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Acid Attack in ConcreteA parabolic-pseudoparabolic model
A.J. Vromans
CASA-Day
April 6, 2016
Acid Attack
Acid transforms concrete (cement) into gypsum, making it brittle.
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.
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
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
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
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.
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
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.
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.
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
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
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.
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.
Future Directions
Implement a stable and well-posed numerical version of thediscrete system, instead of current unstable version.
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.