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Acid Attack in ConcreteA parabolic-pseudoparabolic model
A.J. Vromans
CASA-Day
April 6, 2016
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Acid Attack
Acid transforms concrete (cement) into gypsum, making it brittle.
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Acid attack chemical reaction
We assume a slaked lime based cement reacting with sulfuric acidto create gypsum.
slaked lime (s) + sulfuric acid (f) → gypsum (s)Ca(OH)2 + H2SO4 → CaSO4 · 2H2O
How to model such a reaction?We apply a Continuum Mixture Theory with inter-linkedmechanical, flow, diffusion and reaction effects.
The model will result from mass & momentum conservation laws.
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Mass Conservation
Mass conservation,
∂
∂t
∫g
ραdτ
︸ ︷︷ ︸Mass change
+
∫∂g
ραvα · n dσ
︸ ︷︷ ︸Mass flux
−∫∂g
δα(grad ρα) · n dσ
︸ ︷︷ ︸Fick’s law (diffusion)
=
∫g
Rαdτ
︸ ︷︷ ︸chemical production
self diffusion at temperature T α of molecules of size dα,
δα =2
3
√k3BT αMα
π3
1
Rd2αρ
α=δα
ρα
volume fraction φα = ρα/ρα, incompressibility condition ˙ρα = 0and divergence theorem yield volume fraction conservation
∂φα
∂t+ div (φαvα)− δ div(grad φα) =
Rα
ρα
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Mass Conservation
The volume fraction condition∑α
φα = 1
together with a no net diffusion constraint implies a total fluxcondition
div
(∑α
φαvα
)=∑α
Rα
ρα
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Momentum conservation
Stress is due to internal forces.
stress flux︷ ︸︸ ︷div Tα =
sum of internal forces︷ ︸︸ ︷∑β
Bαβ
Stress contains pressure and Kelvin-Voigt viscoelasticity
Tα︸︷︷︸stress
= −φαpI︸ ︷︷ ︸pressure
+ EαDs︸ ︷︷ ︸Elastic deformation
+ γαLαs︸ ︷︷ ︸Viscous deformation
Dαs︸︷︷︸Elastic deformation
= (grad uα + (grad uα)>)/2︸ ︷︷ ︸Symmetric part of position gradient
Lαs︸︷︷︸Viscous deformation
= (grad vα + (grad vα)>)/2︸ ︷︷ ︸Symmetric part of velocity gradient
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Momentum conservation
Internal forces due to Stokes drag of solid α through fluid β.
en.wikipedia.org/wiki/Stokes law
Stokes drag force depends linear on relativevelocity
Bαβ = χα(vα − vβ)
Newton’s 3rd law (action causes equal andopposite reaction) implies antisymmetry
Bβα = −Bαβ
All other tensor entries are 0.
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Boundary conditions
Flux condition due to Fick’s laws with boundary velocity U.
φα((
U − vα)· n)
︸ ︷︷ ︸flux across boundary
= Jα(φ+α − φα)︸ ︷︷ ︸
concentration difference
Volume fraction insulation condition
grad φα · n = 0︸ ︷︷ ︸no interaction = no change
Clamped boundary conditions
uα = 0︸ ︷︷ ︸fixed position = no displacement
vα = 0︸ ︷︷ ︸no displacement = no movement
Transversal stress free moving boundaries
Tα · n = 0︸ ︷︷ ︸movement without external force = no stress in movement direction
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Plate-layer of cement
Plate layer G (t) = {x , y , z |x , y ∈ R, 0 < z < h(t)} ⊂ R2 × [0,H].Assume only inward flow at z = 0 and z = H and{
φ2(x , t) = φ−2 , φ1(x , t) = φ3(x , t) = 0 for z < 0
φ3(x , t) = φ+3 , φ1(x , t) = φ2(x , t) = 0 for z > h(t)
One expects a growing layer due to influx of reaction components.
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System of 1D-model
We have functions φ1, φ3, v3,w1,w2 and φ2 = 1− φ1 − φ3.
∂φ1
∂t+
∂
∂z
(φ1∂w
1
∂t
)−δ1
∂2φ1
∂z2= εκ1F
∂φ3
∂t+
∂
∂z(φ3v3)−δ2
∂2φ3
∂z2= −εκ3F
∂
∂z
(φ1∂w
1
∂t+ φ2∂w
2
∂t+ φ3v3
)= F
− ∂
∂z(φ1p) + E1
∂2w1
∂z2+γ1
∂3w1
∂z2∂t= χ1
(∂w1
∂t− v3
)− ∂
∂z(φ2p) + E2
∂2w2
∂z2+γ2
∂3w2
∂z2∂t= χ2
(∂w2
∂t− v3
)p = E1
∂w1
∂z+ E2
∂w2
∂z.
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Boundary conditions & reaction term
Chemical reaction F = K (φ1,sat − φ1) · K (φ3 − φ3,thr ).Boundary conditions:
At z = 0 At z = H
w1 = v3 = 0 w1 = w2 = h(t)∂φα
∂z = 0 ∂φα
∂z = 0
−φ2 ∂w1
∂t = J−K(φ−2 − φ2
)φ3(v3 − ∂h(t)
∂t
)= J+K
(φ+
3 − φ3)
−φ2p + E2∂w2
∂z = 0 −φ3p = 0
with K (x) = xH(x) for H(x) the Heaviside and height function
h(t) =
∫ t
0
[∫ H
0F(z , s)dz
−J+(φ+
3 − φ3(H, s)
)H(φ+
3 − φ3(H, s)
)−J−
(φ−2 − φ
2(0, s))H(φ−2 − φ
2(0, s)) ∫]
ds
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Weak solutions: research list
- Apply Rothe method (time discretization) to 1D system- Determine weak system of time discretized 1D system.blaResearch list
Existence and uniqueness of discrete time weak system:Lax-Milgram Theorem
∆t independent bounds of solutions at fixed times
Weak/strong convergence for limit ∆t ↓ 0
Weak limit is a weak solution of continuous time weak system
Uniqueness of this weak solution
Continuous dependence on parameters of weak solution
Numerical approximation of weak limit
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Discrete system of 1D-model
Use discrete functions φk1 = φ1(tk), . . . and φk2 = 1− φk1 − φk3 .
φk1 − φk−1
1
∆t+
∂
∂z
(φk−1
1
w k1 − w k−1
1
∆t
)−δ1
∂2φk1
∂z2= εκ1Fk−1
φk3 − φk−1
3
∆t+
∂
∂z(φk−1
3 v k−13 )−δ2
∂2φk3
∂z2= −εκ3Fk−1
∂
∂z
(φk−1
1
w k1 − w k−1
1
∆t+ φk−1
2
w k2 − w k−1
2
∆t+ φk−1
3 v k3
)= Fk−1
− ∂
∂z(φk−1
1 pk−1) + E1∂2w k
1
∂z2+γ1
∆t
∂2w k1
∂z2− γ1
∆t
∂2w k−11
∂z2= χ1
(w k
1 − w k−11
∆t− v k−1
3
)− ∂
∂z(φk−1
2 pk−1) + E2∂2w k
2
∂z2+γ2
∆t
∂2w k2
∂z2− γ2
∆t
∂2w k−12
∂z2= χ2
(w k
2 − w k−11
∆t− v k−1
3
)pk = E1
∂w k1
∂z+ E2
∂w k2
∂z.
Every equation is linear at t = tk , when functions at t = tk−1 are known.
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Weak solution results
All equations are of the form uk − Γ∆uk = F .Boundary conditions imply coercive and bounded weak bilinearform. ⇒ Uniqueness of uk ∈ H1([0, 1]) follows from Lax-Milgram.
One can show‖uk‖, ‖∇uk‖, ‖(uk − uk−1)/∆t‖, ‖(∇uk −∇uk−1)/∆t‖are bounded independent of ∆t.⇒ Weak convergence of uk in H1([0,T ],H1([0, 1])) for ∆t ↓ 0
Weak ∆t ↓ 0 limit u of uk is weak solution of continuous system⇐ strong convergence of uk in L2([0,T ], L2([0, 1])) for ∆t ↓ 0
Then uniqueness of and continuous dependence on parameters ofweak solution u are almost trivial.
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Future Directions
Implement a stable and well-posed numerical version of thediscrete system, instead of current unstable version.
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Future Directions
Near future choices
Incorporate moving boundary (growth of gypsum layer).
Find weak solutions for the 2D system
Investigate limits of vanishing diffusion, viscosity and/orboundary parameters. (viscosity solution)
Distant future choices
(Stochastic) Homogenization
Find weak solutions for system with arbitrary dimensions
Incorporate temperature and heat transfer effects.