The upstream flow rate per unit width (q) is:
Q 5,000
q=
=
= 50 cfs / ft
W
100
The average velocity (V) is:
Q 5,000
V=
=
= 5.00 ft / s
A 1,000
The specific head (H) is:
2
2
H = V + y = 5 + 10 = 10.39 ft
2g
64.4
According to Section 2.6.2 the maximum unit discharge for a given specific head (specific head
equals to a constant) occurs at the critical depth yc where the Froude number is 1. Equation
2.149 the critical depth yc for this specific head is:
1/ 3
2
2
qmax
2 Vc
2
yc =
= H=
g
2g
3
Solving for qm
1/ 2
1/ 2
2 3
3
2
= g H
= 32.2 x 10.39
= 103.4 cfs / ft
qmax
3
3
Therefore, the width of the channel can be contracted until the unit discharge q is a maximum.
The minimum width (maximum constriction) is:
Q
5,000
W min =
=
= 48.3 ft
qmax
103.4
and the maximum constriction is 100 48.3 = 51.7 ft
This contraction causes the flow to go to critical. This results in an undulating hydraulic jump
downstream. Also, when energy losses are considered there will be some backwater at the
constriction.
2.15.5 PROBLEM 5 Maximum Elevation of a Grade Control Structure
Without Backwater (Neglecting Energy Losses)
A low grade control structure (check dam) is to be placed across a stream downstream of a
highway bridge. The stream is degrading. The purpose of the check dam is to maintain the
2.86
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