Flow Routing
 Flow routing is a procedure to determine the time and magnitude of
flow (i.e. the flow hydrograph) at a point on a watercourse from
known or assumed hydrographs at one or more points upstream.
 We want to predict what a downstream
hydrograph would be if we know the
upstream hydrograph
 If the flow is a flood, the procedure is
specifically called flood routing
 Lumped flow routing: flow is
calculated as a function of time alone
at a particular location. Also called
hydrologic routing.
 Distributed flow routing: flow is calculated as a function of space and
time throughout the space. Also called hydraulic routing.
Hydrologic Routing
 Continuity equation:
dS
 I (t )  Q (t )
dt
 It is not enough to know inflow hydrograph, I(t) to solve for the
outflow hydrograph; Q(t) as S(t) are also unknown.
 A second relationship or storage function is needed: S = f (I,Q)
 In reservoir routing level pool routing is commonly employed where
S=f(Q).
 Q and S are related to reservoir water level, h
 Runge-Kutta method is an alternative to level pool routing. It is a
little bit more complicated, but does not require special computation
of S=f(Q) and is more physics based
 In channel flow routing Muskingum Method is commonly used where
S is linearly related to I and Q.
Flow Routing
 The relationship between outflow and storage could be either invariable or
variable.
 Invariable storage function applies to
reservoirs with a horizontal water
surface. Such reservoirs are wide and
deep with low flow velocities
 When a reservoir has a horizontal
water surface there is a unique S=f(Q)
relationship. For such reservoirs Qpeak
occurs when I=Q, because Smax occurs
when dS/dt =I-Q =0
 A variable S-Q relationship applies to long, narrow reservoirs and to open
channel or streams, where the water surface profile could be significantly curved
due to backwater effects.
 The amount of storage due to backwater depends on the time rate of change of
flow through the system, thus S-Q is no longer unique.
 Due to retarding effect of backwater, Qpeak usually occurs after I=Q
Level Pool Routing
 Let’s reconsider the continuity equation
dS (t )
 I (t )  Q (t )
dt
 In discrete form, for a time interval Dt:
DS
 I (t )  Q (t ) 
Dt
S j 1  S j
Dt

I j  I j 1
2

Q j  Q j 1
2
 The values Ij and Ij+1 are known. The values Qj and Sj are known at the
jth time interval from calculation during previous time interval.
 Hence, two unknowns, Qj+1 and Sj+1. Rearranging:
 2S j 1

 2S j




 Q j 1   I j  I j 1   
 Q j 
 Dt

 Dt

 2S j 1

 2S j


 Q j 1   I j  I j 1   
 Q j 
 Dt

 Dt

 In order to calculate Qj+1 a Q-S
function relating (2S/Dt+Q)
and Q is needed.
 S-h: topographic maps or field
survey.
 Q-h: hydraulic equations
 For Dt use time interval of I(t)
 To set up the data required for the
next time interval, the value of
2Sj+1/Dt +Qj+1 is calculated by
 2S j 1
  2S j 1




 Q j 1   
 Q j 1   2Q j 1
 Dt
  Dt

Level Pool Routing
 2S j 1

 2S j


 Q j 1   I j  I j 1   
 Q j 
 Dt

 Dt

1. Find Ij + Ij+1, 2Sj/Dt - Qj
2. Add (Ij + Ij+1) and (2Sj/Dt - Qj) to find (2Sj+1/Dt + Qj+1)
3. Use the (2S/Dt+Q) vs. Q relation to find the corresponding Qj+1
4. Find (2Sj+1/Dt - Qj+1) by subtracting 2Qj+1 from (2Sj+1/Dt + Qj+1)
5. Repeat calculations for next discharge value
 The inconvenience of this method is that (2S/Dt+Q) vs. Q
relationship is dependent on Dt, i.e. for different Dt’s the relationship
must be reconstructed.
Example
 A detention pond has vertical sites
and is 1 acre in area. The Q-h and
S-h relationships are given in the
table. Assume S(t=0) = 0
 Use level pool routing method to
calculate the outflow hydrograph
from the inflow hydrograph given
in the table on next slide.
 For all elevations:
A = 1 acre = 43,560 ft2
S = 43,560H
 2S j 1

 2S j


 Q j 1   I j  I j 1   
 Q j 
 Dt

 Dt

* Dt = 10 min = 600 sec
 2S j 1

 2S j


 Q j 1   I j  I j 1   
 Q j 
 Dt

 Dt

 S1= Q1= 0
 The value of Qj+1 is found
by linear interpolation. If
(x,y) is between (x1,y1)
and (x2,y2):
y  y1 
Q2  0 
( y2  y1 )
( x  x1 )
( x2  x1 )
(3  0)
(60  0)  2.4
(76  0)
River Routing
 Routing in natural channels is complicated by the fact that storage is not a
function of outflow alone.
 The storage in a stable river reach can be expected to depend primarily on the
inflow to and outflow from the reach, and secondarily on the hydraulic
characteristics of the channel cross section.
 Two types of storages can be defined in natural channels:
 Prism storage: Storage beneath a line
parallel to the streambed drawn from the
downstream end of the reach
 Wedge storage: Storage between this
parallel line and the actual profile.
Wedge storage is (+) during raising flow,
and (-) during falling.
Muskingum Method
 Total storage = S
= KQ + KX(I-Q)
= K[XI + (1-X)Q]
X: dimensionless constant between 0
and 0.5 with typical value of 0.2
K:travel time of a flood wave
through channel
 Sj= K[XIj + (1-X)Qj]

and Sj+1= K[XIj+1 + (1-X)Qj+1]
Sj+1 - Sj = K[XIj+1 + (1-X)Qj+1 - XIj - (1-X)Qj]
 Continuity: dS/dt = I – Q  Sj+1- Sj = {(Ij+Ij+1) - (Qj+Qj+1)}Dt/2
  Qj+1=C1Ij+1 + C2Ij + C3Qj where
C1 
Dt  2 KX
2 K (1  X )  Dt
C2 
Dt  2 KX
2 K (1  X )  Dt
C3 
2 K (1  X )  Dt
2 K (1  X )  Dt
Muskingum Method
 Note that C1 + C2 + C3 =1. Therefore it provides a check
 If X = 0, then S = KQ, i.e. storage is not a function of inflow and becomes
reservoir routing.
 If X = 0.5, then S = K(I+Q)/2, i.e. inflow and outflow have equal weights. In
other words, no change in flood hydrograph.
 Great accuracy in determining X may not be necessary, because results are
relatively insensitive to the value of this parameter.
 K = L/ck, where ck is wave celerity given by dQ/dA or dx/dt.
 Using Manning’s equation for a wide channel, it can be shown that ck = 5u/3
where u is average flow velocity.
3  nB
K
5  S0
23
 K can be approximated as
35

 Q 2 5 L


Example
Qj+1=C1Ij+1 + C2Ij + C3Qj
K= 2.3 hr, X= 0.15, Dt=1 hr
Dt  2 KX
2 K (1  X )  Dt
1  2( 2.3)(0.15)

2( 2.3)(1  0.15)  1
 0.063
C1 
C2 
Dt  2 KX
 0.344
2 K (1  X )  Dt
C3 
2 K (1  X )  Dt
 0.593
2 K (1  X )  Dt
C1  C2  C3  1
Q2  C1 I 2  C2 I1  C3Q1
 0.063(137)  0.344(93)
 0.593(85)  91 cfs