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 produce alternative motor vehicle fuels, fuel additives and number of petrochemicals. Conventionally, methanol is produced from synthesis gas (gas mixtureof carbon monoxide and hydrogen produced 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.
CO + 2H 2 → CH3 OH In this project you are asked to design the reactor(s) to be used for a methanol production at a rate of 400 tons/day: (A) Seaıch the literature for methanol production.Discuss the operating conditions of the process and the critical points for pressure selection. (B) Considering a feed composition of 30% CO and 70% H2 ,examine the thermodynamics of methanol synthesis reaction in order to decide on the operating pressure. Plot equilibrium conversion versus temperature graphs in a pressure range of 50-120 bars. In the equilibrium calculations you should use the fugacities. (C) Taking the pressure as 100 bars and considering a target conversion of CO as 55% decide about 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 of aluminum 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. r = k(PCO P 2H 2 −
PCH3OH ) (mol / kgcat min) Ke
k = 7.6 × 10−6 mol(kgcat)−1 min−1 atm−3 at 250 °C E = 80 kJ mol −1 (activation energy) 90.13kJ.mol−1 −12 K e = 3.567 × 10 exp( )atm−2 RT The catalytic packed bed bulk density was given as 1120 kg m −3
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 Freezing point oC Boiling point oC Critical temperature oC Critical pressure kPa Critical volume mL/mol Critical compressibility factor z in PV=znRT Heat of formation(liquid) at 25oC kj/mol Free energy of formation(liquid) at 25oC kj/mol Heat of fusion J/g Heat of vaporization at boiling point J/g Heat of combustion at 25oC J/g Flammable limits in air Lower, vol % Upper, vol % Autoignition temperature, oC Flash point, closed cup, oC Surface tension, mN/m (dyn/cm) Specific heat of vapor at 25 oC, J/(g.K) of liquid at 25 oC, J/(g.K) Vapor pressure at 25 oC, kPa Solubility in water Density at 25 oC, g/m3 refractive index, nD20 viscosity of liquid at 25 oC, mPa.s(cP) Dielectric constant at 25 oC Thermal conductivity at 25 oC, W/(m.K)
Value -97,68 64,70 239,43 8096 118 0,224 -239,03 -166,81 103 1129 22662 6 36 420 12 22,6 1,370 2,533 16,96 miscible 0,78663 1,3284 0,541 32,7 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
•
Steam reforming of CH4: CH4 + H2O
•
nCH4
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 Temperature , o C 200 250 300 350 400
CO conversion, % 5MPa 10MPa 30MPa
CO2 conversion, % 5MPa 10MPa 30MPa
Exit CH3OH, vol % 5MPa 10MPa 30MPa
95.6 72.1 25.7 -2.3 -12.8
44.1 18.0 14.3 19.8 27.9
27.8 16.2 5.6 1.3 0.3
99.0 90.9 60.6 16.9 -7.2
99.9 98.9 92.8 73.0 38.1
82.5 46.2 24.6 23.6 30.1
99.0 91.0 71.1 52.1 44.2
37.6 26.5 14.2 4.8 1.4
42.3 39.7 32.2 21.7 11.4
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 CuZn-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 oxidechromium 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 CuZnO 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 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
∆H °298 (kj/mol)
∆G °298 (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)
C P,CO (T, K) = 27,113 + 0,655 X 10−2 T – 0,1 X 10−5 T 2 [j /mol.K] C P,H 2 (T, K) = 26,113 + 0,435 X 10−2 T – 0,033 X 10−5 T 2 [j /mol.K] C P,CH3OH (T, K) = 19,038 + 9,146 X 10−2 T – 1,218 X 10−5 T 2 - 8,034 X 10−9 T 3 [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
CO
H2
513.2 K 133 K 33.3 K
79.54 bar 34.96 bar 12.97 bar
7
3.CALCULATIONS 3.1. Obtaining equilibrium constant as a function of temperature, Kf (T) : + ∆H °298 and ∆G°298 values are given at thermodynamics data part.
∆ Η
∑=
H
∆
p
r o d
u
−∑
c t s
t a n
∆
r e a
Hc
t s
∆H °298 = (-201,2)-(-110,52)
= -90,68 kj/mol = -90680 j/mol
∆G
∑=
∆
p
r o
d
u
c t s
G
−∑
t a n
∆
r e a
c
t s
G
∆G °298 = (-162,0)-(-137,2)
= -24,8 kj/mol = -24800j/mol ∆G° = −RT ln K 298 = −8,314
J × 298K × ln K298 =-24800 j/mol mol × K
ln K 298 = 10,0097 K 298 = 22243,38
dInK f ∆ H ° = dT R T2
T
∆H ° dT 2 T 0 RT
.... Van't Hoff Equation ( InKf )T = InK + ∫
∆C P = ∆CP,CHOH − 2∆CP,H 2 − ∆CP,CO
T
T
TR
TR
∫ d∆H = ∫ ∆cpdT
+ Cp values of substances are given at thermodynami properties part. ∆C P = -60,301 + 7,621X 10−2 T – 1,052 X 10−5 T 2 - 8,034 X 10−9 T 3 (J/mol.K)
∆H = ∆H298 +
T
∫∆Cp
dT
298
∆H = - 75985,54 + −(60,301× T) + (0, 0381× T 2 ) − (3,5067 × 10−6 × T3 ) − (2, 0085 ×10−9 × T4 )
8
T
∫
d ln K f =
298
1 T ∆H(T)dT R ∫298 T2
R=8,3145 J/mol.K; ln K 298 = 10,0097
(InKf)= InK 298 + (9,32935 +
9139, 46 − 7, 2529 ln(T) + 4,5826 × 10−3 T − 4, 2178 × 10−7 T2 − 2, 415 × 10−10 T3 ) T
K f = exp(19,33905 +
9139, 46 − 7, 2529 ln(T) + 4,5826 ×10−3 T − 4, 2178 ×10−7 T2 − 2, 415 ×10−10 T3 ) T
Table 3.1 Equilibrium constant versus temperature data T 400 420 440 460 480 500 520 540 560 580 600
Kf 1,619805 0,415906 0,119779 0,03814 0,013267 0,00499 0,002012 0,000863 0,000392 0,000187 9,33E-05
9
3.2 Plotting equilibrium conversion versus temperature graphs in a pressure range of 50-120 bars K T = K Φ × K y × P ∆n
Φ = f / P ( fugacity coefficient)
Basis: 100 moles/s Feed composition enter the reactor CO + 2 H 2 ⇔ CH 3OH
H2 → B ,
CO → A ,
CH3OH → C
A + 2B ⇔ C 30 70 -a -2a a ---- ---- ---30-a 70-2a a nT = 30 – a + 70 – 2a + a = 100 – 2a a=CA0 . XAe
Ky = Kφ =
CA0 = 30 (due to basis 100 moles reactant)
a = 30 XAe
yc y A .yB
2
θic ( θia .θib 2 )
∆n = (1 −1 − 2) = −2
+ 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( X Ae ) versus temperature functions by pressures. Tr = T / TC Pr = P / PC 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) 0,628614534 1,43020595
Pr(H2) 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 400 420 440 460 480 500 520 540 560 580 600
Kf 1,619805 0,415906 0,119779 0,03814 0,013267 0,00499 0,002012 0,000863 0,000392 0,000187 9,33E-05
K(fugacity coefficient) 0,15049 0,23764 0,35310 0,49655 0,70088 0,74634 0,78482 0,81485 0,84074 0,86378 0,88197
Ky 26909,64 4375,284 848,0635 192,0247 47,32129 16,71539 6,409849 2,649156 1,164987 0,541032 0,264582
Sample calculation : At T=400K (P=50bar)
Kf =
30X Ae 100-60X Ae 30-30X Ae 100 − 60X Ae
1, 619805 =
26909,64 =
70 − 60XAe × 100 − 60XAe
2
×
θic × 50−2 2 ( θia .θib )
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
×
0,1538 × 50−2 (1, 017) 2 ×1, 012
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
An equation Ky = f( X Ae ) such as:
Ky =
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
We calculate Xae value using a matlab function (See Appendix A.2) 11
Xae 0,99940 0,99650 0,98380 0,95751 0,86850 0,76720 0,63150 0,47240 0,31600 0,19120 0,10870
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) 0,9429218 2,145308924
Pr(H2) 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 400 420 440 460 480 500 520 540 560 580 600
K(fugacity coefficient) 0,092810483 0,156250745 0,231447055 0,325141594 0,434626865 0,557073141 0,642367583 0,685524544 0,722445487 0,752806781 0,778071256
Kf 1,619805 0,415906 0,119779 0,03814 0,013267 0,00499 0,002012 0,000863 0,000392 0,000187 9,33E-05
Sample calculation : At T=400K (P=75bar)
Kf =
30X Ae 100-60X Ae 30-30X Ae 100 − 60X Ae
1,619805 =
98172,15 =
70 − 60XAe × 100 − 60XAe
2
×
θic × 75−2 2 ( θia .θib )
30X Ae 100-60X Ae 30-30X Ae 100 − 60X Ae
70 − 60XAe × 100 − 60XAe
2
× 0, 09281× 75−2
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
An equation Ky = f( X Ae ) such as:
12
Ky 98172,15 14972,54 2911,074 659,8286 171,6992 50,38741 17,62038 7,085048 3,050439 1,396763 0,6748
Xae 0,9998 0,9989 0,9948 0,9845 0,9427 0,8733 0,7735 0,6476 0,4993 0,3496 0,2237
Ky =
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
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) 1,508674881 3,432494279
Pr(H2) 9,252120278
Table 3.7 Equilibrium constant and Xae values for temperature range of 400-600K, P=120bar T 400 420 440 460 480 500 520 540 560 580 600
K(fugacity coefficient) 0,063209575 0,099519313 0,14732625 0,207338687 0,278338222 0,359381778 0,44432146 0,523088292 0,581157042 0,627223183 0,665798469
Kf 1,619805 0,415906 0,119779 0,03814 0,013267 0,00499 0,002012 0,000863 0,000392 0,000187 9,33E-05
Ky 369013,7 60179,7 11707,52 2648,886 686,3599 199,9485 65,21412 23,77009 9,707646 4,291649 2,018792
Sample calculation : At T=400K (P=120bar)
Kf =
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
1,619805 =
2
×
θic × 120−2 2 ( θia .θib )
30X Ae 100-60X Ae 30-30X Ae 100 − 60X Ae
70 − 60XAe × 100 − 60XAe
2
× 0,063209575 × 75−2
13
Xae 0,99999 0,9997 0,9987 0,9943 0,9806 0,9486 0,8915 0,8065 0,6956 0,5624 0,42
369013,7 =
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
An equation Ky = f( X Ae ) such as:
Ky =
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
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 : r = k(PCO P 2H 2 −
PCH3OH ); Ke
Dalton’s law presents an expression about relation between ya (molar fraction) and pressure as PA = PT × ya . Total amount of A in the total mixture (ya = FA / FT ) can also be defined. yA =
FA PA = FT PT
PA =
FA × PT …(3.1); FT
A + 2B ⇔ C Fa0
Fb0
-Fa0Xa -2Fa0Xa
Fa0Xa
------------------------------------
FA = FA0 − FA0 XA FB = FB0 − 2FA0 XA FC = FA0 X A FT = FA0 (1 − 2XA ) + FB0 Using formula 3.1 and expressions above, these are derived : r=
ra ; r=-ra ; FA = CA × V0 ; −1
FA =
CA0 (1 − X A ) × V0 ; 1+ ∈ X A
FB =
(C B0 − 2CA0 XA ) × V0 ; 1+ ∈ X A
FC =
C A0 X A × V0 ; 1+ ∈ X A
FT = FA + FB + FC
FT =
CA 0 + CB0 − 2CA0 XA × V0 ; 1+ ∈ X A
15
CA 0 (1 − X A ) × V0 1+ ∈ X A (1 − XA ) PA = × 100 = ×100 (Carbon Monoxide) CA 0 + CB0 − 2CA0 XA 7 (1 − 2X A ) + × V0 3 1+ ∈ X A (CB0 − 2CA0 XA ) 7 × V0 − 2X A 1+ ∈ X A PB = ×100 = 3 × 100 (Hydrogen gas) C A0 + CB0 − 2CA0 XA 7 (1 − 2X A ) + × V0 3 1+ ∈ X A CA0 X A × V0 1+ ∈ X A XA PC = ×100 = × 100 (Methanol) C A0 + CB0 − 2CA0 XA 7 (1 − 2X A ) + × V0 3 1+ ∈ X A All of the partial pressure expressions’ numerators and denominators are divided by C A0 , (C B0 / CA0 = 70 / 30) The rate expression is obtained as a function of temperatures(T) and molar fractions (Xa). XA
× 100) 7 (1 − 2XA ) + (1 − X A ) 3 −ra = (k) × (( ×100) × ( ×100)2 − ) 7 7 3,567E − 12 × exp(90130 / 8,314 × T) (1 − 2X A ) + (1 − 2XA ) + 3 3 (
7 − 2X A 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 T 462 463 465 466 467 475 476 478 479 481 482 502 504 553 554 555 559 561 567 569 613 667
Xa 0 0,03 0,12 0,15 0,18 0,38 0,4 0,43 0,45 0,48 0,49 0,66 0,67 0,64 0,64 0,63 0,62 0,61 0,58 0,57 0,35 0,14
r=0,35 T 492 494 498 505 507 509 512 513 514 522 530 539 548 559 583 598 600 602 604 606
Xa 0,01 0,08 0,2 0,35 0,38 0,41 0,45 0,46 0,47 0,54 0,58 0,6 0,6 0,58 0,49 0,42 0,41 0,4 0,39 0,38
r=0,5 T 502 506 507 515 517 523 524 525 536 539 543 558 567 581 586 588 597 599 618 627 663
Other constant rate datas given at appendix B.2. 17
Xa 0,04 0,16 0,19 0,35 0,38 0,45 0,46 0,47 0,54 0,55 0,56 0,56 0,54 0,49 0,47 0,46 0,42 0,41 0,32 0,28 0,15
r=0 T 400 401 408 409 410 411 434 435 436 437 504 505 506 508 533 538 548 598 600 660 696
Xa 0,98 0,98 0,97 0,97 0,97 0,97 0,95 0,95 0,95 0,95 0,83 0,83 0,83 0,82 0,74 0,72 0,68 0,43 0,42 0,16 0,08
Figure 3.3 Constant rate curves at 100 bar 3.4 Energy Balance Inlet stream : FCO, F H 2 Outlet stream : F CH 3OH , FCO, F H 2 General Energy Balance equation : TR
∫
T0
Tf
Σ FC i Pi dT +
i(inlet )
∫
TR
Σ FC i Pi dT − QRemovedbythewalls = −( ∆HR )FA0 XA
i(outlet )
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 ∆H R ° |298 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 FT0 = 100mol / s = 70molH2
2g 28g + 30molCO = 980g / s ; molH 2 molCO
FA0 = 30mol / s ; M methanol = 32g / mol F(CH3OH ) = F(CH3OH)0 + F(CO)0 × XCO = 0 + 30 × 0,55 = 16,5mol / s methanol = 528 g/s methanol Daily production =
528g (60 × 60 × 24)s 1ton × × 6 = 45,61 ton/day methanol s 1day 10 g
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
∫
Q R byflows =
Tf
(FA0C PA + FB0C PB)dT +
T0 TR
Q R byflows =
∫
∫ (F C A
PA
+ F BC PB + F BC PB)dT
TR Tf
(FA0C PA + FB0C PB)dT +
T0
∫ (F
C PA − F A 0X A1C PA + F B0C PB − 2F A 0X A1C PB + F A0X A1C PC)dT
A0
TR
Tf
∫ (F
A0
X A1 =
CPA + FB0 CPB )dT
T0 Tf
∫ (F
A0
CPA + 2FA0 CPB − FA 0 CPC )dT + (−∆HR )FA0
TR
Tf
Tf
F ∫T0 (CPA + FA0B0 CPB )dT
X A1 =
∫ (C
= T0
Tf
∫ (C
PA
FB0 CPB )dT FA0 −(∆H R |T )
PA
+ 2CPB − CPC )dT + (−∆HR )
+
TR
X A1 =
88, 0433(Tf − T0 ) + 0, 0167(Tf 2 − T0 2 ) − 1, 77 ×10 −6 (Tf 3 − T0 3) 90680 + (60,301(298 − T0 ) − 0, 038105(298 2 − T0 2) + 3,507 ×10 −6( 2983 − T0 3 )− 2, 0085× 10−9 (2984 − T0 4 ))
….. 3.5.1
V1 = FAo
0,1593
∫ o
dX A ………… 3.5.2 RA
3.5.1 and 3.5.2 solved simultaneously. XA
× 100) 7 (1 − 2XA ) + (1 − X A ) 3 −ra = (k) × (( ×100) × ( ×100)2 − ) 7 7 3,567E − 12 × exp(90130 / 8, 314 × T) (1 − 2X A ) + (1 − 2XA ) + 3 3 (
7 − 2X A 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 = 0.0879 V1 = 30*0,0879 = 2,637 3.5.1 Reactor 2 0,2863
V2 dX A = ∫ FA 20 0,1593 R A 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
∫
Σ FC i Pi dT +
i(inlet)
T0
TR
∫
∫
TR
Σ FC i Pi dT − QRe movedbythewalls = − (∆HR )FA0 XA
i(outlet)
TR
FA0C PA dT −
T0
∫ FA0 XA1CPA dT +
T0
TF
− ∫ FA0 X A 2 CPA dT + TR
Tf
XA2 =
∫ (CPA +
T0
TR
∫
TR
TR
∫ FB0 CPB dT − 2 ∫ FA0 XA1 CPB dT +
T0
T0
TF
FB0 CPB dT − 2 ∫ FA0 XA2 CPB dT + TR
TR
TR
∫ FA0 XA1 CPC dT +
T0
Tf
∫F
A0
CPA dT
TR
TF
∫F
A0
XA 2 CPC dT = (−∆HR )FA0 XA2
TR
TR
TR
FB0 CPB )dT − XA1 × ( ∫ 2CPB dT − ∫ CPC dT + ∫ CPA ) FA0 T0 T0 T0 −(∆H R |T )
A = ( ∫ 2CPB dT − T0
TF
TR
TR
∫
TR
CPC dT +
T0
∫C
PA
)
T0
A = (60,301(298 − T0 ) − 0, 038105(2982 − T02 ) + 3,507 × 10− 6 (2983 − T03) −2, 0085 ×10 − 9(298 4 −T 04))
X A2 =
88, 0433(Tf − T0) + 0, 0167(T f 2 − T 0 2) − 1,77 × 10 −6(T f 3− T 0 3) + X A1× A 90680 + (60,301(298 − T 0) − 0,038105(298 2− T 0 2) + 3,507× 10 −6( 298−3 T0 3−) 2, 0085 × 10− 9 (298−4 T0 4))
Table 3.5.2 Reactor-2 data To
Tf
497
497
497
507
497
517
497
527
497
537
497
547
497
557
497
567
497
577
497
587
497
597
497 497
607 617
Xa 0,1833 2 0,1946 59 0,2060 16 0,2173 9 0,2287 81 0,2401 9 0,2516 17 0,2630 61 0,2745 22 0,2860 01 0,2974 96 0,3090 1 0,3205
k 2,90614E06 4,25746E06 6,14567E06 8,74859E06 1,22912E05
Pco 27,528 28 27,355 19 27,179 12 27,000 02 26,817 81 26,632 1,7055E-05 39 2,33885E- 26,443 05 7 3,17187E- 26,251 05 64 4,25638E- 26,056 05 13 5,65477E- 25,857 05 06 7,44142E- 25,654 05 35 9,70439E- 25,447 05 9 0,0001254 25,237
22
Ph2 66,292 43 66,032 78 65,768 68 65,500 03 65,226 71 64,948 59 64,665 55 64,377 46 64,084 19 63,785 59 63,481 53 63,171 85 62,856
Pch3oh 6,1792 89 6,6120 37 7,0521 93 7,4999 43 7,9554 83 8,4190 14 8,8907 46 9,3708 94 9,8596 83 10,357 34 10,864 12 11,380 25 11,906
Ke 0,0105 86 0,0068 85 0,0045 53 0,0030 58 0,0020 85 0,0014 42 0,0010 1 0,0007 17 0,0005 15 0,0003 74 0,0002 74 0,0002 03 0,0001
(-ra) 0,3498 83 0,5037 3 0,7129 89 0,9919 56 1,3554 96 1,8164 35 2,3803 87 3,0361 86 3,7389 6 4,3812 23 4,7448 36 4,4230 31 2,6964
1/-ra 2,8581 1,9851 89 1,4025 46 1,0081 09 0,7377 37 0,5505 29 0,4201 0,3293 61 0,2674 54 0,2282 47 0,2107 55 0,2260 89 0,3708
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 FA 20 = FA0 (1 − XA1 ) =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. r = k(PCO P 2H 2 −
PCH3OH ) (mol / kgcat min) Ke
k = 7.6 × 10−6 mol(kgcat)−1 min−1 atm−3
at 250 °C
V1 dX A =∫ FA0 [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
662, 43kgcat 1m3 catalyst 1m3 reactor × × =1,4786 m 1 1120kgcat (1 − 0.6)m3 catalyst
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 400 420 440 460 480 500 520 540 560 580 600
Tr(metanol) 0,77942323 0,81839439 0,85736555 0,89633671 0,93530787 0,97427903 1,01325019 1,05222136 1,09119252 1,13016368 1,16913484
Tr(CO) 3,007518797 3,157894737 3,308270677 3,458646617 3,609022556 3,759398496 3,909774436 4,060150376 4,210526316 4,360902256 4,511278195
Tr(H2) 12,01201201 12,61261261 13,21321321 13,81381381 14,41441441 15,01501502 15,61561562 16,21621622 16,81681682 17,41741742 18,01801802
Table App.A.1.2 Fugacity coefficients of substances at P = 50 bar T,K 400 420 440 460 480 500 520 540 560 580 600
θ (CO)
θ (methanol)
1,012 1,014 1,015 1,017 1,018 1,018 1,019 1,019 1,02 1,02 1,02
0,1538 0,2424 0,3591 0,504 0,7093 0,7553 0,7919 0,8222 0,8475 0,869 0,8873
θ (H2) 1,017 1,017 1,016 1,016 1,015 1,015 1,014 1,014 1,014 1,013 1,013
Table App.A.1.3 Fugacity coefficients of substances at P = 75 bar T 400 420 440 460 480 500
θ (CO)
θ (methanol)
θ (H2)
1,0900 1,0220 1,0240 1,0260 1,0280 1,0290
0,1067 0,1681 0,2490 0,3498 0,4685 0,5999
1,0270 1,0260 1,0250 1,0240 1,0240 1,0230
27
520 540 560 580 600
1,0290 1,0300 1,0300 1,0310 1,0310
0,6904 0,7375 0,7757 0,8075 0,8346
1,0220 1,0220 1,0210 1,0200 1,0200
Table App.A.1.4 Fugacity coefficients of substances at P = 120 bar T 400 420 440 460 480 500 520 540 560 580 600
θ (CO) 1,03400 1,03900 1,04200 1,04500 1,04700 1,04800 1,05000 1,05000 1,05100 1,05100 1,05100
θ (methanol) 0,07151 0,11270 0,16700 0,23480 0,31520 0,40580 0,50170 0,58950 0,65430 0,70480 0,74670
θ (H2) 1,04600 1,04400 1,04300 1,04100 1,04000 1,03800 1,03700 1,03600 1,03500 1,03400 1,03300
Appendix – A.2 Ky =
30X Ae 100-60X Ae 30-30X Ae 70 − 60XAe × 100 − 60X Ae 100 − 60XAe
2
By expanding this equation ; X Ae3 × (−36 − 36 × Ky ) + XAe 2 × (120 + 84 × Ky + 36 × Ky ) + XAe × ( −100 − 49 × Ky − 84 × Ky ) + 49 × Ky = 0
For Solving this equation on MATLAB R2007a, we defined these parameters: a = (−36 − 36 × K y ) b = (120 + 84 × K y + 36 × K y ) c = (−100 − 49 × K y − 84 × K y ) d = 49 × K y 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
29
Appendix B.1 The rate expression is derived as a function of T and Xa. XA
×100) 7 (1 − X A ) 3 r = ρ× (k) × (( ×100) × ( ×100) 2 − ) 7 7 3,567E − 12 × exp(90130 / 8,314 × T) (1 − 2X A ) + (1 − 2X A ) + 3 3 (
7 − 2X A 3
(1 − 2X A ) +
The execution of program is based on scanning T and Xa values at the range of (400
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 T 520 523 524 532 534 538 539 540 541 541 542 546 553 553 556 573 577 581 590 605 621 659 679
Xa 0,01 0,1 0,13 0,29 0,32 0,37 0,38 0,39 0,4 0,4 0,41 0,44 0,47 0,47 0,48 0,48 0,47 0,46 0,43 0,37 0,3 0,16 0,11
r=2 T 541 542 542 543 543 544 545 545 546 546 547 547 548 549 550 550 552 553 560 561 561 564 566 569 569 572 573 588 588 593 597 597 615
Xa 0,03 0,06 0,06 0,08 0,09 0,11 0,13 0,13 0,15 0,15 0,17 0,17 0,19 0,2 0,22 0,22 0,25 0,26 0,33 0,34 0,34 0,36 0,37 0,38 0,38 0,39 0,39 0,39 0,39 0,38 0,37 0,37 0,31
r=5 T 569 570 571 572 579 580 581 582 586 596 614 624 632 643
Xa 0 0,02 0,04 0,06 0,16 0,17 0,18 0,19 0,22 0,26 0,26 0,24 0,22 0,19
r=8 T 590 597 600 604 607 627
Xa 0,07 0,14 0,16 0,18 0,19 0,2
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 olacak h=(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)); end for i=2:2:n-2 x(i)=a+h*i; S=S+2*feval(f,x(i)); end S=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 0,3863
V3 dX A = ∫ FA30 0,2863 R A 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
∫
T0
Tf
Σ FC i Pi dT +
i(inlet)
TR
∫
∫
TR
Σ FC i Pi dT − QRe movedbythewalls = − (∆HR )FA0 XA
i(outlet)
TR
FA0 C PA dT −
T0
∫
FA0X A 2C PAdT +
T0
TF
TF
TR
TR
− ∫ FA0 X A3CPA dT +
∫
TR
TR
TR
Tf
T0
T0
T0
TR
∫ FB0C PBdT − 2 ∫ FA0X A 2C PBdT + TF
TF
TR
TR
FB0 CPB dT − 2 ∫ FA0 XA3 CPB dT +
Tf
X A3
TR
TR
∫F
A0
∫ FA0X A 2C PCdT +
∫F
C PAdT
A0
XA3 CPC dT = (−∆HR )FA0 XA3 TR
F ∫T0 (CPA + FA0B0 CPB )dT − XA 2 × (T0∫ 2CPB dT − T0∫ CPC dT + T0∫ CPA ) = −( ∆H R |T ) TR
A = ( ∫ 2CPB dT − T0
TR
∫
T0
TR
CPC dT +
∫C
PA
)
T0
A = (60,301(298 − T0 ) − 0, 038105(298 2 − T0 2) + 3,507 ×10 −6(298 3 − T0 3 )− 2, 0085× 10−9 (2984 − T0 4 ))
X A3 =
88, 0433(Tf − T0 ) + 0, 0167(Tf 2 − T0 2 ) − 1, 77 ×10 −6 (Tf 3 − T0 3 ) − X A 2 × A 90680 + (60,301(298 − T0 ) − 0, 038105(298 2 − T0 2 ) + 3,507 ×10 −6 ( 2983 − T0 3 ) − 2, 0085× 10−9 (2984 − T0 4 ))
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 FA30 = FA0 (1 − XA3 ) =30*(1-0,3863) = 18,411 mol/s V3 = 18,411 * 0.0757= 1,394 Reactor 4 0,4663
V4 dXA = ∫ FA40 0,3863 R A Tf
XA4
TR
TR
TR
F ∫T0 (CPA + FA0B0 CPB )dT − XA3 × (T0∫ 2CPB dT − T0∫ CPC dT + T0∫ CPA ) = −(∆H R |T ) TR
A = ( ∫ 2CPB dT − T0
TR
∫
T0
TR
CPC dT +
∫C
PA
)
T0
A = (60,301(298 − T0 ) − 0, 038105(2982 − T02 ) + 3,507 × 10− 6 (2983 − T03) −2, 0085 ×10 − 9(298 4 −T 04)) XA4 =
(88, 0433(Tf − T0 ) + 0, 0167(Tf 2 − T0 2 ) − 1, 77 ×10 −6 (Tf 3 − T0 3 )) − X A3 × A 90680 + (60,301(298 − T0 ) − 0, 038105(298 2 − T0 2 ) + 3,507 ×10 −6 (2983 − T0 3 ) − 2, 0085 × 10 −9 (2984 − T0 4 ))
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 FA 40 = FA0 (1 − XA4 ) =30*(1-0,4663) =16,011 mol/s V4 = 16,011 * 0,1004= 1,6075 Reactor 5 0,5243
V5 dX A = ∫ … C.5.1 FA50 0,4663 R A 2 2 3 (88, 0433(T T ) 1,− −77 6 103× (T X4 −A f − 0T )+ 0, 0167(T f 0− f 0 T ))− A XA 5 = 2 2 − 6 3 3 90680+ (60, 301(2980 T− ) 0,−038105(298 0 T ) − 3, 298 507 + 10T )( 2, × 00085 10 (298 −9 −
…C.5.2
× T− )) 4
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
4 0
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 FA50 = FA0 (1 − XA5 ) =30*(1-0,5243) = 14,271 mol/s V5 = 14,271 * 0,0888 = 1,267 Reactor 6 0,5533
V6 dX A = ∫ ….C.6.1 FA60 0,5243 R A X A5 =
(88, 0433(Tf − T0 ) + 0, 0167(Tf 2 − T02 ) − 1, 77 × 10− 6 (Tf 3 − T03 )) − XA4 × A 90680 + (60,301(298 − T0 ) − 0, 038105(2982 − T02 ) + 3,507 × 10− 6 (298 3 −T03 ) −2, 0085 ×10− 9 (2984 −T04 ))
…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 FA60 = FA0 (1 − XA6 ) =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
θ
: Fugacity coefficient
,
J/mol.K
∆G °298 : Standart Gibbs energies of formation , ∆H °298 : Standart Enthalpies
,
J/mol
J/mol
Kf
: Equilibrium constant
Ky
: Equilibrium constant (molar fractions)
Kθ
: Equilibrium constant(fugacity coefficients)
Tr
: Reduced Temperature
Pr
: Reduced Pressure
P
: Pressure
,
bar
R
: Ideal gas law constant
,
8,314 J/mol.K
T
: Temperature
,
K
XA
: Fractional conversion of Carbon monoxide
X Ae
: Equilibrium conversion of Carbon monoxide
40
41
