Chapter 21 performance of distillation columns and absorbers

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

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• Distillation columnsDistillation columns– column pressure– scale‐up/downp/– reflux ratio– flooding/weepingg/ p g

• Absorbers/strippers– operationp– performance


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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


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


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



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?



What does ΔPtray depend on?

Often this is the dominant term and ∴ΔPtray is not a f (operating

ΔPtray = ρlg(hw+ hcr) + kρgvo2

trayconditions)



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


Column Pressure ‐ exampleColumn Pressure ‐ example

Acrylic Acid Example

Wh i l i d ?Why is column operating under vacuum?


47°C and 0.07 bar

11

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


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


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

QQ

+=

++

=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


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

Copyright R.Turton and J.A. Shaeiwitz -2012 15

47°C and 0.07 bar

16

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


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)


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)


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)



scale down

scale up

Taken from J D Seader and E J Henley Separation Process Principles John


Taken from J.D. Seader and E.J. Henley, Separation Process Principles, John Wiley and Sons, NY 1998


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


necessary condenser and reboiler duties

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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


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


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


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




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?


What is the new outlet fraction of acetone in the air?


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



Interpolating we getA = 1.11



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


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



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!)



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




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 ↑


( ) g p T ↓ or P ↑



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




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