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
=
+ Body forces:
Gravity force
Electromagnetic forse
Centrifugal force
Coriolis force
(
2)
Surface forces:
Pressure forces
Viscous forces
We cosider the x-component of (Eq 2).
Since
=
and = ( , , ) we have
∙
We denote the stress tensor
=
,
=
,
+ ,
( pressure forces+ viscous forces)
(
3)
the viscous stress tensor
=
where
and strain ( deformation) rate tensor
⎡
∂u
⎢
∂x
⎢
⎢ 1 ⎛ ∂v ∂u ⎞
+ ⎟
= ⎢ ⎜⎜
2 ⎝ ∂x ∂y ⎟⎠
⎢
⎢ 1 ⎛ ∂w + ∂u ⎞
⎢ 2 ⎜⎝ ∂x ∂z ⎟⎠
⎣
=
Let
=( ,
,
) , = ( ,
perpendicular to the coordinate axes.
,
1 ⎛ ∂u ∂w ⎞⎤
⎜ +
⎟⎥
2 ⎝ ∂z ∂x ⎠⎥
1 ⎛ ∂v ∂w ⎞ ⎥
⎟
⎜ +
2 ⎜⎝ ∂z ∂y ⎟⎠ ⎥
⎥
∂w
⎥
⎥
∂z
⎦
1 ⎛ ∂u ∂v ⎞
⎜ + ⎟
2 ⎜⎝ ∂y ∂x ⎟⎠
∂v
∂y
1 ⎛ ∂w ∂v ⎞
+ ⎟
⎜
2 ⎜⎝ ∂y ∂z ⎟⎠
),
=(
,
,
) be stress vectors on the planes
y
z
x
yz- plane
Then the stress vector
xz-plane
xy-plane
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=
−(
−
3=
−(
−
5 = −(
−
)
2
)
2
)
2
=(
+
(
+
4=
6=
(
)
)
2
+
2
)
Thus
=
,
1
+
2
+
3
+
4
+
5
+
6
=(
+
+
)
(
3)
If we assume that the only body force is the gravity force, we have
=
,
Now from (
∙g =
∙
∙g
3)
∙
=
,
+ ,
we have ∙
We divide by
∙
=
∙
∙ g + (
+
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
+
eq z
z component:
∙(
+
+
+
) = g +
+
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
+
eq C
z component:
∙(
+
+
+
) = g −
+
+
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
+
+
),
= (
+
)
(*)
,
= (
+
)
(**)
= (
+
) ,
= (
We substitute this values
+
)
=2
+
(***)
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
+
= 0 and hence from (*), (**) and (***)
Remark: For an incompressible flow we have
=2
where
r
is the strain rate tensor for the velocity field V = (u, v, w) in Cartesian coordinates:
⎡
∂u
⎢
∂x
⎢
⎢ 1 ⎛ ∂v ∂u ⎞
τ ij = 2 με ij = 2 μ ⎢ ⎜⎜ + ⎟⎟
2 ∂x ∂y ⎠
⎢ ⎝
⎢ 1 ⎛ ∂w + ∂u ⎞
⎢ 2 ⎜⎝ ∂x ∂z ⎟⎠
⎣
1 ⎛ ∂u ∂v ⎞
⎜ + ⎟
2 ⎜⎝ ∂y ∂x ⎟⎠
∂v
∂y
1 ⎛ ∂w ∂v ⎞
+ ⎟
⎜
2 ⎜⎝ ∂y ∂z ⎟⎠
1 ⎛ ∂u ∂w ⎞⎤
⎜ +
⎟⎥
2 ⎝ ∂z ∂x ⎠⎥
1 ⎛ ∂v ∂w ⎞ ⎥
⎟
⎜ +
2 ⎜⎝ ∂z ∂y ⎟⎠ ⎥
⎥
∂w
⎥
⎥
∂z
⎦
⎡
⎛ ∂u ∂v ⎞
∂u
⎛ ∂u ∂w ⎞⎤
μ ⎜⎜ + ⎟⎟ μ ⎜ +
⎟⎥
⎢ 2μ
∂x
⎝ ∂z ∂x ⎠⎥
⎝ ∂y ∂x ⎠
⎢
⎛ ∂v ∂w ⎞ ⎥
∂v
⎢ ⎛ ∂v ∂u ⎞
μ ⎜⎜ +
= ⎢ μ ⎜⎜ + ⎟⎟
2μ
⎟ .
∂x ∂y ⎠
∂y
∂z ∂y ⎟⎠ ⎥
⎝
⎝
⎢
⎥
∂w ⎥
⎢ μ ⎛ ∂w + ∂u ⎞ μ ⎛⎜ ∂w + ∂v ⎞⎟
2μ
⎜ ∂y ∂z ⎟
⎢ ⎜⎝ ∂x ∂z ⎟⎠
∂z ⎥⎦
⎠
⎝
⎣
In the case when we consider an incompressible , isothermal Newtonian flow (density ρ =const,
r
viscosity μ =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
+u
+v
+ w ⎟⎟ = −
+ ρg x + μ ( 2 + 2 + 2 )
∂x
∂y
∂z ⎠
∂x
∂x
∂y
∂z
⎝ ∂t
ρ ⎜⎜
y- component:
⎛ ∂v
∂v
∂v
∂v ⎞
∂P
∂ 2v ∂ 2v ∂ 2v
+ ρg y + μ ( 2 + 2 + 2 )
ρ ⎜⎜ + u + v + w ⎟⎟ = −
∂x
∂y
∂z ⎠
∂y
∂x
∂y
∂z
⎝ ∂t
z component:
⎛ ∂w
∂w
∂w
∂w ⎞
∂P
∂2w ∂2w ∂2w
+u
+v
+ w ⎟⎟ = −
+ ρg z + μ ( 2 + 2 + 2 )
∂x
∂y
∂z ⎠
∂z
∂x
∂y
∂z
⎝ ∂t
ρ ⎜⎜
[ The vector form for these equations:
r
r
r
DV
ρ
= −∇ P + ρ g + μ ∇ 2 V ]
Dt