12

Canal Falls
12.1. Definition ancl Location of Canal ~alls
12.1.1. Definition. Whenever the available natural ground slope is steeper than the
designed bed slope of the chan-
nel, the difference is adjusted by
constructing vertical 'falls' or
'drops' in the canal bed at
suitable intervals, as shown in
VERTICAL
Fig. 12.1. DROP

Such. a drop in a natural

canal bed will not be stable and,
therefore, in order to retain this Fig. 12.1
drop, a masonry structure is constructed. Such a pucca structure is called a canal fall or
a canal drop.
12.1.2. Proper location. The location of a fall in a canal depends upon the topography
of the country through which the canal is passing. In case of the main canal, which does not
directly irrigate any area, the site of a fall is determined by considerations of economy in
'cost of excavation and filling' versus 'cost of fall'. The excavation and filling on two sides
of a fall should be tried to be balanced, because the unbalanced earthwork is quite costly.
By providing-a larger drop in one_step;_the quruitity o{i.mbalanced earth work increases, but
at the same time, the number of fall reduces. An economy between these two factors has to
be worked out before deciding the locations and extent of falls.
In case of branch canals and distributary channels, the falls are located with con-
. sideration to commanded area. The procedure is to fix the FSL required at the head of
the off-taking channels and outlets and mark them on the L-section of the canal. The-
FSL of the canal can then be marked, as to cover all the commanded points, thereby
deciding suitable locations for falls in canal FSL, and hence, in canal beds.
The location of the falls may also be influenced by the possibility of combining it
with a bridge, regulator, or some other masonry work, since such combinations often
result in economy and better regulation. When a fall is combined with a regulator and
a bridge, it is called a~faH-regulator-with roa:d-:cbridge.- - -- - -- ~
12.2. Types of Falls
Various types of falls have been designed and_ tried since the inception of the
idea of 'falls construction' came into being. The important types of such falls, which
were used in olden days and those which are being used in modern days, are
described below:

639
 640 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

(1) Ogee Falls. The 'Ogee type fall' was constructed in olden days on projects like
Ganga canal. The water was DRAW DOWN ..
gradually _led down. by ~ > .,
providing convex and ccin- - ----: =-.,..:t:f:f
WATER -SURFACE
' .:.~
cave curves, as shown in Fig. u/s BED
12.2. ',
,,.,., --
\... ,' D/S BEu
The performance of
such a fall was found to have
the following major defects :
(i) There was heavy
. Fig. 12.2. Ogee fall.
draw-down on the upstream
side, resulting in lower depths, higher velocities and consequent bed erosion. Draw-
down would also affect the supply in a distributary, situated just upstream of fall.
(ii) Due to :smooth transition, the kinetic energy of the flow was not at all dissipated,
causing erosion of downstream bed and banks.
A 'raised crest' was soon added to restrict the draw-down and a long protection was
provided on the downstream side. Later, it was converted into a much better type of fall,
called a 'Vertical Impact type'.
(2) Rapids. In Western Yamuna c-anal, long rapids at slopes of 1 : 15 to 1 : 20 (i.e.,
gently sloping .glacis) with boulder facings, were provided. They worked quite satisfac-
torily, but were very expensive, and hence became obsolete.
_ (3) Trapezoidal Notch Falls. The trapez9_idal n9tch fall was designed by Ried in
1894. It consists of a number of trapezoidal notches constructed in a high crested wall
across the channel with a smooth entra:nce and a flat.circular Hp projecting downstream
from each notch to spread out the falling jet (See Fig. 12.3).

FORONT ELEVATION
B FOUNDATION WALL

Fig. 12.3. Trapezoidal Notch fall.

I
1 The notches.could be designed to maintain the normal water depth in the upstream
channel at any two_discliarges, as the variation at intermediate values is small. Hence,
the depth discharge relationship of the channel remains pr11ctically unaffected by the
'introduction of the fall. In other words, there wotild neither
. .be drawdown
~.. . nor h~ading
,,_
 . (lr\NALFALLS 641
p of water, as the channel approaches the fall. These falls remained quite popular, till
~impler, economical, and better modern falls were developed.
(4) Well Type Falls or Cylinder Fall_s, or Syphon Well Drops. This type of a fall_
onsists of an inlet well with a pipe at its l?ottom, carrying water from the inlet well to
~ downstream well or a cistern. The downstream well (as shown in Fig. 12A) is
necessary in the case of falls
r: greater than 1.8 m and for dis- Ol g'
c-
-.i:::
'f charges greater than 0.29 -:::- u
-~E.
I: curnecs. The water falls into the Vl
~'O
().>

[__ inlet well, througb a trapezoidal--- a:-

ri notch constructed in the steining
of the well, from where it emer-
---
<: Vl
<: 0
s:: a.
..J
- .......
0
ges near the bottom, dissipating
its energy in turbulence inside
the well.
This type of .falls _are very
useful for affecting larger drops
for smaller discharges. They are
commonly used as tail escapes
for small canals, or where high
levelled smaller drains do..outfall
f into a low levelled bigger drain.
I
. (5) Simple Vertical Drop
Type and Sarda Type Falls. A
raised crest fall with a vertic:al
impact (Fig. 12.5) was first of all
introduced on Sarda. Canal Sys-
tem in U.P~. owfrig i6 it.sc.
economy and simplicity. The
necessity for economic falls.
arose because of the . need of
construction of a large number c:
1
of smaller falls on the Sarda ~ Cl
0
.s::
Canal System. In thai area, a .o-;; Ill

thin layer of sandy clay overlied g 2

-0 :::>
u
.s:: 1-
. a stratum of pure sand. If the uCl -a.-
OJ 0
OJ

Cl 3
canal _bed was to be.cut deep and
up. to the ,,,...,.sand _j
, .s;tratum, the
vi
seepage losses would have been IL:
-'-tremendous. Hence, tne depth or.
cutting had t.o be kept low,
necessitating the construction of -o .
a large number of smaller falls. 0
0
.0
In this type of a high crested <:
<:
fall, the nappe impinges. into the ....
u..
Water cushion below. There is
642 IRRIGATION ENGINEERING AND HYDRAULIC STRUCllJRES,

- -- ', I

''
', _- -'--- - - - c/:..H~
FLOOR OF WATER CUSHION

Fig. 12.5. Simple Vertical Drop fall.

no clear hydraulic jump and the energy dissipation is brought about by the turbulent
diffusion, as the high velocity jet enters the deep pool of water downstream.
Two types of crests which are used in Sarda type falls are shown a little later in
Figs. 12.10 (a) and (b ).
Sarda type fall is a high crested fall, and if the discharge in the canal varies (say between
50 to 100% ), the water will head up on the upstream side at low discharges. The reach
upstream of the crest will silt up as the clearer water will pass downstream of the crest. Due
to reduction in silt in the dis discharge, there may be a tendency of scouring on dis, so as
to make up the silt loss. Hence, this type of fall is not quite suitable for canals in which
discharge varies within a wide range. A trapezoidal notch fall, although costlier than Sarcia
type or glacis type fall, is free from such troubles and, therefore, preferred for canals where
the discharge is very small and also varies over a wide range.
- . {6) Straight Gl~cis Faj.ls.111 this type of a modern fall, a 'straight glacis' (generally
a
sloping 2 : 1) .is provided -after 'raised crest' (see Fig~ 12.6). The hydraulic jump is
D/S wrNG WALL TOP OF PITCHING
LOPE PITCHING
D/S
H FL
~=-=--:::--7~ PROF l LE
WALL

SECTION A-A
DEFLECTOR WALL OR
DIS CURTAIN WALL
PROFl.LE WALL OR
OHAMALl
RETURN
WALL FSL
01S WING WALL -
A
+ .' -- ____...o_IS_G_LA_ss CISTERN
'-7 OEfLECTOR
::Cl WALL

u/s WING
WALL
Fig. 12.6. 'Straight Glacisfalf (without fluming), without Regulator and Bridge Details.
 CANAL FALLS 643
made to occur on the glacis, causing sufficient energy dissipation. This type of falls give
very good performance if not flumed, although they may be flumed for economy. They
are suitable up to 60 cumecs discharge and 1.5 m drop.
(7) Montague Type Falls. The energy dissipation on a straight glacis remain
incomplete due to vertical component of velocity remaining unaffected. An improve-
ment in energy dissipation may be brought about in this type of fall [see Fig. 12.7 (a)],
by replacing the straight glacis by a parabolic glacis', commonly known as 'Montague
Profile'.
U/S WI NG WALL

--
D/S WING WALL
_ ~/..FiL-::1 - - - - - - -

Fig.12.7. (a);Montague Type fall.

~~~u~~M"'!"'l,.._.- _LH~R~O~T~ __ --x

I Y I HALF GR~VI TY
. l_PARABOLA
I '
'---x -
'D1S ED GE ___A.I
i{MONTAGUE
I PROFILE!
OF CREST I
I

t
y ""'FREE FALL
UN DER GRAVITY
.

Fig. 12.7. (b) M?ntague Profile.

The Montague profile is given by the equation.

' - f4Y
X= u-\Jg +Y ... (12.1)

where X = The horizontal ordinate of any point of

---- -- the-proflie measuredfrom the dis edge of
crest.
Y = Vertical ordinate measured from the crest
level.
U = Initial velocity of water leaving the crest.
The curved glacis is difficult to construct and thereby1rendering it costlier. This type
of fall is, therefore, generally not adopted in India ..
 644 . 1:.>
IRRIGA TIONENGINEERING AND HYDRAULIC STRUCTURES

(8) Inglis Falls or Baffle ... Falls. A straight

glacis type fall when added with a baffle platform
and a baffle wall as shown in Fig. 12.8, was
developed by Englis, and is called 'Englis Fall' or
'Baffle Fall'. They are quite suitable for all dis-
charges and for drops of more than 1.5 m. They
can be flumed easily as to affect economy.
The baffle wall is provided at a calculated
height and a calculated distance from the toe of
the glacis, so as to en~ure the formation of the
. jump on the baffle platform, as shown in Fig.
,,, ,12.8. -
Comparison of Different Types of Falls.
Extensive testing of various modern types of falls .
" was- carried ouf in 1952 at Research Station
Poona, and the following recomm_endations were
\
made: .....
0
(i) 'Vertical drop' falls are quite suitable for
discharges upto 15 cumecs and drops upto. 1.5 :.o.::f.'~
metres. But this type of fall should not be flumed.
..... ~

(ii) for 'Straight glacis' type falls, the con-

clusion was, that they work satisfactorily for all
conditions, if unflumed ; but in that case they
become costly. Even then, they can be adopted
suitably for discharges upto 60 cumecs and drops
upto 1.5 metres, t:md_ can eve'! ~e flumed. - Q::-
a.

(iii) 'Baffle fall' or 'English fall' may be used ,,.,I

' ' for all discharges when drop is more than 1.5 ii
Zz
metres. This type offall jUnctions very satisfac- a:"' :c
=> :::>
........ vd
! torily, either fl!fined or unflumed, so long as it is
1111
.......... =>I
undrowned.
1
"'"'
(iv) Well type falls are suitable and e_conomi-
' I cal for high drops and very low discharges. They
i can hence be easily used, as tail escapes of small

channels.
Meter and Non-Meter Falls. Meter falls are
those which can be used to measure th~ cl!s_c:_l11irg~ci>fJb~ canal. -If the discharge_ca:nnot
. be measured accurately at the sfte- of the fall; then ,it is called a non~meter fall: Vertical
' drop fall is not suitable as_ a meter due to the formation of partial vacuum under the
i nappe. Glacis type fall is quite suitable as a metering-device. Since a sharp .crest does
1111

' not give a consta~t coefficient of discharge with varying heads, while a broad crest does
so reasonably; a fall to be used as meter must be provided with a broad crest. Generally,
.. a flumed glacis fall or a flumed baffle fall, is used as meter, while an unflumed glacis
fall is used as a non-meter fall. -
 CANAL FALLS 645 ''
DESIGN PRINCIPLES Ofi' VARIOUS TYPES OF FALLS
12.3. Design of a Trapezoidal Notch Fall
As pointed out earlier, a notch fall provides a proportionate fall, in the sense that
there is no heading up or drawdown of water level in the canal near the fall. The whole
width of the channel is divided into a number of notches. The
crest (i.e. the sill level or the level of the bottom of notch) may
be kept higher than the bed level of the canal, which will tend
to increase the length of the weir, but in no case, the total length
of the weir openings should exceed the bed "width of the canal -
upstream, and may well be reduced to about ith of the bed width.
Discharge Formula. The discharge passing through one
notch of a notch fall can be obtained by adding the discharge of
a rectangular notch and a V-notch .
.:. The discharge passing .through a trapezoidal notch such
Fig. 12.9
as shown in Fig. 12.9 is given by Trapezoidal notch
- 2 _~ _ 3;2 8 . -_~ ex s12
Q = 3 Cd '12g l . H + lS Cd . '12g tan J H

= ~3 Cd &g [lH312 +tan~

5 2H ]
512

=~Cd. ~[LH312 +% (2 tan I )H512] ... (12.2)

_If 2 tan I is represented by n, then

Q = ~ Cd . :V2I[LH3;2 + 0..4 . nHs12J ... (12.3)
. ' 1 where, Cd= Coefficient of discharge ""0. 75

Q=~x0.75-V2x9.81 [tH312 +0.4nH512]

or Q = 2.22H312 [z + 0.4nHJ . :(12.4)
_ The above discharge equation contains two unknowns land n. For solving this
equation,- two values of Q and corresponding values of H must be assumed. It is a
common practice to design notches for full supply discharge (Q 100) and half supply
discharge (Q50) with values of H equal to the normal water depths in the channel in the
'respective cases. Let the normal water depths in the channel at full discharge and half
discharge be represented by Yioo andy~o respectively. Then H100 =Y100, an_d Hso = Yso
-----.:.,._ _:_" .. ---- ~--- ~~- - - -- - - - ------- - - --- -- --~ --- --- -------------- --~. --~--- -------

The depth of water in the channel at 50% discharge (i.e. y50~ can be approximately
evaluated in terms of foll supply depth (y 100) as follows :
Let V= C. y.64 ... (Kennedy's Eq. for Vel. in channels)
Now Q=A.V.
... Q = B . y . C . y.64 Using A"" B.y (neglecting sy2)
or Q:::: C . B . YI.64
Q100 = C. B yj1
 646 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

Qso =C . B Y1~
4
and
1.64
Qso Yso
or
Q100 = [ Y100 )
.....22..
Y
=
[ Q )_! = (0.5)
1.64 - I
i. 64 = 0.66
or _2Q..
Y100 Q100
Yso = 0.66' Y100 ... (12.5)
Number of Notches. The number. of riotches should be so adjusted by hit and trial
method that the top width of the notch lies between ~ to full water depth above the ~ill
of the notch. This hit and trial procedure would become clear when we solve a numerical
example.
Notch Piers. The thickness of notch piers should not be less than half th~ water
depth and may be kept more if they have to carry a heavy super structure. The top length
of piers should not be less than their thickness.
In plan, the notch profile is set back by 0.5 m from the downstream face of the notch .
fall for larger canals, and by 0.25 m for distributaries. All curves are circular arcs, and
all centres lie in the plarie of the profile. The splay upstream from the notch section is
45, and the downstream splay is kept 22.5. The lip is circular and is corbelled 'out by
0.8 m on larger canals, and by o;6 m ,on distributaries.
Example 12.1. Design the size and number of notches required for a canal drop
with the following particulars :
Full supply discharge =4cumecs
Bed width =6.0m
' F.S. depth. =I.Sm
.Half supply depth =I.Om
Assume any other data if required. (Madras University I975)
. Solution. The bed width of the canal is 6 m. Each notch at top should be roughly
equal to F.S."depth i.e. 1.5 m: So let us, in the first trial, provide 3 notches.
:. Full supply discharge through e.ach notch=~= 1.33 cmecs
From Eq. (12.4) we have
Q = 2.2'2H312 [i + 0.4 nHJ ..
3/2 .
Using. Q100=2.22()'1oo) [l+OAnY100]
where Q100 = 1.33 cumecs
. _ ...... . . . . . . .. . Y100= 1.5 m
------;.~-we have--1~33';;2.i:f. (t5)312-ll+;OAnxf.sr-''-
or 1.33 =2... 22 .x 1.84 [l + 0.6nl
or l + 0:6n =0.326 . :.. (i)
Now, using
. 3/2 . . .
Qso =2.22. <Yso) [l + OAn . Yso]
. . .. 1 33 . i
.where Q50 =~=0.67cumecs

Yso= I.Om
.. cbNALFALLS 647
0.67 =2.22. (1.0) [l + 0.4n x 1]
312

l+ 0.4n =0.3 .. .(ii)

or
Subtracting (ii) from (i) we get
0.2n=0.026
n=0.13.
putting the value of n in (ii) we get
/+0.4X0.13=0.3
_ l =0.248; say. 0.25.m._ .
By this trial, we find the top width
= 0.25 + 2 tan a H = 0.25 r1- n. JI
=0.25 + 0.13 x 1.5 =0.25 + 0.195
=0.445; say Q.45 m, which is much less than the frill depth of 1.5 m.
In order to increase the top width, and to make it near 1 to ~th FSD, it is necessary
-to-increase l and n :which can be done by reducing the number of notches.
The values of l and n obtained for 3 notches will increase in direct proportion, when
number of notches are reduced.
In other words, the values l of and. n will become 3. times, when number of notches
are reduced 3 times. Thus, when we provide only one notch instead of 3 notches, the
not~_hes against 3, i.e., ~
values of n and l will triple. Similarly, when we use _2
1 5 times
the values n and l will become 1.5 times of those obtained for 3 notches.
Hence when we use 2 notches, values will be
n =1.5 x 0.13 =0.20
l =J..Sx 0.25 =0.3m_
and Top width =1.5 x 0.45 =0.68 m.
Since the width is still quite low, we may use only one notch ..
When we use only one notch, the values will be
n = 3 x 0.13 = 0.39
l = 3 x 0.25 = 0.75m
Top width= 3 x 0.45 =1.35 ""FSD (O.K.)
a _1 n _ 1 0.39
2=tan 2=tan T;::: 110
Since this condition
gives us top width =1.35 m,
which is O.K., we may pro-
vide one notch, centrally
placed in the given channel
of 6 m width. The section of
the. notch to be adopted is
also shown in Fig. 12.10.
Check for raised crest
if possible. It has also been
noticed that when lesser Fig. 12.JO
 648 IRRIGATION EN9INEERING AND HYDRAULIC STRUCTU

number of notches are PEovided,' with their. bottoms kept at U/S DBL of canal th 0

concentration.of flow gets increased considerably.~: avoid.such. an eventuality,'itr~- .

pref~r~ble to increase ~he num~er of notches, and this may sometimes be. achieved b ' i.
prov1dmg the notches m the raised crest. In other words, the bottom of notch open y : .
will be kept higher than UIS DBL of canal. This raising may be between 10% to 3~i ~
of full depth. The design calculations are hence to be repeated to compute n and z with
a raised crest, whenever a detailed designing is being done, and number of notches '
determined are low.
These calculations for the above question 12.1 wlll be-as follows :
Let us assume a raised crest equal to 20% of FSD
= 20% x 1.5 m=0.3 m.
Using Eqn.'(12.4) as
Q = 2.22 H 312 [l + 0.4nH], we have
QIOO = 2.22 (1.5 - 0.3) 312 (l + 0.4n (1.5 - 0.3)]
or +
Q100 = 2.92 (l 0.48n) ... (iii) .
Also Q50 = 2.22 (l.0- 0.3) 312 [l + 0.4n (LO- 0.3)]
[ . FSD at~ discharge= 1.0 m (given)]_
or Qso = 1.3 (l .+ 0.28 n) ... (iv)
But Q100=2Q50
2.92 (l + 0.48 n) = 2 x 1.3 (l + 0.28 n) .)

This gives a negative value ofn, which is not feasible, and hence such a raised crest
may not be feasible in this particular case. Hence, the design made earlier, and shown
in Fig. 12.10, holds good.
12.4. Design of a Syphon Well Drop
A syphon well drop, such as shown in Fig. 12.4, is generally adopted for smaller
discharges and larger drops. The main features of the design involves determining the
size of the inlet well and that of the hume pipe. A suitable size for the outer well, a
proper provision of water cushion at the bottom of the inlet well, the bed and side slope
pitchings iri the canal TRAPEzoro AL
. upstream as well as NOTCH
downstream for suitable .1 y
lengths, are also ~---.---~""-l... 3 100
provided. The size of the -+--.,..___..,,_.,--- R.L. OF C.R
inlet well and that of the
syphon pipe are deter-
. mined--on- the foflowfi1g
conside.rations w .r. to
Fig. 12.11.
First of 11.ll, the size
of the trnpczoidal notch
is determined so as to
pass. the designed dis-
charge by using eqn.
(12.4) in the same way as. Fig'. 12.11
 CANAL FALLS 649
is done for a trapezoidal notch. Then let V1 be the velocity over the notch, V2 be the
velocity of entry in the pipe, and V3 be the velocity through the pipe. All these values
of velocities can be determined easily as below :
V _ Fullsupply discharge
1
- Area of flow over the notch
v = Full supply discharge . .
2
Area of opening at entry (for assumed dia of opening)
V = Full supply discharge ___________ _
3
Areaofpipe (forassumeddia)
The head loss between the inlet well and the d/s FSL is then given by HL 1 as
2 2 .
HL
V:z .
= 0.5 - . (i.e. loss due to entry)+
(V2 -
.
V3) . .
(i.e. the Joss due to sudden
I 2g 2g
. f'LV2 . . . .
3
enlargement)+ gd (i.e. the loss in the assumed pipe length L)
2
v2 .
+ ; (loss due to exit). ..:(12.6)
2
Knowing all the above values, HL1 can be determined, and thus the R.L. of water surface
inlet well (i.e. dis FSL + HL) can be determined .

Now, approximate R.L. of centre of pressure (C.P,) of the trapezoidal waterway

through the notch
.. l
= u/s canal bed level+ 3 FSD.
_ . -=(which can-be calc:u-lated)
Then, the height (Y) of the centre of pressure above the water level in the inlet well
= R.L. of C.P. - R.L. of waterlevei in inlet well
'/ =(Known)
Now using the eqn.

V1 =~* ...(12.7)

where X and Y are the coordinates of the jet (i~suing from centre of pressure) ..;..r. t. the
water surface level in the infot well.

*Tneeqn~cr2:75canoe4eifveiias'6eJ.Ow--w:r.i:-bFig~12.12: ~_c:___c _ _ -'----C---- - - - -

x = v1 t (after a.time t)
2 2
[ : usingS=ut+t gt andu=O, wehaveS=t gr ]
t-- x --j
or
~i.Y1
.or JE:T
I j
Fig. 12.12
650 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

The value of X can be determined. Final.ly the dia of the inlet well may be kept at
about 1.5 times the value of X. The entire procedure will become more clear when we
solve. the following numerical example.
Example 12.2. Design the salient dimensions of a syphon well drop for the follow-
ing particulars :
Fall =3.Bm
General ground level =+ 163.36 m
Full supply depth =75 cm
Bed level upstream + 162.83 ._=
Discharge =1 cumec
Bed width upstream and downstream =2.4 m (Madras University 1972)
Solution. For a trapezoidal notch, we have the discharge eqn. (12.4) as
Q = 2.22 H312 [l + 0.4 n H]
At full supply d~~charge, we have
312
Qioo = 2.22 (y100) [l + 0.4 n Y100]
where y 100 =:= F.S.D. = 0.75 m
Qioo= F.S.Q= 1 cumec
312
1 = 2.22 (0.75) [l + 0.4n (0.75)]
or 0.71=l+0.3n ... (i)
At 50% full discharge, we have
3/2 .
Qso = 2.22 Yso [l + 0.4n Yso]
where Yso"" 0.66 y100 (eqn. 12.5)
=0.66x0.75

Q50 =0.5 cumec

312
0.5 = 2.22 (0.5) [l + 0.4n (0.5)]
or 0.64= l + 0.2 n ... (ii)
Subtracting (ii) from (i) we get
0.07=0.l n
or n=0.1

2 tan 2a = 0.7, or 2a = 19.3

Substituting this value of n in (ii), we get
l=0.64-0.14=0.50 .
Hence~rovide-a-trapez0idal-n0tch-in- the- steining._of the_ -r---1.0 2s.m.
inlet well, with 0.5 m bottom width and each side inclined to an ::'-::.. =-= ~- =-
angle of 19.3 with the vertical. 1
1
r
Now, the ~idth of water (atFSL) flowing over notch : ;
1
= 0.5 + 0.7 x (0.75) = 0.5 + 0.525 = 1.025 m. J 075m
1
Velocity (V1) over the notch
F.S.Q.
=~~~~~-=-~~~-

Area of flow over the notch 1-o.sm -l

Fig. 12.13
I- p.LFALLS 651

r
r
~
=0.5 +21.025 x 0.75 m/sec =0.76 ~ 0.75 m/sec =1.75 m/sec
b_. _ Let us now assume that the diameter of the pipe used be 1 m
:. Velocity Y3 through the pipe
1

t
.. --
= - - m/sec. = 1.27 m/sec.
~ (1)2
4
Let us assume that the diameter of the opening at the inlet of pipe be 0.5 m
:. The velocity of entry into the pipe (V2)
1
= mis = 5 .1 m/sec.
~ (0.5) 2
Now
Loss of head between the inlet well and the dis FSL is given by Eqn. (12.6).
. Vi <V2 - V3)2 f' L v~ v~
= 0 5 . 2g + 2g .+ 2gf1 + :1g

tooef~e~: 0~=~Jtth:,'.~':~,:~=~2
H _ (5.1)
2 .
(5.10-1.27)
2
0.012x 12x (1.27)
m .:"d f' =Da<cey's
. 2'
(1.27)
2
05
L i - x2x9.81+ 2x9.81 + 2x9.81xl.O +2x9.81
= 0.66 + 0.77 + 0.01+0.08 = 1.52 m.
:. R.L. of water surface in the inlet~weH
=dis FSL + 1.52 . .
d/s FSL =u/s FSL - fall ]
[ = (162.83 +0.75)- 3.8 = 159.78)
= 159.78+ 1.52= 161.30.
Approximate R.L. of the centre of pressure (C.P.) of the trapezoidal waterway
through notch

=u!s canal bed level + 31 FSD

0 5
= 162.83 + ; = t6Z,83+0.25=163.os
.. Height Y of" C.I> .. abovecwater level-in-thecin.letc..weU~- - ___..-e;:..__..cc:~"--- --

= 163.08-161.30= 1.78 m.
Now using Eqn. (12.7), we have

f8.'X2
vI =-'12r

or X=-
. -\J~
{Vf2Y
652 I_RRIGATION.ENGINEERING AND HYDRAULIC STRUCTUREs.

= ~.-(1-.7-5....,)2,....9~-82-lx-1-.7-8-= 1.75 x 0.6 = 1.05 m.

Now, the dia. of ~e inlet wen may be kept at about 1.5 X, i.e. l.~ x 1.05 = 1.575 tn,
say 1.6 m. Keep the dia of the dis outlet wen, as say 1.2 m. Also provide a water cushio
at the bottom of the inlet welL Bed and sides of channel for suitable lengths on the u/n.
as well as dis side are protected by dry brick pitching. The complete details are show~ 1
in Fig. 12.14.

i
f
b
F

: . : ...,
~
..
-

. . ';:tz
l.!J
.. ~~
u
<(..;.
""=>
.... ~3, :.:: __,

._;
0.
. . : .wu.. E
,..., z ~ ""
~
"'
d

o,
tl
a E
ct:"'
<( 0 fc
.. - . -- - - ---- -- ---'-------'-al-~p~~g
~
...., E bi
er ""
u...-:-
E ;i; CI
(~

0.
ac
 653
iz.5. pesign of Simple Vertical drop. Fall
In a vertical drop fall, the u;s TEL
gy of the flowing water is -- - - 4-. - - -
di
e~e~pated by means of impact u;s FS L H -===. .:.:::: .:: - - - - f
'ss
'. d by sudden d e fect10n
l. of 1 ' , HL
. ~Jocity from vertical to
:~orizontal dire~tion. J_ - ~~
\
A water -u- - -E-....,..,...,..,.,... ...___...=<""....... - - _ !F.!:_ -
15 8 0
cushion is provided at the toe
. i the drop, so as to reduce the
' ~mpact. of falling jet and thus
, : to save the downstream fl?or
: ., from scour. The water cush10n
; is formed by depressing the
floor below the downstream
Fig. 12.15
bed of the canal, as shown in
, Fig.12.15.

The following dimensions for the .cistern hav~ .been suggested by U.P. Irrigation
.; Research Institute :
Le= 5 -.JH. HL , ... (12.8)
X=~'(H.HJv3 . (12.9)

where Le= The length of the cistern in metres.

X = Cistern depression below the downstream
bed in. metres.
ff= Head of water over the crest, i.nduding
velocity head, in metres, i.e.
=(u/s TEL - Crest level).
12.6. Design of a Sarda Type Fall .
The design criteria for various components of such a fall, based on. the recommen-
dations of Bahadarabad Research Station, are given below :
Length of the Crest. Since fluming is not permissible in this type of falls, the length
of the crest is kept eqr-tal to the.bed width of the canal. Sometimes, for future eX:p11-nsion,
the crest length may be kept equal to (bed w}dth +depth).
Shape of the Crest. A rectangular crest with both faces vertical has been suggested
. for discharges under_ 14 ~~c_s,._Jbe_tQp__widthisJcepLequaLto.0.55:ltiand-theminimumc...c._------ --
base width is kept equal to h ~ d (Take G =2 for masonry) where, dis the height of the
crest above the downstream bed level and h is the head over the crest [See Fig. 12.16
(a)].

For discharges over 14 cumecs; a trapezoidal crest with top width equal to
o. 55 ~H + d with upstream side slope of 1 : 3 and downstream side slope of 1 : 8 is
adopted [See Fig. 12.16 (b)].
 .~
:3
~
654 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTUREs: .'
/S WING WALL

DIS WING
\WALL BRICK SI DE
PITcHING.

U/S CURTAIN D/S CURTAIN

BRICK ON EDGE
WALL WALL
FLOORING

Q = upto a maximum of 14 cumecs

B, =Top width of crest= 0.55 {J
B asew1"dth =h+d
-- .
2

(~r
6

Q=1.84. LH312
Fig. 12.16. (a) Rectangular Crest;for Sarda Type fall ..

U/S Wf NG WALL
_LU/S TE.L .

U/S ~:-1- -+-- - -... D/S \o/fN'G WALL

H h '
t t l-Bt-! ~ - - - __?J~SL__~
.t d
I:;....:_ WAR PED
.
wrNG ...._ - - . Y J

Q = for 14 cumecs and over

= =
B: To.p width of cre8t 0.$5 -../H + d
Bas~ width.= as determined by the batters
. . 116

______ __ ____ Q=L99 H ;(~l__,, _____

31
c_

Fig. 12.16. (b) Trapezoidal crest for a Sarda Type fall.

...
. ;

Crest level. The. following discharge formula is used to determine the height of the;;;
crest.

Q=C,."'2g.L.H'"({f . .,.(12.10),

where Cd= 0.415 for rectangular crest

= 0.45 for trapezoidal crest
 655
L = Length of the crest
Bt = Top width of crest.
Height of the crest above bed =y - h
""y- H (assuming h ""H z.e. neglecting velocity of
approach)
where y is the normal depth of channel
(upstream).
Upstream Wing Wall. For
trapezoidal crest, the upstream wing ~
walls are kept segmental with radius ::g -----~
equal to 5 to 6 times H and subtending
an angle of 60 at centre, and then car~
ried tangential into the berm as shown
in Fig. 12.17. The foundations of the ---1-- CREST
wing walls are laid on the impervious WIDTH OF CANAL: LINE
-concrete floor itself.
For rectangular crest (i.e. discharge
Jess than 14 cumecs), the approach
-___ wmTH OFCREST

l_~-====
wings may be splayed straight at an
angle of 45.
Upstream Protection. Brick ~
pitching in a length equal to upstream ~
water depth may be laid on the
upstream bed, sloping towards the Fig. 12.17. Upstream wing walls for
crest at a slope of 1 : 10. Drain pipes Trapezoidal crest of Sarda Type fall.
should also be provided at the u/s bed
level in-tliecresrso as to drain out theufs bed during the closer of the channel.
t
Upstream Curtain Wall. 1 brick thick upstream curtain wall is provided, having
a depth equal to trd of water depth.

Impervious Concrete Floor. The total length of impervious floor can be. deter-
mined by Bligh's theory for small works and by Khosla's theory for large works. The
minimum length of floor on dis of the. toe of the crest wall should be = [2(water depth
+ 1.2 m) + drop]. The balance can be provided under the crest and on upstream.
The floor thickness required on tl;ie downstream side can be worked out for uplift
pressures (using minimum thickness of0.4 m to 0.6 metre) and only a nominal thickness
of 0.3 metre is provided on the upstream side. The maxiinum seepage head will occur
- when water is-stored-upto-topof-erest-0n-u/s-side-andcthereisno-flow-0n the-downstream----
side.
Cistern. The length and depth of cistern can be worked out from equations (12.8)
and (12.9).
Downstream Protection. The dis bed may be protected with dry brick pitching,
about 20 cm thick resting on 10 cm thick baHast. The length of the dis pitching is given
by the values of Table 12.1; or 3 times the depth of downstream water, whichever is
more. The pitching may be provided between two or three curtain walls. The curtain
 656 IRRIGATION ENGINEERINGAND HYDRAULIC STRUCTURES

walls may be 1~ brick thick and of depth equal to~ the downstream depth; or as given
in Table 12.1 (minimum= 0;5 m).
Table 12.1

Head over the crest Total length of Curtain walls

Remarks
H (metres) dis pitching (metres) No. Depth in metres-
Upto 0.3 m 3.0 All sloping down 1 0.30
atlinlO
0.3 to 0.45 3.0 +Twice HL Horizontal up to end 1 0.30
of masonry wings
0.45 to 0.60 4.5 + Twice HL 1 0.45
and then sloping
0.60to 0.75 6.0 +Twice HL down at 1 : 10 1 0.60

0.75 to 0.90 9.0+TwiceHi " 1 0.75

0.90 to 1.05 13.5 +Twice Hi " 2 0.90
1.05 to 1.20 18.0 +Twice Hi " 2 1.05
1.20 to 1.50 22.5 + Twice Hi " 3 1.35

Slope Pitching. After the return wing, the sides of the channel are pitched with one
. 1 brick on edge. The pitching should rest on a toe wall 1 ~ brick thick and of depth equal
to half the downstream water depth. The side pitching may be curtailed at an angle of
45 from the end of the bed pitching, or extended sttaight from the end of the bed
pitching.
Downstream Wings. Downstream wings are kept straight for a length of 5 to 8
time ...JH. HL and :inay then be gradually wrapped. They should be taken upto. the end
of the pucca floor.
All wing walls must be designed as retaining walls, ~:icibjected to full pressure of
submerged soil at their back when the channel is closed. Such a wall generally has a
base width equal to trd its height.
Example 12.3. Design a 1.5 metres Sarcia typefallfor a canal having a discharge
of 12 c.umecs, with the following data :
Bed level upstream =103.0 m
Side slopes of channel =1 : 1 m .
Bed level downstream =101.5 m
Full supply level upstream =104.5 m
Bed width u/s and dis =1.0 m
Soil = Goodlo(lm
Assume Bligh's Coefficient =-6-- -- - -------~- - - -- ----- ---- ----- ------- - -
Solution.
Length of crest. Same as dis bed width = 10 m
Crest level. A rectangular crest is provided, since the discharge is le.ss than 14
cumecs. The discharge formula is given by

Q =L84. L . H
312
r
[:.
 657
Assume top width of the crest as 0.8 m.
H116
12 = 1.84 x 10 x H312 x---,-=6
. . . (0.8)11
t or
JI513 = 12 x 0.964 = 0 628
I l.84xl0
1;
iI or H = (0.628) 315 = 0.755 m; Say H = 0.76m.
1, Velocity of approach
!:
I
-v -
,... a-
Discharge..:.. -- - -12 - --- CCC--~ '(::-Depth of water ; 1.5
Area - (10 +LS) 1.5
m) .
~ = lLSl~ l.S = 0.696 m/sec.
'l . . v~2 .
:l Velocity head = g = 0.025 m.
2
!lIf'. u/s.TEL = u/s FSL +Velocity Head

r~ RL of the orest
= 104.5 + 0.025 = 104.525 m

= (u/s TEL -H)

' =104.525-0.755=103.77m.
~ Use crest level of 103.77 metres
Height of the crest above dis floor
j
r = 103.77-103.0=0.77m.
'
Shape of the crest.
Width of the crest (Br)
----;;;-0 .ss . -rcr
where d ~.Height of the crest above d/s bed
= 103.77....: 101.5 = 2.27 m
Br= 0.55 ...fd =0.55 'Y2.27 =0.825 rii.
Keep 0.85 m width of the crest
Thickness at base =h ; d

= (0.755-:- 0.025) + 2,27 = 0.73 + 2.27 = 1 5

2 2 . m.
The top shall be capped with 20 cm thick C.C. 1 : 2: 4
. . .
UpstrearrCwlhg Wazi:Iislialloe .sp!ayetstrhlgfilafan"'angfoofLJ.5--from the-ii?secfge-
of the crest and shall be embedded by 1.0 m into the berm. On the dis side, wing walls
are kept straight and parallel up to the end of the floor and joined-to-return walls, as
shown in Fig. 12.19.
Upstream protection . .1.5 m long brick pitching (equal to u/s water depth) ts laid_ on
the u/s bed, sloping down towards the crest at 1 : 10, and three drain pipes of 15 cm
diameter at the ills bed level should be provided in the crest so as to drain out the uh
bed during the closure of the canal. . I I
 658 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

~

Upstream curtain wall._ Maximum depth of u/s curtain wall

Yu 1.5 .
=3=3=0.5m.
Provide 0.4 m x 0.8 m deep curtain wall on the u/s.
Cistern. Depth of cistern, is given by Eq. 12.9 as
3
X= [H Hi)v = [0.76x l.5]v3 =x (1.14) 0667 ... (12.9)
1
= 4 x 1.091 = 0.273 m ; Say0.3 m deep ... _-:
:. R.L. of cistern= 101.5- 0.3=101.2 m.
Length of cistern = 5 -firH;,
= 5 x ..J0.76 x 1.5 = 5 x ..JTl4 = 5.34m; say 5.5 m.
Provide 5.5 m long cistern at R.L. 101.2 m.
Impervious floor.
Maximum Static Head
=(Crest level - dis bed level)
= 103. 77 - 101.5 = 2.27 m.
Total floor length required
=C.H.; where C is Bligh's coefficient
= 6 x 2.27= 13.62m.; say 13.7m.
': ! Minimum d/s floor l~ngth required
= (2 (Water depth + 1.2) + HL)
= 2 (1.5 + 1.2) + 1.5 = 2 (2.7) + 1.5 = 5.4 + 1.5 = 6.9 m; say 7 m.

up UFT

1015

------6-7~------~i----t----~ ~SS
-----------13-7 ------------+t
.Fig. 12.18
. ,. -~____ . FIQ_or TI~knesses._H.G.-1ine_foLthe maximum static-head-is-shown in Fig.12.18.
Maximum
.
unbalanced uplift at the dis toe of the crest
.

(l0 3 . 77 ~/0l. 3) X7=0.3+ 1.16~ l.46m.

. ::::0.3+
3
Thickness required ~ :~~ = 1.29 m ; say 1.3 m.
'11.!

I' Provide 1.1 m thick concrete overlain with .0.2 m thick brick pitching.
 CANAL FALLS
659
Unbalanced head at 3 m from the toe of the crest
2.27 .
= 0.3 + 13.7 x 4 = 0.3 + 0.67 = 0.97

Thickness required=~:;~= 0.78 m; say 0.8 m.

Use 0.6 m thick concrete with 0.2 m brick layer.
Unbalanced head at 5 m from the toe

= 0.3 + ~3~~ x 2 = 0.3 '+- 0.33 = 0.63 m.

Thickness required

= ~:~! = 0.51m;Say0.55 m.
Use 0.35 m thick concrete with 20 cm thick brick layer, as shown in Fig. 12.19.
U/SWING
_.;,.;_ _ _WALL _ _---..,!'.,.,- o.s
_ _1051 PROFILE WALL
103.1 D/S WING WALL 103.6
1045 U/S f.S.L. o.s__;,LE---~:..!......-...z......:.;. _____h..-----1
-"'!"-~-----

1: 10

SLOPE PITCHFffo - - -- PRDflU-, ..- .-----------

WA LL

RL=1012 TOE WALL

CISTERN
BRICK PITCHING 20cm
THICK LAlD OVER 10cm
10m THICK BALLAST
__,_.._......____ - - - - ---. -
CUT OFF WALL

TOE WALL

1.S RETURN SLOPE PITCHING

_ .... c .......... .c.c...--'- -'--~~ -~-AJ_\.
.. _ .. .,_..,...,..____-. ..-_=.-:__-_-
__-__.,......___-_,,-__-__ """""I r - - - - - - i

Fig. 12.19. Details of the Sarda Type fall (rectangular crest) 6fox:ample 12.3.
[)Is Cu,rt(lin Wall. The curtain wall at the dis end of the floor should be 0.75 m deep
(for H = 0,76 min Table 12.1)
:f>rovid~ 0.4 m x L65 m deep curtain wall at dis end of floor, i.e. upto a level of
101.5- 1.65 = 99.85 metres, i.e. the deepest foundation level.
 660 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

Downstream pitching. From 'Fable 12.1,

' Total length of dis pitching
!',, =9+2x 1.5= 12metres.
:I',;I
'1
Pitching is kept sloped at 1 : 10. A curtain wall of 0.4 m x 0.75 m shall be provided
at the end of the pitching, as shown in Fig. 12.19.
Example 12.4. Design a 1.5 metres Sarda Type fall for a canal carrying a discharge
of 40 cumecs with the following data :
Bed level upstream =1os:om.
Bed level downstream =103.5m.
Side slopes of channel =I: 1
Full supply l_evel upstream =106.8 m.
Full supply level downstream = 105.3 m.
Berm level u/s =107.4.m
Bed width u/s and d/s =30m.
Safe exit gradient for Khosla '.s Theory = 115.
Solution.
Length of the crest. Is kept equal to bed width = 30 metres:
Crest level. A trapezoidal crest is provided, since the discharge is more than 14
cumecs.

r
The discharge formula is given by
I
2
Q = 1.99 LU/ (:,

Assume B = 1.0 m.
1
--- - , - - -c--c:--- -. -- H312 -~-'jif/6

40 = 1.99 x 30. 1/6

(1) .
or HS/3=. 40 0.796
1.99 x30
or H = (0.796)3/S = Q.862 m; say 0.865 m.
Velocity of approach

.. = Va= (30 + ~~ 8) 1.8 = 0.7 mlsec ( . Full supply depth = 1.8 m)

2
Velocity head = (OZ;) = 0.025 m.
u!s TEL= u/s FSL +Velocity head __ ____ _ _ __:..__-=---- --
.'!.J_-______ --:-- = roo.-s+o~o2s = 106.82Sm.- -------------
_:,,:..__~~-----

:. R.L. of the crest

I"::-
=u/sTEL -H
= 106.825 - 0.862 = 105.963 m.
Adopt a crest level of 105.97 m.
Shape of the crest. Adopt crest width at top
B1 =0.55.YH + d
cp.NALFALLS 661
! i
where H = 0.865 m
d =Height of the crest above dis bed
= 105.97- 103.5 = 2.47 m.
B 1 =0.55 ..J0.865 + 2.47 = 0.55"13.335=1.0 m.
Adopt a.trapezoidal crest with top width of 1.0 m and u/s slope 1 : 3 and dis slope
1 : 8.
Upstream Wing Walls. Radius of the wings should be 5 to 6 times the head over
the crest= 5 x().865=4.325, to 6 x 0.865 =5.19 m. Use 5.0m:radius for the wings. U/s
wing walls shall be. kept segmental with 5 m radius subtending an angle of60 at centre
and then carried tangentially into the berm.
Downstrea_m Wing Walls. The downstream wings shall be kept straight up to a
distance of say 6 .'-iH Hv i.e. 6 "10.865 x f.5 = 6.8 m; say 7 m, and then warped in a
slope of 1 : 1 and shall _be taken upto the end of pucca floor.
Upstream Protec_tion. Brick pitching equal to u/s water depth i.e. 1.8 m is laid on
the u/s towards the crest at .1 : 10 slope. Provide 20 cm drain holes in the entire length
at 3 m clc to drain out the uls bed during the elo~ure of the canal.
Upstream Curtain Wall. The minimum depth of curtain wall= ~rd water depth, i.e .
.!. x 1.8 = 0.6 m. Provide 0. 7 m deep masonry wall over 0.3 m thick concrete.
3 .
Thus, provide a curtain wall 0.4 m x 1.0 m deep on the u/s.
Downstream Curtain Wall. Minimum thickness
= Depth = _!.! = 0 9 m
2 2 . .
or from Table
.
12.1, it is equal to 0.75 m. .

Provide a dis curtain wall 0.4 m x 1.4 m over 0.3 m cement concrete~ Thus, total
depth of dis curtain wall shall be 1.7 m with its bottom level at 101.8 m.
Cistern. Depth of cistern
~ x = .!: (H. HL)213 ... (12.9)
4

R.L. of Cistern= 103.5- 0.3=103.2 m.

Length of cistern = 5 '-iH HL = 5 "10.865 x 1.5 = 5 x 1.14 = 5.7m.
Provide 5. 7 m fong Cistern
Total EloorLength andExit Gradient-
H 1
GE=d ~:\fr
Maximum static head (H) is caused when water is stored upto th~ crest level and
there is no water dis.
H = 105.97 - 103.5 = 2.47 m.
d = 1) m(i.e. Depth of dis curtain wall)
GE= 115 (given)
 'fl
uIi ______
ii
!rl

662 . IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

1 2.47 1
s=u. 7t.YA.
1 1.7 1
or 1t .,,ff= 2.47 x 5 = 0.137
From plate 11.2
a= 10
or E.:::: 10
d
or b= 10xl.7=17m; UselSm.
Minimum floor length required on the dis
= 2 (Water depth + L2) + fh
= 2 (1.8 + 1.2) + 0.865 = 6.865 m; say7 m.
Provide the balance length of 11 m under and upstream of the crest, as shown in
Fig. 12.20.

Fig. 12.20
b_plift Pressure Calculations
Assume u/s floor thickness as 0.5 m, and dis floor thickness as 0.8 m, and floor
thickness at toe of the crest as 1.3 m.
(1) Upstream Wall.
b = 18 m.
d= I.Om.
_!_ = !!:_ = _hQ_ = 0.056
a b 18.0
From Plate 11.1 (a)
<l>E I =0%
<l>D 1 = 100-<l>D = 100-16 = 84%
;.r :'----'---'---'----~--'<!>-cl= 100-<!>-= roo-.:2s~77%:--=c."-;-'-"--"'--'-''-"--- --'-'-'--.-------

_<;or_r_e_ction to <l>c1 for depth of floor

84%-77% 7 .
= . _ .0 x 0.5 = l x 0.5 = 3.5%(+ ve)
103 0 102
Correction due to influence of other wall is very small and neglected.
<Pc1 (corrected)= 77% + 3.5% = 80.5%
 ,1111
"il""1_,

II

CANAL FALLS 663 Iii

(2) Toe of Crest. i~ I

b 1 =11 m
b= 18m
!!.!.b = .!l
18
= 0.61

d= 103.2- 101.9 = 1.3 m

b 18
a=d=LJ= 13.9
From Plate 11.l(b)
<l>v2 = 36%
(3) Downstream Curtain Wall.
d=l.7m
b= 18m
~= ~; =0.094
From Plate 11.1 (a)
<l>El =<!>=29%
<l>vl=<!>v=20%
<!>cl =0%
Correction for depth to <l>E2
29%:....20% 9
=
1. x0.8=.ux0.8=4.2%(-ve)
7
<l>E (corrected)=
_ _ .. _3. .. .
29-4.2 = 24.8%
---cc-- ..

The levels of H.G. line for maximum static head are worked out in Table 12.2 and
plotted in Fig." 12.20.
Table 12.2

u/s dis Head Height/Elevation of H. G. line above datum

Condition offlow W.L.in W.L. in Hin <Pe1 <l>vl <Pc1 <Pv2 <Pel <l>vl <Pel
metres metres metres 100% 84% 80.5% 36% 24.8% 20% 0%
Maximum static 2.47 2.08 1.99 0.89 0.61 0.50 0.90
head, i.e. Water 105.97 103.50 2.47
up to crest level 105.97 105.58 105.49 104.39 104.11 104.00 103.5
on u/s and no
water
downstream
- ------ .. -
----------~------ ~

Floor Thicknesses
Proyide a nominal thickness of0.4 m under u/s floor.
Unbalanced head at dis toe of giacis = 104.39 - 103 .2 = 1.19 m

Thickness required= !:;: = 0.97 m; Use 1.2 m.

Provide 1.0 m thick C.C. overlain by 0.2 m thick brick pitching.

Unbalanced head at dis end of floor= 104.11-103.5 = 0.61 m.
 666 IRR~GATIONENGINEERING AND HYDRAULIC STRUCTURES

If the heightof the crest works out to be more they 0.4 times the upstream water
depth, the fall may1be flumed or fluming ratio increased, so as to increase Hand to lower
the crest. ,
Hump. For an unflumed non-meter ordinary fall, t}iebed approach may have slope a
- -of~: I (~ H :-IV);-joirr;d tarrgentially to-thew~-edgeor-t:lfe crest with a radius of H/2,
as shown in Fig. 12.23.
2:1
2 ,_
R=H

Fig. 12.23
For a flumed meter fall, the curve should start at the same cross-section as the side
, . - - - U+~ - .
.curves. The radius of hump or bed curve is e,qual to ~(see Fig. 12.24).

Upstream Protection. No upstream protectfon isrequiredfor an ordinary-unflumed

fall. However, for a flumed fall, dry brick pitching on edge may be laid both in bed and
sides for a length equal to u/s water depth (in a slope of 1 : 10 in the bed).

I
I.JP STREAM I
WI NG WALL
1
R=: S TO 6 TIMES H

v:' s ''
"-I - FLOW
....
CD
<
z
<
..... _ _ _1_ .c LINE
--------
Fig. 12.24
Downstream Glacis. A straight glacis with. 2 : 1 slope has been found to be quite
suitable for this type of falls. The glacis may be joined to the. crest at the u/s end, and
to the floor at the dis end, with a radius= H, as shown in Fig. 12.23.
Length an.d Thickness of Floor. The floor is designed as per Khosla's theory and
the total length of the floor is determined from exit gradient considerations, as was done
for Trapezoidal Sarda type fall.
mF"
~

. CANALFALLS 667
. . I .

Downstream Protection. With the provision of a deflector waU, no dis bed pitching .
is provided, and only side slope pitching is provided for a length equal to 3 times the
dis water depth and should rest on a toe wall.
Upstream Wing Walls. For an unflumed non-meter fall, the side walls may -be
splayed at 45~ from u/s edge of crest and carried into the berm for about one metre [as
was done in rectangular crest-Sarda type faU (see Fig. 12.19)].
For a flumed meter.fall, curved wings with a radius of 5 to 6 times H subtending
an angle of 60 at centre, and therefore, carried tangentially into tl:ie berm_[as was done
in Trapezoidal crest- Sarda type fall (see Fig. 12.17)] may be provided.
Downstream Wing Walls. For an unflumed fall, the walls are taken straight up to
the d/s end of the floor and then EN 0 0 F EX PAN s1~N
joined with ret1:1rn walls (as was done
in a rectangular crest - Sarda type
fall-Fig. 12.19,. -
For a flumed fall, the d/s wings
: shall. have to be expanded up to the
START OF
EXPANSION

T
s,
x- TB2

l
1
Bx
normal width of the river. The expan-
sion can be achieved straight with 1 ---t
in 3 slope for smaller works, and a
hyperbolic 'expansion (Fig. 12.25),
given by equation (12.11), may be i------L
used for large works. The hyperbolic
expansion is given by equation : Fig. 12.25. Hyperbolic Expansion.
. B 1 B2 L
B = ---'-----'"---- ... (12.11)
- x L,B2""(B2~B1)_x
where Bx is the width of expanding flume at any
distance x from: the oegirining 6f expan-
sion.
The wings shall be embedded into the berms by at least .1 metre.
Roughening Devices or Energy Dissipators
In Sarda type fall, where energy is dissipated by impact, roughening devices can.
serve as an additional source of energy dissipation. Similarly, in Glacis fall, roughening ~
devices may be employed to arrest the energy left after jump formation. They may help .
in jump formation also. Various types of roughening devices used i~ falls are given
below: ;,
- (1) Friction Blocks. These.are the most simpleccand iluseful of all such devices, They
consist of rectangular concrete blocks securely anchored into the floor. Their height is
approximately th water depth. The spacing between the blocks is about 'twice the height
of the blocks .. The specific recommendation's for their use in the modern falls are as.
given below :
(a) For Vertical Drop Fall or Sarda Type Fall with Trapezoidal crest. Two rows
of friction blocks staggered in plan may be provided, as sho"'.n in Fig. 12.26.
 668 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

The first row may start from a distance

1.5 Ye (where Ye= critical depth) from the --UT
dis toe of the crest. Spacing between the o/s TOE OF
,
rows is kept eqwll to Ye
, THE'CREST

-~
'o-- ~
C.C. ~OCK 0 F SIZE
,
- ~2Yc x Ye xyc
- --- ,.----S-pacing between"tnebTocks =-2 y-;_--~------
i_Yc LJ \ ,
The dimensions of the blocks
=2~eXYeXYe
In addition, cube blocks of size ~th to Fig. 12.26. Provision of friction blocks in
Sarda type fall.
1
16
th of water depth may be provided at the
end of the imperv~ous floor.
(b) For Glacis Fallt Four rows of
, I
staggered blocks are generally
provided hi case of flumed falls only.
Nothing is provided for unflumed
falls. The first row may be provided
''I at a dis~llce equal -to 5 x height of
I block from -the toe of the glacis, as
shown in Fig. 12.27. The height of the
block is generally kept I/8th of the
water depth. The distance between
- two rows is equal :oh (where his the
height o_f block). -
_T_he _dimensions of the bl~c.ks ~e -- . Fig: 12._~? Proyision_of_(riction blocks
3hx hx h. - in flumed glacis fall.

(c) For Baffle Fall. Two rows of friction blocks staggered in plan may be proyided
only when the drop is more than 2 metres. The suitable dimensions are
I 'i

Height (h) = 0.26 Y2

where y2 is the sub-critical depth in the canal dis,
required for-fortnation of jump for paral-
lel downstream sides, in metres. '
Length
Top width.
__ __._.:_DisJa._nJ:_e_b_et.w~e_en.__i:.ows..:.:.....:._- :...;::Ef::h.. _::::: ___ :...: __ :_:. -_::.:_ - - _::-_ - . - ~ ------ ---------- ---

The first row may be placed at a distance trd of the cistern length from the upstr~am
end of-. the
. cistern., .
(2) Glacis Blocks. A single i:ow of blocks called Glacis blocks and of the same
dimensions as friction blocks may be provided just at the dis-toe of the glacis, in case
of flumed falls with drops more than 2 metres, as shown in Fig. 12.28. It helps in
reducing turbulence in flow, which in turn, reduces wave wash, thus ensuring uniform
flow. -
 !:
CANAL FALLS 669 1

ii

DEFLECTOR
WALL

END OF GLACIS ie Cl STERN

D/S TOE

Fig. 12.28. Provision of glads blocks in flumed Fig. 12.29. Provision of a deflector walL
falls with drops more than 2 metres.

(3) Biff Wallor Deflector Wall. A deflector wall of height approximately 1 th of

10
water depth may be provided anhe downstream end Of concrete floor (irfcast w:ith d/s
cutoff) as shown in Fig. 12.29.
Example 12.3. Design a straight flumed meter glacis fall with the following data :
Full supply discharge of the canal . = 120 cumecs.
Bed level of the canal upstream " = 107.5 m
Bed level of the canal downstream = 106.0 m
Drop (HJ ::: 1.5 m.
FSL of the canal upstream = 109. 7 m
FSL of the canal downstream = 108.2 m
Bed width of the canal uls and dis = 60 m
1
Safe Exit Gradient for cancit material = S
5
Solutio
Length of Crest. From Table 12.3, for HL = 1,.5,
flumingratio =75%
:. Length of the crest
=75% of Bed width of canal
= 75%x.60 ril=45m.
Adopt crest length =45 m.
Crest Lev.el. Sine~ the fall is to be used as a meter, a broad crest shall be provided.
:. Dischar:ge Q is given by
. Q = l.7~tH312.
120 =:.1'7
''. ... ..,. 312
''4:C:H
.~~..

or . 1'(,i F:e ~ := 1.57

~ . Vet~#Y~:;~::s:876
/ [assurriill.g l: l siit:~:b:f' canal and taking depth =2.2 as given]
,, ~.i.
 ~
i

670 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

2
ve l oc1ty
u . head = 2<0x879;6)81 =0.039 m; say
. 004 m.

u/s TEL = u/s FSL +Velocity head

= 109.7+0.04= 109.74m.
----erest-Level---=-u/s.'fEb--- H--'-'--------
. = 109.74-1.35 = 108.39m.
. Adopt crest leyel = f 08.39 m.
Ijeight of, the crest above u/s bed
= 108.39- 107.5 = 0.89 m.
wbk):i is approximately equal to 0.4 times the upstream water depth (2.2 x 0.4 = 0.88 m)
. and herice, we can go ahead with this fluming ratio and the present crest level.
Crest Width. Adopt crest width for broad crest
= 2.5H = 2.5 x 1.35 =3.375 m.
Provide crest width = 3.38 m.
- .Hump: The crest shall be joined with u'p~tream~oea withactirve ofi'adiU-s-
L2 + h2
R=-a_ _
2h
where La= 2 m (assume).
h = 0.89 (ht. of crest above u/s bed);
:. Radius of the curve
2
R = 2 + (0.89)2 =4.79 = 2.69 m
2 x 0.89 1.78
Cistern. The dis glacis is provided in a slope of 2 : 1. The cistern level is worked
outas below: . ---- --- -------- . ----
Q = 120 cumecs.
Width provided = 45 m.
1
q =Discharge intensity = ;
5
o= 2.27 cumecs/metre
HL=l.5m._
From plate 10.1, Eli = 1.85 m.
Level at which jump will form
= d/s TEL-12 = 108.24- 1.85=106.39 m.

__-~:2~~!2 .=:= 2~~ ~-Ill - -.

R.L. of Cistern =dis TEL - 1.25 E12
= (108.2 + 0.04) - 2.31.
= 108.24- 2.31=105.93m. '
which is lower than the dis bed level, and hence, adopt cistern level= 1~5.93 m.
Length of Cistern. 5 to 6 times Ef2, i.e. 5 x 1.85 = 9.25 m; or 6 x 1.85 = 11.10 m.
Hence, use 10 m length of cistern joined to d/s bed in a slope of 5 : 1 in a length of
5 (106--105.93) = 0.35 m.
 CANAL FALLS 671
. .
Downstream Curtain Walls. Depth of dis curtam wall below bed = water2depth
::::
2
~ :;;;: 1.1 m; or from Table 10.2, it is equal to 1.35 m, therefore, adopt a curtain wall
of total depth equal to 1.4 m, i.e. 1.1 m masonry laid over 0.3 1!1 cc,mcrete.
Total Floor Length and Exit Gradient. Maximum static head (H) is caused when
water is stored up to crest level on the u/s and there is no water dis ..
Maximum static head (H) = 108.39-106.0=2.39m
Depth of d/s curtain wall (cl)= 1.4m.
Safe Exit gradient Ge=
5~5
H 1
But Ge=d n:\lr
!
1 2.39 1
5.5 = 1.4 . 1t "Ii:
1 1 1.4
or 7t :,rr= . x
5 5 239
=0.107
From Plate 11.2,
ex= 17
Total floor. length required
=a d= 17 x l.4=23.8 m. ;.say24m.
The distribution of floor length is shown in Fig. 12.29. 20.65 mis already provided
and the balance of 3.35 m is provided as u/s floor length.

Upstream C~:tain Wall. A curtain wall of minimum depth ~u =

22
= 0.73 m must 3
-be provided at the end ofpucci floor. Let us provide an u/s curtain wall (cutoff) of
0.4 m x 0.8 m deep masonry wali laid over 0.3 m thick cement concrete, thus giying a
total depth of cutoff as 1. i m.
Uplift' Pressures. Refer Fig. 12.30. Assume floor thickness on the u/s = 0.3 m
' . ' \

and thickness on dis := 0.8 m

Assumed 108.39
1075
03m
I
~f.~~L' I I I I
C"-:t'+',.+-:~--?-5~2 : ;_~-=-~-. :. -=-~ -=-_;_,_o_;_ :.:. - : :.-: :.:-=-= =~=: . ________ _
..

Fig. 12.30
Vis Curtain Wall
b=24m
d= 1.1 m
1 d 1.1
a.=b= 24 =0.046
IiI.If'

I
672 tRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

From P_fate 11.l (a).

<l>E1=100%
<!>c1= 100 - <!>c = 100 - 20 = 80%

------------
<!>v = 100-<!>v=l00-13=87%
- -1---- --- --- ----- -- -- ..

Correction
. .
for depth. to. <!>cI

= 37 %- SO% x 0.3 = 1.91% (+ ve)

I - 1.1 ' .. .
I i'I
<!>cI (corrected)= 86 + 1.91=81.91 %
-

Dis Curtain Wall

I b=24m

I
d= l.4m
1 d 1.4
<X=z;= 24 =o.os83
From Plate 11.1 (a)
<1>2= <l>E= 23%
<l>v2=<!>v=15%
<!>c2 =0%
Correction for. ::"depth
..
to <l>E2

, =
23
% 1~41 ~% x 0.8 = 4.6% (-ve)
<l>E2 (corrP.cted)
...
= 23% + 4.6%
-
= 27.6%
- 'cT11e'-1-eyels 6f'f!.G. lines for maximum static head and flow at FSL are worked out
'
'in Table
..
12.4-
-
and plotted in Figs . .12.31 (a) and-(b), respectively. -
'./-

Table 12.4
--
. iieight1eievatio~ of ii.a. ii~eabov~ datu,;.-
\

Condition
offlow
Iinulsmetres
W.L. dis W.L.
in metres
Head in
metres qlEI qlDI qic1 . q,2 qlD2 qic2
100% 87% 81-91% 27.6% 15% 0%
Maximum 2.39 2.08 1.96 0.61' 0.36 0
Static Head, 108.39 106.0 2.39
i.e_ Water '
upto Crest 108.39 - 108.08 io7.96 106.61 106.36 106.0
.level on u/s
and no
_ Y1atecon__dls,, ----~--- -------- --- -- .. I- --
1.5 1.30 1.23 0.41 0.22 0
Flow atFSL 109.7 108.2 1.5
l 109.7 109.50 109.43 108.61 108.42 108.2

Floor Thicknesses. Provide 0.3 m thicknes_s under upstream floor.

At Toe of Glacis. Level of H.G. line (for static head) at toe of glacis.
= 106.61+ 107.96;106.61x 10.35.
CANAL FALLS 673
107.96 108.39 108.39 SUB SOIL H.G.LJNE
10 7-5 '-.,..c:...:_~----::...
_-_--.-_-- FROM maxm STATIC HEA-0
107.21
- - - - - 106.61106 0 -.1.- -
105.83 - - - ..
1064 ..
r:-3.35_-\-2 +3.38 ~ 4.92 - - - - 1 0 . 3 5
. 240 -------~;::;::;
104.6
Fig. 12.3L-(a) Maximum static. l)ead condition.
109. 70 H.G. LINE FlDW
109.43....:;._ -
- 109 ,~ _{AT F.S.L

2.43 r - 108.61
10 a. 22

. 106-0
---10593
-----1035
9.6
-1---11.27 _ _ _ _ ___...,.

Fig. 12.31. (b) Flow at FSL condition.

= 106.61+
1 5
2! x 10.35
I = 106.61+0.59= 107.20m.
. I
Unbalance.cl head at the toe due to static head
). = 107.20- 105.93 = 1.27 m
Ju,;,,p formation
q,=.~2..2}c_umecstmetre,,-_. __ ..... .
H=L5m.
E~12 =
' .
1~85 m.

. . . E11 (12 +Hi= 1.85 +L50 = 3.35 m.

-From PI'~te 10.2
. y 1 (for E1 ) = 0.3 m.
' I I/
Y2 (for ~2 ) = 1.7 m.

The maximum ordinate at point of jump; from Fig. 12.31 (b) = 2.43 m.

~rd of this ordinate = ~ x 2.43 = 1.62 m, which is ll}Ore than that for static head.
- ... .
-------- . -- - ----- ---
-----.-------
- - -- . " ... .
------ ". - ..
-- - - -- . --- ------------
-------- .. - - -
-- .
. -
- - - . -' -
--------
Hence, the floor thickness at point of jump must not be less than

= ~:~~ = 1.31 m.
Provide 1.5 m at this point and extend upto 2 m beyond toe
Level of H.G. line at 2 m beyond toe (due to static head)
1.35 . .
= 106.61 + x 8._35 = 106.61+0.47=107.08m;
24
 IRRIOATION ENGJNEERINO ANP HYDRAULIC STRUCTURES

Uri!:>al;mceg h.e!J.g = 107.08- 105.93 = l.l5 m.

TP.ic:\(ness reqired = i:~! = 0.925 m.., PFQVicJe 1.0 m.
Level of I,I.O. Hn~ at J m beyond toe {due to static head)
-- -~= io6~pr~\~5 >c535=Too:6r+o.3o= 106:91
Unbalanced head due to static head at this point
= 1Q6.91 - 105.93 = 0.98 m.
Thickness required = ~:;! = 0.79 m., Use 0.8 m.
Upstream ,Wing Walls. Curved wing walls with. radius equal to 6H = 6 x 1.35
= 8.10 m subtending an angle of 60 at centre and then carrieq tangentially into the berm
for a suitable length shall be provided as shown in the attached chart Fig. 12.32.
Downstream Expansion. The width of 45 m shall be expanded to 60 m in a splay
of3: I.

) = ~2.;m.
3 60 45
Minimum length of expansion required= < -
2
The wings can be splayed straight from the dis toe of the glacis in Ii length equal
, to 22.5 m. The bfd shall have to be pitched upto the end point of wings. Slope pitching
shall be provided in a length 3 x 2.2 =,6.6 m beyond this point.
Upstream pitching in a minimum length of 2.2 m (eq1,1a,l to u/s water deptP.) is
required.It shall, however; be provided as shown in Fig. 12.32.
Energy Dissipators. Four rows of St0;ggered friCtion blocks' shall pe provjqed. The
height of the block is tth of water depth, i.e. ix 2.2= 0.275 m. The first,row shall b~ at
a distance 5 x 0.215 = 1.37 5 m from the dis toe ofglacis. The-size--ef the blocks sha)l be
0.825 m x 0.275 m x 0.275 m. Distance between rows = 0.275 m.
A deflector wall at end point of floor is provided and is to have a h,eight equal to

w= .
Yd . 2.2 O22
W' i.e. m.
The detailed arrangement is shown in attached chart Fig. 12.32.
12.8. Design of a Bame Fall or Inglis Fall
Certain flumed type ordinary straight Glacis falls constructed in Punjab were later
found to give some serious troubles, which gave rise to the conclusion that considerable
surplus energy might remain in water even after the jump formation. One major cause
of these troubles was found to be, too, rapid expansion after fluming, which may generate
_CO.!~Sf~W_ eddi~s causing cd~-~P-J!~ours. R,esearch was carried out to eliminate these
defects and Baffle fall w11s
evolved. '
A baffle fall makes use of
the principle of horizontal im- PLAT. BAFFLE
pact for . energy dissipation. The
FD RM Ii
Cl STERN.
jump is held stable on .a horizon-
tal platform by means of a baffle
wall (called baffle).
Fig. 12.n Baffle fall.
 1,
CANAL FALLS 675
I'
I
Baffie Platform. The horizontal pJatforrp. is pn:ivideq at the level at which tqe jump
would normally form. This can pe determined. by Blench curves in case them js no
expansion of the.wings in the region of supercritical jet. If the supercritigaJ jet is splayed,
the optimum level at which the baffle platform shoulq be provided, can be determined
by designer's curves (given in C.B.I. publication No, 10). In the absence of curves, tpe
values can be determined by using the formulas given below. '' !i
Subcritical depth {y2) require<J. for jump formatipn in ordimiry c~ses without flming
is very nearly given by
y 2 =0.98q- 5-iHiJ.L - -- ... (12.10)
where q = the discharge intensity in cumecs/metre
Hi= Drop in metres.
y2 = the subcritical depth in metres.
The subcritlcal depth [y2 (flu~ned)] required for jump formation, in case there is a I'
fluming, is given by
y 2 (flumed) =y2 + (Hx-Hv \.'.\ ... (12.11)
where Hx is the calculated drop in metres given by
Hi
Hx= K 152 .-.(12.12)

where K is the fluming ratio (more than

1), i.e.
_ Actual width of canal before fluming
- Flumed width of canal
The R.~. of the baffle platfQ..rm will then be given by
(i) for ynfll.lmed fall= dis FSL- - h
(ii) For flmed fall =dis FSL -y 2 (flumed)
B~ffie Wall. Height of the baffle wall =hb ='yc __: y1 ... (12.13)
where Ye is the critieal depth, given by
I
.Y =[!L2]1/3. ... (12.14)
1

,,
11

c g
and y 1 is prejump super critical depth, given by
Y1=0.183 q 89 Hx
0 35
. (12.15)
Hx =Hi. when there is no fluming

Thickngs~ q[ th.e b_af[l_~ W(lll =~ hb ... (12.16)

Length of baffle plq.tform = 5.25 hb. ...(12.17)
Cistern. A cistern of length equal to 5 y2 (flumed) (equal. to 5 y2, for without
fluming) may be provided after. the baffle wall. The depth of the cistern below the
downstream bed should be 10% of the downstream water depth (yd), subject to a
minimum of 15 cm for distributaries and minors and 30 cm for main canals and branches.
Upstream Wings. The upstream wings can be curved with a radius equal to
3.6 H.312 (where His the total head over the crest) when His more than 0.3 m, and equal
676 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

to 2H when His less than 0.3 m. The circular wings are continued till they subtend an
angle of 60 at the centre, and afterwards they can be extended tangentially for the
required length into the berm or bank.
Downstream Wings. The _downstream divergence_ for flumed meter falls can be
provided at a slope of 1 in 3 to 1 in 4.
A milder divergence is preferabk for straight Glacis fall as well as for Baffle fall
(1 in 4 to 1 in IO dep~nding upon the bed width depth ratio oft.he downstream channel);
but that will make t~e structure costlier. Hence, adivergence of 1 in 3 is generally used
for meter falls of bo'th types. The dis wings of the unflumed baffle fall are kept as are
kept for unflumed $traight Glacis fall.
Downstream ,Glacis. The dis glacis for unflumed baffle falls is kept at 2/3 : 1; but
for flumed meter falls, it is kept as 2 : 1. The glacis is joined to crest at the u/s end and
to the floor at the dis end with a radius equal to H, as was done in the case of a glacis
fall.
All other details of pitching, etc. rernain the same as for straight glacis fall.
The friction blocks are either not provided (upto 2 m fall); or provided as explained
earlier (for drops of more than 2 m.)
Example 12.6. Design an unflumed non-meter baffle fall for the canal having the
following data :
Full supply discharge = 30 cumecs
Bed level uls =
203.0 m
Bed level dis = 201.2 m
FSL uls = 204.3 m
FSL dis = 202.5 m
Bed width = 28m
Drop (Hu = 1.8m
Side slopes of channel =1: 1.
Solution. "

Crest Length. Equal to bed width : Provide 28 m. crest length

Crest level. A sharp narrow crest is provided, for which
Q = 1.84 LH
312

30 = 1.84 x 28 H312
3/2 . 30
or H = 1.84 x 28 =0.5 82
or H = (0.582)213 =0.697 n,i; say 0.7 m.
Velocity of approach ..:_..:.....:.
_:_ ------- -_
30
=Va= (28 + 1. 3) 1.3 . 0.787 m/sec.
Velocity
. .Hc~ad -_ v~g -_(0.787)
x _
2
0~0315 m; say 0.03 m.
2 2 9 81
Now, u/s TEL= u/s FSL +Velocity head
= 204.3 + 0.03 = 204.33 m.
Crest Level = u/s TEL - H
= 204.33 - 0.7 =203.63 m. f
 PROFILE WALL
~LED DhAMAL!S

R.L.1060

.
,~:
r

+-~-9_2 __

SECTION A-A

..J
UJ
>
UJ
....
~
a:
~k-~~~~---,,.-~~~~~~~~~~~~77'~
- - - - I - - - - - - - - - - - - - - - - - - - - - - - - - - --

SEO LI.NE
---=~-, - - - -:/ / ;~ff~
<:-I
,/
~I I
81
..:;t Lt>
r..
IJ1
~
...J
u.1 20cm DRY BRICK PITCHING,IN BED,
I OVER !Ocm THICK BRICK BALLAST
...J 0
I ;)
UJ
>
6 :::;
UJ
...J
"' BED - I
"'
~
c URVE..:..,.~ REST
UJ
1- - al ! - - - - - - - - - - - - - - - .. - -------

~
A
I

TOP PLAN
FLUMED GLACIS FALL. (FIG. 12.23 )
Fig.1232
r~~F~
I Adopt crest level = 20~.6 m.
677

I Width of the crest is kept equal to t H =t x 0. 7 =0.47 m

Provide crest width = 0.47 m.

U/s Glacis. Glacis of I: 1 joined tangentially to the crest with a ra~ius equal to
!f.2 =0.35 m shall pe provided:
.

Dis Glacis. Glacis oft: 1 joined tangentially to the baffle platform with a radius.
equal to H = 0.70 m.shall be provided.
Upstream Wings. The u/s wing walls shall be splayed at an angle of 45 from the
u/s end of the floor and shall be embedded i~to the bank by 1.0 m beyond FSL line.
i

Downstream Wings. Parallel vertical sides up to the end of pucca floor shall be
provided, which shall be connected with the return walls at 90.
Upstream Protection. No pitching is required in bed and on sides. Depth of u/s
curtain wall required_is
Yu 1.3
=3=3=0.43m.
Provide 0.4 m x 0.6 m deep curtain wall over 0.3 m foundation concrete, thus
making its overall depth as 0.9 m.
Baffle Platform and Raffle Wall
--Baffle Platform. Since there is no fluming,
Hi=Hx=l.8m
Now y2 = 0.98 q 52 H2 21 . (i.e. Eq. 12.10)
. 30
where q = = 1.07 cumecs/rrietre
28
Y2 = 0.98. (l.07)0.s2. (1.8)0.21
= 0.98 x 1.0358 x 1.133 = l.'15 m.
R.L. of baffle platform
= d/s FSL -y2 = 202.5- 1.15 =201.35 m.
Provide Baffle platform at R.L. = 201.35 in
- -. - -- -- ----
.
----- - -
------ --- ---~---
. -
------- --------- --- -- - - -- ------. ------- ------- -------
" - --
Baffle Wall. Height of the baffle wall
=hb=Yc-Yl

where y,=[1r =[<~~7'r =(0.117)'"=049m

Y1 = 0.183 . q0.89 . (Hx)-0.35 (i.e. Eq. 12.15)
= o.l 83 (i .o7)089 (1.8)- 035 = 0.16 ~
 678 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES

:. Height of the baffle wall

hb=Yc-YI
=0.49-0.16= 0.33m.

Thickness of baffle_wall =.~hb :::::_0.22m. _____ _

Length of baffle platform
= 5.25 hb = 5.25 x 0.33 = l.'.73 m. ; say 1.8 m.
Cistern. Depth of cistern below dis bed
= 0.1Yd=0.1x1.3 = 0:13 m.
R.L. of cistern= 201.2- 0.13 =.20l.07m~
Length of cistern = 5 y2
=5X l.15=5.75m;say5.8m.
Dis Curtain Wall. Depth of the downstream curtain wall required
Yd 1.3 ....
=1=2=0.65m; .

or from Table 12.1, the depth for the c.urtain wall is equal to 0.6 m.
Provide 0.4 m x 1.0 m deep .curt~in wall, over 0.3 m thick foundation concrete, thus
making a total depth of curtain wall= 1.3 m
Height of deflector above dis bed
. Yd l.3
=10=10=0.l3m.
Hence, the dis curtain wall shall be rais".d by 0.13 m above dis bed.
Dis Bed.Pitching. No pitching is req_yir~9-:_ ..... ___ : _
a
Dis Side Slope Pitching. Is required {n length equal to .
3 Yd= 3 x 1.3 =.3.9 m.
Provide 0.2 m thick dry brick pitching over 0.1 m thick !;>rick ballast in a length equal to
3.9 m. The slope pitching shall rest on a toe wall 0.4 m thick and 0.8 m deep (overall)
constructed in the bed at the junction of bed and sides. A solid profile wall called
'Dhamali' shall be constructed at the end of pitching. It shall be 0.4 m thick and plastered
in ceme11t mortar. '
Friction Blocks. Not required.
Total Floor Length from Exit Gradient Considerations
Safe exit gradient = 1/5
:--'--'------Maximum,..static--head-fffl-is exerted-when water is stored upto ciest level on u/s
and there is no water on dis. '
H = 203.6- 201.2 = 2.4 m.
Depth of dis curtain wall = d = 1.3 m
. H 1
_,Now, GE=d 1t 'fr
1 2.4 1
s=u1t'!r
CANAL FALLS 679
1 1 1.3
or ;::Jr= 5 x 2 .4 = 0.108
From Plate 11.2,
a= 16.5
:. Total floor length required= ad= 16.5 x 1.3 = 21.45 m;
Provide 22 m overall length.
The floor length already provided (12.365 m) is shown in Fig. 12.34 ; the balance,
i.e. 9.635 mis now provided .on the u/s, as shown in Fig. 12.34. .

Fig. 12.34. Dimensions of the baffle fall of example 12.6.

Calculations for Uplift Pressures
(i) Upstream curtain wall
b=22m
d=0.9m
1 d 0.9
-=-=-=0.041
a b 22 .
From Plate 1 L1 (a),
<l>E1 =100%
<j>D1 =-
100-<j>D=
- .
100%-14%
.
=86%
<J>c1 = 100-<j>E= 100%-20% =80%
Assume upstream floor thickness.= 0.3 m.
Correction to <l>cl for depth
. 86%-80%
= _ x 0.3 = 2.0% (+ ve)
09
<l>c1 (Corrected.) = 80% + 2% =82%
(ii) Downstream curldin wall
b=22m
d= l.3m
_!_ = ~ =ll. = 0.059
a b 22
From Plate 11.1 (a)
<j>E.2 =<j>E,;,,23%
 680 IRRIGA TIOf/ ENGINE.J::RING AND HYDRAULIC STRUCTURES

<i>D2 =<j>D=16%
<Pc2 =0%
Assume dis floor thickness near d/s curtain wall= 0.7 m
Correction to <l>E;for flvvrthiT:kness - - -- - -

= 23% - 16% x 0 .7 = 3 .8o/co (~, ve)

1. 3
<1>2 (Corrected) =2~% - 3.~% = 19.2'fo.
Uplift pressure at the toe of the glacis
'= 19 .2o/c 80%- 19.2% 8 22
' 0 + 22 ' x, .
= 19.2% + 22.7% = 41.9% '
Floor Thicknesses
U/s Floor. Provide a nominal thickness .of 0.3 m on the upstream sid.e and extend
it up to .dis end of crest. Its bottom fovel shall be at R.L. 202.7m. .
Toe of Glacis
Level of H.G. line at toe of glacis
= 201.2+ 41.9% x 2.4 :'.''
= 201.2 + 1.0 = 202.2 m
:. Unbalanced. head due to this maximum static head of 2.4.m
= 202.2:.... 201.35 = 0.85'm
Unbalanced head due to dynamic condition may be taken as
= 50% (y 2 -y1) +%pressure HL (Refer page 558 Chapter 11)
=50% (l.15-"-0.16)"+4L9%x 1.8

l !
= 0.495 + 0. 754 = 1.2.49 m ; say 1.25 m.
Thus, at toe of glacis, the head due to dynamic condition is more than that due to
static condition. Hence, minimum thickness _required at toe = ~:~; = 1.01 m. Provide
I.! '
t1 1.35 m thickness in the entire length of baffle platform, thus keeping its bottom at R.L.
201.0m.
Thickness at the start of cistern
Percentage pressure at 2.02 m from toe of glacis (i.e. start poi.nt of cistern)

=192o/c 80%-19.2% 62 .
. o+ 22 X .
"-- . .__________________ = 19:2-o/o+:T1:2 ;;;3"6~4% -
Level of H.G. line at this point
,;,, 201.2+ 36.4% (2.4)
= 201.2 + 0.87 = 202.07 m.
Maximum unbalanced head at this point
= ,202.07 - 201.07 = 1.0 m
 R.L.20165
R.L. 2049 u/s WING WALL
,RETURN
\ WALL

R.L 2049

-------
9235

1378
220
SECTIONAL ELEVATION
104 104
-j1 j- TOP OF -+j1 ~ PROFILE
PITCHING WALL RL 202.1

OZ THICK BRICK _F__g_ - - SECTION 1-1 ....

r 03U/S GLACIS PITCHING SECTION 2-2
R.L.2031
SED LINE BED LINE

0-4 CUT OFF

"""'f------235------
022
18-11!'4--..---58----l"'-
"'6 "'6
u/s FLOOR AT R.L .2031 O/S CISTERN
BAFFLE WALL

BAFFLE PLA1'FORM

:' <1'
"'6
."cc: ,:
.'; ... : . -~ !". ~ ,,
R.~.20000

_____________SE_C_UQN._3_::3 .. ----,-:::.: :-:-,-- SEC-Tl-ON-4'":4.:e---: :---::

.... ~
--!041-- :

PLAN 2~1 2C
"lf 4 4'
T:~f Wt.l L

UNFLUMED BAF' FLE FALL

SECT)ON 5-5 SECTION 66
Scale 1i:m = lm
Fig.1235. SCi>.~E 1 c m=05 M
 cANALFALLS 681
Floor thickness required at this pofot
., 0
= 1'~ 4 =0.81 m
,L,

Provided thickness 0-= 201.07 - 200.0 = 1.07 m.

Thickness required at 3 m beyond the baffle wall
= 10 ')~ + 80%-19.2%
-'- 0 22
32
x .
=19.2% + 8.85% =28.05%
Level of H.G. line at this point
= 201.2+ 28.05% (2.4)
=201.2+ 0.67 =201.87m
Maximum unbalanced head at this point
= 201.87 - 201.07 0.80 m =
Floor thickness required at this point
=.Q:_SO
1.24
=0.64<:_, m
Hence, provide 0.7 m thickness in the remaining portion, as shown in Fig. 12.34.
Full details of the fall are shown in the attached chart Fig. 12.35.

PROBLEMS
1. (a) What is meant by 'Canal drops' ? Why are canal drops constructed in a canal system ?
(b) Enumerate Ll-ie various types of canal drops which have been used since olden days. Explain in
details the design principles go_verning any one of he modem types.
2. Why are "drops" constructed in an irrigation canal ?
Draw a neat sketch of a syphonweH drop and-explain brief!y:its eompenents .... ---- --- ....
(Madras Universiry, 1973)
3. (a) What is meant by "falls" and where are they located ?
(b) Discuss briefly the components of various types of falls with neat sketches. Also discuss the
suitability of each type.
4. Sketch a syphon well drop to carry 0.3 cumec of water from the following data :
Ground level = 30.00 m.
B.L. of channel above. drop =28.00m.
F.S.L. of channel above drop =30.00m.
F.S.L. of channel below drop = 28.50 m.
B.L. Of channel below drop = 26.50 m.
Provide a 4 m cart track over the well drop. The earthwork connections should be clearly shown.
Assume any other data that you may require. (Madras University, 1976)
5. (a) Explain- why trapezt)id:i.1 notches are preferred<o rectangular notchesiilthedesign of canal
drops. (Madras University, 1974)
(b) Design the size and number of notches required for a canal drop with the following particulars:
Full supply discharge = 20 cumecs
Bed width = 14 m.
F.S. depth = 1.9 m.
Assume any other data if required. .
Ans. [6 Notches as with l = 0.83 m and n = 0.37, provided in a crest raised above the bed level by
0.6m.)
 682 IRRIGATION ENGINEERING AND HYDRAULIC STRUCTURES
6. (a) Discuss the comparative merits and demerits of Notch falls and Sarda type falls.
(b) Design 1 m Sarda type fall on a channel carrying 20 cumecs discharge whose bed width and
water depth are 14 m and 1.9 m respectively.

_______________ 7. (a) What_ are_canaUalls_and why are_they_constructed_1 ___ , ____ _

(b) Design an unflumed non-meter hydraulic jump type fall (i.e. glacis fall) on a ciistributary carrying
10 cumecs discharge. The drop to be affected is 2 m, and the depth of flow and the bed width of channel
are 1.4 m and 8.0 m respectively.

8. (a) Discuss the principle advantage offered by a baffle fall in comparison to a straight glacis fall.
(b) Design a flumed baffle fall for a canal having the following data:
Full supply discharge. of the canal = 120 cumecs.
Bed level of canal upstream = 107.Sm.
Bed level of canal downstream = 105.5 m.
Drop (HJJ =2m.
F.S.L. of canal upstream = 109.7m
F.S.L. of canal downstream = 107.7m.
Bed width of canal u/s and dis - - =60m.
Bligh's safe hydraulic gradient for the soil =C= 12.

9. (a) What are "canal falls" and where are they located ?
(b) What is meant by "flumed falls" and what are theifadvantages ?
(c) Discuss how fluming is done for flumed falls. Write down the equations etc. that you will adopt
for designing the upstream as well as downstream wing walls of flumed falls. -
(d) What are "roughening devices" ? Discuss their use in 'falls construction' ? What roughening
device would you recommend for a straightglacis fall ? (a) when flumed (b) when not flumed.
10. (a) State briefly how you will. decide the location of a canal drop. Explain .how a trapezoidal
notch type fall helps-to maintain depth-discharge-relation-in the canal:
(b) Following data: were observed in a canal fall:
Full supply level = 115 m.
Bed level of canal = 112m.
Bed width of canal =15m.
Full supply discharge =30 cumecs
Side slope of canal =2 (H): 1 CV)
Length of crest of the fall = 10 m (crest section is rectangular)
1/2
Coefficient of discharge over crest = l.70m /sec.
Calculate the crest level.
[Solution. Q = 30 cumecs
y=3m.
-- ~B-=--15 m ..
A= (B + 2y) y (with 2H: 1 V slopes) :::63 sq. m.
Va = velocity in channel

= velocity of approach = q !~
= = 0 .48

~ .
=0.0117;say0.012m
28

-
:. Uls T.E.L. = uls F.S.L. +-
~
g = 115 + 0:012.= 115.012 m.
2
CANAL FALLS 683
Now, Q = 1.7 L ~ H 312

30 = 1.7 x 10 H 312
H= l.462m.
(
:. Crest level = uls T.E.L. - H
= 115.012- 1.462 = 113.65 m. Ans.]
U. Write short notes on any three of the following :
(i) Syphon well drop.
(ii) Roughening Devices.
(iii). Wing w;tli for flum~ falls.
(iv) Trapezoidal notch fall.
(v) Simple vertical drop type fall.
(vi) Sarda type fall.
(vii) Straight glacis fall
(viii) Baffle or Inglis. fall.
(ix) Different types of falls and their suitability for a particular project.

- -- --- -- -'---~--'-