Transient Heat Transfer

Chapter 3: Heat Conduction

Advanced Heat and Mass Transfer by Amir Faghri, Yuwen Zhang, and John R. Howell

3.3 Unsteady State Heat Conduction

1

� For many applications, it is necessary to consider the

variation of temperature with time.

� In this case, the energy equation for classical heat

conduction, eq. (3.8), should be solved.

� If the thermal conductivity is independent from the

temperature, the energy equation is reduced to eq.

(3.10).

3.3 Unsteady State Heat Conduction3.3 Unsteady State Heat Conduction




2

Consider an arbitrarily shaped object with volume V,

surface area As, and a uniform initial temperature of Ti as

shown in Fig. 3.15. At time t = 0, the arbitrarily shaped

object is exposed to a fluid with temperature of T∞ and

the convective heat transfer coefficient between the fluid

and the arbitrarily shaped object is h.

3.3.1 Lumped Analysis3.3.1 Lumped Analysis

T(t)

V

h,

T∞

As

Figure 3.15 Lumped capacitance method




3

The convective heat transfer from the surface is

Similar to what we did for heat transfer from an extended

surface, the heat loss due to surface convection can be

treated as an equivalent heat source

Since the temperature is assumed to be uniform, the

energy equation becomes

(3.188)

which is subject to the following initial condition

(3.189)

( )s

q hA T T∞= −

( )s

hA T Tqq

V V

∞−′′′ = − = −

( )s

p

hA T TTc

t Vρ ∞−∂

= −∂

, 0iT T t= =




4

Introducing excess temperature, , eqs. (3.188) and

(3.189) become

(3.190)

(3.191)

Integrating eq. (3.190) and determining the integral

constant using eq. (3.191), the solution becomes

(3.192)

where

(3.193)

is referred to as the thermal time constant.

T Tϑ ∞= −

s

p

hAd

dt Vc

ϑϑ

ρ= −

, 0i tϑ ϑ= =

/ tt

i

eτϑ

ϑ−=

p

t

s

Vc

hA

ρτ =




5

The cooling process requires transferring heat from the

center of the object to the surface and the heat is further

transferred away from the surface by convection. When

the lamped capacitance method is employed, it is

assumed that the conduction resistance within the object

is negligible compared with the convective thermal

resistance at the surface, therefore, the validity of the

lumped analysis depends on the relative thermal

resistances of conduction and convection. The

conduction thermal resistance can be expressed as

where Lc is the characteristic length and Ac is the area of

heat conduction.

c

cond

c

LR

kA=




6

The convection thermal resistance at the surface is

Assuming , one can define the Biot number, Bi, as

the ratio of the conduction and convection thermal

resistances

(3.194)

If the characteristic length is chosen as , the

lumped capacitance method is valid when the Biot

number is less than 0.1, or the conduction thermal

resistance is one order of magnitude smaller than the

convection thermal resistance at the surface.

1conv

s

RhA

=

s cA A=

Bi= cond c

conv

R hL

R k�

/c sL V A=




7

For the case that the Biot number is greater than 0.1, the

temperature distribution can no longer be treated as

uniform and the knowledge about the temperature

distribution is also of interest. We will now consider the

situation where temperature only varies in one spatial

dimension. Both homogeneous and nonhomogeneous

problem will be considered.

3.3.2 One Dimensional Transient

Systems

3.3.2 One Dimensional Transient

Systems




8

Homogeneous Problems

Figure 3.16 shows a finite slab with thickness of L and a

uniform initial temperature of Ti. At time t = 0, the left

side of the slab is insulated while the right side of the

slab is exposed to a fluid with temperature of T∞. In

contrast to the lumped capacitance method that

assumes uniform temperature, we will present a more

generalized model that takes non-uniform temperature

distribution into account.

h, T∞

Figure 3.16 Transient conduction in a finite slab0 xL

adia

bat

ic




9

The energy equation for this one-dimensional transient

conduction problem is

(3.195)

subject to the following boundary and initial conditions

(3.196)

(3.197)

(3.198)

This is a nonhomogeneous problem because eq. (3.197) is

not homogeneous. By introducing the excess

temperature, , the problem can be

homogenized, i.e.,

(3.199)

2

2

1, 0 , 0

T Tx L t

x tα

∂ ∂= < < >

∂ ∂

0, 0T

xx

∂= =

∂

( ), T

k h T T x Lx

∞

∂− = − =

∂

, 0 , 0iT T x L t= < < =

T Tϑ ∞= −

2

2

1, 0 , 0x L t

x t

ϑ ϑ

α

∂ ∂= < < >

∂ ∂




10

(3.200)

(3.201)

(3.202)

To express our solution in a compact form so that it can be

used for all similar problems, one can define the

following dimensionless variables

(3.203)

and eqs. (3.199) – (3.202) will be nondimensionalized as

(3.204)

(3.205)

0, 0xx

ϑ∂= =

∂

, k h x Lx

ϑϑ

∂− = =

∂

, 0 , 0i iT T x L tϑ ϑ ∞= = − < < =

2, , Fo

i

x tX

L L

ϑ αθ

ϑ= = =

2

2, 0 1, Fo 0

FoX

X

θ θ∂ ∂= < < >

∂ ∂

0, 0XX

θ∂= =

∂




11

(3.206)

(3.207)

This problem can be solved using the method of separation

of variables. Assuming that the temperature can be

expressed as

(3.208)

where are functions of X and Fo, respectively, eq.

(3.204) becomes

Since the left hand side is a function of X only and the right-

hand side of the above equation is a function of Fo only,

both sides must be equal to a separation constant, , i.e.,

Bi , X 1X

θθ

∂− = =

∂

1, 0 1, Fo 0Xθ = < < =

( ,Fo) ( ) (Fo)X Xθ = Θ Γ

and Θ Γ

( ) (Fo)

( ) (Fo)

X

X

′′ ′Θ Γ=

Θ Γ




12

(3.209)

The separation constant µ can be either a real or a

complex number. The solution of from eq. (3.209) will

be . If is a positive real number, we will have

when , which does not make sense, therefore,

cannot be a positive real number. If is zero, we will

have , and is a linear function of X only. The

final solution for will also be a linear function of X only,

which does not make sense either. It can also be shown

that the separation constant cannot be a complex

number, therefore, the separation variable has to be a

negative real number.

( ) (Fo)

( ) (Fo)

X

Xµ

′′ ′Θ Γ= =

Θ Γ

ΓFo

eµΓ = µ Γ → ∞

Fo → ∞ µµ

constΓ = Θ

θ




13

If we represent this negative number by , eqs. (3.209)

can be rewritten as the following two equations

(3.210)

(3.211)

The general solutions of eqs. (3.210) and (3.211) are

(3.212)

(3.213)

where C1, C2, and C3 are integral constants.

Substituting eq. (3.208) into eqs. (3.205) and (3.206), the

following boundary conditions of eq. (3.210) are obtained

(3.214)

(3.215)

2µ λ= −

20λ′′Θ + Θ =

2 0λ′Γ + Γ =

1 2cos sinC X C Xλ λΘ = +2Fo

3C e λ−Γ =

(0) 0′Θ =

(1) Bi (1)′−Θ = Θ




14

Substituting eq. (3.212) into eq. (3.214) yields

Since cannot be zero, C2 must be zero and eq. (3.215)

becomes

(3.216)

Applying the convection boundary condition, eq. (3.215),

one obtains

(3.217)

where n is an integer.

The dimensionless temperature with eigenvalue can be

obtained by substituting eqs. (3.216) and (3.213) into eq.

(4.208), i.e.,

(3.218)

1 2 2(0) sin(0) cos(0) 0C C Cλ λ λ′Θ = − + = =

λ

1 cosC XλΘ =

cotBi

n

n

λλ=

nλ

( )2Fo

cos n

n n nC X e

λθ λ −=




15

where .

Since the one-dimensional transient heat conduction

problem under consideration is a linear problem, the sum

of different for each value of n also satisfies eqs.

(3.204) – (3.206).

(3.219)


Multiplying the above equation by and integrating

the resulting equation in the interval of (0, 1), one obtains

(3.220)

1 3nC C C=

nθ

( )2Fo

1

cos n

n n

n

C X eλθ λ

∞−

=

=∑

( )1

1 cosn n

n

C Xλ∞

=

=∑( )cos m Xλ

1 1

0 01

cos( ) cos( )cos( )m n m n

n

X dX C X X dXλ λ λ∞

=

=∑∫ ∫




16

The integral in the right-hand side of eq. (3.133) can be

evaluated as

(3.221)

Equation (3.217) can be rewritten as

Similarly, for eigenvalue , we have

Combining the above two equations, we have

or

2 21

0

sin cos cos sin,

cos( )cos( )sin 21

, 2 2

m m n n m n

m n

m n

m

m

m

m n

X X dX

m n

λ λ λ λ λ λ

λ λλ λ

λλ

λ

−≠ −

= + =

∫

Bi tann n

λ λ=

mλ

Bi tanm mλ λ=

tan tan 0m m n n

λ λ λ λ− =

sin cos cos sin 0m m n n m nλ λ λ λ λ λ− =




17

therefore, the integral in eq. (3.221) is zero for the case that

, and the right hand side of eq. (3.220) becomes

(3.222)

Substituting eq. (3.222) into eq. (3.220) and evaluating the

integral at the left-hand side of eq. (3.220), we have

i.e.,

Changing notation from m to n, we get

(3.223)

m n≠1

01

sin 2cos( )cos( )

2 2

m m

n m n m

mn

CC X X dX

λλ λ λ

λ

∞

=

= +

∑ ∫

sin 21sin

2 2

m m

m m

m m

C λλ λ

λ λ

= +

4sin

2 sin 2

mm

m m

Cλ

λ λ=

+

4sin

2 sin 2

n

n

n n

Cλ

λ λ=

+




18

The dimensionless temperature, therefore, becomes

(3.224)

If the Biot number becomes infinite, the convection

boundary condition becomes

(3.225)

which is an isothermal condition at the right-hand side of

the wall. Equation (3.217) becomes

(3.226)

and the eigenvalue is therefore

(3.227)

( )2Fo

1

4sincos

2 sin 2nn

n

n nn

X eλλ

θ λλ λ

∞−

=

=+∑

0, X 1θ = =

cos 0nλ =

( )/ 2 , 1,2,3,...n n nλ π= − =




19

The temperature distribution for this case is then

(3.228)

When Fourier’s number is greater than 0.2, only the first

term in eq. (3.219) is necessary and the solution

becomes

(3.229)

[ ]2( / 2) Fo

1

4sin( / 2)cos ( / 2)

2( / 2) sin 2( / 2)

n

n

nn X e

n n

ππθ π

π π

∞− −

=

−= −

− + −∑

( )2

1 Fo

1 1 1cosC X eλθ θ λ −= =




20

� Example 3.3

A long cylinder with radius of ro and a uniform initial

temperature of Ti is exposed to a fluid with temperature

of T∞. The convective heat transfer coefficient between

the fluid and cylinder is h. Assuming that there is no

internal heat generation and constant thermophysical

properties, obtain the transient temperature distribution

in the cylinder.




21

� Solution

Since the temperature changes along the r-direction

only, the energy equation is

(3.230)


(3.231)

(3.232)

(3.233)

Defining the following dimensionless variables

(3.234)

2

2

1 1, 0 , 0o

T T Tx r t

r r r tα

∂ ∂ ∂+ = < < >

∂ ∂ ∂

0, 0 (axisymmetric)T

rr

∂= =

∂

( ), o

Tk h T T r r

r∞

∂− = − =

∂

, 0 , 0i oT T r r t= < < =

2, , Fo , Bi o

i o o

hrT T r tR

T T r r k

αθ ∞

∞

−= = = =

−




22

eqs. (3.230) – (4.233) will be nondimensionalized as

(3.235)

(3.236)

(3.237)

(3.238)

Assuming that the temperature can be expressed as

(3.239)

and substituting eq. (3.239) into eq. (3.235), one obtains

(3.240)

2

2

1, 0 1, Fo 0

FoR

R R R

θ θ θ∂ ∂ ∂+ = < < >

∂ ∂ ∂

0, 0RR

θ∂= =

∂

Bi , R 1R

θθ

∂− = =

∂

1, 0 1, Fo 0Rθ = < < =

( ,Fo) ( ) (Fo)R Rθ = Θ Γ

21 1

Rλ

′Γ ′′ ′Θ + Θ = = − Θ Γ




23

which can be rewritten as the following two equations

(3.241)

(3.242)

Equation (3.101) is a Bessel’s equation of zero order and

has the following general solution

(3.243)

where J0 and Y0 are the Bessel’s function of the first and

second kind, respectively. The general solution of eq.

(3.242) is

(3.244)


210

Rλ′′ ′Θ + Θ + Θ =

2 0λ′Γ + Γ =

1 0 2 0( ) ( ) ( )R C J R C Y Rλ λΘ = +

2Fo

3C eλ−Γ =




24

The boundary conditions for eq. (3.241) can be obtained by

substituting eq. (3.239) into eqs. (3.236) and (3.237), i.e.,

(3.245)

(3.246)

The derivative of is

(3.247)

Since , C2 must be zero. Substituting

eqs. (3.243) and (3.247) into eq. (3.246) and considering

C2 = 0 we have

(3.248)

where n is an integer.

(0) 0′Θ =

(1) Bi (1)′−Θ = Θ

Θ

1 1 2 1( ) ( ) ( )R C J R C Y Rλ λ λ λ′Θ = − −

1 1(0) 0 and (0)J Y= = −∞

1 0( ) Bi ( ) 0n n nJ Jλ λ λ− + =




25

The dimensionless temperature with eigenvalue is

(3.249)

where .

For a linear problem, the sum of different for each value

of n also satisfies eqs. (3.235) – (3.237).

(3.250)


Multiplying the above equation by and integrating

the resulting equation in the interval of (0, 1), one obtains

nλ

( )2Fo

0n

n n nC J R eλθ λ −=

1 3nC C C=

nθ

( )2Fo

0

1

n

n n

n

C J R eλθ λ

∞−

=

=∑

( )0

1

1 n n

n

C J Rλ∞

=

=∑( )0 mRJ Rλ

1 1

0 0 00 0

1

( ) ( ) ( )m n m n

n

RJ R dR C RJ R J R dRλ λ λ∞

=

=∑∫ ∫




26

According to the orthogonal property of Bessel’s function,

the integral on the right-hand side equals zero if

but it is not zero if m = n Therefore, we have

Changing notation from m to n, we get

(3.251)

thus, the dimensionless temperature becomes

(3.252)

m n≠

1

00 1

1 2 22 0 10

0

( ) ( )2

( ) ( )( )

mm

m

m m mm

RJ R dR JC

J JRJ R dR

λ λ

λ λ λλ= =

+

∫∫

1

2 2

0 1

( )2

( ) ( )

n

n

n n n

JC

J J

λ

λ λ λ=

+

( ) 21 0 Fo

2 2

0 11

( )2

( ) ( )nn n

n n nn

J J Re

J J

λλ λθ

λ λ λ

∞−

=

=+∑




27

One-dimensional heat conduction in a spherical coordinate

system can be solved by introducing a new dependent

variable. Consider a sphere with radius of ro and a

uniform initial temperature of Ti. It is exposed to a fluid

with a temperature of T∞ and the convective heat transfer

coefficient between the fluid and finite slab is h.

Assuming that there is no internal heat generation and

constant thermophysical properties, the governing

equation is

(3.253)


(3.254)

2

2

1 ( ) 1, 0 , 0o

rT Tx r t

r r tα

∂ ∂= < < >

∂ ∂

0, 0 (axisymmetric)T

rr

∂= =

∂




28

(3.255)

(3.256)

By using the same dimensionless variables defined in eq.

(3.234), eqs. (3.253) – (3.256) can be

nondimensionalized as

(3.257)

(3.258)

(3.259)

(3.260)

Defining a new dependent variable

(3.261)

( ), o

Tk h T T r r

r∞

∂− = − =

∂, 0 , 0i oT T r r t= < < =

2

2

1 ( ), 0 1, Fo 0

Fo

RR

R R

θ θ∂ ∂= < < >

∂ ∂

0, 0RR

θ∂= =

∂

Bi , R 1R

θθ

∂− = =

∂

1, 0 1, Fo 0Rθ = < < =

U Rθ=




29

eqs. (3.257) – (3.260) become

(3.262)

(3.263)

(3.264)

(3.265)

This problem can be readily solved by using the method of

separation of variables. After the solution is obtained,

one can change the dependent variable back to and

the result is

(3.266)

2

2, 0 1, Fo 0

Fo

U UR

R

∂ ∂= < < >

∂ ∂

0, 0RR

θ∂= =

∂

(Bi-1) , R 1U

Rθ

∂− = =

∂, 0 1, Fo 0U R R= < < =

θ

2Fo

1

4[sin( ) cos( )]1sin( )

[2 sin(2 )]nn n n

n

n n nn

R eR

λλ λ λθ λ

λ λ λ

∞−

=

−=

−∑




30

where the eigenvalue is the positive root of the following

equation

(3.267)1 cot Bin n

λ λ− =




31

Nonhomogeneous Problems

The solution of a nonhomogeneous problem can be

obtained by superposition of a particular solution of the

nonhomogeneous problem and the general solution of

the corresponding homogeneous problem.

Consider a finite slab with thickness of L and a uniform

initial temperature of Ti as shown in Fig. 3.17.

T=Ti

Figure 3.17 Heat conduction under boundary condition of the first kind

0 xL

T=T0

t = 0

t > 0

t → ∞




32

Assuming that there is no internal heat generation in the

slab and the thermophysical properties of the slab are

constants, the energy equation is

(3.268)


(3.269)

(3.270)

(3.271)

By defining the following dimensionless variables

(3.272)

eqs. (3.268) – (3.271) will be nondimensionalized as:

2

2

1, 0 , 0

T Tx L t

x tα

∂ ∂= < < >

∂ ∂

0 , 0T T x= =

, i

T T x L= =

, 0 , 0iT T x L t= < < =

2

0

, , Foi

i

T T x tX

T T L L

αθ

−= = =

−




33

(3.273)

(3.274)

(3.275)

(3.276)

If the steady state temperature is represented by , it must

satisfy the following equations:

(3.277)

(3.278)

(3.279)

which have the following solution:

(3.280)

2

2, 0 1, Fo 0

FoX

X

θ θ∂ ∂= < < >

∂ ∂

1, 0Xθ = =

0, X 1θ = =

0, 0 1, Fo 0Xθ = < < =

sθ

2

20, 0 1s X

X

θ∂= < <

∂

1, 0s

Xθ = =

0, X 1s

θ = =

1s Xθ = −




34

To obtain the generation of the problem described by eqs.

(3.273) – (3.276), a method of partial solution will be

employed. In this methodology, it is assumed that the

solution of a nonhomogeneous problem can be

expressed as

(3.281)

where represent the solution of a homogeneous

problem. Substituting eqs. (3.273) – (3.276) and

considering eqs. (3.277) – (3.279), we have

(3.282)

(3.283)

(3.284)

(3.285)

( ,Fo) ( ) ( ,Fo)s hX X Xθ θ θ= +

hθ

2

2, 0 1, Fo 0

Fo

h h XX

θ θ∂ ∂= < < >

∂ ∂

0, 0h Xθ = =

0, X 1hθ = =

1, 0 1, Fo 0h X Xθ = − < < =




35

which represent a new homogeneous problem. This

problem can be solved using the method of separation of

variables and the result is

(3.286)

The solution of the nonhomogeneous problem thus

becomes

(3.287)

2( ) Fo

1

2 sin( ) n

h

n

n Xe

n

ππθ

π

∞−

=

= − ∑

2( ) Fo

1

2 sin( )1 n

n

n XX e

n

ππθ

π

∞−

=

= − − ∑




36

If the steady-state solution does not exist, we can use the

method of variation of parameter to solve the problem.

Consider a finite slab with thickness of L and a uniform

initial temperature of Ti. At time t = 0, the left side is

subject to a constant heat flux while the right side of the

slab is adiabatic (see Fig. 3.18).

Figure 3.18 Heat conduction under boundary condition of the second kind

0 xL

adiabatic




37

Assuming that there is no internal heat generation in the

slab and the thermophysical properties of the slab are

constants, the energy equation is

(3.288)


(3.289)

(3.290)

(3.291)

By defining the following dimensionless variables

(3.292)

eqs. (3.268) – (3.271) will be nondimensionalized as:

2

2

1, 0 , 0

T Tx L t

x tα

∂ ∂= < < >

∂ ∂

0 , 0T

k q xx

∂′′− = =

∂

0, T

x Lx

∂= =

∂

, 0 , 0iT T x L t= < < =

2

0

, , Fo/

iT T x tX

q L k L L

αθ

−= = =

′′




38

(3.293)

(3.294)

(3.295)

(3.296)

We will use the method of variation of parameters to solve

this problem. This method requires the following steps:

1. Set up a homogeneous problem by dropping the

nonhomogeneous terms,

2. Solve the homogeneous problem to get eigenvalue

and eigenfunctions

3. Assuming the solution of the original nonhomogeneous

problem has the format of

2

2, 0 1, Fo 0

FoX

X

θ θ∂ ∂= < < >

∂ ∂

1, 0XX

θ∂= − =

∂

0, 1XX

θ∂= =

∂

0, 0 1, Fo 0Xθ = < < =

( )n XΘnλ

1

( ,Fo) (Fo) ( )

n

n n

n

X A Xθ=

= Θ∑




39

4. Solve for An(Fo) using orthogonal property of

5. Obtain an ordinary differential equation (ODE) for An

(Fo) and solve for An (Fo) from the ODE

6. Put together the final solution.

We will solve this nonhomogeneous problem be following

the above procedure. The corresponding homogeneous

problem is:

(3.297)

(3.298)

(3.299)

(3.300)

nΘ

2

2, 0 1, Fo 0

Fo

h h XX

θ θ∂ ∂= < < >

∂ ∂

0, 0h XX

θ∂= =

∂

0, 1h XX

θ∂= =

∂

0, 0 1, Fo 0h Xθ = < < =




40

Assuming the solution of the above homogeneous problem

is

(3.301)

eq. (3.297) becomes

(3.302)

The eigenvalue problem is

(3.303)

(3.304)

(3.305)

Solving eqs. (3.303) – (3.305) yields the following

eigenvalues and eigen functions

(3.306)

(3.307)

( ) (Fo)h Xθ = Θ Γ

2( ) (Fo)

( ) (Fo)

X

Xλ

′′ ′Θ Γ= = −

Θ Γ

2 0λ′′Θ + Θ =

(0) 0′Θ =

(1) 0′Θ =

n nλ π=

( ) cos( ), 0,1,2,...n X n X nπΘ = =




41

Now, let us assume that the solution of the original

nonhomogeneous problem is

(3.308)

Multiplying eq. (3.308) by and integrating the

resulting equation in the interval of (0, 1), one obtains

(3.309)

The integral on the right-hand side of eq. (3.309) can be

evaluated as

(3.310)

thus, eq. (3.309) becomes

(3.311)

( )0

( ,Fo) (Fo)cosn

n

X A n Xθ π∞

=

=∑( )cos m Xπ

1 1

0 01

( ,Fo)cos( ) cos( )cos( )n

n

X m X dX A m X n X dXθ π π π∞

=

=∑∫ ∫

1

0

0,

cos( )cos( ) 1/ 2, 0

1, 0

m n

m X n X dX m n

m n

π π

≠

= = ≠ = =

∫

1

00

(Fo) ( ,Fo) , 0A X dX mθ= =∫




42

(3.312)

Differentiating eq. (3.311) with respect to Fo, one obtains:

(3.313)

Substituting eq. (3.293) into eq. (3.313) and integrating with

respect to X yield

(3.314)

Integrating eq. (3.314) with respect to Fo, we have

(3.315)

When Fo = 0, eq. (3.315) becomes

(3.316)

1

m0

(Fo) 2 ( ,Fo)cos( ) , 0A X m X dX mθ π= ≠∫1

0

0dFo Fo

dAdX

θ∂=

∂∫

210

20

1 0

1dFo X X

dAdX

X X X

θ θ θ

= =

∂ ∂ ∂= = − =

∂ ∂ ∂∫

1

0 10

(Fo) Fo ( ,Fo)A C X dXθ= + = ∫1

0 10

(0) ( ,0) 0A C X dXθ= = =∫




43

thus, we have

(3.317)

Differentiating eq. (3.312) and considering eq. (3.293) yield

(3.318)

Using integration by parts twice, the following ODE is

obtained:

(3.319)

Multiplying eq. (3.319) by an integrating factor , we

have

(3.320)

which can be integrated to get

(3.321)

0 (Fo) FoA =

21 1m

20 0

2 cos( ) 2 cos( )Fo Fo

dAm X dX m X dX

d X

θ θπ π

∂ ∂= =

∂ ∂∫ ∫

2m 2 ( )Fo

m

dAm A

dπ= −

2( ) Fome

π

2 2( ) Fo ( ) Fo

m 2Fo

m mdA e e

d

π π =

2( ) Fo

22

2

( )

m

mA C em

π

π−= +




44

where C2 is an integral constant that needs to be

determined by an initial condition. For Fo = 0, eq. (3.312)

becomes

(3.322)

Substituting eq. (3.321) into eq. (3.322), one obtains

therefore, we have

Changing m back to n for notation,

(3.323)

1 1

m0 0

(0) 2 ( ,0)cos( ) 2 cos( ) 0A X m X dX m X dXθ π π= = =∫ ∫

2 2

2

( )C

mπ= −

2( ) Fo

2 2

2 2

( ) ( )

m

mA em m

π

π π−= −

2( ) Fo

2 2

2 2

( ) ( )

n

nA en n

π

π π−= −




45

Substituting eqs. (3.317) and (3.323) into eq. (3.308), the

solution becomes

(3.324)

When the time (Fourier number) becomes large, the last

term on the right-hand side will become zero and the

solution is represented by the first two terms only. To

simplify eq. (3.324), let us assume the solution at large

Fo can be expressed as

(3.325)

which is referred to as asymptotic solution and it must

satisfy eqs. (3.293) – (3.295). Substituting eq. (3.325)

into eqs. (3.293) – (3.295), we have

(3.326)

( ) ( ) 2( ) Fo

2 2 2 2

1 1

cos cos2 2( ,Fo) Fo+ n

n n

n X n XX e

n n

ππ πθ

π π

∞ ∞−

= =

= −∑ ∑

( ,Fo) Fo+ ( )X f Xθ =

( ) 1f X′′ =




46

(3.327)

Integrating eq. (3.326) and considering eq. (3.327), we

obtain

(3.328)

where C cannot be determined from eq. (3.327) because

both boundary conditions are for the first order

derivative. To determine C, we can expand f(X) defined

in eq. (3.328) into cosine Fourier series, i.e.

(3.329)

After determining a0 and an, and considering the f(X) is

identical to the second term on the right-hand side of eq.

(3.324), we have

(3.330)

(0) 1, (1) 0f f′ ′= − =

2

( )2

Xf X X C= − +

2

0

1

( ) cos( )2

n

n

Xf X X C a a n Xπ

∞

=

= − + = +∑

( )2

2 2

1

cos1 2

2 3n

n XXX

n

π

π

∞

=

− + = ∑




47

Substituting eq. (3.330) into eq. (3.324), the final solution

becomes

(3.331)( ) 2

2( ) Fo

2 2

1

cos1 2( ,Fo) Fo+

2 3

n

n

n XXX X e

n

ππθ

π

∞−

=

= − + − ∑




48

Transient Heat Conduction in a Semi-Infinite

Body

Consider heat conduction in a semi-infinite body (x > 0)

with an initial temperature of Ti. The temperature near

the surface of the semi-infinite body will increase

because of the surface temperature change, while the

temperature far from the surface of the semi-infinite body

is not affected and remains at the initial temperature Ti.

The physical model of the problem is illustrated in Fig.

3.19

T=Ti

Figure 3.19 Heat conduction in a semi-infinite body0 x

T=T0

t = 0




49

The governing equation of the heat conduction problem

and the corresponding initial and boundary conditions

are:

(3.332)

(3.333)

(3.334)

(3.335)

which can be solved by using the method of separation of

variables or integral approximate solution.


(3.336)

where L is a characteristic length.

2

2

( ) 1 ( )0 0

T x t T x tx t

x tα

∂ , ∂ ,= > , >

∂ ∂

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

( ) 0iT x t T x t, = → ∞, >

( ) 0 0i

T x t T x t, = > , =

0

2

0

, , Foi

T T x tX

T T L L

αθ

−= = =

−




50

Eqs. (3.332) – (3.335) will be nondimensionalized as

(3.337)

(3.338)

(3.339)

(3.340)

Assuming that the temperature can be expressed as

(3.341)

and substituting eq. (3.341) into eq. (3.337), one obtains

(3.342)

whose general solutions are:

2

2, 0, Fo 0

FoX

X

θ θ∂ ∂= > >

∂ ∂

0, 0Xθ = =

1, Xθ = → ∞

1, 0, Fo 0Xθ = > =

( ,Fo) ( ) (Fo)X Xθ = Θ Γ

2( ) (Fo)

( ) (Fo)

X

Xλ

′′ ′Θ Γ= = −

Θ Γ




51

(3.343)

(3.344)


Substituting eq. (3.341) into eq. (3.338), the following

boundary condition of eq. (3.343) is obtained

(3.345)

Substituting eq. (3.343) into eq. (3.345) yields and eq.

(3.343) becomes

(3.346)


solution becomes

(3.347)

1 2cos sinC X C Xλ λΘ = +2Fo

3C e λ−Γ =

(0) 0′Θ =

2 sinC XλΘ =

( )2Fo

( )sinC X eλ

λθ λ λ −=




52

where . The general solution for the problem can

be obtained by using linear combination of eq. (3.347)

for all possible , i.e.,

(3.348)

Substituting eq. (3.348) into eq. (3.340), one obtains

If we solve the problem by using Laplace transformation,

we have

Comparing the above two equations, an expression of C is

obtained:

2 3C C C=

λ

( )2Fo

0

( )sinC X e dλθ λ λ λ

∞−= ∫

( )0

1 ( )sinC X dλ

λ λ λ∞

== ∫

( ) ( )0 0

21 sin sin

X

X X dX dλ

λ λ λπ

∞ ∞

′= =

′ ′= ∫ ∫

( )0

2( ) sin

X

C X dXλ λπ

∞

′=

′ ′= ∫




53

The temperature distribution, eq. (3.348), becomes

which can be rewritten as

Evaluating the integral with respect to yields

thus

(3.349)

( ) ( )2Fo

0 0

2sin sin

X

e X dX X d dXλ

λθ λ λ λ

π

∞ ∞−

′= =

′ ′ ′= ∫ ∫

[ ]2Fo

0 0

2cos ( ) cos ( )

X

e X X X X d dXλ

λθ λ λ λ

π

∞ ∞−

′= =

′ ′ ′= − − +∫ ∫

22

Fo

0

( )cos ( ) exp

4Fo 4Fo

X Xe X X d

λ

λ

πλ λ

∞−

=

′ −′− = −

∫

22

Fo

0

( )cos ( ) exp

4Fo 4Fo

X Xe X X d

λ

λ

πλ λ

∞−

=

′ +′+ = −

∫

2 2

0 0

1 ( ) ( )exp exp

4Fo 4Fo4 Fo X X

X X X XdX dXθ

π

∞ ∞

′ ′= =

′ ′ − + ′ ′= − − −

∫ ∫




54

Let us define a new variable

The first integral in eq. (3.349), becomes

Similarly, the second integral can be evaluated by following

a similar procedure:

Substituting the above two equations into eq. (3.349), we

have

4Fo

X Xη

′−=

22

0 / 4Fo

( )exp 4Fo exp( )

4FoX X

X XdX dη η

∞ ∞

′= −

′ −′− = −

∫ ∫

22

0 / 4Fo

( )exp 4Fo exp( )

4FoX X

X XdX dη η

∞ ∞

′=

′ +′− = −

∫ ∫

/ 4Fo2 2 2

/ 4Fo / 4Fo 0

1 2exp( ) exp( ) exp( )

X

X X

d d dθ η η η η η ηπ π

∞ ∞

−

= − − − = − ∫ ∫ ∫




55

which can be written as

(3.350)

where erf in eq. (3.350) is the error function defined as:

(3.351)

Equation (3.350) can also be rewritten as dimensional

form:

(3.352)

The surface heat flux can be obtained by applying the

Fourier’s law

(3.353)

erf4Fo

Xθ

=

2

0

2erf ( )

zz

z e dzπ

−= ∫

0

0

erf4i

T T x

T T tα

− =

−

00

0

( )( ) i

x

k T TTq t k

x tπα=

−∂′′ = − =

∂




56

Periodic boundary conditions can be encountered in

various applications ranging from heat conduction in a

building during day and night to emerging technologies

such as pulsed laser processing of materials. Let us

reconsider the problem described by eqs. (3.332) –

(3.335) but replace eq. (3.333) by

(3.354)

where A is the amplitude of oscillation, is the angular

frequency, and is the phase delay. Introducing excess

temperature , the governing equation and

corresponding boundary and initial conditions become

(3.355)

(3.356)

( ) cos( ), 0 0i i

T T f t T A t x tω β= + = + − = , >

ω

β

iT Tϑ = −

2

2

( ) 1 ( )0 0

x t x tx t

x t

ϑ ϑ

α

∂ , ∂ ,= > , >

∂ ∂( ) ( ) cos( ) 0 0x t f t A t x tϑ ω β, = = − = , >




57

(3.357)

(3.358)

. Instead of solving eqs. (3.355) – (3.358) directly, we will

start with a simpler auxiliary problem defined below:

(3.359)

(3.360)

(3.361)

(3.362)

where in eq. (3.360) is treated as a parameter, rather

than time.

( ) 0 0x t x tϑ , = → ∞, >

( ) 0 0 0x t x tϑ , = > , =

2

2

( ) 1 ( )0 0

x t x tx t

x tα

∂ Φ , ∂Φ ,= > , >

∂ ∂

( ) ( ) cos( ) 0 0x t f A x tτ ωτ βΦ , = = − = , >

( ) 0 0x t x tΦ , = → ∞, >

( ) 0 0 0x t x tΦ , = > , =

τ




58

The Duhamel’s theorem stated that the solution the original

problem is related to the solution of auxiliary problem by

(3.363)

which can rewritten using Leibniz’s rule

(3.364)

The second term on the right hand side is

(3.365)

therefore, eq. (3.364) becomes

(3.366)

0

( , ) ( , , )t

x t x t dt τ

ϑ τ τ τ=

∂= Φ −

∂ ∫

0

( , ) ( , , ) ( , , )t

tx t x t d x t

t ττ

ϑ τ τ τ τ τ=

=

∂= Φ − + Φ −

∂∫

( , , ) ( ,0, ) 0t

x t xτ

τ τ τ=

Φ − = Φ =

0

( , ) ( , , )t

x t x t dtτ

ϑ τ τ τ=

∂= Φ −

∂∫




59

The solution of the auxiliary problem can be expressed as

(3.367)

The partial derivative appearing in eq. (3.366) can be

evaluated as

(3.368)

Substituting eq. (3.368) into eq. (3.366), the solution of the

original problem becomes

(3.369)

2

/ 4

2 ( )( , , ) ( ) 1- erf

4 x t

x fx t f e d

t

η

α

ττ τ η

α π

∞−

Φ = =

∫

2

3/ 2( , , ) ( ) exp

4 ( )4 ( )

x xx t f

t ttτ τ τ

α τπα τ

∂Φ − = −

∂ −−

2

3/ 20

( )( , ) exp

( ) 4 ( )4

tx f xx t d

t tτ

τϑ τ

τ α τπα =

= −

− − ∫




60

Introducing a new independent variable

eq. (3.369) becomes

(3.370)

For the periodic boundary condition specified in eq. (3.356),

we have

(3.371)

which can be rewritten as

(3.372)

4 ( )

x

tξ

α τ=

−

22

2/ 4

2( , ) exp( )

4x t

xx t f t d

αϑ ξ ξ

αξπ

∞ = − −

∫

22

2/ 4

2( , ) cos exp( )

4x t

A xx t t d

αϑ ω β ξ ξ

αξπ

∞ = − − −

∫

22

20

2/ 42

20

2( , ) cos exp( )

4

2 cos exp( )

4

x t

A xx t t d

A xt d

α

ϑ ω β ξ ξαξπ

ω β ξ ξαξπ

∞ = − − −

− − − −

∫

∫




61

Evaluating the first integral on the right hand side of eq.

(3.372) yields

(3.373)

It can be seen that as , the second term will become

zero and the first term represents the steady oscillation.

(3.374)

where represents the amplitude of

oscillation at point x, and in the cosine

function represents the phase delay of oscillation at point

x relative to the oscillation of the surface temperature.

1/ 2 1/ 2

2/ 42

20

( , ) exp cos2 2

2 cos exp( )

4

x t

x t A x t x

A xt d

α

ω ωϑ ω β

α α

ω β ξ ξαξπ

= − − −

− − − −

∫

t → ∞

1/ 2 1/ 2

( , ) exp cos2 2

s x t A x t xω ω

ϑ ω βα α

= − − −

( )1/ 2

exp /(2 )A x ω α −

( )1/ 2

/(2 )x ω α−




62

It is useful here to introduce a concept similar to the

thermal boundary layer for convective heat transfer –

thermal penetration depth. Assuming the thickness of the

thermal penetration depth at time is the temperature

of the semi-infinite body at will be affected but the

temperature at will remain unchanged (see Fig. 3.20).

According to the definition of the thermal penetration depth,

the temperature at the thermal penetration depth should

satisfy

(3.375)

(3.376)

Integrating eq. (3.332) in the interval (0, ) one obtains

(3.377)

t δ ,

x δ>

( )0 ( )

T x tx t

xδ

∂ ,= =

∂

( ) ( )iT x t T x tδ, = =

δ( )

0( ) 0

1 ( )t

x t x

T T T x tdx

x x t

δ

δ α= =

∂ ∂ ∂ ,− =

∂ ∂ ∂∫




63

The right-hand side of eq. (3.377) can be rewritten using

Leibnitz’s rule, i.e.,

(3.378)

which represents the energy balance within the thermal

penetration depth.

δ(t)

Ti

T0

T

x

Figure 3.20 Heat conduction in a semi-infinite body with constant wall temperature.

0( ) 0

1x

x t x

T T d dTdx T

x x dt dt

δ

δδ

δ

α == =

∂ ∂ − = − ∂ ∂ ∫




64

Substituting eqs. (3.375) and (3.376) into eq. (3.378) yields

(3.379)

where

(3.380)

Assume that the temperature distribution in the thermal

penetration depth is a third-order polynomial function of

x, i.e.,

(3.381)

where A0, A1, A2, and A3 and are four constants to be

determined using the boundary conditions.

The surface temperature of the semi-infinite body, T0, is not

a function of time t, so

(3.382)

0

( )i

x

T dT

x dtα δ

=

∂− = Θ −

∂( )

0

( ) ( )t

t T x t dxδ

Θ = ,∫

2 3

0 1 2 3( )T x t A A x A x A x, = + + +

( )0 0

T x tx

t

∂ ,= =

∂




65


(3.383)

Substituting eq. (3.381) into eqs. (3.333), (3.375), (3.376)

and (3.383) yields four equations for the constants in eq.

(3.381) . Solving for the four constants and substituting

the results into eq. (3.381), the temperature distribution

in the thermal penetration depth becomes

(3.384)

where the thermal penetration depth, is still unknown.

2

2

( )0 0

T x tx

x

∂ ,= =

∂

3

0

( ) 3 11

2 2

i

i

T x t T x x

T T δ δ

, − = − +

−

δ ,




66

Substituting eq. (3.384) into eq. (3.379), an ordinary

differential equation for is obtained:

(3.385)

Since the thermal penetration depth equals zero at the

beginning of the heat conduction, eq. (3.385) is subject

to the following initial condition:

(3.386)

The solution of eqs. (3.385) and (3.386) is

(3.387)

which is consistent with the result of scale analysis,

4 0d

tdt

δα δ= >

δ

0 0tδ = =

8 tδ α=

tδ α∼




67

Consider transient heat conduction in a rectangular bar

with dimensions of 2L1×2L2 and an initial temperature of

Ti (see Fig. 3.21). At time t = 0, the rectangular bar is

immersed into a fluid with temperature T∞.

3.3.2 Multidimensional Transient

Heat Conduction Systems

3.3.2 Multidimensional Transient

Heat Conduction Systems

Figure 3.21 Two-dimensional transient heat conduction

–

L1

L1 x

h1,T∞

y

0

h1,T∞

h2,T∞L2

h2,T∞-L2




68

The energy equation for this problem is

(3.388)

with the following boundary and initial conditions:

(3.389)

(3.390)

(3.391)

(3.392)

(3.393)

2 2

1 22 2, 0 , 0 , 0

T T Tx L y L t

x y t

∂ ∂ ∂+ = < < < < >

∂ ∂ ∂

20, 0, 0 , 0T

x y L tx

∂= = < < >

∂

1 2( ), , 0 , 0T

k h T T x L y L tx

∞

∂− = − = < < >

∂

10, y 0, 0 , 0T

x L ty

∂= = < < >

∂

2 1( ), y , 0 , 0T

k h T T L x L ty

∞

∂− = − = < < >

∂

1 2, 0 , 0 , 0iT T x L y L t= < < < < >




69


(3.394)

where both Fo1 and Fo2 are Fourier numbers but based on

different characteristic lengths. The dimensional time will

be related to both Fourier numbers, i.e.,

(3.395)

therefore, the right-hand side of eq. (3.388) becomes

(3.396)

1 22 2

1 2 1 2

, , , Fo , Foi

T T x y t tX Y

T T L L L L

α αθ ∞

∞

−= = = = =

−

1 2(Fo ,Fo )t t=

1 2

1 2

2 2

1 1 2 2

Fo Fo( ) ( )

Fo Fo

( )Fo Fo

i i

i

TT T T T

t t t t

T TL L

θ θ θ

θ α θ α

∞ ∞

∞

∂ ∂∂ ∂ ∂ ∂= − = − +

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂= − +

∂ ∂




70

Nondimensionalizing the left-hand side of eq. (3.388) and

considering eq. (3.396), the dimensionless energy

equation of the problem becomes

(3.397)

The boundary and initial conditions, eqs. (3.389) – (3.393)

are nondimensionalized as

(3.398)

(3.399)

(3.400)

(3.401)

2 22 2

1 1

2 2

2 1 2 2Fo Fo

L L

X L Y L

θ θ θ θ ∂ ∂ ∂ ∂+ = +

∂ ∂ ∂ ∂

1 20, X 0, 0 1, Fo 0, Fo 0YX

θ∂= = < < > >

∂

1 1 2Bi , X 1, 0 1, Fo 0, Fo 0YX

θθ

∂− = = < < > >

∂

1 20, Y 0, 0 1, Fo 0, Fo 0XY

θ∂= = < < > >

∂

2 1 2Bi , 1, 0 1, Fo 0, Fo 0Y XY

θθ

∂− = = < < > >

∂




71

(3.402)

where

(3.403)

are Bio numbers for different surfaces.

The idea of product solution is that the solution of the two-

dimensional problem can be expressed as the product of

two one-dimensional problem, i.e.,

(3.404)

where is the solution of the following problem

(3.405)

(3.406)

1 21, 0 1, 0 1, Fo Fo 0X Yθ = < < < < = =

1 1 2 21 2Bi , Bi

h L h L

k k= =

1 2( ,Fo ) ( ,Fo )X Yθ ϕ ψ=

ϕ2

2

1FoX

ϕ ϕ∂ ∂=

∂ ∂

10, X 0, Fo 0X

ϕ∂= = >

∂




72

(3.407)

(3.408)

and satisfies

(3.409)

(3.410)

(3.411)

(3.412)

It can be demonstrated that eqs. (3.397) – (3.402) can be

satisfied by eq. (3.404) if and are solutions of eqs.

(3.405) – (3.408) and (3.409) – (3.4120), respectively.

The solution of eqs. (3.405) – (3.408) is:

(3.413)

1 1Bi , X 1, Fo 0X

ϕϕ

∂− = = >

∂

11, 0 1, Fo 0Xϕ = < < =

ψ 2

2

2FoY

ψ ψ∂ ∂=

∂ ∂

20, 0, Fo 0YY

ψ∂= = >

∂

2 2Bi , 1, Fo 0YY

ψψ

∂− = = >

∂

21, 0 1, Fo 0Yψ = < < =

ϕ ψ

( )2

1Fo

1

4sincos

2 sin 2nn

n

n nn

X eλλ

ϕ λλ λ

∞−

=

=+∑




73

Where

(3.414)

and the solution of eqs. (3.409) – (3.412) is

(3.415)

where

(3.416)


solution of the two-dimensional problem is obtained.

(3.417)

1

cotBi

n

n

λλ=

( )2

2Fo

1

4sincos

2 sin 2mn

m

m mm

Y eνν

ψ νν ν

∞−

=

=+∑

2

cotBi

m

m

νν=

( ) ( ) 2 21 2( Fo Fo )

1 1

16sin sin cos cos

(2 sin 2 )(2 sin 2 )n mn m n m

n n m mn m

X Ye

λ νλ ν λ νθ

λ λ ν ν

∞ ∞− +

= =

=+ +∑∑

Documents

Transient Heat Transfer