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posted:
11/12/2011
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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

Flow is unsteady and nonuniform, so we ask.

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









144a9c49-ce13-41d4-9dd2-45e61ddfd8d8.xlsx, 11/12/2011



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