CH302L

Distillation (Part-1)
Objectives

 Sketch a continuous distillation column and identify various components

 Label the liquid and vapor flowrates and composition in a continuous distillation column
 Write mole balances over the entire distillation column and identify degrees of freedom
 Write the mole balances in the rectifying and stripping sections of a distillation column
 Define the reflux ratio
 Explain when constant molal overflow condition is valid
 Define q for the feed
 Given a reflux ratio, feed condition, bottom and distillate mole-fractions, calculate the
number of ideal stages using McCabe-Thiele method and identify the location of the feed
tray
 Calculate the heating and cooling requirements in the reboiler and condenser for a
continuous distillation column.

Summary

 Material balances for rectifying section 21.10

= − = − 21.11 (25)
= − = − 21.12 (26)
 Material balances for stripping section
= − = − 21.13 (27)
= − = − 21.14 (28)
 Operating lines

= + 21.17 (29)
+ +
= − 21.20 (30)
− −
 Reflux Ratio
moles of liquid returned to the column 21.21 (31)
= =
moles of product withdrawn
 For constant molal overflow
= = =. . = = =⋯=
= = =. . = = =⋯=
= ..= = =⋯=
= ..= = =⋯=

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The operating lines are

= +
+1 +1
21.22 (32)
= − 21.26 (33)
− −

 Cold-liquid feed
( − ) 21.27 (34)
=1+
 Superheated vapor
21.28 (35)
( − )
=−
 Mixed-feed
− (36)
=1− =
−
where and are the equilibrium liquid and vapor mole-fractions at
the temperature of the feed, TF
 Relation between the flows in rectifying and stripping section
= + 21.29 (37)
= + (1 − ) (38)
 Equation of feed line: =− + 21.34 (39)

Feed Condition Feed Temperature q-value

Cold feed < > 1, (34)
Feed at bubble point = =1
Two-phase feed < < 0 < < 1, (36)
Feed at dew point = =0
Superheated vapor > < 0, (35)

 Heat required in the reboiler =

= latent heat of mixture in the reboiler 21.35 (40)
If saturated steam is used as heating medium and all the steam used for
heating is returned from reboiler as liquid water at saturation, then
Heat supplied by steam = ̇
21.35 (41)
 Cooling required in the condenser = 21.36 (41)
= latent heat of mixture in the condenser
Heat removed by cooling water = ̇ ( − ) (42)

Reading Assignment: pp 666-686 (Chapter 21)

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CH302L

Distillation (Part-2)
Objectives

 Explain the concept(s) of minimum reflux ratio and minimum number of stages
 Calculate the minimum reflux ratio and minimum number of stages for a binary distillation
 Explain how reflux ratio affects the fixed and operating costs of a distillation column
 Identify the condition when reboiler and condenser heat duties will be equal
 Solve material and energy balances around an equilibrium stage in a distillation when
constant molar overflow conditions do not apply
 Define overall efficiency of a distillation column
 Define Murphree tray efficiency of a distillation column
 Calculate the total number of actual stages required for a binary distillation, given the
Murphree tray efficiency
 Solve a binary distillation problem using DSTWU in ASPEN
 Solve a binary distillation problem using RADFRAC in ASPEN

Summary

 If the pinch point is at ( , ) then the minimum reflux ratio is

−
=
− 21.47 (43)
 Total enthalpy balance for the column
+ = + + 21.48 (44)
 Overall efficiency
number of ideal plates needed for a desired separation
=
number of actual plates needed for the same separation
 Murphree efficiency, : If ∗ denotes the vapor composition in 21.69 (45)
equilibrium with , and are the vapor compositions
leaving and entering plate n respectively, then
−
= ∗
−

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 The effective equilibrium curve is using Murphree tray efficiency is

calculated by
= + ( − ) 21.72 (46)
: ordinate value at a desired x on the effective equilibrium curve
: ordinate value at x on the actual equilibrium curve
: ordinate value at x on the operating curve

Reading Assignment: pp 687-701, 712-717 (Chapter 21)

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CH302L

Multi-component Distillation Shortcut Method

 The Fenske equation used in the shortcut method assumes that
relative volatility is constant and does not change with temperature.
So, the choice of temperature does not matter; however we will use
the feed temperature to calculate K-factors, identify the HK and LK
and also calculate . Another approach is to calculate at top, feed
and bottom temperatures and take a geometric mean. For now we will
simply use the feed temperature. Fill in the Table given on the next
page. Arrange in increasing order of K-values (or or their boiling
points).

 Identify LK__________ and HK____________.

Calculate the relative volatility of all components w.r.t HK i.e. ,

 Choose a basis for the calculations.

Say, feed flow in F = ____________kmol/hr

 Based on this feed flow and recoveries given calculate the number of
moles of LK and HK in the distillate ( , & , ) and bottoms
( , & , ). Fill in the table.

 At total reflux, calculate the minimum number of plates excluding the

reboiler using the suitable form of Fenske equation, by choosing i=LK
and j=HK. If mole-fractions are known
, / , / , / ,
= −1
,
(22.13 A) (47)
Alternately, if recoveries (and hence moles are known)
, / , / , / ,
= −1
, (22.13 B) (48)
Choose i to be LK and j to be HK. For, example 22.13 B becomes
, / , / , / ,
= −1 (22.13 C
,
(22.13 C) (49)

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 For all non-key components (NK) the Fenske equation still applies.
However, we will use it to calculate the moles of the component
, and , , by choosing j = HK.
, / , / , / , (22.13 D) (50)
= −1
,

 A mass balance for component i yields

(51)
, + , = ,
 Eqs. (50) and (51) can be solved to get , and ,
 Repeat this step for all non-key components in the stream. Add and
find, total flow rates of D and B. Finally, calculate the mole-fractions
of each component in the D and B streams.

 Assuming, same relative volatility in the upper and lower sections,

constant relative volatility and constant molal overflow, one gets the
Underwood equation.
,
1− =∑ (22.29) (52)

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The summation is over all the components. The value of is to be

determined from the above equation based on the feed condition, q. If
there are multiple roots, the root for the equation between and
must be used.

 Calculate the minimum reflux ratio using,

,
, +1=∑ (22.30) (53)
This summation is performed only over the components appearing in
the distillate (the RHS term for other components will go to zero
anyway!).
 The number of trays at desired reflux are then calculated using the
Gilliland correlation (empirical, can be grossly erroneous for non-ideal
systems that deviate from Raoult’s law significantly.

Fig 22.5 from Unit

Operations, McCabe,
Smith and Harriott, ©
McGraw Hill. These
materials are made
available through fair use
act and further copying
and redistributing the
material is a violation of
the copyright law

 The location of the feed tray is estimated from Kirkbride’s correlation.

1+ (54)
=
1+

.
where = , ,
(55)
, ,

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Distillation in a Packed Column

 Obtain N the number of equilibrium stages, location of feed plate and

average liquid and vapor L and V on each tray considering as if it
were a tray tower.
 Obtain HTU or from suitable correlations at the given conditions
for each tray.
ln (56)
= × ×
−1

(57)
= =

(58)
Note that, when the operating and equilibrium lines are parallel only then the
height of the transfer unit (HTU) HOy is same as the height equivalent to one
theoretical plate (HETP). When this is true then, number of transfer units
(NTU) will also be equal to the number of equilibrium stages.

Proof
Note that for distillation the mass transfer is from liquid to the vapor phase,
so the equilibrium line is above the operating line unlike the case of absorption
(where it is below). The driving forces accordingly are ∗ − for the vapor
phase and − ∗ for the liquid phase.

When the lines are linear, from Eq 20.22, the number of equilibrium stages
for distillation is
∗
/( ∗ )
= ( / )
.
On the other hand, analogous to Eq. (18.19) for distillation = ,
∗ ( ∗ )
where Δ = ∗ .
/( ∗ )

From these two relations, =− ∗ ( ∗ )

ln = ∗ ∗ =

∗ ∗ = =

= × = × × = × ⇒ = ×

Calculate for each equilibrium stage and add for all the tray to get total
packed height for the column.

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Batch Distillation
Consider a batch distillation unit as shown below. In a time dt, a distillate of an average
composition y (for the more volatile component) be collected. During this time let the amount
amount of liquid evaporated from the still be dn moles (this will show up as distillate).

A mole balance for more volatile component yields,

= ( )= +
Rearranging the terms and integrating between intial and final number of
moles (n0 & n1) and initial and final concentrations (x0 & x1) we get the
Rayleigh equation
ln =
− (59)

The integral is evaluated (by analytical, graphical or numerical means) using

the equilibrium data.

For the special case of constant relative volatility, the above equation reduces
to
/
= (60)

/
where is the relative volatility of A w.r.t. B. i.e. = at
/

equilibrium. The , are moles of the two components A & B initially

present in the still, while , are the amounts in the still after certain time
has elapsed.

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CH302L

Distillation (Part-3)
Objectives

 Calculate the vapor velocity and hence the diameter of a distillation column
 Calculate the pressure drop and downcomer liquid level
 Choose optimal tray spacing
 Use ASPEN for the above design

 For a given tray spacing find and Kv from Fig 21.26

, are mass flow rates in , , are densities in and

is the surface tension in dyne/cm . Use the lower value; if
unavailable for most organic liquids it is 20; for water it is 72
dyne/cm. is maximum permissible velocity in ft/s.
21.68 (61)
− .
=
20

 Find bubbling area in

Q is the vapor velocity in and is in (multiply with 0.3048
to convert from ft/s to m/s).
(62)
=

 Calculate column area in

(63)
=
fraction of bubbling area
The fraction of bubbling area is usually between 70-90 % of column
area. It accounts for area occupied by downcomers on both the sides.

Shaded area: bubbling area

Unshaded area: occupied by two downcomers

 Column diameter in meters

4
= (64)
π

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 Fractional area occupied by the holes (with respect to bubbling area)

 In triangular pitch
(65)
fractional area occupied by the holes =
2√3
 In square pitch

fractional area occupied by the holes = (66)

4

 Velocity through the holes

= (67)
fractional area occupied by the holes

 Use Fig 21.25 to get the orifice coefficient . The abscissa is

 (68)
ℎ
= fractional area occupied by the holes × fraction of bubbling area
column area

 Pressure drop through the holes, ℎ in mm of liquid. is in m/s

2
ℎ = 51 2
21.62 (69)

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 Length of the weir (Refer Perry’s handbook, Section 1.26, 6th Ed.)

Fraction of bubbling area

0.70 0.8110
0.72 0.7966
0.74 0.7819
0.76 0.7644
0.78 0.7466
0.80 0.7286
0.82 0.7055
0.84 0.6826
0.86 0.6578
0.88 0.6264
0.90 0.5933

 Height of liquid over the weir ℎ in mm

is the liquid flow in / , is in m
/
21.64 (70)
ℎ = 43.4

 Liquid holdup on the weir in mm

ℎ is the height of the weir
ℎ = (ℎ + ℎ ) 21.63 (71)

Use = 0.6, if data is unavailable

 The total pressure drop per plate in mm
ℎ = ℎ +ℎ 21.61 (72)
 The total height of clear liquid in the downcomer is
= 2 (ℎ + ℎ ) + ℎ + ℎ , 21.66 (73)
For frictional loss use ℎ , = 10 , if data is unavailable
 The actual level of aerated liquid in the downcomer is

= 21.67 (74)

For average volume fraction of liquid use = 0.5 , for conservative

design.
 Tray spacing should be MORE than Z

Reading Assignment: pp 703-711

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