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7/24/2019 Hogy Szmol a Klfldi Extrudert
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Extrusion and Injection Molding -Analysis
ver. 1
ME 6222: Manufacturing Processes and SystemsProf. J.S. Colton GIT 2011
7/24/2019 Hogy Szmol a Klfldi Extrudert
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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 SystemsProf. J.S. Colton GIT 2011
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Extrusion schematic
ME 6222: Manufacturing Processes and SystemsProf. J.S. Colton GIT 2011
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Injection molding schematic
ME 6222: Manufacturing Processes and SystemsProf. J.S. Colton GIT 2011
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Schematic
hopper
heaters
barrel
screw
nozzleclamp
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/cosqH
vx vz
v=pDN
q
w is like normal pitch
w/cosqis like axial pitch
ME 6222: Manufacturing Processes and SystemsProf. J.S. Colton GIT 2011
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Flow analysis
Barrel slides across channel at the helixangle
vz= pumping
vx= stirring
w/cosqH
vx vz
v=pDN
q
w is like normal pitch
w/cosqis like axial pitch
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Flow rate
vzshows 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 parallelplates, end effects negligible
vo
y H
H
yvv0
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
dydv mm
Newtonian fluid
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Pressure flow in rectangular channel
Eliminating t
Integrating and noting
@ y = +/- H/2, v= 0
dyy
dz
dpdv
m
1
28
1 22 yH
dz
dpv
m
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Total pressure flow (Qp)
dz
dpwH
dyvwQ
H
Hp
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 zpD
m122
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
flowrate
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
pp + 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
drrrd
drrdrdr
dzdp
ttt
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Flow in round die or runner
drCrrd t
drr
rd
L
p
dz
dp
tC
drCrrdt
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Flow in round die or runner
r
L
pr
C
22
t
2
2rCrt
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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
4 RrL
p
u
m
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Flow in round die or runner
L
pRdrurQ
R
p
0
4
8
2
m
pp
224
RrL
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) = 103N-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
smAvQproduct /1096.1
4
005.01.0 36
2
p
dz
dpHHvwQ zscrew
m122
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 83.4
15sin
25.1
sin
q
83.4
101.5
1012
004.0
2
004.0021.01096.1
6
3
36 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 freezingof plastic and degradation of material
Injection molding cycle
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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 (screwrotates)
Determine the injection pressure tomake the part in 2 seconds
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polymer density (r= 980 kg/m3
polymer viscosity (m) = 103N-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
smmtime
lr
time
volQ /927,33
1505 322
pp
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Injection Molding - Ex. 2-4
Screw RPM calculation
dz
dpHHvwQ zscrew
m122
3
qcosscrewz vv
60DNvscrew p
qsin
ldz
heightchanneldiameterbarrelD
mmD
2
204228
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Injection Molding - Ex. 2-5
Substituting values, solving
15sin
250,11015
10124
2415cos602021927,3
6
3
3
N
p
N = 101 RPM
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Injection Molding - Ex. 2-6
Injection pressure calculation
Part injection is pressure driven
LpRQmold
mp8
4
smm
time
volQ /891,5
2
1505 32
p
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Injection Molding - Ex. 2-7
Substituting, equating, solving
150108
5
891,5 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-11
log
log
m
n
k 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
y
vv
1
0
2
1
dyH
y
wvQ
H
n
n
2
0
1
0
2
12
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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
nn
nn
n
R
rR
Lk
p
n
nu
111
121
n
nnR
Lk
p
n
nQ
131
213
p
n
nn
ave RLk
p
n
n
R
Qv
11
2 213
p
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Example3-1
Compare Newtonian and Non-Newtonian, pressure driven fluid flow ina rectangular channel
Given
H = 2 mm, w = 15 mm, L = 50 mm
Q = 60 cm3/s= 6 x 10-5m3/s
m= 100 Pa-s @ d/dt = 3000/s (Newtonian
viscosity) k = 12198, n = 0.4
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Example3-2
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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Example3-3
28
1 22 yH
L
pv
m
s
m
L
Hpvvy
305.01008
002.01030
8
262
0max
m
(max at y=0 because this gives the greatest value)
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Example3-4
For non-Newtonian flow, determine thep needed for Q = 6 x 10-5m3/s and vave= 2 m/s.
n
n
nave
Hnn
Lkp
wHQv
11
212
4.0
14.0
4.0
1
2002.0
14.024.0
05.0121982
p
sm
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Example3-5
solving
MPap 3.23
and
78.030
3.23
Newtonian
Newtoniannon
p
p
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Example3-6
For non-Newtonian flow, determine Q
for p = 30 MPa.
s
m
vave 77.32
002.0
14.02
4.0
05.012198
1030 4.014.0
4.0
16
s
mHwvQ ave
35103.11002.0015.077.3
88.1106
103.115
5
Newtonian
Newtoniannon
Q
Q
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Example3-7
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 thanplastic (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
TTTT
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) forcenter 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 ~ 10-7m2/s
2l= plate thickness ~ 3 x 10-3m
TW= mold wall temperature ~ 50oC
TM= melt temperature ~ 250oC TE= ejection temperature ~ 100
oC
Minimum cooling time for the center line toreach TE
tc~ 15 sec.
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Minimum cooling time - tc
Approximation for cylinder (radius = r), solvedsimilarly 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/s10 cm
3 mmXY
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
T1
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Ejection force
t
d
t
d
TEp
42
ApFejection m
m
td
td
TAEFejection
42
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Nomenclature
A = area
d = core diameter
E = Youngs
modulus
p = pressure
t = part thickness
= thermal
expansion
coefficient
T = temperaturedifferential
= Poissons ratio
m= frictioncoefficient
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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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