Mandeep Singh

38
[Project#02: 2D Steady State Heat Conduction] [MAE] [542] [Engineering Applications of Computational Fluid Dynamics] [ 22 nd March 2011] By Mandeep Singh Person # 3721 2672

Transcript of Mandeep Singh

Page 1: Mandeep Singh

[Project#02: 2D Steady State Heat

Conduction]

[MAE]

[542] [Engineering Applications of Computational Fluid Dynamics]

[ 22nd March 2011]

By Mandeep Singh

Person # 3721 2672

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Contents 1 Introduction .................................................................................................................................... 4

2 Problem definition / Problem statement ....................................................................................... 4

3 Method of Solution : ...................................................................................................................... 6

4 Discussion of results ....................................................................................................................... 8

4.1 Temperature at the steady state(for mesh size 11 x 6): ........................................................ 8

4.2 Heat balance (Top Vs Bottom) ............................................................................................. 10

4.3 Optimum Relaxation parameter .......................................................................................... 11

4.4 Effect of finer Mesh(for mesh size 21 x 11) :........................................................................ 12

4.4.1 Temperature at the steady state(for mesh size 21 x 11) ............................................. 12

4.4.2 Heat balance (Top Vs Bottom) for Mesh size 21 x 11 .................................................. 14

4.4.3 Optimum Relaxation parameter for finer mesh ........................................................... 15

4.5 Effect of Lower Conductivity(k = 5 5 W/m-oC) : .................................................................. 16

4.5.1 Temperature at the steady state(for mesh size 11 x 6) : ............................................. 16

4.5.2 Heat balance (Top Vs Bottom) for Mesh size 21 x 11 .................................................. 19

4.5.3 Optimum Relaxation parameter for reduced conductivity (K=5W/m- °C)................... 20

4.6 Design Considerations for the Combustion chamber .......................................................... 20

5 Summary and Conclusion ............................................................................................................. 22

6 Appendix....................................................................................................................................... 23

6.1 Matlab Codes written for solving the iterations .................................................................. 23

6.2 Boundary Condition Calculations ......................................................................................... 27

6.2.1 Boundary 1 (Top left corner) ........................................................................................ 27

6.2.2 Boundary 2 (Top Right corner) ..................................................................................... 28

6.2.3 Boundary 3 (Bottom Right corner) ............................................................................... 30

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6.2.4 Boundary 4 (Bottom Left corner) ................................................................................. 31

6.2.5 Boundary 5 ( Left Adiabatic Wall ) ................................................................................ 32

6.2.6 Boundary 6 ( Top convective boundary wall ) .............................................................. 33

6.2.7 Boundary 7 ( Right hand Side Adiabatic boundary wall ) ............................................. 34

6.2.8 Boundary 8 ( Bottom wall , left hand Side Adiabatic boundary wall ) ......................... 35

6.2.9 Boundary 9 ( Bottom wall , Right hand Side convective boundary wall ) .................... 36

6.2.10 Boundary 10 ( Bottom wall , MidPoint ) ....................................................................... 37

6.3 Heat Calculation methodology ............................................................................................. 38

7 References .................................................................................................................................... 38

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1 Introduction In this project we study the basic 2-D Steady state heat conduction (Laplace)

equation. In steady state condition the derivative of the intensive quantity –

Temperature does not depend on the time variable. Hence the rate of change of

temperature with respect to time is zero for the steady state. Generally the steady

state conditions are achieved in those systems where the boundary conditions

remain stagnant with their parameter that leads to a steady state condition of the

system.

The 2D- Steady state equation is given by equation

0yT

xT

2

2

2

2

=∂∂

+∂∂

(1)

where T is the Temperature on the x-y Cartesian plane.

2 Problem definition / Problem statement

In this project we need to solve system which is has steady state temperature

distribution. We have a 2D plate. The length and width of the plate is 10cm and

20cm respectively. On the top of the plate we have Hot gas which is at

temperature(Tg) of 2000 °C and convection co-efficient (hg) 1000 W/m2-°C. The

side walls and the bottom left half of the system have adiabatic boundaries. The

bottom right half of the system have the cooling arrangement with the coolant

temperature maintained at (T1) 60°C and its convection coefficient (h) 8000 W/m2-

°C. Fig 1 illustrates the the boundary in detail.

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Figure 1 (Combustion chamber arrangement)

We need to solve the given system using Gauss-Siedel iteration algorithm with Point

Successive Over Relaxation (PSOR) method. The thermal conductivity of the metal (K) is

20W/ m2-°C .

We have to perform the following cases to analyze the system behavior:

1. By computing the steady state temperature distribution using L2 norm with an error

tolerance of 1 x 10-6 .

2. By computing the rate of heat transfer at the top and compare it to the heat removed

by the coolant at the bottom.

3. Understanding the convergence criteria by varying the relaxation parameter and

estimating the value of optimum relaxation parameter.

4. Effect of changing the mesh size from 11x6 to 21x11(Repeating 1-3).

5. Effect of changing the thermal conductivity of metal by decreasing it to 20W/ m2-°C

(Repeating 1-3).

6. Considerations to be taken while designing the combustion chamber.

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3 Method of Solution :

In this problem we will utilize the Neumann and Dirichlet’s boundary conditions to start our

iteration. Whole system is divided into grid. After that the boundary conditions are applied

on the nodes that are on the surface. After plugging the boundary conditions over the

peripheral nodes, iterations are run using the Gauss- Seidel algorithm till it doesn’t

Gauss Seidel algorithm using PSOR method can be written as

[ ])()1(2

)1( 11,1,

21,1,12

1,

+−+

+−+

+ ++++

+−= kji

kji

kji

kji

kij

kji TTTTTT β

βωω

(2)

Here the function (temperature T) can be calculated for the iteration K+1 , ω is the

relaxation parameter which converges the solution at a faster rate hence increasing the

computation efficiency. We have used Gauss-Seidel PSOR method with five point grid. The

grid used is shown in the figure 2 below

Figure 2(Grid Point arrangement for Gauss-Seidel algorithm)

For solving the boundary conditions we use the ghost nodes technique. This technique can

be understood by the diagram shown in the figure 3.

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Figure 3 (Ghost points)

Here we use the ghost points that are assumed and lie outside the domain boundary. Heat

balance or boundary conditions are used to calculate the value of these nodes. By using the

heat balance equation we calculate the expression for these ghost points and plug these

obtained expressions into the steady state equation. The heat balance equation uses the

temperature and convective coefficients given as the initial and essential boundary

conditions. By use of the ghost points , the boundary conditions gets inserted into the

steady state equation and we run the Gauss Siedel algorithm for getting the desired

temperature distribution.

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4 Discussion of results

4.1 Temperature at the steady state(for mesh size 11 x 6): The computed temperature at the steady state using the L2 norm with an error tolerance of

1 x 10-6 is shown in the Fig 4. The value shows that the temperature drops from the top

edge and attains minimum temperature at the coolant touching boundary.

Figure 4

The matrix for the values of temperature is shown in table 1 on the next page :

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From the Table 1, we can observe that there is a sharp temperature drop on the bottom

edge. On the centre node the temperature is 227°C. At the same time on the left node of

this center node the temperature in 788°C and on the right of it the temperature is 118°C.

Hence there is a drastic temperature fall of around 600°C over the bottom edge.

The contour plot for the temperature distribution over the surface is shown below (fig 4).

From the contour plot we can see that the maximum temperature occure at the extreme

left corner and the minimum temperature occurs at the bottom extreme right corner of the

surface.

Figure 5

4.2 Heat balance (Top Vs Bottom)

When we calculated the heat either through the conduction or convection at the top edge,

the heat input was coming out to be same upto 3rd place decimal and equal to

47442.432225 W/m (calculated from conduction). Likewise we calculated the heat carried

away by the coolant using the convection relation and it comes out to be equal to

47442.432061 W/m. If we take the difference of these two values and divide by the ½ times

the sum of these heat values, we get the error percentage of 3.467 x 10-7 %, which is

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extremely negligible. Hence the heat of the equation is considerably balanced from both the

top (input) and the bottom(output).

We have used the following relations to compare the heat input to the heat carried away by

the coolant.(See Appendix)

Heat Difference in % = (Qin – Qout) / ½ (Qout+Qin)

4.3 Optimum Relaxation parameter Optimum relaxation parameter will yield the values of the Temperature with minimum

iteration cycles. To calculate the approximate optimum value of relaxation parameter, we

changed its value from 0.6 to 2.0 with the increment of 0.1 and the following graph is

plotted for the number of iterations Vs relaxation parameter. Here it is important to

mention that the initial temperature is assumed 60°C to start the iteration. If we change the

value of this initial assumed temperature, the convergence criteria will change. The better

approximation of the initial temperature can also lead to reduce the number of iteration.

Figure 6

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Hence we can observe from Fig. 6 that the value of ω (ωopt )comes out to be around 1.9. We

had also observe while calculating ωopt that the system becomes unstable after the value of

ω equal near about to 2.05. At the value of ω = 1.8 we get the number of iterations equal to

122. At the value of ω = 1.9 we get the number of iterations equal to 112. At the value of ω

= 1.91, the number of iterations again starts increasing and becomes 118. For ω = 2.0, the

number of iterations shoots up to 345.

4.4 Effect of finer Mesh(for mesh size 21 x 11) : Now we inspect the effect of finer mesh on the cases discussed in section 4.1 to 4.3.

4.4.1 Temperature at the steady state(for mesh size 21 x 11) For the finer mesh, we get the fig 7 for temperature distribution over the given domain. The

fine mesh smoothens the plot and give more detailed(with more nodal temperatures) and

accurate interpolated graph for the temperature profile.

Figure 7

The temperature matrix for the new mesh size is shown in the table 2.

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The temperature distribution obtained using the finer mesh (21 x 11) appears higher than

those obtained in the 1st case. For example, the temperature on the right top corner has

increased to approx 1845°C compared to 1839°C as calculated in the first case. Hence there

is a 0.3 % change in the new temperature obtained.

Figure 8

The contour plot is shown if fig 8. We can clearly visualize that because of the finer mesh the

resolution and smoothness of the contour curves has increased since we have obtained the

temperature at higher number of nodes.

4.4.2 Heat balance (Top Vs Bottom) for Mesh size 21 x 11

By using the finer Heat input Qin calculated comes out to be 46488.054401 W/m. At the

same instance the heat taken out by the coolant is 46488.054258 W/m. Hence we can see

that we get considerable heat balance with the error of order 3.0617E-07 % which is

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considerable negligible and smaller than that obtained in the first case with coarse mesh.

Hence the smaller size of mesh size can increase the accuracy for the heat balance.

4.4.3 Optimum Relaxation parameter for finer mesh

For the finer mesh the relaxation factor in slightly more than what we obtained in case 3. In

21 x 11 mesh we get the optimum value of the relaxation parameter ωopt = 1.95 instead of

1.9 as in the former case.

The Relaxation Factor Vs Number of iterations as shown below in fig 9. shows that the

number of iterations required for the finer mesh (21 x 11) for any given relaxation factor is

almost always higher than that required for the mesh size (11 x 6), even for the optimal

values of relaxation factor ω. For the mesh size of 11 x 6 the number of iterations require for

convergence = 112 whereas for 21 x 6 mesh size the number of iterations required for

convergence = 247.

Figure 9

Since we have already compared the coarse mesh given in the case 1 to the finer mesh but

for simple to brief up the whole summary we can say that by refining the mesh size we have

observed that the accuracy of the computation has increased. For example for the top left

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corner node we can see that the temperature was greater than 0.3 %. Also the heat balance

was more accurate compared to the coarse mesh size. But by increasing the mesh size

computational time also increases which will increase the cost for the project.

4.5 Effect of Lower Conductivity(k = 5 5 W/m-oC) :

4.5.1 Temperature at the steady state(for mesh size 11 x 6) :

In this case we obtained the new temperature distribution for the system under

consideration. The temperature plot for this case is shown in figure 10.

Figure 10

By decreasing the conductivity for the metal, we observed that the metal temperature on

the hot gas side has increased. This can be compared by two cases analyzed above. In the

First case when we had the thermal conductivity of 20 W/m-°C, the temperature of the top

left corner node was observed to be approx. 1839°C. For the finer mesh of grid size 21 x 11

we obtained the temperature of the same node to be 1845°C. But when we reduced the

thermal conductivity of the metal, we observed that the temperature has increased to

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around 1958°C, which is 119°C higher compared to the case 1 with mesh size 11 x 6 and

conductivity of 20 W/m-°C.

The contour plot for this condition is shown in the figure 11. Contour plot shows that the

total top edge of the metal exposed to the hot gas has the temperature greater than

1900°C. At the same time if we compare the temperature of the top edge in case 1, the

highest temperature was calculated to be 1839°C.

Figure 11

Also other thing to be noticed in this case that the temperature at the bottom right half

edge has reduced to an average of approximate 73°C. That means since the conductivity has

decreased, the heat is not flowing through the metal and the coolant has less amount on

work to do to reduce the temperature at its end. But at the same time the total

temperature of the system has raised by an average of more than 70°C when compared to

the 1st case. This is basically due to the fact that heat conduction is now happening at lower

rate, the average temperature of the system has increased.

The Matrix for the temperature distribution is shown in the table 3 attached on the next

page.

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We also observed (Fig. 12) that the temperature distribution over the right hand side

vertical adiabatic boundary. For lower thermal conductivity i.e. 5W/m-°C, though the

temperature is high over the node proximal to the hot gas but the temperature at the node

near the coolant is at lower temperature compared to the case 1. The temperature gradient

for lower conductivity is greater than that compared to conductivity of 20W/m-°C.

Figure 12

4.5.2 Heat balance (Top Vs Bottom) for Mesh size 21 x 11

For the heat balance, when we calculate the heat using the convection method for heat

input to the system at the top from the hot gas, we get the heat input equal to approx.

13627.2154248 W/m. When the same heat calculated due to conduction at the top surface,

the value of heat in comes out to be 13627.2154248 W/m. For the bottom coolant section

we calculated the heat taken out by the coolant equal to 13627.2154081 W/m. Hence if we

compare the heat taken away by the coolant to the heat conducted at the top, we get the

percentage difference of 9.97881E-08 % which is extremely small. At the same time if we

compare with the convection heat transfer at the top to the convection heat transfer at the

bottom, we get the percentage difference to be 1.226E-07%. Other important thing to

observe here is that when the heat conduction or heat transfer from any system has

decreased by 4 times than case 1. This is because our conduction coefficient has decreased

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by 4 times. Since the heat transfer has decreased, we get elevated temperatures at the top

and lower temperature at the bottom coolant exposed edge.

4.5.3 Optimum Relaxation parameter for reduced conductivity (K=5W/m- °C)

The graph for the relaxation parameter has been plotted(Fig. 13) for the reduced

conductivity and we found that the number of iterations for the case 1 with 11 x 6 mesh and

for the case of reduced thermal conductivity are considerably similar. Though the optimum

relaxation parameter for it comes out to be 1.85 compared to the case 1 where the

relaxation parameter comes out to be 1.9. For the values of ωopt 1.85 or less, the number of

iterations required for the reduced conductivity case are less than that required for the

conductivity case of 20 W/m-C.

Figure 13

4.6 Design Considerations for the Combustion chamber

As far as the design considerations are concerned, if the thermal conductivity of the metal is

high we get reduced temperatures at the surface exposed to hot gas. This is actually not

desired in combustion chamber. There should be minimum heat loss to the combustion

chamber walls so that more energy is utilized in producing mechanical work. This design

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consideration can be achieved using the wall material with low thermal conductivity. But by

doing this we get elevated temperature on the wall exposed to hot gas. Hence in this case,

the material should be able to withstand higher temperature. At the same time, thermal

expansion should also be considered at elevated temperature for the design. If the

conductivity is high, coolant will get more heat input hence we will require more energy to

cool it down for reuse. By using the lower conductive material we have observed that the

temperature of the metal wall exposed to the coolant have lower temperature. Since the

heat transfer was reduced by approximately 4 times, the coolant mass flow rate required

will also reduce, hence increasing the efficiency of the system. Also we can see that at the

bottom boundary center of the system there is a drastic change in metal temperature. Such

junctions should be avoided because these abrupt changes can lead to thermal stresses in

the material.

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5 Summary and Conclusion

In this project we calculated the temperature distribution for the 2D steady state condition

by the use as Gauss Siedel (PSOR) algorithm. We included the boundary conditions into the

system using the ghost point method (see appendix for detail). We evaluated the

temperature distribution for three cases i.e., using the grid size of 11 x 6 , using the grid size

21 x 11 and finally for the reduced thermal conductivity of 5W/m-°C. We observed that the

temperature accuracy increases using the finer mesh. For comparison we had seen that at

the extreme top right node there was a temperature rise of 0.3% due to finer mesh.At the

same time the computational cost for finer mesh since the convergence is then achieved

using higher number of iterations because of more number of nodes. By reducing the

conductivity we observed that the temperature over the surface exposed to the hot gas is

higher when compare to other cases. The temperature reaches approximately 1950° C in

this case, while the temperature in the case 1 reaches the maximum value of 1840°C. For

better design of the combustion chamber low thermal conductivity will be better. We also

calculated the optimal value of the relaxation factor. For the mesh size of 11 x 6 ,we get the

ωopt = 1.9, for the mesh size of 21x11 we get ωopt = 1.95 and for lower conductivity case we

get ωopt = 1.85 .The numbers of iterations are least for the optimal value of relaxation

parameter.

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6 Appendix

6.1 Matlab Codes written for solving the iterations %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Name : Mandeep Singh % % Project 2 (2-D steady Heat Conduction) % % Assigned: 2/10/11 Due: 2/24/11 % % MAE 542 Engineering Applications of Computational Fluid Dynamics % % Gauss-Siedel(PSOR) method (Explicit) for solving Heat Equation % % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %------------------------------------------------------------------% % Given parameters : % delta x , delta y , beta , WALL LENGTH L = 20 cm % WALL WIDTH W = 10 cm % Tg = 2000°C , hg = 1000 W/m2 - 24°C % WALL % /------------------------------------ % / / % / /ADIABATIC WALL % / / % / / % ///////////////////------------------/ % ADIABATIC T1 = 60°C,h=8000 W/M % WALL %% clear all close all clc %% % Given Parameters Tg=2000; % Temperature at the top hg=1000; % Convection coefficient at top T1=60; % Temperature at Bottom h=8000; % Convection coefficient at bottom L=20/100; % Length in meters W=10/100; % Width in meters Cerror=10^-6; % Acceptable error iter=1000; % number of iteration dx=2/100; % Step in X- Direction dy=2/100; % Step in Y- Direction B=(dx/dy);% Ratio k=20; % Given Conductivity BT=(hg*dy/k);% Beta For Top for Y-direction BB=(h*dy/k); % Beta for Bottom in Y- Direction Z=(1+B*B+B*B*BT); Y=(1+B*B+B*B*BB); %% nx=(L/dx)+1; % Number of nodes in X-Dir ny=(W/dy)+1; % Number of nodes in Y-Dir T=60*ones(ny,nx); % Initializing the matrix om=1.88; % Value of Omega- Relaxation Factor

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s=1; Told=zeros(ny,nx); %% for k = 1:2000% number of iteration m=0.0; %% for l=1:ny for h=1:nx Told(l,h)=T(l,h); end end %% for i=1:ny for j=1:nx if(i==1&&j==1) T(i,j)=(T(1,j+1)+BT*B^2*Tg+B*B*T(i+1,1))/Z; else if(i==1&&j==nx) T(i,j)=(T(1,j-1)+BT*B^2*Tg+B^2*T(2,j))/Z; else if(i==ny&&j==1) T(i,j)=(T(i,j+1)+B^2*T(i-1,1))/(1+B*B); else if(i==ny&&j==nx) T(i,j)=(T(i,j-1)+(B*B*BB*T1)+(B*B*T(i-1,j)))/Y; else if(i>=2 &&i<=ny-1 &&j==1) T(i,j)=(((T(i-1,j)+T(i+1,j))*B*B/2)+T(i,j+1))/(1+B*B); else if(i==1&&j>=2 &&j<=nx-1) T(i,j)=(1/Z)*((T(i,j+1)+T(i,j-1))/2+B*B*(T(i+1,j)+BT*Tg)); else if(i>=2 &&i<=ny-1 &&j==nx) T(i,j)=(((T(i-1,j)+T(i+1,j))*B*B/2)+T(i,j-1))/(1+B*B); else if(i==ny&&j>=2 &&j<=(((nx-1)/2))) T(i,j)=(((T(i,j-1)+(T(i,j+1)))/2)+B*T(i-1,j))/(1+B*B); else if(i==ny&&j>=(((nx-1)/2)+2)&&j<=(nx-1)) T(i,j)=(((T(i,j-1)+T(i,j+1))/2)+(B*B*BB*T1)+(B*B*T(i-1,j)))/Y; else if(i==ny&&j==((nx+1)/2)) T(i,j)=((T(i,j-1)+T(i,j+1)+B*B*(2*T(i-1,j)+B*B*BB*T1))/((2+2*B*B+B*B*BB))); else T(i,j)=(1-om)*T(i,j)+((om/2)*(1/(1+B*B))*(T(i,j+1)+T(i,j-1)+B*B*(T(i+1,j)+T(i-1,j)))); end end end end end end end end end end

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end end %% for l=1:ny for h=1:nx ej(l,h)=T(l,h)-Told(l,h); m=m+(ej(l,h)^2); error=sqrt(m); end end %% if(error < Cerror) break; end %% s=s+1; end Tend=T'; %% figure(1) %mx=0:1:20; %my=0:1:10; %surfc(0:dx*100:x*100,0:dy*100:W*100,T,'EdgeColor','none') x = linspace(0,20,nx); y = linspace(0,10,ny); surfc(y,x,Tend,'EdgeColor','none') %surfc(Tend(1:nx,1:ny)); figure(gcf) %colormap hsv colorbar title('Gauss Seidel (PSOR) to 2D Steady State heat Equation','fontWeight','bold',... 'fontsize',10); xlabel('Width (cm)','fontWeight','bold','fontSize',10); ylabel('Length(cm)','fontWeight','bold','fontSize',10); zlabel('Temperature(C)','fontWeight','bold','fontSize',10); figure(2) imagesc(y,x,Tend); colormap hsv colorbar title('Gauss Seidel (PSOR) to 2D Steady State heat Equation','fontWeight','bold',... 'fontsize',10); xlabel('Length(cm)','fontWeight','bold','fontSize',10); ylabel('Width (cm)','fontWeight','bold','fontSize',10); zlabel('Temperature(C)','fontWeight','bold','fontSize',10); figure(3) %depth = [0:dx:(nx-1)*dx]; contourf(y,x,Tend,'EdgeColor','none'); title('Temperature plot (contourf)') colorbar title('Gauss Seidel (PSOR) to 2D Steady State heat Equation - Temperature plot (contourf)',...

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'fontWeight','bold','fontsize',10); xlabel('Width (cm)','fontWeight','bold','fontSize',10); ylabel('Length (cm)','fontWeight','bold','fontSize',10); zlabel('Temperature(C)','fontWeight','bold','fontSize',10);

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6.2 Boundary Condition Calculations

The boundary conditions are calculated using the ghost point method. Now we will see all

the boundary conditions derivations used for Gauss Seidel PSOR algorithm.

6.2.1 Boundary 1 (Top left corner)

Figure 14 below shows the boundary considered. There are two ghost points considered

outside the system . The values of these ghost points are calculated using the boundary

conditions.

Figure 14

For X-Direction the Boundary condition is adiabatic hence is given by

02

,1,1 =

−=

∂∂ −+

x

TT

xT jiji

Hence we get

jiji TT ,1,1 −+ =

For Y-Direction using the heat balance equation , we get

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−−=− −+

y

TTkTTh jiji

jigg 2)( 1,1,

,

Now we calculate the value of 1, +jiT as it is the ghost node and substitute the value in the 2D

heat equation for the steady state.

1,,1, )(2

−+ +−∆

= jijigg

ji TTTk

yhT

Steady State equation for the 2D is given by

022

21,,1,

2,1,,1 =

+−+

+− −+−+

y

TTT

x

TTT jijijijijiji

Substituting the value of jiji TT ,11, & −+ in the 2D Steady State equation we get :

02)(

22

2

1,,1,,

2,1,,1 =

+−+−∆

+

+− −−++

y

TTTTTk

yh

x

TTT jijijijigg

jijiji

Solving for jiT , , For this node we get the expression to be :

[ ])()1(

11,

2,122, gijiji

iji TTTT ββ

βββ++

++= −+

yx

k

yhWhere g

i ∆∆

=∆

= ββ ,

6.2.2 Boundary 2 (Top Right corner) Figure 15 below shows the top right hand side boundary .

The calculation for this boundary has been done using the same approach as done in section

6.6.1.

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Figure 15

From the Adiabatic wall condition we get

02

,1,1 =

−=

∂∂ −+

x

TT

xT jiji

From which we get

jiji TT ,1,1 −+ =

Also from heat balance in the Y-Direction we get,

−−=− −+

y

TTkTTh jiji

jigg 2)( 1,1,

,

From above equation, solving for 1, +jiT we get the expression

1,,1, )(2

−+ +−∆

= jijigg

ji TTTk

yhT

Substituting the value of jiji TT ,11, & ++ in the 2D Steady State equation we get :

02)(

22

2

1,,1,,

2,1,,1 =

+−+−∆

+

+− −−−−

y

TTTTTk

yh

x

TTT jijijijigg

jijiji

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Solving for jiT , , For this node we get the expression to be :

[ ])()1(

11,

2,122, gijiji

iji TTTT ββ

βββ++

++= −−

6.2.3 Boundary 3 (Bottom Right corner) For this boundary we used the following figure16 where we have the ghost points at i+1,j

and I,j-1.

Figure 16

In the X-Direction we have Adiabatic condition, therefore we get

jiji TT ,1,1 −+ =

In the Y-Direction by balancing the convection heat equal to heat conducted, we get

1,,11, )(2

+− +−∆

= jijiji TTTk

yhT

Substituting these values in the 2D steady state heat equation , we get

02)(

22

2

1,,1,,1

2,1,,1 =

+−+−∆

+

+− ++−−

y

TTTTTkyh

x

TTT jijijijijijiji

Solving for jiT , , For this node we get the expression to be :

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31 | P a g e

[ ])()1(

111,

2,122, TTTT ijiji

iji ββ

βββ++

++= +−

yx

k

yhWhere i ∆

∆=

∆= ββ ,

6.2.4 Boundary 4 (Bottom Left corner)

Below shown(Fig 17) is the boundary under concern where i-1,j and i,j-1 are the ghost points

for which the expressions are calculated .

Figure 17

Since both side wall and the bottom wall boundary are adiabatic in nature, we get

1,1,,1,1 & −+−+ == jijijiji TTTT

Plugging these values of the ghost points into the 2D steady state heat equation , we get

022

21,,1,

2,1,,1 =

+−+

+− ++++

y

TTT

x

TTT jijijijijiji

Solving for jiT , , For this node we get the expression to be :

[ ])()1(

11,

2,12, ++ +

+= jijiji TTT β

β

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32 | P a g e

6.2.5 Boundary 5 ( Left Adiabatic Wall )

For this boundary and configuration(fig 18) we need to calculate the only one ghost node on

the left side of the boundary.

Figure 18

Now we calculate the value of jiT ,1− as it is the ghost node and substitute the value in the 2D

Heat Equation for the Steady State.

Since the side wall is adiabatic

02

,1,1 =

−=

∂∂ −+

x

TT

xT jiji

Hence we get

jiji TT ,1,1 −+ =

Now using 2D Steady state heat equation we get

022

21,,1,

2,1,,1 =

+−+

+− +−++

y

TTT

x

TTT jijijijijiji

Solving for jiT , , For this node we get the expression to be :

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33 | P a g e

++

+= −+

+ 2)1(1 1,1,2

,12,jiji

jiji

TTTT β

β

6.2.6 Boundary 6 ( Top convective boundary wall )

For the top edge (Fig 19) , we have hot gas which transfer the heat to the system through

convection.

Figure 19

From heat balance equation we get

−−=− −+

y

TTkTTh jiji

jigg 2)( 1,1,

,

From this equation , rearranging we get

1,,1, )(2

−+ +−∆

= jijigg

ji TTTk

yhT

Plugging the value calculated above into the 2D heat equation for steady state, we get the

02)(

22

2

1,,1,,

2,1,,1 =

+−+−∆

+

+− −−+−

y

TTTTTk

yh

x

TTT jijijijigg

jijiji

Rearranging the equations and solving for jiT , , we get the expression :

Page 34: Mandeep Singh

34 | P a g e

++

+

++= −

−+ )(2)1(

11,

2,1,122, giji

jiji

iji TT

TTT ββ

βββ

yx

k

yhWhere i ∆

∆=

∆= ββ ,

6.2.7 Boundary 7 ( Right hand Side Adiabatic boundary wall ) The figure 20 for right side adiabatic wall is shown in the figure below

Figure 20

Since the side wall is adiabatic

02

,1,1 =

−=

∂∂ −+

x

TT

xT jiji

Hence we get

jiji TT ,1,1 −+ =

Now using 2D Steady state heat equation we get

022

21,,1,

2,1,,1 =

+−+

+− +−−−

y

TTT

x

TTT jijijijijiji

Solving for jiT , , For this node we get the expression to be :

++

+= −+

− 2)1(1 1,1,2

,12,jiji

jiji

TTTT β

β

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35 | P a g e

6.2.8 Boundary 8 ( Bottom wall , left hand Side Adiabatic boundary wall ) The grid boundary for such condition is shown in figure 21 below

Figure 21

Since the bottom half left hand side wall is adiabatic

02

1,1, =

−=

∂∂ −+

y

TT

yT jiji

Hence we get

1,1, −+ = jiji TT

Now using 2D Steady state heat equation and using the value evaluated above to eliminate

the ghost point, we get

022

21,,1,

2,1,,1 =

+−+

+− −−−+

y

TTT

x

TTT jijijijijiji

Re-arranging and solving for jiT , , we get the following expression :

+

+

+= +

+− )(2)1(

11,

2,1,12, ji

jijiji T

TTT β

β

yx

Where∆∆

Page 36: Mandeep Singh

36 | P a g e

6.2.9 Boundary 9 ( Bottom wall , Right hand Side convective boundary wall )

Figure 22

From heat balance equation for the Y direction (Fig 22) ,we get

−−=− −+

y

TTkTTh jiji

jigg 2)( 1,1,

,

From this equation , rearranging we get

1,,11, )(2

+− +−∆

= jijiji TTTkyh

T

Plugging the value calculated above into the 2D heat equation for steady state, we get the

02)(

22

2

1,,1,,1

2,1,,1 =

+−+−∆

+

+− +++−

y

TTTTTkyh

x

TTT jijijijijijiji

Rearranging the equations and solving for jiT , , we get the expression :

++

+

++= +

−+ )(2)1(

111,

2,1,122, TT

TTT iji

jiji

iji ββ

βββ

yx

k

yhWhere i ∆

∆=

∆= ββ ,

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37 | P a g e

6.2.10 Boundary 10 ( Bottom wall , MidPoint )

For this boundary we use the heat balance control volume condition where we balance the

heat on the middle node coming from top through conduction equal average of the heat

coming out from the left + heat coming out from the right.

Figure 23

Mathematically we can write it as :

[ ]",

",

"

21

RightoutLeftoutcond qqq +=

)(21

2 1,1,1, TTh

y

TTk ji

jiji −=

−−− −+

)(21

2 1,1,1, TTh

y

TTk ji

jiji −=

−−− −+

)( 1,1,1, TTTT jiijiji −−= +− β

Now the value of 1, −jiT as it is the ghost node is substitutes in the 2D Heat Equation for the

Steady State.

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38 | P a g e

0)(22

21,1,,1,

2,1,,1 =

−−+−+

+− +++−

y

TTTTT

x

TTT jiijijijijijiji β

Rearranging, we get

[ ] 0)(22 1,1,,1,2

2

,1,,1 =−−+−∆∆

++− +++− TTTTTyx

TTT jiijijijijijiji β

[ ] 0)(22 1,1,,1,2

,1,,1 =−−+−++− +++− TTTTTTTT jiijijijijijiji ββ

Now rearranging the equation and solving for value of jiT ,

We get ,

[ ])2()22(

111,

2,1,122, TTTTT ijijiji

iji ββ

βββ+++

++= +−+

6.3 Heat Calculation methodology

For calculating the heat at the top , we used the

nx

nx

i yTxk

yTxk

xyT

k

∂∂∆

+

∂∂∆

+∆∂∂

≈∑−

= 22 1

1

2

For comparison for the heat we used the following relation

100)(

.%21

×+−

=inout

inout

QQQQ

differenceHeat

7 References 1. Lecture notes by Prof. Desjardin

2. Hoffman & Chian, Computational Fluid Dynamics Vol-I

dxyT

kdlyT

kdlqqincond ∫∫∫ ∂∂

=∂∂

=−= ."