Answer / kangchuentat

QUANTUM COMPUTING - ANSWER 32.5 : (a) Ba + Be = (00 + 11) + (00 - 11) = 2 (00), 00 = (Ba + Be) / 2. Bc - Bm = (10 + 01) - (10 - 01) = 2 (01), 01 = (Bc - Bm) / 2. Bc + Bm = (10 + 01) + (10 - 01) = 2 (10), 10 = (Bc + Bm) / 2. Ba - Be = (00 + 11) - (00 - 11) = 2 (11), 11 = (Ba - Be) / 2. (b) (C0 + D1) (00 + 11) = C000 + C011 + D100 + D111 = (00) C0 + (01) C1 + (10) D0 + (11) D1 = (1/2) (Ba + Be) C0 + (1/2) (Bc - Bm) C1 + (1/2) (Bc + Bm) D0 + (1/2) (Ba - Be) D1 = (1/2) [ (Ba + Be) C0 + (Bc - Bm) C1 + (Bc + Bm) D0 + (Ba - Be) D1 ] = 0.5 ( Ba C0 + Be C0 + Bc C1 - Bm C1 + Bc D0 + Bm D0 + Ba D1 - Be D1) = 0.5 [ Ba C0 + Ba D1 + Bc C1 + Bc D0 + Be C0 - Be D1 + Bm (-C1) + Bm D0 ] = 0.5 [ Ba (C0 + D1) + Bc (C1 + D0) + Be (C0 - D1) + Bm (-C1 + D0) ] (Proven). 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.

is extractive distillationӞH

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UNIT OPERATION - EXAMPLE 9.4 : Acetone and ethanol are separated using a distillation column with a partial condenser and partial reboiler. An equimolar, sub-cooled liquid feed enters at 100 kmol / hr and condenses 1 mole of vapor for every 6 moles of feed. The separation requires a distillate vapor that is 95 mol % acetone and bottoms liquid that is 5 mol % acetone. The reflux is returned from the condenser to the column as a saturated liquid and the operation is run at (L / V) = 1.4 * (L / V) min. Assume constant overflow conditions. (a) Feed operating line is y = [ q / (q - 1) ] x - z / (q - 1) where z = 0.5 for equimolar liquid mixture of 2 components, q = (L'- L) / F where L' = L + F + (F / 6) for condensation of 1 mole of vapor / 6 moles of feed. What is y = f(x)? (b) The rectifying operating line is y = (L / V) x + (D / V) (xd) where (L / V) min goes through the points A (0.95, 0.95) and B (0.53, 0.69), V = L + D. What is y = f(x)? Let xd = 0.95. (L / V) min is the slope of the 2 points A and B.

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Question 76 - 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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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.6 : In a random walk, a stochastic process starts off with a score of A. This is a score for a chemical engineering test. At each discrete event, there is probability p chance you will increase your score by B and a (1 - p) chance you will decrease your score by C. The event happens T times. Let D = expected value of score. (a) Form an equation of D as a function of A, B, C, T and p. (b) Find the value of D if A = 60, B = 1, C = -1, T = 50, p = 0.5.

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