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Published byWalter Ritchey Modified about 1 year ago

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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).

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S=K*(x*I m +(1-x)*Q m

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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

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Estimation of K and x (I 1 +I 2 )/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*I 1 +(1-x)*Q 1 S2=K*(x*I 2 +(1-x)*Q 2 S2-S1=K(x*(I 2 -I 1 )+(1-x)(Q 2 -Q 1 ) The above equation can be simplified as Q 2 = Q 1 +B 1 *(I 1 -Q 1 )+B 2 (I 2 -I 1 ) B 1 = ∆t/(K*(1-x)+∆t/2) B 2 = (∆t/2-K*x)/(K*(1-x)+∆t/2) Q2 =C 0 *I 2 +C 1 *I 1 +C 2 *Q 1

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Step 2: Plot Σ∆S vrs (x*I+(1+x)*Q. The slope of the equation is K Q2 =C 0 *I 2 +C 1 *I 1 +C 2 *Q 1 Q 2 = Q 1 +B 1 *(I 1 -Q 1 )+B 2 (I 2 -I 1 ) B 1 = ∆t/(K*(1-x)+∆t/2) B 2 = (∆t/2-K*x)/(K*(1-x)+∆t/2)

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Channel flow example t(h)I ( inflow (cume c) Q (Out flow) (cume c) I-Qaverag e (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= ∆t6 Estimation of x X=0.25 K= Estimation of X and K

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Time hrsinflow( I ) cumec C0*I2C1*I1C2*Q1O(cume c) x0.25 k *k*x6.64C0= C1 hrs6C2 Numerator for (C0) Numerator for (C1) Denominator (D) B1= B2= Q 2 = Q 1 +B 1 *(I 1 -Q 1 )+B 2 (I 2 -I 1 ) B 1 = ∆t/(K*(1-x)+0.5*∆t) B 2 = (0.5*∆t-K*x)/(K*(1-x)+0.5*∆t) Q2 =C 0 *I 2 +C 1 *I 1 +C 2 *Q

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Modified Pulse Method Goodrich Method

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10 Goodrich Method Continuity Equation Rewritten 2S 2 /∆t+Q 2 =(I 1 +I 2 )+((2S 1 /∆t)-Q 1 ) (2S 1 /∆t)-Q 1 = [(2S/∆t)+Q) -2Q] ElevationStorage mcm Outflow (cumec) Q Basic Data Time hrs inflow( I ) cumec Actual Data ∆t6 hrs For t=0, reservoir level is 100.6

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Elevatio n Storage mcm Outflo w (cumec ) Q 2S/∆t+Q Cumec_hr ∆t6 hrs For t=0, reservoir level is Basic Data Plot Elevation vrs to [2S/Δt + O]. Plot Discharge vrs to Elevation ,=6*60*60/

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Time hrs inflow( I ) cumec I1+I2(2S/∆t)-Q (col5 (pre)- 2*col7 (pre) (2S/∆t)+Q (col3+col4) elevationO (Outflow) From Graph(1) From Graph (2) Col 6 and read from graph Col 7 and read from graph S 2 /∆t+Q 2 =(I 1 +I 2 )+((2S 1 /∆t)-Q 1 )(2S 1 /∆t)-Q 1 = [(2S/∆t)+Q) -2Q]

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Modified Pulse Method (I 1 +I 2 ) Δt/2+S 1 -(Q 1 *Δt)/2= S 2 +(Q 2 *Δt)/2 Re arranging the above equation, we get

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Modified Pulse Method ElevationStorage mcm AverageOutflow (cumec) (S+Q∆t)/2 mcm delta t0.0216=6*60*60/ (I 1 +I 2 ) Δt/2+S 1 -(Q 1 *Δt)/2= S 2 +(Q 2 *Δt)/2 Re arranging the above equation, we get

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Elevati on Storage mcm delta t Outflow (cumec) Time hrs inflow I cumec I¯I¯*∆tS-(Q∆t/2)S+(Q∆t /2) Elevati on (Read from graph) Q( Read from Graph (I 1 +I 2 ) Δt/2 +S 1 -(Q 1 *Δt)/2= S 2 +(Q 2 *Δt)/2

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