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.
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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
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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
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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
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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,
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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.
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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$"mol )from graph*
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6ote' This point could also be obtained analytically using an energy
balance around the entire tower to obtain Q
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If you continue the construction, nine equilibrium stages )or equilibrium
trays* are required to surpass a bottoms composition of C.CD mole fraction.
This is the minimum number required. If the tray efficiencies are less thanunity, meaning that equilibrium is not achieved on each tray, then the
number of actual stages required will be greater than nine.
b* simplified graphical solution )4c%abe#Thiele*
The first step is to construct ay#xdiagram for the ben!ene 0 toluene system at2 atm pressure. This can be obtained from the T#y#xdiagram discussed above.
nother way is to use separator calculations in HB&B& to produce data points
that can be used to generate the required y#x plot )e.g., in 9xcel*. diagram is
shown below with the graphical solution to this example.
The feed, overhead, and bottoms compositions are found )by material balances ifnecessary*. In this problem, they arex-7 C.DC,xL7 C.MD, andx7 C.CD )mole
fractions of ben!ene*. The feed line is drawn vertically atx-7 C.DC, from the @Do
line upward )for saturated liquid feed*.
The enriching operating line is drawn from its upper end aty27xC7xL)on the @D
o
line*, with a slopeL$V7R$)R02* 7 1.C$@.C 7 C.JD, terminating on the feed line.
-x !iagram for Ben"ene # Toluene
at $ atm
&
&)(
&)*
&)+
&)%
$
& &)( &)* &)+ &)% $
x .Ben"ene0
.
Ben"ene0
$(
3
*
4
+
5
%
'
x!
xB
xFFeed 9ine
Striing Oerating 9ine
Enriching
Oerating
9ine
Equilibrium /urve
Tra
>umbers
R
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The operating line for the stripping section is drawn from where the operating line
for the enriching section terminates on the feed line, down to the bottoms point on
the @Doline )atx*.
&tages are stepped off from the top of the tower, between the operating line and
equilibrium curve, switching over to the stripping section operation line once thefeed line is passed. ll trays are located as points on the equilibrium curve. The
dotted lines indicate the step#wise solution up to tray 1. Bou should complete the
construction on your copy of the notes, to ma"e sure you understand how this isdone.
-rom this diagram, it is concluded that 2C equilibrium trays are needed to achieve
the desired separation. This is one more tray than we found using the 5onchon#&ararit method, but the agreement is pretty good, when comparing two graphical
methods.
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tower with > trays, a partial reboiler )a Mthequilibrium stage*, and a total
condenser. If we had not done the 5onchon#&avarit method, we would need to
guess a number of trays, perform the simulation, and then modify the number oftrays to get the desired separation.
feed stream, an overhead liquid product stream, and a bottoms stream wereplaced on the flowchart.
The feed stream rate )2CC.C "mol$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.
top related