Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Lectures on Heat Transfer --
NUMERICAL METHODS IN STEADY
STATE 1D and 2D HEAT
CONDUCTION – Part-II
by
Dr. M. Thirumaleshwa...
Preface
• This file contains slides on NUMERICAL
METHODS IN STEADY STATE 1D and 2D
HEAT CONDUCTION – Part-II.
• The slides...
• It is hoped that these Slides will be useful
to teachers, students, researchers and
professionals working in this field....
References:
• 1. M. Thirumaleshwar: Fundamentals of Heat &
Mass Transfer, Pearson Edu., 2006
• https://books.google.co.in/...
References… contd.
• 4. Incropera , Dewitt, Bergman, Lavine:
Fundamentals of Heat and Mass Transfer, 6th
Ed., Wiley Intl.
...
NUMERICAL METHODS IN STEADY STATE 1D
and 2D HEAT CONDUCTION- Part-II
• Methods of solving a system of
simultaneous, algebr...
Methods of solving a system of
simultaneous, algebraic equations:
• We shall briefly present a few methods:
• Relaxation m...
• (i) Relaxation method:
• This is basically a trial and error solution and does not
require a computer. But, it is practi...
• Then, the ‘Relaxation technique’
consists of the following steps:
• To start with, assume values for x, y
and z.
• Since...
• 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....
• Set up a ‘Relaxation table’ wherein you begin with the
initially assumed values of x, y and z and the resultant
residual...
Direct methods:
• (a) Gaussian elimination method:
• Here, one of the unknowns is eliminated
systematically in each step, ...
• Now, we ‘triangularize’ the given set of
equations by repeated application of three
basic row operations:
• (i) multipli...
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. ...
• Above set of equations is known as
‘triangularized set’ of equations.
• Having obtained the value of z, now back-
substi...
• Coefficients constitute a square matrix
called ‘coefficient matrix’ and the constant
terms are stored in a vector called...
• Above procedure is done by the computer program to
eliminate the terms below the main diagonal of the
augmented matrix.
...
• Last row means that z = 7. Now, back-
substitution is done to get values of y and
x.
• Gaussian elimination method is
co...
• (b) Matrix inversion method:
• In this method, the set of equations is written in
the following matrix form:
[A] [T] = [...
• Matrix inversion is performed generally by
using readily available computer
subroutines.
• In Mathcad, inverse of a matr...
And, A
1
0.132
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...
(iii) Gauss–Siedel iteration method:
• Iteration methods are used when the number
of algebraic equations to be solved is r...
• (ii) Assume guess values for all
unknowns, and from the equations
developed in step (a), compute the
unknowns, each time...
• To illustrate this procedure, let us consider the example
given below. We have a set of equations as follows:
3 x. y 3 z...
• Now, assume guess values for x, y and z.
Say, x = 1, y =1 and, z = 1. These are the
‘zeroth’ iteration values.
• With th...
x 1 y 1 z 1 ....initial guess values
x
y 3 z.
3
i.e. x 1.333= ..with y = 1, z = 1
y
3 x z( )
2
i.e. y 1.667= ....with x = ...
x 1.333 y 1.667 z 1 ....next guess values from previous iteration
x
y 3 z.
3
i.e. x 0.444= ..with y =1.667, z =-1
y
3 x z(...
• A simple Mathcad program to perform the above iteration is shown
below. It does the iteration 100 times. Final values of...
• In the above program:
• LHS defines a vector R.
• On the RHS, there are 10 lines.
• First three lines assign the initial...
• 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...
• i.e. even with only 10 iterations we are very
close to the final result. By the time 20 iterations
are over, solution ha...
• Above program in Mathcad is shown
only to illustrate the procedure of
iterative solution.
• While actually using Mathcad...
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
Aug. 2016 MT/SJEC/M.Tec...
Solution with ‘Engineering Equation
Solver (EES)’
• Above mentioned 3 equations are solved very easily
using EES, as follo...
Aug. 2016 MT/SJEC/M.Tech 35
Accuracy of the solutions:
• Some comments on the accuracy of finite
difference solutions are appropriate:
• We noted earl...
• Therefore, one should start with a
coarse mesh and then gradually
refine it depending upon the accuracy
of final results...
One-dimensional, steady state
conduction in cylindrical systems:
• We shall now develop finite difference
formulation by e...
Aug. 2016 MT/SJEC/M.Tech 39
• Writing an energy balance for the volume
element around node ‘m’, remembering that all
heat flows are into the volume, w...
• 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...
• L is the length of the cylinder and qm is the
heat gen. rate per unit volume for the
elemental volume ( = qg, generally)...
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:
...
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:
...
• At the periphery: i.e. at node ‘M’:
• Here too, finite difference equation is obtained by
applying the energy balance to...
• Simplifying:
1
1
T. 1
1 ∆ r h.
T. ∆ r h.
T.
∆ r( )
2
q
M
.
0 ....(8.34)
In the above eqn., first term is the heat conduc...
One-dimensional, steady state
conduction in spherical systems:
• Consider a solid sphere of radius R in which the
heat flo...
Aug. 2016 MT/SJEC/M.Tech 48
• Writing an energy balance for the volume
element around node ‘m’, remembering
that all heat flows are into the volume, w...
• 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...
• qm is the heat gen. rate per unit volume for
the elemental volume ( = qg, generally).
• Simplifying the above equation, ...
• 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
...
• The nature of relation obtained will depend
upon the boundary condition i.e.
prescribed temperature, prescribed heat
flu...
• 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...
• the third term is the heat generation term.
• qM is the heat generation rate per unit
volume at node ‘M’ ( = qg, general...
Aug. 2016 MT/SJEC/M.Tech 56
Aug. 2016 MT/SJEC/M.Tech 57
Aug. 2016 MT/SJEC/M.Tech 58
Aug. 2016 MT/SJEC/M.Tech 59
Aug. 2016 MT/SJEC/M.Tech 60
Aug. 2016 MT/SJEC/M.Tech 61
Aug. 2016 MT/SJEC/M.Tech 62
Two-dimensional, steady state
conduction in cartesian coordinates:
• Examples where temperature gradients are
significant ...
Two-dimensional, steady state
conduction in cartesian coordinates:
• Then, the nodes are numbered with a
double subscript ...
Aug. 2016 MT/SJEC/M.Tech 65
• We see that there are basically three types of nodes:
internal nodes, surface nodes, and cor...
• Difference equations for different nodes
are written by making an energy balance
for the elemental volume around the nod...
• Difference equations for internal
nodes:
• Consider a typical internal node, Tm,n in
the x-y plane, with unit depth
perp...
• It is observed that heat flows into the node
from all the four directions, i.e. left, right,
up and down.
• In addition,...
Q left Q right Q up Q down ∆ V q g
. 0 .....(8.40)
i.e.
k ∆ y.
Tm 1 n,
Tm n,
∆ x
. k ∆ y.
Tm 1 n,
Tm n,
∆ x
. k ∆ x.
Tm n ...
• 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...
• i.e., when there is no heat generation, and
a square mesh is used in the analysis,
temperature of an internal node is gi...
• 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 vo...
Aug. 2016 MT/SJEC/M.Tech 73
• Example: Develop finite difference
equations for an interior corner node with
convection conditions, using the energy
ba...
i.e.
k ∆y.
Tm 1 n,
Tm n,
∆x
. k
∆y
2
.
Tm 1 n,
Tm n,
∆x
. k ∆x.
Tm n 1,
Tm n,
∆y
. k
∆x
2
.
Tm n 1,
Tm n,
∆y
. q g
3
4
∆x....
Summary of steady state, finite difference
equations for different boundary conditions:
(q = heat flux, h = conv. heat tr....
(4) Node at a plane surface with uniform heat flux (Fig. (8.7,d):
2 Tm 1 n,
. Tm n 1,
Tm n 1,
2 q. ∆x.
k
4 Tm n,
. 0 ....(...
• 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
...
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...
• For this elemental volume, considering unit depth, heat transfers
into the volume are:
• From left surface, there is no ...
• There is no heat generation term in this problem.
• So, heat balance on the elemental volume for node 1 gives:
h
∆ x
2
1...
• 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...
• 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 ...
• For node 7:
• This is a corner node with conduction from left, convection on the
top, insulated on the right, and conduc...
• We use 'solve block'of Mathcad to solve this set of equations.
• Start with guess values for all unknown temperatures an...
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, ...
• Example 8.9: A very long bar of square cross-section
has its four sides held at constant temperatures as
shown in Fig. E...
• There are 9 internal nodes. Difference eqns. for these nodes are
obtained by applying eqn.(8.42), viz.
Tm 1 n,
Tm 1 n,
T...
• By solving these eqns. simultaneously, we get the temperatures at
nodes 1 to 9.
• We use 'solve block'of Mathcad to solv...
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
176.3...
• Comparison with analytical solution:
• Analytical solution for this problem is a little complicated and is given
in term...
• Above eqn. is solved very easily in Mathcad:
• Let us re-define as a function of (x, y), and consider
only 6 terms of th...
Aug. 2016 MT/SJEC/M.Tech 93
• We make following important
observations:
• (i) Even with a crude mesh of 4 x 4, we get
values of temperatures at the no...
• (iii) Numerical method of formulating
difference eqns. by energy balance
method is easy and straight forward, only
labor...
Upcoming SlideShare
Loading in …5
×

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

3,514 views

Published on

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. students in Mechanical Engineering Dept. of St. Joseph Engineering College, Vamanjoor, Mangalore, India, during Sept. – Dec. 2010.
Contents: Methods of solving a system of simultaneous, algebraic equations - 1D steady state conduction in cylindrical and spherical systems - 2D steady state conduction in cartesian coordinates - Problems

Published in: Engineering
  • Be the first to comment

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

  1. 1. Lectures on Heat Transfer -- NUMERICAL METHODS IN STEADY STATE 1D and 2D HEAT CONDUCTION – Part-II by Dr. M. ThirumaleshwarDr. M. Thirumaleshwar formerly: Professor, Dept. of Mechanical Engineering, St. Joseph Engg. College, Vamanjoor, Mangalore, India
  2. 2. 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
  3. 3. • 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
  4. 4. 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.
  5. 5. 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 Aug. 2016 MT/SJEC/M.Tech 5
  6. 6. 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 Aug. 2016 MT/SJEC/M.Tech 6 spherical systems - 2D steady state conduction in cartesian coordinates - Problems
  7. 7. 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 Aug. 2016 MT/SJEC/M.Tech 7 elimination, and (b) Matrix inversion • Iterative methods: e.g. Gauss – Siedel iteration method
  8. 8. • (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: Aug. 2016 MT/SJEC/M.Tech 8 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
  9. 9. • 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 Aug. 2016 MT/SJEC/M.Tech 9 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:
  10. 10. • 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 Aug. 2016 MT/SJEC/M.Tech 10 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.
  11. 11. • 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 Aug. 2016 MT/SJEC/M.Tech 11 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.
  12. 12. 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’. Aug. 2016 MT/SJEC/M.Tech 12 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)
  13. 13. • 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. Aug. 2016 MT/SJEC/M.Tech 13 • (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:
  14. 14. 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`): Aug. 2016 MT/SJEC/M.Tech 14 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
  15. 15. • 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 Aug. 2016 MT/SJEC/M.Tech 15 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.
  16. 16. • 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 Aug. 2016 MT/SJEC/M.Tech 16 combine these two into a single ‘augmented matrix’.
  17. 17. • 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 Aug. 2016 MT/SJEC/M.Tech 17 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
  18. 18. • 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 Aug. 2016 MT/SJEC/M.Tech 18 and ready subroutines are available to solve a set of N linear algebraic equations simultaneously.
  19. 19. • (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. Aug. 2016 MT/SJEC/M.Tech 19 • Solution of this system by matrix inversion method is given by: [T] = [A]-1 [B], where [A]-1 is the inverse of matrix [A].
  20. 20. • 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 =. Aug. 2016 MT/SJEC/M.Tech 20 A =. • For the problem illustrated above, we have: A 1 1 3 2 4 1 3 1 1 B 33 11 18
  21. 21. And, A 1 0.132 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 Aug. 2016 MT/SJEC/M.Tech 21 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.
  22. 22. (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 Liebmann iteration) method is one of the most popular Aug. 2016 MT/SJEC/M.Tech 22 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
  23. 23. • (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 Aug. 2016 MT/SJEC/M.Tech 23 unknowns in each equation • (iii) Repeat this procedure until the successive values of an unknown converge to a specified accuracy.
  24. 24. • 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. Aug. 2016 MT/SJEC/M.Tech 24 Now, write each eqn. for one of the unknowns. i.e. x y 3 z. 3 y 3 x z( ) 2 z 2 2 x. y
  25. 25. • 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. Aug. 2016 MT/SJEC/M.Tech 25 latest values of unknowns as available. • So, after ‘first’ iteration we have:
  26. 26. x 1 y 1 z 1 ....initial guess values x y 3 z. 3 i.e. x 1.333= ..with y = 1, z = 1 y 3 x z( ) 2 i.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 Aug. 2016 MT/SJEC/M.Tech 26 Now, for the ‘second’ iteration, continue the procedure, with the latest values of unknowns. We get:
  27. 27. x 1.333 y 1.667 z 1 ....next guess values from previous iteration x y 3 z. 3 i.e. x 0.444= ..with y =1.667, z =-1 y 3 x z( ) 2 i.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 Aug. 2016 MT/SJEC/M.Tech 27 -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.
  28. 28. • 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 Aug. 2016 MT/SJEC/M.Tech 28 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.
  29. 29. • 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, Aug. 2016 MT/SJEC/M.Tech 29 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.
  30. 30. • 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 Aug. 2016 MT/SJEC/M.Tech 30 iterations R 1.755 2.718 1.208 = iterations R 1.987 2.982 1.008 = After 20 iterations R 2 3 1 = After 50 iterations R 2 3 1 = After 100 iterations R 2 3 1 =
  31. 31. • 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 Aug. 2016 MT/SJEC/M.Tech 31 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).
  32. 32. • 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 Aug. 2016 MT/SJEC/M.Tech 32 also follows an iterative algorithm), as follows:
  33. 33. 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 Aug. 2016 MT/SJEC/M.Tech 33 Find x y, z,( ) 2 3 1 = You may put any guess value to start with; it makes no difference on the final result. However, it is essential that each unknown is assigned some guess value to start with.
  34. 34. 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 Aug. 2016 MT/SJEC/M.Tech 34 • 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:
  35. 35. Aug. 2016 MT/SJEC/M.Tech 35
  36. 36. 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 Aug. 2016 MT/SJEC/M.Tech 36 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.
  37. 37. • 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 Aug. 2016 MT/SJEC/M.Tech 37 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.
  38. 38. 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). Aug. 2016 MT/SJEC/M.Tech 38 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.
  39. 39. Aug. 2016 MT/SJEC/M.Tech 39
  40. 40. • 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 Aug. 2016 MT/SJEC/M.Tech 40 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
  41. 41. • 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. Aug. 2016 MT/SJEC/M.Tech 41 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.
  42. 42. • 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: 1 1 2 m. Tm 1 . 2 Tm . 1 1 2 m. Tm 1 . ∆ r( ) 2 q m . k 0 .....(8.32) Aug. 2016 MT/SJEC/M.Tech 42 2 m 2 m k Eqn. (8.32) is the finite difference eqn. for internal nodes i.e. for nodes 1, 2, ….(M-1), with constant thermal conductivity and internal heat generation.
  43. 43. 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 Aug. 2016 MT/SJEC/M.Tech 43 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 . k 0 ......(8.33) Eqn. (8.33) gives the finite difference eqn. for the centre node ‘0’.
  44. 44. 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 Aug. 2016 MT/SJEC/M.Tech 44 In the above, first term is the heat conduction rate from 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 . k 0 ......(8.33) Eqn. (8.33) gives the finite difference eqn. for the centre node ‘0’.
  45. 45. • 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 Aug. 2016 MT/SJEC/M.Tech 45 from the periphery is with an ambient at temperature Ta with 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. qM . 0
  46. 46. • Simplifying: 1 1 T. 1 1 ∆ r h. T. ∆ r h. T. ∆ r( ) 2 q M . 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 the convective heat transfer between the periphery and the ambient, and the third term is the heat generation term. Aug. 2016 MT/SJEC/M.Tech 46 1 1 2 M. TM 1 . 1 1 2 M. ∆ r h. k TM . ∆ r h. k T a . M 2 k. 0 ....(8.34) Eqn. (8.34) gives the finite difference eqn. for the boundary node ‘M’, with convection conditions, constant thermal conductivity and internal heat generation.
  47. 47. 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 Aug. 2016 MT/SJEC/M.Tech 47 • 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.
  48. 48. Aug. 2016 MT/SJEC/M.Tech 48
  49. 49. • 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. ∆ r 2 . k. Tm 1 Tm ∆ r 4 π. m ∆ r. ∆ r 2 . k. 4 π. m ∆ r.( ) 2 . ∆ r. q m . 0 ....(8.35) Aug. 2016 MT/SJEC/M.Tech 49 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)’.
  50. 50. • 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. Aug. 2016 MT/SJEC/M.Tech 50 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.
  51. 51. • qm is the heat gen. rate per unit volume for the elemental volume ( = qg, generally). • Simplifying the above equation, we get: 1 1 2 m. 2 Tm 1 Tm . 1 1 2 m. 2 Tm 1 Tm . ∆ r( ) 2 q m . k 0 .....(8.36) Aug. 2016 MT/SJEC/M.Tech 51 Eqn. (8.36) gives finite difference equations for internal nodes, i.e. nodes 1, 2, ….(M-1).
  52. 52. • 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 3 π. ∆ r 2 3 . q 0 . 0 Aug. 2016 MT/SJEC/M.Tech 52 Simplifying, 6 T 1 T 0 . ∆ r( ) 2 q 0 . k 0 ......(8.37) Eqn. (8.37) gives finite difference eqn. for the centre i.e. node ‘0’.
  53. 53. • 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’: Aug. 2016 MT/SJEC/M.Tech 53 • 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 Ta with a heat transfer coeff. of ‘h’.
  54. 54. • 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 . qM . 0 ...(8.38) •In the above, the first term is the heat Aug. 2016 MT/SJEC/M.Tech 54 •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
  55. 55. • 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: 1 1 2 T. 1 1 2 ∆ r h. T. ∆ r h. T. ∆ r( ) 2 q M . 0 ....(8.39) Aug. 2016 MT/SJEC/M.Tech 55 1 1 2 M. TM 1 . 1 1 2 M. ∆ r h. k TM . ∆ r h. k T a . M 2 k. 0 ....(8.39) Eqn. (8.39) gives the difference eqn. for the boundary node ‘M’ when convection conditions prevail at the boundary.
  56. 56. Aug. 2016 MT/SJEC/M.Tech 56
  57. 57. Aug. 2016 MT/SJEC/M.Tech 57
  58. 58. Aug. 2016 MT/SJEC/M.Tech 58
  59. 59. Aug. 2016 MT/SJEC/M.Tech 59
  60. 60. Aug. 2016 MT/SJEC/M.Tech 60
  61. 61. Aug. 2016 MT/SJEC/M.Tech 61
  62. 62. Aug. 2016 MT/SJEC/M.Tech 62
  63. 63. 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 Aug. 2016 MT/SJEC/M.Tech 63 • 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.
  64. 64. 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). Aug. 2016 MT/SJEC/M.Tech 64 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.
  65. 65. Aug. 2016 MT/SJEC/M.Tech 65 • 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).
  66. 66. • 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, Aug. 2016 MT/SJEC/M.Tech 66 • Elemental volumes for the internal node, surface node and corner nodes are shown by dotted lines around the nodes, in the above fig.
  67. 67. • 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 , Aug. 2016 MT/SJEC/M.Tech 67 • 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 .
  68. 68. • 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. Aug. 2016 MT/SJEC/M.Tech 68 g generation rate in the system. • Writing the energy balance, in steady state:
  69. 69. Q left Q right Q up Q down ∆ V q g . 0 .....(8.40) i.e. k ∆ y. Tm 1 n, Tm n, ∆ x . k ∆ y. Tm 1 n, Tm n, ∆ x . k ∆ x. Tm n 1, Tm n, ∆ y . k ∆ x. Tm n 1, Tm n, ∆ y . q g ∆ x. ∆ y. 0 Simplifying, we get, Aug. 2016 MT/SJEC/M.Tech 69 Tm 1 n, 2 Tm n, . Tm 1 n, ∆ x( ) 2 Tm n 1, 2 Tm n, . Tm n 1, ∆ y( ) 2 q g k 0 ....(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).
  70. 70. • 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. k 0 .....(8.42) Eqn.(8.42) is the finite difference eqn. for the internal nodes, with ∆x = ∆y . Aug. 2016 MT/SJEC/M.Tech 70 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)
  71. 71. • 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. Aug. 2016 MT/SJEC/M.Tech 71 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.
  72. 72. • 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. Aug. 2016 MT/SJEC/M.Tech 72 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.
  73. 73. Aug. 2016 MT/SJEC/M.Tech 73
  74. 74. • 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. Aug. 2016 MT/SJEC/M.Tech 74 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)
  75. 75. i.e. k ∆y. Tm 1 n, Tm n, ∆x . k ∆y 2 . Tm 1 n, Tm n, ∆x . k ∆x. Tm n 1, Tm n, ∆y . k ∆x 2 . Tm n 1, Tm n, ∆y . q g 3 4 ∆x. ∆y.. 0 Remembering that ∆x = ∆y =(∆x, say), we get on simplification, Aug. 2016 MT/SJEC/M.Tech 75 Tm n 1, 2 Tm 1 n, . 2 Tm n 1, . Tm 1 n, 6 2 h. ∆x. k Tm n, . 3 2 ∆x( ) 2 . q g k . 2 h. ∆x. k T a . 0 And, if there is no internal heat generation, Tm n 1, 2 Tm 1 n, . 2 Tm n 1, . Tm 1 n, 6 2 h. ∆x. k Tm n, . 2 h. ∆x. k T a . 0 ....(8.44)
  76. 76. 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, 6 2 h. ∆x. k Tm n, . 2 h. ∆x. k T a . 0 ....(8.44) -------------------------------------------------------------------------------------------------- Aug. 2016 MT/SJEC/M.Tech 76 (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. k T a . 2 h ∆x. k 2. Tm n, . 0 ....(8.45) (3) Node at an external corner with convection (Fig. (8.7,c): Tm n 1, Tm 1 n, 2 h. ∆x. k T a . 2 h ∆x. k 1. Tm n, . 0 ....(8.46)
  77. 77. (4) Node at a plane surface with uniform heat flux (Fig. (8.7,d): 2 Tm 1 n, . Tm n 1, Tm n 1, 2 q. ∆x. k 4 Tm n, . 0 ....(8.47) Note: In eqns.(8.42) and (8.44), put h = 0 or q = 0, to get difference equations for an insulated surface or a surface with thermal symmetry. -------------------------------------------------------------------------------------------------- Aug. 2016 MT/SJEC/M.Tech 77
  78. 78. • 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. Aug. 2016 MT/SJEC/M.Tech 78
  79. 79. 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 Aug. 2016 MT/SJEC/M.Tech 79 Nodes are represented by numbers 1, 2,.....7. Elemental volume pertinent 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.
  80. 80. • 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 2 1.. T a T 1 . From right, there is conduction from node 3 through surface a-b: Aug. 2016 MT/SJEC/M.Tech 80 From right, there is conduction from node 3 through surface a-b: i.e. Q right k ∆ y 2 1.. T 3 T 1 ∆ x . From down, there is conduction from node 2 through surface b-c: i.e. Q down k ∆ x 2 1.. T 2 T 1 ∆ y .
  81. 81. • There is no heat generation term in this problem. • So, heat balance on the elemental volume for node 1 gives: h ∆ x 2 1.. T a T 1 . k ∆ y 2 1.. T 3 T 1 ∆ x . k ∆ x 2 1.. 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. Aug. 2016 MT/SJEC/M.Tech 81 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:
  82. 82. • 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, Tm 1 n, 2 h. ∆x. k T a . 2 h ∆x. k 1. Tm n, . 0 ....(8.46) Aug. 2016 MT/SJEC/M.Tech 82 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 be considered is 3/4 volume, g-f-e-b-d-4. Again, we can directly apply eqn. (8.44), viz.
  83. 83. • 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, 6 2 h. ∆x. k Tm n, . 2 h. ∆x. k T 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. Aug. 2016 MT/SJEC/M.Tech 83 • Again, we can directly apply eqn. (8.45), viz. 2 Tm 1 n, . Tm n 1, Tm n 1, 2 h. ∆x. k T a . 2 h ∆x. k 2. 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, we can write: 2 150. T 4 T 6 0.05 T a . 4.05 T 5 . 0 ....(e)
  84. 84. • 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) Aug. 2016 MT/SJEC/M.Tech 84 • Writing the energy balance: k ∆ y 2 1.. T 6 T 7 ∆ x . h ∆ x 2 1.. T a T 7 . 0 k ∆ x 2 1.. 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 7 equations (a) to (g).
  85. 85. • 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 temperatures T 1 50 T 2 50 T 3 50 T 4 50 T 5 50 T 6 50 T 7 50 Given Aug. 2016 MT/SJEC/M.Tech 85 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)
  86. 86. 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 Aug. 2016 MT/SJEC/M.Tech 86 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
  87. 87. • 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. Aug. 2016 MT/SJEC/M.Tech 87
  88. 88. • There are 9 internal nodes. Difference eqns. for these nodes are obtained by applying eqn.(8.42), viz. Tm 1 n, Tm 1 n, Tm n 1, Tm n 1, 4 Tm n, . q g ∆ x( ) 2. k 0 .....(8.42) In the present case, there is no internal heat generation. So, the last term 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) Aug. 2016 MT/SJEC/M.Tech 88 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)
  89. 89. • 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) Aug. 2016 MT/SJEC/M.Tech 89 150 200 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)
  90. 90. 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 Aug. 2016 MT/SJEC/M.Tech 90 153.571 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
  91. 91. • 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 c 2 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: Aug. 2016 MT/SJEC/M.Tech 91 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
  92. 92. • 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 c 2 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 Aug. 2016 MT/SJEC/M.Tech 92 L Now, substitute (x,y) corresponding to different nodes and get the analytical temps. at those nodes immediately:
  93. 93. Aug. 2016 MT/SJEC/M.Tech 93
  94. 94. • 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, Aug. 2016 MT/SJEC/M.Tech 94 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.
  95. 95. • (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. Aug. 2016 MT/SJEC/M.Tech 95

×