Transcript Navier-Stokes - Northern Illinois University

Navier-Stokes
Eulerian View
 
r (r0 , t )
 In the Lagrangian view each
body is described at each point
in space.
 Difficult for a fluid with many
particles
 In the Eulerian view the points
in space are described.

 (r , t )
 
v (r , t )
 Bulk properties of density and
velocity
Streamlines
 A streamline follows the
tangents to fluid velocity.
 Lagrangian view
 Dashed lines at left
 Stream tube follows an area
 A streakline (blue) shows the
current position of a particle
starting at a fixed point.
 A pathline (red) tracks an
individual particle.
Wikimedia image
Fluid Change
 A change in a property like
pressure depends on the view.
 In the Lagrangian view the
total time derivative depends
on position and time.
dp p p dx p dy p dz




dt t x dt y dt z dt
dp p 

 v  p
dt t
d
F  2k  l  x
  v 
dt t
2
 The Eulerian view uses just
the partial derivative with time.
2
l
 Points in space are fixed
dp
p

dt r const t

x
l 2  x2
Jacobian Tensor
 A general coordinate
transformation can be
expressed as a tensor.
 Partial derivatives between
two systems
 Jacobian NN real matrix
 Inverse for nonsingular
Jacobians
 Cartesian coordinate
transformations have an
additional symmetry.
 Not generally true for other
transformations
xi 
xi
qm
qm
 x 
J i 
 qm 
xi
0
q m
xi  J imqm
qm 
xi  
 ei  e j  cosij
x j
J ij 
xi x j

x j xi
qm
xi
xi
Volume Element
V   q1q2q3
x3
V  x1x2x3
 An infinitessimal volume
element is defined by
coordinates.
 dV = dx1dx2dx3

  xi

dx1  
q1 
 q1

x2
x1



V  x1  (x2  x3 )
V   ijk
x j
xi
x
q1
q2 k q3
q1
q2
q3
V  J q1q2q3  J dV 
 Transform a volume element
from other coordinates.
 components from the
transformation
 The Jacobian determinant is
the ratio of the volume
elements.
Compressibility
 A change in pressure on a fluid
can cause deformation.
V
p
V
 p
V
B
Vp 1

V 
 Compressibility measures the
relationship between volume
change and pressure.
 Usually expressed as a bulk
modulus B
 Ideal liquids are
incompressible.
Volume Change
 Consider a fixed amount of
fluid in a volume V.
 Cubic, Cartesian geometry
 Dimensions x, y, z
 The change in V is related to
the divergence.
 Incompressible fluids - no
velocity divergence
v
d
x  x x
dt
x
v y
d
y 
y
dt
y
d
v
z  z z
dt
z
 vx v y vz 
d
xyz
V  


dt
 x y z 

d
V    v V
dt
Balance Equations
 The equation of motion for an arbitrary density in a
volume is a balance equation.
 Current J through the sides of the volume
 Source s inside the volume


d
    v  s    J
dt
 

   ( v  J )  s
t
 Additional balance equations describe conservation of
mass, momentum and energy.
 No sources for conserved quantities
Mass Conservation
 A mass element must remain
constant in time.
 Conservation of mass
 Combine with divergence
relationship.
 Write in terms of a point in
space.


   ( v )  0
t
m  V
d
d
m   V   0
dt
dt
d
dV
V  
dt
dt

d

V    v V  0
dt

d
   v  0
dt

 
 v      v  0
t
Pressure Force
 Each volume element in a fluid
is subject to force due to
pressure.
 Assume a rectangular box
 Pressure force density is the
gradient of pressure
z
p

p
p
p
F   xˆ V  yˆ V  zˆ V
x
y
z

F  pV
V
x
p
 p 
x yz   V
x
 x 
Fx   
y
Equation of Motion
 A fluid element may be subject
to an external force.
 Write as a force density
 Assume uniform over small
element.
 The equation of motion uses
pressure and external force.
 Write form as force density
 Use stress tensor instead of
pressure force
 This is Cauchy’s equation.
 
F  fV


dv
m   F
dt

dv 
 V
 fV  pV
dt


dv
  p  f
dt


dv
   P  f
dt
Euler’s Equation
 Divide by the density.
 Motion in units of force density
per unit mass.
 The time derivative can be
expanded to give a partial
differential equation.
 Pressure or stress tensor
 This is Euler’s equation of
motion for a fluid.


dv 1
f
 p 
dt 



 1
v 
f
 v   v  p 
t




 1
v 
f
 v  v    P 
t


Momentum Conservation

v V


dv
 V
 pV  fV
dt

d 
v V   f  p V
dt


d
v dV  

dt V
V

 The momentum is found for a
small volume.
 Euler equation with force
density
 Mass is constant
 Momentum is not generally
constant.

fdV   pdV
V


d
v dV   fdV    nˆpdS

dt V
V
S
 Effect of pressure
 The total momentum change is
found by integration.
 Gauss’ law
Energy Conservation
 The kinetic energy is related to
the momentum.
d
dt
 Right side is energy density

1
2


 
v V  v  f  p V
2

d
 pV   dp V  p dV
dt
dt
dt

p 

 v    pV  p  v V
t

 v    pV 
 Some change in energy is
related to pressure and
volume.
 Total time derivative
 Volume change related to
velocity divergence

d
dt

1
2
d
 pV   p  p  v V
dt
t
 

p
v V  pV  v  fV   p  v V
t
2

Work Supplied
 The work supplied by
expansion depends on
pressure.

dW
dV
p
 p  v V
dt
dt
 Potential energy associated
with change in volume
dW
d
d
  um   u V
dt
dt
dt
 This potential energy change
goes into the energy
conservation equation.

d
p   v V   u V 
dt
d
dt

1
2
 
p
v V  pV  u V  v  fV 
t
2

Bernoulli’s Equation
 Gravity is an external force.

f  gzˆ  
 

v  fV  v  V
 Gradient of potential
 No time dependence
 d  

   V
 dt t 
d

 V   
V
dt
t
 The result is Bernoulli’s
equation.
 Steady flow no time change
 Integrate to a constant
d 1 2
p

V   V
2 v V  pV   V  u V 
dt
t
t
2
 1 p 
d 1 2 p
v
p
 2 v     u  

  gz  u  k
dt 

2 
  t t


Strain Rate Tensor
 Rate of strain measures the
amount of deformation in
response to a stress.
 Forms symmetric tensor
 Based on the velocity gradient

v x

x

 1  v v 
y

E    x 
 2  y x 
 1  v v 
  x z
 2  z x 

1  v x v y 



2  y x 
v y
y
1  v y v z 



2  z y 
1  v x v z  



2  z x  
1  v y v z  

 

2  z y  

v z


z

Stress and Strain
 There is a general relation
between stress and strain
 Constants a, b include
viscosity
P  aE  b1
a  2
 An incompressible fluid has no
velocity divergence.

2
b   p    v
3

2

P  2E   p    v 1
3


P  2E  p1
Navier-Stokes Equation
 The stress and strain relations can be combined with the
equation of motion.
 Reduces to Euler for no viscosity.


 f 1
v 
 2
 v  v     P   v
t
 

Bernoulli Rederived
1

 P  

v  0
p

 0

v  

v 


   
t t
t
f  gz


 f 1
v 
 v  v     P  0
t
 
2

v
p

   ( gz)    0
t
2

 v 2
p
  gz   c(t )
t 2

 Make assumptions about
flow to approximate fluid
motion.




Incompressible
Inviscid
Irrotational
Force from gravity
 Apply to Navier-Stokes
 The result is Bernoulli’s
equation.