Sympy is a Python package used for solving equations using symbolic math.

Let's solve the following problem with SymPy.

Given:

The density of two different polymer samples $\rho_1$ and $\rho_2$ are measured.

$$ \rho_1 = 0.904 \ g/cm^3 $$$$ \rho_2 = 0.895 \ g/cm^3 $$

The percent crystalinity of the two samples ($\%c_1 $ and $\%c_2$) is known.

$$ \%c_1 = 62.8 \% $$$$ \%c_2 = 54.4 \% $$

The percent crystalinity of a polymer sample is related to the density of 100% amorphus regions ($\rho_a$) and 100% crystaline regions ($\rho_c$) according to:

$$ \%crystallinity = \frac{ \rho_c(\rho_s - \rho_a) }{\rho_s(\rho_c - \rho_a) } \times 100 \% $$

Find:

Find the density of 100% amorphus regions ($\rho_a$) and the density of 100% crystaline regions ($\rho_c$) for this polymer.

Solution:

There are a couple functions we need from Sympy. We'll need the symbols function to create our symbolic math variables and we need the nonlinsolve function to solve a system of non-linear equations.



In [1]:

    
from sympy import symbols, nonlinsolve

We need to define six different symbols: $$\rho_c, \rho_a, \rho_1, \rho_2, c_1, c_2$$



In [2]:

    
pc, pa, p1, p2, c1, c2 = symbols('pc pa p1 p2 c1 c2')

Next we'll create two expressions for our two equations. We can subtract the %crystallinity from the left side of the equation to set the equation to zero.

$$ \%crystallinity = \frac{ \rho_c(\rho_s - \rho_a) }{\rho_s(\rho_c - \rho_a) } \times 100 \% $$$$ \frac{ \rho_c(\rho_s - \rho_a) }{\rho_s(\rho_c - \rho_a) } \times 100 \% - \%crystallinity = 0 $$

Sub in $\rho_s = \rho_1$ and $\rho_s = \rho_2$ to each of the expressions.



In [3]:

    
expr1 = ( (pc*(p1-pa)   ) / (p1*(pc-pa)) - c1)
expr2 = ( (pc*(p2-pa)   ) / (p2*(pc-pa)) - c2)

Now we'll substitue in the values of $\rho_1 = 0.904$ and $c_1 = 0.628$ into our first expression.



In [4]:

    
expr1 = expr1.subs(p1, 0.904)



In [5]:

    
expr1 = expr1.subs(c1, 0.628)



In [6]:

    
expr1









    Out[6]:





$\displaystyle \frac{1.10619469026549 pc \left(0.904 - pa\right)}{- pa + pc} - 0.628$

Now we'll substitue our the values of $\rho_2 = 0.895$ and $c_2 = 0.544$ into our second expression.



In [7]:

    
expr2 = expr2.subs(p2, 0.895)



In [8]:

    
expr2 = expr2.subs(c2, 0.544)



In [9]:

    
expr2









    Out[9]:





$\displaystyle \frac{1.11731843575419 pc \left(0.895 - pa\right)}{- pa + pc} - 0.544$

To solve the two equations for the to unknows $\rho_a$ and $\rho_b$, use SymPy's nonlinsolve() function. Pass in a list of the two expressions and followed by a list of the two variables to solve for.



In [10]:

    
nonlinsolve([expr1,expr2],[pa,pc])









    Out[10]:





$\displaystyle \left\{\left( 0.840789786223278, \  0.946134313397929\right)\right\}$

We see that the value of $\rho_a = 0.840789$ and $\rho_c = 0.94613431$.

The solution is a SymPy FiniteSet object. To pull the values of $\rho_a$ and $\rho_c$ out of the FiniteSet, use the syntax sol.args[0][<var num>].



In [11]:

    
sol = nonlinsolve([expr1,expr2],[pa,pc])



In [12]:

    
type(sol)









    Out[12]:





sympy.sets.sets.FiniteSet



In [13]:

    
sol.args









    Out[13]:





((0.840789786223278, 0.946134313397929),)



In [14]:

    
sol.args[0]









    Out[14]:





$\displaystyle \left( 0.840789786223278, \  0.946134313397929\right)$



In [15]:

    
sol.args[0][0]









    Out[15]:





$\displaystyle 0.840789786223278$



In [16]:

    
pa = sol.args[0][0]
pc = sol.args[0][1]
print(f' Density of 100% amorphous polymer, pa = {round(pa,2)} g/cm3')
print(f' Density of 100% crystaline polymer, pc = {round(pc,2)} g/cm3')









    



 Density of 100% amorphous polymer, pa = 0.84 g/cm3
 Density of 100% crystaline polymer, pc = 0.95 g/cm3



In [ ]: