Lecture10 Momentum Principle

8/12/2019 Lecture10 Momentum Principle

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The Momentum Principle

Hydromechanics VVR090

Hydraulic Jump

(photo taken 2008-02-24 in M. Larson s

kitchen sink)


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Specific Momentum

( )' ' ' ' ' '1 3 2 2 2 1 1fF F F P Q u ug

+ =

Horizontal momentum equation is applied to a controlvolume in an open channel (x-direction):

Small s lopes:

( )1 1 2 2 2 1fz A z A P Q u ug

=

2 2

1 1 2 2

1 2

1 2

2

f

f

P Q Qz A z A

gA gA

PM M

QM zA

gA

= + +

=

= +

Continuity equation:

1 2

1 2

,Q Q

u uA A

= =

Rearranging the momentum equation:


3/11

Specific Momentum Curve

Analogous to the speci fic energy curve.

For a given value on M there are in general two

possib le water depths (sequent depths of ahydraulic jump.

The minimum value of M corresponds to the

minimum value of E.

The Hydraulic Jump

A hydraulic jump results when there is a confl ict between

upstream and downstream control.

Example: upstream control induce supercritical flow and

downstream control subcritical flow.

=> flow has to pass from one regime to another

creating a hydraulic jump

Significant turbulence and energy dissipation occurs in a

hydraulic jump.


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Appl ications of the Hydraulic Jump

1. Dissipation of energy in flows over dams, weirs, and otherhydraulic structures

2. Maintenance of high water levels in channels for water

distribution

3. Increase of discharge of a sluice gate by repelling the

downstream tailwater

4. Reduction of uplift pressure under structures by raising

the water depth

5. Mixing of chemicals

6. Aeration of flow

7. Removal of air pockets

8. Flow measurements

Hydraulic Jump Dependence on Froude Number


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Hydraulic Jump in a Sink

Governing Equation for a Hydraulic Jump I

Apply the momentum equation for a fi ni te control

volume encompassing the jump and let Pf = 0:

1 2

2 2

1 1 2 2

1 2

M M

Q Qz A z A

gA gA

=

+ = +

For a rectangular section:

1 1 2 2

1 1 2 2

1 1 2 2

,

/ 2, / 2

Q u A u A

A by A by

z y z y

= =

= =

= =


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Governing Equation for a Hydraulic Jump II

Momentum equation (rectangular channel):

( )2 2

1 22 1

2 2

y y qu u

g

=

Momentum equation for rectangular section:

( )2

2 2

2 1

1 2

1 1 1

2

qy y

g y y

=

Solutions:

( )

( )

221

1

21

2

2

11 8 1

2

11 8 1

2

yFr

y

yFr

y

= +

= +


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Downstream Froude number (Fr2) usually small so:

2

21 8 1 0Fr+

Sometimes useful to do a Taylor series expansion:

2 2 4 6

2 2 2 21 8 1 4 8 32 ...Fr Fr Fr Fr + = + + +

Substituting into the hydraulic jump equation:

2 4 61

2 2 2

2

2 4 16 ...y

Fr Fr Fr y

= + +

2 31 1 11 1 ...

2 8 16x x x x+ = + + +( )

Submerged Jump

If the downstream water level is high enough (> sequent

depth y2) the jump might be submerged. The depth of

submergence (y3) is then a crucial unknown.

Could occur downstream gates etc.


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Submergence of Hydraulic Jump

Govinda (1963) proposed the foll owing formula for

submerged jumps (horizontal, rectangular channels):

Chow (1959):

( )( )

( )

1/ 22

2 2 23 11

1

4 1

2

221

1

21 2

1

11 8 1

2

y FrS Fr

y S

y yS

y

yFr

y

= + +

+

=

= = +

1/ 2

23 4

4

4 1

1 2 1y y

Fry y

= +


9/11

Energy Loss in a Hydraulic Jump I

Dissipation in a horizontal channel:

1 2E E E =

Rectangular section:

( )

( ) ( )

( )( )

3

2 1

1 2

22

2 1 1 1 2

2

1 1

3/ 22 2

1 11

2 22 1 1

4

2 2 / 1 /

2

8 1 4 1

8 2

y yE

y y

y y Fr y yE

E F

Fr Fr E

E Fr Fr

=

+ =+

+ +=

+

Energy Loss in a Hydraulic Jump II

( )

2 2

1 21 2

2 2

1 11 22

2

22 12 1 1 2

2 1

2 2

12

44

u uy y E

g g

u yE y y

g y

y yE y y y yy y

+ = + +

= +

= +

Energy equation across the jump:

( )3

2 1

1 24

y yE

y y

=


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Hydraulic Jump Length I

Jump length hard to derive through theoretical

considerations.

Length of the jump (Lj) is defined as the distance from the

front face of the jump to a poin t on the surface immediately

downstream the roller associated with the jump.

Hydraulic Jump Length II

Empirical equation for jump length g iven by Silvester (1964)

for rectangular channels,

( )1.01

1

1

9.75 1j

LFr

y=

and for triangular channels:

( )0.695

1

1

4.26 1jL

Fry =

Submerged jump:

2

4.9 6.1j

LS

y= +


11/11

Hydraulic Jump in Sloping Channels

A number of di fferent cases have to be considered.

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Lecture10 Momentum Principle