Computational Rheology (4K430)
dr.ir. M.A. [email protected]
Website: http://www.mate.tue.nl/~hulsenunder 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)
Top view
Bottom view
Side view
Flow Mark
M. Bulters & A. Schepens, DSM-Research
Introduction (Simulation flow front)
0 0.5 1−1
−0.5
0
0.5
1
2xH
___
2y H___
Steady Perturbed
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 ComputationalRheology 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
particle P
reference configuration
time τregion Vτ present configuration
time t
region VtP
d~x
path ofd ~X
P
~X~e3
~e1 Or~e2
~x
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):
d~x = F · d ~X, F =∂~x
∂ ~X, Fij =
∂xi∂Xj
deformation of local volume:
J = detF =dV
dVτ> 0
Deformation (2)
Deformation tensors:C = F T · F “Green”
B = F · F T “Finger”
Rates (1)
material derivative:
DT
Dt=∂Tm( ~X, t)
∂t=∂T
∂t
∣∣∣~X=constant
= T
local derivative:∂T
∂t=∂Ts(~x, t)
∂t=∂T
∂t
∣∣∣~x=constant
velocity:
~u =D~x
Dt= ~x
acceleration:
~a = ~u =D2~x
Dt2= ~x
DT
Dt=∂T
∂t+ ~u · ∇T
Rates (2)
velocity gradient tensor:
D
Dt(d~x) = L · d~x with L = F · F−1 = (∇~u)T , Lij =
∂ui∂xj
rate-of-deformation tensor:
D =12
(L+LT )
vorticity tensor:
W =12
(L−LT )
volume-rate-of-deformation:
J
J= ∇ · ~u = divergence of velocity ~u
Balance (conservation) laws in Eulerian frame (1)
Conservation of mass
ρ+ ρ∇ · ~u = 0
for constant density fluids: ρ = 0:
∇ · ~u = 0
Linear momentum balanceCauchy 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 balanceρε = ∇ · ~q + σ : D + ρr
with
ε internal energy per unit mass
~q heat flux vector: amount of energy flowing through a surface with anormal ~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
withλ5τ +τ = 2ηD
where5τ= τ −L · τ − τ ·LT
ηs = 0 ⇒ upper-convected Maxwell (UCM) fluid
Linear viscoelastic fluid
t
modulus G(t)
G0
τ
1
t
step strain
γ
γ
τ
Linear response theory (Boltzmann superposition):
τ(t) =∫ t
−∞M(t− t′)[γ(t)− γ(t′)] dt′, M(t) = −dG
dt(t)
Elastic reponse at t = 0+:
τ(0+) =(∫ 0
−∞M(−t′) dt′
)γ(0+) = G0γ(0+)
Non-linear viscoelastic fluid (integral model)
Neo-Hookean elastic modelτ = G(B − I)
Viscoelastic (Lodge rubber like liquid)
τ (t) =∫ t
−∞M(t− t′)[Bt′(t)− I] dt′
Spectrum with single relaxation time
G(t) = G0e−t/λ, M(t) =
G0
λe−t/λ
and
τ (t) =∫ t
−∞
G
λe−(t−t′)/λ[Bt′(t)− I] dt′
G0 → G
Non-linear viscoelastic fluid (differential model)
Differentiating to time t
τ = 0︸︷︷︸upper boundary
−∫ t
−∞
G
λ2e−(t−t′)/λ[Bt′(t)− I] dt′ +
∫ t
−∞
G
λe−(t−t′)/λBt′(t) dt′
With B = F · F T from t′ to t and F = L · F
B = F · F T + F · F T = L ·B +B ·LT
we get
τ = −τλ
+L · τ + τ ·LT +∫ t
−∞
G
λe−(t−t′)/λ dt′(L+LT )
= −τλ
+L · τ + τ ·LT + 2GD
Non-linear viscoelastic fluid (Oldroyd-B, UCM)
⇒λ5τ +τ = 2ηD
where
5τ = τ −L · τ − τ ·LTη = Gλ
properties:
B constant steady state viscosity η
B single relaxation time λ
B steady state first-normal stress difference N1 = 2ηλγ2
B no steady state elongation viscosity for ε > 12λ
B second-normal stress difference N2 = 0.
Exercise 1
The linear viscoelastic properties of a particular fluid can be modeled by
G(t) = G1e−t/λ1 +G2e
−t/λ2
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 correspondingdifferential 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 tensorNewtonian:
σ = −pI + τ , τ = 2ηDViscoelastic: (Oldroyd-B/UCM)
σ = −pI + 2ηsD + τ , λ5τ +τ = 2ηD,
5τ= τ −L · τ − τ ·LT
Energy equationρcT = ∇ · (k∇T ) + σ : D + ρr
Boundary and initial conditions
Convection-diffusion-reaction equation
~n Ω
ΓD
~n: outside normal
ΓN
u(~x, t = 0) = u0(~x) in Ω
u = uD on ΓD
−A∂u∂n
= −A~n · ∇u = h on ΓN
∂u
∂t+ ~a · ∇u−∇ · (A∇u) + bu = f
Energy equation:
u = T, ~a = ~u, A =k
ρcwith A ≥ 0, b = 0, f =
1ρcσ : D +
r
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 + 2ηsD + τ , λ5τ +τ = 2ηD,
5τ= τ −L · τ − τ ·LT
Boundary and initial conditions
Set of equations (flow of a visco-elastic fluid) (2)
Rewrite: Find (~u, p, τ ) such that,
ρ(∂~u
∂t+ ~u · ∇~u)−∇ · (2ηsD) +∇p−∇ · τ = ρ~b, in Ω
∇ · ~u = 0, in Ω
λ(∂τ
∂t+ ~u · ∇τ −L · τ − τ ·LT ) + τ = 2ηD, in Ω
Boundary and initial conditions
Convection-diffusion-reaction equation
~n Ω
ΓD
~n: outside normal
ΓN
u(~x, t = 0) = u0(~x) in Ω
u = uD on ΓD
−A∂u∂n
= −A~n · ∇u = h on ΓN
∂u
∂t+ ~a · ∇u−∇ · (A∇u) + bu = f
Finite Element Method (FEM)
Approximation method:
− d
dx(Adu
dx) = f ⇒ K
¯u˜
= f˜
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)
V
w
v
u
linear space V
u ∈ V, v ∈ V,w ∈ Vλ ∈ R,µ ∈ R
B u+ v ∈ VB (u+ v) + w = u+ (v + w)B ∃0 such that u+ 0 = u
B ∃ − u such that u+ (−u) = 0
B λu ∈ VB λ(u+ v) = λu+ λv
B λ(µu) = (λµ)uB 1 · u = u
Linear spaces (2)
Elements of a linear space V : linear combination of independent base vectors.With N base vectors: N -dimensional space.
u =N∑i=1
uiei
Independence:N∑i=1
αiei = 0 ⇒ αi = 0
Examples (1)
B 3D (N = 3) physical spaceAny 3 non-zero vectors not in a plane can act as a base.
B All periodic functions on (−π, π)
π−π
f
g
Fourier expansion:
f(x) =12a0 +
∞∑k=1
(ak cos(kx) + bk sin(kx)
)
Examples (2)
with
ak =1π
∫ π
−πf(x) cos(kx) dx, bk =
1π
∫ π
−πf(x) sin(kx) dx
Base:1, cosx, sinx, cos 2x, sin 2x, . . .
N =∞
B All continuous function on (a, b): C0(a, b).
B All square integrable functions on (a, b): L2(a, b):
f ∈ L2(a, b) then
∫ b
a
f2 dx <∞
Note: δ(x) 6∈ L2(a, b):
Examples (3)
f(x)h
x2/h∫
f(x) dx = 1∫f2(x) dx =
2h3→∞ for h→∞
Inner product and norm
Inner product (u, v):
B (u, v) = (v, u) for all u, v ∈ VB (αu+ βv,w) = (αu,w) + (βv,w) for all u, v, w ∈ V, α, β ∈ RB (u, u) ≥ 0 for all u, v ∈ VB (u, u) = 0 implies u = 0
u and v are orthogonal is (u, v) = 0.
Norm ‖u‖ = (u, u)12.
Distance between u and v: ‖u− v‖.Series uk → u, k = 1, . . . ,∞, converges if ‖uk − u‖ → 0 for k →∞.
Linear space with inner product: Hilbert space.
Example
3D physical space:(~a,~b) = ~a ·~b = |~a||~b| cosφ‖~a‖ =
√|~a|2 = |~a|
orthogonal (orthonormal) base (~e1, ~e2, ~e3):
~ei · ~ej = δij
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
(f, g) =∫ π
−πfg dx
and thus
‖f‖2 =∫ π
−πf2 dx
Notes:
B (f, f) =∫ π−π f
2 dx ≥ 0
B (f, f) = 0 ⇒ ∫ π−π f
2 dx = 0 ⇒ f = 0
B Base:1, cosx, sinx, cos 2x, sin 2x, . . .
are orthogonal functions.
B Expansion:
f(x) =∞∑k=1
αkek(x)
inner product
(f, ei) = (∞∑k=1
αkek, ei) =∞∑k=1
αk(ek, ei)
orthogonal base: (ek, ei) = 0 for k 6= i:
αk =(f, ek)(ek, ek)
⇒ Fourier expansion.Spectral convergence.
B (u, v) = 0 for all v ∈ “periodic functions on (−π, π)” ⇒ u = 0
Example
All square integrable functions on (a, b): L2(a, b).Inner product:
(f, g) =∫ b
a
fg dx
with ‘induced’ norm:
‖f‖2 =∫ b
a
f2 dx
Notes:
B (f, f) =∫ baf2 dx ≥ 0
B (f, f) = 0 ⇒ ∫ baf2 dx = 0 ⇒ f = 0
B (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) are the Legendre polynomials:
-1
-0.5
0
0.5
1
1.5
-1 -0.5 0 0.5 1
x
P0(x)P1(x)P3(x)
P0(x) = 1, P1(x) = x, P2(x) =12
(3x2 − 1), . . .
with
(Pm, Pn) =∫ 1
−1
Pm(x)Pn(x) dx =
0 m 6= n2
2n+ 1m = n
Exercise 2
a. Show that the base:
1, cosx, sinx, cos 2x, sin 2x, . . .
are orthogonal functions.
b. Show thatP0(x), P1(x), P2(x),
are orthogonal and construct P3(x). Hint: write P3(x) as:
P3(x) = a0P0(x) + a1P1(x) + a2P2(x) + a3x3
Example
All square integrable functions u ∈ L2(a, b) with dudx also square integrable
(dudx ∈ L2(a, b)) on (a,b): H1(a, b). H1 is a Hilbert space with inner product:
(f, g)1 =∫ b
a
(df
dx
dg
dx+ fg) dx
B H1 is called a Sobolev space
B Generalizable to Hm(a, b)
When we restrict u: u(a) = 0, u(b) = 0 (zero on the boundary) the space iscalled H1
0(a, b) and an alternative inner product is:
a(f, g) = [f, g] =∫ b
a
df
dx
dg
dxdx
Projections
~p
~q
~r
P
R
OQ
~e
~r0
line ℓ spannedby vector ~e:~r = ~r0 + ~v with
~v = λ~e
Point Q is closest to P of all points on the line `, with ~p− ~q orthogonal to thisline 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~r∈`‖~p− ~r‖
Projections in function spaces
Inproduct and norm gives smallest distance between functions. For example:
B function uN(x) ∈ VN , where VN is a finite approximation space
B function f(x), is the “exact” function that needs to approximated and is notin VN .
The best approximation is the function uN with f − uN orthogonal to VN :
(v, f − uN) = 0 for all v ∈ VN
⇒∫ b
a
v(x)(f(x)− uN(x)) dx = 0 for all v ∈ VN
Note, uN(x) is the function that minimizes the distance (least squares):
minz∈VN
‖f − z‖2 = minz∈VN
∫ b
a
(f − z)2 dx
Example
A periodic function f(x) on [−π, π]. Approximate (space GN):
gN(x) =N∑k=1
αkek
with N finite, ek the orthogonal base 1, sinx, cosx, . . . . Then∫ π
−πv(f − gN) dx = 0, for all v ∈ GN
and thus:
αk =(f, ek)(ek, ek)
Truncated Fourier expansion.
Shape functions
The shape functions φi(x) are the “base vectors” of the approximation space:
uN(x) =N∑i=0
uiφi(x) = φ˜
T (x)u˜
withφ˜
T = (φ0(x), φ1(x), . . . , φN(x)), u˜T = (u0, u1, . . . , uN)
For a N th-order polynomial we could use
φ˜
T (x) = (1, x, x2, . . . , xN)
however it is more practical to use another base, such as Lagrangian interpolants.
Lagrangian interpolation
x0 = a
u2
u3
u4 u5
u1
x2 x3 x4 x5 = bx1
N = 5
u0
xi: nodal pointsui: nodal values
uN(x) =N∑i=0
uiφi(x) = φ˜
T (x)u˜
with
φi(x) =(x− x0) · · · (x− xi−1)(x− xi+1) · · · (x− xN)
(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) shapefunctions on an interval [−1, 1] with equidistant spaced nodal points. Are theshape functions obtained orthogonal?
Linear system for approximate solution
∫ b
a
v(x)(f(x)− uN(x)) dx = 0 for all v ∈ VN
Fill in uN(x) = φ˜
T (x)u˜
, v(x) = φ˜
T (x)v˜
:
∫ b
a
v˜Tφ
˜[f − φ
˜
Tu˜] dx = 0 for all v
˜
⇒ [∫ b
a
φ˜φ˜
T dx]u˜
=∫ b
a
φ˜f dx
or K¯u˜
= f˜
with K¯
=∫ baφ˜φ˜
T dx, f˜
=∫ baφ˜f dx
or Kij =∫ baφiφj dx, fi =
∫ baφif dx.
1D diffusion equation
1D diffusion Eq.: find u(x) such that for x ∈ (a, b)
− d
dx(Adu
dx) = f
and
u = ua, at x = a (ΓD)
−Adudx
= hb at x = b (ΓN)
Notes:
B Strong form; Classical (strong) solution u(x)
B f(x) ∈ C0(a, b) (continuous) then u ∈ C2(a, b) (twice continuouslydifferentiable)
Residual
Residual
r(x) = − d
dx(Adu
dx)− f
should be zero for the exact solution:
r(x) = 0
For an approximate solution uN(x) ∈ VN we get
rN(x) 6= 0
and rN 6∈ VN . We want the approximate solution uN(x) to be such that theresidual is as small as possible or
rN ⊥ VN
Weighted residuals
Therefore we multiply r(x) = 0 with a test function or weighting function v
(v, r) =∫ b
a
v(x)r(x) dx = 0 for all v
Reversely, if this is true for the complete space we get r(x) = 0 again.
⇒Method of weighted residuals
Substitute r(x):
(v, r) =∫ b
a
v(x)(− d
dx(Adu
dx)− f
)dx = 0 for all v
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
dx,Adu
dx)− v(A
du
dx)∣∣∣ba
= (v, f) for all v
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.
B at x = b, −Adu/dx = hb (Neumann condition)
Weak form: Find u ∈ S such that
(dv
dx,Adu
dx) + v(b)hb = (v, f) for all v ∈ V
with S = u ∈ H1(a, b) with u = ua at ΓDand V = v ∈ H1(a, b) with v = 0 at ΓD.
Galerkin approximations
Key elements for an approximate solution using the weak form:
1. expansion/approximation of the trial solution u:
uN(x) =N∑i=0
uiφi(x) = φ˜
T (x)u˜
Forms a finite dimensional subspace of S: SN and must converge to anyfunction in S for N →∞.
2. choice of the test functions vN , which form the finite dimensional space VNto which the residual is orthogonal. In the Galerkin method we take VN=SN :
vN(x) =N∑i=0
viφi(x) = φ˜
T (x)v˜
Linear system of equations for the approximation
Substitution into the weak form:
v˜TK
¯u˜
= v˜Tf
˜for all v
˜
orK¯u˜
= f˜
with
K¯
= (dφ
˜dx,Adφ
˜
T
dx) =
∫ b
a
dφ˜dxAdφ
˜
T
dxdx
(Kij = (
dφidx
,Adφjdx
))
f˜
= (φ˜, f)− hbφ
˜(b) =
∫ b
a
φ˜f dx− hbφ
˜(b)
(fi = (φi, f)− hbφi(b)
)Note: No Dirichlet conditions yet.
“Distance” of approximate to exact solution
Exact u(x), approximate uh(x). We get, with a(v, u) = (dv
dx,Adu
dx)
a(v, u) + v(b)hb = (v, f) for all v ∈ Va(vh, u) + vh(b)hb = (vh, f) for all vh ∈ Vha(vh, uh) + vh(b)hb = (vh, f) for all vh ∈ Vh
a(vh, u− uh) = 0 for all vh ∈ Vh
u − uh is orthogonal to Vh with inner product a(u, v) or uh has a minimumdistance to u with respect to the energy norm:
‖u‖e = a(u, u)12 with a(u, u) =
∫ b
a
du
dxAdu
dxdx
Exercise 4
a. Explain the weighted residual method. Will the residual be zero after applyingthe method?
b. Explain the Galerkin method. Will the residual be zero after applying themethod?
