PTFE4775_PracticeProbsExam3_Solution

# PTFE4775_PracticeProbsExam3_Solution - 1) Fal s e . Wh e n...

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Unformatted text preview: 1) Fal s e . Wh e n a c ro ss lin ke d rubb e r i s st r e t c h e d , i ts e n t ropy in c r e a s e s and t h e r e f or e i t f ee l s warm t o t h e t ou c h . When the rubber is stretched, the entropy decreases because it becomes more ordered / aligned. Tru e . In t e rm s o f probing t h e st orag e modulu s o f a polym e r m e l t , h e a t ing up t h e m e l t ha s t h e s am e e ff ec t a s low e ring t h e ra t e . Fal s e . Ela st i c d e f orma t ion i s t h e d e f orma t ion o f c ro ss lin ke d t h e rmo s e ts wh e r e a s pla st i c d e f orma t ion i s t h e d e f orma t ion o f un c ro ss lin ke d t h e rmopla st i c s . Elastic deformation refers to deformation that is reversible. Plastic deformation is permanent deformation. Either one can occur in both thermosets and thermoplastics. 2) WLF equation relates the viscosity of a polymer melt to temperature (using T g as reference point). ) T (T K . ) T (T . ) (T (T)- ) (T (T) g g g g 6 51 44 17 log log log . We have two data points and therefore two equations: At 150 o C (423 K), )K T (423K K 51.6 ) T 17.44(423K log ) 10 log( g g 5 ) (T- g At 200 o C (473 K), ) T (473K K 51.6 ) T 17.44(473K log ) 10 log( g g 2 ) (T- g We have two unknowns: (T g ) and T g , which we can find using the two equations above. e.g. we can subtract one equation from the other, eliminating the (T g ) term then solve for T g 101 o C (Since the equation is quadratic, there is another solution to the equation but the value for T g is higher than 200 o C, which is not physical because the polymer is supposed to be in the melt at 150 o C). 3) Data from 5 kg/mol: T g = 60.0 o C For M = 5 kg / mol , T = 100 o C, = 12,770 Poise. Lets find out when T = 125 o C. Use WLF: 61 . 7 60)K (100 K 51.6 K 60) 17.44(100 )] (T [ log- C)] (100 [ log ) (T C) (100 log g g 72 . 9 60)K (125 K 51.6 K 60) 17.44(125 )] (T [ log- C)] (125 [ log ) (T C) (125 log g g Taking the difference between the two equations above: 11 . 2 72 . 9 61 . 7 C) (125 C) (100 log C)] (125 [ log- C)] (100 [ log or 11 . 2 10 C) (125 C) (100 We know (100 o C) = 12,770 Poise. Thus (125 o C) = 12,770 Poise / 10 2.11 = 100 Poise. Now at 125 o C, we can compare the two molecular weights and two melt viscosities: M 1 = 5 kg/mol, = 100 Poise M 2 = 20 kg/mol, = 1,600 Poise....
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## PTFE4775_PracticeProbsExam3_Solution - 1) Fal s e . Wh e n...

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