CHAPTER 9: STEADY RAPIDLY VARIED FLOW 
9.1 THE SHARPCRESTED WEIR
Rapidly varied flow differs from gradually varied flow (Chapter 7) 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.
The following characteristics describe rapidly varied flow:
The curvature
of the streamlines is so pronounced that the pressure
distribution cannot be assumed to be hydrostatic.
Unlike gradually varied flow,
changes in flow variables take place
within a relatively short distance (Fig. 91).
The boundary friction is small compared to the other forces,
and often negligible.
The flow characteristics are fixed by the usually rigid
boundary geometry.
The velocity distribution coefficients α
and β are much greater than unity (1) and cannot be determined accurately.
Separation zones, eddies, and rollers tend to complicate
the flow pattern, with flow often confined to separation zones.
To date, there is no theoretical solution for rapidly varied flow. In lieu of theory,
empirical relations are used in many applications.
Fig. 91 A canal drop, a typical example of rapidly varied flow.


The sharpcrested weir
The sharpcrested weir is a measuring device in openchannel flow (Chapter 4). It is also the simplest form of overflow
spillway. The profile of the spillway can be made to match the shape of the lower surface of the flow nappe.
The shape of the flow nappe may be interpreted by the principle of the
projectile (Chow, 1959) (Fig. 92).
According to this principle, the horizontal velocity component of the flow is constant, and the only
force acting on the nappe is the gravitational force. In time t, a particle of water in the lower surface of the nappe
will travel a horizontal distance x (from the face of the weir) equal to:
Fig. 92 Nappe profiles over sharpcrested weir by the principle of the projectile (Chow, 1959).


In the same time, the particle will travel a vertical distance y equal to:
y =  v_{o} t sin θ + (1/2) g t^{ 2} + C'
 (92)

in which g = gravitational acceleration, and C' is the value of y at x = 0. The constant C' may be taken as the vertical distance between the highest point of the nappe
and the elevation of the crest.
Eliminating t from Eqs. 91 and 92:
y x x
^{____} = A (^{____})^{2} + B (^{____}) + C
H H
H
 (93) 
in which:
g H
A = ^{ ________________ }
2 v_{o}^{2} cos^{2} θ
 (94) 
C'
C = ^{ ______ }
H
 (96) 
Assuming the vertical thickness of the nappe T, an additional term may be added to Eq. 95 to yield the nappe equation:
y x x
^{____} = A (^{____})^{2} + B (^{____}) + C + D
H H
H
 (97) 
in which D = T / H .
Discharge of the sharpcrested weir
A common formula for the discharge over a sharpcrested weir is:
in which C = discharge coefficient, L = effective length of the weir crest, and H = measured head
above the crest, excluding the velocity head.
Discharge coefficients for various types of sharpcrested weirs are given in Chapter 4.
9.2 CREST SHAPE OF OVERFLOW SPILLWAY
The spillway shape shown in Fig. 93 is referred to as an ogee shape, because it resembles
an ogee curve, shaped somewhat like an S, consisting of two archs
that curve in opposite sides, so that the ends
are approximately parallel.
In practice, notwithstanding the weir shape, negative pressures may develop in an ogee spillway.
Excessive negative pressures may lead to cavitation damage and are, therefore, to be controlled.
Fig. 93 (a) Emergency spillway of Turner reservoir, in San Diego, California, not spilling.


Fig. 93 (b) Emergency spillway of Turner reservoir, in San Diego, California, spilling on February 24, 2005.


WES spillway shapes
The U.S. Army Corps of Engineers has developed several standard shapes at its Waterways
Experiment Station, in Vicksburg, Mississippi. These shapes, shown in Fig. 94,
are referred to as the WES standard spillway shapes.
[Click here to display]
Fig. 94 The WES standard spillway shapes (Chow, 1959).


The WES shapes follow the equation:
X^{ n} = K H_{d}^{ n1} Y
 (99)

in which X and Y are the coordinates of the crest profile, with the origin of coordinates at the highest
point of the crest; H_{d} = design head excluding the velocity head
of the approach flow; and K and n are parameters which depend on the slope of the
upstream face. The values of K and n are shown in Table 91. For intermediate values,
interpolation is possible.
Table 91 Values of K and n (Eq. 99).

Slope on upstream face
 K
 n

Vertical
 2.000
 1.850

3 V : 1 H
 1.936
 1.836

3 V : 2 H
 1.939
 1.810

3 V : 3 H
 1.873
 1.776

The upstream face of the spillway may sometimes be designed to set back, as shown by the dashed lines of Fig. 9.4.
The set back will not affect the shape of the crest, provided the modification begins with at least onehalf of the
total head H_{e} below the origin. Below this depth, the vertical velocities are small, and the
corresponding effect on the nappe profile is negligible.
For WES shapes, the discharge over the spillway is:
Q^{} = C L H_{e}^{ 3/2}
 (910)

in which H_{e} = total energy head on the crest, including the velocity head in the approach channel.
Model test have shown that the effect of the approach velocity is negligible when the height h of the spillway
is greater than 1.33 H_{d}, where H_{d} = design head, excluding the approach velocity.
Under this condition, the coefficient of discharge, in U.S. Customary units, is C_{d} = 4.03.
For low spillways, where h/H_{d} < 1.33, the approach velocity will have appreciable effect on the discharge and, therefore, on the nappe profile. Figure 95 shows the effect of the approach
velocity on the relationship between (H_{e} /H_{d}) and (C /C_{d})
for WES spillways with a vertical upstream face.
From Fig. 95, for h /H_{d} ≥ 1.33
and H_{e} > H_{d}, the ratio
C/C_{d} > 1.
It is seen that the ratio C /C_{d} varies from 0.70 near crest level (H_{e} /H_{d} ≅ 0)
to 1.05 for H_{e} /H_{d} > 1.3. Thus, effectively, C varies from 2.82 to 4.23 (U.S. Customary units).
For a sloping upstream face, C is multiplied by the correction
factor shown in the topleft side of Fig. 95.
Fig. 95 Headdischarge relation for WES standard spillway shapes (Chow, 1959).


WES spillway: Design example
Determine the crest elevation and the shape of an overflow spillway section (WES standard spillway shape) having a vertical upstream face and a crest length
L = 250 ft. The design discharge is Q = 75,000 cfs.
The upstream water surface, at the design discharge, is at elevation 1,000 ft, while
the channel floor is at elevation 880 ft (Fig. 96).
Fig. 96 Design of an overflow spillway section (Chow, 1959).


Solution
Assume a high overflow spillway, for which the effect of the approach
velocity is negligible. Then: H_{d} = H_{e} , and C = 4.03.
h + H_{d} = 1000  880 = 120 ft.
From Eq. 910:
Q = C L H_{e}^{3/2}
75,000 = (4.03) (250) H_{e}^{3/2}
Therefore, the total energy head, including the velocity head, is: H_{e} = 17.7 ft.
The approach velocity is: V_{a} =
Q / [L (h + H_{d} ) ] = 75,000 / [(250) (120)] = 2.5 fps.
The velocity head is:
H_{a} = V_{a}^{ 2} / (2 g ) = 2.5^{2} / (2 × 32.17) = 0.097 ≅ 0.1
The design head, excluding the velocity head, is:
H_{d} = H_{e}  H_{a} = 17.7  0.1 = 17.6 ft.
The height of the dam is: h = 120  17.6 = 102.4 ft.
h = 102.4 > 1.33 H_{d} = 23.4
Therefore, the effect of the approach velocity is negligible.
The crest elevation is: 1000  17.6 = 982.4 ft.
From Eq. 99 and Table 91:
X^{ 1.85} = 2 H_{d} ^{ 0.85} Y
Therefore, the nappe equation is:
Y = X^{ 1.85} / (22.89)
To calculate the coordinates of the point of tangency (Fig. 96):
Y' = (1.85 / 22.89) X^{ 0.85} = 0.081 X^{ 0.85}
Y' = = ΔY /ΔX = 1 / 0.6 = 1.667
1.667 = 0.081 X^{ 0.85}
Solving for the abscissa of the point of tangency: X = 35.09 ft. ANSWER.
Solving for the ordinate of the point of tangency (Fig. 96): Y = X^{ 1.85} / (22.89) = 31.54 ft. ANSWER.
The length of the base, projected from the point of tangency at a slope 1 V : 0.6 H to the base of the dam
(see Figs. 94 and 96) is:
L_{b} = 0.282 H_{d} + X + (0.6) (h  Y )
L_{b} = [ 0.282 (17.6) ] + 35.09 + [ (0.6) (102.4  31.54) ]
L_{b} = 82.6 ft. ANSWER.
0.282 H_{d} = 4.96 ft. ANSWER.
0.175 H_{d} = 3.08 ft. ANSWER.
0.2 H_{d} = 3.52 ft. ANSWER.
0.5 H_{d} = 8.80 ft. ANSWER.

Example 91.
Solve the WES spillway design example using the calculator ONLINE WES SPILLWAY.
 
ONLINE CALCULATION. Use ONLINE WES SPILLWAY,
the results are the same as the hand calculation. For example, point 10 (the point of tangency) has X(10) = 35.182 ft
(compare to 35.09 ft), and Y(10) = 31.696 ft (compare to 31.54 ft).



Ogeetype spillways
Figure 97 shows several ogeetype spillways.
This type of spillway is preferred for high dams, because their discharge coefficient (C = 4.03 in U.S. Customary units and C = 2.22 in SI units)
is substantially higher than the theoretical value for broadcrested
weirs (C = 3.087 in U.S. Customary units and C = 1.704 in SI units) (Chapter 4 Critical Flow
Control).
Fig. 97 (a) Emergency spillway at El Capitan Dam,
on the San Diego river, San Diego, California.


Fig. 97 (b) Emergency spillway at Mangla Dam,
on the Jhelum river, Pakistan.


Fig. 97 (c) Emergency spillway at Oroville Dam,
on the Feather river, near Oroville, California.


Click here to view a video on the Arizona spillway of
Hoover Dam. 
Broadcrested spillways
Figure 98 shows two broadcrested spillways.
The 8000ft long relief weir (emergency spillway) of the Boerasirie Conservancy, shown in Fig. 98 (a),
has been designed with discharge coefficient varying from C = 1.45 at spill level to C = 1.78 at the design head H_{d} = 0.215 m (discharge coefficient
in SI units). The Valle Grande dam, shown in Fig. 98 (b), is an offstream storage reservoir, with a small contributing drainage area;
thus, no major floods are expected and the broadcrested weir is considered
sufficient to handle the design peak flow.
Fig. 98 (a) The 8000ft long relief weir,
Boerasirie Conservancy, Guyana.


Fig. 98 (b) Overflow spillway at Villa Grande Dam, Cuajone, Peru.


Labyrinth spillways
The labyrinth spillway is used to increase the effective length of the weir crest,
when warranted due to an increased risk of hydrologic failure.
Figure 99 shows three examples of labyrinth spillways. The labyrinth works well
at the design stage above spillway crest.
However, for stages that are higher than the design stage,
the effective length eventually reduces to the actual length
and the labyrinth ceases to provide the desired advantage.
Fig. 99 (a) Valentine Mill pond, Valentine, Nebraska.



Fig. 99 (b) Lower Las Vegas Wash detention basin, Las Vegas, Nevada.



Fig. 99 (c) Ute Dam, New Mexico.


Fuse spillways
A fuse spillway or fuse plug is an embankment designed to wash out in a predictable and controlled manner
when a capacity in excess of the normal capacity of the service spillway and
outlet works is required (Pugh and Gray, 1984). According to the
U.S. Bureau of Reclamation,
fuse plug designs have been selected for the Bartlett and Horseshoe dams, on the Verde River, Arizona.
The fuse plug for Bartlett Dam
is designed with an erosion resistant invert and abutment structure to pass 10,100 m^{3}/s.
Three erodible embankment sections will operate in sequence.
The Horseshoe Dam fuse plug is designed to pass 6,850 m^{3}/s
through three 44 to 52mlong openings 6.0m to 7.9m high.
Fig. 910 Fuse spillway, La Leche river, Lambayeque, Peru.


Other spillways features
Other spillway features include rubbergated spillways (Fig. 911), skyjump spillways (Fig. 912), and baffledissipation structures (Fig. 913).
Fig. 911 Rubbergated
emergency spillway, Arroyo Pasajero retention basin, near Coalinga, California.


Fig. 912 Skyjump spillway,
Oros dam, Jaguaribe river, Ceara, Brazil.


Fig. 913 Baffledissipation structure, Gallito Ciego dam, La Libertad, Peru.


Lake Oroville Spillway Incident
In February 2017, both flood control and emergency spillways at Lake Oroville, in northern California, failed
after 12.8 in of rain fell between February 6 and 10 on the Feather river basin.
The timeline of the events that led to the failure is documented in the video
Lake Oroville Spillway
Incident Timeline.
By September 2018,
the cost of repairs and other issues stemming from the failure had climbed to $1.1 billion.
Fig. 914 Aftermath of the failure of the flood control spillway at Lake Oroville.


9.3 RATING OF SPILLWAYS
At the design discharge, the pressure on the spillway crest is close to atmospheric.
The design head H_{D}, which includes the velocity head,
corresponds to the design discharge.
Flow rates less than the design flow rate will produce pressures on the spillway face above the
atmospheric pressure, while flow rates greater than the design flow rate will produce
subatmospheric pressures.
The discharge over an ungated ogee
spillway may be given as follows (Roberson et al., 1998):
Q_{D} = C_{D} (2 g)^{1/2} L H_{D}^{ 3/2}
 (911)

in which Q_{D} = design discharge, C_{D} = dimensionless design discharge coefficient,
L = crest length perpendicular to the flow, and H_{D} = total design head on the crest,
including the approach velocity head h_{a}.
Figure 915 shows values of C_{D} as a function of P/H_{D},
in which P = height of the spillway crest measured from the channel bed.
Note that for large values of P/H_{D}, the value of C_{D}
approaches asymptotically a value of C_{D} = 0.492.
In general, the discharge over an ogeetype spillways is:
Q^{} = C (2 g)^{1/2} L H^{ 3/2}
 (912)

in which Q = discharge, C = dimensionless discharge coefficient,
L = crest length perpendicular to the flow, and H = head on the crest,
including the approach velocity head.
Fig. 915 Variation of discharge coefficient C_{D} with relative dam height P/H_{D} (Roberson et al., 1998).


The ratio C/C_{D} varies as a function of H/H_{D} as shown in Fig. 916.
Figures 915 and 916 may be used to develop a spillway rating curve, i.e., a relation between flow rate Q
and head above spillway crest H. If the actual head H exceeds the design head H_{D}, subatmospheric
pressures will develop on the spillway, and this may lead to cavitation damage. To prevent cavitation damage, the negative pressure head
should be kept within 20 ft.
Fig. 916 Variation of dimensionless discharge coefficient C/C_{D}
with relative stage above crest level H /H_{D} (Roberson et al., 1998).


Spillway rating: Example
Determine the rating curve for an ogeetype spillway with length L = 30 m.
The design head is: H_{D} = 10 m.
The freeboard is: F_{b} = 5 m.
The spillway crest elevation is at Elev. 1,000 m.
The riverbed elevation is at Elev. 970 m. Neglect the approach velocity.
Solution
L = 30 m.
P = 1000  970 = 30 m.
P /H_{D} = 30 / 10 = 3
From Fig. 915, for
P /H_{D} = 3: C_{D} = 0.493
The spillway rating equation is (Eq. 912):
Q = C (2 g )^{1/2} L H^{ 3/2}
Q = C (2 × 9.806)^{1/2} (30) H^{ 3/2}
Q = 132.856 C H^{ 3/2}
The spillway rating equation is:
Q = 132.856 C H^{ 3/2}
ANSWER.
The calculations are summarized in Table 92.
Column 1 shows postulated watersurface elevations,
from crest elevation (1,000 m) through design watersurface elevation (1,010 m) to a watersurface
elevation
that includes the freeboard (1,015 m).
Column 2 shows the actual head over the spillway crest.
Column 3 shows the H /H_{D} ratios.
Column 4 shows C /C_{D} ratios,
obtained from Fig. 916. Column 5 shows values
of C = C_{D} (C /C_{D})
= 0.493 (C /C_{D})
Column 6 shows the values of Q
calculated with the spillway rating equation:
Q = 132.856 C H^{ 3/2}
The spillway rating curve is shown in Fig. 917.
Table 92 Calculation of spillway rating.

[1] 
[2] 
[3] 
[4] 
[5] 
[6] 
Elevation (m) 
H (m) 
H /H_{D} 
C /C_{D} 
C 
Q (m^{3}/s) 
1000 
0 
0.0 
0.780 
0.385 
0.000 
1001 
1 
0.1 
0.811 
0.400 
53.119 
1002 
2 
0.2 
0.842 
0.415 
155.986 
1003 
3 
0.3 
0.869 
0.428 
295.584 
1004 
4 
0.4 
0.895 
0.441 
468.987 
1005 
5 
0.5 
0.915 
0.451 
670.047 
1006 
6 
0.6 
0.935 
0.461 
900.052 
1007 
7 
0.7 
0.951 
0.469 
1153.604 
1008 
8 
0.8 
0.967 
0.477 
1433.146 
1009 
9 
0.9 
0.984 
0.485 
1739.271 
1010 
10 
1.0 
1.000 
0.493 
2071.234 
1011 
11 
1.1 
1.013 
0.499 
2419.431 
1012 
12 
1.2 
1.025 
0.505 
2790.775 
1013 
13 
1.3 
1.037 
0.511 
3182.097 
1014 
14 
1.4 
1.048 
0.517 
3595.692 
1015 
15 
1.5 
1.059 
0.522 
4029.600 
Fig. 917 Calculated spillway rating curve.



Example 92
Solve the spillway rating example using an online calculator.
 
ONLINE CALCULATION. Use ONLINE OGEE RATING, with L = 30 m, H_{D} = 10 m,
approach velocity V_{a} = 0, dam height P = 30 m,
freeboard F_{b} = 5 m, and
spillway crest elevation E = 1000 m.
The result of ONLINE OGEE RATING is the same of Table 92.



9.4 THE HYDRAULIC JUMP
The hydraulic jump is an openchannel flow phenomenon where
the flow changes suddenly from supercritical to subcritical (Fig. 918).
The equation for the hydraulic jump is developed for a horizontal channel,
for which the jump becomes stationary, i.e., it occurs in a specific place in the channel.
A nonstationary hydraulic jump may occur in a channel of finite (nonzero) slope.
In practice, the
stationary hydraulic jump is preferred over the nonstationary or "moving" jump.
Fig. 918 A hydraulic jump downstream of a sluice gate.


The hydraulic jump is used in the following applications:
To dissipate energy in
water flowing over dams, weirs, and other hydraulic structures, to prevent
downstream scouring of the structures.
To recover hydraulic head on the downstream side of a measuring flume.
To increase weight on an apron and
reduce uplift pressures under a masonry structure.
To aerate water for purposes of water purification.
Hydraulic jump equation
A hydraulic jump will form in a rectangular channel if the following equation is satisfied (See Example 31):
y_{2} 1 ^{____} = ^{____} [
(1 + 8 F_{1}^{ 2} )^{1/2}  1 ]
y_{1} 2
 (913) 
where y_{1} = upstream flow depth (supercritical), y_{2} = downstream flow depth (subcritical), and F_{1} =
Froude number of the upstream flow. The depth y_{1} is the initial depth
and y_{2} is the sequent depth.
The hydraulic jump relation (Eq. 913) is shown in Fig. 919.
Note that for Froude number F_{1} > 2 the equation is almost linear.
Fig. 919 The hydraulic jump relation.


Example 93
Given y_{1} = 0.1 m and v_{1} =
5 m/s, calculate
the sequent depth y_{2}. Confirm the result using ONLINE CHANNEL 11.
The upstream flow Froude number is: F_{1} = 5.049. Using Eq. 913: y_{2} = 0.665 m.
 
ONLINE CALCULATION. Using
ONLINE CHANNEL 11,
with y_{1} = 0.1 m, and v_{1} = 5 m/s, the Froude number is:
F_{1} = 5.049. The sequent depth
is y_{2} = 0.665 m, confirming the hand calculation.



Types of hydraulic jump
Hydraulic jumps
are classified as shown in Table 93 (Chow, 1959).
Table 93
Types of hydraulic jump.

Upstream Froude number F_{1}
 Type of jump
 Graphical description

1.0  1.7
 Undular jump
 
1.7  2.5
 Weak jump
 
2.5  4.5
 Oscillating jump
 
4.5  9.0
 Steady jump
 
> 9.0
 Strong jump
 
Characteristics of the hydraulic jump
The energy loss is:
(y_{2}  y_{1})^{3}
ΔE = E_{1}  E_{2} = ^{ ______________ }
4 y_{1} y_{2}
 (914) 
The relative energy loss is:
ΔE E_{2}
^{ ______} = 1  ^{ ______ }
E_{1} E_{1}
 (915) 
The efficiency of the jump is (see box below):
E_{2} ( 1 + 8 F_{1}^{2} )^{3/2}  4 F_{1}^{2} + 1
^{ ______} = ^{ _______________________________ }
E_{1}
8 F_{1}^{2} ( 2 + F_{1}^{2} )
 (916) 
The height of the jump is:
h_{j} = y_{2}  y_{1}
 (917) 
The relative height of the jump is:
h_{j} y_{2} y_{1}
^{ _____} = ^{ _____ }  ^{ _____ }
E_{1} E_{1} E_{1}
 (918) 
where y_{1} /E_{1} = relative initial depth, and
y_{2} /E_{1} = relative sequent depth.
In terms of the Froude number of the upstream flow,
the relative height of the jump is:
h_{j} ( 1 + 8 F_{1}^{2} )^{1/2}  3
^{ ______} = ^{ _______________________ }
E_{1} 2 + F_{1}^{2}
 (919) 
Figure 920 is a graphical portrayal of the
characteristics of the hydraulic jump. Note the following:
The relative sequent depth reaches a maximum value
y_{2} /E_{1} = 0.8 for F_{1} = 1.73.
The relative height of the jump reaches a maximum value
h_{j} /E_{1} = 0.507 for F_{1} = 2.77.
It is confirmed that for
F_{1} = 1, the initial depth y_{1} is equal to 2/3 (0.667) of the specific energy E_{1}.
For F > 3, the changes in all characteristics become gradual.
Fig. 920 Characteristics of the hydraulic jump.


Efficiency of the hydraulic
jump
With
reference to Fig. 921, the efficiency of the hydraulic jump is:
E_{2} ( 1 + 8 F_{1}^{2} )^{3/2}  4 F_{1}^{2} + 1
^{ ______} = ^{ _______________________________ }
E_{1}
8 F_{1}^{2} ( 2 + F_{1}^{2} )
 (916) 

Fig. 921 Definition sketch for a hydraulic jump.


F_{1} = v_{1} / (gy_{1})^{1/2}
F_{2} = v_{2} / (gy_{2})^{1/2}
v_{1}y_{1} = v_{2} y_{2}
v_{1}^{2}y_{1}^{2} = v_{2}^{2} y_{2}^{2}
F_{1}^{2} y_{1}^{3} = F_{2}^{2} y_{2}^{3}
F_{2}^{2} = F_{1}^{2} / (y_{2}
/ y_{1})^{3}
The hydraulic jump equation is (Eq. 913):
y_{2} / y_{1} = (1/2) [ (1 + 8 F_{1}^{2})^{1/2}  1 ]
N^{ 2} = 1 + 8 F_{1}^{2}
y_{2} / y_{1} = (1/2) [ N  1]
2 (y_{2} / y_{1}) = N  1
(y_{2} / y_{1})^{3} = (1/8) [ N  1]^{3}
2 (y_{2} / y_{1})^{3} = (1/4) [ N  1]^{3}
N = ( 1 + 8 F_{1}^{2})^{1/2}
N^{ 3} = ( 1 + 8 F_{1}^{2})^{3/2}
N^{ 2}  1 = 8 F_{1}^{2}
F_{1}^{2} = ( N^{ 2}  1) / 8
4 F_{1}^{2} = ( N^{ 2}  1) / 2
The efficiency of the hydraulic jump is:
E_{2}/E_{1} = [ y_{2} + v_{2}^{2}/(2g) ] / [ y_{1} + v_{1}^{2}/(2g) ]
E_{2}/E_{1} = [ y_{2}(1 + F_{2}^{2}/2) ] / [ y_{1}(1 + F_{1}^{2}/2) ]
E_{2}/E_{1} = 2 (y_{2}/y_{1}) {1 + F_{1}^{2} / [ 2 (y_{2}/y_{1})^{3}] } / (2 + F_{1}^{2})
E_{2}/E_{1} = (N  1) {1 + (N^{ 2}  1) / [ 2 (N  1)^{3} ] } / (2 + F_{1}^{2})
E_{2}/E_{1} = (N^{ 2}  1)(N  1) {1 + (N^{ 2}  1) / [ 2 (N  1)^{3} ] } / [8 F_{1}^{2}(2 + F_{1}^{2}) ]
E_{2}/E_{1} = [ (N^{ 2}  1)(N  1) + (1/2)(N + 1)^{2} ] / [ 8 F_{1}^{2}(2 + F_{1}^{2}) ]
E_{2}/E_{1} = { (N^{ 3}  N^{ 2}  N  1) + [ (N^{ 2}/2) + N + (1/2)] } / [ 8 F_{1}^{2}(2 + F_{1}^{2}) ]
E_{2}/E_{1} = { (N^{ 3}  [(N^{ 2}  1)/2] + 1} / [ 8 F_{1}^{2} (2 + F_{1}^{2}) ]
E_{2}/E_{1} = [ (1 + 8 F_{1}^{2})^{3/2}  4F_{1}^{2} + 1] / [ 8 F_{1}^{2}(2 + F_{1}^{2}) ] ANSWER.

Example 94
Given q = 0.5 m^{2}/s and ΔE = 0.678 m, calculate
the sequent depths y_{1} and y_{2}. Confirm the result using ONLINE CHANNEL 16.
The calculation proceeds by trial and error.
Assume a low value of F_{1}, say 2.
Use Eq. 913 to calculate y_{2} / y_{1}.
Use Eq. 914 to solve for y_{1}.
Solve for the usptream velocity: v_{1} = q / y_{1}
Calculate the value of F_{1}.
If the value calculated in Step 5 is the same as that assumed in Step 1 (within a certain tolerance), stop and report the sequent depths.
Otherwise, increase the value of F_{1} by a suitable increment, and return to Step 1.
 
ONLINE CALCULATION. Use
ONLINE CHANNEL 16,
with q = 0.5 m^{2}/s and ΔE = 0.678 m. The sequent depths are: y_{1} = 0.1 m and y_{2} = 0.665 m.



Example 95
Given y_{1} = 0.1 m and v_{1} =
5 m/s, calculate
the efficiency of the hydraulic jump. Confirm the result using ONLINE CHANNEL 18.
The upstream flow Froude number is: F_{1} = 5.049. Using Eq. 916: E_{2} / E_{1} = 0.505.
 
ONLINE CALCULATION. Using
ONLINE CHANNEL 18,
with y_{1} = 0.1 m, and v_{1} = 5 m/s, the Froude number is:
F_{1} = 5.049. The efficiency of the jump
is: E_{2} / E_{1} = 0.505, confirming the hand calculation.



Length of the hydraulic jump
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 (see insert of Fig. 921). The relative length of the jump L/y_{2} has been determined experimentally.
It varies from about 5.0 to 6.15 for a wide range of upstream Froude numbers.
Within the range 4.5 ≤ F_{1} ≤ 13,
the relative length of the jump is slightly greater than 6 (Fig. 922).
Fig. 922 Length of the hydraulic jump (U.S. Bureau of
Reclamation).


Profile of the hydraulic jump
Experimental measurements have shown
that the actual surface profile of the hydraulic jump varies
somewhat with the upstream Froude number. The profiles
may be plotted as dimensionless curves, as shown in Fig. 923. A closeup
of a hydraulic jump is shown in Fig. 924.
Fig. 923 Dimensionless profiles of the hydraulic jump (Chow, 1959).


Fig. 924 Closeup of a hydraulic jump.


QUESTIONS
How does rapidly varied flow differ from gradually varied flow?
Is there a theoretical solution of onedimensional
rapidly varied flow?
What is the exponent of the rating in a broadcrested weir?
What is the coefficient of the rating for high
spillways in U.S. Customary units?
What does "ogee" stand for?
What is the rationale for the use of a labyrinth spillway?
What is the risk when flow over a spillway exceeds the design stage?
What flow condition produces a hydraulic jump?
Is the hydraulic jump equation linear or nonlinear?
How is the length of the hydraulic jump measured?
PROBLEMS
An emergency spillway is being considered for the
Demerera Water Conservancy to safeguard the integrity of the dam under conditions of climate change (similar to Fig. 925).
Assume that the existing relief sluices would be inoperable during a major flood due to high tailwater.
Determine the length of the freeoverflow spillway
required to pass the Probable Maximum Flood (PMF). The following data is applicable:
 PMF 1day duration: 428 mm
 Hydrologic abstraction: 18 mm
 Contributing drainage area: 582 km^{2}
 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 triangularshaped flood hydrograph. Use all the freeboard to contain the PMF.
Fig. 925 The 8000ft weir of the
Boerasirie Water Conservancy.


Design an overflowspillway section having a vertical
upstream face and a crest length L = 150 ft.
The design discharge is Q = 50,000 ft^{3}/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. 96 for graphical example).
Use ONLINE OGEE RATING to
determine the rating for an ogee spillway with length L = 15 m,
design head H_{d} = 2 m, spillway crest elevation = 1045 m,
river bed elevation = 1000 m, and freeboard F_{b} = 1 m.
Neglect the approach velocity.
What should be the spillway length to pass the Probable Maximum Flood
Q_{PMF} = 250 m^{3}/s while taking all the freeboard?
Express spillway length to nearest 0.1 m by excess.
Prove Eq. 914.
Prove Eq. 919.
Calculate the energy loss in a
hydraulic jump, given the sequent depths
y_{1} = 0.58 m and y_{2} = 2.688 m.
Calculate the relative
height of the hydraulic jump h_{j} /E_{1} for F_{1} = 3.
Use ONLINE CHANNEL 11
to calculate the sequent depth y_{2} through a hydraulic jump when the
discharge is q = 5 m^{2}/s and the initial depth y_{1} = 0.58 m.
Use ONLINE CHANNEL 12
to determine the sequent depth y_{2} and energy loss ΔE
through a hydraulic jump when the
discharge is q = 10 ft^{2}/s and the upstream flow depth is y_{1} = 0.5 ft.
Use ONLINE CHANNEL 16
to calculate the sequent depths through a hydraulic jump when the
discharge is q = 10 ft^{2}/s and the energy loss is ΔE = 3.287 ft.
Use ONLINE CHANNEL 16
to calculate the sequent depths through a hydraulic jump when the
discharge is q = 5 m^{2}/s and the energy loss is ΔE = 1.5 m.
Use ONLINE CHANNEL 18 to
calculate the efficiency of the hydraulic jump E_{2}/E_{1} for
q = 10 ft^{2}/s and
y_{1} = 0.5 ft.
REFERENCES
Chow, V. T. 1959. Openchannel Hydraulics. McGraw Hill, New York.
Roberson, J. A., J. J. Cassidy,
and M. H. Chaudhry. 1998. Hydraulic Engineering. John Wiley and Sons, New York, Second edition.
Pugh, C. A., and E. W. Gray. 1984. Fuse Plug Embankments in Auxiliary Spillways  Developing Design Guidelines and Parameters,
United States Committee on Large Dams
(as cited in U.S. Bureau of Reclamation).
http://openchannelhydraulics.sdsu.edu 

210913 17:00 
