CENG 3210 Separation Processes Tutorial 7 Solutions 1

CENG 3210 Separation Processes Tutorial 7 Solutions 1

CENG 3210 Separation Processes
Tutorial 7 Solutions
1. Multicomponent Distillation – Fenske equation
A distillation column with a partial reboiler and a total
condenser is being used to separate a mixture of benzene,
toluene, and cumene. The feed is 40 mole % benzene, 30
mole % toluene, and 30 mole % cumene and is input as a
saturated vapor. We desire 95% recovery of the toluene in
the distillate and 95% recovery of the cumene in the
bottoms. The reflux is returned as a saturated liquid, and
constant molal overflow (CMO) can be assumed. Pressure
is 1 atm.
Equilibrium can be represented as constant relative
volatilities. Choosing toluene as the reference component,
αbenz-tol = 2.25 and αcumene-tol = 0.21. Find the number of
equilibrium stages required at total reflux.
Define. A = toluene (LK), B = cumene (HK), C = benzene
(LNK).
αCA = 2.25, αBA = 0.21, αAA = 1
xFA = 0.3, xFB = 0.3, xFC = 0.4
Material balance:
A: FxFA = DxDA + BxBA
B: FxFB = DxDB + BxBB
Fractional recovery of A in the distillate = DxDA/FxFA =
0.95
Fractional recovery of B in the bottoms = BxBB/FxFB = 0.95
Hence,
1
DxDA = 0.95FxFA = 0.95 F(0.3) = 0.285F
BxBA = FxFA - DxDA = F(0.3) – 0.285F = 0.015F
BxBB = 0.95FxFB = 0.95F(0.3) = 0.285F
DxDB = FxFB - BxBB = F(0.3) – 0.285F = 0.015F
 LK , HK  1 / 0.21  4.76
 Dx D, LK / Bx B, LK
ln
Dx D, HK / Bx B, HK

Nm 
ln  LK , HK

 0.285F / 0.015F 

ln



0
.
015
F
/
0
.
285
F
 1  
 1
ln 4.76
 3.8  1  2.8
So the minimum number of ideal stages is 2.8 plus a
reboiler.
2. Multicomponent Distillation – Underwood equation
For the distillation problem given in question 1, use the
Underwood method find the minimum reflux ratio. Use a
basis of 100 kmol/h of feed. The mole fractions of benzene,
toluene and cumene in the distillate are 0.571, 0.408 and
0.021, respectively.
The feed is saturated vapor, q = 0. F = 100 kmol/h.
The  value between 0.21 and 1 is found by trial.
 x
 i Fi  1  q  1  0  1
i  
2
A
B
C
i
xFi
 i x Fi
, =0.5454

i  
1
0.21
2.25
0.3
0.3
0.4
0.6599
-0.1878
0.5280
1.0001
By trial and error, =0.5454, so
 x
1(0.408)
0.21(0.021)
2.25(0.571)
RDm  1   i Di 


 i   1  0.5454 0.21  0.5454 2.25  0.5454
= 0.897 - 0.013 + 0.754 = 1.638
RDm = 0.638
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