Question 5 of 6: Nylon 6,6 Fibre Spinning — Stretch Rate, Stress and Draw Force
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Open-book, 3 hours; six numbered problems of equal value (20 marks each), of which five constitute a complete paper (only the first five in the answer book are marked). All six problems are solved below so the set is complete for study.
Reference texts: Odian, Principles of Polymerization (4th ed., Wiley) — chain- and step-growth kinetics, emulsion polymerization, molecular-weight distributions; Rudin & Choi, The Elements of Polymer Science and Engineering (3rd ed., Academic Press) — dilute-solution properties, osmometry, viscoelasticity; Sperling, Introduction to Physical Polymer Science (4th ed., Wiley) — linear viscoelasticity, terminal-zone moduli; Bird, Armstrong & Hassager, Dynamics of Polymeric Liquids, Vol. 1 — non-Newtonian tube flow; supporting polyolefin process detail from Young & Lovell, Introduction to Polymers (3rd ed.).
Question 5: Nylon 6,6 Fibre Spinning — Stretch Rate, Stress and Draw Force (20 marks)
Find. (a) ε̇max; (b) ratio to die wall shear rate; (c) validity of Vr ≈ 0; (d) σmax and draw force F.
Fig. 5 — isothermal Newtonian spinline: melt leaves the die, swells, then is drawn to take-up speed VL. For a constant-force isothermal filament the velocity rises exponentially, V(z) = V0ekz with k = ln(DR)/L.
Approach. For an isothermal Newtonian filament (gravity, inertia and air drag neglected) the tensile force is constant, which makes the velocity increase exponentially along the spinline; the stretch rate dV/dz is therefore largest where the velocity is largest (take-up). The die shear rate is the ordinary Newtonian wall value, and the Trouton relation (extensional viscosity = 3μ) converts stretch rate to tensile stress.
(a) Maximum stretching rate. With V(z) = V0ekz and V(L)/V(0) = DR, the rate constant is k = ln(DR)/L, and dV/dz = kV is maximal at V = VL:
$$\dot\varepsilon_{max}=\frac{V_L}{L}\ln(DR)=\frac{16.67}{4}\ln(100)=4.167\times4.605=\boxed{19.2\ \text{s}^{-1}}$$
(b) Die wall shear rate. For Newtonian laminar tube flow, $\dot\gamma_w=4Q/\pi R_{die}^3$ with Rdie = 0.8 mm:
$$\dot\gamma_w=\frac{4(1\times10^{-7})}{\pi(8\times10^{-4})^3}=249\ \text{s}^{-1}$$
$$\frac{\dot\gamma_w}{\dot\varepsilon_{max}}=\frac{249}{19.2}\approx\boxed{13}$$
The in-die shear rate is roughly an order of magnitude larger than the spinline stretch rate.
(c) Validity of Vr ≈ 0. Incompressibility with a nearly flat axial profile gives $V_r=-\tfrac12 r\,(dV_z/dz)$, so at the filament surface
$$\frac{|V_r|}{V_z}=\frac{\tfrac12 R\,\dot\varepsilon}{V_z}=\tfrac12 R(z)\,k,\qquad k=\frac{\ln(DR)}{L}=\frac{\ln 100}{4}=1.151\ \text{m}^{-1}$$
since $\dot\varepsilon=kV_z$ everywhere on an exponential spinline. The ratio therefore depends only on the local radius and is largest where the filament is thickest — at the spinline entry — so that is where the approximation must be tested, not at take-up. Continuity at the entry speed $V_0=V_L/DR=0.167$ m/s gives $R_0=\sqrt{Q/\pi V_0}=4.4\times10^{-4}$ m, so
$$\left.\frac{|V_r|}{V_z}\right|_{z=0}=\tfrac12(4.4\times10^{-4})(1.151)=\boxed{2.5\times10^{-4}}$$
while at take-up, where $R_f=\sqrt{Q/\pi V_L}=4.4\times10^{-5}$ m, it is ten times smaller, $2.5\times10^{-5}$. Even bounding the entry by the quoted die-swell diameter (3 × 0.16 cm = 0.48 cm, R = 2.4 mm — the widest the melt ever is) gives only $\tfrac12(2.4\times10^{-3})(1.151)=1.4\times10^{-3}$. The radial velocity is at worst a few parts per thousand of the axial velocity, so the thin-filament (1-D) approximation Vr ≈ 0 is well justified over the entire spinline.
(d) Maximum tensile stress and draw force. By the Trouton relation the extensional viscosity is 3μ, so the tensile stress is greatest where the stretch rate is greatest:
$$\sigma_{max}=3\mu\,\dot\varepsilon_{max}=3(250)(19.2)=\boxed{1.44\times10^{4}\ \text{Pa}}$$
The draw force equals this stress times the cross-section at take-up, $A_L=Q/V_L=6.0\times10^{-9}$ m² (equivalently the constant filament force $F=3\mu Qk$):
$$F=\sigma_{max}A_L=(1.44\times10^{4})(6.0\times10^{-9})=\boxed{8.6\times10^{-5}\ \text{N}}$$
