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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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.
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