dam Note.pdf,lecturenote on dam engineering

COLLEGE OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
WATER RESOURCES ENGINEERING AND MANAGEMENT PROGRAMME
HYDRAULIC STRUCTURES, RESERVOIR ENGINEERING
LECTURE NOTES
F. O. K. Anyemedu
KUMASI
MARCH - 2007
1

Chapter 1; Hydraulic Structures
Hydraulic Structures
Hydraulic structures are engineering structures constructed for the purposes of
harnessing and using water resources (groundwater, surface water, lakes, sea,
etc) or for the prevention of the negative and destructive actions (floods, shore
erosion, etc) of water on the surrounding environment.
There are a large variety of hydraulic structures to serve the many purposes for
which water resources are put to use. There are several classifications of
hydraulic structures, however, the most important is the classification by function
as given below.
2

Table 1. Classification of Hydraulic Structures by function
Type Purpose Structure
1. Storage structures To store water Dams, tanks
2. Flow control
structures
To regulate the
quantity and pass
excess flow
Spillways, outlets, gates,
valves
3. Flow measurement
structures
To determine
discharge
Weirs, orifices, flumes
4. Division structures To divert the main
course of water
Coffer dams, weirs, canal
headworks, intake works
5. Conveyance
structures
To guide flow from
one location to
another
Open channel, pressure
conduits, pipes, canals and
sewers
6. Collection structures To collect water for
disposal
Drain inlets, infiltration
galleries, wells
7. Energy dissipation
structures
To prevent erosion
and structural
damage
Stilling basins, surge tanks,
check dams
8. Shore protection
structures
To protect banks Dikes, groins, jetties,
revetments
9. River training and
waterway stabilization
structures
To maintain river
channel and water
transportation
Levees, cutoffs, locks, piers,
culverts
10. Sediment and
quality control
structures
To control or remove
sediments and other
pollutants
Racks, screens, traps,
sedimentation tanks, filters,
sluiceways
11. Hydraulic machines To convert energy
from one from to
another
Pumps, turbines, rams.
1. Storage structures - Dams
Dam: Dam is any artificial barrier and its appurtenant works constructed for the
purpose of holding water or any other fluid.
There are three common classification schemes for dams. According to the
function performed, dams are classified into:
i) storage dams for impounding water for developmental uses.
ii) Diversion dams for diverting streamflow into canals or other
conveyance system and
iii) Detention dams to hold the water temporary to retard flood flows
From hydraulic design considerations, dams are classified as:
i) overflow dams to carry discharge over their crests, and
3

ii) ii) non-overflow dams, which are not designed to be overtopped.
The most common classification is one based on the materials of which dams are
made. This classification makes further sub-classification by recognizing the
basic type of design, such as concrete gravity or concrete arch dams. Types of
dams include:
i) Earthfill dams
ii) Rockfill dams
iii) Concrete dams
a) concrete gravity dams
b) concrete arch dams
c) concrete buttress dams
iv) Stone masonry
a) stone-masonry gravity dams
b) stone-masonry arch dams
v) Timber dams
vi) Steel coffer dams
SELECTION OF TYPE OF DAM
There are many factor involved in the selection but at the beginning, physical
factors play important role. Topographically, a narrow stream section with high
rocky walls suggest a suitable site for a concrete dam. Where the wall are strong
enough to resist arch thrust, a concrete arch dam is adaptable. Low-rolling plains
suggest an earthfill or rockfill dam.
When the geologic characteristics of foundation are comprised of solid rock, any
type of dam can be constructed, although concrete gravity or arch dams are
favourable. Gravel foundation are suitable for earthfill, rockfill, and low concrete
gravity dams. Silt and fine sand foundations are used to support earthfill and low
concrete gravity dams but not suitable for rockfill dams.
Availability of certain materials close to the sitewill effect considerable reduction
in cost if the type of dam selected utilizes these materials in sufficient quantity.
Size, type and natural restrictions in location of a spillway influence the choice of
dam. A large spillway requirement indicates the adoption of a concrete gravity
dam. A small spillway requirement favours the selection of rockfill dam. When
the excavated material from a site channel spillway can be used in a dam
embankment, an earthfill dam is advantageous.
Apart from the above factors, others, such as the cost of diverting the stream,
availability of labour, and traffic requirements on top of the dam will favour one
type or the other.
4

Chapter 2: EARTHFILL DAMS
Earthfill dams
Earthfill dams are the most common type of dam built to any height. They are
designed as a non-overflow section with separate spillway. The reason for such
wide spread use of earthfill dams are:
• The foundation requirements are not as rigorous as other dams
• Local available soil is the main construction material
• High skill not required
• No special plants are required. Most earth-moving machines can be used
•
Fig. 1. General view of an Earthfill
dam
Figure 2: General view of an Earthfill
dam
Classification of earthfill dams
Earthfill dams are classified by many factors.
1. Based on the method of construction
• Rolled fill earth dams
• Hydraulic fill dam
2. Based on mechanical characteristics of earth materials making the section of
the Dam
• Homogeneous earth dams
• Non-Homogeneous (zoned) earth dams
i) non-homogeneous with inclined impervious zone (ekran) of
artificial material
ii) with impervious zone of soil with low permeability
iii) with central core soil material of low permeability
iv) with a central thin diaphragm of impervious material
Rolled Fill Earth dams
5

In this type of dams, successive layers of moistened or damp soils are laid one
over the other. Each layer not exceeding 20 cm in thickness is properly
consolidated at optimum moisture content, only then is the next layer laid.
Hydraulic Fill dams
In this type of dams, the construction, excavation, transportation of the earth is
done by hydraulic methods. Outer edges of the embankments are kept slightly
higher than the middle portion of each layer. During construction, a mixture of
excavated materials in slurry condition is pumped and discharged at the edges.
This slurry of excavated materials and water consists of coarse and fine
materials. When it is discharged near the outer edges, the coarser materials
settle first at the edges, while the finer materials move to the middle and settle
there. Fine particles are deposited in the central portion to form a water tight
central core. In this method, compaction is not required.
Homogeneous Earthen Dam
These dams are constructed with uniform and homogeneous materials. It is
suitable for low height dams (up to 10m). These dams are usually constructed
with soil and grit mixed in proper ratios. The seepage action of such dams are
not favourable, therefore, for safety in case of rapid drawdown, the upstream
slope is kept relatively flat (3:1)
Homogeneous section is modified by constructing rock toe at the downstream
lower end and providing horizontal filter drain.
6

Zoned Earth Dams
These are dams with the central portions called core or hearting made from
materials which are relatively impervious. The thickness of the core wall is made
sufficiently thick to prevent leakage of water through the body of the dam.
Dam with a Diaphragm
This type of dam is constructed with pervious materials, with a thin impervious
diaphragm in the central part to prevent seepage of water. The thin impervious
diaphragm may be made of impervious clayey soil, cement concrete or masonry
or any impervious material The diaphragm can be constructed in the central
portion or on the upstream face of the dam. The main difference in zoned and
diaphragm type of dams depend on the thickness of the impervious core or
diaphragm. The thickness of the diaphragm is not more than 10 m.
7

The criteria for the design of earth dams are:
1. Sufficient spillway capacity and freeboard are provided so that there is no
danger of overtopping of the dam
2. Seepage flow through the embankment is controlled so that the amount lost
does not interfere with the objective of the dam and there is no erosion or
sloughing of soil. In this respect, seepage line should remain well within the
downstream face of the dam and the portion of the dam on downstream side of
the impervious core should be well drained.
3. Uplift pressure due to the seepage underneath is not enough to cause piping
4. The slopes of the embankment are stable under all conditions of reservoir
operation, including rapid drawdown and during steady seepage under full
reservoir.
5. The stresses imposed by the embankment upon the foundation are less than
the strength of material in the foundation with a suitable factor of safety.
6. The upstream face is properly protected ((stone pitching, riprap, revetment)
against erosion caused by wave action, and the downstream face is protected
(counter-booms, turfs) against the action of rain
Embankment Materials
a. Earth-fill materials.
(1) While most soils can be used for earth-fill construction as long as they are
insoluble and substantially inorganic, typical rock flours and clays with liquid
limits above 80 should generally be avoided. The term “soil” as used herein
includes such materials as soft sandstone or other rocks that break down into soil
during handling and compaction.
(2) If a fine-grained soil can be brought readily within the range of water contents
suitable for compaction and for operation of construction equipment, it can be
used for embankment construction. Some slow-drying impervious soils may be
unusable as embankment fill because of excessive moisture, and the reduction
of moisture content would be impracticable in some climatic areas because of
anticipated rainfall during construction. In other cases, soils may require
additional water to approach optimum water content for compaction. Even
ponding or sprinkling in borrow areas may be necessary. The use of fine-grained
soils having high water contents may cause high porewater pressures to develop
in the embankment under its own weight. Moisture penetration into dry hard
borrow material can be aided by ripping or plowing prior to sprinkling or ponding
operations.
(3) As it is generally difficult to reduce substantially the water content of
impervious soils, borrow areas containing impervious soils more than about 2 to
5 percent wet of optimum water content (depending upon their plasticity
8

characteristics) may be difficult to use in an embankment. However, this depends
upon local climatic conditions and the size and layout of the work, and must be
assessed for each project on an individual basis. The cost of using drier material
requiring a longer haul should be compared with the cost of using wetter
materials and flatter slopes. Other factors being equal, and if a choice is possible,
soils having a wide range of grain sizes (well-graded) are preferable to soils
having relatively uniform particle sizes, since the former usually are stronger, less
susceptible to piping, erosion, and liquefaction, and less compressible. Cobbles
and boulders in soils may add to the cost of construction since stone with
maximum dimensions greater than the thickness of the compacted layer must be
removed to permit proper compaction. Embankment soils that undergo
considerable shrinkage upon drying should be protected by adequate
thicknesses of non-shrinking fine-grained soils to reduce evaporation. Clay soils
should not be used as backfill in contact with concrete or masonry structures,
except in the impervious zone of an embankment.
(4) Most earth materials suitable for the impervious zone of an earth dam are
also suitable for the impervious zone of a rock-fill dam. When water loss must be
kept to a minimum (i.e., when the reservoir is used for long-term storage), and
fine-grained material is in short supply, resulting in a thin zone, the material used
in the core should have a low permeability. Where seepage loss is less
important, as in some flood control dams not used for storage, less impervious
material may be used in the impervious zone.
Some causes of failure of earthfill dams
Like most of engineering structures, earth dams may fail due to faulty design,
improper construction and poor maintenance practices, etc
The various causes of failure may be classified as:
a) Hydraulic failure
b) Seepage failure
c) Structural failure
a) Hydraulic failure:
Hydraulic accounts for over 40% of earth dam failure and may be due to one or
more of the following:
9

.
i) By overtopping: When free board of dam or capacity of spillway is insufficient,
the flood water will pass over the dam and wash it downstream.
ii). Erosion of downstream toe: The toe of the dam at the downstream side may
be eroded due to i) heavy cross-current from spillway buckets, or ii) tail water.
When the toe of downstream is eroded, it will lead to failure of dam. This can be
prevented by providing a downstream slope pitching or a riprap up to a height
above the tail water depth Also, the side wall of the spillway should have
sufficient height and length to prevent possibility of cross flow towards the earth
embankment.
iii) Erosion of upstream surface: During winds, the waves developed near the
top water surface may cut into the soil of upstream dam face which may cause
slip of the upstream surface leading to failure. For preventing against such
failure, the upstream face should be protected with stone pitching or riprap
iv). Erosion of downstream face by gully formation: During heavy rains, the
flowing rain water over the downstream face can erode the surface, creating
gullies, which could lead to failure. To prevent such failures, the dam surface
should be properly maintained; all cuts filled on time and surface well grassed.
Berms could be provided at suitable heights and surface well drained.
F.W.L
Body of dam after
failure
Fig. Dam failure by overtopping
10

Figure 3. Beginning of downstream failure
b). Seepage failure:
Seepage always occurs in the dams. If the magnitude is within design limits, it
may not harm the stability of the dam. However, if seepage is concentrated or
uncontrolled beyond limits, it will lead to failure of the dam. Following are some of
the various types of seepage failure.
i) Piping through dam body. When seepage starts through poor soils in the body
of the dam, small channels are formed which transport material downstream. As
more materials are transported downstream, the channels glow bigger and
bigger which could lead to wash out of dam
.
Failure of dam due to piping through dam
body
H.F.L
.
Channel formation
11

ii) Piping through foundation: When highly permeable cavities or fissures or
strata of gravel or coarse sand are present in the dam foundation, it may lead to
heavy seepage. The concentrated seepage at high rate will erode soil which will
cause increase flow of water and soil. As a result, the dam will settle or sink
leading to failure.
iii) Sloughing of downstream side of dam:
The process of failure due to sloughing starts when the downstream toe of the
dam becomes saturated and starts getting eroded, causing small slump or slide
of the dam. The small slide leaves a relative steep face, which also becomes
saturated due to seepage and also slumps again and forms more unstable
surface. The process of saturation and slumping continues, leading to failure of
dam.
c) Structural Failure:
About 25% of failure is attributed to structural failure, which is mainly due to
shear failure causing slide along the slopes. The failure may be due to:
i) Slide in embankment: When the slopes of the embankments are too steep, the
embankment may slide causing failure. This might happen when there is a
sudden drawdown, which is critical for the upstream side because of the
development of extremely high pore pressures, which decreases the shearing
strength of the soil. The downstream side can also slide especially when dam is
full. Upstream embankment failure is not as serious as downstream failure.
ii) Foundation slide: When the foundation of an earthfill dam is composed of fine
silt, clay, or similar soft soil, the whole dam may slide due to water thrust. If
seams of fissured rocks, such as soft clay, or shale exist below the foundation,
the side thrust of the water pressure may shear the whole dam and cause its
piping
Failure due to piping
12

failure. In such failure the top of the dam gets cracked and subsides, the lower
slopes moves outward and forms large mud waves near the dam heel.
iii) Faulty construction and poor maintenance: When during construction, the
compaction of the embankment is not properly done, it may lead to failure.
iv) Earthquake may cause the following types of failure to earthfill dams;
1 cracks may develop in the core wall, causing leakages and piping failure.
2. slow waves may set up due to shaking of reservoir bottom, and dam may
fail due to overtopping
3. settlement of dam which may reduce freeboard causing failure by
overtopping
4. slidding of natural hills causing damage to dam and its appurtenant
structures
5. Fault movement in the dam site reducing reservoir capacity and causing
overtopping.
6. Shear slide of dam
7. The sand below foundation may liquefy
8. Failure of slope pitching.
Some elements of earthfill dams
1.Crest of dam:
The crest width of dams should be sufficient to keep the seepage line within the
dam, when the reservoir is full. The crest width of the dam if road is not
envisaged should not be less than 3m for low and medium head dams and 6m
for high head dams. If road is envisaged, then the width of the dam is determined
according to the class of road and determined by the road code.
Top width could be determined by the following recommended formulae:
a) For very low dams top width is given by
B = H/5 + 3
b) For dams lower than 30m
B = 0.55(H)1/2
+ H/5
For dams higher than 30m,
13

B = 1.65(H + 1.5)1/3
B = 1.67(H)1/2
Balustrades are provided at the end of the roads to prevent car falling off the
slopes.
2. Side slopes of dam: side slope of dams must satisfy the static stability.
However, since the stability computations can be done only after defining the
profile of the dam and determining the seepage line, it becomes necessary to
give an initial side slopes. Initial slope could be taken from the tables below.
(Taken from “hydraulic structures” –N. P. Rosanova)
Slope Material
of dam
Side slopes depending on height of dam
Less than 5m From 5m -
10m
From 10-15m
Upstream Clayey
Sandy
2
2.5 – 2
2.5
3 – 2.5
3
3
Downstream
With filter
Clayey
Sandy
1.5
2
1.75
2
1.75
2
Downstream
without filter
Clayey
Sandy
1.75
2
2
2.25
2.25
2.25
In low head dams, usually one and constant side slope is used, however in
medium and high head dams, different side slopes are usually adopted to reduce
the volume of the dam.
Side Slope According to the recommendations of Terzaghi
No. Type of material Upstream
slope
(H :V)
Downstream
slope (H:V)
1 Well graded homogeneous soil 2.5:1 2:1
2 Homogeneous coarse silt 3:1 2.5:1
3 Homogeneous silt clay:
i) for dam height less than 15m
ii) for dam height more than
15m
2.5:1
3:1
2:1
2.5:1
4 Sand or sand and gravel with
clay core
3:1 2.5:1
5 Sand or sand and gravel with
reinforced concrete core wall
2.5:1 2:1
3. Berms: Berms are constructed at both the upstream and downstream side of
the dam for the purposes of observing the conditions of protections at the
slopes and their repairs and also for increasing the width of dam at the base
14

with aim of increasing seepage length. It is also done when constructions
coffer dams are made part of the body of the dam. At the downstream side,
berms are done at an interval of 10 – 15m high. Width of berm is taken
between 1 – 2 m.
4. Free Board
Normal freeboard is the vertical distance between the normal pool level and the
crest of the dam. Minimum freeboard is the vertical distance between the high
flood level and the crest of dam.
The minimum height of freeboard is taken as 1.5 hw where hw is given by:
hw = 0.032 (V.F)1/2
+ 0.763 – 0.271(F)1/4
for F, 32 km -----------(X)
and hw = 0.032 (V.F)1/2
for F . 32km -------------------------------------(Y)
where hw = wave height ( height of water from top to trough of waves in meters)
V = velocity of wind in km/hr
F = fetch or straight length of water expanse in km.
Free board values as recommended by U.S.B.R are given in table below.
Free Board by USBR
Spillway Type Dam Height in m Minimum freeboard
over M.W.L
Free spillway Any height 2 m to 3 m
Controlled
spillway
Up to 60 m 2.5 m above top of
gate s
Controlled
spillway
More than 60 m 3.0 m above top of
gates
5. Slope Protection (Revetment)
15

Upstream side protection: For protecting the upstream slope from deterioration
and damage from wave action, the slope is covered with different protective
material.
Rock riprap, either dry dumped stone boulders or hand-packed stone boulders
can be made. Stone pitching provided at slope of 1.5 : 1 to 2 : 1 for ordinary soil
material of dam and 3 : 1 for poor soil material. The revetment stones are fixed
at the toe of the dam to prevent slipping of the embankment. The thickness of the
Figure 3. Upstream protection with hand-
layed rip-rap
Figure 4. Upstream protection
stone pitching is usually more than 60 cm. In most cases the stone pitching is
placed over gravel then sand cushion. Big sizes stones with their broader face
downwards are packed with each other by means of hammer.
Concrete, reinforced concrete slabs, steel plates, bituminous material pavement,
brick tile pavement could also be used. However, extensive survey done by US
Corps of Engineers in the 1940s o over 100 dams show that dry dumped riprap
stone pitching has the best efficiency in terms of failure rate.
16

Downstream protection
One of the cheapest, simple and effective methods of protecting the downstream
from rainfall and wind action is by planting green grass (turfs) on the slope.
Counter-boom could also be done.
17

Figure 4. Downstream protection
6. Drainage: Drainage in earth dams is meant for lowering the seepage curve;
prevent seepage water from flowing onto the downstream slope, and
conveying seepage water through the body of the dam to the downstream
part of the dam. By its function, dam drainage must have two parts; an intake
structure (drainage trench) that allows seepage water from the body and
foundation of the dam, while at the same time preventing deformation due to
seepage and conveyance structure that transports the seepage water from
the dam. However, in most drainage, it is difficult to see clearly these two
parts.
18

Hydrotechnical construction practice has work out many drainage systems
depending on the type of dam, materials of the foundation and body of dam.
Among some of the most commonly used drainage systems are:
i) Drainage prism: with many positive sides (advantages) but requires the use of
large quantity of stones
ii) A type of drainage prism in which the filter material of the drainage system is
laid to extend to a certain height on the downstream side. Such drainage
system is used when there could be rise of the tail water above the crest of
the prism.
iii) Flat horizontal drainage: It requires much smaller quantity of stones and
simplifies construction. It has the advantage of draining both the foundation
and body of dam and it is used mainly when the foundation is made up of
saturated material
iv) Combination of horizontal drain with the prism.
v) Horizontal piped drainage: consists of a pipe (tube), laid parallel to base of
the slope of dam.
vi) Horizontal stone drainage: a type of horizontal piped drain in which instead of
the pipe, a stone prism is used.
19

Seepage through dam
Filtration (seepage) computation through dams are carried out with the aim of:
• determining the position of the seepage depression curve
• finding the value of the gradient and velocity of filtration
• determine the filtration flow (discharge)
The seepage pattern through a dam is shown above. The pattern is the same
irrespective of the material (sand, clay, loam) of the dam, though the rate of
seepage will depend on soil type. The emergence of seepage lines on the
downstream slope tends to make the downstream slope unstable. Either the
downstream slope has to be made very flat or the seepage must be diverted
away from the downstream slope. The second alternative is favoured because it
is economical.
The diagram below shows the seepage lines for a homogeneous earthfill dam
with a horizontal gravel filter on the base of the dam at the downstream side. The
flow lines
20

enter the blanket vertically. Cassagrande (1937) has shown that the phreatic
line, which is the topmost seepage line, quite closely approximates a parabola.
The parabola
Seepage through a homogeneous dam with horizontal drainage blanket
(filter)
Insert drawing
21

intersects the water surface at A such that AB = 0.3 CB. Near the upstream
face, the phreatic line diverges from the parabola and join B perpendicularly.
The focus of the parabola is at F and the directrix passes through D. Every point
on the parabola with origin at the focus F can be given by
0
2
0
2
2y
y
y
x
−
= ----------------1
At the point A, x = d and y = H. Substituting in equation 1 yields
d
d
H
y −
+
= 2
2
0
------------(2)
y0 , which is also the distance FD, is determined from eq. (2)
How to get the point D graphically. With A as the centre and AF as radius, draw
an arc to cut the water surface at J. From J drop a perpendicular to the base at
D. D is the directrix. Locate the mid-point of FD at E which is the vertex of the
parabola.
The phreatic line can be drawn by equation (1).
For an approximate estimate of seepage, the Darcy’s law at point G , per unit
length of dam
0
y
dx
dy
K
KiA
q 





=
= -------------- (3)
From eq. (1) 2
0
0
0
2 y
xy
y
dx
dy
+
= --------------------(4)
Since at G, x= 0, and y = y0 then
dy/dx = 1 ------------------------------(5)
Hence q = K (1) (y0) ------------------------------(6)
The flow through the section at G is the same as at any other section.
Consequently, the total seepage through the length L of the dam is given by:
Q = K(y0)(L) ------------------------------(7)
Worked example
A homogeneous earthfill dam has a top width of 30ft and a height of 100ft with a
freeboard of 10ft. The side slopes are 1V: 2.5H. It has a horizontal drainage
blanket at the base that extends from the downstream toe to a distance of 100ft.
The embankment has a permeability of 1.5 x 10-5
ft/sec. Determine the seepage
through the dam.
Solution
H = 90 ft ; CB = 90 (2.5) = 225ft ; AB = 0.3(225) = 67.5ft and CA = 157.5 ft
d = base length – blanket length – CA
22

= 530 -100 – 157.5 = 272.5 ft.
From eq. (2) d
d
H
y −
+
= 2
2
0
= [(90)2
+ (272.5)2
]1/2
-272.5 = 14.5 ft
From eq. 6 q = K (1) (y0) = (1.5 x 10-5
)(14.5) (1) = 2.18 x 10-4
cfs or 18.8ft3
/day
per foot of dam.
Seepage line of a homogeneous dam without filter
For a homogeneous dam without filter, the focus point F of the parabola is at the
downstream toe of the dam. The base parabola cuts the downstream slope of the
dam and extends beyond the dam.
The seepage line emerges out at point C meeting the downstream face
tangentially. The portion CF of the dam is known as the discharge face and
always remains saturated.
The portion DC (Δa) and CF (a) are inter-related as follows:
( ) 






 −
∆
+
=
∆ 0
0
400
180 α
a
a
a
The table below gives the values of Δa/(a + Δa) for various values of α (angles
which the discharge face makes with the horizontal) as per Cassagrande
α in
degree
Δa/(a + Δa)
30 0.36
60 0.32
90 0.26
120 0.18
135 0.14
150 0.10
180 0.0
Seepage rate calculation from flow net
Flow net is a network form by streamlines (flow lines and equi potential (head
lines) which are mutually perpendicular. In the network, the portion between any
two successive flow lines is called flow channel and the portion between any two
successive equi potential lines called flow field.
Then from Darcy’s law of flow through porous medium, considering unit
thickness,
( )
1
.
. b
l
h
k
q
∆
=
∆ --------------------a)
23

If Δh = drop of head through a field,
h = total head causing flow
Δq = discharge passing through the field
If the total number of potential drops in the flow net = Nd
Then Δh = h/Nd ----------------------b)
Putting b) into a), we obtain 





=
∆
l
b
N
h
k
q
d
. ------------c)
Or Total flow 





=






=
∆
∑
=
l
b
N
N
kh
N
l
b
N
h
k
q
q
d
t
t
d
.
.
. ----------- d)
Where Nt = total number of flow channels in the net.
When the flow field is square, then b = l and the discharge becomes
d
t
N
N
kh
q = ---------------------f)
Seepage under the Dam
If the foundation material is alluvial sand or gravel, seepage could occur
underneath the dam. If the upward seepage pressure of water near the toe is
greater than the effective weight of the soil, the surface of the soil will rise at a
point of least resistance, and water and soil will start flowing away from dam.
This phenomena is known as piping and can result in the sliding of the toe or the
settling of the whole dam.
l
b
equipotential
lines
Flow lines
24

The submerged unit weight soil is given by
w
s
sub
e
G
γ
γ
+
−
=
1
1
----------(Y)
where Gs = specific gravity of soil; e = void ratio.
For a seepage line at a gradient, I, the upward seepage force per unit volume is
iγw . When the two forces are in balance,
e
i
G
i s
+
−
=
1
-----------------(Z)
Eq. (Z) is known as the critical gradient and equal to unity. A gradient of slightly
higher than unit value will cause piping or sand particles to be in an unstable
condition known as quicksand. The actual gradient at the downstream end of the
dam is evaluated from the flow net by dividing the head difference between the
lst two potential lines by the distance between these potential lines. This should
be less than unity.
An empirical approach, the creep ratio, L/H is computed; here L is the length
along the surface of contact between the soil and the base of the structure. This
ratio is kept at 4 for gravels and 18 for sand and silt.
Measures against seepage. Seepage Control.
The following are some common measures for controlling seepage through the
dam and embankments.
1. Prevention of Seepage through Foundation
i) By providing drainage trenches
ii). By providing downstream seepage berms
iii). By providing impervious blanket layer on upstream slope
iv). By providing impervious cutoff.
2. Prevention of Seepage through Embankment (Dam)
i). By providing horizontal drainage filter
25

ii). By providing toe filter
iii). By providing filter downstream of toe
iv). By providing downstream coarse section
v). By providing chimney drains extending upwards into the embankment
Design of Filter
The design of filters should be done in such a way that all the seepage water
through the dam is effectively drained off. The filter consists of several layers.
The first layer of the filter which comes in contact with the seeping water consists
of fine sand material. Subsequent layers of filter are made of sand of increased
fineness. The last layer of the filter is made of gravels. The soil of the earth dam
and the foundation material surrounding the filter are known as the base material.
The filters of filter drains are known as reverse or inverted filter.
The U.S. Bureau of reclamation, Washington 1960 recommended the following
for materials to be used for filters;
i) Filter material should be fine and poorly graded so that the voids in
the filter are small and thus prevent base material from entering the
filter
ii) The filter material should be coarse and pervious in relation to the
base material. This aspect facilitates rapid removal of seeping
water without building up any seepage forces within the filter.
iii) The filter material should be coarser than the perforations of
openings in the drain pipes, so that filter material is not lost in the
drains. The perforations (openings in the pipes drains should be
adequate to admit all seeping water safely.
iv) The thickness of filter material should be sufficient to provide a
good distribution of all particle sizes, also throughout the filter. The
thickness should be adequate to provide safety against piping.
Terzaghi has recommended the following two requirements which should be
fulfilled by the filter:
a). The D15 size of the filter material must not be more than 4 to 5 times the D85 of
the base material. This prevents the foundation material from carrying through
the pores of the filter material.
Chimney drain
26

b). The D15 size of the filter material should be at least 4 to 5 times the D16 of the
base material. This keeps seepage forces within the filter to permissible levels.
The above criteria can be expressed as follows:
D15 of filter < 4 to 5 < D15 of filter
D85 of base material D15 of base material.
The above criteria has been modified by USBR in “ Design of small Dams ” as:
a)
a) D15 of filter material = 5 to 40
D15 of base material
b). D15 of filter material = 5 or less
D85 of base material
c). D85 of filter materials . = 2 or more
Max opening of perforations of pipes
d). The grain size curve of the filter material should be about parallel to the curve
of the base material.
6. Stability of Earthfill dams
Stability computation for earthfil dams are done to check on the following:
2. stability of downstream slope during steady seepage
3. stability of upstream slope during drawdown
4. stability of downstream and upstream slopes during and
immediately after construction
5. Stability of foundation against shear.
Earthfill dams usually fail due to the sliding of large soil mass along a curved
surface.
The most common method used for examining the stability of earthfill dam
embankment slopes is the Swedish slip circle or (the slice) method.
Swedish slip Circle Method (Slide Method)
In this method, the failure of the embankment surface is assumed to be
cylindrical. The factor of safety against sliding, which is the ratio of average
shearing stress as determined by the Coulomb equation s = (c + σ tan φ) to the
average shearing stress determined by statics on a potential sliding surface.
For testing the stability of a slope, the centre of the possible arc is assumed. It is
necessary first to locate the centre of the Critical Circle to locate the Centre of
Critical Slope.
Fellenious has given the following method to locate the locus on which the
probable centre line may lie.
27

For a homogeneous soil, the centre of critical slip centre lies on the line PQ. The
coordinates of point Q is H downwards from toe and 4.5 H horizontal away from
toe as shown above. The location of point P is done with the help of directional
angles i, α and β given in the table below.
Slope angle
(i)o
Slope Directional angles
α β
11.3 5 : 1 25 35
18.4 3 : 1 25 35
26.6 2 : 1 25 35
33.8 1.5 : 1 26 35
45 1 : 1 28 37
60 0.58 : 1 29 40
First extend the surface of the fluid WL. Draw angle i from the toe of dam to cut
the surface of fluid at S. From S draw the angle β and from the toe draw angle α
to intercept angle β at the point P. Now by joining QP, we obtain the line on which
the centre of critical slip centre lies.
Now on line QP we assume point O and an arc AB is drawn with radius OA. The
earth mass above the arc is divided into a number of vertical segments known as
slices. The number of slices may be from 6 to 12. Neglecting the forces between
slices, each slice is assumed to act independently as a column of soil of unit
thickness and width. The weight of each slice is assumed to be acting at its
centre. The weight is resolved into two components N and T (ie. normal
P
S
R
Q
4.5 H
T
H
α
i
β
WL
28

component passing through the centre of the arc O, and will not cause any
driving moment on the slice. But the tangential component T causes a driving
moment of T x R). Near the toe of the dam, some of the slices may cause
resisting moment and in such cases, T is negative.
Let thew slices be numbered 1, 2, …6 and their weights be numbered W1, W2…
W6
Now N = W cos α and T = W sin α
From Coulomb’s equation, the resisting force
FR = c x ΔL + N tan φ ……….1)
where c = unit cohesion ; ΔL = curved length of slice ;
φ = angle of internal friction of soil.
The driving moment
MD = R x ∑T …………………..2)
The resisting moment
MR = R[c∑ΔL + tan φ x ∑N
∑ΔL = sum of curved length of all components = 2πθ/360 - length of arc AB
The factor of safety against sliding
[ ]
[ ] )
3
.......(
..........
tan
.
tan
.
∑
∑ ∑
∑
∑ ∑
+
∆
=
+
∆
=
=
T
N
L
c
S
F
T
R
N
L
c
R
M
M
S
F
D
R
φ
φ
The values of ∑N and ∑T are generally calculated in a tabular form.
T
W
N
R
O
O
R
A
B
α
29

Now if w1, w2, …….wn are the weight of slices, then
∑N = cos α (∑w) and ∑T = sin α∑(w)
No. of slice Weight of
slice
N = w x cos α T = w x sin α c x ΔL
1 w1 N1 T1 c x ΔL1
2 w2 N2 T2 c x ΔL2
3 w3 N3 T3 c x ΔL3
4 w4 N4 T4 c x ΔL4
.. .. .. .. ..
.. .. .. .. ..
n wn Nn Tn c x ΔLn
∑N ∑T c x ∑ ΔL =
c x 2πθ/360
The factor of safety is computed for the point O with equation (3). Different
points on the line QP are chosen and the procedure gone through to deyermine
the factor of safety. Various factor of safety so obtained are plotted as ordinates
on the corresponding centres and a smooth curve of F.S drawn. The centre
corresponding to the lowest factor of safety is the required centre of critical slip
circle.
Downstream Slope Stability during Steady Seepage
The critical condition for downstream slope occurs when the reservoir is full and
seepage is at its maximum rate. The seeping water below the phreatic surface
exerts a pore pressure on the soil mass. Therefore, the shearing strength of the
slices of the critical arc within the range of pore-pressure is reduced.
The net shear strength of the affected slice is
c x ΔL = (N –U) tan φ ………………(5)
where U = the pore pressure
The factor of safety (F.S) for the entire slip circle is
F.S. = c∑ΔL + tan φ (∑N - ∑U) = c∑ΔL + tan φ ∑N’ …….(6)
∑ T ∑ T
where N’ = normal components, calculated on the buoyant unit weight
γ’=(ρs – ρw)g of the dam.
Values of ∑T is calculated on the basis of its saturated weight.
Slope Stability during Sudden Drawdown
If dam is suddenly emptied, it may cause critical conditions for the stability of the
upstream slope. In this condition, the soil pores remain filled with water causing
30

the water level to remain the same as before the sudden drawdown. The water
weight within the soil tends to slide the upstream slope along a circular arc. The
hydrostatic force acting along the upstream slope when reservoir is full is also
removed with the sudden drawdown. Therefore, the tangential component of the
saturated weight is the main force causing disturbance to the upstream slope.
The shear resistance is considerably reduced due to development of pore
pressure on the likely slip surface.
The factor of safety can be calculated by the formula (6) above.
Stability of Upstream and Downstream Slopes during and immediately after
Construction
During construction of the dam and embankment with relatively impervious soil,
8excess pore pressure is developed in the air and water entrapped in the pore
space. This is due to the fact that the soil mass undergoes a change in volume
during compaction during and after construction. With time, this initial excess
pore pressure gets gradually dissipated.
Stability of Foundation against Shear.
At dam site, the silt and clay or fine, loose cohesionless material that form the
foundation may have good imperviousness, but are weak in shear and always
require checking.
For increasing the shear area and keeping the shear stress within permissible
limits, the embankment slopes are flattened or berms are added on both sides of
the dam.
The method for determining the factor of safety, which is approximate is based
on the assumption that earthfill materials have an equivalent liquid weight, which
would produce the same shear stress as the material will develop itself.
The horizontal shear on left is given by:
Toe
Foundation material
B
0.4B
h
h1
h2
L
J
Firm stratum
31

weight
unit
liquid
equivalent
where
h
h
P
o
m
o
m
.
.
.
.....
..........
..........
2
'
45
tan
.
)
7
.........(
..........
2
'
45
tan
.
2
2
2
2
2
2
1












−






−
−
=
φ
γ
φ
γ
φ’
= equivalent angle of friction
( )
1
1
.
tan
.
.
'
tan
h
c
h
m
m
γ
φ
γ
φ
+
= …………………(8)
where φ = angle of repose of foundation material
c = unit cohesion of foundation material
γm = mean unit weight of the dam and foundation weighted in proportion
to the
depth of each.
( )
1
2
2
1 .
h
h
h
h f
d
m
γ
γ
γ
+
−
=
where γd = density of the dam; γf = density of foundation material
The average unit shear = Sav = P/b
Hence the maximum unit shear = Smax = 1.4 x Sav
The maximum unit shear occurs at 0.4 B from point J.
The unit shear strength below toe K
S1 = c + γf . h2 tan φ
The unit shear strength at point J
S2 = c + γf . h1 tan φ
Therefore the average shear strength
S = (S1 + S2)/2
The factor of safety against shear
F.S. = S/Sav
The factor of safetyso obtained must be more than 1.5 for stability of foundation
against shear.
The F.S. at maximum shear, point L, let
S = c + γav . h . tan φ = (γd . h + γf . h2)/(h + h2)
And factor of safety (F.S.) = S/Smax
For stability the value of F.S. so obtained must be more than 1.0.
Chapter 3
GRAVITY DAMS
32

Basically, gravity dams are solid concrete structures that maintain their stability
against design loads from the geometric shape and the weight and strength of
the structure. Generally, they are constructed on a straight axis, but may be
slightly curved or angled to accommodate the specific site conditions. Gravity
dams typically consist of a non-overflow section(s) and an overflow section or
spillway. They are constructed with masonry or concrete but of late conventional
concrete or roller-compacted concrete are popular.
The two general concrete construction methods for concrete gravity dams are
conventional placed mass concrete and roller-compacted concrete (RCC).
a. Conventional concrete dams.
(1) Conventionally placed mass concrete dams are characterized by construction
using materials and techniques employed in the proportioning, mixing, placing,
curing, and temperature control of mass concrete. Construction incorporates
methods that have been developed and perfected over many years of designing
and building mass concrete dams. The cement hydration process of conventional
concrete limits the size and rate of concrete placement and necessitates building
in monoliths to meet crack control requirements. Generally using large-size
coarse aggregates, mix proportions are selected to produce a low-slump
concrete that gives economy, maintains good workability during placement,
develops minimum temperature rise during hydration, and produces important
properties such as strength, impermeability, and durability. Dam construction with
conventional concrete readily facilitates installation of conduits, penstocks,
galleries, etc., within the structure.
(2) Construction procedures include batching and mixing, and transportation,
placement, vibration, cooling, curing, and preparation of horizontal construction
joints between lifts. The large volume of concrete in a gravity dam normally
justifies an onsite batch plant, and requires an aggregate source of adequate
quality and quantity, located at or within an economical distance of the project.
Transportation from the batch plant to the dam is generally performed in buckets
ranging in size from 4 to 12 cubic yards carried by truck, rail, cranes, cableways,
or a combination of these methods. The maximum bucket size is usually
restricted by the capability of effectively spreading and vibrating the concrete pile
after it is dumped from the bucket. The concrete is placed in lifts of 5- to 10-foot
depths. Each lift consists of successive layers not exceeding 18 to 20 inches.
Vibration is generally performed by large one-man, air-driven, spud-type
vibrators. Methods of cleaning horizontal construction joints to remove the weak
laitance film on the surface during curing include green cutting, wet sand-
blasting, and high-pressure air-water jet.
3) The heat generated as cement hydrates requires careful temperature control
during placement of mass concrete and for several days after placement.
Uncontrolled heat generation could result in excessive tensile stresses due to
extreme gradients within the mass concrete or due to temperature reductions as
33

the concrete approaches its annual temperature cycle. Control measures involve
precooling and postcooling techniques to limit the peak temperatures and control
the temperature drop. Reduction in the cement content and cement replacement
with pozzolans have reduced the temperature-rise potential. Crack control is
achieved by constructing the conventional concrete gravity dam in a series of
individually stable monoliths separated by transverse contraction joints. Usually,
monoliths are approximately 50 feet wide.
b. Roller-compacted concrete (RCC) gravity dams.
The design of RCC gravity dams is similar to conventional concrete structures.
The differences lie in the construction methods, concrete mix design, and details
of the appurtenant structures. Construction of an RCC dam is a relatively new
and economical concept. Economic advantages are achieved with rapid
placement using construction techniques that are similar to those employed for
embankment dams. RCC is a relatively dry, lean, zero slump concrete material
containing coarse and fine aggregate that is consolidated by external vibration
using vibratory rollers, dozer, and other heavy equipment. In the hardened
condition, RCC has similar properties to conventional concrete. For effective
consolidation, RCC must be dry enough to support the weight of the construction
equipment, but have a consistency wet enough to permit adequate distribution of
the past binder throughout the mass during the mixing and vibration process and,
thus, achieve the necessary compaction of the RCC and prevention of
undesirable segregation and voids.
Site Selection
a. General. During the feasibility studies, the preliminary site selection will be
dependent on the project purposes. Purposes applicable to dam construction
include navigation, flood damage reduction, hydroelectric power generation, fish
and wildlife enhancement, water quality, water supply, and recreation. The
feasibility study will establish the most suitable and economical location and type
of structure.
b. Selection factors.
(1) A concrete dam requires a sound bedrock foundation. It is important that the
bedrock have adequate shear strength and bearing capacity to meet the
necessary stability requirements. The foundation permeability and the extent
and cost of foundation grouting, drainage, or other seepage and uplift control
measures should be investigated.
(2) The topography is an important factor in the selection and location of a
concrete dam and its appurtenant structures. Construction at a site with a narrow
canyon profile on sound bedrock close to the surface is preferable, as this
34

location would minimize the concrete material requirements and the associated
costs.
(3) The criteria set forth for the spillway, powerhouse, and the other project
appurtenances will play an important role in site selection. The relationship and
adaptability of these features to the project alignment will need evaluation along
with associated costs.
(4) Additional factors of lesser importance that need to be included for
consideration are the relocation of existing facilities and utilities that lie within the
reservoir and in the path of the dam. Included in these are railroads, powerlines,
highways, towns, etc. Extensive and costly relocations should be avoided.
(5) The method or scheme of diverting flows around or through the damsite
during construction is an important consideration to the economy of the dam. A
concrete gravity dam offers major advantages and potential cost savings by
providing the option of diversion through alternate construction blocks, and
lowers risk and delay if overtopping should occur.
Forces on Gravity Dams
3-3. Loads
a. General. In the design of concrete gravity dams, it is essential to determine the
loads required in the stability and stress analysis. The following forces may affect
the design:
(1) Dead load. (2) Water Pressure (Headwater and tailwater pressures).
(3) Uplift. (4) Temperature. (5) Earth and silt pressures. (6) Ice pressure.
(7) Earthquake forces. (8) Wind pressure. (9) Subatmospheric pressure.
(10) Wave pressure. (11) Reaction of foundation.
b. Dead load.
Dead load comprises the major resisting force.
The dead loads considered should include the weight of concrete, superimposed
backfill, and appurtenances such as gates and bridges.
In the computation of the dead load, relatively small voids such as galleries are
normally not deducted except in low dams, where such voids could create an
appreciable effect upon the stability of the structure. The cross section of the dam
is divided into several triangles and rectangles. Weight of each triangle and
rectangle and their points of application at respective centre of gravity are
computed. The resultant of all these downward forces is thus found by talking
moments of the component forces which constitute the total weight of the dam
acting at its centre of gravity. Unit weight of concrete and masonry is taken as
2400kg/m3
and 2300kg/m3
.
c. Water Pressure; (Headwater and tailwater).
35

Water pressure is the major external force acting on the dam. As the water is
stored in the reservoir, and stands against the body of the dam, it exerts
horizontal pressure on the dam.
Fig. G.D. 1 Headwater pressure with vertical upstream face
Fig. G.D. 2 Water pressure for slanted upstream face and water at tailrace
When the upstream face of the dam is vertical, the water pressure P = ½ γh2
and
acting at h/3 from base.
When the upstream face is slanted, the water pressure is resolved in two
components:
Horizontal oressure P = ½ γh2
and acting at h/3 from base.
Vertical pressure Ww = weight of water on slanted side and acting at centre of
gravity of volume of water.
Uplift Pressure.
It is the second major external force acting upwards on the dam.
h
P=1/2γh
h/3
Max. flow level
Dam
P2
Tail water
h
P1
Ww
Ww
Head water
36

Uplift pressure resulting from headwater and tailwater exists through cross
sections within the dam, at the interface between the dam and the foundation,
and within the foundation below the base. This pressure is present within the
cracks, pores, joints, and seams in the concrete and foundation material. Uplift
pressure is an active force that must be included in the stability and stress
analysis to ensure structural adequacy. These pressures vary with time and are
related to boundary conditions and the permeability of the material.
(1) Along the base.
(a) General. The uplift pressure will be considered as acting over 100 percent of
the base. A hydraulic gradient between the upper and lower pool is developed
between the heel and toe of the dam. The pressure distribution along the base
and in the foundation is dependent on the effectiveness of drains and grout
curtain, where applicable, and geologic features such as rock permeability,
seams, jointing, and faulting. The uplift pressure at any point under the structure
will be tailwater pressure plus the pressure measured as an ordinate from
tailwater to the hydraulic gradient between upper and lower pool.
Fig. G.D.3 Uplift distribution without foundation drainage
(b) Without drains. Where there have not been any provisions provided for uplift
reduction, the hydraulic gradient will be assumed to vary, as a straight line, from
headwater at the heel to zero or tailwater at the toe. Determination of uplift, at
any point on or below the foundation, is demonstrated in Figure G.D.3 above.
(c) With drains. Uplift pressures at the base or below the foundation can be
reduced by installing foundation drains. The effectiveness of the drainage system
will depend on depth, size, and spacing of the drains; the character of the
37

foundation; and the facility with which the drains can be maintained. This
effectiveness will be assumed to vary from 25 to 50 percent, and the design
memoranda should contain supporting data for the assumption used.
Along the base, the uplift pressure will vary linearly from the undrained pressure
head at the heel, to the reduced pressure head at the line of drains, to the
undrained pressure head at the toe, as shown in Figure G.D.4
Figure G.D. 4 Uplift distribution with drainage gallery
Where the line of drains intersects the foundation within a distance of 5 percent
of the reservoir depth from the upstream face, the uplift may be assumed to vary
as a single straight line, which would be the case if the drains were exactly at the
heel. This condition is illustrated in Figure G.D.5
If the drainage gallery is above tailwater elevation, the pressure of the line of
drains should be determined as though the tailwater level is equal to the gallery
elevation.
38

Figure G.D.5. Uplift distribution with foundation drains near upstream face
(d) Grout curtain. For drainage to be controlled economically, retarding of flow to
the drains from the upstream head is mandatory. This may be accomplished by a
zone of grouting (curtain) or by the natural imperviousness of the foundation. A
grouted zone (curtain) should be used wherever the foundation is amenable to
grouting. Grout holes shall be oriented to intercept the maximum number of rock
fractures to maximize its effectiveness. Under average conditions, the depth of
the grout zone should be two-thirds to three-fourths of the headwater-tailwater
differential and should be supplemented by foundation drain holes with a depth of
at least two-thirds that of the grout zone (curtain).
Figure G.D.6. Uplift distribution cracked base with drainage, zero
compression zone not extending beyond drains (3-4)
Where the foundation is sufficiently impervious to retard the flow and where
grouting would be impractical, an artificial cutoff is usually unnecessary. Drains,
however, should be provided to relieve the uplift pressures that would build up
over a period of time in a relatively impervious medium. In a relatively impervious
foundation, drain spacing will be closer than in a relatively permeable foundation.
(e) Zero compression zones. Uplift on any portion of any foundation plane not in
compression shall be 100 percent of the hydrostatic head of the adjacent face,
except where tension is the result of instantaneous loading resulting from
earthquake forces. When the zero compression zone does not extend beyond
the location of the drains, the uplift will be as shown in Figure G.D.6. For the
condition where the zero compression zone extends beyond the drains, drain
effectiveness shall not be considered. This uplift condition is shown in Figure
G.D.7
39

Figure G.D.7. Uplift distribution cracked base with drainage, zero
compression zone extending beyond drains(3-5)
.
When an existing dam is being investigated, the design office should submit a
request to CECW-ED for a deviation if expensive remedial measures are
required to satisfy this loading assumption.
(2) Within dam.
(a) Conventional concrete. Uplift within the body of a conventional concrete-
gravity dam shall be assumed to vary linearly from 50 percent of maximum
headwater at the upstream face to 50 percent of tailwater, or zero, as the case
may be, at the downstream face. This simplification is based on the relative
impermeability of intact concrete which precludes the buildup of internal pore
pressures. Cracking at the upstream face of an existing dam or weak horizontal
construction joints in the body of the dam may affect this assumption.
(b) RCC concrete. The determination of the percent uplift will depend on the mix
permeability, lift joint treatment, the placements, techniques specified for
minimizing segregation within the mixture, compaction methods, and the
treatment for watertightness at the upstream and downstream faces. A porous
upstream face and lift joints in conjunction with an impermeable downstream face
may result in a pressure gradient through a cross section of the dam
considerably greater than that outlined above for conventional concrete.
Wave Pressure
40

The portions of the dam is subjected to the impact of waves, which is produced in
water surface during wind
Figure G.D. 8
The wave pressure is tdetermined by the following formula developed by D. A.
Molitor
Wave height, hw = 0.032(V.F.)1/2
+0.763 – 0.271(F)1/4
for F<32 km -------- (A)
and hw = 0.032(V.F)1/2
for F>32 km ---------(B)
where hw = height of wave in metres from top of crest to bottom of trough.
F = fetch or straight length of water expanse in km
V = velocity of wind in km.hr
The maximum pressure intensity due to wave pressure is given by
Pw = 2.4 x γ x hw and act at hw/2m above still water.
Total force due to wave pressure
Pw = ½(2.4 w. hw) x 5/3 . hw = 2 . w . h2
= 2 . 1000 .hw
2
kg/m
This force act at 3/8hw above the still water level of the reservoir.
Wind Load
Wind load in stability analysis is usually ignored
Earth and Silt
Earth pressures against the dam may occur where backfill is deposited in the
foundation excavation and where embankment fills abut and wrap around
concrete monoliths. The fill material may or may not be submerged. Silt
pressures are considered in the design if suspended sediment measurements
indicate that such pressures are expected. Whether the lateral earth pressures
will be in an active or an at-rest state is determined by the resulting structure
lateral deformation.
2.4γh
5/3hw
hw
Water surface
Pw
41

Earthquake Forces
(1) General.
(a) The earthquake loadings used in the design of concrete gravity dams are
based on design earthquakes and site-specific motions determined from
seismological evaluation. As a minimum, a seismological evaluation should be
performed on all projects located in seismic zones.
(b) The seismic coefficient method of analysis should be used in determining the
resultant location and sliding stability of dams. In strong seismicity areas, a
dynamic seismic analysis is required for the internal stress analysis.
(c) Earthquake loadings should be checked for horizontal earthquake
acceleration and, if included in the stress analysis, vertical acceleration. While an
earthquake acceleration might take place in any direction, the analysis should be
performed for the most unfavorable direction.
(2) Seismic coefficient. The seismic coefficient method of analysis is commonly
known as the pseudostatic analysis. Earthquake loading is treated as an inertial
force applied statically to the structure. The loadings are of two types: inertia
force due to the horizontal acceleration of the dam and hydrodynamic forces
resulting from the reaction of the reservoir water against the dam (see Figure
G.D.9). The magnitude of the inertia forces is computed by the principle of mass
times the earthquake acceleration. Inertia forces are assumed to act through the
center of gravity of the section or element. The seismic coefficient is a ratio of the
earthquake acceleration to gravity; it is a dimensionless unit, and in no case can
it be related directly to acceleration from a strong motion instrument. The
coefficients used are considered to be the same for the foundation and are
uniform for the total height of the dam.
42

Figure G.D. 9. Seismically loaded gravity dam, nonoverflow monolith
(a) Inertia of concrete for horizontal earthquake acceleration. The force required
to accelerate the concrete mass of the dam is determined from the equation:
Pe = Max =( W/g)αg = W α
Where Pex = horizontal earthquake force
M = mass of dam
ax = horizontal earthquake acceleration = g
W = weight of dam
g = acceleration of gravity
a = seismic coefficient
(b) Inertia of reservoir for horizontal earthquake acceleration. The inertia of the
reservoir water induces an increased or decreased pressure on the dam
concurrently with concrete inertia forces. FigureG.D.9 shows the pressures and
forces due to earthquake by the seismic coefficient method. This force may be
computed by means of the Westergaard formula using the parabolic
approximation:
Pew = 2/3 Ce (α) y (hy)1/2
where
Pew = additional total water load down to depth y (kips)
Ce = factor depending principally on depth of water and the earthquake vibration
period, t , in seconds e
h = total height of reservoir (feet)
43

Westergaard's approximate equation for Ce, which is sufficiently accurate for all
usual conditions, in pound-second feet units is:
2
000
,
1
72
.
0
1
51








−
=
e
t
h
Ce
where t is the period of vibration.
COMBINATION OF FORCES FOR DESIGN
The design of a gravity dam is performed through an interative process involving
a preliminary layout of the structure followed by a stability and stress analysis. If
the structure fails to meet criteria then the layout is modified and reanalyzed. This
process is repeated until an acceptable cross section is attained.
Analysis of the stability and calculation of the stresses are generally conducted
at the dam base and at selected planes within the structure. If weak seams or
planes exist in the foundation, they should also be analyzed.
Basic Loading Conditions
Dams are designed for the most adverse combination of load conditions as have
reasonable probability of simultaneous occurrence. The following basic loading
conditions are generally used in concrete gravity dam designs (see Figure G.D
10).
44

(1) Load Condition No. 1 - unusual loading condition - construction.
(a) Dam structure completed. (b) No headwater or tailwater.
(2) Load Condition No. 2 - usual loading condition - normal operating.
(a) Pool elevation at top of closed spillway gates where spillway is gated, and at
spillway crest where spillway is ungated. (b) Minimum tailwater. (c) Uplift. (d) Ice
and silt pressure, if applicable.
(3) Load Condition No. 3 - unusual loading condition - flood discharge.
(a) Pool at standard project flood (SPF). (b) Gates at appropriate flood-control
openings and tailwater at flood elevation. (c) Tailwater pressure. (d) Uplift. (e)
Silt, if applicable. (f) No ice pressure.
(4) Load Condition No. 4 - extreme loading condition - construction with operating
basis earthquake (OBE).
(a) Operating basis earthquake (OBE). (b) Horizontal earthquake acceleration in
upstream direction.
(c) No water in reservoir. (d) No headwater or tailwater.
(5) Load Condition No. 5 - unusual loading condition - normal operating with
operating basis earthquake.
(a) Operating basis earthquake (OBE). (b) Horizontal earthquake acceleration in
downstream direction.
(c) Usual pool elevation. (d) Minimum tailwater. (e) Uplift at pre-earthquake level.
(f) Silt pressure, if applicable. (g) No ice pressure.
(6) Load Condition No. 6 - extreme loading condition - normal operating with
maximum credible earthquake.
(a) Maximum credible earthquake (MCE). (b) Horizontal earthquake acceleration
in downstream direction. (c) Usual pool elevation. (d) Minimum tailwater. (e) Uplift
at pre-earthquake level. (f) Silt pressure, if applicable. (g) No ice pressure.
(7) Load Condition No. 7 - extreme loading condition - probable maximum flood.
(a) Pool at probable maximum flood (PMF). (b) All gates open and tailwater at
flood elevation. (c) Uplift. (d) Tailwater pressure. (e) Silt, if applicable. (f) No ice
pressure.
b. In Load Condition Nos. 5 and 6, the selected pool elevation should be the one
judged likely to exist coincident with the selected design earthquake event. This
means that the pool level occurs, on the average, relatively frequently during the
course of the year.
45

Stability Considerations
a. General requirements. The basic stability requirements for a gravity dam for all
conditions of loading are:
(1) That it be safe against overturning at any horizontal plane within the structure,
at the base, or at a plane below the base.
(2) That it be safe against sliding on any horizontal or near-horizontal plane
within the structure at the base or on any rock seam in the foundation.
(3) That the allowable unit stresses in the concrete or in the foundation material
shall not be exceeded.
Characteristic locations within the dam in which a stability criteria check should
be considered include planes where there are dam section changes and high
concentrated loads. Large galleries and openings within the structure and
upstream and downstream slope transitions are specific areas for consideration.
b. Stability criteria. The stability criteria for concrete gravity dams for each load
condition are listed in Table G.D -1.
Overturning Stability
a. Resultant location.
The overturning stability is calculated by applying all the vertical forces (SV) and
lateral forces for each loading condition to the dam and, then, summing moments
(SM) caused by the consequent forces about the downstream toe. The resultant
location along the base is:
Resultant location =
∑
∑
V
M
b. Criteria. When the resultant of all forces acting above any horizontal plane
through a dam intersects that plane outside the middle third, a noncompression
zone will result.
For usual loading conditions, it is generally required that the resultant along the
plane of study remain within the middle third to maintain compressive stresses in
the concrete. For unusual loading conditions, the resultant must remain within the
middle half of the base. For the extreme load conditions, the resultant must
remain sufficiently within the base to assure that base pressures are within
prescribed limits.
46

Sliding Stability
a. General. The sliding stability is based on a factor of safety (FS) as a measure
of determining the resistance of the structure against sliding.
The horizontal components of the loads acting on a dam are resisted by frictional
or shearing forces along horizontal or nearly horizontal planes in the body of the
dam, on the foundation or on horizontal or nearly horizontal seams in the
foundation. It follows that the total magnitude of the forces tending to induce
sliding shall be less than the minimum total available resistance along the critical
path of sliding. The sliding resistance is a function of the cohesion inherent in the
materials and at their contact and the angle of internal friction of the material at
the surface of sliding.
Definition of sliding factor of safety.
(1) The sliding FS is conceptually related to failure, the ratio of the shear strength
(tF), and the applied shear stress (t) along the failure planes of a test specimen
according to Equation 4-2:
( ) ( )








+
−
=
+
=
=
c
F
F
CA
F
u
w
P
c
S
F
φ
φ
τ
φ
σ
τ
τ tan
1
tan
.
.
where tF = s tan f + c, according to the Mohr-Coulomb Failure Criterion; w = total
weight of dam; u = total upthrust force; tan φ = coefficient of internal friction of
material; c = cohesion of the material at the plane considered; A = area under
consideration for cohesion; Fφ =partial factor of safety in respect of friction; Fc =
partial factor of safety in respect of cohesion and P = total horizontal force.
Table G.D.2: Partial factors of safety against sliding (taken from Textbook of
Water and Power Engineering, R.K. Sharma & T.K. Sharma)
Sl. No. Loading
conditions
Fφ Fc
For dams and the
contact plane with
foundation
For foundation
Thoroughly Others
investigated
(i) 1, 2, 3 1.5 3.6 4.0 4.5
(ii) 4, 5 1.2 2.4 2.7 3.0
(iii) 6, 7 1.0 1.2 1.35 1.5
Safety against Crushing
Safety against crushing is ensured if the compressive stresses produced are
within the allowable stresses.
47

Maximum compressive stress = Direct stress + Bending moment
Bending moment MB






+
=
+
=
B
e
A
W
B
A
e
W
A
W .
6
1
.
.
.
6
where W = weight of dam, A = area of dam section, e = eccentricity
ELEMENTARY PROFILE OF A GRAVITY DAM
The elementary profile of the gravity dam to bear only the water pressure, will be
triangular in section as shown below. The width of the profile will be zero at the
water surface, where the pressure is zero and maximum at the base, where the
pressure is maximum. Thus the shape of the elementary profile is the same as
that of the hydrostatic pressure distribution. When the reservoir is empty, the
only force acting is self weight (W) of the dam acting at a distance B/3 from the
heel. It is the maximum possible inner-most position of the resultant so that no
tension develops and provides the maximum possible stabilizing force against
overturning without causing tension at toe under empty dam condition. If any
triangular profile other than the right-angled is provided, its weight will act closer
to the upstream face to provide a higher stabilizing force but will cause tension to
develop at the toe.
Vertical stresses developed when dam is empty will be:
heel
at
B
e
A
W
P
toe
at
B
e
A
W
P
..
....
.
6
1
..
...
.
6
1
min
max






−
=






+
=
Now when reservoir is full and downstream empty, forces acting on elementary
profile will be:
P
R
W
e
h
h/3
P
B
2
B
2
x
48

a) Dam weight W = ½ B. h.sg. γw
where sg = specific gravity of dam material (for concrete, = 2.4)
γw = unit weight of water (9.81 x 1000 kg/m3
)
b). Water Pressure, P = ½.γw.h2
acting at 1/3 from the base.
c). Uplift pressure u = 1/2.c. γw .B.h
where c = uplift pressure intensity coefficient.
The base width, ‘B’ of the elementary profile is determined by the following two
vriteria:
a) Stress criteria
b) Stability or sliding criteria.
a) Stress criteria. When the reservoir is empty, there is no tension in the dam,
the resultant is acting at the inner 1/3rd
point J. When the reservoir is full, for no
tension, the resultan must pass at outer 1/3rd
point K.
Now taking moment of all forces about point K,
½. . γw h2
.h/3 + ½.c. γw.B.h.B/3 - ½.B.h. sg. γw. B/3 =0
Multiplying both sides by 6/ γw.h
h2
+ c.B2
– B2
sg = 0 or
B2
(sg – c) = h2
from whence
B = h/( sg-c)1/2
------------------------------------- (B)
h
3
P
P
R
u
h
W
B
3
B
3
γh
Uplift pressure
J
K
49

By considering the force triangle, using similar triangles, we have:
(W –u )/P = (h/3)/(B/3)
or
( )
( )
c
sg
h
B
h
c
sg
B
B
h
h
h
B
c
Bhsg w
−
=
=
−
=












−
2
2
2
2
1
.
.
.
2
1
2
1
γ
γ
γ
When the uplift force is not considered, c = 0 or B = h/ (sg)1/2
Stability or Sliding Criteria
For no sliding of the dam, the horizontal forces causing sliding should be equal to
the frictional forces, i.e.
P = μ(W – u)
or 1/2γw.h2
= μ(1/2.B.h.sg. γw – 1/2. c.B. γw.h
B = h/μ(sg –c)
And neglecting uplift pressure,
B = h/μsg ----------------------------------------------- ©
The base width ‘B” of the elementary profile should be greater of the widths
obtained in equation (B) or (C)
Stresses in the elementary profile
The normal stressing the dam is given by:
)
(
6
1 D
B
e
B
u
W
p −
−
−
−
−
−
−
−
−
−






+





 −
=
when the reservoir is full, the normal stress at toe is
( ) ( )
[ ] ( ) )
(
.
.
.
.
.
2
/
1
.
.
2
/
1
2
2
1
1
E
c
sg
h
h
B
c
sg
h
B
B
p
B
u
W
B
u
W
p
w
w
w −
−
−
−
−
−
−
=
−
=
−
=
+





 −
=
γ
γ
γ
The corresponding stress at the heel is:
50

( ) )
(
0
1
1 F
B
u
W
p −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
=
−





 −
=
When the reservoir is empty, the only force acting on the elementary profile is its
weight, acting through J. In this case, the maximum compressive stress at the
heel = W/B(1+1) = 2W/B and the corresponding normal stress at toe is
W/B(1-1) =0
The Practical Dam Profile
The elementary profile of a gravity dam is a triangle with maximum water surface
at its apex. This profile is only theoretical one. For meeting the practoical
requirements certain changes have to be made namely: i) for communication,
road has to be provided and therefore a top width; ii) for wave action, free board
above the high flood level must be provided.
The addition of the above will cause the resultant force to shift towards the heel.
Earlier, when the reservoir was empty, the resultant was passing through the
inner middle third. The above changes will shift it towards the heel, crossing the
inner middle third point; this will create tension in the toe. To prevent this tension,
some concrete is added in the dam body towards upstream side.
Figure G.D. : Practical dam profile
Free board
a
h
Dam
a
16
51

Principal and Shear Stress in Dam
Check drawing from main notes
The figure above shows the dam with possible pressure distribution on it. The
maximum normal stress in the dam is the major principal stress which will be
generated on the major principal plane. When the dam is full, the vertical direct
stress is the maximum at the toe as the resultant is near the toe. The principal
stresses near the toe is shown on the second diagram with a small element ABC.
Let dr, ds and db be the lengths of AB, AC and BC respectively, and let
p = be intensity of water pressure
σ1 = principal stress on plane AB
τ = shear stress
Now considering unit length of the element ABC of the dam, the normal forces on
the planes AB, AC and BC are σ1.dr σ1.dr, pv.db and p.ds sinα respectively.
Resolving all the forces in the vertical direction,
p v.db = p.ds sinα + σ1.dr.cosα ----------------- (G)
But dr = db.cosα and ds = db.sinα
Therefore pv .db = p.ds.sin 2
α + σ1.db.cos 2
α
Or pv = p.sin 2
α + σ1.cos 2
α
Therefore σ1 = (p v –p.sin 2
α)/cos 2
α
Need to redraw
52

σ1 = p v. sec 2
α –p.tan 2
α ---------------------- (H)
Equation (H) is known as the principal stress relationship, which is applicable to
both upstream and downstream faces. For the downstream side the worst
condition will be when there is no tail water, and hence p will be zero. In this case
the major principal stress σ1 is given by
σ1 = p v. sec 2
α -----------------------------(I)
If pe’ is the intensity of hydrodynamic pressure of tail water due to an earthquake,
the principal stress at the downstream is given by:
σ1 = p v. sec 2
α –(p – pe.’) tan 2
α
In the same way,considering the hydrodynamic pressure for the upstream side,
in the horizontal direction;
Τ.db = σ1.dr.sin α + p.ds cos α
τ = σ1..sin α(dr/db) + p.(ds/db)cos α
τ = (σ1-p).sin α.cos α.
Now substituting the value of σ1 from eq. (H)
τ = (p v sec 2
α - p.sec 2
α) cos α .sin α
or τ = (p v - p) tan α
The shear stress for the upstream side has the same value but with reversed
direction
For the upstream side, τ = - (p v - p) tan α
STABILITY ANALYSIS OF GRAVITY DAMS
The stability analysis of a gravity dam section can be done by any one of the
following methods:
a) Gravity method of two-dimensional method
b) Slab analogy method
c) Trial load twist method
d) Lattice analogy method
Gravity Method or Two-dimensional Method.
Being an approximate method, it is used for the preliminary calculations.
The gravity method can be carried out by:
i) Graphical method or
53

ii) Analytical method.
We shall consider only the analytical method.
For the analytical method, the following steps are carried out:
a) Considering unit length of the dam, all vertical loads are determined the
algebraic sum of all vertical forces ∑V is calculated
b) Considering unit length of the dam, all horizontal forces are determined
and their algebraic sum ∑H calculated.
c) The sum of the righting moments (∑MR) and the sum of overturning
moments ∑MO) at the toe of the dam are calculated.
The difference between the algebraic sum of the overturning and righting
moments is determined i.e ∑M = ∑MR - ∑MO
d) The location of the resultant force FR from the toe of the dam is also
calculated by the following:
)
(J
V
M
xe −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
=
∑
∑
e) Now the eccentricity of the resultant force is determined by:
)
(
2
K
x
B
e e −
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−






=
f) The normal stress at the toe of the dam is determined by:
)
(
.
6
1 L
B
e
B
V
N T −
−
−
−
−
−
−
−
−
−
−
−
−






+
=
∑
σ
g) The normal stress at the heel is determined by
)
(
.
6
1 M
B
e
B
V
N H −
−
−
−
−
−
−
−
−
−
−
−
−
−
−






−
=
∑
σ
h) The principal and shear stresses at the toe and heel are determined from
σ1 = p v. sec 2
α -----------------------------(I)
τ = - (p v - p) tan α
i) The factor of safety against overturning is calculated by:
54

∑
∑
=
O
R
M
M
S
F .)
.
(
j) The factor of safety against sliding is calculated by:
∑
∑
∑
∑
+
=
−
=
−
H
q
B
V
S
F
factor
friction
Shear
H
V
S
F
factor
Sliding
.
)
.
(
.
.
)
.
(
.
µ
µ
LOW AND HIGH GRAVITY DAMS
Low Dam: is one of limiting height such that the resultant of all forces passes
through the middle third and the maximum compressive stress at the toe does
not exceed the permissible limit i.e
σ1 =γwH(s-c+1) = fc
or limiting height
( )
( )
1
.
1
.
+
=
−
−
−
+
−
=
g
s
f
H
or
c
g
s
f
H
w
w
γ
γ
The limiting height, for the usual stress of dam material i. e. γw = 9.81 x1000
kg/m3
, ρ = 2.4 and f = 30 kg/cm2
(or 300tonnes /m2
) gives:
( )
m
x
x
H 99
.
8
1
4
.
2
1000
81
.
9
1000
300
=
+
=
s = Specific gravity of material; c = coefficient of uplift pressure, γw = specific
weight of water; f = permissible compressive stress of material.
High Dam: A dam the height of which exceeds the limiting height of low dams is
termed as high. In a high dam, the allowable stress are often exceeded if the
resultant of all the forces were to pass through the middle third; to avoid
excessive stresses the resultant is maintained still near the centre of the base for
which purpose the downstream slope is flattened and the upstream slope is also
provided with a batter.
55

Design of Gravity Dams
Before starting, one must establish whether it is low or high dam using the above
relationship.
.
Top width is chosen according to Creager must be about 14%of the dam’s
height.
B
Limiting height
of low dam
Height of
high dam
Upstream
batter
Figure G.D. : Low dam and high dam
Reservoir level
H1
Figure G.D. 13. Economic section of Low Gravity Dam








−
=
c
S
h
B
s
1
B
c
S
a
h s −
= 2
1
h
a
J S
K
Q
O
56

Free Board: is provided on the basis of height of waves and other practical
considerations. In practice a free board of between (3 – 5)% of dam height is
used
Base width B1 I is given by
( )
c
S
h
B
or
c
S
h
B
s
s
−
=
−
−
−
−
=
µ
1
1
WORKED EXAMPLES
The diagram below shows the cross section of a masonry dam. Determine the
stability of the dam. Also determine the principal stress at the toe and heel of the
dam. Take unit weight of dam material as 2250 kg/m3
, density of water 1000
kg/m3
and the permissible shear stress of joint = 15 kg/cm2
. Assume value of
coefficient of friction μ = 0.75
SOLUTION
Stability of dam is tested assuming no free board.
A. Vertical forces
i) Self weight of dam = (12 + 2.5)/2 x 15 x 1 x 2250 = 244,687.5 kg
ii) Weight of water in column DD’A = (1 x 15)/2 x 1000 = 7500 kg
iii) Uplift force on dam =(15 x 12)/2 x 1000 = 90,000kg
D
A B
15m
C
1 2.5 8.5
D
β
α
57

iv) Therefore ∑V = 244687.5 + 7500 – 90,000 = 162.187.5 kg
v) Horizontal water pressure = (γw x h2
)/2 = (1000 x 152
)/2 = 112,500 g
Calculation of moments due to various forces about toe of dam
vi) Moment of self weight
[(1x 15)/2](1 + 2.5 + 8.5) + (2.5x 15 x 2250)[(2.5/2)+ 8.5] + [(8.5x15)/2](2/3x8.5)
= 194,062.5 + 822,656.25+ 406,406.25 = 1,423,124.9kg-m (+ve)
vii) Moment due to weight of water in DD’A
7500 x (1 +2.5 + 8.5) = 78,750kg-m (+ve)
viii) Moment due to uplift force
82,500 x 2/3 x 12 = 660,000 kg-m (-ve)
ix) Moment due to horizontal water pressure
112,500 x 1/3 x 15 = 562,500 kg-m (-ve)
Therefore ∑M =1,423,124.9 + 78750-660000-562500 = + 279,374.9 kg-m
Factor of safety calculation
x) Factor of safety against overturning = Resisting Moments =
Overturning Moments
(1423124.9 + 78750) = 1.228 < 2 unsafe.
(660,000 + 562,500)
xi) Factor of safety against sliding = μ∑V = 0.75 x 162,187.5 = 1.08 >1.0 safe
∑H 112,500
xii) Shear friction factor =μ∑V + b.q = 0.75x 162,187.5 + 12 x 15 x 104
= 17.08
∑H 112,500
Stress calculation
Let the resultant be acting at xav from the toe
Xav = ∑M = 279,374.9 = 1.72m
∑H 162,187.5
Distance of resultant force from centre of dam, (eccentricity, e)
e = B/2 – xav = 12/2 – 1.72 = 4.28
Compressive stress at toe
ft =∑V(1+6.e/B)=162,187.5(1+6x 4.28/12) = 509,268.7 > 50kg/cm2
unsafe
58

B 12
Tensile stress at heel,
fh = ∑V(1– 6.e/B)=162,187.5(1–6x4.28/12)= -15407.81 =-1.54 kg/cm2
unsafe
B 12
In masonry dams, there should not be any tensile stress, therefore the sectionis
not safe.
Calculation of principal stresses
From the diagram, tan β = 8.5/15 = 0.567
Sec α = (1 + 152
)/15 = 1.002
Sec β = (8.52
+ 152
)/15 = 1.149
tan α = 1/15 = 0.067
Principal stress at toe
σ = pn sec2
β = 509,268.7 x 1.149 = 672,336 kg/m2
Shear Stress at toe
Τ = pn tan β = 509,268.7 x 0.567 = 288,755 kg.m2
Principal stress at heel
σh = pn sec2
α – p. tan 2
α = -15407.81 x (1.002)2 -1000 x 15 x (0.067)2
= (15469.5 - 67.33) = 15402.16 kg/m2
Shear stress at heel
= - (pn-p) tan α = - [ - 15407.81 – 10,000] x 1/15 = -(25407.81)/15 = 1693.8 kg/m2
Example 2: From the data given below, design a stone masonry gravity dam of
practical profile.
Ground level, R.L = 1130.5m
R.L of HFL = 1155.5m
Wave height = 1.0m
Specific gravity of masonry = 2.5
Permissible compressive stress for stone masonry 125t/m2
Solution
Free board height = 1.5 x height of wave = 1.5 x1.0 1.5 m
Therefore required level of top of dam = 1155.5 + 1.5 = 1157m
Height of dam = 1157 – 1130.5 = 26.5m
Limiting height of dam =
( )
m
x
S
f
s
w
c
71
.
35
1
5
.
2
(
1000
1000
125
1
=
+
=
+
γ
Therefore the dam is a low gravity dam.
59

The design of the dam can be done with respect to the details.
Depth of water = 1155.5 – 1130.5 = 25m
Top width of dam = 14% height of dam = 14?100 x 26.5 = 3.71m
Assume a roadway width of 4.5 m
Therefore provide top width MN of dam = 4.5m
Base width of dam PS = B1 =
( )
m
c
Ss
h
81
.
15
0
5
.
2
25
=
−
=
−
; assume 16m
Extra width JS = MN/16 = 4.5/16 = 0.281m (assume 0.3m)
Vertical distance LK = 23
.
14
5
.
2
5
.
4
2
2 =
= x
Ss
a assume 14m
Vertical distance LR = m
x
Ss
a 22
5
.
2
5
.
4
1
.
3
1
.
3 =
=
Example 3
What should be the maximum height of elementary profile of a dam, it the safe
limit os stress on the masonry should not exceed 350 tonnes/m2
. Assume weight
of masonry 2.4 tonnes/m3
. Determine the base width also. Determine H and B if
uplift intensity factor is 0.67 and factor of safety is 2
Solution: The limiting height of elementary profile of a masonry dam
B1
= 16m
0.3
22m
14m
25m
1130.5
4.5
1.5
1155.5
1157.0
J S
P
K
R
Q
O
N
M
L
DAM
60

H =
( )
m
xg
xg
x
sg
f
w
c
103
)
1
4
.
2
(
1000
1000
350
1
=
+
=
+
γ
Base width B = m
c
Ss
H
66
4
.
2
103
=
=
−
ii) F.S =2, c = 0.67: H =
( ) ( )
m
xg
x
xg
x
c
Ss
Sx
F
f
w
c
64
1
67
.
0
4
.
2
1000
0
.
2
1000
350
1
.
=
+
−
=
+
−
γ
m
c
Ss
H
B 49
67
.
0
4
.
2
64
=
−
=
−
=
Example 4.
A concrete gravity dam has maximum water level 305.0m, bed level 225.0m, top
required level of dam 309.0m, downstream face slope starts at required level
300.0m, downstream slope 2:3, tail water is nil, upstream face of dam is vertical,
centre line of drainage gallery is 8m downstream of upstream face, uplift
pressure is 100% at heel, 50% at line of gallery and zero at toe, specific gravity
of concrete is 2.4. Considering only weight, water pressure and uplift, determine
i) Maximum vertical stresses at toe and heel of dam
ii) major principal stresses at toe of dam and
iii) Intensity of shear stress on a horizontal plane near the toe
Solution:
Height of dam H = 309.0 – 225.0 = 84m
Depth of water h = 305.0 – 225.0 = 80m
1. Top width of dam = 14% of height = 0.14 x 84 = 12m
2. Bottom width of dam = 12 + (300 – 225) x2/3 = 62m
Calculating weight and moments by considering unit length of dam
61

Designation Force Moment arm Moment about toe
Weight of dam,
w1=12 x84 x 1 x2.4 x γw
w2=50x75/2x1x2.4 γw
w = w1+w2
=2419 γw
=4500 γw
= 6919
50 + 12/2 = 56
50 x 2/3 = 33.33
13546 γw
149985 γw
∑M1=(+)285449 γw
Uplift
U1=40 x 8 x 1/2 γw
U2=40 x 8 γw
U3=54 x40/2
∑u
=160 γw
=320 γw
=1080 γw
=1560 γw
54+2/3x8=59.33
54 + 4 =58
54x2/3=36
9493
18560
38880
∑M2= (-) 66933 γw
Water pressure
P = γwh2
/2=80 x80/2 γw 3200 γw 80 x 1/3=26.67 ∑M3= (-) 85344 γw
∑V = W – U = 6919 – 1560 = 5359 γw ∑M = M1-M2-M3 = (+) 133172 γw
Position of resultant from toe xav = M/V= 133172 γw /5359 γw=24.85
Eccentricity, e = B/2 –xav 62/2 – 24.85=6.15m
Normal compressive stress at toe pn=V/B(1+6e/B)= 5359 γw /62(1+6x6.15/62)=
138 γwN/m2
Normal compressive stress at heel pn
’
= V/B(1-6e/B)= 5359 γw /62(1-6x6.15/62)=
35 γwN/m2
Uplift pressure diagram
80
300
W1
W2
50
62
8
309
305
80
225
40
8 54
U1
u2
u 1
α
62

Principal stress at toe σ = p n sec 2
α – p’ tan 2
α p’ = zero because tail wate is
zero; tan α = 2/3 sec 2
α = 1 + tan 2
α = 1 +(2/3)2
=13/9
Therefore σ = p n sec 2
α= 138 γw x13/9=199 γw N/m2
Intensity of shear stress on a horizontal plane near toe
Τ τo =( pn – p’)tan α = (138 γw – 0)x2/3 = 92 γw N/m2
Example 5. A concrete gravity dam has maximum reservoir level 150.0m, base
level of dam = 100.0m, tail water elevation 110.0m, base width of dam 40m,
location of drainage gallery 10m from upstream face which may be assumed as
vertical. Compute the hydrostatic thrust and the uplift force per metre length of
dam at its base level. Assume 50 % reduction in net seepage head at the
location of the drainage gallery.
Solution:
Free board = 5% of dam height = 0.05x50= 2.5 adopt 3m
Dam height h = 50 + 3 =53m
Top width a = 14%xh = 0.14 x 53 = 7.5m adopt 7.5m
Designation x γw Force x γw Moment arm Moment about toe
X γw
50
143
W1
W2
32
40
1
1
1
50
1
α
Uplift pressure diagram
30
1 30
U3
u2
u 1
U4
63

Weight of dam
W1=7.5x53x1x2.4
W2 = 32.5 x43/2 x1 x2.4
Weight of tail water
W3 = 8x10/2x1x1
=954
=1677
40
∑V1 =2671
32.5+7.5/2=36.25
32.5x2/3=21.67
8 x 1/3 = 2.67
34583
36341
107
∑M1 (+) 71031
Uplift pressure
U1=20 x 10/2
U2=10 x30 x1
U3=10 x30 x 1
U4=20 x 30/2 x 1
100
300
300
300
∑V2= 1000
30 + 10 x2/3=36.67
(30 +10/2) = 35
(30/2) = 15
(30 x 2/3) = 20
3667
10500
4500
6000
∑M2 =24,667
Water pressure
U/S=γwh2
/2=γwx1.10.10/2
D/S=γwh2
/2=γw.1.10.10/2
1250
50
∑H3= 1200
50/3
10/3
(-) 20,833
(+)167
∑M3 (-) 20,666
Position of resultant from toe xav = M/V=25698/1671=15.38m
Eccentricity, e =B/2-xac=40/2 – 15.38 = 4.62m
Normal compressive stress at toe, pn= V/B(1 + 6e/B)
= 1671/40(1 +6.4x62/40)=70.73 γw
Normal compressive stress at heel p’ =V/B(1 – 6e/B)
= 1671/40(1-6x4.62/40)= 12.82 γw
Maximum principal stress at toe σ = pn sec2
α – p’ tan 2
α p’ = 10 γw
tan α = 40/50, sec 2
α= 1.64
σ=70.73 x 1.64 – 10(40/50)2
=110 γw
Intensity of shear stress on a horizontal plane near toe
Ττ0 = (pn – p’)tan α
τ0 = 70.73 -10)40/50 = 49 γw
64

Chapter 4
SPILLWAYS AND GATES
Spillway is a passageway to convey past the dam flood flows that cannot be contained in
the allotted storage space or which are in excess of those turned into the diversion
systems. Spillways function infrequently, at times of flood or sustained high runoff, when
other facilities are inadequate. However its ample capacity is of prime importance for the
safety of the dam and other hydraulic structures. Hydraulic aspects of spillway design
relates to design of the three spillway components: control structure, discharge channel
and terminal structure. The control structure regulates outflows from the reservoir and
may consist of a sill, weir section, orifice, tube or pipe. Design problems here relate to
determining the shape of the section and computing discharge through the section. The
flow released from the control structure is conveyed to the streambed below the dam in a
discharge channel. This can be the downstream face of the overflow section, a tunnel
excavated through an abutment, or an open channel along the ground surface. The
channel dimensions are fixed by the hydraulics of channel flow. Terminal structures are
energy dissipating devices that are provided to return the flow into the river channel
without serious scour or erosion at the toe of the dam.
Spillways are usually located at the normal flood level and the water just overflows the
crest when it gets to that level. However, some spillways are equipped with gates, which
are temporary barriers installed over the permanent crest of the spillway for storing
additional water during period of low water seasons. All small flows exceeding the
barrier top level are allowed to pass over the barrier, but during large flood flows, the
barrier is removed and full spillway capacity is used to discharge the flood water.
Types of Spillways
Spillways may be classified, depending upon the type of structure as:
i) Side channel spillway
ii) Straight drop spillway
iii) Overflow or Ogee spillway
iv) Chute or Trough spillway
v) Shaft (Morning glory) spillway
vi) Siphon spillway
Side channel spillway
This type of spillway is most suited for earthfill and rockfill dams, in narrow canyons,
where construction of other types of spillway is not possible. Also when a long overflow
crest is required for limiting the surcharge head, this type is suited. It is also useful when
its discharge is to be connected to a narrow discharge channel or tunnel. In this type the
control weir is kept along the side and approximately parallel to the upper portion of the
spillway. The discharge after passing over the crest turns at about 90o
before flowing into
a trough to be discharged.
Straight Drop (free overfall) Spillway. In this type, water is allowed to fall freely from a
low weir and vertical fall structure. In some cases the crest of the spillway is extended in
65

the shape of overhanging lip which keeps the discharge away from the straight drop
section. Ogee (Overflow) Spillway: is a special form of a weir whose shape is made to
conform to the profile of lower nappe of a ventilated sheet of water falling from a sharp-
crested weir. The profile is so shaped that the discharging water always remains in touch
with the spillway surface.
Earthfill dam
Side channel spillway
Upstream
downstream
66

Accordingly, the profile of the ogee spillway is made to the shape of the lower nappe of a
free falling jet.
The downstream curve of the ogee has the equation:
( ) 85
.
0
85
.
1
2 h
y
x = --------------------------- (S-1)
where x and y are the co-ordinates of the crest profile measured from the apex of the
crest, and h is the design water head.
The upstream curve may be approximate to the following
Straight drop spillways
With downstream protection
spillway
Spillway
Without downstream protection
O
x
r1
Origin and apex of crest
y
b
r2
Design head h
a = 0.175h
b = 0.292h
r1
= 0.50h
r2
= 0.20h
a
Fig. S-1 Definition sketch of overflow spillway
67

( )
( ) 85
.
0
85
.
1
27
.
0
724
.
0
h
h
x
y
+
= ------------------------ (S-2)
For larger discharges, (eg flows beyond H.F.L), the nappe may leave the ogee profile and
will cause negative pressure resulting in cavitation and increase of discharge.
Discharge over an Overflow Spillway
The discharge equation of the ogee-shaped spillway is given by
Q = C.Le.(he)3/2
------------------------------(S-3)
Where, Q = discharge over the ogee spillway
Le = effective length of the crest
he = h + v2
/2g = total water head at crest including the velocity approach
C = a variable coefficient of discharge, whose value varies from 2.1 to 2.5
depending on various factors.
The effect length of the crest is given by the equation:
Le = L – 2(NKp – Ka) ---------------------------(S-4)
Where L = Total clear length of crest
N = No. of piers in the spillway
Kp = pier contraction coefficient
Ka = abutment contraction coefficient
Table S-2 Pier Contraction Coefficient Kp
Table S-3 Abutment Contraction Coefficient (Kp)
Condition of Abutment Value of Ka
1 Rounded abutment with head wall as 90o
to the direction
of flow, when 0.5h > r > 0.15h
2 Rounded abutment where r> 0.5Ho and the angle of head
wall with direction of flow is <45o
0.00
3 Square abutment with head wall at 90o
to the direction of
flow of water
0.20
Condition of Pier Value of Kp
1 Pointed nose piers 0.00
2 Round nose piers 0.01
3 Square-nose pier with corners rounded on a
radius 0.1 times pier thickness
0.02
68

Example 9.10
Design an overflow spillway section for a design discharge of 50,000cfs. The upstream
water surface is at Elv. 800 and the channel floor is 680. The spillway, having a vertical
face, is 180ft long.
Solution
1. Assume C = 3.95
2. from the discharge equation ( )
1
.
17
32
.
70
180
95
.
3
000
,
50
2
/
3
=
=
=
=
e
e
h
CL
Q
h
3. Depth of water upstream = 800 – 680 = 120 ft.
Velocity of approach vo = 50,000/120(180) = 2.31ft/sec
Velocity head = v2
/2g = (2.31)2
/(32.2) = 0.08 ft
4. Maximum water head = 17.1 – 0.08 = 17.0ft
5. Height of crest, P = 120 – 17.0 = 103ft.
6. Since he <30 ft, design head, hd = 17.1/1.42 = 12.0ft
7. P/hd = 103/12 = 8.58 > 1.33, high overflow section
8. Downstream quadrant of the crest shape
y/12 = ½ (x/12)1.85
or y = 0.06x1.85
X (select)ft Y (computed) ft
5 1.18
10 4.25
15 9.00
20 15.30
30 32.40
9. Point of tangency assume a downstream slope of 2:1
(XDT)/h = 0.485(Kα)1.
176
where XDT = horizontal distance from the apex to the downstream tangent point
α – slope of the downstream face
XDT = 0.485[2(2)]1.176
(12) = 30 ft
10. Upstream quadrant. A/h = 0.28, B/h = 0.165
A = 0.28(12) = 3.36; B = 0.165(12) = 2.00ft
(x2
)/(3.36)2
+ (2.0 – y)2
/(2.0)2
=1
X (selected) ft Y (computed) ft
1.0 0.09
2.0 0.39
3.0 1.10
3,36 2.00
69

CHUTE OR TROUGH SPILLWAY
Chute spillway is a type of spillway in which the discharge is conveyed fro a reservoir to
the downstream river level through a steep open channel placed either along the dam
abutment or through a saddle. In a chute spillway, the velocity of flow ia always greater
than the critical. The name chute applies, regardless of the control device used at the
head of the chute, which can be an overflow crest, a gated orifice or a side channel crest.
The chute spillway consists of four parts: an entrance channel, a control structure or crest,
the sloping chute and a terminal structure. The entrance structure is usually an open
channel of sub-critical flow. The critical velocity occurs when the water passes over the
control.
Flow in the chute is maintained at supercritical stage until the terminal structure DE is
reached. It is desirable that from B to C, where a heavy cut is involved, the chute is
placed on a light slope. From C to D it follows the steep slope and ends with an energy
dissipating device placed at the bottom of the valley D. The axis of the chute is kept as
straight as possible; otherwise, the floor has to be super-elevated to avoid the piling up
high-velocity flow around the curvature.
It is preferable that the width of the control section, the chute, and the stilling basin are
the same.
To prevent hydrostatic uplift under the chute, a cutoff wall is provided under the control
structure and a drainage system of filters and pipes is provided. When the stilling basin is
operating, there is substantial uplift under the lower part of the chute and upstream part of
the stilling basin floor. The floor must be made sufficiently heavy or anchored to the
foundation.
Slope of Chute Channel
Chute section
Terminal Structure
Entrance cha. Control structure
Fig. S – 3 Chute Spillway Section
C
B
A
D E
F
70

It is important that the slope of the chute in the upstream section BC chute, should be
sufficiently steep to maintain a supercritical flow to avoid the formation of a hydraulic
jump. Therefore slope should always be above the critical slope.
Fro the Manning’s formula;
2
.
1
3
/
2
1
I
AR
n
Q = -------------------------- ( S-5 )
For a rectangular channel under critical flow conditions,
h3
= q2
/g ----------------------------------(S-6 )
By putting this into Manning’s formula, the critical slope could be determined. The slope
to be given to the chute must be greater than this critical slope.
A review of existing spillways indicate that the actual slopes of the upstream section of
the chute are 1 or 2 % or more.
Chute Sidewalls
Except for the diverging or converging sections, chute channels are designed with parallel
vertical sidewalls, commonly of reinforced concrete 30 to 45cm thick; designed as
retaining wall. The height of the walls is designed to contain the depth of flow for the
spillway design flood by gradually flows hydraulics equation. Free board or allowance is
made for pier end waves, roll waves and air entrainment. In view of uncertainties
involved in the evaluation of surface roughness, pier end waves, roll waves and air
entrainment, the following empirical equation is added to the computed depth of water
surface profile
Freeboard (ft) = 2.0 + 0.002V h1/3
--------------------- (S-7)
where V - mean velocity in chute section under consideration, h – mean depth.
SIDE CHANNEL SPILLWAYS
In side-channel spillway, the overflow weir is placed along the side of the discharging
channel, so that the flow over the crest falls into a narrow channel section (trough)
opposite the weir, turns through a right-angle, and continues I the direction approximately
parallel to the weir crest
71

This type of weir is adoptable to certain special conditions, such as when a long overflow
crest is desire but the valley is narrow, or where the overflow are most economically
passed through a deep narrow channel or a tunnel. The crest is similar to an overflow or
ordinary weir section. A control section downstream from the trough is achieved by
constricting the channel or elevating the channel bottom to induce critical flow.
Downstream from the control section functions as a chute spillway. Thus the side-
channel design is concerned only with the hydraulic action in the trough upstream of the
control section, where varied flow takes place. The water surface profile in the trough is
determined from the momentum equation applied to gradual varied flow. If flow rate and
velocity in left is Q1 and V1 while those at the right is Q2 and V2, then by applying the
momentum equation, the rate of change of momentum in the reach dx is equal to the
external forces acting in the reach.
Rate of momentum at left(upstream) = ρQ1V1
Rate of momentum at downstream = ρQ2V2
Change in momentum = ρ(Q2V2 – Q1V1) ------------------------------------------(S -8)
Due to smallness of friction forces and weight component in the direction of flow, the
only force acting is the hydrostatic pressure force.
The resultant hydrostatic force Fp1 –Fp2 = γ.(A1.h1 – A2h2) = γ(Q1/V1 – Q2V2)dy ---(S-9)
Trough
Side-channel spillway: plan and section
A
Dam
Side channel crest
Control section
Side channel trough
Chute
A
72

Equating equations (S-8) and (S-9) and rearranging
( ) ( )





 −
+
−








+
=
2
1
2
1
2
1
2
2
1
1
2
Q
Q
Q
V
V
V
Q
V
Q
V
g
Q
dy ---------------------------(S-10)
Equation (S-10) is solved by a trial and error procedure .
Worked example:
Design a side-channel trough for a spillway of 100ft length for a maximum discharge of
2500 cfs. The side-channel trough has a length of 100ft and a bottom slope of 1ft in 100
ft. A control section of 10 ft width is placed downstream from the trough with the bottom
of the control at the same elevation as the bottom of the trough floor at the downstream
end.
Solution
1/. Critical depth at the control section, yc = (q1
2
/g)
2. q1 = 2500/10 = 250cfs/ft.
3. yc = (2502
/32.2)1/3
= 12.44ft
4. vc = q1/yc = 250/12.44 = 20.1ft/sec
5. Velocity head, hc = vc
2
/2g = (20.1)2
/2(32.2) = 6.27ft
6. For a side-channeltrough, assume a trapezoidal section with 2V:1H slope and a 10ft
bottom width. Also, assume that the transition loss from the end of side-channel trough to
the control section is equal to 0.2 of the difference in velocity heads between the end of
the transition.
dx
P2
P1
dy
Crest
73

7. The following energy equation may be written between the trough end and the control
section;
y100 + h100 = yc + hc + 0.2(hc – h100) or
y100 + 1.2h100 = 12.44 + 1.2(6.27) = 19.96 --------- (S-11)
where the subscript c refers to “critical” and “100” refers to the distance of the trough
from the upstream end of the spillway. Equation (S-11) is solved by trial and error.
Assume y100 = 19.2; then A = 376.3 ft2
, v = Q/A = 2 500/376.3 = 6.64 ft/sec, and h100 =
6.64)2
/2(32.2)0.68. Thus Eq. (S-11) is satisfied.
8. The known values above relate to section 1 at the downstream end of the trough.
Section 2 is taken at the upper end of a selected increment dx, 25 ft in this case. A value
of the change in water level, dy, is assumed for each reach, all terms of equation (S-10)
are evaluated, and dy is computed, as in table below. The assumed and computed values
of dy should match; otherwise, a new value is assumed for dy.
9. The process is repeated until the upstream end of the channel is reached.
1 2 3 4 5 6 7 8
dx
(selected
)
Bottom
Level
dy
(Assume)
Water
Level
y
(col 4-col 2)
A Q =
q(L-dx)
v
= Q/A
D/S end 100.0 119.2 19.20 376.3 2500 6.64
25 100.25 1.0 120.2 19.95 398.5 1875 4.71
0.62 119.82 19.57 387.2 4.84
25 100.50 0.50 120.32 19.82 394.6 1250 3.17
0.40 120.22 19.72 391.6 3.19
25 100.75 0.25 120.47 19.72 391.6 625 1.60
15 100.90 0.10 120.57 19.67 390.2 250 0.64
0.07 120.54 19.64 389.3 0.64
9 10 11 12 13 14
Q1 + Q2 Q2 – Q1 v1 + v2 v2 – v1 dy
computed
Remarks on
assumed dy
4375 -625 11.35 -1.93 0.63 High
4325 -625 11.48 -1.80 0.61 Ok
3125 -625 8.01 -1.67 0.41 High
3125 -625 8.03 -1.65 0.41 Ok
1875 -625 4.79 -1.59 0.24 Ok
875 -375 2.24 -0.96 0.07 High
875 -375 2.24 -0.96 0.07 ok
Note: 1. q = Q/L = 2500/100= 25cfs/ft ; bottom slope = 1 in 100ft given
2. Bottom level (col. 2) = slope x channel length + datum
3. Water level (col. 4) = final water level at the preceding station (section) + assumed dy.
4. A (col. 6 = area of cross-section of trough computed for depth y in col.5
5. Q (col.7) = q(L - ∑dx), L = crest length
74

6. Q1 + Q2 (col. 9) = Col. 7 + value in col. 7 at the preceding station (section)
MORNING GLORY OR SHAFT SPILLWAYS
In this type of spillway, water enters over a horizontal circular crest and then drops
through a circular shaft, and then through a horizontal tunnel or conduit. This type of
spillway is suitable under the following situations:
1. For damsites with steeply rising abutments particularly where a diversion tunnel
can be utilized as discharge carrier.
2. For sites in narrow canyons.
3. For sites where there is inadequate space for locating other type of spillway.
Advantages: The main advantage of such type of spillways are as follows:
1. The nearly maximum capacity may be attained at relatively low heads.
2. It is ideal for sites where maximum spillway overflow is to be limited.
This type of spillway consists of four parts: 1) a circular weir at the entry, 2) a flared
transition conforming to the shaft of the lower nappe of a sharp-crested weir, 3) a vertical
drop shaft, and 4) a horizontal or near-horizontal outlet conduit or tunnel. As the head
increases, the control shifts from weir crest, to drop shaft, and to outlet conduit.
Table S-3. Discharge characteristics of shaft spillway
Control Point Condition Characteristics Discharge relations
Weir crest Unsubmerged flow Weir flow Q = CLh3/2
--------(S-12)
Throat of drop
shaft
Partially
submerged
Orifice flow 95
.
0
...
2
1 =
= d
a
d C
gH
A
c
Q
----------------- (S-13)
Downstream of
outlet conduit
Submerged flow Pipe flow
K
gH
A
Q T
Σ
=
2
2 -----(S-14)
∑K = loss coefficient
through pipe
Condition 1: a free-discharging weir prevail as long as the nappe forms to converge into
the shape of a solid jet.
Condition 2: weir crest is drowned out. The US Bureau of Reclamation (1977) indicated
that this condition is approached when Hd/Rs >1, where Hd is the design head and Rs is
the radius of the crest.
Condition 3: Spillway is flooded out, showing only a slight depression and eddy at the
surface. Under condition 3, the head rises rapidly for a small increase in discharge. Thus
the design is not recommended under this condition (i.e., under the design head, the outlet
conduit should not flow more than 75% full).
The U.S. Bureau of Reclamation suggested that the following weir formula may be used
for flow through the shaft spillway entrance regardless of the submergence, by adjusting
the coefficient to reflect the flow conditions.
Q = C(2πRs)H3/2
----------------------------- (S-15)
75

Where C = discharge coefficient related to Hd/Rs and P/Rs , where Hd = design head and
P = crest height from the outlet pipe; Rs = radius to the circular crest; H = head over the
weir ( Gupta page 495)
Equation (S-15) can be used to determine the crest size (radius), Rs for a given discharge
under the maximum head.
Equation (S-13) can be used to determine the shape of the transition (drop shaft) that is
required to pass the design discharge with the maximum head over the crest.
Worked Example: A shaft spillway is to discharge 2000 cfs under a design head of 10 ft.
Determine the minimum size of the overflow crest. Also determine the shape of the
transition if the control section is 4 ft below the crest level.
Solution
1. Since the coefficient C is related to P and Rs, assume that P/Rs > 2 and determine
Rs by trial and error.
2. Try Rs = 7 ft.
Hd/Rs = 10/7 = 1.43
From the curve of C as a function of Hd/Rs, we choose C = 1.44
3. From equation (S-15)
Q = C(2πRs)H3/2
= 1.44(2π) (7) (10)3/2
= 2002cfs
This ois practically the same as the required discharged. Hence the crest radius = 7 ft.
4. Depth of the control section from the water surface;
Ha = 10 + 4 = 14 ft.
5. From equation (S-13), 95
.
0
...
2
1 =
= d
a
d C
gH
A
c
Q
R2
= 2000/{0.95π(2gHa)1/2
}
or R = 9.14/(Ha)1/4
SIPHON SPILLWAYS
A siphon spillway is a short enclosed duct whose longitudinal section is curved. When
flowing full, the highest point in the spillway lies above the liquid level in the upstream
reservoir, and the pressure at that point must therefore be sub-atmospheric. This is the
essential characteristics of a siphon. Siphon spillways can be saddle type or volute type
but the volute is not very common. The siphon has usually three parts, 1) the inlet or
mouth; 2) throat, and 3) lower limb.
The siphon spillway functions as follows:
When the water level exceeds the crest level, the water commences to spill and flows
over the downstream slope in much the same way as a simple Ogee spillway. As the
water rises further, the entrance is sealed off fro the atmosphere. Air is initially trapped
Ha (select), ft R (ft)
14 4.73
16 4.57
18 4.44
20 4.32
76

within the spillway, but the velocity of flow of water tends to entrain the air (giving rise
to aeration of the water) and draws it out through the exit. When all the air has been
expelled, the siphon is primed and is therefore acting as a simple pipe. There are thus
three possible operating conditions depending on upstream depth.
1. Gravity spillway flow
2. Aerated flow (during self priming)
3. Pipe flow (after priming).
Operational problems with siphon spillways. The aerated condition is unstable and is
maintained while the siphon begins to prime. In a simple siphon, a small change in H
produces a sharp increase or decrease in the discharge through the spillway. Depending
on the discharge entering the reservoir, the siphon could go through the following cycle:
a) if the spillway is initially operating with gravity flow, then the upstream
(reservoir) level must rise,
b) when the upstream level has risen sufficiently, the siphon primes and the
spillway discharge increases substantially,
c) the upstream level falls until the siphon de-primes and its discharge drops.
The cycle (a) to (c) is then repeated. This obviously can give rise to radical surges and
stoppages in the downstream flow.
Other potential problems encountered with siphon spillway are:
i) blockage of spillway entrance by debris ( this problem could be overcome by
submerging the inlet of the hood into the water or installing a trash-
intercepting grid in front of the intake).
ii) Substantial foundations required to resist vibrations during operation of
siphon.
iii) Waves arriving in the reservoir during storms may alternately cover and
uncover the entry, thus interrupting smooth siphon action
The discharge of a saddle siphon can be calculated from the following formula;
gH
CA
Q 2
= --------------------------- (S-16)
where A = area of cross section at crown; A = L x b
L = length of hood (going into the paper)
b = height of throat
H = operating head
H = Reservoir level – downstream tail water level if outlet is submerged
H = Reservoir level - downstream centre of outlet if the outlet is discharging
freely.
C = Coefficient of discharge. Its average value may be taken as 0.65.
77

SPILLWAY OR CREST GATES
Spillway gates are the temporary barrier installed over the permanent crest of the
spillway, for storing additional water during dry weather season. The small flows in
excess above the spillway gates is allowed to pass over the gates, but in case of large
flood, the spillway gates are opened and the full capacity is used to remove excessive
flood water.
Spillway gates can be provided on all types of spillways except siphon spillway.
Following types of spillway gates are commonly used:
a) Flash board gates
b) Stop logs or needle gates
c) Radial gates
d) Drum gates
e) vertical lift gates
H2
Crown
Throat
Crest
De-primer
Air vent
Hood
H1
Mouth
Figure S- . Siphon Spillway
78

Flash Board Gate
Flash Board
Rubber stopper
Spillway crest
Stop logs or needle gates
Piers
Groove
Stop log
79

Chapter 5
RESERVOIRS
Reservoirs are the most important man-made storage elements in water systems
because their capacity and operational schedules determine the rates and
volumes of flow in streams. They have enabled humans to make the desert
bloom and flourish and provide water supplies for large and concentrated
populations. A reservoir is usually created by the construction of a dam across a
flowing stream. Storing water is necessary because when water occurs naturally
in streams and is not stored, it is sometimes not available when needed.
Reservoirs solve this problem by capturing water when it occurs and making it
available at latter times. However, with today’s environmental awareness,
reservoirs are sometimes negative symbols of “man taking dominion over nature”
and work against sustainable development as they interfere with natural
ecosystems.
While the concept of a reservoir may bring to mind a large body of water, many
small reservoirs are also in service. These include urban water tanks, farm
ponds, regulating lakes, and small industrial or recreational facilities. These
small reservoirs can have important cumulative effects especially in rural areas
A reservoir is created with the impounding of runoff from the catchment upstream
by the construction of a dam across a river or stream. Storage is done during the
period the flow is in excess of the demand for release during the lean period.
Storage reservoirs are constructed for the purposes of 1) flood control or 2)
conservation of water. Reservoir construction may be contemplated under one of
the following conditions:
Inactive Storage
Active or Live storage
HydroPower
Buffer
Normal. Res. Level
Dead storage
Food Control
High flood level
Max. Elevation
Normal full elev.
Surcharge storage
80

i) Running water is available in sufficient quantity, but running to waste or causing
damage. In this case, an attempt is made to utilize the available water for
developmental purposes of the area.
ii) Water is needed all the time either for the generation of hydro-power to meet
the requirement of power for development or for irrigation , however the flow in
the stream is unable to meet the demand all the time.
Once it is decided to construct a reservoir, it is important to know that it is
cheaper to construct a larger reservoir and combine several functions than to
construct series of separate dams for each purpose. The following factors must
be investigated before deciding to put up a reservoir:
a) Availability of water
b) Availability of suitable site for the construction of a dam to create the
reservoir.
c) Availability of construction material in sufficient quantity, equipment,
labour, etc.
Investigations if carried out in details, will be time consuming and expensive.
Sometimes, it may happen that the investigations made may reveal that the
project is not economical or technically sound. In such a case, all the labour and
resources spent is a waste. Therefore an investigation should be planned and
executed such that the soundness of the project will be determined as early and
as cheap as possible. To achieve this aim, investigation may be divided as
follows:
1. Reconnaissance or preliminary investigations: which seeks to collect and
analyze the following information: i) A not very precise topographic map of the
site; ii) Some information of over burden; iii) Characteristics of the foundation
(few samples are needed); iv) A preliminary geological survey of site; v)
Hydrological studies; vi) Investigation of the available construction materials; vii)
Checking of high flood marks and their use to determine spillway capacity.
The purpose of this preliminary investigation is to obtain sufficient data to carry
out the office studies and estimate cost sufficiently accurate to select the most
economical and suitable site amongst the several options that may be available.
2. Final investigations: Is limited to the point that is necessary to confirm the
relative merits of few selected options and working out details. The investigations
should include: i) to select the final option; ii) to determine the nature of
foundation; iii) to determine availability and quantity of construction materials; iv)
T obtain all other necessary information useful for the design purpose.; v)
requirement for coffer dam, pumping and other provisions for dewatering the site;
vi) transport facilities and accessibility to site; vii) suitable site for construction
camp and repair of construction equipments.
81

Investigations for Reservoir Planning
1. Engineering Surveys; 2) Geological Surveys; 3) Hydrological surveys
1. Engineering surveys of the reservoir and work areas are required to
determine the capacity and spread of the reservoir at various elevation; for laying
out the lines of communications and other various works and for estimating
quantities of materials and their cost.
The contour interval is 2.5 m but at the site of dam an interval of 1.5 m is used
and a map on a scale of 1/250 to 1/500.
Area Capacity Curves.
From the contour map of the reservoir area, the water spread of the reservoir at
any elevation may be directly determined by measuring the area at that contour
with a planimeter. From these measurements, an area – elevation curve can be
drawn. The capacity of the reservoir may be determined by taking contour areas
at equal interval and summing up these areas by any of the following methods:
2. Geological Surveys is essential to determine 1) the suitability of foundation; 2)
the tightness of the reservoir basin and 3) location of quarry sites.
Hydrological investigation is necessary to 1) have accurate estimate of the run-
off pattern at the proposed dam site for the purposes of determining the height of
Volume ( Km2
–m)
Area (Km2
)
Elevation, H(m)
V = f(H)
A = f(H)
82

dam, reservoir size, capacity of power house and other installations and 2) to
determine the hydrograph of the worst flood to estimate the spillway size.
Site Selection for Reservoir
The storage capacity of site should be sufficient to the meet the requirement for
which it is designed.
i) Site where the width of river is narrow but rapidly widening upstream
ii) Site where sufficient quantity of water is available. This will depend on the
intensity of rainfall, run-off and catchment area.
iii) As much as possible, site should be such that water can flow under gravity
from reservoir to demand points
iv) Site must be close to demand points
v) The geological conditions at the site should permit minimum percolation
losses, with maximum run-off.
vi)Site must be close to useful construction materials.
vii) The topography should be favourable for site for spillway.
viii) Site should be such that the run-off water has minimum percentage of
sediment
ix) Site must be free from such minerals and salts, which may make the water
unfit for the purpose, for which the dam built.
Classification of Reservoirs
a) Storage reservoir
b) Flood Control (flood mitigation) Reservoir
c) Distribution Reservoir
d) Single Purpose reservoir,
e) Multipurpose Reservoir.
Storage Reservoir: are small storage capacity reservoir, which are constructed
to store water for meeting the demand for water supply, irrigation, power
generation etc.
Flood Control (Flood Mitigation) Reservoir: The main function of such
reservoirs is to temporary store flood waters and release it at such a safe rate
that it may not flood downstream side.
Distribution Reservoir: These are small capacity reservoirs used for water
supply of towns. These reservoirs may store raw water for treatment. As the
water is required at varying rate during the day, they permit the water treatment
plant to function at a constant rate
83

Single Purpose Reservoir: It is a reservoir to serve only one purpose which
may be: a) Municipal Water Supply, b) Power generation; c) Flood Control;
d) irrigation; e) Navigation; f) Recreation.
Multipurpose Reservoir: In this reservoir, the storage and release cater for a
combination of two or more purposes.
STORAGE ZONES AND LEVELS OF A RESERVOIR
1. Dead Storage: us about 10 – 25 % of the gross storage and it is provided to
cater for sediment deposition by the impounded sediment –laden waters. Usually
this volume is below sluiceway, and therefore below it, the reservoir is not
susceptible to release water by the built in outlet means. The dead storage is
equivalent to the volume of sediment expected to be deposited in the reservoir
during the design life of the reservoir.
2. Live (Active or Useful) Storage: It is the storage capacity above the inactive
storage, which constitutes useable portion of the total storage.It is thus the
difference of gross storage capacity and the sum of dead storage capacity and
inactive capacity Live storage has to be sufficient so that the project is successful
for i) 75% of its life period in an irrigation project; ii) 90% for hydropower and iii)
100% for water supply.
3. Flood Storage: is the storage contained between the normal reservoir level
and the full reservoir level. It is the storage space provided in a reservoir for
storing flood water temporary to moderate the releases downstream.
4. Valley Storage: is the storage in the river in floods after it gas overflowed its
banks. It is important element in the design of large size flood control reservoirs
where it could be a significant proportion in respect to the reservoir storage
volume.
5. Surcharge Storage: It is the storage between the Normal full reservoir level
and the maximum possible level in the reservoir. This storage is usually difficult to
control since it depends on maximum floods, rains and the resulting run-offs.
6. High Flood Level (Maximum Pool Level): The level to which the water will rise
during the design flood
7. Full Reservoir Level The level to which water will rise during ordinary
conditions of operation of reservoir.
Minimum Pool Level: The level to which water from the reservoir in ordinary
conditions may be drawn.
Storage Capacity and Yield
Yield: It is the volume of water that can be supplied from the reservoir I a specific
interval of time. The time interval may vary from a day to a year.
Dependable or Firm Yield: The maximum guaranteed supply of water during the
worst dry period.
84

Secondary Yield: The quantity of water available in excess of dependable yield
during the flood period.
Designed Yield: It is the quantity of water for which the project is designed after
ascertaining the availability of water. It is usually kept lower than firm yield.
Average Yield: It is the arithmetic average of safe yield and secondary yield over
a long period.
Reservoir Storage Capacity Determination
The storage capacity of a reservoir to meet the demand of continuous supply is
determined with the help of observed discharge data of a stream on which the
dam is to be constructed. The flow values for the driest years in as long a period
as is available, eg, 25 to 30 years are used.
There are two approaches for the determination of the size of reservoir storage.
The simplified method, which are commonly used in planning stage studies,
comprise mass curve analysis.
The detail method, used at the time of developing reservoir operating plans,
performs a sequential reservoir routing of the historical flows, which will not
be treated here.
A third, by Bar Graph Method. In this method, the inflows of the driest year are
plotted as ordinates against time as abscissa to get a stepped graph as bar
graph. The area under the bar graph represents total volume of inflow into
reservoir. The average demand is likewise plotted. The area between the two
plots indicates either surplus or deficit volume. The area of maximum deficit
between the demand and inflow represents the minimum storage required.
Surplus
Deficit
Surplus
Max. deficit
STORAGE
Bar Graph Method
85

Simplified Procedure for Reservoir Storage Capacity.
There are two methods of analyses: i) the sequential mass-curve method and ii)
the non-sequential mass-curve method.
Mass curves are useful in reservoir studies since they provide a ready means of
determining storage capacity necessary for a particular average drawoff.
Properties of the mass curve
1. Any point on the curve indicate the total inflow from the beginning of the
period up to the given point.
2. The slope of the tangent to the mass curve at any time gives the inflow
rate at that time
3. The slope of a line joining any two points on the mass curve gives the
average inflow rate withi the period
4. Dry spells (periods) are indicated as hollows on the curve.
A sequential mass-curve method (Rippl Method) considers the most critical
period of recorded flow, which might be a severe drought period. The cumulative
difference between the inflow and outflow to the reservoir during successive
periods are evaluated, the maximum value of which is the required storage.
S = maximum ∑(It –Ot) -------------------------R-1
Where S – required storage capacity
It – inflow during period dt, Ot – outflow (draft) during period dt
Equation (R–1) can be solved either analytically or graphically.
Analytical Method: In this method, the inflow and demand values in each month
are determined. The demand includes prior rights, if any, evaporation, seepage,
etc. The deficit and surplus of water which is the departure of inflow volume from
86

demand volume is determined. The maximum value of cumulative excess of
demand over inflow represents the minimum storage necessary to meet the
demand. The cumulative excess inflow volume starting from each demand
withdrawal from the storage is also determined which indicates the filling up of
the reservoir and the volume in excess of storage to be spilled over.
Example: In the table below, (col. 2) gices the yield of a river, (col. 3 & 4) gives
total losses including evaporation, percolation, etc., and total estimated
consumption of water. Determine the storage capacity of the proposed reservoir.
1 2 3 4 5 6 7
Month Yield of
River
Total
Losses
Total
consumptio
n
Total
(3 + 4)
Surplus
(2 – 5)
Total
surplus
June 0.0385 0.0863 0.1883 0.2746 - -
July 0.4013 0.0784 0.1851 0.2633 0.1378 -
Aug 0.4327 0.0678 0.1782 0.2460 0.1863 -
Sept. 0.4832 0.0615 0.1513 0.2128 0.2704 -
Oct 0.3384 0.0482 0.1482 0.1964 0.2902 -
Nov. 0.3156 0.0257 0.1215 0.1472 0.1684 -
Dec. 0.2983 0.0173 0.0987 0.1160 0.1823 -
Jan 0.2357 0.0195 0.1142 0.1337 0.1020 -
Feb. 0.2013 0.0202 0.1348 0.1550 0.0463 1.3837
Mar. 0.1605 0.0823 0.1562 0.2085 - -
April 0.1114 0.0845 0.1732 0.2577 - -
May 0.0713 0.1032 0.1784 0.2816 - -
June 0.0401 0.0975 0.1895 0.2870 - -
July 0.3247 0.0767 0.1883 0.2650 0.0597 -
Aug 0.3652 0.0681 0.1790 0.2471 0.1181 -
Sep 0.2830 0.0600 0.1551 0.2151 0.0679 -
Oct 0.2218 0.0515 0.1503 0.2018 0.0200 -
Nov 0.1943 0.0301 0.1284 0.1585 0.0358 -
Dec 0.1814 0.0165 0.1010 0.1175 0.0639 -
Jan 0.1703 0.0183 0.1186 0.1369 0.0334 -
Feb 0.1415 0.0197 0.1207 0.1404 0.0011 03997
87

Mar 0.1137 0.0515 0.1531 02046 - -
April 0.0945 0.0726 0.1802 0.2528 - -
May 0.0687 0.0985 0.1854 0.2839 - -
June 0.0312 0.1013 0.1903 0.2916 - -
July 0.4215 0.0695 0.1900 0.2595 0.1620 -
Aug 0.4183 0.0603 0.1805 0.2408 0.1775 -
Sept. 0.3967 0.0587 0.1570 0.2157 0.1810 -
Oct 0.3818 0.0494 0.1525 0.2019 0.1799 -
Nov 0.2323 0.0286 0.1312 0.1598 0.0725 -
Dec 0.1738 0.0108 0.1135 0.1243 0.0495 -
Jan 0.1621 0.0143 0.1218 0.1361 0.0260 0.8484
Feb 0.1358 0.0195 0.1256 0.1451 - -
Mar 0.1148 0.0613 0.1583 0.2196 - -
April 0.0887 0.0688 0.1831 0.2519 - -
May 0.0595 0976 0.1867 0.2843 - -
8 9 10 11 12 13
Deficit
(5-2)
Total deficit Cumulative
yield
Cumulative
Draft
Difference
(10-11)
Carry over
0.2361 - 0.0385 0.2746 0.2361 Emptying
- - 0.4398 0.5381 0.0983 of reserv.
- - 0.8725 0.7481 0.0884 Filling of
- = 1.3557 1.9969 0.3588 Reservoir
- - 1.6941 1.1933 0.5008 Overflow
- - 2.0097 1.3405 0.6692 Overflow
- - 2.3080 1.4565 0.8515 Overflow
- - 2.5437 1.5902 0.9535 Overflow
- - 2.7450 1.7452 0.9998 Emptying
0.048 - 2.9055 1.9537 0.9518 Emptying
0.1463 - 3.0169 2.2114 0.8055 Emptying
0.2103 - 3.0882 2.4930 0.5952 Emptying
88

0.2469 0.8876 3.1283 2.7800 0.3483 Emptying
- - 3.4530 3.0450 0.4080 Emptying
- - 3.8182 3.2921 0.5261 Emptying
- - 4.1012 3.5072 0.5940 Emptying
- - 4.3230 3.7090 0.6140 Emptying
- - 4.5173 3.8675 0.6498 Emptying
- - 4.6987 3.9850 0.7137 Emptying
- - 4.8690 4.1219 0.7471 Emptying
- - 5.0105 4.2623 0.7482 Emptying
0.0909 - 5.1242 4.4669 0.6573 Emptying
0.1583 - 5.2187 4.7197 0.4990 Emptying
0.2125 - 5.2874 5.0036 0.2838 Emptying
0.2604 0.7221 5.3185 5.2952 0.0234 Emptying
- - 6.7401 5.5547 0.1854 Filling
- - 6.1584 5.7955 0.3629 Filling
- - 6.5551 6.0112 0.5439 Filling
- - 6.9369 6.3121 0.7238 Filling
- - 7.1692 6.3729 0.7963 Overflow
- - 7.3430 6.4972 0.8458 Overflow
- - 7.5051 6.6333 0.8718 Overflow
0.0093 - 7.6409 6.7784 0.8625 Overflow
0.1048 - 7.7557 6.9980 0.7577 Emptying
0.1632 - 7.8444 7.2499 0.5945 Emptying
0.2248 0.5021 7,9039 7.5342 0.3697 Emptying
Graphical Method: there are two graphical procedures. In the first method,
(Mass Curve method), It is accumulated separately as a mass inflow curve and
Ot as a mass yield curve. For a constant draft rate, the yield curve is a straight
line having a slope equal to the draft rate. At each high point on the mass inflow
curve, a line is drawn parallel to the yield curve and extended until it meets the
inflow curve. The maximum vertical distance between the parallel yield line and
the mass inflow curve represents the required storage. Assuming the reservoir is
full at A; and going from A to F along the inflow curve. From A to B, the draft is
more than inflow resulting in a lowering of reservoir; from B to C, the inflow is
89

higher than the draft, but not enough to refill the reservoir; from C to D, the draft
is more, once again causing a drop in reservoir level; from D to E, however, the
inflow is very high, thus filling the reservoir at E.
The second graphical procedure plots the difference of successive accumulated
values of inflow and yield ∑(It - Ot) against time. The maximum vertical difference
is the storage.
A
O
B
C
D
E
F
G
Cumulative volume
Time
Storage
Inflow curve
Yield/demand
Curve
Time
∑(I –O)
Storage volume
90

RESERVOIR SEDIMENTATION
Every stream (river) carries some sediment load along with the flow. The larger
solid particles roll along the bed as bed load. Smaller particles are kept in
suspension by the upwards component of the turbulent forces and will only settle
by gravitational force when velocity reduces. Such particles are called
suspended load.
When sediment laden water reaches a reservoir, in the vicinity of a dam, the
velocity and the turbulence are greatly reduced due to enlarged area of the
channel. The larger suspended particles and most of the bed load gets deposited
in the head reaches of the reservoir. The smaller particles remain in suspension
for longer periods and are deposited further down in the reservoir. Very fine
particles may remain in suspension for still longer period and some may pass
over the dam along with the water discharged through spillway or sluiceways.
If the water stored in the reservoir is clear and the inflow is charged with
sediment or is muddy, the heavier water with sediment will flow along the channel
bottom towards the dam under the influence of gravity and the clear light water
will flow on the upper surface of the turbid water. This condition is known as
stratified flow and the under flow of sediment laden water is known as density
current. The process whereby sediments are deposited at the bed of the
reservoir due to reduced turbulence and velocity is known as reservoir
sedimentation. The deposition of sediments will reduce the water storing capacity
of the reservoir and if this process continues for a long time, a stage will be
reached when the whole reservoir will be silted and would be rendered useless.
Mechanism of Sedimentation in Reservoirs
Silt deposition in reservoirs follow a typical pattern just like the formation of deltas
at the confluence of a river at the sea or lake. There are four patterns:
1. Top set bed; 2) Fore set bed; 3) Bottom set bed; and density currents.
1. Top set bed: The velocity of a river is reduced considerably as it
approaches the reservoir even before the river enters the reservoir due to
back water effect. Due to the reduction of velocity of flow, the silt carrying
capacity of the river reduces and it deposits the heavy sediment particles
in the channel above the highest level, just at the entry into the reservoir.
This is called the top set bed.
2. Fore set bed: As the river water enters the relatively calmer waters of the
reservoir, the coarser particles and most medium size sediment particles
settle down at the toe of the top set bed. This is called the fore set bed.
3. Bottom set bed: The fine particles (silt) do not settle till they have moved
through a sufficient distance into the reservoir where they may settle and
get deposited in thin layers as bottom set bed.
4. Density current: is the turbid water which flows under the clear water of
the reservoir due to its relative dense nature
In reservoir engineering, a special storage is allocated to sediment called the
dead storage under the lowest sluice-way. The dead is can be as much as a
fourth of the total capacity of the reservoir.
91

Estimation of the Rate of Sedimentation
The amount of sediment load carried by a stream is determined by taking
samples of water carrying silt at various depths. The samples are filtered,
sediment removed and dried. The sediment load per day may be computed by:
1000
86400
CxQ
x
Wd = tones per day
where Wd = dry weight of sediment
Q = mean daily discharge in cumecs
C = dry sediment weight in grams per litre of water.
Khosla’s Studies.
In the absence of any information on sediments, one could use the studies of Dr.
A. N. Khosla. On the basis of studies on some catchment areas, Khosla
suggested the rate of silting per 100 square kilometer of catchment to be 0.036
million cubic metres.
Recent studies have shown that Khosla’s value is on the low side. The average
sediment rate of reservoirs in USA is given below:
No. Area of watershed
(Km2
)
No. of
measurement
Average annual sediment rate
(in mil m3
per 100 sq. km area).
1 Under 25 650 0.180
2 25- 250 205 0.076
3 250 – 2500 123 0.048
4 Above 2500 118 0.029
TRAP EFFICIENCY AND LIFE OF A RESERVOIR
Trap efficiency η of a reservoir is defined as the ratio of sediments retained in
reservoir to the total sediment brought in the reservoir by the stream.
92

Trap efficiency (η) = Sediment deposited in reservoir .
Total sediment brought into reservoir by river
It has been observed that most reservoirs trap about 95 – 100% sediment
brought to it. It is not possible to reduce it below 90 % irrespective of sediment
control measures taken.
Capacity-Inflow Ratio
Capacity inflow ratio is the ratio of the capacity (volume) of a reservoir to the total
inflow (volume) of water in a given time (usually one year).It is an important factor
as detailed observations of reservoir sedimentation has shown that trap
efficiency is a function of this factor.
Thus: η
trap efficiency(η) = f (Reservoir capacity)
Total inflow
The figure below shows a graph of trap efficiency and capacity inflow ratio based
on observations of existing reservoirs. From the figure, it will be noted that for a
given inflow rate, the trap efficiency decreases with age of reservoir as the
capacity of the reservoir reduces due to sediment deposition. Hence, the rate of
silting is higher in the initial stages and it decreases as silting takes place.
Complete filling of the reservoir may take a long time, but actually the useful life
of the reservoir is terminated when its capacity is reduced to 20 % of the
designed capacity or sediment is so much as to prevent the from serving
intended purpose.
For small reservoirs on large rivers having large inflow rates, the trap efficiency
trap efficiency is low, because the capacity inflow ratio is very small. The silting in
such reservoirs is low, because most of the sediments are passed to the
downstream.
93

The procedure for determination of the life of a reservoir is as follows:
i) The capacity of the reservoir is determined. Next, the capacity inflow ratio and
the trap efficiency are determined for the full capacity of the reservoir with the
help of the curve.
ii) Divide the total capacity into 10 parts. Assuming that 10% capacity has been
reduced due to deposits of sediments, determine the trap efficiency for the
reduced capacity (90% of the total capacity) and the inflow ratio.
iii) For the above 10% interval of capacity, determine the average trap efficiency
by taking into account the trap efficiency (η) obtained in the above steps.
iv) By collecting water samples and drying the sediments, determine the
sediment inflow rate. The total annual sediment collected in the reservoir per
year is multiplied with the trap efficiency determined in step (iii) above. The
sediment quantity so obtained is converted into hectare-meter units deposit per
year.
v) Now the reservoir volume interval ( i.e. 10% of the capacity is divided by the
sediment deposited per year, obtained in step (iv) above. It will give the number
of years required to fill this volume interval of 10% capacity.
vi) Now repeat the procedure (i) to (v) above for further capacity intervals i.e.
80%, 70%, 60%etc of the total reservoir capacity. The total life of the reservoir
will be the total number of years required to fill each of the volume intervals.
Examples
The following data is available from a reservoir regarding its trap efficiency and
capacity inflow ratio;
Cap
Inflow
Ratio
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Trap
efficiency
87 92 93 95.5 96 96 96.5 97 97.5 98
Find the probable life of reservoir with an initial reservoir capacity of 40 million
cubic meters and with an annual average flood inflow of 80 million cubic meters.
The annual sediment inflow is found to be 22 x 104
tonnes. The specific weight of
sediment may be taken as 1.15 tones per m3
. The useful life of reservoir may be
assumed when its capacity is reduced to 20% of its designed capacity.
Solution.
Initial reservoir capacity = 40 x 106
m3
Average annual inflow = 88 x 106
m3
Therefore capacity inflow ratio = 40/80 = 0.5
Average annual sediment inflow = 22 x 104
tonnes
Let’s assume that 20%of the initial capacity is filled in the first interval, i.e. 20/100
x 40 x 106
= 8 x 106
m3
capacity is filled at a time.
94

For 0.5 C-I-R the trap efficiency is 96%.
At the end of the first interval, the capacity of reservoir = (40 – 8) x 106
= 32 x
106
m3
At the end of the first interval the C-I-R = 32/80 = 0.4
From the curve, this correspond to a trap efficiency of 95.5%
Therefore the average trap efficiency for the entire period = (96 + 95.5)/2 =
95.75%
Weight of sediment deposited annually = 22 x104
x 0.975 = 0.2107 x 106
tonnes
Volume of sediment deposited annually = (0.2107 x 106
)/1.15 = 0.1823 x 106
m3
Therefore No. of years to fill 20% (8 x 106
m3
) capacity = (8 x106
)/0.1823 x106
=
43.9 years.
Similarly, the silting period for the other intervals can be calculated. The rest of
the calculations are done in the table below
Capacit
y
x 106
m3
Capacit
y Inflow
Ratio
(C-I-R)
%
Trap Efficiency Sediment
Trapped
At
indicate
d
volume
Average
for
incremen
t
Tonne
s
x 106
tonnes
Volum
e
x 106
m3
Incrementa
l Volume
(x 106
m3
)
Years to
fill
(Col.7)/
(Col. 6)
1 2 3 4 5 6 7 8
40 0.5 96.0
32 0.4 95.5 95.75 2.107 0.1831 8 43.70
24 0.3 95.0 95.25 2.095 0.1822 8 43.93
16 0.2 92.5 93.75 2.062 0.1793 8 44.61
8 0.1 87.5 90.0 1.98 0.1722 8 46.46
178.9
179year
s
Example 2.
Below are data regarding trap efficiency and capacity inflow ratio of a reservoir
Capacity
Inflow ratio
0.
8
0.
1
0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Trap
efficiency
85 86 92 94 95 9
5.5
96 9
6.5
97 9
7.5
9
7.7
Determine the probable life of the reservoir with an initial reservoir capacity of 25
million m3
, if the annual inflow is 50 million m3
and the average annual sediment
inflow is 300,000 tonnes. Assume the density of the sediment as 1250kg/m3
. The
95

useful life of the reservoir will terminate when 85 % of the initial capacity is filled
with sediment.
Solution:
Average annual sediment inflow = 300,000 x 1000 kg
Volume of sediment inflow = 300,000,000/1250 = 0.24 x 106
Initial reservoir capacity = 25 x 106
m3
Annual inflow = 55 x 106
m3
Therefore Initial capacity = 25 x 106
= 0.5
Capacity 50 x 106
The capacity inflow ratio goes on decreasing as the sediment takes place and
trap efficiency also decreases. The volume of interval chosen is 5 x 106
i.e. 20%
of initial capacity. To obtain the 85% filling of initial capacity the volume interval in
last two readings is taken as 5%.
Similarly, for the calculation of the years to fill the last 5% capacity is determined
by taking 1 = . 1 . = 4.87 years
St 0.2.52
Capacity
% Volume
x
106
m3
Cap.
Inflo
w
Ratio
Trap Efficiency η Sediment Trapped
@
indicate
d
Volume
Averag
e for
interval
Tonnes
x 106
=300,00
0
x col(5)
Volum
e
x106
(m3
)
=col(6)
/
1.25
Incre-
menta
l
Vol.
x 106
(m3
)
Years
to fill
col 8
/col 9
1 2 3 4 5 6 7 8 9
10
0
25 0.5 0.955
80 20 0.4 0.95 0.9525 2.8575 0.2286 5 21.7
60 15 0.3 0.94 0.945 2.835 0.2268 5 22.04
40 10 0.2 0.92 0.93 2.79 0.2232 5 22.40
20 5 0.1 0.86 0.89 2.67 0.2136 5 23.40
15 4 0.08 0.85 0.855 2.565 0.2052 121.7 4.87
Total 94.58
Example 3. A reservoir has a drainage area of 49,000 acres. The reservoir has a
capacity of 10,000 acre-ft. Streamflow runoff averages 15.5 in. per year. The
annual sediment production is 9.5 x 106
ft3
. Determine the life of the reservoir,
assuming that the life is over when 80% of the original capacity is lost
Solution
1. Annual inflow I = (49,000) (15.5/12) = 63,290 acre-ft
96

2. Annual sediment production = 9.5 x 106
ft3
or 218.5 acre-ft
3. Unfilled capacity (20%) = (20/100) (10,000) = 2,000 acre-ft
4.
Capacity
% Volum
e
Acre-ft
Cap.
Inflow
Ratio
=col2
/63,29
0
Trap Efficiency η Sediment
Trapped
@
indicate
d
Volume
Averag
e for
interval
Tonnes Volume
Acre-ft
=col(5)
x
218.5
Incre-
menta
l
Vol.
x 106
(m3
)
Years
to fill
col 8
/col 9
1 2 3 4 5 6 7 8 9
10
0
10,000 0.16 0.92
80 8,000 0.13 0.88 0.90 197 2,000 10.1
60 6,000 0.09 0.85 0.865 189 2,000 10.6
40 4,000 0.06 0.80 0.825 180 2,000 11.1
20 2,000 0.03 0.68 0.74 162 2,000 12.3
Total 44.1
Control of Silting in Reservoirs
It is utmost necessary to reduce the depositions of sediments in the reservoir bed
for increasing its life. Sediment control methods may be classified as:
Pre-construction methods
Post-construction methods.
1. Pre-construction Methods includes:
i) Site Selection: For silting control, the site of reservoir should be such that the
catchment area should have firm soil which would not be easily erodable.
ii) The ratio of reservoir capacity and drainage size: If the storage capacity of the
reservoir is much less than the annual inflow into reservoir, large quantity of
water will flow out of reservoir and with it large volume of sediment.
iii) Design of Reservoir: For the release of density currents, adequate outlets
such as sluices should be provided in the design of the dam. I this way, sufficient
quantity of sediment may be flashed fro the reservoir.
iv) Vegetation Screen: If the ground around reservoir is covered by vegetation,
grass, etc., it will prevent the erosion of soil particles as well as it will trap large
amount of sediments. Therefore, if vegetations are grown in the catchment area,
the silting will be reduced to considerable extent. The vegetations covering the
ground are known as vegetation screens and are the cheap and effective means
of silting control.
97

Post-Construction Methods:
i) Mechanical Stirring of Sediment: The sediments which are deposited in the
reservoir bed are scrapped and stirred by mechanical equipments and kept in
moving state. The sediments mixed water is pushed towards the sluices and is
flashed out.
ii) Excavation: It is a very expensive method. Iy is expensive to excavate and
also expensive to dispose of excavated material.
iii) Dredging: It is also expensive as it require the purchase and maintenance of
costly machines as dredger.
Watershed Erosion Control
a) By afforestation: In this method, vegetation is planted to provide a cover to
soil, which reduces the scoring effect of the soil in three different ways:
i) It provides a cushion to the falling rain drops and reduces their impact
and consequent damage due to erosion.
ii) The roots of the plants penetrate deep into the soil and bind soil
particles enabling it resist water and wind erosion.
iii) Plants offer resistance to flow and thereby reduces the velocity of flow.
b) Controlling Overgrazing
OPERATION OF RESERVOIRS
After a dam is built, it must be operated correctly. The key person is the operator,
who makes decisions about when to release or store water. The operator may be
a part-time worker who lives near a lake and who occasionally operates a gate,
or a highly trained engineer working at a remote location who makes system
decisions based on computer forecast. In the past, reservoir operating decisions
were made by RULES CURVES, which provided the operator with simple
guidelines about how much water to release and what lake levels to maintain. As
the science of forecasting and the use of computers has become more complex,
however, reservoir operation has become more sophisticated. It is not
uncommon to have a reservoir control center where operators use computers to
monitor weather forecasts furnished from satellite data and simulate future
demands for water to make decisions about water releases. They may also be
bound by legal requirements to release water for downstream users, including
fish and wildlife.
Basically what is required in the operations problem is an analytical tool to
illustrate the time rate of inflows, outflows, and change in storage. In other
words, it is necessary to consider the storage equation both numerically and
graphically. Recall that the storage equation relates inflow, outflow and change
in storage as a function of time, where for any time period such as an hour, day
month or year.
98

Qi – Q0 = DS
Where Qi – is the rate of inflow; Q0 – is the rate of outflow; and DS – change in
storage.
A simple way to view the operation plan is to consider a rule curve such as one
shown below, which is taken fro the corps of engineers operating plan for lake
Lanier in Georgia. A rule curve illustrates a typical operations time period,
usually a year, and shows the boundaries within which operations should occur.
Rule curve for Lake Lanier, Geopgia
8.1.2 Reservoir Operations
Reservoir operation in the CRYSTAL model is based on the Snohomish County PUD
Jackson Operations model (JKSOPR). Release policies depend on the "state" of both
Spada Reservoir and Lake Chaplain. The state of the reservoir is determined by
comparing the storage to maximum storage capacity, rule curves, and minimum storage
capacity. When water is in abundance, releases are dependent on the capacity of Jackson
Powerhouse transmission line. Otherwise, releases are functions of instream flows,
demands, and relative storage.
99

Snohomish County PUD No. 1 (SnoPUD) operates Spada Reservoir using two rule
curves based on storage (Figure 8.3). In addition,”poweroff" storage may be used during
extreme conditions (Spada reservoir has yet to reach the poweroff storage). Comparing
the reservoir storage to the Full Storage condition (FS), Upper normal rule curve, and
Lower normal rule curve determine the state of the reservoir. If reservoir storage is
between the rule curves (State 3), then a variable amount of flow (based on a ratio of
storage to lower rule curve and time of year) is conveyed through the Powerhouse
pipeline. If storage in Spada is below the Lower normal rule curve (State 4) then
minimum flow to meet instream flow requirements and M&I demand is conveyed. The
maximum flow of 1300 cfs is conveyed if the reservoir is above the Upper normal rule
curve (States 1&2). No flow is conveyed if storage is below the "Power off" level.
FLOOD ROUTING
Flood routing is the process whereby the shape of a flood hydrograph is modified
as it passes through a hydraulic system (such as a river channel, reservoir, lake,
etc). The hydrograph resulting after passing through the hydraulic system is one
with considerably reduced peak and enlarged time base. The process can be
represented as:
100

Flood routing in fact is a technique of determining the extent of change in the
inflow hydrograph as it passes through a hydraulic system and it is done for the
purposes of:
1. Studying the effect of a hydraulic system on the modification of a flood peak
2. Determining the design elevations of flood walls and levees
3. Determining the site for a spillway
4. Derivation of unit hydrograph and synthetic hydrographs
5. For flood forecasting
6. Any other flood flow-related objectives.
In flood routing, all the components of the routing process (inflow, storage and
outflow) are connected by the storage (continuity) equation in the form:
[ Inflow volume in time Δt] + [Change in volume of water stored by the hydraulic
system during time Δt] = [Outflow volume in time Δt]
t
O
S
t
I ∆
=
∆
+
∆ ------------------------- (1)
where I and O are the average inflow and outflow rates within the time Δt.
The hydraulic system can be a reservoir (lake) or a section of a river channel,
and so the routing process is, accordingly classified into two namely: reservoir
routing and channel (streamflow) routing.
In the above equation (1) I- the inflows are known quantities but the outflows, O
and the storage, S are unknown parameters. To solve the equation, either both O
and S have to be related to a common unknown parameter or S has to be
defined in terms of d to a common unknown parameter or S has to be defined in
terms of O. The former approach is applicable to reservoir routing whiles the
latter to channel routing.
Equation (1) when expressed in the differential form can be integrated, however,
the terms in the equation have a form that is not amenable to direct solution.
Therefore, the numerical solution expressed as numerical approximation can be
written as :
2
2
2
1
2
1
2
1 O
O
t
S
S
I
I +
=
∆
−
+
+
------------------ (2)
101

where subscript 1 and 2 denote beginning and end of routing period Δt
RESERVOIR FLOOD ROUTING
Reservoir routing is the process of determining the reservoir stage, storage
volume and outflow rate corresponding to a particular hydrograph of inflow. The
flood routing procedure is used in the detention and storage reservoir for studies
related to the variations in reservoir levels with inflow and outflow discharge with
time with a view to deciding the location and capacity of proposed reservoirs,
determining the spillway outlet capacities, height of dam, extent of land
submergence in the reservoir area.
In reservoir routing, the volume of storage can be expressed as a function of
water surface elevation in the reservoir, so also is the outflow, which can be
expressed as a function of elevation.
In routing through a reservoir, equation (2) is re-organised to the form:
( ) 





+
∆
=
−






+
∆
+
+ 2
2
1
1
1
2
1
2
2
2
O
t
S
O
O
t
S
I
I -------------- (3)
Since the outflow and the storage are both functions of water surface elevation,
the storage equation becomes a relation between the known inflow and the
unknown elevation, from which elevation can be computed as a function of time.
Procedure
1. What is available at the beginning.
i) An inflow hydrograph I =f(t)
ii) An elevation –capacity curve H =f(S) or elevation – area curve of the
reservoir
H =f(A)
iii) An elevation – outflow curve H = f(O)
2. From the above, the routing curve is prepared and plotted
( )
O
f
H
O
t
S
f
H
=






+
∆
=
2
3. At the initial time, t = 0 (start of the routing, just before flood arrives, we
assume a steady state condition, so that I1 = I2 = O1 and S1 corresponds to the
storage corresponding to elevation of O1.
4. Select the time interval Δt for the routing. This is usually taken as the interval
of time given for the inflow hydrograph. Otherwise, the time interval is so chosen
so that one does not miss the peak values.
5. At the beginning of the routing, I1 = I2 =O1. With O1 known, read off H from H
=f(O) curve. With H now known, read off 2S/ Δt +O from the H = f(2S/ Δt + O)
curve. So at time t = 0, the following quantities are known; H, O2 and (2S2 + O2)
and they will become the quantities at the end of a fictitious previous period.
102

Time Inflow (I1 + I2) (2S/ Δt +
O)
-2O1 (2S2/ Δt+O2) H O
0 I1 - - - X*
H*
O*
6 I2 I1+I2 X*
-2O*
X H O
16 I3
18 I4
26 I5
30 I6
6. The values in row corresponding to time t = 0, being the values at the end of a
routing period is transferred to become the initial values on the left hand-side of
the routing equation (3) for the next time step of the routing period.
7. The left hand side of the routing equation now has known values that yield a
value for (2S2 + O2). With this known, H is read from the plot and O is also
determined.
8. The procedure is repeated until all the inflow values are used up.
Example: Given an inflow hydrograph and storage vs elevation data for a
reservoir below. The spillway discharge Q = 3LH3/2.
The crest height of the
spillway is 50 ft and the length of the spillway is 35ft
Time(days) 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0
Flows(cfs) 70 185 360 480 300 165 80 0
Storage data
Elevation(ft) 50 50.5 51.0 51.5 52.0 52.5 53.0 53.5
Storage(acre-ft) 231 247 277 313 353 400 452 509
Solution:
1. The discharge data, which is the outflow, is computed from the given equation
Q = 3 x (35) x H2/3
and the storage data are listed in tabular form.
Water surface
Elevation (ft)
Head on crest
(ft)
Storage
(acre-ft)
Outflow (cfs)
Q=3 x (35) x
H3/2
(2S/ Δt + O)
Δt = 0.5 day
50 0 231 0 465.6
50.5 0.5 247 37.1 536.4
51.0 1.0 277 105.0 664.4
51.5 1.5 313 192.9 825.3
52.0 2.0 353 297.0 1009.4
52.5 2.5 499 415.0 1224.3
53.0 3.0 452 545.6 1459.5
53.5 3.5 509 687.5 1715.7
103

The routing is done in the table below
1 2 3 4 5 6 7 8
Time Inflow I1 + I2 (2S1+O1) -2O1 (2S2+O2) H O
0 0 - - - 465.6 50 0
0.5 70 70 465.6 0 535.6 50.45 35
1.0 185 255.0 535.8 -70 720.6
1.5 360 545.0 720.6
2.0 480
2.5 300
3.0 165
3.5 80
4.0 0
4.5 0
i). Column 1 & 2 are the given inflow hydrograph
ii) Column 3 is obtained by adding the two successive values in column (2)
iii)Column 4 is obtained by transferring col. 6 of the previous routing period
iv)Col. 5 is obtained by transferring 2 times col. 8
v)Col. 6 is obtained by adding columns (3)+(4)+(5) to obtain the right hand side
of eq.(3)
vi. Col. 7 is obtained by reading off H from the plot of H = f(2S/Δt + O)
vii Col 8 is obtained by reading off O from the plot of H = f(O)
5. After the computation, a plot of the inflow and the outflow hydrographs are
made from which the extent of peak reduction can be seen and when the peak
occurs.
Example: An impounding reservoir enclosed by a dam has a surface area that
varies with elevation as given below. The dam is equipped with two circular gated
discharge ports each of 2.7 m diameter, whose centres are at elevation 54.0 with
discharge coefficient cd=0.8 and a free overflow spillway 72.5 m long with crest
level at elevation 66.0. Live storage at elevation 54 is 5.5 x 106
m3
discharge
gates are opened and the surface water level is at elevation 63.5 at time t=o. The
flood hydrograph table below is forecast.. What will the maximum reservoir level
be and when will it occur.
Elevation (m) 54 56 58 60 62 64 66 68 70 72
Surface area
(km2
)
4.12 4.19 4.28 4.36 4.45 4.58 4.71 4.88 5.05 5.22
104

Chapter 6
DIVERSION HEAD WORKS
The construction work done at the river or canal for diversion of water to the off-
taking canal is known as head-work. Depending on the purpose, head-works
may be:
1. storage head-work or
2. diversion head-work
Storage head-work is constructed to store water for some purpose
The main purpose of diversion head-work is to divert the required quantity of
water into the off-taking canals for irrigation purpose. A diversion head-works
may also serve the following purposes:
105

i) to raise the water level in the river for increasing its command
ii) to regulate the intake of water into the canal
iii) to control silt entry into the canal
iv) to store water for meeting emergency requirements
v) to prevent fluctuations in the level of supply of the river.
i). Diversion head-work provides an obstruction across a river, so that the water
level is raised and water is diverted to the channel at required level. The increase
water level helps the flow of water by gravity and results in increasing the
commanded area and reducing the water fluctuations in the river.
ii). Diversion head-work may serve as silt regulator into the channel. Due to the
obstruction, the velocity of the river decreases and silt settles at the bed. Clear
water with permissible percentage of silt is allowed to flow through the regulator
into the channel.
iii).To prevent the direct transfer of flood water into the channel.
6.1.1. Headwork for direct river offtake
In rivers with a stable base flow and a high enough water level throughout the
year in relation to the bed level of the intake canal, one can resort to run-off-river
water supply (Figure 38 and Example 1 in Figure 37).
106

A simple offtake structure to control the water diversion is sufficient. The offtake
should preferably be built in a straight reach of the river (Figure 39). When the
water is free from silt, the centre line of the offtake canal could be at an angle to
the centre line of the parent canal. When there is a lot of silt in the system, the
offtake should have a scour sluice to discharge sediments or should be put at a
90° angle from the parent canal. If it is not possible to build the offtake in a
straight reach of the river, one should select a place on the outside of a bend, as
silt tends to settle on the inside of bends. However, erosion usually takes place
on the outside of the bend and therefore protection of the bank with, for example,
concrete or gabions might be needed.
The offtake can be perpendicular, at an angle or parallel to the riverbank,
depending on site conditions, as illustrated in Figure 40.
107

The functions of the offtake structures are:
To pass the design discharge into the canal or pipeline
To prevent excessive water from entering during flood
108

Considering these functions, the most important aspect of the structure is the
control arrangement, which can be a gate, stop logs, or other structures. When
the gate is fully opened, the intake behaves like a submerged weir (Figure41)
and its discharge is given by equation 23
Where: Q = Discharge in intake (m3/sec) C = Weir coefficient B = Width
of the intake (m)
h = Difference between river water level and canal design water level (m) hd =
Difference between canal design water level and sill level of the intake (m)
River offtake using a weir
Figure 42 shows an example of a river diversion structure, in this case a weir
(Example 2 in Figure 37). Structures constructed across rivers and streams with
an objective of raising the water level are called cross regulators (see Section
6.4).
109

A weir should be located in a stable part of the river where the river is unlikely to
change its course. The weir has to be built high enough to fulfill command
requirements. During high floods, the river could overtop its embankments and
change its course. Therefore, a location with firm, well defined
banks should be selected for the construction of the weir. Where possible, the
site should have good bed conditions, such as rock outcrops. Alternatively, the
weir should be kept as low as possible. Since weirs are the most common
diversion structures, their design aspects will be discussed below.
Diversion head-works may be a weir or a barrage.
Weirs are solid walls constructed across a river for the purpose of raising the
water level
Barrage is a structure constructed either to store water or raise the level of water.
In case of barrages, no wall is constructed across the river, but there is an
arrangement of gates which can be used to store water to the required level.
Diversion Head-Works consists of:
1. A weir or barrage; 2. Under sluices; 3. Divide wall; 4. Fish ladder; 5. Canal
head regulator; 6. Silt controlling devices; 7. Guide bunds; 8. Approach channel.
1. Weir (Barrage) may be masonry; rockfill or cement concrete. All weirs normally
have the following components: i) body wall usually with shutters; ii) upstream
rough stone or boulder pitching; iii) upstream curtain wall; iv) upstream
impervious apron; v) crest shutters; vi) downstream impervious apron; vii)
downstream curtain wall; vii) downstream apron for channel bed.
Head Regulator
CANAL
Weir
u/s GUIDE BUND
d/s GUIDE BUND
Under sluices
Fish ladder
Divide wall
River
Marginal Bunds
110

Masonry (or concrete)
Rockfill
Shutters
Rockfill weir
111

General Equation of Weirs.
The flow over any type of weir is given by
Figure 44. C2 coefficient for different types of weirs in relation to crest
shape
112

Where: Q = Discharge (m3/sec) C1 = Coefficient related to condition of
submergence and crest shape (Figure 43) C2 = Coefficient related to crest
shape (Figure 44)
B = Weir length, i.e. the weir dimension across the river or stream (m) H = Head
of water over the weir crest (m)
113

Design of weir
a) Design of Weir Walls. Weir walls are design as a retaining wall or masonry
dam of solid gravity type. The two main loads to be considered are the horizontal
water thrust, P, and the self – weight of the weir body wall, W. The resultant of
forces P and W should pass through the middle third of the base of the weir, for
the weir to be safe.
Bligh has recommended the following:
If b = bottom width of the body wall (to be designed);
a = top width ( to be fixed)
H = height of weir above the floor or the height of the body wall, and
d = depth of water above the weir crest or height of weir crest wall,
S = specific gravity of the body wall, then
b
H
d
a
115

−
−
−
−
−
−
−
−
−
+
=
−
−
−
−
−
−
−
−
−
−
−
−
−
+
=
)..
(
552
.
0
.. d
H
a
and
S
d
H
b
(D.H -1)
From the above recommendations, stability are checked under the following
conditions:
I) The water in upstream is at crest level or the crest of shutter and there
is no flow over the weir
II) The weir is submerged and water is passing over it.
III) Water is passing over weir crest and the weir is discharging with a
clear overfall.
Design of Impervious Apron
a)The upstream Apron
The length of the upstream impervious apron according to Bligh is:
CH
L s
u 208
.
2
/ = -----------------------( )
where C = creep coefficient
H = head of water stored against the body wall
The thickness of this apron is about 30 cm placed over 30 to 50 cm thick
concrete foundation.
b). Downstream Apron.
The length of this apron is:
13
208
.
2
/
H
C
L s
d = ---------------- (D.H - 2)
u/s apron
u/s curtains d/s boulder pitching
Body of weir
d/s apron
Shutter
Well
u/s boulder pitching
116

Thickness of apron: 





−
−
=
1
3
4
/
SG
h
H
t s
D -------------------(D.H - 3)
Where tD/s = is in metres; H = total head causing seepage.
h = head lost by the creeping water u to the point where thickness is to be
calculated.
SG = specific gravity of material of the apron.
c). Upstream Curtain Wall: This is usually 2.0 to 2.5 m in depth.
d) Downstream Curtain Wall: usually 2.5 to 3.0 m depth
Two major causes of failure of weirs
i) Piping (or Undermining): Since there is always a differential head between
upstream and downstream, water is constantly moving form upstream to
downstream from under the base of weir. However, if the hydraulic gradient
becomes big, greater than the critical value, then at the point of exist of water at
the downstream end, it begins to dislodge the soil particles and carry them away.
In due course, when this erosion continues, a sort of pipe or channel is formed
within the floor through which more particles are transported downstream which
can bring about failure of weir.
The main preventive measure of piping is to reduce the hydraulic gradient by
increasing the length of travel of the fluid by either increasing the length of the
impervious floor or providing curtains or piles at both upstream and downstream.
2.Uplift Pressure Rupturing the Floor of the Weir
The floor of the weir may burst due to inadequate weight of the weir and the
heavy uplift pressure. This bursting of the floor reduces the effective length of
the impervious floor, which will resulting increasing exit gradient, and can cause
failure of the weir.
Rapture of floor could be prevented by:
i) providing adequate length of impervious floor
ii) providing sufficiently thick impervious apron
iii) by providing a pipe at the upstream end to reduce the uplift pressure at the
downstream end.
BLIGH’S CREEP THEORY
Water from upstream percolates and creeps (or travel) slowly through weir base
and the subsoil below it. The head lost by the creeping water is proportional to
the distance it travels (creep length) along the base of the weir profile. The
creep length must be made as big as possible so as to prevent the piping action.
This can be achieved by providing deep vertical cut-offs or sheet
piles.
According to Bligh’s theory, the total creep length for first drawing: L = B and for
second drawing: L = B + 2(d1 + d2 + d3)
117

If H is the total loss of head, then the loss of head per unit length of the creep ©
shall be
( )
[ ]
3
2
1
2
'
d
d
d
B
H
L
H
C
+
+
+
=
=
Bligh called the loss of head per unit length of creep as Percolation coefficient.
The reciprocal, (L/H) of the percolation coefficient is known as the coefficient of
creep C.
.
Table D-1. Coefficient of creep C for various soils.
No. Type of soil Coefficient of
creep
1 Boulder or shingle, gravel and sand
mixed
5 – 9
2 Coarse grained sand 12
3 Fine micaceous sand 15
4 Light sand and mud 18
Design Criteria.
For design, Bligh proposes the following two important criteria:
B = L
d1
d2
d3
Method of increasing creep length
H
118

1. Safety Against Piping: For safety against, the creep length L should be
sufficient to provide a safe hydraulic gradient, depending on the type of soil.
Safe creep length, L = CH
Where L =safe creep length; H = H1 – H2 Total loss of head for the length of
creep;
H1 = upstream head; H2 = downstream head; C = creep coefficient
2. Safety against Uplift: The uplift pressure = γwH!
; where H!
= the uplift pressure
head at any point of the apron.
The downward resisting force per unit length of apron FR = t. γw . S
where t = thickness of the apron floor; S = specific gravity of floor material.
Equating the two forces, we determine the limiting floor thickness
γw H!
= t. γw .S
H!
= t . S
Or H!
– t = t . S – t = t(S – 1)
Or
1
1
!
−
≅
−
−
=
S
h
S
t
H
t - the limiting thickness
where h is the ordinate of the hydraulic gradient line measured above the top of
the floor.
Now applying a safety factor of 4/3






−
=
1
3
4
S
h
t
Divide Wall
This is a long solid wall of stone masonry built at right angles to the axis of the
body of the weir. Its function is to divide the river channel into two portions and to
create a still pond near the canal head regulator end. It is usually built in stone
masonry but it could be done in earth with stone pitching or in concrete. Because
of the divide wall a still pond is created. This helps the deposition of silt before
the water is allowed to pass through the head regulator. It also protects the
regulator from the direct turbulence of river flow.
119

6.1.4. Scour gates for sedimentation control
Many rivers carry substantial sediment loads, especially during the rainy season,
in the form of sand, silt, weeds, moss and tree leaves. Approximately 70% of all
suspended and bed load sediments travel in the lower 25% of the flow profile.
While suspended silt can be beneficial to the scheme by adding nutrients to the
farmland, coarse sediments usually cause problems once they are blocked by a
weir or other diversion structure. Headworks have to be adapted to these
sediment loads to avoid silting of canals and structures. A properly-designed
intake should divert only the relatively clean upper part of the water flow into the
canal and dispose of the lower part down the river. A sluice should
therefore be incorporated into the diversion structure design. It should be placed
in line with the weir near the canal intake (Figure 51). Its seal level is generally
placed at the river bed level while the floor to the intake gate should be located
higher (Figure 52).
The control arrangement in the scour sluice generally consists of a series of stop
logs (timber, concrete) or a sluice gate. This arrangement allows the water to be
raised when there are very few or no sediments in the water. During the flood
season, the sluice is permanently open or opened at
regular intervals so that depositions of sediments can be flushed away. The guide
wall prevents lateral movement of sediments deposited in front of the weir and
separates the flow through the sluice and the flow over the weir.
Under Sluices (Scour Sluices): These are openings provided in the body wall of
the weir. Their main function is to prevent the obstructions to the flow of water
through the main sluice. They also transport deposited silt in front of the head
regulator at the upstream side, to the downstream side, thus preventing the entry
of bed silt in the canal. They create a clear, unobstructed river channel at the
approach portion of the head regulator. They also help to reduce maximum flood
level. They are provided near the wing walls of the weir. They are operated by
Concrete slab
Concrete foundation
Divide Wall
1.5
Stone pitching
Stone Masonry
120

means of gates provided for this purpose. The gates are operated by levers
provided at the top of the weir.
In designing the scour sluices, the following are to be noted:
i) the design flow should be at least twice the discharge in the off-take
channel
ii) the crest of the under sluice should be one metre lower than the crest
of the head regulator.
iii) The downstream portion of the under sluice must be well protected to
prevent downstream erosion.
Canal Head Regulator
A canal head regulator (or head sluice) is a structure constructed at the head of
an off-take channel of a weir. This may consist of a number of spans created by a
number of piers and covered by a number of gates. The functions of canal head
regulator are:
i) to regulate the quantity of water passing into the canal
ii) to control the amount of silt entering the channel
iii) for shutting out flood flows.
The head regulator consists of a raised crest with abutments on both sides. Piers
may be constructed on the crest to create a number of spans and to support a
roadway and a platform at its top to operate the gates.
Marginal Bunds: are earthen embankments constructed parallel to the river along
the banks. They are located near the approaches of the diversion head works.
Their main function is to control the floods and prevent submergence of the areas
behind the hydraulic structures. Their design and construction are similar to earth
dams.
Guide Banks (or Bell’s bunds): are constructed almost parallel to the flow of the
river, near the hydraulic structure. They are bunds with a straight length and a
curved end at their flanks. Their function is to guide the flow into and protect the
hydraulic structure by controlling and regulating the flow of the river.
121

Marginal Bunds(embankment)
Stone pitching
Marginal Embankment
122

Chapter 7
DISCHARGE MEASUREMENT
Discharge measurement in irrigation schemes is important for the following
reasons:
To ensure the maintenance of proper delivery schedules
To determine the amount of water delivered for water pricing, where it is
applicable
To detect the origin of water losses and to estimate the quantity
To ensure efficient water distribution
To conduct applied research
Almost any kind of obstacle that partially restricts the flow of water in an irrigation
canal and provides a free fall, to ensure that upstream and downstream flow are
independent, can be used as a measuring device, provided that it can be
calibrated. Standard structures, which have already been accurately described
and calibrated, exist. Weirs, flumes and orifices are the devices that are normally
used for discharge measurement.
6.6.1. Discharge measurement equations
The three fundamental equations used to solve discharge problems in canals are
based on the principles of conservation of mass, energy and momentum. For our
purposes, only the conservation of mass and energy equations will be dealt with.
Conservation of mass
Conservation of mass leads to the Continuity Equation 12 to be constant:
Q = A x V = Constant
Conservation of energy
Conservation of energy applied along a streamline results in the Bernoulli
Equation:
Equation 38
Where: P = Pressure (kgf/m2) γ = Density of water (kg/m3) V = Water
velocity (m/sec) g = Gravitational force (9.81 m/sec2) z = Elevation above
reference line (m)
123

Equation 38 sums up the pressure head, velocity head and gravitational head to
give the total head. For an open canal, the pressure head equals the water depth
h. When there is frictional loss along the flow path, an expression for frictional
head loss must be included. Thus applying the Bernoulli Equation to two
successive cross sections along a flow path results in:
Equation 39
The numbers 1 and 2 refer to the first and second cross section. HL is the
frictional head loss.
Specific energy
The concept of specific energy is used in the analysis of critical flow. At any
cross-section of a canal, the energy with respect to the canal bed is referred to
as specific energy. It is derived from the Bernoulli Equation according to the
following equation:
124

Where: E = Specific energy (m), h = Depth of flow (m) g = Gravitational
force (9.81 m/sec2)
V = Water velocity (m/sec)
Assuming a uniform velocity distribution, the specific energy is constant across
the section. Combining the above equation and the Continuity Equation gives:
The cross-sectional area varies with the depth of flow only if the geometry of the
canal is constant. Therefore, for a given discharge the specific energy is a
function of depth alone
Specific energy can be determined for different structures:
Rectangular canal
125

Plotting E against h for different values of (Q/b) gives curves as shown in
diagram above.
The curves show that, for a given discharge and specific energy, there are two
alternate depths of flow, which coincide at a point where the specific energy is a
minimum for a given discharge. Below this point, flow is physically not possible.
At this point flow is critical and it occurs at critical depth and velocity. At a greater
depth, the velocity is low and flow is sub-critical. At the lesser depth, the velocity
is high and flow is super-critical. For sub-critical flow, the mean velocity is less
than the
velocity of propagation of stream disturbances such as waves. Thus, stream
effects can be propagated both upstream and downstream. This means that
downstream conditions affect the behaviour of flow. When flow is supercritical,
the velocity of flow exceeds the velocity of propagation. Consequently, stream
effects (for example, waves) cannot be transmitted upstream, and downstream
conditions do not affect the behaviour of the flow. For critical flow, the specific
energy is a minimum for a given discharge. In this case, a relationship exists
between the minimum specific energy and the critical depth. This relationship is
found by differentiating Equation 41 with respect to h, while Q remains constant.
This gives:
Froude Number
The Froude Number is calculated according to:
Where: Fr = 1 for critical flow, Fr = > 1 for super-critical flow, Fr = < 1 for sub-
critical flow
If a structure is built in a canal which has sub-critical flow, it may cause the flow
to pass through the critical to the super-critical state. This means that the state
126

upstream of the structure becomes independent of the state downstream. This
can either be achieved if the structure narrows the canal, which means
increasing the (Q/b)- ratio without altering the specific energy, or if it raises the
canal bed, which means reducing the specific energy without altering the
discharge per unit width. That is how critical flow is obtained with a measuring
device. A control section in a canal is a section that produces a definitive
relationship between water depth and discharge.
Hydraulic jump
If, through a structure, super-critical flow is introduced in a canal where the
normal flow is sub-critical, flow adjusts back to the sub-critical state through a
hydraulic jump in which the water level rises over a short distance with much
visible turbulence. This situation occurs, for example, downstream of a sluice
gate or a flume. It is undesirable to have a hydraulic jump in an unlined canal
because of the risk of scour. In such cases, a jump is usually induced over a
concrete apron by means of a sill or baffle blocks set in the floor, as shown in
Figure 67.
The relationship between depths just upstream and downstream of a hydraulic
jump is found by the application of the momentum theory to the simplified
situation shown in Figure 68. It is assumed that boundary frictions are negligible
over the length of the jump. For a rectangular
canal it can be shown that:
Equation 43
6.6.2. Weirs
FLOW OVER WEIRS AND NOTCHES
127

A weir in general is any regular obstruction over which flow occurs. Weirs, especially the
sharp-crested ones are commonly used for measuring large and small open flows in the
field or laboratory.
The zone of fluid before the wall of the weir is called the head race while the zone of
water after the weir is called the tail race
The distance LB from the wall of the weir to the point where a fall in the level of stream is
noticeable, known as the drawdown section is given by:
LB = (3- 5)H
H - measured at the section B-B is called the geometric head which is the rise of fluid
above the crest of the weir
B – width of weir going into the paper
δ – thickness of the wall of the weir
z – geometric drop in the level of fluid at the weir. It is the difference between the head
race and the tail race.
vo – approach velocity
Ho – The total head which is given by the sum of the geometric head and the velocity
head.
zo – total drop in head over the weir which is the sum of the geometric drop plus the
velocity head.
LB
H
vo
z
P
δ
Tail race
Head race
B
B
Ho
128

Classification of weirs
1. By form of the opening
a) Rectangular
b) Triangular
c) Trapezoidal
d) Circular
e) Parabolic
f) With inclined crest
2. By the thickness of the cross-section of weir wall
a) Sharp-crested weir δ ≤ (0.1 – 0.5)H: In sharp-crested weirs, the nappe springs
free as it leaves the upstream face
b) Broad – crested weir 2H ≤ δ < 8H: In broad-crested weir, the falling nappe is
supported on the crest and does not allow the nappe to fall free.
c) Weir with practical profile (Orgee weir)
3. By width of obstacle in relation to width of stream
a) Suppressed weir: - When the length of the weir is the same as the width of the
channel (stream) It is so called because in this case, the effect of sides or ends of the weir
on contraction of the nappe is eliminated or suppressed. The width of the nappe is the
same as the width of the stream.
b) Weirs with end contractions: - In this case, the length of the weir is less than
the width of the channel. Therefore there is a lateral contraction of the nappe so that the
length of the nappe is less than the length of the weir crest.
4. By the level of downstream water level relative to the crest of the weir.
a) Un-submerged weir: - When the water level on the downstream side of the weir
is below the crest of the weir, it is said to be un-submerged.
b) Submerged weir: When the water level on the downstream side of the weir
rises above the level of the crest, the weir is said to be submerged.
The flow of fluid over weirs is extremely complex phenomenon and virtually impossible
of exact analytical solution. As result, all weir formula are derived by making lots of
129

simplification of the problem which leads to an approximate solution which is somehow
rectified by the use of experimental coefficients derived from experimental set ups.
To derive a simplified relationship for a weir, the following assumptions are made:
i) velocity distribution upstream from the weir is uniform
ii) all fluid particles move horizontally as they trave pass the weir crest
iii) thew pressure in the nappe is zero; ie atmospheric
iv) the influence of viscosity, surface tension, turbulence are all neglected
After simplification, the sharp-crested weir is considered as a large orifice of rectangular
shape placed in the channel such that the head on its upper edge is zero. As a result, the
upper edge is eliminated leaving only the lower edge or the crest. By considering the weir
as a large orifice and by considering an elementary strip of of area Bdh below the free
surface, the velocity of flow is given by:
.
.
2gh
c
u v
=
The flow rate through the strip
gh
Bdhc
dQ v 2
=
The theoretical flow passing over the weir is obtained by integrating the above equation
over the geometric head.
v1
H
v1
2
/2g
P
Fig. Simplified flow over weir with the above assumptions
1 1
2
2
h
dh
H
Weir crest
B
130

2
/
3
0
2
3
2
2 H
g
Bc
dh
gh
Bc
Q v
H
v
∫ =
=
To obtained the actual flow rate over the weir due to the above assumptions, we introduce
an experimental coefficient of discharge
2
/
3
2
3
2
H
g
B
C
Q D
=
If the approach velocity vo is appreciable (>1.0), then instead of geometric head one has
to use the total head Ho
As can be seen, the formula for flow over weir seems simple and easy to use however, the
difficulty is with the correct determination of the discharge coefficient. In the laboratory,
this could be determined quite easily and accurately, however, it becomes extremely
difficult to determine this accurately in the field. A few empirical formulae exist for the
determination of the discharge coefficient for a rectangular weir.
CD = 0.40 + 0.05 H/P ------------Chugaev R. R. (2/3 in front of flow rate absorbed)
CD = 0.405 + 0.003/H -----------Basen (2/3 in front of formula absorbed)
CD = 0.397 + 0.00015/H3/2
Fteley and Stearn (2/3 in front of formula absorbed)
CD = 605 + 0.08H/P + 1/1000H ------Rehbock
CD = 0.611 + 0.075H/P -----------taken from A. K. Jain.
Contracted Weirs
For weir with end contractions, Francis observed that the effect of the contraction is to
reduce the length of nappe by 0.1H. Thus if there are n end contractions, then the length
of the nappe will be reduced by 0.1nH and will be equal to (B – 0.1nH) and the discharge
equation will then be given as:
2
/
3
2
)
1
.
0
(
3
2
H
g
nH
B
C
Q D −
=
Flow over a Triangular Weir (Notch)
A triangular weir (notch) is particularly useful where the discharge is to vary over a large
range and the same accuracy is desired for both small and large discharges because of the
following advantages:
131

i) The coefficient of discharge for a triangular weir is practically independent of the head.
This is because, for all heads the ratio of the head to the wetted length of crest is constant.
But in a rectangular weir, the ratio of head to the wetted length of crest is not constant.
ii) When the discharge is small, a triangular notch provides a greater head than the
rectangular notch. Hence head measurement can be done more accurately over the
triangular notch than over the rectangular notch.
iii) When the flow rate is small, there is the tendency of a clinging nappe in a rectangular
notch. However, for the same flow rate over a triangular notch, the head will be greater
and clinging nappe will be avoided.
The equation for the discharge over the vee-notch can be obtained just like we did in the
rectangular weir by considering an elementary strip of width b and height dh.
Considering the notch as a large orifice, the velocity of flow is given by:
.
.
2gh
c
u v
=
The flow rate through the strip
gh
bdhc
dQ v 2
=
But b = 2(H – h)tan θ/2
Putting b into the discharge equation, dh
gh
c
h
H
dQ v 2
2
tan
)
(
2
θ
−
=
b
H
h
θ
dh
132

( ) 2
/
5
0
2
/
3
2
/
1
2
tan
2
15
8
2
2
tan
2 H
g
c
dh
h
Hh
g
c
Q v
H
v
θ
θ
=
−
= ∫
The above equation represents the theoretical discharge without consideration of the
contraction of the flow area. The actual discharge is obtained by applying the coefficient
of discharge CD. to obtain:
2
5
.
2
tan
2
15
8
H
g
C
Q D
θ
=
By representing k= (8/15)CD(2g)1/2
and taking the average value of CD =0.60
k = 8/15 x 0.60 x 4.43 = 1.42 and hence
Q = 1.42 x tan(θ/2) H5/2
For a right angled Vee-notch, θ = 90o
and hence
Q = 1.42 H5/2
Empirical formulae 1) By Thompson
Q = 1.4 H5/2
2. By King
Q = 1.343 H2.47
By Grave When 22o
≤ θ ≤ 118o
Q = 1.331 (tan θ/2)0.996
H2.47
The coefficient of discharge is made up mainly of the contraction coefficient and the
velocity coefficient. The contraction coefficient of a notch depends on the wetted
perimeter. For a rectangular notch, the wetted perimeter does not vary with head because
the base, which forms part of the wetted perimeter remains constant for all heads.
Consequently, the coefficient of contraction and hence the coefficient of discharge will
not be constant for all heads. In a triangular notch, there is no base and therefore the
contraction is due to the sides only. The wetted perimeter depends on the length of the
sides which in turn depend on the head. Consequently, the coefficient of discharge is
fairly constant in a triangular notch.
Trapezoidal Weir
A trapezoidal weir has an opening of a trapezoidal shape and may be considered as a
combination of a rectangular and a triangular weirs.
133

The trapezoidal notch is equivalent to two notches; one rectangular with length b and the
other a triangular with an apex angle of θ. The total discharge over the trapezoidal notch
would be equal to the sum of discharges over the equivalent rectangular and triangular
notches.; thus
Qtrap = Q1 + Q2 =






+
=
+
2
tan
15
8
3
2
2
2
tan
2
15
8
2
3
2 2
/
3
*
2
5
2
/
3 θ
θ
H
b
H
g
C
H
g
C
H
g
b
C D
D
D
Cippoletti Weir
The principle behind the Cippoletti weir is based on the fact that in a rectangular weir,
due to end contractions, the effective length of the weir is reduced and consequently the
discharge. On the other hand, by the addition of side slopes to a rectangular weir, gives an
increase in discharge. Therefore if the side slopes of a trapezoidal weir are so adjusted
such that the reduction in discharge due to end contractions is just equal to the increase in
discharge due to the addition of side slopes, the net effect I discharge is zero and
therefore the standard weir formula without end contraction can be used.
The Cippoletti Weir is a trapezoidal weir with side slopes 1 horizontal to 4 vertical such
that the discharge in effect equal to that of a rectangular weir without end contraction.
For Cippoletti weir, 2
3
2
3
2
H
g
b
C
Q D
=
Cipolletti gave the equation for the discharge through such a weir as:
Q = 1.86 b H3/2
Submerged Weir
b
θ/2
θ/2
H
Trapezoidal Weir
134

A submerged weir (notch) is one in which the level of fluid in the downstream side of the
weir is above the crest of the weir.
The flow over such a weir can be obtained by dividing the flow into two portions:
i) Flow over the upper part of the section (above line AB) of depth (H1 – H2)
may be considered as a free from a weir into the atmosphere and
ii) Flow through the lower part of depth (below line AB) may be considered as a
discharge through a submerged orifice.
The discharge over the upper portion behaving as a free weir is given by
( ) 2
/
3
2
/
3
2
1
1 2
3
2
2
3
2
H
g
b
C
H
H
g
b
C
Q D
D =
−
=
and the lower portion acting as an orifice is given by:
( ) gH
bH
C
H
H
g
bH
C
Q D
D 2
2 2
2
1
2
2 =
−
=
Therefore the total flow is given by:
gH
bH
C
H
g
b
C
Q D
D 2
2
3
2
2
2
/
3
+
=
Experimental works by Herschel, Villemonte and Mavis on the discharge characteristics
of a number of sharp-crested submerged weirs on rectangular, triangular. Parabolic and
proportional weirs showed the results for all types could be represented by the equation:
Flow over submerged weir
H1
H2
1
2
A
B
H
135

385
.
0
1
2
1 1
















−
=
n
H
H
Q
Q
where Q = discharge for submerged condition
Q1 = free discharge given by Q = K H1
n
n = exponent in the free discharge equation
n = 1.44 for contracted weir
n = 1.5 for suppressed weir
n = 2.50 for 90o
notch.
Broad-Crested Weir
Broad-crested weirs (long-based weirs) are weirs that have crest lengths that are
sufficiently longer (δ > 0.67H) to prevent the nappe from springing free. These weirs are
usually constructed of concrete, have rounded edges, and are capable of handling much
larger discharges than sharp-crested weirs.
Some advantages of broad-crested weirs.
i) It is cheaply and easily fabricated (can be done in-situ or off-site)
ii) It is easy to install
iii) It possesses a wide modular range;
iv) It produces minimum afflux(increases in upstream depth due to weir
installation)
v) It requires a minimum of maintenance.
Flow over a broad-crested weir must pass through the critical depth somewhere near the
downstream corner. Broad-crested weirs are characterized by the presence of two drops
in the water level over the weir.
δ
ZB
1 2
H
P
Ho
h
ZB(o)
v
Zc
136

By taking Bernoulli’s equation in sections 1 and 2 , the velocity of flow on the weir is
obtained as
( )
h
H
g
c
gZ
c
v o
v
B
v −
=
= 2
2 0
For a rectangular weir, the discharge is given by:
Q = Av = b x h x v = ( )
h
H
g
bhc o
v −
2 ----------(*)
( ) ( )
h
f
h
H
g
hc
q o
v =
−
= 2 ---------------------(**)
As can be seen from the above equation, in order to determine the discharge, there is the
need to first find h, the depth of flow on the weir. There are many methods of finding h
and Q but the Belanch’s method based on the principle of maximum discharge is much
more appealing.
As can be seen from the above equation, the discharge is only a function of h and the
value of h varies between 0 ≤ h ≤ Ho however the discharge goes to zero as h = 0 and as
h = Ho
Since the function (**) is a continuous function and the function goes to zero at the two
limits, then it follows that there must be a turning point (maximum or minimum) within
the limits. This is the basis of Belanch’s principle which states that: for a given head Ho
on a broad-crested weir, the depth that will be established on the weir is such as to result
in the maximum discharge on the weir.
With the above postulate, the required depth is obtained by:
( )
[ ]
0
2
=
−
=
dh
h
H
g
h
c
d
dh
dq o
v
Neglecting the constants and differeniating
( ) 0
2
1
=
−
−
−
=
−
h
H
h
h
H
dh
h
H
h
d
o
o
o
which gives h = 2/3Ho
Substituting this value of h into (*) we
2
/
3
2
/
3
2
/
3
2
/
3
2
2
2
385
.
0
2
27
4
3
2
2
3
2
.
o
D
o
D
o
v
o
v
o
o
v
o
H
g
C
q
H
g
b
C
Q
H
g
b
c
H
g
bc
H
H
g
c
H
b
Q
=
=
=
=






−
=
where CD = 0.385 Cv , CD ranges between 0.32 to 0.36.
137

Ventilation of Weirs
When flow takes place over a suppressed weir with wing walls on the downstream side,
the free access of air below the nappe is prevented. The air entrapped between the nappe
and the weir face is gradually evacuated by the dynamic action of the flowing fluid. This
eventually leads to subatmospheric pressure in the underside of the nappe bringing about
increases in the discharge.
Investigations by Bazin have shown that if the flow is sufficient to prevent the air having
access to the underside of the nappe, it may assume one of the forms shown below
Fully aerated (free, ventilated) nappe: In this type, the weir discharges free as the
pressure below the nappe is maintained at atmospheric by proper ventilation.
Drowned (depressed) nappe: When the underside of the nappe is poorly aerated, the air
enclosed between the nappe and the downstream face of the weir gradually mixes up with
the flowing fluid and is carried away. This causes the pressure under the napper to fall
below atmospheric. The pressure difference on the two sides of the nappe-(the upper side
exposed to atmospheric while the bottom side has sub-atmospheric pressure) –causes the
nappe to deflect towards the weir.
Clinging nappe: This type occurs when there is no ventilation and all the air under the
napper has been carried away by the dynamic action of the flowing fluid. The nappe is
forced to adhere to the downstream face of the weir by the atmospheric pressure on the
upper side. In this type, the discharge is 20 to 30 per cent more than in free nappe.
Knowledge of these types of nappes is necessary especially when the weir is used for floe
(discharge) measurement. All the discharge formulae have been derived for free nappe
Fully aerated nappe Drowned nappe Clinging nappe
Atmospheric
Pressure
Sub-atmospheric
pressure
138

and therefore some adjustment in the measured flow must be made when other forms of
nappe occurs during discharge measurement. For discharge measurement it is always
advisable to ensure that the nappe is fully aerated.
FLUMES
Although weirs are the cheapest and simplest structures for flow measurement in open
channels, however the relatively high losses caused by weirs and the tendency of
sedimentation of suspended particles due to reduced velocities in the vicinity of the weir,
could in certain cases pose important limitations Flumes provide a convenient alternative
to weirs for flow measurement in open channels where high head losses and
sedimentation are of concern. Such cases include flow measurement in wastewater
treatment plants and irrigation channels with suspended particles
Flumes are devices in which the flow is locally accelerated by a streamlined lateral
contraction in the channel sides. A flume has: a convergent section, in which the
flow accelerates, a throat section and a divergent section, in which the
flow returns to normal
Flumes are of two types: non-modular or the venturi flume and the modular or the
standing wave flume.
In the non-modular flume, the velocity at the throat is maintained below the critical value
so that no standing wave is produced. However, in the modular type, the flume is
designed such that the velocity of flow at the throat is greater than the critical velocity
thereby resulting in a standing wave within the flume.
Non-Modular Flume (Venturi Flume)
b2
b1
v1
2
/2g
h1
h2
v2
2
/2g
Critical depth
ELEVATION
PLAN
1
1
2
2
139

When the width of a channel is reduced while the bed remains flat, the discharge per unit
width increases. If the losses are neglected, the specific energy remains constant. If the
conditions are made such that the free surface does not pass through the critical depth, the
arrangement forms a Venturi flume.
Referring to the diagram above, for continuity of flow,
h1 x b1 x v1 = h2 x b2 x v2 --continuity equation for section 1-1 and 2-2.
Applying Bernoulli’s equation to sections 1-1 and 2-2
h1 + v1
2
/2g = h2+v2
2
/2g
Substituting v1 for continuity equation,
h1-h2 = v2
2
/2g[1-(h2b2/h1b2)2
]
( )
( ) e
disch
l
Theoretica
h
b
h
b
h
h
g
h
b
A
v
Q
b
h
h
b
h
h
g
v
arg
..
..
1
2
1
2
2
1
2
2
1
2
2
2
2
2
1
2
2
2
1
2
−
−
−
−
















−
−
=
=
















−
−
=
Introducing the discharge coefficient, CD = (Actual discharge)/(Theoretical discharge)
( ) ( )
2
1
2
2
2
1
2
1
2
1
1
2
2
2
1
2
2 2
1
2
h
h
g
A
A
A
A
C
h
b
h
b
h
h
g
h
b
C
Q D
D −
−
=
















−
−
=
CD = 0.95 to o.99
Therefore to measure the flowrate, water depth are measured at two locations, one at
section 1-1 and the other at section 2-2.
Modular Flume or Standing Wave Flume (eg. Parshall Flume)
As already mentioned, modular flumes are designed to make the flow pass through the
critical depth in some section of the throat.
By considering a section upstream with depth h1 and another section at the throat with
depth h2, then applying Bernoulli’s equation,
h1 + v1
2
/2g = h2 + v2
2
= H
140

v2 = (2g(H-h2)1/2
Therefore the discharge Q = b2 h2v2 = b2h2[2g(H-h2)]1/2
( ) ( ) 2
/
1
3
2
2
3
2
2
2 2
2 h
Hh
g
b
h
Hh
g
b
Q −
=
−
=
For the maximum discharge, the quantity (Hh2
-h3
)1/2
should be maximum
Therefore
( )
H
h
h
Hh
dh
h
Hh
d
3
2
0
3
2
0
2
2
2
2
3
2
2
=
=
−
=
−
Thus for the flow to be maximum, the depth at the throat section should be two-thirds the
total energy.
For this condition, the velocity ( ) 2
2
2
2
2
2
3
2
2 gh
h
h
g
h
H
g
v =






−
=
−
=
This implies that for maximum discharge, the depth at the throat must be equal to the
critical depth and the velocity critical velocity.
Substituting h2 = 2/3H in the discharger expression, we shall obtain
2
/
3
2
1
1
2
2
/
3
2
2
/
3
2
2
/
1
2 )
2
/
(
705
.
1
705
.
1
2
3
3
2
3
2
3
2
2 g
v
h
b
H
b
H
gb
H
H
H
g
b
Q +
=
=
=






−
=
Although flat bed modular flume is easier to construct, it is sometimes necessary to raise
the invert in the throat to attain critical conditions. For the flat or raised floor, the throat
length should ideally be sufficient to ensure that the curvature of water surface is small to
make the water surface parallel to the invert. Such ideal flumes become relatively long
and expensive. A more compact, short and less expensive groups of flumes have been
constructed. However, the surface water profile varies rapidly and as such the theoretical
analysis for ordinary flumes is not wholly applicable. Therefore empirical relationship are
usually developed for such flumes (e.g; Parshall Flume), which make each such flume
standard with its unique calibration curve.
The Parshall Flume
The Parshall flume is a widely-used discharge measurement structure. Figure 76 shows
its general form. The characteristics of Parshall flumes are:
i) Small head losses, ii) Free passage of sediments, iii) Reliable measurements even
when partially submerged and iv) Low sensitivity to velocity of approach
141

The Parshall flume consists of a converging section with a level floor, a throat section
with a downward sloping floor and a diverging section with an upward sloping floor.
Flume sizes are known by their throat width and each size has its own characteristics,
which is unique (see Table 31).
The flow through the Parshall flume can occur either under free flow or under submerged
flow conditions. Under free flow the rate of discharge is solely dependent on the throat
width and the measured water depth, ha. The water depth is measured at a fixed point in
the converging section.
The upstream water depth-discharge relationship, according to empirical calibrations, has
the following general form:
Q = K x (ha)u
Where: Q - Discharge (m3
/s); ha Water depth in converging section (m); K -A fraction,
which is a function of the throat width; u Variable, lying between 1.522 and 1.60
Table 32 gives the values for K and u for each flume size.
When the ratio of gauge reading hb to ha exceeds 60% for flumes up to 9 inches, 70% for
flumes between 9 inches and 8 feet and 80% for larger flume sizes, the discharge is
reduced due to submergence. The upper limit of submergence is 95%, after which the
flume ceases to be an effective measuring device because the head difference between ha
and hb becomes too small, such that a slight inaccuracy in either head reading results in a
large discharge measurement error
142

The discharge under submerged conditions is:
Qs = Q - Qc
Where: Qc – Reduction of the modular discharge due to submergence
Figure 77 gives the corrections Qc for submergence for flumes with 6 inch,
9 inch and 1 foot throat width. The correction for the 1 foot flume is made
applicable to other sizes by multiplying the correction Qc for the 1 foot by
the factors given in Figure 77 (1 foot flume).
144

The weir is the most practical and economical device for water measurement.
Weirs are simple to construct, easy to inspect, robust and reliable. Discharge
measurement weirs can either be sharp-crested (Figure 69, 70, 71) or
broadcrested (Figure 72).
149

Sharp-crested weirs
Sharp-crested weirs, also called thin plate weirs, consist of a smooth, vertical, flat
plate installed across the channel and perpendicular to the flow (Figure 69). The
plate obstructs flow, causing water to back up behind the weir plate and to flow
over the weir crest. The distance from the bottom of
the canal to the weir crest, p, is the crest height. The depth of flow over the weir
crest, measured at a specified distance upstream of the weir plate (about four
times the maximum h1), is called the head h1. The overflowing sheet of water is
known as the nappe.
Thin plate weirs are most accurate when the nappe springs completely free of
the upstream edge of the weir crest and air is able to pass freely around the
nappe. The crest of a sharp-crested weir can extend across the full width of
channel or it can be notched. The most commonly used notched ones are:
Rectangular contracted weir
Trapezoidal (Cipoletti) weir (Figure 70)
Sharp sided 90° V-notch weir (Figure 71)
The type and dimensions of the weir chosen are based on the expected
discharge and the limits of its fluctuation. For example, a V-notch weir gives the
most accurate results when measuring small discharges and is particularly
adapted to the measurement of fluctuating discharges.
Calibration curves and tables have been developed for standard weir types. The
conditions and settings for standard weirs are as follows:
i. The height of the crest from the bottom of the approach canal (p) should
preferably be at least twice the depth of water above the crest and should in no
150

case be less than 30 cm. This will allow the water to fall freely, leaving an
airspace under and around the jets.
ii. At a distance upstream of about four times the maximum head a staff gauge is
installed on the crest with the zero placed at the crest elevation, to measure the
head h1.
iii. For the expected discharge, the head (h1) should not be less than 6 cm and
should not exceed 60 cm.
iv. For rectangular and trapezoidal weirs, the head (h1) should not exceed 1/3 of
the weir length.
v. The weir length should be selected so that the head for the design discharge
will be near the maximum, subject to the limitations given in (ii) and (iii).
vi. The thickness of the crest for sharp-crest weirs should be between 1-2 mm.
In sediment-laden canals, a main disadvantage of using weirs is that silt is
deposited against the upstream face of the weir, altering the discharge
characteristics. Weirs also cannot be used in canals with almost no longitudinal
slopes, since the required difference in elevation between the water
levels upstream and downstream side of the weir is not available. Discharge
equations for weirs are derived by the application of the Continuity and Bernoulli
Equations (Equation 12 and 38 respectively). In each case, a discharge
coefficient is used in order to adjust the theoretical discharge found by laboratory
measurement
Rectangular contracted weir
A rectangular contracted weir is a thin-plate weir of rectangular shape, located
perpendicular to the flow. To allow full horizontal contraction of the nappe, the
bed and sides of the canal must be sufficiently far from the weir crest and sides.
Many practical formulae have been developed for
computing the discharge, amongst which are the following
Where:
151

Q = Design discharge over weir (m3/sec) b = Length of weir crest (m) h =
Design water depth measured from the top of the weir crest (m) Table 28 gives
discharge data related to length of crest, b, and water head, h, over a weir.
Trapezoidal (Cipoletti) weir
The trapezoidal weir has a trapezoidal opening, the base being horizontal. The
Cipoletti weir is a trapezoidal weir with the sides having an outward sloping
inclination of 1 horizontal to 4 vertical (Figure 70). This side slope is such that the
water depth-discharge relationship is the same as
that of a full width rectangular weir.
The discharge equation for a Cipoletti weir is:
Where:
Q = Design discharge over weir (m3
/sec) b = Length of weir crest (m) h =
Design water depth measured from the top of the weir crest (m)
Table 29 shows discharge data, related to the design water
depth, h, and weir length, b.
152

V-notch weir
A V-notch weir has two edges that are symmetrically inclined to the vertical to
form a notch in the plane perpendicular to the direction of flow. The most
commonly used V-notch weir is the one with a 90° angle. Other common V-
notches are the ones where the top width is equal to the vertical depth (1/2 x 90°
V-notch) and the one where the top width is half of the vertical depth (1/4 x 90° V-
notch) (Figure 71). The V-notch weir is an accurate discharge-measuring device,
particularly for discharges less than 30 l/sec, and it is as accurate as other types
of sharp-crested weirs for discharges from 30 to 300 l/sec (U.S. Department of
Interior, 1975). To operate properly, the weir should be installed so that the
minimum distance from the canal bank to the weir edge is at least twice the head
on the weir. In addition, the distance from the bottom of the approach canal to the
point of the weir notch should also be at least twice the head on the weir (U.S.
Department of Interior, 1975).
The general and simple discharge equation for a V-notch
153

Where:
Q = Design discharge over the weir (m3/sec) θ = Angle included between the
sides of the notch (degrees)
h = Design water depth (m)
Table 30 gives discharge data for the three common V-notches related to water
depth (head) and angle°.
154

Broad-crested weir
A broad-crested weir is a broad wall set across the canal bed. The way it
functions is to lower the specific energy and thus induce a critical flow (Figure 72
Flumes
Discharge measurement flumes are extensively used in irrigation schemes
mainly because they:
Can be used under almost any flow condition
155

Have smaller head-losses than weirs, thus are more
accurate over a large flow range
Are insensitive to the velocity of approach
Are relatively less susceptible to sediment and debris transport
However, major disadvantages of flumes include the relative large sizes and the
accurate manufacturing/construction workmanship required for optimum
performance (James, 1988).
A canal section that causes flow to pass from sub-critical through critical to the
super-critical state forms a control and the discharge is a single valued function
of the upstream water level. Critical flow can be achieved by raising the canal
bed, thereby reducing the specific energy, or by decreasing the canal width,
thereby increasing the discharge per unit width (see Section 6.6.1). This latter
technique is the one used by flumes. A flume has:
A convergent section, in which the flow accelerates
A throat, in which critical flow occurs
A divergent section, in which the flow returns to normal
Super-critical flow passing from the throat will return to sub-critical flow
downstream of the flume. This occurs due to the development of a hydraulic
jump, which is induced within the divergent section by a sill or other barrier.
Where there is sufficient head available, the divergent section of the
flume could be avoided as the flow could fall freely in a stilling basin. In this case,
weirs could also be used. However, if canals are expected to carry a lot of
sediment, the flume should be the better choice. Flumes are most commonly
rectangular or trapezoidal in cross-section. The former type is the most simple to
construct, but if the canal cross-section is not rectangular there is a risk that
unpredictable flow patterns will result from an abrupt change of cross-section.
The most commonly used flumes are:
Parshall flume
Trapezoidal flume
Cut-throat flume
Trapezoidal flume
Whenever the canal section is not rectangular, trapezoidal flumes such as those
shown in Figure 79, are often preferred, especially for measuring smaller
discharges. A typical trapezoidal flume has an approach, a converging section, a
throat, a diverging and an exit section. A minimum transition will be required. An
additional advantage is the flat bottom, which allows sediment to pass through
fairly easily. Furthermore, the loss in head may be less for comparable
discharges. Trapezoidal flumes are particularly suited for installation in concrete-
lined canals. The flume should normally be put on top of the lining, thus
constricting the flow section to the extent required for free flow conditions over a
156

whole range of discharges up to the canal design discharge. As a rule of thumb,
one can say that the lower the canal gradient the higher the elevation of the
flume above the canal bed level. The flow characteristics of the flume can be
determined experimentally. This allows for the calibrations of the flume.
As an example, a flume with dimensions such as those given in Figure 79 can be
located in a canal with a bed width of 0.30 m (1 foot), having side slopes of 1:1.
The range of calibrated water depth is 6-37 cm and the range of calibrated
discharge is 1.4-169 l/sec. This will suit most conditions in a typical small-scale
irrigation canal.
Cut-throat flume
The cut-throat flume has a converging inlet section, throat and diverging outlet
section. The flume has a flat bottom and vertical walls (Figure 80). It is preferable
157

to have the cut-throat flume operating under free flow conditions. This facilitates
measurements and ensures a high degree of accuracy.
Free flow conditions through the cutthroat flume are described by the following
Where:
Q = Discharge (m3/sec) C = Free flow coefficient ha = Upstream water depth
(m),
K = Flume length coefficient W = Throat width (m)
158

Orifices
Orifices, such as gates and short pipes, are also used as water measuring
devices (Figure 82). However, they do not offer any advantage over the use of
weirs or flumes. Furthermore, their calibrations are not as accurate nor as stable
as other types of measuring devices. For weirs the discharge is proportional to
the head above the crest raised to the power 3/2 (Equations 44, 45, 46,
48). Therefore, they are sensitive to the fluctuations in the upstream water level.
For orifices, including gates and short pipes, the discharge is proportional to the
head of water above the crest raised to the power 1/2, as shown by Equation 34
(see Section 6.1.3). Therefore, they are less sensitive to small fluctuations of the
upstream water level. Under submerged conditions both the upstream and
downstream sides of the structure need water level recordings. For free flow
conditions, the discharge is a function of the upstream water depth alone.
160

The general discharge equation for a free flow orifice is (Equation 34):
Where:
161

Q = Design discharge through orifice (m3/sec) C = Design coefficient
(approximately 0.60)
A = Cross-sectional area of the orifice (m2) g = Gravitational force (9.81 m/sec2)
h1 = Water depth upstream of orifice over reference level (m) (Figure 83)
Partially-opened sluice gates could be used for discharge measurements, in
which case they will be acting like submerged orifices (Figure 84). For partially-
opened sluice gates and submerged orifices
the discharge equation reads:
Where:
Q = Design discharge through orifice (m3/sec), C = Discharge coefficient, which
is 0.63 for sluice gates and submerged orifices and 0.85 for short pipes, A =
Cross-sectional area of the orifice (m2) g = Gravitational force (9.81 m/sec2), h1
= Water depth upstream of orifice over reference level (m), h2 = Water depth
downstream of the structure (m)
Current meters
Current meters are used to measure the velocity in a canal, from where the
discharge can be calculated using the Continuity Equation 12 (see Section 5.1).
Most current meters have a propeller axis in the direction of the current. The
flowing water sets the propeller turning. On a meter, forming part of the
equipment, the number of revolutions per time unit can be read and, by means of
162

a calibrated graph or table, the velocity can be determined. A well known type of
current meter is the Ott instrument C31 for velocities up to 10 m (Figure 85).
Propeller meters are reliable and accurate, but rather expensive. In measuring
the velocities, the number of points per vertical and the number of verticals per
cross-section should be determined. For this purpose, the quantity of work and
the time required should be weighed against the degree of accuracy (Euro-
consult, 1989). For example, measurements can be taken at 10 cm horizontal
distance over the cross-section and at 0.2h and 0.8h depth at each 10 cm (h is
the water depth). The velocity is the average of the velocity at 0.2h and 0.8h
depth. If the water depth is less than 0.5-0.6 m, one reading can be done at 0.6h.
Then, for each vertical the flow per unit width can be calculated according to q =
v av x h (Figure 86a). These qs are distributed over the total width (Figure 86b)
and the area between the q-line and the water surface gives the total discharge.
It is also possible to establish the discharge per section and to consider the sum
of the discharges in the sections as the total discharge.
163

Discharge measurement in pipelines
Several types of devices can be used to measure the discharge in pipelines. This
section will discuss differential pressure and rotating mechanical meters, as they
are the ones commonly used.
Differential pressure flow meters
Differential pressure flow meters create a pressure difference that is proportional
to the square of the
discharge. The pressure difference is created by causing flow to pass through a
contraction. Manometers, bourdon gauges, or pressure transducers are normally
utilized to measure the pressure difference. One good example of a differential
pressure flow meter is the Venturi tube (Figure 87).
Venturi tube
The pressure drop between the inlet and throat is created as water passes
through the throat. In the section downstream of the throat, the gradual increase
in cross-sectional area causes the velocity to decrease and the pressure to
increase. The pressure drop between the Venturi inlet and the throat is related to
the discharge, as follows
164

Where:
Q = Discharge (l/min), C = Flow coefficient, D = Diameter of upstream section
(cm), d = Diameter of contraction (cm)
P1 = Pressure in upstream section (kPa), P2 = Pressure in contraction (kPa), K
= Unit constant (K is 6.66 for Q in l/min, d and D in cm, and P1 and P2 in kPa)
The flow coefficient C for a Venturi metre is 0.97.
Rotating mechanical flow meters
There are many types of rotating mechanical flow meters used in pipelines.
These flow meters normally have a rotor that revolves at a speed roughly
proportional to the discharge and a device for recording and displaying the
discharge and total volume. The rotor may be a propeller or axial flow turbine, or
a vane-wheel with the flow impinging tangentially at one or more points.
Calibration tests are usually needed to accurately relate rotor revolutions to the
flow. The lowest discharge that can be accurately measured by a rotating
mechanical flow meter depends on the amount of bearing friction that can be
tolerated while the occurrence of cavitation often establishes the largest flow rate
that can be measured (see Module 5). Head loss through most rotating
mechanical discharge meters is moderate.
165

Chapter 8
DESIGN OF WEIRS AND BARRAGES. THEORY OF SEEPAGE
Weir or Anicut
It is a solid wall of masonry or concrete constructed along the width of the river
from one bank to the other to raise the water level on the upstream side so that
proper supplies of water may be assured to the canals. Usually water flows over
the crest of the weir but sometimes shutters 1m or more high are provided on the
crest of the weir. During the dry period the shutters are raised to raise the water
level and during rainy season, the shutters are dropped to allow flood waters to
pass without causing damage on the upstream side.
Barrage
The function of a barrage is the same as that of a weir. In a barrage, a low height
weir is constructed first and then according to design piers are constructed at
regular intervals along the width of the river. The gaps between the piers are
closed by means of gates, which can be lifted or lowered down mechanically or
manually or both. Barrages are preferred now-a-days as they offer greater
flexibility in operation and better control. In this case, water level can be raised
many times more than weirs and at will, more discharge can be passed by lifting
the gates, but in case of weirs, the crest is fixed and only a fixed amount of
discharge can be passed.
When the foundation of a hydraulic structure is pervious, seepage will take place
as long as differential head exists across the structure.
Seepage of water below the structure may result in failure due to piping or
pressure uplift.
BLIGH’S CREEP THEORY
Bligh stated that: creep is caused by differential head across a structure and the
loss of head is proportional to the length of creep.
Length of creep: is the total distance travelled by seeping (creeping) water.
If H = total head across structure,
And L =total length of travel of seepage water (creep length,
then the head lost per unit creep length = H/L referred to as hydraulic gradient.
Referring to the diagram below, assume that there are three cut-offs down
beneath the foundation of a weir with depths y1, y2, and y3 at A B and C. Then the
total seepage length (creep length)
Lc = L1 + L2 + 2(y1 + y2 + y3)
If H is the total head difference downstream and upstream, then the hydraulic
gradient is given by:
Hydraulic gradient = H / Lc = head lost per unit length travelled by seepage
water.
Head loss across cut-off A = (H/Lc) x 2y1
166

B = (H/Lc) x 2y2
C = (H/Lc) x 2y3
Based on the above, Bligh suggested the following to prevent failure by seepage.
1. Safety against Piping: If sufficient creep length is provided by providing vertical
cut-off, the head causing creep can be destroyed before the water exits on the
downstream side and piping (undermining) would be prevented. Accordingly
Bligh gave the following criteria.
Lc = C.H where H is the head across structure; and C = a constant
depending on
soil type and given by Bligh below.
No. Type of soil Value of
C
Limiting value of
safe hydraulic
gradient
1 Fine micaceous sand, alluvial soil,
etc
15 1/15
2 Coarse grained sand 12 1/12
3 Sand with boulders and gravel 5 – 9 1/9- 1/5
4 Light sand and mud (silt?) 8 1/8
Safety against Uplift
The ordinate at any point on the bottom of the floor to the hydraulic gradient
represent the uplift at that point. This uplift must be balanced by the weight of
the floor if uplift is to be prevented. Therefore the floor must be of sufficient
thickness to provide the necessary resisting force to this uplift force.
If we let t be the required floor thickness to balance uplift force , then
H
A
B
Hydraulic gradient
Weir or barrage
L1
L2
y3
h Q
y1
C
y2
167

Weight of floor = γ . SG. t
Uplift force = γ.h
where γ = weight density of water, SG specific gravity of material of floor.
In the limiting condition,
.γ. SG. t = γ.h
or t = h/SG.
If h’ is the ordinate to the hydraulic gradient line considered from the top of the
floor, then: h = h’ + t
And t = h/SG = (h’ + t)/SG
Or t = h’/(SG-1)
This is the limiting thickness of the floor to withstand uplift.
Limitations of Bligh’s Theory
1. Bligh made no distinction between horizontal and vertical creep
2. Bligh did not explain exit gradient
3. Bligh did not make any distinction between outer and inner faces of sheet piles
4. Bligh did not explain any effect of sheet piles length and their distances on exit
gradient.
5. Head loss is not directly proportional to creep length; Also uplift pressure
distribution is not linear.
6. Bligh did not specify the necessity of providing downstream end sheet pile.
A. N. Khosla’s Theory
After studying a lot of dam failure constructed based on Bligh’s theory, Khosla
came out with the following;
Seeping water below a hydraulic structure does not follow the bottom profile of
the impervious floor as stated by Bligh but each particle traces its path along a
series of streamlines.
H1
H2
H1
H2
Uplift pressure on foundation
168

For steady flow, seepage in a homogeneous soil obeys the Laplacian equation :
0
2
2
2
2
=
+
dz
Q
d
dx
Q
d
---------------------- ( )
where Q = K . h - flow potential
h = residual head at any point within the soil and K coefficient of
permeability as defined by Darcy.
The above equation represents two sets of curves STREAMLINES and
EQUIPOTENTIAL (VELOCITY POTENTIAL) LINES which intercept each other
orthogonally.
Streamline
A streamline is the path traced out by particles of water seeping through the
subsoil. Every particle traces out its own path which represents the streamline for
that particle.
The first streamline, immediately below the hydraulic structure follows the bottom
profile of the structure and will be almost the same as Bligh’s creep path. The
others streamlines will be curves as they proceed through the pervious soil. If an
impervious boundary intervenes, then the last streamline near the boundary
follows the boundary.
Equipotential Lines.
Let us consider that there is no water on the downstream side. If we assume the
downstream bed as the datum, then the on the upstream side is h. All the floor at
the upstream side AB and every particle of water entering the subsoil on the
upstream side will be having a head h. This head will be gradually utilize as the
particle traces out the streamline and when it emerges at the downstream end,
the entire head h will have been utilized so that the head at the exit end will be
Streamlines
A B C D
Equipotential
lines
Flow network
h
169

zero. At any intermediate position along the streamline, the particle will be
having a head h1 which is the residual head; which will be dissipated by the
particle while traveling the remaining path on the streamline. Thus every
streamline moving from a head of h to a head of zero will pass through a residual
head of h1. If all points on all streamline with a residual head of h1 are joined, we
shall obtain a curve called an equipotential line. Thus equipotential line is a line
joining points of equal residual head.
Flow Nets
The streamlines and equipotential lines always intercept at right angles. In a flow
field, the combination of the two groups of lines form a network in the flow field
called the flow net. The space enclosed between any two adjacent streamlines
and equipotential lines is known as a field. Though it is possible to draw an
infinite streamlines and equipotential lines, for the sake of practical convenience
and easy interpretation, only a limited number is usually drawn in such a way that
every field becomes an elementary square.
By drawing the flow net, all characteristics such as flow rate, velocity etc can be
obtained.
Exit and Critical Gradient
Every particle of water while seeping through the sub-soil, at any position will
exert a force f, which will be tangential to the streamline at any point. As the
streamlines bend upward, the tangential force f will be having a vertical
component f1. Also at that point, there will be a downward force W due to the
submerged weight of the soil particle. Thus at that point there will be two forces
on the particle; one upward vertical component of f, and the other, the
submerged weight. It is evident that if the soil particle is not to be dislodged, then
the submerged weight must be greater than the upward vertical component of f.
h
170

The upward vertical component force at any point is proportional to the water
pressure gradient dp/dx.
Hence for stability of the soil and for the prevention of erosion and piping, the
seeping water when it emerges at the downstream side, at the exit position, the
force f1 should be less than the submerged weight W. In other words the exit
gradient at the downstream end must be safe.
If at the exit point at the downstream side, the exit gradient is such that the force
f1is just equal to the submerged weight of the soil particle, then that gradient is
called critical gradient. Safe exit gradients = 0.2 to 0.25 of the critical exit
gradient.
Values of safe exit gradient may be taken as:
0.14 to 0.17 for fine sand
0.17 to 0.20 for coarse sand
0.20 to 0.25 for shingle
Method of Independent Variable of Khosla.
For the determination of seepage below the foundation of hydraulic structure
developed the method of independent variable.
In this method, the actual profile of a weir which is complex, is divided into a
number simple profiles, each of which cab be solved mathematically without
much difficulty.
The most useful profile considered are:
i) A straight horizontal floor of negligible thickness provided with a
sheet pile at the upstream end or a sheet pile at the downstream
end.
ii) A straight horizontal floor depressed below the bed, but without any
vertical
cut-off.
C
H
d
D
E
b
Sheet pile at u/s end
E C
d
D
b
H
Sheet pile at d/s end
b1
171

ii) A straight horizontal floor of negligible thickness with a sheet pile at
some
intermediate point
The mathematical solution of the flow-nets of the above profiles have been given
in the form of curves. From the curves, percentage pressures at various key
points E, C, E1, C1 etc) be determined. The important points to note are:
i) Junctions of pile with the floor on either side{E, C (bottom), E1, C1
(top) }
ii) Bottom point of the pile (D), and
iii) Junction of the bottom corners (D, D’) in case of depressed floor
The percentage pressures at the key points of a simple forms will become valid
for any complex profile, provided the following corrections are effected:
i) correction for mutual interference of piles
ii) correction for the thickness of floor
iii) correction for slope of the floor.
Sheet pile at intermediate position
Depressed floor
b
D1
D
d
b1
b
d
H
E C
H
172

Correction for Mutual Interference of Piles
Let b1 = distance between the two piles 1 and 2, and
D = the depth of the pile line (2), the influence of which on the neighbouring
pile (1)
of depth d must be determined
b = total length of the impervious floor
c = correction due to interference.
The correction is applied as a percentage of the head
174






 +
=
b
D
d
x
b
D
C
1
19 --------------------------- ( )
This correction is positive when the point is considered to be at the rear of the
interfering pile and negative for points considered in the forward or flow direction
with the interfering pile.
For example, correction for pressure at C1 for pile line (1) by the interference of
pile line (2) is positive as pile line (1) is to the rear of the interfering pile line (2).
Similarly, correction for pressure at E2 for pile line (2) due to the interference of
pile line (1) is negative, because E2 is in the forward or flow direction of
interfering pile line (1).
The interference effect will not be present on the intermediate pile if:
i) the outer pile is equal to or longer than the intermediate pile and
175

ii) if the distance between the intermediate pile and outer interfering
pile is less than twice the length of the outer pile
Correction for Floor Thickness
Standard profiles assuming the floors as having negligible thickness. Hence the
values of the percentage pressures computed from the curves corresponds to
the top levels (E1*, C1*) of the floor. However, the junction points of the floor and
pile are at the bottom of the floor (E1, C1)
The pressures at the actual points E1 and C1 are interpolated by assuming a
straight line variation in pressures from the points E1* to D1 and from D1 to C1*
b1
bs
E1
D1
C1
d
1 2 3
D3
b
D
D2
C3
D3
E1
* C1
*
E1
C1
D1
Depressed floor
176

The corrected pressures at E1 should be less than the computed pressure t E1*.
Therefore the correction for the pressure at E1 will be negative. And so also is for
pressure at C1.
Correction for Slope of Floor.
A correction for a sloping impervious floor is positive for the down slope in the
flow direction and negative for the up slope in the direction of flow.
No. Slope =Ver:Horiz Correction
as % of
pressure
1 1:1 11.2
2 1:2 6.5
3 1:3 4.5
4 1:4 3.3
5 1:5 2.8
6 1:6 2.5
7 1:7 2.3
8 1:8 2.0
The correction factor must be multiplied by the horizontal length of the slope and
divided by the distance between the two poles between which the sloping floor
exists.
In the diagram above, correction for slope can be applied only to point E2. As the
point E2 is terminating at the descending slope in the direction of flow, the
correction will be positive. The value of correction will be:
C.F. x bs/b1
Where C.F. =correction factor
bs = horizontal length of sloping floor
b1 = horizontal distance between the pile lines
Exit Gradient
H
d
b
177

For the standard form consisting of a floor of a length b, and a vertical cut-off of
depth d, the exit gradient at its downstream end is given by:
exit gradient GE = (H/d) x
λ
π
1
By referring to plate (17.3 ), for any value of α = b/d, the corresponding value of
λ
π
1
can be read off. When H and d are given, GE should be easily calculated.
The value of exit gradient GE should be within safe limits as given below.
No. Type of soil Safe exit Gradient
1 Fine sand 1/7 to 1/6
2 Coarse sand 1/6 to 1/5
3 Shingle 1/5 to 1/4
Use of Khosla’s Pressure Curves.
First consider plate 17.1. Values of 1/α = d/b are plotted as abscissa and
percentage pressure φ = P/H x 100 are plotted as ordinates. There are three
curves, one for QD and one for QE for sheet piles fixed at the ends of the floor and
the last one for φD for depressed floor
To find the percentage pressure at the points C1 and D1 of the upstream pile,
calculate
Αα = b/d and then 1/α = d/b
For this value of 1/α determine the value of φD from the curve
Then subtract this value of φD from 100 to get φD1
ΦD1 = 100 – φD
Similarly, determine the value of φE for the value of 1/α. Subtract this φE from
100 to get φC1 φC1 = 100 – φE
Example: Let b = total length of floor = 60.0 m and d = depth of u/s pile = 6.0
Then 1/α = d/b = 6/60 = 0.1
For this value of 1/α, we read the values of φD = 20 and φE = 28.
Then φD1 = 100 – 20 = 80
φC1= 100 – 28 = 72
These percentages pressures must be corrected for mutual interference.
Next, let us consider plate 17.2. This gives pressure curves for φC, φD for sheet
pile not at end condition. The ratios b1/b are plotted as abscissa and φC = Pe/H x
178

100 plotted as ordinates on the left portion of graph and φD = Pe/H x 100 plotted
on the right side of graph for different values of α.
To find φE for any value of α and base ratio b1/b, read φC for base ratio (1-b1/b) for
that value and subtract this value from 100.
Thus for finding φE for base ratio b1/b = 0.3 and α = 3.0
(1 – b1/b) = 1 – 0.3 = 0.7 for α = 3.0
Qc = 20
ΦE = 100 – Qc = 100 – 20 = 80%
ΦD can be calculated for values of b1/b less than 0.5.
Similarly finding φD for b1/b = 0.13 and α = 2.0
(1 – b1/b) = 0.87
Corresponding to 0.87 and α = 2.0 value of φD’ = 40
ΦD = 100 – 40 = 60%
Example. Determine the percentage pressures for various key points shown in
diagram below. Calculate the exit gradient and plot the hydraulic gradient line for
the pond level on the u/s side with no water on downstream side.
Solution:
1. Upstream pile No. 1
Total length of floor b = 60 m
Depth of u/s pile below the top of floor = 105 – 99 = 6.0 m
α = b/d = 60/6 = 10.0
1/α = 0.1
▼107
▼100
60m
(3)
▼102.0
▼94.0
45m
6 6
6 2 25
(1)
(2)
▼110
▼105.0
▼99
14
D1
D2
D3
▼104
0.5m
0.5m
179

From plate 17.1, φC1 = 100 – φE = 100 – 28 = 72 %
ΦD1 = 100-20 = 80 %
For these values, three corrections must be applied .
i) Correction for mutual interference of piles, for pressure at C1
φC1 is affected by the presence of intermediate pile No. (2)
Let D = depth of pile No. (2) = 104.00 – 99.00 = 5.0 m
d = depth of pile No. (1). = 104.00 – 99.00 = 5.0 m
b1 = distance between the two piles = 14.0 m
b = total length of floor = 60.0 m
Then the correction =
)
%(
9
.
1
1
.
0
6
.
0
19
60
5
5
.
14
5
19
.
19
1
ve
x
x
b
D
d
b
D
+
=
=





 +
=





 +
Point C1 is to the rear of the next pile No. (2). Hence correction is positive.
ii) Correction at C1 due to thickness of floor.
φC1 calculated from curves corresponds to point C1
!
At the top of the floor. But we
want the pressure at C1 at the bottom of the floor. Pressure atC1 will be more
than that at C1
!
, therefore the correction will be positive.
Pressure at C1
!
= φC1 = 72%
Pressure at D1 = φD1 = 80%
Distance from C1
!
To D1 = 1050. – 99.0 = 6.0 m
Hence pressure correction at C1 = (80 – 72) x (105.0-104.0)
(105.0 – 99.0)
= 8/6 x 1 = 1.67 % (+ve)
iii) Correction due to slope at C1
As C1 is neither the starting point nor terminating point of a slope, slope
correction is zero.
Hence, corrected pressure at C1
ΦφC1 = 72% + 1.9% + 1.67% = 75.57%
C1
!
E1
C1
1.0
D1
D2
▼99.0
▼105.0
E2
C2
E2
!
C2
!
180

ΦD1 = 80%
2. Intermediate Pile Line No. 2
d = 105 – 99 = 6 m
b = 60 m
α = b/d = 60 / 6 = 10
b1 = 0.5 + 14 = 14.5
b1/b = 14.5 /60 = 0.24
(1 – b1/b) = 1 – 0.24 = 0.76
For b1/b = 0.76, and α = 10, φC = 27% and 100 – 27 = 73% = φE2
For base ratio 0.24 and α = 10,
ΦC2 = 60%
For base ratio 0.76 and α = 10,
ΦD = 34% and φD2 =100 – 34 = 66%.
Corrections for φE2
i) Correction for sheet pile lines at E2.
Pile No. (1) affects the pressure at E2 (φE2). As E2 is in a forward direction with
respect to pile (1), the correction will be negative.
D = depth of pile No.(1), the effect of which on E2 is considered.
= 104 – 99.00 = 5.0 m
d = depth of pile No. (2) which is affected = 104 -99.0 5.0 m
b1 = distance between the two pile lines = 14m
b = Total floor length = 60.0m
Correction
)
%(
9
.
1
60
5
5
14
5
19
19
1
ve
b
D
d
b
D
−
=





 +
=





 +
=
ii) Correction at E2 due to floor thickness
Correction = observed value of φE2 – observed φ D2 x Thickness of floor
Distance E2D2
= 73 – 66 x 1.0 = 7 x 1.0 = 1.17 % (-ve)
105.0.99.0 6
Pressure is observed at E2
!
and the direction of flow is from E2
!
To E2. Hence,
pressure at E2 will be less than that at E2
!
. Correction is therefore negative.
iii) Correction at E2 due to slope is zero as E2 is neither the beginning nor the end
of a slope.
Hence corrected pressure at E2 = (73% - 1.9% - 1.17%) = 69.93%
Corection for φC2
181

i) Correction at C2 due to pile interference. Pressure at C2 is affected by
pile line (3) and as C2 is to the rear of pile line (3), correction will be
positive. The amount of this correction is given by:
Correction 




 +
=
b
d
D
x
b
D
1
Where D = depth of pile No. 3, the effect of which is considered below
the level at which the interference is desired = 104.0 – 99.0 = 10 March
2007
d = depth of pile No. (2), which is affected = 104.0 – 99.0 =
5.0m
b1 = distance between pile (2) and pile (3) = 45m
b = total floor length = 60.0m
Correction
)
%(
28
.
2
60
15
9
2
19
60
5
10
45
10
19
ve
+
=






=





 +
=
ii) Correction at C2 due to floor thickness. From the diagram, it can be
observed that the pressure at C2
!
is less than that at C2 due to the
direction of seepage
flow. The amount of correction is the same as that for E2 and equal to
1.17%
+ve
iii) Correction at C2 due to slope. C2 is at the commencement of a up
slope of 3:1 in the direction of flow and the correction will be negative.
Correction factor 3:1 slope = 4.5 (see table of correction for slope
above)
Horizontal slope length = 6m
Distance between the two pile lines (2) and (3) between which the
sloping floor exists = 45.0 m
Correction = correction factor x horizontal slope length
distance between piles
= 4.5 x 6/45 = 0.6% (-ve)
Hence corrected pressure φC2 = (60% + 2.28% + 1.17% - 0.6% =
62.85%
3. Downstream Pile Line.
d = 102.00 – 94.00 = 8.00m
b = 60m
1/α = d/b = 8/60 = 0.133
Referring to curves of plate 17.1, we find
ΦD3 = 22%
ΦE3 = 32%
182

Note that for downstream pile at end, the values of φD and φE are taken directly
from the curves for the required value of 1/α.
a) Correction for φE3
i) Correction due to mutual interference of pile. The point E3 gets affected by
pile No.2. E3 is in the forward direction of flow with reference to pile No. 2,
correction is negative. Value of correction 




 +
=
b
d
D
x
b
D
1
where D = depth of pile No. 2 which influences point E3 of pile No. 3, the depth
being considered below the level at which interference is desired = 100.0 –
99.0 = 1.0m
d = depth of pile No. 3 which is affected = 100.0 – 94.0 = 6.0m
b1 = distance between the two piles = 45.0m
b = total floor length = 60 m
therefore correction = )
%(
262
.
0
60
6
1
45
0
.
1
ve
x −
=





 +
=
iii) Correction due to floor thickness. From the direction of flow of
seepage water, it is clear that pressure at E3 at the bottom of the
floor is less than that at E3
!
, top of the floor which is the value got by
Khosla’s curves. Hence, correction to be applied for φE3 should be
negative.
Value of this correction = 32% - 22% x 2.00
(102.0-94.0)
= (10 x 2)/8 = 2.5% (-ve)
iv) Correction due to slope. Point E3 is neither starting point nor the
terminus of a slope. Hence, slope correction = 0
Hence, corrected pressure at E3 = φE3=(32%-0.262%-2.5%)=
29.238%
The corrected pressures at all the key points can be shown in the table below
Upstream pile No.
1.
Intermediate pile No.
2.
Downstream pile No.
3.
ΦE1 = 100.0% ΦE2 = 69.93% ΦE3 = 29.24%
ΦD1 = 80.0% ΦD2 = 66% ΦD3 = 22.0%
ΦC1 = 75.57% ΦC2 = 62.85% ΦC3 = 0 %
Exit Gradient
Pond level = 110.0
With no water on downstream, maximum head causing seepage H = (110.0–
102)= 8.0m
Depth of downstream cut–off = d = (102.0 – 94.0) = 8.0m
Total floor length = b = 60m
α = b/d = 60/8 = 7.5
Referring to curve of plate 17.3, for α = 7.5,
183

16
.
0
1
=
λ
π
Hence, exit gradient GE = (H/d) x
λ
π
1
= (8/8) x 0.16 = 0.16
This exit gradient is safe even for fine sand.
184

Chapter 9
CROSS DRAINAGE WORKS
Whenever a canal intercepts a natural or artificial drainage in its passage, cross
drainage works have to be provided.
Types of Cross-Drainage Works
Three main possibilities may occur when a canal has to cross a natural drainage:
1. When the canal bed level is higher than the bed level of the river, the
cross-drainage work constructed is known as aqueduct
2. When the bed level of the natural drainage is higher than that of the canal,
the cross-drainage work constructed is called a super passage
3. When the bed levels of the canal and the natural drainage are almost the
same, and the waters of both the canal and natural drainage are mixed up
while crossing, such cross-drainage works are known as level crossing.
If the bed level of the canal is not very much higher than that of the drainage,
then the cross-drainage work is called siphon aqueduct
Similarly, if the difference between the bed level of the canal and natural drainage
in not very much, the natural drainage being at a higher level, then the cross-
drainage work is called siphon super passage.
The selection of any particular type of cross-drainage work depends on the
following:
1. Bed level of canal and bed level of natural drainage;
2. Discharge of the canal and the drainage
3. The foundation conditions of the site
4. Availability of materials and labour
5. availability of existing modes of communication.
Aqueduct
Canal
Drain
Aqueduct
Canal
Drain
Super passage
185

An aqueduct is a structure constructed at the position, where a canal has to
cross a drainage at a sufficiently high level. It is very similar to a bridge but
instead of a roadway or railway, canal water flows above a natural drainage.
Irrigation water may be taken through a pipe over the drainage water if the
section of the irrigation canal is very small. The pipe is supported in position at
the ends by masonry walls and in between by supports
There are three types of aqueducts:
Type I. In this type of aqueduct, the sides of the aqueduct are completely
earthen embankment with full earthen slopes. The length of the culvert through
which the discharge is passing should be adequate to accommodate the water
way of the canal and also the bottom width of the side embankments. In this
case the original canal section is retained and no fluming of canal section is
done.
Type II.
Type II is very much like type one except that the outer section is reduced by
supporting it by masonry or concrete wall.
Type III.
In this type of aqueduct, the canal banks are discontinued at the aqueduct
portion and the canal water is transported in a masonry or concrete trough
constructed above the drainage culvert. In this case the length of the culvert is
reduced because fluming is done. This type becomes necessary where big
canals have to cross big natural drainage with large catchment areas.
Selection of a suitable type. In order to select a suitable type, it is necessary to
understand the following terms:
Pipe
Tail wall
Head wall
Support
186

1. Culvert length: is the width of the aqueduct, measured along the drain.
The culvert depends upon the shape and size of the canal section
2. Length of aqueduct: is the length measured perpendicular to the drain. It
is evidently equal to the width of the drain
3. Bank connection: consist of masonry (concrete) wings, etc required to
connect the regular section of the canal to its modified section over the
aqueduct.
In the type I, since the canal is fully in earthen section, the culvert length is
maximum. Hence the cost per unit length of aqueduct will be maximum.
However, bank connections are not require. Hence, the selection of this type
depends on the relative cost of bank connections and that of the aqueduct
proper. In all cases the cost of bank connection is independent of the length of
the aqueduct. Therefore, the type I is suitable when the length of the aqueduct is
small and the cost of bank connections would be large in comparison to the
saving from the reduction in the width of work if type III were adopted. On the
contrary, in type III, the culvert length is minimum. Hence the cost per unit length
of the aqueduct is minimum. However, the cost of bank connections will be
additional. Therefore, type III is suitable where the length of culvert is large. Type
two is suitable for intermediate conditions.
Design of Cross-Drainage Works
The following are some of the important features of design of cross-drainage
works:
A. From hydraulic consideration, the following are important
i) determination of the maximum flood discharge and the high flood level
(H.F.L)
ii). Determination of the stable waterway of drain
iii). Contraction of the canal waterway.
iv). Discharge and head losses through the cross drainage works
v). Determination of uplift pressure on the roof of trough
vi). Determination of uplift pressure on the floor.
vii). Design of bank connections
B. Structural design.
1) Design of cross-section of aqueduct trough.
ii). Design of piers and abutments
iii). Design of foundation.
i) Maximum Flood Discharge of Drain: This is determined by any one of the
known methods depending on size of catchment and other available data
ii) Determination of the Waterway of drain: This can be determined by using
Lacey’s regime perimeter equation
Pw = 4.75 Q ½
where Pw = wetted perimeter of the drain (m)
Q = total discharge in drain (cumecs)
In large drains, Pw can be taken as equal to width of river, and hence equal to the
waterway required.. In small drains, a contraction of up to 20% of the waterway is
187

permissible i.e the length between abutments may be increased by 20% to
compensate for contraction of waterway due to the width of the piers.
iii) Velocity of flow through the barrel. This may be taken between 2.0 -3.0 m/s.
Lower velocities may result in silting in the barrel and higher velocities will cause
abrasions of
barrel surface by rolling grit.
Iv0. Height of Opening: The depth of flow in the barrel can be calculated as the
waterway discharge and velocity are fixed. Sufficient margin (free board) should
be provided between the H>F>L and the bottom of canal bed.
Total height of opening = (Depth of flow in barrel + clearance (freeboard)
v). Number of Spans: The total width of the culvert or total length of aqueduct
already been fixed. The number of spans for the culvert should be fixed
considering structural strength and economy of design. More spans may be used
in case of arched culverts.
vi). Contraction of canal waterway: Canal waterway in the aqueduct portion is
reduced in type III by fluming. Fluming ratio is the ration between the restricted
width of canal in the aqueduct portion and the normal width of the canal. This is
taken as ½. However, this reduction in width should not results in the velocity in
the trough exceeding 3m/s or exceeding the critical value bringing the flow to
supercritical flow, which could result in the formation of hydraulic jump in trough –
not desirable.
vii). Length of Contraction: This is the horizontal distance in which the normal
width of the canal is gradually reduced to the contracted width in the aqueduct
portion. A convergence of 2 horizontal to 1 lateral could be assumed.
Viii). Length of expansion: the expansion ration is assumed as 3 horizontal to 1
lateral. This decides the length of the expansion downstream side of the
aqueduct.
ix). Bank connections : these consist of two sets of wing walls for the canal called
canal wings and two sets of wing walls for the drainage called the drainage
wings.
Canal wing walls. These are to be provided on the upstream and downstream
sides of the aqueduct. They protect and retain the earth in the canal banks. The
wings should be constructed on sound foundation.
The Drainage wing are provided on the upstream and downstream sides of the
barrel. They protect and retain the natural slopes of the drain and increase the
seepage path and control the exit gradient. The drainage wings should be taken
deep into the foundation, well below the maximum scour depth. They must be
taken back well into the top of the guide banks.
Example. Design a suitable cross drainage for the following particulars:
a) Drainage particulars:
Catchment area for the drain = 20.0 km2
; value of “C” in Dicken’s formula = 20
Gauge reading at the site of intersection = 1.0m during maximum flood; Bed
level of drainage = 100.00m
b) Channel Particulars
188

Full supply discharge = 20cumecs; full supply depth = 1.5m; Bed width = 10.0m;
Side slopes = 1.5:1. The canal has to be provided with inspection paths of 3.0 m
width on both the banks. The channel is unlined. Bed level of the channel =
103.00m. Good foundation is available at site. Missing data may be suitably
assumed.
Solution:
The bed level of the canal = 103.0m
The bed level of the drainage = 100.0m
The High Flood Level in the drainage = 1.0m so that level of H.F.L = 101.0
The difference between canal bed level and H.F.L = 103.0 – 101.0 = 2.0m
Hence an aqueduct would have to be designed.
Determination of flood discharge: by Dicken’s formula
Q = C . M3/4
Where C = is a constant = 20; M = catchment area = 20.0km2
CULVERT LENGTH
DRAIN
Canal
Road
H.F.L
TYPE I AQUEDUCT
189

Q = 20 x 203/4
= 189.2cumecs ≈ 190 cumecs
Waterway for drainage:
By Lacey’s equation, the wetted perimeter Pw = 4.75 Q1/2
Pw = 4.75 x 1901/2
= 65.47m
iv) Smaller spans can be adopted, as the foundation is good
Assuming a trapezoidal drain of side slopes of 1.5 :1 ( 1.5 horizontal to 1.0
vertical ) the bed width of drain b = 65.47- 2 x 1.5 x1= 62.47m
Provide 12 spans of 5.5m each , to give a total lineal waterway of 12 x 5.x =
66.0m
iv) Thickness of pier at the springing level of the arch:
This can be calculated by t = 0.552 x s1/2
= 0.552 x (5.5)1/2
= 1.294 = 1.3
m
Where s = span of arc (m); t = thickness (m)
A batter of 1 in 15 is given to the pier
v) Total height of the pier
Height of pier above the drainage bed = 1.0
The springing level of the arch will be at H.F.L.
To determine the depth of pier below the bed level of the drainage, the scour
depth is worked out.
Scour depth as per Lacey’s formula
R1 = 1.35 (q2
/f)1/3
Where R1 = scour depth (m), q = discharge per metre width of the drain (m3
/s /m);
f = Lacey’s silt factor, taken as 1.6
R1 = 1.35 {(190/66)2
x 1/1.6}1/3
= 2.4 m
Road
H.F.L
CULVERT LENGTH
DRAIN
Canal
TYPE II AQUEDUCT Masonry wall
190

The pier must be taken 1.5 times the scour depth below the bed of drain,
which is 1.5 x 2.4 = 3.6 m.
therefore total height of pier hp = 3.6 + 1.0 = 4.6 m
vi). Total length of the aqueduct between the abutments LAB.
The pier pose the obstruction to the flow of water and reduce the waterway.
Hence, clear waterway must be increased. This is increased by 20% or the total
thickness of all the piers, whichever is more.
a) 20% of waterway = 20.100 x 66.5 = 13.3 m
b) Total thickness of piers = 11 x 1.3 = 14.3 m (12 spans, so there are 11
piers)
Hence we take 14.3 m for compensating for reduction of waterway by piers.
Therefore, total length of aqueduct between the abutments LAB = 66 + 14.3 =
80.3 m
vii) Rise of arch:
Rise of arch r = span / 5 = 5.5 / 5 = 1.1 m
viii) Radius of arch
Considering the triangle AOB, AO = (R-r); AB = s/2 where s is the span and OB
= R
Therefore using the Phytagoras theorem
AO2
+ AB2
= OB2
(R-r)2
+ (s/2)2
= R2
Knowing r = 1.1 and solving the equation, R = 4.0 m
ix) Thickness of the arch at the springing level. By Trantwine’s formula;
ta = 0.22R + 0.11r + 0.5 = 0.22 x 4.0 + 0.11 x 1.1 + 0.5 = 1.5 m
The water face of the abutment may be kept vertical while the back face (earth
face ) may be given a slope or batter of 1:5 (1 horizontal to 5 vertical).
The canal over the drain is flumed with a fluming ratio of 1:2
Hence, bed width of the canal over the drain bc/d = 10 x ½ = 5.0 m
r
A
B
O
R
s/2
191

The section over the drain will be rectangular and is to be built in R.C.C. The
height of side walls is kept 0.5 m higher than the full supply discharge F.S.D
x) Transitions
Upstream transition is done in 2:1 splay in contraction
Therefore length of contraction transition = (10 – 5)/2 x 2 = 5 m
Upstream transition is done in 3:1 splay expansion
Length of expansion = (10 – 5)/2 x 3 = 7.5 m
xi) An inspection path of 3.0 m width is provided on both sides of the aqueduct.
FLUMING OF THE CANAL
The reduction in waterway of the canal at the aqueduct is known as fluming of
the canal. Fluming reduces the barrel length or the width of aqueduct, and
thereby makes it economical. Fluming is only done for aqueduct type III.
The maximum amount of fluming is dictated by the velocity which can be kept in
trough, because it must not exceed the critical value.
After fixing the canal section and the flumed canal section, the designing of the
transition is done for making smooth entry and exit, and avoiding the formation of
eddiesThe slope of the upstream side should not be more than 30o
(ie 2:1 splay)
and that of the downstream side must not be more than 22.5o
(i.e 1:3 splay). The
normal canal section is trapezoidal, whereas flumed section is rectangular.
Design of Channel Transition
Any of the following methods may be applied for the design:
1. Chaturvedi’s method (when the water depth remains constant)
2. Hind’s method ( when the water depth may or may not vary)
3. Mitra’s method (when water depth remains constant).
1. Chaturvedi’s Method:R. S. Chaturvedi (1963) proposed the following
equation for the design of transition with constant water depth.
















−
−
=
2
/
3
2
/
3
2
/
3
2
/
3
1
.
x
f
f
o
o
B
B
B
B
B
L
x --------------------- ( )
where Bx = channel width at any section X-X at a distance x from the flumed
section.
L = length of transition
Bo = Bed width of the normal channel section
Bf = bed width of the flumed section.
192

ii) Hind’s Method.
This method can be used when the water depth in the trough and the normal
section of the canal vary.
Let V1, V2, V3, and V4 be the velocities of the canal at sections A, B, C. D. Let y1,
y2, y3, y4 be the depths of canal at different sections.
Step I. Let the bed level and cross-section of the canal at section DD be known.
Water surface level (W.S.L.) at section DD = Bed level at section DD + y4)
Total energy line (T.E.L) at section DD = (water surface elevation at DD + V4
2
/2g)
Step II. The energy loss due to expansion of section between section CC and
DD may be taken as equal to 0.3(V3
2
-V4
2
)/2g
Step III. Total energy line at section CC = {T.E.L. at section DD + 0.3(V3
2
– V4
2
) /
2g}
Water surface elevation at section CC, = {T.E.L. at section CC – V3
2
/2g}
Bed level at section BB = {T.E.L at section BB – V2
2
/2g}
Step IV The channel section in the trough from section CC to BB remains
constant. The only loss of head in this section is due to pipe friction, which can
be computed by the Manning’s formula
Q = i/n . A R2/3
(I)1/2
Therefore T.E.L at section BB = (T.E.L at section CC + loss of head)
W.S.E at section BB = (T.E.L at section BB – V2
2
/2g)
Bed level at section BB = (W.S.E. at section BB – y2)
Bf
Bo
Bx
Contraction
Transition
Expansion
Transition
C
B
B
v1
2
/2g
y1
y2
v2
2
/2g v3
2
/2g
v4
2
/2g
y4
L
PLAN
SECTION
A
A
D
D
C
193

Step V. The loss of energy between section AA and BB due to contraction
= 0.2(V2
2
– V1
2
) / 2g
T.E.L at section AA = [T.E.L. at section BB + 0.2(V2
2
– V1
2
) / 2g]
W.S.E at section AA = [T.E.L at section AA – V1
2
/2g]
Bed level at section AA = W.S.E. – y1]
Step VI After calculating the T.E.L and W.S.E and bed levels of all sections, the
total energy line may be drawn and the bed line also.
Step VII. The drop in the W.S.L. between the two adjacent sections is due to:
i)drop in the energy line between the two sections’
ii) increase velocity head for contraction and decreased velocity head for
expansion.
Step VIII) When the water surface profile has been plotted over the whole length,
the velocity head can be determined by measuring the vertical distance between
the T.E.L and the water surface line at any point.
The velocity head can be converted into equivalent velocity by V = (2gh)1/2
Step IX. Discharge at any point,
Q = A.V.
With the velocity known, and the flow also available, the dimensions of the
section can easily be calculated.
If the section is rectangular, then A = b x h
If the section is trapezoidal, A = (b + mh)h, where m = side slope =cot θ; θ =
angle of drain.
SIPHON AQUEDUCT
In siphon aqueduct, the difference between the bed level of the drainage and the
bed level of the canal is not much so the bed level of the drainage is depressed
at the site of the crossing so that there is sufficient clearance between H.F.L of
the drainage and the bed of the canal. The drainage water passes below the
canal through the depressed portion which makes it works like siphon.
Design consideration for a siphon aqueduct
Siphon aqueduct is a bit different from ordinary aqueducts. Therefore in addition
to the design considerations of ordinary aqueduct, the following additional
considerations become necessary:
1. Discharge through the siphon: The head causing flow through the siphon
portion of the barrel can be obtained by Unwin’s formula as follows:
g
V
g
V
R
L
f
f
h a
2
2
.
.
1
2
2
2
1 −






+
+
=
194

where h = head causing flow of the loss of head in the barrel
L = length of barrel,
V = velocity of flow through the barrel.
Va = approach velocity,
R = barrel radius
f1 = entry loss coefficient of barrel; = 0.505 for unshaped mouth and 0.08
for bell
mouth.
f2 = pipe friction loss coefficient given by f2 = a(1 + b/R) where the values
of a and b for different materials may be taken from the table below
Nature of barrel surface Value of “a” Value of “b”
1. Smooth iron pipe 0.00497 0.025
2. Encrusted pipe 0.00996 0.025
3. Smooth cement plaster 0.00316 0.030
4. Brick work 0.00401 0.070
5. Rubble masonry or stone
pitching
0.00507 0.250
After fixing the velocity through the barrel of the siphon the head ‘h’ required to
generate that much velocity can be determined by equation above. As the
downstream H.F.L of the drain remains unchanged, the upstream H.F.L. can be
obtained by adding ‘h’ to the downstream H.F.L. ‘h’ is known as the afflux.
2. Uplift Pressure on the floor of the barrel:
d/s H.F.L u/s
F.S.L
Cross – section of typical siphon aqueduct
u/s H.F.L
d/s H.F.L.
Retaining walls
Canal banks
F.S.L
Arch thickness
195

Since the barrel is depressed below the bed level of the drainage, it will be below
the normal water table in the surrounding area. Due to this, a static uplift
pressure will be exerted on the barrel.
3.Uplift Pressure on the roof of the barrel:
When the barrel is running full, the water in the barrel exerts an upward thrust on
the roof of the barrel. The pressure head on the downstream side of the barrel
will be equal to the height of the water level above the roof of the barrel. The
pressure head on the upstream side of the barrel will be equal to the sum of the
loss of head in the barrel and the pressure head on the downstream side.
Worked Example: Design a siphon aqueduct with the following data:
i) Canal discharge = 30 cumecs
ii) Canal bed width = 22 m
iii) Water depth = 1.5 m
iv) High flood drainage discharge = 420
cumecs
v) Bed level of drainage = 92.50 m
vi) Bed level of canal = 94.50 m
vii) Ground level = 94.50 m
viii) High flood level of drainage = 94.50 m
ix) Canal bank side slope =1.5 : 1
Solution:
The flow is large and therefore the drainage size is assumed large and therefore
we choose type III aqueduct which might be economical. The canal at the
aqueduct shall be flumed and taken in concrete trough.
1. Design of drainage water-way:
Wetted perimeter of the drainage P = 4.75 (Q)1/2
= 4.75 x (420)1/2
= 97.35
m
Provide 12 spans of 6.5 each, separated by 11 No. piers of 1.5 m thick.
Length of clear water way = 12 x 6.5 = 78 m
Length occupied by piers = 11 x 1.5 = 16.5 m
Total length of water way = 78 + 16.5 = 94.5 m
Keeping velocity through siphon barrel = 2.0m/s (concrete material)
Height of barrel required H = Q/(V x B) = 420/(2 x 78) = 2.69 m
Provide size of barrel 6.5m x 2.7 m
Actual velocity through barrel, = 420/(6.5 x 2.7 x 12) = 1.99m/s
Design of Canal Water-Way.
Bed width of canal = 22m
Let the width be reduced to 12 m in trough
Providing a splay of 2:1 in contraction, the length of contraction transition
196

= (22 – 12)/2 x 2 = 10 m
Providing a splay of 3 : 1 in the expansion the length of the expansion transition
= (22 – 12) /2 x 3 = 15 m
Length of the flumed portion from abutment to abutment
= total drainage water-way = 94.5 m
Design of Levels at Various Sections {refer to diagram on page (8)}
a) Section DD. Cross-section area A = B + 1.5h)h = (22+1.5x1.5)x1.5
=36.75 m2
Canal velocity = Q/A = 30/36.75 = 0.82m/s
Velocity head = V2
/2g = (0.82)2
/2x9.81 = 0.034m
Therefore Relative level (R.L) of water surface
= R.L of bed + water depth = 94.5 +1.5 = 96.0 m
Relative level of the Total Energy Line (T.E.L) = 96.0 + 0.034 = 96.034 m
b) Section CC. Cross section area of canal = 12 x 1.5 = 18 m2
Velocity = Q/A 30/18 = 1.67 m/s
Velocity head = V2
/2g = (1.67)2
/2 x 9.81 = 0.142 m
Loss of head in expansion from section CC to DD
= 0.3(V3
2
– V4
2
)/2g = 0.3(0.142 – 0.034) = 0.032
Therefore Level of T.E.L at section CC = T.E.L at section DD + Head loss
= 96.034 + 0.032 = 96.066 m
R.L of water surface at CC = T.E.L – velocity head = (96.066 –
0.142)=95.924 m
Therefore R.L. of canal bed for constant depth,
= R.L. of water surface – water depth = (95.924 – 1.5) = 94.424 m
c) Section BB. Hydraulic mean depth in trough
R = A/P = 18/(10 + 2 x 1.5) = 1.384 m
From Manning’s formula, S = V2
n2
/R4/3
Where V = velocity in the trough
S = required slope
N rougosity coefficient = 0.016
S = (1.67)2
(0.016)2
/(1.384)4/3
=4.63 x 10-4
Length of flumed portion of the trough = 94.5 m
Head loss in trough = 94.5 x 4.63 x 10-4
= 0.044 m
Therefore R. L. of T.E.L at section BB = R.L. of T.E.L. at CC + Head loss in
trough =
96.066 + 0.044 = 96.11 m
.. of water surface at BB = (96.11 – 0.142) 95.968
R.L. of bed for maintaining constant depth = (95.968 – 1.5) = 94.468 m
d) Section AA
Loss of head in contraction transition from section AA to BB
= 0.2(V2
2
-V1
2
)/2g
where V1 = V4
and V2 = V3
= 0.2[(1.67)2
– (0.82)2
]/2 x 9.81 = 0.0215 m
R.L of T.E.L at section AA
197

= R.L of T.E.L at BB + head loss = (96.11 + 0.0215) = 96.325m
R.L. of water surface
= R. L. of T.E.L – V1
2
/2g = 96.325 – (0.82)2
/2 x 9.81= (96.325 – 0.034) = 96.291 m
Design of Contraction Transition: The design of the transition will be done on the
basis of Chaturvedi’s formula
















−
−
=
2
/
3
2
/
3
2
/
3
2
/
3
1
.
x
f
f
o
o
B
B
B
B
B
L
x
here Bo = 22 m; L = 10m; Bf = 12 m
or x =
















−
=
















−
−
2
/
3
2
/
3
2
/
3
2
/
3
2
/
3
12
1
74
.
16
12
1
)
12
(
)
22
(
)
22
.(
10
x
x
B B
The values of x for various values of Bx are calculated from the above equation,
and are tabulated below. The distance x is measured from flumed section BB as
shown in page 8.
Bx 12 13 14 15 16 17 18 19 20 21 22
x 0 1.89 3.38 4.76 5.86 6.81 7.62 8.34 8.96 9.5 10
Design of Expansion Transition
















−
−
=
2
/
3
2
/
3
2
/
3
2
/
3
1
.
x
f
f
o
o
B
B
B
B
B
L
x
here, L = 15m; Bf =12; B0 =22
















−
=
















−
−
=
2
/
3
2
/
3
2
/
3
2
/
3
2
/
3
12
1
11
.
25
12
1
)
12
(
)
22
(
)
22
(
15
x
x
B
x
B
x
x
The values of x, for various values of Bx are calculated in the table below.
Bx 12 13 14 15 16 17 18 19 20 21 22
x 0 2.84 5.18 7.14 8.8 10.21 11.44 12.5 13.44 14.26 15
Design of Trough Aqueduct:
Flumed water way of canal = 12.0m
198

The trough shall be divided into two equal compartments by 30 cm thick concrete
wall. The road way 6.0 m wide shall be carried over one of the compartments
The height of trough = 1.5 + 0.5 m free board = 2.0 m.
The entire trough section shall be constructed in monolithic RCC by usual
structural methods. The outer and flow slab of the trough may be made
tentatively equal to 40 cm thick.
The outer width of the trough = 2 x 6 + 2 x 0.4 + 0.30 = 13.1 m
Loss of head through Siphon Barrel.
The loss of head through the siphon barrel by Unwin’s formula
g
V
R
L
f
f
h
2
.
1
2
2
1 





+
+
=
V = velocity through barrel ( = 1.99m/sec)
f1 = coefficient of loss of head at entry ( 0.505)
f2 = a(1 + b/R) a and b are picked from table given above.
a = 0.00316 and b = 0.030
The hydraulic radius, R, = A/P = (6.5 x2.7)/2(6.5 +2.7) = 0.953m
f2 = 0.00316( 1 + 0.030/0.953) = 0.00415
L = 13.1m
Substituting into head loss equation,
H = [1 + 0.505 + 0.00415 x 13.1/0.953](1.992
)/2 x9.81 = 0.311 m
Downstream H. F. L. = 94.5 m,
Therefore upstream H.F.L = d/s H.F.L + head loss = 94.5 + 0.311 = 94.811
Uplift pressure on the barrel roof
R.Lof trough bottom = R.L. of canal bed – thickness of slab = 94.5-0.40= 94.10m
Loss of head at entry of barrel = 0.505 V2
/2g = 0.505 (1.99)2
/2x9.81 = 0.10 m
Therefore uplift on the roof = 94.811 – (94.1 +0.10) = 0.611m
Uplift pressure = 0.611 ton/m2
Dead weight of slab = 0.4x 24 = 0.96 tons/m2
As the dead weight is greater than the uplift pressure, nominal reinforcement is
required in the slab..
When the water leveling the drainage is low, the trough slab is to be designed for
full water load of the canal.
Uplift pressure on the floor of the barrel
a) Static Head
Barrel floor R. L.
= R.L. of bottom of trough slab – height of barrel = 94.1 – 2.7 = 91.4 m
Assuming tentative thickness of the floor = 0.90 m
R.L. of bottom of the floor = 91.4 – 0.90 = 90.5 m
Bed level of drainage = 92.5m (given)
199

Assuming the subsoil water level is up to the bed level of the drain, the static
uplift on the floor, will be = (92.5 -90.5) = 2.0 m
To be continued on 475
200

Chapter 10
DESIGN OF CHANNELS
Channels must be designed with non-silting and non-scouring velocity. Why?
- When channel is silted up, its carrying capacity is reduced and therefore will
irrigate less area
- If channel sides and bed are eroded away, the cross-section increases and the
full supply depth decreases which reduces its command area.
- Bed and side erosion also causes various types of damages to the canal
structures on it as well as neighbouring areas.
Aim of channel design: is to arrive at a cross-section, which can carry the design flow
without either scouring or silting problems.
Design Parameters
For the design of channels, the following data shall be available:
i) Design discharge Q; ii) Surface and soil properties (roughness coefficient); silt factor f
The design: consists of determining the following four factors:
i) Area of cross-section, A of the channel
ii) Hydraulic mean depth or hydraulic radius R
iii) Velocity of flow, v
iv) Longitudinal slope of the bed, S.
To begin the design, the following two equations are considered (available)
a) Q = Av - continuity equation
b) V = f(n, R, S) – Flow equation, Manning, Chezy, Kutter.
In the above, there are four unknowns and therefore to solve them, we require two
additional equations.
To get the additional two equations, the following procedures may be adopted
i) By providing a channel of best discharging section, an equation between area
A and hydraulic mean depth may be obtained.
ii) Based on consideration of scouring and silting, a limiting equation for velocity
of flow may be obtained.
iii) Fixing the longitudinal slope based on available ground slope.
iv) Based on experience, fixing a suitable width-depth (B/D) ratio not necessary
that which gives best section.
KENNEDY’S THEORY (METHOD)
1) To keep silt in suspension, the silt supporting power is directly proportional to the
bed width of the stream and does not depend on the wetted perimeter.
2) The limiting velocity which does not produce silting or scouring may be called
critical velocity. The relation between critical velocity and depth is
vc = 0.55xD0.64
And after modification we will arrive
v = 0.55 x m x D0.65
where m = v/vc
201

v - velocity of flow in channel, m = critical velocity ratio (C.V.R.), m varies from 0.7
for fine sand to 1.3 for very coarse sand
To determine the mean velocity of the flow in the channel, Kennefdy used the Kutter’s
formula: v = C (RS)1/2
where


















+
+
+
+
=
R
n
S
S
n
C
00155
.
0
23
1
00155
.
0
1
23
The difficulty with Kennedy’s formula is that he did not give any formula to fix the slope
S but has to be fixed based on the natural ground slope which makes it difficult to obtain
the best section for the given discharge.
Design of channel by Kennedy’s Method.
The design procedure:
i) Assume a trial value of depth, D, in metres and determine Kennedy’s velocity
v = 0.55 x m x D0.64.
----------- (K-1)
ii) Calculate the cross-sectional area, A, from the continuity: A = Q/v
iii) Calculate the bed width, b, since A and D are known. For eg for a trapezoidal
channel, one selects the side slope of channel based on material of channel. Eg. Assume
1/2H:1V; then the cross-sectional area is given by:
A = bxD + 1/2xD2
from this, b can be determined.
iv) Calculate the wetted perimeter and the hydraulic radius:
The wetted perimeter, P = b + 2x[D2
+ (D/2)2
]1/2
= b + D(5)1/2
The hydraulic radius R = A/P = (bD +1/2D2
)/(b + D(5)1/2
v) Calculate the actual mean velocity by using the Kutter’s formula;
i.e v = C(RS)1/2
where C is computed by the formula above. ----- (K-2)
The value of v in (K-2) must tarry with the v in (K-1) for the assumed depth. If the
two values do not tarry, the trial D must be changed and the whole steps repeated till
the two values of v are the same.
The design is done for a given value of S.
Worked example
Design an irrigation channel to carry a discharge of 50m3
/s at a slope of 1/5000. Take
Kutter’s n = 0,0225 and m = 0.9.
Solution: i). Assume a depth of 2.0 m and determine Kennedy’s velocity
v = 0. 55mD0.64
= 0.55 x 0.9 x 20.64
= 0.77m/s
ii) Determine the cross-sectional area; A = Q/v = 50/0.77 = 65m2
iii) Assume side slope of 0.5H:1V
then A = bxD + 1/2D2
by putting D = 2.0, b = 31.5m
iv) Determine the wetted perimeter P = b + D(5)1/2
= 35.97m
202

v The hydraulic radius R = A/P = 65/35.97 = 1.806m
The Kutter’s constant 63
.
49
806
.
1
0225
.
0
5000
1
00155
.
0
23
1
5000
1
00155
.
0
0225
.
0
1
23
=




























+
+
+
+
=
C
vi) The actual velocity v = C (RS)1/2
= 49.63(1.806 x1/5000)1/2
=
0.94m/s
The actual velocity is far greater than the Kennedy velocity. Hence we assume a new
depth of 2.6 and repeat the whole procedure.
Kemmedy’s velocity v = 0. 55mD0.64
= 0.55 x 0.9 x 2.60.64
= 0.91m/s
ii) Determine the cross-sectional area; A = Q/v = 50/0.91 = 54.94m2
iii) Assume side slope of 0.5H:1V
then A = bxD + 1/2D2
by putting D = 2.6, b = 19.83m
iv) Determine the wetted perimeter P = b + D(5)1/2
= 25.65m
v The hydraulic radius R = A/P = 54.94/25.65 = 2.14m
05
.
51
14
.
2
0225
.
0
5000
1
00155
.
0
23
1
5000
1
00155
.
0
0225
.
0
1
23
=




























+
+
+
+
=
C
The actual velocity v = C (RS)1/2
= 51.05(2.14 x1/5000)1/2
= 1.06m/s
This value is near to the Kennedy’s critical velocity, however more accuracy could be
achieved by choosing another trial depth. The channel dimensions may be taken as
Bed width = 19.83; Depth D = 6.6
USE OF GARRET’S DIAGRAM FOR DESIGN OF CHANNEL BY KENNEDY’S
METHOD.
The computation that were done previously could be done by using Garret’s
diagrams, which give a graphical solution for Kennedy’s and Kutter’s equation.
Garret’s diagram has discharge plotted as abscissa. The ordinates on the left side
indicate the slope and those on the right side, the depths in the channel and critical
velocity vc. Bed width lines are shown as dotted.
The procedure for using the Garret’s diagrams is as follows:
A) Follow the discharge curve for a given discharge and note its point of intersection
with the given slope line (which are horizontal).
203

B) From this point of intersection, draw a vertical line. This will intersect many bed
width curves.
C) For any bed width, consider the point of intersection of the vertical line and read
out the correspoiding water depth D and critical velocity vc on the right end of
diagram.
D) Calculate the velocity of flow v for the assumed bed width and its corresponding
water depth.
E) Determine the critical velocity ratio (C.V.R = v/vc)
F) The value of C.V.R should be the same as one given. Otherwise, the procedure
must be repeated with other values of bed widths got by the intersection of the
vertical on the bed width curves for the given slope, till the value of C.V.R. is very
near the given value
Garret’s diagram are drawn for side slopes of ½:1. The curves can be used for any
other value of roughness n by making use of the small nomogram provided at the top
of curves. When any other value of n is to be made use of, see whether the value is to
the left or right of the arrow shown on the nomogram and note the shift. When the
vertical line is drawn through the intersection of the slope line and discharge curve, it
is with respect to n = 0.0225 which corresponds to the arrow on the nomogram. The
vertical line has to be shifted either to the left or to the right by the same amount as
the required value of n is shifted on the nomogram, and then the intersection points on
the bed width curves considered.
Example of the use of the Garret’s diagram
Determine the channel dimensions by using Garret’s diagrams. The following data
are available. Discharge in the channel Q = 2.5 m3
/s; slope S = 1/5000; n = 0.0225;
critical velocity ratio m, = 0.99.
Solution:
Refer to the Garret’s diagram. For a slope 1/5000, it corresponds to 0.2 as the slope is
marked in 1/1000 units.
The horizontal from slope 0.2 cuts the discharge curve for 2.5 m3
/s at some point.
From that intersection point, draw a vertical line to cut many bed widths curves. Let
us consider the bed widths curves of b = 6.0; b = 7.0m; b = 8.0m; b = 9.0m;
b = 10m. From these points of intersection, horizontal lines to the right may be
drawn and the corresponding values “D” and vc read out. The rest of the computation
are done in the table below.
No. Discharge
Q (m3
/s)
Bed
width
B (m)
Channel
depth
D (m)
Area A
=bD+D2
/2
(m2
)
Actual
velocity
v=Q/A
Vo
(m/s)
v/v0 =
CVR
Remarks
1 2.5 6 0.77 4.917 0.508 0.485 1.05 Large
2 2.5 7 0.74 5.454 0.458 0.470 0.974 Large
3 2.5 8 0.67 5.58 0.448 0.455 0.985 Suitable
4 2.5 9 0.62 5.772 0.433 0.44 0.984 Less
5 2.5 10 0.58 5.97 0.418 0.43 0.97 Less
204

From the table, we see that No. 3 gives very near value to the one given. Hence
channel dimensions should br: B = 8.0m; D = 0.67m
SILT SUPPORTING CAPACITY OF FLOW (FROM KENNEDY’S THEORY)
The amount of silt held in suspension according to Kennedy’s theory is proportional
to the upwards force of vertical eddies, as it varies as the bed width, b and some
power of the velocity of flow in the channel. This can be formulated as:
Qt α b vc
n
or
Qt = a . b . vc
n
-------------------(A)
Where Qt = quantity of silt transported by channel; a = constant.
Now Qt/Q = p = % of silt in water.
Assuming a trapezoidal channel; Q = b . D. vc
Therefore Qt = p .Q = p . b . D .vc ----------(B)
Equating (A) and (B)
A . b . vc
n-1
= 1/a . p .D
Therefore vc = (p/a)1/(n-1)
. D1/(n-1)
or
Vc = C . D1/(n-1)
where C = const. -----------©
Compare equation © to Kennedy’s critical velocity
Vc = C . D0.64
-----------------------(D)
Equating the indices in © and (D)
1/(n-1) = 0.64
from whence n = 2.56 ≈ 5/2 = 2.5
therefore silt quantity Qt = a . b . vc
5/2
.
205

LACEY’S REGIME THEORY
If for Kennedy’s theory, a channel is said to be in regime, when there is neither
scouring nor silting, then for Lacey, a different view was held.. Lacey observed that
even when there is neither silting nor scouring, the channel may or may not be in
regime. Lacey defined two regime conditions:- i) initial regime and ii) final regime.
i) Initial Regime: refers to the state of the channel that has formed its cross-
section but not formed its longitudinal slope. Such a channel appears to be in
regime outwardly, as there may be no visible silting or scouring. Lacey’s
regime theory does not apply to such channels.
ii) Final Regime: when a channel is constructed with an improper slope, it tries to
remove the incoherent silt on its bed and increase its slope. It first forms its
section and then it forms its final slope so that a stable condition is
established. When the channel has finished its task of forming a stable cross-
section and stable slopes, it is said to be in “final regime”.
iii) Permanent Regime: This condition exists in a channel which is protected both
on its slopes by suitable protecting materials. Due to these protections, the
channel cannot change its cross-section or slope. Lacey’s regime theory is not
applicable to channels in permanent regime.
iv) True regime conditions: A channel will be in regime only when there is no
silting or scouring. To satisfy this condition, the silt load in the channel water
must be efficiently transported by the channel cross-section. There can be only
one channel with a particular cross-section and slope for a particular silt load,
which can produce regime conditions. For a channel to be in true regime, for
any sediment load brought to it, it must adjust its cross-section and slope to be
able to transport the given load.
v) However artificially constructed channels with fixed cross-section and slope
can attain regime conditions when the following conditions are satisfied::
i) flow in channel is uniform
ii) discharge in the channel is constant
iii) channel is flowing through incoherent alluvium. Incoherent alluvium
is that type of soil which can be scoured and deposited with equal ease.
Also the material (silt) transported by the channel should be the same
as the material through which the channel is flowing.
iv) The amount and type of silt in the channel (silt charge and grade) are
constant.
Since all the above conditions cannot usually be satisfied by artificially constructed
channel, it follows that it cannot be in “true regime”. Either in initial or final regime
SHAPE OF REGIME CHANNELS
There is always only one cross-section and one slope for a channel with a particular
discharge carrying a particular grade of silt. Natural silt transporting channels assume a
semi-elliptical cross-section. The coarser the silt particles, the flatter will be the semi-
ellipse and the larger will be the width of water surface. As the grade of silt becomes finer
206

and finer, the shape of the section becomes narrower. Therefore when a channel is
constructed with small cross-section, and steeper slope than necessary, scour starts and
continues till final regime is established. Similarly, in a channel with a bigger cross-
section and flatter slope, silting starts and continues till final regime is established.
LACEY’S REGIME EQUATIONS
According to Lacey, silt is kept in suspension by vertical forces, which are the vertical
component of the forces generated by eddies from the bed and sides of the channel. As
such the hydraulic radius R becomes a variable.
Lacey gave two equations, one of which relates velocity to the hydraulic radius and the
other which relates velocity to area. Silt grade is considered important and therefore
becomes a function in both regime equations.
Lacey’s regime equations:
)
2
(
0
.
140
)
1
(
.
5
2
5
2
−
−
−
−
−
−
−
−
−
−
=
−
−
−
−
−
−
−
−
−
−
−
=
L
v
Af
L
R
f
v
where A = area of cross-section (in m2
)
f = silt factor
R = hydraulic radius (in m)
v = velocity of flow (m/s)
Perimeter – Discharge (P-Q) Relation
From equation (L-1) v4
= (4/25) x f2
x R2
or
f2
= (25/4) x (v4
/R2
) ---------------(L-3)
By putting (L-3) into (L-2) we obtain
A[(25/4) x (v4
/R2
)] = 140 v5
Or (25/4R2
) x A = 140v ----------------(L-4)
By multiplying (L-4) by A, we have
(25/4R2
) x A2
= 140 v x A = 140Q.
Also (A2
/R2
) =P2
-------wetted perimeter of section
(25/4) x P2
= 140Q
hence P2
= (4 x 140/25) Q
P = 4.75 (Q)1/2
----------------------(L-5)
Relation between velocity, discharge and silt factor
From equation (L-2) Af2
= 140 v5
By multiplying by v, A x v xf2
= 140 v6
f2
xQ = 140 v2
or
v = (f2
Q/140)1/6
---------------(L-6)
Lacey gave a relation between v, R and S as follows
v = 10.8 R2/3
S1/3
--------------(L -7)
Regime Slope Equations
207

By using equation (L-1), (L-2) and (l-7) ie the fundamental Lacey’s equations, we shall
obtain
S = (f3/2
)/(4980R1/2
) ---------------(L-8)
S = 0.000178 (f5/3
)/q1/3
------------(L-9)
S = (f5/3
)/(3340Q1/6
) ---------------(L-10)
Regime –Scour Depth Relations
From equation (L-1)
v = (2/5 xf x R)1/2
v2
= 2/5 x f x R
R = 5/2 (v2
/f) -----------------------(L-11)
From equation (L-6)
v = (f2
Q/140)1/6
v2
= (f2
Q/140)1/3
---------L*
By putting L* into (L-11) we obtain
R = 5/2(f2
Q/140) x 1/f
R = 0.47 (Q/f)1/3
-------------------------(L-12)
Similarly
q = 0.21 Q1/2
where q silt load, Q flow rate -----------(L-13)
R = 1.5 (q2
/f)1/3
-------------------------(L-14)
Silt factor – Grain Relationship
If mr = mean diameter of the silt particle (in mm), then the silt factor f = 1.76x(mr)1/2
No. Type of material Size of grain
(mm)
Silt factor (f)
1 Silt
a) very fine
b) fine
c) medium
d) coarse
0.052
0.120
0.158
0.323
0.40
0.60
0.70
1.0
Sand
a)medium
b) coarse
0.505
0.725
1.25
1.50
Gravel
a) medium
b) heavy
7.28
26.10
4.75
9.00
Boulders
a) small
b) medium
c) large
50.10
72.50
188.80
12.00
15.00
24.00
208

Design of Channels by Lacey’s Theory
The following formulae shall be employed:
a) Data necessary
i) Channel discharge
ii) the silt factor
b) Formulae to be used
i) v = (f2
Q/140)1/6
ii) R = 5/2 (v2
/f)
iii) Q = v.A
iv) P = 4.75 (Q)1/2
v) S = (f5/3
)/(3340Q1/6
)
Procedure for design
1). Calculate the velocity from (i) above
2. Calculate the hydraulic radius R from (ii) above
3. Calculate A
4. Calculate wetted perimeter (P)
5. Calculate b and h
6. Calculate longitudinal slope.
Example: Design a regime channel for a discharge of 50m3
/s, with silt factor f = 1.0
using Lacey’s theory.
Solution: Given: Q = 50m3
/s, f = 1.0
1) Calculate velocity , v = (f2
Q/140)1/6
= {50 x (1.0)2
/140}1/6
= 0.842 m/s
2. Calculate the hydraulic radius, R = 5/2 (v2
/f) = 5/2(0.8422
/1.0 = 1.774
3. Calculate the area A = bxD + ½ xD2
= Q/v = 50/0.842 = 59.38 m2
therefore A = b x D + 0.5 D2
= 59.38 (**)
4. Calculate the wetted perimeter P = 4.75 (Q)1/2
= 4.75 (50)1/2
= 33.6
But P = b + 2D [1 + (1/2)2
]1/2
= b + (5)1/2
D =33.6 (***)
Solving equations (**) and (***)
B = 29.34 and D = 1.968m
5. Slope S = (f5/3
)/(3340Q1/6
) = (15/3
)/[3340(50)1/6
] = 1/6409
S = 1: 6409
Therefore channel parameters: b = 29.34m; D = 1.97m; Q = 50 m3
/s
V = 0.842m/s; S = 1:6409
209

Design of channel by use of Lacey’s diagram
Example: determine the channel dimensions and slope for a channel with a discharge of
53m3
/s and silt factor of 1.0 using Lacey’s diagram
Solution: Given: Q = 53m3
/s; f = 1.0
Referring to plate 16.5, there are discharge curves for 50 m3
/s and 55m3
/s. By
interpolation, the discharge curve for 53 m3
/s can be drawn. This curve intersects the
curve for silt factor f = 1.0 at some point. For that point, the abscissa will be 30.0m. This
gives the bed width b = 30.0m
The ordinate for the intersection point = 2.0m. This gives the depth.
Now to determine the regime slope, refer to plate 16.6. Get the intersection point of the
curve for silt factor f = 1.0 and the discharge curve for Q = 53 m3
/s. The ordinate of the
intersection point represents the slope.
The ordinate in this case = 0.15 representing a slope of 0.15 in 1000. t
Therefore S = 0.15/1000 = 1/ 6667
Channel parameters Q = 53m3
/s; f = 1.0; b = 30.0m; D = 2.0; S = 1: 6667
210

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