Answer / kangchuentat

ENVIRONMENTAL ENGINEERING - ANSWER 22.2 : (a) BOD = (8.8 - 5.9) (300 / 30) = 29 mg / L. (b) BOD = (8.8 - 4.2) (600 / 100) = 27.6 mg / L. (c) BOD = (29 + 27.6) / 2 = 28.3 mg / L. (d) NBOD = 45 - 28.3 = 16.7 mg / 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.

Define condensate lift?

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distillation types (vacuum, azeotropic, steam, extractive)

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Thyristor related applications

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.22 : An engineering company intends to produce a small piece of biochemical instrument for sales. Let A = overall fixed cost of production, B = variable cost of production per unit, C = selling price per unit, D = quantity of unit produced. Breakeven Analysis is used where revenue = cost. (a) Explain the role of Breakeven Analysis by using A, B, C and D. (b) Find the value of Contribution Margin in term of A, B, C and / or D. (c) At the breakeven point where revenue = cost, derive an equation of D as a function of A, B and C.

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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.

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which accomplishments in your present position are you proud of and why?

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What is the angle of repose and what are its applications in the chemical industry?

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How can one determine if a particular solid can be fluidized as in a fluidized bed?

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Sir, what is the significant of mmWC what is the definition? How can we Establish the DEMO apparatus in Plant/process lab. ? Arun Kumar Lupin Ltd Bhopal. 9926433779

2 Answers   Lupin,

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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.5 : The following formula is used as a reference : (analyte signal) / (internal standard signal) = (f-factor) x (concentration of analyte) / (concentration of internal standard). A solution containing 3 mM of analyte and 4 mM of internal standard gave peak signals of 2 and 3 mamps respectively. Another similar solution containing 2 mM of analyte and 1 mM of internal standard gave peak signals of 1 and 4 mamps respectively. Find the average f-factor.

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sir/mam . I am a fresh student of chemical engineeing. i've somuch fear to face intervies. I become neverse when entered into company. what shall I do? please suggest me some tips to face the interview.

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ENVIRONMENTAL ENGINEERING - QUESTION 22.3 : A well delivers 225 US-gallons per minute of water to a chemical plant during normal system operation. (a) Calculate its flowrate in the unit of mega US-gallon per day or MGD. (b) The following formula is written next to the chlorine feed point : (chlorine feed rate, lb / day) = (flowrate, MGD) X (dose, mg / L) x (8.34). If this formula is correct, then what should the chlorine feed rate to be in pounds per day (lb / day) if the desired dose is 2 mg / L. (c) Prove by calculations that the constant 8.34 in the formula next to the chlorine feed point is correct. Let 1 US-gallon = 3.78541 L and 1 mg = 0.0000022046 pound.

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