Answer / kangchuentat

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.

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.3 : (a) In the M / M / 1 queue that happens with randomness, let State 0 = the queue and server are empty, State 1 = the server is in use and the queue is empty, State 2 = the server is in use and 1 is in the queue, State 3 = the server is in use and 2 in the queue. Let P (0) = probability of State 0, P (1) = probability of State 1, P (2) = probability of State 2, P (3) = probability of State 3 and so on. If c = constant, P (1) = c P (0), P (2) = c [ c P (0) ], P (3) = c { c [ c P (0) ] }, write an equation that involves P (N), P (N + 1) and c. (b) Let L = market price of risk, r = riskless rate, m = expected return, s = volatility. Given that L = (m - r) / s related to oil prices, expected return = 12 %, s = 20 %, riskless rate = 8 %, calculate the market price of risk.

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Explain the largest application for surfactants?

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Question 53 - In the purchase of a machine with a period n = 8.5 years, the minimum attractive rate of return, i = 12 %, the cost P = $55000, F = $4000 is the salvage, annual maintenance A = $3500. The return of the investment or equivalent uniform annual benefit is $15000. The equivalent uniform annual cost is P (A / P, i, n) + A - F (A / F, i, n). The investment is considered acceptable only when equivalent uniform annual benefit is greater than the equivalent uniform annual cost. From the compound interest table, (A / P, i = 12 %, n = 8 years) = 0.2013, (A / P, i = 12 %, n = 9 years) = 0.1877, (A / F, i = 12 %, n = 8 years) = 0.0813, (A / F, i = 12 %, n = 9 years) = 0.0677. Prove by calculations whether the investment above is acceptable.

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QUANTUM COMPUTING - EXAMPLE 32.10 : In quantum computing, the conversion of Control Not (CNOT) gate in two input quantum bit gate could be decribed as : | 00 > --> | 00 >, | 01 > --> | 01 >, | 10 > --> | 11 >, | 11 > --> | 10 >. If | P > = 0.707 ( | 01 > - | 11 > ), find the value of CNOT | P >.

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please send me Ongc Exam Previous Papers... To MU MAil Id: muyyaboina.kalyani@gmail.com

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What significance the angle of repose holds in the chemical industry, explain?

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How to TDS removed found in lab apparatus.

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QUANTUM BIOLOGY - EXAMPLE 33.8 : (a) Let ^ be the symbol of power where 1 ^ 2 = 1 x 1 = 1, 2 ^ 2 = 2 x 2 = 4. Let the number of electrons in a human body to be 10 ^ 28 = A, the number of all of the grains of sand on Earth planet to be 7 x (10 ^ 20) = B, the number of all the stars in the visible sky to be 8 x (10 ^ 3) = C. By assuming that every star in the visible sky has the same number of grains of sand as on Earth planet, prove by mathematical calculations that there are more electrons in one human body compared to the number of all of the grains of sand on the stars in the visible sky. (b) The incoming solar radiation to the Earth's surface is mainly from sun. Around 51 % of the radiation is absorbed by Earth's surface. Around 19 % is absorbed by atmosphere and clouds. In term of reflection, 4 % of the radiation is from surface of Earth, 6 % is reflected by atmosphere and the rest is reflected by clouds. Find the percentage of radiation absorbed by and reflected by biological beings on Earth, with reason for your response.

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At 150 degree Celsius, a mixture of 40 wt % Sn and 60 wt % Pb present, forming phases of alpha and beta. Chemical composition of Sn at each phase : CO (overall) : 40 %, CA (alpha) : 11 %, CB (beta) : 99 %. (a) State 2 reasons for the existences of alpha and beta phases for the mixture of Sn – Pb at 150 degree Celsius. (b) By using Lever Rule, calculate the weight fraction of each phase for alpha, WA = Q / (P Q) and beta, WB = P / (P Q) where Q = CB – CO and P = CO – CA.

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what is designing step of heat exchanger

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Question 35 – A mixture consists of benzene (B), toluene (T) and xylene (X). At a temperature of 353 K, the data of vapor pressures : B : 754.12, T : 289.71, X : 91.19. Unit is mm Hg. The pressure P is 0.5 atm. The value of k for each substance is k = (vapor pressure) / P. (a) Calculate k for B, T and X. Let L / V = 0.65. (b) By using the equation V = F / [ (L / V) + 1 ], find the value of V when F = 100, then what is the value of L?

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What is the most common carrier gas used in pneumatic conveying?

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