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Hydrology: Muskingum Method Guide

The document provides information about routing hydrographs through river reaches using the Muskingum method. It includes examples of determining K and x values from observed inflow and outflow data, as well as examples of applying the Muskingum method to route given hydrographs through reaches with specified K and x values. Peak lag and attenuation are determined by comparing the routed inflow and outflow hydrographs.

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789 views8 pages

Hydrology: Muskingum Method Guide

The document provides information about routing hydrographs through river reaches using the Muskingum method. It includes examples of determining K and x values from observed inflow and outflow data, as well as examples of applying the Muskingum method to route given hydrographs through reaches with specified K and x values. Peak lag and attenuation are determined by comparing the routed inflow and outflow hydrographs.

Uploaded by

goo od
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Prob: The following inflow and outflow hydrographs were observed in a river

reach. Estimate the values of K and x applicable to this reach for use in the
Muskinghum equation.

Time (h) Inflow (m3/s) Outflow (m3/s)

0 5 5
6 20 6
12 50 12
18 50 29
24 32 38
30 22 35
36 15 29
42 10 23
48 7 17
54 5 13
60 5 9
66 5 7
Prob: Route the following hydrograph through a river reach for which K=
12.0 h and x=0.20, At the start of the inflow flood, the outflow discharge is
10 m3/s.

Time 0 6 12 18 24 30 36 42 48 54
(h)

Inflow 10 20 50 60 55 45 35 27 20 15
(m3/s)

Solution:

Since K=12h and 2Kx=2x12x0.2=4.8h, t should be such that


12h> t>4.8h. In the present case t=6h is selected to suit the given inflow
hydrograph ordinate interval.
 Kx  0.5t
12  0.20  0.5  6 0.6 C0 
C0    0.048 K  Kx  0.5t
12  12  0.2  0.5  6 12.6 Kx  0.5t
12  0.20  0.5  6 C1 
C1   0.429 K  Kx  0.5t
12.6 K  Kx  0.5t
12  12  0.20  0.5  6 C2 
C2   0.523 K  Kx  0.5t
12.6 Note : C0  C1  C2  1

For the first time interval, 0 to 6h,


I1=10.0 C1I1=4.29
I2=20.0 C0I2=0.96
Q1=10.0 C2Q1=5.23
From Eqn. (3), Q=C0I2+C1I1+C2Q1=10.48 m3/s
For the next time step, 6 to 12 h, Q1 = 10.48 m3/s
Similarly, the same procedure was repeated entire duration of the inflow
hydrograph.
NOTE: By plotting the inflow and outflow hydrographs the attenuation and
peak lag are found to be 10 m3/s and 12 h respectively.
Muskingum Method of Routing, t=6h
Time(h) I (m3/s) 0.048I2 0.429I1 0.523Q1 Q (m3/s)

1 2 3 4 5 6
0 10 10.00
0.96 4.29 5.23
6 20 10.48
2.4 8.58 5.48
12 50 16.46
2.88 21.45 8.61
18 60 32.94
2.64 25.74 17.23
24 55 45.61
2.16 23.6 23.85
30 45 49.61
1.68 19.3 25.95
36 35 46.93
1.3 15.02 24.55
42 27 40.87
0.96 11.58 21.38
48 20 33.92
0.72 8.58 17.74
54 15 27.04
Fig. Hydrographs in a channel routing
PROB: Observed values of inflow and outflow hydrograph at the ends of a reach in
a river are given below. Determine the best values of K and x for use in the
Muskingum method of flood routing. (Try x=0.3)

Time (h) 0 6 12 18 24 30 36 42 48 54 60 66
3
Inflow (m /s) 20 80 210 240 215 170 130 90 60 40 28 16
Outflow (m3/s) 20 20 50 150 200 210 185 155 120 85 55 23

PROB: Route the following flood through a reach for which K=22h and x=0.25.
Plot the inflow and outflow hydrographs and determine the peak lag and
attenuation. At t=0 the outflow discharge is 40 m3/s.

Time (h) 0 12 24 26 48 60 72 84 96 108 120 132 144


Inflow 40 65 165 250 240 205 170 130 115 85 70 60 54
(m3/s)

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