CLC with Copper and Nickel as Oxygen Carriers

Depart e t of Che ical E gi eeri g
Research Project ChEN-5
2015
Chemical looping
combustion
CHEMICAL LOOPING COMBUSTION WITH COPPER AND NICKEL
AS OXYGEN CARRIER
VENKATA RATNAKUMAR KAPPAGANTULA
16907427

Chemical Loop Combustion
Page1
Contents
List of Tables.............................................................................................................................................2
List of Figures ...........................................................................................................................................3
Executive Summary..................................................................................................................................4
Acknowledgements..................................................................................................................................4
1 Introduction ..........................................................................................................................................5
1.1 Introduction about fluidized bed reactor .....................................................................................5
1.2 Carbon formation..........................................................................................................................6
1.3 Power plant technology based on fluidization technology ..........................................................7
1.4 Scope of this project .....................................................................................................................8
2 CHEMICAL LOOPING COMBUSTION OF METHANE USING CU-BASED OXYGEN CARRIER ....................9
2.1 Oxygen carrier...............................................................................................................................9
2.2 Hydrodynamic Model..................................................................................................................12
2.3 MASS TRANSFER AND RATE CONSTANT .....................................................................................23
2.4 Material balance .........................................................................................................................24
2.4.1 Material balance calculations .............................................................................................24
2.4.2 Mass rate calculation..........................................................................................................28
2.4.3 Mass loading or the solid circulation:.................................................................................30
2.5 Bubbling bed model:...................................................................................................................32
2.5.1 Evaluating the rate constant:..............................................................................................32
2.5.2 Determination of reaction time and space time of the Fuel Reactor:................................35
2.5.3 Evaluation of the concentrations........................................................................................37
2.5.4 Mole balance of a Continuous Stirred Tank Reactor (CSTR)...............................................38
2.5.5 Mole balance of Plug Flow Reactor (PFR)...........................................................................40
2.6 SIMULATION................................................................................................................................45
2.6.1 The Fuel Reactor .................................................................................................................46
2.6.2 Reduction and oxidation kinetics........................................................................................47
3 CHEMICAL LOOPING COMBUSTION OF METHANE USING NICKEL BASED OXYGEN CARRIER............49
3.1 Air Reactor ..................................................................................................................................49
4 COMBUSTION WITH PURGE GAS........................................................................................................52
5 SIMULATION FOR NATURAL GAS REFORMATION...............................................................................57
6 ECONOMIC ANALYSIS..........................................................................................................................61

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7 CONCLUSIONS.....................................................................................................................................66
8 APPENDICES........................................................................................................................................67
8.1 Optimum Oxygen Supply ............................................................................................................67
8.2 Efficiency of the combustion: .....................................................................................................68
8.3 NOMENCLATURE.........................................................................................................................70
8.4 REFERENCES................................................................................................................................73
8.5 Material and composition streams...............................................................................................0
List of Tables
Table 1 Properties of the Cu14Al-I Oxygen carrier .....................................................................................10
Table 2 Kinetic parameters for reduction of Cu14Al-I with CH4, H2 & CO.................................................10
Table 3 Main operating and modeling parameters of the FR.....................................................................10
Table 4 Algorithm for Hydrodynamic model ..............................................................................................11
Table 5 Bubble dia Vs Bed height................................................................................................................19
Table 6 bed velocities with differential pressure........................................................................................20
Table 7 List of parameters evaluated in the hydrodynamics......................................................................22
Table 8 Algorithm for rate kinetics .............................................................................................................23
Table 9 Mass rate of Copper.......................................................................................................................28
Table 10 Mass loading of Cu .......................................................................................................................31
Table 11 List of parameters evaluated in mass transfer & rate constant section......................................43
Table 12Reaction selection in Hysys...........................................................................................................45
Table 13 Thermodynamic data input..........................................................................................................46
Table14 Hydrodynamic & Kinetic data input..............................................................................................46
Table 15Purge gas composition..................................................................................................................53
Table 16 Thermodynamic data input..........................................................................................................55
Table 17 Hydrodynamics and Kinetics data input for purge gas modelling ...............................................56
Table 18 Oxygen supply ..............................................................................................................................67
Table 19 Material stream for Methane combustion on Copper catalyst .....................................................0
Table 20 Composition of material for Methane combustion on Copper oxygen carrier .............................1
Table 21 Methane combustion on Nickel as Oxygen carrier material stream .............................................2
Table 22 Material composition for methane combustion on Nickel Oxygen carrier ...................................3
Table 23 Material stream for Purge gas combustion on Copper catalyst ....................................................4
Table 24 Composition of material for Purge gas combustion on Copper oxygen carrier.............................5
Table 25 Mass & Heat flow for Partial Oxidation of Methane in chemical loop reforming .........................6
Table 26 Component balance for Methane partial oxidation in chemical loop reforming..........................7
Table 27 Calculated mass flow rate for Methane Combustion on Copper Oxygen carrier..........................8
Table 28 Calculated mass flow rate for Methane Combustion on Copper Oxygen carrier..........................9
Table 29 Calculated component balance for Methane combustion on the Nickel Oxygen carrier............10

Page3
List of Figures
Figure 1 Schematic diagram of chemical loop combustion..........................................................................6
Figure 2 CLC process for power generation..................................................................................................8
Figure 3 Wake parameter and particle diameter relation Adapted from Kunni & Levenspiel, Fluidized
Engineering, 2nd
Edition ..............................................................................................................................16
Figure 4 Bubble diameter at various heights..............................................................................................18
Figure 5 Bed velocities with differential pressure ......................................................................................21
Figure 6 Flow chart for material balance....................................................................................................26
Figure 7 Material balance ...........................................................................................................................27
Figure 8 Mass rate of Copper......................................................................................................................29
Figure 9 Mass loading of Cu........................................................................................................................31
Figure 10Mass transfer resistance for the starting material ......................................................................35
Figure 11 Natural Gas combustion on Copper catalyst ..............................................................................48
Figure 12 Methane combustion on Nickel Oxygen carrier .........................................................................52
Figure 13 Purge gas combustion on Copper catalyst..................................................................................57
Figure 14 Steam Reforming of Methane - ThyssenKrupp technology........................................................59
Figure 15Partial Oxidation of Methane ......................................................................................................60
Figure 16 Chemical Looping Reforming......................................................................................................61
Figure 17 $/ton of Co2 released .................................................................................................................64
Figure 18 O2 requirement for varying concentration of Copper Oxides....................................................68
Figure 19 Combustion efficiency for the various fuel flow rates Cu oxygen carrier...................................70

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Executive Summary
Chemical looping combustion is the novel technology to meet the demand of the control of Carbon-
dioxide emissions. Experiments are being conducted on large scale for the application of this technology
in various cases.
In this project, a basic model was developed with Copper as Oxygen carrier, derived from the previous
work of Abad et al and compared the simulated model for Nickel as Oxygen carrier with the work of
Lyngfelt et al. Tests have been conducted on simulated model for Purge gas composition and touched
the base of partial oxidation of methane for synthesis gas production. Hysys was used for the
simulations. All the models were tested with Equilibrium Reactions only.
All the Hysys models are validated with the material balance sheets are provided at the end of this
document. Few recommendations were made for process optimization in the Appendices.
Acknowledgements
This proje t ould ’t e possi le ithout the technical support from Dr. Vuthaluru H. & Dr. Ingram G. of
Curtin University of Technology. I would like to thank my colleagues Dowker J., Van Smaalen S. &
Hermus H. for proof reading and valuable suggestions throughout the project.

Page5
1 Introduction
1.1 Introduction about fluidized bed reactor
Chemical-looping combustion (CLC) is a combustion technology with inherent separation of the
greenhouse gas (CO2) (A.Lyngelt, 2008)1
. Fossil fuel based power plants are an effective industry to
implement CO2 capturing technologies. The current technologies for CO2 capturing are energy
intensive. Much research effort has been put in to the search for low cost technologies, CLC has
now emerged as the solution for low cost power production and CO2 removing efficiency
(Toftegaard et al, 2010)2
. This technology involves the use of metal oxides such as an oxygen
carrier, where oxygen transfers from the combustion air reactor to the fuel reactor. The air
reactor and fuel reactor are inter-connected fluidized beds.
Different reactor concepts have been developed including alternating fixed bed reactors,
interconnected fluidized bed reactors and rotating reactors. However these systems are not
continuous processes. For continuous power production fluidized bed technology was
conceptualized, where particles are continuously fluidized between two reactors (Lyngfelt,
2001)3
(Adanez, 2006)4
(Proll, 2009)5
(Sridhar, 2012)6
The fuel is introduced into the fuel reactor, where in the fuel and metal oxide reacts in the fuel
reactor as per the following equation
CnH2m + (2n+m) MexOy = (2n+m) MexOy-1 + mH2O + nCO2 (Equation 1-1)
The Schematic diagram of the fluidized bed of chemical loop combustion is shown in the Figure
1 Schematic diagram of chemical loop combustion.
The exit gas stream from the fuel reactor contains CO2 and H2O, and the exit gas stream from the
fuel reactor contained CO2 and H2O, and the CO2 (g) is obtained when H2O (aq) water is condensed.
The reduced metal oxide, MexOy-1 transferred to the air reactor where it is re-oxidized as per
the following reaction
2MexOy-1 + O2 = 2MexOy (Exothermic) (Equation 1-2)
Based on the oxygen carrier, (Equation 1-1) represents the most frequent reaction and is endothermic
and (Equation 1-2) is always an exothermic reaction.

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The amount of heat associated with both reactions (Equation 1-1) and (Equation 1-2) is same as
for normal combustion reaction.
All the oxides have predominantly exothermic reactions in both reactors, if the fuel is H2 or CO
but endothermic when the fuel is CH4. Oxygen carrier as CuO is an exception for this. As the
methane fuel reduction reaction is exothermic for oxygen carrier CuO. However the melting
point of the Cu is major disadvantage for the process (Lyngfelt, 2008)1
Figure 1 Schematic diagram of chemical loop combustion
1.2 Carbon formation
Undesired by-product formation:
Some of the side reactions in the CLC are
CH4 + MeO = CO + 2H2 + Me (Syngas formation) (Equation 1-3)
CH4 = C + 2H2 (Carbon formation) (Equation 1-4)

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CO + H2 = H2O + C (Carbon formation) (Equation 1-5)
2CO = CO2 + C (Carbon formation) (Equation 1-6)
Carbon formation: Carbon formation is a very undesirable in the CLC process as coking may reduce the
activity of the solid oxygen carriers. At lower temperature (under 7000
C) the possibilities of Carbon
formation is higher. At increased temperatures, carbon formation decreases, except for CuO & Ca2SO4 as
it was zero for all temperatures. Deposited carbon can be separated from the gas stream, but cannot be
separated from the oxygen carrier stream. It transports to air reactor with oxygen carrier and oxidizes to
CO2 and contaminates the N2 product. Combustion of carbon generates energy, but increases the air
requirements in the air reactor. The selection of the oxygen carrier at 6000
C with a decreasing yield of
carbon is Mn2O3 < Fe2O3 < CoO < NiO < Na2SO4 < CaSO4 < CaO (Rutuja Bhoje, 2013)7
SO2 & H2S formation: The possibility of SO2 & H2S formation in the fuel reactor is directly related to the
usage of sulfates as Oxygen carriers. H2S formation decreases with increase in temperature for all
sulfates. The SO2 formation increases with increase in the temperature for all the sulfates. (Rutuja Bhoje,
2013)7
, (www.imperial.ac.uk , N/A) 8
1.3 Power plant technology based on fluidization technology
Figure-2 (Yaser Khojasteh, 2015)7
shown with the CLC shown as an integrated process within the total
power plant. The solid metal particles reacting with the air in the air reactor to produce metal oxides,
the metal oxides are then reacted with the fuel in the Fuel Reactor, produces heat, exhaust gases and
metal for re-use. The hot Air /Nitrogen exiting the Air Reactor drive the turbine for power production.
Additional waste heat from the hot Air /Nitrogen used for steam production, Exhaust gases leaving from
the Fuel Reactor used to produce steam for power generation.
Nickel Oxide and Iron Oxide are tested as oxygen carrier. Experimental results obtained by Mattison
(Mattison, 2004)12
& Adanaz (Adanaz, 2012)25
shown that NiO has life span and low circulation rate. On
the other hand Iron Oxide is less expensive than any other catalyst (Corbella, 2007)26
. Lyngfelt tested
both Ni & Fe catalyst for 4000h with 100% CO2 capture (Yaser Khojasteh, 2015)7
& (Lyngelt, 2007)27
.
Linder Holm et al (Linder Holm, 2011)28
has tested and analyzed 10KW plant, Kolibitsh (Kolibitsh, 2008)29
has tested 120KW Ni based plant
From the Figure 2 CLC process for power generation (Yaser Khojasteh, 2015)9
shown with the chemical
looping combustion integrated with the total power plant. The solid metal particles are reacting with the
air in the air reactor to produce metal oxides, the metal oxides are then reacted with the fuel in the Fuel
Reactor to produce heat, exhaust gases and metal for re-use. The hot air /Nitrogen exiting the Air
Reactor drive the turbine for power production. Additional waste heat from the hot air /Nitrogen is used
for steam production, Also exhaust gases exiting Fuel Reactor is used to produce steam for power
generation.
Both Nickel Oxide and Iron Oxide have been tested as the oxygen carrier. Experimental research results
obtained by Mattison (Mattison, 2004)12
& Adanaz (Adanaz, 2012)11
shown that NiO has better life span
and low circulation rate. However Iron Oxide is less expensive than any other catalyst (Corbella, 2007)12
.

Page8
Lyngfelt tested both Ni & Fe catalyst for 4000h with 100% CO2 capture (Yaser Khojasteh, 2015)9
&
(Lyngelt, 2007)13
. Linder Holm et al (Linder Holm, 2011)14
has tested and analyzed 10KW plant, Kolibitsh
(Kolibitsh, 2008)15
has tested 120KW Ni based plant
Figure 2 CLC process for power generation
1.4 Scope of this project
In this project, the following topics are discussed
Section 2 (Copper as Oxygen carrier): In this section Copper is chosen as the Oxygen carrier.
Hydrodynamic model and bubbling bed model are developed. With the data obtained from these
models, few case studies were carried out at the end of each sub-section. Hysys model is developed
based on the data obtained in the section
Section -3 (Nickel as Oxygen carrier): In this section Nickel catalyst is chosen for the same data obtained
in the section-2 and compared the performance of Nickel catalyst over the Copper Catalyst.
Section -4 (Purge gas comparison): Purge-gas (Hydrogen & Carbon-monoxide) is tested with Copper
catalyst and results are analyzed.

Page9
Section -5 (Partial Oxidation of CH4): Tested the model for Partial oxidation of CH4 over Nickel catalyst
for syngas production and results are discussed
Section-6 (Operational costs and plant life cycle): This section is about economic analysis.
2 CHEMICAL LOOPING COMBUSTION OF METHANE USING CU-BASED
OXYGEN CARRIER
2.1 Oxygen carrier
Selection of Oxygen carrier: The selection criteria of the Oxygen Carrier is based on the following points:
 High reactivity under oxidation and reduction reactions
 Able to convert fuel completely in to the products (Carbon dioxide and water)
 Long life expectation without attrition
 Resistance to chlorine and Sulphur degradation (Alstom Inc., 2007)16
A CuO based oxygen carrier (Cu14Al-I) is selected for the study based on the experiment results of the
previous work (Alberto Abad, 2009)17
. Two different particle size ranges were used, namely 0.1-0.3mm
& 0.2-0.5mm. Table 1(Properties of the Cu14Al-I Oxygen carrier) Shows the main properties of fresh
oxygen carrier (Alberto Abad, 2009)17
For determination of kinetics, the shrinking core model with control by the chemical reaction in the
grain was used. The equations describing this model under chemical reaction control in the catalyst
grain are the following (Alberto Abad, 2009)17
t/ = 1-(1-X)1/3
Equation 2-1
Where
 = rg, .CuO/(b.VM.CuO.kCn
) Equation 2-2
The kinetic parameters for the reduction of the oxygen carrier with CH4, H2, &CO are shown in Table 2
(Kinetic parameters for reduction of Cu14Al-I with CH4, H2 & CO) (Alberto Abad, 2009)17

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Table 1 Properties of the Cu14Al-I Oxygen carrier
Description Dp = 0.1-0.3mm Dp = 0.2-0.5mm
Active material xCuO (%) 14 14
ROC (%) 2.7 2.7
Density (kg/m3
) 1500 1560
Table 2 Kinetic parameters for reduction of Cu14Al-I with CH4, H2 & CO
Variables Symbols Units CH4 CO H2
Grain radius rg m 2.10-7
2.10-7
2.10-7
Molar volume
of CO
VM,CuO Cm3
/mole 12.4 12.4 12.4
Stoichiometric
coefficient
b Mole
CuO/mole gas
4 1 1
Reaction order n 0.5 0.6 0.8
Pre
exponential
factor
K0 Mol1-n
mol3n-2
s-1
30.0 0.01 1.0
Activation
energy
E kJ/mole 106 25 60
Table 3 Main operating and modeling parameters of the FR
Variables Symbol Value Units
Reactor geometry
Height Hr 1.2 m
Bottom bed height hb 0.5 m
Diameter Dr 0.1 m
Operational
conditions
Temperature t 973-1073 K
Pressure drop P 4500 Pa
FR solids inventory ms 4 kg
Solids circulation rate Qs 80-250 Kg/h
Inlet fuel flow Qf 0.6-1.2 Nm3
/h
Inlet gas velocity ug 0.1 m/s
Oxygen carrier to fuel
ratio
 0.7-2.2
AR solids conversion Xo,in 1
From the Equation 2-2

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Characteristic time  = rg, .CuO/(b.VM.CuO.kCn
)
Grain radium, rg, CuO = 2*10-7
m
Stoichiometric coefficient, b = 4 mole CuO/mole gas
Molar volume of CuO VM, CuO= 12.4cm3
/mole = 12.4*10-6
m3
/mole
Reaction order of methane, n = 0.5
Pre-exponential factor = 30 mole(1-0.5)
m(3*0.5-2)
s-1
t/ = 1-(1-X)1/3
Concentration of Active material ROC = 14% = 0.14
Table 4 Algorithm for Hydrodynamic model
C: TO EVALUATE THE HYDROYNAMIC PARAMETERS OF THE GIVEN MODEL
READ (dP ,s, g, dr, Qg, )
DIAMETER OF PARTICLE : 2*10-4
m
DENSITY OF THE CATALYST (S) : 1500kg/m3
DENSITY OF THE GAS (g) :0.191kg/ m3
INLET GAS FLOW (Qg) :0.9 m3
/h
VSCOSITY OF THE FLUID () :2.7*10-4
kg/m.s
PRESSURE DROP (P) :4500Pa
WAKE PARAMETER () :0.4
CALCULATE
10 ARCHIMEDES NUMBER (Ar), 20 SUPERFICIAL VELOCITY (u0), REYNOLDS NUMBER (Re)
POROSITY AT MINIMUM FLUIDIZATION (mf), MINIMUM FLUIDIZAATION VELOCITY (umf):
60 IF (Re) < 0.4 GO TO 80
70 TERMINAL VELOCITY(ut) : 1.78*10
-2
*( g(s-g))
2
/(g*))
1/3
*dP [EQUATION 2.6] GO TO 90
80 TERMINAL VELOCITY (ut) : g(s-g)*dP
2
/(18) EQUATION [2.5]
90 HEIGHT OF THE BED AT MIN FLUIDIZATION (hmf): P/hmf = (1-mf)*(s-g)*g EQUATION [2.8]
100 MAXIMUM DIAMETER OF THE BUBBLE (dbm): 0.652(A0(u0-umf))
0.4
EQUATION [2.9]
110 INITIAL DIAMETER OF THE BUBBLE (dbo): db0=0.376(u0-umf)
2
EQUATION [2.10]
120 BUBBLE DIAMETER (db) : (dbm-db)/(dbm-db0) = e
-0.3h/Dr
EQUATION [2.7]
130 BUBBULE RISE VELOCITY (ubr): 0.71(g*db)
0.5
EQUATION [2.11]
140 BUBBLE VELOCITY (ub): ub = u0 –umf + ubr EQUATION [2.12]
150 IF ub>>umf GO TO 170
160 BUBBLE FRACTION () :=(u0-umf)/ub EQUATION [2.14] GO TO 180
170 BUBBLE FRACTION () := (u0-umf)/(ub-umf (1+ EQUATION [2.13]
180 VELOCITY OF THE SOLIDS(us):FROM THE MATERIAL BALANCE Acc(1--us =
ubcAc[2.15]
190 VELOCITY OF THE GAS IN EMULSION (ue): ue = (umf/mf)-us EQUATION [2.16]
200 EMULSION PHASE POROSITY (e):mf + 0.2 – 0.059*exp(-(U0-Umf)/0.429)) EQUATION [2.17]
210 BUBBLE PHASE POROSITY (b): 1-0.146*exp (-(U0-Umf)/4.439) EQUATION [2.18]
220 FLUIDIZATION BED POROSITY (f):*b + (1-)*e EQUATION [2.19]
230 GAS FLOW IN THE BUBBLES (Fb) : A0**ub EQUATION [2.20]
240 GAS FLOW IN THE WAKES (Fw) : A0*mf*ub EQUATION [2.21]
250 GAS FLOW IN THE EMULSION (Fe) : A0*mf*(1- EQUATION [2.22]
260 VOLUME OF THE REACTOR (Vr):/4*dr
2
*h EQUATION [2.23]

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270 VOLUME OF THE BUBBLE PHASE (Vb): Vi* EQUATION [2.24]
280 VOLUME OF THE EMULSION PHASE(Ve):Vi*(1-) EQUATION [2.25]
290 WEIGHT OF THE BED (Ws): hmf(1-mf) EQUATION [2.26]
WRITE Fb, Fw, Fe, Vb, Ve
GAS FLOW IN THE BUBBLES (Fb) : 0.000089m3
/s
GAS FLOW IN THE WAKES (Fw) : 0.000016m3
/s
GAS FLOW IN THE EMULSION (Fe) : 0.00014m3
/s
VOLUME OF THE BUBBLE PHASE (Vb): 0.00026m3
VOLUME OF THE EMULSION PHASE(Ve): 0.00914m3
2.2 Hydrodynamic Model
Modular approach for fluidized bed reactor: Two types of phenomena coexist in the fluidized beds. The
physical phenomenon corresponds to the bed hydrodynamics, the chemical phenomenon corresponds
to the chemical changes occurring in each phase.
In the hydrodynamic model category the fluidized bed is divided into two sections. One is rich in gas
(bubble phase) and rich in solids (emulsion phase). The whole reactor could be axially divided into
several sections, each section itself may be consider to be consisting of two ideal reactors : a PFR to
represent the gas flow through the bubbles and a CSTR to represent the gas flow through the emulsion.
[Sotudeh, 1997)18
, (Rosario Porrazzo, 2014)19
Bubbling bed proposed by Kunni &Levenspiel (1969) is the modified version of the two phase theory. In
addition to the bubble phase and emulsion phase cloud phase (wake) phase is considered. A key
difference between this theory and the other two phase theory is the internal mass transfer coefficient
between bubble and cloud and then cloud and wake phase is considered.
Davidson and Harrison proposed that in the bubble phase, gas stay with the bubble, raise like smoke and
penetrate small distance into the emulsion. This zone of penetration is called clouds. All related
quantities like thickness of the cloud, velocity are simple functions of the bubble diameter. Each bubble
drags substantial amount of wake in the solids
Assumptions of the bubbling bed model:
 Bubbles are of one size and evenly distributed
 Flow of the gas in the vicinity of the bubble follows the Davidson model
 Each bubble drag along with wake of solids, recirculate the solids in the bed, with the up flow
behind the bubbles and down flow with the emulsions
 The emulsion stays at minimum fluidization velocity (Wen & Yu)20
Viscosity of the gas, g = 0.000027Pa-s (Source: Hysys)
Density of the fluid, g = 0.191kg/m3
(Source: Hysys)
Density of the catalyst, s=1500kg/m3
(Table 1)
Diameter of the particle, dP = 0.2mm =2.0*10-4
m (Table 1)
Diameter of the orifice, do = 3.0*103
m (Assumption)
Number of orifices No=68 (Assumption)
Inlet gas flow, Qg=0.9m3
/h (Table 3)

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Diameter of the reactor, Dr=0.1m (Table 3)
Wake parameter,  =0.4 (Kunni Levenspiel, 1991)21
Calculate minimum fluidization porosity (mf):
Void fraction at minimum fluidization or porosity at minimum fluidization require in many equations
mf = 0.586r]-0.029
(g/s) 0.021
(Equation 2-3)
The Equation 2-3  suggested by Wen & Yu in Handbook of fluidization and fluid particles
Ar = Archimedes number and it is expressed as
Ar = gg (s-g) dp3)/
Equation 2-4)
Ar = Archimedes number g*dp
3
*(s-g) g/
= 0.191*(2.0*10-4
)3
(1500-0.191)*9.81/ (2.7*10-5
)2
Archimedes number, Ar = 30.84
Substituting the values in equation [2.1]
mf = 0.586(30.84)-0.0029
(0.191/1500)0.021
=0.439
Therefore, porosity at minimum fluidization velocity, mf = 0.439
Calculate minimum Fluidization Velocity (umf):
Fluidization will be considered to begin at the gas velocity at which the weight of the solids gravitational
force exerted on the particles equals the drag on the particles from the rising gas
umf= 9.23*10-3
*dP
1.801
*(g/g) 0.88
*(s/g) 0.94
(Equation 2-5)
Minimum fluidization velocity, umf
= 9.23*10-3
*(2.0*10-4
)1.801
*(0.191/2.7*10-5
)0.88
*(1500/0.191)0.94
=0.0225m/s
Reynolds number at minimum Fluidization & superficial velocity
ReP = gdPu0/Equation 2-6)
Superficial velocity of the gas, is the velocity of the gas in the same bed without any internals
Inlet gas flow, Qg=0.9m3
/h=2.5*10-4
m3
/s
Cross-sectional area, Ac = *0.12
=0.00785m2
Superficial velocity, u0=Qg/Ac=0.032m/s
Substituting these values in Equation 2-6 for Reynolds Number
ReP = 0.191*2.0*10-4
*0.032/2.7*10-5
= 0.045
Therefore it is laminar flow
Maximum Fluidization Velocity (ut):
If the gas velocity is increased to a sufficiently high value, however, the drag on an individual particle will
surpass the gravitational force on the particle, and the particle will be entrained in a gas and carried out
of the bed. The point at which the drag on an individual particle is about to exceed the gravitational

Page14
force exerted on it is called the maximum fluidization velocity. These following equations were
presented by Kunnii & Levenspiel (1969)
Ut = g (s-g)*dP
2
/ (18) for ReP<0.4 (Equation 2-7)
ut = (1.78*10-2
*(g (s-g)) 2
/ (g*))1/3
*dP for 0.4<Re<500 (Equation 2-8)
As the Reynolds number falls in the first category of the above set of equations, substituting the values,
As the Re number falls under the first category of the above equation, substituting the values, produces
the following;
ut = 9.81(1500-0.191)*(2*10-4
)2
/ (18*2.7*10-5
) =1.21m/s
Now that the maximum and minimum superficial velocities have been determined, operating values for
the bed are known have the maximum and minimum superficial velocities at which we may operate the
bed. The entering superficial velocity, u0 must be above the minimum fluidization velocity umf, but below
the terminal velocity ut.
Both of these conditions must be satisfied for bed operation
The ratio of terminal velocity to minimum fluidizing velocity for spherical particles according to
Bourgeois and Grenier, should be in between 10& 20 for Archimedes number between 104
&106
The ration of terminal velocity to minimum fluidizing velocity for spherical particles according to
Bourgeois and Grenier, should be in between 10 & 60 for Archimedes number less than 104
R = ut/uemf = 1.21/0.0207 = 58.5
Therefore above condition is satisfied
Bubble Diameter (db):
Mori & Wen evaluated the following equation for the bubble diameter with maximum and minimum
bubble diameter & height of the bed
(dbm-db)/ (dbm-db0) = e-0.3h/Dr
(Equation 2-9)
Diameter of the reactor Dr=0.1m
Dbm & Db0 are taken from Equation 2-11&Equation 2-12 substituting in the Equation 2-9
Bubble diameter, db = Dbm – (Dbm-Dbo) exp (-0.3*hmf/Dr)
Fluidized bed height at minimum fluidizing velocity = hmf
P/hmf = (1-mf)*(s-g)*g (Equation 2-10)
According to Amir Farshi (Amir, 2013)22
the minimum height of the fluidized bed is function of catalyst
weight, bed diameter and fluidized bed hydrodynamics. The relation is given in the Equation 2-10
P = 4500Pa, g= 9.81m/s2
mf=0.439
Substituting the values, 4500/hmf = (1-0.439)*(1500-0.191)*9.81
Therefore, bed height at minimum fluidizing velocity, hmf = 0.545m

Page15
Bubble diameter, db = Dbm – (Dbm-Dbo) exp (-0.3*hmf/Dr)
Bubble diameter, db =4.9*10-2
– (4.9*10-2
-0.36*10-2
)*exp (-0.3*0.545/0.1) =4.1*10-2
m
dbm/dr = 4.9*10-2
/0.1=0.49
If dbm >0.3 dr ed is ot freel u li g. Ho a a d Da idso sho ed that . ≤ dbm/dr≤ . u les are
in transitory state between bubbles and slug flow
Maximum bubble diameter (dbm):
In the Equation 2-11&Equation 2-12 u0, umf are in cm/s, A0 is in cm2
and db is in cm
The maximum bubble diameter has been observed to follow the relationship
dbm =0.652(A0 (u0-umf))0.4
(Equation 2-11)
dbm = 0.652(78.54(0.032*102
-0.0225*102
))0.4
= 4.9cm = 4.9*10-2
m
Initial bubble diameter (dbo):
For all the beds initial bubble diameter depends on the type of distributor used. For a porous distributor,
initial bubble diameter follows the equation
db0=0.376(u0-umf)0.4
(Equation 2-12)
Substituting in the above equation, db0=0.376(0.032*102
-0.0225*102
)0.4
=0.36cm = 0.36*10-2
m
Bubble velocity and cloud size:
Davidson found that the velocity of the rise of the single bubble related with the diameter of the bubble.
ubr = 0.71(g*db)0.5
(Equation 2-13)
Substituting the values in the above equation, substituting the values into the above equation generates
the following
Bubble velocity, ubr = 0.71*(9.81*0.041)0.5
= 0.45m/s
Velocity of the bubble rise (ub):
Davidson proposed that rate of bubble rise in a fluidized bed expressed as following equation
ub = u0 –umf + ubr (Equation 2-14)
Velocity of the bubble rise, ub = 0.032-0.0225+0.45 =0.458m/s
Volume fraction bubble ():
The wake parameter (is a function of particle size. The value of  has been observed experimentally
to vary in between 0.25 and 1, with typical values close to 0.4 as shown in the Figure 3 Wake parameter
and particle diameter relation Adapted from Kunni & Levenspiel, Fluidized Engineering, 2nd
Edition

Page16
Figure 3 Wake parameter and particle diameter relation Adapted from Kunni & Levenspiel, Fluidized Engineering, 2
nd
Edition
Expression for the fraction of the bed () occupied by the bubbles proposed by Kunni & Levenspiel
= (u0-umf)/ (ub-umf (1+Equation 2-15)
= (u0-umf)/ub (Equation 2-16)
Equation Equation 2-16 is valid only, if ub>>umf
Substituting the values in the Equation 2-16
= (0.032-0.0225)/0.458 = 0.0204
Velocity of the solids (us):
Using the Kunni & Levenspiel (1969) model, the fraction of the bed occupied by the bubbles and wakes
can be estimated by material balances on the solid particles and the gas flows. The parameter is the
fraction of the total bed occupied by the part of the bubbles that does not include the wake, and is
the fraction of the total bed occupied by the part of the bubbles that does not include the wake, and  is
the volume of wake per volume of bubble. The bed fraction in the wakes is therefore . The bed
fraction in the emulsion phase is (1-. Ac &c represent the cross sectional area of the bed and the
density of the solid particles.
Material balance equation of the solids:
Acc (1--us = ubcAc (Equation 2-17)
Substituting the values in the above equation,
Acc (1--0.4*0.0204us = 0.4*0.0204*0.458cAc
Therefore velocity of the solids, us=0.0038m/s

Page17
Velocity of the gas in emulsion phase (ue):
ue = (umf/mf)-us (Equation 2-18)
Substituting the values in the above reaction, ue = (0.0225/0.439)-0.0038=0.0474m/s
Emulsion phase porosity (e):
e = mf + 0.2 – 0.059*exp(-(U0-Umf)/0.429)) (Equation 2-19)
= 0.439 +0.2-0.059*exp (-(0.032-0.0225)/0.429) =0.582
Bubble phase porosity (b):
b = 1-0.146*exp (-(U0-Umf)/4.439) (Equation 2-20)
= 1-0.146*exp (-(0.032-0.0225)/4.439) = 0.854
Fluidized bed voidage (f):
f = *b + (1-)*e (Equation 2-21)
= 0.0276*0.854 + (1-0.0204)*0.582 = 0.587
Gas flow in the bubbles (Fb):
(Fb) = A0**ub (Equation 2-22)
Substituting the values Fb = /4(0.1)2
*0.0204*0.458 = 0.000073m3
/s
Gas flow in the wakes (Fw):
(Fw) = A0*mf*ub (Equation 2-23)
Substituting the values Fw = /4(0.1)2
*0.439*0.40*0.0204*0.0458 = 0.000013m3
/s
Gas flow in the emulsions (Fe):
(Fe) = A0*mf*(1-ubEquation 2-24)
Substituting the values in the above equation, Fe = 0.00016m3
/s
Volume of the ith
stage (Vi):
Vi= Volume of the reactor =/4*dr
2
*h (Equation 2-25)
Considering the whole reactor as 1 stage
dr= diameter of the reactor = 0.1m
h = height of the reactor = 1.2m
Substituting these values in the Equation 2-25 equation
Vi=/4*0.12
*1.2=0.0094m3

Page18
Volume of the bubble phase (Vb):
Vb=Vi*Equation 2-26)
Vi= 0.0094m3
 = bubble fraction=0.0276 from Equation 2-16
Substituting these values in Equation 2-26
Vb=0.0094*0.0204=0.00019m3
Volume of the emulsion phase, Ve & volume of the cloud phase (Vw):
Ve=Vi*(1-) & Vc = Vb*3*(umf/mf)/ (ub-(umf/mf)) (Equation 2-27)
Vi=0.0094m3
from equation [2.23]
 = bubble fraction=0.0204 from Equation 2-16
Substituting the values in Equation 2-27
Ve=0.0094(1-0.0204) = 0.0092m3
Vc= Volume of the cloud = 0.00019*3*(0.0225/0.439)/ (0.458 -(0.0225/0.439)) = 0.000064m3
/s
Substituting the values Fe = /4(0.1)2
*0.439*(1-0.0204-0.0204*0.4) *0.0474=0.00016m3
/s
Total gas flow, Qg =Fb +Fw +Fe = 0.000073 +0.000013 +0.00016 =0.00025m3
/s =0.9m3
/h
Weight of the bed (Ws):
Ws = hmf (1-mf) (Equation 2-28)
Where hmf =0.545m from the equation [2.8]
Weight of the bed, Ws=1500*/4*0.12
*0.545* (1-0.439) = 3.61kg
Calculation of the bubble diameter at various heights of the reactor:
Taking the values of h from 0.1m to 1.2m in 0.1 intervals, the results are displayed in the following Table
5 Bubble dia Vs Bed height and in the Figure 4 Bubble diameter at various heights. It is apparent that
bubble size is growing up along with the height of the bed
Figure 4 Bubble diameter at various heights
0.00
10.00
20.00
30.00
40.00
50.00
0 0.5 1 1.5
Bubble dia(mm) Vs Bed height
(m)
Bubbledia(mm)

Page19
Table 5 Bubble dia Vs Bed height
Calculating height of fluidized bed at various velocities:
As the fraction of the voids increases, bed height also increases. The correlation used in this study is
h/hmf = mf) = Equation 2-29
Where, height of the fluidized bed at minimum fluidization hmf =0.545m from the Equation 2-10
Porosity of the bed at minimum fluidization, mf= 0.431 from the Equation 2-3
Height of the bed at the porosity,  = h
According to the Kozeny –Carman equation (Mc Cabe, 2005)24
, fluidization velocity, uf is defined as
uf= (s-g) gdP2/ (150*(
Equation 2-30
Density of the solid, s = 1500kg/m3
Density of the fluid, g = 0.191kg/m3
(s & g are from the table)
Acceleration due to gravity, g=9.81m/s2
Diameter of the particle, dP= 2*10-4
m from table
Substituting the values in the equation with different uf values from 0-0.12m in the Equation 2-30, a 3rd
order polynomial expression will be derived as shown below for the velocity of 0.02m/s
0.02= (1500-0.191)9.81*(2*10-4
)2
/ (150*0.000027*(



Solving the above equation for porosity, =0.429
Substituting the correlation for the bed height, porosity in terms of bubble fraction, given in Equation
2-29 . By solving, Height of the bed can be obtained at a velocity and corresponding porosity. Solving the
Equation 2-29 for a velocity of 0.02m/s
Height
(m)
Bubble dia
(mm)
0.1 13.75
0.2 20.91
0.3 26.22
0.4 30.16
0.5 33.07
0.6 35.23
0.7 36.83
0.8 38.01
0.9 38.89
1 39.54
1.1 40.02
1.2 40.38

Page20
Substituting the values in the equation h/hmf = mf)
h/ 0.545 = 0.429/ (1-0.439)
Height of the bed at velocity 0.02m/s and porosity=0.429m, h= 0.417m
Substituting the values in the (Equation 2-10) for differential pressureP
P/hmf = (1-mf)*(s-g)*g
Solving the equation for P by substituting the values
P/0.545 = (1-0.439)*(1500-0.191)*9.81
P=3503.298 Pa
P has been calculated in the above method for all the pressures and tabulated in the Table 6 bed
velocities with differential pressure
Table 6 bed velocities with differential pressure
velocity Diff pre voidage
u m/s P Pa 
0 0 
0.005 2920.026 0.2912
0.01 3267.585 0.3544
0.015 3420.257 0.3962
0.02 3503.298 0.429
0.025 3542.715 0.454
0.03 3562.769 0.477
0.04 3568.463 0.512
0.05 3669.086 0.525
0.06 3642.955 0.549
0.07 3610.248 0.568
0.08 3417.05 0.603
0.09 3372.249 0.618
0.1 3320.826 0.633

Page21
Figure 5 Bed velocities with differential pressure
Bed Velocity in m/s→
DifferentialpressureinPa→

Page22
Table 7 List of parameters evaluated in the hydrodynamics
HYDRODYNAMIC DATA
description symbol unit Ref
Archimedes number Ar 30.83903563 2.4
minimum fluidization porosity mf 0.439458104 2.3
minimum Fluidization Velocity umf 0.022512327 m/s 2.5
inlet gas flow Qg 0.00025 m3
/s
Cross-sectional area Ac 0.00785 m2
Superficial velocity u0 0.031847134 m/s
Reynolds number Re 0.045057797 2.6
Maximum Fluidization Velocity ut 1.210956896 m/s 2.7
maximum bubble diameter dbm 0.049791978 m 2.11
Initial bubble diameter dbo 0.003657884 m 2.12
bed height at minimum fluidizing velocity hmf 0.545631548 m 2.1
bubble diameter db 0.040815051 m 2.9
Bubble velocity ubr 0.449265456 m/s 2.13
Velocity of the bubble rise ub 0.458600262 m/s 2.14
bubble fraction  0.020354996 m/s 2.16
velocity of the solids us 0.00381403 m/s 2.17
velocity of the gas in emulsion ue 0.047413441 m/s 2.18
Emulsion phase porosity e 0.581728045 2.19
Bubble phase porosity b 0.854306702 2.2
Fluidized bed voidage f 0.587276382 2.21
Gas flow in the bubbles Fb 7.32782E-05 m3
/s 2.22
Gas flow in the wakes Fw 1.28811E-05 m3
/s 2.23
Gas flow in the emulsions Fe 0.000158903 m3
/s 2.24
Volume of the reactor Vr 0.00942 m3
2.25
Volume of the bubble phase Vb 0.000191744 m3
2.26
Volume of the emulsion phase Ve 0.009228256 m3
2.27
Volume of the cloud Vc 6.42557E-05 m3
2.27
Weight of the bed W 3.601376006 kg 2.28

Page23
2.3 MASS TRANSFER AND RATE CONSTANT
Table 8 Algorithm for rate kinetics
C: TO EVALUATE THE MATERIAL BALANCE AND KINETIC PARAMETERS OF THE GIVEN MODEL
READ (Fb,Fw, Fe, Vb,Ve, XCUO, XCUO, MWCUO, MWCU2O, Đ,ROC, K0, E,R)
GAS FLOW IN THE BUBBLES (Fb) : 0.000073m3
/s
GAS FLOW IN THE WAKES (Fw) : 0.000013m3
/s
GAS FLOW IN THE EMULSION (Fe) : 0.00016m3
/s
VOLUME OF THE BUBBLE PHASE (Vb) : 0.00019m3
VOLUME OF THE EMULSION PHASE(Ve) : 0.00914m3
MOLAR FRACTION OF CUO AT FR EXIT (XCUO) : 0.3 (ASSUMED)
DIFFERENCE IN THE MOLAR FRACTION (XCUO):0.45(ASSUMED)
MOLECULAR WEIGHT OF CUO :79.545g/mole
MOLECULAR WEIGHT OF CU2O :143.09g/mole
CATALYST % IN OXYGEN CARRIER(ROC) :14%
DIFFUSIVITY, (Đ) : 1.8*10-5
m2
/s
PRE-EXPONENTIAL FACTOR (K0) :30m/s
ACTIVATION ENERGY (E) : 106KJ/mole
UNIVERSAL GAS CONSTANT (R) :8.314J/mole.K
CALCULATE
10 MOLE FRACTION OF CUO (Xcuo) : Ṅcuo/ (ṄCuO +2 Ṅcu2o) [3.1]
20 MOLE FRACTION OF CU ṄCH4 : Ṅcu (Xcuo AR - Xcuo FR)/4 [3.3]
30 RATIO OF MOLAR FLOW OF METHANE TO COPPER ṄCH4 /Ṅcu : 4/ (Xcuo AR - Xcuo FR) [3.4]
40 MOLE FRACTION OF OXYGEN ṄO2 = (Ṅcuo AR - Ṅcuo FR)/4 [3.2]
50 XS:XCUO-AR-XCUO-FR
60 INTERCHANGE OF THE GAS BETWEEN BUBBLE AND CLOUD
Kbc: 4.5(umf/db)+5.85(Đ0.5
g0.25
/db
1.25
)[3.14]
70 INTERCHANGE OF THE GAS BETWEEN CLOUD AND EMULSION Kce:6.78((mf Đub)/db
3
)0.5
[3.15]
80 BUBBLE TO EMULSION MASS TRANSFER COEFFICIENT Kbe : (KbcKce)/(Kbc+Kce) [3.16]
90 RATE CONSTANT (k) : k0e-(E/RT)
100 VOLUME FRACTION OF SOLIDS IN BUBBLE PHASE = 0.0055
110 VOLUME FRACTION OF SOLIDS IN THE CLOUDE c :(1-mf){3(umf/mf)/(ubr-[ umf/mf])+
120 VOLUME FRACTION OF SOLIDS IN THE EMULSION mf)(1-)/b-c [3.23]
130 TOTAL RESISTANCE Ktot=b+1((kcat/Kbc)+1/(c+1/((1/e)+(kcat/Kce)))) [3.25]
WRITE Fb, Fw, Fe, Vb, Ve
RATE CONSTANT (K) : 0.95m/s
INTERCHANGE OF THE GAS BETWEEN BUBBLE AND CLOUD (Kbc) :4.88s-1
INTERCHANGE OF THE GAS BETWEEN CLOUD AND EMULSION (Kce) :1.57s-1
BUBBLE TO EMULSION MASS TRANSFER COEFFICIENT ( Kbe): 1.19s-1
TOTAL RESISTANCE Ktot :4.92s
-1

Page24
Determination of optimum circulation rate:
The oxygen carrier circulation rate along with the total mass of the oxygen carrier is an important
variable in the economic design of chemical looping combustion. The circulation rate for chemical
looping is governed by the requirement to supply the fuel reactor with the oxygen needed to consume
the fuel.
Ṅcu = molar flow of the Cu metal circulating in the system
ṄCH4 = molar flow of the methane in the system
ṄO2 = moles of Oxygen required for combustion
Molar ratio of X is defined as
Xcuo = Ṅcuo/ (ṄCuO +2 Ṅcu2o) (Equation 2-31)
From the chemical equation, CuO→ Cu2O +O2
ṄO2 = (Ṅcuo AR - Ṅcuo FR)/4 (Equation 2-32)
ṄCH4 = Ṅcu (Xcuo AR - Xcuo FR)/8 (Equation 2-33)
ṄCH4 /Ṅcu = 4/ (Xcuo AR - Xcuo FR) (Equation 2-34 )
Combustion reaction in the fuel reactor is
CH4+2O2→CO2+2H2O (Equation 2-35)
Reduction reaction in the Fuel Reactor is
CuO→ Cu2O+O2 (KJ/mole) (Equation 2-36)
Oxidation reaction in the Air Reactor is
2Cu2O+O2→ CuO (Equation 2-37)
2.4 Material balance
2.4.1 Material balance calculations
From Equation 2-35 moles of Oxygen required for complete combustion of methane = 2 moles of
O2/mole of methane
From the Table 3 Main operating and modeling parameters of the FR Methane to the reactor = 0.9m3
/h
= 0.172kg/h = 0.011kg-mole/h
Therefore Oxygen required for combustion=2*0.011=0.022 mole of O2/mole of CH4=0.172kg/h
Mass flow of the Oxygen= 0.022*32=0.7kg/h
CO2 produced from the reaction = mole of CO2 produced/mole of CH4 consumed =
0.011kg-mole/h of CO2 = 0.484kg of CO2
H2O produced from the reaction =2* mole of H2O produced/mole of CH4 consumed=
2*0.011=0.022kg-mole/h of H2O = 0.396kg/h of H2O
From Equation 2-31 XCuO = mole of Cu in CuO/total mole of Cu in CuO & Cu2O
XCuO-FR is calculated at the exit of the fuel reactor and XCuO-AR is calculated at the exit of the air reactor.
According to Adel (Adel F. Sarofim et al, 2011)25
a value of 0.3 of XCuO-FR and 0.45 for XS gives a

Page25
circulation rate of 135kg/MWt. The value of XS is similar to the value 0.2 to 0.4, proposed by Abad
(Abad et al in 2007)26
.Taking the value of XCuO=0.3 and XS=0.45, XCuO-AR=0.3+0.45=0.75
CuO balance at Air reactor:
With reference to the Equation 2-31
XCuO-AR= Ṅcuo-AR / (ṄCuO-AR +2 Ṅcu2o-AR)= 0.75
Simplifying the above equation
Ṅcuo-AR= 6* Ṅcu2o-AR (Equation 2-38)
XCuO-FR= Ṅcuo-FR / (ṄCuO-FR +2 Ṅcu2o-FR) =0.75, by simplifying this equation
Ṅcu2o-FR= 1.167*Ṅcuo-AR (Equation 2-39)
From the given conversion rates, the following two expressions can be obtained
Ṅcu2o-AR= 0.36*Ṅcu2o-FR (Equation 2-40)
Ṅcuo-FR= 0.4*Ṅcuo-AR (Equation 2-41)
Determination of mass flow circulation at XCuO = 0.3 & XS = 0.45
From Equation 2-33= (ṄCuO-AR - Ṅcuo-FR)/8= 0.0107kg-mole
Substituting Equation 2-41 in the above equation,
ṄCuO-AR =0.1427 kg-mole/h & ṄCuO-FR=0.0571kg-mole/h
Substituting the value of ṄCuO-AR in the Equation 2-31, XCuO-AR= Ṅcuo-AR / (ṄCuO-AR +2 Ṅcu2o-AR)= 0.75
Ṅcu2o-AR = 0.0238kg-mole/h
Substituting the value of Ṅcuo-FR in XCuO-FR= Ṅcuo-FR / (ṄCuO-FR +2 Ṅcu2o-FR) =0.3 Ṅcuo2-FR = 0.03325kg-mole/h
Substituting Ṅcu2o-FR in the Equation 2-31 Ṅcu2o-FR = 0.067kg-mole/h
Molar flow of the CuO at the exit of the Fuel Reactor, NCuO-FR = 0.0571kg-mole/h
Mass flow of the CuO at the exit of the Fuel Reactor = NCuO-FR*Molar mass of CuO = 0.0571*79.535
= 4.56kg/h
Molar flow of the CuO at the exit of the Air Reactor, NCuO-AR=0.1427kg-mole/h
Mass flow of the CuO at the exit of the Air Reactor = NCuO-AR* Molar mass of CuO = 0.1427*79.535
= 11.4kg/h
Molar flow of the Cu2O at the exit of the Fuel Reactor, NCu2O-FR=0.067kg-mole/h
Mass flow of the Cu2O at the exit of the Fuel Reactor= NCu2O-FR*Molar mass of Cu2O = 0.067*143.09
=9.57kg/h
Molar flow of the Cu2O at the exit of the Air Reactor, NCu2O-AR = 0.0238kg-mole/h
Molar flow of the Cu2O at the exit of the Air Reactor, NCu2O-AR*Molar mass of Cu2O=0.0238*143.09
=3.42kg/h
Calculated values of the component balance & flow rates are plotted in Table 27 Calculated mass flow
rate for Methane Combustion on Copper Oxygen carrier &Table 28 Calculated mass flow rate for
Methane Combustion on Copper Oxygen carrier

Page26
Figure 6 Flow chart for material balance

Page27
Air requirement for air reactor:
Moles of O2 required for oxidation of Cu2O in air reactor from Equation 2-37=
1 mole of O2 required / 2 mole of Cu2O
From the Equation 2-35, Oxygen required for combustion = 0.0214kg-mole/h=0.6856kg/h
Nitrogen in the exhaust = 0.0214*(0.79/0.21) = 0.0807kg-mole/h
Oxygen in the exhaust gas is taken as 3% as other boilers
Let the oxygen in the exhaust= xO2
Nitrogen because of excessive O2 = x2*(0.79/0.21) =3.77xO2
Balancing the oxygen
xO2/ (xO2+0.0704+3.77xO2) =3%=0.03
Solving the above equation, O2 in the exhaust xO2=0.002826kg-mole/h=0.0904kg/h
Nitrogen in the exhaust = 0.0805+3.76*0.002826=0.0911kg-mole/h=2.55kg/h
Therefore air flow to the reactor = Reaction O2 + Excessive O2 in exhaust + Exhaust N2 due to excessive
O2 + Nitrogen in the exhaust
0.6856 + 0.0904 + 0.2975 + 2.25 =3.328kg/h
All these material balance flow are shown in the Figure 7 Material balance
Figure 7 Material balance

Page28
Lower Heating value of the gas, Qc = 50MJ/kg (from Hysys)
Higher Heating value of the gas, QC=55.15MJ/kg
Average heating value = 52.5MJ/kg
Energy input from the fuel gas = 52.5MJ/kg= 52.5KJ/g =52.5*16.02 =841KJ/mole
2.4.2 Mass rate calculation
The equation can be used to determine the proportionality between the energy in the fuel introduced to
the fuel reactor and mass rate of copper circulation
Mcu = 4(mwcu)/XS (mwCH4) QC (Equation 2-42)
Molecular weight of CuO at the exit of the Fuel Reactor = 0.3*79.545+0.7*143.09=124g/mole
Molecular weight of methane = 16.02g/mole
Molecular weight of the Copper = 63.546g/mole
Energy input from the fuel gas = 52.5MJ/kg= 52.5KJ/g =52.5*16.02 =841KJ/mole
XS=0.45
Substituting these values in Equation 2-42
Mcu = 4*63.546/ (0.45*16.02*841)=0.0419g/KW = 0.0419kg/MWt of solid circulation
The mass rates of the copper are calculate at different values of XCuO & XS and shown in the Table 9
Mass rate of Copper & Figure 8 Mass rate of Copper
Table 9 Mass rate of Copper
mass rate
XCuO=0.1 XCuO=0.2 XCuO=0.3 XCuO=0.4 XCuO=0.5 XCuO=0.6 XCuO=0.7 XCuO=0.8 XCuO=0.9 XCuO=0.10
X=0.05 0.377 0.377 0.377 0.377 0.377 0.377 0.377 0.377 0.377 0.377
X=0.15 0.125 0.125 0.125 0.125 0.125 0.125 0.125 0.125 0.125 0.125
X=0.25 0.075 0.075 0.075 0.075 0.075 0.075 0.075 0.075 0.075 0.075
X=0.35 0.053 0.053 0.053 0.053 0.053 0.053 0.053 0.053 0.053 0.053
X=0.45 0.042 0.042 0.042 0.042 0.042 0.042 0.042 0.042 0.042 0.042
X=0.55 0.034 0.034 0.034 0.034 0.034 0.034 0.034 0.034 0.034 0.034
X=0.65 0.029 0.029 0.029 0.029 0.029 0.029 0.029 0.029 0.029 0.029
X=0.75 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025 0.025
X=0.85 0.022 0.022 0.022 0.022 0.022 0.022 0.022 0.022 0.022 0.022

Page29
Figure 8 Mass rate of Copper
Character circulation rate:
mOC =10-3
br*MCuOFfXf/ (xCuOXs-FR) (Equation 6-4)
mOC = circulation rate of fully oxidized oxygen carrier (kg of CuO/s)
br = stoichiometric factor of the solid component = 4
MCuO = molecular weight of the solid component = 79.545g/mole
Ff = molar flow of the fuel = 0.0107kg-mole/h = 0.00297mole/s
Xf = gas conversion = 1
xCuO=mass fraction of CuO in total oxygen carrier = 0.63
Xs-FR = variations of the solids in the fuel reactor = 0.45
Substituting in the above equation
MOC = 10-3
*4*79.545*0.00297*1/ (0.63*0.45) = 0.00633kg/s =12kg/h
Characteristic circulation rate, ṁC = br*MCuO/ (XCuO*H0
C) (Equation 6-5)
br = stoichiometric factor of the solid component = 4
MCuO = molecular weight of the solid component = 79.545g/mole
xCuO=mass fraction of CuO in total oxygen carrier = 0.63
H0
C = standard heat of combustion of fuel =-841KJ/mole
Substituting in the above equation, ṁC = 4*79.545/ (0.63*841)=0.6g/KW =0.6kg/MW
Circulation rate of fully oxidized oxygen carrier, ṁC = ṁC/XCuO-FR kg of CuO/MW
At X=0.45, ṁC =0.6/0.45 = 1.33 kg of CuO/MW
0
1
2
3
4
5
6
7
0 0.2 0.4 0.6 0.8 1
Massrate
XCuO
Mass rate kg/MW
XCuO
XCuO
XCuO
XCuO
XCuO
XCuO
XCuO
XCuO
XCuO
XCuO
MassofCUcirculatinginkg/MWts

Page30
2.4.3 Mass loading or the solid circulation:
Rate constant values are taken from the CLCRS Journal (Adel F. Sarofim, 2008)25
from the TGA
experimental method. These rate constants are used for obtaining the optimum circulation rate and
simulation development discussed in the later sections of the report.
Rate constant, Kr, CuO =0.0222s-1
Kr, CuO is the pseudo first order decomposition rate determined experimentally by TGA experimental
method (Adel F. Sarofim, 2008)25
Rate constant, Kr, Cu2O (TGA)=0.0129s-1
Kr, Cu2O is the pseudo first order oxidation rate constant determined experimentally by TGA method (Adel
F. Sarofim, 2008)25
The pseudo first order rate constant for oxidation from the inlet concentration of 21% of Oxygen.
Therefore, it is needed to be calculated a log mean value of inlet concentration of 21% to outlet
concentration of 3% of oxygen in the exhaust gas.
Kr, Cu2O = Kr, Cu2O (TGA)*[pO2, inlet/pO2, log mean]0.5
PO2, inlet =0.21& PO2, outlet = 0.03
pO2, log mean = [PO2, inlet - PO2, outlet]/ [ln PO2, inlet -ln PO2, outlet]
Substituting the values in the above equation
pO2, log mean = [0.21– 0.03]/ [ln0.21 – ln0.03] =0.0925
Kr, Cu2O = 0.0129*[0.21/0.0925] 0.5
=0.0195s-1
Reaction time in Fuel Reactor = FR = (1/ Kr, CuO)*(1/ (1-(XS/ XCuO-AR) = (1/0.022)*(1/(1-(0.45/0.75)
=41.27s
Reaction time in Air Reactor = AR = (1/ Kr, Cu2O)*(1/ (1-(XS/ (1- XCuO-FR 0) = (1/0.0195)*(1/(1-(0.45/0.3)
=52.8s
Mass loading of the oxygen carrier = mCu*(FR + AR) = 0.641*(41.27 + 52.8) = 60.3kg/MWt
The mass loading of the copper are calculate at different values of XCuO & XS and shown in the Table 10
Mass loading of Cu & Figure 9 Mass loading of Cu
Reaction time calculated in the section 2.5.2 of this paper is 40.6s, which is within 1.5% of the value
obtained from the assumed value

Page31
Table 10 Mass loading of Cu
mass loading
XCuO X=0.05 X=0.15 X=0.25 X=0.35 X=0.45 X=0.55 X=0.65 X=0.75 X=0.85
0.1 122 97 84 76 72 70 69.5 72.5 84
0.2 77 69 64 62 61 63 67 82
0.3 62 59 57 58 60 66 83
0.4 56 56 57 60 67 87
0.5 56 58 62 71 94
0.6 60 66 76 105
0.7 72 85 122
0.8 101 152
0.9 219
The mass loading of copper in to the system is from 70kg to 220kg. It is matching with the data obtained
from the literature (80 – 250kg) as in Table 3 Main operating and modeling parameters of the FR
Figure 9 Mass loading of Cu
0
50
100
150
200
250
0 0.2 0.4 0.6 0.8 1
Massloading
Xs
Solid circulation
DX=0.05
DX=0.15
DX=0.25
DX=0.35
DX=0.45
DX=0.55
DX=0.65
DX=0.75
DX=0.85

Page32
2.5 Bubbling bed model:
This model was proposed by Kunii & Levenspiel (1969) and it is the development of two phase model. In
addition of bubble and emulsion phase, wake phase is also considered. A key difference between this
phase and other two phase models is interphase mass transfer consider two resistances. One from the
bubble phase to cloud-wake phase and the other cloud-wake phase to the emulsion phase.
The derivation of model involves following background theory and observation reported by Davidson
and Harrison (1963) & Rowe and Patridge (1962)
1. Bubbles gas stays with the bubbles, recirculating very much like smoke rising and only penetrating a
small distance into the emulsions
2. All related quantities like velocity rise, the cloud thickness and recirculation rate are simple functions of
the size of the rising bubble
3. Each bubble of the gas drags a substantial wake of solids up the bed
2.5.1 Evaluating the rate constant:
Based on the above observations bubbling bed model assumes that
1. Bubbles are of one size and uniformly distributed across the bed
2. The flow of the gas in the vicinity of the bubble follows the Davidson model
3. Each bubble drag along with it a wake of solids, creating a circulation of solids in the bed with up flow
behind the bubbles and down flow in the rest of emulsions
4. The emulsion stays at minimum fluidizing conditions, thus the relative velocity of the gas and solids
remain unchanged.
With the above assumptions, material balances for solids and for gas give in turn
Up flow of solids in the bubble = down flow of the solids in the emulsion (Equation 2-43)
Total through flow of the gas = up flow in the bubble + up flow in the emulsion (Equation 2-44)
We have
Bubble velocity, ubr = 0.71*(9.81*0.0408)0.5
= 0.45m/s from Equation 2-13
Velocity of the bubble rise, ub = 0.032-0.0225+0.45 =0.458m/s from Equation 2-14
Bubble fraction, = (0.032-0.0225)/0.458 = 0.0204 from Equation 2-16
The bed fraction in the clouds (
=3Umf/mf)/ {ubr-Umf/mf)} (Equation 2-45)
Substituting the values of umf=0.0225m/s and mf=0.439 from the Equation 2-5 &Equation 2-3
=3/0.439)/{0.45-/0.439)} = 0.0079
Bed fraction in wakes (
=Equation 2-46

Page33
Taken the value of =0.4 and substituting in the above equation
=
Bubble fraction in down flowing emulsion including clouds (ῶ):
ῶ= (1-Equation 2-47
ῶ=1-
The down flow velocity of the emulsion solid, us=0.0038m/s from Equation 2-17
Rise velocity of the emulsion gas, ue= =0.0474m/s Equation 2-18
Usi g Da idso ’s theoreti al e pressio for u le loud e pressio a d Hig ie’s 9 theor of loud
– emulsion diffusion, the interchange of the gas between bubble and cloud found to be
Kbc=4.5(umf/db) +5.85(Đ0.5
g0.25
/db
1.25
) (Equation 2-48)
Đ=1.8*10-5
m2
/s (AP sinha, parameswar) 27
Diameter of the bubble, db= 4.1*10-2
m from Equation 2-9
Minimum fluidization velocity, umf = 0.0225m/s from Equation 2-5
Therefore, interchange of the gas between bubble and cloud, Kbc
Kbc=4.5(0.0225/0.041) +5.85((1.8*10-5
)0.5
*(9.81)0.25
/0.0411.25
) = 4.88s-1
The interchange of the gas between cloud and emulsion found to be
Kce=6.78((mf Đub)/db
3
)0.5
(Equation 2-49)
Đ=1.8*10-5
m2
/s (AP sinha, parameswar)27
Diameter of the bubble, db= 4.1*10-2
m from Equation 2-9
Substituting of umf=0.0225m/s and mf=0.439 from the Equation 2-5 & Equation 2-3
Velocity of the bubble rise, ub = 0.458m/s from equation
The interchange of the gas between cloud and emulsion, Kce
Kce=6.78((0.439*1.8*10-5
*0.458)/0.0413
)0.5
=1.57s-1
Bubble to emulsion mass transfer coefficient, Kbe
Kbe = (KbcKce)/ (Kbc+Kce) (Equation 2-50)
Substituting the values in the above equation bubble to emulsion mass transfer coefficient, Kbe
Kbe = (4.88*1.57)/ (4.88+1.57) = 1.18s-1
Model expression for first order kinetics:
As per Arrhenius theory, rate constant, (k)
k = k0e-(E/RT)
(Equation 2-51)
Where k0 = pre exponential factor = 30mol1-n
m3n-3
s-1
from Table 2 (Kinetic parameters for reduction of
Cu14Al-I with CH4, H2 & CO)
E = activation energy =106KJ/mole from the Table 2 (Kinetic parameters for reduction of Cu14Al-I with
CH4, H2 & CO) = 106*10-3
KJ/kg-mole
R = universal gas constant = 8.314 J/mole.K = 8.314KJ/kg-mole.K

Page34
T= temperature of the reactor = 973K from Table 3 (Main operating and modeling parameters of the FR)
Let us assume n=1
k0 = pre exponential factor = 30mol1-n
m3n-3
s-1
= 30s-1
Rate constant, k = 30*e-(106*0.001 / (8.314 *973)
=30ms-1
For first order kinetic reaction occurring in a gas-solid fluidized bed with A=0, the rate equation may be
expressed as
-rA, s= (1/Vs) (dNA/dt) =coca (Equation 2-52)
If the gas flow is very high with vigorous rising of the bubbles, then gas flow through the emulsion and
cloud become so small and can be ignored. The disappearance of A in rising bubble phase can be
expressed as
Disappearance from bubble phase = Reaction in bubble + transfer to cloud and wake
-rA, b= (-1/Vb) (dNA/dt) =bkCA, b+ Kbc (CA, b-CA, c) (Equation 2-53)
Vb = volume of the bubble phase = 0.00026m3
from Equation 2-26
k= reaction rate constant s-1
CA, b= concentration of A in bubble phase kg-mole/m3
CA, c = concentration of A in cloud phase kg-mole/m3
Kbc= gas interchange coefficient between bubble and cloud phase s-1
=6.16s-1
b=Volume fraction of solids in bubble in a fluidized bed (Experimental value of b=0.001 to 0.01)
(Thomas) 28
Transfer to cloud and wake = Reaction in cloud an wake + Transfer to emulsion
Kbc (CA, b-CA, c) = ckcatCA, c + Kce (CA, c-CA, e) (Equation 2-54)
Kce=gas interchange coefficient between cloud and emulsion=2.12s-1
CA, e = Concentration of A in emulsion phase kg-mole/m3
c=Volume fraction of solids in cloud in a fluidized bed
c= (1-mf) {3(umf/mf)/ (ubr-[umf/mf]) +Equation 2-55
Substituting the values in equation [3.12]
c= 0.204*(1-0.439) {3(0.0225/0.439)/ (0.45-[0.0225/0.439])+0.4
Transfer to emulsion = Reaction in emulsion
Kce (CA, c-CA, e) =ekCn
A, e (Equation 2-56)
e=Volume fraction of solids in emulsion in a fluidized bed
mf)*(1-) b-c (Equation 2-57)
) (1-0.204) -



Page35
b + c + f = 0.406
Figure 10Mass transfer resistance for the starting material
Accounting these five resistances, reaction rate is
-dCA/dt=kcatCAbKtot (Equation 2-58)
Ktot = Total resistance
Ktot=b+1/ ((kcat/Kbc) +1/ (c+1/ ((1/e) + (kcat/Kce)))) (Equation 2-59)
e=Volume fraction of solids in emulsion in a fluidized bed=0.54 from Equation 2-57
c=Volume fraction of solids in cloud in a fluidized bed=0.0089 from Equation 2-55
b=Volume fraction of solids in bubble in a fluidized bed =0.001 (Thomas)28
2.5.2 Determination of reaction time and space time of the Fuel Reactor:
Kcat =30s-1
(Equation 2-51)
Ktot=0.001+1/ ((30/6.16) +1/ (0.125+1/ ((1/0.28) + (30/2.12))))=0.15s-1
From Equation 2-58 -dCA/dt=kcatCAbKtot
Where CAb = CAb, 0(1-X)
Therefore, -d (CAb, 0(1-X))/dt=kcat (CAb, 0(1-X)) Ktot
Put 1-X = X’ taki g partial deri ati es o oth sides
-dX = dX’
-dX’/X’ = kcatKtot dt
Integrating both sides
-l X’ = -ln (1-X) = ln (1/ (1-X)) = kcatKtot t
Where t = HBFB/ub
HBFB = bed height in fluidized condition
ub = bubble velocity =0.458 (From Equation 2-12)

Page36
HBFB =Hfb/ (1-f)
Hfb = height of the bed = 1.2 (From Table 3 (Main operating and modeling parameters of the FR)
f = porosity of the bed = 0.589 (From Equation 2-19)
Therefore, HBFB =1.2/ (1-0.589) = 2.92m
Grain radius of the CuO catalyst as per Table 1 (Properties of the Cu14Al-I Oxygen carrier) = 2*10-7
m
Time for total combustion, = rg, .CuO/(b.VM.CuO.kCn
) (Christina Dueso, N/A)29
With reference to the Table 2 (Kinetic parameters for reduction of Cu14Al-I with CH4, H2 & CO)
Molar volume of CuO = 12.4cm3
/mole = 12.4*10-9
m3
/kg-mole
Stoichiometric coefficient, b=4
Rate constant, k = 30s-1
(Equation 2-51)
Assumed reaction order, n= 1
Inlet concentration, CAb =molar-flow/volume = 0.0107/0.9 =0.012kg-mole/m3
Substituting in the Equation 2-2
Time for total combustion,  = 2*10-7
/ (4*12.4*10-9
*30*0.0121
) = 11.2s
As per the Equation 2-1
Reaction time, t/ = 1-(1-Xs)1/3
(Christina Dueso, N/A)29
Xs = Solid conversion =0.45
Reaction time, t =2s for 0.45 solid conversion
Considering the reaction order, n= 0.5 as given in the Table 2 (Kinetic parameters for reduction of
Where k0 = pre exponential factor = 30mol1-n
m3n-3
s-1
from Table 2 (Kinetic parameters for reduction of
E = activation energy =106KJ/mole from the Table 2 (Kinetic parameters for reduction of Cu14Al-I with
CH4, H2 & CO) = 106*10-3
KJ/kg-mole
T= temperature of the reactor = 973K from Table 3 (Main operating and modeling parameters of the FR)
Let us assume n=0.5
k0 = pre exponential factor = 30/10000.5
mol1-n
m3n-3
s-1
= 0.95(mole/m)0.5
s-1
Rate constant, kcat = 0.95*e-(106*0.001 / (8.314 *973)
=0.95 (mole/m)0.5
s-1
Substituting the k value in the following equation for conversion
Ln (1/ (1-X)) = kcatKtot (HBFB/ub) = 0.95*0.15*(2.92/0.458)
Conversion, X =0.908 =90.8%

Page37
2.5.3 Evaluation of the concentrations
We simply express the concentration of methane in emulsion phase, cloud phase in terms of
concentration of methane in the bubble phase. Initially starting with concentration of methane in
emulsion phase in terms of concentration in the cloud phase and then in terms of concentration of
methane in the bubble phase
Transfer to emulsion = Reaction in emulsion
Kce (CA, c-CA, e) =ekcatCn
A, e (Equation 2-56)
The interchange of the gas between cloud and emulsion, Kce =1.57s-1
(Equation 2-49)
Volume fraction of solids in emulsion in a fluidized bed, Equation 2-58
Rate constant, kcat =0.95 (mole/m)0.5
s-1
(Equation 2-51)
Reaction order n = 0.5
Substituting the values in the (Equation 2-56)
1.57*(CA, c-CA, e) =0.54*0.95*C0.5
A, e
3.1CAc – 3.1CAe = C0.5
Ae
Dividing the whole equation with CAe
3.1(Z-1) = 1/ C0.5
Ae Where
Z=CAc/CAe (Equation 2-61)
C0.5
Ae = 0.327/ (Z-1)
Squaring on both sides
CAe =0.107/ (Z-1)2
(Equation 2-62)
Kbc (CA, b-CA, c) = ckcatCA,c
0.5
+ Kce (CA, c-CA, e) (Equation 2-54)
The interchange of the gas between bubble and cloud, Kbc= 4.886s-1
(Equation 2-48)
(Equation 2-49)
c=Volume fraction of solids in cloud in a fluidized bed, c=Equation 2-55
s-1
(Equation 2-51)
CAe =0.107/ (Z-1)2
(Equation 2-62)
4.88(CA, b-CA, c) = CA,c
0.5
+ 1.57(CA, c- 0.107/ (Z-1)2
)
Re-arranging the equation,
4.88CAb + 0.168/ (Z-1)2
= 6.45CAc +0.0089CAc
0.5
Assuming the value of Z=2, above equation can be simplified as
4.88CAb + 0.168 = 6.45CAc +0.0089CAc
0.5
CA, b= concentration of Methane in bubble phase kg-mole/m3
=molar-flow/volume = 0.0107/0.9 =
0.012kg-mole/m3
Put y= CAc
0.5
, above equation can be written as
6.45y2
+0.0089y – 0.2266 =0
By solving the above equation y= CAc
0.5
=0.1869:

Page38
Squaring both sides CAc = 0.0349kg-mole/m3
Assumption was Z = CAc/CAe =2
Therefore, CAe = 0.0349/2 =0.01745kg-mole/m3
CA, b= concentration of Methane in bubble phase kg-mole/m3
= 0.012kg-mole/m3
CA, c= concentration of Methane in cloud phase kg-mole/m3
= 0.0349kg-mole/m3
CA, e= concentration of Methane in bubble phase kg-mole/m3
= 0.01745kg-mole/m3
Space time () = Volume of the reactor / Flow rate entering the reactor
Volume of the reactor = /4*dr
2
*H = /4*0.12
*1.2 = 0.0094m3
Gas flow to the reactor = 0.9m3
/h
Space time () = 0.0094/0.9=0.0105h=38s
Space time () = 38s
2.5.4 Mole balance of a Continuous Stirred Tank Reactor (CSTR)
Considering the Fuel Reactor is a Continuous Stirred Tank Reactor:
Accumulation of species A = Inlet molar flow of species-A – Out let molar flow of species-A +
disappearance / reaction of species-A
0= FA0 – FA + rAV
V = (FA0 – FA)/-rA
Substituting FA = FA0 – FA0X
Equation 2-63
CSTR Volume, V=FA0X/ (-rA)exit
From the Equation 2-25 geometric volume of the reactor, V = 0.0094m3
From the Figure 7 (Material balance), inlet molar flow NCH4 = 0.172kg/h = 0.0107kg-mole/h
Volumetric flow of methane, = 0.9m3
/h
Concentration of CH4 at the inlet of the Fuel Reactor, CAb = 0.0107/0.9 = 0.012kg-mole/m3
Concentration of CH4 at the exit of the Fuel Reactor, CAc = 0.0104kg-mole/m3
(From the section 2.5.3)
Conversion of CH4 in the Fuel Reactor, X = 0.9082
Rate constant from the Equation 2-55 K = 0.95 (mole/m)0.5
s-1
VCSTR = (0.0107*0.9082)/ (0.95*0.0120.5
) = 0.0934m3
With reference to the Figure 10 (Mass transfer resistance for the starting material) the fluid first pass
through the bubble bed, the fluid is not converted pass through the wakes. The fluid is which is not
converted to the products in the wakes transferred to the cloud phase.
Assume that all the reactors are CSTR
Bubble phase reaction:
rA, b= (-1/Vb) (dNA/dt) =bkcat CA,b
n
+Kbc (CA, b-CA, c) (Equation 2-53)
Volume of the bubble phase, (Vb) = 0.00019m3
(Equation 2-26)

Page39
b=Volume fraction of solids in bubble in a fluidized bed =0.001 (Thomas) 28
s-1
(Equation 2-51)
= 0.012kg-mole/m3
= 0.0349kg-mole/m3
(Equation 2-48)
rA, b= (-1/Vb) (dNA/dt) =0.001*0.95*0.0120.5
+4.88*(0.012-0.0349) = 0.0001 +0.1118 =0.1119
But reaction is happening in the bubble phase and the products are exiting the reactor, remaining fluid is
transferring to the cloud phase. First term of the above reaction is only considered in the rate constant
at this instance.
rA, b= (-1/Vb) (dNA/dt)=0.001*0.95*0.0120.5
= 0.0001
Fluid flow through the bubble phase, Fb0 = 0.000073m3
/s [Equation 2-22] = 0.00000087kg-mole/s
CSTR Volume, Vb =FB0X/ (-rA)exit
0.00019 = 0.00000087*Xbubble-CSTR/0.0001
Therefore conversion in the bubble phase reactor, Xbubble-CSTR = 0.0218 =2.18%
Space time in the bubble reactor, bubble = V/vb = V Cb0/Fb0 = 2.92s
Wake phase reaction:
0.5
Volume of the cloud phase, Vc =0.0001m3
/s
(Equation 2-48)
= 0.012kg-mole/m3
= 0.0349kg-mole/m3
c=Volume fraction of solids in cloud in a fluidized bed, c=Equation 2-55
s-1
(Equation 2-51)
(Equation 2-49)
= 0.01745kg-mole/m3
Substituting the values in the Equation 2-54 for rAc
-rAc=0.0089*0.95*0.03490.5
+1.57*(0.0349 -0.01745) =0.0015 +0.0274 = 0.02589
at this instance.

Page40
rAc =0.0089*0.95*0.03490.5
=0.0015
Fluid flow through the cloud phase, Fc =vc*CAc =0.000013*0.0349= 0.000000454kg-mole/s
CSTR Volume, Vc =FB0Xcloud-CSTR/ (-rAC) exit
0.000064 = 0.000000454*Xcloud-CSTR/0.0015
Therefore conversion in the emulsion phase reactor, Xcloud-CSTR = 0.2116 =21.16%
Emulsion phase reaction:
Fluid flow through the emulsion phase Fe = 0.00016m3
Volume of the emulsion phase, Ve= 0.0092m3
[Equation 2-27]
Kce (CA, c-CA, e) =ekcatCn
A, e (Equation 2-56)
(Equation 2-49)
= 0.0349kg-mole/m3
= 0.001745kg-mole/m3
Volume fraction of solids in emulsion in a fluidized bed, Equation 2-58
s-1
(Equation 2-51)
-rAe = 0.54*0.95*0.0019450.5
=0.00214
Fluid flow through the emulsion phase, Fe =ve*CAe =0.00016*0.25= 0.00004kg-mole/s
CSTR Volume, Vc =FB0Xemulsion-CSTR/ (-rAC) exit
0.0092 = 0.0000019*Xemulsion-CSTR/0.00214
Therefore conversion in the emulsion phase reactor, Xemulsion-CSTR = 0.8 =80%
Space time in the emulsion reactor, cloud = V/vc = V Cb0/Fb0 = 10s
2.5.5 Mole balance of Plug Flow Reactor (PFR)
General equation for the PFR is
-dFA/dV = -rA
Expressing the molar flow in terms of the conversion
FA = FA0 – FA0X
Differentiating the above equation
dFA = -FA0dX
Substituting in the general equation for the PFR

Page41
FA0dX/dV = -rA
Separating the differentiate terms
FA0dX/-rA = dV. Integrating on both sides
V = FA0∫X
dX/-rA
Substituting rA =KCA = KCA0 (1-X) in the above equation
V = FA0∫X
dX/-(KCA0 (1-X)) = -FA0/ (KCA0) ∫X
dX/ (1-X)
Equation 2-64 Volume PFR
V= FA0/ (KCA0) ln (1-X)
From the Figure, inlet molar flow FCuO-AR = 11.4kg/h = 0.143kg-mple/h
FCuO-FR = 4.56kg/h = 0.00573kg-mole/h
Concentration of CuO at the inlet of the Fuel Reactor, XCuO-AR = 0.75
Concentration of CuO at the exit of the Fuel Reactor, XCuO-FR = 0.3
Conversion of CuO in the Fuel Reactor, X = 0.6
Rate constant from the Equation 2-55 Ktot = 19.35s-1
Substituting the values in the Equation 2-63 Volume PFR
Volume of the PFR, VPFR = 0.143/ (19.35*0.75) ln (1-0.6) = 0.009m3
It seems to be Volume of the PFR is matching with the geometric volume of the reactor
Fluid flow through the bubble phase, Fb = 0.000089m3
/s [Equation 2-22]
Fluid rising through the wakes, Fw = 0.000016m3
/s [Equation 2-23]
/s [Equation 2-24]
Volume of the bubble phase, Vb=0.00026m3
[Equation 2-26]
[Equation 2-27]
For reaction order n=0.5
V = FA0∫X
dX/-rA = V = FA0∫X
dX/kcatCA
0.5
V = FA0∫X
dX/kcatCA
0.5
= V = FA0∫X
dX/kcat (CAo (1-X))0.5
V = FA0∫X
dX/kcat (CAo(1-X))0.5
= FA0/( kcat CAo
0.5
∫X
dX/(1-X))0.5
= FA0/( kcat CAo
0.5
)[(1-X 0.5
)/0.5]
Equation 2-65
V = FA0/ (kcatCAo
0.5
) [(1-X 0.5
)/0.5]
Bubble phase reaction:
rA, b= (-1/Vb) (dNA/dt) =bkcat CA,b
n
+Kbc (CA, b-CA, c) (Equation 2-53)
Volume of the bubble phase, (Vb) = 0.00026m3
(Equation 2-26)
b=Volume fraction of solids in bubble in a fluidized bed =0.001 (Experimental value of b=0.001 to
0.01(Thomas)28
s-1
(Equation 2-51)
= 0.012kg-mole/m3
= 0.0114kg-mole/m3
(Equation 2-48)

Page42
rA, b= (-1/Vb) (dNA/dt) =0.001*0.95*0.01140.5
+6.16*(0.012-0.0114) = 0.0001 +0.0037 =0.0038
at this instance.
rA, b= (-1/Vb) (dNA/dt)=0.001*0.95*0.01140.5
= 0.0001
Fluid flow through the bubble phase, Fb0 = 0.000089m3
Substituting the values in the Equation 2-63 Volume PFR
V = FA0/ (kcatCAo
0.5
)*[(1-X 0.5
)/0.5]
0.00026= 0.000089/ (0.95*0.01140.5
)*[(1-Xbubble-PFR
0.5
)/0.5]
Therefore conversion in the bubble phase reactor, Xbubble-PFR = 0.7225 =72.25%
Space time in the bubble reactor, bubble = V/vb = V Cb0/Fb0 = 2.92s
Wake phase reaction:
0.5
Volume of the cloud phase, Vc =0.0001m3
/s
(Equation 2-48)
= 0.012kg-mole/m3
= 0.0114kg-mole/m3
c=Volume fraction of solids in cloud in a fluidized bed, c=Equation 2-55
s-1
(Equation 2-51)
(Equation 2-49)
= 0.0052kg-mole/m3
Substituting the values in the Equation 2-54 for rAc
-rAc=0.125*0.95*0.00520.5
+2.12*(0.0114 -0.0052) =0.0126 +0.013 = 0.0257
at this instance.
rAc = 0.125*0.95*0.00520.5
=0.009
Fluid flow through the cloud phase, Fc =vc*CAc =0.000016*0.0114= 0.00000102kg-mole/s
V = FA0/ (kcatCAo
0.5
)*[(1-X 0.5
)/0.5]

Page43
0.0001 = 0.000016/ (0.95*0.00520.5
)* [(1-X 0.5
)/0.5]
0.0001 = 0.00000102*Xcloud-CSTR/0.009
Therefore conversion in the emulsion phase reactor, Xcloud-CSTR = 0.617 =61.7%
Space time in the emulsion reactor, cloud = V/vc = V Cb0/Fb0 = 1.12s
Emulsion phase reaction:
[Equation 2-27]
= 0.0114kg-mole/m3
= 0.0052kg-mole/m3
Volume fraction of solids in emulsion in a fluidized bed, Equation 2-58
s-1
(Equation 2-51)
Fluid flow through the emulsion phase, Fe =ve*CAe =0.000014*0.25= 0.0000035kg-mole/s
V = FA0/ (kcat CAo
0.5
)*[(1-X 0.5
)/0.5]
0.00914 = 0.00014/ (0.95*0.250.5
)* [(1-X 0.5
)/0.5]
Therefore conversion in the emulsion phase reactor, Xemulsion-CSTR = 0.8 =80%
Space time in the emulsion reactor, cloud = V/vc = V Cb0/Fb0 = 10s
Table 11 List of parameters evaluated in mass transfer & rate constant section
Parameter Value Units Source
XCuO 0.3 - Adel F. Sarofim et al
XS 0.45 - Adel F. Sarofim et al
CuO flow to fuel reactor (FCuO-AR) 83.5 Kg/h CuO balance at air reactor
Cu2O flow to fuel reactor (FCu2O-AR) 25.07 Kg/h CuO balance at air reactor
CuO flow to air reactor (FCuO-FR) 33.5 Kg/h CuO balance at fuel reactor
Cu2O flow to air reactor (FCu2O-FR) 69.5 Kg/h CuO balance at fuel reactor
Air flow to the reactor 5.28 Kg/h Air requirements for reactor
Bed fraction in the clouds ( 0.0112 - [3.11]
Bed fraction in the wakes ( 0.011 - [3.12]
Bubble fraction in down flowing
emulsion including clouds (ῶ)
0.0425 - [3.13]
the interchange of the gas
between bubble and cloud(Kbc)
6.16 1/s [3.14]
interchange of the gas between
cloud and emulsion (Kce)
2.12 1/s [3.15]

Page44
Bubble to emulsion mass transfer
coefficient( Kbe)
1.58 1/s [3.16]
Rate constant (ks) 0.95 m/s [3.17]
Volume fraction of solids in
bubble in a fluidized bed (b)
0.001 - Thomas C.Ho
Volume fraction of solids in cloud
in a fluidized bed (c)
0.452 - [3.21]
Volume fraction of solids in
emulsion in a fluidized bed (e)
19.31 - [3.23]
Total resistance (Ktot) 19.35 1/s [3.25]

Page45
2.6 SIMULATION
Hysys does have five basic reactions and reactors. Selection of the reaction is vital as it will impact the
thermodynamic package, which can give errors in the calculation. The Table 12 (Reaction selection in
Hysys) shows the list of reactions and suitable reactors and thermodynamic data collected from the
Hysys
Table 12Reaction selection in Hysys
Conversion Do not require thermodynamic
knowledge
 Define conversion
 Other conversion reactions
may be ranked from 0 and
1 to set the priorities
Equilibrium  Have Hysys determine
Keq from ideal Gibbs free
energy coefficients
 Equilibrium reactions
also cannot be used with
PFR or CSTR
 Equilibrium reactions
can be used with
equilibrium reactors or
General Reactors
 Kinetic reaction set with
equilibrium reactor acts
like CSTR
 Define equilibrium constant
Keq or Ln(Keq) as constant
 Define Keq as function of
temperature
 Tabulated data of Keq Vs T
Kinetic  Deal with the expression
for the rate of the
reactions
 The first term on the
right side of the rate
reaction is forward and
the latter is reverse
reaction
 R=k*f(basis) –k’*f’ asis
 K=A*exp{-E/RT}
 K’=A’e p{-E’/RT}
Kinetic (rev)  The difference between
kinetic & kinetic (rev) is
information about rate
constant, Keq=forward*krev
 R=k*[f(basis) –f’ asis /K’]
 K=A*exp{-E/RT}
L K’ =A’+B’/T+C’L T+D’T
Langmuir-Hinshelwood  Model heterogeneous
catalysis
 Rate of reaction slow
down when a finite
number of active sites
on the catalyst
 Standard equation rate
is added a denominator
 Rate =
numerator/denominator
 Numerator =k*f(basis) -
k’f’ asis
 Denominator
=1+K1*f1(basis) +K2f2(basis)
 K=A*exp(-E/RT)
 K’=A’*e p -E’/RT
 K1 =A1*exp(-E1/RT)

Page46
Hysys does not have few solids in the library, it is required to import the components. For Copper (I)
Oxide, and Copper (II) Oxide the properties shown in the Table 13 (Thermodynamic data input) following
figure are entered as hypothetical solids. They will be shown with (*) mark throughout the model in
Hysys
Table 13 Thermodynamic data input
Component Copper (I) Oxide Copper (II) Oxide
Chemical Formula Cu2O CuO
Molecular Weight (g/mole) 143.09 79.545
Solid density (kg/m3
) 6000 6315
Boiling point (0
C) 1800 2000
Melting point (0
C) 1322 1326
Enthalpy of formation (KJ/mole) -170 -156
2.6.1 The Fuel Reactor
In the bubbling regime superficial gas velocity (U0), higher than fluidization velocity (Umf) and lower than
terminal velocity (Ut). The bubbling bed can be modelled according to the two phase theory [Kunni D,
Levenspiel , 1991]30
with low solids content in the bubble phase, perfect mixing of gas and solids in the
emulsions and the emulsion phase, are within the bed.
According to Davidson [Davidson JF& Harrison D, 1963]31
, In the emulsion, the gas with the excess
pressure than minimum fluidization velocity (umf) is transferred to the bubble phase the emulsion phase
is at minimum fluidization conditions and the gas in excess with respect to the minimum fluidization
velocity (U0 >Umf) is transferred to the bubble phase.
Assumptions: 1) The bubble diameter, db is constant along the bed height 2) the reactor operates at
isothermal conditions 3) the radial mass solid gradient within the bed is neglected
Table14 Hydrodynamic & Kinetic data input
Components Pure components: CH4, O2, N2, H2O, CO2
Hypothetical solids: CuO, Cu2O
Package Peng Robinson
Reactions: Set-1; & Rxn-1:
Reaction: Equilibrium: 4CuO→ 2Cu2O +O2
(KJ/mole)
Fixed Keq = 0.95
Basis: Molar concentration
Set-2: & Rxn-1:
Reaction: Conversion: CH4 +2O2→ CO2 +2H2O
Basis: Methane, CO =100%
Simulation Composition:
CuO: 0.1427kg-mole/h: Cu2O: 0.0476 kg-mole/h

Page47
CH4: 0.9m3
/h, temperature: 7000
C, pressure
100kpa
Reactor:
Set-1&Rxn-1: Equilibrium reactor
Set-2 & Rxn-1 : Conversion reactor
2.6.2 Reduction and oxidation kinetics
Pure methane and CuO / Cu2O oxygen carrier supported by Aluminum (20m diameter) are chosen as
fuel and solid reactant, respectively. Pure air is used in the air reactor to oxidize Cu2O metal. Fuel reactor
model is only considered in this simulation
The heterogeneous non-catalytic reaction occurring in both the riser and fuel reactor is
Within the fuel reactor the reduction reaction is:
CH4 +4CuO +O2→ CO2 + 2H2O + 2Cu2O
While within the riser the oxidation reaction is:
O2 +2Cu2O→ CuO
Various studies have been carried out to characterize the reduction and oxidation behavior of metal
oxides and diverse gases have been considered including CO, H2, and CH4 [Mattisson T, et al 2001]32
,
[Ryu HY et al, 2001]33
, [Ishida M, 1996]34
. Many authors have used kinetic models, based on shrinking
core and changing grain size, to represent the chemical kinetics of the metal oxides [Hossain MM et al
2008]35
. Garcia – Labiano F et al 2005&2006]36-37
, [Abad A et al, 2007]38
. The heterogeneous reaction
proceeds via three steps, external mass transfer, internal mass transfer (internal diffusion within the
particle), chemical reaction.
Within the fuel reactor the reduction reaction is:
CH4 +4CuO +O2→ CO2 + 2H2O + 2Cu2O
While within the riser the oxidation reaction is:
O2 +2Cu2O→ CuO
The following assumptions are made: 1) the particles are spherical, 2) the external mass transfer step is
fast compared to the internal diffusion and reaction steps, 3) The reaction is first order with respect to
the concentration of the reactant gas 4) The particle volume remains constant 5) The reaction is
isothermal
Ruy et al. demonstrate that the reduction rate for CuO /Cu particles supported by bentonite is
controlled studied the CLC kinetics and the variations in the structure of the oxygen carrier were
considered together with various geometries; the changing grain size model was utilized. Small particles
(30-70m) were selected to minimize mass transfer limitations. The shrinking core model with the
reaction being the controlling step describes well the experimental data [Hossain MM 2008]35
. Indeed,

Page48
the oxygen carrier particles used in CLC have small diameter and high internal porosity, consequently
the assumption is shown to be a reasonable one [Abad A et al, 2007]38
Figure 11 Natural Gas combustion on Copper catalyst
Material stream flow rates and composition are shown in the Table 19 Material stream for Methane
combustion on Copper catalyst & Table 20 Composition of material for Methane combustion on Copper
oxygen carrier in the section 8.5

Page49
3 CHEMICAL LOOPING COMBUSTION OF METHANE USING NICKEL
BASED OXYGEN CARRIER
Hysys does not have few solids in the library, it is required to import the components. For Copper (I)
Oxide, and Copper (II) Oxide the properties shown in the following table. They are entered as
hypothetical solids. They will be shown with (*) mark throughout the model in Hysys
Component Nickel Nickel Oxide
Chemical Formula Ni NiO
) 8908 6670
Boiling point (0
C) 2730 N/A
Melting point (0
C) 1455 1955
Enthalpy of formation (KJ/mole) -240
Components Pure components: CH4, O2, N2, H2O, CO2
Reaction: Equilibrium: 2NiO→ 2Ni +O2
Fixed Keq = 0.2045
Set-2: & Rxn-1:
Basis: Methane, CO =100%
NiO: 0.0684kg-mole/h: Cu2O: 0.0114 kg-mole/h
CH4: 0.9m3
C, pressure
100kpa
Reactor:
3.1 Air Reactor
Superficial velocity of the inlet gas (U0), greater than the terminal velocity of an isolated particle (Ut)
[Kunni D, Levenspiel O, 1991]8
. In fast fluidization, perfect mixing of the gas and the solid is assumed.
The solid volume fraction is assumed to remain constant in the bed radial direction, while two zones
along the bed height are identified: the dense and the lean phases: the latter is divided into lower
acceleration region, upper acceleration region and completely fluidized region [Kunni D, Levenspiel O,
1991}8
, [Sotudeh – Gharebaagh R et al, 1998]10
, (Liu B et al 2012)11
. The relative height of those regions
varies with the inlet superficial gas velocity. The solid voidage of the dense phase is assumed constant
whereas the solid voidage of the lean phase decreases along the bed height. The lean phase is usually

Page50
referred to as the transport disengagement height (TDH) and an exponential change in the solids loading
is assumed to describe the variation of voidage, along the TDH [Kunni D, Levenspiel, 1991]8
.These
hydrodynamic parameters are evaluated in the same way as for Copper Oxide catalyst with the only
difference in the variables are the density of the Ni and NiO catalyst. For the same amount of fuel gas
combustion more CuO flow is required to Fuel Reactor than Nickel Oxide flow. This is because of the
better oxygen carrying capacity of the Nickel than Copper. The parameters of the hydrodynamic model
are presented in the following table.
description symbol unit
Archimedes number Ar 16.82615172
minimum fluidization porosity mf 0.442269631
minimum Fluidization Velocity umf 0.020875962 m/s
inlet gas flow Qg 0.000277778 m3
/s
Cross-sectional area Ac 0.00723456 m2
Superficial velocity u0 0.038395946 m/s
Reynolds number Re 0.033243061
Maximum Fluidization Velocity ut 1.079683987 m/s
maximum bubble diameter dbm 0.059030058 m
Initial bubble diameter dbo 0.004705452 m
bed height at minimum fluidizing velocity hmf 0.373877205 m
bubble diameter db 0.042141928 m
Bubble velocity ubr 0.456509743 m/s
Velocity of the bubble rise ub 0.474029728 m/s
bubble fraction  0.036959675 m/s
velocity of the solids us 0.007287686 m/s
velocity of the gas in emulsion ue 0.039914202 m/s
Emulsion phase porosity e 0.585630601
Bubble phase porosity b 0.854575102
Fluidized bed voidage f 0.595570702
Gas flow in the bubbles Fb 0.000126749 m3
/s
Gas flow in the wakes Fw 2.2423E-05 m3
/s
Gas flow in the emulsions Fe 0.000121102 m3
/s
Volume of the reactor Vr 0.001085184 m3
Volume of the bubble phase Vb 4.0108E-05 m3
Volume of the emulsion phase Ve 0.001045076 m3
Volume of the cloud Vc 1.19814E-05 m3
Weight of the bed W 3.318853513 kg

Page51
description symbol unit
Molar flow of methane NCH4 0.010262 kg-mole/h
O2 required NO2 0.020524 kg-mole/h
CO2 produced NCO2 0.010262 kg-mole/h
H2O produced NH2O 0.020524 kg-mole/h
molar flow of Nio at exit of AR NNiO-AR 0.068414 kg-mole/h
molar flow of NiO at exit of FR NNiO-FR 0.027366 kg-mole/h
molar flow of Ni at exit of AR NNi-AR 0.011402 kg-mole/h
molar flow of Ni at exit of FR NNi-FR 0.031927 kg-mole/h
mass flow of NiO at exit of AR mNiO-AR 4.015479 kg/h
mass flow of NiO at exit of FR mNiO-FR 1.606191 kg/h
mass flow of Ni at exit of AR mNi-AR 0.851678 kg/h
mass flow of Ni at exit of FR mNi-FR 2.384699 kg/h
N2 with O2 for combustion 0.077211 kg-mole/h
excessive oxygen in flue gas 0.002702 kg-mole/h
N2 in the exhaust 0.087377 kg-mole/h
Air required 3.186484 kg/h
mass rates of the copper mNi 0.592122 kg/MWt
bed fraction in the clouds b 0.012787
bed fraction in the wake w 0.014784
Bubble fraction in down flowing emulsion ῶ 0.948256
interchange of the gas between bubble
and cloud
Kbc 4.529646 s-1
interchange of the gas between cloud and
emulsion
Kce 1.522442 s-1
Bubble to emulsion mass transfer
coefficient
Kbe 1.139462 s-1
Pre-exponential factor k0 0.948683
Rate constant k 0.948673 s-1
Volume fraction of solids in cloud gc 0.015377
Volume fraction of solids in emulsion ge 0.52074
total resistance Ktot 4.450963 s-1

Page52
Figure 12 Methane combustion on Nickel Oxygen carrier
Material composition and flow rates are shown in the Table 21 Methane combustion on Nickel
as Oxygen carrier material stream & Table 22 Material composition for methane combustion on
Nickel Oxygen carrier.
The calculated molar compositions are tabulated in Table 29 Calculated component balance for
Methane combustion on the Nickel Oxygen carrier
4 COMBUSTION WITH PURGE GAS
Purge gases are the waste gases generated in the process. Very often they do have heating value
because of the Hydrogen content in the purge gas. These purge gases are used as the secondary fuel in
the furnaces. Purge gas do have low heating value and higher in volumetric flow. They do contain
Hydrogen, Carbon-monoxide, Carbon-dioxide, Nitrogen and Argon. Because of their lower heating value,
purge gas are used as secondary fuel, they need supporting gas with higher heating value to support the
flame. Table 15Purge gas composition shows the purge gas composition of the gas as a fuel supply to
the boiler. The composition of the gas taken from Dyno Nobel – Moranbah facilities

Page53
Table 15Purge gas composition
Component Mole%
CH4 1.2
CO 0.8
CO2 42
H2 17
N2 38
Ar 1
Chemical reactions involved are
CH4+2O2→CO2+2H2O (Equation 2-35)
2H2 + O2 → 2H2O (Equation 4-1)
2CO+O2 →2CO2 (Equation 4-2)
Basis: 100kg-mole of purge gas
H2 in 100 kg-mole of purge gas = 17kg-mole of H2
From the Table 2 Kinetic parameters for reduction of Cu14Al-I with CH4, H2 & CO,
Molar volume of CuO for Hydrogen, VM, CuO = 12.4 Cm3
/mole
Stoichiometric coefficient, b = 1
Reaction order, n= 0.8
Activation energy, E = 60KJ/mole = 60*10-3
KJ/kg-mole
Pre-exponential factor, k0 =30 Mol1-n
m3n-3
s-1
=0.119s-1
Rate constant, k = k0e-(E/RT) (
Equation 2-51)
Temperature, T = 973K
KH2 = 0.119*e-(60*10-3)/(8.314*973)
=0.119s-1
Molar volume of CuO for Carbon monoxide, VM, CuO = 12.4 Cm3
/mole
Stoichiometric coefficient, b = 1
Reaction order, n= 0.6
Activation energy, E = KJ/mole = 25*10-3
KJ/kg-mole
Pre-exponential factor, k0 =1 Mol1-n
m3n-3
s-1
=0.475s-1
Rate constant, k = k0e-(E/RT) (
Equation 2-51)
Temperature, T = 973K

Page54
KCO = 0.475*e-(25*10-3)/(8.314*973)
=0.475s-1
=0.475s-1
From the section 2.5.2 for n=0.5,
KCH4 = 0.95s-1
Oxygen required for 2 mole of Hydrogen =1 mole of Oxygen (Equation 2-35)
For 17kg-mole H2 =8.5 kg-mole of O2
Oxygen required for 2 mole of Carbon monoxide =1 mole of Oxygen (Equation 4-1)
For 0.8kg-mole of CO =0.4 kg-mole of O2
Oxygen required for 1 mole of Carbon Methane =2 mole of Oxygen ( (Equation 4-2)
For 1.2 kg-mole of CH4 = 2.4kg-mole of O2
Therefore total O2 required = 11.3kg-mole of O2
According to the Equation 2-36
CuO→ Cu2O+O2 (KJ/mole)
CuO required to produce 1kg-mole of O2 =4 kg-mole of CuO
Therefore CuO required for 11.3kg-mole of O2 = NCuO-AR= 45.2kg-mole of CuO
Assuming XCuO-FR & Xs are 0.3 & 0.45 as in section 2.5
XCuO-AR = Ṅcuo-AR / (ṄCuO-AR +2 Ṅcu2o-AR) =0.3+0.45 =0.75 Equation 2-31
Substituting the value of NCuO-FR in the above equation and solving, NCu2O-AR =7.533kg-mole/h
From the Equation 2-41, Ṅcuo-FR= 0.4*Ṅcuo-AR
Substituting NCuO-AR= 45.2kg-mole, NCuO-FR =18.08kg-mole
Substituting the value of NCuO-FR in the equation
XCuO-FR = Ṅcuo-FR / (ṄCuO-FR +2 Ṅcu2o-FR) =0.3 Equation 2-31, NCu2O-FR =21.09kg-mole
Molar flow of the CuO at the exit of the Fuel Reactor, NCuO-FR = 18.08kg-mole/h
Mass flow of the CuO at the exit of the Fuel Reactor = NCuO-FR*Molar mass of CuO = 18.08*79.535
= 1438kg/h
Molar flow of the CuO at the exit of the Air Reactor, NCuO-AR=45.02kg-mole/h
Mass flow of the CuO at the exit of the Air Reactor = NCuO-AR* Molar mass of CuO = 45.02*79.535
= 3581kg/h

Page55
Molar flow of the Cu2O at the exit of the Fuel Reactor, NCu2O-FR=21.09kg-mole/h
Mass flow of the Cu2O at the exit of the Fuel Reactor= NCu2O-FR*Molar mass of Cu2O = 21.09*143.09
=3018kg/h
Molar flow of the Cu2O at the exit of the Air Reactor, NCu2O-AR = 7.533kg-mole/h
Molar flow of the Cu2O at the exit of the Air Reactor, NCu2O-AR*Molar mass of Cu2O=7.533*143.09
=1078kg/h
Air requirement for air reactor:
Moles of O2 required for oxidation of Cu2O in air reactor from Equation 2-37 =
1 mole of O2 required / 2 mole of Cu2O
From the Equation 2-32, Oxygen required for combustion = 11.3kg-mole/h=362kg/h
Nitrogen in the exhaust = 0.0214*(0.79/0.21) = 42.51kg-mole/h =1191kg/h
Oxygen in the exhaust gas is taken as 3% as other boilers
Let the oxygen in the exhaust= xO2
Nitrogen because of excessive O2 = x2*(0.79/0.21) =3.77xO2
Balancing the oxygen
xO2/ (xO2+42.51+3.77xO2) =3%=0.03
Solving the above equation, O2 in the exhaust xO2=1.49kg-mole/h=47.68kg/h
Nitrogen in the exhaust = 42.51+3.76*1.49=48.11kg-mole/h=1347kg/h
Therefore air flow to the reactor = Reaction O2 + Excessive O2 in exhaust + Exhaust N2 due to excessive
O2 + Nitrogen in the exhaust
362 + 47.68 + 156.87 + 1191 =1758kg/h
Table 16 Thermodynamic data input
Component Copper (I) Oxide Copper (II) Oxide
Chemical Formula Cu2O CuO
) 6000 6315
Boiling point (0
C) 1800 2000
Melting point (0
C) 1322 1326
Enthalpy of formation (KJ/mole) -170 -156

Page56
Table 17 Hydrodynamics and Kinetics data input for purge gas modelling
Components Pure components: CH4, O2, N2, H2O, CO, CO2 H2, Ar
Reaction: Equilibrium: 4CuO→ 2Cu2O +O2
Fixed Keq = 0.95
Set-2: & Rxn-1:
Basis: Methane, C0 =100%
Rank-1
Set-2: & Rxn-2:
Reaction: Conversion: 2H2 +O2→ 2H2O
Basis: Hydrogen, C0 =100%
Set-2: & Rxn-3:
Reaction: Conversion: 2CO +O2→ 2CO2
Basis: Carbon-monoxide, C0 =100%
CuO: 45.02 kg-mole/h: Cu2O: 7.533 kg-mole/h
CH4: 0.9m3
C, pressure
100kpa
Purge gas data: as per Table 4-1Purge gas
composition
Reactor:

Page57
Figure 13 Purge gas combustion on Copper catalyst
Material flow and component balance are tabulated in Table 23 Material stream for Purge gas
combustion on Copper catalyst &Table 24 Composition of material for Purge gas combustion on Copper
oxygen carrier in the section-8.5
5 SIMULATION FOR NATURAL GAS REFORMATION
Component Nickel Nickel Oxide
Chemical Formula Ni NiO
) 8908 6670
Boiling point (0
C) 2730 N/A
Melting point (0
C) 1455 1955
Enthalpy of formation (KJ/mole) -240

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