ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.4 : A university is enrolling new students of biochemical engineering degree. A long queue is formed during registration. Let L = rate of newcomers to a queue, m = number of clients served at a certain time, T = time in system. In M / M / 1 queue, let T = time waiting in a queue + service time, L = 2 / second, m = 3 / second. (a) Find the service time, A = 1 / m. (b) Calculate time waiting in a queue, B = A (L / m) / (1 - L / m). (c) What is the value of T?
ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.4 : (a) A = 1 / m = 1 / 3 second. (b) B = A (L / m) / (1 - L / m) = (1 / 3) (2 / 3) / (1 - 2 / 3) = 2 / 3 second. (c) T = A + B = 1 / 3 + 2 / 3 = 1 second. 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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ENVIRONMENTAL ENGINEERING - QUESTION 22.3 : A well delivers 225 US-gallons per minute of water to a chemical plant during normal system operation. (a) Calculate its flowrate in the unit of mega US-gallon per day or MGD. (b) The following formula is written next to the chlorine feed point : (chlorine feed rate, lb / day) = (flowrate, MGD) X (dose, mg / L) x (8.34). If this formula is correct, then what should the chlorine feed rate to be in pounds per day (lb / day) if the desired dose is 2 mg / L. (c) Prove by calculations that the constant 8.34 in the formula next to the chlorine feed point is correct. Let 1 US-gallon = 3.78541 L and 1 mg = 0.0000022046 pound.
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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.
PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1 with y(0) = 1. (a) The Laplace transformation, L for given terms are : L (dy / dt) = sY(s) - y(0), L(y) = Y(s), L(1) = 1 / s. Use such transformation to find Y(s). (b) The initial value theorem states that : When t approaches 0 for a function of y(t), it is equal to a function of sY(s) when s approaches infinity. Use the initial value theorem as a check to the answer found in part (a).
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