Answer / kang chuen tat (malaysia - pen

Answer 45 - (a) For component 1, ln (PSAT) = 15 - 3010 / (150 + 250) = 7.475, PSAT = 1763 kPa. For component 2, ln (PSAT) = 14 - 2700 / (150 + 205) = 6.394, PSAT = 598 kPa. Let x(1) PSAT(1) = y(1) P, x(2) PSAT(2) = y(2) P, then x(1) PSAT(1) + x(2) PSAT(2) = [ y(1) + y(2) ] P = P, x(1) (1763) + [ 01 - x(1) ] (598) = 1210, x(1) (1763 - 598) = 1210 - 598, x(1) = 0.525, x(2) = 1 - x(1) = 0.475. (b) y(1) = x(1) PSAT(1) / P = 0.525 (1763) / 1210 = 0.765, y(2) = 1 - y(1) = 0.235. Overall mass balance is assumed 1 mole for F = V + L where F is incoming mole, V is flashed vapor in mole and L is outgoing liquid in mole. For component 1, zF = xL + yV then 0.6(1) = 0.525 L + 0.765 (1 - L), then L = 0.6875, V = 1 - L = 0.3125. Let z to be 60 mole %. 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.

In differential centrifugation of cells with diameter D in centimeter, the square of D is given by D x D = [18n ln (RF / RI) ] / [ (RP – RFF) Wt ] where n is the fluid viscosity (poise), RF is the final radius of rotation (cm), RI is the initial radius of rotation (cm), RP is cell density (g/ml), RFF is the fluid density (g/ml), W the square for the rotational velocity in (radians / s) (radians / s), t is the time required to sediment from RI to RF (s). Derive an equation for W as a function for D, n, RF, RI, RP, RFF and t, with the stated units above, in radian

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For cement company what type of questions come in written exam?

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IS BTech chemical distance education degree from JRN Rajasthan vidyapeth by SSK engineering college Tamilnadu aprooved from UGC or AICTE?

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Explain how does a tank-blanketing valve operate?

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What are the disadvantages of using gear pumps?

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Explain what type of pump may be appropriate for a liquid near saturation, a low flow rate, and very limited npsha (net positive suction head available)?

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What is the maximum recommend pipe velocity for dry and wet gases?

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we have 2 O2/Comb Analyzers Servomex Model 2700 in service for one of the furnaces in our chemical plant. It is giving High Temperature Alarm. Checked Temperature and found it to be 800 Deg C whereas Set Point is 700 Deg c. It is strange why is temperature not at SP. Also response is 23 % with 21 % Inst Air and with N2 reading is not coming down. please clarrify this issuses very urgent

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What's -74C, dew point is better the -70C dew point In draying unit .

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Question 104 - In photoelectrical effect analysis of quantum chemistry, let E = kinetic energy of electron, p = intensity of UV light, f = frequency of UV light. According to Classical Theory, E = c for all values of f, E = mp. According to Quantum Theory, E = c for all values of p, E = mf + c. In a graph, m and c are constants where m is slope and c is y intercept. If m = 2 and c = 3 with similar value of E : (a) find the value of p according to Classical Theory; (b) find the value of f according to Quantum Theory.

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Question 103 - (a) Let | - > = 1 | x > + 0 | y >, | | > = 0 | x > + 1 | y >. Find the value of 2 | x > + 3 | y > in term of | - > and | | >. (b) Let m to be the reduced mass. Find the value of m in term of Ma and Mb where 1 / m = 1 / Ma + 1 / Mb.

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CHEMICAL ENERGY BALANCE - EXAMPLE 11.5 : According to Margules Equation, P = x(1) p(1) g(1) + x(2) p(2) g(2) for a two-component mixture where P is bubble pressure, x is mole fraction, p is saturation pressure, g is constant given by ln g(1) = x(2) A x(2). Find the value of A as a constant when P = 1.08 bar, p(1) = 0.82 bar, p(2) = 1.93 bar in a 50 : 50 mole fraction mixture. Estimate the pressure required to completely liquefy the 30 : 70 mixture using the same equation, by proving P = 1.39 bar. Take note that ln g(2) = x(1) A x(1), ln g(1) = x(2) A x(2).

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