QUESTIONS

1. 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.

2. What is critical flow in terms of specific energy?

Critical flow corresponds to minimum specific energy.

3. 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.

4. 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.

5. What causes a hydraulic jump?

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

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

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

The gravitational force.

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

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

9. What is the difference between alternate depths and sequent depths?

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

10. 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).

11. 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

1. 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.

2. 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.

3. 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 = bc / 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) ]

yc = (2/3) E1 = (2/3) y1 [ 1 + (F12/2) ]

yc / y1 = (2/3) [ 1 + (F12/2) ] = (1/3) (2 + F12)

From water continuity:

Q = v1 y1 b1 = vc yc bc

σ = bc / b1 = (v1 y1) / (vc yc)

σ = v1 / [ vc (yc / y1) ]

σ = v1 / [ ( g yc ) 1/2 (yc / y1) ]

σ = v1 / [ ( g y1 ) 1/2 (yc / y1) 3/2 ]

σ = F1 / (yc / y1) 3/2

σ = F1 / [ (1/3) (2 + F12) ] 3/2

σ = (27)1/2 F1 / (2 + F12) 3/2  ANSWER.

4. Use ONLINE CHANNEL 17 to calculate the required throat width bc 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 bc = 1.445 m.  ANSWER.

Q = v1 y1 b1 = 1.2 × 2.2 × 2.2 = 5.808 m3/s

Run ONLINE CHANNEL 17 to calculate bc = 0.994 m.  ANSWER.

5. Using the specific force principle, show that the force fo (in kN/m) exerted by a blunt obstruction at the bottom of a wide rectangular channel is:

fo = - γ y12 { [ (1 - α 2) / 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 fo. Assuming the flow is from left to right, in what direction is the force fo acting?

The conservation of specific force leads to the force ON the obstruction:

fo, ON = f1 - f2

fo, ON = γ [ q2/(gy1) + (y12/2) ] - γ [ q2/(gy2) + (y22/2) ]

fo, ON = γ [ q2/(gy1) + (y12/2) - q2/(gy2) - (y22/2) ]

fo, ON = γ y12 [ q2/(gy13) + 1/2 - q2/(gy2y12) - y22/(2y12) ]

fo, ON = γ y12 [ q2/(gy13) + 1/2 - q2/(g y13α) - (α2/2) ]

fo, ON = γ y12 [ F12 + 1/2 - (F12 / α) - (α2/2) ]

fo, ON = γ y12 { [ (1 - α 2) / 2 ] +   [ 1 -   (1/α ) ] F12 }

The force exerted by the obstruction is:   fo = - fo, ON

fo = - γ y12 { [ (1 - α 2) / 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

fo = - 9.806 (2.25) [ (1 - 0.8281)/2 + 0.068 (1 - (1/0.91) ) ] = - 1.748 kN/m  ANSWER.

Because fo is negative, it acts in a direction opposite to the flow, i.e., from right to left.  ANSWER.

6. 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 = bc / 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  =  Mc (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  =  vc yc bc (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:

 qc2             yc2    Mc  =  _______  +  _______              g yc             2 (8)

Replacing the discharge Q:

 Q2                yc2    Mc  =  ___________  +  ______              g yc bc2             2 (9)

From continuity:

 v32 y32 b32            yc2    Mc  =  _______________  +  ______                  g yc bc2                2 (10)

Or:

 v32 y33 b32            yc2    Mc  =  _______________  +  ______                g yc bc2 y3              2 (11)

Replacing Eq. 6 in Eq. 11:

 F32 y33            yc2    Mc  =  ___________  +  ______                σ2 yc               2 (12)

By definition, the critical depth is:

 q2                   yc  =  ( ____ ) 1/3               g (13)

Or:

 Q2                   yc  =  ( ________ ) 1/3                bc2g (14)

From continuity:

 v32 y32 b32                    yc  =  ( _____________ ) 1/3                    bc2g (15)

Or:

 v32 y33 b32                    yc  =  ( _____________ ) 1/3                  bc2g y3 (16)

Replacing Eq. 6 in Eq. 16:

 F32 y33                    yc  =  ( _________ ) 1/3                  σ2 (17)

Replacing Eq. 17 in Eq. 12:

 F32 y32 σ2/3                     F32 y33 Mc  =  ______________  +  (1/2) ( _________ )2/3                σ2 F32/3                            σ2 (18)

Reducing terms:

 F34/3 Mc  =  (3/2) ( ________ ) y32                          σ4/3 (19)

Equating Eqs. 7 and 19:

 1                                   F34/3 ( ___  +  F32 )  =  (3/2) ( ________ )     2                                    σ4/3 (20)

Reducing:

 33/4 F3 σ  =  _____________________             ( 1  +  2 F32 ) 3/4 (21)

7. 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.

8. 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.

9. 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 ys to tailwater depth y2 is:

 ys                                         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:

 V1 2y1           ys 2         V2 2y2          y2 2 _________  +  ______  =  ________  +  ______           g                2               g                2 (23)

 ys 2         V2 2y2          y2 2         V1 2y1 ______  =  ________  +  ______  -  ________        2                 g               2               g (24)

 ys 2          2 V2 2         2          2 V1 2y1 ______  =  _______  +  _____  -  __________       y2 2           gy2            2            g y2 2 (25)

 ys 2          2 V2 2         2          2 V1 2y1 V2 2 ______  =  _______  +  _____  -  _______________       y2 2           gy2            2           g y2 2 V2 2 (26)

From continuity:

 V1 y1  =  V2 y2 (27)

 ys 2                                         y2 ______  =  2 F22  +  1  -  2 F22 _____       y2 2                                         y1 (28)

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