NATURAL GAS ENGINEERING - QUESTION 26.2 : (a) The Hyperion sewage plant in Los Angeles burns 8 million cubic feet of natural gas per day to generate power in United States of America. If 1 metre = 3.28084 feet, then how many cubic metres of such gas is burnt per hour? (b) A reservoir of natural gas produces 50 mole % methane and 50 mole % ethane. At zero degree Celsius and one atmosphere, the density of methane gas is 0.716 g / L and the density of ethane gas is 1.3562 mg / (cubic cm). The molar mass of methane is 16.04 g / mol and molar mass of ethane is 30.07 g / mol. (i) Find the mass % of methane and ethane in the natural gas. (ii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the gases are ideal where final volume of the gas mixture is the sum of volume of the individual gases at constant temperature and pressure. (iii) Find the average density of the natural gas mixture in the reservoir at zero degree Celsius and one atmosphere, by assuming that the final mass of the gas mixture is the sum of mass of the individual gases. Assume the gases are ideal where mole % = volume % at constant pressure and temperature.
NATURAL GAS ENGINEERING - ANSWER 26.2 : (a) 1 foot = (1 / 3.28084) metre = 0.3048 metre, 1 cubic foot = 0.3048 x 0.3048 x 0.3048 cubic metre = 0.0283 cubic metre, 8 million cubic feet = 8 million cubic feet x (0.0283 cubic metre / cubic feet) = 0.2264 million cubic metres. If 0.2264 million cubic metres of gas is burnt per day, then 0.2264 million / 24 = 9433 cubic metres of gas is burnt per hour. (b)(i) Let 50 mole of methane and 50 mole of ethane in the gas. Mass (g) = Mole (mol) x Molar Mass (g / mol), then mass of methane = 50 x 16.04 = 802 and mass of ethane = 50 x 30.07 = 1503.5. Total mass of natural gas = 802 g + 1503.5 g = 2305.5 g, then mass % of methane = (802 / 2305.5) x 100 = 34.786 % and mass % of ethane = 100 - 34.786 = 65.214 %. (ii) Density of methane gas = 0.716 g / L, density of ethane gas = 1.3562 mg / (cubic cm) = 1.3562 mg / mL = 1.3562 g / L. Volume (V) = Mass (m) / Density (r). Then V = m / r = 0.34786 m / 0.716 + 0.65214 m / 1.3562 = 0.9667 m, average density, r = m / (0.9667 m) = 1.0344 g / L. (ii) Mass (m) = Volume (V) x Density (r). Then m = Vr = 0.5V x 0.716 + 0.5V x 1.3562 = 1.0361V, average density, r = 1.0361V / V = 1.0361 g / L. 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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QUANTUM CHEMISTRY AND CHEMICAL ENGINEERING - EXAMPLE 31.5 : In a wavefunction, let P(x) = A cos kx + B sin kx. By using the boundary conditions of x = 0 and x = l, where P(0) = P(l) = 0, prove by mathematical calculation that P(x) = B sin (npx / l) where p = 22 / 7 approximately, n is a rounded number. A, B and k are constants.
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FOOD ENGINEERING - QUESTION 23.1 : (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calorie, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?
ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.35 : A voluntary accountant is performing Du Pont Analysis for a charitable engineering organization. References for 3-step Du Pont formula are : (ROE) = (Net Income $) / (Equity $), where ROE is Return On Equity. (ROS) = (Net Income $) / (Sales $), where ROS is Return On Sales. (Asset Turnover) = (Sales $) / (Assets $). (Leverage) = (Assets $) / (Equity $). From the balance sheet and income statement, the company has Assets of $3753.19 this year and $3216.86 last year. Sales for this year is $2292.34. Let ROE = 0.37 and ROS = 0.21. (a) Calculate the Asset Turnover of the organization. (b) Find the value of the organizational Leverage. (c) Calculate the Net Income $ of the organization.
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