MODULE
HYDRAULICS 2
CODE
CIVIL 0018
SEMESTER
SUMMER-20
SUBMITTED BY
18S17731
STUDENT NO 03
Introduction
As we know that there is a complete circulatory process for water in this world. Water from the ocean get
evaporated and get rise to high level in the atmosphere and when get cool causes raining. Similarly, after
raining some of the water seep into to the soil and in this way water ground storage capacity of a earth
remain constant. The water quantity which does not seep become a part of runoff. The runoff generated
from an area depends on several condition of the site. For example, soil properties, topography of the
area, and development in the region, paved and unpaved area of the region. As we know that with the
development in an area the paved area will be increased, so the amount of runoff generated will also
increase in that area. Hence in order to cope with this situation there is a need of systematic system which
should be installed in the area to safely remove excess water generated to a particular place specified for
the water produced as a result of run off event. So, for designing such system there is need of calculation
the discharge generated from rain fall event. Hence to solve this situation several studies has been
conducted be civil engineers and as a result of study different methodology has been developed which can
be used to estimate the runoff generated from a rainfall event. Here in this report we will focus on
rationale method of calculating runoff for a region.
In rational method we used an equation consist of different parameters which purely depends on the
topography and environmental condition of the region. And depend on the rain fall pattern in the area. the
equation used for the calculation of runoff is given below. The rational formula is given below
Q=CiA
The catchment area can we calculate from the gis and also we can use google earth for the data. In the
rational equation the parameters are define below
Q = peak rate of flow
Cf = frequency factor
C = runoff (rational) coefficient
i = intensity of precipitation for a duration equal to time of concentration, t , and a return
period, T
A = drainage area
The above parameter should be identified first for a region then based on the above data
runoff should be calculated. And based on the discharge calculated drainage system
should be designed. While designing each parameter should be considered for having a
sustainable design.
Task 1
Required:
I. Determine the peak runoff for the catchment area which is selected.
II. Design best hydraulic section for the given discharge(triangular, rectangular etc)
III. Differentiate or the feasibility between the two section
IV. In design consider the maximum and minimum velocity that does not cause erosion
Solution
Part(1)
The selected catchment area is given below, location(wadi Min’a, Oman)
Catchment area in Oman wadi
Figure 1 catchment path
Given data;
A= 2100000 m 2 = 2.1 km 2
P= 6220 meter
RL2= 2090 meter
RL1= 699 meter
Solution:
(STEP1) Determine the slope of the catchment path
The elevation of highest and lowest point is given below in figure;
RL2=2090 m
Figure 2 cross section of catchment area RL1=699 m
The calue of C is taken from the following table
The condition of the soil of wadi is sandy so for that we take the value from the above table.
C= 0.15
n= 0.015
Rational Methode
Qmax=0.278 CAI
:
mm mm
i=16 =0.67
day h
The rainfall intensity from the above location is 12 to 20 mm/day so we take the average value which is
16mm/day
Qmax=0.278 x 0.15 x 2.1 x 0.67
Qmax=0.0586 m 3/sec
Part 2
Design the given discharge for the best hydraulic section
GIVEN DATA:
A=2100000 m2=2.1k m 2
P=6220m
R L2=2090 m
R L1=699 m
Solution:
Trapezoidal Section
Qmax =0.278 CAI
mm mm
i=16 =0.67
day h
C = 0.15
n=0.015
m3
Q=0.278× 0.15 ×2.1 ×0.67=0.0586
sec
S=¿)
2090−699
S= =0.32
4340
5
3
1 (A )
Q=
n ¿¿
5
2 1
1 (1.75 D ) 3 2
Q= (0.32)
n 2
(3.5 D)3
5
2 1
1 (1.75 D ) 3
0.0586= (0.32) 2
0.015 2
(3.5 D) 3
(1.75 D 2 )5
(0.0090)3=
(3.5 D)2
(1.75)5 10−2
0.0000007329= (D )
(3.5)2
D8=0.000000547
D=0.165 m
Triangular Section B
Q=0.34 cumecs(m3/sec)
S= 0.32
1
Z= = 0.707
√2
Required Data
Design most efficient hydraulic section
1
Z= = 0.707
√2
Solution:
1
A=2( zd×d)
2
A=ZD2
P=2 (ZD)2 + D 2
√
P=2 D √ 1+ Z 2
1
P=2D 1+(
√ √ 2
1
¿ ) ¿2
√
P=2 D 1+
P=2 D √ 15
2
P=2.449D
1 2/3 1/2
Q=R S
n
1 A
Q= ( ) 2/3 S1/2
n B
1 ZD 2/3 1/2
Q= ( ) S
n 2.499
1 D
Q= ( )2/3 S1/2
n 2
1 D 2/3 1
0.34 = ( ) (0.32) 2
0.015 2
Solving
D=0.171m
Part c
Feasibility study and cost analysis of two hydraulic sections
Ans
The selected section is trapezoidal and rectangular
The trapezoidal section which is selected is more efficient and feasible because of less depth and
more smooth flow passing the=rough it and its design is alos easily compare to the rectangular
section
It is clear that that the depth of trapezoidal section is lower so it cost will also be lower. The
trapezoidal section is frequently used and it required less labor and energy than triangular
section so in for sustainability consideration the trapezoidal section will be good.
Part d
Consider the maximum and minimum velocity that wil cause erosion in design
Solution;
Now to determine the erosion velocity wee can determine from the following equation
Q= AV
Q max
V=
A
0.34
V=
12200000
V =0.27 ×10−7 m/s
So this is the maximum velocity. The velocity of flow should be with this limit. Above this value it will
cause erosion.
Task 2
Part 01:
i. using the given data from excel and determine yo and also comment
ii. comment the type of flow. After the sluice gate(sub-critical or super-critical)
GIVEN DATA:
V = 0.015m 3
T=23 sec
y g= 12mm = 0.012m
t (cd) = 0.75
b=75mm = 0.075m
Required Data:
i) For the given Discharge Coefficient (Cd) determines y 0 and
comments.
ii) Comment on the type of flow after the Sluice Gate.
Solution:
Part (1)
Step 01 Determine Q
V
Q=
t
0.015
Q=
23
Q= 0.000652 m 3/sec
Step 02 Determine y c
1
Q2
( )
yc= 2
b g
3
Putting all the values
1
( 0.00652)2
yc=
((0.075)2∗9.81 )
3
y c =0.091 m
Part (1), Step 03
Determine y 0
We know that
a=cd . b . y . √2 g . y 0
V
= cd . b . y . √ 2 g . y 0
t
0.015
= 0.75 ×0.075 × 0.015 √ 2∗9.81 . y 0
23
0.7729=√ 2 × 9.81. y 0
( 0.7729)2
y 0=
2 ×9.81
y 0=0.030 m
y g < y c So the type of flow after the sluice gate is super critical flow
Part 02:
(I) Evaluate the friction factor
(II) Comment on the flow type
(III) Pressure variation from section 1 to 2
(IV) Kinetic energy variation from section 1 to 2
GIVEN DATA:
V= 0.005m3
T= 15.5sec
L=912 mm= 0.912m
D= 17 mm= 0.017m
h1= 250mm = 0.250m
h2 = 323mm = 0.323m
Required Data
Solution: (i)
For pipe flow we know for two section the Bernoulli’s equation is
P 1 v2 P 2 v2
+ +h1=¿ + +h +h
r 2g r 2g 2 e
h e=h1−hr
h e = 0.323 – 0.250
h e= 0.073m
Now from Doriy Equation
2 gD . he
f= (i)
LV 2
To determine velocity first we find discharge
Vd
Q=
time
0.005
Q=
23
Q=0.000217 m3 / sec
π∗d 2
A=
4
3.1416∗( 0.017)2
A=
4
A=0.000227 m 2
Q 0.000217
V= =
A 0.000227
V =0.956 m/sec
Using equation
2× 9.81× 0.017 ×0.073
f= 2
0.912 ×(0.956)
f =0.0292
(ii) Comment on flow
ξDV 1000 ×0.017 × 0.956
Re = =
μ 0.912 ×10−4
Re =78413.48
(a) Flow is turbulent because the Reynolds’s No is greater then 4000
(b) Pressure is decreased from section 01 to section 02 because piezometric height decreased from
01 to 02 and we know the pressure is directly to the piezometric height
(c) Kinetic energy is same from 01 to 02 because the velocity is same and Kinetic energy is directly
related to the square of velocity
Discussion:
in this report I have discussed the methodology I have adopted to solve the task assigned to us. The report
clearly explained what have been requested to solve. In this report I have clearly explained each and
everything related to each task assigned. I have introduced the rationale method which I have used to
perform the task 1. To perform this task first I have selected a basin area and calculated the related
parameter used in the rationale method. The topography of the area was identified and made some
assumption to consider the other parameter like rain intensity coefficient. And I have used different table
available in the literature to select the soil parameter for the region. The basin area has been calculated
using google map. The other task in this report has been solved using the standard procedure we have
studied in hydrology, fluid mechanic and open channel flow.
Conclusion:
From the study I have conducted to perform the assigned task have helped me to improve my concept in
the filed of hydrology. I have studied different topic and articles to solve the assigned tasks. Different
method of calculation of runoff has been studied along with the one discussed above. Based on my
studies I have solve the given task. The result obtained are well valid and can be used for any purpose
having topographic condition similar to the ones assumed here in performing the first task. Similarly, the
other tasks have been performed based on different techniques available in the literature. If condition
matched, then we can also use the other valued calculated above will be used to design the open channel
flow. The result obtained from the above analysis are valid and can be used for design purpose or
assessment of any facility if condition are similar as given here in this task.
References
1. https://www.hydrocad.net/rational.htm
2. https://www.lmnoeng.com/Hydrology/rational.php
3. https://www.samsamwater.com/catchments/
4. http://www.fsl.orst.edu/geowater/FX3/help/8_Hydraulic_Reference/Open_Channel_Flow.htm
5. https://www.engineeringtoolbox.com/sluice-gate-flow-measurement-d_591.html
6. https://www.researchgate.net/figure/Rainfall-intensity-in-Oman-This-figure-is-available-in-
colour-online-at_fig7_230325195
