고분자의 Rheological Property 측정방법 i) Rotational type viscometer 종류 - RMS (Rheometrics Mechanical Spectrometer) - RDS (Rheometrics Dynamic Spectrometer) - RSR (Rheometrics Stress Rheometer) - ARES (Advanced Rheometric Expansion System) - Weissenberg Rheogoniometer etc

측정가능한 물성 - Steady shear viscosity (η 0 η ) - Synamic viscosity(η ) - Storage modulus (Grsquo) - Loss modulus(Grdquo) - First normal stress difference 등 (a) dynamic test(η ) ndash

oscillatory shear using the coneampplate fixtures is recommended for materials which are very strain sensitive (ie exhibit linear visco-elastic behavior only for small deformations) since the strain is essentially constant throughout the sample volume

- oscillatory parrel plate measurements

useful for materials in approximately the same range of modulus also preferred when making measurements over a wide range of temp since the plate separation may be varied and the use of a relatively wide plate separation minimizes errors due to thermal expansion of the fixture 주로 temp sweep 때 사용함

(b) steady shear test (η ) ndash

the most widely used fixture if the coneampplate which has the advantage that each volume element of fluid within the sample experiences the same shear rate ndash This is especially important for fluids which are shear thinning

- steady shear viscosity(η ) 와 first normal stress difference 측정가능 - viscosity 측정시 사용하는 transducer 가 10 gr-cm이면 torque 는 1dynecm2

까지 측정가능 이것은 약 10sec-1 shear rate 에서 점도 10 poise 에 해당됨

- 예로서(단위환산) poise equiv dyneseccm2 = gcms 1 Pasec = 10 poise = 10gcmsec 1 Pa = 10gcmsec2 = 10 (gcmsec2)(1cm2) = 10(dynecm2) 참고로 dyne = gcmsec2 N = kgmsec2 Pa = Nm2

- steady shear can also be performed between parallel plates if a sample is to be tested at a variety of temp

- when fluids are manipulated in processes of interest such as extrusion fiber

spinning etc they are almost invariably subjected to a shear field involving continuous deformation

- For this reason it is often desirable to examine the behavior of fluids in steady

shear rather than oscillatory shear

P 499 Chap 16 Rheolocical Properties of Polymer solutions A Dilute Polymer solutions

c

c spηη )0lim(][ rarr=

p 512 The glass transition temperature of polymer solutions - the Tg of a plasticized polymer

s

sgpgsgpg K

TKTTT

θθ

)1(1)(

minus+minus+

=

k constant asymp gppl

gsls

αααα

minusminus

lα volume coefficient of expansion above Tg

gα volume coefficient of expansion below Tg

)11(1

1

KTTTT

p

p

gsgp

ggp

minusminus

minus=

minus

minus

φ

φ (1624)

pφ volume fraction of polymer

p 515 (Ex 162) Estimate the Tg for a solution of PS in benzyl alcohol with pφ = 085

(sol) According to Fig 165 at pφ =085 and K=25 (from page 513 bottom )

290=minusminus

gsgp

ggp

TTTT

(1624)

PS 의 Tg = 373K Benzyl alcohol 의 Tg는 Tm으로부터 estimation

(From Table 162) 32=

ms

gs

TT

there4Tgs = 23 times (258K) = 172K

식 (1624) 에 대입 Tgest = 315K Tgexp = 324K

P 515 A new method for estimating the viscosity of concentrated polymer solutions

P 516 (Ex 163) For concentrated solutions of polyisobutylene in decalin( pφ gt01) estimate

the viscosity as a function of the volume fraction of polymer at a temp of 20degC The

viscosity of the bulk polymer at this temp is 91056 times=pη ( 2mSN sdot )

(sol) (1) Tgp=198K Tgs = 23Tms = 23(230K) = 150K

From K=25

KTTTT

pp

p

gsgp

ggp

φφ

φ

+minus

minus=

minus

minus

1

1 로부터 Tg를 계산할 수 있슴

(2) 그림(154)와 )21( gpp Tηη vs TgT

PIB의 A = 125 (P 469 Table)

6760293198 ==

TTg

53))21(log( minus=gpp Tηη

plogη =98 3135389)21(log =+=there4 gp Tη

(3) From Eq(1626)

pηη = at 01=pφ

pp φηη log5loglog += (1626)

여기서 ηlog 를 구할 수 있슴 (see Table 163) (η 의 unit 은 NmiddotSm2) 여기서 η = viscosity of diluted polymer

η = viscosity of undiluted polymer

Chap 17 Transport of Thermal Energy P 530 (Ex 171) Estimate the Heat conductivity of amorphous PMMA (a) at room temp (b) at 200degC

(Sol) From Eq(174) with Cp=1380(JkgmiddotK)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

여기서 λ heat conductivity Cp specific heat capacity ρ density

L distance between the molecules in ldquoadjacent isothermal layersrdquo U velocity of elastic wave(sound velocity)

UR = 6131 ])1(3

1[VVUV L minus

+sdot (P 443)

Molar Sound velocity function or Rao function

Ulog = 213 )1

)1(3()(

vv

VU R

+minus (173)

P 525 uLC p ρλ asymp (172)

λ heat conductivity

Ulong = 213 ]1

)1(3[)(

vv

VU R

+minus

(73)

v poisson ratio (p374) chap 13 - 먼저 chap 14 (p 447 Table 142) 로부터 Rao function 을 구한다

2 ndashCH3 1400times 2=2800 1 -COO- 12254945

So UR V = )(851711101

4945 smMUR ==

ρ

(UR V)3 = 197times 103 (msec) P 531 According to eq(174)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

=gt Cp = 1380(Jkg∙K) (From P112)

ρ = 1170(Kgm3) L = 5times 10-11 (m) (From Table 171 )

(UR V)3 = 197times 103 (msec) v = 040

)(18001311039711110511701380 KmsJ sdotsdot=timestimestimesminustimestimestimes=there4λ

P 527의 Table 171과 비교

)(1930exp KmsJ sdotsdot=λ

(b) at 200degC 먼저 25degC에서 TTg = 298387 = 077 from fig 172

λ (T) λ (Tg) = 096

λ (Tg) = 9601 (0180) = 0188

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

- 예로서(단위환산) poise equiv dyneseccm2 = gcms 1 Pasec = 10 poise = 10gcmsec 1 Pa = 10gcmsec2 = 10 (gcmsec2)(1cm2) = 10(dynecm2) 참고로 dyne = gcmsec2 N = kgmsec2 Pa = Nm2

- steady shear can also be performed between parallel plates if a sample is to be tested at a variety of temp

- when fluids are manipulated in processes of interest such as extrusion fiber

spinning etc they are almost invariably subjected to a shear field involving continuous deformation

- For this reason it is often desirable to examine the behavior of fluids in steady

shear rather than oscillatory shear

P 499 Chap 16 Rheolocical Properties of Polymer solutions A Dilute Polymer solutions

c

c spηη )0lim(][ rarr=

p 512 The glass transition temperature of polymer solutions - the Tg of a plasticized polymer

s

sgpgsgpg K

TKTTT

θθ

)1(1)(

minus+minus+

=

k constant asymp gppl

gsls

αααα

minusminus

lα volume coefficient of expansion above Tg

gα volume coefficient of expansion below Tg

)11(1

1

KTTTT

p

p

gsgp

ggp

minusminus

minus=

minus

minus

φ

φ (1624)

pφ volume fraction of polymer

p 515 (Ex 162) Estimate the Tg for a solution of PS in benzyl alcohol with pφ = 085

(sol) According to Fig 165 at pφ =085 and K=25 (from page 513 bottom )

290=minusminus

gsgp

ggp

TTTT

(1624)

PS 의 Tg = 373K Benzyl alcohol 의 Tg는 Tm으로부터 estimation

(From Table 162) 32=

ms

gs

TT

there4Tgs = 23 times (258K) = 172K

식 (1624) 에 대입 Tgest = 315K Tgexp = 324K

P 515 A new method for estimating the viscosity of concentrated polymer solutions

P 516 (Ex 163) For concentrated solutions of polyisobutylene in decalin( pφ gt01) estimate

the viscosity as a function of the volume fraction of polymer at a temp of 20degC The

viscosity of the bulk polymer at this temp is 91056 times=pη ( 2mSN sdot )

(sol) (1) Tgp=198K Tgs = 23Tms = 23(230K) = 150K

From K=25

KTTTT

pp

p

gsgp

ggp

φφ

φ

+minus

minus=

minus

minus

1

1 로부터 Tg를 계산할 수 있슴

(2) 그림(154)와 )21( gpp Tηη vs TgT

PIB의 A = 125 (P 469 Table)

6760293198 ==

TTg

53))21(log( minus=gpp Tηη

plogη =98 3135389)21(log =+=there4 gp Tη

(3) From Eq(1626)

pηη = at 01=pφ

pp φηη log5loglog += (1626)

여기서 ηlog 를 구할 수 있슴 (see Table 163) (η 의 unit 은 NmiddotSm2) 여기서 η = viscosity of diluted polymer

η = viscosity of undiluted polymer

Chap 17 Transport of Thermal Energy P 530 (Ex 171) Estimate the Heat conductivity of amorphous PMMA (a) at room temp (b) at 200degC

(Sol) From Eq(174) with Cp=1380(JkgmiddotK)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

여기서 λ heat conductivity Cp specific heat capacity ρ density

L distance between the molecules in ldquoadjacent isothermal layersrdquo U velocity of elastic wave(sound velocity)

UR = 6131 ])1(3

1[VVUV L minus

+sdot (P 443)

Molar Sound velocity function or Rao function

Ulog = 213 )1

)1(3()(

vv

VU R

+minus (173)

P 525 uLC p ρλ asymp (172)

λ heat conductivity

Ulong = 213 ]1

)1(3[)(

vv

VU R

+minus

(73)

v poisson ratio (p374) chap 13 - 먼저 chap 14 (p 447 Table 142) 로부터 Rao function 을 구한다

2 ndashCH3 1400times 2=2800 1 -COO- 12254945

So UR V = )(851711101

4945 smMUR ==

ρ

(UR V)3 = 197times 103 (msec) P 531 According to eq(174)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

=gt Cp = 1380(Jkg∙K) (From P112)

ρ = 1170(Kgm3) L = 5times 10-11 (m) (From Table 171 )

(UR V)3 = 197times 103 (msec) v = 040

)(18001311039711110511701380 KmsJ sdotsdot=timestimestimesminustimestimestimes=there4λ

P 527의 Table 171과 비교

)(1930exp KmsJ sdotsdot=λ

(b) at 200degC 먼저 25degC에서 TTg = 298387 = 077 from fig 172

λ (T) λ (Tg) = 096

λ (Tg) = 9601 (0180) = 0188

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

P 499 Chap 16 Rheolocical Properties of Polymer solutions A Dilute Polymer solutions

c

c spηη )0lim(][ rarr=

p 512 The glass transition temperature of polymer solutions - the Tg of a plasticized polymer

s

sgpgsgpg K

TKTTT

θθ

)1(1)(

minus+minus+

=

k constant asymp gppl

gsls

αααα

minusminus

lα volume coefficient of expansion above Tg

gα volume coefficient of expansion below Tg

)11(1

1

KTTTT

p

p

gsgp

ggp

minusminus

minus=

minus

minus

φ

φ (1624)

pφ volume fraction of polymer

p 515 (Ex 162) Estimate the Tg for a solution of PS in benzyl alcohol with pφ = 085

(sol) According to Fig 165 at pφ =085 and K=25 (from page 513 bottom )

290=minusminus

gsgp

ggp

TTTT

(1624)

PS 의 Tg = 373K Benzyl alcohol 의 Tg는 Tm으로부터 estimation

(From Table 162) 32=

ms

gs

TT

there4Tgs = 23 times (258K) = 172K

식 (1624) 에 대입 Tgest = 315K Tgexp = 324K

P 515 A new method for estimating the viscosity of concentrated polymer solutions

P 516 (Ex 163) For concentrated solutions of polyisobutylene in decalin( pφ gt01) estimate

the viscosity as a function of the volume fraction of polymer at a temp of 20degC The

viscosity of the bulk polymer at this temp is 91056 times=pη ( 2mSN sdot )

(sol) (1) Tgp=198K Tgs = 23Tms = 23(230K) = 150K

From K=25

KTTTT

pp

p

gsgp

ggp

φφ

φ

+minus

minus=

minus

minus

1

1 로부터 Tg를 계산할 수 있슴

(2) 그림(154)와 )21( gpp Tηη vs TgT

PIB의 A = 125 (P 469 Table)

6760293198 ==

TTg

53))21(log( minus=gpp Tηη

plogη =98 3135389)21(log =+=there4 gp Tη

(3) From Eq(1626)

pηη = at 01=pφ

pp φηη log5loglog += (1626)

여기서 ηlog 를 구할 수 있슴 (see Table 163) (η 의 unit 은 NmiddotSm2) 여기서 η = viscosity of diluted polymer

η = viscosity of undiluted polymer

Chap 17 Transport of Thermal Energy P 530 (Ex 171) Estimate the Heat conductivity of amorphous PMMA (a) at room temp (b) at 200degC

(Sol) From Eq(174) with Cp=1380(JkgmiddotK)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

여기서 λ heat conductivity Cp specific heat capacity ρ density

L distance between the molecules in ldquoadjacent isothermal layersrdquo U velocity of elastic wave(sound velocity)

UR = 6131 ])1(3

1[VVUV L minus

+sdot (P 443)

Molar Sound velocity function or Rao function

Ulog = 213 )1

)1(3()(

vv

VU R

+minus (173)

P 525 uLC p ρλ asymp (172)

λ heat conductivity

Ulong = 213 ]1

)1(3[)(

vv

VU R

+minus

(73)

v poisson ratio (p374) chap 13 - 먼저 chap 14 (p 447 Table 142) 로부터 Rao function 을 구한다

2 ndashCH3 1400times 2=2800 1 -COO- 12254945

So UR V = )(851711101

4945 smMUR ==

ρ

(UR V)3 = 197times 103 (msec) P 531 According to eq(174)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

=gt Cp = 1380(Jkg∙K) (From P112)

ρ = 1170(Kgm3) L = 5times 10-11 (m) (From Table 171 )

(UR V)3 = 197times 103 (msec) v = 040

)(18001311039711110511701380 KmsJ sdotsdot=timestimestimesminustimestimestimes=there4λ

P 527의 Table 171과 비교

)(1930exp KmsJ sdotsdot=λ

(b) at 200degC 먼저 25degC에서 TTg = 298387 = 077 from fig 172

λ (T) λ (Tg) = 096

λ (Tg) = 9601 (0180) = 0188

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

P 515 A new method for estimating the viscosity of concentrated polymer solutions

P 516 (Ex 163) For concentrated solutions of polyisobutylene in decalin( pφ gt01) estimate

the viscosity as a function of the volume fraction of polymer at a temp of 20degC The

viscosity of the bulk polymer at this temp is 91056 times=pη ( 2mSN sdot )

(sol) (1) Tgp=198K Tgs = 23Tms = 23(230K) = 150K

From K=25

KTTTT

pp

p

gsgp

ggp

φφ

φ

+minus

minus=

minus

minus

1

1 로부터 Tg를 계산할 수 있슴

(2) 그림(154)와 )21( gpp Tηη vs TgT

PIB의 A = 125 (P 469 Table)

6760293198 ==

TTg

53))21(log( minus=gpp Tηη

plogη =98 3135389)21(log =+=there4 gp Tη

(3) From Eq(1626)

pηη = at 01=pφ

pp φηη log5loglog += (1626)

여기서 ηlog 를 구할 수 있슴 (see Table 163) (η 의 unit 은 NmiddotSm2) 여기서 η = viscosity of diluted polymer

η = viscosity of undiluted polymer

Chap 17 Transport of Thermal Energy P 530 (Ex 171) Estimate the Heat conductivity of amorphous PMMA (a) at room temp (b) at 200degC

(Sol) From Eq(174) with Cp=1380(JkgmiddotK)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

여기서 λ heat conductivity Cp specific heat capacity ρ density

L distance between the molecules in ldquoadjacent isothermal layersrdquo U velocity of elastic wave(sound velocity)

UR = 6131 ])1(3

1[VVUV L minus

+sdot (P 443)

Molar Sound velocity function or Rao function

Ulog = 213 )1

)1(3()(

vv

VU R

+minus (173)

P 525 uLC p ρλ asymp (172)

λ heat conductivity

Ulong = 213 ]1

)1(3[)(

vv

VU R

+minus

(73)

v poisson ratio (p374) chap 13 - 먼저 chap 14 (p 447 Table 142) 로부터 Rao function 을 구한다

2 ndashCH3 1400times 2=2800 1 -COO- 12254945

So UR V = )(851711101

4945 smMUR ==

ρ

(UR V)3 = 197times 103 (msec) P 531 According to eq(174)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

=gt Cp = 1380(Jkg∙K) (From P112)

ρ = 1170(Kgm3) L = 5times 10-11 (m) (From Table 171 )

(UR V)3 = 197times 103 (msec) v = 040

)(18001311039711110511701380 KmsJ sdotsdot=timestimestimesminustimestimestimes=there4λ

P 527의 Table 171과 비교

)(1930exp KmsJ sdotsdot=λ

(b) at 200degC 먼저 25degC에서 TTg = 298387 = 077 from fig 172

λ (T) λ (Tg) = 096

λ (Tg) = 9601 (0180) = 0188

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

Chap 17 Transport of Thermal Energy P 530 (Ex 171) Estimate the Heat conductivity of amorphous PMMA (a) at room temp (b) at 200degC

(Sol) From Eq(174) with Cp=1380(JkgmiddotK)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

여기서 λ heat conductivity Cp specific heat capacity ρ density

L distance between the molecules in ldquoadjacent isothermal layersrdquo U velocity of elastic wave(sound velocity)

UR = 6131 ])1(3

1[VVUV L minus

+sdot (P 443)

Molar Sound velocity function or Rao function

Ulog = 213 )1

)1(3()(

vv

VU R

+minus (173)

P 525 uLC p ρλ asymp (172)

λ heat conductivity

Ulong = 213 ]1

)1(3[)(

vv

VU R

+minus

(73)

v poisson ratio (p374) chap 13 - 먼저 chap 14 (p 447 Table 142) 로부터 Rao function 을 구한다

2 ndashCH3 1400times 2=2800 1 -COO- 12254945

So UR V = )(851711101

4945 smMUR ==

ρ

(UR V)3 = 197times 103 (msec) P 531 According to eq(174)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

=gt Cp = 1380(Jkg∙K) (From P112)

ρ = 1170(Kgm3) L = 5times 10-11 (m) (From Table 171 )

(UR V)3 = 197times 103 (msec) v = 040

)(18001311039711110511701380 KmsJ sdotsdot=timestimestimesminustimestimestimes=there4λ

P 527의 Table 171과 비교

)(1930exp KmsJ sdotsdot=λ

(b) at 200degC 먼저 25degC에서 TTg = 298387 = 077 from fig 172

λ (T) λ (Tg) = 096

λ (Tg) = 9601 (0180) = 0188

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

2 ndashCH3 1400times 2=2800 1 -COO- 12254945

So UR V = )(851711101

4945 smMUR ==

ρ

(UR V)3 = 197times 103 (msec) P 531 According to eq(174)

213 ]1

)1(3[)(

υυ

ρλ

+minus

=sdot V

UL

CR

p

=gt Cp = 1380(Jkg∙K) (From P112)

ρ = 1170(Kgm3) L = 5times 10-11 (m) (From Table 171 )

(UR V)3 = 197times 103 (msec) v = 040

)(18001311039711110511701380 KmsJ sdotsdot=timestimestimesminustimestimestimes=there4λ

P 527의 Table 171과 비교

)(1930exp KmsJ sdotsdot=λ

(b) at 200degC 먼저 25degC에서 TTg = 298387 = 077 from fig 172

λ (T) λ (Tg) = 096

λ (Tg) = 9601 (0180) = 0188

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

p 531 at T = 200 + 273 = 473K

221387473 ==

gTT

From Fig 172

950)()473(

=gTλ

λ

)(1780)950(1880)473( KmsJ sdotsdot==there4λ

P 532 (Ex 172)

Calculate the λ of PET at 40 crystallinity rArr λ (at roomp) = 0218 ( KmsJ sdotsdot )

(Sol) 4651=cρ 3351=aρ Eq(177)로부터

)1(851 minus=minusa

cc

ρρ

λλ

34005612180))133514651(851(2180 =times=minus+=cλ

여기서 58121803400 ==a

c

λλ and Xc = 04

(see Fig174 in P531)

360asympminusminus

ac

a

λλλλ or (From P531 Fig 174 )

36021803400

2180 =minus

minusλ

λ = 0044 + 0218 = 0262 )( KmsJ sdotsdot

expλ = 0272 )( KmsJ sdotsdot

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

Chap18

Properties Determining Mass Transfer in Polymeric Systems A Permeation of simple Gases P (permeability) S (solubility) D (diffusivity) P = Stimes D

P 536 - permeation is a sequential process starting with solution of the gas on the outer surface of the polymer followed by slow inward diffusion

P 542 Ex 181) Estimate the solubility and the heat of solution (sorption) of oxygen in PET

both in the quenched amorphous glassy state and semi-crystalline state (Xc = 045)

Sol) i) Tg(PET) = 345K For O2 εk=107 From Eq(187b) ε(potential energy constant) k(Boltzman const) log S(298) = -74+0010εk plusmn 06 = -633 plusmn 06

there4 S = 56times 10-7 SExp = 99times 10-7

ii) For semicrystalline state

Ssc(298) = Sa(298)(1-Xc) = 56times 10-7(055) = 31times 10-7

Ssc(Exp) = 74times 10-7

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

P 552 (Ex 182) Estimate the diffusivity at 298K and the activation energy of diffusion for oxygen in PET both in the glassy and in the semicrystalline state (sol) For the derivation of ED (activation energy of diffusion ) we use Fig 183 Tg = 345K P = 675 plusmn 15 (from Eq1812b p 552)

25165)51756(830)51756()38

347()66

(10 22

2

23 plusmn=plusmn=plusmnsdot=sdot=sdot minus PNO

RED (KJmol)

10-3 times ED = (83)56 plusmn (83)125 = 465 plusmn 105 (KJmol) EDExp = 461 ~ 485

From Eq(1813b) (p552) log D0 = 10-3 ED R ndash 50 plusmn 08 = 06 plusmn 08

From Eq(1815) log D(298) = log D0 ndash 146times 10-3EDR = 76 plusmn 08

there4 Da(298) = 25times 10-8 (cm2sec)

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

P 560 Ex 183 Estimate the permeability P(298) for oxygen of two polymer films one a neoprene rubber film and the other a PVC film

P 571 Ex 184 Estimate the moisture content of nylon-6 at 25degC and a relative humidity of 07 The crystallinity is 70

P 581 Ex 185 Estimate the rate of dissolution of PS in toluene at 35degC(308K)

(a) at a very low Reynolds number (NRe asymp 0 ) (b) at a Reynolds number of 100D PS(Mw)=150000 D = toluene in PS at 35degC is 15times 10-6

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

Chap 19 Crystallization and Recrystallization A1 Crystallinity - the change from a random liquid structure to a well ordered periodic crystalline

structure can occur this transformation is called crystallization the reverse is called melting

- Many polymeric solids consistlargely of folded chain lamellae

A2 Nucleation and Growth

P 586 - the theory is based on the assumption that in supercooled melts there occur

fluctuations leading to the formation of a new phase The phase transformation begins with the appearance of a number of very small

particles of the new phase (nucleation) (Ex 191) Estimate for isotactic PS a the temperature of maximum crystallization velocity b the linear growth rate at this temp c the probable(maximum) degree of crystallization a) Two methods of estimation are available eq(198) and Fig(196b) i) From eq(198) Tk cong 05 (Tm + Tg ) = 05 (513 + 373 ) = 443 K ii) using Fig(196b) with TgTm = 373513 = 07

여기서 860max =m

x

TT

so Txmax = 441K Texp = 449 K

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement

b) applying Eq(1916) with T=Txmax= 442K

i) log 50320 TTTT

TTT

vv

mgm

mm

minus+

minusminusasymp (1916)

여기서 v0 = 1012 (nmsec) (universal const For crystalline polymer)

P 661 log v= 12-23 )7150

373513513(

442513 +

minus = 12 - 47 = 03 maxv = 20

ii) From Fig 196-a TgTm = 0725 그림으로부터 log maxv = 05

the average log maxv =2

5030 + =04

there4 maxv = 1004 = 25(nmsec) expmaxv = 42(nmsec)

c) Maximum degree of crystallinity Applying Fig 196c TgTm = 0725 Xcmax = 03 Xcmaxexp = 034 in good agreement