Answer / kangchuentat

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.17 : (a) For Fisher Effect, 1 + R = 1.1 (1 + r) = 1.1 + 1.1 r, 0.1 = R - 1.1 r (Equation 1). For General Inflation Effect, r = R - 0.1, 0.1 = R - r (Equation 2). Equation 2 - Equation 1 : r = 0, then R = 0.1. (b) (1 + R) = (1 + r) (1 + I) = 1 + I + r + rI, then R = (I + r) + rI (Equation 3). Let R = I + r (Equation 4). Equation 3 - Equation 4 : 0 = Ir, where (I = 0) or (r = 0) or (I = 0 and r = 0). If I = 0, r = R. If r = 0, I = R. If I = 0 and r = 0, R = 0 = 0 R = 1 R. 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.

How can the particle size distribution be determined in a given bulk solid?

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GENETIC ENGINEERING - EXAMPLE 27.4 : 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 of 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?

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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).

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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.4 : 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.

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