4.2 HYDRAULIC JUMP .1 Presentation
4.2.2 Basic equations
For a steady flow in a horizontal rectangular channel of constant channel width, the three fun- damental equations become:
(A) Continuity equation
Q = V^d^B = ^2^2^ (4-2) where V^ and dx are, respectively, the velocity and flow depth at the upstream cross-section (Fig.
4.4), V2 and ^2 are defined at the downstream cross-section, B is the channel width and Q is the total flow rate.
Fig. 4.2 Hydraulic jump in a natural waterway: Bald Rock Creek at the Junction Qld,Australia (9 November 1997)-flowfrom the left to the right.
4.2 Hydraulic jump 55
Fig. 4.3 Photographs of hydraulic jump in a rectan- gular channel (a) Undular hydraulic jump: Fr^ = 125, d^ = 0.081m, 5 = 0.5m, flow from the left to the right (courtesy of Ms Chantal Donnely); (b) Steady/strong jump: Fr^ = 9.4, d^ = 0.013 m, B = 0.5m, flow from the left to the right;
(c) Strong jump: Fr^ = 13.2, c/i = 0.013m, e = 0.5m, flow from the left to the right.
(B) Momentum equation (Belanger equation)^
The momentum equation states that the sum of all the forces acting on the control volume equals the change in momentum flux across the control volume. For a hydraulic jump, it yields:
\pgd'x - \pg4 \B - FfHe = pQ{V^ - n) (4.3) where Ff^^ is a drag force exerted by the channel roughness on the flow (Fig. 4.4).
(C) Energy equation
The energy equation (2.24) can be transformed as:
H^=H2-\- MI (4.4a) where AH is the energy loss (or head loss) at the jump, and Hi and H2 are upstream and down-
stream total heads, respectively. Assuming a hydrostatic pressure distribution and taking the channel bed as the datum, equation (4.4a) becomes:
^ + . . = ^
2g 2g -\-d2+MI (4.4b)
Note that equations (4.1)-(4.4) were developed assuming hydrostatic pressure distributions at both the upstream and downstream ends of the control volume (Fig. 4.4). Furthermore, the upstream and downstream velocity distributions were assumed uniform for simplicity.
Notes
1. In the momentum equation, (pV) is the momentum per unit volume.
2. In the simple case of uniform velocity distribution, the term (pW) is the momentum flux per unit area across the control surface. (pWA) is the total momentum flux across the control surface.
Roller
Hydrostatic pressure distribution
Hydraulic jump
^fric
Bottom friction
Fig. 4.4 Application of the momentum equation to a hydraulic jump.
^ Jean-Baptiste Belanger (1789-1874) was the first to suggest the application of the momentum principle to the hydraulic jump flow (Belanger, 1828). The momentum equation applied across a hydraulic jump is often called the Belanger equation.
4.2 Hydraulic jump 57
Neglecting the drag force on the fluid, the continuity and momentum equations provide a rela- tionship between the upstream and downstream flow depths as:
* = j i i r + ^2'
X^lif' or in dimensionless terms:
i^U.-^^-i
(4.5a)
(4.5b) where Frx is the upstream Froude number: Fri = Yi/Jgdi. It must be noted that Fri> \. The depths di and t/2 ^re referred to as conjugate depths (or sequent depths). Using equation (4.5) the momentum equation yields:
Fr, = 2"'Fn
^ 1 + 8FK
3/2 (4.6)
where Fr2 is the downstream Froude number. The energy equation gives the head loss:
A / / = ( ^ 2 - ^ 1 ) -\3
4(^1^2
(4.7a) and in dimensionless terms:
d, 16
1 + SFrj - 3 1 + %Fr{ - 1
(4.7b)
Equations (4.5)-(4.7) are summarized in Fig. 4.5. Figure 4.5 provides means to estimate rapidly the jump properties as functions of the upstream Froude number. For example, for Fr^ = 5, we can deduce Fr2 ~ 0.3, t/2/^1 ~ 6.5, AH/di ~ 7.
/ LM d2/d^ (equation 4.5)
—•— Fr2 (equation 4.6) A/-//c/i (equation 4.7) Lr/c/i (equation 4.8)
Fig. 4.5 Flow properties downstream of a liydraulic jump 11 Fri in a rectangular horizontal channel.
Notes
1. In a hydraulic jump, the downstream flow depth J2 is always larger than the upstream flow depth di.
2. The main parameter of a hydraulic jump is its upstream Froude number Fri.
3. If only the downstream flow conditions are known, the solution of the continuity and momentum equations is given by:
i4(>^-')
-*24. A hydraulic jump is a very effective way of dissipating energy. It is also an effective mixing device. Hydraulic jumps are commonly used at the end of spillway or in dissipation basin to destroy much of the kinetic energy of the flow. The hydraulic power dissipated in a jump equals;
pggAif where AH is computed using equation (4.7).
5. Hydraulic jumps are characterized by air entrainment. Air is entrapped at the impingement point of the supercritical inflow with the roller. The rate of air entrainment may be approximated as:
^ « 0.018 (Frj - 1)^^^^ (Rajaratnam, 1967)
^ « 0.014(Fri - 1)^ ^ (Wisner, 1965)
Wood (1991) and Chanson (1997) presented complete reviews of these formulae.
If the jump is in a closed duct then, as the air is released, it could accumulate on the roof of the duct. This phenomena is called 'blowback' and has caused failures in some cases (Falvey, 1980).
6. Recent studies showed that the flow properties (including air entrainment) of hydraulic jumps are not only functions of the upstream Froude number but also of the upstream flow conditions: i.e.
partially developed or fiilly developed inflow The topic is currently under investigation.
Application
Considering a hydraulic jump in a horizontal rectangular channel, write the continuity equation and momentum principle. Neglecting the boundary shear, deduce the relationships d2/di =f(Fri) and Fr2=f(Fr,).
Solution
The continuity equation and the momentum equation (in the flow direction) are respectively:
q = V^d^ = ^2^2 [C]
1 9 1 2
-Pgdi - -Pgd2 = pq (V2 - Fj) [M]
where q is the discharge per unit width.
[C] implies V2 = Vidi/d2. Replacing [C] into [M], it yields:
-Pgdi - -pgdl = pV^^d^ d,
yd2
1 _ . T / 2
- P ^ l ^ l
4.2 Hydraulic jump 59 Dividing by (pgdi) it becomes:
1 I d^ 2 _ Z7.2 d,
= Fr'\-^\-Fn 2 lyd^
After transformation we obtain a polynomial equation of degree three in terms ofdildi'.
I do
or
21 ^1
ih^-i
+ Fn'
d, + Fn^ =0
do I , \ dj
2 d,
The solutions of the momentum equation are the obvious solution d2 = d\ and the solutions of the polynomial equation of degree two:
d, - IFn = 0
The (only) meaningful solution is:
^ = 1
Ji 2
^ - - ' ^ ' l + 8F/i^ - 1
Using the continuity equation in the form V2 = Vxdxldi, and dividing by igd2, it yields:
F., = ^ ^ - - ^ f A
[gd~2 ^x\^2
yi >3/2
= F^-
i
1 + 8Fri - 13/2
>App//cat/on
A hydraulic jump takes place in a 0.4 m wide laboratory channel. The upstream flow depth is 20 mm and the total flow rate is 311/s. The channel is horizontal, rectangular and smooth. Calculate the downstream flow properties and the energy dissipated in the jump. If the dissipated power could be transformed into electricity, how many 100 W bulbs could be lighted with the jump?
Solution
The upstream flow velocity is deduced from the continuity equation:
- 3
Vx = 31 XIO"
Bdx 0.4 X (20 X 10"^)
= 3.875 m/s
The upstream Froude number equals Frx = 8.75 (i.e. supercritical upstream flow). The downstream flow properties are deduced from the above equation:
^ = i(^7T^-i) = n.
,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