1 mech 221 fluid mechanics (fall 06/07) tutorial 7

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MECH 221 FLUID MECHANICS(Fall 06/07)Tutorial 7

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Outline

1. Conservation of Momentum(Navier-Stokes Equation)

2. Dimensional Analysis1. Buckingham Pi Theorem2. Normalization Method

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1. Conservation of Momentum (Navier-Stokes Equations)

In Chapter 3, for inviscid flows, only pressure forces act on the control volume V since the viscous forces (stress) were neglected and the resultant equations are the Euler’s equations. The equations for conservation of momentum for inviscid flows were derived based on Newton’s second law in the Lagrangian form.

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1. Conservation of Momentum (Navier-Stokes Equations)

Euler’s equation:

This is the Integral Form of Euler’s equation.

dVpdddV)(t S )t(VS)t(V

gss vvv

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1. Conservation of Momentum (Navier-Stokes Equations)

This is the Differential Form of Euler’s equation.

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1. Conservation of Momentum (Navier-Stokes Equations)

Here we should include the viscous stresses to derive the momentum conservation equations.

With the viscous stress, the total stress on the fluid is the sum of pressure stress( , here the negative sign implies that tension is positive) and viscous stress ( ), and is described by the stress tensor given by:

τ

ISIτσσ ))(( v 3

2pp

Iσ pp

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1. Conservation of Momentum (Navier-Stokes Equations)

Here, we generalize the body force (b) due to all types of far field forces. They may include those due to gravity , electromagnetic force, etc.

As a result, the total force on the control volume in a Lagrangian frame is given by

)( g

V(t))t(S

dVd b sσF

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1. Conservation of Momentum (Navier-Stokes Equations)

The Newton’s second law then is stated as:

Hence, we have

S)t(V

)t(V

ddV)(t

dVdt

d

dt

d

s

MF

vvv

v

S)t(VS )t(V

ddV)(t

dVd sb sσ vvv

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1. Conservation of Momentum (Navier-Stokes Equations)

By the substitution of the total stress into the above equation, we have

which is integral form of the momentum equation.

dVdpdddV)(t S )t(V)t(SS)t(V

b s ss τvvv

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1. Conservation of Momentum (Navier-Stokes Equations)

For the differential form, we now apply the divergence theorem to the surface integrals to reach:

Hence, V→0, the integrands are independent of V. Therefore,

which are the momentum equations in differential form for viscous flows. These equations are also named as the Navier-Stokes equations.

dVdVdVpdVdV)(t )t(V)t(V)t(V)t(V)t(V

b τ)()( vvv

bτ

)()( p)(t

vvv

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1. Conservation of Momentum (Navier-Stokes Equations)

For the incompressible fluids where = constant.

If the variation in viscosity is negligible (Newtonian fluids), the continuity equation becomes , then the shear stress tensor reduces to .

)(

0 vSτ

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1. Conservation of Momentum (Navier-Stokes Equations)

The substitution of the viscous stress into the momentum equations leads to:

where is the Laplacian operator which in a Cartesian coordinate system reads

b

vvvv 2pt

2

2

2

2

2

2

22

zyx

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2.1 Buckingham Pi Theorem

Given the quantities that are required to describe a physical law, the number of dimensionless product (the “Pi’s”, Np) that can be formed to describe the physics equals the number of quantities (Nv) minus the rank of the quantities, i.e., Np=Nv –Nm,

),...,,(

)u,...,u,u(fu

mv

v

N-N321

N321

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2.1 Example

Viscous drag on an infinitely long circular cylinder in a steady uniform flow at free stream of an incompressible fluid.

Geometrical similarity is automatically satisfied since the diameter (R) is the only length scale involved.

D

D: drag (force/unit length)

, ,

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2.1 Example

n-h-n-3k-hgnk201

11312

113

12

v 5

tlmtlm

)tml()ml()lt(lmt

URD

),,U,R(DD

tml;ml

lR;ltU;mtD

tl,m,

N,,R,U,D:

nhg

nkhg

:dimensions lFundamenta

Quantities

Dynamics similarity

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2.1 Example

Both sides must have the same dimensions!

n-h

n-g

n-k:

n- h-:

n-3k-hg:

nk:

ntlm

2

1

1

2-

0

1

of terms in Solving

of Exponent

of Exponent

of Exponent

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2.1 Example

where is the kinematic viscosity.

The non-dimensional parameters are:

n

n

nnnn

UR

v

RU

D

)RU/(RUD

)UR(RUD

2

2

2

So,

/v

number Reynolds the

and tcoefficien drag the

/URRe

RU/DCD2

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2.1 Example

Therefore, the functional relationship must be of the form:

The number of dimensionless groups is Np=2

(Re)FCD

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2.1 Example

The matrix of the exponents is

The rank of the matrix (Nm) is the order of the largest non-zero determinant formed from the rows and columns of a matrix, i.e. Nm=3.

tdererminan zero-non

μ ρ U R D

t

l

m

3)(3

10102

13110

11001

235mvp NNN

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2.1 Example

Problems:

No clear physics can be based on to know the involved quantities

Assumption is not easy to justified. nkhgURD

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2.2 Normalization Method

The more physical method for obtaining the relevant parameters that govern the problem is to perform the non-dimensional normalization on the Navier-Stokes equations:

where the body force is taken as that due to gravity.

g

vvvv

v

2

0

pt

t

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2.2 Normalization Method

As a demonstration of the method, we consider the simple steady flow of incompressible fluids, similar to that shown above for steady flows past a long cylinder.

Then the Navier-Stokes equations reduce to:

g

vvv

v2

0

p

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2.2 Normalization Method

If the proper scales of the problem are:

Here the flow domain under consideration is assumed such that the scales in x, y and z directions are the same

LPU :Length:Pressure :Velocity

U

LP:

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2.2 Normalization Method

Using these scales, the variables are normalized to obtain the non-dimensional variables as:

Note that the non-dimensional variable with “*” are of order one, O(1).

The velocity scale U and the length scale L are well defined, but the scale P remains to be determined.

LPppU / /vv

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2.2 Normalization Method

The Navier-Stokes equations then become:

where is the unit vector in the direction of gravity which is dimensionless.

*g

0

2*2

2ivvv

v

*** gL

Up

L

P

L

U

L

U

*gi

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2.2 Normalization Method

The coefficient of in the continuity equation can be divided to yield

Dividing the momentum equations by in the first term of left hand side gives:

*

U

gL

ULp

U

P *

g2

2

2ivvv ***

0 v

L/U

L/U 2

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2.2 Normalization Method

Since the quantities with “*” are of O(1), the coefficients appeared in each term on the right hand side measure the ratios of each forces to the inertia force. i.e.,

where Re is called as Reynolds number and Fr as Froude number.

2Fr

Re

1force inertia

force nalgravitatio

1force inertiaforce viscous

force inertiaforce pressure

2

2

U

gL

UL

U

P ,

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2.2 Normalization Method

Conservation of mass:

Conservation of momentum:

*g2

1

Re

1 2*2

ivvv ***

Frp

U

P

0 v

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Given that: Velocity: U = U∞cos(ωt) Pressure: P Length: L

2.2 Example

U

LP:

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2.2 Example

)(

or

2

2

ggpt

pt

i

v

v

vvv

gvvv

Momentum conservation equation with viscous effect:

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2.2 Example

Normalized parameters:

scale) time(transient /2 where

2//

/

/

T

*

ttt

L

Ppp

U

T

vv

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2.2 Example

ggU

LPp

L

UL

Ut

U

ggp

t

*T

i

i

)()())((

)()()()(

)(

2**

*

2

**

***

v

vvv

vvv v

Normalization:

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2.2 Example

gU

gL

LUp

U

P

tU

L

gg

L

Up

L

P

L

U

t

U

*T

*T

i

i

2

2**2

*

2*2

*

*2

**

***

**

***

v

)v(vv

v

)v(vv

Normalization:

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2.2 Example

gp

U

Pt

gL

ULU

L

U

*

T

i2

2**2

*

Fr

1

Re

1

Th

1

Fr ;Re ;Th

**

***

v

)v(vv

Since:

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The End