kang chuen tat


{ City } penang
< Country > malaysia
* Profession * biochemical engineer
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Answers / { kang chuen tat }

Question { 1945 }

DIFFERENTIAL EQUATIONS - EXAMPLE 20.2 : During the landing process of an airplane, the velocity is constant at v. (a) If the displacement of the plane is x at time t, find the differential equation that relates t, x and v. (b) The plane has 2 parts of wheels - the front and the back, separated by a distance L. The front part of the wheel touches the land first, that allows the straight body of the plane to form an angle T with the horizontal land. If the vertical distance between the back part of the wheel and the horizontal land is y, find the equation of y as a function of L and T. (c) Find the differential equation that relates dy as a function of dt, v and sin T. (d) Find the differential equation that consist of dy as a function of y, L, v and dt. (e) Find the equation of y as a function of v, L, t and C where C is a constant. (f) When t = 0, prove that y = exp C as the initial value of y.


Answer

DIFFERENTIAL EQUATIONS - ANSWER 20.2 : (a) The kinematic relationship is dx = v dt. (b) The trigonometric relationship is sin T = y / L. (c) y = -vt sin T, then dy = -v dt sin T since y / x = sin T and x = -vt for y is decreased with an increasing x. (d) dy = -v dt sin T = -v dt (y / L) = -y (v / L) dt. (e) dy / y = (-v / L) dt. Integrate both sides of equation will produce ln y = -vt / L + C, then y = exp (-vt / L + C) = (exp C) exp (-vt / L). (f) When t = 0, y = (exp C) exp (-vt / L) = (exp C) exp (0) = exp C. 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.

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Question { 1388 }

DIFFERENTIAL EQUATIONS - EXAMPLE 20.3 : A differential equation is given as y” + 5y’ + 6y = 0, y(0) = 2 and y’(0) = 3. By using Laplace transform, an engineer has correctly produced the equation L {y} = (2s + 13) / [(s + 2)(s + 3)] = A / (s + 2) + B (s + 3). (a) Find the values of A and B. (b) The inversed Laplace transform of 1 / (s + a) is given by exp (-at) where a is a constant. If the statement : L {y} = 9 L { exp (-2t) } - 7 L { exp (-3t) } is correct, find the equation of y as a function of t as a solution to the differential equation stated in the beginning of this question. When L {d} = 9 L {b} - 7 L {c}, then d = 9b - 7c with b, c and d are unknowns.


Answer

DIFFERENTIAL EQUATIONS - ANSWER 20.3 : (a) Let (2s + 13) / [(s + 2)(s + 3)] = [A (s + 3) + B (s + 2)] / [(s + 2)(s + 3)]. Then 2s + 13 = As + 3A + Bs + 2B = (A + B)s + (3A + 2B), then A + B = 2 and 3A + 2B = 13. Let 2A + 2B = 4 as first equation by doubling A + B = 2 and 3A + 2B = 13 as second equation, the difference of first and second equations will produce A = 9. When A = 9, B = 2 - A = 2 - 9 = -7. (b) Let d = y, b = exp (-2t) and c = exp (-3t), then y = 9 exp (-2t) - 7 exp (-3t). 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.

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Question { 1492 }

PROCESS DESIGN - EXAMPLE 21.1 : According to rules of thumb in chemical process design, consider the use of an expander for reducing the pressure of a gas when more than 20 horsepowers can be recovered. The theoretical adiabatic horsepower (THp) for expanding a gas could be estimated from the equation : THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] where 3 ^ 3 is 3 power 3 or 27, Q is volumetric flowrate in standard cubic feet per minute, Ti is inlet temperature in degree Rankine, a = (k - 1) / k where k = Cp / Cv, Po and Pi are reference and systemic pressures respectively. (a) Assume Cp / Cv = 1.4, Po = 14.7 psia, (temperature in degree Rankine) = [ (temperature in degree Celsius) + 273.15 ] (9 / 5), nitrogen gas at Pi = 90 psia and 25 degree Celsius flowing at Q = 230 standard cubic feet per minute is to be vented to the atmosphere. According to rules of thumb, should an expander or a valve be used? (b) Find the outlet temperature To by using the equation To = Ti (Po / Pi) ^ a.


Answer

PROCESS DESIGN - ANSWER 21.1 : (a) a = (k - 1) / k = (1.4 - 1) / 1.4 = 0.286. Ti = 25 degree Celsius to degree Rankine = (25 + 273.15) (9 / 5) = 536.67 degree Rankine. THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] = 230 [ 536.67 / (8130 x 0.286) ] [ 1 - (14.7 / 90) ^ 0.286 ] = 21.469 horsepowers which is more than 20 horsepowers for expander to be used. (b) To = 536.67 (14.7 / 90) ^ 0.286 = 319.629 degree Rankine. 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.

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Question { 1641 }

Question 107 - In N + 1 Rule in Quantum Chemistry, whenever a spin 1 / 2 nucleus is adjacent to N other nuclei, it is split into N + 1 distinct peaks. In 1 peak or singlet, there is only 1 magnitude. In 2 peaks or doublet, the ratio of magnitude of each peak is 1 : 1. In 3 peaks or triplet, the ratio of magnitude of each peak is 1 : 2 : 1. In 4 peaks or quartet, the ratio of magnitude of each peak is 1 : 3 : 3 : 1. In 5 peaks or quintet, the ratio of magnitude of each peak is 1 : 4 : 6 : 4 : 1. (a) By using binomial coefficients or Triangle of Pascal find the ratio of magnitude of each peak if 6 peaks exists. (b) How many adjacent nuclei are available in a spin 1 / 2 nucleus in such situation of 6 peaks?


Answer

Answer 107 - (a) 1 : 5 : 10 : 10 : 5 : 1, since (0 + 1) : (1 + 4) : (4 + 6) : (6 + 4) : (4 + 1) : (1 + 0) with reference to the ratio of magnitude of each peak in quintet. (b) Number of adjacent nuclei = Number of peaks - 1 = 6 - 1 = 5. 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.

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Question { 1710 }

PROCESS DESIGN - EXAMPLE 21.2 : The names of the flow streams could be represented by : H1 for first hot stream, H2 for second hot stream, C1 for first cold stream, C2 for second cold stream. Data of supply temperature Ts in degree Celsius : 150 for H1, 170 for H2, 30 for C1, 30 for C2. Data of target temperature Tt in degree Celsius : 50 for H1, 169 for H2, 150 for C1, 40 for C2. Data of heat capacity Cp in kW / degree Celsius : 3 for H1, 360 for H2, 3 for C1, 30 for C2. (a) Find the enthalpy changes, dH for all streams of flow H1, H2, C1 and C2 in the unit of kW. Take note of the formula dH = (Cp) (Tt - Ts). (b) Match the hot streams H1 and H2 with the suitable cold streams C1 and C2 to achieve the maximum energy efficiency.


Answer

PROCESS DESIGN - ANSWER 21.2 : (a) dH for streams H1 : 3 (50 - 150) = -300 kW. H2 : 360 (169 - 170) = -360 kW. C1 : 3 (150 - 30) = 360 kW. C2 : 30 (40 - 30) = 300 kW. (b) dH (H1 + C2) = 0 kW; dH (H2 + C1) = 0 kW where all heats from hot utilities are supplied to cold utilities. Then H1 is matched to C2 and H2 is matched to C1. 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.

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Question { 1781 }

PROCESS DESIGN - EXAMPLE 21.3 : According to a heuristic of chemical engineering plant design, assume a pressure difference dP = 4 psi for each 10-ft rise in elevation. A pump is needed to pump liquid from a storage tank at a lower elevation to a heating tank at a higher elevation with the elevation difference of 60 ft. (a) Find the pressure loss due to such elevation. (b) If the required minimum inlet pressure to the heating tank is 9 psi, with 1 control valve is installed between pump and heating tank, what is the dP minimum required for the control valve and the entrance to the heating tank when the heuristic mentions that at least 10 psi is required for the control valve? (c) The pressure at the inlet of the pump is 8 psi and the flowrate of the liquid produces pressure head of 50 psi. What is the total pressure produced by the pump? (d) Assume a pipeline dP of 2 psi / 100 ft for liquid flow in a pipe according to heuristic, what is the approximate maximum length of the pipe in ft that can be installed between the pump and the heating tank?


Answer

PROCESS DESIGN - ANSWER 21.3 : (a) dP = (4 psi / 10 ft) x 60 ft = 24 psi. (b) dP = 9 + 10 = 19 psi minimum. (c) P of pump = 8 + 50 = 58 psi. (d) dP for pipe = Answer of (c) - Answer of (a) - Answer of (b) = 58 - 24 - 19 = 15 psi. Maximum length of pipe = (100 ft / 2 psi) x 15 psi = 750 feet. 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.

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Question { 1609 }

ENVIRONMENTAL ENGINEERING - QUESTION 22.1 : In order to predict the wastewater production, the population number has to be understood. The population data is : 72000 (for year 1961 or P-1961), 85000 (for year 1971 or P-1971), 110500 (for year 1981 or P-1981). (a) Find the average population increase, or [ (P-1981 - P-1971) + (P-1971 - P-1961) ] / 2. (b) Find the average percentage population increase, or [ (P-1981 - P-1971) / P-1971 + (P-1971 - P-1961) / P-1961 ] / (2) X 100. (c) Find the incremental increase or P-1981 - 2 (P-1971) + P-1961. (d) Let Po = P-1981. After 2 decades or n = 2, the population is P-2001. By using arithmetical increase method, find P-2001 = Po + n (Answer for a). (e) By using incremental increase method, find P-2001 = (Answer of d) + n (n + 1) (Answer of c) / 2. (f) By using geometrical increase method, find P-2001 = Po [ 1 + (Answer of b) / 100 ] ^ n where ^ is power sign, or 1 ^ 2 = 1 x 1 = 1. (g) If the actual P-2001 = 184000, which method of estimation is more accurate, based on your answer in (d), (e) and (f)?


Answer

ENVIRONMENTAL ENGINEERING - ANSWER 22.1 : (a) Average population increase = [ (110500 - 85000) + (85000 - 72000) / 2 = 19250. (b) Average percentage population increase = [ (85000 - 72000) / 72000 + (110500 - 85000) / 85000 ] x 100 / 2 = 24.025. (c) Incremental increase = 110500 - 2 (85000) + 72000 = 12500. (d) P-2001 = 2(19250) + 110500 = 149000. (e) P-2001 = 149000 + 2 (3) (12500) / 2 = 186500. (f) P-2001 = 110500 (1 + 0.24025) ^ 2 = 169973. (g) Errors of answers : (d) 184000 - 149000 = 35000, (e) 184000 - 186500 = -2500, (f) 184000 - 169973 = 14027. Answer (e) or incremental increase is more accurate to such data, or the most accurate prediction among the three available formulae. 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.

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Question { 2299 }

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).


Answer

ENVIRONMENTAL ENGINEERING - ANSWER 22.2 : (a) BOD = (8.8 - 5.9) (300 / 30) = 29 mg / L. (b) BOD = (8.8 - 4.2) (600 / 100) = 27.6 mg / L. (c) BOD = (29 + 27.6) / 2 = 28.3 mg / L. (d) NBOD = 45 - 28.3 = 16.7 mg / L. 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.

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Question { 1468 }

ENVIRONMENTAL ENGINEERING - QUESTION 22.3 : A well delivers 225 US-gallons per minute of water to a chemical plant during normal system operation. (a) Calculate its flowrate in the unit of mega US-gallon per day or MGD. (b) The following formula is written next to the chlorine feed point : (chlorine feed rate, lb / day) = (flowrate, MGD) X (dose, mg / L) x (8.34). If this formula is correct, then what should the chlorine feed rate to be in pounds per day (lb / day) if the desired dose is 2 mg / L. (c) Prove by calculations that the constant 8.34 in the formula next to the chlorine feed point is correct. Let 1 US-gallon = 3.78541 L and 1 mg = 0.0000022046 pound.


Answer

ENVIRONMENTAL ENGINEERING - ANSWER 22.3 : (a) Flowrate = (225 US-gallon / minute) / [ (hour / 60 minute) x (day / 24 hour) ] = 324000 or 0.324 MGD. (b) Chlorine feed rate = 0.324 MGD x 2 mg / L x 8.34 = 5.40432 lb / day. (c) The constant is derived from 0.0000022046 pound / mg x 1000000 US-gallon / M US-gallon x 3.78541 L / US-gallon = 8.345 with minor error (proven). 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.

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Question { 1680 }

FOOD ENGINEERING - QUESTION 23.1 : (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 calorie, 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?


Answer

FOOD ENGINEERING - ANSWER 23.1 : (a) Energy gained = 150 g (77 kcal / 100 g) for potatoes + 200 g (18 kcal / 111 g) for tomatoes = 147.93 kcal. (b) 1 food Calorie = 1 Calorie = 1000 calories, then 9600 calories is 9600 calories (1 Calorie / 1000 calories) = 9.6 Calories. (c) Calories per serving size in actual experiment = 34 g (9.6 Calories / 1.75 g) = 186.51 Calories. Then difference of Calories = 210 - 186.51 = 23.49. (d) Temperature difference T in degree Celsius is same as Kelvin. Then c = 1 = 9600 calories / [ (150 g) T ] or T = 9600 / 150 = 64 degree Celsius. 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.

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Question { 1875 }

FOOD ENGINEERING - QUESTION 23.2 : (a) A dryer reduces the moisture content of 100 kg of a potato product from 80 % to 10 % moisture. Find the mass of the water removed in such drying process. (b) During the drying process, the air is cooled from 80 °C to 71 °C in passing through the dryer. If the latent heat of vaporization corresponding to a saturation temperature of 71 °C is 2331 kJ / kg for water, find the heat energy required to evaporate the water only. (c) Assume potato enters at 24 °C, which is also the ambient air temperature, and leaves at the same temperature as the exit air. The specific heat of potato is 3.43 kJ / (kg °C). Find the minimum heat energy required to raise the temperature of the potatoes. (d) 250 kg of steam at 70 kPa gauge is used to heat 49,800 cubic metre of air to 80 °C, and the air is cooled to 71 °C in passing through the dryer. If the latent heat of steam at 70 kPa gauge is 2283 kJ / kg, find the heat energy in steam. (e) Calculate the efficiency of the dryer based heat input and output, in drying air. Use the formula (Ti - To) / (Ti - Ta) where Ti is the inlet (high) air temperature into the dryer, To is the outlet air temperature from the dryer, and Ta is the ambient air temperature.


Answer

FOOD ENGINEERING - ANSWER 23.2 : (a) The mass of dried material is constant throughout the drying process, or 100 kg x 0.2 = 20 kg. Mass of water = mass of dried material (% of water / % of dried material). Initial mass of water = 20 (80 / 20) = 80 kg. Final mass of water = 20 (10 / 90) = 2.222 kg. Mass of the water removed = Final mass - Initial mass = 80 - 2.222 = 77.778 kg. (b) Heat energy required to evaporate the water only = Mass of the water removed x Latent heat of vaporization = 77.778 kg x 2331 kJ / kg = 181300.518 kJ. (c) Minimum heat energy required to raise the temperature of the potatoes = Mass of potatoes of 80 % moisture x specific heat of potato x temperature changes = 100 kg x 3.43 kJ / (kg °C) x (71 - 24) °C = 16121 kJ. (d) Heat energy in steam = mass of steam x latent heat of steam = 250 kg x 2283 kJ / kg = 570750 kJ. (e) Efficiency of the dryer based heat input and output = (Ti - To) / (Ti - Ta) = (80 - 71) / (80 - 24) = 0.1607. 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.

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Question { 1375 }

Question 108 - (a) The correct statement about both the average value of position () and momentum (

) of a 1-dimensional harmonic oscillator wavefunction is =

= 1 - x. Find the value of x. (b) The probabilities of finding a particle around points A, B and C in the wavefunction y = f(x) are P(A), P(B) and P(C) respectively. Coordinates are A (3,5), B (4,-10) and C (6,7). Arrange P(A), P(B) and P(C) in term of a < b < c, when | y-coordinate | signifies the probability.


Answer

Answer 108 - (a) Let =

= 1 - x = 0, then x = 1 - 0 = 1. (b) With reference to the | y-coordinate | for each point in A, B and C coordinates, if 5 < 7 < |-10| then a < b < c = P(A) < P(C) < P(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.


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Question { 1494 }

FOOD ENGINEERING - QUESTION 23.3 : (a) In the measurement of the browning (optical density) of fruit juice at 10 day interval, the following pairs of data are obtained with time t in days and browning or optical density (OD) : t = 0, OD = 0.05; t = 10, OD = 0.071; t = 20, OD = 0.089; t = 30, OD = 0.11; t = 40, OD = 0.128; t = 50, OD = 0.149; t = 60, OD = 0.17. (i) By using Excel or other programs, determine if such browning reaction can be characterised by pseudo zero order kinetics, with strong correlations between the data pairs of t and OD. (ii) When OD = 0.24, the shelf life T of such juice is expired. Calculate the value of T. (b) Let food cost percentage = ( food cost / food sales ) x 100. If total food cost of bread and butter are $25, food cost percentage of bread is 50 % and for butter is 50 %, find the total food sales of bread and butter.


Answer

FOOD ENGINEERING - ANSWER 23.3 : (a) (i) Excel program is used for quick graph plotting where x-axis is t and y-axis is OD. A linear graph with equation y = 0.002 x + 0.0501 with r squared value of 0.9996 could be obtained. High r squared value approaching 1 signifies strong correlations between the data pairs of t and OD for pseudo zero order kinetics. (ii) By using the equation y = 0.002 x + 0.0501, substituting y for OD and x for T, then T = (OD - 0.0501) / 0.002 = (0.24 - 0.0501) / 0.002 = 94.95 days. (b) Total food cost = c + d = $25 = 0.5 (C + D) and total food sales = C + D = (c + d) / 0.5 = 25 / 0.5 = $50 where c = food cost of bread, d = food cost of butter, C = food sales of bread, D = food sales of butter. 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.

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Question { 1719 }

POLYMER ENGINEERING - QUESTION 24.1 : The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18. The fraction of polymer chain X of 3 different monomers a, b and c in a polymer are Xa = 0.5, Xb = 0.3 and Xc = 0.2. (i) Calculate number average molecular weight by using the formula Ma Xa + Mb Xb + Mc Xc. (ii) Calculate weight average molecular weight by using the formula (Ma Xa Ma + Mb Xb Mb + Mc Xc Mc) / (Ma Xa + Mb Xb + Mc Xc). (iii) Calculate the polydispersity by using the answer in (ii) divided by answer in (i). (iv) If the molecular weight of repeat unit is 12, calculate the degree of polymerization by using the formula (Ma Xa + Mb Xb + Mc Xc) / (molecular weight of repeat unit).


Answer

POLYMER ENGINEERING - ANSWER 24.1 : (i) Number average molecular weight = Ma Xa + Mb Xb + Mc Xc = 14 x 0.5 + 16 x 0.3 + 18 x 0.2 = 15.4 kg / mol. (ii) Weight average molecular weight = (Ma Xa Ma + Mb Xb Mb + Mc Xc Mc) / (Ma Xa + Mb Xb + Mc Xc) = (14 x 0.5 x 14 + 16 x 0.3 x 16 + 18 x 0.2 x 18) / 15.4 = 15.558 kg / mol. (iii) Polydispersity = [ answer in (ii) ] / [ answer in (i) ] = 15.558 / 15.4 = 1.0103. (iv) Degree of polymerization = (Ma Xa + Mb Xb + Mc Xc) / (molecular weight of repeat unit) = (15.4 kg / mol) / (12 g / mol) = 1.283 k = 1283. 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.

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Question { 1431 }

POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs = density of sample, Ra = density of amorphous form and Rc = density of crystalline form. In a polymer, these unknowns could be related by the equation C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra). (a) Find the equation of Rc as a function of C%, Rs and Ra. (b) Two samples of a polymer, C and D exist. For sample C, C% = 0.513 when Rs = 2.215 unit. For sample D, C% = 0.742 when Rs = 2.144 unit. Both samples C and D have the same values of Ra and Rc. Find the values of Ra and Rc in 6 decimal places.


Answer

POLYMER ENGINEERING - ANSWER 24.2 : (a) C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra), C% (Rc - Ra) Rs = Rc Rs - Rc Ra = Rc C% Rs - Ra C% Rs, Rc C% Rs - Rc Rs + Rc Ra = Ra C% Rs = Rc (C% Rs - Rs + Ra), Rc = Ra C% Rs / (C% Rs - Rs + Ra). (b) For sample C, Rc = Ra C% Rs / (C% Rs - Rs + Ra) = Ra (0.513) (2.215) / [ (0.513) (2.215) - 2.215 + Ra ] = 1.136295 Ra / (Ra - 1.078705). For sample D, Rc = Ra C% Rs / (C% Rs - Rs + Ra) = Ra (0.742) (2.144) / [ (0.742) (2.144) - 2.144 + Ra ] = 1.590848 Ra / (Ra - 0.553152). Then Rc = 1.136295 / (Ra - 1.078705) = 1.590848 / (Ra - 0.553152), 1.136295 (Ra - 0.553152) = 1.590848 (Ra - 1.078705), 1.136295 Ra - 0.628544 = 1.590848 Ra - 1.716056, Ra (1.590848 - 1.136295) = 1.716056 - 0.628544 = 1.087512 = 0.454553 Ra. Then Ra = 1.087512 / 0.454553 = 2.392487. When Ra = 2.392487, Rc = 1.136295 / (Ra - 1.078705) = 1.590848 / (Ra - 0.553152) = 0.864904 with 6 decimal places. 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.

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