ENGINEERING ECONOMY - EXAMPLE 7.3 : There are 2 alternatives of investment. Choice 1 : A trader offers you an investment opportunity where your investment of A$15000 presently will be A$18000 after 4 years. Choice 2 : A bank offers you 5 % annual return for your initial investment of A$15000. Question a : What is the equivalent bank payment after 4 years? Question b : By using the concept of equivalence in engineering economy, which is the better choice, between 1 and 2, that will be more profitable after 4 years?
ENGINEERING ECONOMY - ANSWER 7.3 : (a) For Choice 2, A$15000 x 1.05 = A$15750 after 1 year, A$15750 x 1.05 = A$16537.50 after 2 years, A$16537.50 x 1.05 = A$17364.375 after 3 years, A$17364.375 x 1.05 = A$18232.6 after 4 years. (b) The gain of Choice 2 > Choice 1 with A$18232.6 > A$18000, then Choice 2 is preferred. 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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ENGINEERING PHYSICS - EXAMPLE 30.4 : (a) Time evolution in Heisenberg picture, according to Ehrenfest theorem is m (d / dt) < r > = < p >, where m = mass, r = position, p = momentum of a particle. If v = velocity, prove that m < v > = < p >. (b) Lande g-factor is given by Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. If Gl = 1 and under approximation of Gs = 2, prove by calculation that Gj = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ].
Hi.. Me going to tke test for management trainee(technical) Post I am an chemical engineering student . Can u guide how to prepare for this test.Can u send me test paper pattern..?
Question 45 - 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.
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ENGINEERING MATHEMATICS - EXAMPLE 8.3 : Solve the first order differential equation : (Z + 1)(dy/dx) = xy in term of ln |y| = f(x). Z = (x)(x).
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.3 : (a) In the M / M / 1 queue that happens with randomness, let State 0 = the queue and server are empty, State 1 = the server is in use and the queue is empty, State 2 = the server is in use and 1 is in the queue, State 3 = the server is in use and 2 in the queue. Let P (0) = probability of State 0, P (1) = probability of State 1, P (2) = probability of State 2, P (3) = probability of State 3 and so on. If c = constant, P (1) = c P (0), P (2) = c [ c P (0) ], P (3) = c { c [ c P (0) ] }, write an equation that involves P (N), P (N + 1) and c. (b) Let L = market price of risk, r = riskless rate, m = expected return, s = volatility. Given that L = (m - r) / s related to oil prices, expected return = 12 %, s = 20 %, riskless rate = 8 %, calculate the market price of risk.
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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