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 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 ( pressure forces+ viscous forces)

= , the viscous stress tensor = and strain ( ) 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 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 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