FLOOD ROUTING
Flow Routing
Q
t


Procedure to
determine the flow
hydrograph at a
point on a
watershed from a
known hydrograph
upstream
As the hydrograph
travels, it


attenuates
gets delayed
Q
t
Q
t
Q
t
Why route flows?
Q
t


Account for changes in flow hydrograph as a flood wave
passes downstream
This helps in


Calculate for storages
Studying the attenuation of flood peaks
Types of flow routing

Lumped/hydrologic



Flow is calculated as a function of time alone
at a particular location
Governed by continuity equation and
flow/storage relationship
Distributed/hydraulic


Flow is calculated as a function of space and
time throughout the system
Governed by continuity and momentum
equations
Lumped flow routing

Three types
1.
Level pool method (Modified Puls)

2.
Muskingum method

3.
Storage is nonlinear function of Q
Storage is linear function of I and Q
Series of reservoir models

Storage is linear function of Q and its time
derivatives
S and Q relationships
Level pool routing

Procedure for calculating outflow
hydrograph Q(t) from a reservoir with
horizontal water surface, given its
inflow hydrograph I(t) and storageoutflow relationship
Wedge and
Prism Storage
• Positive wedge
I>Q
• Maximum S when I = Q
• Negative wedge
I<Q
Hydrologic River Flood
Routing
Basic Equation
dS
t I O
t
t
dt
Hydrologic river routing
(Muskingum Method)
Wedge storage in
reach
S Prism  KQ
S Wedge  KX ( I  Q)
Advancing
Flood
Wave
I>Q
K = travel time of peak through the reach
X = weight on inflow versus outflow (0 ≤ X ≤ 0.5)
X = 0  Reservoir, storage depends on outflow,
no wedge
X = 0.0 - 0.3  Natural stream
S  KQ  KX ( I  Q)
Receding
Flood
Wave
Q>I
I
Q
I Q
Q
Q
I
Q
QI
S  K [ XI  (1  X )Q]
I
I
Continuity Equation in
Difference Form

Referring to figure, the continuity
equation in difference form can be
expressed as DS  S2  S1  _I  O_  (I1  I2 )  (O1  O2 )
Dt
t t
2 1
2
2
Derivation of Muskingum Routing Equation
• By Muskingum Model,
at t = t2, S2 = K [X I2 + (1 - X)O2]
at t = t1, S1 = K [X I1 + (1 - X)O1]
• Substituting S1, S2 into the continuity equation and after some
algebraic manipulations, one has
O2 = Co I2 + C1 I1 + C2 O1
• Replacing subscript 2 by t +1 and 1 by t, the Muskingum routing
equation is
Ot+1 = Co It+1 + C1 It + C2 Ot, for t = 1, 2, …
KX  0.5Dt

KX

0.5
D
t
C

where C 
;
; C2 = 1 – Co – C1
o K  KX  0.5Dt 1 K  KX  0.5Dt
Note: K and Dt must have the same unit.
Routing
Muskingum Routing
Equation
Q2  C0 I 2  C1 I1  C2Q1
Qt 1  C0 I t 1  C1 I t  C2Qt
where C’s are functions of x, K, Dt and sum to 1.0
Muskingum Equations
where
C0 = (– Kx + 0.5Dt) / D
C1 = (Kx + 0.5Dt) / D
C2 = (K – Kx – 0.5Dt) / D
D = (K – Kx + 0.5Dt)
Repeat for Q3, Q4, Q5 and so on.
Estimating Muskingum
Parameters, K and x


Graphical Method:
Referring to the Muskingum Model, find X
such that the plot of XIt+ (1-X)Ot (m3/s) vs
St (m3/s.h) behaves almost nearly as a single
value curve. The assume value of x lies
between 0 and 0.3.
The corresponding slope is K.
Example 8.4: Estimating the
value of x and K.

Try and error to get the nearly straight
line graph.
Muskingum Routing
Procedure
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Given (knowns): O1; I1, I2, …; Dt; K; X
Find (unknowns): O2, O3, O4, …
Procedure:
(a) Calculate Co, C1, and C2
(b) Apply Ot+1 = Co It+1 + C1 It + C2 Ot
starting from t=1, 2, … recursively.

Example 8.5
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
Given K and x.
Initial outflow, Q also given.
Solution:
Calculate Co, C1, and C2
C0 = (– Kx + 0.5Dt)/ D
C1 = (Kx + 0.5Dt)/ D
C2 = (K – Kx – 0.5Dt)/ D
D = (K – Kx + 0.5Dt)
Solution:

Time
(hr)
0
Inflow 10
(m3/s)
Route the following flood hydrograph through
a river reach for which K=12.0hr and X=0.20.
At the start of the inflow flood, the outflow
flood, the outflow discharge is 10 m3/s.
6
12
18
24
30
36
42
48
54
20
50
60
55
45
35
27
20
15
Reservoir Routing
• Reservoir acts to store water
and release through control
structure later.
•
Max Storage
Inflow hydrograph
• Outflow hydrograph
• S - Q Relationship
• Outflow peaks are reduced
• Outflow timing is delayed
Inflow and Outflow
dS
I Q
dt
Inflow and Outflow
I1 + I2 – Q1 + Q2
2
2
=
S2 – S1
Dt
Inflow & Outflow Day 3
= change in storage / time
S3  S2
I 2  I 3 / 2  Q2  Q3 / 2  dt
Repeat for each day in progression
Determining Storage
• Evaluate surface area at several different depths
• Use available topographic maps or GIS based DEM
sources (digital elevation map)
• Outflow Q can be computed as function of depth for
either pipes, orifices, or weirs or combinations
Q  CA 2gH for orifice flow
Q  CLH
3/2
for weir flow
Typical Storage -Outflow
• Plot of Storage in vs. Outflow in Storage is
largely a function of topography
• Outflows can be computed as function of
elevation for either pipes or weirs
Combined
S
Pipe
Q
Comparisons:
River vs.
Reservoir
Routing
Level pool reservoir
River Reach
Flood Control
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Structural Measures
Non-structural Methods