NUMERICAL METHODS IN STEADY STATE, 1D and 2D HEAT CONDUCTION- Part-II

Lectures on Heat Transfer --NUMERICAL METHODS IN STEADY STATE 1D and 2D HEAT CONDUCTION – Part-II

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 STEADY STATE 1D and 2D HEAT CONDUCTION – Part-II.

• The slides were prepared while teaching Heat Transfer course to the M.Tech. 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-II, Bookboon, 2013

• http://bookboon.com/en/software-solutions-problems-on-heat-transfer-cii-ebook


NUMERICAL METHODS IN STEADY STATE 1D and 2D HEAT CONDUCTION- Part-II

• Methods of solving a system of simultaneous, algebraic equations - 1D steady state conduction in cylindrical and spherical systems - 2D steady state


spherical systems - 2D steady state conduction in cartesian coordinates -Problems

Methods of solving a system of simultaneous, algebraic equations:

• We shall briefly present a few methods:• Relaxation method• Direct methods: (a) Gaussian

elimination, and (b) Matrix inversion


elimination, and (b) Matrix inversion• Iterative methods: e.g. Gauss – Siedel

iteration method

• (i) Relaxation method:• This is basically a trial and error solution and does not

require a computer. But, it is practicable to use only when the number of equations is small, say, less than 10.

• Consider an example of a set of following three algebraic equations:


a 1 x. b 1 y. c 1 z. 0

a 2 x. b 2 y. c 2 z. 0

a 3 x. b 3 y. c 3 z. 0

• Then, the ‘Relaxation technique’ consists of the following steps:

• To start with, assume values for x, y and z.

• Since the assumed values are certainly likely to be in error, each of


certainly likely to be in error, each of the above equations will not be zero, but equal to some residual values R1, R2 and R3:

• Our aim is to reduce R1, R2 and R3 to zero by suitably varying the assumed

a 1 x. b 1 y. c 1 z. R 1

a 2 x. b 2 y. c 2 z. R 2

a 3 x. b 3 y. c 3 z. R 3


zero by suitably varying the assumed values of x, y and z, by trial and error. This is done systematically, by first setting up a ‘unit change table’. i.e. a table showing the change in the values of residuals for unit change in x, y and z.

• Set up a ‘Relaxation table’ wherein you begin with the initially assumed values of x, y and z and the resultant residuals. Then, start ‘relaxing’ the largest residual by

a 1 x. b 1 y. c 1 z. R 1

a 2 x. b 2 y. c 2 z. R 2

a 3 x. b 3 y. c 3 z. R 3


residuals. Then, start ‘relaxing’ the largest residual by suitably changing the value of x, y or z, taking guidance from the ‘unit table’ already set up.

• Continue the procedure till all the residuals are relaxed to zero.

• This procedure is slow and time consuming and can not be used when the number of equations to be solved is large.

Direct methods:• (a) Gaussian elimination method:• Here, one of the unknowns is eliminated

systematically in each step, and at the end of the elimination process, the last equation involves only one unknown, and then the remaining unknowns are obtained one by one by ‘back substitution’.


obtained one by one by ‘back substitution’.• Consider an example of solving the following 3 algebraic

equations: x 2 y. 3 z. 33= ....(a)

x 4 y. z 11= .... b( )

3 x. y z 18= .....(c)

• Now, we ‘triangularize’ the given set of equations by repeated application of three basic row operations:

• (i) multiplication of a row by a constant • (ii) adding one row to another row, and • (iii) interchange of two rows.


• (iii) interchange of two rows.• In the above, use eqn. (a) to eliminate x

from eqns. (b) and (c), by adding –1 times (a) to (b) and by adding –3 times (a) to (c). We get:

x 2 y. 3 z. 33= ....(a')

6 y. 2 z. 44 ....(b')

5 y. 8 z. 81 ....(c')

Next, eliminate y from eqn. (c )̀ by multiplying eqn. (b )̀ by –5/6 and adding to eqn. (c )̀:


eqn. (b )̀ by –5/6 and adding to eqn. (c )̀:We get:

x 2 y. 3 z. 33=

6 y. 2 z. 44

z 7

• Above set of equations is known as ‘triangularized set’ of equations.

• Having obtained the value of z, now back-substitute in the previous eqn. to get value of y as y =5, and one more ‘back-substitution’ in the preceding eqn. gives


substitution’ in the preceding eqn. gives the value of x as x=2.

• Gaussian elimination method for a system of large number of equations is done with a computer, using matrix notation to represent the equation.

• Coefficients constitute a square matrix called ‘coefficient matrix’ and the constant terms are stored in a vector called ‘right hand side vector’.

• Computation sub-routines normally combine these two into a single


combine these two into a single ‘augmented matrix’.

• Above procedure is done by the computer program to eliminate the terms below the main diagonal of the augmented matrix.

• This results in a matrix of ‘upper diagonal form’. Then, back – substitution is performed by the program systematically to get the solution.

• So, for the above set of equations, the augmented matrix will be: 1 2 3 33


will be: 1

1

3

2

4

1

3

1

1

33

11

18

The ‘upper diagonal’ form of the matrix is obtained as:1

0

0

2

6

0

3

2

1

33

44

7

• Last row means that z = 7. Now, back-substitution is done to get values of y and x.

• Gaussian elimination method is conveniently programmed in a computer and ready subroutines are available to


and ready subroutines are available to solve a set of N linear algebraic equations simultaneously.

• (b) Matrix inversion method:• In this method, the set of equations is written in

the following matrix form:[A] [T] = [B], where

[A] is the coefficient matrix, [T] is the vector of temperatures to be found out, and [B] is the vector of constants (RHS) of the equations.


• Solution of this system by matrix inversion method is given by:

[T] = [A]-1 [B], where [A]-1 is the inverse of matrix [A].

• Matrix inversion is performed generally by using readily available computer subroutines.

• In Mathcad, inverse of a matrix A is obtained in a single step by the command A-1 =.


A =.• For the problem illustrated above, we

have:

A

1

1

3

2

4

1

3

1

1

B

33

11

18

And, A 10.132

0.053

0.342

0.026

0.211

0.132

0.368

0.053

0.158

=

Therefore,

T A 1 B. ...T is the vector containing x, y, z as its elements

2


i.e. T

2

5

7

=

i.e. x = 2, y = 5 and z = 7. This result is the same as that obtained earlier.

Once again, when the number of equations is relatively large,this is not a preferred method, from the point of view of computer memory and storage.

(iii) Gauss–Siedel iteration method:

• Iteration methods are used when the number of algebraic equations to be solved is relatively large.

• Gauss-Siedel iteration (also called Liebmanniteration) method is one of the most popular


iteration methods because of its simplicity.

• The method involves the following steps:• (i) Solve each equation for one of the

unknowns, i.e. write each unknown in terms of other unknowns

• (ii) Assume guess values for all unknowns, and from the equations developed in step (a), compute the unknowns, each time using the most recently computed values for the unknowns in each equation


unknowns in each equation• (iii) Repeat this procedure until the

successive values of an unknown converge to a specified accuracy.

• To illustrate this procedure, let us consider the example given below. We have a set of equations as follows:

3 x. y 3 z. 0= ....(a)

x 2 y. z 3= .... b( )

2 x. y z 2= .....(c)

Now, write each eqn. for one of the unknowns. i.e.


Now, write each eqn. for one of the unknowns. i.e.

xy 3 z.

3

y3 x z( )

2

z 2 2 x. y

• Now, assume guess values for x, y and z. Say, x = 1, y =1 and, z = 1. These are the ‘zeroth’ iteration values.

• With these guess values, begin the iteration and in each equation, use the latest values of unknowns as available.


latest values of unknowns as available. • So, after ‘first’ iteration we have:

x 1 y 1 z 1 ....initial guess values

xy 3 z.

3i.e. x 1.333= ..with y = 1, z = 1

y3 x z( )

2i.e. y 1.667= ....with x = 1.333, z = 1

z 2 2 x. y i.e. z 1= ...with x = 1.333, y = 1.667


Now, for the ‘second’ iteration, continue the procedure, with the latest values of unknowns. We get:

x 1.333 y 1.667 z 1 ....next guess values from previous iteration

xy 3 z.

3i.e. x 0.444= ..with y =1.667, z =-1

y3 x z( )

2i.e. y 1.778= ....with x=-0.444, z=-1

z 2 2 x. y i.e. z 4.667= ...with x=-0.444, y=1.778

For the ‘third’ iteration, take x = -0.444, y = 1.778 and z = -4.667, and continue. This process is programmed easily


-4.667, and continue. This process is programmed easily in a computer and the results normally converge within about 100 iterations.

Of course, we can also instruct the program to stop when the difference between successive values of unknowns converge to a pre-determined accuracy.

• A simple Mathcad program to perform the above iteration is shown below. It does the iteration 100 times. Final values of x, y and z are returned as a vector R.

R x0

1

y0

1

z0 1

xi 1

yi

3 zi

.i 0 100..∈for

2


xi 1 3

yi 1

3 xi 1 zi

2

zi 1 2 2 xi 1. yi 1

xi 1

yi 1

zi 1

R

2

3

1

=And,

i.e. x = 2, y = 3 and z = -1.

• In the above program: • LHS defines a vector R. • On the RHS, there are 10 lines. • First three lines assign the initial guess values

for x, y and z. • Next 4 lines show the ‘ for loop’, for 100

iterations, wherein x, y and z are calculated,


iterations, wherein x, y and z are calculated, each time using the latest available values of unknowns.

• Next 3 lines constitute the latest values of x, y and z which are stored as the elements of the vector R.

• It is interesting to note that in the above program, if iteration is carried out only for 5,10, 20, 50 and 100 loops (by changing the 4th line), following are the results:

After 5 After 10 After 20 After 50 After 100


iterations

R

1.755

2.718

1.208

=

iterations

R

1.987

2.982

1.008

=

After 20iterations

R

2

3

1

=

After 50 iterations

R

2

3

1

=

After 100 iterations

R

2

3

1

=

• i.e. even with only 10 iterations we are very close to the final result. By the time 20 iterations are over, solution has already converged to the final result.

• It is stated that for steady state heat conduction problems, Gauss–Siedel iteration process is inherently stable and always converges into a


inherently stable and always converges into a solution.

• Note: Of course, above program can be further refined to stop when the successive values of x, y and z differ by a pre-determined small value ε. (say, ε = 0.001).

• Above program in Mathcad is shown only to illustrate the procedure of iterative solution.

• While actually using Mathcad, we would use the ‘Solve block’ (which also follows an iterative algorithm), as


also follows an iterative algorithm), as follows:

x 0 y 0 z 0 ...guess values

Given

3 x. y 3 z. 0 ....(a)

x 2 y. z 3 .... b( )

2 x. y z 2 .....(c)

2


Find x y, z,( )

2

3

1

=

You may put any guess value to start with; itmakes no difference on the final result.However, it is essential that each unknown isassigned some guess value to start with.

Solution with ‘Engineering Equation Solver (EES)’

• Above mentioned 3 equations are solved very easily using EES, as follows:

• Enter the 3 equations in the ‘Equations Window’ as shown.

• Press F2


• Press F2• The equations are solved and the results appear in the

‘Solutions Window’• A screen–shot containing both the Equations Window

and the Solutions Window is shown below:

Accuracy of the solutions:• Some comments on the accuracy of finite

difference solutions are appropriate:• We noted earlier that, accuracy improves

as the number of nodes is made larger. • However, this would mean that a larger

number of algebraic equations have to be


number of algebraic equations have to be solved simultaneously.

• This situation has following inherent draw backs: the computer memory required increases and also, more importantly, the round off errors in successive calculations increase since they are cumulative.

• Therefore, one should start with a coarse mesh and then gradually refine it depending upon the accuracy of final results required.

• Note that for the normal problems encountered in practice, a coarse mesh generally gives results of


mesh generally gives results of acceptable accuracy.

• Remember that, anyway, there are uncertainties in the values of thermal properties and heat transfer coefficients available to the designer.

One-dimensional, steady state conduction in cylindrical systems:• We shall now develop finite difference

formulation by energy balance method:• Consider a long, solid cylinder of radius R in

which the heat flow is only in the radial direction. Let the rate of internal heat generation be qg(W/m3).


g(W/m3).

• The region from r = 0 to r = R is divided into M sub-regions, each of thickness ∆r = R/M.

• Therefore, there are (M + 1) nodes, numbered as 0, 1, 2,….M.

• See Fig. 8.5.

• Writing an energy balance for the volume element around node ‘m’, remembering that all heat flows are into the volume, we get:

Tm 1 Tm

∆ r

2 π. m ∆ r. ∆ r

2. L. k.

Tm 1 Tm

∆ r

2 π. m ∆ r. ∆ r

2. L. k.

2 π. m. ∆ r. ∆ r.( ) L. q m. 0

Tm 1

Tm

∆ r

2 π. m ∆ r. ∆ r

2. L. k.

Tm 1

Tm

∆ r

2 π. m ∆ r. ∆ r

2. L. k.

2 π. m. ∆ r. ∆ r.( ) L. q m. 0


2 2

First term in the above eqn. is the heat flowing into node ‘m’ from node ‘(m – 1)’. Denominator of the first term is the thermal resistance between ‘m’ and ‘(m – 1)’;

2 2

• It is written in the form (L/k A) where A is the mean area i.e. area of the plane mid-way between nodes ‘m’ and ‘(m – 1)’.

• This form of thermal resistance ( as if for a plane wall), is alright for the cylindrical system when ∆r << R, which is generally the case.


the case. • Second term in the above eqn. is the heat

flowing into node ‘m’ from node ‘(m + 1)’.• The third term gives the heat generated in

the elemental volume.

• L is the length of the cylinder and qm is the heat gen. rate per unit volume for the elemental volume ( = qg, generally).

• Simplifying the above equation, we get:

11

2 m.Tm 1

. 2 Tm. 1

1

2 m.Tm 1

.∆ r( )

2q m

.

k0 .....(8.32)


2 m 2 m k

Eqn. (8.32) is the finite difference eqn. forinternal nodes i.e. for nodes 1, 2, ….(M-1),with constant thermal conductivity andinternal heat generation.

At the centre: i.e. at r = 0:Writing the energy balance for the half-volume(of thickness ∆r/2) around node ‘0’, we get:

T 1 T 0

∆ r

2 π. ∆ r

2. L. k.

π ∆ r

2

2. L. q 0

. 0

In the above, first term is the heat conduction rate from node ‘1’ to node ‘0’ and the second term is the heat


node ‘1’ to node ‘0’ and the second term is the heat generation term.

Simplifying the above equation, we get:

4 T 1 T 0.

∆ r( )2

q 0.

k0 ......(8.33)

Eqn. (8.33) gives the finite difference eqn. for the centre node ‘0’.

At the centre: i.e. at r = 0:Writing the energy balance for the half-volume (of thickness ∆r/2)around node ‘0’, we get:

T 1 T 0

∆ r

2 π. ∆ r

2. L. k.

π ∆ r

2

2. L. q 0

. 0

In the above, first term is the heat conduction rate from node ‘1’ to


In the above, first term is the heat conduction rate from node ‘1’ tonode ‘0’ and the second term is the heat generation term.

Simplifying the above equation, we get:

4 T 1 T 0.

∆ r( )2

q 0.

k0 ......(8.33)

Eqn. (8.33) gives the finite difference eqn. for the centre node ‘0’.

• At the periphery: i.e. at node ‘M’:• Here too, finite difference equation is obtained by

applying the energy balance to the half-volume around node ‘M’.

• Of course, the nature of equation depends on the boundary condition.

• For convection boundary conditions, where heat transfer from the periphery is with an ambient at temperature Ta


from the periphery is with an ambient at temperature Tawith a heat transfer coeff. of h, energy balance around node ‘M’, gives:

TM 1

TM

∆ r

2 π. M ∆ r. ∆ r

2. L. k.

2 π. M ∆ r. L.( ) h. T a T M. 2 π. M. ∆ r. ∆ r

2. L. q

M. 0

• Simplifying:

11

T. 11 ∆ r h.

T. ∆ r h.T.

∆ r( )2

qM

.0 ....(8.34)

In the above eqn., first term is the heat conduction rate from node‘(M-1)’ to node ‘M’ and the second term is theconvective heat transfer between the periphery and theambient, and the third term is the heat generation term.


11

2 M.TM 1

. 11

2 M.∆ r h.

kTM

. ∆ r h.

kT a

. M

2 k.0 ....(8.34)

Eqn. (8.34) gives the finite difference eqn. for theboundary node ‘M’, with convection conditions, constantthermal conductivity and internal heat generation.

One-dimensional, steady state conduction in spherical systems:

• Consider a solid sphere of radius R in which the heat flow is only in the radial direction.

• Let the rate of internal heat generation be qg(W/m3).

• The region from r = 0 to r = R is divided into M


• The region from r = 0 to r = R is divided into M sub-regions, each of thickness ∆r = R/M.

• Therefore, there are (M + 1) nodes, numbered as 0, 1, 2,….M.

• See Fig. 8.5.

• Writing an energy balance for the volume element around node ‘m’, remembering that all heat flows are into the volume, we get:

Tm 1 Tm

∆ r

4 π. m ∆ r. ∆ r2

. k.

Tm 1 Tm

∆ r

4 π. m ∆ r. ∆ r2

. k.

4 π. m ∆ r.( )2. ∆ r. q m

. 0 ....(8.35)


4 π. m ∆ r. ∆ r

2. k. 4 π. m ∆ r. ∆ r

2. k.

First term in the above eqn. is the heat flowing into node ‘m’ from node ‘(m – 1)’. Denominator of the first term is the thermal resistance between ‘m’ and ‘(m – 1)’.

• It is written in the form (L/k A) where A is the mean area i.e. area of the plane mid-way between nodes ‘m’ and ‘(m – 1)’.

• This form of thermal resistance ( as if for a plane wall), is alright for the spherical system when ∆r << R, which is generally the case.


the case. • Second term in the above eqn. is the heat

flowing into node ‘m’ from node ‘(m + 1)’. • The third term gives the heat generated in

the elemental volume.

• qm is the heat gen. rate per unit volume for the elemental volume ( = qg, generally).

• Simplifying the above equation, we get:

11

2 m.

2T

m 1T

m. 1

1

2 m.

2T

m 1T

m.

∆ r( )2

q m.

k0 .....(8.36)


Eqn. (8.36) gives finite difference equationsfor internal nodes, i.e. nodes 1, 2, ….(M-1).

• At the centre, r = 0:• Applying the energy balance to the half-volume

around node ‘0’: T 1 T 0

∆ r

4 π. ∆ r

2

2. k.

4

3π. ∆ r

2

3. q 0

. 0


Simplifying, 6 T 1 T 0.

∆ r( )2

q 0.

k0 ......(8.37)

Eqn. (8.37) gives finite difference eqn. for the centrei.e. node ‘0’.

• The nature of relation obtained will depend upon the boundary condition i.e. prescribed temperature, prescribed heat flux, or convection boundary condition.

• Let us write the difference eqn. for node

For the boundary node ‘M’:


• Let us write the difference eqn. for node “M’ when convection conditions prevail at the boundary:

• Let there be heat transfer at the boundary with a fluid flowing at a temperature of Tawith a heat transfer coeff. of ‘h’.

• Then, writing an energy balance for the half-volume around node ‘M’, we get:

TM 1

TM

∆ r

4 π. M ∆ r. ∆ r

2

2. k.

4 π. M ∆ r.( )2

h. T a T M. 4 π. M ∆ r.( )

2. ∆ r

2. q

M. 0 ...(8.38)

•In the above, the first term is the heat


•In the above, the first term is the heat conduction rate from node ‘(M-1)’ to node ‘M’.•The second term is the convective heat transfer between the outer surface and the ambient, and

• the third term is the heat generation term. • qM is the heat generation rate per unit

volume at node ‘M’ ( = qg, generally). • Simplifying the above equation, we get:

11 2

T. 11 2 ∆ r h.

T. ∆ r h.T.

∆ r( )2

qM

.0 ....(8.39)


11

2 M.TM 1

. 11

2 M.∆ r h.

kTM

. ∆ r h.

kT a

. M

2 k.0 ....(8.39)

Eqn. (8.39) gives the difference eqn. forthe boundary node ‘M’ when convectionconditions prevail at the boundary.

Two-dimensional, steady state conduction in cartesian coordinates:• Examples where temperature gradients are

significant in more than one direction: large chimneys and L–shaped bars etc.

• Consider a two-dimensional system in which


• Consider a two-dimensional system in which temperature gradients are significant in the x and y directions.

• Let the x-y plane be subdivided into rectangular mesh of nodes, with spacing of ∆x and ∆y in x and y directions respectively.

Two-dimensional, steady state conduction in cartesian coordinates:

• Then, the nodes are numbered with a double subscript notation. i.e. a typical node, T m,n is the node with a x-coordinate of (m.∆x) and y-coordinate of (n.∆y).


of (m.∆x) and y-coordinate of (n.∆y). • Node count is m = 0, 1, ..M in the x

direction and n = 0, 1,….N in the y direction. See Fig. 8.6.


• We see that there are basically three types of nodes: internal nodes, surface nodes, and corner nodes, marked 1, 2 and 3 respectively in the fig. 8.6(a).

• Difference equations for different nodes are written by making an energy balance for the elemental volume around the node in question, with all the heat flow lines going into the volume.

• Elemental volumes for the internal node,


• Elemental volumes for the internal node, surface node and corner nodes are shown by dotted lines around the nodes, in the above fig.

• Difference equations for internal nodes:

• Consider a typical internal node, Tm,n in the x-y plane, with unit depth perpendicular to the plane of paper, as shown in Fig. 8.6(b).

• It is surrounded by 4 nodes: T , T ,


• It 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 .

• It is observed that heat flows into the node from all the four directions, i.e. left, right, up 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 volumetric heat generation rate in the system.


ggeneration rate in the system.

• Writing the energy balance, in steady state:

Q left Q right Q up Q down ∆ V q g. 0 .....(8.40)

i.e.

k ∆ y.T

m 1 n, Tm n,

∆ x. k ∆ y.

Tm 1 n, T

m n,

∆ x. k ∆ x.

Tm n 1, T

m n,

∆ y. k ∆ x.

Tm n 1, T

m n,

∆ y. q g ∆ x. ∆ y. 0

Simplifying, we get,


Tm 1 n, 2 T

m n,. T

m 1 n,

∆ x( )2

Tm n 1, 2 T

m n,. T

m n 1,

∆ y( )2

q g

k0 ....(8.41)

Eqn. (8.41) gives the difference eqn. for internal nodes, i.e. for m = 1,2,…(M-1), and n = 1,2,….(N-1).

• Now, generally a square mesh is used i.e. ∆x = ∆y = (∆x , say). Then, the eqn.(8.41) simplifies to:

Tm 1 n, Tm 1 n, Tm n 1, Tm n 1, 4 Tm n,.

q g ∆x( )2.

k0 .....(8.42)

Eqn.(8.42) is the finite difference eqn. for theinternal nodes, with ∆x = ∆y .


internal nodes, with ∆x = ∆y .

When there is no heat generation in the body, the difference eqn. for the node reduces to:

Tm n,

Tm 1 n, Tm 1 n, Tm n 1, Tm n 1,

4.....(8.43)

• i.e., when there is no heat generation, and a square mesh is used in the analysis, temperature of an internal node is given as the arithmetic average of the surrounding four temperatures.

• Difference equations for boundary nodes:• Boundary nodes may be on the surface or

on the corners.


on the corners.• Difference equations are developed for

boundary nodes in a similar manner as for interior nodes, i.e. by making an energy balance on the elemental volume surrounding the node.

• In Fig. 8.6(a), we can see that the surface node 2 is surrounded by a half-volume and the corner node 3 has a quarter volume attached to it.

• Exact form of the difference equation will depend upon the boundary conditions i.e.


depend upon the boundary conditions i.e. prescribed temperature, prescribed heat flux, insulated, convection or radiation boundary conditions.

• Fig. (8.7) shows some common boundary conditions encountered in practice.

• Example: Develop finite difference equations for an interior corner node with convection conditions, using the energy balance method. See Fig. 8.7(a).

• As shown in the fig., elemental volume around the node is ¾ of full volume.


around the node is ¾ of full volume. Writing an energy balance for this volume, we apply eqn. (8.40):

Q left Q right Q up Q down ∆V q g. 0 .....(8.40)

i.e.

k ∆y.T

m 1 n, Tm n,

∆x. k

∆y

2.

Tm 1 n, T

m n,

∆x. k ∆x.

Tm n 1, T

m n,

∆y. k

∆x

2.

Tm n 1, T

m n,

∆y. q g

3

4∆x. ∆y.. 0

Remembering that ∆x = ∆y =(∆x, say), we get onsimplification,


Tm n 1, 2 Tm 1 n,. 2 Tm n 1,

. Tm 1 n, 62 h. ∆x.

kTm n,. 3

2∆x( )

2.q g

k. 2 h. ∆x.

kT a. 0

And, if there is no internal heat generation,

Tm n 1, 2 Tm 1 n,. 2 Tm n 1,

. Tm 1 n, 62 h. ∆x.

kTm n,. 2 h. ∆x.

kT a. 0 ....(8.44)

Summary of steady state, finite difference equations for different boundary conditions:(q = heat flux, h = conv. heat tr. coeff., k = thermal cond., no int. heat gen., and ∆∆∆∆x = ∆∆∆∆y)

(1) Node at an internal corner with convection (Fig. (8.7,a):

Tm n 1, 2 Tm 1 n,. 2 Tm n 1,

. Tm 1 n, 62 h. ∆x.

kTm n,. 2 h. ∆x.

kT a. 0 ....(8.44)

--------------------------------------------------------------------------------------------------


(2) Node at a plane surface with convection (Fig. (8.7,b):

2 Tm 1 n,. Tm n 1, Tm n 1,

2 h. ∆x.

kT a. 2

h ∆x.

k2. Tm n,

. 0 ....(8.45)

(3) Node at an external corner with convection (Fig. (8.7,c):

Tm n 1, T

m 1 n,2 h. ∆x.

kT a. 2

h ∆x.

k1. T

m n,. 0 ....(8.46)

(4) Node at a plane surface with uniform heat flux (Fig. (8.7,d):

2 Tm 1 n,

. Tm n 1, T

m n 1,2 q. ∆x.

k4 T

m n,. 0 ....(8.47)

Note: In eqns.(8.42) and (8.44), put h = 0 or q = 0, to get differenceequations for an insulated surface or a surface with thermal symmetry.

--------------------------------------------------------------------------------------------------


• Example 8.8: For the two-dimensional region shown in Fig. Ex. 8.8, with constant k ( = 20 W/(m.C)) and no internal heat generation, and with the indicated boundary conditions, formulate the finite difference equations and solve for unknown temperatures. Use ∆∆∆∆x = ∆∆∆∆y = 1 cm.


Data:

∆ x 0.01 m

∆ y 0.01 m

T a 20 C.....ambient temp.

h 50 W/(m2.C).....heat transfer coeff.

k 20 W/(m.C)....thermal cond.

Nodes are represented by numbers 1, 2,.....7. Elemental volume


Nodes are represented by numbers 1, 2,.....7. Elemental volumepertinent to each node is also marked around it and numbered a, b,....r.

For node 1:Elemental volume to be considered is 1/4 volume, 1-a-b-c-1.

• For this elemental volume, considering unit depth, heat transfers into the volume are:

• From left surface, there is no heat transfer, since it is insulated.

i.e. Q left 0

From top, i.e surface 1-a: there is convection:

i.e. Q top h∆ x

21.. T a T 1

.

From right, there is conduction from node 3 through surface a-b:


From right, there is conduction from node 3 through surface a-b:

i.e. Q right k∆ y

21..

T 3 T 1

∆ x.

From down, there is conduction from node 2 through surface b-c:

i.e. Q down k∆ x

21..

T 2 T 1

∆ y.

• There is no heat generation term in this problem.• So, heat balance on the elemental volume for node 1 gives:

h∆ x

21.. T a T 1

. k∆ y

21..

T 3 T 1

∆ x. k

∆ x

21..

T 2 T 1

∆ y. 0

i.e. 0.25 T a T 1. 10 T 3 T 1

. 10 T 2 T 1. 0 ....(a)

Eqn. (a) is the difference eqn. for node 1.


For node 2:Here, elemental volume to be considered is 1/2 volume, c-b-e-r. and, energy balance can be written as we did for node 1.However, since the surface is insulated, it is easier to use the mirror image concept and consider the node 2 as an internal node. So, to the left of node 2, we have T4, mirror image of temp. of node 4. Then, considering 2 as internal node, we get difference eqn for node 2:

• For node 3:• This is a corner node with convection . Elemental volume to be

considered is 1/4 volume, a-3-d-b.• We can directly apply eqn. (8.46), viz.

T 4 T 1 T 4 150 4 T 2. 0 .....(b)

Tm n 1, T

m 1 n,2 h. ∆x.

kT a. 2

h ∆x.

k1. T

m n,. 0 ....(8.46)


k k

i.e. T 4 T 1 0.05 T a. 2.05 T 3

. 0 ....(c)

For node 4:

This is an internal corner node with convection . Elemental volume to beconsidered is 3/4 volume, g-f-e-b-d-4.

Again, we can directly apply eqn. (8.44), viz.

• For node 5:• This is a surface node with convection . Elemental volume to be

considered is 1/2 volume, g-f-i-h-g.

Tm n 1, 2 Tm 1 n,. 2 Tm n 1,

. Tm 1 n, 62 h. ∆x.

kTm n,. 2 h. ∆x.

kT a. 0 ....(8.44)

i.e. 150 2 T 2. 2 T 3

. T 5 6.05 T 4. 0.05 T a

. 0 .....(d)

Eqn.(d) is the difference eqn. for node 4.


• Again, we can directly apply eqn. (8.45), viz.

2 Tm 1 n,. Tm n 1, Tm n 1,

2 h. ∆x.

kT a. 2

h ∆x.

k2. Tm n,

. 0 ....(8.45)

Remembering that eqn. (8.45) was developed for a vertical surface,and in the present case, we are dealing with a horizontal surface, wecan write:

2 150. T 4 T 6 0.05 T a. 4.05 T 5

. 0 ....(e)

• For node 7:• This is a corner node with conduction from left, convection on the

top, insulated on the right, and conduction from down. Elemental volume to be considered is 1/4 volume, k-7-p-j.

• Writing the energy balance:

For node 6:This is identical to node 5. So, we get:

2 150. T 5 T 7 0.05 T a. 4.05 T 6

. 0 ....(f)


• Writing the energy balance:

k∆ y

21..

T 6 T 7

∆ x. h

∆ x

21.. T a T 7

. 0 k∆ x

21..

150 T 7

∆ y. 0

i.e. 10 T 6 T 7. 0.25 T a T 7

. 10 150 T 7. 0 ....(g)

Temperatures at nodes 1 to 7 are obtained by simultaneously solving 7equations (a) to (g).

• We use 'solve block' of Mathcad to solve this set of equations. • Start with guess values for all unknown temperatures and

immediately below 'Given', type the constraint equations. Then, the command 'Find(T1,...T7)' gives the temperatures immediately:

...guess values of temperaturesT 1 50 T 2 50 T 3 50 T 4 50 T 5 50 T 6 50 T 7 50

Given


0.25 T a T 1. 10 T 3 T 1

. 10 T 2 T 1. 0 ....(a)

T 4 T 1 T 4 150 4 T 2. 0 .....(b)

T 4 T 1 0.05 T a. 2.05 T 3

. 0 ....(c)

150 2 T 2. 2 T 3

. T 5 6.05 T 4. 0.05 T a

. 0 .....(d)

2 150. T 4 T 6 0.05 T a. 4.05 T 5

. 0 ....(e)

2 150. T 5 T 7 0.05 T a. 4.05 T 6

. 0 ....(f)

10 T 6 T 7. 0.25 T a T 7

. 10 150 T 7. 0 ....(g)

Temp Find T 1 T 2, T 3, T 4, T 5, T 6, T 7, ....node temps. are stored in vector 'Temp'.

i.e. Temp

138.552

142.929

137.139

141.582=

T1

138.552 C

T2

142.929 C

T 137.139 C


i.e. Temp 141.582

145.437

146.438

146.636

=

i.e. The node temperatures are:

T3

137.139 C

T4

141.582 C

T5 145.437 C

T6 146.438 C

T7

146.636 C

• Example 8.9: A very long bar of square cross-section has its four sides held at constant temperatures as shown in Fig. Ex. 8.9. Determine the temperatures at the internal nodes. Compare the results with analytical solution.


• There are 9 internal nodes. Difference eqns. for these nodes are obtained by applying eqn.(8.42), viz.

Tm 1 n, T

m 1 n, Tm n 1, T

m n 1, 4 Tm n,

.q g ∆ x( )

2.

k0 .....(8.42)

In the present case, there is no internal heat generation. So, the lastterm of the above eqn. will be zero. Therefore, we get:

Node 1: 150 200 T 2 T 4 4 T 1. 0 .....(a)

Node 2: T 1 200 T 3 T 5 4 T 2. 0 .....(b)


1 3 5 2

Node 3: T 2 200 150 T 6 4 T 3. 0 .....(c)

Node 4: 150 T 1 T 5 T 7 4 T 4. 0 .....(d)

Node 5: T 4 T 2 T 6 T 8 4 T 5. 0 .....(e)

Node 6: T 5 T 3 150 T 9 4 T 6. 0 .....(f)

Node 7: 150 T 4 T 8 150 4 T 7. 0 .....(g)

Node 8: T 7 T 5 T 9 150 4 T 8. 0 .....(h)

Node 9: T 8 T 6 150 150 4 T 9. 0 .....(i)

• By solving these eqns. simultaneously, we get the temperatures at nodes 1 to 9.

• We use 'solve block' of Mathcad to solve this set of equations. • Start with guess values for all unknown temperatures and immediately

below 'Given', type the constraint equations. Then, the command 'Find(T1,...T9)' gives the temperatures immediately:

T 1 50 T 2 50 T 3 50 T 4 50 T 5 50...guess values of temperaturesT 6 50 T 7 50 T 8 50 T 9 50

Given

150 200 T 2 T 4 4 T 1. 0 .....(a)


150 200 T 2 T 4 4 T 1. 0 .....(a)

T 1 200 T 3 T 5 4 T 2. 0 .....(b)

T 2 200 150 T 6 4 T 3. 0 .....(c)

150 T 1 T 5 T 7 4 T 4. 0 .....(d)

T 4 T 2 T 6 T 8 4 T 5. 0 .....(e)

T 5 T 3 150 T 9 4 T 6. 0 .....(f)

150 T 4 T 8 150 4 T 7. 0 .....(g)

T 7 T 5 T 9 150 4 T 8. 0 .....(h)

T 8 T 6 150 150 4 T 9. 0 .....(i)

Temp Find T 1 T 2, T 3, T 4, T 5, T 6, T 7, T 8, T 9, ....node temps. are stored in vector 'Temp'.

i.e. Temp

171.429

176.339

171.429

159.375

162.5

159.375

153.571

154.911

153.571

= T1 171.429 C

T2 176.339 C

T3 171.429 C


153.571

i.e. The node temperatures are:

T4 159.375 C

T5 162.5 C

T6 159.375 C

T7 153.571 C

T8 154.911 C

T9 153.571 C

• Comparison with analytical solution:• Analytical solution for this problem is a little complicated and is given

in terms of an infinite series, as follows:

θ θ c2

π.

1

∞

n

1( )n 1 1

n

sinh n π. y

L.

sinh n π. W

L.

. sin n π. x

L..

=

.

Nomenclature for the above eqn. for the present problem is as follows:


Nomenclature for the above eqn. for the present problem is as follows:

θ T 150 ....T = temp. at the desired point; 150 C is the const. temp. on three sides

θ c 200 150 ..temp. difference between the temp. of fourth side and the const. temp. of three sides.

n...no. of terms considered in the infinite series

x, y ....coordinates of the point where temp. is desired

L 2 m...length along x-axis

W 2 m....length along y-axis

• Above eqn. is solved very easily in Mathcad:• Let us re-define � as a function of (x, y), and consider

only 6 terms of the infinite series (n = 6) as shown below, for convenience:

θ x y,( ) θ c2

π.

1

6

n

1( )n 1 1

n

sinh n π. y

L.

sinh n π. W

L.

. sin n π. x

L..

=

. .(A)....define θ as a function of x and y


L

Now, substitute (x,y) corresponding to different nodes and get the analytical temps. at those nodesimmediately:

• We make following important observations:

• (i) Even with a crude mesh of 4 x 4, we get values of temperatures at the nodes very close to the analytical results.

• (ii) Note that the analytical relation to find the temp. at any point is very complicated,


the temp. at any point is very complicated, and to solve it without a computer is rather laborious and time consuming. But, with Mathcad, even this analytical solution is easy to perform.

• (iii) Numerical method of formulating difference eqns. by energy balance method is easy and straight forward, only labor being in solving the set of simultaneous equations. But, with Mathcad, this is also very easy.


NUMERICAL METHODS IN STEADY STATE, 1D and 2D HEAT CONDUCTION- Part-II

Engineering

Transcript of NUMERICAL METHODS IN STEADY STATE, 1D and 2D HEAT CONDUCTION- Part-II