111812009 navier stokes derivation in cylindrical coordinates
TRANSCRIPT
NAVIER-STOKES
Derivation of Navier-
Stokes Equation Using cylindrical co-ordinates (r, Ɵ, z)
Year 2012
PRAXIE
This document provides a step-by-step guide to deriving the NS equation using
cylindrical co-ordinates. The steps have been collected from different documents
available on the web; frankly speaking, this document just assembles them into a
single file. Although the derivation is not the most detailed one can go for, it
certainly helps the user to have a feel of the derivation process and proceed in a
more detailed fashion if one decides to. Thank You :-)
STEP I (Pgs. 3-5)
Using the figures highlighting stresses in r, Ɵ and z directions, we find out the equilibrium equations
in terms of normal and shear stresses.
STEP II (Pgs 6-8)
The material derivative or acceleration terms are derived
in terms of cylindrical coordinates (r, Ɵ, z).
STEP III (Pg. 9)
Substituting the acceleration terms from step II in the
equilibrium equations calculated in step I, we find out the r, Ɵ and z components of the momentum
equation.
STEP IV (Pg. 10)
The normal and shear stresses are shown in cylindrical
coordinates; no derivation has been done over here!
STEP V (Pg. 11)
Substituting the normal and shear stresses from step IV into the momentum equation derived in step III and using the continuity
equation (in cylindrical coordinates) for simplification, we finally get the Navier-Stokes equation in r, Ɵ and z directions
NOTE: The derivation has been divided into five steps, with each
step collected from a different pdf, the necessary pages of which
have been attached herein and the required figures and terms
either clouded out or highlighted in yellow.
Remember, no detailed calculations have been shown in this
document, only the superficial things necessary for one to derive the
NS equation!!! U r highly encouraged to do the detailed derivation
and share it on the web for others to learn and use. Cheers ☺
16 2. Governing Equations
δθ
θθσ
rrσ
rθτ
rθτ
rrrr r
r∂σ
σ + δ∂
rr
θθ
∂ττ + δθ
∂θ
rr r
rθ
θ∂τ
τ + δ∂
θθθθ
∂σσ + δθ
∂θ
zrzr z
z∂τ
τ + δ∂
FIGURE 2.2. Stresses in the r and θ Directions.
of stress, strain, and displacement in cylindrical coordinates. The follow-ing sections provide a succinct review of essential topics needed for theestablishment of the governing elasto-dynamic equations.
2.1 State of Stresses at a Point
A three dimensional state of stress in an infinitesimal cylindrical elementis shown in the following three figures. Figure 2.1 depicts such an elementwith direct stresses, dimensions, and directions of the cylindrical coordi-nate. Figure 2.2 represents the direct and shear stresses in the radial andtransverse directions (r and θ), and the variation of direct and shear stressesin these two directions. Figure 2.3 shows direct and shear stresses associ-ated with the planes perpendicular to the r and z directions, as well astheir variations along these directions.
In the above graphical representations the changes in direct and shearstresses are given by considering the first order infinitesimal term used inTaylor series approximation. The series approximation has been truncatedafter the second term. Further terms within the series representation con-tain terms of an infinitesimal length squared. Assuming that the second
Page 3
2. Governing Equations 17
zδ
rrσ
rθτ
rrrr r
r∂σ
σ + δ∂
zzzz z
z∂σ
σ + δ∂
θθσ
z
Fr
r
rδ
zθτ
zrzr r
r∂τ
τ + δ∂
zz
θθ
∂ττ + δθ
∂θ
δθ
FIGURE 2.3. Stresses in the plane perpendicular to r and z direction.
order terms are very small, they can be neglected. Therefore, the changein stress across the element is considered very small.
2.2 Equilibrium Equations in Terms of Stress
Utilizing Newton’s second law and the graphical representation of the stateof stress, the equilibrium equations for an infinitesimal element in a cylin-drical coordinates will be developed. By examining the state of stress onthe element shown in section 2.1, the following equilibrium equation in ther direction is given.µ
σrr +∂σrr∂r
δr
¶(r + δr) δθδz +
µτrθ +
∂τrθ∂θ
δθ
¶δrδz cos
δθ
2
+
µτrz +
∂τrz∂z
δz
¶µr +
δr
2
¶δrδθ + Frδrδθδz
= σrrrδθδz + δτrθδrδz cosδθ
2+ τrz
µr +
δr
2
¶δrδθ +
+
µσθθ +
∂σθθ∂θ
δθ
¶δrδz sin
δθ
2+ σθθδrδz sin
δθ
2(2.1)
Page 4
18 2. Governing Equations
Canceling appropriate terms from both sides of the equation and aftersimplifying, it yields:
∂σrr∂r
+1
r
∂τrθ∂θ
+∂τrz∂z
+σrr − σθθ
r+ Fr = 0 (2.2)
Similarly, the equilibrium equation for the θ direction yields:
∂τ rθ∂r
+1
r
∂σθθ∂θ
+∂τθz∂z
+2
rτrθ + Fθ = 0 (2.3)
and finally, for the z direction one may write:
∂τrz∂r
+1
r
∂τθz∂θ
+∂σzz∂z
+1
rτ rz + Fz = 0 (2.4)
In the above simplifications, due to very small angle of δθ, the followingapproximations were used:
cosδθ
2≈ 1 sin
δθ
2≈ δθ
2(2.5)
In addition to the stresses, body forces acting throughout the element havebeen considered for each direction. These are denoted by Fr, Fθ, and Fzwhich are introduced as forces in the r, θ, and z direction per unit ofvolume. Due to the cancellation of the moments about each of the threeperpendicular axes, the relations among the six shear stress componentsare presented by the following three equations:
τrθ = τθr τθz = τzθ τzr = τrz (2.6)
Therefore, the stress at any point in the cylinder may be accurately de-scribed by three direct stresses and three shear stresses.
2.3 Stress-Strains Relationships
The constitutive relation between stresses and strains for a homogeneousand isotropic material can be expressed by Hooke’s law. By definition, ahomogeneous and isotropic material has the same properties in all direc-tions. From this, the following three equations for direct strain in terms ofstress are presented:
errE = σrr − ν (σθθ + σzz) (2.7)
eθθE = σθθ − ν (σzz + σθθ) (2.8)
ezzE = σzz − ν (σrr + σθθ) (2.9)
Page 5
2. The Material Derivative in Cylindrical Coordinates This one a little bit more involved than the Cartesian derivation. The reason for this is that the unit vectors in cylindrical coordinates change direction when the particle is moving. In the Lagrangian reference, the velocity is only a function of time. When we switch to the Eulerian reference, the velocity becomes a function of position, which, implicitly, is a function of time as well as viewed from the Eulerian reference. Then
(Eq. 1) and the material derivative is written as (with the capital D symbol to distinguish it from the total and partial derivatives)
(Eq. 2)
Special attention must be made in evaluating the time derivative in Eq. 2. In dynamics, when differentiating the velocity vector in cylindrical coordinates, the unit vectors must also be differentiated with respect to time. In this case, the partial derivative is computed at a fixed position and therefore, the unit vectors are "fixed" in time and their time derivatives are identically zero. Then, we have
(Eq. 3)
we can now evaluate the remaining terms in Eq. 2 as follows
Page 6
(Eq. 4)
and
(Eq. 5) finally
Page 7
(Eq. 6)
When these are put together, the material derivative in cylindrical coordinates becomes
Page 8
5.7 Basic Equations in different Coordinate Systems 143
x2−Component: ρ
(∂U2
∂t+ U2
∂U2
∂x1+ U2
∂U2
∂x2+ U3
∂U2
∂x3
)
= −∂P
∂x2
+ µ
(∂2U2
∂x21
+∂2U2
∂x22
+∂2U2
∂x23
)
+ ρg2
(5.110)
x3−Component: ρ
(∂U3
∂t+ U1
∂U3
∂x1+ U2
∂U3
∂x2+ U3
∂U3
∂x3
)
= −∂P
∂x3
+ µ
(∂2U3
∂x21
+∂2U3
∂x22
+∂2U3
∂x23
)
+ ρg3
(5.111)
– Momentum Equations in Cylindrical Coordinates
- Momentum equations with τij−terms:
r−Component: ρ
(
∂Ur
∂t+ Ur
∂Ur
∂r+
Uϕ
r
∂Ur
∂ϕ−
U2ϕ
r+ Uz
∂Ur
∂z
)
= −∂P
∂r
−
(1
r
∂
∂r(rτrr) +
1
r
∂τrϕ
∂ϕ−
τϕϕ
r+
∂τrz
∂z
)
+ ρgr
(5.112)
ϕ−Component: ρ
(∂Uϕ
∂t+ Ur
∂Uϕ
∂r+
Uϕ
r
∂Uϕ
∂ϕ+
UrUϕ
r+ Uz
∂Uϕ
∂z
)
= −1
r
∂P
∂ϕ−
(1
r2
∂
∂r(r2τrϕ) +
1
r
∂τϕϕ
∂ϕ+
∂τϕz
∂z
)
+ ρgϕ
(5.113)
z−Component: ρ
(∂Uz
∂t+ Ur
∂Uz
∂r+
Uϕ
r
∂Uz
∂ϕ+ Uz
∂Uz
∂z
)
= −∂P
∂z
−
(1
r
∂
∂r(rτrz) +
1
r
∂τϕz
∂ϕ+
∂τzz
∂z
)
+ ρgz
(5.114)
- - Navier-Stokes equations for ρ and µ equally constant:
r−Component: ρ
(
∂Ur
∂t+ Ur
∂Ur
∂r+
Uϕ
r
∂Ur
∂ϕ−
U2ϕ
r+ Uz
∂Ur
∂z
)
(5.115)
= −∂P
∂r+ µ
[∂
∂r
(1
r
∂
∂r(rUr)
)
+1
r2
∂2Ur
∂ϕ2−
2
r2
∂Uϕ
∂ϕ+
∂2Ur
∂z2
]
+ ρgr
Page 9
26 FLUID MECHANICS
+12
(∂vx
∂z+∂vz
∂x
)2
+12
(∂vy
∂z+∂vz
∂y
)2 . (2.141)
In the above,γ, µ, κ, andM are treated as uniform constants.
2.19 Fluid Equations in Cylindrical Coordinates
Let us adopt the cylindrical coordinate system,r, θ, z. Making use of the results quoted in Section C.3, the componentsof the stress tensor are
σrr = −p+ 2µ∂vr
∂r, (2.142)
σθθ = −p+ 2µ
(1r∂vθ
∂θ+vr
r
), (2.143)
σzz = −p+ 2µ∂vz
∂z, (2.144)
σrθ = σθr = µ
(1r∂vr
∂θ+∂vθ
∂r− vθ
r
), (2.145)
σrz = σzr = µ
(∂vr
∂z+∂vz
∂r
), (2.146)
σθz = σzθ = µ
(1r∂vz
∂θ+∂vθ
∂z
), (2.147)
whereas the equations of compressible fluid flow become
DρDt
= −ρ∆, (2.148)
Dvr
Dt−v 2θ
r= −1
ρ
∂p∂r− ∂Ψ∂r
+µ
ρ
(∇2vr −
vr
r2− 2
r2
∂vθ
∂θ+
13∂∆
∂r
), (2.149)
DvθDt+vr vθ
r= − 1
ρ r∂p∂θ− 1
r∂Ψ
∂θ
+µ
ρ
(∇2vθ +
2r2
∂vr
∂θ− vθ
r2+
13r
∂∆
∂θ
), (2.150)
Dvz
Dt= −1
ρ
∂p∂z− ∂Ψ∂z+µ
ρ
(∇2vz +
13∂∆
∂z
), (2.151)
1γ − 1
(DρDt− γ p
ρ
DρDt
)= χ +
κMR ∇
2
(pρ
), (2.152)
where
∆ =1r∂(r vr)∂r
+1r∂vθ
∂θ+∂vz
∂z, (2.153)
DDt
=∂
∂t+ vr
∂
∂r+vθ
r∂
∂θ+ vz
∂
∂z, (2.154)
∇2 =1r∂
∂r
(r∂
∂r
)+
1r2
∂2
∂θ2+∂2
∂z2, (2.155)
Page 10
Navier - Stokes equation:
We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field ))()()(( x,y,z, w x,y,z, vx,y,zuV =
r
Incompressible continuity equation:
0=∂∂
+∂∂
+∂∂
zw
yv
xu eq1.
Navier - Stokes equation:
vector form: VgPDt
VD rrr
2∇++−∇= μρρ
x component:
)( 2
2
2
2
2
2
zu
yu
xug
xP
zuw
yuv
xuu
tu
x ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq2.
y component:
)( 2
2
2
2
2
2
zv
yv
xvg
yP
zvw
yvv
xvu
tv
y ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq3.
z component:
)( 2
2
2
2
2
2
zw
yw
xwg
zP
zww
ywv
xwu
tw
z ∂∂
+∂∂
+∂∂
++∂∂
−=⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+∂∂
+∂∂
+∂∂ μρρ eq4.
Cylindrical coordinates ),,( zr θ : We consider an incompressible , isothermal Newtonian flow (density ρ =const, viscosity μ =const), with a velocity field .uuuV zr ),,( θ=
r
Incompressible continuity equation:
0)(1)(1
=∂∂
+∂
∂+
∂∂
zuu
rrru
rzr
θθ eq a)
r-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
−∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+−∂∂
+∂∂
+∂∂
2
2
22
2
22
2
211zuu
ru
rru
rur
rrg
rP
zuu
ruu
ru
ruu
tu
rrrrr
rz
rrr
r
θθμρ
θρ
θ
θθ
eq b)
θ -component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+∂∂
+−⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
++∂∂
+∂∂
+∂∂
2
2
22
2
22
2111zuu
ru
rru
ru
rrr
gPr
zu
uruuu
ru
ru
ut
u
r
zr
r
θθθθθ
θθθθθθ
θθμρ
θ
θρ
eq c)
z-component:
⎥⎦
⎤⎢⎣
⎡∂∂
+∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
∂∂
++∂∂
−=
⎟⎠⎞
⎜⎝⎛
∂∂
+∂∂
+∂∂
+∂∂
2
2
2
2
2
11zuu
rrur
rrg
zP
zuuu
ru
ruu
tu
zzzz
zz
zzr
z
θμρ
θρ θ
eq d)
Page 11