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MICROBIOLOGICAL ENGINEERING - ANSWER 28.1 : (a) Dilution = (final volume of mixture) / (initial volume of mixture). First dilution = ( 0.2 + 1.8 ) L / (0.2 L) = 10. Second dilution = first dilution x ( 0.2 + 1.8 ) L / (0.2 L) = 10 x 10 = 100. (b) 100 microlitres = 100 microlitres x 1 mL / (1000 microlitres) = 0.1 mL. CFU / mL = (number of colonies x dilution) / (amount plated, in unit mL). In the first plate, CFU / mL = 250 x 10 / 0.1 = 25000. In the second plate, CFU / mL = 23 x 100 / 0.1 = 23000. Average CFU / mL = (first plated value + second plated value) / 2 = (25000 + 23000) / 2 = 24000. 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.

Heat transfer: 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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Explain some of the consequences of an undersized kettle type reboiler?

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How is waste heat boilers categorized?

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Question 72 - (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calory, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?

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What is difference between Piping Standard & Piping Specification ?

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Explain how can viscosity affect the design of a mixer?

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Explain what are the affinity laws associated with dynamics pumps?

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Question 37 - Calculate the bubble temperature T at P = 85-kPa for a binary liquid with x(1) = 0.4. The liquid solution is ideal. The saturation pressures are Psat(1) = exp [ 14.3 - 2945 / (T + 224) ], Psat(2) = exp [ 14.2 - 2943 / (T + 209) ] where T is in degree Celsius. Please take note that x(1) + x(2) = 1. Please take note that y(1) + y(2) = 1, y(1) = [ x(1) * Psat(1) ] / P, y(2) = [ x(2) * Psat(2) ] / P, * is multiplication. P is in kPa.

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QUANTUM COMPUTING - EXAMPLE 32.4 : A system of linear congruences consists of 3 equations : X ≡ 1 (mod 2), X ≡ 3 (mod 3), X ≡ 4 (mod 5). X has positive values. (a)(i) List the values of these equations from 1 to approximately 40. (ii) Find the first smallest value and second smallest value of X. (iii) Guess the third smallest value of X. (b) Let X ≡ Aa (mod Ma), X ≡ Ab (mod Mb), X ≡ Ac (mod Mc). According to Chinese remainder theorem, X ≡ (Aa x Ya x Md + Ab x Yb x Me + Ac x Yc x Mf) [ mod (Ma x Mb x Mc) ]. (i) Show that Ma, Mb and Mc have the greatest common divisor of Ma x Mb x Mc. (ii) Find the values of Md, Me and Mf if Md = Mb x Mc, Me = Ma x Mc and Mf = Ma x Mb. (iii) Find the values of Ya, Yb and Yc if Ya = Remainder of (Md / Ma), Yb = Remainder of (Me / Mb) and Yc = Remainder of (Mf / Mc). (iv) Use Chinese remainder theorem to find X.

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Question 47 - In a cylinder with a hollow, let a is outside radius and b is the inside radius. In a steady state temperature distribution with no heat generation, the differential equation is (d / dr) (r dT / dr) = 0 where r is for radius and T is for temperature. (a) Integrate the heat equation above into T(r) in term of r. (b) At r = a, T = c; at r = b, T = d. Find the heat equation of T(r) in term of r, a, b, c, d.

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Chemical Engineering Unit Operation - Which of the sequence below represent a feasible flows of ethanol processing plants using cellulose as starting material? A. raw material --> heat exchanger --> distillation column --> reactor. B. reactor --> distillation column --> raw material --> heat exchanger. C. heat exchanger --> raw material --> distillation column --> reactor. D. raw material --> heat exchanger --> reactor --> distillation column. E. distillation column --> raw material --> reactor --> heat exchanger.

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REACTION ENGINEERING - EXAMPLE 13.2 : A batch reactor is designed for the system of the irreversible, elementary liquid-phase hydration of butylene oxide that produces butylene glycol. At the reaction temperature T = 323 K, the reaction rate constant is k = 0.00083 L / (mol - min). The initial concentration of butylene oxide is 0.25 mol / L = Ca. The reaction is conducted using water as the solvent, so that water is in large excess. (a) Let the molecular weight of water is 18 g / mol and the mass of 1 kg in 1 L of water, calculate the molar density of water, Cb in the unit of mol / L. (b) Determine the final conversion, X of butylene oxide in the batch reactor after t = 45 min of reaction time. Use the formula X = 1 - 1 / exp [ kt (Cb) ] derived from material balance. (c) Find the equation of t as a function of X.

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