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- Find the storm hydrograph for the following data using time area method. Given rainfall excess ordinate at time is 0.5 in./hr
A B C D Area (ac) 100 200 300 100 Time to gage G (hr) 1 2 3 4 A
B
C
D
G
Time area histogram method uses Qn = Ri A 1 + Ri-2 A 2 +…….+ Ri Aj For n = 5, i = 5, and j = 5 Q 5 = R 5 A 1 + R 4 A 2 + R 3 A 3 + R 2 A 4 (0.5 in./ hr) (100 ac) + (0.5 in./hr) (200 ac) + (0.5 in./hr) (300ac) + (0.5 in./hr) (100) Q5 = 350 ac-in./hr Note that 1 ac-in./hr ≈ 1 cfs, hence Q5 = 350 cfs
Example Problem
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Example Problem Contd…
Tim e (hr)
Hydrograp h Ordinate (R1:Rn)
Basi n No.
Time to gage
Basin area A1:An (ac)
R1:An R2:An R2:An R2:An R2:An Storm hydrograph
0 0 1 0.5 A 1 100 *^50 2 0.5 B 2 200 100 50 + 150 3 0.5 C 3 300 150 100 50 300 4 0.5 D 4 400 50 150 100 50 350 5 50 150 100 50 350 6 50 150 100 300 7 50 150 200 8 50 50 9 0
Excel spreadsheet calculation
- (^) =(R1A1) = (0.5100) and + (^) = (adding the columns from 6 to 10) (^) Module 3 Docsity.com
- Rainfall of magnitude 3.8 cm and 2.8 cm occurring on two consecutive 4-h durations on a catchment area 27km^2 produced the following hydrograph of flow at the outlet of the catchment. Estimate the rainfall excess and φ-index Time from start of rainfall (h) -6^0 6 12 18 24 30 36 42 48 54 60 Observed flow (m 3 /s) 6 5 13 26 21 16 12 9 7 5 5 4.5 4.
Example Problem-
Baseflow separation:
Using Simple straight line method,
N = 0.83 A0.2^ = 0.83 (27) 0.
= 1.6 days = 38.5 h
So the baseflow starts at 0 th^ h and ends at the point (12+38.5)h Docsity.com
Hydrograph
(^6 )
13
26
21
16 12 9 7 (^5 5) 4.5 4.
0
5
10
15
20
25
30
-10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
Discharge(m
3 /s)
Time (hr)
50.5 h ( say 48 h approx.)
Constant baseflow of 5m 3 /s
Example Problem-1 Contd…
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Hydrograph
(^6 )
13
26
21
16
12 9 7 (^5 5) 4.5 4.
0
5
10
15
20
25
30
-10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
Discharge(m
3 /s)
Time (hr)
Area of Direct runoff hydrograph
Example Problem-1 Contd…
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Area of DRH = (66060)[1/2 (8)+1/2 (8+21)+
1/2 (21+16)+ 1/2 (16+11)+
1/2 (11+7)+ 1/2 (7+4)+ 1/2 (4+2)+ 1/2 (2)]
= 1.4904 * 10 6 m 3 (total direct runoff due to storm)
Run-off depth = Runoff volume/catchment area
= 1.4904 * 10 6 /27* 10 6
= 0.0552m = 5.52 cm = rainfall excess
Total rainfall = 3.8 +2.8 = 6.6cm
Duration = 8h
φ-index = (P-R)/t = (6.6-5.52)/8 = 0.135cm/h
Example Problem-1 Contd…
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Time from start of storm(h)
Time interval ∆t
Accumulated rainfall in ∆t (cm)
Depth of rainfall in ∆t (cm)
φ ∆t (cm) ER (cm)
Intensity of ER (cm/h) 0 _ 0 _ _ _ _ 2 2 0.6 0.6 0.8 0 0 4 2 2.8 2.2 0.8 1.4 0. 6 2 5.2 2.4 0.8 1.6 0. 8 2 6.7 1.5 0.8 0.7 0. 10 2 7.5 0.8 0.8 0 0 12 2 9.2 1.7 0.8 0.9 0. 14 2 9.6 0.4 0.8 0 0
Example Problem-2 Contd…
- Total effective rainfall = Direct runoff due to storm = area of ER hyetograph = (0.7+0.8+0.35+0.45)*2 = 4.6 cm
- Volume of direct runoff = (4.6/100) * 5.0*(1000)^2 = 230000m 3 Docsity.com
