ENGIN5201 Hydrology Assignment: Problems 3, 4, and 5 Solution

This document presents a comprehensive solution to a hydrology assignment, focusing on key concepts in surface water hydrology. The solution addresses three problems: calculating groundwater velocity and daily water flow through an aquifer given hydraulic gradient, hydraulic conductivity, and porosity; determining stream discharge using velocity measurements at different depths; and analyzing the water surface profile based on given data related to concrete structures. The solution includes detailed calculations, application of relevant formulas, and graphical representations to illustrate the concepts. The assignment covers topics such as groundwater flow, stream discharge calculations, critical and normal depth, and water surface profile analysis. The document provides a complete, step-by-step solution to the assignment problems.

HYDROLOGY

Problem 3:
According to the question the given value is:
Hydraulic gradient is 0.08 m/m
Hydraulic conductivity = 120 m/day
Porosity=25%
Area=87500
Velocity= Q*porosity/area
=120*0.08*0.25/87500 m/s
=2.74 m/s
Problem 4:
Distance across
stream (m)
Velocity at 80% of
water depth (m/s)
Velocity at 20% of
water depth (m)
Total water
depth (m)
0.00 0.00 0.00 0.00
0.5 0.25 0.19 0.21
1.0 0.33 0.18 1.47
1.5 0.45 0.27 0.94
2.0 0.55 0.22 0.83
2.5 0.67 0.45 1.02
3.0 0.93 0.69 1.12
3.5 1.27 0.84 1.14
4.0 0.73 0.60 0.89
4.5 0.44 0.34 0.56
5.0 0.00 0.00 0.00
Table 1: stream depth and velocity measurement

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
Velocity at 80% of water depth (m/s) Total water depth (m)
Figure 1: Stream depth and velocity measurement
The total steam discharge can be calculated through critical evaluation of above gathered.
Q =?
It is assumed that the total depth of water is D
Therefore, at distance 1 m, the velocity of the water is 0.33 m/s at 80% depth and at 20%
depth the velocity of water is 0.18 m/s. The total water depth is 1.47 m
In order to obtain necessary value, float method has been used to measure the steam float
rate, which can be represented mathematically below are as follow:
Flow rate (Q): width (b) * depth (D) * (distance (d) / time (t)) ---- (1)
By putting the value in equation (1)
Q = 2.5 * 1.47 * (1/t) ---- (2)
In order to determine the at distance 1 m with respect to velocity continous formula has been
used, which can be mathematically represented as follow:
D = sped * time
1 = 0.33 *t
t =1/0.33
t = 3.03 s ---- (3)

By putting the equation (3) in equations (2), we get
Q = 2.5 * D * (1/3.03)
Q = 2.5*1.47*0.33
Q = 1.21 m3/s
Problem 5:
The above data have been to obtain different value in respect to concrete structure
significantly.
Given data,
Q = 1.21 m3/s -------- [Problem 4]
Width (b) = 2.5 m
The total depth flow over the hydraulic control is 1.85 m
Height of the hydraulic control Ps = 0.8 m
Roughness is estimated n = 0.06
Slope is estimated = 0.0022 m/m
The above data have been used to obtain below rate
1:
The normal depth of water (Yn) can be determined as follow:
Let the Yn be the height under uniform flow
According to continuity flow:
Q = A*V --- (6)
Where,
Q = flow rate
A= Area
V= volume
Q = (b*Yn)*c * (slope of bed)1/2

Q = (2.5*Yn)* m1/6/N*m1/2*ib
1.21 = (2.5 *Yn)* m2/3*0.0022
0.06
1.21*0.06 /2.5 = Yn*m2/3 ------ (7)
0.02904 = Yn*m2/3 ------ (7)
The value of m can be determined as follow:
m = A/ps
m = (2.5*Yn) / 0.8 ----- (8)
By putting the equation (8) into equation (7), we get
0.02904 = Yn *(3.125Yn)2/3
0.0294 = Yn (9.765Yn2)1/3
0.0294 = Yn*2.13Yn1/3
0.0294/2.13 =Yn4/3
0.0138 = Yn4/3
Yn = (2.62 *106)1/4
Yn = 0.040 m
2:
The critical depth (Yc) for the stream flow can be determined as follow:
Yc = (q2 / g)1/3 ------ (4)
In order to obtain the value of q, below formula has been used as follow
q = Q/b
q = 1.21/2.5
q = 0.484 m2/s ---- (5)
By putting the equation (5) into equation (4), we get the value of critical depth

Yc = ((0.484)2/ 9.81)1/3
Yc = (0.234/ 9.81)1/3
Yc = (0.024)1/3
Yc = 0.28 m
From the above obtain data, it has been evident that
Y >Yc > Yn ------- (9)
Through critical analysis of equation (9), it can clearly state the surface profile in the stream
is Mild Slope Curve
3:
Depth
(m)
Area (m) Paramete
r
Hydraulic
radius (m
=N/p)
Velocity V2/2g E=
h+v2/2g
Slope
(ie)
0.5 0.02 24 1.6 1.625 0.134 2.1346 0.00053
1 0.04 24.4 1.8 0.11 0.11 2.311 0.00039
1.5 0.06 24.8 1.9355 1.35 0.0934 2.4934 0.00034
Table 2: Water surface profile
4:
0%
20%
40%
60%
80%
100%
0.5 0.02
24 1.6
1.625
0.134 2.1346
0.000534
1 0.04
24.4 1.8 0.11
0.11 2.311 0.00039
Series1 Series2 Series3
Figure 2: Water surface profile