Final project for haydroulic engginering

Final project design of diversion structure
FACULITY OF CIVIL AND WATER RESOURCE ENGINEERING
DEPARTMENT OF HYDRAULIC & WATER RESOURCE ENGINEERING
PROJECT TITLE: Weyizero wuhaDiversions Weir Small scale
Irrigation Project
SubmissionDate 24/09/2009 EC

i
DECLARATION
we are a fifth year Hydraulic and Water Resources engineering students and this is the final year
project under the academic advisor supervision. Certified further, that to the best of our
knowledge the work reported here is not from part of any other project report which mean that
this project is worked by us successfully.
Students name ID number Signature
1. Kedir Jemal 0501610
2. Mehari Kiros 0501803
3 .Rabia Ali 0502185
4. Melat Gete 0501849
5. Matiwos Mekonnen 0501774
6. Kindu Melese 0501660

ii
Approval of the Advisor
I approve that those students has done the final project report by themselves.
Name of Advisor Signature
Misbah Abdela ……………

iii
ACKNOWLEDGEMENTS
First and above all, we praise God, the almighty for providing us this opportunity and granting us
the capability to proceed successfully. We would like to express our gratitude to all those who
gave us the possibility to complete final project. Especially we are deeply indebted to express our
thanks to our advisor instructor Misbah Abdela who gave us a complete and series comments
and suggestion how to proceed. He also gave countless help and advices that encourage us to
finalize this project. Also we wish to thanks Amhara Design and Supervision Works Enterprise
for providing us the necessary data for our project and also thanks to tour parents for their
tremendous contributions and support both morally and financially towards the completion of
this project. Finally Our thanks and appreciations go to our friends in developing the project and
people who have willingly helped us with their abilities.

iv
Executive Summery
The design of Weyizero wuha small scale diversion headwork irrigation project will enable the
farmers to use the available water and land resources efficiently and get themselves food
secured. Weyizero wuha diversion irrigation development project area is found in South Gondar
Administrative zone of Amhara National Regional State. It is located in Nifasemewcha woreda,
Keble 03. The specific location of the project site is called Weyizero wuha.
This report contains six different chapters. The first chapter, is about introduction of the project
which includes the Back ground, Objective, methodology, which includes physical feature like
location, climate, rainfall. The second chapter is discussed in detail about the hydrological
analysis which includes outlier test, checking consistency of the given hydrologic data and
estimating the design rainfall using normal, gamble, person type 3, log person type 3, and log
normal method.
The third chapter is mainly about peak discharge determination using peak flood analysis by SCS
unit hydrograph method.
The fourth chapter is mainly about the hydraulic design of the weir starting from the weir type
selection up to the determination of weir height, calculation and determination of U/s and D/s
HFL, hydraulic jump computation, and design of impervious floor and pervious Apron, about the
structural design of the weir which includes the stability of the weir, design of divide wall,
retaining wall, under sluice and head regulator. The fifth chapter is discusses about the cost
estimation of the project (bill of quantity) and the last chapter includes general conclusion and
recommendation of the project.

v
Abbreviations
AMC…………………………………..anticipated moisture condition
A.M.S.L……………………………….above mean sea level
BM-……………………………………bench mark
CN…………………………………….curve number
D/S HFL………………………………downstream high flood level
D/S -……………………………………down stream
FAO…………………………………… international food aid organization
F.S.L-…………………………………...full supply level
H.F.L-…………………………………..high flood level
JHC……………………………………..jump height curve
LRBL -………………………………….lowest river bed level
MM/DAY………………………………millimeter per day
NGOS…………………………………..non-governmental organization
OGL-…………………………………....original ground level
PH. ……………………………………..hydro static pressure
PS……………………………………....soil pressure
R.F……………………………………...rainfall
TWRC……………………………….…. tail water rating curved
TAW - ………………………………….total available water
USCS……………………………………united soil conservation system
U/S - …………………………………….up stream
UTM - ………………………………...universal transverse Mercator
WC………………………………………weight of concrete
X-SECTION……………………………..cross section

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Contents
CHAPTER ONE: INTRODUCTIO1
1.1. Back ground ..................................................................................................................1
1.2. Description of the project..............................................................................................1
1.2.1. Location......................................................................................................................... 1
1.2.2. Watershed characteristics .............................................................................................. 2
1.2.3. River Geomorphology................................................................................................... 3
1.2.4 .Hydro-meteorological data availability ......................................................................... 4
1.3. Objective .......................................................................................................................6
1.3.1. General Objective.......................................................................................................... 6
1.3.2. Specific objective ........................................................................................................... 6
1.4. Methodology.................................................................................................................6
CHAPTER TWO: Design rainfall and design flood estimation..................................................... 7
2.1. Data quality...................................................................................................................7
2.1.1 Checking Data Reliability............................................................................................... 7
2.1.2 Outlier test ...................................................................................................................... 7
2.2. Determination of return Period .....................................................................................9
2.3. Design Storm Computation.........................................................................................10
CHAPTER THREE: PROJECT DESIGN FLOOD ..................................................................... 13
3.1. General............................................................................................................................ 13
3.3. Direct Run off Analysis .................................................................................................. 17
3.4. Rational method.............................................................................................................. 19
3.5. Flood mark Method ........................................................................................................ 20
3.6. Selected Design flood ..................................................................................................... 22
CHAPTER FOUR: HEAD WORK DESIGN............................................................................... 23
4.1.Introduction...................................................................................................................... 23
4.2. Weir type selection ......................................................................................................... 23
4.3. Weir Cross section.......................................................................................................... 23
4.4. Irrigation water requirement/management ..................................................................... 24
4.5. Determination of the weir height.................................................................................... 26
4.6. Hydraulic Jump Calculation........................................................................................... 29
4.7. Stability of the weir structure ......................................................................................... 31
4.8. Design of cutoff and impervious floor............................................................................ 38
4.9. Design of under sluice .................................................................................................... 46

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4.10 .Design of operation slab and Breast wall .................................................................49
4.11. Design of retaining wall............................................................................................50
4.12. Design of Divide Wall ..............................................................................................53
CHAPTRE FIVE: BILL OF QUANTITY AND COST ESTIMATION ..................................... 56
CHAPTER SIX: CONCLUSION AND RECOMMENDATION................................................ 61
6.1. Conclusion ..................................................................................................................61
6.2. Recommendations.......................................................................................................62
7. REFFERENCES. ...................................................................................................................... 63
8. Appendix................................................................................................................................... 64
List of Figures
Figure 1.location of Weyizero wouha............................................................................................. 2
Figure 2. Nifasemewch station annual RF ...................................................................................... 5
Figure 3. Unit hydro graph............................................................................................................ 19
Figure 4, River cross section......................................................................................................... 20
Figure 5stage discharge curve....................................................................................................... 21
Figure 6.tail water depth vs ,y2..................................................................................................... 31
Figure 7 .x- section of broad crested weir..................................................................................... 33
Figure 8 x .section of weir ............................................................................................................ 37
Figure 9. detail x section of head work ......................................................................................... 42
Figure 10. Gate for under sluice.................................................................................................... 48
Figure 11. x- section of retting wall.............................................................................................. 52
Figure 12. X-Section of divide wall.............................................................................................. 55
List of Tables
Table 1.Data availability and adequacy for Nifasemewcha............................................................ 5
Table 2.Hydro climatic Data Availability and Its Quality .............................................................. 9
Table 3.Determination of return Period ........................................................................................ 10
Table 4.Test for goodness to fit using D-index............................................................................. 12
Table 5.Estimating Time of Concentration................................................................................... 14
Table 6.Antecedent Rainfall Conditions and Curve Numbers...................................................... 15
Table 7.Design Rainfall Arrangement .......................................................................................... 16
Table 8.Direct Runoff analysis ..................................................................................................... 17
Table 9, Hydrograph coordinates.................................................................................................. 18
Table 10.tail water depth............................................................................................................... 21
Table 11weir height determination ............................................................................................... 27
Table 12.Hydraulic Jump Calculation ................................................................... 31

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Table 13 . Stability analysis at static condition............................................................................. 34
Table 14.Checked by Dynamic stability ....................................................................................... 36
Table 15.U/S and D/S cutoff depth calculation ............................................................................ 38
Table 16.Bligh’s Creep Coefficient .............................................................................................. 41
Table 17.Floor thickness determination........................................................................................ 42
Table 18 .stability analysis of u/s wall.......................................................................................... 52
Table 19.Downstream Retainer walls (Both left and Right Side .................................................. 52
Table 20 .stability analysis of divide wall .................................................................................... 55

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CHAPTER ONE: INTRODUCTION
1.1. Back ground
Irrigation has been considered as an engine for agricultural growth all over the world. In
Ethiopia, irrigation has to be considered not only as engine for agricultural development but also
as a crucial factor for the overall economic growth. About 85% of the total populations directly
depend on agriculture for their livelihood.
Ethiopian government is running to develop small and large scale irrigation schemes to alleviate
the impact of recurrent draught in the whole country. This can be achieved by working together
with the community, local and international NGOs and the government organizations to use all
the available resources efficiently and bring significant change. Weyizero wuha small scale
irrigation project is part of the development strategy carried out by the regional Bureau of Water
Resource Development. <<ADSWE>>
The design and study of Weyizero wuha irrigation project under modern irrigation scheme will
enable the farmers to use the available water and land resources efficiently. In addition to this
they will save time and money for which they will lose for temporary diversion of the project
every catastrophe flood event. Weyizero wuha irrigation project will enable the irrigation water
users of the project area to positive economic change and improve their life standard by
producing excess production and livestock feed for their live stokes using advanced irrigation.
1.2. Description of the project
1.2.1. Location
The project area is located in South Gondar Administrative zone of Amhara National Regional
State. It is locate1d in Nifasemewcha woreda, Keble 03.The specific location of the project site is
called Weyizero wuha. It can be accessed by all-weather gravel road along the route which. The
project site is 4.0km from the main road and 15kms from the city of Nifasemewcha .The location
map of the Weyizero wuha project is shown in the figure below.

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Figure 1.location of Weyizero wuha
<< Source from feasibility study report Amara Design and Supervision Enterprise >>
1.2.2. Watershed characteristics
In any small scale modern irrigation system, most of the headwork component structures are to
be designed considering the magnitude of flood produced by a fifty years return period design
rainfall. Once the rainfall is determined the next step is to investigate about the characteristics of
the watershed.
Determination of catchment area, main stream length and the vertical elevation difference are the
major and the primary activity for watershed runoff simulation using various accepted models.
Weyizero wuha irrigation project has a total catchment area of 1.71km2 for Rivers having the
main stream lengths of 2364m. The watershed has an average main stream bed slope of 0.0738.
The average Curve Number in Antecedent Moisture Condition II is found to be 73.89.

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1.2.3. River Geomorphology
The river bed sediments are dominated with cobbles and boulders at the top with an average
thickness of 1m. At some portions of the stream bed, especially both left and right ends, there are
outcrops of the bedrock (agglomerate rock). On the other hand, the banks of the stream at the
headwork site are made up of different geologic materials; most part of the right and left bank is
made up of old alluvial sediment of low plastic clay. It is dense due to consolidation of the
sediment.
 Left Bank conditions
At the headwork site/axis, the left bank have similar geological formation with the right bank. It
is characterized by relatively moderate to gentle slope, having about 2m height from stream bed.
It reveals nearly genteel section within this height. From visual observation of the natural cuts at
the bank, there are two distinct geological materials forming the bank section from top to bottom,
these are:-
a. Low to Medium Plastic Clay brown to reddish color (CL), with some silt
b. Silty Gravel(GM) old alluvium, and
These two units have variable thickness/depth at the area. Just at the intake axis, the top clay soil
has about 40cm, whereas the middle Silty Gravel old alluvial sediment possesses 1.6m thickness,
both of which increase towards upstream. These overburden soil materials have been affected by
erosion/ flood under cutting which is widening the bank by forming nearly vertical slope. Such
vertical slope configuration observed at upstream bank part forms instability or collapse.
 Right Bank condition
At the headwork site/axis, the right bank is characterized by relatively moderate to gentle slope,
having about 3m height from stream bed. It reveals nearly genteel section within this height.
From visual obsetion of the natural cuts at the bank, there are two distinct geological materials
forming the bank section from top to bottom.
These are:-
a. Low to Medium Plastic Clay brown to reddish color (CL), with some silt
b. Silt Gravel(GM) old alluvium, and
These two units have variable thickness/depth at the area. Just at the intake axis, the top clay soil
has about 1.6m, whereas the middle salty Gravel old alluvial sediment possesses 1.4m thickness,
both of which increase towards upstream. These overburden soil materials have been affected by
erosion/ flood under cutting which is widening the bank by forming nearly vertical slope. Such
vertical slope configuration observed at upstream bank part forms instability or collapse. Here, it
is important to design the bank slope to stable configuration, just by providing appropriate slope
(indifferent from the present vertical slope).

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 Stream Bed
At the proposed headwork site the stream bed or course is undefined, nearly zigzag shape or
channel, and shows rough surface due to recent sediment. Along the intake axis, the bed is made
up of two basically different geologic materials, as seen from surface observation. These are thin
layer of reddish color low plastic Clay with some block of rocks, and underling bedrock.
The stream areas of the bed are mostly covered with thin layer of reddish to brown color low
plastic Clay (CL) which is eroded from the top of both banks intercalated with some angular
shape block of rock.. At the head work site and nearby the stream bed (bedrock) is covered with
these materials. At about 100m upstream and 800m in downstream, there is clearly exposed
slightly weathered and fractured bed rock. It is dark gray color coarse grained basaltic
agglomerate. Hence at the head work axis the nearby bedrock surface forms the OGL of the
stream bed, having about 1 to 1.5m.
As described above, the foundation area of the headwork structure is characterized by non-
uniform geologic materials of the stream bed; the block of angular shape aphantic and bedrock.
The former is irregular and pervious, while the bedrock is strong and impervious. It is therefore
better to incorporate a positive cut-off masonry wall at the central portion of the bed that
anchored to the bedrock after intercepting the 1to1.5m thickness thin layer of CL soil and the top
most weathered part of Basaltic agglomerate rock layer. This will help for both seepage barrier
and also stability conditions.
(Source: from ADSWE Feasibility study report)
1.2.4 .Hydro-meteorological data availability
 Climate
Data for the hydro-meteorological analysis for this project D/tabor & Nifasemewcha but due to
nearest station from the project area all available elements the selected station have been taken
from Nifasemewcha station.

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Table 1.Data availability and adequacy for Nifasemewcha
climate data
Elements No. years
Range of
Data Years Missing Data Adequacy Remark
Minimum
temperature - - - -
Maximum
temperature - - - -
Daily heaviest
Rainfall 42 1970-2012 23 Adequate
monthly
rainfall 19 1954-2010 Adequate
wind speed - - - -
sunshine
hours - - - -
Source: <<ADSWE>>
 Daily Heaviest Rainfall Data
In order to compute the design flood for design of the diversion structure, the daily maximum
rainfall is collected from Nifasemewcha (Nifasemewcha) Metrological stations with a record
period of 18 years. Nifasemewcha station is selected because it is the nearest one as compared to
other Debretabor and Gondar
Figure 2. Nifasemewcha station annual RF
0
20
40
60
80
1990 1995 2000 2005 2010
RF(MM)
year
Nifasemewch station anual RF.

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 River flow data
The source of water i.e. proposed for the project is Weyizero Wuha River. It is a perennial river,
which flows throughout the year are 20l/sec.
Source :<< ADSWE feasibility study report >>
1.3. Objective
1.3.1. General Objective
The general objective of our project is to design a diversion weir to upgrade the existing
traditional farming system in order to increase the living standards of the local people. The main
objectives that enforce us to study this project are:
To assist the project area farmers by upgrading the existing traditional irrigation practices to
modern irrigation.
To design permanent diversion structures at the proposed river.
Promote the crop production per hectare of land by improving water resources utilization
efficiency.
1.3.2. Specific objective
Specific objective of the project is designing a stable and economical design of head work
structure that can resist the anticipated loads over the weir structure. It includes;
Determination of annual rainfall and peak flood.
 Flood analysis
Selection of weir type
Hydraulic design of component parts of head work
1.4. Methodology
while designing this final year project on small scale diversion headwork weir (broad crest
weir), to have well organized structure, we have used the following procedures: Primary data is
obtained through our Advisor from Amhara Design an Supervision Works Enterprise which have
recorded data in the previous many years and we follow those steps.
Determination of maximum daily rainfall.
Design rainfall or storm: to design rainfall checking and also we use some soft weir like excel,
Auto cad, Auto cad civil 3D.

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CHAPTER TWO: Designrainfall and designflood estimation
2.1. Data quality
Although daily rainfall data is obtained from 1992-2009 are used for analysis ommiting some
years with poor registration.Finally daily rainfall in that particular date for 18 years and then the
daily heaviest rain fall is selected . These data should be checked for their consistency or outlier
test to assure the reliability of data for further design flood simulation.
2.1.1 Checking Data Reliability
A18 year’s data checking using arithmetic mean and outlier test is shown below.
Number of data = 18
Standard deviation, 1n 12.3873
Mean, X= 56.00 mm
Standard error of mean, =
n
n
n
1


 =12.387 ÷ √18 = 2.91972
Relative standard, 100*
X
n
=(2.91972÷ 56.00) ∗ 100
=5.21379 % < 10 %
Hence, the data series is regarded as reliable and adequate since the value of "Relative standard"
is relatively small enough. Now, let us check the data outlier test.
2.1.2 Outlier test
This is done to check whether the adopted data are within a limited range or not. Outliers are
data points that depart significantly from the trend of the remaining data. The retention or
deletion of these outliers can significantly affect the magnitude of statistical parameters as mean
and standard deviation that are computed from the data, especially for small samples. Procedures
for treating outliers require judgment involving both mathematical and hydrologic
considerations. However, here simple mathematical approaches are practiced to sort out the data
that seem reliable of the trend of the parent. Input data: Summation of the daily maximum
rainfall data records of 18years
Arithmetic mean of the data, X,∑ 𝑅𝐹 /𝑁 = 56
Summation of common logarithms of the data, ∑ 𝑌 = 31.3033

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Mean value of common logarithms of the data,
∑ 𝑌
𝑁
=
31 .3033
18
= 1.7391
1323.0)( 2


YY
002016.0)( 3


YY
Standard deviation of the common logarithms, √
∑( 𝑦−𝑦𝑚)
( 𝑛−1)
^2 = 0.0896
Skew ness of the common logarithms of the daily maximum rainfall data, Cs
7737.0
*2)1)(18(18
0.00840*18
2)S1)(N(N
Y)(YN
(0.0896)
33
y
3
i








sC
Consideration of the outliers depends on the value of skew ness coefficient. If the value is b/n -
0.4 and +0.4, we consider both the Higher and the Lower outliers; if the value is < -0.4, and if
skew ness coefficient is >+0.4 consider the higher outlier first; based on this we consider the
Lower And higher outlier .so based on this The value of coefficient of scewness (Cs)=0.7737 is
greater than 0.4 entail the data shall be checked for higher outlier only
higher outlier 𝑦 𝐻 =𝑦̅+𝑘 𝑛*𝑠 𝑦 where 𝑘 𝑛=2.33512from table for sample size N=18
hence 𝑦 𝐻 =1.7391+2.335*0.0896
𝑦 𝐻 =1.9483
Y=101.948316
=88.758
But the highest record value is 85mm in the year 1992 which is lower than the threshold value
(88.758).Hence there is no omitting of data from the data set.

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Hydro climatic Data Availability and Its Quality
Table 2. Hydro climatic Data Availability and Its Quality
S.No. Year Max. RF
Descending
Order Rank
Logarithmic
Value/Yo/ (Yo-Ym)2 (Yo-Ym)3
1 1992 72 85.00 1 1.9294 0.0362328 0.0068969
2 1993 48 79.00 2 1.8976 0.0251404 0.0039862
3 1994 33 72.00 3 1.8573 0.0139861 0.0016540
4 1995 51.2 64.10 4 1.8069 0.0045952 0.0003115
5 1996 37.9 63.00 5 1.7993 0.0036326 0.0002189
6 1997 42.5 60.00 6 1.7782 0.0015274 0.0000597
7 1998 54.5 56.00 7 1.7482 0.0000831 0.0000008
8 1999 39.5 55.70 8 1.7459 0.0000460 0.0000003
9 2000 46.5 51.70 9 1.7135 0.0006543 -0.0000167
10 2001 44.4 50.00 10 1.6990 0.0016080 -0.0000645
11 2002 44.3 49.90 11 1.6981 0.0016785 -0.0000688
12 2003 45.6 49.00 12 1.6902 0.0023886 -0.0001167
13 2004 35.3 48.90 13 1.6893 0.0024762 -0.0001232
14 2005 56.7 46.30 14 1.6656 0.0054006 -0.0003969
15 2006 44.2 46.20 15 1.6646 0.0055395 -0.0004123
16 2007 56.5 46.00 16 1.6628 0.0058235 -0.0004444
17 2008 44.9 45.00 17 1.6532 0.0073715 -0.0006329
18 2009 56.5 40.20 18 1.6042 0.0181828 -0.0024518
SUM 1008.00 31.3033 0.1363671 0.0084000
MEAN 56.00 1.7391 0.0075760 0.0004667
STANDARD DEVATION 12.39 0.0896
SKEWNESS COEFICIENT 1.122 0.7737
2.2. Determination of return Period
Selection of the design return period, also called recurrence interval, depends on economic
balance between the cost of periodic repair or replacement of the facility and the cost of
providing additional capacity to reduce the frequency of repair or replacement vegetated control
and temporary structures are usually designed for a runoff that may be excepted to occur once in

10
10 years; expensive permanent structures will be designed for runoffs expected only once in 50
or 100 years.
For the small – scale irrigation project, it would be recommended that the project design flood
once in 100 years be used for design of storage dams, the flood once in 50 years for design of
diversion weirs, and the flood once in 10-20 years for design of drainage structures. However,
for the case that the downstream damage potential by resulting from failure of the structure may
dictate the choice of the design frequency, the flood once in 200 years should be selected. The
following table shows safety factor for the different return period of the project design flood.
Table 3.return Period
Type of
Structure
Project Life
(Years)
Return Period
( Years)
Safety Factor
(Percent)
Storage dams 30
200
100
50
86
74
54
Diversion weir and
drainage structures
15 50
20
74
54
<<Source from Amhara Design and Supervision Works Enterprise manual>>
2.3. Design Storm Computation
After checking consistency (reliability and outliers) test, the rainfall data are obtained as
representative for the analysis. The magnitude of the design rainfall of 50 years of return period
is estimated by the recommended distributions such as Gamble, EVI, Log Pearson and Log
Normal distributions. The best fitting distribution to be used can be done by using D-index.
Design rain fall analysis
Gumball’s Method
return period T yrs. 50.00
Standard vitiate,Sn 1.06
standard mean, Yn 0.52
Yt 3.90

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frequency factor , KT 3.19
Y=X mean + Z*X Standard deviation 95.49
Gumball’s EVI Method
return period T yrs. 50.00
Standard variant,Sn 1.06
standard mean, Yn 0.52
Yt 3.90
frequency factor , KT 2.59
Y=Xmean + Z*XStandared deviation 88.11
Log Pearson Type 3 Method
Design Period, T 50.00
Probability,P 0.02
K=(Cs/6) 0.129
W=(Ln(1/P2))0.5 2.80
Frequency Factor,
KT=(w((2.515517+0.802853*w+0.010328*w2)/(1+1.432788*w+0.189269*w2+0.001308*w3))) 2.05
Standared Normal Variance, Z=KT+(KT2-1)*K+1/3*(KT3-6*KT)*K2-(KT2-
1)*K3+KT*K4+1/3*K5 2.44
Y=Ymean + Z*YStandared deviation 1.958
Design Rainfall, X50 = 10Y 90.75
Log Normal Method
Probability,P 0.02
K=(Cs/6) 0.000
W=(Ln(1/P2))0.5 2.80
Frequency Factor,
KT=(w((2.515517+0.802853*w+0.010328*w2)/(1+1.432788*w+0.189269*w2+0.001308*w3))) 2.05
Standard Normal Variance, Z=KT+(KT2-1)*K+1/3*(KT3-6*KT)*K2-(KT2-
1)*K3+KT*K4+1/3*K5 2.05
Y=Ymean + Z*σy
1.923
Design Rainfall, X50 = 10Y 83.76

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Pearson Type 3 Distribution
Probability,P 0.02
K=(Cs/6) 0.129
W=(Ln(1/P2))0.5 2.80
Z=(w((2.515517+0.802853*w+0.010328*w2)/(1+1.432788*w+0.189269*w2+0.001308*w3))) 2.05
Kt=Z+(Z2-1)*K+1/3*(Z3-6*Z)*K2-(Z2-1)*K3+Z*K4+1/3*K5 2.44
X50=Xmean + KT*σx 86.26
Normal
Probability,P 0.02
K=(Cs/6) 0.000
W=(Ln(1/P2))0.5 2.80
KT=(w((2.515517+0.802853*w+0.010328*w2)/(1+1.432788*w+0.189269*w2+0.001308*w3))) 2.05
X50=Xmean + KT*σx 81.45
Test for goodness to fit using D-index
Table 4.Test for goodness to fit using D-index
Rank XI
Normal
Log
Pearson
Type III
Log
Normal
Pearson Type
III
Gumball
EVI Gumball
XI -'XI' XI -'XI'
XI -
'XI' XI -'XI' XI -'XI' XI -'XI'
1 85.00 8.930 8.331 8.408 8.635 6.396 4.539
2 79.00 7.488 7.980 8.005 7.388 7.363 13.593
3 72.00 3.574 4.561 4.561 3.577 4.565 23.215
4 64.10 1.864 0.617 0.631 1.795 0.255 33.718
5 63.00 0.845 0.536 0.513 0.731 1.118 37.632
6 60.00 1.935 0.502 0.531 1.790 0.217 43.911
Sum 24.636 22.527 22.649 23.916 19.915 156.608
Sum/Mean 0.440 0.402 0.404 0.427 0.356 2.797
Point Rainfall 90.75 90.75 83.76 86.26 88.11 95.49
Design Point
Rainfall = 88.11
Based on D-index the minimum error was Gumball EVI distribution however, in our project the
design rain fall was found to be 88.112mm.

13
CHAPTER THREE: PROJECT DESIGN FLOOD
3.1. General
The flood used for design against failure is termed the “Project design flood” can usually be
determined by estimating the runoff that results from an occurrence of design storm based on
meteorological factors. This hydro meteorological approach is necessary because stream flow
records often are not available.
Flood formulas primarily have been derived from and are directed toward peak discharge
computation. However, the volume of runoff associated with peak discharge and its time
distribution is of vital concern to the designers, who usually need a hydrograph of the inflow
design flood for computing flood routing.
Rational method formula is the simplest but the reliability of the results decrease with increase in
size of catchment area. Hydrograph analysis method is applicable to watersheds of any size
where flow originates as direct runoff from rainfall.
The watershed of given small scale irrigation project is less than 5km2 therefore we can use
rational method without computing the peak discharge computation by SCS method but, it
doesn’t mean that SCS method invalid in this limited catchment area therefore it is better to
compute SCS method in addition to rational method and flood mark method.
For ungagged stream the design flood can be simulated by using SCS unit hydrograph method.
The computation is done using design rainfall or storm estimated earlier, In the hydrologic
analysis of flood using SCS method, rainfall amount and storm distribution; catchment area,
shape and orientation; ground cover; type of soil; slopes of terrain and stream(S); antecedent
moisture condition; Storage potential (over bank, ponds, wetlands, reservoirs, channel, etc.) can
be used and all such data shall be carefully determined before proceeding to SCS simulation..
3.2.1 Estimating Time of Concentration
The time of concentration, Tc, is the time required for a drop of water falling on the most remote
part of the drainage basin to reach the basin out let or the at the point of the diversion .It includes
the time required for all portion of the drainage basin to contribute runoff to the hydrograph and
this time represents the maximum discharge that can occur from a given storm intensity over
drainage basin.
Time of concentration has been calculated by taking the stream profile of the longest streamline
and dividing it in to different elevation ranges. Kirpich formula is adopted for computation.
 TciTc
385.0
3
948.0










H
L
T ic

14
Table 5.Estimating Time of Concentration
Partial Elevation Elevation
Difference
in dm
Length(m) in m Slope of river, Decimal
0
3215 0 0 0
555.00 3160 55.00 9.91 0.10
1079.00 3100 60.00 5.56 0.21
696.00 3051 49.00 7.04 0.14
2330.00 164.00 7.50 0.46
Total Tc,in
hr. 0.46
The formula is,
𝑇𝑐 = ∑ 0.948 {(
𝐿1
3
𝐻1
)
0.385
+ (
𝐿2
3
𝐻2
)
0.385
+ ⋯ + (
𝐿 𝑛
3
𝐻 𝑛
)
0.385
}
I.Tc = 0.46hr
Hence the total time of concentration as computed above is 0.46hr and as it is less than 3hr the
recommended practical time increment is taken as 0.5hr
Having determined time increment of 0.5hr, the point rainfall is applied over the entire area (1.71
km2) and the design rain fall arrangement is shown in the following table.
II.Time to peak,
Tp=
D
2
+ 0.6 ∗ 1.2Tc
III.Base time,
𝑇𝑏 = 2.67 ∗ 𝑇𝑝
IV.Recession time,
𝑇𝑟 = 1.67 ∗ 𝑇𝑝=
Where, L = water course (stream) length in (km)
H = Elevation difference
Tci =Time of concentration for each divided stream length (hr.)
TC= total time of concentration (hr).












 385.0
155.1
*
3000
1
H
L
Tc

15
3.2.2. Curve number (CN)
Curve number (CN) is achieved based on SCS method by watershed characterization in terms of
land cover, treatment, hydrologic condition and soil group. Curve number at condition II =73.89,
since peak rainfall is found at an antecedent moisture condition III state, this value has to be
changed to antecedent moisture condition III (worst condition).
 Conversion factor = 1.18 which is found from table by using interpolation technique
 CN Condition (III) = (Factor from Table x CN condition II) = 73.89*1.18= 87.222
Antecedent Rainfall Conditions and Curve Numbers (for Ia=0.25)
Table 6.Antecedent Rainfall Conditions and Curve Numbers
Curve Number
for
Condition II
Factor to Convert Curve number for
Condition II to
Condition I Condition III
10
20
30
40
50
60
70
80
90
100
0.40
0.45
0.50
0.55
0.62
0.67
0.73
0.79
0.87
1.00
2.22
1.85
1.67
1.50
1.40
1.30
1.21
1.14
1.07
1.00
<<Source: U.S Soil Conservation Service. National Engineering Handbook Hydrology, Section 4(1972)
and U.S Dept. Agr. ARS 41-172(1970)>>.
3.2.3. Run off coefficient
Run off coefficient (c) is achieved based on by watershed characterization in terms of land cover,
treatment, hydrologic condition and soil group. From this consideration the run off coefficient is
0.4.

16
3.2.4. Areal Rainfall
As the area of the catchment gets larger, coincidence of all hydrological incidences becomes less
and less. This can be optimized by changing the calculated point rainfall to aerial rainfall. The
conversion factor is taken from standard table or curves that relate directly with the size of
watershed area and type of the gauging station (IDD manual). But in this project a table was
used.
3.2.5. Rainfall Profile
Rainfall profile is the distribution of the proportion of design rainfall during every incremental
time on the watershed area during the 24 hours duration. Well-developed models are needed to
determine such an event for the selected basin area. But there are no sufficient modeling studies
in the vicinity and adaptation of standard curves has been taken as the only option. For designing
this project has we adopted the standard curve from Design Guidelines for Small Scale Irrigation
Projects in Ethiopia. With the aid of rainfall profile versus duration curve the percentages of
design rainfall distribution on the catchment area are computed for the first most intensive storm
duration.
3.2.6. Design Rainfall Arrangement
Table 7.Design Rainfall Arrangement
Duration
(hr.)
Design
Rainfall
Rainfall Profile
Area to
Point
Ratio %
Areal
Rainfall
(mm)
Incremental
Rainfall
(mm)
Descending
order Rank
% Mm
0.50
88.112
30
26.4
63
16.7 16.65 16.65
1
1 45
39.7
71.65
28.4 11.76 11.76
2
1.5
52.28 46.1
74.8
34.5 6.05 6.05
3
2 59
52.0
78
40.5 6.09 6.09
4
2.5
62.5 55.1
80
44.1 3.51 3.51
5
3 67 3.0 82
48.4 4.35 4.35
6

17
3.3. Direct Run off Analysis
Input data:
Curve number at antecedent moisture condition III =87.222
Catchment Area, A = 1.71Km2
Direct run-off,
Q =
(p − 0.2 ∗ S)2
(p + 0.8 ∗ S)
Where, p = Rearranged cumulative run-off depth (mm
S = Maximum run off potential difference,
𝑆 = (
25400
𝐶𝑁
) − 254
Peak run-off for incremental;
𝑄 𝑝 = 0.21 ∗
(𝐴 ∗ 𝑄)
𝑇𝑝
Where, A=Catchment area (Km2)
Tp=Time to peak (hr)
Q = Incremental run-off (mm)
Direct Runoff analysis
Table 8.Direct Runoff analysis
Duration cumulative
RF
Incremental
RF
accumulative
runoff
incremental
runoff
runoff
in
unit hydrograph time
Remark
Hr Mm Mm Mm Mm m3/s Beginning Peak End
0.50
4.35 4.35 0.00 0.00
0.00 0.00 0.53 1.42 H1
1.00
7.86 3.51 0.00 0.00
0.00 0.50 1.03 1.92 H2
1.50
13.91 6.05 0.96 0.95
0.52 1.00 1.53 2.42 H3
2.00
30.56 16.65 8.86 7.90
6.99 1.50 2.03 2.92 H4
2.50
42.32 11.76 16.87 8.01
3.23 2.00 2.53 3.42 H5
3.00
48.41 6.09 21.47 4.60
2.05 2.50 3.03 3.92 H6

18
Procedure to prepare the above Table
(i) Calculate the direct runoff using accumulated rainfall amounts by progressive time
increments, and determine accumulated direct runoff for respective progressive time
increments.
(ii) Tabulate incremental rainfall and respective incremental runoff, and subtract
incremental runoff from incremental rainfall to determine incremental loss.
(iii) When incremental loss rate reaches the limit minimum infiltration rate, the direct
runoff equation is no longer used. The incremental runoff is then computed by
subtracting the limiting loss rate amounts from the incremental rainfall.
Hydrograph coordinates
Table 9, Hydrograph coordinates
HYDROGRAPH
TIME H1 H2 H3 H4 H5 H6 HT
Beginning 0.00 0.00
0.00 0.00 0.00 0.00
0.50 0.00 0.00 0.00
1.00 0.00 0.00 0.00 0.00
1.04 0.00 0.00 0.05 0.00 0.00 0.05
1.50 0.00 0.00 0.65 0.00 0.00 0.65
1.53 0.00 0.00 6.99 0.40 0.00 7.39
1.90 0.00 0.00 3.23 5.34 0.00 8.57
2.00 0.00 3.15 6.99 0.00 0.00 10.15
2.03 0.00 0.00 6.94 0.19 0.00 7.12
2.20 H5 5.58 1.23 0.18 7.00
2.50 0.00 3.18 3.23 1.26 7.68
2.53 0.00 2.94 3.21 1.37 7.52
3.00 0.00 0.00 1.47 3.15 4.62
3.03 0.00 1.36 3.13 4.49
3.42 0.00 1.72 1.72
3.92 0.00 0.00

19
Figure3.Unithydrograph
From the analysis, the 50 years return period design flood is 10.15m 3/s at 2.0hr peak time. This
implies that for this watershed the peak flood rate per km2 area of the watershed is about
5.93m3/s/km2.
3.4. Rational method
The rational method or CIA method can be compute the design peak discharge, however this
method is limited to watershed of less than 5km2, therefore our small scale irrigation project
catchment area is less than 5km2 i.e. The catchment area is 1.71km2.so it is possible to compute
the design peak discharge by using rational method then after select the maximum peak
discharge by comparing the other method of computing peak discharge (SCS method & Flood
mark method
𝑄 = (
1
3.6
) 𝐶𝐼𝐴
Where=Design peak discharge (m3/s)
C=runoff Coefficient
I=rainfall intensity (mm/h)
𝐼 = (
𝑃
6
)(
𝑇+1
𝑇𝐶+1
) = 𝐼 = (
88.112
6
)(
6+1
0.46+1
) =70.409mm/hr.
𝑄 = (
1
3.6
) 0.4 ∗ 70.409 ∗ 1.712=13.36 m3/s
-2
0
2
4
6
8
10
12
0.00 1.00 2.00 3.00 4.00
dis

20
3.5. Flood mark Method
This is just to check the design flood what we have determined using rough simulation methods
such as SCS for engaged catchments. That was the main purpose of taking flood marks during
field assessment. Also Stage- discharge analysis for tail water depth determination helps to fix
the head of expected peak flood at the proposed weir axis cross section, the stage discharge
analysis can be done after obtaining the weir axis cross section before construction data, river
bed slope and the Manning’s roughness coefficients of the river channel as follows: From the
weir axis cross section we can compute the wetted area, wetted perimeter, hydraulic radius and
depth of flood at different points to both sides of the river banks starting at the center of the river.
On the other hand, the bed slope of the river (So) and Manning’s roughness coefficients of the
river channel (n) are obtained by using best fit curve and considering the river channel
characteristics respectively. Having all those values and adopting the Manning’s formula the
flow velocity of the expected flood (V) and corresponding discharge (Q) are computed for a
particular stage and this stage is considered as the flood level at the intake axis .And that will be
checked using the stage discharge analysis in the following section.
Figure 4, River cross section
The river cross sections and bed profiles are used to determine the hydraulic parameters such as
area, wetted perimeter, hydraulic radius and the bed slope. These data enable to calculate the
approaching flow depth and the flow velocity which are the basic input parameters to determine
the scour depth using those selected equations.
 River bed slope = 0.061 from the weir tope data
3056.50
3057.00
3057.50
3058.00
3058.50
3059.00
3059.50
3060.00
3060.50
3061.00
0.00 10.00 20.00 30.00 40.00 50.00 60.00
elivation
commuladis
River cross section
ELEVATION
y = -0.0626x + 3062.6
3053
3054
3055
3056
3057
3058
3059
3060
3061
3062
3063
3064
0 100 200
AxisTitle
Axis Title
Chart Title
Series1
Linear (Series1)

21
 River bed material manning coefficient, n= 0.035
then from manning’s equation: q=
𝐴𝑅
(
2
3
) 𝑆0.5
𝑛
,𝑄 = 𝐴 ∗ 𝑉 and 𝑣 =
𝑅^ (
2
3
) 𝑆^0.5. so, we can obtain the data values in table below
Table 10.tail water depth
ELEVATIO
N
DEPT
H
WET
AREA
TOTAL
PERIM
TOP
LENGT
H
WET
PERI
M
Hydrauli
c radius
R (m)
Velocit
y V
(m/sec.)
Discharge
Q
(m^3/sec.)
3056.980 0.00 0 0 0 0.00 0 0 0
3057.230 0.25 0.362 5.853 2.901 2.95 0.12 1.74 0.63
3057.480 0.50 1.25 8.541 4.195 4.35 0.29 3.07 3.84
3057.730 0.75 2.494 11.749 5.755 5.99 0.42 3.93 9.80
3057.740 0.76 2.554 11.913 5.835 6.08 0.42 3.95 10.10
3057.741 0.761 2.564 11.94 5.848 6.09 0.42 3.96 10.15 scs
3057.742 0.762 2.566 11.946 5.852 6.09 0.42 3.96 10.16
3057.754 0.77 2.64 12.144 5.949 6.20 0.43 3.99 10.54
3057.83 0.850 2.992 12.778 6.567 6.21 0.48 4.33 12.96
3057.840 0.86 3.188 13.648 6.685 6.96 0.46 4.19 13.35
CI
A
3057.850 0.87 3.22 13.721 6.725 7.00 0.46 4.20 13.53
3057.86 0.880 3.289 13.895 6.806 7.09 0.46 4.22 13.89
3057.880 0.900 3.462 14.222 6.968 7.25 0.48 4.30 14.90
Figure 5stage discharge curve
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0 2 4 6 8 10 12
depth
discharg(m^3/s)
depth

22
3.6. SelectedDesignflood
Based on the stage analysis result of the flood mark elevation & CIA method the amount of
flood computed in that elevation has been found to be which is higher than the computed flood
amount using SCS method. The overflow over the length becomes minimum and therefore for
the case of designing the protection walls of a Intake it is advisable to take the design flood
obtained by using CIA method ( higher value ) and hence design flood is taken as 13.365m3/s.
So From the above stage discharge table and curve the maximum flood level corresponding to
the computed design peak discharge is 3057.84 m.a.s.l (0.86m)from the river bed) and it is
considered as the d/s high flood level. I.e. expected at the weir axis before construction of the
weir d/s hfl =3057.84 a.m.s.l.

23
CHAPTER FOUR: HEAD WORK DESIGN
4.1.Introduction
Weir or barrage are relatively low level dams constructed across a river to rise the river level
sufficiently and to divert the floe in full, or in part into a supply canal or conduit for the purposes
of irrigation, power generation, domestic and industrial uses. These diversion structures usually
provided a small storage capacity.
4.2. Weir type selection
As the proposed project has a catchment area of a relatively plain topography and it has no any
transported boulder materials (ADSWE feasibility report) the expected flood from u/s catchment
is not so much; availability of construction materials and for simplicity of workmanship broad
crested weir having an external R.C.C capping is preferably recommended to be designed.
4.3. Weir Cross section
 Referring from the top map of head work x-section and existing traditional canal profile
of irrigable land the following data are obtained:
 Optimum irrigable command level = 3057.02m.a.s.l.
 River bed level at the weir cross-section = 3056.98 m.
 The actual site conditions of the river banks and average width of the river channel is
8m.
4.3.1. Hydraulic design consideration of weir structure
 The following hydrological and topographical data must be collected before designing
the weir.
 High flood levels for river at the weir site.
 Maximum flood discharge for the river at the weir site.
 River cross section at the river at the weir site.
 The stage discharge curve for the river at the weir site.
 In addition to above, there are also some basic factors, which have to be considered:
 Crest level
 Afflux
 Waterway and the discharge per meter.
 Pond level
Appropriate Weir site should be investigated on field and hence bed and abutments data,
Bed level (deepest point or center) of the river, the existing commanding sill level at d/s,
Canal length b/n outlet and command sill level, Water depth required in the canal, the
selection of best canal slope and other required datasets must be investigated for weir
height determination.
4.3.2 Weir crest length determination
In fixing the weir crest length Laceys flow regime width may be used when the actual width of the river
is higher. However, the Lacey’s crest length value is most of the time higher and hence weir
crest length Selection depends according to the actual river section width and check with Lacey’s
regime width by using the following equation.

24
L = 4.75 ∗C0.5Where L= Lacey regime length in meter (m), Q= the design flood (m3/s)
L = 4.75 ∗ (13.35) ^.5=17.35m, but the lacey’s water way is higher than the actual conditions of
the river which is 8m .There for we should take the actual width of the river as the crest length.
There for the crest length of the weir is 8m.
4.3.3. Discharge and Head Over the weir
When a weir is constructed across the river, head is produced above the crest of the weir. This
head is an important factor in the design of hydraulic structures. Discharge over the weir is
generally expressed as: 𝑄 = 𝐶𝐿𝐻𝑒^(
3
2
)
Where; Q = Design discharge (m3/s)
C = Discharge coefficient (Usually C = 1.7, is used to broad crested type)
He = Height of energy line above the datum (m)
He = (Qd/Cd*L) 2/3 L = length of the weir (m)
He=0.987m
The velocity head, ha is computed from the approach velocity as shown below
ℎ𝑎 = 𝑣𝑎2
/2𝑔
Where g: acceleration due to gravity = 9.81m/sec2
Va is Approach velocity determined by
𝑉𝑎 =
𝑄
𝐿∗( 𝑃∗𝐻𝑑)
but Hd=He-
𝑽 𝟐
𝟐∗𝟗.𝟖𝟏
, He-Hd= 0.987-Hd
Expressing the above two equations interms of Hd
(
𝑸
𝑳∗(𝒑+𝑯𝒅)
)2 *
𝟏
𝟐∗𝟗.𝟖𝟏
=0.987-Hd using iteration Hd=0.378m
Hence velocity head Ha=He-Hd
Ha=0.987m-0.378m
ha =0.609
V=√(𝐻𝑎 ∗ 2 ∗ 9.81
= 3.4567m/s
4.4. Irrigation water requirement/management
Interims of water management this would mean that water allocations of a controlled but limited
supply would be directed toward meeting the full water requirements of the crop during the most
sensitive growth periods for water deficit rather than spreading the available limited supply to the
crop equally over the total growing period.

25
4.4.1. Irrigation Duty:
Irrigation duty is the volume of water required per hectare for the full flange of the crops; and it
is also the relationship between the volume of water and the area of the crop matures. It helps in
designing an efficient irrigation canal system. The area, which will be irrigated, can be calculated
by knowing the total available water at the source and the overall duty for all crops required to be
irrigated in different seasons of the years.
The proposed cropping pattern of the project has showed a maximum irrigation water
requirement (IWR) in the month of October (ADSWE, feasibility report). IWR has to be taken
for designing of the irrigation water application and the flows in the entire canal system.
However, here for the convenience of the designing and operation of the project, from all the
proposed crops the potato crop peak net irrigation water requirement (NIWR) has taken for the
irrigation project duty calculation. The potato peak NIWR is 6mm/day in the months of March &
two decades in April.
The gross irrigation water requirement (GIWR) is calculated by NIWR and irrigation efficiency:
GIWR = NIWR x IE
Where; GIWR – Gross irrigation water requirement [mm/day]
NIWR – Net irrigation water requirement [mm/day]
IE – Irrigation Efficiency [%]
e = ec × ea/100 = (60 × 80)/100=48% ≈ 50% Where, ea = scheme irrigation efficiency (%), e =
conveyance efficiency (%) and ea = field application efficiency (%)
The GIWR for the design of the project is given for the selected irrigation method (i.e., surface
irrigation) as follows:
GIWR = 6/0.5 = 12[mm/day]
The GIWR represents the daily quantity of water that is required to be applied. This water
quantity is also used for determination of the canal discharge in consideration of the time of flow
and is defined as the duty, expressed as l/s/ha. The duty is calculated by:
DUTY (D)= GIWR x 1000 x 10 / (t x 60 x 60)
Where; Duty – the duty [l/s/ha]
GIWR – Gross Irrigation Requirement [mm/day]
t – Daily irrigation or flow hours [hrs.]. Since farmers are well aware of the irrigation
technology in the project area (ADSWE feasibility report)we have selected 7 days irrigation days
per week and 24hrs of irrigation time per day b/c the base flow of the spring is very small so it is

26
better to use effective water use. Using this time input data and the crop watt based maximum
duty, the design duty has been calculated
The duty for the GIWR of 6 mm/day (standard) and 24 hours of daily irrigation time (t = 24) is
supported to be used with furrow irrigation method. Hence, Duty for 24 working hours is
computed as follows:
D = 12 x 1000 x 10 / (24 x 3600) = 1.4 l/s/ha
The NIWR and GIWR can be expressed as the duty for the net water requirement and for the
gross water requirement of the proposed cropping pattern.
4.4.2. Water Supply and Demand Analysis
Though excess amount of irrigable area is available the water source is limited to command
large area. Hence the size of the irrigable area should be planned in accordance of the supply
amount. Therefore the designed flow to pass through the canal should accommodate the need of
the planned irrigation area during the driest season and this is optimized by supply and demand
analysis.
Moreover in water resource projects the downstream release is a must and considered as part of
the project and 20l/s water is released for this purpose. Therefore the size of irrigable area is
limited to the supply 20 L/S.
Duty as calculated in the agronomic report=1.4l/s/ha
Irrigation area= 20l/s)/1.4 l/s/ha
=14.4ha
Hence 14.4 ha of land would be irrigated.
4.5. Determination of the weir height
The height of weir should be sufficient in order to attain the full supply level (FSL) of the canal
at the dry season, so we can fix the based on full supply level and crest level.
4.5.1. Weir Crest level determination based on full supply level (FSL)
The weir height is fixed based on the maximum elevation of the command area to be irrigated,
different losses and outlet position (level).The detail considerations and calculations are
summarized with the following table.

27
Table 11weir height determination
Table: weir height determination Remarks
1. River Bed Level at weir axis 3056.98 From River x-section
2. Canal length from outlet to the maximum command area 820 From top map
3. Average level of the highest field of the command area
3057.02
Top map &
observation
4. Water depth required at canal outlet 0.4 Hydraulic computation
5. Free board at canal outlet 0.30 Assumed
6. Head loss across the field 0.05 Estimated Loss
7. Head loss at the turnout 0.05 Estimated Loss
10. Head loss across head regulator 0.10 Estimated Loss
8. Canal slope 0.001 Hydraulic computation
9. (Canal slope) * (Canal Length) 0.82 Canal slope Loss
Total Loss 1.02 Sum of all Losses
11. Crest level of the weir = Command area Level + Total
Loss+ canal depth+ free boar
3058.74
12. Weir height = Crest level – River bed level 1.8
Full supply level (FSL) =optimum irrigable command +water depth in canal +head regulator
water depth in the field+ total loss.
=3057.02+0.4+0.15+1.02
=3058.59
Pond level=Full supply level (FSL) +modular head (head loss in head regulator)
=3058.59+0.10
=3058.69
From the above two calculation of the weir crest level determination is little difference and the
height of weir due to U/s TEL is 1.5m and due to full supply level is 1.8m, thus we can fix the
height of the weir is larger of the two which is 1.8m . Now the crest level of the weir is
3058.69ma.s.l

28
4.5.2. Top and bottom width
According to the Bligh’s formula, top and bottom width of the weir body is determined as
follows
Input Data:
P: Height of weir (m) = 1.8m
He: specific energy head (over flow depth + approaching velocity head (m), 0.987m the
above calculation.
The top width is fixed as the larger of the two values obtained from the following relations based on no
tension and no sliding criteria
Top width, no tension criteria 𝑏 =
ℎ𝑒
(−1)0.5 =0.87 let’s take 1
No sliding condition criteria 𝑏 =
2
3
∗ (
𝐻𝑒
(−1)0.5)=0.58 let’s take 0.6
Therefore top width is 1m
 =Specific weight of weir body (2.3 for cyclopean concrete)
Bottom width
allowable limits and the tension does not develop. For preliminarily design, the base width may
be taken as:
Bottom width 𝑏′
=
0.378+𝑝
(2.3−1)0.5= , =
0..378+1.76
(2.3−1)0.5 =1.88m lets take 2m.
For preliminary design the top and bottom widths are calculated to be 1m and 2m respectively.
These values are to be checked for stability requirements later and readjusted dimensions are to
be set. Then we need to analysis the stability of the weir considering the weight itself, sediment
load and upstream horizontal water load and uplift pressure.
4.5.3. U/S and D/S HFL Calculation & Determination
From the stage –discharge curve prepared the high flood level before construction (i.e. D/s HFL)
corresponding to the design flood is 3057.840ma.s.l.
D/s HFL = 3057.84ma.s.l
D/S TEL= D/s HFL+ velocity head at D/s=3057.84+ (4.19)2/19.62=3058.7348m.a.s.l
U/s HFL = U/s bed level + weir height + HD=weir crest level +HD
HD is the depth of water over the weir crest is 0.378m previous calculation.
U/S HFL=3056.98+1.76+0.378=3056.98+2.138=3059.118m.a.s.l
U/S TEL=U/s HFL+ ha=weir crest level +He, where ha=He-Hd=0.987-0.378=0.609m

29
= 3059.118+0.609=3059.727 m.a.s.l
Head loss=U/s TEL-D/s TEL=3059.727- 3058.7348=0.992m
Afflux
The rise of the maximum level of river U/S of the weir after construction is known as afflux. The
amount of afflux will determine the top level of guide banks and marginal banks.by providing a
higher afflux, the waterway and, therefore the length of the weir can be reduced, but it will
increase the cost of training works and the risk of failure by outflanking. Generally afflux is
directly related to the guide banks advice versa of the waterway.
Afflux = U/s HFL- D/s HFL = 3059.727-3057.84m.a.s.l = 1.8875m,
This calculated value of afflux is larger than limit of consideration which is afflux is between 1-
1.2m. However in steep reaches with rocky bed, a higher value of afflux may be permitted. From
the flood level analysis, it is seen that the flood overtops the banks of the river u/s of the
structure. This condition is allowed to take place as it doesn’t bring pronounced negative impacts
on the structures, rather than constructing bulky structures to confine it.
4.6. Hydraulic Jump Calculation
By constructing the head work across the weir, there is rise of the water level on the U/S and
there will be jump at the D/s to dissipate the energy. For diversion head works constructed in
pervious foundation, the length of the jump is an important and should be determined hydraulic
using jump equation as follow.
The jump length is very essential whether to construct or not energy dissipating structure at the
downstream of hydraulic structures. Retaining walls at upstream right and left sides are mainly
needed to confine the peak flood within the river channel/ weir. To keep the downstream banks
from erosion, retaining walls are extended downstream for the same length with the downstream
impervious apron. The length of wing walls is determined based on the length of Jump, and it is
calculated as shown below.
 Weir crest length =8m
 Weir height = p = 1.8m
 Pre-jump depth = y1=0.238
 Post -jump depth =y2=1.429m
Neglecting losses between U/s and D/s points and considering Similar datum,
𝑝 + 𝐻𝑒 = 𝑦1 + ℎ𝑎
𝑞 = 𝑄/𝐿
=13.35/8=1.668
But He=0.987
ℎ𝑎 = 𝑞^2/(2 ∗ 𝑔 ∗ 𝑦1^2)

30
After iteration y1 is 0.238
𝑉1 = 𝑞/𝑦1=1.668/0.238
=7.00m/s
𝐹𝑟 = 𝑣/(𝑔𝑦1)^0.5=7/(9.81*0.238)^0.5=4.586
𝑦2 =
𝑦1
2
∗ ((1 + 8 ∗ 𝑓𝑟2)0.5 − 1)
=0.238/2*((1+8*4.586^2)^0.5-1)=1.429m
𝑉2 = 𝑞/𝑦2=1.668/1.429=1.167m/s
To find the jump (basin) Length=6 ∗ ( 𝑦2 − 𝑦1) = 6 ∗ (1.429 − 0.238) = 7.128𝑚, 𝑡𝑎𝑘𝑒 7
Specific energy before the jump is given by
𝐸𝑓1 = 𝐸𝑓2 + 𝐻𝐿, 𝑤ℎ𝑒𝑟𝑒 𝐻𝐿 𝑖𝑠 ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑗𝑢𝑚𝑝 𝑖𝑠,
𝐻𝐿 =
(𝑌2 − 𝑌1)^3
4 ∗ 𝑌2 ∗ 𝑌1
=
(1.429 − 0.238)^3
4 ∗ 0.9635 ∗ 0.116
= 1.248𝑚
The specific energy after the jump is given by
𝐸𝑓2 = 𝑦2 +
𝑣22
2 ∗ 𝑔
= 1.429 +
1.1672
2 ∗ 9.81
= 1.498𝑚
𝐸𝑓1 = 𝐸𝑓2 + 𝐻𝐿 = 1.248 + 1.498 = 2.746𝑚
Check whether the flow is free (modular) or submerged (non-modular).For the flow to be
modular, i.e.
not affected by submergence, the ratioH2/H1, where H1 and H2 are the upstream and
downstream heads
above the weir crest, is less than 0.75 (BSI, 1969; Boss, 1976)
𝐻2 = 𝑌2 − ℎ𝑤 = 1.429 − 1.76 = −0.334
H1=HD+hv=He=0.987m
Where, HW = Weir height So, H2/H1 =-0.33/0..987=-0.334
H2/H1= 0.334<0.75, the flow is free (modular)
so the length of the stilling basin, L is equal to => 7m

31
Table 12.Hydraulic Jump Calculation
Figure 6.tail water depth vs ,y2
The above graph shows that the tail water rating curve is lower than the jump height curve depth
which means that need to constructs the energy dissipater structure. For the formation of the
jump, the horizontal apron maybe depressed by excavating the river bed D/s of the toe of the
weir to increase the tail water depth. The depth is depression can be taken as the difference of
between the tail water depth and post jump depth. Other option we can provide chute blocks at
the weir toe and wall (sill) at the end of impervious floor. Now the wall of the height is the
difference between the post jump depth and tail water depth.
Height of wall=Y2-TWD=1.403-0.861=0.569m
4.7. Stability of the weir structure
4.7.1 .Acting Forces on Weirs
Stability analysis is carried out to see the already determined weir/intake section is safe against
Overturning, sliding, tension. The stability analysis is carried out considering the effect of the
following forces.
 Weight of the over flow weir section
 Water pressure
 Sediment load
 uplift pressure
A .Self-weight of the structure
for the ease of calculating moment arm for each section of the curved profile of the broad
crested, the curved surface was assumed to be linear at proper intervals so that a trapezium
section can be obtained. Now the total section of the weir was divided in to sub sections as
shown in figure below.
Weight (w) =rmas *Ac
Q tail
water
depth
y2
0 0 0
0.07875 0.25 0.333572
0.48 0.5 0.811271
1.225 0.75 1.250452
1.26875 0.761 1.26749
1.27 0.762 1.267973
1.3175 0.77 1.289482
1.66875 0.86 1.430061
1.73625 0.88 1.455102
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 0.5 1 1.5 2
Chart Title
tail water depth y2

32
Where, mas= unit weight of masonry =23KN/m3
Ac= area of the concrete
B=2m, bottom width
B1=1, top width
B2=1.m, triangle part of bottom width.
P = 1.8m, weir height
B. Water pressure (Hydrostatic pressure (PH)
These are the forces acting on the weir due to the reservoir created upstream of the overflow
section and the dynamic pressure created at the toe due to change in the momentum of the flow.
The external water pressure on the upstream face of the weir is calculated for sever case i.e. for
the design discharge level. It has the four components Pw1, Pw2, Pw3, and Pw4 as shown in fig
below. The water pressure that could be exerted on the weir body due to a change in momentum
as the water flows over the curved toe surface was also calculated and incorporated in the
analysis. This is calculated based on the following formula
Ph1=1/2*w*h acting on h/3 KN/m where h is weir height.
Ph2=1/2*w*Y22
Ph3= 1/2*yc*B1*w
Ph4=(Hd-yc)*B1*w
C. Uplift water pressure
Pu = γw*h1∗B 2
where B= bottom width of the weir,
3. Dynamic
PU1= γw*h2*B acts at B/2 from the toe
PU 2 =0.5*B* γw*(h1-h2)
Pu1=B*w*h toe
Pu2=0.5*w*B*(Hheel-Htoe)
D. Silt pressure
the gradual accumulation of significant deposit silt, against the face of the weir generates a result
of horizontal pressure Ps on the upstream section of the weir. Its magnitude is a function of the
sediment depth at worst condition with a height equals to silt height (hs).
Ps = 1.8*hs2 acting on h/3 KN/m, where hs silt height

33
4.7.2. Stability analysis consideration
For diversion weir stability, the critical load case may be the pond level case (i.e. the water level
is up to the crest level of the weir and no over flow) or the high flood level case (i.e. when there
is over flow and the weir is submerged).therefore it is necessary considering the two cases to
check whether the preliminary section of the designed weir is stable or not. The designed should
be safe against sliding overturning and tension crack.
Figure 7 .x- section of broad crested weir
i. Weir stability analyses with static condition
 Overturning
To prevent overturning, the sum of the stabilizing moments must exceed the sum of the
overturning moments on the structure. Maximum upstream and minimum downstream
water surfaces subject these structures to unsymmetrical loads which tend to cause
overturning. The resultant of all forces acting on the structure should fall within the
middle third of the structure base to provide safety against overturning. This location of
the resultant also provides a more uniform bearing pressure on the foundation.
 Sliding
the weir should be stable against sliding at the toe base for different conditions and it is
the function of the shear strength of the construction materials.
 No tension criteria
for no tension on the base of head work structure, for critical section, the resultant (R)
should be act as the middle third part of the critical section. In the computation process

34
the structure considering monolithic section & a unit length of the weir & earth quake
force is assumed to be negligible.
Note: sign convention
 Vertical forces downward is positive and upward is negative
 Horizontal forces towards upstream positive and towards downstream negative
 Moments clock wise moment negative and anticlockwise moment positive
 Summation of all moments about at toe must be equal to zero: ΣMtoe=0
 Summation of all horizontal forces must be equal to zero: ΣFv=0
 Summation of all vertical forces must be equal to zero: ΣFh=0
for a structure to remain stable, the moments which tend to topple it must be equal to
the moments which balance it. In practice, this condition does not satisfy design
engineers, since unpredictable situations are likely to occur and cause the toppling
moment to exceed the balancing one and hence the structure fails. The load
combination on the weir stability is checked for both cases (Static and dynamic):
The preliminary section of the weir dimensions is checked for its stability in both static and
dynamic cases and the computations are tabulated below.
Table 13 . Stability analysis at static condition
Dimension
Height,H
Triangle
,B1 Rectangle,B2
Bed width,
B
1.80 1.00 1.00 2.00
Stability analysis
Code Load
Lever
Arm, R Moment (about toe)
Vertical Horizontal Positive Negative
W1 41.400 1.500 62.100
W2 20.700 0.667 13.800
Ps -5.832 0.600 -3.499
Ph -16.200 0.600 -9.720
SUM 62.100 -22.032 3.367 75.900 -13.219
Factor of safety against,
overturning test Fo 5.742 >1.5 OK
Sliding test Fs 2.114 >1.50 OK
Tension test X 1.009
B/6= 0.333 E 0.009 <B/6 OK

35
Factor of Safety
I. Factor of safety against overturning (FO): the factor of safety against overturning should not
be less than 1.5.
fso = ∑𝑠𝑡𝑎𝑏𝑙𝑖𝑧𝑖𝑛𝑔
𝑚𝑜𝑚𝑒𝑛𝑡
∑overturning
moment ⃒ = 5.742 ≫ 1.5 SAF
ii. Stability against Sliding
The weir should be stable against sliding at the base for different conditions and it is the function
of the shear strength of the construction materials. It is given by:
Safe
F
F
F
H
V
s 5.1114.2
032.22
100.62
*75.0 


Where;  VF and  HF is summation of vertical and horizontal forces respectively and μ is
coefficient of friction b/n the material and the horizontal section and its value varies b/n 0.65 to
0.75 up on the materials used ( here 0.75 is taken ) . Fs should be greater than or equal to 1.5.
Safety against Tension
For no tension on the base of the head work structure, for critical section, the resultant (R) should
act as the middle third part of the critical section. This implies that the eccentricity (e) should be
less than or equal to one-sixth (1/6) of the base width (b) of the weir at the critical section.
009.1
100.62
219.13900.75






  
VF
MM
X
And the eccentricity, SafeB
B
Xe 313.06/009.0
2
2
009.1
2


The resultant lays out of the middle third implying that there is tension developed at the weir
body at the toe.
Conclusion: From stability analysis, the designed weir section it is safe for two conditions but it
is had beater to increase the weir dimensions to minimize tension
ii. Checked by Dynamic stability

36
Table 14.Checked by Dynamic stability
Dimension
Height,H
Triangle
,B1
Rectangle,B
2
Bed width,
B
Thickness
, t Yc
1.76 1.00 1.00 2.00 0.45 0.521
Stability analysis
Code Load
Lever Arm,
R Moment (about toe)
Vertical Horizontal Posetive Negative
w1 40.480 1.5 60.720
w2 20.240
0.66666666
7 13.493
Ps -5.576
0.58666666
7 -3.271
ph1 -18.814
0.71266666
7 -13.408
ph2 10.225
0.47333333
3 4.840
ph3 3.780 1.500 5.670
pu1 -28.580 1.000 -28.580
pu2 -3.310 0.667 -2.207
SUM 32.610 -14.166 84.723 -45.259
Factor of safety against,
overturing test Fo 1.872 >1.5 OK
Sliding test Fs 1.727 >1.50 OK
Tension test X 1.210
B/6= 0.333 E 0.210 <B/6 OK

37
 Factor of Safety
Factor of safety against overturning (FO): the factor of safety against overturning
should not
be less than 1.5.
fso = ∑𝑠𝑡𝑎𝑏𝑙𝑖𝑧𝑖𝑛𝑔
𝑚𝑜𝑚𝑒𝑛𝑡
∑overturning
moment ⃒ = 1.872 ≫ 1.5 SAFE
 Stability against Sliding
The weir should be stable against sliding at the base for different conditions and it is the
function of the shear strength of the construction materials. It is given by:
Safe
F
F
F
H
V
s 5.1727.1
166.14
610.32
*75.0 


Where;  VF and  HF is summation of vertical and horizontal forces respectively and μ is
coefficient of friction b/n the material and the horizontal section and its value varies b/n 0.65 to
0.75 up on the materials used ( here 0.75 is taken ) . Fs should be greater than or equal to 1.5.
 Safety against Tension
For no tension on the base of the head work structure, for critical section, the resultant (R) should
act as the middle third part of the critical section. This implies that the eccentricity (e) should be
less than or equal to one-sixth (1/6) of the base width (b) of the weir at the critical section.
210.1
61.32
253.45773.84






  
VF
MM
X
And the eccentricity, SafeB
B
Xe 313.06/210.0
2
2
210.1
2


The resultant lays out of the middle third implying that there is tension developed at the weir
body at the toe.
Conclusion: From stability analysis, the designed weir section it is safe for three conditions. But
the factor safety is greater than 50% so it is had better to minimize its bottom width
Figure 8 x .section of weir

38
4.8. Designof cutoff and impervious floor
Hydraulic structures such as dams and weirs may be founded on an imperious solid rock
foundation or on a pervious foundation. Whenever, such a structure is founded on a pervious
foundation, it is subjected to seepage of water beneath the structure, in addition to all other forces
to which it will be subjected when founded on a impervious rock foundation. The water seeping
below the body of the hydraulic structure endangers the stability of the structure and may cause
its failure either by piping or direct uplift. Hence seepage and uplift calculations are required to
determine the lengths of upstream and downstream cut-offs required (subject to scour
considerations) in relation to the length of the structure, and to determine the floor thicknesses
required at various places. The primary purpose of cutoff walls is to increase the percolation path
to prevent piping of foundation material and reduce percolation. Cutoffs also protect a structure
from undermining, if excessive erosion should occur in a structure
4.8.1 U/S and D/S cutoff depth calculation
Table 15.U/S and D/S cutoff depth calculation
SCOURING DEPTH
DETERMINATION
specific gravity 2.3
hd 0.378
bottom width 1.88
Q 13.35
Crest length 8
weir heght 1.76
Unit design flood 1.66875
average river bed
material diammeter, mr
12
Laceys silt factor, f 6.096818843
Laceys scouring depth ,
R
1.039692397
u/s scouring depth 1.299615496
Bottom level of u/s
scouring depth
3057.818
u/s cutoff depth , du -0.838 1
d/s scouring depth 1.819
Bottom level of d/s
cutoff
m 3056.02
d/s cutoff depth C2 m 0.959 1
Impervoius Apron
creep coefficient,C 12
creep length, L m 21.120
D/s Length of
impervoius Apron Ld
m 11.12576426
Length of u/s
impervious floor
m 4.114235739

39
pervious apron design
length of pervious
apron total for
downstream
6.433174818
Length of d/s pervious
protection
-4.69258944
Length of filter -1.25757676
Length of Launching
Apron
2.398654236
Length of pervious
apron d/s
1.14107748
Floor thickness
determination
Location creep
Length
pressure
head
tickness recomenede floor thickness
at the heel 6.11 1.25 1.28 1.3
at the toe 7.99 1.09 1.12 1.20
at 1m from the toe 10.99 0.84 0.87 0.9
at 3m from the toe 12.99 0.68 0.69 0.7
at 6m from the toe 15.99 0.43 0.44 0.5
at 11.22m from the toe 21.21 0.00 0.00 0
Cut off Depth Calculation
the primary purpose of cutoff is to increase the percolation path to prevent piping of the
foundation material and reduce percolation. Cutoffs also protect a structure from undermining if
excessive erosion should occur at the end of the structure.
 Depth of u/s pile
U/s pile level = u/s HFL-1.25R, R= 1.35(q^2/f) ^1/3=1.04m
= 3059.118m a.s.+ 1.299
=3060.477m.a.s.l
Depth of u/s pile (d1) = river bed level-U/s pile level =U/S scour depth-(p+Hd )
=3056.988m.a.s.l-3060.477m.a.s.l =1.299-(1.76+0.378)= -0.839m, this indicates there is no need
to provide cutoff at the upstream of the weir .But a nominal of 05-1 m cutoff should provide to
be safe,
take the nominal depth 1m is provided. Therefore d1=1m
 Depth of d/s pile
The downstream cutoff is designed to have a depth of d/s pile,
d2 = D/S Scour depth – Tail water depth
Bottom level of D/S cutoff=D/S HFL-D/S cutoff =3057.84-1.819=3056.02m.a.s.l
Hence D/S cutoff depth =river bed level-bottom level of D/S cutoff
=3056.988-3056.02= 0.959 take 1m

40
4.8.2. Impervious floor depth
Design of Impervious floor thickness the Seepage head should be cheeked designing the
impervious floor using different theories. It may occur under a no flow condition, where the head
difference is the difference between the weirs crest level and the downstream bed level or under
a full discharge condition with a hydraulic jump in the stilling basin.
The main purpose of u/s apron is to protect the channel bed from the impact of the flow against
the weir, and to protect the upstream bed against cross currents flow along the face of the weir,
particularly when the scouring sluices are in operation. The upstream apron also provides extra
length to the structure and hence reduces the under-floor pressure and exit gradient of seepage
flow. The upstream apron generally is set at the minimum bed level of the channel at the site.
The purpose of downstream apron is to resist uplift pressure, reduce the exit gradient of seepage
and to dissipate the energy over the weir
1. D/S impervious floor length (Ld.)
The basic probable seepage heads are considered for the two cases i.e. at pond level and at
maximum flood level. The main assumption here is there is no significant tail water for the case
of pond level and it exists for high flood level case with significant depth/level/.
1) Pond level case: Thus, Hs = P = The head difference between the U/S&D/S
Hs = crest level –bed level
Hs =3058.44-3056.98=1.76 m
2) Maximum flood case:
Hs = U/s HFL- D/s HFL
Hs = 3059.118-3057.84m=1.278m
Therefore maximum seepage head occurs when water is stored up to the pond level and there is
no water on the d/s.Bligh constant, Cb depends on the type of the foundation. dHence the
downstream apron length is
Ld = 2.21dd * Cb * (Hs/10)0.5, Cb 12 for Coarse grained sand.
Ld = 2.21 *12 * (2.1/10)0.5 = 11.125m, and compare the jump length which is 7.128m and take
the larger.
Therefore Ld is 11.125 m, take 12m.
B. U/S impervious floor, Lu
Ld= 11 m
B= 1.88m
d1 = 1m
d2 =1m
Lu =L - (Ld + B + 2d1 + 2d2) = 4.11m
Take, Lu 4m
Hence total creep length ,LT = 2*d1 + L1+ B1 + L2 + 2*d2 =21.12m

41
2. .U/s impervious floor length (Lu)
The u/s impervious floor length, Lu =L - (Ld + B + 2d1 + 2d2)=21.12-
(11+1.88+2*1+2*1)=4.11m ,but take 4.2m
the nominal length in the upstream side 4m far from the heel of the weir in the left side.
From stability analysis, top width, a=1m, bottom width, B=2m
Total creep length, Lc=C*Hs=9*2.1=21.12m and total creep length after providing u/s
and D/S cutoff and U/s and D/S impervious floor, total creep length is,
Lc=2d1+Lu+B+Ld+2d2=2+1+1.88+11+2=21.12m
 Bligh’s theory
to check the safety of hydraulics structure on pervious foundation, the following two
criteria should be satisfied
1. The sub soil hydraulic gradient should be less than the permeable value to prevent
piping failure i.e. In Bligh’s creep theory weir height over creep length must less than
one over creep. Thus for a safe design, i ≤1/C; C (Bligh creep coefficient and is a
function of soil property); i<1/C, i=H/L, H/L =1.76/21.12=0.083, 1/c=1/12=0.083
therefore 0.083=0.083 safe.
Table 16.Bligh’s Creep Coefficient
The floor should be sufficient thickness to prevent its rupture due to uplift pressure To improve
the safety of floor the thickness should be provide by this; T=4/3*(h/G-1), where the, h=H-
(H/L)Leq, Material of specific gravity (G) for concrete=2.3
The thickness of u/s impervious pronominal thickness of the u/s impervious apron= 0.5m
Nominal length of u/s impervious =4m for the u/s, as the upward and downward forces are
balanced, nominal thickness,(0.5m) masonry may be enough for the downstream, the floor
should be enough to resist up lift pressure developed due to the seepage water. Hence, using

42
Bligh’s theory, the thickness of the floor can be calculated as shown below table. The
unbalanced pressure head at any point is given by: h=H-(H/L)* Lequ and the floor thickness is
given by T=4/3(h/G-1) and the values are tabulated on table below.
Table 17.Floor thickness determination
4.8.3. Designof Protection Work
As the geological investigation shows that the foundation is pervious which is made up of
alluvial soil. Therefore protection works should be made to prevent the migration of
particles and erosion. This purpose will be achieved by providing inverted filter and
launching apron block protection detail arrangement is shown in the drawing.
 U/S Protection Works
In the upstream side of the weir the provided protection work is not much because at
short time accumulation of silt, therefore we provided 1.5m with the thickness of 0.5m
stone blocks is enough.
 D/S Protection Works
After the end of impervious concrete floor an inverted filter; 1.5 to 2.5D long is
generally provided, where D is depth of U/s and D/S cutoff. Length of the inverted filter
= 2* d2 =2m Thickness of the inverted filter is usually provided of 50 to 70 cm. Take
60 cm. The inverted filter and the length of the launching apron is taken to be the same
as the length of the block stone protection it. Generally 1 to 1.2m stone deep concrete
blocks width open joints laid over 0.6m thick grade filter material.
Figure 9. detail x section of head work
Floor thickness determination
Location creep Lengthpressure headtickness recomenede floor thickness
at the heel 6.11 1.25 1.28 1.3
at the toe 7.99 1.09 1.12 1.20
at 1mfromthe toe 10.99 0.84 0.87 0.9
at 3mfromthe toe 12.99 0.68 0.69 0.7
at 6mfromthe toe 15.99 0.43 0.44 0.5
at 11.22mfromthe toe 21.21 0.00 0.00 0

43
Designof impervious floor thickness
From practical point of view, the u/s apron (impervious floor) mostly covered by river deposit,
one thickness cover of the structure, and uplift pressure is also counter balanced by the weight of
the standing water. Hence provide nominal thickness of 0.5m u/s of the weir.
Thickness of d/s impervious apron
The thickness of concrete at the particular point under consideration resisting the uplift pressure
under no flow condition (case (a)) is determined from:
t=
4
3
(
𝐻𝑟
𝐺−1
), where, Hr=is the residual head remaining at a point
Hr=HW -
𝐻𝑤
𝐿
(𝐿𝑝) where HW=Percolation head
L=Total creep length
Lp =Length at a point where to
Calculate the thickness
G=unit weight of floor material =2.3
Point G Hw Lp L Hr T
A 2.3 1.76 6.11 21.21 1.25 1.29
B 2.3 1.76 7.99 21.21 1.10 1.13
C 2.3 1.76 10.99 21.21 0.85 0.87
D 2.3 1.76 12.99 21.21 0.68 0.70
D 2.3 1.76 15.1 21.21 0.51 0.52
When a hydraulic jump forms in the basin under the maximum flow condition (case (b)) the
thickness of concrete is determined from:
t=
4
3
(
𝐻𝑟
𝐺−1
), where, H r = u/s HFL-RBL-y1
=is the uplift head at the point of the hydraulic jump on the stilling basin

44
Point G Hw Lp L Hr T
A 2.3 1.9 6.11 21.21 1.35 1.39
B 2.3 1.9 7.99 21.21 1.18 1.21
C 2.3 1.9 10.99 21.21 0.92 0.94
D 2.3 1.9 12.99 21.21 0.74 0.76
E 2.3 1.9 15.1 21.21 0.55 0.56
The concrete thickness to be adopted for the structure is the greater of the two cases. Hence
adopt the second case.
Check for the exit gradient
B=Total length of impervious apron
d2=d/s cutoff depth
GE=
𝐻𝑤
𝑑2
∗
1
𝜋∗√λ
Where 𝜆 =
1
2
(1 + √1 + 𝛼2) 𝛼 =
𝑏
𝑑2
=
19.7
1
= 19.7
𝜆 =
1
2
(1 + √1 + 19.7) = 10.36
GE=
1.9
1
∗
1
𝜋∗√3.95
= 0.1879
The maximum permissible exit gradient for mixture of gravel, boulder, cobble and sand is 0.25
which is greater than the GE=0.1879 then the structure is safe against piping.
Checking the thickness of the impervious floor by khoslas
 Pressure at key points of u/s cutoff
𝜆 =
1
2
(1 + √1 + 𝛼2 ), where, 𝛼 =
𝑏
𝑑1
= 19.7/1 = 19.7
=
1
2
(1 + √1 + 𝛼2) =
1
2
(1 + √1 + 20.52) = 10.36
фC1=100- фE, Where фE=
100
𝜋
cos−1
(
𝜆−2
𝜆
)
=100-22.3433
=77.6567 =
100
180
∗ 40.2179
=22.3433
фD1=100- фD, Where фD=
100
180
cos−1
(
𝜆−1
𝜆
)

45
=100-15.665
=84.3345 =
100
180
∗ 28.197786
=15.665
Ct=correction for thickness фC1
= (
ф𝐷1−ф𝐶1
𝑑1
) ∗ 𝑡 =
84.3345 −77.6567
1
∗ 0.5 = 3.3389 where 0.5 is the nominal thickness of u/s
apron.
Cif=correction for interference of d/s cutoff on фC1
Cif=19*(
𝑑+𝐷
𝑏
)∗ √
𝐷
𝑏′
, Where D=depth of pile whose influence has to be
Determined on the adjacent pile depth d.
b’=distance b/n two piles
d= depth of pile on which the effects of another
Depth (D) is to be calculated
b=Total floor length
Cif=19*(
𝑑+𝐷
𝑏
)∗ √
𝐷
𝑏′
=19*(
1+1
19.7
)∗ √
1
16.2
= 0.479
Corrected фC1=0.479+3.3389+77.6567
= 81.4748
The residual pressure head at C1=Hw* Corrected фC1
=1.9*0.814748=2.714748
Floor thickness at point c1=
2.714748
2.3−1
= 2. 𝑚
From practical point of view, the u/s apron (impervious floor) mostly covered by river deposit,
one thickness cover of the structure, and uplift pressure is also counter balanced by the weight of
the standing water. Hence provide nominal thickness of 0.5m.

46
4.9. Designof under sluice
Under sluice is used to maintain a deep channel in front of the head regulator and dispose of
heavy silt and a part of flood discharge on the D/S side of the weir. Sluice gate refers to a
movable gate allowing water to flow under it. When a sluice is lowered, water may spill over the
top, in which case the gate operates as a weir. Usually, a mechanism drives the sluice up or
down. It is used to maintain a deep channel in front of the head regulator and dispose of heavy
silt and a part of flood discharge on the d/s side of the weir. The under sluice or scouring sluice is
a comparatively less turbulent pocket of water is created near the canal head regulator by
constructing under sluice portion of the weir. A divide wall separates the main weir portion from
the under sluice portion of the weir. The crest of the under sluice portion of the weir is kept at a
lower level than the crest of the normal proportion of the weir. The purpose of the weir sluice is
to prevent the entrance of the silt loads in to the off take canals. The under sluice located to the
same side of the off take canal. To maintain well defined water flow towards the canal head
regulator and to remove the silt deposit on the riverbed near the head regulator
4.9.1. Functions:
Preserve a clear and defined river channel approaching the regulator.
Scour the silt deposited in the river bed above the approach channel.
Pass low floods without dropping the shutter of the main weir; and
Provide greater waterway for floods, thus lowering flood levels.
4.9.2. Designconsideration of under sluice
The capacity of under sluice is determined considering the following points.
The capacity should be at least two times the head regulator discharge
Capacity of passing about 10% to 20% of the maximum flood discharge at high floods.
During construction, it should be able to pass the prevailing (at least base flow) discharge
of the river.
 Capacity:
From stated above two times the canal discharge is taken to be the discharging capacity
of the under sluice. The reason is at the time of raining season the head regulator is closed
so the coming flood should in the head regulator is back into the under sluice. Therefore
the size of under sluice must be fix this principle. But if we take 10-20% maximum flood
discharge the size of under sluice which is the height and length is larger compared to the
above two criteria. We are going to design for second therefore the size of under should
be easily moveable without crane. The dimensions of under sluice are determined by
using orifice flow formula.
𝑄 = 2/3 ∗ 𝐶𝑑 ∗ 𝐿 ∗ 𝐻3/2 ∗ (2𝑔)^0.5

47
Where
Qd = Discharge of the under-sluice portion (m3/s)
Cd = Coefficient of discharge = 0.62
L=Width of the under-sluice portion (m)
H = Height of under crest (m)
g = 9.81m/s2
The sluice way gate should have a capacity of passing about two times head regulator
discharge which is 0.02 m3/s OR 10% OF 13.35M^3/sec know let’s take 10%of
13.35=1.335.
1.335 = 2/3 ∗ 𝐶𝑑 ∗ 𝐿 ∗ 𝐻3/2∗ (2𝑔)^0.5
1.335 = 2/3 ∗ 0.62 ∗ 𝐿 ∗ 𝐻3/2(2 ∗ 9.81)0.5
Take Width (L) =0.9 and height (H)=0.9 and now check this size to pass the required discharge .
𝑄 = 2/3 ∗ 0.62 ∗ 0.9 ∗ 0.9^(3/2)∗ (19.62)^0.5 = 1.335𝑚3/s
Which is more than two times the head regulator discharge. Hence during non-rainy time, it is
possible to flush the silt easily when required.
The gate for under sluice is to be vertical sheet metal of size 0.9m x 0.9m for the closure of the
opening space providing some extra dimensions for the groove insertion 5cm provided. Gross
area of sheet metal for the gate will be 0.95m x 0.95m. The grooves are to be provided on the
walls using angle iron frames at the two sides of the gate opening.
Crest level: should be lower than the crest of head regulator by at least 1 to 1.2m if special silt
exclusion mechanism is not provided. The silt level of under sluice is consider with the river bed
level by plastering the bed. Impervious floor: thickness and length of impervious floor should be
designed on the same line as the floor of the weir portion. (HS-2 hand out).
Design of under sluice gate thickness
Hydrostatic water pressure, Pa=0.9*10=14KN/m2
Hydro Static water Pressure for head of 1.4m at the bottom of the gate=9kN/m2=0.98N/cm2
The allowable tensile and bending stress of the steel during wet condition=0.45*300=135
N/mm2=13500N/cm2
Hence bending stress in flat plate should be, δ =
K∗P∗a2
100 ∗S2
Where S=thickness of the sheet metal (cm)
P=Hydrostatic pressure (N/Cm)=0.9N/cm2
K=Non-dimensional factor
a =minor support length which related with KFor 𝑏/𝑎 =
0.9
0.9
= 1, K=28.7 from the
table for different supporting condition.

48
S = (
K ∗ P ∗ a2
100 ∗ δ
)0.5
= 0.44 = cm
Hence considering incoming boulders and transported materials, take S=6 mm
Weight of gate= gsteel *s*a*b, Where s=thickness (m) =0.006 m
h=gate height (m)=0.9 ,b=width=0.9m ,gs=Density of steel =7800kg/m3
Weight of gate=7800*0.006*0.9*0.9=37.908kg.
Hence the weight of the sheet metal gate is light; we can use stiffening materials for further safety.
 Know let as take plastering thickness 0.5m’ so the floor level of under sluice gate is
3056.98+0.5=3057.48 m.s.l.
Figure 10. Gate for under sluice
 Canal outlet level
it is a structure constructed at the heads of a canal taking off from a reservoir behind the
weir. The head regulator is provided on both side of the river in reference to the flow
direction. The silt level of this head regulator is fixed from different angle of
observations. Hence this level is fixed based on the optimum route alignment and the
maximum irrigated command level including minor and major losses criteria. Based on
this condition, the silt level is fixed to be 3058.04m a.s.l.(give)
Outlet sill level=river bed level +overall losses=3056.98+1.02=3058.04m a.s.l.
 Outlet capacity
the minimum command area is determined by the minimum flow of the river. But the
canal capacity should be determined for maximum command area and the corresponding
discharge. In this case the outlet capacity is fixed considering maximum duty and
command area. Outlet capacity = Duty x command area x correction factor (when
necessary)
See on the previous title

49
 Outlet size
From the weir discharge formula the outlet size is determined as follows
Qd =2/3* Cd*L*H (3/2) *(2g) 0.5
Where; Cd = Coefficient of discharge = 0.62
L = Length of water way (m)
H =height water (m)
Now by trial and error fix the dimension of head regulator and checking this size
is the capacity to pass
the required discharge. Take H=0.4m and L=0.4m
Checking, Q=2/3*0.62*0.4* (0.4)3/2*(19.62) =0.185m3/s which is greater than discharge in
head regulator. Therefore the size of the gate it is ok. Provided 6cm for groove. The area of the
metal plate is 0.46 by 0.46 m.
The gate of Water pressure is, 𝑝 = ᵞ𝑤 ∗ 𝑤𝑎𝑡𝑒𝑟 = 10 ∗ 0.4 = 4𝐾𝑁/𝑀^2
Design of head regulator gate thickness
Hydrostatic water pressure, Pa=0.4*10=4KN/m2
Hydro Static water Pressure for head of 0.6m at the bottom of the gate=4kN/m2=0.4N/cm2
The allowable tensile and bending stress of the steel during wet condition=0.45*300=135
N/mm2=13500N/cm2
Hence bending stress in flat plate should be, δ =
K∗P∗a2
100 ∗S2
Where S=thickness of the sheet metal (cm)
P=Hydrostatic pressure (N/Cm)=0.4N/cm^2
K=Non-dimensional factor
a =minor support length which related with K
For
𝑏
𝑎
=
0.4
0.4
= 1, K=28.7 from the table for different supporting condition.
S = (
K∗P∗a2
100∗δ
)0.5
= (
28.7∗0.4∗402
100∗13500
)0.5
= 0.12cm
Hence considering incoming boulders and transported materials, take S=6mm
4.10 .Design of operation slab and Breast wall
To avoid spilling of water during HFL over the canal regulator gate, a R.C.C wall is provided
from the gate top level up to the HFL (i.e. known as breast wall). A vertical raised gate is
designed for the head regulator. These gates are slides over the breast wall-using spindle during
opening and closing. The thickness of the breast wall is simply determined from
recommendations (point of construction) rather than the imposed load. The thickness required
for the imposed load is less than this nominal value taken 0.2m.
For the breast wall, the minimum reinforcement area is taken as 0.15% along the respective
direction. Hence A steel=0.0015*1000*200=300
A steel=300mm2, Provide  12 @C/c 200 mm

50
Considering cover thickness of 50 mm, effective depth, de=D-(50+12/2) =200-(50+12/2)=144
Hence spacing of reinforcement=200mm <3*de=432mm A steel=3.14*12^2/4*5=565.2 mm2
Therefore the actual provided steel area per meter width is 565.2 mm2/m>300 mm2/m Ok!
Hence, provide t = 0.20m = 20cm thickness for the breast wall work. And provide the
reinforcement bar of 12mm @200mm c/c spacing in all directions with reinforcement covers of
50mm for the breast wall.
Weir, Apron and sluice Protection Work (Capping)
In order to avoid cracking and shearing of the weir, apron and under sluice during overflowing
and incoming of boulders, RCC of thickness 200 mm is provided with proper capping. The
nominal reinforcement is taken as 0.13% of the concrete cross sectional area per meter width.
Hence, A Steel=0.0013*1000*200=260mm2
Thus, Provide  14 @ C/C 300mm.
Actual area of steel=3.14*14^2/4*1/0.3 =512.35mm2
Since A steel=260 mm2<512.35 mm2 it is ok!
Covering of the reinforcement=50mm+14/2=57mm. 57 mm as gross covering depth. But it is
easy for fixing 50 mm gross covering thickness.
The spacing of the reinforcement bar should be less than three times the effective depth or 450
mm, which is smaller of the two. Effective depth, de=300-50=250mm
Hence the actual spacing, 300 mm<3*250=>300mm<750mm.Hence it is ok.
The spacing @ c/c should be account the diameter of the reinforcement when
Intermediate smaller or larger size of the reinforcement is applied during actual operation. This
capping should also apply for divide wall with reinforcement size of  14 mm with C/c spacing
of 300 mm.
This capping detail is provided for the weir, apron, under sluice and Divide wall and check the
design drawing for further information.
4.11. Design ofretaining wall.
Retaining walls have been provided to safeguard the structure from scour of banks at the
ends and also as a facility to the canal outlet operation and maintenance at the canal outlet
portion. The walls are basically provided to keep the highest flood flow within the weir
crest section and to safeguard areas out of the river bank. The bottom level of the
retaining wall should start from sound strata. The common concern in design of retaining
wall is that the masonry section of the retaining wall must have sufficient
 Self-weight to resist the thrust due to earth pressure occurs at the back without
overturning, sliding, tension and compressive stress developed within the body of
the structure.

51
 The maximum design flood and the flood jump height govern the height of the
retaining wall with some free board provided to protect overtopping of flood and
scouring of the banks.
 The triangular wedge of the retained soil is assumed to assist the stabilizing effect.
The loads considered are
Dead weight.
Pressure due to back fill soil.
Hydrostatic pressure.
A. U/S right and left retaining wall height fixation
The existing topographical condition at the weir axis is considered to be governing parameters to
fix the wall height.
The HFL level after construction of the weir (U/s HFL) =3059.118m a.s.l.
River bed level (RBL) =3056.98 m a.s.l.
Wall height = U/S HFL –RBL+Free board or
U/s wall height (H) = weir height + Hd + Free board (Fb), minimum free board assume 0.3m
H = 1.76 +0.378+0.4 = 2.538, take 2.6m
Top elevation of the U/S retaining wall is = RBL +H =3056.98+2.6=3059.58m a.s.l.
Assume top width T=0.6m
Bottom width of wall (u/s) B=(50%TO 70%) of H, Lets take 60%of H
B=70% OF H=0.7*2.6=1.82 take 2.3m .
Self-weight of retaining wall
W1 =B1*H*rmas= 0.6*2.6*23=38.889KN/m
W2=1/2*(B-B1)*H*rmas=0.5*(2.3-0.6)*(2.6-0.5)*23=41.06KN/m
Back fill load
W3=1/2*(B-B1) *H*rdry*k=0.5*2.6^2*(19-10)*0.33=10.039KN/m
Earth pressure load (P)=1/2*rdry*(H-0.5)^2*k=0.5*9.8*(2.6-0.5) ^2*0.33=21.609
Where angle internal friction of soil assume,= 0.33

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