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MAE 5130: VISCOUS FLOWS

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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

- 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

- Tensors are often displayed as a matrix
- The transpose of a tensor is obtained by interchanging the two indicies
- Transpose of Tij is Tji

- 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

- 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

- Recall that in White’s nomenclature:
- x1, y1, and z1 are principal axes
- x, y, and z are arbitrary axes

- 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

- 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

- Satisfied for all gases and most common liquids
- Fluid is continuous, and its stress tensor tij is at most a linear function of strain rates eij
- 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

- When strain rates are zero (for example if fluid is at rest, V=0), deformation law must reduce to hydrostatic pressure condition, tij = -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

- Using the principal axes the deformation law could involve 3 linear coefficients
- Isotropic condition requires that e22 = e33 (cross-flow terms) be equal
- -p is added to satisfy hydrostatic condition
- Re-write with gradient of velocity
- Try to write t22 and t33 terms

- Examples of general deformation law
- Comparing with shear flow between parallel plates
- 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

- Now substitute into Newton’s 2nd Law
- Note that shear stresses are expressed as velocity derivatives as desired

- 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