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.
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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Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli’s equation where P Rgy 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid’s velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.
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QUANTUM BIOLOGY - EXAMPLE 33.6 : The time to maintain quantum coherence in a biological system is t. Let t = k / (MT) where k = constant, M = mass, T = temperature. For living farm animals, normal rectal temperatures ranges for pig and goat are 38.7 - 39.8 and 38.5 - 39.7 in degree Celsius respectively. Let a pig has a mass of 100 kg and a goat has a mass of 300 pounds. (a) Find the mass of goat in the unit of kilogram, when 1 pound = 0.4536 kg. (b) Find t (p) / t (g) for living animals when t (p) = t for pig and t (g) = t for goat. (c) State the assumption of your calculation in question (b).
Please help to get IOCL/HPCL or other chemical engineering paper pls send kathir_skv@yahoo.com Regards, kathirvel
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