Heat Transfer

Assumtions

1. Equations

heat equation
ρCpTt(kT)=Q, in ΩρCpTt(kT)=Q, in Ω

which is completed with boundary conditions and initial value

at t=0,T(x,0)=T0(x)at t=0,T(x,0)=T0(x)

1.1. Convective heat transfer

convective heat equation
ρCp(Tt+uT)(kT)=Q, in Ω

1.2. Steady case

steady heat equation
(kT)=Q, in Ω
steady convective heat equation
ρCpuT(kT)=Q, in Ω

1.3. Multi-materials

Given a domain ΩRd,d=1,2,3, Ω is partitioned into Nr regions Ωi,i=1,,Nr corresponding to different materials (solid or fluid). We consider ρi, Cp,i and ki the material properties defined in each regions Ωi. We define also ni the outward unit normal vector associated to the boundary Ωi.

ρiCp,iTt(kiT)=Q, in Ωi,i=1,,NrT|Ωi=T|Ωj, on ΩiΩj=Γij,ijkiTni=kjTnj, on ΩiΩj=Γij,ij

We assume the operator L tel que LT=ρiCp,iTt(kiT) is elliptical.

We multiply Lu=Q by a function test vV and integrates by part on Ωi. Which give:

ρCp,iΩiTtvΩi[kiT]v=ΩiQv,vH10(Ω)fori=1,,Nr

By the formula of Green, we get

ρCp,iΩiTtv+Ωiki(y)TvΩkiTniv=ΩiQvvV

Additivity of the integral, we have

Nri=1(ρCp,iΩiTtv+ΩikiTvΩikiTniv)=Nri=1(ΩiQv)vV

Note that

Ni=1Ωi=i,jΓijΩ

Use the conditions in the interfaces, we get

Nri=1(ρCp,iΩiTtv+ΩikiTvΩkiTnv)=Nri=1(ΩiQv)vV

Using the implicit Euler method for the time term:

Tt(tk+1)T(tk+1)T(tk)dttkR+ et kN

Denoting Tk=T(tk), we write the formula in tk+1, we obtain:

Nri=1(ρCp,iΩiTk+1dtv+ΩikiTk+1vΩkiTk+1nv)=Nri=1(ΩiTkdtv+ΩiQv)vV

So, the weak wording becomes:

The weak formulation
 On cherche TH telle que:a(Tk+1,v)=l(v)vV anda(Tk+1,v)=Nri=1(ρCp,iΩiTk+1dtv+ΩikiTk+1vΩkiTk+1nv)l(v)=Nri=1(ΩiTkdtv+ΩiQv)

So we have a(uk+1,v) a continuous bilinear form coercive in vV and l(ϕ) a continuous linear form . We are in a Hilbert space, so we have all the conditions for the application of the Lax-Milgram theorem. So this problem is well posed.

Correct approximation:

We use the Galerkin approximation method:

Let {Th} a family of meshes of :Ω.

Let {K,P,} a finite element of Lagrange of reference of the degree k1.

Let Pkc,h the conforming approximation space defined by

Pkc,h={vC0(Ω),KTh,v|KPk(K)}

To obtain a conformal approximation in V, we add the boundary conditions

Vh=Pkc,hV

Discrete problem is written:

Problème discrète
 Find ThVh such thata(Th,vh)=l(vh)vhVh

Let {φ1,φ2,...,φN} the base of Vh. An element ThVh is written as

Th=Nl=1Tlφl

Using v as a basic function of Vh, our problem becomes

Nri=1(ρCp,iΩiNl=1Tk+1lφldtφj+ΩikiNl=1Tk+1lφlφjΩkiNl=1Tk+1lφlnφj)=Nri=1(ΩiNl=1Tklφldtφj+ΩiQφj)

The variational problem of approximation is then equivalent to a linear system

Algebraic problem
Determine Tl satisfyingNl=1a(φl,φj)Tk+1l=l(φj)j=1,,N

Introduce

A=(a(φi,φj)),1i,jN,Uk+1=(Tk+11,Tk+12,...,Tk+1N)RN,F=(l(φ1),l(φ2),...,l(φN))RN

We write the system in matrix form

AU=F