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.
BIOCHEMICAL ENGINEERING INSTRUMENTATION - ANSWER 29.4 : 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.
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ENGINEERING PHYSICS - EXAMPLE 30.4 : (a) Time evolution in Heisenberg picture, according to Ehrenfest theorem is m (d / dt) < r > = < p >, where m = mass, r = position, p = momentum of a particle. If v = velocity, prove that m < v > = < p >. (b) Lande g-factor is given by Gj = Gl [ J (J + 1) - S (S + 1) + L (L + 1) ] / [ 2J (J + 1) ] + Gs [ J (J + 1) + S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ]. If Gl = 1 and under approximation of Gs = 2, prove by calculation that Gj = (3/2) + [ S (S + 1) - L (L + 1) ] / [ 2J (J + 1) ].
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Question 49 - According to rules of thumb in chemical process design, consider the use of an expander for reducing the pressure of a gas when more than 20 horsepowers can be recovered. The theoretical adiabatic horsepower (THp) for expanding a gas could be estimated from the equation : THp = Q [ Ti / (8130a) ] [ 1 - (Po / Pi) ^ a ] where 3 ^ 3 is 3 power 3 or 27, Q is volumetric flowrate in standard cubic feet per minute, Ti is inlet temperature in degree Rankine, a = (k - 1) / k where k = Cp / Cv, Po and Pi are reference and systemic pressures respectively. (a) Assume Cp / Cv = 1.4, Po = 14.7 psia, (temperature in degree Rankine) = [ (temperature in degree Celsius) + 273.15 ] (9 / 5), nitrogen gas at Pi = 90 psia and 25 degree Celsius flowing at Q = 230 standard cubic feet per minute is to be vented to the atmosphere. According to rules of thumb, should an expander or a valve be used? (b) Find the outlet temperature To by using the equation To = Ti (Po / Pi) ^ a.
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Question 51 - A batch reactor is designed for the system of the irreversible, elementary liquid-phase hydration of butylene oxide that produces butylene glycol. At the reaction temperature T = 323 K, the reaction rate constant is k = 0.00083 L / (mol - min). The initial concentration of butylene oxide is 0.25 mol / L = Ca. The reaction is conducted using water as the solvent, so that water is in large excess. (a) Let the molecular weight of water is 18 g / mol and the mass of 1 kg in 1 L of water, calculate the molar density of water, Cb in the unit of mol / L. (b) Determine the final conversion, X of butylene oxide in the batch reactor after t = 45 min of reaction time. Use the formula X = 1 - 1 / exp [ kt (Cb) ] derived from material balance. (c) Find the equation of t as a function of X.
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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.
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