Channel Routing Methods

Channel RoutingChannel Routing
• Simulate the movement of water through a channel
• Used to predict the magnitudes, volumes, and
temporal patterns of the flow (often a flood wave)
as it translates down a channel.
• 2 types of routing : hydrologic and hydraulic.
• both of these methods use some form of the
continuity equation.
Continuity equation
Hydrologic Routing
Hydraulic Routing
Momentum Equation

Continuity EquationContinuity Equation
•The change in storage (dS) equals the difference
between inflow (I) and outflow (O) or :
O-I=
dt
dS
•For open channel flow, the continuity equation
is also often written as :
q=
x
Q
+
t
A
∂
∂
∂
∂ A = the cross-sectional area,
Q = channel flow, and
q = lateral inflow
Continuity equationContinuity equation
Hydrologic Routing
Hydraulic Routing
Momentum Equation

Hydrologic RoutingHydrologic Routing
• Methods combine the continuity equation with some
relationship between storage, outflow, and possibly
inflow.
• These relationships are usually assumed, empirical,
or analytical in nature.
• An of example of such a relationship might be a
stage-discharge relationship.
Continuity equation
Hydraulic Routing
Momentum Equation

Use of Manning EquationUse of Manning Equation
• Stage is also related to the outflow via a relationship
such as Manning's equation
2
1
3
249.1
fh
SAR
n
Q =
Continuity equation
Hydraulic Routing
Momentum Equation

Hydraulic RoutingHydraulic Routing
• Hydraulic routing methods combine the continuity
equation with some more physical relationship
describing the actual physics of the movement of the
water.
• The momentum equation is the common relationship
employed.
• In hydraulic routing analysis, it is intended that the
dynamics of the water or flood wave movement be
more accurately described
Continuity equation
Hydrologic Routing
Hydraulic RoutingHydraulic Routing
Momentum Equation

Momentum EquationMomentum Equation
• Expressed by considering the external forces acting on a
control section of water as it moves down a channel
)S-Sg(=
A
vg
+
2x
A)y(
A
g
+
x
v
V+
t
v
fo
∂
∂
∂
∂
∂
• Henderson (1966) expressed the momentum equation as :
t
v
g
1
-
x
v
g
v
-
x
y
-S=S of
∂
∂
∂
∂
∂
∂
Continuity equation
Hydrologic Routing
Hydraulic Routing
Momentum EquationMomentum Equation

Combinations of EquationsCombinations of Equations
• Simplified Versions :
t
v
g
1
-
x
v
g
v
-
x
y
-S=S of
∂
∂
∂
∂
∂
∂
x
v
g
v
-
x
y
-S=S of
∂
∂
∂
∂
x
y
-S=S of
∂
∂
Sf = So
Unsteady -Nonuniform
Steady - Nonuniform
Diffusion or noninertial
Kinematic
Continuity equation
Hydrologic Routing
Hydraulic Routing
Momentum Equation

Routing MethodsRouting Methods
• Modified Puls
• Kinematic Wave
• Muskingum
• Muskingum-Cunge
• Dynamic
Modified Puls
Kinematic Wave
Muskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Modified PulsModified Puls
• 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.
Kinematic Wave
Muskingum
Muskingum-Cunge
Dynamic
Modeling Notes

t
S-S
=
2
O+O(
-
2
I+I 122121
∆
Continuity Equation
O+
t
S2
=O-
t
S2
+I+I 2
2
1
1
21
∆






∆
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.
Kinematic Wave
Muskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Modified Puls ExampleModified 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:
Hydrograph For Modified Puls Example
0
30
60
90
120
150
180
0 2 4 6 8 10
Time (hr)
Discharge(cfs)

2S/∆t + O curve for Modified Puls
example
0
200
400
600
800
1000
1200
0 10 20 30 40 50 60 70
Outflow (cfs)
2S/t+O(cfs)
•The following 2S/∆t + O curve is also given:

•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

Hydrograph For Modified Puls Example
0
30
60
90
120
150
180
0 2 4 6 8 10
Time (hr)
Discharge(cfs)
•Next, using the hydrograph and interpolation, insert the Inflow
(discharge) values.
•For example at 1 hour, the inflow is 30 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

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
•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.

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
•This is then repeated for the rest of the values in the column.

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
O+
t
S2
=O-
t
S2
+I+I 2
2
1
1
21
∆






∆
•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

2S/∆ t + O curve for Modified Puls
example
0
200
400
600
800
1000
1200
0 10 20 30 40 50 60 70
Outflow (cfs)
2S/t+O(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
•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.

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
•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

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
•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

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
•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.

MuskingumMuskingum MethodMethod
Sp = K O
Sw = K(I - O)X
Prism Storage
Wedge Storage
CombinedS = K[XI + (1-X)O]
Modified Puls
Kinematic Wave
MuskingumMuskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Muskingum,Muskingum, cont...cont...
O2 = C0 I2 + C1 I1 + C2 O1
Substitute storage equation, S into the “S” in
the continuity equation yields :
S = K[XI + (1-X)O] O-I=
dt
dS
t0.5+Kx-K
t0.5-Kx
-=C0
∆
∆
t0.5+Kx-K
t0.5+Kx
=C1
∆
∆
t0.5+Kx-K
t0.5-Kx-K
=C2
∆
∆
Modified Puls
Kinematic Wave
MuskingumMuskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Muskingum Notes :Muskingum Notes :
• The method assumes a single stage-discharge
relationship.
• In other words, for any given discharge, Q, there can
be only one stage height.
• This assumption may not be entirely valid for certain
flow situations.
• For instance, the friction slope on the rising side of a
hydrograph for a given flow, Q, may be quite different
than for the recession side of the hydrograph for the
same given flow, Q.
• This causes an effect known as hysteresis, which can
introduce errors into the storage assumptions of this
method.
Modified Puls
Kinematic Wave
MuskingumMuskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Estimating KEstimating K
• K is estimated to be the travel time through the reach.
• This may pose somewhat of a difficulty, as the travel
time will obviously change with flow.
• The question may arise as to whether the travel time
should be estimated using the average flow, the peak
flow, or some other flow.
• The travel time may be estimated using the kinematic
travel time or a travel time based on Manning's
equation.
Modified Puls
Kinematic Wave
MuskingumMuskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Estimating XEstimating X
• The value of X must be between 0.0 and 0.5.
• The parameter X may be thought of as a weighting coefficient
for inflow and outflow.
• As inflow becomes less important, the value of X decreases.
• The lower limit of X is 0.0 and this would be indicative of a
situation where inflow, I, has little or no effect on the storage.
• A reservoir is an example of this situation and it should be
noted that attenuation would be the dominant process
compared to translation.
• Values of X = 0.2 to 0.3 are the most common for natural
streams; however, values of 0.4 to 0.5 may be calibrated for
streams with little or no flood plains or storage effects.
• A value of X = 0.5 would represent equal weighting between
inflow and outflow and would produce translation with little
or no attenuation.
Modified Puls
Kinematic Wave
MuskingumMuskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Did you know?
Lag and K is a special case of
Muskingum -> X=0!

More Notes - MuskingumMore Notes - Muskingum
• The Handbook of Hydrology (Maidment, 1992)
includes additional cautions or limitations in the
Muskingum method.
• The method may produce negative flows in the initial
portion of the hydrograph.
• Additionally, it is recommended that the method be
limited to moderate to slow rising hydrographs being
routed through mild to steep sloping channels.
• The method is not applicable to steeply rising
hydrographs such as dam breaks.
• Finally, this method also neglects variable backwater
effects such as downstream dams, constrictions, bridges,
and tidal influences.
Modified Puls
Kinematic Wave
MuskingumMuskingum
Muskingum-Cunge
Dynamic
Modeling Notes

Muskingum Example ProblemMuskingum Example Problem
Time Inflow C0I2 C1I1 C2O1 Outflow
0 3 3
1 5
2 10
3 8
4 6
5 5
•A portion of the inflow hydrograph to a reach of channel is given
below. If the travel time is K=1 unit and the weighting factor is
X=0.30, then find the outflow from the reach for the period shown
below:

•The first step is to determine the coefficients in this problem.
•The calculations for each of the coefficients is given below:
t0.5+Kx-K
t0.5-Kx
-=C0
∆
∆
C0= - ((1*0.30) - (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.167
t0.5+Kx-K
t0.5+Kx
=C1
∆
∆
C1= ((1*0.30) + (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.667

C2= (1- (1*0.30) - (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.167
t0.5+Kx-K
t0.5-Kx-K
=C2
∆
∆
Therefore the coefficients in this problem are:
•C0 = 0.167
•C1 = 0.667
•C2 = 0.167

0 3 0.835 2.00 0.501 3
1 5
2 10
3 8
4 6
5 5
•The three columns now can be calculated.
•C0I2 = 0.167 * 5 = 0.835
•C1I1 = 0.667 * 3 = 2.00
•C2O1 = 0.167 * 3 = 0.501

•Next the three columns are added to determine the outflow at time
equal 1 hour.
•0.835 + 2.00 + 0.501 = 3.34
0 3 0.835 2.00 0.501 3
1 5 3.34
2 10
3 8
4 6
5 5

•This can be repeated until the table is complete and the outflow at
each time step is known.
0 3 0.835 2.00 0.501 3
1 5 1.67 3.34 0.557 3.34
2 10 1.34 6.67 0.93 5.57
3 8 1.00 5.34 1.49 8.94
4 6 0.835 4.00 1.31 7.83
5 5 3.34 1.03 6.14

Muskingum Example - Tenkiller
Look at R-5

Reach at Outlet
K = 2 hrs
X = 0.25

Let’s Alter X
K = 2 hrs.
K = 2 hrs
X = 0.25
K = 2 hrs
X = 0.5

And Again....
K = 2 hrs
X = 0.5
K = 2 hrs
X = 0.05

Muskingum-CungeMuskingum-Cunge
• Muskingum-Cunge formulation is similar to the
Muskingum type formulation
• The Muskingum-Cunge derivation begins with the
continuity equation and includes the diffusion form
of the momentum equation.
• These equations are combined and linearized,
Modified Puls
Kinematic Wave
Muskingum
Dynamic
Modeling Notes

“working equation”“working equation”
where :
Q = discharge
t = time
x = distance along channel
qx = lateral inflow
c = wave celerity
m = hydraulic diffusivity
Latcq
x
Q
x
Q
t
Q
+
∂
∂
=
∂
∂
+
∂
∂
2
2
µ
Modified Puls
Kinematic Wave
Muskingum
Dynamic
Modeling Notes

Muskingum-Cunge, cont...Muskingum-Cunge, cont...
• Method attempts to account for diffusion by taking
into account channel and flow characteristics.
• Hydraulic diffusivity is found to be :
O
BS
Q
2
=µ
• The Wave celerity in the x-direction is :
dA
dQ
C =
Modified Puls
Kinematic Wave
Muskingum
Dynamic
Modeling Notes

Solution of Muskingum-CungeSolution of Muskingum-Cunge
• Solution of the Method is accomplished by
discretizing the equations on an x-t plane.
QC+QC+QC+QC=
1+j
1+n
Q L4
n
1+j3
1+n
j2
n
j1
x)-2(1+
k
t
2x+
k
t
=C1
∆
∆
x)-2(1+
k
t
2x-
k
t
=C2
∆
∆
x)-2(1+
k
t
k
t
-x)-2(1
=C3
∆
∆
x)-2(1+
k
t
k
t
2
=C4
∆





 ∆
X
t
Modified Puls
Kinematic Wave
Muskingum
Dynamic
Modeling Notes

Calculation of K & XCalculation of K & X
c
x
=k
∆






∆
−=
xcBS
Q
X
O
1
2
1
Estimation of K & X is more “physically based”
and should be able to reflect the “changing”
conditions - better.
Modified Puls
Kinematic Wave
Muskingum
Dynamic
Modeling Notes

Muskingum-Cunge - NOTESMuskingum-Cunge - NOTES
• Muskingum-Cunge formulation is actually considered an
approximate solution of the convective diffusion equation.
• As such it may account for wave attenuation, but not for
reverse flow and backwater effects and not for fast rising
hydrographs.
• Properly applied, the method is non-linear in that the flow
properties and routing coefficients are re-calculated at
each time and distance step
• Often, an iterative 4-point scheme is used for the solution.
• Care should be taken when choosing the computation
interval, as the computation interval may be longer than
the time it takes for the wave to travel the reach distance.
• Internal computational times are used to account for the
possibility of this occurring.
Modified Puls
Kinematic Wave
Muskingum
Dynamic
Modeling Notes

Muskingum-Cunge ExampleMuskingum-Cunge Example
• The hydrograph at the upstream end of a
river is given in the following table. The
reach of interest is 18 km long. Using a
subreach length ∆x of 6 km, determine the
hydrograph at the end of the reach using the
Muskingum-Cunge method. Assume c =
2m/s, B = 25.3 m, So = 0.001m and no lateral
flow.
Time (hr) Flow (m3
/s)
0 10
1 12
2 18
3 28.5
4 50
5 78
6 107
7 134.5
8 147
9 150
10 146
11 129
12 105
13 78
14 59
15 45
16 33
17 24
18 17
19 12
20 10

• First, K must be determined.
• K is equal to :
c
x
K
∆
=
seconds3000
/2
/10006
=
×
=
sm
kmmkm
K
• ∆x = 6 km, while c = 2 m/s
Is “c” a
constant?

• The next step is to determine x.






∆
−=
xcBS
Q
x
O
1
2
1
•All the variables are known, with B =
25.3 m, So = 0.001 and ∆x =6000 m,
and the peak Q taken from the table.
253.0
/)6000)(2)(001.0)(3.25(
/150
1
2
1
3
3
=





−=
sm
sm
x
Time (hr) Flow (m3
/s)
0 10
1 12
2 18
3 28.5
4 50
5 78
6 107
7 134.5
8 147
9 150
10 146
11 129
12 105
13 78
14 59
15 45
16 33
17 24
18 17
19 12
20 10

• The coefficients of the Muskingum-Cunge method can now
be determined.
)1(2
2
1
x
K
t
x
K
t
C
−+
∆
+
∆
=
7466.0
)253.01(2
3000
7200
)253.0(2
3000
7200
1 =
−+
+
=C
Using 7200 seconds or 2 hrs for
timestep - this may need to be
changed

be determined.
)1(2
2
2
x
K
t
x
K
t
C
−+
∆
−
∆
=
4863.0
)253.01(2
3000
7200
)253.0(2
3000
7200
2 =
−+
−
=C

be determined.
)1(2
)1(2
3
x
K
t
K
t
x
C
−+
∆
∆
−−
=
232.0
)253.01(2
3000
7200
3000
7200
)253.01(2
3 −=
−+
−−
=C

be determined.
)1(2
2
4
x
K
t
K
t
C
−+
∆





 ∆
=
233.1
)253.01(2
3000
7200
3000
7200
2
4 =
−+






=C

• Then a simplification of the original formula can be made.
L
n
j
n
j
n
j
n
j QCQCQCQCQ 413
1
21
1
1 +++= +
++
+
• Since there is not lateral flow, QL = 0. The simplified
formula is the following:
n
j
n
j
n
j
n
j QCQCQCQ 13
1
21
1
1 +
++
+ ++=

• A table can then be created in 2 hour time steps similar to the
one below:
Time (hr) 0 km 6 km 12 km 18 km
0 10
2 18
4 50
6 107
8 147
10 146
12 105
14 59
16 33
18 17
20 10
22 10
24 10
26 10
28 10

• It is assumed at time zero, the flow is 10 m3
/s at each distance.
0 10 10 10 10
2 18
4 50
6 107
8 147
10 146
12 105
14 59
16 33
18 17
20 10
22 10
24 10
26 10
28 10

• Next, zero is substituted into for each letter to solve the
equation.
n
j
n
j
n
j
n
j QCQCQCQ 13
1
21
1
1 +
++
+ ++=
0
13
1
02
0
01
1
1 QCQCQCQ ++=

• Using the table, the variables can be determined.
0
13
1
02
0
01
1
1 QCQCQCQ ++=
0 10 10 10 10
2 18
4 50
6 107
8 147
10 146
12 105
14 59
16 33
18 17
20 10
22 10
24 10
26 10
28 10
=
=
=
0
1
1
0
0
0
Q
Q
Q 10
18
10

• Therefore, the equation can be solved.
0
13
1
02
0
01
1
1 QCQCQCQ ++=
smQ
Q
/89.13
)10)(2329.0()18)(4863.0()10)(7466.0(
31
1
1
1
=
−++=
0 10 10 10 10
2 18 13.89
4 50
6 107
8 147
10 146
12 105
14 59
16 33
18 17
20 10
22 10
24 10
26 10
28 10

• Therefore, the equation can be solved.
1
13
2
02
1
01
2
1 QCQCQCQ ++=
smQ
Q
/51.34
)89.13)(2329.0()50)(4863.0()18)(7466.0(
31
1
1
1
=
−++=
0 10 10 10 10
2 18 13.89
4 50 34.51
6 107
8 147
10 146
12 105
14 59
16 33
18 17
20 10
22 10
24 10
26 10
28 10

• This is repeated for the rest of the columns and the subsequent
columns to produce the following table. Note that when you
change rows, “n” changes. When you change columns, “j”
changes. Time (hr) 0 km 6 km 12 km 18 km
0 10 10 10 10
2 18 13.89 11.89 10.92
4 50 34.51 24.38 18.19
6 107 81.32 59.63 42.96
8 147 132.44 111.23 88.60
10 146 149.91 145.88 133.35
12 105 125.16 138.82 145.37
14 59 77.93 99.01 117.94
16 33 41.94 55.52 73.45
18 17 23.14 29.63 38.75
20 10 12.17 16.29 21.02
22 10 9.49 9.91 12.09
24 10 10.12 9.70 9.30
26 10 9.97 10.15 10.01
28 10 10.01 9.95 10.08

Look again at the Discretization
X
t

Things to Consider
• As the wave comes into a reach:
• the flow and depth go up and then down
• the wave speed and ALL coefficients would
then change...
• What effect would this have?

Full Dynamic Wave EquationsFull Dynamic Wave Equations
• The solution of the St. Venant equations is known as
dynamic routing.
• Dynamic routing is generally the standard to which
other methods are measured or compared.
• The solution of the St. Venant equations is generally
accomplished via one of two methods : 1) the method
of characteristics and 2) direct methods (implicit and
explicit).
• It may be fair to say that regardless of the method of
solution, a computer is absolutely necessary as the
solutions are quite time consuming.
• J. J. Stoker (1953, 1957) is generally credited for
initially attempting to solve the St. Venant equations
using a high speed computer.
Modified Puls
Kinematic Wave
Muskingum
Muskingum-Cunge
DynamicDynamic
Modeling Notes

Dynamic Wave SolutionsDynamic Wave Solutions
• Characteristics, Explicit, & Implicit
• The most popular method of applying the implicit
technique is to use a four point weighted finite
difference scheme.
• Some computer programs utilize a finite element
solution technique; however, these tend to be more
complex in nature and thus a finite difference technique
is most often employed.
• It should be noted that most of the models using the
finite difference technique are one-dimensional and that
two and three-dimensional solution schemes often revert
to a finite element solution.
Modified Puls
Kinematic Wave
Muskingum
Muskingum-Cunge
DynamicDynamic
Modeling Notes

Dynamic Wave SolutionsDynamic Wave Solutions
• Dynamic routing allows for a higher degree of accuracy
when modeling flood situations because it includes
parameters that other methods neglect.
• Dynamic routing, when compared to other modeling
techniques, relies less on previous flood data and more
on the physical properties of the storm. This is
extremely important when record rainfalls occur or
other extreme events.
• Dynamic routing also provides more hydraulic
information about the event, which can be used to
determine the transportation of sediment along the
waterway.
Modified Puls
Kinematic Wave
Muskingum
Muskingum-Cunge
DynamicDynamic
Modeling Notes

Courant Condition?Courant Condition?
• If the wave or hydrograph can travel through the
subreach (of length Δx) in a time less than the
computational interval, Δt, then computational
instabilities may evolve.
• The condition to satisfy here is known as the Courant
condition and is expressed as :
c
dx
dt ≤
Modified Puls
Kinematic Wave
Muskingum
Muskingum-Cunge
Dynamic
Modeling NotesModeling Notes

Some DISadvantagesSome DISadvantages
• Geometric simplification - some models are designed to use
very simplistic representations of the cross-sectional
geometry. This may be valid for large dam breaks where very
large flows are encountered and width to depth ratios are
large; however, this may not be applicable to smaller dam
breaks where channel geometry would be more critical.
• Model simulation input requirements - dynamic routing
techniques generally require boundary conditions at one or
more locations in the domain, such as the upstream and
downstream sections. These boundary conditions may in the
form of known or constant water surfaces, hydrographs, or
assumed stage-discharge relationships.
• Stability - As previously noted, the very complex nature of
these methods often leads to numeric instability. Also,
convergence may be a problem in some solution schemes.
For these reasons as well as others, there tends to be a
stability problem in some programs. Often times it is very
difficult to obtain a "clean" model run in a cost efficient
manner.
Modified Puls
Kinematic Wave
Muskingum
Muskingum-Cunge
Dynamic
Modeling NotesModeling Notes

Channel Routing Methods

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