From Polymers to Plastics Problems and Solved Exercises

7/28/2019 From Polymers to Plastics Problems and Solved Exercises

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ExercisesFrom polymers to

plastics

A.K. van der Vegt

DUP Blue Print


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NUGI 83 1Key word : polymers


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Contents

1. INTRODUCTION 5

2. M OLECULAR COMPOSITION 7

3. G LASSY STATE AND GLASS -RUBBER TRANSITION 13

4. C RYSTALLINE POLYMERS 15

5. R UBBERY AND LIQUID PHASES 19

6. V ISCO -ELASTICITY 23

7. M ECHANICAL PROPERTIES 268. F URTHER PROPERTIES 30

9. P OLYMERIC COMPOUNDS AND COMPOSITES 32

11. P RINCIPLES OF PROCESSING 36

ANSWERS 411. Introduction 412. Molecular composition 44

3. Glassy state and glass-rubber transition 524. Crystalline polymers 565. Rubbertoestand en vloeibare toestand 636. Visco-elasticity 677. Mechanical properties 738. Further properties 789. Polymeric compounds and composites 8111. Processing techniques 86


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1Introduction

1.1. Give twenty high molecular materials from a natural (both from vegetable andanimal) origin, that are used as technical materials

1.2. Give ten technically used materials based on half synthetic polymers andexplain the addition half to it.

1.3. Why are hydrocarbons, obtained from a cracking process, very suitable for theproduction of synthetic polymers?

1.4. Why are polymers, built up from dienes, very suitable for application astechnical rubbers?

1.5. Which kind of chainstructures can be formed in the polymerisation of butadiene(CH 2=CHCH=CH 2)?

1.6. Why is it not possible to vulcanise ethylene-propylene copolymers and poly-isobutylene in the usual way. In what way are these polymers modified in order tomake vulcanisation possible anyway?

1.7. In what way can polymers be formed from saturated monomers?

1.8. In which two manners can a three dimensional network be formed?

1.9. What is the meaning of the name thermosets? Why can this name lead to

misunderstanding?

1.10. What is the difference between thermoplastics on the one side and thermosetsand rubbers on the other side?

1.11. What is the difference between thermosets and technical rubbers?

1.12. What is the difference between thermoplastic rubbers and ordinary rubbers?

1.13. What is the difference between thermoplastic rubbers and thermoplastics?

1.14. Give the most important thermoplastics, thermosets and synthetic rubbers.

1.15. Give the three maintypes of PE. What are the differences between these types.

1.16. Give a number of polymers (thermoplastics, thermosets and rubbers) in which


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6 Problems Polymers

styrene plays a role.

1.17. Which polymer cannot be manufactured to an end product in a normalproduction process? In what way is this polymer being processed? Which similarpolymers are more suitable for production processes?

1.18. Give a polymer with an extremely high temperature resistance and somemembers of the same family.

1.19. Give a number of composite plastics

1.20. When and for what purpose are the following materials added to a polymer?

a. fine carbon black b. chalk c. glassfibresd. plasticizer e. mica f. acceleratorsg. lubricants h. sulphur i. pentane

j. wood flour k. anti-oxidants l. silicium carbidem. UV-stabilisers n. antistatic agents o. rubber particles.


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2Molecular composition2.1. In what aspects can the mainchains of polymers differ?

2.2. a . Give the name of four polymers with sidegroups only consisting of hydrogen.b. Give the name of some vinylpolymers of the type -CH 2-CHR-.c. Give the name of some vinylidenepolymers, -CH 2-CR 2-.

2.3. Of three polymers, PE, PS and PA-12 the degree of polymerisation is 500. Calculate their molar masses. Calculate for each polymer the contour length. For which types of PE and PS the contour length calculation is valid?

2.4. If we lay a chain with molar mass of 80,000 kg/kmol next to a chain with molarmass of 120,000 kg/kmol, why is the average molar mass not 100,000 kg/kmol?What is the real meaning of the value 100,000 kg/kmol and what is the real value of M w?

2.5. Give a definition of n i, w i, z i, and

M n ,

M w en

M z. What is the value of ni, wiand zi?

2.6. Show thatM n can be calculated from the weight fractions w i through

1/

M n = (wi /M i).

2.7. Show that

M w

M n = niM i2 .

2.8. Show that M w M z = wi M i2 .

2.9. a . Give a description of a monodisperse polymer.b. Give the definition of the degree of polymerisation.

2.10. How will an originally very narrow molar mass distribution change if thepolymer is being broken down gradually, for example under the influence of oxygen?

2.11. We mix two polydisperse polymers, A and B with average molar masses: A:(

M n)A, (

M w)A and (

M z)A; B: (

M n)B, (

M w)B and (

M z)B. The mass fractions of A and Bare wA and wB. Give a formula from which the

M w can be calculated.

2.12. We have three batches, A, B and C of a polymer, each with a

M w of of 60,000kg/kmol, while the heterogenity indices ( D =

M w /

M n) are 6.3 and 2, respectively. 10


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8 Problems Polymers

kg of A is mixed with 10 kg of B. How many kg of C must be added to this mixtureto obtain a mixture with D= 4?

2.13. Equal weight quantities of two polydisperse polymers with M n of 15,000 and10,000 respectively,

M w of 50,000 and 60,000 respectively and

M z of 100,000 and

200,000 kg/kmol are mixed. Calculate the

M n ,

M w and

M z of the mixture.

2.14. Three monodisperse polystyrene fractions A, B and C have molar masses of 76,000 , 250,000 and 850,000 kg/kmol respectively. The intrinsic viscosities [ ], asmeasured in toluene, are 0.382, 0.91 en 2.21 dl/g. Determine the values of theconstants k and a in the Mark-Houwink equation from these data.

2.15. From the fractions in 2.14, 37 grammes of A, 18 grammes of B and 45grammes of C are mixed. Calculate

M w and

M n of the mixture.

2.16. From the mixture of 2.15 solutions in toluene are being made withconcentrations of 0, 0.1, 0.2, 0.3 and 0.4 g/dl. These fluids need 95.0, 107.7,121.4, 136.3, and 152.8 seconds to run through an Ubbelohde viscosimeter.Determine the intrinsic viscosity of the mixture and from the intrinsic viscosity thethe average molar mass. Compare the value with the values calculated in 2.15.

2.17. Calculate the M v of the mixture, without the determination of the [ ].

2.18. Eight fractions of polystyrene with very narrow molar mass distributions aremixed

fraction nr 1 2 3 4 5 6 7 8mass (g) 10 18 25 17 12 8 6 4

M (kg/mol) 15 27 39 56 78 104 120 153

a . To which particular fraction can the mixture be compared best with respect tomelt-viscosity?b. What is the difference between the injection moulding behaviour of this

particular fraction in comparison with that of the mixture?c. To which particular fraction can the mixture be best compared with respect to

impact strength?d . To which particular fraction can the mixture be compared best with respect to

die-swell?e. To which particular fraction can the mixture be compared best with respect to

modulus of elasticity in the solid state?

2.19. 50 kg of a monodisperse polymer with M = 60,000 kg/kmol is mixed with acertain quantity of a comparable, also monodisperse polymer with M = 600,000kg/kmol


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Molecular composition 9

a . How many kg of the latter polymer are needed to obtain a number average molarmass of 100,000?

b. What will then be the values of weight average and z-average molar masses.2.20. For the end-to-end distance in a coil of a polystyrene with M = 1000,000kg/kmol it can be determined that ( r 0

2 )1/2 = 73,5 nm. Calculate the coil density andcompare this value with the density of polystyrene 1050 kg/m 3 .

2.21. Calculate the degree of dilution for a polystyrene with M= 10,000 kg/kmol.

2.22. What are the consequences from the results found in 2.20 and 2.21 for the so-called dilute solution (a solution in which the coils do not penetrate into each other)?

2.23. In what way will this influence the measurement of the intrinsic viscosity?

2.24. Two batches of PE, A and B, with M w = 200,000 and 30,000 kg/kmol

respectively are being mixed with the intention to obtain a mixture with

M w =81,000. Both batches have the same form of molar mass distribution, with D =

M w /

M n = 5, and

M z /

M w = 3.Calculate how many kg of each component is needed to obtain 1000 kg of thedesired mixture. Calculate the

M n and

M z of the mixture. To what extent will the

properties of this mixture differ from a directly produced batch with M w = 81,000and the same distribution width as A and B?

2.25. The intrinsic viscosity is related to the molar mass by the Mark-Houwink equation: [ ] = kM a.a . In what range can the value of a usually be found?b. What is the relation of

M v of a polydisperse polymer, calculated from the Mark-

Houwink equation, to

M n and

M w?

c. In which case can

M v be higher than

M w?

2.26. We have two white cubes. One cube is a polymer. The other one is a lowmolecular substance. Both cubes can be solved in the same solvent. Give somesimple ways how to determine in the laboratory which of the two cubes is made of the polymer?

2.27. Explain why the heterogenity index D =

M w /

M n is approximately 2 for anideal polycondensation process.

2.28. Equal molar quantities of 1,2 ethanediole and benzene-1,4-dicarbonic acid aremixed in order to react to a polymer.a. What is the mass ratio of the components?b. Give the equation of reaction. What is the name of the polymer?c. What is the meaning of the symbols in the formula P = 1/(1 p)?


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10 Problems Polymers

d. Which conversion is needed to reach a

M n of 40,000 kg/kmol? What will then bethe value of

M w?

e. Which circumstances can prevent that the desired conversion will be reached?2.29. Which type of polymerisation process will result in a polymer with a D-valueclose to 1?

2.30. Which methods can be use for the measurement of

M n , of

M w and

M z andgive a description of the principles on which these methods are based.

2.31. For a polymer in solution the following data were found from a viscometerexperiment:

run time solvent 100 secrun time solution 0.05 g/dl 106.25 sec

0.1 g/dl 115 sec0.15 g/dl 126.25 sec0.2 g/dl 140 sec

Calculate the intrinsic viscosity from these data.

2.32. If the value of [ ] for the same type of polymer with M = 50,000 and M =800,000 is 0.5 and 2.0 dl/g respectively, what is the value of M of the sample underinvestigation?

2.33. Calculate the values of the constants in the Mark-Houwink equation for thepolymer solvent combination from 2.31.

2.34. a . A monodisperse polymer with M = 20,000 is contaminated with 2% (w/w)polymer of M = 200. Calculate the values of M n ,

M w and

M z.

b. Now the monodisperse polymer contains 2% (w/w) chains with M = 2,000,000.Calculate also the values of M n ,

M w and

M z for this contaminated polymer.

2.35. Osmotic pressures are measured at 25 C from a dissolved polymer at variousconcentrations c and the following fluid heights were found:

c (g/dl) h (mm)0.12 6.500.18 9.940.25 14.050.33 18.80.45 26.6

Determine

M n from these data. The solvent density is 903 kg/m 3, R = 8.314 J/mol.K


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Molecular composition 11

and g = 9.81 m/s 2.

2.36 . The two figures give imaginary results of osmotic pressure measurements( /c versus c) and solution viscosity measurements ( red versus c), each with twodifferent solvents A and B. Explain why in the case of the osmotic pressuremeasurements extrapolation leads to the same value and in the case of the viscositymeasurements extrapolation does not lead to the same value.

c

c

c

red

Figure 2.1. Figure for problem 2.36

2.37. Why does the sensitivity of the determination of the intrinsic viscosity

increase with increasing molar mass but decrease with the determination of theosmotic pressure?

2.38. Nylon 6.6 is produced twice by an ideal polycondensation process from adipicacid, HOOC-(CH 2)4-COOH, and hexamethylenediamine H 2N-(CH 2)6-NH 2. In thefirst case (A) the conversion p = 0.99, in the second case (B) the conversion is 0.995.Calculate

M n and

M w for both cases.

2.39. 30% of A and 70% of B are mixed. Calculate

M n and

M w for this mixture.

2.40. How can, without measuring the viscosities of solutions at variousconcentrations, a simple comparative measure be found for the average molar mass?

2.41. For PVC k-values are used, like k = 55, k = 70. What is the meaning of thesevalues both molecular and technological?

2.42. Why is the melt-index, as used for e.g. PE, PP and ABS a useful andfunctional measure for the molar mass?

2.43. A light scattering experiment applied to two types, A and B, of the samepolymer results in the same value for M for both types. The value of the intrinsicviscosity of type A is however higher than that of type B. What can be said about theratio of the osmotic pressures of A and B?


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2.44. Which principle is the basis of GPC (Gel Permeation Chromatography orSEC, Size Exclusion Chromatography) How can the various average molar masses

like

M n ,

M w and

M z be determined from the result of a GPC-measurement?2.45. In what aspects can a polymer chain be built up irregularly? Give examples of each aspect.

2.46. Give schematically the difference in the structure of an isotactic, an atacticand a syndiotactic polymer formed from the monomer CH 2=CHX.

2.47. What are the consequences of the steric configuration of the chain for practicalusability and for the properties if X is CH 3 or C 6H5? (T g of PP and PS are 15 Cand 95 C respectively)

2.48. Give two factors which cause a high chain stiffness. Give an example of eachfactor.

2.49. Give two strong and one weak interaction force between chains and giveexamples of polymers in which these forces occur.

2.50. What is being understood by the expression cross-link density? Give another

measure to characterize the degree of crosslinking. How can these properties bedetermined?

2.51. What is the difference between chemical and physical cross-linking. Giveexamples of both types. In what way does the difference show up in the propertiesof the network?


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3Glassy state and glass-rubbertransition3.1. Give differences and similarities between ordinary anorganic glass and apolymeric glass like polystyrene.

3.2. Give three essential differences between a glass-rubber transition and a meltingpoint, in particular concerning the temperature dependency of volume V , enthalpy Hand entropy S around these transition points..

3.3. Why can, just above the T g , complete chains not yet move with respect to eachother? When are they able to do so?

3.4. Give examples of the phenomenon that enlarging of the distance between thechains leads to lower T g-values.

3.5. Why can the glass-rubber transition not be considered as thermodynamic second-order transition?

3.6. Polyhexene-1 and polymethylenepentene both have a C 4H9-sidegroupconnected to the main chain. The T g-values are 50 C and +30 C respectively.Explain the difference.

3.7. If we look at the following series:

PE (-CH 2-CH 2-) T g = 120 CPP (-CH 2-CHCH 3-) = 15 CPVC (-CH 2-CHCl-) = 90 C

a strong rise of the glass-rubber transition temperatures can be seen. Explain thedifferences.Let us now consider the following series.

BR (-CH 2-CH=CH-CH 2) T g = 90 CIR (-CH

2-CCH

3=CH-CH

2-) = 73 C

CR (-CH 2-CCl=CH-CH 2-) = 50 C

Explain why the differences between the glass-rubber transition temperatures arenow considerably smaller.


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14 Problems polymers

3.8. We compare the T g of PMMA (110 C) first with the T g of poly methyl acrylate(5 C), then with the T g of poly (n-butyl methacrylate)(25 C) and finally with the T gof poly(n-octylmethacrylate)(-20 C). Explain the differences.

3.9. An amorphous plastic sometimes shows two transitions, within or from theglassy state. Give three possible causes for this phenomenum.

3.10. The T g of an SBR is 65 C. Which factors determine this value?SBS shows two glass-rubber transitions, one at 90 C and one at +85 C.Explain these values if you know that the T g of PS is 95 C.

3.11. What analogy is there, with respect to glass-rubber transitions, betweencopolymers and polymerblends?

3.12. If a single glass-rubber transition is present does this mean that the polymer iscompletely mixed on a molecular scale?

3.13. Why does crosslinking influence the value of T g?

3.14. A polybutadiene rubber (BR) with molar mass of 200,000 kg/kmol is beingvulcanised with sulphur in such a way that the average number of connections that

each chain has with another one is four. Explain why the T g of this vulcanised rubberdoes not differ significantly from the T g of the one that does not contain crosslinks.

3.15. Why is the T g of a polyamide variable?


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4Crystalline polymers4 . 1. Give a reason for the fact that some polymers with regular chains, like PC, PPE,PIB, cis-polyisoprene and cis-polybutadiene do not crystallize.

4 . 2. Why is not desirable that these polymer should crystallize?

4 . 3. In the production of IR and BR one aims at the highest possible cis-content or inother words, a maximum in chain regularity, although crystallization (see Question4.2) is not desired! Explain.

4 . 4. Give the thermodynamic formula for the melting point and describe the symbolsin this formula. Describe also the relation between these quantities and the chainproperties.

4 . 5. Estimate the melting point of PVC (with a regular chain) when T g = 85 C.

Would a stereospecific PVC be a useful thermoplastic?4.6. For the melting points of some members of the alkane-range (C nH2n+2 ),unbranched, the following values were found :

n = 8: 56.8 n = 10: 29.7 n = 15: 10.0 n = 20: 36.66 n = 32: 69.7

Estimate from these values the melting point of PE with the formula:

1

T m= a +

b

M

Discuss the question whether in practice the melting point of PE will be dependenton the chain length.

4 . 7. What is the reason for the fact that polymers cannot be entirely in a crystallinestate? Give an example where, exceptionally, a level of almost entire crystallisationis achieved. How is this being realised?

4 . 8. Two samples of PP from the same batch have melting points of 161 C and 165C respectively. What is the reason for this difference?

4 . 9. How would the width of the melting-range differ for the two samples from 4.8?

4 . 10. The melting points of PETB and PBTB are 260 C and 210 C respectively.What causes this difference? Why is PBTB preferred to PETB for a number of


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applications like injection moulding products?

4 . 11. How will it be possible to make semi-crystalline products anyway , in injectionmoulding products, from PETP?

4 . 12. In which processes can PETP be brought to rapid crystallisation in anotherway?

4 . 13. With respect to problem 4.12 which analogy can be found with rubbers?

4 . 14. What is the critical nucleus size r c? Which, counteracting, phenomenadetermine r c? Which factor in the formula for r c is the most important? In what way

is this property present in the formula?

4 . 15. The maximum crystallisation speed v cr for a polymer with M = 87,000appeared to be 98 m/min. and with M = 143,000 it is 23 m/min. Explain thedifference in a quantitave way.

4 . 16. Polyamides (nylons) should crystallize. Why? As an exception, alsoamorphous polyamides exist. What are the requirements for the chain structure toachieve this?

4 . 17. Why do the values of T g and T m of polyamides (PA-s) decrease with increasingn?Why does the curve of T m as a function of n show a zig-zag form?

4 . 18. A non-crystallizing polymer is being :

A. Slowly cooled down from T 1 = T g + 50 K to T 2 = T g 50 K,B. Subsequently heated slowly to T 1 .C. Subsequently cooled down rapidly to T 2 .

D. Held at T 2 for a long time.

- Show in a schematic diagram how the volumes changes as a function of T for A,B, C and D.

- Do the same for a crystalllizable polymer, for which now T 1 = T m + 50 K and T 2= T m 50 K.

4.19. Polyethylene-terephtalate is being treated from the melt in the followingmanner:

A. Cooled down rapidly to room temperature: It appears to be amorphous now andto stay amorphous for an unlimited period of time.

B. Raised to a higher temperature, e.g. 80 C and kept at this value for a longerperiod of time; it now crystallizes partly.

C. Cooled down to20 C,D. Heated above the melting point.


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Crystalline polymers 17

a . What conclusion can be drawn from the observations about the value of T g?b. Estimate in which temperature range T m will be.c. Explain the observations under B.d. Explain why at a certain temperature the crystallization process proceeds at a

maximum rate. What is the postion of this temperature compared to thecharacteristic phase-transistion temperatures of the polymer?

e. Draw in a volume-temperature diagram the volume change during A, B,C and D. f. Draw in a log E-temperature diagram the temperature dependency of E for the

situations reached after A and C respectively.g. Give several technological consequences of the behaviour of PETP as mentioned

above.

4.20. The dimensions of the orthorombic unit cell of crystalline PE are 0.740, 0.493and 0.253 nm. Calculate the density of crystalline PE from these data.

4.21. Explain the formula for the crystalline volume fraction:

= a

c a

What is the formula for the crystalline massfraction?

4.22. A quantity of 0.1 gramme of PP (melting point 165 C) is being melted in adilatometer and subsequently cooled down rapidly to 150 C. At 150 C thefollowing data were found for the volume as a function a time:

time (sec) 0 500 1000 1500 2000 2500vol (mm 3) 120.00 119.16 114.97 110.94 110.04 110.00

a. Investigate if this behaviour can be described with the Avrami-equation: V = ( V ) [1 exp( ct )], and determine the values of c and .

b. Which conclusions can be drawn from the value of ?c. How would the values of c and have changed if the experiment would have

been performed at 155 C?d. Determine the degree of crystallization in the end situation, assuming that

crystalline PP has a density of 937 kg/m 3 at 150 C.

4.23. Calculate the crystalline volume fractions for a LDPE with = 0.91 g/cm 3 andfor a HDPE with = 0.965 g/cm 3. ( am = 0.855, cr = 1.00 g/cm 3)

4.24. From the answer to problem 4.23 it appears that HDPE has a crystallinevolume fraction twice as large as the crystalline volume fraction of LDPE. From thisone could expect that the stiffness of HDPE would be twice as high as LDPE(the


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stiffness of the amorphous fraction, in the rubbery state is negligibly small).However the E-modulus of HDPE appears to be 10 times the value of the E-modulusof LDPE.

What is the cause of this difference?

4.25. Why is a semi-crystalline polymer not transparant in most cases? Under whichcircumstances can it be transparant?

4 . 26. Why is a PE with a molar mass as high as possible used for the production of an extremely strong PE fibre?

4 . 27. When is the statement valid that semi-crystalline polymers show more creepthan amorphous ones and when is it not valid?

4 . 28. What is the difference between lyotropic and thermotropic LCP-s (liquidcrystalline polymers). In which cases is it essential to use either one type or theother?

4 . 29. To what extent can the three characteristic temperatures of a polymer T g, T mand T v be regarded as material constants? If this is not the case, what can cause

changes to these temperatures?

4 . 30. Give a kind of polymers in which all three temperatures T g, T m and T v areimportant.

4 . 31. Give two sorts of polymers in which only two transitions occur.

4 . 32. Give two sorts of polymers in which only one transition occurs.

4 . 33. Why is a semi-crystalline polymer impact resistant between T g and T m?


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5Rubbery and liquid phases5 . 1 . The free energy of a stretched polymer bar is given by F = U TS . Explain thesymbols used.

5 . 2. In what way will F change if the bar is being stretched? Describe the way d F /d llooks for a polymer in the glassy state and for a polymer in the rubbery phase.

Explain the difference.

5 . 3. Why is d S /d l independent of the temperature?

5 . 4. A polyisoprene rubber IR is being vulcanised with 0.5 % sulphur (S = 32,C = 12, H = 1 g/mol). All sulphur is being used and the cross-links contain 8 S-atomseach.

a. What is the average number of monomer units present between two cross-links.

b. Show that the E -modulus at 25 C should have the value of 264 kPa if E iscalculated from the formula E = 3 RT/M c ( = 0.91 g/cm 3 , R = 8.31 J/molK, beaware of the units!)

c. If the real E -value of the vulcanised rubber appears to be 1.5 MPa, how can thedifference with the value calculated under b be explained?

d . Give some general argumentation for the position of T and M c in the formula.

5 . 5 . A nonvulcanised butadiene rubber shows at 20 C an E -value of 2.2 MPa in afast stretching experiment. Its density = 0.91 g/cm 3 .

a . Calculate M C from the formula E = 3 RT/M c ( R = 8.314 J/molK). What is themeaning of M C in this case?

b. What would the value of E , according to the formula, be at 100 C? Why is thereal E -value found significantly lower, even lower than the value at 20 C, thanthe calculated one?

5 . 6 . Draw the log E vs T curve for :a. a nonvulcanised rubber,

b. for the same rubber, but now lightly vulcanised.c. for the same rubber, but now strongly crosslinked.d . Discuss the differences in T g in the three cases above.e. Discuss the differences in the E -level above T g in the three cases above.

f. What is the shape of the curve above T g for case b and c?


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5.7. A small rubber balloon (ideal vulcanisate, ideal rubber behaviour, without leaks)is blown up and put in store for a long time at -20 C. The diameter is now 20 cm.Subsequently it is brought to a room where the temperature is 25 C. What willhappen to the diameter?

5.8. Why is the flow temperature not a real, for a certain polymer characteristic,transition temperature from the rubbery state to the fluid state?

5.9. In figure 5.1 the apparent viscosity, , has been plotted against the shear stress , both on a logarithmic scale.

10 5

10 4

10 3

10 2

0,1 1 10 100 1000

Pas

(Pa)

Figure 5.1. Figure for problem 5.9.

a . Draw schematically the shape of the curve for a comparable polymer with a molarmass twice as large as the original polymer. Which average is the most importantproperty? What is de range of shift of the curve?

b. Do the same for a polymer type with the same molar mass as in the originalcurve, but with a broader molar mass distribution.

c. The so-called Power Law is often used to describe Non-Newtonian behaviour: =k (

. )n . On which part of the original curve in Figure 5.1 can this approximation be

applied and what will be the slope of that part of the curve for n = 0.5?d . Give some consequences of the width of the molar mass distribution for

processing. Distinguish between various type of processes.

5.10. Why is plotting of log versus log more useful than log versus log. ?

5.11. a . What restriction has the value of the melt-index in predicting the processingbehaviour of a polymer?b. In what way can the conditions in the determination of the melt-index be altered

in order to obtain a better prediction?c. What does the ratio of the two melt-indices, measured as above, say about the

chain structure of the polymer?


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Rubbery and liquid phases 21

5.12. In Figure 5.2 published values of the impact strength are plotted against themelt-index for a number of PP-s (homopolymers). There is a clear trend of decreaseof impact strength with increasing melt-index (not taking into account two types thathave, as reported, a narrower molar mass distribution. Explain these data.

1

2

3

4

1 10 100

m.i.

IZOD


5.13. The same source as problem 5.12 gives the bending modulus as a function of the melt-index (see Figure 5.3). Here also a trend of increasing E with increasingmelt-index can be observed, with the two types of narrower molar mass distributionshowing a lower stiffnes. Explain these observations.

1 10 100

m.i.

E MPa

2500

2000

1500

1000

Figure 5.3. Figure for problem 5.13

5.14. Can these trends, that is to say a changing impact strenght and stiffness withchanging molar mass, also be expected with PE?

5.15. Give four effects of the elastic behaviour of fluid polymers on the processing


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behaviour and the properties of the end product.


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23

6Visco-elasticity6.1. To what extent can a Maxwell-element (spring and dashpot in series) be used asa model for the real behaviour of a polymer?

6.2. A fluid polymer behaves like a Maxwell-element: a spring with E = 10 5 Pa inseries with a viscosity = 10 7 Pas. At t = 0 a constant stress of 10 4 Pa is applied to

the material during a period of 500 s. From t = 500 s the deformation is keptconstant, while the stress relaxation is recorded as a function of time.a. What is the value of the deformation immediately after applying the stress at t =

0?b. What is the value of the deformation at t = 500 s?c. What is the value of the stress at t = 800 s?d. Draw the curves of and as a function of time between t = 0 and t = 800 s.e. To what extent will a real polymer deviate from this behaviour?

6 . 3 . To what extent can a Kelvin-Voigt element (spring and dashpot parallel) be usedas a model for the real behaviour of a polymer?

6 . 4 . A force of 600 N is applied in the length direction of a small plastic bar with alength of 10 cm, a width of 1 cm and a thickness of 1 mm. Under the influence of this force the bar shows an elongation as a function of time as given in figure 6.1

a . Which spring-dashpot model should be able to describe this behaviour?b Determine or estimate the values of parameters for this particular model ( E

moduli and viscosity).c. Draw the elongation as a function of time when the force is withdrawn after 250

seconds.d . Give a number of reasons why a polymer in reality will not behave like this

pattern.

6.5. A polymer behaves like a Kelvin-Voigt element (spring E 1 = 310 8 Pa parallel toviscosity = 310 14 Pas) connected in serie with a spring E 2= 310 9 Pa. At t = 0 a

stress = 610 6 Pa is applied to the system; After 3.10 6 seconds the stress isremoved.Draw the curve of the deformation as a function of time and calculate thedseformation (%)


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Visco-elasticity 25

6.11. A visco-elastic system is subjected to an instantaneous deformation of 100 %.The stress, necessary for this deformation, is 2 MPa. After the deformation the stressdecreases as a function of time as displayed in the table below.

t (sec) 0.01 0.10 0.20 0.50 1 2 5 10 20 50 100 (MPa) 1.99 1.90 1.82 1.61 1.37 1.14 1.01 1.00 1.00 1.00 0.99

t (sec) 200 500 1,000 2,000 5,000 10,000 20,000 50,000 (MPa) 0.98 0.95 0.90 0.82 0.61 0.37 0.14 0.00

Give a model for this behaviour.

6.12. The modulus of elasticity of a polymer is often represented by a complexquantity: E * = E 1 + iE 2.a . What is the meaning of E 1 and E 2 in this formula?b. In what way can these quantities be measured, e.g. around 1 Hz?c. In what way does the dampingfactor, tan , depend on these quantities?d . Give a number of cases in which tan has a practical technological meaning.

6.13. In figure 6.2 the damping factor is plotted against the temperature for twosituations ( p and q).a. p and q are valid for which two types of polymers?b. Where is the first maximum for both of these two curves situated?c. Which value will the curve of p reach with a further increase of temperature?.b. Explain why the curve of situation q does not approach to zero, but to a finite

value.

tan p

q

T



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26

7Mechanical properties7.1. A polymer bar, width 1 cm, thickness 1 mm and length 10 cm is beingelongated by 2 mm. The necessary force is 600 N. What is the value of the modulusof elasticity ( E )?

7.2. In general a value for the modulus of elasticity ( E ) of 3 GPa is found for hard ,

amorphous polymers in the glassy state. Why is the value for E for polycarbonateonly 2.1 GPa?

7.3. Why do semi-crystalline polymers differ considerably in stiffness? Examples :PEEK; E = 4 GPa, PP: E = 1.3 GPa, LDPE even down to 0.15 Gpa.

7.4. In which two manners can the stiffness of a polymer be increased?

7.5. We have two equally formed bars; the first one is ordinary PP, the other one

contains a small amount of copolymer (to increase the impact strength). What is themost simple way to determine the material which the bars are made of? How can theamount of dispersed rubber be estimated easily?

7.6. Why are fibres stiffer than the polymers that they are made of? What order of magnitude can the stiffness increase be? Give a number of extreme examples.

7.7. Why is it, practically spoken, not possible to say that a stiffer polymer chainleads to a higher stiffness of the polymer? Explain why extremely stiff chains , like

aromatic polyamides, result in a high stiffness?7.8. Why is it not possible to predict the creep of an amorphous glass-like polymerafter a long time by shifting creep curves at higher temperatures along the log t - axisas suggested by the principle of time-temperature equivalenc?

7.9. Draw schematically a number of creep isochrones of :a. A polymer with a high short-term modulus of elasticity and much creep.b. A polymer with the same high short-term modulus of elasticity but little creep.

c. A polymer with a short-term modulus of elasticity three times lower as in a and band much creep.

For which kinds of thermoplastic polymers can the cases of a, b and c be valid?


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Mechanical properties 27

7.10. The creep formula of Kohlrausch: D(t ) = D 0 exp( t/t 0)1/3 appears to describe thecreep behaviour surprisingly well for a large number of materials, includingpolymers. What is the meaning of D

0in the formula? Why can this formula not be

applied for the practical behaviour of polymers?

7.11. A glass-like polymer can show physical ageing.a . What is the real cause of this phenomenon?b. What is the influence on the creep behaviour?c. Draw bundles of isochrones with and without physical ageing.

7.12. Figure 7.1 shows published creep isochrones of Noryl (PPE/PS) and PBTP,

both at 23 C. Explain the difference in shape of both isochrone bundles.

20

15

10

5

0 0.5 1.0 1.5%

PPO/PS(Noryl)23 C

MPa

hours

20

15

10

5

0 0.5 1.0 1.5%

PBTP23 C

MPa

hours

1 10 10 2 10 3 10 4 1 1010 2

10 3

10 4


7.13. In figure 7.2 creep isochrones are displayed at five different temperatures;Each temperature has its own vertical axis for the stress.a. What will be the deformation at 20 C under a stress of 8 MPa after 100 hours?b. What will be the deformation at 70 C under a stress of 5 MPa after 10 hours?c. What is the value of the short-term modulus of elasticity at 20 C?

d. What is the value of the short-term modulus of elasticity at 80 C?e. What stress is allowed at 20 C, if the deformation after 1 year must not exceed

2%? f. What stress is allowed at 100 C, if the deformation after 1000 hours must not

exceed 3%?


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100 80 60 40 20 C 1 2 3 4 %

PP

MPa

10 2 hours

1

10 2

10 4

2

4

6

8

10

12

14

16

18

1

11

1

2

2

2

2

3

3

3

3

4

4

4

4

5

5

5

5

6

6

6

7

8

8

10


7.14. Breaking stress experiments on PB (poly-1-butene) pipes gave the followingbreaking times t b at various wall stresses and temperatures T :

(MPa) 189 183 180 175 163 157 151 135 128 123 119 107 105T (C) 20 20 20 20 40 40 40 60 60 60 60 80 80t b (hours) 0.02 1.1 12.9 850 0.15 40 3000 0.3 95 2000 8100 0.2 3

100 80 80 77 72 60 43 58 53 42 26.7 20

T 80 80 100 100 100 100 100 110 110 110 110 110t b 600 12000 0.01 20 1100 3800 9800 0.03 20 2000 5000 7500

Estimate from these data the safe wall stress for a period of use of 10 years at 80 C.

7.15. a . Why is the safe wall stress, as determined in 7.14, not safe enough forcertain applications?b . Which information is needed in order to construct in such a way that never an

irreversible material damage can occur?

7.16. Why is a standard impact test often performed with notched test bars?


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Mechanical properties 29

7.17. a . How can a polymer with a low impact strenght as obtained from a testdescribed in 7.16 be improved?

b. Which mechanisms are involved in this improvement?c. To what property must be given in with this improvement?d . Give examples of this principle.

7.18. Why can the impact strenght of a polymer not be considered as a materialconstant?

7.19. a . Why is there a rather good correlation between the Shore-D hardness of polymers and their modulus of elasticity?

b. Why can this correlation not be perfect?c. Why are polymers (like TPE-s) often characterised by their hardness (Shore-D)instead of their modulus of elasticity?

7.20. Why is the damping of of a vulcanised rubber lower with a lower T g?

7.21. By which factors is the tan of a rubber vulcanisate, far above T g , controlled?


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30

8Further properties8 . 1 . Where would the brittleness temperature approximately lie for:- an amorphous thermoplastic?- a semi-crystalline thermoplastic?- a rubber-modified thermoplastic?

8 . 2 . Why is the softening behaviour of an amorphous thermoplast different of that of a semi-crystalline thermoplast?

8 . 3 . Look at the data for Vicat B and ISO/A (Chapter 10). Assume that thecharacteristic E-moduli for these determinations are 200 MPa and 1000 MParespectively. Try to verify these values for PBTB, PTFE, PP, PMMA, PA 6 . 6, POMand HDPE from the E(T) curves (see end of chapter 10). What are the reasons for thedifferences found?

8 . 4 . Two different softening temperatures of two different polymers may show anapparent contradiction, like (C):

ISO/A Vicat BABS 95 100PBTP 60 180

a. What causes this contradiction?b. Which softening temperature should be used in practice?

8 . 5 . Manufacturers of semi-crystalline polymers sometimes claim that the additionof short glassfibres not only causes an increase in stiffness, but also cause aconsiderable rise in the softening temperature. Is this an extraordinary achievementor is it just in the nature of things?

8 . 6 . What are the disadvantages of the relatively high expansion coefficient of polymers?

8 . 7 . Give both advantages and disadvantages of the low thermal conduction of polymers.

8 . 8 . The heat of conduction of a polymer foam is composed of four components.a . Which are these components?b . Point out the way changes with the cell size at constant density.


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Further properties 31

c . Point out the way changes with the foam density.d . Point out the way changes with the type of gas.e . Which factor determines the maximum insulation value possible for a foam?

8 . 9 . Why does a high softening temperature of a polymer not guarantee sufficientlythat the polymer can be used in practice at high temperatures?

8 . 10 . a. What is understood by stress corrosion?b . Why is this name in general not correctc . When is this name correct?d . Give a number of examples.

8 . 11 . For what reason would one lower the excellent high electrical resistance? Inwhat manners can this be done?

8 . 12 . Which property influences the tendency to electrostatic charge? How can thistendency be surpressed? What is the importance of the time scale in this respect?

8 . 13 . What are the advantages and disadvantages of the extremely low dielectriclosses of PE?

8 . 14 . a . Why are polymers not or hardly suitable for fine optical applications?b . Polymers are frequently used in the manufacture of spectacles. What are theadvantages and disadvantages in this application?

8 . 15 . Why would one choose a PC with a molar mass as low as possible for themanufacturing of compact discs?

8 . 1 6 . The nitrogen-permeabilities of HDPE, PP and IIR are about equal.Nevertheless IIR is considered as very gastight, while PP and PE are considered asrather permeable. Explain this.

8 . 17 . Why is the permeability of LDPE higher than of HDPE?


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32

9Polymeric compounds andcomposites9.1. The Flory-Huggins formula is:

G mV = kT [

1V

1

ln 1 + 2V

2

ln 2 +V

s

1 2]

Show from this formula that the miscibility of two polymers decreases withincreasing molar masses.

9.2. Which role does the interaction parameter play?

9.3. Simplify the Flory-Huggins equation for a 50/50 (volume) mixture of macromolecules of equal size and derive for this case a simple condition for the signof the free enthalpy of mixing.

9.4. Show from the formula that, independent of the parameters, there will alwaysbe values for 1 and 2 for which the free energy of mixing is negative.

9.5. If after mixing in the melt a homogeneous mixture is obtained, it can segregateduring cooling to the solid state. When will this be the case?

9.6. The reverse: One has mixed at a not too high temperature and obtains, aftercooling, particles consisting of a homogeneous mixture. During processing in an

injection moulding process segregation occurs resulting in an endproduct being adispersion. Explain when this can happen.

9.7. Why does the postion of the G- curve not always guarantee, that there willindeed result a homogeneous mixture from a mixing process?

9 . 8 . How can it simply be explained that segregation in a block copolymer willbecome more perfect with increasing length of the blocks?

9 . 9 . Which properties govern the mimimum droplet size in a mixing process?9 . 10 . Why do we find a much larger droplet size in real dispersions?

9 . 11 . Which two processes contribute to the reduction of the droplet size in amixing process?


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Polymeric compounds and composites 33

9 . 12 . Why is it not possible to obtain the properties of a dispersion from a rule of mixing following from a serial or parallel arrangement?

9 . 13 . Referring to problem 4.24 we look at the bending modulus E of a polyethenewith density 0.935. The degree of crystallization, (volume fraction) can becalculated with :

=d d am

d cr d am

with d cr =1.00 and d am = 0.855 (see maintext 4.4.1) and so = 0.552. The E -modulus is 660 Mpa.

Can we determine the morphology of PE from these data (either amorphousdispersed in crystalline,the reverse or co-continuous)? Use the formulae of Kernerand Nielsen ( 9.1.6 of the main text).

9.14. A three block copolymer SBS is built up a polybutadiene middle section of 2000 butadiene-units -(CH 2-CH=CH-CH 2)- with two end sections, each consisting of 150 styrene units -(CH 2-CHC 6H5))-. Complete segregation occurs; the PS domainsare spherically shaped with a diameter of 25 nm.a. What is the mass fraction of polystyrene?b . How many PS chains are present in a domain ( = 1.05 g/cm 3 and N A = 610 23?c. Why does SBS lose its properties already around 10 C below the glasspoint of

PS?d . Estimate with the Kerner-formula the modulus of elasticity of SBS, assuming that

the E-moduli of polybutadiene and polystyrene are 3 MPa and 3 GPa respectivelyand the density of BR = 0.91 g/cm 3.

9.15. Three foams of a glasslike thermoplastic polymer, each with a density of =

0.5 g/cm 3 (density of the solid polymer = 1 g/cm 3 , E modulus is 3200 Mpa) show E -moduli of 1100, 300 and 100 MPa respectively. Which conclusions can be drawnfrom these data about about the structure of these foams?

9.16. Show that the stiffness of a heavy foam (little foaming) is, roughly estimated,proportional to its density.

9 . 17 . Why is the expression reinforcing fillers in general not really correct? Howshould these be called in a correct way?

9 . 18 . Why is carbon black in particular a reinforcing filler and why is a technicalSBR-vulcanisate not possible without a carbon black reinforcement and can a naturalrubber vulcanisate do without?

9 . 19 . Give a number of cases in which it is very desirable or even essential for the


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properties of a heterogenous mixture that the smallest phase is the continuous phase.

9.20. Two polymers A and B are mixed in two mixing ratios to M1 and M2. For theISO heat deflection temperatures of A, M1, M2 and B values of 50, 80, 110 and140 C are found respectively. The Vicat softening temperature however is more orless the same for all four materials, all values are around 145 C. Give a schematicdrawing of log E versus T for all four materials. In what respect do A and B differ?

9.21. Now two polymers C and D are mixed in two mixing ratios to M3 and M4. Forthe ISO heat deflection temperatures of C, M3, M4 and D values 90, 130, 170 and210 C are found. The Vicat softening temperatures are 95, 135, 180 and 220 C.

Give also a schematic drawing of log E

versusT

for these four materials. What is thenature of the polymers C and D and what can be said about their miscibility?

9.22. The addition of glass fibres to a polymer increases the impact strenght of certain polymers (the brittle ones), but decreases the impact strenght of others (thetough ones) (see main text Figure 9.1.7). How can these contradicting observationsbe explained?

9.23. It is desired to enhance the impact strength of a glassy thermoplastic ( E = 3

GPa, p = 1.4 g/cm3) by adding a dispersed rubbery phase ( E = 2 MPa, r =

0.9 g/cm 3). However the stiffness should not be reduced by more than 20% as aresult of this addition. How many %(weight) rubber may be added? The Poissonnumber of the matrix is = 0.33.

9.24. The mixture made in problem 9.23 should be returned to its original stiffnesswith glassfibres ( E = 75 GPa, g = 2.4 g/cm 3). How many %(weight) of glassfibresmust be added?

9.25. Why is the velocity of sound in a composite important?

9.26. In a certain composite glass fibres are dispersed in a parallel alignment in thepolymer. The volume fraction of the fibres is 0.25. Imagine a number of cases inwhich the lenght of the fibres increases more and more. Calculate (without using theFormula of Kerner) to which limit the E- modulus of the composite will approach inthe direction of the glassfibres. (the E -moduli of the polymer and the glass are 3 and75 GPa respectively).

9.27. Estimate the value of the E -modulus in the direction perpendicular to thedirection of the glass fibres with the aid of the Einstein formula. Why is it allowed touse this formula here?

9.28. Estimate, without too detailed calculations, the E -modulus of a similarcomposite in which the glass fibres are now dispersed at random.


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Polymeric compounds and composites 35

9.29. To a semi-crystalline thermoplastic short glass fibres are added in one case andin another case rubber particles are added. In what way and why does each of thefollowing properties alter for both cases?a. Transition temperature from glassy tot rubbery phaseb. Vicat softening temperature.c. Melting point.d. ISO-HDT.e. Modulus of elasticity.f. Tensile strength.g. Conduction of heat.h. Impact strength.


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36

11Principles of processing11.1. Mention four different principles to heat-up a polymer in a processingoperation. Give an example of each of these. Which material properties are importantfor each of these principles?

11.2. Mention a few examples of thermosets and thermoplastics where casting is a

possible shaping technique.

11.3. Which processing techniques are not- or less suited for semi-crystallinepolymers? Why?

11.4. Why is compression moulding for rubbers and thermosets more attractive thanfor thermoplastics?

11.5. a . Give two different routes to manufacture PS (or TPS) cups from the raw

material.b. How can these two cups be distinguished in a non-destructive way?c. How does their behaviour difference upon heating?d . Which factors govern the choice of the processing methods?

11.6. Figuur 11.1 gives, schematically, two "short-shot" lines for an injectionmoulding process.

T

A B

p

Figure 11.1. Figure to problem 11.6.

a. Explain the shape of the curves.b. If A and B hold for the same injection moulding machine and the same polymer,

but for different moulds, in which respects do the moulds then differ?


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Principles of processing 37

c. If A and B now hold for the same mould, but for different grades of the samepolymer, which conclusion could, at first sight, be drawn about the differencebetween A and B?

d. If the conclusion from the above question would appear to be wrong, what wouldthen be the following conclusion about the difference between A and B?

e. What can happen when, at a certain temperature, the pressure p is being set at amuch higher value

f. The same when, at a certain pressure, the temperature T is strongly increased?g. The same when T is strongly decreased?

11.7. In an injection moulding process an HDPE sheet of 10 cm 10 cm 3 mm is

being produced. The injection moulding pressure is 800 bar, the melt temperature is220 C.

a. What is the minimum required clamping force on the mould?b. How many grammes of the polymer are present in the mould, immediately after

filling?c. What would the dimensions of the product be if the filling gate would be closed at

this moment?d. How can it be prevented that the end-product is too small?

The compressibility of molten HDPE is 7.5 105 /bar. The volume-temperaturerelation at atmospheric pressure is given in figure 11.2.

1.4

1.3

1.2

1.1

1.0

0.9

0 50 100 150 200 250T C

specific volume

HDPE

Figure 11.2. Figure to problem 11.7.

11.8. Why is the mould shrinkage for PE, PP and PA greater than for PS, PVC and


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PC?

11.9. Why is the density in the interior of a casted, compression-moulded, orinjection-moulded block smaller than at the outside?

11.10. Describe the competition between flow and curing of powder resins, used fore.g. bicycle frames,

11.11. Why exists the risk in a calender process of uneven thickness over the widthof the sheet? What can be done to prevent this?

11.12. Mention some processing techniques where the elastic behaviour of the

molten polymer plays an essential part. Why is the word molten not appropriate?

11.13. In which two respects is a broad molar mass distribution, e.g. with PE, of importance when making shrink films?

11.14. In which two respects is a narrow MMD, e.g. with PE, important in theproduction of an oil tank in a rotational moulding process?

11.15. When a wire coating is extruded around a copper wire, a high throughput is

required. This involves the risk of melt fracture, resulting in a rough surface. Howcan melt fracture be avoided?

11.16. Mention a few cases where the meeting of two streams of molten polymermay result in weak spots in the end-product.

11.17. What is meant by an "adiabatic" extruder? Why is this to be recommended?

11.18. Why can an extruder be used as a mixer?

11.19. The figure gives, for a certain combination of an extruder, a die and apolymer, two lines for Q (throughput) as a function of the pressure p , resp. Q = aN b p/ and Q = cp/ .

a. explain the symbols in these equations.b. What is, at the given combination of extruder screw and polymer, independent of

the die, the maximum throughput and the maximum pressure? When are thesemaxima being reached?

c. How do the throughput and the pressure change when the rotational speed of the

screw is halved?d. Estimate the rate of throughput and the pressure when the die has a twice as large

flow resistance.e. The temperature is lowered to such an extent that the viscosity becomes three

times higher. How does the rate of throughput and the pressure change?


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Principles of processing 39

Illustrate the points b. to e. with extra lines in the diagram. f. Why are the lines in the diagram not straight lines in reality? Sketch (only for the

die) a more realistic curve.

p (bar)

Q (dm 3 /h)80

70

60

50

40

30

20

10

00 100 200 300

Figure 11.3. Figure to roblem 11.18.

11.20. In an extruded pipe internal as well as frozen-in stresses may be present.a. Describe the difference and explain the origin of both.b. Give a simple detection method for each.c. What are the consequences of each of these kinds of stresses for the properties in

practice?d. Why are internal stresses more pronounced with thick-walled pipes, and frozen-in

stresses with thin-walled pipes?

11.21 . When making PET bottles, the first step is the injection moulding of anamorphous preform, which is later blown-up.a. Why is this pre-form amorphous?b. Why is this necessary for the further fabrication of the bottle?c. What happens more with the blowing-up?d. Why should the final bottle be (semi)-crystalline?e. Why does the material crystallise during the formation of the bottle?

f. Why is the bottle beautifully transparant?g. What is the analogy with the manufacture of PETP (polyester) fibres?

11.22 . How and why does, with extrusion, the die-swell change:


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a. with a polymer with a higher melt-flow-index m.f.i.?b. with a higher ratio of m.f.i.(10) / m.f.i.(2.16)?c with a higher rate of throughput?d. at a higher melt temperature ?e. at a larger l/d ratio of the exit channel ?

11.23 . Why does the extrusion of PVC require a longer compression zone than withPA?

11.24 . What is a structural foam? Mention four advantages in its manufacturing andits use.

11.25 . Which processing techniques could be applied to manufacture (from the rawmaterial) of the following articles:a. 2-litre bottle of PE,b. ship propellor of PA,c. butter cup of PP,d. bottle crate of PP,e. floor tiles of PVC-compound,

f. rubber ball,g. broad PVC film,h. tyre tread of SBR,i. waste bag of LDPE,

j. packaging case of PS foam,k. rubber O-ring,l. rubber shoe-solem. small rowing-boat of PVC,n. thick PMMA-sheets ,

o. wire-coating of PVC, p. sewer pipe of PVC,q. sewer pipe of epoxy/glass,r. 10 m 3 vessel of PE,s. switch housing of UF,t. light armature of PMMA,u. toothrad of PA,v. thin-walled PP pipe of 3 m diameter,

w. PU thermal insulation of a storage tank, x. rowing-boat of UP/glass, y. PET-bottle. z. thin flat sheets of PS foam?


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41

Answers

1. Introduction

1.1. Vegetable: timber, cotton, cork, sisal, hemp, straw, reed, cane, capok, rattan, etc.Animal: wool, silk, fur, mohair, horse-hair, intestine, down, ivory, horn, sponge,beeswax etc.

1.2. With half-synthetic plastics the polymer chain has been formed in a livingtissue, but it has been chemically modified afterwards.Cellophane is, after chemical modification, obtained from the cellulose in wood, justas paper (from cellulose and lignin), cellulose fibres (rayon), and cellulose plastics.Leather is made from animal hides in a tanning process.Natural rubber has to be vulcanised (with sulphur) to obtain a technically usableproduct.Ebonite is also made from natural rubber, but with a much higher amount of sulphur;it belongs, therefore, to the family of thermosets.Linen is obtained from flax after a chemical treatment (retting).The boundaries between half-synthetic and natural polymers are sometimes vague.

1.3. In a cracking process the long, saturated, hydrocarbon chains in the crude oil aretransformed into shorter ones, which, for lack of hydrogen atoms, contain doublebonds, (-C=C-), which provide polymerisability.

1.4. When a diene, C=C-C=C, is polymerised, only one of the two double bonds isused; an unsaturated chain results, -C-C=C-C-, which offers the possibility of vulcanisation with sulphur.

1.5. a . 1.4-polybutadiene, -CH 2-CH=CH-CH 2-b . 1.2- of 3.4-polybutadiene, -CH-CH 2

|CH||CH 2

-CH-CH 2|

CH||CH 2

-

in which under a. two different structures, cis and trans can be formed.

-C C- \ / C=C

or C=C /

-C

/ C-

1.6. Ethylene-propylene copolymers, and also polyisobutylene, are rubbers withsaturated chains, which cannot be vulcanised with sulphur. They are, therefore,


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copolymerised with a small amount of diene: EP with a diene in a sidegroup (e.g.dicyclopentadiene), PIB with isoprene (IIR = isobutylene-isoprene rubber = butylrubber).

1.7. a . Polycondensation is possible with saturated monomers, e.g. a diol with abifunctional carbon acid, by ester formation on two sides. In this reaction a smallmolecule, e.g. water is also formed.b . After ring opening of saturated monomers these can also polymerise (e.g.caprolactam to PA-6).

1.8. a . By a reaction between tri-functional and bi-functional components, such as

with the formation of phenol-formaldehyde or a polyurethane from glycerol and di-isocyanate.b . By cross-linking polymer chains containing double bonds: with rubbers usingsmall amounts of sulphur; with unsaturated polyesters using (poly)styrene.

1.9. Thermosets are cured into networks at elevated temperature from smallmolecules or from single chains.The name could suggest that thermosets become harder at temperature increase; onthe contrary: they soften just as all polymers at their glass-rubber transition, thoughtheir stiffness remains much higher than that of a rubber.Moreover: curing is not always carried out at a high temperature; also cold-curingresins are used (e.g. in two-component adhesives or in large ship decks).

1.10. In thermoplasts the parts of different chains are bound to each other by weak secondary bonds, which are broken at higher temperatures. In thermosets andvulcanised rubbers the chains are permanently linked by primary chemical bonds,which, also at higher T , do not allow flow.

1.11. Thermosets have a tight network; with rubbers it is much looser; the cross-links are just sufficient to connect all chains at a few points.

1.12. In thermoplastic rubbers the chains are not connected by primary bonds as isthe case with sulphur-vulcanised rubbers, but by weaker secondary bonds (e.g.within a PS domain in SBS), which are loosened upon temperature increase).

1.13. Thermoplastic rubbers (TPEs) show rubbery behaviour at ambienttemperature because the rubber phase plays the dominant part. In thermoplastics thebehaviour of the glassy phase or of the crystalline part is dominant.

1.14. Thermoplasts: PE, PS, PVC, PP. Thermosets: PF, UP. Rubbers: SBR.


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Answers 43

1.15. LDPE, HDPE and LLDPE. They differ in density: LDPE between 0.91 and0.92, HDPE between 0.94 and 0.96, and LLDPE between 0.92 and 0.95 g/cm 3. Sincethe density reflects the crystallinity, they also differ considerably in stiffness,between as well as within types.

1.16. thermoplasts: PS, TPS, SAN, ABS en ASA. Thermosets: UP (is cross-linked with styrene). Rubbers: SBR and SBS.

1.17. PTFE with its very high melting point (327 C) can only be transformed intoan end product as a powder in a sintering process, since as a melt it would degrade at

the high temperatures involved in other processing operations. Other fluor polymerssuch as FEP, PVDF and EFTE are suitable for the usual processing techniques.

1.18. PI, polyimide, can be used short-time up to 500 C. Related to PI are: poly-etherimides, polyesterimides en polyamidimides.

1.19. a . a. Blends of polymersb. Particle- and short fibre- reinforced polymersc. Composites

d. Foams

1.20. a . Carbon black is added to thermoplasts to decrease the electric resistance, inparticular to prevent electrostatic charge, and also to shield the polymer fromlight, thus preventing ultraviolet degradation. Carbon black is added to rubbers toincrease strength and abrasion resistance,

b. Chalk is a reinforcing filler in thermoplasts and thermosets.c. Glass fibres are added to thermoplasts and thermosets to increase stiffness and

strength, more than particles do.d. Plasticisers are added to a polymer to reduce the glass rubber transition

temperature drastically (e.g. with PVC), so that the polymer behaves as a rubberat ambient temperature rather than as hard glassy thermoplast.

e. Mica is a reinforcing filler which, due to its platelet structure, increases thestiffness more than spherical particles do.

f. Accelerators are added to thermosets and rubbers to speed-up the curing orvulcanisation process to such an extent that an end product can be formed underoptimal conditions.

g. Lubricants are sometimes used to reduced the friction resistance in a shapingprocess.

h. Sulphur is essential in the usual processing technology of rubbers; it brings aboutthe formation of cross-links, which provide the necessary shape stability.

i. Pentane is a blowing agent, applied in a.o. PS, to form foam.


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j. Wood flour is a cheap filler, which brings about some stiffening (e.g. in PVC),and which reduces the price.

k. Antioxidants can, in the first place, protect a polymer against degradation at thehigh temperatures during processing; in the second place they can (mostlydifferent types) prevent chain rupture during long-term use at elevatedtemperatures.

l. SiC is a very hard filler, mostly used to increase the abrasion resistance of apolymer.

m. UV-stabilisers protect a polymer against chain break under the influence of theultraviolet components of daylight, often in combination with pigments (e.g. withPP).

n. Anti-static agents bring about a very thin layer of a conducting (hydrophilic)substance at the surface of the product, which allows electric charges, resultingfrom friction or contact, to be dissipated quickly.

o. Rubber particles are blended into a brittle polymer to increase its impact strength,e.g. butadiene rubber in PS, etc.

2. Molecular composition

2.1. C-atoms only, or also other ones such as N, O, S or Si, saturated or unsaturated chains simple chains or chains containing rings, even composite rings (examples in MT

2.1.1)

2.2. a . PE -CH 2-CH 2-BR -CH 2-CH=CH-CH 2polyacetylene -CH=CH-CH=CH-POM -CH 2-O-CH 2-O-

b. PP, PVC, PS, PB, PAN.c. PIB, PVDC, PVDF (see MT 2.1.2).

2.3. PE, -CH 2-CH 2-, monomer mass = 2C + 4H = 2 12 + 4 1 = 28 g/mole,PS, -CH 2-CH(C 6H5)-, monomer mass = 8C + 8H = 8 12 +8 1 = 104 g/mole;PA-12, -(CH 2)11 -CO-NH-, monomer mass = 12C + O + N + 23H =12 12 + 1 16 + 1 14 + 23 1 = 197 g/mole.M(PE) = 500 28= 14,000 g/mole,M(PS) = 500 104 = 52,000 g/moleM(PA-12)= 500 197 = 98,500 g/mole.-C-C-C-C-C- chains lie zig-zag shaped, with C-C distances of 0.154 nm and with anangle between the bonds of 109.5. The contribution of a C-C link to the contourlength is, therefore, 0.154 sin 54.2 = 0.126 nm. for PE the stretched length is thus 1000 0.126 nm = 126 nm (since the monomer


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Answers 45

contains two C-atoms), for PS also 126 nm, for PA-12: 13 links (for simplicity taking the same length for CO-NH-), so 500

13 0.126 nm = 820 nm.The calculation is only valid for strictly linear PE, without branches. As to PS, itholds only for the syndio-tactic structure, since otherwise the neighbouring phenylrings prevent a stretched conformation. (MT 2.5).

2.4. In this case equal numbers (both 1) are mixed. The number fractions are both1/2 and the number average is

12 80,000 +

12 120,000 = 100,000. The weight

fractions are, respectively, 80/200 and 120/200 or 0.4 and 0.6, so that the weight

average is: 0.4 80,000 + 0.6 120,000 = 104,000.

2.5. n i = N i / N i = number of the species i divided by the total number = numberfraction, wi = n iM i / niM i = weight fraction, zi = w iM i / w iM i = z-fraction

M n =

niM i ,

M w = wiM i,

M z = ziM i , ni = wi = zi = 1 (all normalised).

2.6. M n = niM i; ni = ( wi /M i)/ (wi /M i),

M n = ((wi /M i))M i)/ (wi /M i)) = w i / (wi /M i)) = 1/ (wi /M i))

2.7.

M w = wiM i; wi = ( nM i)/ (niM i) = niM i /

M n ; M w = (ni M i2)/

M n ; M w

M n = niM i2 .

2.8. M z = ziM i; zi = ( wiM i)/ (wiM i) = wiM i /

M w;

M z = (wiM ii2)/

M w;

M z

M w = wiM i i2 .

2.9. a . In a mono-disperse polymer all chains are equally long (exception). Then

M n=

M w =

M z etc..

b. The degree of dispersion or heterogeneity index, D, is defined as

M w /

M n ,, and is,

therefore, D = 1 for a mono-disperse polymer.

2.10. The chains break in a statistical way, by which shorter ones of various lengthsare formed. The molar mass distribution is broadened in the direction of the low-molecular side.

2.11. In the blend every ( wi) A is replaced by: w A(wi) A and every ( wi)B byw B(wi)B. So

M w of the blend is:

M w = (w A(wi) A( M i) A) + (w B(wi)B( M i)B)

= w A(wi)A( M i)A + w B(wi)B( M i)B = w A(

M w)A + w B(

M w)B

2.12. The number of chains in one gramme of polymer is N / M n ( N = Avogadronumber). One gramme of the mixture contains w A grammes of A, w B grammes of Band wC grammes of C, so that:

N /

M n = w A N /(

M n)A + w BN /(

M n)B + wC N /(

M n)C


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of 1/

M n = w A /(

M n)A + w B /(

M n)B + wC /(

M n)C.The

M n values of A, B and C are 10, 20 and 30 kg/mole respectively 10, 20 en 30

kg/mole; w A

= w B

= 10/(20 + x ), wC

= x /(20 + x ),

M n

= 15 kg/mole, so:1/15 = 10/(20 + x )10 + 10/(20 + x )20 + x /(20 + x )30, which results in: x = 5.

2.13.

M z = wi M i2 / wi M i of wi M i2 /

M w. Further evaluation is analogous to 2.11 andresults in:

M z

M w = w1 (

M z)1 (

M w)1 + w 2(

M z)2 (

M w)2 . So:

M z = 154.5 kg/mole.

2.14. 0.382 = k (76,000) a; 0.91 = k (250,000) a; 2.21 = k (850,000) a; we can plotlog[ ] versus log M to determine the slope a ; we can also consider ratios of [ ]:0.91/0.382 = (250/76) a; a = 0.729; 2.21/0.91 = (850/250) a; a = 0.725; 2.21/0.382 =

(850/76)a

; a = 0.727; on average a = 0.727. With 0.91 = k 250,0000,727

we find: k =10.810 5.

2.15. w A = 0.37; w B = 0.18; wC = 0.45.

M w (in kg/mole) = 0.3776 + 0.18250 +0.45850 = 455.62 = 455,620 g/mole.1/

M n = 0.37/76 + 0.18/250 + 0.45/850;

M n = 163.457 = 163,457 g/mole.

2.16. c t t t o (t t o)/t o (t t o)toc = red0 95.0 0

0.1 107.7 12.7 0.1337 1.3370.2 121.4 26.4 0.2779 1.3900.3 136.3 41.3 0.4347 1.4490.4 152.8 57.8 0.6084 1.521

When we plot red against c, it is clear that extrapolation of red to leads to 1.30 dl/g,so [ ] = 1.30 dl/g. From Mark-Houwink it follows (see 2.14): 1.30 =10.810 5M 0.727 , from which

M v = 410,000 g/mole, indeed slightly below

M w.

red(dl/g)1.55

1.50

1.45

1.40

1.35

1.30

1.250 0.1 0.2 0.3 0.4 c (g/dl)

Figure A.1. Figure to the answer of problem 2.16.


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Answers 47

2.17. With

M v = (wi M ia)1/a :

M v = (0.3776,000 0,727 + 0.18250,000 0,727 + 0.45850,000 0,727 )1/0,727 = 409,340,close enough to the answer on 2.16 .

2.18. We first calculate the various averages:w1 = 0.10, w2 = 0.18, w3 = 0.25, w4 = 0.17,w5 = 0.12, w6 = 0.08, w7 = 0.06, w8 = 0.04.

M 1 = 15, M 2 = 27, M 3 = 39, M 4 = 56, M 5 = 78, M 6 = 104, M 7 = 120, M 8 = 153.

M n = 1/ (wi /M i)) = 1/(0.10/15 + 0.18/27 + ) = 38.7 kg/mole.

M w = (wi M i) = 0.1015 + 0.1827 + = 56.6 kg/mole.

M z = (wi M i2)/

M w = (0.1015 2 + 0.1827 2 + )/56.6 = 78.8 kg/mole.

M z+1 = (wi M i3)/

M z

M w = (0.1015 3 + 0.1827 3 + )/(78.856.6) = 99.2 kg/mole

a. The melt viscosity (zero-shear) is governed by

M w, so as far as this is concernedthe blend resembles fraction 4 .

b. The blend has a broader molar mass distribution; it behaves, therefore, strongernon- Newtonian (see MT 5.3.2 ) and flows better at high rates of shear.

c. The impact strength is mainly dependent on

M n , so the blend resembles fraction 3.d. The die-swell is a consequence of melt-elasticity which is governed by higher

averages such as

M z

and

M z+1

; here the fractions 5 and 6 are of importance.e. The Youngs modulus in the glassy state is independent of M.

2.19. In kg/mole: M 1 = 60, M 2 = 600,

M n =1/ (wi /M 1) = 1/((1 w2)/60 + w2 /600) =100, so w2 = 4/9 and w1 = 5/9; so 40 kg are needed.

M w = (5/9)60 + (4/9)600 = 300

kg/mole = 300.000 g/mole. M z = [(5/9)60 2 + (4/9)600 2)]/300 = 540 kg/mole =

540.000 g/mole.

2.20. If we consider the coil as a sphere with a radius r = ( r o2 )1/2 = 73.5 nm then the

volume of such a sphere is V = (4/3) r3

= 1.6631021

m3

. The mass m of the chain is1000/610 23 = 1.66710 21 kg. The density, , is then m/V = 1.00 kg/m 3 . When wecompare this with the density of PS, 1050 kg/m 3, then the coil is 1050 times diluted

2.21. With M = 10,000 r o2 = nb 2 is 100 times smaller; so the radius of the sphericalcoil is 10 times smaller and the volume V is 1000 times smaller. The chain mass is100 times smaller, so the density, r , is 10 times greater. The dilution is then 105times.

2.22. In a tightest array of spheres the volume fraction is about 0.65; if we consider amaximum volume fraction of 0.3 to exclude all interaction effects, we find in thefirst case ( M = 10 6) a maximum concentration of 0.3/1050 = 0.29 g/l = 0.028 g/dl(the densities of solvent and polymer considering as equal for convenience), and inthe second case ten times 10 higher, so 0.28 g/dl.


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2.23. The slope of sp versus c is greater as M is higher (at lower concentrationsmore effect of entanglements).

2.24. For convenience in kg/mole: w A200 + (1 w A)30 = 81 ; w A = 0.3; w B = 0.7,so 300 kg of A and 700 kg of B.(

M n)A = 40, (

M n)B = 6, (

M z)A = 600, (

M z)B = 90 kg/mole.1/

M n = w A /

M n)A + w B /(

M n)B = 0.3/40 + 0.7/6;

M n = 8.054 = 8054 g/mole.

M z = [w A (

M z)A(

M w)A + w B (

M z)B(

M w)B]/

M w = (0.3600200 + 0.79030)/81 =

467.8 = 467,800 g/mole.A directly produced batch would show:

M n = 81/5 = 16.2 = 16,200 g/mole, and

M z =

381 = 243 = 243,000 g/mole. So the blend, though on-spec as regards its

M w, has

a two times lower

M n and a two times higher

M z as a user would expect. A blownbottle is, therefore, less resistant against stress-corrosion and impact failure at lowtemperatures. As a result of the stronger pronounced melt-elasticity the die-swell ishigher , so the bottle is too thick and its manufacture requires more material!

2.25. a . a is usually between 0.5 and 1.0.b.

M v is usually between

M n and

M w, but closer to

M w as a approaches to a = 1.

c. When a = 1,

M v = M w ; when a > 1,

M v >

M w (this can occur with very stiff

chains).

2.26. When in solution, both substances will cause an increase in boiling point. Atthe same concentration this increase will, for a low-molecular substance, beconsiderably higher than for a polymer (readily a factor 100 of 1000 times greater),since the increase in boiling point is proportional to the number of molecules solved.An even simpler criterion is the viscosity of the solution: with a low-molecularsubstance hardly any change is noticed, where a polymer solution exhibits asignificant increase in viscosity.

2.27. The number average degree of polymerisation P n = 1/(1 p), in which p is thedegree of conversion. The weight average degree of polymerisation

P w =

(1 + p)/(1 p) (see MT 2.2.2). so the heterogeneity index

M w /

M n =P w /

P n = 1 + p ,

close to 2.

2.28. a . HO-C 2H4-OH = 62; HOOC-C 6H4-COOH = 166; the mass ratio is 62:166 =1:2.68.

b. HO-C 2H4-OH + HOOC-C 6H4-COOH = -O-CH 2-CH 2-O-CO-C 6H4-CO- + 2H 2O;

the polymer formed is polyethylene terephtalate (PETP).c. P = number average degree of polymerisation, p = degree of conversion.d . The molar mass of one unit is 192 (= 166 + 62 218), so 40.000/192 =

1/(1 p) and p = 0.9952.

M w is 192(1 + p)/(1 p) = 79,800 g/mole, nearly twice

M n .


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Answers 49

e. Growing chains can close to rings; small acid or basic contamination canterminate chain growth, etc.

2.29. With anionic polymerisation (living polymers) all chains start at the sametime and they grow at the same rate until the monomer is exhausted; the chains are,therefore, equally long.

2.30. M n: ebulliometry (boiling point increase); measurement of end groupconcentration (both for relatively low molar masses); osmometry for higher, but nottoo high molar masses.

M w: light scattering.

M z: sedimentation in solution by ultracentrifugeing.

Further: see MT 2.2.3.

2.31. c (g/dl) t (sec) ( t t o)to (t t o)/ t oc (dl /g )0 1000.05 106.25 0.0625 1.250.1 115 0.15 1.50.15 126.25 0.2625 1.750.2 140 0.40 2.0

In this oversimplified case the limit for c = 0 can easily be determined withoutdrawing a graph: [ ] = 1,0 dl/g.

2.32. [] = kM a, so: 0.5 = k (510 4)a (1)2 = k (810 5)a (2)1 = k ( x )a (3)

From (1) and (2) it follows: 4 = 16 a, so a = 0.5, and from (3) and (1): ( x /510 4)0.5 = 2so x = 200,000.

2.33. a is already known ( a = 0.5); k follows e.g. from: 0.5 = k(510 4)0.5 ,so k = 22.36 10 4.

2.34. a . M n = 1/(0.98/20.000 + 0.02/200) = 6,711

M w = 0.9820,000 + 0.02200 = 19,604

M z = (0.9820,000 2 + 0.02200 2)/19,604 = 19.996so: very strong effect on

M n , small effect on

M w, and hardly any effect on

M z.

b. Here it is easier in kg/mole: M n = 1/(0.98/20 +0.02/2000) = 20,4 = 20,400 g/mol

M w = 0.9820 + 0.022000 =59,6 = 59,600 g/mol

M z = (0.9820 2 + 0.022000 2)/59,6 = 138,490 = 1.384,900 g/molso: hardly any influence on

M n , considerable effect on

M w, very large effect on

M z.


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2.35. We calculate / c, for convenience first in mm/(g/dl):

c (g/dl) (mm) / c0.12 6.50 54.20.18 9.94 55.20.25 14.05 56.20.33 18.8 57.00.45 26.6 59.1

Now we plot / c versus c. Extrapolation to c = 0 gives 52.8 mm/(g/dl).1mm fluid pressure = gh = 903 kg/m 3 9.81 m/sec 2 10 3 m = 46.7 Pa and

1 g/dl = 10 kg/m3. It follows that: /c = 46.77 m

2 /sec

2. This equals RT/

M n =8.314298/

M n , so

M n = 52.97 kg/mole or 52,970 g/mole.

2.36. Measurements of osmotic pressure provide an absolute determination (withoutcalibration) of the number-average molar mass. This is independent of the type of solvent; for each solvent the extrapolation to zero concentration results in the samevalue: / c = RT/ M n . With viscosimetry the determination of M is not absolute;dependent on the solvent and the temperature one finds a value for the intrinsicviscosity, [ ], which is not unique but which has to be calibrated.

c

60

58

56

54

52

500 0.1 0.2 0.3 0.4 0.5 c

Figure A.2. Figure to the answer of problem 2.35.

2.37. The increase of viscosity from a certain concentration becomes morepronounced with increasing M (see Mark-Houwink), but the osmotic pressure isinversely proportional to the molar mass.

2.38. with p = 0.99, P n = 1/(1 p)= 100, with p = 0.995,P n = 200. The mass of the

unit link is: (-CO-(CH 2)4-CO-NH-(CH 2)6-NH-) = 1212 + 214 + 216 + 221 = 226g/mole. For A, therefore:

M n = 100226 = 22,600 g/mole, for B:

M n = 200226 =


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Answers 51

45,200 mole. The weight-average degree of polymerisation isP w = (1 + p)/(1 p) =

199 for A and 399 for B. So the weight average molar masses are 199226 = 44,970(for A) and 399226 = 90,170 (for B). (end groups disregarded for convenience).

2.39. w A = 0.3; w B = 0.7: (

M w)A = 44,970; (

M w)B = 90,170;

M n = 1/(0.3/22.600 + 0.7/45,200) = 34,770 g/mole (see 2.12),

M w = 0.344,970 + 0.790,170 = 76,610 g/mole,

2.40. We can measure rel = / o or sp = ( )/ o at a single concentration c(from the run times t o and t),and calculate red = ( o)/ oc. rel or sp are in itself already convenient data to rank polymers of the same kind after their molar mass. If

a quantitative estimation of [ ] is desired, then these values can be corrected by theknown general dependence of red on c.

2.41. The principle of Problem 2.40 is the basis of the k -value, used with PVC. Ameasurement of the running time through a viscometer at a fixed concentration,compared with that of the pure solvent, results, after a standard calculation, in the k-value, which is, for that polymer, a simple index for the molecular mass. Intechnological terms one knows that a PVC with k = 70 (high M ) does not flow easily,but it provides excellent end-use properties, e.g. in the long-term strength of pipes. Ininjection moulding a better flow is required; then a PVC with k = 55 is moreappropriate.

2.42. The melt flow index is a useful indication of the molar mass, since it is areciprocal measure of the melt viscosity . depends very strongly on: (:) M w3,4 ,(doubling of

M w results in a 10.6 times higher !). This relation is valid for the

zero-shear viscosity; the melt index is measured at a shear stress where the non-Newtonian behaviour, and thus the width of the molar mass distribution, is already

playing a part (see MT 5.3.2). The melt index is a functional measure for the molarmass, because for a producer of end products the processability is often of primaryimportance.

2.43. From the results of the light scattering experiments it follows that A and Bhave the same

M w. The viscosity average,

M v , is for A higher than for B, and is thus

closer to

M w. A has, therefore, a narrower molar mass distribution and a higher

M n .The osmotic pressure is therefore lower.

2.44. Large coils diffuse through the porous gel via the broadest channels, thusfollowing the shortest way, and they arrive as the first at the end. Smaller coils areable to follow all side paths and have a longer travelling time.The result of a GPC-measurement is a curve of wi against M i from which

M w can be

calculated. Conversion to w i /M i) against M i and to wiM i against M i gives thepossibility to calculate

M n and

M z as well.


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2.45. head-head and head-tail polymerisation in irregular sequences (exception). atactic instead of iso- or syndiotactic structure (example: PP). irregular branching (example: PE). with unsaturated chains: cis-, trans- and 1.2 configurations distributed at random

(examples: BR and IR). with copolymers: random distribution of building blocks (examples: SBR and

E/P).

2.46. Isotactic: all R-groups are at the same side of the main chain; syndiotactic: inalternating positions; atactic: randomly situated. All of this has to be considered in3D space due to the tetraedric structure round a C-atom (see MT 2.3).

2.47. An atactic structure is in both cases not crystallisable. Atactic PP is because of its glass-rubber transition temperature ( T g = 15 C) rubbery and technically of nouse. Isotactic PP is able to crystallise and can, therefore, be used in practice. For PSatacticity is no objection; its properties as a glassy polymer are retained up to its T g(95 C).

2.48. by a stiff main chain, e.g. with rings (example: aromatic polyamides). by big side groups (example: PS).

2.49. strong interactions: dipole forces, e.g. in PVC (-Cl) and PMMA(-COOCH 3).

also: hydrogen bridges, e.g. in PA and PU (NH....OC). weak interactions: dispersion forces, e.g. in PE and PP.

2.50. cross link density = number of connected units / total number of units. Anothermeasure is the molar mass between cross-links, M c, The degree of cross-linking canbe determined experimentally from the swelling in a solvent or from the E-modulus.vernettingsdichtheid = aantal verbonden eenheden / totaal aantal eenheden.

2.51. Chemical cross linking is the formation of primary chemical bonds betweenchains, such as sulphur bridges (rubbers) or polystyrene bridges (UP), both betweenunsaturated chains. In saturated chains cross linking can be brought about afterradical formation by irradiation or peroxides. Physical cross linking is based onsecondary bonds (e.g. domains in thermoplastic elastomers) or entanglements.

3. Glassy state and glass-rubber transition

3.1. Similarities: Both are not able to crystallise because of their irregular molecularstructure. Both will, therefore, reach the glassy condition upon cooling down fromthe liquid phase.


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Answers 53

Differences: When a low-molecular glass is heated it will, at its T g , immediately

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