help with equation substituting

hi i want to substitute in an equation that has 2 variables (q) and (pwf), given that the user has to input several values for (pwf) to substitute into the equation and solve for (q)

Thank you

Answer 1

[Walter Roberson]

Example:

new_pwf = rand();
new_equation = subs(TheEquation, pwf, new_pwf);
sol = solve(new_equation, q)

[mohamed]

ok this is what i did

clc
clear
new_pwf=[0:500:2500]
new_equation=subs('q=((1-(0.2*(pwf/pr)))-(0.8*(pwf/pr)^2))*(1067)',pwf,new_pwf);
sol=solve(new_equation,q)

but it gave me an error

Undefined function or variable 'pwf'.

Error in Untitled2 (line 4) new_equation=subs('q=((1-(0.2*(pwf/pr)))-(0.8*(pwf/pr)^2))*(1067)',pwf,new_pwf);

[Walter Roberson]

syms pwf pr q
expression = ((1-(0.2*(pwf/pr)))-(0.8*(pwf/pr)^2))*(1067);
new_expression = subs(expression, pfw, 0:500:2500);

There is no point solving for a variable (q) that appears isolated on one side of a comparison.

Solving a vector of expressions can only work when the number of elements in the vector is the same as the number of variables indicated to solve for. With 5 values for the new pwf you would need to be solving for 5 variables.

[mohamed]

ty very much you really helped me if i may ask another question whne i substitute into an equation this is the result

q =

   1.0e+03 *

    1.0670    0.9902    0.8451    0.6317    0.3500         0

i need the results to be

1067 990 845 632 350 0 i dont want them multiplyed by 1.0e+03 and i tried the double() code and it didnt work thank you

[Walter Roberson]

format long g