Question

Consider a distillation column that receives a feed of 500 kg/min of a mixture of benzene and cyclohexane that is 55 mole % benzene. We wish to design the column so that the distillate has a composition that is 95 mole % benzene and contains 90 % of the benzene fed to the column. Calculate the distillate and bottoms flow rates (in kg/min) and the composition of the bottoms product in mole %.

Answer 1

a) 3.2279 kmol/min

b) 0.3407 kmol/min

c) 0.1148

Step-by-step explanation:

molecular weight of Benzene = 78

molecular weight of cyclohexane = 84

mass flow rate of the feed = 500kg/min

molecular weight of mixture = 0.55 Mb + 0.45 Mc

Mmix = 0.55 ( 78 ) + 0.45 ( 84 ) = 80.7

Given that The molar flow rate of mixture feed ( F ) = 6.195 kmol/min

Molar flow rate of Benzene in the feed = 6.195 * 055 = 3.4072kmol/min

a)Determine the Distillate flow rate

Molar flow rate of Benzene in distillate = 0.9 * 3.4072 = 3.0665 kmol/min

mol % of Benzene in distillate = 0.95

therefore the molar flow rate of Distillate = 3.0665 / 0.95 = 3.2279 kmol/min

b)Determine the Bottom flow rate

molar flow rate of Benzene in the Bottom = 3.4072 - 3.0665 = 0.3407 kmol/min

c) composition of the bottoms product in mole %

composition of the bottoms product in mole %. = molar flow rate of Benzene in the Bottom / overall mol balance

overall mol balance ( w ) = F - D = 6.195 - 3.2279 = 2.967 kmol/min

composition of the bottoms product in mole %. = 0.3407 / 2.967 = 0.1148