5. Transient and multidimensional conduction - EPFL

5. Transient and multidimensional conduction

John Richard Thome

5 avril 2008

John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Conduction 5 avril 2008 1 / 60

5.2 Lumped capacitance method - Dimensional analysisWe consider a clearly representative problem of one-dimensional transient heatconduction, without internal heat conduction and constant thermal conductivity

∂2T∂x2 =

1α

∂T∂t with

i.c. : T (t = 0) = Tib.c. : T (t > 0, x = 0) = Tl

b.c. : −k ∂T∂x

∣∣∣∣x=L

= h (T − Tl)x=L

The solution of this problem must take the form of the following dimensionalfunctional equation

T − Tl = fn[(Ti − Tl) , x , L, t, α, h, k

]Nondimensionalizing :

Θ ≡ T − TlTi − Tl

ξ ≡ xL

Fo ≡ αtL2


5.2 Lumped capacitance method - Dimensional analysiswhere Fo = Fourier number (dimensionless time). Then

∂2Θ

∂ξ2 =∂Θ

∂Fo

Θ(ξ, 0) = 1∂Θ

∂ξ

∣∣∣∣ξ=0

= Tl

∂Θ

∂ξ

∣∣∣∣ξ=1

= −BiΘ(1, Fo)

Finally, in dimensionless form

Θ = fn [ξ, Fo, Bi ]

Notice, that t/T can be viewed as

tT =

hk (V /A) tρc (V /A)2 k

= Bi · Fo (5.7)


5.2 Lumped capacitance method

We turn our attention to an analysis of unsteady heat transfer, beginning with amore detailed consideration of the lumped-capacity system.Basis of method : we assume that the solid is uniform in temperature and that its

mean temperature changes with time, i.e. negligible temperaturegradients.

SettingΘ ≡ T − T∞ =⇒ dΘ

dt =dTdt

The problem is that of predicting the transient cooling of a convectively cooledobject (cooling of a hot metal forging, for example). With reference to figure nextslide, we apply our familiar first law statement

Q =dUdt

−hA (T − T∞) = ρcV dTdt



The cooling of a body for which the Biot number is small :



Then, first law becomes−Θ =

ρcVhA

dΘ

dtSeparating variables and integrating from time = 0, where T (0) = Ti

ρcVhA

∫ Θ

Θi

dΘ

Θ= −

∫ t

0dt

where Θi = Ti − T∞. ThenρcVhA

ln ΘiΘ

= t

OrΘ

Θi=

T − T∞Ti − T∞

= exp[−

(hAρcV

)t]


5.2 Electrical analogy

The term capacitance is adapted from electrical circuit theory to the heattransfer problem. We can defined a thermal time constant

τt =

(1

hA

)(ρcV ) = RtCt

wherethermal resistance : Rt =

1hA

lumped thermal capacitance : Ct = ρcV

Electrical solution for E (t) is analogous to the thermal response. Total heattransfer up to time t is

Q = (ρcV ) Θi

[1− exp

(− t

τt

)]


5.2 Electrical analogy

A simple resistance-capacitance circuit (Fig 5.1) :


5.2 Validity of method

The cooling of a body for which the Biot number is large :


5.2 Validity of method

Normally, in a heat conduction problem the thermal capacitance is distributed inspace. But when the Biot number is small, T (t) is uniform in the body (almostconstant within the body at any time) and we can lump the capacitance into asingle circuit element. Thus

Bi ≡ hLk � 1 =⇒ Tb (x , t) ' T (t) ' Tsurface

and the thermal conductivity, k, becomes irrelevant to the cooling process. Utilizelumped capacitance method when

Bi ≡ hLk < 0.1

Where L = V /A, equal to :L for wall 2L thickr0/2 for long cylinder of radius r0

r0/3 for a sphere of radius r0


5.2 Transient cooling of a slabFigure 5.6 : Transient cooling in a one-dimensional slab.


5.6 Transient conduction - Semi-infinite solid

Constant surface temperature

The calculation of the temperature distribution in a semi-infinite region poses adifficulty in that we can impose a definite b.c. at only one position - the exposedboundary. We shall be able to get around that difficulty in a nice way with thehelp of dimensional analysis.



The dimensional function equation is

T − T∞ = fn [t, x , α, (Ti − Tinfty )]

We have two dimensional groups

T − T∞Ti − T∞

= fn(

x√αt

)Θ = fn (ζ)

Thus, we transform each side of

∂2T∂x2 =

1α

∂T∂t



as follows∂T∂t = ∆T ∂Θ

∂ζ

∂ζ

∂t = ∆T(− x

2t√

αt

)∂Θ

∂ζ;

∂T∂x = ∆T ∂Θ

∂ζ

∂ζ

∂x =∆T√

αt∂Θ

∂ζ;

∂2T∂x2 =

∆T√αt

∂2Θ

∂ζ2∂ζ

∂x =∆Tαt

∂2Θ

∂ζ2 .

We getd2Θ

dζ2 = −ζ

2dΘ

dζ



If we call dΘ/dζ ≡ χ, then equation becomes the first-order equation

dχ

dζ= −ζ

2χ

which can be integrated twice to get

Θ = C∫ ζ

0e−ζ2/4dζ + Θ(0)

Thus the solution to the problem of conduction in a semi-infinite region, subjectto a b.c. of the first kind is

Θ =1√π

∫ ζ

0e−ζ2/4dζ =

2√π

∫ ζ/2

0e−s2

ds ≡ erf (ζ/2)



Error function and complementary error function :



Temperature distribution in a semi-infinite region (Fig 5.15) :


5.6 Transient conduction - Surface convection

If repeated the problem with a boundary condition of the third kind, we wouldexpect to get Θ = Θ(Bi , ζ), except that there is no length, L, upon which tobuild a Biot number. Therefore, we must replace L with

√αt, so

Θ = Θ

(ζ,

h√

αtk

)≡ Θ(ζ, β)

withβ ≡ Bi

√Fo

The solution of this problem is

Θ = erf(

ζ

2

)+ exp

(βζ + β2) [

erfc(

ζ

2 + β

)]


5.6 Transient conduction - Surface convection

Fig 5.16 : The cooling of a semi-infinite region by an environment at T∞, througha heat transfer coefficient, h. (Θ function of βζ).


5.6 Example 5.4

For what maximum time can a samurai sword be analyzed as a semi-infiniteregion after it is quenched, if it has no clay coating and hexternal = infinity ?

Temperature distribution in a semi-infinite region.


5.6 Example 5.4

Solution : First, we must guess the half-thickness of the sword (say, 3 mm) andits material (probably wrought iron with an average α around 1.5x10−5m2/s).The sword will be semi-infinite until δgg equals the half-thickness. We find

δ2gg

3.642α=

0.0032

13.3(1.5)(10−5)= 0.045s

Thus the quench would be felt at the centerline of the sword within only 1/20 s.The thermal diffusivity of clay is smaller than that of steel by a factor of about 30,so the quench time of the coated steel must continue for over 1 s before thetemperature of the steel is affected at all, if the clay and the sword thicknesses arecomparable.


5.7 Two-dimensional conduction

The two-dimensional flow of heat between two isothermal walls (Fig 5.20) :



The general equation for T(−→r )

withsteady conduction.constant thermal conductivity.without internal heat generation.

Is called Laplace’s equation :∇2T = 0

It looks easier to solve than it is, since the Laplacian is a sum of several secondpartial derivatives. Faced with a steady multidimensional problem, four routes areopen to us

Find out whether or not the analytical solution is already available in a heatconduction text or in other published literature.Solve the problem analytically (chapter 4).Obtain the solution graphically.Solve the problem numerically.



The method of flux plotting will solve all steady planar problems in which allboundaries are held at either of two temperatures or are insulated. We identify aseries of channels, each which carries the same heat flow, ∂Q W/m. We alsoinclude a set of equally spaced isotherms, ∂T apart, between the walls. Since theheat fluxes in all channels are the same,∣∣∣∂Q

∣∣∣ = k ∂T∂n ∂s

Notice that ∂s/∂n must be the same for each rectangle. We therefore arbitrarilyset the ratio equal to unity, so all the elements appear as distored squares. Theobjective then is to sketch the isothermal lines and the adiabatic, or heat flow,lines which run perpendicular to them. This sketch is to be done subject to twoconstraints :

Isothermal and adiabatic lines must intersect at right angles.They must subdivide the flow field into elements that are nearly square.



Steps in constructing a flux plotIdentify all lines of symmetry (from thermal and geometrical considerations)Note that lines of symmetry are adiabatic and no flux crosses themIdentify all isotherms at boundaries and then try to sketch in the isothermlines within the system (with all isotherms normal to the adiabatic lines)Heat flow lines are then drawn trying to create curvilinear squares (with heatflow lines and isotherms intersecting at right angles and trying to keep allsides of the square about the same lenght)If you goof, erase and go back to start !



A first rough sketching :



The evolution of the flux plot :



A flux plot with no axis of symmetry to guide construction :



Heat transfer through a wall with isothermal ribs :



Once the grid has been sketched, the temperature anywhere in the field can beread directly from the sketch. And the heat flow per unit depth into the paper is

Q = Nk∂T ∂s∂n =

NI k∆T

where N is the number of heat flow channels and I is the number of temperatureincrements, ∆T/∂T .



A heat conduction shape factor S may be defined for steady problems involvingtwo isothermal surfaces as follows

Q ≡ Sk∆T

where for a flux plotS =

NI

It follows that the thermal resistance of a two-dimensional body is

Rt =1

kS where Q =∆TRt

The virtue of the shape factor is that it summarizes a heat conduction solution ina given configuration. Once S is known, it can be used again and again.



The shape factor for two similar bodies of different size (Fig 5.24) :

Some tables in the course include a number of analytically derived shape factorsfor use in calculating the heat flux in different configurations.


5.7 Two-dimensional conduction : example 5.9

Calculate the shape factor for a one-quarter section of a thick cylinder.

Solution : We already know Rt for a thick cylinder. From it we compute

Scyl =1

kRt =2π

ln(r0/ri)

so on the case of a quarter-cylinder,

S =π

2ln(r0/ri)

The quarter-cylinder is pictured in Fig. 5.24 for a radius ratio, ro/ri =3, but fortwo different sizes. In both cases S = 1.43. (Note that the same S is also given bythe flux plot shown.)



Calculate S for a thick hollow sphere (picture 5.25)

The general solution of the heat diffusion equation in spherical coordinates forradial heat flow is

T =C1r + C2



The boundary conditions are Ti and To and substituting the general solution weget

Ti =C1ri

+ C2 and T0 =C1r0

+ C2

ThereforeC1 =

Ti − T0r0 − ri

r0ri and C2 = Ti −Ti − T0r0 − ri

r0

In the general solution we get

T = Ti + ∆T[

ri r0r (r0 − ri)

− r0r0 − ri

]Then

Q = −kAdTdr =

4π (ri r0)

r0 − rik∆T

S =4π (ri r0)

r0 − ri[m]



A spherical heat source of 6 cm in diameter is buried 30 cm below the surface of avery large box of soil and kept at 35C. The surface of the soil is kept at 21C. Ifthe steady heat transfer rate is 14 W, what is the thermal conductivity of thissample of soil ?

Solution :Q = Sk∆T =

(4πR

1− R/2h

)k∆T

where S is that for situation 7 in Table 5.4 (see book). Then

k =14

(35− 21)

1− (0.06/2)/2(0.3)

4π(0.06/2)= 2.545 W /mK


5.comp. Numerical method

Nodal Network :



Another approximate solution is use of numerical techniques : finite-difference,finite-element, and boundary-element methods. We will present thefinite-difference method.

Each node represents a small zone with an average temperature of the zoneassigned as the node’s temperature. The nodes and mesh are set to the user’sconvenience, the finer the mesh the more accurate the calculation (but thisincreases the computation time).


5.comp. Numerical methodThe value of the second derivative, ∂2T/∂x2 is

∂2T∂x2

∣∣∣∣∣m,n

=

∂T/∂x∣∣∣m+1/2,n

− ∂T/∂x∣∣∣m−1/2,n

∆x

The temperature gradients in terms of nodal temperatures are

∂T∂x

∣∣∣∣∣m+1/2,n

=Tm+1,n − Tm,n

∆x

∂T∂x

∣∣∣∣∣m−1/2,n

=Tm,n − Tm−1,n

∆x

Substituting equations

∂2T∂x2

∣∣∣∣∣m,n

=Tm+1,n + Tm−1,n − 2Tm,n

(∆x)2



Similarly, in the y-direction, the analogous expression is

∂2T∂y2

∣∣∣∣∣m,n

=

∂T/∂y∣∣∣m,n+1/2

− ∂T/∂y∣∣∣m,n−1/2

∆y =Tm,n+1 + Tm,n−1 − 2Tm,n

(∆y)2

For a mesh in which ∆x = ∆y , then the heat conduction equation ?? is

∂2T∂x2 +

∂2T∂y2 = 0

⇒ Tm,n+1 + Tm,n−1 + Tm+1,n + Tm−1,n − 4Tm,n = 0

This is an approximate algebraic equation for the heat conduction in this node.



Energy balance approach - First approach


5.comp. Numerical methodThe modal equation from an energy balance : here assuming all heat flows intothe node, steady-state conditions and internal heat generation

Q(m−1,n) 7−→(m,n) = k (∆y · 1)Tm−1,n − Tm,n

∆x

Q(m+1,n) 7−→(m,n) = k (∆y · 1)Tm+1,n − Tm,n

∆x

Q(m,n+1) 7−→(m,n) = k (∆x · 1)Tm,n+1 − Tm,n

∆y

Q(m,n−1) 7−→(m,n) = k (∆x · 1)Tm,n−1 − Tm,n

∆yWith equilibrius equation

Ein + Eg = 0 ⇒4∑

i=1Q(i) 7−→(m,n) + Q (∆x∆y1) = 0

we obtain (∆x = ∆y)

Tm,n+1 + Tm,n−1 + Tm+1,n + Tm−1,n +Q (∆x)2

k − 4Tm,n = 0



Energy balance approach - Second approach, with convection



Q(m−1,n) 7−→(m,n) = k (∆y · 1)Tm−1,n − Tm,n

∆x

Q(m+1,n) 7−→(m,n) = k(

∆y2 · 1

)Tm+1,n − Tm,n

∆x

Q(m,n+1) 7−→(m,n) = k (∆x · 1)Tm,n+1 − Tm,n

∆y

Q(m,n−1) 7−→(m,n) = k(

∆x2 · 1

)Tm,n−1 − Tm,n

∆yAnd for convection

Q(∞) 7−→(m,n) = h(

∆x2 · 1

)(T∞ − Tm,n) + h

(∆y2 · 1

)(T∞ − Tm,n)

Then, for ∆x = ∆y , we obtain

Tm−1,n + Tm,n+1 +12 (Tm+1,n + Tm,n−1) +

h∆xk T∞ −

(3 +

h∆xk

)Tm,n = 0


5.comp. Numerical method ∆x = ∆y

Case 1 : Interior node

Tm,n+1 + Tm,n−1 + Tm+1,n + Tm−1,n − 4Tm,n = 0



Case 2 : Node at an internal corner with convection

2 (Tm−1,n + Tm,n+1) + (Tm+1,n + Tm,n−1) + 2h∆xk T∞ − 2

(3 +

h∆xk

)Tm,n = 0



Case 3 : Node at a plane surface with convection

(2Tm−1,n + Tm,n+1 + Tm,n−1) +2h∆x

k T∞ − 2(

h∆xk + 2

)Tm,n = 0



Case 4 : Node at an external corner with convection

(Tm,n−1 + Tm−1,n) + 2h∆xk T∞ − 2

(h∆x

k + 1)

Tm,n = 0



Case 5 : Node at a plane surface with uniform heat flux

(2Tm−1,n + Tm,n+1 + Tm,n−1) +2Q′′∆x

k − 4Tm,n = 0



Depending upon your mathematical background and the specific problem, thenumerical solution can be found with

Matrix inversion.Gauss-Seidel iteration....

The reader who wishes to study such analyses in depth should refer to specificpublications.


5.comp.2 Finite-difference methods for transient heatconductionFor transient conditions with two-dimensional effects, constant properties and nointernal heat generation, the general expression

∂2T∂x2 +

∂2T∂y2 +

∂2T∂z2 +

Qk =

1α

∂T∂t

reduces to∂2T∂x2 +

∂2T∂y2 =

1α

∂T∂t

To obtain the finite-difference form, we can use the central-difference form of

∂2T∂x2

∣∣∣∣∣m,n

=Tm+1,n + Tm−1,n − 2Tm,n

(∆x)2

∂2T∂y2

∣∣∣∣∣m,n

=Tm,n+1 + Tm,n−1 − 2Tm,n

(∆y)2


5.comp.2 Finite-difference methods for transient heatconduction

We discretize in time using the integer p as : t = p∆t and obtain

∂T∂t

∣∣∣∣∣m,n

=T p+1

m,n + T pm,n

∆t

Hence, the time derivative is in terms of the difference in temperatures at time(p+1) new and (p) previous, separated by the time interval ∆t.

Explicit method : the temperatures are evaluated at (p)Implicit method : the temperatures are evaluated at (p+1)


5.comp.2 Finite-difference methods : Explicit

It’s a forward-difference method. Evaluate the terms on the right-hand side ofequations at p.

1α

T p+1m,n − T p

m,n∆t =

T pm+1,n + T p

m−1,n − 2T pm,n

(∆x)2 +T p

m,n+1 + T pm,n−1 − 2T p

m,n

(∆y)2

Solving for new nodal temperature at p+1 for ∆x = ∆y :

T p+1m,n = Fo

(T p

m+1,n + T pm−1,n + T p

m,n+1 + T pm,n−1

)+ (1− 4 Fo) T p

m,n

withFo =

α∆t(∆x)2

For a one-dimensional transient heat conduction, the expression becomes

T p+1m = Fo

(T p

m+1 + T pm−1

)+ (1− 2 Fo) T p

m


5.comp.2 Finite-difference methods : Explicit

Equations are explicit since the unknown nodal temperatures at time p+1 aredetermined with known temperatures at time p in each time step.

Initial condition must be known so that the temperature of each node is known attime t = 0 when p = 0. Then, the temperatures at t = ∆t for p = 1 arecalculable and the calculations proceed for t = 2∆t for p = 2 and so forth.

Accuracy is increased by decreasing the size of the time step ∆t and the size of∆x , at the expense of increasing calculation time.


5.comp.2 Finite-difference methods : Stability ofcalculation

Stability criterion for 1-D interior node : Fo ≤ 12

Stability criterion for 2-D interior node : Fo ≤ 14


5.comp.2 Finite-difference methods : Stability ofcalculationEquations may also be derived from an energy balance. For example, for a surfacenode with a convection boundary 1-D. Starting with

Ein + Eg = Est

We obtain

hA (T∞ − T p0 ) +

kA∆x

(T 0

1 − T p0)

= ρcA∆x2

T p+10 − T p

0∆t

And solving for the surface temperature at t + ∆t

T p+10 =

2h∆tρc∆x (T∞ − T p

0 ) +2α∆t(∆x)2 (T p

1 − T p0 ) + T p

0

since2h∆tρc∆x = 2h∆x

kα∆t

(∆x)2 = 2 Bi Fo


5.comp.2 Finite-difference methods : Stability ofcalculation

ThenT p+1

0 = 2Fo (T p1 + Bi T∞) + (1− 2Fo − 2Bi Fo) T p

0

The finite-difference form of the Biot number is

Bi =h∆x

k

For stability, we require that the coefficient for T p0 ≥ 0, so

1− 2 Fo − 2 Fo Bi ≥ 0

orFo (1 + Bi) ≤ 1

2Note : The stability limit for the most restrictive requirement must be used !


5.comp.2 Finite-difference methods : Implicit

Implicit method, obtained using

∂T∂t

∣∣∣∣∣m,n

=T p+1

m,n + T pm,n

∆t

to approximate the time derivative and evaluating all other temperatures at timep+1 rather than p. This gives a backward-difference method, which intwo-dimensional form is

1α

T p+1m,n − T p

m,n∆t =

T p+1m+1,n + T p+1

m−1,n − 2T p+1m,n

(∆x)2 +T p+1

m,n+1 + T p+1m,n−1 − 2T p+1

m,n

(∆y)2

or for ∆x = ∆y it becomes

T pm,n = (1− 4 Fo) T p+1

m,n − Fo(

T p+1m+1,n + T p+1

m−1,n + T p+1m,n+1 + T p+1

m,n−1

)


5.comp.2 Finite-difference methods : Implicit

Notice :

Thus, the new temperature at node m,n depends on the new unknowntemperatures at the other adjacent nodes. Consequently, a simultaneous solutionis required using Gauss-Seidel iteration or matrix inversion.

The implicit solution scheme is implicitly unconditionally stable. Hence, we canchoose time steps ∆t and node spacings ∆x and ∆y to our own advantage.


5.comp.2 Finite-difference methods : Energy balancemethod

Energy balance method :

Surface node :

(1 + 2Fo + 2Bi Fo) T p+10 − 2Fo T p+1

1 = 2Fo Bi T∞ + T p0

Interior node :

(1 + 2Fo) T p+1m − Fo

(T p+1

m−1 + T p+1m+1

)= T p

m


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