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for Nth boundary. Where:
gives the fluxes at cell boundaries.
It is assumed that constant temperature distribution for each triangular cell.
Solution domain and unstructured domain is needed. Sample of the solution domain is as follows:
Tbc(1)= 1200
Tbc(2)= 200
Tbc(3)= 200
Tbc(4)= 200
For the solution flux terms are based on cell averaged variables and the FORTRAN
program is written as follows:
c..mxc : Max number of cells
c..mxn : Max number of nodes
parameter (mxc=5001,mxn=3001)
common /grid/ ncell,nnode,node(3,mxc),neigh(3,mxc),
> xy(2,mxn),area(mxc)
common /var/ time,dt,Tcell(mxc),Tbc(10)
common /grad/ dTdx(mxc),dTdy(mxc)
data mxstep,iostep/7000,1000/ dt,delTallow/0.1,0.01/
c..Read the input data and initialize the solution
call INIT
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c..Start the solution loop
delTmax = 1.
nstep = 0
DO WHILE (nstep .lt. mxstep .and. delTmax .gt. delTallow)
nstep = nstep + 1
time = time + dt
c..Evaluate temperature gradients for each cell
call GRADIENT
c..Sweep all the cells and solve for T^n+1
delTmax = 0.
do n = 1,ncell
c..Evaluate temperature change for the cell
delT = dt/area(n) * FLUX(n)
Tcell(n) = Tcell(n) - delT
delTmax = max(abs(delT), delTmax)
enddo
print*, ' Nstep, Time, DelTmax :',nstep,time,delTmax
c..Output the intermediate solutions
if( mod(nstep,iostep) .eq. 0 .and. nstep .ne. mxstep )
> call TECout(nstep)
ENDDO
c..Output the final solution
call TECout(nstep)
stop 'FINE'
end
subroutine INIT
parameter (mxc=5001,mxn=3001)
common /grid/ ncell,nnode,node(3,mxc),neigh(3,mxc),
> xy(2,mxn),area(mxc)
common /var/ time,dt,Tcell(mxc),Tbc(10)
character fn*16
logical ok
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enddo
call TECout(0)
return
end
subroutine GRADIENT
parameter (mxc=5001,mxn=3001)
common /grid/ ncell,nnode,node(3,mxc),neigh(3,mxc),
> xy(2,mxn),area(mxc)
common /var/ time,dt,Tcell(mxc),Tbc(10)
common /grad/ dTdx(mxc),dTdy(mxc)
DO n = 1,ncell
dTdx(n) = 0.
dTdy(n) = 0.
do nf = 1,3
n1 = node(nf,n)
if(nf .lt. 3) then
n2=node(nf+1,n)
else
n2=node(1,n)
endif
dx = xy(1,n2)-xy(1,n1)
dy = xy(2,n2)-xy(2,n1)
ne = neigh(nf,n)
if(ne .gt. 0) then !..real neighbor
Tneigh = Tcell(ne)
else !..walls
Tneigh = Tbc(-ne)
endif
Tface = 0.5*(Tcell(n)+Tneigh)
dTdx(n) = dTdx(n) + Tface*dy
dTdy(n) = dTdy(n) - Tface*dx
enddo
dTdx(n) = dTdx(n)/area(n)
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dTdy(n) = dTdy(n)/area(n)
ENDDO
return
end
function FLUX(n)
parameter (mxc=5001,mxn=3001)
common /grid/ ncell,nnode,node(3,mxc),neigh(3,mxc),
> xy(2,mxn),area(mxc)
common /var/ time,dt,Tcell(mxc),Tbc(10)
common /grad/ dTdx(mxc),dTdy(mxc)
data alpha /22.5E-6/
FLUX = 0.
c..Sum surface fluxes over the cell faces
do nf = 1,3
n1 = node(nf,n)
if(nf .lt. 3) then
n2=node(nf+1,n)
else
n2=node(1,n)
endif
dx = xy(1,n2)-xy(1,n1)
dy = xy(2,n2)-xy(2,n1)
ne = neigh(nf,n)
if(ne .gt. 0) then !..real neighbor
flux_x = (dTdx(n)+dTdx(ne))*0.5
flux_y = (dTdy(n)+dTdy(ne))*0.5
else !..walls
flux_x = dTdx(n)*0.5
flux_y = dTdy(n)*0.5
endif
FLUX = FLUX + (flux_x*dy - flux_y*dx)
enddo
FLUX = -alpha*FLUX
return
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do n=1,nnode
Tnode(n) = 0.
npass(n) = 0
enddo
c..Find the contribution of cells to the node temperatures
do n=1,ncell
do nf=1,3
nn = node(nf,n)
Tnode(nn)=Tnode(nn)+Tcell(n)
npass(nn)=npass(nn)+1
enddo
enddo
c..Average the total node temperature with # of contributing cells
do n=1,nnode
Tnode(n)=Tnode(n)/npass(n)
enddo
return
end
RESULTS & DISCUSSION
With one cooling hole
We plotted the graph while our mx step 7000 and iostep 1000. This means we will get 7
temperature distributions with one initial state situation. As we can see blade cooling hole is not
enough to provide keep all blade cool. While time is increasing, blade is getting hotter from
starting from aft. Nevertheless we can say cooling hole is enough to keep their neighbor
environment cool.
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It can be observed with considering the temperature scale that due to hot gases turbine blades getshotter and firstly blade tips are effected.
When a second cooling hole is added temperature distributions are as follows:
We changed the given blade.d file to put second cooling hole as follows. First hole has been kept
constant to be able to add a second cooling hole onto turbine blade. Secondly we changed the
first holes place on coordinate system by changing x and y points of first hole. By adding all
these information into blade.d we obtained this illustration as shown in below.
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With the second hole temperature distrubutions seems more uniform and the blade affected from
hot gases less when the temperature scale is considered. Especially on the middle part of turbine
blade temperature seems low if we compare with one cooling hole. Therefore, we can briefly say
second cooling hole is usefull to keep temperature less in middle section.
After many step solution steady-state condition has been reached. Then the temperature
distrubition keeps its stability over the turbine blade surface.
In steady state temperature distrubition turbine blades seem in their neutral situation. In this case
we couldnt observe too much difference between one hole and double hole blades.
REFERENCES
http://www.mathematik.uni-dortmund.de/~kuzmin/cfdintro/lecture5.pdf
Lecture notes
Steady State Temperature distribution
http://www.mathematik.uni-dortmund.de/~kuzmin/cfdintro/lecture5.pdfhttp://www.mathematik.uni-dortmund.de/~kuzmin/cfdintro/lecture5.pdfTop Related