Linear Viscoelasticity — η 0 , J e 0 and Longest Relaxation Time: 23-Chem-B8 Polymer Engineering, May 2016, Q6

Question 6: Linear Viscoelasticity — η₀, J_e⁰ and Longest Relaxation Time (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.

Find. Zero-shear viscosity η₀, steady-state recoverable compliance J_e⁰, and the terminal (longest) relaxation time τ.

Approach. In the terminal zone (low ω) the storage and loss moduli follow $G''\to\eta_0\omega$ and $G'\to J_e^0\eta_0^2\omega^2$, so η₀ is the low-ω limit of G''/ω, J_e⁰ is the low-ω limit of G'/G''², and τ = η₀J_e⁰.

Fig. 6 — log–log dynamic moduli. At low ω (terminal zone) G'' has slope 1 and G' slope 2; G'' stays above G' across the whole measured window, the two converging toward a crossover at ω ≈ 1/τ ≈ 1.7×10² rad/s beyond the data. The plateau of G''/ω gives η₀ and of G'/ω² gives J_e⁰η₀².

Zero-shear viscosity from the low-ω plateau of G''/ω:

ω 9.12 5.75 3.63

G''/ω (Pa·s) 1000 1000 1025

$$\eta_0=\lim_{\omega\to0}\frac{G''}{\omega}\approx\boxed{1.0\times10^{3}\ \text{Pa}\cdot\text{s}}$$
Steady-state compliance from the low-ω plateau of G'/G''²: $$J_e^0=\lim_{\omega\to0}\frac{G'}{(G'')^2}\approx\frac{200}{5750^2}\approx\frac{83.2}{3720^2}\approx\boxed{6.0\times10^{-6}\ \text{Pa}^{-1}}$$
Longest relaxation time as the product of the two terminal parameters: $$\tau=\eta_0 J_e^0=(1.0\times10^{3})(6.0\times10^{-6})=\boxed{6.0\times10^{-3}\ \text{s}}$$ As a consistency check, the terminal forms $G'=J_e^0\eta_0^2\omega^2$ and $G''=\eta_0\omega$ would cross at $\omega=1/(\eta_0J_e^0)=1/\tau\approx1.7\times10^{2}$ rad/s — beyond the measured window, which is why G'' exceeds G' at every tabulated frequency (the curves are still converging at ω = 57.5 rad/s, where G''/G' = 2.8).

ω	9.12	5.75	3.63
G''/ω (Pa·s)	1000	1000	1025

Quantity	Value
Zero-shear viscosity, η₀	≈ 1.0×10³ Pa·s
Steady-state compliance, J_e⁰	≈ 6.0×10^-6 Pa^-1
Longest relaxation time, τ	≈ 6.0×10^-3 s

Topic: Terminal-zone linear viscoelasticity of a monodisperse melt.

Key relations: Terminal limits $G'(\omega)=J_e^0\eta_0^2\omega^2$, $G''(\omega)=\eta_0\omega$; steady-state compliance $J_e^0=\lim_{\omega\to0}G'/G''^2$; terminal time $\tau=\eta_0 J_e^0$.

Why this works: As ω → 0 the melt approaches its slowest (longest) relaxation mode, where the response is dominated by viscous flow (G'' ∝ ω) with a small in-phase elastic storage (G' ∝ ω²). The ratio of these two terminal coefficients isolates the recoverable elasticity J_e⁰, and its product with η₀ is the terminal relaxation time. For a monodisperse polymer J_e⁰ is essentially independent of M while η₀ ∝ M^3.4 above entanglement.

Common pitfall: Reading η₀ or J_e⁰ off high-ω data still in the transition/plateau zone (where the slope-1/slope-2 scaling fails); the single lowest point (ω = 2.91) here scatters low in G', so use the consistent plateau of several terminal points.

Source: Sperling, Introduction to Physical Polymer Science, 4th ed., Ch. 8 (viscoelasticity); Ferry, Viscoelastic Properties of Polymers, 3rd ed., Ch. on terminal zone.

Question 6 of 6: Linear Viscoelasticity — η 0 , J e 0 and Longest Relaxation Time

Question 6: Linear Viscoelasticity — η0, Je0 and Longest Relaxation Time (20 marks)

Question 6: Linear Viscoelasticity — η₀, J_e⁰ and Longest Relaxation Time (20 marks)