Question { 1666 }

HEAT TRANSFER - EXAMPLE 5.3 : 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.

Answer

HEAT TRANSFER - ANSWER 5.3 : Let r (dT / dr) = e, then dT / dr = e / r. T(r) = e ln r + f by integrating both sides where e and f are constants. (b) Let c = e ln a + f and d = e ln b + f. Subtracting both equations gives c - d = e (ln a - ln b) = e ln (a / b). Then e = (c - d) / [ ln (a / b) ] and f = d - e ln b. T(r) = e ln r + d - e ln b = e ln (r / b) + d = (c - d) [ ln (a / b) ] ln (r / b) + d. 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 { 1617 }

Question 91 - In the application of Theory of Spectrometry in spectrophotometer, let n = N x C x V, V = A x t, e = a x N where n = number of molecules, N = Avogadro's number, V = volume of cuvette, A = area of cuvette, t = thickness of cuvette, C = concentration of fluid in the cuvette, e = extinction coefficient, a = effective area of molecule. (a) By using calculus in dI = -I x a x N x C x dt, prove that ln (I / Io) = -a x N x C x t, where dI is the small difference in I and dt is the small difference in t. I = intensity of light. Io = initial intensity of light. (b) Show by calculations that ln (Io / I) = e x C x t based on the answer in the previous question (a). (c) Find the equation of log (Io / I) as a function of e, C and t based on the answer in the previous question (b).

Answer

Answer 91 - (a) Let dI = -I x a x N x C x dt, then dI / I = -a x N x C x dt. Integrate both sides of the equation gives ln (I / Io) = -a x N x C x t. (b) When ln (I / Io) = -a x N x C x t and e = a x N, then ln (I / Io) = -e x C x t. If ln (I / Io) = - ln (Io / I), then ln (Io / I) = e x C x t. (c) Let ln (Io / I) = log (Io / I) / log E, where E = 2.718 approximately. Then log (Io / I) = log E x ln (Io / I) = log E x e x C x t. 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 { 1983 }

ENGINEERING ECONOMY - EXAMPLE 7.2 : In the purchase of a machine with a period n = 8.5 years, the minimum attractive rate of return, i = 12 %, the cost P = $55000, F = $4000 is the salvage, annual maintenance A = $3500. The return of the investment or equivalent uniform annual benefit is $15000. The equivalent uniform annual cost is P (A / P, i, n) + A - F (A / F, i, n). The investment is considered acceptable only when equivalent uniform annual benefit is greater than the equivalent uniform annual cost. From the compound interest table, (A / P, i = 12 %, n = 8 years) = 0.2013, (A / P, i = 12 %, n = 9 years) = 0.1877, (A / F, i = 12 %, n = 8 years) = 0.0813, (A / F, i = 12 %, n = 9 years) = 0.0677. Prove by calculations whether the investment above is acceptable.

Answer

ENGINEERING ECONOMY - ANSWER 7.2 : (A / P, i = 12 %, n = 8.5 years) = (0.2013 + 0.1877) / 2 = 0.1945. (A / F, i = 12 %, n = 8.5 years) = (0.0813 + 0.0677) / 2 = 0.0745. Equivalent uniform annual cost = P (A / P, i = 12 %, n = 8.5 years) + A - F (A / F, i = 12 %, n = 8.5 years) = $55000 (0.1945) + $3500 - $4000 (0.0745) = $13899.5 < $15000 (equivalent uniform annual benefit). The investment is proven acceptable. 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 { 1774 }

ENGINEERING ECONOMY - EXAMPLE 7.3 : There are 2 alternatives of investment. Choice 1 : A trader offers you an investment opportunity where your investment of A$15000 presently will be A$18000 after 4 years. Choice 2 : A bank offers you 5 % annual return for your initial investment of A$15000. Question a : What is the equivalent bank payment after 4 years? Question b : By using the concept of equivalence in engineering economy, which is the better choice, between 1 and 2, that will be more profitable after 4 years?

Answer

ENGINEERING ECONOMY - ANSWER 7.3 : (a) For Choice 2, A$15000 x 1.05 = A$15750 after 1 year, A$15750 x 1.05 = A$16537.50 after 2 years, A$16537.50 x 1.05 = A$17364.375 after 3 years, A$17364.375 x 1.05 = A$18232.6 after 4 years. (b) The gain of Choice 2 > Choice 1 with A$18232.6 > A$18000, then Choice 2 is preferred. 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 { 1669 }

Question 92 - (a) In order to determine the slurry volume (L) in chromatography, arrange the formula using the following parameters : slurry concentration (%), slurry volume (L), settled chromatography medium volume (L), 100. (b) In order to determine the chromatography medium volume (L) in gel filtration chromatography, arrange the formula using the following parameters : sample volume to be processed (L), chromatography medium volume (L), sample volume in percent of total column volume (%), 100.

Answer

Answer 92 - (a) By fraction of concentration, [ settled chromatography medium volume (L) ] / [ slurry volume (L) ] = slurry concentration (%) / 100. Then [ slurry volume (L) ] = [ settled chromatography medium volume (L) ] / [ slurry concentration (%) / 100 ]. (b) By fraction of volume, [ sample volume to be processed (L) ] / [ chromatography medium volume (L) ] = sample volume in percent total volume (%) / 100. Then chromatography medium volume (L) = [ sample volume to be processed (L) ] / [ sample volume in percent of total column volume (%) / 100 ]. 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 { 2152 }

Question 93 - In thin layer chromatography (TLC), let retention factor Rf = (distance traveled by solute) / (distance traveled by solvent). Silica gel is used as stationary phase which is more polar than the hexane solvent as mobile phase. If the Rf values of 3 solutes A, B and C are 1.5, 0.75 and 1 respectively, compare the polarities of A, B and C with reasons.

Answer

Polarities : A < C < B. A is least polar since it travels farthest with the mobile phase, whereas B is most polar as it is attracted to the stationary phase and travels least. 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 { 2429 }

ENGINEERING MATHEMATICS - EXAMPLE 8.1 : A local utility burns coal having the following composition on a dry basis : Carbon (C) 83.05 %, hydrogen (H) 4.45 %, oxygen (O) 3.36 %, nitrogen (N) 1.08 %, sulfur (S) 0.7 % and ash 7.36 %. Calculate the ash free composition of the coal with reference to C, H, O, N and S.

Answer

ENGINEERING MATHEMATICS - ANSWER 8.1 : Total percentage without ash is 83.05 + 4.45 + 3.36 + 1.08 + 0.7 = 92.64 %. For ash free composition percentage calculations, C = 83.05 / 92.64 x 100 % = 89.65 %. H = 4.45 / 92.64 x 100 % = 4.80 %. O = 3.36 / 92.64 x 100 % = 3.63 %. N = 1.08 / 92.64 x 100 % = 1.17 %. S = 0.7 / 92.64 x 100 % = 0.76 %. 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 { 1771 }

Question 94 - The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.

Answer

Answer 94 - Reference formula Rs = (difference in retention time) / (average width at the base) is used. (a) First equation : Rs = (Tb - Ta) / [ (Wa + Wb) / 2 ] = 2 (Tb - Ta) / (Wa + Wb). Second equation : Rs = (Tc - Ta) / [ (Wa + Wc) / 2 ] = 2 (Tc - Ta) / (Wa + Wc). (b) Substitute Ta = Tc - Tb, Wa = 20 s, Wb = 40 s and Wc = 30 s into first equation and second equation. First equation : Rs = 2 = 2 [ Tb - (Tc - Tb) ] / (20 + 40) = (2 Tb - Tc) / 30, 2 Tb - Tc = 60. Second equation : Rs = 4 = 2 [ Tc - (Tc - Tb) ] / (20 + 30) = Tb / 25, Tb = 100 s. Substitute Tb = 100 s into first equation gives 2 Tb - Tc = 2 x 100 - Tc = 200 - Tc = 60, then Tc = 200 - 60 = 140 s. Then Ta = Tc - Tb = 140 - 100 = 40 s. 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 { 1605 }

ENGINEERING MATHEMATICS - EXAMPLE 8.2 : In the US (United States), one barrel is equal to 42 US gallons and equivalent to 0.15898 cubic metres. One British barrel is equal to 36 Imperial gallons and equivalent to 0.163659 cubic metres. If an oil refinery in England sells 42 Imperial gallons of crude oil to a company in US, how much US gallons of crude oils has been sold?

Answer

ENGINEERING MATHEMATICS - ANSWER 8.2 : Volume of crude oil sold in US = 42 x 42 (US gallons) / 0.15898 (cubic metres) x 0.163659 (cubic metres) / 36 (Imperial gallons) = 50.4421 US gallons. 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 { 1678 }

ENGINEERING MATHEMATICS - EXAMPLE 8.3 : Solve the first order differential equation : (Z + 1)(dy/dx) = xy in term of ln |y| = f(x). Z = (x)(x).

Answer

ENGINEERING MATHEMATICS - ANSWER 8.3 : Rearranging the equation (1/y)(dy) = [ x / (Z+1) ] (dx). Use the additional equation u = Z + 1, du = 2x dx, 0.5 du = x dx. Then the equation becomes (1/y)(dy) = (0.5)(1/u) du. Integrating both sides of the equation gives ln |y| = (0.5) ln |Z+1| + c where Z = square of x. 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 { 1428 }

Question 95 - The following formula is used as a reference : (analyte signal) / (internal standard signal) = (f-factor) x (concentration of analyte) / (concentration of internal standard). A solution containing 3 mM of analyte and 4 mM of internal standard gave peak signals of 2 and 3 mamps respectively. Another similar solution containing 2 mM of analyte and 1 mM of internal standard gave peak signals of 1 and 4 mamps respectively. Find the average f-factor.

Answer

Answer 95 - Let (f-factor) = (analyte signal) (concentration of internal standard) / [ (internal standard signal) (concentration of analyte) ]. For first solution, (f-factor) = (2 mamps) (4 mM) / [ (3 mamps) (3 mM) ] = 8 / 9. For second solution, (f-factor) = (1 mamps) (1 mM) / [ (4 mamps) (2 mM) ] = 1 / 8. Average : Average f-factor = (8 / 9 + 1 / 8) / 2 = (64 + 9) / (72 x 2) = 73 / 144. 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 { 1457 }

Question 96 - In infrared spectrum, one of the factors affecting peak location is the mass of the atoms. The stretching frequency of a bond connected to a lighter atom will be greater than the same bond connected to a heavier atom. (a) For halogens like florine (F), chlorine (Cl), bromine (Br), iodine (I) and astatine (At), what is their IUPAC group number? Hint : The proton numbers for F, Cl, Br, I and At are 9, 17, 35, 53 and 85 respectively. (b) For the compounds of H-F, H-Cl, H-Br, H-I and H-At, which one has the lowest stretching frequency and which one has the highest stretching frequency? State the reasons.

Answer

Answer 96 - (a) 17. (b) Lowest stretching frequency : H-At. Highest stretching frequency : H-F. Reasons : F has the lowest proton number 9 and At has the highest proton number 85. For the compound of H-X, the stretching frequency of the bond decreases when the mass of X increases and vice versa. 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 { 1605 }

Question 97 - Two viruses a and b with masses of Ma and Mb are moving at velocities of Va and Vb respectively, facing towards each other and collide. After collision both masses of Ma and Mb disappear. (a) Find the total momentum available for both a and b. Hint : momentum = mass x velocity = M x V. (b) Guess the total energy E generated from the disappearance of a and b. Let c to be the velocity of light. Hint : E is equal to M c square.

Answer

Answer 97 - (a) Momentum of a = Ma x Va. Momentum of b = Mb x Vb. Total momentum of a and b = Ma x Va + Mb x Vb. (b) After mass disappearance : Energy of a = Ma x c x c. Energy of b = Mb x c x c. Then E = Ma x c x c + Mb x c x c = (Ma + Mb)(c x 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.

Question { 1814 }

Question 98 - The Planck-Einstein relation connects the particulate photon energy E with its associated wave frequency f to produce E = hf. Let h to be the Planck constant. The frequency f, wavelength L and speed of light c are related by E = hc / L. With p denoting the linear momentum of a particle, the de Broglie wavelength L of the particle is given by L = h / p. (a) Find the equation of E as a function of p and c. (b) If E has a unit of electron-volt and f has a unit of 1 / second, then what is the unit of h?

Answer

Answer 98 - (a) Let first equation : E = hc / L, second equation : L = h / p. Substitute the second equation into first equation will produce E = hc / L = hc / (h / p) = pc. (b) E = hf then h = E / f. Dimensional analysis of h will produce E / f of electron-volt / (1 / second) or electron-volt-second. 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 { 1529 }

Question 99 - (a) The quantum number m is given by m = -s, -s + 1. If s = 0.5, find the values of m. (b) | T > = (cos T) | V > + (sin T) | H >. The V and H states form a basis for all polarizations. Let cos T = 0.8. (i) If (sin T)(sin T) + (cos T)(cos T) = 1, find the value of sin T. (ii) For | T > = a | V > + b | H >, where a x a represents the probability of | V > and b x b represents the probability of | H >. Which one is more abundant, | V > or | H >? (iii) Find the value of T without using any mathematical tools.

Answer

Answer 99 - (a) When s = 0.5, m = -s = -0.5 and m = -s + 1 = -0.5 + 1 = 0.5. (b)(i)(sin T)(sin T) = 1 - (cos T)(cos T) = 1 - 0.8 x 0.8 = 0.36. Then sin T = 0.6. (ii) Let a = cos T = 0.8, probability of | V > = a x a = 0.8 x 0.8 = 0.64 (more abundant) and b = sin T = 0.6, probability of | H > = b x b = 0.6 x 0.6 = 0.36 (less abundant). (iii) cos T = 0.8, then T = arc cos 0.8. 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.