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.