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06 Newtonian Fluids - 6 Newtonian Fluids and Navier-Stokes...

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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)
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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 = + 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 4 42 4 43 where μ is the first coefficient of dynamic viscosity λ is the second coefficient of dynamic viscosity 2 N s m . 6.1.2 Stokes Hypothesis Recall, Stokes Hypothesis: t m p p =
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t p : thermodynamic pressure; function of the thermodynamic state m p : mechanical pressure, due to translational motion only 1 3 m ii p T = - Recall that t m p p = when normal viscous stresses can be neglected ( 29 ii τ .
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