Training Services

Binary Fractionation

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 Binary Fractionation

n Heat and Material Balance

n Equilibrium
n Concept of Equimolar Overflow
n McCabe - Thiele Graphical Method
n Analytical Methods
n Tray Efficiency

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 BF-R00-01
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 Binary Distillation

ENVELOPE (1) - Overall Material and Heat Balance

Mass Balance F = D+ B
X iF F = X iD D + X iB B
(i = 1 to N components)

Heat Balance hF F + QR = hD D + hB B + Qc

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 Binary Distillation

ENVELOPE (2) - Rectifying Section

Mass Balance V n + 1 = Ln + D
Yin + 1V n + 1 = X in Ln + X iD D

Heat Balance hvn + 1V n + 1 = hLn Ln + hD D + Qc

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 Binary Distillation

ENVELOPE (3) - Stripping Section

Mass Balance Ln' = V n + 1' + B

xin Ln' = yin + 1V n + 1' + X iB B

Heat Balance QR + hL Ln ' = hvV n + 1' + hb B

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 Binary Distillation

ENVELOPE (4) - A single tray

Mass Balance V n + 1 + Lo = V 1 + Ln

Heat Balance hvn + 1V n + 1 + hLo Lo = hv1V 1 + hLn Ln

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 Binary Distillation

ENVELOPE (4)
Rearranging the mass balance yields

V n + 1 = V 1 + Ln − Lo
Inserting this into the heat balance

hVn + 1V 1 + hVn + 1 Ln − hVn + 1Lo + hLo Lo = hV1 V 1 + hLn Ln

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 Binary Distillation

ASSUME
Sensible Heat Changes are Negligible

i) hLo = hL1 = hLn = hL

Molal Heats of Vaporization are Constant

ii) λ0 = λ1 = ... = λn
n n n
Since λ = hV − hL

iii) hVo = hV1 = ... = hVn = hV

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 Binary Distillation

Heat Balance

hV V 1 + hV Ln − hV Lo + hL Lo = hV V 1 + hL Ln

(hv − hL )Ln = (hv − hL )Lo

Therefore Ln = Lo = Constant

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 Binary Distillation
Constant Molal Overflow

Ln = Constant

n Sensible Heat Changes are Negligible

n Molal Heats of Vaporization are Constant

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 Binary Distillation
Constant Molal Overflow
VALIDITY?

n Boiling Point Range of Components is Narrow

n Molecular Characteristics of Components are

Similar
– For example: all paraffinic hydrocarbons or all
aromatic hydrocarbons not mixture of
paraffins and aromatics

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 Binary Distillation

ENVELOPE 2 - Rectifying Section

Mass Balance yin + 1V n + 1 = xin Ln + xid D

Equimolal Overflow
L = Constant
V = Constant
yin + 1 = ( L V )xin + ( D V )xiD

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 Binary Distillation

BINARY Separations
xi x1, x2

x 2 = 1 - x1

Used convention of dropping subscript i

x = x1 Mole fraction of more volatile

y = y1 (lower boiling point) component

yn+1 = (L/V) xn + (D/V) xD Trays above feed

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 Binary Distillation

ENVELOPE 3 - Stripping Section

Mass Balance xin Ln = yin + 1V n + 1' + xiB B

Equimolal Overflow
L' = Constant ≠ L
V ' = Constant ≠ V

yn+1 = (L' /V' ) xn + (B/V ' ) xB Trays below feed

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 Binary Distillation

L
V

FV
F V' L
FL
L'
V'

BF-R00-02
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 Binary Distillation

Feed Stage

Mass Balance F + V ' + L = V + L'

Heat Balance hF F + h'V V ' + hL L = hV V + h'L L'

V = L+ D

V ' = L' − B
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 Binary Distillation

ASSUME h'V − hV and h'L = hL

Then hF F + hV ( L' − B )+ hL L = hV ( L + D )+ hL L'

Rearranging (hV − hL )L' = (hV − hL )L + hV ( D + B )− hF F

Since D + B = F

hV − hF 
Then L' = L + 
h − h  F
 V L

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 Binary Distillation

hFV − hF
Define: q= V
hF − hFL

Then: L' = L + q F

n+ 1  q  n  1 
Also: y = x −  xF At Feed Tray
(q − 1) (q − 1)

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 Q Values

Condition Value of q
BP Liquid 1.0
DP Vapor 0
Sub Cooled Liquid >1.0
Superheated Vapor <0
Partly Vapor Mol Frac. Liquid

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 Equilibrium K Value Definition

y
K=
x

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 T-x Diagram
Dew Point Curve,
Saturated Vapor

T3

Bubble Point Curve,

T2 Saturated Liquid

T1
Temperature

x3 x2 x1 y3 y2 y1

Mole Fraction (x or y)
Vapor or Liquid Phase

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 Binary Distillation

EQUILIBRIUM

y = K 1x

(1 − y )= K 2 (1 − x )

1− K 2
x=
( K1 − K 2 )

y = K1 x

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 Binary Distillation

EQUILIBRIUM
y = K 1x
(1 − y )= K 2 (1 − x ) Define
K K1
α=
=
y 1 x
K2
( 1− y) K (1 − x )
2

α x
y=
1 + (α − 1)x

y
x=
K1
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 Equilibrium Curve or x-y Diagram
α x
y= Equilibrium Curves
1 + (α − 1)x
α =5
0.9
α = 2. 5
0.8 α = 1. 5
0.7 α =1
0.6

y, composition in 0.5
the vapor phase 45o line
0.4

0.3

0.2

0.1

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

x, composition in the liquid phase

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 Binary System: Propane/Isobutane

n Specified
– Feed = 200 mol/h C3 + 200 mol/h iC4 at bubble
point
– Distillate = 196 mol/h
– Reflux = 400 mol/h at 100ºF
– Column pressure = 250 psia
– Number of trays = 24, feed on 13
n Problem
– Find xD and xB
464
– L = 1.16 R = 464 mol/h L F= = 1.16
400

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 Binary System: Propane/Isobutane

Rigorous Heat and Weight Equimolal Overflow

Balanced Tray-to-Tray Simplified Tray-to-Tray
Method Method

xD = 0.927 0.917
xB = 0.090 0.098

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 Internal vs. External Reflux

BF-R00-03
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 Internal vs. External Reflux

hV 2 V2 + hR R = hV 1 V1 + hL1 L1
V1 = R + D
V2 = L1 + D

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 Internal vs. External Reflux

ASSUME: hV 1 = hV 2

(hV − hR ) hV = saturated vapor enthalpy @ π

L= R hL = saturated liquid enthalpy @ π
(hV − hL ) hR = liquid enthalpy @ T

Receiver Temp = 100ºF

Receiver Press = 250 psia

Assume: Receiver liquid is 100% propane

hV = 168 Btu/lb; hL = 47 Btu/lb; hR = 27 Btu/lb
L = 1.16 R

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 Binary System: Propane/Benzene

n Specified
– Feed = 50 mol/h C3 + 50 mol/h Bz at bubble point
– Distillate = 50 mol/h
– xD = 0.99
– Column pressure = 215 psia
– Number of trays = 10, Feed on 6
n Problem
– Find reflux rate (temp = 100ºF)

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 Binary System: Propane/Benzene

Equimolal Overflow Rigorous Heat and Weight

Simplified Tray-to-Tray Balance Tray-to-Tray
Method Method

R = 5 Mol/hr R = 14 Mol/h !!

If 5 Mol/h, xD = 0.967

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 Equilibrium
Pressure Constant

T y

x, y x
Ideal Vapor/Liquid Equilibria:
Systems that conform to Raoult's Law
(i.e. p* = P vx, ∴ α = Pv1 = constant)
P v2
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 Equilibrium
Pressure Constant

T y

x, y x
Large Deviation from Ideality:
e.g. Minimum boiling azeotrope

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 Water K Values

Y
K=
X
P o (1 − X HW )
Y=
π
and
X = X wh

(1 − X hw )P o
Kw =
X whπ

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 Water Equilibrium Curve

α X
Y=
1 + (α − 1)X

As X goes to 0
Y=αX
and, therefore: ln(Y )= ln(α )+ ln( X )

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 Summary of Equations

Equilibrium curve

n(1 − K 2 )
x = , y n = K 1 xn
(K 1 − K 2 )
or

n α xn
y = α = K1 K 2
1 + (α − 1)x n

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 Summary of Equations

Rectifying Section - upper operating line

y n + 1 = ( L V )x n + ( D V )xD V = L+ D

Stripping Section - lower operating line

y n + 1 = ( L' V ' )x n − ( B V ' )xB V = L− B

Feed Tray - q-line

hVF − hF
y n+ 1 = q (q − 1)x n − 1 (q − 1)xF q=
λF

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 McCabe-Thiele

BF-R00-04
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 Problem

n Binary System: Propane-Normal Butane

– System pressure = 200 psia
– Feed = 50 mol/h C3 + 50 mol/h nC4 at bubble
point
– Desired purities top and Bottom
xD = 0.95 xB = 0.05
– Reflux rate (internal)
R = L = 100 mol/h

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 Problem

Question: How Many Trays?

x F = 50 100 = 0.50

x F F = xD D + xB B

B=F− D

(0.5)(100 )= (0.95)D + 0.05(100 − D )→ D = 50 mol h

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 Propane - Normal Butane
π= 200 psia

ºF K1 K2 α x1 y1
110 1.058 0.4098 2.58 0.911 0.963
120 1.151 0.4622 2.49 0.781 0.899
130 1.249 0.5180 2.41 0.659 0.824
140 1.350 0.5769 2.34 0.547 0.739
150 1.456 0.6389 2.28 0.442 0.643
160 1.566 0.7036 2.23 0.344 0.538
170 1.681 0.7710 2.18 0.252 0.423
180 1.801 0.8408 2.14 0.166 0.299
190 1.925 0.9130 2.11 0.086 0.165
200 2.051 0.9875 2.08 0.102 0.024

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 Rectifying Section

y n + 1 = ( L V )x n + ( D V )x D , V = L+ D

Note: When xn = xD

 L D 
y n+ 1 =  + xD = xD
L + D L + D 

can plot equation as a straight line with slope equal to

L/(L + D) that passes through the point (xD, xD)

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 Rectifying Section

For This Problem

L = 100 Mol/h

D = 50 Mol/h

L L
Slope = =
V L+ D

100
Slope = = 0.67
100 + 50

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 Feed Q Line

n+ 1 q n 1 hVF − hF
y = x − xF q= V
(q − 1) (q − 1) hF − hFL
Note: When xn = xF

n+ 1  q 1 
y = − xF = x F
q − 1 q − 1
can plot equation as a straight line with slope equal to
q/(q-1) that passes through the point (xF, xF)

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 Feed Q Line

For This Problem

hF = hFL (bubble point liquid feed)

Therefore,

q =1
1
q Line Slope = =∞
1− 1

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 McCabe Thiele

BF-R00-05
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 McCabe Thiele
Effect of Feed Enthalpy

BF-R00-14
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 McCabe Thiele

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 McCabe Thiele

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 McCabe-Thiele
Limiting Conditions

n Total reflux - minimum trays

– maximum separation but no feed or products

n Minimum reflux - infinite trays, minimum duty

– pinch points
– adjusting feed location and condition
– operating reflux and duty

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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 McCabe Thiele
Minimum Trays

BF-R00-15
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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 McCabe Thiele

BF-R00-11
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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 McCabe Thiele
Minimum Reflux

n At minimum, the slope of the upper operating

line is:
– slope = L/V
– slope = (xD - yF*) / (xD - xF)
– where yF and xF are the compositions where
the q-line meets the equilibrium line
– (R/D)min = L/V / (1 - L/V)

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 More Reflux or More Trays

Reflux

Rmin

Nmin Trays

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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 McCabe Thiele

BF-R00-12
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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 McCabe Thiele

BF-R00-13
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 Minimum Trays

Fenske Equation

Log (rD rB )
nm =
Log α

XD
rD =
1− X D

XB
rB =
1− X B

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 Feed Tray Location

For Bubble Point Feed

R LOG (rD rF ) Fenske Equation

nm =
LOG α { rectifying section

Then
R
LOG (rD rF ) Assume ratio holds
nm
=
nm LOG (rD rB )
{ for any reflux ratio

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 Minimum Reflux

Underwood Method
1 − xF α xF
1) + = 1 − q,1 < φ< α
1− φ α − φ

(1 − xD ) α xD
2) (L D )min = + − 1
(1 − φ) (α − φ)
Solve equation 1 for the proper root of the
quadratic for φ.
Solve equation 2 for (L/D)min using the φfrom
equation 1.
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 Minimum Reflux

Note: If terms are multiplied out, equation 1

becomes

(1 − q )φ2 + (q + αq + α xF − α − xF )φ− qα = 0

A φ2 + B φ+ C = 0

− B ± B 2 − 4 AC
φ=
2A

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 Minimum Reflux

Underwood Method

If feed is bubble point liquid:

q =1

The Underwood equations reduce to

1 xD α (1 − xD )
(L D )m =  −
α − 1 xF (1 − xF ) 

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 Minimum Reflux

Note: For perfect separation

( xD = 1)
 
(L D )m =  1  1 
α − 1  xF 
Since xF = D F
Then x F F = xD D + xB B When
xF = D F xD = 1
Therefore 1
(L F )m =
α− 1
For perfect separation with bubble point feed.

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 Analytical Techniques
1.0
The Gillliland Relation
0.9 Trays and reflux as a function of their minimal
IEC, 1940, (p.1220)
0.8
N = Theoretical Plates (Design)
0.7 Nm = Min. Theor. Plates (L/D = ∞ )
R = L/D (Design)
Rm = L/D Min. (N = ∞ )
0.6
0.5
0.4
R - Rm
X= R+1
0.3 N - Nm
Y= N+1

0.2
0.1
0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
X
BF-R01-16
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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 Analytical Techniques

R − Rm Rm + X
X= R=
R+ 1 1− X

N − Nm Nm + Y
Y= N=
N+ 1 1− Y
X + Y are parameters, not compositions

Determine:
Rm from Underwood Equation
Nm from Fenske Equation

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 Example

XF = 0.5 rF = 1.0
XD = 0.927 rD = 12.7 D/F = 0.49
XB = 0.090 rB = 0.0989
α = 1.76 q = 1.0

Given n = 24, calculate needed reflux

UNDERWOOD

(L D )m = 1 X D − α (1 − X D )
 
α − 1 X F (1 − X F ) 
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 Example

FENSKE
LOG (rD rB )
nm = = 8.59
LOGα

R
nm LOG (rD rF )
= = 0.523
nm LOG (rD rB )

n (real in rectifying section) = 0.523 x 24 = 12.6

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distributed for any purpose whatsoever except by written permission of UOP LLC and except as authorized under agreements with UOP LLC.
 Example

GILLILAND

N − Nm
Y= = 0.616 → X = 0.055
N+ 1

( L D )m + X
L D= = 2.28 → R F = 1.12
1− X

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 Example

N = 24 L/D = 2.28

nm = 8.59 L/Dm = 2.10

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 Smoker’s Equation

n Simplified form

ln (S )
N=
 R+ q 
ln α 1 − 
 ( R + 1)( RxF + q) 

n where S is a separation parameter

S = ( xLK / xHK ) D ( xHK / xLK ) B

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 Equilibrium Stage

n How most methods see it

v n lin− 1
i

V nhvn Ln− 1 hln− 1

Stage n

vin+ 1 lin

V n+ 1 hvn+ 1 Ln hln

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 Conventional Tray

Vn? Vn?
Ln-1

Ln

Vn+1

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 Efficiency

n Deviation from ideal

efficiency = N ideal / N actual

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 Typical Observed Tray Efficiency
(Simplistic Overall)

Xylene Isomer Fractionators 80-100

Deisopentanizers/Deisobutanizers 80-100
Benzene/Toluene/Xylene 75-80
Depropanizers/Debutanizers 75-80
Naphtha Fractionators 65-85
High Pressure Deethanizers 50-60
Low Pressure Drop Columns 40-60
Distillation Dryers 15
Gas Strippers 7-10
Gas Con Absorbers
Primary 30-35
Sponge 20-25
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 General Guide for Tray Efficiency
as Function of α Alone

Alpha Tray Efficiency

1.2 90
2.0 70
3.0 50
5.0 20
15.0 10

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 Tray Efficiency

Glitsch Bulletin 4900 Fifth Edition

BF-R00-17
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 Designing a Column

n Define Feed
n Define Product Specification
n Set Column Pressure
n Optimize Column Design
n Set Composition Control

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 Define Feed

n Composition
n Flow Rate
n Temperature
n Pressure
n Enthalpy

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 Define Product Specifications

n Receiver Temperature

n Top Tray Vapor Temperature

n Product Purities and Recoveries

n Zero Purity Spec is Not Acceptable

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 Set Column Pressure

n Maximize Alpha Value

n Minimize Column Cost
n Keep Flare Material Out of Overhead
n Totally Condense Overhead Products
n Prevent Need for Net Gas Compression

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 Set Column Pressure

n Minimize Net Overhead Vapor

n Use Condenser as Heat Source
n Use Bottoms as Hot Oil
n Limit Bottom Temperature
– Cracking
– Polymerization
– Approach to critical

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 Optimize Column Design

n Best Trays vs. Reboiler

n Best Feed Tray Location

n Feed Preheat

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 Optimum Trays
Trays vs. Reboiler Duty
15
Reboiler Duty

10

0
0 10 20 30 40 50 60
Number of Theoretical Stages
BF-R01-19
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 Best Feed Tray
38 Total Stages
15

10

0
0 10 20 30 40
Feed Stage BF-R01-19
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 Feed Preheat Efficiency
60

50

40

30

20

10

0
0 10 20 30 40 50 60
Preheat Duty, MBtu/h
BF-R00-20
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