Chap.3 Viscoelastic Response of Polymeric Fluids

Design of a Parison die for a Viscoelastic Fluids

Design a die that will allow one to extrude at the highest rate possible for

LDPE (NPE953) at 170℃ with (Fig 3.1)

Diameter D = 5.13cm

Thickness H = 0.565cm

Specify die dia D using rheology data in Appendix A.1

Gap H ρ = 772kg/cm@ 170℃

Assume L=10H Maximum feed to die, m = 300cc/h @melt fracture, τ = 1.13 × 10Pa In your calculation consider the extrudate swell.

Die swell ratio =D/D

Fig 3.1 here.

3.1 Material Functions for Viscoelastic Fluids

3.1.1 Kinematics

Flows Shear flow

Shear-free flow (Elongational flow)

For rectilinear shear flow

v = γ(t)y v = v = 0 ∙∙∙∙∙∙∙∙∙∙(1)

0

γ = γ = + = = v = y + c @ y=0, v =0 →c=0

v = y = γy

For shear-free flow:

v = − ϵ(1 + b)x v = − ϵ(1 − b)y v = +ϵz (2)

ϵ=extrusion rate , b=0 or 1

b=0, ϵ > 0 => Uniaxial extension

b=0, ϵ < 0 => Equibiaxial extension

b=1, ϵ > 0 => Planar extrusion

Note γ(t − t) = γ Shear strain

ϵ(t − t) = ε Elongational strain

More generally:

γ(t, t) = ∫ γ(t)dt

reference time

ϵ(t, t) = ∫ ϵ(t)dt

Note for simple elongation:

v = =

v = − ϵx v = ϵz

= − ϵx = ϵz

= − ϵdt

= ϵdt

lnx = − ϵt + c lnz = ϵt + c @ t=0, x = x c = lnx

→ln = − ϵt Likewise,z = ze

x = xe = e(t2−t1) = √ xyz = 1

y = ye = e(t2−t1) = √ (Mass conservation)

z = ze = eϵ(t2−t1) = l xyz = 1

Fig 3.2

Components of rate of deformation tensor (Table 2.10)

Simple shear flow : v = γy v = v = 0

γ = + γ = γ = + =

Other γ = 0 γ = γ(t) 0 1 01 0 00 0 0 ∙∙∙∙∙∙∙∙∙∙∙∙ (3) Shear-free flow :

v = − ϵ(1 + b)x = − ϵ(1 + b) γ = −ϵ(1 + b)

v = − ϵ(1 − b)y = − ϵ(1 − b) γ = −ϵ(1 − b)

v = ϵz = ϵ γ = 2ϵ

γ = + So,

γ = −ϵ(1 + b) 0 00 −ϵ(1 − b) 00 0 2ϵ ∙∙∙∙∙∙∙∙∙∙∙∙ (4)

3.1.2 Stress tensor component

State of stress

τ + P τ ττ τ + P ττ τ τ + P ∙∙∙∙∙∙∙∙∙∙∙∙ (5) Shear stress Normal stress (diagonal)

P= Isotropic pressure

τ , τ … = Extra stress or molecular stress π = τ π = τ + P = Total stress

→π = τ + Pδ ..... (7)

δ =1 for i = j = 0 for i ≠ j (Kronecker Delta) (8)

Newtonian shear flow (incompressible):

τ = −2μ + μ(∇ ∙ v) (3.2-11)BSL τ = −μ(∂v∂x + ∂v∂x)

τ = τ = τ = 0 (See Table 2.10)

State of stress (5) → P τ 0τ P 00 0 P ∙∙∙∙∙∙∙∙∙∙∙∙ (9)

Viscoelastic shear flow

τ + P τ 0τ τ + P 00 0 τ + P ∙∙∙∙∙∙∙∙∙∙∙∙ (10) (τ + P)hasno rheological significance. So we difine three independent quantities

of stress of rheological significance which are directly measured: τ = τ , N1, N2

Shear stress: τ = τ (symmetric) ∙∙∙∙∙∙∙∙∙∙∙∙ (11) First normal stress difference: N = π − π = τ − τ ∙∙∙∙∙∙∙∙∙∙∙∙ (11)

Second normal stress difference:N = π − π = τ − τ ∙∙∙∙∙∙∙∙∙∙∙∙ (11) Definition of directions:

Direction 1= flow direction

Direction 2= normal to the flow directio

Direction 3= neutral direction

There is no way to separate τ and P!

→So, N1 and N2 are a way around to understand the normal stress.

N, N ∶ Related to die swell, elastic recoil, rod climbing (Elasticity) → N = N = 0 (For Newtonian)

Shear-free flow, the form of extra stress:

τ + P 0 00 τ + P 00 0 τ + P ∙∙∙∙∙∙∙∙∙∙∙∙ (12)

Two normal stress differences:

τ − τ ,τ − τ ∙∙∙∙∙∙∙∙∙∙∙∙ (13)

For b=0 in (2) (uniaxial, equal biaxial), τ − τ = 0

So, the only nonzero normal stress difference is:

τ − τ (14)

3.1.3 Material Functions for Shear Flows

Various types of shear flows used in the characterizations of polymeric fluids

(Fig 3.3)

Steady shear flows :

γ = γ = constant

τ = −η(γ)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (15) τ − τ = −ψ(γ)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (16) τ − τ = −ψ(γ)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (17)

η ∶ Shear viscosity

ψ ∶ Primary normal stress difference coefficient

ψ ∶ Secondary normal stress difference coefficient

See Fig. 3.4 for steady data as well as dynamic ones.

Fig. 3.4 here Note:

ψ is more shear dependent than η. ψ > 0, large ψ < 0 small => Neglected in processing

=> −ψ ψ = 0.1~0.2

There are numerous transient shear flows (unsteady) where shear rate varies in a

specific way with time. One most frequently used is γ = γ cosωt (18).

------------------------Briefly discussed or skipped--------------------------------------

Constitutive Eq.

100% elastic solid τ = Gγ 100% viscous fluids τ = −ηγ

More commonly γ andτ are given in terms of complex variables to make

up for the phase lag between γ andτ.

So,

γ∗ = γe = γ(cosωt + sinωt) = γ + γ′′ τ∗ = τe() = τ[cos(ωt + δ) + sin(ωt + δ)] Then

Complex modulus (G∗) G∗ = ∗∗ = () = e = (cos δ + sin δ) = G + G′′

Storage modulus G = cos δ Loss modulus G = sin δ Complex viscosity

η∗ = ∗∗ = ∗ = ∗∗ η∗ = e = − (cos δ + sin δ) = sin δ − cos δ = − () = η − η′′

So G = ωη′′ G′′ = ωη′ tanδ = G′′G′ = η′η′′ ( )

G*=G’+iG” =iω η* (24)

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Back to the text:

Viscous point of view

γ = γ cosωt ∙∙∙∙∙∙∙∙∙∙∙∙ (18) => τ = −B(ω)γ0 cos(ωt − ϕ) = −B(ω)γ0cosϕ cosωt − B(ω)γinsϕ sin ωt

This allows us to define a complex viscosity as

η*= η’- iη”

where

η’=−B(ω)γ0cosϕ

(in phase with γ(t)=dynamic viscosity=viscosity contribution) η” = B(ω)γinsϕ

(90o out of γ(t)=loss part of complex viscosity=elastic

contribution associated with energy storage per cycle)

→See close correlation of η* and η in Fig 3.4.

Stress growth (Fig 3.3c)

Suddenly set in motion and stress is measured as a function of time (t)..

γ(t) = γH(t) Unit step function H(t) = 0; t < 0andH(t) = 1; t > 0

Stress growth materials function (Fig 3.7)

τ = −η(t, γ)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (26) τ − τ = −ψ1(t, γ)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (27) τ − τ = −ψ2(t, γ)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (28)

Stress relaxation after cessati

Deformation history (

γ(t) = γ[ H(t) = 0; t <

τ = −η( τ − τ = τ − τ = τ relaxes faster than

As γ increases, relaxation time becomes shorter

3.1.4 Shear Free Flow Material

=elongational or extensional flow

τ − τ = τ − τ =

Steady uniaxial extrusion

τ − τ = => η(ϵ) = η(ϵ, 0)

cessation of a steady shear flow (Fig 3.3d)

eformation history (See Fig 3.7) [1 − H(t)] ∙∙∙∙∙∙∙∙∙∙∙∙ (30) < 0andH(t) = 1; t > 0 γ, t)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (31) −ψ1(γ, t)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (32) −ψ2(γ, t)γ ∙∙∙∙∙∙∙∙∙∙∙∙ (33)

relaxes faster than τ − τ.(Note from the figure)

increases, relaxation time becomes shorter (Observed)

aterial Functions

elongational or extensional flow −η(ϵ, b)ϵ ∙∙∙∙∙∙∙∙∙∙∙∙ (34) −η(ϵ, b)ϵ ∙∙∙∙∙∙∙∙∙∙∙∙ (35) 0 => η = 0(Leaves one viscosity function) ) ∙∙∙∙∙∙∙∙∙∙∙∙ (36) See Fig 3.9

figure)

(Observed).

ne viscosity function)

At steady state (Fig 3.9)

η = 3η at low shear (

η (shear viscosity) shows shear thinning.

η increases slightly w/ stress and then decrease

Stress growth

Two viscosity functions are defined

τ − τ = τ − τ =

Uniaxial flow defines

η =η η = 0

Linear polymer tends

w/ branching or w/

at low shear (Trouton viscosity).

(shear viscosity) shows shear thinning.

increases slightly w/ stress and then decrease

Two viscosity functions are defined for stress growth. −ηϵ ∙∙∙∙∙∙∙∙∙∙∙∙ (37) −ηϵ ∙∙∙∙∙∙∙∙∙∙∙∙ (38) defines only one viscosity function. ,

tends to reach steady value (Fig 3.10), but

w/ HMW tail →to increase w/o bound (Fig 3.11

).

Fig 3.11)

Fig 3.11 here

3.2 Nonlinear Constitutive Eq.

Constant viscosity (μ) does not describe shear dependence.

GNF ηdoes not describe time dependence (nonelastic).

Nonlinear constitutive eq describe time dependence + nonlinear behavior

Maxwell model-Review

τ = Gγ γ = (Reversible deformation-Elastic)

τ = ηγγ = (Dissipation-Viscous)

Stress τ = τ = τ(Needs a mechanic model for Maxwell here)

Strain

γ = γ + γ

= (γ + γ) = + => τ + = η = ηγ

Define λ = ηG mLtmLt2L2= t: Relaxation time

Then τ + λ = η = ηγ (One dimensional model)

τ + λ = −ηγ (Tensor form)

τ + λ = −ηγ (Component form) ∙∙∙∙∙∙∙∙∙∙∙∙ (39) For stress relaxation

γ = = 0, = − lnτ = − + c @t = 0, τ = τ, c = lnτ => τ = τe/

Generalized Maxwell model-linear viscoelastic model

(See LN 184 for the solution of (39) as)

GNF (power-law model etc) and linear viscoelastic models (generalized

Maxwell model ) do not predict the normal stress difference. For example

for steady SSF, v = γy,v = v = 0

τ τ ττ τ ττ τ τ = −ηγ γ γγ γ γγ γ γ = −η0 γ 0γ 0 00 0 0

γ = γ = So, τ = τ = −ηγ τ − τ = 0 → Do not predict N, N τ−τ = 0

How can the model predict

→Cast the time derivative (39) into codeforming or corotating coordinates

i) Codeformation model

Cast the time derivative

=>

τ + λτ = where,

τ = ∂∂t +

Recognize the first two terms in RHS are

This CE is called UCM

White-Metzner (WM

See Table 3.1 for other models

ii) Corotational Model-CEF

Tadmor and Gogos (Ex 3.5

The CEF constitutive equation in

vz (r)=vc (1-(r/R)2).

ow can the model predict , ? ast the time derivative (39) into codeforming or corotating coordinates

Codeformation model

the time derivative (39) into codeforming (and translating

= −ηγij ∙∙∙∙∙∙∙∙∙∙∙∙ (40) (v ∙ ∇τ) − (∇v) τ + τ (∇v)

Recognize the first two terms in RHS are substantial time derivative.

UCM (upper convected Maxwell model) or

(WM) model (μ → η) for other models.

CEF (Criminale-Ericksen-Filbery) Eq

Ex 3.5)

quation in steady, fully developed flow in tube

ast the time derivative (39) into codeforming or corotating coordinates!

and translating) coordinate!

∙∙∙∙∙∙∙∙∙∙∙∙ (41)

or

ube:

τ = −ηγ − ψ + ψ {γ ∙ γ} + ψ γ CEF (E3.5-1)

=Corotational derivative.

----------------------------------------------------------------------------------------

Rate-of-deformation tensor is obtained from : vz(r)=vc(1-(r/R)2).

= + () = 0 0 γ0 0 0γ 0 0 (E3.5-2)

{γ ∙ γ} = γ 0 00 0 00 0 γ (E3.5-3)

Next we calculate

γ = γ + {v ∙ ∇γ} + ({ω ∙ γ} −{γ ∙ ω}) (E3.5-4)

γ = 0(ss)

The components of ∙∇γ is obtained from Table (Table 3.2, T&G)

(v ∙ ∇γ) = (v ∙ ∇)γ −vr γ = v + +v γ − γ(E3.5-5)

Since v = 0, v = 0,and ∂v ∂z = 0,⁄ (v ∙ ∇γ) = 0.

The vorticity tensor can be obtained as

= − () = 0 0 γ0 0 0−γ 0 0 (E3.5-6)

Then,

{ω ∙ γ} = 0 0 γ0 0 0−γ 0 0 0 0 γ0 0 0γ 0 0 = γ 0 00 0 00 0 −γ (E3.5-7)

And {γ ∙ ω} = −γ 0 00 0 00 0 γ (E3.5-8)

Thus, Eq. E3.5-4 reduces to

γ = ({ω ∙ γ} − {γ ∙ ω}) = γ 0 00 0 00 0 −γ (E3.5-9)

Finally, we substitute Eqs. E3.5-2, E3.5-3, and E3.5-9 into Eq. E3.5-1(CEF) to obtain

the stress field

τ τ ττ τ ττ τ τ =−η(γ) 0 0 γ0 0 0γ 0 0 − Ψ(γ) + Ψ(γ)γ 0 00 0 00 0 γ

+ Ψ(γ)γ 0 00 0 00 0 −γ (E3.5-10)

τ =τ =−ηγ τ =−12Ψ +Ψ γ +12ψγ = −ψγ τ = 0 τ =− 12Ψ +Ψ γ −12ψγ =−(Ψ +ψ)γ

(E3.5-11) τ =τ =−ηγ (E3.5-12) τ −τ =τ −τ =−Ψγ (E3.5-13) τ −τ =τ −τ =−Ψγ (E3.5-14)

• Viscoelastic model is used to predict die swell (Tanner)

= 0.1 + [1.0 + (, )] ∙∙∙∙∙∙∙∙∙∙∙∙ (3.89)

Dp=Extrudate diameter, Do=Capillary diameter

• Cases where nonlinear viscoelastic model is used = Unsteady flow

•Viscoelastic effect is also important in steady flow in capillary flow, slit die flow if

residence time (t) < longest relaxation time (λ), or Deborah number

D ≡ ≫ 1→ Elastic effect is important.

•For example in injection molding if the melt stress relaxes slowly relative to the

heat transfer rate => Residual stress are frozen into the part

3.3 Rheometry

3.3.1 Shear flow measurement

Low shear property => rotatry rheometry

ie cone-and-plate (Fig 3.18) and parallel plate (Fig 3.19)

High shear property => capillary rheometer

Rotatry Rheometry

Fig 3.19 here

Steady property =η(γ), G(γ)…

Dynamic property =η∗(η, η′′), G∗(G, G ), tanδ = = (= f(ω)) η = ψ = γ = vθ= = (cone and plate)

cone angle

Capillary Rheometry

τ = (− ) ∙∙∙∙∙∙∙∙∙∙∙∙ (91) Pressure drop across the capillary+end effect (Fig 3.21)

−() = − ΔP = ΔP + ΔP ∙∙∙∙ (92) Capillary length

→ τ = ∆ (Wall shear stress)

End effect and wall shear rate corrections

▶End effect correction

1) Plot ΔP vs L D at various γ for several L D capillaries

=>GetΔP = ΔP at L D = 0 =>τ = ( ) (See Fig 3.22)

2) Correction by equivalent dia length required to produce P

Extrapolation of ΔP to 0 => Read L D value

(L D ) = L + Nentx(D 2 ) D ∙∙∙∙∙∙∙∙∙∙∙∙ (94)

Corrected

=> τ = = ∙∙∙∙∙∙∙∙∙∙∙∙ (95)

▶Wall shear rate correction ∫ udv = uv − ∫ v du Volumetric flow rate

Q = ∫ vds = ∫2πr vdr = v(πr)| − ∫πrdv (dv = 2πrdr, v = πr) Q = −π∫ rd v = +π∫ r(− ) dr = +π∫ rγd r (γ = − > 0) Chang of variable:

τ = r

r = τ,dr = dτ

∴ Q = π∫ ( τ) γ( )dτ = ∫ τ γdτ → ( )τ ≡ Γτ = 4∫ τ γdτ Γ≡ = Newtonian shear rate @ wall

Differentiation wrt τ 0

(Γ) = 4 ∫ (τγ)dτ + τγ| − τγ

= 4τγ ←true shear rate @ wall

Leibnitz formula applied

∫ f(x, t)()() dx = ∫ f(x, t)dx + f(a, t) − f(a, t)

∴ γ = (Γ)

= 3τ Γ + τ Γ = Γ + Γ Γ = Γ ΓΓ = τ Γ = Γ

where n′ = Γ ←From pressure drop measurement

i.e., γ = Γ( ) Newtonian shear rate @ wall = (Measured)

True shear rate @ wall

From Newtonian’s law of viscosity

τ = ηγ

→ η = (Calculated)

3.3.2 Shear free flow measurement

Measurement of uniaxial viscosity (Fig 3.24)

Ballmen method (a)

Polymer melt is either glued or clamped at both ends, and then one end is

moved in such a way as to generate a uniform extension rate (ϵ). That is, the

length of sample should increase exponentially with time as L=Lo exp(ϵt).

τ − τ = −η(ϵ, b)ϵ → η =

For uniaxial extrusion

b = 0,ϵ > 0,η = 0 -> η(ϵ) = η(ϵ, 0)

Viscosity measured by this method is the startup viscosity. Π = −(F(t) A(t)) + P⁄ ∙∙∙∙∙∙∙∙∙∙∙∙ (102) Π = Pa Π − Π = τ − τ = −(F(t) A(t)⁄ ) ∙∙∙∙∙∙∙∙∙∙∙∙∙ (103)

Volume=A (t) L= Ao Lo(= 1), L=Lo exp(ϵt)=e → A = Ao /e η = − = () ∙∙∙∙∙∙∙∙∙∙∙∙∙ (104)

Meissner method (b)

Both ends of the materials are pulled at a constant velocity or a constant stress.

3.4 Useful Relationships for Material Functions

▶η ∝ MforM < M ↑ critical mol.wt η ∝ M.forM > M

▶The power is

3.4 for nonpolar polymer (PE,PP)

>3.4 for polar polymer (SAN…)

▶For γ → 0 ψ, ∝ η ∝ M

▶Cox-Merz rule η(γ) = η∗(ω)| ω[=] , γ[=]1/s (N/2)| = G′ N = 2ωη′′(ω) (1 + ( )).

Solution to the design problem?

Design a die that will allow one to extrude at the highest rate possible for

LDPE (NPE953) with (Fig 3.1), diameter D = 5.13cm, thickness H = 0.565츠.

Specify die dia D and Gap Husing rheology data in Appendix A.1.

Assume L=10H Maximum feed to die, m = 300cc/h @melt fracture, τ = 1.13 × 10Pa

In your calculation consider the extrudate swell:

Die swell ratio =D/D

Die swell is given by Tanner eq as = 0.1 + [1.0 + (, )] ∙∙∙∙∙∙∙∙∙∙∙∙ (3.89) Shear stress of annular flow is derived in class as τ == P0′−PL′2L R rR− (1−k2)2ln1k Rr (2.29) or Table 2.7

Maximum stress occurs at r=R, i.e.,

τ (R) = τ, = (P − P)2L R 1 − (1 − k)2ln 1k Wallshearrateis then obtained by τ, = ηγω →γ=τ,/η Then, N1,w is obtained from the rheology data(ψ(γ)) → Dofrom(89), Hofrommasslow rate(m) data.

Now you are able to design an annular die with desired product dimension.