Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
System Identification to Optimize Inputs
Conditions For Plant Design of Ethanol
Farah Wahidah Mohd Azman1and Ahmmed Saadi
Ibrahem1*
Article Info
Received:10th April 2012
Accepted: 24th April 2012
Published online: 1st May
1
Faculty of Chemical Engineering, UiTM, Universiti Teknologi
Mara , Shah Alam, 40450 Selangor, Malaysia.
1*
[email protected]
ISSN: 2232-1179
© 2012 Design for Scientific Renaissance All rights
reserved
ABSTRACT
The purpose of this research project is to study the effect of inputs variables on the system by using
system identification method for the production for ethanol by using HYSYS software. One of the
important parts in the study is modeling, simulation and system identification to reach to the optimum
conditions. In this work, system identification method was used to capture the reactor characteristics of
production rate of ethanol based on mathematical model by using HYSYS software. The identification
method was used to measure the percentage effect on the production rate of ethanol by measuring the
effect of inputs factors like temperature of reaction and reactants concentrations have a big effects on the
output of the system .All these results depend on model of HYSYS software and these results are very
important in industrial plants.
Keywords: System identification, reaction, temperature, concentration
1. Introduction
Alcohol can be defined as an organic derivative of water with one of the water oxygen is
replaced by an organic group. It present widely in nature and have many applications in the
industrial and pharmaceutical sectors.
Today, approximately 4 billion gallons of ethanol is produced annually in United State (US)
by fermentation of corn, barley and sorghum and production is expected to double by 2012.
Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
Essentially, the entire amount is used to make E85 automobile fuel, a blend of 85% ethanol
and 15% gasoline. Besides that, about 110 million gallons of ethanol a year is produced in US
for use as a solvent or as a chemical intermediate in other industrial reactions (John McMurry,
2008).
The increase in the world ethanol production is mainly due to the economic and
environmental security concern. At certain countries, like Europe, India, China and Brazil are
aiming on reduction of petroleum import and increase the consumption and production of
renewable fuel (Ethanol 2020: Global Market Survey, Next Generation Trends, and Forecasts).
There are several ways of ethanol production for industrial such as fermentation of ethanol,
indirect hydration (esterification-hydrolysis) process and direct hydration of ethylene. Besides
that, there is also another method have been investigated but still not become commercial such as
the hydration of ethylene in the presence of dilute acid; the conversion of acetylene to
acetaldehyde, followed by hydrogenation of the aldehyde to ethyl alcohol; and the FischerTropsch hydrocarbon synthesis (John McMurry, 2008)
Fermentation can be categorized as one of the oldest chemical processes knows in producing
variety of products including production of fuel. There are 3 types feedstock that can be used in
the fermentation process such as sugar, starch and cellulose materials (DiPardo, 2007., Balat &
Balat, 2009., Sanchez & Cardona, 2008., O'Brien et al., 2000., Mussatto et al., 2010., KirkOthmer, 2005).
There are two main processes of the synthesis of ethanol which is the indirect hydration and
direct hydration process. Indirect hydration process also called as strong sulfuric acid-ethylene
process. It still use in Russia. The different between indirect hydration and direct hydration
process, the sulfuric acid is not use in direct hydration. Today, only Dow Chemical, Texas City,
Texas uses the direct hydration process.
Indirect hydration also called as esterification hydrolysis process. There are steps that
involved in this process such as absorption, hydrolysis and re-concentration of dilute sulfuric
acid. In the absorption step, the feedstock (ethylene) is contacted with organic acid which act as
esterification agent (Meadows et al., 1989).
Direct hydration of ethylene to produce ethanol is by contacting the ethylene mixture with
water vapor. This process does not required sulfuric acid. There are two main process categories
in the direct hydration of ethylene which are the vapor-phase processes contact a solid or liquid
catalyst with gaseous reactants and the mixed-phase processes contact a solid or liquid catalyst
with liquid and gaseous reactants. But, normally ethanol will be produced by a vapor-phase
process (Kirk-Othmer, 2005). The catalyst used in this process is the sesqui-phosphoric acid
catalyst and the conditions of the process are at the elevated temperature and pressure (Dreyfus,
2003).
2. Methodology
In this study, the ethanol plant production system was modeled by using HYSYS software
after completed the mass and energy balance calculations of the 8800000 L/year ethanol
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
production. After completing the simulation system by HYSYS, the system identification
method was applied to the simulation system. The process used in this study is direct hydration
of ethylene.
Without use of any laws concerning on the fundamental nature and properties of the
nonlinear system the system identification are involving with the building of a dynamic model
from an input/output data. In a process can involve many variables and can make it as a SingleInput Single-Output (SISO) process. One of the variables is input or manipulated variable (MV).
Manipulated variable is a variable that can have direct effect on the process performance and
practically easy to actuate. In this study the MV are the input temperature and concentration of
reactants while the controlled variables (CV) are the output stream of the reactor in the ethanol
production system (Ahmmed S. Ibrahem, 2011)
The steps involve in the system identification method (Ahmmed S. Ibrahem, 2011) were
define as follow:
a) Make the actual calculations for the system from the nominal conditions xn,i in order to
find the yn,i. this is done by calculate the mass and energy balance for the production
system.
b) Step a) is repeat by increase of inlet temperature of the system.
c) Each of the element from the first matrix in the first step from the corresponding element
in the second matrix and divide the difference by yn,i.
-
(1)
d) The similar step is used to calculate the change in parameters
-
(2)
e) Then, result from c) is divided by result obtain form d). This is to produce the sensitivity
matrix k that depends completely on scale matrix without using any proper factor.
f) All steps above are repeated by decreasing the value of the temperature as the same
amount of the increased value made.
g) Lastl k average slope s calculated
order to f d the average a gle (θ) wh ch w ll
represent the overall effect of each parameter on the measured output.
θ1: effect of temperature.
θ2: effect of concentrations.
From the average analysis, a preliminary partition into different groups depends on the slope
a gle (θ) of k average a d these groups are class f ed as follows:
a) (θ) ≥ 20˚: large effect on the system.
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b) 20˚ ≥ θ ˃ 15˚: middle effects on the system.
c) 15˚ ≥ θ ≥ 10˚: weak effects on the system.
d) 10˚ > θ: ca ot be establ shed.
Note: θtotal θoutput + θinteraction
(3)
All of the steps mentions before are being repeated for other MV such as the concentration of the
reactants and also the inlet flow rate of reactor. The result then will be record.
3
Mass Balance Calculation
3.1 Introduction
After finding details or data from the previous inventions such as from journals, patents and
articles, mass balances of the process is done by based on the production rate required. Basic of
this technique is referring to the conservation law which is no matter can disappear or be created
spontaneously. Mass balance also called as material balance. In mathematically, the general mass
balance or material balance can be written as below:
Input +(4
Generation – Output – Consumption = Accumulation …….
(4)
There few rules can be followed to simplified material balance equation (4). The rules are as
follow (Felder & Rouseeau, 2005):
a) If the balance quantity is total mass, set generation and consumption equal to zero.
Except for nuclear reactions, mass neither be created nor destroyed.
b) If the balance substance is a non-reactive species (neither reactant nor product), set
generation and consumption equal to zero.
c) If the system is at steady state, set accumulation equal to zero, regardless of what
being balanced.
Based on those rules and assumptions mentioned before, Equation 1 can be simplified to be:
Input – Input = 0
(5)
By referring the mass balance done, the amount of the raw material needed to produce the
required production of one material can be determined. The value obtained then will be put into
the simulation system (HYSYS), so that the simulation can calculate the energy needed for
process to be achieve and to determined whether the process are feasible in real world or not.
For this research project, the ethanol production has been specified as 8800000 L/yrs or
840.78 kg/hrs. The required amount of raw materials needed to produce the required amount of
ethanol has been calculated by using the stoichiometry equation of the process. From the mass
balance done the mass flow rate for produce the required production was 2047.65 kg/hrs (73
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
kmol/hrs) of ethylene (C2H4) and 1315.46 kg/hrs (73 kmol/hrs) of water (H2O). The details of the
calculations are shown in the next section.
3.2 Overall mass balance for Ethanol production by direct hydration of Ethylene
Some of the details required in the mass balance are as follow:
b) Molecular weight of
c) Molecular weight of
kg
.
a) Molecular weight of ethanol =
k ol
kg
ethylene = 28.05 k
kg
water = 18.02 k ol
kg
d) Density of ethanol = 789
e) Required production
=
L
ear
ear
L
da
da s
kg
hrs
k ol
.
kg
.
k ol
hrs
f) Conversion of the ethylene to ethanol: 4-24 %
g) Process conditions:
i.
Temperature: 300 oC.
ii.
Pressure: 6.8 MPa.
Based on the literature review section, overall reaction equation for direct hydration
process:
+
(6)
Table 1 below shows the molecular balance of equation 3. The molecular balance is made
by balancing each of the molecular at the reactants side (left hand side, LHS) and at the product
side (right hand side, RHS).
Table 1 Molecular balance.
LHS
RHS
C=2
C=2
H=6
H=6
O=1
O=1
Conversion of Ethylene to Ethanol = 24% = 0.24
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
Therefore,
Un-reacted ethylene = 76% = 0.76
By consider the conversion factor; equation 6 will become equation 7.
+
.
+ .
+ .
(7)
Molecular balance for equation 4 is then being calculated. Table 2 shows the molecular balance
made and the molecular for LHS and RHS are both balanced.
Table 2 Molecular balance
LHS
RHS
C=2
C = (0.24 × 2) + (0.76 × 2) = 2
H=6
H = (0.24 × 6) + (0.76 ×6) = 6
O =1
O = 0.24 + 0.76 = 1
After balancing the molecular at the both sides; reactants and products, the correction factor
is calculated. Correction factor is a value that suppose to be multiply with equation 7 in order to
obtain the require amount of reactants needed to produce required production of ethanol.
Correction factor = required production × conversion
= 18.25 ÷ 0.24
= 76.04167
Multiply equation 7 with correction factor,
.
+
.
.
+ 57.7917
+ 57.7917
(8)
By referring to equation 8, the amount of ethylene and water needed to produce the required
production of ethanol is .
kmol/hrs each respectively or 2132.97 kg/hrs of ethylene and
1370.27 kg/hrs of water. Table 3 shows the molar and mass flow rate for each of the reactants
and products in the process.
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Table 3 Mass flow rate and mass fraction for reactants and products.
Reactants
Products
Molar flow
rate,
kmol/hrs
Mass flow
rate, kg/hr
Ethylene
76.0417
2132.97
Water
76.0417
1370.27
Total flow
rate
152.083
Molar flow
rate,
kmol/hrs
Mass flow
rate, kg/hr
Ethylene
57.7917
1621.06
Water
57.7917
1041.41
Ethanol
18.25
840.78
133.8334
3503.25
3503.24
3.3 Mass balance for mixer
Before entering reactor, ethylene and water are mix in a mixer at temperature of 30 oC and at
atmospheric pressure. Fig. 1 below shows the mass balance around the mixer unit.
m1 = 2132.97 kg/hrs
m3 = 3503.24 kg/hrs
m2 = 1370.27kg/hrs
Fig. 1 Mass balance at mixer.
m1: mass flow of ethylene.
m2: mass flow of water.
m3: mass flow of water and ethylene mixture.
From the overall mass balance, the mass flow rate for ethylene and water entering the mixer are
2132.97 kg/hrs and 1370.27 kg/hrs respectively. By using equation 17, mass flow rate at the
outlet of the mixer is calculated.
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
Mass in –mass out = 0
(9)
2132.97 +1370.27 = 3503.24
Then, the mass fraction at the outlet of the mixer is calculated. The calculation is as below:
ass flow rate of each of co po e t
total
(10)
ass flow rate at outlet strea
By using equation 10, the mass fractions for each of the component in the outlet stream of the
mixer are as follow:
Table 4 Mass fraction at the outlet stream of mixer.
Component
Mass fraction
Ethylene
0.6089
Water
0.3911
3.4 Mass balance for reactor
Fig. 2 represents the operation condition of reactor. The inlet of the reactor is from the
cooler. The reaction of producing the ethanol is happens here.
m3 = 3503.24 kg/hrs
m4 = 3503.24 kg/hrs
xC2H4 = 0.6089
xC2H4 = 0.4628
xH2O = 0.3911
xH2O = 0.2973
xC2H5OH = 0.24
Fig. 2 Mass balance at reactor unit.
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m3: mass flow of water and ethylene mixture.
m4: mass flow out from reactor.
xC2H4: mass fraction of ethylene.
xH2O: mass fraction of water.
xC2H5OH: mass fraction of ethanol.
The inlet of the reactor is the outlet of the mixer unit. Based on the overall mass balance the
mass flow rate for ethylene, water and ethanol at the outlet of the reactor are 1621.06 kg/hrs,
1041.41 kg/hrs and 840.78 kg/hrs respectively. By using equation 10, the mass fractions of the
outlet of reactor are shows as in table 5 follow:
Table 5 Mass fraction of each component at the outlet stream of reactor.
Component
Ethylene
Water
Ethanol
4
Mass fraction
0.4628
0.2973
0.24
Energy Balance Calculations
4.1 Introduction
Energy balance on the process is one of the jobs of chemical engineer. Engineers can known
the flow in and out of energy around the system and also the total energy required by the system
based on the energy balance. Energy balance is much like the mass balance only that mass
balance is represents the mass flow in and out around the system.
The principles that can be related to the energy balance is the law of conservation of energy
or also know as first law of thermodynamics which state that energy neither be created nor
destroyed.
Energy balance can be done either by closed system or open system at steady state. Closed
system can be defined as no mass is transferred across the system boundaries while process is
taking place. The energy may be transferred between the system and its surrounding in two
ways; as heat and as work. By definition, an open system has mass crossing its boundaries as the
process occurs. For this system, it required work done on the system to push mass in and it also
has work done on the surroundings by mass that emerges. Both work terms must be included in
the energy balance.
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4.2 Assumptions
To make the calculation of the energy balance easier few assumptions before start the
calculation is made. The assumptions are as below:
a) The system is an open system at steady state.
b) Since the effect of pressure difference to the energy balance in the process give a very
small value as compared to the value contributed by the sensible heat and the heat of
formation, heat obtained from the pressure difference is assumed to be negligible.
4.3 Equation used for calculation
4.3.1General equation
Q-Ws =
+
k+
(11)
p
However, based on the assumption made, equation 1 is reduced to the form of:
(12)
p
4.3.2 Equation for reactive process
 


H   ni   H i   H
Outlet 

f

 





n

H


H
i
  i 
Inlet



f






(13)
4.3.3 Equation for process involving phase change


T
 Tbp

H i =  C p (l ) dT   H v   C p ( v ) dT 
Tref
Tbp


(14)
4.3.4Equation for non-reactive system
T
H i  ni   C p dT 
 Tref

or
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
T
 Tbp

H i  ni  C p (l ) dT   H v   C p ( v ) dT 
Tref
Tbp


(15)
4.3.5Equation heat capacity
(16)
(17)
4.3.6Equation for heat of vaporization
(18)
4.3.7Total heat for energy balance (non-reactive system)
i
H Total   (H Outlet stream  H Inlet stream )
i 1
(19)
4.3.8Constants value for gas heat capacity equation
Table 6 and are the constant values that needed to calculate the heat capacity equation both
for gas and liquid. All of these values were obtained from Elementary Principles of Chemical
Processes.
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Table 6 Represent the constant values for heat capacity equation of gas.
Component
C1 × 10-3
C2 × 10-5
C3 × 10-8
C4 × 10-12
Ethylene
40.75
11.47
-6.891
17.66
Water
33.46
0.688
0.7604
-3.593
Ethanol
61.34
15.72
-8.749
19.83
Table 7 Constant values for heat capacity equation of liquid.
Component
C1 × 10-3
C2 × 10-5
C3 × 10-8
C4 × 10-12
Water
75.4
-
-
-
Ethanol (at
temperature, T ≤ 0)
103.1
-
-
-
Ethanol (at
temperature, T ≤
100)
158.8
-
-
-
4.3.9Constant values for heat of vaporization
Table 8 shows the constant values that require performing calculation of latent heat of
vaporization for each of component. All of these values were obtained from Elementary
Principles of Chemical Processes.
Table 8 Latent heat of vaporization constant for each component.
Component
ΔĤv (kJ/mol)
Ethylene
13.54
Water
40.656
Ethanol
38.58
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4.3.10Energy balance around mixer
Fig. 3 shows the mixer unit operation of the process. The inlets components of this mixer are
ethylene and water. Ethylene is enter by first stream while water in second stream. The outlet
mixer is a mixture of water and ethylene with equal molar composition of the mixture.
P = 101.3 kPa
P = 101.3 kPa
P = 101.3 kPa
Fig. 3 Mixer unit operation condition.
The reference point is chose as 303 K and 101.3 kPa. Based on the reference point, the
enthalpy at the inlet of the mixer is equal to zero.
By using equation 17, the outlet enthalpy of the mixer is calculated. The calculations are as
follow:
Enthalpy at the outlet stream is calculated as below:
a) Enthalpy of ethylene.
Δ
ethylene
=
Δ
ethylene
=
.
= -0.17858 kJ/mol
= -178.58 kJ/kmol
165
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
b) Enthalpy of water.
Δ
water =
Δ
water =
.
pd
.
-
(
.
-
)
= -0.31065 kJ/mol
= -310.65 kJ/kmol
c) Total heat flow of mixer unit operation.
Δ
Ĥ Ĥ
.
out
= - 37201.05 kJ/hrs
(-
.
) +
.
(-
.
)
4.3.11Energy balance around first heater
Fig. 4 shows the first heater condition in the process. The inlet of this heater is from the
mixer outlet stream. The mixture of ethylene and water is heat up to 373 K within this unit
operation.
T= 293.88 K
T= 373 K
Fig. 4 First heater unit operation condition.
Reference point: 298 K, 101.3 kPa.
By using equation 17, the enthalpy of inlet stream for this heater can be calculated. The
calculations are shows as follow:
a) Enthalpy of ethylene.
Δ
ethylene
=
Δ
ethylene
=
.
pd
= -0.17858 kJ/mol
= -178.58 kJ/kmol
b) Enthalpy of water.
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Δ
water =
Δ
water =
.
pd
.
-
(
.
)
-
= -0.31065 kJ/mol
= -310.65 kJ/kmol
Then, by using the same equation, the enthalpy at the outlet stream is calculated as follow:
a) Enthalpy of ethylene.
Δ
ethylene
=
Δ
ethylene
=
= 3.57174 kJ/mol
= 3571.74 kJ/kmol
b) Enthalpy of water.
Δ
water =
Δ
water =
pd
-
.
(
-
)
= 2.54416 kJ/mol
= 2544.16 kJ/kmol
c) Heat of vaporization for water.
ΔĤv = 40.656 kJ/mol
ΔĤv = 40656 kJ/kmol
Based on the enthalpy calculation, the heat flow for first heater is:
–
= 3.59 × 106 kJ/hrs
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4.3.12Energy balance around compressor
The operation condition for compressor is shows in figure 5 below. The inlet of this unit
operation is from the heater. In this unit the mixture ethylene-water is compress until 6800 kPa
which the pressure of the reaction required.
P = 101.3 kPa
P = 6800 kPa
T = 373 K
T = 585.4 K
Fig. 5 Compressor unit operation condition.
Reference point: 373 K, 101.3 kPa.
Energy balance calculations for inlet and outlet stream of the compressor are as below:
The compressor is assuming to operate in adiabatic condition at all the time. There are no
s g f ca t vert cal d sta ce separates the let a d outlet parts so Δ p ≈ 0. There are also nonegl g ble te peratures cha ges occur so Δ k ≈ 0. Therefore, equation 11 can be simplified as
follow:
s
-
-
out
-
Table 9 shows the simplify energy balance for the compressor by referring to the reference point
and all the assumptions has been made before.
Table 9 Energy balance around compressor.
Component
nin (kmol/hr)
Ĥin ( J/kmol)
nout (kmol/hr)
Ĥout ( J/kmol)
Ethylene
76.04
0
76.04
Ĥout, ethylene
Water
76.04
0
76.04
Ĥout,water
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Ĥout,ethylene: ethylene (gas, 373 K, 101.325 kPa)
kPa)
ethylene (gas, 585.4 K, 7000
Ĥout,ethylene: ΔĤout,ethylene
=
= 176839.5571 kJ/kmol
Ĥout,water: water (gas, 373 K, 101.325 kPa)
Ĥout,water: ΔĤout,water
= 8216.5767 kJ/kmol
= -13.51 × 106 kJ/hr
169
water (gas, 585.4 K, 7000 kPa)
Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
4.3.13Energy balance around first cooler
After exit compressor, the mixture is cool down to the reaction temperature which is 573 K.
this is done by cooler. Figure 6 shows the operation condition of the cooler.
T = 585.4 K
T = 573 K
Fig. 6 First cooler unit operation condition.
Reference point: 298 K, 6800 kPa.
By using equation 17, the enthalpy of inlet stream for this cooler can be calculated. The
calculations are shows as follow:
a) Enthalpy of ethylene.
Δ
ethylene
=
Δ
ethylene
=
= 38.3647 kJ/mol
= 38364.7 kJ/kmol
b) Enthalpy of water.
Δ
Δ
water =
water =
= 20.3306 kJ/mol
= 20330.6 kJ/kmol
Then, by using the same equation, the enthalpy at the outlet stream is calculated as follow:
a) Enthalpy of ethylene.
Δ
ethylene
=
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Δ
ethylene
=
= 21.6707 kJ/mol
= 21670.7 kJ/kmol
b) Enthalpy of water.
Δ
Δ
water =
water=
= 10.1416 kJ/mol
= 10141.6 kJ/kmol
Based on the enthalpy calculation, the heat flow for cooler is:
Δ
Ĥ –
.
. + .
6
= -2.04 × 10 kJ/h
Ĥ
out
(
. ) -( .
. + .
. )
4.3.14Energy balance around reactor
After being cooled, the mixture enters the reactor. Here the reaction will take place to
produce ethanol. The operation conditions of the reactor are shown in figure 7.
xethylene= 0.5
xethylene= 0.4286
xwater= 0.5
xwater= 0.4286
xwater= 0.1429
Fig. 7 Reactor unit operation
condition.
Reference point: 573 K, 6800 kPa.
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By using equation 17, the enthalpy of inlet stream for this cooler can be calculated. The
calculations are shows as follow:
a) Enthalpy of ethylene.
Δ
ethylene
=
Δ
ethylene
=
= 21.6707 kJ/mol
= 21670.7 kJ/kmol
b) Enthalpy of water.
Δ
Δ
water =
water=
= 10.1416 kJ/mol
= 10141.6 kJ/kmol
Then, by using the same equation, the enthalpy at the outlet stream is calculated as follow:
a) Enthalpy of ethylene.
Δ
ethylene
=
Δ
ethylene
=
= 30.6224 kJ/mol
= 30622.4 kJ/kmol
b) Enthalpy of water.
Δ
water =
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Δ
water =
= 14.284 kJ/mol
= 14284 kJ/kmol
c) Enthalpy of ethanol.
Δ
Δ
ethanol =
ethanol =
= 44.4513 kJ/mol
= 44451.3 kJ/kmol
Heat of reaction is calculated as below:
Multiply the stoichiometry constant of the reaction equation with their respective heat of
formation. Table 10 shows the heat of reaction for each of the component that involve in the
reaction to produce ethanol.
Table 10 Heat of reaction for all components involved in ethanol production system.
Component
Ethylene
Water
Ethanol
Stoichiometry
constant
1
1
1
Heat of formation
(kJ/kmol)
52280
241830
235310
Heat of reaction:
= 235310 – (52280 + 241830)
= -58800 kJ/kmol
+
Equal to:
173
+
(20)
Heat of reaction
(kJ/kmol)
52280
241830
235310
Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
P = 6800 kPa ; T =300 ˚C
Mole fraction of A = 0.5
Mole fraction of B = 0.5
FAo = 0.5 FTo = 0.5 (152.08) = 76.04 kmol/hr
FBo = 0.5 FTo = 0.5 (152.08) = 76.04 kmol/hr
Table 11 Conversion table for each of component in the ethanol production.
Species
Ethylene
Water
Ethanol
Symbol
A
B
C
o(
v
- )
Initial
FAo
FBo
0
Change
- FAo X
-FAo X
+ FAo X
Remaining
FA = FAo (1-X)
FB = FAo (1-X)
FC = FAo X
(21)
v
(22)
When neglecting the pressure drop across the system, equation will become:
(23)
Substitute equation 23 into equation 21:
o(
v
vo ( -
)( )
o
o(
v
( -
- )
- )
)( )
(24)
o
CAo = yAo CTo
= yAo
= .
.
= 0.7137 kmol/m3
ε
Ao δ
( . ( -1-1)) = -0.5
By using equation 39
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CA =
= 0.3633 kmol/m3
CB = 0.3633 kmol/m3
CC =
= 0.1147 kmol/m3
-rA = k CACB = k CA2
= 25.69 × (0.3633)2
= 3.3907 kmol/m3 hr
Based on the enthalpy calculation, the heat flow for reactor is:
Δ
Ĥ –
Ĥ
out
= (57.79(30622.4) + 57.79(14284) + 18.25(44451.3) – (76.04(21670.7) + 76.04(10141.6)) + (58800 × 3.3907 × 5)
= -9496 kJ/hr
5
Result and Discussion
Fig. 8 below shows the manipulated variable (MV) involved in this study which is the inlet
temperature, composition and also the flow rate of the mixture of ethylene and water of the
reactor. The controlled variable (CV) that is being concentrated in this study is the ethanol
production by the reactor at every changing of the MV done every time.
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
Temperature
Composition
Reactor
Ethanol
Flow rate
Fig. 8 manipulated and controlled variables the process of producing ethanol by direct hydration
of ethylene method.
Each time the changes done toward to any of the manipulated variable, the result of the
controlled variable were recorded into a tabulated form as shows in the next section of this
chapter.
5.1 Results
5.1.1Changes of inlet temperature
After changes has made to the inlet temperature of the reactor, the result of changes occurred
towards the controlled variable were recorded as in table 12. Each of the controlled variable
values was obtained from the HYSYS simulation system.
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Table 12 Effect of changes of inlet temperature.
Temperatur Temperature
e inlet (˚C) outlet (˚C)
Outlet composition
Ethylene
Water
Ethanol
Slope (k)
200
218.5
0.3485
0.3485
0.3030
0.3333
1.2117
3.6350
260
371.3
0.4249
0.4249
0.1520
0.1333
0.1095
0.8212
270
378.5
0.4251
0.4251
0.1467
0.1000
0.0708
0.7080
280
385.7
0.4282
0.4282
0.1434
0.0667
0.0467
0.7007
290
392.9
0.4299
0.4299
0.1401
0.0333
0.0226
0.6788
300
400.0
0.4315
0.4315
0.1370
0
0
0
310
407.6
0.4331
0.4331
0.1339
0.0333
0.0226
0.6788
320
415.1
0.4345
0.4345
0.1309
0.0666
0.0445
0.6678
330
422.6
0.4360
0.4360
0.1280
0.1000
0.0656
0.6569
340
430.2
0.4374
0.4374
0.1252
0.1333
0.0861
0.6459
400
400.0
0.4450
0.4450
0.1099
0.3333
0.1978
0.5934
k ave
0.8897
θ1 ave
41.66˚
% effect
46.29
5.1.2Changes of Ethylene and water composition
The results for the changes made to the inlet composition which is the ethylene and water
mixture into the reactor was being recorded into tabulated form as shows in table 13. The result
for the controlled variable was also obtained from the HYSYS simulation system.
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
Table 13 Effect of changes of the inlet composition.
Inlet composition
Outlet
Outlet composition
temperature
Ethylene Water Ethanol
(˚C)
Slope
(k)
Ethylene
Water
0.1
0.9
348.6
0.0562
0.8951
0.048
0.8
0.6496
0.8120
0.2
0.8
378.7
0.1311
0.7828
0.0862
0.6
0.3708
0.6180
0.3
0.7
395.6
0.2202
0.6658
0.114
0.4
0.1679
0.4197
0.4
0.6
402.1
0.3212
0.5475
0.1313
0.2
0.0416
0.2080
0.5
0.5
400.2
0.4315
0.4315
0.137
0
0
0
0.6
0.4
391.1
0.5479
0.3219
0.1302
0.2
0.0496
0.2482
0.7
0.3
375.7
0.6666
0.2221
0.1113
0.4
0.1876
0.4689
0.8
0.2
355
0.7836
0.1346
0.0818
0.6
0.4029
0.6715
0.9
0.1
329.7
0.8956
0.0604
0.044
0.8
0.6788
0.8485
k ave
0.4772
θ2 ave
25.51
%effect
28.34
5.1.3Changes of inlet flow rate of the reactor
The last manipulated variable that being used in this study was the inlet flow rate of the
reactor. Each time the inlet flow rate was changed, the result of the ethanol produced by the
reactor was recorded also in a tabulated form as shows by table 14.
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Table 14 Effect of flow rate changes.
Molar
flow rate
(kmol/hr)
Temperature
outlet (˚C)
112.1
Outlet composition
Slope (k)
Ethylene
Water
Ethanol
481.3
0.4163
0.4163
0.1674
0.2630
0.2228
0.8472
122.1
413.1
0.4208
0.4208
0.1584
0.1972
0.1570
0.7962
132.1
408.4
04248
04248
0.1505
0.1315
0.0993
0.7555
142.1
404.1
042.84
042.84
0.1433
0.0657
0.0467
0.7111
152.1
329.7
0.4315
0.4315
0.1369
0
0
0
162.1
396.6
0.4344
0.4344
0.1311
0.0657
0.0424
0.6444
172.1
393.3
0.4371
0.4371
0.1258
0.1315
0.0811
0.6166
182.1
390.2
0.4395
0.4395
0.1210
0.1972
0.1161
0.5888
192.1
387.4
0.4417
0.4417
0.1165
0.2630
0.1490
0.5666
k ave
0.6141
θ3 ave
31.55
% effect
35.06
5.2 Discussion
Based on the result obtained in table 12, the convergence of the controlled variables has been
calculated by using equation and all of the values were recorded. Then, the average slope, k ave,
was calculated by taking the ratio of summation of each of the slope to total number of
manipulated variable being changed in the this system identification method and the k ave or the
average slope was .
a d the average θ was . ˚. Base on the classification done before
for the θ correspo d g to the k ave prev ous sect o th s ake the var able have large effect
on the system with 46.29% effect on the system.
Same steps taken in order to obtain the same result for the other manipulated variables that
involved in this system identification method which were the changes of inlet composition and
the inlet flow rate of the reactor. The results for the other two manipulated variables used in this
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
method were shown by table 13 and 14 which is the changes of inlet composition (ethylenewater) and inlet flow rate of the reactor respectively.
Based on the result from table 13, the ethanol production was decreased when ethylene
composition ratio at the inlet stream was too high or too low. This is shows when ethylene
composition at the inlet stream was 0.1 and 0.9; the ethanol mole fraction at the outlet stream
was approximately to just 0.04. The optimum composition condition for ethanol production by
direction hydration of ethylene was 1:1 ratio of ethylene to water. By this ratio, the ethanol
produce was the highest which is 0.137.
Besides that, based on the result in table 13, k ave when changed the composition was 0.4772
a d the average θ for th s a pulated var able was . .
us g th s value the perce tage of
effect toward the s ste ca be calculated a d the value was . . ased o the average θ value
and the percentage of effect value, it can be said that the manipulated variable have a large effect
toward the system.
It is same as the first manipulated variable (inlet temperature). However, the first
manipulated variable have a larger effect toward the system which is 41.66 compared to the
second manipulated variable which is only 25.51.
For the third manipulated variable; inlet flow rate of the reactor, the amount of ethanol
produce was increased as the molar flow rate of the inlet stream of the reactor was decreased at a
constant volume of reactor. This was shown in the result tabulated in table 14. When the flow
rate of the inlet stream was decreased to 112.1 kmol/hr, the mole fraction of ethanol produced by
the system was 0.1674. By comparing this result to the initial state of the system which was the
flow rate is 152.1 kmol/hr, the mole fraction of ethanol produced by the system was less which is
0.1369.
The amount of continued decreasing as the molar flow rate of the inlet stream of the reactor
increased. As the molar flow rate was increased up to 192.1 kmol/hr, the mole fraction was
decreased to 0.1165. The decreased of the ethanol mole fraction in the system as the molar flow
rate is increased may be due to the space velocity of the mixture in the reactor. When the flow
rate increased the space velocity will decreased. Therefore, there will be not enough time for the
ethylene-water mixture to undergo the reaction to produce ethanol in the reactor.
Based on result in table 14, the k ave value for this manipulated variable is 0.6141 and the
respective θ value for th s k ave s . . It also has a large effect o the s ste wh ch s arou d
35.06. The effect toward the system is bigger compare the effect of changing the inlet
composition of the reactor. However, the effect is smaller if it is compared to the changing in the
inlet temperature of the reactor. Therefore, the effect of the manipulated variable can be arranged
in the following manner:
Composition < Flow rate < Temperature
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Journal of Purity, Utility Reaction and Environment Vol.1 No.3, May 2012, 153-183
he b us g the equat o
is 98.72˚.
θtotal
the total value of θ (θtotal) was calculated.
he value for θtotal
θ1 + θ2 + θ3 = 41.66˚ + 25.51˚ + 31.55˚ = 98.72˚
fter that ew relat o sh p betwee each of the ajor a pulated var able θ w th the θoutput
a d θinteraction was form as represent by equation 24 until 26.
θ1 = 41.66˚
θ1,output + θ1,interaction
(25)
θ2 = 25.51˚
θ2,output + θ2,interaction
(26)
θ3 = 31.55˚
θ3,output + θ3,interaction
(27)
The maximum effect on the output is 90˚ and the relationship is shows as the equation 28.
Maximum effect on output: 90˚ = θ1,output + θ2,output + θ3,output
(28)
relat o sh p for θinteraction for each of the manipulated variables is as equation 29.
θtotal –
a . effect o output
θ1,interaction + θ2,interaction + θ3,interaction
8.72˚ = θ1,interaction + θ2,interaction + θ3,interaction
(29)
(30)
herefore for θ1,output (x1) was 58.45˚ and θ1,interaction (x2) was -16.79˚. Based on this result, it
shows that the θ1,output have a large effect o the s ste wh le θ1,interaction the effect cannot be
establish on the system because it is too small (θ < ˚). By referring to table 6.1, it shows that as
inlet temperature increased the production of ethanol will decreased. This table also shows that,
the optimum inlet temperature for this process was 200˚C which ethanol composition at the
outlet stream was 0.3030.
Besides that, θ2,output also will not have any effect on the system because the value for this
co sta t was zero. owever the θ2,interaction having a large effect on the system because its value
was equal 25.51˚.The mixture ethylene and water must be in equal composition because if one
of the components is too high or too low, the production of ethanol will be decrease. This is
proven by the result in table 13.
or θ3,output, value was equal 31.55˚. Based on the rules of the system identification method
proposed by Ahmmed S. Ibrahem, it means that θ3,output have large effect on the system. This
shown in table 14, as the inlet flow rate of the reactor was increased; the production of ethanol
will decrease. This is due to the decreasing of the space velocity of the ethylene-water mixture in
the reactor as the inlet flow rate is increased. Therefore, there will be not enough time for the
reaction of converting the ethylene-water
ture to etha ol the reactor. owever θ3,interaction
have not effect at all toward the system because the value was zero (θ < ˚).
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6
Conclusion
By referring to result and discussion section, it can be conclude that by using system
identification method, the relationship of different variables in the system and the optimum
condition for a process can be obtained. This information is important to industry because
optimum condition of a process can minimum the operation cost at the same time maximum the
production. For ethanol production by using direct hydration of ethylene, the variable that has
biggest effect on the system was the inlet temperature compared to other variables. The second
variable that has big effect on the system was the inlet flow rate of the reactor and lastly followed
by the inlet composition stream of the reactor.
Composition < Flow rate < Temperature
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