distil performance

8/3/2019 Distil Performance

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Column Pressure

How do we choose the operating pressure

of a column?

Usually based on:

(i) Temperature of overhead condenser

(ii) Temperature at bottom of column

Turton, 3rd edition

Column Pressure

1. If possible use cheapest coolant streamavailable cooling water (or air) in thecondenser

2. If by following (1), the bottom temperature is

too high then reduce pressure so that bottomtemperature is acceptable and choosewarmest coolant (refrigerant)

Turton, 3rd edition


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Column Pressure - example

A column separates A (light) from B (heavy). At whatpressure would you operate this column?

log P*(mmHg) =a -b/(c+T)

A a = 6.80398 b = 803.810

c = 246.990

B a = 6.80896 b = 935.860

c= 238.730

Turton, 3rd edition

Vapor pressure

Turton, 3rd edition

Distilllation ColumnOverhead Condenser

Alog P*(mmHg) = 6.80398 - 803.810 /(246.990 + 49)

= 4.0975

P* = 12,516 mmHg = 242 psi

What is pressure and temperature at bottom of the column?

Q

T

32

40-45

49

Top product is pure A

Assume vapor condenses to 49 C (120 F)

T-Q diagram

Condensing media:For water inlet 32 C (90 F)Maximum outlet: 45 C (115 F)

X

water

3240-45


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What does DPtraydepend on?

DPtray= rlg(hw+ hcr) + krgvo2

Often this is the dominantterm

DPtray is not a function of theoperating conditions.

Turton, 3rd edition

Weir height Contribution from gas flowthrough tray orifices

DPtray

Assume DPcol= 5 psi : Pressure drop per tray is on the order of 7.6cm (3 in) of water = 0.0007 bar (0.1 psi)

Pbot= 242 + 5 = 247 psia = 12,770 mmHg

For B bottom product:

a - b/(c+T) = log(12,770)

T=b/[a- log(12,770)]-C= 935.860/[6.80896 log(12,770)] - 238.730

T = 107.5C

Top product is pureA

Bottom product is pure B

DPcol= nDPtray Pbot= Ptop+ DPcol

Turton, 3rd edition

Distillation Column


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R.Turton and J.A. Shaeiwitz -

Copyright 2008

7

Number oftrays, n= 31

Turton, 3rd edition

Column PressureAcrylic Acid from Propylene

68

0.16

0.07

47

Why is column operatingunder vacuum?

Acrylic Acid production

Why is column operating under vacuum?

Polymerization of acrylic acid above 90C

Must reduce pressure so that Pbot:P*(acrylic acid) =a-b/(c+ 89C)P=0.16 bar

What about DPcol?DPcol= Pbot Ptop=0.16 - 0.07 = 0.09 bar

Number of trays, n= 31DPtray= 0.09/31 = 0.003 bar (= 1.2 of acrylic acid)

Turton, 3rd edition


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McCabe-Thiele Analysis

y

x

xD/(1+R)

xD

Feed line slope

= -q/(1-q)

xB xF

Turton, 3rd edition, ch. 18

1

RR

xyyx

LDL

VL

FD

FD

xxy

11

FD

FD

xy

yxR

min

R

R

V

L

achial

1

Column Pressure and

TemperatureReboiler temp ~ boiling point of

heavy component

Condenser temp ~boiling pointof light component

Increasing column pressure:

increases both temp.decreases relative volatility and

hence make separation more

difficult

Slope = L/V

Slope = L/V V

D

B

V

F

L = B + V

L = V - D

R = L/D

R = L/B

L

L

reflux

Boil-up

Distillation of Benzene from Toluene

Flows in kmol/hComposition in benzene mole fractionReflux ratio=2.38Pressure = 2 bar

Benzene recovery:G=DXD/(FxF) = 30(0.885)/(100(0.321))=0.827

Diameter = 0.83 mContains 20 sieve plates with 0.61 m plate spacing

Feed is introduced in plate 13

Tower operates with a reflux ratio of 2.38

= 70 (mass balance)XB=0.079

1

N

Analyze this operation: Number of theoretical plates?

where is the feed added? Column oversized/undersized? Effect ofchanging L/V?

McCabe-Thiele construction:

Speed of computation usingMcCabe-Thiele method is far

slower than for process

simulators.However, it remains an important

analytical tool.


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Distillation of Benzene from Toluene


Benzene recovery:

G=DXD/(FxF) = 30(0.885)/( 100(0.321))=0.827

Diameter = 0.83 mContains 20 sieve plates with 0.61 m plate spacingFeed is introduced in plate 13Tower operates with a reflux ratio of 2.38



McCabe-Thiele construction yields 13 THEORETICAL stages (12 + reboiler).Average plate efficiency is 12/20=0.6 (60%)

Feed is added in plate 13 (actual). Equivalent to theoretical plate number 8: 13x0.6=8

R=L/D

q=fraction of feed that is liquid


1

N

XB=0.079

XD=0.885

ZF=0.321

Some observations from previous process:

-Feed is not being introduced at optimum location inthe column. The optimal location is plate 6(theoretical) or 10 (6/0.6) actual

-Separation steps near the feed plate are small


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Flows in kmol/hComposition in benzene mole fraction

Reflux ratio=2.38Pressure = 2 bar

G=DXD/(FxF) = 30(0.885)/ (100(0.321))=0.827




R=L/Dq=fraction of feed that is liquid

Current situation (feed at plate 13 vs optimum atplate 10): The tower appears to

be oversized for the

present separation. Itcould process a

lower quality of feedand improve

benzene recovery


There are more than the optimum # of stages in therectifying (top) section: This means that the tower

was designed to process a lower concentration offeed or to produce a higher concentration distillatestream (benzene)


G=DXD/(FxF) = 30(0.885)/ (100(0.321))=0.827




Operating line moves away from equilibrium this increases separation ofeach stage (larger step size).


Increasing L/V:

The concentration of benzene in the bottom stream (xB) will be lowered

If xDremains the same:

Increasing L/V:


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Scale-up/down of ColumnMcCabe-Thiele Analysis

y

x

xD/(1+R)

Feed line slope= -q/(1-q)

xF xD

xB

What happenswhen refluxratio increasesbut number oftrays stays thesame?


Flows in kmol/hComposition in benzene mole fraction

Reflux ratio=2.38Pressure = 2 bar

G=DXD/(FxF) = 30(0.885)/ (100(0.321))=0.827





Computer simulation:Flowrate(kmol/h)

Mole fraction Concentration

Inputs

F 100 zF 0.321

Outputs from Simulation

L0 71.5 X0 0.885

VN+1 96.4 yN+1 0.079

V1 101.5 Y1 0.885

Ln 166.4 xN 0.079Turton, 3rd edition, ch. 18


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Mill Situation

(see working example)

You are told that benzene in feed will varybetween 25 and 40%. However, you arerequired to keep Dand xDconstant.

You are asked to asses the effect of suchchanges in feed concentration on theoperation of the distillation process

Performance

diagrams


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Mill Situation:

You are told that benzene in feed will vary between 25 and45%. However, you are required to keep Dand xDconstant.

Reflux is chosen as dependentvariable (most common way to

regulate the performance of adistillation column)

Observations (after running anumber of simulations):

For constant recovery G:

As zFincreases, the reflux L0decreases.

For a constant feed

concentration zF:As Gincreases, the refluxincreases

You can set up a number of other situations

(estimate reflux rate to obtain desireddistillate conditions, etc.)

Example: Find changes needed to get 90%

recovery at same zF.

Refluxrate

Feed mole fraction, zF

Constant recovery, G

Benzenerecovery:

Base Case

Weeping gas velocity: 0.35 m/s

Flooding gas velocity: 1.07 m/s

Scale-up/down of ColumnOperation

What other factors must we consider whencolumn 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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J.D. Seader and E.J. Henley, Separation Process Principles, John Wiley and Sons, NY 1998

scaledown

scaleup

Benzene

recovery:

Base Case



Example: Findchanges needed toget 90% recoveryat same zF (0.321)


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Benzene

recovery:

Base Case




Example: Findchanges needed toget 90% recoveryat same zF (0.321)

New Feed rate:F=xDD/zFG= 91.9

New reflux L0(from figure):

L0=81 (point b)

The distillation can be

operated at a recoveryof 0.9; this reduces the

feed that can be

processed by 8.1%

keep D and xDconstant

Benzene

recovery:



Example: Find the

maximum recoverypossible for thedistillation equipment for

a feed concentrationzF=0.275


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Scale-up/down of Column

What if we want to change the feed rate to the

column?

If we want tokeep all the purities the samethen the reflux ratio must stay the same.

Scale-down

How do we adjust the condenser and reboilerso as to keep the reflux ratios the same? (tokeep all the product purities the same

V

D

B

V

F

L = B + V

L = V - D

R = L/D

R = L/B

L

L

reflux

Boil-up


Top of column

QD= VlD=D(1+R)lD

Bottom of column

QB=VlB=B(R-1)lB


What if we do not adjust any of the utilityflows while increasing/reducing the feed tothe column?

Scale-down case

F = B + D

Fxf= BxB+ DxD

Top of columnQD= VlD=D(1+R)lD

Bottom of columnQB=VlB=B(R-1)lB

V

B

V

F

L = B + V

L = V - D

R = L/D

R = L/B

L

L

reflux

Boil-up



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R.Turton and J.A. Shaeiwitz -

Copyright 2008

31

47C and 0.07 bar

Turton, 3rd edition


What if we do not adjust any of the utility flowswhile increasing/reducing the feed to the column?

Scale-down case

F = B + D

Fxf= BxB+ DxD

Top of columnQD= VlD=D(1+R)lD

Bottom of columnQB=VlB=B(R-1)lB

Material balance control willadjust D+ Bto equal F

Same cooling water flowand steam pressure causessame amount ofcondensation and reboil what happens to Rand R?

Rand Rwill increase what happens to theseparation?

V

B

V

F

L = B + V

L = V - D

R = L/D

R = L/B

L

L



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For Scale-up must be aware of flooding limit 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 flooding

For Scale-down

weeping becomes an issue at 30-40% of flooding,

depending on type of tray

generally can avoid weeping problems by using higherthan necessary condenser and reboiler duties

Turton, 3rd edition

Number of Sequences of Ordinary Distillation Columns

Benzene, Tb=80.1 CToluene, Tb 110.8 CBiphenyl, Tb=254.9 C

Most volatile,taken asoverhead

Toluene, Tb 110.8 CBiphenyl, Tb=254.9 C

Least volatile

Pure products produced (largedifferences in volatility)

Other separation sequences arepossible!

ethylenzenep-Xylenem-Xyleneo-Xylene

p-Xylenem-Xylene

o-Xylene

Non-pure products produced(p- and m- xylene Tbs areseparated by only 0.8 C)

Seider, Ch. 7


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Synthesis of all possible ordinary distillation sequences for multicomponent feed that is to

be separated into Pfinal products (nearly pure components or multicomponent mixtures)Lets order the feed components in decreasing volatilities and assume that this correlates with the normal

boiling point (near ideal liquid solution).

We come up with an equation for the number of different sequences of ordinarycolumns Nsto produce a number of products P:

For example, for P=5 (A, B, C, D, E), the number of separation points is 5-1=4:A-B, B-C, C-D, D-E

j= # of final products that must be developed from the distillate of 1st columnThus P-jis the # of final products that must be developed from the bottoms of the firstcolumn.

Let Ni= the # of different sequences for I final products (thus, the number of sequences fora given separation point in the 1st column is NjNP-j. Note that in the first separator P-1different separation points are possible.

Thus, the number of different sequences for Pproducts is:

)!1(!

)!1(21

1

PP

PNNN

P

j

jPjs

Seider, Ch. 7

)!1(!

)!1(21

1

PP

PNNN

P

j

jPjs

Number of different sequences for P products:

Number ofProducts, P

Number ofseparators in thesequence, (P-1)

Number ofdifferentSequences, Ns

2 1 1

3 2 2

4 3 5

5 4 14

6 5 42

7 6 132

8 7 429

9 8 1430

10 9 4862

Seider, Ch. 7


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Number ofProducts, P

Number ofseparators inthe sequence,

(P-1)

Number ofdifferentSequences,

Ns

2 1 1

3 2 2

4 3 5

5 4 14

6 5 42

7 6 132

8 7 429

9 8 1430

10 9 4862 Direct sequence

(commonly used in

industry)

pure" distillates (free of high-boiling compounds)

If purity of D iscritical: addrerun(or

finishing)column

Sequence 2

Sequence 3

Sequence 4 Sequence 5: Indirect Sequence(least desirable due to difficulties in achieving

purity for bottom products) Seider, Ch. 7

Heuristics for Determining Favorable Sequences

For P=3 or 4: designing and costing all sequences can be best to choose the mosteconomical one. However, in most case the direct sequence is often the choice.

Otherwise, use following heuristics:

1 Remove thermally unstable, corrosive, or chemically reactivecomponents early in the sequence

2 Remove final products one by one as distillates (directsequence)

3 Sequence separation points to remove, early in the

sequence, those components of greatest molar % in the feed4 Sequence separation points in the order of decreasing

relative volatility so that the most difficult splits are made inthe absence of the other components

5 Sequence separation points to leave last those separationsthat give the highest-purity products

6 Sequence separation points that favor near equimolaramounts of distillate and bottoms in each column

When energy costs are high,this heuristics often leads tothe most economicalsequence.

Note the marginal vapor Rate Method, Chapter 7 Seider

Seider, Ch. 7


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Example:

Large variations in relative volatility andmolar % in process feed. Heuristics 4dominates over heuristic 3 and leads tothis sequence.

H4: Sequence separation points in theorder of decreasing relative volatility sothat the most difficult splits are made inthe absence of the other components

Seider, Ch. 7

distil performance

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