Floodplain Hydraulics! Hydrology and Floodplain Analysis Dr. Philip Bedient


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1 Floodplain Hydraulics! Hydrology and Floodplain Analysis Dr. Philip Bedient
2 Open Channel Flow 1. Uniform flow  Manning s Eqn in a prismatic channel  Q, V, y, A, P, B, S and roughness are all constant 2. Critical flow  Specific Energy Eqn (Froude No.) 3. Nonuniform flow  gradually varied flow (steady flow)  determination of floodplains 4. Unsteady and Nonuniform flow  flood waves
3 Uniform Open Channel Flow Chezy and Manning s Eqn. 1. Must use results from Fluid Mechanics 2. Derivation of these eqns requires a force balance (x) F = ρq[ V V ]= 0 x Actual forces (F = hydrostatic) are summed across C.V. [ F 1 F 2 ] τ w Pl + W sinθ = 0
4 Chezy and Manning s Eqn. 1. Since hydrostatic forces are equal, and sinθ = S 0 τ w = W sinθ = WS 0 Pl Pl W = γ Al Slopes are very mild 2. Define R = A/P, the hydraulic radius Now τ w = Kρ V 2 2 forturbulent flows
5 Chezy and Manning s Eqn. Finally, we can equate the two eqns for shear stress γ RS 0 = V 2 Kρ 2,solving forv = C RS 0 C = Chezy Coefficient (1768) in Paris Manning was an Irish Eng and 1889 developed his EQN.
6 Uniform Open Channel Flow Manning s Eqn for velocity or peak flow rate v = 1 n R 2 /3 S S.I. units v = 1.49 R 2 /3 S English units n where!!!n = Manning s roughness coefficient!!r = hydraulic radius = A/P!!S = channel slope!! Q = V A = flow rate (cfs)
7 Uniform Open Channel Flow Brays B. Brays Bayou" Concrete Channel"
8 Normal depth is function of flow rate, and geometry and slope. One usually solves for normal depth or width given flow rate and slope information B b
9 Normal depth implies that flow rate, velocity, depth, bottom slope, area, top width, and roughness remain constant within a prismatic channel as shown below UNIFORM FLOW Q = C V = C y = C S 0 = C A = C B = C n = C
10 Optimal Channels  Max R and Min P
11 Uniform Flow Energy slope = Bed slope or dh/dx = dz/dx Water surface slope = Bed slope = dy/dz = dz/dx Velocity and depth remain constant with x H
12
13 Critical depth is used to characterize channel flows  based on addressing specific energy E = y + v 2 /2g :!!E = y + Q 2 /2gA 2 where Q/A = q/y and q = Q/b!!!!!Take de/dy = (1 q 2 /gy 3 ) and set = 0. q = const! E = y + q 2 /2gy 2! y Min E Condition, q = C E
14 !!!Solving de/dy = (1 q 2 /gy 3 ) and set = 0.! For a rectangular channel bottom width b,! 1.!E min = 3/2Y c for critical depth y = y c! 2. y c /2 = V c2 /2g! 3. y c = (Q 2 /gb 2 ) 1/3 Froude No. = v/(gy) 1/2 We use the Froude No. to characterize critical flows
15 Y vs E E = y + q 2 /2gy 2! q = const!
16 ! Critical Flow in Open Channels In general for any channel shape, B = top width!! (Q 2 /g) = (A 3 /B)!at y = y c!!! Finally Fr = v/(gy) 1/2 = Froude No.!! Fr = 1 for critical flow! Fr < 1 for subcritical flow! Fr > 1 for supercritical flow
17
18 NonUniform Open Channel Flow With natural or manmade channels, the shape, size, and slope may vary along the stream length, x. In addition, velocity and flow rate may also vary with x. Nonuniform flow can be best approximated using a numerical method called the Standard Step Method.
19 NonUniform Computations Typically start at downstream end with known water level  y o. Proceed upstream with calculations using new water levels as they are computed. The limits of calculation range between normal and critical depths. In the case of mild slopes, calculations start downstream. In the case of steep slopes, calculations start upstream. Calc. Q Mild Slope
20 NonUniform Open Channel Flow Let s evaluate H, total energy, as a function of x. H = z + y + ( α v 2 / 2g) Take derivative, dh dx = dz dx + dy dx + α 2 g dv 2 dx Where!H = total energy head!!!z = elevation head,! αv 2 /2g = velocity head
21 Replace terms for various values of S and S o. Let v = q/y = flow/unit width  solve for dy/dx, the slope of the water surface S = S + dy o dx 1 q 2 gy 3 since v = q / y 1 2g d [ dx v 2 ]= 1 2g d dx q 2 y 2 = q2 g 1 y 3 dy dx
22 Given the Froude number, we can simplify and solve for dy/dx as a fcn of measurable parameters Fr 2 = v 2 / gy ( ) dy dx = S S o 1 v 2 / gy = S o S 1 Fr 2 *Note that the eqn blows up when Fr = 1 and goes to zero if S o = S, the case of uniform OCF. where S = total energy slope S o = bed slope, dy/dx = water surface slope
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24 Y n > Y c Uniform Depth Mild Slopes where  Y n > Y c
25 Now apply Energy Eqn. for a reach of length L y + v g L = = y + v g 2 y + v g 2 y 2 + v 2 2g S S 0 + ( S S )L o This Eqn is the basis for the Standard Step Method Solve for L = Δx to compute water surface profiles as function of y 1 and y 2, v 1 and v 2, and S and S 0 2
26 Backwater Profiles  Mild Slope Cases x
27 Backwater Profiles  Compute Numerically Compute y 3 y 2 y 1
28 Routine Backwater Calculations 1. Select Y 1 (starting depth) 2. Calculate A 1 (cross sectional area) 3. Calculate P 1 (wetted perimeter) 4. Calculate R 1 = A 1 /P 1 5. Calculate V 1 = Q 1 /A 1 6. Select Y 2 (ending depth) 7. Calculate A 2 8. Calculate P 2 9. Calculate R 2 = A 2 /P Calculate V 2 = Q 2 /A 2
29 Backwater Calculations (cont d) 1. Prepare a table of values 2. Calculate V m = (V 1 + V 2 ) / 2 Energy Slope Approx. 3. Calculate R m = (R 1 + R 2 ) / 2 nv S = m R 3 m 4. Calculate Manning s 5. Calculate L = X from first equation 2 6. X = X i for each stream reach (SEE SPREADSHEETS) L = 2 y 1 + v 1 2g y + v g S S 0
30 100 Year Floodplain" Floodplain Tributary" D" Bridge Q D" C" Q C" Bridge Section" Main Stream" B" Q B" A" Cross Sections" Q A" Cross Sections"
31 The Floodplain" Top Width"
32 Floodplain Determination"
33 The Woodlands" v The Woodlands planners wanted to design the community to withstand a 100year storm." v In doing this, they would attempt to minimize any changes to the existing, undeveloped floodplain as development proceeded through time.
34 HEC RAS (River Analysis System, 1995)" HEC RAS or (HEC2)is a computer model designed for natural cross sections in natural rivers. It solves the governing equations for the standard step method, generally in a downstream to upstream direction. It can Also handle the presence of bridges, culverts, and variable roughness, flow rate, depth, and velocity.
35 HEC  2 Orientation  looking downstream
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37
38
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40 Multiple Cross Sections River
41 HEC RAS (River Analysis System, 1995)"
42 HEC RAS Bridge CS"
43 HEC RAS Input Window"
44 HEC RAS Profile Plots"
45 3D Floodplain"
46 HEC RAS Cross Section Output Table"
47 Brays BayouTypical Urban System" Bridges cause unique problems in hydraulics" Piers, low chords, and top of road is considered" Expansion/contraction can cause hydraulic losses" Several cross sections are needed for a bridge" 288 Bridge causes a 2 ft " Backup at TMC and is being replaced by TXDOT" 288 Crossing"
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