Spinning a Polymeric Fiber A Newtonian polymeric liquid of viscosity μ is being “spun” (drawn into a fiber or filament of small diameter before solidifying by pulling it through a chemical setting bath) in the apparatus shown in Fig. E6.8. The liquid volumetric flow rate is Q, and the filament diam

Question

Spinning a Polymeric Fiber

A Newtonian polymeric liquid of viscosity [latex]\mu[/latex] is being "spun" (drawn into a fiber or filament of small diameter before solidifying by pulling it through a chemical setting bath) in the apparatus shown in Fig. E6.8.

The liquid volumetric flow rate is [latex]Q[/latex], and the filament diameters at [latex]z=0[/latex] and [latex]z=L[/latex] are [latex]D_{0}[/latex] and [latex]D_{L}[/latex], respectively. To a first approximation, neglect the effects of gravity, inertia, and surface tension (this last effect is examined in detail in Problem 6.26). Derive an expression for the tensile force [latex]F[/latex] needed to pull the filament downward. Assume that the axial velocity profile is "flat" at any vertical location, so that [latex]v_{z}[/latex] depends only on [latex]z[/latex], which is here most conveniently taken as positive in the downward direction. Also derive an expression for the downward velocity [latex]v_{z}[/latex] as a function of [latex]z[/latex]. The inset of Fig. E6.8 shows further details of the notation concerning the filament.

Accepted Answer

It is first necessary to determine the radial velocity and hence the pressure inside the filament. From continuity:

[latex]\frac{1}{r} \frac{\partial\left(r v_{r} ight)}{\partial r}+\frac{\partial v_{z}}{\partial z}=0.[/latex] (E6.8.1)

Since [latex]v_{z}[/latex] depends only on [latex]z[/latex], its axial derivative is a function of [latex]z[/latex] only, or [latex]d v_{z} / d z=[/latex] [latex]f(z)[/latex], so that Eqn. (E6.8.1) may be rearranged and integrated at constant [latex]z[/latex] to give:

[latex]\frac{\partial\left(r v_{r} ight)}{\partial r}=-r f(z), \quad r v_{r}=-\frac{r^{2} f(z)}{2}+g(z).[/latex] (E6.8.2)

But to avoid an infinite value of [latex]v_{r}[/latex] at the centerline, [latex]\mathrm{g}(\mathrm{z})[/latex] must be zero, giving:

[latex]v_{r}=-\frac{r f(z)}{2}, \quad \frac{\partial v_{r}}{\partial r}=-\frac{f(z)}{2}.[/latex] (E6.8.3)

To proceed with reasonable expediency, it is necessary to make some simplification. After accounting for a primary effect (the difference between the pressure in the filament and the surrounding atmosphere), we assume that a secondary effect (variation of pressure across the filament) is negligible; that is, the pressure does not depend on the radial location (see Problem [latex]6.27[/latex] for a more accurate investigation). Note that the external (gauge) pressure is zero everywhere, and apply the first part of Eqn. (5.64) at the free surface:

[latex]\sigma_{r r}=-p+2 \mu \frac{\partial v_{r}}{\partial r}-\frac{2}{3} \mu abla \cdot \mathbf{v},[/latex] (5.64)

[latex]\sigma_{r r}=-p+2 \mu \frac{\partial v_{r}}{\partial r}=-p-\mu f(z)=0, \quad ext { or } \quad p=-\mu \frac{d v_{z}}{d z}.[/latex] (E6.8.4)

The axial stress is therefore:

[latex]\sigma_{z z}=-p+2 \mu \frac{d v_{z}}{d z}=3 \mu \frac{d v_{z}}{d z}.[/latex] (E6.8.5)

It is interesting to note that the same result can be obtained with an alternative assumption. [latex]{ }^{3}[/latex] The axial tension in the fiber equals the product of the cross-sectional area and the local axial stress:

[latex]F=A \sigma_{z z}=3 \mu A \frac{d v_{z}}{d z} .[/latex] (E6.8.6)

Since the effect of gravity is stated to be insignificant, [latex]F[/latex] is a constant, regardless of the vertical location.

At any location, the volumetric flow rate equals the product of the crosssectional area and the axial velocity:

[latex]Q=A v_{z} .[/latex] (E6.8.7)

A differential equation for the velocity is next obtained by dividing one of the last two equations by the other, and rearranging:

[latex]\frac{1}{v_{z}} \frac{d v_{z}}{d z}=\frac{F}{3 \mu Q} .[/latex] (E6.8.8)

[latex]{ }^{3}[/latex] See S. Middleman's Fundamentals of Polymer Processing, McGraw-Hill, New York, 1977, p. [latex]235 .[/latex] There, the author assumes [latex]\sigma_{r r}=\sigma_{ heta heta}=0[/latex], followed by the identity: [latex]p=-\left(\sigma_{z z}+\sigma_{r r}+\sigma_{ heta heta} ight) / 3[/latex].

Integration, noting that the inlet velocity at [latex]z=0[/latex] is [latex]v_{z 0}=Q /\left(\pi D_{0}^{2} / 4 ight)[/latex], gives:

[latex]\int_{v_{z 0}}^{v_{z}} \frac{d v_{z}}{v_{z}}=\frac{F}{3 \mu Q} \int_{0}^{z} d z, \quad ext { where } \quad v_{z 0}=\frac{4 Q}{\pi D_{0}^{2}},[/latex] (E6.8.9)

so that the axial velocity increases exponentially with distance:

[latex]v_{z}=v_{z 0} e^{F z / 3 \mu Q}.[/latex] (E6.8.10)

The tension is obtained by applying Eqns. (E6.8.7) and (E6.8.10) just before the filament is taken up by the rollers:

[latex]v_{z L}=\frac{4 Q}{\pi D_{L}^{2}}=v_{z 0} e^{F L / 3 \mu Q}.[/latex] (E6.8.11)

Rearrangement yields:

[latex]F=\frac{3 \mu Q}{L} \ln \frac{v_{z L}}{v_{z 0}},[/latex] (E6.8.12)

which predicts a force that increases with higher viscosities, flow rates, and drawdown ratios [latex]\left(v_{z L} / v_{z 0} ight)[/latex] and that decreases with longer filaments.

Elimination of [latex]F[/latex] from Eqns. (E6.8.10) and (E6.8.12) gives an expression for the velocity that depends only on the variables specified originally:

[latex]v_{z}=v_{z 0}\left(\frac{v_{z L}}{v_{z 0}} ight)^{z / L}=v_{z 0}\left(\frac{D_{0}}{D_{L}} ight)^{2 z / L},[/latex] (E6.8.13)

a result that is independent of the viscosity.

Question 6.8: Spinning a Polymeric Fiber A Newtonian polymeric liquid of v...

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