IMA Annual Program Year Tutorial An Introduction to Funny (Complex) Fluids: Rheology, Modeling and Theorems September 12-13, 2009 Understanding silly putty, snail slime and.
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Transcript IMA Annual Program Year Tutorial An Introduction to Funny (Complex) Fluids: Rheology, Modeling and Theorems September 12-13, 2009 Understanding silly putty, snail slime and.
IMA Annual Program Year Tutorial
An Introduction to Funny (Complex) Fluids: Rheology, Modeling and Theorems
September 12-13, 2009
Understanding silly putty, snail slime
and other funny fluids
Chris Macosko
Department of Chemical Engineering and Materials Science
NSF- MRSEC
(National Science Foundation sponsored Materials Research Science and Engineering Center)
IPRIME
(Industrial Partnership for Research in Interfacial and Materials Engineering)
What is rheology?
rein (Greek) = to flow
ta panta rei = every thing flows
rheology =
study of flow?, i.e. fluid mechanics?
honey and mayonnaise
stress=
f/area
honey and mayo
viscosity =
stress/rate
rate of deformation
rate of deformation
What is rheology?
rein (Greek) = to flow
ta panta rei = every thing flows
rheology =
study of flow?, i.e. fluid mechanics?
rubber band and silly putty
honey and mayo
modulus
= f/area
viscosity
rate of deformation
time of deformation
4 key rheological phenomena
rheology = study of deformation of complex materials
fluid mechanician: simple fluids
complex flows
materials chemist: complex fluids complex flows
rheologist:
complex fluids simple flows
rheologist fits data to constitutive equations which
- can be solved by fluid mechanician for complex flows
- have a microstructural basis
from: Rheology: Principles, Measurement and Applications, VCH/Wiley (1994).
ad majorem Dei gloriam
Goal: Understand Principles of Rheology:
(stress, strain, constitutive equations)
stress = f (deformation, time)
Simplest constitutive relations:
Newton’s Law:
Hooke’s Law:
d
t
dt
t G
Key Rheological Phenomena
•
•
•
•
shear thinning (thickening)
time dependent modulus
normal stresses in shear
extensional > shear stress
G(t)
N1
u >
1
k1
ELASTIC SOLID
k2
L´
The power of any spring
is in the same proportion
with the tension thereof.
Robert Hooke (1678)
k1
f L
f kL
k2
f
L´
Young (1805)
modulus
stress
t
t G
strain
1-8
Uniaxial Extension
Natural rubb
G=3.9x105Pa
T11
f
a
extension a
or strain e
a = area
L
L
L - L
a 1
L
1
T11 G a 2
a
natural rubber G = 400 kPa
1-9
Shear gives different stress response
=0
= -0.4
T21 G
= 0.4
Goal: explain different results in extension and shear
obtain from Hooke’s Law in 3D
If use stress and deformation tensors
T11 T22 G 2
Silicone rubber G = 160 kPa
1-10
Stress Tensor - Notation
Txx
T Tyx
Tzx
ˆ ˆ Txx xy
ˆ ˆ Txy xz
ˆ ˆ Txz
T xx
ˆ ˆ Tyx yy
ˆ ˆ Tyy yz
ˆ ˆ Tyz
yx
ˆ ˆ Tzx zy
ˆ ˆ Tzy zz
ˆ ˆ Tzz
zx
dyad
ˆ x yt
ˆ y zt
ˆ z
T xt
ˆ ˆ Txy
xy
direction of stress on plane
plane stress acts on
Txz
Tyz
Tzz
Txy
Tyy
Tzy
ˆ ,y
ˆ ,z
ˆ x
ˆ1 ,x
ˆ 2 ,x
ˆ3
x
T11
Tij
T21
T31
3
T
i 1
T13
T23
T33
T12
T22
T32
3
xˆ xˆ T
j 1
i
j
ij
Other notation besides Tij: sij or Pij
Tij
1-11
Rheologists use very simple T
1. Uniaxial Extension
T11 0 0
T 0 0 0
0 0 0
T22 = T33 = 0
or
T22
T33
0
0
T 0 T22
0
0
0
0 T11 = 0
T33
T22 = T33
T11 -T22 causes deformation
1-12
Consider only normal stress components
T11 0
Tij 0 T22
0
0
Hydrostatic Pressure
p
Tij 0
0
T pI
0
0
T33
T11 = T22 = T33 = -p
0
1
p 0 p 0
0 p
0
1 0 0
I 0 1 0
0 0 1
0
TI T
0
1
0
0
0
1
Then only t the extra or viscous stresses
cause deformation
T = -pI + t
If a liquid is
incompressible
and only the normal stress differences
cause deformation
G ≠ f(p) ≠ f(p)
T11 - T22 = t11 - t22 ≡ N1 (shear)
1-13
Rheologists use very simple T
2. Simple Shear
x̂2
x̂2
T12
T21
x̂1
x̂3
x̂1
T21
0 0 0
Tij T21 0 0
0 0 0
But to balance angular momentum
in general
x̂3
0 T12 0
Tij 0 0 0
0 0 0
T12
T21 T12
T TT
T T i j (xi x jTij )T i j xi x jT ji
tn nˆ T TT nˆ
Stress tensor for simple shear
T11 T12 0
Tij T12 T22 0
0 0 T33
Only 3 components:
T12
T11 – T22 = t11 – t22 ≡ N1
T22 – T33 = t22 – t33 ≡ N2
1-14
Stress Tensor Summary
1.
t n nˆ T stress at point
n
T11
T11
on any plane
2. in general T = f( time or rate, strain)
3. simple T for rheologically complex materials:
- extension and shear
4. T = pressure + extra stress = -pI + t.
5. τ causes deformation
6. normal stress differences cause deformation, t11-t22 = T11-T22
7. symmetric T = TT i.e. T12=T21
Hooke
(1678)
→
Young
(~1801)
→
Cauchy
(1830’s)
→
Gibbs
Einstein
(1880,~1905)
2
Deformation Gradient Tensor
Q
Q
s’ is a vector connecting
s
Motion
s'
two very close points in the
material, P and Q
xˆ 1
P
P
Rest or past state at t'
Deformed or present state at t
xˆ 3
Q
Q
s′
w′
y′
w
P
s′ = w′ - y′
w’ = y’ + s’
s
y
P
s=w–y
w=y+s
x = displacement function
describes how material points move
w x( w) x( y s)
x
2
s Os
x y
x(y) F s
y F s
x y
a new tensor !
x
F=
x
or Fij
xi
xj
s w - y F s
1-16
Apply F to Uniaxial Extension
Displacement functions describe how coordinates of P in undeformed state, xi‘
have been displaced to coordinates of P in deformed state, xi.
x1
x1
x1 a1 x1
x1
x 2
x2
x2 a 2 x2
x2
x3
x3
x3 a 3 x3
x3
Fij xi x j
x1 x1 a1 x1 x2 0 x1 x3 0
x x 0 x x a x x 0
2
2
2
2
3
2 1
x3 x1 0 x3 x2 0 x3 x3 a 3
a1 0 0
Fij 0 a 2 0
0 0 a 3
1-17
a1 0
Fij xi xj 0 a
2
0 0
0
0
a 3
Assume:
1) constant volume V′ = V
x1x2 x3 x1x2 x3 or a1a 2a 3 1
x
x2
2
2) symmetric about the x1 axis
x3
a2 a3
x3
0
0
a1
a1a 1 a2 1 2
Fij 0 a11 2
0
a1
0
0
a11 2
1-18
Can we write Hooke’s Law as τ G F - I ?
2
2
1
Can we write Hooke’s Law as τ G F - I ?
Solid Body Rotation – expect no stresses
x1 x1 cos θ x2 sin θ
x2 x1 sin θ x2 cos θ
x3 x3
Fij xi x j
cos
Fij sin
0
sin
cos
0
0
0
1
For solid body rotation,
expect F = I
t=0
But
F≠I
F ≠ FT
Need to get rid of rotation
1-19
create a new tensor!
Finger Tensor
F·FT V·R·V·R
T
FV R
V·R·RT ·VT
V stretch
V·I·VT
R rotation
BF F
T
V2
xi x j
or Bij Fik Fjk
xk xk
Solid Body Rotation
cos
Bij sin
0
sin
cos
0
0 cos
0 sin
1 0
sin
cos
0
0 1 0 0
0 0 1 0
1 0 0 1
Bij gives relative local change in area within the sample.
1-20
Neo-Hookean Solid
τ GB - I
or
T pI GB
1. Uniaxial Extension
0
0
a1
Fij 0 a11 2
0
1 2
0
a1
0
0
0 a1
0
0 a12 0
0
a1
Bij Fik F jk 0 a11 2
0 0 a11 2
0 0 a11 0
1 2
1 2
0
a1 0
0
a1 0
0 a11
0
t 11 G (a12 1)
t 22 G (a11 1)
t 11 t 22 G (a12
t 22 t 33 0
1
a1
) T11 T22 T11
since
T22 = 0
f1
a1
1-21
2. Simple Shear
τ G B - I
x1 x1
x2 x2
x3 x3
s
x2 x1 x2
x '2
t 21 G
t 11 t 22 G 2
agrees with experiment
1 0
Fij 0 1 0
0 0 1
2 0
1 0 1 0 0 1
Bij 0 1 0 1 0
1 0
0 0 1 0 0 1 0
0
1
1-22
Silicone rubber G = 160 kPa
Finger Deformation Tensor Summary
1. B F F area change around a point
on any plane
T
2. symmetric
3. eliminates rotation
4. gives Hooke’s Law in 3D
fits rubber data fairly well
predicts N1, shear normal stresses
for uniaxial tension B11 a1
for a1 1 e
f1
1
2
T11 T22 T11
G(a1 )
a1
a1
(1 e )3 1
T11 G
1
e
for e
T11
e
T11 3Ge
tensile modulus 3G
Course Goal: Understand Principles of Rheology:
(constitutive equations)
stress = f (deformation, time)
Simplest constitutive relations:
Newton’s Law:
Hooke’s Law:
d
t
dt
t G
τ G B - I
Key Rheological Phenomena
•
•
•
•
shear thinning (thickening)
time dependent modulus
normal stresses in shear
extensional > shear stress
G(t)
N1
u >
2
VISCOUS LIQUID
The resistance which arises
From the lack of slipperiness
Originating in a fluid, other
Things being equal, is
Proportional to the velocity
by which the parts of the
fluids are being separated
from each other.
Isaac S. Newton (1687)
t yx
d x
dy
dv
t
dy
t yx
Bernoulli
Newton, 1687
dvx
dy
Stokes-Navier, 1845
Material
Glass
Molten Glass (500C)
Asphalt
Molten polymers
Heavy syrup
Honey
Glycerin
Olive oil
Light oil
Water
Air
measured in shear
1856 capillary (Poiseuille)
1880’s concentric cylinders
(Perry, Mallock,
Couette, Schwedoff)
Approximate
Viscosity
(Pa - s)
1040
1012
108
103
102
101
100
10-1
10-2
10-3
10-5
Familiar materials have a wide range in viscosity
Adapted from
Barnes et al.
(1989).
dv
t
dy
Newton, 1687
t yx
Bernoulli
dvx
dy
Stokes-Navier, 1845
measured in shear
1856 capillary (Poiseuille)
1880’s concentric cylinders
(Perry, Mallock,
Couette, Schwedoff)
measured in extension
1906 Trouton
u = 3
“A variety of pitch which gave by the traction method
l = 4.3 x 1010 (poise) was found by the torsion
method to have a viscosity m = 1.4 x 1010 (poise).”
F.T. Trouton (1906)
To hold his viscous pitch samples, Trouton forced a
thickened end into a small metal box. A hook was
attached to the box from which weights were hung.
polystyrene 160°C
Münstedt (1980)
Goal
1.Put Newton’s Law in 3 dimensions
• rate of strain tensor 2D
• show u = 3
recall Deformation Gradient Tensor, F
xˆ 2
Q
Q
s
Motion
s'
xˆ 1
P
P
Rest or past state at t'
Deformed or present state at t
xˆ 3
Separation and displacement
of point Q from P
Q
w′
y′
s′ = w′ - y′
s′
P
Q
w
w x ( y, t) F s
s
y
P
s=w-y
s = w - y = F s
Viscosity is “proportional to the velocity by which the parts of the
fluids are being separated from each other.” —Newton
Velocity Gradient Tensor
L is the velocity gradient tensor.
v
dv
dx or dv L dx
x
s F s
rate of separation
s F
s
s F
t t
t
s
dv = F s F dx
t
dv F dx L dx
dv F dx L dx
F LF
lim F I
x x
or
t t
lim F L
x x
Alternate notation:
v LT xˆ i xˆ j
i
j
vj
xi
Can we write Newton’s Law for viscosity as t = L?
F VR
v θ Ωr
•
vr vz 0
•
•
F = V R+V R
solid body rotation
lim V t lim R (t ) I
x x
x x
•
•
•
lim F L V + R
0 Ω 0
Lij Ω 0 0
0 0 0
x x
t2 ≠ t2
•
•
T
•
R is anti-symmetric
Rate of Deformation Tensor D
•
2 V 2D L LT (v)T + v
Vorticity Tensor W
•
Other notation:
•
2 R 2W L LT
L D+W (v )
T
•
L (V R ) V R
T
•
2D
Example 2.2.4 Rate of Deformation Tensor is a Time Derivative of B.
Show lim
t t
dB
2D
dt
Thus
t t
lim B F+ FT
dB
T
B F F F FT F FT
dt
lim F lim FT I
t t
recall that lim F L
t t
x x
lim B L + LT 2D
t t
Show that 2D = 0 for solid body rotation
0 0 0 0
2 D L LT 0 0 0 0 0
0 0 0
0 0
0
0 0
2W L - LT 2 0 0
0 0
0
Newtonian Liquid
t = 2D
or
T = -pI + 2D
Steady simple shear
Here planes of fluid slide over each other like cards in a deck.
x1 x1 γx2
x2 = x2
x3 = x3
Time derivatives of the displacement functions
for simple, shear
dx1 dγ
x2 v 1
x2 x2 dt
dt
lim
v 1 x2
0 0
Lij 0 0 0
0 0 0
0 0 0
L ji 0 0
0 0 0
1 0 0
0 0
Tij p 0 1 0 0 0
0 0 1
0 0 0
dv1
x2
dx2
dx2
0 v2
dt
and
v2 v3 0
0 0
2Dij 0 0
0 0 0
T12 t12 t 21
(2.2.10)
Newtonian Liquid
Steady Uniaxial Extension
time derivatives of the displacement functions at x1 x1
x1 α1 x1
dx1 dα1
dα
x1 or v 1 x1 a1 x1
dt
dt
dt
Similarly
v2 a 2 x2
v3 a 3 x3
a1 0 0
Lij 0 a 2 0
0 0 a 3
incompressible fluid (1.7.9)
v 0
or a1 a 2 a 3 0
symmetric, υ2 υ3 and thus
a 2 a3
a 2 a3
a1
2
a1
Lij 0
0
0
0
e
a1 2
0 0 e 2
0
0
a1 2 0
0
e
0
0
2e
2Dij (Lij L ji ) 0
0
0
e
0
0
0
e
2
Newtonian Liquid
Apply to Uniaxial Extension
2e
2Dij 0
0
0
e
0
0
0
e
t = 2D
t 11 2e
t 22 t 33 e
From definition of extensional viscosity
t 11 t 22
u
3
e
Newton’s Law in 3 Dimensions
•predicts 0 low shear rate
•predicts u0 = 30
but many materials show large deviation
Polystyren
e melt
Summary of Fundamentals
1.
t n nˆ T stress at point on plane
n
T11
simple T - extension and shear
T = pressure + extra stress = -pI + t.
symmetric T = TT i.e. T12=T21
2. B F F area change around a point on plane
symmetric, eliminates rotation
T
gives Hooke’s Law in 3D, E=3G
T
T
2
D
L
L
(
v
)
+ v rate of separation of particles
3.
symmetric, eliminates rotation
gives Newton’s Law in 3D,
u 3
T11
Course Goal: Understand Principles of Rheology:
stress = f (deformation, time)
NeoHookean:
τ G B - I
Newtonian:
t = 2D
Key Rheological Phenomena
•
•
•
•
shear thinning (thickening)
time dependent modulus
normal stresses in shear
extensional > shear stress
G(t)
N1
u >