PHYSICS POLYMER - Spring 2005
EMA 6165 - EXAM 2
STUDENT UFID#:______________________
(use the NEW UF ID No. )
INSTRUCTIONS: Read the entire exam carefully. Answer every question to the best of your ability, based
upon the available information. State all assumptions you use in answers.
SECTION I: Polymer Structures/Properties.
State the polymerization mechanism, i.e., condensation, addition, ring opening..
State the relative Tg and/or Tm for each polymer (24pts)
Polymer
1. poly(propylene)
2. poly(ethylene
terephthalate)
3. poly(caprolactam)
Nylon 6
4. poly(ethylene
oxide)
5. poly(styrene)
6. poly(phenyl
terepthalamide)
Kevlar
Polymerization Mechanism
(2pt)
Tg (1pt)
Tm (1pt)
POLYMER PHYSICS - Spring 2005
EMA 6165 - EXAM II
Concepts. Answer the following statements with TRUE or FALSE. If the statement is
false, state the reason(s). If you believe a statement can be made TRUE by a simple
qualifier, then state it clearly. (Normally only a single word change is required.) (12 pts)
7.
The retractive force of an ideal elastomer is predominantly controlled by the entropy at
extension ratios below ca. 3.0
8.
Single chains of an elastomer have a higher modulus for a phantom network compared to
a Gaussian like network.
9.
Increasing a solvent concentration in an ideal rubber reduces the recovery stress by 1/3 for
a fixed temperature and elongation.
10. The entropy of a chain mixture with a solvent increases with increasing temperature above
the LCST.
11. The differences in the principle recovery stress of an ideal elastomer scales directly with
the square root of the product of the biaxial strains under biaxial loading conditions.
12. The osmotic pressure equals zero for conditions where the Flory interaction parameter,
i.e., chi, equals .
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POLYMER PHYSICS - Spring 2005
EMA 6165 - EXAM II
II.
Graphical Relationships: You must provide the appropriate equations for answering each
question. You must label axes appropriately and provide appropriate units and orders of magnitude.
STATE ALL ASSUMPTIONS required. SHOW ALL Equations required. It is preferred that the
relationships are linearized.
13.
Sketch the relationship between the Flory interaction parameter chi versus the (4pts)
radius of gyration of a polymer at the theta condition.
14. Plot the relationship between intrinsic viscosity and the shift factor
(4pts)
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POLYMER PHYSICS - Spring 2005
15. Plot the Osmotic Pressure versus radius of gyration.
EMA 6165 - EXAM II
(4pts)
16. Plot the recovery stress of an ideal elastomer versus the root mean square radius of (4pts)
gyration.
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POLYMER PHYSICS - Spring 2005
EMA 6165 - EXAM II
SECTION II: Definitions/Concepts:
Give a BRIEF description for each of the following terms in the SPACE provided. 16 pts,
Use equations or words or figures.
17. Extension ratio:
18. Fictive temperature:
19. Spinodal decomposition:
20. LCST:
21. Gaussian Chain:
22. Physical aging:
23. Bindoal decomposition:
24. Characteristic ratio:
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POLYMER PHYSICS - Spring 2005
25.
EMA 6165 - EXAM II
SECTION III: Engineering Problems
CLEARLY state any and all assumptions applied developing your the solution.
Given the following information calculate the average molar mass of chains 12 pts
between cross-link junctions. State all assumptions necessary.
Density = 1.143 g/cm2; volume fraction of polymer in equilibrium = 0.56; Chi is
estimated = 0.0082. and molar
volume of 8.11 1 1005 m3 mol01 ,
1
2
Gmix = RT ln (1 2 ) + 1 2 + 1 2
x
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POLYMER PHYSICS - Spring 2005
EMA 6165 - EXAM II
HONOR CODE: We, the members of the University of Florida community, pledge to hold ourselves and our peers to
the highest standards of honesty and integrity.
Signature:___________________________________________________________
26.
Compare, i.e., estimate the number of polymer chains that occupy the cross sectional
area of one cylinder defined in the referenced paper. Use the radius of the cylinders
which were generated by annealing for 60 minutes. Compare to the ideal dimensions
of the poly(styrene-block-butadiene-block-styrene) using the information contained in
the reference the on course web page listed as PaperEleven Hashimoto. Note that
there are specific distributions provided for the configurations of the butadiene block.
You will have to determine the appropriate value to calculate for the block copolymer.
This paper is located on the course webpage at:
http://brennan.mse.ufl.edu/dileup/ema6165/References/SBS_SB_Morphology_Macro
molecules_1993_Hashimoto.pdf
This is due by Friday, February 11 at 9AM. Limit response to one page. Show all
work. (20 pts)
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POLYMER PHYSICS - Spring 2005
EMA 6165 - EXAM II
Rubber Elasticity:
= NV kT ( 2 )
=
=
Thermodynamics of Mixing
2
RT
2 MC
1
( 2 )
MC Mo
RT
MC
H m = N 1 2 1, 2 k T
1
3
( )
2
(
0 = N v kT x x 3y 2
x
0x =
0x =
E 1 / 2 E 1 / 2
H m = Vm 1 + 2 (1 2 )
V
V1
2
RT
Mc
1/ 3 x
)
1
x 2
RT
Mc
1
x 2
x
f * 2
=
N V RT( 2 )
*
f
1
2
G mix = RT ln (1 2 ) + 1 2 + 1 2
x
1, 2 = B1 +
C
S
E
f =
l T l
T ,V
T ,V
E
f
+ T
f =
l
T
T ,V
l,V
2 (1 2 ) C
2
1
1
=
1+
1
2
X2
RT
V1
Nk
S = c [
2
2
( x + y 2 + z 2 3 ) + (ln( x y z )) ]
2
C
x (t ) =
2
C
1
2
ln(1 2 ) + 1 x 2 + 12
t
a
1
c = 1
2
T
Crystallization & Kinetics
4d N kT
1
P = 0 v3 * 7
R0
2G
1
1
=
+ 1 + 2
22
(1 2 )C X
2
=
f * 2
ri 2
=
N v RT 2 ( 2 )
*
r0
f
1
=
C
V1
( 1 2 )2
RT
1 1 = R V u
T
T o H V 1
f
f
f 1
T
f
= h
Tm o
Energy
f
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POLYMER PHYSICS - Spring 2005
EMA 6165 - EXAM II
v = 2b l + 2v f ab e
General Thermodynamics
GI = boao nsi
Q=
G = Go expQ * RT *exp K * RT
I
D
gI
G = b S g aN
I
T
1 V
V T P ,l
P=
1 2g
SK =
=
Nk i
0.5
n 1
1 V
V P l ,T
Chain Dimensions/Molecular Weight
Tm
T
(t ) = 1 exp( kt
n
n 1
)
Physical Aging Behavior/Viscoelasticity
ln t =
=
05
.
G n
C Tm
=
kTC TC TC
m1 mo 1
mo s
(B
f o )(T To )
( f o f ) + (T To )
(B f o )(T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) ( f o f ) + (T To )
Mn =
n
p( r , n) = 2 l 2
3
n
p ( x , y , z , n ) = 2 l 2
3
2
g
r
3
2
( )
4 r 2 e
(
2
e
3 x 2 + y 2 + z 2
2 nl 2
3 r 2
2 nl 2
dr
)
dxdydz
nl 2
=
=
6
6
r2
f = f o + T To
f
3
ro2
ln1 (t1 , T1 ) 17.4 (T To )
=
ln 2 (t 2 , T2 ) 51.6 + (T To )
= Ae
i
2
Kc
1
1 16 2 r
sin 2 + 2 A2 c
=
+
2
R M W M W 3
2
6
r 2 = nl 2
i
i
8.86 (T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) 101.6 + (T To )
Vo
B
Vf
N M
N
1 + cos(180 )
1 cos(180 )
1 + cos (180 ) 1 + cos
= nl 2
1 cos (180 ) 1 cos
[ ] = 2 .5 4
3
2
R eo
M
3
2
M
1
2
3
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POLYMER PHYSICS - Spring 2005
r2
[ ] =
M
3
2
M
1
2
EMA 6165 - EXAM II
ln1 (t1 , T1 ) 17.4 (T To )
=
ln 2 (t 2 , T2 ) 51.6 + (T To )
3
f = f o + T To
f
[ ] =
[ o ]
5 3 = 2Cm 1 1 M
T
1
2
(t ) o e i
=
G=
(t ) o
Gf
= n
=
K
M
*
De =
=
Go
P R 4
8LQ
= A e E / RT
= 1 + 2.5 + 14.1 2
(t ) = o cos t
w = 0.5 max min
= A eV/V
0
Fluid Dynamics (Rheology)
Viscoelasticity and Mechanics
EYoung 's = 2 G Shear (1 + ) = 3 B (1 2 )
ln t =
(B
f o )(T To )
( f o f ) + (T To )
(B f o )(T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) ( f o f ) + (T To )
8.86 (T To )
ln1 (t1 , T1 )
=
ln 2 (t 2 , T2 ) 101.6 + (T To )
R 4 P
Q=
8 L
Wh 2 h
P
Q =
2 (s + 2 ) 2m * L
S
Dh 3 sin 2 P
*
QP =
12
L
Dh 2 sin 2
=
QP
2 (s + 2 )
h
P
2m * L
S
2
E h
2
&
m = bulk D 2 tan D B D S
4
sin
Q D = 1/2 2 D 2 Nh cos sin
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POLYMER PHYSICS - Spring 2005
F2
P=
2
9 2 r1 r 2
2
8LQ
P=
R4
2
F
P=
222
9 r1 r 2
P=
8LQ
R4
h0
= B R DR
hf
x2
Bt = B
2
p
vf
DR =
BR =
X=
v0
Rf
R0
Z
R0
yield = 0.028 ETensile
F=
t
F0 =
Re =
EMA 6165 - EXAM II
R0 fz
Q
E = 2 (1 ) G
E = 2 (1 2 ) B
DV
R0 p
Q
3
P=
Q =V A
= m (T )& (n 1 )
= (T )
=
E = E f v f + E m vm
vf
v
1
=
+m
E
Ef
Em
k
Cp
4 h2
t cooling =
Miscellaneous
8 (Tm T w )
ln 2
(T D TW )
Blow Molding
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