UNIVERSITY OF ANBAR
COLLEGE OF ENGINEERING
DAM & WATER RESOURCES ENG.
DEPARTMENT
Chapter 3
Estimating of flood :part II
Lecture Notes Prepared
by :
Ahmed Sh. Moneer Al-Salmani
Dams &Water Resources Engineering
Department
Master Students
١
outlines
1.Normal
method
2.Log‐
7.Stochastic normal
method method
Flood
frequency
analysis 3.Log‐
6.Powel person
method method
5.Gumbel 4.Chow
method method
1.Normal distribution method
• This method based on normal probability law.
• The flood of any return period which follows the
normal probably law is computed by flowing steps:
∑
Compute mean flood
∑
Find standard deviation
Determine from normal probability table.
Find the required flood
• Example 1
From the annual flood series of river Barnaby , a tributary of the
Brahmaputra, following data are given in cumec . Estimate flood
discharge for return period 100 years.
Year Flow year flow year flow
1978 2949 1987 4798 1996 6599
1979 3521 1988 4290 1997 3700
1980 2399 1989 4651 1998 4175
1981 4124 1990 5050 1999 2988
1982 3496 1991 6960 2000 2709
1983 2947 1992 4366 2001 3873
1984 5060 1993 3380 2002 4593
1985 4903 1994 7826 2003 6761
1986 3751 1995 3320 2004 1971
Solution
∑ ⋯
.
∑
Non‐ exceeding probability .
From normal probability table .
the required flood
. .
Normal probability table
2‐Log normal distribution method
• This method based on log normal probability law ,,the
flood of any return period is computed by flowing steps.
• Compute ,
∑
Compute
∑
and from given flood series
Determine K from normal probability table.
Fine the required flood
Example 2 Resolve Example 1 by log‐normal method
• Solution
Compute
Q Log Q Q Log Q Q Log Q
2949 3.469 4798 3.368 6599 3.819
3521 3.546 4290 3.632 3700 3.568
2399 3.38 4651 3.667 4175 3.620
4124 3.625 5050 3.703 2988 3.478
3496 3.543 6960 3.838 2709 3.432
2947 3.469 4366 3.640 3873 3.588
5060 3.704 3380 3.528 4593 3.662
4903 3.690 7826 3.893 6761 3.830
3751 3.574 3320 3.521 1971 3.294
∑
Compute = 3.607 ,
∑
0.1472
From probability table = 2.327
. . . .
Take antilog .
3‐Log Pearson Type III Distribution
It was Pearson (1930) who developed this model , the
procedure to computed ;
Compute ,
∑
Estimate standard
∑
Compute stranded deviation
∑
Compute skew coefficient
Then
• Example 3
From the annual flood series of river Barnaby , a tributary
of the Brahmaputra, following data are given . Estimate
flood discharge for return period 100 years.
Year flow year flow year flow
1978 2949 1987 4798 1996 6599
1979 3521 1988 4290 1997 3700
1980 2399 1989 4651 1998 4175
1981 4124 1990 5050 1999 2988
1982 3496 1991 6960 2000 2709
1983 2947 1992 4366 2001 3873
1984 5060 1993 3380 2002 4593
1985 4903 1994 7826 2003 6761
1986 3751 1995 3320 2004 1971
• Estimate standard
∑ . . … .
.
• find standard deviation of log Q
∑ .
.
• Compute skew coefficient
∑ ∑
.
.
• from table (15.2 ) K=2.35 at Cs=0.043 , T=100 yrs.
• Then
. . . .
.
• Take antilog for both side → . /
4‐Chow method
Another modification of the Gumbel’s method was made by V.T.
Chow by using the frequency factor.
To find flood of return period( ) :
Sort the data
compute plotting position
Find
By least squares method , the following equation are used and
solve to obtain two parameters a and b.
∑ and ∑ ∑ ∑
Then
• Example 4
Resolve example 1 by using Chow method.
m Q
1 7826 28 ‐1.801 ‐14094.6 3.24
2 6960 14 ‐1.492 ‐10384.3 2.22
3 6761 9.33 ‐1.308 ‐8843.4 1.71
4 6599 7 ‐1.174 ‐7747.2 1.38
5 5060 5.6 ‐1.068 5404.1 1.14
6 5050 4.66 ‐0.978 ‐4938.9 0.95
7 4903 4 ‐0.903 ‐4094 0.82
8 4798 3.5 ‐0.835 ‐4006.3 0.68
9 4651 3.11 ‐0.773 ‐3595.2 0.6
10 4593 2.8 ‐0.717 ‐3293.2 0.51
11 4366 2.54 ‐0.663 ‐2894.6 0.44
12 4290 2.33 ‐0.613 ‐2629.8 0.37
m
13 4175 2.15 ‐0.566 ‐2363 0.32
14 4124 2 ‐0.521 ‐2148.6 0.27
15 3873 1.87 ‐0.478 ‐1851.3 0.23
16 3751 1.75 ‐0.434 ‐1627.9 0.19
17 3700 1.65 ‐0.393 ‐1454.1 0.15
14 3521 1.55 ‐0.346 ‐1218.3 0.12
19 3496 1.47 ‐0.305 ‐1066.3 0.09
20 3380 1.4 ‐0.264 ‐892.3 0.07
21 3320 1.33 ‐0.218 ‐723.8 0.05
22 2988 1.27 ‐0.172 ‐513.9 0.03
23 2949 1.22 ‐0.128 ‐377.5 0.016
24 2947 1.16 ‐0.065 ‐191.6 0.01
25 2709 1.12 ‐0.0132 ‐35.8 0.005
26 2399 1.08 0.053 127.15 0.003
27 1971 1.04 0.151 297.6 0.02
115160 ‐16.024 ‐84552.3 15.634
• from previous table we obtained;
• ∑ 115160 , ∑ ‐16.024 , ∑ 15.634
∑ 84552.3 , n=27
• ∑ → . … . 1
• ∑ ∑ ∑ → . . . … 2
• Solve (1) and(2) . , .
• Then . .
• .
• Then . . .
Table 1. flood corresponding to return period in cumec for example 1
by log‐person ,log‐normal and Chow method respectively
Return probability Normal Log‐person Log‐normal Chow
period method
2 50% 4265 4035 4046 4073
5 20% 5475 5332 5382 5376
10 10% 6107 6177 6165 6238
25 4% 6789 7228 7190 7329
50 2% 7217 8013 7954 8136
100 1% 7608 8757 8902 8941
200 0.5% 7965 9506 9429 9740
Flood frequency curve for example 1 by four method
10000
9000
8000
7000
Log‐person 6000
Log‐normal
charegedis
Chow method 5000
normal method
4000
3000
2000
1000
0
1000 100 10 1
return period
Flood frequency distribution for example 4 by three method
12000
10000
Log‐person
Log‐normal
Chow method
8000 normal method
Flow
6000
4000
2000
0
٢٠٠ ١٠٠ ٥٠ ٢٥ ١٠ ٥ ٢
return period
5‐Gumbel method when sample is not infinite
In practical field , sample is not infinite
Procedure to compute :
Find the and from given data.
Find vlaue of
Compute the reduced variate by . .
Find frequency factor
Then
Example 5
The Annual Peak discharges in Ganga river at Hardwar for the
period 1885‐1971 (87 yrs.) is given in flowing table. find the annual
flood which has a return period of 1000,500,200,100,50 yrs.
SI.No. year Ann. peak ) .SI.No year Ann. peak )
Q( Q(
1. 1885 7241 11. 1895 9680
2. 1886 9164 12. 1896 14336
3. 1887 7404 13. 1897 8174
4. 1888 6870 14. 1898 8953
5. 1889 9855 15. 1899 7546
6. 1890 11887 16. 1900 6652
7. 1891 8827 Cont.. Cont.… Cont..
8. 1892 7546 85. 1969 4546
9. 1893 8494 86. 1970 3842
10. 1894 16767 87. 1971 4542
Solution
∑
∑ 6635.63 , 3130.8 ,
for n=87 from table 1.1 then 0.55815, 1.1987
T‐yrs. . .
1000 ‐3.361 6.907 5.29 23185
500 ‐3.060 6.213 4.7 21335
200 ‐2.662 5.295 3.95 19005
100 ‐2.360 4.6000 3.36 17155
50 ‐2.056 3.901 2.79 15365
5 ‐1.013 1.497 1.714 12003
Table 1.1 Reduced mean( ) and reduced standard deviation (
as functions of sample size n
Size of Size of
sample sample
10 0.4952 0.2457 55 0.5504 1.1681
15 0.5128 1.0206 60 0.5521 1.1747
20 0.5236 1.0628 65 0.5536 1.1803
25 0.5309 1.0915 75 0.5549 1.1898
30 0.5362 1.1124 85 0.5578 1.1973
35 0.5403 1.1283 90 0.5553 1.2038
40 0.5436 1.1413 100 0.5600 1.265
45 0.5436 1.1518 200 0.5672 1.2359
50 0.5465 1.1608 500 0.5724 1.2588
1000 0.5745 1.2685
6‐Powell Method
• Powel made a modification of Gumball method when
sample is not infinite (when sample is not infinite)
Compute
γ ln ln , 0.5772
Simplifying, . .
Then
Example 6
Resolve example 5 by using Powel method .
Solution
6635.63 , 3130.8
T‐yrs.
. .
1000 ‐3.361 4.933 22079.8
500 ‐3.060 4.393 20389.2
200 ‐2.662 3.678 18150.7
100 ‐2.360 3.136 16453.8
50 ‐2.056 2.591 14747.5
5 ‐1.013 0.719 8876.14
7.Stochastic method
• The basic approach to stochastic generation applies to stream flow
,evaporation, precipitation and other hydraulic factors.
• The flood . ,Her , is
number of recorded floods ,counting only one for same flood peak occurring in
different years.
• Example7 ; resolve example 5 by using stochastic method
• 6635.63 , 3130.8 , 2341 , 77
• 2341
77 T‐yrs. .
2.3 665.63 2341 log
87
1000 2.947 315141
2341 9890 log 0.885
500 2.646 28541
200 2.242 24514
100 1.947 21541
50 1.616 18601
5 0.646 8729
Table 2. flood in cumec corresponding with return period
for example 1 by three method respectively
Return period P=1/T % Gumbel Powel stochastic
T
5 20% 23185 22079.8 315141
50 2% 21335 20389.2 28541
100 1% 19005 18150.7 24514
200 0.5% 17155 16453.8 21541
500 0.2% 15365 14747.5 18601
1000 0.1% 12358.3 8876.14 8729
Flood frequency curve for example 1 by three method
35000 Gumbel
Powel
stochastic
30000
25000
flood in cumec
20000
15000
10000
5000
0
1000 100 10 1
return period in year
Thank you
٣٠
