Question 4 of 7: Ethylene–Propylene Copolymer Molecular Weight; Fastener Stress Relaxation Life

Question 4: Ethylene–Propylene Copolymer Molecular Weight; Fastener Stress Relaxation Life (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) 1 kg C$_2$H$_4$ + 3 kg C$_3$H$_6$, degree of polymerization $\overline{DP}=4500$. (b) $\sigma_0=4000$ psi at $t=0$; $\sigma=3500$ psi at $t=100$ hr; failure limit $\sigma_{\min}=2500$ psi.

Find. (a) Copolymer molecular weight. (b) Fastener life (time to reach 2500 psi).

Approach

The copolymer's average mer molecular weight is the mole-fraction-weighted average of the two mers' molecular weights (found by converting the given mass feed to moles first), then multiplied by the degree of polymerization. The stress-relaxation problem is the standard Maxwell-model exponential decay at constant strain; two data points fix the relaxation time, and the same model then gives the time to reach the failure stress.

1. (a) Mer molecular weights and mole fractions. $$M_{\text{C}_2\text{H}_4}=2(12.01)+4(1.01)=28.06\ \text{g/mol},\qquad M_{\text{C}_3\text{H}_6}=3(12.01)+6(1.01)=42.09\ \text{g/mol}.$$ Moles per kg fed: $n_{\text{eth}}=1000/28.06=35.64$ mol; $n_{\text{prop}}=3000/42.09=71.28$ mol; total $=106.92$ mol, so $$x_{\text{eth}}=0.3335,\qquad x_{\text{prop}}=0.6665.$$
2. Average mer weight and polymer molecular weight. $$\overline M_{\text{mer}}=x_{\text{eth}}M_{\text{eth}}+x_{\text{prop}}M_{\text{prop}} =0.3335(28.06)+0.6665(42.09)=37.41\ \text{g/mol},$$ $$\overline M=\overline{DP}\times\overline M_{\text{mer}}=4500\times37.41=\boxed{168{,}300\ \text{g/mol}}.$$
3. (b) Relaxation time from the two given points. Stress relaxation at constant strain follows $\sigma(t)=\sigma_0e^{-t/\tau}$. From $\sigma(100\,\text{hr})=3500$ psi, $$\tau=\frac{100}{\ln(4000/3500)}=\frac{100}{0.1335}=\boxed{749\ \text{hr}}.$$
4. Life to the 2500 psi limit. Using the same $\tau$, $$t_{\text{life}}=\tau\ln\!\left(\frac{\sigma_0}{\sigma_{\min}}\right)=749\times\ln\!\left(\frac{4000}{2500}\right) =749\times0.4700=\boxed{352\ \text{hr}}.$$ (about 14.7 days of continuous service before the fastener stress falls below its functional minimum.)