Computational Heat Transfer ME673 Mini Project 2 Revised

8/12/2019 Computational Heat Transfer ME673 Mini Project 2 Revised

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Numerical Heat Transfer ME673

Mini-Project 2

Second Semester 1435

Due date Wednesday April 23rd

Consider the two-dimensional steady state heat conduction in an isotropic medium,

Defined over the domain .Subject to the following boundary conditions:

Where,

1. Obtain an expression for evaluating the boundary condition at x=0.2. Discretize the PDE using a five-point stencil3. Use Gauss-Seidel iterative scheme to obtain the steady-state temperature distribution.

Use the convergence criteria below. Develop a computer code to do the calculations.

4. Plot temperature contours at steady state.5. Investigate over-relaxation on convergence acceleration.

Convergence CriteriaConvergence can be assumed when the convergence parameter, p, is less than .Wherepis defined as,

[ ]

Note: All submitted work should be entirely yours. Include source code in your report.


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In order to evaluate the boundary condition at x = 0, the following condition is used.

(1)

The R.H.S of Eq. (1) will be discretized by using a ghost node/point (a point outside the

domain) as shown inFigure 1.Furhtermore, a second order central difference scheme will be

used on the R.H.S of this equation (see Eq.(2) and Figure 1).

( )

(2)

Then, by means of this point, the final difference equation for the boundary condition

can be constructed in conjunction with the five-stencil FDE of the corresponding PDE that

will be developed later. For this reason, the discussion on the expression for the convection

boundary condition will be postponed until the five-point stencil has been developed in the

next discussion/question. It should be mentioned that in this report the index count starts from

(i , j) = (1 , 1) on the lower-left-corner of the domain.

Figure 1: Showing the points and indices used to develope the expression for the convection boundary

condition, Eq.(1).Note: The total number of nodes is only an example.


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In the next question (question no.2), it is asked to discretize the PDE considered using

a five-point stencil. The PDE is shown below, Eq.(3).

(3)

Figure 2: Showing the points and indices used in developing the FDE of the correspondingPDE, Eq.(3).Note: The total number of nodes in this figure is only an example.

Based onFigure 2,we will develop the five-point stencil finite difference equation (FDE) as

follows.

(4)

(5)


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Substitute Eq.(4),and(5) into Eq.(3)

(6)

The truncation error of the FDE, Eq.(6),is O[,]. Collecting and re-arranging termswe obtain the following.

(7)

In order to simplify, Eq.(7) can be written as shown below.

(8)

where

(9)

Continuing the previous discussion regarding question no.1, i.e., obtaining expression

for evaluation the convection boundary condition, the FDE of Eq.(8) will be utilized together

with Eq.(2).In the following derivation, points and indices inFigure 1 are used.

Solving Eq.(2) for the ghost point, , Eq.(10) will be obtained.

{( ) } (10)

Now, Eq. (8) will be applied on point (i,j) in Figure 1 in order to get the other equation.

Afterwards, Eq.(10) will be substituted into this equation to get:


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{( ) }

(11)

After collecting terms and solving for, , we obtain the final expression, Eq. (12), fortreating the convection boundary condition.

(12)

Next, the above equation are written into FORTRAN code to be solved using

Successive Over Relaxation (SOR) method. The code can be seen at the end of this report. In

order to verify the code, a test was conducted by comparing the results of a simple (low

number of segments) mesh with 2x2 segments (Figure 3) obtained by means of the

FORTRAN code with that of manual calculation. The results show a very good agreement

between both calculation methods as can be seen inTable 1.

Figure 3: Schematic of the mesh used for verification of the FORTRAN code

Table 1: Comparison between the results obtained by manual calculation andFORTRAN code for 2x2 segments mesh

Nodes Manual calculation (C) FORTRAN code (C)

(i,j) = (1,2) 287.9 287.9(i,j) = (2,2) 172 172


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In task no.4 and 5, it is required that the temperature contours be plotted and over-

relaxation on convergence acceleration/speed be investigated. We, firstly, will address the

latter, i.e., the effect of relaxation factor, , on the speed of convergence.Calculations were done for several segments of mesh with various relaxation factors

() as can be seen in theTable 2.It can be seen clearly from the table that increasing the to

a value of greater than one will increase the convergence speed. This founding is consistent

with the rule that . However, it was also found that there is a limit whereincreasing beyond that value will decrerase the convergence speed. This can be seen in thetable for 10x10, 20x20, and 200x200 segments. Hence, an optimum value ofexists. Also,it is found that at the calculation is not converging even up to 500 iterations for 20x20mesh.

Table 2: The various calculation parameter settings

Next, the contour plot of the temperature is presented with constant omega ( but changing the number of segments. FromFigure 4 -Figure 7,it can be observed that the

Number of segments Omega No. of iterations

1 95

1.2 65

1.4 41

1.6 21

1.8 52

1 321

1.2 225

1.4 151

1.6 90

1.8 55

1.9 121

2Still not converge

for 500 iterations

50 x 50 1.6 470100 x 100 1.6 1528

1.8 2453

1.95 697

1.97 401

1.98 630

1.99 1201

10 x 10

20 x 20

200 x 200


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trend of the temperature distribution for all number of segments presented are the same. The

difference is only that the least number of segments (10x10) gives coarser temperature

contour. It becomes smoother when the number of segments increased.

Figure 4: Temperature contour plot for mesh of 10x10 segments



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The FORTRAN Code

PROGRAM Steady_2D_heat_equation

IMPLICIT NONE

INTEGER :: i,j,iter,m,n,max_iter

!m is number of segments in x direction

!n is number of segments in y direction grid points

!iter is for iteration number (count from 0)

!max_iter is the possible maximum iteration

REAL, PARAMETER :: h=250.d0, Tamb=300.d0, k=5.d0

!h is the convection HTC (W/m^2.C)

!Tamb is the ambient temperature (C)

!k is the thermal conductivity of the material (W/m.C)

REAL, PARAMETER :: T_bottom = 50.d0, T_right = 200.d0, T_top = 150.d0

!The Ts above are the temp. of boundaries

REAL :: xsize, ysize, p, omega

!xsize is size of the domain in x dir

!ysize is size of the domain in y dir

!omega is the over/under relaxation factor

REAL :: dx, dy, a, b, c

!dx is the size of the segment in x direction

!dy is the size of the segment in y direction


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DOUBLE PRECISION, ALLOCATABLE :: T(:,:), T_old(:,:)

!T is the temperature (C). It is designated to be

!allocatable, i.e., 2D matrix with adjustable size.

WRITE(*,10,ADVANCE='no')

READ(*,*) xsize


READ(*,*) ysize


READ(*,*) m


READ(*,*) n


READ(*,*) omega


READ(*,*) max_iter

WRITE(*,*)

WRITE(*,*) 'The values that you have entered are:'

WRITE(*,*)

WRITE(*,*) 'x =', xsize

WRITE(*,*) 'y =', ysize

WRITE(*,*) 'm =', m

WRITE(*,*) 'n =', n

WRITE(*,*) 'omega =', omega

WRITE(*,*) 'max_iter =', max_iter

WRITE(*,*)

!Calculate dx, dy, a, b, and c

dx = xsize/m

dy = ysize/n

a = 1/(dy*dy)


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b = 1/(dx*dx)

c = -2*(a + b)

WRITE(*,*) 'Thus, the value of dx, dy, a, b, and c obtained are as follows'

WRITE(*,*)

WRITE(*,*) 'dx=',dx

WRITE(*,*) 'dy=',dy

WRITE(*,*) 'a=',a

WRITE(*,*) 'b=',b

WRITE(*,*) 'c=',c

WRITE(*,*)

WRITE(*,*) 'Where a = 1/dx^2, b = 1/dy^2, and c = -2(a+b)'

WRITE(*,*)

OPEN (unit = 1 , file = "result_2D_steady_cond")

WRITE(1,*) 'The values that you have entered are:'

WRITE(1,*)

WRITE(1,*) 'x =', xsize

WRITE(1,*) 'y =', ysize

WRITE(1,*) 'm =', m

WRITE(1,*) 'n =', n

WRITE(1,*) 'omega =', omega

WRITE(1,*) 'max_iter =', max_iter

WRITE(1,*)

WRITE(1,*) 'Thus, the value of dx, dy, a, b, and c obtained are as follows'

WRITE(1,*)

WRITE(1,*) 'dx=',dx

WRITE(1,*) 'dy=',dy

WRITE(1,*) 'a=',a

WRITE(1,*) 'b=',b

WRITE(1,*) 'c=',c

WRITE(1,*)


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WRITE(1,*) 'Where a = 1/dx^2, b = 1/dy^2, and c = -2(a+b)'

WRITE(1,*)

!Allocate the size of T and T_old

ALLOCATE(T(0:m+1,0:n+1))

ALLOCATE(T_old(0:m+1,0:n+1))

!initialize

T(:,:) = 0.d0

T_old(:,:) = 0.d0

!Define the boundary conditions

T(:,1) = T_bottom

T(:,n+1) = T_top

T(m+1,:) = T_right

!Calculate the Temperature distribution

iter = 0 !iteration counter

p = 1 !first guess for p (any number > 1e-5)

DO WHILE (p >= 1.d-5 .AND. iter


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T(i,j) = T(i,j) + omega*(1/c*(-a*T(i,j-1) - b*T(i-1,j) - b*T_old(i+1,j) -

b*T_old(i,j+1)) - T(i,j))

ENDIF

ENDDO

ENDDO

p = ABS(MAXVAL(T(1:n,2:m)-T_old(1:n,2:m))/MAXVAL((T(1:n,2:m))))

iter=iter+1

ENDDO

WRITE(1,*)'The temperature distribution:'

WRITE(1,*)

WRITE(1,*)' i=1 i=2 i=3 ...(so on if available)'

DO j=1,n+1

WRITE(1,22,ADVANCE='no') 'j =',j

WRITE(1,22,ADVANCE='no')' '

DO i=1,m+1

WRITE(1,21,ADVANCE='no') T(i,j)

ENDDO

WRITE(1,*)

ENDDO

WRITE(1,*)

WRITE(1,*) 'The residual =', p

WRITE(1,*)

WRITE(1,*) 'Number of iteration =', iter

10 FORMAT('The size of the domain in x direction, x = ')


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11 FORMAT('The size of the domain in y direction, y = ')

12 FORMAT('The number of segments in x direction, m = ')

13 FORMAT('The number of segments in y direction, n = ')

14 FORMAT('Maximum iteration, max_iter = ')

15 FORMAT('Over/under relaxation factor, omega = ')

21 FORMAT(F6.1)

!FORMAT no.21 is to let the value using this format to be Real decimal form (for 'F') and

!the last digit is written at the 6th space (for '6') and consist of 1 decimal number after point

(for .1)

22 FORMAT(A3,I4)

END PROGRAM Steady_2D_heat_equation

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Computational Heat Transfer ME673 Mini Project 2 Revised