navier-stokes equation

IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Navier-Stokes Equation

Newtonian Fluid Constant Density, Viscosity Cartesian, Cylindrical, spherical coordinates


Cartesian Coordinates gVPDt

DV 2

xxxxx

zx

yx

xx g

z

V

y

V

x

V

x

P

z

VV

y

VV

x

VV

t

V

2

2

2

2

2

2

yyyyy

zy

yy

xy g

z

V

y

V

x

V

y

P

z

VV

y

VV

x

VV

t

V

2

2

2

2

2

2

zzzzz

zz

yz

xz g

z

V

y

V

x

V

z

P

z

VV

y

VV

x

VV

t

V

2

2

2

2

2

2


Cylindrical Coordinates gVPDt

DV 2

rrr

r

rz

rrr

r

gV

rz

VV

rrV

rrr

r

P

z

VV

r

VV

r

V

r

VV

t

V

22

2

2

2

2

2

211

zzzzz

zzz

rz g

z

VV

rr

Vr

rrz

P

z

VV

VV

r

VV

t

V

2

2

2

2

2

11

gV

rz

VV

rrV

rrr

P

rz

VV

r

VVV

r

V

r

VV

t

V

r

zr

r

22

2

2

2

2

211

1

Centrifugal force

Coriolis force


Spherical Coordinates gVPDt

DV 2

rrr

r

rrrr

r

gV

r

V

rVr

rr

r

P

r

VVV

r

VV

r

V

r

VV

t

V

2

2

222

2

2

2

22

sin

1sin

sin

11

sin

gV

r

V

r

V

r

Vr

Vrrr

P

rr

VVVV

r

VV

r

V

r

VV

t

V

r

rr

sin

cot22

sin

1

sinsin

111

1cot

sin

222

2

2

22

2

2

2

2


Spherical Coordinates gVPDt

DV 2

gV

r

V

r

V

r

Vrr

Vr

rr

P

rr

VVVVV

r

VV

r

V

r

VV

t

V

r

rr

sin

cot2

sin

2

sin

1

sinsin

111

sin

1cot

sin

222

2

2

22

2

BSL has g here instead of g


Spherical Coordinates (3W) gVPDt

DV 2

rrr

rrrr

r

gV

rV

r

V

rV

rV

r

P

r

VVV

r

VV

r

V

r

VV

t

V

sin

2cot

222

sin

22222

22

2

2

222

22

sin

1sin

sin

11

rrr

rrr

3W &R have the formula in terms of However, the expression for is incorrect in the book2 2


Continuity 0.

Vt

0

zyx Vz

Vy

Vxt

011

zr Vz

Vr

Vrrrt

0sin

1sin

sin

11 22

V

rV

rVr

rrt r


Newton’s law of viscosity

VVV .3

2

Constant Density, zero dilatational viscosity

y

V

x

Vxy

xyyx

rrr

V

rr

V

rr

1

z

VVr z

zz

r

V

z

V zrrzrz

Cylindrical coordinates


Newton’s law of viscosity

VVV .3

2

Spherical coordinates:Constant Density, zero dilatational viscosity

rrr

V

rr

V

rr

1


N-S Equation: Examples

•ODE vs PDE•Spherical and cylindrical coordinates•Eqn for pipe flow (Hagen Poiseulle)•Flow between rotating cylinders (not solved in class)•Thin film flow with temp variation (not solved in class, steps were discussed briefly. BSL ‘worked out’ example)•Radial flow between circular plates (BSL 3B.10)


Example problems

1. Pressure driven steady state flow of fluid

between two infinite parallel platesinside a circular tube

2. Steady state Couvette flow of a fluidbetween two infinite parallel plates with top plate moving

at a known velocitybetween two circular plates of finite radius, with the top

plate rotating at a known angular velocitybetween two circular cylinders with outer cylinder

rotating at a known angular velocity (end effects are negligible)between a cone and plate (stationary plate and cone is

rotating at a known angular velocity). Angle of cone is very small (almost a parallel plate with almost zero gap)3. Coutte Poisseuille flow

between two parallel plates


N-S Equation: Examples

•PDE• Please refer to the book “Applied Mathematical Methods for Chemical Engineers” by Norman W Loney (CRC press), pages 330 to 342 for “worked out” examples for Momentum Transfer problems involving PDE.• Either multi dimensional or time dependent (however multidimensional and time dependent cases are not discussed in detail)• Steady state in Rectangular channel: pressure driven , coutte flow• Plan suddenly moving with constant velocity (or stress) from time t=0 (Stokes problem)• Sudden pressure gradient in a cylindrical tube (unsteady flow , converging to Hagen-Poisseuille’s flow (Bessel functions)• Flow between two (non rotating) cylinders, caused by boundary movement (coutte flow). Unsteady vs steady (not discussed in class or covered in tutorial)


Guidelines for solving PDE in Momentum Transfer

Method: If the problem involves finite scales, “separation of variable” method should

be tried If the problem involves infinite (or semi-infinite) distances, “combination of

variables” method should be tried



Solution forms, for finite scales: Applying the separation of variables directly may not always give proper

results If the equation is non-homogenous

For time dependent problems, first try to get steady state solution (and try that as the ‘particular solution’ for the equation). Unsteady state solution may be the ‘general solution’ for the corresponding homogenous equation

For multi dimensional problems, first try to get solution for ‘one dimensional’ problem and try that as particular solution. The ‘correction term’ may be the ‘general solution’ for corresponding homogenous equation.

Even if the equation is homogenous, you can try the above methods of obtaining ‘steady state’ or ‘one dimensional’ solution. The ‘complete solution’ will be the sum of ‘steady state + transient’ solution OR ‘one dimensional solution + correction for presence of plates’ (for example).

Always make sure that the ‘correction term’ goes to zero in the appropriate limit (eg time --> infinity, or the ‘width of the channel --> infinity)



Other relevant Information: Problems in Cartesian coordinates tend to give Cosine/ Sine series

solution. In cylindrical coordinates, Bessel functions. In spherical coordinates, Legendre functions

When you attempt a ‘complete solution’ as ‘steady state+ transient’ (OR ‘one dimensional + correction’), make sure that you also translate the boundary conditions correctly

While solving for the ‘transient’ or ‘correction’ terms, you may encounter a situation where you have to choose an arbitrary constant (either positive or negative or zero). Usually the constant will not be zero. Choose the constant as positive or negative, depending on the boundary conditions (otherwise, you will proceed only to realize that it will not work).


Stoke’s first problem (Please refer to BSL for solution)


xxxxx

zx

yx

xx g

z

V

y

V

x

V

x

P

z

VV

y

VV

x

VV

t

V

2

2

2

2

2

2

N-S Equation: Example: Steady state flow in Rectangular channel

• Steady state in Rectangular channel: pressure driven flow, incompressible fluid

h2

h

hbb2

h2• Vy = Vz =0•Vx is function of y and z• gravity has no component in x direction

x

y

z

2

2

2

2

0z

V

y

V

x

P xx

• Method employed: Find a particular solution satisfying above equation; then find a general solution satisfying following differential eqn

2

2

2

2

0z

V

y

V generalxgeneralx particularxgeneralxx VVV

1

23


N-S Equation: Example; Rectangular channel

• Hint: To obtain a physically meaningful format, we can take particular solution to resemble one dimensional flow (when b goes to infinity)

h2

h

hbb2

h2x

y

z

2

22

12 h

yh

x

PV particularx

• Note: Check that the above solution is a valid particular solution• Before trying to get general solution, write down the boundary conditions for the over all solution Vx

0, zhyVx

0, bzyVx

000

z

x

y

x

z

V

y

V

4

5


N-S Equation: Examples; rectangular channel

• Translate that to get the boundary conditions for Vx-general

0, zhyV generalx

particularxgeneralx VbzyV ,

000

z

generalx

y

generalx

z

V

y

V

particularxxgeneralx VVV

2

22

12 h

yh

x

PV particularx

• We know

•Hence, from equation 5,

• Use separation of variables method

zgyfVAssume generalx

2

2

2

2

0z

V

y

V generalxgeneralx implies 0 fggf

3

2

Eqn

6

7b

7a



• Since LHS is only a function of y and RHS is fn of z, both must be equal to a constant•We say

• Note: Why do we say , why not ? What will happen if you try that? Or if we say ?

implies0 fggfg

g

f

f

2

g

g

f

f

• In any case, the chosen constant leads to

zz eCeCzg 43 )sin()cos( 21 yCyCyf

00

z

generalx

z

V implies43 CC

implies 02 C00

y

generalx

y

V

8

22

0

9

From 6

and



Hence, substituting in

zzgeneralx eeyCCzgyfV )cos(31

Using superposition principle

zz

ngeneralx eeyCV )cos(

0, zhyV generalx

2

22

12 h

yh

x

PV particularx

Now, from

)cos(12 2

22

yeeCh

yh

x

P bbn

9 7a

6

particularxgeneralx VbzyV ,

h

n2

12

Again, from 6

implies

implies



• Using Fourier cosine expansion for an even function

we can find Kn

2

22

12 h

yh

x

PV particularx

)cos(12 2

22

yKh

yh

x

Pn

•Equating the co-efficients, we get

nbb

n KeeC bbn

n ee

KC

Hence, general solution part is

zz

bbn

generalx eeyee

KV

)cos(



2

22

12 h

yh

x

PV particularx

“Complete” solution for the original problem is given by

• Note: When “b” goes to infinity, the ‘correction’ part goes to zero

zzbb

nx eey

ee

K

h

yh

x

PV

)cos(12 2

22


N-S Equation: Other examples

• To determine the velocity profile in a rectangular channel, where the top plate is moving at a constant velocity of V-zero, under steady state conditions

• Try out a solution of the form “V-parallel-plate + V-correction”• Use separation of variable techniques, to determine V-correction• What happens if you try separation of variable in the first place?

• To determine the unsteady state solution for a flow in a cylindrical pipe, caused by sudden application of pressure

• Try a solution of the form ‘Steady state + Transient’, just like the one we saw for flow between parallel plates• You will get Bessel Equations. Just like we represented functions in rectangular coordinates by sine and cosine functions, we can represent functions in cylindrical co ordinates by Bessel functions, because they are orthogonal.

navier-stokes equation

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