What function (if any) can I use to simplify symbolic equation in terms of another variable?

I would like dH3 in terms of Cv

Edit: I see that I have caused a bit of confusion here. What I meant to say was, I would like dH3 in terms of both Cp and Cv. I can achieve this by hand manipulation. For example, if we multiply the coefficient of dV1 by beta/beta then we can rewrite the coefficient as Cp/(Cv*beta). This produces a simpler expression by simply multiplying the numerator and denominator by beta and substituting in Cv. I hope this helps.

[Star Strider]

If you post (or attach the .m file of) your complete code, we could experiment with it.

[Miguel Ramirez]

Sounds good, I have attached the files. Let me know if you or anyone need me to clarify the code.

[Walter Roberson]

Note: you cannot pass a cell array as the second parameter of collect(): you need to change those {} list of variables to [] lists.

[Miguel Ramirez]

I changed the list of variables within the braces {}, to brackets [], but I didn't see any difference. Would you mind clarifying what you meant? Thanks in advance.

Answer 1

[Joshua]

If you define your variables using the 'syms' function, you can create equations that will simplify. For example, the code

syms a b c d e
b=d*e
a=b+c

will give the following output

b = 
d*e

a =
c + d*e

showing that the equation b=d*e was automatically plugged into the equation a=b+c. You can do something similar for your two equations. Just make sure to put the second equation in the form cp= and put it before your larger equation.

[Miguel Ramirez]

I think for the expression dH3, it will need some algebraic manipulation to achive dH3(Cv, alpha, V etc..). What I mean by this is, for the coefficient of dV1, I would need to multiply the numerator and denominator by beta, so that I may substitute Cv in the denominator (this would mean that I end up with [Cp/(beta*Cv)]dV1.

I see what you mean by the code you wrote, but this is assuming we do not need algebraic manipulation for the substitution.

Answer 2

[Star Strider]

Looking at your equations (specifically in ‘T4_5.m’), the only way I can see to cast them in terms of ‘Cv’ is to solve the last equation for ‘Cp’, the substitute in the rest of them:

Eqn = Cv == Cp - (T*V*alpha^2)/beta;
Cpv = solve(Eqn, Cp);

sys = [dV; dS; dU; dH; dF; dG];
sys = simplify(subs(sys, Cp, Cpv), 'Steps',10)

This gives:

sys =
                                             V*(alpha*dT - beta*dP)
                      (dT*((T*V*alpha^2)/beta + Cv))/T - V*alpha*dP
 dT*((T*V*alpha^2)/beta - P*V*alpha + Cv) + V*dP*(P*beta - T*alpha)
                  dT*((T*V*alpha^2)/beta + Cv) - V*dP*(T*alpha - 1)
                                   P*V*beta*dP - dT*(S + P*V*alpha)
                                                        V*dP - S*dT

leaving you with the slightly boring task of having to re-define ‘dV’*, ‘dS’, ‘dU’, ‘dH’, ‘dF’, and ‘dG’ in terms of the elements of ‘sys’. This would give your final result as a function of ‘Cv’ (and other variables).

If I understand correctly what you want to do, this could work. If not, at least it may suggest an approach to a solution.

[Miguel Ramirez]

It would be great to have the final expression in terms of Cp and Cv, I've tried altering the code to see if I could get sys in terms of Cp and Cv and it does not work out unless I were to perform some algebraic manipulation/'trickery' by hand.

In the expression dH3, I can recognize that if I multiply the coefficient of dV1 by beta/beta, then I can substitute in Cv all while leaving in Cp (i.e., simplify the expression into a simpler one or one with a lower amount of variables).

Answer 3

[John D'Errico]

I would just use subs. Without running your scripts, this will work:

dH3 = subs(dH3,Cp,Cv + T*V*alpha/beta)

Now the entire expression will be in terms of Cv. You might want to do some simplification, but that comes after the substitution. I could have picked some other variable to use for the substitution.

[Miguel Ramirez]

I've tried this in the past and it did not work for me. See my comment above.

Answer 4

[Walter Roberson]

In R2017a or later, instead of

solve(dH3, Cp)       %implicit is dH3 == 0

you can https://www.mathworks.com/help/symbolic/isolate.html

isolate(dH3 == 0, Cp)       %the == 0 must be made explicit

"The result is similar to solving eqn for expr. If isolate cannot isolate expr, it moves all terms containing expr to the left side."

isolate() is a bit more flexible in cases where getting a complete solution might be difficult