Hydraulic Jumps[1]1

Momentum Applications in Open Channel Flow

Momentum

ss mvF

Basic relationship in mechanics:

Sum of forces in the s direction

Change in momentum in the s direction

mass

Velocity in the s direction

Momentum cont.

• For a constant mass and a per unit width consideration: (rectangular channel)

12 vvqmvs

Momentum forces – Open Channel Application

v2

v1

P1

P2

W

Wsin

Rf

L

f21s RPsinWPF

Momentum Forces cont.

• Rf is the frictional resistance.

• P1 and P2 are pressure forces per unit width given by:

2yP

2

Momentum contd.

• Combining terms we get:

)vv(qRsinW2y

2y

12f

22

21

Momentum cont.

• Considering a short section so that Rf is negligible and the channel slope is small so that sin is near zero the equation can be written as:

2

22

1

21 qv

2yqv

2y

or

Mg

qv2y

gqv

2y 2

221

21

Momentum cont.

• M is called the momentum function or the specific force plus momentum.

• For a constant q, M can be plotted against depth to create a curve similar to the specific energy curve.

• Under steady conditions, M is constant from point to point along a channel reach.

Specific force plus momentum curve.

Mc

y

yc

MM

y1

y2

q1

q2

y = yc

Momentum cont.

v2

v1

y1

y2

1 2

Mg

qv2y

gqv

2y 2

221

21

q = q1 = q2

Hydraulic Jump as an application of Momentum Equation (p. 458-465 text)

Lab Jump in Flume (Right to Left)

Hydraulic Jump in a Sink?• http://www.eng.vt.edu/fluids/msc/gallery/waves/sink.htm

hydrojump.mov

Hydraulic Jumps

• Occurs when there is a sudden transition from supercritical (y < yc) to subcritical (y > yc) flow.

• Examples of where this may occur are :– At the foot of a spillway– Where a channel slope suddenly turns flat.

• In analyzing hydraulic jumps we assume there is conservation of momentum, i.e. :

2

222

1

221

gyq

2y

gyq

2y

We can algebraically manipulate this to find:

1F8121

yy 2

11

2

1F8121

yy 2

22

1

• y1 is known as the initial depth and is < yc (supercritical flow).

• y2 is known as the sequent depth and is > yc (subcritical flow).

• The energy loss in a hydraulic jump can be found by:

21

312

1 yy4yyE

gyVyy

y 12

12

112

222

Example

Example 3.18, Streeter, et al.

If 12 m3/sec of water per meter of width flows down a spillway onto a horizontal floor and the velocity is 20 m/sec, determine (a) the downstream depth required to cause a hydraulic jump, (b) the loss in energy head, and (c) the losses in power by the jump per meter of width.

Solution for depth, losses, power lost

kWmmmNlossesQmpower

NmNlosses

m

1659)1.14sec)(/12)(/9806()(/

/1.14)7.6)(6.0(4

)6.07.6(

7.6806.9

)6.0)(20(23.03.0

33

3

22