Download PHY 711 -- Assignment #24

PHY 711 Classical Mechanics and
Mathematical Methods
11-11:50 AM MWF Olin 107
Plan for Lecture 32:
Effects of viscosity in fluid motion –
Chap.12 in Fetter & Walecka
1. Navier-Stokes equation
2. Terminal velocity of a sphere
moving with constant applied force
in a viscous medium
3. Stokes’ viscosity relation
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PHY 711 Fall 2016 -- Lecture 32
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11/22/2013
PHY 711 Fall 2013
2016 -- Lecture 34
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2
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Brief introduction to viscous effects in incompressible fluids
Stokes' analysis of viscous drag on a sphere of radius R
moving at speed u in medium with viscosity  :
FD   6Ru 
u
FD
Plan:
1. Consider the general effects of viscosity on fluid
equations
2. Consider the solution to the linearized equations
for the case of steady-state flow of a sphere of
radius R
3. Infer the drag force needed to maintain the
steady-state flow
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Newton-Euler equation for incompressible fluid,
modified by viscous contribution (Navier-Stokes equation):
v
p  2
  v    v  f applied 
  v
t
 
 Kinematic viscosity
Typical kinematic viscosities at 20o C and 1 atm:
Fluid
1.00 x 10-6
Water
14.9 x 10-6
Air
Ethyl alcohol
Glycerine
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n (m2/s)
1.52 x 10-6
1183 x 10-6
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Stokes' analysis of viscous drag on a sphere of radius R
moving at speed u in medium with visc osity  :
FD   6Ru 
u
F
FD
Effects of drag force on motion of
particle of mass m with constant force F :
du
F  6Ru  m
with u (0)  0
dt
6R

t

F
m
1  e

 u (t ) 

6R 

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Effects of drag force on motion of
particle of mass m with constant force F :
du
F  6Ru  m
with u (0)  0
dt
u
6R

t


F
F
1  e m 
 u (t ) 
FD

6R 

u
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t
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Effects of drag force on motion of particle of mass m
with an initial velocity with u (0)  U 0 and no external force
du
6 R u  m
dt
 u (t )  U 0e

u
6 R
t
m
FD
u
t
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Recall: PHY 711 -- Assignment #24
Oct. 28, 2015
Determine the form of the velocity potential for an
incompressible fluid representing uniform velocity in the z
direction at large distances from a spherical obstruction
of radius a. Find the form of the velocity potential and the
velocity field for all r > a. Assume that for r = a, the
velocity in the radial direction is 0 but the velocity in the
azimuthal direction is not necessarily 0.
 0
2
3

a 
  r ,   v0  r  2  cos
2r 

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Newton-Euler equation for incompressible fluid,
modified by viscous contribution (Navier-Stokes equation):
v
p  2
  v    v  f applied 
  v
t
 
Continuity equation:   v  0
v
Assume steady state: 
0
t
Assume non-linear effects small
Initially set f applied  0;
  p   v
2
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 p   2 v
Take curl of both sides of equation:
    p   0      v 
2
Assume (with a little insight from Landau):
v      f  r  u   u
where
f (r ) 
0
r 
Note that:
    A      A   2A
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Digression
Some comment on assumption: v       f  r  u   u
    A      A   2A
Here A  f  r  u
  v          A        2 A 
Also note :
p   v
2
   p     2 v
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or  2   v   0
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v      f r u   u
u  uzˆ
2
ˆ
ˆ
    f r z     f r z    f (r )zˆ
  2   v   0
 v  0
 4   f (r )zˆ   0
  4 f (r )  zˆ   0
  4 f (r )  0
C4
f (r )  C1r  C2 r  C3 
r
2C2 2C4 
 2 df 

vr  u cos 1 
 3 
  u cos  1  4C1 
r
r 
 r dr 

2
 d 2 f 1 df 
C2 C4 

v  u sin  1  2 
 3
  u sin  1  4C1 
dr
r dr 
r
r 


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Some details:
2
d
2 d 
 f (r )  0
 2 
 f (r )  0
r dr 
 dr
C4
2
f (r )  C1r  C2 r  C3 
r
v  u      f  r  zˆ   zˆ
2
4

 


=u     f  r  zˆ    2 f (r ) zˆ  zˆ

Note that: zˆ  cos  rˆ  sin  θˆ
  df


2
ˆ
v  u    cos      f (r )   1 cos rˆ  sin  θ 

  dr


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2C2 2C4 
 2 df 

vr  u cos 1 
 3 
  u cos  1  4C1 
r
r 
 r dr 

 d 2 f 1 df 
C2 C4 

v  u sin  1  2 
 3
  u sin  1  4C1 
dr
r dr 
r
r 


To satisfy v(r  )  u :
 C1  0
To satisfy v( R)  0
solve for C2 , C4
 3R R 3 
vr  u cos 1 
 3
2r 2r 

 3R R 3 
v  u sin  1 
 3
4r 4r 

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 3R R 3 
vr  u cos  1 
 3 
 2r 2r 
 3R R 3 
v  u sin  1 
 3 
 4r 4r 
Determining pressure :

 3R  
p   v   u cos   2  
 2r  

 3R 
 p  p0  u cos   2 
 2r 
2
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 3R 
p  p0   u cos  2 
 2r 
Corresponds to:
FD cos   p ( R )  p0  4 R
2
 FD =   u  6 R 
u
FD
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