1. 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).

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

3. 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

4. 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

5. 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)

6. 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

7. 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=
B2=
-0.025
0.488
0.537
 -0.32
 6.32
 12.96

9. Modified Pulse Method
Goodrich Method

10. 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

11. 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/

12. (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)

14. 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

15. 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/

17. 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

19. End