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28.4 General Hydraulic Jump Equation

The internal equation of motion for a hydraulic jump can be integrated in two ways

whether the control volume is defined on a microscopic scale. The hydraulic jump takes

place over a short distance (of the order of five times the sequent depth) the transition is

dominated by initial momentum flux and pressure forces due to sequent depth.

Boundary shear forces are secondary in nature. Consider the situation shown in

Figure 28.1 (unsubmerged, forced hydraulic jump in a radially diverging sloping

channel). The macroscopic approach is as follows:

Ps/2

V2V1

Ps/2Plan

__

__

z_

y2y1

++

1 z__

2

Section 1Section 2

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y1

y

Wcos

Ff

V1P1 V2

x

y2

FD

1 2

Figure 28.1 - Schematic diagram of a hydraulic jump in free surface flows

P2

Longitudinal Section

__

__

Using the Greens theorem, the Reynolds equation for turbulent flow can be integrated

over the control volume V to obtain.

xx xji i d + d= - p d + x d Vi

i j i ju u u u

( ) ( ) ( ) ( ) ( )

u ji + d (28.1)j

1 2 3 4 5

x

x

in which is outwardly directed normal.

The first term represents the net flux of momentum through the boundary , due to

mean flow. The second term represents the net momentum transfer through the

boundary due to turbulence. The third term represents the pressure force (resultant

mean normal) exerted on the fluid boundary . The fourth term represents the net

weight of the fluid within the control volume V and the fifth term represents mean

tangential force exerted on the boundary .

A macroscopic momentum equation is obtained if the above equation is applied to the

control volume V , shown in Fig 1.

Consider the momentum in the x direction, then it can be written as

{ } ( )2 12 1 2 1 1 2V V (28.2)2

+ = +

s D fQ Q I I P P P sin F W sin F

in which, pressure force, is force on the side wall.

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( )2dA/ QV , I = v' dA, P = ' gAz cosv= , pressure force sP , is force on the side

wall.

The continuity equation is 1 21 2Q = V A = V A (28.3)

Equations 2 and 3 are to be solved simultaneously to determine the sequent depth,

velocity ( )2 2v , y for given initial condition ( )1 1v , y . If 0 0S = , rectangular channel without

baffles, and no side thrust, then it simplifies to the standard format (equation 4)

2 22 2 2 21 21 1 2 2V V

2 2

y y y y

g g+ = + (28.4)

1 21 2V Vy y= (28.5)

When solved results in

( )3 2 20 1 0 12 1 2 0y F y F + + = or ( )20 11 1 8 12

y F= + (28.6)

in which 0y is the sequent depth ratio2

1

y

y.

Bed friction decreases the ratio by about 4% at 1 10 0F .= . It is to be noted that the

macroscopic approach yields only sequent depth ratio and no information regarding

surface profile or the length of the jump. In radial stilling basins, sloping basins, forced

hydraulic jump even the sequent depth ratio depends on the internal flow and hence the

physical model is used for determining.