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+HO=3H+CO HO=H+O CO+O=CO CO=C(solid)+O So, the following components are involved. CH4,HO,CO,H,CO,O,C(solid) Feed Composition CH4 : 1 kg-molh HO : 1 kg-molh Temperature: 600 degc Pressure: kg/cm Q-1 Do we expect carbon deposit in the reactor under the above conditions? 108
1. Preparation 1)UOM: Metric )Components Select the following components. CH4 HO CO H CO O C Select SIMSCI as the databank. 3) Thermo set Select SRK.. Flow sheet setting 1) Addition of unit operation Add a Gibbs reactor. 109
S1 S R1 ) 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. 110
3.Run the simulation Add stream property list (Material balance list). Carbon appears in the outlet stream. S1 S Stream Name Stream Description Phase Temperature Pressure Flowrate Composition CH4 HO CO H CO O C R1 C KG/CM KG-MOL/HR S1 Vapor 600.000.000.000 0.500 0.500 0.000 0.000 0.000 0.000 0.000 S Mixed 600.000.000.749 0.8 0.18 0.06 0.454 0.058 0.000 0.016 4.Results Q-1:Carbon deposition is expected. Needs some measure to avoid the deposition. -Increase in HO/CH4 ratio -Increase in Temperature 111
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.) CH 3 COOH+ CH 3OH CH 3COOCH 3 + H O r = k HOAC f C HOAC C MeOH MeOH C C K MeOAc MeOAc eq H O = 8 1493. 6 k f 9.73 10 exp RT 1555. 78 K eq =.3 exp RT The above equitation is supposed to be derived from the following equations for forward and back reactions. r = k f f C HOAC C MeOH rb = kbcmeoacc H O k f = = 8 14049.4 kb 4.195 10 exp K eq RT Equilibrium constant Keq becomes as follows; ( ) 78. 98 ln K eq = 0.84156 + T where C:[kg-mol/m 3 ],R: 1.987[kcal/kg-mol K],r:[kg-mol/m 3 h] Q-1. Calculate the conversion of methanol at a Plug Flow Reactor under the following conditions. FEED Temp.:77deg C, Press.:1kg/cm acetic acid: 80kg-mol/h, methanol:80kg-mol/h Reactor ID: 1000mm, length:0m, Temp.:77deg C Reaction Operation Phase;Liquid Reaction Kinetics 1)Power low )Procedure Q-. 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,37,1917-198(1998) 11
. Preparation 1)UOM: Metric )Components Select the following components in exactly the same order. HOAC MEOH MEAC HO Select SIMSCI as the databank. Set component phases as vapor-liquid for all the components. 113
3) Property estimation method Select NRTL (single phase). Push Modify. Vapor fugacity: Hyden- O Connel Vapor enthalpy: SRK-Modified- Panag. Check calculate transport property. 114
4)Reaction Stoichiometric Data input Make reaction sets SET1 and SET. SET1 includes reactions R1(forward reaction) and R(back reaction). Stoichiometric data for R1 is set as HOAC+MEOH ->MEAC+HO. Stoichiometric data for R is set as MEAC+HO ->HOAC+MEOH. So that the GUI appearance finally becomes as follows. 115
5)Input power law constants for reaction kinetics The following equation is set for R1 and R. R1: = 8 1493. 6 k f 9.73 10 exp RT k f R: = = 8 14049.4 kb 4.195 10 exp K RT eq 116
6)Reaction equilibrium data For reaction set SET, only R1;HOAC+MEOH ->MEAC+HO is set. The following equation is set for R1. ( ) 78. 98 ln K eq = 0.84156 + T 117
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 ) 118
.Process flow diagram Make a flow diagram that consists of two PFRs, an Equilibrium Reactor and a Gibbs Reactor. FEED FEED PFR1-POWER PFR1-OUT PFR-OUT Appropriately name each stream. PFR-PROCE FEED3 CONR-OUT CON-RX FEED4 GIBBS-OUT GIBBS-RX 1)Set the contents of PFR-1 This reactor uses power-law as the kinetics. FEED PFR1-OUT Temp.:77Cdeg Press.:1Kg/cm HOAC:80kg-mol/h MEOH:80kg-mol/h PFR1-POWER 119
)Set the contents of the PFR- This reactor uses Procedure for the kinetics. FEED PFR-OUT Define to FEED PFR-POWER 3) Set the contents of the Equilibrium Reactor FEED3 CONR-OUT CON-RX 10
4)Set the contents of Gibbs Reactor FEED4 GIBBS-OUT GIBBS-RX 3.RUN the simulation FEED PFR1-OUT PFR1-POWER FEED PFR-PROCE PFR-OUT Stream Name Stream Description Phase Fluid Rates HOAC MEOH MEAC HO KG-MOL/HR PFR1-OUT Mixed 50.355 50.355 9.7645 9.7645 PFR-OUT Mixed 50.355 50.355 9.7645 9.7645 CONR-OUT Mixed 49.4770 49.4770 30.530 30.530 GIBBS-OUT Mixed 57.4798 57.4798.50.50 FEED3 CON-RX CONR-OUT Rate Temperature Pressure Enthalpy Molecular Weight Mole Fraction Vapor Mole Fraction Liquid KG-MOL/HR C KG/CM M*KCAL/HR 560.000 77.0000 1.0000 4.4535 46.0474 0.7386 0.614 560.000 77.0000 1.0000 4.4535 46.0474 0.7386 0.614 560.000 77.0000 1.0000 4.4648 46.0474 0.7409 0.591 560.000 77.0000 1.0000 4.3444 46.0474 0.7163 0.837 FEED4 GIBBS-OUT GIBBS-RX 11
4.Results Q-1. PFR 1) Conversion of methanol at PFR with Power-law Recovery =( 1-49.445/80)x100 =8.341 % 5.Discussion ) Conversion of methanol at PFR with Procedure Recovery =( 1-49.445/80)x100 =8.341 % Because the exactly the same equations are used, the result must be exactly the same. Q-. Equilibrium Reactor Recovery =( 1-49.477/80)x100 =8.39 % 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 =( 1-57.48/80)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. 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 9.3%, 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. CH 3 COOH+ CH 3OH CH 3COOCH 3 + H O HOAC MeOH MeOAc Reactive distillation in the next session uses this principle. 1
WS-5-0 Production of ethylene oxide Problem According to a literature, the reaction kinetics for oxidation of ethylene on catalyst is described as follows; 1 C H 4 + O C H 4O C H 4 + 3O CO + H O 5 C H 4O + O CO + H O K K K 1 3 = 1.36x10 15 = 9.59x10 = 1.35x10 4 4.59x10 exp RT 4 5.7x10 exp RT 4 4.43x10 exp RT 17 13 Unit of each variable is; kg - mol kg - mol r K kg - cat s kg - cat s K1A1 AC1C r1 = ( 1+ A C + A C + A C ) r r 3 1 1 1 K A1 AC1C = ( 1+ A C + A C + A C 1 1 K 3 A1 ACC3 = ( 1+ A C + A C + A C 1 3 4.91x10 A = 1.9 exp RT 4 11 =.65x10 A 3.00 10 exp RT 3 3.37x10 A = 3 1.3 10 exp RT m 3 kg - mol cal A C kg - mol 3 R m mol K 3 3 3 3 3 3 ) ) 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: 50 degc(fixed) Pressure 10Kg/cm Feed flow rate:10kg-mol/h composition: ethylene 30mol%,O 7mol%,N 63mol% Temp.: degc, Press.:10Kg/cm ID: 1000mm, Length: 10m Q-: 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. 13
1.Preparation 1)UOM: Metric ) Components Set the following components just in the same order. ETHYLENE O EO CO HO N Set component phase as Vapor-Liquid. 3)Thermo system Select NRTL (single liquid). Modify it to calculate transport property. 14
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=.9*EXP(4.91E+3/(RGAS*RTABS)) A=3.00E-11*EXP(.65E+4/(RGAS*RTABS)) A3=1.3E+*EXP(3.73E+3/(RGAS*RTABS)) AK1=1E+15*PREEXP(1)*EXP(-ACTIVE(1)*1000/(RGAS*RTABS)) AK=1E+17*PREEXP()*EXP(-ACTIVE()*1000/(RGAS*RTABS)) AK3=1E+13*PREEXP(3)*EXP(-ACTIVE(3)*1000/(RGAS*RTABS)) C1=XVCONC(1) C=XVCONC() C3=XVCONC(3) SS=(1.0+A1*C1+A*C+A3*C3)** R1=AK1*A1*A*C1*C/SS R=AK*A1*A*C1*C/SS R3=AK3*A*A3*C*C3/SS RRATES(1) = R1*3600.0*800. RRATES() = R*3600.0*800. RRATES(3) = R3*3600.0*800. ISOLVE=1 RETURN 15
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 K = 1.36x10 1 15 4.59x10 exp RT 4 K = 9.59x10 17 5.7x10 exp RT 4 K = 1.35x10 3 13 4.43x10 exp RT 4 16
.Flow sheet setting 1)Addition of unit operations PLUG-IN PLUG-OUT R1 CSTR-IN CSTR-OUT R )Setting Plug Flow Reactor 17
3) Setting CSTR 4) Setting feed stream Feed for Plug Flow Reactor flow rate:10kg-mol/h composition: ethylene 30mol%,O 7mol%,N 63mol% Temp.: degc, Press.:10Kg/cm Feed for CSTR Reference to the feed stream for the Plug Flow Reactor 18
3.RUN the simulation PLUG-IN PLUG-OUT R1 CSTR-IN CSTR-OUT R Stream Name Stream Description Phase PLUG-IN Vapor PLUG-OUT Vapor CSTR-IN Vapor CSTR-OUT Vapor Fluid Rates ETHYLENE O EO CO HO N KG-MOL/HR 3.0000 0.7000 6.3000.647 0.064 0.1737 0.367 0.367 6.3000 3.0000 0.7000 6.3000.781 0.111 0.1307 0.84 0.84 6.3000 Rate KG-MOL/HR 10.000 9.913 10.000 9.935 Temperature C Pressure KG/CM Enthalpy M*KCAL/HR Molecular Weight Mole Fraction Vapor Mole Fraction Liquid 10 1 0.0135 8.3045 1.0000 5 1 0.0389 8.555 1.0000 10 1 0.0135 8.3045 1.0000 5 1 0.0380 8.4907 1.0000 19
4.Chage the flow sheet 1) Addition of calculator and optimizer PLUG-IN PLUG-OUT R1 CSTR-IN CSTR-OUT R CA1 OP1 ) Setting the content of calculator 130
3) Setting the content of Optimizer 5.RUN the simulation Right click the optimizer and execute RUN RESULT. 131
6.RESULT Q-1 Conversion at Plug Flow Reactor Conv.=(0.1737/3.0)x100=5.79% Q- Conversion at CSTR Conv.=(0.1307/3.0)x100=4.36% Q-3 Kinetic parameter (pre-exponential factor) 4.35 kg-mol/(kg-cat sec) 13
WS-5-03 Heat balance of Reactor Theme Q-1 Check the heat of reaction at the PFR of WS-5-0 is correctly calculated. Q- Check the heat duty of the PFR of WS-5-0 is correctly calculated. 133
1.Change of flow sheet Open the file made at WS-5-0. 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..run the simulation PLUG-IN PLUG-OUT R1 CSTR-IN CSTR-OUT R Stream Name Stream Description Phase Fluid Rates ETHYLENE O EO CO HO N KG-MOL/HR PLUG-IN Vapor 3.0000 0.7000 6.3000 PLUG-OUT Vapor.647 0.064 0.1737 0.367 0.367 6.3000 CSTR-IN Vapor 3.0000 0.7000 6.3000 CSTR-OUT Vapor.781 0.111 0.1307 0.84 0.84 6.3000 Rate KG-MOL/HR 10.000 9.913 10.000 9.935 Temperature Pressure Enthalpy Molecular Weight Mole Fraction Vapor Mole Fraction Liquid C KG/CM M*KCAL/HR 10 1 0.0135 8.3045 1.0000 5 1 0.0389 8.555 1.0000 10 1 0.0135 8.3045 1.0000 5 1 0.0380 8.4907 1.0000 134
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. 135
From PLUG FLOW HEAT BALANCE retrieve enthalpy of reactant and product both at operating condition and reference condition. H1 H4 H H3 4.Check 1)Heat of formation Component A: HEAT FORM. B: Change AxB kcal/kg-mol kg-mol/h kcal/h ETHYLENE 1541.8-0.3573-4481.18514 O 0-0.6376 0 EO -1570.46 0.1737-183.48890 CO -93988.6 0.367-3451.48907 HO -57756.9 0.367-108.10969 N 0 0 0 Total -6385.78 Agrees with 0.064 M*Kcal/h in the output file ) Duty Duty = (H-H1)+ DH + (H4-H3) = -0.046 M*Kcal/h Agrees with -0.046 M*Kcal/h in the output file 5.Results Q-1 Heat of reaction at the PFR of WS-3-0 is correctly calculated. Q- Heat duty of the PFR of WS-3-0 is correctly calculated. 136