MAE 5130: VISCOUS FLOWS
Momentum Equation: The Navier-Stokes Equations, Part 2
September 9, 2010
Mechanical and Aerospace Engineering Department
Florida Institute of Technology
D. R. Kirk
1
GOAL: INCOMPRESSIBLE, CONSTNAT m N/S EQUATION
 
ma  F

 
DV

 g  f surface
Dt
•
Start with Newton’s 2nd Law for a fixed mass
•
•
•
•
Divide by volume
Introduce acceleration in Eulerian terms
Ignore external forces
Only body force considered is gravity
•
Express all surface forces that can act on an element
– 3 on each surface (1 normal, 2 perpendicular)
– Results in a tensor with 9 components
– Due to moment equilibrium only 6 components are
independent
•
Employ Stokes’ postulates to develop a general
deformation law between stress and strain rate
– White Equation 2-29a and 2-29b
Assume incompressible flow and constant viscosity


DV

 g    ij
Dt



DV
2

 g  p  m V
Dt
•
2
TENSOR COMMENT
 T11 T12 T13 


Tij   T21 T22 T23 
T T

 31 32 T33 
• Tensors are often displayed as a matrix
• The transpose of a tensor is obtained by
interchanging the two indicies
– Transpose of Tij is Tji
 T11 T21 T31 


T ji   T12 T22 T32 
T T

 13 23 T33 
Qij  Q ji
Rij   R ji
Tij 
1
Tij  T ji   1 Tij  T ji 
2
2

 1
 1 2 3 

 42
4
0
5


 2 1 3  2
23



 2
24
2
0
1 5
2
3 2  
  1
2  
5 1   4  2

2   2
  23
3  
  2
24
2
0
1 5
2
3 2

2 
5 1 
2 

3 

• Tensor Qij is symmetric if Qij = Qji
• Tensor is antisymmetric if it is equal to
the negative of its transpose, Rij = -Rji
• Any arbitrary tensor Tij may be
decomposed into sum of a symmetric
tensor and antisymmetric tensor
 1 0.5 
 1 2 3   1 3 2.5   0

 
 

0
2 
 4 0 5   3 0 3    1
 2 1 3   2.5 3 3    0.5  2 0 

 
 

3
EXAMPLES OF TENSOR PROPERTIES
  xx

 ij    yx

 zx
 xy  xz 

 yy  yz 
 zy  zz 
I1   xx   yy   zz
I 2   xx yy   yy  zz   zz xx      
2
xy
 xx  xy  xz
I 3   yx  yy  yz
 zx  zy  zz
0 
  11 0


0

0


22
0
0  33 

2
yz
2
zx
•
Although component magnitudes vary with
change of axes x, y, and z, the stress and
strain-rate tensor follow the transformation
laws of symmetric tensors
•
3 invariants are particularly useful
•
I3 is the determinant
•
•
Another property of symmetric tensors is that
there exists one and only one set of axes for
which the off-diagonal terms (the shear-strain
rates in this example) vanish.
These are called the principal axes
•
Invariants for principal axes
I1   11   22   33
I 2   11 22   22 33   33 11
I 3   11 22 33
4
COMMENT ON NOTATION
• Recall that in White’s nomenclature:
– x1, y1, and z1 are principal axes
– x, y, and z are arbitrary axes
 xx   11l12   22m12   33n12
 xx   11l12   22m12   33n12
 xy   11l1l2   22m1m2   33n1n2
 xy   11l1l2   22m1m2   33n1n2
• With respect to principal axes
• x-axis has directional cosines:
l1, m1, and n1
• y-axis has directional cosines:
l2, m2, and n2
• z-axis has directional cosines:
l3, m3, and n3
• Using tensor transformation from
principal to arbitrary axes we arrived at
general expressions for diagonal and
off-diagonal terms for shear stress and
strain in arbitrary orientation
5
COMMENTS FROM SECTION 2-4.2
•
Simplest assumption for variation between viscous stress and strain rate is a linear law
– Satisfied for all gases and most common liquids
Stokes’ 3 postulates
1. Fluid is continuous, and its stress tensor ij is at most a linear function of strain rates ij
2. Fluid is isotropic
– Properties are independent of directions (no preferred direction)
– Deformation law is independent of coordinate system choice
– Also implies that principal stress axes be identical with principal strain-rate axes
3. When strain rates are zero (for example if fluid is at rest, V=0), deformation law must
reduce to hydrostatic pressure condition, ij = -pdij
•
Begin derivation of deformation law with element aligned with principal axes
– White notation for principal axes: x1, y1, z1
– Axes where shear stresses and shear strain rates are zero
6
FORMULATING THE DEFORMATION LAW
 11   p  C1 11  C2 22  C3 33
C 2  C3

If V  0,  11   p
 11   p  K 11  C2  11   22   33 

 11   p  K 11  C2  V
K  C1  C2
• Using the principal axes the
deformation law could involve
3 linear coefficients
• Isotropic condition requires that
22 = 33 (cross-flow terms) be
equal
• -p is added to satisfy hydrostatic
condition
• Re-write with gradient of
velocity
• Try to write 22 and 33 terms
7
FORMULATING THE DEFORMATION LAW

• Examples of general deformation law
  p  K xx  C2  V
 xx
 xy  K xy
•
Comparing with shear flow between
parallel plates
•

 ui u j 
  d ij   V
 ij   pd ij  m 



x

x
j
i


Often called the ‘second coefficient of
viscosity’ or coefficient of bulk
viscosity or Lamé’s constant (linear
elasticity)
– Only associated with volume
expansion through divergence of
velocity field


DV

 g    ij
Dt
Now substitute into Newton’s 2nd Law
Note that shear stresses are expressed
as velocity derivatives as desired
K  2m
C2  
•
•
8
THE NAVIER-STOKES EQUATIONS
9
N/S EQUATION FOR INCOMPRESSIBLE, CONSTANT m FLOW
 
ma  F

 
DV

 g  f surface
Dt
•
Start with Newton’s 2nd Law for a fixed mass
•
•
•
•
Divide by volume
Introduce acceleration in Eulerian terms
Ignore external forces
Only body force considered is gravity
•
Express all surface forces that can act on an element
– 3 on each surface (1 normal, 2 perpendicular)
– Results in a tensor with 9 components
– Due to moment equilibrium (no angular rotation of
element) 6 components are independent)
•
Employ a Stokes’ postulates to develop a general
deformation law between stress and strain rate
– White Equation 2-29a and 2-29b
Assume incompressible flow and constant viscosity


DV

 g    ij
Dt



DV
2

 g  p  m V
Dt
•
10