A model for calcium carbonate neutralization in the

presence of armoring

L. Fusia, M. Primicerioa, A. Montia,b

a Dipartimento di Matematica e Informatica “U.Dini”, Universita degli Studi di Firenze,Viale Morgagni 67/A, 50134 Firenze, Italia.

bI2T3 - Innovazione Industriale Tramite Trasferimento Tecnologico Associazione (Onlus)- Polo Scientifico Sesto Fiorentino (FI) - Via Madonna del Piano 6, c/o Incubatore

Universitario 50019 Sesto Fiorentino (Fi)

Abstract

In this paper we present a mathematical model for the reaction betweencalcium carbonate (CaCO3) and a solution containing sulfuric acid (H2SO4).We assume that Ca2+ ions (liberated on the reaction surface) react with SO2−

4

ions to form solid gypsum crystals (CaSO4 ) which, in turn, may accumulateon the reaction surface, hindering the neutralizing process (armoring). Wesuppose that neutralization is governed by a first order reaction, with thereaction rate depending on the superficial density of solid CaSO4. We provethat the problem is multiscale in time and study the problem in fast timescale (order of hours), showing numerical and analytical results. In particularwe determine the conditions that guarantee complete neutralization.

Keywords: Neutralization, Reaction Kinetics, Armoring, PartialDifferential Equations, multi-scale modelling

1. Introduction

The problem of acid mine drainage (AMD) treatment is certainly one ofthe most challenging and critical in mining industry, since it encompasses aseries of complex phenomena that may lead to dramatic pollution problems,[6]. Among the various processes occurring in a treatment plant, neutraliza-tion is surely one of the most significant, since it allows to buffer the wasteeffluent to a pH which is not environmentally-threatening. Limestone (i.e.CaCO3) is definitely the least expensive neutralizing agent and it is widelyused for this purpose (especially in pulverized form), even if some concerns

Preprint submitted to Applied Mathematical Modelling June 7, 2013

are expressed about its relatively slow reaction rates and possible coating(armoring) of the limestone surface by solid precipitates. A predictive modelthat evaluates and quantifies the efficiency of a neutralizing system (account-ing for phenomena like precipitation and armoring) is undoubtedly a usefultool for designing neutralizing devices (e.g. cartridges).

To this aim, in the last few years, we have developed a series of mathemat-ical models that describe the evolution of a neutralizing system and wherethe neutralizing agent is calcium carbonate (see [3], [4], [5]). These mod-els have been proposed for different geometrical settings and both in staticand dynamical regimes. In this paper we propose a mathematical model forthe neutralization of a sulfuric acid solution (H2SO4) by means of marble(CaCO3). Differently from some previous papers on this topic , here we alsotake into account the so-called “armoring” phenomenon, i.e. the formationof an insoluble precipitate (calcium sulfate, CaSO4) that settles (because ofgravity) on the reacting surface, hindering the neutralization process. Thearmoring phenomenon is indeed a very challenging one, since it involvesmany parameters such as solubility product constant (Kps), ions diffusivity,turbulence of the chemical environment and interactions between the insol-uble formed species. The armoring can be due to many chemical speciesamong which there are iron hydroxides and oxyhydroxides, metal hydroxidesin general (Aluminium et al) and insoluble salts (see [10]). The formationmechanisms are multiple, depending on the precipitating chemical speciesand environmental conditions and may include different states as gels, col-loids et al. After the first crystal/molecule is formed, the species can evolvein different ways, some of which bring to dispersions that do not affect theneutralization process, while others generate a precipitate that coat the otherchemical species present inside the system, making them incapable to react.The precipitation times can be various, ranging from seconds to day or evenweeks and they mainly depend on ions concentrations, solubility product andturbulence. The present work focuses on calcium sulphate (CaSO4) chieflybecause of the high concentration of calcium and sulphates that can be foundin the acid mine drainage to which the present model is aimed. We stressalso that this is the simplest species to model, especially for what concernsthe parameters required by the model, that are unavailable for more complexcompounds like oxyhydroxides (i.e. lepidocrocite, feroxyhyte, goethite et al).

The sulfuric acid H2SO4 - which is dissociated in SO2−4 and 2H+ - reacts

with CaCO3 liberating Ca2+ ions in the solution. The latter in turn react withavailable SO2−

4 ions to form solid CaSO4 (calcium sulfate). The precipitation

2

rate depends on the solubility product of the calcium ions Ca2+ and sulfateions SO2−

4 and in particular on how this product deviates from the equilibriumvalue Kps.

The problem is proposed in a simple 1-D geometry and in a static setting(see Fig. 1), assuming that the precipitate motion is driven only by gravitywhereas the motion of the ions is driven only by diffusion. We suppose thatthe incoming flux of 2H+ and SO2−

4 ions at the “top” free surface x = L isproportional to the concentrations of such ions at x = L (Robin boundarycondition). We also assume that the only source of Ca2+ ions is at x = 0(where they are produced by the reaction between H2SO4 and CaCO3).

We suppose that the thickness of the “coating” layer is by far smallerthan the other characteristic lengths of the system so that we can neglectthe problem of its growth (i.e. a free boundary problem). Further we assumethat the consumption of the marble slab occurs on a time scale scale whichis substantially larger than those of the other phenomena and therefore weneglect this phenomenon as well. Finally we suppose that the neutralizationreaction rate γ depends on the superficial density S of precipitated calciumsulfate deposited on the reacting surface (armoring) and we postulate theexistence of a threshold beyond which neutralization ceases. In particular wesuppose that γ is a decreasing function of S, since an increase of the precipi-tated CaSO4 on the reacting surface entails less efficiency in the neutralizingprocess.

We demonstrate that the system is multi-scale in time and space and wesee that there exists a “fast” time scale in which all the diffusive phenomenacan be safely neglected. This time scale is characterized by the presence of“thin” boundary layers outside which the variables can be considered spa-tially homogeneous (bulk). We prove that in the bulk the system reduces toa nonlinear system of differential equations whose solution depends on theefficiency of the armoring phenomenon (parameter Υ) and on the the rate atwhich ions H+ and SO2−

4 are provided on the free surface (parameter ω). Inparticular we study the case in which the solution is slowly stirred.

We solve the problem numerically and we determine the “critical” valuesfor Υ, ω for which armoring/neutralization is activated/inhibited.

2. The physical model

Assume to have a beaker filled with a solution of H2SO4 (sulfuric acid)and assume that a slab of CaCO3 (marble) is placed horizontally inside the

3

Figure 1: Sketch of the system

beaker as depicted in Fig. 1. In a neighbourhood of the marble surface thefollowing reaction1 occurs

CaCO3 + 2H+ + SO2−4 ⇋ Ca2+ + SO2−

4 + H2O + CO2. (1)

In the solution the Ca2+ and SO2−4 ions react to form CaSO4, i.e.

Ca2+ + SO2−4 ⇋ CaSO4. (2)

In practice the sulfuric acid H2SO4 dissolved in the solution reacts withmarble, neutralizing the solution and liberating calcium ions Ca2+, that isreaction (1). The sulfate ions SO2−

4 dissolved in the solution then find avail-able Ca2+ ions and react with them to form calcium sulfate CaSO4, that isreaction (2). If the solubility product of Ca2+ and SO2−

4 exceeds the Kps

value (for calcium sulfate Kps = 4.9 · 10−11 (moℓ/cm3)2), i.e.

[Ca2+] · [SO2−4 ] > Kps,

then CaSO4 precipitates in the form of small solid particles (average diameter50 µm) which are transported to the bottom surface because of gravity. The

1Sulfuric Acid is written considering its ionic dissociation.

4

solid CaSO4 that reaches the marble surface forms a film that hinders theneutralization process (armoring). To describe the system we suppose thatthe variables depend only on the vertical coordinate x and on time t, withthe spatial domain represented by the interval [0, L] as depicted in Fig. 1.We denote by x = 0 the position of the reacting surface and with x = Lthe free surface of the solution. We suppose that the thickness of the film ofCaSO4 formed on the slab is so small that it can be neglected and we alsoneglect the problem of the consumption of the slab. The system is static anddiffusion is taken as the only driving force for the displacement of the ions.We introduce the following molar concentrations

(i) c(x, t) moles of H+ ions per unit volume, [c] = moℓ/cm3;

(ii) g(x, t) moles of SO2−4 ions per unit volume, [g] = moℓ/cm3;

(iii) h(x, t) moles of Ca2+ ions per unit volume, [h] = moℓ/cm3;

(iv) s(x, t) moles of solid CaSO4 per unit volume, [s] = moℓ/cm3;

We define the solubility product

K(x, t) = h(x, t) · g(x, t), [K] = moℓ2/cm6,

so that when K(x, t) > Kps precipitation occurs. The molar mass balanceequations for c, g and h are

∂c

∂t−Dc

∂2c

∂x2= 0, t > 0, 0 < x < L, (3)

∂g

∂t−Dg

∂2g

∂x2= −q t > 0, 0 < x < L, (4)

∂h

∂t−Dh

∂2h

∂x2= −q t > 0, 0 < x < L, (5)

where the diffusivities Dc, Dh, Dg are supposed to be constant (see [9]). Wenotice that the sink/source term for H+ ions is null since these ions are notadded or removed inside the solution. The sink term for g and h may be,on the other hand, different from zero, since these ions react to form theprecipitated calcium sulfate. Of course q > 0 assuming the same values inboth (4), (5) because of reaction (2).

5

A simple choice for q can be made assuming that this rate is proportionalto the difference between the solubility product K and the Kps (i.e. the moreK exceeds Kps, the faster precipitation occurs), namely

q(x, t) = α [K(x, t) − Kps]+ , [α] = cm3/(moℓ · s),

where we have taken the positive part because precipitation does not occurwhen K(x, t) ≤ Kps. Evidently other choices are possible and actually qshould be determined experimentally. In general one can write q as a functionof K(x, t) − Kps, that is

q = Q(K(x, t) − Kps)H(K(x, t) − Kps),

where the Heaviside step function H guarantees that precipitation does notoccur for K(x, t) ≤ Kps. The concentration of precipitated calcium sulfates(x, t) satisfies a transport equation (we assume diffusion is negligible for s)of the form

∂s

∂t−

∂

∂x(s v(s)) = q t > 0, 0 < x < L, (6)

where v(s) is the modulus of the velocity (which depends on gravity and issuch that v > 0) of the precipitated particles. We assume thermodynamicalequilibrium so that the source term for s is equal to q.

3. Initial and boundary conditions

Let us assume that the initial conditions for problems (3)-(5) are givenby the non-negative constants

c(x, 0) = cin, g(x, 0) = gin, h(x, 0) = 0, x ∈ [0, L].

The last condition comes from the assumption that initially there are noCa2+ ions dissolved in the solution. From reaction (1) we observe that theratio between the initial molar concentrations

cingin

=1

2, (7)

since every mole (p.u.v.) of H2SO4 is made up of one mole (p.u.v.) of SO2−4

and two moles of H+. Taking cin =√

Kps/2 (which corresponds to a highly

acid solution, pH≈ 1) we get gin =√

Kps = O(10−6) moℓ/cm3, that is a

6

typical concentration of a strongly acid solution (see [9]). Hence the initialboundary conditions can be rewritten as

c(x, 0) =

√Kps

2, g(x, 0) =

√

Kps, h(x, 0) = 0, x ∈ [0, L]. (8)

Let us now consider the boundary conditions for c, h and g. From reaction(1) we see that the rate at which Ca2+ ions p.u.v. are liberated on thereacting surface must be equal to twice the rate at which H+ ions p.u.v. areconsumed in the solution, that is

d

dt

[

−2

∫ L

0

c(x, t)dx

]

,

while the rate at which Ca2+ ions are consumed because of precipitation mustbe equal to the rate at which SO2−

4 are consumed, i.e.

d

dt

[∫ L

0

g(x, t)dx

]

.

Hence the overall rate of change of Ca2+ is given by

d

dt

[∫ L

0

h(x, t)dx

]

=d

dt

[

−2

∫ L

0

c(x, t)dx

]

+d

dt

[∫ L

0

g(x, t)dx

]

. (9)

On the boundary x = L we assume mixed (Robin) boundary conditions forc and g

Dc∂c(L, t)

∂x= −ω[c(L, t) − cin], Dg

∂g(L, t)

∂x= −ω[g(L, t) − gin], (10)

where ω represents the injection rate of “fresh” solution at x = L. In practicewe are assuming that at x = L the beaker is continuously fed with a new acidsolution (with c = cin and g = gin) that replaces the existing one. Recalling(8) we get

2Dc∂c(L, t)

∂x= −ω[2c(L, t)−

√

Kps], Dg∂g(L, t)

∂x= −ω[g(L, t)−

√

Kps].

(11)On x = L we also write the homogeneous Neumann boundary condition

Dh∂h(L, t)

∂x= 0

7

where the flux of Ca2+ is set to zero because these ions are produced only onx = 0. On x = 0 we write

Dg∂g(0, t)

∂x= 0, (12)

because SO2−4 are not produced/removed on the bottom surface. Recalling

(3)-(5) and recalling (11), equation (9) can be rewritten as

Dh∂h(0, t)

∂x= −2Dc

∂c(0, t)

∂x− ω[2c(L, t) − g(L, t)].

The boundary condition for problem (6) is s(L, t) = 0 (see [1]) and, assumingthat initially the solution does not contain precipitated CaSO4, we write theinitial condition as s(x, 0) = 0. The overall rate of change of H+ ions is dueto the reaction with the marble surface and the injection of fresh solution,namely

d

dt

[∫ L

0

c(x, t)dx

]

= − γ(S)(c− co)+|x=0︸ ︷︷ ︸

reaction

+ω[cin − c(L, t)]︸ ︷︷ ︸

injection

,

which recalling (3) and (10)1 yields

Dc∂c(0, t)

∂x= γ(S)(c− co)+|x=0 , (13)

where γ(S) is the reaction rate ([γ] = cm/sec )which is assumed to be a nonincreasing function of the superficial density S of solid calcium sulfate thathas deposited on x = 0 (see [3] for a derivation of the reaction kinetics). In(13) co > 0 is the concentration of ions H+ at neutralization. The quantityS is obtained integrating w.r.t. time the flux of precipitated calcium sulfateon x = 0, namely

S(t) =

∫ t

0

s ·v(s)|(0,τ) dτ, [S] =moℓ

cm2(14)

The function γ must be non-increasing, whence γ′

(S) ≤ 0. A simple choicecan be

γ(S) = Γ

(

1 − ΥS

S∞

)

+

(15)

8

where S∞ > 0 is a typical value for the superficial density of deposited CaSO4

and Γ is the rate constant of neutralization when armoring is not effective (i.e.when S = 0: see [3] where a model to infer the value of Γ is presented). In(15) the parameter Υ is an adimensional parameter related to the efficiencyof armoring (it must be determined experimentally). We have

i) When Υ = 1 neutralization stops when S = S∞;

ii) When Υ > 1 neutralization stops when S = S∞/Υ < S∞;

iii) When Υ < 1 neutralization stops when S = S∞/Υ > S∞;

Remark 1. If we assume that the solution is stirred so that there is notransport of “precipitated” CaSO4, then the function S cannot be evaluatedby means of the integral (14). In this particular case the function S must bereplaced by λs, where [λ] = cm. Typically λ = O(100 µm).

4. Sedimentation

In this section we show how to write the transport term s ·v(s) appearingin (6). Let us consider the representative elementary volume VT centered atsome x ∈ [0, L]. Such a volume is decomposed in a solid and a liquid part:VT = Vs + Vl. We have

φs + φl = 1,

where φs and φl are the solid and liquid volume fractions respectively. ClearlyφsVT = Vs and

Vs =Ns

ρs,

where Ns is the number of solid CaSO4 moles in VT and ρs is the molardensity of solid CaSO4. Recalling that s represents the number of moles ofprecipitate per unit volume of solution we have

s =Ns

VT=VsρsVT

= ρsφs.

In terms of the solid volume fraction φs the transport equation (6) is typi-cally written using the Richardson-Zaki empirical formula (see [11]) for theconvective term, namely

∂φs∂t

−∂

∂x

[

v∞φs

(

1 −φsφso

)kφso]

=q

ρs, (16)

9

where φso < 1 is the volume fraction at dense packing (colloidal gelification),k is an empirical parameter and where v∞ is Stokes velocity

v∞ =2R2g(s − l)

9µ. (17)

In (17) we are tacitly assuming that precipitated CaSO4 is formed by solidspheres of uniform radiusR, g being gravity, s, l being the “densities” (massper unit volume) of the calcium sulfate and of the solution respectively andµ being the dynamic viscosity of the solution. In the dilute limit (φs ≪ φo)the convective term in (16) can be expanded, so that, neglecting o(φ3

s) terms,we get the following Burgers’ equation (see [1])

∂φs∂t

−∂

∂x[v∞φs (1 − kφs)] =

q

ρs. (18)

When written for s(x, t) equations (16) and (18) become

∂s

∂t−

∂

∂x

s · v∞

(

1 −s

so

)kso/ρs

︸ ︷︷ ︸

=v(s)

= q, (19)

∂s

∂t−

∂

∂x

s · v∞

(

1 − ks

ρs

)

︸ ︷︷ ︸

=v(s)

= q, (20)

where so = ρsφso.

5. Non dimensional problem

In this section we write the mathematical problem in a non-dimensionalformulation, showing that there exists a peculiar time scale in which diffusiveprocesses can be safely neglected. We adopt the following rescaling

c = c

√Kps

2, g = g

√

Kps, h = h√

Kps, x = xL, t = tT,

10

K = KKps, γ = γΓ, s = s√

Kps, v = vv∞, S = SS∞

where T is a characteristic time to be specified. Omitting the tilda to keepnotation simple we get the following problem for c(x, t)

(Pc)

∂c

∂t−

(T

Tc

)∂2c

∂x2= 0 x ∈ [0, 1], t > 0,

c(x, 0) = 1, x ∈ [0, 1],

∂c(1, t)

∂x= −

(TcTω

)

(1 − c), t > 0,

∂c(0, t)

∂x=

(TcTγ

)

(1 − ΥS)+[c(0, t) − δ]+, t > 0,

(21)where

S =

(T

Ts∞

) ∫ t

0

s ·v(s)|(0,τ) dτ, δ =2co

√Kps

,

and where we have set

Tc =L2

Dc, Tγ =

L

Γ, Ts∞ =

S∞

v∞√

Kps

, Tω =L

ω.

Remark 2. In the case of a dilute solution (see (20)) the velocity v(s) islinear in s

1 − ΥS = 1 − Υ

(T

Ts∞

) ∫ t

0

s (1 − ζs)dτ, v(s) = (1 − ζs), (22)

where

ζ =k√

Kps

ρs.

11

The problem for h(x, t) is

(Ph)

∂h

∂t−

(T

Th

)∂2h

∂x2= −

(T

Tα

)

[K − 1]+ , x ∈ [0, 1], t > 0,

h(x, 0) = 0, x ∈ [0, 1],

∂h(1, t)

∂x= 0, t > 0,

∂h(0, t)

∂x= −

(ThTc

)∂c(0, t)

∂x−

(ThTω

)

(c− g), t > 0,

where we have set

Th =L2

Dh, Tα =

1

α√

Kps

.

The problem for g(x, t) is

(Pg)

∂g

∂t−

(T

Tg

)∂2g

∂x2= −

(T

Tα

)

[K − 1]+ , x ∈ [0, 1], t > 0,

g(x, 0) = 1, x ∈ [0, 1],

∂g(0, t)

∂x= 0, t > 0,

∂g(1, t)

∂x= −

(TgTω

)

(g − 1), t > 0,

where

Tg =L2

Dg.

The problem for s(x, t) becomes

(Ps)

∂s

∂t−

(T

Ts

)∂

∂x(v(s)s) =

(T

Tα

)

[K − 1]+ , x ∈ [0, 1], t > 0,

s(x, 0) = 0, x ∈ [0, 1],

s(1, t) = 0, t > 0,(23)

12

where

Ts =L

v∞.

We make use of typical values

Parameter Value Unit Parameter Value Unit

Dc 3 · 10−5 cm2/s R 50 µm

Dg 1 · 10−5 cm2/s s 2.96 gr/cm3

Dh 8 · 10−6 cm2/s l 1.86 gr/cm3

L 3 · 10 cm µ 0.27 gr/(cm · s)

g 9.8 · 102 cm/s2 Γ 9 · 10−4 cm · s−1

co 1 · 10−10 moℓ/cm3 α 6.4 cm3/(moℓ · s)

Kps 4.9 · 10−11 moℓ2/cm6 ρs 2.1 · 10−2 moℓ/cm3

S∞ 6.1 · 10−4 moℓ/cm2 k 0.65

λ 200 µm

taken from [7], [8], [9], [3]. A few words must be spent on parameter α.From [8] we know that a typical value for the precipitaion rate at 22 o C forparticles of average grain size 50 µm is 10−9.5 mol/(cm3 · s). Therefore fromthe knowledge of Kps we have

α =10−9.5

Kps

moℓ

cm3 · s≈ 6.4

cm3

moℓ · s

With these parameters we see thatTc Th Tg

3.5 · 102 gg 1.3 · 103 gg 1 · 103 gg(Diffusive characteristic times)

Tγ

3.8 · 10−1 gg(Neutralization characteristic time)

Tα

2.5 · 10−1 gg(Precipitation characteristic time)

Ts

1.5 · 10−2 gg(Sedimentation characteristic time)

Ts∞4.4 · 10−2 gg

(Armoring characteristic time)

13

δ ζ

10−5 2.0 · 10−4 (Parameters)

v∞

0.022 cm/s(Stokes velocity)

We immediately notice that the diffusive characteristic times Tc, Th and Tghave order of magnitude O(102 ∼ 103) gg, while the characteristic times Tγ ,Tα are O(10−1) gg. The characteristic times and Ts, Ts∞ are O(10−2) gg.Moreover we observe that δ and ζ can be safely neglected as long as c = O(1)and s = O(1) respectively. Therefore the problem is multiscale in time andcan be simplified according to which time scale we decide to focus on. Inconclusion we can assert that the problem has substantially two time scales

(I) Transport, armoring, neutralization, precipitation: O(hours)

(II) diffusion: O(years)

In this paper we will focus only on the “fast” time scale (I), showing that inthis case the system of PDEs reduces to a system of nonlinear ODEs that canbe solved numerically. We conclude this section with the following remarks

Remark 3. Notice that so far we have not specified the values of ω and Υ,since we will use these parameters to determine the “most efficient” neu-tralizing system. In particular we will fix the value Υ - which correspondsto select a particular choice of the minimum superficial density of CaSO4

required to inhibit neutralization - and determine the maximum values of ωthat guarantees complete neutralization.

Remark 4. We remark that, since δ, ζ ≪ 1, we have that (21)4 becomes(see also (22))

∂c(0, t)

∂x=

(TcTγ

) [

1 − Υ

(T

Ts∞

) ∫ t

0

s(0, τ)dτ

]

+

c(0, t)+,

Moreover equation (23)1 becomes linear

∂s

∂t−

(T

Ts

)∂s

∂x=

(T

Tα

)

[K − 1]+

14

6. The problem in the fastest time scale

Let us select T = Ts∞ as characteristic time, i.e. the fast time scale. Wenotice that the coefficients of the second derivatives of the PDE’s in (Pc),(Ph), (Pg) becomes

(Ts∞Tc

)

,

(Ts∞Th

)

,

(Ts∞Tg

)

(24)

These coefficients define the following adimensional boundary layers

L

√

Ts∞Tc

≈ 3.3·10−1 cm, L

√

Ts∞Th

≈ 1.7·10−1 cm, L

√

Ts∞Tg

≈ 1.9·10−1 cm.

Hence, in the characteristic time scale Ts∞, we have three boundary layerswhose width is O(1 mm ). Outside these boundary layers, that is in the bulkof the solution, the ions concentrations c, h and g will therefore depend ontime only, meaning that no significant variation along the spatial coordinateis observed in the time scale Ts∞ (or analogously Ts,Tα, Tγ). To write theproblem for this special case let us go back for a moment to dimensionalvariables. It is easy to see that in the fast time scale the molar mass balancefor the molar concentrations c, h, g reduces to the following system

c = −Γ

L

(

1 − ΥS

S∞

)

+

(c− co)+ −ω

2L(2c−

√Kps), t > 0,

c(0) =

√Kps

2,

h = −α [hg − Kps]+ + 2Γ

L

(

1 − ΥS

S∞

)

+

(c− co)+, t > 0,

h(0) = 0,

g = −α [hg − Kps]+ −ω

L(g −

√Kps), t > 0,

g(0) =√

Kps.

(25)

Remark 5. We notice that:

15

(i) the first term on the r.h.s. of (25)1 represents the consumption rateof H+ because of neutralization, while the second term the rate of H+

production because of the injection of “fresh” solution from the uppersurface;

(ii) the first term on the r.h.s. of (25)3 represents the consumption rate ofCa2+ because of precipitation, while the second term is the productionrate of Ca2+ because of reaction (1) on the bottom surface;

(iii) the first term on the r.h.s. of (25)5 represents the consumption rate ofSO2−

4 because of precipitation, while the second term the rate of SO2−4

production because of the injection of “fresh” solution from the uppersurface;

Returning to non-dimensional variables we see that problem (25) becomes

c = −σ[

1 − Υ∫ t

0s(0, τ)dτ

]

+c+ − (c− 1)θ, t > 0,

c(0) = 1,

h = −β [hg − 1]+ + σ[

1 − Υ∫ t

0s(0, τ)dτ

]

+c+, t > 0,

h(0) = 0,

g = −β [hg − 1]+ − (g − 1)θ, t > 0,

g(0) = 1,

(26)

where

σ =

(Ts∞Tγ

)

, β =

(Ts∞Tα

)

, θ =

(Ts∞Tω

)

,

are non dimensional quantities. The problem for s becomes

∂s

∂t− ψ

∂s

∂x= β [hg − 1]+ , x ∈ [0, 1], t > 0,

s(x, 0) = 0, x ∈ [0, 1], s(1, t) = 0, t > 0,

(27)

where

ψ =

(Ts∞Ts

)

.

16

We notice that, using the typical values introduced in Section 5, σ, β, ψ =O(1), while θ and Υ have to be specified (see Remark 3). In practice wehave shown that in the fast time scale the problem reduces to a system of 3ODE’s coupled with a linear first order PDE.

6.1. The case of a slowly stirred solution

Suppose that the solution is gently stirred so that the concentration ofsolid particles of CaSO4 does not depend on x and the neutralization rate γin dimensional variables is given (see Remark 1)

γ(s) = Γ

(

1 − Υλs

S∞

)

+

,

so that (26)1 becomes

c = −σ [1 − Υξs]+ c+ + (1 − c)θ, ξ =λ√

Kps

S∞

.

In this case we can get rid of the convective term in (27)1, so that in conclusionproblem (26)-(27) reduces to

c = −σ [1 − Υξs]+ c+ + (1 − c)θ, c(0) = 1,

h = −β [hg − 1]+ + σ [1 − Υξs]+ c+, h(0) = 0

g = −β [hg − 1]+ + (1 − g)θ, g(0) = 1.

s = β [hg − 1]+ , s(0) = 0.

By the sign preserving theorem for continuous function (which holds assum-ing that g and h are at least continuous), there exists a time interval [0, t)in which hg < 1. This occurs because initially the term [hg − 1]+ = 0, sinceh(0) = 0. In this time interval precipitation (and hence armoring) does notoccur and the the problem reduces to

c = −σc + (1 − c)θ,

h = σc,

g = (1 − g)θ.

(28)

17

Integrating we get

c(t) =1

σ + θ{σ exp [−(σ + θ)t] + θ} , (29)

h(t) =σ

σ + θ

{

θt+σ

σ + θ[1 − exp [−(σ + θ)t]]

}

, (30)

g(t) = 1. (31)

This solution is valid in a time interval [0, t] with t such that h(t) · g(t) =h(t) · 1 = 1. We define

Σ(t; θ) =:σ

σ + θ

{

θt+σ

σ + θ[1 − exp [−(σ + θ)t]]

}

− 1 = 0,

and notice that∂Σ

∂t> 0,

and that Σ(0) = −1, Σ(∞) = ∞. Therefore there exists only one timet(θ) > 0 Σ(t; θ) = 0, or equivalently such that h(t) = 1. In particular weobserve that (29) can be used to give an estimate of time t. Indeed, from thedefinition of Σ we know that at time t = t

−σ

σ + θexp [−(θ + σ)t] =

σ + θ

σ−

σ

θ + σ− θt,

that recalling (29) yields

c(t) = θ

(

t−1

σ

)

.

Now, since c ∈ (0, 1) we get

1

σ< t <

1

σ+

1

θ. (32)

Therefore we have proved the following

Proposition 1. There exists a finite time interval (0, t) such that c, h,g aregiven by (29)-(31) respectively, with t satisfying inequalities (32).

Remark 6. In the limit θ → ∞, which corresponds to an infinite (instanta-neous) rate of replacement of “fresh” solution, we get c ≡ g ≡ 1 and h = σtin (0, t). In this case precipitation starts at time t = σ−1.

18

6.2. The problem in the fast time scale without stirringIf we now consider the problem without stirring we have to take into

account the transport problem (27). We begin by studying the PDE fors(x, t). From (26)5 and (27)1

∂s

∂t− ψ

∂s

∂x= −g + (1 − g)θ. (33)

The characteristics of (33) are the straight lines ψt+ x = const so that

s(x, t) =

∫ t

0

[(1 − g)θ − g] dτ, x < 1 − ψt,

∫ t

(x−1)+ψtψ

[(1 − g)θ − g] dτ, x ≥ 1 − ψt.

Therefore

s(0, t) =

∫ t

0

[(1 − g)θ − g] dτ, ψt < 1,

∫ t

ψt−1ψ

[(1 − g)θ − g] dτ, ψt ≥ 1.

Recalling that S = s(0, t) we get

S = H(1 − ψt)

∫ t

0

[(1 − g)θ − g]|τ dτ + H(ψt− 1)

∫ t

t−ψ−1

[(1 − g)θ − g]|τ dτ,

or equivalently

S =

∫ t

H(ψt−1)(t−ψ−1)

[(1 − g)θ − g]|τ dτ.

The system to be solved now becomes

c = −σ [1 − ΥS]+ c+ + (1 − c)θ, c(0) = 1,

h = −β [hg − 1]+ + σ [1 − ΥS]+ c+, h(0) = 0,

g = −β [hg − 1]+ + (1 − g)θ, g(0) = 1,

S = β

∫ t

H(ψt−1)(t−ψ−1)

[hg − 1]+ dτ, S(0) = 0,

(34)

19

which is a nonlinear integro-differential system.

Remark 7. We notice that Proposition 1 holds true also for problem (34).Indeed it is easy to show that there exists a time interval [0, t) in whichhg < 1 and that this interval is exactly the same we have found in the casewith stirring.

7. Some qualitative results

In this section we prove some analytical results for problem (34). Let usmultiply (34)2 by g and (34)3 by h. Summing up we get

hg = −β[(hg − 1)(h+ g)]+ + σg[1 − ΥS]+c+ + (h− hg)θ. (35)

For t = t we have (recall that h = 1, g = 1 and S = 0 in t = t)

hg(t) = σc(t) > 0,

where we recall that

c(t) = θ

(

t−1

σ

)

> 0.

Hence there exists a time interval (t, ¯t), finite or infinite, in which hg > 1 (¯t(see Fig. 2 (A)). Differentiating (34)3 we get

g = −βhg − βgh− gθ in (t, ¯t). (36)

where we have omitted the positive part because hg > 1 in (t, ¯t). We haveg(t) = 0 and g(t) = 1 since g(t) ≡ 1 in [0, t] (see Fig. 2 (B)), so that

g(t) = −βh(t) < 0,

since h(t) = σc(t) > 0. Therefore in a right neighbourhood of t the functiong is decreasing. Now, suppose there exists a time t > t such that g(t) = 1,i.e. t is the first instant at which g attains again the value 1 (see Fig. 2 (B)).At instant t we get g = −β[h−1]+ ≤ 0, which can be true only if g = 0 (thisbecause g < 1 in (t, t) and the derivative in t = t must be non-negative, seeagain Fig. 2 (B)). Hence in t = t we have g = 1 and g = 0, implying, from(34)3

[h(t) − 1]+ = 0, ⇐⇒ h(t) ≤ 1.

20

Figure 2: The functions h(t)g(t) and g(t).

Therefore, from (34)2,

h(t) = σ[c(t)]+[1 − ΥS(t)]+ ≥ 0, (37)

Now h(t) ≥ 0 and h(t) ≤ 1 implies that, in a left neighbourhood of t, h(t) ≤ 1(case I), unless h(t) = 0 and h(t) = 1 (case II).(CASE I) In this case we have h(t) ≤ 1 in (t− ε, t) with g satisfying

g = (1 − g)θ, t < t

g(t = 1 ,

whose unique solution in (t− ε, t) is g = 1. This is a contradiction, since wewere supposing that t is the first time at which g = 1. Therefore the onlypossible case is Case II.(CASE II) In this case we have g(t) = h(t) = 1 and h(t) = 0 so that, from(34)2

σ[c(t)]+[1 − ΥS(t)]+ = 0.

If we assume that c(t) = 0 then, by (34)1, we get c(t) = θ. This leads to acontradiction since, by the continuity of c, there should exist a time t∗ < twhere c(t∗) = 0 with c(t∗) ≤ 0, which is impossible since c(t∗) = θ > 0.Hence we conclude that

[1 − ΥS(t)]+ = 0,

21

which means that for t ≥ t neutralization is not effective (the coating filmdue to armoring has completely inhibited neutralization). The equation forc then becomes

c = (1 − c)θ, t ≥ t,

which yields

c(t) = 1 − [1 − c(t)]e−θ(t−t) =⇒ c(t)t→∞→ 1. (38)

The problem for h and g then becomes

h = −β [hg − 1]+ , h(t) ≤ 1,

g = −β [hg − 1]+ + (1 − g)θ, g(t) = 1,

(39)

which admits a unique solution h = h(t) ≤ 1 and g(t) ≡ 1 for all times.Therefore we have prove the following

Proposition 2. If at some time t > t we have g(t) = 1, then c(t) is givenby (38), g(t) = 1 and h = h(t) ≤ 1 for all t ≥ t. As a consequence g(t) ≤ 1for all times.

Moreover suppose that the time ¯t <∞, i.e. suppose that there exists a finitetime ¯t such that h(¯t)g(¯t) = 1 (see Fig. 2 (A)). Then, from (35)

hg = (h− 1)θ + σg[1 − ΥS]+c+, in t = ¯t. (40)

From Proposition 2 we know that g ≤ 1 for all times so that h(¯t) ≥ 1, since

hg = 1 for t = ¯t. Hence hg(¯t) ≥ 0 which can be true only if hg(¯t) = 0 (recallthat hg > 1 in (t, ¯t), see again Fig. 2 (A)). Therefore, from (40), we have

θh = θ−σg[1−ΥS]+c+, =⇒ h = 1−σg

θ[1−ΥS]+c+ ≤ 1, in t = ¯t.

(41)We have found that h(¯t) ≤ 1 (whereas we were supposing h(¯t) ≥ 1) and hencewe conclude that h(¯t) = 1 that, in turn, implies g(¯t) = 1, since h(¯t)g(¯t) = 1.Hence we are once again in the hypotheses of Proposition 2 with only onedifference: the system (39) has now h(¯t) = 1 has initial conditions, whichleads to h(t) = g(t) ≡ 1 as solution. Therefore we can state the following

22

Figure 3: Υ = 0.01

Figure 4: Υ = 0.1

23

Figure 5: Υ = 1

Figure 6: Υ = 10

24

Proposition 3. If at some time ¯t > t we have h(¯t)g(¯t) = 1, then c(t) isgiven by

c(t) = 1 − [1 − c(¯t)]e−θ(t−¯t), t ≥ ¯t

and g(t) ≡ h(t) ≡ 1 for all t ≥ ¯t. As a consequence g(t)h(t) ≥ 1 for all t ≥ t.

Coupling Proposition 2 and 3 we get the following theorem

Theorem 1. For all times t > t we have g(t) ≤ 1 and h(t)g(t) ≥ 1, implyingh(t) ≥ 1.

As a consequence we have

Corollary 1. If there exists a time t such that g(t) = 1 then g(t) ≡ h(t) ≡ 1for all t > t.

The proof of this corollary is trivial.

8. Simulations

In this section we present and comment some numerical simulations forsystem (34). In particular we show how the solutions depend on the param-eters Υ and ω and how these parameters depend on each other. We performsome numerical simulations of system (34) and plot the evolution of c(t) (i.e.the evolution of pH). We have selected four values of Υ (ranging from 0.01 to10 as shown in Figures 6-??) that correspond to different “effectiveness” ofthe armoring. Then we have solved numerically system (34) for some valuesof ω starting from ω = 10 cm/s and progressively decreasing the injectionrate. For each simulation (and thus for each Υ) we have determined thegreatest value of ω for which complete neutralization is achieved. We havecalled such a value ωN = ωN(Υ), so that ωN represents the maximum rateat which we can supply “fresh” solution and obtain complete neutralizationof the waters (of course within the time scale of neutralization). This meansthat, for a fixed Υ and for ω > ωN(Υ), the solution will return to its initialpH after a finite time and complete neutralization is never achieved. Weshow (see the Table below) that the function ωN is a decreasing function ofΥ and this means that as armoring becomes more effective (increasing Υ) themaximum flux that guarantees utter neutralization decreases (as physicallyexpected).

Υ = 0.01 Υ = 0.1 Υ = 1 Υ = 10ωN = 0.3 ωN = 0.05 ωN = 0.01 ωN = 0.005

25

9. Conclusion and Perspective

We have proposed a one-dimensional model for the neutralization of anacid solution (H2SO4) by means of a reacting material (CaCO3). We haveconsidered the possible formation of a coating thin film (armoring) and wehave evaluated the effects produced on the neutralizing process. We havefound that the problem has a multi-scale structure in time and space andin particular we have seen that the diffusive time scales are by far largerthan the other time scale of the system (armoring, neutralization, etc.). Wehave focussed on the “fast” time scale showing that in this case the set ofPDEs that form the mathematical problem reduces to a nonlinear integro-differential system. For this peculiar problem we have proved some analyticalresult and performed some numerical simulations showing the evolution ofpH.

The problem considered is clearly a simplification in which some impor-tant phenomena have been neglected. A first extension of the model studiedhere is to study the system in the diffusive time scale, in which the varia-tion along the spatial coordinates cannot be neglected. Moreover one couldanalyze the system in a dynamical setting (as we did for the sole neutraliza-tion in [4]) and take into account the evolution of the coating layer and theconsumption of the slab (free boundary problem).

References

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[3] Fusi L., Monti A., Primicerio M., Determining calcium carbonate neu-tralization kinetics from experimental laboratory data, J. Math. Chem.2492–2511(2012).

[4] Fusi L., Monti A., Farina A., Primicerio M., Mathematical Model forCalcium Carbonate Acid Mine Drainage Reaction: a Multiple TimeScale Approach, Rend. Lincei Mat. Appl. 24, 1–24 (2013).

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[5] Fusi L., Farina A., Primicerio M., A Free Boundary Problem for CaCO3

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[7] Lee S.H., Rasaiah J.C., Proton transfer and the mobilities of the H+and OH ions from studies of a dissociating model for water, JournalChem. Phys. 135, 124505, (2011)

[8] De Meer S., Spiers C.J., Peach C.J., Kinetics of precipitation of gypsumand implications for pressure-solution creep, Journal Geol. Soc., London,157, 269–281 (2000).

[9] Samson E., Marchand J., Snyder K.A., Calculation of ionic diffusion co-efficients on the basis of migration test results, Materials and Structures/ Matriaux et Constructions, 36, N. 257, 156–165, (2003).

[10] Sun Q., McDonald L., Skousen J., Effects of Armoring on LimestoneNeutralization of AMD, West Virginia Mine Drainage Task Force Sym-posium, Division of Plant and Soil Sciences, West Virginia University(2000).

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