06 Newtonian Fluids

06 Newtonian Fluids - 6 Newtonian Fluids and Navier-Stokes Equations 6.1 Newton’s Viscosity Law Derivation Need a relationship between ij T and

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6 Newtonian Fluids and Navier-Stokes Equations 6.1 Newton’s Viscosity Law Derivation Need a relationship between ij T and ij ε to evaluate the governing equations 6.1.1 Development of ij T For a Newtonian fluid we can decompose ij T into { { fluid @ fluid in rest motion ij ij ij T A d = + where ij d deviatoric stress tensor The deviatoric stress tensor is non-isotropic and is only concerned with the stresses due to fluid motion Definition: isotropic => transforms onto itself or put another way, the components are always the same regardless of rotation, e.g. , ij ijk δ ε ij A = ”isotropic” t ij p δ = - 6.1.1.1 Newtonian Fluid Assume a linear relationship between the surface stress tensor, ij T , and the velocity gradient, { l k V du dx ur Thus, the deviatoric stress tensor is defined as follows to ensure a linear relationship l ij ijkl k u d B x ∴ = Most general form of a linear relationship l ijkl k u B f x Note: ijkl B is a rank-4 tensor with 81 components, obviously we need to simplify! Assumptions for the constitutive relationship between the surface stress tensor, ij T , and the velocity gradient 1. linear relationship between deviatoric stress tensor and velocity gradient (Newton) 2. ij ε is symmetric by definition 3. ij T is symmetric via angular momentum equation 4. fluid is isotropic (no preferred direction) We can decompose deviatoric stress tensor { { i j ij ijkl kl ijkl symmetric antisymmetric d d B B dt ε Ω = + Recall assumptions #2 and #3, so ijkl B must be symmetric ij ijkl kl d B ε = Since the fluid is isotropic (without preferred direction (assump. #4)), ijkl B is also isotropic, so it can be represented as a “general rank-4, isotropic tensor” (see web handout), ' '' ijkl ik jl il jk ij kl B μ δ δ μ δ δ λδ δ = + + , where ', '', μ μ and λ are thermodynamic variables. Apply symmetry (assumptions #2 and #3) • ij d : i and j can switch so that ik jl δ δ can be factored from the first two terms of ijkl B . Set ' '' μ μ μ = = . ( 29 ( 29 ' '' 2 ij ik jl il jk ij kl kl ik jl ij kl kl d μ δ δ μ δ δ λδ δ ε μδ δ λδ δ ε = + + = + For the first term to be nonzero, , i k j l = = , and for the second term to be nonzero, , i j k l = = . By applying the definition of the strain rate tensor and , i j k l = = to the second term such that 1 or 2 k l k l kl kl kl l k k l u u u u x x x x δ ε δ = + = Now the deviatoric stress is 2 k ij ij ij k u d x με λδ = + Now the stress tensor can be written such that { 2 ij ij k ij ij ij ij k A d u T p x δ με λδ = - + + 1 442 4 43 where • μ is the first coefficient of dynamic viscosity • λ is the second coefficient of dynamic viscosity 2 N s m ....
View Full Document

This note was uploaded on 01/17/2012 for the course EGM 6812 taught by Professor Renweimei during the Fall '09 term at University of Florida.

Page1 / 15

06 Newtonian Fluids - 6 Newtonian Fluids and Navier-Stokes Equations 6.1 Newton’s Viscosity Law Derivation Need a relationship between ij T and

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online