ENVIRONMENTAL ENGINEERING - QUESTION 22.1 : In order to predict the wastewater production, the population number has to be understood. The population data is : 72000 (for year 1961 or P-1961), 85000 (for year 1971 or P-1971), 110500 (for year 1981 or P-1981). (a) Find the average population increase, or [ (P-1981 - P-1971) + (P-1971 - P-1961) ] / 2. (b) Find the average percentage population increase, or [ (P-1981 - P-1971) / P-1971 + (P-1971 - P-1961) / P-1961 ] / (2) X 100. (c) Find the incremental increase or P-1981 - 2 (P-1971) + P-1961. (d) Let Po = P-1981. After 2 decades or n = 2, the population is P-2001. By using arithmetical increase method, find P-2001 = Po + n (Answer for a). (e) By using incremental increase method, find P-2001 = (Answer of d) + n (n + 1) (Answer of c) / 2. (f) By using geometrical increase method, find P-2001 = Po [ 1 + (Answer of b) / 100 ] ^ n where ^ is power sign, or 1 ^ 2 = 1 x 1 = 1. (g) If the actual P-2001 = 184000, which method of estimation is more accurate, based on your answer in (d), (e) and (f)?
ENVIRONMENTAL ENGINEERING - ANSWER 22.1 : (a) Average population increase = [ (110500 - 85000) + (85000 - 72000) / 2 = 19250. (b) Average percentage population increase = [ (85000 - 72000) / 72000 + (110500 - 85000) / 85000 ] x 100 / 2 = 24.025. (c) Incremental increase = 110500 - 2 (85000) + 72000 = 12500. (d) P-2001 = 2(19250) + 110500 = 149000. (e) P-2001 = 149000 + 2 (3) (12500) / 2 = 186500. (f) P-2001 = 110500 (1 + 0.24025) ^ 2 = 169973. (g) Errors of answers : (d) 184000 - 149000 = 35000, (e) 184000 - 186500 = -2500, (f) 184000 - 169973 = 14027. Answer (e) or incremental increase is more accurate to such data, or the most accurate prediction among the three available formulae. 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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Question 81 - (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.10 : Let D be the random outcome of rolling a dice once. A new dice has values of D* = D - 3.5. There is a total of n rolls of a dice. (a) Find the variance for D* by using the formula 6 V = [ D* (D = 1) ] [ D* (D = 1) ] + [ D* (D = 2) ] [ D* (D = 2) ] + [ D* (D = 3) ] [ D* (D = 3) ] + [ D* (D = 4) ] [ D* (D = 4) ] + [ D* (D = 5) ] [ D* (D = 5) ] + [ D* (D = 6) ] [ D* (D = 6) ]. (b) Calculate the standard deviation of D* as a square root of V. (c) Another new dice has values of D** = kD*. (i) Find the value of k so that D** has a standard deviation of 1. (ii) Find the values of D** for each outcome of D = 1, 2, 3, 4, 5 and 6, when the standard deviation is 1. (iii) Given that the average score of a dice is 3.5, find the equivalent, new and improved model of a dice, Sn in term of n and D**. (iv) Find the expected value of D** as the average of D**.
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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.
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