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Case Studies in Thermal Engineering 34 (2022) 102035
Available online 13 April 2022
2214-157X/© 2022 Published by Elsevier Ltd. This is an open access article under the CC BY-NC-ND license
(http://creativecommons.org/licenses/by-nc-nd/4.0/).
Dynamic simulation of multiple-effect evaporation
Rubens E.N. Castro
*
, Rita M.B. Alves, Claudio A.O. Nascimento
Department of Chemical Engineering, Escola Polit´
ecnica, Universidade de S˜
ao Paulo, Av. Prof Luciano Gualberto 380, tr 3, S˜
ao Paulo, SP, 05508-
900, Brazil
ARTICLE INFO
Keywords:
Multiple-effect evaporation
Batch simulation
Sugar crystallization
Quasi-steady state
Non-steady state
ABSTRACT
The main purpose of this work is to simulate multiple-effect evaporation and its variation as a
result of the interconnection between batch and continuous unit operations. Therefore, the
challenges posed by the hybrid continuous-discrete character of the model are addressed. The
computer code is based on detailed fundamental material and energy balance equations, physical
and thermodynamic properties, and well-proven correlations for the heat transfer coefcients. A
sugarcane industry was used to demonstrate the effectiveness of the model; in this industry, there
is an interconnection between a batch and a continuous process. The steam consumed in the batch
process changes in each batch step. The results showed that changing the steam ow rate changes
the pressure of the steam produced in a multiple-effect evaporation. Additionally, when a bypass
valve and a relief valve are used to maintain constant steam pressure, it increases the overall
steam consumption.
1. Introduction
The multiple-effect evaporation system, MEE, was rst adopted in the sugar mill industry in Florida in 1830. Invented by Norbert
Rillieux [1], it has since become widely used in many industries, such as pulp and paper [2], salt production [3], and water desali-
nation [4]. An evaporator is essentially a heat exchanger in which vapor is used to boil a liquid to produce vapor. It is hence possible to
treat it as a low-pressure boiler, in which the steam thus produced is used in the following evaporator, and/or it can be used as a heat
utility in the industrial process. Similarly, MEE is a thermal unit operation used in industrial processes to produce a concentrated
stream. Therefore, MEE plays two roles in the industry: producing utility steam and concentrating a given stream. The higher the
number of effects, the more efcient the evaporation process is, which means that the less energy is used to concentrate it.
A dynamic process is one that constantly changes, i.e., a non-steady state. To simulate the multiple-effect evaporation in a non-
steady state, we assume a quasi-steady state: this is the situation where all parameters and variables change slowly enough for
them to be considered constant for a short period. As an example to illustrate MEE in dynamic behaviour, we used a sugarcane industry
process.
In a sugarcane industry, not all processes operate continuously. In some operations, such as sugar crystallization, the main steps
operate discontinuously. A batch process delivers its product in discrete amounts. This means that heat, mass, temperature, con-
centration, and other properties vary with time. Most batch processes are made up of a series of batches and semi-continuous steps. A
semi-continuous step runs continuously with periodic start-ups and shutdowns. A liquid stream from a continuous process in-
terconnects to a batch process using buffers, i.e., storage containers. For example, sugar crystallization occurs in a batch vacuum pan;
when the sugar crystals reach the required size, the syrup stops being fed and the massecuite is discharged into the massecuite tanks.
* Corresponding author.
E-mail address: rubenscastro@usp.br (R.E.N. Castro).
Contents lists available at ScienceDirect
Case Studies in Thermal Engineering
journal homepage: www.elsevier.com/locate/csite
https://doi.org/10.1016/j.csite.2022.102035
Received 24 January 2022; Received in revised form 4 April 2022; Accepted 10 April 2022
Case Studies in Thermal Engineering 34 (2022) 102035
2
The massecuite tanks vary their level to ensure the continuous supply of massecuite to centrifugation. Not only do liquid streams, such
as syrup and massecuite, vary during the crystallization but also the steam consumed in the vacuum pan. Steam consumption varies
because the crystallization process stops to discharge the massecuite and ll with magma before the start of the next crystallization;
additionally, steam consumption also varies during the crystallization process itself: the crystallization starts with a high steam
consumption and ends with a lower one due to the increase in the viscosity of the massecuite [5,6]. The steam used in this process is
bled from the multiple-effect evaporation and this is the reason why MEE operates in a non-steady state.
MEE allows many congurations to be designed; for example, Kaya [7] has modelled many systems, such as counter-current,
co-current, and parallel-current; Ahmetovi´
c [8,9] used vapor recompression to recycle steam. In the sugarcane industry, the
co-current conguration is preferred to avoid sucrose degradation [10]. In addition to different congurations, many thermal in-
tegrations with different streams within the main industrial process are also possible [11] or even with some by-processes, such as
vinasse concentration [12]. The integration of MEE into the process has also produced many studies on process optimization [13,14]
and the application of the pinch analysis technique [11,15]. The complexity level of the mathematical model depends on the
assumption each author made, which results in a number of possibilities in modeling, such as Heluane [16], who considered fouling in
their modeling, and Westphalen [17], who considered boiling point elevation. The models used by the authors cited above usually
consider the temperature (or saturation point pressure) as a constant parameter. This assumption simplies the resolution of the model
since the enthalpy, heat capacity, and heat transfer coefcient are calculated using equations that depend on the temperature and/or
pressure. However, these assumptions would produce an imprecise answer and would not be able to simulate changes in steam bleed,
such as those caused by the batch crystallization process in the sugarcane industry.
Fig. 1. –Steam diagram.
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
3
Unlike the models aforementioned, this study proposes a model in which the temperature and pressure of the produced steam are
estimated based on the amount of steam consumed: i.e., different consumption of steam results in different steam temperatures. When
the temperature of the steam produced in the MEE body increases near the temperature of the fed steam (steam from the previous body
in the MEE or exhaustion for the rst effect), the production of steam decreases due to the small temperature difference, δT. The steam
produced, ˚
m in the ith effect, is obtained by Equation (1).
˚mi=Ui·Ai·δT
ΔHi
(1)
Therefore, changes in the steam consumption in the process, ˚
m, result in a change in the temperature difference to produce more or
less steam in the MEE. Since the fed steam temperature cannot be modied by the evaporator, the temperature that changes close to or
far from the fed steam temperature is that inside the evaporator (T
i
in δT =T
i
–T
i-1
): therefore, even an evaporation body with a large
evaporation area would not produce a large amount of steam when T
i
approaches T
i-1
also, a small area evaporator can produce a large
amount of steam when the temperature of the produced steam is far from the inlet steam temperature. Additionally, if the temperature
is not constant, the heat transfer coefcient, U, and the enthalpy, ΔH, are not constant either, since they are obtained using equations
that use the temperature or pressure as a variable. As shown in Equation (1), both parameters are used to nd the evaporation rate.
Therefore, the temperature and pressure of the steam produced in the MEE are a function of the evaporation area (which is constant)
and the rate of steam bleed (which varies due to the batch process). The evaporation rate is the sum of the steam bleed and the steam
consumed in the next evaporation body. The model proposed in this work also includes the elevation of the boiling point and the sugar
concentration in solution. The following hypotheses to reduce the complexity of the mathematical model were used in this study:
•The steam produced in the evaporator is free from sugar or other solutes;
•No degradation of sucrose during the evaporation process;
•No pressure drop in steam pipes.
•Heat dissolution was neglected
1.1. Process overview
The application of the proposed model is illustrated through a typical sugarcane industry, which uses batch crystallizers to produce
sugar. We have considered that the sugar mill has a ve-effect evaporation train, a cogeneration system, and an annexed ethanol
process that uses the nal molasse and part of the sugarcane juice.
Multiple-effect evaporation (MEE) is responsible for providing thermal energy for all industrial processes, such as the crystalli-
zation of sugar, distillation, and juice heating. Therefore, the total thermal energy input for the whole renery is the exhaust steam,
from the turbo-generator or straight from the boiler via a bypass valve, which enters the rst-effect evaporator. As an example, Fig. 1
shows a steam diagram. The blocks identied as EVAP
i
represent the ith evaporation body. In this gure, vapor V
i
is the vapor from the
ith evaporation body. Part of the vapor produced by the rst and second evaporation bodies is used as a process utility heat.
In general, evaporators consist of a heating section and an expansion vessel. The Robert evaporator type, which is the most used in
sugar factories, consists of a calandria at the bottom and a ash space at the top. Heat is transferred through the tube wall in the
calandria: the hotter uid (fed steam) goes outside the tube and the cold (liquor), inside; thus, the steam condenses outside the tube,
Fig. 2. Temperature distribution in a ve-effect evaporation train.
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
4
and the liquid evaporates inside. The driving force for heat transfer is based on the temperature difference between the steam and the
liquor, as shown by Equation (2). The greater the temperature difference, the greater the heat transfer, and thus the greater the
production of steam.
qi=Ui·Ai·δT,i=1,…,N(2)
Fig. 2, shown here for explanatory purposes only, present two hypothetical scenarios of the temperature distribution in the same
multiple-effect evaporator operating with different temperature distributions. Only the rst and last temperatures are the same across
the two distributions, because the rst is the temperature of the steam from the turbine or boiler, and the last depends on the cooling
water and on the efciency of the barometric condenser. If the area and overall heat coefcient are assumed constant, the increase in
the steam bleed results in an increase in the temperature difference. In the example shown in Fig. 2, the distribution b) produces more
steam in the 2nd effect than a); however, the temperature (and pressure) of the steam produced in b) is lower than a). Note, however,
that this also results in a different concentration in the syrup, which causes a different steam consumption in the sugar crystallization
process; besides, the different temperature of the steam affects the juice heaters and thus the temperature of the juice that enters the
MEE.
Thus, when sugar is crystallized in a batch vacuum pan instead of a continuous vacuum pan, the demand for energy (steam) is
different during the batch cycle. To cope with this change in the energy demand, two alternatives are possible: rst, one can let the
temperature of the MEE oscillate to produce more or less steam as aforementioned; or second, they can keep the temperature of the
steam constant by bypassing the higher pressure steam to boost the lower pressure steam when the demand is high or release the steam
to the atmosphere when the steam demand is low. Here, we investigate these two possibilities, which require a mathematical model to
nd the temperature distribution. To our knowledge, there are no models in the literature that allow us to nd the temperature
distribution in multiple-effect evaporation: the previously mentioned papers use the temperature as a xed input parameter; to do that,
most of them considered the steam consumption as continuous, using the average steam consumption in batch processes.
2. Nomenclature
A Area, m
2
bpe Boiling point elevation, K
Brix Soluble solid concentration in kg of soluble solids per kg of solution
C Concentration in kg of solute per kg of solution
Cp Specic heat capacity, kJ/(kg⋅K)
CV Condensate from a multiple-effect evaporator, kg/s
CVF Condensate from a ash tank, kg/s
EV Steam evaporated from a multiple-effect evaporator, kg/s
F Streamow rate, kg/s
H Enthalpy, kJ/kg
ΔH Enthalpy of vaporization, kJ/kg
J Juice ow rate, kg/s
LMTD Logarithmic mean temperature difference, K
m Mass, kg
Mass ow rate, kg/s
P Purity in kg of sucrose per kg of solids in solution
q Heat transfer rate, W
t Time, s
T Temperature, ◦C
V Feed steam in a multiple-effect evaporator, kg/s
VS Steam bleed from MEE, kg/s
VFC Steam ashed, kg/s
δT Temperature difference,
U Overall heat coefcient, W/(m
2
⋅K)
KG Crystallization rate, m/h
kg Kinetic crystallization parameter
ρ
Density, kg/m
3
Subscript
c Crystal
cv Condensate
E or 0 Exhaust steam: in equation (3) trough 7 the subscript ‘0’ is used instead of ‘E’
heat Used to specify sensible heat when both latent and sensible heat, are exchanged
I Impurities
i Evaporation body effect
in Vacuum pan feed
J Juice
MC Massecuite
s Sucrose
st Steam
T Total
(continued on next page)
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
5
(continued )
vap_out Steam withdrawn from massecuite
vap_in Steam consumed in the vacuum pan calandria
w Water
3. Method
In this section, we describe the model used to simulate MEE and sugar batch crystallization. Other unit operations, such as heaters,
clariers, distillation, extraction, fermentation, boiler, etc. are not shown in this article; these unit operations are described in Castro
et al. [18].
3.1. Modeling of the evaporator
The multiple-effect evaporator train is composed of four different unit operations: a) the evaporator itself, identied as EVAP
i
in
Fig. 3; b) the condensate tank, Tk
i
, which extracts the condensate from the evaporation body and allows it to ash in the lower pressure
steam; c) the mix/splitter, Sp
i
, that plays three roles, which are to incorporate the steam ashed from the condensate Tk
i
, divert steam
to the industrial process (steam bleed) as a thermal utility, and feed the next effect evaporator EVAP
i+1
with the remaining steam.
Therefore, the mathematical model is obtained by balancing the mass and energy in each unit operation and applying the design
equation to the evaporator bodies.
The expressions of the mass balance around the ith effect evaporator (EVAP
i
) are given by Equations (3) and (4). There are 2 N
equations according to the number of evaporation effects, where N is the number of evaporation effects. In these equations, J
1
is a
known variable that represents the juice stream that enters the MEE; all other variables are calculated.
Ji−Ji+1−EVi=0,i=1,…,N(3)
Vi−1−CVi−1=0,i=1,…,N(4)
The expressions of the energy balance around the ith effect evaporator (EVAP
i
) are given by Equations (5) and (6). Once again, there
are 2 N equations according to the number of evaporation effects. In this case, the latent heat q
i
produced by the condensation of the fed
steam is assumed to be transferred entirely to the liquor to produce low-pressure steam. The boiling point elevation caused by the sugar
concentration in the juice leads to the production of superheated steam, which quickly cools down and becomes saturated steam before
entering the next evaporation body. We assume that this energy of cooling the steam until its saturation point is lost to the surrounding.
Hst,i−1·Vi−1−Hcv,i−1·CVi−1−qi=0,i=1,…,N(5)
HJ,i·Ji−HJ,i+1·Ji+1+Hst,i·EVi+qi=0,i=1,…,N(6)
The heat transferred from the higher-pressure steam condensation to the liquor evaporation is given by Equation (7), which is the
design equation for a heat transfer unit in the MEE. This equation is the same as Equation (1); however, it is rearranged. Thus, the
amount of heat transferred depends on the evaporation area, the heat transfer coefcient, and the temperature difference. In this
model, the temperature of the steam from the turbine, T
st,0
, and the last effect evaporator, T
st,N
, are known; the intermediary steam
temperatures T
st,i
to T
st,N-1
are calculated. The term q
heat
represents the heat necessary for the liquor to raise its temperature to the
Fig. 3. Multiple effect evaporation representing the steam bleed in the Sugarmill.
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
6
boiling point; q
heat
is given by Equation (8). The liquor needs to be heated to its boiling point only in the rst-evaporation effect: In a
feedforward multiple-effect evaporation, this term is equal to zero from the second to the last evaporation body because the liquor
enters the evaporation body above its boiling point.
Ui·Ai·Tst,i−1−Ui·Ai·Tst,i−qi=Ui·Ai·bpei+qheat ,i=1,…,N(7)
qheat =Ui·Ai·LMTD,i=1qheat =0,i=2,…,N(8)
The unit operation Mixer/Splitter (Sp
i
) is responsible for incorporating steam ashed from the condensate tank and diverting steam
between the next evaporation body and the process. The expression of the mass balance around the Mixer/Splitter is given by Equation
(9). There are N–1 equations according to the number of evaporation bodies; the steam from the last evaporation body is not used as
thermal energy. For the evaporation bodies where no steam is bled, the term VS is null.
EVi−Vi+VFCi−VSi=0,i=1,…,N−1(9)
The condensate from the EVAP
i
evaporator goes to the Tk
i
ash tank, where the condensate ashes into the lower pressure steam.
The expressions of the mass and energy balance around the ith ash tank are given by Equations (10) and (11). Usually, the condensate
from the rst tank is not allowed to ash on the lower pressure steam, because this condensate is used to feed the boiler in a closed
circuit: if the condensate were allowed to ash, it would increase the amount of make-up water in the boiler, which would demand an
additional amount of energy to heat this water. Moreover, the last ash tank does not ash in the lowest pressure steam, since this
steam is not used as thermal energy: ashing this condensate would hinder the barometric condenser and increase the amount of
cooling water. Therefore, there are 2N–2 equations according to the number of evaporation effects.
CVi−1+CVFi−1−VFCi−CVFi=0,i=2,…,N−1(10)
Hcv,i−1·CVi−1−Hst,i+1·VFCi+Hcv,i−1·CVFi−1−Hst,i−1·CVFi−1=0,i=2,…,N−1(11)
3.2. Thermodynamic properties and heat transfer coefcient
Thermodynamic properties of water and steam were obtained adjoining packages from the International Association for the
Properties of Water and Steam, IAPWS IF – 97. The quality or range of the data is available in their release guidelines on their website
[19]. The Matlab® correlation written for it was created by Holmgren and is available online [20].
The specic heat and enthalpy of the juice are given by Equation (12a) and b adapted from Hugot [5].
CpJ= (1− (0.6−0.0018 ·TJ+0.08 · (1−P)) · Brix ) · 4.1868 (12a)
HJ=CpJ·TJ(12b)
As the concentration of dissolved solids increases, the boiling temperature of the juice increases above the temperature of saturated
Table 1
Stages and steps for crystalizaiton B.
Stage Step B-pan 1 B-pan 2
1 1 A-molasse feed Stand-by
2 Concentration Stand-by
3 Seeding and Crystallization Stand-by
2 1 Magma transfer (cut) Magma transfer (cut)
2 Crystal washing Crystal washing
3 Crystallization Crystallization
3 1 Magma transfer (cut) Magma transfer (cut)
2 Crystal washing Crystal washing
3 Crystallization Crystallization
4 Tightening Tightening
5 Discharging Discharging
4 1 Stand-by Magma transfer (cut)
2 Stand-by Crystal washing
3 Stand-by Crystallization
4 Stand-by Tightening
5 Stand-by Discharging
5 1 Stand-by Magma transfer (cut)
2 Stand-by Crystal washing
3 Stand-by Crystallization
4 Stand-by Tightening
5 Stand-by Discharging
6 1 Stand-by Magma transfer (cut)
2 Stand-by Crystal washing
3 Stand-by Crystallization
4 Stand-by Tightening
5 Stand-by Discharging
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
7
steam at the same pressure. The degree of boiling point elevation is determined as a function of the concentration of soluble solids, Brix,
by the following Equation (13) [21].
bpe =2·Brix
1−Brix (13)
Finally, the correlation of the heat transfer coefcient for conventional Robert evaporators is given by Equation (14), adapted from
Wright [22].
U=0.00049 ·((1.10 −Brix ) · 100)1.1616 ·T1.0808
J·δT0.266 (14)
3.3. Sugar batch crystallization process
In any crystallization process, the goal is to transfer sucrose from the mother liquor to the surface of the sugar crystal. This may take
many steps and will depend on many factors, such as initial crystal size, sucrose concentration, impurities, machinery size, etc. Most
Brazilian sugar mills use a batch vacuum pan to crystallize the sugar in a two-boiling pan scheme [23]. Thus, the batch processes that
consume steam are A-Pan, B-Pan, and Graining-Pan. In this section, we briey describe the sugar process using a batch vacuum pan.
We will not detail the kinetic crystallization model because it is already detailed in the literature [21,24,25]. Since the focus of this
study is MEE, we describe only the crystallization operation schedule that occurs in the boiling pans with a focus on steam consumption
during different batch operation steps. In Section 3.3.1, we describe the steps and stages in crystallization B, which are summarized in
Table 1, and crystallization A, summarized in Table 2; Section 3.3.2 presents the mathematical model.
3.3.1. Sugar process recipe
In a two-boiling pan scheme, the crystals start growing in crystallization B, then continue growing in crystallization A: at the end of
crystallization A, the crystals reach the desired size. The syrup, which contains a higher purity of sucrose, feeds crystallization A; A-
molasse, which contains a lower purity of sucrose, feeds crystallization B, while B-molasse is the exhausted molasse used to produce
ethanol. The term massecuite describes the mixture of crystals, which is the solid fraction, and mother liquor, which is the liquid
fraction. The term magma describes the stream rich in crystals that provide the nuclei to initiate the growing process. Both crystal-
lizations, A and B, occur in multiple vacuum pans; we use the term stage when massecuite or magma is transferred from one vacuum
pan to another; thus, there are many steps in each stage. The ow of magma and massecuite among the vacuum pan is shown in Fig. 4.
In crystallization B, the crystal growth process begins in graining (stage 1). The rst step in this stage is to partially ll the vacuum
pan with A-molasse. In the second step, the uid is concentrated: the sucrose concentration must be above the solubility curve and
below the labile curve. In this region, namely the metastable region, sugar crystals grow, but no new nuclei form. Crystallization is
initiated by adding very ne seeds in the form of a slurry (about 4-
μ
m size ground sugar crystals suspended in alcohol). Adding seeds
starts the third step: as sucrose leaves the mother liquor and deposits on the crystal surface, more A-molasse is added to increase the
amount of sucrose in the mother liquor; then, steam is used to withdraw the water and adjust the sucrose concentration. This stage ends
when the volume of the graining pan reaches its maximum level. The graining pan product is the magma that contains the crystals used
to start the B-crystallization in stage 2.
The amount of magma necessary to start the crystallization in each B-Pan is the volume that covers the calandria: typically, the
graining-pan produces an amount of magma to start many B-Pans. After having the calandria lled with magma, which is the rst step
of stage 2, the second step of stage 2 starts, namely crystal washing: water is added to the massecuite; then, steam is fed to evaporate
the water added; this step aims to start the circulation of the massecuite by convection forces. In the third step, the sugar crystals start
to grow and molasse from A-Centrifuge is added as a source of sucrose to continue growing the crystals. Steam is used during crys-
tallization to evaporate the water from the mother liquor. As crystallization occurs, the viscosity of the massecuite increases due to the
growth in the size of the crystals and the concentration of impurities: the higher the viscosity of the massecuite, the lower the heat
transfer coefcient [26]. Therefore, the steam consumption is higher at the beginning of batch crystallization and decreases as the
crystal concentration increases and the concentration of mother liquor impurities increases. This stage stops when the maximum
volume of the pan is reached; the magma is then partially transferred to the next vacuum pan, in a process, namely cut, to start the next
stage.
Stage 3 repeats steps 1 to 3 described in stage 2; however, in stage 3, the crystals reach the desired size to be discharged and
centrifuged. Thus, in step 4, namely tightening, the massecuite reaches the desired concentration of crystals, and then in step 5, the
massecuite is discharged into massecuite buffer tanks. From these buffer tanks, massecuite goes to a centrifuge to separate the B-
Table 2
Stages and steps for crystalization A.
Stage Step A-pan 3 A-pan 5 A-pan 4 A-pan 6
1 1 Magma transfer (cut) Magma transfer (cut) Stand-by Stand-by
2 Crystal washing Crystal washing Stand-by Stand-by
3 Crystallization Crystallization Stand-by Stand-by
2 1 Magma transfer (cut) Magma transfer (cut) Magma transfer (cut) Magma transfer (cut)
2 Crystal washing Crystal washing Crystal washing Crystal washing
3 Crystallization Crystallization Crystallization Crystallization
4 Tightening Tightening Tightening Tightening
5 Discharging Discharging Discharging Discharging
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8
molasse from the crystals. The B-molasse is the nal molasse in the two-boiling sugar production scheme; it goes to the ethanol
fermentation process; the crystals from the B-centrifuges are mixed with water to produce magma for the A-pan. Stages 4, 5, and 6
repeat the steps from stage 3.
In crystallization A, the magma is the sugar produced in B and the sucrose source is the syrup from MEE. Step 1 in stage 1 of
crystallization A is the magma transfer; the amount of magma necessary to start the crystallization in each A-Pan is the volume that
covers the calandria. After having the calandria lled with magma, water is added to the massecuite (step 2 of stage 1), steam is then
fed to evaporate the water added; this step aims to start the circulation of the massecuite by the convection forces (crystal washing). In
step 3, the sugar crystals start growing and syrup from MEE is added as a source of sucrose to continue growing the crystals. Steam is
used during crystallization to withdraw the water from the mother liquor in the vacuum pan. This step ends when the vacuum pan
reaches its maximum level. Afterward, half of the massecuite is transferred to another vacuum pan, i.e., stage 2 and step 1. The next
steps, steps 2 and 3, are the same as described in stage 1; then, when the vacuum pan reaches the maximum volume, before discharging
the massecuite, the massecuite is tightened (step 4), then discharged (step 5).
In all the steps that involve heat transfer, the temperature of evaporation is controlled by maintaining the vacuum inside the
heating chamber. The vapor, withdrawn from the product inside the vacuum pan, is conveyed through a condenser to a vacuum
system, usually a water ejector or a vacuum pump.
3.3.2. Sugar crystallization mathematical model
In the recipe for sugar production, the steps named crystallization, crystal washing, concentration, and tightening are solved using
the system of ordinary differential equations, ODE: the difference in each step is the fed stream as described above. These equations
consist of an overall mass balance on the massecuite side of the vacuum pan, Equation (15), and on its components crystals, sucrose,
impurities, and water, Equations (16)–(19), respectively; the energy balance on the calandria side and on the massecuite side is given
Fig. 4. Sugar crystallization stages.
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
9
by Equations (20) and (21), respectively. The kinetic crystallization is given by Equation (22) adapted from Castro et al. [24].
dmT
dt =Fin −dmvap out
dt (15)
dmc
dt =KG ·Ac·
ρ
c(16)
dms
dt =Fin ·Cs,in −dmc
mt=Fin ·Cs,in −KG ·Ac·
ρ
c(17)
dmI
dt =Fin ·CI,in (18)
dmW
dt =Fin ·CW,in −dmvap out
dt
dmW
dt =Fin ·CW,in −U·A·Tvap in −TMC+Fin ·Cpin ·Tin −Fin ·CpMC ·TMC
Hvap out −CpMC ·TMC
(19)
dmvap in
dt =U·A·Tvap in −TMC
ΔHvap in
(20)
dmvap out
dt =U·A·Tvap in −TMC+Fin ·Cpin ·Tin −Fin ·CpMC ·TMC
Hvap out −CpMC ·TMC
(21)
KG =kg ·exp−57000
R·(TMC +273)(P−1)exp−8mI
ms+mI1+2
ρ
mc
ρ
c
mc
mT (22)
Fig. 5. Algorithm to solve multiple-effect evaporation in a Sugarmill industry.
R.E.N. Castro et al.
Case Studies in Thermal Engineering 34 (2022) 102035
10
3.4. Interactive algorithm
The sugar crystallization model is represented by the ODE system, which is solved using Runge-Kutta for a small time interval (1
min). At this time interval, the entire sugarcane industrial process is calculated taking into account the result from the sugar crys-
tallization step. Depending on the step of the sugar crystallization process recipe, a different amount of steam is required from MEE.
The calculation of the mathematical model of the sugar mill requires solving the equations of mass and energy balance, thermody-
namic properties, and heat transfer coefcient. The MEE mass and energy balance equations can be rearranged into a set of linear
equations. The thermodynamic properties and the heat transfer coefcient are a set of non-linear equations. To solve this set of linear
and non-linear equations, an iterative method combining the Gaussian elimination for the matrix of linear equations and the xed-
point method is used. The calculation is repeated until all the steam temperature and the syrup concentration become constant.
To solve the MEE model, the mass and energy balance and the evaporator design equation are rearranged into 8N–5 linearly in-
dependent equations, where N is the number of effects. The xed parameters are (a) the mass ow rate and temperature of the juice
that enters the evaporation train; (b) the temperature and pressure of the exhaust steam from the turbine; (c) the temperature of the
steam from the last evaporation body withdrawn by the barometric condenser; (d) the evaporation area of each body. The parameters
calculated previously are (a) the overall heat coefcient; (b) the juice boiling point elevation, which depends on the concentration of
the juice; and (c) the steam bleed.
The intermediate temperature of the juice and steam streams and the concentration of the juice, in the multiple-effect evaporator,
are results from the set of linear equations; however, before solving the mass and energy balance, a rst guess of the temperature and
juice concentration is made. Using this rst guess, the enthalpy of the liquor, condensate, and steam; the boiling point elevation; and
the heat transfer coefcient are estimated. Then the set of linear equations is solved and new values of temperature and concentration
are obtained. Again, the thermodynamic properties are calculated using these obtained values of temperature and juice concentration.
This is repeated until the values of the temperature and juice concentration become constant. A summary of this interaction is shown in
Fig. 5 and is represented by the dashed line identied as Interaction 1.
After nding the intermediate steam temperature, which balances the energy for a given amount of steam bleed, it is necessary to
recalculate the steam bleed for juice heating. Changing the temperature of the steam utility changes the heat transferred to a given
area. Additionally, changing the temperature distribution in MEE changes the syrup concentration; thus, different syrup concentra-
tions require a different amount of steam to crystalize the sugar. Therefore, a new interaction is necessary to recalculate the tem-
perature of the steam using the new values of steam bleed. In Fig. 5, it is represented by the dashed line identied as interaction 2.
By applying this algorithm, it is possible to nd the steam consumption for a given time considering a continuous process operation;
however, the crystallization process is not continuous. Thus, the steam consumption in the crystallization process will depend on the
batch crystallization stage. The steps and stages for sugar crystallization are represented by the dashed line identied as interaction 3
in Fig. 5. Therefore, interaction 3 represents the small time interval at which the ODE for sugar crystallization is solved.
4. Results
In this section, we show the result of a 48-h simulation of a sugarcane mill. We simulate two boundary scenarios and between these
two scenarios. In the rst scenario, the steam pressure changes according to the steam consumption on sugar crystallization. In the
second, the pressure and thus the temperature of the steam are xed; therefore, a by-pass valve from the higher-pressure steam to the
lower one is used when the steam consumption rises above the average; analogously, a relief valve releases steam to the atmosphere
Fig. 6. Steam consumed during the crystallization process, steam fed in MEE and steam pressure produced at each multiple-effect evaporation body.
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Case Studies in Thermal Engineering 34 (2022) 102035
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when the steam consumption decreases below the average; in this way, the amount of steam bleed from the MEE is constant and
corresponds to the average steam consumption in batch process. Since the steam bleed ow remains constant, the temperature and
pressure of all of the vapors produced in MEE are constant as well. Then, in the third part (Section 4.3), we set the lower and upper
boundaries; in this way, part of the steam is bypassed only when the steam consumption in sugar crystallization is above an upper limit,
i.e., the average plus a delta; analogously, steam is released into the atmosphere when steam consumption is below a lower limit, i.e.,
the average minus a delta. Here, this delta is a range around the average steam consumption.
4.1. Flexible steam pressure
The energy input for the whole sugarcane mill is produced by the boiler that feeds the rst evaporation body of the MEE. Changing
the steam bled from the MEE in any of the evaporation bodies changes the fed steam required in the rst body. Fig. 6 shows (during 48-
h simulation) the steam consumed in the crystallization process and the total steam from the boiler consumed by the entire industrial
process. Depending on each crystalization step, different consumptions are expected. The steam used by the vacuum pan is bled from
the second evaporator body. The higher the amount of steam required by the vacuum pan, the lower the pressure of the steam; this
change in the steam pressure guarantees the amount of steam to feed the sugar process. The pressure of the steam from MEE is shown
on the right-hand side of Fig. 6; the steam extracted from the ith body evaporator is identied as V
i
and the steam that enters the rst
body of evaporation is identied as V
E
. In this simulation, no bypass from the higher-pressure steam was considered to boost the lower-
pressure steam. The total amount of V
E
consumed during this operation interval is 1.2210 ×10
7
kg and the average steam ow rate
used in the crystallization process (V2) is 28.425 ×10
3
kg/h.
4.2. Fixed steam pressure
In this simulation, the steam pressure is maintained unchanged; for this, a bypass valve is used to increase the pressure of the steam
when the pressure drops, and a relief valve is used to relieve steam into the atmosphere when the pressure increases. Thus, the pressure
of the steam remains constant during all the batch operation steps. We do not assess the control: we assume that the bypass and the
relief valve can control the pressure with an immediate response; nonetheless, this might be an issue for a future study. Additionally, a
controller study is a matter of seconds, and in this article, we assess 48-h of operation aiming at the total steam consumption; i.e., the
energy requirement. Fig. 7 shows the steam consumed by the crystallization process; the exhausts steam, which is the total steam from
the boiler consumed at the evaporation in the rst body; and the amount of steam bypassed, V
E
to V
1
, V
1
to V
2
, and V
2
to the at-
mosphere. The steam pressure remains constant during the simulation (V
1
=1.857 bar(a); V
2
=1.450 bar(a); V
3
=1.085 bar(a); V
4
=
0.721 bar(a)). The total amount of V
E
consumed during this operation interval is 1.2498 ×10
7
kg.
4.3. Lower and upper boundaries condition
Industrial processes, in general, do not let the steam pressure oscillate freely (as described in Section 4.1), as well as heat exchange
devices are not strict with a xed steam pressure point (as described in Section 4.2). Therefore, the usual is that boundaries are created
Fig. 7. Steam consumed at the crystallization process.
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Case Studies in Thermal Engineering 34 (2022) 102035
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around an average steam consumption; then a control system, such as a bypass valve and a relief valve, actuate only when these
boundaries are surpassed: knowing the average steam consumption, the bypass valve or relief valve actuates to increase or decrease the
steam pressure. This boundary, namely Delta in this article, represents the amount of steam ow rate that is allowed to oscillate freely
around the average steam ow rate. For example, when Delta is 5000 kg/h and the required steam is 37000 kg/h while average steam
consumption is 28425 kg/h, only 3575 kg/h is by-passed; analogously, if steam consumed is 23000 kg/h, then 425 kg/h is released into
the atmosphere. In this example, no bypass or relief valve is used when the steam consumed is between 23450 kg/h and 33450 kg/h.
Fig. 8 shows the total exhaust steam consumed during the 48-h of simulation against the boundary limit, Delta. The red line, at the
top of the graph, considered the upper and lower limits, i.e., bypass and relief valve; while the blue line considered only the lower limit,
i.e., only the bypass valve. In the second case, when the steam consumed is higher than the average, the steam is not released into the
atmosphere; consequently, the steam pressure increases freely above the average plus Delta. Steam with a higher pressure should not
be a problem in an industrial process since the heat exchange area reduces when temperature increases; conversely, a lower steam
pressure would require a higher heat exchange area.
4.4. Electricity production
Changing steam consumption changes the consumption of bagasse, which is considered a feedstock with many applications in the
sugarcane industry [27]. Here, we assess the production of electricity using a condensation turbine and a cogeneration system. A
cogeneration system is dened as an energy system that produces two useful forms of energy simultaneously. Thus, the cogeneration
system consists of a boiler and a backpressure turbine: the backpressure turbine produces electricity and steam (thermal energy), while
the condensation turbine produces only electricity. Using less exhaustion steam results in using less high-pressure steam. In this
simulation, the surplus high-pressure steam is used to produce electricity using a condensation turbine, which produces more elec-
tricity per kg of steam than the backpressure turbine, 267,8 W/kg and 187,7 W/kg, respectively.
We considered that all bagasse is burned to produce high-pressure steam and high-pressure steam is diverted to feed the back-
pressure turbine or condensation turbine, as shown schematically in Fig. 1. Thus, the amount of steam diverted to the backpressure
turbine depends on the amount of exhaustion steam required on MEE. Fig. 9 shows the power during the 48-h simulation for the two
scenarios described above; therefore, the area below the curve represents the total energy produced. The energy produced is 2200.9
MW h and 2219.2 MW h for the process that uses and does not use the bypass valve, respectively; that is, it is produced a surplus of 18.2
MW h considering the scenario when no bypass valve is used or the steam pressure changes as a function of the crystallization process.
5. Conclusion
The proposed algorithm was able to nd the steam temperature when the amount of steam produced in a MEE changes. To our
knowledge, there are no other works with this approach. Since the output of the simulation was the temperature or pressure, a much
higher computational effort is required compared to the algorithms that use this variable as input. This is because the constitutive and
thermodynamic equations which are used to calculate properties such as enthalpy, heat capacity, density, etc. require the temperature
as an input variable. Thus, these properties were obtained initially by guessing the temperature and then solving them until they
converge to a value that represents the heat transferred in both the thermodynamic and constitutive equations.
For the case in which this algorithm was used, it was possible to assess the consumption of exhaust steam due to the dynamics of
batch sugar production: saving exhaust steam means saving bagasse. Surplus steam (produced using surplus bagasse) was used in a
condensation turbine to produce electricity. The scenarios studied in this paper, in which steam is released or bypassed, increased the
overall steam consumption (exhaust steam). Only bypassing steam reduces the overall consumption compared to that also releases
steam. The scenario that saved more steam was the one that did not bypass or release steam into the atmosphere. The drawback of not
releasing or bypassing steam is the changes in steam temperature; steam with a lower temperature requires more area to perform the
same heat exchange.
This algorithm was shown to be a powerful tool for assessing the energy consumption of the whole sugar mill considering a dynamic
process. Besides this case study, the proposed model can be used to investigate different scenarios, such as the change in steam pressure
Fig. 8. Total exhaust steam consumed in a 48-h simulation and Delta, lower and upper limit around average steam consumption.
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Case Studies in Thermal Engineering 34 (2022) 102035
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due to fouling or due to a change in feed ow rate; additionally, it can be used to simulate MEE in different applications such as pulp
and paper, desalinization, etc.
Author statement
Rubens E. N. de Castro: Conceptualization; Methodology; Software; Investigation; Validation; Data curation; Writing - original
draft; Rita M. B. Alves: Writing - review & editing; Validation; Visualization; Formal analysis. Claudio A.O. Nascimento: Validation;
Funding acquisition; Supervision; Resources; Project administration.
Declaration of competing interest
The authors declare that they have no known competing nancial interests or personal relationships that could have appeared to
inuence the work reported in this paper.
Acknowledgments
This work has been supported by the following Brazilian research agencies: CNPq (National Council for Scientic and Techno-
logical Development) process number 306115/2019-0; and FAPESP-NERC (S˜
ao Paulo Research Foundation and National Environment
Research Council) process number 2015/50684-9.
The authors also gratefully acknowledge the support of the RCGI – Research Centre for Gas Innovation, hosted by the University of
S˜
ao Paulo (USP) and sponsored by FAPESP – S˜
ao Paulo Research Foundation (2014/50279-4 and 2020/15230-5) and Shell Brasil, and
the strategic importance of the support given by ANP (Brazil’s National Oil, Natural Gas and Biofuels Agency) through the R&D levy
regulation. This study was nanced in part by the Personnel Coordination of Improvement of Higher Level - Brazil (CAPES) - Finance
Code 001.
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