Answer / kangchuentat

MASS TRANSFER - ANSWER 4.2 : (a) S / G minimum = (0.086957 - 0.0304) / (0.0000488 - 0) = 1160. (b) S / G actual = 1.5 (1160) = 1740. (c) Let x = y / 1640 = 0.08 / 1640 = 0.0000487805. The value of y corresponds to 8 % or 0.08. 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.

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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I am writing the chemical engineering subject test for Reliance India Limited. Could someone who has taken the test please list some of the questions from that paper?

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What are the effects of oils on the properties of Polyolefins?

0 Answers   IOCL,

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Explain the working of a spray condenser?

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How can you estimate the efficiency of a pump?

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Explain how is waste heat boilers categorized?

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Question 101 - (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb Cc). Find | A > + | C > - | B > in term of Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb and Cc. (b) Let d | E > = d (Ea Eb Ec) = (d Ea d Eb d Ec). If | E > = (6 7 8), find the value of 10 | E >.

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Question 71 - (a) The byte is the basic building block of computer data used in chemical engineering process simulation where 16 bits make a word, 4 bits make a nibble, 32 bits make a quad word and 8 bits make a byte. Then how many nibbles are there in a megabytes? (b) In computer data items, let : 1 bit - counts from 0 to 1, 8 bits - counts from 0 to 255, 16 bits - counts from 0 to A. What is the value of A? (c) In a binary system of 4 bits, if 1100 = 12, 1101 = 13, 1110 = 14, 1111 = 15, B = 16, then guess the value of B. (d) By using any form of tools, find the exact value of 2 power 64 or 2^64.

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what is difference between urea and NPK fertiliser?

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Question 94 - 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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For a Vernier scale of representative fraction (RF) = 1 / 25, calculate the length of the scale for the reading up to 4 metres.

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REACTION ENGINEERING - EXAMPLE 13.1 : In a furnace, 2 chemical reactions are happening - 1 mole of solid carbon reacts with 1 mole of oxygen gas to generate 1 mole of carbon dioxide gas; 1 mole of solid carbon reacts with 0.5 mole of oxygen gas to generate 1 mole of carbon monoxide gas. In a given process, 100 kmol of carbon is burned in a furnace. (a) Calculate the theoretical oxygen gas needed by assuming that all the carbon is burned completely to carbon dioxide gas. (b) Calculate the theoretical air needed by assuming that all the carbon is burned completely to carbon dioxide gas and there is only 21 % of oxygen gas. (c) Determine the amount of air required (in kmol) if 50 % excess oxygen gas must be satisfied for (a) and (b). (d) It has latter been found that 20 % of the carbon undergoes incomplete combustion resulting to carbon monoxide gas production. The rest of the carbon undergoes complete combustion. Calculate the total oxygen gas required stoichiometrically based on the actual process.

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