Derivation of the NAVIER-STOKES EQUATION (For Compressible

DERIVATION OF THE NAVIER STOKES EQUATION  DV      P   g    2 V Dt

The motion of a non turbulent Newtonian fluid ( i. e. liquid or gas) is governed by the Navier-Stokes equation (N-S eq).

When we derive N-S eq. we usually begin with the Cauchy momentum equation.

1. CAUCHY’S MOMENTUM EQUATION

The Cauchy momentum equation is a vector partial differential equation  DV     σ  f Dt where  is the density of the fluid,   V  V (t, x, y, z)  u(t, x, y, z),v(t, x, y, z),w(t, x, y, z) is the velocity vector field.

σ is the Cauchy stress tensor  and f denotes the sum of body forces per unit volume.

METOD 1.

We consider a differential fluid element as a material element and apply Newton’s second law

∙

Since ( denotes acceleration and ,, velocity of the element) we have

∙ 1 The total force we can express 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

Now we cosider the x-component of (Eq 2).

Since and ,, we have

∙ 3 , ,

Stress tensor. We denote the stress tensor σ ( pressure forces+ viscous forces)

σ ,

Let ,, , ,,, ,, be stress vectors on the planes perpendicular to the coordinate axes. (Note that we can write stress vectors as column vectors)

x z

yz‐ plane xz‐plane xy‐plane

Then the stress vector at any point associated with a plane with a unit normal vector

,,,

according to Cauchy, can be expressed as

 ,, = nσ ,

The matrix σ = is a symmetric matrix ,

i.e. , .

Remark 1. If we denote all vectors as column vectors then  Remark 2. For the stress vector and related" stress force F ” on the small (imaginary) surface S , passing through an internal material poiint P (x,y,z),

we have the following relation    F  T  area(S)  (nσ)area(S) . [R1]

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In order´to derive Cauchy’s equation we use a box-shaped small (infinitesimal) control volume with dimensions dx, dy and dz . WWe consider the x-component off the resultant surface force ∑ , .

dx Let P(x,y,z) be the center of the box, and W(x  , y, z) be the point on the ”west side” of 2 the box.  To find the force F(W ) we use the relation [R1]     F(W )  T (W )  area(SW )  (n(W )σ(W ))area(SW )  i σ(W )dydz

1,0,0 dydz

,,dydz  If F1, x denote the x coordinate of F(W ) then we have

F1, x

Using Taylor’s formula, f (x, y, z) f (x, y, z) f (x, y, z) f (x  x, y  y, z  z)  f (x, y, z)  x  y  z x y z applied to the function at the center P(x,y,z) of the box, with

dx x   , y  0, z  0 2 we get

F1, x 2

or simply F1, x .

Similarly we calculate F2, x, …, F6, x and get

F1, x F2, x F3, x F4, x 2 2

F5, x F6, x 2 2 Thus

F1 F2 F3 F4 F5 F6 ,

If we assume that the only body force is the gravity force, we have

, ∙g ∙ ∙ g

Now from

∙ 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, called Cauchy momentum equations can be written as one vector equation:  DV     σ  g Dt

METOD 2. (Derivation using volume and surface integrals)

We apply Newton’s second law on a control volume K . The time rate of change of the momentum of K is equal the sum of external forces:

( The rate of change of the momentum of K ) = (*)

6 i) To find the rate of change of the momentum, we make a partition of K by dividing K in small volumes K k . Thus K   Kk and the mass( Kk ) = mk  k xk yk zk . k

The rate of change of the momentum for K k is approximately mk ak and consequently the rate of change of the momentum for the whole K is the limit of the Riemann sums, that is the following triple integral:   DV m a   a x y z    a dxdydz   dxdydz lim k k lim k k k k k   k k K K Dt ii) For the resulting surface force we have:

Tk Fk    Fsurface   Fk  Tk area(Sk ) K k k Sk

   F  T area(S )  TdS  nσdS  (modified Gauss div.th.) surface lim k k   k K K

 σ dxdydz . K

Remark. We applied Gauss divergence theorem on every coordinate of         nσ  (n  k1,n  k2 ,n  k3 ) , where k1,2,3 are columns of σ . Thus

             nσdS  (n  k ,n  k ,n  k )dS  ( n  k dS, n  k dS, n  k dS)    1 2 3  1  2  3 K K K K K     ( k dxdydz, k dxdydz, k dxdydz )  1  2  3 K K K

    (k ,k ,k )dxdydz ,  1 2 3 K  σ dxdydz K    where σ = (k1 ,k2 ,k3 ) =          ( xx  yx  zx , xy  yy  zy , xz  yz  zz ) . x y z x y z x y z

iii) For the resulting body force we have:   F   g dxdydz body  K iv) Now we account all external forces and substitute in the relation:

(The rate of change of the momentum of K ) = .

We have  DV   dxdydz =  g dxdydz + σ dxdydz K Dt K K or  DV   (  g  σ)dxdydz  0 (**) K Dt  DV   Since (**) is valid for an arbitrary control volume K we have   g  σ  0 Dt or  DV    g  σ ( Cauchy’s equation, vector form). Dt

2. NAVIER STOKES EQUATIONS

The stress tensor σ is usualy expressed as the sum of pressure tensor p and viscous tensor τ :

0 0 σ  p  τ where σ , 00 , 00

the viscous stress tensor τ .

Let ε denotes the strain ( deformation) rate tensor,

 u 1  u v  1  u w         x 2 y x 2  z x       1  v u  v 1  v w  ε        2  x y  y 2  z y       1  w u  1  w v  w        2 x z 2  y z  z        ------

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 (by 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 , thus div( V )  0 , and hence from (*), (**) and (***) 2 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    2  2       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   2        x  y x   z x    v u  v  v w      2    .   x y  y  z y         w u   w v  w        2    x z   y z  z   Thus, for an incompressible ( div( V )  0) , isothermal Newtonian flow (density  =constant,  viscosity  =constant), with a velocity field V  (u(x,y,z), v(x,y,z), w (x,y,z)) , we can simplify  u 2u 2u the Navier‐Stokes equations. Using that  is a constant we write 2 ( )     x x x2 x2 and rearrange x component:  u u u u  P 2u 2u 2u  u v w   u  v  w     g  (   )   (   ) .   x 2 2 2  t x y z  x x y z x x y z  u v w Again since div( V )     0 we have x y z x component:  u u u u  P  2u  2u  2u   u  v  w     g  (   ) .   x 2 2 2  t x y z  x x y z In the same way we get y- component:  v v v v  P  2v  2v  2v   u  v  w     g  (   )   y 2 2 2  t x y z  y x y z and z-component:  w w w w  P  2 w  2 w  2 w   u  v  w     g  (   )   z 2 2 2  t x y z  z x y z We can express these N-S equations as one vector-equation  DV      P   g    2 V Dt