Download NUMBER OF EQUILIBRIUM STAGES IN BINARY DISTILLATION

V1
qD
Over-all material balance:
D, xD, hD
L0
x0
h0
F=D+B
(1)
Component material balance:
F
xF
hF
qB
B, xB, hB
F x F = D xD + B x B
(2)
F xF = D xD + (F – D) xB
(3)
F x F  x B 
D
xD  xB
(4)
ASSUMPTIONS:
 Equimolal overflow
V1  V2    Vn
and
VF1  VF2  . . .  Vm
(5)
L 0  L1  . . .  Ln
and
LF  LF1  . . .  Lm
(6)
 The two components have equal and constant Hvap.
 The sensible enthalpy changes of the vapor and
liquid are negligible compared to the latent heats.
 The binary mixture behaves as an ideal solution.
 The stages are adiabatic except at designated
locations (condenser and reboiler).
 The pressure is constant throughout the column.
ENRICHING SECTION
D is calculated using eq. (4)
V1
V1
V2
V3
V4
V5
L0
L1
L2
L3
L4
L5
Providing that R is fixed, L0 is
calculated using the equation
D
defining reflux ratio:
L0
R
(5)
D
Material balance around condenser:
V1 = L 0 + D
(6)
Since V1 is in equilibrium with L1, the
composition (x1) of L1 is obtained
from equilibrium data.
Material balance around envelope A:
A
V2 = L 1 + D
V1
V2
V3
V4
V5
L0
L1
L2
L3
L4
L5
D
(7)
Component balance:
V2 y2 = L1 x1 + D xD
(8)
L1
D
y2  x1  xD
V2
V2
(9)
A
D
xD
Vn+1
F
xF
vF
n
Ln
The composition of V2, V3,. ..,
Vn is calculated by
performing material balance
in envelope A:
Vn+1 yn+1 = Ln xn + D xD
(10)
Ln
D
y n 1 
xn 
xD
Vn1
Vn1
(11)
n+1
LF-1
L'F
Since the molar liquid overflow is constant, Ln = L
and Vn+1 = V:
L
D
y n 1  x n  x D
V
V
(12)
Eq. (12) relates the composition of the vapor rising
to a plate to the composition of the liquid on the
plate.
Equilibrium equation:
Vn, yn
Stage n
Ln, xn
y 1 ,n
K1 
x 1 ,n
y 2 ,n 1  y 1 ,n 
K2 

x 2 ,n 1  x 1 ,n 
K2 1  x1,n   1  y1,n 
K2 1  x1,n   1  K1 x1,n 
x1,n K1  K2   1  K2
1  K2
x 1 ,n 
K1  K 2
FEED PLATE
Material balance around feed plate:
F  VF1  LF1  VF  LF
vF
(13)
LF-1
F
xF
Component balance around feed plate:
VF1
LF
FxF  VF1 yF1  LF1xF1  VFYF  LFxF
(14)
VF+1 and LF are calculated using eqs. (13) along with the
information about thermal condition of the feed.
yF+1 is calculated using eq. (14)
STRIPPING SECTION
The composition of Vm is calculated
by performing material balance in
envelope A:
F
xF
Vm1
m
Lm
m+1
A
B
xB
Vm1  Lm  B
(15)
Vm1 ym1  Lm xm  B xB
(16)
Lm
B
y m 1 
xm 
xB
Vm1
Vm1
(17)
Lm
B
y m 1  x m  x B
Vm
Vm
(18)
EXAMPLE 1
Pertinent data on the binary system heptane-ethyl benzene at
760 mm Hg are as follows.
t, C
136.2
129.5
122.9
119.7
116.0
110.8
106.2
103.0
100.2
98.5
xH
0.000
0.080
0.185
0.251
0.335
0.487
0.651
0.788
0.914
1.000
yH
0.000
0.233
0.428
0.514
0.608
0.729
0.834
0.904
0.963
1.000
H
-1.23
1.19
1.14
1.12
1.06
1.03
1.00
1.00
1.00
EB
1.00
1.00
1.02
1.03
1.05
1.09
1.15
1.22
1.27
--
A feed mixture composed of 42 mole % heptane, 58
mole % ethyl benzene is to be fractionated at 760 mm Hg
to produce distillate containing 97 mole % heptane and a
residue containing 99 mole % ethyl benzene. Using (L/D)
= 2.5, determine the number of equilibrium stages
needed for a saturated liquid feed and bubble-point
reflux.
SOLUTION
Since xn is calculated using equilibrium relationship, it is
necessary to develop an equation correlating xn and yn.
Based on the available data, an equation is established:
x  0.771 y3  0.104y2  0.334y
Assume 100 mole of feed is introduced:
F x F  x B  100  0.42  0.01
D

 42.71
xD  xB
0.97  0.01
B = F – D = 100 – 42.71 = 57.29
L0 = R D = (2.5) (42.71) = 106.77 
L1 = L2 = . . . . = L0 = 106.77
V1 = L0 + D = 106.77 + 42.71 = 149.48 
V1 = V2 = . . . . = 149.48
ENRICHING SECTION
Ln
D
106.77
42.71 

y n 1  x n  x D 
x n 1  
 0.97
Vn
Vn
149.48
 149.48 
yn1  0.7143 xn  0.277
3
xn  0.771 yn
2
 0.104yn
 0.334yn
n
yn
xn
1
0.970
0.930
2
0.941
0.865
3
0.895
0.769
4
0.826
0.640
5
0.734
0.494
6
0.630
0.362
FEED PLATE
Feed is a saturated liquid:
FL = F
FV = 0
LF  L5  FL  106.77  100  206.77
5
V6
L5
F
L6  L7  206.77
6
V7'
A
L'6
Material balance around feed plate (envelope A):
F  VF1  LF1  VF  LF
F  V7  L5  V6  L6
100  V7  106.77  149.48  206.77
V7  149.48
7
Component balance around feed plate:
F xF  VF1 yF1  LF1 xF1  VF yF  LF xF
F xF  V7 y 7  L5 x 5  V6 y 6  L6 x 6
V6 y 6  L 6 x 6  F x F  L 5 x 5
y7 
V7
y7 
149.48 0.63  206.770.362
149.48

1000.42  106.770.494
y7 = 0.497
149.48
STRIPPING SECTION
Lm
B
206.77
57.29 

y m1  x m  x B 
xm  
 0.01
Vm
Vm
149.48
 149.48 
ym1  1.3833 xm  0.0038
1.383261975
3
xm  0.771 ym
m
7
8
9
10
11
2
 0.104ym
 0.334ym
ym
0.497
0.321
0.165
0.073
0.030
xm
0.235
0.122
0.056
0.024
0.010
Number of equilibrium stages = 11
0.5729