How to solve 4 equations with 4 unknowns using matlab

I'm attempting to use matlab to solve a set of 4 nonlinear equations with 4 unknowns. This is what I've entered so far, but it keeps saying it cant find a solution. Any suggestions?

mf=.12;
Yfin=.0826;
Kg=(6.19*10^9)*exp(-15098/298);
MW=29;
V=.000268;
P=101325;
R=8315;
M=.1;
N=1.65;
Yoxin=.9174;
AF=16;
Hf=40000000;
Cp=1200;
Tin=298;
eqn1='mf*(Yfin-R1)-Kg*MW*V*((P/R*R2)^(M+N))*(((R1^M)*((.233*R3)^N))/1)'
eqn2='mf*(Yoxin-R1)-(AF)*Kg*MW*V*((P/R*R2)^(M+N))*(((R1^M)*((.233*R3)^N))/1)'
eqn3='1-R1-R3-R4'
eqn4='(R1-Yfin)*Hf+Cp*(R2-Tin)'
sol=fsolve(eqn1,eqn2,eqn3,eqn4,'R1','R2','R3','R4')

[Azzi Abdelmalek]

We can't test your code. Posting an image instead of a text is not a good idea

[Josh]

sorry, just uploaded

[Walter Roberson]

The fsolve() syntax you are using would be the syntax for solve(), a symbolic solver, rather than fsolve(), a numeric solver. If you want to use the symbolic solver, you should use

syms R1 R2 R3 R4

and then when you define your equations, do not put in the quotation marks,

eqn1 = mf*(Yfin-R1)-Kg*MW*V*((P/R*R2)^(M+N))*(((R1^M)*((.233*R3)^N))/1);

and so on. And change to solve()

Answer 1

[Roger Stafford]

I suggest you try to solve it numerically using 'fsolve' rather than symbolically.

[Josh]

Can you explain a little bit. I'm sorry, I'm a beginner at using matlab. Thank you for your answer

[Roger Stafford]

The 'fsolve' function is found in the Optimization Toolbox. You can read about it at:

 http://www.mathworks.com/help/optim/ug/fsolve.html

It requires that you make an initial estimate of a solution and it attempts to converge to an accurate solution. If there are many roots, it will be necessary to use many initial estimates in order to find them all. It uses matlab's usual double precision numbers rather than the symbolic forms used by 'solve'. It does require that you specify the numerical values of all parameters involved but you have satisfied that requirement.

[Josh]

I think I've managed to do it , can you please check ? Thank you very much for your time!!!

Answer 2

[Walter Roberson]

You will probably not be able to find a meaningful numeric solution. The gradient of the function is very very steep, and outside a very narrow valley of R3 values it goes complex. The width of the range of R3 values that lead to non-complex values is less than eps() of the representable value, so numerically you are always going to end up with complex results if you using "double". You can get meaningful results if you work symbolically with more than 150 digits of precision.

R1 is on the order of 8 * 10^(-12), R2 is on the order of 3051 1/3, R3 is on the order of 8336.5; and because of the constraint that 1-R1-R3-R4 = 0, R4 comes out approximately -8335.5

Answer 3

[Alex Sha]

r1: 0.0269466666666667

r2: 2153.11111111111

r3: 2523.94560037811

r4: -2522.97254704477