chapter 2. thermophysical properties of polymersoecomposites/notes/527 u3040/ch2.pdf · chapter 2....
TRANSCRIPT
1
1
Chapter 2. Thermophysical Properties of Polymers
Thermophysical properties include:
1. Volumetric properties2. Calorimetric properties3. Transition temperatures4. Cohesive properties and solubility5. Interfacial energy properties6. Transport properties (diffusion et al.)
2
2
3
4
3
5
6
• Physical aging of glassing polymers
4
7
I. Volumetric properties
1. Specific volume (=volume per unit weight)2. Molar volume
8
5
9
10
• Enthalpy relaxation
6
11
)/()( ∞∞ −−= υυυυδ iLet
τδδ // −=dtdSo, [ ]τδ /)(exp itt −−=
12
7
13
PT⎟⎠⎞
⎜⎝⎛∂∂
=υ
υα 1
14
8
15
16
II. Calorimetric properties
1. Heat capacity
9
17
18
10
19
20
11
21
22
m
mm T
HS ∆=∆ , ∑ ∆=∆ iim snS
12
23
24
13
25
26
III. Transition Temperature
14
27
28
15
29
M
IYY
MY
T ixiggi
gg
∑ +==
)(
where Ix is
5.022
2=
+=I
30
16
31
32
M
ODDYIYY
MYT i
mxmmim
m
∑ ∑ ∑++==
)()(
17
33
34
18
35
IV. Cohesive properties and solubility
36
Cohesive energy density of polymers can be determined by swelling or dissociation experiments.
19
37
38
20
39
• The solubility parameter
40
21
41
42
V. Interfacial energy properties
22
43
44
where κ is the compressibility
1.
2.
23
45
46
• Interfacial tension between a solid and a liquid
Young’s equation svsllv γγθγ =+cos
24
47
48
• How to estimate the surface tension of solid polymers
1. Measure the contact angle with the liquid with known surface tension.
2. , where γ expressed in mJ/m2 and ecoh in mJ/m3.
3. Using group contribution to Parachor for estimation
3/275.0 cohe≈γ
25
49
50
26
51
52
27
53
54
28
55
56
29
57
VI. Transport properties (Diffusion)
58
J
J+ㅿJΔx
the flux xcDJ∂∂
−=
tc
xJJJ
x ∂∂
=∆
∆−−→∆ 0
lim
So,xcD
xtc
∂∂
∂∂
=∂∂
If D is independent of c and x, 2
2
xcD
tc
∂∂
=∂∂
R=nl6
22 nlS =
lu oφ=If ,6
2lD oφ= where Φo is the jump frequency
From the Fick’s first law
(Fick’s second law)
30
59
B.C. C=Ci for 0<x<h at t≤0 and C=Cm at x=0, x=h , and t>0
By using the separation method, we can obtain
∑∞
=⎥⎦
⎤⎢⎣
⎡ +−++
−=−−
02
22)12(exp)12(sin12
141jim
i
hDtj
hxj
jcccc ππ
π
Let ∫=hcdxAm
0, where A is the conversion factor
[ ]∑∞
= ++−
−=−−
=0
2
222
2 )12()/()12(exp81
jim
it
jhDtj
mmmmG π
π
In the environment test, M%=[(weight of moist sample-weight of dry sample)/weight of dry sample]x100%=
%100%100 ×=×−
dd
d
Wm
WWW
,im
it
MMMMG
−−
=
60
31
61
• For the specimens with an infinitive thickness
2
2
xcD
tc
∂∂
=∂∂ B. C. C=Ci for 0<x<∞ at t≤0
and C=Cm at x=0, and t>0
txerf
CCCC
im
it
21−=
−−
πDtCCAdt
xCDAm im
t
xx )(2
00
−=⎟⎠⎞
⎜⎝⎛∂∂
−= ∫=
• Since %%100%dd
d
Wm
WWWM =×
−=
for the initial stage of moisture absorption
tDhMM m
π4
= ,22
12
12
4 ⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛
−−
=mM
htt
MMD π
,
62
• For the three dimensional consideration,
2
2
2
2
2
2
zCD
yCD
xCD
dtdC
zyx ∂∂
+∂∂
+∂∂
=
( ) ⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡+−
+⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
−−
= ∑∞
=02
22
2222 )12(exp12
1414181iim
it
hDti
inDt
lDt
MMMMG π
πππ
In the initial stage of moisture absorption,
πtDhlDnhDnlCCm zyxim ))((4 ++−=
tDnhD
lhD
hMM zyx
m ⎟⎠⎞
⎜⎝⎛ ++=
π4
tDhM m
π4
=
where 2)1(x
z
x
yx D
Dnh
DD
lhDD ++=
If zyx DDD ==2
1 ⎟⎠⎞
⎜⎝⎛ ++=
nh
lhDD x
32
63
DT
DLFor the unidirectional composites,
ffmfL DVDVD +−= )1(0
( ) ( ) ⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
+
−
−−+−= −
π
π
πππ
/1
)/(1tan
/1
4/212
1
2fD
fD
fDD
mmfT VB
VB
VBBDDVD
where ⎟⎟⎠
⎞⎜⎜⎝
⎛−= 12
f
mD D
DB . When Df 0, ( ) mfT DVD π/21−≅
Since yxT DDD == , zL DD =2
/211
1 ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
−
−++=
πf
fT V
Vnh
lhDD
From the slope of experimental data,we can obtain D, DT,, Dm.and DL.