5 Analysis of polymer melt flow

5 Analysis of polymer 5.1

melt flow

Introduction In general, a l l polymer processing methods involve three stages-ie heating, shaping and cooling of a plastic. However, t h i s apparent simplicity can be deceiving. Most plastic moulding methods are not straightforward and the practical know-how can only be gained by experience,often using t r i a l and error methods. In most cases plastics processing has developed from other technologies (eg metal and glass) as an art rather than as a science. ~his is p r i n c i p a l l y because in the early days the flow of polymeric materials was not understood and the rate of increase in the usage of the materials was much greater than the advances in the associated technology. Nowadays the position is changing because, as ever increasing demands are being put on materials and moulding machines i t is becoming essential to be able to make reliable quantitative predictions about performance. In Chapter 4 i t was shown that a simple Newtonian approach gives a useful f i r s t approximation to many of the processes but unfortunately the assumption of constant viscosity can lead to serious errors in some cases. For this reason a more detailed analysis using a Non-Newtonian model is often necessary and this w i l l now be i l l u s t r a t e d . Most processing methods involve flow in c a p i l l a r y or rectangular sections, which may be uniform or tapered. Therefore the approach taken w i l l be to develop f i r s t the theory for Newtonian flow in these channels and then when the Non-Newtonian case is considered i t may be seen that the steps in the analysis are identical although the mathematics is a l i t t l e more complex. At the end of the chapter a selection of processing situations are analysed quantitatively to i l l u s t r a t e the use of the theory.

223

224

5.2

R . J . CRAWFORD

GeneralBehaviourof Polymer Melts In a f l u i d under stress, the r a t i o of the shear stress, z, to the rate of strain, #, is called the shear viscosity, n, and is analogous to the modulus of a solid. In an ideal f l u i d the viscosity is a material constant. However,for plastics the viscosity varies depending on the stress, strain rate, temperature etc. A typical relationship between shear stress and shear rate for a plastic is shown in f i g . 5.1. In analysing the flow of polymer melts i t is convenient to have an equation for this nonlinear behaviour and although many have been suggested, the most satisfactory is a power law relation of the form (5.1)

z = 3 0 '-'ty) n

where T and n a r e c o n s t a n t s . O

Newtonion

shear

ftuid

-[[=ll~

stress %

/

•,•fLPuower

shear

0 Fig 5-1

RELATIONS

law

BETWEEN

--

id ][=-C,~ n

SHEAR

strain r a t e ,

STRESS/SHEAR

RATE

As a starting point i t is useful to consider the relationship between shear stress and shear rate as shown in f i g . 5.1 since this is similar to the stress-strain characteristics for a solid. However, in practice i t is often more convenient to rearrange the variables and plot viscosity against stress as shown in f i g . 5.2. Logarithmic scales are also common so that several decades of stress and viscosity can be included. Lines of constant shear rate may also be drawn for quick reference. When a f l u i d is flowing along a channel which has a uniform cross section, then the f l u i d w i l l be subjected to shear stresses only. I f , on the other hand, the channel section changes then tensile stresses w i l l also be set up in the f l u i d and i t is often

PLASTICS ENGINEERING

225

log viscosit,j

IU

Fig 5.2

IU

TYPICAL

IU

VARIATION

STRESS

FOR

OF A

log stheor s t r e s s

VISCOSITY

POLYMER

WITH

SHEAR

MELT

necessary to determine the t e n s i l e v i s c o s i t y , x, f o r use in f l o w calculations. I f the t e n s i l e s t r e s s i s ~ and the t e n s i l e s t r a i n r a t e i s E then o = XE

(5.2)

For many p o l y m e r i c m e l t s the t e n s i l e v i s c o s i t y i s f a i r l y c o n s t a n t and a t low stresses i s a p p r o x i m a t e l y t h r e e times the shear viscosity. To add to t h i s p i c t u r e i t should be r e a l i s e d t h a t so f a r only the viscous component o f behaviour has been r e f e r r e d t o . Since p l a s t i c s are v i s c o e l a s t i c there w i l l also be an e l a s t i c component which w i l l i n f l u e n c e the behaviour o f the f l u i d . This means t h a t t h e r e w i l l be a shear modulus, G, and, i f the channel s e c t i o n i s not uniform, a tensile modulus, E, to consider. I f yp and ~R are the recoverable shE:ar and tensile strains respectivelp then G : ! YR

(5.3)

E =~ ~R

(5.4)

These two moduli are not m a t e r i a l constants and t y p i c a l

variations

are

R. J. CRAWFORD

226

shown in f i g 5.3. As with the viscous components, the tensile modulus tends to be about three times the shear modulus at low stresses. Fig 5.3 has been included here as an introduction to the type of behaviour which can be expected from a polymer melt as i t flows. The methods used to obtain this data w i l l be described later, when the effects of temperature and pressure w i l l also be discussed.

I I

log scale

-

I

~ I i ~ !!i t e n s i [ e modutus

ii

ii

i

II

.,,,r

i

II

II II

lr '

i i iiJi

I Ill]i I 11 I I I

, . VlSCOSlt~

[',~"" shear I modulus

~,, ,

I\l

,

,

9

[ I~lll ' J.~l~ I

sheQr vl s c o s d y / , III

I

5.3

PROPERTIES

I

i

, l l l l [I] [11

III III

NI

Ill

Irl,

III

\

Ill loc

TYPICAL FLOW

ill

II E

Fig5.3

L'~ , , ,III

I,,

.--

Irl

i

'tensde-

I

stress

FOR A POLYMER

Isothermal Flow in Channels : Newtonian Fluids In the analysis of flow in channels the following assumptions are made : I.

There is no s l i p at the wall.

2.

The melt is incompressible.

3.

The flow is steady, laminar and time independent.

4.

Fluid viscosity is not affected by pressure changes along the channel.

5.

End effects are negligible.

The steady isothermal flow of incompressible fluids through straight horizontal tubes is of importance in a number of cases of practical interest

PLASTICS ENGINEERING

(a)

227

Flow of Newtonian Fluid along a Channel of Uniform C i r c u l a r CrossSection

Consider the forces acting on an element of f l u i d as shown in f i g 5.4. Bp F1 : ~r2(p -~-~ dz) F2 = ~r2p F3 = 2~rdz.T Since the flow is steady zFz = 0 ~r2p = ~r2(p - ~P dz) + 2~rdz.T

: ~r ( ~P ~)

(5.5)

In many cases the pressure gradient is uniform, so that f o r a pressure drop, P, over a length, L, the shear stress is given by Pr (5 6) T = -~ 3V Also since z = n y = n~-~ then in equation (5.5)

~v

r (~P,

n.~

: ~

low

~

P- ~zPdz

P

I/

I

pressure

/ z / z / / / /

I l l / I l l

I

,

~ ~ ~ high V ~ s s u r e

. I Z direclion

L F

" F~] dz

z / I / / / / / / / //I

I

/ / / / /

/

/ / I I

Fig 5.4 ELEMENT OF FLUID IN A CAPILLARY

Integrating this gives v r 1 f dv = f 0 R 2n

l V =~

dP dz

. r dr

dP r 2 R2 ~ (~---T-)

228

R. J. CRAWFORD At r = O, V = Vo So

1 dP R2 Vo = " 4~ " dz "

(5.7)

and

V = V0 (I - (~)2)

(5.8)

The volume f l o w r a t e , Q, may be o b t a i n e d from R Q = ; 2~rVdr o R

Q= f

2~r

2 Vo (I - ( ~ ) )

. dr = ~VO R2

o This can be rearranged using (5.7) to the form Q =

~R4 dP 8n " dz

(5.9)

Once again i f the pressure drop is uniform t h i s may be expressed in the more common form Q=

~R4P 8qL

(5.10)

I t is a l s o convenient to d e r i v e an expression f o r the shear r a t e y. since ~ = ~_~V then

2 ~ =-~-~ ~ ' [ Vo(l - ( ~ ) ) ]

This may be rearranged using the r e l a t i o n Vo to give the shear rate a t r = R as

(5.11) between f l o w r a t e and

y =

(5.12)

The negative signs f o r v e l o c i t y and f l o w r a t e i n d i c a t e t h a t these are in the o p p o s i t e d i r e c t i o n t o the chosen z - d i r e c t i o n . (b)

Flow o f Newtonian F l u i d Between P a r a l l e l Plates

Consider the forces a c t i n g on u n i t w i d t h o f an element as shown in f i g 5.5. F 1 = (P - BP . dz) 2y F2 = P.2y F3 =

T dz

For steady f l o w there must be e q u i l i b r i u m o f forces so P2y = (P - - ~@P . dz) 2y + 2T dz = y

~P

. ~

(5.13)

PLASTICS ENGINEERING low pressure

ap dz P-~-~. / / / /

(

K / K K

/

/

,/

,/

/

/

/

/

Z direction

-Iiiiiii#i!iiiiiiiiiiiiiiiii#iiiiiiiiiiiiiil-F --w,

I= /

/

/

L

./

high pressure

P

Ti .Yl ]ii i ji i i Ji j#i i i ji i i i i i i i i i iJ_

,

/

229

/

/

Fig, 5.5. ELEMENT

/

-

F3 dz /

/

_ /

/

/

/

/

/

/

/

/

OF FLUID BETWEEN PARALLEL PLATES.

In many cases the pressure g r a d i e n t is uniform so t h a t the shear stress a t the w a l l is given by PH T : 2--L

(5.14)

For a Newtonian F l u i d the shear s t r e s s , ~, is also given by av T = n~ = n - ~ So using (5.13) av

aP

nT# =YT~ I n t e g r a t i n g t h i s gives v

y 1 Y n H/2

% dV = o

dP dz.y dy

1 dP V : 2--n d--z (y2 _ ( ) 2 ) Now a t y = 0

V = Vo dP ~ ) 2

1

So

V = - 2n o

Substituting

dP ~-

dz (

(5.]5)

in the e x p r e s s i o n f o r V g i v e s

V = Vo (1 - ( ~ ) 2 )

(5.16)

230

R. J. CRAWFORD

The volume flow r a t e , Q, is given by

H/2 Q= 2

f

TV dy

0

H/2 = 2 %

)2 TVo (I - (H 2-~

) dy

0

Q = 2T Vo (~) Using equation (5.15) t h i s may be expressed i n the form T H3 dP Q: - ~ 9~ (5.17) which f o r a uniform pressure g r a d i e n t reduces to T P H3 Q = 12nL

(5.18)

An expression f o r the shear rate, y, may also be derived from ~v

[Vo

2

)l

= - 8VoY/H2

(5.19)

but as shown above Vo = 3Q So

-12120v Y = T~

At the wall y =

H

= - ~H

(5.2O)

I t is worth noting t h a t the equations f o r flow between p a r a l l e l plates may also be used w i t h acceptable accuracy f o r flow along a c i r c u l a r annular s l o t . The relevant terms are i l l u s t r a t e d in f i g 5.6. 5.4

Isothermal Flow i n Channels : Non-Newtonian F l u i d s The Non-Newtonian f l u i d which is most amenable to simple a n a l y s i s is one in which the r e l a t i o n s h i p s between shear s t r e s s , T, shear rate, #, and v i s c o s i t y are given by a power law of the form .n-I n-I n = no [ " .]C_yo ]

= no [~-oo] n

(5.21)

PLASTICS ENGINEERING

231

f

//

F.Lg.5.6. FLOW IN AN ANNULAR SLOT

where y^ and To.represent values of shear rate and shear stress in some a r b i t r a r i l y chosen standard state and no is the viscosity in this state. For convenience Yo is often taken as l s-lo Note that from (5.21)

or

~_=

[~] 1/n

u

L%J

~

:

~o

~o

and i f Yo is taken as ] .n

T :'COY

which is the more common form of the Power Law. (a)

Flow of Power Law Fluiq Alon9 a Channel of Uniform Circular Cross Section

Referring to f i g 5.4 and remembering that equilibrium of forces was the only condition necessary to derive (5.5) then this equation may be used as the starting point for the analysis of any f l u i d . r aP T : 2 (Tz)

(5.5) av

Now

T : n~ = n T ~

so

av _ r (B--P: ar 2~ "az'

(5.22)

R. J, CRAWFORD

232

but from (5.21)

=

n-l

r ~ l n-I

hi@V/Dr]

~

:

j

So i n (5.22)

DV

-Dr- :

r

@P

"(Ti) 2no rDVlDrln_l L__~o J

(~V) n = r ( i o )n-I Dr"

2n ~

Integrating and

~p '(Tz )

t h i s between the ] i m i t s

V e l o c i t y : V at radius = r V e l o c i t y = 0 at r a d i u s = R n I / n ~- n+___11

n~lFDp/D:1

v: (~T)#o At r = 0

L2-~-oJ

V = Vo

-

1

n-I

.-~-F

Vo:

L rn

n+l]

Rn

BP

-

l~z] n

n+l (5.23)

R n

The equation f o r v e l o c i t y at any radius may then be expressed as r n+l

V : Vo [1 - ( ~ ) - - ~ ]

(5.24)

The volume flow rate, Q, is given by

R Q= I

2~r V dr

0

n+l 'n

R = 27 V o I

r (I

- (~)

) dr

0

{n+l Q : ,-~-~, ~R2 Vo Also Shear Rate, y : DV Dr

(5.25)

n+l n

= LDr [Vo (l - (~) Y= -

n+l l --n--. Vo I/n R " (~)

)1 (5.26)

The shear rate at the wall is obtained by l e t t i n g r = R and the equation is more convenient i f Vo is replaced by Q from (5.25). In

PLASTICS ENGINEERING

233

this case

3n+l, Shear Rate at wall ~ = - ( T )

n

I t is worth noting t h a t since a Newtonian F l u i d is a special case of the Power Law f l u i d when n : l , then the above equations a l l reduce to those derived in section 5.3(a) i f t h i s s u b s t i t u t i o n is made. (b)

Flow of Power Law Fluid Between Parallel Plates

Referring to f i g 5,5 and recognising t h a t equation (5.13) is independent of the f l u i d then BP ~ =YTz but

~ = n~

So

~V

n-I (5.13)

and n = no [ ~ o ]

y pv/,,lol no --=-_

~V _ 9 n ~y Yo

yn

~P

I-,z)

(I__ ~P)-n n~ Tz"

I n t e g r a t i n g t h i s equation between the l i m i t s v e l o c i t y = V at distance = X velocity 0 at distance H/2 1 n+l n+l ~ P ~ n [ y n _ (H) n ]

n

v = (~TT) ~o ( ~ x o T/, At y = 0

V = Vo n

( 1

Vo = - (-~-~)"~o

'~-0%

l ~P~n

~'

n+l n

(~)

And using t h i s to s i m p l i f y the expression f o r V n+___Z n V = Vo (1 - (2H-~) )

(5.27)

(5.28)

The flow rate, Q, may then be obtained from the expression 1112 Q = 2Tf

V dy o

= 2T ~/2 Vo (I - (2y) H o

n+l n)dy

234

R. J. CRAWFORD

{ n+l Q = ~-~TTj T Vo H

(5.29)

And as before ~ : a.~V ay

n+__l :

[ V0 (l

~-~

(2H_~Z)n )1

-

n+l = . Vo (~) n

1 (n+l) "-T" y

(5.30)

At the wall y = ~ and s u b s t i t u t i n g f o r V~ from (5.29) then =

_

(2n+l)

" n

" 2T-~H

I t can be seen that in the special case of n = 1 these equations reduce to those for Newtonian Flow.

5.5

Isothermal Flow in Non-Uniform Channels In many p r a c t i c a l s i t u a t i o n s i n v o l v i n g the flow of polymer melts through dies and along channels the cross sections are tapered. In these circumstances, t e n s i l e stresses w i l l be set up in the f l u i d and t h e i r effects superimposed on the effects due to shear stresses as analysed above. Cogswell has analysed t h i s problem f o r the flow of a power law f l u i d along c o n i - c y l i n d r i c a l and wedge channels. The flow in these sections is influenced by three factors (I) entry effects given s u f f i x 0 (2) shear effects given s u f f i x S (3) extensional effects given suffix E.

Each of these w i l l since each r e s u l t s dominate depending more c l e a r l y l a t e r (a)

contribute to the behaviour of the f l u i d although from a d i f f e r e n t deformation mode, one e f f e c t may on the geometry of the s i t u a t i o n . This w i l l be seen when s p e c i f i c design problems are tackled.

Flow in C o n i - C y l i n d r i c a l Dies

Consider the channel section shown in f i g 5.7(a) from which the element shown in f i g 5.7(b) may be taken. Pressure Drop Due to Shear~ Ps

Assuming t h a t the d i f f e r e n t modes of deformation are separable then considering e q u i l i b r i u m of forces in regard to the shear stress

235

PLASTICS ENGINEERING

~

F19" 5.7.

ANALYSIS

OF

CONI-CYLINDRICAL

meni

of f l u ' l d .

DIE

onl~ gives dP$ .~r 2 = 2~r dl sec e.~ cos e 2T

dP = - s r

(5.31)

. dl

but f o r a power law f l u i d "c :

.n ToY

and ~

dr

and using dl = t a n ' e

3n+l = (T)

~

236

R . J . CRAWFORD

then (5.31) becomes 2~o dPs = tan e

c3n+l) ~)n dr -Trl+3n

(

Integrating 2To (3n+l) Q n Ps =tan e ( 9 n - ~) fRo Rl

dr l+3n r 3n

3n+l R ~ l n [

Ps 3n tan e =

2~~

( ( T )

[

2T, PS = 3n tan e

)

1 -

Rl (~o)

~Io) 3n ]

l - (

(5.32

Pressure Drop Due to Extensional Flow, PF Considering equilibrium of the extensional forces dPE.~r2 = o [ ~(r+dr)2 - ~r 2] dPE =

2o dr

Integrating PE = f

R o 2o dr r

(5.33

R1

In order to integrate the right hand side of this equation i t is necessary to get an equation for o in terms of r. This may be obtained as follows. Consider a converging annulus of thickness h and radius a, within the die. I f the angle of convergence is @then simple geometry indicates that for uniform convergence a dh da tan ~ = ~ tan e and-h-- = ~ Now i f the simple tensile strain, ~, is given by dA =~-then the strain rate, ~, w i l l be given by

l Ea =area 2 da a dt

dar~

1

d(2.ah)

dt

= 2xah "

dt

PLASTICS ENGINEERING but

237

da dl a - ~ = tan @ . - ~ =-~ tan 0 . V

wh'ere V is the v e l o c i t y o f the f l u i d p a r a l l e l to the centre l i n e of the annulu.s. From (5.24) and (5.25) this velocity is given by

,3n+l, V = ~ n+lJ ~ r

n+l [ a--n-- i] (-r) n+l

So

~ a ) - - n - n + l[J Ea = ~2 tan e ~I3n+l~-~r

i]

The average extensional stress is obtained by integrating across the element so r

a ~r 2 = /

~a 2~a da

0

and since by d e f i n i t i o n oa = xE a r

o ~r 2 = f

~Ea 2~a da

0

r

=/

n+l

X. ~ tan e '-n~-T'

(r) -~-- 1

2~a da

0

(5.34)

= X (2-~r) tan e

So in (5.33)

PE = ~R~ ~X'2 (2--~r) tan o.dr I

_ ~ tan83 (~4-~R I) [ I -

(~)3

]

Using (5.34) PE = 3 ~I

(5.35)

Pressure Drop at Die Entrx~ Po When the fluid enters the die from a reservoir i t will conform to a streamline shape such that the pressure drop is a minimum. This will tend to be of a coni-cylindrical geomet~ and the pressure drop, Po, may be estimated by considering an i n f i n i t e number of very short frustrums of a cone.

238

R . J . CRAWFORD

Consider a coni-cylindrical die with outlet radius, Ro, and inlet radius, RI, and an included angle 2%. From (5.32)

PS = 3n2~9 tan ao

From (5.35)

PE : ~ tan 3 e

[

I - xn ]

(xR4_~o ) [l-x]

where x = (~)3-- for convenience The pressure drop, PI' for such a die is PI = PS + PE =a tan so

+ b tan %

2T O

where a : ~ -

[ I - xn];

For minimum pressure drop the d i f f e r e n t i a l to the angle ~o should be zero

dPl

xl of pressure drop with respect

a

-

d(tan So) = tan-~% + b = 0

tan ~o = (-~)89 So the minimum value of pressure is given by p

I

: a89 ~ + a89 ~ : 2a89 89

I f t h i s procedure is repeated f o r other short c o n i - c y l i n d r i c a i i t is found that l+n

P2 = x-2- PM l+n P3 = ( x T ) 2 PI and so on. So the total entry pressure loss, Po, is given by i =| Po = Limit z Pi x§ i=I =T

(

) (-~n~)89 L i m i t f ( x )

dies then

PLASTICS ENGINEERING

239

1+n 1+n l+n where f(x) = [(~-xn)c1-x)] 89 [ I + x 2 + (x2--~2+ (x-'2-) 3 + . . . . ] and n = viscosity corresponding to shear rate at die entry.

4V-~

PO = 3-'~'-~-~-

(~-~o) ( nx )~

(5.36)

F 1-xn ] 3 .~R 3 Limit L ~ -~-

2T0 Also tan 2 ~o = 3n tan ~o = (~-~) So that from (5.34) 8~ P o : 3-~-~TT (b)

(5.37)

Flow in Wedge Shaped Die

Referring to the terminology in f i g 5.8 and using analysis similar to that for the coni-cylindrical die, i t may be shown that the shear, extensional and die entry pressure losses are given by I r PS = 2n tan =

82- (

)

-

0

Po:~

(5.38) (5.3g

9

(

) (nx)1

(5.40

where ncorresponds to shear rate at die entry Also V = .~2n+~ n " (2T_~H) J

.

E = ~ y tan ~ and tan B0 = ~ (~)~

(5.41) (5.42 (5.43)

where B is the half angle of convergence to the die. o 5.6

Elastic Behaviour of Polymer Melts As discussed e a r l i e r , polymer melts can also exhibit e l a s t i c i t y . During flow they have the a b i l i t y to store strain energy and when the stresses are removed, this strain is recoverable. A good example of elastic recovery is post extrusion swelling. After extrusion the dimensions of the extrudate are larger than those of the die, which may present problems i f the dimensions of the extrudate are c r i t i c a l . In these c i r cumstances some knowledge of the amount of swelling l i k e l y to occur is

240

R, J. CRAWFORD

T

Fig 5.8 ANALYSIS

OF FLOW IN TAPERED WEDGE

swelling rQtio = Bs ~

I

R~

-~

I''

l R'/

D

-r--~

,

i

K

Long die

C~ ' ' ~ / / ' ' / / / / / / / E D

drI

C

-

dr R

9 / annuLc1/r/eLement Fig 5.9 POLYMER MELI:

EMERGING

FRO~vl A LONG DfE

essential f o r die design. I f the die is of a non-uniform section (tapered f o r example) then there w i l l be recoverable t e n s i l e and shear s t r a i n s . I f the die has a uniform cross-section and is long in r e l a t i o n to i t s transverse dimensions then any t e n s i l e stresses which were set up at the die e n t r y f o r example, normally relax out so that only the shear component contributes to the s w e l l i n g at the die e x i t . I f the die is very short ( i d e a l l y of zero length) then no shear stresses w i l l be set up and the

PLASTICS ENGINEERING

241

swelling at the die e x i t w i l l be the result of recoverable tensile strains only. In order to analyse the phenomenon of post extrusion swelling i t is usual to define the swelling r a t i o , B, as dimension of extrudate B = dimension of die Swellin9 Ratios Due to Shear Stresses (a)

Lon9 Capillary

Fig 5.9 shows an annular element of f l u i d of radius r and thickness dr subjected to a shear stress in the capillary. When the element of f l u i d emerges from the die i t w i l l recover to the form shown by ABCD. I f the shear strain at radius r is Yr then ED Yr = tan ~ = A--E Also area of swollen annulus 2~rd# AD i n i t i a l area of annulus =2~--~-r-dr- =A-E

=

[AE 2+ ED2) 89 AE

~ED~2 )~

= (l + ~AEj =

(I + yr2) 89

So from the definition of swelling ratio and using the subscripts S and R to denote shear swelling in the radial direction then R f (I + ~r2) 892~rdr BSR2 = area of swollen extrudate = o area of capillary iR2~rdr 0

Assuming that the shear strain, ~r' varies l i n e a r l y with radius, r then r

Yr = R YR where YR is the shear strain at the capillary wall R r2 So % (l + ~ u189 2~rdr 0

BSR2 =

~R2

3 ie

BSR = [~ YR { (l + yR'2)-~ - yR'3}] 89

(5,44)

242

R.J. CRAWFORD

(b)

Lon9 Rectangular Channe!

When the polymer melt emerges from a die with a rectangular section there will be swelling in both the width (T) and thickness (H) directions. By a similar analysis to that given above; expressions may be derived for the swelling in these two directions. The resulting equations are 1

BST =

(5.45)

[89 + yR2)89+2-lyRln {xR + (l + ~R2)89 T

BSH = BST2

(5.46)

Equations (5.44), (5.45) and (5.46) can be cumbersome i f they are to be used regularly so the relationships between swelling ratio and recoverable strain are often presented graphically as shown in fig 5.10.

2.4

i

/

4 = thickness R = radial T = width

2'2

/

2.0

m

o"

1.8

O

1.6 E

Ill

/zJ

1.4

,.2

1 . 0 / " ~/ 0

, 1

i

2

3

recoverable

F j g . 5.10

I

4

I

5

I

6

strainYR

VARIATION OF SWELLING RATIO WITH SHEAR RECOVERABLE STRAIN FOR s AND SLOT DIES.

243

PLASTICS ENGINEERING

Swellin 9 Ratio Due to Tensile Stresses (a)

Short Capillary (zero length)

Consider the annular element of fluid shown in fig 5.11. tens,~le strain ~R in this element is given by ER

In (l + ~)

where r is the nominal strain (extension - original length) SR : ]n (I + (d~r dr) ) ~R e =l+

( ~ )

swelling RI ratio =" BE =

~shor t 1

Rll dr7F-'~C

D C. dr

IR I

/

. ~

polymer melt

////

annulare emen

I

1

Fig 5.11. POLYMER MELT EMERGING FROM A SHORT DIE, (eER - 1) dr + dr = d~

2~rdrI drt Now area of swollen annulus original area of annulus : 2~--~r-d~- : d--~ :

(eER

=

e

ER

I) dr + dr dr

The true

244

R . J . CRAWFORD

So from the definition of swelling ratio and using the subscript, E, to denote extensional stresses then

2R2~reeR dr 2 area of swollen extrudate BER = area of capillary

o

R f 2xr dr o

~R

BER = ( e ) (b)

(5.47)

Short Rectangular ChanneI

By similar analysis i t may be shown that for a short rectangular s l i t the swelling ratios in the width (T) and thickness (H) directions are given by

BET

=

~R

(e)I

(5.48)

~R 89 BEH = ( e )

(5.49)

Although these expressions are less d i f f i c u l t to use than the expressions for shear, i t is often convenient to have the relationships available in graphical form as shown in fig 5.12. 5.7

Residence and Relaxation Times (a)

Residence (or Dwell) Time

This refers to the time taken for the polymer melt to pass through the die or channel section. Mathematically i t is given by the ratio Volume of channel Residence time, t R = Volume flow rate

For the uniform channels analysed earlier, i t is a simple matter to show that the residence times are : Newtonian Flow Circular Cross Section

tR =8_8~pR ~

(5.50)

Rectangular Cross Section

tR= ~

(5.51)

Power Law Fluid = 13n+l~ L ~ n+l j V-o

Circular Cross Section

tR

Rectangular Cross Section

f2n+l~ tR = ~ n+l j

L V-o

(5.52) (5.53)

PLASTICS ENGINEERING

245

2-61

H =thickness R = radial 2.~

T = width

BEH 0

r

J

16

~~

J

recoverable

F~g 5.12

Y j-'J

strain

E:r

VARIATION OF SWELLING RATIO WITH TENSILE RECOVERABLE STRAIN FOR CAPILLARY AND SLOT

(b)

Y

1,8

DIES.

Relaxation (or Natural) Time

In Chapter 2 when the Maxwell and Kelvin models were analysed, i t was found that the time constant for the deformations was given by the ratio of viscosity to modulus. This ratio is sometimes referred to as the Relaxation or Natural time and is used to give an indication of whether the elastic or the viscous response dominates the flow of the melt. I f the relaxation time is small compared to the residence time then the response of the f l u i d w i l l be largely plastic. 5.8

Power Used to Extrude a Fluid Through a Die The Power is the work done per unit time, where work in the simplest sense is defined as Work : (force) x (distance)

R.J. CRAWFORD

246

Therefore Power = (force) x (distance per unit time) = (Pressure drop x area) x (velocity) But the volume flow rate, Q, is given by Q = area x velocity So

Power = P.Q

(5.54)

where P is the pressure drop across the die. Using this expression i t is possible to make an approximation to the temperature rise of the f l u i d during extrusion through a die. I f i t is assumed that a l l the work is changed into shear heating and that a l l the heat is taken up evenly by the polymer, then the work done may be equated to the temperature rise in the polymer. or

Power = Heat required to change temperature = mass x s p e c i f i c heat x temperature rise

So

PQ = pQ x Cp x AT

P (5.55) AT pC where p is t~e density of the f l u i d and Cp is i t s specific heat. =

5.9

E~erimental Methods Used to Obtain Flow Data In section 5.10 design examples r e l a t i n g to polymer processing w i l l be i l l u s t r a t e d . In these examples the flow data supplied by material manufacturers w i l l be referred to, so i t is proposed in t h i s section to show how t h i s data may be obtained. The equipment used to obtain flow data on polymer melts may be divided into two main groups. (a)

Rotational Viscometers - these include the cone and plate and the concentric c y l i n d e r .

(b)

C a p i l l a r y Viscometer - t h e extruder.

main example of t h i s is the ram

Cone and Plate Viscometer In t h i s apparatus the p l a s t i c to be analysed is placed between a heated cone and a heated plate. The cone is truncated and is placed above the plate in such a way that the imaginary apex of the cone is in the plane of the plate. The angle between the side of the cone and the plate is small ( t y p i c a l l y <5o).

PLASTICS ENGINEERING

247

The cone i s r o t a t e d r e l a t i v e t o the p l a t e and the t o r q u e , T, necessary to do t h i s i s measured over a range o f r o t a t i o n a l r a t e s , 6. Referring to fig.

5.13

Area o f annulus Force Torque

= 2~r.dr : 2xr T . d r = 2xr 2 ~ . d r r~

So

Total Torque, T T = ~ J

Jo

2~r 2 ~ . d r

(5.56)

nR3

,

heated cone

oct_ 5~

I

I I

h e a t e d plate (fixed)

R

Fig, 5,13 CONE AND

.

mohen jplastic

PLATE VISCOMETER.

248

R.J.

Also

Shear Strain, v

X _

CRAWFORD

r0

= ~

r~

0

Strain Rate,~ = And since Viscosity n = n

3T 2~Ra

(5.57)

T

Y

(5,58)

0

The disadvantage of this apparatus is that i t is limited to strain rates in the region 10-3 s- I t o l s-I whereas in plastics processing equipment the strain rates are in the order of 103 s-l to 104 s-Z. Concentric Cylinder Viscometer In this apparatus the polymer melt is sheared between concentric cylinders. The torque required to rotate the inner cylinder over a range of speeds is recorded so that viscosity and strain rates may be calculated. Referring to fig. 5.14 Torque, T = 2~RL.R.~ = 2~R2L ~

(5.59)

Strain Rate, # ~ ~du = T2~RN

(5.60)

Viscosity, n = ~ TH - = 4~-~

(5.61)

As in the previous case, Lilis apparatus is usually restricted to relatively low strain rates.

H~

heated I concentric cylinders

~

~

molten plastic

~

Fi_g. 5.14. CONCENTRIC CYLINDER VISCOMETER.

PLASTICS ENGINEERING

249

Ram Extruder In t h i s apparatus the p l a s t i c to be t e s t e d i s heated in a barre# and then forced through a c a p i l l a r y d i e as shown in f i g . 5 . 1 5 . Normally the ram moves a t a c o n s t a n t v e l o c i t y to g i v e a constant volume f l o w r a t e , Q. From t h i s i t i s c o n v e n t i o n a l to c a l c u l a t e the shear r a t e from the Newtonian f l o w e x p r e s s i o n .

4Q

i" = --~R

piston

/// \\'.~

/ / /

///

\

pressure transducer

,\

/ /

.

i

/

/

x

I

\

///

~

/

/

/

" " " i/.-

Fi_g 515

SECTION

molten nifty,asit

j

YZ,

/ / /

die

heater

-it:::: / / /

/ / /

\

/

~

/ / / ,

/ /

/ / /

/

/// ~

// / / / /

/

/

i / /

\

/// /

/

I:.~!!

THROUGH

/

/

/

/

/

/

/

insulation

extrudate

RAM

EXTRUDER 9

Since i t i s recognised t h a t the f l u i d i s Non-Newtonian, t h i s i s o f t e n r e f e r r e d to as the apparent shear r a t e to d i f f e r e n t i a t e i t from the t r u e shear r a t e . I f the pressure d r o p , P, across the d i e i s a l s o measured then the shear s t r e s s , T, may be c a l c u l a t e d from PR

R. J. CRAWFORD

250

This leads to a d e f i n i t i o n o f apparent v i s c o s i t y as the r a t i o of shear stress to apparent shear rate q

=

~pR 4 I 8LQ

A p l o t of apparent v i s c o s i t y against shear stress produces a unique flow curve f o r the melt as shown in f i g . 5.17. Occasionally t h i s information may be based on the true shear rate. As shown in section 5.4 (a) t h i s is given by

However, t h i s then means that the true shear rate must always be used in the flow s i t u a t i o n being analysed. Experience has shown that t h i s a d d i t i o n a l complexity is unnecessary because i f the Newtonian shear rate is correlated with flow data which has been calculated using a Newtonian shear rate, then no error is involved. Note t h a t r o t a t i o n a l viscometers give true shear rates and i f t h i s is to be used with Newtonian based flow curves then, from above, a correction f a c t o r of (4n/3n+l) needs to be applied to the true shear rate. There are two other points worth noting about t h i s t e s t . F i r s t l y the flow data is produced using a c a p i l l a r y die so that i t s use on channels of a d i f f e r e n t geometry would require a correction f a c t o r . However, in most cases of practical i n t e r e s t , the f a c t o r is not s i g n i f i c a n t l y d i f f e r e n t from 1 to j u s t i f y the additional complication by i t s i n c l u s i o n . For example, from the analysis e a r l i e r i t may be shown t h a t , f o r n = 0.3: the correction f a c t o r necessary when using c a p i l l a r y flow data on a rectangular s l i t is given by =

1.035

Secondly, the pressure drop, P, in the above expression is the pressure drop due to shear flow along the die. i f a pressure transducer is used to record the pressure drop as shown in f i g . 5.16, then i t w i l l also pick up the pressure losses at the die entry. This problem may be overcome by carrying out f u r t h e r tests using e i t h e r a series of dies having d i f f e r e n t lengths or a die with a very short ( t h e o r e t i c a l l y zero) length. In the former case, the pressure drops f o r the various lengths of die may be extrapolated to give the pressure drop f o r entry i n t o a die of zero length (See f i g . 5.16). In the second case t h i s pressure is obtained d i r e c t l y by using the so called zero length die. This is then subtracted from the measured pressure loss, PL' ~n the long die being considered, so that
PLASTICS ENGINEERING

251

~ o

,o

.J

l/1

o

13-

J

J J Die Length

F_ig.5.16. VARIATION OF PRESSURE " DROP WITH DIE LENGTH In addition i f the swelling of the extrudate is measured in each of these two tests then the swelling r a t i o using the long die w i l l be B~p and the swelling r a t i o using the short die w i l l be ~ (see sectTBn 5.6). Using equation (5.44) and (5.47) t h i s enables shear and t e n s i l e components of the recoverable s t r a i n s to be calculated and from them the shear and t e n s i l e moduli. From t h i s r e l a t i v e l y simple t e s t , t h e r e f o r e , i t is possible to obtain complete flow data on the material as shown in f i g . 5.17. Note t h a t shear rates s i m i l a r to those experienced in processing equipment can be achieved. Variations in melt temperature and h y d r o s t a t i c pressure also have an e f f e c t on the shear and t e n s i l e v i s c o s i t i e s of the melt. An increase in temperature causes a decrease in v i s c o s i t y and an increase in h y d r o s t a t i c pressure causes an increase in v i s c o s i t y . T y p i c a l l y , f o r low density polythene an increase in temperature of 40~ causes a v e r t i c a l s h i f t of the v i s c o s i t y curve by a f a c t o r of about 3. Since the p l a s t i c w i l l be subjected to a temperature rise when i t is forced through the die, i t is u s u a l l y worthwhile to check (by means of equation 5.55) whether or not t h i s is s i g n i f i c a n t . In addition to t h i s , a change in pressure from atmospheric (= 0.I 14Pa) to about I00 MPa (a pressure l i k e l y to be experienced during processing) causes a v e r t i c a l s h i f t of the v i s c o s i t y curve by a f a c t o r of about 4 f o r LDPE. This e f f e c t may be important during processing because the material can be subjected to large changes in pressure in sections such as nozzles, gates, etc. However, i t should be noted t h a t in some cases the increase or decrease in pressure r e s u l t s i n , or is associated w i t h , an increase or decrease in temperatures so that the net e f f e c t on v i s c o s i t y may be n e g l i g i b l e .

252

R. J. C R A W F O R D

106 i

X = 11

Tensile v i s c o s i t y Apparent viscosity

E

=

Tensile

G

=

Shear

2

3i

4

modulus modulus

5 67891

2

3 4 567891

2

3 4 5 67891

.

T i i hG

'

I~111

/

345

/

/

/

~

67 8911 9 8 7 6 5 4 3

/ /

]i /

i

U./!

10' ~

Z ~E

4

/

3

/

Illlil

,,

Z

I"

L

2 1

104

non laminar flow in unstrearnlined die:_ I

7

ID Z

7 g: 5

,/

/

14

,

/

!3

/

/ / I

10:

\/,.

I

lo:

1 103

2

/

3"~4 5 6 7 8 9 1 104

2

3 4 $67891 105

4

;3

I

2

3 106

4

5

7891 107

s t r e s s (Nm-2)

FigI 5.17

FLOW DATA FOR A

GRADE OF POLYETHYLENE AT 170~163

Melt Flow Index The Melt Flow Index Test is a method used to characterise polymer melts. I t is in e f f e c t a single point ram extruder test using standard t e s t i n g conditions (BS 2782) as i l l u s t r a t e d in f i g . 5.18. The polymer sample is heated in the barrel and then extruded through a standard die using a standard weight on the piston, and the weight (in gms) of polymer extruded in I0 minutes is quoted as the melt flow index (MFI) of the polymer.

PLASTICS ENGINEERING

I' .IA

FII . "-

oil-filled thermometer wells (.

4-

253

w~

|

::::-.iffs

!

~]--:::.- !if I

barrel

+~]i-iii~:i'i'~ii[

heater

- - insutatian

~ i

polymer melt

die Fig. 5.18 DIAGRAM OF APPARATUS FOR MEASURING MELT FLOW INDEX. Flow Defects When a molten plastic is forced through a die i t is found that under certain conditions there w i l l be defects in the extrudate. In the worst case this w i l l take the form of gross distortion of the extrudate but i t can be as s l i g h t as a dullness of the surface. In most cases flow defects are to be avoided since they affect the quality of the output and the efficiency of the processing operation, iiowever, in some cases i f the flow anomaly can be controlled and reproduced, i t can be used to advantage - for example, in the production of sheets with matt surface f i n i s h . Flow defects result from a combination of melt flow properties, die design and processing conditions but the exact causes and mechanisms are not

completely understood.

The two most common defects are

(a) Melt Fracture When a polymer melt is flowing through a die, there is a c r i t i c a l shear rate above which the extrudate is no longer smooth. The defect may take the form of a s p i r a l l i n g extrudate or a completely random configuration. With most plastics i t is found that increasing the melt temperature or the L/D r a t i o of the die w i l l increase the c r i t i c a l value of shear rate. I t is generally believed that the distortion of the extrudate is caused by s l i p - s t i c k mechanism between the melt and the die wall due to the high shear rates. I f there is an abrupt entry to the die then the t e n s i l e /

254

R . J . CRAWFORD

shear stress history which the melt experiences is also considered to contribute to the problem. (b) Sharkskin Although this defect is also a visual imperfection of the extrudate i t is usually differentiated from melt fracture because the defects are perpendicular to the flow direction rather than helical or irregular. In addition experience has shown that this defect is a function of the linear output rate rather than the shear rate or die dimensions. The most probable mechanism of sharkskin relates to the velocity of skin layers of the melt inside and outside the die. inside the die the skin layers are almost stationary whereas when the extrudate emerges from the die there must be a rapid acceleration of the skin layers to bring the skin velocity up to that of the rest of the extrudate. This sets up tensile stresses in the melt which can be s u f f i c i e n t to cause fracture. 5.10 Analysis of Flow in Some Processing Operations Design methods involving polymer melts are d i f f i c u l t because the flow behaviour of these materials is complex. In addition flow properties of the melt are usually measured under well defined uniform conditions whereas unknown effects such as heating and cooling in processing equipment make service conditions less than ideal. However, s u f f i c i e n t experience has been gathered using the equations derived e a r l i e r that melt flow problems can be tackled quantitatively and with an accuracy which compares favourably with other engineering design situations. In this section a number of polymer processing methods w i l l be analysed using the tools which have been assembled in this chapter. In some cases the heating and cooling of the melt may have an important effect on the flow behaviour and methods of allowing for this are developed. A summary of the equations which are useful in polymer flow analysis are given at the end of this chapter, EXAMPLE (1) In a plunger type injection moulding machine the torpedo has a length of 40 mm, a diameter of 23 mm and is supported by three spiders. I f , during moulding of polythene at 170~ the plunger moves forward at a speed of 25 mm/s, estimate the pressure drop across the torpedo and the shear force on the spiders. The barrel diameter is 25 mm. SOLUTION : Assume that the flow is isothermal. Volume flow rate, Q = area x velocity 2 (25 x lO-3) x 25 x I0-3 = 12.27 x 10-6 m3/S. The gap between the torpedo and the barrel ma be coqsidered as a rectangular s l i t with T = (~ x 24 x I0-3 m) and H = # x lO- j m. So apparent strain rate, ~ =-~H = 6 x 12.27 x lO"6 x 24 x IO-3x lO -6 = 976.6 S- l

PLASTICSENGINEERING

255

From f i g 5.17 at this strain rate PH Shear stress, T = 2.7 x 105 = 2-L p:

2LT : 2 x 40 x I0 "3 x 2.7 x 105 lO "3

= 21.6 MPa The force on the spider results from (a)

(b)

the force on the torpedo due to the pressure difference across i t an the force on the torpedo due to viscous drag. Cross-sectional area of torpedo = ~r 2

So force due to pressure : ~r2P

~ ( l l . 5 x lO-a)

=

2

x 21.6 x 106

= 8.97 kN Surface area of torpedo

=

Viscous drag force

= ~DLT

xDL

: ~(23 x 1 0 - 3 ) ( 4 0 x 1 0 - 3 ) ( 2 . 7

x IO s )

= 0.78 kN So T o t a l Force

: 9.75 kN

EXAMPLE (2) Fhe output of polythene from an extruder is 30 x 10-6 m3/S. I~ the breaker plate in this extruder has 80 holes, each being 4 mmdiameter and 12 mm long, estimate the pressure drop across the plate assuming the material temperature is 170~ at this point. SOLUTION: Assumethe flow is isothermal. Flow rate through each hole

30 x 10-G m3/s = 80 4Q So Apparent shear rate, # = - - ~ 4x30x10-6 x 80 x 23 x 10-9 = 59.6 s-Z From f i g . 5.17 the shear stress, T, is 1.2 x I0 S Nm-2 and since PR T=~T

256

R. J. CRAWFORD P=

2 x 12 x lO-a x 1.2 x IOs 2 x lO-~

= 1.44 MPa EXAMPLE ~-3(3)5 A power law f l u i d with constants n_ = 1.2 x I04 Ns/m2 and n = is injected through a centre gate ~nto a disc cavity which has a depth of 2 mm and a diameter of 200 mm. I f the injection rate is constant at 6 x IO-s m3/s, estimate the time taken to f i l l the cavity and the minimum injection pressure necessary at the gate. SOLUTION: I f the volume flow rate is Q, then for any increment of time, dt, the volume of material injected into the cavity w i l l be given by (Qdt). During this time period the melt front w i l l have moved from a radius, r , to a radius (r+dr). Therefore a volume balance gives the relation Qdt = 2~rHdt where H is the depth of the cavity. Since the volume flow rate, Q, is constant this expression may be integrated to give

t

dt - 2~H Q C rdr t - ~R2H

For the conditions given t = ~(lOOxlO-3)2 (2xi0-3)

6xlO-S = 1.05 seconds I t is now necessary to derive an expression for the pressure loss in the channel. Since the cavity f i l l s very quickly i t may be assumed that effects due to freezing-off of the melt may be ignored. In section 5.4 (b) i t was shown that for the flow of a power law f l u i d between parallel plates q = (T~F) TVoH Now for the disc, T = 2~r and substituting for Vo from (5.27) ~ n

Q = (~-~F) (2~rH)

1 i / n dp I / n

n+--~T)(~)

dP = Qn ~2n+l~ n no (2)n+l

(~r) dr

n+l]

(H/2) -~--

PLASTICS ENGINEERING This may be integrated to give the pressure as

Qn ,2n+l, n no (2)n+l

p

RTM

For the conditions given

P : (6xi0-5) ~

(~_~T) ~

35

( i . 2 x i 0 4 ) ( 2 ) 1"35 IlOOxlO-3) ~ (2xi0-3) 1"7 (0.65)

= 12.42 MPa EXAMPLE (4) In a polythene f i l m blowing die the geometry of the die l i p s Is as shown in f i g . 5.19. I f the output is 300 kg/hour and the density o f the polythene is 0.76 g/cm3, estimate the pressure drop across the die l i p s . SOLUTION: Volume flow rate =

300 x 103 x I0 -6 = 109.6 x I0 -6 m3/s 60 x 60 x 0.76

The die l i p s may be considered in three sections A, B and C and the t o t a l pressure drop is the sum of the losses in each. Section A Since t h i s section is of a uniform section there w i l l only be a pressure drop due to shear. For a rectangular s l o t , as shown e a r l i e r , the apparent shear rat e , ~ at A1 is given by = _6--~:

6 x 109.6 x 10-6 ~(260 x 10-3)(0.7 x 10-3) 2

=

103 s - I

1.64

So front f i g . 5.17, ~

x

= 3.2 x IO s Nm-2

2LTj PSi 2

:

TI 2 x 4 x 3.2 x 105

=

0.7

= 3.66 MPa Section B This section is tapered so there w i l l be pressure losses due to both shear and extensional flows. From simple geometry :IH31: ~HP + (2L tan ~) where ~ is h a l f the included angle and L

257

R. J. CRAWFORD

258

H2 : 0.7 mm, H3 = 1.276 mm Now a t B2,# = 1.64 x 103 S-z I & = ~#

I tan ~ = ~x

1.64 x 103 t a n ( l . 5 o )

= 13.96 s -1 At B3, 6q = 6 x 109.6 x 10 -3 = TH32 ~ x 260 x I 0 "3 (1.276 x 10-3) 2

= 494.5 s -z

tu be

tube

1 I_

ncluded B

1

260

=

11

C die

poty

Fig 5-19

DIE

E=~ 1 (494.5)

USED

FOR

er mett

TUBE

EXTRUSION

t a n 1.5 o

= 4.3 s- I From f i g . 5.17 t h e f l o w curve can be t a k e n as a s t r a i g h t l i n e f o r t h e shear r a t e range 500 S- I t o 1600 s -z and t h e power law i n d e x n may

PLASTICS ENGINEERING

be taken as 0.3. ~2 So from (5.38) PS23 =~-6 tan ~

[

H22n] 1 - (~33)

3.2 x IOs 2(0.3) tan 1.5 = 6.16 MPa Also from (5.39)

PE23 =-~- 9

1 - (

)

Now ~ = 13.96 s- I so from f i g . 5.17 o2 = 3.6 x lO 6 Nm-2

PE~3 =

0x,o 2

[,

= 1.26 MPa Section C Although t h i s section i s tapered on one side only, since the angle is small the e r r o r in assuming i t is tapered on both sides w i l l be n e g l i g i b l e . The included angle is taken as 80.. H4 : H3 + L tan 8o = 1.276 + 40 tan 8o = 6.9 mm At C3 = 494.5 ~ i = l (494.5) tan 40 = II.52 s-I At C4

6Q = 6 x I09.6 x lO-6 = TH4Z ~ x 260 x lO -3 x (6.9 x I0-3) z = 16.9 s-1 = 1 (16.9) tan 40 = 0.39s-I

259

260

R. J. CRAWFORD

As before T3 PS34 = ~-n tan from f i g .

[

i

(

H~ 2n ] )

5.17, T3 = 2.3 x IO s Nm-2 at # = 494.5 s-I

2.3 x I0:~ I 1.276, ~ PS34 = 0.6 tan 1 -( 6.9 J = 3.49 MPa Also #3 PE34 = T

s : 11.52 PE34 =

[ ' S- I

H32] 1 - (-~4) SO

from f i g .

2

1 -

5.17 ~3 = 2.9 x 106 Nm-2 .

= 1.4 MPa In this case there w i l l also be a pressure loss due to flow convergence at the die entry. This may be obtained from (5.40). i.e.

Po = ~Yn~F)-' (

) ' (nx)89

6Q = 16.9, so n = 4.4 x 103 Nsm-2, ~ = 2.5 x I0 s Nsm-2 At C4, # = T--~4 Po =T~-g (16.9) (4.4 x 103 x 2.5 x IOS) 89 : 1.18 MPa So the total

pressure drop through the die lips is given by

P = 3.66 + 6.16 + 1.26 + 3.49 + 1.4 + 1.18 : 17.15 MPa EXAMPLE (5) Estimate the dimensions of the tube which w i l l be produced by the die in example 4, assuming that there is no draw down. SOLUTION: Since the f i n a l

section of the die l i p s is not tapered

PLASTICS ENGINEERING

i t may be assumed t h a t the s w e l l i n g a t the d i e e x i t i s due to shear effects only. At e x i t , #

6Q = 1.64 x 103 s_ 1 =~-~z

From f i g .

5.17, x = 3.2 x IO s Nm-z and G = 5 x 104 Nm-2

T 3.2 x IOS = 6.4 So r e c o v e r a b l e s t r a i n , y R = ~ = 5 . 0 x I04 Then using equations (5.45) and ( 5 . 4 6 ) , or more c o n v e n i e n t l y f i g . 5.10, the s w e l l i n g r a t i o s may be o b t a i n e d a s BSH = 2.25 and BST = 1.51 Therefore swollen thickness o f f i l m = HI x BSH = 0.7 x 2.25 = 1.575 mm and swollen d i a m e t e r o f bubble = D x BST = 260 x 1.51 = 392.6 mm EXAMPLE (61 in many p r a c t i c a l s i t u a t i o n s i t i s the dimensions o f the o u t p u t which are known o r s p e c i f i e d and i t i s necessary to work back ( o f t e n using an i t e r a t i v e procedure) to design a suitable die. Design a d i e which w i l l produce p l a s t i c f i l m 0.523 mm t h i c k a t a l i n e a r v e l o c i t y o f 20 mm/s. The l a y - f l a t w i d t h o f the f i l m i s to be 450 mm and i t i s known t h a t a blow-up r a t i o o f I . g i w i l l g i v e the necessary o r i e n t a t i o n i n the f i l m . Assume t h a t t h e r e i s no draw-down. SOLUTION: Using the t e r m i n o l o g y from the a n a l y s i s o f blow moulding in section 4.2.5 2(LFW) = ~Db 2 x 450 Db = ~ =

286.5 mm

Now i t makes the s o l u t i o n s i m p l e r to up r a t i o is g i v e n by Db/D I ( i e r a t h e r than seems p r a c t i c a l because tile change from DI by i n f l a t i o n whereas the change from Dm tO effects. .'.

D l = 1.91 (286.5) = 150 mm

assume t h a t the blow Dh/D_ ) . Also t h i s t~ D~ i s caused s o l e l y Db iMcludes d i e s w e l l

261

R.J. CRAWFORD

262

then f o r constant volume xDih I

hl

=

=

xDbh b

286.5 x 0.523 150

= i mm

So having obtained the dimensions of the tube to be produced, the procedure is to s t a r t by assuming t h a t there is no die swell. This means t h a t the die dimensions w i l l be the same as those of the tube.

6Q

Apparent shear r a t e , ~ = H~-~T but volume flow r a t e , Q = VTH /

where V = v e l o c i t y of p l a s t i c melt. 6V

Y=-FF _ 2 6 _ _x~ 0

=

I

120 s- I

from f i g . 5.17, z = 1.5 x IO s Nm-2 and G = 3 x lO 4 Nm-2 so recoverable shear s t r a i n , YR' is given by T _ 1 . 5 x IO s

YR = G

= 5

3 x 104

Assuming t h a t swelling results from shear effects only then from f i g . 5.10 at YR = 5 and the swelling r a t i o s are BSH = 2 and BST = 1.42 The i n i t i a l swelling

die dimension must now be adjusted to allow f o r this

New die gap = H' = ~ : 0.5 mm New die circumference = T' = 3 T ~ The volume flow r a t e , Q, w i l l be constant so the v e l o c i t y of the melt in the die w i l l also be adjusted. Q = VTH = V'T'H' V'

VTH =1~-~r = V.BsH.BsT

New s h e a r r a t e ,

6V' # = ~

From f i g .

9 = 2 . 5 x IO s Nm- 2 and G = 4 . 2 x 104 Nm- 2

5.17,

YR =5.95

=

6 x 20 x 2 x 1 . 4 2 0.5

= 681.6

s -I

PLASTICS ENGINEERING and from f i g . 5.10

BSH = 2.175 and BST = 1.485 I So new die gap = H " = ~ / ~ . = new d i e

circumference

:

T"

0.459 mm

=

I f this i t e r a t i v e procedure is repeated again the next values obtained are die gap, H ' "

= 0.455 mm

die circumference, T " '

~ : T

x 150 mm

These values are s u f f i c i e n t l y close to the previous values so the die e x i t dimensions are Gap :

0 . 4 5 5 mm

Diameter

x 150 : T-.Tg-~-T = 100.67

The shear rate, #, in the

mm

die land is

6 x 20 x 1.49 x 2.2 =

0.455

= 864.5 s-I From f i g . 5.17 i t may be seen that this exceeds the shear rate for non-laminar flow (approximately 30 s- I ) so that the entry to this region would need to be streamlined. Fig. 5.17 also shows that the extensional strain rate, ~, in the tapered entry region should not exceed about 15 s-I i f turbulence is to be avoided. I

Therefore since ~ = ~ k tan = tan_I (3_~) = tan-1 ( ~ g ~ ) = 2.97~ So the f u l l angle of convergence to the die land must be less than (2 x 2.97) o = 5.940. To complete the die design i t is l i k e l y that a third tapered section w i l l be necessary as shown in f i g . 5.19. This w i l l be designed so that the entry to the die is compatible with the outl e t of the die body on the extruder. The maximumangle of convergence of this section can be estimated as shown e a r l i e r and

263

264

R. J. CRAWFORD

the lengths of the two tapered sections selected to ensure the angle is not exceeded f o r the f i x e d die i n l e t dimension. I f the length of the 5.940 tapered region is chosen as lO mm i t may be shown t h a t the m~ximum angle o f convergence f o r the entry to t h i s taper is 58.5 . I f the length o f t h i s section is f i x e d a t 40 mm and the annular gap a t the e x i t from the extruder is 16 mm, then a s u i t a b l e angle o f convergence would be about 400 . EXAMPLE (7) i t is desired to blow mould a p l a s t i c b o t t l e with a dimater of 60 mm and a wall thickness of 2 mm. I f the extruder die has an annular s l o t o f outside diameter = 32 mm and inside diameter = 28 mm, c a l c u l a t e the output r a t e needed from the e x t r u d e r and recommend a s u i t a b l e i n f l a t i o n pressure. Use the f l o w c h a r a c t e r i s t i c s given in f i g . 5.17. Density o f polythene = 760 kg/m~. SOLUTION: In the analysis of blow moulding in section 4 . 2 . 5 , i t was shown t h a t the thickness, h, o f the i n f l a t e d b o t t l e is given by Dd h = BSTa hd (~--) m where hd = die gap = 2 mm D = mould diameter = 60 mm m Dd = die diameter = 89 So

= 30 mm

3 2 x 60 BST =Z~-O-~-Z BST = l .26

From f i g . So

5.10 a t BST = 1.26, YR = 3.15 3.15 = ~/G

From f i g . 5.17 i t is now necessary to determine the combination of 9 and G to s a t i s f y t h i s equation. For example a t 9 = 105 N/m2

G = 2.1xlO ~

so YR = 4.76

a t T = 5xlO 4 N/m~ G = 1.7xlO 4

so ~R = 2.94

a t 9 = 6xlO 4 N/m2

so YR = 3.15

G = 1.9xlO4

Therefore the l a t t e r combination is c o r r e c t and from f i g . a t T = 6xlO 4 N/m2, # = IO s - l

5.17,

6q # = I0 = T-~T

and

Q=

lO x ~ x 30 x 22 x lO -9 = 1.26 x 10-6 m3/s 6

= 1.26 x 10-6 x 760 x 3600 : 3.45 kg/hour

265

PLASTICS ENGINEERING

To calculate the i n f l a t i o n pressure i t is necessary to get the melt fracture stress. From f i g . 5.17 i t may be seen that this is 4 x lO6 N/mz. Therefore since the maximum stress in the inflated bubble is the hoop stress, a, then the i n f l a t i o n pressure, P, is given

by

2ha 2 x 2 x 4 x IOG P =ml~- = 60 = 0.13 MPa EXAMPLE (8) During the blow moulding of polythene bottles the parisonisO.3 m long and is l e f t hanging for 5 seconds. Estimate the amount of sagging which occurs. The density of polythene is 760 kg/m3. Consider the small element of parison as shown

SOLUTION: I

i I I

I

- - ~, y

(a)

From the relationship between stress, strain and modulus

,I

Fdx 6L =-AZ-

I 1

Elastic Strain

I

pxDl hI xdx

x' I I

_L'

xDlhIE So total extension = ~ I

i I

xdx = P ~ 0

gl

(b)

L

So Elastic Strain = ER = ~1~

Viscous Strain force _ weight below YY The stress on the element - -area area PxDlhldX

So

total stress, a = I

pdx : pL

Now the viscous strain on the element is given by

~'V

cT

: ~dt = ~ dt ev =

~ dt =

- - dt = x

= pdx

266

R.J. CRAWFORD Therefore the total strain, ~ = ~R + ~V = pL pLt ~E + X

= pL[-~-E +t] Note that this solution applies for values of t less than the characteristic time (X/E) for the process. This is generally the case for blow moulding. For the situation given stress, ~ = pL = 760 x 9.81 x 0.3 = 2.L4 x lO3 N/m2 From f i g . 5.17, x = 2.2 x IOs Ns/m2, E = 1.9 x lO4 N/m2 So characteristic time = _k = l l . 6 s E

Therefore since the parison sag time is less than this the above expression may be used to calculate the amount of stretching = 760 x 9.81 x 0.3

2xl.9x104 +

O.ll

=

So extension

= O.ll x 0.3m 33 mm

=

Analysis of Heat Transfer during Polymer Processing Most polymer processing methods involve heating and cooling of the polymer melt. So far the effect of the surroundings on the melt has been assumed to be small and experience in the situations analysed has proved this to be a reasonable assumption. However, in most polymer flow studies i t is preferable to consider the effect of heat transfer between the melt and i t s surroundings. I t is not proposed to do a detailed analysis of heat transfer techniques here since these are dealt with in many standard texts, instead some simple methods which may be used for heat flow calculations are demonstrated. Fouriers equation for non-steady heat flow in one dimension, x, is

B2T 1 .aT aRT

~ at

where T is temperature and ~ is the thermal d i f f u s i v i t y defined as the r a t i o of thermal conductivity, K, to the heat capacity per unit volume K

pCp

PLASTICS ENGINEERING

267

where p is density and Cp is specific heat. Host materials manufacturers supply data on the thermal d i f f u s i v i t y of their plastics but in the absence of any information a value of IxlO-? m2/s may be used for most thermoplastics. Solutions to FourieHs equation are in the form of i n f i n i t e series but are often more conveniently expressed in graphical form. In the solution the following dimensionless groups are used (1)

st Fourier Number, F~ = x-T

(5.62)

where t is time and X is the radius of the sphere or cylinder (or half thickness of the sheet) considered. In the case of a f l a t sheet, i f the heating/cooling is from one side only then the dimension, X, is taken as the f u l l thickness. (2)

T3 - T2 Temperature Gradient, AT = I " ~

(5.63)

where Tl = I n i t i a l uniform temperature of the melt T2 = Temperature of heating or cooling medium T3 = Temperature at time t. Figures 5.20 and 5.21 show the solution to Fourier~ equation in terms of the temperature gradient at centre l i n e of section considered and the Fourier Number for the cases of a f l a t sheet and a cylinder respectively. In each case i t is assumed that there is no resistance to heat transfer at the boundary between the melt and the heatingZcooling medium. The way in which this data may be used is i l l u s t r a t e d by the following example. EXAMPLE (9) An injection moulding is in the form of a f l a t sheet I00 mm square and 4 mm thick. I f the melt temperature is 180oc, the mould temperature is 60oc and the melt freezes o f f at a temperature o f llOOC estimate (a)

the temperature of the material at the centre of the moulding after 7 seconds

(b)

the time taken for the moulding to s o l i d i f y .

SOLUTION:

(a)

Fourier st Fo : x-~

Number, Fo, is

1 xlO-Tx7 Fo = (2 x 10-3) 2 = 0.175

268

R. J. C R A W F O R D

l

0'8

0.6

\

0.4

0.2

0

0'2

0"4

0'6

0'8

1-0

1"2

14

Fo Fi_g. 5.20.

TEMPERATURE FOR A FLAT

GRADIENT SHEET.

AGAINST

FOURIER NUMBER

1.0

0.8

0.6

i i

0,4

0,2

0

0'2

0'4

0'6

0.8

1.0

Fo Fig 5.21 TEMPERATURE

NUMBER

GRADIENT

FOR

A

AGAINST

CYLINDER

FOURIER

PLASTICS ENGINEERING

From f i g . 5.20 the temperature gradient is 0.82 T3 - 60 So 0.82 : TB-O--Z-~O ie T3

(b)

158,4~ For freeze-off to occur the temperature gradient is AT =

~

=

0.42

From f i g . 5.20, F~ = 0.455 ~t So 0.455 = x-~ ort:

0.455 x (2 x 10-3) 2 l x l O -7 = 18.2

s

EXAMPLE ~0) Estimate the heat transfer from the surroundings t o the melt as i t passes through the die land in Example (4) SOLUTION: Volume of die land = ~DHL = (~ x 260 x 0.7 x 4 x lO-9) m3 Volume flow rate Q = I09.6 x lO-s m3/s So Residence time'= ~ x 260 x 0.7 x 4 x lO-9 I09.6 x lO-s = 2.08 x 10 -2 s ~t So F o u r i e r number =x-~Z

F = o

1 x 10-7 x 2.08 x 10 -2 (0.35 x I 0 - ~ ) z

= 0.017 From f i g . 5.20 i t may be seen t h a t the c e n t r e l i n e t e m p e r a t u r e g r a d i e n t a t t h i s F o u r i e r Number i s a l m o s t I . T3 - T2 Since AT = T - ~ = 1

then T3 = Tl

=IreS 7 -

R

269

270

R. J. CRAWFORD

ie the melt temperature after 2.08 x lO-2 seconds is the same as the i n i t i a l melt temperature (TI) so that as the melt passes through the die land i t is r e l a t i v e l y unaffected by the temperature of the die. Any temperature rise which occurs in the melt as i t passes through this section w i l l be as a result of the work done on the melt and may be estimated using (5.55). P Temperature rise = p Tn where the specific heat, Cp, may be taken as 2.5 x lO3 J/kg~ 3.66 x IOt So Temperature rise = 760 x 2.5 x lO3 : l .9oc

As an additional aid to assessing the effect of surroundings on melt temperature (5.63) may be rearranged to the form Tl - T3 T-~ = (I - AT)

(5.64)

This gives the r a t i o of the actual temperature change (TI - T3) to the possible temperature change (Tl - Tp) and may be expressed in terms of Fo as shown in the following table.

Actual temperature change Possible temperature change Fourier Number Fo 0.01 0.03 0.I 0.2 0.3 0.5 1.0 3.0

Flat sheet

0 0 0.05 0.23 0.39 0.63 0.89 0.99

Cylinder 0 O.Ol 0.15 0.53 0.72 0.91 0.98 0.99

Note that this table refers to the temperature change at the centre l i n e of the melt. At a point half way between the centre line and the surface the effects are given in the following table.

PLASTICS ENGINEERING

271

Actual temperature change Possible temperature change

Fourier Number Fo

Flat sheet

Cylinder

0.02 0.05 0.27 0.45 0.57 0.73 0.92 0.99

0.05 O.l 0.39 0.66 0.81 O.94 0.99 l.O

0.01 0.03 0.I 0.2 0.3 0.5 1,0 3.0

EXAMPLE i l l ) Derive an expression for the flow length of a power law fluid when i t is injected at constant pressure into a rectangular section channel assuming (a) (b)

the flow is isothermal there is freezing off as the melt flows.

SOLUTION: Ca) As shown earlier the flow rate of a power law fluid in a rectangular section is given by Q = (zn+l)n+l TH (n+-~F) (__p_p)lnno~ (~) n+l n

(5.64)

During any increment of time, dt, the volume of the fluid injected into the channel is Qdt. This will be equal to the increase in volume of the fluid in the channel

Qdt= THd~ So from above

1/n

n+l n

TH ~

L 1/n 1/n n+l it I ~ d~= (~T) (n~) (~) n 0

dt

0

n

,~+I n

(n-~F) L

,=

~

= (~)

n

l/n (

)

n+l (~)-n-. t

1 n (PI R+-Ti )t

(5.65)

no

The volume flow rate at any instant in time may be determined by substituting for L in equation (5.64).

272

R. J. C R A W F O R D

(b) I f the melt is freezing o f f as i t flows then the e f f e c t i v e channel depth w i l l be h instead of H as shown in f i g . 5.22. Therefore the above expression may be w r i t t e n as i/n 1/n n+l d~ = (~-~T) (@oo)

(~)

n

dt

(5.66)

This expression cannot be integrated j u s t as simply as before because h is now a f u n c t i o n of time. I t is necessary therefore to make an assumption about the rate at which the channel thickness changes. Barrie has investigated t h i s problem in d e t a i l and concluded t h a t the freezing o f f could be described by a r e l a t i o n of the form Ay = Ct 1/3

(5.67)

where C is a constant and ay is the thickness of the frozen layer as shown in f i g . 5.22.

cool

mould

~, R ~, \ \ \ "-..~.. \ \ \

,.frozen layer \ \ \ \ \

iil!

-T-

..5.,... ~......X...5...?~... 5...h,....~....'5.....?~...?~...,.---k

I A,, "

iJ•iiii•iii!•!iiii7iii7iiiiiii•iii7iiiiiiiiiiiiiiiiii!i!iii•i•ii!i•ii!•7 i::iii::i::iii::;;::i m oi i e"~ ::iii::iiiii::iii::iiiiT#::ilili::i::Tiiiiiiiiii#~ii;::; iiii~iil;!;! p~ a s t ~c. i',ili',iiiiiiiTiiiiiiii#~iiiiiiiiiiiiiiiii!!~iiii . frozen

layer~'\~

lay

coot mould Fig. 5.22 FLOW OF MOLTEN POLYMER INTO A COLD MOULD. From (5.67) using the boundary condition that t = t f o f f time) at Ay = H/2

(freeze-

then C : (H/2tf z/3) also 2Ay = (H-h) so in (5.67)

( t ~I13

ie

h : H(I -

(4) 113)

So s u b s t i t u t i n g t h i s expression f o r h i n t o (5.66), then i l n dl~

(5.68)

PLASTICS ENGINEERING

273

this may then be integrated to give n+l

(~) n

Ln

=

(~-6~-) (-~-P)I/n H n+l [6 )] (g)--n- tf (TFfiD$~F) (3-~T) (T6~F qo

1

n

no Similar expressions may also be derived for a circular section channel and for the situation where the injection rate is held constant rather than the pressure (see questions end of the chapter). In practical injection moulding the injection rate would probably be held constant until a pre-selected value of pressure is reached. From this point on the pressure would be held constant and the injection rate would decrease. Note that for a Newtonian f l u i d , n = l , so for the isothermal case, equation (5.65) becomes

L = 0.408 ( ~ )

H

and for the non-isothermal case, equation

(5.69) becomes

Ptf 89 L = 0.13 ( ~ ) H For the non-isothermal cases the freeze-off time, t f , may be estimated by the method described in Example 9. EXAMPLE (12) During injection moulding of low density polythene, 15 kg of material are plasticised Rer hour. The temperature of the melt entering the mould is 190uC and the mould temperature is 40~ I f the energy input from the screw is equivalent to l kW, calculate (a) (b)

the energy required from the i~eater bands the flow rate of the circulating water in the mould necessary to keep i t s temperature at 40 • 2~

SOLUTION: The steady flow energy equation may be written as

q - W = Ah

(5.70)

where q is the heat transfer per unit mass Wis the work transfer per unit mass h is enthalpy Enthalpy is defined as the amount of heat required to change the temperature of unit mass of material from one temperature to another. Thus the amount of heat required to change the temperature of a material between specified l i m i t s is the product of i t s mass and the enthalpy change.

274

R . J . CRAWFORD

The enthalpy of plastics is frequently given in graphical form. For a perfectly crystalline material there is a sharp change in enthalpy at the melting point due to the latent heat. However, for semicrystalline plastics the rate of enthalpy change with temperature increases up to the melting point after which i t varies l i n e a r l y with temperature as shown in figure 5.23. For amorphous plastics there is only a change in slope of the enthalpy line at glass transition points. Figure 5.23 shows that when LDPE is heated from 20oc to 190oc the change in enthalpy is 485 kJ/kg. (a)

In equation (5.70) the sign convention is important. Heat is usually taken as positive when i t is applied to the system and work is positive when done by the system. Hence in this example where the work is done on the system by the screw i t is regarded as negative work. So using (5.70) for a mass of 15 kg per hour q -

(-4)

: 15 (485

q = 1.02 kW i.e.

(b)

the heater bands are expected to supply t h i s powe~

At the mould there is no work done so in terms of the total heat absorbed in cooling the melt from 190~ to 40~ q = mAh

= 15 = 1.85

] kW

This heat must be removed by the water circulating in the mould at a rate, Q q = Qah or by d e f i n i t i o n of enthalpy q = Q CpAT where Cp is the specific heat (= 4.186 kJ/kg~ for water) and AT is the temperature change (=4oc i . e . • 2oc) So 1.85 = Q x 4.186 x 4 or Q = O.ll kg/s = O.ll l i t r e s / s I t is also possible to estimate the number of cooling channels required. I f the thermal conductivity of the mould material is K then the heat removed through the mould per unit time w i l l be given by K.A (AT) q = y

PLASTICS ENGINEERING

275

0

o

\

~g ~'~ c

i

&

\ '--

CS"

,3E d~ LZl

o

0 0

0 CO 0

0 LeO

0 ~1.0

0 Oe. ~

(6~IIr~) AdloUIlUe

0 O o

0 0

0

276

R.J. CRAWFORD

where AT is the temperature between the melt and the circulating fluid Y is the distance of the cooling channels from the mould and A is the area through which the heat is conducted to the coolant. This is usually taken as half the circumference of the cooling channel multiplied by i t s length.

The K Value f o r s t e e l i s 11.5 c a l / m . s . ~ so assuming t h a t the cooling channels have a diameter of lO rmn and they are placed 40 mm from the c a v i t y , then

L =

2x40x10-3xl'85x10a ll.5x4.2x~xl0xl0-ax(190-40)

= 0.65 m I f the length of the cavity is 130 mm then f i v e cooling channels would be needed to provide the necessary heat removal. 5.11

Su~ (])

of Melt Flow Equations Uniform Circular Die (Long) 2LT PS = T PE = 0

Po : TCT~TT (

) (nx)~ 3

:r : (-~---)3n+l ~ Q

(= ~-R-4Q3for Newtonian Fluid) O

(2)

Uniform Circular Die (Zero length) PS = 0 PE : 0

Po

4 V-~ =

BER2 = eCR

9 4Q )

(n~) 89

277

PLASTICSENGINEERING

(3)

Uniform Rectangular S l i t (long) 2LT PS = T PE = 0 PO = - ~

4

(T-~o Q ) (n~) ~

1 BST = [89 { (I + yR2)89 + I__ YR In (YR + (I + yR2) 89 } ] # BSH = BST2 (2n+l)

(4)

2Q

(= ~

for Newtonian Flev)

Uni form Rectangular S l i t (Zero length) PS = 0 PE = 0 4

(=f~z) (nx) 89

Po : TT6-~T

o

C

BET = (e R)~ C

BEH = (e R)89 (5)

Coni-Cyl indrical

2~] [

Die

RI 3n ] PS = 3n tan o 1 - (~oo)

[

PE = 3 o] 4J2

PO : ~ BSR BER

=

1 - (--)3

wh~r~ o I

=

(

) (nl)89 3 YR (I + y~IR )2 - ~

ER ( e ) 89

3n+l "{ = (--~---)

q3 (: 4_q_ ~ 3 for Newtonian flow) ~R

2

(

)

278

R. J. C R A W F O R D

(6)

Wedge Shaped Die

"~I

I

PS = 2n tan ~

PE

~

-

HI

1 - (~o)

2n]

(no)2

I9 892) BST=~I + yR2)~ + l-- In8('YR+9(I +YR BSH = BST2 BET = (eER)88

BEH= BET2 ( = ~ H for Newtonian flow) 1 . = ~ y tan

FURTHER READING I.

J.L. Throne, "Plastics Process Engineering", Marcel Dekker, !979.

2.

R.T. Fenner, "Principles of Polymer Processing", Macmillan, 1979.

3.

Z. Tadmor and C.G. Gogos, "Principles of Polymer Processing", Wiley Interscience, 1979.

4,

J.A. Brydson, "Flow Properties of Polymer Melts", George Godwin~ 1981.

5.

H. Grober, "Fundamentals of Heat Transfer", McGraw-Hill, 1961.

6.

F.N. Cogswell, "Polymer Melt Rheology", George Godwin, 1981.

PLASTICS ENGINEERING

5.0

Questions

5.1 In a p a r t i c u l a r type of cone and p l a te rheometer the torque is applied by means of a weight suspended on a piece of cord. The cord passes over a p u l l e y and is wound around a drum which is on the same axis as the cone. There is a d i r e c t drive between the two. During a t e s t on polythene at 190oc the f o l l o w i n g results were obtained by applying a weight and, when the steady state had been achieved, noting the angle of r o t a t i o n of the cone in 40 seconds. I f the diameter of the cone is 50 mm and i t s angle is 5o , estimate the v i s c o s i t y of the melt at a shear stress of 104 N/m2. Weight (g)

50

Angle (i~~

0.57

I00

200

500

I000

2000

1.25

2.56

7.36

17.0

42.0

5.2 Derive expressions f o r the v e l o c i t y p r o f i l e , shear stress, shear rate and volume flow rate during the isothermal flow of a power law f l u i d in a rectangular section s l i t of width W, depth H and length L. During tests on such a section the f o l l o w i n g data was obtained.

Flow rate (kg/min)

0.21

0.4

0.58 0.8

1.3

2.3

Pressure drop (MPa) 1.8

3.0

4.0

7.6

12.0

5.2

I f the channel has a length of 50 mm, a depth of 2 mm and a width of 6 mm, establish the a p p l i c a b i l i t y of the power law to this f l u i d and determine the relevant constants. The densifv of the f l u i d is 940 kg/m3. 5.3 The v i s c o s i t y c h a r a c t e r i s t i c s of a polymer melt are measured using both a c a p i l l a r y rheometer and a cone and plate viscometer at the same f(~nperature. The c a p i l l a r y is 2.0 mm diameter and 32.0 mm ong. For volumetric flow rates of 70 x 10-9 m3/s and 200 x lG 9 m3/s, the pressures measured j u s t before the entry to the c a p i l l a r y are 3.9 MPa and 5.7 MPa, r e s p e c t i v e l y . The angle between the cone and the p l a te in the viscometer is 3o and the diameter of the base of the cone is 75 mm. When a torque of 1.18 Nm is applied to the cone, the steady rate of r o t a t i o n reached is observed to be 0.062 rad/s. Assuming t h a t the melt v i s c o s i t y is a power law function of the rate of shear, c a l c u l a t e the percentage d i f f e r e n c e in the shear stresses given by the two methods of measurement at the rate of shear obtained in the cone and plate experiment.

279

280

R: J. CRAWFORD

5.4 The correction f a c t o r f o r converting apparent shear rates at the wall of a c i r c u l a r c y l i n d r i c a l c a p i l l a r y to true shear rates is (3n + l ) / 4 n , where n is the power law index of the polymer melt being extruded. Derive a s i m i l a r expression f o r correcting apparent shear rates at the walls of a die whose cross-section is in the form of a very long narrow s l i t . A s l i t die is designed on the assumption that the material is Newtonian, using apparent viscous properties derived from c a p i l l a r y rheometer measurements,at a p a r t i c u l a r wall shear stress,to calculate the volumetric flow rate through the s l i t f o r the same wall shear stress. Using the correction factors already derived, Qbtain an expression f o r the e r r o r involved in t h i s procedure due to the melt being non-Newtonian. Also obtain an expression f o r the e r r o r in pressure drop calculated on the same basis. What is the magnitude of the e r r o r in each case f o r a typical power law index n = 0.37? 5.5 An a c r y l i c moulding material at a temperature of 230~ passes through a c y l i n d r i c a l die of radius 3 mm and length 37.5 mm at a rate of 2.12 x 10-6 m3/s. Using the flow curves supplied, calculate the natural time of the process and comment on the meaning of the value obtained. 5.6 Polythene is passed through a rectangular s l i t die 5 mm wide, 1 mm deep at a rate of 0.75 x 10-9 m~/s. I f the time taken is 1 second, c a l c u l a t e the natural time and comment on i t s meaning. 5.7 In a plunger type i n j e c t i o n moulding machine the torpedo has a length of 30 mm and a diameter of 23 mm. I f , during the moulding of polythene at 170~ (flow curves given), the plunger moves forward at a speed of 50 mm/s estimate the pressure drop along the torpedo. The barrel diameter is 25 mm. 5.8 An a c r y l i c moulding material at 230oc (flow curves supplied) is injected i n t o a mould at a pressure of I00 MPa. I f the mould cavity has the form of a long channel with a rectangular cross-section 6 mm x 1 mm deep, estimate the length of the flow path a f t e r 1 second. The flow may be assumed to be isothermal and over the range of shear rates experienced (I03-I0 s s-z) the material may be considered to be a power law f l u i d . 5.9 Repeat the previous question f o r the s i t u a t i o n in which the mould temperature is 60oc and the f r e e z e - o f f temperature f o r a c r y l i c is 165~ What difference would i t make i f i t had been assumed that the material was Newtonian with a v i s c o s i t y of 1.2 x 102 Ns/mz. 5.10 During the blow moulding of polypropylene b o t t l e s , the parison is extruded at a temperature of 230oc and the mould temperature is 50~ I f the wall thickness of the b o t t l e is

PLASTICS ENGINEERING

281

1 mm and the b o t t l e s can be ejected at a temperature of 120oc estimate the coqling time in the mould. 5.11 For a p a r t i c u l a r polymer melt the power law constants are x~ = 40 kN.sn/m 2 and n = 0.35. i f the polymer flows through an i n j e c t i o n nozzle of diameter 3 mm and length 25 mm at a rate of 5 x I0 -s m3/s, estimate the pressure drop in the nozzle. 5.]2 Polythene at 170~ is used to i n j e c t i o n mould a disc with a diameter of 120 mm and thickness 3 n~. A sprue gate is used to feed the material into the centre of the disc. I f the i n j e c t i o n rate is constant and the c a v i t y is to be f i l l e d in ] second estimate the minimum i n j e c t i o n pressure needed at the nozzle. The flow curves f o r this grade of polythene are given in Fig. 5.17. 5.13 During the i n j e c t i o n moulding o f a polythene container having a volume of 4 x 10-5 m3, the melt temperature is 210oc, the mould temperature is 50oc and rectangular gates with a land length of 0.6 mm are to be used. I f i t is desired to have the melt enter the mould at a shear rate of 6 x 103 s -1 and f r e e z e - o f f at the gate a f t e r 3 seconds, estimate the dimensions of the gate and the pressure drop across i t . It may be assumed that f r e e z e - o f f occurs at a temperature of 135~ The flow curves in Fig. 5. i7 should be used. 5.14 An a c r y l i c moulding powder at a temperature o f 230~ passes through the annular die shown,at a rate of 50 x 10-6 m3/s. Using the flow curves provided and assuming the power law index n = 0.25 over the working section o f the curves, c a l c u l a t e the t o t a l pressure drop through the die. Also estimate the dimensions o f the extruded tube. I

--

-

40mm

--,,....--3'Smm 6mm

15 m m

5.15 A polythene tube of outside diameter 40 mmand wall thickness 0.75 mm is to be extruded at a l i n e a r speed of 15 mm/s. Using the 170oc polythene flow curves supplied, calculate s u i t a b l e die e x i t dimensions.

282

R.J. CRAWFORD

5. i6 An a c r y l i c moulding m a t e r i a l a t 230~ passes along the channel shown a t a r a t e o f 4 x 10-6 m3/s. Using the f l o w curves and assuming n : 0.25 c a l c u l a t e the pressure drop along the channel. all

dimensions

in

mm

..J

5.17 A power law p l a s t i c is i n j e c t e d i n t o a c i r c u l a r s e c t i o n channel using a c o n s t a n t p r e s s u r e , P. Derive an expression for the flow length assuming t h a t (a) the flow i s isothermal (b) the melt is f r e e z i n g off as i t flows along the channel. 5.18 A polymer m e l t i s i n j e c t e d i n t o a c i r c u l a r s e c t i o n channel under c o n s t a n t pressure. ~Jhat i s the r a t i o o f the maximum n o n - i s o t h e r m a l f l o w l e n g t h to the isothermal f l o w l e n g t h i n the same time f o r (a) a Newtonian m e l t and (b) a power law m e l t w i t h i n d e x , n = 0.3. 5.19 A power law f l u i d w i t h the constants n_ = I04 Ns/m 2 and n = 0.3 i s i n j e c t e d i n t o a c i r c u l a r s e c t i o n ~hannel o f diameter I0 mm. &how how the i n j e c t i o n r a t e and i n j e c t i o n pressure vary w i t h time i f I~ I the i n j e c t i o n the i n j e c t i o n

pressure is held constant a t 140 MPa r a t e i s held c o n s t a n t a t 10-3 m3/S.

:he f l o w in each case may be considered to be i s o t h e r m a l . 5.20 An a c r y l i c moulding m a t e r i a l a t 230~ i s used to i n j e c t i o n mould a f l a t plaque measuring 50 mm x I00 mm x 3 mm. A r e c t a n g u l a r gate which i s 4 mm x 2 mm w i t h a land l e n g t h o f 0.6 mm i s s i t u a t e d in the centre o f the 50 mm s i d e . The runners are 8 mm diameter and 200 mm long. The m a t e r i a l passes from the b a r r e l i n t o t h e runners i n 1 second and the pressure losses i n the nozzle and sprue may be taken as the same as those i n the runner. I f the i n j e c t i o n r a t e is f i x e d a t I0 -s m3/s, e s t i m a t e (a) the pressure losses in the runner and gate and (b) the i n i t i a l packing pressure on the moulded plaque. Flow curves f o r a c r y l i c a t 230oc are s u p p l i e d . 5.21 I t i s d e s i r e d to blow mould a c y l i n d r i c a l p l a s t i c cont a i n e r o f diameter I00 mm and w a l l thickness 2.5 mmo I f the e x t r u d e r d i e has an average diameter o f 40 mm and a gap o f

PLASTICS ENGINEERING

283

2 mm, calculate the output rate needed from the extruder. Comment on t h e ~ s u i t a b i l i t y of an i n f l a t i o n pressure in the region of 0.4 MPa. The density of the molten p l a s t i c may be taken as 790 kg/m3. Use the flow curves in Fig. 5.17. 5.22 During the blow moulding of polypropylene at 230oc the parison is 0.4 m long and is l e f t hanging f o r l second. Estimate the natural time f o r the process and the amount of sagging which occurs. The density of the melt may be taken as 730 kg/m 3. 5.23 The v i s c o s i t y , n, of p l a s t i c melt is dependent on temperature, T, and pressure, P. The v a r i a t i o n s f o r some common p l a s t i c s are given by equations of the form n/n R = I0 AAT and n/np = IoBAH where ~T = T-T R (oc), ~P = P-P~ (MPa), and the"subscript R s i g n i f i e s a reference value~ Typical values of the constants A and B are given below.

;,crylic

Polypropylene

A (xlO -3)

-28.32

-7.53

B (xlO -3)

9.54

6.43

LDPE -ii.29 6.02

Nylon

Acetal

-i2.97

-7.53

4.22

3.89

Durin 9 flow along a p a r t i c u l a r channel the temperature drops by 40oC and the pressure drops by 50 MPa. Estimate the overall change in viscosity of the melt in each case. Determine the r a t i o of the pressure change to tile temperature change which would cause no change in v i s c o s i t y f o r each of the above materials.

284

R.J.

10 6

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CRAWFORD

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i

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i_

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!

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---

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I

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@

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7

~

.......

10 a

i flill"~

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,i

IIIII

I

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.......

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@

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10 ~ "

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~:ryLir

(230~C,

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0

\ ~Iiiii

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. . . . .

~ I~IIIIII

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i

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Stress (NIm 2)

A

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io~

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io'

lo 2 Io~

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Io~ (NIm

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