Chap12_7pgs - bll hnse P $trettgtheltirt0[ispersiott nnd...

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Trattsfomations nnd ]|eat Trenlmenl l2-2 Determine the constants c and n in Equation l2-2 frrat describe the rate of crystal- lization ofpolypropylene at 140"C. (See Figure 12-30) Solution: f : I - exp(-ctn) T - 140'C : 413 K We can rearrange the equation and eliminate the exponential by taking natu- ral logarithms of both sides of the rearranged equation. We can then note that ln(l - f) versus r is a power equation; if these terms are plotted on a log-log plot, we should obtain a linear relationship, as the graph of the data below indicates. Note that in setting up the equation for plotting, we switch the minus sign from the right hand to the left hand side, since we don't have negative numbers on the log-log paper. I - f : exp(-ctn) ln(l - f): -ctn ln[-ln(l --l)] : ln(ct") ln[ -ln(l - f)] : ln(c) * n ln(t) A log-log plot of "-ln(l - f)" versus "t" is shown. From the graph, we find that the slope n - 2.89 and the constant c can be found from one of the points from the curve: iff:0.5, t:S1.Then I - 0.5 : exp[-c(55)1$1 c:6.47 X 10-6 f r(min) -ln(l - f) [ispersiott $trettgtheltirt0 bll Phnse 0.1 0.22 0.36 0.51 0.69 0.92 1.20 r.61 2.302 28 37 M 50 55 60 67 73 86
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t{6 Tfie science and Englneering of Materlals fnstructor's Sofutions Manuaf 10 f (min)
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Determine the activation energy for crystallization of polypropylene, using the curves in Figure l Z-30. solutlon: we can determine how the rate (equar to r/r)changes with temperature: rate:: l/r:cexp(-Q/n4 r/r (r-r) ur(K-') L/(9 rnin)(60 s/min) = 1.85 x t0-3 r/(SS min)(60s/min) : 3.03 x t0-4 t/(lt6 minx6o vmin) : 5.2,1 x lo-5 r/(130 + 273):2.48 x t0-3 1/(140 + 273) : 2.42 x t0-3 l/(150 + 273) :2.36 x l0-3
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This note was uploaded on 02/01/2012 for the course MSEG 302 taught by Professor Snively during the Spring '08 term at University of Delaware.

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Chap12_7pgs - bll hnse P $trettgtheltirt0[ispersiott nnd...

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