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Derivation of the Navier Stokes Equation DERIVATION OF THE NAVIER STOKES EQUATION 1. CAUCHY’S EQUATION First we derive Cauchy’s equation using Newton’s second law. We take a differential fluid element. We consider the element as a material element ( instead of a control volume) and apply Newton’s second law ∙= () or since () = ∙ = ( 1) We express the total force as the sum of body forces and surface forces ∑ = ∑ + ∑ . Thus ( 1) can be written as ∙ = + ( 2) Body forces: Surface forces: Gravity force Pressure forces Electromagnetic forse Viscous forces Centrifugal force Coriolis force We cosider the x-component of (Eq 2). Since = and =(,,) we have ∙ = + ( 3) , , We denote the stress tensor ( pressure forces+ viscous forces) = , the viscous stress tensor = and strain ( deformation) rate tensor where ∂ ∂ ∂ ∂ ∂ ⎡ u 1 ⎛ u + v ⎞ 1 ⎛ u + w ⎞⎤ ⎢ ⎜ ⎟ ⎜ ⎟⎥ ∂x 2 ⎝ ∂y ∂x ⎠ 2 ⎝ ∂z ∂x ⎠ ⎢ ⎥ ⎢ 1 ⎛ ∂v ∂u ⎞ ∂v 1 ⎛ ∂v ∂w ⎞⎥ = = ⎜ + ⎟ ⎜ + ⎟ ⎢ 2 ⎜ ∂x ∂y ⎟ ∂y 2 ⎜ ∂z ∂y ⎟⎥ ⎢ ⎝ ⎠ ⎝ ⎠⎥ 1 ⎛ ∂w ∂u ⎞ 1 ⎛ ∂w ∂v ⎞ ∂w ⎢ + ⎜ + ⎟ ⎥ ⎢ ⎜ ∂ ∂ ⎟ ⎜ ∂ ∂ ⎟ ∂ ⎥ ⎣2 ⎝ x z ⎠ 2 ⎝ y z ⎠ z ⎦ Let =(,,) , =(,,), =(,,) be stress vectors on the planes perpendicular to the coordinate axes. y x z yz- plane xz-plane xy-plane Then the stress vector at any point associated with a plane of unit normal vector =(,,) can be expressed as = + + =(,,) . We consider the x-component of the net surface force ∑ , using the figure below. Using Taylor’s formula we get 1=−( − ) 2 =( + ) 3=−( − ) 4=( + ) 2 2 5=−( − ) 6=( + ) 2 2 Thus =1 +2 +3 +4 +5 +6 =( + + ) , If we assume that the only body force is the gravity force, we have , =∙g =∙∙g Now from ( 3) ∙ = + ( 3) , , we have ∙ ∙ = ∙ ∙ g +( + + ) We divide by and get the equation for the x-component: ∙ =g + + + or ∙( + + + )=g + + + eq x In the similar way we derive the following equations for y component: ∙( + + + )=g + + + eq y z component: ∙( + + + )=g + + + eq z Equations eq x,y,z, are called Cauchy’s equations. THE NAVIER STOKES EQUATION When considering ∑ , we can separate x components of pressure forces and viscous forces: =− + , = , = In the similar way we can change y-component and z-component Thus Cauchy’s equations become ∙( + + + )=g − + + + eq A In the similar way we derive the following equations for y component: ∙( + + + )=g − + + + eq B z component: ∙( + + + )=g − + + + eq C According o the NEWTON’S LOW OF VISCOSITY the viscous stress components are related ( throw a linear combination) to the ( first) dynamic viscosity and the second viscosity . =2 + , =( + ) , =( + ) (*) =( + ) , =2 + , =( + ) (**) =( + ) , =( + ) =2 + (***) We substitute this values in to Cauchy’s equations eq A, B, C and get THE NAVIER STOKES EQUATIONS for the compressible flow: x-component: ∙( + + + ) =g − + 2 + + ( + ) + ( + ) y-component: ∙( + + + ) =g − + ( + ) + 2 + + ( + ) z-component: ∙( + + + ) =g − + ( + ) + ( + ) + 2 + Remark: For an incompressible flow we have =0 and hence from (*), (**) and (***) =2 r = where is the strain rate tensor for the velocity field V (u,v, w) in Cartesian coordinates: ⎡ ∂u 1 ⎛ ∂u ∂v ⎞ 1 ⎛ ∂u ∂w ⎞⎤ ⎜ + ⎟ + ⎢ ∂ ⎜ ∂ ∂ ⎟ ⎜ ∂ ∂ ⎟⎥ ⎢ x 2 ⎝ y x ⎠ 2 ⎝ z x ⎠⎥ ⎢ 1 ⎛ ∂v ∂u ⎞ ∂v 1 ⎛ ∂v ∂w ⎞⎥ τ = 2με = 2μ ⎜ + ⎟ ⎜ + ⎟ ij ij ⎢ 2 ⎜ ∂x ∂y ⎟ ∂y 2 ⎜ ∂z ∂y ⎟⎥ ⎢ ⎝ ⎠ ⎝ ⎠⎥ 1 ⎛ ∂w ∂u ⎞ 1 ⎛ ∂w ∂v ⎞ ∂w ⎢ + ⎜ + ⎟ ⎥ ⎢ ⎜ ∂ ∂ ⎟ ⎜ ∂ ∂ ⎟ ∂ ⎥ ⎣2 ⎝ x z ⎠ 2 ⎝ y z ⎠ z ⎦ ⎡ ∂u ⎛ ∂u ∂v ⎞ ⎛ ∂u ∂w ⎞⎤ 2μ μ⎜ + ⎟ μ + ⎢ ∂ ⎜ ∂ ∂ ⎟ ⎜ ∂ ∂ ⎟⎥ ⎢ x ⎝ y x ⎠ ⎝ z x ⎠⎥ ⎢ ⎛ ∂v ∂u ⎞ ∂v ⎛ ∂v ∂w ⎞⎥ = μ⎜ + ⎟ 2μ μ⎜ + ⎟ . ⎢ ⎜ ∂x ∂y ⎟ ∂y ⎜ ∂z ∂y ⎟⎥ ⎢ ⎝ ⎠ ⎝ ⎠⎥ ⎛ ∂w ∂u ⎞ ⎛ ∂w ∂v ⎞ ∂w ⎢μ + μ⎜ + ⎟ 2μ ⎥ ⎢ ⎜ ∂ ∂ ⎟ ⎜ ∂ ∂ ⎟ ∂ ⎥ ⎣ ⎝ x z ⎠ ⎝ y z ⎠ z ⎦ In the case when we consider an incompressible , isothermal Newtonian flow (density ρ =const, μ r = viscosity =const), with a velocity field V (u(x,y,z), v(x,y,z), w (x,y,z)) we can simplify the Navier-Stokes equations to his form: x component: ⎛ ∂u ∂u ∂u ∂u ⎞ ∂P ∂ 2u ∂ 2u ∂ 2u ρ⎜ + u + v + w ⎟ = − + ρg + μ( + + ) ⎜ ⎟ x 2 2 2 ⎝ ∂t ∂x ∂y ∂z ⎠ ∂x ∂x ∂y ∂z y- component: ⎛ ∂v ∂v ∂v ∂v ⎞ ∂P ∂ 2v ∂ 2v ∂ 2v ρ⎜ + u + v + w ⎟ = − + ρg + μ( + + ) ⎜ ⎟ y 2 2 2 ⎝ ∂t ∂x ∂y ∂z ⎠ ∂y ∂x ∂y ∂z z component: ⎛ ∂w ∂w ∂w ∂w ⎞ ∂P ∂ 2 w ∂ 2 w ∂ 2 w ρ⎜ + u + v + w ⎟ = − + ρg + μ( + + ) ⎜ ⎟ z 2 2 2 ⎝ ∂t ∂x ∂y ∂z ⎠ ∂z ∂x ∂y ∂z r DV r [ The vector form for these equations: ρ = −∇ P + ρ gr + μ ∇ 2 V ] Dt.
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