3. Routing- is the process of
prediction temporal and spatial
variation of a flood wave as it
travels through a river or
channel reach or reservoir.
4. Hydrologic routing
Lumped/hydrologic
Flow is calculated as a function of time alone at
a particular location
Governed by continuity equation and
flow/storage relationship
Hydraulic routing
Distributed/hydraulic
Flow is calculated as a function of space and
time throughout the system
Governed by continuity and momentum
equations
5. Flood routing- procedure to
compute output hydrograph
when input hydrograph and
physical dimensions of the
storage are known.
6. Reservoir -usually means an
enlarged natural or artificial
lake, storage
pond or impoundment created
using a dam or lock to store
water.
8. Reservoir routing- used to
determine the peak-flow
attenuation that a hydrograph
undergoes as it enters a
reservoir or other type of
storage pool.
9.
10. The modified puls routing method is
probably most often applied to reservoir
routing
The method may also be applied to river
routing for certain channel situations.
The modified puls method is also referred
to as the storage-indication method.
The heart of the modified puls equation is
found by considering the finite difference
form of the continuity equation.
11. Continuity
Equation
Rewritten
The solution to the modified puls method is
accomplished by developing a graph (or
table) of O -vs- [2S/Δt + O]. In order to do
this, a stage-discharge-storage relationship
must be known, assumed, or derived.
t
S
-
S
=
2
O
+
O
(
-
2
I
+
I 1
2
2
1
2
1
O
+
t
S
2
=
O
-
t
S
2
+
I
+
I 2
2
1
1
2
1
12. Given the following hydrograph and the 2S/t + O
curve, find the outflow hydrograph for the
reservoir assuming it to be completely full at the
beginning of the storm.
The following hydrograph is given:
0
30
60
90
120
150
180
0 2 4 6 8 10
Discharge
(cfs)
Time (hr)
Hydrograph For Modified Puls Example
14. A table may be created as follows:
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0
1
2
3
4
5
6
7
8
9
10
11
12
15. • Next, using the hydrograph and interpolation, insert the
Inflow (discharge) values.
• For example at 1 hour, the inflow is 30 cfs.
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0
1 30
2 60
3 90
4 120
5 150
6 180
7 135
8 90
9 45
10 0
11 0
12 0
Hydrograph For Modified Puls Example
0
30
60
90
120
150
180
0 2 4 6 8 10
Time (hr)
Discharge
(cfs)
16. • The next step is to add the inflow to the inflow in the next
time step.
• For the first blank the inflow at 0 is added to the inflow at
1 hour to obtain a value of 30.
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30
1 30
2 60
3 90
4 120
5 150
6 180
7 135
8 90
9 45
10 0
11 0
12 0
17. This is then repeated for the rest of the values in the
column.
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30
1 30 90
2 60 150
3 90 210
4 120 270
5 150 330
6 180 315
7 135 225
8 90 135
9 45 45
10 0 0
11 0 0
12 0 0
18. The 2Sn/t + On+1 column can then be calculated using the
following equation:
Note that 2Sn/t - On and On+1 are set to zero.
30 + 0 = 2Sn/t + On+1
O
+
t
S
2
=
O
-
t
S
2
+
I
+
I 2
2
1
1
2
1
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30 0 0
1 30 90 30
2 60 150
3 90 210
4 120 270
5 150 330
6 180 315
7 135 225
8 90 135
9 45 45
10 0 0
11 0 0
12 0 0
19. • Then using the curve provided outflow can be determined.
• In this case, since 2Sn/t + On+1 = 30, outflow = 5 based on
the graph provided.
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30 0 0
1 30 90 30 5
2 60 150
3 90 210
4 120 270
5 150 330
6 180 315
7 135 225
8 90 135
9 45 45
10 0 0
11 0 0
12 0 0
20. • To obtain the final column, 2Sn/t - On, two times the
outflow is subtracted from 2Sn/t + On+1.
• In this example 30 - 2*5 = 20
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30 0 0
1 30 90 20 30 5
2 60 150
3 90 210
4 120 270
5 150 330
6 180 315
7 135 225
8 90 135
9 45 45
10 0 0
11 0 0
12 0 0
21. • The same steps are repeated for the next line.
• First 90 + 20 = 110.
• From the graph, 110 equals an outflow value of 18.
• Finally 110 - 2*18 = 74
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30 0 0
1 30 90 20 30 5
2 60 150 74 110 18
3 90 210
4 120 270
5 150 330
6 180 315
7 135 225
8 90 135
9 45 45
10 0 0
11 0 0
12 0 0
22. • This process can then be repeated for the rest of the
columns.
• Now a list of the outflow values have been calculated and
the problem is complete.
Time In In+In+1 2Sn/t - On 2Sn/t + On+1 On+1
(hr) (cfs) (cfs) (cfs) (cfs) (cfs)
0 0 30 0 0
1 30 90 20 30 5
2 60 150 74 110 18
3 90 210 160 224 32
4 120 270 284 370 43
5 150 330 450 554 52
6 180 315 664 780 58
7 135 225 853 979 63
8 90 135 948 1078 65
9 45 45 953 1085 65
10 0 0 870 998 64
11 0 0 746 870 62
12 0 0 630 746 58
23. In the above equation the starting inflow and
end inflow at time period t is known (read it
from the inflow hydrograph), and the initial
storage and discharge is also known
Then estimate the value remember
both are unknown quantities
2
2
1
1
2
1
2
2
Q
t
S
Q
t
S
I
I
2
2
2
Q
t
S
24. To know the discharge, we need a graph
between elevation Vs
Thus called as semi graphical method
This quantity is called storage-elevation-
discharge data
The graph gives the relationship between
discharge and elevation
From graph estimate the elevation
From elevation estimate the discharge
Q
t
S
2
25. The storage-elevation-discharge data is as follows:
The initial conditions are when t = 0, the reservoir
elevation is 100.60 m.
Time (h) 0 6 12 18 24 30 36 42 48 54 60 66
Inflow (m3/s) 10 30 85 140 125 96 75 60 46 35 25 20
Elevation Storage (106 m3) Outflow discharge (m3/s)
100.00 3.350 0
100.50 3.472 10
101.00 3.880 26
101.50 4.383 46
102.00 4.882 72
102.50 5.370 100
102.75 5.527 116
103.00 5.856 130
26. Assume a time period of
6 hr (t )
Equal to time of
discharge measurement
in the inflow hydrograph
Estimate the values of
Plot a graph
elevation-Vs-discharge
Elevation-Vs-
For initial time period
t=0 find the Q2 and
From the graph
Elevation Storage
(10
6
m
3
)
Outflow
dischar
ge
(m
3
/s)
(m3/s)
100 3.35 0
310.19
100.5 3.472 10
331.48
101 3.88 26
385.26
101.5 4.383 46
451.83
102 4.882 72
524.04
102.5 5.37 100
597.22
102.75 5.527 116
627.76
103 5.856 130
672.22
34. 1. The peak discharge magnitude is
reduced, this is called attenuation.
2. The peak of outflow gets shifted
and is called as lag
3. The difference in rising limb
shows the reservoir is storing the
water
4. The difference in receding limb
shows the reservoir is depleted.
5. When the outflow is through
uncontrolled spillway, the peak of
outflow always occurs at point of
inflection of inflow hydrograph and
also is the point at which the inflow
and outflow hydrograph intersect.
0 10 20 30 40 50 60 70 80
Time in hrs
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
160.00
Inflow/outflow
in
cu.m/s
Inflow hydrograph
Outflow hydrograph
