[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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[ ])()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
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 :
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 :
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
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∆∆
=β
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 ,
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.
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 ∫∫∫ ∂∂
=∂∂
=−= ."
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