Polymer Crystallinity

4.1. Polymer Crystallinity#

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

Sympy can be installed with the Anaconda Prompt or a terminal using the command below:

conda install sympy

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.

from sympy import symbols, nonlinsolve

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

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.

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.

expr1 = expr1.subs(p1, 0.904)

expr1 = expr1.subs(c1, 0.628)

expr1
\[\displaystyle \frac{1.10619469026549 pc \left(0.904 - pa\right)}{- pa + pc} - 0.628\]

Now we’ll substitue our the values of \(\rho_2 = 0.895\) and \(c_2 = 0.544\) into our second expression.

expr2 = expr2.subs(p2, 0.895)

expr2 = expr2.subs(c2, 0.544)

expr2
\[\displaystyle \frac{1.11731843575419 pc \left(0.895 - pa\right)}{- pa + pc} - 0.544\]

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.

nonlinsolve([expr1,expr2],[pa,pc])
\[\displaystyle \left\{\left( 0, \ 0\right), \left( 0.840789786223278, \ 0.946134313397929\right)\right\}\]

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>].

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

type(sol)

sympy.sets.sets.FiniteSet

sol.args

((0.840789786223278, 0.946134313397929), (0, 0))

sol.args[0]
\[\displaystyle \left( 0.840789786223278, \ 0.946134313397929\right)\]
sol.args[0][0]
\[\displaystyle 0.840789786223278\]
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
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