Institute for Chemical Technology and

[email protected] http://www.itcp.kit.edu/wilhelm/

Introduction to

Prof. Dr. Manfred Wilhelm

private copy 2019

KIT – Universität des Landes Baden-Württemberg und nationales Forschungszentrum in der Helmholtz-Gemeinschaft www.kit.edu

Contents (overview)

 Motivation, Literature, Journals

 First principles

 Simple models: Maxwell, Voigt, Burger, Carreau, Ostwald - de Waele

 Glossary

 Rheological hardware

 Examples: Dispersions (response and phenomena), Polymer melts, ...

 Fourier-Transformation

 FT-Rheology

Contents

Literature: ...... 1 Books ...... 1 Journals ...... 2 Internet ...... 2

Definition of the term “Rheology” ...... 3 Typical examples of daily live: ( motivation) ...... 3 1) Brush with on a wall ...... 3 2) Piston in an engine ...... 4 Why can we assume that Hooke’s law could be correct? ...... 5 Hooke for (rubber ) ...... 8 Why can we assume that Newton’s law could be correct? ...... 10 Gedankenexperiment ...... 11 Linear models: Hooke, Newton, Maxwell, Kelvin-Voigt … ...... 14 Detailed analysis of Maxwell model ...... 16 Without any mathematics: step experiments (step in or step in strain) ...... 27 Memory (Gedächtnis) ...... 28 Multimode models ...... 29 Glossary ...... 31 a) Lamellar flow ...... 31 b) ...... 31 c) Cox-Merz-rule ...... 32 d) Lissajous figures ...... 33 e) ...... 34 Ostwald-de Waele (example for 2 parameter model) ...... 34 Carreau (example for 3 parameter model) ...... 35 4 parameter models:...... 35 shear thinning + long memory ( Hysteresis) ...... 36 Shear thickening ( rheopex  dilatancy) ...... 36 Anti-thixotropy shear thickening + memory ( Hysteresis) ...... 36 Rheopexy ...... 37 Dilatancy ...... 37

I Bingham ...... 37 Dimensionless groups ...... 38 ...... 39 Péclet number ...... 40 Taylor vortex ...... 40

What do we expect for (p,T)? ...... 42 ...... 42 of , dependence ...... 44 Stress-strain tensor and normal ...... 46 Definition of the extra stress tensor (right handed system!) ...... 48 Properties of the extra stress tensor ...... 48 What do normal stress differences mean? ...... 49

What do we expect for N1,2 γ, γ0 ? ...... 50 Phenomena where we can directly “see” normal forces ...... 51 a) Rod-climbing ...... 51 b) Secondary flow for rotating disc ...... 52 c) Extrudate swell ...... 52 Possible measurements (for oscillatory ) and hardware ...... 53 1) Detection of onset of non-linearity at fixed frequency ...... 53

2) Measurement of G’, G” at T = const., : variable, 0: parameter ...... 53 3) Temperature dependent measurement ...... 54 4) Shear rate dependent viscosity ...... 55 Hardware: ...... 55 Couette geometry ...... 55 Hardware ...... 58 Stress and strain , typical types of construction: ...... 58 Typical hardware specifications (ARES) ...... 59 Typical pathway of a signal from the transducer to G’, G” ...... 60 Vane rheometer ...... 61 Melt-flow index ...... 62 Capillary rheometer ( high shear rates) ...... 62 Elongational rheology, viscosity ...... 63

II Rheology on two specific examples: polymers and dispersions ...... 66 Polymers ...... 66 Reptation theory ...... 66 Typical shape for G’(), G”() for monodisperse linear polymer melts ...... 69 Time-Temperature-Superposition (TTS) and the Williams-Landel-Ferry (WLF) equation 73 Dispersions ...... 77 Fourier-Transform-spectroscopy ...... 88 Problem of discretisation (ADC, analogue digital converter) ...... 89 Some important mathematical relations ...... 90

Appendix A ...... 93 Appendix B ...... 115 Appendix C ...... 124

III Literature:

Books

Einführung in Rheologie und Rheometrie (also available in English) Gebhard Schramm, Gebr. Haake GmbH, Karlsruhe (easy; book to start with)

Das Rheologie Handbuch (also available in English) Thomas Mezger, Vincentz Verlag, 2000 (easy, covers lots of practical problems, nice hardware section)

Rheology for , an Introduction J. W. Goodwin and R. H. Hughes, Royal Society of Chemistry 2000 (easy)

A Handbook of elementary Rheology Howard A. Barnes, University of Wales, Institute of Non-Newtonian Mechanics, Aberystwyth 2000, (good overview, very elaborated literature at the end)

The structure and rheology of complex Ronald G. Larson (Head of society of rheology), Oxford University Press 1999 (more advanced)

Rheology Principles, Measurements and Applications Ch. W. Macosko, Wiley-VCH 1994 (more advanced)

Engineering Rheology R. J. Tanner, Oxford University Press 2000 (for mechanical )

1 Rheological measurements A. A. Collyer and D.W. Clegg, Chapman & Hall 1995 (Hardware)

Rheology: A Historical Perspective R. I. Tanner and K. Walters, Elsevier 1998 (lots about people and phenomena)

Journals

Journal of Applied Rheology http://www.ar.ethz.ch/ (incl. Jobs!, Hardware guide, reports about upcoming and previous conferences)

Rheologica Acta (Springer) http://www.springerlink.com/

Journal of Rheology (The Society of Rheology) http://scitation.aip.org/joro/

Journal of Non-Newtonian http://www.elsevier.com/

Internet www.rheologie.de www.rheology-esr.org www.rheology.org

2 Definition of the term “Rheology”

Rheology is the of and flow of matter.

Side conditions: conservation of , conservation of , constraints, incompressibility;

Analysis of: Deformations: strain (shear), stretch (elongation); stress (torque); normal forces;

Typical examples of daily live: ( motivation)

1) Brush with paint on a wall v

v = 1 m/s = 1000 mm/s d = 0.2 mm d

What is the relevant quantity?

Assumption: layered structure vn

n layers vi

di

v i = constant for all i! di

3 v v v 1000 mm/s 1  i  n    5000 di d n d 0.2 mm s v  shear rate γ  [1/s] d Why is this a rate and not a frequency? Frequency is only used with respect to periodic phenomena, otherwise: rate! both: [1/s] !

2) Piston in an engine d frequency: ω 1 s v 1  6000 2π min (=rpm, rotations per minute) ω 1 1  100 2π s stroke (german: Hub): s = 10 cm = 0.1 m ω m m 1  v  1 s2 1000.12  20 2π  s   s  m m v  ( / 2)20  30 why  / 2 ? max  s   s  d  20 μm  20106 m

30 m vmax s 6 1  γ max   1.510 d 20106 m s

think about: - shower lotion - lipstick - coating of paper - extrusion of fibres (clothing)

4 unit of γ : [1/s] = inverse time v

d 1 comparison:  λ γ p

Polymer with relaxation time p γ  λp is a unitless quantity

Why can we assume that Hooke’s law could be correct?

Do we “buy” this law? Hooke: F  k  x F

0x Possible reasons:

! a) F  F(x,...)  F(x) (assumption)

Taylor-expansion (Taylor around x = 0  MacLaurin series)

 F  1  2F  F  F     x     x 2 (x) (x0) x 2! 2   x 0  x x 0 0, k vanishing for because small x at equil.

linear nonlinear part b) C C C C C C C C C

Interaction Potential for vibrational (IR) spectroscopy (beside Hooke): Morse-Potential U(x) (Potential, not !) U  F  U  Fdx x 

5 Potential has !

2 Morse: U(x)  A(1 exp(β(x  x0 )))

U/A U/A

1

1-exp(-(x-x0)) exp(-(x-x0))

x

1

x X0=0 X0=0

U/A

1

2 (1-exp(-(x-x0)))

Note: A: Dissoziation energy x X0=0

set: x0 = 0 β2x 2 β3x3 Taylor: eβx  1 βx    ... 2! 3!

2 2 2 2  U (x)  A(1 (1β  x ...))  A(β  x ...)  Aβ x U F   2Aβ2x  k  x x k a) + b)  no proof, but we “buy” Hooke’s law exercise: prove Hooke’s law for the finite extendable nonlinear elastic interaction (FENE),

frequently used in computer simulation, for x  R0

2   x   U  A  ln1    (x)   R     0   6 Remark: If we remember typical force-constant from IR ( spectroscopy books)

k k = 500 N/m ( ω  ) m and we remember typical area needed for a chain, e.g. : orthorhombic a = 7.5 Å 5Å b = 5 Å c = 2.5 Å

7.5 Å

2 chains per unit cell

o 2 o 2 A 7.55A 37.5A o 2    20A  201020 m2 chain 2 2

F

different A F

different x and L

F

L

Renormalization to area + relative change in length F x  σ  E   + E: unit: [ 1Pa = 1 N/m2 ] A L

stress

E-module 7 upper limit: F k  x x   σ  E  A A L k 5001020 1010  N m   L  E   250109 Pa  250 GPa A 20 mm 2 

C in unit cell  10-10 m (one bond) Tungsten (W): 150 GPa

C But: modes are weaker  only several GPa C C

Hooke for polymers (rubber elasticity)

Start of chain in coordinate origin, where is end?

W(r)  Gauss

1  (x  μ)2  W  exp  , here  = 0 (r)  2  2π σ  2σ 

Boltzmann: S  k  ln(W)  C  k  ln(exp(-x2 ))  C  k  x 2 with G  H - T S, H  0

G  - T (C - k  x 2 )

G: units of energy: ΔG F, (W Fdx)  F  T k 2 x (temp. + elongation!) x 

only needed: H  0 ; W(r)  Gauss

See analogy for Gauss in crystallography!  Debye-Waller factor!

8

9 Why can we assume that Newton’s law could be correct?

x F v Newton:  σ  η  η γ rough surface A d

Note: In Rheology Newton’s law is associated with σ  η γ ,

not with his other law: F  mx  ma .

F Why not:  x ? A F  a, a   x ? Do we “buy” this? 1 π F  v 3  a ?? 2 why?

We need proportionality between viscous force and : Fviscous  v

Remember: Law from Stokes: F  6π ηr  v (F: e.g. gravity)

F  Sphere in viscous media 2r v

Note: unit of : F pd Pa ms η  A   η Pa s , old: Poise: 1Pa s = 10 P; 1 cP = 1 mPas (Poiseuille); v v  m  d typical values: blood 100 – 4 mPas (thicker than water!; shear thinning) Glycerin 0°C 10,000 mPas 20°C 1,400 mPas T   60°C 60 mPas Oil, SAE 10 30°C 200 mPas

H2O 1 mPas ( memorize!) Air 0.02 mPas

10 Gedankenexperiment (  Prof. Sillescu, article Lord Rayleigh 1891! see Appendix A, p. 93-114)

Tube; big mass M; lots of particles with small mass m strike on mass M; M is moved with vM; What force is needed?

m m M vm = v vM - vm = - v

For M: ΔlM  vM Δt For m: Δl  vΔt , same velocity for all small particles!

After time t: lM

particles lM l

l

l - lM

N Average of particles m: ρ  , number density, not mass density! Δl

l

The mass M is hit by the following number of particles during t: 1 1 N   N   ρΔl  Δl  ρΔl  Δl  ρΔl  N 2 M 2 M

50:50 probability that particles fly in correct direction

direction of particles but N- > N+ !!

11 if we define a clash-rate: N N Δt N Z Z   ρ     Z  ρ v Δt ΔlΔt vΔt v

1 1 v  v Z  ρv  v  Z M 2 M 2 v 1 1 v  v Z  ρv  v  Z M 2 M 2 v

Z  Z  Z  ρ v

  each particle transfers elastic impact onto mass M with relative p  2mv

p  2mv  v mv M  p  2mv  vM mv

In one time unit t, this balances the outer force F needed to push mass M with velocity vM.

F  Z p  Z p

1 v  v 1 v  v F  Z M 2mv  v  Z M 2mv  v 2 v M 2 v M Zm   v  v v  v  v  v v  v v M M M M

Zm 2 2 2 2  v  2vvM  vM  v  2vvM  vM v Zm   4vv  4mZ v v M M

 F  vM is proportional to the velocity of the mass M.

12

Cf tance and to treat a case of motion in a viscous fluid.” magnitude of the force required “Newton was the first to formulate a hypothesis regarding the Princi pia i iSc I Sect ii Lib Sir 1642 – 1727 – 1642 mlHatschek Emil X to overcome viscous resis-

13 Linear models: Hooke, Newton, Maxwell, Kelvin-Voigt … incl. oscillatory excitation and response

Hooke – spring

,  d dt

σ  G  γ  σ  G  γ

Pa no unit

Newton – dash-pot

dγ σ  η  γ, γ  dt

σ,γ  γ

Math. def. of linear models: A linear model is a mathematical description of the relation between stress and strain (respective: ) where only linear terms of γ1 or γ 1 are used. Further more G and  are constant.

Experimental def.: Linear response can be assumed if the response (stress, strain, strain rate) is large enough to be detected but still in a regime where G and  are not affected by the measurement.

14 The non-linear regime should be avoided for linear response measurements:

 asymptotic deviation!

rate sweep: η γ  linear regime

γ

G G’ strain sweep: Gγ ,Gγ  G’, G”  later 0 0 linear regime G’’ fixed frequency

0

Dash-pot (DP) and spring (S) can be arranged in series or in parallel:

σ  σ γ  γ S DP S DP viscosity with a elasticity with bit of elasticity a bit of viscosity (long term) (long term) γ  γ  γ σ  σ  σ S DP S DP γ  γ  γ σ  Gγ  ηγ S DP

γ, γ, G, η

σ Maxwell model Kelvin-Voigt model (for liquids with some (for with some γ, γ, G 0 , η elastic response) viscous response) σ

15 Detailed analysis of Maxwell model

σ σ  G  γ  γ  S G σ σ  η  γ  γ  DP η σ σ γ   (1) G η

1. step-experiment 

0

0 t at time t > 0, γ  0 (not in the dash-pot, but overall system!)

σ σ using (1): 0   G η 0 σ σ  G 0 η dσ G   0 dt (see: first order kinetic, or Lambert-Beer) σ η σ(t) 1 G t  dσ   0 dt σ(0) σ η 0

G 0 ln(σ(t) )  ln(σ(0) )   t η σ(t) G ln   0 t relaxation time σ(0) η

G 0 t η Pa s  σ(t)  σ(0) exp( t)  σ(0) exp( ); τ   s η τ G  Pa  (for short time force is fully in spring) σ(0)  G 0  γ G σ G σ  G  γ exp( 0 t)  (t)  G  G exp( 0 t) (t) 0 η γ (t) 0 η

limG (t)  G 0 t0  G exp( 0 t) η

G 0 

Memory!

0 t 16 Oscillatory response:

Hooke

, 

γ  γ0 sin(ω t)

0

t

  distinguish: amplitude  elongation! System has  memory ( stored energy  storage modulus G’)

Newton

, , γ phase shift  γ  γ sin(ω t)  0 γ  γ ωcos(ω t) γ 0 0

t

distinguish: 2π  [rad/s]  , f [1/s = Hz] ω   2π  T System has no memory ( energy is lost  loss modulus G’’)

17 Maxwell:

γ(t)  γ0 exp(iω t)

γ  γ0 (iω)exp(iω t)  i ω γ(t) after initial time we reach dynamic steady state:

σ(t)  σ0 exp(i(ω t  δ)

σ  σ0 (iω)exp(i(ω t  δ)  iωσ(t) σ σ eq. (1): γ   G η

iωσ(t) σ G iω γ(t)    G η iωσ(t)

G  γ(t) G γ(t) 1 1 , def.:  * σ(t) i ωη σ(t) G ( ) G 1 η * 1 , τ  G ( ) i ω τ G G 1 i ω τ  G* iω τ iω τ G*  G  sep. into real and imag. part :(1-iω τ) 1 iω τ iω τ(1-iω τ) G*  G  iG  G  ( ) ( ) (1 i ω τ)(1-i ω τ) ω2  τ2  iω τ  ω2  τ2 ω τ   G   G   i  2 2  2 2 2 2  1 ω  τ 1 ω  τ 1 ω  τ  ω2  τ2 G  G  1 ω2  τ 2 ω τ G  G  Note: (a+b)(a-b)=a2-b2 1 ω2  τ 2 Plot: G’, G” (linear scale) G’, storage G

G/2 G”, loss module Why G’ = storage ?

Why G” = loss ?

log() [rad/s]  = 1

18 ω2  τ2 limG  lim  limω2  τ2  ω2 ω0 ω0 1 ω2  τ 2 ω0  1

G  ω2 for small 

ω τ limG  lim  limω τ  ω1 ω0 ω0 1 ω2  τ 2 ω0  1

G  ω1 for small  trick to memorize: G a  ωb , a + b = 3

log G

1 G” 2

G’

 = 1 log 

Note for “NMR-People”: t FID: M  exp( )  Lorentz  shape in  T2

Lorentz: Re (ˆ G ) τ Re  G” ( ) ω2  τ2 1 Factor  - ω τ2 missing G’ Im( )  2 2  ω  τ 1 0

Im (ˆ G ) in Rheology:   0 !!

19 * G ω0  Gω0  G (module of spring within Maxwell-element) width of relaxation spectrum for G”: G” ω τ  G  G  2 2 ω  τ 1

 = 1 log 

Set  to  = 1: ω G  G  ; G = 1 ω2 1

dG 1 ω Maximum at: ( )  0   2ω  0 dω ω2 1 2 2 ω 1 2 1 2ω 2 2  2 ω 1 ω 1 ω2 1  2ω2 1  ω2 1 ω2 1  2ω2 1  ω2 ω  1 physically meaningful: ω  1 1 1 G(ω1)  G  G 11 2

full width at half maximum, 1, 2 ? fwhm 1 G 2 1 G fwhm 4

 log 

1 ω  4 ω2 1

1 1 ω2   ω 4 4 1 1 ω2  ω   0 4 4 20 remember: ax 2  bx  c  0  b  b2  4ac x1/2  2a

1 1 4 1  1 ω  4 4 1/2 1 2    21 3   4 

 2  3

ω1  3.732

ω2  0.268

ω ratio : log 1 1.14 10 ω 2

The full width at half maximum (fwhm) of a single exponential relaxation is 1.14 decades in frequency space.

* * σ G ( )   G  iG γ* ( ) ( )

* * *  σ  G  γ ; γ  γ0 exp(iω t), γ  iω γ0 exp(iω t)

* G( )  iG( )   σ     γ  iω 

 η*

 iωη*  G* units: [1/s  Pa  s = Pa] [to memorize: “iong”, i omega n equals G]

η*  η  iη η*  η2  η2

ω η*  G*

21 G Phase lag : tanδ  G experimental advantage: G”, G’ extensive quantities tan  intensive quantity if, e.g. filling factor is “bad”, G’ is wrong, G” is also wrong, but G”/G’ is still accurate  tan  is generally very reproducible

G  G  10%  typical error margin for rheological measurements G  G

22

23

24

25

26 Without any mathematics: step experiments (step in stress or step in strain)

0 t

σ γ (t0)  η σ γ  (ta ) G

 F, , 

0 ta t Kelvin-Voigt model

 2nd possibility: strain step:

0 t  t

 t  ~ exp  spring τ elastic part   memory?  spring, viscous part  elastic part t

0 t F, , 

Maxwell - model

27 More complex models (but still linear models!):

G1

 0 t G2 1

2 1, G2 1 2

2

 G1

0 t

Burger - model

Memory (Gedächtnis)

The memory of the system might be defined for a step strain experiment as follows: dσ dG  γdt dt

dG (t) M   , minus sign, so that M(t) is positive; (t) dt  dσ  M  γ dt

σ tt  σt   dσ  -  M(t-t )γ t  dt 0 t- Memory depends only on elapsed time: t – t’ = s, dt’ = -ds

 σ  - M γ ds (t)  (s) (t-s) 0 exchange of limits and: dt’ = -ds

28 for infinite small motion: dσ  Gdγ dγ dσ  G dt  Gγdt dt t σ  G γ dt, s  t  t  (t-t) t -  σ  G γ ds  (s) (t-s) 0 if we have a modulus function with an exponential memory:

t  - t   - (t - t)  monomodal: G (t)  G 0 exp   σt  G0exp  γ t dt τ    -  τ  Improvement of this model: several relaxation times

Multimode models N   t   G(t)   Gk exp  , N - mode model k1  τk 

t N   t    σ    G k exp  γ (t ) dt  k 1  τk  picture for multimode Maxwell-model: 1 2 3 4

 = 1 = 2 = 3 = 4

 = 1 + 2 + 3 + 4

F, , 

Also possible: multimode Kelvin-Voigt (several Kelvin-Voigt models in series);

Under oscillatory shear, a multimode Maxwell-model will respond as follows (see next page) remember: fwhm for single Maxwell: 1.14 decades  spacing ?! in  not uncommon for polymers: 5 – 7 decades relaxation time distribution

29

H. M.Laun

30 To reduce the need of maths for a while, a glossary on important rheological terms is inserted:

Glossary

a) Lamellar flow

x γ  x h

1 dx 1 h γ     v h dt h flux r0 2 2 4 For this model: γ  γ (h) , in contrast: tube 0 vr  r0  r  I  r0 !! Hagen-Poiseuille r0

Shear deformation is equally distributed throughout the sample. For sliding plate geometry: the points of similar elongation amplitude form lamellae. For high shear rates, generally instabilities can occur (Reynolds number) and the lamellar flow profile is disrupted. Other possibility: plug-flow (tooth !) v = const.

b) Reynolds number

The Reynolds number describes the ratio between the of a system and the energy lost by viscous flow. E Re  kin , for Re > 2000 we find transition between lamellar ( F  γ ) and turbulent E viscous ( F  γ 2 ) flow! 2rρ v For a capillary (diameter: circle) we find: Re  η

r: radius, : density, v: avg. velocity, : viscosity

31 example: Aorta (main blood vessel close to the heart): r = 1 cm = 0.01 m  = 4 mPas = 0.004 N/m2  s ; 1 N = 1 kgm / s2  = 1000 kg/m3 v = 0.3 m/s

2 10-2 103 3101 m s2 m2 kg m Re  -3  3  1,500 ,close to transition: lamellar  turbulent! 410  kg m sm s 

c) Cox-Merz-rule

The Cox-Merz-rule is an empirical rule that connects the shear rate dependent viscosity with the absolute value of the frequency dependent complex viscosity, as calculated by:

* * iωη ( )  G ( )

* η(γ)  η (aω) , a  1 (experimentally) via: *  η(γ)  η ω ; note : ω  frequency ν ; ω  2π  ν

This rule holds only for rheologically simple materials!

 2  2 * Gω  G ω ηγ  η ω  ω ω2  τ2  G ( )  G  2 2 e.g. Maxwell-model: 1 ω  τ ω τ G  G  ( ) 1 ω2  τ2

G’ G0

G /2 0 G”

 = 1 log()

G” leading term G’ leading term 32 2 2 2 * G ω  τ lim η   lim   const.  η0 ω0 ω0 ω ω

2 * G const. 1 lim η   lim   ω ω ω ω ω

* log( ( ) ) 0 η for single Maxwell-model

-1

 = 1 log()

d) Lissajous figures

oscillatory shear different representations σ (t) σ max  1 

t 1 γ (t) γ max

 ~ cos (t),  ~ sin (t) vector description of circle general for ellipse: y

x  cos(ω t)  b      a  y cos(ω  t  δ) x δ b phase lag tan  2 a

33 Linear response: ( ellipse) Contains symmetry elements for Lissajous figure: 2 mirror plains + point symmetry

In case of non-linearity: only point symmetry ( I(31), I(51), …)

Note: deviations < 2-3% of sinusoidal response can generally not be seen in Lissajous figures!!  much less sensitive compared to FT-Rheology (see later)

e) Shear thinning

(deutsch auch: Strukturviskos), pseudoplastic monotonically decaying viscosity flow curves; viscosity as a function of shear rate in steady state, so no implicit memory involved.

log η η 0 = Newton η 0 2

-a η η  0 1 β  γ c η  b  γ a

1 log γ

β

ηγ 1  η γ 2 for γ 1  γ 2

To describe shear-rate dependent viscosity, empirical equations with 1, 2, 3, 4 parameters are used, e.g.:

1 parameter: Newtons law!   = 0

Ostwald-de Waele (example for 2 parameter model) η  b  γ a a: scaling parameter, shear thinning exponent; a[0, 1] : 0  Newton 1  max. shear thinning exponent

34 if a = 1: F  η  γ  b  γ -1  γ  b force independent of γ , force is constant! at γ  1  η  b

Carreau (example for 3 parameter model) η η η  0 also other def. for Carreau: η  0 (not equal!) 1 β  γ c 1 β  γ c c: scaling parameter c[0, 1] : pivot point (knee), 1 η η η  0 0 0 if γ   η  c  c  β  1  11 2 1 β    β 

4 parameter models: - further parameter needed to: e.g. model the width of the knee as the next parameter

log η η e.g.: η(γ)  0 d  1 β  γ c

Polymers: where cd < 1 increase in Mw/Mn! WHY??

log γ

- introduction of “second Newtonian plateau” for high shear rates

log η st η 0 , 1 Newtonian plateau

η  η e.g.: η(γ)  η  0   1 β  γ c

η , 2nd Newtonian plateau  log γ

35 Thixotropy shear thinning + long memory ( Hysteresis)

A decrease of apparent viscosity under constant shear rate, followed by a gradual recovery when the stress or shear rate is removed. The effect is time dependent. greek: thixis: shake trepo: changing in principle we can have two types of hysteresis:

start of shear η η time after shear γ = const. γ increase  Plateau  τc steady state t t or:

σ Newton

lim thixotrop  shear thinning τc 0

γ

Shear thickening ( rheopex  dilatancy)

ηγ 1  η γ 2 for γ 1  γ 2

Anti-thixotropy shear thickening + memory ( Hysteresis)

η γ = const. η time after shear

t t

36 or: σ

Newton

γ

Rheopexy Structure is generated without shear so that viscosity or module increases as a function of time only (not as a result of applied shear).

Dilatancy Why is wet sand “dry” for a few seconds when we walk barefoot on the beach?  dilatancy! Application of shear changes (reduces) level of in packed spheres (granula). This can cause shear thickening.

Experiment: “dry” particles on top

shear Vsand = const. !  sand level Vwater = const. !  water level or shake

ordered disordered spheres spheres need ”sand” more in beaker to pack

Dilatation: Ausweitung

Bingham plastic (the “evil” in the bottle!) (deutsch: strukturviskose Flüssigkeit mit Fließgrenze = plastisches Fluid)

σ critical stress γ = const. σ  η  γ  σ

below: elastic, -like σ yield  G  γ yield σ above: liquid-like yield

G σ yield Ketchup  20 Pa

γ γ yield 37

σ shear thinning (pseudoplastic) Bingham σ yield γ  0 for σ  σ yield Newton σ  η γ  σ yield for σ  σ yield η

γ

Extension of Bingham-model: Herschel-Bulkley

n include: powerlaw for viscosity σ  σ HB  k  γ for measurements  vane rheometer (see later)

Dimensionless groups

Reynolds (already covered), Deborah, Péclet, Taylor For several phenomena in nature only unitless quantities seem to play the important role: e.g.: 1) Arrhenius group - E  a  k r  A  exp   RT 

unitless

if Ea << RT  kr  A

if Ea >> RT  kr << A, slow down

2) kinetics

A(t)  A (0)  exp- k r  t

kr  t >> 1 basically complete reaction

kr  t << 1 just started

38 Deborah number [book of judges 5.5, song of Deborah: “Even the mountains flowed before the Lord ...”] remember:  G    step in strain σ (t)  G  γ exp  t  η 

dimensionless

1 G η   t,  τ De η G η τ  internal relaxation time De   G  t t  observation time

De 1, short observation time  solid like De  1, viscoelastic reponse De  1, long observation time  liquid response

1 if we take:  γ t γ  γ  cos(ω  t) under oscillatory shear: 0

γ  ω  γ 0   sin(ω  t)

De  γ  τ  ω γ 0  τ e.g. longest relaxation time in polymer

Note: Generally Deborah-nr. is not precisely defined ( γ  ω  γ 0 , or γ  ω), and there is (osc. shear) (Cox-Merz) confusion with Weissenberg-nr.: Wi  γ  τ Weissenberg normally used in the context of:  γ =const., steady shear

 with respect to normal forces

Pipkin diagram:

γ 0 N l yield ( ) e i non linear response w q s t u o

o i γ 0 ω = const. l n d i d viscoelastic

1 De, ω γ 0  τ, or ω τ

39 Péclet number

F  6π  η  r  v Stokes: 2r F F  ξ  v

k T k T Stokes-Einstein for diffusion coefficient D: D   ξ 6π  η  r

r 2 6π  η  r  r 2 6π  η Time needed to displace object by distance r: t     r 3 D k T k T

6π  η 6π  r 3  σ Pe  t  γ   r 3  γ  k T k T

σ  η  γ frequently used in context with ;

Taylor vortex moving bob

Ro secondary flow caused by generates vortices Ri in addition to shear

moving cup

more sensitive less sensitive to Taylor vortices to Taylor vortices

ρ 2 Ω 2 R  R 3  R Ta  o i i  3400 ηγ 2 ρ : density Ω : angular velocity

40 Units for Ta (only check, no proof):    2 4 2 4 2   kg  m kg  m  kg  m  1  Ta  2  2 4 2   2   2 1  6 2  N  m  s  N  s  N  m  s   2  s    m    unitless quantity

-=- END OF GLOSSARY -=-

41 What do we expect for (p,T)?

Gases Mean free path length: l 2r

v L >> r mean distance

Cross-section A  π2r 2  4π  r 2

in :  (confusing for rheology)

one particle 1 1 1 ρ    ΔV L  A L  4π  r 2

volume 1 1  L   ρ  A N 2  4π  r V normal conditions (, 1 bar, 300 K): pV  nRT  1 Mol ˆ 22.4 l n 61023 ρ    31025 m3 V 22.4103 m3 2 A  4π 10-10 m2 1019 m2

o r  1A

1 m3 1 1  L   m  μm  300 nm 31025 1019 m2 3106 3 mean free path length clash rate: 1 Δt L m 1 3 v  , v  330 more precise : mv2  RT Δt s 2 2

L 310-6 m 1 Δt    s v m 9 330 10 s 1 1 typical clash  rate : 109 Δt s

42 velocity distribution (Maxwell-Boltzmann) (see Physical Chemistry books for details)

3 2 2  m   mv  2 P(v)dv  4π   exp   v dv  2π  kT   2kT  1 1 1  8kT  2  kT  2  3kT  2 v     2.54  simple picture : v     π  m   m   m  1 ___ 3 ___ m v 2  kT note : v 2  v 2 , generally true for distributions 2 2 1st moment 2nd moment model: L: mean free path length

unit area A x

 dvs  vs  L    xi+1  dx 

L vs xi

vs: shear velocity y v: particle velocity z

e.g. xi: bottom layer, xi+1: top layer, gap of width L

 dv  momentum transport if particle leaves layer xi+1 to go to layer xi : mL s   dx 

Number of particles n leaving layer xi+1 in unit time in direction xi : N N N density of particle ρn     ρn  A  v  t  N V0 A  L A  v  t

volume average velocity! only half fly in correct direction: 1 1 1 N

 ρ n  A  v  t  n   ρ n  v    v 2 2 2 V0

unit time, unit area

43   in unit time, unit area the following momentum p is transferred:

Δp 1 N  dv     v  m  L  s  Δt 2 V0  dx   dv  this must be equal to the force F  η   s   dx  1 N 1  η    m  v  L  ρ  v  L 2 V0 2

V using : L  0 N  4π  r 2

1 N  m  v  V 1 m  v  η   0  2 2 2 V0  N  4π  r 8 π  r 1  3kT  2 using : v     m  1 η   m  3kT  T  m 8 π  r 2

Viscosity of gases is: - independent of density!

- therefore independent of pressure! ηp,T  η T !

- a function of mass and temp. of particles! η  T !

Viscosity of liquids, temperature dependence

- no shear: Boltzmann distribution for particles making transition from left to right ; typically 1vacancy per shell (= 12 neighbours)  5-10% free volume no shear (density diff. amorphous  crystall!) E*

kT  E *  N   exp  , E*: activation energy h  RT 

44 r - shear: force on single molecule, typical distance r F  σ  A  σ  r 2

r/2 apply this force for half distance r r σ  r 3 σ  V E  σ  r 2    m σ  r3 2 2 2 E * 2

σ  r3 Vm: average occupied volume per molecule, E *  2 VM: volume per NL

So effective jumps N are jumps to the right N minus jumps to the left N

  σ  VM   σ  VM    E *    E *   k B  T 2 2 (1) N  N N  exp   exp    h  RT   RT           

The shear rate is the effective number of jumps in one second divided by layer thickness in unit time:

v r  N (2)  γ   N d r

(2) in (1):

k B  T  E *   σ  VM   σ  VM  γ   exp exp   exp  h  RT   RT 2   RT 2 

remember : sinh(x)  ex  ex /2 limsinh(x)  limex  ex /2  lim1 x 1 x /2  x x0 x0 x0

k  T  E *  V  γ  B  exp   2  M  σ h  RT  RT 2 1 V V  N V γ   σ M  m L  m η R R k B

h  E *   η(γ 0)   exp  Vm  RT 

45 - Arrhenius for T-dependence - increase of free volume reduces viscosity (hopping probability  )

- Ea     ; Ea     - Pressure dependence via average volume per molecule  weak p-dependence

(T)

 E  exp a  liquid RT   gases

 T T

Stress-strain tensor and normal forces (Why might we need a tensorial property?!)

   So far we have used v,x and F as collinear (parallel) vectors  scalar description   If we would like to extend this, what happens if x and F are not parallel?

 F

x

  We need a transformation between x  F . This transformation should: 1. transform a vector into a vector 2. transform a plane into a plane 3. have a fixed origin in both systems

1-3 define an affine coordination transformation. This transformation is linear if the new   system y  (y1,y2 , y3 ) is generated out of the old system x  (x1,x 2 ,x 3 ) by a linear set of equations:

46 y1  a11  x1  a12  x 2  a13  x 3

y2  a 21  x1  a 22  x 2  a 23  x 3

y3  a 31  x1  a 32  x 2  a 33  x 3 if we introduce matrix (3 by 3 matrix, second rank tensor)  a a a   11 12 13    A  a 21 a 22 a 23  , we can write: y  A  x    a 31 a 32 a 33 

 Example for a simple rotation of a vector x in 2 dimensions:

x2 y2  Rotation around y origin by angle   in math. positive sense x  (counterclockwise)

α

x1 y1

 cos α  cos (α ) x    , y     sin α   sin (α )  use of addition theorems: cos (α )  Reeiα  Reeiα ei  Recos α  isin αcos   isin   Recos α cos   sin α sin   i ...  cos α cos   sin α sin  analogue for the sine (using the imaginary part): sin (α )  ...  cos αsin   sin αcos 

 cos (α ) cos   sin   cos α  y         sin (α )   sin  cos    sin α 

 x A

47 Definition of the extra stress tensor (right handed system!)

2

22 indices ij :

21 i: the force acts on a plane that is

12 normal to the basis vector i 11 23 j: the force acts in the direction of

13 1 the basis vector j

3

This results in the following extra stress tensor:  τ τ τ   11 12 13  τ   τ21 τ22 τ23     τ31 τ32 τ33  The stress-tensor σ is the sum of the extra stress tensor plus the hydrostatic pressure. The hydrostatic pressure acts equally along the τ11,τ22 and τ33 components.

1 0 0   σ  -p E  τ , p : pressure, E : unit tensor  0 1 0   0 0 1

Properties of the extra stress tensor

- The tensor is symmetric (like many in , see e.g. Fermi’s golden rule):

τij  τ ji  reduction from 9 variables to 6 variables - forces that pull have positive prefactor - forces that push have negative prefactor

The tensor has properties that are under transformation of coordinates:

1st invariant: Trace of the tensor A

n I1  tr A  a ii  a11  a 22  a 33 i1

48  (see also quantum mechanic books   αi αi ) i1

2 nd 1  2  2 invariant: I2  tr A  tr A  2  

a11 a12 a13 rd 3 invariant: I3  det A  a 21 a 22 a 23

a 31 a 32 a 33

Due to the first invariant τ11  τ 22  τ33  0 the trace of the extra stress tensor has only two variables.  N1 = 11 - 22 normal stress differences N2 = 22 - 33

What do normal stress differences mean?

- assume along 21

2

22

21

11

1 33

3

22 : force that pushes plates apart

33 : force that pushes material into plate-plate geometry

N1 = 11 - 22 : first normal stress difference, generally positive

N2 = 22 - 33 : second normal stress difference, generally negative N1  N2

to memorize: Na = aa - a+1, a+1

49 What do we expect for N1,2 γ, γ0 ?

- N1,2 should only be a function of γ due to kinetic nature of the phenomenon, e.g.:

2 N1,2  a  b γ  c γ ... a,b,c : constant

- if we do not apply a shear rate N1,2 should be 0  a = 0

- if we apply a shear rate, the force N1,2 should be independent of the direction

 N1,2 γ  N1,2 - γ 

 even function with respect to γ n

2 We expect equation like: N1,2  c γ as first approximation N τ  τ ψ  1  11 22 1 γ 2 γ 2

N τ  τ ψ  2  22 33 2 γ 2 γ 2

1 : first normal stress coefficient

2 : second normal stress coefficient

1 : generally positive, ψ1  ψ2 (typical factor: 10);

N1 can be as high or even higher than 12 !

2 : generally small and negative

1 + 2 can be measured separately using both:

γ N1 γ N1 + N2

&

cone-plate plate-plate

Information: N1 Information: superposition N1 and N2

50 Typical examples for extra stress tensor: a) ideal viscous fluid b) viscoelastic liquid  0 τ 0  τ τ 0   12   11 12  τ   τ21 0 0 τ   τ21 τ22 0  5 unknown      0 0 0  0 0 τ33 

12 = 21 12 = 21 , 11 + 22 + 33 = 0

3 degrees of freedom  , 1 , 2

The first normal stress coefficient can be estimated using:

2G( ) limψ1 γ  for γ  ω ω0 ω2

Phenomena where we can directly “see” normal forces a) Rod-climbing

Parabola, f(r) r 2 leading term, f(r) r 4 !

centrifugal forces f(R,r,ψ ,ψ ,ρ,ω,...) 1 2 (e.g. water) Non- Newtonian fluid climbing effect is called Weissenberg effect

51 b) Secondary flow for rotating disc up!

centrifugal see: cover page forces book: Tanner Newton or not?

Newtonian fluid Non-Newtonian fluid

c) Extrudate swell

parabola

die swell  film blowing (e.g. plastic bags)

Newtonian fluid Non-Newtonian fluid

52 New chapter:

Possible measurements (for oscillatory rheometers) and hardware

1) Detection of onset of non-linearity at fixed frequency

log G linear regime 1 = const., fixed

a b G’

c G”

log 0 a: Problem: torque too low, hard to get sensitivity b: onset of non-linearity, but: depends on accuracy of detection! c: in filled materials sometimes an overshoot in G” is detected: Payne-effect (name!: not pain) typical values: polymer melts: 0 < 0.05 – 0.3

solutions: 0 < 0.1 – 1

cross-linked rubber: 0  0.01

 we know linear regime for a fixed frequency ( γ max  γ0  2π  ω1 )

we can assume: linear if 2 < 1

perhaps non-linear if 2 > 1

2) Measurement of G’, G” at T = const., : variable, 0: parameter

difference! log G

Frequency dependent module 1  distribution of relaxation times G” e.g. via Multimode-Maxwell models 2

 see section about polymers (later) G’

(reptation, rubber plateau, TTS)  = 1 log 

53 Typical range: 10-2 <  < 60, dynamic range: 4 decades of hardware due to mechanical device!

log G adjust 0 !,

torque will change for an increase

in  by 104 by up to 104!

log 

γ(t)  γ0 sin(ω t) γ  γ ωcos(ω t) 0  1

γ max  γ0 ω

 adjust 0 every 1 - 2 decades for best performance

3) Temperature dependent measurement

0: fixed but parameter, T: variable, : fixed

G’ G’ or or 9 G” 10 G” [Pa] if 1/T   105 - 106 Arrhenius

T log 

Instrument for this: DMTA, (Dynamic mechanical thermo analyser) cheap due to limited -range (sometimes also E-module measurements)

54 4) Shear rate dependent viscosity

log  asymptotic behaviour  in principle no linear regime!

0 experimental linear regime

e.g. 10% reduction relative to  0

log γ

Fit with: 1 - 4 parameter model (see before)

Hardware:

Couette geometry preferentially: static bob, moving cup (because of Taylor vortices!) inside moving: Searle-type, outside moving: Couette-type

N(t): torque

bob

or or

cup

(t) (t) (t)

Mooney-Ewart double couette Haake-type

for low viscosity Air bubble 

materials, low friction at lower end

e.g. water

55 To prevent evaporation of water: saturated H2O vapor

H2O Dodecane, C12H26 H2O

or water trap H2O sample

If plate - geometries are used: 2 Area  r0 r0 Torque at infinitesimal area:    ΔN  ΔF r  ΔN  r1

 total torque N  r 3 0 e.g. change from 50 mm plate-plate geometry to plate-plate 8 mm plate-plate geometry : torque reduction by  3 non homogeneous γ0 ,γ !  50     244 , 2.5 decades reduction  8 

r0 3 N  r0

 typical values for : 0.02 - 0.1 rad h  1.14 –5.73°  very small!

cone-plate

homogeneous γ0 ,γ !

or

Advantage of plate-plate (cone-plate) vs. Couette: - less sample volume (e.g. 0.1ml vs. 10ml) Disadvantage of plate-plate: truncated cone - leakage (low viscosity material) easier to manufacture - less area  less sensitivity for low viscosity materials - heterogeneity of shear rate

56 5) Creep experiment

0

t

G(t)

t

 G(0,t) measured

Overlay via h() log G(0,t) 1 < 2 < 3 < 4 3 (modulus measured, not stress!) 1

2 4

log t

Wagner-Ansatz (Manfred Wagner, Prof. in Berlin): Gγ,t  G t h γ

damping function

1. limhγ 1, linear response γ0 2. decreasing strictly monotonic as a function of time, limhγ  0 γ typical examples: hγ  exp- n  γ, e.g. n  0.18 for PE  melt

hγ  f1  exp - n1  γ  f2  exp - n 2  γ  , f1  f2  1 1 hγ  Doi-theory 1 a  γ2

57 Hardware

Stress and strain rheometer, typical types of construction:

motor, Volt + Amp.  torque, (t)

optical encoder, (t) disc or position (capacity) stress-rheometer, stress is given, strain measured air bearing strain is measured via optical encoder controlled stress is imposed sample geometry, e.g. plate-plate

frame

rigid spring, deflection < 1°, otherwise problem with Bingham fluid

disc optical encoder or position sensor

air bearing strain-rheometer (A), strain is given, stress measured plate-plate

disc position sensor, (t) nominal actual value comparison, feedback loop motor, (t)

magnetic , seal + normal forces stress detection

position sensor force rebalance transducer (FRT) feedback: “stand still” e.g. our ARES: 2K FRT N1 (2K = 2000 gcm, motor acts as rigid spring N1 = normal forces can be measured)

strain-rheometer (B), air bearing

ARES-type sample ball (cheap) or air bearing (expensive); for normal forces air bearing needed

position sensor strain application

feedback  (t) motor

58 Typical hardware specifications (ARES)

Magnets, transducer: Al Ni Co - alloy 0.01% / °C

Magnets, motor: Nd (Neodym) 0.1% / °C (compare: Cu: 0.39% / °C)

Optical encoder: 30,000 lines + interpolation 0.0810-6 radian resolution = 810-8 rad

810-8 rad 0.08 mm ! 1 km 1 1 Alternative: capacitive encoding: ┤ ├ C  ω  d LC LC d dynamic range of transducers: newest: 1K FRT N1 (Rheometrics, also Haake) 6 -1 -7 Nmax / Nmin = 10 , e.g. 10 Nm to 10 Nm!

Typical prices (2002): Stress rheometer: 15k – 40k € (Haake, Bohlin, TA, Rheometrics, …)

ARES (strain): 60k – 100k € (3 types of motors, diff. types of temp. control, 7 diff. transducers, …)

+ cooling (N2): 10k € + dielectric option: 30k € + birefringence, dichroism option: 30k €  up to 180k €

Geometry: 1.5 – 3k €

59 Typical pathway of a signal from the torque transducer to G’, G”

Torque transducer: analogue filter “smoothing” V average V “low-pass filter”

0 e.g. integration or 0 t t “oversampling”

Dynamic range of V V “zero-frequency-artefact” ADC adjusted

autorange Autobias a 0 a 0 t t “NMR: RGA” (receiver gain autorange)

Imax dynamic range

4 bit ADC, Imin

0 24 = 16 slots t

dwell-time

ADC: discrete in time ( dwell-time) and in intensity (k-bit ADC, 2k slots)

60 Typical acoustic ADC’s: 16 bit = 216 = 65,536  dynamic range: 1: 65,536 (remark: limit for S/N in FT-Rheology!) dwell-time: 10s, sampling rate: 100 kHz  105  6.55  104 = 6.55  109 decisions per second

position signal

corrections + geometric factors G’, G” cross correlation or Fourier-Transformation torque signal

corrections: e.g. inertia of geometry, stress transducer, motor, phase lag due to lumped circuit

Vane rheometer (German: Schaufel, Flügelrad) useful for determination of yield stress Rv ( Bingham fluid) in concentrated suspensions, greases or food (yoghurt!); especially if the history of loading should be avoided. Lv Approximation:

3  L v 2 Tm  2π R v    σ y R v 3

Tm: torque maximum

y: yield stress

61 Melt-flow index cheap + robust version of a capillary rheometer (see later) - uncontrolled, non-homogeneous flow M - relative measurement (“index”) typical parameters: 9.57 mm T = 190°C condition “E” M = 2.16 kg 0.8 mm  pressure  3  105 Pa MFI: flow of polymer in [g] per 10 min 0.209 mm

 rough measure of average MW

Capillary rheometer ( high shear rates)

Model system for e.g. polymer extrusion process (see also: melt-flow index) important shear rates:

oscillatory / vibrational

rotational capillary

elongational

processing

γ [s-1] -4 -2 0 2 4

62 Set-up: constant force or constant velocity M, v

d pressure L d:L  1:30 for steady state, developed streamlines

ds die swell, extrudate swell  normal forces

m(t) mass

Dominantly viscous properties of the material are determined, pressure loss at entrance can be corrected  “Bagley-Correction”

Information: m(t), ds, pn, …, p1 for different T, M, v, d, L

Elongational rheology, viscosity important for: fibre spinning, blow moulding, flat film extrusion, film blowing

L0 2

3 1

L(t)

dx  v(t)  ε  x , ε :stretch rate dt 1

63 if specimen is stretched with constant rateε dx  ε x1 dt L 1 t dx  ε dt ; ε  const. x L0 0 L ln  ε  t L0

Hencky-strain, sample length L  et !

[Hencky worked for many years in Mainz-Gustavsburg! See Appendix B, p. 115 - 123]

σ tensile viscosity: η  E E ε η without proof: lim E  3 for simple liquids ε0 η0 Trouton’s ratio

experimental apparatus: B A

M sample

oil: + compensates gravity sample thickness + elongation + temp. control monitored via camera - can act as plasticizer in the sample Prof. Meissner Prof. Münstedt Zürich Erlangen ex BASF ex BASF

64 Muenstedt, Laun, Rheol. Acta, 18, 492, 1979

65 Rheology on two specific examples: polymers and dispersions

Polymers

 End-to-end distance R , bond length b, N monomers  n  R   ri i1 Gauß:   R R  R 2  b N  R 2  b2  N contour length: L = Nb (“odometer”)

e.g. high Mw-PE, N = 100,000, b = 1.5 Å contour length 15 m (in principle visible!), R = 47 nm

R simplified model: R  g 6

Reptation theory basic idea: one-dimensional stochastic process of chain along contour (reptate: reptile)

simplified tube with diameter d and other chains are static, s typical distance of other chains: s  d typical d  30 - 80 Å d one-dimensional Fick-equation, for chain distribution probability P  2P  D t 1d x 2

66 Solution for P(x, t): Gauß-statistics x 2 Gauß: 2σ 2 1  x 2  Px,t  exp  σ 2  r 2  2nDt 4π D1d  t  4D1d  t  n: dimensionality Mean square displacement ( second moment)

 x 2  x 2P x,t dx  2D  t   1d 

If we assume stochastic friction coefficient ’, where this friction coefficient ’ is proportional to N, therefore also M ’ = N : friction per monomer unit

Using the Einstein-relation for the 1-d. diffusion:

kT kT D    M 1 1d ξ ξ  N

The time  needed to diffuse along L will allow a fully different conformation, so that all memory of the other chains (static) is erased 

2 L  λ 2D1d L2 M 2 λ    M3 1 2D1d M

λ  M3

The self-diffusion coefficient Ds is given by the time t to move the center of mass by a typical coil diameter R (3-dimensional problem!).

r2  2nDt , n : dimensionality, here : 3 t

2 R M 2 Ds    M 6  λ M3 R

2 Ds  M

67 assuming a Maxwell-model: η λ  G η  λ G with λ  M3 G  M 0 molecular weight independent given by temporary entanglements, “mesh-length”

Ml

3 3.4 ηPolymer  M , DeGennes 1971, exp.: ηPolymer  M

1 for non-entangled: ηPolymer  M friction of polymer-contour

log  Mc  3 Me 3.4 entangled

“3 fingers needed to hold a stick” 1

log M 1/10 Mc Mc

Rule of thumb for flexible monomers with 2 carbons per polymer backbone (so not true for PPP, poly-paraphenylene  persistence length)

 ne  100 – 200 monomers contourlength between entanglements: 150  3Å = 45 nm, R e  n 3Å  3 - 4 nm

examples, Me: PE: 828 g/mol PS: 13 kg/mol PDMS: 12.3 kg/mol PIB: 7.3 kg/mol PMMA: 10 kg/mol 1,4 PBd: 1.8 kg/mol 1,4 PI: 5.4 kg/mol might differ depending on lit. sources

68 Typical shape for G’(), G”() for monodisperse linear polymer melts

I II III IV log G

G’p

1 G” 2 tan  G’ minimum

 = 1 log  Maxwell-model related length scales:

Rg, 10-50 nm 5 - 10 nm 2-3 nm R , distance between e entanglements related time scales:

d R e s

(d: disengagement, R: Rouse time, e: entanglement, s: segmental motion)

Zone I: G  ω2 see Maxwell-model G  ω1 flow-zone, viscosity and the dominates response length scale probed  Rg longest relaxation time:  = 1 for tan  = 1 (Maxwell-model) or

extrapolated crossing point log G’, for G  ω2 and G  ω1 log G”

d = 1 log 

69 Zone II: (rubber-plateau)

After G’ exceeds G”, G’ levels off. Response is dominated by elastic spring (G’!) of physically cross linked entanglements (see motivation for Hooke-solid!). Maximum of relative elastic response is reached for tan  = Minimum, corresponding G’p (p: plateau). length scale probed:

5-10 nm

It is possible to calculate from G’p the entanglement molecular weight: ρR T M e  Gp : density R: Gas constant T: temperature

Assuming typical values for a polymer melt:

3  = 1,000 kg/m , R = 8.3 J/(mol K), T = 450 K, Me  150  70 g/mol = 10.5 kg/mol

1,0008.3450  kg J K mol  G  , 1J 1Nm p 10.5 m3 molK kg   5  Nm  Gp  3.510  Pa  m3  typical plateau value: 105-106 Pa ( memorize!)

(high Tg + low Me  increase in rubber plateau modulus)

 for higher cross link density in chemical cross linked systems we expect higher modules

3 3  Mn does not affect G’p, but η  M n , λ  M n ; so increase in molecular weight by factor 10  103 shift in  for plateau length;

70 Strobl, The Physics of polymers

at  3 Me we start to see plateau

increase in Mn by  100  shift in  by 1003 = 106

71 plateau length (width)

log G’ Mn2

log G” Mn1

105

Mn2  10  Mn1

3 decades

22 = 1 11 = 1 log 

Zone III:

G” exceeds again G’. Strong increase as a function of frequency. Transition zone towards glass plateau;

Zone IV: (glass plateau)

High torque and high frequency regime, experimentally difficult to obtain. e.g. small sample diameter ( 5 mm), use of TTS (see later) Length scale probed in dimension of typical length scale of polymer glasses, e.g. 2-3 nm. shear rate dependent viscosity (or measured by ηγ  η*ω , Cox-Merz), typical shape:

log  3.4 η0  M n

typical slope for linear polymers: - 0.8  0.1

γ  λ  1 log γ or

: longest relax- log  ation time

72 Time-Temperature-Superposition (TTS) and the Williams-Landel-Ferry (WLF) equation

Assumption: The internal mobility of a polymer is monotonically (+ continuously) changed via the temperature. The changes keep the ratio (not the difference!) between the different relaxation time distributions and relative strength. This is related to the concept of the “internal clock” that is only affected by the temperature (McKenna). Obviously this assumption must fail if phase transitions (e.g. first order: crystallisation, second order: glass transition or TODT) are involved. If we set a reference temperature T2, where we know or have measured G’((T2)),

G”((T2)), we can predict G’((T1)), G”((T1)).

Maths: Modification of Arrhenius law  Vogel-Fulcher equation:  E  (1) η T  η  exp a   0    RT  TVF  note: Ea for flow of linear polymer melts  25-30 KJ/mol (typical value)

no information about TVF yet, except:

- if TVF = 0  Arrhenius

- if T = TVF  singularity in  therefore we expect: T a) VF  1 T for typical T  300-500 K, because it is only a correction!

b) fixed difference of TVF relative to Tg due to the assumption of similar mobility of

different polymers at Tg; using (1): η T η 1 1 :  ; T1  T2 , T2 : ref. temp. ηT2 η2 η ω 1 η 1  2 ; η  τ  ; τ  Maxwell η2 ω1 ω G : characteristic “frequency” of motion

73 η 1  f  η1 T1 η2

2: fixed value at reference temperature T2 (not defined yet)

ω E log 2  log f  a  loge ω1 RT1  TVF

=: - C1 (no units!) 0.434

E a  loge :  C  C ; C : unit of temp. ! R 1 2 2

TVF :  T2  C2 ; choice of T2 will change C1 and C2!

ω C  C - C  T  T  2 1 2 1 1 2 log  -C1   ω1 T1  T2  C2 C2  T1  T2

only diff. to ref. T2 important!

ω - C  T  T 2 1 1 2 WLF-equation log :  log a T  ω1 C2  T1  T2 shift factor

If we choose the reference temperature as T2 = Tg (other choices also possible!): ωT  - C  T  T  log g  1 1 g ω1 C2  T1  Tg

For these conditions (T2 = Tg) and for typical polymers it is found:

C1  17.4 C2  51.6 K

(C1  7.6 C2  100 K for T2 = Tg + 50 K ; rem.: C1  C2  const. ( 900 K))

C  C  R  1  apparent activation energy: E  1 2  17.5 kJ/mol for  0 a loge  T 

74

Prefactor  A – B stretch, IR frequency  typically 1012 - 1014 1/s log 

Arrhenius

slope ~ apparent activation energy,

differs as a function of temperature! -1

0 1/Tg 1/T

singularity at: Tg - C2

If we assume (Tg)  0.1 rad/s ( 0.01 Hz) as the typical jump rate (motion) at the glass transition temperature for a spatial entity of several monomerunits (e.g.  100)  - relaxation. We do not look at side chain motion  -relaxation (typically pure Arrhenius) or

12 –CH3  10 Hz (at room temperature)

 0.1 17.4T  T  17.4T limlog   g   17.4 T   ω 51.6  T  Tg 51.6  T T2 Tg  

 C1 is related to prefactor

0.1 1017.4 ω ω ω 1016.4 ;  ν  2π 15.6 ν  10

further: TVF = T2 – C2 T2 = Tg

TVF = Tg – 51.6 K

75 WLF curve defined via: 1) axis intercept  Arrhenius log  16.4 1 Ea = 17.5 kJ/mol 2) slope lim  E a T T

3) singularity at TVF

-1 C2!

1 1 1 1 T Tm T T g VF if crystalline

singularity

so: Tm > Tg > TVF rule of thumb for polymers: T 2 g  (in Kelvin!, absolute energy scale) Tm 3

Tg – C2 = TVF (C2  50 K)

0.1 17.4ΔT log a T  log   ω(T) 51.6  ΔT

T2 = Tg:

T aT (T) [rad/s] 0 1 0.1 3 deg  1 decade change in 5 10-1.5 3.5 mobility, close to Tg 10 10-2.8 60 20 10-4.8 7103 30 10-6.4 2.5105 50 10-8.6 3.6107 15 deg  1 decade change 100 10-11.5 31010

don’t take this table to literally!

76 Glossary: (to relax from maths for a second)

Boger fluid: To study the relaxation of high Mn polymers (e.g. normal forces, G’, G”) the

very long relaxation times are shifted to more “practical” values via low Mn solvents.

Dispersions

Definition: lat.: dispersio, fragmentation A system built of several phases where one is a continuous and at least one more phase is fine fragmentated within the continuous phase. If the size of the dispersed phase is < 0.2 m (visibility!) they might be classified as colloids or colloidal dispersions.

continuous phase dispersed phase name example solid solid vitreosol ruby glass solid liquid solid butter solid gas solid pumic-stone (Bims)

liquid solid colloidal of Au, S in H2O liquid liquid emulsion milk, pharmaceutic or cosmetic emulsion liquid gas foam -foam

gas solid smoke NH4Cl, carbon black smoke gas liquid fog, mist natural mist gas gas ------why?!

77 Zero-shear viscosity as a function of solid content: Einstein 1906: (see Appendix C, p. 124-144 for original )

η  ηs 1 2.5  for   0.1 Idea:

s : viscosity solvent  : volume fraction lit.: A. Einstein, Ann. Physik, 1906, 10, 289 1911, 34, 591 flow field, hard, rotating particle General behaviour for higher concentrations:

2 2 η  ηs 1 2.5  O  ... , O : not defined yet

Intrinsic viscosity (relative change of viscosity normalised to solvent viscosity): η  η η  s ηs limη  2.5 , Einstein coefficient  0

Extension of Einstein, O(2)  Batchelor (1977) η shear  1 2.5  6.2  2  O 3  ηs ηt extension  1 2.5  7.6  2  O 3  ηs shear + extension already anisotropic!

Shear can deform liquid particles to prolate or oblate shape if , mobility and shear rate are sufficient (e.g. blood):

a c a c b b

a = b < c prolate a < b = c oblate a In both cases: aspect ratio:  1 c

78 High aspect ratio  more excluded volume (see liquid crystals: Onsager theory)

Zero-shear viscosity as a function of volume fraction (no information about ηγ !) for higher fractions:  d  η  ηs  2.5 ηs     d 

(1) dη  2.5 ηs  d At certain volume fraction the addition of d leads to an increase in d that is expected to be: dη  2.5 η  d η dη   2.5 d  η  ηs 0 η  ln  2.5  0 ηs

η  ηs  exp2.5 Ball, Richmond 1980

 5 25 2  2 Taylor: η  ηs 1     ... too low increase in  !  2 2!4 

Other idea: Addition of small amount of particles d to the volume fraction (1-) of remaining fluid, raises volume fraction by: d

1 in analogy to (1): 5 d dη    η 2 1 dη 5 d   η 2 1    2.5  2.5 η  1   1  2.5 ln  ln    ln   ln1 ηs  1   1    0 2.5 η  1  2.5     1 , singularity at   1  not physical ηs 1 

79 If we assume a maximum filling fraction m, we find:

-2.5 η    m  1  , Krieger-Dougherty (1959) ηs  m 

Maximum filling factor ( crystallography, inorganic chemistry), examples:

Simple cubic, sc m = 0.52

Hexagonal packed sheet m = 0.605 (colloids high shear rates)

random close packing m = 0.637

body-centered cubic, bcc m = 0.68

face-centered cubic (fcc) / m = 0.74 hexagonal close packing

Max. possible value for monodisperse!

80 Best theoretical equation for ():

1   3 9  η m   Frankel and Acrivos, with m  0.62 - 0.64 1 ηs      3  81     m     experimental: η  1 2.5  10.05 2  2.7 103  exp16.6  ηs

Zero-shear-(rate)-viscosity can drastically be influenced by multimodal distribution:

e.g. r1 : r2 = 5 : 1 1 η   η !   60% pure2 50 50:50 total

ηpure2, 60%  η 60% 215% 1 15% increase in solid content, same viscosity picture:

filling of voids

Viscosity as a function of shear-rate for colloids:

st 0, 1 Newtonian plateau log  scaling law Shear thickening, layered structure

log γ 2nd Newtonian plateau

γ c : critical shear stress

81 Scaling-law behaviour (+2-Newton) can be described by: - Ostwald de Waele ηγ  A  γ -B , B[0,1]

- e.g. Ellis-model, using the Péclet-number as universal parameter

η  η 1  p η0  η 1 b  Pe

6  σ  r3 η  γ  r3 Pe   s kT kT

The critical shear-rate can be estimated for a 0.50 =  mixture:

nm2 d2  γ  107 ; d [nm] , γ [1/s] c s c

γ c [1/s]

5 10 e.g.: d = 100 nm, γ  103 1/s c 103

101

10-1

10-3

101 102 103 104 105 particle diameter [nm]

Understanding of related forces F(x) and potentials ( ˆ )  Fdx  Vx for colloidal particles:

Vtotal = Vvan-der-Waals + Velectrostatic + Vdepletion + Vsteric

DLVO-theory

DLVO: Derjaguin – Landau – Verwey – Overbeek (1941 + 1948)

82 Van-der-Waals:

Attractive force between atoms, molecules and particles caused by induced electrical dipols of cloud. 1 Permanent dipol interaction Vr   , r3

2  1  1 induced Vr      , minus prefactor, because attractive!  r3  r6

For two quadratic surfaces with side length L one finds: A Vr    L2 12π  r2 A: Hamaker constant [energy]; typical value: 0.4 - 4  10-19 J

For Lennard-Jones potential, also short range repulsion  Vr f r , for hard spheres, one typically finds: 1 1 Vr  a  b r6 r12

Velectrostatic:

q1  q2 q1 r q2 Fr  2 4π  ε0  εr  r Coulomb

ρ 1st - Maxwell-equation: E  , : charge density ε0  εr  E  Ve , Ve: electric potential ( F  E  q; W  Ve  q ) in spherical coordinates: 1 d2 1 (1) ΔV  2 r  Ve   ρ r dr ε0  εr

83 Approximation of  via a Boltzmann-distribution of screened Coulomb potential (single ions):

 q  Ve r   q  Ve r ρr  ρ0  exp   ρ0 1   kT   kT  homo q  V  ρ  ρ  e  V using (1) eff 0 kT e

2 1 d 2 ΔVe  r  Ve  χ  Ve Eigenvalue problem r dr2 ( Quantum mechanics Hˆ  ψ  E  ψ ) solution: q Ve r   exp χ  r 4π  ε0  εr  r no screening screened potential, screening length1/: Debye length

exp(χ 2 ) ρr    exp χ  r 4π  r

1/ : equivalent to Bohr-radius in H-atom, first Laguerre polynom

1 0.304 rD   [nm] for 1:1 electrolyte, c in mol/litre χ c

c = 1 mol/l  rD = 3 Å

c = 0.01 mol/l  rD = 30 Å

Bjerrum length:

What is the distance l b, where the electrostatic energy of ion is equivalent kT?? (“electrostatic yardstick”) + l b

84 W   Fdx l b e  e kT  dx , RT = 2.4 KJ/mol , r = 80  2  4π  ε0  εr  x

 l b = 7 Å for distances smaller 7 Å  Manning , opposite charges bound to each other

In case of identical spheres at constant surface potential 0, radius a of spheres:

4π  ε  a 2  ψ 2 V  0  exp χ  r for   a < 5 e r

2 Ve  2π  ε  a  ψ0 ln1 exp χ  r for   a > 5

Vdepletion:

In bimodal systems with large size difference, e.g. polymeric solution plus particle, polymer does not bind to particle. Potential caused by osmotic pressure.

Vd  - , : osmotic pressure

forces acting!

areas not accessible for polymer

mixed flocculated

85 Vsteric:

particle particle

non charged or polymers

Influenced by: - number of chains per area

- layer thickness (Mn!) - solvent quality - anchor strength systems: e.g. block copolymers (5 - 50% as anchor) triblocks: bridging  flocculation (e.g. sewage water treatment)

86

Book: Israelachvili

87 Fourier-Transform-spectroscopy

[Joseph Baron de Fourier (1768-1830), and ]

free induction decay (FID)

pulse time

Ta

FT, Fourier transform

1 0 ν  a Ta

In the past (in ESR till today): CW (continuous wave) excite with single frequency

measure the resonance

change frequency

FT: all signals are acquired simultaneously (“multiplex advantage”)

In words: A Fourier transform analyses the corresponding frequencies of a given timesignal with respect to amplitude, frequency and phase (i.e. full information).

88 Note: This method is of special importance in NMR, IR, X-ray, neutrons, QM, ... and Rheology!

Math:

 Fω   f t exp  i ω t  dt  complex! complex (can be separated in cos + sin)

real + imaginary part or magnitude + phase

This operation is reversible! (one-to-one) 1  f t   Fω exp  i ω t  dω 2π 

In NMR we use a single-sided, complex, discrete Fast-Fourier transform (FFT) (special algorithm (“butterfly-algorithm”), which needs 2N datapoints)

Problem of discretisation (ADC, analogue digital converter)

FID

FID

t scan rate of the signal (dwell-time, DW)

Signals are not distinguishable! Nyquist-frequency (cp. solid state-modes, Einstein, Debye model)

Frequency regime in which the signal can be assigned unambiguously (= spectral width, SW): 1 SW  2  DW

89 Some important mathematical relations

1) FT is linear a f t  b g t  a  F ω  b Gω i.e. the different signals can be detected independently! (“proof”: A  BC   AC  BC , linearity of the integral)

2) The point in time t = 0 is proportional to the whole integral of the absorptive spectrum.

 Proof: f t   F ω exp  i ω t  dω 

  f t  0   F ω  exp  i  ω  0  dω   F ω dω  

  A FT A B   B

t 0  b   a

3) Timesignal and spectrum are inverse to each other with respect to the full width at half maximum. (units: s, 1/s !)

pulse t

FT same integral, because  FID(t=0) is the same!

0 

Heisenberg’s : t  E  h E = h   t    1

90 4) important Fourier-pairs:

a) exponential FT Lorentzian

exp(-t/) (cp. Relaxation etc.)

  t     1    exp   exp  i  ω  t dt   exp  i  ω    t dt 0  τ  0   τ      1  i  ω  1   1   1   exp i  ω    t   τ  1  1 1 i  ω    τ   i  ω   i  ω  τ 0 τ τ 1 i  ω  τ  1 1 ω2  ω2  τ2 τ2

real part imaginary part

absorptive dispersive

real (absorptive)

1

T2 π

 0

imaginary (dispersive)

Magnitude: Re2  Im2  “broader feet” ( test it! MC)

b) box FT sinc

1

FT

 -t0 0 t0 cp. single slit diffraction pattern -x x q-vector 91 proof:

t t0  1  0 exp i  ω  t dt   exp i  ω  t  i  ω  t0   t0 1  exp i  ω  t  exp i  ω  t i  ω 0 0 1  cosω  t  i sin ω  t  cos ω  t  i sin ω  t i  ω 0 0 0 0 1 sinω  t   2   i sinω  t  2 0 i  ω 0 ω sinc()

remark : expi  ω  t  cos ω  t  i sin ω  t ; exp i  π 1  0 (Euler)

FT c) Gaussian Gaussian (without proof)

  t 2   ω2  σ 2  exp   exp i  ω  t  2π  σ  exp t    2  t  2    2σt   

5) convolution

Multiplication of a timesignal t(t) with a function g(t) corresponds to a convolution in Fourierspace. In NMR the measured timesignals are often multiplied with exp(-kt) or exp(-k’t2). (This is equal to a convolution with a lorentzian or a gaussian.)

Example: Convolution

with a box- S/N  5/(1/3) =15 S/N  5/1 function but: broader peak! S

N N  1/3

  averaging, new function

92 Appendix A Phil. Mag. 32 (5. Series), 424 (1891)

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114 Appendix B

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123 Appendix C

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144