Open Channel Flow -
Combination Application
Example: (Continued) The surface profile looks
approximately:
|
For
the case of a mild slope, y3 = yN
= 2.85 ft Also q
= y3V3, so V3
= 16/2.85 = 5.6 ft/sec Fr3 = V3
/ √ gy3 = 5.6 / √ (32.2)(2.85) =
0.34 Now
for a hydraulic jump, y2 /
y3 = [ √ ( 1 + 8 Fr32
) ˗ 1 ] / 2 So
y2 = 1.3 ft
(result) |
The
specific energy plot is as follows: |
The
energy dissipated across the hydraulic jump is E2 ˗ E3 =
0.3 ft (result) Click
here go to a summary of equations for open channel flow. |
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