Analysis of a plastic extruder An extruder screw with a thread angle of 20° has a melt-pumping zone that is 1 m long, with a channel depth of 7 mm for a 50-mm-diameter barrel. The screw is used to extrude 5-mm-diameter circular nylon bars through a die with a 20-mm land length at 573 K. If the

Question

Analysis of a plastic extruder

An extruder screw with a thread angle of 20^{°} has a melt-pumping zone that is 1 m long, with a channel depth of 7 mm for a 50-mm-diameter barrel. The screw is used to extrude 5-mm-diameter circular nylon bars through a die with a 20-mm land length at 573 K. If the extruder is operated at 50 rpm =0.833 rev/s, determine the extruder and die characteristics and obtain the operating flow rate. What is the speed of material leaving the extruder? Ignore die swell.

Accepted Answer

Note from Fig. 10.12 that the viscosity of nylon at 573 K is around [latex]300 Ns/m^{2}[/latex] and that the final bar cross-sectional area is [latex]A=\pi D^{2}_{d} /4 =1.96 imes 10^{-5}m^2 . [/latex] The extruder characteristic is given by Eq. (10.23) as

[latex]Q=\frac{\pi ^2HD^2N\sin heta \cos heta }{2} -\frac{p\pi DH^3\sin ^2 heta }{12\eta l} [/latex] (10.23)

[latex]=\frac{\pi ^{2}(0.007)(0.050)^2(0.833)\sin 20^{\omicron }\cos 20^{\omicron }}{2}-\frac{\pi (0.050)(0.007)^3\sin ^2 20^{\omicron }}{12(300)(1)}p [/latex]

[latex]=2 .31 imes 10^{-5}-(1.75 imes 10^{-12})p [/latex]

where Q is in [latex]m^{3}[/latex]/s and [latex]p[/latex] is in N/[latex]m^2. [/latex]From Eq. (10.26),

[latex]K=\frac{\pi D^{4}_{d} }{128\eta l_{d}} [/latex] (10.26)

[latex]K=\frac{\pi D^{4}_{d} }{128\eta l_{d}} =\frac{\pi (0.005)^4}{128(300)(0.020)} =2.56 imes 10^{-12}, [/latex]

so that the die characteristic is given by Eq. (10.25) as

[latex]Q_{die}=Kp[/latex] (10.25)

[latex]Q=Kp= (2.56 ×10^{-12})p .[/latex]

The extruder and die characteristics are plotted in Fig. 10.24. Note that for both to be valid, the extruder must operate at the intersection of the two lines. This can be obtained from the graph or directly from the foregoing equations as [latex]p = 5.4[/latex] MPa. At this pressure, the flow rate can be calculated as [latex]Q = 1.37 × 10^{-5} m^{3}[/latex]/s. Based on the cross-sectional area of the product, the final velocity is

[latex]Q=vA;[/latex] [latex]v=\frac{Q}{A} =\frac{1.37 imes 10^{-5}}{1.96 imes 10^{-5}} =0.70 \ m/s.[/latex]

Question 10.5: Analysis of a plastic extruder An extruder screw with a thre...

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