Answer / kangchuentat

QUANTUM COMPUTING - ANSWER 32.4 : (a)(i) 1 (mod 2) = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39. 3 (mod 3) = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39. 4 (mod 5) = 4, 9, 14, 19, 24, 29, 34, 39. (ii) By observation on X, 3 equations have common values of 9 and 39. First smallest value = 9, second smallest value = 39. (iii) Third smallest value = second smallest value + (second smallest value - first smallest value) = 39 + (39 - 9) = 69. (b)(i) Let Ma = 2, Mb = 3, Mc = 5 where they are prime numbers. Their greatest common divisor is 2 x 3 x 5 = Ma x Mb x Mc (shown). (ii) Md = Mb x Mc = 3 x 5 = 15, Me = Ma x Mc = 2 x 5 = 10, Mf = Ma x Mb = 2 x 3 = 6. (iii) Ma = 2, Mb = 3, Mc = 5, Md = 15, Me = 10, Mf = 6. Md / Ma = 15 / 2 = 7 remain 1, Ya = 1. Me / Mb = 10 / 3 = 3 remain 1, Yb = 1. Mf / Mc = 6 / 5 = 1 remain 1, Yc = 1. (iv) Let Aa = 1, Ab = 3, Ac = 4, Ya = 1, Yb = 1, Yc = 1, Ma = 2, Mb = 3, Mc = 5, Md = 15, Me = 10, Mf = 6. X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ] = (1 x 1 x 15 + 3 x 1 x 10 + 4 x 1 x 6) [ mod (2 x 3 x 5) ] = 69 mod 30 = 39 mod 30 = 9 mod 30. 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.

Question 51 - A batch reactor is designed for the system of the irreversible, elementary liquid-phase hydration of butylene oxide that produces butylene glycol. At the reaction temperature T = 323 K, the reaction rate constant is k = 0.00083 L / (mol - min). The initial concentration of butylene oxide is 0.25 mol / L = Ca. The reaction is conducted using water as the solvent, so that water is in large excess. (a) Let the molecular weight of water is 18 g / mol and the mass of 1 kg in 1 L of water, calculate the molar density of water, Cb in the unit of mol / L. (b) Determine the final conversion, X of butylene oxide in the batch reactor after t = 45 min of reaction time. Use the formula X = 1 - 1 / exp [ kt (Cb) ] derived from material balance. (c) Find the equation of t as a function of X.

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MASS TRANSFER - EXAMPLE 4.3 : According to Adolf Eugen Fick (1829 - 1901) : rate of diffusion v increases with less wall thickness t, increased area A and decreased molecular weight of a fluid M. The diffusion constant D decreased with increasing M. (a) By assuming v, t, dP, A, M and D changes proportionally of each other, find the equation of v as a function of t, dP, A and D. (b) The ratio of self diffusion constant D, at T = 273 K and P = 0.1 MPa, for gases B and C are 1.604 : 0.155. If only 2 gases exist in such a system : hydrogen and nitrogen, find the type of gas for B and C with reference to their molecular weights M. (c) By using the equation of kinetic energy 0.5 MV = constant where V = square of v, find the ratio of V for B and V for C, or V(B) / V(C), as a function of M(B) and M(C), where M(B) is molecular weight of B and M(C) the molecular weight of C : Graham's Law of Diffusion.

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MASS TRANSFER - EXAMPLE 4.1 : A concentric, counter-current heat exchanger is used to cool lubricating oil. Water is used as the coolant. The mass flow rate of oil into the heat exchanger is 0.1 kg / s = FO. For oil, the inlet temperature TIO = 100 degree Celsius and the outlet temperature TOO = 55 degree Celsius. For water, the inlet temperature TIW = 35 degree Celsius and the outlet temperature TOW = 42 degree Celsius. What is the mass flow rate of water in kg / s, FW needed to maintain these operating conditions? Constant for heat capacity of oil is CO = 2131 J /(kg K) and for water is CW = 4178 J /(kg K). Use the equation (FO)(CO)(TIO ?TOO) = (FW)(CW)(TOW ?TIW).

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