Question

In melt spinning of a monofilament, a polymer is extruded at a volumetric flow rate of $5\times10^{-5}\ \text{m}^3\!/\text{s}$ 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

Correct Answer: 5.6

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