WS-4-03 Prediction of the carbon deposition in steam ...web.nchu.edu.tw/pweb/users/cmchang/lesson/3638.pdf · ©Invensys Systems Japan, Inc. Thermo Workbook - 108 WS-4-03 Prediction

PRO/II Advanced Training for Customers in Taiwan

©Invensys Systems Japan, Inc. Thermo Workbook - 108

WS-4-03

Prediction of the carbon deposition

in steam reforming unit (Equilibrium reaction calculation in Gibbs Reactor)

Problem

Steam reformer is often used in refineries or chemical plants. Design and operation of steam reformer need to avoid deposition of solid carbon that can plug the reactor.

Some authors reported that reactions in steam reformer are well described by equilibrium assumption. We predict the deposition of carbon using Gibbs reactor under the following condition. The following reactions are involved.

CH4+H2O=3H2+CO 2H2O=2H2+O2 2CO+O2=2CO2 2CO=2C(solid)+O2

So, the following components are involved. CH4,H2O,CO2,H2,CO,O2,C(solid)

Feed Composition

CH4 : 1 kg-molh H2O : 1 kg-molh

Temperature: 600 degC Pressure: 2 kg/cm2 Q-1 Do we expect carbon deposit in the reactor under the above conditions?



1. Preparation 1）UOM: Metric

2）Components Select the following components.

CH4 H2O CO2 H2 CO O2 C

Select SIMSCI as the databank.

3) Thermo set Select SRK.

2. Flow sheet setting 1) Addition of unit operation

Add a Gibbs reactor.



R1

S1S2

2) Setting the feed stream

3) Setting the Gibbs Reactor

At first select SET1 and click OK.

After that open the Gibbs reactor again and select “None”.



3.Run the simulation

Add stream property list (Material balance list). Carbon appears in the outlet stream.

4.Results

Q-1:Carbon deposition is expected.

Needs some measure to avoid the deposition. -Increase in H2O/CH4 ratio -Increase in Temperature

R1

S1S2

Stream NameStream Description

PhaseTemperaturePressure

FlowrateComposition CH4 H2O CO2 H2 CO O2 C

CKG/CM2

KG-MOL/HR

S1

Vapor600.000

2.000

2.000

0.5000.5000.0000.0000.0000.0000.000

S2

Mixed600.000

2.000

2.749

0.2280.1820.0620.4540.0580.0000.016



WS-5-01

Production of Methyl Acetate

Problem Reaction kinetics for synthesis of methyl acetate from methanol and acetic acid is described as follows. (Song et al.)

OHCOOCHCHOHCHCOOHCH 23333 +⇔+MeOAcMeOHHOAC

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

eq

OHMeOAcMeOHHOACf K

CCCCkr 2

⎟⎠⎞

⎜⎝⎛−×=

RTk f

6.12493exp10732.9 8

⎟⎠⎞

⎜⎝⎛=

RTKeq

78.1555exp32.2

The above equitation is supposed to be derived from the following equations for forward and back reactions.

MeOHHOACff CCkr =

OHMeOAcbb CCkr 2=

⎟⎠⎞

⎜⎝⎛−×==

RTKk

keq

fb

4.14049exp10195.4 8

Equilibrium constant Keq becomes as follows;

( )T

K eq98.78284156.0ln +=

where C:[kg-mol/m3],R: 1.987[kcal/kg-mol K],r:[kg-mol/m3 h]

Q-1. Calculate the conversion of methanol at a Plug Flow Reactor under the following conditions.

FEED Temp.:77deg C, Press.:1kg/cm2、acetic acid: 280kg-mol/h, methanol:280kg-mol/h

Reactor ID: 1000mm, length:20m, Temp.:77deg C、 Reaction Operation Phase;Liquid

Reaction Kinetics １）Power low ２）Procedure Q-2. Calculate the conversion of methanol at a Equilibrium Reactor under the same condition (Reaction Operation Phase;Liquid) Q-3. Calculate the conversion of methanol at a Gibbs Reactor under the same condition (Reaction Operation Phase;Liquid)

[reference] Song,W.,G. Venimadhavan,J.M.Manning,M.F.Malone, and M.F.Doherty “Measurement of Residue Curve Map and Heterogeneous Kinetics in Methyl Acetate Synthesis”, Ind.Eng.Chem.Research,37,1917-1928(1998)



2. Preparation 1）UOM: Metric

2）Components Select the following components in exactly the same order.

HOAC MEOH MEAC H2O

Select SIMSCI as the databank.

Set “component phases” as “vapor-liquid” for all the components.



3) Property estimation method

Select NRTL (single phase). Push “Modify”. Vapor fugacity: Hyden- O’Connel Vapor enthalpy: SRK-Modified- Panag. Check “calculate transport property”.



4）Reaction Stoichiometric Data input

Make reaction sets SET1 and SET2. SET1 includes reactions R1(forward reaction) and R2(back reaction). Stoichiometric data for R1 is set as HOAC+MEOH ->MEAC+H2O.

Stoichiometric data for R2 is set as MEAC+H2O ->HOAC+MEOH. So that the GUI appearance finally becomes as follows.



5)Input power law constants for reaction kinetics

The following equation is set for R1 and R2.

R1: ⎟⎠⎞

⎜⎝⎛−×=

RTk f

6.12493exp10732.9 8

R2: ⎟⎠⎞

⎜⎝⎛−×==

RTKk

keq

fb

4.14049exp10195.4 8



6)Reaction equilibrium data

For reaction set “SET２”, only R1；HOAC+MEOH ->MEAC+H2O is set. The following equation is set for R1.

( )T

K eq98.78284156.0ln +=



7)Input Procedure for the reaction kinetics Procedure set “PSET1” is declared. The code is input as follows; (copy and paste from attached “MethylAcetate.txt”)



2.Process flow diagram

Make a flow diagram that consists of two PFRs, an Equilibrium Reactor and a Gibbs Reactor. Appropriately name each stream.

１）Set the contents of PFR-1

This reactor uses power-law as the kinetics.

Temp.:77CdegPress.:1Kg/cm2HOAC:280kg-mol/hMEOH:280kg-mol/h

PFR1-POWER

FEED PFR1-OUT

PFR1-POWER

FEED PFR1-OUT

PFR1-POWER

FEED PFR1-OUT



PFR1-POWER

FEED PFR1-OUT

PFR1-POWER

FEED PFR1-OUT

PFR1-POWER

FEED PFR1-OUT

PFR1-POWER

FEED PFR1-OUT


PFR1-POWER

CON-RX

GIBBS-RX

PFR2-PROCE

FEED PFR1-OUT

FEED3 CONR-OUT

FEED4GIBBS-OUT

FEED2 PFR2-OUT

PFR1-POWER

CON-RX

GIBBS-RX

PFR2-PROCE

FEED PFR1-OUT

FEED3 CONR-OUT

FEED4GIBBS-OUT

FEED2 PFR2-OUT



２）Set the contents of the PFR-2

This reactor uses Procedure for the kinetics.

PFR2-POWER

FEED2 PFR2 -OUT

Define to FEED PFR2-POWER

FEED2 PFR2 -OUT

PFR2-POWER

FEED2 PFR2 -OUT


FEED2 PFR2 -OUT

PFR2-POWER

FEED2 PFR2 -OUT


FEED2 PFR2 -OUT

PFR2-POWER

FEED2 PFR2 -OUT

Define to FEED

3) Set the contents of the Equilibrium Reactor

CON-RX

FEED3CONR-OUT

CON-RX

FEED3CONR-OUT

CON-RX

FEED3CONR-OUT

CON-RX

FEED3CONR-OUT

CON-RX

FEED3CONR-OUT

CON-RX

FEED3CONR-OUT



4)Set the contents of Gibbs Reactor

GIBBS-RX

FEED4GIBBS-OUT

GIBBS-RX

FEED4GIBBS-OUT

GIBBS-RX

FEED4GIBBS-OUT

3.RUN the simulation

PFR1-POWER

CON-RX

GIBBS-RX

PFR2-PROCE

FEEDPFR1-OUT

FEED3 CONR-OUT

FEED4GIBBS-OUT

FEED2 PFR2-OUT

Stream NameStream DescriptionPhase

Fluid Rates HOAC MEOH MEAC H2O

Rate

TemperaturePressureEnthalpyMolecular WeightMole Fraction VaporMole Fraction Liquid

KG-MOL/HR

KG-MOL/HR

CKG/CM2M*KCAL/HR

PFR1-OUT

Mixed

50.235550.2355

229.7645229.7645

560.000

77.00001.00004.4535

46.04740.73860.2614

PFR2-OUT

Mixed

50.235550.2355

229.7645229.7645

560.000

77.00001.00004.4535

46.04740.73860.2614

CONR-OUT

Mixed

49.477049.4770

230.5230230.5230

560.000

77.00001.00004.4648

46.04740.74090.2591

GIBBS-OUT

Mixed

57.479857.4798

222.5202222.5202

560.000

77.00001.00004.3444

46.04740.71630.2837



4.Results Q-1. PFR

１） Conversion of methanol at PFR with Power-law Recovery =( 1-49.4452/280)x100 =82.341 %

２） Conversion of methanol at PFR with Procedure

Recovery =( 1-49.4452/280)x100 =82.341 % Because the exactly the same equations are used, the result must be exactly the same.

Q-2. Equilibrium Reactor Recovery =( 1-49.477/280)x100 =82.329 %

Since the result of PFR is closed to that of Equilibrium Reactor, it is presumed that increase in the volume of PFR dose not significantly increase the recovery.

Q-3.Gibbs Reactor

Recovery =( 1-57.48/280)x100 =79.47 % Since the result of Gibbs Reactor is closed to that of Equilibrium Reactor, the literature value for equilibrium equation is in accord with thermodynamic requirements.

5.Discussion

All the problems above regards the reactor is full with liquid. PFR. Users need to select one of vapor or liquid for PFR, Equilibrium Reactor and CSTR. Gibbs reactor can handle phase equilibria simultaneously with reaction

equilibria.

Conversion of methanol increases to 92.3%, if the Reactor Operation Phase is “Calculated” at the Gibbs Reactor in Q-3. This increase is caused by the vaporization of MeOAc. So the rate of back reaction decreases.

OHCOOCHCHOHCHCOOHCH 23333 +⇔+MeOAcMeOHHOAC

Reactive distillation in the next session uses this principle.



WS-5-02

Production of ethylene oxide Problem

According to a literature, the reaction kinetics for oxidation of ethylene on catalyst is

described as follows;

OHC O21 HC 42242 →+

2332211

212111

)1( CACACACCAAK

r+++

=

OH22CO 3O HC 22242 +→+ 2

332211

212122

)1( CACACACCAAK

r+++

=

OH22CO O25 OHC 22242 +→+

2332211

322133

)1( CACACACCAAK

r+++

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RTxxK

415

11059.4exp1036.1 ⎟

⎟⎠

⎞⎜⎜⎝

⎛=

RTxA

3

11091.4exp92.2

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RTxxK

417

21027.5exp1059.9 ⎟

⎟⎠

⎞⎜⎜⎝

⎛×= −

RTxA

411

21065.2exp1000.3

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RTxxK

413

31043.4exp1035.1 ⎟

⎟⎠

⎞⎜⎜⎝

⎛×=

RTxA

32

31037.3exp1032.1

Unit of each variable is;

⎥⎦

⎤⎢⎣

⎡• scat -kg

mol-kgr 、 ⎥

⎦

⎤⎢⎣

⎡• scat -kg

mol-kgK 、

⎥⎥⎦

⎤

⎢⎢⎣

⎡

mol-kgm3

A 、 ⎥⎦

⎤⎢⎣

⎡3mmol-kg

C 、 ⎥⎦⎤

⎢⎣⎡

•KmolcalR

Q-1: Calculate the conversion of ethylene to ethylene oxide at a Plug Flow Reactor under the following conditions. Reactor Catalytic bulk density: 800kg-cat/m3-bed Reaction temperature: 250 degC(fixed) Pressure 10Kg/cm2 Feed flow rate:10kg-mol/h composition: ethylene 30mol%,O2 7mol%,N2 63mol% Temp.: degC, Press.:10Kg/cm2 ID: 1000mm, Length: 10m Q-2: Calculate the conversion of ethylene to ethylene oxide at a CSTR with the same temperature, pressure ,feed condition and the reaction volume. Q-3: Recent development of catalyst increased the conversion to 80% at the same condition with Q-1. This is mainly due to an increase in the rate of the first reaction (K1). Determine the pre-exponential factor for K1 for the new catalyst.



1.Preparation

1)UOM: Metric

2) Components Set the following components just in the same order. ETHYLENE O2 EO CO2 H2O N2

3)Thermo system

Select NRTL (single liquid). Modify it to calculate transport property.

Set component phase as “Vapor-Liquid”.



4)Reaction stoichiometry

Set as follows.

5)Procedure data

Declare new procedure “PSET1”.

Input Procedure Data Code as follows; (The code can be transferred from the file “EO.txt” in attached disk by copy & paste.)

A1=2.92*EXP(4.91E+3/(RGAS*RTABS)) A2=3.00E-11*EXP(2.65E+4/(RGAS*RTABS)) A3=1.32E+2*EXP(3.73E+3/(RGAS*RTABS)) AK1=1E+15*PREEXP(1)*EXP(-ACTIVE(1)*1000/(RGAS*RTABS)) AK2=1E+17*PREEXP(2)*EXP(-ACTIVE(2)*1000/(RGAS*RTABS)) AK3=1E+13*PREEXP(3)*EXP(-ACTIVE(3)*1000/(RGAS*RTABS)) C1=XVCONC(1) C2=XVCONC(2) C3=XVCONC(3) SS=(1.0+A1*C1+A2*C2+A3*C3)**2 R1=AK1*A1*A2*C1*C2/SS R2=AK2*A1*A2*C1*C2/SS R3=AK3*A2*A3*C2*C3/SS RRATES(1) = R1*3600.0*800. RRATES(2) = R2*3600.0*800. RRATES(3) = R3*3600.0*800. ISOLVE=1 RETURN



7) Reaction rate constant

The values of pre-exponential factors (PREEXP) in the code in procedure are retrieved from “Kinetic Reaction Data” of which data entry window is accessed by clicking “K” in the reaction definition window. Since the power part (1E+15,1e+17,1E+13) is multiplied in the procedure, only the figures (1.36,9.56 and 1.35) need to be set. Input

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RTxxK

415

11059.4exp1036.1

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RTxxK

417

21027.5exp1059.9

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

RTxxK

413

31043.4exp1035.1



2.Flow sheet setting

1)Addition of unit operations

R1

R2

PLUG-IN PLUG-OUT

CSTR-IN CSTR-OUT

2）Setting Plug Flow Reactor



3) Setting CSTR

4) Setting feed stream

Feed for Plug Flow Reactor

flow rate:10kg-mol/h composition: ethylene 30mol%,O2 7mol%,N2 63mol% Temp.: degC, Press.:10Kg/cm2

Feed for CSTR

Reference to the feed stream for the Plug Flow Reactor




R1

R2

PLUG-IN PLUG-OUT

CSTR-IN CSTR-OUT


Fluid Rates ETHYLENE O2 EO CO2 H2O N2

Rate


KG-MOL/HR

KG-MOL/HR

CKG/CM2M*KCAL/HR

PLUG-IN

Vapor

3.00000.70000.00000.00000.00006.3000

10.000

100.000010.00000.0135

28.30451.00000.0000

PLUG-OUT

Vapor

2.64270.06240.17370.36720.36726.3000

9.913

250.000010.00000.0389

28.55251.00000.0000

CSTR-IN

Vapor

3.00000.70000.00000.00000.00006.3000

10.000

100.000010.00000.0135

28.30451.00000.0000

CSTR-OUT

Vapor

2.72810.21110.13070.28240.28246.3000

9.935

250.000010.00000.0380

28.49071.00000.0000



4.Chage the flow sheet

1) Addition of calculator and optimizer

R1

R2

CA1 OP1

PLUG-IN PLUG-OUT

CSTR-IN CSTR-OUT

2) Setting the content of calculator



3) Setting the content of Optimizer


Right click the optimizer and execute “RUN RESULT”.



6.RESULT Q-1 Conversion at Plug Flow Reactor

Conv.=(0.1737/3.0)x100=5.79% Q-2 Conversion at CSTR

Conv.=(0.1307/3.0)x100=4.36% Q-3 Kinetic parameter (pre-exponential factor)

24.35 kg-mol/(kg-cat・sec)



WS-5-03 Heat balance of Reactor

Theme

Q-1 Check the heat of reaction at the PFR of WS-5-02 is correctly calculated. Q-2 Check the heat duty of the PFR of WS-5-02 is correctly calculated.



1.Change of flow sheet Open the file made at WS-5-02. Save with different name using “Save as” from the file menu. Open the Print option of Plug Flow Reactor. Check to print out enthalpy balance.


R1

R2

PLUG-IN PLUG-OUT

CSTR-IN CSTR-OUT


Fluid Rates ETHYLENE O2 EO CO2 H2O N2

Rate


KG-MOL/HR

KG-MOL/HR

CKG/CM2M*KCAL/HR

PLUG-IN

Vapor

3.00000.70000.00000.00000.00006.3000

10.000

100.000010.00000.0135

28.30451.00000.0000

PLUG-OUT

Vapor

2.64270.06240.17370.36720.36726.3000

9.913

250.000010.00000.0389

28.55251.00000.0000

CSTR-IN

Vapor

3.00000.70000.00000.00000.00006.3000

10.000

100.000010.00000.0135

28.30451.00000.0000

CSTR-OUT

Vapor

2.72810.21110.13070.28240.28246.3000

9.935

250.000010.00000.0380

28.49071.00000.0000



3.Retrieve data Generate output. From the output file retrieve the data of Heat of Formation (at the last of P-1)

From PLUGFLOW REACTOR SUMMARY (P5) retrieve the data on Duty and Heat of reaction.

From PLUGFLOW REACTOR SUMMARY (P5) retrieve the data on flow rate change for each component.



From PLUG FLOW HEAT BALANCE retrieve enthalpy of reactant and product both at operating condition and reference condition.

4.Check 1)Heat of formation

Agrees with –0.0624 M*Kcal/h in the output file

2) Duty Duty = (H2-H1)+ DH + (H4-H3) = -0.0426 M*Kcal/h Agrees with -0.0426 M*Kcal/h in the output file

5.Results Q-1 Heat of reaction at the PFR of WS-3-02 is correctly calculated. Q-2 Heat duty of the PFR of WS-3-02 is correctly calculated.

H1

H2

H4

H3

Component A: HEAT FORM. B: Change AxBkcal/kg-mol kg-mol/h kcal/h

ETHYLENE 12541.8 -0.3573 -4481.18514O2 0 -0.6376 0EO -12570.46 0.1737 -2183.488902CO2 -93988.26 0.3672 -34512.48907H2O -57756.29 0.3672 -21208.10969N2 0 0 0

Total -62385.2728

Documents

WS-4-03 Prediction of the carbon deposition in steam ...web.nchu.edu.tw/pweb/users/cmchang/lesson/3638.pdf · ©Invensys Systems Japan, Inc. Thermo Workbook - 108 WS-4-03 Prediction