solve failing on some values.

hi and thank you for the attention.

i'm trying to use Matlab to write a code on waste waters pipes sizingfor my college thesis. i have started since one year ago and i did more tries

i have this problem that i will write and i don't know how to fix it.

it consist in it : i use GDRC = solve( eqn == value , GDR ), it works for more values that i give in input passing them as datas to the function that contains GDRC , but failes for some of them, saying me

" Warning: Explicit solution could not be found. > In solve at 179 In GDRC at 17 In Main at 45 Error using sym/eval (line 11) Error: This statement is incomplete.

Error in GDRC (line 17) GDRC = eval(solve ( Q_Prog == Q , GDR ) ) ;

Error in Main (line 45) [Val_Comm] = GDRC ( GS, Q_Prog, j, Ris_Prog(5) )"

my code ( and functions) ( see below ) is based on no-linear problem. for same datas, excell solver and maple one find a real value of GDR and it is the same for both, why matlab not?

Main

clear all

% Dati di prova ----------------------------------------

Q_Prog = 0.030 % changing  and encreasing this value  Dimensionamento sizes the diameter but GDRC gives error 8 it recalculates idraulic parameters for common pipes ( bigger than calculated teorical diameter in dimensionamento)
% 
% 

% Q_Prog = 0.04
% 
% GS = 70 ;
% 
% L = 370 ;
% 
% z_i = 53 ;
%  
% z_f = 52 ;

%
%
GS = 70 ;

L = 370 ;

z_i = 53 ;

z_f = 52 ;

j = ( z_i - z_f) / L ;

GDR_Prog = 0.75 ;

%-------------------------------------------------------

% Diametro teorico e relativi valori idraulici ed inviduazione 
% diametro commerciale --------------------------------

[Ris_Prog] = Dimensionamento( GDR_Prog, Q_Prog, GS, j )

% valori idraulici relativi diametro commeciale 
% ed eventuale cambio pendenza --------------------------

   [Val_Comm] = GDRC ( GS, Q_Prog, j, Ris_Prog(5) )

dimensionamento ( sizing )

function [ Ris_Prog ] = Dimensionamento( GDR_Prog, Q_Prog, GS, j )

clear beta A R chi D DC

syms DT real

beta = 2 * acos( 1 - 2 * GDR_Prog ) ;

A = ( DT ^ 2 / 8) * ( beta - sin( beta ) ) ;

R = ( DT / 4 ) * ( 1 - ( sin( beta ) / beta ) ) ;

chi = GS * R ^ ( 1 / 6 ) ;

Q = chi * A * R ^ ( 1 / 2 ) * j ^ ( 1 / 2) ;

DT = eval( solve ( Q_Prog == Q , DT ) ) ;

D = [ 200: 50 : 2000 ] ;

A = eval ( subs ( A , DT , DT ) ) ;

R = eval ( subs ( R , DT , DT ) ) ;

chi = eval ( subs ( chi , R , R ) ) ;

for a = 1 : length ( D )

      if D(a) / 1000 >= DT

          DC = D(a) / 1000 ;

          break

      end

end

Ris_Prog = [ DT, A, R, chi, DC ] ;

end

GDRC ( values for real diameters in shops )

function [Val_Comm] = GDRC ( GS, Q_Prog, j, DC )

clear  GDR beta A R chi AC chiC RC GDRC

syms GDR real

beta = 2 * acos( 1 - 2 * GDR ) ;

A = ( DC ^ 2 / 8 ) * ( beta - sin(beta) ) ;

R = ( DC / 4 ) * ( 1 - ( sin(beta) / beta )) ;

chi = GS * R ^ ( 1 / 6 ) ;

Q = chi * A * R ^ ( 1 / 2 ) * j ^ (1 / 2 ) ;

GDRC = eval(solve ( Q_Prog == Q , GDR ) ) ;

betaC = eval( subs(beta, GDR, GDRC )) ;

AC = eval( subs(A, GDR, GDRC )) ;

RC = eval( subs(R, GDR, GDRC  )) ;

chiC = eval( subs(chi, GDR, GDRC  )) ;

v_c = Q_Prog / AC ;

Val_Comm = [ GDRC, betaC, AC, RC, chiC, v_c ] ;

end

[Ecor]

i successed to find solutions using vpasolve.

GDRC = eval(vpasolve( Q_Prog == Q , GDR,[0.0001 , 0.75] ) )  ;

at this point i think there is a bug in solve script and i will try to report this error to MATLAB report bug.

thank you again for your attention

Answer 1

[Walter Roberson]

Never eval() a symbolic formula. Symbolic formula are written in a language which is not exactly the same as MATLAB, so the results are not going to be what you want.

If you have a symbolic formula that you want to evaluate into a number, then use double() on the formula.

[Ecor]

hi and thanks for your answer. the problem persist also using double, as you wrote, and don't using eval( solve (..., gdr)

if i change gdr from "syms GDR real" to " syms GDR " it works but it finds complex numbers.

also if i encrease the diameter, GDRc fuction works. this truely weird for me.

thank you again.