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# Flood

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• pg 1
```									Given a rectangular channel:                 The discharge at pt a increases
Channel width b      25     (m)         from Q1 to Q2 over Dt                          Flood Water       Normal flow                      Note that a number of cells
Channel slope So    0.002                          Q1      10      (m^3/s)                                before flood                     are named and that those
Hydraulic roughness n    0.03                          Q2      20      (m^3/s)                                                                 names are used in formulas
Dt                          a
600      (s)
Enter Data only in green cells
          dh U dU 1 dU          Dh U DU 1 DU                                     To make approximatio
t   gR  S                         
 t   gR  S        
Dx g Dx g Dt 
which terms in the St Venant Eqn are significant? 0                   0                           0               0
          dx g dx g dt 
How does the shear stress change, approximately, over this reach of nonuniform flow?                              
>>> we need to estimate the magnitude of the different terms in the St Venant Eqn.
The length of the flood wave Dx is found from Uf = Dx/Dt, where Uf is the speed of the flood wave and is approximately 1.5 times the mean velocity in the reach.
Dx    902      (m)
We need to determine depth and velocity at the upstream and downstream end of the reach                       S0 2 / 3
UPSTREAM                                       DOWNSTREAM                                            U            R         and Q  UA
hu    0.70     (m)                               hd     0.46     (m)                                             n
Ru
Uu
0.67
1.14
(m)
(m/s)
Rd
Ud
0.44
0.87
(m)
(m/s)
so, for a rectangular channel (A  bh, P  b  2h)
 
We also need the mean velocity and hydraulic radius for the reach                                                            2/3
R    0.56     (m)                                                                       S0       bh
U    1.00     (m/s)                                                            Q                              bh or
Dh -0.243 (m)
DU    -0.27     (m/s)

n b  2h 2 / 3     
Now we can compute all of the components in the brackets of the St Venant eqn                                           3/ 5
Channel slope So      0.00200
1  nQ 
Dh/Dx -0.00027
drop if any of these terms are more than an              h                       
b  2h
2/5

(U/g)*(DU/Dx) -0.00003
(1/g)*(DU/Dt) -0.00005
b  S0 
         
S   0.00235
which can be solved iteratively for h
Mean to in nonuniform reach, approximated from St Venant eqn.
to    12.77 Pa
to in steady uniform flow before flood
8.72     Pa

f62f3a68-cbf0-4afb-935a-d675dd3cff5a.xlsx, 10/14/2011

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