Question 83 - The United States of America Energy Information Administration reports the following emissions in million metric tons of carbon dioxide in the world for year 2012 : Natural gas : 6799, petroleum : 11695, coal : 13787. Coal-fired electric power generation emits around 2000 pounds of carbon dioxide for every megawatt hour generated, which is almost double the carbon dioxide released by a natural gas-fired electric plant per megawatt hour generated. If 1 metric ton = 1000 kg and 1 pound = 0.4536 kg, estimate the total energy generated by natural gas in the world for year 2012, in gigawatt hour.
Answer 83 - Mass of carbon dioxide emitted by natural gas-fired electric power generation for every megawatt hour generated = 2000 / 2 = 1000 pounds = 1000 pounds x (0.4536 kg / pound) x (metric ton / 1000 kg) = 0.4536 metric tons. Total energy generated by natural gas in the world for year 2012 = 6799 million metric tons x (megawatt hour / 0.4536 metric tons) = 14988.977 million megawatt hour = 14988977 gigawatt hour, approximately. 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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MICROBIOLOGICAL ENGINEERING - QUESTION 28.1 : In the calculation of the growth of bacteria, colony forming unit (CFU) in serial dilution is used. In a laboratory, viable count assay is used to estimate CFU. Formula applied is CFU / mL = (number of colonies x dilution) / (amount plated, in unit mL). Acceptable plate count is either between 20 and 200 or between 30 and 300 according to 2 different references. A wastewater sample of 200 ml is added to and mixed with 1.8 L of sterile water. Another 200 ml of the mixture is added to and mixed with 1.8 L of sterile water. (a) Calculate the dilution of first mixture and the dilution of the second mixture. (b) 100 microlitres of wastewater samples from the first mixture and the second mixture are placed separately on 2 different alga plates. The first plate has 250 colonies and the second plate has 23 colonies. Calculate the average CFU / mL.
Explain the procedure to estimate the friction factor involved in heat exchanger tubes?
Calculate the cooling duty, H required to condense and cool acetone from 100 degree Celsius to 25 degree Celsius at atmospheric pressure. The heat of vaporization for acetone at its normal boiling point is 30.2 kJ / mol. The boiling point of acetone at atmospheric pressure is 56 degree Celsius. The flowrate of acetone through the condenser is 100 mol / s = N. Value of sensible heat needed to increase the temperature of acetone in liquid form from 25 to 56 degree Celsius is 4.06 kJ / mol. Value of sensible heat needed to increase the temperature of acetone in vapor form from 56 to 100 degree Celsius is 3.82 kJ / mol. Unit of H is kJ / s.
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A distillation column separates 10000 kg / hr of a mixture containing equal mass of benzene and toluene. The product D recovered from the condenser at the top of the column contains 95 % benzene, and the bottom W from the column contains 96 % toluene. The vapor V entering the condenser from the top of the column is 8000 kg / hr. A portion of the product from the condenser is returned to the column as reflux R, and the rest is withdrawn as the final product D. Assume that V, R, and D are identical in composition since V is condensed completely. Find the ratio of the amount refluxed R to the product withdrawn D. Hint : Solve the simultaneous equations as follow in order to find the answer (R / D) : 10000 = D + W; 10000 (0.5) = D (0.95) + W (0.04); 8000 = R + D.
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How is plate heat exchangers used in an ammonia refrigeration system?
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can u tel me d question paper of gail?
On what basis are materials to be crushed evaluated apart from size?
In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI – TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI – TB); second evaporator : q(2) = UA (TB – TC); third evaporator : q(3) = UA (TC – TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI – TF).
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