THE PARAMETERS OF A P.E.M. FUEL CELL
ANCA C ˘AP ˘AT¸ˆIN ˘A, HORIA ENE, GELU PAS¸A, DAN POLISEVSCHI and RUXANDRA STAVRE
We use a mathematical model of the combustion gas behavior during the passage through the anode of a PEM fuel cell. We study here the influence of the electro chemic parameters and of the fluid dynamics upon the qualitative and quantitative properties of the fields that characterize the problem. First, we briefly describe the physical problem and establish the mathematical model. We present the relations which insure the existence, uniqueness and regularity of the weak solution of our problem. Finally, we consider a specific optimal control problem; the main result consists of the necessary conditions of optimality that are obtained here. These are the relations that the free parameters (the difference of the concentration and the pressure between the entry and exit borders) have to satisfy in order to obtain a certain thermodynamic configuration.
AMS 2000 Subject Classification: 76S05, 80A20, 76R50.
Key words: PEM fuel cell, variational problem, existence of a weak solution, con- trol problem.
1. INTRODUCTION
The anode of the PEM fuel cell is represented by a three-dimensional porous medium consisting of a solid parallelepiped crossed by thin channels.
The process is modeled by using the Boussinesq approximation of the Darcy law (for details see [2], [4], [6], [7]) and the boundary conditions which describe the absorption and the emission of hydrogen on the different parts of the anode surface (see also [3]).
It is well known that the efficaciousness of the PEM fuel cells depend strongly on the temperature distribution in the domain. The variation of the temperature in the anode of the PEM fuel cell is influenced by many param- eters. Among them, the main part is played by the concentration and the pressure imposed on the absorption and the emission of hydrogen boundaries.
The domain in which we study the phenomenon is the anode of the PEM fuel cell, represented by the parallelepiped Ω = (0, l)×(0, a)×(0, b). The absorption of the hydrogen is made through the part of the lateral surface of
MATH. REPORTS10(60),4 (2008), 299–308
the anode Σ1 defined by the equationx1= 0. During the oxidation process, in the presence of an electrolyte, the hydrogen is dissociated in ions and electrons.
Then, the emission is made through the part of the lateral surface Σ2 defined by the equation x1=l.
The unknown functions of our problem are: the velocity,uthe pressurep and the temperatureT of the hydrogen-electrolyte mixture in the anode while C is the concentration of the hydrogen in the mixture. After homogenizing the boundary conditions and denoting Σ = Σ1∪Σ2 si Γ =∂Ω−Σ, we obtain a system of equations and boundary conditions (one can find details of these procedures in [1]), namely,
(1)
µm
k u =−∇p+ [ah−αC−βT]f in Ω,
div u= 0 in Ω,
div (Tu−D1∇T) +u∇Th=Qh in Ω, div (Cu−D2∇C) +C2−C1
l u1= 0 in Ω,
C = 0 on Σ,
∂C
∂n = 0 on Γ,
T = 0 on∂Ω,
p=πi on Σi, i= 1,2,
u·n= 0 on Γ.
where the functions ah(x) = 1−αC2−Cl 1x1−βTh and Qh are bounded with respect to hwhile Th∈H2(Ω) has (see [5]) the property
kθ∇Thk(L2(Ω))3 ≤hk∇θk(L2(Ω))3, ∀θ∈H01(Ω).
Denoting δ = π2−πl 1, d = C2−Cl 1 and takingp = (1− xl1)π1+ xl1π2+q withq = 0 on Σ, after dividing equation (1)1 µm
k and by keeping the notation for pand f, we obtain for (u, T, C, p) the system
(2)
u =−∇p+ [ah−αC−βT]f−δ(e)1 in Ω,
div u= 0 in Ω,
div (Tu−D1∇T) +u∇Th =Qh in Ω, div (Cu−D2∇C) +d(u)1 = 0 in Ω,
C= 0 on Σ,
∂C
∂n = 0 on Γ,
T = 0 on∂Ω,
p= 0 on Σ,
u·n= 0 on Γ.
2. EXISTENCE AND UNIQUENESS OF THE SOLUTION
In this section we give a variational formulation of problem (2). This formulation allows us to choose the parameters of the problem in order to get the existence and uniqueness of the solution. Denote
X1 ={v∈[L2(Ω)]3; divv= 0 in Ω,v·n= 0 on Γ}, X2=H01(Ω), X3 ={ϕ∈H1(Ω); ϕ= 0 on Σ},
Y1 =X1, Y2=W01,4(Ω), Y3 ={ϕ∈W1,4(Ω);ϕ= 0 on Σ}, X=X1×X2×X3, Y =Y1×Y2×Y3.
We then obtain the following variational formulation of problem (2):
(VP)d,δ
( Find (u, T, C)∈X such that
hGd,δ(u, T, C),(v, θ, ϕ)iY0,Y = 0, ∀(v, θ, ϕ)∈Y,
where for real parametersη andλthe operator Gd,δ :X→Y0 is defined by
(3)
hGd,δ(u, T, C),(v, θ, ϕ)iY0,Y
= Z
Ω
u·vdx+δ Z
Ω
(v)1dx− Z
Ω
(ah−αC−βT)f·vdx +η2D1
Z
Ω
∇T · ∇θdx+η2 Z
Ω
u· ∇T θdx+η2 Z
Ω
u· ∇Thθdx
−η2 Z
Ω
Qhθdx+λ2D2 Z
Ω
∇C∇ϕdx+λ2 Z
Ω
u· ∇Cϕdx +λ2d
Z
Ω
(u)1ϕdx, ∀((u, T, C),(v, θ, ϕ))∈Y×Y.
We begin our study with the a priori estimation given below.
Proposition 1. If (u, T, C) is a solution of problem (VP)d,δ, then
(4)
maxn
kuk2(L2(Ω))3, η2D1k∇Tk2(L2(Ω))3, λ2D2kCk2(L2(Ω))3
o
≤4
max
x∈Ω¯
|ah(x)| kfk(L∞(Ω))3 +C(Ω)δ 2
+4η2 D1
C2(Ω)kQhk2L2(Ω), where C(Ω)is a positive constant only depending on Ω.
Proof. By using the same technique as in [5], we can prove that the solution of problem (2) has the regularity p ∈ H2(Ω),u ∈ (H1(Ω))3. This allow us to set (v, θ, ϕ) = (u, T, C) in (VP)d,δ, where (u, T, C) is a solution of
problem (VP)d,δ. So, we obtain
kuk2(L2(Ω))3+η2D1k∇Tk2(L2(Ω))3 +λ2D2kCk2(L2(Ω))3 =
−α Z
Ω
Cf·udx−β Z
Ω
Tf·udx−η2 Z
Ω
Tu· ∇Thdx
−λ2d Z
Ω
C(u)1dx+ Z
Ω
ahf·udx−δ Z
Ω
(u)1dx+η2 Z
Ω
QhTdx
≤C(Ω)kfk(L∞(Ω))3(αk∇Ck(L2(Ω))3 +βk∇Tk(L2(Ω))3)kuk(L2(Ω))3
+η2hk∇Tk(L2(Ω))3kuk(L2(Ω))3 +λ2dC(Ω)k∇Ck(L2(Ω))3kuk(L2(Ω))3
+ max
x∈Ω¯
|ah(x)| kfk(L∞(Ω))3kuk(L2(Ω))3 +C(Ω)δkuk(L2(Ω))3
+η2C(Ω)kQhkL2(Ω)k∇Tk(L2(Ω))3.
It follows that there existd0 >0 andh0 >0 such that ford≤d0,h≤h0 and λand µproperly chosen we have
kuk2(L2(Ω))3 +η2D1k∇Tk2(L2(Ω))3+λ2D2kCk2(L2(Ω))3
≤ 1
2kuk2(L2(Ω))3+ 2
max
x∈Ω¯
|ah(x)| kfk(L∞(Ω))3 +C(Ω)δ 2
+η2D1
2 k∇Tk2(L2(Ω))3 +2η2 D1
C2(Ω)kQhk2L2(Ω). Hence
maxn1
2kuk2(L2(Ω))3, η2D1
2 k∇Tk2(L2(Ω))3, λ2D2
2 kCk2(L2(Ω))3
o
≤2 max
x∈Ω¯
|ah(x)| kfk(L∞(Ω))3 +C(Ω)δ2
+2η2
D1C2(Ω)kQhk2L2(Ω). that is, (4).
We are now in a position to prove the following result of existence and uniquenes.
Theorem 1. (1) There existh0 >0 and d0 >0 such that for all h >0 andd∈R withh < h0 and|d|< d0one can chooseλ=λ(h, d)andη=η(h, d) such that problem (VP)d,δ has at least one solution for all δ >0.
(2) If the solution of problem (VP)d,δ is regular then, for d, h, λ and η chosen as above, there exists δ0 such that the problem (VP)d,δ has a unique solution for all δ≤δ0.
Proof.(1) In order to obtain this result we use a generalization of Gossez’
theorem as in [1] (see also [5]). We have to look for λ, η andr >0 such that (5) hGd,δ(v, θ, ϕ),(v, θ, ϕ)iY0,Y ≥0, ∀(v, θ, ϕ)∈Y,k(v, θ, ϕ)kX =r.
Taking into account that the terms of order 3 vanish (see [1]), we have
(6)
hGd,δ(v, θ, ϕ),(v, θ, ϕ)iY0,Y =kvk2(L2(Ω))3+η2D1k∇θk2(L2(Ω))3
λ2D2kϕk2(L2(Ω))3 +α Z
Ω
ϕf·vdx+β Z
Ω
θf·vdx +η2
Z
Ω
θv· ∇Thdx+λ2d Z
Ω
ϕ(v)1dx− Z
Ω
ahf·vdx +δ
Z
Ω
(v)1dx−η2 Z
Ω
Qhθdx.
Using the same technique as in [1] we obtain (5) by taking h0 = D1
βkfk(L∞(Ω))3
, d0 = D2
αkfk(L∞(Ω))3
.
(2) To prove the uniqueness of the solution, we consider two solutions (ui, Ti, Ci),i= 1,2, of problem (VP)d,δ. Let (u, T, C) = (u1−u2, T1−T2, C1− C2). By subtracting the equations satisfied by the two solutions and by setting (v, θ, ϕ) = (u, T, C) we obtain
Z
Ω
u2dx+η2D1
Z
Ω
∇T2dx+λ2D2
Z
Ω
∇C2dx=− Z
Ω
(αC+βT)f·udx +η2
Z
Ω
∇T1u· ∇Tdx−η2 Z
Ω
∇Tu· ∇Thdx+λ2 Z
Ω
C1u· ∇Cdx
−λ2d Z
Ω
C(u)1· ∇Cdx
≤C(Ω)kfk(L∞(Ω))3(αk∇Ck(L2(Ω))3 +βk∇Tk(L2(Ω))3)kuk(L2(Ω))3
+η2hk∇Tk(L2(Ω))3kuk(L2(Ω))3 +λ2dC(Ω)k∇Ck(L2(Ω))3kuk(L2(Ω))3
+η2 Z
Ω
∇T1u· ∇Tdx+λ2 Z
Ω
C1u· ∇Cdx .
For d0, h0 chosen as in Proposition 1, for d ≤ d0, h ≤ h0 and for λ, µ properly chosen, we have
kuk2(L2(Ω))3 +η2D1k∇Tk2(L2(Ω))3+λ2D2kCk2(L2(Ω))3
≤2η2 Z
Ω
T1u· ∇Tdx+ 2λ2 Z
Ω
C1u· ∇Cdx .
One can prove (by using the same technique as in [5]) thatT1 ∈H2(Ω), C1 ∈H2(Ω), henceT, C ∈L∞(Ω). From the above relation we get
kuk2(L2(Ω))3 +η2D1k∇Tk2(L2(Ω))3+λ2D2kCk2(L2(Ω))3
≤(kT1kL∞(Ω)k∇Tk(L2(Ω))3 +kC1kL∞(Ω)k∇Ck(L2(Ω))3)kuk2(L2(Ω))3. By using the estimates from Proposition 1 we obtain that (u, T, C) = (0,0,0) and the proof of the theorem is complete.
3. THE CONTROL PROBLEM
In this section we will consider a problem which allow us to determine the parameters dand δ in order to obtain an optimal temperature at the porous anode of the PEM fuel cell. Denote
(7) D= [d1, d2]×[δ1, δ2],
where di ∈(0, d0], δi ∈(0, δ0], i= 1,2 with d0 and δ0 given by Proposition 1.
In the sequel we suppose that the data and parameters of the problems are such that for all (d, δ)∈Dproblem (VP)d,δhas a unique solution denoted by (ud,δ, Td,δ, Cd,δ) (i.e. the assumptions of Theorem 2.1 are satisfied).
Define the functionalJ :D→ Rby
(8) J(d, δ) = 1
2 Z
Ω
(Td,δ−T0)2dx,
where T0 ∈X2 is the desired configuration for the temperature. Consider the optimal control problem
(CP) Find (d∗, δ∗)∈Dsuch thatJ(d∗, δ∗)≤J(d, δ), ∀(d, δ)∈D.
Proposition 2. The control problem (CP)has at least one solution.
Proof. We shall prove that the functional J is continuous. Indeed, let {(dn, δn)}n⊂Dand (d∗, δ∗)∈Dbe such that lim
n→∞dn=d∗ and lim
n→∞δn=δ∗. Let (un, Tn, Cn) be the unique solution corresponding to problem (VP)(dn,δn). It follows from Proposition 1 that the sequence {(un, Tn, Cn)}n is bounded
in X, hence there exist a subsequence {(unp, Tnp, Cnp)}np and (u, T, C) ∈X such that
unp*u in (L2(Ω))3, Tnp* T inH01(Ω), Cnp * C inH1(Ω).
By passing to the limit in (VP)dnp,δnp, we obtain that (u, T, C) is the unique solution of problem (VP)d∗,δ∗, which imply that Tn * T = Td∗,δ∗ in H1(Ω) so Tn→T inL2(Ω). Therefore,
n→∞lim J(dn, δn) = 1 2
Z
Ω
(T−T0)2dx=J(d∗, δ∗).
The proof is now complete by taking into account thatDis compact.
In order to get the necessary conditions of optimality, we shall first prove the next result.
Lemma 1. Let (d∗, δ∗) be an optimal control for problem (CP). Then (9)
Z
Ω
(T∗−T0)Td,δ∗ dx≥0, ∀(d, δ)∈D,
where T∗ is such that(u∗, T∗, C∗)is the unique solution of problem(VP)(d∗,δ∗)
and, for all d ∈ [d1, d2] and δ ∈ [δ1, δ2], (u∗d,δ, Td,δ∗ , Cd,δ∗ ) ∈ X is the unique solution of the problem
(Q∗d,δ)
Z
Ω
u∗d,δ·vdx+ (δ−δ∗) Z
Ω
(v)1dx+α Z
Ω
Cd,δ∗ f·vdx
+β Z
Ω
Td,δ∗ f·vdx+η2D1 Z
Ω
∇Td,δ∗ · ∇θdx+η2 Z
Ω
u∗d,δ· ∇T∗θdx
+η2 Z
Ω
u∗· ∇Td,δ∗ θdx+η2 Z
Ω
u∗d,δ· ∇Thθdx+λ2D2 Z
Ω
∇Cd,δ∗ · ∇ϕdx
+λ2 Z
Ω
u∗d,δ· ∇C∗ϕdx+λ2 Z
Ω
u∗· ∇Cd,δ∗ ϕdx+λ2d∗ Z
Ω
(u∗d,δ)1ϕdx
+λ2(d−d∗) Z
Ω
(u)∗1ϕdx= 0 ∀(v, θ, ϕ)∈Y.
Proof. Lett∈[0,1], d∈[d1, d2] and δ∈[δ1, δ2]. Putdt=d∗+t(d−d∗), δt=δ∗+t(δ−δ∗) and let (udt,δt, Tdt,δt, Cdt,δt) be the unique solution of problem (VP)d
t,δt. An easy computation yields udt,δt = u∗ +tu∗d
t,δt, Tdt,δt = T∗+ tTd∗
t,δt, Cdt,δt =C∗+Cd∗
t,δt, where (u∗d
t,δt, Td∗
t,δt, Cd∗
t,δt) is the unique solution
(uniqueness is obtained in the same way as in the proof of Theorem 1 of the problem
(Q∗d
t,δt)
Z
Ω
u∗dt,δt ·vdx+ (δ−δ∗) Z
Ω
(v)1dx+α Z
Ω
Cd∗t,δtf·vdx +β
Z
Ω
Td∗t,δtf·vdx+η2D1
Z
Ω
∇Td∗
t,δt· ∇θdx +η2t
Z
Ω
u∗dt,δt· ∇Td∗
t,δtθdx+η2 Z
Ω
u∗dt,δt· ∇T∗θdx +η2
Z
Ω
u∗· ∇Td∗t,δtθdx+η2 Z
Ω
u∗dt,δt · ∇Thθdx +λ2D2
Z
Ω
∇Cd∗t,δt · ∇ϕdx+λ2t Z
Ω
u∗dt,δt· ∇Cd∗t,δtϕdx +λ2
Z
Ω
u∗dt,δt· ∇C∗ϕdx+λ2 Z
Ω
u∗· ∇Cd∗
t,δtϕdx +λ2t(d−d∗)
Z
Ω
(u∗dt,δt)1ϕdx+λ2d∗ Z
Ω
(u∗dt,δt)1ϕdx +λ2(d−d∗)
Z
Ω
(u)∗1ϕdx= 0 ∀(v, θ, ϕ)∈Y.
By using the method in the proof of Proposition 1 we deduce the bounded- ness of the solution of this problem with respect tot. Hence, lettingt→0, we deduce that the limit (u∗d,δ, Td,δ∗ , Cd,δ∗ ) is the unique solution of problem (Q∗d,δ).
Using these facts we get
(10)
J0(d∗, δ∗)·(d−d∗, δ−δ∗) = lim
t→0
J(dt, δt)−J((d∗, δ∗)) t
= 1 2lim
t→0
Z
Ω
[(Tdt,δt−T0)2−(T∗−T0)2] dx t
= 1 2lim
t→0
2
Z
Ω
Td∗t,δt(T∗−T0) dx+t Z
Ω
(Td∗t,δt)2dx
= Z
Ω
Td,δ∗ (T∗−T0) dx.
Taking into account that (d∗, δ∗) is a minimum point for J on D, rela- tion (9) follows from (10).
The above result allows us to obtain the necessary conditions of optima- lity for problem (CP).
Theorem 2. Let (d∗, δ∗) be an optimal control for problem(CP). Then there exist unique elements (u∗, T∗, C∗) ∈ X and (ν∗, τ∗, γ∗) such that the system below is verified:
(11) hGd∗,δ∗(u∗, T∗, C∗),(v, θ, ϕ)iY0,Y = 0, ∀(v, θ, ϕ)∈Y,
(12)
Z
Ω
ν∗·vdx+α Z
Ω
ν∗·fϕdx+β Z
Ω
ν∗·fθdx +η2D1
Z
Ω
∇τ∗· ∇θdx+η2 Z
Ω
τ∗∇T∗·vdx+η2 Z
Ω
τ∗u∗· ∇θdx +η2
Z
Ω
τ∗∇Th·vdx+λ2D2 Z
Ω
∇γ∗· ∇ϕdx+λ2 Z
Ω
γ∗∇C∗·vdx +λ2
Z
Ω
γ∗u∗· ∇ϕdx+λ2d∗ Z
Ω
γ∗(v)1dx− Z
Ω
(T∗−T0)θdx= 0,
∀(v, θ, ϕ)∈Y,
(13) λ2(d−d∗) Z
Ω
γ∗(u∗)1dx+ (δ−δ∗) Z
Ω
(ν∗)1dx≤0, ∀(d, δ)∈D.
Proof. The existence and uniqueness of the element (u∗, T∗, C∗) was proved in Theorem 1. By computations similar to those in the proof of Theo- rem 1, one can prove that problem (12) has a unique solution.
In order to obtain (13), we take (v, θ, ϕ) = (u∗d,δ, Td,δ∗ , Cd,δ∗ ) (the unique solution of problem Q∗d,δ) in problem (12) and get
(14)
Z
Ω
ν∗·u∗d,δdx+α Z
Ω
ν∗·fCd,δ∗ dx+β Z
Ω
ν∗·fTd,δ∗ dx +η2D1
Z
Ω
∇τ∗· ∇Td,δ∗ dx+η2 Z
Ω
τ∗∇T∗·u∗d,δdx+η2 Z
Ω
τ∗u∗· ∇Td,δ∗ dx +η2
Z
Ω
τ∗∇Th·u∗d,δdx+λ2D2 Z
Ω
∇γ∗· ∇Cd,δ∗ dx+λ2 Z
Ω
γ∗∇C∗·u∗d,δdx +λ2
Z
Ω
γ∗u∗· ∇Cd,δ∗ dx+λ2d∗ Z
Ω
γ∗(u∗d,δ)1dx− Z
Ω
(T∗−T0)Td,δ∗ dx= 0.
Next, by taking in problem (Q∗d,δ) as test function the unique solution (ν∗, τ∗, γ∗) of problem (12), we obtain
(15)
Z
Ω
u∗d,δ·ν∗dx+ (δ−δ∗) Z
Ω
(v)1dx+α Z
Ω
Cd,δ∗ f·ν∗dx +β
Z
Ω
Td,δ∗ f ·ν∗dx+η2D1
Z
Ω
∇Td,δ∗ · ∇τ∗dx+η2 Z
Ω
u∗d,δ· ∇T∗τ∗dx +η2
Z
Ω
u∗· ∇Td,δ∗ τ∗dx+η2 Z
Ω
u∗d,δ· ∇Thτ∗dx+λ2D2
Z
Ω
∇Cd,δ∗ · ∇γ∗dx +λ2
Z
Ω
u∗d,δ· ∇C∗γ∗dx+λ2 Z
Ω
u∗· ∇Cd,δ∗ γ∗dx+λ2d∗ Z
Ω
(u∗d,δ)1γ∗dx +λ2(d−d∗)
Z
Ω
(u)∗1γ∗dx= 0.
By substrating (15) from (14) we conclude that Z
Ω
(T∗−T0)Td,δ∗ dx=−λ2(d−d∗) Z
Ω
γ∗(u∗)1dx−(δ−δ∗) Z
Ω
(ν∗)1dx . The proof is now complete on account of the last equation and Lem-
ma 1.
Acknowledgements. This work was done with the support of Contract CEx 189/2006.
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Received 28 January 2008 Romanian Academy
“Simion Stoilow” Institute of Mathematics Calea Grivitei 21
014700 Bucharest, Romania Anca.Capatina@imar.ro