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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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1

Rheological behaviour of molten polymer, models and experiments

Rudy VALETTE Cemef – Mines Paristech

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2

Cemef – Mines Paristech

Center for Material Forming Polymer, pastes, composites, metals Industrial partnership Polymer processing

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3

Examples of polymers

Polystyrene Polypropylene Polyethylene

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4

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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6

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 ?

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SLIDE 7

Polymer processing flows: extrusion

flow instabilites interfacial instabilities

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8

Which material ?

macromolecule : ... close view ... 0.2 nm

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9

Which material ?

macromolecule : ... further away. 20 nm

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10

Which material ?

macromolecule : ... further away : free segments 20 nm

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11

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

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SLIDE 12

Entropic elasticity

Energy kT Relaxation: diffusion in the space of R

=

i

r R

  • ( )

0 R

R R kT R F

12

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SLIDE 13

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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14

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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15

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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16

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

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SLIDE 17

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)

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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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19

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

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Viscosity function : experiment 2 on viscoelasticity

Response to a continuous macroscopic deformation

macro

γ

  • micro

γ

Stress = ,

τ = G0 γmacroθ γmicro = γmacroθ

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SLIDE 21

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 =

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SLIDE 22

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 γ τ

t

e G

− =

  • G

θ G

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23

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

= =

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Long chains are entangled

Entanglements : topological constrains between chains

Courtesy K. Kremer, from Everaers et al., Science, 2004

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a

Long chains are entangled

Entanglements: topological constrains between chains

Courtesy K. Kremer, from Everaers et al., Science, 2004

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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 ≅ ≅

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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

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Long chains are entangled

Entanglements: topological constrains between chains Relaxation: diffusion out of a « tube », reptation dynamics Tube survival probability

Doi-Edwards 1986

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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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30

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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31

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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32

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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33

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

Spectrum fitted using conventional rheometers

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34

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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35

Rheometric tools

Cone-plate rheometer: mechanical spectroscopy (linear regime) Capillary rheometer: steady shear viscosity

Piston Pressure transducer Die

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36

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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37

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

  • =

( )

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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38

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

  • =

( )

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

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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

39

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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

40

2

. R R R trB

  • =
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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

  • 41

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

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SLIDE 42

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

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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

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SLIDE 44

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

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SLIDE 45

Valette, Mackley, Hernandez, JNNFM 2006

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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

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SLIDE 47

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

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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 >>

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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
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Planar elongational flow Coventry, Mackley, Valette 2004

Piston 1

« Purely » elongational :

  • 3 s-1
  • birefringence
  • polystyrene
  • compare different models

Piston 2

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Planar elongational flow Coventry, Mackley, Valette 2004

Same physics, different expressions (closure approximation) :

  • Rolie-Poly
  • Pompom
  • DXPompom

Too many adjustable parameters !!!!

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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

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  • 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
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Computation Velocity +gradients Stress ?

Rheometry : Linear regime: spectrum Capillary: steady shear

Optical rheometer

Boukellal, Durin, Valette, Agassant 2011

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Optical rheometer: on symmetry axis get velocity

Boukellal, Durin, Valette, Agassant 2011

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Optical rheometer: on symmetry axis deduce strain rate

Boukellal, Durin, Valette, Agassant 2011

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Optical rheometer: compute stress and compare

Boukellal, Durin, Valette, Agassant 2011

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SLIDE 58

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

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Polymer processing flows: instabilities

Very strong flows

r

θ γ / 1 >>

  • Combeaud, Demay, Vergnes 2004
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Polymer processing flows: instabilities

flow instabilites interfacial instabilities

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Polymer processing flows: coextrusion instabilities

interfacial instabilities

Valette, Laure, Demay, Agassant 2003

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Polymer processing flows: coextrusion instabilities

interfacial instabilities inside flow cell / after cooling

Valette, Laure, Demay, Agassant 2004a

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Polymer processing flows: coextrusion instabilities

interfacial instabilities: solve direct model -> convective instability forced system: Level-set + SUPG method

Valette, Laure, Demay, Fortin 2002

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Polymer processing flows: coextrusion instabilities

interfacial instabilities: solve direct model -> convective instability wavepacket: Discontinuous Galerkin method

Valette, Laure, Demay, Fortin 2001

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Polymer processing flows: coextrusion instabilities

interfacial instabilities: linear stability of 2 layer viscoelastic Poiseuille flow

Valette, Laure, Demay, Agassant 2004b

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SLIDE 66

Polymer processing flows: instabilities

flow instabilites interfacial instabilities