Answer / kangchuentat

ELECTRICAL TECHNOLOGY - ANSWER 16.3 : (a) Input = Output / Efficiency = 100 watts / 0.5 = 200 watts. (b) Number of inverter = Input / Rated power = 200 / 300 = 2 / 3 < 1. Practically only 1 inverter is needed. (c) I = Pin / V = 200 / 12 = 16.667 A. (d) Number of charge controllers = Input / Rated current = 16.667 / 10 = 1.6667 < 2. Practically only 2 are needed. (e) Energy consumption = 0.1 kilowatt x 2 hours / day = 0.2 kilowatt hour per day. (f) Input energy = Energy consumption / Efficiency = 0.2 / 0.5 = 0.4 kilowatt hour per day. (g) Energy generation per day per solar panel = 2.88 hours x 0.02 kilowatt = 0.0576 kilowatt hour. (h) Number of solar panels needed = Answer for (f) / Answer for (g) = 0.4 / 0.0576 = 6.944 < 7. Practically only 7 are needed. (i) Ampere hour per day to run the load = 0.4 kilowatt hour per day / 12 V = 1 / 30 k. Total ampere hour needed for 3 days = 1 / 30 k x 3 = 0.1 k. Number of batteries needed = 0.1 k / 5 = 100 / 5 = 20. 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.

ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.10 : Let D be the random outcome of rolling a dice once. A new dice has values of D* = D - 3.5. There is a total of n rolls of a dice. (a) Find the variance for D* by using the formula 6 V = [ D* (D = 1) ] [ D* (D = 1) ] + [ D* (D = 2) ] [ D* (D = 2) ] + [ D* (D = 3) ] [ D* (D = 3) ] + [ D* (D = 4) ] [ D* (D = 4) ] + [ D* (D = 5) ] [ D* (D = 5) ] + [ D* (D = 6) ] [ D* (D = 6) ]. (b) Calculate the standard deviation of D* as a square root of V. (c) Another new dice has values of D** = kD*. (i) Find the value of k so that D** has a standard deviation of 1. (ii) Find the values of D** for each outcome of D = 1, 2, 3, 4, 5 and 6, when the standard deviation is 1. (iii) Given that the average score of a dice is 3.5, find the equivalent, new and improved model of a dice, Sn in term of n and D**. (iv) Find the expected value of D** as the average of D**.

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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.5 : 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.

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Question 42 - 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).

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.8 : Cash flow agreement where the sign + or - of the amounts are start negative and at a point switch to positive and stay positive is referred to has an investment. The cash flow could be denoted by the symbol of [a, b, c ...]. Consider the following two cases in a biochemical factory : (a) You pay $1 today. You receive $1.05, one year from now. (b) You pay $3 today. You receive $5, one year from today. You pay $2, two years from today. Determine if cases (a) and (b) are an investment or a business transaction with cash flow symbol. Find the interest rate r in the cash flow in both cases (a) and (b).

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