numerical methods in transient-heat-conduction

Lectures on Heat Transfer --NUMERICAL METHODS IN

TRANSIENT HEAT CONDUCTION

by

Dr. M. ThirumaleshwarDr. M. Thirumaleshwarformerly:

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. Tmi

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 Tmi+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 V

element. C p

.T m

i 1( ) T mi

∆τ. .......(8.48)

• In the above eqn., as already mentioned, Tmi 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 - Tmi )/∆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 lefti Q up

i Q righti Q down

i Q gi ρ Velement

. C p.

T mi 1( ) T m

i

∆τ. .......(8.49)

• Implicit method:

Q lefti 1 Q up

i 1 Q righti 1 Q down

i 1 Q gi 1 ρ Velement

. C p.

T mi 1( ) T m

i

∆τ. ...(8.50)


In the explicit method, time derivative is calculated in‘forward difference’ form, and in implicit method, the timederivative 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 mi

∆τ. ...(8.51)

• Simplifying,

Tm 1 2 Tm. Tm 1

q m ∆x( )2.

k

∆x( )2

α ∆τ.T m

i 1 T mi. .....(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

Tmi 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 mi ∆x( )

2. Tmi 1 Tm

i

.....(8.54)


Tm 1i 2 Tm

i. Tm 1i q m x( )

k

Tm Tm

Fo.....(8.54)

Now, the new temperature Tmi+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 :

Tmi 1 Fo Tm 1

i Tm 1i. 1 2 Fo.( ) Tm

i. Foqm

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 1q

mi 1 ∆x( )

2.

k

Tm

i 1 Tm

i

Fo.....(8.57)

• Eqn. (8.57) is simplified to:

1 2 Fo.( ) Tmi 1. Fo Tm 1

i 1 Tm 1i 1 qm

i 1 ∆x( )2.

k. Tm

i 0 ....(8.58)

Eqn.(8.58) is the implicit difference eqn. valid for all interiornodes,1,2….(M-1), when there is internal heat generation.

When there is no heat generation, eqn. (8.58) reduces to:


1 2 Fo.( ) Tmi 1. Fo Tm 1

i 1 Tm 1i 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 T0i. k A.

T1i T0

i

∆x. q0

i A. ∆x

2. ρ A. ∆x

2. C p

.T0

i 1 T0i

∆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 Bih ∆x.

k= Biot number

When there is no heat generation, eqn. (8.61) for explicit formulation becomes:

T0i 1 1 2 Fo. 2 Fo. Bi.( ) T0

i. Fo 2 T1i. 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 Tmi+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. Foα ∆τ.

2

1

2for 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 morerestrictive 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 bedetermined 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 5 106. 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.

Tmi 1 Fo Tm 1

i Tm 1i. 1 2 Fo.( ) Tm

i. Foqm

i ∆x( )2.

k. ....(8.55)


We get:

Node 1: T 1i 1 Fo T 0

i T 2i. 1 2 Fo.( ) T 1

i. Foqg

i ∆ x( )2.

k. ...(b)


i T 3i. 1 2 Fo.( ) T 2

i. Foqg

i ∆ x( )2.

k. ...(c)

• Difference eqns. for boundary nodes:• For node '0':


i T 4i. 1 2 Fo.( ) T 3

i. Foqg

i ∆ x( )2.

k. ...(d)


i T 5i. 1 2 Fo.( ) T 4

i. Foq

gi ∆ x( )

2.

k. ...(e)


• Node '0' is on the left surface, subjected to convection. Applying the eqn. (8.61) directly:

T0i 1 1 2 Fo. 2 Fo. Bi.( ) T0

i. Fo 2 T1i. 2 Bi. Ta

.q0

i ∆x( )2.

k. .....(8.61)

wherewherewherewherewhere Bih ∆ x.

kBi

h ∆ x.

kBi

h ∆ x.

kBi

h ∆ x.

kBi

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 5i. k A.

T 4i T 5

i

∆ x. q g A. ∆ x

2. ρ A. ∆ x

2. C p

.T 5

i 1 T 5i

∆τ.

i.e. T 5i 1 1 2 Fo. 2 Fo. Bi.( ) T

5i. Fo 2 T

4i. 2 Bi. T

a.

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. Fo1

2 1h ∆ x.

k.

i.e. ∆τ ∆ x( )2

2 α. 1h ∆ 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, Foα ∆τ.

∆ x( )2

i.e. Fo 0.1563=

Substituting all relevant numerical values in eqn. (a) to (f), we get theexplicit difference equations as:


T 0i 1 0.353T 0

i. 0.1563 2 T 1i. 71.429. .....(a)

T 1i 1 0.1563 T 0

i T 2i. 0.688T 1

i. 11.161 ...(b)

T 2i 1 0.1563 T 1

i T 3i. 0.688T 2

i. 11.161 ...(c)

T 3i 1 0.1563 T 2

i T 4i. 0.688 T 3

i. 11.161 ...(d)

T 4i 1 0.1563 T 3

i T 5i. 0.688 T 4

i. 11.161 ...(e)

T 5i 1 0.353 T5

i. 0.1563 2 T4i. 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 00 T 1

0 T 20 T 3

0 T 40 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 1.25 103 169.388 314.878 387.761 387.761 314.878 169.388=

Temp 260( ) 260 1.3 103 169.395 314.891 387.778 387.778 314.891 169.395=


• 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 asStep1, 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

400Transient temp. distr. in a plate

Tem

pera

ture

(deg

. C)


0 1 2 3 4 550

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.( ) Tmi 1. Fo Tm 1

i 1 Tm 1i 1 qm

i 1 ∆x( )2.

k. Tm

i 0 ....(8.58)


m m 1 m 1 k m

Node 1 : 1 2 Fo.( ) T1i 1. Fo T0( )i 1 T2( )i 1 q g ∆ x( )2.

k. T1i 0 ....(b)


k. T2i 0 ....(c)

• Difference eqns. for boundary nodes:• Nodes 0 and 5 are boundary nodes, with convection

conditions.


k. T3i 0 ....(d)


k. T4i 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.

kT a T0( )i 1. 2 Fo. T1( )i 1 T0( )i 1.

Fo q g. ∆ x( )

2.

kT0( )i 1 T0( )i

....(a)

Eqn. (a) is the implicit finite difference formulation for node '0', withconvection 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 T5i 1. k A. T4i 1 T5i 1

∆ x. q g A. ∆ x

2. ρ A. ∆ x

2. C p

.T 5

i 1 T 5i

∆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.

kT a T5( )i 1. 2 Fo. T4( )i 1 T5( )i 1.

Fo q g. ∆ x( )

2.

kT5( )i 1 T5( )i

....(f)


• Let us choose:

∆τ 10 s

Therefore, Foα ∆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)

1.625T1i 1. 0.3125 T0( )i 1 T2( )i 1 71.429. T1i 0 ....(b)

1.625T2i 1. 0.3125 T1( )i 1 T3( )i 1 71.429. T2i 0 ....(c)


1.625T3i 1. 0.3125 T2( )i 1 T4( )i 1 71.429. T3i 0 ....(d)

1.625 T4i 1. 0.3125 T3( )i 1 T5( )i 1 71.429. T4i 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 subsequenttime 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.

Aug. 2016 MT/SJEC/M.Tech.58

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 mi 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, T

m 1 n, Tm n 1, T

m n 1, 4 Tm n,

.q g ∆x( )

2.

k

T mi 1 T m

i

Fo.....(8.66)

where Foα ∆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. Foq g ∆x( )

2..


Tm n, Fo T

m 1 n, Tm 1 n, T

m n 1, Tm n 1,

. 1 4 Fo.( ) Tm n,

. Fok

.

.....(8.68)

Eqn. (8.68) is valid for all interior nodes, when there is heatgeneration.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:

Foα ∆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 implicitformulation of finite difference eqn. forinterior 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.