Chapter 21Chapter 21 Performance Curves for Individual
i O iUnit Operations(Distillation Columns and (Absorbers/Strippers)
Department of Chemical Engineering
West Virginia UniversityWest Virginia University
Copyright R.Turton and J.A. Shaeiwitz - 2012 1
OutlineOutline
• Distillation columnsDistillation columns– column pressure– scale‐up/downp/– reflux ratio– flooding/weepingg/ p g
• Absorbers/strippers– operationp– performance
Copyright R.Turton and J.A. Shaeiwitz - 2012 2
OutlineOutline
• Distillation columnsDistillation columns– column pressure– scale‐up/downp/– reflux ratio– flooding/weepingg/ p g
• Absorbers/strippers– operationp– performance
Copyright R.Turton and J.A. Shaeiwitz - 2012 3
Column PressureColumn Pressure
How do we choose the operating pressureHow do we choose the operating pressure of a column?
Usually based on:Usually based on:
(i) Temperature of overhead condenser
(ii) Temperature at bottom of column
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Column PressureColumn Pressure
Algorithm:Algorithm:
1. If possible – use cheapest coolant stream available – cooling water (or air) in the condenserg ( )
2. If by choosing (1) above, the bottom temperature is too high – then reduce pressure so that bottom g ptemperature is acceptable and choose warmest coolant (refrigerant) for condenser
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Column Pressure ‐ exampleColumn Pressure example
A depropanizer is a column that separates propane from n‐p p p p pbutane. At what pressure would you operate this column?
log P*(mmHg) = A ‐ B/(C+T)
Propane A = 6.80398 B = 803.810
C 246 990C = 246.990
n‐butane A = 6.80896 B = 935.860
C = 238 730C 238.730
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Column Pressure ‐ exampleColumn Pressure example
Depropanizer Column – Overhead Condenserp p
Top product is ≅ pure propaneT
40
50
PropaneQ
30
Propanelog P*(mmHg) = 6.80398 ‐ 803.810 /(246.990 + 50)
= 4.0975P* = 12 516 mmHg = 242 psiaP = 12,516 mmHg = 242 psia
What is pressure and temperature at bottom of the column?
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Column Pressure ‐ exampleColumn Pressure example
What does ΔPtray depend on?
Often this is the dominant term and ∴ΔPtray is not a f (operating
ΔPtray = ρlg(hw+ hcr) + kρgvo2
trayconditions)
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Column Pressure ‐ exampleColumn Pressure example
Depropanizer ColumnTop product is ≅ pure propane
ΔPcol = nΔPtray Pbot = Ptop + ΔPcol
Say ΔPcol = 5 psiPbot = 242 + 5 = 247 psia = 12,770 mmHgFor butane bottom product
Bottom product is ≅ pure butane
pA ‐ B/(C+T ) = log(12,770)T = B/[A‐ log(12,770)]‐C = 935.860/[6.80896 – log(12,770)] ‐ 238.730T = 107.5°CT 107.5 C
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Column Pressure ‐ exampleColumn Pressure ‐ example
Acrylic Acid Example
Wh i l i d ?Why is column operating under vacuum?
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Column Pressure ‐ exampleColumn Pressure ‐ example
Acrylic Acid Example
Why is column operating under vacuum?
Polymerization of acrylic acid above 90°C
Must reduce pressure so that P*(acrylic acid) = A‐B/(C+89°C)Must reduce pressure so that P (acrylic acid) = A‐B/(C+89 C)
What about ΔPcol?ΔPcol = Pbot – Ptop = 0.16 ‐ 0.07 = 0.09 barcol bot top
Number of trays, n = 31ΔPtray = 0.09/31 = 0.003 bar = 1.2” of acrylic acid
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Scale‐up/down of ColumnScale‐up/down of Column
What if we want to change the feed rate to the column?
If we want to keep all the purities the same then
V
DR = L/D
the reflux ratio must stay the same.
Look at scale-down
How do we adjust the condenser and reboiler to
F L = V - D
How do we adjust the condenser and reboiler to keep the reflux ratios the same? This will keep all the product purities the same
Top of columnB
V’ L’ = B + V’
F = B + DTop of column
QD = VλD=D(1+R)λD
Bottom of columnR’ = L’/B
Fxf = BxB + DxD
QD and QB scale linearly with F if purities all
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QB = V’λB = B(R’-1)λBwith F if purities all remain the same
Scale‐up/down of ColumnScale‐up/down of Column
( ) ( )[ ]1pl MRDTcmTAUQ +ΔΔ λ&
2,2, cwp mM
mM
&&==
( )( )
( )[ ]( )[ ]
1,1,2
1
2
1,
2,
1,
2,
1
2
1
2
1
2
11
11
11
oi
D
D
cwp
cwp
lm
lm
hhUU
RDMRD
Tcm
Tcm
TT
AA
UU
+=
++
=Δ
Δ=
Δ
Δ=
λλ
&V
DR = L/D
process
1,1, cwcw
p mM
mM
&&==
1,8.0
1,
1 11ocwi hMh
U +
F L = V - D
Reduce cw flowrate and
( )( )
( )[ ]( )[ ]
1 22,2,222 −′==
Δ= Bsteamlm MRBmTAUQ λλ&B
V’ L’ = B + V’Reduce cw flowrate and reduce steam pressure
( ) ( )[ ]
constant11
11
1
11
1,1,
1
2
11,1,111
=+
+=
−′Δ
i
oi
Bsteamlm
hh
hhUU
RBmTAUQ λλ&
R’ = L’/B process
1,1, oi hh
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Scale‐up/down of ColumnScale up/down of Column
V
DR = L/D
What if we do not adjust any of the utility flows while increasing/reducing the feed to the column?
Look again at scale-down case
F L = V - DF = B + D
Fxf = BxB + DxD
B
V’ L’ = B + V’ Top of columnQD = VλD=D(1+R)λD
B tt f l
R’ = L’/B
Bottom of columnQB = V’λB = B(R’-1)λB
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Scale‐up/down of ColumnScale up/down of Column
V
DR = L/D
What if we do not adjust any of the utility flows while increasing/reducing the feed to the column?
Look again at scale-down case
F L = V - DF = B + D
Fxf = BxB + DxD
Material balance control will adjust D + B to based on F
B
V’ L’ = B + V’ Top of columnQD = VλD=D(1+R)λD
B tt f l
Same cooling water flow and steam pressure causes same amount of
R’ = L’/B
Bottom of columnQB = V’λB = B(R’-1)λB
same amount of condensation and reboil –what happens to R and R’ ?
R and R’ will increase – what happens to the
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R and R will increase what happens to the separation?
Scale‐up/down of Column – McCabe‐h l lThiele Analysis
y Feed line slope = -q/(1-q)
What causes a high reflux ratio?
1.High purity products1.High purity products
2.Difficult separation
3.Impure feed
xD/(1+R)
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xxDxB xF
Scale‐up/down of Column – McCabe‐h l lThiele Analysis
y Feed line slope = -q/(1-q)Back to scale-
down analysis
What happens when reflux ratio increases but number of trays stays the same?
xD/(1+R)
Copyright R.Turton and J.A. Shaeiwitz - 2012 19
xxF xD⇒⇐ xB
Scale‐up/down of ColumnOperation
What other factors must we consider when column operations change?column operations change?
Capacity of the column to process more or less material is limited (bounded) by:
Flooding (scale-up)
Weeping (scale-down)
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Scale‐up/down of ColumnOperation
scale down
scale up
Taken from J D Seader and E J Henley Separation Process Principles John
Copyright R.Turton and J.A. Shaeiwitz - 2012 21
Taken from J.D. Seader and E.J. Henley, Separation Process Principles, John Wiley and Sons, NY 1998
Scale‐up/down of ColumnOperation
For Scale-up • must be aware of flooding limit
l d i i b t 75 85% f fl di l i d• normal design is between 75-85% of flooding, so scale-up window is small
• for vacuum operations – changes in column pressure have a large effect on floodingeffect on flooding
For Scale-down
• weeping becomes an issue at 30-40% of flooding depending on• weeping becomes an issue at 30 40% of flooding, depending on type of tray
• generally can avoid weeping problems by using higher than necessary condenser and reboiler duties
Copyright R.Turton and J.A. Shaeiwitz - 2012 22
necessary condenser and reboiler duties
OutlineOutline
• Distillation columnsDistillation columns– column pressure– scale‐up/downp/– reflux ratio– flooding/weepingg/ p g
• Absorbers/strippers– operationp– performance
Copyright R.Turton and J.A. Shaeiwitz - 2012 23
Stripper and Absorber OperationStripper and Absorber Operation
xA,in yA,out
GL For dilute systems, L and G do not change in the column ⇒ operating lines are straightTower can contain trays or packing –but the approach is very similar
For absorbers – we are interested in reducing the amount of A in the entering gas stream by transfer to the liquid
xA,out yA,in
For strippers – we are interested in reducing the amount of A in the incoming liquid stream by transfer to the gas
Copyright R.Turton and J.A. Shaeiwitz - 2012 24
the gas
Stripper and Absorber OperationStripper and Absorber Operation
yyA in
Absorptiony
Stripping
yA,inol
y
el
l
yA,out
ol
xxA,out
yA,out
xA,in
el: y = mx
xxA,in
yA,inxA,out
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Strippers and Absorber OperationStrippers and Absorber Operation
For dilute systems we have
Packed Beds – Colburn Equation (18.23) in text
* [1 1/ ], ,
*[1/ ]
1 [1/ ]
OGN AA in A out
A out A out
y y e AAy y
−− −=
−−
Packed Beds Colburn Equation (18.23) in text
Written in absorber form, , [ ]A out A outy y
* 1y y ATray Columns – Kremser Equation (18.22) in text
absorber form
A=(L/mG)
Slope of el, ,
* 1, ,
11
A out A outN
A in A out
y y Ay y A +
− −=
− −
Slope of el
L and G are the molar flowrates
Copyright R.Turton and J.A. Shaeiwitz - 2012 26
y*A,out = mxA,in
Stripper and Absorber OperationFrom Analysis Synthesis andFrom Analysis, Synthesis and Design of Chemical Processes, 2nd ed., Turton, Bailie, Whiting, and Shaeiwitz, Prentice-Hall, Upper Saddle River, NJ, 2003
Copyright R.Turton and J.A. Shaeiwitz -2012 27
Stripper and Absorber OperationFrom Analysis Synthesis andFrom Analysis, Synthesis and Design of Chemical Processes, 2nd ed., Turton, Bailie, Whiting, and Shaeiwitz, Prentice-Hall, Upper Saddle River, NJ, 2003
Copyright R.Turton and J.A. Shaeiwitz -2012 28
Performance of Strippers and Absorberspp
For dilute systems, we can use Figures 21.14 an 21.15 (or Eqs. 21.22 and 21.23) to solve most problems.
Example
An absorber with 10 stages is used to reduce the acetone concentration in 80 kmol/h of air from 0.03 to 0.0015. Theconcentration in 80 kmol/h of air from 0.03 to 0.0015. The acetone is absorbed into 50 kmol/h of pure water.
During a process upset, it is found that the water available for this service is reduced to 40 kmol/h.
What is the new outlet fraction of acetone in the air?
Copyright R.Turton and J.A. Shaeiwitz - 2012 29
What is the new outlet fraction of acetone in the air?
Performance of Strippers and Absorberspp
Example - Solution
Information given – drop the subscript A for the solute
L = 50 kmol/h xin = 0
G = 80 kmol/h yin = 0.03 yout = 0.0015
Find y*out= mxin = 0
∴Y ordinate for Fig 18.14 – Kremser Plot is
Y = (yout – y*out)/(yin – y*out) = (.0015 – 0)/(0.03 – 0) = 0.05
Locate intersection of Y and N = 10 on figure
Copyright R.Turton and J.A. Shaeiwitz - 2012 30
Stripper and Absorber OperationFrom Analysis Synthesis andFrom Analysis, Synthesis and Design of Chemical Processes, 2nd ed., Turton, Bailie, Whiting, and Shaeiwitz, Prentice-Hall, Upper Saddle River, NJ, 2003
Interpolating we getA = 1.11
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Performance of Strippers and Absorberspp
Example - Solution
Now look at new condition
L2 = 40 kmol/h
A1 = 1.11 = L1/mG1 = (50)/(m)(80) ⇒ m = (50)/(1.11)(80) = 0.563
A2 = L2/mG2 = (40)/(0.563)(80) = 0.888
Find intersection point of A2 = 0.888 and N = 10 on figure
Copyright R.Turton and J.A. Shaeiwitz - 2012 32
Stripper and Absorber OperationFrom Analysis Synthesis and
Y = 0.15
From Analysis, Synthesis and Design of Chemical Processes, 2nd ed., Turton, Bailie, Whiting, and Shaeiwitz, Prentice-Hall, Upper Saddle River, NJ, 2003
Copyright R.Turton and J.A. Shaeiwitz -2012 33
Performance of Strippers and Absorberspp
Example – Solution
Does not change because x remains at 0
Y = 0.15 = (yout – y*out)2/(yin – y*out)2 = (yout,2 – 0)/(0.03 – 0)
Does not change because xin remains at 0
Yout,2 = (0.15)(0.03) = 0.0045
(originally 0.0015 – up by a factor of 3!)
Copyright R.Turton and J.A. Shaeiwitz - 2012 34
Performance of Strippers and Absorberspp
Example – continued
What can we do to bring back the exit concentration of vapor g pto 0.0015 under the constraint that the water flowrateremains at 40 kmol/h?
Discuss and find all possible strategies
Copyright R.Turton and J.A. Shaeiwitz - 2012 35
Performance of Strippers and Absorberspp
Example – continued
Discuss and find all possible strategies
1. Reduce the gas flow to [(40)/(50)](80) = 64 kmol/h – A remains at 1.11 and we are at the same original operating point on Figure.
the problem is that this is the process stream so not a viable option!the problem is that this is the process stream, so not a viable option!
2. Change m to have the same value of A = 1.11 – m decreases to (0.563)(40/50) = 0.45
t t d Atwo ways to decrease A
(i) Change temperature
(ii) Change pressureyP = xp* ⇒ m = y/x = p*/P
T ↓ or P ↑
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( ) g p T ↓ or P ↑
Performance of Strippers and Absorberspp
Example – continued
Discuss and find all possible strategiesp g
3. Reduce yin – but this is the process variable we are trying to reduce so not an option
4. Increase the number of trays – could be costly and in this case it will not work – see graph – we prefer solution not requiring new/modified equipment
Copyright R.Turton and J.A. Shaeiwitz - 2012 37
Stripper and Absorber OperationFrom Analysis Synthesis andFrom Analysis, Synthesis and Design of Chemical Processes, 2nd ed., Turton, Bailie, Whiting, and Shaeiwitz, Prentice-Hall, Upper Saddle River, NJ, 2003
Copyright R.Turton and J.A. Shaeiwitz -2012 38
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