Sugar Cane Process Performance
Transcript of Sugar Cane Process Performance
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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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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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H. Heluane et al. / Chemical Engineering and Processing 46 (2007) 198209 207
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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208 H. Heluane et al. / Chemical Engineering and Processing 46 (2007) 198209
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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