Flood Routing In channel routing the storage is a function of both inflow and outflow. In reservoir routing Storage is a unique function of the outflow discharge S= f(Q).

S=K*(x*Im+(1-x)*Qm

S= Storage in the channel I= Inflow Q= out flow K and x are constants S=K*(x*Im+(1-x)*Qm S= Storage in the channel I= Inflow Q= out flow K and x are constants m= 0.6 to 1, 0.6 is for rectangular channel and 1.0 is for natural channels. If m =1 , the above equation changes to X= is a weighing factor varying from 0 to 0.5 When x=0, the equation becomes S=K * Q When x=0.5, both input and output are important

S2-S1=K(x*(I2-I1)+(1-x)(Q2-Q1) Estimation of K and x (I1+I2)/2*∆t-(Q1+Q2) /2* ∆t =∆s Step1 : Plot S vrs (x*I+(1-x)*Q and by trial and error we can get x. We should select x value so that the above curve obtained is far as possible as straight line. S1=K*(x*I1+(1-x)*Q1 S2=K*(x*I2+(1-x)*Q2 S2-S1=K(x*(I2-I1)+(1-x)(Q2-Q1) The above equation can be simplified as Q2= Q1+B1*(I1-Q1)+B2(I2-I1) B1= ∆t/(K*(1-x)+∆t/2) B2= (∆t/2-K*x)/(K*(1-x)+∆t/2) Q2 =C0*I2+C1*I1+C2*Q1

B2= (∆t/2-K*x)/(K*(1-x)+∆t/2) Step 2: Plot Σ∆S vrs (x*I+(1+x)*Q . The slope of the equation is K Q2 =C0*I2+C1*I1+C2*Q1 Q2= Q1+B1*(I1-Q1)+B2(I2-I1) B1= ∆t/(K*(1-x)+∆t/2) B2= (∆t/2-K*x)/(K*(1-x)+∆t/2)

Estimation of X and K Estimation of x X=0.25 K= 13.28 t(h) Channel flow example t(h) I ( inflow (cumec) Q (Out flow) (cumec) I-Q average (I-Q) ∆S= col5 * ∆t ( cumec.Hr) S=Σ∆S ( cumec.Hr) x*I+(1-x)*Q x=0.4 x*I+(1-x)*Q x=0.25 1 2 3 4 5 6 7 8 10   0.4 0.25 20 14 42 11.6 9.5 12 50 38 26 156 198 27.2 21.5 18 29 21 29.5 24 32 -6 7.5 30 22 35 -13 -9.5 36 15 -14 -13.5 23 48 17 -10 -11.5 54 13 -8 -9 60 9 -4 66 -2 -3 ∆t Estimation of x X=0.25 K= 13.28 177 375 37.4 34.25 45 420 35.6 36.5 -57 363 29.8 31.75 -81 282 23.4 25.5 201 17.8 19.75 -69 132 13 14.5 -54 78 9.8 11 -36 42 7.4 8 -18 24 6.2 6.5

Q2 =C0*I2+C1*I1+C2*Q1 Time hrs inflow( I ) cumec C0*I2 C1*I1 C2*Q1 O(cumec) 5   5.00 6 20 -0.49 2.44 2.69 4.63 12 50 -1.23 9.75 2.49 11.00 18 24 32 30 22 36 15 42 10 48 7 54 x 0.25 k 13.28 1 2*k*x 6.64 C0= C1 hrs C2 Numerator for (C0) Numerator for (C1) Denominator (D) -1.23 24.38 5.91 29.06 -0.79 15.61 39.20 -0.54 15.60 21.05 36.11 -0.37 10.73 19.39 29.75 -0.25 7.31 15.98 23.05 -0.17 4.88 12.38 17.08 -0.12 3.41 9.17 12.46 Q2 =C0*I2+C1*I1+C2*Q1 Q2= Q1+B1*(I1-Q1)+B2(I2-I1) B1= ∆t/(K*(1-x)+0.5*∆t) B2= (0.5*∆t-K*x)/(K*(1-x)+0.5*∆t) B1= 0.462963 B2= -0.02469 -0.025 0.488 0.537  -0.32  6.32  12.96

Modified Pulse Method Goodrich Method

Goodrich Method Basic Data 2S2/∆t+Q2=(I1+I2)+((2S1/∆t)-Q1) Actual Data Continuity Equation Time hrs inflow( I ) cumec 10 6 30 12 85 18 140 24 125 96 36 75 42 60 48 46 54 35 25 66 20 Basic Data Elevation Storage mcm Outflow (cumec) Q 100 3.35 100.5 3.472 10 101 3.88 26 101.5 4.383 46 102 4.882 72 102.5 5.37 102.75 5.527 116 103 5.856 130 Rewritten 2S2/∆t+Q2=(I1+I2)+((2S1/∆t)-Q1) (2S1/∆t)-Q1= [(2S/∆t)+Q) -2Q]  ∆t 6 hrs 0.0216 For t=0, reservoir level is 100.6

Basic Data ,=6*60*60/1000000 Plot Elevation vrs to [2S/Δt + O]. Plot Discharge vrs to Elevation Basic Data Elevation Storage mcm Outflow (cumec) Q 2S/∆t+Q Cumec_hr 100 3.35 310.2 100.5 3.472 10 331.5 101 3.88 26 101.5 4.383 46 102 4.882 72 102.5 5.37 102.75 5.527 116 103 5.856 130    ∆t 6 hrs 0.0216 For t=0, reservoir level is 100.6 385.3 451.8 524.0 597.2 627.8 672.2 ,=6*60*60/1000000

(2S/∆t)-Q (col5 (pre)-2*col7 (pre) Time hrs inflow( I ) cumec I1+I2 (2S/∆t)-Q (col5 (pre)-2*col7 (pre) (2S/∆t)+Q (col3+col4) elevation O (Outflow)   1 2 3 4 5 6 7 10 340. 100.6 12 From Graph(1) From Graph (2) 30 40 316 356 100.75 17 1. Col 6 and read from graph 2 85 115 322 437 101.35 2. Col 7 and read from graph 1 18 140 225 24 125 265 96 221 36 75 171 42 60 135 48 46 106 54 35 81 25 66 20 45 357 582 102.5 95 392 657 102.92 127 403 624 102.7 112 400 571 102.32 90 391 526 102.02 73 380 486 101.74 57 372 453 101.51 46 361 421 101.28 37 347 101.02 27 338   (2S1/∆t)-Q1= [(2S/∆t)+Q) -2Q] 2S2/∆t+Q2=(I1+I2)+((2S1/∆t)-Q1)

Modified Pulse Method (I1+I2) Δt/2+S1-(Q1*Δt)/2= S2+(Q2*Δt)/2 Re arranging the above equation, we get (I1+I2) Δt/2+S1-(Q1*Δt)/2= S2+(Q2*Δt)/2

Modified Pulse Method (I1+I2) Δt/2+S1-(Q1*Δt)/2= S2+(Q2*Δt)/2 Re arranging the above equation, we get (I1+I2) Δt/2+S1-(Q1*Δt)/2= S2+(Q2*Δt)/2 Elevation Storage mcm  Average Outflow (cumec) (S+Q∆t)/2 mcm 100 3.35   100.5 3.472 3.411 10 3.58 101 3.7 3.586 26 3.98 101.5 4.383 4.0415 46 4.88 102 4.882 4.6325 72 5.66 102.5 5.37 5.126 6.45 102.75 5.527 5.4485 116 6.78 103 5.856 5.6915 130 7.26 delta t 0.0216 =6*60*60/1000000

Elevation (Read from graph) Storage mcm 100 3.35 100.5 3.472 101 3.7 101.5 4.383 102 4.882 102.5 5.37 102.75 5.527 103 5.856 delta t 0.0216 Outflow (cumec) 10 26 46 72 100 116 130 Time hrs inflow I cumec I¯ I¯*∆t S-(Q∆t/2) S+(Q∆t/2) Elevation (Read from graph) Q( Read from Graph 10   100.5 6 20 15 0.324 3.364 3.7 100.70 15.00 12 55 37.5 0.810 3.36 4.2 101.12 30.00 18 80 67.5 1.458 3.53 5.0 101.60 50.00 24 73 76.5 1.652 3.90 5.6 101.95 72.00 30 58 65.5 1.415 4.00 5.4 101.80 61.00 36 46 52 1.123 4.10 5.2 101.70 55.00 42 41 0.886 4.03 4.9 101.45 43.00 48 27.5 31.75 0.686 3.99 4.7 101.40 41.00 54 23.75 0.513 3.79 4.3 101.20 34.00 60 17.5 0.378 3.57 3.9 100.90 22.00 66 13 14 0.302 3.47 3.8 100.80 18.50 72 11 0.259 3.37 3.6 100.50 10.00 (I1+I2) Δt/2 +S1-(Q1*Δt)/2= S2+(Q2*Δt)/2

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