Solutions Manual for
Polymer Science and
Technology
Third Edition
Joel R. Fried
Upper Saddle River, NJ • Boston • Indianapolis • San Francisco
New York • Toronto • Montreal • London • Munich • Paris • Madrid
Capetown • Sydney • Tokyo • Singapore • Mexico City
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
The author and publisher have taken care in the preparation of this book, but make no
expressed or implied warranty of any kind and assume no responsibility for errors or
omissions No liability is assumed for incidental or consequential damages in connection
with or arising out of the use of the information or programs contained herein.
Visit us on the Web: InformIT.com/ph
Copyright © 2015 Pearson Education, Inc.
This work is protected by United States copyright laws and is provided solely for the use
of instructors in teaching their courses and assessing student learning. Dissemination or
sale of any part of this work (including on the World Wide Web) will destroy the
integrity of the work and is not permitted. The work and materials from it should never
be made available to students except by instructors using the accompanying text in their
classes. All recipients of this work are expected to abide by these restrictions and to
honor the intended pedagogical purposes and the needs of other instructors who rely on
these materials.
ISBN-10: 0-13-384559-1
ISBN-13: 978-0-13-384559-4
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
SOLUTIONS TO PROBLEMS IN POLYMER SCIENCE AND TECHNOLOGY,
3RD EDITION
TABLE OF CONTENTS
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 7
Chapter 11
Chapter 12
Chapter 13
1
5
14
24
28
36
40
51
52
CHAPTER 1
1-1 A polymer sample combines five different molecular-weight fractions, each of equal weight. The
molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000.
Calculate M n , M w , and M z . Based upon these results, comment on whether this sample has a
broad or narrow molecular-weight distribution compared to typical commercial polymer samples.
Solution
Fraction #
1
2
3
4
5
Σ
=
Mn
5
Wi N
∑=
i =1
Mi (×10-3)
20
40
60
80
100
300
Wi
1
1
1
1
1
5
Ni = Wi/Mi (×105)
5.0
2.5
1.67
1.25
1.0
11.42
5
=
43,783
1.142 × 10−4
5
=
Mw
∑W M
i
i
=
5
∑Wi
i =1
300,000
= 60,000
5
i =1
5
Mz
=
∑W M
i
2
i
=
∑Wi M i
i =1
5
4 × 108 + 16 × 108 + 36 × 108 + 64 × 108 + 100 × 108
= 73,333
3 × 105
i =1
M z 60,000
=
= 1.37 (narrow distribution)
M n 43,783
1-2 A 50-gm polymer sample was fractionated into six samples of different weights given in the table
below. The viscosity-average molecular weight, M v , of each was determined and is included in the table.
Estimate the number-average and weight-average molecular weights of the original sample. For these
calculations, assume that the molecular-weight distribution of each fraction is extremely narrow and can
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
1
be considered to be monodisperse. Would you classify the molecular weight distribution of the original
sample as narrow or broad?
Fraction
1
2
3
4
5
6
Weight
(gm)
1.0
5.0
21.0
15.0
6.5
1.5
Mv
1,500
35,000
75,000
150,000
400,000
850,000
Solution
Let M i ≈ M v
=
Mn
6
W N
∑=
i
i =1
Fraction
Wi
Mi
1
2
3
4
5
6
Σ
1.0
5.0
21.0
15.0
6.5
1.5
50.0
1,500
35,000
75,000
150,000
400,000
850,000
Ni = Wi/Mi
(×106)
667
143
280
100.
16.3
1.76
1208
WiMi
1500
175.000
627,500
2,250,000
2,600,000
1,275,000
7,929,000
50.0
= 41,322
1.21 × 10−3
6
=
Mw
∑W M
i
i
=
6
∑Wi
i =1
7,930,000
= 158,600
50.0
i =1
M w 158, 600
=
= 3.84 (broad distribution)
Mn
41,322
1-3 The Schultz–Zimm [11] molecular-weight-distribution function can be written as
=
W (M )
a b +1
M b exp ( − aM )
Γ ( b + 1)
where a and b are adjustable parameters (b is a positive real number) and Γ is the gamma function (see
Appendix E) which is used to normalize the weight fraction.
(a) Using this relationship, obtain expressions for M n and M w in terms of a and b and an expression for
M max , the molecular weight at the peak of the W(M) curve, in terms of M n .
Solution
Mn =
∫
∞
0
∞
WdM
∫ (W
0
M ) dM
let t = aM
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
2
∫
∞
0
∞
1 ∞ b
1
a b +1
a b +1
b
d
t
a
t exp ( −=
t ) dt
−
=
Γ ( b=
+ 1) 1
exp
t
a
t
(
)
(
)
(
)
∫
Γ ( b + 1) 0
Γ ( b + 1) a b +1 ∫0
Γ ( b + 1)
=
WdM
∞
∫0 (W
M=
) dM
∞
a b +1
a b +1 1
b −1
−
=
d
t
a
t
a
t
exp
(
)
(
)
(
)
Γ ( b + 1) ∫0
Γ ( b + 1) a b
∫
∞
0
−t ) dt
t b −1 exp (=
a b +1 1
=
Γ (b)
Γ ( b + 1) a b
a
a
Γ (b) =
bΓ ( b )
b
1
b
=
ab a
M
=
n
∫
M w=
∞
0
∫
WMdM
∞
0
=
WdM
∫
∞
0
WMdM =
b +1
∞
a b +1
a b +1 Γ ( b + 2 )
t
exp
−
=
=
t
d
t
a
a
(
)
(
)
(
)
Γ ( b + 1) ∫0
Γ ( b + 1) a b + 2
( b + 1) Γ ( b + 1) = b + 1
aΓ ( b + 1)
a
(b) Derive an expression for Mmax, the molecular weight at the peak of the W(M) curve, in terms of M n .
Solution
dW
a b +1
bM b −1 exp ( − aM ) + M b ( − a=
=
) exp ( −aM ) 0
dM Γ ( b + 1) 
bM b − a = aM b
b
a
= M
=
M n (i.e., the maximum occurs at M n )
a
(c) Show how the value of b affects the molecular weight distribution by graphing W(M) versus M on the
same plot for b = 0.1, 1, and 10 given that M n = 10,000 for the three distributions.
Solution
b
a=
10,000
b
a
=
W
0.1
1×10-5
1
1×10-4
10
1×10-3
a b +1
M b exp ( − aM ) dM
Γ ( b + 1)
where
=
Γ ( b + 1)
∞
∫ ( aM )
0
b
exp ( − aM ) dM .
Plot W(M) versus M
Hint:
∫
∞
0
x n exp ( − ax ) dx =
Γ ( n + 1) a n +1 =
n ! a n +1 (if n is a positive interger).
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
3
1-4 (a) Calculate the z-average molecular weight, M z , of the discrete molecular weight distribution
described in Example Problem 1.1.
Solution
3
=
Mz
∑W M
i
2
i
=
∑Wi M i
i =1
3
1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 )
= 80,968
1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 )
2
2
2
i =1
(b) Calculate the z-average molecular weight, M z , of the continuous molecular weight distribution
shown in Example 1.2.
Solution
M dM ( M 3)
∫=
=
MdM
( M 2)
∫
105
Mz
=
2
3
105
2
103
105
3
10
105
103
66,673
103
(c) Obtain an expression for the z-average degree of polymerization, X z , for the Flory distribution
described in Example 1.3.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
4
Solution
∞
∑1 X 2W ( X )
Xz =
=
∞
∑ XW ( X )
1
∞
∑X
3
p x −1
∑X
2
p x −1
1
∞
1
Let
∞
1
A =∑ Xp x −1 =1 + 2 p + 3 p 2 +  =
1− p
1
(geometric series)
∞
1 + 22 p + 32 p 2 + 
B=
∑ X 2 p x −1 =
1
∞
C=
1 + 23 p + 32 p 2 + 
∑ X 3 p x −1 =
1
Can show that B (1 − p ) = A (1 + p )
Therefore B =
1+ p
(1 − p )
Write C (1 − p ) =
3
∞
=
x 1
Therefore C =
∞
∞
∑ 3 X 2 p x −1 − ∑ 3 Xp x −1 + ∑ p x −1 = 3B − 3 A2 +
=
x 1=
x 1
1
1 + 4 p + p2
=
3
1− p
(1 − p )
1 + 4 p + p2
(1 − p )
4
∞
∑X
p x −1
3
2
C 1 + 4 p + p (1 − p )
1
and finally X=
= =
=
z
∞
4
B
2 x −1
p
1
−
1
+
p
(
)
(
)
∑X p
3
1 + 4 p + p2
1 + 4 p + p2
=
1 − p2
(1 − p )(1 + p )
1
Mz = Mo X z
CHAPTER 2
2.1 If the half-life time, t1/2, of the initiator AIBN in an unknown solvent is 22.6 h at 60°C, calculate its
dissociation rate constant, kd, in units of reciprocal seconds.
Solution
=
[ I] [ I]o exp ( −kd t )
[ I]=
[ I]o
1
= exp ( −kd t )
2
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
5
− kd t =
ln (1 2 ) =
−0.693
=
kd
0.693 0.693 h
=
= 8.52 × 10−5 s -1
t
22.6 h 3600 s
2.2 Styrene is polymerized by free-radical mechanism in solution. The initial monomer and initiator
concentrations are 1 M (molar) and 0.001 M, respectively. At the polymerization temperature of 60°C, the
initiator efficiency is 0.30. The rate constants at the polymerization temperature are as follows:
kd = 1.2 × 10-5 s-1
kp = 176 M-1 s-1
kt = 7.2 × 107 M-1 s-1
Given this information, determine the following:
(a) Rate of initiation at 1 min and at 16.6 h
Solution
Ri =
2 fkd [ I ] =
2 ( 0.30 ) (1.2 × 10−5 ) [ I ] =
7.2 × 10−6 [ I ]
=
[ I] [ Io ] exp ( −kd t )
at 1 min:
=
=
( 0.9993) 0.0009993 M
[ I] 0.001
Ri =
7.19 × 10−9 M s -1
( 7.2 ×10−6 ) ( 0.0009993) =
at 16.6 h:
=
=
( 0.488) 0.000488 M
[ I] 0.001
Ri =
3.51 × 10−9 M s -1
( 7.2 ×10−6 ) ( 0.000448) =
(b) Steady-state free-radical concentration at 1 min
Solution
12
 fk 
[ IM x ⋅] = d 
 kt 
at 1 min:
[ I]
12
 ( 0.30 ) (1.2 × 10−5 ) 
12

 ( 0.0009993=
) 7.08 × 10−9 M
[ IM=
]
x
7
7.2
×
10


(c) Rate of polymerization at 1 min
12
Solution
=
Ro kp [ IM x ⋅][ M ]

M ] [ M ]o exp ( − kp [ IM x =
⋅] t ) (1) exp  −176 ( 7.08 × 10−9 ) 60
=
[=

−9
−6
−1
Ro =×
176 ( 7.08 10 ) ( 0.9999 ) =
1.24 × 10 M s
0.9999 M
(d) Average free-radical lifetime, τ, at 1 min, where τ is defined as the radical concentration divided by
the rate of termination
Solution
τ
=
[ IM x ⋅]
1
1
=
=
= 0.981 s
2
7
2k t [ IM x ⋅] 2 ( 7.2 × 10 )( 7.08 × 10−9 )
2k t [ IM x ⋅]
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
6
(e) Number-average degree of polymerization at 1 min
Solution
Rp 1.24 × 10−6
1.24 × 10−6
1.24 × 10−6
Xn =
=
=
==
= 172
Rt 2k t [ IM x ⋅]2 2 ( 7.2 × 107 )( 7.08 × 10−9 )2
7.22 × 10−9
2.3 It has been reported that the rate of a batch photopolymerization of an aqueous acrylamide solution
using a light-sensitive dye is proportional to the square of the monomer concentration, [M]2, and the
square root of the absorbed light-intensity, I1/2. Note that, although this polymerization is free radical, the
apparent kinetics appear not to be typical of usual free-radical polymerization for which the rate of
polymerization is proportional to the first power of monomer concentration and to the square root of the
initiator concentration (eq. (2.25)). The following polymerization mechanism has been proposed to
explain the observed kinetics:
Initiation
k ,hν
1
M + D 
→ R
k
2
R + M →
RM1 
Propagation
k
3
RM1 + M 
→ RM 2 
. .
.
. .
.
. .
.
k
3
RM n + M 
→ RM n +1 
Termination
k
4
RM n  + RM n  →
P
k
5
R  
→S
where
M, monomer
D, dye
P, terminated polymer
S, deactivated initiator
Show that this mechanism appears to be correct by deriving an equation for the rate of propagation in
terms of [M], I, and the appropriate rate constants. The following assumptions may be made:
1. Equal reactivity in the propagation steps
2. Steady-state concentration of R• and RMn•
3. k2 << k5
4. The concentration of dye, [D], that has been activated by light and thereby contributes to the first
initiation step is proportional to the absorbed light intensity.
Solution
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
7
=
Ro
−d [ M ]
= k3 [ M ][ RM n ⋅]
dt
12
 k2

steady state for RM n ⋅ : k2 [ R ⋅][ M=
] k4 [ RM n ⋅] or [ RM=
 [ R ⋅][ M ]  (eq. 2.3.1)
n ⋅]
 k4

k1 [ M ][ D ]
(eq. 2.3.2)
steady state for [ R ⋅] : k1 [ M ][ D ] = k2 [ R ⋅][ M ] + k5 [ R ⋅] or [ R ⋅] =
k 2 [ M ] + k5
substituting eq. 2.3.2 into eq. 2.3.1, gives
2
 k k1 [ M ]2 [ D ] 

[ RM n ⋅] = 2
k k M +k 
 4 2[ ] 5 
12
 k k1 [ M ][ D ]2
and Ro = k3 [ M ]  2
 k 4 k 2 [ M ] + k5

12




12
kk 
2
2
Letting [ D ] = I and k5 >> k2 [ M ] , we have Ro = k3  1 2  [ M ] I1 2 or Ro ∝ [ M ] I1 2
 k 4 k5 
See G.K. Oster, G. Oster, & G. Prati, JACS 79, 595 (1957).
2.4 Reactivity ratios for styrene and 4-chlorostyrene are given in Table 2-6.
(a) Using these values, plot the instantaneous copolymer composition of poly(styrene-co-4-chlorostyrene) as a function of comonomer concentration in the copolymerization mixture.
Solution
Styrene, 1; 4-chlorostyrene, 2
Q1
1.00
=
r1
exp −
e1 ( e1 −=
e2 ) 
exp 0.8 ( −0.8 + 0.33
=
)  0.667

Q2
1.03
=
r2
1.03
exp 0.33 ( −0.33 +=
0.8 )  1.203
1.00
(b) Comment on the expected monomer sequence distribution in the resulting copolymer.
Solution
F1 =
r1 f12 + f1 f 2
r1 f12 + 2 f1 f 2 + r2 f 2 2
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
8
1
0.9
r1 = 0.667; r2 = 1.203
0.8
ideal
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
f1
(c)
Near ideal with copolymer enrichment in the more reactive monomer, 4ClS.
2.5 If the number-average degree of polymerization for polystyrene obtained by the bulk polymerization
of styrene at 60°C is 1000, what would be the number-average degree of polymerization if the
polymerization were conducted in a 10% solution in toluene (900 g of toluene per 100 g of styrene) under
otherwise identical conditions? The molecular weights of styrene and toluene are 104.12 and 92.15,
respectively. State any assumptions that are needed.
Solution
[SH ]
1
1
=
+C
Xn ( Xn )
[M]
o
=
C 1.25 × 10−5 (Table 2.4)
( X n )o = 1000
=
[M]
=
[SH ]
100
= 0.9604
104.12
900
= 9.767
92.15
1
1
9.767
=
+ 0.125 × 10−4
=1.127 × 10−3 or X n = 887
X n 1000
0.9604
2.6 Assume that a polyesterification is conducted in the absence of solvent or catalyst and that the
monomers are present in stoichiometric ratios. Calculate the time (min) required to obtain a numberaverage degree of polymerization of 50 given that the initial dicarboxylic acid concentration is 3 mol L-1
and that the polymerization rate constant is 10-2 L mol-1 s-1.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
9
Solution
Xn =
[ A − A ]o kt + 1
t
=
X n −1
50 − 1
1633 s or 27.2 min
= =
[ A − A ]o k 3 × 10−2
2.7 Show how the assumption of steady-state free radical concentration,
M1 or
M2 , can be used
to obtain the instantaneous copolymerization equation in the form of eq. (2.45) starting with eq. (2.39).
Solution
d [ M1 ] [ M1 ] k11 [ ~M1 ⋅] + k21 [ ~M 2 ⋅]
=
d [ M 2 ] [ M 2 ] k12 [ ~ M1 ⋅] + k22 [ ~M 2 ⋅]
Steady state in ~ M1 ⋅ gives
[ M1 ] k11 [ ~M1 ⋅] [ ~M 2 ⋅] + k21
[ M 2 ] k12 [ ~M1 ⋅] [ ~M 2 ⋅] + k22
[ ~ M1 ⋅] = k21 [ M1 ]
[ ~ M 2 ⋅] k12 [ M 2 ]
Substitution in the equation given above, rearranging, and introducing the definitions of the reactivity
ratios gives
d [ M1 ] [ M1 ]  r1 [ M1 ] + [ M 2 ] 
=


d [ M 2 ] [ M 2 ]  [ M1 ] + r2 [ M 2 ] 
2.8 Show that the ceiling temperature in a free-radical polymerization can be obtained as
Tc =
−∆H p
Rln ( Ap [ M ] Adp )
.
Solution
Rp = Rdp
kp [ M ][=
M x ⋅] kdp [ M x ⋅]
 Ep 
 Edp
AP exp  −
 [ M ] =Adp exp  −
 RTc 
 RTc
Ep − Edp −∆H p
ln { Ap [ M
=
] Adp } =
RTc
RTc



and Tc =
{
−∆H p
}
R ln ( Ap [ M ] Adp )
2.9 Find the azeotropic composition for the free-radical copolymerization of styrene and acrylonitrile.
Solution
Substituting f1 = F1 and rearranging eq. 2.45 (using f 2 = 1 − f1 ) gives
( r1 − 2 + r2 ) f12 + ( 2 − 2r2 − r1 + 1) f1 + r2 − 1 =0
Using reactivity ratios given in Table 2-6 (r1 = 0.290 and r2 = 0.020) and solving the quadratic equation,
gives F=
f=
0.580 .
1
1
2.10 Describe the copolymer composition that would be expected in the free-radical copolymerization
of styrene and vinyl acetate.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
10
Solution
Since the reactivity ratio for vinyl acetate is not given in Table 2-6, try Q-e scheme using values given in
Table 2-7
 1.00 
=
=
r1 
+ 0.88 )  41.0 (styrene)
 exp 0.80 ( −0.80
 0.026 
 0.026 
=
=
r2 
+ 0.80 )  0.0242 (vinyl acetate)
 exp 0.88 ( −0.88
 1.0 
Values of 55 and 0.01 are sometimes reported; experimental values range from 18.8 to 60 for r1 and –0.04
to 0.16 for r2 (Polymer Handbook, 4th ed)
k11
k22
, there is a tendency for consecutive homopolymerization since M1 (styrene) will
Since =
r1
>>=
r2
k12
k21
polymerize until it is completely consumed and then M2 (vinyl acetate) will polymerize.
2.11 Explain why high pressure favors the propagation step in a free-radical polymerization. How would
the rate of termination be affected by pressure?
Answer
It would be expected that pressure would increase the rate of propagation but decrease the rate of
termination. This is supported by studies of styrene polymerization (Ogo, Macromol. Sci.-Rev. Macromol.
Chem. Phys. C24, 1 (1984)). One argument that can be made is that pressure increases viscosity and,
therefore, the diffusion of long-chain radicals is reduced (i.e., the rate of termination decreases). Kiran &
Saraf (J. Supercritical Fluids 3, 198 (1990) discuss volume production arguments. Net volume decreases
during propagation and is, therefore, favored at high pressure. On the other hand, net volume increase
during termination and unfavored at high pressure.
2.12 From data available in Section 2.2.1, calculate the activation energy for propagation for the freeradical polymerization of styrene. Do you expect the activation energy to be dependent upon solvent in a
solution polymerization?
Solution
Using data for styrene bulk polymerization in Table 2-3 and
=
kp Ap exp ( − Ep RT )
R = 8.3144 J mol-1 K-1
Plot gives slope = –3.925×103; using this value gives Ep = 32.6 kcal mol-1
Polymer Handbook, 4th ed., cites a value of 31.5 kcal mol-1
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
11
6
5
4
3
2
1
0
0.003
0.0031
0.0032
0.0033
0.0034
-1
1/T (K )
2.13 Draw the chemical structures of the two ends of a terminated polystyrene chain obtained by the
atom transfer radical polymerization of styrene using 1-phenylethyl chloride (1-PECl) as the initiator,
CuCl as the catalyst, and 2,2’-bipyridine as the complexing agent.
Solution
1-phenyl chloride initiator
Cl
CH3
CH2
CH
See Wang & Matyjaszewski, Macromolecules 28, 7901 (1995) or Coessens et al., Progr. Polym. Sci. 26,
337 (2001).
2.14 Show that the rate of polymerization in atom transfer radical polymerization is proportional to the
equilibrium constant defined in eq. (2.50).
Solution
Rp = kp [ Pn ⋅][ M ] = K e
[ Pn − X ][Cu(I)X ] M
[ ]
[Cu(II)X 2 ]
2.15 Show that azeotropic copolymerization occurs when the feed composition is given as
f1 =
Solution
At azeotrope:
d [ M1 ]
1 − r1 .
2 − r1 − r2
[ M1 ]
d [M2 ] [M2 ]
=
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
12
From eq. 2.42: r1 [ M1 ] + [ M 2 ] = [ M1 ] + r2 [ M 2 ]
Dividing both side by [M2] gives:
1− r
[ M1 ] = 2 [ M 2 ]
1 − r1
Dividing by [M1]+[M2] gives
1 − r2
=
f1
(1 − f1 )
1 − r1
Rearrangement gives
1 − r2
2 − r1 − r2
Substituting values of r1 and r2 from problem 2.9 gives the same result, f1 = 0.580 at the azeotrope.
f1 =
2.16 Methyl methacrylate is copolymerized with 2-methylbenzyl methacrylate (M1) in 1,4-dioxane at
60°C using AIBN as the free-radical initiator.
(a) Draw the repeating unit of poly(2-methylbenzyl methacrylate).
Solution
CH3
CH2
C
C
O
O CH2
H3 C
(b) From the data given in the table below, estimate the reactivity ratios of both monomers.
f1
F1*
0.10
0.25
0.50
0.75
0.90
0.14
0.33
0.52
0.70
0.87
* From 1H-NMR measurements
Solution
Data can be fitted to Eq. 2.45 using nonlinear regression analysis. Alternately (and less preferred) is the
traditional linearization of the instantaneous copolymerization equation in the form (see Flory, Principle
of Polymer Chemistry, pp. 185–189)
=
G r1 H − r2
where
f1  F2 
=
G
1 − 
f2 
F1 
and
2
 f  F
H = 1  2
 f 2  F1
A plot of G versus H gives r1 from the slope and r2 from the intercept.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
13
Solutions reported in I. Erol, C. Soykan, J. Macromol. Sci.: Part A – Pure & Applied Chemistry A39,
953 (2002) give average values of r1 = 1.03 and r2 = 0.77. A nonlinear regression (Matlab) gives r1 =
0.6311 and r2 = 0.5328.
CHAPTER 3
3.1 Polyisobutylene (PIB) is equilibrated in propane vapor at 35°C. At this temperature, the saturated
vapor pressure (p1o) of propane is 9050 mm Hg and its density is 0.490 g cm-3. Polyisobutylene has a
molecular weight of approximately one million and a density of 0.915 g cm-3. The concentration of
propane, c, sorbed by PIB at different partial pressures of propane (p1) is given in the following table.
Using this information, determine an average value of the Flory interaction-parameter, χ12, for the PIB–
propane system.
p1 (mm
Hg)
496
941
1446
1452
c (g propane/g
PIB)
0.0061
0.0116
0.0185
0.0183
Solution
CH3
CH2
C
n
CH3
;r
M o = 56.11=
1
106
−
0.9999 ≈ 1
= 1.78 × 104 and 1 =
r
56.1
w1
; w1 = ρ1V1 ; w2 = ρ 2V2
w2
ρV
V
ρ
c = 1 1 or 2 = 1
ρ 2V2
V1 ρ 2 c
ρ1
V1
V
1
=+
φ1 =
1 2 =+
1
or
φ1
ρ2c
V1 + V2
V1
p1
p1
=
a1 =
o
p1 9050
c
c=
0.0061
0.0116
0.0183
0.0185
φ1
0.01126
0.02120
0.03304
0.03339
p1
496
941
1452
1446
lna1
-2.9039
-2.2636
-1.8298
-1.8340
ave.
χ
12
0.6075
0.6381
0.6559
0.6410
0.64
See S. Prager, E. Bagley, and F. A. Long, JACS 75, 2742 (1953).
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
14
3.2 The following osmotic pressure data are available for a polymer in solution:
c (g dL-1)
0.32
0.66
1.00
1.40
1.90
h (cm of solvent)
0.70
1.82
3.10
5.44
9.30
Given this information and assuming that the temperature is 25°C and that the solvent density is 0.85 g
cm-3, provide the following:
(a) A plot of Π/RTc versus concentration, c
Solution
=
Π ρ=
gh 0.85 ( 980.665 ) h
c (g dL-1)
h (cm)
0.32
0.66
1.00
1.40
1.90
0.70
1.82
3.10
5.44
9.30
Π(
RTc ) ×106
7.310
9.216
10.360
12.985
16.358
18
16
14
12
10
8
6
4
2
0
0
0.2
0.4
0.6
0.8
c (g
1
1.2
1.4
1.6
1.8
2
dL-1 )
(b) The molecular weight of the polymer and the second virial coefficient, A2, for the polymer solution
Solution
Π
1
≈
+ 2 A2 c
RTc M n
Least squares fit of data ( R 2 = 0.9884 ) gives slope = 5.654 × 10-6 ( M n = 176,866 ) and intercept = 5.276
A
(=
2
2.64 × 10−6 ) .
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
15
3.3 (a) What is the osmotic pressure (units of atm) of a 0.5 wt % solution of poly(methyl methacrylate) (
M n = 100,000) in acetonitrile (density, 0.7857 g cm-3) at 45°C for which [η] = 4.8×10-3 M0.5?
Solution
The form of the Mark-Houwink-Sakurata equation shown above (see eq 3.101) indicates that for PMMA
in acetonitrile a = 0.5 and, therefore, A2 = 0 (i.e., θ solvent) and from eq. 3.85 we can write
RTc
atm cm3
0.5 g mol
=
Π
= 82.057
= 1.305 × 10−3 atm
318 K
3
5
Mn
mol K
100 cm 10 g
(b) What is the osmotic head in units of cm?
Solution
From eq 3.88, we have
h
=
1 kg
1000 g m 2
Π 1.305 × 10−3 atm cm3 s 2
=
= 1.716 cm
ρg
0.7857 g 9.80665 m 9.869 × 10−6 atm m s 2 kg 104 cm 2
(c) Estimate the Flory interaction parameter for polysulfone in methylene chloride.
Solution
2
χ12 ∝ (δ1 − δ 2 )
Table 3-3: δ PSF = 9.92 ( cal cm -3 )
12
and δ CH Cl = 9.92 ( cal cm -3 )
12
3
and therefore χ12 ≈ 0
(d) Based upon your answer above, would you expect methylene chloride to be a good or poor solvent for
polysulfone?
Answer
Good solvent (solubility parameters match).
3.4 The osmotic pressure of two samples, A and B, of poly(vinyl pyridinium chloride)
CH2
CH
N
n
Cl
were measured in different solvents. The following data were obtained:
Osmotic Pressure Data in Distilled Water
Sample c (g mL-1) Π (atm × 103)
A
0.002
29
A
0.005
50
B
0.002
31
B
0.005
52
Osmotic Pressure Data in 0.01 N Aqueous NaCl
Samplee c (g mL-1) Π (atm × 103)
A
0.002
5
A
0.005
13
B
0.002
2
B
0.005
5.5
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
16
Discuss these results and account for any features that you consider anomalous.
Solution
 1

Π
= RT 
+ A2 c +  
c
 Mn

Distilled Water
Sample
A
A
B
B
c ( g mL-1)
0.002
0.005
0.002
0.005
Π (atm × 103) Π
29
14.5
50
10.0
31
15.5
52
10.4
0.01 N aq NaCl
Sample
A
A
B
B
c ( g mL-1)
0.002
0.005
0.002
0.005
Π (atm × 103) Π
5
2.5
13
2.6
2
1.0
5.5
1.1
As shown by the data for the aqueous solution, Π c increases with dilution which is opposite to expected
behavior of a polymer in solution. This may be attributed to dissociation of the chlorine substituent and,
thereby, an increase in the effective number of osmotic units at high dilution. When chloride anions are
added, dissociation is inhibited and, as shown by the data given in the second table, Π c is essentially
independent of concentration ( A2 ≈ 0 ).
See, for example, the discussion of osmotic pressure of polyelectrolytes given in Flory, Principles of
Polymer Chemistry, pp. 633-635.
3.5
The following viscosity data were obtained for solutions of polystyrene (PS) in toluene at 30°C:
c (g dL-1)
0
0.54
1.08
1.62
2.16
t (s)
65.8
101.2
144.3
194.6
257.0
Using this information, please do the following:
(a) Plot the reduced viscosity as a function of concentration
Solution
ηi
= [η ] + kH [η ] c where
c
η − η s t − ts
=
≈
ηi
ηs
ts
ts = 65.8 s
2
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
17
c (g dL-1)
0.54
1.08
1.62
2.16
η
i
0.538
1.193
1.957
2.906
η
i
0.996
1.105
1.208
1.345
(b) Determine the intrinsic viscosity of this PS sample and the value of the Huggins constant, kH
Solution
From plot (R2 = 0.9958); we can obtain the following:
From eq 3.103
ηi
2
= [η ] + k H [η ] c
c
the intercept gives [η] = 0.8760 dL g-1
and the slope (kH[η]) is 0.2130 which gives kH = 0.278
(c) Calculate the molecular weight of PS using Mark–Houwink parameters of a = 0.725 and K = 1.1 × 10dL g-
4
Solution
[η ] = KM v a
Substituting values, we have
= (1.1 × 10−4 ) M v 0.725 or M v = 240,350
0.876
3.6 Given that the molecular weight of a polystyrene (PS) repeating unit is 104 and that the carboncarbon bond distance is 1.54 Å, calculate the following:
(a) The mean-square end-to-end distance for a PS molecule of 1 million molecular weight assuming that
the molecule behaves as a freely-rotating, freely-jointed, and volumeless chain. Assume that each link is
equivalent to a single repeating unit of PS.
Solution
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
18
106
= 9.62 × 103
104
=
l 2=
(1.54 Å ) 3.08 Å
=
n
2
2
r=
nl=
9.13 × 104 Å 2
(b) The unperturbed root-mean-square end-to-end distance, ⟨r2⟩o1/2, given the relationship for intrinsic
viscosity, [η], of PS in a θ solvent at 35°C as
[η ]=
8 × 10−4 M 0.5
where [η] is in units of dL g-1.
Solution
M [η ]θ
2
r=
Φ
o
r2
12
o
106 ( 8 × 10−4 )(106 )0.5 
= 5.26 × 105 Å 2
= 


2.1 × 1021


2/3
23
= 725 Å
(c) The characteristic ratio, CN, for PS.
Solution
r2
5.26 × 105
=
= 5.76
nl 2 9.13 × 104
=
CN
3.7 The use of universal calibration curves in GPC is based upon the principle that the product [η] M,
the hydrodynamic volume, is the same for all polymers at equal elution volumes. If the retention volume
for a monodisperse polystyrene (PS) sample of 50,000 molecular weight is 100 mL in toluene at 25°C,
what is the molecular weight of a fraction of poly(methyl methacrylate) (PMMA) at the same elution
volume in toluene at 25°C? The Mark–Houwink parameters, K and a, for PS are given as 7.54 × 10-3 mL
g-1 and 0.783, respectively; the corresponding values for PMMA are 8.12 × 10-3 mL g-1 and 0.71.
Solution
1+ a
η  M=KM v
7.54 × 10−3 ( 5 × 104 )
1.783
=
8.12 × 10−3 M1.71
or M V = 75,988
3.8 Show that the most probable end-to-end distance of a freely-jointed polymer chain is given as
( 2n 3)
12
2
.
Solution
 b 
3
ω=
( r )  1 2  exp ( −b 2 r 2 ) 4πr 2
π 
The maximum value occurs at
dω ( r )
dr
dω ( r )
dr
=0
3
 b 
8πr  1 2  (1 − b 2 r 2 ) exp ( −b 2 r 2 ) =
0
=
π 
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
19
or b 2 r 2 = 1 and
12
1  2nl 
r= = 

b  3 
3.9 The (reduced or excess) Rayleigh ratio ( Rθ ) of cellulose acetate (CA) in dioxane was determined
as a function of concentration by low-angle laser light-scattering measurements. Data are given in the
following table. If the refractive index ( no ) of dioxane is 1.4199, the refractive-index increment ( dn dc )
for CA in dioxane is 6.297 × 10-2 cm3 g-1, and the wavelength (λ) of the light is 6328 Å, calculate the
weight-average molecular weight of CA and the second virial coefficient (A2)
c × 103
(g mL-1)
0.5034
1.0068
1.5102
2.0136
2.517
R(θ) × 105
(cm-1)
0.239
0.440
0.606
0.790
0.902
Solution
2
2π 2 (1.4199 ) ( 6.297 × 10−2 )
2π 2 no2  dn 
=
=
= 1.63 × 10−8
K
 
4
−8 4
23
N A λ 4  dc 
6.023 × 10 ( 6328 ) (10 )
2
c × 103
0.5034
1.0068
1.5102
2.0136
2.517
R ( θ ) ×105
0.239
0.440
0.606
0.790
0.902
2
Kc R ( θ ) ×106
3.43
3.73
4.06
4.15
4.55
Least squares (R2 = 0.9758) fit of data gives (from the intercept)
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
20
=
Mn
1
=
313,873
3.186 × 10−6
and (from the slope)
1
A2 = 5.284 × 10−4 =2.64 × 10−4 mL-mol g -2
2
(
)
3.10 Chromosorb P was coated with a dilute solution of polystyrene in chloroform, thoroughly dried,
and packed into a GC column. The column was then heated in a GC oven and maintained at different
temperatures over a range from 200°C to 270°C under a helium purge. At each temperature, a small
amount of toluene was injected and the time for the solute to elute the column was recorded and
compared to that for air. From this information, the specific retention volume was calculated as given in
the table below. Using this data, plot the apparent Flory interaction parameter as a function of
temperature.
T
°C
200
210
220
230
240
250
260
270
Vg
mL/g-coating
6.55
5.58
4.66
4.07
3.38
2.87
2.88
2.38
Solution
0.46
0.44
0.42
0.4
0.38
0.36
0.34
0.32
0.3
0.0018
0.0019
0.002
0.0021
0.0022
1/T (K-1 )
See J. R. Fried and A. C. Su, "Poly(2,6-dimethyl-1,4-phenylene oxide) Blends Studied by Inverse Gas
Chromatography," Adv. Chem. Ser. 211, 59–66 (1986).
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
21
3.11 (a) Derive eq (3.72) and (b) develop an expression that can be used to obtain the exchange
interaction parameter, X 12 , appearing in the Flory equation of state from inverse gas chromatography
measurements.
Solution
(a) The F-H equation, eq 3.37
ln a1 = ln (1 − φ2 ) + φ2 + χ12φ22
Rearrangement of eq. 3.37 gives
a 
2
ln  1 = ln a1 − ln φ1= ln a1 − ln (1 − φ2 )= 1 − φ1 + χ12 (1 − φ1 )
 φ1 
and it follows that
a 
lim ln  1  = 1 + χ12
 φ1 
φ1 → 0
Substituting eq. 3.70 (below) into the LHS of the equation above
 273.16 Rv2  p1o ( B11 − V1 )
 a1 
 a1 
ln γ
ln=
lim
ln
=
=
−

  φ →0  
 V p o 
1
RT
g 1
 φ1 
 φ1 


∞
∞
1
finally gives the required result eq. 3.72
 273.16 Rv2  p1o ( B11 − V1 )
−
−1
 V p oV 
RT
 g 1 1 
χ12= ln 
(b) For the Flory EOS, write eq. 3.61 as
ln a1=

 v* 
 v1 3 − 1  1 1  
∆µ1
1 θ 22 v1* X 12
= ln φ1 + 1 − 1*  φ2 +
+ v1* p1* 3T1 ln  11 3  + −  

RT
RT  v
 v − 1  v1 v  
 v2 

and then rearrangement gives
ln

 v* 
 v1 3 − 1  1 1  
1 θ 22 v1* X 12
= 1 − 1*  φ2 +
+ v1* p1* 3T1 ln  11 3  + −  

φ1  v2 
RT  v
 v − 1  v1 v  

a1
φ2 1;=
φ1 1; θ 2 =1; θ1 =1;
=
v v2=
; T T2 ; α =α 2 and using eq. 3.70 (part a), we have
Noting that=
13
 273.16 Rv2  p1o ( B11 − V1 )  v1*  1 v1* X 12 v1* p1* 

 a1 
 ln  v1 − 1  + 1 − 1 
T
+
+
−
=
−
lim ln   = ln 
1
3




1
o
*
13


φ1 → 0
RT
RT 
 φ1 
 v2  RT v2
 v2 − 1  v1 v2 
 Vg p1

X 12
 v    273.16 Rv2
RT  2*  ln 
o

 v1    Vg p1
 p1o ( B11 − V1 )  v1*  v1* p1* 
 v11 3 − 1  1 1  
− 1 − *  −
 −
3T1 ln  1 3
 + − 
RT
 v2  RT 
 v2 − 1  v1 v2  

This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
22
and finally
 v   273.16 Rv2 
 v 
 v*  v 
 v  
 v1 3 − 1  1 1 
X 12 RT  2*  ln 
=
− p1o ( B11 − V1 )  2*  − RT 1 − 1*  2*  − v1* p1*  2*  3T1 ln  11 3

+ − 
o


 v1   Vg p1
 v1 
 v2  v1 
 v1  
 v2 − 1  v1 v2 

3.12 Derive the expression for osmotic pressure given by eq. (3.83).

1  Mν 3  2
Π RT   Mν 2   1

=
1 + 
  − χ12  c + 
 c + 
c
M   V1   2
3  V1 


Solution
From eq. 3.80, we can write
Π =−
RT
ln a1
V1
Substituting eq. 3.36 for lna1 in the Flory-Huggins model gives
Π =−
RT
V1

 1
2
ln (1 − φ2 ) + 1 − r  φ2 + χ12φ2 




Substituting the Taylor series (Appendix E) in the form
ln (1 − φ2 ) ≈ −φ2 −
φ2 2
2
−
φ23
3
+
gives
Π≈
3
 RT  φ2  1

φ2 2 φ23  1 
RT 
 2 φ2
+
− 1 −  φ2 − χ12φ2 2 + 
=
+ 
φ2 +

 +  − χ12  φ2 +
V1 
2
3  r
3

 V1  r  2

Since r =
vM
and φ2 = cv (eq. 3.82)
V1
we can write
φ2 cv
1
=
=
V1
cV1
r vM
M
and finally, we obtain eq. 3.83 as

 1 1
 RT   Mv 2   1
c 2 v3
1  Mv 3  2
Π


χ
1
c
= RT  +  − χ12  cv 2 +
+ =
+
−
+




 c + 
12 

c
3
3  V1 


M  2
 M   V1   2

3.13 Show how eq. eq. 3.78 for the relationship between the Flory-Huggins interaction parameter and
the solubility parameters of polymer and solvent was derived.
Eq. 3.78
V
2
χ12 ≅ 1 (δ1 − δ 2 )
RT
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
23
Eq. 3.34
kT χ12 n1φ2
∆H m =
Eq. 3.77
∆H m = V (δ1 − δ 2 ) φ1φ2
Equating equations 3.34 and 3.77 and noting that k = R/NA and
φ1 = ν 1 V
V
v
V
V
2 N
2 N
2
χ12 = (δ1 − δ 2 ) A φ1 = (δ1 − δ 2 ) A 1 =1 (δ1 − δ 2 )
RT
n1
RT
n1 V RT
2
CHAPTER 4
4.1 Show that σ= σ (1 + ε ) for an incompressible material.
T
Solution
F
F  ∆L  F  Lo + L − Lo  F  L 
σ T = = σ (1 + ε ) = 1 +
= 
=  
A
Ao 
Lo  Ao 
Lo
 Ao  Lo 
For an incompressible material, AL = Ao Lo
or Ao = A ( L Lo )
Substituting this expression for Ao gives
σT
=
F  L  F  Lo   L  F
=
=
 
 
Ao  Lo  A  L   Lo  A
4.2 A tensile strip of polystyrene that is 10 cm in length, 5 cm in width, and 2 cm in thickness is
stretched to a length of 10.5 cm. Assuming that the sample is isotropic and deforms uniformly, calculate
the resulting width and the % volume change after deformation.
Solution
Vo = 10 × 5 × 2 = 100 cm3
eq. 4.45, v = − ε T ε L
=
ε L ln=
=
10 ) 0.04879
( L Lo ) ln (10.5
Table 4.13, v = 0.35 for PS
Assuming isotropic,
εT =
ln (W Wo ) =
ln (T To ) =
−vε L =
−0.01708
=
W 5=
( 0.9831) 4.92 cm
=
T 2=
( 0.9831) 1.97 cm
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
24
V = 10.5 × 4.92 × 1.97 = 101.77 cm3
∆
=
V 101.77 − 100
= 1.77 cm3 or 1.77%
4.3 A polymer has a crystalline growth parameter (n) of 2 and a rate constant (k) of 10-2 s-2 at 100°C.
The polymer is melted and then quenched to 100°C and allowed to crystallize isothermally. After 10 s,
what is the percent crystallinity of the sample?
Solution
φ=
1 − exp ( − kt n )
2
φ =1 − exp  −102 (10 )  =1 − exp ( −1) =1 − 0.3679 =0.6321 or 63.2% crystalline


4.4 What is the % volume change that is expected at 100% elongation of natural rubber, assuming that
no crystallization occurs during deformation?
Solution
=
ε ln ( L L=
ln=
2 0.6931
o)
∆v = (1 − 2v ) ε Vo where v = 0.49 (Table 4.13)
∆V
=
0.0139 or 1.39%
(1 − 0.98)( 0.6931) =
Vo
4.5 Give your best estimate for the weight fraction of plasticizer required to lower the Tg of poly(vinyl
chloride) (PVC) to 30°C. Assume that the Tg of PVC is 356 K and that of the plasticizer is 188 K. No
other information is available.
Solution
Eq. 4.33
 303  (1 − W1 ) ln ( 356 188 )
ln 
=
 188  W1 ( 356 188 ) + 1 − W1
solve for W1 gives W1 = 0.151 or 15.1 wt%
4.6 Show that the inverse rule of mixtures given by eq.4.34 can be obtained from the generalized
relationship given by eq.4.32 when Tg,1 ≈ Tg,2
Solution
Substituting ∆Cp,1 =
constant Tg ,2 into eq. 4.32 and multiplying numerator and
constant Tg ,1 and ∆Cp,2 =
denominator by Tg,2 gives eq. 4.33.
Next, let ln (1 + x ) ≈ x where
=
x
(T
g
Tg,1 ) − 1
Substitute into eq. 4.33 gives
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
25
T

W2  g,2 − 1


Tg
 Tg,1

−1 =
Tg,1
W1 (Tg,2 Tg,1 ) + W2
Tg
=
W2 (Tg,2 − Tg,1 ) + W1Tg,2 + W2Tg,1
W2Tg,2 + W1Tg,2
Tg,2
(W1 + W2 ) Tg,2
= = =
W1 (Tg,2 Tg,1 ) + W2
W1 (Tg,2 Tg,1 ) + W2 W1 (Tg,2 Tg,1 ) + W2 W1 (Tg,2 Tg,1 ) + W2
1 W1 W2
=
+
Tg Tg,1 Tg,2
4.7 Polytetrafluoroethylene has been reported to exhibit a negative Poisson ratio. Explain why this
polymer exhibits this unusual behavior.
Teflon thickens upon elongation due to a rotation of crystals perpendicular to the draw direction.
Solution
See B. W. Ludwig and M. W. Urban, Polymer 35, 5130 (1994).
4.8 A sample of poly(ethylene terephthalate) is reported to be 20% crystalline.
(a) What is the expected density of this sample?
Solution
Using eq. 4.6 and densities give in Table 4-5
ρ= φ ( ρc − ρ a ) + ρ a= 0.2 (1.396 − 1.280 ) + 1.280= 1.303 g cm -3
(b) What is the expected specific heat increment of this semicrystalline sample?
Solution
From eq. 4.26
∆Cp =(1 − φ ) ( ∆Cp ) =0.8 ( ∆Cp )
am
am
Simha–Boyer rule: ( ∆Cp )
am
≈ 115 J g -1 Tg
From Table 4-3, Tg = 342 K
=0.336 J g -1 K -1
( ∆Cp )am =115
342
-1
=
∆Cp ( 0.8=
K -1 0.269 ( 0.2387
=
) 0.336 0.269 J g=
) 0.06416 cal g -1 K -1
(d) What is the expected heat of fusion of this sample?
Solution
From eq. 4.25 and Table 4-4
∆Q =φ∆H f =0.2 ( 6.431) kcal mol-1 =1.286 kcal mol-1 =1.286 ( 0.2387 ) =0.307 kJ mol-1
4.9 Twenty wt % of a styrene oligomer having a number-average degree of polymerization of 7 is mixed
with a commercial polystyrene sample having a number-average molecular weight of 100,000.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
26
(a) What is the Tg (K) of the styrene oligomer?
Solution
Use of the Fox-Flory relation, eq 4.27, and parameters given in Table 4-11
M n =×
7 104 =
728
Tg,1 = 373 −
1.2 × 105
= 373 − 165 = 208 K
728
(b) What is the Tg (K) of the polystyrene mixture.
Solution
1.2 × 105
=
372 K
105
Use eq. 4.34 as best approximation since Tgs are far apart
Tg,2 =−
373
0.8ln ( 372 165 )
 Tg 
=
ln 
 = 0.5199
 165  0.2 ( 372 165 ) + 0.8
=
Tg 1.6819
=
(165) 277.5 K
4.10
The 1% secant modulus of a polystyrene sample is 3 GPa.
(a) What is the nominal stress (MPa) of this sample at a nominal strain of 0.01?
Solution
σ = Eε = 0.01(3 GPa) == 0.03 GPa=30 MPa
(b) What is the true stress (MPa) of this sample at a nominal strain of 0.01?
Solution
σ T = σ (1 + ε ) = 30(1.01) = 30.3 MPa
(c) What is the percent change in volume of this sample at the nominal strain of 0.01?
Solution
From eq. 4.44 and Table 4-13,
∆V
=
(1 − 2v ) ε T =
(1 − 0.70 ) ε T
Vo
From eq. =
4.43 ε T ln=
(1.01 1) 0.00995
∆V
= 0.3
=
( 0.009950 ) 0.002985 or 0.299%
Vo
If the Young’s modulus of a sample of polystyrene is determined to be 3 GPa at room temperature,
calculate its shear modulus.
Solution
From eq. 4.56:=
G E=
3 1 GPa
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
27
4.12 Isotactic poly(methyl methacrylate) has a much lower Tg than the corresponding syndiotactic
polymer. How can isotactic and syndioactic PMMA be polymerized? Explain why the isotactic polymer
has the lower Tg
Solution
Syndiotactic PMMA can be obtained by free-radical polymerization at low temperatures (T. G. Fox et al.,
JACS 80, 1768 (1958); F. A. Bovey, JPS 46, 69 (1960)). Both syndiotactic and isotactic polymers can be
obtained from anionic polymerizations. Recent molecular modeling (A. Soldera, Polymer 43, 4269
(2002)) indicate that differences in both intermolecular interactions and bending angle energy along the
backbone can be correlated with the higher Tg of the syndiotactic polymer.
CHAPTER 5
5.1 Show that E * = σ o ε o and D * = 1 E * .
Solution
( E ')
=
E*
σo
E'= o
ε
2
+ ( E ")
2

σo 
;
=
E
"
cos
δ
 o  sin δ

ε 

12
 σ o  2

σo
Therefore E=
*  o  ( sin 2 δ + cos 2 δ ) = o
ε 
ε


Similarly D * ×D* =
D*
2
where D=
* D '+ iD "
* D '− iD " and D=
 εo
D' =  o
σ

 εo
and
cos
D
"
=
δ

 o

σ

 sin δ

12
 ε o  2

therefore D * =
 o  ( sin 2 δ + cos 2 δ ) 
 σ 

εo
1
=
=
o
σ
E*
5.2 Show that the work per cycle per unit volume performed during dynamic tensile oscillation of a
o o
viscoelastic solid may be given as πσ ε sin δ (eq. (5.30)).
Solution
2π
W = ∫ σ *d ε *
0
=
σ * σ o sin (ωt + δ )
ε * = ε o sin ωt and d ε * = ε o cos ωtd (ωt )
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
28
=
W σ oε o ∫
2π
ωt = 0
sin (ωt + δ ) cos ωtd (ωt )
Since sin (ω=
t + δ ) sin ωt cos δ + cos ωt sin δ , write
W
σ ε
o
o
2π
2π
0
0
cos δ ∫ sin ωt cos ωtd (ωt ) + sin δ ∫ cos 2ωtd (ωt )
Note ∫ sin ω x cos ω xdx =
Therefore
∫
2π
0
1 1
sin 2 ω x
and sin 2 θ=
− cos 2θ
2 2
2ω
t) ω
sin ωt cos ωtd (ω=
2π
sin 2 ωt
1 1 1
1 1
=
− ( cos 4π ) − + ( cos 0=
) 0

2ω 0
2 2 2
2 2

This is the elastic response (no work expanded).
Next,
2π
W
σ ε
o
o
= sin δ ∫ cos 2ωtd (ωt )
0
Note ∫ cos 2udu=
Then
u sin 2u
+
2
4

sin ( 4π )
sin 0 
 ωt sin 2ωt 
= sin δ  +
−0−
 = π sin δ and
 = sin δ  π +
o o
4 0
4
4 
σ ε
 2

2π
W
W = πσ oε o sin δ
5.3 Given the expression G (=
t ) Go exp ( −t τ ) + G1 , show that the compliance function, J(t), can be
written in the form J (t ) =
A − B exp ( −C t τ ) , where A, B, and C are constants.
Solution
L  J ( t )  L G ( t )  =
L G=
( t )
1
p2
Go
G
+ 1
p +1 τ
p
1
p +1 τ
=
L  J ( t ) 
=
2
p L G ( t )  p ( Go + G1 ) p + ( G1 τ ) 
Rearrange in form
p+a
bp ( p + c )
where
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
29
a=
1
τ
;=
b Go + G1 ; c = G1 a b = G1 τ ( Go + G1 ) 
 p+a 
Then J ( t ) = L−1 

 bp ( p + c ) 
Use partial fractions, to find inverse
p ( A + bB ) + Ac
p+a
A
B
p+a
= +
=
=
bp ( p + c ) bp p + c
bp ( p + c )
bp ( p + c )
Then A + bB =
1
a = Ac or =
A a=
c
=
B
Go + G1
G1
1− A c − a
=
b
cb
  G1  t 
 A
 B  A
1
J ( t ) = L−1   + L−1 
 
 = + B exp ( −ct ) = + B exp  − 
G1
 bp 
 p+c b
  Go + G1  τ 
=
B
−Go
1− A
=
b
G1 + ( Go + G1 )
J (=
t)
Go
1
−
G1 G1 ( Go + G1 )
5.4 For a Maxwell model, show the following:
(a) The equation for complex modulus E* (eq. (5.57)) can be obtained from the Fourier transform of the
stress-relaxation modulus, Er (eq. (5.49)).
(b) A maximum in the loss modulus plotted as a function of frequency occurs at ω = 1 / τ .
Solutions
 t
(a) eq. 5.49:
Er E exp  − 
=
 τ
∞
∞
0
0
E ′ ω ∫ sin (ω=
s )E ( s ) ds ω E ∫ exp ( − s τ ) sin (ω s ) ds
=
Fourier transform f =
( x ) exp ( −bx ) ; FS =
α
α + b2
2
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
30
(ωτ )
E′ = E
2
(ωτ ) + 1
2
Therefore
∞
Similarly
E ′′ ω E ∫ exp ( − s τ ) cos (ω s )ds
=
0
Fourier transform f =
( x ) exp ( −bx ) ; Fc (α ) =
b
α + b2
2
Get eq. 5.59:
1τ
ωτ
=
E ′′ ω=
E 2
E
2
2
ω + (1 τ )
(ωτ ) + 1
dE ′′
(b)
=
dω
(ω τ
2 2
ω τ + 1) Eτ − Eτω 2ω wτ
(=
(ω τ + 1)
2 2
2
2 2
2
0
+ 1) Eτ − Eτω 2ω wτ 2 =
0
or ω 2τ 2 = 1 and ω = 1 τ at maximum
5.5 Given the four-element model illustrated below, derive an analytical solution for the strain behavior
and sketch ε ( t ) versus time under the following stress conditions:
t<0
σ =0
0 ≤ t < t1
σ = σ o (creep)
t1 ≤ t < t2 σ = 0
(creep recovery)
Solution
Four-element model is a series combination of a Voigt element. Dashpot, and spring and, therefore,
strains are additive
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
31
=
ε
σo
EM
+
or D ( t ) ≡
σo
σ
1 − exp ( −t τ )  + o t ; where τ = η V E V
EM 
ηM
ε (t ) 1
1
t
t
=
+
1 − exp ( − t τ )  + = D1 + D2 ( 1 − exp ( − t τ )  ) +
EM EV
σo
ηM
ηM
Boltzmann Superposition Principle says that removal of stress at t1 is equivalent to applying negative
stress (i.e., −σ o ) at t1
Therefore, t1 < t < t2
σ
σ t
ε= σ o  D ( t ) − D ( t − t1 ) = o exp ( - ( t -t1 ) τ ) − exp ( −t τ )  + o 1
EV
ηM
σo / EM
0
t1
5.6 In the case of the expression for the dielectric loss constant given by eq. (5.112), the relaxation time
can be assigned a temperature dependence of the form
τ = τ o exp ( H RT )
where H is the activation energy. If ε ′′ assumes its maximum value at that temperature (Tmax) where
ω = 1 / τ , how can the value of H be determined given data in the form of a plot of ε ′′ versus T at
different frequencies, f ?
Solution
ωτ ( ε R − ε U )
ε ′′ =
2
1+ (ωτ )
′′ , ω = 2πf =
at ε max
=
f
1
τ
=
1
τo
exp ( − H RTmax )
1
H
1
−
+ ln
exp ( − H RTmax ) and ln f =
RTmax
2πτ o
2πτ o
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
32
Given data in the form of ε ′′ versus temperature at different frequencies, find the temperature
corresponding to the maximum in ε ′′ at a given frequency (i.e., Tmax). A plot of lnf versus 1/Tmax should
then give a straight line with slope –H/R, from which H can be obtained.
5.7 Calculate the WLF shift factor, aT, for polystyrene (PS) at 150°C given that the reference
temperature is taken to be the Tg of PS, 100°C, and using (a) the "universal" values of the WLF
parameters, C1 and C2, and (b) the reported values of C1 = 13.7 and C2 = 50.0 K for PS.
Solutions
(e)
log aT =
aT 2.61 × 10−9
=
log aT =
(f)
−17.44 ( 423 − 373)
51.6 + 50
−13.7 ( 50 )
50 + 50
= −8.583
= −6.85
=
aT 1.41 × 10−7
5.8 If the maximum in the α−loss modulus of polystyrene at 1 Hz occurs at 373 K, at what temperature
would the maximum occur at 110 Hz if the activation energy for this relaxation is 840 kJ mol-1?
Solution
From eq. 5.37, write
 f 
 E  1 1 
ln  1  =
− a  − 
 R   T1 T2 
 f2 
1
 1 
 840,000   1
ln 
−
− 
=

 110 
 8.3144   373 T2 
T2 = 379.6 K
5.9 (a) Calculate the relaxation modulus, in SI units of GPa, at 10 seconds after a stress has been applied
to three Maxwell elements linked in parallel using the following model parameters:
E1 = 0.1 GPa
τ 1 = 10 s
E2 = 1.0 GPa
τ 2 = 20 s
E3 = 10 GPa
τ 3 = 30 s
(b) Does this model give a realistic representation of stress relaxation behavior of a real polymer?
Explain.
Solutions
(a)
E=
(t )
3
τ )
∑ E exp ( −t=
i =1
i
i
9.983 GPa
(b)
The value is a bit high. In general, The Maxwell–Wichert model is a good representation of the
behavior of stress relaxation modulus, particularly as the number of Maxwell elements increases.
5.10 An elastomeric cube, 2 cm on a side, is compressed to 95% of its original length by applying a
mass of 5 kg. What force is required to stretch a strip of the same elastomer by 50%? The initial length of
the strip is 2 cm and its original cross-sectional area is 1 cm2.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
33
Solution
∆P
K=
∆V Vo
The hydrostatic pressure is obtained as follows:
∆=
P 5 kg
9.80665 m 104 cm 2
= 2.043 × 104 kg m -1 s -2
s 2 24 cm 2 m 2
Vo = 8 cm3 ; V = 6.859 cm3 ; ∆V = 8 − 6.859 =1.141 cm3
K=
2.043 × 104
= 14.32 × 104 = 0.143 MPa
1.141 8
From eqs. 4.54 and 4.56, we have
K=
2 (1 + v ) G
3 (1 − 2v )
or G =
3 (1 − 2v ) K
2 (1 + v )
Table 4-13, v = 0.49 for NR
=
G
=
λ
3 (1 − 0.98 )
=
0.143 0.00288 MPa
2 (1.49 )
L
= 1.5
Lo
1 
1 


f * = Go  λ − 2  = 0.00288 1.5 − 2  = 0.00288 (1.0556 ) = 0.00304 MPa
λ 
1.5 


or unit conversion
kg
m2
f*=
3.04 × 103
1 cm 2 4 2 A =
0.304 kg f
2
ms
10 cm
5.11 If the stress at 23°C of an ideal rubber is 100 psi when stretched to twice its original length, what
would be the stress at a 50% extension?
Solution
1 

Eq. 5.157=
f * Go  λ − 2 
λ 

λ=
100 psi
L
=2; Go = 57.14 psi
=
1
Lo

2− 
4

This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
34
=
λ
1.5
= 1.5
1
f=
* 57.14 (1.5 − 0.444=
) 60.3 psi
5.12 The length of an ideal rubber band is increased 100% to 12.0 cm at 23°C. Stress on this rubber
band increases by 0.2 MPa when it is heated to 30°C at 100% elongation. What is its tensile modulus in
MPa at 23°C when it is stretched 2%?
Solution
Neglecting any changes in volume with temperature
 296 
 303 
 296 
 303 
σ1 =

 σ 2 =+

 (σ 1 0.2 MPa ) =0.9769σ 1 + 0.1954 MPa
0.02310σ 1 = 0.1954 MPa
σ 1 = 8.4589 MPa
=
G1
8.4589
= 4.8336 MPa
2 −1 4
1 

σ 3 4.8336 1.02 − =
=
0.2844 MPa
2 
1.02


=
E
σ
3
=
0.02
0.28436
= 14.2 MPa
0.02
5.13 Calculate the shear modulus (GPa) of a polymer sample in a torsion pendulum with a period of 1.0
sec. The specimen is 10 cm long, 2 cm wide, and 7.5 mm thick and the moment of inertia is 5000 g cm2.
Solution
 1  64π2 L 1  1 
1
=
G ′ FS I  =
I =
3.743 × 107 g cm -1 s -2
2 
3
2 
Wt u  p 
u
p 
See ASTM D 4065 (ref. in Appendix C)
16
t 
t4 
u =− 3.36 1 −
4.075
=
3
W  12W 4 
G′ =
3.743 × 107
1
Pa
g cm -1 s -2 =×
0.9185 107 g cm -1s −2 =
9.185 × 106 dynes cm -2
4.075
10 dynes cm -2
=0.9185 MPa
5.14 For a two-phase system for which the complex shear modulus follows a log rule of mixtures, show
that
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
35
G "
G " G1 "
≅
φ1 + 2 φ2
G ' G1 '
G2 '
Solution
Write log rule of mixtures as
=
ln G* φ1 ln G1 * +φ2 ln G2 *
Therefore ln ( G '+ iG "=
) φ1 ln ( G1 '+ iG1 ") + φ2 ln ( G2 '+ iG2 ")


G1 "  
G2 "  

G"  
or ln 1 + i
 G1 ' + φ2 ln 1 + i
 G2 '
 G ' = φ1 ln 1 + i
G'  
G1 '  
G2 '  



and


G1 " 
G2 " 
G" 

ln 1 + i
 + φ1 ln G1 '+ φ2 ln 1 + i
 + φ2 ln G2 '
 + ln G ' = φ1 ln 1 + i
G' 
G1 ' 
G2 ' 



Applying the binomial expansion that ln (1 + x ) ≈ x , gives
G"
G "
G"
+ ln G ' ≅ φ1i 1 + φ1 ln G1 '+ φ2i 2 + φ2 ln G2 '
G'
G1 '
G2 '
Equating imaginary terms gives
G "
G "  G1 "
≅  φ1
+ φ2 2  i
i
G '  G1 '
G2 ' 
i
or
G ′′
G ′′ G1′′
≅
φ1 + 2 φ2
G ′ G1′
G2′
See L. E. Nielsen, Trans. Soc. Rheol. 13, 141 (1969).
CHAPTER 7
7.1 Poly(2,6-dimethyl-1,4-phenylene oxide) (PPO) is blended with polystyrene. Compare the
predictions of the inverse rule of mixtures and the logarithmic rule of mixtures (see Section 4.3.4) by
plotting the calculated Tg of the blend against the weight fraction of PS. Tg values obtained from DSC
measurements are as follows:
Wt %
PPO
0
20
40
60
80
100
Tg
(°C)
105
121
140
158
185
216
Solution
1, PS
W1
0
Tg (K)
exp.
489
Tg (K)
inverse rule
Tg (K)
log rule
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
36
0.2
0.4
0.6
0.8
1.0
458
431
413
394
378
461.9
437.6
415.7
396.0
464.5
441.1
419.0
398.0
Both equations overpredict blend Tg; however, the inverse rule of mixtures (Fox equation) provides the
best estimate in this case.
490
470
450
430
410
390
Experimental
Inverse rule of mixtures
370
Log rule of mixtures
350
0
0.1
0.2
0.3
0.4
0.5
W
0.6
0.7
0.8
0.9
1
1
7.2 For a graphite-fiber composite of polysulfone containing 40 vol % filler, what are the maximum
modulus and maximum strength that can be expected?
Solution
The maximum modulus and strength of the fiber-reinforced composite is achieved in the longitudinal
direction for an uniaxially oriented fiber composite by eq. (7.11):
EL =
(1 − φ f ) Em + φ f Ef
For the fiber, use the upper bound for the modulus of high-modulus graphite (Table 7-3): for PSF, use 2.5
GPa for modulus (Table B-2). 1
EL =
208.3 GPa
(1 − 0.4 ) 2.5 + 0.4 ( 517 ) =
Note that in this case, the modulus of the composite is dominated by the fiber modulus.
For strength, use eq. 7.14,
σ L =−
1.17 GPa
(1 φf )σ m + φf σ f =−
(1 0.4 )( 0.065) + 0.4 ( 2.83) =
7.3. Draw the chemical structures for the following plasticizers:
(a) TCP
(b) TOTM
(c) DOA
(d) DIOP
1
Rows for polystyrene and polysulfone are interchanged in first printing of second edition.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
37
Solution
(g) TCP, tricresyl phosphate or tritolyl phosphate (a mixture of isomers)
O
RO
P
OR
OR
where
R=
CH3
(h) TOTM, trioctyl trimellitate or tris(2-ethylhexyl trimellitate)
O
C O R
R
O C
C O R
O
O
where
R=
CH2 CHCH2 CH2 CH2 CH3
CH2 CH3
(i)
DOA, dioctyl adipate or di-2-ethylhexyl adipate (derivative of adipic acid)
O
R O C
O
(CH2 )4
C O R
where
R=
CH2 CHCH2 CH2 CH2 CH3
CH3 CH3
(d)
DIOP, diisoctyl phthalate
O
C O R
C O R
O
where
R=
CH2 CH2 CH2 CH2 CH2 CH CH3
CH3
7.4 Give your best estimate for the Tg of PVC plasticized with 30 phr of TOTM (Tg = -72°C).
Solution
Tg ,1 = 201 K ; Tg ,1 = 354 K ( Table B-1, Appendix B )
With the information available, the accurate expression would be eq. 7.3 since the Tgs are widely
separated.
30
=
W1 = 0.231
100 + 30
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
38
W2 ln (Tg ,2 Tg ,1 )
 Tg 
0.769ln ( 354 201)
 Tg 
ln
=
=
= 0.3699

 ln=


 201  W (Tg ,2 Tg ,1 ) + W21 0.231( 354 201) + 0.769
 Tg ,1 
=
Tg 1.448
=
( 201) 291 K or 18°C
7.5 Explain why the curve representing the Tg of miscible TMPC/PS blends appearing in Figure 7.4
appears to diverge at low TMPC compositions.
Solution
At low TMPC concentration, PS is the major component and the overall Tg of the blend is reduced by the
lowest-molecular-weight PS ( M n = 42,000 ) component; the actual reduction in Tg is small. From the
Fox–Flory equation (eq. 4.27)
Tg =−
373
1.2 × 105
=
370 K .
42,000
7.6 Show that the Halpin–Tsai equation (eq. 7.9) reduces to the simple law of mixtures
=
M φ1M 1 + φ2 M 2
when the Halpin–Tsai parameter A approaches infinity and reduces to the inverse rule of mixtures
φ
φ
1
= 1 + 2
M M1 M 2
when A approaches zero.
Solution
As A → ∞ (corresponding to the case where fibers are aligned in a parallel direction)
=
M φ1M 1 + φ2 M 2
As A → 0 , B =
( M1
M 2 ) −1
M1 M 2
= 1−
M2
M1
M
1
=
M 2 1 − ψφ1 (1 − M 2 M 1 )
 M 
M
1 − ψφ1 1 − 2 
=
M1
 M1 
φ
1
1
=
(1 −ψφ1 ) + ψ 1
M M2
M1
When ψ ≈ 1 , the inverse rule of mixture
and
φ
φ
1
= 1 + 2
M M1 M 2
is obtained.
(See L. E. Nielsen, “Mechanical Properties of Polymers and Composites,” Dekker, Vol. 2.)
7.7 Inverse gas chromatography (see Section 3.2.5) can be used to determine an apparent Flory
interaction, χ 23,app′ , using a solvent probe through the relationship
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
39
Vg,b


 Vg,3 
 Vg,2 
χ 23,app′φ=
ln 
 − φ2 ln 

 − φ3 ln 
2φ3
 v2 
 w2 v2 + w3v3 
 v3 
where w and v are weight fraction and specific volume, respectively. Specific retention volumes for
polystyrene (2), poly(2,6-dimethyl-1,4-phenylene oxide) (PPO) (3), and a 50/50 blend of PPO and
polystyrene using toluene as the probe at 270°C are given below. From this data, calculate χ 23 for the
PPO/PS blend. Is this value consistent with the reported miscibility of this blend?
Coating
PS
PPO
PPO/PS
Vg
mL/g-coating
2.38
2.96
2.42
Solution
Rearranging, we have
  Vg,blend 
Vg,2
V 
χ 23,app ' ln 
=
− φ3 ln g,3  φ2φ3
 − φ2 ln
v2
v3 
  w2 v2 + w3v3 
where
ρ 2 ( PS) = 1.05 g cm-3 (Tables 9-3 and B-1, Appendix B)
v2 = 0.9524 cm3 g-1
w2 = 0.5
ρ3 ( PPO ) = 1.06 g cm-3 (Table 4-5)
v3 = 0.9434 cm3 g-1
w3 = 0.5
φ2 =
w2 ρ 2
w2 ρ 2 + w3 ρ3
φ3 =
w3 ρ3
w2 ρ 2 + w3 ρ3
w2 v2 +=
w3v3 0.5 ( 0.9524 ) + 0.5 ( 0.9434
=
) 0.9479
0.4762
= 0.5024
0.9479
0.4717
= 0.4976 or φ3 = 1 − φ2
φ3 =
0.9479
  2.42 
 2.38 
 2.96  
χ 23,app ' =
−0.367
ln  0.9479  − 0.5024ln  0.9524  − 0.4976ln  0.9434   0.5024 ( 0.4976 ) =





 
=
φ2
The negative value is consistent with a compatible blend. The value reported for this blend at 270°C from
IGC measurements using a toluene probe is -0.31 and -0.35±0.03 as an average value for five solvent
probes (A. C. Su and J. R. Fried, in D. R. Lloyd, T. C. Ward, and H. P. Schreiber, eds., ACS Symp. Series
No. 390, ACS, Washington, DC, 1989, pp. 155–165).
CHAPTER 11
11.1 Poly(vinyl acetate) (PVAc) is extruded at 180°C at constant temperature through a capillary
rheometer having a ram (reservoir) diameter of 0.375 in. and a capillary with an inside diameter of 0.041
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
40
in. and length of 0.622 in. The data provided give the efflux time to extrude 0.0737 in.3 at different ram
loads. Using the following data:
Ram Load
(lbf)
Efflux
Time
(min)
97.5
145
217
250
5.32
1.58
0.31
0.17
(a) Determine the power-law parameters n and m for PVAc and state all assumptions used to obtain your
results.
Solution
ram load ( lb f ) ram load
in psi
=
∆p
=
2
0.110
π ( 0.375 2 )
eq. 11.38
τw
=
Q=
=
φ
R∆p
=
2L
∆p
( 0.041 2 )=
2 ( 0.622 )
0.0165 ∆p in psi
0.0737
in unit of in3 min-1
efflux time
4Q
4Q
=
=
1.48 × 105 Q in units of min-1
3
3
πR
π ( 0.0205 )
A plot of logτ w versus log φ will yield n as the slope and log m′ as the intercept where
n
 4n 
m = m′ 
 (eq 11.40)
 3n + 1 
Δ p
(
886
1318
1973
2273
)
τ w ( psi )
logτ w
Q ( in3 min-1 )
14.6
21.7
32.5
37.5
1.16
1.34
1.51
1.57
0.0139
0.0466
0.2377
0.4335
φ ( min-1 )
2050
6890
35,100
64,100
logφ
3.31
3.84
4.55
4.81
A least squares fit of the plot gives a slope (n) of 0.2694 and an intercept of 0.2829 ( m′ = 1.92 ) .
 4 ( 0.269 ) 
=
m 1.92  =
1.67
=
psi-min 0.269 34,600 Pa-s 0.269

 3 ( 0.269 ) + 1 
0.269
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
41
The usual assumptions are fully-developed, isothermal, laminar, steady-state, and incompressible flow;
negligible body forces, no slip at the wall, no viscous heating, and viscosity that is independent of
pressure.
(b) Plot the apparent viscosity, η, in units of Pa-s versus the nominal shear rate at the wall, γw (s-1), using
logarithmic coordinates.
Solution
Eq. 11.40,
γw
=
η=
τw
γw
3n + 1
φ 1.68 φ
=
4n
γw ( s -1 )
η (Pa - s )
57.3
193
983
1800
1750
774
228
144
Since η = mγw n −1 (eq. 11.14), logη =
−0.728 log γw + 4.53 using values obtained from a least-squares fit
2
(R = 0.9993) of the four data points.
The experimental data shows shear-thinning behavior over the range of wall shear rates from 57 to 1800
s-1.
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
log dγw/dt
11.2 Plot the dimensionless velocity profile for polystyrene flowing in a capillary at 483 K.
Solution
Get power-law index (n = 0.25) for PS at 483 K (210°C) from Table 11-1.
Eq. 11-275
uz ( r )
u z max
r
1−  
=
R
1+ n
n
r
1−  
=
R
5
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
42
11.3 As illustrated, two capillaries of identical length are connected to the same liquid reservoir in
which a power-law fluid is held. The tubes differ in radii by a factor of 2. When a pressure is applied to
the reservoir, the volumetric flow rates from the two tubes differ by a factor of 40. What is the value of n?
How different are the nominal shear rates in the two cases?
Reservoir
Solution
Eq. 11.28
nπR 3  R∆p 


1 + 3n  2mL 
1n
Q=
Q2 R23 R21 n  R2 
=
=  
Q1 R13 R11 n  R1 
40 = 2
3+
3+
1
n
1
n
1

log 40
=  3 +  log 2
n

gives n = 0.43.
3n + 1  4Q 
γw =


4n  πR 3 
3
γw ,1  R2   Q1 
Therefore
=  =
  
γw ,2  R1   Q2 
 
2)  
(=
3
1
0.2
 40 
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
0.4
0.5
r/R
0.6
0.7
0.8
0.9
1
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
43
11.4 Molten polystyrene flows through a circular tube at 210°C under a pressure drop of 1,000 psi.
Given that the inside diameter of the tube is 0.25 in. and that the tube is 3 in. in length, calculate the
following:
(a) the (nominal) shear stress at the wall in units of N m-2
Solution
Eq. 11.38, τ =
w
∆
=
p
R∆p 0.125
=
∆=
p 2.08 × 10−2 ∆p
2L
2 ( 3)
1000 lb f
1N
in 2
104 cm 2
= 6.895 × 106 N m -2
2
2
2
in
0.2248 lb f ( 2.54 cm )
m
τ w = 2.08 × 10−2 ( 6.895 × 106 ) =1.43 × 105 N m -2
(b) the (nominal) shear rate at the wall in s-1
Solution
Get values of power-law parameters m and n for PS (210°C) from Table 11-1
4
 1.43 × 105 
τ 
=
= 1.30 × 103 s -1
γw  w = 
4 
×
m
2.38
10
 


Note that this value is within the allowable range for the parameters used (i.e., 100–4500 s-1).
1n
the volumetric flow rate in cm3 s-1
4
3
1n
0.25π ( 0.125 )  0.125∆p 
nπR 3  R∆p 
 =
 19.1 cm3 s -1
=
Q =


4
1 + 3n  2mL 
1 + 0.75
 2 ( 2.38 × 10 ) 3 
Assume that flow is isothermal, steady, and fully developed.
11.5 (a) Given that tensile (Trouton's) viscosity is defined as
ηT =
σ
ε
where σ and ε are the true tensile stress and true strain, respectively, show that
 1 
=
ln L   σ t + ln Lo
 ηT 
when viscosity is independent of ε and Lo is the initial length of the sample.
(a)
Solution
dε
or σ dt =
ηT d ln ( L Lo ) 
σ = ηT
dt
 1
integrating gives σ t = ηT ln ( L Lo ) =
or ln L 
 ηT

 σ t + ln Lo

This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
44
(b)
Solution
σ =F A
Assuming that PIB is incompressible (a reasonable assumption for an elastomer)
=
V 6.0
=
( 0.699 )( 0.155) 0.650 cm3
Therefore, A = 0.650 L at any time.
Force (F) is 75 gm × go=9.81 × 102 cm s-2
t (min)
1
2
3
6
12
15
18
21
24
L (cm)
6.90
7.00
7.10
7.25
7.48
7.60
7.69
7.79
7.90
σ × 10-5
A (cm2)
ln L
1.932
1.946
1.960
1.981
2.013
2.029
2.040
2.053
2.067
0.0942
0.0929
0.0915
0.0897
0.0869
0.0855
0.0845
0.0834
0.0823
dynes cm
7.81
7.92
8.04
8.21
8.47
8.61
8.71
8.82
8.94
σt × 10-7
-2
g cm-1 s-1
4.69
9.51
14.48
29.54
60.98
77.47
94.07
111.2
128.8
2.1
2
1.9
1.8
1.7
1.6
1.5
0
20
40
60
80
100
120
140
σt X 10 -7 (g cm-1 s -1 )
Note that the data provides a linear fit only from values of =
σ t 2.95 × 108 g cm -1 s -1 and larger. A least
squares fit (R2 = 0.9927) of the six data points from σ t =
2.95 × 108 to 1.28 × 108 g cm -1 s -1 gives a slope
of 8.54 × 10-11 cm s g-1 and an intercept of 1.959.
1
= 8.54 × 10−11 cm s g -1
ηT
Pa-s
=
1.17 × 109 Pa-s
10 poise
Note that extrapolation of the linear portion gives an intercept of 1.959 or an apparent Lo of 7.09 cm
versus an actual Lo of 6.00 cm. The difference is due to initial elastic deformation (stretching of coiled
chains) prior to viscous flow (i.e., chain slippage).
ηT =
1.17 × 1010 g cm -1 s -1 =1.17 × 1010 poise
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
45
(b)
Solution
A strip of polyisobutylene (800,000 molecular weight) is subjected to a fixed tensile load at ambient
conditions. Initially, the sample is 0.699 cm wide, 6.0 cm long, and 0.155 cm thick. The strip is hung
vertically and a mass of 75 g is attached to the bottom of the strip. The sample length is then recorded as a
function of time with the following measurements:
Time (min)
1
2
3
6
12
15
18
21
24
Length (cm)
6.9
0
7.0
0
7.1
0
7.2
5
7.4
8
7.6
0
7.6
9
7.7
9
7.9
0
Plot the data given in the form of ln L versus σ t and determine the value of ηT in SI units. Comment on
the probable phenomenological significance of the actual intercept of the plot obtained by extrapolating
the linear portion of the data.
11.6 Show that eq. 11.37, that defines shear stress in pressure flow through a capillary, is correct by
balancing pressure force and shear force in a cylindrical element.
Solution
The balance of forces gives
r ∆p
πr 2 ∆p = τ 2πrL or τ =
2L
11.7 A 2-in. melt extruder is pumping a Newtonian fluid through a slit die for which the form factor, Fp,
is 0.5. The viscosity, µ, of the fluid at operating conditions is 0.2 lbf-s in.-2. The dimensions of the slit die
are 1 in. in width, 0.8 in. in height, and 3 in. in length. The geometric parameters for the extruder are
given in the following table.
Extruder Geometry
Extruder length, Lextr
Diameter of the screw,
D
Channel depth, B
Flight angle, θ
Channel width, W
14.75 in.
1.982 in.
0.166 in.
30°
11 in.
If the extruder is rotating at 60 rpm under isothermal conditions, determine the following:
(a) pressure drop, ∆p, in psi
Solution
µ AN
Eq. 11.60, ∆p =
k +C
Eq. 11.55:
A=
1
1
πDWB cos θ=
π (1.982 )( 3.11)( 0.166 ) cos30°= 1.392 in 3
2
2
From Table 11-4
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
46
WH 3 Fp 1( 0.8 ) 0.5
k
=
=
= 7.11 × 10−3 in 3
12L
12 ( 3)
3
Eq. 11.57,
=
Z
L
14.75
=
= 29.5 in
sin θ sin 30°
Eq. 11.56,=
C
WB 3 3.11( 0.166 )
=
= 4.019 × 10−5 in 3
12 Z
12 ( 29.5 )
3
∆p
0.2 lb f s (1.392 in 3 ) 60 min
= 38.9 psi
in 2 min ( 7.11 × 10-3 in 3 + 4.019 × 10−5 in 3 ) 60 sec
(b) Volumetric flow rate, Q, in units of in.3 min-1
Solution
Eq. 11.54, Q =
k
µ
∆p =
7.11 × 10−3
38.9 = 83.0 in 3 min -1
0.2
(c) power, P, required to operate the extruder in hp (1 hp = 550 ft-lbf s-1)
Solution
 π3 (1.982 )3 D 3 

 sin 30° (1 + 3sin 2 30° )
0.166


1 1
1 1
Note: sin 2 θ= − cos 2θ (Appendix E); sin 2 30°= − cos 60°= 0.25
2 2
2 2
 π3 D 3 
Eq.
11.62, E 
1 + 3sin 2 θ )
=
 sin θ (=
B


E = 1454.3(0.5)(1+ 0.75) = 1273 in
2
Next eq. 11.61,
1
+ 1.392 ( 60 ) 38.9
60
lb -in. Hp s min ft
= 4.506 × 105 + 3.25 × 103 = 4.54 × 105 f
= 1.146 Hp
min 550 ft-lb f 60 s 12 in
=
=
∆p 1273 ( 0.2 )( 60 ) ( 29.5 )
P E µ N 2 Z + AN
2
11.8 Using the dynamic equations for cylindrical coordinates given in Appendix A.2.2 of this chapter,
show how eq. 11.23 can be obtained making the usual assumptions of isothermal, steady, fully developed,
laminar flow through a capillary. State any additional assumptions necessary to obtain eq. 11.23.
Solution
The assumption of fully developed flow means that the flow is simple shear (pressure) flow and the only
component of the stress tensor that needs to be considered in the dynamic equations is τ rz ( ≡ τ zr ) .
Neglecting all inertial terms and body forces results in the following components of the dynamic equation
in cylindrical coordinates:
∂τ rz ∂p
r:
=
∂z
∂r
but fully developed flow means that ∂ ∂r =0 and, therefore, ∂p ∂r =0
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
47
θ:
∂p
=0
∂θ
Finally
z:
1 ∂
∂p
( rτ rz ) = which is eq. 11.23.
r  ∂r
 ∂z
11.9 Derive eq. 11.25 for the velocity profile of a power-law fluid for pressure flow through a capillary.
Solution
Rearrangement of eq. 11.23 gives
∂
∆p
( rτ rz ) = r
∂r
∆z
Integration then gives ∫ d ( rτ rz ) =
r 2  ∆p 
∆p
which
yields
τ
=
r
rdr
rz

 + C1
2  ∆z 
∆z ∫
Note that as r → 0,
r  ∆p 
∆p
→ 0 and, therefore, C1 = 0 and τ rz = 

2  ∆z 
∆z
n
 du 
Substituting the PLF constitutive equation (eq. 11.24) =
τ rz m=
γ m  z  into the above equation gives
 dr 
n
du z  1 ∆p  1 n
=
 r
dr  2m ∆z 
1n
 1 ∆p 
Integrating ∫ du z = 

 2m ∆z 
1n
r dr gives u
∫=
1n
z
1+
1
n
 1 ∆p  r
+ C2


 2m ∆z  1 + 1
n
1n
1+ n
n  1 ∆p 
n
The no-slip assumption, u z ( R ) = 0 , gives C2 = −

 R
1 + n  2m ∆z 
1n
Letting the length of the capillary L = ∆z , finally yields gives eq. 11.23.
11.10
Derive eq. 11.46 for the velocity uθ of a power-law fluid in a Couette rheometer.
Solution
This problem provides an opportunity for the instructor to go deeper into the fundamentals of rheology
including tensor representation of rate of strain in cylindrical coordinates.
Eq. A.14 should read
∂u
∂u
u ∂u
uu
 ∂uθ
+ ur θ + θ θ + r θ + u z θ
∂r
∂z
r ∂θ
r
 ∂t
ρ
1 ∂p  1 ∂ 2
1 ∂τ θθ ∂τ θ z

−
+  2 ( r τ rθ ) +
+
=
∂z
r ∂θ  r ∂r
r ∂θ


 + ρ gθ

Eliminating all zero terms (viscometric flow, uθ ( r ) only) gives
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
48
1 ∂ 2
( r τ rθ ) = 0
r 2 ∂r
from which
a
τ rθ = 2
r
where a is a constant of integration. The constitutive equation for a PLF can be written as (see textbooks
by Middleman or McKelvey)
n
 d  uθ  
C1
τ rθ m=
=
 r dr  r  
r2
 

or
1n
2
d  uθ  1  C1 
 C1  −1− n
=
=
r


 
dr  r  r  mr 2 
m
where C1. Integration gives
( C m ) r −2 n + C
uθ
=
− 1
2
2n
r
or
( C m ) r1− 2 n + C r
uθ =
− 1
2
2n
where C2 is another constant of integration. Using the no-slip boundary conditions:
(1) uθ ( r= Ri )= Ri Ω and
1n
(2) uθ =
0 gives
( r R=
o)
gives the two constants of integration as
1n
nC 
C2 =  1  Ro − 2 n
2 m 
and
Ω
C2 =
−2 n
1 − ( Ri Ro )
Substitution and using κ = Ri Ro gives, upon rearrangement, eq. 11.46
 r  1 − ( Ro r )
uθ
= 
Ri Ω  Ri  1 − κ −2/ n
2/ n
11.11 Derive eqs. 11-72 and 11-74 for the axial annular Couette flow of a Newtonian fluid in a wire
coating die.
Solution
Eq. 11.72
Q =−2πR ( R − Ri )Vz
2κ 2 ln κ − κ 2 + 1
4 (1 − κ ) ln κ
Put eq. 11.30 in form (where R = D/2)
Q = 2π ∫ ru z ( r ) dr
R
Ri
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
49
Use eq. 11.32 in form
uz ln ( r R )
=
U
ln κ
or
Vz
ln ( r R )
ln κ
where eq. 11.73 should read
R
κ= i
R
Then
uz =
Q=
2πVz
ln κ
∫
R
Ri
r ln ( r R )dr
Note
ln xdx
∫ x=
x2 
1
 ln x − 
2
2
Therefore
R
 r2 

1   r2 
ln
ln
ln
ln
ln
r
r
R
dr
r
rdr
r
rdr
r
R
=
−
=
−
−
=
(
)






∫Ri
∫Ri
∫Ri
2  2 
2
 Ri
R
R
R
1 1
1 1
1
1
1
1 2

R  ln R −  − Ri 2  ln Ri −  − R 2 ln R − R12 ln R =
− ( R 2 − Ri 2 ) − Ri 2 ln κ
2 2
2
4
2
2 
2 2 
Next
 R 2 − Ri 2 + 2 Ri 2 ln κ 
Q = −2π Vz 

4ln κ


Multiply numerator and denominator by ( R − Ri ) R gives
 R 2 − Ri 2 + 2 Ri 2 ln κ 
Q=
−2π Vz R ( R − Ri ) 

 4 R ( R − Ri ) ln κ 
Division of the numerator and denominator in brackets by R2 and substitution of κ = Ri R gives eq.
11.68.
Next eq. 11.70
m c = m d
2
−2πρ ′ ( R − Ri )Vz F (κ )
Q=
Qd + Qp =
Vzπ ( Ri + h ) − Ri 2  ρ =


where F (κ ) is defined by eq. 11.75. Rearrangement gives
0
ρ h 2 + 2 Ri ρ h + 2 ρ ′R ( R − Ri ) F (κ ) =
Using the quadratic formula (Appendix E), we have
 2 ρ ′ ( R − Ri ) F (κ ) 
h
h′ = =−1 + 1 −

Ri
ρ Ri


12
12
 2ρ ′  1 

=1 −
 − 1 F (κ ) 

 κρ  κ

−1
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
50
11.12 A Newtonian fluid having a viscosity of 15,000 poise is to be coated on a wire having a diameter
of 0.06 inch through an annular die of 0.08-inch inside diameter. The length of the die is 1.25 inch.
Assuming isothermal flow, calculate the required pressure drop (in psi) across the die to produce a
uniform coating having a thickness of 0.06 inch. The wire is moving at a velocity of 100 ft min-1. What is
the nominal shear rate in the die?
Solution
This is a problem not specified developed in Section 11.5.2, but can be solved with the information
provided. For a Newtonian fluid subject to both isothermal drag and pressure flows,
Q
= Qd + Qp ( see eq. 5-29 in Middleman )
and
m=
m d + m p .
c
Assuming ρ = ρ ′ , we get
Vz π ( Ri + h )

2
2
2
2
4 
1−κ 2 ) 
(
pR
2
ln
1
κ
κ
κ
−
+
π∆
4
1 − κ −

− Ri  =−2πR ( R − Ri )Vz
+

2 (1 − κ ) ln κ
8µ L 
ln (1 κ ) 


2
where κ = Ri R . Need to substitute the parameters given in the problem and solve for the pressure drop,
∆p . Given and calculated parameters are µ = 15,000 poise ; h = 0.06 in; R = 0.04 in; L =1.25 in;
κ = 0.75; Vz = 100 ft min-1; and Ri = 0.03 in. Note 10 poise = 1 Pa-sec and 1 Pa =1.45×10-4 psi.
Substitution gives ∆p =
0.00609 psi .
11.13
Derive eq. 11-76 for a PLF in a wire coating operation.
Solution
No pressure flow, PLF. Start with eq. 11.36 for isothermal axial annular Couette flow of a PLF. Put this
expression for Q into eq. 11.72 for m d and equate with eq. 11.70 for m c . Rearrangement and solution of
the quadratic gives the expression for the dimensionless thickness, eq. 11.76.
CHAPTER 12
12.1 An asymmetric hollow fiber of polysulfone has a surface pore area, A3/A2, of 1.9 × 10-6 and an
effective skin thickness of 1000 Å. If the fiber is coated with a 1-µm layer of silicone rubber, calculate the
effective P  for the coated membrane for CO2 and the permselectivity for CO2/CH4.
Solution
Use eq. 12.25 with permeability data given in Table 12-4.
For CO2,
−1

10−7
P  10−6

 =
=
+
4.86 × 107 barrer cm -1
−6
  4553 4.9 + 4553 (1.9 × 10 ) 


For CH4,
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
51
−1

10−7
P  10−6

 =
=
+
2.15 × 106 barrer cm -1
 1339 0.213 + 1339 (1.9 × 10−6 ) 


7

P
( )CO2 4.86 × 10
CO2
α CH4
=
=
= 23
( P  )CH4 2.15 × 106
This value matched the reported permselectivity of polysulfone (see Table 12-4).
12.2 What pore size is required for Knudsen flow of oxygen and nitrogen through a porous membrane?
What would be the ratio of diffusion coefficient, D(O2)/D(N2), for the permeation of air through this
membrane. How does this ratio compare to that for the permeation of air through a polysulfone membrane
hollow fiber membrane. How does this ratio of diffusion coefficients compare to the ideal
permselectivity, α(O2,N2), reported for oxygen/nitrogen through the polysulfone membrane.
Solution
Using eq. 12.8
12
D ( O2 )  M ( N 2 ) 
 14.0067 
=
=
=
0.936




D ( N 2 )  M ( O 2 ) 
 15.9994 
The Knudsen number is the ratio of mean free path to pore diameter (see for example J. Gilron, A. Soffer,
J. Membr. Sci. 209, 339 (2002). The mean free path is given as
12
3η ( πRT )
2 P 2M
Typical diameters are in the area of 5 Å.
D(O2)/D(N2) = 3.6 while P(O2)/P(N2) = 5.6 for PSF due to higher O2 solubility. See Koros et al., Annu.
Rev. Mater. Sci. 22, 47 (1992).
12
λ=
12.3 What are the limitations, if any, to the statement that P = SD (eq. 12.7)?
Solution
Equation 12.9 requires sufficiently low gas solubility that Henry’s law is valid and the diffusion
coefficient is independent of concentration (usually a good assumption for supercritical gases in rubbery
polymers). See the excellent review by Stern (J. Membr. Sci. 94, 1 (1994)).
CHAPTER 13
13.1 Poly(2,6-dimethyl-1,4-phenylene oxide) (PDMPO) can be partially crystallized in solution (i.e.,
solvent-induced crystallization). (a) Calculate the density of 100% crystalline PDMPO using the groupcontribution parameters given in Table 13-1. (b) If the crystallinity of a semicrystalline sample of
PDMPO is 8%, estimate its crystallinity using the relationship
Vsc= xcVc + (1 − xc )Va
where xc is the degree of crystallinity, Vc is the molar specific volume of a 100% crystalline polymer, and
Va is the molar specific volume of the totally amorphous polymer.
Solution
For Va, see Example Problems 13.1
(a)
Vc = 7.1 + 94 = 101.1
cm3 mol
cm3
= 0.8411
mol 120.2 g
g
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
52
(b)
Vsc = 0.08 ( 0.8411) + 0.92 ( 0.9318 ) = 0.0673 + 0.8573 = 0.9246
=
ρsc
cm3
g
1
g
= 1.082 3
0.9246
cm
13.2 Using the values of molar attraction constants given by van Krevelen in Table 3-2, calculate the
solubility parameters, units of (MPa)1/2, at 25°C of the following polymers whose densities are given
within parentheses:
(a) Polyisobutylene (ρ = 0.924 g cm-3)
(b) Polystyrene (ρ = 1.04 g cm-3)
(c) Polycarbonate (ρ = 1.20 g cm-3)
(a)
CH3
CH2
C
n
, Mo= 56.11
CH3
∑ Fi
Fi
δ=∑
1
C(CH3 )2
840
840
1
CH2
280
280
1,120 (MPa)1/2 cm3 mol-1
F1
V
M o 56.11
=
= 60.73 cm3 mol−1
V =
ρ
0.924
1.937
12
=
δ = 19.4 ( MPa )
100.1
(b)
104.12
= 100.1 cm3 mol-1
1.04
∑ Fi = 280 + 140 + 1517 = 1,937
=
V
=
δ
1,937
12
= 19.4 ( MPa )
100.1
(c)
254.3
= 211.9 cm3 mol-1
1.20
=
F
= 4,361
∑ i 2 (1377 ) + 767 + 840
=
V
=
δ
4361
12
= 20.6 ( MPa )
211.9
13.3 Using UNIFAC-FV, estimate the activity of toluene in a 50 wt % solution of polydimethylsiloxane
in toluene at 298 K.
Prob. 12-1 (2nd ed)
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
53
See J. R. Fried, J. S. Jiang, and E. Yeh, Comput. Polym. Sci. 2, 95 (1992).
a1 = 0.9579
13.4 Based upon their calculated Permachors, order the following polymers in terms of their expected
performance as oxygen barriers: (a) amorphous Teflon; (b) polyisobutylene; (c) polychloroprene; (d)
polybutadiene; (e) nitrile rubber; (f) silicone rubber; and (g) butyl rubber. Does your result give the
correct order based upon experimental permeability values?
Write eq. 12.2 as
=
log P ( 298 ) log P * ( 298 ) −
sπ
2.303
(a) Amorphous Teflon
For the purposes of these calculations, assume amorphous Teflon (see Section 10.1.9) is primarily noncrystalline polytetrafluoroethylene with the repeat unit
CF2
CF2
Since the Permachor for the comonomer is not available in Table 12-2.
1
240
π =  2 (120 )  =
= 120
2
2
First calculate P for N2 where logP*(298) = -12 and s = 0.12:
0.12 (120 )
log P ( 298 ) =
−12 −
=
−12 − 6.2527 =
−18.2527
2.303
P ( 298=
) N2 5.59 × 10−19 cm3 (STP) cm ( cm 2 s Pa )
−1
For O2,
P ( 298 )O2 =3.8 × P ( 298 ) N2 =3.8 × 5.59 × 10−19 =2.12 × 10−18 cm3 (STP) cm ( cm 2 s Pa )
Note (from Appendix D), 1 Pa=7.5 × 10-3 mmHg=7.5 × 10-3 mmHg
P ( 298=
)O2 2.12 × 10−18 cm3 (STP) cm ( cm 2 s Pa )
−1
= 2.8 × 10−15 cm3 (STP) cm ( cm 2 s cmHg )
−1
cm
=
7.5 × 10-4 cmHg
10 mm
1 Pa
7.5 × 10−4 cmHg
−1
(b) Polyisobutylene
CH3
CH2
C
CH3
1
2
π=
−2.5
(15 − 20 ) =
log P ( 298 ) N2 =
−12 +
0.12 ( 2.5 )
2.303
=
−11.8697
P ( 298=
) N2 1.35 × 10−12 cm3 (STP ) cm ( cm 2 s Pa )
−1
P ( 298 )O2 =3.8 (1.35 × 10−12 ) =5.13 × 10−12 cm3 ( STP ) cm ( cm 2 s Pa )
= 6.84 × 10−9 cm3 ( STP ) cm ( cm 2 s cmHg )
−1
−1
(c) Polychloroprene
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
54
CH2
C(Cl)
CH
CH2
1
 2 (15 ) + 33 = 21
3
0.12 ( 21)
log P ( 298 ) N2 =
−12 −
=
−13.0942
2.303
π=
P ( 298=
) N2 8.05 × 10−14 cm3 (STP ) cm ( cm 2 s Pa )
−1
P ( 298 )O2 =3.8 ( 8.05 × 10−14 ) =3.06 × 10−13 cm3 ( STP ) cm ( cm 2 s Pa )
= 4.08 × 10−10 cm3 ( STP ) cm ( cm 2 s cmHg )
(d) Polybutadiene
CH2 CH CH
−1
−1
CH2
1
 2 (15 ) + 33 = 22
3
0.12 ( 22 )
log P ( 298 ) N2 =
−12 −
=
−13.1463
2.303
π=
P ( 298=
) N2 7.14 × 10−14 cm3 (STP ) cm ( cm 2 s Pa )
−1
P ( 298 )O2 =3.8 ( 7.14 × 10−14 ) =2.71 × 10−13 cm3 ( STP ) cm ( cm 2 s Pa )
= 3.62 × 10−10 cm3 ( STP ) cm ( cm 2 s cmHg )
−1
−1
(e) Nitrile rubber is a copolymer of butadiene (part d) and 15% to 40% acrylonitrile. For comparison,
calculate the oxygen permeability of polyacrylonitrile.
CH2
CH
C
N
1
( 205 + 15=) 110
2
0.12 (110 )
log P ( 298 ) N2 =
−12 −
=
−17.73
2.303
π=
P ( 298=
) N2 1.86 × 10−18 cm3 (STP ) cm ( cm 2 s Pa )
−1
P ( 298 )O2 =3.8 ( 2.71 × 10−18 ) =7.05 × 10−18 cm3 ( STP ) cm ( cm 2 s Pa )
= 2.48 × 10−15 cm3 ( STP ) cm ( cm 2 s cmHg )
−1
−1
Estimate the permeability of the copolymer using the inverse rule of mixtures. For 15% and 40% AN
content, permeabilities are estimated as
1 W W
= 1+ 2
P P1 P2
1
0.15
0.85
= −15 +
=
6.05 × 1013
−10
P 2.48 × 10
3.62 × 10
−14
P = 1.65 × 10
1
0.4
0.6
= −15 +
=
1.61 × 1014
P 2.48 × 10
3.62 × 10−10
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
55
=
P 6.21 × 10−15
Note: results show that the permeability of the copolymers is dominated by the high barrier properties of
the AN component.
(f) Silicone rubber is crosslinked polydimethylsiloxane which will be used for calculations.
CH3
Si
O
CH3
1
2
π=
−23
( 70 − 116 ) =
log P ( 298 ) N2 =
−12 +
0.12 ( 23)
2.303
=
−10.8016
P ( 298=
) N2 1.58 × 10−11 cm3 (STP ) cm ( cm 2 s Pa )
−1
P ( 298 )O2 =3.8 (1.58 × 10−11 ) =6.00 × 10−11 cm3 ( STP ) cm ( cm 2 s Pa )
= 8.00 × 10−8 cm3 ( STP ) cm ( cm 2 s cmHg )
−1
−1
(g) Butyl rubber is polyisobutylene with 0.5 to 2% isoprene comonomer for crosslinking. Look at
polyisoprene
CH2
C(CH3 )
CH
CH2
for comparison with PIB (part b)
1
π =  2 (15 ) − 30  = 0 (essentially natural rubber)
3
log P ( 298 ) N2 = −12
P ( 298 ) N2
= 1.0 × 10−12 cm3 ( STP ) cm ( cm 2 s Pa )
−1
P ( 298 )O2 =
3.8 (1.0 × 10−12 ) =
3.8 × 10−12 cm3 ( STP ) cm ( cm 2 s Pa )
= 5.07 × 10−9 cm3 ( STP ) cm ( cm 2 s cmHg )
−1
−1
This value for the O2 permeability of polyisoprene is very close to that of polyisobutylene so little effect
of copolymer concentration on permeability would be expected in this case.
The order of increasing barrier properties: silicone rubber<butyl rubber<polybutadiene<polychloroprene
<polyisobutylene<nitrile rubber<amorphous Teflon.
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
56