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ERT 246 Hydrology & Water Resources Eng.

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Presentation on theme: "ERT 246 Hydrology & Water Resources Eng."— Presentation transcript:

1 ERT 246 Hydrology & Water Resources Eng.
FLOOD ROUTING ERT 246 Hydrology & Water Resources Eng.

2 Flow Routing Q t Q Procedure to determine the flow hydrograph at a point on a watershed from a known hydrograph upstream As the hydrograph travels, it attenuates gets delayed t Q t Q t

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

4 Types of flow routing Lumped/hydrologic Distributed/hydraulic
Flow is calculated as a function of time alone at a particular location Governed by continuity equation and flow/storage relationship Distributed/hydraulic Flow is calculated as a function of space and time throughout the system Governed by continuity and momentum equations

5 Lumped flow routing Three types Level pool method (Modified Puls)
Storage is nonlinear function of Q Muskingum method Storage is linear function of I and Q Series of reservoir models Storage is linear function of Q and its time derivatives

6 S and Q relationships

7 Level pool routing Procedure for calculating outflow hydrograph Q(t) from a reservoir with horizontal water surface, given its inflow hydrograph I(t) and storage-outflow relationship

8 Wedge and Prism Storage
Positive wedge I > Q Maximum S when I = Q Negative wedge I < Q

9 Hydrologic River Flood Routing
Basic Equation

10 Hydrologic river routing (Muskingum Method)
Wedge storage in reach Advancing Flood Wave I > Q K = travel time of peak through the reach X = weight on inflow versus outflow (0 ≤ X ≤ 0.5) X = 0  Reservoir, storage depends on outflow, no wedge X =  Natural stream Receding Flood Wave Q > I

11 Continuity Equation in Difference Form
Referring to figure, the continuity equation in difference form can be expressed as 2 ) O 1 (O I (I _ t S + - = D

12 Derivation of Muskingum Routing Equation
By Muskingum Model, at t = t2, S2 = K [X I2 + (1 - X)O2] at t = t1, S1 = K [X I1 + (1 - X)O1] Substituting S1, S2 into the continuity equation and after some algebraic manipulations, one has O2 = Co I2 + C1 I1 + C2 O1 Replacing subscript 2 by t +1 and 1 by t, the Muskingum routing equation is Ot+1 = Co It+1 + C1 It + C2 Ot, for t = 1, 2, … where ; ; C2 = 1 – Co – C1 Note: K and t must have the same unit. Routing

13 Muskingum Routing Equation
where C’s are functions of x, K, Dt and sum to 1.0

14 Muskingum Equations where C0 = (– Kx + 0.5Dt) / D
C2 = (K – Kx – 0.5Dt) / D D = (K – Kx + 0.5Dt) Repeat for Q3, Q4, Q5 and so on.

15 Estimating Muskingum Parameters, K and x
Graphical Method: Referring to the Muskingum Model, find X such that the plot of XIt+ (1-X)Ot (m3/s) vs St (m3/s.h) behaves almost nearly as a single value curve. The assume value of x lies between 0 and 0.3. The corresponding slope is K.

16 Example 8.4: Estimating the value of x and K.
Try and error to get the nearly straight line graph.

17 Muskingum Routing Procedure
Given (knowns): O1; I1, I2, …; t; K; X Find (unknowns): O2, O3, O4, … Procedure: (a) Calculate Co, C1, and C2 (b) Apply Ot+1 = Co It+1 + C1 It + C2 Ot starting from t=1, 2, … recursively.

18 Example 8.5 Given K and x. Initial outflow, Q also given. Solution:
Calculate Co, C1, and C2 C0 = (– Kx + 0.5Dt)/ D C1 = (Kx + 0.5Dt)/ D C2 = (K – Kx – 0.5Dt)/ D D = (K – Kx + 0.5Dt)

19 Solution: Route the following flood hydrograph through a river reach for which K=12.0hr and X=0.20. At the start of the inflow flood, the outflow flood, the outflow discharge is 10 m3/s. Time (hr) 6 12 18 24 30 36 42 48 54 Inflow (m3/s) 10 20 50 60 55 45 35 27 15

20 Reservoir Routing Reservoir acts to store water and release through control structure later. Inflow hydrograph Outflow hydrograph S - Q Relationship Outflow peaks are reduced Outflow timing is delayed Max Storage

21 Inflow and Outflow

22 Inflow and Outflow I1 + I2 – Q1 + Q S2 – S1 = 2 2 Dt

23 Inflow & Outflow Day 3 = change in storage / time
Repeat for each day in progression

24 Determining Storage Evaluate surface area at several different depths
Use available topographic maps or GIS based DEM sources (digital elevation map) Outflow Q can be computed as function of depth for either pipes, orifices, or weirs or combinations

25 Typical Storage -Outflow
Plot of Storage in vs. Outflow in Storage is largely a function of topography Outflows can be computed as function of elevation for either pipes or weirs Combined S Pipe Q

26 Comparisons: River vs. Reservoir Routing
Level pool reservoir River Reach

27 Flood Control Structural Measures Non-structural Methods


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