NAVIER-STOKES

Derivation of Navier-

Stokes Equation Using cylindrical co-ordinates (r, , 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, 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, , 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, 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, 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

rr

r

r

rrrr rr

+

rr

+

rr rr

+

+

zrzr zz

+

FIGURE 2.2. Stresses in the r and 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 ), 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

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2. Governing Equations 17

z

rr

r

rrrr rr

+

zzzz zz

+

z

Fr

r

r

z

zrzr rr

+

zz

+

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 Newtons 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.

rr +rrr

r(r + r) z +

r +

r

rz cos

2

+

rz +

rzz

z

r +r2

r + Frrz

= rrrz + rrz cos2+ rz

r +

r2

r +

+

+

rz sin

2+ rz sin

2

(2.1)

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18 2. Governing Equations

Canceling appropriate terms from both sides of the equation and aftersimplifying, it yields:

rrr

+1

rr

+rzz

+rr

r+ Fr = 0 (2.2)

Similarly, the equilibrium equation for the direction yields:

rr

+1

r

+zz

+2

rr + F = 0 (2.3)

and finally, for the z direction one may write:

rzr

+1

rz

+zzz

+1

r rz + Fz = 0 (2.4)

In the above simplifications, due to very small angle of , the followingapproximations were used:

cos2 1 sin

2 2

(2.5)

In addition to the stresses, body forces acting throughout the element havebeen considered for each direction. These are denoted by Fr, F, and Fzwhich are introduced as forces in the r, , 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:

r = r z = z zr = rz (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 Hookes 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 = rr ( + zz) (2.7)eE = (zz + ) (2.8)ezzE = zz (rr + ) (2.9)

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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

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(Eq. 4)

and

(Eq. 5) finally

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(Eq. 6)

When these are put together, the material derivative in cylindrical coordinates becomes

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5.7 Basic Equations in different Coordinate Systems 143

x2Component:

(U2t

+ U2U2x1

+ U2U2x2

+ U3U2x3

)=

P

x2

+

(2U2x21

+2U2x22

+2U2x23

)+ g2

(5.110)

x3Component:

(U3t

+ U1U3x1

+ U2U3x2

+ U3U3x3

)=

P

x3

+

(2U3x21

+2U3x22

+2U3x23

)+ g3

(5.111)

Momentum Equations in Cylindrical Coordinates

- Momentum equations with ijterms:

rComponent:

(Urt

+ UrUrr

+Ur

Ur

U2r

+ UzUrz

)=

P

r

(1

r

r(rrr) +

1

r

r

r

+rzz

)+ gr

(5.112)

Component:

(Ut

+ UrUr

+Ur

U

+UrUr

+ UzUz

)

= 1

r

P

(1

r2

r(r2r) +

1

r

+zz

)+ g

(5.113)

zComponent:

(Uzt

+ UrUzr

+Ur

Uz

+ UzUzz

)=

P

z

(1

r

r(rrz) +

1

r

z

+zzz

)+ gz

(5.114)

- - Navier-Stokes equations for and equally constant:

rComponent:

(Urt

+ UrUrr

+Ur

Ur

U2r

+ UzUrz

)(5.115)

= P

r+

[

r

(1

r

r(rUr)

)+

1

r22Ur2

2

r2U

+2Urz2

]+ gr

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26 FLUID MECHANICS

+12

(vx

z+vz

x

)2+

12

(vy

z+vz

y

)2 . (2.141)In the above, , , , and M are treated as uniform constants.

2.19 Fluid Equations in Cylindrical Coordinates

Let us adopt the cylindrical coordinate system, r, , z. Making use of the results quoted in Section C.3, the componentsof the stress tensor are

rr = p + 2 vr

r, (2.142)

= p + 2 (

1r

v

+vr

r

), (2.143)

zz = p + 2 vzz

, (2.144)

r = r =

(1r

vr

+v

r v

r

), (2.145)

rz = zr =

(vr

z+vz

r

), (2.146)

z = z =

(1r

vz

+v

z

), (2.147)

whereas the equations of compressible fluid flow become

DDt

= , (2.148)

DvrDt

v2

r= 1

pr

r

+

(2vr vr

r2 2

r2v

+

13

r

), (2.149)

DvDt

+vr v

r= 1

r

p

1r

+

(2v + 2

r2vr

v

r2+

13r

), (2.150)

DvzDt

= 1

pz

z

+

(2vz + 13

z

), (2.151)

1 1

(DDt

p

DDt

)= +

MR

2(

p

), (2.152)

where

=1r

(r vr)r

+1r

v

+vz

z, (2.153)

DDt

=

t+ vr

r+v

r

+ vz

z, (2.154)

2 = 1r

r

(r

r

)+

1r2

2

2+

2

z2, (2.155)

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Navier-StokesHighlight

Navier-StokesHighlight

Navier - Stokes equation:

We consider an incompressible , isothermal Newtonian flow (density =const, viscosity =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++= x component:

)( 22

2

2

2

2

zu

yu

xug

xP

zuw

yuv

xuu

tu

x +

+++

=

+

++

eq2. y component:

)( 22

2

2

2

2

zv

yv

xvg

yP

zvw

yvv

xvu

tv

y +

+++

=

+

++

eq3. z component:

)( 22

2

2

2

2

zw

yw

xwg

zP

zww

ywv

xwu

tw

z +

+++

=

+

++

eq4. Cylindrical coordinates ),,( zr : We consider an incompressible , isothermal Newtonian flow (density =const, viscosity =const), with a velocity field .uuuV zr ),,( =

r

Incompressible continuity equation:

0)(1)(1 =+

+

zuu

rrru

rzr

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

eq b)

-component:

+

++

++

=

++

++

2

2

22

2

22

2111zuu

ru

rru

rur

rrgP

r

zuu

ruuu

ru

ruu

tu

r

zr

r

eq c)

z-component:

+

+

++

=

+

++

2

2

2

2

2

11zuu

rrur

rrg

zP

zuuu

ru

ruu

tu

zzzz

zz

zzr

z

eq d)

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