Flood Routing
Introduction
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Flood routing is an analytical technique of
determining the flood hydrograph at a particular
location in a channel or a reservoir resulting from a
known flood at some other location upstream.
Routing techniques may be classified in two major
categories
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Hydrologic routing (based on the equation of continuity and
empirical equation. This involves the balancing of inflow, outflow
and volume of storage through use of continuity)
Hydraulic routing (based on equations of continuity and
momentum)
Continued…..
Reservoir Routing
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The main components of Dam are inflow channel,
storage reservoir and outflow structures like
spillways, tunnels etc.
Once a flood enters a reservoir, part of it may be
stored in the reservoir and balance safely passes
through or over outflow structures.
The main function of a reservoir is to store water
from which releases are made according to water
demands on downstream of reservoir.
A multipurpose hydroelectric project has storage of
water as well as generation of electricity.
Continued…..
Reservoir Routing
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The reservoirs may be small or large. An example of
small reservoir is pond of a barrage. A small reservoir
has small capacity and hence water levels in barrage
pond are sensitive.
The outflow from a small reservoir is solely a function
of pond elevation if the outflow is not controlled. In
case of large reservoirs the moderate inflow may not
have large impact on reservoir elevation however
large floods need to be negotiated keeping in view the
operational rules.
Continued…..
Reservoir Routing
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The reservoir routing may be classified according to
outflow control at a particular reservoir e.g.
Flood routing in reservoirs with uncontrolled outflow
Flood routing in reservoirs with controlled outflow
The basic equation applied is storage equation given as
Inflow-Outflow = Rate of Change of Storage
I - O = ds / dt - - - - - - - - - - - - (1)
Equation (1) shows that if inflow is assumed constant the
reservoir storage is a simple function of outflow.
Continued…..
Reservoir Routing
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If average values of inflow and outflow are considered for
opted time interval ∆t then equation (1) can be written as
[(I1+I2)/2]-[(O1+O2)/2]=[(S2-S1)/∆t] - - - - - - - - (2)
Where 1 to 2 are the time step values of I, O and S
Equation (2) can be rearranged as
(I1+I2)+[(2S1/∆t)-O1]=[(2S2/∆t)+O2] - - - - - -(3)
As mentioned above the subscripts 1 and 2 denote values
of Inflow, Outflow and Storage at beginning and end of ∆t
say from 1 to 2. The time ∆t is known as routing period.
This period should not be so large that peak of inflow
hydrograph is not intercepted.
Flood routing in reservoirs with
uncontrolled outflow
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The following steps explain procedure of
reservoir routing
1. The Elevation vs Storage of reservoir
information should be known. Here storage
means volume of water that a reservoir can
accommodate at certain elevation.
This elevation vs storage information may
either be in the form of table or graph. A
typical elevation vs storage graph is shown in
Figure 1.
Continued…..
Flood routing in reservoirs with
uncontrolled outflow
Elevation vs Surface Area Realationship
124
122
120
Elevation (m)
118
116
114
112
110
108
106
104
102
40000
42000
44000
46000
48000
50000
52000
54000
Surface Area (m²)
Fig. 1. Variation of Surface Area of a Reservoir with Elevation
Flood routing in reservoirs with uncontrolled
outflow
Flood routing in reservoirs with uncontrolled
outflow
Flood routing in reservoirs with
uncontrolled outflow
2. The discharge capacity of overflow structure
with change in water level should be calculated.
For this purpose the applicable discharge formula
need to be applied. The well known weir equation
is:
Q = Cd BH3/2 - - - - - - - - - -(3A)
Where,
Q
= Total Discharge
Cd
= Coefficient of Discharge
B
= Width of Weir
H
= Differential head over the crest of the
weir neglecting velocity of approach
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Continued…..
Flood routing in reservoirs with
uncontrolled outflow
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The coefficient of discharge depends on
degree of submergence of the weir. Its value
is determined experimentally e.g. by model
tests.
The value can also be determined from
Gibson’s curve. Its value generally ranges
from 1.6 to 2.2. A mean value of 1.70 is often
used in SI units.
Continued…..
Flood routing in reservoirs with
uncontrolled outflow
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For other types of outflow structures like Sluice
Gates, Pipes etc. different equations are used for
calculations of discharge and can be found in books
of hydraulics.
Once the outflow is determined for different reservoir
elevations, a graph is plotted between storage and
outflow. A typical such graph is shown in figure (2).
Please note that outflow is taken along y-axis and
[(2S/∆t)+O] is taken along x-axis. The quantity
[(2S/∆t)+O] is called ‘Storage Indication’.
Continued…..
Flood routing in reservoirs with
uncontrolled outflow
300
Outflow (m³/s)
250
200
150
100
50
-
500
1,000
1,500
Storage Indication
2,000
2,500
[(2S/∆t)+O]
3,000
3,500
4,000
(m³/s)
Figure 2 Outflow and Storage Indication Relationship for certain
reservoir
Continued…..
Flood routing in reservoirs with
uncontrolled outflow
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3. The inflow hydrograph should be known. It
may be actual or forecasted flood. The inflow is
added for successive values to get I1+I2.
Corresponding to the initial outflow value storage
indication [(2S/∆t)+O] is found from storage
indication curve.
To this value double of outflow is subtracted to
get [(2S/∆t)-O]. To this value of [(2S/∆t)-O], I1+I2
is added to get next value of [(2S/∆t)+O]. Read
out next outflow from storage indication curve
and repeat the procedure till whole of inflow
hydrograph is used to get outflow values.
Continued…..
Flood routing in reservoirs with
uncontrolled outflow
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Now the inflow and outflow hydrographs are plotted.
The difference in peak of inflow and outflow
hydrograph is known as attenuation and time
between two peaks is known as reservoir lag.
Stream Channel Routing or River
Routing
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The routing in channels involves solution of storage equation as
was done in case of reservoir routing. The storage is function of
both inflow and outflow.
The method of channel routing is known as Muskingum Method.
Consider a channel reach having prismatic cross section as
shown in Figure 5.
Let,
S = Storage
I = Inflow
O = Outflow
The storage in the channel reach consists of two parts:
❑ Prism storage equal to KO.
❑ Wedge storage equal to K (I-O).
Stream Channel Routing or River
Routing
Wedge Storage
=K(I-O)
I
O
Prism Storage
=KO
Figure 5. Prism and Wedge Storage in Channel
Stream Channel Routing or River
Routing
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Then total storage ‘S’ is therefore sum of
prism and wedge storage. That is:
S = K [XI + (1-X)O] - - - - - - (6)
Where ‘X’ is a dimensionless constant for
certain reach or segment of channel. ‘K’ is
storage constant having dimensions of time.
Both X and K are determined from inflow and
outflow hydrographs for reach under
consideration.
Stream Channel Routing or River
Routing
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These constants vary from reach to reach and are
determined as follows.
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The inflow and outflow hydrographs are known for the reach.
Find values of (I-O) for each time interval.
Find the mean and cumulative mean values of (I-O) which is
storage.
Assume value of ‘X’ and find the term [XI + (1-X) O] for each
time interval. The storage value is already calculated against
time as explained.
Plot [xI + (1-x)O] values against storage. Inspect if data
plotted nearly fits a straight line. If not assume new value of x
and repeat steps 1-4.
The best-fit straight line corresponds to required value of ‘x’.
The slope of this straight line is our required value of ‘K’.
Stream Channel Routing or River
Routing
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Now we proceed for channel flow routing once
values of ‘x’ and ‘K’ are known. Routing means
finding outflow hydrograph for given inflow
hydrograph.
The linear relationship is expressed as for
Muskingum Method
◼ S = K [(xI) + (1-x) O]
Since rate of change of storage in a particular
channel reach is given as
[I2+I1/2]Δt – [O1+O2/2] Δt = S2–S1 -----------(1)
Stream Channel Routing or River
Routing
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For if Muskingum Theorem is applied to that
change; the equation can be written as in terms of
storage
◼ S2–S1 = K [ x (I2+I1) + (1-x)(O1+O2)]---------(2)
Now equating equations 1 and 2 we get
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(I1/2)Δt+(I2/2)Δt–(O1/2)Δt-(O2/2)Δt = K{(xI2-xI1)+(1-x)(O2–O1)}
rearranging the terms we get
(I1/2)Δt+KI1x+(I2/2)Δt-KI2x-(O1/2)Δt+O1K-xKO1 = (O2/2)Δt+O2K-xKO2
(0.5Δt+Kx)I1=(0.5Δt-Kx)I2+(K-Kx-0.5Δt)O1 = (0.5Δt+K-Kx)O2 ---------(3)
Stream Channel Routing or River
Routing
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O2= [(0.5Δt+Kx)/(0.5Δt+K-xK)]I1+
[(0.5Δt-Kx)/(0.5Δt+K-xK)]I2+
[(K-Kx-0.5Δt/0.5Δt+K-Kx)]O1
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OR O2 = C0I2+C1I1+C2O1
Where
Co = [(0.5Δt-Kx)/(0.5Δt+K-xK)]
C1 = [(0.5Δt+Kx)/(0.5Δt+K-xK)]
C2 = [(K-Kx-0.5Δt/0.5Δt+K-Kx)]
It may be noted that sum of weighing coefficients
Co + C 1 + C2 = 1
Knowing values of ‘x’ and ‘K’ these coefficients are
determined simply by substitution of values in
equations
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