CHEMICAL PROCESS SIMULATION OF HEXANE 11111:.
ISOMERIZATION IN A FIXED-BED AND
A 9~:!'<2~ REACTOR .. ----
By
TAl-CHANG KAO II
Bachelor of Science
Tunghai University
Taiwan, Republic of China
1986
Submitted to the Faculty of the Graduate College of the
Oklahoma State University in partial fulfillment of
the requirements for the Degree of
MASTER OF SCIENCE May, 1991
~ . '' l •, '·····
r,
i :
Oklahoma State Univ. Library
CHEMICAL PROCESS SIMULATION OF HEXANE
ISOMERIZATION IN A FIXED-BED AND
A CSTCR REACTOR
Thesis Approved:
Dean of the Graduate College
ii
1393133
PREFACE
In recent years, Chemical Reaction Engineering has
developed to a science that uses complicated theoretical
apparatus and sophisticated mathematical models to describe
the behavior of reacting system. It is not simple to find a
realistic approach to the application of the theory in
practical technological research. The main purpose of this
study is to develop a more reliable model to simulate
chemical reactions which proceed in reactors. For this
research work, the ideal plug flow reactor and CSTCR are
chosen.
Many people have aided me in this research work, and it
is impossible to adequately acknowledge their efforts except
in a general way. I am deeply indebted to Dr. Arland H.
Johannes who offered me numerous valuable suggestions and,
who was the main promoter of this study. Also, Dr. Robert
L. Robinson, Jr. and Dr. Khaled A. M. Gasem are most
generous in the encouragement and cooperation. Financial
support from the School of Chemical Engineering is appreci
ated.
Finally, I should like to acknowledge the help of my
parents, Mr. Kao, Ming-Pan and Mrs. Kao, Cheng Su-Chu for
their moral encouragement and constant support throughout
iii
this endeavor. All these are gratefully acknowledged.
iv
TABLE OF CONTENTS
Chapter Page
I. INTRODUCTION 1
II. LITERATURE REVIEW ........ ..... ... ... .. . . . .. . . . . . 3
Importance of Isomerization................ 3 Hexane Isomerization Kinetics . .. .. . . . . ..... 7 Catalytic Reactors . . . . . . . . . . . . . . . . . . . . . . . . . 10 LHSV and Hydrogen-to-Hydrocarbon Ratio ..... 15
LHSV ....... •. . . . . . . . . . . . . . . . . . . . . . . . . . 15 Hydrogen-to-Hydrocarbon Ratio ......... 15
Process Descriptions ....................... 16 Penex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 I s orne rat e . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 6 BP . ................... I • • • • • • • • • • • • • • • • 17
Catalysts .. ·................................ 17 Deactivation . . . . . . . . . . . . . . . . . . . . . . . . . . 19
III. FIXED-BED REACTOR AND CSTCR DESIGN PRINCIPLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Fixed-Bed Reactors ......................... 21 Derivation of Mass Balance
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Derivation of Energy Balance
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Pressure Drop Prediction . . . . . . . . . . . . . . 26 Model Simulation . . . . . . . . . . . . . . . . . . . . . . 27 Numerical Approach . . . . . . . . . . . . . . . . . . . . 28
Design Equations for a CSTCR ............... 28 Derivation of Mass Balance
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 9 Derivation of Energy Balance
Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 Model Simulation . . . . . . . . . . . . . . . . . . . . . . 32 Numeri ca 1 Approach . . . . . . . . . . . . . . . . . . . . 3 2
IV. PROGRAM DESCRIPTIONS AND TESTING ................ 33
Software Applied 33
v
Chapter Page
Program Organization and Subroutine Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
fixed-bed reactor ..................... 34 DATAN . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 ARRAY ... ·- ... I •••••••••• I • • • • • • • • • 3 4 FNC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 PROP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 EQCON . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2 SLTRES . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3 RXN . • . . . . . . . . . • . . . . • . . . . . . . . . . . . . 4 5 LINPAC. ~ ...... I • • • • • • • • • • • • • • • • • • • 46
ester . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 NEWTN . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4'6 ENGBALS . . . . . . . . . . . . . . . . . . . . . . . . . . 48 THEQ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Testing and Results ........................ 50 Overview ............................ -. . SO
Testing . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Optimized Model.............. 52 Fixed-Length Model .......... 55 HEXCR . . . . . . . . . . . . . . . . . . . . . . . 68
V. CONCLUSIONS AND RECOMMENDATIONS ................. 72
LITERATURE
APPENDIXES
Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 Recommenda t i ens . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 3
CITED ................................. · · · · · · 75
77
APPENDIX A- COMPUTER PROGRAM FOR HEXFI ......... 78 APPENDIX B- COMPUTER PROGRAM FOR HEXCR ......... 99 APPENDIX C- LISTING OF CONTROL PANELS .......... 120 APPENDIX D - EFFECT OF PRESSURE UPON
EQUILIBRIUM CONSTANT ............... 131
vi
LIST OF TABLES
Table Page
I. Isomerization Catalysts .......................... 5
II. Octane Numbers of Selected Hydrocarbons and Refinery Blend Stocks ....................... 8
III. Heat Capacities of Hexane Isomers ................ 39
IV. Gibbs Energy of Formation of Hexane Isomers ...... 40
v. Results of the Comparison of Experimental Data with Model Predictions for an Isothermal Fixed-Bed Reactor ................... 54
VI. Results of the Comparison of Isothermal and Adiabatic Operation of a CSTCR ............. 71
vii
LIST OF FIGURES
Figure Page
1. Sketch of Fixed-Bed Reactor ....................... 12
2 . Sketch of CSTCR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3 . Typical Isomerization Process 18
4. Program Organization of HEXFI 35
Sa. Program Organization of CSTCR 36
5b. Program Organization of Energy Balance ............ 37
5c. Program Organization of Newton's Method
6.
7.
8.
9.
10.
11.
12.
13.
Mole Fraction Distributions of Hexane Isomerization (35 ATM, 408 K, Isothermal)
Mole Fraction Distributions of Hexane Isomerization (61 ATM, 408 K, Isothermal)
Mole Fraction Distributions of Hexane Isomerization (35 ATM, 366 K, Isothermal)
Mole Fraction Distributions of Hexane Isomerization (61 ATM, 366 K, Isothermal)
Mole Fraction Distributions of Hexane Isomerization (35 ATM, 366 K, Adiabatic)
Mole Fraction Distributions of Hexane Isomerization {61 ATM, 366 K, Adiabatic)
Mole Fraction Distributions of Hexane Isomerization {35 ATM, 408 K, Adiabatic)
Mole Fraction Distributions of Hexane. Isomerization {61 ATM, 408 K, Adiabatic)
14. Temperature Profile vs Reactor Length of
38
56
57
58
59
60
61
62
63
Hexane Isomerization {35 ATM, 366 K) ............ 64
viii
Figure Page
15. Temperature Profile vs Reactor Length of Hexane Isomerization (61 ATM, 366 K) ............ 65
16. Temperature Profile vs Reactor Length of Hexane Isomerization (35 ATM, 408 K) ............ 66
17. Temperature Profile vs Reactor Length of Hexane Isomerization (61 ATM, 408 K) ............ 67
18. Control Panel 1 for Fixed-Bed Reactor Initial Parameter Settings ...................... 121
19. Control Panel 2 for Fixed-Bed Reactor Initial Parameter Settings ...................... 122
20. Control Panel 1 for Fixed-Bed Reactor Optimized Model, Isothermal Reaction ............ 123
21. Control Panel 2 for Fixed-Bed Reactor Optimized Model, Isothermal Reaction ............ 124
22. Control Panel 1 for Fixed-Bed Reactor Fixed-Length Model, Isothermal Reaction ......... 125
23. Control Panel 2 for Fixed-Bed Reactor Fixed-Length Model, Isothermal Reaction ......... 126
24. Control Panel 1 for CSTCR Initial Parameter Settings ...................... 127
25. Control Panel 2 for CSTCR Initial Parameter Settings ...................... 128
26. Control Panel 1 for CSTCR Isothermal Reaction ............................. 129
27. Control Panel 2 for CSTCR Isothermal Reaction ............................. 130
ix
CHAPTER I
INTRODUCTION
Present isomerization applications in petroleum
refining are used to provide additional feedstock for
alkylation units or high-octane fractions for gasoline
blending. Straight chain paraffins, such as n-butane,
n-pentane or n-hexane can be converted to isomers by
continuous, catalytic (aluminum chloride, antimony
trichloride, etc.) processes.
Isomerization found initial commercial application
during World War II for making high-octane aviation
gasoline. Atlas Processing Company of Shreveport, La., was
the first to install a hexane isomerization process (Penex)
for the production of a motor-fuel blending component (1).
Licensed by the Pure Oil Company, a division of the
Union Oil Company of California, Isomerate is another
continuous isomerization process designed to convert
pentanes and hexanes into highly branched isomers. A rugged
dual-function catalyst is used in a fixed-bed reactor system
(2). Another process licensed by British Petroleum Company,
BP is a two fixed-bed-reactor (one for Cs feed) process
using high activity Platinum catalyst and external hydrogen
(3). These processes will be described in more detail.
1
2
This study is devoted to model design and neglects
the mechanical design and stability study of hexane
isomerization reactors. To model a reactor it is necessary
to write a set of mathematical equations which express the
behavior of the reacting system under various operating
conditions. For the fixed-bed reactor model, two ordinary
differential equations (ODEs) are required to describe the
reactor system. One is a reactor mass balance and the other
is energy balance. For the perfectly mixed reactor model,
two linear equations are also required to describe the
reactor system. One is a reactor material balance and the
other is a reactor energy balance. The two ODEs describing
catalytic fixed-bed and the two linear equations describing
perfectly mixed reactors must be solved simultaneously and
can be solved by means of a computer.
There are various numerical methods which have been
developed to solve systems of simultaneous ODEs. A modified
fourth-order, Runge-Kutta algorithm will be utilized to
obtain the solution of the initial-value ODEs encountered in
fixed-bed reactor problems. Newton's method was used to
solve the linear equations in the Continuously Stirred Tank
Catalytic Reactor (CSTCR) model.
The fixed-bed reactor program is called HEXFI. The
CSTCR model is called HEXCR. Both programs are written in
the FORTRAN language and are listed in Appendix A and B.
CHAPTER II
LITERATURE REVIEW
The literature review will cover the following
subjects:
1. Importance of Isomerization
2. Hexane Isomerization Kinetics
3. Catalytic Reactors
4. Process Descriptions and Catalysts
Importance of Isomerization
The demand of today's automobiles for high-octane
gasolines has stimulated the use of catalytic reforming.
Catalytic reformate furnishes approximately 45-55\ of the
United States gasoline requirements and with the increasing
utilization of low-lead and lead-free gasolines, this can be
expected to increase (4).
Catalytic reforming is a continuous process to upgrade
low-octane virgin, or heavy catalytically cracked naphthas
into high-octane components for motor or aviation fuel
blending or petrochemical usage. The commercial processes
available for use today can be broadly classified as moving
bed, fluidized-bed or fixed-bed types. The primary reaction
mechanisms include in the followings (2):
3
(1) dehydrogenation of naphthenes
(2) dehydrocyclization of paraffins
{3) paraffin isomerization
(4) dehydroisomerization of naphthenes
(5) paraffin hydrocracking
{6} desulfurization
{7) olefin saturation
The petroleum processing industry is without doubt the
largest user of catalysts in the chemical industry. The
catalytic materials include both solids and liquids and
range all the way from common clay to precious metals.
Table I lists various catalytic processes commonly used in
petroleum processing and the materials employed.
4
To understand the significance of catalytic reforming
in refinery operations, the use of the octane number as a
standard for gasoline quality must be understood. Octane
rating has been used for years to measure the antiknock
performance of gasoline. The higher the octane number, the
less the tendency for a gasoline to produce a knocking sound
in an automobile engine.
In 1923, a standard was established for measuring the
octane number of gasoline. The straight-chain paraffin,
n-heptane, was assigned an octane number of zero, and a
branched-chain paraffin, iso-octane {2,2,4-trimethyl
pentane), was assigned an octane number of 100. The octane
number of a gasoline is determined by comparing its anti-
TABLE I
AISOMERIZATION CATALYSTS
Low-temperature processes Catalyst Selectivity
Vapor phase: Anglo-Jersey . . . . . . . . . Impreg. bauxite 95 Phi 11 ips . . . . . . . . . . . . . Impreg. bauxite 95 Shell . . . . . . . . . . . . . . . . Impreg. bauxite 95
Liquid phase: UOP . . . . . . . . . . . . . . . . . . Complex on quartz 97 Standard Oil . . . . . . . . . Liquid complex 97
High-temperature processes Catalyst Regeneration
Butamer . . . . . . . . . . . . . . . Platinum None required Iso Kel . . . . . . . . . . . . . . . Precious metal Regenerable Isomerate . . . . . . . . . . . . Nonnoble metal Infrequent Penex . . . . . . . . . . . . . . . . Platinum None required
AREFERENCE (2)
(11
knock engine performance with various blends of n-heptane
and iso-octane under specified laboratory conditions.
Automotive engineers fix compression ratios for particular
engine designs. Engines with higher compression ratios
require higher-octane-number fuel than those with lower
ratios.
6
Two methods of determining motor fuel octane number are
now in use (5):
1) the research method, ASTM D-2699, a laboratory
simulation of engine performance at low speed
(reported as RON, research octane number).
2) the motor method, ASTM D-2700, a laboratory
simulation of engine performance at high speed
(reported as MON, motor octane number).
Road testing a number of different autos under varying
conditions and gasolines have shown that the average of the
RON and MON, (R+M}/2, gives an acceptable number for rating
gasolines. This average is now a specification on gasoline
and is the octane number displayed on the pumps at service
stations. In the U.S., most service stations offer three
gasoline choices: leaded regular, unleaded regular and
unleaded premium. Leaded regular at service stations has an
octane rating of 88-89. Unleaded regular has an octane
rating of 87-88 and unleaded premium is 91-92.
A few examples of octane number of individual
hydrocarbons and some selected refinery motor fuel blend
stocks are shown in Table II. Note Cs+ reformate
7
(pentane and heavier) from a reformer is the only gasoline
stock that varies in octane number. Reformate octane number
can be varied from 1 to 25 or more. That flexibility is
what makes the catalytic reformer so useful to the petroleum
refiner.
There is one thing I would like to mention here about
tetraethyl lead (TEL) in TABLE II. First introduced in
1922, its effect in improving the octane number of motor
gasoline is well established, the response varying with the
hydrocarbon composition of the gasoline. In spite of much
research work, the exact mechanism by which TEL works to
suppress knock is not known. It is visualized that the
compound is decomposed by heat in the combustion chamber.
This gives rise to particles which then influence the
chemical reactions involved in the combustion of the fuel.
This promotes smooth combustion to the exclusion of knock.
However, TEL has certain well-recognized disadvantages such
as tending to increase deposits in the combustion chamber,
tending to inverse exhaust valve burning, and tending to
foul spark plugs (6).
Isomerization Kinetics
The isomerization of n-hexane to its isomeric forms has
been the subject of a great deal of study (7-10). Various
postulations related to the reaction paths have been made
and rate constants for the mathematical model have been
determined. There are some possible models which describe
TABLE II
OCTANE NUMBERS OF SELECTED HYDROCARBONS AND REFINERY BLEND STOCKSa
Research Motor ml TEL/galb ml TEL/gal
Octane Rating 0.0 3.0 0.0 3.0 (R + M)/2
n-Butane 94.0 104.0 89.0 104.7 91.5 i-Butane 102.0 118.0 97.0 - 99.5 n-Pentane 61.8 84.6 83.2 84.8 72.5 i-Pentane 93.0 104.9 89.7 107,. 3 91.4 n-Octane - 24.8 - 28.1 2 I 2 I 4-TMPC 100.0 115.5 100.0 115.5 100.0 Cyclohexane 84.0 96.6 77.6 87.4 80.8 Alkyl ate 93.0 104.0 92.0 106.0 91.0 Csf- reformate 90.0 98.0 81.0 89.0 85.5 Csf- reformate 95.0 101.0 85.0 93.0 90.0 Csf- reformate 100.0 104.5 90.0 94.0 95.0
a REFERENCE (5)
b To convert milliliters of tetraethyl lead per gallon(ml TEL/gal) to grams Pb/gall multiply ml TEL/gal by 1.057
c Trimethylpentane
00
9
the reaction kinetics:
Frolich (11) and Evering (9) deduced from experimental
data that the isomerization of n-hexane proceeded stepwise
according to the scheme.
Model I:
2-MP
~ ll ~DMB <---=> 2,2-DMB
3-MP
Note that for this model, they used AlCl3-HCl catalyst.
Frolich and Evering also found that the rate determining
step for isomerizing n-hexane to 2,2-dimethylbutane (2,2-
DMB) was the last step in Model I. Later on, Evering
proposed another mathematical model (9) based on graphical
analysis of his experimental data.
Model II:
2-MP
<=> J t <====> 2,3-DMB <====> 2,2-DMB
3-MP
To simplify calculation of the reaction kinetics, a
slightly different reaction mechanism was proposed by Cull
and Brenner (12). First-order reactions were assumed and
the system was described by differential equations given in
the following:
Model III:
n-C6 --- 3-MP --- 2-MP --- 2,3-DMB --- 2,2-DMB
d(n-C6)
dt
d(3-MP)
dt
ka
10
d(2-MP)
dt = k3{3-MP) + k6(2,3-DMB) - (k4+ks)(2,3-DMB)
d(2,3-DMB) = ks(2-MP) + ka(2,2-DMB) - (k6+k7)
dt {2,3-DMB)
d(2,2-DMB) = k7{2,3-DMB) - ka(2,2-DMB)
dt
To test this model, data were obtained from batch runs
and a nonlinear regression technique was applied to
determine the rate constants.
Catalytic Reactors
Choosing a suitable reactor for a gas-solid reaction is
a question of matching the characteristics of the reaction
system, especially the reaction kinetics, with the charac-
teristics of the reactors under consideration.
There is a wide choice of contacting methods and
equipment for gas-solid reaction. These reactors include
fixed-bed reactor, Carberry reactor and fluidized-bed
11
reactor. Each of these reactors has different key features
and are discussed below.
Fixed-bed reactors consist of one or more tubes packed
with catalyst particles and are typically operated in a
vertical position. The catalyst particles may be a variety
of sizes and shapes: granular, pelleted, cylinders, spheres,
etc .. Because of the necessity of removing or adding heat,
it may not be possible to use a single large-diameter tube
packed with catalyst. In this event the reactor may be
built of a number of tubes encased in a single body, such as
is illustrated in Figure 1. The energy exchange with the
surroundings is obtained by circulating, or boiling, a fluid
in the space between the tubes. For an exothermic reaction,
heat evolved due to reaction is much often greater than that
can be transferred to the cooling fluid. This leads to a
maximum temperature somewhere in the reactor, and is called
a "hot spot" (13).
The construction of this type of reactor is
straightforward. In general, unsteady operation results due
to catalyst aging. The reactor is, not very useful for
gathering kinetic data when the catalyst decays rapidly
(14). Leva {15) reported the calculation of pressure drop
along the fixed bed, and Nauman (16) suggests using the
Ergun equation to calculate the pressure drop in this type
of reactor.
The rotating-basket reactor (often known as the
Carberry reactor) has been widely used for gas-solid
Heotinq (or coo!inq) __ !-Fluid in
Tubes pocked witll coto!yst-~
--v
Product stream
t r-----r-r-
~---~--- .... ~....--
t Feed
stream
Heotinq (or coolinq)
----+- Flutd out
Figure l. Multitube Reactor, Fixed-Bed; adapted from Smith (1970).
12
13
catalytic reactions. The construction is not very
difficult, but it is more complex and expensive to build
than a batch or fixed-bed reactor. The catalyst baskets can
be attached to a stirrer or they can be used as the stirrer
paddles. The reactor is operated under transient conditions
if the catalyst decays rapidly. Otherwise, steady-state
operation is obtained. Baffles can be installed to obtain
better contact (17). Figure 2 sketches the main features of
an experimental reactor.
This type of reactor has several disadvantages.
Erosion of the catalyst may occur under sever agitation and
it can be a problem to keep powered catalyst in the baskets.
The surface temperature of the catalyst is very difficult
to measure and it is often erroneously assumed to be equal
to the bulk temperature. For these reasons, the use of very
small catalyst particle size is not recommended.
Fluidized-bed reactors are catalyst particles supported
by an upflow of gas as a fluid bed (18). A mechanical
advantage is also gained by the relative ease with which
solids may be conveyed and, because of solids mixing, the
gas in the reactor is at approximately the same temperature.
Another important advantage of the fluidized-bed reactor
over the fixed-bed type is that the catalyst can be
externally regenerated without disturbing the operation of
the reactor. A disadvantage of the fluidized-bed reactors
is that the equipment is large. To avoid the solid
particles from being blown out the top of the reactor, the
Composition is uniform at XA, out• CA. out
Four rapidly spinning wire baskets containing catalyst pellets, W
-..:::::::::::> ~
Spinning shaft
Figure 2. CSTCR; adapted from Levenspiel (1972).
Fluid out, XA, out• CA, out
14
15
gas velocity must be low. This means that we need to design
large-diameter vessels and this increases the initial cost . . There are also losses of catalyst fines from the reactor,
necessitating expensive dust-collection equipment in the
exit streams.
LHSV and Hydrogen-to-Hydrocarbon Ratio
Space velocity is an important variable in refinery
because it is interchangeable with reaction temperature.
Space velocity has to do with the length of time of contact
between the reactants and the catalyst. Refiners choose an
easily accessible parameter of residence time in either
liquid hourly space velocity (LHSV) or in weight hourly
space velocity (WHSV).
LHSV is the volume per hour of reactor charge per
volume of catalyst. The higher the LHSV, the greater the
volume of feed charge per hour over a given amount of
catalyst. Therefore, contact time with catalyst is less.
Normally, in most isomerization process, LHSVs are
controlled between 1.5- 2.5 (19).
Hydrogen-to-Hydrocarbon Ratio
The main purpose of hydrogen recycle is to increase
hydrogen partial pressure in the reactors. The hydrogen
react with co*ke precursors, removing them from the catalyst
before they can form polycyclic aromatics which ultimately
deactivate the catalyst. Also, hydrogen can inhibit side
reactions such as cracking. Most of the present processes
control the ratio between 2 - 1 (3,20).
Process Descriptions
16
Many papers have published information about their
isomerization processes and operating conditions. These
processes such as Penex, Isomerate and BP are now applied in
commercial processes.
Pen ex
Licensed by Universal Oil Products, Penex is a
nonregenerative Cs and/or C6 isomerization process. The
reaction takes place in the presence of hydrogen and a
platinum catalyst. The Penex process may be applied to many
feedstocks by varying the fractionating system. Mixed feed
may be split into pentane and hexane fractions, and
respective isofractions separated from each other. Reactor
temperatures range from 500 - 900 oF; pressures from 300 -
1000 psig. Hydrogen requirements are low-- (49 scf/bbl)
for pentane isomerization and slightly higher for hexane
isomerization (21,22,23).
Isomerate
Licensed by the Pure Oil Company, Isomerate is a
continuous isomerization process designed to convert
17
pentanes and hexanes into highly branched isomers. A rugged
dual-function catalyst is used in a fixed-bed reactor
system. Operating conditions include reactor temperatures
and pressures which are less than 750 oF and 750 psig,
respectively (24,25).
Licensed by British Petroleum Company, BP is a two
fixed-bed-reactor isomerization process. It uses very high
activity, regenerable Platinum catalyst and hydrogen.
Pentanes, hexanes or mixtures of the two from catalytic
reforming or solvent extraction may be processed. Operating
temperatures are typical less than 320 OF , pressure is
about 250 psig, LHSV is around 1 to 2 and hydrogen to
hydrocarbon mole ratio is 2:1.
A typical simplied flow diagram is shown in Figure 3
( 1 1 3 ) o
Catalysts
The platinum-based catalysts used for isomerization are
similar to those used in catalytic reforming but the
conditions are much less severe. A catalyst promoter such
as hydrogen chloride is added continuously to maintain high
catalyst activity but catalyst deactivation occurs so slowly
that catalyst regeneration is not necessary except at long
intervals (generally greater than one year). Hydrogen is
used to minimize carbon deposits on the catalyst but
DESULFURIZED FEED
ORGANIC CHLORIDE MAKEUP I (
LEAD TAIL ISOMERIZATION ISOMERIZATION
REACTOR REACTOR
MAKEUP H2
STABILIZER
PRODUCT TO STORAGE
Figure 3. Hexane Isomerization Flow Diagram
1-' (X)
19
hydrogen consumption is negligible {19). Table I lists the
catalysts used in some commercial processes. Supported
metal catalysts have been developed for use in high
temperature processes which operate in the range 700 - 900
OF and 300 to 750 psig. Aluminum chloride plus hydrogen
chloride are universally used in the low-temperature
processes.
Deactivation
Most often catalysts are employed to speed up
reactions that are sluggish or will not proceed at all.
They may also allow operation at a lower operating
temperature level, influence the product distributions, or
more rarely, slow down a reaction.
During the chemical process the properties of the
catalyst gradually deteriorate
reasons for this {27):
There may be several
Catalytic Poisoning Catalysts become poisoned when
feed stream contains impurities which are deleterious to the
activity of the catalyst. Particularly strong poisons are
substances whose molecular structure contains lone electron
pairs capable of forming covalent bonds with catalyst
surfaces. For instance, catalytic poisons for metals are
compounds containing sulphur, arsenic and nitrogen. Acidic
catalyst poisons are all normally basic compounds.
Catalytic poisons most often come from impurities present in
raw materials, but sometimes may be present in the material
20
used for preparation of the catalyst itself.
Catalyst Fouling Reactions involving organic compounds
are inevitably accompanied by decomposition of the materials
to carbon or possibly the formation of high-molecular weight
compounds. These gradually cover the surface of the pellets
and block access the active surface (20).
CHAPTER III
FIXED-BED REACTOR AND CSTCR
DESIGN PRINCIPLES
Fixed-Bed Reactors
Beecher, and Voorhies (28) reported that hexane
isomerization was obtained under plug flow (tubular flow)
conditions in a fixed-bed reactor. In a plug-flow reactor
specific assumptions are usually made about the extent of
mixing: no mixing in the axial direction, complete mixing in
the radial direction and uniform velocity across the radius.
The absence of longitudinal mixing is the special character
istic of this type of reactor (29).
Derivation of Mass Balance Equation
The performance equation for a steady state plug flow
reactor is:
Rp dW = F dx
where
W = mass of catalyst, kg
F = molar flow rate of reactant, kgmole/hr
x = conversion of reactant
21
(3-1)
Rp = global rate of reaction per unit mass of
catalyst, kgmole/(kg of cat -hr)
22
Equation (3-1) is based on a material balance and is derived
by several authors (29,30).
Although equation (3-1) is the general form for
tubular-flow reactor packed with catalyst pellets, it does
not meet our requirements. Therefore, it was converted into
the following form which can be utilized for this study:
I (-rij Rj} dW = Fio dXi
Rearranging,
dW
Integrating,
w
Since,
=I __ d_x1 __ I( -n jRj)
= I
F1 = Fio ( 1-Xi )
then,
dF1 = and,
dXi =
Therefore,
w
-Fio dXi
dFi
-Fio
= I ( -dFi /Fi o >
I(-rtjRj}
(3-2)
23
or,
w
1
taking derivatives of both sides, it becomes
dFi dW =
or,
(3-3)
If we assume the reactor tube has a diameter, D, then:
dZ (3-4) 4
Substituting equation (3-4) into equation (3-3) gives:
xD2 dFi = ( pb dZ) I: ( rt j Rj )
4 or,
dFi xD2 = pb t( n j Rj) (3-5)
dZ 4
This is the final mass balance equation utilized in my
design.
where
Fi = the molar flow rate of species i, kgmole/hr
rij = stoichiometric coefficient of the ith
component in the jth reaction.
Rj = reaction rate of the jth reaction,
kgmole/(kg of cat -hr)
A, = the bulk density of the bed, kgjm3
D = the diameter of the tube, m
z = the length of the reactor, m
Derivation of EnergY Balance Equation
If the enthalpy of the reaction per unit mass
above a base state is H at the entrance to an element and
H+~H at the exit, a standard energy balance can be written
as:
FtHilt- Ft(H+t1H) t + U(~Ab.)(T.-T)Llt = 0
or
24
-Ft.dH + U(ilAb)(T.-T) = 0 (3-6)
where
Ft = total molar-flow rate, kgmole/hr
u = overall heat transfer coefficient, kj/(m2-hr-K)
nH = enthalpy change due to reaction,
11 Ab = heat transfer area, m2
but,
llt = time interval, hr
T. = surrounding temperature, K
T = bulk temperature in reactor, K
ilH = Cpt6T + l: (L\Haj} Xij
Fi
Ft
kj/kgmole
25
Using this expression for fj H in equation (3-6) and
simplifying yields
If we divide each term by ~W, and take the limit as
~W ---> 0, we obtain;
dW dW dW
Combining with,
gives,
dT Ft Cp t = U(Ta -T)
dW dW
or,
and,
U{Ta-T):~tDdZ + l:(-~HRj )RjfJDAcdZ dT =
since,
dAh = 7tDdZ
dW = PbAcdZ
Substituting into equation {3-7) gives equation (3-8)
1tD2
dT U(T.-T):JtD + 4 Pbl: Rj(-~HRj) =
dZ
{3-7)
(3-8)
26
Equations (3-5) and (3-8) are the two basic equations
which will be used to model a fixed-bed reactor.
Pressure Drop Prediction
For the calculation of the pressure drop for a catalyst
bed, Ergun (31) recommends the following equation:
d~
where
dr
L
es (1-e) p = 150 + 1.75
1-e dr Go
e = void fraction of the bed, dimensionless
dr = effective diameter of particles, m
L = height of the bed, m
dP = pressure drop, Pa
p = viscosity of fluid, Pa-s = kg/(m-s)
Go = superficial mass velocity, kg/(m2-s)
p = density of fluid, kg/m3
For turbulent region, characterized by:
~ = > 100 p(1-e)
Hence, equation (3-9) may be simplified to:
dP =1.75LpVo2(1-e)/(dre3)
(3-9)
As I mentioned before, there are numerous correlations
in the literature for the calculation of pressure drop (15),
the Ergun equation is probably the best.
Model Simulation
In Chapter 2, it was mentioned that there were three
existing types of theoretical models. The mechanism
proposed by Cull and Brenner (12) is one of the simplest
possible mechanisms involving all five hexane isomers and
requires estimation of a minimum number of kinetic
parameters, ie., eight rate constants. This mechanism can
be written as:
27
n-C6<=====> 3-MP <=====> 2-MP <=====> 2,3-DMB <=====>2,2-DMB
The following assumptions were made to simplify the
kinetic model:
1. Reactions from n-C6 to 2,3-DMB occur very fast and
were modeled as being equilibrium controlled.
2. Total moles of hexane in the reactor are conserved
down the reactor. That is, no appreciable side
reactions such as hydrocracking occur.
3. The reactor operates in plug flow.
4. The reaction of 2,3-DMB to nee-hexane is the rate
controlling step.
With these assumptions, we can eliminate rate constants
k1r through k3r and k1t through k3t. Reaction coordinates
can be used to solve for the equilibrium mole fractions
of n-hexane, 3-MP and 2-MP.
28
Numerical Approach
Computer-implemented numerical methods are now commonly
used for solving system of ordinary differential equations.
In order to calculate the equilibrium mole fraction of n-c,,
3-MP, 2-MP, and 2,3-DMB, three linear simultaneous equations
were required to solve for the reaction coordinates. A
numerical method was used to solve these equations, followed
by a fourth order Runge-Kutta method to approximate the two
ordinary differential equations (equations {3-5) and {3-8)).
The detailed approach will be discussed in the following
chapter.
Design Equations for a CSTCR
Another approach uses a Continuously Stirred Tank
Catalytic Reactor (CSTCR) to carry out the isomerization.
Carberry (17) introduced this type of reactor, which
consists of a rotating basket of catalyst particles. This
reactor also accommodates commercial-size pellets and
extruded catalysts. Levenspiel (30) also refers to this
kind of catalytic reactor as a basket-type mixed reactor.
In the theory of continuous stirred tank reactors, an
important basic assumption is that the contents of the tank
are well mixed (32). This means that the compositions in
the tank are everywhere uniform and that the product stream
leaving the tank has the same composition as the mixture
within the tank.
29
Derivation of Mass Balance Equation
Assume the following hypothetical reaction occurs in a
CSTCR
where
kt A<----> B
kr
kt = forward rate constant
kr = reverse rate constant
A CSTCR is used for this reaction system. The volu
metric feed to the reactor is Q, the catalyst weight W, and
the total flow rate is Fto. Since a CSTCR is designed to
operate at steady-state, a steady-state mole balance defines
the performance of the system. The following mole balance
can be constructed:
(component A)
(component B)
OUT = IN +
CaQ = CaoQ +
CsQ = CsoQ +
GENERATION
RrW
RtW
Assuming simple first order reaction,
Rt = kt Ca
Rk = kr C:a
CONSUMPTION
RtW
RrW
Substituting the rate expressions and rearranging
results in the following set of linear equations:
w
Q ( kr Cs - kt Ca ) = 0 (3-10)
30
w F2 = -Ca + Cao + ---- ( ktCA - krCa) = 0 (3-11)
Q
Note that the only unknowns are C& and Ca. These
equations can be solved using Newton's method and the final
concentrations of C& and Ca can be calculated.
The performance equation of a mixed reactor can then be
utilized to calculate the conversion. It appears below:
W Xaou'l' = (3-12)
F&o -R 'a ouT
where
F&o = the molar flow rate of species i, kgmole/hr
Xaou'l' = conversion of reactant
R'&ou'l' = reaction rate, kgmole/(kg of cat -hr)
W = weight of catalyst, kg
Equation (3-12) is derived by Levenspiel (30}.
Derivation of Energy Balance Equation
Consider a mixed flow reactor, in which conversion is
X&, and T1 is the a temperature on which the enthalpies and
heats of reactions are based.
enthalpy of entering feed:
H' 1 = Cp I ( Tl - Tl ) = 0
enthalpy of leaving system:
H" 2 Xa + H' 2 ( 1-XA) = Cp" (T2 -Tl )XA + Cp' (T2 -Tl) { 1-XA)
energy released by reaction:
at T1
31
At steady state the energy balance is:
input = output + accumulation + released by reaction
or,
0 = [ Cp" (T2 -Tl )XA + Cp' (T2 -Tl) (1-XJL) ] + AHRlXA
rearranging,
0 = (Cp"T2-Cp"T1 )XA + Cp' (T2-T1-T2XA+T1XJL) + tJIR1XA
= XJLCp"T2 - XACp"Tl + Cp'T2 - Cp'T1 - Cp'T2XA
+Cp' T1 XA + ltHR 1 XA
= T2 (XaCp" + Cp'- Cp'Xa) - XaCp"T1 - Cp'T1 + Cp'T1Xa
+ ~HR 1 XA
-AH~tlXA + T1 (Cp"XA + Cp' - Cp' XA)
Cp"XA + Cp' - Cp'XA (3-13)
where subscripts 1,2 refer to temperatures of entering
and leaving streams and,
Cp' , Cp" = mean specific heat of unreacted feed stream
and of completely converted product stream
per kgmole of entering reactant A.
H' , H" = enthalpy of unreacted feed stream and of
completely converted product stream per
kgmole of entering reactant A.
~HR = heat of reaction per kgmole of entering
reactant A.
Equations (3-12) and (3-13) are the two basic equations
which will be used for modeling a CSTCR.
Model Simulation
For the CSTCR it was also assumed that the model
proposed by Cull and Brenner (12) was valid and the
assumptions made previously were still valid.
Numerical Approach
32
A program for simulating CSTCR performance called
"HEXCR" was written in the FORTRAN language. Newton's
method was applied to solve two linear equations containing
two unknowns. The detailed procedures will be discussed in
the next chapter.
CHAPTER IV
PROGRAM DESCRIPTIONS AND TESTING
The main purpose of this research is to simulate
chemical reactions in a fixed bed reactor and a CSTCR.
HEXFI and HEXCR are the names of programs developed for
simulating the fixed-bed reactor and the CSTCR, respective
ly. The HEXFI program offers users two kinds of simulation,
one is optimized model, the other is fixed-length model.
The optimized model calculates the equilibrium mole
fractions of hexane isomers at specified condition and
outputs the required reactor length. However, in the fixed
length model, it calculates mole fractions at specified
reactor length. Both programs provide users the option to
simulate isothermal or adiabatic operation. Both models
allow users to see the effects of changing the reactor
operating conditions, such as temperature, pressure,
catalyst weight, etc.. All models were using the FORTRAN
language.
Software Applied
In order to simulate real control panels in the
chemical industry, IBM software (EZVU) was used. This
software can create panels which connect design variables
33
and users together. Users can easily manipulate different
operating conditions from the panels and outputs will be
shown on the panels simultaneously. This software is very
user friendly to the persons who are undertaking
simulations.
Program Organization and Subroutine Descriptions
34
Figures 4,5 show the FORTRAN flow diagrams for the
fixed-bed and the CSTCR, respectively. A short description
of the program subroutines follows.
Fixed-Bed Reactor
DATAN
This subroutine is used in HEXFI and HEXCR. The
function of DATAN is to defined stoichiometric coef
ficients, feed conditions, and it is called at the beginning
of each simulation run.
ARRAY
Subroutine ARRAY also appears in both HEXFI and HEXCR.
It calculates the heat capacities of each isomer, the heats
of reactions, and the Gibbs free energies of reactions at T
K. Heat capacities and Gibbs free energies can be expressed
as polynomials in terms of temperature. Table III and IV
which appear on the next pages are the data sources. A
Cubic Spline polynomial approximation was applied to
' r-- --, RUNGE-KUTTA I
COEFF.S L-- __ _.
I
Figure 4. Program Organization of HEXFI
35
....
CAL. MOL FC LINPAC
r-- --, I NEWTON'S 1 I. -~HOD-- .I
I
Figure Sa. Program Organization of HEXCR
36
N
MAIN PROGRAM
ASSUME
OUTPUT TEMP .
CALCULATE
E.B. EQUATION
NEW OUTPUT TEMP
CONTINUE
Figure Sb. Program Organization of Energy Balance
37
38
N
Figure Sc. Program Organization of Newton's Method
*TABLE I I I
HEAT CAPACITY FOR THE IDEAL GAS STATE
Compound 298.15
Name
n-Hexane 142.59
•2-M.P. 142.21
b3-M.P. 140.12
c 2, 2-DMB 141.46
d 2, 3-DMB I 139.41
•2-methylpantane b3-methylpantane c2,2-dimethylbutane d2,3-dimethylbutane
*TRC TABLE (1985)
Temperature in K
400 500 600 700 800
Heat Capacity Cp 0 (T) in J K-1 mol-l
181.54 217.28 248.11 274.05 296.23
183.51 219.83 251.04 277.40 300.41
181.17 217.48 248.95 275.73 298.74
183.13 220.33 253.13 281.58 306.69
181.71 218.36 250.20 277.40 301.67
1000
331.37
337.23
335.98
348.11
340.58
w \0
"'TABLE IV
GIBBS ENERGY OF FORMATION FOR IDEAL GAS STATE
Compound
I 298.15 Name
n-Hexane I 0.15
a2-M.P. -5.14
b3-M.P. -3.17
c 2 I 2-DMB -8.52
d 2, 3-DMB I -2.90
a2-methylpentane b3-methylpentane c2,2-dimethylbutane d2,3-dimethylbutane
*TRC TABLE {1985)
Temperature in K
400 500 600 700 800
Gibbs Energy of Formation AG(T), in KJ mol-l
58.87 118.96 180.51 243.15 306.28
54.42 115.20 177.52 241.25 304.56
56.15 116.88 179.14 242.25 305.71
53.30 116.41 180.99 246.28 312.59
58.21 120.61 184.56 249.47 314.70
1000
433.73
433.09
434.67
445.06
446.49
""' 0
41
fit the data. Heat capacities and the Gibbs free energies
at any specified temperature can then be easily
approximated. Function CP and function GF are included in
this subroutine predict these values. To reduced program
execution time, ninety points between 298 and 1000 K were to
evaluate heat capacities and Gibbs free energies and the
data was stored in vector form. Therefore, in order to
predict the heat capacities and the Gibbs free energies at
any temperatures, subroutine PROP was used to look up a pre
calculated value.
Subroutine FNC is the heart of HEXFI and it includes
several additional subroutines. Basically, it contains
two differential equations (O.D.E.s). One is the mass
balance equation, the other is the energy balance equation.
In order to evaluate these two equations, it also needs to
call additional subroutines. These subroutines are PROP,
EQCON, SLTRES, RXN. These subroutines have different
functions to calculate the terms appearing in the
differential equations. After finishing calculations, this
subroutine will transfer the values of the two differential
equations to the main program and use the Runge-Kutta method
to evaluate the function values.
Subroutine PROP calculates the heats of reactions, heat
42
capacities and Gibbs free energies at any temperatures
between 298 and 1000 K. As mentioned previously in
subroutine ARRAY, ninety points were calculated by Cubic
Spline approximation in terms of temperature. Other points
between any two known points were evaluated by linear
interpolation.
EQCON
Subroutine EQCON handles the calculation of equilibrium
constants and forward and reverse rate constants. From
subroutine PROP, we can get Gibbs free energies of each
isomers at specified temperatures. Then the following
equations applied:
~Ga o = -R9 T 1 n K
-6GR 0
In K = RvT
-L'lGa o
K = exp( ) RvT
where
R9 = gas constant, 8.314 kj/kgmole-K
T = temperature, K
~GR 0 =standard Gibbs free energy, kj/kgmole
Equilibrium constants for all the reactions in the
mechanism can be evaluated.
By definition, the equilibrium constant is the ratio of
43
forward rate constant to reverse rate constant. That is:
K =
or,
kt = K * kr
where
kr = reverse rate constant, m'/(kg of cat -hr)
kt = forward rate constant, m'/(kg of cat -hr)
The reverse rate constant it is always expressed in
Arrhenius form:
E k = ko exp( --- )
R9 T
where
E = activation energy, kj/kgmole-K
ko = frequency factor
Rg = gas constant, 8.314 kj/kgmole-K
T = temperature, K
If the reverse rate constant has been determined from
experimental data, the forward rate constant is fixed.
SLTRES
Subroutine SLTRES calculates the thermodynamic equi-
librium mole fractions of the following reactions:
44
( 1 ) ( 3)
n-C6 <----> 3-MP <----> 2-MP <----> 2,3-DMB
(2)'
Assume the reaction coordinates for the first, second and
third reactions are EI, EII, EIII, r•spectively. For
example, assume the initial feed compositions is 50% n-
hexane and 50% 3-MP. Nauman (17) proposed an equation to
solve reaction coordinates; i.e.,
N - No = 11 e (4-1)
where N and No are vectors ( N*1 matrices ) giving the final
and initial number of moles of each component, 11 is the
matrix of stoichiometric coefficients, and e is the reaction
coordinate vector ( M*1 matrix ). In more explicit form,
=
llA , I
llB t I
\lA ti I
llB ti I
converting equation (4-1) into our case, it has the
following form:
n-c, 0.5 -1 0 0
3-MP 0.5 1 -1 0 - + + 2-MP 0.0 0 1 -1
2,3-DMB 0.0 0 0 1
EI
EI I
EI I I
or
Nn- c 6 = 0. 5 - EI
N3-HP = 0.5 + EI - En·
N2-KP = 0.0 + EJJ - EJJI
N2,3-DHB = 0.0 + EIII
45
Using equilibrium constants to solve reaction coordinates
for reaction I, it becomes:
( 0. 5 + EI - eu ) * Por (4-2)
( 0 • 5 - e I ) * Por
for reaction II, it becomes:
( EI I - EI I I ) * Por (4-3)
( 0. 5 + EI + EI I ) * Por
for reaction III, it becomes:
( en I + 0. 0 ) * Por K3 = (4-4)
( EI I - EI I I ) * Por
From subroutine EQCON, we have evaluated equilibrium con
stants K1, K2, K3. Hence, equations (4-2), (4-3) and (4-4)
will turn out to be three simultaneous equations with three
unknowns EI I EII and EIII. Subroutine LINPAC then solves
handle these equations for specified condition.
Subroutine RXN.calculates the globe rate of the final
reaction.
kt 2,3-DMB <====-=> 2,2-DMB
kr
Rate2,2-nMB = kt * C2,3-»MB - kr * C2,2-nMB
46
When the rate is determined it is substituted into equations
(3-5) and (3-8) for calculating the value of the two
differential equations.
LINPAC
This subroutine was written by individuals at Argonne
National Laboratory. It uses partial pivoting and matrix
decomposition with Gaussian elimination to very efficiently
solve large sets of linear equations (33).
CSTCR
NEW TN
Subroutine Newtn is one of the biggest difference
when comparing CSTCR with the fixed-bed design. This
subroutine employs Newton's method in order to solve a set
of two linear equations containing two unknowns. It
includes subroutines DER, FUNC, ADER and FADI. First of
all, let me explain its algorithm and all the functions of
the subroutines.
The algorithm for this case is a two-dimensional
problem and may be represented as simultaneous solution of
the following equations:
47
'b.h (X) 'bf1 (X) dl ( j) + d2 ( j) + f1 ( x< j > ) =
"bX1 x< n bX2 x< j >
"bf2 (X) bf2 (X) dl ( j) + d2 ( j) + f2 ( x< n ) =
OXl x< n 'bX2 x< j >
where
Xl ( j + l ) = Xl ( j ) + d1 ( j )
and
X2 ( j + 1 ) : X2 ( j ) + d2 ( j )
·Note that the superscript (j) or (j+1) indicates the number
of linear approximations that have been used in searching
for the roots. The coefficient matrix for this system of
two linear equations contains all the possible combinations
of partial derivatives of functions, fk(X), with respect to
each independent variable, Xi. This coefficient matrix is
called the Jacobian matrix. For the isothermal case, the
two linear equations are equations (3-10) and (3-11).
Hence, the partial derivatives of these two equations are:
"bil w = -1 - kt
"bC& Q
'bfl w = kr
'bCu Q
"bj 2 w = kt
"bC& Q
48
= -1 - kr 'bCa Q
the above equations are inputed in subroutine DER.
ENGBALS
Subroutine ENGBALS performs the energy balance in a
CSTCR. The energy balance equations are constructed for the
following reactions:
As I mentioned previously, it was assumed to reach
thermodynamic equilibrium very fast. So the energy balance
equations are:
INPUT = OUTPUT + ACCUMULATION + DISAPPEARANCE
The reference temperature was picked to be the same as
the feed temperature.
input term:
Fto * t [YI(i) * Cp(i) * (To-To)]= 0
energy released by reaction:
Ft o * t [ EXC ( i) * fl H ( i) ]
output term:
assuming an output temperature of T1 , the equation
becomes:
Fto * t {YF(i) * Cp(i) * (Tl-To)}
Replacing these quantities in the energy balance gives,
~ {YF(i) * Cp(i) * (Tl-To)}
= - E [EXC(i) * AH(i)]
49
(4-5)
In the above equation, To, EXC(i) and also dH(i), Cp(i) are
function of temperature. In order to satisfy both sides in
equation (4-5), a trial and error method is applied to find
the temperature T1 . Another energy balance for the final
reaction is then required.
2,3-DMB < > 2,2-DMB (I) (II)
OUT = IN + GENERATION CONSUMPTION
(I) H2C1Q = H1 C1 o Q + H2RrW
(I I) H t 2 CI I Q = H t 1 CI I 0 Q + H '2 Rt W
converting (a) and (b) into the following set of linear
equations:
w .h = -H2 CI + Hl c, 0 + (c)
Q
w (d)
Q
Substituting the rate expressions into (c) and (d) and
taking derivatives of each equations with respect to each
variable is done in subroutine FAD!.
w = -H2 - (4-6)
oc, Q
50
"bf 1 w = ( H2 kr ) (4-7)
bCtt Q
'hf2 w = ( H '2 kt ) (4-8)
'hCt Q
'hf2 w = -H '2 - ( H '2 kr ) (4-9)
'hCt I Q
Equations {4-6) to (4-9) are the coefficients of Jocobian
matrix for adiabatic case. After setting up the Jocobian
matrix, we can apply LINPAC to solve for the roots (i.e.,
concentrations at some temperature).
Subroutine THEQ is similar to subroutine FNC in HEXFI.
It also contains several subroutines, such as PROP, EQCON,
SLTRES and ENGBALS. The main difference between THEQ and
FNC is the subroutine ENGBALS. However, the basic function
of this subroutine are similar to what was calculated in
FNC.
Testing and Results
Overview
The programs developed in this study were tested using
the model proposed by Cull and Brenner (12). However, their
experimental data was based on results from data in a batch
reactor and were not suitable verification use. Other
51
literature, listed in references did not supply parameters
and constants related to their experiments. Because of the
above reasons, it was impossible to proceed using this data.
Finally, a copy of operating data from a proprietary source
was obtained and this data was used to validate the model.
The results of the numerical solutions are close to the
proprietary data.
Testing
In industry, there are several isomerization processes
licensed by the Pure Oil Company, British Petroleum and
Phillips Petroleum Company. Generally speaking, their
operating conditions are:
1. isothermal reaction
2. LHSV = 1 - 20
3. reactor charge = 3000 barrels per stream day,
(bpsd)
4. bulk density of catalyst = 40 lb/ft3
5. hydrogen-to-hydrocarbon ratio = 2 : 1
Basically, HEXFI was designed for industrial
simulation. However, adiabatic operation was added in this
study to compare the benefits and shortcomings of each.
Another option supplied allows users to select one of two
objective functions, optimized or fixed-length models. A
few examples are shown below to demonstrate design. Units
used in HEXFI were all converted into the metric system.
52
Control panels of some cases are all listing in Appendix C.
Optimized Model
Figures 18 and 19 in Appendix C show the control panels
for fixed-bed reactor design. Original inputs were set to
be zero. The following operating conditions were then used
as inputs:
Case I
1. isothermal reaction
2. flow rate = 153.5 kgmole/hr
3. feed temperature = 408 K
4. system pressure = 35 ATM
5. tube diameter = 0.05 m
6. number of tubes = 150
7 . bulk density of the catalyst = 640 kg/m3
8. opz = I y I
9. pure n-hexane as feedstock
The final results from the computer monitor are shown
in Figures 20 and 21 in Appendix C. Figure 6 shows that
hexane isomers mole fract~on distributions with respect to
reactor length. From this figure, it is evident that if the
reaction reaches equilibrium conditions, it needs 12.1 m of
reactor length. Comparison of the simulation outputs with
the proprietary source, shows that both sets of data have
similar trends. The equilibrium prediction from HEXFI model
for final mole percent of nee-hexane is 37.69% which is
53
very close to the experimental value of 36.25 \.
Case II
operating conditions:
1. system pressure = 61 ATM
2. other factors are the same as Case I
For this case the pressure was increased to 61 atm and
results are shown in Figure 7. The equilibrium value from
the proprietary data for neo-hexane is 36.25 \ which is
close to the model prediction of 37.69 %. Comparing flow
trends of both sets of data shows that the simulation curves
closely resemble the experimental curves. From Case I and
II, we find that pressure effect does not influence the mole
fraction of neo-hexane. Slightly lower temperature cases
were also tested and shown in Figures 8 and 9. The results
are summaried in Table V. From this table, it can be
summaried that higher temperatures do not favor
isomerization reaction.
Besides the isothermal reaction, HEXFI can also
simulate adiabatic reactors. Case III is a typical case of
adiabatic operations.
Case III
operating conditions:
1. adiabatic reaction
2. other factors are the same as Case I
Results are shown in Figure 10. Comparing Case III
TABLE V
RESULTS OF THE COMPARISON OF EXPERIMENTAL DATA WITH MODEL PREDICTIONS FOR AN ISOTHERMAL FIXED-BED REACTOR
CASE NO. T (K) P (ATM) MODEL PREDICTION! EXPERIMENTAL DATA
1. 408.0 35.0 37.69% 36.25 %
2. 408.0 61.0 37.69% 36.05 %
3. 366.5 35.0 46.84 % 42.75 %
4. 366.5 61.0 46.84 % 41.25 %
---------- - -------------- ----
1. product mole percentage of nee-Hexane
01 ~
with Case I, we notice that the adiabatic reactor has a
lower 2,2-DMB formation than that of the isothermal
reactor. This is because hexane isomerization equilibrium
is favored at lower temperatures.
55
Other adiabatic conditions were also tested and are
plotted in Figures 11 - 13. From these figures, it can be
concluded that although adiabatic reactors give lower mole
fractions of 2,2-DMB, they also require a shorter reactor
length than for isothermal conditions. In general the
initial cost for building an adiabatic reactor is less than
that required to build an isothermal reactor. Furthermore,
Figure 14 through 17 show the difference in temperature
profiles inside a reactor during adiabatic operation. At
the beginning of the reactors, the slopes of the curves are
quite steep. This means that the reactions release a large
amount of heat. After some reactor length, reactions
gradually approach equilibrium and then the temperatures
remain constant.
Fixed-Length Model
In addition to the optimized-model, HEXFI can also
simulate a Fixed-Length reactor. The reason for this model
is sometimes room is available to build a long reactor or
information is required to know the conversion of a reactor
of specified length.
Case IV
1. specified reactor length = 3.0 m
c 0. ~~ 0
\. ·--+-' u ~ 0 -~--1 LL •
Q) -·~ 0. ~~ I
0. 0
Isothermal Reaction 35 ATM, 408 K
/ ------
------
2
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
4 6 8 10 Reactor Length, m
Figure 6. Mole Fraction Distributions of Hexane Isomerization
12 14
01 0\
c 0. ~~ 0
\. ·--+-' 0
~ 0 -~-1 LL •
(J) -
~ 0. ~-l I
0. 0
Isothermal Reaction 61 ATM, 408 K
/ ------
-------
1 2 3 Reactor
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
4 5 6 Length, m
Figure 7. Mole Fraction Distributions of Hexane Isomerization
7 8
01 ..,J
c 0. 0
-+-' ()
0 ~ 0.
(])
0 2 0.
0. 0 2
Isothermal Reaction 35 ATM, 366.5 K
4 6 8 10 Reactor Length, m
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
12
Figure 8. Mole Fraction Distributions of Hexane Isomerization
14 16
01 00
c 0. 0
+-' ()
0 ~ 0.
())
0 2 0.
0. 0 1
Isothermal Reaction 61 ATM, 366.5 K
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
2 3 4 5 6 Reactor Length, m
7
Figure 9. Mole F roction Distributions of Hexane Isomerization
8 9
U'l
"'
§ 0. ·--+-' 0 0 ~
l.L
(])
0 2 0.
0. 0
Adiabatic Reaction Pressure - 35 ATM Initial Temperature - 366.5 K
2
2.2-DMB
2-MP
3-MP
2,3-0MB
n-Hexane
4 6 8 10 Reactor Length, m
Figure 1 0. Mole Fraction Distributions of Hexane Isomerization
12 14
0\ 0
c 0. 0
-+-' u 0 ~ 0.
<l)
0 2 0.
0. 0
Adiabatic Reaction Pressure - 61 ATM Initial Temperature - 366.5 K
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
1 2 3 4 5 Reactor Length, m
Figure 11. Mole Fraction Distributions of Hexane Isomerization
6 7
0'1 ......
0.
c 0. 0
-+-' (.)
0 ~ 0.
Q)
0 2 0. 2
0. 0 1
Adiabatic Reaction Pressure - 35 ATM Initial Temperature - 408 K
2 3 4 5
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
6 Reactor Length, m
Figure 12. Mole Fraction Distributions of Hexane Isomerization
7 8
0\
"'
c 0 _j ~ -
-+-' u 0 ~ 0. 91
(]) -~ 0. ~~ I
0. 0
Adiabatic Reaction Pressure - 61 ATM Initial Temperature -408 K
~
-------
1 2 3 Reactor Length, m
2,2-DMB
2-MP
3-MP
2,3-DMB
n-Hexane
Figure 13. Mole Fraction Distributions of Hexane Isomerization
4 5
0\ w
40
39
39 ~
Q)- 38 L :::::;
-+-'
'2 38 Q) Q_
E 37 Q)
1---37
Pressure - 35 ATM Initial Temperature - 366.5 K
36t~~~~~~~~~~~~~~~~~~~~~~~~~~
0 2 4 6 8 10 Reactor Length, m
Figure 14. Temperature Profile vs Reactor Length of Hexane Isomerization
12 14
0\ ~
40,~------------------------------------------------~
39
39 ~
v"' 38 L
::J -t-J 0 38 L Q) Q_
E 375 Q)
1-37
Pressure - 61 ATM Initial Temperature - 366.5 K
36r~~~~~~~~~~~~~~~~~~~~~~~~~~
0 1 2 3 4 5 Reactor Length, m
Figure 15. Temperature Profile vs Reactor Length of Hexane Isomerization
6 7
0'\ 0'1
43
Y: 42 -
Q) L
::::1
0 415 L Q)
0..
~ 41 ~
40 0 1
Pressure - 35 ATM Initial Temperature - 408 K
2 3 4 5 6 Reactor Length, m
Figure 16. Temperature Profile vs Reactor Length of Hexane Isomerization
7 8
0\ 0\
43
42
~ 42 ...
Q) L ::J
0 41 L Q) a..
~ 41 I-
40 0
Pressure - 61 ATM Initial Temperature - 408 K
1 2 3 4 Reactor Length, m
Figure 17. Temperature Profile vs Reactor Length of Hexane Isomerization .
5
0\ ....:I
68
2. opz = 'N'
3. other factors are the same as Case I
Final results are shown in Figures 22 and 23 in
Appendix c. The mole percentage of 2,2-DMB is 35.64 % at
3.0 m compared with a mole percentage of 37.69% at 12.1 m.
Comparing the outputs for these cases, the following
conclusions can be reached:
HEXCR
1. At the same initial temperature and system
pressure, isothermal reaction can reach higher nee
Hexane mole fractions than the adiabatic case.
2. Higher temperature does not favor the yield of
neo-hexane. For isothermal conditions a reactions,
a reactor temperature decrease of 1 K, increases
the mole percentage of 2,2-DMB by 0.2 %. For
adiabatic operation, an initial temperature
decrease of 1 K, increase the mole percentage of
2,2-DMB by 0.16 %.
3. Pressure has a very little influence on the
conversion of nee-hexane but does affect the
reactor length. Comparing Case I with Case II,
the reactor length in Case II decreases almost 42 %
over that of Case I.
At present, CSTCRs still are used mainly in the
laboratory. This is because it is more complex and
69
expensive to build than a batch or fixed-bed reactor.
Further, the mole fraction of neo-hexane is generally lower
than that of the fixed-bed reactor. The program for
simulating a CSTCR was called HEXCR and supplies two choices
to the user, isothermal or adiabatic operation. At this
time, no published papers and no data is available that
shows that hexane isomerization in a CSTCR. Two basic
simulation runs are illustrated below.
Figures 24 and 25 in Appendix c are the control panels
for the CSTCR.
Initial values of the parameters are set to zero.
Case I.
operating conditions:
1. isothermal reaction
2. reactor charge = 153 kgmole/hr
3. feed temperature = 408 K
4. volumetric flow rate = 125 m3/hr
5. weight of catalyst = 2200 kg
6. pure n-hexane as feedstock
For these parameters~ results are shown in Figures 26
and 27 in Appendix C. The mole fraction of neo-hexane is
significantly lower than that of a fixed-bed reactor using
same amount of catalyst.
Case II.
operating conditions:
70
1. adiabatic reaction
2. other conditions are the same as Case I
Results for the two cases are shown in Table VI. From
this table, we find that isothermal reaction gives higher
mole fractions of nee-hexane than that of adiabatic
operation. This is because higher temperature does not
favor the formation of nee-hexane.
made:
From the above cases, the following conclusions can be
1. For isothermal operation, a temperature decrease of
1 K, increases the mole fraction of nee-hexane
by 0.04 %. For adiabatic operation, a temperature
decrease of 1 K, increases the mole fraction of
nee-hexane by 0.03 %.
2. A CSTCR requires more catalyst than fixed-bed
reactor.
3. If we have equal catalyst weight a large number of
CSTCRs connected in series will behave as a plug
flow reactor.
4. If the volumetric flow rate increases, the final
mole fraction of nee-hexane decreases. This is
a function of reactor residence time.
TABLE VI
RESULTS OF THE COMPARISON OF ISOTHERMAL AND ADIABATIC OPERATION OF A CSTCR
CASE Ti (K) Tt (K) MODEL PREDICTION!
Isothermal 366.5 366.5 13.14 \
Isothermal 408.0 408.0 11.37 \
Adiabatic 366.5 435.1 11.39 \
Adiabatic 408.0 469.1 10.07 \
1. product mole percentage of nee-hexane
-...1 .....
CHAPTER V
CONCLUSIONS AND RECOMMENDATIONS
Conclusions
The purpose of this study is to simulate an ideal plug
flow reactor and a CSTCR for hexane isomerization. An
isomerization catalyst, Pt-Al203 and proprietary data were
used to validate the model.· In actual operation, hydrogen
and chloride are added to the reactors in order to prevent
hydrocracking and keep the activity of the catalyst.
In summary, the models developed in this study can
perform the following:
1. The optimized-length model of a fixed-bed reactor
predicts that the distribution of hexane isomers
optimal and the optimal length under specified
operating conditions.
2. The fixed-length model of a fixed-bed reactor
predicts the distributions of hexane isomers at
specified reactor length.
3. The CSTCR model evaluates the distribution of hexane
isomers at different input conditions.
In this study several conclusions can be made from
model output data.
72
73
1. High temperature does not favor the isomerization
process and isothermal operation gives higher yields
of neo-hexane than that for adiabatic reaction for
both optimized and fixed-length models.
2. Pressure does not affect the yield of neo-Hexane,
but does influence the optimal reactor length.
3. Pressure drop in most fixed-bed reactors will be
small compared to total system pressure and
therefore can be neglected.
Recommendations
1. The ideal gas law was used in both reactor models to
evaluate concentrations, however, since this system
operates at high pressure, the gases do not behave
as ideal gases. Therefore, it is suggested that a
more accurate equation of state such as Redlich
Kwong or Peng-Robinson be used instead of the ideal
gas law. Generally speaking, equilibrium constants
are defined in terms of fugacities or activities and
are not dependent upon the pressure. The behavior
is explained in Appendix D.
2. This study did not consider any side reactions. It
is recommended that these reactions be included in
future work to more realistically model the hexane
isomerization process.
3. It is recommended that additional experimental data
be obtained to test the model for other catalysts.
This model can not be generalized until additional
kinetic data is generated for a specific catalyst.
4. The equilibrium assumptions of the first three
reactions are probably reasonable, but should be
verified with experimental data.
74
LITERATURE CITED
1. Belden, D. H., Haensel V., Starnes, W. G. and Zabor, R. C. J. Oil & Gas, 55{20), 142 {1957).
2. Bland, W. F. and Davidson, R. L., "Petroleum Processing Handbook", McGraw-Hi 11 Book Company, Inc. , New York {1967).
3. Hydrocarbon Processing, "Isomerization", 62{9), 120, (1984).
4. Gary, J. H. and Handwerk, G. E., "Petroleum Refining", Marcel Dekker, Inc., New York (1975).
5. Little, D. M., "Catalytic Reforming", PennWell Publishing Company, Oklahoma {1985).
6. Guthier, Virgil, "Petroleum Products Handbook", McGraw-Hill Book Co., New York {1960).
7. Brooks, B. T., Boord, C. E., Kurtz, S. S.and Schmerling, L., "The Chemistry of Petroleum Hydrocarbons", Reinhold, New York (1955).
a.- Evering, B. L., d'Ouville, E. L., Lien, A. P. and Waugh,R. C., Ind. Eng. Chern., 45{3), 582 {1953).
9. Evering, B. L. and Waugh, R. C., Ibid., 43{8), 1820 (1951).
10. McCaulay, D. A., J. Am. Chern. Soc. 81, 6437 {1959).
11. Scheider, V. and Frolich, PerK., Ind. Eng. Chern., n_(12), 1405 {1931).
12. Cull, N. L. and Brenner, H. H., Ind. Eng. Chern., ~{10), 833 (1961).
13. Smith, J.M., "Chemical Engineering Kinetics", McGrawHill, New York {1956).
14. Shah, Y. T., "Gas-Liquid-Solid Reactor Design", McGrawHill International Book Company, New York (1979).
15. Leva, M., Chemical Engineering, ~{5), 115 {1949).
75
76
16. Nauman, E. B., "Chemical Reactor Design", John Wiley & Sons, Inc., New York (1987).
17. Carberry, J. J., Ind. Eng. Chern., 56(11), 39 (1964).
18. Tarhan, M. Orhan, "Catalytic Reactor Design", McGraw-Hill, New York (1983).
19. Hydrocarbon Processing., "Isomerization", 49(9), 195, (1970).
20. Prasher, B. D., et. al., Ind. Eng. Chemical, Process Div. Develop., 17, 266 (1978).
21. Oil & Gas J., "Penex", 63(14), 135 (1965).
22. Pet. Engr., "Penex Catalytic Isomerization", 32(2), 19 (1960).
23. Grote, H. W., Pet. Ref., 35(7), 148 (1956).
24. Oil & Gas J., "Isomerate Process", 63(14), 136 (1965).
25. Oil & Gas J., "New Process Expands High Octane PoolPure Oil Company's Isomerate Process", 54(52), 86 (1956).
26. O'May, T. C. and Knights, D. L., Hydrocarbon Processing & Petroleum Refiner, 41(11), 229 (1962).
27. Butt, J. B., Catalyst Deactivation, Adv. Chern. Ser., 148 (1975).
28. Beecher, R. and Voorhies, A., I & EC Product Research and Development, ~(4), 366 (1969).
29. Smith, J.M., "Chemical Engineering Kinetics", McGrawHill, New York (1970).
30. Levenspiel, 0., "Chemical Reaction Engineering", 2nd ed., John Wiley & Sons, Inc., New York (1972).
31. Ergun, S., Chemical Engineering Process, 48(2), 89 (1952).
32. Coulson, J. M. and Richardson, J. F., "Chemical Engineering", Vol 3, Page Bros Ltd., Great Britain (1979).
33. Riggs, J. B., "An Introduction to Numerical Methods for Chemical Engineers", Texas Tech University Press, Texas (1988).
APPENDIXES
77
APPENDIX A
COMPUTER PROGRAM FOR HEXFI
78
$debug
c~--------------ABSDAC! _________ _
c c c c c c c c c c
!HIS PROGRAM CAl BE USED !0 DESIGJf A FIXED BED REActOR II NHICH fBIRE AR! MOL!IPLE R!AC!IOifS OCCORIIG UIDBR ISO!JIERMAL COIDifiOif OR ADIABAtiC COifDITION. !HIS MODEL ASSIJM!S PLUG PLOW AID I!GL!C'lS AXIAL DISPIRSIOI AID RADIAL !EMPERATORE GRADIA!f!S WI'lHIIf !HE BED. THE DESIGN IQOATIOIIS ARE IlfTIGRAT!D OSIIIG A POOR!Il ORDER ROIIGE·KUHA METHOD WI!R A VARYIIG STEP SIZE. THE STEP SIZE IS S!'l' SUCH !RAT 'lHBR! IS A C!R!AII CRAIG! II fBE T!MPERA'l'I1RE OR MOLE PRACTIOIIS OF !HE RBACI'Ain'S.
C, _______ _ IOM!ICLATURE -------
c~------------------------------------c c c c c c c c c c c c c c c c c c c c
D - !HE DIAMETER OP !BE REACTOR TOB!, M P - 'I'll! SYS!!M PRESSOR!, A!M Q - !HI VOLUME!RIC PLOW RAT!, M• /HR T - !!MP!RATORE, K If - CATALYS! WEIGHT I KG Zl - 'I'll! LDGTH OP !HE REActOR !UBE, M ZSP - REAC!OR LDG'l'll SPECIFIED BY USER, M ID - FLAG NHICH DETERMIIIES REACTIOII TYPE NC - THE NUMBER OF CHEMICAL COMPOmrJ'S IR - !HE lUMBER or CHEMICAL RBAC!IOifS DZ - THE AXIAL STEP SIZE, M DZT - !HE AXIAL STEP SIZE BASED UPOII A 10 K T!MPERATORE PTO - THE !O'l'AL MOLAR PLOW RATE A'!' 'l'HE IlfLET PLOW RATE,
PTRTifT -F(I)
KGMOLES/HR PLOW RATE Ill A SIIIGL! TOB!, KGMOLE/HR lfO OP !UB!S Ilf REACTOR
- FOR I=S, THE MOLAR FLOW RAT! OF THE I-TH COMPOif!IT, KGMOLES/HR) ; FOR I= 6, fBE REACTOR TEMPERATURE, K
C R(I) c
- fBE RATE OF THE I-TH REACTIOif, KGMOLES/(KG OF CA'l'-HR)
c c
B(I) - THE COifSTANT TERM Ilf THE ITH EQUATION FMPl - PEED MOL! PERCENTAGE OF n-HEXAIIE
C FHP2 - PEED MOLE PERCDTAGE OF 3-METHYLPANTAIE C FMP3 - FEED MOLE PERCEif'l'AGE OF 2-METHYLPANTANE C PMP4 - PEED MOLE P!RCDTAGE OF 2,3 DIMETHYLBUTANE C PMPS - PEED MOLE PERCENTAGE OF 2,2 DIMETHYLBUTANE C FMP6 - PEED MOLE PERC!lfTAGE OF IlfERT GAS C PMPl - PRODUCT MOLE PERCENTAGE OF n-HEXAifE C PMP2 - PRODUCT MOLE PERC!IfTAGE OF 3-ME'l'HYLPANTANE C PMP3 - PRODUCT MOLE PERCEif'l'AGE OF 2·M!'l'HYLPAN'l'Aif! C PMP4 - PRODUCT MOLE PERCENTAGE OF 2,3-DIMETHYLBUTANE C PMP5 - PRODUCT MOLE PERCENTAGE OF 2,2-DIMETHYLBUTAME C PMP6 - PRODUC! MOL! or INERT GAS C THCAP - THE TOTAL HEAT CAPACITY OF THE REACTION MIXTURE C ,KJ/K C YO(I) - THE MOLE FRACTION OF THE I-TR COMPONENT IN THE C PEED
79
C CP( J) - !HI HEAT CAPACITY OP THE J-TH COMPOif!lfT AT C TEMPERA!URI T, KJ/(KGMOLE-K) C PP(I) - THE DERIVATIVE OP P(I) HI!H RESPECT TO Zl C CA( I) - THE BEAT CAPACITY OF THE I -TH COMPOm'l', C KJ/(KGMOL!-K) C BULD!If - !BE BULK DDSITY OP '!'BE BED, KG/M' C TDHRXIf - 'I'll! I!T HEAT OP R!ACTIOJ, KJ/MS-HR) C EXC(I) - REACTION COORDINATE, I=1,3 C A(I,J) - 'I'll! COEPICIDT OP THE J'l'H VARIABLE II 'l'HE ITH C EQUATION C CAYl(I)]-C CAY2(I) > 'l'HE K'S USED IN THE RUIGE-KUTTA M!THOD C CAY3(I) C CAY4(I) C PSAVE(I) - A VECTOR WHICH SAVE '!'BE VALUES OF P(I) C GAM(I,J) - THE STOICHIOMETRIC COEPFICI!IfT OF THE I-TH C COMPOIEKT IN '!'HI J-THE REACTION C GP!RV(I) - THE GIBBS PRE! BJERG! OF THE I-TH R!ACTIOif, C KJ/KGMOLE C CPV(I,J) - A VECTOR COJTAIIIIG HEAT CAPACITIES OF C COMPOK!IfT I AT DiSCRETE VALUE OF C TEMP!RA'lURE, KJ/(KGMOLE-K) C DHRXIf( I) - THE HEAT OF REACTION OF THE I -TH REACTION C KJ/KGMOLE C DBRXV(I,J) - A VICTOR CONTAINING THE HEAT OF REACTION C OF THE I-TH REACTIOif AT DISCRETE VALUES C OF TEMPERATURE, KJ/KGMOLE c----------------------------------------------------------c INPUT FORMAT DESCRIPTION------, c C INPUT DATA FOR HEXANE ISOMERIZATION AND SET UP CPV C AND BlAT OF REACTION. c ~----------------~ c C THIS PILE CONTAINS ALL THE SCREEN C INPUT VAULES AKD DEPINITIOIS c
$S'l'ORAGE:2 INTEGER RC DOUBLE PRECISION PP(lO),PSAV(lO),TS DOUBLE PRECISION CAY1(10),CAY2(10),CAY3(10),CAY4(10) COMMON /DHRX/ DHRXN(4),T,YO(l0) COMMOif /DATA6/ P,GAM(5,4),D COMMON /DATAS/ JC,PTO,P(lO),IR COMMON /CATAP/ BULDEN DIM!IfSION YP(lO)
c ~-------------------------------~ C !HIS PROGRAM APPLIES SOPTHARE EZVU DEVELOPED BY IBM C DEFIKE IKPUT AND OUTPUT VARIABLES FOR SCREEN C HEX1,HEX2,HEX3 c ~----------------------------~
80
RC=O CALL ISPPPV(4,'T P7',RC,T,4) CALL ISPPPV(4,'P PS',RC,P,4) CALL ISPPPV(4,'D P5',RC,D,4) CALL ISPPPV(S,'OPZ C' ,RC,OPZ,4) CALL ISPPPV(S,'Zl P7' ,RC,Zl,4) CALL ISPPPV(S,'AI Pl',RC,AI,4) CALL ISPPPV(6,'0PZ1 C',RC,OPZ1,4) CALL ISPPPV(6,'ZSP P7',RC,ZSP,4) CALL ISPPPV(6,'FTO P7' ,RC,Pf0,4) CALL ISPPPV(7,'T!MP P7',RC,TIMP,4) CALL ISPPPV(7,'PMP1 P6',RC,PMP1,4) CALL ISPPPV(7,'PMP2 P6',RC,PMP2,4) CALL ISPPPV(7,'PMP3 P6',RC,PMP3,4) CALL ISPPPV(7,'PMP4 P6',RC,PMP4,4) CALL ISPPFV(7,'PMP5 P6',RC,PMP5,4) CALL ISPPPV(7,'PMP6 P6',RC,PMP6,4) CALL ISPPPV(7,'PMP1 P6',RC,PMP1,4) CALL ISPPPV(7,'PMP2 P6',RC,PMP2,4) CALL ISPPFV(7,'PMP3 P6',RC,PMP3,4) CALL ISPPPV(7,'PMP4 P6',RC,PMP4,4) CALL ISPPPV(7,'PMP5 P6',RC,PMP5,4) CALL ISPPFV(7,'PMP6 P6',RC,PMP6,4) CALL ISPFFV(9,'BULDEN F6' ,RC,BULDEN,4)
~ I SliT PUIICTIOII KEYS
ZFlO='QUI'l'' ZCMD=' I
ZATR='HRI' c .----------, C SE'l' INITIAL VALUES OF C PUICTION KEYS c '----------'
CALL ISPFPV(6,'ZATR C',RC,ZATR,4) CALL ISPFPV(6,'ZF01 C' ,RC,ZF01,4) CALL ISPFPV(6,'ZF02 C',RC,ZF02,4) CALL ISPFPV(6,'ZF03 C' ,RC,ZF03,4) CALL ISPFPV(6, 'ZFlO C' ,RC,ZF10,"4) CALL ISPFFV(6,'ZCMD C' ,RC,ZCMD,4)
c r-------------------, C INPU'l' VARIABLES FOR SCREEN HEX1,HEX2,HEX3 C GET DEFAULT VALUES PROM PROFILE c '-----------------~
CALL ISPFP(S,'VGET D P',RC) CALL ISPFF(S,'VGE'l' P P',RC) CALL ISPPF(S,'VGET T P',RC) CALL ISPFF(9,'VGE'l' Zl P' ,RC)
81
CALL ISPFF(lO, 'VGET OPZ P' ,RC) CALL ISPFF(lO, 'VG!! P"1'' P' ,RC) CALL ISPFF(lO,'VG!T ZSP P',RC) CALL ISPFF(ll,'VGET TEMP P' ,RC) CALL ISPFF(ll,'VGET PMPl P',RC) CALL ISPFF(ll,'VG!T PMP2 P',RC) CALL ISPFF(ll,'VGET PMP3 P',RC) CALL ISPFF(ll,'VGET FMP4 P' ,RC) CALL ISPFF(ll,'VGET FMP5 P',RC) CALL ISPFF(ll,'VGET PMP6 P',RC) CALL ISPPF(ll,'VG!T PMPl P',RC) CALL ISPFP(ll,'VGET PMP2 P',RC) CALL ISPFF(ll,'VGET PMP3 P',RC) CALL ISPFF(ll,'VG!f PMP4 P',RC) CALL ISPFP(ll,'VGET PMPS P',RC) CALL ISPPP(ll,'VGET PMP6 P',RC) CALL ISPPP(ll, 'VGET OPZl P' ,RC) CALL ISPPF(13,'VGET BOLDEN P' ,RC)
~ I sr.IR! SCHill IIPU!S I CALL ISPFF(14,'DISPLAY OPTION' ,RC) IP((OPZl.!Q. 'a').OR.(OPZl.!Q. 'A')) GOTO 777 IF((OPZl.!Q. 'b').OR.(OPZl.!Q. 'B')) GOTO 101
101 ZCMD=' CALL ISPFF(lO,'VG!T ZSP P' ,RC) ZFOl='H!Xl' ZF02='HEX2' ZP03='HEX3' CALL ISPFP(13,'DISPLAY KEYS!' ,RC) CALL ISPFP(12,'DISPLAY H!Xl',RC)
IP{ZCMD.!Q.'QUI!') CALL EXIT IP(ZCMD.EQ. 'HEX2') GOTO 200 IF{ZCMD.EQ.'HEX3') GO'l'O 300 IF(OPZ.EQ. 'N') GOTO 777 IF(OPZ.EQ.'n') GOTO 777 GOTO 102
200 ZCMD=' CALL ISPPP{13,'DISPLAY KEYS2',RC) CALL ISPPP{l2,'DISPLAY HEX2',RC)
IP{ZCMO.EQ.'QUIT') CALL EXIT IP{ZCMD.EQ.'HEXl') GOTO 101 IP(ZCMO.EQ. 'HEX3') GOTO 300 GOTO 102
300 ZCHO=' CALL ISPPP{13,'DISPLAY KEYS3' ,RC) CALL ISPPP{l2,'0ISPLAY HEX3',RC)
IF(ZCMO.EQ.'QUIT') CALL EXIT
82
IP{ZCMD.EQ. 'HEX2') GO TO 200 IP{ZCMD.EQ.'HEX1') GO TO 101 GOTO 102
777 COift'llftJE CALL ISPPP{9,'VGET Z1 P',RC)
ZF01='HEP1' ZP02= 'HEX2 I ZP03='HEP3' ZCMD=' I
CALL ISPPF{13,'DISPLAY KEYS1',RC) CALL ISPFF{12,'DISPLAY HEF1',RC)
IP{ZCMD.EQ. 'QUIT') CALL EXIT IF{ZCMD.EQ.'HEX2') GOTO 201 IF{ZCMD.EQ.'HEP3') G0'1'0 301 IP{OPZ.BQ.'Y') GOTO 101 if{OPZ.EQ.'y') GOTO 101 GOTO 102
201 ZCMD=' CALL ISPPF{13,'DISPLAY KEYS2' ,RC) CALL ISPFF{12,'DISPLAY HEX2' ,RC)
IF{ZCMD.EQ.'QUIT') CALL EXIT IF{ZCMD.EQ.'HEF1') GOTO 777 IP{ZCMD.EQ.'HEF3') GOTO 301 GOTO 102
301 ZCMD=' CALL ISPFF{13,'DISPLAY KEYS3' ,RC) CALL ISPFF{12,'DISPLAY HEF3',RC)
IP{ZCMD.!Q.'QUIT') CALL EXIT IF{ZCMD.!Q.'HEF1') GOTO 777 IF{ZCMD.EQ.'HEX2') GOTO 201 GOTO 102
102 ZCMD=' c ++++++++++++++++++++++++++++++++++++++++ C + STARTING MAIN PROGRAM + c ++++++++++++++++++++++++++++++++++++++++ c ~-------------------------, c c c
SET UP INITIAL DATA INFORMATION 1. ADIABATIC {ID=1) 2. ISOTH!RMAL{ID=2)
c ~------------------------~
ID=l IF {AI.GT.l.S) ID=2
T=T!MP
83
~ I PLOI Rl!l II A SIBGLE IVBBI
F'l'R = rro /m Cr-------------., C IKPOT IKITIAL MOLE FRACTION C OP 1-HIXAKE ••. !0(1) C 3-MP ....... Y0(2) C 2-MP •••.••• Y0(3) C 2,3-DMB •••• Y0(4) C 2,2-DMB ..•• YO(S) C IM!RT ••..•. Y0(6) C.__ _________ __,
Y0(1)=PMP1/100. Y0(2)=1!'MP2/100. Y0(3)=1!'MP3/100. Y0(4)=1!'MP4/100. YO(S)=I!'MPS/100. Y0(6)=1!'MP6/100.
SET UP DATA POR REACTOR MODEL c~ I ~---------~
CALL DATAlf c r----------------------------~ C CALCULATE HEAT CAPACITIES AID REACTION HEATS c ~--------------------------~
CALL ARRAY
g I IJI!IALIZE REACTOR DIS!!ICE, II (M!!ER)
Z1=0.0 c r--------------------------~ C INITIALLY ASSUME MOLE FRACTION OF C N!O-HEXAlfE c ~------------------------~
COC1=0.0 c ************************************ C * BEGIN INTEGRATION LOOP * c ************************************
1 CALL FMC (F,FP,YO,ID) TS=O.ODO
c~ I CALCULATE NEW STEP SIZE ~--------~
84
Cr--------------. C ISOTHERMAL CASE C --> USING CONCEKTRATION CHANGES C'---------------J
RT=ABS(FP(5)/FTR) IF (RT.GT.TS) TS=RT DZ=2.5E-4/TS IF (ID.EQ.2) GOTO 35
Cr-----------, C ADIABATIC CASE C -> USING !BMP!RA'rUR! CHANGE C'-------------1
DZT=l0.0/ABS(FP(6)) IF(DZT.LT.DZ) DZ=DZT IF(DZ.Gf.l) DZ=O.OlO
c >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> C > STARTIKG RUKG!-KUT!A > c >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
35 DO 3 1=5,6 3 FSAV(I)=F(I)
CCC I _ SET UP CAYl(I)
DO 4 1=5,6 4 CAYl(I)=PP(l)
DO 5 1=5,6 5 P(I)=FSAV(I)+O.SO*DZ*CAYl(I)
CALL FNC(F,FP,YO,ID)
CCC I _ SET UP CAY2(I)
DO 6 I=5,6 6 CAY2(I)=FP(I)
DO 7 I=5,6 7 F(I)=FSAV(I)+0.50*DZ*CAY2(I)
CALL FNC(F,FP,YO,ID)
CCC I _ S!'l' UP CAY3(I)
DO 8 1=5,6 8 CAY3(I)=PP(l)
DO 9 1=5,6 9 P(I)=PSAV(I)+DZ*CAY3(I)
CALL FNC(F,FP,YO,ID)
CCC I _ SEf UP CAY4(I)
85
DO 10 1=5,6 10 CAY4(I)=FP(I)
c r----------. C CALCULATE THE HEN VALUES C OF F{I) AT Z1+DZ c L...----------1
DO 11 1=5,6 11 F(I)=FSAV(I)+DZ/6.0*(CAY1(I)+2.0*CAY2{I)
&+2.0*CAY3(I)+CAY4(I)) C<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< C END OF RUNGE-KUTTA METHOD!!! C<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
C\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ C IF USER CHOOSE OPTIMIZED MODEL C CHECK WHETHER REACTION REACHES EQUILIBRIUM OR NOT C\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
IF((OPZ.!Q.'y').OR.(OPZ.EQ. 'Y')) THEM COC2=F(S)/PTR !RLIM=(COC2-COC1)*FTR COC1=COC2 IF(ERLIM.LT.O.OOOl) GOTO 100
EHDIF c r--------------~ C CALCULATE MEN MOLE PRACTIOI OF lEO-HEXANE AID C HEN HOLE FRACTION OF 2,3-DIHETHYLBUTAHE c L...--------------~ ~ l1111 MOLE PIW:'IIOR or REO-HWIII j
YP(S)=F{5)/FfR
g lm MOLE PIW:'IIOII or 2, 3-DIIB I
Y0(4)=Y0(4)-(YP(5)-Y0(5))
~ I ASSIGR Yr 10 YO I YO(S)=YF(S)
~ I MOLE PIW:'IIOII or IIIIR'I CAS I YF{6)=Y0(6)
~ I CALCULATE !'liE BI!ACI'OR LI!IIGtll, -I
86
Zl=Zl+DZ c ~---------------------, C DE'l'ERMIKE WHETHER THE REACTOR C LEKGTH OVER THE SPECIFIED C LEKGTH OR NOT c ~--------------------~
I P ( ( OPZ . EQ. 1 n 1 ) • OR • ( OPZ . EQ. 1 lf 1 ) ) THEif ERLilf=ZSP-Zl IF{ERLIN.LT.O.O) GOTO 100
ElfDIP
CCC I . ITERATE RUlfGE-KUTTA
GOTO 1
CCC I *** PINAL OUTPUTS ~--------------~
***
100 CONTINUE c ~--------~--------~ c c c c c c c
MOLE PERCENTAGE OF I-HEXANE ..••• PMP1
3-MP ......•.. PMP2 2-MP ..•.•.••. PMP3 2,3-DMB ...•.. PMP4 2,2-DMB •..•.. PMPS INERT •••..••• PMP6
c ~----------------------~ PHPl=Y0(1)*100. PMP2=Y0{2)*100. PMP3=Y0{3)*100. PMP4=Y0{4)*100. PHP5=Y0{5)*100. PMP6=Y0{6)*100.
~ I PIIAL RIIIC'IOR I'EIIPIRA!URE, I
'f=P{6)
IP({OPZ.EQ. 'y 1 ).0R.{OPZ.EQ.'Y')) GO!O 101 IP{{OPZ.EQ. 1n').OR.(OPZ.EQ.'I')) GOTO 777
STOP ElfD
c ++++++++++++++++++++++++++++++++++++++ C + END ! of MAIN Program + c ++++++++++++++++++++++++++++++++++++++
87
c r---------------------------------~ C THIS SUBROUTIME SUPPLIES THE MAJORITY OF THE DATA C FOR THE FIXED-BED REACTOR MODEL. THIS SUBROUTINE IS C USER SUPPLIED AND PROVIDES DATA, FEED CONDITIOBS, C STOICHIOMETRIC COEFFICIENTS, HEATS OF REACTIONS, c~-------------------------------'
SUBROUTINE DATAN COMMON /DHRX/ DHRXH(4),T,Y0(10) COMMON /DATAS/ NC,FTR,F(10),NR COMMON /DATA6/ P,GAM(5,4),D COMMON /CATAP/ BULDEH
CCC I _ THE lfUMB!R OF REACTIONS
NR=4 Cr-------------------~ c HUMBER OF COMPOif!N'l'S I C (EXCEPT THE INERT GAS BECAUSE IT C DOESN'T REACT WITH OTHER REAC'l'ANTS) c~-----------------------'
NC=S Cr----------------------~ C THE StOICHIOMETRIC COEFICIEI'l'S C ** IBITIALIZE GAM(I,J) ** C'-------------------------'
DO 1 I=1,NC DO 1 J=1,1fR
1 GAM(I,J)=O.O Cr----------------------, c c c
(1)n-H!XAif! <--> (1) 3-methylpantane
l____>GAM(1,1) l_____>GAM(2,1) C'-------------------'
GAM(l,1)=-l.O GAM(2,1)=1.0
c r---------------------, c (1) 3-methylpantane <---> (1) 2-methylpantane
~ l_____>GAM(2,2) I >GAM(3,2) c .__ ____________________________________ ~
GAM(2,2)=-l.O GAM(3,2)=1.0
c r------------------------------------~ C (1) 2-methylpantane <--> (1) 2,3-dimethylbutane
~ l_____>GAM(3,3) l__>GAM(4,3) c~------------------------------------~
88
GAM(3,3)=-l.O GAM(4,3)=1.0
c .----------------------------------, c (1) 2,3-dimethylbutane <------> (1) 2,2-dimethylbutane
~ l______>GAM(4,4) l_____>GAM(5,4) c ~--------------------------------~
GAM(4,4)=-l.O GAM( 5 ,4)=1.0
c ~----------------~ C CALCULATE THE INLET C MOLAR FLOW RATES, (KGMOLE/HR) c ~------------------~
DO 2 I=1,lfC 2 F(I)=YO(I)*FTR
c ~------------...., C SET F(6) STANDS FOR C 'l'HE T!MP Ilf THE SYST!M c ..._ ______________ __.
F(NC+l )=T
RETURN !KD
c ******************************************************* C * THIS SUBROUTINE CALCULATES THE HEAT CAPACITIES OF * C * EACH SPECIES AND THE HEATS OF REACTIOlfS AT T (K). * c *******************************************************
SUBROUTINE ARRAY COMMOlf /DATAS/ lfC,FTR,F(10),lfR COMMOlf /DATA6/ P,GAM(5,4),D COMMON /DHRX/ DHRXlf(4),T,Y0(10) COMMON /VECTR/ DHRXV(4,1SO),CPV(5,1SO),GFERV(4,150) DOUBLE PRECISION SUM
c ~----------------------~ C DECIDE THE IlfTEGRATIOlf INTERVAL, C ( 'l'!MP RANGE FROM 298.15 to 1000 II:) c ..._ ______________________ ~
TI=298.15 DT=(1000.-TI)/90.
c ~------------------------~ C REACTIOK HEAT (KJ/KGMOL!) AT 298.15 K c C n-H!XAK! <------> 3-M!THYLP!KTAJIE (3-MP) c L-------------------------~
DHRXV{1,1)=-5.050*1000.
89
c r-------------------------~ C REACTION HEAT (KJ/KGMOLB) AT 298.15 K C 3-MBTHYLPEMTAME <------> 2-METHYLPEHTANB c ~------------------------------------~
DHRXV(2,1)=-2.580*1000. c r-----------------------------------------~ C REACTION HEAT (KJ/KGMOLE) AT 298.15 K C 2-ME'l'HYLPEM'l'AME <-> 2,3-DIMETHYLBUTAME c ~----------------------------------------------~
DHRXV(3,1)=-2.250*1000. c r-----------------------------------------------------~ C REACTION HEAT (KJ/KGMOLE) AT 298.15 K C 2,3-DIME'l'HYLBUTAJE < > 2,2-DIMETHYLBUTANE c ~----------------------------------------------------~
DHRXV(4,1)=-7.880*1000. c----------------------------------------------------------c C CALCULATE REACTION HEAT AT EVERY DT, (KJ/KGMOLE) C---------- From T=298.15 K TO 1000 K c C For e1ample: 1st reaction c c c c c c c c c c c c c c c c c c c c c c c c c c
n-Heiane <--> 3-Methylpantane
IT2 (Cp'-Cp)dT = heat of reaction
Tt
where Cp'= heat capacity of 3-MP Cp = heat capacity of n-Hexane
p.s. 1st reaction: I=l TI=298.15 K (referance temp) sum=O.O K=2
we want to evaulate reaction heat at 305.98 K (TI+DT) SUIFO.O+DT*GAM(l,1)*Cp(1,305.98)=-DT*Cp(1,305.98) sum=sum+DT*GAM(2,l)*Cp(2,305.98)=-DT*Cp(l,305.98)
+DT*l*Cp(2,305.98) sum=sum+DT*GAM(3,1)*Cp(3,305.98}=-DT*Cp(1,305.98)
+DT*Cp(2,305.98)+DT*O.O*Cp(3,305.98) =-DT*Cp(1,305.98)+DT*Cp(2,305.98)
SUIFSum+DT*GAM(4,l)*Cp(4,305.98) =-DT*Cp(1,305.98)+DT*Cp(2,305.98)
sum=sum+DT*GAM(S,l)*Cp(5,305.98) =-DT*Cp(1,305.98)+DT*Cp(2,305.98)
C Therefore, heat of reaction with respect to referance C temp is delta H= DT*( Cp(2,305.98)-Cp(1,305.98)) c-----------------------------------------------------------
90
DO 4 I=l,KR TI=298.15 SOM=O.ODO DO 5 K=2,91 TI=TI+DT DO 6 J=l,NC SOM=SUM+DT*GAM(J,I)*CP(J,TI)
6 CONTINUE
IF (TI.GT.lOOO.) GOTO 4 c ~-----------------------, C NEW VALUE OF REACTION HEAT c AT (T+dT) with respect to 298.15 K c ~----------------------~
DHRXV(I,K)=DHRXV(I,K-l)+SUM
5 SUM=O.ODO 4 CONTINUE
c --------------------------------------------------------c +CALCULATE THE GIBBS ENERGY OF FORMATION OP EACH REACTION+ C ------------- GF!RV(I,J), (KJ/KGMOLE) -------------C For example: 1st reaction C referance temp=298.15 K C 0'1'=7.8 K C new temp to evaulate : 305.9 K c suml=O.O c c c c c c c c c c c c c c c c c
n-Hexane <--> 3-methylpantane
show steps: suml=GAM(l,l)*GF(l,305.9)+suml =-l*GF(l,305.9)+0.0
suml=GAM(2,l)*GF(2,305.9)+suml= =l*GF(2,305.9)-GF(1,305.9)
suml=GAM(3,1)*GP(3,305.9)+suml=GP(2,305.9)GF(l,305.9)
suml=GAM(4,l)*GP(4,305.9)+suml=GP(2,305.9)GP(l,305.9)
suml=GAM(5,l)*GP(5,305.9)+suml=GP(2,305.9)GF(l,305.9)
Therefore, Gibbs energy of formation of 1st reaction at t+dt GP!RV(l,305.9)=suml
c-----------------------------------------------------------
DO 34 I=l,KR '1'1=298.15 SUMl=O.O DO 35 K=2,91 TI=TI+DT DO 36 J=l,KC
91
SUM1=GAM(J,I)*GF(J,TI)+SUM1 36 CONTIKUE
IF (TI.GT.1000.) GO TO 34 GFERV(I,K)=SUMl
35 SUM1=0.0 34 CONTIKUE
c ~-----------------------, C SET UP CPV(I,J} -->> CONVERT EVERY C VALUE OF Cp(i,ti) INTO VECTOR IN C ORDER TO USE LINEAR INTERPOLA'l'ION c ~----------------------~
DO 10 I=1,NC '1'1=298.15 DO 11 J=1,90 CPV(I,J}=CP(I,TI)
11 TI=TI+D'l' 10 CONTINUE
RETURN ElfD
c ***************************************** C * THIS FUNCTION CALCULATES THE HEAT * C * CAPCAITY OF EACH SPECIES * C * FROM 200-1000 (K} (KJ/KGMOLE-K} * c *****************************************
f*ckCTIOB CP(I,T} IF (I.EQ.1) GOTO 1 IF (I.EQ.2) GO'l'O 2 IF (I.EQ.3) GOTO 3 IF (I.EQ.4) GO!O 4 IF {I.EQ.5) GOTO 5
1 CP=(-1.0540+1.39E-1*T-7.449E-5*T2+1.551!-8*TS)*4.184 RETURN
2 CP=(-0.570+1.359E-1*T-6.854E-5*T2+1.202E-8*T')*4.184 RETURlf
3 CP=(-2.5240+1.477E-1*T-8.533E-5*T2+1.931E-8*'1'')*4.184 RmJRI
4 CP=(-3.489+1.469E-1*'1'-8.063E-5*T2+1.629E-8*TS)*4.184 RETORI
5 CP={-3.973+1.503E-1*T-8.314E-5*T2+1.636E-8*T')*4.184 RETURlf EHD
c ******************************************* C * THIS FUICTIOI CALCULATE THE GIBBS FREE * C * ElfERGY OP PORMATIOif OP HUAMI ISOMERS * C * FROM 200-1000 K, KJ/KGMOL! * c *******************************************
PUICTIOK GP{I,T) IF (I.EQ.l) GOTO 1 IF (I.EQ.2) GO!O 2
92
IF (I.EQ.3) GOTO 3 IP (I.EQ.4) GOTO 4 IF (I.EQ.S) GOTO 5
1 GP=-0.000087*TS+0.219336*T2+453.119366*T-151738.80896 RETURN
3 GF=-0.000092*TS+Q.225563*T2+458.412884*T-158998.25354 RETURN
2 GP=-0.000085*TS+0.216435*T2+460.331617*T-156934.60254 RETURN
5 GF=-0.000092*TS+0.224945*T2+481.313618*T-169166.74548 RETURN
4 GF=-0.000091*TS+0.22493*T2+473.536175*T-161194.52295
RETURN END
c ******************************************************** C * THIS SUBROUTINE CALCULATE THE DERIVATIVE OF P(I) * C * WITH RESPECT TO Z. THE DERIVATIVES ARE CALCULATED * C * PROM MATERIAL BALANCE WHEN P(I) IS THE MOLAR PLOW * C * RATE OF A COMPOlfm AND PROM ElfERGY BALANCE WHEJ * C * P(I) IS THE TEMPERATURE. * c ********************************************************
SUBROUTINE PNC(P,PP,YO,ID) COMMON /CATAP/ BULDEN COMMOI /DATAS/ NC,FTR,Y(lO),NR COMMON /DATA6/ P,GAM(5,4),D DIMENSIOM P(lO),EK(4),FK(4),RK(4),YO(lO) DOUBLE PRECISIOlf PP(lO),CA(S),DHRXN(4),R(4),GPERN(4) DOUBLE PRECISION TDHRXN,THCAP,CONC(S),EXC(3)
c~ I _ REACTOR PRESSURE, (ATM)
PO=P
REACTOR TEMPERATURE, K c~ J ~----------------------------~
T=P(6) c r---------------------------------------------------------~ c_ CALCULATE THE VOLUMETRIC PLOW RATE, (MS/HR) C Using Ideal Gas Law, PV=nRT c ~--------------------------------------------------------~
QO=PT0*22.4*T/273.15/PO c r-----------------------------------------------------~ C CALCULATE 'l'HB HEATS OP RBAC'I'IOifS, HEAT C CAPACITIES AID GIBBS EIERGIBS AT AIY TEMP C B!TWE!If, 298 -1000 K C Using linear interpolation c ~----------------------------------------------------~
93
CALL PROP(T,DHRXN,CA,GFERN) c r-----------------------~ C CALCULATE THE EQUILIBRIUM CONSTANTS C ,FORWARD AND REVERSE RATE CONSTAM'l'S c ~----------------------~
CALL EQCON(EK,FK,RK,F,GFERN) c r-----------------------------~ C CALCULATE THE EQUILIBRIUM MOLE C ,FRACTIONS OF HEXANE ISOMERS. C Using 'LINPAC' To Solve Reaction Coordinates c ~----------------------------~
CALL SLTRES(EK,YO,EXC) c r----------------------, C CALCULATE THE CONCENTRATION OF C 2,3-DMB AKD NEOHEXAKE, KGMOLE/M' c ~--------------------~
DO 80 I=4,5 80 CONC(I)=FTO*YO(I)/QO
~ I CALCULATE Till! GLOBE RAI'I or FORTI! REACTION I CALL RXH(R,CONC,FK,RK) R(4)=R(4)*BULD!N
~ I ID,2, ie. ISDTB!RMAL COIDI!IOI I IF (ID.EQ.2) GOTO 75
c ~--------------------------------~ C CALCULATE THE NET HEAT DUE TO R!ACTIOif, KJ/KGMOLE c ~--------------------------------~ C by energy balance equation: c c dT/dZ = (xOZ/4)* t(bulk(-dH)) I t(Fi*Cpi) c I< >1 1<->1 c part A part B c f< >I C part C c ~--------------------------------~ c C part A c
SRXN=O.O DO 77 1=1,3
77 SRXK=!XC(I)*FTR*DHRXIf(I)+SRXN SRXI=SRXI*BULD!I TDHRXN=SRXK+R(4)*DHRXIf(4)
94
c c c
part B
THCAP=O.ODO DO 3 I=l,lfC
3 THCAP=THCAP+FTR*YO(I)*CA(I) Cr---------------. C PERFORM THE ENERGY BALANCE (K/M) C part C c ~-------------~
FP(6)=(-TDHRXN*3.14159DO*D*D/4.DO)/THCAP c ~------------------------~ C PERFORM THE MATERIAL BALANCE, (KGMOLE/HR/M) c C dFi/dZ= BULK*(,DI/4)* tRi c ~----------------------------~
75 FP(S)=R(4)*3.14159*D*D/4.0DO
RETURJf END
c ********************************************************** C * THIS SUBROUTIIE CALCULATES THE HEAT CAPACITY AID * C * THE HEATS OF REACTIONS FOR A GIVEN TEMPERATURE BY * C * LINERALY INTERPOLATING BETWEEN VALUES OF CPV( I, J) AlfD* C * DHRXV(I,J), RESPECTIVELY. * c **********************************************************
SUBROUTINE PROP(T,DHRXK,CA,GPERI) DOUBLE PRECISIOK DHRXK(4),CA(5),GFERB(4) COMMON /VECTR/ DHRXV(4,1SO),CPV(5,150),GFERV(4,150) COMMON /DATA5/ KC,FTR,F(10),BR
DT=(1000.-298.15)/90. • I=IFIX((T-298.15)/DT)+1
PRO=(T-298.15-DT*FLOAT(I-1))/DT Cr----------------. C CALCULATE THE HEATS OP REAC'l'IOKS C AT SPECIFIED T!MP, KJ/KGMOLE C'---------------A
DO 1 J:l,lfR 1 DHRXI(J)=DHRXV(J,I)+PRO*(DHRXV(J,I+1)-DHRXV(J,I))
c r---------------------------~ C CALCULATE THE HEAT CAPACITIES OP C ISOMERS AT SPECIFIED T!MP, KJ/KGMOLE-K c .__ _______________________ ~
DO 2 J:l,KC 2 CA(J)=CPV(J,I)+PRO*(CPV(J,I+l)-CPV(J,I))
95
c r----------------------, C CALCULATE THE GIBBS EHERGI!S C AT SPECIFIED TEMP, (KJ/KGMOLE) c ~--------------------~
DO 3 J=l,IfR 3 GFERK(J)=GFERV(J,I)+PRO*(GFERV(J,I+l)-GFERV(J,I))
RETURN END
c ********************************************************** C * THIS SUBROUTINE CALCULATES THE EQULIIBRIUM * C * CONSTANTS OF EACH REACTIOif AND THE FORWARD AND REVERSE* C * REACTION RATE CONSTANTS. * c **********************************************************
SUBROUTINE EQCON(EK,FK,RK,F,GFERN) DIMENSION EK(4),FK(4),RK(4),F(l0) DOUBLE PRECISION GFgRN(4)
C-- REACTOR TEMPERATURE
T=F(6) c r---------------------------, C CALCULATE EQUILIBRIUM COISTAlfT OF NORMAL C HEXANE TO IT'S ISOMERS C delta G=-RT*lnK c ~------------------------~
DO 5 1=1,4 5 EK(I)=EXP(-GF!Rif(I)/(8.314*T))
c r---------------------------, C CALCULATE REVERSE REACTIOI RATE CONSTANTS, C (FTS/LB of CAT-HR) C kr=k·elp(-E/RT) c ~------------------------~
TR=9./5.*(T-273.15)+32.+459.67 RK(4)=5.24*1000.* EXP(-9.55*1000./10.7302/TR)
c r-------------------------------------, C CALCULATE FORWARD REACTION C RATE CONSTANTS, (FTS/LB-HR) c K=kf/kr -----> kf=K*kr c ~-------------------------------------'
FK(4)=RK(4)*EK(4) c r-----------------------, C COIIV!RT FORWARD AID REVERSE C RATE CONSTANTS TO, MI/KG-HR) c ~------------------~
FK(4)=FK(4)*0.06243 RK(4)=RK(4)*0.06243
96
RETURN DD
c ********************************************************** C * THIS SUBROUTINE CALCULATES THE RATES OF EACH REACTION * C * GIVEN THE COMPOSITION AND TEMPERATURE. THE DEFINITION * C * AND UNITS OF THE VARIABLES CAN BE FOUND IN THE * C * NOMENCLATURE SECTION OF THE MAIN PROGRAM. * c **********************************************************
SUBROUTINE RXN(R,CONC,FK,RK) COMMON /DATAS/ NC,FTR,Y(10),NR DOUBLE PRECISION R(4),CONC(S) DIMENSION FK(4),RK(4)
c r-----------------------------~ C CALCULATE THE GLOBE REACTION RATE C OF FORTH REACTION, KGMOLE/(KG OF CAT-HR) c ~----------------------------~
R(4)=FK(4)*CONC(4)-RK(4)*CONC(5)
RETURN END
c ********************************************************** C * THIS SUBROUTINE SOLVE 3 EQUATIONS SIMULTANEOUSLY IN * C * ORDER TO FIND THE THERMODYNAMIC EQUILIBRIUM MOLE * C * FRACTION OF N-HEXANE, 3-MP, 2-MP, 2,3-DMB. * c **********************************************************
SUBROUTINE SLTRES(EK,YO,EXC) DOUBLE PRECISION A(3,3),B(3),EXC(3) DIMENSION EK(4),Y0(10)
INTEGER IPVT(3)
C*** NOTE THAT A,B,X,IPVT MUST BE DOUBLE PRECISIONED AND DIMENSIONED BY c C---- NUMBER OF UKKHOWKS (REACTION COORDINATES) C
N=3 c---------------------------C ZERO B(I) AND A(I,J)
DO 1 I=1,N B(I)=O.O DO 1 J=l,N
1 A(I,J)=O.O
C SET THE NONZERO VALUES OF A(I,J) AND B(I)
A(l,l)=EK(l)+l. A(1,2)=-l. A(2,1)=EK(2) A(2,2)=-EK(2)-l.
97
A(2,3)=1. A(3,2)=EK(3) A(3,3)=-EK(3)-l. B(l)=YO(l)*EK(l)-Y0(2) B(2)=Y0(3)-Y0(2)*!K(2) B(3)=Y0(4)-!K(3)*Y0(3)
C SOLVE REACTION COORDINATES EXC(l),EXC(2),EXC(3)
C CALL LINPAC c
CALL LIMPAC(N,A,B,EXC,IPVT)
YO(l)=YO(l)-EXC(l) Y0(2)=Y0(2)+EXC(l)-EXC(2) Y0(3)=Y0(3)+EXC(2)-EXC(3) Y0(4)=Y0(4)+EXC(3) YO(S)=YO(S)
RETURN END
98
APPENDIX B
COMPUTER PROGRAM FOR HEXCR
99
$debug
C ************************ ABSTRACT *********************** c * * C * This program calculates the performance of an * C * CSTCR . Using a steady state mole balance for each * C * species, a system of two linear equations containing * C * two unknown is generated. This system of equations * C * is solved using the 11 Newton's Method 11 • * c * * C * ****************** NOMENCLATURE ********************* c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c * c *
W - weight of catalyst, kg Q- volumetric flow rate, ~/hr N - no. of linear equations TM - average temperature, K NC - no. of components NR - no. of reactions ID - flag which determine reaction type CDO - initial concentration of 2,3-DMB, kgmole/~ CEO - initial concentration of 2,2-DMB, kgmole/ms FTO - total flow rate, kgmole/hr HHl- enthalpy of 2,3-DMB at temp tz, kj/kgmole HH2- enthalpy of 2,2-DMB at temp t2, kj/kgmole HGl - enthalpy of 2,3-DMB at temp t1, kj/kgmole HG2- enthalpy of 2,2-DMB at temp t1, kj/kgmole TIN - feed input temperature, K TOUT - product output temperature, K CONS - conversion PMPl - feed mole percentage of n-hexane FMP2 - feed mole percentage of 3-MP FHP3 - feed mole percentage of 2-MP FMP4 - feed mole percentage of 2,3-DMB FHP5 - feed mole percentage of 2,2-DMB FMP6 - feed mole percentage of inert gas PMPl - product mole percentage of n-hexane PMP2 - product mole percentage of 3~MP PMP3 - product mole percentage of 2-MP PMP4 - product mole percentage of 2,3-DMB PMP5 - product mole percentage of 2,2-DMB PMP6 - product mole percentage of inert gas YO(I) - mole fractions of hezane isomers, I=l,S Y0(6) - mole fraction of inert gas PX(I) - value of linear equation RK(4) - reverse rate constant, ~/(kg of cat-hr) FK(4) - forward rate constant, ~/(kg of cat-hr) EK(I) - equilibrium constants, dimensionless CP(I) - heat capacity of the i-th component at
temperature t, kj/(kgmole-K) EXC(I) - reaction coordinates, I=l,3 GPERK(I) - gibbs free energy, kj/kgmole CPV(I,J) - a vector containing heat capacities of
component i at discrete value of
100
C * temperature, kj/{kgmole-K) C * DHRXN(I) - the heat of reaction of the i-th reaction c * ,kj/kgmole c ********************************************************* c c c c c c c c c c
I INPUT DESCRIPTION I The initial guesses are specified in the main program
as well as the error criteria and the number of linear equations. The functions are specified in subroutine ~UNC AND ~ADI, the partial derivatives of the functions with respect to the independent variables are specified in subroutine DER and ADER.
C***********************************************************
$storage:2
INTEGER RC DIMENSION X{2),FX{2) COMMON /DHRX/TIN,Y0{6),yd{6) COMMON /ONE/ Q,CDO,CEO,HHl,HH2,HGl,HG2 COMMON /DATA5/ NC,PTO,NR DIMENSION RK(4),PK{4) DOUBLE PRECISION CA{5),GFERN{4),DHRXN{4)
~ I set up data for EZVU I RC=O CALL ISPPFV{S,'AI Pl',RC,AI,4) CALL ISPFFV(6, 'TIN F5' ,RC,TIN,4) CALL ISPFFV{7,'TOUT F5',RC,TOUT,4) CALL ISPPPV( 4, I Q PSI ,RC,Q,4) CALL ISPPFV{6,'FTO F5',RC,PT0,4) CALL ISPFPV{4,'W F5',RC,W,4) CALL ISPFFV{7,'FMP1 F6',RC,FMP1,4) CALL ISPPPV(7,'PMP2 F6' ,RC,PMP2,4) CALL ISPPFV{7,'FMP3 P6',RC,FMP3,4) CALL ISPPPV(7,'PMP4 F6',RC,FMP4,4) CALL ISPPPV(7,'FMP5 F6',RC,FMP5,4) CALL ISPFPV{7,'FMP6 F6',RC,FMP6,4) CALL ISPFFV(7,'PMP1 F6',RC,PMP1,4) CALL ISPPPV{7,'PMP2 F6',RC,PMP2,4) CALL ISPFFV{7,'PMP3 F6',RC,PMP3,4) CALL ISPFFV{7,'PMP4 P6',RC,PMP4,4) CALL ISPFFV{7,'PMP5 F6',RC,PMP5,4) CALL ISPFPV(7,'PMP6 F6',RC,PMP6,4)
~ set function keys I
101
c
ZF10='QUIT' ZCMD=' I
ZATR='WRI I
ZF01='CSR1' ZF02='CSR2'
C set initial values of function keys c
c
CALL ISPFFV(6,'ZATR C',RC,ZATR,4) CALL ISPFFV(6,'ZF01 C' ,RC,ZF01,4) CALL ISPFFV(6,'ZF02 C' ,RC,ZF02,4) CALL ISPFFV(6,'ZF10 C' ,RC,ZF10,4) CALL ISPFFV(6,'ZCMD C',RC,ZCMD,4)
c data for screens csl,cs2 C get default values from profile c
c
CALL ISPFF(lO,'VGET TIN P' ,RC) CALL ISPFF(S, 'VGET Q P',RC) CALL ISPFF(lO,'VGET FTO P' ,RC) CALL ISPFF(S, 'VGET W P' ,RC) CALL ISPFF(ll,'VGET FMPl P',RC) CALL ISPFF(ll,'VGET FMP2 P',RC) CALL ISPFF(ll,'VGET FMP3 P',RC) CALL ISPFF(l1,'VGET FMP4 P',RC) CALL ISPFF(l1,'VGET FMPS P',RC) CALL ISPFF(11,'VGET FMP6 P' ,RC)
C start screen inputs c
101 ZCMD=' CALL ISPFF(l3,'DISPLAY KEYC1' ,RC) CALL ISPFF(12,'DISPLAY CSR1',RC)
IF(ZCMD.EQ.'QUIT') CALL EXIT IF(ZCMD.EQ. 'CSR2') GOTO 202 GOTO 303
202 ZCMD=' CALL ISPPP(13,'DISPLAY KEYC2' ,RC) CALL ISPPP(12,'DISPLAY CSR2',RC)
IF(ZCMD.EQ. 'QUIT') CALL EXIT IF(ZCMD.EQ.'CSR1') GOTO 101 GOTO 303
303 ZCMD=' c c id=1 for adiabatic condition C id=2 for isothermal condition c
102
ID=1 IF(AI.GT.1.5) ID=2
~ I assume isothermal case
TOUT= TIN c C input initial mole fraction c of n-hexane ----> Y0(1) C 3-HP > Y0(2) C 2-HP > Y0{3) C 2,3-DHB --> Y0(4) C 2,2-DHB --> YO(S) c
c
Y0(1)=FHP1/100. Y0(2)=FHP2/100. Y0{3)=FHP3/100. Y0(4)=FHP4/100. YO(S)=FHPS/100. Y0{6)=FHP6/100.
C let yd(i) be yo(i) c
yd=yo c c check initial condition c
IF((W.EQ.O.).AND.(ID.EQ.l)) GOTO 555 IF((W.EQ.O.).AND.(ID.EQ.2)) GOTO 320
c c set up data for reaction model c
CALL DATAlf c C calculate heat capacities C and reaction heats c
CALL ARRAY c C make initial guesses of Newton's method C x(1) = concentration of 2,3-DMB C x(2) = concentration of nee-hexane c C unit : (kgmole/~) c
X(l)=lO.O
103
X(2)=10.0 c C calculate the thermo equilibrium c mole fraction of hexane isomers c
c c
TINN=TIN CALL THEQ{YD,FK,RK,TINN,ID)
C convert mole fractions into concentrations, C (kgmole/ml) c C cdo = new cone. of 2,3-DMB C ceo = new cone. of nee-hexane c
c
CDO=YD(4)*FTO/Q CEO=YD(S)*FTO/Q
IF(ID.EQ.l) GOTO 139
C perform isothermal condition c c call Newton's method C calculate the output cone. of 2,3-DMB & neo-hexane c
CALL NEWTN(X,FX,K,FK,RK,ID,TINN,TOUT) c C From Newton's method, we can find the final C conc.s of 2,3-DMB [x(1)] and nee-hexane [x(2)]. c Then convert them into mole fraction, YD(4) and C YD(S). c
c
YD(S)=X(2)*Q/PTO YD(4)=X(l)*Q/FTO GOTO 224
C adiabatic case (id=1) c
139 COift'INU!
~ I assumed the final output teoperature
TOUT=TINN+lOO. 145 CALL KEWTJ(X,FX,W,PK,RK,ID,TINN,TOUT)
104
c c c c c c c
calculate the conversion the following reaction for adiabatic case:
2,3-DHB <--==> neo-hexane
CONS=W*{FK{4)*X{l)-RK{4)*X{2))/{YD{4)*FTO) c c using average temperature to estimate the C heat capacities of 2,3-DMB and neo-hexane c
TM={TINN+TOUT)/2. CALL PROP(TM,DHRXN,CA,GFERN)
~ I check convergence I USUM=CA{S}*TINN*CONS+CA(4)*TINN-CA{4)*TINK*CONS-
Hl*CONS DSUM=CA{S)*CONS+CA(4}-CA(4)*CONS TF2=USUM/DSUM TERML=ABS{TF2-TOUT) TOUT=TF2 IF{TERML.LT.l.O) GOTO 542 GOTO 145
~ end of adiabatic case I c C print out results of c isothermal condition c
224 CONTINUE c C calculate the conversion of the following reaction C for isothermal case c C 2,3-DMB <~> neo-Helane c
CONS=W*{PK{4)*X(l)-RK(4)*X(2))/{YD(4)*PTO)
~ I **** final outputs to BZVU ****
320 CONTINUE
105
C > Isothermal condition <-----. c C moles of hexane isomers at the end of the C reaction (convert into percentage form) c C PMPl ----> n-hexane C PHP2 -> 3-HP C PMP3 ----> 2-MP C PMP4 ----> 2,3-DHB C PMPS ----> 2,2-DMB c PHP6 -> inert gas c
c
PMP1=YD(1)*100. PMP2=YD(2)*100. PHP3=YD{3)*100. PMP4=YD(4)*100. PHP5=YD(5)*100. PHP6=YD(6)*100.
GO'i'O 101
C adiabatic condition c C final outputs to EZVU c
542 COifTINUE
YD(S)=YD(S)+YD(4)*CONS YD{4)=YD(4)*(1.-CONS) CONS=CONS*lOO.
555 COif'l'IlfUE C > Adiabatic condition <------. c C moles of hexane isomers at the end of the C reaction (convert into percentage fora) c c PHP1 -> n-hexane C PMP2 ----> 3-MP C PHP3 ----> 2-MP C PMP4 -> 2,3-DMB C PHP5 -> 2,2-DHB c PMP6 -> inert gas c
PHP1=YD(1)*100. PHP2=YD(2)*100. PHP3=YD(3)*100. PHP4=YD(4)*100. PHP5=YD(5)*100. PHP6=YD(6)*100.
106
107
GOTO 101 666 STOP
END c c end of the main program!!! c
c ******************* ABSTRACT ************************ c * * c * THIS SUBROUTINE CALCULATES THE PARTIAL DERIVATIVES* c * OF THE FUNC'l'IONS WITH RESPECT TO THE INDEPENDENT * c * VARIABLES. A(I,J) REPRESENTS THE PARTIAL OF THE ith* c * FUNCTION WITH RESPECT TO THE JTH VARIABLE. * c * (ISOTHERMAL CASE) ·t
c ******************************************************* c
SUBROUTINE DER(N,A,FK,RK,W) DIMENSION A{2,2) COMMON /ONE/Q,CDO,CEO,HH1,HH2,HG1,HG2 DIMENSION RK(4),FK(4)
DO 1 I=1,N DO 1 J=1,N
1 A(I,J)=O.O A(1,1)=-1.-W/Q*FK(4) A(1,2)=RK(4)*W/Q A{2,1)=W/Q*FK(4) A(2,2)=-1.-W/Q*RK(4)
RETURN END
c ******************* ABSTRACT ************************ c * * c * THIS SUBROUTINE CALCULATES THE PARTIAL DERIVATIVES* c * OF THE FUNCTIONS WITH RESPECT TO THE INDEPENDENT * c * VARIABLES. A{I,J) REPRESENTS THE PARTIAL OF THE ith* c * FUNCTION WITH RESPECT TO THE JTH VARIABLE. * c * (ADIABATIC CASE) * c ******************************************************* c
SUBROUTINE AD!R(N,A,FK,RK,W) DIMENSIOI A(2,2),RK(4),FK(4) COMMOI /ONE/ Q,CDO,CEO,HRl,HH2,HGl,HG2
DO 1 I=1,N DO 1 J=1,lf
1 A(I,J)=O.O
A{l,1)=-HR1-(HH1*W/Q)*FK(4) A(1,2)=(HHl*N/Q)*Rk(4) A(2,l)={HH2*W/Q)*FK(4) A{2,2)=-RH2-(HH2*N/Q)*RK(4)
RETURN END
c * ***************************************************** C * THIS SUBROUTINE CALCULATES THE VALUES OF EACH * C * LINEAR EQUATION GIVEN THE VALUE OF X(I) AND N. * C * THESE VALUES ARE SUPPLIED TO THIS SUBROUTINE WHEN * C * IT IS CALLED BY N!HTN. (ISOTHERMAL CASE) * c *******************************************************
SUBROUTINE FUNC(X,FX,FK,RK,W) COMMON /ONE/Q,CDO,CEO,HHl,HH2,HGl,HG2 DIMENSION RK(4),FK(4),X(2),FX(2)
FX(l)=CDO-X(l)+W/Q*(RK(4)*X(2)-FK(4)*X(l)) FX(2)=-X(2)+CEO+W/Q*(FK(4)*X(l)-RK(4)*X(2))
RETURN END
c * ***************************************************** C * THIS SUBROUTINE CALCULATES THE VALUES OF EACH * C * LINEAR EQUATION GIVEN THE VALUE OF X(I) AND H. * C * THESE VALUES ARE SUPPLIED TO THIS SUBROUTINE WHEN * C * IT IS CALLED BY NEWTN. (ADIABATIC CASE) * c *******************************************************
SUBROUTINE FADI(X,FX,FK,RK,W,Tl,T2) COMMON /OK!/ Q,CDO,CEO,HHl,HH2,HGl,HG2 DIMENSIOK RK(4),FK(4),X(2),FX(2) DOUBLE PRECISION CA(5),DHRXN(4),GFERK(4)
CALL PROP(Tl,DHRXN,CA,GFERN) CA3=CA(4) CA4=CA(5) Hl=DHRXN(4) TM=(Tl+T2)/2. CALL PROP(TM,DHRXN,CA,GF!RK) CAl=CA(4) CA2=CA(5) HHl=-CAl*(T2-298.15)-176.80*1000. HH2=-CA2*(T2-298.15)-184.68*1000. HG1=-(Tl-298.15)*CA3-176.8*1000. HG2=-(Tl-298.15)*CA4-184.68*1000. FX(l)=-HHl*X(l)+CDO*HGl+(W/Q)*(RK(4)*X(2)*HHl-
FK(4)*X(l)*HHl) FX(2)=-HH2*X(2)+CEO*HG2+(W/Q)*(FK(4)*X(l)*HH2-
RK(4)*X(2)*HH2)
RETURN END
C ttttttttttttttttttt ABSTRACT tttttttttttttttttttttttt
c * * C * THIS SUBROUTIKE EMPLOYES NEWTOK'S METHOD IN * C * ORDER TO SOLVE A SET OF H LINEAR EQUATIONS * C * CONTAINING N UHKNOHHS. THIS SUBROUTINE IS *
108
c * CALLED BY THE MAIN PROGRAM AMD IS CCSUPPLIED * c * THE VALUES OF THE INITIAL GUESS FOR X(I)'S AS * c * WELL AS THE VALUE OF K. THIS SUBROUTINE USES * c * THE VALUES OF THE FUNCTION FROM FUNC AND THE * c * VALUES OF THE PARTIAL DERIVATIVES OF THE * c * FUNCTION IN ORDER TO DETERMINE THE SOLUTION. * c * THIS C METHOD USES THE LIBRARY ROUTINE LINPAC * c * TO SOLVE THE C SYSTEM OF LINEAR EQUATION USED * c * BY NEWTON'S METHOD. * c ****************************************************** c
c
SUBROUTINE NEWTN(X,FX,W,FK,RK,ID,T,T2) DIMENSION A{2,2) ,X(2) ,FX{2) ,B{2) ,RAT{-2) ,FK{4) ,RK( 4)
DOUBLE PRECISION AA(2,2),BB{2),XX{2) INTEGER IPVT{2)
N=2 ERLIH=l.OE-3
1 CONTINUE ITEST=O
C HAKE FUNCTION EVALUATIONS c
IF {ID.EQ.1) GOTO 10 CALL FUNC{X,FX,FK,RK,W) GOTO 4
10 CALL FADI{X,FX,FK,RK,W,T,T2) c c set up constant terms c in jacobian matrix c
4 DO 3 I=1,N 3 B(I)=-FX(I)
~ I evaluate JACOBIIX matri1 I IF(ID.EQ.1) GOTO 15 CALL DER(N,A,PK,RK,H) GOTO 20
15 CALL ADER(N,A,FK,RK,W) c C establish coefficients C in jacobian matrix c
20 DO 32 I=1,N
c
DO 32 J=l,N 32 AA(I,J)=A(I,J)
DO 35 I=l,N 35 BB(I)=B(I)
C Using linpac to solve jacobian matrix c C CALL LINEAR EQUATION SOLVER c
CALL LINPAC(N,AA,BB,XX,IPVT)
~ I make an improved value for x(i)
c
DO 5 I=l,N RAT(I)=XX(I)/X(I)
5 X(I)=X(I)+XX(I)
C CHECK FOR CONVERGENCE c
c
DO 125 I=l,N 125 IF(ABS(RAT(I)).GT.ERLIM)ITEST=ITEST+l
IF(ITEST.NE.O)GO TO 1 RETURN END
C This subroutine supplied the majority of the C data for the CSCTR reactor model. Feed conditions, C stoichiometric coefficients of reaction model. c
SUBROUTINE DATAN COMMON /DHRX/ TIN,Y0(6),YD(6) COMMON /DATA5/ NC,FTO,NR COMMON /DATA6/ GAM(5,4)
~ I The nuober of reactions
MR=4 c C THE NUMBER OF COMPONENTS, c (except the inert qas because it C dosen't react with other reactants) c
NC=5
110
c c c c
1 c c c c c
c c c c c
c c c c c
c c c c c
The stoichiometric coefficients ** INITIALIZE GAM(I,J) **
DO 1 I=1,NC DO 1 J=1,NR GAM(I,J)=O.O
(1) n-HEXANE <-------> (1) 3-methylpantane
l____>GAM(1,1) l_____>GAM(2,1)
GAM(1,1)=-l.O GAM(2,1)=1.0
(1) 3-methylpantane <--> (1) 2-methylpantane
L_____>GAM(2,2) I >GAM(3,2)
GAH(2,2)=-l.O GAM(3,2)=1.0
(1) 2-methylpantane <--> (1) 2,3-dimethylbutane
l_____>GAM(3,3) l______>GAM(4,3)
GAM( 3 I 3) =-1. 0 GAM( 4,3)=1.0
(1) 2,3-dimethylbutane <----> (1) 2,2-dimethylbutane
l______>GAM(4,4) l_____>GAM(5,4)
GAM(4,4)=-l.O GAM(S,4)=1.0
RETURN END
c ******************************************************* C * THIS SUBROUTINE CALCULATES THE HEAT CAPACITIES * C * OF EACH SPECIES AND THE HEATS OF REACTION AT T * C * (K). * c *******************************************************
SUBROUTINE ARRAY COMMON /DATAS/ NC,FTO,NR COMMON /DATA6/ GAM(5,4) COMMON /DHRX/ TIN,Y0(6),YD(6)
111
c
COMMON /VECTR/ DHRXV(4,150),CPV(5,150),GFERV(4,150) DOUBLE PRECISION SUM,SUM1
c decide the integration interval C (temp range from 298.15 to 1000 K) c
c
TI=298.15 DT=(1000.-TI)/90.
C REACTION HEAT (KJ/KGMOLE) AT 298.15 K c C n-HEXANE <-.-> 3-METHYLPANTANE (3-MP) c
DHRXV(1,1)=-5.050*1000. c C REACTION HEAT (KJ/KGMOLE) AT 298.15 K c C 3-M!THYLPAlf'l'Alf! <--> 2-M!THYLPANTANE c
DHRXV(2,1)=-2.580*1000. c C REACTION HEAT (KJ/KGMOLE) AT 298.15 K c C 2-M!THYLPAlfTAlf! <--> 2,3-DIMETHYLBUTAifE c
c c c c c
DHRXV(3,1)=-2.250*1000.
REACTION HEAT (KJ/KGMOL!) AT 298.15 K
2,3-DIMETHYLBUTAN! <--------> 2,2-DIMETHYLBUTAlf!
DHRXV(4,1)=-7.880*1000. C++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ C CALCULATE REACTION HEAT AT EVERY T+DT, (KJ/KGMOLE) C--------- From '1'=298.15 K TO 1000 K c For e1ample: lst reaction C n-Hez:ane --> 3-Methylpantane c c c c c c c c c c
IT2 (Cp'-Cp)dT = heat of reaction
T1
where Cp'= heat capacity of 3-MP Cp = heat capacity of n-Hexane
112
C p.s. 1st reaction: 1=1 C TI=298.15 K {referance temp) C sum=O.O C K=2 C we want to evaluate reaction heat at 305.98 K {TI+DT) C sum=O.O+DT*GAM(1,1)*Cp{1,305.98)=-DT*Cp(1,305.98) C sum=sum+DT*GAM(2,1)*Cp(2,305.98)=-DT*Cp(1,305.98) C +DT*1*Cp(2,305.98) C sum=sum+DT*GAM(3,1)*Cp(3,305.98)=-DT*Cp{1,305.98) C +DT*Cp(2,305.98)+DT*O.O*Cp(3,305.98) C =-DT*Cp(l,305.98)+DT*Cp(2,305.98) C sum=sum+DT*GAM(4,1)*Cp(4,305.98) C =-DT*Cp{l,305.98)+DT*Cp(2,305.98) C sum=sum+DT*GAM(5,1)*Cp(5,305.98) C =-DT*Cp(1,305.98)+DT*Cp{2,305.98) c C Therefore, heat of reaction with respect to reference C temp is delta H= DT*{ Cp(2,305.98)-Cp{l,305.98)) c C+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
c
DO 4 I=1,NR TI=298.15 SUM=O.ODO DO 5 K=2,91 TI=TI+DT DO 6 J=1,HC SUM=SUM+DT*GAM{J,I)*CP{J,TI)
6 CONTINUE
IF (TI.GT.1000.) GOTO 4
c new value of reaction heat C at (t+dt) with respect to 298.15 K c
DHRXV(I,K)=DHRXV(I,K-l)+sum
5 SUM=O.ODO
4 CONTINUE c C CALCULATE THE GIBBS ENERGY OF FORMATION C OF EACH REAC'l'IOH C "PERV(I ,J), (KJ/KGMOLE)---t c C For example: 1st reaction C reference temp=298.15 K C DT=7.8 K C new temp to evaluate : 305.9 K c suml=O.O c C n-Hexane <--> 3-methylpantane
113
C Show steps: C suml=GAM(l,l)*GF(1,305.9)+suml C =-1*GF(1,305.9)+0.0 c c sum1=GAM(2,1)*GF(2,305.9)+sum1 C = l*GF(2,305.9)-GF(l,305.9) C sum1=GAM(3,1)*GF(3,305.9)+sum1 C =GF(2,305.9)-GF(l,305.9) C suml=GAM(4,1)*GF(4,305.9)+suml C =GF(2,305.9)-GF(l,305.9) C suml=GAM(5,1)*GF(5,305.9)+suml C =GF(2,305.9)-GF(l,305.9) c C Therefore, Gibbs energy of formation C of 1st reaction at t+dt C GFERV(1,305.9)=suml c c
DO 34 I=1,lfR TI=298.15 SUMl=O.O DO 35 K=2,91 TI=TI+DT DO 36 J=l,NC SUM1=GAM(J,I)*GF(J,TI)+SUM1
36 CONTINUE IF (TI.GT.1000.) GO TO 34 GFERV(I,K)=SUMl
35 SUM1=0.0
34 CONTINUE c C SET UP CPV(I,J) ---->> convert every value of C Cp(I,TI) into vector in order to use linear C interpolation c
c
DO 10 I=l,lfC TI=298.15
DO 11 J=l,90 CPV(I,J)=CP(I,TI)
11 TI=TI+DT 10 CONTINUE
RE'l'URlf EHD
C This function subroutine calculates the heat C capacity of each species from 200 - 1000 (K) C (KJ/KGMOLI!!-K) .C
FUNCTION CP(I,T)
114
c
IF (I.EQ.1) GOTO 1 IF (I.EQ.2) GOTO 2 IF (I.EQ.3) GOTO 3 IF (I.EQ.4) GOTO 4 IF (I.EQ.5) GOTO 5
1 CP=(-1.0540+1.39E-1*T-7.449E-5*T2+1.551E-8*TS)*4.184 RETURN
2 CP=(-0.570+1.359E-1*T-6.854E-5*T2+1.202E-8*TS)*4.184 RETURN
3 CP=(-2.5240+1.477E-1*T-8.533E-5*T2+1.931E-8*TS)*4.184 RETURN
4 CP=(-3.489+1.469E-1*T-8.063E-5*T2+1.629E-8*TS)*4.184 RETURN
5 CP=(-3.973+1.503E-1*T-8.314E-5*T2+1.636E-8*TS)*4.184 RETURN END
C This function calculates the gibbs free C energy of formation of hexane isomers C from 200-1000'K, KJ/KGMOLE c
c
FUNCTION GF(I,T) IF (I.EQ.1) GOTO 1 IF (I.EQ.2) GOTO 2 IF (I.EQ.3) GOTO 3 IF (I.EQ.4) GOTO 4 IF (I.EQ.S) GOTO 5
1 GF=-0.000087*T'+0.219336*T2+453.119366*T-151738.80896 RETURN
3 GF=-0.000092*T'+0.225563*T2+458.412884*T-158998.25354 RETURN
2 GF=-0.000085*TS+0.216435*T2+460.331617*T-156934.60254 RETURN
5 GF=-0.000092*TS+0.224945*T2+481.3136l8*T-169166.74548 RETURN
4 GF=-0.00009l*TS+0.22493*T2+473.536175*T-161194.52295 RETURN END
c This subroutine calculates the rate constants, c heat capacitites and also the output temperature of C reaction n-hexane -----> 2,3-DMB c
SUBROUTINE THEQ(YO,FK,RK,T,ID) COMMON /DATA5/ NC,FTO,NR COMMON /DATA6/ GAM(5,4) DIMENSION EK(4),FK(4),RK(4),Y0(6) DOUBLE PRECISION CA(5),DHRXN(4),GFERN(4)
115
DOUBLE PRECISION EXC(3) c C Calculate the heats of reactions, heat c capacities and gibbs energies at any temp C between (298 -1000 K) C Using linear interpolation c
CALL PROP(T,DHRXH,CA,GFERN) c C Calculate the equilibrium constants, C forward and reverse rate constants c
CALL EQCON(EK,FK,RK,GFERN,T) c C calculate the equilibrium mole fractions C of hexane isomers C Using 'LINPAC' To Solve Reaction Coordinates c
c
CALL SLTRES(EK,YO,EXC) IF(ID.EQ.2) GOTO 100
C do energy balance c
c
c c c
c
CALL EKGBALS(YO,EXC,DHRXN,T)
100 RETURN END
SUBROUTINE ENGBALS(YO,EXC,DHRXH,T) COMMON /DATAS/ NC,FTO,KR DOUBLE PRECISION EXC(3),DHRXN(4),CA(S),GFERN(4) DIMENSION Y0(6),EK(4),FK(4),RK(4)
input term =0
C disappearance term c
SDIS=O.O DO 20 I=1,3
20 SDIS=EXC(I)*FTO*DHRXH(I)+SDIS
c C input=output+acc+disapp c
116
DIF=-SDIS
~ I output term
'l'M=T TE=T INX=1
50 CALL PROP(TH,DHRXN,CA,GFERN) SOUT=O.O DO 25 !=1,4
25 SOUT=SOUT+(TE-T)*CA(I)*YO(I)*FTO ERH1=ABS(DIF-SOUT) IF(INX.EQ.1) GOTO 77 ERH=ERH2-ERH1 IP(ERM.LT.O.) GOTO 100
77 ERH2=ERH1 TE=TE+0.2 INX=INX+1 CALL PROP(TE,DHRXN,CA,GFERN)
. CALL EQCON(EK,FK,RK,GFERN,TE) CALL SLTRES(EK,YO,EXC) TH=(TE+T)/2. GOTO 50
100 T=TE
110 RETURN END
c ********************************************************** C * THIS SUBROUTINE CALCULATES THE HEAT CAPACITY AND * C * THE HEATS OF REACTIONS FOR A GIVEN TEMPERATURE BY * C * LINEARLY INTERPOLATING BETWEEN VALUES OF CPV(I,J) AND* C * DHRXV(I,J), RESPECTIVELY. * c **********************************************************
SUBROUTINE PROP(T,DHRXN,CA,GFERN) DOUBLE PRECISION DHRXN(4),CA(5),GFERK(4) COMMON /VECTR/ DHRXV(4,150),CPV(5,150),GFERV(4,150) COMMON /DATA5/ NC,FTO,MR
DT=(1000.-298.15)/90. I=IFIX((T-298.15)/DT)+1 PRO=(T-298.15-DT*FLOAT(I-1))/DT
c..---------------, C CALCULATE THE HEA'l'S OF REACTIONS C AT SPECIFIED TEMP, KJ/KGMOLE C'--------------'
DO 1 J=1,MR 1 DHRXN(J)=DHRXV(J,I)+PRO*(DHRXV(J,I+1)-DHRXV(J,I))
117
c .--------------------------, C CALCULATE THE HEAT CAPACITIES OF C ISOMERS AT SPECIFIED TEMP, KJ/KGMOLE-K c ~------------------------~
DO 2 J=l,NC 2 CA(J)=CPV(J,I)+PRO*(CPV(J,I+l)-CPV(J,I))
c .----------------------, C CALCULATE THE GIBBS ENERGIES C AT SPECIFIED TEMP, (KJ/KGMOLE) c ~--------------------~
DO 3 J=l,lfR 3 GFERN(J)=GFERV(J,I)+PRO*(GFERV(J,I+l)-GFERV(J,I))
RETURN END
c ********************************************************** C * THIS SUBROUTINE CALCULATES THE EQUILIBRIUM * C * CONSTANTS OF EACH REACTION AND THE FORWARD AND REVERSE* C * REACTION RATE CONSTANTS. * c **********************************************************
SUBROUTINE EQCON(EK,FK,RK,GFERN,T) DIMENSION EK(4),FK(4),RK(4) DOUBLE PRECISION GFERN(4)
c .---------------------------, C CALCULATE EQUILIBRIUM CONSTANT OF NORMAL C HEXANE TO IT'S ISOMERS C delta G=-RT*lnK c ~------------------------~
DO 5 I=l,4 5 EK(I)=EXP(-GFERN(I)/(8.314*T))
c .-------------------------~ C CALCULATE REVERSE REACTION RATE CONSTANTS, C (FT'/LB of CAT-HR) C kr=k·elp(-E/RT) c ~------------------------~
TR=9./5.*(T-273.15)+32.+459.67 RK( 4) =5. 24*1000. * !XP( -'1. 55*1000. /1.987 /TR)
c .--------------------, C CALCULATE FORWARD REACTION C RATE CONSTANTS, (FTS/LB-HR) C K=kf/kr -----> kf=K*kr c ~------------'
FK(4)=RK(4)*EK(4) c .-----------, C CONVERT FORWARD AJID REVERSE C RATE CONSTANTS TO, MS /KG-HR) c .__ _________ _,
FK(4)=FK(4)*0.06243 RK(4)=RK(4)*0.06243
118
RETURN END
c ********************************************************** C * THIS SUBROUTINE SOLVE 3 EQUATIONS SIMULTANEOUSLY IN * C * ORDER TO FIND THE THERMODYNAMIC EQUILIBRIUM MOLE * C * FRACTION OF N-HEXANE, 3-MP, 2-MP, 2,3-DMB. * c **********************************************************
SUBROUTINE SLTRES(EK,YO,EXC) DOUBLE PRECISION A(3,3),B(3),EXC(3)
DIMENSION EK(4),Y0(10) INTEGER IPVT(3)
C*** NOTE THAT A,B,X,IPVT MUST BE DOUBLE PRECISIONED AND DIMENSIONED BY c C---- NUMBER OF UNKNOWNS (REACTION COORDINATES) C
N=3 c---------------------------C ZERO B(I) AND A(I,J)
DO 1 I=l,N B(I)=O.O DO 1 J=l,N
1 A(I,J)=O.O
C SET THE NONZERO VALUES OF A(I,J) AND B(I)
A(1,l)=EK(1)+1. A(l,2)=-l. A(2,l)=EK(2) A(2,2)=-EK(2)-l. A(2,3)=1. A(3,2)=EK(3) A(3,3)=-EK(3)-l. B(l)=YO(l)*EK(l)-Y0(2) B(2)=Y0(3)-Y0(2)*EK(2) B(3)=Y0(4)-EK(3)*Y0(3)
C SOLVE REACTION COORDIMATES EXC(l),EXC(2),EXC(3)
CALL LINPAC(H,A,B,EXC,IPVT) YO(l)=YO(l)-EXC(l) Y0(2)=Y0(2)+EXC(l)-EXC(2) Y0(3)=Y0(3)+EXC(2)-EXC(3) Y0(4)=Y0(4)+EXC(3) YO(S)=YO(S)
RETURN END
119
APPENDIX C
LISTING OF CONTROL PANELS
120.
************************ Hexane Isomerization
Fixed-Bed Reactor 300 -- 1000 K
REACTOR CONDITION
1= ADIABATIC RX PLEASE PICK EITHER 1 OR 2 2= ISOTHERMAL RX [ 0] .
FEED ->~ ~
0 0 0 0 o.l
opz:N
0 0 0 0
0 0 0 0 0
o o o o o.oo m 0000011
0 0 0 0 u L»- PRODUCT
F2 HEX SCREEN 2
I Final Mol% NEO
I 0.000
F3 HEX SCREEN 3
NO OF TUBES o.o TUBE D (M) 0.000
FEED CONDITIONS
FLOW RATE (KGMOLE/HR) 0.0000
TEMP ( K) 0.0000
PRESSURE 0.0000
BULK DEN OF CAT. (KG/M~3) 0.0000
FlO QUIT
Figure 18. Control Panel 1 for Fixed-Bed Reactor Initial Parameter Settings
ATM
I-' IV ......
INPUT. REACTANTS MOLE %
HEXANE ISOMERIZATION 300-1000 K
OUTPUT PRODUCTS MOLE %
N-HEXANE [100.0] % N-HEXANE 0.000
3 - MP o.ooo % 3 - MP 0.000
2 - MP 0.000 % 2 - MP 0.000
2,3-DMB 0.000 % 2,3-DMB 0.000
NEO-NEXANE 0.000 % NEO-HEXANE 0.000 (2,2-DMB) (2,2-DMB)
INERTS. 0.000 % INERTS 0.000
Fl HEX SCREEN 1 F3 HEX SCREEN 3 FlO QUIT
Figure 19. Control Panel 2 for Fixed-Bed Reactor Initial Parameter Settings
%
%
%
%
%
%
..... 1\.) 1\.)
************************ Hexane Isomerization
Fixed-Bed Reactor 300 -- 1000 K
FEED ->>~ I I Final Mol% NEO
I .... ··l
opz:Y
0 0 0 0
0 0 0 0 0
o o o o 12.1 m 0000011 .... u
L»- PRODUCT
F2 HEX SCREEN- 2
37.69
F3 HEX SCREEN 3
REACTOR CONDITION
1= ADIABATIC RX 2= ISOTHERMAL RX [2]
NO OF TUBES 150. TUBE D (M) 0.050
FEED CONDITIONS
FLOW RATE (KGMOLE/HR) 153.50
TEMP ( K) 408.00
PRESSURE 35.000
BULK DEN OF CAT. (KG/M-3) 640.00
FlO QUIT
Figure 20. Control Panel 1 for Fixed-Bed Reactor Optimized Model, Isothermal Reaction
ATM
---
I-' IV w
HEXANE ISOMERIZATION 300-1000 K
~ I INPUT REACTANTS MOLE % f;~::;.
:-:\'::{
.;:-:-::: :-:-:-:·:~-
~?{
f N-HEXANE (100.0) % 0
=~ :-:-:-:~-:-1 ~~~;~::~E 0.000 % I I INERTS 0.000 %
n I OUTPUT PRODUCTS MOLE % q ~
~ ~
~ N-HEXANE 7.825 % 0 I ~
I :.:_::8 :~::: :I ~!}!-------- ------.....::: I NEO-HEXANE 37.69 % ~ ftl (2,2-DMB)
-------<.:H if"( INERTS 0.000 % /:j
Ft HEX SCREEN 1 F3 HEX SCREEN 3 FlO QUIT
Figure 21. Control Panel 2 for Fixed-Bed Reactor Optimized Model, Isothermal Reaction
...... N ,j:>.
****~******************* Hexane Isomerization
Fixed-Bed Reactor 300 -- 1000 K
FEED ->~ I 1 Final Mol% NEO
I 35.64 .... ·•l 0 0 0 0
0 0 0 0 0
o o o o 3.00 m
OOOOOJ 0 0 0 0
L»- PRODUCT '· ~
opz:N
F2 HEX SCREEN 2 F3 HEX SCREEN 3
REACTOR CONDITION
1= ADIABATIC RX 2= ISOTHERMAL RX. [2]
NO OF TUBES 150. TUBE D (M) 0.050
.FEED CONDITIONS
FLOW RATE (KGMOLE/HR) 153.50
TEMP ( K) 408.00
PRESSURE 35.000
BULK DEN OF CAT. (KG/M- 3) 640.00
FlO QUIT
Figure 22. Control Panel 1 for Fixed-Bed Reactor Fixed-Length Model, Isothermal Reaction
I
ATM
1-' 1\J ()1
INPUT REACTANTS MOLE %
HEXANE ISOMERIZATION 300-1000 K
OUTPUT PRODUCTS MOLE %
N-HEXANE [100.0] % N-HEXANE 8.086
3 - MP 0.000 % 3 - MP 17.40
2 - MP 0.000 % 2 - MP 29.60
2,3-DMB 0.000 % 2,3-DMB 9.252
NEO-NEXANE 0.000 % NEO-HEXANE 35.64 (2,2-DMB) (2,2-DMB)
INERTS 0.000 % INERTS 0.000
Fl HEX SCREEN 1 F3 HEX SCREEN 3 FlO QUIT
Figure 23. Control Panel 2 for Fixed-Bed Reactor Fixed-Length Model, Isothermal Reaction
%
%
%
%
%
%
...... IV 0'1
*********************** HEXANE ISOMERIZATION
CSTCR 300 1000 K
"1 11 or "2" only
•• •• •• ••
Reactor condition
1. Adiabatic Rx 2. Isothermal Rx (OJ
>> Feed condition <<
Flow rate (kgmolefhr) 0.0000
Volumetric flow rate 0.0000 (m-3/hr)
=>~ spinning shaft
Tout= o.oooo >>=
Product out
K
Temperature ( K) 0.0000
weight of cat. (kg) 0.0000
Feed in Neo mol% = o.ooo %
F2 HEX SCREEN 2 FlO QUIT
Figure 24. Control Panel 1 for CSTCR · Initial Parameter Settings
I
1-' IV -....]
HEXANE ISOMERIZATION 300-1000 K
INPUT"REACTANTS MOLE% OUTPUT PRODUCTS MOLE %
N-HEXANE [100.0) % N-HEXANE
3 - MP 0.000 % 3 - MP
2 - MP 0.000 % 2 - MP
2,3-DMB 0.000 % 2,3-DMB
NEO-HEXANE 0.000 % NEO-HEXANE (2,2-DMB) (2,2-DMB)
INERTS 0.000 % INERTS
Fl HEX SCREEN 1 FlO QUIT
Figure 25. Control Panel 2 for CSTCR Initial Parameter Settings
0.000
o.ooo
o.ooo
o.ooo
o.ooo
0.000
%
%
%
%
%
%
...... 1\J 00
*********************** HEXANE ISOMERIZATION
CSTCR 300
•• •• •• ••
=>~ Feed in
spinning shaft
1000 K
~ II out= 408.00 ~>>= Product out
Neo mol% = 11.37 %
F2 HEX SCREEN 2
K
Reactor condition
1. Adiabatic Rx 2. Isothermal Rx [2]
>> Feed condition <<
Flow rate (kgmole/hr) 153.00
Volumetric flow rate 125.00 (mA3/hr)
Temperature · ( K) 408.00
weight of cat. (kg) 2200.0
-------- --- ----
FlO QUIT
Figure 26. Control Panel 1 for CSTCR Isothermal Reaction
I
...... "-> 1.0
HEXANE ISOMERIZATION 300-1000 K
INPUT REACTANTS MOLE % OUTPUT PRODUCTS MOLE %
N-HEXANE [100.0] % N-HEXANE
3 - MP 0.000 % 3 - MP
2 - MP 0.000 % 2 - MP
2,3-DMB 0.000 % 2,3-DMB
NEO-HEXANE o.ooo % NEO-HEXANE ( 2, 2-DMB) (2, 2-DMB)
INERTS o.ooo % INERTS
Fl HEX SCREEN 1 FlO QUIT
Figure 27. Control Panel 2 for CSTCR Isothermal Reaction
12.56
27.03
45.99
3.038
11.37
0.000
%
%
%
%
%
%
...... w 0
APPENDIX D
EFFECT OF PRESSURE UPON THE EQUILIBRIUM CONSTANT
This appendix explains that the effect of pressure on
the equilibrium constant.
As mentioned earlier, ~Go is based upon a fixed initial
and final state and is not influenced by the conditions at
any intermediate point. In fact, pressure does affect
equilibrium yield for a gas phase reaction. This effect of
pressure can be accounted for in the relationship between
Ky, K. The detailed steps are shown below.
For reaction aA + bB ---> cC + dD
fi v = flli v Yi P
where
fiv = the fugacity of components
fl!iv = mixture fugacity coefficients
Yi = mole fraction in the gaseous mixture
P = Total pressure
Using this expression for the fugacity, K becomes
c d
[fl!P]c [fl!P]D K =
• b
[fl!P]A [fl!P]B
c d
Yc YD
• b
YA YB
131
(D-1)
where Ky = equilibrium constant in terms of
mole fractions.
132
Assuming the mixture fugacity coefficients are equal to
unity is equivalent to assuming that the gas phase behaves
as an ideal solution. With this simplification, equation
(D-1) becomes
K = [p(c+d)-(a+b)] Ky; ( •:· c+d-a-b = 0 )
= Ky
Therefore, pressure does not affect the equilibrium yield if
ideal gas behavior is assumed.
Tai-Chang Kao
Candidate for the Degree of
Master of Science
Thesis: CHEMICAL PROCESS SIMULATION OF HEXANE ISOMERIZATION IN A FIXED-BED AND A CSTCR REACTOR
Major Field: Chemical Engineering
Biographical:
Personal Data: Born in Taipei, Taiwan, Republic of China, August 3, 1963, the son of Ming-Pan & SuChu Kao.
Education: Graduated from Cheng-Kwo High School, Taiwan, R.O.C., in June 1981; received the Bachelor of Science degree in Chemical Engineering from Tunghai University, Taiwan, in June 1986; completed requirements for the Master of Science degree at Oklahoma State University in May, 1991.
Professional Experience: R & D assistant engineer of Shin-Kwong Synthetic Fibers Company in Taiwan, 1988. Employed as a teaching assistant at Oklahoma State University during the Fall semester of 1990 and Spring semester of 1991.