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HW11 Solutions - ChE 178 Problem Set#11 SOLUTIONS Due...

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ChE 178: Problem Set #11 SOLUTIONS Due Monday, December 8, 2004 1. The G(t, T) for a certain polymer with a density equal to 1x10 6 g/m 3 at a temperature T = 350 K is given by G(t,T) = 1000exp[-(t/ τ 1 o ) 0.333 ] + exp[-(t/ τ 2 o ) 0.45 ]       where   τ 1 o =1sec and τ 2 o = 10 6 sec and the units of G are MPa. (30 pts total) a.) Plot log G(t) versus log t from t=10 -10 to 10 +10 . (3 pts) Using Excel, produce the following plot : G(t,T) 1E-28 1E-26 1E-24 1E-22 1E-20 1E-18 1E-16 1E-14 1E-12 1E-10 0.00000001 0.000001 0.0001 0.01 1 100 10000 1.00E-10 1.00E-08 1.00E-06 1.00E-04 1.00E-02 1.00E+00 1.00E+02 1.00E+04 1.00E+06 1.00E+08 1.00E+10 t G(t,T G(t,T) b.) Identify the transition zone and the terminal zone on the plot. (4 points total – 2 for transition zone and 2 for terminal zone) The transition zone is the decrease that occurs around t=10 to t=1000 sec and the terminal zone is the decrease that occurs above around t=10 8 . c.) Find the rubbery plateau modulus G N o for this polymer. (3 points) The plateau modulus is the area of 0 slope between the transition and terminal zone. It is more easily identified in the region of the plot blown up below. From the plot G N 0 = 0.9 Mpa
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G(t,T) 0.1 1 10 1.00E+02 1.00E+03 1.00E+04 1.00E+05 1.00E+06 t G G(t,T) d.) Describe the mechanism by which the stress relaxes in the terminal zone. (3 points) Here the stress is mainly relaxed by disentanglement through reptation type movement of the chains. We know this because there is a rubbery plateau, meaning we can use the Reptation model to describe the entangled polymer melt. e.) Compute the entanglement molecular weight M e of this polymer from the plot. (3 points) M e = ( ρ R T)/G N 0 =(1x10 6 g m -3 )(8.314 m 3 Pa mol -1 K -1 )(350 K)/(0.9x10 6 Pa) = 3233 g/mol f.) Estimate the critical molecular weight for entanglement of this polymer. (3 points) M c = 2 M e = 6466 g/mol Recall that the molecular weight of the polymer must be GREATER than the
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