polymers: calculate degree of crystallinity?

polymers: calculate degree of crystallinity?
in [1[2,3]] an equation for quiscent crystallisation kinetics,
calculating "material derivative" or relative degree of crystallisation
is given:
(1*) (D/Dt)theta = n * K(T) * (1-theta) * (-ln(1-theta))^((n-1)/n)
(D/Dt) is "material derivative" - what is its meaning?
theta is the relative degree of crystallisation
(theta=theta_abs/theta_max)
is my interpretation right, that (D/Dt)theta is the speed of change in
crystallinity degree? my problem is, that if initially theta=0 then
(D/Dt)theta=0 everytime, because
ln(1-(theta=0))=ln(1)=0? there is obviously something wrong
is this rearrangement of (1*) allowed?
(2) (D)theta = (Dt) *
n * K(T) * (1-theta) * (-ln(1-theta))^((n-1)/n)
to calculate theta(t)?
(3) theta(t+Dt) = theta(t) + (D)theta(t)
in [1[4,5,6,7]] this equations are given for K(T), Delta_T, f and
T_inf:
(4*) K(T) = (ln2)^(1/n) * (1/t_12)
* exp(-U_/(R*(T-T_inf)))
*
exp(-K_k/(T* Delta_T *f))
Delta_T = T_m_0 - T
f = 2*T / (T + T_m_0)
T_inf = T_g - 30
R = 8.314 [J/mol K]
for profax PP-6723 (IPP) [1] states:
n = 3 avrami-index
(1/t_12) = 3.858E8 [1/s] factor: influcences crystallisation
independent of T
K_k = 3.930e5 [1/K] nucleation exponent
U_ = 6284 [J/mol] activation energy for semental jump
i took missing data from [8]:
Tg = -3 [K] glass-temperature
T_m_0 = 187 [K] equilibrium melting point
xi_inf = 0.460 maximum crystallisation degree
substituting with those values leads to huge K(T):
T = 250
(ln2)^(1/n) ~ 1
(1/t_12) ~ 1E8
exp(-U_/(R*(T-T_inf))) ~ exp(-6284/8.314*288) ~ 1E-3
exp(-K_k/(T* Delta_T *f)) ~ exp(3.9E5/(250*63*1)) ~ 1E10
K(T) ~ 1E15 ?
using a simple simulation K(T) is increasing to infinity (for float
numers in c).
where is my thinking-error?
I would be glad to get some hints.
Thank you,
Bernd Schmitt
[1] x. guo, a.i. isayev* and i. guo, "crystallinity and microstructure
..."
polymer engineering and science, oct 1999, vol. 39, no. 10, p2096ff
http://130.83.61.160/~robert/temp/ISA99a_CrystallinityAndMicroStructurePartII.pdf
[2] k. nakamura, t. watanabe, k. katayama and t. amano,
j. appl. polym. sci, 16, 1077 (1972)
[3] k. nakamura, k. katayama and t. amano,
j. appl. polym. sci., 17, 1031 (1973)
[4] t.w. chan, a.i. isayev*,
polym. eng. sci., 34, 461 (1994)
[5] t.w. chan, g.d. shyu, a.i. isayev*
,
polym. eng. sci., 35, 733 (1995)
[6] b. wunderlich, "crystal nucleation, ...",
academic press, new york (1976)
[7] b. wunderlich,
macromolecular phzsics, vol. 2, academic press, london (1976)
[8] m. moneke, "die kristallisation von verstaerkten thermoplasten
...",
dissertation, tu-darmstadt (2001)
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Reply to
Bernd.Schmitt.News
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The material (or substantial) derivative D/Dt is defined for transport problems as d/dt + v dot del (just google it), where del is the gradient. It is useful for calculations when the point of reference is moving with the flow or whatever is transporting.
If you have zero degree of crystallization, why would you expect D(theta)/Dt to be other than zero? If you set theta equal to a constant, it won't change...you need to solve for theta as a function using theta = 0 as an initial condition AT t = 0.
This equation is not analytically solvable unless you make some serious assumptions about what directions can be eliminated from del. D/Dt contains 4 independent variables (t, x, y, z), so it is a nasty PDE. Equation (4) must incorporate some of these assumptions so you should find out what they are, exactly.
Reply to
rekuci
Thank you. You mean:
(D/Dt)A(t,x) = (Partial / Partial t)A(t,x) + (Partial /Partial x)A(t,x)*(dx/dt)
?
Hm. So (3) is not allowed? I mean, if focusing on one small cell, i thought (4*) and (1) ((3)) would describe how much nucleation happens / crystalls grow and (2) would give me the amount of how much crystallinity has changed (which results in fusion heat). Like:
theta(0) = 0 theta(0+dt) = theta(0) + (D/Dt)theta(0)*dt ... theta(t+dt) = theta(t) + (D/Dt)theta(t)*dt ?
What is my missing point?
If starting like theta(0) = theta_0 > 0 theta(t+dt) = theta(t) + (D/Dt)theta(t)*dt this seems to work ...
Reply to
Bernd.Schmitt.News
All partials for any multivariate function. I don't know what you're doing above...Dq/Dt = dq/dt + v.(dq/dx i + dq/dy j + dq/dz k) = dq/dt + v.del(q). All lower case d's are partial by necessity. This is for Cartesian coordinates. You might be more interested in something like spherical or cylidrical coordinates to describe a polymer nucleation phenomena, so use the appropriate del.
What on earth are you trying to do? Finite differences?
Reply to
rekuci
I do not know this "v dot del", "v.del(q)" ... is this div or grad you mean? I did not find any hints of this kind of mathematical notation, not in sci.polymers faq nor googling for usenet mathematical notation nor ascii mathematical notation .... sorry - can you give me keywords to use for googling?
AFAIK I do not need coordinates, because I first look at a small homogenous block of material at a homogenous temperature T. In a certain temperature range nucleation should start. Neglegting morphology I am only interested in T(t) and theta(t) for that cell.
So I thought a first simple numerical approach could be like that (Q=heat conduction, H=reaction/nucleation heat, c = heat capacity of that cell): -> dT = dQ/c + dH/c | T(t+dt) = T(t) + dt | dQ= ... - dH= ... I was looking for a simple equation for dH=dH(dtheta).
A very simple numerical iteration assuming _quiescent_ melt (only heat conduction and reaction heat): T(t,x) -> dQ(t,x), dH(t,x) -> dT(t,x) -> T(t+dt,x) -> ...
Why is that wrong?
Reply to
Bernd.Schmitt.News
sorry I meant
-> dQ= ... | dH= ... | dT= dQ/c + dH/c - T(t+dt) = T(t) +dT
Reply to
Bernd.Schmitt.News

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