A FemVariational approach to the droplet spreading over dry surfaces

S.Manservisi

Nuclear Engineering Lab. of Montecuccolino University of Bologna, Italy Department of mathematics Texas Tech University, USA

Simulations of droplets impacting orthogonally over dry surfaces at Low Reynolds Numbers

OUTLINE OF THE PRESENTATION

- Introduction to the impact problem- Front tracking method- Variational formulation of the contact problem- Numerical experiments

Depostion

Prompt Splashd

Corona Splashd

INTRODUCTION

Depostion

Partial reboundd

Total reboundd

INTRODUCTION

An experimental

An experimental investigation .....C.D. Stow & M.G. Hadfield

Spreading smooth surfacev=3.65 m/sr=1.65mm

INTRODUCTION

An experimental

An experimental investigation .....C.D. Stow & M.G. Hadfield

Splashing rough surfacev=3.65 m/sr=1.65mm

INTRODUCTION

An experimental

INTRODUCTION

1) Problem : Numerical Representation of Interfaces• Impact Dynamics : solid surface + liquid interface = drop surface • Splash Dynamics : liquid interface -> more liquid interfaces

2) Problem : Correct Physics•Impact Dynamics : solid surface + liquid interface = drop surface

• Splash Dynamics : liquid interface -> more liquid interfaces

Hypoteses:No simulation of the impactNo splash o total rebound (low Re numbers, no rough surfaces) Axisymmetric simulation

Numerical Representation of Interfaces -> okCorrect Physics ?

Some features:

• Behavior of the impact for: Wettable-P/Wettable N/Wettable surfaces • •Deposition – Partial rebound – total rebound

• Surface capillary waves

• Spreading ratio and Max spreading ratio

• Static/Dynamic/apparent Contact angle

INTRODUCTION

D=1.4mmv=0.77m/s

Re=1000We=10

Wettable Partially Wettable Non-Wettable

Deposition

Partially Wettable

Non-Wettable

INTRODUCTION

τ= τ(μ) = Stress tensor

Dynamics (incompressible. N.S.eqs)

incompressible

u = velocity p=pressure

f_s = Surface tension f = Body force

μ =viscosity = μ1 χ + (1-χ) μ2

ρ =viscosity = ρ1 χ + (1-χ) ρ2

Kinematics (Phase eq.)

Equation for χ (phase indicator)χ =0 phase 1χ =1 phase 2

Solution:1) Weak form (method of characteristics)2) Geometrical algorithm

Boundary conditions

Static cos() =cos(s) v=0 no-slip boundary condition

Non Static cos() =cos(s) ? v=0 no-slip boundary condition ?

V. FORM OF THE STOKES PROBLEM

2

0,||min 1

0uSS

VuHu

0

0

up

upuu

V

VV

gives

20

10

Lp

Hu

20

10

Lp

Hu

CONTACT PROBLEM (NO INERTIAL FORCES)

dAdAdAF

dVuS

FS

gsls

gsls

V

uHu

lg

10

2

0,

||2

1

)min(

10

10

Hu

Hu

F = Shape derivative in the direction u

CONTACT PROBLEM (NO INERTIAL FORCES)

un

dAdAdAdt

dF

gsls

gsls

lg

lg

0

0lg

up

unupuu

V

VV

Minimization gives

20

10

Lp

Hu

20

10

Lp

Hu

10Hu

No angle condition

)2

1(min 2

20,10

s

zdAuFS

uHu 1

0zHu

dAussun

utunF

sc

cs

s

2))cos()(cos()(

))cos()(cos(

lg

lg

MINIMIZATION WITH PENALTY

10

10 HuHu z

Remarks:s

dAu222

1 Is a dissipation term

Contact angle condition

CONTACT PROBLEM WITH PENALTY

0

0))cos()(cos(lg

2

up

utun

uuupuu

V

cs

VV s

Minimization gives

20

10

Lp

Hu z

20

10

Lp

Hu z

10zHu

0))cos()(cos()(

2

ssc

sss

uss

uuupuu

s

sss

Boundary condition over the solid surface

)(10 ss Hu

02 u

0),,,,( suf Boundary condition

0 Full slip boundary cond

V.F.OF THE CONTACT PROBLEM

0

0))cos()(cos(

)(

lg

2

up

utun

uuupuu

uuuut

u

V

cs

VV

VV

s

20

10

Lp

Hu z

20

10

Lp

Hu z

0 Near the contact point

otherwise

Numerical solution

Fem solution

•Weak form -> fem•Advection equation -> integral form•Density and viscosity are discountinuous -> weak f.•Surface term singularity-> weak form

ADVECTION EQUATION

0

ut

1

0

0t

t

udtxx

10 tt Surface advection

Integral form

Advection equation

(2D)

ReconstructionAdvection

ADVECTION EQUATION

Markers= intersection (2markers) Conservation (2markers)

Fixed mrks (if necessary)

VORTEX_SQUARE.MPEG

ADVECTION EQUATION

Vortex testsADVECTION EQUATION

ADVECTION EQUATION

ADVECTION EQUATION

Fem surface tension formulationSurface form

Volume formc

hhhh

hh uds

dxdA

ds

u

ds

dxdAun

lglg

dVudVu

dVudAun

V

hh

V

hh

V

hhhh

lglg

lglg

Is extended over the droplet domain

Static: Laplace equationSolution for bubble v=0, p=p0

Spurious Currents

Fem surface tension formulation

Static: Laplace equationSolution for bubble v=0, p=p0

1) Computation of the curvature2) Computation of the singular term

Solution v=0, v=0p=0 outside p=P0=a/R inside

Fem surface tension formulation

Casa A: exact curvature

SolutionCurvature=1/RSurface tens=σV=0; p=p0

No parassitic currents

Fem surface tension formulation

Case B: Numerical curvatureWith exact initial shape

A t=0 B t=15 C t=50Curvature

Initial velocity

Final velocity

Fem surface tension formulation

Case C: Numerical curvature (ellipse)

Shape

time

Fem surface tension formulation

Steady solution

angle=120

angle=60

angle=90

Boundary condition over the solid surface

Boundary condition over the solid surface

1

)(10 ss Hu

02 u

0),,,,( suf

02 Full slip boundary cond

Re=100 We=20 =60 Deposition

t=0

t=2.5

t=4

t=15

t=50

t=0t=0

t=0

t=0.5

t=3

t=1.5

t=1

Re=100 We=20 =60 Deposition

Re=100 We=20 =90 partial rebound

t=4

t=5

t=0t=0t=0t=0t=0t=0

t=6

t=0t=0

t=3

t=2

t=1.5

t=1

t=0.5

t=0

t=7

t=9

t=8

t=10

t=11

t=14

Re=100 We=20 =90 partial rebound

Re=100 We=20 =120 total rebound

t=.5

t=1.5 t=3t=0

t=2 t=4

t=7

t=2.5t=1

Re=100 We=20 =120 total rebound

DIFFERENT WETTABILITY

Wettable (60) A Non-wettable (120) C

partially wettable (90) B

Re=100 We=100 =120

Re=100 We =120 u0 =120

We= 100 A 50 B 20 C 10 D

u0= 2 A 1 B .5 C

Different impact velocityDifferent We

DYNAMICAL ANGLE

cWeaD

D)(Re 5.0

max0

Glycerin droplet impact v=1.4m/s D=1.4mm

Wettable (18) Partially wettable (90)

DYNAMICAL ANGLE

0))cos()(cos()(

)(

2

ssc

ss

SS

uss

uuupuu

uuuut

u

s

sss

ss

))cos())(cos((' sdcss

' dFriction over the solid surface Friction over the rotation

DYNAMICAL ANGLE MODEL

Cox

mds ACa )cos()cos(

)96.4tanh()cos()cos( 706.0CaAds

)sinh()cos()cos( BCaAds Blake

Power law

Jing

Non-Wettable

Wettable

D=1.4mm u0=1.4m/s glycerin droplet

A=1B=2C=10

A=1B=2C=10

D/D0

h angle

Conclusions

- Variational contact models can be used

- Open question: Can we simulate large classes • of droplet impacts with a unique setting of• boundary conditions ?

Thanks