chapter 2. thermophysical properties of polymersoecomposites/notes/527 u3040/ch2.pdf · chapter 2....

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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.)

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• Physical aging of glassing polymers

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I. Volumetric properties

1. Specific volume (=volume per unit weight)2. Molar volume

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• Enthalpy relaxation

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11

)/()( ∞∞ −−= υυυυδ iLet

τδδ // −=dtdSo, [ ]τδ /)(exp itt −−=

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13

PT⎟⎠⎞

⎜⎝⎛∂∂

=υ

υα 1

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II. Calorimetric properties

1. Heat capacity

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22

m

mm T

HS ∆=∆ , ∑ ∆=∆ iim snS

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III. Transition Temperature

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M

IYY

MY

T ixiggi

gg

∑ +==

)(

where Ix is

5.022

2=

+=I

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M

ODDYIYY

MYT i

mxmmim

m

∑ ∑ ∑++==

)()(

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IV. Cohesive properties and solubility

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Cohesive energy density of polymers can be determined by swelling or dissociation experiments.

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• The solubility parameter

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V. Interfacial energy properties

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where κ is the compressibility

1.

2.

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• Interfacial tension between a solid and a liquid

Young’s equation svsllv γγθγ =+cos

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• 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

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VI. Transport properties (Diffusion)

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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)

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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

−−

=

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• 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

,

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• 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

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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.