15.55: PROBLEM DEFINITION
Situation:
A hydraulic jump is described in the problem statement.
γ=9,810 N/m2,B
3=5m.
y1=40cm =0.40 m, V=10m/s.
Find:
Depth of ow downstream of jump (m).
SOLUTION
Check Fr upstream to see if the ow is really supercritical ow. Then apply the
momentum principle.
Then
Then
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Also
Flow rate equation
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15.56: PROBLEM DEFINITION
Situation:
A hydraulic jump occurs in a wide rectangular channel.
The upstream depth is y1=0.5ft.
The downstream depth is y2=10ft.
Find:
Discharge per foot of width of channel ¡ft3/s/ft¢.
PLAN
Apply the Hydraulic jump equation to solve for the Froude number. Next, use the
value of the Froude number to solve for the discharge q.
SOLUTION
Hydraulic jump equation
Solve the above equation for Froude number.
Froude number
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15.57: PROBLEM DEFINITION
Situation:
A rectangular channel has three dierent reaches–additional details are provided
in the problem statement.
Find:
(a) Calculate the critical depth and normal depth in reach 1 (ft).
(b) Classify the ow in each reach (subcritical, critical or supercritical).
(c) For each reach, determine if a hydraulic jump can occur.
PLAN
Apply the critical depth equation. Determine jump height and location by applying
the hydraulic jump equation.
SOLUTION Critical depth equation
Thus one concludes that the normal depth in each reach is
Hydraulic jump equation
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15.58: PROBLEM DEFINITION
Situation:
Water ows out a sluice gate and then over a free overfall–additional details are
provided in the problem statement.
Find:
(a) Determine if a hydraulic jump will form.
(i) If a jump forms, locate the position.
(ii)If a jump does not form, sketch the full prole and label each part.
(b) Sketch the EGL
PLAN
Check Froude numbers. Then determine y1for a y2of 1.1 m by applying the hydraulic
jump equation.
SOLUTION Froude number
A hydraulic jump will form because ow goes from supercritical to subcritical.
Hydraulic jump equation
Therefore the jump will start at about the 29 m distance downstream of the sluice
gate. Prole and energy grade line:
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15.59: PROBLEM DEFINITION
Situation:
Water ows out of sluice gate and then through a hydraulic jump–additional
details are provided in the problem statement.
Find:
Horsepower lost in hydraulic jump (hp).
Assumptions:
Negligible energy loss for ow under the sluice gate.
PLAN
Apply the Bernoulli equation from a location upstream of the sluice gate to a location
downstream. Then, calculate the Froude number and apply the equations that govern
a hydraulic jump. Calculate the power using P=hL/550, where the number
“550” is a unit conversion.
SOLUTION
Bernoulli equation
V1=64 ×64.4=64.2ft/s
Froude number
Hydraulic jump equations
Power equation
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15.60: PROBLEM DEFINITION
Situation:
Water ows in a rectangular channel.
A sill installed on the bottom of the channel forces a hydraulic jump to occur.
Additional details are provided in the problem statement.
channel width = 8 m
y1=32cm
Find:
Estimate the height of hydraulic jump (the height is the change in elevation of the
water surface) (m).
Assumptions:
n=0.012.
PLAN
Calculate Froude number in order to apply the Hydraulic jump equation.
SOLUTION
Froude number
Hydraulic jump equation
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15.61: PROBLEM DEFINITION
Situation:
Water ows in a rectangular channel.
A sill installed on the bottom of the channel forces a hydraulic jump to occur.
Additional details are provided in the problem statement.
Find:
(a) Estimate the shear force associated with the jump (N/m2).
(b) Calculate the ratio Fs/FH,whereFsis shear force and FHis the net hydrostatic
force acting on the jump.
Assumptions:
(a) The shear stress will be the average of τ01(associated with uniform ow ap
proaching the jump), and τ02(associated with uniform ow leaving the jump).
(b) The ow may be idealized as normal ow in a channel.
PLAN
Apply the local shear stress Equation (10.11, EFM10e) and calculate the Reynolds
numbers. Then nd V2by applying the same solution procedure from Problem 15.60
(EFM10e). Then estimate the total shear force by using an average shear stress.
SOLUTION
Local shear stress
From solution to Prob. 15.60
Assume ks=3×103m
Then
Thus
REVIEW
The above estimate is probably inuenced too much by τ01because shear stress will
not be linearly distributed. A better estimate might be to assume a linear distribution
of velocity with an average fand then integrate τ0dA from one end to the other.
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15.62: PROBLEM DEFINITION
Situation:
Water ows out of a sluice gate–additional details are provided in the problem
statement.
Find:
(a) Determine the type of water surface prole that occurs downstream of the sluice
gate.
(b) Calculate the shear stress on bottom of the channel at a horizontal distance of
0.5 m downstream from the sluice gate (N/m2).
Assumptions:
The ow can be idealized as a boundary layer ow over a at plate, with the leading
edge of the boundary layer located at the sluice gate.
PLAN
Apply the hydraulic jump equation by rst calculating qapplying the ow rate equa
tion. Then apply the local shear stress equation.
SOLUTION
Flow rate equation
s
Hydraulic jump equation
Then we have y<y
n<y
c; therefore, the water surface prole will be an S3.
Reynolds number
The local shear stress coecient is
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Local shear stress
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15.63: PROBLEM DEFINITION
Situation:
Water ows in a rectangular channel
Q=100ft3/s
yn=2ft yactual =4ft width = 10 ft
Find:
Classify the water surface prole as
a.) S1
b.) S2
c.) M1
d.) M2
PLAN
Use principles of water classication for gradually-varied ow
SOLUTION
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15.64: PROBLEM DEFINITION
Situation:
Water surface is labeled with a question mark in gure in text.
Find:
Classify the water surface prole as one of the following:
a.) M2
b.) S2
c.) H2
d.) A2
PLAN
Use principles of water classication for gradually-varied ow
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15.65: PROBLEM DEFINITION
Situation:
The problem statement shows a partial sketch of a water-surface prole.
Find:
(a) Sketch the missing part of the water prole.
(b) Identify the various types of proles.
PLAN
Check the Froude number at points 1 and 2. Apply the Broad crested weir—Discharge
equation to calculate y2for the second Froude number.
SOLUTION
Froude number
Broad crested weir—Discharge equation
Solving by iteration gives H=0.917 m. Depth upstream of weir =0.917 + 1.6=2.52
m
Therefore a hydraulic jump forms.
Hydraulic jump equation
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