) No consolidation or expansion of the soil takes place, (iv) The soil and water are incompressible, and

(v) The flow obeys Darcy’s law and is steady. For two-dimensional flow, Eq. (9.44) reduces to

∂	2h	+	∂	2h	= 0	(9.45)
∂x^2	∂y^2

Equation (9.45) can be solved by graphical, analytical, numerical or some other suitable method, such as analogue method.
9.3.2. Graphical Solution of Seepage Equation
A graphical solution of Laplace equation results in two families of curves which intersect at right angle and form a pattern of ‘square’ figures known as flownet (Fig. 9.21). One set of lines is called the streamlines or flow lines along which water can flow through a cross-section. The other set of lines comprise what are called equipotential lines which are lines of equal head or energy level. Any flownet must satisfy the following basic requirements:

Water surface

q

a # q

q

dl	dp
	+
p

H

# q

Streamlines

#I # q
p

Equipotential

lines

Fig. 9.21 Flownet for seepage under a weir
(i) Flow lines and equipotential lines must intersect at right angle to form areas which are approximately squares. Most flownets are composed of curves and not straight lines. As such, the “square” figures are not true squares but curvilinear squares.The requirement of square figures is met, if the average width of any squire is equal to its average length.

SURFACE AND SUBSURFACE FLOW CONSIDERATIONS FOR DESIGN OF CANAL STRUCTURES

339

(ii) Certain entrance and exit requirements must be satisfied.

(iii) All pairs of adjacent equipotential lines must have equal head losses between them. (iv) The same amout of seepage flows between all pairs of adjacent flow lines.

Using Darcy’s law, a simple expression for seepage discharge can be obtained. Referring to Fig. 9.21, let the number of flow channels be N_f , the number of equipotential drops be N_d, and the seepage quantity flowing between any two adjacent flow lines be equal to ∆q. Thus, the total seepage quantity,

	q = Nf ∆q
or	q = Nf k	∆h	a
		∆l
or	q = Nfk∆ h
as ∆ l = a, since the flownet is composed of squares.
Since	∆ h = H/Nd,
	q = Nf k	H	= kH	Nf	(9.46)
	Nd	Nd

Equation (9.46) enables computation of the seepage quantity.The ratio N_f / N_d is called the shape factor. The seepage quantity is, therefore, the product of the coefficient of permeability, the net head, and the shape factor.
Considering an elementary cylindrical element of soil of cross-sectional area dA and length dl along any flow line, shown in Fig. 9.21, one can formulate the expression for the net seepage force dF in the direction of flow as
dF = pdA – (p + dp) dA or dF = – dp dA

Therefore, seepage force per unit volume of soil,

dF	= − ^dp	= − ρ g ^dH
dA dl
	dl	dl

As the head H decreases in the direction of flow, seepage force is positive in the flow direction.
(9.47) dH/dl is negative and, hence, the

Obviously, at the exit end (Fig. 9.21), the seepage force is vertical and may cause the lifting of soil particles resulting in piping failure. Hence, to provide safety against piping failure, the seepage force at the exit end must be less than the submerged weight of the soil particles. At the critical condition, the two forces will just balance each other. Thus,

− ρ g ^dH	= (1 – n)(ρs – ρ)g	(9.48)
dl

Here, n is the porosity of the soil and (ρ_s – ρ)g is the submerged weight of unit volume of soil particles. Dividing Eq. (9.48) by ρ g, one gets

		dH		F	ρs		I
	−		= ( 1	− n) G		− 1J	(9.49)
	dl	ρ
					H		K
or			− dH	= (1 − n) (G − 1)		(9.50)
			dl

where, G is the relative density of soil.

340 IRRIGATION AND WATER RESOURCES ENGINEERING
The quantity dH/dl represents the hydraulic gradient at the exit (or, simply, the exit gradient) which is negative. The value of dH/dl given by Eq. (9.50) is termed the critical gradient which should not be exceeded in order to prevent failure by piping.
Assuming G = 2.65 (true for most of the river sands) and n = 0.4, the value of the critical gradient is approximately 1.0. In practice, however, the actual gradient at the exit is kept around 1/4 to 1/6 depending on the safety requirements.
9.3.3. Method for Determination of Seepage Pressure
The seepage equation (or the Laplace equation), Eq. (9.45), cannot be solved exactly for usual canal structures having complex boundary conditions. Khosla, et al . (12) used the method of independent variables and obtained solutions of Laplace equation for a number of simple profiles. This solution is commonly known as Khosla’s solution. The following forms of these simple profiles (Fig. 9.22) are very useful in the design of weirs and barrages on permeable foundations:
(i) A straight horizontal floor of negligible thickness with a sheet pile at either end [Fig. 9.22 (i) and (ii)].
(ii) A straight horizontal floor of negligible thickness with an intermediate sheet pile [Fig. 9.22 (iii)].
(iii) A straight horizontal floor depressed below the bed but with no vertical cutoff [Fig. 9.22 (iv)]. This arrangement is useful for very small structures where no cutoff is provided.
The solutions for these simple profiles have been obtained (12) in terms of the pressure head ratio φ at ‘key’ points. These key points are the junction points of sheet piles with floor, i.e., E, C, D, E₁, C₁, and D₁ in case of floors of negligible thickness and D′ and D₁′ in case of depressed floor. The pressure head ratio φ at any key point, say E, is thus H_E/H. Here, H_E is the uplift pressure head at E. The mathematical expressions for the pressure head ratio φ at the key points are as follows:

	H
E1	C1
d
	b

D₁
(i) Sheet pile at the upstream end
^H b₂

E C

^b1 d

D
b (iii) Intermediate sheet pile

H
E ^C

b d
D
(ii) Sheet pile at the downstream end


H

d

D	D
1

b

(iv) Depressed floor

Fig. 9.22 Simple standard profiles of weir floors.

SURFACE AND SUBSURFACE FLOW CONSIDERATIONS FOR DESIGN OF CANAL STRUCTURES

341

(i) For sheet piles at either the upstream end [Fig. 9.22 (i)] or the downstream end [Fig. 9.22 (ii)],

	1			−1	F	λ − 2I
φE =			cos		G			J
π		λ
						H		K
			1			−1	F	λ − 1I
	φD =				cos		G			J
π				λ
						H	K

φC = 100 – φ

1

E

φ D

= 100 – φ

1

D

where,

λ =

1

[1 +

1 + α ^2 ]

2

and

α =

b

d

(ii) For sheet piles at the intermediate point [Fig. 9.22 (iii)],

1

−1

F

λ 1 − 1I

φE

=

cos

G

J

π

λ 2

H

K

1

−1

F

λ 1 I

^φD

=

cos

G

J

π

H

λ 2 K

1

−1

F

λ 1 + 1I

φC

=

cos

G

J

π

λ 2

H

K

Here,

λ

1

=

1

[ 1 + α 1^2 −

1 + α2^2 ]

2

λ

2

=

1

[ 1 + α 1^2 + 1 + α2^2 ]

2

where,	α	=	^b1	and	α =	^b2
d	d
			1				2

(iii) In the case of a depressed floor [Fig. 9.22 (iv)],

(9.51)
(9.52)

(9.53)
(9.54)
(9.55)

(9.56)

(9.57)

(9.58)
(9.59)

(9.60)

		φ	= φ D − ^2	( φ E − φ D ) +	3		(9.61)
		α ^2
			D′	3
			^φ D1 ′ = 100 – φ	D′				(9.62)
							b
where, φ		and φ are given by Eqs. (9.52) and (9.51), respectively, and α =	.
D
		E					d

The above solutions have also been presented in the form of curves of Fig. 9.23 which too can be used for determining the pressure head ratio φ at the key points. One can directly obtain, from Fig. 9.23, the values of φ_C , and φ_D (for b₁ /b > 0.5) for known values of α (= b/ d) and b₁/b of intermediate pile [Fig. 9.22 (iii)]. The value of φ_E for given values of α (=b/d) and b₁/b of intermediate pile is obtained by subtracting the value of φ_C (for 1 – b₁/b and given α) from 100. For example, φ_E [for b₁/b = 0.4 and α = 4] = 100 – φ_C (for b₁/b = 0.6 and α = 4) = 100 – 29.1 = 70.9. Likewise, to obtain φ _D (for b₁/b < 0.5 and given α), subtract φ_D (for 1 – b₁/b and given α) from 100. For example, φ_D [for b₁/b = 0.4 and α = 4] = 100 – φ_D (for b₁/b = 0.6 and α = 4) = 100 – 44.8 = 55.2.

342 IRRIGATION AND WATER RESOURCES ENGINEERING
Similarly, while φ_E and φ_D for the sheet pile at the downstream end (Fig. 9.22 (ii) can be read from Fig. 9.23, the values of φ_C and φ_D for the sheet pile at the upstream end (Fig. 9.22 (i)) are obtained as follows:
φ_C1 = 100 – φ_E (for the sheet pile at the downstream end) φ _D1 = 100 – φ_D (for the sheet pile at the downstream end)
And for the depressed floor (Fig. 9.22 (iv)),

^φ D₁ ′ = 100 – φ_D′
The uplift pressures obtained for the simple forms from either the above-mentioned mathematical expressions or the curves of Fig. 9.23 are corrected for (i) the floor thickness, (ii) mutual interference of sheet piles, and (iii) the slope of the floor.

Fig. 9.23 Variation of φ and	1
	(12)

π λ

SURFACE AND SUBSURFACE FLOW CONSIDERATIONS FOR DESIGN OF CANAL STRUCTURES

343

The corrected values of the pressures are valid for actual profiles. The pressures at intermediate points between the adjacent key points are assumed to vary linearly. This assumption causes only negligible error.