Chapter 1 Computational modelling of

steady diffusion process

Lecture 1: Solving 2D Steady Heat Conduction Equation

Prof. Nilanjan Chakraborty E-mail: nilanjan.chakraborty@ncl.ac.uk

Computational Modelling : MEC3014/MEC8043

1

Governing law of conduction heat transfer

xTkqx ∂∂

−=

)(TgradkTkq ×−=∇−=

xTkAQx ∂∂

−=

Conductive heat transfer rate in x direction is given by Fourier’s law of heat conduction

where T is temperature, k is the thermal conductivity and A is the area normal to the direction of heat transfer and qx is the heat flux in the x direction.

or

Heat flux is a vector. The generalised expression for conductive heat flux is given by:

kzTj

yTi

xTTgradT ˆˆˆ)(

∂∂

+∂∂

+∂∂

==∇where

SI Units: k → W/m.K qx and q → W/m2 Qx → Watt .

2

General Heat Conduction Equation

Cartesian co-ordinates

3

Energy balance of the control volume gives us

gzzyyxxzyxst EQdQQdQQdQQQQE ++−+−+−++= )()()(

where tTzyxCEst ∂∂

∆∆∆= )(ρ and zyxqEg ∆∆∆=

=q volumetric heat generation

4

xTzykqzyQ xx ∂∂

∆∆−=∆∆= )()(

))(( xxxx dqqzyQdQ +∆∆=+

yyq

qqdqq yyyyyy ∆

∂

∂+==+ ∆+

Similarly in y and z directions

yTzxkqzxQ yy ∂∂

∆∆−=∆∆= )()(

))(( yyyy dqqzxQdQ +∆∆=+

xxqqqdqq x

xxxxx ∆∂∂

+==+ ∆+

zTyxkqyxQ zz ∂∂

∆∆−=∆∆= )()(

))(( zzzz dqqyxQdQ +∆∆=+

zzqqqdqq z

zzzzz ∆∂∂

+==+ ∆+

Let us consider x direction

5

Energy balance of the control volume gives us

gzzyyxxzyxst EQdQQdQQdQQQQE ++−+−+−++= )()()(yxqzxqzyqE zyxst ∆∆+∆∆+∆∆=

gzzyyxx Eyxdqqzxdqqzydqq +∆∆+−∆∆+−∆∆+− )()()(

zyxqzyxzqzyx

yq

zyxxqzyx

tTC zyx ∆∆∆+∆∆∆

∂∂

−∆∆∆∂

∂−∆∆∆

∂∂

−=∆∆∆∂∂

ρ

tTzyxCEst ∂∂

∆∆∆= )(ρ

zyxqEg ∆∆∆=

6

Energy balance of the control volume gives us

qzq

yq

xq

tTC zyx +

∂∂

−∂

∂−

∂∂

−=∂∂ρ

According to Fourier’s law of heat conduction

xTkqx ∂∂

−= ; yTkqy ∂∂

−= ; zTkqz ∂∂

−=

qzTk

zyTk

yxTk

xtTC +

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=∂∂ρ

Generalised conduction equation in Cartesian co-ordinates.

7

If the thermal conductivity k remains constant

Generalised conduction equation in Cartesian co-ordinates.

qzTk

zyTk

yxTk

xtTC +

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=∂∂ρ

In vector notation the same equation is given by:

qTktTC +∇∇=∂∂ ).(ρ or qTgradkdiv

tTC +×=∂∂ ))((ρ

qzT

yT

xTk

tTC +

∂∂

+∂∂

+∂∂

=∂∂

2

2

2

2

2

2

ρ

tT

kq

zT

yT

xT

∂∂

=+

∂∂

+∂∂

+∂∂

α1

2

2

2

2

2

2

The above equation can be rewritten as:

where

Ckρ

α =

=α Thermal diffusivity

8

Generalised conduction equation in Cartesian co-ordinates.

For steady state

tT

kq

zT

yT

xT

∂∂

=+

∂∂

+∂∂

+∂∂

α1

2

2

2

2

2

2

0=∂∂

tT

For 2D problem 0=∂∂

zT

Under this condition the governing equation becomes

02

2

2

2

=+

∂∂

+∂∂

kq

yT

xT

or 02 =+∇kqT

2

2

2

2

2

22 (_)(_)(_)(_)

zyx ∂∂

+∂∂

+∂∂

=∇ Laplace operator

In 2D Laplace operator becomes

2

2

2

22 (_)(_)(_)

yx ∂∂

+∂∂

=∇

9

The analytical solutions of Poisson and Laplace equations are often mathematically involved and are beyond the scope of the present module. Interested readers should consult the following text for analytical solutions.

Conduction of Heat in Solids: H.S. Carslaw & J.A. Jaeger (OUP) We will opt for Numerical Solution Technique for 2D heat conduction problems.

kq

yT

xT

−=

∂∂

+∂∂

2

2

2

2or

kqT

−=∇2

2D steady heat conduction equation with internal heat generation

The above partial differential equation is known as Poisson Equation

2D steady heat conduction equation without internal heat generation

02

2

2

2

=

∂∂

+∂∂

yT

xT or 02 =∇ T

The above partial differential equation is known as Laplace Equation

10

P E W

N

S

Finite Difference Technique

Δx

Δy

Tw

Tp TE

Δx Δx

TN Tp TS

Δy Δy

11

Examples of meshing of engineering structures are shown below. The figures shown below demonstrate Cartesian adaptive grid for a full scale motorcycle and an IC engine. It can be seen from the figures shown below that the geometrical complexities can be handled satisfactorily if the cells are made sufficiently small (i.e. If the mesh is sufficiently refined).

Courtesy: Profs. W.N. Dawes and R.S. Cant, Cambridge University

12

According to Taylor series expansion

)()(61)(

21)()( 43

3

32

2

2

xOxxTx

xTx

xTxTxxT

xxx

∆+∆∂∂

+∆∂∂

+∆∂∂

+=∆+

(1)

(2)

ETxxT =∆+ )(

WTxxT =∆− )(

PTxT =)(

P E W

Δx Δx

x

From eqs. (1) and (2) one gets:

)()(

)(2)()( 222

2

xOx

xTxxTxxTxT

x

∆+∆

−∆−+∆+=

∂∂

)()(

2 222

2

xOx

TTTxT PWE

P

∆+∆

−+=

∂∂

)()(61)(

21)()( 43

3

32

2

2

xOxxTx

xTx

xTxTxxT

xxx

∆+∆∂∂

−∆∂∂

+∆∂∂

−=∆−

13

According to Taylor series expansion

)()(61)(

21)()( 43

3

32

2

2

yOyyTy

yTy

yTyTyyT

yyy

∆+∆∂∂

+∆∂∂

+∆∂∂

+=∆+

(3)

NTyyT =∆+ )(

STyyT =∆− )(

PTyT =)(

P S

From eqs. (3) and (4) one gets:

)()(

)(2)()( 222

2

yOy

yTyyTyyTyT

y

∆+∆

−∆−+∆+=

∂∂

)()(

2 222

2

yOy

TTTyT PSN

P

∆+∆

−+=

∂∂

)()(61)(

21)()( 43

3

32

2

2

yOyyTy

yTy

yTyTyyT

yyy

∆+∆∂∂

−∆∂∂

+∆∂∂

−=∆−

(4) N Δy Δy

y

14

P E W

N

S Δx

Δy

kq

yT

xT

−=

∂∂

+∂∂

2

2

2

2

Discretised form of

kqyOxO

yTTT

xTTT PSNPWE

−=∆+∆+∆

−++

∆−+ )()(

)(2

)(2 22

22

)()(

2 222

2

yOy

TTTyT PSN

P

∆+∆

−+=

∂∂

)()(

2 222

2

xOx

TTTxT PWE

P

∆+∆

−+=

∂∂

For small values of Δx and Δy: kq

yTTT

xTTT PSNPWE

−=∆

−++

∆−+

22 )(2

)(2

For Δx = Δy = Δ : 04 2 =∆+−+++kqTTTTT PSNWE

Truncation error

Truncation error

15

P E W

N

S Δx

Δy

Discretised form of Poisson equation for equal grid spacing:

The general form of the discretised equation can be written as:

bTaTa nbnbpP +=∑

Where subscript ‘nb’ refers to neighbouring points

04 2 =∆+−+++kqTTTTT PSNWE

Which can be rewritten as:

24 ∆++++=kqTTTTT SNWEP

bTaTaTaTaTa SSNNWWEEPP ++++=

1,1,1,1,4 ===== SNWEP aaaaa and 2∆=kqb

16

P E W

N

S Δx

Δy

02

2

2

2

=

∂∂

+∂∂

yT

xT

Discretised form of

For small values of Δx and Δy:

0)(

2)(

222 =

∆−+

+∆

−+y

TTTx

TTT PSNPWE

For Δx = Δy = Δ :

04 =−+++ PSNWE TTTTT

The above discretised equation can be written as:

bTaTa nbnbpP +=∑

where 1,1,1,1,4 ===== SNWEP aaaaa and 0=b

17

P E W

N

S Δx

Δy

Golden rules of the discretised equation:

bTaTa nbnbpP +=∑

Rules 1. The discretisation coefficients aP, aE, aW, aN, aS have to be positive or zero. 2. The discretistaion coefficients will be such

that it will satisfy :

Explanation: 1. If temperature increases in the neighbouring nodes, temperature will increase in node P. 2. The second criterion satisfies isothermal condition. The inequality holds

when there is a source term which depends on temperature.

SNWEP aaaaa +++≥ or ∑≥ nbP aa

18

Flux Boundary Condition Specification in Finite Difference Method

P E q

)()(21)()( 32

2

2

xOxxTx

xTxTxxT

xx

∆+∆∂∂

+∆∂∂

+=∆+

)()()( xOx

xTxxTxT

x

∆+∆

−∆+=

∂∂

)( xOxTT

xT PE

P

∆+∆−

=∂∂

Boundary condition for node P is given by:

PxTkq∂∂

−= )( xOxTTkq PE ∆+

∆−

−=

For small Δx the boundary condition is given by:

Δx

∆−

=xTTkq EP

19

Convective Boundary Condition Specification in Finite Difference Method

P

S

h T∞

)()(21)()( 32

2

2

yOyyTy

yTyTyyT

xy

∆+∆∂∂

+∆∂∂

−=∆−

)()()( yOy

yyTyTyT

y

∆+∆

∆−−=

∂∂

)( yOyTT

yT SP

P

∆+∆−

=∂∂

Δy

Boundary condition for node P is given by: )( ∝−=∂∂

− TThyTk P

P

Discretised boundary condition is given by:

)()( yOyTTk

yTkTTh SP

PP ∆+

∆−

−=∂∂

−=− ∝

For small Δy the boundary condition is given by:

∆−

=− ∝ yTTkTTh PS

P )(

20

Solving 1D Heat Conduction problem using Finite Difference Method

50 oC -20 oC

L= 0.25m

Δx= 0.05m

T1 T2 T3 T4 T5 T6

Discretised form

2002020202

50

6

654

543

432

321

1

−==+−=+−=+−=+−

=

TTTTTTTTTTTTT

T1 0 0 0 0 0 1 -2 1 0 0 0 0 1 -2 1 0 0 0 0 1 -2 1 0 0 0 0 1 -2 1 0 0 0 0 0 1

T1

T2 T3

T4

T6

=

50 0 0 0 0

-20

Tri-diagonal matrix

T5

21

Solving Tri-diagonal system of equations by Tri-Diagonal Matrix Algorithm (TDMA)

iiiiiii bTuTaTl =++ +− 11

iiiii fTeTd =+ +1

1111 −−−− =+ iiiii fTeTd 111 /)( −−− −= iiiii dTefT

iiiiii

iiii bTuTa

dTefl =++

−+

−

−−1

1

11

(i)

(ii)

From eq. (ii) one gets: (iii)

Substituting eq. (iii) in eq. (i) one gets:

−=+

−

−

−+

−

−

1

11

1

1

i

iiiiii

i

iii d

flbTuTdela

−=

−

−

1

1

i

iiii d

elad ii ue =; ;

−=

−

−

1

1

i

iiii d

flbf

0

0

0 0

22

Solving 1D Heat Conduction problem using Finite Difference Method

50 oC -20 oC

L= 0.25m

Δx= 0.05m

T1 T2 T3 T4 T5 T6

Discretised form

=

50 36 22 8 -6 -20 20

02020202

50

6

654

543

432

321

1

−==+−=+−=+−=+−

=

TTTTTTTTTTTTT

T

6

5

4

3

2

1

TTTTTT

Analytical solution

50280)( +−= xxT

=

50 36 22 8 -6 -20

The numerical scheme is second order accurate and the actual profile is linear so the numerical scheme here gives exact solution because there is no error due to truncation of higher order terms.

6

5

4

3

2

1

TTTTTT

23

Solving 2D Heat Conduction problem using Finite Difference Method

0 0 0 0

0

0

0

100

100

100

100 100 100 100

1 2 3

4 5 6

For each nodes with unknown temperature we can write the discretised equation in the form :

04 =−+++ PSNWE TTTTT

040100 142 =−+++ TTTFor point 1:

For point 2: 040 2513 =−+++ TTTT

For point 3: 0400 362 =−+++ TTT

For point 4: 04100100 415 =−+++ TTT

For point 5: 04100 5246 =−+++ TTTT

For point 6: 041000 635 =−+++ TTT

We have a closed set of 6 equations with 6 unknowns

24

Solving 2D Heat Conduction problem using Finite Difference Method

0 0 0 0

0

0

0

100

100

100

100 100 100 100

1 2 3

4 5 6

040100 142 =−+++ TTT

040 2513 =−+++ TTTT

0400 362 =−+++ TTT

Set of equations:

04100100 415 =−+++ TTT

04100 5246 =−+++ TTTT

041000 635 =−+++ TTT

One can solve the above set of equations by calculator. However we have 6 equations but in the actual problem we might have 1000 equations. As direct Matrix inversion is a computationally expensive process the system of equation is solved by iterative numerical methods. One such method is Gauss Seidel Iteration scheme.

25

Gauss Seidel Iteration 1. Equations should be constructed in the following manner

It is necessary to make sure that coefficients aP, aE, aW, aN, aS follow the “Golden rules” of discretisation. 2. Make an initial guess about the unknowns. 3. Recast the equation in the following manner:

4. Check whether a prescribed convergence criterion is satisfied:

5. If not carry out steps 3 and 4 until convergence.

bTaTa nbnbpP +=∑

)1()1()()( −− ++++= kS

P

SkE

P

EkN

P

NkW

P

W

P

kp T

aaT

aaT

aaT

aa

abT

where k and k -1 are iteration counts and are not exponents. For initial guess k = 0.

ε<− − )1()( kP

kP TT ε is a small number

26

Solving 2D Heat Conduction problem using Finite Difference Method

0 0 0 0

0

0

0

100

100

100

100 100 100 100

1 2 3

4 5 6

421 1004 TTT ++=

51324 TTTT ++=

6234 TTT +=

Set of equations:

2004 154 ++= TTT

1004 2465 +++= TTTT

1004 356 ++= TTT

Initial guess based on the boundary condition & common sense 70,80,50,50,60 )0(

5)0(

4)0(

3)0(

2)0(

1 ===== TTTTT and 50)0(6 =T

5.574/)1008050(4/)100( )0(4

)0(2

)1(1 =++=++= TTT

4.444/)705.5750(4/)( )0(5

)1(1

)0(3

)1(2 =++=++= TTTT

The process is repeated till T6 using most recent values. When improved values of T1 to T6 are obtained the whole process is repeated until convergence

…….

27

Solving 2D Heat Conduction problem using Finite Difference Method

Initial guess 1st iteration 2nd iteration 3rd iteration 4th iteration

57.5 56.6 54.7 54.0

44.4 37.3 35.7 35.1 23.6 21.4 20.7 20.4 81.9 81.4 80.4 80.0 69.0 66.7 65.8 65.4 48.2 47.0 46.6 46.5

Actual solution of discretised equation:

601 =T

502 =T

503 =T

804 =T705 =T506 =T

2.65,7.79,3.20,8.34,6.53 54321 ===== TTTTT and 4.466 =T

It is worth doing a couple of iterations before the convergence

The actual solution of discretised equation may not be the exact solution of the physical problem. In general it is appropriate to solve the problem with finer mesh. If the solution does not change appreciably with finer mesh then grid independence is said to be achieved. At that point the solution can be taken as the exact solution of the problem.

28

E W

N

S

Energy Balance Method

Δx

Δy

P

NQ

EQWQ

SQ

According to Fourier’s law of heat conduction

)1.( xyTTkQ PN

N ∆∆−

= )1.( xyTTkQ PS

S ∆∆−

=; ; )1.( yxTTkQ PE

E ∆∆−

=

)1.( yxTTkQ PW

W ∆∆−

=

q

From energy balance: 0)1.( =∆∆++++ yxqQQQQ WESN

For Δx = Δy = Δ : 04 2 =∆+−+++kqTTTTT PSNWE

Δy / 2

Δx / 2

29

Application of Energy balance method in constructing boundary conditions

Case 1: Boundary with convection

P

S

E W

NQ

Δy / 2

Δx

∝T

WQ EQ

SQ

From energy balance method:

01.2

=

∆∆

++++yxqQQQQ WESN

))(1.( PN TTxhQ −∆= ∝

)1.( xyTTkQ PS

S ∆∆−

=

∆

∆−

= 1.2y

xTTkQ PE

E

∆

∆−

= 1.2y

xTTkQ PW

W

For Δx = Δy = Δ :

0242)2( 2 =∆+

∆

+−∆

+++ ∝ kqT

khT

khTTT PSWE

Δx / 2

h

30

Application of Energy balance method in constructing boundary conditions

Case 2: External corner

P

S

W

NQ

Δy / 2

Δx

∝T

WQEQ

SQ

From energy balance method:

01.4

=

∆∆

++++yxqQQQQ WESN

)(1.2 PN TTxhQ −

∆= ∝

∆

∆−

= 1.2x

yTTkQ PS

S

∆

∆−

= 1.2y

xTTkQ PW

W

For Δx = Δy = Δ :

02

222)( 2 =∆+∆

+

∆

+−+ ∝ kqT

khT

khTT PSW

Δx / 2

)(1.2 PE TTyhQ −

∆= ∝

h Δy

31

Application of Energy balance method in constructing boundary conditions

Case 3: Internal corner

P

S

W

NQ

Δy / 2

Δx ∝T

WQEQ

From energy balance method:

01.4

3=

∆∆

++++yxqQQQQ WESN

∆−

∆+−

∆= ∝ y

TTxkTTxhQ PNPN 1.

2)(1.

2

For Δx = Δy = Δ :

023232)22( 2 =∆+

∆+

∆+−+++ ∝ k

qTk

hTk

hTTTT PWSEN

Δx / 2

∆−

∆+−

∆= ∝ x

TTykTTyhQ PEPE 1.

2)(1.

2

h

SQE Δy

N

)1.( yxTTkQ PW

W ∆∆−

=

)1.( xyTTkQ PS

S ∆∆−

=

32

EXAMPLE

The cross-section of a water-cooled component is illustrated below. The wall is symmetrical about AB. The faces AF and EF are maintained at 200oC whilst face ED is well insulated. For faces BC and CD, which are exposed to water at 20oC, the convective heat transfer coefficient is 500 W/m2K. As shown in the figure a 10 mm square mesh of nodal points is employed in a finite difference analysis of the two-dimensional, steady-state, temperature distribution. The thermal conductivity of the material is 0.25 W/mK. (a) Derive discretised equations for nodes 1 to 7 based on energy balance method. (b) Determine the temperatures at nodes 1 to 7. (c) Calculate the rate of convective heat transfer from the component per metre depth into the page.

A

B C

D E

F

1 2 3

4 5

6 7

30mm

10mm 20mm

30mm

200oC

200oC

h = 500 W/m2K T∞= 20oC

33

Epilogue

You will need to understand this part to understand the rest of the module. Please devote time on this material.

And

Most of the practical heat transfer problems in engineering analysis are solved using the numerical techniques described in this lecture.

Moreover

The numerical techniques described above are also valid for 3D problems

The next lecture will be on solving unsteady (transient) heat conduction problems.

34