Numerical methods in Transient-heat-conduction

Lectures on Heat Transfer --
NUMERICAL METHODS IN
TRANSIENT HEAT CONDUCTION
by
Dr. M. ThirumaleshwarDr. M. Thirumaleshwar
formerly:
Professor, Dept. of Mechanical Engineering,
St. Joseph Engg. College, Vamanjoor,
Mangalore,
India

Preface
• This file contains slides on Numerical
methods in Transient heat conduction.
• The slides were prepared while teaching
Heat Transfer course to the M.Tech.
students in Mechanical Engineering Dept.
of St. Joseph Engineering College,
Vamanjoor, Mangalore, India, during Sept.
– Dec. 2010.
Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful
to teachers, students, researchers and
professionals working in this field.
• For students, it should be particularly
useful to study, quickly review the subject,useful to study, quickly review the subject,
and to prepare for the examinations.
•
Aug. 2016 3MT/SJEC/M.Tech.

References:
• 1. M. Thirumaleshwar: Fundamentals of Heat &
Mass Transfer, Pearson Edu., 2006
• https://books.google.co.in/books?id=b2238B-
AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false
• 2. Cengel Y. A. Heat Transfer: A Practical
Approach, 2nd Ed. McGraw Hill Co., 2003
Aug. 2016 MT/SJEC/M.Tech. 4
Approach, 2nd Ed. McGraw Hill Co., 2003
• 3. Cengel, Y. A. and Ghajar, A. J., Heat and
Mass Transfer - Fundamentals and Applications,
5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.
• 4. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.
• 5. M. Thirumaleshwar: Software Solutions to• 5. M. Thirumaleshwar: Software Solutions to
Problems on Heat Transfer – CONDUCTION-
Part-III, Bookboon, 2013
• http://bookboon.com/en/software-solutions-problems-on-heat-
transfer-ciii-ebook

NUMERICAL METHODS IN TRANSIENT
HEAT CONDUCTION
• Finite difference eqns. by energy balance
– Explicit and Implicit methods – 1-D
transient conduction in a plane wall –
stability criterion – 2-D transient heat
conduction – Finite diff. eqns. for interior
conduction – Finite diff. eqns. for interior
nodes – Explicit and Implicit methods -
stability criterion – difference eqns for
different boundary conditions – Accuracy
considerations – discretization error and
round–off error

Numerical methods in Transient
heat conduction:
• In transient conduction, temperature varies
with both position and time.
• So, to obtain finite difference equations for
transient conduction, we have to discretize
transient conduction, we have to discretize
both space and time domains.
• This scheme is illustrated in Fig. 8.9.

• Starting from initial temperature at τ = 0, at
each node we calculate the temperature at
a successive time interval of ∆τ till we
reach the desired time at which
temperature has to be calculated.
• Time step is shown in superscript, i.e. Tm
i
is the temperature of node ‘m’ at time step
is the temperature of node ‘m’ at time step
‘i’ ( at time = i.∆τ from start up);
• The notation Tm
i+1 means the temperature
of node ‘m’ at the time step (i + 1) ( at
time = (i + 1)∆τ from start up).

• Formulation of finite difference equations
in transient conduction is done by an
energy balance on the elemental volumes
containing the nodes, just as was done in
the case of steady state conduction;
• however, now, on the RHS, there appears
a term representing the change in energy
a term representing the change in energy
content of the elemental volume, with time.
• We write, for a given volume element:
Q left Q up Q right Q down Q g ρ Velement
. C p
.
T m
i 1( )
T m
i
∆τ
. .......(8.48)

• In the above eqn., as already mentioned, Tm
i is
the temperature of node ‘m’ at time step ‘i’ ( i.e.
at time = i.∆τ from start up) and Tm
i+1 is the
temperature of node ‘m’ at the time step (i + 1) (
i.e. at time = (i + 1)∆τ from start up). Cp is the
specific heat and ρ is the density of the medium.
(Tm
i+1 - Tm
i )/∆τ is the finite difference
approximation of the term dT/dτ.
approximation of the term dT/dτ.
• Now, regarding the terms on the LHS of eqn.
(8.45), the question arises as to whether we
should consider the temperatures of the nodes
at step ‘i’ or step ‘(i + 1)’. In fact, both the
methods are adopted in practice.

• While applying eqn. (8.45) to write the
finite difference eqn. for a node, if the
terms on the LHS of the eqn. are
considered at time step ‘i’, then, the
method is known as explicit method of
approach;
• if the terms on the LHS of the eqn.
• if the terms on the LHS of the eqn.
are considered at time step ‘(i + 1)’,
then, the method is known as implicit
method of approach.

• To summarize:
• Explicit method:
Q left
i
Q up
i
Q right
i
Q down
i
Q g
i
ρ Velement
. C p
.
T m
i 1( )
T m
i
∆τ
. .......(8.49)
• Implicit method:
Q left
i 1
Q up
i 1
Q right
i 1
Q down
i 1
Q g
i 1
ρ Velement
. C p
.
T m
i 1( )
T m
i
∆τ
. ...(8.50)
In the explicit method, time derivative is calculated in
‘forward difference’ form, and in implicit method, the time
derivative is in the ‘backward difference’ form.

• Explicit method is called so, because temperature of the
node ‘m’ at time step (i+1) is calculated explicitly in terms
of the temperatures calculated at the previous time step
‘i’; therefore, the calculations are quite straight forward;
• However, it suffers from a serious limitation that the time
increment can not be independently fixed, but has an
upper limit because of stability considerations.
• But in case of implicit method, this limitation on time
• But in case of implicit method, this limitation on time
duration is not there and we can choose any time step;
• But, the implicit method requires that at each time step,
nodal temperatures have to be solved simultaneously.

One-dimensional Transient heat
conduction in a plane wall:
• Consider one-dimensional, transient heat
conduction in a plane wall of thickness L, with
heat generation rate qg(x,τ) and constant
thermal conductivity k.
• Now, let us divide the region 0 < x < L into M
• Now, let us divide the region 0 < x < L into M
sub-regions.
• Then, thickness of each sub-region is:
∆x = L/M. So, there are totally (M+1) nodes,
starting from m = 0 to m = M, as shown in Fig.
8.10.

• It is clear that interior nodes 1,2…M-1
represent full sub-volumes whereas
boundary nodes 0 and M represent half
volumes (of thickness ∆x/2).
• Volume of element surrounding node ‘m’ is
A.∆x.
A.∆x.
• To get the finite difference formulation, we
apply the general energy balance, viz.
eqn. (8.48):
k A.
Tm 1
Tm
∆x
. k A.
Tm 1
Tm
∆x
. q m A ∆x.( ). ρ A. ∆x. C p
.
T m
i 1
T m
i
∆τ
. ...(8.51)

• Simplifying,
Tm 1
2 Tm
. Tm 1
q m ∆x( )
2
.
k
∆x( )
2
α ∆τ.
T m
i 1
T m
i. .....(8.52)
where, α
k
ρ C p
.
= thermal diffisivity of the material.
Now, the term
α ∆τ.
is the finite difference form of the Fourier number, Fo
So, eqn. (8.52) reduces to:
Tm 1
2 Tm
. Tm 1
q m ∆x( )
2
.
k
Tm
i 1
Tm
i
Fo
.....(8.53)
Now, the term
α ∆τ
∆x( )
2
is the finite difference form of the Fourier number, Fo

• In LHS, if we use temperatures at time
step ‘i’, it is the ‘explicit method’ and if the
temperatures at time step ‘i+1’ are used,
then, it is the ‘implicit method’.
• Explicit method:
T
i
2 T
i. T
i
q m
i
∆x( )
2
. Tm
i 1
Tm
i
.....(8.54)
Tm 1
i
2 Tm
i. Tm 1
i
q m x( )
k
Tm
Tm
Fo
.....(8.54)
Now, the new temperature Tm
i+1 can be explicitly solved since the
Other terms involved at the previous time step ‘i’, are already known.
So, we write for Tm
i+1 :
Tm
i 1
Fo Tm 1
i
Tm 1
i. 1 2 Fo.( ) Tm
i. Fo
qm
i
∆x( )
2
.
k
. ....(8.55)

• Eqn.(8.55) is the explicit difference eqn.
valid for all interior nodes, 1,2….(M-1),
when there is internal heat generation.
• When there is no heat generation, eqn.
(8.55) reduces to:
T
i 1
Fo T
i
T
i. 1 2 Fo.( ) T
i. ....(8.56)
Tm
i 1
Fo Tm 1
i
Tm 1
i. 1 2 Fo.( ) Tm
i. ....(8.56)
Implicit method:
If in the LHS of eqn. (8.53), we use the values at time step (i + 1),
we get the implicit relation for the node temperatures:
i.e. Tm 1
i 1
2 Tm
i 1. Tm 1
i 1
qm
i 1
∆x( )
2.
k
Tm
i 1
Tm
i
Fo
.....(8.57)

• Eqn. (8.57) is simplified to:
1 2 Fo.( ) Tm
i 1. Fo Tm 1
i 1
Tm 1
i 1
qm
i 1
∆x( )
2.
k
. Tm
i
0 ....(8.58)
Eqn.(8.58) is the implicit difference eqn. valid for all interior
nodes,1,2….(M-1), when there is internal heat generation.
When there is no heat generation, eqn. (8.58) reduces to:
1 2 Fo.( ) Tm
i 1. Fo Tm 1
i 1
Tm 1
i 1. Tm
i
0 ....(8.59)
With the use of either the explicit or the implicit eqns. given above,
we get M-1 nodal equations.
We need two more equations for the boundary nodes ‘0’ and ‘M’. These
are obtained by applying the energy balance for the half-volumes around
these nodes. See Fig. 8.10.

• For node ‘0’ with convection boundary condition:
• Explicit formulation:
h A. Ta
T0
i. k A.
T1
i
T0
i
∆x
. q0
i
A. ∆x
2
. ρ A. ∆x
2
. C p
.
T0
i 1
T0
i
∆τ
. ....(8.60)
Simplifying:
T0
i 1
1 2 Fo. 2 Fo. Bi.( ) T0
i. Fo 2 T1
i. 2 Bi. Ta
.
q0
i
∆x( )
2
.
k
. .....(8.61)
0 0 1 a
k
where Bi
h ∆x.
k
= Biot number
When there is no heat generation, eqn. (8.61) for explicit formulation
becomes:
T0
i 1
1 2 Fo. 2 Fo. Bi.( ) T0
i. Fo 2 T1
i. 2 Bi. Ta
.. .....(8.62)

• For other types of boundary conditions,
difference equations are developed in a
similar manner, by applying the energy
balance on the elemental volume
containing the node and considering all
the heat flows to be into the volume.
• Next step is to choose a suitable time
increment ∆τ; then, starting with the initial
• Next step is to choose a suitable time
increment ∆τ; then, starting with the initial
conditions at τ = 0, explicitly solve the
difference equations for the temperatures
Tm
i+1 at all the nodes at the next time step
τ = ∆τ .

• Now, using these values of temperatures
as ‘previous values’, again get the nodal
temperatures at the next time step τ =
2. ∆τ, using the same difference
equations.
• Thus, continue to march in time till the
• Thus, continue to march in time till the
solution is obtained for the desired time
interval.

Stability criterion:
• Suitable time interval ∆τ has to be chosen;
• Explicit method is not unconditionally
stable, and above a certain value of ∆τ,
the solution will not converge.
• This limit on ∆τ is determined as follows:
• This limit on ∆τ is determined as follows:
• “Coefficients of all Tm
i in the Tm
i+1
expressions (called ‘primary
coefficients’) must be greater than or
equal to zero for all nodes ‘m’”.

• Considering eqn.(8.55) for interior nodes,
we see that coeff. of Tm
i is (1-2.Fo) and
applying the above mentioned criterion for
stability, we get:
1 2 Fo. 0
i.e. Fo
α ∆τ.
2
1
2
for interior nodes, one-dimensional conduction.............(8.63)
∆x( )
2 2
Now, ∆τ must be fixed from eqn. (8.63).
However, generally, boundary nodes with convection conditions are more
restrictive and in such cases, coeff. of Tm
i from the most restrictive eqn.
must be considered for the stability criterion and the time step ∆τ∆τ∆τ∆τ must be
determined with respect to that coefficient.

• Example 8.10:
• A large Uranium plate of thickness L = 10 cm, (k
= 28 W/(m.C), a = 12.5 x 10-6 m2/s) is initially at
an uniform temperature of 100 C. Heat gen. rate
in the plate is 5 x 10^6 W/m3. At time t = 0, both
the left and right sides of the plate are subjected
to convection with a fluid at at temperature of 0
C and a heat transfer coeff. of 1500 W/(m2.C).
Using a uniform nodal spacing of 2 cm, develop
Using a uniform nodal spacing of 2 cm, develop
the explicit finite difference formulations for all
nodes, and determine the temperature
distribution in the plate after 5 min. Also, find out
how long it will take for steady conditions to be
reached in the plate.
• (b) Also, solve this problem by implicit finite
difference formulation.

Data:
L 0.1 m....thickness of plate
k 28 W/(m.C)..thermal cond. of plate
α 12.5 10
6. m2/s....thermal diffusivity of plate
q g 5 10
6. W/m3...heat gen. rate in the plate
T 100 C...initial temp. of plate
T a 0 C....temp. of ambient fluid
h 1500 W/(m2.C)....heat tr. coeff. between the ambient fluid and the plate surface.
∆ x 0.02 m....nodal spacing
M 5 no. of equal spacings, i.e. nodes 0,1,2....5
τ 300 s...time after which temp. distribution in plate is desired

• Difference eqns. for interior nodes:
• Nodes 1,2,3 and 4 are interior nodes. Finite difference
equations for these nodes by explicit method are
obtained from eqn. (8.55), by setting m = 1,2,3,4. i.e.
Tm
i 1
Fo Tm 1
i
Tm 1
i. 1 2 Fo.( ) Tm
i. Fo
qm
i
∆x( )
2
.
k
. ....(8.55)
We get:
Node 1: T 1
i 1
Fo T 0
i
T 2
i. 1 2 Fo.( ) T 1
i. Fo
qg
i
∆ x( )
2
.
k
. ...(b)
Node 2: T 2
i 1
Fo T 1
i
T 3
i. 1 2 Fo.( ) T 2
i. Fo
qg
i
∆ x( )
2
.
k
. ...(c)

• Difference eqns. for boundary nodes:
• For node '0':
Node 3: T 3
i 1
Fo T 2
i
T 4
i. 1 2 Fo.( ) T 3
i. Fo
qg
i
∆ x( )
2
.
k
. ...(d)
Node 4: T 4
i 1
Fo T 3
i
T 5
i. 1 2 Fo.( ) T 4
i. Fo
qg
i
∆ x( )
2.
k
. ...(e)
• Node '0'is on the left surface, subjected to convection.
Applying the eqn. (8.61) directly:
T0
i 1
1 2 Fo. 2 Fo. Bi.( ) T0
i. Fo 2 T1
i. 2 Bi. Ta
.
q0
i
∆x( )
2.
k
. .....(8.61)
wherewherewherewherewhere Bi
h ∆ x.
k
Bi
h ∆ x.
k
Bi
h ∆ x.
k
Bi
h ∆ x.
k
Bi
h ∆ x.
k
= Biot number= Biot number= Biot number= Biot number= Biot number

• For node 5:
• This is a node with convection boundary
condition. So, applying the energy balance to
the half-volume around node 5, with all the heat
lines flowing into the element, we get:
i.e. T0
i 1
1 2 Fo. 2 Fo. Bi.( ) T0
i. Fo 2 T1
i. 2 Bi. T a
.
q g ∆ x( )
2
.
k
. .....(a)
lines flowing into the element, we get:
h A. T a T 5
i. k A.
T 4
i
T 5
i
∆ x
. q g A. ∆ x
2
. ρ A. ∆ x
2
. C p
.
T 5
i 1
T 5
i
∆τ
.
i.e. T 5
i 1
1 2 Fo. 2 Fo. Bi.( ) T5
i. Fo 2 T4
i. 2 Bi. Ta
.
qg
∆ x( )
2.
k
. .... f( )

• Now, we have to fix the upper limit of t from stability
criterion.
• To do that, we observe that in eqns. (a) to (f), the smaller
coeff. of Tm
i is in eqn. (f), i.e. (1 - 2. Fo -2.Fo.Bi) must be
greater than or equal to zero. Putting this condition, we
get:
1 2 Fo. 2 Fo. h ∆ x.
k
. 0
i.e. Fo
1
2 1
h ∆ x.
k
.
i.e. ∆τ
∆ x( )
2
2 α. 1
h ∆ x.
k
.
i.e. ∆τ 7.724 s

• This means that a time step less than 7.724 s has to be
employed from stability criterion.
• Let us choose: ∆τ 5 s
Then, Fo
α ∆τ.
∆ x( )
2
i.e. Fo 0.1563=
Substituting all relevant numerical values in eqn. (a) to (f), we get the
explicit difference equations as:
T 0
i 1
0.353T 0
i. 0.1563 2 T 1
i. 71.429. .....(a)
T 1
i 1
0.1563 T 0
i
T 2
i. 0.688T 1
i. 11.161 ...(b)
T 2
i 1
0.1563 T 1
i
T 3
i. 0.688T 2
i. 11.161 ...(c)

T 3
i 1
0.1563 T 2
i
T 4
i. 0.688T 3
i. 11.161 ...(d)
T 4
i 1
0.1563 T 3
i
T 5
i. 0.688T 4
i. 11.161 ...(e)
T 5
i 1
0.353 T5
i. 0.1563 2 T4
i. 71.429. .... f( )
Initial temp. of the plate at τ = 0 and i = 0, is given as 100 C.
• Therefore, at the next time step i = 1, i.e. at ∆τ =
5 s, temperatures at nodes 0 to 5 can be
explicitly calculated from eqns. (a) to (f).
Initial temp. of the plate at τ = 0 and i = 0, is given as 100 C.
i.e. T 0
0
T 1
0
T 2
0
T 3
0
T 4
0
T 5
0
100 C

• Then, calculate temperatures at the nodes for
next time step of ∆τ = 10 s, using the same
eqns. (a) to (f), since the temperatures at the
previous time step are already calculated.
• Thus, march in time till we reach the time limit
specified in the problem, viz. 5 min i.e. there are
60 time steps of 5 s each.
• In the small Mathcad program given below, LHS
defines a function 'Temp(n)'where n is the no. of
time steps, which we can specify. Output is a
vector containing step no., total time elapsed,
and node temperatures T0, T1,...T5.

Temp n( ) T00
100
T10
100
T20
100
T30
100
T40
100
T50
100
T0i 1
0.353T0i
. 0.1563 2 T1i
. 71.429.
T1i 1
0.1563 T0i
T2i
. 0.688T1i
. 11.161
i 0 n..∈for
T2i 1
0.1563 T1i
T3i
. 0.688T2i
. 11.161
T3i 1
0.1563 T2i
T4i
. 0.688T3i
. 11.161
T4i 1
0.1563 T3i
T5i
. 0.688T4i
. 11.161
T5i 1
0.353T5i
. 0.1563 2 T4i
. 71.429.
i 5 i. T0i
T1i
T2i
T3i
T4i
T5i
Temp 0( ) 0 0 100 100 100 100 100 100= .....starting at time = 0

• i = step no.; t = one time step = 5 s; t = time
duration from beginning = i. t, s
-------------------------------------------------------------------------------------------
i τ T0 T1 T2 T3 T4 T5
-------------------------------------------------------------------------------------------
Temp 2( ) 2 10 73.369 117.213 122.449 122.449 117.213 73.369=
Temp 4( ) 4 20 75.447 127.666 142.472 142.472 127.666 75.447=
Temp 12( ) 12 60 95.576 169.805 202.468 202.468 169.805 95.576=
Temp 12( ) 12 60 95.576 169.805 202.468 202.468 169.805 95.576=
Temp 18( ) 18 90 108.991 196.256 236.479 236.479 196.256 108.991=
Temp 24( ) 24 120 120.036 217.958 264.179 264.179 217.958 120.036=
Temp 36( ) 36 180 136.45 250.194 305.285 305.285 250.194 136.45=
Temp 48( ) 48 240 147.409 271.714 332.724 332.724 271.714 147.409=
Temp 60( ) 60 300 154.724 286.079 351.041 351.041 286.079 154.724=

Temp 120( ) 120 600 167.466 311.102 382.946 382.946 311.102 167.466=
Temp 180( ) 180 900 169.155 314.419 387.176 387.176 314.419 169.155=
Temp 250( ) 250 1.25 10
3
169.388 314.878 387.761 387.761 314.878 169.388=
Temp 260( ) 260 1.3 10
3
169.395 314.891 387.778 387.778 314.891 169.395=
• Temp. distribution after 5 min.:
• Temp(60) corresponds to 60th time step, i.e. 300 s from
beginning.
• We note that after 5 min. the node temps. are:
T0 T5 154.724 C; T1 T4 286.079C; T2 T3 351.041 C.

• Time to reach steady state:
• It may be seen from the Table that from
about 240th step, the temperatures at the
nodes do not vary much as we advance in
time. i.e. steady state is reached at about
20 min. from start up.
20 min. from start up.
• To draw the temperatures at the nodes
at different times:
• First represent the node temperatures at
different time steps as vectors:

Step0
100
100
100
100
100
100
....initial temp. distribution in nodes 0,1,...5
Similarly, temp. distributions after 1, 5, 10, 20 and 30 min. are given as
Step1, Step5,...etc., below:
Step1
95.58
169.81
202.47
202.47
169.81
95.58
Step5
154.72
286.08
351.04
351.04
286.08
154.72
Step10
167.47
311.1
382.95
382.95
311.1
167.47
Step20
169.38
314.86
387.74
387.74
314.86
169.38
Step30
169.41
314.92
387.82
387.82
314.92
169.41

150
200
250
300
350
400
Transient temp. distr. in a plate
Temperature(deg.C)
0 1 2 3 4 5
50
100
Initial temp. distribn.
After 1 min.
After 5 min.
After 10 min.
After 20 min.
After 30 min.
Node number
It is seen from the graph that steady state is reached at about 20 min. from start up.

• (b) Implicit method:
• Difference eqns. for interior nodes:
• Nodes 1,2,3 and 4 are interior nodes. Finite difference
equations for these nodes by implicit method are
obtained from eqn. (8.58), by setting m = 1,2,3,4. i.e.
1 2 Fo.( ) Tm
i 1. Fo Tm 1
i 1
Tm 1
i 1
qm
i 1
∆x( )
2
.
k
. Tm
i
0 ....(8.58)
m m 1 m 1
k m
Node 1 : 1 2 Fo.( ) T1
i 1. Fo T0( )
i 1
T2( )
i 1
q g ∆ x( )
2
.
k
. T1
i
0 ....(b)
Node 2 : 1 2 Fo.( ) T2
i 1. Fo T1( )
i 1
T3( )
i 1
q g ∆ x( )
2.
k
. T2
i
0 ....(c)

• Difference eqns. for boundary nodes:
• Nodes 0 and 5 are boundary nodes, with convection
conditions.
Node 3 : 1 2 Fo.( ) T3
i 1. Fo T2( )
i 1
T4( )
i 1
q g ∆ x( )
2
.
k
. T3
i
0 ....(d)
Node 4 : 1 2 Fo.( ) T4
i 1. Fo T3( )
i 1
T5( )
i 1
q g ∆ x( )
2
.
k
. T4
i
0 ....(e)
conditions.
• For node '0':
• Writing the energy balance for the half-volume around
node '0', with all heat flow lines going into the volume
element, with the LHS of eqn. (8.60) expressed at time
step (i + 1), we get:

• For node '5':
h A. T a T0( )
i 1. k A. T1( )
i 1
T0( )
i 1
∆ x
. q g A. ∆ x
2
. ρ A. ∆ x
2
. C p
. T0( )
i 1
T0( )
i
∆τ
.
i.e.
2 Fo. h. ∆ x.
k
T a T0( )
i 1. 2 Fo. T1( )
i 1
T0( )
i 1.
Fo q g
. ∆ x( )
2.
k
T0( )
i 1
T0( )
i
....(a)
Eqn. (a) is the implicit finite difference formulation for node '0', with
convection conditions.
• For node '5':
• Writing the energy balance for the half-volume around
node '5', with all heat flow lines going into the volume
element, with the LHS of energy balance eqn. expressed
at time step (i + 1), we get:
h A. T a T5
i 1. k A. T4
i 1
T5
i 1
∆ x
. q g A. ∆ x
2
. ρ A. ∆ x
2
. C p
.
T 5
i 1
T 5
i
∆τ
.

• Eqn. (f) is the implicit finite difference formulation for
node 5, with convection conditions.
• Now, we can choose any ∆τ, since there is no problem
of stability in implicit formulation.
• Let us choose:
i.e.
2 Fo. h. ∆ x.
k
T a T5( )
i 1. 2 Fo. T4( )
i 1
T5( )
i 1.
Fo q g
. ∆ x( )
2.
k
T5( )
i 1
T5( )
i
....(f)
• Let us choose:
∆τ 10 s
Therefore, Fo
α ∆τ.
∆ x( )
2
i.e. Fo 0.3125=
Inserting numerical values, eqns. (a) to (f) are written as:

0.67 T a T0( )
i 1. 0.625 T1( )
i 1
T0( )
i 1. 22.321 T0( )
i 1
T0( )
i
....(a)
1.625T1
i 1. 0.3125 T0( )
i 1
T2( )
i 1
71.429. T1
i
0 ....(b)
1.625T2
i 1. 0.3125 T1( )
i 1
T3( )
i 1
71.429. T2
i
0 ....(c)
1.625T3
i 1. 0.3125 T2( )
i 1
T4( )
i 1
71.429. T3
i
0 ....(d)
1.625T4
i 1. 0.3125 T3( )
i 1
T5( )
i 1
71.429. T4
i
0 ....(e)
0.67 T a T5( )
i 1. 0.625 T4( )
i 1
T5( )
i 1. 22.321 T5( )
i 1
T5( )
i
....(f)

• Now, to start with, i.e. at τ = 0, all the node
temperatures T0, T1,....T5 are known.
• Then, at the next time step, solve eqns. (a)
to (f) simultaneously to get the node
temperatures at that time step.
• Using these results, solve the eqns. (a) to
(f) at the next time step, etc. till you reach
(f) at the next time step, etc. till you reach
the given time limit.
• A Mathcad program to perform these
calculations is shown below:

• This calculation is easily done in Mathcad. We slightly
change the notation for convenience in calculation: we
write the superscripts as subscripts to work with matrix
notation, as shown below:

i.e. Temperatures at the end of first time step are as shown in the vector
i.e. Temperatures at the end of first time step are as shown in the vector
on the RHS above.
Now, to proceed with the next time step, re-set T00 = 84.57,
T10 = 114.826… etc. and use the above Function to calculate
temperatures at the end of second time step.
Then, repeat the procedure to get temperatures at the subsequent
time steps.
Mathcad program to do this is shown below:

• We note that after 5 min. i.e. after 300 s, i.e. after 30th time step, the
node temperatures are:
Similarly, after 200th time step, i.e. after 2000 s, i.e. after about 33 min.
Similarly, after 200th time step, i.e. after 2000 s, i.e. after about 33 min.
the node temperatures have stabilised; compare with the temperatures
after 220th time step :

• Exercise:Write a computer program to
accomplish this task. Use the Gauss –
Siedel iteration technique for the
solution of simultaneous equations.
solution of simultaneous equations.

Results:

Two-dimensional Transient heat
conduction:
• Fig. 8.11 shows a rectangular region where the heat
transfer in x and y directions are significant, and heat
transfer in the z direction is negligible.
• Divide the rectangular region into a nodal network of
thicknesses ∆x and ∆y as shown.
thicknesses ∆x and ∆y as shown.
• Let the thickness in the z direction be unity.
• Finite difference equations are developed by writing the
energy balance for an elemental volume surrounding the
node under consideration.
• All heat flows are considered to be flowing into the
volume.

• Difference equations for interior nodes:
• A typical interior node, Tm,n and the elemental volume
surrounding it, and immediate neighbours of this node
are shown in Fig. (8.11, b).
• Node Tm,n is surrounded by 4 nodes: Tm-1,n , Tm,n+1 ,
Tm+1,n , and Tm,n-1.
• Let us make an energy balance on the elemental volume
surrounding the node Tm,n .
surrounding the node Tm,n .
• It is observed that heat flows into the node from all the
four directions, i.e. left, up, right and down.
• In addition, let there be heat generation in the volume at
a rate of (∆V.qg) , W, where qg , (W/m3), is the uniform
heat generation rate in the system.

• Writing the energy balance,
Q left Q up Q right Q down ∆V q g
. m C p
. dT
dτ
. .....(8.64)
i.e.
k ∆y.
Tm 1 n,
Tm n,
∆x
. k ∆x.
Tm n 1,
Tm n,
∆y
. k ∆y.
Tm 1 n,
Tm n,
∆x
. k ∆x.
Tm n 1,
Tm n,
∆y
. q g ∆x. ∆y.
= ρ ∆x. ∆y. C p
.
T m
i 1
T m
i
. ....(8.65)
= ρ ∆x. ∆y. C p
.
∆τ
. ....(8.65)
For ∆x = ∆y (i.e. a square mesh), we get:
Tm 1 n,
Tm 1 n,
Tm n 1,
Tm n 1,
4 Tm n,
.
q g ∆x( )
2
.
k
T m
i 1
T m
i
Fo
.....(8.66)
where Fo
α ∆τ.
∆x( )
2
= Fourier number, and α is thermal diffusivity.

• Now, on the LHS of eqn. (8.66), if we use the ‘previous’ time step
‘i’, we get the explicit formulation of finite difference eqn. for
interior nodes:
i.e.
Tm 1 n,
i
Tm 1 n,
i
Tm n 1,
i
Tm n 1,
i
4 Tm n,
i.
q g ∆x( )
2
.
k
Tm n,
i 1
Tm n,
i
Fo
.....(8.67)
i.e.
T ,
i 1
Fo T ,
i
T ,
i
T ,
i
T ,
i. 1 4 Fo.( ) T ,
i. Fo
q g ∆x( )
2
.
.
Tm n,
Fo Tm 1 n,
Tm 1 n,
Tm n 1,
Tm n 1,
. 1 4 Fo.( ) Tm n,
. Fo
k
.
.....(8.68)
Eqn. (8.68) is valid for all interior nodes, when there is heat
generation.
If there is no heat generation, eqn. (8.68) simplifies to:
Tm
n,
i 1
Fo Tm 1 n,
i
Tm 1 n,
i
Tm n 1,
i
Tm n 1,
i. 1 4 Fo.( ) Tm n,
i. .....(8.69)

• Stability criterion in the explicit method requires
the coefficient of (Tm,n) i to be positive and this
condition gives the upper limit on the time
increment ∆τ, as follows:
Fo
α ∆τ.
∆x( )
2
1
4
....stability criterion for interior nodes.....(8.70)
∆x( ) 4
Now, on the LHS of eqn. (8.66), if we use the
‘future’ time step ‘i+1’, we get the implicit
formulation of finite difference eqn. for
interior nodes.

• Difference equations for boundary nodes:
• Boundary nodes may be on the surface or on the corners.
• Fig. (8.12) shows some common boundary conditions encountered
in practice:

• Finite difference equations for the boundary situations shown in Fig.
(8.12) are given in Table 8.2.

Accuracy considerations:
• Numerical methods yield approximate values as
compared to ‘exact analytical solutions’.
• This is due to the following errors inherent in
numerical methods:
• 1. Discretization error:
This is due to the error involved in writing the
This is due to the error involved in writing the
derivatives in terms of differences. Discretization error
is cumulative; but if the function changes sign, it is
possible that the errors may cancel. Discretization
error is proportional to the square of the time step ∆τ
(or ∆x).
Therefore, smaller the mesh size, smaller the
discretization error.

• 2. Round - off error:
This is due to the fact that computer retains
only 15 digits accuracy in a calculation and the
rest of the digits are either chopped off or
rounded off.
When this is done continuously for a large
number of calculations, error is carried over to
number of calculations, error is carried over to
successive calculations and the cumulative
error can be significant.
Obviously, the round-off error is proportional to
the total number of computations performed,
and reduces as the mesh size increases.

• We note that we have to deal with two
opposing effects: if the mesh size ∆x (or
time step size ∆τ) is decreased,
discretization error is reduced, but the
round off error increases since the total
number of calculations increases.
• So, practical way of approaching the
solution is to start with a coarse mesh and
then gradually refine the mesh size and
observe if the results converge.

Numerical methods in Transient-heat-conduction

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