Fundamentals of Polymer Engineering Part 15 doc

Polymer melts and solutions, however,have a steady shear viscosity that depends on the shear rate.. Unlike with Newtonian liquids, the shear stress takes some time to reach a steady valu

Flow Behavior of Polymeric Fluids

In order to use polymers (whether available in the form of pellets or as melt from

a polymerization reactor), the material has to be converted into useful shapes such

as fibers, films, or molded articles This is done using unit operations such as fiberspinning and injection molding, which are analyzed in detail inChapter 15 Here,

we simply mention that flow is an integral part of any shaping operation, and,very frequently, it is useful to know quantities such as the pressure drop needed topump a polymeric fluid at a specified flow rate through a channel of a givengeometry The answer is easy to obtain if we are working with low-molecular-weight liquids that behave in a Newtonian manner; all we need, by way ofmaterial properties, is information about the temperature dependence of the shearviscosity If the process is isothermal, the shear viscosity is a constant and it can

be measured in any one of several ways Polymer melts and solutions, however,have a steady shear viscosity that depends on the shear rate Therefore, it is amaterial function rather than a material constant For polymeric fluids the typicalshape of the steady shear viscosity curve as a function of shear rate is shown inFigure 14.1 At steady state, the viscosity is constant at low shear rates It usuallydecreases with increasing shear rate and often becomes constant again at highshear rates There is, therefore, a lower Newtonian region characterized by thezero shear rate viscosity Z0 and an upper Newtonian region characterized by an

infinite shear viscosity Z1 Between these two regions, the viscosity versus shearrate behavior can usually be represented as a straight line on logarithmiccoordinates—this is the power-law region.

In the foregoing, we have been careful to append the word ‘‘steady’’ to theshear viscosity Unlike with Newtonian liquids, the shear stress takes some time

to reach a steady value upon inception of shear flow at a constant shear rate This

is sketched in Figure 14.2, which shows that the shear stress can also overshootthe steady-state value Polymeric fluids are therefore non-Newtonian in the sensethat the shear viscosity depends on both shear rate and time Obtaining thepressure drop corresponding to a given flow rate is, consequently, a slightly morecomplicated process

FIGURE14.1 Qualitative behavior of the steady shear viscosity of polymeric fluids

FIGURE14.2 Start-up and shutdown of shearing at constant shear rate

A shear-thinning viscosity is not the only non-Newtonian feature of thebehavior of polymeric fluids; several other unusual phenomena are observed If,

in the situation depicted inFigure 14.2, the shear rate is suddenly reduced to zeroafter the attainment of a steady state, low- and high-molecular-weight liquidsagain behave differently The stress in the Newtonian fluid goes to zero instantly,but it takes some time to disappear in the polymer The time scale over which thisstress relaxation occurs is known as the relaxation time and is denoted by thesymbol y Additionally, if a small-amplitude sinusoidal strain is imposed on thepolymer, the resulting stress is neither in phase with the strain nor out of phasewith the strain: There is an out-of-phase component representing energy dissipa-tion and an in-phase component representing energy storage (see Sect 12.4).Both stress relaxation and the phase difference in dynamic experiments are elasticeffects; we say that the polymers are both viscous and elastic (i.e., viscoelastic)

In time-dependent flow, the relative extent of these two effects depends on thevalue of the dimensionless group known as the Deborah number (De) and defined

FIGURE14.3 The die-swell phenomenon

various polymer processing operations Some of these models are discussed inChapter 15, and these are based on the conservation principles of mass,momentum, and energy, together with appropriate constitutive equations andboundary conditions They are useful for process optimization and for determin-ing the effect of the various material, geometrical, and processing variables on theproperties of the polymeric product The models also allow us to relateperformance variables to machine variables They are also useful for predictingthe onset of flow instabilities In this chapter, though, we describe methods ofmeasuring the stress response of polymeric fluids in well-characterized flowsituations, present the associated methods of data analysis, and give typicalresults This naturally leads to a discussion of theories available to explain theobserved behavior in terms of material microstructure and to methods ofmathematically representing the stress–deformation relations or constitutivebehavior This is the realm of rheology, wherein we examine both polymersolutions and polymer melts Note that, at a less fundamental level, thesemeasurements can be employed for product characterization and quality-controlpurposes A succinct treatment may be found in Ref [1].

The flow field that is generated in most standard instruments used to measurerheological or flow properties is a particular kind of shear flow called viscometricflow All of the motion in a viscometric flow, whether in Cartesian or curvilinearcoordinates, is along one coordinate direction (say, x1 inFig 14.5), the velocityvaries along a second coordinate direction (say, x2), and the third direction isneutral An illustration of this, shown in Figure 14.5, is the shearing of a liquidbetween two parallel plates due to the motion of one plate relative to the other.The velocity gradient or shear rate, _gg, then is dv1=dx2, where v1 is the onlynonzero component of the velocity vector If the shear rate is independent ofposition, the flow field is homogeneous and the components of the extra stress[see Eq (10.5.6) ofChap 10] are also independent of position This is intuitivelyobvious (the extra stress depends on the rate of shear strain), and because theshear rate is the same everywhere, so must be the extra stress This fact is used to

FIGURE14.4 Extrudate melt fracture

great advantage in several viscometers or rheometers, as such instruments arecalled.

As discussed in Section 10.5, there are, in general, only six independentstress components For a Newtonian liquid, these can be evaluated usingNewton’s law of viscosity,

from a measurement of Tii However, p can be eliminated if we take stressdifferences We can, therefore, define the first and second normal stress differ-ences as follows:

N1ð_ggÞ ¼ T11
T22¼ t11
t22 ð14:2:3Þ

N2ð_ggÞ ¼ T22
T33¼ t22
t33 ð14:2:4Þand these must depend uniquely on the shear rate _gg, because each of the extra-stress components is a unique function of _gg

The existence of a positive first normal stress difference during shear flowcan be used to explain die swell If the fluid being sheared between parallel plates

in Figure 14.5 were to emerge into the atmosphere, T11 would obviously equal

FIGURE14.5 Viscometric flow in rectangular Cartesian coordinates

pa, where pais atmospheric pressure and is a compressive stress A positive firstnormal stress difference would then imply that T22is negative (compressive) andgreater than pain magnitude In other words, the upper plate pushes down on theliquid being sheared and the liquid pushes up on the plate with a stress thatexceeds pa in magnitude Because only atmospheric pressure acts on the outside

of the upper plate, it has to be held down by an externally applied force to preventthe fluid from pushing the two plates apart When the fluid emerges into theatmosphere, there is no plate present to push down on it, and it, therefore,expands and we observe die swell

Because the shear stress is an odd function of the shear rate and the normalstress differences are even functions, it is customary to define the viscosityfunction and the first and second normal stress coefficients as follows:

by Eqs (14.2.5)–(14.2.7) Note that F equals zero for a Newtonian liquid.This flow is a viscometric flow when viewed in a spherical coordinatesystem, and there is only one nonzero component of the velocity This component

is vf, which varies with both r and y; the streamlines are closed circles If werotate the plate at an angular velocityO, the linear velocity on the plate surface atany radial position is Or On the cone surface at the same radial position,however, the velocity is zero If the cone angle is small, we can assume thatvf

varies linearly across the gap between the cone and the plate The shear rate _gg atany value of r is then given as follows:

is oscillated or given a step strain

If we integrate the shear stress over the cone surface, we can get anexpression for the torque M as follows:

M ¼

ðR 0

In order to obtain the normal stress functions, we need to solve theequations of motion in spherical coordinates [3] An examination of the ycomponent of this equation shows @p=@y ¼ 0 Thus, p depends on r alone,because derivatives with respect to f are zero Further, because most polymerfluids are fairly viscous, we can neglect inertia and, as a result, the r component ofthe equation of motion yields the following [3]:

½
pðrÞ þ tyy2pr dr ¼ 0

or

F þ pR2paþ pR2tyy ¼

ðR 0

Clearly, the first normal stress difference in shear is a positive quantity

Finally, if we were to use a pressure transducer to measure Tyyon the platesurface, we would have

Shown in Figure 14.7 are data for the shear viscosity of a low-densitypolyethylene sample (called IUPAC A) as a function of the shear rate over a range

of temperatures [4] These data were collected as part of an international studythat involved several investigators and numerous instruments in different labora-tories For this polymer sample,
Mnis 2  104and
Mwexceeds 106 It is seen thatthe shear rates attained are low enough that, at each temperature, the zero shearrate viscosity can be identified easily; these Z0 values are noted in Figure 14.7itself Clearly, the viscosity values increase with decreasing temperature If weplot Z0 as a function of the reciprocal of the absolute temperature, we get astraight line, showing that the Arrhenius relation is obeyed; that is,

612, 1975.)

Note, though, that close to the glass transition temperature, the WLF equation[see also Eqs (12.5.4) and (13.6.13) ofChapters 12and 13, respectively]

In addition to temperature, the zero-shear viscosity is also influenced by thepressure, especially at high pressures and especially close to the glass transitiontemperature The pressure dependence is again of the Arrhenius type,

and is the result of the tendency of the free volume to decrease on the application

of a large hydrostatic pressure

The first normal stress difference in shear, N1, on the IUPAC A sample andmeasured at 130C is displayed on logarithmic coordinates in Figure 14.8 If wecompare this figure to Figure 14.7, we determine that N1 is comparable to theshear stress at low shear rates but significantly exceeds this quantity at highershear rates This happens because of the stronger dependence of N1 on _ggcompared with the dependence of tfy on _gg Indeed, straight lines typically

FIGURE14.8 First normal stress difference in shear for IUPAC A LDPE at 130C.(From Ref 4.)

result when these two functions are plotted in terms of the shear rate onlogarithmic coordinates Although the slope of the N1 plot usually lies between

1 and 2, the maximum value of the slope of the shear stress plot is unity.Although all of the data discussed so far have been melt data, the shearbehavior of polymer solutions is similar to that of polymer melts, except that thestress levels are lower The same viscometers are used for both fluids, and thecone-and-plate viscometer has become a standard laboratory tool Indeed, it is theinstrument of choice for making simultaneous measurements of the viscosity andthe first normal stress difference This is because data analysis involves fewassumptions There are limitations, though; we can have viscous-heating effects ifthe viscosity of the fluid is high but, more importantly, centrifugal forces andelastic instabilities cause the sample to be expelled from the gap at high rates ofrevolution This limits the instrument to shear rates less than about 100 sec
1.Difficulties also arise during measurements on filled polymers if the size of theparticulates is comparable to the gap size In such a situation, we use two parallelplates with a large gap This situation can also be analysed [5], but the analysis issignificantly more involved Also, N1 and N2 cannot be obtained separately

Example 14.1: If ln N1 is plotted versus ln tfy, a straight line of slope 2frequently results Furthermore, the same plot is obtained irrespective of thetemperature of measurement If data are obtained with the help of a cone-and-plate viscometer, how would M be related to F for this relationship to hold?Solution: We know that ln N1¼ 2 ln tfyþ ln c, where c is a constant Conse-quently,

N1

t2

fy

¼ cand using Eqs (14.3.3) and (14.3.9) gives

time-shear rate but also at specified values of the time-shear stress; the time-shear rate thenbecomes the dependent variable Other popular rotational viscometers are theparallel-plate viscometer and the Couette viscometer These are described inProblems 14.4 and 14.5, respectively.

The need to measure fluid properties at shear rates higher than those accessiblewith rotational viscometers arises because deformation rates can easily reach

105–106sec
1in polymer processing operations To attain these high shear rates,

we use flow through capillaries or slits and calculate the viscometric functionsfrom a knowledge of the pressure drop-versus-flow rate relationship

Consider the steady flow through a horizontal capillary of circular crosssection, as shown in Figure 14.9 In most of the capillary in a region of length L,away from the exit or inlet, the pressure gradient is constant and there is fullydeveloped flow Thus,Dp=L equals dp=dz In the entry and exit regions, though,there are velocity rearrangements and the actual pressure drops exceed thosecalculated based on fully developed flow To neglect these extra pressure losses,

we need to use capillaries for which L=D exceeds 50, where D is the capillarydiameter and its value is usually greater than about 0.025 cm For polymer melts,though, short capillaries are used because of thermal degradation problems andbecause long capillaries require very high pressures However, methods exist thatcorrect for entrance effects, which are the larger of the two losses [6] Of course,

FIGURE14.9 Schematic diagram of a capillary viscometer

to obtain consistent results, the ratio of the reservoir diameter to the capillarydiameter should also be large—a value of 12–18 is safe.

The fully developed flow region is again a region of viscometric flowbecause only vz is nonzero and it varies only with r Thus, the shear rate _gg is

dvz=dr However, the deformation is not homogeneous because the shear ratevaries from zero at the capillary axis to a maximum at the tube wall To obtain theviscosity, we need to be able to measure the shear stress and the shear rate at thesame location Although the shear stress can be obtained at any radial positionfrom a macroscopic force balance, the shear rate is most easily calculated at thetube wall Note that in the analysis that follows,@=@z is taken to be zero for theextra stresses due to fully developed flow and@=@y is zero due to symmetry From

a macroscopic force balance on the control volume shown in Figure 14.9, wehave

Eliminating Dp in Eq (14.4.5) through the use of Eq (14.4.6) andrearranging the result, we have

d ln Q=d ln tw, which, with the help of Eq (14.4.9), gives _ggðtwÞ Dividing tw

by this value of _ggwgives Zð_ggwÞ By repeating this process, we can get Z at othervalues of _ggw Because it does not matter where the viscosity is measured so long

as the shear stress and shear rate are determined at the same location, we can dropthe subscript w, resulting in the term Zð_ggÞ

For a Newtonian liquid the fully developed velocity profile is parabolic andthe wall shear rate is as follows (see Problem 14.6):

in which K and n are constants Equation (14.4.12) is known as the power-lawequation, K is called the consistency index, and n is called the power-law index.Usually, n 1.

Equation (14.4.12) obviously cannot represent the viscosity over the entirerange of shear rates, because it suggests the following:

Z ¼ K

which predicts a meaningless increase in the viscosity as the shear rate isdecreased A popular alternative model that does not have this shortcoming isthe Carreau model [7],

Z
Z1

Z0
Z1¼ ½1 þ ðlc_ggÞ2ðn
1Þ=2 ð14:4:14Þwhich has four constants: Z0, Z1, lc, and n Although this model can accom-modate a limiting shear viscosity at both high and low shear rates, the flexibilitycomes at the expense of greater complexity Note that the parameter lc

determines the point of onset of shear thinning; the onset of shear thinning isfound to move to larger values of the shear rate as the temperature of measure-ment is increased (see Fig 14.7)

FIGURE14.10 Viscosity functions for samples A, B, and C at 150C Ia, IVa:Weissenberg rheogoniometer (Ia measures at 130C and shifts the data); II: Kepesrotational rheometer (cone-and-plate); Ib, IVb: capillary viscometers MF denotes theonset of melt fracture (From Ref 4.) (Reprinted with permission from Meissner, J.: ‘‘BasicParameters, Melt Rheology, Processing and End-Use Properties of Three Similar LowDensity Polyethylene Samples,’’ Pure Appl Chem., vol 42, pp 553–612, 1975.)

Example 14.2: The (unstable) melt fracture behavior illustrated inFigure 14.4isfound to occur in polymer melts when the wall shear stress is about 106dyn=cm2

If we were to extrude IUPAC A LDPE through a long capillary 0.025 cm diameter

at 150C, at what flow rate might we expect to observe melt fracture?

Solution: By trial and error, we find that the shear rate at which Z_gg equals

106 dyn=cm2

inFigure 14.10is about 100 sec
1 (as indicated in the figure, meltfracture actually occurs at slightly lower shear rates)

If we represent this polymer as a power-law fluid, the shear rate at the wall

is given by the following (see Problem 14.7):

_ggw¼ 4Q

pR3

3n þ 14n

a pressure-driven instrument Of course, it must be ensured that viscous tion does not lead to nonisothermal conditions; this computation is easy to carryout [8] The lower limit of operation of a plunger-type viscometer is set by theamount of friction between the plunger and the barrel As the plunger speeddecreases, this friction can become a significant part of the totalDp and make theviscosity measurements unreliable More details about capillary viscometry areavailable in the article by Kestin et al [9] The theory for flow through slits issimilar to that for circular tubes [10] An advantage of slit viscometry, as shown

dissipa-by Lodge, is that the technique can also be used to measure normal stressdifferences [10]

14.5 EXTENSIONAL VISCOMETERS

Besides viscometric flow, the other major category of flow that can be generated

in the laboratory is extensional flow In mathematical terms, extensional flow may

be represented in a rectangular Cartesian coordinate system xi by the set ofequations for the three components of the velocity vector

In physical terms, the distance between material planes that are cular to the flow direction increases in extensional flow This is illustrated foruniaxial extension in Figure 14.11a A material plane is a surface that alwayscontains the same material points or particles In viscometric flow, the fluid onany surface for which the coordinate x2is a constant forms a material plane (seeFigure 14.11b) Here, each material plane moves as a rigid body and there is

perpendi-FIGURE14.11 (a) Increase in distance between material planes in a uniaxial sional flow; (b) Sliding of material planes in viscometric flow (From Ref 11.)

exten-relative sliding between neighboring material planes In extensional flow, asopposed to shear flow, polymer molecules tend to uncoil and ultimately there caneven be stretching of chemical bonds, which results in chain scission Therefore,stresses in the flow direction can reach fairly large values.

Although it is somewhat difficult to visualize how a rod of polymer might

be stretched in the manner ofFigure 14.11a, this can be done for both polymermelts [11] and polymer solutions [12] For polymer melts, a cylindrical sample isimmersed in an oil bath with one end attached to a force transducer and the otherend moved outward so that the stretch rate is maintained constant Similarly, forpolymer solutions, we merely place the liquid sample between two coaxial disks,one of which is stationary and connected to a microbalance, and the other diskmoves outward, generating the stretching The filament diameter thins progres-sively but remains independent of position Simultaneously, the filament length lincreases exponentially as follows:

ln l

l0

 

which follows directly from Eq (14.5.1)

In this flow field, there is no shear deformation, and the total stress tensor aswell as the extra stress tensor are diagonal As a consequence, there are only threenonzero stress components, but, due to fluid incompressibility, we can measureonly two stress differences Further, in uniaxial extension, the two directions thatare perpendicular to the stretching direction are identical, so there is only onemeasurable material function: the net tensile stress sE, which is the difference

where Z0 is zero-shear rate viscosity

Laun and Munstedt have obtained tensile stress growth data on the IUPAC

A LDPE sample at 150C [13,14], and these are shown inFigure 14.12 At agiven stretch rate, the stress increases monotonically and, ultimately, tends to alimiting value even for stretch rates as high as 10 sec
1 The extensionalviscosity, calculated using these steady-state data, is displayed in Figure 14.13

as a function of the stretch rate, and at low stretch rates, Eq (14.5.6) is found tohold As the stretch rate is increased, the extensional viscosity increases, goesthrough a maximum, and finally decreases to a value even below that of the zero-shear value The behavior of polymer solutions is qualitatively similar to that ofpolymer melts, except that the extensional viscosity of polymer solutions canexceed the corresponding shear viscosity by a far wider margin than does theextensional viscosity of polymer melts [15].

Example 14.3: How is the extensional viscosity of a Newtonian fluid related tothe shear viscosity?

Solution: Using Eqs (14.2.2), (14.5.1), and (14.5.2) gives

For uniaxial extension, however,_ee1¼
2_ee2 ¼_ee Therefore sE ¼ 2Z½_ee þ ð_ee=2Þ ¼3Z_ee and ZE ¼ 3Z.

Even though commercial instruments are available for making extensionalviscosity measurements on polymer melts, this is not a routine measurement Thestretch-rate range of these extensional viscometers is such that the maximumstretch rate that can be achieved is of the order of 1 sec
1; in polymer processingoperations, a stretch rate of 100 sec
1 is commonplace Also, not every polymerstretches uniformly, and, even when it does, steady-state stress levels are notalways attained For all of these reasons, extensional viscometry is an area ofcurrent research Additional details regarding extensional and other viscometersmay be found in the book by Dealy [16]

Experimental data obtained in viscometric flow or extensional flow are obviouslyuseful for predicting material behavior in a flow field that is predominantly shear

or extension Thus, the shear viscosity function is needed to compute the pressure

FIGURE14.13 Steady-state viscosity in shear and elongation as a function ofdeformation rate (From Ref 14.) Reprinted from Rheol Acta, vol 17 Laun, H M.,and H Munstedt: Elongational behavior of an LDPE melt: I Strain rate and stressdependence of viscosity and recoverable strain in steady state Comparison with shear data.Influence of interfacial tension, 415–425, 1978 Used by permission of SteinkopffPublishers, Darmstadt (FRG)

be employed to calculate the force of peel adhesion in a pressure-sensitiveadhesive In addition, the data reveal material behavior and can be used formaterial characterization and product comparison However, it is logical to askwhether it is necessary to conduct a new experiment every time we wantinformation in a different flow field or if we want transient data in the sameflow field The answer, we hope, is that behavior in one flow situation can bepredicted using data generated in a different flow situation In other words, astress constitutive equation for non-Newtonian fluids is sought This equationwould involve a limited number of constants or functions that would be specific

to the material being examined and whose numerical values could be determined

by conducting one or two experiments in idealized flow fields In this section, weshow how this goal can be partially achieved

As seen earlier in Problem 12.3 ofChapter 12and more recently in Figure14.2, the stress in a polymeric fluid does not decay to zero once deformation ishalted Instead, stress relaxation occurs Thus, upon imposition of a shear strain g,

we expect the shear stress t to be

in which GðtÞ is a modulus function that decays with time

If a series of strains g1, g2; is imposed on the material at times t1, t2;

in the past, the stress at the present time t is a combination of the stresses resultingfrom each of these strains If this combination can be taken to be a linearcombination (which is the thesis of the Boltzmann superposition principle), wecan write the following expression:

tðtÞ ¼ Gðt
t1Þg1þ Gðt
t2Þg2þ    ð14:6:2Þwhere t
ti is the time elapsed since the imposition of strain gi Converting thesum in Eq (14.6.2) into an integral gives

tijðtÞ ¼

ðt

with the deformation rate components _ggijbeing ð@vi=@xjþ @vj=@xiÞ

Example 14.4: Is a shear-thinning viscosity consistent with the predictions of

Solution: If the shear rate is held constant, the shear viscosity according to Eq.(14.6.3) is given by

Gðt0Þ dt0¼ constantand, not surprisingly, we find that a linear superposition of stresses does not allowfor nonlinear effects such as shear thinning Shear thinning is a nonlinear effectbecause the shear stress is less than doubled when we double the shear rate Thus,the viscosity calculated in this example is the zero-shear viscosity

The Boltzmann superposition principle is the embodiment of the theory oflinear viscoelasticity, and it is valid for both steady and transient deformations,provided that the extent of deformation is low A specific form of the stress–relaxation modulus may be obtained by permitting the stress response in apolymer to be made up of an elastic contribution and a viscous contribution.Thus, if we again use the Maxwell element encountered previously in Example12.2 and Figure 12.4 ofChapter 12, the total strain g in the spring and dashpotcombination is (at any time) a sum of the individual strains; that is,

where the subscripts s and d denote the spring and dashpot, respectively.Because the applied stress t equals the stress in both the spring and thedashpot, Eq (14.6.5) implies (see Example 12.1) the following:

dg

dt¼dgs

dt þdgd

dt ¼ 1G

Comparing Eqs (14.6.3) and (14.6.7) yields

Gðt
sÞ ¼ G exp
t
s

y

ð14:6:8Þwhich, when introduced into Eq (14.6.1), reveals that

tðtÞ

and it is seen that the stress decays over a time scale of the order of y.Consequently, y is called a relaxation time Although a single relaxation timecan be expected to fit data on monodisperse polymers, most polymer samples arepolydisperse It is for this reason that we modify the mechanical analog to make itconsist of N Maxwell elements in parallel; each spring has a modulus Giand eachdashpot has a damping constant Zi By repeating the analysis presented here, wefind that

in which yi¼ Zi=Gi The constants Giand yiare usually obtained from the results

of dynamic mechanical experiments rather than by conducting a stress–relaxationexperiment

As discussed in Section 12.4 of Chapter 12, dynamic mechanical testing isconducted by subjecting a polymer sample to a sinusoidal strain of amplitude g0and frequency o Because the strain amplitude is usually of infinitesimalmagnitude, we can legitimately apply the Boltzmann superposition principle tothis flow situation Before doing this, though, we change the independent variable

in Eq (14.6.3) from s to t0, where t0¼ t
s Thus, we have

tðtÞ ¼

ð1 0

Gðt0Þ cosðot0Þ dt0þ o sinðotÞ

ð1 0

Gðt0Þ sinðot0Þ dt0

ð14:7:2Þwhich when compared to Eq (12.4.4) yields the following expressions for thestorage and loss moduli, respectively:

G0ðoÞ ¼ o

ð1 0

Solution: As o ! 0, sinðot0Þ ! ot0and cosðot0Þ ¼ 1, with the result that

lim

o!0 G0ðoÞ ¼ ½

ð1 0

lim

o!0 G00ðoÞ ¼ ½

ð1 0

In order to accomplish these objectives, we need methods of interrelatingthe various linear viscoelastic functions The general techniques of obtaining onefunction from another have been discussed by Ferry [17] In the present case,Baumgaertel and Winter have proposed a particularly simple method [18] If weintroduce Eq (14.6.10) into Eqs (14.7.3) and (14.7.4) and carry out theintegrations, we get (see also Example 12.2)

The terms Giand yiare found by simply fitting Eqs (14.7.9) and (14.7.10)

to measured G0ðoÞ and G00ðoÞ data using a nonlinear least-squares procedure[18] In doing so, the choice of N is crucial; a small value of N can lead to errors,whereas a large value of N cannot be justified given the normal errors associatedwith measuring G0and G00 The set of relaxation times yiand moduli Giis called

a discrete-relaxation time spectrum, and we typically choose between one and tworelaxation modes per decade of frequency Figure 14.14 shows a master curve of

G0and G00 values in a temperature range of 130–250C on an injection-moldinggrade sample of polystyrene; all of the data have been combined by means of ahorizontal shift using time–temperature superposition with a 150C referencetemperature according to the procedure of Section 12.5 The stress–relaxationmodulus (calculated in the manner of Baumgaertel and Winter) is displayed inFigure 14.15 along with actual stress–relaxation data at 150C; the agreementcould not be better

The utility of dynamic data appears to go beyond the theoretical tions considered in this section We find, for example, that the modulus of thecomplex viscosity Z*, defined as

applica-jZ*j ¼ ½ðZ0Þ2þ ðZ00Þ21 =2 ð14:7:11Þwhere Z0¼ G00=o and Z00¼ G0=o, when plotted versus frequency, often super-poses with the steady shear viscosity as a function of the shear rate [19] This isknown as the Cox–Merz rule, and it provides information about a nonlinearproperty from a measurement of a linear property Note that Z0is generally calledthe dynamic viscosity

FIGURE14.14 Master curve of storage and loss moduli of polystyrene

Another useful empiricism, Laun’s rule, relates the first normal stresscoefficient to dynamic data [20]:

The shear viscosity of polymer melts depends primarily on the molecular weight,the temperature, and the imposed shear rate; for polymer solutions, the concen-tration and nature of solvent are additional variables In one of the earliesttheories, [21,22], the shear viscosity was calculated by determining the amount ofenergy dissipated due to fluid friction in a steady laminar shearing flow at aconstant shear rate_gg For this flow situation, depicted inFigure 14.16, the energydissipated per unit time, per unit volume, P, is given as follows [23]:

Because ðdv1=dx2Þ is_gg, the viscosity is known if P can be computed P, in turn, iscalculated as the dot product, or scalar product, of the drag force with the velocityvector.

If, in this flow field, we consider the end-to-end vector of the typicalpolymer molecule, the two ends move at different velocities in the x1 directiondue to the presence of a velocity gradient The analysis is simplified if we bringthe center of mass of the molecule to rest by imposing the average velocity in thenegative x1 direction In this case, as is clear from Figure 14.16, the moleculerotates in the clockwise direction At steady state, the torque due to the imposedshear flow balances the opposing torque due to fluid friction and there is noangular acceleration The polymer molecule, as represented by the end-to-endvector, rotates at a constant angular velocity o The linear velocity at any radialposition r is or The drag force F acting on any segment of polymer (say, amonomer unit) is as follows:

For all of the n segments taken together,

F  v ¼n2l2z_gg2

To obtain the energy dissipated per unit volume, we need to know thenumber of polymer molecules per unit volume For melts, this is rNA=M and forsolutions it is cNA=M, where r is density, c is the concentration in mass units, NA

is Avogadro’s number, and M is the molecular weight (assuming monodispersepolymer) Multiplying the right-hand side of Eq (14.8.7) by either rNA=M or

nl2

M

 

and Z is predicted to be proportional to M or cM because nl2=M is not a function

of M Although the dependence on temperature can enter through z, it is obviousthat Eq (14.8.9) cannot predict shear thinning, which is almost always observed

at high enough shear rates

When experimental data for the zero-shear viscosity of polymer melts andsolutions are examined in light of Eq (14.8.9), it is found that melts exhibit theexpected proportionality with molecular weight (at least for low molecularweights), whereas solutions show a much stronger dependence of the viscosity

on molecular weight The unexpected behavior of solutions apparently resultsfrom a dependence of z on M and c When data are corrected to give viscosities at

be stretched in the manner ofFigure 14.11a, this can be done for both polymermelts [11] and polymer solutions [12] For polymer melts, a cylindrical... expense of greater complexity Note that the parameter lc

determines the point of onset of shear thinning; the onset of shear thinning isfound to move to larger values of the