Biological fluid mechanics at the
micro‐ and nanoscale
Lecture 2:
Some examples of fluid flows
Anne Tanguy
University of Lyon (France)
Some reminder
I. Simple flows
II.Flow around an obstacle
III.Capillary forces
IV.Hydrodynamical instabilities
REMINDER:
The mass conservation:
, for incompressible fluid:
The Navier-Stokes equation:
with
 2 e  (  
2
 )  . v I with e   v for a « Newtonian fluid ».
S
3
Thus:
for an incompressible and Newtonian fluid.
Claude Navier
1785-1836
Georges Stokes
1819-1903
(Giesekus, Rheologica Acta, 68)
Non-Newtonian liquid
Different regimes:
Born: 23 Aug 1842
in Belfast, Ireland
Died: 21 Feb 1912
in Watchet, Somerset,
England
Re = 5.7 10-4
Re = 1.25 10-2
(Boger, Hur, Binnington, JNFM 1986)
Re << 1 Viscous flow (microworld)
Re >> 1
Ex. perfect fluids (=0)
or transient response t<<tc ,at large scales L>Lc
and

v
t
2
   v diffusive transport of momentum
needs a time
tc  L
tc=10-6 s (L=10-6m)
L  Lc 
T

2


to establish
tc=106s (L=1m)
Lc=0.1mm for w=20 Hz
Lc=10mm for w=20 000 Hz
Bernouilli relation when viscosity is negligeable (ex. Re >>1):
  v2 

  v    v    p   U pot

    
t
  2 

v
Along a streamline (dr // v), or everywhere for irrotational flows (   v  0 ),
For permanent flow :

v
2
2
 p   U pot  cste
For « potential flows » ( with v   
): 

t
Daniel Bernouilli
1700-1782

v
2
2
 p   U pot  cste
How solve the Navier-Stokes equation ?
Non-linear equation. Many solutions.
• Estimate the dominant terms of the equation (Re, permanent flow…)
• Do assumptions on the geometry of the flows (laminar flow …)
• Identify the boundary conditions (fluid/solid, slip/no slip, fluid/fluid..)
Ex. Fluid/Solid: rigid boundaries
  0  v // solid surface
  0  v fluid  v solid in general, at the solid surface
(see lecture 5 !)
Ex. Fluid/Fluid: soft boundaries
(see lecture 3 !)
 1
1 
 with  , surface
p I  p II   

R
R
2 
 1
tension
I. Simple flows
Flow along an inclined plane:
Assume: a flow along the x-direction: v ( x , z )  v x ( x , z ) e x
Continuity equation:
v x
x
 0  v ( z ), then ( v . ) v  0
Boundary conditions: v ( z  0 )  0
P ( x , z  h )  P0
 ' (x, z  h)  0  
P
Navier-Stokes equation:
x
P
z
v
z
(z  h )  0
 v
2
  g sin   
   g cos 
z
2
(z )  v (z ) 
g

sin  z ( h  z / 2 )
 P ( z )  P0   g cos  ( z  h )
Flow along an inclined plane:
Flow rate: Q 
  v ( z ) dz . L
z0
 g sin  L y h
2
h
y

3
3

test for rheological laws
L x L y  g sin  h
Force applied on the inclined plane: F  L x L y  . n 
0
 P(z )
Friction and pressure compensate the weight of the fluid (stationary flow).
Planar Couette flow:
Assume: a flow along the x-direction: v ( x , z )  v x ( x , z ) e x
Continuity equation:
v x
Boundary conditions:
P ( x  0 , z )  P ( x  L , z ) no pressure
x
 0  v ( z ), then ( v . ) v  0
gradient
v (z  0)  0
v (z  h )  U e x
Navier-Stokes equation:
P
x
P
z
 v
2

z
2
(z )  v (z )  U
h
  g
Force applied on the upper plane: Fx  
z
 P ( z )  P0   gz
U
h
per unit surface
Fx=106 Pa U=1 m.s-1 h=1 nm
Cylindrical Couette flow:
Assume: symetry around Oz + no pressure gradient along Oz:
v ( r ,  , z )  v r ( r ) e r  v  ( r ) e  and p ( r )
Continuity equation:  v r  v r  1  ( rv )  0  rv ( r )  cst
r
r
r
r
r r
Boundary conditions: v ( R 1 )   1 e 
v ( R 2 )   2 e

Navier-Stokes equation:
v
r
2
 
1 P
 r
 rv r ( r )  0
radial gradient compensates radial inertia
  2v
1 v 
v 
 2
r  r   0 no torque
0   

 2  
2
r r
r 
r
 r
Cylindrical Couette flow:
Friction force on the cylinders:
 1   2 R 1 R 2 1
v 
 v
         2
. 2
2
2
r 
r
R 2  R1
 r
2
 r
torque 1 / Oz 
 dz . 2 R
1
2
. R 1 . r  ( R 1 )   (  2   1 ) / d
z
Couette Rheometer:
Rotation is applied on the internal cylinder, to limit v .
Taylor-Couette instability:
Planar Poiseuille flow:
z
Assume: a flow along the x-direction: v ( x , z )  v x ( x , z ) e x
Continuity equation:
v x
Boundary conditions:
x
 0  v ( z ), then ( v . ) v  0
 P  P ( x  0 , z )  P ( x  L , z ) Pressure
gradient
v (z  0)  0
v (z  h )  0
Navier-Stokes equation:
P
x
P
z
Flow rate Q 

 v
2

z
2
(z )  v (z ) 
0
L y v x ( z ) dz 
P
2 L
. z .( z  h )
 P (x )  P (0) 
 P.L y
12  . L x
small Force exerted on the upper plane:
P
.x
L
.h
3
Fx  
P
2L
. h per unit surface
Poiseuille flow in a cylinder (Hagen-Poiseuille):
Assume: flow along Oz+ rotational invariance: v ( r ,  , z )  v z ( r , z ) e z
Continuity equation:
v z
 0  v z (r)
z
Boundary conditions: v ( r  R ,  )  0
P ( z  0 )  P ( z  L )   P Pressure
P
Navier-Stokes equation:
r

P
r
 P   gL
8 L
 0  P(z )
2
  2v z
1  v z 1 v z  P
 
 
 2

 g
2
2
r 
r r  z
 r
v z(r) 
Flow rate: Q 
Gradient
. . R
4
 P   gL
4 L
.( R  r )
2
Friction force:
2
Fz  (  P   gL ) . R
Total pressure force:
2
Fr  P  2  . R . L
Jean-Louis Marie Poiseuille
1797-1869
(1842)
(2010)
Rheological properties of blood
Elasticity of the vessel
Bifurcations
Thickening
Non-stationary flow…
Other example of Laminar flow with Re>>1:
Lubrication hypothesis (small inclination)
v ( y  0)  U ,
p ( x  0 )  p ( x  L )  P0
v(y  h)  0
  ( v . ) v  m  v
2
 vx
2
and m
y
2

P
x
 vy
2
 m
y
2
cf. planar flow with x-dependence
Poiseuille
+ Couette

P
y
0
 Q / Lt
6mL  U  h(x )


p ( x )  P0 
 1  
2

h0  h L  h(x )  h L
h
(
x
)


with the flow rate Q/L t  U.
h 0h L
h 0  h1
  h(x ) 2


  1  

 hL 



=1.2kg.m-3 =2.10-5 Pa.s L ~ 1m, h ~ 1 cm, U ~ 0.1m/s Re ~ 6000< (L/h)2 = 10000
xM ~ e1.L/h ~ 10 cm
Supporting pressure PM ~ 10-1Pa
Flow above an obstacle: hydraulic swell
Froude 
U
gh

fluid velocity
gravitatio nal waves
1
Mass conservation: U.h=U(x).h(x)
(I)
Bernouilli along a streamline close to the surface:
P0 
U
2
2
  gh  P0 
U ( x )
2
2
  g h ( x )  e 0 ( x ) 
then
(II)
Case (I): dU/dx(xm)=0 then U2(x)-gh(x)<0
then
U(x)
and h(x)
Case (II): dU/dx(x) >0 then U2(xm)-gh(xm)=0
then
U(x)
and h(x)
U2(x)-gh(x) <0 becomes >0
low velocity of surfaces waves
Hydraulic swell
End of Part I.