Sugar Cane Process Performance

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H. Heluane et al. / Chemical Engineering and Processing 46 (2007) 198209 199

scheduling of continuous processes [68]. Jain and Grossmann

[9] studied the scheduling of multiple feeds on parallel units

and developed a mixed integer nonlinear programming model

(MINLP). Georgiadis and Papageorgiou [2] considered a cyclic

cleaning scheduling on heat exchanger networks and proposed

a mixed integer linear programming model (MILP). Alle et al.

[10] addressed the cyclic scheduling of cleaning and produc-

tion operations in continuous plants. In all cases performance

decay with time was considered, but the additional complex-

ity of multiple-unit parallel evaporation lines in the sugar cane

industry has not been contemplated yet.

Otherwise, a wide range of chemical engineering prob-

lems can be framed as mixed integer nonlinear programming

(MINLP) like process synthesis problems (e.g., heat recovery

networks, separation systems, reactor networks) and process

operations problems (e.g., scheduling and design of batch pro-

cesses) [912].

The objective of this work is to address the scheduling of

production and cleaning operations in a sugar plant with perfor-

mance decay. A detailed mixed integer nonlinear programming(MINLP) model including the effect of fouling on the overall

heat-transfer coefficient is presented.Multiple-unit parallel lines

are modeled for the evaporation section. The cyclic nature of

the cleaning operations is also taken into account. The objective

function to be minimized considers the costs of the evapora-

tion and the crystallization sections and other facilities (i.e. heat

exchangers) that require vapor (or eventually steam) to oper-

ate. The problem solution provides the following information:

the cleaning (maintenance) frequency, the mass flow to be pro-

cessed by each line, vapor bleed as energy source for external

heat requirements and the starting time (scheduling) for each

cleaning (maintenance) task in each line.

1.1. Problem statement

The evaporation and crystallization sections of the typical

sugar caneplant consideredin this workare shown schematically

in Fig. 1.

As seen in Fig. 1, the evaporation system, the crystalliza-

tion stage and other operations (i.e. heat exchangers) are steam

consumers. Heat exchangers are used for pre-heating the juice

before being fed to the first unit of the evaporation line. The

so called other operations can be operated with either vapor

generated at the evaporation and/or steam depending on plant

availability.

This paper seeks enhanced process integration in sugar plants

by considering the simultaneous roles of the evaporation and

crystallization sections as material processors as well as energy

suppliers.

In particular, the objective of this work is to determine the

optimal production schedule that minimizes the plant cost asso-

ciated to cleaning and steam consumed by the evaporation and

crystallization sections, and by other steam-consuming opera-

tions (other operations).

The problem can be formally stated as follows:

given:

(i) the amount of material to be processed during a certain

time period

(ii) the equipment models, parameters and initial status

(iii) the individual equipment performance as a time function

(iv) product (sugar) concentration(v) other steam related requirements (other operations heat

requirements)

determine:

(i) the cleaning (maintenance) frequency

(ii) the mass flow to be processed by each line

(iii) starting time for each cleaning (maintenance) task

(iv) flows of vapor extracted from the evaporation (bleeds).

1.2. Cost considerations

As explained by Heluane et al. [13], the aim of evaporation

and crystallization processes at a sugar factory is to eliminatewater from the juice and, thus, to obtain crystals of sucrose.

The evaporation process is economically more effective than

the crystallization process due to the multiple-effect scheme

employed (several evaporators working in series). In multiple-

effect evaporation withIunits, the water extracted from the juice

is approximately I times the steam used in the process. Other-

wise, at the crystallization stage the water is extracted roughly

in a proportion 1:1 with the consumed steam. Therefore, the

objective function (to be minimized) has to take into account

not only the additional cost due to the evaporator fouling, but

Fig. 1. Use of steam and extracted vapor for the system studied.


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200 H. Heluane et al. / Chemical Engineering and Processing 46 (2007) 198209

also the crystallization cost, a necessary although economically

less efficient operation. So it is important to consider the neces-

sary trade-off between evaporation and crystallization sections

in the overall cost study.

The evaporation process leads to the formation of fouling

on the inner surface of the evaporator tubes. The rate of

fouling formation is dependent on the nature of the feed, and

is particularly significant for the case of liquid feeds. Fouling

deposits inside the tubes act as insulation thus causing higher

heat-transfer resistance. It is convenient to clean the equipment

periodically in order to restore conditions of high heat-transfer

rate. If a high concentration of the product is desired, then

the evaporators have to be cleaned frequently which would

increase costs. Thus, there is also a compromise between juice

concentration and cleaning costs.

Special consideration must be given to the vapor produced

by an evaporation unit, which is mainly used for two purposes

(a) vapor source for the following evaporator unit, and

(b) vapor source for heating purposes other than theevaporationunits. This vapor is named as bleed in the sugar industry.

If the bleed is not enough to meet heating targets then more

steam must be generated at the boiler with the consequent

increase of operating costs. Usually, vapor produced by the

last units of evaporation lines and the vapor produced at the

crystallization stage are not used as energy source but they are

condensed in the so called barometric condenser to maintain

appropriate vacuum conditions in the system.

Hence, total operating cost can be expressed as

C = Cevaporator cleaning + Csteam evaporation + Csteam crystallization

+ Csteam other operations (1)

2. Mathematical formulation

2.1. Objective function

The objective is to minimize the steam cost at the evaporation

and crystallization sections as well as in other operations, i.e.

heat exchangers, and the evaporator cleaning costs. Eq. (1) will

be used as objective function.

2.2. Model assumptions

The following assumptions have been considered to formu-

late the model presented in this work:

i. Negligible sensitive heat and boiling point increase at the

evaporator units.

ii. No sugar loss during juice processing.

iii. Constant fouling factor during evaporation.

iv. Fixed operating conditions for the equipment units.

2.3. Fouling model for evaporation units

In order to calculate the global heat-transfer coefficient, an

empirical expression that depends on juice temperature () and

juice concentration (x)isused [14]. This expression is frequently

used in sugar industry calculations within the typical range of

temperature andjuiceconcentration andis known as theSwedish

formula

U =

x(2)

where is a proportionality constant.

During evaporator operation, and as a consequence of foul-

ing, global heat transfer coefficient gradually decreases with

time. In order to handle this situation, time dependence of U

must be taken into account in Eq. (2). The model that better

describes the decreasing behavior of U is given by the follow-

ing expression, which was determined fitting several models to

experimental data from a local sugar plant

U =

x(1 + bt)1/2(3)

The temperature () of the boiling juice inside the evaporator

is a parameter of the problem and is usually maintained constant

during equipment operation. In Eq. (3) juice concentration (x)

must be expressed as Brix (Bx) defined as grams of solids per

100 g of water.

As sugar mass remains constant at every evaporator (no sugar

loss), the outlet juice concentration for unit j can be obtained

from the mass balance under a pseudo steady state condition

(see Fig. 2)

xj = xj1 +xj1Vj

Fj(4)

If sensible-heat is neglected from the evaporator energy balance,Eq. (5) is obtained

jVj = UjAjj (5)

For a given operating time t, by substituting Ufrom Eq. (3) and

Vj from Eq. (5) into Eq. (4) the following expression for the

outlet juice concentration from unit j is derived

xj = xj1 +jAjj

jFj1(1 + bjt)1/2

= xj1 +j

Fj1(1 + bjt)1/2

(6)

Fig. 2. Scheme of two evaporation units working in series.


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Fig. 3. Scheme of multiple-effect evaporation line i and the following crystallization stage.

For sake of simplicity, a new variable j is introduced, as follows

j =jAjj

j(7)

By operating with sugar mass balance for unit j and xj given

by Eq. (6), the following expression is derived for the outlet

concentration of the juice leaving unit Mat operating time t:

xM = x0

Mj=1

1 +

j

F0x0(1 + bjt)1/2

(8)

where F0 and x0 are the mass flow and concentration, respec-tively of the juice fed to the evaporation system.

2.4. Steam for the evaporation section

As seenin Fig. 3, for each evaporation line i steam is fed only

to the first unit (j = 1) while for the j following units the energy

is provided by the vapor produced at the previous one, j 1.

Under the hypothesis mentioned above, the total steam, SE,

required for the evaporation system withNlines for an operating

time ti is given by Eq. (9)

SE =

Ni=1

sei =

Ni=1

Vi1ti (9)

By substituting the corresponding conservation balances for the

first evaporation unit of a line i in Eq. (9), the steam required by

the whole evaporation stage can be expressed as

SE =

Ni=1

Fi0

1

xi0

xi1

ti (10)

where Fi0 is the flow with a concentration xi0 fed to the first unit

of line i and xi1 is the average concentration of the flow leaving

the unit.

2.5. Steam for the crystallization section

Under the hypothesis mentioned above, the total steam

required for the crystallization section is given by Eq. (11)

SC =

Ni=1

sci =

Ni=1

VCiti (11)

By substituting the mass balances for crystallization and evap-

oration sections in Eq. (11), the steam required for the crystal-

lization stage for Nlines can be expressed as

SC =

Ni=1

Fi0xi0

xT xi

xTxi

ti (12)

where xi is the average concentration of the juice leaving unit

Mfor a line i.

Eq. (8) can be adapted to express the concentration of the

outlet flow of an evaporation line i with Mi units as follows:

xi = xi0

Mij=1

1 +

ij

Fi0xi0(1 + bijti)1/2

(13)

Sugar concentration of the juice leaving the evaporator decays

with time dueto thefouling of theheat-exchangesurface.Hence,the average concentration of the concentrated juice is given by

xi =

t2t1

xi dt

t2 t1(14)

Note that for calculating the average concentration of the con-

centrated juice when the evaporation line starts operating clean

(maximum heat exchange capacity) t1 = 0.

2.6. Steam requirements for other operations

Many heating operations are met by making use of the vapor

produced by the evaporators. As different vapors have differ-


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ent temperature levels, those vapors are selectively used (i.e.

vapors from first units are used exclusively as heating supply

for requirement E1). All vapors except those from the last unit

may be used for heating operations. Last unit vapors are sent to

a barometric condenser to assure vacuum conditions in the units

of the line (see Fig. 3). When vapor is not enough as heating

supply, steam is used.

Let us assume thatEj is the energy demand during the operat-

ingtimeofline i, ti, by operationsclassified as other operations

that can be supplied with vapor from thermal levelj. This is total

vapor produced by the units in the jth position in each line. Thus

Ej =

Ni=1

ijVBijti + ssrj (15)

where VBij represents the flow of vapor (bleed) extracted from

the evaporator j on line i, s is the heat of vaporization of steam

and srj is the amount of steam used when vapor from units j is

not enough to supply energy demand. Due to temperature levels,

when vapor is used, only that from unit j can be used to supplyEj requirements.

Hence, steam requirements can be obtained from Eq. (15)

srj =Ej

s

Ni=1

ij

sVBijti, j = 1, 2, . . . , M 1 (16)

when

Ni=1

ijVBijti Ej (17)

srj must be set equal to zero because bleed is enough to supply

other heating requirements.

Therefore, for a certain operation time ti, the total steam

consumption (sr) for other operations in a system withNevap-

oration lines will be expressed as

sr

0 if

Ni=1

ijVBijti Ej, j = 1, 2, . . . , M 1

M1j=1

srj =

M1j=1

Ej

s

Ni=1

ij

sVBijti

otherwise

(18)

2.7. Cleaning costs

Evaporators are cleaned by line, thus, all evaporators belong-

ing to a certain line are stopped at the same time and cleaning

operations are performed. For a certain time period the clean-

ing costs (Cc) for N evaporation lines can be calculated as

follows:

Cc = cc

Ni=1

ni (19)

where cc is the cost of cleaning one evaporation line; ni the

number the of cleanings of a line i during a certain period.

2.8. Cycle

Thecyclic nature of the scheduling may be taken into account

by the mathematical model. The model allows determining the

operation schedule for one cycle ofTC hours [9]. This cycle can

be repeated until the desired production level is achieved. If H

is the time horizon, then the number of evaporation cycles can

be calculated by the following equation:

=H

TC(20)

The steam consumed during a time horizon Hwill be

steam =

N

i=1

Vi1 +

Ni=1

VCi

ti + SR (21)

where

SR

0 if

Ni=1

ijVBijti EHj , j = 1, 2, . . . , M 1

M1j=1

SRj =

M1j=1

EHj

s

Ni=1

ij

sVBijti

otherwise

(22)

EHj is the energy demand (obtained from vapor from units j

and/or steam) for the time horizon H.

Therefore the objective function canbe expressed as follows:

minFO = csu

N

i=1

Vi1 +

Ni=1

VCi

ti

+

M1j=1

j(TC, VBij, ti)

+ cc N

i=1

Ni (23)

with

j(TC, VBij, ti) = max

0,

EHj

s

Ni=1

ij

sVBijti

,

j = 1, 2, . . . , M 1 (24)


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The equivalent mathematical representation of Eq. (24) is the

following:

LO1(1 zj) EHj

s

Ni=1

ij

sVBijti UP1zj,

j = 1, 2, . . . , M 1 (25)

0 j(TC, VBij, ti)

EHj

s

Ni=1

ij

sVBijti

UP2(1 zj), j = 1, 2, . . . , M 1 (26)

0 j(TC, VBij, ti) UP3zj, j = 1, 2, . . . , M 1 (27)

where LO1, UP1 are lower and upper bounds onEHjs

N

i=1

ijs VBijti, where j =1, 2, . . ., M 1, respectively. UP2 is

upper bound on j(TC, VBij, ti)

EHjs

N

i=1ijs

VBijti

,

where j = 1, 2, . . ., M 1. UP3 is upper bound on j(TC, VBij,

ti) j =1, 2, . . ., M 1. zj is a binary variable. If zj =1 then,

j(TC, VBij, ti) =

Ejs

N

i=1ijs

VBijti

, and if zj =0 then

j(TC, VBij, ti) = 0 .

2.9. Integrality constraints for the number of subcycles

Each evaporation line may be cleaned many times during one

cycle time (TC). This fact determines subcycles (Ni) for eachline

Ni =

Kk=1

kyik, i (28)

Kk=1

yik = 1, i (29)

If the number of subcycles for evaporation line i is k then the

binary variable yik is one. Note that for any evaporation line

the number of subcycles will be at least one, therefore all the

evaporators will operate during the cycle time.

2.10. Last evaporation unit outlet flow concentration

Given the operation time for each subcycle (ti/Ni) Eq. (14)

needs to be solved for each particular case (the setMi) for obtain-

ing outlet juice average concentration

xi =x0

t

t2t1

M

j=1

1 +

j

F0x01 + bj tiNi1/2

(30)

2.11. Mass balance

For a system of N lines, the total mass flow of juice

(F) fed to the evaporation system must be processed in the

evaporators

FTc =

Ni=1

Fi0ti (31)

Mass and energy balances for each unit can be expressed by the

following equations:

xij = xij1 +

2ij

1 + b ti

Ni

1/2 1

bFij1

tiNi

, i, j (32)

Vij = Fij1

1

xij1

xij

, i, j (33)

Fij = Fij1 Vij, i, j (34)

If a unit does not exist for an evaporator system ij will be zero

(see Eq. (7)). On the other hand, when ij is not zero an amount

of vapor is generated in unit (i, j) and is available to be used at

the next unit of the line. Therefore

If ij = 0 then VPij = Vij+1, i,j = 1, 2, . . . , M 1

(35)

The bleed can be calculated as follows:

VBij = Vij VPij, i, j (36)

If for a given unit (i, j) no bleed is required, VBij will be

equal to zero for that unit. An additional constant is usedBij.This

constant takes the value 1 when the unit has a bleed, otherwise

the constants value is 0

If Bij = 0 then VBij = 0, i,j (37)

2.12. Storage tank

The implementation of the results of this model will require

a storage tank because the inlet flows (Fi0) to the evaporationsystem remain constant during TC (operation + cleaning times).

Therefore, when line i is shut down to be cleaned, the corre-

sponding Fi0 is diverted to a storage tank until the operation of

the evaporation line i is re-established. As operating times are

longerthan cleaning times,it is possible to implementa sequence

of cleaning in such a way that no overlapping of cleaning oper-

ations occurs. Hence, the minimum desirable tank volume is

given by the following equation:

vol = maxi

i

F

l=iFl0

(38)


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2.13. Processing and cleaning time

The total time (operating and cleaning time) for line i is given

by the following equations:

ttoti = Nii + ti, i (39)

TC

= Ni

i+ t

i+ sl

i, i (40)

sli UP ysli 0, i (41)

Ni=1

(1 ysli ) 1 (42)

The above constraints ensure that Eq. (39) will accomplish for

at least one line.

Being L a number large enough, the following constraints

ensure that the processing time for every line is longer than the

cleaning time

ti LiNi, i (43)

2.14. Bounds

Ni 1, i; Fi0 > 0, i; Tc > 0;

yik {0, 1}, i, k; ysli {0, 1}, i; zj {0, 1}, j

(44)

The MINLP model has Eq. (23) as objective function and the

above constraints.

The formulation is flexible enough to model multiple unit Munits) and parallelNlines) evaporator systems. It can also model

situationswhere extraction of vapor (bleed) from the evaporation

units is needed to supply other operations.

2.15. Case study I

The following example is based on a sugar plant located in

Tucuman, Argentina. Five parallel evaporation lines are consid-

ered and each line is a quintuple effect system. Concentration

of the juice leaving evaporation line i will be expressed by Eq.

(45) which was obtained by integrating Eq. (13) with time for a

line with five evaporation units and assuming identical fouling

coefficient (b) for all evaporators

xi = xi0 +2

b

1 + b ti

Ni

1/2 1

ti

Ni

Mij=1ij

Fi0

+ln

1 + b tiNi

ti

Ni

Mij=1

r>j(ijir)

bF2i0xi0

+2

b

1 + b ti

Ni

1/2 1

ti

Ni1 + b ti

Ni1/2

Mij=1

r>j

s>r(ijiris)

F3i0x2i0

+

Mij=1

r>j

s>r

t>s(ijirisit)

F4i0x3i0

1 + b ti

Ni

+2

3b

Mij=1j

F5i0x4i0

1 + b ti

Ni

3/2 1

1 + b tiNi3/2

(45)

For this particular case also the operating time is imposed to be

at least six times the cleaning time of each line

ti 6iNi, i (46)

It is desired to determine a configuration and cycle schedule

to process 800 t/h of 16 Bx juice. The final concentration of the

sugar (xT) must be 99 Bx. As vapors extracted from units 1, 2,

3, and 4 have different enthalpy conditions, they will be used

at different stages of the process. The energy demand of each

type of vapor (EHj ) is 42,200 MW h for a time horizon (h) of

720 h. If any vapor is not enough to meet the requirements, steam

will be used. Cost of cleaning one evaporation line (cc) wasassumed $4500 and steam cost per mass unit (csu) 8.386 $/t.

Some parameters of the problem are shown in Table 1.

The case was implemented in GAMS [15] using DICOPT++

as a solver. The results are given in Tables 2 and 3.

The optimal value obtained for the objective function is

$4,076,400 for the time horizon and the cycle time is 154h.

Table 1

Parameters of the problem case study I

Line i1 i2 i3 i4 i5 i (h) b

1 1025 770 621 503 450 18 0.01

2 649 550 449 349 343 16 0.013 848 688 602 517 386 18 0.01

4 977 843 749 667 550 16 0.01

5 908 721 645 531 458 19 0.01

Table 2

Results of the case study I

Line ti (h) Fi0 (t/h) Ni xi (Bx) SRj (t) FPi (t/h)

1 136 189 1 35.7 32634 84.6

2 138 150 1 32.6 48424 73.7

3 136 172 1 35.6 53653 77.2

4 138 209 1 36.2 58804 92.3

5 135 182 1 36.0 0 80.7

Table 3

Bleed (t/h) from the units for case study I

Line Evaporator units

1 2 3 4 5

1 14.2 7.6 4.9 2.5 0

2 7.4 5.4 4.0 1.1 0

3 10.9 5.8 4.2 4.2 0

4 10.9 6.9 4.9 4.4 0

5 12.1 5.7 5.0 3.0 0

Nos. 15 refers to lines.


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Fig. 4. Gantt chart of the optimal cleaning time distribution.

Cleaning cost is $105,194. As seen from Table 2 the amount ofvapor generated in the evaporation stage is not enough to supply

other energy requirements (SR1, SR2, SR3 and SR4 >0).

The cleaning policy showed in Fig. 4 will be adopted in order

to avoid the following inconveniences:

1. Under-utilization of cleaning resources, in particular man-

power.

2. The superposition of the cleaning times at the end of the

schedule yields a great mass accumulation that would require

storage until the beginning of a new cycle.

This sequence allows using the same manpower for cleaningoperation and, on the other hand, avoids the storage of juice for

a long time that would cause a decrease in sugar yield.

2.16. Storage requirements

The Gantt chart shown in Fig. 4 presents 10 time intervals

where the variations of total processed mass flow are due to

the cleaning policy and different feed conditions. If the pro-

cessed flows are analyzed at each interval, two situations are

observed. When one line is being cleaned, the mass flow arriv-

ing to the evaporation section exceeds the flow being processed,

and when the five lines are operating simultaneously the oppo-

site occurs. Therefore, it is necessary to contemplate the storageof juice so that the evaporation section could be operated contin-

uously (which is not explicitly taken into account in the MINLP

model).

As shown in Fig. 5, the accumulation of juice is 13,742 m3

per each cycle when all units are stopped and cleaned at the same

time.

When the cleaning task for a unit starts immediately after

the previous one is finished, the volume of juice accumulated is

6840 m3 (see Fig. 5). In the situation shown in Fig. 4, the juice

is accumulated while one line is being cleaned and immediately

used in the next time interval where the five lines are work-

ing together. In that case, the storage requirement is reduced

Fig. 5. Storage of juice per cycle for different situations.

to 1716 m3 as shown in Fig. 5. Any storage tank of a volume

of 1716 m3 or higher will allow the operation of the proposed

optimum scheduling but higher storage capacities gives more

operational flexibility.

Fig. 6 shows the variation of the value of the optimum (mini-

mal) costs defined by Eq. (1) with storage tank volume available

for that purpose at the plant.

2.17. Case study II

The same problem was considered when no tank is avail-

able (zero storage). Then, the Gantt chart shown in Fig. 7 is

obtained. The optimal value obtained for the objective functionis $4,185,186, a cycle time of 95 h, and flows fed to the evapo-

ration lines of 200 t/h.

2.18. Computational statistics

The GAMS modeling system was used to implement the

mathematical model as mentioned above. The NLP subproblem

was solved using Minos5.

The resulting MINLP for the case study had 60 binary vari-

ables, 227 continuous variables and 263 equations. The solution

was obtained in 0.70 CPU seconds on a Pentium I.

Fig. 6. Variation of total operation cost with available storage tank volume.


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Fig. 7. Gantt chart with zero storage.

The sensitivity of the solution to the initial point provided to

the solver is shown in Fig. 8. A sample of 96 points was cho-

sen at random. The minimum value obtained was $4,076,400(solution of the problem) and the maximum was $4,120,600

(range: $44,200). Only 22.9% of the initial points led to the

minimum cost value but for the rest of the initial points, costs

were below 1% of difference from the minimum cost value

($4,076,400) although the schedules were different. This gives

a remarkable flexibility because the system can be operated

with different cleaning schedules and costs remain practically

constant.

2.19. Case study III

Case study III has the same parameters than case study I, thedifference between them is the evaporator scheme. The evap-

oration system was considered as follows: five parallel lines,

three of them with quintuple units, one with quadruple units and

another one with 3 units. Values of ij are the same as in case

study I except 15 = 0, 54 =0, and 55 = 0. Therefore, line 1 is

a quadruple unit line while line 5 is a triple unit line. The opti-

mal value for the objective function is $4,295,100. The lack of

three units in the new evaporation system causes an increase of

water to be extracted at the crystallization section and hence an

increase of costs.

Fig. 8. Sensitivity of the solution to the initial point.

2.20. Heuristic case

For comparison purposes, a heuristics based case study is

used to schedule the five lines. A 7-day cycle is considered.

During the first 87 h of the cycle (time devoted to cleaning

sequentially every line) each operating line is fed at a rate of

200 t/h of juice. At the following period (87168 h), all lines are

fairly clean and operate with the same mass flow of 160 t/h. The

cost calculated in this case was $4.618E+6 which imply savings

of $542,000 for the time horizon of 720 h.

2.21. Sensitivity analysis

The influence of the different parameters and variables on

the objective function, once the optimum has been achieved,

was determined.

The relative influence of the main parameters of the evapora-

tor model such as cleaning costs per unit, steam costs per mass

unit, body temperature, driving force, area, fouling factor, inlet

juice concentration, and final sugar concentration on the objec-tive function has been studied and the corresponding results for

the case study are shown in Fig. 9.

Theparameter used for such purposes is Sp defined according

to

Sp =Z

p

p

Z(47)

The influence ofA, and is the same, in agreement with

Eq. (7). It should be also noted that even when the influence of

inlet juice concentration and sugar concentration are significant,

their values could be obtained with relative accuracy from plant

information. More significant is the relatively small dependence

Fig. 9. Sensitivity to parameters of the model.


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Fig. 10. Sensitivity to an increase of 10% in variable values.

on b, which is usually the most uncertain parameter. Otherwise,

the model had shown a strong influence of steam cost per mass

unit, which is another uncertain parameter.

In order to reflect the loss of performance when an optimaloperating condition cannot be implemented, the sensitivity due

to the decision variables, Sv, has been calculated according to

Sv =Z

v

v

Z(48)

where v has been set to 10%.

According to Fig. 10 (v = +10%), the objective function

shows to be especially sensitive to the cycle time Tc) and the

flows for every evaporator (Fi0). A similar result is obtained for

(v = 10%).

2.22. Technical objective function

If cleaning costs are neglected, minimum costs are obtained

maximizing the outlet juice concentration at the evaporation

stage [13]. Therefore, a new MINLP problem was solved. In this

case the constraints were maintained but the following equation

was considered as objective function:

max xT =

Ni=1Fi0tiNi=1

Fi0tixei

(49)

where xT is the average outlet juice concentration of all evapo-

ration lines.

2.23. Case study IV

The same parameters as in case study I were used. The opti-

mal value obtained for the objective function is 35.6 Bx. Thecal-

culated costs using optimum values obtained from the problem

solved with the technical objective function has no significant

difference with the cost obtained with Eq. (23) as objective func-

tion. Table 4 shows the results obtained for the different cases

studied. The optimization of a technical objective function with

economical background is useful in cases where uncertain cost

parameters are involved. For instance, in this work, accurate val-

ues of cleaning and steam costs may be difficult to determine,

Table 4

Results of the optimization studies

Variable Case study Heuristic

I II III IV

t1 (h) 136 77 135 99 150

t2 (h) 138 79 137 101 152

t3 (h) 136 77 135 99 150t4 (h) 138 71 137 101 152

t5 (h) 135 76 134 98 149

F1 (t/h) 189 200 176 198 200/160

F2 (t/h) 150 200 150 150 200/160

F3 (t/h) 172 200 192 178 200/160

F4 (t/h) 209 200 233 221 200/160

F5 (t/h) 182 200 150 192 200/160

x1 (Bx) 35.7 36.5 33.5 35.9 36.9

x2 (Bx) 32.6 28.8 32.6 33.7 28.9

x3 (Bx) 35.6 33.9 32.9 35.9 34.0

x4 (Bx) 36.2 40.5 33.5 36.0 39.9

x5 (Bx) 36.0 35.7 31.1 35.9 35.5

FP1 (t/h) 84.6 87.6 84.0 88.0FP2 (t/h) 73.7 111.1 73.7 71.3

FP3 (t/h) 77.2 94.2 93.3 79.4

FP4 (t/h) 92.3 78.9 111.6 98.4

FP5 (t/h) 80.7 89.5 77.0 85.3

C($) 4.07E6 4.18E6 4.29E6 4.09E6 4.62E6

C(%) 0 2.7 5.4 0.49 13.5

hence, to find an alternative objective function is a valuable tool

for process optimization.

3. Conclusions

Efficient process integration in sugar cane plants and

enhanced operation performance may be achieved by consider-

ing the combined operation of the evaporationand crystallization

sections, along with the appropriate management of their asso-

ciated steam bleeds for satisfying energy demands from other

plant operations. A common cost objective allows formulating a

problem for determining the optimal operating conditions under

different scenarios.

Aimed at a practical application of the results this work seeks

to evaluate different operating conditions of multiple evapo-

ration systems working in parallel in order to choose those

conditions leading to minimum operating costs.

A MINLP model was developed to determine an optimalschedule for the evaporator system. The formulation is flexible

enough to model multiple units (Munits) and parallel (Nlines)

evaporator systems, as well as network arrangements arising

from the combination of these basic cases. The formulation may

also consider bleed at any unit. Results show that significant

savings of steam could be achieved just operating the evapora-

tion section in a different way and with no additional investment

needed.

Although the solution of the MINLP model is sensitive to

the initial point, most of the times, costs were only about 1%

higher than minimum cost (optimal solution). This situation is

remarkable because it gives operational flexibility because the


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evaporation system may be operated with different schedules

without sensible cost increase.

Identical costs were determined with an alternative technical

objective function which is convenient because the objective

function based on costs is very sensitive to a parameter (steam

cost per unit mass) that has a fairly uncertain value.

Acknowledgements

This work was partially supported by Consejo de Investi-

gaciones de la Universidad Nacional de Tucuman (Argentina).

Support received by the European Commission is alsothankfully

acknowledged (Project no. MRTN-CT-2004-512233).

Appendix A. Nomenclature

A heat-exchange area (m2)

b Fouling coefficient for the evaporator

Bij bleed constantcc cost of cleaning one evaporation unit ($/unit)

csu cost of steam per mass unit ($/t)

C total cost ($)

Ccleaning costs of the cleaning operation ($)

Csteam crystallization cost of steam of the crystallization section

($)

Csteam evaporation cost of steam of the evaporation section ($)

Csteam other uses cost of steam used as supply for other operations

of the process ($)

E energy required for other operations (MW h)

F total mass flow of fed juice (t/h)

Fi0 mass flow of juice fed to line i (t/h)FPi juice flow leaving evaporation line i (t/h)

H time horizon (h)

K maximum expected number of cleaning tasks during

TcNi number of subcycles in line i

RQ mass of vapor and/or steam required for other opera-

tions (t/h)

sci steam condensed at the crystallizer in the line i (t/h)

sei steam condensed in the first evaporator of each line i

(t/h)

sl slack variable

srt steam required for other operations of the process for t

(t/h)Sp sensitivity to parameters

Sv sensitivity to variables

SC total steam condensed atthe crystallization section (t/h)

SE total steam condensed at the evaporation section (t/h)

SRH steam required for other operations of the process for

H (t/h)

ti total operation time of line i (h)

ttoti processing and cleaning time of line i in Tc (h)

TC cycle time (h)

U global heat-transfer coefficient (kW/m2 C)

UP upper bound

vol storage tank volume (m3

)

V total water removed as vapor from an evaporator (t/h)

VB vapor removed as bleed (t/h)

VC water removed as vapor from crystallization section

(t/h)

VP vapor removed from an evaporator and derived to the

following one (t/h)

xij outlet juice concentration at evaporation unit (i, j) (Bx)

xi0 inlet juice concentration at evaporation unit j = 1 (Bx)

xT sugar concentration of the product obtained at the crys-

tallization section (Bx)

x0 concentration of the juice fed to an evaporator (Bx)

xi average concentration of the concentrated juiceat evap-

oration line i (Bx)

X average sugar concentration obtained at evaporator

(Bx)

yik binary variable (yi,k= 1 if unit i operates ksubcycles in

Tc)

ysli , zj binary variable

Indicesi evaporation line

j evaporation unit

Greek letters

proportionality constant (kWBx/(m2 C2))

number of evaporation cycles in the time horizon

juice temperature in the evaporator (C)

driving force (C)

heat of vaporization of water (kWh/t)

i time devoted to clean line i (h)

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Sugar Cane Process Performance

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