Science of Extruder #1

2012
MA201
The Art & Science of Extrusion
for Wire & Cable – Part I
Instructor: Dr. Stephan Puissant
Professional Development Courses

Iwcs 2012/11/11
MA201:
THE ART AND SCIENCE OF EXTRUSION
FOR WIRE AND CABLE – Part I
Stéphan Puissant
E-mail: stephan.puissant@brugg.com; stephan_puissant@yahoo.fr
 Now
 Middle XIX’s
20111111 IWCS – The art and science of extrusion for wire and cable Part 1 – S. Puissant - 1 -
Single screw extrusion. An Introduction.
Presentation Overview.
 0 Presentation overview

 1 Polymers, their viscosity and thermal behavior (11 slides).

 2 Single Screw Extruder (55 slides)

 3 Extrusion Crosshead (19 slides)

 4 Cooling bath (7 slides)

 5 References (1 slide)

1 Polymers mechanical and thermal characteristics.
1.1 Archimedes screw vs. plasticizing screw.
 The extruder is a machine to transport and process materials .


 287-212 before J. C.(Archimedes).
 Transport screw.
 1845 (Bewley, Faraday).

 First extruder (Gutta Percha, submarine cable)
Extruder screw
 Now.
 An extruder screw difference with
these screws is related to the
polymer characteristics.
1.2 Introduction.
 Polymers are materials with quite different properties depending on:

Chemical composition
Molecular structure
Fillers and additives
 Most of the practical polymers have two properties in common:

Low heat conduction
High viscosity
 They are important for the processing
1.3 Mechanics: definitions of terms used for polymer flows.
W L S=W.L
The shear rate is defined as 
V/h (Velocity/height)
m  V (velocity of upper plate F (pull force)
V
γ = in  s =s −1 (S.I. units)  plate)
h m 
 
The shear stress is defined as  h
F/S (Force/Surface)
F N 
τ= in  2 = Pa (S.I. units) 
S m 
The (dynamic) viscosity is defined as


shear stress/ shear rate (F/S)/(V/h)
F⋅ h  N 
η= in  2 .s = Pa.s(S.I units) = 0.1Poises(c.g.s. units) 
S⋅ V  m 
Do not confound with the kinematical viscosity ! (often used for lubricants): 
dynamic viscosity/density

η  Pa.s
ν = in  =
N 2 .s
m
( )
=
m2
(SI units) = 1.10 4 cm
2

= 1.10 Stokes (c.g.s. units) 
4
ρ kg kg s s
 m 3 m3 
1.4 Mechanics: order of magnitude of viscosity.
The viscosity is very high for polymer melts (ca. 1 million higher than water)
η (Pa.s) ν (m2/s)
Air (0°C) 0.000017 (1.7 10-5 ) 1.33 10-5
Water (0°C) 0.0018 (1.8 10-3) 1.79 10-6
Water (20°C) 0.001 (10-3) 1.0 10-6
Mercury (20°C) 0.0016 (1.6 10-3) 1.2 10-7
Oils 0.01(1.0 10-2) to 1.0 1.0 10-5 to 1.0 10-3
Molten Polymers 100 (102) to 10 000 (104 ) 1.0 10-1 to 1.0 100
Molten Glass 100 (102) to 10 000 (104 ) 1.0 10-1 to 1.0 100
1.5 Mechanics: influence of high viscosity on flow.
Consequence of the high viscosity:

 The number of Reynolds is defined as the ratio of the inertia forces (acceleration) to
the viscous forces.   
− ∇ p+ η∆v + F − ργ = 0
 From Navier-Stokes equation (Dynamic equilibrium + continuity +conservation of mass):
ρ .V .h
V
ρ
Re = ρ ⋅ γ acceleration ρV h 2
Re = = t = V = ρ ⋅ V = ρ .V .h
η
η .∆ v η .V 2 η .V 2 η .V / h η
h h
 In Practical polymer processing:

 ρ ~ 103 kg.m-3, η>102 Pa.s,,
 h <20mm, V < 1m/s (majority of extruders work with a maximum peripheral speed of 0.5 m/s).
 Re < 2.10-1, which is much less than 2100, the critical number of Reynolds under which the flow
is laminar and becomes turbulent above it.
The flow of polymers inside an extruder or a X-head is always laminar.
1.6 Mechanics: influence of high viscosity on pressure levels.
 High pressures are required for the extrusion

(example of a flat die):
 Expression of output vs Pressure gradient: y W
z
Q.L  12η 
.  = ( p1 − p 2 )
W  h3  pressure P1 pressure P2
 Pressure order of magnitude
h
 P2=0 MPa,
 L=0.1m, W=0.1m, h=0.001m, 0 u(z) L x
 η=103,
 Q=60kg/h=1kg/min=10-3 m3/min (for ρ ~ 1000kg/m3)
 P1(h=1mm)~(10-3/60)x0.1/0.1x(12 .103/ 10-9)
 P1(h=1mm)~2. 108Pa=2000 bar
 P1(h=2mm)=250bar
Question: For same flowrates, what would be the pressure levels for water ?
(viscosity water~ viscosity plastic/1'000'000)
1.7 Mechanics: dependence of viscosity to shear rate and temperature.
The viscosity is also

 temperature dependent (η decreases with increasing temperature)
 shear rate dependent (η decreases with increasing shear rate) η(γ,T)
 Example of viscosity of a LDPE 22D760 of BAYER.
V is c o s ity o f L D P E 2 2 D 7 6 0 ( B a y e r )
100000
10000
170
E ta (P a .s )
1000 210
250
100
10
0 ,1 1 10 100 1000
S h e a r ra te (s -1 )
1.8 Thermals: low heat conduction.
 Thermal conductivity: λ = 0.1 - 0.3 W/mK (1000 x lower than

copper)
λ m 2
 Thermal diffusivity: a= . − 6.5 ⋅ 10− 7
a = 13
ρ ⋅c s
 A heat wave penetrates to a depth of:

 1 cm in 17 min
 1 mm in 10 sec
 0.1 mm in 0.1 sec
 Consequence: The material has to be melted as a constantly
renewed thin layer in contact with the barrel.
1.9 Thermals: low heat conduction vs. high viscosity (1/2).
Another consequence of high viscosity: Tw U
 Brinkman number (Dissipation energy/heat

h V Tav
exchange):
W ηV
2
 Br = 2.  =
Q k (Tp − T )
 Mechanical energy dissipated (V average
 In the case of extrusion, with a peripheral speed velocity of the melt=Uupper wall/2=U/2:
( )
2
 
W = η.γ 2 = η. V  = 4.η. V 2
of 0.2m/s:




h
2
( ) 

h2
Br ≥ 2
 Heat exchanged with the 2 walls at the center
of flow (k is conductivity of melt, Tw, wall
temperature, Tav average temperature of 
melt): Q ≈ 2. − k .
(T − Tw )  = −8.k . (T − Tw )
 ( h 2 ) 2  h2
 Shear (friction) heating is important and is 
 Ratio dissipation/conduction 2
4.η V 2
directly used in the melting process  W
= h =
ηV 2 h 2
=
ηV 2
 Q 8k (Tp − T ) 2k (T p − T )h 2 2k (Tp − T )
h2

Watt 
 In the case of extrusion η ≈10 3 Pa.s, k = 0,2 
m.°C  ⇒ Br ≥ 2
T p −T ≤100°C , V = 0.2m / s 

Question: Majority of extruder have a maximum speed of 0.5m/s (30m/min or 251,3 rpm for
a D=38mm):
Will be dissipation or conduction the main phenomenon?
1.10 Thermals: low heat conduction vs. high viscosity (2/2).
Example: Evolution of temperature in a slit die.

T function of output (height 10mm)
 Case of the slit die (width=10cm,height=10mm) 350
Temperature of mass outside (ºC)h=10mm
with which we measured viscosity (slide 7) 330
310 DeltaTheoretical=10mm
 Measurement on a 60-30D 290
The 6 first values are in agreement with the increase

270
T(°C)
250
of temperature obtained by viscous dissipation: 230
210
 i.e. increase of temperature in function of the 190
square of velocity. 170
( h ).h
W = η .γ 2 .h = η . V
2
2
150
10 100
l/h
Viscosity of LDPE 22D760 (Bayer)
1000
100000
10000 170
Question: only the first point are in accordance with this law 210
Eta (Pa.s)
How to explain it with the viscosity curve on the right? 1000
250
Average shear rate (s-1)

100 in flat die
10
0,1 1 10 100 1000
Shear rate (s-1)
1.10 Thermals: Necessary melting Heat.
Heat necessary to heat up a semi

crystalline material (PBT).
 From [G.E.]
 Crystalline part increases
Surface under the peak
increases ∆H (Enthalpy of
fusion increases.
 Energy necessary to heat from
20°C to 280°C 1 kg of PBT:
Q = C p .∆T + ∆H
Q = C p .( 280 − 20) + ∆H
Q =1700.290 + 41500
kJ
Q = 534.50
kg
2 The single screw extruder.
2.1 Generalities on extruder.
2.1.1 The role of the extruder.
 The extruder must prepare the raw material (powder, pellets,

stripes) for the subsequent shaping operations. Its task is therefore
to
 plasticize the solid material
 homogenize (Mixing of pigments, minimize temperature fluctuations ∆T)
 build up pressure (pressure consumption of tool, minimize pressure
fluctuations ∆P)
2.1.2 Screw and barrel.
 The heart of the extruder is the screw and barrel

 Barrel: smooth or grooved over a fraction of its length
 Practical length of screw and barrel: 20-35 times the diameter
 one or several flights
 Division of screw into functional zones
 feed zone
 melting zone
 pumping zone
 mixing zone
2.1.3 Screw and barrel functional zones.
Feed zone Melting zone Pumping and mixing zone
2.1.4 Other components.
Hopper Zone 1 Zone 2 Zone 3 Zone 4 Zone 5
Water cooled
Feeding flange Barrel Heating
zone Cover
Gearbox
X-Head
Water cooling Clamp
sleeves
Pulley and Cooling thins

Belt Cover
Heater Band
Motor Cooling fan cover (1 per Zone) Frame

2.1.5 Geometry of screw used in simulation and practical demo.
 Screw 38-25D showing high compression:

 Compression ratio=Hfeed/Hmetering=3.
Hopper Feeding Compression Metering

Length 1.3D 14D 3D 8D
Depth 6.5mm 6.5mm 2.16mm
2.2 Feed zone.
2.2.1 Diagnosis: pressure and rotation speed dependence of output.
 Specific throughput decreases with head pressure: Higher screw speed

required for given output => Increased melt temperature
Throughput of barrier screw with smooth barrel,
D=63.5 mm (Han, Lee and Wheeler, Polym. Eng. & Sc.,
Vol. 31, No. 11, p. 837)
1.2
Specific flow rate
0.8
[kg/h/rpm]
50 rpm
75 rpm
0.4
100 rpm
0
50 100 150 200 250
Head pressure [bar]
Question: Why are outputs decreasing with head pressure?
2.3 Feed zone.
2.3.2 Description of feed mechanism.
Direction of
screw rotation
Friction based mechanism.
Pitch=1D x An increasing solid conveying
z
angle ϕ increases the
y
velocity of the solid vs.
Ff
ϕ
W e screw
Vb/g
Vb/s Approximation:
Fna
ϕ Fp2 Geometry considered in
ϕ

ϕ+θ first approximation flat.

Ffa
Fp1 θ  Granules considered as a
Fr Fnp Vg/s solid plug.
Ffp
Velocities of :
 Barrel moving screw
• barrel vs. screw fixed
Enumeration of forces acting on • granules vs. screw
the solid plug of granules. • barrel vs. solid
Question: What is the reason for the polymer to move forward?

2.3 Feed zone.
2.3.3 Friction coefficients.
 Physical interpretation of Coulombs law: (Ref.

[J.F.A.] p.158).
Coefficient of friction of HDPE. ([G.A.M.], p.24):

*independant on velocity
*diminishes a lot with temperature
*diminishes also with pressure
2.3 Feed zone.
2.3.4 Flat plate model: influence of channel depth.
Results based on a flat plate model Ou t p u t a n d co n v e y in g a n g le in f u n ct ion of ch a n n e l d e p t h
(D=38mm=1.5”): ( P/ P0 = 1 , f f = 0 , 3 6 ) ,
e x t r u d e r 3 8 m m , f lig h t w id t h = 3 , 8 m m , f lig h t p it ch = 3 8 m m , PE.
45 27,5
 Solids conveying is governed by difference 40 q(° ) 25
Ou t p u t ( k g / h )
Q(kg/h)
of friction forces between screw and barrel. 35
30
22,5
20
 Throughput depends on geometry 25

20
17,5
15
 Optimum feeding channel depth: 15 12,5

10 10
 6,5mm (screw real value) for P/P0=1, for 5 7,5
ff=0,36. 0
0 1 2 3 4 5 6 7 8
5
Ch a n n e l d e p t h ( m m )
 5mm for P/P0=100, for ff=0,36.
Ou t p u t a nd con v e y in g a n g le in f u n ct ion of ch a n n e l d e p t h
( P/P0 = 1 0 0 , f f = 0 , 3 6 ) ,
e xt r u d e r 3 8 m m , f lig ht w id t h = 3 , 8 m m , f lig h t p it ch = 3 8 m m , PE.
40 27,5
 Question: 35 q(° )
Q(kg/h)
25
22,5
Ou t p u t ( k g / h )
30 20
 Why is the output decreasing above 25 17,5
this optimum channel depth?

15
20
12,5
15 10
10 7,5
5
5
2,5
0 0
0 1 2 3 4 5 6 7 8
Ch a n n e l d e p t h ( m m )
2.3 Feed zone.
2.3.4 Flat plate model: influence of flight pitch.
Results based on a flat plate Ev olu t ion of ou t p u t in f u n ct ion of f lig h t p it ch ,

model D=38mm=1.5”: f or d if f e r e n t coe f f icie n t of f r ict ion s, P/P0 = 1 0 0 .
 For a Channel depth = 70
6,5mm, N=100 rpm
60 Q(kg/h) ff= 0,5,
 Dependence to
Out p ut ( kg /h)
fv= 0.2
50
 flight pitch. Q(kg/h)
 Barrel friction 40 ff= 0.45,
fv= 0.2
coefficient
30 Q(kg/h) ff= 0.4,
 Max outputs may vary fv= 0.2
20
from 20kg/h to about 60 Q(kg/h)
ff= 0.33,
kg/h. 10
fv= 0.2
0
0 0,5 1 1,5 2 2,5
Fligh t Pit ch ( * D )
 Question: Where does the optimum flight pitch lies?
2.3 Feed zone.
2.3.5 Flat plate model: influence of pressure.
Dependance on coefficient of friction and back pressure at the end of the feeding zone
(D=38mm=1.5”).
 Higher back pressure reduces output.
 Higher coefficient of friction against barrel reduces the pressure dependence of the output.
Out put in funct ion of ba ck pr e ssure at t he e nd of t he fe e ding zone

f or dif fere nt coe fficie nt of f rict ion on bar r el f f
 Question: Ext ruder dia m e t e r 3 8 , f light pit ch= 3 8 m m , flight w idt h= 3 , 8 m m , 1 f light , PE.
60
How to increase
50
Out put /kg/h)
machine performance
40
(coefficient of friction)?
30
20
10
0 P/P0
1 10 100 1000
Q(kg/h), ff= 0.33 Q(kg/h), ff= 0.35 Q(k g/h), ff= 0.4 Q(kg/h), ff= 0,45 Q(kg/h), ff= 0.5
Q(kg/h), ff= 0,6
2.3 Feed zone.
2.3.6 Grooved barrel.
 Increase apparent barrel friction (polymer-polymer friction is higher

than polymer-metal friction). Can build up pressure and override
subsequent zones. Output becomes virtually independent of
backpressure
 Problems:
 frictional heat generation may become excessive
 wear and abrasion at end of grooves
2.3 Feed zone.
2.3.9 Feed zone with helical grooves: its limit.
Straight grooved feedings depend also on internal coefficient of friction.

Example of a 100-20D with TPU (Elastollan 1195):
 specific outputs at 25 rpm,
 for two feeding some temperatures Tz1=155°C, and Tz1=130°C.
Increase of 44.8% of output!-->TPU has a coefficient of friction with metal varying a lot!
Zone 1 varying between the two temperatures about all 10 minutes:
 Low T: high friction= high friction energy= T increases after some time
 High T: low friction=low friction energy=T decreases after some time
N(rp Tz1(° g/36s g/36s g/36s g/36s Q(kg/

m) C) h)
25 155 870 870 840 900 87
25  Question:
130 Any 1210 1270
idea how 1260things?
to improve 1300 126
2.3 Feed zone.
2.3.7 Feed zone with helical grooves: how it works.
 The nut-on-screw operation is achieved through geometrical

interlocking of solids. Hence throughput is governed by geometry
(section S0) rather than by frictional force balance
 Virtually independent of back pressure.
 Sections must be carefully compensated to prevent wear
S0
2.3 Feed zone.
2.3.8 Feed zone with helical grooves: example of results.
 Independance against counter pressure for feed zone with helical

groove:
Example: Specific throughput vs. head pressure
(NMA-80,D=31/4 “Alain- Aline II, LDPE (13.9.1991))
Specific throughput vs. head pressure
2.5
Rate (kg/h/rpm)
2
30 rpm
1.5
60 rpm
1
0.5
0
0 100 200 300 400 500
Pressure (bar)
2.3 Feed zone.
2.3.9 Feed zone with helical grooves: advantages and disadvantages.
Advantages:
 Output depends on screw speed only
 Product dimensions are easier to keep under control during ramps
 Transient effects weak and only over a short period of time
 pressure build-up capability
but:
 Screw sections have to be carefully matched (melting capacity etc)
 Feed section may not be adapted for some materials, particularly materials
with low 'internal' or intergranular coefficient of friction(but: Is there any
universal high-performance screw anyway?)
2.3 Feed zone.
2.3.10 Comparison of the different feeding zones.
Output and specific output on a 150-24D (courtesy Maillefer) with PVC

ρ=1.41kg/l:
Spiraled feeding zone more linear
Out pu t in f unct ion of scre w sp e e d Sp e cif ic ou t p u t in f u n ct io n o f scr e w r ot a t ion
1600 30
1400 28
1200 26
Qs ( k g/h/r pm )
1000 24
Q(k g/h)
800 22
600 20
400 18
200 16
0 14
5 10 15 20 25 30 35 40 45 50 55 60 5 10 15 20 25 30 35 40 45 50 55 60
N ( rpm ) N( rpm )
Q(kg/h) sm oot h Q(kg/h) grooved Q(kg/h) spiraled Qs(kg/h/rpm ) Qs(kg/h/rpm ) Qs(kg/h/rpm )

barrel barrel (BASF) barrel sm oot h grooved (BASF) spiraled
Question: Advantages of linear output (constant

specific output) vs. rpm?
2.4 Melting zone.
2.4.1 Description of melting mechanism.
The solid material is compacted and gradually melted. The molten material accumulates into
a melt pool. (Pull-out tests by Maddock 1960)
 Photo from [J.F.A.]p.172
2.4 Melting zone.
2.4.2 Melting zone unwrapped.
 Unwrap the screw into a plane. Solid and molten material separates
according to the scheme below. Material that melts early or late
does not have the same history!
Metering zone
Melting zone Solid
Relative motion of barrel Feed zone
Liquid
2.4 Melting zone
2.4.3 Viscous dissipation against conduction (again!
.
Basic melting mechanism: "Tadmor model” (1968). More elaborate
models (Lindt, Han) can be traced back here
Barrel Melt film
Melt pool Solid bed

 Viscous dissipation in the melt film above the solid bed provides the heat at
the interface to the solid bed for melting.
 Heat conduction is almost irrelevant except at low screw speeds, or for some
highly filled materials
2.4 Melting zone.
2.4.4 Energy balance.
Energy Balance of the Melting Process at the interface between melt

film and solid bed
Eth = λ ⋅ S ⋅
Tw − Tm
δ
δ
E th U Eth
ηU 2
Ediss = S ⋅
2δ
Eth: Heating power Ediss: Dissipation power
λ: Thermal conductivity Tw: Barrel temperature
Tm: Melting temperature S: Surface of melt film
δ: Melt film thickness U: Relative barrel velocity
Stability: If δ increases, Eth and Ediss decrease, which tends to make δ
decrease again
2.4 Melting zone.
2.4.5 Melting energy necessary for a given polymer.
Melting energy necessary to bring the material from the solid into the
liquid phase:
(
E p = Q ⋅ cs ( Tm − Ts ) + ∆hm )
Ep: Melting power Q: Mass throughput
cs: Heat capacity ∆hm: Heat of fusion
Tm: Melting temperature Ts: Solid bed temperature
Theoretical energy balance:

Eth + Ediss = E p
used to calculate melting rates and melting length along the screw
2.4 Melting zone.
2.4.6 Melting energy necessary for a given polymer.
Thermal parameters and viscosity used for calculation of melting.

For FEP the shadowed values are estimated.
Calorific Solid Melt Fusion

Conductivity capacity density density Enthalpy Melting Zone T
Material (W.m .K-1) (J.kg-1.K-1)
-1
(kg.m ) (kg.m ) Viscosity (Pa.s) (J.kg-1) T. (°C) (°C)
-3 -3
PE Bayer 0.20 2500 920 1800 1000 15000 110 150

PB GE 0.25 1600 1310 1080 500 48800 225 265
FEP Daikin 0.25 1163 2120 1800 5000 to 60000 31900 270 390
FEP Dupont 0.195 1172 2140 263 390
2.4 Melting zone.
2.4.7 Simple Tadmor melting model: screw pitch and number of flights.
Influence of pitch and number of flights on solid bed Axial m elt ing length versus pitch and num ber of flights
length (D=38mm=1.5”): 4,2

4,1
 Above a pitch of 1D to 2D, the melting length is only 4
decreasing a little (results for 1 flight) 3,9

3,8
 Decrease of 5% of length with increase of pitch from 3,7
0.6D to 1D 3,6
Decrease of 4% of length by increase of pitch from 1D to
Axial m elt ing lengt h (*D)

 3,5
3,4
2D. 3,3
 Decrease of 1% of length with increase of pitch from 2D 3,2
to 3D. 3,1
3
 The melting length is smaller with increasing 2,9
number of flights. 2,8
 Decrease from 3.9D for 1flight, to 3.3D for two flights. 2,7
2,6
0,5 0,75 1 1,25 1,5 1,75 2 2,25 2,5 2,75 3 3,25 3,5 3,75 4 4,25 4,5
Pitch (*D)
Lp(D), b= 3,8m m, Lp(D), b= 3,8mm ,nflight= 2 Lp(D), b= 3,8mm ,

nflight= 1 nflight= 3
 Question: Why are most of screw at a pitch of 1D, when the figure above shows that we
could further reduce the melting length?
2.4 Melting zone.
2.4.8 Simple Tadmor melting model: screw pitch and flight width.
Influence of flight width on melting

length (D=38mm=1.5”, pitch=1D): Melting length in function of pitch
 For an decrease of flight width from for different flight widths
7.6mm to 3.8mm, decrease from 4,7
Lp(D), b=0mm,
4.5D to 4.4D. 4,65 p=1
4,6 Lp(D), b=3,8mm,
 For an decrease of flight width from 4,55
p=1
Lp(D), b=7,6mm,
Melting length (/D)

3.8mm to 0 mm (theoretical!), length 4,5 p=1
decreases from 4.4D to 4.3D. 4,45
4,4
4,35
4,3
4,25
4,2
4,15
 Question: Why not decrease flight 4,1

0,5 1 1,5 2 2,5 3 3,5 4 4,5
width to decrease plasticizing Flight pitch (/D)
length?
2.4 Melting zone.
2.4.9 Simple Tadmor melting model: influence of polymer.
The melting length for a D=38mm=1.5”:

changes with Plastification length in function of screw speed
 increasing rotation speed (specific 4,5
volume output considered constant) 4,25

4
 For PE, above 60 rpm, and for PBT, 3,75
Plastification length (/D)

above 80 rpm melting length decreases 3,5
again. 3,25 Lp(D) FEP
 Different materials (Here PE, PBT 3 Lp(D) PBT
Lp(D) PE
and FEP) 2,75
2,5
 Due to higher energy needed for 2,25
melting PBT 2
 Due to different viscosity 1,75
1,5
0 25 50 75 100 125 150 175
N(rpm)
 Question 1: Decrease of length above a certain speed: realist?

 Question 2: Do you think the same screw should be used for PE and PBT?
2.4 Melting zone.
2.4.10 Simple Tadmor melting model: its limits.
In reality, the melting process is complex because:

 The local viscosity varies with local shear rate and local temperature.
 The melting rate depends on local viscosity: Mathematical models
become involved
 Temperature peaks in the melt film can be considerably higher than
the average melt temperatures: Degradation!
 Leakage flow across flight clearance adversely affects melting rates:
worn out screws perform badly
 Melting length increases with screw speed, if throughput is
proportional to screw speed: Danger of incomplete melting at high speeds =>
Barrier screws / Mixing elements!
2.4 Melting zone.
2.4.11 Improvement: the barrier zone.
 Invented in 1959 by Charles Maillefer. A secondary flight with

increased pitch and increased flight clearance acts as a "filter". It
retains unmolten pellets in the extruder. Output and melt homogeneity
are increased compared to a 3-zone standard screw.
 Other designs derived from Maillefer's initial concept:

 Barr, Kim, Ingen-Housz etc. They all take advantage from the
mechanical separation between solid and melt.
2.4 Melting zone.
2.4.12 Improvement: description of the barrier zone.
Barrel Melt film
Main flight
Melt pool
Barrier flight
Solid bed
Primary Secondary
channel channel
Coffee Break
2.5 Melt conveying.
2.5.1. Simplification of flow matter.
 A complex flow field appears in all melt-filled zones. In a simplified

manner, it may be thought of as a superposition of drag and
pressure flow in downchannel direction along the screw:
Downchannel flow Apparent downchannel

flow
Flow pattern
Is decomposed into:
Drag flow Pressure flow
2.5 Melt conveying.
2.5.2. description of simplified flow.
W ⋅U z ⋅ h W ⋅ h 3 ∆P
Q = Qd − Qp Qd = Qp =
2 12 ⋅ η Z
Qd: Drag flow Qp: Pressure flow
W: Channel width h: Channel height
Uz: Rel. peripheral velocity ∆P: Pressure difference
η: Viscosity Z: Channel length
In reality, additional effects are present:

 Non-Newtonian behavior of molten polymer
 Coupling of flow and temperature fields
 cross-channel flow and leakage flow across flights
 channel curvature
2.5 Melt conveying.
2.5.3 Simplified flow first analysis.
Influence of depth on flow (C.Rauwendahl,

‘Polymer extrusion’ Hanser Verlag, 3rd
edition 1994)
 By using the ratio rd between pressure
flow and drag flow:
Qp ∆P h 3
rd = =
Qd z 6η.u fz h
 Rauwendahl (Ref. [C.R.], p283) shows
that optimal mixing is for rd=0.5
 Question: How can it be that at rd=1

we are in close discharge?
2.5 Melt conveying.
2.5.4 Simple complete screw model: Interaction with feed zone.
Output of a D=60mm=2.5” Out put of a 3 8 m m ( 1 0 0 rpm ) in funct ion of backpre ssure

for different frict ion coefficient s on barre l.
extruder is shown, in De pt h of scre w in m et e ring se ct ion= 2 ,1 6 m m .
function of coefficient of 40
friction on barrel. 35
In fact the extruder screw 30
Out p ut ( k g/h)
Out put (kg/h)ff= 0,325
Out put (kg/h) ff= 0,350
performance is based on the 25 Out put (kg/h) ff= 0,375
Out put (kg/h) ff= 0,400
interaction between: 20 Xhead out put (kg/h) L= 5,
R= 20
 the 3 zones, 15 Xhead out put (kg/h) L= 3,

R= 20
 and the Xhead. 10
0
0 200 400 600 800
Ba ckp re ssure @ X he a d ( ba r)
 Question: To what do the different curves (extruder output for different friction
coefficients) correspond in reality?
2.5 Melt conveying.
2.5.5 Simple complete screw model: Influence of metering flight depth.
To improve mixing, metering zone depth can be increased:

Output in function of counter pressure
for different melt conveying zone depth.
38-25D, ff=0.35, PE.
50 Output (kg/h),
Hp=1,4mm
45 Output (kg/h),
Output (kg/h)
40 Hp=2,16mm
Output (kg/h),
35 Hp=3,00 mm
30 Output (kg/h),
Hp=5,0 mm
25
Xhead output (kg/h)
20 L=5, R=20
15 Xhead output (kg/h)
L=3, R=20
10
5
0
0 100 200 300 400 500 600 700 800
Pressure end metering zone (bar)
 Question: any idea why the curve for depth=5.0mm stops in the region of low pressure,
high output?
2.5 Melt conveying.
Pressure along the screw (D=38mm=1.5”, N=100rpm):
Case of real metering depth =2.16mm.
 For P<10 bar (Q=18.9kg/h) metering zone consuming pressure

Evolution of pressure along screw (Hm=2.16mm)
for different outputs
500
450
400
Pressure (bar)
350
18,85
300 17,46
250 16,08
13,68
200
150
100
50
0
0 5 10 15 20 25
Axial length(/D)
 Question: What means a negative pressure gradient in the ast zone?

2.5 Melt conveying.
Pressure along screw (D=38mm=1.5”, Evolution of pressure along screw (Hm=4.2mm)
N=100rpm) for high metering depth 200
175
Pressure (bar)
 If:
150
125 35,62
24,12
100
16,57
P<10bar(Hm=4,2mm), 75
50
11,49
P<50bar(Hm=4,6mm), 25
P<150bar(Hm=5mm)
0 5 10 15 20 25
Axial length (/D)
Evolution of pressure along screw (Hm=4.6mm)

 Pumping phenomenon: 400
 output of metering zone increases 350
dramatically (to reach the counter
Pressure (bar)
300
35,63
250 23,82
pressure) 200 15,81
11,09
150
 But, as this output is higher than the one 100
given by the feeding, the screw empties 50
itself.
0 2,5 5 7,5 10 12,5 15 17,5 20 22,5 25
Length along screw (/D)
 Output decreases dramatically until the Evolution of pressure along screw (Hm=5mm)
screw is again full. 600
550
500
450
32,72
Pressure (bar)
400
 Question: How to solve pumping? 350

300
250
19,73
12,42
10,76
200
150
100
50
0
0 2,5 5 7,5 10 12,5 15 17,5 20 22,5 25
Axial length (/D)
2.6 Mixing zones.
2.6.1 Type of mixing.
Two types of mixing:

 Dispersive mixing. Application of shear stress to agglomerates to break them
up. Ex: Dispersion of fillers in a matrix
 Distributive mixing. Longitudinal mixing (reduce variations in time) and
crossmixing (reduce variations in space). Distributive mixing is achieved by
increasing interface area between “flow layers” and splitting and
recombination of flow paths. Ex: homogenizing of colors
2.6 Mixing zones.
2.6.2 Description of the two type of mixing.
3.0 mm
Which of the photo correspond to a lack of dispersive (distributive) mixing?
2.6 Mixing zones.
2.6.3 Evaluation criteria of mixing and solutions.
Evaluation criteria for mixing devices:

 Mixing effect
 Process compatibility (Viscous heating, pressure drop, stagnation, self-
cleaning etc)
 Handling (Assembly, dismantling, cleaning)
 Cost
Types of mixing devices:
 Static (Kenics, Sulzer SMX, Elongational mixer)
 Dynamic (TransferMix, Pineapple, Rapra (CTM), Saxton, Stat-Dyn,
Maddock, Egan.....)
 Other (Planetary gear, “Chaotic mixer”,...)
2.6 Mixing zones.
2.6.4 Some different mixing zones.
Maddock mixing zone
Second barrier zone
Multi flights zone (1, 2 and 4 flights)
Stat-dyn device
Saxton device
2.6 Mixing zones.
2.6.5 Conclusion.
 Problem: Mixing action is achieved by shear. This leads to

dissipative heating and generation of new thermal inhomogenities
 Attractive designs for mixing elements:
 Dispersive: Barrier screw sections, Maddock, Egan (self-cleaning, shear

stress controlled through flight clearance)
 Distributive: Saxton, Stat-Dyn (self cleaning, flow reorientation)
2.7 General relations for Extruder.
2.7.1 Output range for standard extruder.
Ref. [G.M.A.], p. 170, gives following specific output range .
Specific output.
Spiral grooved feeding zone. Smooth feeding zone.

Screw diameter
2.5”

Specific output is proportional to
 D3 at same L/D.
 L/D
Qsp(38-25D)=Qsp(60-24D)*(38/60)3 *(25/24)=1.1*0.254*25/24=0.29kg/h/rpm.
2.7 General relations for Extruder.
2.7.2 Needed power of motor (for different polymers).
As we have seen above (in §2) the melting of a polymer needs a certain energy.
Q = C p .∆T + ∆H
From this energy, one can determine the necessary power of motor needed, knowing that
the plasticizing effect is mainly done by the screw.
From Ref. [G.B.A.], p171,following practical values are proposed
Polymer HDPE LDPE PP PB PS ABS PC PMMA PA6 PA11

E (kWh/kg) 0.20-0.25 0.18-0.22 0.23-0.28 0.18-0.23 0.17-0.22 0.17-0.22 0.30-0.35 0.24-0.29 0.27-0.32 0.24-0.29
E (kJ/kg) 720-900 648-792 828-1008 648-828 612-792 612-792 1080-1260 864-1044 972-1152 864-1044
For example a 60-24D with a maximum output of 120kg/h of HDPE will need a motor of:
Pmin=0.20(kWh/kg)*120(kg/h)=24kW, or Pmax=0.25*120=30kW.
 Question: What will happen when running PA6 on a extruder designed for
HDPE with corresponding motor (30kW)?
2.8 Conclusions for the extruder part: screw design criteria.
 The design of extruder screws must account for the most important
thermodynamic and rheological properties of the material.
 The different zones (Feeding, Melting, Mixing) must be matched in order to

guarantee a stable operation.
 Screw design is a matter of experience and understanding. Computer simulations

are of some help for the design of extruder screws, but experimental data are of
prime importance.
 En extruder designed for a given output of a given polymer may be not capable of
working with another material or with higher output.
3 The extrusion head.
3.1 Generalities on Cross head.
3.1.1 Description.
 The extrusion head is where the shaping operation of the

plastic material takes place (wire coating, film blowing,
pipe, profile or sheet extrusion)
Extruder
Crosshead
Wire
Insulation
3.1 Generalities on Cross head.
3.1.2 Construction with two main components.
 The extrusion head has two main components

 Melt distributor. The distributor has the task of transforming the
melt flow along the axis of an extruder into an annular flow, and to
distribute it uniformly over the circumference or width. There are
two basic distributor concepts:
 “Coathanger” flow distributors
 “Spiral mandrel” flow distributors
 Tooling. The tooling has the task of shaping the extrudate, resp. of
applying the insulation on the core.
3.2 The Spider die.
3.2.1 Construction and applications.
 Advantage simplicity but risk of stagnation and weld lines appear

after the spider legs.
 Used mainly for tubing in PVC (Battenfeld drawing) et for profile
die.
3.2 The spider die.
3.2.2 Velocity repartition.
 Velocity repartition difference may

appear.
 Geometry optimized with the help of
softwares like Polyflow or Rem3D.
3.3 The coathanger distributor.
 Popular design for wire coating and side-fed applications

 This type of geometry can be of cylindrical, conical or even radial
configuration
3.3 The coathanger distributor.
3.3.2 The drawbacks of the coathanger distributor.
 Narrow processing window (material-specific design)

 For flat die, Coathanger-type flow divider generate an intrinsic weak
spot at the weld line (weaker cohesion of polymer molecules across weld
line). This may lead to unacceptable results in some applications
3.4 The helical flow distributor.
 Flow leaks from helical channel into slit

 Overlapping flows eliminate weld lines
 Needs primary splitting of flow in side-fed application
 From left to right: Maillefer for tube, Battenfeld 5 layers for tube,
Cincinnatti 8 layers for film.
3.4 The helical flow distributor.
3.4.2 Distributed weld lines.
 The helical flow distributor generates a layered structure. Weld

lines are distributed over the circumference. No specific weak spot
is created.
Manifold Distributed
weld lines
Slit
3.5 The distributor.
3.5.1 The flow distributor design and analysis.
 Melt flow distributors are numerically optimized for specific

materials or material classes (Design and analysis software, support
from experimental data)
 But: The result can not be better than the available material data
and operating practice!
3.5 The distributor.
3.5.1 Simulation and comparison to experiments.
Simulation vs. experiment (conical distributor)
1.10
Measurement
Relative
Flowrate Calculation
1.05
Distributor
1.00
Flux separator
Sector 1 Sector 5 0.95
Sector Sector Sector Sector Sector

1 2 and 8 3 and 7 4 and 6 5
Sector 2 Sector 4 0.90
Sector 3
3.6 The extrusion tool.
3.6.1 Two types of extrusion tools.
Pressure tool Tubing tool

 insulation  sheathing
 good adhesion of insulation  good centricity
 no air inclusion  takes shape of
conductor
3.6.2 Tube tools parameters: DDR definition.
 Two important design parameters in tubing tool:

 1) Draw Down Ratio DDR De Di
de di
 D −D
2 2
DDR = e i
d −d
2
e i
2
 High values of the DDR permit to keep shear stresses low inside the
tool (avoid melt fracture), but yet maintain high production speeds.
Condition: High elongational strength of the melt acceptable.
 The DDR may reach values up to 200 in practice (TEFLON 340)
3.6.3 Tube tools parameters: Some practical values of DDR.
The values given below may still depend on the grade of the polymer used (from Ref.
[B.B.], pp. 50-51).
Polymer PVC PVC PE, all PE, PP PA12 PA6 PBT TPE PVDF ETFE FEP PFA
pure with types with
50% 50%
chalk ATH
DDR < 4 2 4 2 10 10 40 16 20 25 25 150 200
 Some points have to be specified:

 DDR considers only the elongation in the cone. The higher the speed of elongation (variation of
speed of elongation in direction of extrusion), the lower the admissible DDR.
 Melt Temperature increases admissible DDR increases
 Big tools and low viscosity melts gravity may deform cone admissible DDR smaller (not true
if vertical direction of extrusion)
 Higher DDR higher molecular orientation different mechanical properties along and
perpendicular to extrusion direction orientation of molecules in extrusion direction time and
polymer-dependent reorientation will mean a shrinkage (for example PE tube from 1m to 80
cm) shrinkage has to be taken as low as possible.
3.6.4 Tube tools parameters: DRB.
 2) Draw Ratio Balance DRB: Compares aspect ratio of sections

De Di
 de di
De di
DRB = ⋅
Di d e
 Practical range of DRB : 0.9 - 1.15
 DRB < 0.9: out-of-round insulation, fold-over
 DRB > 1.15: Melt tears
 Consequence: The tool must be designed according to the material
and the production conditions
3.6.5 Pressure tool.
 The geometry of the tool influences the pressure in the extrusion

head, but also the mechanical stress of the wire (example:
Elongation of wire in telephone wire insulation).
 Very high shear rates are present in the melt flow in the tool, and
accordingly high local temperatures build up close to the die walls.
For this reason, an extrusion head may be heated up to e.g. 220°C,
although the bulk temperature of the melt does not exceed e.g.
200°C.
3.6.6 Pressure tool: simulation result.
Head profile Pressure profile

4 176
radius [mm]
3 132
1000 m/min
p [bar]
2 88
1 44
0 0
0 4 7 11 15 0 4 7 11 15
x [mm] x [mm]
Temperature profile Shear stress

232 11.0
Wire
224 5.5 Die
tau [bar]
T (°C)
216 0.0
208 -5.5
200 -11.0
0 4 7 11 15 0 4 7 11 15
x [mm] x [mm]
3.6 General relation between pressure and melt temperature in Xhead.
A high pressure needed in the cross head will mean an increase in

mass temperature (demo in small characters beneath).
∆P H = U + PVwhich can be written dH = dU + PdV + VdP
∆T = with dU = δQ + δW with δW = − PdV work of pressure forces
ρ .c p We obtain dH = δQ + VdP and as we are Adiabatic ⇒ δQ = 0
With the basic thermal properties of polymers, we can evaluate the We have m.c p .dT = VdP with m = ρ .V
heat generated by pressure. which, at least gives ρ .V .c p .dT = VdP

∆P
∆T =
ρ .c p
Polymer PS PVC PMMA SAN ABS PC LDPE LLDPE HDPE PP PA6 PA66 PET PBT PVF2 FEP
Specific heat 1.20 1.10 1.45 1.40 1.40 1.40 2.30 2.30 2.25 2.10 2.15 2.15 1.55 1.25 1.38 1.18
Cp
(kJ/kg/°C)
Density ρ 1.06 1.40 1.18 1.08 1.02 1.20 0.92 0.92 0.95 0.91 1.13 1.14 1.35 1.35 1.76 2.15
(gr/cc)
ρ.Cp 1272 1540 1711 1512 1428 1680 2116 2116 2137.5 1911 2429 2451 2092 1687 2429 2537
(kJ/m3/°C)
∆T /100bars 7.86 6.49 5.84 6.61 7.00 5.95 4.73 4.73 4.68 5.23 4.12 4.08 4.78 5.93 4.12 3.94
(°C)
3.8 Conclusions for the crosshead part.
 Melt distributor and tool together determine the function of the

extrusion head
 They both must be carefully designed accordingly to the material

properties and the operating conditions
4 Cooling bath.
4.1 Heat transfer on a piece of cable.
 Heat transfer during cooling.



 Heat transfer on ∂ 2 +
 
 Q conduction z 
dz
2 
= ∫ k ∇T 
n d ∂ 2 +
t  2 -
 Heat transfer on ∂ 2 -
  Q  2  = ∫ k ∇ T n d ∂ 2
z −
dz
conduction
-
HEat transfer on ∂ 1
 Q convection
= ∫ −hTsurface−Twater d ∂ 1
t t
  2-  1
 Heat balance:
∫ ρC p w⃗ . ∇ Td Ω= ∫ k ∇ .T . ⃗n . d ∂Ω= ∫ −h(Tsurf.−Twat. ). d ∂ Ω 1 + ∫ k ∇ .T . ⃗
n. d ∂ Ω 2 +
−
∫ k ∇ .T .n
+
⃗ . d ∂Ω 2
Ω ∂Ω ∂ Ω1 ∂ Ω2 − ∂Ω2
+
4 Cooling bath.
4.2 Heat transfer general assumptions.
 Convection:
 Coefficient of convection in water h is depending on the type of flow inside the
water:
 A value of 1000 W.m-2.K-1 is often used.
 Simplification:
 We can show that the conduction in wire direction is negligible against the
convection in the same direction (number of Peclet).
P =
Q
=
 R1 . ρ . C . R2 −R1  . ρ . C  w . L
conv
2
Cu pCu
2 2
Isol pIsol
e 
Q cond 2R1 . k Cu 2  R2 −R1  . k Isol
2 2 2
Numerical application:
kg J W
Polyethylene insulation: ρ PE ≈1000 3
, C p−PE ≈2500 , k PE ≈0 . 33 ,
m  kg.° C   m.° C 
kg J W
Core conductor  in cupper: ρ Cu≈8940 3
, C p−Cu≈385 , k Cu≈389 ,
m  kg . °C   m.° C 
Conductor radius: R1=0 . 15mm ,Insulation radius R2=0 .3 mm
m
Line speed w=20 ,Unit cooling through length L=1 m, P e≈1,4. 104 >>1
s
4 Cooling bath.
4.3 Heat transfer analytical solution.
4.3.1 Validity conditions.
 We use the ratio of the two heat transfers in radial direction, the one
by conduction (in the polymer) and the other by convection (with
water):

 
h . ΔT h
water .R water
Bi= =
 k .
ΔT
R k
solid
solid
 If this ratio is small, we can neglect the difference of temperature in

the radial direction in the wire.
W
 Polyethylen wire in water convection coefficient h=1000
m² . ° C
h .R k
 Bi=
k
eau
<<1 which corresponds to
solide h
solide
>> R or R <<0 . 33. 10 or R <<0 . 33. mm
eau
−3
 This is true for small external diameters(D<0.6mm)
4 Cooling bath.
4.3 Heat treansfer analytical solution.
4.3.2 Example.
 Example PE wire:
 Analytical solution:
 Length to reach a certain temperature
  R1 . ρ . C .  R2 −R1  . ρ .C  .w . Log Tm  L −Twater
2 2 2
L=
Cu pCu
−2 . h. R2
Isol pIsol
 Tm  z =0 −Twater  


Numerical application:
 Polyethylene insulation: ρ PE ≈1000
kg
3
, C p−PE ≈2500
J
 kg.° C 
, k PE ≈0 . 33
W
 m.° C 
,
m
 Core conductor  in cupper: ρ Cu≈8940
kg
m 3
, C p−Cu≈385
J
 kg . °C 
, k Cu≈389
W
 m.° C 
,
 Conductor radius: R1=0 . 15mm ,Insulation radius R2=0 .3 mm , h=1000

W
m² . °C
m
Line speed w=20
s
To reach a cable temperature of 100 °C , we need: L=6,6 m
4 Cooling bath.
4.4 Heat transfer numerical solution.
4.4.1 Numerical resolution.
 The heat balance reduces to:

 At any point in the volume:
 ρC p w
 . ∇ T =kΔT
 With boundary conditions

 at interface between copper and insulation
 k Copper ∫ ∇ . T . n . d ∂ =k insulation ∫ ∇ . T . n . d ∂ 
∂ ∂
 at outside insulation surface
 ∫
∂1
−h  Tsurface−Twater . d ∂  1=k insulation ∫
∂1
∇ .T .n
 . d ∂ 1
 at the center of the wire

 k ∇ .T .n
 =0 Copper
 Cylindrical coordinates
 At any point (conduction <<convection in axis dir.
 ρC p w .
∂T
∂z
=k
1 ∂
r ∂r   
r
∂T
∂r
 With same boundary conditions as above
4 Cooling bath.
4.4.2 Finite difference method of solution.
 Implicit finite difference method:

 i axis dir., j radial dir., nbpc=number of point in a layer, n=current layer number,
nmax=total layer number.
At any point T i , j −T i−1, j 1 ∂ T ∂2 T
  
1 T i , j −T i , j−1 T i , j−1−2 . T i , j T i , j1


ρC p w . =k  2 =k k .  2
Δz r ∂r ∂ r rj Δr j
  Δr j 
 With boundary conditions
 at interface between copper and insulation (j=n*nbpc)
T i , j −T i , j−1 T i , j1−T i , j
 k copper
Δr j
=k insulation
Δr j1
 at outside insulation surface (j=nmax*nbpc)
T i , j −T i , j−1
 −h T i , k∗nbpc −Twater =k insulation
Δr j
 at the center of the wire
T i ,1−T i,0
 Δr
=0 or: T =T k copper
1
i,1 i ,0
 Matricial system to solve in each slab (i) along axis .

4 Cooling bath.
4.4.2 Comparison with analytical resolution.
 Analytical (simplified) vs numerical (PE wire):

 Analytical temperature at a certain length
 Tm  z =  Tm  z =0  −Twater  . exp
 −2. h . R2 . z
R1 . ρ Cu . C pCu .  R2 2−R12  . ρ Isol .C pIsol .w
2
  Twater
 Graphical comparisons
Cooling case Bi=0.6 Cooling case Bi=0.9 Cooling case Bi=3

(Rcu=0.05mm RPE=0.1mm W=20 m/s) (Rcu=0.15mm RPE=0.3mm W=20 m/s) (Rcu=0.5mm RPE=1mm W=20 m/s)
200 200 200
175 TmCu-TmPE (°C) 175 175

Tm numerical(°C)
150 Tm analytical (°C) 150 150
125 125 125

Temperature (°C)
TmCu-TmPE (°C)
Temperature(°C)
Temperature (°C)
100 100 100 Tm numerical(°C)
TmCu-TmPE (°C) Tm analytical (°C)
75 75 Tm numerical(°C)
75
Tm analytical (°C)
50 50 50
25 25 25
0 0
0
0 2,5 5 7,5 10 12,5 15 17,5 20 0 2,5 5 7,5 10 12,5 15 17,5 20 0 2,5 5 7,5 10 12,5 15 17,5 20
Abscisse(m) Abscisse(m) Abscisse(m)

5 References.
 Ref
 [J.F.A]: J.F.Agassant, “Polymer Processing, Principles and Modelling”, Hanser
Verlag, 1991, Münich,Vienna, New-York.
 [G.A.M.]: G.A. Martin et al., “Der Einschnecken Extruder- Grundlagen und
Systemoptimierung”, VDI-Veralg, Düsseldorf 2001.
 [C.R.]: C. Rauwendahl, “Polymer extrusion” , Hanser Verlag, 3rd. Edition,1994,
Münich,Vienna, New-York.
 [B.B.]: B. Buluschek, “The Art and Science of Extrusion for Wire & Cable 1”,
52nd. IWCS professional Development Course, Nov. 2003.
 [G.E.]: General Electric’s Web site: www.geplastics.com
 [D.S.]:D. Schlaefli: « Plastics extrusion: a short Overview », International
Technical Symposium 2000, Ecublens (CH).
 [S.P.]: S. Puissant: « DC32: Etude d’outillage en extrusion », Cours FITI3 GIP-
InSIC, Ecole des Mines de Nancy, 2002-2011, Saint Dié des Vosges, France.
Thanks
 Thanks to all who helped draft this course:

 Maillefer extrusion for all the data's and and experiences gathered
 Habia Cable, for accepting to make a course trying to bind theory
and practice, and helping me to streamline it.
 Brugg Cables Industry AG for allowing me to add some concrete
examples






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2012

Professional Development Courses

 The extruder is a machine to transport and process materials .

 1845 (Bewley, Faraday).

 Polymers are materials with quite different properties depending on:

Fillers and additives

 Most of the practical polymers have two properties in common:

 They are important for the processing

The shear stress is defined as  h

The (dynamic) viscosity is defined as

Consequence of the high viscosity:

 In Practical polymer processing:

 High pressures are required for the extrusion

The viscosity is also

 Thermal conductivity: λ = 0.1 - 0.3 W/mK (1000 x lower than

 A heat wave penetrates to a depth of:

Another consequence of high viscosity: Tw U

 Brinkman number (Dissipation energy/heat

Example: Evolution of temperature in a slit die.

with which we measured viscosity (slide 7) 330

 Measurement on a 60-30D 290

The 6 first values are in agreement with the increase

of temperature obtained by viscous dissipation: 230

Average shear rate (s-1)

Heat necessary to heat up a semi

 The extruder must prepare the raw material (powder, pellets,

 The heart of the extruder is the screw and barrel

Feed zone Melting zone Pumping and mixing zone

Hopper Zone 1 Zone 2 Zone 3 Zone 4 Zone 5

Pulley and Cooling thins

Motor Cooling fan cover (1 per Zone) Frame

 Screw 38-25D showing high compression:

Hopper Feeding Compression Metering

 Specific throughput decreases with head pressure: Higher screw speed

ϕ+θ first approximation flat.

Question: What is the reason for the polymer to move forward?

 Physical interpretation of Coulombs law: (Ref.

Coefficient of friction of HDPE. ([G.A.M.], p.24):

Results based on a flat plate model Ou t p u t a n d co n v e y in g a n g le in f u n ct ion of ch a n n e l d e p t h

 Throughput depends on geometry 25

 Optimum feeding channel depth: 15 12,5

this optimum channel depth?

Results based on a flat plate Ev olu t ion of ou t p u t in f u n ct ion of f lig h t p it ch ,

 Question: Where does the optimum flight pitch lies?

Out put in funct ion of ba ck pr e ssure at t he e nd of t he fe e ding zone

 Increase apparent barrel friction (polymer-polymer friction is higher

Straight grooved feedings depend also on internal coefficient of friction.

N(rp Tz1(° g/36s g/36s g/36s g/36s Q(kg/

 The nut-on-screw operation is achieved through geometrical

 Independance against counter pressure for feed zone with helical

Output and specific output on a 150-24D (courtesy Maillefer) with PVC

Q(kg/h) sm oot h Q(kg/h) grooved Q(kg/h) spiraled Qs(kg/h/rpm ) Qs(kg/h/rpm ) Qs(kg/h/rpm )

Question: Advantages of linear output (constant

Melting zone Solid

Relative motion of barrel Feed zone

Melt pool Solid bed

Energy Balance of the Melting Process at the interface between melt

Theoretical energy balance:

Thermal parameters and viscosity used for calculation of melting.