Computational method for engineers

8/8/2019 Computational method for engineers

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1Statement of the problem

1. Statement of the problem

Figure 1 shows a laminar incompressible flow in a four-element and three-node network of circular

pipes. If a quantity Q m3/s of fluid enters and leaves the pipe network, it is necessary to compute the

fluid nodal pressures and the volume flow rate in each pipe.

Figure 1: Fluid flow network.

2. Mathematical modelling

The fluid resistance for an element is given as

(1)Where L is the length of the pipe section;D, the diameter of the pipe section and, the dynamic

viscosity of the fluid and the subscript k, indicates the element number. The mass flux rate entering

and leaving an element can be written as

(2)And

(3)Where pis the pressure, q is the mass flux rate and the subscripts i andj indicate the two nodes of an

element.

(a) Using the equations given above, show that the characteristics of the element can be written as and

based on Eq. 4, construct the characteristics for each of the four elements in Figure 1.

(4) (5) (6)


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2Solution to the problem

Now if we want to make a matrix of coefficients, unknowns and constants (we assume qi and qJ to be

constant) if would be like following equation

As you can see this is the same equation as equation number (4).in this equation is the pressure atthe starting point of the element, is the pressure at the end point of the element, is the enteringmass flux rate of each element and

is the leaving mass flux rate of each element.

For element number 1 we have

(7)For element number 2 we have

(8)For element number 3 we have

(9)

For element number 4 we have

(10)These equations show that the mass flux rate entering and leaving an element are equals. And the

mass flux rate of an element in the forward cycle is equal to the mass flux rate in the backward cycle

of that element but off course in opposite direction.

Table 1: Details of pipe network


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3S

to the proble

(b) From t

element equations in Part a, write t

e nodal equations and t

ence construct t

e matri

form of a system oflinear equations fort

e flow network.

hat would you assume to sol e forp and q?

3. Solu ion to the problem

3.1analyti al solution1. (11 2. (12 3. (13)4. (14)5.

(15)

6. (16)As you can see here we have si equations but we have seven unknowns, it means that we should add

an extra equation to these equations. If we do so then we will have seven equations and seven

unknowns and we can solve these equations using a matrix. For doing so we assume thatthe pressure

in nod numberthree is equalto 1atm.so the last equation is

7. (17)

For making these equations in the standard form we should calculate the fluid resistance for all the

elements. So we rewrite equation (1) to calculate

(1)Where

is the diameter of element k; L is the length of pipe section and, the dynamic viscosity ofthe fluid equals to and subscript k, indicates the element number. We shouldcalculate R for all ofthe four elements, so we have

(18) (19) (20) (21)


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4Solution to the proble

So we write those 7 equations again but this time with these numerical findings, and we also have

Q=0.1

1. (22)2. (23)3. (24)4. (25)5. (26)6. (27)7.

Now we can make a matrix of coefficients, known constants and unknowns.

Afterthis step we use Matlab 7.0 to solve this equation and find allthe unknowns. You can see the

process bellow

A=

b=

and x=


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Solving this equation in Matlab 7.0, we find these amounts for unknowns.

3. Numeri al solution

(c) If water enters the network, shown in Figure 1, at a rate of ( ) with a viscosity of , write a FORTRAN program to iteratively determine the pressure values at all nodes.Write a MATLAB M-file to solve the same and compare the results.

3. .1 Program code

A: First solution

If we want to determine the value of something iteratively one choice is to use Gauss-Siedel or

Jacobian to do this. But for doing this we must at first write the equations in the standard form that

can be used in these methods. You can see the standard form of writing the equations bellow. Theseequations arrive from equation 22 to 27. In this part because the main idea is to find the value of

pressure in every node iteratively and we assume we only need two equations foruse in Gauss-Sidel or Jacobian. You can see these two equations bellow

1.

If we put the values of all the fluid resistance and the assuming value of we willhave the following equations

(28)


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2. If we putthe values of allthe fluid resistance and the assuming value of we willhave the following equations

( (29)So we can rewrite our equations as follow

MATLAB

For using these equations in Gauss-Siedel method ofMATLAB we should write them in matrix formso we will have

, We putthese matrixes in our program and solve the equations iteratively with Gauss-Siedel you can

see the program and the result bellow. Firstitis an M-file to callthe GaussSiedel program.

% A = known coe

matrix

% b = known constant vector

% xg = initial (guessed) values of x

% tol = tolerance

% maxit = maximum number of iteration allowed

A=[33.28951435 19.48335369;-20.45307739 39.9364310]

b=[5347210.954 1974150.813]'

xg = [00]'

tol = 1e-8;

maxit = 100;

% call gauss-siedel

X = GaussSiedel(A,b,xg,tol,maxit)

And here is the Gauss-Siedel program

function X=GaussSiedel(A,B,P,delta, max1)

% Input - A is an N x N nonsingular matrix

% - B is an N x 1 matrix


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% - P is an N x 1 matrix; the initial guess

% - delta is the tolerance for P

% - max1 is the maximum number of iterations

% Output - X is an N x 1 matrix: the Gauss-Siedel approximation to

% the solution of AX = B

N = length(B);

for k=1:max1

for j=1:N

if j==1

X(1)=(B(1)-A(1,2:N)*P(2:N))/A(1,1);

elseif j==N

X(N)=(B(N)-A(N,1:N-1)*(X(1:N-1))')/A(N,N);

else

%X contains the kth approximations and P the (k-1)st

X(j)=(B(j)-A(j,1:j-1)*X(1:j-1)'-A(j,j+1:N)*P(j+1:N))/A(j,j);

endend

err=abs(norm(X'-P));

relerr=err/(norm(X)+eps);

P=X';

if (err


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xg =

0

0

------------------------------------------- -------------------iter no. p1 p2

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

17 1.0132500246e+005 1.0132500147e+005

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

Nima Fouladinejad***[email protected]

X =

1.0e+005 *

1.01325002463959

1.01325001473900

FORTRAN

For using equations 28 and 29 in Gauss-Siedel method ofFORTRAN we should write them in a

different mode. You can see new equations bellow

(30) (31)

And here is the Gauss-Siedel program write in FORTRAN

c

c Gauss-Seidel method

c

c

c file: GaussSiedel.f

c

dimension x(2)

data x/0.,0./

c

print *

print *,' Gauss -Seidel method 'print *,' Nima Fouladinejad [email protected] '

print *

c

print *,'iteration p1 p2'

print 3,0,x

do 2 k=1,13

x(1) = (5347210.954-19.48335369*x(2))/33.28951435

x(2) = (1974150.813+20.45307739*x(1))/39.93643108


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print 3,k,x

2 continue

c

3 format (3x,i2,2x,2(e23.15,2x))

stop

end

And as you can see here are the results ofthis program

Gauss-Seidel method

Nima Fouladinejad [email protected]

iteration p1 p2

0 0.000000000000000E+00 0.000000000000000E+00

1 0.160627500000000E+06 0.131696234375000E+06

2 0.835496328125000E+05 0.922215078125000E+05

3 0.106653007812500E+06 0.104053687500000E+06

4 0.997279843750000E+05 0.100507101562500E+06

5 0.101803695312500E+06 0.101570156250000E+06

6 0.101181523437500E+06 0.101251515625000E+06

7 0.101368015625000E+06 0.101347031250000E+06

8 0.101312109375000E+06 0.101318398437500E+06

9 0.101328867187500E+06 0.101326976562500E+06

10 0.101323851562500E+06 0.101324414062500E+06

11 0.101325351562500E+06 0.101325179687500E+06

12 0.101324898437500E+06 0.101324945312500E+06

13 0.101325039062500E+06 0.101325015625000E+06

Press any key to continue

Comparing of results

If you mention the results that gain from MATLAB and FORTRAN more precisely and compare

them with our findings in analytical solution, you can see clearly that MATLAB results is much more

closer to these results and it is interesting that MATLAB is much more easierto program. But off

course for learning computer programming FORTRAN

is a better guide because it has all thefundamentals of programming.


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B: Second solution

For solving this part we can make a function with one variable then we can solve it by bisection

method atthis part again we assume that .then we should find a way to write interms of or vase versa, doing this at last we would have a equation just in terms ofor. Thecomputational will be in this order

By looking in the figure we can see that solving this equation we can have in termsof

(35) (36) (37) (38)Now we have

in terms of

.if we look atthe figure we can see thatin nod number 1 we have this

equation and so (39)And from our previous calculation we have

So we can write the equation (28) this way

(41) (42)In this step we should write in terms of, so we will have (43)


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(44)And atlast we can show in terms oflike this (45)In the part (c) of the question it says that we should find the pressure values at all the nodesiterativelyto manage this we can have an objective function like this

(46) (47)Our objective function is now and we should try to find the zero of this objective function by amethod that we have learned in computational method, maybe bisection method is a proper way to do

this. Forthis project we use bisection method to solve for

, and , by this method we can findfrom the function and putting this in equation (40) we can find a corresponding to this. Butat first we should give a boundary to the variable ofthe function, as you will see in our program we

use 101,324 forlower bound and 101,326 for our upper bound, and for ourtolerance we use 0.0001.

MATLAB

function [p1,p2,p3,err]=bisect1(f,a,b,delta)

%producer Nima Fouladine jad MM081053

%Input - f is the function input as a string 'f'

% - a and b are the left and right endpoints

% - delta is the tolerance

%Output - p2 is the value for pressure in node number 2

% - p1 is the value for pressure in node number 1% - p3 is the value for pressure in node number 3

% - err is theerror estimate for p2

ya=feval(f,a);

yb=feval(f,b);

if ya*yb > 0,return,end

max1=1+round((log(b-a)-log(delta))/log(2));

fprintf('---------------------------------------------------------------------- \n');

fprintf('iter no. p1 p2 p3 err \n');

fprintf('--------------------------------------------------------------- -------\n');

for k=1:max1

p2=(a+b)/2;

yp2=feval(f,p2);if yp2==0

a=p2;

b=p2;

elseif yb*yp2>0

b=p2;

yb=yp2;

else

a=p2;


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ya=yp2;

end

yp2=feval(f,p2);

p1=1.952587883*p2-0.952587883*101325;

p3=101325;

err=abs(b-a);

fprintf('%.2d %.10e%.10e%.6e%.6e\n', k, p1, p2, p3, err);if b-a < delta, break,end

end

fprintf('---------------------------------------------------------------------- \n');

fprintf(' Nima Fouladinejad***[email protected]')

And with this program in MATLAB we achieve the results bellow for our problem

>> format long

>> f=@myfun;

>> [p1,p2,p3,err]=bisect1(f,101324,101326,.0001)

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

iter no. p1 p2 p3 err----------------------------------------- ------------------------------------------------------------

01 1.0132500000e+005 1.0132500000e+005 1.013250e+005 1.000000e+000

02 1.0132597629e+005 1.0132550000e+005 1.013250e+005 5.000000e-001

03 1.0132548815e+005 1.0132525000e+005 1.013250e+005 2.500000e-001

04 1.0132524407e+005 1.0132512500e+005 1.013250e+005 1.250000e-001

05 1.0132512204e+005 1.0132506250e+005 1.013250e+005 6.250000e-002

06 1.0132506102e+005 1.0132503125e+005 1.013250e+005 3.125000e-002

07 1.0132503051e+005 1.0132501563e+005 1.013250e+005 1.562500e-002

08 1.0132501525e+005 1.0132500781e+005 1.013250e+005 7.812500e-003

09 1.0132500763e+005 1.0132500391e+005 1.013250e+005 3.906250e-003

10 1.0132500381e+005 1.0132500195e+005 1.013250e+005 1.953125e-003

11 1.0132500191e+005 1.0132500098e+005 1.013250e+005 9.765625e-004

12 1.0132500286e+005 1.0132500146e+005 1.013250e+005 4.882813e-004

13 1.0132500238e+005 1.0132500122e+005 1.013250e+005 2.441406e-004

14 1.0132500215e+005 1.0132500110e+005 1.013250e+005 1.220703e-004

15 1.0132500226e+005 1.0132500116e+005 1.013250e+005 6.103516e-005

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

Nima Fouladinejad *** [email protected]

p1 =

1.013250022643536e+005

p2 =

1.013250011596680e+005

p3 =

101325

err =


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6.103515625000000e-005

Now we have all the data from , and in a table. We can save this data in a file in MATLAB by the suffix. dat . We did this and name the file rasm.dat. After that by using these data in

MATLAB we can plot

,

,

and

versus the iteration number. You can see these processes

bellow

>> load rasm.dat

>> x=rasm(:,1)

>> p1=rasm(:,2)

p1 =

1.0e+005*

1.01325000000000

1.01325976290000

1.01325488150000

1.01325244070000

1.01325122040000

1.01325061020000

1.01325030510000

1.01325015250000

1.01325007630000

1.01325003810000

1.01325001910000

1.01325002860000

1.01325002380000

1.01325002150000

1.01325002260000

>> plot(x,p1)

>>xlabel(iter. Number)

>>ylabel(p1)

>>

rid

Figure 3-1: p1 versus iteration number (MATLAB)


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>> p2=rasm(:,3)

p2 =

1.0e+005*

1.01325000000000

1.01325500000000

1.01325250000000

1.013251250000001.01325062500000

1.01325031250000

1.01325015630000

1.01325007810000

1.01325003910000

1.01325001950000

1.01325000980000

1.01325001460000

1.01325001220000

1.01325001100000

1.01325001160000

>> plot(x,p2)>>xlabel(iter. Number)

>>ylabel(p2)

>>!

rid

Figure 3-2: p2 versus iteration number (MATLAB)

>> p3=rasm(:,4)

p3=101325

101325

101325

101325

101325

101325

101325

101325


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101325

101325

101325

101325

101325

101325

101325>> plot(x,p3)


>>ylabel(p3)

>>"

rid

Figure 3-3 p3 versus iteration number (MATLAB)

>> err=rasm(:,5)err =

1.00000000000000

0.50000000000000

0.25000000000000

0.12500000000000

0.06250000000000

0.03125000000000

0.01562500000000

0.00781250000000

0.00390625000000

0.00195312500000

0.000976562500000.00048828130000

0.00024414060000

0.00012207030000

0.00006103516000

>> plot(x,err)


>>ylabel(error)


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>>#

rid

Figure 3-4: error versus iteration number (MATLAB)

FORTRAN

And here is the same program but this time writes with FORTRAN

C B $ SECTION METHOD

C producer Nima Fouladinejad MM081053

CInput - f is the function input as a string % F'

C - A and B are the left and right endpoints

C - N is the number of iterations

COutput - p2 is the value for pressure in node number 2

C - P1 is the value for pressure in node number 1C - P3 is the value for pressure in node number 3

C - err is the error estimate for p2

C

EXTERNAL F

DATA AF&

101324.0/, BF/101326.0/, N/15/

CALL BISECT(F,AF,BF,N)

STOP

END

FUNCTION F(X)

F=X-101325.00118

RETURN

ENDSUBROUTINE BISECT(F,A,B,N)

FA =F(A)

FB=F(B)

IF( SIGN(1.0,FA) .EQ. SIGN(1.0,FB) ) THEN

PRINT5,A,B

RETURN

ELSE

PRINT3


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WIDTH = B - A

DO 2 I = 1,N

p2 = A + (B - A)*0.5

Fp2 = F(p2)

WIDTH = WIDTH/2.0

IF( SIGN(1.0,FA) .' (

.SIGN(1.0,Fp2) ) THEN

A = p2FA = Fp2

ELSE

B = p2

FB = Fp2

ENDIF

p1=1.952587883*p2-(0.952587883*101325)

p3=11325.0000

PRINT 4,I,p1,p2,p3,WIDTH

2 CONTIN)

E

RET)

RN

ENDIF

3 FORMAT(//4X,'STEP',10X,'p1',19X,'p2',18X,'p3',11X,'ERROR',//)4 FORMAT(2X,I5,E21.13,E21.13,E21.13,E10.3)

5 FORMAT(2X,'F)

NCTION HASSAME SIGN AT',2E22.14)

END

And here are the results ofthe problem

STEP p1 p2 p3 ERROR

1 0.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.100E+01

2 0.1013240156250E+06 0.1013245000000E+06 0.1132500000000E+05 0.500E+00

3 0.1013245078125E+06 0.1013247500000E+06 0.1132500000000E+05 0.250E+00

4 0.1013247500000E+06 0.1013248750000E+06 0.1132500000000E+05 0.125E+00

5 0.1013248750000E+06 0.1013249375000E+06 0.1132500000000E+05 0.625E-01

6 0.1013249296875E+06 0.1013249687500E+06 0.1132500000000E+05 0.313E-01

7 0.1013249609375E+06 0.1013249843750E+06 0.1132500000000E+05 0.156E-01

8 0.1013249765625E+06 0.1013249921875E+06 0.1132500000000E+05 0.781E-02

9 0.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.391E-02

100.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.195E-02

110.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.977E-03

120.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.488E-03

13 0.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.244E-03

14 0.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.122E-03

15 0.1013249921875E+06 0.1013250000000E+06 0.1132500000000E+05 0.610E-04

And if we enter these data in MATLAB like the previous program we can plot,, and versus the iteration number as before. So we saved this data in a file fortplan with the suffix of.dat. And again for drawing the plots we have

>> format long

>> load fortplan.dat

>> x=fortplan(:,1)

x =

1

2


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3

4

5

7

8

9

1112

13

14

15

>> p1=fortplan(:,2)

p1 =

1.0e+005*

1.01324992187500

1.01324015625000

1.01324507812500

1.01324750000000

1.0132487

50000001.01324929687500

1.01324960937500

1.01324976562500

1.01324992187500

1.01324992187500

1.01324992187500

1.01324992187500

1.01324992187500

1.01324992187500

1.01324992187500

>> plot(x,p1)

>>xlabel(iteration Number)>>ylabel(p1)

>>grid

Figure 3-5: p1 versus iteration number (FORTRAN)


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>> p2=fortplan(:,3)

p2 =

1.0e+005*

1.01325000000000

1.01324500000000

1.01324750000000

1.013248750000001.01324937500000

1.01324968750000

1.01324984375000

1.01324992187500

1.01325000000000

1.01325000000000

1.01325000000000

1.01325000000000

1.01325000000000

1.01325000000000

1.01325000000000

>> plot(x,p2)>>xlabel(iteration Number)

>>ylabel(p2)

>>grid


>> p3=fortplan(:,4)

p3=

1132511325

11325

11325

11325

11325

11325

11325

11325


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11325

11325

11325

11325

11325

11325

>> plot(x,p3)>>xlabel(iteration Number)

>>ylabel(p3)

>>grid


>> error=fortplan(:,5)error =

1.00000000000000

0.50000000000000

0.25000000000000

0.12500000000000

0.06250000000000

0.03130000000000

0.01560000000000

0.00781000000000

0.00391000000000

0.00195000000000

0.000977000000000.00048800000000

0.00024400000000

0.00012200000000

0.00006100000000

>> plot(x,error)

>>xlabel(iteration Number)

>>ylabel(error)

>>grid


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Figure 3-8: error versus iteration number (FORTRAN)

Comparing of results

If you mention the results that gain from MATLAB and FORTRAN more precisely and compare

them with our findings in analytical solution, you can see clearly that MATLAB results is much more

closer to these results and it is interesting that MATLAB is much more easier to program. But off

course for learning computer programming FORTRAN is a better guide because it has all the

fundamentals of programming.

Documents

Computational method for engineers