Problem 4' Solve the conduction heat transfer problem in a square of length 1 with the following BCs

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Advanced Topics in Heat, Momentum and Mass
Transfer
Lecturer Payman Jalali, Docent Faculty of
Technology Dept. Energy Environmental
Technology Lappeenranta University of Technology
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Postprocessing of numerical solution Using PDE
toolbox, we solved the following problem. Here,
we perform some postprocessing of the results of
the problem.
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(No Transcript)
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We export data from PDE, Mesh and Solve menus.
Then they will be known in MATLAB environment. As
already mentioned, these information are related
to the points and triangles (meshes) with the
solution. For example, we would like to see the
variation of heat flux over the two left and
right boundaries. The exported solution is only u
(temperature). We can calculate the flux vector
components in the center of triangles in
MATLAB cux,cuy pdecgrad(p,t,c,u) cux-cux
cuy-cuy
p data for points (x,y) t data for triangles
(indices) c data for conductivity u solution
(at nodes)
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After exporting data from 'mesh', 'PDE' and
'Solve' menus we run this code. Assume that we
have already saved the exported data in a file.
So, we load them first. load sol1 cux,cuypdec
grad(p,t,c,u) cux-cux cuy-cuy We specify
x,y coordinates of the vertices of triangles for
i13 for j1length(t)
tx(i,j)p(1,t(i,j)) ty(i,j)p(2,t(i,j))
end end
Here, we determine the location of interest for
analysis and corresponding calculations ii0 fo
r i1length(tx) aatx(,i) pb1 aa
contains some key information of the boundary
of interest. pb represents
the position of
boundary. For various boundaries these must
change accordingly. Current
setting is for the
right boundary This is the place we specify
the right boundary (x1) as our boundary of
interest if (aa(1)pbaa(2)pb)
(aa(1)pbaa(3)pb) (aa(2)pbaa(3)pb)
iiii1 wall(ii,)cux(i),cuy(i),me
an(tx(,i)),mean(ty(,i))
end end figure,plot(wall(,4),wall(,1),'ko'),
xlabel('y'),ylabel('q_x-kgrad_x(T)') figure,plot
(wall(,4),wall(,2),'rs'), xlabel('y'),ylabel('q_
y-kgrad_y(T)')
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mean(wall(,1)) ans -19.4305
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If we change pb to 0 in the code, it will
represent the left boundary and we get the
following results
mean(wall(,1)) ans 119.3735
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For the analysis of heat flux around the circle,
we may need to make changes in the previous code
load sol1 cux,cuypdecgrad(p,t,c,u) cux-cux
cuy-cuy for i13 for j1length(t)
tx(i,j)p(1,t(i,j)) ty(i,j)p(2,t(i,j))
end end ii0 for i1length(tx)
aasqrt((tx(,i)-0.5).2(ty(,i)-0.5).2)
pb0.20.005 if (aa(1)ltpbaa(2)ltpb)
(aa(1)ltpbaa(3)ltpb) (aa(2)ltpbaa(3)ltpb)
iiii1 wall(ii,)cux(i),cuy(i),me
an(tx(,i)),mean(ty(,i))
end end tetaangle(wall(,3)-0.5sqrt(-1)(wall(,
4)-0.5)) Qsqrt(wall(,1).2wall(,2).2) figur
e,polar(teta,Q,'ro')
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Angular position
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• Assignment Using the PDE toolbox for the
  above-mentioned problem, investigate how the
  following parameters affect on the mean heat flux
  across the right boundary.
• The temperature of the fluid inside the internal
  circular cylinder,
• the conductivity of the right half of the domain
  as other conditions are fixed.
• Calculate for specific values as given below,
  then plot the results

a)
b)