Data reliability checking
Before proceeding to the other analysis, the adequacy of rainfall data series should be checked and it
should be realized. The data series could be considered and adequate if relative standard error, e10%,
where eis the relative standard error, A 30 years data checking using arithmetic mean and outlier test is
shown below.
S/no. Year Heaviest Rainfall (mm/day)=Xi
1 1953 55.3
2 1954 54.6
3 1955 59
4 1956 41
5 1957 61.9
6 1958 53.6
7 1959 63
8 1960 41.2
9 1961 54.5
10 1962 51.4
11 1963 51.2
12 1964 76
13 1965 34.2
14 1966 41
15 1967 65
16 1968 53.5
17 1969 47.6
18 1970 59.5
19 1971 45.1
20 1972 43.4
21 1973 15
22 1974 58.1
23 1975 94.5
24 1976 39.5
25 1977 60.7
26 1978 48.9
27 1979 56.4
28 1980 38
29 1981 62.3
30 1982 58
31 1983 82.8
32 1984 49.6
33 1985 62.4
34 1986 42.8
35 1987 50.5
36 1988 67.9
37 1989 74.8
38 1990 59
39 1991 42.2
40 1992 52.4
41 1993 52.4
42 1994 58.5
43 1995 33.8
44 1996 42.7
45 1997 54.3
46 1998 68.4
47 1999 61.2
48 2000 50.6
49 2001 71.6
50 2002 64.4
51 2003 60.9
52 2004 46.8
53 2005 70.4
54 2006 64.8
Sum 2968.6
Mean 54.97407
Standard deviation 13.23526
Statistical series analysis
For normal series
x=
∑ xi =2968 .6 =54 . 97407
Mean = n 54
Standard deviation, Sx
√
S x= ∑
n−1
=
54−1 √
( xi−x ) 2 9284 . 124
=13 . 23526
For log transformed series
y=
∑ yi =1 .725756
Mean n
Standard deviation, Sy= √ ∑ ( yi− y )2 =
n−1 √ 0 .767642
53
=0 . 120348707
Se Sx
e= ∗100 ,Where Se= , standard error
Xm √N
13.23526 1.8011
Se= =1.8011, e= ∗100 =3.27627 % < 10% ok!
√ 54 54.97407
Therefore the data is reliable and adequate.
Data consistency testing
Outliers are data points that depart significantly from the trend of the remaining data. Procedures for
treating outliers require judgment involving both mathematical and hydrologic consideration. The
retention and deletion of these outliers can significantly affect the magnitude of statistical parameters
more significantly in small samples. Therefore according to US Water resource council (1981), If the
station skeweness coefficient is greater than +0.4 tests for higher outliers should be done first; If the
station skeweness coefficient is less than -0.4 tests for lower outliers should be done first and if the
station skeweness is between -0.4 and+0.4 tests for both high and lower outliers should be applied before
any elimination from the data set. The station skeweness coefficient is computed as follows:
Skeweness coefficient is given by:
n ∑ ( yi− y )3
3
= 54∗(−0 . 15396 ) =−1 .73065942
Csy= ( n−1 ) ( n−2 ) Sy 53∗52∗0 . 000174311
Where, n=sample size
Yi = log transform value
y = Log transformed mean
Sy = Standard deviation of data log units
Since the station skeweness coefficient is less than -0.4 tests for lower outliers should be done first.
Here table of log transfored
Determination of the threshold value of lower outlier:
Lower outlier is given by: YL = Ym – Kn*Sy
YL= 1.725756- *0.120348707 =
Minimum rainfall(X) = 10YL = 10
Lower outlier is = 10 = mm
The lowest recorded data i.e. 15 mm is not lower than the lower outlier therefore no data will be
eliminated and the fifty four years recorded data is used for storm analysis.
Determination of the threshold value of lower outlier:
Lower outlier is given by: YL = Ym – Kn*Sy
YL= 1.7950- 2.2008*0.1001 =1.5387
Daily heaviest rainfall
Areal rain fall
After checking the consistency of the data for higher and lower outlier, the 30 years data is obtained as
representative for the analysis. The probability of occurrence of maximum probable rain fall is estimated
as follow: We have taken the return period (T) is 50 years, (Subramanian K., 1984)
Statistical Parameters for normal series
x=
∑ xi =2968 .6 =54 . 97407
Mean = n 54
Standard deviation, Sx
√
S x= ∑
n−1
=
54−1 √
( xi−x ) 2 9284 . 124
=13 . 23526
Coefficient of skewness, Csx
n ∑ ( xi−x )3 54∗10680 . 47
3
=
Csx = ( n−1 ) ( n−2 ) Sx 53∗52∗13 .23526 3 = 0.090263
Coefficient of kurtosis, Ckx
n 2 ∑ ( xi−x )4 54 2∗6241584 . 04
Ckx=
( n−1 ) ( n−2 )( n−3 ) Sx 4 =53∗52∗51∗13. 23526 4 = 4.687129
Statistical Parameters for log transferred series
y=
∑ yi =1 .725756
Mean n
Standard deviation Sy= √ ∑ ( yi− y )2 =
n−1 √ 0 .767642
53
=0 . 120348707
Coefficient of skewness ,Csy
n ∑ ( yi− y )3
3
= 54∗(−0 . 15396 ) =−1 .73065942
Csy= ( n−1 ) ( n−2 ) Sy 53∗52∗0 . 000174311
Coefficient of Kurtosis, Cky
( y− y )4
n ∑
2
=
2916∗0 .103106
4 =10 .19661
Cky= ( n−1 ) ( n−2 ) ( n−3 ) sy 53∗52∗51∗0 .00021
The design daily point rainfall of a certain return period in this case, 50 years, is assumed to follow different
types of distribution as discussed below.
Frequency analysis
Frequency method may be used for drainage basin of any size. The reliability of the frequency estimates
depends on the lengths of the observed record rather than the method of probability analysis. The return
period or recurrence interval (T) is the average number of year during which a flood of given magnitude
will be equaled or exceeded once and computed by; Wilbur method (1939) as;
T = n+1/m , where n=number of event ,m=order or rank of the event , T=recurrence interval (year)
For determining extreme events, the following techniques are carried out for analysis
1. Normal distribution
XT = Xm + KT *Sx where, For T = 50 years, p = 1/50 = 0.02
W = (ln(1/p2))0.5 = (ln(1/0.022))0.5 = 2.7971
2
2.51557+ 0.608486∗w+0.01033 ¿ w
Kt =w− 2 3
1+1.143279∗w+ 0.1992 ¿ w +0.00131 ¿ w
KT = 2.054
From the above table we have, Xm = 54.9407 and Sx = 13.23526
By substituting these values on the formula , XT = Xm + KT *Sx
X50 = 54.97407 + 2.054*13.23526
X50 = 82.1259mm
2. Log normal distribution
XT = 10 (Ym + KT *Sy) ,where Ym = 1.725756 , Sy = 0.120348 and KT =2.054 since it is the same as the
calculated value on normal distribution.
XT = 10 (1.725756 + *2.054 0.120348) = 93.962mm
3. Extreme value- I distribution
XT = U + α * YT , U = Xm – 0.5772 α , α = (√6/∏) * Sx , KT = (YT – Yn)/Sn
YT = -ln(-ln(1-1/T))= -ln(-ln(1-1/50)) = 3.902
KT = (3.902 – 0.5485)/1.1607 = 2.8892
U = 54.9407- 0.5772*10.3195 = 48.9843, α = (√6/∏) *13.23526 = 10.3195
X50= U + α * YT = 48.9843 + 10.3195*3.902 = 89.251mm
4. Pearson type III
XT = Xm + KT *Sx ,where Xm = 54.9407 , Sx = 13.23526
KT= z+(z2-1)k +1/3(z3-6z)k2 –(z2-1)k3 + zk4+1/3k5 , where k = Csx/6 = 0.09026/6 = 0.01504 and z is
obtained from the table values of Cs and return period. But Csx = 0.09026251 and a return period of 50
years then we obtain z value through interpolation. 0→2.054
0.09→z
0.1→2.107 from this we get z = 2.1017
KT= z+(z2-1)k +1/3(z3-6z)k2 –(z2-1)k3 + zk4+1/3k5 , by substituting z= 2.1017 and k = 0.01504
KT= 2.1528
XT = Xm + KT *Sx
X50 = 54.9407+ 2.1528 *13.23526 = 83.4335mm
5. Log pearson type III distribution
XT = 10(Ym + KT *Sy) , where Ym = 1.725756 , and Sy = 0.120349
KT= z+(z2-1)k +1/3(z3-6z)k2 –(z2-1)k3 + zk4+1/3k5 , where k = Csy/6 = -1.730659/6 = -0.28844 and z is
obtained from the table values of Csy and return period. But Csy = -1.730659 and a return period of 50
years then we obtain z value through interpolation. -1.7→1.116
-1.730659→z
-1.8→1.069 from this we get z = 1.10159
KT= z+(z2-1)k +1/3(z3-6z)k2 –(z2-1)k3 + zk4+1/3k5 , , by substituting z= 1.10159 and k = -0.28844
KT = 1.0614
XT = 10(Ym + KT *Sy)
X50 = 10(1.725756 + 1.0614 *0.120349) = 71.366mm
Normal distribution 82.1259mm
Log normal distribution 93.962mm
Extreme value- I distribution 89.251mm
Pearson type III 83.4335mm
Log pearson type III distribution 71.366mm
Therefore, from the above calculation Log normal distribution (type II) gives the highest value of point
rainfall depth.
Testing the goodness of fit
The D-index tests for the comparison of the fit of various distributions in the upper tail are given by:
D-index = (1/Xm)* ∑Abs (Xi-Xical)
The validity of a probability distribution function proposed to fit the empirical frequency distribution of a
given sample may be tested graphically or analytically. Graphical methods are usually based on visual
comparison which may make decisions a bit subjective and the procedure require a specially designed paper
such as Gamble, log- log etc. However, the paper is not available for all distributions in practice. Graphical
approach for dialoging, how well distribution fits the data are subjective. A number of analytical tests are in
practice now a day at global level. Some of the commonly used tests are :- Dindex test, K-s test , X2 test
How ever, for the sake of simplicity and for power and flexibility of the test Dindex test is applied in our case.
Dindex = / where: - is mean of normal data
Xi is the ith observed data
Xic is the ith computed data
The distribution giving the least D-index is considered to be the best fit distribution.
Table . statistical results for both nomal and log transformed data
Statistical parameter Normal data Log transformed data
Mean 54.97407 1.725756
Standard deviation 13.23526 0.120348707
Skwness.coefficient 0.090263 -1.73065942
Variation coefficient 0.24090085 0.0697368
Normal distribution
Xical=Xi+Kt*σx
2
2.51557+ 0.608486∗w+0.01033 ¿ w
Kt =w− 2 3 The answer is found in the table 3.2 below.
1+1.143279∗w+ 0.1992 ¿ w +0.00131 ¿ w
W= (ln (1/pr) ^2) ^0.5
Table 3.2 Normal distribution
Xi rank P=m/n+1 KT Xi cal Abs(Xi – X
ical)
94.5 1 0.142857 1.043192 108.3069 13.8069
82.8 2 0.285714 0.525359 89.75327 6.95327
76 3 0.428571 0.126265 77.67115 1.67115
74.8 4 0.571429 -0.24479 71.5601 3.2399
71.6 5 0.714286 -0.63756 63.16177 8.438234
70.4 6 0.857143 -1.12864 55.46219 14.93781
SUM 49.047264
1
D-index = ∗∑ |( xi− Xical )|
Xm
= 49.047264/54.97407 = 0.8922
2. Log normal distribution
Yical=Yi+Kt*σy
Table 3.3 Log normal distribution
Yi Rank P=m/n+1 Kt Yi cal Abs(Yi-Yi cal)
1.975432 1 0.142857 1.043192 1.851303 0.124129
1.91803 2 0.285714 0.525359 1.788982 0.129048
1.880814 3 0.428571 0.126265 1.740952 0.139862
1.873902 4 0.571429 -0.24479 1.696296 0.177606
1.854913 5 0.714286 -0.63756 1.649026 0.205887
1.847573 6 0.857143 -1.12864 1.589926 0.257647
SUM 1.034179
1
D-index= ∗∑ |( Yi−Yical )|
Ym
=1.034179/1.725756 = 0.5993
Pearson Type III Distribution Method
Kt=Z+(Z^2-1)*k+1/3*(z^3-6*Z)*K^2-(Z-1)*k^3+Z*K^4+1/3*K^5
Where K=Cs/6 =0.0529
2
2.51557+ 0.608486∗w+ 0.01033¿ w
Z=w− 2 3
1+ 1.143279∗w+0.1992∗w +0.00131 ¿ w
Where W= (ln(1/pr)^2)^0.5
Table 3.4 Pearson Type III Distribution Method
Xi Rank P=m/n+1 Kt X ical Abs(Xi-Xical)
94.5 1 0.142857 1.044133 108.3194 13.8194
82.8 2 0.285714 0.514245 89.60617 6.80617
76 3 0.428571 0.111411 77.47455 1.47455
74.8 4 0.571429 -0.25881 71.37452 3.425483
71.6 5 0.714286 -0.64621 63.04722 8.552785
70.4 6 0.857143 -1.12411 55.52208 14.87792
SUM 48.956
1
D-index= ∗∑ |( xi− Xical )|
Xm
= 48.956/54.97407 = 0.8905
Gambel EVI Distribution method
Table 3.6 Gambel EVI Distribution method
xi Rank P=m/n+1 kt Xi cal Abs(Xi-Xical)
94.5 1 0.142857 1.138387 68.27997 26.22003
82.8 2 0.285714 0.465875 60.22472 22.57528
76 3 0.428571 0.027575 54.97483 21.02517
74.8 4 0.571429 -0.3298 50.69426 24.10574
71.6 5 0.714286 -0.66671 46.65878 24.94122
70.4 6 0.857143 -1.04612 42.1143 28.2857
SUM 147.1531
XT = U + α * YT , U = Xm – 0.5772 α , α = (√6/∏) * Sx ,
KT = (YT – Yn)/Sn
YT = -ln(-ln(1-1/T))
U = 54.9407- 0.5772*10.3195 = 48.9843, α = (√6/∏) *13.23526 = 10.3195
Xi cal = U + α * YT
1
D-index = ∗∑ |( xi− Xical )|
Xm
= 147.1531/54.97407 = 2.6768
Log Pearson Type III Distribution Method
Kt=Z+(Z^2-1)*k+1/3*(z^3-6*Z)*K^2-(Z-1)*k^3+Z*K^4+1/3*K^5
Where K=Cs/6 =0.0529
2
2.51557+ 0.608486∗w+ 0.01033¿ w
Z=w− 2 3
1+ 1.143279∗w+0.1992∗w +0.00131 ¿ w
Where W= (ln(1/pr)^2)^0.5
Yi Rank P=m/n+1 Kt Yi cal Abs(Yi-Yical)
1.975432 1 0.142857 1.044133 2.`83125 0.20769
1.91803 2 0.285714 0.514245 2.125723 0.20769
1.880814 3 0.428571 0.111411 2.088507 0.20769
1.873902 4 0.571429 -0.25881 2.081595 0.20769
1.854913 5 0.714286 -0.64621 2.062606 0.20769
1.847573 6 0.857143 -1.12411 2.055266 0.20769
SUM 1.24617
1
D-index = ∗∑ |( Yi−Yical )|
Ym
= 1.24617/1.725756 = 0.7221
Dindex is minimum in case of log normal distribution hence on the bases of this test log normal
distribution best fits the data, as result X50=93.962mm is taken as design point rainfall .therefore, the
average depth over the catchment area should be adjusted for the whole size of the watershed.
Areal Rainfall
A point rainfall X=93.962mm is taken as a point design rain fall and it has to be distributed over the area
using area reduction factors (ARF) given as;
The area reduction factor (ARF) is introduced to account for the spatial variability of point rainfall over
the catchment. This is not significant for small catchments but becomes so as catchment size increases.
ARF = 1- 0.044(A) 0.275 , where ARF= areal reduction factor, A = catchment area(km2)
= 1- 0.044(73)0.275 = 0.8568
As it is known for catchment greater than 25 square km, the point rain fall has to be multiplied with areal
reduction factor.
Then the areal rainfall =0.8568 * 93.962mm = 80.51mm, therefore areal rainfall value of 80.51mm is
used to determine the peak discharge.
Peak discharge determination
Step Designation/formula Symbol Unit Results
1 Area of the catchment(This can be determined from A Km2 73
1:50,000 scale topographical maps or aerial
photographs)
2 Length of main watercourse from watershed divide L m 17000
to proposed diversion or storage site(topographical
map)
3 Elevation of watershed divide opposite to the head of H1 M 2384.02
the main watercourse (topographical map)
4 Elevation of streambed at proposed or storage site H2 M 1199.26
(topographical map)
5 Slope of the main watercourse, S=(H1-H2)/L S m/m 0.0697
6 Time of concentration ,Tc = 1/3000(L/S)0.77 Tc hr. 1.681
7 Rainfall excess duration D=Tc/6 if Tc<3hrs D hr. 0.28
D= 1 , if Tc ≥3hrs
8 Time to peak ,tp=0.5D+0.6Tc tp hr. 1.1486
9 Time base of hydrograph, tb= 2.67tp tb hr. 3.067
10 Lag time,te=0.6Tc te Hr 1.0086
11 Peak rate of discharge created by 1mm runoff excess qp m3/s/mm 13.347
on whole of the catchment,qp=(0.21A)/tp
12 13 14 15 16 17 18 19
Duration Daily point Rainfall Rainfall Area to Areal Incremental Descending
rainfall for profile Profile point rainfall
return rainfall rainfall Order
period 50 ratio
years
Hr mm % mm % mm mm No
0.0 – 0.28 80.51 19 15.297 32.1 4.91 4.91 1. 8.315
0.28 -0.56 28 22.5 58.52 13.167 8.257 2. 8.257
0.56 -0.84 39 31.4 68.413 21.482 8.315 3. 4.91
0.84 -1.12 46 37.035 69.856 25.87 4.388 4. 4.388
1.12 – 1.4 51 41.06 73.282 30.09 4.22 5. 4.22
1.4 – 1.68 54 43.48 71.569 31.12 1.03 6. 1.03
12 Fill in 0-D hr, D-2D hr,… 5D-6D hr.
13 Determine the magnitude of the daily rainfall with the given recurrent interval by applying
statistical method.
14 Read from (Appendix Fig.A-2) the rainfall profile (%) occurring in D, 2D, 3D…6D hours, and
enter in 14.
15 Multiply 13 and 14 to find the rainfall profile (mm) and enter in 15.
16 From (Appendix Table A-3), read areal to point rainfall ratio for different duration and particular
catchment area. The method is based on research conducted in India and influenced by return
period, magnitude of storm shape and orientation of area etc.
17 Multiply 15 and 16 and file in 17
18 Calculate incremental rainfall by deducting the current areal rainfall from the preceding areal
rainfall as listed in (18)
19 Assign order to the rainfall depths in descending order 1 to 6
20 21 22 23 24 25
Rearranged Rearrange Cumulative Times of increamental hydrogragh
incremental rainfall
Order rainfall Time of Time to peak Time to end
beginning
No. mm mm Hr Hr Hr
6 1.03 1.03 0 1.149 3.067
4 4.388 5.418 0.28 1.429 3.347
3 4.91 10.328 0.56 1.709 3.627
1 8.315 18.643 0.84 1.989 3.907
2 8.257 26.9 1.12 2.269 4.187
5 4.22 31.12 1.4 2.549 4.467
20 From 19, mention the rearranged order as 6,4,3,1,2,5(arbitrarily) but considering ascending
and descending feature of the hydrograph or lines, where peak value is around the middle
of the hydrograph.
21 Fill in the corresponding incremental rainfall value to the rearranged order of 20 from 18
22 Fill in the cumulative rainfall values of 21 by adding with the rainfall values in the
preceding duration.
23 Fill in the time of beginning of hydrograph as 0,D,2D,…5Dhr.
24 Fill in the time to peak as tp,D+tp,2D+tp…5D+tp or add tp in every value of 23 and fill in
24.
25 Add tb in every value of 23 and fill in 25.
26 27 28 29 30 31
Land use cover Area ratio Hydrologic Curve WEIGHED Sum of weighed
condition no”CN” “CN” “CN’
Cultivated land Good ,D AMC CN
Forest land Good ,D I
Pasture or range (grazing Good, D II
land)
Shrub grass land Good, D III 92.115
26 Identify all type of land use cover such as cropped area, follow land, pastures,
meadow forest etc. from the catchment map or aerial photo.
27 Find ratio of each type of land use cover to the total catchment area and enter in
27.
28 Ascertain treatment practice of each type of land use cover, hydrologic
condition corresponding to it from the catchment map and enter in 28.
29 Ascertain hydrologic soil groups for each type of land use cover as below:
Group A:low run off potential
Group B:Moderate runoff potential
Group C:soil having runoff potential
Group D:soil having very high runoff potential
Find the corresponding curve number “CN” from (Appendix table A-4)
30 Multiply 27 and 29 and enter in 30.
31 Add 30. This curve number is corresponding to antecedent moisture condition II
(AMC-II). Find ”CN” for AMC III from tables A-1
The curve number of a given land use/ land cover can be calculated through this steps. But in our case it
is given as CN= 92.115
No Descriptions/Formula Symbol Unit Results
32 Find Maximum potential S Mm 21.74
difference between
rainfall(P), which is given by
the following formula:
25400
s= −254 , where
CN
CN= value corresponding to
AMC III as found in 31
33 Substitute the value of “S” in Q Mm = (P-4.348)2/(P+17.392)
the following formula, giving
the relation between Direct
Runoff(Q) and rainfall(P)
Q=(P-0.2S)2/(P+0.8S)
34 Substitute values of P1 as 22 34
mentioned in 22, in the above
formula and find the X(mm) Q(mm)
corresponding values of
1.03 0.598
Q(34).
5.418 0.05
Enter the value of Q in 36
10.328 1.29
18.643 5.671
26.9 11.483
31.12 14.774
35 36 37 38 39 40
Duration Cumulativ Incrementa Peak runoff 23 24 25
e run off l Runoff for
increment Time of Time to Time to Composite
beginnin peak end hydrograph
g
hr mm mm m 3/s hr hr mr t-Q
0-0.28 0.598 0 0.00 0 1.149 3.067 Triangular
0.28-0.56 0.05 -0.548 0.00 0.28 1.429 3.347 Hydrograph
0.56-0.84 1.29 1.24 16.55 0.56 1.709 3.627 Synthesis
0.84 –1.12 5.671 4.381 58.473 0.84 1.989 3.907
1.12 – 1.4 11.483 5.812 77.573 1.12 2.269 4.187
1.4- 1.68 14.774 3.291 43.925 1.4 2.549 4.467
35 Enter the same time as in 12 i.e.
0-D,D-2D,……5D-6D
36 There are the values of Q as found out in 34 corresponding to the value of P.
37 Find incremental runoff by reducing the values of36from preceding values.
38 Multiply 37with peak rate of runoff corresponding to mm runoff excess as found at 11.
39 Plot triangular hydrograph, (fig.2.2) with time of beginning, peak time and time to end as mentioned
in 23, 24, 25 and peak runoff as mentioned in 38.
40 Plot a composite hydrograph,( Fig.2.3) by adding all the triangular hydrographs. The resultant
hydrograph will be composite hydrograph of desired return period. The coordinate of peak of this
hydrograph will give the peak runoff with desired return period.
Plot of triangular hydrographs
Q Q
16.55
58.473
0.56 1.709 3.627 t 0.84 1.989 3.907
Y3 = 14.44(t3 – 0.56), t3 = 0.56 1.709 Y4 = 50.89(t4 – 0.84), t4 = 0.84 1.989
Y3* = 8.63( 3.627 - t3* ), t3* =1.709 3.627 Y4* = 30.49( 3.907 - t4* ), t4* =1.989 3.907
Q Q
77.573 43.925
1.12 2.269 4.187 1.4 2.549 4.467
Y5 = 67.51(t5 – 1.12), t5 = 1.12 2.269 Y6 = 38.23(t6 – 1.4), t6 = 1.4 2.549
Y5* = 40.44( 4.187- t5* ), t5* =2.269 4.187 Y6* = 22.9( 4.467 - t6* ), t6* =2.549 4.467
Figure 1. triangular hydrograph for peak discharge determination
90 Triangular Hydrograph
80
70
Y1
60 Y2
50 Y3
Y4
40 Y5
Y6
30 base flow
20
10
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Table 2. composite hydrograph
Y1 Y2 Y3 Y4 Y5 Y6 Base total
flow
0 0 0.15 0.15
0.28 0 0 0.15 0.15
0.56 0 0 0 0.15 0.15
0.84 0 0 4.403 0 0.15 4.193
1.12 0 0 8.086 14.25 0 0.15 22.486
1.149 0 0 8.505 15.725 1.958 0.15 26.338
1.4 0 0 12.13 28.498 18.903 0 0.15 59.681
1.429 0 0 12.55 29.974 20.861 1.109 0.15 64.644
1.709 0 0 16.55 44.22 39.763 11.83 0.15 112.496
1.989 0 0 14.136 58.473 58.67 22.52 0.15 153.949
2.269 0 0 11.72 49.94 77.573 33.22 0.15 172.603
2.549 0 0 9.303 41.405 66.241 43.925 0.15 161.024
3.067 0 0 4.83 25.61 45.293 32.06 0.15 107.943
3.347 0 2.42 17.074 33.97 25.648 0.15 79.252
3.627 0 8.537 22.646 19.236 0.15 50.569
3.907 0 11.323 12.824 0.15 24.297
4.187 0 6.412 0.15 6.562
4.467 0 0.15 00.15
composite Hydrograph
200
172.603m 3 /s
180
d 160
i
s 140
c
h 120
a
r 100 composite Hydrograph
g
e 80
m
3
/ 60
s
40
20
0
0 0.5 1 1.5 2 2.5duration,
3 3.5 hr 4 4.5 5
Figure 3.Composite hydrograph
