Question

In melt spinning of a monofilament, a polymer is extruded at a volumetric flow rate of \( 5 \times 10^{-5}\ \text{m}^3/\text{s} \) through a circular spinneret. The take-up velocity is \( 100\ \text{m/s} \) with a draw ratio of \( 50 \). The diameter (mm) of the spinneret orifice (rounded off to 2 decimal places) is .................

Hint:

In melt spinning, $\text{DR}=V_{\text{take-up}}/V_0$. Get $V_0$ first, then $A_0=Q/V_0$, and finally $d=\sqrt{4A_0/\pi}$.

Answer 1

Draw ratio $\text{DR}=\dfrac{V_{\text{take-up}}}{V_0}⇒ V_0=\dfrac{100}{50}=2\ \text{m/s}$ (jet exit velocity).
Volumetric flow rate $Q=A_0V_0⇒ A_0=\dfrac{Q}{V_0}=\dfrac{5\times10^{-5}}{2}=2.5\times10^{-5}\ \text{m}^2$.
For a circular orifice, $A_0=\dfrac{\pi d^2}{4}⇒ d=\sqrt{\dfrac{4A_0}{\pi}}=\sqrt{\dfrac{4(2.5\times10^{-5})}{\pi}}=5.643\times10^{-3}\ \text{m}$.
Convert to mm: $d=5.643\ \text{mm}⇒ \boxed{5.64\ \text{mm}}$.