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Problem No .1: Chart Title

i) The ordinates of a 6-hour unit hydrograph for a catchment are provided. This is used to calculate the ordinates of the direct runoff hydrograph (DRH) due to 3.5 cm of rainfall excess occurring in 6 hours. ii) The ordinates of the combined DRH from two sequential storms of 3.0 cm and 2.0 cm rainfall excess each lasting 6 hours are calculated. iii) The flood hydrograph is derived for a storm with time-varying rainfall and a base flow component is included.

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136 views11 pages

Problem No .1: Chart Title

i) The ordinates of a 6-hour unit hydrograph for a catchment are provided. This is used to calculate the ordinates of the direct runoff hydrograph (DRH) due to 3.5 cm of rainfall excess occurring in 6 hours. ii) The ordinates of the combined DRH from two sequential storms of 3.0 cm and 2.0 cm rainfall excess each lasting 6 hours are calculated. iii) The flood hydrograph is derived for a storm with time-varying rainfall and a base flow component is included.

Uploaded by

sirkali1973
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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PROBLEM No .

1
The ordinates of a 6 hr unit hydrograph for a catchment are given below
Time(hr) 0 3 6 9 12 15 18
UHO (Cumecs) 0 25 50 85 125 160 185

i) Compute the ordinates of the DRH due to rainfall excess of 3.5 cm occurring in 6 hr
ii)Two storms each of 6 hr duration and having rainfall excess values of 3.0cm and 2.0 cm respe
The 2 cm rainfall excess follows the 3 cm rainfall. Calculate the resulting DRH

iii) Derive the flood hydrograph due to the storm given below
Time from start of storm(h) 0 6 12 18
Accumulated rainfall (cm) 0 3.5 11 16.5
The storm loss rate for the catchment is estimated as 2.5mm/hr The base flow can be assumed a
and increasing by 2.0Cumecs for every 12 hrs till the end of the DRH.

i) Ordinate
s of 3.5
Cm
6 hr DRH 700
UHO (Cumecs 600
Time hr (Cumecs) )
500
0 0 0
400
3 25 87.5
300
6 50 175
9 85 297.5 200

12 125 437.5 100

15 160 560 0
0 10 20
18 185 647.5
Column B
24 160 560
30 110 385
36 60 210
42 36 126
48 25 87.5
54 16 56
60 8 28
69 0 0

ii)
Ord. of
Ordinate 2.0 Cm Ord. of
s of 3. DRH 5.0 Cm
Cm (Cumecs) DRH
6 hr DRH Col 2 (Cumecs
UHO (Cumecs lagged by) Col
Time hr (Cumecs) ) 6hr*2 3+col 4
0 0 0 0
3 25 75 75
6 50 150 0 150
9 85 255 50 305
12 125 375 100 475
15 160 480 170 650
18 185 555 250 805
21 172.5 517.5 320 837.5
24 160 480 370 850
30 110 330 320 650
36 60 180 220 400
42 36 108 120 228
48 25 75 72 147
54 16 48 50 98
60 8 24 32 56
66 2.7 8.1 16 24.1
72 0 0 5.4 5.4
78 0 0 0 0

iii)
Duration First 6 hSecond 6hr Third 6 hr
Total rainfall depth cm 3.5 7.5 5.5
Loss 1.5 1.5 1.5
Effective rainfall cm 2 6 4

Ord. of
Ord. of 4.0 Cm
Ordinate 6.0 Cm DRH
s of 2 DRH (Cumecs
cmCm (Cumecs) ) Col 2
6 hr DRH Col 2 lagged
UHO (Cumecs lagged by by Base
Time hr (Cumecs) ) 6hr*6 12hr*4 flow
0 0 0 15
3 25 50 15
6 50 100 0 15
9 85 170 150 15
12 125 250 300 0 17
15 160 320 510 100 17
18 185 370 750 200 17
21 172.5 345 960 340 17
24 160 320 1110 500 19
27 135 270 1035 640 19
30 110 220 960 740 19
36 60 120 660 640 21
42 36 72 360 440 21
48 25 50 216 240 23
54 16 32 150 144 23
60 8 16 96 100 25
66 2.7 5.4 48 64 25
69 0 0 32.1 48 25
72 0 0 16.2 32 27
75 0 0 0 21.4 27
78 0 10.8 27
81 0 29
84

PROBLEM No. 2
The runoff data at a stream gauging station for a flood are given below. The drainage area is 40 k
for the basin and plot the same.
Time(hr) 2 5 8 11 14 17 20
Discharge (Cume 50 47 75 120 225 290 270
State the peak of the unit hydrograph.

Assuming a base flow of 47.0 Cumecs from the start of runoff and 60 cumecs at 38 th hr
UHO
Base (Cumec Time
Discharg flow DRH s) from
e (Cum (Cumecs Col4/26 beginning
Time(hr) (Cumecs) ecs) ) .62 hr
2 50 0 0 0

10
9
8
5 47 47 0 0 0 0
8 75 46 29 1.09 3 1.09 10

11 120 45 75 2.82 6 2.82 9


8
14 225 45 180 6.76 9 6.76
7
17 290 45 245 9.2 12 9.2 6
20 270 46 224 8.41 15 8.41 5
23 145 48 97 3.64 18 3.64 4

26 110 50 60 2.25 21 2.25 3


2
29 90 53 37 1.39 24 1.39
1
32 80 54 26 0.98 27 0.98
0
35 70 57 13 0.49 30 0.49 0 5 10

38 60 60 0 0 33 0
41 55
44 51
47 50
986
Volume of runoff = 986 *60*60*3
1E+07
Effective rainfall depth 986*60*60*3/(40*10^6)
Effective rainfall depth = Volume / Area of catchment
0.2662 m
26.622 cm
24 30 36 42 48 54 60 69
160 110 60 36 25 16 8 0

occurring in 6 hr
of 3.0cm and 2.0 cm respectively occur successively.
ing DRH

ase flow can be assumed as 15 Cumecs at the beginning

Chart Title
700

600

500

400

300

200

100

0
0 10 20 30 40 50 60 70 80

Column B Column C Column D Column E


0 0
3 75
6 150
9 305 Chart Title
12 475 900
800
15 650
700
18 805
600
21 837.5
500
24 850
400
30 650
300
36 400
200
42 228 100
48 147 0
0 10 20 30 40 50 60 70
54 98
60 56
66 24.1
75 0

Ord. of
total DRH Ord. of Chart Title
(Cumecs) total DRH
Col (Cumecs) 2500
Col
3+col4+co 3+col4+co
l 5 + col6 Time hr l 5 + col6 2000

1500

1000
2500

2000

15 0 15
1500
65 3 65
115 6 115 1000
335 9 335
567 12 567 500

947 15 947
0
1337 18 1337 0 10 20 30 40 50

1662 21 1662
1949 24 1949
1964 27 1964
1939 30 1939
1441 36 1441
893 42 893
529 48 529
349 54 349
237 60 237
142.4 66 142.4
105.1 69 105.1
75.2 72 75.2
48.4 75 48.4
37.8 78 37.8
29 81 29

. The drainage area is 40 km2. The duration of rainfall is 3 hours. Derive the 3-hour unit hydrograph

23 26 29 32 35 38 41 44 47
145 110 90 80 70 60 55 51 50

cumecs at 38 th hr

Chart T
Chart Title 350
10
300
9
250
8
Chart T
Chart Title 350
10
300
9
250
8
7 200
6 150
5
100
4
50
3
2 0
0 5 10 15 20 2
1
0 Column B
0 5 10 15 20 25 30 35

UNIT HYDROGRAPH

2.091279
1.735762

41.65828
tle

50 60 70 80

Chart Title
20 30 40 50 60 70 80 90

our unit hydrograph

Chart Title
Chart Title

5 10 15 20 25 30 35 40 45 50

Column B Column C

DIRECT RUNOFF HYDROGRAPH

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