1

Transient Heat Transfer

• Having gotten a feel for the steady state heat transfer, let

us get a feel for the transient behavior of heat transfer

• To begin with, let us assume that spatial variation of

temperature is negligible and the temperature of the body

as a whole changes with time

• This implies that the thermal conductivity is very high.

• We shall evolve a criterion for the validity of this

assumption as we go along

• Let us look at the steps in analysis

• This type of problem is called lumped analysis

Lumped Analysis-I

• Let the body temperature be denoted

by T

T

T∞, h

• The body interacts with the

surroundings with the temperature at

T∞ and heat transfer coefficient h

• First law implies that WQE &&& −=

( ) )TT(hAmcTdt

d∞−−=⇒

• For constant specific heat

)TT(hAdt

dTmc ∞−−=⇒

Lumped Analysis-II

• Defining θ = T - T∞, we can write

θ−=θ

⇒ hAdt

dmc θ−=

θ⇒

mc

hA

dt

dLet θ = θo at t = 0

• The solution for the above equation is

tmc

hA

0e−

θ=θ

• If the object is sphere, we have

23 R4A;R3

4m π=ρπ=

R

3

c

h

R3

4

R4

c

h

mc

hA

3

2

ρ=

ρπ

π=⇒

tR

3

c

h

0e ρ−

θ=θ

Lumped Analysis-III

• In generalt

V

A

c

h

0

tcV

hA

0

tmc

hA

0 eee ρ−

ρ−−

θ=θ=θ=θ

• If the object is a cylinder of Radius R and Length L,

we have

( )RL

RL2

LR

R2RL2

V

A2

2 +=

π

π+π=⇒

tLR

)RL(2

c

h

0e

+

ρ−

θ=θ

• If the cylinder has L>>R , then

tR

2

c

h

0e ρ−

θ=θ

Lumped Analysis-III

• This concept can now be extended to any geometry

• To generalize the result, it is customary to introduce

the characteristic length L

• The logic suggests that V/A is the most obvious

choice

22L

t

k

hL

0

tcL

k

k

hL

0

tcL

h

0 eee

α−

ρ−

ρ−

θ=θ=θ=θ⇒

• In the above expression, we have introduced the

property called thermal diffusivity = k/(ρc)

• The non-dimensional parameter hL/k and αt/L2 are

called the Biot Number and Fourier No respectively

Lumped Analysis-IV

• Thus, the temperature variation is a function of two

non dimensional parameters Biot number and Fourier

number.

• We will appreciate these parameters, as we go into

more complex cases

• We can give a physical interpretation for the Biot

number as follows

cetansisReconvection

cetansisReconduction

1

hA

KA

L

K

hLBi ===

• When Bi is very small, it implies that conduction

resistance is very small and hence lumped analysis valid

• The criterion used is Bi < 0.1

2

Transients with spatial effects

• If Bi > 0.1 spatial effects become important and so

more complications are involved

• Exact analytical solutions can be obtained using

separation of variables similar to 2-D steady state

analysis

• Let us look at 1-D transient analysis in a slab

geometry with no heat generation

• The governing equation for this case is

qz

T

y

T

x

Tk

t

)T(c

2

2

2

2

2

2

′′′+

∂

∂+

∂

∂+

∂

∂=

∂

∂ρ

2

2

x

T

t

T1

∂

∂=

∂

∂

α⇒

1-D Transient in a Plate-I • The governing equation

• Let us define

• Boundary conditions

T(0,x) = Ti ; [∂T/∂x](t,0) = 0; [-k∂T/∂x](t,L) = h(T(t,L)- T∞)

∞−=θ TT

[∂θ/∂x](t,0) = 0; -k[∂θ/∂x](t,L) =hθ(t,L);

θ(0,x) = θ0

x

0 L

h, T∞ h, T∞0 ≤ x ≤ L; 0 ≤ t2

2

x

T

t

T1

∂

∂=

∂

∂

α

0 ≤ x ≤ L; 0 ≤ t2

2

xt

1

∂

θ∂=

∂

θ∂

α⇒

1-D Transient in a Plate-II• The solution for θ(t,x) is assumed of the form:

θ(t,x) = X(x)T(t)

• Substituting this in the governing equation, we get

0dx

)x(Xd)t(T

dt

)t(dT1)x(X

2

2

==α

⇒2

2

dx

Xd

X

1

dt

dT

T

11=

α⇒

• Since LHS is only a function of t and RHS is only

a function of x and yet they be equal, would

require that both sides be equal to a constant

0Tdt

dT;0X

dx

Xd 22

2

2

=αλ−=λ+⇒2

2

2

dx

Xd

X

1

dt

dT

T

11λ−==

α⇒

1-D Transient in a Plate-II• The solution for X and T are;

)xsin(C)xcos(CX 21 λ+λ= t3

2

eCT αλ−=

• The BC at x = 0; ∂θ/ ∂x = 0, requires a symmetric

solution and hence C2 = 0

( )( )t321

2

eC)xsin(C)xcos(CXT αλ−λ+λ==θ⇒

( )( )t1

2

e)xcos(C αλ−λ=θ⇒

( )( )t21

2

e)xsin(C)xcos(C αλ−λ+λ=C3 is absorbed

in C1 and C2

( )( )t1

2

e)xsin((Cx

αλ−λ−λ=∂

θ∂⇒

BC at x = L ( )( ) ( )( )t1

t1

22

e)Lcos(Che)Lsin()(kC αλ−αλ− λ=λλ−−⇒

( ) ( ))Lcos(k

h)Lsin( λ=λλ⇒

Bi

L

h

k)Lcot(

λ=

λ=λ⇒

1-D Transient in a Plate-III• The solution for the eigen values can be

graphically interpreted as shown

λL

λL/BiCot(λL)

λ1L λ2L λ3L

• Thus we will have infinite

values of eigen values, but

these have to be numerically

determined. Your book

summarizes the first four

values as a function of Bi in

Appendix B3

• Thus the solution for θ can be written as

∑∞

=

αλ− λ=θ⇒1n

nt

n )xcos(eC2

n Still Cn needs to

be determined

1-D Transient in a Plate-IV• The constants have to be determined from

initial condition

• This will be again done by orthogonal functions

• It turns out that

mnfor0dx)xcos()xcos(L

0

mn ≠=λλ∫ Messy to prove

n

nnnL

0

n2

2

)Lcos()Lsin(Ldx)x(cos

λ

λλ+λ=λ∫

n

nL

0

n

)Lsin(dx)xcos(

λ

λ=λ∫

Will be given as home work

3

1-D Transient in a Plate-V• The initial condition is θ(0,x) = θ0

∑∞

=

λ=θ⇒1n

nn0 )xcos(C

dx)xcos()xcos(Cdx)xcos(L

0

n1n

mnm

L

0

0 ∫ ∑∫ λλ=λθ∞

=

• Note that in the RHS, only term n=m will survive

dx)x(cosCdx)xcos(L

0

n2

nn

L

0

0 ∫∫ λ=λθ

n

nnnn

n

n0

2

)Lcos()Lsin(LC

)Lsin(

λ

λλ+λ=

λ

λθ

)Lcos()Lsin(L

)Lsin(2C

nnn

n0n

λλ+λ

λθ=∴

1-D Transient in a Plate-VI

• Thus the solution for θ can be written as

∑∞

=

αλ− λλλ+λ

λ=

θ

θ⇒

1nn

t

nnn

n

0

)xcos(e)Lcos()Lsin(L

)Lsin(2

2n

αλ=

θ

θ⇒

L

x,

L

t,Lf

2n0

∑∞

=

αλ− λλλ+λ

λ=

θ

θ⇒

1nn

t

nnn

n

0

)L

xLcos(e

)Lcos()Lsin(L

)Lsin(2

2n

• But, λnL is determined by Bi

=

α=

θ

θ∴

L

x,Fo,Bif

L

x,

L

t,Bif

20

Heisler Charts-I

• The solution for θ was shown to be

∑∞

=

αλ− λλλ+λ

λ=

θ

θ⇒

1nn

t

nnn

n

0

)xcos(e)Lcos()Lsin(L

)Lsin(2

2n

• In functional form

=

α=

θ

θ

L

x,Fo,Bif

L

x,

L

t,Bif

20

• Heisler showed that when Fo > 0.2, just one term is

adequate to describe the solution

)xcos(eC 1t

10

21 λ=

θ

θ⇒ αλ−

)Lcos()Lsin(L

)Lsin(2CWhere

111

11

λλ+λ

λ=

Heisler Charts-II• The coordinate of the mid-plane is x =0

( ))Bi,Fo(f)L,Fo(feCeC

)0(1

FoL1

t1

0

21

21 =λ===

θ

θ⇒ λ−αλ−

• Thus, the spatial profile at any time can be written as

)L/x,Bi(f)L/x,L(f)xcos()0(

11 =λ=λ=θ

θ⇒

• Heisler presented these results in the form of two

graphs, called Heisler Charts

• For Cartesian system, it is shown in the next slide.

Similar curves for the cylindrical and spherical one

dimensional cases are given in Appendix D of your

book

Heisler Charts-III Criterion for Lumped Analysis• If the variation of temperature within the slab is less

than 5%, we can call it lumped

95.0)Lcos()0(

)Lx(1 =λ=

θ

=θ⇒ radian318.0L1 =λ⇒

1.0)Lcot(

LBi

1

1 =λ

λ=⇒

• Thus, Bi should be < 0.1 for lumped analysis to be

valid. This can also be viewed in the chart

4

Energy Storage-I• As energy storage is one of the main application of

transient, it is useful to get a method to estimate the

total energy stored

• From thermodynamics

∫∫ θ−θρ=−ρ=∆L

0

0

L

0

0 dx)(L

1AcL2dx)TT(Ac2E

∫ −λθρ= αλ−L

0

1t

10 dx)1)xcos(eC(L

1AcL2

21

L

01

1t1

0

x)xsin(

eCL

1

AcL2

E 21

−

λ

λ=

θρ

∆ αλ−

)xcos(eC 1t

10

21 λθ=θ αλ−

−

λ

λθ=

−

λ

λ=

∆

∆ αλ−1

L

)Lsin()0(1

L

)Lsin(eC

E

E

1

1

1

1t1

0

21

Energy Storage-II

)Fo,Bi(f)Fo,L(f1L

)Lsin()0(

E

E1

1

1

0

=λ=

−

λ

λθ=

∆

∆

1-D Transient in a Semi-Infinite Plate-I

• Another case of interest in transient heat transfer is

what is called heat transfer in a Semi-infinite plate

• It is has many useful applications. In fact, all

transient problems start as semi-infinite wall

0

h, T∞

• The governing equation

0 ≤ x ≤ ∞; 0 ≤ t2

2

x

T

t

T1

∂

∂=

∂

∂

α

• Boundary conditions

T(0,x) = Ti ; T(t,0) = Ts ; T(t,x→∞) = Ti

• Mathematically, problems that have boundary at

infinity are often solved by a method called similarity

solution

1-D Transient in a Semi-Infinite Plate-II

• In this method, a new variable, called similarity

variable is introduced

• There are systematic ways by which this can be

derived, but often involves some qualitative

arguments

• Thus the governing equation will be transformed into

an ODE from PDE

• This variable is chosen such that T becomes a

function of only this variable

5.0)t4(

1

d

dT

xd

dT

x

T

αη=

∂

η∂

η=

∂

∂

• Using chain rule

1-D Transient in a Semi-Infinite Plate-III

( ) 5.0t4

x

α=η

• In this course we shall give you the form of the

variable. Note that it will be a combination of x and t

)t4(

1

d

Td

)t4(

1

xd

Td

)t4(

1

d

dT

xx

T

xx

T2

2

5.02

2

5.02

2

αη=

α∂

η∂

η=

αη∂

∂=

∂

∂

∂

∂=

∂

∂

5.05.15.0 )t4(t2

x

d

dT

)t(

5.0

)4(

x

d

dT

td

dT

t

T

α

−

η=

−

αη=

∂

η∂

η=

∂

∂

( ) 5.0t4

1

x α=

∂

η∂⇒

( ) 5.15.0 t

5.0

4

x

t,

−

α=

∂

η∂

1-D Transient in a Semi-Infinite Plate-IV

• Substituting for the partial derivatives in the heat

equation, we get

)t4(

1

d

Td

)t4(t2

x

d

dT12

2

5.0 αη=

α

−

ηα

ηη

−=α

α−

ηα=

η⇒ 2

d

dT

)t4(t2

)t4(x

d

dT1

d

Td5.02

2

ηη−=

η d

dT2

d

Td2

2

• Thus we get an ODE in η

• The boundary conditions

T(t,0) = Ts T(η = 0) = Ts

T(0,x) = Ti ; T(t,x→∞) = Ti T(η = ∞) = Ti

5

1-D Transient in a Semi-Infinite Plate-V

η=

η⇒=

η

++

d

dT

d

TdT

d

dT2

2

• To get the solution, we make the transformation

• The equation in the new variables can be written as

++

η−=η

T2d

dT

ηη−=

η d

dT2

d

Td2

2

• The governing equation is

ηη−=⇒+

+

d2T

dT

C)Tln( 2 +η−=+• Integration gives2

eCT 1η−+ =⇒

2

eCd

dT1

η−=η

⇒ 2

0

u12

0

1 CdueCCdeCT22

+=+η=⇒ ∫∫η

−η

η−u is a

dummy

variable

Error Function

∫η

−

0

u due2

• The integral occurs very frequently in physics

• Though not integrable in explicit form, it has been

integrated with series expansion and tables have been

constructed under what is called Error Function

• It turns out that 2

due0

u2 π=∫

∞−

• Hence erf(∞)=1, erf(0) = 0

∫−

π=

x

0

u due2

)x(erf2

erf(x) is tabulated in

Appendix B of your

book

x=3 is as good as ∞

Erf(3) = 0.99998• A complimentary error function

is also defined as

erfc(x) = 1 – erf(x)

∫∫∞

∞−

+−∞

∞−

= dxdyeI )yx( 22

Consider ∫∫π

−∞

θ=2

0

)r(

0

rdrde2

Put r2 = t π=

−

π=θ=⇒

∞−

π−

∞

∫∫0

t2

0

t

0

e2

2d

2

dteI

Note that dxeIwhere,I4dyedxeI0

x1

21

yx 222

∫∫∫∞

−∞

∞−

−∞

∞−

− ===

From above2

IorI4 12

1

π=π=

Integration

...!4

x

!3

x

!2

xx1e

8642x 2

++−+−=−

Error function can be computed by using the fact that

1-D Transient in a Semi-Infinite Plate-VI

2

0

u1 CdueCT

2

+= ∫η

−• The solution

• The Boundary condition, T(η=0) = Ts implies

21s C)0(erf2

CT +π

= s2 TC =⇒

21 C)(erf2

C +ηπ

=

• The Boundary condition, T(η=∞) = Ti implies

s1i T)(erf2

CT +∞π

= s1i T2

CT +π

= ( )π

−=⇒2

TTC si1

( ) ssi T)(erfTTT +η−=∴

( )θ=η=

−

−⇒ )(erf

TT

TT

si

s

η

θ

00

10.99

1.82

1-D Transient in a Semi-Infinite Plate-VII

• For η = 1.82, θ=0.99; This implies that η = 1.82

is for all practical purposes is ∞

( )82.1

t4

x82.1

5.0=

α⇒=η ( ) 5.0

t64.3x α=⇒

• The above numbers can be interpreted in the

following manner

• x > 3.64 (αt)0.5 can be considered as infinitely thick

• Similarly for t < x2/(13.25 α), the plate can be

considered infinite

• Now we will turn our attention to heat transferred

=∂

∂−=′′

=0xx

Tkq

1-D Transient in a Semi-Infinite Plate-VIII

2

eCd

dT1

η−=η

• We had shown that1

0

Cd

dT=

η⇒

=η

( )π

−=2

TT si

( ) ( )5.0

is

5.0

si

)t(

TTk

2

)t4(

TTkq

πα

−−=

πα

−−=′′∴

• Solutions are available for other boundary conditions and

are summarized in your book

5.00 )t4(

1

d

dTk

αη−

=η

5.0)t4(

1

d

dT

x

T

αη=

∂

∂

6

Application-I

• Semi-infinite wall finds lots of applications, such as

bubble growth, vapor explosion etc.

• In vapor explosion, analysis liquid metal drops falling

in relatively cold water is required

• This can be idealized by two semi-infinite slabs

brought into intimate contact

• We shall see the gist by trying to calculate what is

known as the contact temperature

T1

T2

Application-II

• At the instant they come in contact, thermal

equilibrium would dictate that the contact temperature

Ts would lie in between T1 and T2

• If the contact temperature remains same, thermal

equilibrium demands that the heat flux from one face

should be equal to the heat flux from the other face.

• The solution on the two slabs can now be worked

similar to what we have obtained with T1 and T2

similar to Ti

( ) ( )5.0

2

2s2125.0

1

1s11

)t(

TTkqand

)t(

TTkq

πα

−−=′′

πα

−−=′′∴

Application-III

• Since heat flux coming out from 1 will be getting into

the other,

21 qq ′′−=′′⇒( ) ( )

5.02

2s25.0

1

1s1

)t(

TTk

)t(

TTk

πα

−=

πα

−−⇒

( )( )

( )( ) 5.0

111

5.0222

5.02

5.01

1

2

2s

s1

ck

ck

k

k

TT

TT

ρ

ρ=

α

α=

−

−⇒

• Rearranging, we get

( ) ( )( ) ( ) 5.0

2225.0

111

5.02222

5.01111

Sckck

ckTckTT

ρ+ρ

ρ+ρ==

Finite Difference Method - I

• We found that analytical methods are complex

• These methods are restrictive and are applicable for

simple boundary conditions

• Numerical methods are easy to implement and we can

obtain results quickly

• We will see the gist of the method

Physical Domain:

t

x

Computational

Domain:n+1

n

n

iT

i-1 i i+1 N1

Finite Difference Method - II

Governing Equation:2

2

x

T

t

T

∂

∂α=

∂

∂Governing Equation:2

2

x

T

t

T

∂

∂α=

∂

∂ • One of the FDM approximation is

t

TT

t

T n

i

1n

i

n

i ∆

−=

∂

∂ +

2

n

1i

n

i

n

1i

n

i

2

2

x

TT2T

x

T

∆

+−=

∂

∂ −+

• This leads to the nodal equation

)TT2T(x

tTT

n

1i

n

i

n

1i2

n

i

1n

i −++ +−

∆

∆α+=

Finite Difference Method - III

• This method is called explicit method, as the

values at Tin+1 are readily obtained explicitly, once

the initial and boundary conditions are known

7

Finite Difference Method - IV

• This method suffers from a disadvantage that the

time step ∆t < 0.5α∆t/∆x2

• This is called stability limit. This occurs due to

explosion of errors above the limit

• If we need more accurate results, we need more

nodes, and this implies small ∆x. This will limt ∆t

to be small and takes more computational time

• This can be overcome by choosing a different

method called implicit method

Finite Difference Method - V

t

TT

t

T n

i

1n

i

n

i ∆

−=

∂

∂ +

2

1n1i

1ni

1n1i

n

i

2

2

x

TT2T

x

T

∆

+−=

∂

∂ +−

+++

• This leads to the nodal equation

ni2

1n1i2

1ni2

1n1i T

x

tT

x

t21T

x

tT =

∆

∆α−+

∆

∆α++

∆

∆α− +

−++

+

2

1n1i

1ni

1n1i

ni

1ni

x

TT2T

t

TT

∆

+−α=

∆

− +−

+++

+

Implicit Method

Finite Difference Method - VI

• For the simple case of boundary temperature

known

=

∆

∆α−

∆

∆α+

∆

∆α−

∆

∆α−

∆

∆α+

∆

∆α−

∆

∆α−

∆

∆α+

∆

∆α−

+

+

+

+

+

+

+

1n5

n4

n3

n3

1n1

1n5

1n4

1n3

1n2

1n1

222

222

222

T

T

T

T

T

T

T

T

T

T

10x

t

x

t21

x

tx

t

x

t21

x

tx

t

x

t21

x

t

01

n+1

n2 3 4 51

• The matrix can be inverted in Matlab or any other software