Transcript Potential flow

Biological fluid mechanics at the
micro‐ and nanoscale
Lectures 3:
Fluid flows and capillary forces
Anne Tanguy
University of Lyon (France)
Lecture 3
Some reminder
I. Simple flows
II.Flow around an obstacle
III.Capillary forces
IV.Hydrodynamical instabilities
II. Flow around an obstacle
The case of « Potential flows »
incompress ible fluid.
 2 v
is negligeable.
if   v ( r , t 0 )  0 then   v ( r , t )  0 always
  v ( r , t )  0  v ( r , t )   " Potential flow"
Mass conservati on : .v  0  2  0 (Laplace' s equation)

v2
Bernouilli relation : 

 P  U pot  cst
t
2
Boundary conditions ?
U
« Potential flow » around a fixed cylinder:
Boundary conditions:
Potential :
 lim r v  U

v r ( r  R ,  )  0
 R2 
p cos 
   0  ( r, )  U.r. cos  
 U.r. cos  .1  2 
2 .r
r 

2
Uniform flow + dipole
Stream lines
Velocity v:
Pressure
 R2 

vr 
 U. cos  .1  2 
r
r 

 R2 

v 
  U. sin  .1  2 
r
r 

« Potential flow » around a rotating cylinder: the Magnus force.
U
Boundary conditions: 

lim r  v  U
 v.d r    2R.R


r  R
Potential :
Velocity v:
Stream lines
vorticity

R 2  
2  0  ( r, )  U.r. cos  .1  2  
r  2

Fixed cylinder + vortex


R2 
vr 
 U. cos  .1  2 
r
r 



R2 

v 
  U. sin  .1  2  
r
r  2r

Asymetric flow: arrest points
If ||<4R|U|
sin=/4RU, r=R.
Else r=rP>R.
Magnus Force Fz=-U=-∫P(R).sin.Rd
on the solid.
No viscous dissipation (no drag force).
Pressure
Force
Air foil / birds wing
Conformal mapping
Joukowski’s transform:
Z= g(z) =
f (z ) known
F( Z)  f (g 1 ( Z))
Stream lines
Pressure
Force
Perfect potential flow around a sphere:
Spherical coordinates
2  0   ( r ,  )  U.r. cos  
p cos 
4 .r 2
Uniform flow + 3D dipole
 R3 

vr 
 U. cos  .1  3 
r
r 

velocity decrease ~1/r3.


R3 
v 
  U. sin  .1  3 
r
 2r 
Viscous flow around a sphere: the Stoke’s force
Navier - Stokes equation P  2 v
with .v  0 then P    (  v ) and thus 2 P  0.
Boundary conditions: lim r v  0 and v ( r  R , )  U sphere
 3R R 3 

v r  U cos  

2 r 2 r 3 
3
cos 

P( r ,  )  UR 2 
 3R R 3 
2
r
v   U sin  
 3 
4
r
4r 

 ( r ,  )   P I   2 v 
Low velocity decrease ~1/r.
cos 
R  F   .ndS   6  R U
sphere

3
sin 
 r ( r  R , )   U
2
R
3
2
 rr ( r  R ,  )   P   U
Stoke’s force.
II. Capillary forces
Surface tension
Definition of capillary forces.
At the interface between different phases/different chemical composition
Effective force insuring the equilibrium
Energy per unit surface:  , « surface tension »
E   .A
F.dh  dE   .L.dh
F   .L
« capillary force »
Examples: Liquid/vapor interface
Molecular Dynamics Simulations
at constant T and V
(L. Joly, LPMCN)
cf. lecture 7
Water (20°C)=72.8 mN/m
Ethanol (20°C)=22.10 mN/m
Comparison with gravitational forces:
area A
h
V=h.A
Total Energy:
Einterfaces≈ A.(LV+SL-SV)
Egravity ≈ 0.5 .g.h2.A
Egravity >> Einterfaces for h>> lc
lc 

 .g
« capillary length »
lc=2,7 mm for water et 20°C
Examples: Liquid/solid interfaces (without gravity)
Equilibrium of capillary forces :
  SV .dl   SL .dl   LV . cos  .dl  0
  LV . cos   ( SV   SL )
You ng' s Law
possible if  SV   SL   LV .
Contact angle 
0<<90°: liquid is « partially wetting »
90°<: liquid is « non wetting »
=0°: « complete wetting »
3
Effect of the curvature on the pressure:
Laplace’s law
Equilibrium of forces : pressure  surface tension
 2 .dl 2 . sin( 1 )  2 .dl1. sin(  2 )  Pint .dl1.dl 2  Pext .dl1.dl 2  0
f1
f1
sin( 1 )  1 
dl1
2R1
sin(  2 )   2 
dl 2
2R 2
 1
1 
  0
Pint  Pext   .

 R1 R 2 
Laplace' s Law
(1749-1827)
Δp for water drops of different radii at STP
Droplet radius
1 mm
0.1 mm
1 μm
10 nm
Δp (atm)
0.0014
0.0144
1.436
143.6
Example: Alveoli of the lungs
R ≈ 50 mm DP≈2,8 .103 Pa if water.
DP smaller with a surfactant
≈ 5 to 45.10-3 N.m-1
Allows a common work of all the alveoli.
Else:
PB
PC
PB > PC . The small bubble will lose air
Example 2: droplet between 2 plates.
r
R
V  π.a 2.h
Pressure :
 1 1
Pint  Pext   LV .  
R r
 1 2 cos   V. LV

 Fpressure   .a 2 . LV . 
h 
h2
R
Vertical component of the capillary force :
Fcapillary  2 .a. LV . sin    LV
E. Csapo (2007)
V
h
LV=70 mN.m-1 =130° V=10-1 cm3
h= 100 mm FP= 0,95 N
FC= 6,25.10-3N
h= 1 mm FP= 9500 N ! FC= 6,25.10-2N
h= 1 nm FP= 95.108 N !! FC= 1,98 N
Example 3: ascent of a liquid in a thin tube (d<lc). Jurin’s law
Assume a spherical shape for the meniscus (r  lC )
PA  P0
2
(Laplace' s Law)
r
PD  PB   .g.h (static equilibrium of the fluid)
PB  PA 
PD  PE  P0

h
2.
2. . cos 

 .g.r
 .g.R
Jurin' s Law
For water at 20°C with =0°
R=1mm h=1,46 cm
R=10mm h=1,46 m
R=1mm h=14,6 m !
Sap and trees:
Example 4: Shape of the Meniscus in a free surface
Pliquid( z ( x ))  Pext    .g.z ( x ) and Pliquid( z ( x ))  Pext 
z

r(x)
2
l
 z(x )   c
r(x)
with lc 
Pext

capillary length
g
  dz 2 
1    
  dx  

and r ( x )   
2
dz
dx 2
x
3/ 2
local radius of curvature of the interface
2
dz
2 d z
if
 1 then z ( x )  l c . 2
dx
dx
 x
z ( x )  lc . cot an . exp   
 lc 
Interactions between 2 plates:
d 2 h( x )
h( x )  lc .
dx 2
2
I.
II.
III.
with boundary conditions:
I. h1(-∞)=0
II. h2’(x=0)=-cotan1
III. h3’(x=d)=-cotan2
h1’(x=0)=cotan1
h2’(x=d)=cotan2
h3(+∞)=0
 x
h1 ( x )  lc . cot an1. exp   
 lc 

dx
 x  2lc 2
lc
  cot an 2 . cosh   
h2 (x) 
.cot an1. cosh 
cos  (Jurin) if d  lc
l
l c 
d
d 
c



sinh  
 lc 
 dx

h3 ( x )  lc . cot an 2 . exp  
l
c


Vertical Capillary forces:
T = -2..cos.L ez
Horizontal Pressure forces:
FP = ∫P(z).dz.L ex = 0.5 gL.[ h22(o)-h12(0)] ex
If 1=2
I.
II.
III.
If d>>lc
If d<<lc FP ≈ 2 .lc2.L.(cotan1)2/d2 ex
FP ≈ 2 .L.(cotan1)2.exp(-d/lc) ex ≈ -T. cotan1.exp(-d/lc) ex
If cotan1.cotan2<0
Non wetting
wetting
Attractive forces (either for wetting or non-wetting surfaces)
If d << lc FP ≈ 2 .lc2.L.(cotan1+ cotan2)2/d2 ex Attractive Force
If d=d*
FP ≈ - 0.5 .L.(cotanmax)2 ex
If d > d* FP <0
Max. Repulsive Force
Repulsive Force at large distances.
d*=lc.acosh(|cotanmin/cotanmax|)
Beetle Larva
III. Related instabilities
The Marangoni effect:
Effect of boundary conditions
Gradient of surface tension on the upper free surface (cf. lecture 3)
 S xz ( x, z  h ) 
d
d
v
viscous
( x )   xz
( x, z  h ) 
( x )   x ( x, z  h )  0
dx
dx
z
Ex. Temperature gradient // surface,
Chemical gradient (soap on water, Tears of wine: alcohol in water)
2

Navier-Stokes equation:  v2x ( x, z )  0
z

v x ( x, z ) 
d
.z
dx
Motion in the direction of larger surface tension
(flow from alcohol to water, hot places to cold places..)
The Bénard-Marangoni instability:
Local gradient of temperature (cf. Marangoni)
Flow due to coupling between T and v
T
Fourier’s law
 (v. )T  2T
(cf. lecture 4)
t
d
.DT.h
dt
Marangoni number: Ma 


motion
 Ma c
dissipation
The Taylor-Couette instability:
(Couette 1921, Taylor 1923)
Volumic competition between inertia
and viscous forces
when motion is driven by the internal cylinder.
Taylor number:
12 .  R  .( R 2  R1 )3
inertia
Ta 

 Ta c
dissipation
( /  )2
Next lecture:
From Liquid to Solid,
Rheological behaviour
(Lecture 6)