Answer / kang chuen tat (malaysia - pen

Answer 78 - (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.

what is density

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If sum one has HAL placement paper plz mail me ,I m Chemical Engg.my id is er.kotnala@gmail.com

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Name the three classes of organic solvents?

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CHEMICAL FLUID MECHANIC - EXAMPLE 3.1 : Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli's equation where P + Rgy + 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid's velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.

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The power required for size reduction in crushing is???

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What are the steps taken to operate a tank-blanketing valve?

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1 Answers  

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.34 : A biochemical engineering company has stock P = 20,000 shares of M % preferred stock of Q = $100 par and N number of shares of $20 par common stock. In year 1 : $10,000 is distributed as dividend. In year 2 : $25,000 is distributed as dividend. Let (Total Dividend) = (Dividend for Preferred Stock) + (Dividend for Common Stock). (a) In year 1, calculate the amount distributed as dividend for preferred stock, when no dividend was distributed for common stock. (b) In year 2, the amount distributed for common stock was $5000. Find the values of M and N, when the dividend per share for common stock is $0.05.

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Question 61 – A biochemical trolley of mass 15 kg is towing a trailer of mass 5 kg along a straight horizontal pathway. The trailer and the trolley are connected by a light inextensible tow-bar. The engine of the trolley exerts a driving force of magnitude 100 N. The trailer and the trolley experience resistances of magnitude 10 N and 30 N respectively. (a) Form 2 equations with unknowns T and a, that represents the equilibrium for the 2 systems of the trolley and trailer. (b) Solve the simultaneous equations from the 2 equations that are obtained in part (a) of this question. T is the tension of the tow-bar and a is the acceleration.

1 Answers  

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