Pressure Distribution in a Calendered Sheet To illustrate the preceding analysis, evaluate the point of departure and the pressure distribution for a calendering operation in which R = 0.35 m, H = 0.00025 m, and x1 = −0.01 m. The rolls rotate at 20 rpm, and the viscosity of the melt is 0.5 × 10^4 P

Question

Pressure Distribution in a Calendered Sheet

To illustrate the preceding analysis, evaluate the point of departure and the pressure distribution for a calendering operation in which [latex]R=0.35 \mathrm{~m}, H=[/latex] [latex]0.00025 \mathrm{~m}[/latex], and [latex]x_{1}=-0.01 \mathrm{~m}[/latex]. The rolls rotate at [latex]20 \mathrm{rpm}[/latex], and the viscosity of the melt is [latex]0.5 imes 10^{4} \mathrm{P}[/latex].

Accepted Answer

First, note that [latex]\alpha=1 / 2 H R=1 /(2 imes 0.00025 imes 0.35)=5,714 \mathrm{~m}^{-2}[/latex]. The solution of Eqn. (8.112) can then be effected by any standard numerical rootfinding technique. The Goal-Seek feature of the Excel spreadsheeting application gives the downstream distance before separation as [latex]x_{2}=0.00403 \mathrm{~m}[/latex].

[latex]\frac{x_{1} \sqrt{\alpha}\left[\alpha x_{1}^{2}\left(1-3 \alpha x_{2}^{2} ight)-1-5 \alpha x_{2}^{2} ight]}{\left(1+\alpha x_{1}^{2} ight)^{2}}+\frac{x_{2} \sqrt{\alpha}\left(1+3 \alpha x_{2}^{2} ight)}{1+\alpha x_{2}^{2}}[/latex]

[latex]+\left(1-3 \alpha x_{2}^{2} ight)\left[ an ^{-1}\left(x_{1} \sqrt{\alpha} ight)- an ^{-1}\left(x_{2} \sqrt{\alpha} ight) ight]=0.[/latex] (8.112)

When [latex]x_{2}[/latex] has thus been found, Eqn. (8.111) then gives the complete pressure distribution, which can again be evaluated with the aid of a spreadsheet. The results are shown in Fig. E8.6, which actually shows a closely related value as the ordinate (namely, the expression in the large brackets in Eqn. (8.111)):

[latex]\begin{aligned} p=\int_{0}^{p} d p=& \int_{x_{2}}^{x} \frac{3 \mu \omega}{2 H^{3}} \frac{x^{2}-x_{2}^{2}}{\left(1+\alpha x^{2} ight)^{3}} d x \=& \frac{3 \mu \omega}{2 H^{3}} \frac{1}{8 \alpha^{3 / 2}}\left[\frac{x \sqrt{\alpha}\left[\alpha x^{2}\left(1-3 \alpha x_{2}^{2} ight)-1-5 \alpha x_{2}^{2} ight]}{\left(1+\alpha x^{2} ight)^{2}}+\frac{x_{2} \sqrt{\alpha}\left(1+3 \alpha x_{2}^{2} ight)}{1+\alpha x_{2}^{2}} ight.\ &\left.+\left(1-3 \alpha x_{2}^{2} ight)\left[ an ^{-1}(x \sqrt{\alpha})- an ^{-1}\left(x_{2} \sqrt{\alpha} ight) ight] ight] . (8.111) \end{aligned}[/latex]

[latex]p^{*}=\frac{16 H^{3} \alpha^{3 / 2}}{3 \mu \omega} p .[/latex] (E8.6.1)

Observe that the pressure rises from zero at the inlet to a maximum upstream of the "nip," and then declines to zero as the sheet leaves the calender. In accordance with an earlier observation, note that the pressure gradient is zero at the exit.

The angular velocity of the rolls is [latex]\omega=2 \pi imes 20 / 60=2.09 \mathrm{rad} / \mathrm{s}[/latex]. Since the maximum value of [latex]p^{*}[/latex] is about [latex]0.288[/latex], the maximum pressure between the rolls is:

[latex]p_{\max }=\frac{3 \mu \omega p^{*}}{16 H^{3} \alpha^{3 / 2}}=\frac{3 imes 0.5 imes 10^{4} imes 2.09 imes 0.288}{16 imes 0.00025^{3} imes 5,714^{3 / 2}}=8.38 imes 10^{7} \mathrm{~Pa}=838 \mathrm{bar} .[/latex]

As mentioned previously, the maximum pressure can be simply enormous.

Question 8.6: Pressure Distribution in a Calendered Sheet To illustrate th...

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