Determination of flow surface profile imperfection allowing hydraulic drop at mid section of contraction? When Q = 100 cfs.
Solution:
E = y + V2/2g
E = y + Q2/2gA2
E = 5 + 1002 /2 × 32.2 × 102 × 52
E = 5.06’
Now
Yc =2E/3
Yc = 3.37’
And, E= Yc + Q2/2gA2
5.06 = 3.37 + 1002 /2 × 32.2 × bc2 × 3.372
Bc = 2.84’
Slope;
Before Mid Section
S1 = (10-2.84)/25 = 0.286
After Mid Section
S2 = (8-2.84)/25 = 0.206
Table: Flow Surface Profile
Length (L)
|
Breadth
|
Depth (y)
| |
( Ft )
|
( Ft )
|
( Ft )
|
( Ft )
|
0
|
10
|
5
|
0.589
|
5
|
8.57
|
4.974
|
0.696
|
10
|
7.14
|
4.934
|
0.851
|
15
|
5.71
|
4.858
|
1.096
|
20
|
4.28
|
4.672
|
1.555
|
25
|
2.85
|
3.494
|
3.249
|
30
|
3.88
|
4.565
|
1.77
|
35
|
4.91
|
4.778
|
1.31
|
40
|
5.94
|
4.875
|
1.047
|
45
|
6.97
|
4.928
|
0.873
|
50
|
8
|
4.96
|
0.75
|
Flow Surface Profile:
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