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