Polymer Dynamics and Rheology 1
Polymer Dynamics and Rheology Brownian motion Harmonic Oscillator Damped harmonic oscillator Elastic dumbbell model Boltzmann superposition principle Rubber elasticity and viscous drag Temporary network model (Green & Tobolsky 1946) Rouse model (1953) Cox-Merz rule and dynamic viscoelasticity Reptation The gel point 2
The Gaussian Chain Boltzman Probability Gaussian Probability For a Thermally Equilibrated System For a Chain of End to End Distance R By Comparison The Energy to stretch a Thermally Equilibrated Chain Can be Written Force Force Assumptions: -Gaussian Chain -Thermally Equilibrated - Small Perturbation of Structure (so it is still Gaussian after the deformation) 3
Stoke’s Law F = v ς ς = 6 πη s R F 4
Creep Experiment Cox-Merz Rule 5
6
Boltzmann Superposition 7
Stress Relaxation (liquids) ( ) ( ) = ε t Creep (solids) J t σ Dynamic Measurement δ Harmonic Oscillator: = 90° for all except = 1/ where = 0° ω ω τ δ Hookean Elastic Newtonian Fluid 8
9
Brownian Motion For short times For long times 0 10
11
12
http://mathworld.wolfram.com/First-OrderOrdinaryDifferentialEquation.html 13
14
15
16
The response to any force field 17
18
19
20
21
22
Both loss and storage are based on the primary response function, so it should be possible to express a relationship between the two. The response function is not defined at t = ∞ or at ω = 0 This leads to a singularity where you can’t do the integrals Cauchy Integral 23
24
W energy = Force * distance 25
D = ε 0 E + P = ε 0 ε E Dielectric Displacement Parallel Analytic Technique to Dynamic Mechanical (Most of the math was originally worked out for dielectric relaxation) Simple types of relaxation can be considered, water molecules for instance. ε 0 Free Space ε Material Creep: ε u Dynamic material Instantaneous Time-lag Response Response Dynamic: 26
Rotational Motion at Equilibrium τ relaxation A single relaxation mode, 27
Creep Measurement Response K = 1 τ http://mathworld.wolfram.com/First-OrderOrdinaryDifferentialEquation.html 28
Apply to a dynamic mechanical measurement ( ) d γ 12 t 0 J * ω ( ) exp i ω t ( ) = i ωσ 12 dt ( ) Multiply by i ωτ − 1 Single mode ( ) i ωτ − 1 Debye Relaxation 29
Single mode Debye Relaxation Symmetric on a log-log plot 30
31
Single mode Debye Relaxation More complex processes have a broader peak Shows a broader peak but much narrower than a Debye relaxation The width of the loss peak indicates the difference between a vibration and a relaxation process Oscillating system displays a moment of inertia Relaxing system only dissipates energy 32
Equation for a circle in J’-J” space 33
34
Time Dependent Equilibrium Value Value ( ) d γ dt = − γ t + Δ J σ 0 τ Lodge Liquid Boltzmann Superposition 35
36
37
Boltzmann Superposition 38
Rouse Dynamics Flow 39
40
41
42
Rouse Newtonian Entanglement Behavior Flow Reptation 12 ω ω 2 ω 1 57 ω 43
44
45
46
47
48
49
Rouse Newtonian Entanglement Behavior Flow Reptation 12 ω ω 2 ω 1 57 ω 50
Lodge Liquid and Transient Network Model Simple Shear Finger Tensor Simple Shear Stress First Normal Stress Second Normal Stress 51
For a Hookean Elastic 52
For Newtonian Fluid 53
54
55
Dumbbell Model ( ) ⎛ ⎞ dt 'exp − k t − t ' t ( ) = ( ) ∫ x t ⎟ g t ⎜ ⎝ ξ ⎠ −∞ 56
Dilute Solution Chain Dynamics of the chain Rouse Motion Beads 0 and N are special For Beads 1 to N-1 For Bead 0 use R -1 = R 0 and for bead N R N+1 = R N This is called a closure relationship 57
Dilute Solution Chain Dynamics of the chain Rouse Motion The Rouse unit size is arbitrary so we can make it very small and: With dR/dt = 0 at i = 0 and N Reflects the curvature of R in i, it describes modes of vibration like on a guitar string 58
x, y, z decouple (are equivalent) so you can just deal with z dz l ς R dt = b R ( z l + 1 − z l ) + b R ( z l − 1 − z l ) For a chain of infinite molecular weight there are wave solutions to this series of differential equations ⎛ ⎞ z l ~ exp − t ( ) ⎟ exp il δ ⎜ Phase shift between adjacent beads ⎝ ⎠ τ Use the proposed solution in the differential equation results in: sin 2 δ τ − 1 = b R ) = 4 b R ( 2 − 2cos δ ζ R ζ R 2 59
sin 2 δ τ − 1 = b R ) = 4 b R ( 2 − 2cos δ For N R = 10 ζ R ζ R 2 z l = z l + N R Cyclic Boundary Conditions: N R δ = m 2 π N R values of phase shift δ m = 2 π ⎛ ⎞ m ; m = − N R ⎟ ,..., N R 2 − 1 ⎜ ⎝ ⎠ N R 2 60
sin 2 δ τ − 1 = b R ) = 4 b R ( 2 − 2cos δ For N R = 10 ζ R ζ R 2 z l − z 0 = z N R − 1 − z N R − 2 = 0 Free End Boundary Conditions: dz ) = dz ( ) = 0 ( dl l = 0 dl l = N R − 1 ( ) δ = m π N R − 1 N R values of phase shift π ( ) δ m = m ; m = 0,1,2,..., N R − 1 ( ) N R Rouse Modes of order “m” N R − 1 61
Lowest order relaxation time dominates the response ⎛ ⎞ ζ R ⎜ ⎟ ⎝ ⎠ 2 a R 1 τ R = 4 R 0 3 π 2 kT ⎛ ⎞ ζ R This assumes that ⎜ ⎟ ⎝ ⎠ 2 a R is constant, friction coefficient is proportional to number of monomer units in a Rouse segment This is the basic assumption of the Rouse model, 2 ~ N ζ R ~ a R = n R N R 62
Lowest order relaxation time dominates the response ⎛ ⎞ ζ R ⎜ ⎟ ⎝ ⎠ 2 a R 1 τ R = 4 R 0 3 π 2 kT 2 = a 0 Since 2 N R 0 τ R ~ N 2 kT 63
The amplitude of the Rouse modes is given by: 2 2 = 2 R 0 Z m 3 π 2 m 2 The amplitude is independent of temperature because the free energy of a mode is proportional to kT and the modes are distributed by Boltzmann statistics ⎛ ⎞ ) = exp − F ( p Z m ⎜ ⎟ ⎝ ⎠ kT 90% of the total mean-square end to end distance of the chain originates from the lowest order Rouse-modes so the chain can be often represented as an elastic dumbbell 64
Rouse dynamics (like a dumbell response) Dumbbell Rouse ⎛ ⎞ dU ⎜ ⎟ ζ R ⎝ ⎠ + g ( t ) = − k spr x τ R = dx dx dt = − + g ( t ) 4 b R sin 2 δ ζ ζ 2 π t dt 'exp − t − t ' ⎛ ⎞ ( ) = ( ) ∫ δ = N R − 1 m , m=0,1,2,...,N R -1 ⎜ ⎟ x t g t ⎝ ⎠ τ −∞ τ = ζ k spr 65
Rouse dynamics (like a dumbell response) ( ) g t 2 ( ) = 2 D δ t ( ) where t = t 1 − t 2 and δ ( ) is the delta function whose integral is 1 g t 1 D = kT Also, ζ ⎛ ⎞ kT exp − t ⎜ ⎟ τ = ζ x 2 = kT ⎝ ⎠ τ ( ) x 0 ( ) = For t => 0, x t k spr k spr k spr 66
Predictions of Rouse Model − 1 ( ) ~ t G t 2 1 ( ) ( ) ~ ωη 0 G ' ω 2 π 2 η 0 = kT ρ p τ R 12 ~ N 67
Rouse Newtonian Entanglement Behavior Flow Reptation 12 ω ω 2 ω 1 57 ω 68
Dilute Solution Chain Dynamics of the chain Rouse Motion Predicts that the viscosity will follow N which is true for low molecular weights in the melt and for fully draining polymers in solution Rouse model predicts Relaxation time follows N 2 (actually follows N 3 /df) Diffusion constant follows 1/N (zeroth order mode is translation of the molecule) (actually follows N -1/df ) Both failings are due to hydrodynamic interactions (incomplete draining of coil) 69
Dilute Solution Chain Dynamics of the chain Rouse Motion Predicts that the viscosity will follow N which is true for low molecular weights in the melt and for fully draining polymers in solution Rouse model predicts Relaxation time follows N 2 (actually follows N 3 /df) 70
Hierarchy of Entangled Melts Chain dynamics in the melt can be described by a small set of “ physically motivated, material-specific paramters ” Tube Diameter d T Kuhn Length l K Packing Length p http://www.eng.uc.edu/~gbeaucag/Classes/MorphologyofComplexMaterials/SukumaranScience.pdf 71
Quasi-elastic neutron scattering data demonstrating the existence of the tube Unconstrained motion => S(q) goes to 0 at very long times Each curve is for a different q = 1/size At small size there are less constraints (within the tube) At large sizes there is substantial constraint (the tube) By extrapolation to high times a size for the tube can be obtained d T 72
There are two regimes of hierarchy in time dependence Small-scale unconstrained Rouse behavior Large-scale tube behavior We say that the tube follows a “ primitive path ” This path can “ relax ” in time = Tube relaxation or Tube Renewal Without tube renewal the Reptation model predicts that viscosity follows N 3 (observed is N 3.4 ) 73
Without tube renewal the Reptation model predicts that viscosity follows N 3 (observed is N 3.4 ) 74
Recommend
More recommend
