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1. CHAPTER 6: HYDROGRAPH ROUTING 1

2. 2 Definitions Flood routing is variously defined as follows: Routing, flood- The procedure that determines the timing and magnitude of a flood wave at a point on a stream from the known or assumed data at one or more points upstream . Routing, flow - A mathematical procedure that predicts the changing magnitude, speed, and shape of a flood wave as a function of time at one or more points along a watercourse .

3. 3 Routing, stream channel- Mathematical relations that calculate outflow from a stream channel once inflow , lateral contributions, and channel characteristics are known. Routing, reservoir- This procedure derives the outflow hydrograph from a reservoir from the inflow hydrograph into the reservoir with consideration of elevation , storage , and discharge characteristics of the reservoir and spillways. The conservation of mass equation is solved with the assumption that outflow discharge and volume of storage are directly related.

4. 4 In flood forecasting, hydrologists may want to know how a short burst of intense rain in an area upstream of a city will change as it reaches the city. Routing can be used to determine whether the pulse of rain reaches the city as a deluge or a trickle. Other uses of routing include reservoir and channel design, floodplain studies and watershed simulations.

5. Flow Routing Procedure to determine the flow hydrograph at a point on a watershed from a known hydrograph upstream As the hydrograph travels, it attenuates (peak reduction) gets delayed Q t Q t Q t Q t 5

6. Why route flows? Account for changes in flow hydrograph as a flood wave passes downstream This helps in Accounting for storages Studying the attenuation of flood peaks Q t 6

7. 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/flow routing 7

8. The flood hydrograph is in fact a wave The Stage-discharge relationship represent the passage of waves As this wave 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 8

9. 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 9

10. 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 10

11. Flood Routing Hydrologic Routing Hydraulic Routing (Based on Continuity Equation) Based on Momentum equation Reservoir Routing Channel Routing 11

12. RESERVOIR ROUTING 12

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15. 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 15

16. Flood Routing via Storage A Reservoir sacrifices floodplain and adjacent land to provide a basin that spreads stream floodwaters over a large area, slowing velocity and allowing slow discharge before, during , and after the storm. The height of the flood wave is greatly reduced, because part of the flood is retained in the reservoir, to be released later. Recall that Flow rate Q = Velocity x Area 16

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18. Inflow and outflow hydrographs plotted on same graph. Area A volume of water that fills reservoir, dS / dt > 0. Due to inflow, which is due to precip . At t 1 reservoir is full and inflow = outflow. Area C is the volume of water that flows out. If inflow has slowed (storm is past) outflow can be slow. dS / dt < 0, so water in reservoir drops. Outflow rate mostly depends on reservoir water depth, sqrt (2gh), but the spillway operator can lessen it. Lag due to travel time and filling of reservoir 18

19. Most reservoirs have spillways to slowly lower levels in prep for next storm; slow enough not to flood the downstream valley. 19

20. 20 General configuration of a reservoir As inflow exceeds outflow, the storage increases. If the outflow is greater than the inflow, the storage decreases as long as the water level is higher than the crest of the outlet device.

21. Hydrologic Storage Routing (Level Pool Routing) horizontal water surface is assumed in the reservoir 21

22. 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 22

23. Inflow and outflow from a reservoir Inflow and outflow are equal and maximum storage is reached Inflow exceeds outflow and the reservoir is filling Outflow exceeds inflow and the reservoir empties

24. D S/ D t = Inflow – Outflow - Storage change is equal to the area between the inflow and outflow hydrographs. Example 1: 24

25. 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 ( Δ t) I is average Inflow, Q is average outflow during the time interval If storage at beginning is S 1 and at end is S 2 during time Δ t In this time interval the hydrograph is linear and is smaller than the transit of the flood wave through the reach. …..2 …..3 Basic equations used in hydrologic routing …..1 25

26. Goodrich Method Semi-graphical Method 26 Discharge Time Storage Time Inflow Outflow Unknown Known Need a function relating Storage-outflow function

27. Steps Develop Q versus Q+ 2S/ D t relationship using Q/H relationship Compute Q+ 2S/ D t Use the relationship developed in step 1 to get Q 27

28. ? Route the following flood hydrograph through the reservoir by Goodrich method: Inflow hydrograph: Elevation Storage (10 6 m 3 ) Outflow discharge (m 3 /s) 100.00 3.350 100.50 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) 6 12 18 24 30 36 42 48 54 60 66 Inflow ( m 3 /s) 10 30 85 140 125 96 75 60 46 35 25 20 The initial conditions are when t = , the reservoir elevation is 100.60 m . The storage-elevation-discharge data is as follows: 28

29. 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 Q 2 and From the graph Elevation Storage (10 6 m 3 ) Outflow discharge (m 3 /s) (m 3 /s)   100 3.35 310.19 100.5 3.472 10 331.48 101 3.88 26 385.26 101.5 4.383 46 451.83 102 4.882 72 524.04 102.5 5.37 100 597.22 102.75 5.527 116 627.76 103 5.856 130 672.22 29

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31. Time (h) (m 3 /s) (m 3 /s) (m 3 /s) Elevation (m) Discharge Q (m 3 /s) 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 Solution: 31

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33. Time (h) (m 3 /s) (m 3 /s) (m 3 /s) Elevation (m) Discharge Q (m 3 /s) 10 340 100.6 12 6 30 40 340-2*12=316 =  (2S 1  / Δt + O 1 )  - 2 x  O 1   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 33

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35. Time (h) (m 3 /s) (m 3 /s) (m 3 /s) Elevation (m) Discharge Q (m 3 /s) 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 35

36. Time (h) (m 3 /s) (m 3 /s) (m 3 /s) Elevation (m) Discharge Q (m 3 /s) 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 36

37. What we achieved through this flood routing 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 Reservoir storing Reservoir Depleting 37

38. 38 CHANNEL/RIVER ROUTING

39. Channel/river/stream 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 39

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41. It’s because of storage within the river system itself. When flow is rising, there’s a parcel of storage within the *reach between inflow and outflow, slowed because of friction and large cross sectional area when the floodplain gets some of the water. Suppose, for example, we have gauges both upstream (station 1) , and downstream(station 2). Both have floodplains that store water. We could write a water balance equation with averages: Average inflow minus average outflow = average change in storage * Reach : any portion of a stream.

42. 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 42

43. 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 ) 43

44. When Inflow is greater upstream than in the lower watershed, then inflow > outflow, here I > Q, and the water storage will be a wedge with higher water upstream. The elevated (deeper) reach of water during the flood crest is called prism (=rectangular) storage , it occurs when I = Q, and finally as the inflow falls, there’s again wedge storage while the outflow is greater than the inflow Q>I , with the higher water level close to the outlet. Flood arrives at reach Entire reach flooded Storm over, inflow slows, higher water in lower reach

45. So, if you had a routing method that allows for wedge storage, you could predict the flow at points downstream, and see how the flood wave attenuates. Several such methods exist. We’re going to talk today about the Muskingum method.

46. Muskingum Method The Muskingum method uses the basic hydrologic continuity equation with averages we just saw: and a storage term that depends both on the inflow and outflow: where x is a weighting factor between 0 and 0.5 that says something about how inflow and outflow vary within a given reach, and K is the travel time of the flood wave.

47. Muskingum Method K and x In a perfectly smooth channel, x = 0.5 and S = 0.5 K (I + Q), which results in simple translation of the wave. However , typical streams have values of x=0.2 to 0.3.

48. Muskingum Method Our storage discharge equation is written in a finite difference form: The Muskingum routing procedure itself uses this form combined with in the form To produce the Muskingum outflow equation

49. 49 Estimating 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.

50. Derivation of Muskingum Routing Equation By Muskingum Model, at t = t 2 , S 2 = K [X I 2 + (1 - X)O 2 ] at t = t 1 , S 1 = K [X I 1 + (1 - X)O 1 ] Substituting S 1 , S 2 into the continuity equation and after some algebraic manipulations, one has O 2 = C o I 2 + C 1 I 1 + C 2 O 1 Replacing subscript 2 by t +1 and 1 by t , the Muskingum routing equation is O t+1 = C o I t+1 + C 1 I t + C 2 O t , for t = 1, 2, …   w here C 2 = 1 – C o – C 1 Note: K and  t must have the same unit. 50

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52. Muskingum Routing Procedure Given ( know n s ): O 1 ; I 1 , I 2 , …;  t; K; X Find (unkno w ns): O 2 , O 3 , O 4 , … Procedure: (a) Calculate C o , C 1 , and C 2 (b) Apply O t+1 = C o I t+1 + C 1 I t + C 2 O t starting from t=1, 2, … recursively. 52

53. 53 Muskingum Example Problem 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:

54. 54 Muskingum Example Problem The first step is to determine the coefficients in this problem. The calculations for each of the coefficients is given below: C = - ((1*0.30) - (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.167 C 1 = ((1*0.30) + (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.667

55. 55 Muskingum Example Problem C 2 = (1- (1*0.30) - (0.5*1)) / ((1-(1*0.30) + (0.5*1)) = 0.167 Therefore the coefficients in this problem are: C = 0.167 C 1 = 0.667 C 2 = 0.167

56. 56 Muskingum Example Problem The three columns now can be calculated. C I 2 = 0.167 * 5 = 0.835 C 1 I 1 = 0.667 * 3 = 2.00 C 2 O 1 = 0.167 * 3 = 0.501

57. 57 Muskingum Example Problem Next the three columns are added to determine the outflow at time equal 1 hour. 0.835 + 2.00 + 0.501 = 3.34

58. 58 Muskingum Example Problem This can be repeated until the table is complete and the outflow at each time step is known.

59. Muskingum Routing - Example 59

60. Muskingum - Example Given: Inflow hydrograph K = 2.3 hr, X = 0.15, D t = 1 hour, Initial Q = 85 cfs Find: Outflow hydrograph using Muskingum routing method 60

61. Muskingum – Example (Cont.) C 1 = 0.0631, C 2 = 0.3442, C 3 = 0.5927 61

62. ? 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 m 3 /s. Time (h) 6 12 18 24 30 36 42 48 54 Inflow (m 3 /s) 10 20 50 60 55 45 35 27 20 15 Time (h) I (m 3 /s) 0.429 I 1 0.048 I 2 0.523 Q 1 Q (m 3 /s) 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 62