On Thu, 11 Feb 2021 at 22:54, mpierro3 <[email protected]> wrote: > > Yes, thank you. The rest is as follows: > > from sympy import * > import numpy as np > > M_e = 1 > gamma_1 = 1.667 > gamma_4 = 1.667 > MW_1 = 39.948 > MW_4 = 4.0026 > D_4 = 5 / 39.37 > D_1 = 3 / 39.37 > A_4 = (np.pi / 4) * Pow(D_4, 2) > A_1 = (np.pi / 4) * Pow(D_1, 2) > a_4 = (gamma_4 + 1) / (gamma_4 - 1) > a_1 = (gamma_1 + 1) / (gamma_1 - 1) > A_4_A_1 = A_4 / A_1 > > M_3a = symbols('M_3a', positive=true, nonzero=true) > eqn = nonlinsolve([Eq((M_e / M_3a) * Pow((2 + (gamma_4 - 1) * Pow(M_3a, 2)) / > \ > (2 + (gamma_4 - 1) * Pow(M_e, 2)), a_4 / 2), A_4_A_1), M_3a <= 1], M_3a) > print(eqn)
Your equation to be solved is (truncating the numbers for clarity): Eq(0.56*(0.33*M_3a**2 + 1)**1.99/M_3a, 2.77) I think it is unlikely that you will get analytic expressions for the solutions to an equation like this. Sympy can solve it numerically though: In [31]: nsolve(eq[0], M_3a, 0.2) Out[31]: 0.208399106769846 -- Oscar -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/CAHVvXxST3XuJ_6EZOfOYXr0jAvg28YPPKJJ9pZA2QKpA6Sxn3w%40mail.gmail.com.
