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Flash Flood Risk Management: A Training of Trainers Manual Routing methods Session 13 The basic principle of flood routing is the continuity of flow as expressed by the continuity equation. Several methods can be used to calculate flood routing, these include: modified plus, kinematic wave, Muskingum, Muskingum-Cunge, and dynamic (Chow et al. 1988). The following discussion is limited to the Muskingum method, a commonly used hydrological flood routing method that models the storage volume of flooding in a stream channel by a combination of wedge and prism storage. In the Muskingum method, K and X are determined graphically from the hydrograph, while in the Musking- Cunge method they can be determined using the following equations: ∆X 1 Qmax K = and X = (1 − ) C k 2 BS C x where C k is celerity and B is the width of the water surface. 0 k ∆ The exercise in Box 13 makes the method much clearer. It shows the calculation of peak discharge at 6 km distance from the origin. The calculation can be carried out in a similar way for the remaining hydrographs at 12 and 18 km distance. RM 13.5: Flood frequency and return period Flood frequency Floods are recurring phenomena. The chance of a flood happening at a given location can be calculated using a probability function. One of the simplest probability functions used to determine flood intensity is the return period (T). It is a statistical estimate indicating the average recurrence interval based on data collected over an extended period of time. The return period is an important parameter required for risk analysis. Return period can be determined using the following equation: T = n + 1 m where n is the number of years on record, and m is the rank of the flood being considered (in terms of the flood size in m³/sec). The return period is important in relating extreme discharge to average discharge. The return period has an inverse relationship with the probability (P) that the event will be exceeded in any year. For example, a 10-year flood has 0.1 or 10% chance of being exceeded in any year while a 50-year flood has 0.02 (2%) chance of being exceeded in any year. The term ‘return period’ is actually a misnomer. It does not necessarily mean that the design storm of a 10-year return period will return every 10 years. It could, in fact, never occur, or occur twice in a single year. It is still considered a 10-year storm. The study of return period is useful for risk analysis (such as the natural, inherent, or hydrological risk of failure). For structural design expectations, the return period is useful in calculating the risk to the structure with respect to a given storm return period given the expected design life. The equation for assessing this risk can be expressed as: _ 1 n n R = 1− (1− ) = 1− (1− P ( X ≥ xT )) T 1 where = P ( X ≥ x T ) expresses the probability of the occurrence for the hydrological event in question, and T n is the expected life of the structure in years. 88
Day 3 Box 13: Exercise on flood routing Example: The hydrograph at the upstream end of a river is given in the following table (Shrestha 2008). The reach of interest is 18 km long. Using a sub-reach length Dx of 6 km, determine the hydrograph at the end of the reach using the Muskingum-Cunge method. Assume C k = 2 m/sec, B = 25.3 m, S o = 0.001 m, and there is no lateral flow. Session 13 Time (hour) 0 1 2 3 4 5 6 7 8 9 10 11 12 Flow (m 3 /sec) 10 12 18 28.5 50 78 107 134.5 147 150 146 129 105 Time (hour) 13 14 15 16 17 18 19 20 21 22 23 24 Flow (m 3 /sec) 78 59 45 33 24 17 12 10 10 10 10 10 Solution: Step 1: Determine K ∆x 6,000 K = = = 3,000 sec 2 c k Step 2: Determine X 1 Q X = (1 − 2 BS C max ∆X 0 k ) = 1 2 (1 − 150 ( 25.3) (0.001) (2) (6000) ) = 0.253 Step 3: Determine, C 1 , C 2 , and C 3 ∆t − 2KX 3600 − ( 2)( 3000)( 0.253) C1 = = = 0. 26 2K (1 − X ) + ∆t ( 2)( 3000)( 1 − 0.253) + 3600 C C ∆t + 2KX 2K (1− X ) + ∆t 3600 + ( 2)( 3000)( 0.253) ( 2)( 3000)( 1− 0.253) + 3600 2 = = = 2K (1 − X ) − ∆t 2K (1 − X ) + ∆t ( 2)( 3000)( 1 − 0.253) − 3600 ( 2)( 3000)( 1 − 0.253) + 3600 3 = = = 0.633 0.109 Here Dt is 1 hour= 3,600 sec. If we want our hydrograph to show a 2-hour interval, then we must take Dt = 7,200 sec, and so on. Step 4: Calculate discharge at 6, 12, and 18 km distances. The initial flow at 0 hours is taken as 10 m 3 /sec at all three locations. km The initial flow at 6 km at 0 hours ( Q 6 0 ) is 10 m 3 /sec. The flow at 1 hour at 6 km distance is given by 6km = 0km + 0km + 6km Q 1 C1Q0 C2Q1 C3Q0 = ( 0.26)( 10) + ( 0.6333)( 12) + ( 0.109)( 10) = 11.3 m 3 /sec Similarly, the flow at 2 hours at 6 km distance is given by 6km = 0 km + 0km + 6km = Q 2 C1 Q 1 C2Q 1 C3Q 1 ( 0.26)( 12) + ( 0.633)( 18.0) + ( 0.109)( 18) = 15.7 m 3 /sec The calculations can be carried out in a similar manner for the remaining part of the hydrograph at 6 km distance for the remaining times. The flow at 1 hour at 12 km distance is given by 12 km 6km 6km 12 km Q ( 0.26)( 10) ( 1 = C Q0 + C Q 1 + C Q0 = + 0.633)( 10.9) + ( 0.109)( 10) = 10.8 m 3 /sec 1 For further details, see Shrestha (2008). 2 3 89
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Flash Flood Risk Management A Train
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Foreword The Hindu Kush-Himalayan (
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Acronyms and Abbreviations CFFRMC D
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Day 1 Session 2 Flash Flood Hazards
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Day 1 RM 2.4: Flash floods in the H
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Page 55 and 56: Day 2 Step 4 Step 5 Discuss the dif
Page 57 and 58: Day 2 Figure 10: Framework for comm
Page 59 and 60: Day 2 • First-aid and health: Th
Page 61 and 62: Day 2 Session 8 Gender Perspectives
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Page 81 and 82: Day 3 Geospatial Stream Flow Model
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Page 87 and 88: Day 3 their livelihoods. Watershed
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