Question { 1759 }

Question 83 - The United States of America Energy Information Administration reports the following emissions in million metric tons of carbon dioxide in the world for year 2012 : Natural gas : 6799, petroleum : 11695, coal : 13787. Coal-fired electric power generation emits around 2000 pounds of carbon dioxide for every megawatt hour generated, which is almost double the carbon dioxide released by a natural gas-fired electric plant per megawatt hour generated. If 1 metric ton = 1000 kg and 1 pound = 0.4536 kg, estimate the total energy generated by natural gas in the world for year 2012, in gigawatt hour.

Answer

Answer 83 - Mass of carbon dioxide emitted by natural gas-fired electric power generation for every megawatt hour generated = 2000 / 2 = 1000 pounds = 1000 pounds x (0.4536 kg / pound) x (metric ton / 1000 kg) = 0.4536 metric tons. Total energy generated by natural gas in the world for year 2012 = 6799 million metric tons x (megawatt hour / 0.4536 metric tons) = 14988.977 million megawatt hour = 14988977 gigawatt hour, approximately. 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 { 1667 }

MASS TRANSFER - EXAMPLE 4.3 : According to Adolf Eugen Fick (1829 - 1901) : rate of diffusion v increases with less wall thickness t, increased area A and decreased molecular weight of a fluid M. The diffusion constant D decreased with increasing M. (a) By assuming v, t, dP, A, M and D changes proportionally of each other, find the equation of v as a function of t, dP, A and D. (b) The ratio of self diffusion constant D, at T = 273 K and P = 0.1 MPa, for gases B and C are 1.604 : 0.155. If only 2 gases exist in such a system : hydrogen and nitrogen, find the type of gas for B and C with reference to their molecular weights M. (c) By using the equation of kinetic energy 0.5 MV = constant where V = square of v, find the ratio of V for B and V for C, or V(B) / V(C), as a function of M(B) and M(C), where M(B) is molecular weight of B and M(C) the molecular weight of C : Graham's Law of Diffusion.

Answer

MASS TRANSFER - ANSWER 4.3 : (a) v = (dP) AD / t. (b) Hydrogen has least M among all gases - general knowledge - highest D. Then B = hydrogen and C = nitrogen. (c) Let 0.5 M(B) V(B) = 0.5 M(C) V(C), then V(B) / V(C) = M(C) / M(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 { 1831 }

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).

Answer

HEAT TRANSFER - ANSWER 5.1 : Q = q(1) + q(2) + q(3) = [ UA (TI - TB) ] + [ UA (TB - TC) ] + [ UA (TC - TF) ] = UATI + (UATB - UATB) + (UATC - UATC) - UATF = UATI - UATF = UA (TI - TF) (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.

Question { 1517 }

Heat transfer: In a steady state one dimensional conduction with no heat generation, the differential equation is d / dx (k dT / dx) = 0. Prove that T(x) = ax b, where k, a and b are constants. (b) At x = 0, T = c and at x = L, T = d. Prove that T(x) = (d - c) x / L c for boundary conditions.

Answer

HEAT TRANSFER - ANSWER 5.2 : When d / dx (k dT / dx) = 0, d (dT) / [ (dx) (dx) ] = 0. Integrate both sides gives dT / dx = a. Second integration gives T(x) = ax + b for both sides (proven). (b) T(0) = a(0) + b = b = c. T(L) = d = aL + c then a = (d - c) / L. Substitute in T(x) = ax + b gives T(x) = (d - c) x / L + c (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.

Question { 1638 }

Question 84 - In Mendelian genetics, yellow (Y) is dominant to green (y) and round (R) is dominant to wrinkled (r). (a) What is the probability P of Rr x Rr producing wrinkled seeds? (b) What is the probability P of Yy x yy producing green seeds? (c) What is the probability that RRYy x RrYy would produce RrYy?

Answer

(a) There are 3 types of seeds produced : RR, Rr, rR, rr where RR, Rr, rR (3) are dominantly round and rr (1) is recessively wrinkled. P (wrinkled) = 1 / (3 + 1) = 1 / 4. (b) There are 2 types of seeds produced : Yy, Yy, yy, yy where Yy (2) are dominantly yellow and yy (2) are recessively green. P (green) = 2 / (2 + 2) = 1 / 2. (c) Let RR x Rr produces RR, Rr, RR, Rr, then P (Rr) = 2 / 4 = 1 / 2. Let Yy x Yy produces YY, Yy, yY, yy, then P (Yy) = 2 / 4 = 1 / 2. P (RrYy) = P (Rr) x P (Yy) = 1 / 2 x 1 / 2 = 1 / 4. 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 { 1723 }

Question 85 - (a) Three genes, I, J and K are available. All these genes are linked with respect to one another. If the percent recombination between I and J is 8 %, that between J and K is 10 %, and that between K and I is 18 %, what is the order of the gene? (b) Twenty six genes, a, b, c, d, e, f, ... x, y and z are available. All these genes are linked with respect to one another. If the percent recombination between a and b is 3 %, between b and c is 3 %, between c and d is 3 %, ... between w and x is 3 %, between x and y is 3 %, between y and z is 3 %, then what is the percentage recombination between b and y?

Answer

Answer 85 - (a) From I as origin, J - I = 8 %, K - J = 10 %, K - I = 18 %, then I ... 8 % ... J ... 10 % ... K. (b) There are 22 genes between b and y, separated by 3 % recombination between 2 adjacent genes. Percentage recombination between b and y = (22 + 1) x 3 % = 69 %. 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 { 1764 }

Question 86 - (a) Male with genotype GGmm and phenotype gray wingless mates with female with genotype ggMM and phenotype black winged in fruit flies. G is dominant to g in color. M is dominant to m in wing shape. If the actual distribution of the second generation of the fruit flies was as follow : 890 gray wingless, 900 black winged, 115 gray winged, 95 black wingless, calculate the recombination frequency betwen the two genes and distance in recombination units. Let 1 map unit = 1 % recombination. (b) A DNA molecule has 180 base pairs and 20 % adenine. How many cytosine nucleotides are present in this molecule of DNA?

Answer

Answer 86 - (a) Phenotypes of parents = gray wingless, black winged. Phenotypes of recombinant children = gray winged, black wingless. Number of recombinant children = gray winged + black wingless = 115 + 95 = 210. Recombination frequency between the two genes = Number of recombinant children / Total number of childen = 210 / (210 + 890 + 900) = 0.105 or 10.5 % = distance = 10.5 map units. (b) Base pairs number of adenine = 180 x 0.2 = 36 = base pairs number thymine. Base pairs number of guanine = base pairs number cytosine = (Total number of base pairs - 2 x Base pairs number of adenine) / 2 = (180 - 2 x 36) / 2 = 54. 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 { 1715 }

PROCESS CONTROL - EXAMPLE 6.1 : In a Laplace Transform Table, the Laplace transfer function of f(t) is F(s). When d(t) = f(t) then 1 = F(s). When x(t) = f(t) then X(s) = F(s). If d(t) is the impulse of a spring when d(t) = kx(t), then derive the equation for the impulse of a spring as X(s) in term of k. Next question : A controller has a transfer function a and the other controller has a transfer function b. The overall transfer function of both controllers is ab. What is the transfer function overall when both controllers have similar transfer function 1 / (Cs + k)?

Answer

PROCESS CONTROL - ANSWER 6.1 : For f(t), d(t) = kx(t). Transform such equation to F(t) gives 1 = kX(s) then X(s) = 1 / k. Next answer : Overall transfer function for 2 controllers in combination is 1 / (Cs + k) x 1 / (Cs + k) = 1 / [ (Cs + k) (Cs + k) ]. 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 { 1742 }

Question 87 - According to Hardy-Weinberg Equation, p x p + 2 x p x q + q x q = 1 where p = dominant allele frequency and q = recessive allele frequency. Let p + q = 1. Fraction of population has 2 copies of the p gene = p x p. Fraction of population has 2 copies of the q gene = q x q. Fraction of population has a copy of p gene and a copy of q gene = 2 x p x q. In a small town, the allele frequency is q = 0.2 for a recessive gene, the delta-32 mutation, that gives human protection from HIV infection. (a) Find the allele frequency a dominant gene, p. (b) What percent of the population has at least a copy of the gene that cause the population either immune to HIV or less susceptible to the disease?

Answer

Answer 87 - (a) Let p + q = 1, then p = 1 - q = 1 - 0.2 = 0.8. (b) Percentage of population has 2 copies of the p gene = p x p x 100 = 0.8 x 0.8 x 100 = 64 %. Percentage of population has at least a copy of the q gene = (1 - Fraction of population has 2 copies of the p gene) x 100 = (1 - p x p) x 100 = (1 - 0.8 x 0.8) x 100 = (1 - 0.64) x 100 = 36 %. 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 { 2401 }

Question 88 - In the calculation of the growth of bacteria, colony forming unit (CFU) in serial dilution is used. In a laboratory, viable count assay is used to estimate CFU. Formula applied is CFU / mL = (number of colonies x dilution) / (amount plated, in unit mL). Acceptable plate count is either between 20 and 200 or between 30 and 300 according to 2 different references. A wastewater sample of 200 ml is added to and mixed with 1.8 L of sterile water. Another 200 ml of the mixture is added to and mixed with 1.8 L of sterile water. (a) Calculate the dilution of first mixture and the dilution of the second mixture. (b) 100 microlitres of wastewater samples from the first mixture and the second mixture are placed separately on 2 different alga plates. The first plate has 250 colonies and the second plate has 23 colonies. Calculate the average CFU / mL.

Answer

Answer 88 - (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.

Question { 1552 }

PROCESS CONTROL - EXAMPLE 6.2 : A stream with volumetric flow rate Q enters a cylindrical tank and a stream with volumetric flow rate q exits the tank. The fluid has a constant heat capacity and density. There is no temperature change or chemical reaction occurring in the tank. Develop a model for determining the height of the tank, h. Let V is the volume, A is the cross sectional area, r is the density, m is the mass, where V and A are for the tank, r and m are for the fluid. The rate of mass of fluid accumulation, dm / dt = (Q - q) r. Prove the model to be dh / dt = (Q - q) / A.

Answer

PROCESS CONTROL - ANSWER 6.2 : Mass of fluid in tank, m = Vr = hAr. Then d (hAr) / dt = (Q - q) r. Ar (dh / dt) = (Q - q) r leads to the answer A (dh / dt) = Q - q. Finally dh / dt = (Q - q) / A. 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 { 1705 }

PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1 with y(0) = 1. (a) The Laplace transformation, L for given terms are : L (dy / dt) = sY(s) - y(0), L(y) = Y(s), L(1) = 1 / s. Use such transformation to find Y(s). (b) The initial value theorem states that : When t approaches 0 for a function of y(t), it is equal to a function of sY(s) when s approaches infinity. Use the initial value theorem as a check to the answer found in part (a).

Answer

PROCESS CONTROL - ANSWER 6.3 : (a) For the equation 3 dy / dt + 2y = 1, its Laplace transformation is 3 [ sY(s) - y(0) ] + 2Y(s) = 1 / s, 3 (sY - 1) + 2Y = 1 / s, 3sY - 3 + 2Y = 1 / s, 3sY + 2Y = 1 / s + 3, Y (3s + 2) = (1 + 3s) / s. Y = Y(s) = (1 + 3s) / [ s (3s + 2) ]. (b) Initial value theorem states that y(0) = sY(infinity), then 1 = s (1 + 3s) / [ s (3s + 2) ] = (1 + 3s) / (3s + 2) approaches 1 when the value of s approaches infinity. The checking is correct. 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 { 3677 }

Question 89 - A hemocytometer is a device that is used for counting cells. In an engineering experiment, 100 microlitres of cell suspension is diluted with 50 microlitres of Trypan blue dye. Only death cells appear blue in color when stained with the dye. There are 57 cells detected in a hemocytometer, where 5.3 % of them appear blue when the chamber of the meter is placed under a microscope. Each square of a chamber can contain 0.0001 mL of liquid. (a) Calculate the number of viable cells. (b) The cells ocupied 5 squares. Calculate the average number of viable cells / square. (c) Calculate the dilution factor of the cell suspension by using the formula : Dilution = final volume / initial volume. (d) Calculate the concentration of viable cells / mL by using the formula : Concentration = (Average number of viable cells / square) x dilution x (square / volume).

Answer

Answer 89 - (a) Number of death cells = Total cells x % of death cells / 100 = 57 x 5.3 / 100 = 3.021 rounded to 3 cells. Number of viable cells = Total cells - number of death cells = 57 - 3 = 54 cells. (b) Average number of viable cells / square = 54 / 5 = 10.8 cells / square. (c) Dilution = final volume / initial volume = (100 + 50) microlitres / 100 microlitres = 1.5. (d) Concentration = (Average number of viable cells / square) x dilution x (square / volume) = 10.8 cells x 1.5 x (1 / 0.0001 mL) = 162000 cells / mL. 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 { 1590 }

ENGINEERING ECONOMY - EXAMPLE 7.1 : In engineering economy, the future value of first year is FV = PV (1 + i). For second year it is FV = PV (1 + i) (1 + i). For third year it is FV = PV (1 + i) (1 + i)(1 + i) where FV = future value, PV = present value, i = interest rate per period, n = the number of compounding periods. By induction, what is the future value of $1000 for 5 years at the interest rate of 6 %?

Answer

ENGINEERING ECONOMY - ANSWER 7.1 : By induction, the future value for 5 year = FV = PV (1 + i) (1 + i) (1 + i) (1 + i)(1 + i) = $1000 (1.06) (1.06) (1.06) (1.06)(1.06) = $1338.2256. 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 { 1558 }

Question 90 - In the calculation of the growth of bacteria, absorbance, A in spectrophotometry is used. According to Beer-Lambert Law, A = e x l x c where A is the absorbance of the solution (no unit), l is the distance of light travels through the solution (in cm), e is the molar absorptivity or the molar extinction coefficient [ in L / (mol.cm) ]. For a particular solute and fixed path length : As / Ao = Cs / Co where Ao is the observed signal for a known concentration Co, and As is the observed signal for a sample concentration Cs. (a) For a cell concentration of 560 cells / mL, a spectrophotometre gives an absorbance reading of 1.0. A mixture of concentration 3600000 cells / mL can be diluted in several operations, with each operation having a dilution of 1:20. How many dilutions should be made so that the concentration of this mixture can be calculated within a range of A = 0.0 to 1.0. (b) In another experiment, a sample tube of 1 cm in width is used. Let A = 0.06 and e = 0.0012 ml / (cell.cm). Find the cell concentration of the sample.

Answer

Answer 90 - (a) Let Ao = 1.0, Co = 560 cells / mL, Cs = 3600000 cells / mL. By using the equation As / Ao = Cs / Co, then As = Ao x Cs / Co = 1.0 x 3600000 / 560 = 6428.57. For a dilution of 1:20, first stage dilution gives A = 6428.57 / 20 = 321.43, second stage dilution gives A = 321.43 / 20 = 16.07, third stage dilution gives A = 16.07 / 20 = 0.8, which is within a range of A = 0.0 to 1.0. Total stages of dilution needed = 3. (b) By using the equation A = e x l x c, then c = A / (e x l) = 0.06 / { [ 0.0012 ml / (cell.cm) ] x 1 cm } = 50 cells / mL. 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.