Chapter 11 - Stress, Strain and Deformation in...

2000, W. E. Haisler 1

Chapter 14 - Fluid Mechanics

Fluid mechanics problems, like solid mechanics problems, must satisfy all the conservation principles. The major differences are that in fluid mechanics problems:

1. The continuum (the fluid) generally has a velocity.

2. The constitutive relations relate stress to velocity gradients [instead of stress to displacement gradients (strain) as in a solid].

3. The final solution of the problem (after combining all applicable principles) is in terms of velocities [instead of displacements as in a solid].


Conservation of Mass

t

v ( )

Conservation of Linear Momentum

[ ] vtv v g T

S T PI where P Txx Tyy Tzz tr T 1

3 1 3( ) / ( )


[ ] vtv v g S P

2000, W. E. Haisler 4Incompressible Fluids ( =constant in time and space ). Conservation of mass (continuity equation) becomes

v 0 Continuity equation for =constant

Constitutive equations for Newtonian Fluids

The strain tensor has been defined as 12[( ) ( ) ]u uT . We can define the

rate of deformation tensor by taking the time derivative of the strain tensor to obtain

D v v T 12[( ) ( ) ] =

vx

x

vy

x

vx

y

vz

x

vx

zvx

y

uy

x

vy

y

vz

y

vy

zvx

z

uz

x

vy

z

uz

y

vz

z

1

2

1

21

2

1

21

2

1

2

( ) ( )

( ) ( )

( ) ( )


Fluids are generally isotropic and we can observe experimentally the following linear relation between stress and velocity gradients <-- defined to be a Newtonian fluid.

S tr D I D ( ) 2 <-- defined to be a Newtonian fluid

where and are 2 experimentally observed material constants (similar to E and in isotropic solids).

The above can also be written as

T S PI PI tr D I D ( ) 2


Meaning of and

Consider the case where all velocity gradients are zero except that in the x-y plane (as might occur with steady flow between two parallel plates with vx vx y ( ) only).

x

y

fluid flow

Dxydvxdy

12 where vx vx y ( )

Dxx Dyy Dzz Dxz Dyz 0 (flow only in x direction)

Substituting above into the constitutive equation gives


Txx Tyy Tzz P , Txz Tyz 0 , Txydvxdy

The material constant relates the shearing stress to the velocity gradient in the y direction. is called the coefficient of viscosity.

Txy

dvx/dy

units of MPa secpsi sec


Stokes proposed a relation between and based on the following argument (and physical observation). First take the sum of the diagonal terms of the stress tensor [S]

tr S tr tr D I D tr D tr I tr D( ) ( ( ) ) ( ) ( ) ( ) 2 2

note

tr S Sxx Syy Szz( ) ; tr I( )

3 ;

tr D Dxx Dyy Dzz( )

Thus

Sxx Syy Szz Dxx Dyy Dzz Dxx Dyy Dzz 3 2 ( ) ( )or


( / )( ) ( / )( )13 0 2 3Sxx Syy Szz Dxx Dyy Dzz

tr S Sxx Syy Szz( ) 0 and thus ( / ) 2 3 0. (

/ ) is called the coefficient of bulk viscosity . Thus 2 3/ .


Substitute Stokes’ assumption back into the constitutive relation (stress-velocity gradient relation) to obtain

S D tr D I 2 1 3[ ( / ) ( ) ]

But tr D vx x vy y vz z v( ) / / / 0

from conservation of mass. Thus S D v v T 2 [( ) ( ) ]

Note, we need S in linear momentum equation:

S D v vT

v v v

( ) ( [( ) ( ) ])

[ ( )]

2 2 122 2

Note: ( ) v 0 from incompressibility. Substitute for

S

into Linear Momentum equation to obtain the Navier-Stokes Equation (constant density and viscosity):


[ ] vtv v g v P 2

For body forces which are conservative, it convenient to re-write the body force term in terms of a potential function (means the force is the gradient of a potential function). The body force due to gravity can be written as g g h and h is the distance above a reference in the direction of the gravitational vector. If we assume g is a constant, then g g h gh ( ). Thus the two terms

g P can be

combined and written ( ) ( ) gh P P gh

where P gh is called the modified pressure potential function. With this the Navier-Stokes equation becomes


[ ] vtv v g v P 2

or

[ ] vtv v v 2

2000, W. E. Haisler 13Summary - Navier-Stokes Equations for Newtonian Fluid

Constant Mass (), Constant Viscosity ()

Continuity: ( ) v 0

Deformation Gradient: D v v T 12[( ) ( ) ]

Constitutive (Stress-Velocity): S D v v T 2 [( ) ( ) ]

Linear Momentum:

[ ] vtv v g v P 2


or

[ ] vtv v v 2 where

P gh


Cylindrical (r, , z) coordinates.

x

y

r

vv

rvz

Recall er r

e r ez z

1

2 22

1 12

22

22

rr r

r z


1. Steady-State, Laminar Flow in a Tube of radius R.

zr

axial flow

Continuity: ( ) v 0

Assume no radial flow and that flow is axisymmetric (velocity is independent of ), ie, vr v 0 . Continuity

reduces to vzz0 . Thus vz vz r C ( ) .


Linear Momentum:

[ ] vtv v g v P 2


For steady-state:

vt0. Neglect body forces:

g0.

In cylindrical coordinates, components of v vr v vz[ , , ] . From

continuity, vr v 0. Components of gradient operator are

er re r ez z

1 . Thus [ ] v v becomes in matrix notation

vr v vz

rr

z

vr v vz vz

vzr

1 0 0

0 00 0 00 0 0

0 0 0T

he two remaining terms on the right side of L.M. become:


0 2002

21

1

v P

d vz

drrdvzdr

Pr

rP

Pz

Thus we have 3 equations:

PrrP

01 0 --> Implies that P=P(z)


d vzdr

rdvzdr

dPdz

2

21

To solve the last equation for vz(r), we assume that dP/dz is a given value (and not a function of r). Integrating the above differential equation twice gives

dvzdr

dPdzr C

r 12

1 and vz dP

dzr C r C 1 24 1 2 ln

C1 must be zero since at r=0, the velocity gradient must be finite (not infinity). Assume BC that axial flow velocity is

zero at the outer wall: vz(r=R) = 0. Thus C dPdzR2 1 24 .


The axial velocity solution is thus vz r dPdzR r

R( )

1 24

22 1 .

The velocity distribution is parabolic and the maximum

value occurs as r=0: vz dPdzR

max

24 . The shear stress

distribution (from constitutive equation) is given by

Srzdvzdr

dPdz

r r dPdz

( )2 2 . The shear stress varies

linearly with the radius, is zero at r=0, and is a maximum at the wall where r=R. The volumetric flow rate is given by

Q vzAdA vz rdr

RdPdzR r

R

Rrdr dP

dzR

20

24

22 1

02 4

8


Note: Both flow rate Q and velocity vz are proportional to axial pressure gradient, dP/dz.


2. Plane Couette Flow Between Two Parallel Plates

z

y

d

Continuity: ( ) v 0

Assume fluid flow is only in the z direction, ie, vx vy 0 .

Continuity reduces to vzz0 . Thus vz vz y C ( ) .

Linear Momentum:

[ ] vtv v g v P 2


( )

( )

( )

( )

( ) (

vxt

vxvxx

vyvxy

vzvxz

vyt

vxvyx

vyvyy

vzvyz

vzt

vxvzx

vyvzy

vzvzz

gxvxx

vxy

vxz

Px

gyvyx

vyy

vyz

Py

gzvzx

22

22

22

22

22

22

22

22

22

vzy

vzz

Pz)

For steady-state:

vt0. Neglect body forces:

g0.

Linear Momentum reduces to


00

02

2

PxPyd vzdy

dPdz

The first two equations imply that P=P(z). Solving the third equation for vz(y) and taking dP/dz as a given (and not a function of y), then

vz dPdz

y C y C ( )122 1 2

B.C.: vz(0)= vz(d)=0 -->

C dPdz

d1

12 ( )

Thus vz y dPdz

d yd

yd( ) ( )

1 22

2 which is parabolic in y.

The max velocity occurs at the center:


vz vz d d dPdzmax

( / ) ( ) 22

8 . The flow rate (vol/time) is

Q vzA

dA vzWdy Wd dPdz

d 0

312 ( ) where W is the width in the x

direction of the plates. The shear stress is

Syzdvzdy

dPdz y d ( )( / )2 .

It should be kept in mind that the definition of P is given by P = -(1/3) tr[T] = -(1/3) (Txx+Tyy+Tzz) = - average hydrostatic stress. Thus, a negative dP/dz causes flow in the positive z direction.

Some units:force: 1 N = 1 kg m/s2

pressure or stress: 1 Pa = 1 N/m2 = 1 kg/(m s2),

2000, W. E. Haisler 271 psi = 6,894.76 Pa, 1 Mpa = 145.04 psi

viscosity, : 1 Pa s = 1 N s/ m2 = 1 kg/(m s)


3. Annular Flow between two pipes of radius R2 and R1.

R1R2

flow

flowz

rsolid (no flow)

The general solution for this problem is identical to the first example (laminar flow in a tube) except for geometry and boundary conditions.

vz dPdzr C r C 1 24 1 2 ln . B.C.: vz(R1)= vz(R2)=0


vz R dPdzR

C R C( ) ln1 0 1 12

4 1 1 2

vz R dPdzR

C R C( ) ln2 0 1 22

4 1 2 2

Solving for C1 and C2 gives

C dPdz

R RR R1 1 2

212

412 1

( )ln( / )

C dPdz

RdPdz

R R RR R2 1 1

24

1 22

12

41

2 1

( ) ( )

lnln( / )

Substituting the constants of integration back into vz gives


vz dPdz

RrR r R

( ) ( ln )ln( / )12

4 11

2 2 11

where R R2 1/

This is the same as equation (13-80) except that (13-80) is written in terms of the pressure potential where

dPdz

LL 0 = change in potential (pressure) from z=0 to

z=L. Note: ln()=-ln(1/). Note also that as 1 (thin annulus), the ln term goes to zero and the above solution is similar to plane couette flow.

Chapter 11 - Stress, Strain and Deformation in...

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