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.
PETROLEUM ENGINEERING - ANSWER 25.1 : (a) Ideally (moles of nitrogen) / (moles of oxygen) = (volume of nitrogen) / (volume of oxygen) = 78.09 % V / 20.95 % V = 3.7274 where V is the volume of dry air. (Fact 1, Fact 2). (b) Moles of nitrogen = (moles of oxygen) x (volume of nitrogen) / (volume of oxygen) = 12.5 x 3.7274 = 46.5925 moles. (c) Mass of 1 mole of octane fuel = 1 mole x 114.232 g / mole = 114.232 g. (d) Mass of dry air = mass of oxygen + mass of nitrogen = 12.5 mole x 31.998 g / mole + 46.5925 mole x 28.014 g / mole = 1705.217295 g. (Fact 5) (e) Air-fuel ratio (AFR) = [ Answer in (d) ] / [ Answer in (c) ] = 1705.217295 g / 114.232 g = 14.928. (Fact 3, Fact 4) (f) Fuel-air ratio (FAR) = 1 / [ Answer in (e) ] = 1 / 14.928 = 0.067. 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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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.13 (CORRECTION) : (i) In the Present Value Multiplication Rule, let PV = present value, Ra = interest rate for first discount, A = duration for first discount; Rc = interest rate for second discount, C = duration for second discount. Let PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] where ^ is the symbol of power : 3 ^ 2 = 3 x 3, 2 ^ 3 = 2 x 2 x 2. (a) For discounts involving 8 % / year for 3 years and 10 % / year for 9 years, find the value of PV. (b) If Re = interest rate for third discount, E = duration of third discount, form a mathematical equation of PV as a function of A, C, E, Ra, Rc, Re. Note : Discounts are available in the purchase of certain biochemical engineering instruments. (ii) Let R = nominal interest rate related to growth rate of money, r = real interest rate related to growth rate of purchase power. If I = inflation, where the unit of R, r and I is %, find the mathematical relationship of r as a function of R and I.
Question 33 – By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?
ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.33 : An engineering corporation that produces a type of healthy food has A number of shares of $B preferred stock and C number of shares of $H common stock outstanding. In a typical financial year, let the net income obtained = D. The amount retained by the corporation = E. (a) Calculate the amount of money distributed to the shareholders if : (i) D > E; (ii) D = E; (iii) D < E. (b) Let preferred dividend = F, common dividend = G, CH > G. Find the values of F and G, in term of A, B, D and / or E when : (i) (D - E) > AB; (ii) (D - E) < AB; (iii) (D - E) = AB.
i am a prefinal year student of polymer engineering .could you please send the latest question papers regarding entrace exam.
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hai i A neelakanteswara rao ,selected for the interview on 15th july 2008. I completed my diploma in chemical engineering.let me know how to face the interview& the type of technical questions.
how many TDS & TOTAL HARDNESS, PH of drinking water is recomanded for human being use.
QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.9 : When an algebraic product is defined on the space, the Lie bracket is the commutator [x,y] = xy - yx according to Lie algebra in mathematics. If [p,x] f(x) = px f(x) - xp f(x), p = -ih d / dx, find the value of [p,x] in term of i and h.
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