1 Extrusion and Injection Molding - Analysis ver. 1 ME 6222: Manufacturing Processes and Systems Prof. J.S. Colton © GIT 2011
1
Extrusion and Injection Molding -
Analysis
ver. 1
ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2011
2
Overview
• Extrusion and Injection molding
– Flow in screw
– Flow in cavity or die
• Injection molding
– Clamp force
– Cooling time
– Ejection force
ME 6222: Manufacturing Processes and Systems
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Extrusion schematic
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Injection molding schematic
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Schematic
hopper
heaters
barrel
screw
nozzle clamp
mold
cavity
pellets
motor /
drive
throat
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Flow in screw -
Extrusion and Injection molding
• Understood through simple fluid
analysis
• Unroll barrel from screw
– rectangular trough and lid
w/cosq H
vx vz
v=pDN
q
w is like normal pitch
w/cosq is like axial pitch
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Flow analysis
• Barrel slides across channel at the helix
angle
• vz = pumping
• vx = stirring
w/cosq H
vx vz
v=pDN
q
w is like normal pitch
w/cosq is like axial pitch
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Flow rate
• vz shows viscous traction work against
exit pressure
flow rate = f(exit pressure, vbarrel, m, d, w, l)
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Flow analysis
• Simplify by using Newtonian fluid
• Separate into drag and pressure flows
• Add solutions (superposition)
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Drag flow in rectangular channel (QD)
• Simple viscous flow between parallel
plates, end effects negligible
vo
y H
H
yvv 0
wHvAvQD 2
10
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Pressure flow in rectangular channel
• Assumptions
– no slip at walls
– melt is incompressible
– steady, laminar flow
– end and side wall effects are negligible
p p + dp
dz
y
z
2y
t
t
hopper exit
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Pressure flow in rectangular channel
Equilibrium
p p + dp
dz
y
z
2y
t
t
hopper exit
022 dzydppp t
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Pressure flow in rectangular channel
022 dzydppp t
dz
dpy t
dy
dvγτ mm
Newtonian fluid
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Pressure flow in rectangular channel
• Eliminating t
• Integrating and noting
@ y = +/- H/2, v = 0
dyydz
dpdv
m
1
28
1 22 yH
dz
dpv
m
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Total pressure flow (Qp)
dz
dpwHdyvwQ
H
H
p
2
2
3
12m
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Total flow (Q)
• dp/dz set by
– back pressure on reciprocating screw (injection
molding)
– die resistance (extrusion)
dz
dpHHvwQQQ z
pDm122
3
f(screw speed) f(pressure drop)
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Nomenclature
• dz = helical length = axial length/sinq
• vz = helix velocity = vbarrel*cosq
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Flow rate
flow
rate
output pressure
w
2w
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Schematics
Injection molding
Extrusion
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Flow in round die or runner
dr dz
r z
t + dt
t
p p + dp
dzrddrrdprdrr tttpp 2][ 22
Equilibrium
Same assumptions as above
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Flow in round die or runner
Neglecting HOT
dzdrdrdpdrr ttpp 22
dzrddrrdprdrr tttpp 2][ 22
drr
rd
drr
drdr
dz
dp
ttt
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Flow in round die or runner
drCrrd t
drr
rd
L
p
dz
dp
tC
drCrrd t
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Flow in round die or runner
rL
pr
C
22
t
2
2rCr t
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Flow in round die or runner
• At center, t = 0
• At edge of tube (R), t = max
L
Rp
2max
t
dr
dumt
Newtonian fluid
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Flow in round die or runner
m
L
rp
dr
du
2
finally
22
4Rr
L
pu
m
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Flow in round die or runner
L
pRdrurQ
R
p
0
4
82
m
pp
22
4Rr
L
pu
m
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Flow in rectangular die or runner
• as above
L
pwHQp
m12
3
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Extrusion
• Pressure generated by screw rotation
– flow rate through screw =
flow rate through die
Q(extruder) = Q(die)
– pressure rise in screw = pressure drop in
die
dp(extruder) = p(die)
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Extrusion - Ex. 1-1
• Extrude a polymer through a die with
dimensions diameter 5 mm, length 40
mm at rate 10 cm/s
• Screw is fixed, barrel rotates
• More data on next page
• Calculate barrel RPM
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• polymer density (r = 980 kg/m3
• polymer viscosity (m) = 103 N-s/m2
• barrel diameter (D) = 28 mm
• channel width (w) = 21 mm
• channel height (H) = 4 mm
• helix angle (q) = 15 degrees
• length of screw (L) = 1.25 m
Extrusion - Ex. 1-2
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Extrusion - Ex. 1-3
• First, calculate flow rate
smAvQ product /1096.1
4
005.01.0 36
2
p
dz
dpHHvwQ z
screwm122
3
L
pRQdie
m
p
8
4
pdp
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Extrusion - Ex. 1-4
• Substituting, equating, solving
dieproduct QQ
04.0108
2
005.0
1096.13
4
6 p
p
MPap 1.5
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Extrusion - Ex. 1-5
• Substituting, equating using p, solving
screwproduct QQ
ml
dz 83.415sin
25.1
sin
q
83.4
101.5
1012
004.0
2
004.0021.01096.1
6
3
3
6 zv
smmvz /5.49
solving
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Extrusion - Ex. 1-6
• Solving for RPM
smmv
v zbarrel /2.51
15cos
5.49
cos
q
RPMD
vN barrel 35
28
2.516060
pp
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Injection molding cycle
1. To make a shot: use screw (extruder)
equation for flow rate (Q) to produce a
shot volume (vol = Q*t).
– back pressure gives dp term
– time (t) bounded by cycle time (upper)
and degradation of material (lower)
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2. To inject the plastic: use pressure flow
equations and injection pressure (p)
or injection time (t) and volume to be
filled (shot volume) to determine flow
rate (Q) and hence time (t) or injection
pressure (p) required to fill mold
– injection time (t) will be limited by freezing
of plastic and degradation of material
Injection molding cycle
ME 6222: Manufacturing Processes and Systems
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Injection Molding - Ex. 2-1
• Injection mold a polymer in a steel tool
• Model the sprue, runner and part as a cylinder of diameter 10 mm, length 150 mm
• Determine the screw RPM to make a shot in less than 3 seconds (screw rotates)
• Determine the injection pressure to make the part in 2 seconds
ME 6222: Manufacturing Processes and Systems
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• polymer density (r = 980 kg/m3
• polymer viscosity (m) = 103 N-s/m2
• barrel diameter (D) = 28 mm
• channel width (w) = 21 mm
• channel height (H) = 4 mm
• helix angle (q) = 15 degrees
• length of screw (L) = 1.25 m
Injection Molding - Ex. 2-2
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Injection Molding - Ex. 2-3
• Screw RPM calculation
• Back pressure = 15 MPa
• Assume 3 seconds to make shot
• Calculate Q
smm
time
lr
time
volQ /927,3
3
1505 3
22
pp
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Injection Molding - Ex. 2-4
• Screw RPM calculation
dz
dpHHvwQ z
screwm122
3
qcosscrewz vv
60
DNvscrew
p
qsin
ldz
heightchanneldiameterbarrelD
mmD
2
204228
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Injection Molding - Ex. 2-5
• Substituting values, solving
15sin
250,1
1015
1012
4
2
415cos60
2021927,3
6
3
3
Np
N = 101 RPM
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Injection Molding - Ex. 2-6
• Injection pressure calculation
• Part injection is pressure driven
L
pRQmold
m
p
8
4
smm
time
volQ /891,5
2
1505 3
2
p
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Injection Molding - Ex. 2-7
• Substituting, equating, solving
150108
5891,5
3
4p
p
psiMPap 5226.3
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Power law viscosity
k, n are consistency and
power law index
1 nk m log mo
n-1 1
log
log
m
nk mt
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Non-Newtonian, pressure driven flow
in rectangular channel
• NB: drag flow analysis is similar to the following
n
n
H
yvv
1
02
1
dyH
ywvQ
H
n
n
2
0
1
0
212
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Non-Newtonian, pressure driven flow
in rectangular channel
n
n
n H
n
n
Lk
pwQ
121
212
2
n
n
nave
H
n
n
Lk
p
wH
Qv
11
212
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Non-Newtonian, pressure driven flow in
round channel
n
n
n
nn
R
rR
Lk
p
n
nu
111
121
n
nn
RLk
p
n
nQ
131
213
p
n
nn
ave RLk
p
n
n
R
Qv
11
2 213
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Example – 3-1
• Compare Newtonian and Non-Newtonian, pressure driven fluid flow in a rectangular channel
• Given
– H = 2 mm, w = 15 mm, L = 50 mm
– Q = 60 cm3/s= 6 x 10-5 m3/s
– m = 100 Pa-s @ d/dt = 3000/s (Newtonian viscosity)
• k = 12198, n = 0.4
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Example – 3-2
s
m
wH
Qvave 2
002.0015.0
106 5
MPa
wH
LQp 30
002.0015.0
10605.010012123
5
3
m
First, determine the Newtonian flow properties
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Example – 3-3
28
1 22 yH
L
pv
m
s
m
L
Hpvv
y3
05.01008
002.01030
8
262
0max
m
(max at y=0 because this gives the greatest value)
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Example – 3-4
• For non-Newtonian flow, determine the p needed for Q = 6 x 10-5 m3/s and vave = 2 m/s.
n
n
nave
H
n
n
Lk
p
wH
Qv
11
212
4.0
14.0
4.0
1
2
002.0
14.02
4.0
05.0121982
p
s
m
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Example – 3-5
solving
MPap 3.23
and
78.030
3.23
Newtonian
Newtoniannon
p
p
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Example – 3-6
• For non-Newtonian flow, determine Q
for p = 30 MPa.
s
mvave 77.3
2
002.0
14.02
4.0
05.012198
1030 4.0
14.0
4.0
16
s
mHwvQ ave
35103.11002.0015.077.3
88.1106
103.115
5
Newtonian
Newtoniannon
Q
Q
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Example – 3-7
• One can see the effect of shear-thinning
– reduction in pressure needed to maintain a
flow
– increase in flow with a constant pressure
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Clamp force
• Typically 50 tons/oz of injected material
• Can be approximated by
– injection pressure x projected area of part
at parting line
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Cooling in a mold
• Assume 1-D heat conduction
• Assume mold conducts much better than plastic (Biot > 1)
• Center temperature important
2
2
x
T
t
T
T = temperature
t = time
= thermal diffusivity
=k/rc
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Cooling in a mold
2x
tFo
k
hxBi
WM
WE
TT
TT
Fo
2
2
TE = ejection temp
TM = injection temp
TW = mold wall temp
2l = thickness of part
l
x
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Cooling in a mold
• Solution
– must be approximated or solved numerically
oddn
nFo
n
nFo
,
2
2sin
2exp
14,
pp
p
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Minimum cooling time - tc
Approximation for time taken (tc) for
center of flat sheet (thickness, 2l) to
reach ejection temperature (TE)
WE
WMc
TT
TTlt
pp
4ln
42
2
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Minimum cooling time - Ex. 4-1
• = thermal diffusivity ~ 10-7 m2/s
• 2l = plate thickness ~ 3 x 10-3 m
• TW = mold wall temperature ~ 50oC
• TM = melt temperature ~ 250oC
• TE = ejection temperature ~ 100oC
• Minimum cooling time for the center line to reach TE
– tc~ 15 sec.
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Minimum cooling time - tc
• Approximation for cylinder (radius = r), solved
similarly to the plate
WE
WMc
TT
TTrt 7.1ln
7.12
2
p
2r
tFo
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Non-isothermal flow
• Flow rate characteristic time constant:
– t = 1/t ~ V/Lx
• Heat transfer rate characteristic time constant:
– t = 1/t ~ /Lz2
• Small numbers give short shots
– thick runners needed
– ratio should be greater than one for filling
x
zz
x
z
L
LLV
L
LV
ratetransferHeat
rateFlow
2
~
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25.21.0
0015.0
/10
0015.0/01.0~
27
m
m
sm
msm
ratetransferHeat
rateFlow
Non-isothermal flow
So, the mold should fill.
1 cm/s 10 cm
3 mm X Y
Z
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Limits on ejection temperature
• Plastic must be cool enough to
withstand ejection force from ejection
pins without breaking
• Plastic must be cool enough so that
upon further cooling will not warp
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Ejection force
• Ejection pins force the part out of the
mold after the part has cooled and
solidified enough.
• The part will shrink onto any cores,
leading to an interference fit.
• Model as a thin walled cylinder with
closed ends (plastic part) on a rigid core
(metal mold).
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Thin-walled cylinder with closed ends
12
t
pdt
24
t
pda
30 r
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Biaxial strain
tE
pd
tE
pd
EE 42
211
t
d
t
d
E
p
421
T 1
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Ejection force
t
d
t
d
TEp
42ApFejection m
m
t
d
t
d
TAEFejection
42
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Nomenclature
• A = area
• d = core diameter
• E = Young’s
modulus
• p = pressure
• t = part thickness
• = thermal
expansion
coefficient
• T = temperature
differential
• = Poisson’s ratio
• m = friction
coefficient
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Summary
• Extrusion and Injection molding
– Flow in screw
– Flow in cavity or die
• Injection molding
– Clamp force
– Cooling time
– Ejection force
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71 ME 6222: Manufacturing Processes and Systems
Prof. J.S. Colton © GIT 2011