iteration

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mohamed saber
mohamed saber 2011-11-12
plz .... why there is no answer in this iteration when i use this input by order :
30 , .01 ,1 1000 ,18,.02 ,1.2 ,1200, 2 , .05 , .75 , 1400
clear , clc
fprintf(' \n <<< this program calculate Q in three branched pipes from three tanks >>> \n %12.5f ')
fprintf (' \n 1- Data for highest tank \n %12.5f')
% Data for highest tank
E1=input ('\n Enter tank_ energy (m) : ');
f1=input (' Enter pipe_friction coefficient (m) : ');
d1=input (' Enter pipe_diameter (m) : ');
l1=input (' Enter pipe_length (m) : ');
fprintf (' \n 2- Data for intermediate tank \n %12.5f ')
% Data for intermediate tank
E2=input ('\n Enter tank_ energy (m) : ');
f2=input (' Enter pipe_friction coefficient (m) : ');
d2=input (' Enter pipe_diameter (m) : ');
l2=input (' Enter pipe_length (m) : ');
fprintf (' \n 3- Data for lower tank \n %12.5f ')
% Data for lower tank
E3=input ('\n Enter tank_ energy (m) : ');
f3=input (' Enter pipe_friction coefficient (m) : ');
d3=input (' Enter pipe_diameter (m) : ');
l3=input (' Enter pipe_length (m) : ');
% resistance calculation
R1=(.8*f1*l1)/(9.8*d1^5); % 1st pipe resistance
R2=(.8*f2*l2)/(9.8*d2^5); % 2nd pipe resistance
R3=(.8*f3*l3)/(9.8*d3^5); % 3rd pipe resistance
% junction's energy assumption (Ej)
Ej=(E1+E3)/2; % E1 <Ej< E3
% Q calculation
Q1=((abs(E1-Ej))/(R1))^.5;
Q2=((abs(E2-Ej))/(R2))^.5;
Q3=((abs(E3-Ej))/(R3))^.5;
% Q enter to j is +ve , Q out from j is -ve
if E2<Ej % Q in 2nd pipe is -ve
dQ=Q1-Q2-Q3; % applying sign law
elseif E2>Ej % Q in 2nd pipe is +ve
dQ=Q1+Q2-Q3; % applying sign law
end
if dQ>0 % if dQ +ve ... increase Ej
iter=0;
while abs(dQ)>.0001 % tolerrance
Ej=Ej+.001; % increase Ej
iter=iter+1; % no of try & error
q1=((abs(E1-Ej))/(R1))^.5;
q2=((abs(E2-Ej))/(R2))^.5;
q3=((abs(E3-Ej))/(R3))^.5;
if Ej>E2
dQ=q1-q2-q3;
else
dQ=q1+q2+q3;
end
end
fprintf('\n Q at 1st pipe(m^3/s)= %12.5f\n ',q1)
fprintf('\n Q at 2nd pipe(m^3/s)= %12.5f\n ',q2)
fprintf('\n Q at 3rd pipe(m^3/s)= %12.5f\n ',q3)
fprintf('\n Energy at junction(m)= %12.5f\n ',Ej)
fprintf('\n no.of iteration= %12.5f\n',iter)
elseif dQ<0; % if dQ +ve ... decrease Ej
iter=0;
while abs(dQ)>.0001 % tolerrance
Ej=Ej-.001; % decrease Ej
iter=iter+1; % no of try & error
q1=((abs(E1-Ej))/(R1))^.5;
q2=((abs(E2-Ej))/(R2))^.5;
q3=((abs(E3-Ej))/(R3))^.5;
if Ej>E2
dQ=q1-q2-q3;
else
dQ=q1+q2+q3;
end
end
fprintf('\n Q at 1st pipe(m^3/s)= %12.5f\n ',q1)
fprintf('\n Q at 2nd pipe(m^3/s)= %12.5f\n ',q2)
fprintf('\n Q at 3rd pipe(m^3/s)= %12.5f\n ',q3)
fprintf('\n Energy at junction(m^3/s)= %12.5f\n ',Ej)
fprintf('\n no.of iteration(m^3/s)= %12.5f\n',iter)
end
generally , the program run well when (E1+E3)/2 >E2 but it doesn't run when (E1+E3)/2 >E2
please , help me

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回答（1 个）

Jan
Jan 2011-11-13
You can use the debugger to check, what's going on: Open the file in the editor, set a breakpoint after the input commands and step through the program line by line.
Working with the debugger is more efficient than asking the forum.

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