Answer / kangchuentat

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.10 : (a) 6 V = (1 - 3.5) (1 - 3.5) + (2 - 3.5) (2 - 3.5) + (3 - 3.5) (3 - 3.5) + (4 - 3.5) (4 - 3.5) + (5 - 3.5) (5 - 3.5) + (6 - 3.5) (6 - 3.5) = 17.5. V = 17.5 / 6 = 2.92. (b) Standard deviation = square root of V = (2.92) ^ (0.5) = 1.71. (c) (i) k = 1 / (standard deviation of D*) = 1 / 1.71 = 0.585. (ii) D** (D = 1) = -2.5 / 1.71 = -1.462, D** (D = 2) = -1.5 / 1.71 = -0.877, D** (D = 3) = -0.5 / 1.71 = -0.292, D** (D = 4) = 0.5 / 1.71 = 0.292, D** (D = 5) = 1.5 / 1.71 = 0.877, D** (D = 6) = 2.5 / 1.71 = 1.462. (iii) Sn = 3.5 n + 1.71 nD**. (iv) Expected value of D** = [ D** (D = 1) + D** (D = 2) + D** (D = 3) + D** (D = 4) + D** (D = 5) + D** (D = 6) ] / 6 = (- 1.462 - 0.877 - 0.292 + 0.292 + 0.877 + 1.462) / 6 = 0. 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.

why we make urea prills , why we do not make crystals of urea,

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Which type of drier is used for drying wet material?

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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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Is it possible to compare the resistance to chloride attack of several materials of construction?

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PETROLEUM ENGINEERING - QUESTION 25.1 : Fact 1 : Dry air contains 20.95 % oxygen, 78.09 % nitrogen, 0.93 % argon, 0.039 % carbon dioxide, and small amounts of other gases by volume. Fact 2 : Volume occupied is directly proportional to the number of moles for ideal gases at constant temperature and pressure. Fact 3 : 12.5 moles of pure oxygen is required to completely burn 1 mole of pure octane. Fact 4 : Air-fuel ratio (AFR) is the mass ratio of dry air to fuel present in a combustion process such as in an internal combustion engine or industrial furnace. Fact 5 : Molecular weight of oxygen gas is 31.998 g / mole and molecular weight of nitrogen gas is 28.014 g / mole. (a) Find the molar ratio of nitrogen and oxygen, or (moles of nitrogen) / (moles of oxygen) in dry air, by assuming ideal features of nitrogen and oxygen gases. (b) How many moles of nitrogen are available if dry air is used to completely burn the 1 mole pure octane? (c) Find the mass of fuel of 1 mole of octane with molecular weight of 114.232 g / mole. (d) Find the mass of dry air with 12.5 moles of pure oxygen by assuming only oxygen and nitrogen gases exist in the air. (e) Find the air-fuel ratio (AFR) when octane is used as fuel. (f) Find the fuel-air ratio (FAR) when octane is used as fuel.

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Question 107 - In N + 1 Rule in Quantum Chemistry, whenever a spin 1 / 2 nucleus is adjacent to N other nuclei, it is split into N + 1 distinct peaks. In 1 peak or singlet, there is only 1 magnitude. In 2 peaks or doublet, the ratio of magnitude of each peak is 1 : 1. In 3 peaks or triplet, the ratio of magnitude of each peak is 1 : 2 : 1. In 4 peaks or quartet, the ratio of magnitude of each peak is 1 : 3 : 3 : 1. In 5 peaks or quintet, the ratio of magnitude of each peak is 1 : 4 : 6 : 4 : 1. (a) By using binomial coefficients or Triangle of Pascal find the ratio of magnitude of each peak if 6 peaks exists. (b) How many adjacent nuclei are available in a spin 1 / 2 nucleus in such situation of 6 peaks?

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plz send the interview questions in the written test asked by hindustan petroleum corporation ltd.Iam a b>tech in chemical engineering.My written test is on feb 2008.urgent plz.send the questions to indra_bits10@yahoo.com

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What are the types of clarifier used in water treatement plant?

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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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Question 80 - Liquid octane has a density of 703 kilograms per cubic metre and molar mass of 114.23 grams per mole. Its specific heat capacity is 255.68 J / (mol K). (a) Find the energy in J needed to increase the temperature of 1 cubic metre of octane for 1 Kelvin. (b) At 20 degree Celsius, the solubility of liquid octane in water is 0.007 mg / L as stated in a handbook. For a mixture of 1 L of liquid octane and 1 L of water, prove by calculations that liquid octane is almost insoluble in water.

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