Derivation of the Navier-Stokes (N-S) Equations
in Cylindrical Coordinates
1
General Equation
Consider a Newtonian, incompressible viscous fluid. The Cauchy stress T in this case is
given by
T = −pI + 2µD , D =
1
L + LT , L = ∂v/∂x .
2
(1)
where p is the pressure, µ is the fluid viscosity coefficient, v is the velocity of a material
point in the present configuration which is located by the position vector x at time t
relative to a fixed origin, L is the velocity gradient, and the total deformation rate D is
the symmetric part of L.
The incompressibility constraint yields
div(v) = vk,k = 0 .
(2)
Furthermore, substituting (1) into the balance of linear momentum equation
ρ v̇ = ρ b + div(T ) ,
(3)
with {ρ, b} being the fluid density and body force, respectively, it follows that
ρ v̇ = ρ b + div(−pI + 2µD) .
1
(4)
Expanding the divergence term on the right-hand-side of (4) gives
div(−pI + 2µD) = (−pI + 2µD),i ·ei = −p,i ei + 2µ (D,i ·ei )
= −∇p + 2µ (D,i ·ei ) .
(5)
However,
h
i
2µD,i ·ei = µ L + LT ,i ·ei = µ ∂v/∂x + (∂v/∂x)T ,i ·ei = µ (vm,n + vn,m ) ,n em
h
i
2
= µ (vm,nn em + vn,mn em ) = µ ∇2 v + (v
n,n ),m em = µ ∇ v ,
(6)
where use has been made of (2). Consequently, the N-S equation takes the form
ρ v̇ = −∇p + ρ b + µ ∇2 v .
2
(7)
N-S Equations in Cylindrical Coordinates
The basis vectors {er (θ), eθ (θ), ez } associated with a cylindrical coordinate system are
given by
er (θ) = cos(θ) e1 + sin(θ) e2 , eθ (θ) = − sin(θ) e1 + cos(θ) e2 , ez = e3 .
(8)
Also, the position vector of a point in space can be expressed as
x = r er (θ) + z ez ,
(9)
where {r, z} are the radial and axial distances of the point from the origin, respectively.
Therefore, the convariant gi and contravariant g i vectors simplify to
g1 =
∂x
∂x
∂x
= er (θ) , g2 =
= r eθ (θ) , g3 =
= ez ;
∂r
∂θ
∂z
1
g 1 = er (θ) , g 2 = eθ (θ) , g 3 = ez .
r
(10)
(11)
Next, the gradient of the pressure, ∇p, can expressed as
∇p = p,i g i =
∂p
1 ∂p
∂p
er +
eθ +
ez .
∂r
r ∂θ
∂z
2
(12)
Also, the gradient of the velocity field, ∇v, with v = vr er + vθ eθ + vz ez , becomes
1 ∂v
∂v
∂v
⊗ er +
⊗ eθ +
⊗ ez
∂r
r ∂θ !
∂z
∂vr
1 ∂vr
∂vr
=
er ⊗ er +
− vθ er ⊗ eθ +
er ⊗ ez
∂r
r ∂θ
∂z
!
1 ∂vθ
∂vθ
∂vθ
eθ ⊗ er +
+ vr eθ ⊗ eθ +
eθ ⊗ ez
+
∂r
r ∂θ
∂z
∂vz
1 ∂vz
∂vz
+
ez ⊗ er +
ez ⊗ eθ +
ez ⊗ ez .
∂r
r ∂θ
∂z
∇v = v,i ⊗g i =
(13)
Using this result, the incompressibility constraint (2), vk,k = ∇v · I = 0, yields
1 ∂
1 ∂vθ ∂vz
(rvr ) +
+
=0 .
r ∂r
r ∂θ
∂z
(14)
Now, the Laplacian of the velocity field, ∇2 v, takes the form
∂(∇v)
1 ∂(∇v)
∂(∇v)
· er +
· eθ +
· ez
∂r
r ∂θ
∂z
"
!
#
1 ∂
∂vr
vr
1 ∂ 2 vr
2 ∂vθ ∂ 2 vr
=
r
− 2+ 2
−
+
er
r ∂r
∂r
r
r ∂θ2
r2 ∂θ
∂z 2
"
!
#
1 ∂
∂vθ
vθ
1 ∂ 2 vθ
2 ∂vr ∂ 2 vθ
+
r
− 2+ 2
+ 2
+
eθ
r ∂r
∂r
r
r ∂θ2
r ∂θ
∂z 2
"
!
#
1 ∂
∂vz
1 ∂ 2 vz ∂ 2 vz
+
+
r
+ 2
ez .
r ∂r
∂r
r ∂θ2
∂z 2
∇2 v = ∇ · ∇v = (∇v) ,i ·g i =
(15)
Moreover, the material time derivative of the velocity field v(x, t) is given by
∂v
+ ∇v · v
∂t
∂vr
∂vr vθ
=
+ vr
+
∂t
∂r
r
∂vθ
∂vθ vθ
+ vr
+
+
∂t
∂r
r
∂vz
∂vz vθ
+
+ vr
+
∂t
∂r
r
v̇ =
!
∂vr vθ2
∂vr
−
+ vz
er
∂θ
r
∂z
!
∂vθ vr vθ
∂vθ
+
+ vz
eθ
∂θ
r
∂z
!
∂vz
∂vz
+ vz
ez .
∂θ
∂z
(16)
Substituting {(12), (15), (16)} into (7), it follows that
!
∂vr
∂vr vθ ∂vr vθ2
∂vr
ρ
+ vr
+
−
+ vz
∂t
∂r
r ∂θ
r
∂z
"
!
#
∂p
1 ∂
∂vr
vr
1 ∂ 2 vr
2 ∂vθ ∂ 2 vr
r
− 2+ 2
− 2
+
,
= − + ρ br + µ
∂r
r ∂r
∂r
r
r ∂θ2
r ∂θ
∂z 2
3
(17)
!
∂vθ vθ ∂vθ vr vθ
∂vθ
∂vθ
+ vr
+
+
+ vz
ρ
∂t
∂r
r ∂θ
r
∂z
"
!
#
2
1 ∂
∂vθ
vθ
1 ∂p
1 ∂ vθ
2 ∂vr ∂ 2 vθ
+ ρ bθ + µ
r
− 2+ 2
+
=−
+ 2
,
r ∂θ
r ∂r
∂r
r
r ∂θ2
r ∂θ
∂z 2
∂vz
ρ
+ vr
∂t
"
∂p
1
= − + ρ bz + µ
∂z
r
(18)
!
∂vz vθ ∂vz
∂vz
+
+ vz
∂r
r ∂θ
∂z
!
#
2
∂
∂vz
1 ∂ vz ∂ 2 vz
r
+ 2
+
.
∂r
∂r
r ∂θ2
∂z 2
4
(19)