Chapter7
FLOOD ROUTING
____________________________________________________________________________________________
7.1
Introduction
The flood hydrograph discussed in many hydrology and hydraulics textbooks is in fact a wave.
The stage and discharge hydrographs of unsteady and non-uniform flows represent the passage of
waves of either through a river or a reservoir. As this wave moves down the river, the shape of
the wave gets modified due to various factors, such as storage, frictional resistance, lateral
addition or withdrawal of flows etc. When a flood wave passes through a reservoir, its peak is
attenuated and the time base is enlarged due to the effects of storage. Flood waves passing down
a river have their peaks attenuated due to frictional resistance if there is no lateral inflow. The
addition of lateral inflows can cause a reduction of attenuation or even an amplification of the
flood wave. The aspects of the changes in a flood wave passing through a channel system form
the basis of flood routing.
In general, flood routing is the technique of determining the characteristics of a flood hydrograph
at the outlet of a reservoir or at a downstream section of a river channel by utilizing the data of
flood flow at one or more upstream sections. The hydrologic analysis of problems such as flood
forecasting, flood protection, reservoir design and spillway design invariably require inputs from
flood routing analysis. In these applications two broad categories of routing can be recognized.
These are:
1. Reservoir or storage routing and
2. channel routing
In reservoir routing, which is also occasionally referred to as lumped routing, the effect of a flood
wave entering a reservoir is studied. Knowing the volume-elevation characteristics of the
reservoir and the outflow-elevation relationship for the spillways and other outlet structures in
the reservoir, the effect of a flood wave entering the reservoir is studied to predict the variations
of reservoir elevation and discharge outflow with time. This form of reservoir routing is essential
in:
1. the design of the capacity of spillways and other reservoir outlet structures and
2. the location and sizing of the capacity of reservoirs to meet specific requirements.
In channel routing, which is also referred to as distributed routing, the change in the shape of a
hydrograph as it travels down a channel is studied. By considering a channel reach and an input
hydrograph at the upstream end, this form of routing aims to predict the flood hydrograph at
various sections of the reach. Information on the flood peak attenuation and the duration of high
water levels obtained by channel routing is of utmost importance in flood forecasting operations
and flood protection works.
7-1
A variety of routing methods are available and they can be broadly classified into two categories
namely:
i. hydrologic routing and
ii. hydraulic routing
Hydrologic routing methods employ essentially the equation of continuity. Hydraulic routing
methods, on the other hand, utilize both the continuity equation and the equation of motion
(derived from Newton’s second law of motion) for unsteady flow. The limitations of data
unavailability notwithstanding, the basic differential equations used in the hydraulic routing,
known as the St. Venant equations afford a better description of unsteady flow, than hydrologic
methods.
7.1.1 The Basic Equations
The passage of a flood through a reservoir or a channel reach is an unsteady flow phenomenon. It
is classified in open channel hydraulics as gradually varied unsteady flow.
The lumped form of the equation of continuity, which is used in all hydrologic (lumped) routing
as primary equation states that, the difference between the inflow rate (I) and the outflow rate (O)
is equal to the rate of change of storage (S), i.e.
I O 
dS
dt
------- 7.1
On the other hand, the hydraulic (or distributed) routing requires the distributed form of the
continuity and momentum conservation differential equations. The differential equation of
continuity for unsteady flow in a reach no lateral inflow is given by:
 y
 V  y
V
y

0
Nonconservative form:
------7.2a
 x
 x  t
 Q  A

0
 x  t
Conservative form:
------7.2b
The momentum conservation equation (equation of motion) for a flood wave is derived from the
application of the Newton’s second law of motion and is given as:
Nonconservative form:


y V  V  1  V 
 
 
  S f  S0
x g   x g   t 
------ 7.3a
7-2
Conservative form:
1 Q 1

A  t
A
 Q2 

 A

g
 x

 
y
 g( S0  S f )  0
x
------7.3b
where V is the velocity of flow at a given section, S0 is the channel bed slope, Sf is the slope of
the total energy line, A is the cross-sectional area of flow, Q is the discharge and g is the
acceleration due to gravity. The first to the sixth terms in the flow equation (either 7.3a or 7.3b)
represent the local acceleration term, convective acceleration term, pressure gradient force term,
gravitational force term and the frictional force term respectively. When all the forces relevant to
all six terms in either (7.3a) or (7.3b) are significant, then the corresponding flow is unsteady and
non-uniform (commonly referred to as the dynamic wave). The flow is steady and non-uniform
in flow where the local acceleration term is negligible. On the other hand, the flow is said to be
steady and uniform when the gravitational forces tend to balance the frictional forces. In this
case, the flow characterizes a kinematic wave. A diffusion wave exists in flows where there is no
acceleration.
The continuity equations 7.2a and 7.2b and the equations of motion 7.3a and 7.3b are believed to
have been first developed by A.J.C Barrè de Saint Venant around 1871 and these equations are
commonly known as the St. Venant equations. Hydraulic flood routing involves the numerical
solution of the St. Venant equations. Details about these equations can be found in many standard
hydrology and hydraulics textbooks.
7.2
Hydrologic storage routing
There are essentially two types of storage routing. These are: reservoir hydrologic-routing and
channel hydrologic-routing. In both these methods, the characteristics of an outflow (O), either at
a given channel section or at the outlet of a reservoir are estimated by utilising the information on
inflow (I) at a given upstream section or sections.
7.2.1 Hydrologic reservoir routing
A flood wave I(t), enters a reservoir provided with an outlet flow control such as a spillway. Due
to the passage of the flood wave through the reservoir, the water level in the reservoir changes
with time. Thus, the discharge and storage also change with time. But the outflow is a function of
the reservoir elevation only, i.e. O=O(H). In an uncontrolled reservoir, typically,
O
2
Cd 2 g Le H 3/ 2  O(h)
3
------7.4
where H is the depth of flow over the spillway, Le is the effective length of the spillway crest and
Cd is the coefficient of discharge characteristic to the spillway. Similarly, for other forms of
outlets, such as gated spillways, sluice gates, etc. other relations for O(H) are available.
7-3
Similarly, the storage in the reservoir is also a function of the reservoir elevation and thus
S=S(h). Unlike in an ideal channel for which storage is a function of both inflow and outflow, in
an ideal reservoir storage is a function of outflow only. The relationship between reservoir
storage and outflow can be expressed in the following general form:
S  f ( O)
------7.5
A common relationship between outflow and reservoir storage is the following power function:
S  KO n
------7.6
in which K = storage coefficient and n = exponent. For n = 1, Eq. 7.6 reduces to the linear form
S = KO
------7.7
Reservoir storage routing involves finding the variation of S, H, and O with time, i.e. finding
S=S(t), O=O(t) and H=H(t) given I=I(t).
For the reservoir storage routing, the following data have to be known:
i. a relationship between the reservoir volume and the water level above the crest of the
spillway
ii. a relationship between the reservoir water-surface elevation and the outflow and hence a
relationship between the reservoir storage and the outflow discharge.
iii. the temporal distribution of the inflow hydrograph I=I(t)
iv. The initial values of S, I and Q at time t=0
There are a variety of methods available for the routing of floods through a reservoir. All of them
use the finite difference form of the lumped continuity equation (mass balance) Eq.7.1. The
initial reservoir water elevation is often assumed to be horizontal. Thus, the storage routing is
also known as Level Pool Routing. Many methods have been developed for Level Pool routing.
Two of the most commonly used level pool routing methods are:
 The Coefficient method
 The storage indication (Puls) method
These methods are described in the following sections.
7.2.1.1 The coefficient routing method
The coefficient routing method, which is commonly referred to as the general reservoir routing
method, assumes that the reservoir or channel storage is directly proportional to the outflow from
the reservoir. The storage concept is well established in flow-routing theory and practice.
Storage routing is used not only in reservoir routing but also in stream channel and catchment
routing. Techniques for storage routing are invariably based on the differential equation of
continuity (Eq 7.1).
7-4
Equation 7.1 can be solved by analytical or numerical means. The numerical approach is usually
preferred because it can account for an arbitrary inflow hydrograph and because it lends itself
readily to computer solution. The approach to solving this equation can be visualised by
discretizing Eq. 7.1 on the x-t plane (Fig. 7.1). The x-t plane is a graph showing the values of a
certain variable in discrete points in time and space.
Figure 7.1 shows two consecutive time levels, 1 and 2 separated between them by an interval t,
and two spatial locations or sections depicting inflow (section 1) and outflow (section 2), with
the reservoir located between them. The discretization of Eq. 7.1 on the x-t plane leads to:
I 1  I 2 O1  O2 S 2  S1


---------7.8
2
2
 t
in which I1 = inflow (section 1) at time level 1; I2 = inflow (section 1) at time level 2; O1 =
outflow (section 2) at time level 1; 02 = outflow (section 2) at time level 2; S1 = storage at time
level 1; S2 = storage at time level 2; and t = time interval. Equation 7.8 states that between two
time-levels 1 and 2 separated by a time interval t, average inflow minus average outflow is
equal to change in storage.
time t
Time level 2=t+t
S2
I2
O2
t
Time level 1= t
S1
I1
O1
x
Inflow section 1
Figure 7.1
Outflow section 2
Discretization of storage equation in the x-t plane
For linear reservoirs, the storage-outflow relationship for time-level 1 can be written as:
S1= KO1
------7.9
And for time-level 2, the equation can be written as:
S2 = KO2
-------7.10
Substituting Eq. 7.9 into 7.10, and solving for O2 we get:
7-5
O2 = C0I2 + C1I1 + C2O1
---------7.11
in which CO, C1 and C2 are routing coefficients defined as follows:
C0 
t / K
2  ( t / K )
--------7.12
Cl = C0
C2 
--------7.13
2  ( t / K )
2  ( t / K )
--------7.14
The reader can verify that C0 + C1 + C2 = 1. Thus, the routing coefficients are interpreted as
weighting coefficients. These routing coefficients are a function of t/K, the ratio of time
interval to storage constant. The components C0I2 , C1I1 and C2O1 of Eq. 7.11 are called partial
outflows.
The coefficient (linear reservoir) routing procedure is illustrated by the following example.
Example 7.1
A linear reservoir has a storage constant K = 2 h, and it is initially at equilibrium with inflow and
outflow equal to 100 m'/s. Route the following inflow hydrograph through the reservoir.
Table 7.1
Inflow hydrograph for Example 7.1
Time(hr)
Inflow(m3/s)
0
100
1
150
2
250
3
400
4
800
5
1000
Time(hr)
Inflow(m3/s)
8
550
9
400
10
300
11
250
12
200
13
150
6
900
14
120
7
700
15
100
First it is necessary to select an appropriate time interval. An examination of the inflow
hydrogaph reveals that the time-to-peak is tp = 5 h. A rule-of-thumb for adequate temporal
resolution is to make the ratio tp/t at least equal to 5. Setting t = 1 hr assures that tp/t = 5.
With t =1 h, the ratio t/K = 1/2. From Eqs. 7.12 to 7.14, C0 = C1=1/5, and C2 = 3/5. The
routing calculations are shown in Table 7.2. Column 1 shows the time and Column 2 shows the
inflow hydrograph ordinates. Columns 3 to 5 gives the partial flows while Col. 6 shows the
computed outflow, which is the sum of the partial flows at each time-level. The recursive
procedure continues until the calculated outflow (Col. 6) is within 5 percent of baseflow (100
m3/s). Plotted inflow and outflow hydrographs (Cols. 2 and 6) are shown in Fig. 7.2. The
calculated peak outflow (758 m3/s) occurs at t = 7 hr. However, the shape of the outflow
hydrograph reveals that the true peak outflow occurs somewhere between 6 and 7 hr. The true
peak-outflow is approximated graphically at 765 m3/s, occurring at about 6.6 hr. The peak
7-6
outflow is substantially less than the peak inflow (1000 m3/s), showing the attenuating effect of
the reservoir. Also, the time elapsed between the occurrences of peak inflow and peak outflow
(1.6 hr) is approximately equal to the storage constant.
The reservoir exerts a diffusive action on the flow, with the net result that peak flow is attenuated
and time base is increased. In the linear reservoir case, the amount of attenuation is a function of
t/K. The smaller this ratio, the greater the amount of attenuation exerted by the reservoir.
Conversely, large values of t/K cause less attenuation. Values of t/K greater than 2 can lead to
negative attenuation. This amounts to amplification; therefore, values of t/K greater than 2 are
not used in reservoir routing.
Table 7.2
Col. 1
Time (hr)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
The coefficient method routing for Example 7.1
Col. 2
Col. 3
3
Inflow(m /s) C0I2
100
150
30
250
50
400
50
800
160
1000
200
900
180
700
140
550
110
400
50
300
60
250
50
200
40
156
30
120
24
100
20
100
20
100
20
100
20
100
20
100
20
100
20
Col. 4
Cl Il
20
30
50
80
160
200
180
140
110
50
60
50
40
30
24
20
20
20
20
20
20
Col. 5
C2O1
60
66
87.6
130.6
222.4
349.4
437.6
454.6
422.8
367.7
304.6
248.8
203.3
164.0
131.8
104.9
86.9
76.1
69.7
65.8
63.5
Col. 6
Outflow(m3/s)
100.0
110.0
146.0
217.6
370.6
582.4
729.4
757.6
704.6
612.8
507.7
414.6
338.8
273.3
218.0
174.8
144.9
126.9
116.1
109.7
105.8
103.5
7-7
^ 3 /s )
( m
1200
In flo w H y d r o g r a p h
1000
D
is c h a r g e
O u t flo w H y d r o g r a p h
800
600
400
200
0
0
2
4
6
8
10
12 14 16 18 20
T im e (H r s )
Figure 7.2
Application of the coefficient routing method: Example 7.1
A distinct characteristic of reservoir routing is the occurrence of peak outflow at the time when
inflow equals outflow (see Fig. 7.2). Since outflow is proportional to storage (Eq. 7-7), peak
outflow corresponds to maximum storage. Since storage ceases to increase when outflow equals
inflow, maximum storage and peak outflow must occur at the time when inflow and outflow
coincide.
Another characteristic of reservoir routing is the immediate outflow response, with no apparent
lag between the start of inflow and the start of outflow (see Fig. 7.2). From a mathematical
standpoint, this property is attributed to the infinite propagation velocity of surface waves in an
ideal reservoir.
7.2.1.2 Storage indication method
The storage indication method is also known as the modified Puls routing method. In some
references, it is also called the Goodrich routing method. It is used to route flood wave through
actual reservoirs, for which the relationship between outflow and storage is usually of a nonlinear nature.
The method is based on the differential equation of storage, Eq. 7.1. The discretization of this
equation on the x-t plane (Fig. 7.1) leads to Eq. 7.8. In the storage indication method, Eq. 7.8 is
transformed to its equivalent form:
2S2
2S
 O2  I 1  I 2  1  O1
--------7.15
t
t
in which the unknown values (S2 and 02) are on the left side of the equation and the
known values (inflows, initial outflows and storage) are on the right hand side. The left side of
Eq. 7.15 is known as the storage indication quantity.
7-8
In the storage indication method, it is first necessary to assemble geometric and hydraulic
reservoir data in suitable form. For this purpose, the following curves (or tables) are prepared:
(1) elevation-storage, (2) elevation-outflow, (3) storage-outflow, and (4) storage indicationoutflow. For computer applications, these curves are re-placed by tables of elevation-outflowstorage indication quantities.
The elevation-storage relation is determined based on topographic information. The minimum
elevation is that for which storage is zero, and the maximum elevation is the minimum elevation
of the dam crest.
The elevation-outflow relation is determined based on the hydraulic properties of the outlet
works, either closed conduit, overflow spillway, or a combination of the two. In the typical
application, the reservoir pool elevation provides a head over the outlet or spillway crest, and the
outflow can be calculated using an equation such as Eq. 7.4. When routing floods through
emergency spillways, storage is alternatively expressed in terms of surcharge storage, i.e., the
storage above a certain level, usually the emergency spillway crest elevation.
Elevation-storage and elevation-outflow relations lead to the storage-outflow relation. In turn,
the storage-outflow relation is used to develop the storage indication-outflow relation. To
develop the storage indication-outflow relation it is first necessary to select a time interval such
that the resulting linearization of the inflow hydrograph remains a close approximation of the
actual non-linear shape of the hydrograph. For smoothly rising hydrographs, a minimum value of
tp/t = 5 is recommended, in which tp is the time-to-peak of the inflow hydrograph.
Once the data has been prepared, Eq. 7.15 is used to perform the reservoir routing. The procedure
consists of the following steps:
1.
2.
3.
4.
5.
Set the counter at n = 1 to start.
Use Eq. 7.15 to calculate the storage indication quantity [(2Sn+1/t) +On+1] at time level
n+1.
Use the storage indication quantity versus outflow relation to determine the outflow On+1
at time level n + 1.
Use the storage indication quantity and outflow at time level n + 1 to calculate
[(2Sn+1/t) - O n+1] = [(2S n+1/t) + O n+1]-2(O n+1).
Increment the counter by 1, go back to step 2 and repeat. The recursive procedure is
terminated either when the inflow ceases or when the outflow hydrograph has
substantially receded to baseflow discharge.
The procedure is illustrated in Examples 7.2 and 7.3 using the same data as in Example 7.1, not
only to confirm that the storage indication method is not only applicable to a linear but also to a
non-linear reservoir.
7-9
Example 7.2
Use the data of example 7.1 to perform a reservoir routing by the storage indication method. The
reservoir storage (S) and outflow from the reservoir (O) are related by an equation S=2(O). Use
t = 1 hr.
The required table for outflow (O), the storage (S) and the storage indication quantity
[(2*S/t)+O] is shown below.
Table 7.3
The outflow-storage relationship for Example 7.2
Outflow (m3/s)
Storage (S)
m3s-1-h
100
200
300
400
500
600
700
800
900
1000
1100
Storage indication
quantity
[(2*S/t)+O] m3s-1
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
5500
200
400
600
800
1000
1200
1400
1600
1800
2000
2200
This table can also be replaced by a graph of the storage indication quantity versus outflow. In
this example, this graph is as given below:
6000
Storage indication quantity
3 -1
ms
5000
4000
3000
2000
1000
1100
1000
900
800
700
600
500
400
300
200
100
0
Outflow (O) m 3s-1
Figure 7.3
The outflow-storage relationship for Example 7.2
7-10
The calculations are shown in Table 7.4. At time t=0 the counter is at n=1. The outflow is 100
m3/s (the baseflow) and the corresponding storage indication quantity (from the table or graph
above) is 500 m3/s. The entries in column 3 of the calculations table represent [(2S/t)-O]n which
is equal to [(2S/t)+O]-2O]n the indication quantity minus twice the output at calculation time
step n. At t=1 this is therefore equal to 500-2*100=300 m3/s. For t=2 and other subsequent time
steps, the indication quantity is computed from Eq. 7.15 using the current and previous inflows
as well as the previous entry of [(2S/t)-O] in column 3. For each step-computation of the
storage indication quantity, the estimates of the corresponding outflow are computed by
interpolation using Table 7.3 or the graph on Fig. 7.3. This recursive procedure is continued until
the outflow hydrograph ordinate (column 5) is within 5% of the equilibrium inflow. The
complete routing solution is summarised in Table 7.4 below.
The solution of this problem as summarized in column 5 of Table 7.4 can be confirmed to be the
same as that was obtained in Table 7.2 in Example 7.1. This must be expected because we have
used the same outflow-storage relationship in both examples. Example 7.3 below illustrates the
use of the indication method for a non-linear reservoir.
Example 7.3
Use the data of example 7.1 to perform a reservoir routing by the storage indication method if the
reservoir has an outflow-storage relationship S=kOn where k=0.02 and n=2. Use t =1 hr.
Proceeding as in the above example, it is important to construct the storage-outflow relationship
table and/or graph. The corresponding table and graph for this example are given in Table 7.5
and Figure 7.4 respectively.
Table 7.4
Solution of Example 7.2
Col. 1
Time t (hr)
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Col. 2
Col. 3
Inflow (I) m3/s
[(2S/delt)-O]
100
300.0
150
330.0
250
438.0
400
652.8
800
1111.7
1000
1747.0
900
2188.2
700
2272.9
550
2113.8
400
1838.3
300
1523.0
250
1243.8
200
1016.3
150
819.8
120
653.9
Col. 4
Col. 5
[(2S/delt)+O] Outflow (O) m3/s
500.0
100.0
550.0
110.0
730.0
146.0
1088.0
217.6
1852.8
370.6
2911.7
582.3
3647.0
729.4
3788.2
757.6
3522.9
704.6
3063.8
612.8
2538.3
507.7
2073.0
414.6
1693.8
338.8
1366.3
273.3
1089.8
218.0
7-11
15
16
17
18
19
20
21
Table 7.5
100
100
100
100
100
100
100
524.3
434.6
380.8
348.5
329.1
317.4
310.5
873.9
724.3
634.6
580.8
548.5
529.1
517.4
174.8
144.9
126.9
116.2
109.7
105.8
103.5
The outflow-storage relationship for Example 7.3
Outflow Storage (S) Storage indication Outflow
Storage (S)
Storage indication
(m3/s)
m3s-1-h
quantity
(m3/s)
m3s-1-h
quantity
[(2*S/t)+O] m3s-1
[(2*S/t)+O] m3s-1
1100.0
540
1161.9
100
500.0
2863.8
1323.2
580
1204.2
140
591.6
2988.3
1521.6
620
1245.0
180
670.8
3110.0
1703.2
660
1284.5
220
741.6
3229.0
1872.5
700
1322.9
260
806.2
3345.8
2032.1
740
1360.1
300
866.0
3460.3
2183.9
780
1396.4
340
922.0
3572.8
2329.4
820
1431.8
380
974.7
3683.6
2469.4
860
1466.3
420
1024.7
3792.6
2604.8
900
1500.0
460
1072.4
3900.0
2736.1
500
1118.0
7-12
m
O
]
4 000
3 500
2
(
3 000
2 500
- 1
S
/ d
e
l
3
s)
t
+
S = 0 .0 2 * O
[
2
2 000
O u t flo w [ O ] m
Figure 7.4
0
9
0
6
8
0
2
8
0
8
7
0
4
7
0
0
7
0
6
6
0
2
6
0
8
0
3
5
0
4
5
0
0
5
0
6
4
2
4
0
8
3
0
4
3
0
0
3
0
6
2
0
2
2
0
8
1
0
4
1
0
0
1 000
1
0
1 500
s
-1
The storage indication quantity versus outflow for Example 7.3
Using the same procedure as in Example 7.2, we obtain the solution as summarised in Table 7.6
below.
Table 7.6
Solution of Example 7.3
Col. 1
Time t (hr)
0
1
2
3
4
5
6
7
8
9
10
11
Col. 2
Col. 3
Inflow (I) m3/s
[(2S/delt)-O]
100
900
150
934
250
1056
400
1270
800
1634
1000
1968
900
2092
700
2046
550
1930
400
1790
300
1638
250
1506
Col. 4
Col. 5
[(2S/delt)+O] Outflow (O) m3/s
1100
100
1150
108
1334
139
1706
218
2470
418
3434
733
3868
888
3692
823
3296
683
2880
545
2490
426
2188
341
7-13
12
13
14
15
16
17
18
19
20
21
200
150
120
100
100
100
100
100
100
100
1394
1284
1180
1090
1022
980
954
936
924
916
1956
1744
1554
1400
1290
1222
1180
1154
1136
1124
281
230
187
155
134
121
113
109
106
104
Occasionally, the reservoir elevation-outflow-storage information is also provided especially
when the reservoir water-elevation is pertinent in the reservoir regulation plans. In such cases, the
storage indication method can be extended to provide this required reservoir elevation
information. This is illustrated in the example given below.
Example 7.4
A reservoir has an elevation-storage-outflow relationship as shown on Table 7.7 below:
Table 7.7
Elevation-Outflow-storage relationship for Example 7.4
Elevation (m)
100.00
100.50
101.00
101.50
102.00
102.50
102.75
103.00
S*106 m3
3.35
3.47
3.88
4.38
4.88
5.37
5.53
5.86
(O) m3/s
(2*S/t+O) m3/s
0.00
310.19
10.00
331.48
26.00
385.26
46.00
451.83
72.00
524.04
100.00
597.22
116.00
627.76
130.00
672.22
Route the flood wave given on Table 7.8 through this reservoir if the initial water level in the
reservoir is 100.5 meters. Use a t=6 hrs=21600 sec.
Table 7.8
Flood hydrograph for routing in Example 7.4
Time Hrs
Inflow m3/s
0 6 12 18 24 30 36 42 48 54 60 66
10 30 85 140 125 96 75 60 46 35 25 20
7-14
q u a n t
650
m /s
600
3
550
500
450
400
1 4 0
1 3 0
1 2 0
1 1 0
9 0
8 0
7 0
6 0
5 0
4 0
3 0
2 0
0
300
1 0 0
350
1 0
in d ic a t io n
S t o r a g e
700
3
O u tflo w m /s
Figure 7.5
Storage indication quantity versus outflow for Example 7.4.
As it has been shown in the previous examples, it is important to construct a storage indication
versus outflow graph if any of these types. It has been stressed that such graph is very useful in
these problems for interpolation purposes. The graph for this example is given on Figure 7.5.
With the information provided so far, it is now possible to estimate the required outflow
hydrograph. This has been done using the usual procedure. The computations are shown in
column 5 of Table 7.9.
The reservoir elevations corresponding to these outflows are shown in column 6 of Table 7.9.
These were estimated by interpolation from Figure 7.6.
7-15
Table 7.9
Summary of the computations for Example 7.4
0
6
12
18
24
30
36
42
48
54
60
66
. 0
3
0
1
Elevation-Outflow relationship for Example 7.4.
Time (hr)
. 5
0
1
E le v a t io n ( m )
Figure 7.6
Col. 1
2
. 0
0
1
2
. 5
0
1
1
. 0
0
1
1
. 5
0
1
0
0
1
0
4 0 .0 0
3 0 .0 0
2 0 .0 0
1 0 .0 0
0 0 .0 0
9 0 .0 0
8 0 .0 0
7 0 .0 0
6 0 .0 0
5 0 .0 0
4 0 .0 0
3 0 .0 0
2 0 .0 0
1 0 .0 0
0 .0 0
. 0
m
3
/ s
u t f lo w
O
1
1
1
1
1
Col. 2
Col. 3
Col. 4
Col. 5
Col. 6
Inflow (2S/delt)-O
O
m3/s
m3/s
(2S/delt)+O m3/s m3/s
Elevation m
10
314
340
13
100.6
30
322
354
16
100.72
85
355
437
41
101.4
140
396
580
92
102.35
125
409
661
126
102.93
96
398
630
116
102.75
75
391
569
89
102.3
60
382
526
72
102
46
370
488
59
101.75
35
361
451
45
101.48
25
349
421
36
101.25
20
338
394
28
101.05
Figure 7.7 shows the temporal distribution of the inflow flood wave, the estimated outflow
hydrograph and the corresponding reservoir elevation for Example 7.4.
7-16
Outflow
Elevation m
103.5
103.0
102.5
102.0
101.5
101.0
100.5
100.0
99.5
99.0
160
140
120
100
80
60
40
20
0
0
6
12
18
24
30
36
42
48
54
60
Elevation m
Outflow m 3/s
Inflow
66
Time (hrs)
Figure 7.7
7.2.1.3
A comparison of the temporal distribution of the inflow flood wave, the
outflow hydrograph and the reservoir elevation for Example 7.4.
General remarks on storage routing
Most large reservoirs have some type of outflow control, wherein the amount of outflow is
regulated by gated spillways. In this case, both hydraulic conditions and operational rules
determine the prescribed outflow. Operational rules take into account the various uses of water.
For instance, a multipurpose reservoir may be designed for hydropower generation, flood control,
irrigation, and navigation. For hydropower generation, reservoir pool level is kept within a
narrow range, usually close to the optimum operating level of the installation. On the other hand,
flood-control operation may require that a certain storage volume be kept empty during the floodseason in order to receive and attenuate the incoming floods. Flood-control operations also
require that the reservoir releases be kept below a certain maximum, usually taken as the flow
corresponding to bank-full stage. Irrigation requirements may vary from month to month
depending on the consumptive needs and crop patterns. For navigation purposes, outflow should
be a nearly constant value that will ensure a minimum draft downstream of the reservoir.
Reservoir operational rules are designed to take into account the various water-demands. These
are often conflicting and, therefore, compromises must be reached. Multipurpose reservoirs
allocate reservoir volumes to the different uses. In this way, operational rules may be developed
to take into account the requirements of each use. In general, outflow from a reservoir with gated
outlets is determined by prescribed operational policies. In turn, the latter are based on the
current level of storage, incoming flow, and downstream flow requirements.
The differential equation of storage can be used to route flows through reservoirs with controlled
outflow. In general, the outflow can be either (1) uncontrolled (ungauged), (2) controlled (gated),
or (3) a combination of controlled and uncontrolled. The discretized equation, including
controlled outflow, is given by:
I 1  I 2 O1  O2
S  S1

 Or  2
2
2
t
-------7.16
7-17

In which Qr is the mean regulated outflow during the time interval At. Equation 7.16 can be
expressed in storage indication form:
2S2
2S
 O2  I 1  I 2  1  O1  2Or
t
t
--------7.17

With Qr known, the solution proceeds in the same way as with the uncontrolled out-flow case.
In the case where the entire outflow is controlled, Eq. 7.17 reduces to:
S 2  S1 
t
( I 1  I 2 )  ( t )O r
2
---------7.18
By which the storage volume can be updated based on average inflows and mean regulated
outflow. Other requirements, such as estimates of reservoir evaporation where warranted (i.e., in
semiarid and regions) may be implemented properly to account for the storage volumes.
7.2.2 Hydrologic channel routing
In reservoir routing presented in the previous sections, the storage was a unique function of the
outflow discharge. However, in channel routing, the storage is a function of both the inflow and
the outflow discharges. Hence, a different routing approach is required. It is important to note
that the flow in a river during a flood passage belongs to the category of gradually varied
unsteady flow. The water surface is not only unparallel to the channel bottom but also varies with
time. Thus, considering a channel reach having a flood flow, the total volume in storage can be
considered under two categories as:
1. Prism storage
2. Wedge storage
Prism storage is the volume that would exist if uniform flow occurred at the downstream depth,
i.e. the volume formed by an imaginary plane parallel to the channel bottom drawn at the outflow
section of the flow. On the other hand, wedge storage is a wedge-like volume formed between
the actual water surface profile and the surface of the prism storage. These storages are illustrated
on Figure 7.8 below.
Wedge storage
Inflow (I)
Outflow (O)
Prism storage
(Fig. 7.8a)
7-18
-ve wedge storage
+ve wedge
storage
Prism
Prism
Prism
(Fig. 7.8b)
Figure 7.8
7.2.2.1
Prism and wedge storage in a river channel during flood flow
The Muskingum channel routing method
One of the most commonly used hydrologic channel routing procedure is the Muskingum method.
This method was developed during the period 1934-1935 for studies of the Muskingum
Conservation District Flood Control Project (MCDFCP) of the U.S Army Corps of Engineers.
The method which utilises the concept of both wedge and prism storage is of the form

S  K xI m  1  x O m

---------7.19
Where K is a channel coefficient usually called the storage-time constant. It has the dimension of
time. It is approximately equal to the time of travel of a flood wave through the channel reach.
The coefficient x is a weighting factor that takes values between 0 and 0.5. When x=0 Eq. 7.19
reduces to the linear storage equation (Eq. 7.7). When x=0.5, then the inflow and outflow are
equally important in determining the storage. On the other hand, m is a positive exponent which
varies from 0.6 for rectangular channels to a value of about 1.0 for natural channels. When m=1.0
Eq. 7.19 reduces to the linear form:
S=K[xI+(1-x)O]
---------7.20
Equation 7.20 is commonly referred to as the Muskingum Equation.
Estimation of K and x in the Muskingum equation
Figure 7.9(a) shows a typical inflow and outflow hydrographs through a channel reach. Note that
unlike the case with the reservoir routing, the outflow peak does not occur at the point of
intersection of the inflow and the outflow hydrographs. Using the continuity equation (Eq. 7.8 ),
(I1+I2)t/2 – (O1+O2)t/2 = S
The increment in storage at time t and the time increment t can be calculated. Summation of the
various incremental storage-values enables one to find the channel storage S versus time
relationship (Fig 7.9(a)). If the inflow and outflow data is available for a given reach, values of S
7-19
at various time intervals can be determined by the above approach. By choosing a trial value of x,
values of S at time t are plotted against the [xI+(1-x)O] values. If the value of x is chosen
correctly, a straight-line relationship as given by Eq. 7.20 is obtained. An incorrect value of x
gives a plot of S versus [xI+(1-x)O] values which trace a looping curve. By trial and error, a
value of x can be chosen for which the plot very nearly describes a straight line. The inverse of
the slope of such line gives the value of K. In a given channel reach the values of x and K are
assumed to be constant. This procedure is demonstrated in example 7.5 below.
Example 7.5
Estimate the values of K and x for a channel reach with an inflow and outflow discharge
hydrographs as given in columns 1 to 3 in Table 7.10 below.
Calibration of Muskingum routing parameters for Example 7.5
Time Inflow -I Outflow-O I-O (I-O)/2 S=Col5* S=S
hrs
(m3/s)
(m3/s) (m3/s) (m3/s) t (m3/s- (m3/s-hr)
hr)
0
6
12
18
24
30
36
42
48
54
60
66
(2)
(3)
5
20
50
50
32
22
15
10
7
5
5
5
(4)
5
6
12
29
38
35
29
23
17
13
9
7
D is c h a rg e
(1)
(5)
0
14
38
21
-6
-13
-14
-13
-10
-8
-4
-2
(6)
(7)
0
42
156
177
45
-57
-81
-81
-69
-54
-36
-18
7
26
29.5
7.5
-9.5
-13.5
-13.5
-11.5
-9
-6
-3
xI+(1-x)O] (m 3/s)
x=0.35
x=0.3
(8)
(9)
0
42
198
375
420
363
282
201
132
78
42
24
5
10.9
25.3
36.35
35.9
30.45
24.1
18.45
13.5
10.2
7.6
6.3
5
10.2
23.4
35.3
36.2
31.1
24.8
19.1
14
10.6
7.8
6.4
60
500
50
400
40
300
30
200
20
x=0.2
(10)
5
8.8
19.6
33.2
36.8
32.4
26.2
20.4
15
11.4
8.2
6.6
S t o ra g e
Table 7.10
100
10
0
0
0
6
12
18 24
30
36
42 48
54
60
66
T im e (H rs )
In f l o w
m ^ 3 /s
O u t f lo w
m ^ 3 /s
S to r a g e m ^ 3 /s .h r
(Fig 7.9a)
7-20
m ^ 3 /s
[ x I+ ( 1 - x ) O ]
4 0
3 5
3 0
2 5
2 0
1 5
1 0
x = 0 .2 0
x = 0 .3
x = 0 .3 5
5
0
0
1 0 0
2 0 0
3 0 0
4 0 0
5 0 0
S to r a g e S m ^ 3 /s - h r
Figure 7.9
(Fig 7.9b)
Estimation of the K and x values in the Muskingum equation
As a first trial, x=0.35 is selected and the value of [xI+(1-x)O] evaluated (column 8) and plotted
against S in Figure 7.9(b). Since a looped curve is obtained, further trials are performed with
x=0.3 and 0.20. It is seen that from Figure 7.9(b) that for x=0.2, the data very nearly describes a
straight line and as such, x=0.20 is taken as an appropriate value for the channel reach. The value
of K is then computed from the slope of the x=0.2 line. In this case K=13.2 hrs.
Muskingum method of routing
For a given channel reach, by selecting a routing interval t and using the Muskingum equation,
the change in storage is
S2 – S1 = K [x (I2 – I1) + (1 - x) (O2 - Ol]
-------------7.21a
where suffixes 1 and 2 refer to the conditions before and after the time interval t. The continuity
equation (Eq. 7.8) for the reach is
I I 
 O  O1 
S 2  S1   2 1 t   2
----------7.21b
t
 2 
 2 
From the above equation, O2 is evaluated as
O2 = C0I2+C1I1+C2O1
------------7.22
where C0 = (-Kx + 0.5 t)/( K-Kx+0.5 t)
C1 = (Kx+0.5t)/ (K-Kx+0.5t)
C2 = (K-Kx -0.5 t)/ (K-Kx+0.5t)
Note that C0 + C1 + C2 =1.0, Eq. 7.22 can be written in a general form for the n-th time step as
7-21
On =C0In+C1In-1 +C2On-1
---------7.23
Eq. 7.23 is known as Muskingum Routing Equation and provides a simple linear equation for
channel routing. It has been found that for best results the routing interval t should be so
chosen that K > t > 2Kx. If t < 2Kx, the coefficient C0 will be negative. Generally, negative
values of coefficients are avoided by choosing appropriate values of t
To use the Muskingum equation to route a given inflow hydrograph through a reach, the values
of K and x for the reach and the value of the outflow, Ol, from the reach at the start are needed.
The procedure is indeed simple.
a)
Knowing K and x, select an appropriate value of t
b)
Calculate C0 , C1 , and C2.
c)
Starting from the initial conditions I1 , O1 , and known I2 at the end of the first time step t
calculate O2 by Eq. 7.22 or Eq. 7.23
d)
The outflow that is calculated in step (c) becomes the known initial outflow that is used
for the next time step. Repeat the calculations for the entire inflow hydrograph.
The calculations are best done row by row in a tabular form. Example 7.6 illustrates the
computation procedure. Any computer SpreadSheet is ideally suited to perform the routing
calculations and to view the inflow and outflow hydrographs.
Example 7.6
Route the following hydrograph through a river reach for which K = 12.0 hr and x = 0.20. At the
start of the inflow flood, the outflow discharge is 10 m3/s.
Table 7.11
Time (hr)
Inflow (m3/s
The inflow hydrograph for Example 7.6
0
10
6
20
12
50
18
60
24
55
30
45
36
35
42
27
48
20
54
15
Solution:
Since K = 12 hr and 2Kx = 2 x 12 x 0.2 = 4.8 hr, t should be such that 12 hr >t > 4.8 hr. In
the present case t = 6 hr is selected to suit the given inflow hydrograph ordinate interval.
Using Eqs. 7.22 or Eq. 7.23 the coefficients C0, C1 and C2 are calculated as
C0 = (- 12* 0.20 + 0.5 * 6)/ (12 - 12 * O.2+ 0 5*6 = 0.6/12.6 = 0.048
C1= (12 * 0.2 + 0.5 * 6)/12.6 = 0.429
C2 = (12 - 12 * 0.2 - 0.5* 6)/12.6 = 0.523
For the first time interval, 0 to 6 hr,
I1 = 10.0
C1I1 = 4.29
I2 = 20.0
C0I2 = 0.96
O1 = 10.0
C2O1 = 5.23
7-22
From Eq. 7.22 O2 = C012+Cll+C2O1 = 10.48 m3/s
For the next time step, 6 to 12 hr, O1 = 10.48 m3/s. The procedure is repeated for the entire
duration of the inflow hydrograph. The computations are done in a tabular form as shown in
Table 7.12. By plotting the inflow and outflow hydrographs the attenuation and peak lag are
found to be 10 m3/s and 12 hr respectively.
Table 7.12
Muskingum method of routing-Example 7.6.(t = 6 hrs)
Time (h) I (m3/s)
(1)
(2)
0
10
6
20
12
50
0.048I2
(3)
0.429I1
(4)
0.523O1
(5)
0.96
4.29
5.23
10.48
2.40
18
55
30
45
2.88
21.45
8.61
2.64
25.74
17.23
32.94
45.61
23.60
23.85
49.61
1.68
19.30
25.95
1.30
15.02
24.55
35
42
27
48
20
46.93
40.87
0.96
11.58
21.38
33.92
0.72
54
5.48
16.46
2.16
36
8.58
60
24
O (m3/s)
(6)
10.00
8.58
15
17.74
27.04
Equations 7.21 and 7.8 can be combined in an alternative form of the routing equation as
O2 = O1 + B1 (I1 – O1) + B2 (I2 – I1)
---------7.24
Where B1 = t/([K (1 - x) + 0.5 t]
B2 = (0.5t-Kx)/([K (1 - x) + 0.5 t]
The use of Eq. 7.24 is essentially the same as that of Eq. 7.22.
7-23
7.2.2.2
Muskingum-Cunge routing method
One of the modified methods that has worked well in river routing is the Muskingum-Cunge
method. As shown earlier, the Muskimgum method is based on the storage equation with
coefficients K and x derived by trial and error. Cunge showed that K and x could be derived by
considering the hydraulics of the flow in the channel. Recall that K was shown to be
approximately equal to the time of travel of a flood wave through the reach. This assumption was
used by Cunge to give K=L/c, where c is the average speed of the flood peak and L is the
length of the river reach. Substituting for K=L/c inEq. 7.21a and 7.21b gives:
Lx
L
1
( I 2  I1 ) 
(1  x)(O2  O1 )  (O2  I 2  O1  I1 )  0
ct
ct
2
----7.25
Multiplying through Eq. 7.25 by c/L and rearranging gives:
x( I 2  I )1 (1  x)(O2  O1 )
c


[(O2  I 2 )  (O1  I1 )]  0
ct
t
2L
-------7.26
With K==L/c as an acceptable approximation, a means is required for obtaining the x factor.
Cunge derived the following expression for x from the channel properties:
x
Qp
1

2 2sBc L
----------7.27
where Qp is the mean peak discharge, s is the average bed-slope and B is the mean channel width and L is the
channel reach. In practice, the Muskingum coefficients are evaluated according to each reach forming the
subdivisions of the total length of the river reach being considered. Then the routing of the inflow could proceed as
shown earlier to obtain the outflow by recurrent application of the Muskingum routing equation.
7-24