Hogy Számol a Külföldi Extrudert

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


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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



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Extrusion schematic



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Injection molding schematic



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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



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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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Equilibrium

p p + dp

dz

y

z

2y

t

t

hopper exit

022 dzydppp t



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022 dzydppp t

dz

dpy t

dydv mm

Newtonian fluid



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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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Neglecting HOT

dzdrdrdpdrr ttpp 22

dzrddrrdprdrr tttpp 2][ 22

drrrd

drrdrdr

dzdp

ttt



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drCrrd t

drr

rd

L

p

dz

dp

tC

drCrrdt



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r

L

pr

C

22

t

2

2rCrt



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At center, t= 0

At edge of tube (R), t= max

L

Rp

2max

t

dr

dumt

Newtonian fluid



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m

L

rp

dr

du

2

finally

22

4 RrL

p

u

m



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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

ME 6222: Manufacturing Processes and Systems

Prof. J.S. Colton GIT 2011


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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





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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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Screw RPM calculation

dz

dpHHvwQ zscrew

m122

3

qcosscrewz vv

60DNvscrew p

qsin

ldz

heightchanneldiameterbarrelD

mmD

2

204228




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Substituting values, solving

15sin

250,11015

10124

2415cos602021927,3

6

3

3

N

p

N = 101 RPM




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Injection pressure calculation

Part injection is pressure driven

LpRQmold

mp8

4

smm

time

volQ /891,5

2

1505 32

p




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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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