ENGINEERING MATERIAL - EXAMPLE 12.3 : Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.
ENGINEERING MATERIAL - ANSWER 12.3 : (a) V = 0.3349 x 0.3349 x 0.3349 = 0.0376 cubic nanometers. (b) Mass per unit cell = 2 x 92.9 (g / mol) [ 1 mol / (6.02 x 10 ^ 23 atoms) ] = 3.086 x 10 ^ (-22) g = m. V = 0.0376 x [ 10 ^ (-7) ] ^ 3 = 0.0376 x 10 ^ (-21) cubic cm. Then density is m / V = 8.207 g / cubic cm.
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Question 36 - For a mixture of benzene (B), toluene (T) and xylene (X), the equation applies where x for B, T and X will sum up to 1. The equation of x for each component is x = (L / V + 1) (F) / (L / V + K). The data of F for each component are : 0.5 for B, 0.35 for T, 0.15 for X. The data of K for each component are : 1.98 for B, 0.76 for T, 0.24 for X. When x for B + x for T + x for X = 1, find the values of (a) L / V; (b) x for each component of B, T, X respectively. You may use Excel program - Data : What-If-Analysis for Goal Seek to perform the iterative calculations.
COMPUTER PROGRAMMING FOR ENGINEERS - EXAMPLE 17.2 : There are 5 simultaneous equations with 5 unknowns as follow : 16 V - 4 X - Z = 36, 4 X - 4 W - Y + Z = 3, 8 V + 4 W - X + Y = 10, X - W = 1, V + W = 0. Find the values of V, W, X, Y and Z accurately within 15 minutes. State the computer program that you use to get the answers.
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).
GENETIC ENGINEERING - EXAMPLE 27.3 : (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 between 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?
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