2 1 1 2 2 1 r u r r u r r D \u03a9 r r D \u03a9 r r D r Fluid Mechanics Spring 2019

2 1 1 2 2 1 r u r r u r r d ω r r d ω r r d r fluid

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2 1 1 2 2 1 r u r r u r r D Ω r r D Ω r r D r ------------ Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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Physical meaning of the deformation tensor - shear (2) Then 2 sin cos 1 cos 1 d d d ds d ds d 1 2 r D r = + θ θ θ θ 1 2 2 2 1 1 ds ds dt dt ds dt ds / d ds = r e / e r = ds d e e Let with instantaneously 1 1 , 1 2 2 2 2 1 90 , o θ = cos θ = 0, and sin θ = 1. Let with instantaneously, Then 21 1 2 2 2 D e e dt d = = D θ or dt d D D θ 2 1 21 12 = = (represents a pure shear motion) Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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Example 1 Given velocity field : kx u = 0 v = 0 = w We have : 11 1 , 2 u u u D k x x x = + = = 0 31 23 12 33 22 = = = = = D D D D D A pure straining (if k > 0) motion in the x-direction: Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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Example 2 Given velocity field : , u ky = 0, v = 0. w = We have : 12 21 1 1 1 , 2 2 2 d v u D D k dt x y θ = = − = + = 0 31 23 33 22 11 = = = = = D D D D D (a pure shearing motion in the xy -plane, with k the shear rate) Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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Rigid or deformable motion If D = 0, the motion is called rigid, i.e., the distance between any two fluid particles remains unchanged during the fluid motion. (Example : the steady state fluid motion within a cylindrical tank, which is rotating about its principal axis at a constant rate.) If D 0, we have stretching and shearing motion. However, it is l ibl t fi d th di l di ti ( t i t) always possible to find three perpendicular directions (at a point) along which there is straining and such that the angles between them are instantaneously rigid the axes along these directions are them are instantaneously rigid, the axes along these directions are called the principal axes . We have 1 2 3 ', ', ') (e e e ' ' ' ' ' ' 3 3 3 2 2 2 1 1 1 e e e e e e D λ λ λ + + = Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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To find the principal values and directions Refer to the eigenvalue problem for a square matrix e e D λ = 0 = e I) (D λ or 0 13 12 11 λ λ D D D D D D or 33 23 13 23 22 12 = λ D D D eigenvalues : λ 1 2 3 ( , , ) λ λ λ Principal strain rate eigenvectors (e-functions) : e , 1 2 3 ( ', ', ') e e e Principal axes Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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Vorticity tensor and vorticity vector (1) Vorticity tensor : Ω i j u u 1 Vorticity tensor : = j i ij x x 2 1 u u Vorticity vector : ( ) , = ∇ × ω u i 2 j k ijk j k e x x ω = Relationship : ( ) ( ) ( ) i 1 1 1 2 2 2 j k ijk j k ijk jk u u x x e e ω = = ∇ × + ∇ × = ∇ × = Ω u i u i u i i ω ilm jk ijk ilm e e e = Ω Fluid Mechanics (Spring 2019) – Chapter 2 - U. Lei ( 李雨 )
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Vorticity tensor and vorticity vector (2) , ilm ijk lj mk lk mj e e δ δ δ δ = With We have i 2 ( ) lj mk lk mj jk ilm l l m m m l e δ δ δ δ ω = Ω = Ω = Ω − Ω or 1 1 0 = Ω Ω Ω
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