CHEMICAL MATERIAL BALANCE - EXAMPLE 2.3 : A 1.5 weight % aqueous salt solution i

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.

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CHEMICAL MATERIAL BALANCE - EXAMPLE 2.3 : A 1.5 weight % aqueous salt solution is concentrated to 4 ..

Answer / kangchuentat

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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