solving a system of equation using symbolic expressions returns empty solution

Hello,

I'm trying to solve a system of equation using symbolic expressions but it outputs empty value.

% values are predefined for the following variables: CG0, HC, R, T, K0, Cm0, beta
syms RH q_H2O CG K Cm

eq1 = q_H2O == Cm * CG*K*RH / ((1-K*RH)*(1+(CG-1)*K*RH));
eq2 = CG == CG0*exp(HC/R/T);
eq3 = K == K0*exp(HK/R/T);
eq4 = Cm == Cm0*exp(beta/T);

eqns = [eq1, eq2, eq3, eq4];
[~, ~, ~, q_H2O] = solve(eqns);

Running this code yields:

ans = 

  struct with fields:

       CG: [0×1 sym]
        K: [0×1 sym]
       RH: [0×1 sym]
    q_H2O: [0×1 sym]

I was expecting a result that is a function of 'RH' (q_H2O = f(RH))

But if I predefine 'RH' and run the code (e.g. RH=0.5), this code outputs a value as a result.

I would appreciate any comments!

Answer 1

[Walter Roberson]

syms CG0  HC R T K0 Cm0 beta
syms RH q_H2O CG K Cm HK

eq1 = q_H2O == Cm * CG*K*RH / ((1-K*RH)*(1+(CG-1)*K*RH));
eq2 = CG == CG0*exp(HC/R/T);
eq3 = K == K0*exp(HK/R/T);
eq4 = Cm == Cm0*exp(beta/T);

eqns = [eq1, eq2, eq3, eq4];
sol = solve(eqns, [q_H2O, CG, K, Cm] )

[Walter Roberson]

Remember, when you do not specify which variables to solve for, solve will choose by itself

[Tae Lim]

Hi Walter, thank you for your reponse. the other variables not defined as symbolic variable have predefined values. Please see below.

    CG0 = 7;
    HC = -4;
    K0 = 2;
    HK = -2.5;
    Cm0 = 0.02;
    beta = 1500;
    R = 8;
    T = 300;

    syms RH q_H2O CG K Cm

    eq1 = q_H2O == Cm * CG*K*RH / ((1-K*RH)*(1+(CG-1)*K*RH));
    eq2 = CG == CG0*exp(HC/R/T);
    eq3 = K == K0*exp(HK/R/T);
    eq4 = Cm == Cm0*exp(beta/T);

    eqns = [eq1, eq2, eq3, eq4];
    sol = solve(eqns, [q_H2O])
    
    
sol =
 
Empty sym: 0-by-1

In this case I still have the same issue. I am not sure what went wrong as 'q_H2O' is the only unknown here.

[Walter Roberson]

When you solve(), the number of equality equations cannot exceed the number of unknowns being solved for.

[Walter Roberson]

sol = solve(eqns, [q_H2O, CG, K, Cm] )

You are not required to use all of the components returned.

[Tae Lim]

Hi Walter, Thank you again for your comments. I am still confused since I previously used the code to solve another system of equations without issues. Please see below:

ns0 = 2;
chi = 2;
T0 = 296;
T = 300;
b0 = 2e5;
H_CO2 = 60;
R = 8;
t0 = 1;
alpha = 1;

syms P_CO2 q ns b t

eq1 = q == ns*b*P_CO2 / (1+(b*P_CO2)^t)^(1/t);
eq2 = ns == ns0 * exp(chi * (1-T0/T));          % maximum adsorption capacity
eq3 = b == b0 * exp(H_CO2/R/T0 * (T0/T-1));    % adsorption affinity
eq4 = t == t0 + alpha * (1-T0/T);              % heterogeneity of the adsorption system

eqns = [eq1, eq2, eq3, eq4];
solve(eqns)

ans = 

  struct with fields:

     b: 3434813230887443/17179869184
    ns: 4625311226299041/2251799813685248
     q: (15887080197064170033712929842163*P_CO2)/(38685626227668133590597632*(((3434813230887443*P_CO2)/17179869184)^(76/75) + 1)^(75/76))
     t: 76/75

I do not really see the primary difference between this code and the code above. In essence, I'm just trying to generate an equation in the form y=f(x) so I can either evaluate 'y' in specific values of 'x' or generate plots of 'y' as a function of 'x'. I appreciate your help!

[Torsten]

As Walter already said:

In your previous code, you specified to solve for one solution variable q_H2O with four equations. This is not possible with solve - the number of equations mustn't exceed the number of variables solved for.

In the code above, you didn't specify the variables to solve for. So MATLAB chose b, ns, q and t as sympathic and expressed q in terms of the last of the five symbolic variables, namely P_CO2. If you try harder, you might find the rule MATLAB applied to choose the variables it solved for (maybe alphabetic order or something similar).

[Tae Lim]

Hi Walter and Torsten, thank you for your comments! I ended up simplifying my codes to evaluate just one function with one unknown since I can fix all the other variables into certain values (for now). I don't need to worry about a system of equations in this case. I appreciate all your help!