WATER RESOURCE ENGINEERING

Modified Puls Method :-

A time interval = 6

From the available data the elevation discharge = (S+ 2
) table is prepared

= 60 X 60 X 6 =0.216X106S

=0.0216MS

Elevation discharge(m3/s) Storage

S+ S-
2 2
(m)
100.00 0 3.38 3.38 3.38
100.50 11 3.536 3.654 3.41
101.00 27 3.437 3.728 3.145
101.50 48 4.413 4.931 3.894
102.00 74 4.997 5.796 4.1197
102.50 101 5.436 6.526 4.345
102.85 118 5.68 6.95 4.4
103.25 132 5.934 7.359 4.5

(S+ )2 = (I1+I2) + (S - )2
2 2 2

0.0216
=(10+20)X + 3.38
2

= 3.736
Now from graph (1) the corresponding elevation = 100.6m and discharge 12 m3/sec

Now, (S - )= (S+ )2 - (Q)
2 2

= 3.736 (12X0.0216) = 4.329


(S+ )3 = 0.853 +3.476 = 4.329
2

The process is repeated for the entire duration of the inflow hydrograph in a tabular form

1+2
=
2

Time(h) Inflow Reservoir Q

S+
S-
I(m3/sec) Elevation 3
(m /sec)
(m)
0 11 100.5 11
16.5 0.356 3.38 3.736
6 22 100.6 12
39.5 0.853 3.476 4.329
12 57 101.3 27
70 1.512 3.745 5.257
18 83 101.7 56
78.5 1.695 4.05 5.745
24 74 101.95 70
67.5 1.458 4.233 5.7
30 61 101.9 68
54.5 1.177 4.23 5.41
36 48 101.77 57
42.5 0.918 4.18 5.1
42 37 101.6 52
47 1.015 3.97 4.99
48 57 101.47 48

40 0.864 3.95 4.81

54 23 101.4 42
19.5 0.42 3.9 4.32
60 16 101.05 28
14 0.3 3.71 4
66 12 100.77 18

11 0.237 3.61 3.85

72 10 100.65 14
 103.5
103
102.5
elevation (m)

102
101.5
101
100.5
100
99.5
0 1 2 3 4 5 6 7 8
S+/

103.5
103
102.5
elevation (m)

102
101.5
101
100.5
100
99.5
0 20 40 60 80 100 120 140
outflow (m3/s)

102.2
102
reservoir elevation (m)

101.8
101.6
101.4
101.2
101
100.8
100.6
100.4
0 10 20 30 40 50 60 70 80
time(h)
Good rich Method :-

Atime interval = 6h = 0.021Ms

Using the known storage elevation discharge data,the following table is prepared
2
A graph depiciting Q vs elevation and ( t + ) vs elevation is prepared from this data

Elevation (cm) Storage (10h3) Out flow(m3/S) 2

( + )( m3/S)
t
100 3.380 0 313.0
100.5 3.536 11 338.4
101 3.437 27 345.24
101.5 4.413 48 456.6
102 4.997 74 536.7
102.5 5.436 101 604.3
102.85 5.680 118 644.0
103.25 5.934 132 681.4

At t = 0 ,elevation = 100.60
2
From graph corresponding ,Q = 15m3/s and ( t + ) = 352.5 m3/s

2 2
( )1 = ( + )1 (2 X Q)
t t

= 352.5 (2X15) = 322.5

Now,
2 2
( )2 = ( 1 + 2) + ( + )1
t t

(11+22)+322.5 = 355.5

Find corresponding elevation and Q

2 2
( )2 = ( + )2 (2 X Q) = 355.5 (2X16) = 323.5
t t
And process repeated

Time (h) (m3/S) 1+2 Elevation(m) Discharge

+

0 11 352.5 100.6 15
33 322.5 355.5
6 22 100.625 16
79 323.5 402.5
12 57 101.1 35
140 332.5 472.5
18 83 101.625 16
157 366.5 523.5
24 74 101.225 69
135 385.5 520.5
30 61 101.9 68
109 384.5 493.5
36 48 101.75 59
85 375.5 460.5
42 37 101.52 49

94 362.5 456.5

48 57 101.47 47

80 362.5 442.5
54 23 101.4 45
39 352.5 391.5

60 16 101 32
28 327.5 355.5
66 12 100.56 16
22 323.5 345.5

72 10 100.5 11
 103.5
103
102.5
elevation (m)

102
101.5
101
100.5
100
99.5
0 100 200 300 400 500 600 700 800
(2/t+)( m3/S)

103.5
103
102.5
elevation (m)

102
101.5
101
100.5
100
99.5
0 20 40 60 80 100 120 140
outflow (m3/s)

102
101.8
reservoir elevation (m)

101.6
101.4
101.2
101
100.8
100.6
100.4
0 10 20 30 40 50 60 70 80
time (h)
 80

70

60
discharge (m3/s)

50

40

30

20

10

0
0 10 20 30 40 50 60 70 80
time (h)

OUTFLOW DISCHARGE VS TIME

90

80

70

60
discharge (m3/s)

50

40

30

20

10

0
0 10 20 30 40 50 60 70 80
time (h)

INFLOW DISCHARGE VS TIME