Chapter 5

Transient Heat Conduction:

Analytical Methods

1 Introduction

Many heat conduction problems encountered in engineering applications involve time as inindependent variable. The goal of analysis is to determine the variation of the temperatureas a function of time and position T (x, t) within the heat conducting body. In general,we deal with conducting bodies in a three dimensional Euclidean space in a suitable set ofcoordinates (x ∈ R3) and the goal is to predict the evolution of the temperature field forfuture times (t > 0).

Here we investigate specifically solutions to selected special cases of the following formof the heat equation

ρCp∂T

∂t= ∇ · (k∇T ) + g(r)

Solutions to the above equation must be obtained that also satisfy suitable initial and bound-ary conditions.

2 Fundamental Solutions

Transient problems resulting from the effect of instantaneous point, line and planar sourcesof heat lead to useful fundamental solutions of the heat equation. By considering media ofinfinite or semi-infinite extent one can conveniently ignore the effect of boundary conditionson the resulting solutions.

Let a fixed amount of energy H0 (J) be instantaneously released (thermal explosion) attime t = 0 at the origin of a three dimensional system of coordinates inside a solid body of in-finite extent initially at T (x, 0) = T (r, 0) = 0 everywhere, where x = r =

√x2 + y2 + z2. No

other thermal energy input exists subsequent to the initial instantaneous release. Assumingconstant thermal properties k, ρ and Cp, the heat equation is

∂T

∂t= α

1

r2

∂

∂r(r2∂T

∂r)

1

where α = k/ρCp. This must be solved subject to the initial condition

T (r, 0) = 0

for all r > 0 plus the statement expressing the instantaneous release of energy at t = 0 atthe origin. Since the body is infinitely large, the far field temperature never changes and allthe released energy must dissipate within the body itself.

The fundamental solution of this problem is given by

T (r, t) =H0

(4παt)3/2ρCpexp(− r2

4αt)

where H0/ρCp is the amount of energy released per unit energy required to raise the tem-perature of a unit volume of material by one degree.

This solution may be useful in the study of thermal explosions where a buried explosiveload located at r = 0 is suddenly released at t = 0 and the subsequent distribution oftemperature at various distances from the explosion is measured as a function of time.A slight modification of the solution produced by the method of reflexion constitutes anapproximation to the problem of surface heating of bulk samples by short duration pulses offinely focused high energy beams.

Consider now the situation where heat is released instantaneously at t = 0 at x = y = 0but along the entire z−axis in a circular cylindrical system of coordinates in the amount(per unit length) of H0/L (J/m). The temperature is expected to be independent of z andthe corresponding fundamental solution is

T (r, t) =(H0/L)

4παtρCpexp(− r2

4αt)

where here r =√x2 + y2. This could be a model of heat conduction inside a massive body

heated by a short pulse of current passing through a thin wire embedded within it.Finally, if heat is instantaneously released at t = 0 at x = 0 but on the entire the x− y

plane, the corresponding fundamental solution is independent of x and y and is given as

T (r, t) =(H0/A)

(4παt)1/2ρCpexp(− r2

4αt)

where r = z and H0/A is now the amount of thermal energy released per unit area (J/m2).If the reflexion method is used here, the resulting expression may be a representation of thetemperature distribution resulting from a short pulse of a uniformly distributed energy beamapplied at the surface of a semi-infinite medium.

Another useful fundamental solution is obtained for the case of a semi-infinite solid(x ≥ 0) initially at T = T0 everywhere and suddenly exposed to a fixed temperature T = 0at x = 0. The statement of the problem is

1

α

∂T

∂t=∂2T

∂x2

2

subject to

T (x, 0) = T0

and

T (0, t) = 0

The solution is easily obtained introducing the Laplace transform as follows. Multiplythe heat equation by exp(−st), where s is the parameter of the transform, and then integratewith respect to t from 0 to ∞, i.e.

1

α

∫ ∞

0exp(−st)∂T

∂tdt =

∫ ∞

0exp(−st)∂

2T

∂x2dt

introducing the notation T ∗ =∫ ∞0 exp(−st)Tdt and allowing for the exchange of the integral

and differential operators, the heat equation transforms into

d2T ∗

dx2=s

αT ∗

This is a simple ordinary differential equation which is readily solved for T ∗(x). The desiredresult T (x, t) is finally obtained from T ∗ by an inversion process yielding

T (x, t) = T0erf(x

2√αt

)

where the error function, erf is defined as

erf(z) =2√π

∫ z

0exp(−ξ2)dξ

The above solution is an appropriate mathematical approximation of the problem of quench-ing hot bulk metal samples.

3 Separation of Variables and Integral Transforms in

Transient Heat Conduction Problems

Analytical solutions to boundary value problems in linear ordinary differential equations areusually obtained by first producing a general solution. The general solution is obtained bysimple integration and contains as many independent arbitrary constants as there are deriva-tives in the original differential equation. Incorporation of boundary conditions subsequentlydefine the values of the integration constants.

In contrast, general solutions of linear partial differential equations involve arbitrary func-tions of specific functions. Incorporation of boundary conditions involves the determination

3

of functional relationships and is rarely feasible or practical. An alternative approach tofinding solutions is based instead on first determining a set of particular solutions directlyand then combining these so as to satisfy the prescribed boundary conditions. A specific,useful and simple implementation of the above idea is known as the method of separationof variables. This section illustrates the application of the separation of variables methodfor the determination of analytical solutions of steady and transient one dimensional linearheat conduction problems.

In essence, the method is based on the assumption that if one is looking for a solutionto a transient, one-dimensional heat conduction problem of the form T (x, t) it is possible toexpress it as a product

T (x, t) = X(x)Γ(t)

where the functions X(x) and Γ(t) are each functions of a single independent variable sat-isfying specific ordinary differential equations. One proceeds by first solving the associatedordinary differential equations (ODEs) which are then combined in the product form givenabove.

Regardless of the dimensionality of a problem, the equation for Γ(t) is always of the firstorder and readily solvable by elementary methods.

The equation for X(x) is always of second order and together with the boundary condi-tions leads to an eigenvalue problem (proper Sturm-Liouville system). The required solutionfor X(x) can be formally and very generally expressed by the inversion formula

X(x) =∞∑

m=1

K(βm, x)X(βm)

where the kernel functions K(βm, x) are the normalized eigenfunctions of the associatedSturm-Liouville system and the sum runs over all the eigenvalues of the system. Moreover,the integral transform X(βm) is given by the formula

X(βm) =∫ L

x′=0K(βm, x

′)X(x′)dx′

Therefore, once the boundary conditions are translated from T (x, t) toX(x), the resultingSturm-Liouville system is solved yielding the appropriate eigenfunctions and eigenvalues.From these one then determines the kernel K(βm, x) and the integral transform X(βm). Theinversion formula is next used to obtain the function X(x). Finally, the desired solutionT (x, t) is obtained substituting in the assumed product form.

The transform method is also applicable to problems in semi-infinite domains. In thiscase the inversion formula is given instead by

X(x) =∫ ∞

β=0K(βm, x)X(β)dβ

4

while the integral transform is

X(βm) =∫ ∞

x′=0K(βm, x

′)X(x′)dx′

Note that the sum over the discrete set of eigenvalues in the inversion formula has beenreplaced by an integral over the continuous spectrum of eigenvalues obtained in the semi-infinite case.

4 Separation of Variables in One-Dimensional Systems

In transient, one-dimensional heat conduction problems, the required temperature is a func-tion of distance and time.

4.1 Imposed Boundary Temperature in Cartesian Coordinates

A simple but important conduction heat transfer problem consists of determining the tem-perature history inside a solid flat wall which is quenched from a high temperature. Morespecifically, consider the homogeneous problem of finding the one-dimensional temperaturedistribution inside a slab of thickness L and thermal diffusivity α, initially at some specifiedtemperature T (x, 0) = f(x) and exposed to heat extraction at its boundaries x = 0 andx = L such that T (0, t) = T (L, t) = 0 (Dirichlet homogeneous conditions), for t > 0. Thethermal properties are assumed constant.

The mathematical statement of the heat equation for this problem is:

∂T (x, t)

∂t= α

∂2T (x, t)

∂x2

subject to

T (0, t) = T (L, t) = 0

and

T (x, 0) = f(x)

for all x when t = 0.As indicated before, the method of separation of variables starts by assuming the solution

to this problem has the following particular form

T (x, t) = X(x)Γ(t)

If the assumption is wrong, one discovers soon enough, but if it is correct then one may justfind the desired solution to the problem! Fortunately, the latter turns out to be the case forthis and for many other similar problems.

5

Introducing the above assumption into the heat equation and rearranging yields

1

X

d2X

dx2=

1

αΓ

dΓ

dt

However since X(x) and Γ(t), the left hand side of this equation is only a function of xwhile the right hand side is a function only of t. For this to avoid being a contradiction (forarbitrary values of x and t) both sides must be equal to a constant. For physical reasons(in this case we obviously are after a temperature function that either increases or decreasesmonotonically depending on the initial conditions and the imposed boundary conditions),the required constant must be negative; let us call it −ω2.

Therefore, the original heat equation, (a partial differential equation) is transformed intothe following equivalent system of ordinary differential equations

1

αΓ

dΓ

dt= −ω2

and

1

X

d2X

dx2= −ω2

General solutions to these equations are readily obtained by direct integration and are

Γ(t) = C exp(−ω2αt)

and

X(x) = A′ cos(ωx) +B ′ sin(ωx)

Substituting back into our original assumption yields

T (x, t) = X(x)Γ(t) = [A cos(ωx) +B sin(ωx)] exp(−ω2αt)

where the constant C has been combined with A′ and B ′ to give A and B without loosingany generality.

Now we introduce the boundary conditions. Since T (0, t) = 0, necessarily A = 0. Fur-thermore, since also T (L, t) = 0, then sin(ωL) = 0 (since B = 0 is an uninteresting trivialsolution.) There is an infinite number of values of ω which satisfy this conditions, i.e.

ωn =nπ

L

with n = 1, 2, 3, .... The ωn’s are the eigenvalues and the functions sin(ωnx) are the eigen-functions of the Sturm-Liouville problem satisfied by the function X(x). These eigenvaluesand eigenfunctions play a role in heat conduction analogous to that of the deflection modes

6

in structural dynamics, the vibration modes in vibration theory and the quantum states inwave mechanics.

Note that each value of ωn yields an independent solution satisfying the heat equationas well as the two boundary conditions. Therefore we have now an infinite number ofindependent solutions Tn(x, t) for n = 1, 2, 3, ... given by

Tn(x, t) = [Bn sin(ωnx)] exp(−ω2nαt)

The principle of superposition allows the creation of a more general solution from theparticular solutions above by simple linear combination to give

T (x, t) =∞∑

n=1

Tn(x, t) =∞∑

n=1

[Bn sin(ωnx)] exp(−ω2nαt) =

∞∑

n=1

[Bn sin(nπx

L)] exp(−(

nπ

L)2αt)

The final step is to ensure the values of the constants Bn are such that they satisfy theinitial condition, i.e.

T (x, 0) = f(x) =∞∑

n=1

Bn sin(nπx

L)

But this is just the Fourier sine series representation of the function f(x).Recall that a key property of the eigenfunctions is that of orthonormality property. This

is expressed here as

∫ L

0sin(

nπx

L) sin(

mπx

L)dx =

{

0 , n 6= mL/2 , n = m

Using the orthonormality property one can multiply the Fourier sine series representationof f(x) by sin(mπx

L) and integrate from x = 0 to x = L to produce the result

Bn =2

L

∫ L

0f(x) sin(

nπx

L)dx

for n = 1, 2, 3, ....Finally, the resulting Bn’s can be substituted into the general solution above to give

T (x, t) =∞∑

n=1

Tn(x, t) =∞∑

n=1

[2

L

∫ L

0f(x′) sin(

nπx′

L)dx′] sin(

nπx

L) exp(−n

2π2αt

L2)

Explicit expressions for the Bn’s can be readily obtained for simple f(x)’s, for instance if

f(x) = Ti = constant

then

Bn = −Ti2(−1 + (−1)n)

nπ

7

And if

f(x) =

{

x , 0 ≤ x ≤ L2

L− x , L2≤ x ≤ L

then

Bn =

4Ln2π2 , n = 1, 5, 9, ...− 4L

n2π2 , n = 3, 7, 11, ...0 , n = 2, 4, 6, ...

For instance, for the second specific f(x) given above, the result is

T (x, t) =4Ti

π

∞∑

n=0

1

(2n+ 1)sin(

(2n+ 1)πx

L) exp(−n

2π2αt

L2)

And for the second case, the result is

T (x, t) =4L

π2[exp(−π

2αt

L2) sin(

πx

L) − 1

9exp(−9π2αt

L2) sin(

3πx

L) + ...]

4.2 Convection at the Boundary in Cartesian Coordinates

The separation of variables method can also be used when the boundary conditions spec-ify values of the normal derivative of the temperature (Newmann conditions) or when linearcombinations of the normal derivative and the temperature itself are used (Convective condi-tions; Mixed conditions; Robin conditions). Consider the homogeneous problem of transientheat conduction in a slab initially at a temperature T = f(x) and subject to convectionlosses into a medium at T = 0 at x = 0 and x = L. Convection heat transfer coefficients atx = 0 and x = L are, respectively h1 and h2. Assume the thermal conductivity of the slabk is constant.

The mathematical formulation of the problem is to find T (x, t) such that

∂2T (x, t)

∂x2=

1

α

∂T (x, t)

∂t

−k∂T∂x

+ h1T = 0

at x = 0 and

k∂T

∂x+ h2T = 0

at x = L, with

T (x, 0) = f(x)

8

for all x when t = 0.Assume the solution is of the form T (x, t) = X(x)Γ(t) and substitute to obtain

1

X

d2X

dx2=

1

αΓ

dΓ

dt= −β2

The solution for Γ(t) is

Γ(t) = exp(−αβ2t)

while X(x) is the solution of the following eigenvalue (proper Sturm-Liouville) problem

d2X

dx2+ β2X = 0

with

−kdXdx

+ h1X = 0

at x = 0 and

kdX

dx+ h2X = 0

at x = L.Let the eigenvalues of this problem be βm and the eigenfunctions X(βm, x). Since the

eigenfunctions are orthogonal

∫ L

0X(βm, x)X(βn, x)dx =

{

0 , n 6= mN(βm) , n = m

where

N(βm) =∫ L

0X(βm, x)

2dx

is the norm of the problem.It can be shown that for the above problem the eigenfunctions are

X(βm, x) = βm cosβmx+h1

k1sinβmx

while the eigenvalues are the roots of the transcendental equation

tan βmL =βm(h1

k1+ h2

k2)

β2m − h1

k1

h2

k2

9

Therefore, the complete solution is of the form

T (x, t) =∞∑

m=1

cmX(βm, x) exp(−αβ2mt)

The specific eigenfunctions are obtained by incorporating the initial condition

f(x) =∞∑

m=1

cmX(βm, x)

which expresses the representation of f(x) in terms of eigenfunctions and requires that

cm =1

N(βm)

∫ L

0X(βm, x)f(x)dx

As am example, the case when h1 = 0 and h2 = h yields the eigenfunctions

X(βm, x) = cos(βmx)

and the eigenvalues are the roots of βm tan(βmL) = h/k. Note that the eigenvalues inthis case cannot be given explicitly but must be determined by numerical solution of thegiven transcendental equation. For this purpose, it is common to rewrite the transcendentalequation as

cot(βmL) =βmL

Bi

where Bi = hL/k. This can be easily solved either graphically or numerically by bisection,Newton’s or secant methods. Finally, the norm in this case is

N(βm) =∫ L

0X(βm, x)

2dx =∫ L

0cos(βmx)

2dx =

=1

2

cos(βmL) sin(βmL) + βmL

βm=L(β2

m + (h/k)2) + (h/k)

2(β2m + (h/k)2)

4.3 Imposed Boundary Temperature in Cylindrical Coordinates

Consider the quenching problem where a long cylinder (radius r = b) initially at T = f(r)whose surface temperature is made equal to zero for t > 0.

The heat equation in this case is

1

r

∂

∂r(r∂T

∂r) =

1

α

∂T

∂t

subject to

T = 0

10

at r = b and

∂T

∂r= 0

at r = 0.According to the separation of variables method we assume a solution of the form T (r, t) =

R(r)Γ(t). Substituting this into the heat equation yields

1

rR

d

dr(rdR

dr) =

1

αΓ

dΓ

dt

This equation only makes sense when both terms are equal to a negative constant −λ2. Theoriginal problem has now been transformed into two boundary value problems, namely

dΓ

dt+ αλ2Γ = 0

with general solution

Γ(t) = C exp(−αλ2t)

where C is a constant and

r2d2R

dr2+ r

dR

dr+ r2λ2R = 0

subject to R(b) = 0 and necessarily bounded at r = 0.The last equation is a special case of Bessel’s equation. The only physically meaningful

solution of which has the form

R(r) = A′J0(λr)

where A′ is another constant and J0(z) is the Bessel function of first kind of order zero ofthe argument given by

J0(z) = 1 − z2

(1!)222+

z4

(2!)224− z6

(3!)226+ ...

Since R(b) = 0 this requires that J0(λb) = 0 which defines the eigenvalues and eigenfunc-tions for this problem. The eigenvalues are thus given as the roots of

J0(λnb) = 0

Specifically, the first four roots are: λ1b = 2.405, λ2b = 5.520, λ3b = 8.654, and λ4b = 11.79.A particular solution is then

Tn(r, t) = Rn(r)Γ(t) = AnJ0(λnr) exp(−αλ2nt)

11

where An = A′nC and n = 1, 2, 3, ...

Superposition of particular solutions yields the more general solution

T (r, t) =∞∑

n=1

Tn(r, t) =∞∑

n=1

Rn(r)Γ(t) =∞∑

n=1

AnJ0(λnr) exp(−αλ2nt)

All is left is to determine the An’s. For this we use the given initial condition, i.e.

T (r, 0) = f(r) =∞∑

n=1

AnJ0(λnr)

This is the Fourier-Bessel series representation of f(r) and one can use the orthogonalityproperty of the eigenfunctions to write

∫ b

0rJ0(λmr)f(r)dr =

∞∑

n=1

An

∫ b

0rJ0(λmr)J0(λnr)dr = An

∫ b

0rJ2

0 (λmr)dr =

=b2Am

2[J2

0 (λmb) + J21 (λmb)] =

b2Am

2J2

1 (λmb)

where J1(z) = −dJ0(z)/dz is the Bessel function of first kind of order one of the argument.Therefore,

An =2

b2J21 (λnb)

∫ b

0rJ0(λnr)f(r)dr

so that the required solution is

T (r, t) =2

b2

∞∑

n=1

J0(λnr)

J21 (λnb)

exp(−αλ2nt)

∫ b

0r′J0(λnr

′)f(r′)dr′

An important special case is obtained when f(r) = Ti = constant. The required solutionthen becomes

T (r, t) =2Ti

b

∞∑

n=1

J0(λnr)

λnJ1(λnb)exp(−αλ2

nt)

4.4 Convection at the Boundary in Cylindrical Coordinates

Many problems involving more complex boundary conditions can also be solved using theseparation of variables method. As an example consider the quenching problem where along cylinder (radius r = b) initially at T = f(r) is exposed to a cooling medium at zerotemperature which extracts heat uniformly from its surface.

The heat equation in this case is

1

r

∂

∂r(r∂T

∂r) =

1

α

∂T

∂t

12

subject to

−k∂T∂r

= hT

at r = b, and subject to

∂T

∂r= 0

at r = 0.Separation of variables T (r, t) = R(r)Γ(t) gives in this case

T (r, t) =2

b2

∞∑

m=1

exp(−αβmt)β2

mJ0(βmr)

(β2m + (h

k)2)J2

0 (βmb)

∫ b

0r′J0(βmr

′)f(r′)dr′

A special case of interest is when the initial temperature is constant = Ti and the sur-rounding environment is at a non-zero temperature = T∞. In this case the above equationreduces to

T (r, t) = T∞ +2

b(Ti − T∞)

∞∑

m=1

1

βm

J1(βmb)J0(βmr)

J20 (βmb) + J2

1 (βmb)e−β2

mαt

where the eigenvalues βm are obtained from the roots of the following transcendental equation

βbJ1(βb)− BiJ0(βb) = 0

with Bi = hb/k.

4.5 Imposed Boundary Temperature in Spherical Coordinates

Consider the quenching problem where a sphere (radius r = b) initially at T = f(r) whosesurface temperature is made equal to zero for t > 0.

The heat equation in this case is

1

r2

∂

∂r(r2∂T

∂r) =

1

α

∂T

∂t

subject to

T = 0

at r = b and

∂T

∂r= 0

13

at r = 0.According to the separation of variables method we assume a solution of the form T (r, t) =

R(r)Γ(t). Substituting this into the heat equation yields

1

r2R

d

dr(r2dR

dr) =

1

αΓ

dΓ

dt

Again, this equation only makes sense when both terms are equal to a negative constant−λ2. The original problem has now been transformed into two boundary values problems,namely

dΓ

dt+ λ2αΓ = 0

with general solution

Γ(t) = C exp(−αλ2t)

where C is a constant and

1

r2

d

dr(r2dR

dr) + λ2R = 0

subject to R(b) = 0 and necessarily bounded at r = 0. The general solution of the lastproblem is

R(r) = A′ sin(λr)

r+B ′cos(λr)

r= A′ sin(λr)

r

where A′ and B ′ are constants and B ′ = 0 since the temperature must be bounded at r = 0.Moreover, the boundary condition at r = b yields the eigenvalues

λn =nπ

b

and the eigenfunctions

Rn(r) =A′

n

rsin(λnr) =

1

rsin(

nπr

b)

for n = 1, 2, 3, ....A particular solution is then

Tn(r, t) = Rn(r)Γ(t) =An

rsin(λnr) exp(−αλ2

nt)

where An = A′nC and n = 1, 2, 3, ...

14

Superposition of particular solutions yields the more general solution

T (r, t) =∞∑

n=1

Tn(r, t) =∞∑

n=1

Rn(r)Γ(t) =∞∑

n=1

An

rsin(λnr) exp(−αλ2

nt)

To determine the An’s we use the given initial condition, i.e.

T (r, 0) = f(r) =∞∑

n=1

An

rsin(λnr)

this is again a Fourier series representation with the Fourier coefficients given by

An =2

b

∫ b

0f(r′) sin(

nπr′

b)r′dr′

An important special case results when the initial temperature f(r) = Ti = constant. Inthis case the Fourier coefficients become

An = −Tib

nπcos(nπ) = −Tib

nπ(−1)n

4.6 Convection at the Boundary in Spherical Coordinates

Consider a sphere with initial temperature T (r, 0) = F (r) and dissipating heat by convectioninto a medium at zero temperature at its surface r = b. The heat conduction equation in1D spherical coordinates is

∂2T

∂r2+

2

r

∂T

∂r=

1

α

∂T

∂t

In terms of the new variable U(r, t) = rT (r, t) the mathematical formulation of theproblem is

∂2U

∂r2=

1

α

∂U

∂t

subject to

U(0, t) = 0, r = 0

∂U

∂r+ (

h

k− 1

b)U = 0, r = b

and

U(r, 0) = rF (r), t = 0

15

This is just like heat 1D conduction from a slab so the solution is

T (r, t) =2

r

∞∑

m=1

e−αβ2mt β2

m + (h/k − 1/b)2

b(β2m + (h/k − 1/b)2) + (h/k − 1/b)

sin(βmr)

∫ b

r′=0r′F (r′) sin(βmr

′)dr′

and the eigenvalues βm are the positive roots of

βmb cot(βmb) + b(h/k − 1/b) = 0

Consider now as another example a hollow sphere a ≤ r ≤ b initially at F (r) anddissipating heat by convection at both its surfaces via heat transfer coefficients h1 andh2. Introducing the transformation x = r − a, the problem becomes identical to 1D heatconduction from a slab and the solution is

T (r, t) =1

r

∞∑

m=1

e−αβ2mt 1

N(βm)R(βm, r)

∫ b

r′=ar′F (r′)R(βm, r

′)dr′

where

R(βm, r) = βm cos(βm[r − a]) + (h1/k + 1/a) sin(βm[r − a])

and the eigenvalues βm are the positive roots of

tan(βm[b− a]) =βm([h1/k + 1/a] + [h2/k − 1/b])

β2m − [h1/k + 1/a][h2/k − 1/b]

4.7 Non-homogeneous Problems in One-Dimensional Systems

Consider the problem of finding the temperature field T (x, t) resulting from transient heatconduction in a slab (1D) initially at T (x, 0) = f(x) whose surfaces at x = 0 and x = lare maintained respectively at constant non-zero temperatures T0 and Tl for t > 0. Themathematical formulation is

∂2T

∂x2=

1

α

∂T

∂t

subject to

T (0, t) = T0

T (l, t) = Tl

16

and

T (x, 0) = f(x)

This problem can be transformed into two equivalent but simpler problems. A steadystate non homogeneous problem and a transient homogeneous problem.

Introduce new functions v(x) and w(x, t) such that

T (x, t) = v(x) + w(x, t)

Substituting into the original PDE one gets

d2v

dx2= 0

and

∂2w

∂x2=

1

α

∂w

∂t

subject to

w(0, t) = T0 − v(0)

w(l, t) = Tl − v(l)

and

w(x, 0) = f(x) − v(x)

Now we select v(x) such that the boundary conditions for w become homogeneous, i.e.

v(x) = T0 +x

l[Tl − T0]

With this, the solution for w with F (x) = f(x) − v(x) is readily obtained by separation ofvariables. Finally

T (x, t) = T0 + (x

l)(T1 − T0) +

∞∑

n=1

cne−n2π2αt/l2 sin(

nπx

l)

with

cn =2

l

∫ l

0[f(x)− v(x)] sin(

nπx

l)dx

=2

l

∫ l

0f(x) sin(

nπx

l)dx+

2

nπ[(−1)nT1 − T0]

17

Consider a now long cylinder initially at T = F (r) inside which heat is generated ata constant rate g0 and whose boundary r = b is subjected to T = 0. The mathematicalstatement of the problem consists of the heat equation *

∂2T

∂r2+

1

r

∂T

∂r+

1

kg0 =

1

α

∂T

∂t

with T = 0 at r = b and t > 0 and T = F (r) at t = 0. The problem can be split into asteady state problem giving Ts(r) = (g0/4k)(b

2 − r2) and a homogeneous transient problemgiving Th(r, t). The solution to the original problem becomes

T (r, t) = Ts(r) + Th(r, t).

5 Separation of Variables for the Multidimensional Heat

Equation

The governing equation for three dimensional heat conduction is

∇2T (r, t) =1

α

∂T (r, t)

∂t

This must be solved subject to

ki∂T

∂ni+ hiT = 0

on boundary Si and

T (r, 0) = f(r)

initially.Separation of variables in this case requires a solution of the form

T (r, t) = ψ(r)Γ(t)

which leads to

1

ψ(r)∇2ψ(r) =

1

αΓ(t)

dΓ

dt= −λ2

The solution for Γ is the same as before while ψ must be found to satisfy

∇2ψ(r) + λ2ψ(r) = 0

subject to

ki∂ψ

∂ni+ hiψ = 0.

This is called the Helmholtz equation and can in turn be solved by separating variables aswell.

18

5.1 Separation of Variables for the Helmholtz Equation

Separation of variables of the Helmholtz equation requires assuming the solution has theform

ψ(x, y, z) = X(x)Y (y)Z(z)

substituting this produces

X ′′ + β2X = 0

Y ′′ + γ2Y = 0

Z ′′ + η2Z = 0

where β2 + γ2 + η2 = λ2. Further, the solution for the time dependent part is

Γ = exp(−αλ2t)

Finally, the complete solution is obtained by superposition of the individual solutions.

5.2 Multidimensional Homogeneous Problems

Consider the problem of transient 2D conduction in a plate of dimensions a × b, initiallyat T (x, y, 0) = f(x, y) and subject to an insulating condition at the boundary x = 0, azero temperature condition at the boundary y = 0 and to convective heat exchange witha surrounding environment at zero temperature at boundaries x = a and y = b with heattransfer coefficients h2 and h4, respectively.

Separation of variables starts by assuming a solution of the form

T (x, y, t) = Γ(t)X(x)Y (y)

substituting one gets

X ′′ + β2X = 0

subject to

X ′ = 0

at x = 0 and

X ′ +h2

kX = 0

19

at x = a, and

Y ′′ + γ2Y = 0

subject to

Y = 0

at y = 0 and

Y ′ +h4

kY = 0

at y = b. The complete solution is then

T (x, y, t) =∞∑

m=1

∞∑

n=1

cmne−α(β2

m+γ2n)t cos(βmx) sin(γny)

where cos(βmx) and sin(γnx) are the eigenfunctions of this problem and the Fourier coef-ficients cmn must be obtained by substituting the initial condition T (x, y, 0) = f(x, y) andtaking advantage of the orthogonality of the eigenfunctions to yield

cmn =1

N(βm)N(γn)

∫ a

0

∫ b

0cos(βmx) sin(γnx)f(x, y)dxdy

where the norms N(βm) and N(γn) are

N(βm) =∫ a

0cos2(βmx)dx =

1

2

a(β2m + (h2/k)

2) + h2/k

β2m + (h2/k)2

and

N(γn) =∫ b

0sin2(γny)dy =

1

2

b(γ2n + (h4/k)

2) + h4/k

γ2n + (h4/k)2

While the eigenvalues βm are obtained as the roots of

βm tanβma =h2

k

and the eigenvalues γn are the roots of

γn cot γnb = −h4

k

20

5.3 Quenching of a Billet

Consider a long billet of rectangular cross section 2a× 2b initially at a uniform temperatureTi which is quenched by exposing it to convective exchange with an environment at T∞ bymeans of a heat transfer coefficient h. Since the billet is long it is appropriate to neglectconduction along its axis. Because of symmetry it is enough to analyze one quarter of thecross section using a rectangular Cartesian system of coordinates with the origin at the centerof the billet.

Define first the shifted temperature θ(x, y, t) = T (x, y, t)−T∞. In terms of the shiftedtemperature the mathematical formulation of the problem is as follows.

∂θ

∂t= α(

∂2θ

∂x2+∂2θ

∂y2)

subject to

θ(x, y, 0) = Ti − T∞ = θi

∂θ

∂x= 0

at x = 0,

∂θ

∂y= 0

at y = 0,

∂θ

∂x= −h

kθ(a, y, t)

at x = a, and

∂θ

∂y= −h

kθ(x, b, t)

at y = b.The solution to this problem obtained by separation of variables is

θ(x, y, t)

θi= 4

∞∑

n=1

∞∑

m=1

e−α(λ2n+β2

m)t × sin(λna) cos(λnx) sin(βmb) cos(βmy)

[λna + sin(λna) cos(λna)][βmb+ sin(βmb) cos(βmb)]

where the eigenvalues λn and βm are the roots of the following transcendental equations

λn tan(λna) =h

k

and

βm tan(βma) =h

k

21

5.4 Product Solutions

The solution above can be rewritten as

θ(x, y, t)

θi

= 2∞∑

n=1

sin(λna) cos(λnx)

λna + sin(λna) cos(λna)e−αλ2

nt

×2∞∑

m=1

sin(βmb) cos(βmy)

βmb+ sin(βmb) cos(βmb)e−αβ2

mt

Note that the first sum is precisely the solution to the 1D problem of quenching of a slabof thickness 2a while the second sum is the corresponding solution for a slab of thickness 2b.Therefore

θ(x, y, t)

θi|billet =

θ(x, t)

θi|2a−slab ×

θ(y, t)

θi|2b−slab

The method of obtaining solutions to multidimensional heat conduction problems bymultiplication of the solutions of subsidiary problems of lower dimensionality is called themethod of product solutions.

In general, if the initial temperature of the medium can be expressed as a product ofsingle space variable functions one can produce solutions to multidimensional problems asproducts of single dimensional solutions.

Consider transient homogeneous 2D conduction in a plate with convective exchange onall four sides. Initially, T (x, y, 0) = F (x)F (y). This problem can be regarded as a compositeof two one-dimensional ones with solutions T1(x, t) and T2(y, t). The product solution forthe 2D problem is then

T (x, y, t) = T1(x, t)T2(y, t)

5.5 Non-homogeneous Problems

Consider the non homogeneous problem of transient multidimensional heat conduction withinternal heat generation

∇2T (r, t) +1

kg(r) =

1

α

∂T (r, t)

∂t

subject to non homogeneous time independent conditions on at least part of the boundary.This problem can be decomposed into a set of steady state non homogeneous problems

in each of which a single non homogenous boundary condition occurs and a transient homo-geneous problem. If the solutions of the steady state problems are T0j(r) and that for thetransient problem is Th(r, t), the solution of the original problem is

T (r, t) = Th(r, t) +N

∑

j=0

T0j(r)

22

6 Transient Temperature Nomographs: Heisler Charts

The solutions obtained for 1D non homogeneous problems with Neumann boundary con-ditions in Cartesian coordinate systems using the method of separation of variables havebeen collected and assembled in the form of transient temperature nomographs by Heisler.The given charts are a very useful baseline against which to validate one’s own analytical ornumerical computations. The Heisler charts summarize the solutions to the following threeimportant problems.

The first problem is the 1D transient homogeneous heat conduction in a plate of spanL from an initial temperature Ti and with one boundary insulated and the other subjectedto a convective heat flux condition into a surrounding environment at T∞. (This problemis equivalent to the quenching of a slab of span 2L with identical heat convection at theexternal boundaries x = −L and x = L).

Introduction of the following non dimensional parameters simplifies the mathematicalformulation of the problem. First is the dimensionless distance

X =x

L

next, the dimensionless time

τ =αt

L2

then the dimensionless temperature

θ(X, τ ) =T (x, t)− T∞

Ti − T∞

and finally, the Biot number

Bi =hL

k

With the new variables, the mathematical formulation of the heat conduction problembecomes

∂2θ

∂X2=∂θ

∂τ

subject to

∂θ

∂X= 0

at X = 0,

∂θ

∂X+Biθ = 0

23

at X = 1, and

θ = 1

in 0 ≤ X ≤ 1 for τ = 0.The solutions obtained for one-dimensional nonhomogeneous problems with Neumann

boundary conditions in cylindrical and spherical coordinate systems using the method ofseparation of variables have also been collected and assembled in the form of transienttemperature nomographs by Heisler. As in the Cartesian case, the charts are a useful baselineagainst which to validate one’s own analytical or numerical computations. The Heisler chartssummarize the solutions to the following three important problems.

The first problem is the 1D transient homogeneous heat conduction in a solid cylinderof radius b from an initial temperature Ti and with one boundary insulated and the othersubjected to a convective heat flux condition into a surrounding environment at T∞.

Introduction of the following nondimensional parameters simplifies the mathematicalformulation of the problem. First is the dimensionless distance

R =r

b

next, the dimensionless time

τ =αt

b2

then the dimensionless temperature

θ(X, τ ) =T (r, t)− T∞

Ti − T∞

and finally, the Biot number

Bi =hb

k

With the new variables, the mathematical formulation of the heat conduction problembecomes

1

R

∂

∂R(R

∂θ

∂R) =

∂θ

∂τ

subject to

∂θ

∂R= 0

at R = 0,

∂θ

∂R+Biθ = 0

24

at R = 1, and

θ = 1

in 0 ≤ R ≤ 1 for τ = 0.As the second problem consider the cooling of a sphere (0 ≤ r ≤ b) initially at a uniform

temperature Ti and subjected to a uniform convective heat flux at its surface into a mediumat T∞ with heat transfer coefficient h. In terms of the dimensionless quantities Bi = hb/k ,τ = αt/b2, θ = (T (r, t) − T∞)/(Ti − T∞) and R = r/b, the mathematical statement of theproblem is

1

R2

∂

∂R(R2 ∂θ

∂R) =

∂θ

∂τ

in 0 < R < 1, subject to

∂θ

∂R= 0

at R = 0, and

∂θ

∂R+Biθ = 0

at R = 1, and

θ = 1

in 0 ≤ R ≤ 1, for τ = 0.As before, the solutions to the above problems are well known and are readily available

in graphical form (Heisler charts).

7 Other Analytical Methods

There is a variety of heat conduction problems which although linear, are not readily solvedusing the separation of variables method. Fortunately, a number of methods not based onseparation of variables are available for the solution of these heat conduction problems.

Four such methods are briefly described below, namely: the Laplace transform method,the Duhamel method, the Green’s function method and Goodman’s approximate integralmethod. In each case, examples are used to illustrate the application of the method.

7.1 Laplace Transform Method

The Laplace transform method makes use of the Laplace transform to convert the originalheat equation into an ordinary differential equation which is more readily solved. The so-lution of the transformed problem must then be inverted in order to obtain the solution tothe original problem.

25

7.1.1 Theory

Consider a real function F (t). The Laplace transform of F (t), L[F (t)] = F (s) is defined as

L[F (t)] = F (s) =∫ ∞

t′=0e−st′F (t′)dt′

The inverse transform (to recover F (t) from L[F (t)]) is

F (t) =1

2πi

∫ γ+i∞

s=γ−i∞estF (s)ds

where γ is large enough that all the singularities of F (s) lie to the left of the imaginary axis.For the transform and inverse to exist, F (t) must be at least piecewise continuous, of

exponential order and tnF (t) must be bounded as t → 0+.Laplace transforms have the following properties:Linearity.

L[c1F (t) + c2G(t)] = c1F (s) + c2G(s)

Transforms of Derivatives.

L[F ′(t)] = sF (s) − F (0)

L[F ′′(t)] = s2F (s) − sF (0) − F ′(0)

L[F ′′′(t)] = s3F (s) − s2F (0) − sF ′(0) − F ′′(0)

L[F (n)(t)] = snF (s) − sn−1F (0) − sn−2F (1)(0) − sn−3F (2)(0) − ...− F ′(n−1)(0)

Transforms of Integrals.Let g(t) =

∫ t0 F (τ )dτ (i.e. g′(t) = F (t)).

L[∫ t

0F (τ )] =

1

sF (s)

L[∫ t

0

∫ τ2

0F (τ1)dτ1dτ2] =

1

s2F (s)

L[∫ t

0...

∫ τn

0F (τ1)dτ1...dτn] =

1

snF (s)

26

Scale change.Let a > 0 be a real number.

L[F (at)] =1

aF (s

a)

L[F (1

at)] = aF (as)

Shift.

L[e±atF (t)] = F (s∓ a)

Transform of translated functions.The translated function is

U(t− a)F (t− a) =

{

F (t− a) t > a0 t < a

where

U(t− a) =

{

1 t > a0 t < a

The Laplace transform of the translated function is

L[U(t− a)F (t− a)] = e−asF (s)

Also

L[U(t− a)] = e−as1

s

Transform of the δ function.

L[δ(t)] = 1

Transform of convolution.The convolution of two functions f(t) and g(t) is

f ∗ g =∫ t

0f(t− τ )g(τ )dτ = g ∗ f

The Laplace transform is

L[f ∗ g] = f(s)g(s)

27

Derivatives.If F (s) is the Laplace transform of F (t) then

dF

ds= F ′(s) = L[(−t)F (t)]

dnF

dsn= F ′(s) = L[(−t)nF (t)]

Integrals.

∫ ∞

sF (s′)ds′ = L[

F (t)

t]

Functions of More than One Variable.Let f(x, t) be a function of the two independent variables x and t. The Laplace transform

of f(x, t), L[f(x, t)] = f (x, s) is defined as

f(x, s) =∫ ∞

0e−stf(x, t)dt

With this, the following useful formulae are obtained

L[∂f(x, t)

∂x] =

df(x, s)

dx

L[∂2f(x, t)

∂x2] =

d2f(x, s)

dx2

L[∂f(x, t)

∂t] = sf(x, s) − f(x, 0)

L[∂2f(x, t)

∂t2] = s2f(x, s) − sf(x, 0) − ∂f(x, 0)

∂t

The inverse Laplace transform of L[F (t)] is

F (t) =1

2πi

∫ γ+i∞

s=γ−i∞estF (s)ds

The integration must be carried out in the complex domain and requires a basic understand-ing of analytic functions of a complex variable and residue calculus. Fortunately, inversionformulae for many transforms have been computed and are readily available in tabular form.A short table of useful transforms is included below.

28

F (t) F (s)

1 1s

t 1s2√

t√

π

2√

s3

1√t

√

πs

tn n!sn+1

tn−1

(n−1)!1sn

tneat n!(s−a)n+1

sin kt ks2+k2

cos kt ss2+k2

sinh kt ks2−k2

cosh kt ss2−k2

F (t− a); t ≥ a 0; otherwise e−asF (s)k

2√

πt3e−

k2

4t e−k√

s

erfc( k2√

t) 1

se−k

√s

−eakea2terfc(a√t+ k

s√

t) + erfc( k

2√

t) ae−k

√s

s(a+√

s)

1 − 2∑∞

n=1e−λ2

nt/a2J0(λnx/a)

λnJ1(λn); λn roots of J0(λn) = 0 J0(ix

√s)

sJ0(ia√

s)

7.1.2 Examples

Consider the problem of finding T (x, t) in 0 ≤ x <∞ satisfying

∂2T

∂x2=

1

α

∂T

∂t

and subject to

T (0, t) = f(t)

T (x→ ∞, t) = 0

and

T (x, 0) = 0

Taking Laplace transforms one gets

d2T (x, s)

dx2− s

αT (x, s) = 0

29

T (0, s) = f (s)

T (x → ∞, s) = 0

The solution of this problem is

T (x, s) = f(s)e−x√

s/α = f(s)g(x, s) = L[f(t) ∗ g(x, t)]

Inversion then produces

T (x, t) = f(t) ∗ g(x, t) =∫ t

0f(τ )g(x, t− τ )dτ

Inversion of g(x, s) to get g(x, t) finally gives

T (x, t) =x√4πα

∫ t

τ=0

f(τ )

(t− τ )3/2exp[− x2

4α(t− τ )]dτ

If f(t) = T0 = constant, the solution is

T (x, t) = T0erfc(x

2√αt

)

Consider now the transient problem of quenching a solid sphere (radius b) initially at T0

by maintaining its surface at zero. The Laplace transform method can be used to find andexpression for T (r, t). The transformed problem is

1

r

d2(rT )

dr2− s

αT = −T0

α

the solution of which is

T (r, s) =T0

s− T0b

rs

sinh(r√sα)

sinh(b√sα)

To invert this the trigonometric functions are expressed as asymptotic series and then in-verted term by term. The final result is

T (r, t) = T0 −bT0

r

∞∑

n=0

{erfc(b(2n + 1) − r

2√αt

) − erfc(b(2n + 1) + r

2√αt

)}

7.2 Duhamel’s Method

Duhamel’s method is based on Duhamel’s theorem which allows expressing the solution ofa problem subjected to time dependent boundary conditions in terms of an integral of thesolution corresponding to time independent boundary conditions.

30

7.2.1 Theory

Many practical transient heat conduction problems involve time dependent boundary condi-tions. Duhamel’s theorem allows the analysis of transient heat conduction problems involvingtime varying thermal conditions at boundaries.

Consider the following problem

∇2T (r, t) +1

kg(r, t) =

1

α

∂T (r, t)

∂t

subject to

ki∂T

∂ni+ hiT = fi(r, t)

on boundaries Si for i = 1, 2, ...N and for t > 0, as well as

T (r, 0) = F (r)

Solution by separation of variables is not possible because of the time dependent nonhomo-geneous terms g(r, t) and fi(r, t).

Consider instead an auxiliary problem obtained by introducing a new parameter τ , un-related to t. Find Φ(r, t, τ ) such that

∇2Φ(r, t, τ ) +1

kg(r, τ ) =

1

α

∂Φ(r, t, τ )

∂t

subject to

ki∂Φ

∂ni+ hiΦ = fi(r, τ )

on boundaries Si for i = 1, 2, ...N and for t > 0, as well as

Φ(r, 0, τ ) = F (r)

Since here g(r, τ ) and fi(r, τ ) do not depend on time, the problem can be solved by separationof variables. Duhamel’s theorem states that once Φ(r, t, τ ) has been determined, T (r, t) canbe found by

T (r, t) = F (r) +∫ t

τ=0

∂

∂tΦ(r, t− τ, τ )dτ

If the nonhomogeneity occurs only in the boundary condition at one boundary and theinitial temperature is F (r) = 0, the appropriate form of Duhamel’s theorem is

T (r, t) =∫ t

τ=0f(τ )

∂

∂tΦ(r, t− τ )dτ

31

Sometimes the nonhomogeneous term at a boundary f(t) changes discontinuously withtime. The integral in Duhamel’s theorem must then be done separately integrating by parts.For a total of N discontinuities at times over the time range 0 < t < τN , the temperatureT (r, t) for τN−1 < t < τN is given by

T (r, t) =∫ t

τ=0Φ(r, t− τ )

df(τ )

dτdτ +

N−1∑

j=0

Φ(r, t− τj)∆fj

where ∆fj = f+(τj) − f−(τj) is the step change in the boundary condition at t = τj. If thevalues of f are constant between the steps changes, Duhamel’s theorem becomes

T (r, t) =N−1∑

j=0

Φ(r, t− j∆t)∆fj

for the time interval (N − 1)∆t < t < N∆t.

7.2.2 Examples

Consider the following 1D transient problem in a slab of thickness L. Find T (x, t) satisfying

∂2T

∂x2=

1

α

∂T

∂t

subject to

T (0, t) = 0

T (L, t) = f(t) =

{

bt; 0 < t < τ10; t > τ1

and

T (x, 0) = 0

The appropriate auxiliary problem here is, find Φ(x, t)

∂2Φ

∂x2=

1

α

∂Φ

∂t

subject to

Φ(0, t) = 0

Φ(L, t) = 1

32

and

Φ(x, 0) = 0

The desired function Φ(x, t − τ ) is obtained from the solution to the above problem byreplacing t by t− τ , i.e.

Φ(x, t) =x

L+

2

L

∞∑

m=1

e−αβ2mt 1

βmsin βmx

where βm = mπ/L. So, for t < τ1

T (x, t) =∫ t

τ=0Φ(x, t− τ )

df(t)

dτdτ =

= bx

Lt+ b

2

L

∞∑

m=1

(−1)m

αβ3m

(1 − e−αβ2mt) sin βmx

And for t > τ1

T (x, t) =∫ τ1

τ=0Φ(x, t− τ )

df

dτdτ +

∫ t

τ1

Φ(x, t− τ )df

dτdτ + Φ(x, t− τ1)∆f1

since df/dτ = b when t < τ1, df/dτ = 0 for t > τ1 and ∆f1 = −bτ1, so

T (x, t) =∫ τ1

τ=0{xL

+2

L

∞∑

m=1

e−αβ2m(t−τ ) (−1)m

βmsinβmx}bdτ

−bτ1[x

L+

2

L

∞∑

m=1

e−αβ2m(t−τ1)

(−1)m

βmsinβmx]

Consider now the transient 1D problem in a solid cylinder (radius b) with internal heatgeneration g(t). Initially and at the boundary the temperature is zero. The auxiliary problemis

∂2Φ

∂r2+

1

r

∂Φ

∂r+

1

k=

1

α

∂Φ

∂t

subject to

Φ(b, t) = 0

Φ(r, 0) = 0

The solution to this problem is

Φ(r, t) =b2 − r2

4k− 2

bk

∞∑

m=1

e−αβ2mt J0(βmr)

β3mJ1(βmb)

33

where βm are the roots of J0(βmb) = 0. From Duhamel’s theorem, the desired solution is

T (r, t) =∫ t

τ=0g(τ )

∂Φ

∂tdτ =

=2α

bk

∞∑

m=1

e−αβ2mt J0(βmr)

βmJ1(βmb)

∫ t

τ=0g(τ )eαβ2

mτdτ

7.3 Green’s Function Method

The method of Green’s functions is based on the notion of fundamental solutions of the heatequation described before. Specifically, Green’s functions are solutions of heat conductionproblems involving instantaneous point, line or planar sources of heat. Solutions to otherheat conduction problems can then be expressed as special integrals of the Green’s functions.

7.3.1 Theory

Consider the problem of a 3D body (volume V ) initially at zero temperature. Suddenly, athermal explosion at a point r′ instantaneously releases a unit of energy at time τ while theentire outer surface S of the body undergoes convective heat exchange with the surroundingenvironment at zero temperature. The thermal explosion is an instantaneous point sourceof thermal energy which then diffuses throughout the body. The mathematical formulationof this problem consists of finding the function G(r, t|r′, τ ) satisfying the nonhomogenoeustransient heat equation in V ,

∇2G +1

kδ(r− r′)δ(t− τ ) =

1

α

∂G

∂tsubject to

k∂G

∂n+ hG = 0

on the boundary S. Here, δ(x) is Dirac’s delta function (which is zero everywhere exceptat the origin, where it is infinite). The function G(r, t|r′, τ ) represents the temperature atlocation r and time t resulting from an instantaneous point source of heat releasing a unitof thermal energy at location r′ and time τ and it is called a Green’s function.

Consider now the related problem of a 3D body of volume V , initially at T (r, 0) = F (r)inside which thermal energy is generated at a volumetric rate g(r, t) while its outer surface(s)Si exchange heat by convection with a medium at T∞. The solution to this problem can beexpressed in terms of the Green’s function above as

T (r, t) =∫ ∫ ∫

VG(r, t|r′, τ )|τ=0F (r′)dV ′ +

α

k

∫ t

τ=0dτ

∫ ∫ ∫

VG(r, t|r′, τ )g(r′, τ )dV ′ +

α∫ t

τ=0dτ

N∑

i=1

∫ ∫

Si

G(r, t|r′, τ )|r′=ri

1

kihiT∞dS

′i

34

The terms on the RHS above are: i) the contribution due to the initial condition, (ii) thecontribution due to the internal energy generation, and (iii) the contribution due to thenonhomogeneity in the boundary conditions.

Line sources and plane sources can be similarly defined respectively for two- and one-dimensional systems and nonhomogeneous transient problems in 2D and 1D are the solvablein terms of appropriate Green’s functions. For instance, for a 1D system of extent L,

T (x, t) =∫

LG(x, t|x′, τ )|τ=0F (x′)dx′ +

α

k

∫ t

τ=0dτ

∫

LG(x, t|x′, τ )g(x′, τ )dx′ +

α∫ t

τ=0dτ

2∑

i=1

G(x, t|x′, τ )|x′=xi

1

kihiT∞

Here, the Green’s function G(x, t|x′, τ ) represents the temperature at x, t resulting from theinstantaneous release of a unit of energy at time τ , by a planar source located at x′.

Point, line and surface sources can be instantaneous or continuous. Subscript and su-perscript signs are used to make the distinction explicit. For instance gi

p denotes an in-stantaneous point source of heat. In Cartesian coordinates, the relationship between gi

p andg(x, y, z, t) is

gipδ(x− x′)δ(y − y′)δ(z − z′)δ(t− τ ) = g(x, y, z, t)

Once the Green function associated with a particular problem is known, the actual tem-perature can be calculated simply by integration. So, one needs techniques for Green’sfunction determination.

Consider the following homogeneous problem in a region R. Find T (r, t) satisfying

∇2T (r, t) =1

α

∂T

∂t

subject to

∂T

∂ni+HiT = 0

on all surfaces Si and

T (r, t) = F (r)

Many special cases of this problem have already been solved by separation of variables.In all instances one can symbolically write the solution as

T (r, t) =∫

RK(r, r′, t)F (r′)dv′

35

where K(r, r′, t) is called the kernel of the integration. Now, applying the Green’s functionapproach to this problem yields

K(r, r′, t) = G(r, t|r′, 0)so the kernel is the Green’s function in this case. The associated nonhomogeneous prob-lem can also be solved once G(r, t|r′, 0) is known by simply replacing t by t − τ to obtainG(r, t|r′, τ ).

Consider as an example the 1D transient heat conduction in a cylinder.

1

r

∂

∂r(r∂T

∂r) +

1

kg(r, t) =

1

α

∂T

∂tsubject to

T (b, t) = f(t)

and

T (r, 0) = F (r)

The associated homogeneous problem with homogeneous boundary condition is readily solvedby separation of variables and the solution is

ψ(r, t) =∫ b

r′=0[2

b2

∞∑

m=1

e−αβ2mt 1

J21 (βmb)

r′J20 (βmr)J

20 (βmr

′)]F (r′)dr′ =

∫ b

r′=0r′G(r, t|r′, 0)F (r′)dr′

where βm are the roots of J0(βmb) = 0. Therefore, the Green’s function with τ = 0 is

G(r, t|r′, τ )|τ=0 =2

b2

∞∑

m=1

e−αβ2mt 1

J21 (βmb)

r′J20 (βmr)J

20 (βmr

′)

and replacing t by t− τ , the desired Green’s function is

G(r, t|r′, τ ) =2

b2

∞∑

m=1

e−αβ2m(t−τ ) 1

J21 (βmb)

r′J20 (βmr)J

20 (βmr

′)

For homogeneous 1D transient conduction in an infinite medium, the Green’s function is

G(x, t|x′, τ )|τ=0 =1

(4παt)1/2exp(−(x− x′)2

4αt)

For homogeneous 1D transient conduction in a semiinfinite medium, the Green’s functionis

G(x, t|x′, τ )|τ=0 =1

(4παt)1/2[exp(−(x− x′)2

4αt) − exp(−(x+ x′)2

4αt)]

For homogeneous 1D transient conduction in a slab of thickness L,

G(x, t|x′, τ )|τ=0 = 2∞∑

m=1

e−αβ2mt β2

m +H2

L(β2m +H2) +H

cos(βmx) cos(βmx′)

36

7.3.2 Examples

The problem of finding ψ(x, t) for −∞ < x <∞ satisfying

∂2ψ

∂x2=

1

α

∂ψ

∂t

and subject to

ψ(x, 0) = F (x)

has the solution

ψ(x, t) =∫ ∞

x′=−∞[(4παt)−1/2 exp(−(x− x′)2

4αt]F (x′)dx′ =

=∫ ∞

x′=−∞G(x, t|x′, τ )|τ=0F (x′)dx′

Therefore, the Green’s function of the problem of finding T (x, t) for −∞ < x < ∞satisfying

∂2T

∂x2+

1

kg(x, t) =

1

α

∂T

∂t

and subject to

T (x, 0) = F (x)

is simply

G(x, t|x′, τ ) = (4πα[t− τ ])−1/2 exp(− (x− x′)2

4α[t− τ ]

and the desired solution is

T (x, t) = (4παt)−1/2∫ ∞

x′=−∞exp(−(x− x′)2

4αtF (x′)dx′

+α

k

∫ t

τ=0dτ

∫ ∞

x′=−∞(4πα[t− τ ])−1/2 exp(− (x− x′)2

4α[t− τ ]g(x′, τ )dx′

The problem of finding ψ(r, t) for 0 ≤ r ≤ b satisfying

∂2ψ

∂r2+

1

r

∂ψ

∂r=

1

α

∂ψ

∂t

and subject to

ψ(0, t) = 0

37

and

ψ(r, 0) = F (r)

has the solution

ψ(r, t) =∫ b

r′=0r′[

2

b2

∞∑

m=1

e−αβ2mtJ0(βmr)

J21 (βmb)

J0(βmr′)]F (r′)dr′

=∫ b

r′=0r′G(r, t|r′, τ )|τ=0F (r′)dr′

Therefore, the Green’s function of the problem of finding T (r, t) for 0 ≤ r ≤ b satisfying

∂2T

∂x2+

1

r

∂T

∂r

1

kg(x, t) =

1

α

∂T

∂t

and subject to

T (b, t) = f(t)

and

T (r, 0) = F (r)

is simply

G(r, t|r′, τ ) =2

b2

∞∑

m=1

e−αβ2m[t−τ ] J0(βmr)

J21 (βmb)

J0(βmr′)

and the desired solution is

T (r, t) =∫ b

r′=0r′G(r, t|r′, τ )|τ=0F (r′)dr′ +

α

k

∫ t

τ=0dτ

∫ b

r′=0r′G(r, t|r′, τ )g(r′, τ )dr′

−α∫ t

τ=0[r′∂G

∂r′]f(τ )dτ

7.4 Goodman Approximate Integral Method

The integral method provides approximate solutions to heat conduction problems in whichthe energy balance is satisfied only in an average sense.

The integral method is usually implemented in four steps:

• The heat equation is first integrated over a distance δ(t) called the thermal layer toobtain the heat balance integral. The thermal layer is selected so that the effect ofboundary conditions is negligible for x > δ(t).

38

• A low order polynomial is then selected to approximate the temperature distributionover the thermal layer. The coefficients in the polynomial are in general functions oftime and must be determined from the conditions of the problem.

• The approximating polynomial is substituted in the heat balance integral to obtain adifferential equation for δ(t).

• The resulting δ(t) is substituted in the polynomial formula to obtain the requiredtemperature distribution.

Consider the case of a semiinfinite medium initially at Ti where at the boundary x = 0the temperature is maintained at T0 > Ti. The formulation is, find T (x, t) satisfying

∂2T

∂x2=

1

α

∂T

∂t

subject to

T (x, 0) = Ti

and to

T (0, t) = T0

First integrate the heat equation from x = 0 to x = δ(t) to obtain

∫ x=δ(t)

x=0d(∂T

∂x) =

∂T

∂x|x=δ(t) −

∂T

∂x|x=0 =

=1

α

∫ x=δ(t)

x=0

∂T

∂tdx =

1

α[d

dt(∫ x=δ(t)

x=0Tdx)− T |x=δ(t)

dδ

dt]

where Leibnitz rule has be used on the right hand side. Since (dT/dx)|x=δ(t) = 0 andT |x=δ(t) = Ti

−α∂T∂x

|0 =d

dt(∫ x=δ(t)

x=0Tdx− Tiδ)

this is the heat balance integral equation.Now assume that the temperature distribution in the thermal layer can be represented

as a low order (fourth order) polynomial of x, i.e.

T (x, t) = a+ bx+ cx2 + dx3 + ex4

The coefficients a, b, c, d and e are determined by introducing the conditions

T (0, t) = T0

39

T (δ, t) = Ti

∂T

∂x x=δ= 0

∂2T

∂x2 x=0= 0

and

∂2T

∂x2 x=δ= 0

Substituting into the polynomial, the result is

T (x, t)− Ti

T0 − Ti= 1 − 2

x

δ+ 2(

x

δ)3 − (

x

δ)4

Substituting this last expression for T (x, t) into the heat balance integral results in

20

3α = δ

dδ

dt

which must be solved subject to δ(0) = 0 to give

δ(t) =

√

40

3αt

This expression for δ(t) can now be substituted into the expression for T (x, t) or inq(0, t) = −k(∂T/∂x)|x = 0 to give

q(0, t) =2k

δ(T0 − Ti)

Consider now instead the case of a semiinfinite medium initially at Ti = 0 where theboundary x = 0 is subjected to a prescribed, time dependent heat flux f(t) and in whichthe thermal properties are all functions of temperature.

The formulation is, find T (x, t) satisfying

∂

∂x(k(T )

∂T

∂x) = ρ(T )Cp(T )

∂T

∂t

subject to

T (x, 0) = 0

40

and to

−k∂T∂x

|x=0 = f(t)

Introducing now a new variable U defined as U =∫ T0 ρCpdT the problem transforms into

∂

∂x(α(U)

∂U

∂x) =

∂U

∂t

subject to

U(x, 0) = 0

and to

−(α(U)∂U

∂x)|x=0 = f(t)

Integrating over the thermal layer δ(t) and since (∂U/∂x)|x=δ = U |x=δ = 0, leads to theheat balance integral equation

d

dt

∫ δ

0Udx = f(t)

Now, assuming that the distribution of U is cubic in x and using the conditions of theproblem to define the coefficients leads to

U(x, t) =δf(t)

3αx=0(1 − x

δ)3

Substituting this into the heat balance integral gives

d

dt[δ2f(t)

12αx=0] = f(t)

In this case, however, the solution depends on αx=0 which is unknown but can be determinedfrom the value of U at x = 0. The final result is

(U√α)|x=0 = [

4

3f(t)

∫ t

0f(t′)dt′]1/2

This one can be used to find αx=0(t). From that, δ(t) can be calculated, which in turn givesU , which in turn gives T .

41