Is It possible create a function that, given an equation, finds the value of the (x) so the two members are equal?

Hi,

I have the previous problem and consists in finding the values of (Mach) and (pp1). I thought to find then the values of (Mach) and (pp1) such that the two members of the equation are equal and I used the function "fsolve" but without acceptable results (I have insert my personal script and linked function). Is it possible use the construct "while"?

[dpb]

Did you not have an earlier Q? (or two) on the same problem? What happened to those? Seems like I suggested fzero there??? But, I see no links...

What are knowns and what is/are unknown(s) in the equations figure?

[Gianmarco Moccia]

These are the equations and I need to valuate the quantities (Mp1) and (pp1) and I propose the following script file and linked function file but the result is not acceptable

%DATI_INPUT_GEOMETRICI 
Dt=2.82 %mm 
At=6.2458 %mm^2 
Dp1=5.10 %mm 
Ap1=20.4282 %mm^2 
A3=66.4761 %mm^2 
%DATI_INPUT_TERMODINAMICI 
pev=0.040 %Mpa 
Tev = 8 %°C 
pg=0.604 %Mpa 
Tg=95 %°C 
cp= 0.85 %[kJ/(kg*K)]
 gam=1.2 R=0.143 %[kJ/(kg*K)] 
eta_p=0.95 
Msy=1 
lossp=0.88 
etap = 0.95 
%Algoritmo 
%Calcolo porata massica del primario 
mp=((pg*At)/sqrt(Tg))*sqrt((gam/R)*(2/(gam+1))^((gam+1)/(gam-1)))*sqrt(eta_p) 
%Calcolo della pressione del fluido primario e del numero di Mach del  %primario nella sezione 1  
x0=[1.1,0.1]; xsol = fsolve(@(x)NLfunctions1(x), x0) 
return

Function file 
function F = NLfunctions1(x)
Ap1=20.4282  
At=6.2458 
gam=1.2 
pg=0.604 
%definizione funzioni 
F = (Ap1/At)^2 - ((1/(x(1))^2))*[(2/(gam+1))*(1+((gam-1)/(2)))(x(1))^(2))]^((gam+1)/(gam-1)); 
F = ((pg)/(x(2)))-(1+(((gam-1)/(2))*x(1)^(2))^(gam/(gam-1))); 
end

[dpb]

Sorry, I ran out of time last night...didn't have chance to do anything -- but, one thing I notice that is problem in the NLFunctions1 routine -- you don't return a 2-vector of the the two equations--but only one.

See if

function F = NLfunctions1(x)
  Ap1=20.4282  
  At=6.2458 
  gam=1.2 
  pg=0.604 
  %definizione funzioni 
  F(1) = (Ap1/At)^2 - ((1/(x(1))^2))*[(2/(gam+1))*(1+((gam-1)/(2)))(x(1))^(2))]^((gam+1)/(gam-1)); 
  F(2) = ((pg)/(x(2)))-(1+(((gam-1)/(2))*x(1)^(2))^(gam/(gam-1))); 
end

won't help...

Answer 1

[David Goodmanson]

Hi Gianmarco,

the two equations aren't coupled, and using fzero on Mp^2 seems to work. In the first equation I multiplied by both sides by Mp^2 and took both sides to the (gam-1)/(gam+1) power, so the variations in the functions are smaller and the search limits are easier to guess.

At=6.2458;    % mm^2 
Ap1=20.4282;  % mm^2 
pg=0.604;     % Mpa 
gam=1.2;
fun1 = @(m2) (m2*(Ap1/At)^2).^((gam-1)/(gam+1));   % left
fun2 = @(m2) (2/(gam+1))*(1+((gam-1)/2)*m2);       % right
Mp2 = fzero(@(m2) fun1(m2)-fun2(m2), [.01 1])
pp1 = pg*(1+((gam-1)/2)*Mp2).^(-gam/(gam-1))
% check
(Ap1/At)^2 - (1/Mp2)*((2/(gam+1)*(1+((gam-1)/2)*Mp2))).^((gam+1)/(gam-1))
ans = 4.6185e-14
pg - pp1*(1+((gam-1)/2)*Mp2).^(gam/(gam-1))
ans = 0