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Re: Pump - Problems


From: klausp
Subject: Re: Pump - Problems
Date: Sun, 14 Apr 2013 11:25:17 -0700 (PDT)

Thank you Dmitri!

Well, I understand that the mathematical model wasn´t right. I corrected the
problem but it still not working.
I´ve tried different initial values and some substituctes of the Colebrook
law (like Haaland). And isn´t working. could someone give more one tip?

Thank you!
Klaus Peter

>>>warning: matrix singular to machine precision, rcond = 4.36065e-018 
warning: attempting to find minimum norm solution 
warning: zgelsd: rank deficient 3x3 matrix, rank = 2, tol = -1.000000e+000 
>>>warning: matrix singular to machine precision, rcond = 2.66901e-020 
warning: attempting to find minimum norm solution 
warning: zgelsd: rank deficient 3x3 matrix, rank = 2, tol = -1.000000e+000 
warning: matrix singular to machine precision, rcond = 8.19905e-022 
warning: attempting to find minimum norm solution 
warning: zgelsd: rank deficient 3x3 matrix, rank = 2, tol = -1.000000e+000 
warning: matrix singular to machine precision, rcond = 4.57172e-017 
warning: attempting to find minimum norm solution 
warning: zgelsd: rank deficient 3x3 matrix, rank = 2, tol = -1.000000e+000 
warning: matrix singular to machine precision, rcond = 4.36065e-018 
warning: attempting to find minimum norm solution 
warning: zgelsd: rank deficient 3x3 matrix, rank = 2, tol = -1.000000e+000 
 

The code:
%pump1 
clear all; 

#FUNCTION DEFINITION 
function y=fun(x) 


#VARIABLES 
%x=diam; 

P2=x(1); 
Q= x(2); 
FM=x(3); 

#PARAMETERS DEFINITION AND CONVERTION TO INTERNATIONAL SYSTEM 
P1=14.696;%psi 
P3=14.696;%psi 
P1=P1*6894.75729;%Pa 
P3=P3*6894.75729;%Pa 

a=16.7; %psi 
a=a*6894.75729;%Pa 

b=0.052;%psi/(gpm)^1,5 
b=6894.75*b;%Pa/(gpm)^1,5 
L=50;%ft 
L=0.305*L;%m 
p=62.4;%lbm/ft3 
p=16.02*p;%kg/m3 

vis=0.00003228;%VISCOSITY-lbs/ft2 
vis=0.00102;%m.s/ft2 


E=0.00006;%metros 

D=2.469;%in 
D=D*0.0254;meters 
v=2;%m/s 

#Another equations 
Re=p*v*D/vis; %Reynolds 


#EQUATIONS SET 
y(1)=a-b*Q^1.5-P2+P1; %Empirical/Pump 
y(2)=(8*FM*p*L*Q/(pi*D^5))-P2+P3;%Relation flow and pressure loss 
y(3)=-2*log10((E/(3.7*D))+(2.51/(Re*sqrt(FM))))-1/sqrt(FM);%Colebroock
equation 


endfunction 

#Initial values 

x0=[200105 5 1.5e-2]; 


%Solving 

[x,fval,info]=fsolve(@(x) fun(x),x0); 
 



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