Flood routing. Prof. (Dr.) Rajib Kumar Bhattacharjya Indian Institute of Technology Guwahati

Transcription

1 Flood routing Prof. (Dr.) Rajib Kumar Bhattacharjya Indian Institute of Technology Guwahati Guwahati, Assam Web: Visiting Faculty NIT Meghalaya

2 Q (m 3 /sec) Q (m 3 /sec) Type equation here.? T (hr) T (hr)

3 Flood routing ds dt = I t Q(t) S = f I, di dt, d I dq dt,, Q, dt, d Q dt, Level pool routing S j+1 ds = S j j+1 Δt jδt I t dt j+1 Δt jδt Q t dt S j+1 S j = I j + I j+1 Δt Q j + Q j+1 Δt S j+1 Δt + Q j+1 = I j + I j+1 + S j Δt Q j

4 Discharge (Q) Storage (S) Discharge (Q) S H(m)Q m 3 /s S m 3 Δt + Q Elevation (H) Elevation (H) Storage-Outflow function

5 EXAMPLE index (min) Q m 3 /s H(m)Q m 3 /s S m

6 Discharge (Q) Discharge (Q) S H(m)Q m 3 /s S m 3 Δt + Q Storage-Outflow S function Δt + Q m3 /sec y = 3E-08x 5-1E-06x x x x Storage-Outflow S function Δt + Q m3 /sec

7 Outflow (Q) Discharge (Q) Inflow S j index (min) m 3 /s I j + I j+1 Δt Q S j+1 Outflow j + Q Δt j+1 m 3 /s S j+1 Δt + Q j+1 = I j + I j+1 + S j Δt Q j y = 3E-08x 5-1E-06x x x x Storage-Outflow function (Min)

8 Elevation Elevation Elevation RUNGA-KUTTA METHOD ds = I t Q(H) dt ds = A H dh dh dt = I t Q(H) A(H) H 1 = I t j Q(H j ) A(H j ) H j 1 t j Slope = H 1 t j+1 H 1 H j + H 1 3 t j Slope = H t j + 3 t j+1 H H = I t j + 3 Q H j + H 1 3 A H j + H 1 3 H 3 = I t j + 3 Q H j + H 3 A H j + H 3 H j + H 3 Slope = H 3 H 3 H j+1 = H j + H Where H = H H t j t j + 3 t j+1

9 MUSKINGUM METHOD S = KQ + KX I Q S = K XI 1 X Q The value of storage at time j and time j + 1 S j = K XI j 1 X Q j S j+1 = K XI j+1 1 X Q j+1 Combining S j+1 S j = K XI j+1 1 X Q j+1 XI j 1 X Q j Change in storage can also be written as S j+1 S j = I j + I j+1 Δt Q j + Q j+1 Δt

10 MUSKINGUM METHOD Q j+1 = C 1 I j+1 + C I j + C 3 Q j Where C 1 = C = C 3 = KX K 1 X + + KX K 1 X + K 1 X K 1 X + C 1 + C + C 3 = 1

11 Saint-Venant equations Continuity equations x + A t = q Conservation form V y x + y V x + y t = 0 Non conservation form Momentum equations 1 A t + 1 A x Q A + g y x g S 0 S f = 0 Local acceleration Convective acceleration Pressure force Gravity force Friction force V t + V V x + g y x g S 0 S f = 0 Kinematic wave Diffusion wave Dynamic wave

12 Kinematic Wave model Continuity equations x + A t = q Momentum equations S 0 = S f The momentum equation can also be written as A = αq β Manning s equation with S 0 = S f and R = A/P is Q = S 1/ o np /3 A5/3 A = np/3 S o 1/ 3/5 Q 3/5 α = np/3 S o 1/ 3/5 β = 0.6

13 t Kinematic Wave model A t = αβqβ 1 t j + 1 Linear Model Q i j+1 x j+1 Q Continuity equation x + A t = q Q t x + αβqβ 1 t = q j Q i j j Q x = Q j+1 j+1 Q i x x t = Q j+1 j Q Q = Q i j+1 + Q j i x Distance x i + 1 x

14 Linear Model x + αβqβ 1 t = q x = Q j+1 j+1 Q i x t = Q j+1 j Q Q j+1 j+1 Q i x + αβ Q j+1 j i Q β 1 j+1 j Q Q = q j+1 j + q Q = Q j+1 j i + Q j+1 Solving for Q q = q j+1 j + q Q j+1 = x Q j+1 i + αβ Q j+1 j i Q β 1 x + αβ Q j+1 j i Q + q j+1 j + q β 1

15 Nonlinear Model x + A t = q Q j+1 j+1 Q i x + A j+1 j A = q j+1 j + q A = αq β A j+1 j+1 = α Q β j A j = α Q β x = Q j+1 j+1 Q i x A t = A j+1 j A q = q j+1 j + q x Q j+1 j+1 + α Q β = x Q j+1 j β q j+1 j i + α Q + + q x Q j+1 j+1 + α Q β = C Residual error First Derivative f Q j+1 = x Q j+1 j+1 + α Q β C f Q j+1 = x + αβ Q j+1 β 1

16 Nonlinear Model The objective is to find Q j+1 for f Q j+1 = 0 This can be solved using Newton s method j+1 Q = Q j+1 k+1 f Q j+1 k f j+1 Q k k Convergence criteria for the iterative processes j+1 f Q k+1 ε

17 t Muskingum Cunge Method Q j+1 = C 1 I j+1 + C I j + C 3 Q j j + 1 Q i j+1 x j+1 Q Q j+1 j = C 1 Q + C Q j j i + C 3 Q Q t K = x c k j Q i j x j Q X = 1 1 Q BC k S o x i x i + 1 x Distance x

18 Q = S 1/ o A5/3 np/3 Q = αa m α = S 1/ o np/3 and m = 5/3 = αma m 1 A = αm Am A A = m Q A A = mv A A = mv = c k

19 Example problem K =.3 h X = 0.15 = 1 h Q o = 3.1 m 3 /s C 1 = C = C 3 = Routing period (hr.) Inflow m 3 /s

20 Discharge (Q) Routing period (hr.) Inflow m 3 /s C 1 I j+1 C I j C 3 Q j Outflow m 3 /s Inflow Outflow in hr.

21 α = β = 0.6 Q j+1 = Example problem Width = m Length = 5000 m Slope = 1% n = np/3 S o 1/ 0.6 = x = 1000 m = 3 min = 180 s Initial discharge m 3 /s / / x Q j+1 i + αβ Q j+1 j i Q 0.6 β 1 x + αβ Q j+1 j i Q = q j+1 j + q β 1 index (mm) Inflow m 3 /s index (mm) Inflow m 3 /s index (mm) Inflow m 3 /s

22 index (min) Distance along channel (m) index (min) Distance along channel (m)

23 Discharge (cubic meter per sec) index (min) Distance along channel (m) in hr

24 Discharge in cfs Example: Apply the Muskingum-Cunge method to route the following inflow hydrograph. The peak flow is 680 cfs. The area of cross section of the river the peak flow is 95 sq. ft. and the width is 0 ft. The channel has a bottom slope of and the reach length is 5 mi. Take = 1 h. Sol. x = 5 mi=13000 ft. V = Q p A = = 7.16 ft/sec c k = mv = = ft/sec X = 1 1 Q BC k S o x = K = x 100 = s = 3.070hr c k C 1 = C = C 3 = KX K 1 X + = KX K 1 X + = 0.91 Inflow Outflow K 1 X K 1 X + = in hr I Q

25 Example: By the Muskingum-Cunge method, route the following hydrograph. The peak flow rate is 0 m 3 /sec. The area of cross section of the river the peak flow is 183 m. and the width is 35 m. The channel has a bottom slope of and the reach length is 6 km.take = 1 h. (hr) I (m3/s)