Injection of a Power-Law Melt into a Runner: 23-Chem-B8 Polymer Engineering, May 2017, Q5

Question 5: Injection of a Power-Law Melt into a Runner (20 points: 15 + 5)

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.

Quantity	Symbol	Value
Consistency	m	2.18×10⁴ N·sⁿ/m²
Power-law index	n	0.39
Volumetric injection rate	Q	1.2×10^-6 m³/s
Runner length	L	0.254 m
Runner radius	R	2.54×10^-3 m

Find. (a) ΔP(t) and penetration depth z(t); (b) the pressure to fill the full runner.

Approach. Constant Q fixes the melt-front position linearly in time; the power-law tube-flow law gives the pressure drop over the currently-filled length, so pressure rises linearly with penetration and hence with time.

Penetration depth. A constant volumetric rate fills a constant cross-section, so the melt front advances at a fixed speed:$$z(t)=\frac{Q\,t}{\pi R^2}=\frac{1.2\times10^{-6}}{\pi(2.54\times10^{-3})^2}\,t=\boxed{0.0592\,t\ \text{m}}\quad(t\ \text{in s})$$
Power-law pressure gradient. For a power-law fluid in a tube the throughput and pressure drop over a filled length z are related by $Q=\dfrac{\pi n R^3}{3n+1}\left(\dfrac{R\,\Delta P}{2mz}\right)^{1/n}$; solving for the gradient,$$\frac{\Delta P}{z}=\frac{2m}{R}\left[\frac{Q(3n+1)}{n\pi R^3}\right]^{n}=\frac{2(2.18\times10^4)}{2.54\times10^{-3}}(129.7)^{0.39}=1.145\times10^{8}\ \text{Pa/m}$$where 129.7 s^-1 is the apparent wall shear rate.
(a) Injection pressure vs time. Since ΔP is proportional to the filled length and z ∝ t, the pressure rises linearly:$$\Delta P(t)=\left(\frac{\Delta P}{z}\right)z(t)=1.145\times10^{8}\times0.0592\,t=\boxed{6.78\times10^{6}\,t\ \text{Pa}}$$
Fill time. The front reaches the end when z = L:$$t_{fill}=\frac{L}{0.0592}=\frac{\pi R^2 L}{Q}=\boxed{4.29\ \text{s}}$$
(b) Pressure to fill the runner. Set z = L in the pressure law:$$\Delta P_{fill}=\left(\frac{\Delta P}{z}\right)L=1.145\times10^{8}(0.254)=\boxed{2.91\times10^{7}\ \text{Pa}=29.1\ \text{MPa}\ (\approx4220\ \text{psi})}$$

Fig. 5 — injection pressure (left axis) and penetration depth (right axis) both grow linearly with time; the melt fills the 0.254 m runner at 4.29 s and 29.1 MPa.

Quantity	Value
Penetration depth	z(t) = 0.0592 t m
Apparent wall shear rate	129.7 s^-1
Pressure gradient, ΔP/z	1.145×10⁸ Pa/m
Injection pressure	ΔP(t) = 6.78×10⁶ t Pa
Fill time	4.29 s
(b) Pressure to fill runner	29.1 MPa (≈ 4220 psi)

Topic: Isothermal filling of a tubular runner by a power-law melt.

Key relations: front position $z=Qt/\pi R^2$; power-law tube law $Q=\dfrac{\pi nR^3}{3n+1}\left(\dfrac{R\Delta P}{2mz}\right)^{1/n}$; hence $\Delta P=\dfrac{2mz}{R}\left[\dfrac{Q(3n+1)}{n\pi R^3}\right]^{n}$.

Why this works: At constant Q the shear-rate field is fixed, so the pressure needed per unit filled length is a constant; only the length grows, and it grows linearly in time. The result is the tell-tale linear pressure ramp seen on injection-machine cavity traces during the fill stage, before pack-and-hold.

Common pitfall: Using the Newtonian Hagen–Poiseuille law (n = 1) — which badly over-predicts ΔP for a shear-thinning melt — or forgetting the (3n+1)/n factor in the wall shear rate.

Source: Middleman, Fundamentals of Polymer Processing (power-law tube flow); Tadmor & Gogos, Principles of Polymer Processing, §13 (injection-mold filling).

Question 5 of 6: Injection of a Power-Law Melt into a Runner

Question 5: Injection of a Power-Law Melt into a Runner (20 points: 15 + 5)