CIVL253 –HYDROLOGY
Final Examination (25 October 2006)
Name (English/Chinese)
Student Number:
Problem 1
/ 20
Problem 2
/ 15
Problem 3
/ 20
Problem 4
/ 45
Total Score
/ 100
Declaration on Academic Integrity
I declare that the answers submitted for this examination are my own work.
I understand that sanctions will be imposed, if I am found to have violated the
University regulations governing academic integrity.
Student's Signature:
Note:
(1) This is a close-book examination. No reference material of any kind can be used
during the examination.
(2) All computation steps must be shown clearly - without them, no score will be given.
(3) Assumptions made must be clearly stated and justified.
(4) Proper units must be attached.
Problem-1 (20%)
Consider a rainfall event having 5-min cumulative rainfall record given below:
Time (min)
Cumulative
Rainfall (mm)
Time (min)
Cumulative
Rainfall (mm)
Time (min)
Cumulative
Rainfall (mm)
0
5
10
15
20
25
30
0
7
14
23
34
45
58
35
40
45
50
55
60
65
70
81
91
100
110
119
125
70
75
80
85
90
131
136
140
140
140
(a) What is the duration of the entire rainfall event and the corresponding total rainfall
amount? (5%)
(b) Find the rainfall depth hyetograph (in tabular form) with 10-min time interval for the
storm event. (7%)
(c) Find the maximum 10-min and 20-min average rainfall intensities (in mm/hr) for the
storm event (8%)
Problem-2 (15%)
A 600-hectare farm land receives annual rainfall of 2,500 mm. There is a river flowing
through the farm land with inflow rate of 5 m3/s and outflow rate of 4 m3/s.
It is measured that the annual water storage in the farm land increases by 2.5×106 m3.
Based on the hydrologic budget equation, determine the annual evaporation amount (in
mm). (15%) [Note: 1 hectare = 10,000 m2]
Problem-3 (20%)
You are hired as a consultant to help design a coffer dam for protecting the construction
of bridge piers in a river (see figure). Based on 30 years of flood data in the river, the
largest flood is 7,800 m3/s and the smallest is 2,400 m3/s.
The probability plot indicates that the flood magnitude and the corresponding return
period, estimated by the Weibull plotting position formula, Pr[X≥x(m)]≈m/(n+1), reveal a
linear relation.
Assume that the bridge piers have a construction period of 3 years and the required
overtopping probability over the construction period must not exceed 10 percents.
(a) Determine the required design return period for the coffer dam. (10%)
(b) Determine the magnitude of flood to be protected against for the coffer dam. (10%)
Problem-4 (45%)
Given flood data in a river a shown in the table below.
Peak Discharge (m3/s)
95
300
Year
1963
Date
1 June
3 August
1964
7 June
60
1965
2 August
80
1966
28 May
15 August
100
90
1967
11 July
25 August
4 September
800
700
90
(a) Determine the data set for both annual maximum series and annual exceedance series.
(5%)
(b) Determine the first three sample Lmoments based on the annual exceedance series
using plotting position formula Pr[X≤x(m)]≈(m0.35)/n. (15%)
(c) Suppose the sample L-kurtosis (t4) is 0.05. Identify the best-fit distribution for the
annual exceedance series between two candidate distribution models: lognormal and
generalized Pareto distributions. (15%)
(d) Based on the best-fit distribution identified for the annual exceedance series in Part(c), determine the magnitude of 50-year flood. (10%).
FORMULA SHEETS
L-moments:
Population
Sample

r  E X  F ( X )
r



^
1 n
br   x m  F [ x( m ) ]
n m 1
r
for r  0, 1, ...
1 =  0
2 = 2 1   0
3 = 6  2  6 1 +  0
 4 = 20  3  30  2 + 12  1   0
l1 = b 0
l 2 = 2 b1  b0
l 3 = 6 b2  6 b1 + b0
l 4 = 20 b3  30 b2 + 12 b1  b0
 2 =  2 / 1 ;  3 =  3 /  2 ;  4 =  4 /  2
t 2 = l 2 / l1 ; t 3 = l 3 / l 2 ; t 4 = l 4 / l 2
Sample Product-Moments:
  E( X )
x
  E  X    
2
2

s

2
E  X    



3
2
g

x
i
n
  x  x

2
n 1

n xi  x

x

2
i
i
 nx
2
n 1
2
(n  1)(n  2) s 3

n
  x  3x x  2nx 
3
i
2
i
3
(n  1)(n  2) s 3
Approximating formula for frequency factor of Pearson and log-Pearson type 3 distribution:
g 1
g
KT  zT   z  1     zT3  6 zT   
6 3
6
2
2
T
3
4
g
 g  1 g 
  zT2  1    zT     
6
 6  3 6 
5
where zT is the standard normal quantile corresponding to 1/T exceedance probability and g
is the skewness coefficient.
Approximated 3 4 relations for selected distribution models:
Lognormal distribution:
4 = 0.1228 + 0.7752 32 + 0.1228 34  0.1364 36 + 0.1137 38
Generalized Pareto distribution:
4 = 0.202 3 + 0.959 32  0.201 33 + 0.0406 34
Relations between moments and parameters of selected distribution models
Distribution
Normal
PDF or CDF
f x (x) 
 1  x  
1
exp   

2 
 2   
1
Log-normal
f x (x) 
Rayleigh
f x (x) 
Range
2 ln x
2



 1  ln ( x )   2 
ln x
exp   

 ln x
x
 2 
 
 x    exp   1 

2
x   


 
 2    
2
 1
 x  
1
  ( )   
 ( x  ) / 
Pearson 3
f x (x) 
Exponential
f x (x)  e x /  
Gumbel
(EV1 for maxima)
f x (x) =
Weibull
fx ( x ) =
GEV
(Generalized
Extreme Value)
   (x   )  1/ 
Fx (x) = exp  1 



 
GPA
(Generalized
Pareto)
e

  x   
  x  

exp   
  exp       



 
 

 

1
  x  
   
 1
  x   
exp   
 
    
1/ 

( x   )
Fx (x)  1   1  
 




Product Moments
L-moments
1=; 2=
 < x < 
ln x  ln( x )   ln2 x / 2 ;  ln2 x  ln   x2  1 ;
x>0
 x  3 x   x3
x<

    2  ;   2  
2
>0
for >0: x>
for <0: x<
     ;  2  2 ;   sign( )
x>0
= 
2 
 >0; x>0


x   


<0:


x    



>0:



x    



<0: x<;



;



2 1

1=; 2=/2; 3=1/3; 4=1/6
2 2
1=; 2=
3=0.1699; 4=0.1504
;   1.1396
6

1
 
2

1 


 




1   1     ;
 
2
 
 2     1  2    2 1   
 
  

1

2
2

1
; 2 

1   
2(1   ) (1  2 )1 / 2
(1  3 )
2

1    
3 
2
 (1  2 )
1

1
1 / 

 1  
1       1  ; 2   1  2






;
 
 [1   (1   ) ]
 
  
;
2 
3=0.1140; 4=0.1054
     1   ;  2   2   1     2  1   



>0:
Eq. (2.68); Eq. (2.70)
1      2 ;
    0.5772 ;
 < x < 
; 3=0; 4=0.1226
;
2  1  3 
 1 2 

1   
3 
;
 3 4 

1
1
3
;
2 



 1  2     1    ;





1  5 4  10 3  6 2
; 2 
1  2

(1   )(2   )
(1   )(2   )
4 
(3   )(4   )
