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

Examples of polymers Polyethylene Polystyrene Polypropylene 3

Phenomenology Density : ~water « Melting » temperature : ~150 ° C (PE, PP, PS) Viscosity : ~10 6 water « Internal » structure : remains/relaxes over differents timescales → memory, elasticity → → → « High » temperature : viscous + elastic → visco-elastic fluid → → → 4

« Internal » structure ? Macromolecules , ex PE : -[CH 2 ]- n etc … … 10 000 CH 2 groups ; 10 18 molecules per cm 3 ; very flexible Polydispersity : distribution molecular mass 5

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 ? 6

Polymer processing flows: extrusion flow instabilites interfacial instabilities

Which material ? macromolecule : 0.2 nm ... close view ... 8

Which material ? macromolecule : ... further away. 20 nm 9

Which material ? macromolecule : ... further away : free segments 20 nm 10

Flexible chain Degrees of freedom not dof dof θ φ « Free » equivalent segment = � � = r , r b i i � Conformation = { }= "random flight" of N segments of length b r i � � ∑ = R r i ↔ ↔ ↔ ↔ 11

Entropic elasticity � � � � � kT R ∑ ( ) = ≅ R r F R i Energy kT R 0 R 0 → Relaxation: diffusion in the space of R 12

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

Entropic elasticity � � ∑ = Statistical conformation = = end-to-end vector R r i � Mean value : = R 0 � � � ∑ ∑ ) ( ) ( 2 2 2 Mean square : = = = R r r Nb R i j 0 3  −  2 3 R     � 2 ( ) 3     2   2 R Gaussian density distribution : ψ = R e 0   2  π  2 R 0 � ( ( ) ) Statistical entropy : = Ω ψ S k ln R tot � � � dS 3 kT ( ) � Thermodynamic force : = − = F R T R 2 d R R 0 14 Doi-Edwards 1986

Elastic dumbbell � � � 3 kT ( ) Force: = F R R 2 R 0 kT = 3 3 kT Spring constant: 2 2 Nb R 0 kT kT Einstein relation: = ζ = D ζ N b 2 2 2 ζ R N b b = Relaxation time: 2 0 θ = = θ N 0 D kT 15 Doi-Edwards 1986

Macroscopic scale � � � 3 kT ( ) Force: = F R R 2 R 0 chains per unit volume ν chains per unit surface ν R � � 2 3 kT R Stress: 2 ν = ν R 3 kT 2 2 R R 0 0 Elastic modulus: = ν G kT 0 Relaxation time: 2 θ ∝ N 16 Doi-Edwards 1986

Relaxation function : experiment 1 on viscoelasticity Response to a sudden macroscopic deformation Stress decreases with time because microscopic strain γ micro decreases Deformation a t=0 : stress = G 0 x strain Sample at rest (the arrows are proportional to the chains oriented and stretched in their direction) γ macro γ micro Whatever γ macro : G(t) is constant (same ratio stress/strain)

Relaxation function : experiment 1 on viscoelasticity Stress relaxation is faster when far from equilibrium τ d 1 = − τ θ dt γ d 1 micro = − γ micro θ dt

Linear (separable) response of a simple viscoelastic fluid Step strain : log shear modulus t=0 rest, t>0 constant shear γ macro θ G G 0 0 10 − 2 s log time − θ t / τ γ = = / G ( t ) G e 0 fraction of unrelaxed segments 19

Viscosity function : experiment 2 on viscoelasticity Response to a continuous macroscopic deformation γ micro = � τ = G 0 � γ macro θ γ macro θ Stress = , � γ macro γ micro

Viscosity function : experiment 2 on viscoelasticity Response to a continuous macroscopic deformation drag relax 1 = � γ γ − γ d dt dt micro macro micro θ � γ = θ γ Stationary : micro macro � τ = γ = θ γ G G Stress : 0 micro 0 macro η = θ G Viscosity : 0

Response in steady shear Steady shear: log shear stress t=0 rest, elastic: slope G Viscous: viscosity G θ t>0 onstant shear rate, θ G G � = γ V t − / θ � t τ γ = θ − / G ( 1 e ) log time ~ deformation

Steady shear � γ Shear rate: � Drag (affine deformation)/relaxation balance → stress: τ = γ θ G 0 2 Viscosity: η = θ = ν ζ G Nb N 0 b Weissenberg effect: 23

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

Long chains are entangled Entanglements: topological constrains between chains � � kT R kT ≅ ≅ F R R a a 0 0 Courtesy K. Kremer, from Everaers et al., Science, 2004

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 2 2 ζ L N N Disengagement time : 3 θ = ∝ b ∝ θ N d 0 D kT a = 10 − 9 m Z = 5 − 50 N / Z = 25 θ Z = 1 ≈ 0.1 s a

Macroscopic scale Elastic modulus: = ν G kT N 0 e 3 Relaxation time: θ d ∝ N a 3 kT Tension on each segment: a 30

Linear response of an entangled polymer melt Step strain : log shear modulus t=0 rest, fast retraction along tube contour 9 10 Pa reptation (disengagement) t>0 constant strain γ 6 10 Pa macro 10 − 2 0 s 10 s log time 2 3 ∝ ∝ N N 31

Linear response of an entangled polymer melt Step strain : log shear modulus t=0 rest, retraction 9 10 Pa reptation t>0 constant strain γ 6 10 Pa macro distribution 10 − 2 0 s 10 s log time 2 3 ∝ ∝ N N 32 molecular mass

Linear response of an entangled polymer melt Step strain : log shear modulus t=0 rest, retraction 9 10 Pa reptation t>0 constant strain γ 6 10 Pa macro 1 θ G G 1 1 2 θ G G 2 2 10 − 2 0 s 10 s log time 3 θ G G 3 3 2 3 ∝ ∝ N N G θ 33 ... Spectrum fitted using conventional rheometers i i

Steady shear � γ < θ 1 / Shear rate: d � Drag (affine deformation)/relaxation balance → stress: τ = γ θ G N 0 d η = G θ Viscosity: 0 N d � γ > θ Shear thinning: 1 / rotation of segments d log viscosity θ G N 0 Doi-Edwards 1986 -1 34 log shear rate θ 1 / d

Rheometric tools Cone-plate rheometer: mechanical spectroscopy (linear regime) Capillary rheometer: steady shear viscosity Piston Pressure transducer 35 Die

Flow-induced birefringence: stress measurements Stress-optical law: Sodium source k λ λ = 589nm ∆ σ = Cw polarizer Schweizer et al., Rheo Acta 2005 45 ° λ /4 Robert et al. 2003 w 135 ° λ /4 analyzer Used to test pertinence Picture of constitutive equations 36

Constitutive equation (3D) � � T R R Stress: (autocorrelation) τ = = 3 G 3 G B N 0 N 0 2 R 0 � � � � ( ) ( ) � � T T R R R R ∫ B conformation tensor: 3 = ψ R d R � � � 2 2 R � R R � T � 0 0 R R n = τ = d f n ds 3 G ds 1 N 0 2 At rest: = R B I 0 3 � � � � � � ( ) T T � � ∇ ∇ dB v R R R v R Affine deformation: T = + = ∇ + ∇ v B B v 2 2 dt R R 0 0 � � d ( ) = 0 ( ) d Because when no relaxation R 3 ψ R 37 dt

Constitutive equation (3D) � � T R R Stress: τ = = 3 G 3 G B N 0 N 0 2 R 0 � � � � ( ) ( ) � � T T R R R R ∫ B conformation tensor: 3 = ψ R d R � � � 2 2 R � R R � T � 0 0 R R n = τ = d f n ds 3 G ds 1 N 0 2 At rest: = R B I 0 3 � � � � � � ( ) T T � � ∇ ∇ dB v R R R v R Affine deformation: T = + = ∇ + ∇ v B B v 2 2 dt R R 0 0   � � Upper-convected dB 1 1 Relaxation:   T = ∇ + ∇ − − v B B v B I Maxwell model 38   θ dt 3