The industrial partner manufactures high quality lead crystal glassware. The cutting of decorative features in the glass damages the surface and the cuts are optically opaque; to restore transparency, the glass is polished in a solution of hydrofluoric (HF) and sulphuric acid (H2 SO4 .) The polishing process comprises three stages:

   1. immersion in a polishing tank containing acid;

   2. rinsing in a tank containing water; and

   3. settlement of the solid reaction products in a settlement tank.

The manufacturer hopes to optimise its polishing process to

    • minimise the health/environmental impact of the process;

    • maximise throughput;

    • maintain the sharpness of the cut edges while still polishing to an acceptable level of transparency.



The study group was asked to focus on modelling three aspects of the process:

    • the chemical reactions involved in the etching at the glass-acid solution interface;

    • the removal of reaction products in the settlement tank;

    • flow within the polishing tank

Fowler, Andrew

Gonzalez, Maria

Hegarty, Alan

King, John

Ockendon, Hilary

O’Brien, Stephen

Robinson, Marguerite

Shishkin, G. I.

Tocher, David

Wang, Qi

English

Mathematics in Industry Information Service Eprints Archive

Acid Polishing of Lead Crystal Glass
Contributors:
Andrew C. Fowler (UL), Maria Gonzalez (UL), Alan F. Hegarty (UL),
John King (U. Nottingham), Hilary Ockendon (U. Oxford),
Stephen O’Brien (UL), Marguerite Robinson (UL), G. I. Shishkin (UCC).
David Tocher (UL). Qi Wang (DIT)
January 2008
0.1 Introduction
The industrial partner manufactures high quality lead crystal glassware. The cutting
of decorative features in the glass damages the surface and the cuts are optically
opaque; to restore transparency, the glass is polished in a solution of hydrofluoric
(HF) and sulphuric acid (H2SO4.) The polishing process comprises three stages:
1. immersion in a polishing tank containing acid;
2. rinsing in a tank containing water; and
3. settlement of the solid reaction products in a settlement tank.
The manufacturer hopes to optimise its polishing process to
• minimise the health/environmental impact of the process;
• maximise throughput;
• maintain the sharpness of the cut edges while still polishing to an acceptable
level of transparency.
The study group was asked to focus on modelling three aspects of the process:
• the chemical reactions involved in the etching at the glass-acid solution interface;
• the removal of reaction products in the settlement tank.
• flow within the polishing tank;
0.2 Etching
The polishing process involves the dissolution of cut glass surfaces in a reservoir of
hydrofluoric acid (HF) and sulphuric acid (H2SO4).
1 Lead crystal consists largely of
lead oxide PbO, potassium oxide K2O, and silica SiO2, and these react with the acids
according to the reactions
PbO + H2SO4
r1→ PbSO4 + H2O,
SiO2 + 4HF
r2→ SiF4 + 2H2O,
K2O + H2SO4
r3→ K2SO4 + H2O,
K2O + 2HF
r4→ 2KF + H2O. (2.1)
The potassium salts and the silicon hexafluoride are soluble, but not the lead sulphate,
which precipitates on the cut surface before being washed away in the rinsing bath.
This rinsing action must have a chemical effect, with the water acting to dissolve the
bonds which tie the sulphate crystals to the surface.
1Thanks to J.J. Leahy of the Department of Chemical and Life Sciences (UL) who explained
some chemistry to us.
1
0.2.1 Surface evolution
In general, if a surface is given by F (x, t) = 0, then its velocity v satisfies Ft+v.∇F =
0, whence also Ft + vn|∇F | = 0, where vn = v.n denotes the normal velocity of the
surface, and n =
∇F
|∇F | is the unit normal. For example, if the surface is denoted by
z = s(x, y, t), then (taking F = s− z)
st = −
[
1 + |∇s|2
]1/2
vn, (2.2)
where vn is the normal downward removal rate of the surface. In general, we expect
the reaction rates to increase with the curvature of the surface. Specifically, the mean
curvature of the surface κ is defined by
2κ =∇.n = −∇.
[ ∇s
(1 + |∇s|2)1/2
]
, (2.3)
and thus
st = −
[
1 + |∇s|2
]1/2
vn
{ ∇s
(1 + |∇s|2)1/2
}
, (2.4)
and vn is an decreasing function of its argument; hence (2.4) is a non-linear diffusion
equation for s. As such, the surface will smooth as it is etched, thus explaining simply
enough why polishing works.
0.2.2 Etching rate
We now wish to relate the etching rate vn to the reaction rates of (2.1). We denote
the reaction rates of the four reactions in (2.1) as r1, r2, r3 and r4, respectively,
with units of moles per unit area per unit time. Denote further the densities of lead
sulphate, silica and potassium oxide by ρP , ρS and ρK , respectively, their volume
fractions within the crystal by φP , φS and φK , and their molecular weights by MP ,
MS and MK . Then the density of species i in the crystal is φiρi, and its molar density
is
φiρi
Mi
. Therefore if vn is the rate of removal of the surface, then the rate of removal
of species i from the surface is
φiρivn
Mi
, and this must be equal to the rate of reaction,
thus
φiρivn
Mi
= Ri. (2.5)
In terms of the reaction rates rj, we would have
RP = r1, RS = r2, RK = r3 + r4. (2.6)
The quantity
mj =
φjρj
Mj
(2.7)
2
is the molar density of species j, and if we denote
m =
∑
j
mj (2.8)
as the molar density of the glass, then
fj =
mj
m
(2.9)
is the fraction of sites in the crystal occupied by species j. It is natural to take
Ri = fiFi, (2.10)
where Fi is the effective flux of external (acid) reactant to the surface, and thus (2.5)
and (2.10) imply
mvn = Fi. (2.11)
It was observed that if one of the acids is not present, etching will not occur. For
example, one can store sulphuric acid in a glass jar without damage; the hydrofluoric
acid is also necessary to cause etching. And yet, the sulphuric acid must attack
the lead sulphate. The mathematical part of this conundrum lies in the general
impossibility of satisfying (2.11) for each species, since it would require the specific
effective reaction rates Fi to be related to each other, and this is unrealistic. In order
to determine what the etching rate vn is, we thus need to consider in greater detail
just what the surface reaction process is.
Physically, we can explain this conundrum in the presence of a single acid, say
H2SO4, by means of the following conceptual picture. Imagine the glass as a crystal
lattice — which it is not . but the concept is valid — where lead, silicon and potassium
atoms are distributed at random. The sulphuric acid can pick off the lead atoms, and
we suppose that it can excavate downwards into the lattice until it encounters a
silicon atom. At this point, no further stripping is possible, and reaction at that
horizontal location ceases. This stripping will happen at each point of the surface,
and, supposing only vertical excavation is possible, eventually a molecularly rough
surface will be obtained, in which only silicon atoms are exposed, thus preventing any
further reaction.
In order to describe the surface reaction, we need to account for the molecularly
rough surface, and to do this, we again suppose that the molecules are arranged in
a lattice, with the horizontal layers denoted by an index i, with i = 0 indicating the
initial surface, and i increasing with depth into the lattice. As etching proceeds, the
surface will have exposed sites at different levels. We let ψji denote the fraction of
exposed surface at level i of species j. As before, the specific effective reaction rate
of species j is denoted Fj, and the species is present in a fraction of sites fj in the
crystal. Thus ∑
j
fj = 1. (2.12)
We define
ψi =
∑
j
ψji (2.13)
3
to be the fraction of exposed sites at level i, and thus
∞∑
i=0
ψi = 1. (2.14)
The reaction equations describe the evolution of exposed sites, and are
ψ˙ji = −Ajψji + fj
∑
k
Akψ
k
i−1, i ≥ 1,
ψ˙j0 = −Ajψj0. (2.15)
The negative term in (2.15) represents the reactive rate of removal of exposed j sites,
while the positive term represents the creation of new exposed sites (a fraction fj of
which are j sites). The quantities Aj (units s
−1) are given by
Aj = NFj(∆x)
2, (2.16)
where N is Avogadro’s number (6× 1023 mole−1), and ∆x is the lattice spacing (m).
Note that the molar density is
m =
1
N(∆x)3
, (2.17)
so that (2.16) is
Aj =
Fj
m∆x
. (2.18)
Thus the paradox (2.11) can be written as
vn = Aj∆x, (2.19)
and is a paradox unless the Aj’s are equal.
Ideally, the solution for the total exposed site fraction ψi would tend to a Gaussian
shaped travelling wave, whose speed would then be the effective etching rate, vn. In
practice, numerical solution of these equations indicates a travelling wave which also
diffuses as it propagates (see figures 1 and 2.)
The computations also indicate that the solution varies smoothly with i, and this
suggests examining the continuum limit of (2.15). To do this, we put
x = i∆x, ψj(x, t) = ψji ; (2.20)
Taylor expanding about x, (2.15) becomes
∂ψj
∂t
= −Ajψj + fj
∑
k
Ak
{
ψk −∆x∂ψ
k
∂x
+
(∆x)2
2
∂2ψk
∂x2
. . .
}
, (2.21)
with boundary condition
∂ψj
∂t
= −Ajψj at x = 0. (2.22)
4
Figure 1: Development of φ = φ1 + φ2 with r1 = 1, r2 = 0.1, f1 = 0.1, f2 = 0.9
Figure 2: φ = φ1 + φ2 at fixed t with r1 = 1, r2 = 0.1, f1 = 0.1, f2 = 0.9
5
The initial condition is
ψj = 0 on t = 0, x > 0,
ψj = fj on t = 0, x = 0, (2.23)
and the small diffusion terms induce an extra boundary condition (culled from the
initial condition)
ψj → 0 as x→∞. (2.24)
Summing (2.22) over j, we find that
∂ψ
∂t
+
∑
k
vk
∂ψk
∂x
=
∑
k
Dk
∂2ψk
∂x2
, (2.25)
where
vk = Ak∆x, Dk =
1
2
vk∆x, (2.26)
and
ψ =
∑
j
ψj. (2.27)
We see that if the Ak’s are all equal, say to A, then the apparent paradox disappears,
and the wave speed is vn = A∆x, as suggested by (2.19). In addition there is a set
of small diffusion coefficients Dk. We can see that these are small, since over the
time l/vn during which the front advances a distance l into the solid, the diffusion
acts over a distance
√
Dl/vn ∼
√
l∆x. This indicates that the surface is molecularly
rough over the distance
√
l∆x.
Now let us consider what happens if the Ak’s are unequal. We write (2.21) in the
form
∂ψj
∂t
+ fj
∑
k
vk
∂ψk
∂x
= −Ajψj + fj
∑
k
Akψ
k + fj
∑
k
Dk
∂2ψk
∂x2
, (2.28)
with
ψj = fj exp(−Ajt) at x = 0. (2.29)
At this point it is convenient to scale the equation. We order the k’s so that v1 =
min vk, and we then scale x and t as
x ∼ l, t ∼ l
v1
, (2.30)
and we define
ε =
∆x
l
, Vk =
vk
v1
, (2.31)
and note that V1 = 1, Vk > 1 for k > 1. The dimensionless form of the equation is
then
∂ψj
∂t
+ fj
∑
k
Vk
∂ψk
∂x
= −1
ε
[
−Vjψj + fj
∑
k
Vkψ
k
]
+ εfj
∑
k
1
2
Vk
∂2ψk
∂x2
, (2.32)
6
with
ψj = fj exp
(
−Vjt
ε
)
at x = 0. (2.33)
If we define the matrices B and V by
Bjk = fjVk, V = diag Vk, (2.34)
then (2.32) is written succinctly as the linear equation
ψt +Bψx =
1
ε
[−Vψ +Bψ] + 1
2
εBψxx, (2.35)
where ψ = (ψ1, . . . , ψn)T , if there are n components.
We may conjecture what the solution of (2.35) should be. Physically, the quantities
vk in (2.26) represent the rate of etching of a surface of pure species k. Insofar as it
is necessary to remove all species, it seems natural to suppose that the actual rate
of etching will be vn = min vk = v1. In dimensionless terms, this suggests that the
solution of (2.35) should (rapidly) tend to a solution in which ψ is non-zero, away
from x = st, where we expect s = minVk = V1 = 1 by choice of the scale for vk.
Inspecting (2.35), this suggests that we write2
x− st = εξ, t = ετ, (2.36)
and then (2.35) becomes
ψτ + (B − sI)ψξ = (B − V )ψ + 12Bψξξ, (2.37)
with the appropriate boundary conditions being that
ψ → 0 as ξ → ±∞. (2.38)
The solution is constrained to be non-zero because of the normalising constraint
(2.14), which in the present terms can be approximated as
∑
j
∫ ∞
−∞
ψj dξ = 1. (2.39)
The hyperbolic, diffusionless part of (2.35) has characteristic speeds λ, these being
the eigenvalues of B. What happens along these characteristics is opaque, because of
the linear source term. If we examine the diffusionless version of (2.37), we can see
that there are steady exponential solutions ψ = ceαξ, ξ < 0, ψ = 0, ξ > 0, providing
α(B − sI)c = (B − V )c, (2.40)
2It should be noted that the continuum approximation really loses justification on the ξ scale, in
which the lattice spacings are ∆ξ = 1. However, for a glass, we can reclaim its justification, since
random molecular spacings will yield a continuous variation with ξ when varied horizontally over
the surface.
7
which is some kind of eigenvalue problem for α, given s (or vice versa). Normalisation
of c follows from (2.29).
The question arises, how should both α and s be chosen? We apparently need
α > 0, and we would like for any solution where s > V1 (if this is the actual wave
speed) to have α < 0, which (with ψ → 0 at −∞) would then imply ψ ≡ 0.
In addition, we must have ci ≥ 0 for each i. These constraints appear mutually
contradictory, and it remains to be seen what the structure of the solutions of (2.35)
or (2.37) (with or without the diffusion term) actually is.
The industrial partner has agreed to supply samples of glass at various stages of
the etching process; it is planned to produce scanning electron micrographs of each
sample to validate the proposed model.
0.3 Settlement
For a sediment of uniform size, we define the volume fraction to be φ. Then mass
conservation of the sediment takes the form
φt + (φv)z = 0, (3.1)
where v is the average sediment velocity. Mass conservation of the liquid takes the
form
−φt + [(1− φ)u]z = 0, (3.2)
where u is the fluid velocity. Adding these, we obtain the total flux condition
φv + (1− φ)u = 0. (3.3)
Additionally, we suppose that settling occurs at a rate
v − u = −vs(φ), (3.4)
where vs is the settling velocity. Typical assumptions are
vs = v0(1− φ)ν , (3.5)
where v0 is the Stokes settling velocity,
v0 =
2
9
ρga2
µ
, (3.6)
where µ is the viscosity, g is gravity, and a is particle radius.
From these it follows that u = φvs(φ), and writing z = −Z, so that Z points
downwards, we find
∂φ
∂t
+
∂
∂Z
[φ(1− φ)vs(φ)] = 0. (3.7)
The particle flux is thus
q(φ) = φ(1− φ)vs(φ), (3.8)
8
and is a unimodal (one-humped) function of φ. The characteristic speed is thus
V (φ) = q′(φ), (3.9)
and is a monotonically decreasing function of φ.
It is straightforward to solve (3.7), for example from an initially uniform concen-
tration φ0. The solution then consists of two uniform regions in which φ = 0 and
φ = φ0, separated by a shock at Z = Vst, where the uniform shock speed is given by
Vs =
[q]
[φ]
= {(1− φ)vs}|φ0 . (3.10)
0.3.1 Variable grain size
Now suppose that the particulate has a grain size distribution f(a; z, t), where a is
particle radius; thus the volume fraction of particles is
φ =
∫ ∞
0
f da, (3.11)
and f has units of inverse length. Conservation of particle mass now takes the form
∂f
∂t
+
∂
∂z
(fv) = 0, (3.12)
where we also define the drift velocity
v − u = −vs, (3.13)
where u is the liquid velocity and, for example,
vs(a, φ) = v0(a)(1− φ)ν . (3.14)
Again we might take v0 to be the Stokes settling velocity in (3.6).
Integrating (3.12) using (3.13), we find
φt +
∂
∂z
[
φu−
∫ ∞
0
fvs da
]
= 0. (3.15)
In addition, conservation of liquid mass takes the form
−φt + ∂
∂z
[(1− φ)u] = 0, (3.16)
and from these last two equations we obtain
u =
∫ ∞
0
fvs da, (3.17)
assuming no net volume flux. Using this in (3.15) and putting z = −Z, we obtain
the evolution equation for φ,
φt +
∂
∂Z
[
(1− φ)
∫ ∞
0
fvs da
]
= 0, (3.18)
9
while for f , (3.12) implies
ft +
∂
∂Z
[
f
{
vs −
∫ ∞
0
fvs da
}]
= 0. (3.19)
In general, this equation requires numerical solution. Simplifications are possible for
two grain sizes, or for dilute suspensions (φ  1), for which also fa  1, and the
integral term can be dropped from (3.19); in this case the particles fall independently
of each other, and the time for settlement is determined entirely by particle size.
0.4 Movement of the glass
We finally consider under what conditions the rinsing will cause a glass to lift from
its base.
The glass is sitting on an oscillating slatted base (figure 3.) The flow at ∞ is also
oscillating.
• Glass sits upon an oscillating slatted base
• Flow at infinity is also oscillating
Glass mov ment within the insert
Figure 3: Position of glass in basket
Let v = αeiωt(iω) at ∞. (α could be complex.) We a sume the flow is inviscid
and linearise.
Equation for glass
mz¨ = −mg + L+R (4.1)
where L is fluid lift and (see figures 4 and 5)
R = 0, if z >  cosωt
> 0, if z =  cosωt
10
Figure 4:
Figure 5:
11
φ = (iω)αeiωt(r sin θ +
a2
r
sin θ)− z˙ a
2
r
sin θ (4.2)
Lift on cylinder = −
∫ ipi
0
p sin θa dθ
= +ρ
∫ ipi
0
(φt + g(a+ z + r sin θ)a sin θ dθ
= ρa[−apiz¨ + gapi + 2apiα(iω)2eiωt]
When glass is in contact z = eiωt and so
R = mz¨ +mg + pia2ρz¨ − pia2ρg + 2ρa2piαω2eiωt
= (m+ m¯)z¨ + (m− m¯)g + 2ρa2piαω2eiωt
= −ω2(m+ m¯)eiωt + (m− m¯)g + 2ρa2piαω2eiωt
where m¯ = ρpia2 is the mass of fluid displaced by the object. Thus
R = −ω2(m+ m¯) cosωt+ (m− m¯)g + 2m¯ω2(αr cosωt− αi sinωt) (4.3)
We can now consider two cases.
Figure 6:
0.4.1 Case 1.
First consider α = 0; i.e., there is no flow at ∞.
R = 0 when cosωt = (m−m¯)g
(m+m¯)ω2
; i.e., the glass will rise from the support if
ω2 >
(
m− m¯
m+ m¯
)
g (4.4)
m = mass of glass + water ≈ 920 grams. m¯ = added mass = water displaced by
glass and water ≈ 650 grams; i.e., we need ω2 > 270
1570
g or (using  = 0.05)
ω2 >
270
157
× 20 ≈ 34.39 (4.5)
Thus, if ω > 6 the glass will lift. The actual ω ≈ 40 cycles/min ≈ 4 rad/sec.
12
0.4.2 Case 2.
Now let αr = 1, αi = 0; i.e., flow at ∞ moves with the cage
R = 0 when
ω2(m¯−m) cosωt+ (m− m¯)g = 0;
i.e., if ω2 cosωt = g, which can happen only if ω2 > g (which is less likely).
13


Acid Polishing of Lead Crystal Glass

Acid Polishing of Lead Crystal Glass

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