FLOOD ROUTING

Flood Routing

• We may have to find the magnitude of flood and flood hydrograph parameters at a particular space due to flood at another space or at same space– Of course both the spaces are hydrologically

connected– The space may be very nearer also

• Upstream to downstream of a reservoir – storage is large• One section of the river to another section of the river –

long distance

– This way of finding the hydrologic route of flood from one space to another space is called flood routing

Flood Routing (contd…)

• The flood hydrograph is in fact a wave– The Stage-discharge relationship represent the

passage of waves – As this waves moves down the river

• The shape of wave gets modified– Channel storage – Resistance– Lateral addition/withdrawal

• When the flood passes through a reservoir– The peak is attenuated– Time base is enlarge

» Due to effect of storage

Flood Routing (Contd…)• Thus the main purpose of reservoir flood routing

is to safely dispose the flood magnitude by reducing the peak and increasing the base time.

• Determination of flood hydrograph at a river section– By utilizing the data of flood flow at one or more

upstream sections

• Flood routing is more useful in:– Flood forecasting– Flood protection– Reservoir design– Spillway design

Applications (Types of flood routing)

• Reservoir Routing

• Channel Routing

Reservoir Routing• To predict the variations of reservoir elevation

and outflow discharge with time– Study the effect of a flood wave entering a reservoir

• Volume-elevation characteristic of reservoir• Outflow-elevation relationship for the spillways and other

outlets

• Reservoir Routing is Essential– Design of the capacity of spillway/other reservoir

outlets– Location and sizing of the capacity of reservoirs to

meet specific requirements

Channel Routing

• Change in shape of Hydrograph as it travels down a channel is studied

– To predict the flood hydrograph at various sections of the reach

– Information on the flood-peak attenuation and the duration of high-water levels

• Flood forecasting• Flood protection• Flood operations

Classification of Routing

• Hydrologic Routing – employs the continuity equation

• Hydraulic Routing - employs the continuity equation together with the momentum equation of unsteady flow– St. Venant Equations

Flood Routing

Hydrologic Routing Hydraulic Routing

(Based on Continuity Equation) Based on Momentum equation

Reservoir Routing Channel Routing

• The change in storage is the difference between the inflow and outflow

I is Inflow, Q is outflow and S is the storage

• In a small time interval (Δt)

I is average Inflow, Q is average outflow during the time interval

• If storage at beginning is S1 and at end is S2 during time Δt

In this time interval the hydrograph is linear and is smaller than the transit of the flood wave through the reach.

StQtI

122121

22SSt

QQt

II

…..2

…..3

Basic equations used in hydrologic routing

dt

dSQI …..1

• Differential form of Continuity Equation

• Equation of motion for a flood wave

0

t

yT

x

Q

fSSt

V

gx

V

g

V

x

y

01

…..4

…..5

St. Venant equation (based on application of momentum equation

(used in hydraulic routing)

Hydrologic Storage Routing(Level Pool Routing)

horizontal water surface is assumed in the reservoir

• Uncontrolled spillway provided2/32

3

2HLgCQ ed …..6

Data required for reservoir routing

• Storage volume vs elevation for the reservoir

• Water-surface elevation vs outflow and hence storage vs outflow discharge

• Inflow hydrograph, I = I (t)

• Initial values of S, I and Q at time t = 0

Methods for Flood Routing Through a Reservoir

• Modified Pul’s Method

• Goodrich Method

• Standard Fourth – Order Runge-Kutta Method (SRK)

Goodrich MethodSemi-graphical Method

• On rearranging equation 3

• On collecting known and initial values

• In the above equation the starting inflow and end inflow at time period t is known (read it from the inflow hydrograph), and the initial storage and discharge is also known

• Then estimate the value remember both are unknown quantities

t

S

t

SQQII

12

2121

22

22

11

21

22Q

t

SQ

t

SII

…..7

…..8

222

Qt

S

Contd….• To know the discharge, we need a graph between elevation Vs

– Thus called as semi graphical method

– This quantity is called storage-elevation-discharge data

• The graph gives the relationship between discharge and elevation– From graph estimate the elevation

– From elevation estimate the discharge

Qt

S2

Is flood routing is too confusing

• The following problem will help to understand this method

? Route the following flood hydrograph through the reservoir by Goodrich method:

Inflow hydrograph

Elevation Storage (106 m3) Outflow discharge (m3/s)

100.00 3.350 0

100.50 3.472 10

101.00 3.880 26

101.50 4.383 46

102.00 4.882 72

102.50 5.370 100

102.75 5.527 116

103.00 5.856 130

Time (h) 0 6 12 18 24 30 36 42 48 54 60 66

Inflow (m3/s) 10 30 85 140 125 96 75 60 46 35 25 20

The initial conditions are when t = 0, the reservoir elevation is 100.60 m.

The storage-elevation-discharge data is as follows:

Step 1: Construct the storage-elevation-discharge curve

• Assume a time period of 6 hr (t )– Equal to time of discharge

measurement in the inflow hydrograph

– Estimate the values of

• Plot a graph– elevation-Vs-discharge – Elevation-Vs-

• For initial time period t=0 find the Q2 and

From the graph

Elevation Storage (106 m3)

Outflow dischar

ge (m3/s)

(m3/s)  

100 3.35 0310.19

100.5 3.472 10331.48

101 3.88 26385.26

101.5 4.383 46451.83

102 4.882 72524.04

102.5 5.37 100597.22

102.75 5.527 116627.76

103 5.856 130672.22

Qt

S2

Qt

S2

Qt

S2

Qt

S2

300 350 400 450 500 550 600 650 700Storage -D ischarge cu.m /s

100.00

100.50

101.00

101.50

102.00

102.50

103.00

Res

ervo

ir w

ater

leve

l Ele

vatio

n (m

)

Storage-D ischarge Vs E levation

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

O utflow cu.m /s

D ischarge Vs E levation

Time(h)

(m3/s) (m3/s) (m3/s)

Elevation (m) DischargeQ

(m3/s)

0 10 340 100.6 12

6 30 40 316 =(340-2*12) 356 =(40+316)

Find this from graph

12 85 115

18 140 225

24 125 265

30 96 221

36 75 171

42 60 135

48 46 106

54 35 81

60 25 60

66 20 45

I 21 II

112

Qt

S

222

Qt

S

Solution

300 350 400 450 500 550 600 650 700Storage -D ischarge cu.m /s

100.00

100.50

101.00

101.50

102.00

102.50

103.00

Res

ervo

ir w

ater

leve

l Ele

vatio

n (m

)

Storage-D ischarge Vs E levation

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

O utflow cu.m /s

D ischarge Vs E levation

I 21 II

Qt

S2

Qt

S2Time(h)

(m3/s) (m3/s) (m3/s)

Elevation (m) DischargeQ

(m3/s)

0 10 340 100.6 12

6 30 40 340-2*12=316 40+316 =356

100.74 17

12 85 115 356-2*17=322

322+115 =437

From graph find this

18 140 225

24 125 265

30 96 221

36 75 171

42 60 135

48 46 106

54 35 81

60 25 60

66 20 45

Solution

300 350 400 450 500 550 600 650 700Storage -D ischarge cu.m /s

100.00

100.50

101.00

101.50

102.00

102.50

103.00

Res

ervo

ir w

ater

leve

l Ele

vatio

n (m

)

Storage-D ischarge Vs E levation

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160

O utflow cu.m /s

D ischarge Vs E levation

I 21 II

Qt

S2

Qt

S2Time(h)

(m3/s) (m3/s) (m3/s)

Elevation (m) DischargeQ

(m3/s)

0 10 340 100.6 12

6 30 40 340-2*12=316 40+316 =356

100.74 17

12 85 115 356-2*17=322

322+115 =437

101.38 40

18 140 225 437-2*40 = 357

357+225 = 582

…

24 125 265

30 96 221

36 75 171

42 60 135

48 46 106

54 35 81

60 25 60

66 20 45

Solution

I 21 II

Qt

S2

Qt

S2Time(h)

(m3/s) (m3/s) (m3/s)

Elevation (m) DischargeQ

(m3/s)

0 10 340 100.6 12

6 30 40 316 =(340-2*12) 356 100.74 17

12 85 115 322 437 101.38 40

18 140 225 357 582 102.50 95

24 125 265 392 657 102.92 127

30 96 221 403 624 102.70 112

36 75 171 400 571 102.32 90

42 60 135 391 526 102.02 73

48 46 106 380 486 101.74 57

54 35 81 372 453 101.51 46

60 25 60 361 421 101.28 37

66 20 45 347 392 101.02 27

335

Solution

What we achieved through this flood routing

0 10 20 30 40 50 60 70 80Tim e in hrs

0.00

20.00

40.00

60.00

80.00

100.00

120.00

140.00

160.00

Inflo

w/o

utflo

w in

cu.

m/s In flow hydrograph

O utflow hydrograph

1. The peak discharge magnitude is reduced, this is called attenuation.

2. The peak of outflow gets shifted and is called as lag

3. The difference in rising limb shows the reservoir is storing the water

4. The difference in receding limb shows the reservoir is depleted.

5. When the outflow is through uncontrolled spillway, the peak of outflow always occurs at point of inflection of inflow hydrograph and also is the point at which the inflow and outflow hydrograph intersect.

Attenuation

Lag

Res

ervo

ir st

orin

g

Res

ervo

ir D

eple

ting

Hydrologic Channel Routing

• In reservoir routing storage was a unique function of the outflow discharge, S = f(Q)

• Here, Storage is a function of both outflow and inflow discharges

• Therefore different routing method needed

• River flow during floods belongs to the category of gradually varied unsteady flow

• Water is not only parallel to the channel bottom, but also varies with time

Total Volume of Storage in a channel during flood

• Prism Storage: vol. that would exist if uniform flow occurred at the downstream depth

= function (outflow)

• Wedge storage: wedge like vol. formed between the actual water surface profile and the top surface of the prism storage

= function (inflow)

• The total storage in the channel is given by:

mm QxxIKS 1 …..9

K, and x are coefficients and m-is a constant

Muskingum Equation• One of the most popular channel routing

• Uses the hydrologic spatially lumped form of the continuity equation

• First applied to Muskingum river in Ohio state, USA

• Tributary of Ohio river

• Length 179 km (111 mi )• Basin area 20,852 km² (8,051 mi² )

Muskingum Equation (Contd…)• Using m =1.0, equation 9. reduces to a linear relationship

for S in terms of I and Q as

• x – weighting factor varies bet. 0 to 0.5

• When x = 0, storage function is discharge only– Linear storage or linear reservoir– x = 0.5 both inflow and outflow are equally imp. In determining

storage– K – storage-time constant ~ time of travel of a flood wave

through the channel reach

QxxIKS 1

QKS

…..10

…..11

Estimation of K and x

• Like reservoir routing in channel routing also we can draw inflow-outflow hydrograph through a channel reach

• The increment in storage at any time t due to a small time period t can be calculated.

• The summation of the various incremental storage gives us the channel storage Vs time relationship

St

QQt

II

22 2121

Estimation of K and x (contd…)

• Once this storage Vs time is known for a reach

• Assume a value of x and estimate for various time intervals

• Draw the graph between the storage and if the assumed x is correct we will get a linear line, else a loop will be formed

• By trial and error find the value of x until a straight line is formed

• Inverse slope of the line will give the value of K

Let us see how to estimate this using an example

QxxI 1

QxxI 1

? The following inflow and outflow hydrographs were observed in a river reach. Estimate the values of K and x applicable to this reach for use in the Muskingum equation.

Time (h) 0 6 12 18 24 30 36 42 48 54 60 66

Inflow (m3/s) 5 20 50 50 32 22 15 10 7 5 5 5

Outflow (m3/s) 5 6 12 29 38 35 29 23 17 13 9 7

Time I Q (I-Q) Avg. ΔS= S = ΣΔS

[xI + (1-x) Q] (m3/s)

(h) (m3/s) (m3/s)   (I-Q) Col. 5 x Δt        

          (m3/s.h)   x = 0.35 x = 0.3 x = 0.25

0 5 5 0     0 5 5 5

6 20 6 14 7 42 42 10.9 10.2 9.5

12 50 12 38 26 156 198 25.3 23.4 21.5

18 50 29 21 29.5 177 375 36.35 35.3 34.25

24 32 38 -6 7.5 45 420 35.9 36.2 36.5

30 22 35 -13 -9.5 -57 363 30.45 31.1 31.75

36 15 29 -14 -13.5 -81 282 24.1 24.8 25.5

42 10 23 -13 -13.5 -81 201 18.45 19.1 19.75

48 7 17 -10 -11.5 -69 132 13.5 14 14.5

54 5 13 -8 -9 -54 78 10.2 10.6 11

60 5 9 -4 -6 -36 42 7.6 7.8 8

66 5 7 -2 -3 -18 24 6.3 6.4 6.5

0 100 200 300 400 500Storage from continu ity equation (cu.m /s) hr

5.00

10.00

15.00

20.00

25.00

30.00

35.00

40.00

[x I+

(1-x

)Q] c

u.m

/s

x=0.35

x=0.3

x=0.25

x=0.2

Muskingum Method of Routing• For a given channel reach

– K and x are assumed to be constant

• will not change with respect to time

• But changes when the shape of channel changes in the reach

• For a given channel reach by selecting a routing interval Δt and using the Muskingum equation the change in storage is

• The continuity eqn. for the reach is

121212 1 QQxIIxKSS

tQQ

tII

SS

22

121212

…..12

…..13

0 100 200 300 400 500Storage from continuity equation (cu.m /s) hr

5.00

10.00

15.00

20.00

25.00

30.00

35.00

40.00

[x I+

(1-x

)Q] c

u.m

/s

x=0.35

x=0.2

• On simplifying the equations 12 and 13

• where

1322112 QCICICQ …..14

tKxK

tKxC

5.0

5.02tKxK

tKxC

5.0

5.01

tKxK

tKxKC

5.0

5.03

0.1321 CCC

• In general form for the n th time step

n=2,3,4,……

• This equation is known as Muskingum Routing Equation

• It provides a simple linear equation for channel routing

13211 nnnn QCICICQ …..15

Procedure to use Muskingum Equation to route a given inflow hydrograph through a reach

– Knowing K and x, select an appropriate value of Δt

– Calculate C1, C2 and C3

– Starting from the initial conditions I1, Q1 and known I2 at the end of the first time step Δt calculate Q2 by Muskingum equation (14 or 15)

– The outflow calculated in above step becomes the known initial outflow for the next time step. Repeat the calculations for the entire inflow hydrograph

? Route the following hydrograph through a river reach for which K = 12.0 h and x = 0.20. At the start of the inflow flood, the outflow discharge is 10 m3/s.

Time (h) 0 6 12 18 24 30 36 42 48 54

Inflow (m3/s) 10 20 50 60 55 45 35 27 20 15

Time (h) I (m3/s) 0.429 I1 0.048 I2 0.523 Q1 Q (m3/s)

0 10       10

6 20 4.29 0.96 5.23 10.48

12 50 8.58 2.40 5.48 16.46

18 60 21.45 2.88 8.61 32.94

24 55 25.74 2.64 17.23 45.61

30 45 23.60 2.16 23.85 49.61

36 35 19.31 1.68 25.94 46.93

42 27 15.02 1.30 24.54 40.86

48 20 11.58 0.96 21.37 33.91

54 15 8.58 0.72 17.74 27.04

Flood Control• All the measures adopted to reduce damages

to life and property by floods

• Flood control measures– Structural Methods

• Storage and detention reservoirs• Levees (flood embankments)• Channel improvement• Flood ways (new channels)• Soil conservation

– Non-structural methods• Flood plain zoning• Flood warning evacuation and relocation

Self Study

• Hydrologic routing –other famous methods

• Hydraulic routing

• Flood forecasting

• Solve all the problems in this section

Current research in this area

1. Suitable methods for flood forecasting

2. Estimation of Muskingum parameters using AI (mostly using Genetic Algorithms and Genetic Programming).

3. Use of Nash IUH model for flood forecasting

4. Use of AI for estimation of Nash parameters