POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs

POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs = density of sample, Ra = density of amorphous form and Rc = density of crystalline form. In a polymer, these unknowns could be related by the equation C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra). (a) Find the equation of Rc as a function of C%, Rs and Ra. (b) Two samples of a polymer, C and D exist. For sample C, C% = 0.513 when Rs = 2.215 unit. For sample D, C% = 0.742 when Rs = 2.144 unit. Both samples C and D have the same values of Ra and Rc. Find the values of Ra and Rc in 6 decimal places.

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POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs = density of sample..

Answer / kangchuentat

POLYMER ENGINEERING - ANSWER 24.2 : (a) C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra), C% (Rc - Ra) Rs = Rc Rs - Rc Ra = Rc C% Rs - Ra C% Rs, Rc C% Rs - Rc Rs + Rc Ra = Ra C% Rs = Rc (C% Rs - Rs + Ra), Rc = Ra C% Rs / (C% Rs - Rs + Ra). (b) For sample C, Rc = Ra C% Rs / (C% Rs - Rs + Ra) = Ra (0.513) (2.215) / [ (0.513) (2.215) - 2.215 + Ra ] = 1.136295 Ra / (Ra - 1.078705). For sample D, Rc = Ra C% Rs / (C% Rs - Rs + Ra) = Ra (0.742) (2.144) / [ (0.742) (2.144) - 2.144 + Ra ] = 1.590848 Ra / (Ra - 0.553152). Then Rc = 1.136295 / (Ra - 1.078705) = 1.590848 / (Ra - 0.553152), 1.136295 (Ra - 0.553152) = 1.590848 (Ra - 1.078705), 1.136295 Ra - 0.628544 = 1.590848 Ra - 1.716056, Ra (1.590848 - 1.136295) = 1.716056 - 0.628544 = 1.087512 = 0.454553 Ra. Then Ra = 1.087512 / 0.454553 = 2.392487. When Ra = 2.392487, Rc = 1.136295 / (Ra - 1.078705) = 1.590848 / (Ra - 0.553152) = 0.864904 with 6 decimal places. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.

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