05%20Newtonian%20Fluids%20EGM%206812%20F11

05%20Newtonian%20Fluids%20EGM%206812%20F11 - EGM 6812, F11,...

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EGM 6812, F11, University of Florida, M. Sheplak 1/15 Section 5, Newtonian Fluids 6 Newtonian Fluids and Navier-Stokes Equations 6.1 Newton’s Viscosity Law Derivation Need a relationship between ij T and l k u x (or i u , lk , etc. )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 Recall definition: isotropic => transforms onto itself or put another way, the components are always the same regardless of rotation, e.g. , ij ijk  ij A ”isotropic” t ij p  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 Thus, the deviatoric stress tensor is defined as follows to ensure a linear relationship l ij ijkl k u dB x  Most general form of a linear relationship l ijkl k u Bf 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
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EGM 6812, F11, University of Florida, M. Sheplak 2/15 Section 5, Newtonian Fluids 1. linear relationship between deviatoric stress tensor and velocity gradient (Newton) 2. ij 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 kl ij ijkl kl ijkl symmetric antisymmetric d d B B dt  Recall assumptions #2 and #3, so ijkl B must be symmetric ij ijkl kl dB 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”, ' '' ijkl ik jl il jk ij kl B      , where ', '',  and are thermodynamic variables. Apply symmetry (assumptions #2 and #3) ij d : i and j can switch so that ik jl  can be factored from the first two terms of ijkl B . Set ' ''  .     ' '' 2 ij ik jl il jk ij kl kl ik jl ij kl kl d   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      Now the deviatoric stress is 2 k ij ij ij k u d x   Now the stress tensor can be written such that 2 ij ij k ij ij ij ij k A d u Tp x   where is the first coefficient of dynamic viscosity is the second coefficient of dynamic viscosity 2 Ns m    .
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EGM 6812, F11, University of Florida, M. Sheplak 3/15 Section 5, Newtonian Fluids 6.1.2 Stokes Hypothesis Recall, Stokes Hypothesis: tm pp t p :
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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.

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05%20Newtonian%20Fluids%20EGM%206812%20F11 - EGM 6812, F11,...

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