Computational Rheology (4K430)

dr.ir. M.A. Hulsen [email protected]

Website: http://www.mate.tue.nl/~hulsen under link ‘Computational Rheology’.

– Section Polymer Technology (PT) / Materials Technology (MaTe) – Introduction

Computational Rheology important for:

B Polymer processing

B Rheology & Material science

B Turbulent flow (drag reduction phenomena)

B Food processing

B Biological flows

B ... Introduction (Polymer Processing)

Analysis of viscoelastic phenomena essential for predicting

B Flow induced crystallization kinetics

B Flow instabilities during processing

B Free surface flows (e.g.extrudate swell)

B Secondary flows

B Dimensional stability of injection moulded products

B Prediction of mechanical and optical properties Introduction (Surface Defects on Injection Molded Parts)

Alternating dull bands perpendicular to flow direction with high surface roughness (M. Bulters & A. Schepens, DSM-Research). Introduction (Flow Marks, Two Color Polypropylene)

Flow Mark

Side view

Top view

Bottom view

M. Bulters & A. Schepens, DSM-Research Introduction (Simulation flow front)

1

0.5 Steady Perturbed H 2y 0 ___

−0.5

−1 0 0.5 1 ___2x H Introduction (Rheology & Material Science)

Simulation essential for understanding and predicting material properties:

B Polymer blends (morphology, viscosity, normal stresses)

B Particle filled viscoelastic fluids (suspensions)

B Polymer architecture macroscopic properties (Brownian dynamics (BD), molecular dynamics (MD),⇒ Monte Carlo, . . . )

Multi-scale. ⇒ Introduction (Solid particles in a viscoelastic fluid)

B Microstructure (polymer, particles)

B Bulk rheology

B Flow induced crystallization Introduction (Multiple particles in a viscoelastic fluid) Introduction (Flow induced crystallization) Introduction (Multi-phase flows) Goal and contents of the course

Goal: Introduction of the basic numerical techniques used in Computational Rheology using the Finite Element Method (FEM).

Contents (tentatively):

B Basic equations from Continuum Mechanics and Rheology B Introduction of basic FEM techniques: Galerkin method, mixed methods, Petrov-Galerkin: SUPG, DG, time discretization. B FEM for flow problems. Navier-Stokes, Mixed Stokes. Viscoelastic. B Stabilization techniques for viscoelastic flows. B Benchmarks B Micro-macro methods B Integral models B Suspensions B ... Configurations

reference configuration time τ region Vτ P present configuration time t ~ dX region Vt path of P particle P ~e X~ 3 d~x ~x

~e2 ~e r 1 O Deformation (1)

material description (Lagrangian): T = Tm(X,~ t) spatial description (Eulerian): T = Ts(~x,t) mapping deformation: ≡ ~x = ~x(X,~ t) X~ = X~ (~x,t) ⇔ deformation gradient (local deformation):

∂~x ∂xi d~x = F dX,~ F = ,Fij = · ∂X~ ∂Xj deformation of local volume: dV J = det F = > 0 dVτ Deformation (2)

Deformation tensors: C = F T F “Green” · B = F F T “Finger” · Rates (1) material derivative:

DT ∂T (X,~ t) ∂T = m = = T˙ Dt ∂t ∂t X~ =constant

local derivative: ∂T ∂T (~x,t) ∂T = s = ∂t ∂t ∂t ~x=constant velocity: D~x ~u = = ~x˙ Dt acceleration: D2~x ~a = ~u˙ = = ~x¨ Dt2

DT ∂T = + ~u T Dt ∂t · ∇ Rates (2) velocity gradient tensor:

D 1 T ∂ui (d~x) = L d~x with L = F˙ F − = ( ~u) ,Lij = Dt · · ∇ ∂xj rate-of-deformation tensor: 1 D = (L + LT ) 2 vorticity tensor: 1 W = (L LT ) 2 − volume-rate-of-deformation:

J˙ = ~u = divergence of velocity ~u J ∇ · Balance (conservation) laws in Eulerian frame (1)

Conservation of mass ρ˙ + ρ ~u = 0 ∇ · for constant density fluids: ρ˙ = 0:

~u = 0 ∇ · Linear momentum balance Cauchy stress tensor σ gives the ‘traction’ on surface with normal ~n:

~t = σT ~n = ~n σ · · ρ~u˙ = σ + ρ~b ∇ · constant density fluids:

σ = pI + t, p : hydrostatic pressure, t : extra-stress tensor, − Balance (conservation) laws in Eulerian frame (2)

angular momentum balance

σT = σ symmetric energy balance ρε˙ = ~q + σ : D + ρr ∇ · with

ε internal energy per unit mass ~q heat flux vector: amount of energy flowing through a surface with a normal ~n per unit area q = ~q ~n by conduction · r body heat source. CEs for the stress tensor

Constant density fluids: σ = pI + t − Newtonian fluids: t = 2ηD with viscosity η a constant.

Viscoelastic fluids: for example the Oldroyd-B fluid

t = 2ηsD + τ with λ τ5 +τ = 2ηD where τ5= τ˙ L τ τ LT − · − · η = 0 upper-convected Maxwell (UCM) fluid s ⇒ Linear viscoelastic fluid

γ τ τ G0 γ 1 step strain modulus G(t)

t t

Linear response theory (Boltzmann superposition):

t dG τ(t) = M(t t0)[γ(t) γ(t0)] dt0,M(t) = (t) − − − dt Z−∞ Elastic reponse at t = 0+:

0 + + + τ(0 ) = M( t0) dt0 γ(0 ) = G γ(0 ) − 0 Z−∞  Non-linear viscoelastic fluid (integral model)

Neo-Hookean elastic model τ = G(B I) − Viscoelastic (Lodge rubber like liquid)

t τ (t) = M(t t0)[B (t) I] dt0 − t0 − Z−∞ Spectrum with single relaxation time

t/λ G0 t/λ G(t) = G e− ,M(t) = e− 0 λ and t G (t t0)/λ τ (t) = e− − [B (t) I] dt0 λ t0 − Z−∞ G G 0 → Non-linear viscoelastic fluid (differential model)

Differentiating to time t

t t G (t t0)/λ G (t t0)/λ τ˙ = 0 e− − [B (t) I] dt0 + e− − B˙ (t) dt0 − λ2 t0 − λ t0 upper boundary Z−∞ Z−∞ |{z} T With B = F F from t0 to t and F˙ = L F · · T B˙ = F˙ F T + F F˙ = L B + B LT · · · · we get

t τ T G (t t0)/λ T τ˙ = + L τ + τ L + e− − dt0(L + L ) −λ · · λ Z−∞ τ = + L τ + τ LT + 2GD −λ · · Non-linear viscoelastic fluid (Oldroyd-B, UCM)

⇒ λ τ5 +τ = 2ηD where

τ5 = τ˙ L τ τ LT − · − · η = Gλ properties:

B constant steady state viscosity η B single relaxation time λ 2 B steady state first-normal stress difference N1 = 2ηλγ˙ 1 B no steady state elongation viscosity for ˙ > 2λ B second-normal stress difference N2 = 0. Exercise 1

The linear viscoelastic properties of a particular fluid can be modeled by

t/λ1 t/λ2 G(t) = G1e− + G2e−

Apply the procedure we used to derive the Lodge integral model to propose a non- linear model suitable for large deformation of the fluid. Derive the corresponding differential model. CEs for ε and ~q

We assume (in this course):

ε˙ = cT˙ , c : specific heat (constant),T : temperature

~q = k T, k : thermal conductivity (constant) − ∇ and:

σ does not depend on the thermal history

decoupling ⇒ Set of equations (summary)

Conservation of mass ~u = 0 ∇ · (Linear and angular) Momentum equation

ρ~u˙ = σ + ρ~b, with σ = σT ∇ · CE for the stress tensor Newtonian: σ = pI + τ , τ = 2ηD − Viscoelastic: (Oldroyd-B/UCM) σ = pI + 2η D + τ , λ τ5 +τ = 2ηD, τ5= τ˙ L τ τ LT − s − · − ·

Energy equation ρcT˙ = (k T ) + σ : D + ρr ∇ · ∇ Boundary and initial conditions Convection-diffusion-reaction equation

Γ D u(~x,t = 0) = u0(~x) in Ω

ΓN u = uD on ΓD ∂u ~n Ω A = A~n u = h on Γ − ∂n − · ∇ N ~n: outside normal ∂u + ~a u (A u) + bu = f ∂t · ∇ − ∇ · ∇ Energy equation:

k 1 r u = T, ~a = ~u, A = with A 0, b = 0, f = σ : D + ρc ≥ ρc c Set of equations (flow of a visco-elastic fluid) (1)

Conservation of mass ~u = 0 ∇ · (Linear and angular) Momentum equation

ρ~u˙ = σ + ρ~b, with σ = σT ∇ · Viscoelastic fluid model: (Oldroyd-B/UCM)

σ = pI + 2η D + τ , λ τ5 +τ = 2ηD, τ5= τ˙ L τ τ LT − s − · − · Boundary and initial conditions Set of equations (flow of a visco-elastic fluid) (2)

Rewrite: Find (~u,p, τ ) such that,

∂~u ρ( + ~u ~u) (2η D) + p τ = ρ~b, in Ω ∂t · ∇ − ∇ · s ∇ − ∇ · ~u = 0, in Ω ∇ · ∂τ λ( + ~u τ L τ τ LT ) + τ = 2ηD, in Ω ∂t · ∇ − · − ·

Boundary and initial conditions Convection-diffusion-reaction equation

Γ D u(~x,t = 0) = u0(~x) in Ω

ΓN u = uD on ΓD ∂u ~n Ω A = A~n u = h on Γ − ∂n − · ∇ N ~n: outside normal ∂u + ~a u (A u) + bu = f ∂t · ∇ − ∇ · ∇ Finite Element Method (FEM)

Approximation method:

d du (A ) = f Ku = f −dx dx ⇒ ¯ ˜ ˜ where u: an approximate solution using a finite number of unknowns N. ˜ For N :“u u” → ∞ ˜ →

B quite general distribution of ‘elements’ without losing accuracy

B local refinements near large gradients

B quite general geometries in multiple dimensions Linear spaces (1)

u V linear space V v u V, v V, w V ∈ ∈ ∈ w λ R, µ R ∈ ∈

B u + v V λu V ∈ B ∈ B (u + v) + w = u + (v + w) B λ(u + v) = λu + λv 0 such that u + 0 = u λ(µu) = (λµ)u B ∃ B B u such that u + ( u) = 0 1 u = u ∃ − − B · Linear spaces (2)

Elements of a linear space V : linear combination of independent base vectors. With N base vectors: N-dimensional space.

N

u = uiei i=1 X Independence: N α e = 0 α = 0 i i ⇒ i i=1 X Examples (1)

B 3D (N = 3) physical space Any 3 non-zero vectors not in a plane can act as a base.

All periodic functions on ( π, π) B −

f

π π − g

Fourier expansion:

1 ∞ f(x) = a + a cos(kx) + b sin(kx) 2 0 k k k=1 X
Examples (2)

with 1 π 1 π ak = f(x) cos(kx) dx, bk = f(x) sin(kx) dx π π π π Z− Z− Base: 1, cos x, sin x, cos 2x, sin 2x, . . . N = ∞ 0 B All continuous function on (a, b): C (a, b).

2 B All square integrable functions on (a, b): L (a, b):

b f L2(a, b) then f 2 dx < ∈ ∞ Za

Note: δ(x) L2(a, b): 6∈ Examples (3)

h f(x)

x 2/h

f(x) dx = 1 Z 2h f 2(x) dx = for h 3 → ∞ → ∞ Z Inner product and norm

Inner product (u, v):

(u, v) = (v, u) for all u, v V B ∈ (αu + βv, w) = (αu, w) + (βv, w) for all u, v, w V, α, β R B ∈ ∈ (u, u) 0 for all u, v V B ≥ ∈ B (u, u) = 0 implies u = 0 u and v are orthogonal is (u, v) = 0.

1 Norm u = (u, u)2 . k k Distance between u and v: u v . k − k Series u u, k = 1,..., , converges if u u 0 for k . k → ∞ k k − k → → ∞ Linear space with inner product: Hilbert space. Example

3D physical space: (~a,~b) = ~a ~b = ~a ~b cos φ · | || | ~a = ~a 2 = ~a k k | | | | orthogonal (orthonormal) base (~e1,~e2,~ep3):

~e ~e = δ i · j ij We have, due to orthogonality

~a = ai~ei, with ai = (~a,~ei)

Note: if u is orthogonal to all vectors:

(u, v) = 0, for all v u = 0 ⇒ Example

All periodic functions on ( π, π) that are square integrable, with − π (f, g) = fg dx π Z− and thus π f 2 = f 2 dx k k π Z− Notes:

π 2 B (f, f) = π f dx 0 − ≥ R π 2 B (f, f) = 0 π f dx = 0 f = 0 ⇒ − ⇒ R B Base: 1, cos x, sin x, cos 2x, sin 2x, . . . are orthogonal functions. B Expansion: ∞ f(x) = αkek(x) kX=1 inner product ∞ ∞ (f, ei) = ( αkek, ei) = αk(ek, ei) kX=1 kX=1 orthogonal base: (e , e ) = 0 for k = i: k i 6

(f, ek) αk = (ek, ek)

Fourier expansion. ⇒Spectral convergence.

(u, v) = 0 for all v “periodic functions on ( π, π)” u = 0 B ∈ − ⇒ Example

All square integrable functions on (a, b): L2(a, b). Inner product: b (f, g) = fg dx Za with ‘induced’ norm: b f 2 = f 2 dx k k Za Notes:

(f, f) = b f 2 dx 0 B a ≥ R (f, f) = 0 b f 2 dx = 0 f = 0 B ⇒ a ⇒ R (u, v) = 0 for all v L2(a, b) u = 0 B ∈ ⇒ “Delta functions” are not allowed: [δ(x)]2 dx = B ∞ An orthogonal base for L2( 1, 1) areR the Legendre polynomials: B −

1.5 P0(x) P1(x) P3(x) 1

0.5

0

-0.5

-1 -1 -0.5 0 0.5 1 x 1 P (x) = 1,P (x) = x, P (x) = (3x2 1),... 0 1 2 2 − with 1 0 m = n 6 (Pm,Pn) = Pm(x)Pn(x) dx = 2 1  m = n Z− 2n + 1  Exercise 2

a. Show that the base:

1, cos x, sin x, cos 2x, sin 2x, . . . are orthogonal functions. b. Show that P0(x),P1(x),P2(x), are orthogonal and construct P3(x). Hint: write P3(x) as:

3 P3(x) = a0P0(x) + a1P1(x) + a2P2(x) + a3x Example

All square integrable functions u L2(a, b) with du also square integrable ∈ dx (du L2(a, b)) on (a,b): H1(a, b). H1 is a Hilbert space with inner product: dx ∈ b df dg (f, g) = ( + fg) dx 1 dxdx Za 1 B H is called a Sobolev space

m B Generalizable to H (a, b)

When we restrict u: u(a) = 0, u(b) = 0 (zero on the boundary) the space is 1 called H0 (a, b) and an alternative inner product is:

b df dg a(f, g) = [f, g] = dx dxdx Za Projections

P

~p

~q line ℓ spanned O Q by vector ~e: ~r ~r = ~r0 + ~v with R ~r0 ~v = λ~e

~e

Point Q is closest to P of all points on the line `, with ~p ~q orthogonal to this line or: − (~v, ~p ~q) = 0 for all ~v in ` − Point Q is the orthogonal projection of P on the space spanned by ~e. Note:

~p ~q = min ~p ~r k − k ~r ` k − k ∈ Projections in function spaces

Inproduct and norm gives smallest distance between functions. For example:

function u (x) V , where V is a finite approximation space B N ∈ N N B function f(x), is the “exact” function that needs to approximated and is not in VN .

The best approximation is the function u with f u orthogonal to V : N − N N (v, f u ) = 0 for all v V − N ∈ N b v(x)(f(x) u (x)) dx = 0 for all v V ⇒ − N ∈ N Za

Note, uN (x) is the function that minimizes the distance (least squares):

b min f z 2 = min (f z)2 dx z VN k − k z VN − ∈ ∈ Za Example

A periodic function f(x) on [ π, π]. Approximate (space G ): − N

N

gN (x) = αkek kX=1 with N finite, ek the orthogonal base 1, sin x, cos x, . . . . Then

π v(f gN ) dx = 0, for all v GN π − ∈ Z− and thus: (f, ek) αk = (ek, ek) Truncated Fourier expansion. Shape functions

The shape functions φi(x) are the “base vectors” of the approximation space:

N T uN (x) = uiφi(x) = φ (x)u i=0 ˜ ˜ X with T T φ = (φ0(x), φ1(x), . . . , φN (x)), u = (u0, u1, . . . , uN ) ˜ ˜ For a N th-order polynomial we could use

φT (x) = (1, x, x2, . . . , xN ) ˜ however it is more practical to use another base, such as Lagrangian interpolants. Lagrangian interpolation

N = 5

u4 u5 xi: nodal points u3 u2 ui: nodal values u0 u1

x0 = a x1 x2 x3 x4 x5 = b

N T uN (x) = uiφi(x) = φ (x)u i=0 ˜ ˜ X with (x x0) (x xi 1)(x xi+1) (x xN ) φi(x) = − ··· − − − ··· − (xi x0) (xi xi 1)(xi xi+1) (xi xN ) − ··· − − − ··· − Note: φi(xj) = δij Exercise 3

Construct linear (N = 1), quadratic (N = 2) and third-order (N = 3) shape functions on an interval [ 1, 1] with equidistant spaced nodal points. Are the shape functions obtained orthogonal?− Linear system for approximate solution

b v(x)(f(x) u (x)) dx = 0 for all v V − N ∈ N Za T T Fill in uN (x) = φ (x)u, v(x) = φ (x)v: ˜ ˜ ˜ ˜ b vT φ[f φT u] dx = 0 for all v ˜ − ˜ ˜ Za ˜ ˜ b b [ φφT dx]u = φf dx ⇒ ˜ Za ˜˜ Za ˜ or Ku = f ¯ ˜ ˜

b T b with K = a φφ dx, f = a φf dx ¯ b ˜˜ ˜ b ˜ or Kij = Ra φiφj dx, fi = Ra φif dx. R R 1D diffusion equation

1D diffusion Eq.: find u(x) such that for x (a, b) ∈ d du (A ) = f −dx dx and

u = ua, at x = a (ΓD) du A = h at x = b (Γ ) − dx b N Notes:

B Strong form; Classical (strong) solution u(x)

0 2 B f(x) C (a, b) (continuous) then u C (a, b) (twice continuously differentiable)∈ ∈ Residual

Residual d du r(x) = (A ) f −dx dx − should be zero for the exact solution:

r(x) = 0

For an approximate solution u (x) V we get N ∈ N r (x) = 0 N 6 and rN VN . We want the approximate solution uN (x) to be such that the residual is6∈ as small as possible or

r V N ⊥ N Weighted residuals

Therefore we multiply r(x) = 0 with a test function or weighting function v

b (v, r) = v(x)r(x) dx = 0 for all v Za Reversely, if this is true for the complete space we get r(x) = 0 again.

Method of weighted residuals ⇒ Substitute r(x):

b d du (v, r) = v(x) (A ) f dx = 0 for all v a − dx dx − Z   Space S for u(x): trial space (H2(a, b)). Space V for v(x): test space (L2(a, b)). Rhs f(x) L2(a, b). Already weaker than the strong form. ∈ Weak form of 1D diffusion equation

Partial integration of second-order term:

dv du du b ( ,A ) v(A ) = (v, f) for all v dx dx − dx a

Boundary conditions: B at x = a, u = ua (Dirichlet condition) – restrict trial solutions space S to u(a) = ua – restrict test function space V to v(a) = 0. at x = b, Adu/dx = h (Neumann condition) B − b Weak form: Find u S such that ∈ dv du ( ,A ) + v(b)h = (v, f) for all v V dx dx b ∈

1 with S = u H (a, b) with u = ua at ΓD and V = {v ∈H1(a, b) with v = 0 at Γ .} { ∈ D} Galerkin approximations

Key elements for an approximate solution using the weak form:

1. expansion/approximation of the trial solution u:

N T uN (x) = uiφi(x) = φ (x)u i=0 ˜ ˜ X

Forms a finite dimensional subspace of S: SN and must converge to any function in S for N . → ∞

2. choice of the test functions vN , which form the finite dimensional space VN to which the residual is orthogonal. In the Galerkin method we take VN =SN :

N T vN (x) = viφi(x) = φ (x)v i=0 ˜ ˜ X Linear system of equations for the approximation

Substitution into the weak form:

vT Ku = vT f for all v ˜ ¯ ˜ ˜ ˜ ˜ or Ku = f ¯ ˜ ˜ with

T b T dφ dφ dφ dφ dφi dφj K = ( ˜,A ˜ ) = ˜A ˜ dx Kij = ( ,A ) ¯ dx dx dx dx dx dx Za b   f = (φ, f) hbφ(b) = φf dx hbφ(b) fi = (φi, f) hbφi(b) ˜ ˜ − ˜ a ˜ − ˜ − Z   Note: No Dirichlet conditions yet. “Distance” of approximate to exact solution

dv du Exact u(x), approximate u (x). We get, with a(v, u) = ( ,A ) h dx dx

a(v, u) + v(b)h = (v, f) for all v V b ∈ a(v , u) + v (b)h = (v , f) for all v V h h b h h ∈ h a(v , u ) + v (b)h = (v , f) for all v V h h h b h h ∈ h a(v , u u ) = 0 for all v V h − h h ∈ h u uh is orthogonal to Vh with inner product a(u, v) or uh has a minimum distance− to u with respect to the energy norm:

b 1 du du u = a(u, u)2 with a(u, u) = A dx k ke dx dx Za Exercise 4

a. Explain the weighted residual method. Will the residual be zero after applying the method? b. Explain the Galerkin method. Will the residual be zero after applying the method?