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Problem 4. Problem 4. Hydraulic Hydraulic Jump Jump

Problem 4

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Problem 4. Hydraulic Jump. Problem. When a smooth column of water hits a horizontal plane, it flows out radially. At some radius, its height suddenly rises. Investigate the nature of the phenomenon. What happens if a liquid more viscous than water is used?. Experiment. Obtaining the effect - PowerPoint PPT Presentation

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Page 1: Problem 4

Problem 4.Problem 4.

Hydraulic JumpHydraulic Jump

Page 2: Problem 4

ProblemProblem

When a smooth column of water hits a When a smooth column of water hits a horizontal plane, it flows out radially. At horizontal plane, it flows out radially. At some radius, its height suddenly rises. some radius, its height suddenly rises. Investigate the nature of the phenomenon. Investigate the nature of the phenomenon. What happens if a liquid more viscous What happens if a liquid more viscous than water is used?than water is used?

Page 3: Problem 4

ExperimentExperiment

• Obtaining the effectObtaining the effect

• Parameters:Parameters:

• Liquid densityLiquid density

• Liquid viscosityLiquid viscosity

• Flow rateFlow rate

• Jet heightJet height

Page 4: Problem 4

Experiment Experiment cont.cont.

• Measurements:Measurements:

• Dependence of flow velocity on radiusDependence of flow velocity on radius

• Dependence of jump radius on flow rateDependence of jump radius on flow rate

• Dependence of jump radius on viscosityDependence of jump radius on viscosity

• Dependence of jump radius on jet heightDependence of jump radius on jet height

• Jump structure in dependence on velocityJump structure in dependence on velocity

Page 5: Problem 4

ApparatusApparatus

pump

p l a t e

container

Page 6: Problem 4

Apparatus Apparatus cont.cont.

Page 7: Problem 4

Viscosity variationViscosity variation

• Water was heated from 20Water was heated from 20˚C to 6˚C to 60˚C0˚C

• The achieved viscosity change was over The achieved viscosity change was over 50%50%

• Dependence of viscosity on temperature:Dependence of viscosity on temperature:

Te1730

610693.2

S. Gleston, Udžbenik fizičke hemije, NKB 1967

Page 8: Problem 4

Viscosity variation Viscosity variation cont.cont.

thermometer

heater

Page 9: Problem 4

Velocity measurementVelocity measurement

• A Pitot tube was usedA Pitot tube was used

H

HHgv 2v – flow velocity

H – water height in tube

ΔH – cappilary correction

Page 10: Problem 4

Explanation Explanation

• Hydraulic jump – sudden slow-down and Hydraulic jump – sudden slow-down and rising of liquid because of turbulencerising of liquid because of turbulence

• The turbulence appears when the viscous The turbulence appears when the viscous boundary layer reaches the flow surfaceboundary layer reaches the flow surface

• Boundary layer detachment appears and a Boundary layer detachment appears and a vortex is formedvortex is formed

• The vortex spends flow energy and slows itThe vortex spends flow energy and slows it

Page 11: Problem 4

Explanation Explanation cont.cont.

• Due to turbulenceDue to turbulence energy is lost in the jumpenergy is lost in the jump

Flow before the jump is slower than behindFlow before the jump is slower than behind

Water level is higher due to continuityWater level is higher due to continuity

Boundary layer

jump

˝Nonviscous˝ layer

Page 12: Problem 4

Explanation Explanation cont.cont.

• Tasks for the theory:Tasks for the theory:

• Dependence of jump radius on parametersDependence of jump radius on parameters

• Dependence of flow velocity on radiusDependence of flow velocity on radius

• Jump structureJump structure• Governing equations:Governing equations:

• Continuity and energy conservationContinuity and energy conservation

• Navier – Stokes equationNavier – Stokes equation

Page 13: Problem 4

Critical radiusCritical radius

• Critical radius – jump formation radiusCritical radius – jump formation radius

• Condition for obtaining critical radius:Condition for obtaining critical radius:

krhh – flow height

rk – critical radius

Δ – boundary layer thickness

Page 14: Problem 4

Critical radius Critical radius cont.cont.

• Continuity equation:Continuity equation:

• Energy conservation:Energy conservation:

h

vdzrQ0

2

Q – flow rate

v – flow velocity

r – distance from jump centre

z – vertical axis

0drdJ

drdJ ot

J – kinetc energy pro unit time

Jot – friction power

Page 15: Problem 4

Critical radius Critical radius cont.cont.

• Flow velocity is approximately linear in height Flow velocity is approximately linear in height because of hte small flow height:because of hte small flow height:

zv ξ – constant

z – vertical coordinate

2rhQ

• The constant is obtained from continuity:The constant is obtained from continuity:

Q – flow rate

r – radius

h – flow height

Page 16: Problem 4

Critical radius Critical radius cont.cont.

• Friction force is Newtonian due to flow thinnessFriction force is Newtonian due to flow thinness

flow height equation:flow height equation:

rQr

hdrdh

4

η – viscosity

ρ – density

v0 – initial velocity

0

2

234

rvQr

Qrh

Page 17: Problem 4

Critical radius Critical radius cont.cont.

• Free fall of the liquid causes the existence of Free fall of the liquid causes the existence of initial velocity:initial velocity:

gdv 20 g – free fall acceleration

d – jet height

Page 18: Problem 4

Critical radius Critical radius cont.cont.

• Boundary layer thickness isBoundary layer thickness is

• Inserting:Inserting:

vr

e.g. D. J. Acheson, ˝Boundary Layers˝, in Elementary Fluid Dynamics (Oxford U. P., New York, 1990)

61

31

323

1

2 243

dQ

grc

Page 19: Problem 4

Result comparationResult comparation

• Theoretical scaling confirmedTheoretical scaling confirmed

• Comparation of constant in flow rate Comparation of constant in flow rate dependence:dependence:

η 1.1·10-3 Pas

ρ 103 kg/m3

d 5 cm

constant 41.16 s/m3

Experimental value: Experimental value: 41.0 ± 1.0 s/m3

Page 20: Problem 4

flow rate [m3/s]

2e-5 3e-5 4e-5 5e-5 6e-5 7e-5 8e-5

radi

us [m

]

0,02

0,03

0,04

0,05

0,06

0,07

0,08

r(Q)experiment

Result comparation Result comparation cont.cont.

Page 21: Problem 4

viscosity [Pas]

0,0005 0,0006 0,0007 0,0008 0,0009 0,0010

radi

us [c

m]

2,8

3,0

3,2

3,4

3,6

3,8

4,0

4,2

r()experiment

Result comparation Result comparation cont.cont.

Page 22: Problem 4

jet height [cm]

0 2 4 6 8 10

radi

us [c

m]

4,20

4,25

4,30

4,35

4,40

4,45

r(d)experiment

Result comparation Result comparation cont.cont.

Page 23: Problem 4

Result comparation Result comparation cont.cont.

radius [m]

0,01 0,02 0,03 0,04 0,05 0,06

wat

er le

vel i

n P

itot t

ube

[m]

0,0

0,1

0,2

0,3

0,4

v(r)experiment

Page 24: Problem 4

Jump structureJump structure

• Main jump modes:Main jump modes:

• Laminar jumpLaminar jump

• Standing waves – wave jumpStanding waves – wave jump

• Oscillating/weakly turbulent jumpOscillating/weakly turbulent jump

• Turbulent jumpTurbulent jump

Page 25: Problem 4

Jump structure Jump structure cont.cont.

• Decription of liquid motion – Navier - Stokes Decription of liquid motion – Navier - Stokes equation:equation:

zvvvv ˆ2 gt

Inertial termInertial termConvection Convection termterm

Viscosity Viscosity termterm

Gravitational Gravitational term (pressure)term (pressure)

Page 26: Problem 4

Jump structureJump structure cont.cont.

• laminar jump conditon:laminar jump conditon:

vvvv

t

2

small velocitiessmall velocities

Viscous liquidsViscous liquids

Steady rotation in jump regionSteady rotation in jump region

Page 27: Problem 4

(slika1)

Page 28: Problem 4

Jump Structure Jump Structure cont.cont.

• Stable turbulent jump:Stable turbulent jump:

vvv 2

Large velocitiesLarge velocities

Weakly viscous liquidsWeakly viscous liquids

Time – stable modeTime – stable mode

Page 29: Problem 4

(slika3)

Page 30: Problem 4

Struktura skoka Struktura skoka cont.cont.

• The remaining time – dependent modes areThe remaining time – dependent modes are

• Difficult to obtainDifficult to obtain

• Unstable Unstable • Mathematical cause: the inertial term in the Mathematical cause: the inertial term in the

equation of motionequation of motion

• Observing is problematicObserving is problematic

Page 31: Problem 4

Conclusion Conclusion • We can now answer the problem:We can now answer the problem:

• The jump is pfrmed because of boundary layer The jump is pfrmed because of boundary layer separation and vortex formationseparation and vortex formation

• Energy is lost in the jump, so the flow height is Energy is lost in the jump, so the flow height is larger after the jumplarger after the jump

• The jump in viscous liquids is laminar or The jump in viscous liquids is laminar or wavelike, without turbulencewavelike, without turbulence