Assignment #5
ChE 4X03/6X03 Introduction to Polymer Processing
Due Tuesday April 7
1. [15 Marks] An extrusion line has been set up to prepare two planks at the same time. The line
extrudes a wood composite (WC) resin
...
Assignment #5
ChE 4X03/6X03 Introduction to Polymer Processing
Due Tuesday April 7
1. [15 Marks] An extrusion line has been set up to prepare two planks at the same time. The line
extrudes a wood composite (WC) resin which has a measured melt index of I2.16kg = 0.8 g/10min
and I10kg = 6 g/10min at 180oC. Melt density of WC at this temperature is 1190 kg/m3. The
measurement was determined under the standard conditions of ASTM D-1238 [Lc = 8.000 mm
(capillary length); Dc = 2.0955 mm (capillary diameter); Dp = 9.5504 mm (plunger diameter)].
We are only concerned about the 250 mm, 15 L/D melt extruder (i.e. Db = 250mm with no solids
conveying zone or melting zone) being used in the line, which has a 12mm deep constant depth
square-pitched metering screw (flight thickness, e = 25 mm) rotating at 60 RPM. The die splits
the flow to produce two planks of equal cross-section, 18 cm wide and 5 cm high, and is drawn
in the figure below. The internal length of the die (i.e. length of the flow path in the die) is L =
75 cm.
a. Determine the output rate (m3/s) if
the pressure rise along the screw is
2.961 MPa. [10 Marks]
b. Determine how closely the two
opening in the die can be placed, i.e.
distance S, if the polymer has a
known first normal stress
difference, ?! = 30 ∙ ?"!.$%[units
MPa] (use the Tanner equation to
solve). [5 marks
2. [15 Marks] Determine the diameter of runner B which is attached between sprue A to mold
M1, seen in the figure below (also showing the clamping force direction for the system as a big
arrow). Both runners B and C are 10 cm long (ignore the bend in runners B and C for the
calculations). Two molded sheets are made at the same time, each with 5 mm thick and 15 cm
wide (into the plane of the paper) but the sheet in mold M1 was 8 cm long while the sheet in mold
M2 was 12 cm long. The differences in length for the molded sheets can be seen in the figure
below. The PET resin has a viscosity of η = 12,770*γ-0.48. The elapsed time for each mold to fill
from the time the liquid starts to enter the mold cavities was 9 seconds; runners are not included
in this fill time. The clamping force was equal to the packing pressure, which was 5 times greater
than the filling pressure. Determine the clamping force in tonnes (1 tonne=9810 N).
Sprue A diameter: 4 mm; length: 4 cm
Runner C diameter: 3 mm length: 10 cm
3. [10 Marks] The mold for four (4) cell
phone back plates are shown in the attached
figure (with only two back plates actually
showing and the other two behind). Each
cavity is filled in 5 seconds through two
secondary runners shown (total of 8
secondary runners in the mold). Each runner
has a round cross-section with a 3 mm
diameter and is 40 mm long. The packing
pressure is 6 times larger than the pressure at
the end of filling. Determine in tonnes the
clamping force required. The polymer
viscosity is defined by power-law
parameters, m = 7000 Pa-sn and n = 0.7, and
its melt density is 710 kg/m3. The primary
runner and sprue feeding the secondary
runners have a negligible pressure drop compared to the secondary runners and flow in the mold cavity.
4. [20 Marks] Determine the filling time of each cavity for a mold used for preparing two
polyurethane rubber shoe pads (modelled as a slab of 30.48 cm long x 10.16 cm wide x 1.25 cm thick
for Cavity C, C’). The necessary clamping force is 750 tonnes to balance a packing pressure that is 3
times the filling pressure. The rubber has a viscosity of η = 3,600*γ-0.5. Pipe dimensions [A: 5mm dia.,
8 cm long] [B,B’: 3.5mm dia., 7 cm long]. The direction of the clamping force is out of the page and
ignores the presence of the pipes in the projected area calculation.
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