Equilibrium Stage Processes

7/25/2019 Equilibrium Stage Processes

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Equilibrium Stage Processes

A. Introduction

There are many separations operations used in chemical processing

where the objective is to have two phases form in a unit, come toequilibrium, and leave the unit as two separate streams. These two

phases may be formed from a single inlet stream, or, they may beformed by mixing two inlet streams. In either case, we refer to

such a process as an equilibrium stage. To achieve better

separations of components than can be achieved in a single stage,

chemical engineers use towers or extraction trains containing manyindividual stages where streams can be contacted and brought to

near equilibrium conditions.

flash vapori!ation unit is a process unit where a single inlet stream is brought to

conditions such that a liquid phase and a vapor phase will develop and approach

equilibrium in a vessel, which is commonly called a flash drum or a separator. The vapor

and liquid phases are ta"en off as separate exit streams. The purpose is to separate more#volatile components from less#volatile components. In an ideal separator, equilibrium

will be achieved. In practice, this will not occur, however, if properly designed, a flash

unit may produce streams that are close to equilibrium compositions.

In some cases, a single inlet stream may result in three phases formed and separated, such

as in a crude oil separator that you might see next to an oil well in an oil field. In such athree#phase separator, light hydrocarbon gases form a gas phase, heavy hydrocarbon

liquids form an oil phase, and water drops out and is separated as a second liquid phase.

mixer$settler unit is an example of a process unit where two inlet streams are contactedso that some components are exchanged between them. The two phases are brought to

equilibrium, and then separated as two separate exit streams. This type of unit can be

Feed

Feed

Solvent

Steam

Product

Overhead Raffinate

Product

Bottoms

Extract

One or More

Equilibrium

Stages


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used in liquid#liquid extraction, solid#liquid extraction, distillation, gas absorption, steam

stripping and other operations.

%alculational methods for these multistage operations are always based on the

assumption of equilibrium in every stage. &tage efficiencies are then used to relate real

operations to the ideal calculations based on equilibrium. These efficiencies depend onthe fluid dynamics, heat transfer, and mass transfer in the real processes, and methods

used in the estimation of stage efficiencies are beyond the scope of this class.

B. Processes with a Single Inlet Stream (Two-Phase Separators)

In these processes, we will consider that a single feed stream

enters a unit, where it is allowed to come to equilibrium,forming two phases, which are removed as an overhead product

and a bottoms product. The feed stream may be a single phase,

or may already be in two#phase flow before entering the unit.

The exit streams can be any two phases that can be separated'vapor and liquid, liquid and liquid, liquid and solid, etc.

(e will focus on a separator )flash vapori!ation unit*, where a

liquid )or sometimes mixed liquid and vapor* feed is separated

into an overhead vapor product and a bottoms liquid product,the two product streams assumed to be in equilibrium with each

other. This is usually accomplished by suddenly lowering the pressure on the feed stream

just as it enters the flash drum. In practice, this is done by use of a valve to throttle the

inlet stream. diagram is shown below, although the valve is not explicitly shown.

+verhead )apor*

P VVVi HPTyV ,,,,

apor

-eed T

FFFiF HPTxF ,,,,

ottoms )/iquid*

Q BBBi HPTxB ,,,,

In the variable list for each stream, the first symbol represents the molar )or mass* flow

rate, the second represents the mole )or mass* fractions of the various components )some

could be !ero*, and the last three represent temperature, pressure, and molar enthalpy.

If there are n components in the system, and there are no chemical reactions, then there

are n01 variables for each of the three streams' 2 molar flow rate )-, , or /*, n#2independent mole fractions )remember mole fractions must add up to unity*, and 1 other

stream properties )temperature, pressure, and enthalpy*. (e have to include temperature

and pressure, because they are needed to define the thermodynamic state of the stream.lso, the phase equilibrium relations depend on these variables, in addition to


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compositions of the phases. (e include enthalpies, not because they are needed to

specify the state of a stream, but because they will appear explicitly in the energy balance

equation. 3 is the duty for the flash vapori!ation unit, that is, the heat rate term in theenergy balance. 3 can be positive )energy added*, negative )energy removed*, or !ero

)adiabatic*. 4any separators operate close to adiabatic conditions.

The total number of variables is 1)n01* 0 1, where the last three are 3, T, and 5'

Total 6o. ariables 7 1n028

6ow, lets write all of the equations that we can to provide relations between these

variables. (e can write materials balances, an energy balance, equilibrium relations, and

enthalpy property relations. (e will assume steady#state operation of the flashvapori!ation unit. The system will be the contents of the flash drum, with the throttling

valve outside the system boundary where the feed stream enters.

4aterial balances'

Total'

%omponents' iiiF LxVyFx += )i 7 2 to n#2*

9nergy balance'

6ote' The valve on the feed stream does not affect the energy balance, because

the process of steady#state flow across a valve is considered an isenthalpic process)constant enthalpy*.

9quilibrium relations'

Thermal'

4echanical'

&pecies' ( ) ( )sxPTfsyPTf iL

ii

V

i :,,;:,,; = )i 7 2 to n*

6ote' To use the species equilibrium relations, we will have to provide specificequations, such as


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9nthalpy property relations'

( )( )( )sxPTHH

syPTHH

sxPTHH

iLLLLL

iVVVVV

iFFFFF

:,,

:,,

:,,

=

=

=

6ote' These relations are written in functional form, because they may come

from a table or chart in a reference boo" )e.g., steam tables or an enthalpy#

concentration diagram*, or, more li"ely, they will be ta"en from some

calculational scheme in a process simulation software pac"age.

The total number of equations we have written is 8n0> ?n )material balances* 0 2 )energy

balance* 0 n0@ )equilibrium relations* 0 1 )enthalpy relations*A.

Total 6o. 9quations 7 8n0>

To solve a real problem, the number of equations must equal the number of un"nowns,

that is, the degrees of freedom must be !ero. Thus, we can calculate the number of

variables that must be specified to allow a solution to be obtained'

6o. &pecifications 7 6o. ar. # 6o. 9qns. 7 )1n028* # )8n0>* 7 n0@ .

Thus, we must specify n0@ design variables before we can solve the above equationsfor the remaining un"nown state variables. Bou should remember that not all sets of

variables ma"e valid sets of design variables, because we have to be careful not to violate

any of the equations by our specification of variables.

The most common set of design variables to specify is as follows'

F,xi=s, T-,P-,P, and )either Qor T*

6ote' n adiabatic specification for a separator means Q7 C, whereas, an

isothermal specification means T7 T-.

Eample T!pe I Problem. "lash #apori$ation %or a Ben$ene & Toluene Stream

5roblem' stream containing 1D.C mol E ben!ene and FD.C mol E toluene is flowing at

2CC.C mol$s, 2CC.C o%, and 2C atm. It is flashed across a valve and into a flash drum heldat 2CC.C o% and 2.CC atm. (hat are the compositions and flow rates of the product

streams, and what is the duty for this flash unitG 5repare solutions by a* hand

calculations assuming


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a* analytical solution using


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vapori!ation at the normal boiling points from Table .2 in -elder and


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s can be seen,x2is about C.8D andy2 is about C.@D, in agreement with our

analytical solution.

The amounts of vapor and liquid can be found using the lever rule. Thus, byvisual inspection, the amounts of vapor and liquid product are roughly equal )DC

mol$s each*, which agrees with the analytical results.

complete graphical solution can be obtained by use of aH#x#ydiagram, but first

we must construct one for this system. Three steps are needed to construct such a

diagram' 2* location of the enthalpy line for vapors at their dew points, 8*location of the enthalpy line for liquids at their bubble points, and 1* location of

tie lines connecting equilibrium liquid and vapor compositions at various

temperatures. lthough we need only the location of a single tie line at 2CC o% to

solve the current problem, we will construct the entire diagram, for clarificationand for future reference when multi#stage processes are discussed.

(e will retain the same reference conditions for enthalpy' Hi/7 C for both

components at 2CC

o

%. lso, we will continue to assume that the heats of mixingfor liquid mixtures are small. Thus, the enthalpy of the liquid mixture which has

a bubble point at 2CC o% will have H/7 C )see the analytical solution above*. (ecan calculate the enthalpy for other bubble#point mixtures by selecting a

temperature )T*, reading the liquid composition from the bubble#point curve on

the T#x#ydiagram )or calculating the equilibrium compositions directly fromCC K$mol

atx27 C.8DJ H/7 C )2CC.Co% point*

atx27 2 H/7 #1,CCC

apor enthalpies are found by adding the heats of vapori!ation to the liquid

enthalpies that have just been calculated. &ince the pressure is 2.CC atm, the heats

of vapori!ation for the two components )x27 2 andx27 C* are just the values attheir normal boiling points )>C.C and 22C.F o%, respectively*. The values are

1C,JFD K$mol for ben!ene and 11,@JC K$mol for toluene )from -elder and


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aty27 C.@DF H7 18,CCC )2CC.Co% point*

aty27 2 H7 #1,CCC 0 1C,JFD 7 8J,>CC

The dew# and bubble#point compositions for any tie line )at any desired

temperature* can be found by solving


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gain, note that the enthalpy of the feed is ta"en as !ero at 2CC.C o% and 2C atm.

The results agree very well with the analytical solution.

(hy would a person use a graphical solution, when the analytical solution is so

easyG The answer will become apparent when we consider multistage processes,

where analytical calculations by hand become tedious. efore high#speedcomputers, chemical engineers relied heavily on graphical methods. 6ow, we use

them only to illustrate the principles behind the computer calculations.

c* solution from a HB&B& simulation

%omponents selected' ben!ene, toluene

Thermodynamics pac"age' ntoine )ideal gas, ideal liquid solution*-eed stream properties' T7 2CC o%,P7 2C atm

F7 2CC mol$s,x27 C.1D

-lash unit specifications' T 7 2CC o%,P7 2 atm

x2CFK$s

6ote that the enthalpy for the feed is slightly different from the enthalpy for the

liquid product in the HB&B& simulation. This is one reason for the differencebetween this result and the analytical result )2.@Mx2CFK$s*, but not the major

factor.

'. Processes with Two Inlet Streams (istillation Tower Tra! *ier+Settler)

In these processes, two streams )different phases*

enter the unit, are mixed and allowed to come toequilibrium, and are then separated into two product

,a! .P-/ Plume 'leanup

*ultiphase Etraction 0nit


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streams )different phases*. flow chart for such a unit is shown below, where streamsLCand V8are mixed in the unit &tage 2, brought to equilibrium, and separate streams V2

andL2exit.

P T

V1 V2 Stage 1

L3 L1 Q

The unit name and the numbering of the streams are chosen as indicated, so that we can

later treat countercurrent multistageprocesses without changing the notation. lthough

L)liquid* and V)vapor* are used to represent stream flow rates )as would be mostappropriate in distillation, gas absorption, or stripping*, the phases could equally well be

two liquids )liquid#liquid extraction*, or one a fluid and the other a solid )e.g., solid#liquidextraction or gas adsorption*.

-or each stream, if there are n components in the system, there will be a set of n01 streamvariables )e.g.,LC,xiC)i 7 2,N,n#2*, T/C,P/C,H/Cfor the liquid stream entering &tage 2*.

In addition, we need to count the unit variables T,P, and Q. Thus, the total number of

variables for the entire unit will be given by

Total 6o. of ariables 7 @)n01* 0 1 7 @n 0 2D

6ow, the number of equations that can be written for this unit is the same as for Type 2processes )n material balances, 2 energy balance, n0@ equilibrium relations, 1 enthalpy

relations*, except that now one additional enthalpy relation can be written for the specific

enthalpy of the fourth stream )in terms of its temperature, pressure and composition*'

Total 6o. of 9quations 7 n020)n0@*0@ 7 8n0M

Thus, the number of specifications )design variables* to ma"e a real problem solvable)force the degrees of freedom to !ero* is given by

6o. of &pecifications 7 6o. ar. # 6o. 9qns. 7 )@n02D* # )8n0M* 7 8n0F

The most common set of variables that would be specified when considering a single

stage are the properties of the two feed streams, the unit pressure and either the unittemperature or duty'

%ommon &pecifications' LC,xiC=s, T/C,P/C, V8,yi8=s, T8,P8,P, and )Qor T*


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This does not have to be the list specified, as any 8n0F variables will suffice to allow a

solution, remembering that none of the equations can be violated by the specifications.

However, some sets may be more#easily accommodated in process simulation pac"ages.

Eample T!pe 2 Single-Stage Problem. Ben$ene & Toluene Equilibrium

liquid streamLCcontains D1.C mol E ben!ene and the rest toluene. It is flowing at

1>.C "mol$h and M2.C o%. This stream is mixed with a vapor stream V8containing

8F.C mol E ben!ene, which is flowing at F8.C "mol$h and 2CD.C o%. The mixer$separatoroperates at 2CC.C o%. The pressure is constant throughout this system at 2.CC atm. (hat

are the flow rates and compositions of the product vapor and liquid streamsG (hat is the

duty for this mixer$separatorG %ompare analytical calculations, graphical results, and a

HB&B& simulation.

a* analytical calculation

/ets first ma"e a list of all the variables that are specified'

LC7 1>.C "mol$h x2C7 C.D1C T/C7 M2.Co%

V87 F8.C "mol$h y287 C.8FC T87 2CD.Co%

T 7 2CC.C o% P/C7P87P7 2.CC atm

In this notation,y=s are mole fractions in vapor streams, andx=s are mole fractions

in liquid streams. The first subscript refers to the component and the second

subscript refers to the stream )numbered by the stage from which the stream

originates*. Thus, we have specified 8n0F 7 2C variables, and we should be ableto solve for the rest.

+nce again, we have a binary system )ben!ene)2* 0 toluene)8** which comes tophase equilibrium at 2CC.C o% and 2.CC atm. -rom the


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To solve this for 3 requires us to estimate the four stream enthalpies. gain,ta"ing the references to beH/7 C at 2CC.C

o% for both components, and assuming

negligible heats of mixing, the enthalpy of streamL2will be !ero. (ith these

references, the enthalpies for all four streams can be read from theH#x#ychartconstructed earlier )assuming pressure effects are small*'

Then, the duty becomes

b* graphical solution

-irst, the feed points are located on theH#x#ydiagram, and they are found to lieon the saturated liquid and vapor curves. If the vapor was significantly

superheated, or the liquid greatly subcooled, then we would need enthalpy lines

on the graph for these regions to be able to locate the feed points. The method ofsolution presented below would be exactly the same.

The flow rates )LC7 1>.C "mol$h and V87 F8.C "mol$h* specify a feed mixture

point )by the lever rule*, which is at a flow rate of 2CC.C "mol$h, a composition ofC.1F, a temperature near 2C2 o%, and an enthalpy of 8C,2CC "K$"mol.

The actual mixture in the unit must be at this same composition, but at 2CC.C o%,so, coming straight down from the feed mixture point to the 2CC.C o% tie line, the

actual mixture point in the unit can be found. y reading the difference in

enthalpy between these two points, an enthalpy change of #1,8CC "K$"mol isrequired to bring a mixture of this composition to the proper temperature.2-x- !iagram for Ben"ene # Toluene

at $ atm

-$&&&&

&

$&&&&

(&&&&

3&&&&

*&&&&

& &)$ &)( &)3 &)* &)4 &)+ &)5 &)% &)' $

x$, $

Enthal-.67mol0

$&& /$&% $&* '+ '( %% %*

H8

H9

$$&)+

%&

V(

L &L $

V$


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Thus, the duty for this unit can be calculated from the total amount of mixture and

the required change in enthalpy'

Q7 )2CC.C "mol$h*)#1,8CC "K$"mol* 7 #C.18x2CF"K$h

The end points for the 2CC.C o% tie line tell us the compositions of the equilibrium

streams leaving the unit'

x227 C.8F y227 C.@F

y use of the lever rule on the 2CC.C o% tie line, bothL2and V2can be found'

L27 @J "mol$hV27 D1 "mol$h

c* HB&B& simulation results

L27 @C.M "mol$h V27 DM.2 "mol$h Q7 #C.8Dx2CF"K$h

x227 C.8@8 y227 C.@@F

gain, these results are not in perfect agreement with those above, for the samereason as in the previous example problem. The vapor pressure of ben!ene is

different in HB&B& from that predicted by ntoine equations from the %h9 8C2

and 1C2 texts. (hich is rightG Loes it matterG

. 'ountercurrent *ultistage 4perations (Etraction 5as Absorption Stripping)

To obtain good separations in processes such as liquid#liquidextraction, gas absorption, and stripping, multiple stages are often

used in practice. +ne common arrangement is to have countercurrent

flow through a series of stages. This could be a train of

mixer$settlers, or a series of plates in a vertical tower.

Steam Stripper


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simple flowchart for such a process with m stages is shown below. The advantage of

this arrangement is that the driving force for mass transfer is more nearly constantthroughout this system, the streamsLCand V2being the richest in transferable

component)s*, and streamsLmand Vm02being the most dilute. There are three unit

variables of importance, which are not shown on the figure, Tm,Pmand Qm.

V2 V8 V1 Vm#2 Vm Vm02

2 8 m#2 m

LC L2 L8 Lm#8 Lm#2 Lm

y counting variables and equations, we obtain

6o. ariables 7 &tm ar 0 nit ar 7 )8m08*)n01* 0 1m 7 8mn0Mm08n0F

6o. 9quations 7 4atl al 0 9ner al 0 9qlm


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The analytical solution is now more complicated than for a single stage, because, with the

specifications indicated above, we do not "now values for the flow rate, compositions,

and conditions of two streams for any single stage, which would be needed to get thecalculations started. 6ote that we cannot solve for the exit streamsLmand V2by overall

material balances, because we cannot predict the combined effect of all the equilibria for

the m stages. +n the other hand, if we had specified streamsLCand V2, we couldcalculate stage 2, then proceed to stage 8, etc. (hen we finished calculations for stage

m, we would "now all un"nown stream and unit variables.

(ith the specifications indicated above, the analytical problem is trial and error, in that

assumptions have to be made about stage 2 )e.g., specifying variables for stream V2* to

allow stage#to#stage calculations to be made for stages 2, 8, 1, N.., m. Then, a chec" has

to be made, to see if the "nown properties of stream Vm02are duplicated. If not, then theassumptions for stage 2 need to be modified, and a new set of stage#to#stage calculations

performed. This needs to be repeated until convergence is achieved )hopefully*.

lthough modern computers can do these stage#to#stage calculations very rapidly, theconvergence problem is a difficult one. This is because there are no simple, foolproof

ways to adjust the assumptions about stage 2, to converge on the correct answers forstream Vm02)at the other end of the process*. %hemical engineers are often frustrated

because it is difficult to get computer simulation pac"ages for multicomponent, stage#

wise processes to converge, even when the (egstein method, or other accelerationschemes are used.

The 'oncepts o% ,et "low and the elta Point

%onsider material balances for a system including stages 2, 8, N, j in the flowchart

shown above, where stage j is some arbitrary stage number. (e can write a total material

balance, n#2 component material balances, and an energy balance around this system'

2222CC

2222CC

22C

VLjjVjjL

iijjijji

jj

HVHLHVHL

yVxLyVxL

VLVL

+=+

+=+

+=+

++

++

+

6ote' -low rates can be per mole or per unit mass. If the latter is used, then the

compositions are mass fractions, rather than mole fractions.

These equations can be rearranged to give the equations of net flow between any two

stages'

22CC22

22CC22

2C2

VLVjjLjj

iiijjijji

jj

HVHLHVHLH

yVxLyVxLx

VLVL

==

==

==

++

++

+


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&ince we are assuming steady#state operation, the flow rates and properties of streamsLC

and V2are not changing, thus, we can conclude from these net flow equations that the net

flow to the right )of total mass, mass of component i, or energy* between any two stagesis a constant, independent of where we are in the series of stages. 6ote that net flows

may be positive or negative, depending on whether the / or streams carry more

material )or energy*.

(e can define a fictitious concentration for component i, and a fictitious enthalpy, by

combining the above net flow equations. These will be the definitions of the point,

which will be used in graphical solutions to multistage processes'

2

22

2

22

+

++

+

++

=

=

jj

VjjLjj

jj

ijjijj

i

VL

HVHL

H

VL

yVxLx

These equations may be used on phase diagrams to locate #points, which are constant

for a series of stages, and provide the relationship between streams that are passing

between stages. -or example, the composition of component i for the #point )xi* lies on

a straight line throughxCandy2, and also on a straight line throughxmandym02, the

composition points at the other end of the series of stages.

Lelta points )relating streams passing between stages* and tie lines )relating equilibrium

streams leaving any stage* are all that are needed to step from stage#to#stage on a phase

diagram, as will be shown in the example below.

In the derivation above, the energy balance was written based on the assumption that allof the stages are adiabatic. This is a requirement for a fixed delta point on anH#x

diagram. If there are significant heat losses for each stage, then the delta point will

incrementally change position for each stage, ma"ing a graphical construction more

difficult. If heat effects are small and energy balances are being ignored, such as in theexample below, then this is not a problem.

Eample 'ountercurrent *ultistage Problem. Etraction o% acetic acid %rom water

with isoprop!l ether.

three#stage mixer#settler system is to be used to extract acetic acid from 2CC. lb$h of a

DC. wt E aqueous solution, with the solvent being 8CC. lb$h of pure isopropyl ether. The

feed streams are at F> o- and 2 atm, and the stages are all assumed to be isothermal and

isobaric. 9stimate the flow rates and concentrations of the extract and raffinate streamsleaving this extraction train, using both analytical and graphical solutions. (hat

percentage of the acetic acid is extractedG


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cetic cid )2* 0 Isopropyl 9ther )8* 0 (ater )1*

V2 V8 V1 V@7 8CC. lb$h

y@' y2@7 C.CC

2 8 1 y8@7 2.CC y1@7 C.CC

LC7 2CC. lb$h L2 L8 L3xC' x2C7 C.DC x8C7 C.CC

x1C7 C.DC

9quilibrium data'


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upper or lower part of this equilibrium curve. (e will use the other construction

lines in the graphical solution described below.

a* analytical solution

-irst, lets see if the problem is well posed. There should be 8n08m0@ 7 2Fspecifications. (e "now six flow rates and compositions, plus the temperature

and pressure for these two feed streams. In addition, we "now T7 F> o- )8C o%*

andP7 2 atm for each stage. Thus, there are 2F specifications.

ecause we "now the temperature and pressure in each stage, the energy balance

equations do not have to be solved to find the equilibrium conditions for each

stage. They could be used to find the Qfor each stage, but this is not part of theproblem statement, so we will ignore the energy balances. If the heats of mixing

are small, then the Qvalues will be small as well, and each stage will be nearly

adiabatic.

To start the calculations, we will assume a flow rate and composition for stream

V2)remember that the compositions for this stream must lie on the upper part ofthe equilibrium curve, because it flows out of stage 2*'

V27 8@1 lb$h,y2' y227 C.2>,y827 C.JJ,y127 C.CD

(e can then solve the material balances for stage 2, using the tie lines on the

handout phase diagram to determinex2from they2point assumed'

x2' x227 C.11,x827 C.CD,x127 C.F8

Three material balance equations can be written for stage 2'

8228228888CC

2222222882CC

228C

yVxLyVxL

yVxLyVxL

VLVL

+=+

+=+

+=+

lthough it loo"s li"e there are four un"nowns )V8,L2,y28,y88* , we must

remember that the pointy8must also lie on the upper part of the equilibriumcurve, because it is for a stream leaving stage 8. Thus,y28andy88must be related.

-rom the phase diagram, this relation is nearly a straight line )up toy2of C.8*'

2888 CD.2M>.C yy =

The analytical solution of the above equations for stage 2 results in


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L27 FM lb$h, V87 828 lb$h,y8' y287 C.CJ>,y887 C.>M>,y187 C.C8@

&imilarly, for stage 8, we use a tie line and then solve the balance equations'

x8' x287 C.88,x887 C.C1,x187 C.JD

L87 DD lb$h, V17 2MM lb$h,y1' y217 C.C12,y817 C.M@J,y117 C.C88

-inally, for stage 1 )withy8@7 2, since the point for V@does not have to lie on the

equilibrium curve*,

x1' x217 C.22,x817 C.C8D,x117 C.>FD

L17 @@ lb$h, V@7 2>> lb$h,y@' y2@7 C.CCF

These results are close to, but not in perfect agreement with, the "nown values for

stream V@)see the flowchart above*. If better agreement is desired, we could

choose slightly different values for V2andy22and repeat the stage#to#stage

calculations.

ssuming the above values are close enough, the results are as follows'

-eed' LC7 2CC. lb$h xC' x2C7 C.DC

x8C7 C.CC x1C7 C.DC

&olvent' V@7 2>> lb$h y@' y2@7 C.CC

y8@7 2.CC

y1@7 C.CCF9xtract' V27 8@1 lb$h y2' y227 C.2>

y827 C.JJ

y127 C.CD

acetic acid extracted 7 )C.2>x8@1*x2CCE$)C.DCx2CC* 7 >J E

b* graphical solution

&ince this problem does not involve energy balances, we do not have to use

enthalpies, and the solution can be constructed on the triangular composition

diagram, with mass fraction acetic acid on the hori!ontal axis and mass fractionisopropyl ether on the vertical axis. (e will find this solution relatively easy to

construct, as the equilibrium data are initially available in graphical form.


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The first step is to locate the feed and solvent stream points, and the mixture point

representing the combination of these two streams. The feed pointxCis at C.D,C

and the solvent pointy@is at 2,C. line is drawn between these two points, andthe feed mixture pointz-is located on this line by the lever rule using the feed and

solvent rates )see the medium weight, solid line on the above diagram*'

LC$V@7 2CC$8CC

y an overall material balance for the entire extraction train, this !-mixture pointmust also lie on a line connecting the composition points for the two exit streams'

x1andy2. ssuming a position for the pointy2)on the upper phase boundary*,

immediately defines the pointx1)on the lower phase boundary*. y the lever

rule, the flow ratesL1and V2can be found, since, by overall material balance, theymust add up to 1CC lb$h )the sum of the two feed streams*.

-or example, ify227 C.2> is assumed, thenx217 C.2C,L17 @8 lb$h, and

V27 8D> lb$h are found from the diagram. These are the values shown on thediagram, with a medium weight, solid line connectingy2andx1and passing

throughz-.

6ext, the #point is found by the intersection of two lines representing streams

passing each other' thexC#y2line and thex1#y@line. This construction is shown on

the figure above. The #point is the point that lets us graphically solve the

material balances to relate streams passing each other between extraction units.

The plates are stepped off by starting at pointy2, following a tie line tox2)on the

lower phase boundary*, then following a line toward the #point up toy8)on the

upper phase boundary*, then following a tie line down tox8, then a #point line uptoy1, then a tie line tox1. These construction lines are all shown as dashed lines

on the diagram above.

Ifx17 C.2C, then we assumed the correct y2starting point. In fact, the result is

x1O C.22, which is pretty close, probably as close as we can read this diagram.

E. "eed Streams Side raws 6e%lu and 6eboil (istillation Towers)

If we tipped the m stages in the above figure so they ran

vertically down the page, it would almost loo" li"e a distillationtower with m trays inside. In fact, it could be a tray#type gas

absorber or steam stripper, where the feed and exit streams are all

ta"en off at the top and bottom of the tower.


21/30

distillation tower is usually operated in a different arrangement. The feed is usually

introduced somewhere in the middle, and product streams come off

each end through the use of reflux and reboil. There may be manyother complications in practice, such as multiple side#draw product

streams, pump#arounds where fluid is ta"en from a tray and passed

through a heat exchange and returned to the tower, etc.

elow is a flowchart for a distillation process with the minimum

number of liquid streams' feedF, overhead product streamD, andbottoms product streamB.

%ondenser


22/30

heat losses from the trays in the tower. In counting variables below, it will be assumed

that there is an arbitrary duty for each stage.

/ets once again count variables and equations, so that we can define how many

specifications are needed to form a well#posed distillation tower problem'

6o. ariables 7 )8m0D*)n01* 0 1)m02* 7 8mn 0 Mm 0 Dn 0 2>

6o. 9quations 7 )m02*)n02* 0 m)n0@* 0 8 0 )8m0D* 0 )8n01* 0 )n08*7 8mn 0 Jm 0 @n 0 21

6o. &pecifications 7 8m 0 n 0 D

The difficulty in preparing the above list is counting the equations, as the condenser

equilibrium relations )8*, feed stream mixer relations )8n01* and distillate ta"e#off point

splitter relations )n08* require special consideration to get the number of equations right.

Two typical lists of specifications are as follows'

2* F,xi-=s, T-,P-,P%, T%)exit*, Q%, Q


23/30

separation. The tray efficiency depends on fluid mechanics and mass transfer

considerations, and is beyond the scope of a thermodynamics course. tray efficiency of

unity means that equilibrium was achievedP so actual tray efficiencies are 2.

The Ponchon-Saarit *ethod %or Binar! S!stems

In distillation, energy effects are significant, as materials are being vapori!ed and

condensed. Thus, energy balances cannot be neglected. s is often the case in chemical

processing, analytical calculations are simplified for binary systems, and a graphicalsolution is possible on an enthalpy#concentration diagram.

The only real difference from the countercurrent multistage solutions, given in the

preceding section, is that the net flows are different in the sections of the tower above andbelow the feed location. Thus, two delta#points will be needed to allow stages to be

stepped off throughout the tower.

The upper section of a distillation tower )above the feed* is called the enriching )orrectifying* section, and the lower section )below the feed* is called the stripping section.

In the upper section, the vapor stream is being enriched in the more#volatilecomponent)s* by removal of the heavier component)s*. In the lower section, the more#

volatile component)s* is being stripped out of the liquid stream.

The net flow equations become

9nriching &ection' DVLVL jj === +22C

&tripping &ection' BVLL jjm === +2

The coordinates of the two delta#points become

9nriching &ection' ( ) ( )DVDC

DD HHRHD

QHHxx ++=== 22,

&tripping &ection'B

QHHxx RBB == ,

6ote' Q%will be negative, so the delta#point in the enriching section will be

aboveHL, the enthalpy of the overhead product stream. The second form forH

is useful when the reflux ratio )R7LC$D* is given, but it is only correct for a total

condenser. Q


24/30

6ote' The two delta#points and the feed pointz-must lie on a straight line, so, if

one delta#point and the feed point are "nown, then the other delta#point can be

located on the enthalpy#concentration diagram )e.g., as the intersection of thisstraight line with a vertical line atx*.

+nce the two delta#points have been located, the stages can be stepped off, starting at theoverhead point and using the upper delta#point and the tie lines in the enriching section.

Then, once the feed pointz- has been passed, the lower delta#point and the tie lines are

used to step off stages in the stripping section.

This 5onchon#&avarit method will be illustrated in the example problem below.

The Approimate *c'abe-Thiele *ethod

simplified procedure for binary distillation can be used under the assumption of

constant molal overflow. (hat this means is thatLand V)as molar flow rates* remain

constant throughout the rectifying section of a tower, and constant at different values )L=and V=* throughout the stripping section. -or this to be valid, a number of assumptions

have to hold'

ssumptions'

2. The molar heats of vapori!ation of all mixtures are equal )distance

between the bubble#point and dew#point curves on the H#x#y diagram

everywhere the same*.

8. The heats of mixing, heat losses, and sensible heat effects due totemperature gradients are all negligible.

The assumption of constant molal overflow is not usually made in modern simulationpac"ages. However, in many cases this is a reasonable approximation, and the 4c%abe#

Thiele method is sometimes used as a quic" hand calculation to give a quic", visual

chec" of what is happening in the simulation. It has the advantage of producing a visualpicture of the operating and equilibrium lines, allowing the user to identify situations that

may be close to a pinch point. This will be discussed further below.

This assumption constant molal overflow allows temperatures and enthalpies to beignored, and stages to be stepped off on ay#xdiagram for the binary system at the

pressure of the tower )assumed constant*. The equilibrium curve on they#xdiagram can

be found from experimental data, ta"en from a calculated T#y#xdiagram, or calculateddirectly from the phase equilibrium relations. y tradition, the more volatile component

is chosen as component 2, so that they#xequilibrium is usually above the @Doline )where

y 7x*. +f course, if there is an a!eotrope, then the equilibrium curve will cross the @Doline at the a!eotrope composition.

To finish the construction of this graphical solution, we have to locate operating lines

on they#xdiagram. In this approximate method, operating lines are straight lines,


25/30

representing the relative compositions of streams passing each other between trays.

Thus, operating lines ta"e the place of the delta points used above in the 5onchon#&avarit

method.y material balances around the top end of the column, withLand Vconstant, an

equation can be easily derived for the straight operating line for the enriching section

where y and x are mole#fraction compositions for any two streams passing betweenstages.

&imilarly, for the stripping section the operating line becomes

These two operating lines cross on the feed line, which is a vertical line atx7x-for asaturated liquid feed stream. -or conditions other than saturated liquid for the feed

stream, refer to 5erry=s Handboo" or a nit +perations text for details on construction of

the feed line. -or subcooled liquid feed, the feed line will angle slightly to the right ofvertical. -or a feed with is a mixture of liquid and vapor, the feed line will angle to the

left of vertical, being somewhere between vertical )all liquid* and hori!ontal )all vapor*.

-or a saturated liquid feed at rateF,

L= 7L0F and V= 7 V

&ince, by material balances,

BDF BxDxFx

BDF

DLV

+=

+=

+=

we can locate both operating lines ifF,x-,D,xL, andR7L$Dare "nown. (e could "now

B,xrather thanD,xLand still ma"e the construction.

&tarting at the top of the tower, wherey27xL)on the @Doline*, we can findx2by moving

hori!ontally to the left to the equilibrium curve, theny8by moving vertically down to theoperating line, thenx8hori!ontally on the equilibrium curve, theny1vertically on the

operating line, etc. This procedure is followed until the feed line is passed, then theoperating line for the stripping section is used to step off stages to the bottom of the

tower. The construction can also be made starting at the bottom of the tower.

This 4c%abe#Thiele construction will be illustrated in the example problem below.


26/30

Eample istillation Problem. Separation o% Ben$ene and Toluene.

n equimolar ben!ene )2* 0 toluene )8* liquid stream is flowing at 2CC.C "mol$h,2 atm, and its bubble#point temperature. It is to be separated into two liquid

streams by use of a distillation tower. The overhead product stream is to be DC.C

"mol$h and contain at least MD E of the ben!ene fed to the tower. total condenser)with no subcooling of the liquid*, a reflux ratio of 1.CC, and a partial reboiler are to

be used. (hat is the minimum number of trays requiredG (hat are the duties of

the condenser and reboilerG ssume negligible pressure drop, and no heat lossesfrom the column. se both graphical solution methods indicated above and

compare with a HB&B& simulation.

Is this problem well posedG The number of specifications should be 8m 0 n 0 F,since the number of stages is un"nown. The pressure and duty are specified for

each stage except the reboiler )8)m#2* specs*. In addition, T-,P-,Fandx-are

specified for the feed )@*P TL,PL,DandxLare specified for the overhead stream )@*P

the reflux ratioRis specified )2*P and the pressure of the reboilerP, which is the required number for a binary distillation,

where the number of stages in not "nown. 6ote that the distillate producttemperature is fixed by the specification that there is no subcooling of the exit

stream from the condenser.

a* graphical solution )5onchon#&avarit*

Qiven' D7 DC.C "mol$h xL7 C.MD)C.DC*)2CC.C*$DC.C 7 C.MD

)4ole fractions are for the more volatile component R ben!ene.*

9quilibrium and enthalpy data' H#x#yphase diagram at 2 atm )given above and

reproduced below on a much larger vertical scale, to accommodate delta points*

/ocation of #point' H7HL0)R02*)H2#HL*

7 )#1,CCC* 0 @?8>,CCC#)#1,CCC*A 7 282,CCC "K$"mol

x7xL7 C.MD

6ote' Q%can be obtained from the alternate equation given for H'

Q%7D)HL#H* 7 DC.C)#1CCC#282CCC* 7 #F.8x2CF"K$h

/ocation of feed point' z-7x-7 C.DC

H-7H/7 #2CCC "K$"mol )bubble point liquid*/ocation of =#point' )straight line through #point and feed point !-tox*

B7F#D7 DC.C "mol$h

x 7 C.CD)C.DC*)2CC.C*$DC.C 7 C.CD

H=7 #28@,CCC "K$"mol )from graph*


27/30

6ote' This point could also be obtained analytically using an energy

balance around the entire tower to obtain Q


28/30

If you continue the construction, nine equilibrium stages )or equilibrium

trays* are required to surpass a bottoms composition of C.CD mole fraction.

This is the minimum number required. If the tray efficiencies are less thanunity, meaning that equilibrium is not achieved on each tray, then the

number of actual stages required will be greater than nine.

b* simplified graphical solution )4c%abe#Thiele*

The first step is to construct ay#xdiagram for the ben!ene 0 toluene system at2 atm pressure. This can be obtained from the T#y#xdiagram discussed above.

nother way is to use separator calculations in HB&B& to produce data points

that can be used to generate the required y#x plot )e.g., in 9xcel*. diagram is

shown below with the graphical solution to this example.

The feed, overhead, and bottoms compositions are found )by material balances ifnecessary*. In this problem, they arex-7 C.DC,xL7 C.MD, andx7 C.CD )mole

fractions of ben!ene*. The feed line is drawn vertically atx-7 C.DC, from the @Do

line upward )for saturated liquid feed*.

The enriching operating line is drawn from its upper end aty27xC7xL)on the @D

o

line*, with a slopeL$V7R$)R02* 7 [email protected] 7 C.JD, terminating on the feed line.

-x !iagram for Ben"ene # Toluene

at $ atm

&

&)(

&)*

&)+

&)%

$

& &)( &)* &)+ &)% $

x .Ben"ene0

.

Ben"ene0

$(

3

*

4

+

5

%

'

x!

xB

xFFeed 9ine

Striing Oerating 9ine

Enriching

Oerating

9ine

Equilibrium /urve

Tra

>umbers

R


29/30

The operating line for the stripping section is drawn from where the operating line

for the enriching section terminates on the feed line, down to the bottoms point on

the @Doline )atx*.

&tages are stepped off from the top of the tower, between the operating line and

equilibrium curve, switching over to the stripping section operation line once thefeed line is passed. ll trays are located as points on the equilibrium curve. The

dotted lines indicate the step#wise solution up to tray 1. Bou should complete the

construction on your copy of the notes, to ma"e sure you understand how this isdone.

-rom this diagram, it is concluded that 2C equilibrium trays are needed to achieve

the desired separation. This is one more tray than we found using the 5onchon#&ararit method, but the agreement is pretty good, when comparing two graphical

methods.


30/30

tower with > trays, a partial reboiler )a Mthequilibrium stage*, and a total

condenser. If we had not done the 5onchon#&avarit method, we would need to

guess a number of trays, perform the simulation, and then modify the number oftrays to get the desired separation.

feed stream, an overhead liquid product stream, and a bottoms stream wereplaced on the flowchart.

The feed stream rate )2CC.C "mol$h*, composition )x27x87 C.DC* , and conditions)P7 2.CC atm, bubble point* were set.

-or the tower, the feed was set on tray @, which is the default location in the

simulator for this problem. -or the condenser, the type total was selected, andthe pressure set at 2 atm. partial reboiler was selected, and the pressure set at 2

atm. ll tray efficiencies were set to 2.CCC as default values in HB&B&, and these

were chec"ed but not changed.

Two tower specifications were written. -irst, the overhead rate was set at

DC.C "mol$h. &econd, the reflux ratio was set at 1.CC on a mole basis. 6ote thatthese can be done on the column menu during setup. &ince we set the number of

trays, we could not set the overhead composition as a specification.

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Equilibrium Stage Processes

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