Flood Routing

August 7, 2015

Flood Routing

Flood Routing is the technique of determining the flood hydro-graph at a section of a river by utilizing the data of flood flow ofone or more upstream sections. The hrydrologic annalysis of prob-lems such as flood forecasting, flood protection, reservior designand spillway design invariably include flood routing. In these ap-plications two broad categories of routing can be recognised. Theseare

1. Reservoir Routing and

2. Channel Routing.

Contd...

Reservoir Routing: In reservoir routing the effect of a flood wave enteringa reservoir is studied. This form of reservoir routing is essential (i) in thedesign of the capacity of spillways and other reservoir outlet structures. and(ii) in the location and sizing of the capacity of reservoirs to meet the specificrequirements.

Channel Routing: In channel routing the change in shape of a hydrographas it travels down a channel is studied. Information of the flood-peak at-tenuation and the duration of high water levels obtained by channel routingis of utmost importance in flood-forecasting operations and flood-protectionworks.

A variety of routing procedures are available and they can be broadly classi-fied as i) Hydrologic Routing and ii) Hydraulic Routing.

Basic Equations

The passege of flood hydrograph through a reservoir or a channel reach isan unsteady-flow phenomenon.It is classified in open channel Hydraulics asgradually varied unsteady flow. The equation of continuity used in all hy-drologic routing as the primary equation i.e.

I − Q =dSdt

(1)

where I= inflow rate, Q= outflow rate andS= storage. Alternatively,

I4t − Q4t = 4S (2)

where I =average inflow in time 4t,Q =average outflow in time 4t,and4S =change in storage.or, (

I1 + I22

)4t −

(Q1 + Q2

2

)4t = S2 − S1 (3)

The time interval 4t must be shorter than the time of transit of the floodwave through the reach.

Hydraulic Routing

Flood Routing by St. Venant Equation

The basic differential equation used in hydraulic routing for unsteady flow ina reach is given by

I Continuity Equation

∂Q∂x

+∂A∂t− q = 0 (4)

I Momentum equation

∂Q∂t

+∂

∂x

(Q2

A

)= gA (S0 − Sf )− gA

∂h∂x

(5)

q=lateral inflow; Q=discharge in the channel; A=area of flow in the channel,S0=bed slope; S f = slope of the energy line.

Hydrologic Storage Routing

Level Pool Routing

It is required to find out the variation of S, h, and Q with time, i.e. FindS=S(t), Q=Q(t) and h=h(t)given I= I(t).If an uncontrolled spillway is provided in a reservoir, Typically

Q =23Cd√

2gLeH32 = Q(h)

where H= head over the spillway, Le =effective length of the spillway crestand Cd= Coefficient of discharge.For the reservoir the following data shouldbe known.

I Storage vs Elevation characteristics of thereservoirI Outflow vs Elevation relationship of the spillway and hence outflow vs

storage relationshipI The inflow hydrograph

I Initial values of inflow, outflow, and storage at time t=0

Modified Pul’s Method

Eqn. (3) is rearraged as(I1 + I2

2

)4t +

(S1 −

Q14t2

)=

(S2 +

Q24t2

)(6)

Contd...

1. Prepare a curve of(S + Q4t

2

)vs elevation . Here ∆t is any chosen

interval.2. On the same plot prepare a curve of outflow discharge vs elevation.3. The storage, elevation and outflow discharge at the starting of routing

are known. For the first time interval ∆t,( I1+I2

2

)and

(S1 − Q14t

2

)are

known and hence by Eq (6) the term(S2 + Q24t

2

)is determined.

4. The water-surface elevation corresponding to(S2 + Q24t

2

)is found by

using the plot of step(1). The outflow discharge Q2 at the end of thetime step ∆t is found from plot of step (2).

5. Deducting Q2∆t from(S2 + Q24t

2

)Gives

(S − Q4t

2

)1for the

beginning of the next time step.

6. The procedure is repeated till the entire inflow hydrograph is routed.

Example 1

A reservoir has the following elevation, discharge and storage relationships:Elevation (m) Storage (106 m3) Outflow Discharge (m3/s)

100.00 3.35 0

100.50 3.472 10

101.00 3.88 26

101.50 4.383 46

102.00 4.882 72

102.50 5.37 100

102.75 5.527 116

103.00 5.856 130

When the reservoir level was at 100.50 m the following flood hydrographentered the reservoir.

Time(h) 0 6 12 18 24 30 36 42 48 54 60 66 72

Discharge(m3/s) 10 20 55 80 73 58 46 36 55 20 15 13 11

Route the flood and obtain (i) the outflow hydrograph and (ii) the reservoirelevation vs time curve during the passage of the flood wave.

Solution

A time interval of 4t = 6h is choosen. From the available data the elevation-discharge-

(S + Q4t

2

)is prepared.

4t = 6 x 60 x 60 = 0.0216 x 106s = 0.0216 Ms

Elevation (m) 100.00 100.50 101.00 101.50 102.00 102.50 102.75 103.00

Discharge Q (m3/s) 0 10 26 46 72 100 116 130(S + Q4t

2

) (Mm3

)3.35 3.58 4.16 4.88 5.66 6.45 6.78 7.26

Solution

Contd...

At the start of the routing, elevation= 100.5 m, Q=10 m3/s , and(S − Q4t

2

)= 3.362 Mm3. Starting from this value of

(S − Q4t

2

), Eq (6)

is used to get(S + Q4t

2

)at the end of first time step of 6h as

(S + Q4t

2

)2 = (I1 + I2) 4t

2 +(S − Q4t

2

)1

= (10 + 20) x 0.02162 + (3.362) = 3.686

Now, the water surface elevation corresponding to(S + Q4t

2

)= 3.686 Mm3is

100.62 m and and the corresponding outflow discharge Q is 13 m3/s. For thenext step, initial value of(

S − Q4t2

)=(S + Q4t

2

)of the previous time step -Q4t

= (3.686− 13x0.0216) = 3.405

The process is repeated for the entire duration of inflow hydrograph in atabular form as shown in the following table.

Contd...

Time (h) I(m3/s) (I1 + I2) (I1 + I2).4t [(S-4tQ/2)] [(S+4tQ/2)] Elevation(m) Discharge Q (m3/s)

0 10 100.5 10

15 0.324 3.362 3.636

6 20 100.62 13

37.5 0.81 3.405 4.215

12 55 101.04 27

67.5 1.458 3.632 5.09

18 80 101.64 53

76.5 1.6524 3.945 5.597

24 73 101.96 69

65.5 1.4148 4.107 5.522

30 58 101.91 66

52 1.1232 4.096 5.219

36 46 101.72 57

41 0.8856 3.988 4.874

Contd...

Time (h) I(m3/s) (I1 + I2) (I1 + I2).4t [(S-4tQ/2)] [(S+4tQ/2)] Elevation(m) Discharge Q (m3/s)

42 36 101.48 48

31.75 0.6858 3.902 4.588

48 27.5 101.3 37

23.75 0.513 3.789 4.302

54 20 100.1 25

17.5 0.378 3.676 4.054

60 15 100.93 23

14 0.3024 3.557 3.859

66 13 100.77 18

12 0.2592 3.47 3.729

72 11 100.65 14

3.427

Goodrich MethodAnother Popular method of hydrologic reservoir routing, known as Goodrichmethod utilizes Eq. (3) rearranged as

I1 + I2 − Q1 − Q2 =2S2

4t− 2S1

4twhere suffixes 1 and 2 stand for the values at the beginning and end of atime step respectively. Collecting the known and initial values together,

(I1 + I2) +

(2S1

4t− Q1

)=

(2S2

4t+ Q2

)(7)

For a given time step, the left-hand side of Eqn. (7) is known and theterm

(2S4t + Q

)2 is determined by using Eqn. (7). For the next time step,[(

2S4t

)2− 2Q2

]of the previous time step=

(2S4t − Q

)1 for use at the initial

values.

Example 2

Route the following flood hydrograph through the reservoir of Example 1 byGoodrich method.Elevation (m) Storage (106m3) Outflow Discharge (m3/s)

100.00 3.35 0

100.50 3.472 10

101.00 3.88 26

101.50 4.383 46

102.00 4.882 72

102.50 5.37 100

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 100.60m.

Solution

Elevation (m) 100.0 100.5 101.0 101.5 102.0 102.5 102.75 103.0

Outflow (m3/s) 0 10 26 46 72 100 116 130

2S4t + Q 310.2 331.5 385.3 451.8 524.0 597.2 627.8 672.2

Solution

At time t=0, Elevation=100.6 m and Q = 12m3/s and(2S4t + Q

)= 340m3/s(

2S4t − Q

)1

= 340− 24 = 316m3/s

For the first time interval of 6 h,I 1 = 10, I2 = 30,Q1 = 12and(

2S4t + Q

)2

= (10 + 30) + 316= 356m3/s

From fig the reservoir elevation for this(

2S4t + Q

)2is 100.74 m and discharge

is 17 m3/s.

For the next time increment(2S4t − Q

)1

= 356− 2x17= 322m3/s

Contd...

Time (h) I(m3/s) (I1+I2) [(2S/4t) -Q] [(2S/4t)+Q] Elevation(m) Discharge Q (m3/s)

0 10 340 100.6 12

40 316 356

6 30 100.74 17

115 322 437

12 85 101.38 40

225 357 582

18 140 102.5 95

265 392 657

24 125 102.92 127

221 403 624

30 96 102.7 112

171 400 571

Contd...

Time (h) I(m3/s) (I1+I2) [(2S/4t) -Q] [(2S/4t)+Q] Elevation(m) Discharge Q (m3/s)

135 391 526

42 60 102.02 73

106 380 486

48 46 101.74 57

81 372 453

54 35 101.51 46

60 361 421

60 25 101.28 37

45 347 392

66 20 101.02 27

335

Attenuation

Owing to the storage effect, the peak of the outflow hydrograph will besmaller than that of the inflow hydrograph. This reduction in peak valueis known as attenuation. Further the peak of the outflow occurs after thepeak of the inflow; the time difference between the two peaks is known as lagThe attenuation and lag of a flood hydrograph at a reservoir are two veryimportant aspects of reservoir operating under a flood control criteria. Byjudicious management of the initial reservoir level at the time of arrival ofa critical flood, considerable attenuating of the floods can be achieved. Thestorage capacity of the reservoir

Contd...

lag

Attenuation

Time

Dis

ch

ag

re

lag

Attenuation

Time

Dis

ch

ag

reAc

cum

utat

ed s

tora

ge S

Time

Accumulation of storageRelease from storage

Hydrologic Channel Routing

Considering a channel reach having a flood flow, the total volume in storagecan be considered under two categories as

1. Prism storage: It is the volume that would exist if uniform flowoccurred at the downstream depth,i.e. the volume formed by animaginery plane parallel to the challel bottom drawn at the outflowsection water surface.

2. Wedge storage: It is the wedge-like volume formed between theactual water surface profile at the top surface of the prism storage.

Muskingum Equation

The total storage in the channel reach can be expressed as

S = K [xI + (1− x) Q] (8)

This relationship is known as the Muskingum equation. In this the parameterx is known as weighting factor and takes a value between 0 and 0.5.When x=0,

S = KQ (9)

Such a storage is known as linear storage or linear reservoir . Thecoefficient K is known as storage-time constant and has the dimentions oftime. It is approximately equal to the time of travel of a flood wave throughthe channel reach.

Example 3

The following inflow and outflow hydrographs was observed in a river reach.Estimate the values of K and x applicable to this reach for use in the Musk-ingum equation.

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

Inflow (m3/s) 20 80 210 240 215 170 130 90 60 40 28 16

Outflow (m3/s) 20 20 50 150 200 210 185 155 120 85 55 23

SOLUTION

Time (h) I(m3/s) Q(m3/s) (I-Q) Average (I-Q) 4S S =∑4S [xI+(I-x)Q] (m3/s)

x=0.35 x=0.30 x=0.25

0 20 20 0 0 20 20 20

30 180

6 80 20 60 180 41 38 35

110 660

12 210 50 160 840 106 98 90

125 750

18 240 150 90 1590 181.5 177 172.5

52.5 315

24 215 200 15 1905 205.25 204.5 203.75

-12.5 -75

30 170 210 -40 1830 196 198 200

Solution

Time (h) I(m3/s) Q(m3/s) (I-Q) Average (I-Q) 4S S =∑4S [xI+(I-x)Q] (m3/s)

x=0.35 x=0.30 x=0.25

-47.5 -285

36 130 185 -55 1545 165.75 168.5 171.25

-60 -360

42 90 155 -65 1185 132.25 135.5 138.75

-62.5 -375

48 60 120 -60 810 99 102 105

-52.5 -315

54 40 85 -45 495 69.25 71.5 73.75

-36 -216

60 28 55 -27 279 45.55 46.9 48.25

-17 -102

66 16 23 -7 177 20.55 20.9 21.25

GRAPHS

K = 1905204.5 = 9.315

MUSKINGUM METHOD OF ROUTING

For a given channel reach using the Muskingum equation, the change instorage is

S2 − S1 = K [x (I2 − I1) + (1− x) (Q2 − Q1)] (10)

The continuity equation for the reach is(I1 + I2

2

)4t −

(Q1 + Q2

2

)4t = S2 − S1 (11)

From Eqn. (10) and (11), Q2is evaluated as

Q2 = C0I2 + C1I1 + C2Q1 (12)

whereC0 =

−Kx + 0.54tK − Kx + 0.54t

(13)

C1 =Kx + 0.54t

K − Kx + 0.54t(14)

C2 =K − Kx − 0.54tK − Kx + 0.54t

(15)

Contd...

Note that C0 + C1 + C2 = 1. Eqn (12) can be written in general form

Qn = C0In + C1In−1 + C2Qn−1 (16)

Eqn. (12) is known as Muskingum Routing Equation and provides a simplelinear equation for channel routing. For best results the routing interval 4tshould be so chosen that K > 4t > 2Kx .

Example 4

Route the following hydrograph through a river reach for which K= 12.0 hand x= 0.2. 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 37 27 20 15

SolutionSince K = 12 h and 2Kx = 2x12x0.2 = 4.8h, 4t should be such that 12 >4t > 4.8h.In the present case 4t = 6h is selected to suit the given inflowhydrograph ordinate interval.

Now, C0 = −12x0.2+0.5x612−12x0.2+0.5x6 = 0.6

12.6 = 0.048

C1 = 12x0.2+0.5x612−12x0.2+0.5x6 = 0.429

C2 = 12−12x0.2+0.5x612−12x0.2+0.5x6 = 0.523

For the first time interval 0 to 6h,

I 1 = 10 C1I 1 = 4.29I2 = 20 C0I 2 = 0.96Q1 = 10 C2Q1 = 5.23From Eqn. (12) Q2 = 10.48m3/s

For the next time step, 6 to 12h, Q1 = 10.48m3/s.

Solution

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

0 10 10

0.96 4.29 5.32

6 20 10.48

2.4 8.58 5.48

12 50 16.46

2.88 21.45 8.61

18 60 32.94

2.64 25.74 17.23

24 55 45.61

2.16 23.595 23.85

Contd

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

30 45 49.61

1.68 19.305 25.95

36 35 46.93

1.296 15.015 24.55

42 27 40.87

0.96 11.583 21.38

48 20 33.92

0.72 8.58 17.14

54 15 27.04