Numerical Techniques for Conduction - LTCM | EPFL · The method of ﬂux plotting will solve all...

Numerical Techniques for Conduction

John Richard Thome

March 2010

John Richard Thome (LTCM - IGM - EPFL) Heat transfer - Conduction March 2010 1 / 35

Two-dimensional conduction

The general equation for T(−→x )

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

is called Laplace’s equation :∇2T = 0

The Laplacian is a sum of several second partial derivatives. Faced with a steadymultidimensional problem, four routes are open to us

Find out whether or not the analytical solution is already available.Solve the problem analytically.Obtain the solution graphically.Solve the problem numerically.


Graphical method : flux plot

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 distorted squares. Theobjective then is to sketch the isothermal lines and the adiabatic, or heat flow,lines which run perpendicular to them. This sketch is subject to two constraints :

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 length)If you fail, 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

kΔT

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


Graphical method : The shape factor

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

Q ≡ SkΔT

where for a flux plot

S =NI

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

Rt =1

kSwhere Q =

ΔTRt

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.


Graphical method : The shape factor

The shape factor for two similar bodies of different size :

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


Numerical method

Nodal Network :


Numerical method

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

Each node represents a small zone with an average temperature of that 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 (at increasedcomputational time).


Finite-difference form of heat conduction equation

The value of the second derivative, ∂2T/∂x2 is

∂2T∂x2

∣∣∣∣∣m,n

=

∂T/∂x∣∣∣m+1/2,n

− ∂T/∂x∣∣∣m−1/2,n

Δx

The temperature gradients in terms of nodal temperatures are

∂T∂x

∣∣∣∣∣m+1/2,n

=Tm+1,n − Tm,n

Δx

∂T∂x

∣∣∣∣∣m−1/2,n

=Tm,n − Tm−1,n

Δx

Substituting equations

∂2T∂x2

∣∣∣∣∣m,n

=Tm+1,n + Tm−1,n − 2Tm,n

(Δx)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

Δy=

Tm,n+1 + Tm,n−1 − 2Tm,n

(Δy)2

For a mesh in which Δx = Δy , 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



The nodal equation froms an energy balance : here assuming all heat flows intothe node, steady-state conditions and internal heat generation

Q(m−1,n) �−→(m,n) = k (Δy · 1)Tm−1,n − Tm,n

Δx

Q(m+1,n) �−→(m,n) = k (Δy · 1)Tm+1,n − Tm,n

Δx

Q(m,n+1) �−→(m,n) = k (Δx · 1)Tm,n+1 − Tm,n

Δy

Q(m,n−1) �−→(m,n) = k (Δx · 1)Tm,n−1 − Tm,n

ΔyWith equilibrium equation

Ein + Eg = 0 ⇒4∑

i=1

Q(i) �−→(m,n) + Q (ΔxΔy1) = 0

we obtain (Δx = Δy)

Tm,n+1 + Tm,n−1 + Tm+1,n + Tm−1,n +Q (Δx)2

k− 4Tm,n = 0



Energy balance approach - Second approach, with convection



Q(m−1,n) �−→(m,n) = k (Δy · 1)Tm−1,n − Tm,n

Δx

Q(m+1,n) �−→(m,n) = k(

Δy2· 1

)Tm+1,n − Tm,n

Δx

Q(m,n+1) �−→(m,n) = k (Δx · 1)Tm,n+1 − Tm,n

Δy

Q(m,n−1) �−→(m,n) = k(

Δx2· 1

)Tm,n−1 − Tm,n

Δy

And for convection

Q(∞) �−→(m,n) = h(

Δx2· 1

)(T∞ − Tm,n) + h

(Δy2· 1

)(T∞ − Tm,n)

Then, for Δx = Δy , we obtain

Tm−1,n + Tm,n+1 +12

(Tm+1,n + Tm,n−1) +hΔx

kT∞ −

(3 +

hΔxk

)Tm,n = 0


Summary of nodal finite-difference equations Δx = Δy

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) + 2hΔx

kT∞ − 2

(3 +

hΔxk

)Tm,n = 0



Case 3 : Node at a plane surface with convection

(2Tm−1,n + Tm,n+1 + Tm,n−1) +2hΔx

kT∞ − 2

(hΔx

k+ 2

)Tm,n = 0



Case 4 : Node at an external corner with convection

(Tm,n−1 + Tm−1,n) + 2hΔx

kT∞ − 2

(hΔx

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 ′′Δx

k− 4Tm,n = 0


Finite-difference solutions

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.


Transient conduction - Finite-difference methods fortransient heat conduction

For 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

(Δx)2

∂2T∂y2

∣∣∣∣∣m,n

=Tm,n+1 + Tm,n−1 − 2Tm,n

(Δy)2


Transient conduction - Finite-difference methods fortransient heat conduction

We discretise in time using the integer p as : t = pΔt and obtain

∂T∂y

∣∣∣∣∣m,n

=T p+1

m,n + T pm,n

Δt

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 Δt.

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


Transient conduction - Explicit method

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

Δt=

T pm+1,n + T p

m−1,n − 2T pm,n

(Δx)2 +T p

m,n+1 + T pm,n−1 − 2T p

m,n

(Δy)2

Solving for new nodal temperature at p+1 for Δx = Δy :

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 =

αΔt(Δx)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


Transient conduction - Explicit method

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 = Δt for p = 1 arecalculable and the calculations proceed for t = 2Δt for p = 2 and so forth.

Accuracy is increased by decreasing the size of the time step Δt and the size ofΔx , at the expense of increasing calculation time.


Transient conduction - Stability of calculation

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

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



Equations 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 ) +

kAΔx

(T 0

1 − T p0)

= ρcAΔx2

T p+10 − T p

0Δt

And solving for the surface temperature at t + Δt

T p+10 =

2hΔtρcΔx

(T∞ − T p0 ) +

2αΔt(Δx)2 (T p

1 − T p0 ) + T p

0

since2hΔtρcΔx

= 2hΔx

kαΔt

(Δx)2 = 2 Bi Fo



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 =hΔx

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 !


Transient conduction - Implicit method

Implicit method, obtained using

∂T∂y

∣∣∣∣∣m,n

=T p+1

m,n + T pm,n

Δt

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

Δt=

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

m−1,n − 2T p+1m,n

(Δx)2 +T p+1

m,n+1 + T p+1m,n−1 − 2T p+1

m,n

(Δy)2

or for Δx = Δy 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

)


Transient conduction - Implicit method

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 Δt and node spacings Δx and Δy to our own advantage.


Transient conduction - Energy balance method

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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