modelling the flow via 2d constant source panel method

7/26/2019 Modelling the flow via 2D constant source panel method

1/20

1

I tried to model the flow around symmetric objects via the 2D constant strength source panel

method. I used a circle and a naca foil. All calculations were carried out on Matlab and all codes are

given below this paragraph.

Number of panels were chosen as 8 and 20 for circle and 10 and 20 for the naca profile and the

collocation points were chosen at the half and the quarter of the panelslengths.

Uinf=0

R=1

The program reads the panel coordinates from the excel sheet and then continues the calculation.

THE CODE :

clc;close all;clear allR=1;u=1;

filename='koordinatlar';zz=xlsread(filename); %koordinatlar girildin=length(zz(:,1))-1;Y=zz(:,2);X=zz(:,1); %koordinatlar blnd

subplot(3,1,1)axis square;hold on; grid on;ezplot('x^2+y^2-1');title(['Daire zerinde Panel Yerleimi( 'num2str(n),' PANEL,R=',num2str(R),' m ,U_G= ',num2str(u) ' m/s)'])

%% alarn bulunmas

fori=1:nteta(i,:)=atan2(Y(i+1,:)-Y(i,:),X(i+1,:)-X(i,:));

ifteta(i,:)2*piteta(i,:)=teta(i,:)-2*pi;

elseifteta(i,:)==2*piteta(i,:)=0;

elseteta(i,:)=teta(i,:); %teta degerleri dogru

endend

fori=1:nbeta(i,:)=teta(i,:)+(pi()/2); %beta degerleri dogru

ifbeta(i,:)(2*pi())beta(i,:)=beta(i,:)-2*pi;


2/20

2

elseifbeta(i,:)==2*pibeta(i,:)=0;

elsebeta(i,:)=beta(i,:); %beta degerleri dogru

end

end

%% orta noktalarclear ifori=1:n

x(i,:)=((X(i+1)-X(i))/2)+X(i);y(i,:)=((Y(i+1)-Y(i))/2)+Y(i); % orta noktalar dogru bulunduplot(x(i,1),y(i,1),'.m')

endclear i

fori=1:nS(i,:)=((X(i+1)-X(i))^2 + ((Y(i+1)-Y(i))^2))^0.5; %panel boylar tamam

endclear iclear j

%% abromovitz donsm ve kaynak siddetlerifori=1:n

forj=1:nA(i,j)=-(x(i)-X(j))*cos(teta(j))-((y(i)-Y(j))*sin(teta(j))); B(i,j)=(x(i)-X(j))^2+ (y(i)-Y(j))^2;C(i,j)=sin(teta(i)-teta(j));

D(i,j)=-(x(i)-X(j))*sin(teta(i))+(y(i)-Y(j))*cos(teta(i)); E(i,j)=(x(i)-X(j))*sin(teta(j))-(y(i)-Y(j))*cos(teta(j)); ifi==jI(i,j)=0.5;Iy(i,j)=pi();else

I(i,j)=(1/2*sin(teta(i)-teta(j))*log(1+((S(j)^2+2*A(i,j)*S(j))/B(i,j)))-cos(teta(i)-teta(j))*(atan((S(j)+A(i,j))/E(i,j))-atan(A(i,j)/E(i,j))))/(2*pi());

Iy(i,j)=(D(i,j)-A(i,j)*C(i,j))/(2*E(i,j))*log(1+((S(j)^2+2*A(i,j)*S(j))/B(i,j)))-C(i,j)*(atan2((S(j)+A(i,j)),E(i,j))-atan2(A(i,j),E(i,j)));

endend

end% I ve Iy(J(i,j)) degerleri dogru bulundu.I(i,j)=I(j,i);%Iy(i,j)=Iy(j,i); %matriste satrlar stunlar ters olmustu.clear iclear jfori=1:nsonuc(i,:)=-u*cos(beta(i));end%sonuc matrisi dogru

lamda=inv(I)*sonuc; % lamda degerleri dogru bulundu.


3/20

3

%% hzlar ve basnclarn bulunmas

fori=1:1:nhiz=0;

forj=1:1:nhiz=hiz+Iy(i,j)*lamda(j,1)/(2*pi);

endtop_hiz(i,1)=hiz+u*sin(beta(i,1));Cp(i,1)=1-(top_hiz(i,1)/u)^2;end

fori=1:nV_an(i,:)=2*u*sin(beta(i));

end

subplot(3,1,2)hold on; grid on;

plot(teta,Cp,'+r')

ezplot('1-4*(cos(x)^2)',[0,2*pi])set(gca,'XTick',0:pi/2:2*pi)set(gca,'XTickLabel',{'0','pi/2','pi','3pi/2','2pi'})xlabel('0 \leq \beta \leq 2\pi')ylabel('C_P')title(['Daire Etrafndaki Basn Dalm( 'num2str(n),' PANEL )'])graf=legend('Nmerik zm','Analitik zm')set(graf,'Location','southeast')

hold on; grid on;subplot(3,1,3)hold on; grid on;ezplot('2*1*cos(x)',[0,2*pi])plot(teta,top_hiz,'.m')set(gca,'XTick',0:pi/2:2*pi)set(gca,'XTickLabel',{'0','pi/2','pi','3pi/2','2pi'})xlabel('0 \leq \beta \leq 2\pi')

ylabel('U')title(['Daire Etrafndaki Hz Dalm( 'num2str(n),' PANEL )'])graf=legend('Analitik zm','Nmerik zm')set(graf,'Location','southeast')


4/20

4

For the circle;

Number of panels = 8

Collocation points = 0.5

LAMBDATHETA

Radian Degree

2,3656 1,5708 90

1,6728 0,7854 45

0,0000 0,0000 0

-1,6728 5,4978 315

-2,3656 4,7124 270

-1,6728 3,9270 225

0,0000 3,1416 180

1,6728 2,3562 135

The total of the strength of sources are 0 and this guarantees that the results are true up to now.

(I checked this code with the problem which was solved at the class and the results were the same.)


5/20

5

For the circle;



LAMBDATHETA

Radian Degree

2,1406 1,5708 90

2,0358 1,2566 72

1,7318 0,9425 54

1,2582 0,6283 36

0,6615 0,3142 18

0,0000 6,2832 360

-0,6615 5,9690 342

-1,2582 5,6549 324

-1,7318 5,3407 306

-2,0358 5,0265 288

-2,1406 4,7124 270

-2,0358 4,3982 252

-1,7318 4,0841 234

-1,2582 3,7699 216-0,6615 3,4558 198

0,0000 3,1416 180

0,6615 2,8274 162

1,2582 2,5133 144

1,7318 2,1991 126

2,0358 1,8850 108

Again if we sum the lambdas, we can realize that this summation equals zero.


6/20

6

When we look the graphs we can say that the more number of panels, the nearer results to

the real. But also we can say that it cant answer totally correct even if the number of panels is

too high.


7/20

7

For the circle;



LAMBDATHETA

Radian Degree

2,069878 1,570796 90

1,919748 1,256637 72

1,581699 0,942478 54

1,088822 0,628319 36

0,489364 0,314159 18

-0,158 6,283185 360-0,78989 5,969026 342

-1,34447 5,654867 324

-1,76743 5,340708 306

-2,01739 5,026548 288

-2,06988 4,712389 270

-1,91975 4,39823 252

-1,5817 4,08407 234

-1,08882 3,769911 216

-0,48936 3,455752 198

0,157996 3,141593 180

0,789891 2,827433 162

1,344465 2,513274 144

1,767434 2,199115 126

2,017395 1,884956 108

When we have selected the quarter of the panels as collocation points, the values of

lambdas have changed. But although they have changed, the summation has not changed andit is still zero. So it still makes sense.


8/20

8

ThetaCollocation Points

Exact ResultRelative Error

0.25(col. pt.) 0.5(col. pt.) 0.25(col. pt.) 0.5(col. pt.)

1,5708 1,396856 1,070307 2,0000 0,3016 0,4648

1,2566 1,913575 1,635956 1,9021 0,0060 0,1399

0,9425 2,242979 2,041467 1,6180 0,3862 0,2617

0,6283 2,352825 2,247145 1,1756 1,0014 0,9115

0,3142 2,23236 2,232856 0,6180 2,6120 2,6128

6,2832 1,893376 2 0,0000 too high too high

5,9690 1,369056 1,57137 -0,6180 3,2152 3,5425

5,6549 0,710722 0,988923 -1,1756 1,6046 1,8412

5,3407 -0,01718 0,309674 -1,6180 0,9894 1,1914

5,0265 -0,7434 -0,39989 -1,9021 0,6092 0,7898

4,7124 -1,39686 -1,07031 -2,0000 0,3016 0,4648

4,3982 -1,91357 -1,63596 -1,9021 0,0060 0,13994,0841 -2,24298 -2,04147 -1,6180 0,3862 0,2617

3,7699 -2,35283 -2,24714 -1,1756 1,0014 0,9115

3,4558 -2,23236 -2,23286 -0,6180 2,6120 2,6128

3,1416 -1,89338 -2 0,0000 too high too high

2,8274 -1,36906 -1,57137 0,6180 3,2152 3,5425

2,5133 -0,71072 -0,98892 1,1756 1,6046 1,8412

2,1991 0,017182 -0,30967 1,6180 0,9894 1,1914

1,8850 0,743404 0,399888 1,9021 0,6092 0,7898

Total 21,4512 23,5115


9/20

9

If we want to compare this two situations for the 20 panels, we can clearly say that the results

which belongs to collocation points=0.25 are a little better than the others.

For the circle;



LAMBDATHETA

Radian Degree

2,1794 1,5708 90

1,2513 0,7854 45

-0,4098 0,0000 0

-1,8309 5,4978 315

-2,1794 4,7124 270

-1,2513 3,9270 225

0,4098 3,1416 180

1,8309 2,3562 135

As all the others, the summation of lambdas equals to zero and again as all the others the

values of lambdas make sense. For example they start as being positive like sources and then

they continue as being negative like sinks. When we think about the geometry of the circle, it

is exactly possible.

The influence of changing location of collocation points on the lambdas:

Lambda

0.25(col. pt.) 0.5(col. Pt.)

2,1794 2,3656

1,2513 1,6728

-0,4098 0,0000

-1,8309 -1,6728

-2,1794 -2,3656

-1,2513 -1,6728

0,4098 0,0000

1,8309 1,6728


10/20

10

ThetaVelocities



1,5708 1,4708 1,1828 2,0000 0,2646 0,4086

0,7854 2,2587 2,2506 1,4142 0,5971 0,5914

0,0000 1,7235 2,0000 0,0000 too high too high

5,4978 0,1787 0,5778 -1,4142 1,1263 1,4086

4,7124 -1,4708 -1,1828 -2,0000 0,2646 0,4086

3,9270 -2,2587 -2,2506 -1,4142 0,5971 0,5914

3,1416 -1,7235 -2,0000 0,0000 too high too high

2,3562 -0,1787 -0,5778 1,4142 1,1263 1,4086

Total 3,9761 4,8172

Lastly for this part, we can say that this method with these numbers of panels is not suitable

to model this situation correctly. The results can be better when the number of panels is

increased.

The differences between two situations, which are related to the location of control or

collocation points, increase when the number of panels increases. Although there is a

difference, it is not significant for the circle.


11/20

11

NACA0015

The formula of the Naca profile:

The code from which I calculated the y and x values:

clcclear allformat long%% naca geometri hesaplamat=0.7502;% maksimum kalnlk .c=1; % kord uzunluuN=5 ; %panel saysnn yarsdb=pi/N;u=1;

%% koordinatlarn belirlenmesi

fori=1:2*N+1x(i,:)= c/2*(1-cos((i-1)*db)); % x degerleri dogru .

end

forj=1:2*N+1y(j,:)=(2*c*t)/(10*0.2)*(0.2969*((x(j,1)/c)^0.5)-0.1260*(x(j,1)/c)-

0.3516*((x(j,1)/c)^2)+0.2843*((x(j,1)/c)^3)-0.1015*((x(j,1)/c)^4)); % ydegerleri 0.001 hassasiyetle doru

end

----------------------------------------------------------

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000

Naca 0015


12/20

12

For the naca profile;



The coordinates are got from the upper code and then program continues calculation via the

code which is explained in early parts of the report.

The x and y values of the panels:

x y

0,0000 0,0000

0,0955 0,0576

0,3455 0,0745

0,6545 0,0512

0,9045 0,01741,0000 0,0000

0,9045 -0,0174

0,6545 -0,0512

0,3455 -0,0745

0,0955 -0,0576

0,0000 0,0000

Because the profile is symmetric only for the x axis, the length of the panels is not the

same. So if we want to check the accuracy of the lambdas, we have to multiply each of themwith the panel lengths. Then we have to sum and it should be zero. But when we look at this

summation, we can say that the sensitivity of the system is 10-3

. Because when we calculate

the x and y values, we cant choose more sensitive formula.But it is important, system is still

symmetric.( I mean lambdas and panel lengths.)

Panel length LAMBDATHETA

Radian Degree

0,1115 0,7584 0,5426 31,09

0,2506 0,0604 0,0675 3,870,3099 -0,1080 6,2078 355,68

0,2523 -0,1551 6,1489 352,31

0,0971 -0,1833 6,1030 349,67

0,0971 -0,1833 3,3218 190,33

0,2523 -0,1551 3,2758 187,69

0,3099 -0,1080 3,2169 184,32

0,2506 0,0604 3,0740 176,13

0,1115 0,7584 2,5990 148,91


13/20

13





Radian Degree

0,1115 0,3481 0,5426 31,09

0,2506 -0,1487 0,0675 3,870,3099 -0,2620 6,2078 355,68

0,2523 -0,3150 6,1489 352,31

0,0971 -0,4371 6,1030 349,67

0,0971 0,1115 3,3218 190,33

0,2523 0,0173 3,2758 187,69

0,3099 0,0482 3,2169 184,32

0,2506 0,2515 3,0740 176,13

0,1115 1,0240 2,5990 148,91


14/20

14

ThetaVelocities



0,5426 1,2388 1,3850 2,0000 0,3806 0,30750,0675 1,1242 1,2334 1,9021 0,4090 0,3516

6,2078 0,9621 1,0771 1,6180 0,4054 0,3343

6,1489 0,8284 0,9638 1,1756 0,2953 0,1801

6,1030 0,6308 0,8434 0,6180 0,0207 0,3646

3,3218 -0,8663 -1,0267 0,0000 too high too high

3,2758 -1,0150 -1,1189 -0,6180 0,6422 0,8105

3,2169 -1,0770 -1,1851 -1,1756 0,0839 0,0081

3,0740 -1,0131 -1,1730 -1,6180 0,3739 0,2750

2,5990 -0,3529 -0,6266 -1,9021 0,8145 0,6706

Total 3,4255 3,3024

When we look at the differences of two situations, we can clearly say that for 10 panels,

there is no significant difference. But also we can say that putting the collocation points on the

quarter of the panel breaks the symmetry of the system.


15/20

15





Radian Degree

0,0406 1,1927 0,9235 52,92

0,0754 0,4180 0,3410 19,54

0,1116 0,1420 0,1311 7,51

0,1394 -0,0039 0,0167 0,96

0,1547 -0,0855 6,2295 356,93

0,1552 -0,1272 6,1862 354,44

0,1405 -0,1485 6,1581 352,84

0,1118 -0,1609 6,1374 351,65

0,0720 -0,1652 6,1214 350,73

0,0252 -0,2237 6,0503 346,66

0,0252 -0,2237 3,3745 193,34

0,0720 -0,1652 3,3034 189,27

0,1118 -0,1609 3,2874 188,35

0,1405 -0,1485 3,2666 187,16

0,1552 -0,1272 3,2386 185,56

0,1547 -0,0855 3,1952 183,07

0,1394 -0,0039 3,1249 179,040,1116 0,1420 3,0105 172,49

0,0754 0,4180 2,8006 160,46

0,0406 1,1927 2,2180 127,08

As before, because the profile is symmetric only for the x axis, the length of the panels is

not the same. So if we want to check the accuracy of the lambdas, we have to multiply each

of them with the panel lengths. Then we have to sum and it should be zero. But when we

look at this summation, we can say that the sensitivity of the system is 10-2

. Because when

we calculate the x and y values, we cant choose more sensitive formula. But it is important,

system is still symmetric. (I mean lambdas and panel lengths.)


16/20

16

This graph has some accumulations especially at some degrees. Because of the geometry this is

possible. But also we can clearly say that almost all values are far from the exact values.


17/20

17





Radian Degree

0,0406 0,9131 0,9235 52,92

0,0754 0,2287 0,3410 19,54

0,1116 0,0220 0,1311 7,51

0,1394 -0,0976 0,0167 0,96

0,1547 -0,1668 6,2295 356,930,1552 -0,2041 6,1862 354,44

0,1405 -0,2281 6,1581 352,84

0,1118 -0,2529 6,1374 351,65

0,0720 -0,2892 6,1214 350,73

0,0252 -0,4533 6,0503 346,66

0,0252 0,0537 3,3745 193,34

0,0720 -0,0276 3,3034 189,27

0,1118 -0,0623 3,2874 188,35

0,1405 -0,0651 3,2666 187,16

0,1552 -0,0481 3,2386 185,56

0,1547 -0,0031 3,1952 183,07

0,1394 0,0890 3,1249 179,04

0,1116 0,2559 3,0105 172,49

0,0754 0,5708 2,8006 160,46

0,0406 1,3197 2,2180 127,08

As before, because the profile is symmetric only for the x axis, the length of the panels is

not the same. So if we want to check the accuracy of the lambdas, we have to multiply each

of them with the panel lengths. Then we have to sum and it should be zero. But when we

look at this summation, we can say that the sensitivity of the system is 10-3

. Because when

we calculate the x and y values, we cant choose more sensitive formula. But it is important,

system is still symmetric. (I mean lambdas and panel lengths.)


18/20

18

ThetaVelocities



0,9235 1,3594 1,3388 2,0000 0,3203 0,3306

0,3410 1,3694 1,4012 1,9021 0,2801 0,2633

0,1311 1,2459 1,2969 1,6180 0,2300 0,1985

0,0167 1,1394 1,1981 1,1756 0,0308 0,01916,2295 1,0481 1,1133 0,6180 0,6958 0,8013

6,1862 0,9726 1,0438 0,0000 too high too high

6,1581 0,9063 0,9848 -0,6180 2,4664 2,5934

6,1374 0,8380 0,9283 -1,1756 1,7128 1,7897

6,1214 0,7538 0,8705 -1,6180 1,4659 1,5380

6,0503 0,5517 0,7479 -1,9021 1,2900 1,3932

3,3745 -0,8341 -0,9715 -2,0000 0,5830 0,5142

3,3034 -0,9693 -1,0357 -1,9021 0,4904 0,4555

3,2874 -1,0428 -1,0892 -1,6180 0,3555 0,3268

3,2666 -1,0945 -1,1333 -1,1756 0,0690 0,0360

3,2386 -1,1306 -1,1710 -0,6180 0,8294 0,8948

3,1952 -1,1484 -1,1988 0,0000 too high too high

3,1249 -1,1310 -1,2020 0,6180 2,8300 2,9449

3,0105 -1,0454 -1,1550 1,1756 1,8892 1,9825

2,8006 -0,7991 -0,9832 1,6180 1,4939 1,6077

2,2180 0,1078 -0,1461 1,9021 0,9433 1,0768

Total 17,9758 18,7662


19/20

19

The influence of changing location of collocation points on the lambdas for the naca profie:

For 20 panels:

LAMBDAPanel length

0.5(col. pt.) 0.25(col.pt.)1,1927 0,9131 0,0406

0,4180 0,2287 0,0754

0,1420 0,0220 0,1116

-0,0039 -0,0976 0,1394

-0,0855 -0,1668 0,1547

-0,1272 -0,2041 0,1552

-0,1485 -0,2281 0,1405

-0,1609 -0,2529 0,1118

-0,1652 -0,2892 0,0720

-0,2237 -0,4533 0,0252

-0,2237 0,0537 0,0252

-0,1652 -0,0276 0,0720

-0,1609 -0,0623 0,1118

-0,1485 -0,0651 0,1405

-0,1272 -0,0481 0,1552

-0,0855 -0,0031 0,1547

-0,0039 0,0890 0,1394

0,1420 0,2559 0,1116

0,4180 0,5708 0,07541,1927 1,3197 0,0406

For 10 panels:

LAMBDAPanel length

0.5(col. pt.) 0.25(col.pt.)

0,7584 0,3481 0,1115

0,0604 -0,1487 0,2506

-0,1080 -0,2620 0,3099

-0,1551 -0,3150 0,2523

-0,1833 -0,4371 0,0971

-0,1833 0,1115 0,0971

-0,1551 0,0173 0,2523

-0,1080 0,0482 0,3099

0,0604 0,2515 0,25060,7584 1,0240 0,1115


20/20

20

Conclusions:

1- For better modelling the flow, the number of panels has to be increased. I used 10 and 20

panels for the naca profile and 8 and 20 panels for the circle but they dont enough.

2- Selecting collocation points at half or quarter of the panels length hasno significant effect

on the results for both of geometries.

3- Selecting collocation points at quarter of the panels breaks the symmetry of the lambdas.

So it means that it breaks the symmetry of the geometry for this method.

4- Because of third conclusion for this method the collocation points have to be selected at

the half of the panels length, if we dont want to break the symmetry.

5- Modelling the circle is easier than modelling the naca profiles. Because the former is

symmetric for both x and y axis.

(NOTE: Because Cpvalues are derived from the velocities, naturally, the comparisons about

the Cps are similar to the comparisons about the velocities.)

modelling the flow via 2d constant source panel method

Documents