SECTION VI: FLOOD ROUTING Consider the watershed with 6 sub-basins Q1 = QA + QB (Runoff from A & B) mross/work/coursewk/watres/notes/notes6.bigovh 1 Q2 = (QA + QB)2 (Routed runoff from Q1) + QC+ QD + (Direct runoff from C & D) What causes attenuation? 1) Storage What causes flow lag? 1) Flood wave travel time (celerity) - f(length, depth, friction, slope) Consider a short reach or reservoir (pond) Continuity Eq. Qin - Qout = S/ t mross/work/coursewk/watres/notes/notes6.bigovh 2 and 2) Friction See Appendix Figure 4.2 Bedient Simple Hydrologic Flood Routing Simple methods based on lumped continuity equation f(x)) Qi - Qo = S/ t Problem: Qi is known, need to solve for Qo What about S? mross/work/coursewk/watres/notes/notes6.bigovh 3 ("lumped" not To solve the flood routing problem a relationship between S and Q is needed. Either: (See handout, USGS rating curve) 2) Q = f(S) or S = f(Q) directly - Less common than (1) 3) Solve for Q(y) from momentum equation (e.g., Q = kym, kinematic ), S(A), A = cross-sectional area mross/work/coursewk/watres/notes/notes6.bigovh 4 RESERVOIR ROUTING - simplest hydrologic method - sometimes called "Pond Routing" "Linear" Reservoir Qo = kS k = [1/time] = routing coefficient =f (channel geometry) Qo = outflow linearly related to storage then Qin - Qout = S/ t (Homework #’s 4.1, 3, 4, 5, 7) mross/work/coursewk/watres/notes/notes6.bigovh 5 Qin - Qout = d(k-1Qout)/dt k dQout/dt + k Qout = k Qin(t) Can solve analytically (exponential solution) f(time) Solution: Constant = Qout at t=0 Need functional form for Qi (i.e., Qi(t)) so that it can be integrated over time. For example, could assume Qi = f(sine function) or mross/work/coursewk/watres/notes/notes6.bigovh 6 could use numerical (discrete step) methods Finite Difference Method of Continuity Equation Qin + Qo = ds/dt Subscript 1 implies beginning of a time step Known: Qi1, Qi2, Qo1, S1 Subscript 2 implies end of a time step Unknown: Qo2, S2 mross/work/coursewk/watres/notes/notes6.bigovh 7 Modified Puls Method (Finite Difference Continuity Eq.) - also called Storage Indication Method (book) - collect knowns and unknowns on opposite sides of the equation To solve: use a relationship between S and Q from rating curve (stagedischarge relation), weir equation, uniform flow assumption, or other information. Can construct a table or graph of Qo = f(S + ( t/2) Qo) Book has good example (p. 256 - 260) mross/work/coursewk/watres/notes/notes6.bigovh 8 Consider a river reach shown Consider uniform flow mross/work/coursewk/watres/notes/notes6.bigovh 9 Manning's Equation for uniform flow Seq1 = Sbed A = cross-sectional area L = length of channel reach and S = L A = reach storage Then: R = Hydraulic radius = Area/ Wetted perimeter For rectangular channel (y = depth, b = bottom width), Area = y b WP = 2y + b Rh = yb/(2y + b) y/2 for b = 2y Qo = Constant y2/3 S mross/work/coursewk/watres/notes/notes6.bigovh 10 Muskingum Method For Flood Routing S = prism storage + wedge storage = KQo + Kx (Qi - Qo) Two Parameters K, x x is not distance Can substitute into continuity mross/work/coursewk/watres/notes/notes6.bigovh 11 Can solve analytically as before if x = 0, linear reservoir In general, 0 x 0.5 note x is higher for more regular ("improved") channels x is lower for more natural channel e.g., natural irregular channels x ~ 0.15 concrete lined, trapezoidal channel 0.3 if x = 0.5, pure translation mross/work/coursewk/watres/notes/notes6.bigovh 12 x 0.4 Finite differences applied to continuity Equation Muskingum Eq. For a linear reservoir x = 0 Must have storage and flow data to evaluate parameters Plot graph shown for trial values of x, keep trying with different x, until loop narrows (approximates a line) Slope = K , S = K [ x Qi + (1 - x ) Qo] mross/work/coursewk/watres/notes/notes6.bigovh 13 mross/work/coursewk/watres/notes/notes6.bigovh 14 When storage is a maximum dS/dt = 0 = Qi - Qo, substitute S = K [ Qo + Kx(Qi - Qo)] mross/work/coursewk/watres/notes/notes6.bigovh 15 If we had these at the same time, we could solve Use two slope values and solve for x mross/work/coursewk/watres/notes/notes6.bigovh 16 In the real river reach the parameters are a function of the flow, Q. For this case need parameter estimation for several flow rates (i.e., variable parameter, K(flow), x(flow), Muskingum Method). Probably better to just go to full St. Venant (dynamic) equations. Another method is Muskingum-Cunge Method - A better way for parameter estimation (relate them to physical parameters of the channel) ref Cunge 1969 (Ref 66) let Muskingum parameter, x, now be mross/work/coursewk/watres/notes/notes6.bigovh 17 (x in the following eqs reps. distance) Taylor Series Remember: and Substitute for Q(x + x, t) and dQ(x + x, t)/dt in continuity equation (neglect d3Q/dt dx2 term): Use continuity equation of the form: Define: (Muskingum-Cunge) mross/work/coursewk/watres/notes/notes6.bigovh 18 Solve for: LHS Combine d2Q/dx2 terms: mross/work/coursewk/watres/notes/notes6.bigovh 19 Muskingum - Cunge This is a form of the advection - diffusion equation (the bracketed term represents a sort of flow wave diffusivity, D). [RHS] Likely to be curved: Numerical Diffusivity e.g., c = f(flow) Also, approximate c by flood profiles: Hydraulic Flood Routing Large "Dynamic" Rivers "Dynamic" - subjected to rapid fluctuations in flow requiring inclusion of acceleration terms in equations of flow * Must use St. Venant Eqns to adequately describe flow (p. 237-8) mross/work/coursewk/watres/notes/notes6.bigovh 20