ABSTRACT
In this project we studied on manufacturing methanol. We made investigations for methanol manufacturing process. The flow chart for the process, reactor configurations are asked.Firstly, Using thermodynamic properties of C0,H2 and CH3OH, equilibrium line is plotted and 100 bar is selected. Constant rate curves are obtained by using reaction rate expression which is given at term project part.The operating lines goes through the extremums of these curves.Adiabatic energy balance equation for each reactor gives adiabatic line equations.These adiabatic line equations plotted until the reaching desired conversion of 0.55.This conversion is attained with 6 plug flow reactors.
Reactor Design Project
Methanol is considered as a potential source of energy and as an intermediate to producealternative motor vehicle fuels, fuel additives and number of petrochemicals. Conventionally,methanol is produced from synthesis gas (gas mixtureof carbon monoxide and hydrogenproduced by reforming of natural gas) in a series of fıxed bed catalytic reactors at a relatively high pressure. It is also possible to produce synthesis gas by gasification of biomass or coal.
2 3CO 2H CH OH+ →
In this project you are asked to design the reactor(s) to be used for a methanol productionat a rate of 400 tons/day:
(A) Seaıch the literature for methanol production.Discuss the operating conditions of theprocess and the critical points for pressure selection.
(B) Considering a feed composition of 30% CO and 70% H2 ,examine the thermodynamics ofmethanol synthesis reaction in order to decide on the operating pressure. Plot equilibriumconversion versus temperature graphs in a pressure range of 50-120 bars. In theequilibrium calculations you should use the fugacities.
(C) Taking the pressure as 100 bars and considering a target conversion of CO as 55% decideabout the reactor configurations, operation mode (adiabatic, non adiabatic, isothermal),reactor inlet temperature(s).
(D) Design the reactor(s) to find the catalyst volume and total reactor volume.
In methanol synthesis, a catalyst containing Cu/ZnO system, with the addition ofaluminum is generally used. You may use the rate expression given in the datapage.
DATA
For systems where the synthesis gas is composed of only CO and H2, the following rate equation was proposed for the hydrogenation of CO over a Cu/Zno on alumina catalyst for a temperaturen range between 450-650K and pressures 50-100 bar.
2 CH3OHCO H2
e
Pr k(P P )
K= − (mol / kgcat min)
6 1 1 3k 7.6 10 mol(kgcat) min atm− − − −= × at 250 C°E = 80 kJ 1mol− (activation energy)
112 2
e
90.13kJ.molK 3.567 10 exp( )atm
RT
−− −= ×
The catalytic packed bed bulk density was given as 1120 kg 3m−
2
1.INTRODUCTION
1.1 Methyl Alcohol as an Industrial Chemical
Methanol (methyl alcohol), CH3OH, is clear, water- white liquid with a mild odor at
ambient temperatures. From its discovery in the late 1600s, methanol has grown to become
the 21st largest commodity chemical with over 12x106 metric tons annually produced in the
world. Methanol has been called wood alcohol (or wood spirit) because it was obtained
commercially from the destructive distillation of wood for over a century. However, true
wood alcohol contained more contaminants (primarily acetone, acetic acid, and ally alcohol)
than the chemical- grade methanol avaible today.
Table 1.1 Physical Properties of Methanol
Property ValueFreezing point oC -97,68Boiling point oC 64,70Critical temperature oC 239,43Critical pressure kPa 8096Critical volume mL/mol 118Critical compressibility factor z in PV=znRT 0,224Heat of formation(liquid) at 25oC kj/mol -239,03Free energy of formation(liquid) at 25oC kj/mol -166,81Heat of fusion J/g 103Heat of vaporization at boiling point J/g 1129Heat of combustion at 25oC J/g 22662Flammable limits in air Lower, vol % Upper, vol %
636
Autoignition temperature, oC 420Flash point, closed cup, oC 12Surface tension, mN/m (dyn/cm) 22,6Specific heat of vapor at 25 oC, J/(g.K) of liquid at 25 oC, J/(g.K)
1,3702,533
Vapor pressure at 25 oC, kPa 16,96Solubility in water miscibleDensity at 25 oC, g/m3 0,78663refractive index, nD
20 1,3284viscosity of liquid at 25 oC, mPa.s(cP) 0,541Dielectric constant at 25 oC 32,7Thermal conductivity at 25 oC, W/(m.K) 0,202
For many years the largest use for methanol has been as a feedstock in the production
of formaldehyde, consuming almost half of the entire methanol produced. In the future,
3
formaldehyde’s importance to methanol will decrease as newer uses increase such as the
production of acetic acid and methyl tert-butyl ether (MTBE, a gasoline octane booster).
Methanol’s direct use as a fuel may be significant in special circumstances.
1.2 Manufacturing and Processing
Modern industrial- scale methanol production is based on exclusively on synthesis
from pressurized mixtures of hydrogen, carbon monoxide, and carbon dioxide gases in the
presence of metallic heterogeneous catalysts. The required synthesis pressure is dependent on
the activity of the particular catalyst. By convention, technology is generally distinguished by
pressure as follows; lower pressure processes, 5-10 MPa (50-100atm); medium pressure
processes, 10-25 MPa (100-250 atm); and high pressure processes, 25-35 MPa (250-350 atm).
[1]
In the late 1960a medium and low pressure methanol technology came into use with
the successful development of highly active, durable copper-zinc oxide catalysts. Copper
catalysts’ sensitivity to poisons required careful purification of feed streams. Low and
medium pressure technology has advantages of reduces compression power, good catalyst
life, larger capacity single- train converter designs and milder operating pressures.
Some reactions rate expressions uses for methanol production is listed on appendix
D.1
1.3 Natural Gas
• Hydrocracking of heavy hydrocarbons:
CnH(2n+2) + (n-1)H2 nCH4
• Steam reforming of CH4:
CH4 + H2O CO +3H2
• Water gas shift:
CO + H2O CO2+H2
For low pressure catalysts, the excess hydrogen improves the catalyst effectiveness.
Thus, converter costs are reduced and the necessity of shifting and removing excess hydrogen
from the synthesis feed gas, as commonly practiced with high pressure technology, is
avoided. Excess hydrogen is vented during synthesis and used as fuel in the reforming step.
Thus, a high overall energy efficiency is mainted which makes the process economical. [1]
4
Table 1.2 Equilibrium CO, CO2 Conversion, and Exit CH3OH Concentration vs Pressure and
Temperature
1.4 Catalyst
Methanol, an important industrial chemical is produced on a large scale so called “low
pressure” (50-100 bar) process. The formation of methanol is catalyzed by Cu-Zn-Al or Cu-
Zn-Cr mixed oxides important design factors in modeling a methanol reactor are the values
of equilibrium constants of the following reaction. [2]
CO+2H2 CH3OH
CO2+H2 CO+H2O
Catalyst used in high pressure (25-35 MPa or 250-350atm) synthesis is zinc oxide-
chromium oxide. It is a more robust catalyst than the low pressure copper-based catalyst and
can tolerate higher temperature and sulfur levels. The copper- zinc oxide catalyst, However, is
more attractive and can be operated at lower pressure (5-25 MPa or 50-250 atm) and
temperature (200-300 C). [1]
1.5 Low Pressure Processes
A more active catalyst than the above can be made from a combination of copper and
zinc together with a textural promoter such as chromia or alumina. These permit the use of a
lower pressure in the range of about 5 to 10 MPa, and a temperature of about 240 to 260
centigrade degrees. Recent laboratory studies indicate that the active phase is a solution of Cu
in ZnO and that methanol yield are increased by the presence of CO2, H2O or O2 in the
synthesis gas. If none of these is present, the catalyst gradually loses activity, since the Cu-
ZnO phase apparently may be gradually reduced to inactive copper metal. This process is
irreversible once the crystallites of copper metal have grown. The fact that the copper
produces a chemical effect rather than a physical effect is also shown by the fact that this
catalyst exhibits considerably lower apparent activation energy than the Zno-Cr2O3 catalyst.
Low pressure process utilizes a single bed of catalyst and quench cooling, obtained by
Temperature,oC
CO conversion, %5MPa 10MPa 30MPa
CO2 conversion, %5MPa 10MPa 30MPa
Exit CH3OH, vol %5MPa 10MPa 30MPa
200 95.6 99.0 99.9 44.1 82.5 99.0 27.8 37.6 42.3250 72.1 90.9 98.9 18.0 46.2 91.0 16.2 26.5 39.7300 25.7 60.6 92.8 14.3 24.6 71.1 5.6 14.2 32.2350 -2.3 16.9 73.0 19.8 23.6 52.1 1.3 4.8 21.7400 -12.8 -7.2 38.1 27.9 30.1 44.2 0.3 1.4 11.4
5
lozenge distributors especially designed to obtain good gas distribution and gas mixing and to
permit rapid loading and unloading of catalyst. A low pressure methanol synthesis process is
advantageously combined with production of synthesis pressure, thus avoiding the necessity
of intermediate gas compression. These low pressure processes are usually the process of
choice in new installations.
To produce relatively pure methanol product directly requires care in catalyst
manufacture , and requires procedures to avoid catalyst contamination. [3].
1.5.1 Catalyst Characteristic
Zinc oxide serves several important functions that enhance the stability and life of
the catalysts.
• Its credited with an important role in the proprietary manufacturing produce that
creates a high- surface area of copper
• Along with alumina, it prevents copper agglomeration
• ZnO reacts readily with copper, poisons such as sulfur and chlorine compounds. [4]
1.5.2 Side Reactions
Prior to commercialization of the low-to-medium pressure process using copper
catalysts, the most troublesome side reaction was the reverse of thee steam reforming
reaction. Occurs in high pressure plants above 450 C and causes exit bed temperatures to
exceed 600 C. Such runaway temperatures usually require reactor shutdown to prevent
catalyst and equipment damage. The low pressure copper-based catalysts operate in a lower
temperature range, ie, 200-300 C , where the methanation reaction is unimportant.
Alcohols other than methanol are produced in small quantities with ethanol the chief
impurity. Formation of the higher alcohols can be suppressed by keeping the reaction
temperature as low as possible for the methanol production rate desired. High hydrogen
concentration also suppresses the formation of higher alcohols and the other by products.
Other by products produced is small amounts are aldehydes, ketones, ethers and esters. [1]
6
2.THERMODYNAMIC DATA
Table 2.1 Thermodynamics properties of methanol,carbon monoxide and hydrogen gaseous
Components 298°∆H (kj/mol) 298°∆G (kj/mol)
CH3OH (Methanol) -201,2 -162
H2 0 0
CO -110,52 -137,2( Referrence : Sandler , S.I, Chemical and Engineerig Thermodynamics, Third edition, John Wiley &
Sons Inc. , 1999 , page 759)
P,COC (T, K) = 27,113 + 0,655 X 210− T – 0,1 X 510− 2T [j /mol.K]
P,H2C (T,K) = 26,113 + 0,435 X 210− T – 0,033 X 510− 2T [j /mol.K]
P,CH3OHC (T,K) = 19,038 + 9,146 X 210− T – 1,218 X 510− 2T - 8,034 X 910− 3T [j /mol.K]
( Referrence : Sandler , S.I, Chemical and Engineerig Thermodynamics, Third edition, John Wiley &
Sons Inc. , 1999 , page 745-747)
Table 2.2 Critical tempertaure and pressure of substances
513.2 K 79.54 barCO 133 K 34.96 bar
2H 33.3 K 12.97 bar
7
3.CALCULATIONS
3.1. Obtaining equilibrium constant as a function of temperature, Kf (T) :
+ 298°∆H and 298°∆G values are given at thermodynamics data part.
298°∆H = (-201,2)-(-110,52)
= -90,68 kj/mol = -90680 j/mol
298°∆G = (-162,0)-(-137,2)
= -24,8 kj/mol = -24800j/mol
298 298
JG RT ln K 8,314 298K ln K
mol K∆ ° = − = − × ×
×=-24800 j/mol
298ln K = 10,0097 298K = 22243,38
2RT
H
dT
dInK f °∆= .... Van't Hoff Equation dTRT
HInKInKf
T
T
T ∫°∆+=
0
2)(
P P,CHOH P,H2 P,COC C 2 C C∆ = ∆ − ∆ − ∆
dTcHdT
T
T
T
p
R R
∫ ∫∆=∆
+ Cp values of substances are given at thermodynami properties part.
PC∆ = -60,301 + 7,621X 210− T – 1,052 X 510− 2T - 8,034 X 910− 3T (J/mol.K)
∆H = ∆H298 + ∫∆T
Cp298
dT
∆H = - 75985,54 + 2 6 3 9 4(60,301 T) (0,0381 T ) (3,5067 10 T ) (2,0085 10 T )− −− × + × − × × − × ×
8
t a np r o d u c t s r e a c t sH H∆ Η = ∆ − ∆∑ ∑
t a np r o d u c t s r e a c t sG G G∆ = ∆ − ∆∑ ∑
TT
f 2298298
1 H(T)dTd ln K
R T
∆=∫ ∫
R=8,3145 J/mol.K;
298ln K = 10,0097
(InKf)= 298 InK +
3 7 2 10 39139,46(9,32935 7,2529ln(T) 4,5826 10 T 4,2178 10 T 2,415 10 T )
T− − −+ − + × − × − ×
3 7 2 10 3f
9139,46K exp(19,33905 7,2529ln(T) 4,5826 10 T 4,2178 10 T 2,415 10 T )
T− − −= + − + × − × − ×
Table 3.1 Equilibrium constant versus temperature data
T Kf400 1,619805420 0,415906440 0,119779460 0,03814480 0,013267500 0,00499520 0,002012540 0,000863560 0,000392580 0,000187600 9,33E-05
9
3.2 Plotting equilibrium conversion versus temperature graphs in a pressure range of
50-120 bars
nT yK K K P∆
Φ= × × f / PΦ = ( fugacity coefficient)
Basis: 100 moles/s Feed composition enter the reactor
OHCHHCO 322 ⇔+
CO A→ , 2H B→ , 3CH OH C→
A 2B C+ ⇔
30 70 -a -2a a---- ---- ----30-a 70-2a a
nT = 30 – a + 70 – 2a + a = 100 – 2a CA0 = 30 (due to basis 100 moles reactant)
a=CA0 . XAe a = 30 XAe
2. BA yy
ycKy =
( )ic
2ia ib
K.
φθ=
θ θ 2)211( −=−−=∆n
+ With changing the operating pressure, reduced pressure (Pr) and temperature (Tr) values be
changed . So, fugacities of substances might be changed. According to this change, obtained
different equilibrium conversion( AeX ) versus temperature functions by pressures.
r CT T / T=
r CP P / P=
Reduced temperature and pressure values of substances are shown on Appendix A.1
+ At P = 50 bar, temperature range of 400-600 K ;
Table 3.2 Reduced Pressures at P=50 bar
Pr(metanol) Pr(CO) Pr(H2)0,628614534 1,43020595 3,855050116
10
Fugacity coefficient of substances are read on a graph by parameters Tr,Pr,θ shown on
Appendix A.1
Table 3.3 Equilibrium constant and Xae values for temperature range of 400-600K, P=50bar
T Kf K(fugacity coefficient) Ky Xae400 1,619805 0,15049 26909,64 0,99940420 0,415906 0,23764 4375,284 0,99650440 0,119779 0,35310 848,0635 0,98380460 0,03814 0,49655 192,0247 0,95751480 0,013267 0,70088 47,32129 0,86850500 0,00499 0,74634 16,71539 0,76720520 0,002012 0,78482 6,409849 0,63150540 0,000863 0,81485 2,649156 0,47240560 0,000392 0,84074 1,164987 0,31600580 0,000187 0,86378 0,541032 0,19120600 9,33E-05 0,88197 0,264582 0,10870
Sample calculation : At T=400K (P=50bar)
( )
Ae
2Ae icf 2 2
ia ibAe Ae
Ae Ae
30X
100-60XK 50
.30-30X 70 60X
100 60X 100 60X
−θ= × ×θ θ −× − −
Ae
2Ae2 2
Ae Ae
Ae Ae
30X
100-60X 0,15381,619805 50
(1,017) 1,01230-30X 70 60X
100 60X 100 60X
−= × ×× −× − −
26909,64 =
Ae
Ae2
Ae Ae
Ae Ae
30X
100-60X
30-30X 70 60X
100 60X 100 60X
−× − −
An equation Ky = f( AeX ) such as:
Ae
Aey 2
Ae Ae
Ae Ae
30X
100-60XK
30-30X 70 60X
100 60X 100 60X
= −× − −
We calculate Xae value using a matlab function (See Appendix A.2)
11
By using other Ky values on this matlab function, we get the Xae values on Table 3.3
+ At P = 75 bar, temperature range of 400-600 K ;
Table 3.4 Reduced Pressures at P=75 bar
Pr(metanol) Pr(CO) Pr(H2)0,9429218 2,145308924 5,782575173
Fugacity coefficient of substances are read on a graph by parameters Tr,Pr,θ shown on
Appendix A.1
Table 3.5 Equilibrium constant and Xae values for temperature range of 400-600K, P=75bar
T KfK(fugacity coefficient) Ky Xae
400 1,619805 0,092810483 98172,15 0,9998420 0,415906 0,156250745 14972,54 0,9989440 0,119779 0,231447055 2911,074 0,9948460 0,03814 0,325141594 659,8286 0,9845480 0,013267 0,434626865 171,6992 0,9427500 0,00499 0,557073141 50,38741 0,8733520 0,002012 0,642367583 17,62038 0,7735540 0,000863 0,685524544 7,085048 0,6476560 0,000392 0,722445487 3,050439 0,4993580 0,000187 0,752806781 1,396763 0,3496600 9,33E-05 0,778071256 0,6748 0,2237
Sample calculation : At T=400K (P=75bar)
( )
Ae
2Ae icf 2 2
ia ibAe Ae
Ae Ae
30X
100-60XK 75
.30-30X 70 60X
100 60X 100 60X
−θ= × ×θ θ −× − −
Ae
2Ae2
Ae Ae
Ae Ae
30X
100-60X1,619805 0,09281 75
30-30X 70 60X
100 60X 100 60X
−= × × −× − −
98172,15 =
Ae
Ae2
Ae Ae
Ae Ae
30X
100-60X
30-30X 70 60X
100 60X 100 60X
−× − −
An equation Ky = f( AeX ) such as:
12
Ae
Aey 2
Ae Ae
Ae Ae
30X
100-60XK
30-30X 70 60X
100 60X 100 60X
= −× − −
We calculate Xae value using a matlab function (See Appendix A.2)
By using other Ky values on this matlab function, we get the Xae values on Table 3.5
+ At P = 120 bar, temperature range of 400-600 K ;
Table 3.6 Reduced Pressures at P=120 bar
Pr(metanol) Pr(CO) Pr(H2)1,508674881 3,432494279 9,252120278
Table 3.7 Equilibrium constant and Xae values for temperature range of 400-600K, P=120bar
T KfK(fugacity coefficient) Ky Xae
400 1,619805 0,063209575 369013,7 0,99999420 0,415906 0,099519313 60179,7 0,9997440 0,119779 0,14732625 11707,52 0,9987460 0,03814 0,207338687 2648,886 0,9943480 0,013267 0,278338222 686,3599 0,9806500 0,00499 0,359381778 199,9485 0,9486520 0,002012 0,44432146 65,21412 0,8915540 0,000863 0,523088292 23,77009 0,8065560 0,000392 0,581157042 9,707646 0,6956580 0,000187 0,627223183 4,291649 0,5624600 9,33E-05 0,665798469 2,018792 0,42
Sample calculation : At T=400K (P=120bar)
( )
Ae
2Ae icf 2 2
ia ibAe Ae
Ae Ae
30X
100-60XK 120
.30-30X 70 60X
100 60X 100 60X
−θ= × ×θ θ −× − −
Ae
2Ae2
Ae Ae
Ae Ae
30X
100-60X1,619805 0,063209575 75
30-30X 70 60X
100 60X 100 60X
−= × × −× − −
13
369013,7 =
Ae
Ae2
Ae Ae
Ae Ae
30X
100-60X
30-30X 70 60X
100 60X 100 60X
−× − −
An equation Ky = f( AeX ) such as:
Ae
Aey 2
Ae Ae
Ae Ae
30X
100-60XK
30-30X 70 60X
100 60X 100 60X
= −× − −
We calculate Xae value using a matlab function (See Appendix A.2)
By using other Ky values on this matlab function, we get the Xae values on Table 3.7
Finaly, Xae – T graph plotted on graph 3.1 at selected pressure of 50,75 and 120 bar.
Figure 3.1 Xae versus Temperature(K) graph for methanol production at given conditions
14
3.3 Drawing constant rate curves to get operating line
The rate expression which will be used for obtaining kinetic data is given in the
reactor desing project part as :
2 CH3OHCO H2
e
Pr k(P P )
K= − ;
Dalton’s law presents an expression about relation between ya (molar fraction) and
pressure as A TP P ya= × . Total amount of A in the total mixture A T(ya F / F )= can also be
defined.
A AA
T T
F Py
F P= =
AA T
T
FP P
F= × …(3.1);
A 2B C+ ⇔ Fa0 Fb0
-Fa0Xa -2Fa0Xa Fa0Xa
------------------------------------
A A0 A0 AF F F X= −
B B0 A0 AF F 2F X= −
C A0 AF F X=
T A0 A B0F F (1 2X ) F= − +
Using formula 3.1 and expressions above, these are derived :
rar
1=
−; r=-ra ; A A 0F C V= × ;
A0 AA 0
A
C (1 X )F V
1 X
−= ×+ ∈ ;
B0 A0 AB 0
A
(C 2C X )F V
1 X
−= ×+ ∈ ;
A0 AC 0
A
C XF V
1 X= ×
+ ∈ ;
T A B CF F F F= + +
A0 B0 A0 AT 0
A
C C 2C XF V
1 X
+ −= ×+ ∈ ;
15
A0 A0
A AA
A0 B0 A0 AA0
A
C (1 X )V
1 X (1 X )P 100 100
C C 2C X 7(1 2X )V
31 X
− ×+ ∈ −= × = ×+ − − +×+ ∈
(Carbon Monoxide)
B0 A0 A0 A
AB
A0 B0 A0 AA0
A
(C 2C X ) 7V 2X1 X 3P 100 100C C 2C X 7
(1 2X )V31 X
− × −+ ∈= × = ×+ − − +×+ ∈
(Hydrogen gas)
A0 A0
A AC
A0 B0 A0 AA0
A
C XV
1 X XP 100 100
C C 2C X 7(1 2X )V
31 X
×+ ∈= × = ×+ − − +×+ ∈
(Methanol)
All of the partial pressure expressions’ numerators and denominators are divided by A0C ,
B0 A0(C / C 70 / 30)= The rate expression is obtained as a function of temperatures(T) and
molar fractions (Xa).
A
A A2A
A A
X( 100)
7 72X (1 2X )(1 X ) 3 3ra (k) (( 100) ( 100) )
7 7 3,567E 12 exp(90130 / 8,314 T)(1 2X ) (1 2X )3 3
×− − +−− = × × × × −
− × ×− + − +
k 743,198 exp(-80000 / (8,3145 T))= × ×
The constant rate curves are drawen by cooperation with an C#.NET program and
Excel . (See Appendix B.1 for C# program)
Constant rate curves are drawen for the vaules of r; 0; 0,05, 0,1, 0,35, 0,5, 1, 2, 5 and 8
Kinetics and thermodynamics equilibrium lines (r=0) are shown on figure 3.2
16
Figure 3.2 Equilibrium lines from kinetics and thermodynamics
Table 3.1 Constant rates T,Xa data
r=0,1 r=0,35 r=0,5 r=0T Xa T Xa T Xa T Xa462 0 492 0,01 502 0,04 400 0,98463 0,03 494 0,08 506 0,16 401 0,98465 0,12 498 0,2 507 0,19 408 0,97466 0,15 505 0,35 515 0,35 409 0,97467 0,18 507 0,38 517 0,38 410 0,97475 0,38 509 0,41 523 0,45 411 0,97476 0,4 512 0,45 524 0,46 434 0,95478 0,43 513 0,46 525 0,47 435 0,95479 0,45 514 0,47 536 0,54 436 0,95481 0,48 522 0,54 539 0,55 437 0,95482 0,49 530 0,58 543 0,56 504 0,83502 0,66 539 0,6 558 0,56 505 0,83504 0,67 548 0,6 567 0,54 506 0,83553 0,64 559 0,58 581 0,49 508 0,82554 0,64 583 0,49 586 0,47 533 0,74555 0,63 598 0,42 588 0,46 538 0,72559 0,62 600 0,41 597 0,42 548 0,68561 0,61 602 0,4 599 0,41 598 0,43567 0,58 604 0,39 618 0,32 600 0,42569 0,57 606 0,38 627 0,28 660 0,16613 0,35 663 0,15 696 0,08667 0,14
Other constant rate datas given at appendix B.2.
17
Figure 3.3 Constant rate curves at 100 bar
3.4 Energy Balance
Inlet stream : FCO, F 2H
Outlet stream : F 3CH OH , FCO, F 2H
General Energy Balance equation :
TR Tf
i Pi i Pi Removedbythewalls R A0 Ai(inlet ) i(outlet )
T0 TR
FC dT FC dT Q ( H )F XΣ + Σ − = − ∆∫ ∫Flow reactors are used for methanol production.At PFR reactors adiabatic operations
are easier to control than isothermal operations. So, heat lost by the system is neglected.
Cp(T) functions listed on thermodynamics data chapter and R 298H |∆ ° is calculated in
calculations 3.1.
18
3.4.1 Adiabatic lines to calculate number of reactors to achieve 0.55 conversion of A
Taken basis 100 mol/s feed composition
T0 22
2g 28gF 100mol / s 70molH 30molCO 980g / s
molH molCO= = + = ;
A0F 30mol / s= ;
methanolM 32g / mol=
(CH3OH) (CH3OH)0 (CO)0 COF F F X 0 30 0,55 16,5mol / s= + × = + × = methanol = 528 g/s methanol
Daily production = 6
528g (60 60 24)s 1ton
s 1day 10 g
× ×× × = 45,61 ton/day methanol
Figure 3.4.1 Adiabatic lines and number of reactors
According to figure 3.4 six plug flow reactor must be used to achive 0,55 conversion at the exit.
Figure 3.4.2 6 PFR Reactors
19
3.5 Reactor Volumes Calculation
3.5.1 Reactor 1
Energy balance for reactor 1:
TR Tf
R byflows A0 PA B0 PB A PA B PB B PB
T0 TR
TR Tf
R byflows A0 PA B0 PB A0 PA A0 A1 PA B0 PB A0 A1 PB A0 A1 PC
T0 TR
Q (F C F C )dT (F C F C F C )dT
Q (F C F C )dT (F C F X C F C 2F X C F X C )dT
= + + + +
= + + − + − +
∫ ∫
∫ ∫Tf
A0 PA B0 PB
T0A1 Tf
A0 PA A0 PB A0 PC R A0
TR
(F C F C )dT
X
(F C 2F C F C )dT ( H )F
+=
+ − + −∆
∫
∫Tf Tf
B0 B0PA PB PA PB
A0 A0T0 T0A1 Tf
R TPA PB PC R
TR
F F(C C )dT (C C )dT
F FX
( H | )(C 2C C )dT ( H )
+ += =
− ∆+ − + −∆
∫ ∫
∫
2 2 6 3 3f 0 f 0 f 0
A1 2 2 6 3 3 9 4 40 0 0 0
88,0433(T T ) 0,0167(T T ) 1,77 10 (T T )X
90680 (60,301(298 T ) 0,038105(298 T ) 3,507 10 ( 298 T ) 2,0085 10 (298 T ))
−
− −
− + − − × −=+ − − − + × − − × −
….. 3.5.1
0,15931 A
Ao Ao
V dX
F R= ∫ ………… 3.5.2
3.5.1 and 3.5.2 solved simultaneously.
A
A A2A
A A
X( 100)
7 72X (1 2X )(1 X ) 3 3ra (k) (( 100) ( 100) )
7 7 3,567E 12 exp(90130 / 8,314 T)(1 2X ) (1 2X )3 3
×− − +−− = × × × × −
− × ×− + − +
k 743,198 exp(-80000 / (8,3145 T))= × ×
20
From adiabatic line equation and –ra equation, data on table 3.5.1 obtained
Table 3.5.1 Reactor-1 data
To Tf Xa k Pco Ph2 Pch3oh Ke (-ra) 1/-ra
490 490 0 2,20397E-06 30 70 0 0,014456 0,323983 3,086582
490 500 0,009866 3,2641E-06 29,8809 69,82135 0,297754 0,009287 0,475376 2,103597
490 510 0,019696 4,76028E-06 29,76082 69,64123 0,597958 0,006071 0,686615 1,456419
490 520 0,029492 6,84225E-06 29,63972 69,45958 0,900698 0,004034 0,97692 1,023625
490 530 0,039255 9,70108E-06 29,51757 69,27636 1,206064 0,002722 1,369968 0,729944
490 540 0,048988 1,35777E-05 29,39434 69,09151 1,514147 0,001864 1,894161 0,527938
490 550 0,058692 1,87725E-05 29,26998 68,90498 1,825039 0,001294 2,582348 0,387244
490 560 0,06837 2,56562E-05 29,14447 68,7167 2,138834 0,00091 3,470503 0,288143
490 570 0,078022 3,46819E-05 29,01775 68,52662 2,45563 0,000648 4,594486 0,217652
490 580 0,087652 4,6398E-05 28,88979 68,33469 2,775525 0,000467 5,983472 0,167127
490 590 0,09726 6,14627E-05 28,76055 68,14083 3,098619 0,00034 7,647774 0,130757
490 600 0,106848 8,06591E-05 28,62999 67,94499 3,425015 0,00025 9,557504 0,10463
490 610 0,116418 0,000104912 28,49807 67,74711 3,754818 0,000186 11,60663 0,086158
490 620 0,125971 0,000135305 28,36475 67,54712 4,088137 0,00014 13,55422 0,073778
490 630 0,13551 0,000173099 28,22997 67,34495 4,42508 0,000106 14,93089 0,066975
490 640 0,145035 0,000219751 28,0937 67,14054 4,765762 8,1E-05 14,89287 0,067146
490 650 0,154548 0,000276936 27,95588 66,93382 5,110296 6,24E-05 11,99905 0,08334
490 655 0,1593 0,000310066 27,88638 66,82957 5,284045 5,49E-05 8,788086 0,11379
By -1/ra vs Xa data on table 3.5.1 excel regression gives this equation
-1/ra=y(Xa)= (-157200*x^5) + (85543*x^4) - (18291*x^3) + (1979.3*x^2) - (114.38*x) + (3.0754);
Using simpson’s integration rule the equation above is integrated by simpson matlab
function (See Appendix C.1 for simpson fuction and reactor volume functions)
Simpson('reactorvolume1',0,0.1593,1000) gives
V1/Fa0 = 0.0879V1 = 30*0,0879 = 2,637
3.5.1 Reactor 2
0,28632 A
A20 A0,1593
V dX
F R= ∫
Inlet outlet
A : FA1=FA0-FA0*XA1 A : FA2 = FA0 – FA0XA2
B: FB1=FB0-2*FA0*XA1 B : FB2 = FB0 – 2*FA0*XA2
C: FC1=FA0*XA1 C : FC2 = FA0*XA2
21
TR Tf
i Pi i Pi Removedbythewalls R A0 Ai(inlet) i(outlet)
T0 TR
FC dT FC dT Q ( H )F XΣ + Σ − = − ∆∫ ∫
TR TR TR TR TR Tf
A0 PA A0 A1 PA B0 PB A0 A1 PB A0 A1 PC A0 PA
T0 T0 T0 T0 T0 TR
F C dT F X C dT F C dT 2 F X C dT F X C dT F C dT− + − + +∫ ∫ ∫ ∫ ∫ ∫TF TF TF TF
A0 A2 PA B0 PB A0 A2 PB A0 A2 PC R A0 A2
TR TR TR TR
F X C dT F C dT 2 F X C dT F X C dT ( H )F X− + − + = −∆∫ ∫ ∫ ∫Tf TR TR TR
B0PA PB A1 PB PC PA
A0T0 T0 T0 T0A2
R T
F(C C )dT X ( 2C dT C dT C )
FX
( H | )
+ − × − +=
− ∆
∫ ∫ ∫ ∫
A = TR TR TR
PB PC PA
T0 T0 T0
( 2C dT C dT C )− +∫ ∫ ∫2 2 6 3 3 9 4 4
0 0 0 0A (60,301(298 T ) 0,038105(298 T ) 3,507 10 (298 T ) 2,0085 10 (298 T ))− −= − − − + × − − × −
2 2 6 3 3f 0 f 0 f 0 A1
A2 2 2 6 3 3 9 4 40 0 0 0
88,0433(T T ) 0,0167(T T ) 1,77 10 (T T ) X AX
90680 (60,301(298 T ) 0,038105(298 T ) 3,507 10 ( 298 T ) 2,0085 10 (298 T ))
−
− −
− + − − × − + ×=+ − − − + × − − × −
Table 3.5.2 Reactor-2 data
To Tf Xa k Pco Ph2 Pch3oh Ke (-ra) 1/-ra
497 4970,18332
2,90614E-06
27,52828
66,29243
6,179289
0,010586
0,349883 2,8581
497 5070,194659
4,25746E-06
27,35519
66,03278
6,612037
0,006885
0,50373
1,985189
497 5170,206016
6,14567E-06
27,17912
65,76868
7,052193
0,004553
0,712989
1,402546
497 5270,21739
8,74859E-06
27,00002
65,50003
7,499943
0,003058
0,991956
1,008109
497 5370,228781
1,22912E-05
26,81781
65,22671
7,955483
0,002085
1,355496
0,737737
497 5470,24019 1,7055E-05
26,63239
64,94859
8,419014
0,001442
1,816435
0,550529
497 5570,251617
2,33885E-05
26,4437
64,66555
8,890746
0,00101
2,380387 0,4201
497 5670,263061
3,17187E-05
26,25164
64,37746
9,370894
0,000717
3,036186
0,329361
497 5770,274522
4,25638E-05
26,05613
64,08419
9,859683
0,000515
3,73896
0,267454
497 5870,286001
5,65477E-05
25,85706
63,78559
10,35734
0,000374
4,381223
0,228247
497 5970,297496
7,44142E-05
25,65435
63,48153
10,86412
0,000274
4,744836
0,210755
497 6070,30901
9,70439E-05
25,4479
63,17185
11,38025
0,000203
4,423031
0,226089
497 617 0,3205 0,0001254 25,237 62,856 11,906 0,0001 2,6964 0,3708
22
4 71 6 4 52 41 59
By -1/ra vs Xa data on table 3.5.2 excel regression gives this equation
-1/ra=y(Xa)= (38246*x^4) - (36734*x^3) + (13296*x^2) - (2154.8*x) + (132.5);
Simpson('reactorvolume2',0.1593,0.2863,1000)
V2/Fa20 = 0,0889
A20 A0 A1F F (1 X )= − =30*(1-0,1593) = 25,221 mol/s
V2 = 25,221 * 0,0889 = 2,242
Other Reactors’ volume calculation shown on Appendix C.3
v1,v2,v3,v4,v5 and v6 values are not in unit of volume.
2 CH3OHCO H2
e
Pr k(P P )
K= − (mol / kgcat min)
6 1 1 3k 7.6 10 mol(kgcat) min atm− − − −= × at 250 C°
A
A0
dXV1
F [mol / s] ra[mol / kgcat min]=
− ×∫After making unit correction (seconds convert to minute), kgcat unit is obtained.
Table 3.5.3 Catalyst uses
Reactors Calculated Catalyst mass(kg)
Reactor 1 2,637 158,22
Reactor 2 2,242 134,52
Reactor 3 1,394 83,64
Reactor 4 1,6075 96,45
Reactor 5 1,267 76,02
Reactor 6 0,501 30,06
Total catalyst mass 662.43
23
3 3
3
662,43kgcat 1m catalyst 1m reactor
1 1120kgcat (1 0.6)m catalyst× ×
−=1,4786 m
Table 3.5.4 Reactor volumes
Reactors Volume(lt)
Reactor 1 353,16
Reactor 2 300,26
Reactor 3 186,69
Reactor 4 215.29
Reactor 5 169,68
Reactor 6 67
Total volume 1478,6 lt
24
4.RESULTS & DISCUSSIONS
Firstly, thermodynamics equilibrium line is drawn by using van’t hoff equation for
different pressures (figure 3.1). Plot shows that methanol production is rising by pressure
increasing. But, at very high pressures catalyst lifetime is decreasing and also reactor material
may not resist the high pressures. So, Achieving 0.55 converion of methanol, pressure of 100
bar is selected.
Constant rate curves drawn as figure 4.1 by using –ra=f(Xa,T) formula. The line goes
through the maximum points of these curves is operating line.To close to operating line, first
reactor inlet temperature is selected 490. If more less T0 value is selected, the number of
reactors should be decreased. But, according to the rate equation(the function of temperature
and conversion); reaction rate is decreasing with temperature decreasing. If,T0 value is very
high, the reactor number will increase. So, optimum tempretature should be selected.
Figure 4.1 Adiabatic lines and number of reactors (Xa(y-axis) vs T(x-axis))
Xa=f(T) function is obtained by using adiabatic energy balance. Using these lineer
functions, adiabatic lines for each reactor are drawn(figure 4.1).
–ra=f(T,xa) is also function of temperature.adiabatic line and rate equations are solved
simultaneously and Xa vs -1/ra data are obtained.These data plotted on figure 4.2.Each reactor
25
functions is monitorized with excel.And using these functions in a simpson’s integration rule
matlab function, areas under the curves are calculated.These results equals to Vi/Fa0(i)
Fa0 unit is selected as [mol/s]. It is converted to mol/min. –ra unit is [mol/kgcat*min].
After doing these unit conversions, areas under the curves(at figure 4.2) gives the result
[kgcat]. The reactor volume results are obtained by dividing the kgcat results by catalyst
density and a volume conversion factor (volume of catalyst to volume of reactor).(catalyst
void volume is selected 0,6).
Figure 4.2 -1/ra(y axis) vs Xa(x axis)
26
Appendix - A.1
Table App.A.1.1 Reduced temperatures of substances
T,K Tr(metanol) Tr(CO) Tr(H2)400 0,77942323 3,007518797 12,01201201420 0,81839439 3,157894737 12,61261261440 0,85736555 3,308270677 13,21321321460 0,89633671 3,458646617 13,81381381480 0,93530787 3,609022556 14,41441441500 0,97427903 3,759398496 15,01501502520 1,01325019 3,909774436 15,61561562540 1,05222136 4,060150376 16,21621622560 1,09119252 4,210526316 16,81681682580 1,13016368 4,360902256 17,41741742600 1,16913484 4,511278195 18,01801802
Table App.A.1.2 Fugacity coefficients of substances at P = 50 bar
T,K θ (CO) θ (methanol) θ (H2)
400 1,012 0,1538 1,017
420 1,014 0,2424 1,017
440 1,015 0,3591 1,016
460 1,017 0,504 1,016
480 1,018 0,7093 1,015
500 1,018 0,7553 1,015
520 1,019 0,7919 1,014
540 1,019 0,8222 1,014
560 1,02 0,8475 1,014
580 1,02 0,869 1,013
600 1,02 0,8873 1,013
Table App.A.1.3 Fugacity coefficients of substances at P = 75 bar
T θ (CO) θ (methanol) θ (H2)
400 1,0900 0,1067 1,0270420 1,0220 0,1681 1,0260440 1,0240 0,2490 1,0250460 1,0260 0,3498 1,0240480 1,0280 0,4685 1,0240500 1,0290 0,5999 1,0230
27
520 1,0290 0,6904 1,0220540 1,0300 0,7375 1,0220560 1,0300 0,7757 1,0210580 1,0310 0,8075 1,0200600 1,0310 0,8346 1,0200
Table App.A.1.4 Fugacity coefficients of substances at P = 120 bar
T θ (CO) θ (methanol) θ (H2)
400 1,03400 0,07151 1,04600420 1,03900 0,11270 1,04400440 1,04200 0,16700 1,04300460 1,04500 0,23480 1,04100480 1,04700 0,31520 1,04000500 1,04800 0,40580 1,03800520 1,05000 0,50170 1,03700540 1,05000 0,58950 1,03600560 1,05100 0,65430 1,03500580 1,05100 0,70480 1,03400600 1,05100 0,74670 1,03300
Appendix – A.2
Ae
Aey 2
Ae Ae
Ae Ae
30X
100-60XK
30-30X 70 60X
100 60X 100 60X
= −× − −
By expanding this equation ;3 2
Ae y Ae y y Ae y y yX ( 36 36 K ) X (120 84 K 36 K ) X ( 100 49 K 84 K ) 49 K 0× − − × + × + × + × + × − − × − × + × =
For Solving this equation on MATLAB R2007a, we defined these parameters:
a = y( 36 36 K )− − ×
b = y y(120 84 K 36 K )+ × + ×
c = y y( 100 49 K 84 K )− − × − ×
d = y49 K×
a,b,c,d are polynomial coefficients and there is a function on matlab to find roots of high order
functions by using these polynomial coefficients.
M.File of Matlab is ;
function a=c(A)a = -36 - (36 * A);
28
b = 120 + (84 * A) + (36 * A);c = -100 - (49 * A) - (84 * A);d = 49*A;p=[a b c d];a = roots(p);
and we call the function on command window like ‘ c(Ky) ‘
Example :
c (31217.42) gives the result below
ans =
1.1669 + 0.0075i
1.1669 - 0.0075i
0.9995
Our Xae value is 0,995. Other roots are imaginer and not validating Xae.(Xae must be
0<Xae<1 and real )
By using other Ky values on this matlab function, we get the Xae values on Table 3.3, 3.5,
3.7.
29
Appendix B.1
The rate expression is derived as a function of T and Xa.
A
A A2A
A A
X( 100)
7 72X (1 2X )(1 X ) 3 3r (k) (( 100) ( 100) )
7 7 3,567E 12 exp(90130 /8,314 T)(1 2X ) (1 2X )3 3
×− − +−= ρ× × × × × −
− × ×− + − +
The execution of program is based on scanning T and Xa values at the range of
(400<T<650 T+=0.1) and (0<Xa<1 Xa+=0.01). In the program constant r value is given from
textbox and the program calculate and r value from T and Xa values. If r(given from textbox)
– r(calculated from equation) = 0 (almost 0, its sensivity differs by T and r to obtain more
data).
The variables are defined as;
k = 743.198 * (Math.Exp(-80000 / (8.3145 * T)));Pco = ((1 - Xa)/((10/3)-(2*Xa)))*100;Ph2 = (((7/3)-(2*Xa))/((10/3)-(2*Xa)))*100;Pch3oh = ((Xa)/((10/3)-(2*Xa)))*100;Ke = (0.000000000003567) * ((Math.Exp((90130) / (8.3145 * T))));ra = (k * ro * ((Pco * (Ph2 * Ph2)) - (Pch3oh / Ke)));
Main code block of this program is :
for (T = 400; T < 700; T += 1) { for (Xa = 0; Xa < 1; Xa += 0.01) { k = 743.198 * (Math.Exp(-80000 / (8.3145 * T))); Pco = ((1 - Xa)/((10/3)-(2*Xa)))*100; Ph2 = (((7/3)-(2*Xa))/((10/3)-(2*Xa)))*100; Pch3oh = ((Xa)/((10/3)-(2*Xa)))*100; Ke = (0.000000000003567) * ((Math.Exp((90130) / (8.3145 * T)))); ra = (k * ((Pco * (Ph2 * Ph2)) - (Pch3oh / Ke)));
if (r <= 0.03 && (ra - r) < 0.000015&& (ra - r) > -0.000015) { listBox1.Items.Add(T.ToString() + " " + Xa.ToString() + " " + ra.ToString()); listBox2.Items.Add(T.ToString()); listBox3.Items.Add(Xa.ToString()); } if (T < 470 && r > 0.03 && r <= 0.15 && (ra - r) < 0.00015 && (ra - r) > -0.00015) { listBox1.Items.Add(T.ToString() + " " + Xa.ToString() + " " + ra.ToString()); listBox2.Items.Add(T.ToString());
30
listBox3.Items.Add(Xa.ToString()); } if (T<550 && T > 470 && r > 0.03 && r <= 0.15 && (ra - r) < 0.001 && (ra - r) > -0.001) { listBox1.Items.Add(T.ToString() + " " + Xa.ToString() + " " + ra.ToString()); listBox2.Items.Add(T.ToString()); listBox3.Items.Add(Xa.ToString()); }
if (T > 550 && r > 0.03 && r <= 0.15 && (ra - r) < 0.03 && (ra - r) > -0.03) { listBox1.Items.Add(T.ToString() + " " + Xa.ToString() + " " + ra.ToString()); listBox2.Items.Add(T.ToString()); listBox3.Items.Add(Xa.ToString()); }...
//(other if else loops exist here for other range of T for varying (r-ra) sensibilities.These code block written to get uniform distributed data at temperature range 400 – 700 K)
else { continue; }
All code block and .exe file of this program is given at CD-ROM (attached to report).
Figure A.3 A screenshot from constant rate program
31
Appendix B.2
Table B.2 constant rate data
r=1 r=2 r=5 r=8T Xa T Xa T Xa T Xa520 0,01 541 0,03 569 0 590 0,07523 0,1 542 0,06 570 0,02 597 0,14524 0,13 542 0,06 571 0,04 600 0,16532 0,29 543 0,08 572 0,06 604 0,18534 0,32 543 0,09 579 0,16 607 0,19538 0,37 544 0,11 580 0,17 627 0,2539 0,38 545 0,13 581 0,18540 0,39 545 0,13 582 0,19541 0,4 546 0,15 586 0,22541 0,4 546 0,15 596 0,26542 0,41 547 0,17 614 0,26546 0,44 547 0,17 624 0,24553 0,47 548 0,19 632 0,22553 0,47 549 0,2 643 0,19556 0,48 550 0,22573 0,48 550 0,22577 0,47 552 0,25581 0,46 553 0,26590 0,43 560 0,33605 0,37 561 0,34621 0,3 561 0,34659 0,16 564 0,36679 0,11 566 0,37
569 0,38569 0,38572 0,39573 0,39588 0,39588 0,39593 0,38597 0,37597 0,37615 0,31
For more little scattered constant rate data by different values of r are available in the
Excel file (constant rate curces at 100 bar.xls) on CD-ROM.
32
Appendix C.1
Simpson’s integration M.File on matlab:
function I=Simpson(f,a,b,n)%f nin integrali simpson kuralı% n çift sayı olacakh=(b-a)/n;S= feval(f,a);for i=1:2:n-1 x(i)=a+h*i; S=S+4*feval(f,x(i));endfor i=2:2:n-2 x(i)=a+h*i; S=S+2*feval(f,x(i));endS=S+feval(f,b);I=h*S/3;The function is called from command window like:
V1/Fa01=Simpson(‘reactorvolume1’,0,0.1563,1000)
‘reactorvolume1’ = is another m.file which includes function of -1/ra by Xa;
0 = Xa0;
0.1563 = Xa1;
1000 = Diveding factor (if it is larger, result of the numerical result approaches analytical
result )
Appendix C.2
M.Files of reactor volume equations (-1/ra function by Xa) :
Reactor 1:
function y1=reactorvolume1(x)y1=(-157200*x^5) + (85543*x^4) - (18291*x^3) + (1979.3*x^2) - (114.38*x) + (3.0754);Reactor 2:
function y2=reactorvolume2(x)y2=(38246*x^4) - (36734*x^3) + (13296*x^2) - (2154.8*x) + (132.5);Reactor 3:
function y3=reactorvolume3(x)y3=(65582*x^4) - (91531*x^3) + (48084*x^2) - (11278*x) + (997.57);Reactor 4:
function y4=reactorvolume4(x)y4=(119805*x^4) - (207253*x^3) + (134786*x^2) - (39071*x) + (4261.7);Reactor 5:
33
function y5=reactorvolume5(x)y5=(177966*x^4) - (355243*x^3) + (266463*x^2) - (89032*x) + (11184);Reactor6:
function y6=reactorvolume6(x)y6=(366299*x^4) - (785097*x^3) + (631702*x^2) - (226167*x) + (30405);
Appendix C.3
C.3.1 Reactor 3
0,38633 A
A30 A0,2863
V dX
F R= ∫
Inlet outlet
A : FA2=FA0-FA0*XA2 A : FA3 = FA0 – FA0XA3
B: FB2=FB0-2*FA0*XA2 B : FB3 = FB0 – 2*FA0*XA3
C: FC2=FA0*XA2 C : FC3 = FA0*XA3
TR Tf
i Pi i Pi Removedbythewalls R A0 Ai(inlet) i(outlet)
T0 TR
FC dT FC dT Q ( H )F XΣ + Σ − = − ∆∫ ∫
TR TR TR TR TR Tf
A0 PA A0 A2 PA B0 PB A0 A2 PB A0 A2 PC A0 PA
T0 T0 T0 T0 T0 TR
F C dT F X C dT F C dT 2 F X C dT F X C dT F C dT− + − + +∫ ∫ ∫ ∫ ∫ ∫TF TF TF TF
A0 A3 PA B0 PB A0 A3 PB A0 A3 PC R A0 A3
TR TR TR TR
F X C dT F C dT 2 F X C dT F X C dT ( H )F X− + − + = −∆∫ ∫ ∫ ∫Tf TR TR TR
B0PA PB A2 PB PC PA
A0T0 T0 T0 T0A3
R T
F(C C )dT X ( 2C dT C dT C )
FX
( H | )
+ − × − +=
− ∆
∫ ∫ ∫ ∫
A = TR TR TR
PB PC PA
T0 T0 T0
( 2C dT C dT C )− +∫ ∫ ∫2 2 6 3 3 9 4 4
0 0 0 0A (60,301(298 T ) 0,038105(298 T ) 3,507 10 (298 T ) 2,0085 10 (298 T ))− −= − − − + × − − × −
2 2 6 3 3f 0 f 0 f 0 A2
A3 2 2 6 3 3 9 4 40 0 0 0
88,0433(T T ) 0,0167(T T ) 1,77 10 (T T ) X AX
90680 (60,301(298 T ) 0,038105(298 T ) 3,507 10 ( 298 T ) 2,0085 10 (298 T ))
−
− −
− + − − × − − ×=+ − − − + × − − × −
34
Table C.3.1 Reactor-3 data
To Tf Xa k Pco Ph2 Pch3oh Ke (-ra) 1/-ra
502 502 0,2863 3,52419E-06 25,85183 63,77774 10,37043 0,008519 0,366295 2,730037
502 512 0,2963 5,12434E-06 25,67561 63,51341 10,81098 0,005587 0,520835 1,919994
502 522 0,3063 7,34494E-06 25,4968 63,2452 11,25799 0,003724 0,72688 1,375743
502 532 0,3163 1,03863E-05 25,31535 62,97302 11,71163 0,002521 0,994431 1,005601
502 542 0,3263 1,45004E-05 25,13118 62,69677 12,17204 0,001731 1,330492 0,751602
502 552 0,3363 2,00009E-05 24,94425 62,41638 12,63937 0,001205 1,733817 0,576762
502 562 0,3463 2,72739E-05 24,75449 62,13173 13,11378 0,000849 2,185292 0,457605
502 572 0,3563 3,67905E-05 24,56183 61,84274 13,59543 0,000606 2,631077 0,380073
502 582 0,3663 4,91199E-05 24,3662 61,54931 14,08449 0,000438 2,95395 0,33853
502 592 0,3763 6,4944E-05 24,16755 61,25132 14,58113 0,00032 2,925888 0,341777
502 602 0,3863 8,50728E-05 23,96579 60,94869 15,08552 0,000236 2,131355 0,469185
By -1/ra vs Xa data on table C.3.1 excel regression gives this equation
-1/ra=y(Xa)= (65582*x^4) - (91531*x^3) + (48084*x^2) - (11278*x) + (997.57);
Simpson('reactorvolume3',0.2863,0.3863,1000)gives
V3/Fa30 = 0.0757
A30 A0 A3F F (1 X )= − =30*(1-0,3863) = 18,411 mol/s
V3 = 18,411 * 0.0757= 1,394
Reactor 4
0,46634 A
A40 A0,3863
V dX
F R= ∫
Tf TR TR TRB0
PA PB A3 PB PC PAA0T0 T0 T0 T0
A4R T
F(C C )dT X ( 2C dT C dT C )
FX
( H | )
+ − × − +=
− ∆
∫ ∫ ∫ ∫
A = TR TR TR
PB PC PA
T0 T0 T0
( 2C dT C dT C )− +∫ ∫ ∫2 2 6 3 3 9 4 4
0 0 0 0A (60,301(298 T ) 0,038105(298 T ) 3,507 10 (298 T ) 2,0085 10 (298 T ))− −= − − − + × − − × −
2 2 6 3 3f 0 f 0 f 0 A3
A4 2 2 6 3 3 9 4 40 0 0 0
(88,0433(T T ) 0,0167(T T ) 1,77 10 (T T )) X AX
90680 (60,301(298 T ) 0,038105(298 T ) 3,507 10 (298 T ) 2,0085 10 (298 T ))
−
− −
− + − − × − − ×=+ − − − + × − − × −
35
Table C.3.2 Reactor-4 data
To Tf Xa k Pco Ph2 Pch3oh Ke (-ra) 1/-ra
507 507 0,3863 4,25746E-06 23,96579 60,94869 15,08552 0,006885 0,369699 2,704904
507 517 0,3963 6,14567E-06 23,76086 60,64128 15,59786 0,004553 0,515938 1,938218
507 527 0,4063 8,74859E-06 23,55267 60,329 16,11833 0,003058 0,703841 1,420776
507 537 0,4163 1,22912E-05 23,34115 60,01173 16,64712 0,002085 0,935085 1,069422
507 547 0,4263 1,7055E-05 23,12623 59,68934 17,18443 0,001442 1,201953 0,831979
507 557 0,4363 2,33885E-05 22,90781 59,36171 17,73049 0,00101 1,477438 0,676848
507 567 0,4463 3,17187E-05 22,6858 59,02871 18,28549 0,000717 1,697902 0,588962
507 577 0,4563 4,25638E-05 22,46014 58,6902 18,84966 0,000515 1,733567 0,576845
507 587 0,4663 5,65477E-05 22,23071 58,34606 19,42323 0,000374 1,339535 0,746528
507 587 0,4663 5,65477E-05 22,23071 58,34606 19,42323 0,000374 1,339535 0,746528
By -1/ra vs Xa data on table C.3.2 excel regression gives this equation
-1/ra=y(Xa)= (119805*x^4) - (207253*x^3) + (134786*x^2) - (39071*x) + (4261.7);Simpson('reactorvolume4',0.3863,0.4663,1000)gives
V4/Fa40 = 0.1004
A40 A0 A4F F (1 X )= − =30*(1-0,4663) =16,011 mol/s
V4 = 16,011 * 0,1004= 1,6075
Reactor 5
0,52435 A
A50 A0,4663
V dX
F R= ∫ … C.5.1
2 2 6 3 3f 0 f 0 f 0 A 4
A 5 2 2 6 3 3 9 4 40 0 0 0
(88, 0433(T T ) 0, 0167(T T ) 1, 77 10 (T T )) X AX
90680 (60, 301(298 T ) 0, 038105(298 T ) 3, 507 10 (298 T ) 2, 0085 10 (298 T ))
−
− −
− + − − × − − ×=+ − − − + × − − × −
…C.5.2 Solving simultaneously equation C.5.1 and C.5.2
Table C.3.3 Reactor-5 data
To Tf Xa k Pco Ph2 Pch3oh Ke (-ra) 1/-ra
514 514 0,4663 5,51309E-06 22,23071 58,34606 19,42323 0,005146 0,396415 2,522606
514 519 0,4713 6,60261E-06 22,11455 58,17183 19,71362 0,0042 0,463111 2,159311
514 524 0,4763 7,88028E-06 21,99742 57,99614 20,00644 0,003441 0,537237 1,861376
514 529 0,4813 9,3738E-06 21,87931 57,81896 20,30174 0,00283 0,618375 1,61714
514 534 0,4863 1,11142E-05 21,76019 57,64028 20,59953 0,002336 0,705488 1,417458
514 539 0,4913 1,31362E-05 21,64006 57,46008 20,89986 0,001935 0,79665 1,255256
514 544 0,4963 1,54783E-05 21,5189 57,27835 21,20276 0,001608 0,888692 1,12525
36
514 549 0,5013 1,81837E-05 21,3967 57,09505 21,50825 0,001341 0,976726 1,023828
514 554 0,5063 2,12999E-05 21,27345 56,91017 21,81638 0,001122 1,053527 0,949192
514 559 0,5113 2,48796E-05 21,14913 56,7237 22,12717 0,000942 1,108712 0,901947
514 564 0,5163 2,8981E-05 21,02373 56,5356 22,44067 0,000793 1,127685 0,886772
514 572 0,5243 3,67905E-05 20,82081 56,23122 22,94797 0,000606 1,029673 0,971182
By -1/ra vs Xa data on table C.3.3 excel regression gives this equation
-1/ra=y(Xa)= (177966*x^4) - (355243*x^3) + (266463*x^2) - (89032*x) + (11184);
Simpson('reactorvolume5',0.4663,0.5243,1000) gives
V5/Fa50 = 0.0888
A50 A0 A5F F (1 X )= − =30*(1-0,5243) = 14,271 mol/s
V5 = 14,271 * 0,0888 = 1,267
Reactor 6
0,55336 A
A60 A0,5243
V dX
F R= ∫ ….C.6.1
2 2 6 3 3f 0 f 0 f 0 A4
A5 2 2 6 3 3 9 4 40 0 0 0
(88,0433(T T ) 0,0167(T T ) 1,77 10 (T T )) X AX
90680 (60,301(298 T ) 0,038105(298 T ) 3,507 10 (298 T ) 2,0085 10 (298 T ))
−
− −
− + − − × − − ×=+ − − − + × − − × −
…C.6.2
Solving simultaneously equation C.6.1 and C.6.2
Table C.3.4 Reactor-6 data
To Tf Xa k Pco Ph2 Pch3oh Ke (-ra) 1/-ra
533 533 0,5243 1,07448E-05 20,82081 56,23122 22,94797 0,002426 0,605749 1,650848
533 537 0,5283 1,22912E-05 20,71828 56,07742 23,2043 0,002085 0,664023 1,505973
533 542 0,5333 1,45004E-05 20,5891 55,88365 23,52725 0,001731 0,73527 1,360045
533 547 0,5383 1,7055E-05 20,45878 55,68816 23,85306 0,001442 0,7999 1,250157
533 552 0,5433 2,00009E-05 20,32729 55,49093 24,18178 0,001205 0,850469 1,175822
533 557 0,5483 2,33885E-05 20,19463 55,29194 24,51343 0,00101 0,876376 1,141063
533 562 0,5533 2,72739E-05 20,06078 55,09116 24,84806 0,000849 0,862807 1,159007
By -1/ra vs Xa data on table C.3.4 excel regretion gives this equation
-1/ra=y(Xa)= (366299*x^4) - (785097*x^3) + (631702*x^2) - (226167*x) + (30405)
Simpson('reactorvolume6',0.5243,0.5533,1000) gives
V6/Fa60 = 0.0374
A60 A0 A6F F (1 X )= − =30 *(1-0,5533) = 13,401 mol/s
37
V6 = 13,401 * 0,0374 = 0,501
Appendix D.1
Rate equation uses for methanol productions[7]
38
References
1- Wade L.E. Gengelbach, R.B. Taumbley, J.L. Hallhover W.L. Kırk-Other Encyclopedia of
Chemical Technology, 3rd. Edition, Wiley New York 1981 Vol 15 page 398-415
2- G.H. Graaf, Sıytsema P.J.J., Stamhuıs E.J., Joosten G.E.H. Chem. Eng.Sci. Vol 41, 11,
page 2883 (1986)
3- Satterfield N.D., Heterogeneous Catalysis in Practice, McGraw Hill, 1980
4- Howard F. Rase, Handbook of Commercial Catalysts, Crc Pres,2000, page 429-430
5- Sandler , S.I, Chemical and Engineerig Thermodynamics, Third edition, John Wiley &
Sons Inc. , 1999 , page 759
6- Smith, J.M., VanNess, H.C., “Introduction to Chemical Engineering Thermodynamics”, 3rd
Ed., McGrow Hill, Newyork, 1996.
7- http://www.rajwantbedi.com/dg1_final.pdf
39
SYMBOLS
Cpi : Heat capasity , J/mol.K
θ : Fugacity coefficient
298°∆G : Standart Gibbs energies of formation , J/mol
298°∆H : Standart Enthalpies , J/mol
fK : Equilibrium constant
yK : Equilibrium constant (molar fractions)
Kθ : Equilibrium constant(fugacity coefficients)
rT : Reduced Temperature
rP : Reduced Pressure
P : Pressure , bar
R : Ideal gas law constant , 8,314 J/mol.K
T : Temperature , K
XA : Fractional conversion of Carbon monoxide
AeX : Equilibrium conversion of Carbon monoxide
40
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