Answer / Mrityunjay Kumar

This question pertains to job applications, which falls outside the scope of SAP ABAP, Manual Testing, and Chemical Engineering. It would be best to research online for graduate trainee (chemical engineering) model questions or reach out to specific companies for their application requirements.

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.

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How can I treat a waste stream containing both hexavalent chromium and arsenic?

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ENVIRONMENTAL ENGINEERING - QUESTION 22.2 : Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) + Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).

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Explain the reasons of removal of particles from effluent gas?

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QUANTUM BIOLOGY - EXAMPLE 33.7 : (a) In a DNA of a living cell, the quantum information available in the bases guanine (G) and thymine (T) are | G > = | 110 > and | T > = | 010 > respectively. Calculate | G > - | T >. (b) In a living biological cell, the step time for random walk of an electron is t. The localization time of an electron is T. If i is the geometric average of T and t, find log T as a function of t and i.

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which book is better for written test of a refinery or a chemical industry-o.p gupta or DR.RAM PRASAD?

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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.4 : The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.

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Question 66 – The drag coefficient Cd = 0.05 and lift coefficient Cl = 0.4 for a levelled flow aircraft are measured. The velocity of the aircraft is v = 150 ft / s with its weight W = 2677.5 pound-force. (a) Find the value of the lift of the aircraft, L, when it is also its weight. (b) The drag of the aircraft, D = Cd M, L = Cl M. Find the value of D. (c) The power required is P = Dv. If 1 pound-force x (ft / s) = 1.356 W, find the value of P in the unit of Watt or W.

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Define an additive?

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i need sample question paper of hpcl entrance exam... in chemical engneering category... please mail me to mychitti07@yahoo.co.in

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

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