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Rheological behaviour of molten polymer, models and experiments - - PowerPoint PPT Presentation
Rheological behaviour of molten polymer, models and experiments - - PowerPoint PPT Presentation
Rheological behaviour of molten polymer, models and experiments Rudy VALETTE Cemef Mines Paristech 1 Cemef Mines Paristech Center for Material Forming Polymer, pastes, composites, metals Industrial partnership Polymer processing 2
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Cemef – Mines Paristech
Center for Material Forming Polymer, pastes, composites, metals Industrial partnership Polymer processing
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Examples of polymers
Polystyrene Polypropylene Polyethylene
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Phenomenology
Density : ~water « Melting » temperature : ~150°C (PE, PP, PS) Viscosity : ~106 water « Internal » structure : remains/relaxes over differents timescales → → → → memory, elasticity « High » temperature : viscous + elastic → → → → visco-elastic fluid
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« Internal » structure ?
Macromolecules, ex PE : -[CH2]-n … 10 000 CH2 groups ; 1018 molecules per cm3 ; very flexible Polydispersity : etc …
molecular mass distribution
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Research topics
Link between structure and rheology Viscoelastic flows modelling → Polymer processing flows :
- large deformations
- large deformation rates
- complex geometries
How pertinent are viscoelastic constitutive equations ?
Polymer processing flows: extrusion
flow instabilites interfacial instabilities
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Which material ?
macromolecule : ... close view ... 0.2 nm
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Which material ?
macromolecule : ... further away. 20 nm
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Which material ?
macromolecule : ... further away : free segments 20 nm
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Flexible chain
Degrees of freedom « Free » equivalent segment Conformation = { }= "random flight" of N segments of length b φ θ
dof not dof
=
b r r
i i
=
- ,
i
r
- ↔
↔ ↔ ↔
∑
=
i
r R
Entropic elasticity
Energy kT Relaxation: diffusion in the space of R
∑
=
i
r R
- ( )
0 R
R R kT R F
- ≅
→
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Viscoelasticity
Relaxation of entropic elastic forces Relaxation processes involve:
- which object (chain, subchain) ?
- which time ?
- which distance/which topology ?
Role of the flow ? Overview:
- Scalings
- Linear viscoelasticity/elastic dumbell
- Topological constraints, tube model
- Treatment of non-linear deformation (flow)
- Some examples
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Entropic elasticity
Statistical conformation = = end-to-end vector Mean value : Mean square : Gaussian density distribution : Statistical entropy : Thermodynamic force :
∑
=
i
r R
- =
R
- (
)( )
2 2 2
R Nb r r R
j i
= = =
∑ ∑
- ( )
2 3
2 2
2 3 2 3 2 −
=
R R
e R R π ψ
- ( )
( )
R k S
tot
- ψ
Ω = ln
( )
R R kT R d dS T R F
- 2
3 = − =
Doi-Edwards 1986
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Elastic dumbbell
Force: Spring constant: Einstein relation: Relaxation time:
( )
R R kT R F
- 2
3 =
2 2
3 3 Nb kT R kT =
b
N kT kT D ζ ζ = =
2 2 2 2
θ ζ θ N kT b N D R
b =
= =
Doi-Edwards 1986
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Macroscopic scale
Force: chains per unit volume chains per unit surface Stress: Elastic modulus: Relaxation time:
( )
R R kT R F
- 2
3 =
2
N ∝ θ ν R ν
2 2 2 2
3 3 R R kT R R kT
- ν
ν = kT G ν =
Doi-Edwards 1986
Relaxation function : experiment 1 on viscoelasticity
Response to a sudden macroscopic deformation
macro
γ
micro
γ
Sample at rest
(the arrows are proportional to the chains oriented and stretched in their direction)
Deformation a t=0 : stress = G0 x strain Stress decreases with time because microscopic strain γmicro decreases Whatever γmacro : G(t) is constant (same ratio stress/strain)
Relaxation function : experiment 1 on viscoelasticity
Stress relaxation is faster when far from equilibrium
micro micro
dt d γ θ γ 1 − = τ θ τ 1 − = dt d
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Linear (separable) response of a simple viscoelastic fluid
Step strain :
t=0 rest, t>0 constant shear log time log shear modulus
macro
γ
θ
γ τ
/
) ( /
t
e G t G
−
= = G θ G
s
2
10−
fraction of unrelaxed segments
Viscosity function : experiment 2 on viscoelasticity
Response to a continuous macroscopic deformation
macro
γ
- micro
γ
Stress = ,
τ = G0 γmacroθ γmicro = γmacroθ
Viscosity function : experiment 2 on viscoelasticity
Response to a continuous macroscopic deformation
drag
dt dt d
micro macro micro
γ θ γ γ 1 − =
macro micro
γ θ γ
- =
relax Stationary : Stress :
macro micro
G G γ θ γ τ
- =
=
Viscosity :
θ η G =
Response in steady shear
Steady shear:
t=0 rest, t>0 onstant shear rate, log time ~ deformation log shear stress
t V γ
- =
elastic: slope G Viscous: viscosity Gθ
) 1 ( /
/θ
θ γ τ
t
e G
−
− =
- G
θ G
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Steady shear
Shear rate: Drag (affine deformation)/relaxation balance → stress: Viscosity: Weissenberg effect:
γ
- θ
γ τ
- G
=
b
N Nb G ζ ν θ η
2
= =
Long chains are entangled
Entanglements : topological constrains between chains
Courtesy K. Kremer, from Everaers et al., Science, 2004
a
Long chains are entangled
Entanglements: topological constrains between chains
Courtesy K. Kremer, from Everaers et al., Science, 2004
Long chains are entangled
Entanglements: topological constrains between chains
Courtesy K. Kremer, from Everaers et al., Science, 2004
a
a kT R R R kT F ≅ ≅
Long chains are entangled
Entanglements: topological constrains between chains Relaxation: diffusion out of a « tube », reptation dynamics convenient mean field model (to make self-similar !)
From T.C.B. McLeish group, Leeds University
Long chains are entangled
Entanglements: topological constrains between chains Relaxation: diffusion out of a « tube », reptation dynamics Tube survival probability
Doi-Edwards 1986
Long chains are entangled
Entanglements: topological constrains between chains Tube length L = Za, scales as N Disengagement time:
a
3 2 2
θ ζ θ N kT N N D L
b d
∝ ∝ = a=10−9m Z = 5− 50 N / Z = 25 θZ=1 ≈ 0.1s
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Macroscopic scale
Elastic modulus: Relaxation time: Tension on each segment:
3
N
d ∝
θ a kT 3 kT G
e N
ν =
a
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Linear response of an entangled polymer melt
Step strain :
t=0 rest, t>0 constant strain log time log shear modulus
macro
γ
Pa
9
10 Pa
6
10 s
2
10− s 10
fast retraction along tube contour reptation (disengagement)
2
N ∝
3
N ∝
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Linear response of an entangled polymer melt
Step strain :
t=0 rest, t>0 constant strain log time log shear modulus
macro
γ
Pa
9
10 Pa
6
10 s
2
10− s 10
retraction reptation
2
N ∝
3
N ∝
molecular mass distribution
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Linear response of an entangled polymer melt
Step strain :
t=0 rest, t>0 constant strain log time log shear modulus
macro
γ
Pa
9
10 Pa
6
10 s
2
10− s 10
retraction reptation
2
N ∝
3
N ∝
3
G
3 3θ
G
2
G
2 2θ
G
1
G
1 1θ
G ...
i i
Gθ
Spectrum fitted using conventional rheometers
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Steady shear
Shear rate: Drag (affine deformation)/relaxation balance → stress: Viscosity: Shear thinning: rotation of segments
d
θ γ / 1 <
- d
N
G θ γ τ
- =
d N
G θ η =
d
θ γ / 1 >
- θ
N
G
log viscosity log shear rate
- 1
d
θ / 1
Doi-Edwards 1986
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Rheometric tools
Cone-plate rheometer: mechanical spectroscopy (linear regime) Capillary rheometer: steady shear viscosity
Piston Pressure transducer Die
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Sodium source λ = 589nm Picture polarizer 45° λ/4 135° λ/4 analyzer
Robert et al. 2003
Flow-induced birefringence: stress measurements
Cw kλ σ = ∆
Stress-optical law:
Schweizer et al., Rheo Acta 2005
w
Used to test pertinence
- f constitutive equations
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Constitutive equation (3D)
Stress: (autocorrelation) B conformation tensor: At rest: Affine deformation: Because when no relaxation
3 3
2
B G R R R G
N T N
= =
- τ
( ) ( )
3 2 2
R d R R R R R R R
T R T
- ∫
= ψ
ds R n R R G ds n f d
T N 2
3
- =
=τ
( )
v B B v R R v R R R R v dt dB
T T T
- ∇
+ ∇ = ∇ + ∇ =
2 2
d dt ψ
- R
( )d
- R3
( ) = 0
I B 3 1 =
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Constitutive equation (3D)
Stress: B conformation tensor: At rest: Affine deformation: Relaxation:
3 3
2
B G R R R G
N T N
= =
- τ
( ) ( )
3 2 2
R d R R R R R R R
T R T
- ∫
= ψ
ds R n R R G ds n f d
T N 2
3
- =
=τ
( )
v B B v R R v R R R R v dt dB
T T T
- ∇
+ ∇ = ∇ + ∇ =
2 2
− − ∇ + ∇ = I B v B B v dt dB
T
3 1 1 θ
- I
B 3 1 =
Upper-convected Maxwell model
Why choosing differential models for B ?
Maxwell model: closed form from Smoluchowski equation But general kinetic theory models (Likthman et al., Öttinger & Kröger et al, etc ..) involve too many refinements for polymer processing and require closure approximations Integral models (KBKZ 1962-63, Doi-Edwards 1986, Wagner et al. 1976-2012) require lots of computational ressources to be solved in practice
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Constitutive equation (3D)
Evolution equation for , tube model:
- take Maxwell model
- trace(B) is mean square chain stretch:
- rescale B so that trB=1
− − ∇ + ∇ = I B v B B v dt dB
T
3 1 1 θ
- 2
R R R B
T
- =
( ) ( )
1 1 2 − − ∇ = trB B v tr dt dtrB θ
- (
)
− − ∇ − ∇ + ∇ = I B B B v tr v B B v dt dB
T
3 1 1 2 θ
- Larson 1988
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2
. R R R trB
- =
Test of reptation differential model
Evolution equation for , tube model:
- Marrucci & Ianniruberto (1996): convective constrain release
relaxation is accelerated by retraction
- linear viscoelasticity spectrum + β fitted from steady shear
2
R R R B
T
- =
( )
− − ∇ − ∇ + ∇ = I B B B v tr v B B v dt dB
T
3 1 1 2 θ
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1 θ = 1 θ +2β tr ∇uB
( )
Valette, Mackley, Hernandez 2006
θ
N
G
log viscosity log shear rate slope
1 θ = 1 θd + ∇u
Piston 1 Piston 2
1 2 3 4 5 6 7 8 9 10 11 12 13 0,25 0,5 0,75 1 1,25 1,5 1,75 2 2,25 2,5 2,75 3 3,25 3,5 3,75
Temps (s) Perte de charge (bars)
Flow:
- start
- flow
- relax
Measure:
- birefringence
- pressure drop
- strain rate ~1/θd
Valette, Mackley, Hernandez, JNNFM 2006
Equations : momentum equation incompressibility transport equations for conformation tensors Splitting : Perturbed Stokes problem ← extra-stress Stabilization : « solvent » part (discretization) SUPG (transport) Time : implicit Euler Approximation : P, Conformation (linear continuous) Velocity (quadratic continuous)
Numerics: finite elements
1 2 3 4 5 6 7 8 9 10 11 12 13 0,25 0,5 0,75 1 1,25 1,5 1,75 2 2,25 2,5 2,75 3 3,25 3,5 3,75
Temps (s) Perte de charge (bars)
a : 0.06s b : 0.28s c : 0.84s d : 1.20s e : 2.85s f : 2.89s g : 2.96s h : 3.40s
|
a
|
b
|
c
|
d
|
e
|
f
|
g
|
h
polymer symmetry
½ gap
Valette, Mackley, Hernandez, JNNFM 2006
Valette, Mackley, Hernandez, JNNFM 2006
Valette, Mackley, Hernandez, JNNFM 2006
Solving it with suitable numerical methods, then compute birefringence
0.06s 0.28s 0.84s 1.20s 2.85s 2.89s 2.96s 3.40s
Valette, Mackley, Hernandez, JNNFM 2006
Solving it with suitable numerical methods, then compute pressure drop
1 2 3 4 5 6 7 8 9 10 11 12 13 0,25 0,5 0,75 1 1,25 1,5 1,75 2 2,25 2,5 2,75 3 3,25 3,5 3,75
Temps (s) Perte de charge (bars)
Carreau, compressible Rolie Poly, incompressible Rolie Poly, compressible 1 2 3 4 5 6 7 8 9 10 11 12 13 2,8 2,85 2,9 2,95 3 3,05
Temps (s)
1 2 3 4 5 6 7 8 9 10 11 12 13 0,05 0,1 0,15 0,2 0,25
Temps (s) Perte de charge (bars)
Carreau, compressible Rolie Poly, incompressible Rolie Poly, compressible
- - Viscous compressible
__ VE incompressible __ VE compressible
- - Viscous compressible
__ VE incompressible __ VE compressible
Predictions of tube-based differential models are ≈ ≈ ≈ ≈ OK
Flows were not as strong as polymer processing flows: Birefringence:
- measure of stress
- not linked to a measure of strain (rate)
d
θ γ / 1 >>
Flows stronger than 1/θ θ θ θd
Take chain retraction dynamics into account:
- stress is still
- partition of B:
- not really satisfactory predictions
3GB = τ S B
2
λ =
( )
− − ∇ − ∇ + ∇ = I B B B v v B B v dt dB
d T
3 1 1 : 2 θ
- McLeish & Larson 1998
( ) ( )
1 1 − − ∇ = λ θ λ
r
B v tr dt d
- Koventry, Valette, Mackley 2004
Planar elongational flow Coventry, Mackley, Valette 2004
Piston 1
« Purely » elongational :
- 3 s-1
- birefringence
- polystyrene
- compare different models
Piston 2
Planar elongational flow Coventry, Mackley, Valette 2004
Same physics, different expressions (closure approximation) :
- Rolie-Poly
- Pompom
- DXPompom
Too many adjustable parameters !!!!
Flows stronger than 1/θ θ θ θd
Take chain retraction dynamics and finite extensibility into account:
- stress is still
- finite extensibility: chains are no more Gaussian:
- rest dimension
- unfolded chain length
- max extension square
- modulus
- single equation partition for B:
3GB = τ
Gf trB N N G G = − − → 1
( )I
ftrB I trB B f v B B v dt dB
r T
1 3 1 3 − − − − ∇ + ∇ = θ θ
- Marrucci & Ianniruberto 2003
b N R =
Nb R =
max
( )
N R R =
2 max /
( ) ( )
1 1 1 1 1 1 1 − + − − + = ftrB ftrB
d r d
β β θ θ θ θ
- Measure Vx (axial)
Spatial resolution 35 μ μ μ μm, velocity resolution 50μ μ μ μm/s
Test constitutive equations using LDV
- D. Hertel, PhD, LSP Erlangen
Computation Velocity +gradients Stress ?
Rheometry : Linear regime: spectrum Capillary: steady shear
Optical rheometer
Boukellal, Durin, Valette, Agassant 2011
Optical rheometer: on symmetry axis get velocity
Boukellal, Durin, Valette, Agassant 2011
Optical rheometer: on symmetry axis deduce strain rate
Boukellal, Durin, Valette, Agassant 2011
Optical rheometer: compute stress and compare
Boukellal, Durin, Valette, Agassant 2011
Conclusion
Models are OK for moderately fast flows and simple chain topologies No adjustable parameters (modulo uncertainties) Predictive constitutive models for complex topologies and faster flows remain a challenge
Polymer processing flows: instabilities
Very strong flows
r
θ γ / 1 >>
- Combeaud, Demay, Vergnes 2004
Polymer processing flows: instabilities
flow instabilites interfacial instabilities
Polymer processing flows: coextrusion instabilities
interfacial instabilities
Valette, Laure, Demay, Agassant 2003
Polymer processing flows: coextrusion instabilities
interfacial instabilities inside flow cell / after cooling
Valette, Laure, Demay, Agassant 2004a
Polymer processing flows: coextrusion instabilities
interfacial instabilities: solve direct model -> convective instability forced system: Level-set + SUPG method
Valette, Laure, Demay, Fortin 2002
Polymer processing flows: coextrusion instabilities
interfacial instabilities: solve direct model -> convective instability wavepacket: Discontinuous Galerkin method
Valette, Laure, Demay, Fortin 2001
Polymer processing flows: coextrusion instabilities
interfacial instabilities: linear stability of 2 layer viscoelastic Poiseuille flow
Valette, Laure, Demay, Agassant 2004b