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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.pdf