Mathematical modelling of alumina feeding

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Mathematical Institute Universit y of Oxfo rd

Mathematical modelling of alumina feeding

Attila Kovacs

Chris Breward, James Oliver and Andreas Münch (Oxford) Svenn Halvorsen and Ellen Nordgård-Hansen (Norce)

Eirik Manger (Norsk Hydro)

March 14, 2019

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Mathematical Institute Universit y of Oxfo rd Hall-Héroult cell

Alumina (300K)

Electrolysis in the hot cryolite (1200K)

Cathode Molten aluminium

Anode Anode

Tapping Crust

Current (200 kA)

Insulation

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Mathematical Institute Universit y of Oxfo rd Alumina feeding processes

A

time

C

E D

B

How does the molten cryolite

infiltrate and dissolve a porous

alumina structure?

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Mathematical Institute Universit y of Oxfo rd Previous Analysis: Solid alumina particle

Molten Cryolite Stefan Condition Thermal Contact r

t Alumina particle

Frozen Cryolite

Dissolution

t

D

t

M

t

F

R

f

Freezing time Melting time Dissolution time R

0

t

F

∼

â²_π^c2Âk^ρAÂ

t

M

∼

â²^c_3kÂ^ρÂ^(T^m^−TÂ⁾

c(T_c−Tm)

t

D

∼

^a²_2D(C^ρ^a^(1−C^s^/ρ^c⁾

s−C0)

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Mathematical Institute Universit y of Oxfo rd Alumina feeding: raft problem

Fluid Saturated

Uninfiltrated

Dissolution boundary Infiltration boundary

Front movement Mush?

A

time

C

E D B

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Mathematical Institute Universit y of Oxfo rd Alumina feeding: raft problem

x y

Uninfiltrated

Saturated

Fluid

Mush

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Mathematical Institute Universit y of Oxfo rd 1D Infiltration problem

T

c

T

m

y Front movement

φ

₀

(top)

Solid Alumina Liquid Cryolite

Solid Cryolite

Air 1

Volume fraction Temperature

y = H y = c(t )

y = b(t)

φ

₀

+ ψ

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Mathematical Institute Universit y of Oxfo rd 1D Infiltration problem: uninfiltrated region

c(t)

b(t) H

y 3 For c (t) < y < H :

∂T

3

∂t = k

^∗

ρ

_A

c

A

φ

∂

²

T

3

∂y

²

At y = c (t ) : T

3

= T

m

, ρ

c

ψ c ˙ = − k

3

∂T

3

∂y At y = H : k

3

∂T

3

∂y = h(T − T

e

)

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Mathematical Institute Universit y of Oxfo rd 1D Infiltration problem: mushy region

c(t)

b(t) H

y 2

For b(t) < y < c (t) :

∂ψ

∂t = 0

∂

∂y ((1 − φ − ψ)u

2

) = 0

−K (ψ + φ) µ( 1 − φ − ψ)

∂p

2

∂y + ρ

c

g

= u

2

At y = c (t) : p

2

= p

a

− σ

_c

r

pore

, u

2

= ˙ c

1 + ψρ

1 − φ − ψ

At y = b(t) : [p

i

] = 0 , u

1

− 1 − φ − ψ

1 − φ u

2

= ψ b ˙ 1 − ρ

1 − φ

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Mathematical Institute Universit y of Oxfo rd 1D Infiltration problem: infiltrated region

c(t)

b(t) H

y For 0 < y < b(t): 1

∂T

1

∂t + ∂

∂y

1 − φ

a u

1

T

1

= k

1

aρ

A

c

A

∂

²

T

∂y

²

∂

∂y ((1 − φ)u

1

) = 0

− K (φ) µ(1 − ψ)

∂p

1

∂y + ρ

_c

g

= u

1

At y = b(t) : T

1

= T

m

, ρ

c

ψ b ˙ = − k

1

∂T

∂y , [p

i

] = 0, u

1

− 1 − φ − ψ

1 − φ u

2

= ψ b ˙ 1 − ρ

1 − φ ,

At y = 0 : T

1

= T

h

, p

1

= ρgH

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Mathematical Institute Universit y of Oxfo rd Full dimensionless problem

y

t H

b(t)

∂T₃

a(t)

∂t

=

_γPe¹

T2

∂²T₃

∂y²

∂T₁

∂t

+

_∂y^∂

_1−φ

α

u

₁

T

₁

=

_αγPe¹

T

∂²T₁

∂y²

∂

∂y

(Φu

₂

) = 0

∂

∂y

((1 − φ)u

₁

) = 0 u

₁

=

^−K(φ)_1−φ

_∂p

1

∂y

+ β

∂ψ

∂t

= 0,

u

₂

=

^−K(ψ+φ)_1−φ−ψ

_∂p

2

∂y

+ β T

₂

= 1

1 2 3

(top)

0 T

₁

= 1 + ε T

_i

= 1 T

_i

= 1

^∂T_∂y³

= Nu(T

₃

− θ) p

₁

= 0 [p

_i

]

⁺₋

= 0 p

₂

= −1

γψ c ˙ = −

^∂T_∂y²

γκψ b ˙ = −

^∂T_∂y¹

u

₂

= ˙ c

1 +

^ψρ_Φ

u

₁

−

_1−φ^Φ

u

₂

= ψ b ˙

^1−ρ_1−φ

y = 0 y = b(t) y = c(t) y = 1

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Mathematical Institute Universit y of Oxfo rd Dimensionless parameters

Meaning Symbol Equation Value

Infiltrating force γ KP τ /(µ

c

H

²

) 40–400

Newton cooling Nu hH /k

^∗

3 Condictivity ratio κ k

^∗

/k

m

0.2 Thermal diffusivity Pe

_T

H

²

ρ

c

/(k

m

τ ) 1.3 Thermal diffusivity Pe

_T2

H

²

ρ

A

c

A

φ

τ k

A

(k

air

/k

A

)

^1−φ

1.77 Particle coldness θ T

A

/T

m

0.246 Overheat ε

T

c

/T

m

− 1 0.015

Density ratio ρ ρ

_f

/ρ

_c

1.00966

Importance of gravity β ρ

c

gH/P 0.01

Energy stored in phases α φ

^ρ_ρ^A^c^A

ccc

+ ( 1 − φ) 0.81

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Mathematical Institute Universit y of Oxfo rd Small time asymptotics

• Expect b, c = O ( √

t) as t → 0

⁺

, so need small t asymptotics to initiate numerical solution.

• Seek similarity solution given T

i

∼ f

i

(η), u

i

∼ g √

i

(η)

t , p

i

∼ h

i

(η), ψ ∼ ψ

c

, b ∼ 2λ

_b

√

t, c ∼ 2λ

_c

√ t with η = y/2 √

t = O ( 1 ) as t → 0

⁺

and λ

b

, λ

c

, ψ

c

to be determined.

t

y O( √

t)

c(t) b(t)

1

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Mathematical Institute Universit y of Oxfo rd Transcendential system for λ _b , λ _c , ψ _c

K

2

(ψ

c

) = −2 (λ

b

− λ

c

)g

2

( 1 − φ − ψ

c

) + λ

_b

g

1

( 1 − φ) K

2

(ψ

c

) K

1

γ 2κψ

c

λ

b

= 2 √

Pe

_T1

e

^−Pe^T1^(ag¹^+λ^b⁾²

√ π erf √

Pe

_T1

(ag

1

+ λ

b

)

− erf ag

1

√ Pe

_T₁

2γψ

c

λ

c

= − 2(θ − 1) √

Pe

_T2

e

^−Pe^T2^λ²^c

√ πerfc λ

c

√ Pe

_T2

where g

1

and g

2

constants given by

g

1

= λ

c

+ (λ

c

− λ

b

)(ρ − 1 )ψ

c

1 − φ g

2

= λ

c

1 + ψ

c

ρ 1 − φ − ψ

_c

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Mathematical Institute Universit y of Oxfo rd Constraint surfaces

Two physical solutions with γ = 200 and Pe

T2

= 1/7.

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Mathematical Institute Universit y of Oxfo rd Similarity solution for γ = 200, Pe _T ₂ = 0.14

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0

0.5 1

T

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0

0.1 0.2 0.3 0.4

η

ψ

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Mathematical Institute Universit y of Oxfo rd Sanity check: sublimits

φ

0

Solid Alumina

Liquid Cryolite Air

1 No cooling and heating (ε, θ = 1) φ

0

Solid Alumina Liquid Cryolite Solid Cryolite

Air 1

φ

0

+ ψ

No heating (ε = 0, λ

b

= 0)

y

Volume fraction

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Mathematical Institute Universit y of Oxfo rd Comparison with numerical results

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Mathematical Institute Universit y of Oxfo rd Nonexistence of Solutions

No physical solutions with γ = 75 and Pe

T2

= 0.14 due to no intersection in the physically admissible region.

No physical solutions with γ = 75 and Pe

T2

= 1.7 due to the Stefan condition (blue).

ψ > 1 − φ, meaning freezes more

than the available fluid.

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Mathematical Institute Universit y of Oxfo rd Conclusions and Future work

Conclusions:

• Developed a simple model for the infiltration of molten cryolite into a cold porous alumina structure (with phase change).

• Similarity solution at small times yields nonexistence and nonuniqueness.

• Preliminary numerical simulation suggest one solution is admissible while the other may blow up.

Future work:

• Characterise regions for similarity solution

• Investigate small time behaviour numerically: Alternatives to similarity solution?

• Are we missing physics e.g.

• capillary pressure dependence on porosity

• dropping local thermodynamical equilibrium

• including composition effects