Heat transfer in a material

Introduction

Definitions

Heat is a form of energy

The temperature is the measurable manifestation of the stored heat

If two bodies which have different temperatures are in contact, a heat tranfer occurs, transfering the heat from the warmer to the colder body.

Several heat transfer modes exist:

The heat conduction corresponds to the heat exchange between two points of a motionless and opaque solid, liquid or gaz.

The convection corresponds to a heat exchange between a wall and a fluid with a heat transport done by the moving fluid.

The radiation corresponds to a heat exchange between two walls separated by a transparent environment.

The heat equations

The heat conduction law:

(OverVector[(Subscript[j, T])])=-k(OverVector[grad])(T)

where k is the thermal conduction coefficient and  (OverVector[(Subscript[j, T])]) the diffusive thermal flux.

The heat equation:

(Subscript[ρC, P])((∂T)/(∂t))+(Subscript[ρC, P])(OverVector[v])·(OverVector[grad])(T)+div((OverVector[(Subscript[j, T])]))=(Subscript[Q, T])

where ρ is the density of the medium, (Subscript[C, P]) the heat capacity, (OverVector[v]) the heat flux speed and (Subscript[Q, T]) a heat source term.

Combining both equations, we get:

(Subscript[ρC, P])((∂T)/(∂t))+(Subscript[ρC, P])(OverVector[v])·(OverVector[grad])(T)-div(k(OverVector[grad])(T))=(Subscript[Q, T])

Examples

A wall of a house

As first example, a simple wall of a house, with thickness e is choosen. Its thermal state is stable (((∂T)/(∂t))=0 and (OverVector[v])=0) and there is no calorific power ((Subscript[Q, T])=0). A thin wall is considered, with an unidirectionnal heat flux along x axis. The temperatures on both sides of the wall are different ((Subscript[T, 1])<(Subscript[T, 2])). The temperature profil in the wall is researched.

To solve this problem, it is adapted to use Cartesian system. The temperature only changes along x axis, and  an homogeneous distribution of the temperature is waited in the other directions.

The conditions of this problem allow to simplify the heat equations presented before.

(Subscript[ρC, P])((∂T)/(∂t))+(Subscript[ρC, P])(OverVector[v])·(OverVector[grad])(T)-div(k(OverVector[grad])(T))=(Subscript[Q, T])

become

(((∂^2)T)/(∂(x^2)))=0

The boundary conditions are:

T(x=0)=(Subscript[T, 1])

T(x=e)=(Subscript[T, 2])

The differential equation system is then solved:

ODEandBCs={T''[x]==0, T[0]==T1, T[e]==T2}

Tsol=T[x]/.DSolve[ODEandBCs,T[x],x, Assumptions-> T1∈Reals&&T2∈Reals][[1]]

To be able to have a general solution, non-dependant of the thickness of the wall, a variable change is done: the temperature is not in function of x anymore, but in function of the ratio x/e, which vary between 0 and 1. The differential system of equations and the solution are as following:

ODEandBCsxsure={T''[xsure]==0, T[0]==T1, T[1]==T2}

Tsol=T[xsure]/.DSolve[ODEandBCsxsure,T[xsure],xsure, Assumptions-> T1∈Reals&&T2∈Reals][[1]]

Changing the disposition of the equation, it’s possible to get

T(x)=(Subscript[T, 1])+((Subscript[T, 2])-(Subscript[T, 1]))(x/e)

Tsol=Tsol/.(T2*xsure-T1*xsure)->xsure*(T2-T1)

It’s possible to generalize this equation to every case doing the following transformation:

((T(x)-(Subscript[T, 1]))/((Subscript[T, 2])-(Subscript[T, 1])))=(x/e)

Tsol=Tsol/.Tsol->(Tsol-T1)/(T2-T1)

Thus, the temperature profile for this case looks like:

Plot[Tsol,{xsure,0,1}, AxesLabel->{"x/e", "(T(x)-T1)/(T2-T1)"}]

In the wall, the temperature has the following distribution:

DensityPlot[Tsol,{xsure,0,1},{y,-5,5},ColorFunction->(ColorData["TemperatureMap"]),Axes->True,AxesOrigin->{0,0},  AxesLabel->{"x/e"}]

A new concrete wall

As a second example, a concrete wall with a thickness e is chosen. This concrete wall is hardening due to an exothermic chemical reaction - hydration of the concrete. The temperature is the same at both side of the wall and is equal to (Subscript[T, w]). This case is assumed as a stable and motionless case (((∂T)/(∂t))=0 et (OverVector[v])=0).

To solve this problem, it is adapted to use Cartesian system. The temperature only changes along x axis, and  an homogeneous distribution of the temperature is waited in the other directions.

The conditions of this problem allow to simplify the heat equations presented before.

(Subscript[ρC, P])((∂T)/(∂t))+(Subscript[ρC, P])(OverVector[v])·(OverVector[grad])(T)-div(k(OverVector[grad])(T))=(Subscript[Q, T])

become

(((∂^2)T)/(∂(x^2)))=-((Subscript[Q, T])/k)

The boundary conditions are:

T(x=0)=(Subscript[T, w])

T(x=e)=(Subscript[T, w])

The differential equation system is then solved:

ODEandBCs={T''[x]==-Q/k, T[0]==Tw, T[e]==Tw}

Tsol=T[x]/.DSolve[ODEandBCs,T[x],x,Assumptions->Q∈Reals && k∈Reals && Tw∈Reals && e∈Reals][[1]]

To get a general solution, usable for a large amount of cases, it looks like to be easier to adimensionnalize this result. Thus, a variable change is done: the temperature is not in function of x anymore, but in function of the ratio x/e, which vary between 0 and 1. The differential system of equations and the solution are as following:

ODEandBCsxsure={T''[xsure]==-Q/k, T[0]==Tw, T[1]==Tw}

Tsolxsure=T[xsure]/.FullSimplify[DSolve[ODEandBCsxsure,T[xsure],xsure,Assumptions->Q∈Reals && k∈Reals && Tw∈Reals][[1]]]

Furthermore, all terms that arent dependent of xsure are placed to the left side of the equal sign:

((T(x/e)-(Subscript[T, w]))/((Subscript[Q, T])(e^2)/k))=(1/2)[(x/e)-((((x/e)))^2)]

TsolxsureAdim=Tsolxsure/.Tsolxsure->(Tsolxsure-Tw)/(Q/k)

Thus, the temperature profil looks like:

Plot[TsolxsureAdim,{xsure,0,1}, AxesLabel->{"x/e", "(T(x)-Tw)/(Q*e^2/k)"}]

In the wall, the temperature has the following distribution:

DensityPlot[TsolxsureAdim,{xsure,0,1},{y,-5,5},ColorFunction->(ColorData["TemperatureMap"]),Axes->True,AxesOrigin->{0,0},  AxesLabel->{"x/e"}]

A continuous thermal treatment

For this example, a thin steel bar is considered during a continuous thermal treatment process: the bar is extracted from an oven (heated at a temperature of (Subscript[T, 0])) and then quenched, after a L distance, in a water bath which is at a (Subscript[T, L]) temperature. The temperature profil of the bar between the exit of the oven and the entrance of the water bath is researched, when a steady state is established.

To solve this problem, it is adapted to use Cartesian system. As the bar is thin, a uniform temperature is assumed in its transversal section, i.e. the temperature field only depend on the x position.

The conditions of this problem allow to simplify the heat equations presented before.

(Subscript[ρC, P])((∂T)/(∂t))+(Subscript[ρC, P])(OverVector[v])·(OverVector[grad])(T)-div(k(OverVector[grad])(T))=(Subscript[Q, T])

become

(((∂^2)T)/(∂(x^2)))=(((Subscript[ρC, P])v)/k)·((∂T)/(∂x))

The boundary conditions are:

T(x=0)=(Subscript[T, 0])

T(x=L)=(Subscript[T, L])

The differential equation system is then solved:

ODEandBCs={D[T[x],{x,2}]==rho*Cp*v/k*D[T[x],x],
T[0]==T0,
T[L]==TL}

Tsol=T[x]/.FullSimplify[DSolve[ODEandBCs,T[x],x,Assumptions->rho∈Reals&&rho>0&&Cp∈Reals&&Cp>0&&v∈Reals&&T0∈Reals&&TL∈Reals]][[1]][[1]]

The factor ((rho*Cp)/k)=(1/α), where α is the thermal diffusivity coefficient of the material.

Tsolalpha=Tsol/.Cp*rho/k->1/α

This solution looks like complicated, however it appears simplier rewriting it in the following way

T(x)=(Subscript[T, 0])+(((Subscript[T, L])-(Subscript[T, 0]))/(exp((vL/α))-1))(exp((v/α)x)-1)

As for the previous examples, a general solution is researched. In this case, this solution looks like:

((T(x)-(Subscript[T, 0]))/((Subscript[T, L])-(Subscript[T, 0])))=((exp((v/α)x)-1)/(exp((vL/α))-1))

TAdim=FullSimplify[Tsolalpha/.Tsolalpha->(Tsolalpha-T0)/(TL-T0)]

The ratio (vL/α) corresponds to the adimensional Péclet number, which defines the ration between advection and diffusion phenomena of a process.

TsolPeclet=TAdim/.(v*L/α)->Pe

TsolPecletfinal=TsolPeclet/.(v/α)->(Pe/L)

Finally, in the same way as the previous cases, a variable change is done, remplacing x/L with the variable xsurL.

Tsoladimfinal=TsolPecletfinal/.(x/L)->xsurL

It’s possible to try to understand how the Péclet number works, varying the temperature profil in function of it

Manipulate[Dynamic[Plot[Tsoladimfinal/.Pe->Péclet,{xsurL,0,1}, PlotRange->All,AxesLabel->{"x/L", "(T(x)-T0)/(TL-T0)"}]],{{Péclet,1},0.001,100}]

or showing the graphical solution of the temperature profil for different Péclet number

listpeclet={0.0001,1,5,10,40,100}

TsolseveralPeclet=Tsoladimfinal/.Pe->listpeclet

Plot[TsolseveralPeclet,{xsurL,0,1}, PlotLegends->listpeclet]

It's also possible to present this solution in a wall, allowing a variable Péclet number

Manipulate[Dynamic[DensityPlot[Tsoladimfinal/.Pe->pe,{xsurL,0,1},{y,-5,5}, Axes->True,AxesOrigin->{0,0},PlotRange->All,AxesLabel->{"x/L"},ColorFunction->(ColorData["TemperatureMap"])]],{{pe,1},0.001,100}]

Heat flux

Attention: this example isn’t in the video!

For this last example, an ordinary heat flux is considered on a surface, changing with the time. A simplified version of the heat equation is anyway used.

(Subscript[ρC, P])((∂T)/(∂t))+(Subscript[ρC, P])(OverVector[v])·(OverVector[grad])(T)-div(k(OverVector[grad])(T))=(Subscript[Q, T])

becomes

ΔT(x,y)=((∂T)/(∂t))

The initial condition is the following

T(0,x,y)=0

and the boundary conditions are arbitrary chosen as:

T(t,x,0)=cos(t)

T(t,0,y)=cos(t)

T(t,x,L)=0

T(t,L,y)=0

An analytical solution does exist for such a case. However, it’s probably a complicated one (with some sum where the convergence has to be checked) and which could take some time to calculate, even for computer. Thus, for this case, a numerical solution is researched. That mean, that the length L has to be defined. L is assumed to be equal to 10. Then, the system and its solution are as follow:

ODEandBCs={D[T[t,x,y],t]==Laplacian[T[t,x,y],{x,y}],T[0,x,y]==0,
T[t,x,0]==Cos[t], T[t,0,y]==Cos[t],
T[t,x,10]==0,T[t,10,y]==0}

Tsol=T[t,x,y]/.NDSolve[ODEandBCs,
T[t,x,y],{t,0,20},{x,0,10},{y,0,10}][[1]]

It' s possible to represent graphically this solution, allowing a varying time and to see how it’s changing the heat flux.

Manipulate[Plot3D[Tsol/.t->temps,{x,0,10},{y,0,10}, PlotRange->{{0,10},{0,10}, {-1,1}},ColorFunction->(ColorData["TemperatureMap"][(#3+1)/2]&), ColorFunctionScaling->False, AxesLabel->{"x", "y", "T[temps, x, y]"}], {temps,0,20}]

References

Amstad Esther and Bowen Paul. Transfer phenomena lecture for bachelor 2nd year student in material science and engineering. Ecole polytechnique fédérale de Lausanne (EPFL). Spring 2016.

Desbiolles Jean-Luc. Modelisation of materials lecture for bachelor 3rd year students in material science and engineering. Ecole polytechnique fédérale de Lausanne (EPFL). Spring 2017