100 mol/min of an equimolar mixture of A and B is separated in a two stage evapo
ID: 801395 • Letter: 1
Question
100 mol/min of an equimolar mixture of A and B is separated in a two stage evaporation
process. In the first stage, the liquid and vapor flow rates exiting the stage are each 50
mol/min. The liquid phase effluent from the first stage is fed to a second stage that operates
at the same temperature as the first stage. The effluent flow rates of the liquid and vapor
phases from the second stage are 25 mol/min each. The vapor phase effluent from the
second stage is combined with the vapor phase effluent from the first stage. At the
temperature that both stages are operating at, the vapor pressure of A is 10 kPa and the vapor
pressure of B is 100 kPa. The liquid and vapor phases leaving each unit are in equilibrium. Find the composition of each stream and the pressure at each stage.
Explanation / Answer
Assuming that A and B make an ideal mixture, Raoult's law can be used to solve the riddle.
In any minute considered, there's 100 moles of mixture in the system
Stage 1 :
The mixture is equimolar.
Mole fractions of A and B are each 0.5
total vapor pressure at this stage before exiting = 0.5*10 + 0.5*100 = 5+ 50 = 55kPa
Stage 2:
Only 50mol/min of liquid is entering the second stage.
Stage 1 vapor effluents :
Fraction of A in vapor in stage 1 : 5/55 = 1/11
Fraction of B in vapor = 1- (1/11) = 10/11
Moles of A in vapor = 50/11
Moles of B in vapor = 500/11
Total moles of A = 50 = Total moles of B [Equimola mixture]
Moles of A left in liquid phase = 50- 50/11 = 500/11 mol/min
Moles of B left in liquid phase = 50 - 500/11 = 50/11 mol/min
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Pure liquid entering stage 2 will establish a liquid-vapor equilibrium.
Mole fraction of A = 10/11
Mole fraction of B = 1/11
Total vapor pressure in stage 2 = (10/11)*10 + (1/11)*100 = 200/11 = 18.182 kPa
Vapor effluents of stage 2:
Fraction of A in vapor phase = (100/11)/(200/11) = 0.5
Fraction of B in vapor phase = (100/11)/(200/11) = 0.5
Now as only25 mol/min of liquid gets out, 25mol/min is the vapor effluent rate in this stage.
This vapor effluent has 12.5 mol/min of A and B each respectively
Similarly the fractions of A and B in the liquid effluents of stage 2 are equal.
The final stream has 12.5 mol/min of A nad B each.
The vapor effluents are mixed.
Moles of A = 50/11 + 12.5 =17.045 mol/min
Moles of B = 500/11 + 12.5 = 57.955 mol/min
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