4.2 HYDRAULIC JUMP .1 Presentation
4.2.3 Discussion Types of hydraulic jump
,3/2
Fr2 = Fr^
1 + 8Fri - 1
3/2 0.213
Hence ^2 = 0.238 m and V2 = 0.33 m/s.
The head loss in the hydraulic jump equals:
A / / = ^ ^ ^ ^ ^ = 0.544 m 4d^d2
The hydraulic power dissipated in the jump equals:
pgQ/SH = 99S.2 X 9.8 X 31 X 10"^ X 0.544 = 165 W
where p is the fluid density (kg/w?), Q is in m^/s and A ^ is in m. In the laboratory flume, the dissi- pation power equals 165 W: one lOOW bulb and one 60 W bulb could be lighted with the jump power.
4.2.3 Discussion
4.2 Hydraulic jump 61 1995a). Furthermore, the inflow conditions (uniform velocity distribution, partially developed or
fully developed) modify substantially the flow properties and affect the classification of jumps.
Also, the shape of the channel cross-section affects the hydraulic jump characteristics (Hager, 1992b). Note that the above table is given for a rectangular cross-section only.
A hydraulic jump is a very unsteady flow. Experimental measurements of bottom pressure fluc- tuations indicated that the mean pressure is quasi-hydrostatic below the jump but large pressure fluctuations are observed (see reviews in Hager, 1992b and Chanson, 1995b). The re-analysis of bottom pressure fluctuation records below hydraulic jumps over long periods of time indicates that the extreme minimum pressures might become negative (i.e. below atmospheric pressure) and could lead to uplift pressures on the channel bottom. The resulting uplift loads on the channel bed might lead to substantial damage, erosion and destruction of the channel.
Length of the roller
Chov^ (1973) proposed some guidelines to estimate the length of the roller of hydraulic jump as a function of the upstream flow^ conditions. Hager et al, (1990) review^ed a broader range of data and correlations. For v^ide channel (i.e. dxlB < 0.10), they proposed the following correlation:
^ = 1 6 0 t a n h ^ ^ - 12 2 < Fr. < 16
d^ (, 20 J ^ (4.8)
w^here Lj. is the length of the roller. Equation (4.8) is valid for rectangular horizontal channels with 2 < Fri < 16. Such a correlation can be used when designing energy dissipation basins (see Fig. 4.5).
Notes
1. The hyperbolic tangent tanh is defined as:
tanh(x) =
e'' + e"''
2. Equation (4.8) is an empirical correlation based upon model and prototype data. It fits reasonably well experimental data for hydraulic jumps with partially developed inflow conditions (in rect- angular channels).
Application: energy dissipation basin
Hydraulic jumps are known for their energy dissipation performances. Prior to late 19th century, designers tried to avoid hydraulic jumps whenever possible to minimize the risks of channel destruction. Since the beginning of the 20th century and with the introduction of high-resistance materials (e.g. reinforced concrete), hydraulic jumps are used to dissipate flow energy dovm- stream of supercritical flow structures (e.g. spillways and bottom outlets) (e.g. Henderson, 1966:
pp. 221-225).
In practice, energy dissipation structures^ are designed to induce a steady jump or a strong jump. The lowest design (inflow) Froude number must be above 4.5. The selection of a strong jump requires a careful analysis of the risks of bed erosion.
^Energy dissipation structures are also called stilling basin, transition structures or energy dissipators.
Application
Considering a dissipation basin at the downstream end of a spillway, the total discharge is Q = 2000 m^/s. The energy dissipation structure is located in a horizontal rectangular channel (25 m wide). The flow depth at the downstream end of the spillway is 2.3 m. Compute the energy dissipa- tion in the basin.
Solution
The upstream flow conditions of the jump are:
di = 2.3 m Fi = 34.8 m/s
Fri = 7.3 (i.e. steady jump) The downstream flow conditions are:
^2 = 22.707 m
^2 = 3.52 m/s Fr2 = 0.236 The head loss across the jump equals:
AT/= 40.7 m The power dissipated in the jump is:
pgQ/!iH = 796 X 10^ W (i.e. nearly 800 MW!)
In a dissipation basin with flat horizontal bottom, the location of the jump may change as a func- tion of the upstream and downstream flow conditions. That is, with a change of (upstream or downstream) flow conditions, the location of the jump changes in order to satisfy the Belanger equation (4.1). The new location might not be suitable and could require a very long and uneco- nomical structure.
In practice, design engineers select desirable features to make the jump stable and as short as possible. Abrupt rise and abrupt drop, channel expansion and channel contraction, chute blocks introduce additional flow resistance and tend to promote the jump formation. Hager (1992b) described a wide range of designs.
Application
Considering a hydraulic jump in a horizontal rectangular channel located immediately upstream of an abrupt rise (Fig. 4.6), estimate the downstream flow depth d^ (see Fig. 4.6) for the design flow conditions di = 0.45 m, Vi = 10.1 m/s. The step height equals AZQ = 0.5 m.
Solution
First, we will assume that the complete jump is located upstream of the bottom rise as sketched in Fig. 4.6.
The continuity equation between sections 1 and 3 is
q = Vidi = V^d^ Continuity equation
4.2 Hydraulic jump 63 Hydraulic jump
Roller
Q'l /
Vi / //////
h
Hydrostat distri
Pressure fo
^ p /
W/ // // A c pressure Dution
^ dzl —•
rce F n ^ I
U^t
/>- '-^ ^^7^
C/3
/^PS
<
^^///
A
y ^
yy ^ ^
Fig. 4.6 Sketch of hydraulic jump at an abrupt bottom rise.
The momentum equation between the section 1 (upstream flow) and section 2 (i.e. immediately upstream of the abrupt rise) is:
\pgdl-]^pgd\\ = pq{V^-V,) Momentum equation
where q is the discharge per unit width. The momentum equation implies hydrostatic pressure distribu- tion at section 2.
The momentum equation between the section 2 (taken immediately downstream of the bottom rise) and section 3 is:
Momentum equation
\9g{d^-^,f -\9gdi\=pq^,-V^) The solution of the non-linear system of equations is:
q = 4.54 m^/s Fri = 4.8
^2/^1 = 6.32 (Equation 4.5)
^2 = 2.84 m
F2 = 1.6m/s Continuity equation J3 = 2.26 m
F3 = 2.0 m/s Fr^ = 0.43 Comments
1. (d2 — AZQ) is not equal to d^. At section 2, the velocity V2 is indeed slower than that at section 3.
2. The hydraulic jump remains confined upstream of the abrupt rise as long as J3 > d^ where d^ is the critical flow depth (in this case dc= 1.28 m).
3. Experimental data showed that the pressure distribution at section 2 is not hydrostatic. In practice, the above analysis is somehow oversimplified although it provides a good order of magnitude.
4.3 SURGES AND BORES