CE 356 Elements of Hydraulic Engineering

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CE 356 Elements of Hydraulic Engineering

• Specific Energy
• (Alternate Depths)
• Hydraulic Jump
• (Sequent Depths)

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Energy in Open Channels
Specific Energy
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Specific Energy, E
E energy (head) measured with respect to the
channel bottom
E V2/2g y q2/(2gy2) y
Multiply through by y2 and arrange to find
y3 E y2 q2/2g 0
What kind of equation is this? How many roots?
Significance of roots?
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Specific Energy Diagram
Rectangular channel B 20 ft Q 600 ft3/s.
E y
Fr 1
Fr 2
Fr V/(gy)0.5
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Example Sluice Gate
q 30 ft2/s
y 5.6 ft
y 1.8 ft
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figure 1
Hydraulic jump
Note that there is head loss in an hydraulic jump
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jump
Hydraulic jump on Rattan Creek, TX. July 2, 2002
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classification
Classification
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picture
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figure 2
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jump cv
Now lets consider an hydraulic jump
We know that
or
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momentum balance
Our momentum balance
can be written as
rearranging and dividing by ?g
Using ABy for a rectangular channel
Lets go back to our hydraulic jump
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solvable set
If we have an open channel where we know Q, B,
and y1, we can solve for y2
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rectangular
Starting from
Recalling the definition of the flow rate per
unit width
we can obtain
Solving for y2
y1 solution is similar (see text)
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limits
Now, what are the limitations of this equation?
shear forces are neglected shear causes smaller
head loss than turbulence in the jump requires
jump occurs over a short distance
rectangular channel
effect of gravity downslope is neglected again
requires jump to occur over a short distance
neglect non-hydrostatic pressure
uniform density fluid
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principle
and what is the fundamental principle for this
equation?
conservation of momentum
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EM graphs
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head loss
The more general head loss formula is
Valid for any cross section