Hydraulic Structures: Fourth Edition

WORKED EXAMPLES 263

From equation (5.4), for S50 m,

! 1 0.0155

H

S

 1 0.0155

5.

5

0

0

32

0.846.

Forn0.028,d1/60.028/0.04; thus d0.118 m (Section 8.2).
For the design discharge, the uniform flow depth y 0 from q
y 0 5/3S 0 1/2/n,

y 0 

3/5

6.416 m,



d(2.

R

65

S

0

1)

0.0330.05 (Shields)

(Section 8.2); the river bed is therefore stable.
From equation (5.7) with the datum at river bed level,

50 5.032y 1  ;

by trial and error y 1 0.907 m.
From equation (5.8) for Fr^21 q^2 /gy^31  252 /(9.810.907^3 )85.38,

y 2  [ 1 (1 8 85.38)1/2]11.40 m.

Asy 2 #y 0 a stilling basin is required.

For1.2, from equation (5.9),

y1.211.40 6.4167.264 m.

Assumey7.50 m (will be reduced by lowering the datum) and repeat
the computation:

E 50 5.0327.5062.532, ! 1 (0.015557.5)/5.0320.823.

From 62.532y 1  252 /(19.620.823^2 y^21 ),y 1 0.873 m, Fr^21 95.75, and y 2 
11.65 m, 1.19 (satisfactory).
Check the design for a smaller discharge, say 10 m^3 s^1 :

y 0 

3/5

3.7 m.

0.028 10



(0.001)1/2

0.907



2

252



19.60.846^2 y^21

6.4160.001



0.1181.65

0.028 25



(0.001)1/2