QUESTIONS

1. How does rapidly varied flow differ from gradually varied flow?

Rapidly varied flow differs from gradually varied flow in the greater curvature of the streamlines. In extreme cases, the flow is virtually broken, resulting in high levels of turbulence and associated energy loss.

2. Is there a theoretical solution of one-dimensional rapidly varied flow?

No. Empirical equations are used in practical applications.

3. What is the exponent of the rating in a broad-crested weir?

The exponent of the rating is:  3/2 = 1.5.

4. What is the coefficient of the rating for high spillways in U.S. Customary units?

4.03.

5. What does "ogee" stand for?

An ogee curve is shaped somewhat like an S, consisting of two archs that curve in opposite sides, so that the ends are approximately parallel.

6. What is the rationale for the use of a labyrinth spillway?

The labyrinth spillway is used to increase the effective length of the weir crest, when warranted due to an increased risk of hydrologic failure.

7. What is the risk when flow over a spillway exceeds the design stage?

Flow rates greater than the design flow rate will produce subatmospheric pressures that may lead to cavitation.

8. What flow condition produces a hydraulic jump?

The hydraulic jump is an open-channel flow phenomenon where the flow changes suddenly from supercritical to subcritical.

9. Is the hydraulic jump equation linear or nonlinear?

The hydraulic jump equation is nonlinear, but it approximates to a linear equation for Froude nymbers greater than 2.

10. How is the length of the hydraulic jump measured?

The length of the hydraulic jump is defined as the distance measured from the front face to a point (on the water surface) located immediately downstream of the roller.

PROBLEMS

1. An emergency spillway is being considered for the East Demerera Water Conservancy, in Guyana, to safeguard the integrity of the dam under conditions of climate change (similar to Fig. 9-20). Assume that the existing relief sluices would be inoperable during a major flood due to high tailwater. Determine the length of the free-overflow spillway required to pass the Probable Maximum Flood (PMF). The following data is applicable:

• PMF 1-day duration:   428 mm

• Hydrologic abstraction:   18 mm

• Contributing drainage area:   582 km2

• Time base of the flood hydrograph:   3 days

• Embankment crest elevation:   18.288 m

• Spillway crest elevation:   17.526 m

• Freeboard:   0.3 m

• Weir discharge coefficient:   1.45

For simplicity, assume a triangular-shaped flood hydrograph. Use all the freeboard to contain the PMF.

Fig. 9-20  The 8000-ft weir, Boerasirie Conservancy, Guyana.

The volume of runoff is:

Vr = (428 - 18) mm × 582 km2 × (1 m / 1000 mm) × (1000 m / 1 km)2

Vr = 238,620,000 m3

Time base Tb = 3 d × 86,400 s/d = 259,200 s

Assuming a triangular-shaped flood hydrograph:

Qp = 2 Vr / Tb = (2) × (238,620,000) / (259,200)

Qp = 1841.2 m3

When the entire freeboard is used, the head H is:

H = 18.288 - 17.526 = 0.762 m

L = Q / (C H 3/2) = 1841.2 / (1.45 × 0.7623/2) = 1909 m.  ANSWER.

2. Design an overflow-spillway section having a vertical upstream face and a crest length L = 150 ft. The design discharge is Q = 50,000 ft3/s. The upstream water surface at design discharge is at Elev. 750 ft, and the average channel floor is at Elev. 650 ft (see Fig. 9-6 for graphical example).

Q = 50,000 cfs; L = 150 ft.

h + Hd = 750 - 650 = 100 ft.

Assume h / Hd > 1.33: Cd = 4.03

Q = Cd L He1.5

He = [50000/(4.03 × 150)]2/3 = 18.98 ft.

Va = Q / [L (h + Hd)] = 50000/(150 × 100) = 3.33 fps.

Ha = Va2/(2g) = 3.332/(2 × 32.17) = 0.17 ft.

Hd = He - Ha = 18.98 - 0.17 = 18.81 ft.

h = 100 - Hd = 100 - 18.81 = 81.19 ft.

h/ Hd = 81.19 / 18.81 = 4.31 > 1.33 OK!

Crest elevation:  750 - 18.81 = 731.19 ft.

Crest shape:  X 1.85 = 2 Hd 0.85 Y

X1.85 = 2 (18.81) 0.85 Y

Y = 0.04128 X 1.85

Y' = 1.0/0.6 = 1.85 × 0.04128 X 0.85

1.66667 = 0.0764 X 0.85

21.82 = X 0.85

X = (21.82)1.176 = 37.54 ft.  ANSWER.

Y = 0.04128 X 1.85 = 33.77 ft.  ANSWER.

0.282 Hd = 5.30 ft.  ANSWER.

0.175 Hd = 3.29 ft.  ANSWER.

0.5 Hd = 9.40 ft.  ANSWER.

0.2 Hd = 3.76 ft.  ANSWER.

3. Use ONLINE OGEE RATING to determine the rating for an ogee spillway with length L = 15 m, design head Hd = 2 m, spillway crest elevation = 1045 m, river bed elevation = 1000 m, and freeboard Fb = 1 m. Neglect the approach velocity. What should be the spillway length to pass the Probable Maximum Flood QPMF = 250 m3/s while taking all the freeboard? Express spillway length to nearest 0.1 m by excess.

Run ONLINE OGEE RATING with spillway length L = 15 m, design head Hd = 2 m, dam height P = 45 m, freeboard Fb = 1 m, and spillway crest elevation E = 1045 m, to give: (1) design discharge at Elev. 1047 is Q = 92.647 m3/s; (2) discharge to use all freeboard, at Elev. 1048, is QFb = 180.246 m3/s.  ANSWER.

By trial and error, for spillway length L = 20.9 m, the discharge to use all freeboard, at Elev. 1048, is QFb = 251.143 m3/s.  ANSWER.

4. Prove Eq. 9-14.

 ΔE  =  E1  -  E2  =  y1 + [V1 2/(2g)]  -  y2 - [V2 2/(2g)] (1)

 ΔE  =  y1 + y1 [F1 2/2]  -  y2 - y1 [F1 2/2] (V2 / V1)2 (2)

 ΔE  =  y1 + y1 [F1 2/2]  -  y2 - y1 [F1 2/2] (y1 / y2)2 (3)

The hydraulic jump equation, Eq. 9-13:

 y2         1                                ____  =  ____  [ (1  +  8 F1 2 )1/2  -  1 ]   y1         2 (4)

Combining Eqs. 3 and 4, after some algebraic manipulation:

 (y2  -  y1)3                                 ΔE  =  _____________                  4 y1 y2 (5)

5. Prove Eq. 9-19.

hj / E1 = (y2 - y1) / E1

hj / E1 = [(y2 / y1) - 1 ] / (E1/y1)

hj / E1 = [(2 y2 / y1) - 2 ] / [2 (E1/y1)]

hj / E1 = [(1 + 8F12)1/2 - 3 ] / [2 (E1/y1)]

(E1/y1) = 1 + (1/2)F12

(2 E1/y1) = 2 + F12

hj / E1 = [(1 + 8F12)1/2 - 3 ] / [ 2 + F12 ]

6. Calculate the energy loss in a hydraulic jump, given the sequent depths y1 = 0.58 m and y2 = 2.688 m.

Using Eq. 9-14:  ΔE = 1.5 m.  ANSWER.

7. Calculate the relative height of the hydraulic jump hj / E1 for F1 = 3.

Using Eq. 9-19:  hj / E1 = 0.504  ANSWER.

8. Use ONLINE CHANNEL 11 to calculate the sequent depth y2 through a hydraulic jump when the discharge is q = 5 m2/s and the initial depth y1 = 0.58 m.

Run ONLINE CHANNEL 11 with v1 = 5.0 / 0.58 = 8.621 m2/s and y1 = 0.58 m to give y2 = 2.688 m.  ANSWER.

9. Use ONLINE CHANNEL 12 to determine the sequent depth y2 and energy loss ΔE through a hydraulic jump when the discharge is q = 10 ft2/s and the upstream flow depth is y1 = 0.5 ft.

v1 = q / y1 = 10 / 0.5 = 20 ft/s

Run ONLINE CHANNEL 12 with y1 = 0.5 ft and v1 = 20 ft/s to give: y2 = 3.285 ft and ΔE = 3.287 ft.  ANSWER.

10. Use ONLINE CHANNEL 16 to calculate the sequent depths through a hydraulic jump when the discharge is q = 10 ft2/s and the energy loss is ΔE = 3.287 ft.

Run ONLINE CHANNEL 16 with q = 10 ft2/s and ΔE = 3.287 ft to give: y1 = 0.5 ft and y2 = 3.284 ft.  ANSWER.

11. Use ONLINE CHANNEL 16 to calculate the sequent depths through a hydraulic jump when the discharge is q = 5 m2/s and the energy loss is ΔE = 1.5 m.

Run ONLINE CHANNEL 16 with q = 5 m2/s and ΔE = 1.5 m to give: y1 = 0.58 m and y2 = 2.687 m.  ANSWER.

12. Use ONLINE CHANNEL 18 to calculate the efficiency of the hydraulic jump E2 / E2 for q = 10 ft2/s and y1 = 0.5 ft.

v1 = q / y1 = 10 / 0.5 = 20 ft/s

Run ONLINE CHANNEL 18 with y1 = 0.5 ft and v1 = 20 ft/s to give: E2 / E2 = 0.51.  ANSWER.

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140821 22:30

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