3. What is routing?
• 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. Two types of routing
• 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. What is flood routing?
• Flood routing- procedure to
compute output hydrograph when
input hydrograph and physical
dimensions of the storage are
known.
6. What is reservoir?
• Reservoir -usually means an
enlarged natural or artificial
lake, storage
pond or impoundment created
using a dam or lock to store water.
7. Types of reservoir
•Storage/ conservation reservoir
•Flood control reservoir
•Multi purpose reservoir
•Distribution reservoir
8. What is reservoir routing?
• Reservoir routing- used to
determine the peak-flow
attenuation that a hydrograph
undergoes as it enters a reservoir or
other type of storage pool.
10. Pul’s Method or Inflow-storage-
discharge Method
• 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. Pul’s Method or Inflow-storage-
discharge Method
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. Modified Puls Example
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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Modified Puls Example
• 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. Goodrich Method
• 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. Route the following flood hydrograph through the reservoir by
Goodrich method:
Inflow hydrograph
• The storage-elevation-discharge data is as follows:
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. Step 1: Construct the storage-elevation-
discharge curve
• 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
discharg
e (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. What we achieved
through this flood
routing
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
