Chapter 15 Two situations are of interest: an up step

15.38: PROBLEM DEFINITION

Situation:

Water ﬂows over a broad-crested weir.

Additional details are given in the problem statement.

Find:

The water surface elevation in the upstream reservoir (ft).

PLAN

Apply the Broad crested weir—Discharge equation.

SOLUTION

From Fig. 15.13 (EFM10e), C≈0.85

Broad crested weir—Discharge equation

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15.39: PROBLEM DEFINITION

Situation:

Water ﬂows in a rectangular channel.

Twosituationsareofinterest: anupstepandadownstepinbottomelev,eachof

30 cm.

V=3m/s

Initial depth = 3m

Find:

(a) Change in depth and water surface elevation for upstep of 30cm (m).

(b) Change in depth and water surface elevation for the downstep of 30cm (m).

(c) Maximum size of upstep so that no change in upstream depth occurs (m).

PLAN

Apply the speciﬁcenergyequationandchecktheFroudenumber.

SOLUTION

(a) Speciﬁc Energy Equation prior to upstep

Froude number

Apply upstep

SpeciﬁcEnergyEquation

Solving for y2yields

42

Then, for change in depth

So water surface drops from 3.0 to 2.49, which is a drop of 0.21 m.

For a downstep

Solving for y2gives

Then

where

Maximum size of step

15.40: PROBLEM DEFINITION

Situation:

Water ﬂows in a rectangular channel.

Two situations are of interest: an upstep and a downstep.

Additional details are provided in the problem statement.

Find:

(a) Change in depth and water surface elevation for the upstep (m).

(b) Change in depth and water surface elevation for the downstep (m).

(c) Maximum size of upstep so that no change in upstream depth occurs (m).

PLAN

Apply the speciﬁcenergyequationbyﬁrst calculating Froude number and critical

depth.

SOLUTION

For the upstep

=0.369

SpeciﬁcEnergyEquation

44

Solving, one gets y2=2.24 m. Then

Solving: y2=3.17 mor

Then

45

15.41: PROBLEM DEFINITION

Situation:

Water ﬂows over a gradual upstep, shown in ﬁgure.

Desired: Unit ﬂowrate q=6m2/sec .

Upstream depth y1=3m

Find:

Maximum value of ∆zto permit a unit ﬂow rate of 6 m2/s without increasing the

upstream depth (m).

Assumptions:

No energy loss.

SOLUTION

Critical depth equation

where ycis depth allowed over the hump for the given conditions.

SpeciﬁcEnergyEquation

46

15.42: PROBLEM DEFINITION

Situation:

A rectangular channel has a gradual contraction in width–additional details are

provided in the problem statement.

Find:

(a) Change in depth (m).

(b) Change in water surface elevation (m).

(c) Greatest contraction allowable so that upstream conditions are not altered (m).

SOLUTION

Froude number

SpeciﬁcEnergyEquation

Then

15.43: PROBLEM DEFINITION

Situation:

“Ship squat” problem when the draft of vessels approaches the depth of the ship

channel.

Ship reduces the cross-sectional area available for ﬂow in the channel.

Ship is lower when moving than it would be if it were stationary.

Vship =5kt =2.575m/s; y1= 35m; channel width = 200 m

Draft of ship is 29m when fully loaded; width ship = 63m; length ship = 414m

Find:

The change in elevation or “ship squat” of a fully loaded supertanker (m).

PLAN

Reference the water velocity to the ship, and apply the energy equation.

Apply the speciﬁc energy equation from a section in the channel upstream of the ship

to a section where the ship is located.

Then apply the ﬂow rate equation and solve for y2.

SOLUTION

SpeciﬁcEnergyEquation

Flow rate equation

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15.44: PROBLEM DEFINITION

Situation:

A rectangular channel has a small reach that is roughened with angle irons–

additional details are provided in the problem statement.

Find:

Determine the depth of water downstream of angle irons (ft).

PLAN

Apply the momentum principle for a unit width.

SOLUTION Momentum principle

49

15.45: PROBLEM DEFINITION

Situation:

Water ﬂows out of a reservoir into a steep rectangular channel–additional details

are provided in the problem statement.

Find:

Discharge (m

3/s).

Assumptions:

Negligible velocity in the reservoir and negligible energy loss. Then the channel

entrance will act like a broad crested weir.

PLAN

Apply the Broad crested weir—Discharge equation.

SOLUTION

where L=4mandH=3m. Then

50

15.46: PROBLEM DEFINITION

Situation:

A small wave is produced in a pond.

Pond depth = 6 in.

Find:

Speed of the wave (ft/s).

PLAN

Apply the wave celerity equation.

SOLUTION

Wave celerity

51

15.47: PROBLEM DEFINITION

Situation:

A small wave travels in a pool of water.

Depth of water is constant.

Wave speed = 1.5m/s.

Find:

Depth of water (m).

PLAN

Apply the wave celerity equation.

SOLUTION

Wave celerity

52

15.48: PROBLEM DEFINITION

Situation:

As ocean waves approach a sloping beach, they curve so that they are aligned

parallel to the beach.

Find:

Explain the observed phenomena.

PLAN

Apply the wave celerity equation.

SOLUTION

As the waves travel into shallower water their speed is decreased.

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15.49: PROBLEM DEFINITION

Situation:

For a hydraulic jump, __________. (Select all of the following that are

correct.)

a. the ﬂow changes from subcritical to supercritical.

b. the ﬂow changes from supercritical to subcritical.

c. signiﬁcant energy is lost.

d. the height of the water abruptly increases from the upstream to the downstream

cross-section.

e. the downstream and upstream depth are related quantitatively in terms of the

upstream Fr.

f. the energy equation is a better tool for analysis than the momentum equation.

SOLUTION

a. No

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15.50: PROBLEM DEFINITION

Situation:

Abaﬄed ramp is used to dissipate energy in an open channel–additional details

are provided in the problem statement.

Find:

(a) Head that is lost (ft).

(b) Power that is dissipated ( hp).

(c) Horizontal component of the force exerted by ramp on the water (lbf).

Assumptions:

The kinetic energy correction factors are ≈1.0.xpositive in the direction of ﬂow.

PLAN

Let the upstream section (where y=3ft) be section 1 and the downstream section

(y=2ft) be section 2. Solve for the velocities at 1 and 2 using the ﬂow rate equation.

Then apply the energy equation and power equation. Determine the force of ramp

by writing the momentum equation between section 1 and 2. Let Fxbe the force of

the ramp on the water.

SOLUTION

(a) Flow rate equation

(b) Power equation

55

(c) Momentum principle, for horizontal component

56

15.51: PROBLEM DEFINITION

Situation:

Water ﬂows out a reservoir, down a spillway and then forms a hydraulic jump near

the base of the spillway.

Flow rate is q=2.9m

3/sper mof width.

Additional details are provided in the problem statement.

Find:

Depth downstream of hydraulic jump (m).

PLAN

Apply the speciﬁc energy equation to calculate y1. Then calculate Froude number

in order to apply the Hydraulic jump equation.

SOLUTION

SpeciﬁcEnergy

Froude number

Hydraulic jump equation

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15.52: PROBLEM DEFINITION

Situation:

Water ﬂowsoutasluicegate

Depth = 32 cm

Q=5.2 m3/s per meter

Find:

(a) Determine if a hydraulic jump can exist.

(b) If the hydraulic jump can exist, calculate the depth downstream of the jump

(m).

PLAN

Calculate Froude number, then apply the hydraulic jump equation.

SOLUTION

Froude number

Hydraulic jump equation

58

15.53: PROBLEM DEFINITION

Situation:

A hydraulic jump is described in the problem statement.

Find:

Depth upstream of the hydraulic jump (ft).

PLAN

Apply the hydraulic jump equation.

SOLUTION

Hydraulic jump equation

whereFroudenumber

Then

However

Therefore

59

15.54: PROBLEM DEFINITION

Situation:

An obstruction in a channel causes a hydraulic jump.

On the upstream side of the jump: V1=8m/s,y1=0.40 m.

Find:

Depth of ﬂow downstream of the jump (m).

PLAN

Calculate the upstream Froude number. Then apply the Hydraulic jump equation

to ﬁnd the downstream depth.

SOLUTION

Froude number

Hydraulic jump equation

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