_ class and home prOblems )

The objec t of this column is to enhance our readers ' co llections of interesting and novelproblems in chemical engineeri ng. Problems of the type that can be used to motivate the studentby present ing a particular principle in class, or in a new light , or that can be assig ned as a novelhome problem , are requested , as well as those that are more traditional in nature and whichelucida te difficul t conce pts. Please submit them to Professor James O. Wi lkes (e-mail:wilkes@e ngin.umich.edu) or Mark A. Burns (e-mail: maburns@engin.u mich.edu ). ChemicalEngineering Department , University of Michigan , Ann Arbor, MI 48 109-2 136.

AN INTRODUCTIONTO PROCESS FLEXIBILITY

Part 2: Recycle Loop with Reactor

W . E. J ONES, J.A. WILSONUni versity of N ottingham No tti ngha m, N G7 2RD, England

An earlier article!I ] disc ussed teaching flexibilit y via asimple heat-exchan ge problem. Heat exchange is aconvenient starting point for flexibility teaching be-cause the equations and resulting ca lculations are easy tohandle. Other process operations can rapidly become com-plex and a rigorous treatm ent very demanding.!" Hence it ispleasing to find that a potentiall y complex recycle loop withreactor can be simplified to provide a thought-p rovokin g,process-flowsheet-based exercise that emphasizes understand-ing of the system's operation.

It is important to stress, eve n with simplification, that therecycle loop is still a much broader exampl e than that forheat exchange , makin g the exercise more suitable as a basisfor project work or a discussion question. One of the authorsfirst became interested in the prob lem when attempting toexpl ain the operating bounds of an ammonia or methanolsynthesis loop durin g the course of final-yea r design projects.The reactor used in this exercise is a simpler design thancomm only found in amm onia and methanol plants, but theexerc ise still illustrates many of the issues that make designand operation of such process systems so interesting.

BACKGROUNDThe exerc ise considers a reversible exo thermic reaction

taking place in the gas phase over a catalyst in an adiabatic

reactor. The reactor is part of a recycle loop where product isseparated from the reac tor effluent and unreacted feed isrecycled (see Figure I). It is well known that when designingrecycle loops, an eco nomic optimum exis ts for reactor con-version that balances reactor size against the combination ofprodu ct separa tion load and circulation rate.

Having chose n the design conversion, the circulation rateis an immediate consequence in order to meet the required

Warren Jones holds as c and PhD degrees inchemica l engineering from the University ofNottingham and is a registered Chartered Engi-neer. He has a wide-ranging interest in both front-end process and detailed plant design, devel-oped initially through nine years of experiencewith a major engineering and construction com-pany. Teaching responsibilities include severaldesign courses, and engineering thermodynam-ics.

Tony Wilson holds a s c and PhD degrees inchemical engineering from the University ofNottingham. With industrial and consulting ex-perience in process control and batch processengineering. and with active research in bothfields. he coordinates the department's researchin computer-aided process engineering and isresponsible for process control teaching at theundergraduate level.

224

Copyrigh t ChE Division of ASEE 1998

Chemical Engineering Education

The catalyst can be deactivated in several ways.!"Here, we are more interested in the overall effects ofdeac tivation than we are in the details of one par-ticular mechanism, so we adopt a simple model-namely, uniform deac tivation throughout the cata-lyst inventory. The normal response to loss of pro-duction through catalyst deactivation is to raise thereact ion rate by increa sing the inlet temp erature. Forrever sible exo thermic reactions, however, this ac-tion further limit s the conversion, and unle ss thecirculation rate is also increased , the prod uction tar-get will not be met.

The effects of catalyst deacti vation are demon-strated in Figure 2b, which compares performanceat beginni ng-of-run (BOR) and end-of-run (EOR).Note particularly that the BOR circulation rate is

a relatively low temp erature permits a higher conversion,but this must be balanced against a slower rate of reactionleading to a large catalyst inventory. The design conversionfor reversible exo thermic reactions ofte n approac hes quit eclosely the upper bound imposed by thermodynamics.

Finally, the temp eratu re wil l increase through the catalystbed, but a single tem perature value is needed to characte r-ize operatio n; for th is we wi ll use inlet temperatur e. It isassumed that reactor inl et temperatu re ca n be adju stedinde pe nde ntly .!!'

For a flexi bi lity study. the ca talyst inve nto ry and othe requipme nt will be fixe d; we are interested in investigat-ing how the plant will per form und er alte rna tive operat-ing co ndi tio ns.

From the earlier discussion. two obvious parameters tocharacter ize operation are circulation rate and reac tor inlettemperature. As noted before, the reactor inlet temperaturescan be adjusted to maximi ze production for fixed circ ulationrate (illustrated in Figure 2ap l).

To the left of the maximum, increasing inlet temperatureimproves the rate of reaction, lead ing to more pro-duction. But as the temperature increases, the dete-riorating equilibrium conversion imposed by ther-modynamics "bites," and production peak s. Opera-tion at the optimum inlet temperature ensures thatthe minimum circulation rate to meet production isbeing used ; we assume the optimum temp erature tobe selected in each case.

A typical industrial exa mple is more complex andinvo lves four additional considera tions: Catalyst deactivation Producing circulation around the loop Loop pressure Presence of an inert component in the make-up

gas535

525

Loopci rculation

515

tEOR CatalystAc tiv ity

20% extra flow

Reactor

20% less flow

505

Figure 2.

495

Circulationcompressor

Reactor Inlet Temperature K

Fig ure 1.

510 515 520 525 530Re ac to r In le t Te mperature K

'---- Product

- -------------

---

---

485

..".--------- ..............

...,.-

,/ / k"" 40 t

.-' .-_ .- . ' . _'.... "

....-~ .

30t

Pr od uctsepa ra tion

.--- -----

505

0.08 L- ...... ...... ...... ""- -'-_...I

475

Purge .(only needed ifMUG co ntai ns a ninert co mpo nent)

~ Iake-up ),L---+ r - - ---,gas (MU G) (,.

a0.125

0.12

u 0.115Ql.!!l 0.11"0E""

0.105(J'0 0.1l:0 0.095

:;:;u:::> 0.09"'0ea. 0.085

0.08

0.075500

production. The choice of reactor ope rating temp erature isalso important and may be opt imized by the desig ner as partof the reactor size-separation load/circulation rate trade-off.

In the case of reversible exo thermic reactions. selection of

Summer / 998 225

Parts 1 and 2 of this paper introduce flexibility by providing exercises th at force the studentto consider how the system will actually operate-this is an important first step fordeveloping a robust design. Further, a wide-ranging knowledge of basic chemicalengineering is required, making these exercises (particularly Part 2)ideal as the basis for project work.

15% less than that for Ea R; if the EaR circulation rate wereimplemented at BaR, overpro ductio n wou ld result.

Norm al design practice wou ld be to set the loop processparameters for EaR operation, but producing a design suffi-ciently flexib le to opera te at BaR is clearly very desirable.Loop circulation is normally produced using a var iable speedcentrifuga l compressor, so compressor characteristics be-come important in meeting process flexibility.

Centrifuga l compressor charac teristics are represe nted bya head-capacity curve and, for recycle systems (where theloop pressure drop is 10% or less of the nomin al operatingpressure) the loop can be treated as one of constant density,and the compressor characteristics will have roughly thesame shape as those obtained for centrifugal pump s.

Figure 3 shows the operating point to be defined where thesystem curve intersec ts the compressor characteristic; for arecycle loop the system curve is based solely on fricti onlosses beca use the discharge returns to the suction.

Also show n in Figure 3 is the implementation of reducedcirculation rate via compressor speed reduction. The affinitylawslsi can be applied to relate compressor performance atdifferent speeds . Sometimes the speed reduction requi red isquite large, taking the compresso r close to a critical speed,which induces synchronous whirling of the shaft; this condi-tion must be avoided and hence places a lower bound onoperation.

Typically, critical speeds are found in the range of 60 to80% of normal running speed. Two other constraints thatarise from compresso r opera tion are

loop pressure. Varying loop pressure is helpfu l when match-ing compressor operation to process requ irements because

A reduction in loop pressure increases the circulat-ing gas specific volume; hence. circulating the samequantity of gas would require a faster compressorspeed to cope with the larger volumetric flowrate,i.e., moving operation mvay fro m the critical speed.

If the reaction is one where there is volume reduc-tion as the reaction proceeds, pressure reductionwill give a lower, thermodynamic limited, conver-sion, and this will lead to a larger circulation rateand a move away fro m the critical speed.

The loop pressure cannot be set independently, however.In a simple exa mple where the feed contains no inert compo-nent , the rate of reac tant consumption (in the reactor) mustbalance the rate of make-up gas (MUG) flow into the loop inorder for the pressure to remain steady. In other word s, atsteady-state, the reac tor inlet temperature and circulationrate (combined with the pressure) must give a reaction ratethat just balances the MUG rate. If the circulatio n rate, say ,were smaller, the loop would equilibrate at a higher pressurewith the extent of the pressure increase being limited by therelief valve setting on the loop.

Finally, we need to consider an inert compo nent in theMUG. In many industrial exa mples, the MUG to the synthe-sis loop is not pure reactant, but contains I to 2% by volumeof an inert component. Unless the inert is removed, it willaccumulate in the loop and slow the rate of reaction. In apressurized loop, some inert will dissolve in the product, but

Chemical Engineering Education

800 r----------------------------,

1.41.31.21.1

EOR Operati on

SystemsCurves

0.8 0.90.7

10000 rpm

8420 rpm

Figure 3.

Ac tual Vol umetric Flowrate m A3/sec

400r----2~~--_~200 '----.......::;;'O::'--........ "'--__.......__---''--__.......__.......__--J

0.6

300

700

600

E'g SOOell:I:

Maintenance of sufficient volumetricflowra te through the machine to avoidsurge; for a low-head compressor (asconsidered here) this constraint willlie well to the lef t on the head-capacitycurve (around 50% of normal capac-ity) and hence will not be influential.Overspeed trip. typically set at 10 to15% above the normal running speed;this constraint obviously "bites " fo roperating modes requi ring higher cir-culation rates.

The constraints impo sed by compressoroperation lead to the third parameter thatrequired furth er consideration-namel y,

226

stoichiometric proportions (mixture RMM=50): hence therecycle stream also contains only A and B, and the purgeflowrat e will be zero .

The recycl e loop at end-of-run (EOR) is designed to oper-ate at a nomin al pressure of 50 bar, with a circulation ratethrough the reactor of 2.2 kmol sec' . The EO R loop pressuredrop is 5 bar; this comprises 4 bar for all equipment exclud-ing the reactor and I bar for the reactor. The EO R reactorpressure drop may be furth er decomposed:

Pressure drop (based 0/1 EOR circulation rate) forreactor if catalyst were in good mechanical condition:0.7 bar

Allowan ce (based on experience) fo r catalyst particlebreakdown and loss of voidage: 0.3 bar

Circulation through the reactor is provided by a single-stage centrifugal compressor who se head (He. m) - capacity(Q, rrr' sec') cha rac teristic at 10.000 rpm is represen ted by

He = 530.5 + 149.52 Q - 130.585 Q 2

The exercises below examine the process tlex ibi lity re-quirements for the synthes is loop as the catalyst inventory of35000 kg slowly deacti vate s. Catalyst deactivation is mod-eled by assumi ng EOR k, and kH values are only 30% ofthose applying at BOR (see Tab le I).

You may assume the heat exc hanger arrangements on thereactor feed have sufficient flexibi lity to prod uce a widerange of reactor feed temperatures and the compressorsuction operates at 40C. Also, assume the reacta nts A andB have negligible solubility in the product.

I) Determine the maximum production of C achievableat EO R conditions by varying the reactor inlettemperature .

2) You wish to achieve the same production rate of C asthat calc ulated in ( I). but using BOR catalyst.a) Calc ulate the circulation rate required if the loop

pressure remains at 50 bar.b) If the circulation rate is maintained at 2.2 kmol

sec:' . at what pressure wou ld the loop need tooperate?

c) Use the insight you have gained from solving 2(a)and (b) to exp lain the interrelatio nship betweenloop pressure . circu lation rate. and reactor inlettemp erature. How might you ex ploit this flexibil-ity to deal with variation of catalyst activity?

3) a) Confirm the compressor speed of 10.000 rpm willsatisfy EO R operating condition s.

b) Determine the co mpressor speed for 2(a).c) Determin e the co mpressor speed for 2(b).d) Comment on your answer to 2(c) in light of the

compressor speeds you have just calculated.

PROBLEM STATEMENT

BOR rate constant. krnol sec" bar" (kg of ca talyst)"525000 exp(-108000/8.314 T)partial pressures of A. B. and C. barabsolute temperature of the reactin g mixture. K

(-------------)

k, = BOR rate constant. krnol sec" bar" (kg of catalyst)"k, = 100 exp(-94000/8.3 14 T )

Rate of destruct ion of C by the reverse reaction = kHPcwhere

Specifi c heat capacities: CPA' CpR' Cpc = 30. 40. 70 kl krno l' K"

Heat of reaction =-14000 kJ per kmol of C formed

TAB LE 1Reaction Data for the Exercise

A +B ~ Ctakes place in the gas phase over a fixed catal yst bed in areac tor operated adiabatica lly. The reac tion data are summa-rized in Table I. The reactor form s part of a recycl e loop(show n in Figure I) where C is totally separa ted from thereac tor eftl uent. For the first three parts of the exe rcise, youmay assume the make-up gas co ntains only A and B in

The reversible exot hermic chemical reaction

Rate of production of C by the forwa rd reaction = k.PA PHwhere

most has to be purged from the loop (see Figure I). Arelatively high purge rate ensure s the inert composition inthe recycle loop is low and the rate of reaction high for thepressure beca use the inert has only a small diluti ng effect. Ahigh purge rate, however , results in a large reactant lossand. in des ign mode, se tti ng the purg e rat e is an eco-nomic optimization that add s the third dimension of MUGrat e to the reactor size- separation load/circulati on-ratebalance already di scussed .

From a flexibility viewpoint. recycle loops with purge canfully exp loit the interacting trio of parameters (circulationrate. reactor inlet temperature , and pressure ) becau se pres-sure can now be set indepe nde ntly via the purge rate ratherthan being a consequence, as previously noted.

To illustrat e, if the loop is initially at steady state, reducingthe purge rate causes the inert composi tion to increase, sothe loop press ure must increase to restore the reactant partialpressure s. There will be a compensating reduction in MUGdemand. If the circulation rate is also increased . it will limitthe change in press ure needed.

In view of the complexi ty added by con sidering an inert inthe MUG, the main part of the exercise concentrates on pureMUG; handl ing an inert in the MUG is co nsidered at the endof the exe rcise.

Summer /998 227

Chemical Engineering Education

Solving the quadratic gives N=8420 rpm ; this is shown asthe second compressor curve in Figure 3. (This is quite alarge speed reduction and may bring the compressor qu iteclo se to the critica l speed.)

dent s should be aware that there are three important vari-ables . Reactor inlet tempera ture can and should be opti-mized for all situations, but circulation rate and loop pres-sure are related and cannot be set independently. The aboveexe rcises take the two ex treme posit ions of maintain ing looppressure or maintainin g circulation rate; the correspondingmaximum red uctio n in circulation rate or loop pressure isthen calculated. In practice, an opera tor would use a sma llerchange in circulation rate and allow the loop pressure toequilibrate with a reduction somewhat less than the maxi-mum change previously calculated.

(circulation rate )2loop pressure

1 3(c)1In this case, the sys tem head will be proportional tothe

13(a.)1 Plottin g the head -capacity curve using the given equa-tion produces the 10,000 rpm line in Figure 3. The Ea Rcirculation rate is 2.2 kmol sec" or a co mpressor suctionflowrate of [(2.2)(0.08314)(313)]/50=1.145013 sec", and asmay be seen from the curve or equation, this implies that aHe of 530 .5m will be thrown up. If the loop pressure drop is5 bar, thi s is eq u iva le nt to an H, o f (5)( I05)/[(9.81)(96.07)]=530.5m, i.e., Hs=He and the co mpressorspeed of 10,000 rpm will satisfy Ea R opera ting conditions(96.07 kg m' is the gas density at the compressor suctionand is calculated from the ideal gas law).

13(b).1 In this case, the sys tem head will be proportional tothe (circulation rate r' because the loop pressure is main-tained at 50 bar and thus the gas density can be assum edconstant. We neglect the effect of minor com position andreactor exit temp erature changes. Thu s the sys tem curvepasses through the volumetric tlowrate [(2)(0.95 1)( 1.145)]/2.2=0.9899 rrr ' sec' at a head of [(4 .7)(530.5)(0.951 {2/2.2} r ]/5=372.7m. Stable opera tion requ ires the com pressor speedto be reduced such that the head-capacity curve also passesthrough this point. Let N be the new compressor speed andthe stable operating point must map back onto the 10,000rpm head-capacity curve . Thus

SOLUTION(------ - - -----)The solution of this problem requi res a model for the

chemical reactor, which is eas ily generated by integratingthe differential mass and energy balances thro ugh the reac-tor :

4) The M UG to the synthes is loop contains 2% of aninert co mpound, I, in addition to equal proportions ofA and B. If the circulation rate is 2.4 kmol sec' I andthe purge rate is 0.050 91 kmol sec', de term ine theloop opera ting pressure to meet the production targetfrom part I. What wo uld happen if the purge ratewere reduced further , and how would you mitigatethe co nsequences ?

where w is the mass of catalyst and q is the mult iplyingfactor to acco unt for ca talyst deacti vation.

At the reactor inlet, the tlow rates and tem perat ures areFA=FAO' FB=FBO' Fe=O, T=To, and at any subsequent point ,FA=FAO- Fe, FB=F Bo-Fe. Ideal gas behavior is assumed tocalculate the part ial pressures, e.g., PA=PTFA/(FA+FB+Fe),where PTis the nominal loop pressure. Any one of a numb erof numerical integration packages can be used.

dFc ( )- = q k pPAPB - kBPCdw

dT q 14000 (kp PAPB - kBPC)dw (FACPA + FBCpB + FcCpc)

[!JSetting FAo=Fso=l. l kmol sec:', q=0.3, and Pr50 bar,the model is run for a range of inlet temp eratures To. Fe isnoted after integration thro ugh 35 tons of catalyst and theresult s recorded as a plot of Fe against To(see the middl ecurve in Figure 2a). The peak indicates a maximum produ c-tion of 0.10206 kmol sec' for C at an inlet temp erature of517K. (Note the sensitivi ty to circulation rate changes, butrelative lack of sensitivity to cata lyst inventory changes asindicated by the other curves in Figure 2a.)

12(b).IFor BaR catalyst q= 1.0 and FAo=FBo= 1.1 kmol sec:',the problem requires a trial-and-error solution to find PTthatgive s the requi red production of Fe. A solution is found forPT=43.64 bar with a reactor inlet temperature of 495K.

12(a).IFor BaR catalyst q= 1.0 and Pr50 bar, the prob lemrequ ires a trial-and-error solution to find FAOand FBOthatgive the req uired production of Fe. A solution is found forFAo=FBO=0.95 I kmol sec' with a reactor inlet temp erature of487 K (see Figure 2b). (Note: maintaining the EaR circula-tion rate gives a large overprod uction .)

12(c).1By performing the calculations in 2(a) and 2(b), stu-228

because dropping the loop pres sure has a strong effect on gasdensity. The req uired point on the sys tem cur ve is defined by[( 1. 145)(50 ) ]/43 .64 = 1.3 1 19 m 3se c1 at a head o f[(4.7 )(530.5 )({50/43.64 f) ]=654.6m. Using the same ap-proach as that applied in 3(b), but this time anticipating aspeed increase, we solve the quadrat ic to give N= 11,160 rpm .(This is quite a large speed increase and may brin g themachi ne clo se to overspeed trip.)

13(d).1 Answer 2(c) advocates a midd le path- some redu c-tion in circ ulatio n rate and a corresponding loop pres surereduction. The answers to sec tion 3(b) and (c) show that ifloop press ure or circulation rate is mai ntain ed, the extreme sof com pressor speed are also approached, i.e., 2(a) take sopera tion close the critical speed and 2(b ) results in opera-tion near overspee d trip. Hence, the midd le path requires asteadier machine speed, closer to normal running speed.

2. Swan ey, R.E. , and I.E . Grossmann, "An Index forOperational Flexibility in Che mical Proc ess De-sign ," AlChE J ., 31(4), 621 (1985 )

3. Fromen t , G.F., and K.B . Bischoff, Chem ical Re-actor Ana lys is, 2nd ed., J ohn Wiley, New York ,NY, p. 425 (1990)

4. Ibid . p. 2195. Aer st in , F., and G. Street , Applied Chem ical Pro-

cess Design, Plenum Press, New York , NY, p. 237(1978 ) 0

REFERENCES1. J ones , W.E ., and J.A. Wilson , "An In troduct ion to

Process Flexibility: Part 1," Chem. Eng. Ed., 31(3 ),172 (1997)

X = 0.49M +(X - 0. 1020d 2.144971l 2.19588 )

? 4 - ?X=O O?M (? 4 -? X)( 2.14497 1_. - . - + _. - l Z.19588)

Parts I and 2 of this paper introduce flexibility by provid-ing exercises that force the student to considerhow the sys tem will actually ope rate- this is animportant first step to developing a rob ust de-sign. Furth er, a wide-ranging knowledge of ba-sic chemical engineering is required, makingthes e exerci ses (particularly Part 2) idea l as thebasis for proje ct work .

remain; namel y I) flowrates of A or B in the reactor feed, X,and 2) MUG flow rate. M. Writing component mass bala ncesfor a reactant and the inert at the mixing point ju st before thecompresso r gives

Solving for X and M gives X= 1.089976 kmo l sec " andM=0.255028 kmol sec ", and henc e the reactor feed for thesimulation is A and B, 1.089976 kmol sec' and I is 0.220048kmol sec' . Note particularly that the MUG rate has in-crea sed by 24.9%, compared to an inert free MUG , to ac-count for the react ants lost in the purge. So lving for the looppressure in this part of the exercise is also by trial and error.A loop operating pressure of 55 .62 bar will ensure the targe tproduction is met if the reactor inlet temperature is opti-mized at 5 17K.

Reducing the purge will increa se the inert s com position ofthe gas in the loop. At constant circulation rate, the changemust cause the loop press ure to increase in order to restorethe reac tan t partial pressures. For high-pre ssure systems, therelief valve is typically set at 10% above norm al runningpressure ; hence the calculated loop pressure of 55 .62 isprob ably unacceptable if 50 bar is viewed as norma l opera-tion . A further reducti on in purge flow would cer ta inly causethe relief valve to lift. To limit the pressure increase, the besttacti c would be to increa se the loop circulation rate.

SUMMARY

Reactor fcedCompo nent kmol/see

A X8 X1 2.4-2X

2.4

Reactor effluentComponent kmol/see

A X-0.102068 X-0.10206C 0.102061 2.4-2X

2.29794

2.14497 kmollsec

Figure 4.

L-__.... Pr odu ctComponent kmol/see

C 0.10206

l\Iole frac tion0.490.490,02

2.19588 kmollscc

Purge0.05091 kmol/sec

l\IUG, 1\1 (,L- - _ .- - - --,Co mpo nent

A81

~The equat ions represe nting the reac tor must be mod ifiedto includ e the presence of the inert. Two changes are needed:

I. Inclu sion of the inert component flowrate, F" in thecirc ulation rate (FA+FB+Fc+F,); this reduc es the par-tial pressure of any reactant.

2. Inclusio n of the inert component heat capacity, F,CPI,in the heat capacity (FACPA+FBCPB+FcCpc+F,Cp,);this provides a greater heat sink and so red uces thetemperature rise across the reacto r.

Before the reac tor simulation can be used , the effect of theinert component on the recycle loop material balance mustbe estab lished . In particular we need the reactor feed compo-sition. Figure 4 shows all the information easily deri vablefrom the prob lem sta tements , but two important unknowns

Slimmer 1998 229

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