Green’s Function for Heat Conduction in a Slab X55

with Type 5 Boundary Conditions

Donald E. Amos

Abstract The Green’s function is the principal tool in construction of the general solution to the classical heat conduction problem. The solution is presented in terms of the internal heat generation, initial temperature and integrals which reflect the physical influence of the boundary. In the current literature ( http://Exact.unl.edu ) the common boundary conditions are presented as Types 1,2,3,4, and 5 ranging from a specified temperature (Type 1) to the most general form (Type 5) where input energy (flux), heat loss to the surroundings, heat storage on a boundary layer and conduction into the material are considered. Since the driving energy for the Green’s function is internal, the homogeneous form of the boundary condition is used to define the Green’s function. The thrust of this work is to derive the Green’s function, labeled X55, for a slab with Type 5 boundary conditions on both faces.

In the general solution with a Type 5 boundary, the usual integrals emerge, but an extra term which accounts for the release (or absorption) of energy stored in the boundary layer also appears. The results are used to construct the solution to an X55T0 slab problem where fluxes are the energy sources at the Type 5 boundaries x=0 and x=L. Another problem, describing the cooling of a boundary layer, is constructed to utilize only the extra term where the energy source is the heat stored in a boundary layer and the slab acts as a heat sink. Both solutions agree with the direct Laplace transform solutions. Finally, a closed system where there is no heat loss or gain is considered. The initial temperature differences of the slab and boundary layers provide the driving force for a redistribution of energy. The transient temperature distribution and equilibrium temperatures are calculated and agree with known results.

1. Introduction

The Green’s function is the principal tool in construction of the general solution to the classical heat conduction problem. In [2], the solution is presented in terms of the internal heat generation, initial temperature and integrals which reflect the physical influence of the boundary. In [1, p 49] and

( http://Exact.unl.edu ) the common boundary conditions are presented as Types 1,2,3,4, and 5 ranging from constant temperature (Type 1) to the most general form (Type 5) where input energy (flux), heat loss to the surroundings, heat storage on the boundary and conduction into the material are considered.

In mathematical terms,

(GX.1.1)

where is an input flux, h is the heat transfer coefficient for losses to the exterior, is the product of density, layer thickness and specific heat and the flux derivative is an interior normal derivative. The sub-cases can be summarized by

Type 5 Boundary condition:

Type 4 Boundary condition:

(GX.1.2) Type 3 Boundary condition:

Type 2 Boundary condition:

Type 1 Boundary condition:

In the Type 1 condition, a zero boundary temperature is produced by taking the limit.

The range of problems solved by classical Green’s functions with boundary Types 1-5 is outlined in [2]. For the slab on [0,L] with Type 5 boundaries, the problem to be solved is given by

(GX.1.3) .

(GX.1.4) .

The solution in terms of the corresponding Green’s function is [2]

(GX.1.5)

Notice that the usual integrals emerge, but extra terms in which account for the release (or absorption) of energy stored in the boundary layers also appear. It was noted above that one can have mixed boundary types in one problem. In the general solution [2], integrands over subsets of the boundary may look quite different depending on the type which applies there. This point is illustrated in Section 5, equation (GX.5.3).

The Green’s function corresponding to the Type 5 boundary condition is derived in Section 2. Section 3 is devoted to sub-cases X51 and X15 because the Type 1 condition cannot be obtained by setting h or to zero. In Section 4, the main results of Section 2 are used to construct the solution to an X55T0 slab problem where fluxes are the energy source in the Type 5 boundaries x=0 and x=L. This solution agrees with the direct solution in [3]. In Section 5 a problem is constructed to utilize only one extra term of (GX.1.5) where the energy source is the heat stored in a boundary layer and the slab acts as a heat sink. This solution is compared with the direct (Laplace transform) solution. Finally, a closed system where there is no heat loss or gain is considered in Section 6. The initial temperature differences in the slab and boundary layers provide the driving force for a redistribution of energy. The transient temperature distribution and equilibrium temperatures are calculated and agree with known results.

2. Green’s Function for X55 Slab

The differential equation for the slab is

(GX.2.1)

with 55 boundary conditions expressing how the internal heat generation is distributed to the external region or stored on a surface film with only heat capacity [2]. It is common practice to use the homogenous form of the boundary condition for the Green’s function since the energy source is interior to the material,

(GX.2.2) .

These are the Type 5 boundary conditions in the nomenclature of [1, p. 49] and http://Exact.unl.edu . The initial temperature distribution is taken to be zero

(GX.2.3) ,

except in the neighborhood of where the heat source has the form

(GX.2.4)

The Laplace transform is generally the method of choice for these linear problems with constant parameters where the boundary conditions vary with time. Then

(GX.2.5) ,

and we form the solution from the source term and the solutions of this equation,

(GX.2.6)

(GX.2.7) .

We apply this equation to the boundary conditions at x=0 and x=L

(GX.2.8)

Then

(GX.2.9)

The solution is

(GX.2.10) .

The algebra which follows expands this transform, collects like terms, and rewrites in a more classical form. The final result is presented in (GX.2.31). We start with the coefficients:

(GX.2.11)

(GX.2.12)

(GX.2.13)

For ,

(GX.2.14)

(GX.2.15)

(GX.2.16)

(GX.2.17)

(GX.2.18)

(GX.2.19)

(GX.2.20)

We use on the numerator

(GX.2.21)

(GX.2.22)

(GX.2.23)

Now we can write the transform explicitly by dividing by in the form

(GX.2.24)

(GX.2.25)

Now, is an analytic function in the whole p plane because each of the terms

(GX.2.26)

can be represented as a power series in p which has an infinite radius of convergence( analytic functions) while are only linear functions of p. However the term looks like it may have a pole at p=0 , but the following analysis shows that the term has the estimate which makes the numerator an analytic function of p in the whole p plane. The analysis is

(GX.2.27)

This analysis is further confirmed by the following manipulations

(GX.2.28)

(GX.2.29)

(GX.2.30)

Finally,

(GX.2.31)

We have demonstrated (2 ways) that we have the quotient of two analytic functions. Any singularities can only come from the zeros of the denominator, . Therefore we can apply the residue theorem in the classical way and sum the residues of poles from . We assume from previous history for finite thicknesses that the poles are on the negative real axis at . Therefore,

(GX.2.32)

The residue computation for the ratio of two analytic functions at a simple pole where is . Then, using and the result for is

(GX.2.33)

where is a non-zero solution of , (GX.2.24), with

(GX.2.34)

(GX.2.35)

and

(GX.2.36)

Here in the definition of is either .

The derivative computation can be found in [3]. It looks like is a solution of the eigen-equation and there is a pole at p=0. The analysis presented in [3] shows that

(GX.2.37) .

Therefore, p=0 is a pole if and only if . Then,

(GX.2.38)

and the n=0 term is

(GX.2.39) .

The complete solution has to include the case where . The analysis can be repeated, but the problem for is not essentially different from that for . That is, we map the region onto so that . The mapping y=L-x does the job and the contents of the APPENDIX confirms this approach. Thus we exchange subscripts and replace x and by L-x and L-respectively. Notice that the eigen-equation and the denominator are symmetric in and the exchange does not change any value. Then

(GX.2.40)

where is either . The complete solution is then a combination of (GX.2.33), (GX.2.39) and (GX.2.40).

3. Slabs X51, X15 and Sub-cases

The Type 5 boundary condition

(GX.3.1)

and the corresponding Green’s function boundary condition

(GX.3.2)

has two parameters [1, p 50] which lead to sub-cases of Types 4, 3, and 2 by setting h or or both to zero, making a total of 10 different slab problems. The Green’s function for Type 1 for a zero boundary temperature is derived by setting to zero and taking . We can delineate the cases by considering XI5, I=5,4,3,2,1; then XI4, I=4,3,2,1; etc., making a total of 15 cases. The cases for X5I; I=4,3,2,1; X4I, I=3,2,1; etc., lead to 10 more cases for a total of 25 cases with the subscripts defined according to (GX.1.2)

We consider the Green’s function for X51 because it requires taking a limit and all sub cases X41, X31, X21 can be obtained by setting or or both to zero. X11 is obtained from and with and .

X51 is define as a Type 5 boundary at x=0 and a constant zero temperature at x=L. This means that we set in (GX.2.33) and (GX.2.40). The limit is obtained first by dividing the numerator and denominator by . Then for in (GX.2.33)

(GX.3.4)

and we have

(GX.3.5)

And similarly for ,

(GX.3.6)

If in addition we take , we get the Green’s function for the X11 case

(GX.3.7)

We do the case for X15 so that the sub-cases X14, X13, X12 can be obtained by setting or or both to zero. X15 is obtained from X55 by setting

(GX.3.8)

and

(GX.3.9)

4 Application to the X55T0 Problem

The differential equation for the X55T0 slab on (0,L) is

(GX.4.1)

with 55 boundary conditions expressing how input fluxes are conducted into the region, conducted or radiated into an exterior medium, and stored on a surface layer with only heat capacity

(GX.4.2) .

The initial temperature distribution (T0) is taken to be zero

(GX.4.3) .

This problem was solved in [3]. The application here gives an independent check on both the Green’s function and the solution in [3]. The theory for the solution of this problem using the Green’s function is presented in [2],

(GX.4.4) .

Since on S from (GX.4.3) only the first term is applicable. Then, in one dimension, the surface integral is just sum of the evaluations at the endpoints. From (GX.2.33) and (GX.2.40) we get

(GX.4.5)

where we have used the expression for for and the expression for for . Equation (GX.4.5) agrees with the solution from [3]. If , then this becomes an X44T0 case and the leading terms for n=0 from the two formulations match also.

5. Cooling of a Boundary Layer

In [2], the general theory for the solution of a heat conduction problem was derived using Type 5 boundary conditions. The components of the solution follow the classical forms for volumetric heat generation, initial temperature and input flux on the boundary, but an extra term appears which comes from the release of heat from a boundary layer. This boundary layer has heat storage properties, acting like a source or sink, with infinite conductive properties and assumes the temperature of the slab on the boundary. It is the purpose of this example to construct a problem which uses only this extra term as the solution and then verify the result by direct solution of the problem. The problem is expressed by

(GX.5.1)

with 14 boundary conditions

(GX.5.2) .

This problem is of Type 14 since we have a zero temperature at x=0 and a Type 5 condition with at x=L with no external driving energy . These equations represent the cooling of the boundary layer initially at temperature using the slab as a heat sink. The heat energy contained in the boundary layer is . The zero in this formula is the ambient temperature. We write the formula from [2] for the solution, taking into account the different types at x=0 and x=L.

(GX.5.3) .

When we apply this to an interval, surface integrals are just the integrands at end points. With we get

(GX.5.4) .

In the development we have G for X15 from (GX.3.9)

(GX.5.5)

For the X14 geometry, we set and take the case for since G is evaluated at x=L,

(GX.5.6)

Then, for x=L,

(GX.5.7)

Solution of (GX.5.1) and (GX.5.2) by Laplace Transform

To verify the solution, we solve the initial problem by the Laplace transform.

(GX.5.8)

(GX.5.9)

We take a solution which satisfies the boundary condition at x=0

(GX.5.10)

and apply the boundary condition at x=L,

(GX.5.11)

(GX.5.12)

and

(GX.5.13) .

Now we have the ratio of two analytic functions in the p plane. The zeros of the denominator give contributions to the residues at

(GX.5.14)

The inversion becomes

(GX.5.15)

which agrees with the Green’s function approach in (GX.5.7).

6. Temperatures in a Slab with No Gain or Loss of Heat

In this section we are interested in constructing a problem which will test the validity of and verify transient terms for consistency. This means that we take which implies no loss of energy from the system. Further, if we take the input fluxes at the boundaries to be zero, , there is no gain of energy to the system either. The Green’s function solution from [2], with the only non-zero terms coming from the initial temperatures , is

(GX.6.1) .

Note that with , in the Green’s function from (GR2.33) and (GR2.40) is not zero, but has the value

(GX.6.2) .

Steady State

Since there is no gain or loss of heat, a system with unequal initial temperatures in the slab and boundary layers must come to an equilibrium temperature, for . This energy balance expresses the fact that the initial energy must be equal to the final energy with all components at the equilibrium temperature. Then, if the slab is initially at temperature on (0,L) and the boundary layers initially at temperatures , the energy balance is

(GX.6.3)

Since the transient part vanishes for , in each of the terms of T and we get the steady state or equilibrium temperature form (GX.2.33) and (GX.2.40)

(GX.6.4)

Thus, we have verified that the Green’s function approach using gives the correct result. Further, since is a convex linear combination of the initial temperatures, lies between the minimum and maximum of the three initial temperatures.

Transient Part

The mathematical problem can be stated as

(GX.6.5)

with boundary and initial conditions

(GX.6.6)

The Green’s function solution for this problem is given by (GX.6.1). We continue by investigating the transient part of the solution

To simplify the algebra, we drop the summation in (GX.2.33) and (GX.2.40) and pick up the coefficients which depend only on x and . Then, we have

(GX.6.7)

where we have used the expressions for for the integration. We want to compute the terms of (GX.6.7)

(GX.6.8)

The computation is:

(GX.6.9)

(GX.6.10)

Now we pick out the coefficients of the product of the sines and cosines in each expression to get

(GX.6.11)

Now the terms involving are a form of the eigen equation for this case where and the combination is zero.

(GX.6.12)

Then, the final sum is

(GX.6.13)

and the transient part is

(GX.6.14) .

The final solution with both the steady state and transient terms in (GX2.33) and (GX2.40) is

(GX.6.15)

It is interesting to note that if all three initial temperatures are the same, , we have equilibrium and there is no gradient to change the distribution. Consequently, we get the known result

(GX.6.16)

7. References

[1] Cole, KD, Beck, JV, et. al. (2010), Heat Conduction Using Green’s Functions, 2^nd Ed., CRC Press

Boca Raton, 643pp

[2] Amos, DE (2014) Theory of Heat Conduction with a Type 5 Boundary Condition,

http://nanohub.org/resources/20365

[3] Amos, DE, (2014) Heat Conduction in a Slab X55T0 and Sub-cases,

http://nanohub.org/resources/20381

APPENDIX

Algebra for X55 Green’s Function for

The object here is to do the algebra for the case and compare it to the results obtained in (GX2.33) and (GX2.40) with the mapping of onto the region with . The source term for requires . Then, for , (GX2.14) - (GX2.17) become

(GX2.14A)

(GX2.15A)

(GX2.16A)

(GX2.17A)

Now we add the source term (GX2.14A) to this sum

(GX2.18A)

where

(GX2.19A)

Finally, the analog to (GX2.20) is

(GX2.20A)

Now, if we take the expression for (GX2.20),

(GX2.20)

exchange the subscripts, replace with and x with, we get (GX2.20A). That is, exchanging the subscripts with the mapping gets the expression for . This change carries over to the inversion shown in (GX2.33) and (GX2.40).

Green's Function for Type 5 Boundary Condition on a Slab