131245357 52610363 methanol production from syngas reactor design

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 mixture of 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 CH 3 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) Search 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 data page.

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 temperature range between 450-650K and pressures 50-100 bar.

at 250˚C

The catalytic packed bed bulk density was given as

INTRODUCTION

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 available 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 % 6Upper, vol % 36

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) 1,370of liquid at 25 oC, J/(g.K) 2,533

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

1.3Natural 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

Temperature CO conversion, % CO2 conversion, % Exit CH3OH, vol %, 5MPa 10MPa 30MPa 5MPa 10MPa 30MPa 5MPa 10MPa 30MPaoC200 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

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

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]

1.5.2Side 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) G298 (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)

CP,CO (T,K) 27,113 + 0,655 X102 T – 0,1 X105

T2 [j /mol.K]

CP,H2 (T,K) = 26,113 + 0,435 X102 T – 0,033 X105

T2 [j /mol.K]

CP,CH3OH (T,K) = 19,038 + 9,146 X102 T – 1,218 X105 T2 - 8,034 X 109 T3 [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 barH 2 33.3 K 12.97 bar

7

3.CALCULATIONS

3.1. Obtaining equilibrium constant as a function of temperature, Kf

(T) : + H 298 and G298 values are given at thermodynamics data part.

H p r

od u c t s

t a n

H 298 = (-201,2)-(-110,52)

= -90,68 kj/mol = -90680 j/mol

G p r

o d u c t s

Gt an

G298 = (-162,0)-(-137,2)

= -24,8 kj/mol = -24800j/mol

G RT ln K298 8,314J

298K ln K298 =-24800 j/molmol K

ln K298 = 10,0097 K298 = 22243,38

dInKf H .... Van't Hoff Equation (InKf )T InK T H

dT

dT RT 2 T

0 RT 2

CP C

P,CHOH

2C

P,H2 C

P,CO

T T

dH cpdTTR TR

+ Cp values of substances are given at thermodynami properties part.

CP = -60,301 + 7,621X102 T – 1,052 X105 T2 - 8,034 X 109 T3 (J/mol.K)

T

H = H298 + Cp dT298

H = - 75985,54 + (60,301 T) (0,0381 T2 ) (3,5067 106 T3 ) (2,0085 109 T4 )

8

T 1 T H(T)dT d ln Kf 298 2

R T298

R=8,3145 J/mol.K;

ln K298 = 10,0097

(InKf)= InK298 +

(9,32935 9139,46

7,2529ln(T) 4,5826 103 T 4,2178 107 T2 2,4151010 T3 )

T

Kf exp(19,33905 9139,46

7,2529ln(T) 4,5826 103 T 4,2178 107 T2 2,415 1010 T3 ) T

Table 3.1 Equilibrium constant versus temperature data

T Kf

400 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

K T K K y Pn f / P ( fugacity coefficient) Basis:

100 moles/s Feed composition enter the reactor

CO 2H 2 CH 3OH

CO A , H2 B , CH3OH 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

Ky

yc

yA .yB 2

K

ic

n (1 1 2) 2 ia .ib 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( XAe ) 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) Pr(H2)0,628614534 1,43020595 3,855050116

10

Fugacity coefficient of substances are read on a graph by parameters Tr,Pr,

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)

30XAe ic

Kf

100-60XAe

50

2

30-30XAe 70 60XAe 2 ia .ib 2

100 60XAe 100 60XAe 30XAe

1,619805 100-60XAe 0,1538 502

2 (1,017)2 1,01230-30XAe

70 60XAe

100 60XAe 100 60XAe 30XAe

26909,64 =100-60XAe

30-30XAe

70 60XAe

2

100 60XAe 100 60XAe

An equation Ky = f( XAe ) such as:

30XAe

Ky 100-60XAe

30-30X 70 60X 2

Ae Ae

100 60XAe 100 60XAe 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




Fugacity coefficient of substances are read on a graph by parameters Tr,Pr,

shown on Appendix A.1


K(fugacityT Kf coefficient) Ky Xae400 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


30XAe ic

Kf

100-60XAe

75

2

30-30XAe 70 60XAe 2 ia .ib 2

100 60XAe 100 60X

Ae 30XAe

100-60XAe 2

1,619805 0,09281 7530-30X 70 60X 2Ae Ae

100 60XAe 100 60XAe 30XAe

98172,15 =100-60XAe

30-30XAe

70 60XAe

2

100 60XAe 100 60XAe An equation Ky = f( XAe ) such as:

12

30XAe

Ky 100-60XAe

30-30X 70 60X 2

Ae Ae







K(fugacityT Kf coefficient) Ky Xae400 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


30XAe ic

Kf

100-60XAe

120

2

30-30XAe 70 60XAe 2 ia .ib2

100 60XAe 100 60X

Ae 30XAe

1,619805 100-60XAe 0,063209575 752

230-30X 70 60X Ae Ae

100 60XAe 100 60XAe

13

30XAe

369013,7 =100-60XAe

30-30XAe

70 60XAe

2

100 60XAe 100 60XAe An equation Ky = f( XAe ) such as:

30XAe

Ky 100-60XAe

30-30X 70 60X 2

Ae Ae



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

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 P2H2

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 F A PA PA

FA PT …(3.1);

FT PT FT

A 2B CFa0 Fb0

-Fa0Xa -2Fa0Xa Fa0Xa

------------------------------------

FA

F

A0

F

A0 X

A

FB FB0 2FA0 XA

FC

F

A0 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 C A0

(1 X A

) V0 ;

1 XAF

B (C B0

2C A0

X A

) V0 ;

1 XA

FC C

A0 X

A V0 ;

1 XA

FT FA FB FC

FT C A0

C B0

2C A0

X A V0 ;

1 XA

15

C A0

(1

X

A )

V0

(1 XA )PA

1 X A 100 100 (Carbon Monoxide)C

A0

C B0

2C A0

X A V (1 2X

A) 7

0

31 XA(C B0

2C A0

X A

) V 7 2XA0

3PB

1 XA 100 100 (Hydrogen gas)C

A0

C B0

2C A0

X A V (1 2X

A) 7

0

31 XAC

A0X

A V0 X

A

PC 1 XA

100 100 (Methanol)C A0

C B0

2C A0

X A V (1 2X

A) 7

0

31 XA

All of the partial pressure expressions’ numerators and denominators are divided

by CA0 , (CB0 / CA0 70 / 30) The rate expression is obtained as a function of

temperatures(T) and molar fractions (Xa).

7 (XA

100)(1 XA ) 2XA (1 2XA )

7

ra (k) (( 100) (3

100)2

3

)(1 2XA )

7(1 2XA )

7 3,567E 12 exp(90130 / 8,314 T)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

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 H2

Outlet stream : F CH 3 OH , FCO,

F H2 General Energy Balance equation :

TR Tf

FC dT FC dT Q ( H )F Xi(inlet)

i Pi i(outlet) i Pi Removedbythewalls R A0 AT0 TR

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 HR |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 30molCO 28g 980g / s ;molCOmolH2

FA0 30mol / s ;

Mmethanol 32g / molF F F X 0 30 0,55 16,5mol / s

methanol = 528 g/s methanol(CH3OH) (CH3OH)0 (CO)0 CO

Daily production = 528g (60 60 24)s 1ton6 45,61 ton/day methanols 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 TfQ

R byflows (FA0C PA FB0C PB)dT (F AC PA F BC PB F BC PB)dT

T0 TRTR Tf

QR byflows

(F CPA F C

PB)dT (F C

PA F X C F C

PB 2F X C F X C

PC)dT

A0 B0 A0 A0 A1 PA B0 A0 A1 PB A0 A1

T0 TR

Tf

XA1

(FA0

C PA

F

B0 C

PB )dT

T0Tf

(FA0C PA 2FA0 CPB FA0 CPC )dT ( HR )FA0

TR

Tf FB0

Tf FB0 (C PA CPB )dT

(CPA CPB )dTF

A0XA1 T0

FA0 T0

Tf

( H R |T ) (C PA 2C PB C PC )dT ( HR )TR

XA1 88,0433(T T ) 0,0167(T 2 T 2 ) 1,77 10 6 (T 3 T 3)

f 0 f 0 f 0

90680 (60,301(298 T ) 0,038105(298 2 T 2) 3,507 10 6 3 T 3 ) 2,0085 10 9 4 4 ))( 298 (298 T0 0 0 0

….. 3.5.1

V 0,1593 dX1 A

………… 3.5.2F RAo o A

3.5.1 and 3.5.2 solved simultaneously.

7 2XA (

XA

100)(1 XA )

(1 2XA ) 72 33

ra (k) (( 100) ( 100) )(1 2XA )

7(1 2XA )

7 3,567E 12 exp(90130 / 8,314 T)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

V 0,2863 dXA2

F RAA20 0,1593

Inlet

A : FA1=FA0-FA0*XA1

B: FB1=FB0-2*FA0*XA1

C: FC1=FA0*XA1

outlet

A : FA2 = FA0 – FA0XA2

B : FB2 = FB0 – 2*FA0*XA2 C : FC2 =

FA0*XA2

21

TR Tf

FC dT FC dT Q ( H )F Xi(inlet) i Pi i(outlet) i Pi Removedbythewalls R A0 A

T0 TR

TR TR TR TR TR Tf

F

A0C

PAdT

FA0

X A1

CPA

dT

F

B0 C

PB dT

2

F

A0 X

A1C

PB dT

F

A0 X

A1 C

PC dT

FA0

CPA

dT

T0 T0 T0 T0 T0 TR

TF TF TF TF

F XA2

CPA

dT F C dT 2 F X C dT F X C dT ( H )F X A0 B0 PB A0 A2 PB A0 A2 PC R A0 A2

TR TR TR TR

Tf FB0

TR TR TR

(C PA CPB )dT X A1 ( 2CPB dT C

PC dT

C

PA )

FA0X

A2 T0 T0 T0 T0

( H R |T )TR TR TR

A = ( 2C

PBdT

CPC dT CPA )T0 T0 T0

A (60,301(298 T ) 0,038105(2982 T 2 ) 3,507 10 6 (2983 T 3) 2,0085 10 9(298 4 T 4))0 0 0 0

XA2

88,0433(T T ) 0,0167(Tf

2 T 2) 1,77 10 6(T 3 T 3) X Af 0 0 f 0 A1

2 2 6 3 3 2,0085 9 4 490680 (60,301(298 T ) 0,038105(298 T

0) 3,507 10 ( 298 T ) 10 (298 T ))

0 0 0

Table 3.5.2 Reactor-2 data


497 4970,1833 2,90614E- 27,528 66,292 6,1792 0,0105 0,3498

2,85812 06 28 43 89 86 83

497 5070,1946 4,25746E- 27,355 66,032 6,6120 0,0068 0,5037 1,985159 06 19 78 37 85 3 89

497 5170,2060 6,14567E- 27,179 65,768 7,0521 0,0045 0,7129 1,402516 06 12 68 93 53 89 46

497 5270,2173 8,74859E- 27,000 65,500 7,4999 0,0030 0,9919 1,00819 06 02 03 43 58 56 09

497 5370,2287 1,22912E- 26,817 65,226 7,9554 0,0020 1,3554 0,737781 05 81 71 83 85 96 37

497 5470,2401

1,7055E-0526,632 64,948 8,4190 0,0014 1,8164 0,5505

9 39 59 14 42 35 29

497 5570,2516 2,33885E- 26,443 64,665 8,8907 0,0010 2,3803

0,420117 05 7 55 46 1 87

497 5670,2630 3,17187E- 26,251 64,377 9,3708 0,0007 3,0361 0,329361 05 64 46 94 17 86 61

497 5770,2745 4,25638E- 26,056 64,084 9,8596 0,0005 3,7389 0,267422 05 13 19 83 15 6 54

497 5870,2860 5,65477E- 25,857 63,785 10,357 0,0003 4,3812 0,228201 05 06 59 34 74 23 47

497 5970,2974 7,44142E- 25,654 63,481 10,864 0,0002 4,7448 0,210796 05 35 53 12 74 36 55

497 6070,3090 9,70439E- 25,447 63,171 11,380 0,0002 4,4230 0,22601 05 9 85 25 03 31 89

497 617 0,3205 0,0001254 25,237 62,856 11,906 0,0001 2,6964 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 = 0,0889FA20 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(PCOP2H2

PCH3OH ) (mol / kgcat min)Ke

k 7.6106 mol(kgcat) 1 min1 atm3 at 250 C

V1 dX AF [mol / s] ra[mol / kgcat min]

A0

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 1m 3 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 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 (CO) (methanol) (H2)

400 1,012 0,1538 1,017420 1,014 0,2424 1,017440 1,015 0,3591 1,016460 1,017 0,504 1,016480 1,018 0,7093 1,015500 1,018 0,7553 1,015520 1,019 0,7919 1,014540 1,019 0,8222 1,014560 1,02 0,8475 1,014580 1,02 0,869 1,013600 1,02 0,8873 1,013


T (CO) (methanol) (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


T (CO) (methanol) (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

30XAe

Ky 100-60XAe

30-30X 70 60X 2

Ae Ae

100 60XAe 100 60XAe By expanding this equation ;

XAe

3 ( 36 36 K ) X 2 (120 84 K 36 K ) X ( 100 49 K 84 K ) 49 K 0y Ae y yAe y y y

For Solving this equation on MATLAB R2007a, we defined these parameters:

a = ( 36 36 Ky )

b = (120 84 Ky 36 Ky )

c = ( 100 49 Ky 84 Ky )

d = 49 Ky

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.

7(

XA

100)(1 XA ) 2XA (1 2XA )

7

r (k) (( 100) (3

100)2

3

3,567E 12 exp(90130 /8,314 T))

(1 2XA ) 7

(1 2XA ) 7

3 3

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){


}

if (T > 550 && r > 0.03 && r <= 0.15 && (ra - r) < 0.03 && (ra - r) > -0.03){


}...

//(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 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

V3 0,3863 dXA

FA30 0,2863

RA

Inlet

A : FA2=FA0-FA0*XA2

B: FB2=FB0-2*FA0*XA2

C: FC2=FA0*XA2

outlet

A : FA3 = FA0 – FA0XA3

B : FB3 = FB0 –

2*FA0*XA3 C : FC3 =

FA0*XA3

TR Tf

FC dT FC dT Q ( H )F Xi(inlet) i Pi i(outlet) i Pi Removedbythewalls R A0 A

T0 TR

TR TR TR TR TR Tf

F

A0C

PAdT

FA0

X A2

C PA

dT

F

B0C

PBdT

2

F

A0X

A2C

PBdT

F A0X

A2C

PCdT

F

A0C

PAdT

T0 T0 T0 T0 T0 TR

TF TF TF TF

F XA3

CPA

dT F C dT 2 F X C dT F X C dT (H )F X A0 B0 PB A0 A3 PB A0 A3 PC R A0 A3

TR TR TR TR

Tf FB0

TR TR TR

(CPA CPB )dT XA2 ( 2CPB dT CPC dT

CPA

)F

A0XA3 T0 T0 T0 T0

( HR |T )TR TR TR

A = ( 2CPBdT CPC dT CPA )T0 T0 T0

A (60,301(298 T ) 0,038105(298 2 T 2 ) 3,507 10 6 (2983 T 3 ) 2,0085 109 (2984 T4 ))0 0 0 0

XA3

88,0433(T T ) 0,0167(T 2 T 2 ) 1,77 106 (T 3 T 3) XA2 A

f 0 f 0 f 0

90680 (60,301(298 T ) 0,038105(298 2 T 2 ) 3,507 10 6 3 3 ) 2,0085 10 9 4 T 4 ))( 298 T (298

0 0 0 0

Table C.3.1 Reactor-3 data


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 = 0.0757FA30 FA0 (1 XA3 ) =30*(1-0,3863) = 18,411 mol/s

V3 = 18,411 * 0.0757= 1,394

Reactor 4

V 0,4663 dX4

R AFA40 0,3863 A

Tf FB0

TR TR TR

(C

PA CPB )dT XA3 ( 2CPB dT CPC dT CPA )

XA4 F

A0T0 T0 T0 T0

( HR |T )TR TR TR

A = ( 2CPBdT CPC dT CPA )T0 T0 T0

A (60,301(298 T0 ) 0,038105(2982 T02 ) 3,507 10 6 (2983 T0

3) 2,0085 10 9(298 4 T 04))

XA4

(88,0433(T T ) 0,0167(T 2 T 2 ) 1,77 10 6 (T 3 T 3 )) XA3 A

f0 f 0 f 0

90680 (60,301(298 T ) 0,038105(298 2 T2 ) 3,507 10 6 3 T 3 ) 2,0085 10 9 (298 4 T 4 ))(298

0 0 0 0

35



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 = 0.1004FA40 FA0 (1 XA4 ) =30*(1-0,4663) =16,011

mol/s V4 = 16,011 * 0,1004= 1,6075

Reactor 5

V 0,5243 dXA5

… C.5.1F RA50 0,4663 A

(88, 0433(T T ) 2 2 T ) 6 3 3 T )) X AXA 5

0, 0167(T 1,77 10 (T0f 0 f 0 f A 4

90680 (60, 301(2980 T ) 0, 038105(298 0 T 6 3

)(3

) 3, 298507 10T 2, 00085 10 (298…C.5.2 Solving simultaneously equation C.5.1 and C.5.2



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 0,971182By -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 = 0.0888FA50 FA0 (1 XA5 ) =30*(1-0,5243) = 14,271 mol/s

V5 = 14,271 * 0,0888 = 1,267

Reactor 6

V 0,5533 dX6 R A ….C.6.1F

A60 0,5243 A

XA5

(88,0433(T T ) 0,0167(T 2 T 2 ) 1,77 106 (T3 T3 )) X Af 0 f 0 f 0 A4

90680 (60,301(298 T ) 2 T 2 ) 6 (2983 T 3 ) 9 (298 4 T 4 ))0,038105(298 3,507 10

0

2,0085 10

0 0 0

…C.6.2

Solving simultaneously equation C.6.1 and C.6.2



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 = 0.0374FA60 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]

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 398-4152- 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

q : Fugacity coefficient

G298 : Standart Gibbs energies of formation , J/mol

H 298 : Standart Enthalpies , J/mol

Kf : Equilibrium constant

Ky : Equilibrium constant (molar fractions)

K : 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

XAe : Equilibrium conversion of Carbon monoxide

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131245357 52610363 methanol production from syngas reactor design

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