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1
Distillation
Introduction
Distillation is a method used to separate the components of liquid solution,
which depends upon the distribution of these various components between
a vapor and a liquid phase.
The vapor phase is created from the liquid phase by vaporization at the
boiling point.
Distillation is concerned with solution where all components are appreciably
volatile such as in ammonia-water or ethanol-water solutions, where both
components will be in the vapor phase.
2
Vapor-Liquid Equilibrium Relations
Raoult’s Law
An ideal law, Rault’s law, can be defined for vapor-liquid phases in
equilibrium (only ideal solution e.g. benzene-toluene, hexane-heptane etc.
Where
pA is the partial pressure of component A in the vapor in Pa (atm)
PA is the vapor pressure of pure A in Pa (atm)
xA is the mole fraction of A in the liquid.
AAA xPp =
BA xx +=1
BA yy +=1
Composition in liquid:
Composition in vapor:
(1)
(2)
(3)
3
Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure
of 101.32 kPa.
Boiling-Point Diagrams and xy Plots
Dew point is the temperature at which
the saturated vapor starts to condense.
Bubble-point is the temperature at
which the liquid starts to boil.
The difference between liquid and vapor
compositions is the basis for distillation
operations.
4
Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure
of 101.32 kPa.
Boiling-Point Diagrams and xy Plots
If we start with a cold liquid
composition is xA1 = 0.318 (xB1
= 0.682) and heat the mixture, it
will start to boil at 98ÂșC.
The first vapor composition in
equilibrium is yA1 = 0.532 (yB1 =
0.468).
Continue boiling, the
composition xA will move to the
left since yA is richer in A.
5
Boiling-Point Diagrams and xy Plots
Ppp BA =+
PxPxP ABAA =−+ )1(
P
xP
P
p
y AAA
A ==
The boiling point diagram can be calculated from (1) the pure vapor-
pressure data in the table below and (2) the following equations:
(4)
(5)
(6)
Where
pA, pB are the partial pressure of component A and B in the vapor in Pa (atm)
PA , PB are the vapor pressure of pure A and pure B in Pa (atm)
P is total pressure in Pa (atm)
xA is the mole fraction of A in the liquid.
6
Boiling-Point Diagrams and xy Plots
The boiling point diagram can be calculated from (1) the pure vapor-
pressure data in the table below and (2) the following equations:
1
7
Exmpl Use of Raoult’s Law for Boiling-Point Diagram
Calculate the vapor and liquid compositions in equilibrium at 95ÂșC
(368.2K) for benzene-toluene using the vapor pressure from the
table 1 at 101.32 kPa.
Solution: At 95ÂșC from Table 1 for benzene, PA = 155.7 kPa and PB = 63.3
kPa. Substituting into Eq.(5) and solving,
155.7(xA) + 63.3(1-xA) = 101.32 kPa (760 mmHg)
Hence, xA= 0.411 and xB= 1 – xA = 1 - 0.411 = 0.589. Substituting into eqn.
(6),
PxPxP ABAA =−+ )1(
632.0
32.101
)411.0(7.155
====
P
xP
P
p
y AAA
A
8
The boiling point diagram can be calculated from the pure vapor-pressure
data in the table below and the following equations:
1
9
A common method of plotting the equilibrium data is shown in Fig. 2 where
yA is plotted versus xA for the benzene-toluene system. The 45Âș line is
given to show that yA is richer in component A than is xA.
Fig. 2 Equilibrium diagram for system benzene(A) – toluene(B) at 101.32
kPa (1atm).
10
Vapor-Liquid Equilibrium Relations
Boiling-Point Diagrams and xy Plots
Maximum-boiling azeotropeMinimum-boiling azeotropeIdeal boiling point diagram
An azeotrope is a mixture of two or more liquids in such a ratio that its
composition cannot be changed by simple distillation.
This occurs because, when an azeotrope is boiled, the resulting vapor has
the same ratio of constituents as the original mixture.
11
Boiling-Point Diagrams and xy Plots
Vapor-Liquid Equilibrium Relations
12
Single-Stage Equilibrium Contact for Vapor-Liquid System
V1 V2
L0 L1
MVLVL =+=+ 1120Total mass balance:
AMAAAA MxyVxLyVxL =+=+ 11112200Mass A balance:
Where
V1, V2 is a vapor
L0, L1is a liquid
A single equilibrium stage is
- the two different phases are brought into intimate contact with each other.
- The mixing time is long enough and the components are essentially at
equilibrium in the two phases after separation.
In case of constant molal overflow : V1 = V2 and L0 = L1
13
Exmpl Equilibrium Contact of Vapor-Liquid Mixture
A vapor at the dew point and 101.32 kPa containing a mole fraction
of 0.40 benzene (A) and 0.60 toluene (B) and 100 kg mol total is
contacted with 110 kg mol of a liquid at the boiling point containing a
mole fraction of 0.30 benzene and 0.70 toluene. The two streams are
contacted in a single stage, and the outlet streams leave in
equilibrium with each other. Assume constant molal overflow.
Calculate the amounts and compositions of the exit streams.
Solution: The given values are V2 = 100 kg mol, yA2 = 0.40, L0=110 kg
mol , and xA0 = 0.30.
V1 V2
L0 L1
For constant molal overflow,
V2 = V1 and L0 = L1.
Material balance on component A,
To solve equation above, the equilibrium relation between yA1 and xA1 in
figure below must be used.
First, we assume that xA1 = 0.20 and substitute into equation above to
solve for yA1.
AMAAAA MxyVxLyVxL =+=+ 11112200
1100)2.0(110)40.0(100)30.0(110 Ay+=+
11 100110)40.0(100)30.0(110 AA yx +=+
Assuming that xA1 = 0.20 and solving yA1 = 0.51.
Next, assuming that xA1=0.40 and solving, yA1 = 0.29.
Next, assuming that xA1=0.40 and solving, yA1 = 0.29.
(These point are plotted on the graph.)
At the intersection of this line with the equilibrium curve,
yA1 = 0.455 and xA1 = 0.25.
16
17
Relative Volatility of Vapor-Liquid Systems
)1)(1(
/
/
/
AA
AA
BB
AA
AB
xy
xy
xy
xy
−−
==α
AAB
AAB
A
x
x
y
)1(1 −+
=
α
α
P
xP
y AA
A =
B
A
AB
P
P
=α
P
xP
y BB
B =
Relative volatility
It is a measure of the differences in volatility between 2 components, and hence their boiling
points. It indicates how easy or difficult a particular separation will be.
Where αAB is the relative volatility of A with respect to B in the binary system.
when αAB is above 1.0, a separation is possible.
Raoult’s law:
)( ABα
18
Exmpl Using data from table 1 calculate the relative volatility for the
benzene-toluene system at 85ÂșC (358.2K) and 105ÂșC (378.2K)
Solution: At 85ÂșC, substituting into equation below for a system following
Rault’s law,
Similarly at 105ÂșC,
The variation in α is about 7%. Answer
54.2
0.46
9.116
===
B
A
AB
P
P
α
38.2
0.86
2.204
==α
19
Equilibrium or Flash Distillation
Distillation has two main methods in practice.
Introduction to distillation methods
1. Production of vapor by boiling the liquid mixture to be separated in a
single stage and recovering and condensing the vapors. No liquid
is allowed to return to the single-stage still to contact the
rising vapors.
2. Returning of a portion of the condensate to the still. The vapors
rise through a series of stages or trays, and part of the condensate
flows downward through the series of stages or trays counter
currently to the vapors (“fractional distillation, distillation with reflux,
or rectification”).
There are 3 important types of distillation that occur in a single stage or
still: Equilibrium or flash distillation, Simple batch or differential
distillation and simple steam distillation
20
Equilibrium or Flash Distillation
heater
Separator
xA
yA
Flash distillation is a single stage separation technique.
1. A liquid mixture is pumped through a heater to raise the temperature
and enthalpy of the mixture.
2. It then flows through a valve and the pressure is reduced, causing
the liquid to partially vaporize.
3. Once the mixture enters a big enough volume (the “flash drum”), the liquid and
vapor separate.
4. Because the vapor and liquid are in such close contact up until the “flash” occurs,
the product liquid and vapor phases approach equilibrium.
21
Equilibrium or Flash Distillation
heater
Separator
Total mass balance:
Component A balance:
where
F, V and L are flow rate of feed, vapor and liquid phases.
xF, yA and xA are mole fraction of component A in feed, vapor and liquid.
AAF xfyfx )1( −+=
Where
f = V/F = molal fraction of the feed that is vaporized and withdrawn continuously as vapor.
1-f = one as liquid
Material balance for more volatile component :
xA
yA
AAF x
F
V
F
F
y
F
V
x )()( −+=
AAF LxyVFx +=
LVF +=
22
Exmpl A mixture of 50% mole normal heptane and 50% normal
octane at 30ÂșC is continuously flash distilled at 1 standard
atmosphere so that 60 mol% of the feed is vaporized. What will be
the composition of the vapor and liquid products?
xA 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
yA 0.247 0.453 0.607 0.717 0.796 0.853 0.898 0.935 0.968
Solution: Given: xF = 0.5, f = 0.6
Find: xA, yA
Basis: F = 100 mols
Applying the mass balance yields:
Since ,
LVF +=
60)100(6.0 === fFV
4060100 =−=−= VFL
FVf /=
23
Material balance for more volatile component,
Substituting value of f =0.6 and xF =0.5 we get,
Assuming that xA = 0.5 and solving yA = 0.5.
Next, assuming that xA=0 and solving, yA = 0.83.
(These point are plotted on the graph.)
At the intersection of this line with the equilibrium curve,
yA = 0.58 and xA = 0.39.
AAF xfyfx )1( −+=
AAF x
F
V
F
F
y
F
V
x )()( −+=
AA xy )6.01(6.05.0 −+=
AA xy 4.06.05.0 +=
24
yA
xA
xF =0.5
yF = 0.5
x=0
y-intercept= 0.834
xA =0.39
yA = 0.58
1st
2nd
3rd
25
Simple Batch or Differential Distillation
The pot is filled with liquid mixture and heated.
Vapor flows upwards though the column and
condenses at the top.
Part of the liquid is returned to the column as
reflux, and the remainder withdrawn as
distillate.
Nothing is added or withdrawn from the still
until the run is completed.
26
Simple Batch or Differential Distillation
The total moles of component A left in the still
nA will be
nA = xn
where
n is the moles of liquid left in the still at a given
time
y and x is the vapor and liquid compositions
If a small amount of liquid dn is vaporized, the change in the moles of component A is ydn, or
dnA. Differentiating equation a gives
ydnxdnndx
xdnndxxnddnA
=+
+== )(
xy
dx
n
dn
−
=
27
B
A
AB
B
A
x
x
y
y
α=
dx/(y-x) can be integrated graphically or numerically using tabulated equilibrium data
or an equilibrium curve.
For ideal mixture:
∫∫ =
−
=
1
0 0
1
1
0
ln
x
x
n
n
n
n
xy
dx
n
dn
xy
dx
n
dn
−
=
B
B
AB
A
A
B
A
AB
B
A
B
A
n
dn
n
dn
n
n
dn
dn
dndn
dndn
α
α
=
==
AB
A
A
B
B
n
n
n
n
α/1
00
ïŁ·ïŁ·
ïŁž
ïŁ¶
ïŁŹïŁŹ
ïŁ­
ïŁ«
=
B
B
AB
A
A
n
n
n
n
00
lnln α=
Simple Batch or Differential Distillation
Integrating
28
Exmpl A batch of crude pentane contains 15 mole percent n-
butane and 85 percent n-pentane. If a simple batch distillation at
atmospheric pressure is used to remove 90 percent of butane,
how much pentane will be removed? What will be the
composition of the remaining liquid?
Solution: The final liquid is nearly pure pentane, and its boiling point is
36ÂșC. The vapor pressure of butane at this temperature is 3.4 atm, giving a
relative volatility of 3.4. For the initial conditions, the boiling point is about
27ÂșC, and the relative volatility is 3.6. Therefore, an average value of 3.5 is
used for αAB.
Basis: 1 mol feed
(batane) (pentane)
From equation:
nB = total mole of B left in still, nA = total mole A left in still.
n0B = total initial mole of B in still, n0A = total initial mole A lin still.
15.0=OAn 015.0=An 85.0=OBn
AB
A
A
B
B
n
n
n
n
α/1
00
ïŁ·ïŁ·
ïŁž
ïŁ¶
ïŁŹïŁŹ
ïŁ­
ïŁ«
=
29
( ) 518.01.0
85.0
5.3/1
==Bn
440.0)85.0(518.0 ==Bn
moln 455.0015.044.0 =+=
033.0
455.0
015.0
==Ax
Total mole of liquid left in still:
Mole fraction of butane in
liquid left:
30
Simple Steam Distillation
Note that by steam distillation, as long as water is present, the high-
boiling component B vaporizes at a temperature well below its normal
boiling point without using a vacuum. The A and B are usually
condensed in condenser and the resulting two immiscible liquid
phases separated.
Disadvantage: large amounts of heat must be used to simultaneously
evaporate the water with high-boiling compound.
31
Simple Steam Distillation
When the sum of the separate vapor pressures equals the total pressure,
the mixture boils and
Where
is vapor pressure of pure water A
is vapor pressure of pure B
Then the vapor composition is
The ratio moles of B distilled to moles of A distilled is
PPP BA =+
BP
P
P
y A
A =
AP
P
P
y B
B =
A
B
A
B
P
P
n
n
=
32
Temperature PA(water)
(mm Hg)
PB(ethylaniline)
(mm Hg)K ÂșC
353.8 80.6 48.5 1.33
369.2 96.0 87.7 2.67
372.3 99.15 98.3 3.04
386.4 113.2 163.3 5.33
Exmpl A mixture contains 100 kg of H2O and 100 kg of ethylene
(mol wt = 121.1 kg/kg mol), which is immiscible with water. A very
slight amount of nonvolatile impurity is dissolved in the organic.
To purify the ethyaniline it is steam-distilled by bubbling saturated
steam into the mixture at a total pressure of 101.32 kPa (1 atm).
Determine the boiling point of the mixture and the composition of
the vapor. The vapor pressure of each of the pure compounds is
as follows (T1):
33
Solution:
PPP BA =+
Temperature PA
(water)
(kPa)
PB
(ethylaniline)
(kPa)
P=PA+PB
(kPa)K ÂșC
353.8 80.6 48.5 1.33 49.83
369.2 96.0 87.7 2.67 90.37
372.3 99.15 98.3 3.04 101.34
386.4 113.2 163.3 5.33 169.23
The boiling temperature = 99.15ÂșC since total pressure in this temperature
is equal to atmospheric pressure.
The vapor composition are:
97.0
32.101
3.98
===
kPa
kPa
P
P
y A
A 03.0
32.101
04.3
===
P
P
y B
B
34
Distillation with Reflux and McCabe-Thiele
method
Introduction to Distillation with Reflux
Rectification (fractionation) or stage distillation with reflux is
a series of flash-vaporization stages are arranged in a series which the vapor and liquid
products from each stage flow counter currently to each other.
V2
L1
V1
V2
L0 L1
V3
L2
Vn Vn+1
Ln-1 Ln
The liquid in a stage is conducted or flows to the stage below and the vapor from a stage
flow upward to the stage above.
nnnn LVLV +=+ −+ 11
nnnnnnnn xLyVxLyV +=+ −−++ 1111
A total material balance:
A component balance on A:
35
Introduction to Distillation with Reflux
1. Feed enters the column somewhere in
the middle of the column.
5. The vapor continues up to the next tray
or stage, where it is again contacted with
a downflowing liquid.
2. Feed is liquid, it flows down to a sieve
tray or stage.
4. The vapor and liquid leaving the tray
are essentially in equilibrium.
In a distillation column the stages (referred to as sieve plates or trays) in a distillation tower are
arranged vertically, as shown schematically in figure below.
3. Vapor enters the tray and bubbles
through the liquid on this tray as the
entering liquid flows across.
6. The concentration of the more volatile
component is being increased in the
vapor form each stage going upward and
decreased in the liquid from each stage
going donwards.
36
Introduction to Distillation with Reflux
7. The final vapor product coming
overhead is condensed in a condenser
and a portion of the liquid product
(distillate) is removed, which contains a
high concentration of A.
9. The liquid leaving the bottom tray
enters a reboilier, where it partially
vaporized, and the remaining liquid, which
is lean in A or rich in B, is withdrawn as
liquid product.
In a distillation column the stages (referred to as sieve plates or trays) in a distillation tower are
arranged vertically, as shown schematically in figure below.
8. The remaining liquid from the
condenser is returned (refluxed) as a
liquid to the top tray.
10. The vapor from the reboiler is sent
back to the bottom stage or trays is much
greater.
37
McCabe-Thiele Method of Calculation for Number of
Theoretical Stages
A mathematical – graphical method for determining the number of theoretical trays
or stages needed for a given separation of a binary mixture of A and B has been
developed by McCabe and Thiele.
The method uses material balances around certain parts of the tower, which give
operating lines and the xy equilibrium curve for the system.
Main assumption
1) Equimolar overflow through the tower between the feed inlet and the top tray and
the feed inlet and bottom tray.
2) Liquid and vapor streams enter a tray, are equilibrated, and leave.
A) Introduction and assumptions
38
A total material balance:
nnnn LVLV +=+ −+ 11
A component A balance:
nnnnnnnn xLyVxLyV +=+ −−++ 1111
Where
Vn+1 is mol/h of vapor from tray n+1
Ln is mol/h liquid from tray n
yn+1 is mole fraction of A in Vn+1 and so on.
39
A total material balance:
A component A balance:
Where
F is the entering feed (mol/h)
D is the distillate (mol/h)
W is the bottoms (mol/h)
WDF +=
wDF WxDxFx +=
B) Equation for enriching section
(1)
(2)
40
Material balance over dashed-line section: DLV nn +=+1
A balance on component A: DDxxLyV nnnn +=++ 11 (4)
(3)
41
Solving for yn+1, the enriching-section operating line is
)1/(, 11 +=+= ++ RRVLDLV nnnn
DLR n /=
11
1
+
+
+
=+
R
x
x
R
R
y D
nn
where = reflux ratio = constant.
The eqn. (1) is a straight line on a plot of vapor composition versus liquid
composition.
Since and equation becomes
11
1
++
+ +=
n
D
n
n
n
n
V
Dx
x
V
L
y
(6)
(5)
42
The slope is or . It intersects the y=x line (45Âș diagonal
line) at . The intercept of the operating line at x = 0 is .
1/ +nn VL )1/( +RR
)1/( += Rxy DDxx =
The theoretical stages are determined by starting at xD and stepping off the first
plate to x1. Then y2 is the composition of the vapor passing the liquid x1.
In a similar manner, the other theoretical trays are stepped off down the tower in
the enriching section to the feed tray.
43
C) Equation for stripping section
Material balance over dashed-line section: WLV mm −=+1
A component A balance: wWxxLyV mmmm −=++ 11 (8)
(7)
44
Solving for ym+1, the enriching-section operating line is
11
1
++
+ +=
m
D
m
m
m
m
V
Dx
x
V
L
y (9)
1+= mm VL
Again, since equimolal flow is assumed, = constant and
= constant, eqn. (2) is a straight line when plotted as y versus x,
with a slope of . It intersects the y = x line at x = xw.
The intercept at x = 0 is .
Nm LL =
Nm VV =+1
1/ +−= mW VWxy
45
The theoretical stages for the stripping section are determined by starting at
xW, going up to yW, and then across to the operating line, etc.
46
D) Effect of feed conditions
The condition of feed stream is represented by the quantity q, which is the mole
fraction of liquid in feed.
qFLL nm +=
FqVV mn )1( −+=
Dnn DxxLyV +=
wmm WxxLyV −=
)()()( wDnmnm WxDxxLLyVV +−−=−
The enriching and striping operating-line
equations on an xy diagram can be
derived as follows:
Where the y and x values are the point of
intersection of the two operating lines.
Subtracting eqn.(3) from eqn.(4),
(12)
(13)
(14)
(10)
(11)
47
D) Effect of feed conditions
Substituting eqn.(2), (10), and (11) into eqn.(14) and rearranging,
(15)
11 −
−
−
=
q
x
x
q
q
y F
λ
)(
1
FbpL TTc
q
−
+=
λ
)(
1
dFpV TTc
q
−
+=
Cold-liquid feed Superheated vapor
where
CpL, CpV = specific heats of liquid and vapor, respectively
TF = temperature of feed
Tb, Td = bubble point and dew point of feed respectively
λ = heat of vaporization
48
E) Location of the feed tray in a tower and number of trays.
q = 0 (saturated vapor)
q = 1 (saturated liquid)
q > 1(subcooled liquid)
q < 0 (superheated vapor)
0 < q < 1 (mix of liquid and
vapor)
From eqn.(15), the q-line equation and is the locus of the intersection of the two
operating lines. Setting y = x in eqn(15), the intersection of the q-line equation
with the 45Âș line is y=x=xF, where xF is the overall composition of the feed.
In given below the figure, the q line is plotted for various feed conditions. The
slope of the q line is q/(q-1).
Slope = R/(R+1)
Slope = q/(1-q)
Slope = L/ V
F) Using Operating Lines and the Feed Line in McCabe-Thiele Design
Exmpl A continuous fractioning column is to be designed to separate
30,000 kg/h of a mixture of 40 percent benzene and 60 percent toluene
into an overhead product containing 97 percent benzene and a bottom
product containing 98 percent toluene. These percentages are by weight.
A reflux ratio of 3.5 mol to 1 mol of product is to be used. The molal
latent heats of benzene and toluene are 7,360 and 7,960 cal/g mol,
respectively. Benzene and toluene from a nearly ideal system with a
relative volatility of about 2.5. The feed has a boiling point of 95ÂșC at a
pressure of 1 atm.
a) Calculate the moles of overhead product and bottom product per hour.
b) Determine the number of ideal plates and the position of the feed plate
(i) if the feed is liquid and at its boiling point; (ii) if the feed is liquid and at
20ÂșC (specific heat 0.44 cal/g.ÂșC); (iii) if the feed is a mixture of two-thirds
vapor and one-third liquid.
Solution (a)
The average molecular weight of the feed is
The average of heat vaporization is
The feed rate F is 30,000/85.8 = 350 kg mol/h. By an overall benzene
balance, using Eq. below
440.0
92
60
78
40
78
40
=
+
=Fx 0235.0
92
98
78
2
78
2
=
+
=Bx974.0
92
3
78
97
78
97
=
+
=Dx
8.85
92
60
78
40
100
=
+
gmolcal /696,7)960,7(56.0)360,7(44.0 =+=λ
hkgmolD /4.153
0235.0974.0
0235.0440.0
350 =ïŁ·
ïŁž
ïŁ¶
ïŁŹ
ïŁ­
ïŁ«
−
−
=
hkgmolB /6.1964.153350 =−=
Solution (b) (i),
We determine the number of ideal plates and position of the feed plate.
1) Plot the equilibrium diagram, erect verticals at xD, xF, and xB.
2) Draw the feed line. Here q=1, and the feed line is vertical.
3) Plot the operating lines. The intercept of the rectifying lie on
the y axis is, xD/(R+1) = 0.974/(3.5+1) = 0.216 (eqn (6)). From the
intersection of the rectifying operating line and the feed line, the stripping
line is drawn.
4) Draw the rectangular steps between the two operating lines
and the equilibrium curve. The stripping line is at the seventh step. By
counting steps it is found that, besides the reboiler, 11 ideal plates are
needed and feed should be introduced on the seventh plate from the top.
Solution (b) (ii),
The latent heat of vaporization of the feed λ is 7,696/85.8 = 98.7 cal/g.
The slope of the feed line is -1.37/(1-1.37) = 3.70. When steps are drawn
for this case, as shown in Fig. below, it is found that a reboiler and 10
ideal plates are needed and that the feed should be introduced on the
sixth plate.
37.1
7.89
)2095(44.0
1 =
−
+=q
11 −
−
−
=
q
x
x
q
q
y F
Solution (b) (iii),
From the definition of q it follows that for this case q = 1/3 and the slope
of the feed line is -0.5. The solution is shown in Fig. below. It calls for a
reboiler and 12 plates, with the feed entering on the seventh plate.
58
Total and Minimum Reflux Ratio for McCabe-Thiele Method
A) Total Reflux
One limiting values of reflux ratio is that of total reflux, or R = ∞. Since R = Ln/D
and, by eqn.(16).
DLV nn +=+1
Then Ln is very large, as is the vapor flow Vn. This means that the slope R/(R+1)
of the enriching operating line becomes 1.0 and the operating lines of both
sections of the column coincide with the 45Âș diagonol line, as shown in Fig below.
Minimum number of trays can be obtained by returning all the overhead
condensed vapor V1 from the top of the tower back to the tower as reflux, i.e., total
reflux. Also, the liquid in the bottoms is reboiled.
(16)
59
Minimum number of theoretical steps Nm
when a total condenser is used (α is constant).
av
w
w
D
D
m
x
x
x
x
N
αlog
1
1
log ïŁ·ïŁ·
ïŁž
ïŁ¶
ïŁŹïŁŹ
ïŁ­
ïŁ« −
−
=
( ) 2/1
1 wav ααα =For small variations in α,
where α1 is the relative volatility of the overhead vapor
αw is the relative volatility of the bottoms liquid.
60
B) Minimum reflux ratio
The minimum reflux ratio (Rm) will require an infinite number of trays for the given
separation desired of xD and xW.
If R is decreased, the slope of the enriching operating line R/(R+1) is decreased,
and the intersection of this line and the stripping line with the q line moves farther
from the 45Âș line and closer to the equilibrium line.
xx
yx
R
R
D
D
m
m
â€Č−
â€Č−
=
+1
Two operating lines touch the
equilibrium line (“pinch point”) at y’
and x’ (number of steps required
becomes infinite).
The line passes through the points x’,
y’ and xD (y=xD):
61
C) Operating and optimum reflux ratio
Total reflux = number of plates is a minimum, but the tower diameter is infinite.
This corresponds to an infinite cost of tower and steam and cooling water. This
is the limit in the tower operation.
Minimum reflux = number of trays is infinite, which again gives an infinite cost.
These are the two limits in operation of the tower.
Actual operating reflux ratio to use is in between these two limits. The
optimum reflux ratio to use for lowest total cost per year is between the
minimum Rm and total reflux (1.2Rm to 1.5Rm).
General Design Consideration
1. A tower design is normally divided into two main steps, a process
design followed by a mechanical design. The purpose of the
process design is to calculate the number of required theoretical
stages, column diameter and tower height. On the other hand, the
mechanical design focuses on the tower internals and heat
exchanger arrangements.
2. Many factors have to be considered in designing a distillation column
such as the safety and environmental requirements, column
performance, economics of the design and other parameters, which
may constrain the work.
The first step in distillation column design is to determine the separation
sequences, which depends on the relative volatility and concentration of
each component in the feed. King has outlined a few design rules as
follows:
1) Direct sequences that remove the components one by one in the
distillate are generally favored.
2) Sequences that result in a more equal-molar division of the feed
between distillate and bottoms products should be favored.
3) Separations where the relative volatility of two adjacent components is
close to unity should be performed in the absence of other components; ie,
reserve such a separation until the last column in the sequence.
4) Separations involving high-specified recovery fractions should be
reserved until last in the sequence.
Once the separation sequence is decided, engineering calculations follow
to determine the number of theoretical stages, operating parameters and
tower dimensions. In general, the steps included in distillation calculations
are summarized into the following:
1) Performing a material balance for the column
2) Determining the tower operating pressure (and/or temperature)
3) Calculating the minimum number of theoretical stages using the Fenske
equation
4) Calculating the minimum reflux rate using the Underwood equations
5) Determining the operating reflux rate and number of theoretical stages
6) Selection of column internals (tray or packings)
7) Calculating the tower diameter and height
Some general design rules (from Cheresources.com) that should be
considered are as follows:
1) Distillation is usually the most economical method of separating
liquids.
2) For Ideal mixtures (low pressure, medium temperature, and non-
polar), relative volatility is the ratio of vapor pressures i.e. α = P2/P1
3) Tower operating pressure is determined most often by the temperature
of the available cooling medium in the condenser or by the maximum
allowable reboiler temperature.
4) Tower Sequencing :
A. Easiest separation first – least trays and reflux
B. When neither relative volatility nor feed concentrations vary widely,
remove components one by one as overhead products.
C. When the adjacent ordered components in the feed vary widely in
relative volatility, sequence the splits in order of decreasing volatility.
D. When the concentration in the feed varies widely but the relative
volatilities do not, remove the components in the order of decreasing
concentration in the feed.
5) Economically optimum reflux ratio is about 120% to 150% of the
minimum reflux ratio.
6) The economically optimum number of stages is about 200% of the
minimum value.
7) A safety factor of at least 10% above the number of stages by the
best method is advisable.
8) A safety factor of at least 25% about the reflux should be utilized for
the reflux pumps.
9) Reflux drums are almost always horizontally mounted and designed
for a 5 min holdup at half of the drum's capacity.
10) For towers that are at least 3 ft (0.9 m) in diameter, 4 ft (1.2 m)
should be added to the top for vapor release and 6 ft (1.8 m) should be
added to the bottom to account for the liquid level and reboiler return.
11) Limit tower heights to 175 ft (53 m) due to wind load and foundation
considerations.
12) The Length/Diameter ratio of a tower should be no more than 30
and preferably below 20.
13) A rough estimate of reboiler duty as a function of tower diameter is
given by:
Q = 0.5 D2 for pressure distillation
Q = 0.3 D2 for atmospheric distillation
Q = 0.15 D2 for vacuum distillation
Where,
Q : Energy in Million Btu/hr
D : Tower diameter in feet.
The Selection of Column Internals
The selection of column internals has a big impact on the column
performance and the maintenance cost of a distillation tower.
There are several choices of column internals and the two major
categories are trays and packing. The choice of which to utilize depends
on the
1) pressure,
2) fouling potential,
3) liquid to vapor density ratio,
4) liquid loading, and
5) most importantly the life cycle cost.
Trays can be divided into many categories, such as baffle trays, dual flow
trays, conventional trays, high capacity trays, multiple down comer trays
and system limit trays. According to some rules of thumb, trays should be
selected if:
1) the compounds contain solids or foulants
2) there are many internal transitions
3) liquid loads are high
4) there is a lack of experience in the service
5) vessel wall needs periodic inspection
6) there are multiple liquid phases
On the other hand, packing divisions include grid packing, random
packing, conventional structured packing, and high capacity structured
packing. The rules of thumb for selecting packing are:
1) the compounds are temperature sensitive
2) pressure drop is important (vacuum service)
3) liquid loads are low
4) towers are small in diameter
5) highly corrosive service (use plastic or carbon)
6) the system is foaming
7) the ratio of tower diameter to random packing is greater than 10
Some design guidelines should be considered when designing a tray tower,
such as follows:
1) Tray spacing should be from 18 to 24 inches, with accessibility in mind
(Generally, for a tower diameter of 4 feet and above, the most
common tray spacing is 24 inches to allow easy access for maintenance.
However, for a tower diameter below 4 feet, a tray spacing of 18 inches
is adequate as the column wall can be reached from the man way.)
2) Peak tray efficiencies usually occur at linear vapor velocities of 2 ft/s
(0.6 m/s) at moderate pressures, or 6 ft/s (1.8 m/s) under vacuum
conditions.
3) A typical pressure drop per tray is 0.1 psi (0.007 bar)
4) Tray efficiencies for aqueous solutions are usually in the range of 60-
90% while gas absorption and stripping typically have efficiencies closer
to 10-20%
5) Sieve tray holes are 0.25 to 0.50 in. diameter with the total hole area
being about 10% of the total active tray area. Maximum efficiency is 0.5 in
and 8%.
6) Valve trays typically have 1.5 in. diameter holes each with a lifting cap.
12-14 caps/square foot of tray is a good benchmark.
7) The most common weir heights are 2 and 3 in and the weir length is
typically 75% of the tray diameter.
The packed tower design concepts are listed below:
1) Packed towers almost always have lower pressure drop compared to
tray towers.
2) Packing is often retrofitted into existing tray towers to increase capacity
or separation.
3) For gas flow rates of 500 ft3/min (14.2 m3/min), use 1 in (2.5 cm)
packing, for gas flows of 2000 ft3/min (56.6 m3/min) or more, use 2 in
(5 cm) packing.
4) Ratio of tower diameter to packing diameter should usually be at least
15
5) Due to the possibility of deformation, plastic packing should be limited to
an unsupported depth of 10-15 ft (3-4 m) while metal packing can
withstand 20-25 ft (6-7.6 m).
6) Liquid distributor should be placed every 5-10 tower diameters (along t
he length) for pall rings and every 20 ft (6.5 m) for other types of
random packing.
7) For redistribution, there should be 8-12 streams per sq. foot of tower
area for towers larger than three feet in diameter. They should be even
more numerous in smaller towers.
8) Packed columns should operate near 70% flooding.
9) Height Equivalent to Theoretical Stage (HETS) for vapor-liquid
contacting is 1.3- 1.8 ft (0.4-0.56 m) for 1 in pall rings and 2.5-3.0 ft
(0.76-0.90 m) for 2 in pall rings.
10) Design pressure drops should be as follows:
Table 1: Pressure drop in difference services

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Distillation

  • 1. 1 Distillation Introduction Distillation is a method used to separate the components of liquid solution, which depends upon the distribution of these various components between a vapor and a liquid phase. The vapor phase is created from the liquid phase by vaporization at the boiling point. Distillation is concerned with solution where all components are appreciably volatile such as in ammonia-water or ethanol-water solutions, where both components will be in the vapor phase.
  • 2. 2 Vapor-Liquid Equilibrium Relations Raoult’s Law An ideal law, Rault’s law, can be defined for vapor-liquid phases in equilibrium (only ideal solution e.g. benzene-toluene, hexane-heptane etc. Where pA is the partial pressure of component A in the vapor in Pa (atm) PA is the vapor pressure of pure A in Pa (atm) xA is the mole fraction of A in the liquid. AAA xPp = BA xx +=1 BA yy +=1 Composition in liquid: Composition in vapor: (1) (2) (3)
  • 3. 3 Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa. Boiling-Point Diagrams and xy Plots Dew point is the temperature at which the saturated vapor starts to condense. Bubble-point is the temperature at which the liquid starts to boil. The difference between liquid and vapor compositions is the basis for distillation operations.
  • 4. 4 Boiling-point diagram for system benzene (A)-toluene (B) at a total pressure of 101.32 kPa. Boiling-Point Diagrams and xy Plots If we start with a cold liquid composition is xA1 = 0.318 (xB1 = 0.682) and heat the mixture, it will start to boil at 98ÂșC. The first vapor composition in equilibrium is yA1 = 0.532 (yB1 = 0.468). Continue boiling, the composition xA will move to the left since yA is richer in A.
  • 5. 5 Boiling-Point Diagrams and xy Plots Ppp BA =+ PxPxP ABAA =−+ )1( P xP P p y AAA A == The boiling point diagram can be calculated from (1) the pure vapor- pressure data in the table below and (2) the following equations: (4) (5) (6) Where pA, pB are the partial pressure of component A and B in the vapor in Pa (atm) PA , PB are the vapor pressure of pure A and pure B in Pa (atm) P is total pressure in Pa (atm) xA is the mole fraction of A in the liquid.
  • 6. 6 Boiling-Point Diagrams and xy Plots The boiling point diagram can be calculated from (1) the pure vapor- pressure data in the table below and (2) the following equations: 1
  • 7. 7 Exmpl Use of Raoult’s Law for Boiling-Point Diagram Calculate the vapor and liquid compositions in equilibrium at 95ÂșC (368.2K) for benzene-toluene using the vapor pressure from the table 1 at 101.32 kPa. Solution: At 95ÂșC from Table 1 for benzene, PA = 155.7 kPa and PB = 63.3 kPa. Substituting into Eq.(5) and solving, 155.7(xA) + 63.3(1-xA) = 101.32 kPa (760 mmHg) Hence, xA= 0.411 and xB= 1 – xA = 1 - 0.411 = 0.589. Substituting into eqn. (6), PxPxP ABAA =−+ )1( 632.0 32.101 )411.0(7.155 ==== P xP P p y AAA A
  • 8. 8 The boiling point diagram can be calculated from the pure vapor-pressure data in the table below and the following equations: 1
  • 9. 9 A common method of plotting the equilibrium data is shown in Fig. 2 where yA is plotted versus xA for the benzene-toluene system. The 45Âș line is given to show that yA is richer in component A than is xA. Fig. 2 Equilibrium diagram for system benzene(A) – toluene(B) at 101.32 kPa (1atm).
  • 10. 10 Vapor-Liquid Equilibrium Relations Boiling-Point Diagrams and xy Plots Maximum-boiling azeotropeMinimum-boiling azeotropeIdeal boiling point diagram An azeotrope is a mixture of two or more liquids in such a ratio that its composition cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the resulting vapor has the same ratio of constituents as the original mixture.
  • 11. 11 Boiling-Point Diagrams and xy Plots Vapor-Liquid Equilibrium Relations
  • 12. 12 Single-Stage Equilibrium Contact for Vapor-Liquid System V1 V2 L0 L1 MVLVL =+=+ 1120Total mass balance: AMAAAA MxyVxLyVxL =+=+ 11112200Mass A balance: Where V1, V2 is a vapor L0, L1is a liquid A single equilibrium stage is - the two different phases are brought into intimate contact with each other. - The mixing time is long enough and the components are essentially at equilibrium in the two phases after separation. In case of constant molal overflow : V1 = V2 and L0 = L1
  • 13. 13 Exmpl Equilibrium Contact of Vapor-Liquid Mixture A vapor at the dew point and 101.32 kPa containing a mole fraction of 0.40 benzene (A) and 0.60 toluene (B) and 100 kg mol total is contacted with 110 kg mol of a liquid at the boiling point containing a mole fraction of 0.30 benzene and 0.70 toluene. The two streams are contacted in a single stage, and the outlet streams leave in equilibrium with each other. Assume constant molal overflow. Calculate the amounts and compositions of the exit streams. Solution: The given values are V2 = 100 kg mol, yA2 = 0.40, L0=110 kg mol , and xA0 = 0.30. V1 V2 L0 L1 For constant molal overflow, V2 = V1 and L0 = L1.
  • 14. Material balance on component A, To solve equation above, the equilibrium relation between yA1 and xA1 in figure below must be used. First, we assume that xA1 = 0.20 and substitute into equation above to solve for yA1. AMAAAA MxyVxLyVxL =+=+ 11112200 1100)2.0(110)40.0(100)30.0(110 Ay+=+ 11 100110)40.0(100)30.0(110 AA yx +=+
  • 15. Assuming that xA1 = 0.20 and solving yA1 = 0.51. Next, assuming that xA1=0.40 and solving, yA1 = 0.29. Next, assuming that xA1=0.40 and solving, yA1 = 0.29. (These point are plotted on the graph.) At the intersection of this line with the equilibrium curve, yA1 = 0.455 and xA1 = 0.25.
  • 16. 16
  • 17. 17 Relative Volatility of Vapor-Liquid Systems )1)(1( / / / AA AA BB AA AB xy xy xy xy −− ==α AAB AAB A x x y )1(1 −+ = α α P xP y AA A = B A AB P P =α P xP y BB B = Relative volatility It is a measure of the differences in volatility between 2 components, and hence their boiling points. It indicates how easy or difficult a particular separation will be. Where αAB is the relative volatility of A with respect to B in the binary system. when αAB is above 1.0, a separation is possible. Raoult’s law: )( ABα
  • 18. 18 Exmpl Using data from table 1 calculate the relative volatility for the benzene-toluene system at 85ÂșC (358.2K) and 105ÂșC (378.2K) Solution: At 85ÂșC, substituting into equation below for a system following Rault’s law, Similarly at 105ÂșC, The variation in α is about 7%. Answer 54.2 0.46 9.116 === B A AB P P α 38.2 0.86 2.204 ==α
  • 19. 19 Equilibrium or Flash Distillation Distillation has two main methods in practice. Introduction to distillation methods 1. Production of vapor by boiling the liquid mixture to be separated in a single stage and recovering and condensing the vapors. No liquid is allowed to return to the single-stage still to contact the rising vapors. 2. Returning of a portion of the condensate to the still. The vapors rise through a series of stages or trays, and part of the condensate flows downward through the series of stages or trays counter currently to the vapors (“fractional distillation, distillation with reflux, or rectification”). There are 3 important types of distillation that occur in a single stage or still: Equilibrium or flash distillation, Simple batch or differential distillation and simple steam distillation
  • 20. 20 Equilibrium or Flash Distillation heater Separator xA yA Flash distillation is a single stage separation technique. 1. A liquid mixture is pumped through a heater to raise the temperature and enthalpy of the mixture. 2. It then flows through a valve and the pressure is reduced, causing the liquid to partially vaporize. 3. Once the mixture enters a big enough volume (the “flash drum”), the liquid and vapor separate. 4. Because the vapor and liquid are in such close contact up until the “flash” occurs, the product liquid and vapor phases approach equilibrium.
  • 21. 21 Equilibrium or Flash Distillation heater Separator Total mass balance: Component A balance: where F, V and L are flow rate of feed, vapor and liquid phases. xF, yA and xA are mole fraction of component A in feed, vapor and liquid. AAF xfyfx )1( −+= Where f = V/F = molal fraction of the feed that is vaporized and withdrawn continuously as vapor. 1-f = one as liquid Material balance for more volatile component : xA yA AAF x F V F F y F V x )()( −+= AAF LxyVFx += LVF +=
  • 22. 22 Exmpl A mixture of 50% mole normal heptane and 50% normal octane at 30ÂșC is continuously flash distilled at 1 standard atmosphere so that 60 mol% of the feed is vaporized. What will be the composition of the vapor and liquid products? xA 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 yA 0.247 0.453 0.607 0.717 0.796 0.853 0.898 0.935 0.968 Solution: Given: xF = 0.5, f = 0.6 Find: xA, yA Basis: F = 100 mols Applying the mass balance yields: Since , LVF += 60)100(6.0 === fFV 4060100 =−=−= VFL FVf /=
  • 23. 23 Material balance for more volatile component, Substituting value of f =0.6 and xF =0.5 we get, Assuming that xA = 0.5 and solving yA = 0.5. Next, assuming that xA=0 and solving, yA = 0.83. (These point are plotted on the graph.) At the intersection of this line with the equilibrium curve, yA = 0.58 and xA = 0.39. AAF xfyfx )1( −+= AAF x F V F F y F V x )()( −+= AA xy )6.01(6.05.0 −+= AA xy 4.06.05.0 +=
  • 24. 24 yA xA xF =0.5 yF = 0.5 x=0 y-intercept= 0.834 xA =0.39 yA = 0.58 1st 2nd 3rd
  • 25. 25 Simple Batch or Differential Distillation The pot is filled with liquid mixture and heated. Vapor flows upwards though the column and condenses at the top. Part of the liquid is returned to the column as reflux, and the remainder withdrawn as distillate. Nothing is added or withdrawn from the still until the run is completed.
  • 26. 26 Simple Batch or Differential Distillation The total moles of component A left in the still nA will be nA = xn where n is the moles of liquid left in the still at a given time y and x is the vapor and liquid compositions If a small amount of liquid dn is vaporized, the change in the moles of component A is ydn, or dnA. Differentiating equation a gives ydnxdnndx xdnndxxnddnA =+ +== )( xy dx n dn − =
  • 27. 27 B A AB B A x x y y α= dx/(y-x) can be integrated graphically or numerically using tabulated equilibrium data or an equilibrium curve. For ideal mixture: ∫∫ = − = 1 0 0 1 1 0 ln x x n n n n xy dx n dn xy dx n dn − = B B AB A A B A AB B A B A n dn n dn n n dn dn dndn dndn α α = == AB A A B B n n n n α/1 00 ïŁ·ïŁ· ïŁž ïŁ¶ ïŁŹïŁŹ ïŁ­ ïŁ« = B B AB A A n n n n 00 lnln α= Simple Batch or Differential Distillation Integrating
  • 28. 28 Exmpl A batch of crude pentane contains 15 mole percent n- butane and 85 percent n-pentane. If a simple batch distillation at atmospheric pressure is used to remove 90 percent of butane, how much pentane will be removed? What will be the composition of the remaining liquid? Solution: The final liquid is nearly pure pentane, and its boiling point is 36ÂșC. The vapor pressure of butane at this temperature is 3.4 atm, giving a relative volatility of 3.4. For the initial conditions, the boiling point is about 27ÂșC, and the relative volatility is 3.6. Therefore, an average value of 3.5 is used for αAB. Basis: 1 mol feed (batane) (pentane) From equation: nB = total mole of B left in still, nA = total mole A left in still. n0B = total initial mole of B in still, n0A = total initial mole A lin still. 15.0=OAn 015.0=An 85.0=OBn AB A A B B n n n n α/1 00 ïŁ·ïŁ· ïŁž ïŁ¶ ïŁŹïŁŹ ïŁ­ ïŁ« =
  • 29. 29 ( ) 518.01.0 85.0 5.3/1 ==Bn 440.0)85.0(518.0 ==Bn moln 455.0015.044.0 =+= 033.0 455.0 015.0 ==Ax Total mole of liquid left in still: Mole fraction of butane in liquid left:
  • 30. 30 Simple Steam Distillation Note that by steam distillation, as long as water is present, the high- boiling component B vaporizes at a temperature well below its normal boiling point without using a vacuum. The A and B are usually condensed in condenser and the resulting two immiscible liquid phases separated. Disadvantage: large amounts of heat must be used to simultaneously evaporate the water with high-boiling compound.
  • 31. 31 Simple Steam Distillation When the sum of the separate vapor pressures equals the total pressure, the mixture boils and Where is vapor pressure of pure water A is vapor pressure of pure B Then the vapor composition is The ratio moles of B distilled to moles of A distilled is PPP BA =+ BP P P y A A = AP P P y B B = A B A B P P n n =
  • 32. 32 Temperature PA(water) (mm Hg) PB(ethylaniline) (mm Hg)K ÂșC 353.8 80.6 48.5 1.33 369.2 96.0 87.7 2.67 372.3 99.15 98.3 3.04 386.4 113.2 163.3 5.33 Exmpl A mixture contains 100 kg of H2O and 100 kg of ethylene (mol wt = 121.1 kg/kg mol), which is immiscible with water. A very slight amount of nonvolatile impurity is dissolved in the organic. To purify the ethyaniline it is steam-distilled by bubbling saturated steam into the mixture at a total pressure of 101.32 kPa (1 atm). Determine the boiling point of the mixture and the composition of the vapor. The vapor pressure of each of the pure compounds is as follows (T1):
  • 33. 33 Solution: PPP BA =+ Temperature PA (water) (kPa) PB (ethylaniline) (kPa) P=PA+PB (kPa)K ÂșC 353.8 80.6 48.5 1.33 49.83 369.2 96.0 87.7 2.67 90.37 372.3 99.15 98.3 3.04 101.34 386.4 113.2 163.3 5.33 169.23 The boiling temperature = 99.15ÂșC since total pressure in this temperature is equal to atmospheric pressure. The vapor composition are: 97.0 32.101 3.98 === kPa kPa P P y A A 03.0 32.101 04.3 === P P y B B
  • 34. 34 Distillation with Reflux and McCabe-Thiele method Introduction to Distillation with Reflux Rectification (fractionation) or stage distillation with reflux is a series of flash-vaporization stages are arranged in a series which the vapor and liquid products from each stage flow counter currently to each other. V2 L1 V1 V2 L0 L1 V3 L2 Vn Vn+1 Ln-1 Ln The liquid in a stage is conducted or flows to the stage below and the vapor from a stage flow upward to the stage above. nnnn LVLV +=+ −+ 11 nnnnnnnn xLyVxLyV +=+ −−++ 1111 A total material balance: A component balance on A:
  • 35. 35 Introduction to Distillation with Reflux 1. Feed enters the column somewhere in the middle of the column. 5. The vapor continues up to the next tray or stage, where it is again contacted with a downflowing liquid. 2. Feed is liquid, it flows down to a sieve tray or stage. 4. The vapor and liquid leaving the tray are essentially in equilibrium. In a distillation column the stages (referred to as sieve plates or trays) in a distillation tower are arranged vertically, as shown schematically in figure below. 3. Vapor enters the tray and bubbles through the liquid on this tray as the entering liquid flows across. 6. The concentration of the more volatile component is being increased in the vapor form each stage going upward and decreased in the liquid from each stage going donwards.
  • 36. 36 Introduction to Distillation with Reflux 7. The final vapor product coming overhead is condensed in a condenser and a portion of the liquid product (distillate) is removed, which contains a high concentration of A. 9. The liquid leaving the bottom tray enters a reboilier, where it partially vaporized, and the remaining liquid, which is lean in A or rich in B, is withdrawn as liquid product. In a distillation column the stages (referred to as sieve plates or trays) in a distillation tower are arranged vertically, as shown schematically in figure below. 8. The remaining liquid from the condenser is returned (refluxed) as a liquid to the top tray. 10. The vapor from the reboiler is sent back to the bottom stage or trays is much greater.
  • 37. 37 McCabe-Thiele Method of Calculation for Number of Theoretical Stages A mathematical – graphical method for determining the number of theoretical trays or stages needed for a given separation of a binary mixture of A and B has been developed by McCabe and Thiele. The method uses material balances around certain parts of the tower, which give operating lines and the xy equilibrium curve for the system. Main assumption 1) Equimolar overflow through the tower between the feed inlet and the top tray and the feed inlet and bottom tray. 2) Liquid and vapor streams enter a tray, are equilibrated, and leave. A) Introduction and assumptions
  • 38. 38 A total material balance: nnnn LVLV +=+ −+ 11 A component A balance: nnnnnnnn xLyVxLyV +=+ −−++ 1111 Where Vn+1 is mol/h of vapor from tray n+1 Ln is mol/h liquid from tray n yn+1 is mole fraction of A in Vn+1 and so on.
  • 39. 39 A total material balance: A component A balance: Where F is the entering feed (mol/h) D is the distillate (mol/h) W is the bottoms (mol/h) WDF += wDF WxDxFx += B) Equation for enriching section (1) (2)
  • 40. 40 Material balance over dashed-line section: DLV nn +=+1 A balance on component A: DDxxLyV nnnn +=++ 11 (4) (3)
  • 41. 41 Solving for yn+1, the enriching-section operating line is )1/(, 11 +=+= ++ RRVLDLV nnnn DLR n /= 11 1 + + + =+ R x x R R y D nn where = reflux ratio = constant. The eqn. (1) is a straight line on a plot of vapor composition versus liquid composition. Since and equation becomes 11 1 ++ + += n D n n n n V Dx x V L y (6) (5)
  • 42. 42 The slope is or . It intersects the y=x line (45Âș diagonal line) at . The intercept of the operating line at x = 0 is . 1/ +nn VL )1/( +RR )1/( += Rxy DDxx = The theoretical stages are determined by starting at xD and stepping off the first plate to x1. Then y2 is the composition of the vapor passing the liquid x1. In a similar manner, the other theoretical trays are stepped off down the tower in the enriching section to the feed tray.
  • 43. 43 C) Equation for stripping section Material balance over dashed-line section: WLV mm −=+1 A component A balance: wWxxLyV mmmm −=++ 11 (8) (7)
  • 44. 44 Solving for ym+1, the enriching-section operating line is 11 1 ++ + += m D m m m m V Dx x V L y (9) 1+= mm VL Again, since equimolal flow is assumed, = constant and = constant, eqn. (2) is a straight line when plotted as y versus x, with a slope of . It intersects the y = x line at x = xw. The intercept at x = 0 is . Nm LL = Nm VV =+1 1/ +−= mW VWxy
  • 45. 45 The theoretical stages for the stripping section are determined by starting at xW, going up to yW, and then across to the operating line, etc.
  • 46. 46 D) Effect of feed conditions The condition of feed stream is represented by the quantity q, which is the mole fraction of liquid in feed. qFLL nm += FqVV mn )1( −+= Dnn DxxLyV += wmm WxxLyV −= )()()( wDnmnm WxDxxLLyVV +−−=− The enriching and striping operating-line equations on an xy diagram can be derived as follows: Where the y and x values are the point of intersection of the two operating lines. Subtracting eqn.(3) from eqn.(4), (12) (13) (14) (10) (11)
  • 47. 47 D) Effect of feed conditions Substituting eqn.(2), (10), and (11) into eqn.(14) and rearranging, (15) 11 − − − = q x x q q y F λ )( 1 FbpL TTc q − += λ )( 1 dFpV TTc q − += Cold-liquid feed Superheated vapor where CpL, CpV = specific heats of liquid and vapor, respectively TF = temperature of feed Tb, Td = bubble point and dew point of feed respectively λ = heat of vaporization
  • 48. 48 E) Location of the feed tray in a tower and number of trays. q = 0 (saturated vapor) q = 1 (saturated liquid) q > 1(subcooled liquid) q < 0 (superheated vapor) 0 < q < 1 (mix of liquid and vapor) From eqn.(15), the q-line equation and is the locus of the intersection of the two operating lines. Setting y = x in eqn(15), the intersection of the q-line equation with the 45Âș line is y=x=xF, where xF is the overall composition of the feed. In given below the figure, the q line is plotted for various feed conditions. The slope of the q line is q/(q-1).
  • 49. Slope = R/(R+1) Slope = q/(1-q) Slope = L/ V F) Using Operating Lines and the Feed Line in McCabe-Thiele Design
  • 50. Exmpl A continuous fractioning column is to be designed to separate 30,000 kg/h of a mixture of 40 percent benzene and 60 percent toluene into an overhead product containing 97 percent benzene and a bottom product containing 98 percent toluene. These percentages are by weight. A reflux ratio of 3.5 mol to 1 mol of product is to be used. The molal latent heats of benzene and toluene are 7,360 and 7,960 cal/g mol, respectively. Benzene and toluene from a nearly ideal system with a relative volatility of about 2.5. The feed has a boiling point of 95ÂșC at a pressure of 1 atm. a) Calculate the moles of overhead product and bottom product per hour. b) Determine the number of ideal plates and the position of the feed plate (i) if the feed is liquid and at its boiling point; (ii) if the feed is liquid and at 20ÂșC (specific heat 0.44 cal/g.ÂșC); (iii) if the feed is a mixture of two-thirds vapor and one-third liquid.
  • 51. Solution (a) The average molecular weight of the feed is The average of heat vaporization is The feed rate F is 30,000/85.8 = 350 kg mol/h. By an overall benzene balance, using Eq. below 440.0 92 60 78 40 78 40 = + =Fx 0235.0 92 98 78 2 78 2 = + =Bx974.0 92 3 78 97 78 97 = + =Dx 8.85 92 60 78 40 100 = + gmolcal /696,7)960,7(56.0)360,7(44.0 =+=λ hkgmolD /4.153 0235.0974.0 0235.0440.0 350 =ïŁ· ïŁž ïŁ¶ ïŁŹ ïŁ­ ïŁ« − − = hkgmolB /6.1964.153350 =−=
  • 52. Solution (b) (i), We determine the number of ideal plates and position of the feed plate. 1) Plot the equilibrium diagram, erect verticals at xD, xF, and xB. 2) Draw the feed line. Here q=1, and the feed line is vertical. 3) Plot the operating lines. The intercept of the rectifying lie on the y axis is, xD/(R+1) = 0.974/(3.5+1) = 0.216 (eqn (6)). From the intersection of the rectifying operating line and the feed line, the stripping line is drawn. 4) Draw the rectangular steps between the two operating lines and the equilibrium curve. The stripping line is at the seventh step. By counting steps it is found that, besides the reboiler, 11 ideal plates are needed and feed should be introduced on the seventh plate from the top.
  • 53.
  • 54. Solution (b) (ii), The latent heat of vaporization of the feed λ is 7,696/85.8 = 98.7 cal/g. The slope of the feed line is -1.37/(1-1.37) = 3.70. When steps are drawn for this case, as shown in Fig. below, it is found that a reboiler and 10 ideal plates are needed and that the feed should be introduced on the sixth plate. 37.1 7.89 )2095(44.0 1 = − +=q 11 − − − = q x x q q y F
  • 55.
  • 56. Solution (b) (iii), From the definition of q it follows that for this case q = 1/3 and the slope of the feed line is -0.5. The solution is shown in Fig. below. It calls for a reboiler and 12 plates, with the feed entering on the seventh plate.
  • 57.
  • 58. 58 Total and Minimum Reflux Ratio for McCabe-Thiele Method A) Total Reflux One limiting values of reflux ratio is that of total reflux, or R = ∞. Since R = Ln/D and, by eqn.(16). DLV nn +=+1 Then Ln is very large, as is the vapor flow Vn. This means that the slope R/(R+1) of the enriching operating line becomes 1.0 and the operating lines of both sections of the column coincide with the 45Âș diagonol line, as shown in Fig below. Minimum number of trays can be obtained by returning all the overhead condensed vapor V1 from the top of the tower back to the tower as reflux, i.e., total reflux. Also, the liquid in the bottoms is reboiled. (16)
  • 59. 59 Minimum number of theoretical steps Nm when a total condenser is used (α is constant). av w w D D m x x x x N αlog 1 1 log ïŁ·ïŁ· ïŁž ïŁ¶ ïŁŹïŁŹ ïŁ­ ïŁ« − − = ( ) 2/1 1 wav ααα =For small variations in α, where α1 is the relative volatility of the overhead vapor αw is the relative volatility of the bottoms liquid.
  • 60. 60 B) Minimum reflux ratio The minimum reflux ratio (Rm) will require an infinite number of trays for the given separation desired of xD and xW. If R is decreased, the slope of the enriching operating line R/(R+1) is decreased, and the intersection of this line and the stripping line with the q line moves farther from the 45Âș line and closer to the equilibrium line. xx yx R R D D m m â€Č− â€Č− = +1 Two operating lines touch the equilibrium line (“pinch point”) at y’ and x’ (number of steps required becomes infinite). The line passes through the points x’, y’ and xD (y=xD):
  • 61. 61 C) Operating and optimum reflux ratio Total reflux = number of plates is a minimum, but the tower diameter is infinite. This corresponds to an infinite cost of tower and steam and cooling water. This is the limit in the tower operation. Minimum reflux = number of trays is infinite, which again gives an infinite cost. These are the two limits in operation of the tower. Actual operating reflux ratio to use is in between these two limits. The optimum reflux ratio to use for lowest total cost per year is between the minimum Rm and total reflux (1.2Rm to 1.5Rm).
  • 62. General Design Consideration 1. A tower design is normally divided into two main steps, a process design followed by a mechanical design. The purpose of the process design is to calculate the number of required theoretical stages, column diameter and tower height. On the other hand, the mechanical design focuses on the tower internals and heat exchanger arrangements. 2. Many factors have to be considered in designing a distillation column such as the safety and environmental requirements, column performance, economics of the design and other parameters, which may constrain the work.
  • 63. The first step in distillation column design is to determine the separation sequences, which depends on the relative volatility and concentration of each component in the feed. King has outlined a few design rules as follows: 1) Direct sequences that remove the components one by one in the distillate are generally favored. 2) Sequences that result in a more equal-molar division of the feed between distillate and bottoms products should be favored. 3) Separations where the relative volatility of two adjacent components is close to unity should be performed in the absence of other components; ie, reserve such a separation until the last column in the sequence. 4) Separations involving high-specified recovery fractions should be reserved until last in the sequence.
  • 64. Once the separation sequence is decided, engineering calculations follow to determine the number of theoretical stages, operating parameters and tower dimensions. In general, the steps included in distillation calculations are summarized into the following: 1) Performing a material balance for the column 2) Determining the tower operating pressure (and/or temperature) 3) Calculating the minimum number of theoretical stages using the Fenske equation 4) Calculating the minimum reflux rate using the Underwood equations 5) Determining the operating reflux rate and number of theoretical stages 6) Selection of column internals (tray or packings) 7) Calculating the tower diameter and height
  • 65. Some general design rules (from Cheresources.com) that should be considered are as follows: 1) Distillation is usually the most economical method of separating liquids. 2) For Ideal mixtures (low pressure, medium temperature, and non- polar), relative volatility is the ratio of vapor pressures i.e. α = P2/P1 3) Tower operating pressure is determined most often by the temperature of the available cooling medium in the condenser or by the maximum allowable reboiler temperature. 4) Tower Sequencing : A. Easiest separation first – least trays and reflux B. When neither relative volatility nor feed concentrations vary widely, remove components one by one as overhead products. C. When the adjacent ordered components in the feed vary widely in relative volatility, sequence the splits in order of decreasing volatility. D. When the concentration in the feed varies widely but the relative volatilities do not, remove the components in the order of decreasing concentration in the feed.
  • 66. 5) Economically optimum reflux ratio is about 120% to 150% of the minimum reflux ratio. 6) The economically optimum number of stages is about 200% of the minimum value. 7) A safety factor of at least 10% above the number of stages by the best method is advisable. 8) A safety factor of at least 25% about the reflux should be utilized for the reflux pumps. 9) Reflux drums are almost always horizontally mounted and designed for a 5 min holdup at half of the drum's capacity. 10) For towers that are at least 3 ft (0.9 m) in diameter, 4 ft (1.2 m) should be added to the top for vapor release and 6 ft (1.8 m) should be added to the bottom to account for the liquid level and reboiler return. 11) Limit tower heights to 175 ft (53 m) due to wind load and foundation considerations. 12) The Length/Diameter ratio of a tower should be no more than 30 and preferably below 20.
  • 67. 13) A rough estimate of reboiler duty as a function of tower diameter is given by: Q = 0.5 D2 for pressure distillation Q = 0.3 D2 for atmospheric distillation Q = 0.15 D2 for vacuum distillation Where, Q : Energy in Million Btu/hr D : Tower diameter in feet.
  • 68. The Selection of Column Internals The selection of column internals has a big impact on the column performance and the maintenance cost of a distillation tower. There are several choices of column internals and the two major categories are trays and packing. The choice of which to utilize depends on the 1) pressure, 2) fouling potential, 3) liquid to vapor density ratio, 4) liquid loading, and 5) most importantly the life cycle cost. Trays can be divided into many categories, such as baffle trays, dual flow trays, conventional trays, high capacity trays, multiple down comer trays and system limit trays. According to some rules of thumb, trays should be selected if: 1) the compounds contain solids or foulants 2) there are many internal transitions 3) liquid loads are high 4) there is a lack of experience in the service 5) vessel wall needs periodic inspection 6) there are multiple liquid phases
  • 69. On the other hand, packing divisions include grid packing, random packing, conventional structured packing, and high capacity structured packing. The rules of thumb for selecting packing are: 1) the compounds are temperature sensitive 2) pressure drop is important (vacuum service) 3) liquid loads are low 4) towers are small in diameter 5) highly corrosive service (use plastic or carbon) 6) the system is foaming 7) the ratio of tower diameter to random packing is greater than 10
  • 70. Some design guidelines should be considered when designing a tray tower, such as follows: 1) Tray spacing should be from 18 to 24 inches, with accessibility in mind (Generally, for a tower diameter of 4 feet and above, the most common tray spacing is 24 inches to allow easy access for maintenance. However, for a tower diameter below 4 feet, a tray spacing of 18 inches is adequate as the column wall can be reached from the man way.) 2) Peak tray efficiencies usually occur at linear vapor velocities of 2 ft/s (0.6 m/s) at moderate pressures, or 6 ft/s (1.8 m/s) under vacuum conditions. 3) A typical pressure drop per tray is 0.1 psi (0.007 bar) 4) Tray efficiencies for aqueous solutions are usually in the range of 60- 90% while gas absorption and stripping typically have efficiencies closer to 10-20% 5) Sieve tray holes are 0.25 to 0.50 in. diameter with the total hole area being about 10% of the total active tray area. Maximum efficiency is 0.5 in and 8%. 6) Valve trays typically have 1.5 in. diameter holes each with a lifting cap. 12-14 caps/square foot of tray is a good benchmark. 7) The most common weir heights are 2 and 3 in and the weir length is typically 75% of the tray diameter.
  • 71. The packed tower design concepts are listed below: 1) Packed towers almost always have lower pressure drop compared to tray towers. 2) Packing is often retrofitted into existing tray towers to increase capacity or separation. 3) For gas flow rates of 500 ft3/min (14.2 m3/min), use 1 in (2.5 cm) packing, for gas flows of 2000 ft3/min (56.6 m3/min) or more, use 2 in (5 cm) packing. 4) Ratio of tower diameter to packing diameter should usually be at least 15 5) Due to the possibility of deformation, plastic packing should be limited to an unsupported depth of 10-15 ft (3-4 m) while metal packing can withstand 20-25 ft (6-7.6 m). 6) Liquid distributor should be placed every 5-10 tower diameters (along t he length) for pall rings and every 20 ft (6.5 m) for other types of random packing. 7) For redistribution, there should be 8-12 streams per sq. foot of tower area for towers larger than three feet in diameter. They should be even more numerous in smaller towers. 8) Packed columns should operate near 70% flooding.
  • 72. 9) Height Equivalent to Theoretical Stage (HETS) for vapor-liquid contacting is 1.3- 1.8 ft (0.4-0.56 m) for 1 in pall rings and 2.5-3.0 ft (0.76-0.90 m) for 2 in pall rings. 10) Design pressure drops should be as follows: Table 1: Pressure drop in difference services