Question 2 of 7: Modulus of Semicrystalline Nylon; Porosity of an Oil-Impregnated Bearing

Question 2: Modulus of Semicrystalline Nylon; Porosity of an Oil-Impregnated Bearing (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. (b) cylinder $d=2.00$ cm, $L=6.00$ cm; mass after sintering $m_s=123.85$ g; mass after oil impregnation $m_{oi}=126.80$ g; oil specific gravity $\rho_{oil}=0.90$ g/cm$^3$; true (pore-free) brass density $\rho_{Pb}=8.47$ g/cm$^3$.

Find. (a) Sketch of modulus vs. temperature for partially crystalline nylon and how it shifts with more/less crystallinity. (b) % open porosity and % closed porosity.

Approach

(a) is a qualitative sketch-and-explain question; the shape follows directly from how crystallites behave as reinforcing, physically-crosslinking regions within the amorphous matrix. (b) is a mass/volume porosity problem: the oil that soaks in during impregnation can only reach open (surface-connected) pores, so its volume directly measures the open-pore volume; comparing the sintered part’s mass to its true (pore-free) density gives the total solid volume, and hence the total pore volume by difference from the bulk (measured) volume.

1. (a) Modulus vs. temperature for partially crystalline nylon. Four regions are seen on cooling from the melt (right to left on the sketch): a low-modulus viscous flow region above the melting temperature $T_m$; a rubbery plateau between $T_g$ and $T_m$ whose height is set by the crystalline regions acting as physical crosslinks (tying the amorphous chains together, the same reinforcing role fillers play in a rubber); a sharp modulus drop of 2–3 orders of magnitude at the glass transition $T_g$ as the amorphous fraction becomes rubbery; and a high, roughly temperature-independent glassy plateau below $T_g$.
   
   Increasing crystallinity raises and extends the rubbery plateau (more, and more extensive, crystallites reinforce the structure and resist flow to a higher temperature, closer to $T_m$) and raises the glassy-region modulus slightly; decreasing crystallinity lowers the rubbery plateau and shortens it — in the limit of a fully amorphous polymer, the plateau disappears and the modulus falls almost directly from the glassy value into viscous flow just above $T_g$, with no true rubbery region.
2. (b) Bulk volume of the bearing. $$V_{bulk}=\frac{\pi}{4}d^2L=\frac{\pi}{4}(2.00)^2(6.00)=18.85\ \text{cm}^3$$
3. Open-pore volume, from the absorbed oil. Oil can only enter pores connected to the surface, so the mass gained on impregnation is entirely oil filling open pores: $$m_{oil}=126.80-123.85=2.95\ \text{g}\ \Rightarrow\ V_{open}=\frac{m_{oil}}{\rho_{oil}}=\frac{2.95}{0.90}=3.278\ \text{cm}^3$$ $$\boxed{\%\text{open porosity}=\frac{3.278}{18.85}\times100\approx17.4\%}$$
4. Total pore volume, from the true density of solid brass. The volume of solid brass actually present in the sintered part (before impregnation) is $$V_{solid}=\frac{m_s}{\rho_{Pb}}=\frac{123.85}{8.47}=14.62\ \text{cm}^3$$ so the total (open + closed) pore volume is the remainder of the bulk volume: $$V_{pore,total}=V_{bulk}-V_{solid}=18.85-14.62=4.226\ \text{cm}^3\ \ (\%\text{total}\approx22.4\%)$$
5. Closed-pore volume, by difference. $$V_{closed}=V_{pore,total}-V_{open}=4.226-3.278=0.948\ \text{cm}^3$$ $$\boxed{\%\text{closed porosity}=\frac{0.948}{18.85}\times100\approx5.0\%}$$ (check: $17.4\%+5.0\%=22.4\%$, matching the independently computed total — internally consistent).