Solution: CIVL2104 Hydraulics and Hydrology Assignment - HKU

This document presents a detailed solution to a CIVL2104 Hydraulics and Hydrology assignment from The University of Hong Kong, covering various aspects of hydrology and hydraulics. The solution includes calculations and analyses related to runoff, the rational method, unit hydrographs, and the SCS method. It addresses multiple questions, providing step-by-step solutions, tables, and graphs to illustrate the concepts. Key topics include the determination of the recession constant, baseflow separation, hydrograph sketching, peak runoff calculation using the rational formula, the phi-index method, excess rainfall hyetograph determination, and direct runoff hydrograph computation. The assignment employs formulas and data analysis to solve practical problems in hydrology, supported by references to relevant literature.

Question 2
Applying the rational method (Robert J. Houghtalen, 2010);
Qp=1.008 CiA
Where;
Qp is the peak flow
C is a coefficient
∅ is the rainfall intensity.
A is drainage area.
Making A the subject;
A= Qp
1.008 C ∅

¿ 48
1.008∗0.25∗12∗10−3
¿ 15873 m2
Area of the basin = 15.873 k m3
Developing the unit hydrograph, we formulate a table as shown below.
Time in
min
Base
flow
Discharg
e
Cumulative
discharge
rainfall depth
(mm) runoff
0 2.5 8 10.5 0.5
0.01322
8
5 2.5 13 15.5 6 0.15873
10 2.5 18 20.5 6 0.15873
15 2.5 24 26.5 1
0.02645
5
20 2.5 30 32.5 3.5
0.09259
3
25 2.5 20 22.5 3.5
0.09259
3
30 2.5 10 12.5 3.5
0.09259
3
Plotting the unit hydrograph with base flow of 15 m3 /s, we obtain;
0 5 10 15 20 25 30
0
5
10
15
20
25
30
35
Cumulative discharge
runoff
Time (min)
Discharge (cumec/s)

Question 2(second)
Applying the rational method;
A= Qp
1.008 C ∅
¿ 30
1.008∗0.25∗12∗10−3
¿ 9920.63 m2
The catchment area ¿ 9.921 k m2
Developing the unit hydrograph, we formulate a table as shown below.

Time in
min
Base
flow Discharge
Cumulative
discharge
rainfall depth
(mm) runoff
0 2.5 8 10.5 0.5 0.008267
5 2.5 13 15.5 6 0.0992
10 2.5 18 20.5 6 0.0992
15 2.5 24 26.5 1 0.016533
20 2.5 30 32.5 3.5 0.057867
25 2.5 20 22.5 3.5 0.057867
30 2.5 10 12.5 3.5 0.057867
Graphing the data, we get;
0 5 10 15 20 25 30
0
5
10
15
20
25
30
35
Cumulative discharge
runoff
Time (min)
Discharge (cumec/s)

Question 3
From the rational method, we make ∅ the subject to obtain
A= Qp
1.008 C ∅
∅ = QP
1.008 CA
¿ 17.9
1.008∗0.25∗2.4
¿ 29.60 mm
≈ 30 mm
The deconvolution equation for a linear system is given by

Qp=∑
m=1
n ≤m
Pm U n−m+1
Where Qn is the direct runoff.
Pm is the excess rainfall.
Un−m +1 is the unit hydrograph.
The values are calculated in the attached excel in sheet 2.
Tabulating the data, we get
time Stream flow rainfall
Base
flow Unit hydrograph Excess rainfall Q
0 0.5 0 0.5 1 1.8 1.8
10 1.3 24 0.5 1.8576 1.8 5.14368
20 6.5 66 0.5 7.1584 1.8 18.0288
30 14.9 12 0.5 15.4288 0 44.00064
40 17.9 0 0.5 18.4 0 73.77696
50 5.9 0 0.5 6.4 0.5 72.91184
60 0.5 0 0.5 1 0.5 47.8688
Graphing the data, we get;
0 10 20 30 40 50 60
0
10
20
30
40
50
60
70
80
Time in minutes
Discharge in cumec/s

Question 4
a) Curve number of 75 and a normal antecedent moisture condition, use the SCS method to
compute the cumulative abstractions and cumulative excess rainfall. Determine and plot
excess rainfall hyetograph
Solution
Abstraction FA=S∗( P−Ia
P−Ia+ S ) CITATION SKM13 \l 2057 (S.K. Mishra, 2013)
Where S is total maximum storage.
Ia is the initial abstraction.
P is the total rainfall
S= 1000
CN −100

¿ 1000
75 −100
¿ 3.333∈¿
¿ 0.085 m
Ia=0.2 S=0.2∗0.085
¿ 0.017 m
The calculated values of the cumulative abstractions are in excel file SCS sheet. The cumulative
excess rainfall and excess rainfall hyetograph are also obtained in the attached excel.
Cumulative abstractions
Time
Rainfall
(mm/h)
Cumulative
rainfall Ia Fa
Cum. Excess
rainfall
Excess rainfall
hyetograph
0 0 0 0 0 0 0
0.5 12 12 12 0 0 0
1 90 102 85 0.084577114 16.91542289 16.91542289
1.5 26 128 85 0.084832308 42.91516769 25.99974481
Graphing the data to obtain the excess rainfall hyetograph, we obtain;
0 0.5 1 1.5
0
5
10
15
20
25
30
Excess rainfall hyetograph
Time in hrs
Excess rainfall (mm)

b) A half-hour triangular unit hydrograph is developed for the basin in (a) and the unit
hydrograph has a peak discharge of 120 m3 /s /cm at 60 minutes and a base time of 150
minutes. Determine the river basin area. For the excess rainfall hyetograph obtained in (a)
compute and plot the direct runoff hydrographs.
Solution
Basin area A= Qp
1.008 C ∅ CITATION MRY13 \l 2057 (Putty, 2013)
A= 120
1.008∗0.25∗90 =5.291 k m2
Direct runoff hydrograph is shown in excel SHEET 4. The formulated table is as shown
below.
Excess 1 2 3
time Precipitation (mm/h) 12 90 26 Direct runoff
0 0 0 0
0.5 0.5 6 0 6.5
1 16.92 203.04 45 0 264.96
1.5 42.92 515.04 1522.8 13 2093.76
2 3862.8 439.92 4302.72
3 1115.92 1115.92
This data was used to develop direct runoff hydrograph below.

0 0.5 1 1.5 2 3
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
Time in Hrs
Flow rate in cumec/s
Question 4 (second one)
Basin area A= Qp
1.008 C ∅
A= 116
1.008∗0.25∗6 =76.720 k m2

The hydrograph is obtained in excel file QUESTION 4(SECOND ONE). The tabulated data are as follows
time Stream flow rainfall Unit hydrograph Excess rainfall Q
0 21 1 21.7672 5 108.836
2 43 4.2 46.22224 5 339.9472
4 72 2.2 73.68784 0 599.5504
6 116 0 116 0 948.4392
8 51 0 51 3 900.3016
10 26 0 26 3 588.9683
This data was used to generate the unit hydrograph below.
0 2 4 6 8 10
0
100
200
300
400
500
600
700
800
900
1000
Unit hydrograph
Q
Time in minutes
Discharge in cumec/s
References

Putty, M. R. Y., 2013. Principles of Hydrology. In: s.l.:I.K. International Publishing House Pvt. Limited, pp.
222-226.
Robert J. Houghtalen, A. O. A. N. H. C. H., 2010. Fundamentals of Hydraulic Engineering Systems. In:
s.l.:Prentice Hall, pp. 213-217.
S.K. Mishra, V. S., 2013. Soil Conservation Service Curve Number (SCS-CN) Methodology. In: s.l.:Springer
Science & Business Media, pp. 176-188.