Examples on Method of Characteristics in Open Channel Flow by Ih2uBoX6

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									Examples on Method of Characteristics in Open Channel Flow

Example 1

Water flows at a depth of 2.0m, and with a velocity of 1.0m/s in a rectangular channel
discharging into an estuary. The estuary level, initially the same as that of the river
level at the point of discharge, falls at the rate of 0.25 m/hour over a period of 3
hours. Neglecting bed slope and resistance, determine how long it takes for the river
level to fall by 0.50 m at a section 1km upstream from the mouth of the channel. At
this time, how far upstream will the river level just be starting to fall?

Solution:

1.     Take the origin of spatial co-ordinates to be the mouth of the channel, and the
       +ve x direction to be upstream. The initial conditions may then be expressed
       as:

               v o  1.0m / s         co  gy  2g  4.43m / s
       and
               (v o  c o )  3.43 m / s

2.     The problem definition sketch is shown in Figure 8.4. For the first part of the
       problem we are seeking the point B, 1000m upstream of the estuary at A. We
       are seeking the time at which the depth at this point will be 1.50m.




                                    Problem definition sketch

       At point B, the wave speed:         c  1.5g  3.84m / s

       To solve the problem we are seeking a characteristic along which c is
       constant and equal to 3.84 m/s. This would originate from the estuary mouth
       when the depth at the estuary mouth was 1.50 m, which from the rate of
       decline of estuary levels would occur after 2 hours.

3.     The slope of the characteristic from A to B may now be determined:
             dx
                 3c(t )  v o  2c o  3  3.84  1.0  8.86  1.66m / s
             dt

4.   Knowing that the distance from A to B is 1 km, the time interval is 602
     seconds or 10 mins 2 seconds. The total time for the water level to fall by 0.5
     m at a point 1 km upstream of the estuary is thus 2 hours and 10 minutes and
     2 seconds.

5.   The distance upstream to the zone of quite at this time may be determined
     directly now from the wave speed of the first characteristic, c0:

     Distance = (v o  c o )  (2  3600  602 ) = 26.8 km.
Example 2

A river flows at a depth of 2m and with a velocity of 1m/s in a rectangular channel
discharging into an estuary. The estuary level is initially at the same level as the river,
and then starts to rise at a rate of 1 m/hour for a period of 3 hours. Determine where
and when a surge will first develop. At the time of development of the surge, what
will be the depth of midway between the surge and the river mouth?

Solution:

1.     Take the origin of the spatial co-ordinates to be the mouth of the channel, and
       the +ve x direction to be upstream:

        v o  1.0m / s         and co     gy  2g  4.43m / s

        (v o  c o )  3.43 m / s

2.     The intersection of characteristics is given by:

               (3c(t )  vo  2co ) 2
        xs 
                    3dc(t ) / dt

       Which for the first characteristic, may be written as:

               (c o  v o ) 2
        xs 
                 3dc / dt

                                dc dc dy 1 g dy                      dy   1
       Since, c  gy ,                        , and at the origin,    
                                dt dy dt 2 y dt                      dt 3600

                  dc 1 1            9.81
       Then                              3.08  10  4
                  dt 2 3600          2

                            3.43 2
       and        xs                  12 .73 km
                         9.24  10 4

3.     The time at which this occurs can be determined from the first characteristic:

        dx
            (v o  c o )  3.43m / s
        dt

                           12 .73 10 3
       Travel time =                     3711 s  1 hr 1.8 min
                               3.43

4.     The depth at the midway position can only be computed by trial and error.
       The distance to the point of interest is 6.36 km, and we wish to know the
       depth at this point after 3711 seconds. If we assume that the depth would be
       2.5 m (and therefore that this is the depth at which the characteristic starts at
the origin) the corresponding wave speed would be       2.5g =4.95 m/s. The
corresponding characteristic gradient would be:

dx
    3  4.95  1.0  2  4.43  5.03 m / s
dt

                6.36  10 3
Travel time =                1264 s
                   5.03

Total time=time for depth to get to 2.5m at start plus above travel time

Time for depth to get to 2.5m = half an hour = 1800s

Total time=1800+1264=3064s

This is a bit too short. Need to increase the depth/start time.

Try depth = 2.67m.

Wave speed would be        2.67g =5.11 m/s. The corresponding characteristic
gradient would be:

dx
    3  5.11  1.0  2  4.43  5.49 m / s
dt

                6.36  10 3
Travel time =                1158 s
                   5.49

Total time=time for depth to get to 2.67m at start plus above travel time

Time for depth to get to 2.67m = 2412s

Total time=2412+1158=3570s

Which is nearer. Reiterate ….

								
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