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
INTRODUCTION
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 surface
v=3.65 m/s
r=1.65mm
INTRODUCTION
An experimental
An experimental
investigation .....
C.D. Stow & M.G.
Hadfield
Splashing
rough surface
v=3.65 m/s
r=1.65mm
INTRODUCTION
1) Problem : Numerical Representation of Interfaces
An experimental
•
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 impact
No splash o total rebound (low Re numbers, no rough surfaces)
Axisymmetric simulation
Numerical Representation of Interfaces -> ok
Correct Physics ?
INTRODUCTION
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
D=1.4mm
v=0.77m/s
Re=1000
We=10
Deposition
Partially
Wettable
Wettable
Non-Wettable
Partially Wettable
Non-Wettable
INTRODUCTION
Dynamics (incompressible. N.S.eqs)
incompressible
τ= τ(μ) = Stress tensor 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 1
χ =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
min SuH 1 ,u 0
0
S    | u |
2
V
gives
1


u



u

p



u

0
u  H 0


V
 p  u  0
V
V
p  L
2
0
uH
pL
1
0
2
0
CONTACT PROBLEM (NO INERTIAL FORCES)
min( S  F )uH 1 ,u 0
uH
0
1
0
u  H
1
2
S    | u | dV
2V
F    dA    ls dA    gs dA
lg
F
ls
g s
= Shape derivative in the direction u
1
0
CONTACT PROBLEM (NO INERTIAL FORCES)


d
F     dA    ls dA    gs dA  

dt  lg
ls
g s


u  H
  n  u
1
0
lg
Minimization gives
 u  u   p  u   n  u  0
V
V
lg
 p  u  0No angle condition
V
u  H
p  L
2
0
1
0
uH
pL
1
0
2
0
MINIMIZATION WITH PENALTY
1
2

min uH 1 ,u 0 ( S  F   u2 dA)
0z
2 s
u  H
1
0z
F   n  u   (cos( )  cos(s ))t  uc 
lg
 n  u     (s
lg

sc )(cos( )  cos(s ))u2 dA
s
u  H   u  H
1
0z
Remarks:
1
2

u
dA
2
2 s
1
0
Is a dissipation term
Contact angle condition
CONTACT PROBLEM WITH PENALTY
Minimization gives
 u  u   p  u   u2  u 
V
V
s
u  H
1
0z
  n  u   (cos( )  cos(s ))t  uc  0
lg
 p  u  0
V
u  H
p  L
2
0
1
0z
uH
pL
1
0z
2
0
Boundary condition over the solid surface
 u  u   p  u   u
s
s
s
s
2
s
 us 
us  H (s )
    ( s  sc )(cos( )  cos(s ))us  0
1
0
s
f ( , u,  , ,s )  0
   u2  0
Boundary condition
 0
Full slip boundary cond
V.F.OF THE CONTACT PROBLEM
u  H u  H
u
V  t  u  V  (u  )u  u 
 u  u   p  u   u
V
1
0z
p  L
2
 u 
2
0
pL
s
V
  n  u   (cos( )  cos(s ))t  uc  0
lg
 p  u  0
V
 0
 
Near the contact point
otherwise
1
0z
2
0
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

 u    0
t
Advection equation
t1
x  x0   udt
Integral form
t0
t0  t1
Surface advection
ADVECTION EQUATION
Reconstruction
(2D)
Advection
Markers= intersection (2markers)
Conservation (2markers)
Fixed mrks (if necessary)
ADVECTION EQUATION
VORTEX_SQUARE.MPEG
ADVECTION EQUATION
Vortex tests
ADVECTION EQUATION
ADVECTION EQUATION
Fem surface tension formulation
Surface form
dxh uh
 dxh


n


u
dA




dA



u
h
h

 ds ds
 ds h 
c
lg
lg
Volume form

n


u
dA





u
dV

h
h
h
h


lg
Vlg
  uh   h dV    h  uh dV

Vlg
Vlg
Is extended over the droplet domain
Fem surface tension formulation
Spurious Currents
Static: Laplace equation
Solution for bubble v=0, p=p0
Fem surface tension formulation
1) Computation of the curvature
2) Computation of the singular term
Static: Laplace equation
Solution for bubble v=0, p=p0
Solution
v=0,
v=0
p=0 outside p=P0=a/R inside
Fem surface tension formulation
Casa A: exact curvature
Solution
Curvature=1/R
Surface tens=σ
V=0; p=p0
No parassitic currents
Fem surface tension formulation
Case B: Numerical curvature
With exact initial shape
Curvature
A t=0 B t=15 C t=50
Initial velocity
Final velocity
Fem surface tension formulation
Case C: Numerical curvature (ellipse)
time
Shape
Steady solution
angle=120
angle=90
angle=60
Boundary condition over the solid surface
Boundary condition over the solid surface
us  H (s )
1
0
f ( , u,  , ,s )  0
1   u2  0
2  0
Full slip boundary cond
Re=100 We=20 =60 Deposition
t=0
t=1.5
t=4
t=0
t=0
t=0.5
t=2.5
t=15
t=3
t=50
t=0
t=1
Re=100 We=20 =60 Deposition
Re=100 We=20 =90 partial rebound
t=0
t=1.5
t=0
t=0
t=0
t=0
t=0
t=0
t=0.5
t=1
t=2
t=3
t=4
t=0
t=0
t=5
t=6
t=7
t=8
t=9
t=10
t=11
t=14
Re=100 We=20 =90 partial rebound
Re=100 We=20 =120 total rebound
t=1.5
t=0
t=2
t=.5
t=7
t=1
t=2.5
t=3
t=4
Re=100 We=20 =120 total rebound
DIFFERENT WETTABILITY
partially wettable (90) B
Wettable (60) A
Non-wettable (120) C
Re=100 We=100 =120
Different We
Re=100 We =120
We= 100 A
50 B
20 C
10 D
Different impact velocity
u0 =120
u0=
2A
1B
.5 C
DYNAMICAL ANGLE
Glycerin droplet
impact v=1.4m/s D=1.4mm
Wettable (18)
D
0.5 c
 a(Re We )
D0 max
Partially wettable (90)
DYNAMICAL ANGLE
u
  t  uS    (u  )u  uS 
s
s
 u  u   p  u   u
s
s
s
s
2
 u 
s
    ( s  sc )(cos( )  cos(s ))us  0
s
   ' (s  sc )(cos(d )  cos(s ))
'
Friction over the solid surface
d
Friction over the rotation
DYNAMICAL ANGLE MODEL
Cox
cos( s )  cos(d )  A sinh( BCa ) Blake
m
Power law
cos( s )  cos(d )  ACa
cos( s )  cos(d )  A tanh( 4.96Ca
0.706
)
Jing
D=1.4mm u0=1.4m/s
A=1
B=2
C=10
Non-Wettable
glycerin droplet
A=1
B=2
Wettable C=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