why I get the error of "Unable to find variables in equations." in line 29?

I have the function in line 17 (calculating M2) and it works but the same function in line 29 (calculating Mx) doesnt work and i get the error above?

how can i fix it?

clc
clear
syms M2;
syms Mx;
L=input('enter the L value\n');
D=input('enter the D value\n');
P1=input('enter the P1 value\n');
T1=input('enter the T1 value\n');
M1=input('enter the M1 value\n');
f=0.005;
L1=0.1;
Ls1=(((1-M1^2)/(1.4*(M1^2)))+((2.4/2.8)*log((2.4*(M1^2))/(2+0.4*(M1^2)))))*(D/(4*f));
  Z = sprintf('L* is = %s.',Ls1);
  disp(Z);
if L<Ls1
    Ls2=Ls1-L;
M2=vpasolve((((1-M2^2)/(1.4*(M2^2)))+((2.4/2.8)*log((2.4*(M2^2))/(2+0.4*(M2^2)))))*(D/(4*f))- Ls2,M2);
P2=P1*(M1/M2)*(((2+0.4*(M1^2))/(2+0.4*(M2^2)))^0.5);
T2=T1*((2+0.4*(M1^2))/(2+0.4*(M2^2)));
  X = sprintf('M2 is = %s.',M2);
  Y = sprintf('T2 is = %s and P2 is = %d .',T2,P2);
disp(X);
disp(Y);
else
    M2=1;
    while L1<L
        L2=L-L1;
        Lsx=Ls1-L1;
        Mx=vpasolve((((1-Mx^2)/(1.4*(Mx^2)))+((2.4/2.8)*log((2.4*(Mx^2))/(2+0.4*(Mx^2)))))*(D/(4*f))- Lsx,Mx);

Answer 1

[Walter Roberson]

You define

syms Mx

then you do

Mx=vpasolve(an expression in Mx, Mx);

Before you do that first vpasolve(), Mx holds a reference to the symbolic variable named Mx . After the assignment, Mx holds a reference to the symbolic numeric result of the vpasolve(), roughly sym(1.934) .

The next iteration of the loop, Mx is the sym(1.934), and the expression in Mx evaluates to a symbolic numeric result, and the Mx at the end of the call is sym(1.934)-ish, so you have asked to vpasolve() and passed two symbolic numeric values that have no unresolved symbolic variable names. vpasolve() cannot figure out what you want to solve for.

[fatemeh p]

so what should i do to fix it?

[Walter Roberson]

Ideally you should avoid assignment a value to any variable you have defined as a symbolic variable: there is just too much risk of the kind of error you just encountered, or risk that you will program something like

A = expression in Mx
Mx = new value

and then expect A to operate as if it had the new value of Mx.

For example if you had assigned to

sol_Mx = vpasolve(....,Mx);
    

and had then used sol_Mx in the later lines, then the next iteration through the loop, Mx would still be the symbolic variable it originally was and the problem would not have occurred.

The alternative to the ideal is...

Insert

syms Mx

inside the while loop.

[fatemeh p]

it worked, thanks a lot

Answer 2

[madhan ravi]

syms M2 Mx
L=input('enter the L value\n');
D=input('enter the D value\n');
P1=input('enter the P1 value\n');
T1=input('enter the T1 value\n');
M1=input('enter the M1 value\n');
f=0.005;
L1=0.1;
Ls1=(((1-M1^2)/(1.4*(M1^2)))+((2.4/2.8)*log((2.4*(M1^2))/(2+0.4*(M1^2)))))*(D/(4*f));
Z = sprintf('L* is = %s.',Ls1);
disp(Z);
if L<Ls1
    Ls2=Ls1-L;
    M2=vpasolve((((1-M2^2)/(1.4*(M2^2)))+((2.4/2.8)*log((2.4*(M2^2))/(2+0.4*(M2^2)))))*(D/(4*f))- Ls2,M2);
    P2=P1*(M1/M2)*(((2+0.4*(M1^2))/(2+0.4*(M2^2)))^0.5);
    T2=T1*((2+0.4*(M1^2))/(2+0.4*(M2^2)));
    X = sprintf('M2 is = %s.',M2);
    Y = sprintf('T2 is = %s and P2 is = %d .',T2,P2);
    disp(X);
    disp(Y);
else
    M2=1;
    L2=L-L1;
    Lsx=Ls1-L1;
    Mx=vpasolve((((1-Mx^2)/(1.4*(Mx^2)))+((2.4/2.8)*log((2.4*(Mx^2))/(2+0.4*(Mx^2)))))*(D/(4*f))- Lsx,Mx);    
end

[fatemeh p]

i dont understand

it is because of "while"?

but i need that part of my code.

here is the complete one.

clc
clear
syms M2;
syms Mx;
L=input('enter the L value\n');
D=input('enter the D value\n');
P1=input('enter the P1 value\n');
T1=input('enter the T1 value\n');
M1=input('enter the M1 value\n');
f=0.005;
L1=0.1;
Ls1=(((1-M1^2)/(1.4*(M1^2)))+((2.4/2.8)*log((2.4*(M1^2))/(2+0.4*(M1^2)))))*(D/(4*f));
  Z = sprintf('L* is = %s.',Ls1);
  disp(Z);
if L<Ls1
    Ls2=Ls1-L;
M2=vpasolve((((1-M2^2)/(1.4*(M2^2)))+((2.4/2.8)*log((2.4*(M2^2))/(2+0.4*(M2^2)))))*(D/(4*f))- Ls2,M2);
P2=P1*(M1/M2)*(((2+0.4*(M1^2))/(2+0.4*(M2^2)))^0.5);
T2=T1*((2+0.4*(M1^2))/(2+0.4*(M2^2)));
  X = sprintf('M2 is = %s.',M2);
  Y = sprintf('T2 is = %s and P2 is = %d .',T2,P2);
disp(X);
disp(Y);
else
    M2=1;
    while L1<L
        L2=L-L1;
        Lsx=Ls1-L1;
        Mx=vpasolve((((1-Mx^2)/(1.4*(Mx^2)))+((2.4/2.8)*log((2.4*(Mx^2))/(2+0.4*(Mx^2)))))*(D/(4*f))- Lsx,Mx);
%  [downstream_mach] = flownormalshock(1.4 , Mx)
My=((1+0.2*(Mx^2))/(1.4*(Mx^2)-0.2))^0.5;
Lsy=(((1-My^2)/(1.4*(My^2)))+((2.4/2.8)*log((2.4*(My^2))/(2+0.4*(My^2)))))*(D/(4*f));
error=abs(Lsy-L2);
if (error<=0.05)
    P2=P1*(M1/Mx)*(((2+0.4*(M1^2))/(2+0.4*(Mx^2)))^0.5);
    P3=P2*(1+((2.8/2.4)*((Mx^2)-1)));
    P4=P3*Mx*(((2+0.4*(Mx^2))/2.4)^0.5);
    T2=T1*((2+0.4*(M1^2))/(2+0.4*(Mx^2)));
    T3=T2*(1+((2.8/2.4)*((Mx^2)-1)))*((2+0.4*(Mx^2))/(2.4*(Mx^2)));
    T4=T3*((2+0.4*(Mx^2))/2.4);
    U=sprintf('T2 is = %s and P2 is = %d .',T4,P4);
    disp(U);
    K=sprintf('M2 is = %s and The shock happens in = %d .',M2,L1);
    disp(K);
    break
end
L1=L1+0.05;
    end
end

[Walter Roberson]

What are some sample inputs for us to test with?

[fatemeh p]

of cource

this code producing the ouput mach number,pressure and temperture of a pipe with extra friction in aerodynamics

for a L=1.8, T1=288, P1=1, D=0.1, M1=2 it should go into while loop.