CHEMICAL MATERIAL BALANCE – EXAMPLE 2.6 : According to Raoult's law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
CHEMICAL MATERIAL BALANCE – ANSWER 2.6 : (a) For component 1, ln (PSAT) = 15 - 3010 / (150 + 250) = 7.475, PSAT = 1763 kPa. For component 2, ln (PSAT) = 14 - 2700 / (150 + 205) = 6.394, PSAT = 598 kPa. Let x(1) PSAT(1) = y(1) P, x(2) PSAT(2) = y(2) P, then x(1) PSAT(1) + x(2) PSAT(2) = [ y(1) + y(2) ] P = P, x(1) (1763) + [ 01 - x(1) ] (598) = 1210, x(1) (1763 - 598) = 1210 - 598, x(1) = 0.525, x(2) = 1 - x(1) = 0.475. (b) y(1) = x(1) PSAT(1) / P = 0.525 (1763) / 1210 = 0.765, y(2) = 1 - y(1) = 0.235. Overall mass balance is assumed 1 mole for F = V + L where F is incoming mole, V is flashed vapor in mole and L is outgoing liquid in mole. For component 1, zF = xL + yV then 0.6(1) = 0.525 L + 0.765 (1 - L), then L = 0.6875, V = 1 - L = 0.3125. Let z to be 60 mole %. 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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DIFFERENTIAL EQUATIONS - EXAMPLE 20.3 : A differential equation is given as y” + 5y’ + 6y = 0, y(0) = 2 and y’(0) = 3. By using Laplace transform, an engineer has correctly produced the equation L {y} = (2s + 13) / [(s + 2)(s + 3)] = A / (s + 2) + B (s + 3). (a) Find the values of A and B. (b) The inversed Laplace transform of 1 / (s + a) is given by exp (-at) where a is a constant. If the statement : L {y} = 9 L { exp (-2t) } - 7 L { exp (-3t) } is correct, find the equation of y as a function of t as a solution to the differential equation stated in the beginning of this question. When L {d} = 9 L {b} - 7 L {c}, then d = 9b - 7c with b, c and d are unknowns.
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.5 : In an American style option for share market, 2 persons - A and B agree to the following : B is required to sell 100 shares of IBMS to A for $85 per share anytime that A wants in the next 8 months. A will pay B $2 per share up front, non-refundable for this option. IBMS involves in petrochemical processing. IBMS stock is currently selling for $80 per share. (a) If A did not buy the share of IBMS from B after 8 months, how much will B earn? (b) If the share of IBMS goes up to $100 / share in 6 months later : (i) how much should A pay B for 100 shares according to their optional agreement? (ii) how much will A earn from 100 shares purchased from B when all the 100 shares are sold to the open market? (iii) how much net profit will A earn for selling 100 shares to the open market?
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Question 33 – By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?
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Question 63 – Autocad computer program is used to draw a pipe that is used to transport natural gases for processing. Actual length of the pipe is 10 metres. The pipe has cross sectional area of circle in shape with measured circumference of s = 3 metres. (a) If 10 units are used to represent the actual length of the pipe in computerised engineering drawing, find the diameter of the pipe, d that will be shown in the engineering drawing. (b) If one unit in the program equals to 1 centimetre, find the scale used in the computer program.
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ENGINEERING MECHANIC - EXAMPLE 15.3 : A biochemical trolley of mass 15 kg is towing a trailer of mass 5 kg along a straight horizontal pathway. The trailer and the trolley are connected by a light inextensible tow-bar. The engine of the trolley exerts a driving force of magnitude 100 N. The trailer and the trolley experience resistances of magnitude 10 N and 30 N respectively. (a) Form 2 equations with unknowns T and a, that represents the equilibrium for the 2 systems of the trolley and trailer. (b) Solve the simultaneous equations from the 2 equations that are obtained in part (a) of this question. T is the tension of the tow-bar and a is the acceleration.
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