Mutli Component Distillation
Transcript of Mutli Component Distillation
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Multi-Component Distillation
Phase equilibrium For multi-components
iii
s
i pPyxP == 12
iii
s
i
i xKxP
P
y == 13
Taking 1 as the reference component
i
i
i
iiiii
x
x
y
y
x
x
y
y
xy
xy
K
K
====
1
1
/
/ 1
1
1
1111
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Bubble point calculations
Given the total pressure and liquid mole fraction T and y
i
n
i
iii
n
i
i
s
in
i
i
n
i
xKxKxP
Py
====
====1
1
111
1 15
If vapor pressure data is available solve:
011
== =
i
s
in
i
xP
Pf
If k-values data is available solve:
011
== =
ii
n
i
xKf
If relative volatility data is available solve:
011
1 == =
ii
n
i
xKf
Iterate over temperature till solution is found calculate the vapor mole fraction from
i
s
ii x
P
Py =
iii xKy =
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=
==
n
i
ii
iiiii
x
xxKy
1
1 ; note that iixK = /11
See Example in Geankoplis BookA liquid stream at 405.3 kPa with 40% butane, 25% bentane, 30% hexane and 15%hepatne is fed to a distillation column. Find the bubble point and the corresponding vapor
composition in equilibrium with the liquid.
Using the K-value diagram we build this Table:
Let K-hexane be the reference component
Let T = 65 oC
component xi Ki xi
Butane
Pentane
0.4
0.25
1.68
0.63
6.857
2.571
2.743
0.643
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Hexane
Heptane
0.2
0.15
0.245
0.093
1.0
0.38
0.2
0.057
Total 3.643
According to Equation 18: 0107.0)643.3(245.01111
===
=ii
n
i
xKf
Let T = 70 oC
component xi Ki xi yi
Butane
BentaneHexane
Heptane
0.40.25
0.2
0.15
1.860.71
0.2815
0.11
6.6072.522
1.0
0.391
2.6430.631
0.2
0.059
0.748
0.178
0.0570.017
Total 3.533 1.000
According to Equation 18: 00547.0)533.3(2815.0111
1 === = iin
ixKf
Dew point calculations
Given P and vapor mole fraction Tandx
i
in
is
i
in
i
i
n
i K
y
PP
yx
===
===111 /
1 16
Thus iterate over temperature till the following is solved:
011
== = i
in
i K
yf
17
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Flash calculations
iiiLxVyFz += 18
iii xVFVyFz )( += 19
iii xF
Vy
F
Vz )1( +=
20
iiixffyz )1( += 21
iii xf
fz
fy
)1(1 =
22
F, z
V, y
L, x
P, T
Using the equilibrium relation:
iiiiix
f
fz
fxKy
)1(1 ==
23
Solving forx:
1)1( +=
i
ii
Kf
zx
24
11)1(11 =+= == ii
n
ii
n
i Kf
z
x
25
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iii LxVyFz +=
iiii KLyVyFz /+=
( ) iii yKLVFz /+=
( ) iii VyVKLFz /1+=
i
iii
VKL
FzVyV
/1+==
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Another approach
VLF += 1 L 1
iii VyLxFz += n-1 V 1
= 1ix 1 X n =1iy 1 Y n
iiixKy = n T 1
vLF VHLhFh += 1
2n+3 2n+3
Non-ideal systems (activity coefficient)
For non-ideal systems the phase equilibrium:
i
s
iii
xP
Py
=
where is the liquid phase activity coefficient and depends on temperature, pressure andcomposition.
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Distillation Column design
Stage to stage calculation using computer program Short cut methods
Key component
A single distillation column is designed to separate binary components. Therefore an
important species to be concentrated in the top product Is known as the light key while
the one to be concentrated in the bottom product is the heavy key.
Total reflux minimum number of stages
[ ]log ( / )( / )
log( )av
LD HD HB LB
m
L
x D x D x B x BN
=
LBLDLav=
Bx
Dx
Bx
Dx
HB
HDN
avei
iB
iD m,=
Minimum Reflux
=i
ifixq1
=+i
iDim
xR 1
is the relative volatility at average column temperature
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Erbar-Maddox Chart
Feed position
= 2)()(log206.0log
HD
LB
Lf
Hf
x
x
D
B
x
x
Ns
Ne
Example:
A liquid stream at 405.3 kPa and 100 mole/h with 40% butane (A), 25% bentane (B),30% hexane (C) and 15% hepatne (D) is fed to a distillation column.
90% ofB is recovered in distillate and 90% ofCis the bottom, calculate
(a)mole and composition of distillate and bottom(b)Dew point of the distillate and bubble point of the bottom(c)minimum stages for total reflux(d)minimum reflux ratio(e)number of theoretical stage working at 1.5Rm(f) location of feed tray
(a)solve the material balancemole balance on butane
25 = nBD + nBB
Q nBD = 0.9(25) = 22.5 mole nDD = 2.5 mole
mole balance on hexane
30 = nCD + nCB
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Q ncB = 0.9(20) = 18 mole nCD = 2 mole
Assume allA in distillate nAD= 40 mole nAB = 0Assume allD in bottom nAD= 40 mole nAB = 0
comp nf nD yD nb xB
A 25 40 0.62 0 0
B 25 22.5 0.349 2.5 0.07
C 20 2 0.031 18 0.507
D 15 0 0 15 0.423
total 100 64.5 1.0 35.5 1.0
part (b)
Dew point calculations:i
in
is
i
in
i
i
n
i K
y
PP
yx
===
===111 /
1
start at T = 67oC
comp yD Ki yD/Ki A 0.62 1.75 0.354 6.73
B 0.349 0.65 0.537 2.5
C 0.031 0.26 0.119 1.0D 0 0.1 0 0.385
total 1.0 1.01
Bubble point calculation
i
n
i
iii
n
i
i
s
in
i
i
n
i
xKxKxP
Py
====
====1
1
111
1
start at T = 132 oC
comp xB Ki xBKi A 0 5.0 0 4.384
B 0.07 2.35 0.165 2.043
C 0.507 1.15 0.583 1.0
D 0.423 0.61 0.258 0.53
total 1.0 1.006
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part ( c)
letB light componentlet C heavy component
LD = 2.5LB = 2.04
258.2)043.2(5.2 === LBLDave
[ ] [ ]4.5
)258.2log(
)07.0/507.0)(031.0/349.0(log
)log(
)/)(/(log==
=
avL
HBLBHDLDm
BxBxDxDxN
Distribution of component A:
43.5)384.4(73.6 === ABADave
103218
243.5 4.5 ==
AB
AD
n
n
40 = nAD + nAB = 1033nAB nAB = 0.039 mole/h nAD = 39.961 mole/h
Distribution of component D:
45.0)385.0(53.0 === DBDDave
0015.018
245.0 4.5 ==
DB
DD
n
n
15 = nDD + nDB = 1.0015nDB nDB = 14.97 mole/h nDD = 0.022 mole/h
comp nD yD nb xB
A 40 0.62 0.024 0.0007
B 22.5 0.349 2.5 0.07C 2 0.031 18 0.507
D 0.022 0.00034 15 0.423
total 64.522 1.0 35.5 1.0
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part (d)
=i
ifixq1
=+i
iDim
xR 1
The average tower temperature = (67 + 123)/2 = 99.5oC
comp xB Ki (99.5) (99.5)A 0.4 3.12 5.2
B 0.25 1.38 2.3
C 0.2 0.6 1.0
D 0.15 0.28 0.467total 1.0
For feed at boiling point q =1
+
+
+
===
467.0
)15.0(467.0
0.1
)2.0(0.1
3.2
)25.0(3.2
2.5
)4.0(2.50111 q
Solve by trial-and-error (see book)
= 1.2096
12096.1467.0
)0(467.0
2096.11
)031.0(0.1
2096.13.2
)35.0(3.2
2096.12.5
)62.0(2.51
+
+
+
=
=i
iDim
xR
Rm = 0.395
part (d)
R = 1.5Rm = 1.5(0.395) = 0.593
R/(R+1) = 0.3723
Rm/(Rm+1) = 0.283
Using the chart Nm/N = 0.49 N = Nm/0.49 N = 5.4/0.49 = 11
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part (e)
= 2)()(log206.0log
HD
LB
Lf
Hf
xx
DB
xx
NsNe
07344.0)031.0
0704.0(
5.64
5.35)
25.0
2.0(log206.0log 2 =
=Ns
Ne
Ne/Ns = 1.184 Ne+Ns = 11
Ne = 6Ns = 5
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