WLF shift — extruder temperature for a lower-molecular-weight batch: 23-Chem-B8 Polymer Engineering, December 2015, Q4

Question 4: WLF shift — extruder temperature for a lower-molecular-weight batch (25 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. Same polymer, $T_g=80$ °C. Batch 1: $\bar X_w=750$, gives $\eta=20\,000$ P at $T_1=150$ °C. Batch 2: $\bar X_w=500$, target the same $\eta=20\,000$ P.

Find. The processing temperature $T_2$ for batch 2.

Approach. Melt viscosity factorises into a molecular-weight term ($\eta\propto \bar X_w^{3.4}$ above the entanglement point) and a temperature term given by the WLF equation referenced to $T_g$. Both batches share $T_g$, so equating the two viscosities eliminates the molecular-weight prefactor and leaves a WLF equation for $T_2$.

Viscosity model. Write $\eta(T,\bar X_w)=B\,\bar X_w^{3.4}\,a_T(T)$, where the WLF shift factor (universal constants $C_1=17.44$, $C_2=51.6$) is $$\log a_T(T)=-\frac{C_1(T-T_g)}{C_2+(T-T_g)}$$ Equal viscosity for both batches gives $\bar X_{w,1}^{3.4}a_T(T_1)=\bar X_{w,2}^{3.4}a_T(T_2)$.
Molecular-weight ratio. Solving for the required shift ratio, $$\frac{a_T(T_2)}{a_T(T_1)}=\left(\frac{\bar X_{w,1}}{\bar X_{w,2}}\right)^{3.4} =\left(\frac{750}{500}\right)^{3.4}=1.5^{3.4}=3.97$$ The lower-MW batch is thinner, so its temperature must be reduced to raise $a_T$ (and hence $\eta$) back to target.
Evaluate the reference shift. At $T_1=150$ °C, $T_1-T_g=70$: $$\log a_T(T_1)=-\frac{17.44(70)}{51.6+70}=-10.04$$ so $\log a_T(T_2)=\log a_T(T_1)+\log 3.97=-10.04+0.60=-9.44$.
Solve the WLF equation for $T_2$. With $\theta_2=T_2-T_g$, $$\frac{17.44\,\theta_2}{51.6+\theta_2}=9.44 \;\Longrightarrow\;\theta_2=\frac{9.44\times51.6}{17.44-9.44}=60.9\;\text{K}$$ $$T_2=80+60.9=\boxed{141\ ^\circ\text{C}}$$

Quantity	Value
Required shift ratio $(750/500)^{3.4}$	3.97
New processing temperature $T_2$	$\approx141$ °C

Topic: Time–temperature (WLF) superposition and the $M^{3.4}$ melt-viscosity law.

Key relations: $\eta\propto M^{3.4}$ (above $M_c$); WLF $\log a_T=-C_1(T-T_g)/[C_2+(T-T_g)]$ with universal $C_1=17.44$, $C_2=51.6$ K.

Why this works: Because both batches are the same chemistry with the same $T_g$, their viscosity–temperature curves are identical in shape and differ only by the vertical $M^{3.4}$ scale factor. Dropping molecular weight from 750 to 500 lowers viscosity by $1.5^{3.4}\approx4\times$; to recover the extruder's optimum you slide back down the WLF curve (closer to $T_g$), where viscosity rises steeply. The single unknown $T_2$ then follows from one WLF inversion.

Common pitfall: Using an Arrhenius viscosity law near $T_g$ — WLF governs the $T_g$ to $T_g+100$ window and predicts a much stronger temperature dependence. Also, remember the exponent is $3.4$ (entangled melt), not 1.0.

Source: Sperling, Physical Polymer Science 4th ed., §8.6 (WLF) and §10 ($M^{3.4}$ law); Rudin & Choi, Elements of Polymer Science 3rd ed., Ch. 11.

Quantity	Value
Required shift ratio \((750/500)^{3.4}\)	3.97
New processing temperature \(T_2\)	\(\approx141\) °C

Question 4 of 6: WLF shift — extruder temperature for a lower-molecular-weight batch

Question 4: WLF shift — extruder temperature for a lower-molecular-weight batch (25 marks)