Fundamentals of Open-channel Hydraulics, Chapter 1, Introduction, Victor Miguel Ponce, San Diego State University, 2014

QUESTIONS

What is the difference between total energy and specific energy?

Total energy considers the bed elevation; specific energy does not. Therefore, specific energy is applicable to channels of very mild slope, where the difference in bed elevation can be neglected.

What is critical flow in terms of specific energy?

Critical flow corresponds to minimum specific energy.

What causes a hydraulic drop?

The hydraulic drop is triggered by a sharp depression in the water surface, usually caused by an abrupt change in bottom elevation.

Why is the flow depth at a free overfall less than critical?

Flow in the proximity of a free overfall is usually rapidly varied; therefore, the brink depth is somewhat smaller than the critical depth computed by the theory of parallel flow.

What causes a hydraulic jump?

The hydraulic jump is triggered by an abrupt rise in the water surface as the flow progresses downstream.

Why is momentum unsteady?

Momentum is equal to a force integrated over a period of time; therefore, it is essentially unsteady.

What is the body force considered in open-channel flow?

The gravitational force.

What is the magnitude of the body force in a horizontal channel?

Zero; the gravitational force does not exist in a horizontal channel.

What is the difference between alternate depths and sequent depths?

Alternate depths have the same specific energy; sequent depths do not.

Where does specific force apply?

In a horizontal channel of short length, where the gravitational force is zero and the frictional force can be neglected, compared to the remaining forces (pressure-gradient force and inertia force).

For what type of problems is specific force used?

For those problems where forces can be shown to play a significant role in the solution.

PROBLEMS

Derive the relation for the discharge per unit of width q under a sluice gate, as a function of upstream depth y₁ and downstream depth y₃.

Fig. 3-12 Discharge under a sluice gate.

The conservation of specific energy between the u/s flow and the d/s flow is:
y₁ + v₁²/(2g) = y₃ + v₃²/(2g)
y₁ + q²/(2gy₁²) = y₃ + q²/(2gy₃²)
[q²/(2g)] [ (1/y₁²) - (1/y₃²) ] = y₃ - y₁
q²/(2g) = (y₁² y₃²) / (y₁ + y₃)
q = y₁ y₃ [ 2 g / (y₁ + y₃) ] ^1/2 ANSWER.
Prove that the equation derived in the previous problem (Problem 1) is mathematically equivalent to the equation based on y₁ and y₂ on used in ONLINE CHANNEL 13.

The two equations are mathematically the same.
Using the specific energy principle, derive the formula for the dimensionless throat width of a channel constriction that forces critical flow through it [Henderson (1966), p. 267].
σ = b_c / b₁ = (27)^1/2 F₁ / (2 + F₁²) ^3/2

Fig. 3-13 Critical width constriction using specific energy.

The specific energy for the u/s flow is:
E₁ = y₁ + v₁²/(2g) = y₁ [ 1 + (F₁²/2) ]
y_c = (2/3) E₁ = (2/3) y₁ [ 1 + (F₁²/2) ]
y_c / y₁ = (2/3) [ 1 + (F₁²/2) ] = (1/3) (2 + F₁²)
From water continuity:
Q = v₁ y₁ b₁ = v_c y_c b_c
σ = b_c / b₁ = (v₁ y₁) / (v_c y_c)
σ = v₁ / [ v_c (y_c / y₁) ]
σ = v₁ / [ ( g y_c ) ^1/2 (y_c / y₁) ]
σ = v₁ / [ ( g y₁ ) ^1/2 (y_c / y₁) ^3/2 ]
σ = F₁ / (y_c / y₁) ^3/2
σ = F₁ / [ (1/3) (2 + F₁²) ] ^3/2
σ = (27)^1/2 F₁ / (2 + F₁²) ^3/2 ANSWER.
Use ONLINE CHANNEL 17 to calculate the required throat width b_c for the following upstream conditions: y₁ = 2.2 m, v₁ = 1.2 m/s, and b₁ = 3.2 m. What would be the required throat width if the upstream channel width is b₁ = 2.2 m?

Q = v₁ y₁ b₁ = 1.2 × 2.2 × 3.2 = 8.448 m³/s
Run ONLINE CHANNEL 17 to calculate b_c = 1.445 m. ANSWER.
Q = v₁ y₁ b₁ = 1.2 × 2.2 × 2.2 = 5.808 m³/s
Run ONLINE CHANNEL 17 to calculate b_c = 0.994 m. ANSWER.
Using the specific force principle, show that the force f_o (in kN/m) exerted by a blunt obstruction at the bottom of a wide rectangular channel is:
f_o = - γ y₁² { [ (1 - α²) / 2 ] + [ 1 - (1/α ) ] F₁² }
where γ = unit weight of water, F₁ = upstream Froude number, y₁ = upstream flow depth, and α = y₂/y₁, where y₂ = downstream flow depth (after the obstruction). Given q = 1.5 m²/s, v₁ = 1.0 m/s, and α = 0.91, calculate the force f_o. Assuming the flow is from left to right, in what direction is the force f_o acting?

The conservation of specific force leads to the force ON the obstruction:
f_o, ON = f₁ - f₂
f_o, ON = γ [ q²/(gy₁) + (y₁²/2) ] - γ [ q²/(gy₂) + (y₂²/2) ]
f_o, ON = γ [ q²/(gy₁) + (y₁²/2) - q²/(gy₂) - (y₂²/2) ]
f_o, ON = γ y₁² [ q²/(gy₁³) + 1/2 - q²/(gy₂y₁²) - y₂²/(2y₁²) ]
f_o, ON = γ y₁² [ q²/(gy₁³) + 1/2 - q²/(g y₁³α) - (α²/2) ]
f_o, ON = γ y₁² [ F₁² + 1/2 - (F₁² / α) - (α²/2) ]
f_o, ON = γ y₁² { [ (1 - α²) / 2 ] + [ 1 - (1/α ) ] F₁² }
The force exerted by the obstruction is: f_o = - f_o, ON
f_o = - γ y₁² { [ (1 - α²) / 2 ] + [ 1 - (1/α ) ] F₁² }
y₁ = q/v₁ = 1.5 / 1.0 = 1.5 m.
y₁² = (1.5)² = 2.25 m²
y₁³ = (1.5)³ = 3.375 m³
γ = 9.806 kN/m³
α = 0.91
F₁² = q²/(g y₁³) = (1.5)² / (9.806 × 3.375 ) = 0.068
f_o = - 9.806 (2.25) [ (1 - 0.8281)/2 + 0.068 (1 - (1/0.91) ) ] = - 1.748 kN/m ANSWER.
Because f_o is negative, it acts in a direction opposite to the flow, i.e., from right to left. ANSWER.

Using the specific force principle, derive the formula for the dimensionless throat width of a channel constriction that forces critical flow through it [modified from Henderson (1966), p. 267].

σ = b_c / b₃ = (3)^3/4 F₃ / (1 + 2 F₃²) ^3/4

Fig. 3-14 Critical width constriction using specific force.

The momentum-based equation is derived by equating the specific force at critical flow at the contraction (Section 2, or Section c) with the specific force at the downstream flow section (Section 3).

M₃ = M₂ = M_c (1)

The specific force at the downstream flow section is:

q₃² y₃²
M₃ = ^_______ + ^_______
g y₃ 2 (2)

Replacing the discharge Q:

Q² y₃²
M₃ = ^___________ + ^______
g y₃ b₃² 2 (3)

From continuity:

Q = v₃ y₃ b₃ = v_c y_c b_c (4)

Thus:

v₃² y₃² y₃²
M₃ = ^__________ + ^______
g y₃ 2 (5)

The Froude number at the downstream section is defined as follows:

v₃²
F₃² = ^_______
g y₃ (6)

Thus, Eq. 5 reduces to:

1
M₃ = y₃² ( ^____ + F₃² )
2 (7)

The specific force at the contraction is:

q_c² y_c²
M_c = ^_______ + ^_______
g y_c 2 (8)

Replacing the discharge Q:

Q² y_c²
M_c = ^___________ + ^______
g y_c b_c² 2 (9)

From continuity:

v₃² y₃² b₃² y_c²
M_c = ^{_______________} + ^______
g y_c b_c² 2 (10)

Or:

v₃² y₃³ b₃² y_c²
M_c = ^{_______________} + ^______
g y_c b_c² y₃ 2 (11)

Replacing Eq. 6 in Eq. 11:

F₃² y₃³ y_c²
M_c = ^___________ + ^______
σ² y_c 2 (12)

By definition, the critical depth is:

q²
y_c = ( ^____ ) ^1/3
g (13)

Or:

Q²
y_c = ( ^________ ) ^1/3
b_c²g (14)

From continuity:

v₃² y₃² b₃²
y_c = ( ^{_____________} ) ^1/3
b_c²g (15)

Or:

v₃² y₃³ b₃²
y_c = ( ^{_____________} ) ^1/3
b_c²g y₃ (16)

Replacing Eq. 6 in Eq. 16:

F₃² y₃³
y_c = ( ^_________ ) ^1/3
σ² (17)

Replacing Eq. 17 in Eq. 12:

F₃² y₃² σ^2/3 F₃² y₃³
M_c = ^{______________} + (1/2) ( ^_________ )^2/3
σ² F₃^2/3 σ² (18)

Reducing terms:

F₃^4/3
M_c = (3/2) ( ^________ ) y₃²
σ^4/3 (19)

Equating Eqs. 7 and 19:

1 F₃^4/3
( ^___ + F₃² ) = (3/2) ( ^________ )
2 σ^4/3 (20)

Reducing:

3^3/4 F₃
σ = ^{_____________________}
( 1 + 2 F₃² ) ^3/4 (21)

ANSWER.

Using ONLINE LIMITING CONTRACTION, calculate the limiting contraction ratios using both energy and momentum principles, for a Froude number F = 0.3.

Using the specific energy principle, for F = 0.3: σ = 0.516 ANSWER.
Using the specific force principle, for F = 0.3: σ = 0.604 ANSWER.
Using ONLINE LIMITING CONTRACTION SET, calculate the limiting contraction ratios using both energy and momentum principles, for Froude numbers is the range 0.1 ≤ F ≤ 2.0, at intervals of 0.1. Discuss the results.

Using the specific energy principle, σ varies between 0.182 and 1.0, for F = 0.1 to F = 1.0; then, it reduces to σ = 0.707 at F = 2. ANSWER.
Using the specific force principle, σ varies between 0.225 and 1.0, for F = 0.1 to F = 1.0; then, it reduces to σ = 0.877 at F = 2. ANSWER.

A submerged hydraulic jump occurs immediately downstream of a sluice outlet in a rectangular channel. Using the momentum principle, prove that the ratio of submerged depth y_s to tailwater depth y₂ is:

y_s y₂
^_____ = [ 1 + 2 F₂² ( 1 - ^_____ ) ]^1/2
y₂ y₁ (22)

Fig. 3-15 A submerged hydraulic jump at a sluice outlet.

The specific force principle applied between the upstream and downstream sections is:

  V₁²y₁           y_s²         V₂²y₂          y₂²
^_________ + ^______ = ^________ + ^______
      g                2               g                2 (23)

  y_s²         V₂²y₂          y₂²         V₁²y₁
^______ = ^________ + ^______ - ^________
   2                 g               2               g   (24)

  y_s²          2 V₂²         2          2 V₁²y₁
^______ = ^_______ + ^_____ - ^__________
  y₂²           gy₂            2            g y₂² (25)

  y_s²          2 V₂²         2         2 V₁²y₁ V₂²
^______ = ^_______ + ^_____ - ^{_______________}
  y₂²           gy₂            2           g y₂² V₂² (26)

From continuity:

V₁ y₁ = V₂ y₂ (27)

  y_s²                                         y₂
^______ = 2 F₂² + 1 - 2 F₂² ^_____
  y₂²                                         y₁ (28)

y_s y₂
^_____ = [ 1 + 2 F₂² ( 1 - ^_____ ) ]^1/2
y₂ y₁ (29)

ANSWER.

http://openchannelhydraulics.sdsu.edu

141110 13:30

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