Extrusion and Injection Molding - Analysislibvolume2.xyz/.../polymerextrusion/polymerextrusionnotes2.pdf · Extrusion and Injection Molding - Analysis ver. 1 ... or injection pressure

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



3

Extrusion schematic



4

Injection molding schematic



5

Schematic

hopper

heaters

barrel

screw

nozzle clamp

mold

cavity

pellets

motor /

drive

throat



6

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



7

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



8

Flow rate

• vz shows viscous traction work against

exit pressure

flow rate = f(exit pressure, vbarrel, m, d, w, l)



9

Flow analysis

• Simplify by using Newtonian fluid

• Separate into drag and pressure flows

• Add solutions (superposition)



10

Drag flow in rectangular channel (QD)

• Simple viscous flow between parallel

plates, end effects negligible

vo

y H

H

yvv 0

wHvAvQD 2

10



11

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



12


Equilibrium

p p + dp

dz

y

z

2y

t

t

hopper exit

022 dzydppp t



13


022 dzydppp t

dz

dpy t

dy

dvγτ mm

Newtonian fluid



14


• Eliminating t

• Integrating and noting

@ y = +/- H/2, v = 0

dyydz

dpdv

m

1

28

1 22 yH

dz

dpv

m



15

Total pressure flow (Qp)

dz

dpwHdyvwQ

H

H

p

2

2

3

12m



16

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)



17

Nomenclature

• dz = helical length = axial length/sinq

• vz = helix velocity = vbarrel*cosq



18

Flow rate

flow

rate

output pressure

w

2w



19

Schematics

Injection molding

Extrusion



20

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



21


Neglecting HOT

dzdrdrdpdrr ttpp 22

dzrddrrdprdrr tttpp 2][ 22

drr

rd

drr

drdr

dz

dp

ttt



22


drCrrd t

drr

rd

L

p

dz

dp

tC

drCrrd t



23


rL

pr

C

22

t

2

2rCr t



24


• At center, t = 0

• At edge of tube (R), t = max

L

Rp

2max

t

dr

dumt

Newtonian fluid



25


m

L

rp

dr

du

2

finally

22

4Rr

L

pu

m



26


L

pRdrurQ

R

p

0

4

82

m

pp

22

4Rr

L

pu

m



27

Flow in rectangular die or runner

• as above

L

pwHQp

m12

3



28

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)



29

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



30

• 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



31

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



32

Extrusion - Ex. 1-4

• Substituting, equating, solving

dieproduct QQ

04.0108

2

005.0

1096.13

4

6 p

p

MPap 1.5



33

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



34

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



35

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)



36

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



37

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



38

• 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




39


• 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



40


• Screw RPM calculation

dz

dpHHvwQ z

screwm122

3

qcosscrewz vv

60

DNvscrew

p

qsin

ldz

heightchanneldiameterbarrelD

mmD

2

204228



41


• Substituting values, solving

15sin

250,1

1015

1012

4

2

415cos60

2021927,3

6

3

3

Np

N = 101 RPM



42


• Injection pressure calculation

• Part injection is pressure driven

L

pRQmold

m

p

8

4

smm

time

volQ /891,5

2

1505 3

2

p



43


• Substituting, equating, solving

150108

5891,5

3

4p

p

psiMPap 5226.3



44

Power law viscosity

k, n are consistency and

power law index

1 nk m log mo

n-1 1

log

log

m

nk mt



45

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



46

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



47

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

pME 6222: Manufacturing Processes and Systems


48

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



49

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



50

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)



51

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



52

Example – 3-5

solving

MPap 3.23

and

78.030

3.23

Newtonian

Newtoniannon

p

p



53

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



54

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



55

Clamp force

• Typically 50 tons/oz of injected material

• Can be approximated by

– injection pressure x projected area of part

at parting line



56

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



57

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



58

Cooling in a mold

• Solution

– must be approximated or solved numerically

oddn

nFo

n

nFo

,

2

2sin

2exp

14,

pp

p



59

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



60

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.



61

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



62

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

~



63

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



64

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



65

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).



66

Thin-walled cylinder with closed ends

12

t

pdt

24

t

pda

30 r



67

Biaxial strain

tE

pd

tE

pd

EE 42

211

t

d

t

d

E

p

421

T 1



68

Ejection force

t

d

t

d

TEp

42ApFejection m

m

t

d

t

d

TAEFejection

42



69

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



70

Summary

• Extrusion and Injection molding

– Flow in screw

– Flow in cavity or die

• Injection molding

– Clamp force

– Cooling time

– Ejection force



71 ME 6222: Manufacturing Processes and Systems


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Extrusion and Injection Molding - Analysislibvolume2.xyz/.../polymerextrusion/polymerextrusionnotes2.pdf · Extrusion and Injection Molding - Analysis ver. 1 ... or injection pressure