chapter 11 - stress, strain and deformation in...
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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].
2000, W. E. Haisler 2
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( ) / ( )
2000, W. E. Haisler 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
( ) ( )
( ) ( )
( ) ( )
2000, W. E. Haisler 5
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
2000, W. E. Haisler 6
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
2000, W. E. Haisler 7
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
2000, W. E. Haisler 8
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
2000, W. E. Haisler 9
( / )( ) ( / )( )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/ .
2000, W. E. Haisler 10
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):
2000, W. E. Haisler 11
[ ] 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
2000, W. E. Haisler 12
[ ] 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
2000, W. E. Haisler 14
or
[ ] vtv v v 2 where
P gh
2000, W. E. Haisler 15
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
2000, W. E. Haisler 16
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 ( ) .
2000, W. E. Haisler 17
Linear Momentum:
[ ] vtv v g v P 2
2000, W. E. Haisler 18
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:
2000, W. E. Haisler 19
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)
2000, W. E. Haisler 20
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 .
2000, W. E. Haisler 21
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
2000, W. E. Haisler 22
Note: Both flow rate Q and velocity vz are proportional to axial pressure gradient, dP/dz.
2000, W. E. Haisler 23
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
2000, W. E. Haisler 24
( )
( )
( )
( )
( ) (
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
2000, W. E. Haisler 25
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:
2000, W. E. Haisler 26
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)
2000, W. E. Haisler 28
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
2000, W. E. Haisler 29
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
2000, W. E. Haisler 30
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.
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