CHEMICAL MATERIAL BALANCE - EXAMPLE 2.3 : A 1.5 weight % aqueous salt solution is concentrated to 4 weight % in a single-effect evaporator. The feed rate to the evaporator is F = 7500 kg / h and the feed is at 85 degree Celsius. The evaporator operates at 1 bar. By assuming that only pure solvent of water exists in the form of vapor from the feed, calculate the flow rate of such vapor V.
CHEMICAL MATERIAL BALANCE - ANSWER 2.3 : Mass balance of salt : Mass of salt in feed = Mass of salt in concentrated liquid. (0.015) (7500 kg / h) = 0.04L where L = flowrate of concentrated liquid = (0.015) (7500) / (0.04) = 2812.5 kg / h. Let F = V + L as overall mass balance, then V = F - L = 7500 - 2812.5 = 4687.5 kg / h.
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While there, are many tests available to detect leaks on vessels, is there a technology available to quantify the leak, or measure the flow through a leak?
ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.13 (CORRECTION) : (i) In the Present Value Multiplication Rule, let PV = present value, Ra = interest rate for first discount, A = duration for first discount; Rc = interest rate for second discount, C = duration for second discount. Let PV = [ 1 / (1 + Ra) ^ A ] [ 1 / (1 + Rc) ^ C ] where ^ is the symbol of power : 3 ^ 2 = 3 x 3, 2 ^ 3 = 2 x 2 x 2. (a) For discounts involving 8 % / year for 3 years and 10 % / year for 9 years, find the value of PV. (b) If Re = interest rate for third discount, E = duration of third discount, form a mathematical equation of PV as a function of A, C, E, Ra, Rc, Re. Note : Discounts are available in the purchase of certain biochemical engineering instruments. (ii) Let R = nominal interest rate related to growth rate of money, r = real interest rate related to growth rate of purchase power. If I = inflation, where the unit of R, r and I is %, find the mathematical relationship of r as a function of R and I.
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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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