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).
ENGINEERING MATHEMATICS - ANSWER 8.3 : Rearranging the equation (1/y)(dy) = [ x / (Z+1) ] (dx). Use the additional equation u = Z + 1, du = 2x dx, 0.5 du = x dx. Then the equation becomes (1/y)(dy) = (0.5)(1/u) du. Integrating both sides of the equation gives ln |y| = (0.5) ln |Z+1| + c where Z = square of x. 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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CHEMICAL FLUID MECHANIC - EXAMPLE 3.2 : The terminal velocity of a falling object, v is given by v = sqrt [ 4g (R - r) D / (3Cr) ] where sqrt is the square root of, g = 9.81, D = 0.000208, R = 1800, r = 994.6, m = 0.000893. The Reynold number, L is given by L = rD (v) / m. The C for various conditions are : C = 24 / L for L < 0.1; C = 24 (1 + 0.14 L^0.7) / L for 0.1 <= L <= 1000; C = 0.44 for 1000 < L <= 350000; C = 0.19 - 80000 / L for 350000 < L. Find the value of v for the situation above by trial and error, ^ is power, <= is less than or equal to.
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ENVIRONMENTAL ENGINEERING - QUESTION 22.2 : Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) + Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
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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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