Question { 1478 }

CHEMICAL ENERGY BALANCE - EXAMPLE 11.4 : 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.

Answer

CHEMICAL ENERGY BALANCE - ANSWER 11.4 : By trial and error, T = 84.37 degree Celsius. Excel program may be used, either by Solver or Goal Seek. Overall equation : y(1) + y(2) = 1 = { 0.4 exp [ 14.3 - 2945 / (T + 224) ] + 0.6 exp [ 14.2 - 2943 / (T + 209) ] } / 85. Solve the equation by computer iteration with one unknown will produce T = 84.37 degree Celsius. Let x(2) = 1 - x(1) = 1 - 0.4 = 0.6, P = 85. 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 { 1660 }

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).

Answer

CHEMICAL ENERGY BALANCE - ANSWER 11.5 : Let x(1) = x(2) = 0.5 for 50 : 50 mole fraction mixture, then P = x(1) p(1) g(1) + x(2) p(2) g(2) or 1.08 = 0.5 (0.82) exp [ (0.5) A (0.5) ] + 0.5 (1.93) exp [ (0.5) A (0.5) ]. Then exp [ (0.5) A (0.5) ] = 1.08 / [ 0.5 (0.82 + 1.93) ] = 0.785, A = - 0.97. P for 30 : 70 mixture is P = x(1) p(1) g(1) + x(2) p(2) g(2) = (0.3) (0.82) exp [ (-0.97) (0.7) (0.7) ] + (0.7) (1.93) exp [ (-0.97) (0.3) (0.3) ] = 1.39 bar (proven) where g(1) = exp [ A x(2) x(2) ], g(2) = exp [ A x(1) x(1) ]. 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 { 1574 }

ENGINEERING MATERIAL - EXAMPLE 12.1 : In crystal material, hexagonal crystal system could form 4-digit index in certain direction of solid. For [1(-1)0] direction in the hexagonal crystal systems of particular catalyst applied in fume removal of incinerator, what is the four-digit index for this direction? Hint : The transformation equations between the 3-digit [h'k'l'] and the 4-digit [hkil] indices are : h = (1/3) (2h'-k'); i = - (h + k); k = (1/3) (2k'-h'); l = l' A. [(-1)100] B. [1(-1)00] C. [(-1)000] D. [00(-1)(-1)] E. [(-1)0(-1)0]

Answer

ENGINEERING MATERIAL - ANSWER 12.1 : h = (1/3) [ 2 - (-1) ] = 1; k = (1/3) ( - 2 - 1 ) = -1; i = - ( 1 - 1 ) = 0; l = i = 0. Solution is [1(-1)00]. Answer is B. 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 { 1684 }

ENGINEERING MATERIAL - EXAMPLE 12.2 : At 150 degree Celsius, a mixture of 40 wt % Sn and 60 wt % Pb present, forming phases of alpha and beta. Chemical composition of Sn at each phase : CO (overall) : 40 %, CA (alpha) : 11 %, CB (beta) : 99 %. (a) State 2 reasons for the existences of alpha and beta phases for the mixture of Sn - Pb at 150 degree Celsius. (b) By using Lever Rule, calculate the weight fraction of each phase for alpha, WA = Q / (P + Q) and beta, WB = P / (P + Q) where Q = CB - CO and P = CO - CA.

Answer

ENGINEERING MATERIAL - ANSWER 12.2 : (a) Reasons : (1) atomic radii difference; (2) different crystal structure. (b) P + Q = CO - CA + CB - CO = CB - CA, then WA = Q / (P + Q) = (CB - CO) / (CB - CA) = (99 - 40) / (99 - 11) = 0.6705. WB = P / (P + Q) = (CO - CA) / (CB - CA) = (40 - 11) / (99 - 11) = 0.3295. 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 { 1550 }

ENGINEERING MATERIAL - EXAMPLE 12.3 : Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.

Answer

ENGINEERING MATERIAL - ANSWER 12.3 : (a) V = 0.3349 x 0.3349 x 0.3349 = 0.0376 cubic nanometers. (b) Mass per unit cell = 2 x 92.9 (g / mol) [ 1 mol / (6.02 x 10 ^ 23 atoms) ] = 3.086 x 10 ^ (-22) g = m. V = 0.0376 x [ 10 ^ (-7) ] ^ 3 = 0.0376 x 10 ^ (-21) cubic cm. Then density is m / V = 8.207 g / cubic cm.

Question { 1962 }

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.

Answer

REACTION ENGINEERING - ANSWER 13.1 : (a) Theoretical oxygen gas = 100 kmol C (1/1) = 100 kmol of oxygen gas. (b) Theoretical air = (100 kmol) (1 / 0.21) = 476.2 kmol. (c) Amount of air required 50 % excess = 476.2 kmol (1.5) = 714.3 kmol. (d) Carbon consumption to produce carbon dioxide gas = 0.8 x 100 = 80 kmol = oxygen consumption to produce carbon dioxide gas. Carbon consumption to produce carbon monoxide gas = 0.2 x 100 = 20 kmol = 2 times oxygen consumption to produce carbon monoxide gas, then oxygen consumption to produce carbon monoxide gas = 20 / 2 = 10 kmol. Total oxygen gas required = 80 + 10 = 90 kmol. 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 { 1522 }

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.

Answer

REACTION ENGINEERING - ANSWER 13.2 : (a) Cb = (1000 g / L) / (18 g / mol) = 55.556 mol / L. (b) X = 1 - 1 / exp [ kt (Cb) ] = 1 - 1 / exp (0.00083 x 45 x 55.556) = 0.8744. (c) X = 1 - 1 / exp [ kt (Cb) ], 1 / exp [ kt (Cb) ] = 1 - X, exp [ kt (Cb) ] = 1 / (1 - X), (Cb) kt = ln [ 1 / (1 - X) ], t = 1 / [ k (Cb) ] ln [ 1 / (1 - X) ] = ln [ 1 / (1 - X) ] / [ k (Cb) ]. 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 { 1438 }

REACTION ENGINEERING - EXAMPLE 13.3 : The half-life for first order reaction could be described in the differential equation dC / dt = -kC where k is a constant, C is concentration and t is time. (a) Find the equation of C as a function of t. (b) Find the half life for such reaction or the time required to reduce 50 % of the initial concentration, where k = 0.139 per minute. (c) When the initial concentration Co is 16 mol / cubic metre, how long does the reaction required to achieve the final concentration of 1 mol / cubic metre?

Answer

REACTION ENGINEERING - ANSWER 13.3 : (a) dC / dt = -kC, dC / C = -k dt. Integrate both sides gives ln C - ln Co = ln (C / Co) = -kt, C = Co exp (-kt). (b) When C / Co = 0.5, t = [ ln (C / Co) ] / (-k) = [ ln (0.5) ] / (-0.139) = 4.9867 minutes. (c) Step 1 : 16 to 8 mol per cubic metre. Step 2 : 8 to 4. Step 3 : 4 to 2. Step 4 : 2 to 1. Each step takes t minutes. There are 4 steps needed from 16 to 1 mol / cubic metre or 4t = 19.9468 minutes. 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 { 1643 }

BIOPROCESS ENGINEERING - EXAMPLE 14.1 : 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 & degree.

Answer

BIOPROCESS ENGINEERING - ANSWER 14.1 : By algebraic formula, W = [18n ln (RF / RI) ] / [ (D x D) (RP - RFF) t ] where W is in (radians / s) (radians / s). One radian is approximately 57.288 degrees, then the W (radian) = (57.288) (57.288) W (degree) or W (in radian) = 3281.96 W (in degree). 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 { 1791 }

BIOPROCESS ENGINEERING - EXAMPLE 14.2 : An aqueous solution with 2.5 g of a protein dissolved in 600 cubic centimeters of a solution at 20 degree Celsius was placed in a container that has a water-permeable membrane. Water permeated through the membrane until the h - level of the solution was 0.9 cm above the pure water. (a) Calculate the absolute temperature of the solution, T in Kelvin, where T (Kelvin) = T (degree Celsius) + 273.15. (b) Calculate the osmotic pressure, P of the solution by using the formula P = hrg where h is level of the solution, r is density of water with 1000 kg per cubic meter, g = 9.81 N / kg as gravitational acceleration. (c) Calculate the concentration of the protein solution, C in kg / cubic meter. (d) Calculate the molecular weight of the protein, (MW) = CRT / P where R = 8.314 Pa cubic meter / (mol K) as ideal gas constant.

Answer

BIOPROCESS ENGINEERING - ANSWER 14.2 : (a) T = 20 + 273.15 = 293.15 K. (b) P = hrg = (0.9 / 100) (1000) (9.81) = 88.29 Pa. (c) C = (2.5 / 1000) / (600 / 1000000) = 4.167 kg / cubic meter. (d) (MW) = CRT / P = (4.167) (8.314) (293.15) / 88.29 = 115.03 kg / mol = 115030 g / mol. 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 { 1567 }

ENGINEERING MECHANIC - EXAMPLE 15.1 : What is the meaning – strain? In a distillation column, if a block 10 cm on a side is deformed so that it becomes 9 cm long, what is the strain involved in percentage?

Answer

ENGINEERING MECHANIC - ANSWER 15.1 : Strain is defined as the amount of deformation an object experiences compared to its original size and shape. Calculation of strain = (original length – final length) / (original length) = (10 – 9) / 10 = 0.1 or 0.1 x 100 % = 10 %. 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 { 1807 }

BIOPROCESS ENGINEERING - EXAMPLE 14.3 : The kinetic behavior of an enzyme could be described using Michalis - Menten equation : Vo = Vmax [S] / (Km + [S]). Derive this equation from [ES] = [E]total [S] / (Km + [S]), Vmax = Kcat [E]total, Vo = Kcat [ES].

Answer

BIOPROCESS ENGINEERING - ANSWER 14.3 : Vo = Kcat [ES] = Kcat [E]total [S] / (Km + [S]) where [ES] = [E]total [S] / (Km + [S]). When Vmax = Kcat [E]total, Vo = Kcat [E]total [S] / (Km + [S]) = Vmax [S] /(Km + [S]) (Derived). 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 { 1489 }

ENGINEERING MECHANIC - EXAMPLE 15.2 : A cantilever beam of length L = 3 m carries a uniformly distributed load (UDL) of W = 20 N / m throughout the length. Calculate the bending moment, BM of the beam near the fixed end. What is the shear force, SF at this point? Let SF = -WL, BM = -LWL/2.

Answer

ENGINEERING MECHANIC - ANSWER 15.2 : BM near the fixed end = -LWL/2 = -3 x 20 x 3 / 2 = -90 Nm. SF near the fixed end = -WL = -20 x 3 = -60 N. 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 { 1432 }

ENGINEERING MECHANIC - EXAMPLE 15.3 : A biochemical trolley of mass 15 kg is towing a trailer of mass 5 kg along a straight horizontal pathway. The trailer and the trolley are connected by a light inextensible tow-bar. The engine of the trolley exerts a driving force of magnitude 100 N. The trailer and the trolley experience resistances of magnitude 10 N and 30 N respectively. (a) Form 2 equations with unknowns T and a, that represents the equilibrium for the 2 systems of the trolley and trailer. (b) Solve the simultaneous equations from the 2 equations that are obtained in part (a) of this question. T is the tension of the tow-bar and a is the acceleration.

Answer

ENGINEERING MECHANIC - ANSWER 15.3 : (a) Trolley : 100 - 30 - T = 70 - T = 15a as Equation (1). Trailer : T - 10 = 5a as Equation (2). (b) Trailer : 3 times Equation (2) : 3T - 30 = 15a as Equation (3). Equation (1) - Equation (3) = 100 - 4T = 0 then T = 100 / 4 = 25 N. From Equation (2), 5a = T - 10 then a = (T - 10) / 5 = (25 - 10) / 5 = 3 N / kg. 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 { 1919 }

Question 105 - In a rigid rotor model in quantum chemistry, the moment of inertia I is given by an Equation E as I = Ma x La x La + Mc x Lc x Lc = m x L x L, where m = (Ma x Mc) / (Ma + Mc) and L = La + Lc, m is the reduced mass, Ma is the mass of a, Mc is the mass of c, La is the radius of a from point O, Lc is the radius of c from point O. Prove by simplest method that Equation E is wrong.

Answer

Answer 105 - Let Ma = 0 and Mc = 1 as assumption. Substitute them into Equation E as I = Ma x La x La + Mc x Lc x Lc = 0 + Lc x Lc = Lc x Lc. However, m = (0 x 1) / (0 + 1) = 0, then I = Lc x Lc, which is not equal to 0 but I = m x L x L = 0 x L x L = 0. Equation E is proven wrong. 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.