MATHEMATICAL MODEL FOR THE P.E.M. FUEL CELLS USING SULFURETTED HYDROGEN

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CELLS USING SULFURETTED HYDROGEN

ANCA C ˘AP ˘AT¸ˆIN ˘A, HORIA ENE, GELU PAS¸A, DAN POLISEVSCHI and RUXANDRA STAVRE

In this paper we present a mathematical model of the combustion gas behav- ior, that is the sulphuretted hydrogen, during the passage through the Proton Exchange Membrane (P.E.M.) fuel cell. The anode is represented by a three dimensional porous medium, formed by a system of thin channels which cross a solid parallelepiped. The main mathematical contribution is the proof of the existence of a weak solution of the model problem.

AMS 2000 Subject Classification: 76S05, 80A20, 76R50.

Key words: PEM fuel cell, sulphuretted hydrogen, variational formulation, weak solution

1. INTRODUCTION

In order to model the process we use the Boussinesq approximation of the Darcy law, the heat equation, the diffusion equation and the boundary conditions which describe the absorption and the emission of hydrogen on the different parts of the anode surface. More discussions regarding the basic equations that we use here can be found in [1], [2], [4], [5], [7] and [10].

The characteristic feature of the present study is that by considering the nonlinear terms in the heat and diffusion equations we try to describe more realistically the influence of the gas absorbed by the anode. This is a refinement with respect to other papers which studied similar phenomena (see [3], [6] and [8]).

We find the variational formulation of the model problem. The main contribution of this paper is the proof of the existence of a weak solution, based on a generalization of the Gossez’ theorem, which can be found in [9].

We study the miscible displacement of the sulphuretted hydrogen by the electrolyte, considered as incompressible fluids, in the anode of the PEM fuel cell, considered as a porous medium placed in the parallelepiped Ω = (0, l)×(0, a)×(0, b).

The absorption of the sulphuretted hydrogen takes place through Σ1, the lateral surface of the anode; then, during the oxidation process it is dissociated

MATH. REPORTS11(61),1 (2009), 1–10

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in ions and electrons. We denote by Σ2 the part of the boundary where the migration of the ions into the free electrolyte of the PEM fuel cell takes place.

The rest of the boundary of the anode is denoted by Γ.

The unknowns are~u, S, pandB; they stand for the velocity, the temperature and pressure of the hydrogen-electrolyte mixture and for the concentration of the hydrogen in the mixture, respectively.

The heat and mass transport are due to a thermal and diffusive flow, for which we adopt the Fick law, denoting by −D₁_∂x^∂S

i and −D₂_∂x^∂B

i the thermal and diffusive fluxes,D1 andD2 standing for the thermal conductivity and the molecular diffusivity.

The process is governed by the system below. It consists of the Darcy law, the continuity equation, the diffusion equation of the gas and the heat equation:

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









div~u= 0 in Ω,

K~u=−∇p+ (1−αB−βS)f~ in Ω, div(S~u−D1∇S) =Q in Ω, div(B~u−D₂∇B) = 0 in Ω.

Here, f~ denotes a molecular attraction force,K is the permeability tensor of our porous medium, Q is the radiant source, while α > 0 and β > 0 stand for the volumetric coefficients of diffusive and respectively, thermal expansion, specific to the so-called Boussinesq approximation of the Darcy law. With a slightly loss of generality we considerKij =−^µ_k^mδij, wherekis the permeability of the medium, µ_m the viscosity of the mixture and δ_ij is the Kronecker symbol in R³.

The temperature and the concentration are given on Σ₁ and Σ₂. As the channels that cross the anode do not touch Γ, the normal velocity, the diffusive flow and the heat flux are zero on that part of the boundary. The pressure is imposed on the transfer surfaces Σi. Thus, the corresponding boundary conditions are

(2)











B =B_i on Σ_i, S=Si on Σi,

p=p0 on Σ = Σ1∪Σ2,

~

u·~n= 0 on Γ,

∂B

∂n = 0 on Γ,

∂S

∂n = 0 on Γ,

where Bi and Si are the traces of some smooth functions defined in Ω.

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2. THE VARIATIONAL FORMULATION

In order to obtain a simplified variational formulation of the problem, we introduce new functions that satisfy homogeneous boundary conditions.

LetC₀ andT₀ be two regular functions in Ω such that T₀ =S_i,C₀=B_i on Σ_i, ^∂T_∂n⁰ = 0, ^∂C_∂n⁰ = 0 on Γ (T₀ =S₁+ (S₂−S₁)^x_l¹, C₀=B₁+ (B₂−B₁)^x_l¹, ifS₁,S₂, respectively,B₁,B₂ are constants). Define the functions

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T =S−T₀, C =B−C₀.

It is obvious that the pair (~u, S, B) verifies system (1) and the boundary conditions (2) iff the pair (~u, T, C) verifies the system

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







 µm

k ~u=−∇p+ [(1−αC0−βT0)−αC−βT]f~ in Ω,

div~u= 0 in Ω,

div(T ~u−D₁∇T) +~u∇T₀ =Q₁ in Ω, div(C~u−D2∇C) +~u∇C₀= 0 in Ω,

T = 0 on Σ,

C= 0 on Σ,

p=p0 on Σ,

~

u·~n= 0 on Γ,

∂T

∂n = 0 on Γ,

∂C

∂n = 0 on Γ,

where Q1=Q+^S²^−S_l ¹.

In the sequel we shall rewrite system (4) in dimensionless variables for obtaining better coefficients in variational formulation. Let L be a characteristic length, U a characteristic velocity, P = ^µ_k^mU L a characteristic pressure, γ a characteristic concentration and τ a characteristic temperature. Consider the dimensionless variables

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









x⁰ = x

L, ~u⁰ = ~u

U, p⁰= p

P, C⁰ = C

γ, T⁰ = T τ, f~⁰ =

f~ kf~k_(L∞(Ω))³

, C₀⁰ = C0

γ , T₀⁰ = T0

τ . By using (5) in (4)₁ one obtains

~

u⁰ =−∇p⁰k+ kf

µU[(1−αγC₀⁰ −βτ T₀⁰)−αγC⁰−βτ T⁰]f~⁰, where f =kfk~ _(L∞(Ω))³.

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Analogously, equations (4)3 and (4)4 can be written as div⁰

T⁰~u⁰− D1

LU ∇⁰T⁰

+~u⁰· ∇⁰T₀⁰=Q⁰₁, div⁰

C⁰~u⁰− D2

LU ∇⁰C⁰

+~u⁰· ∇⁰C₀⁰= 0, where Q⁰₁= ^Q_θU¹^L. By denoting

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









a(x) = kf

µU(1−αγC₀⁰ −βτ T₀⁰) α1= kf

µUαγ β₁ = kf

µUβτ

and omitting the symbol ⁰, we obtain the system

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









~u =−∇p+ (a(x)−α₁C−β₁T)f~in Ω, div~u= 0 in Ω,

div

T ~u− D₁ LU ∇T

+~u· ∇T₀ =Q₁ in Ω, div

C~u− D₂ LU ∇C

+~u· ∇C₀ = 0 in Ω, T = 0 on Σ, C = 0 on Σ, p=p₀ on Σ,

~u·~n= 0 on Γ, ∂T

∂n = 0 on Γ, ∂C

∂n = 0 on Γ.

For obtaining a weak formulation of problem (6) we introduce the spaces of functions X=X1×X2×X3 and Y=Y1×Y2×Y3,where

X₁ ={~v∈[L²(Ω)]³; div ~v= 0, ~v·~n= 0 on Γ}, X2 =X3 ={θ∈H¹(Ω); θ= 0 on Σ}, Y1 =X1,

Y₂=Y₃={θ∈W^1,4(Ω);θ= 0 on Σ}.

By multiplying equations (7)1, (7)3, (7)4 with test functions fromY and taking into account condition (7)₂ and the boundary conditions (7)₅–(7)₁₀, we obtain

Z

Ω

~

u·~vdx=−hγ_ν~v, p₀i

H⁻¹²(Σ)×H¹²(Σ)+ (8)

+ Z

Ω

(a(x)−α1C−β1T)f~·~vdx= 0, ∀~v∈Y1,

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D1

LU Z

Ω

∇T · ∇θdx− Z

Ω

T ~u· ∇θdx+

(9)

+ Z

Ω

θ~u· ∇T₀dx= Z

Ω

Q₁θdx, ∀θ∈Y₂, D2

LU Z

Ω

∇C· ∇ϕdx− Z

Ω

C~u· ∇ϕdx+ Z

Ω

ϕ~u· ∇C₀dx= 0, ∀ϕ∈Y3. (10)

For, η, λ >0 arbitrarily chosen, we define the operator hG(~u, T, C),(~v, θ, ϕ)i=

Z

Ω

~

u·~vdx+hγ_ν~v, p₀i

H⁻¹²(Σ)×H¹²(Σ)− (11)

− Z

Ω

((a(x)−α₁C−β₁T)f~·~vdx+ηD₁ LU

Z

Ω

∇T · ∇θdx−

−η Z

Ω

T ~u· ∇θdx+η Z

Ω

θ~u· ∇T₀dx−η Z

Ω

Q1θdx+

+λD₂ LU

Z

Ω

∇C· ∇ϕdx−λ Z

Ω

C~u· ∇ϕdx+λ Z

Ω

ϕ~u· ∇C₀dx.

The variational formulation of problem (7) is (12)

Find (~u, T, C)∈X such that

hG(~u, T, C),(~v, θ, ϕ)i_Y⁰_×Y = 0∀(~v, θ, ϕ)∈Y.

Indeed, we have the following result.

Theorem 1. If(u, T, C)is a solution of(12), then there existsp∈ D⁰(Ω) such that (u, p, T, C) is a weak solution of (7).

Proof. By takingθ=ϕ= 0 in (12), we obtain Z

Ω

~

u·~vdx+hγ_ν~v, p0i

H⁻¹²(Σ)×H¹2(Σ)− (13)

− Z

Ω

(a(x)−α₁C−β₁T)f~·~vdx= 0, ∀~v∈Y₁, which implies

Z

Ω

[~u−(a(x)−α₁C−β₁T)f~]·~vdx= 0, ∀~v∈ {w~ ∈(D(Ω))³/ divw~ = 0}.

Hence, by applying De Rham’s theorem (see [11]), there follows the existence of an element p₁∈ D⁰(Ω) such that

(14) ~u−(a(x)−α1C−β1T)f~=−∇p₁ in (D⁰(Ω))³.

Moreover, since (~u, T, C) ∈X, this relation implies ∇p₁ ∈ (L²(Ω))³. It then follows that (see [11]) p1 ∈ H¹(Ω), hence γ0p1 ∈ H¹²(∂Ω). By multiplying

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relation (14) by~v∈X1, we obtain (15)

Z

Ω

~

u·~vdx− Z

Ω

(a(x)−α₁C−β₁T)f~·~vdx=−hγ_ν~v, γ₀p₁i

H⁻¹²(Σ)×H¹²(Σ). Relations (13) and (15) yield

hγ_ν~v, γ0p1−p0i

H⁻¹²(Σ)×H¹2(Σ)= 0, ∀~v∈X1,

so γ₀p₁−p₀ =ct. on Σ. It is obvious that p=p₁−ctis the solution to (7)₁ and (7)7.

Now, by taking~v=~0 and ϕ= 0 in (12), we obtain D1

LU Z

Ω

T ~u· ∇θdx+ Z

Ω

θ~u· ∇T₀dx−

(16)

− Z

Ω

Q₁θdx= 0, ∀θ∈ D(Ω), thus

(17) div

T ~u− D1

LU ∇T

+~u· ∇T₀ =Q1 inD⁰(Ω),

which is (7)3. Multiplying (17) byθ∈Y2 and integrating on Ω, we get D₁

LU Z

Ω

T ~u· ∇θdx+ Z

Ω

θ~u· ∇T₀dx+

(18)

+D

γ_ν~u− D1

LU

∂T

∂n, θE

H⁻¹²(Γ)×H¹²(Γ) = Z

Ω

Q₁θdx.

Recalling that~u∈X₁, relations (16) and (18) imply ^∂T_∂n = 0 inH⁻¹²(Γ) so we get condition (7)9.

Similarly, to obtain the equation in concentration we take~v =~0,θ = 0 in (12) and get

(19) D₂ LU

Z

Ω

∇C· ∇ϕdx− Z

Ω

C~u· ∇ϕdx+ Z

Ω

ϕ~u· ∇C₀dx= 0, ∀ϕ∈ D(Ω), that is, relation (7)4.

Relations (7)₂, (7)₅, (7)₆, (7)₈ follow from the definitions of the spaces X1,X2,X3. Relation (7)10 follows similarly to (7)9.

3. THE EXISTENCE OF THE WEAK SOLUTION

From the previous theorem we deduce that the existence of a weak solution for problem (7) is a consequence of the existence of a solution to the variational problem (12).

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Theorem 2. In the adimensional relations (5) one can choose L such that problem (12) has at least one solution.

Proof. We shall prove that the operator G defined by (11) satisfies the hypothesis of the following generalization of Gossez’ theorem (see [9]):

Theorem 3. Let Y be a separated locally convex space, continuously embedded in the reflexive Banach space X, and let G : X 7→ Y^∗ (the dual of Y) with the properties

a) G is weakly continuous, that is, continuous between the weak topolo- gies;

b) G is coercive, that is, (∃)r > 0 such that hG(y), yi ≥ 0, (∀)y ∈ Y, kyk_X =r.

Then(∃)x₀∈B_r={x∈X | kxk_X ≤r} such that Gx₀ = 0.

First we prove thatG is weakly continuous.

Let{(~u_n, T_n, C_n)}_n be a weakly convergent sequence inXto an element (~u1, T1, C1) ∈ X. We shall prove that {G(~un, Tn, Cn)}_n is weakly convergent toG(~u1, T1, C1) in Y^∗, that is,

n→∞limhy^∗∗, G(~u_n, T_n, C_n)i_Y^∗∗_×Y^∗ =hy^∗∗, G(~u₁, T₁, C₁)i_Y^∗∗_×Y^∗, ∀y^∗∗∈Y^∗∗, or, equivalently, because the space Y is reflexive,

n→∞limhG(~u_n, T_n, C_n),(~v, θ, ϕ)i_Y^∗_×Y =hG(~u₁, T₁, C₁),(~v, θ, ϕ)i_Y^∗∗_×Y^∗

∀(~v, θ, ϕ)∈Y. We have

hG(~un, Tn, Cn),(~v, θ, ϕ)i_Y^∗_×Y = Z

Ω

~

un·~vdx+hγ_ν~v, p0i

H⁻¹²(Σ)×H¹2(Σ)−

− Z

Ω

((a(x)−α₁C_n−β₁T_n)f~·~vdx+ηD₁ LU

Z

Ω

∇T_n· ∇θdx−

−η Z

Ω

Tn~un· ∇θdx+η Z

Ω

θ~un· ∇T₀dx−η Z

Ω

Q1θdx+

+λD2

LU Z

Ω

∇C_n∇ϕdx−λ Z

Ω

C_n~u_n· ∇ϕdx+λ Z

Ω

ϕ~u_n· ∇C₀dx.

It follows from the weak convergence of the sequence{(~u_n, T_n, C_n)}_nthat the only terms which may produce difficulties are

Z

Ω

T_n~u_n· ∇θdx and Z

Ω

C_n~u_n· ∇ϕdx.

Their convergence is obtained by taking into account that ~u_n_n→∞^−→ ~u₁ weakly in (L²(Ω))³,T_n_n→∞^−→ T₁ strongly in L⁴(Ω) (respectivelyC_n _n→∞^−→ C₁ strongly in L⁴(Ω)), and ∇θ (respectively ∇ϕ) ∈(L⁴(Ω))³. Consequently, the operator G is weakly continuous.

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In order to verify the coercivity condition (b) from Theorem 3, let (~v, θ, ϕ)

∈Y withk(~v, θ, ϕ)k_X =r, where

k(~v, θ, ϕ)k²_X=k~vk²_(L2(Ω))³+k∇θk²_(L2(Ω))³+k∇ϕk²_(L2(Ω))³

We have

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









hG(~v, θ, ϕ),(~v, θ, ϕ)i_Y^∗_×Y = Z

Ω

~

v²dx+α1

Z

Ω

ϕ ~f ·~vdx+

+hγ_ν~v, p0i

H⁻¹²(Σ)×H¹²(Σ)− Z

Ω

a(x)f~·~vdx+

+β₁ Z

Ω

θ ~f·~vdx+ηD₁ LU

Z

Ω

|∇θ|²dx−η Z

Ω

θ~v· ∇θdx+

+η Z

Ω

θ~v· ∇T₀dx−η Z

Ω

Q₁θdx+λD₂ LU

Z

Ω

|∇ϕ|²dx−

−λ Z

Ω

ϕ~v· ∇ϕdx+λ Z

Ω

ϕ~v· ∇C₀dx.

Let us examine the terms of (20). We have

• −η Z

Ω

θ~v· ∇θdx=−η 2 Z

Ω

~v· ∇(θ²) dx=−η

2hγ_ν~v, γ₀θ²i

H⁻¹²(Σ)×H¹²(Σ)= 0;

• −λ Z

Ω

ϕ~v·∇ϕdx=−λ 2 Z

Ω

~v·∇(ϕ²) dx=−λ

2hγ_ν~v, γ0ϕ²i

H⁻¹²(Σ)×H¹2(Σ)= 0;

• α₁ Z

Ω

ϕ ~f·~vdx≥ −α₁kϕk_L2(Ω)k~vk_(L2(Ω))³ ≥

−α₁a0k∇ϕk_(L2(Ω))³k~vk_(L2(Ω))³;

• β₁ Z

Ω

θ ~f ·~vdx≥ −β₁kθk_L2(Ω)k~vk_(L2(Ω))³ ≥

−β₁a0k∇θk_(L2(Ω))³k~vk_(L2(Ω))³,

where one takes into account that kf~k_(L∞(Ω))³ = 1, and a0 is the constant which appears in the inequalitykφk_L2(Ω)≤a₀k∇φk_(L2(Ω))³ ∀φ∈H¹(Ω),φ= 0 on Γ,

• η Z

Ω

θ~v· ∇T₀dx≥ −ηkθk_L4(Ω)k~vk_(L2(Ω))³k∇T₀k_(L4(Ω))³ ≥

−ηa₁k∇θk_(L2(Ω))³k~vk_(L2(Ω))³k∇T₀k_(L4(Ω))³;

• λ Z

Ω

ϕ~v· ∇C₀dx≥ −λa₁k∇ϕk_(L2(Ω))³k~vk_(L2(Ω))³k∇C₀k_(L4(Ω))³;

• hγ_ν~v, p₀i

H⁻¹²(Σ)×H¹²(Σ)≥ −a₂kp₀k

H¹²(Σ)k~vk_X₁ =

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−a₂kp₀k

H¹²(Σ)k~vk_(L2(Ω))³;

• −η Z

Ω

Q1θdx≥ −ηa₀kQ₁k_L2(Ω))k∇θk_(L2(Ω))³. By using these estimates in (20), we get

hG(~v, θ, ϕ),(~v, θ, ϕ)i_Y^∗_×Y ≥A1+A2, where

A₁=

2k~vk²_(L2(Ω))³ + ηD1

2LUk∇θk²_(L2(Ω))³ + ηD1

2LUk∇ϕk²_(L2(Ω))³

−a₂kp₀k

H¹²(Σ)k~vk_(L2(Ω))³ −ηa0kQ₁k_L2(Ω))k∇θk_(L2(Ω))³, A₂=

2k~vk²_(L2(Ω))³ + ηD₁

2LUk∇θk²_(L2(Ω))³ + λD₂

−(β₁a₀+ηa₁k∇T₀k_(L4(Ω))³)k∇θk_(L2(Ω))³k~vk_(L2(Ω))³

−(α₁a₀+λa₁k∇C₀k_(L4(Ω))³)k∇ϕk_(L2(Ω))³k~vk_(L2(Ω))³. Then

A₁≥m(k~vk²_(L2(Ω))³ +k∇θk²_(L2(Ω))³+k∇ϕk²_(L2(Ω))³)−

−M(k~vk_(L2(Ω))³ +k∇θk_(L2(Ω))³ +k∇ϕk_(L2(Ω))³)≥r(mr−M

√ 3), where

m= min n

2, ηD1

2LU, ηD2

2LU o

, M = max

n

a2kp₀k

H¹²(Σ), ηa0kQ₁k_L2(Ω)), λD2

LUk∇C₀k_(L4(Ω))³

o . By choosingr(, η, λ, L, U) = ^M

√3

m , we have A1 ≥0, (∀), η, λ > 0 and (∀)r≥r(, η, λ, L, U).

Next, to prove that the termA2 is positive, let us we remark that it can be rewritten as

A₂=h

4k~vk²_(L2(Ω))³−(β₁a₀+ηa₁k∇T₀k_(L4(Ω))³)k∇θk_(L2(Ω))³k~vk_(L2(Ω))³+ +ηD1

2LUk∇θk²_(L2(Ω))³

i +

h

4k~vk²_(L2(Ω))³−

−(α₁a0+λk∇C₀k_(L4(Ω))³)k∇ϕk_(L2(Ω))³k~vk_(L2(Ω))³ + ηD2

i . Hence, by choosing =η =λ= 1 and

L≤min

n D1

U(β₁a₀+ηa₁k∇T₀k_(L4(Ω))³)² , D2

(α₁a₀+λk∇C₀k_(L4(Ω))³)² o

,

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one can see that A2≥0, (∀)U >0, (∀)r >0.

Consequently, for = η = λ = 1 and for (∀)U > 0, (∃)L > 0 and (∃)r >0 such that

hG(~v, θ, ϕ),(~v, θ, ϕ)i_Y^∗_×Y ≥0, ∀(~v, θ, ϕ)∈Y, k(~v, θ, ϕ)k_X=r, which complete the proof.

Acknowledgements. This work was done with the support of GRANT CEx 320/2006.

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Received 28 January 2008 Romanian Academy

“Simion Stoilow” Institute of Mathematics Calea Grivitei 21

010702 Bucharest, Romania Anca.Capatina@imar.ro