Gibbs vs Equilibrium Reactors

[colancto]

Henk,

are you using HSC Chemistry software? If yes, do you see an opportunity and a need to make such a software somehow CAPE-OPEN compliant?

Michel

[HenkF]

Michel,
yes, I've been using HSC Chemistry for many years now. I love the way it has grown to the current state, including the SIM simulator, and the rather free way you can define and build your models (in an EXCEL kind of environment), but this also presents the weakness. Creating models is a messy process, making it hard(er) to reuse models. Some parts will easy integrate with CAPE-OPEN. In the passed I defined a wrapper to use it more integrated. Currently HSC Chemistry is shipped with a maintained GIBBS routine, marketed by Outotec and an unmaintained (?) SOLGASMIX executable. The latter may be wrapped to live within a CAPE-OPEN environment. Not too hard to implement, I guess.

regards, Henk

[nrgeng]

I am now able to meet your original request, from April 20, for the fsd file for your analysis. I have sent the three (3) fsd files, that I used in my comparison, containing the pure data from my thermodynamic calculations. The data have not been altered, no fictitious data have been used.

One file is for the Gibbs Minimization Reactor Unit and the other two are for the Equilibrium Reactor Unit.
One Equilibrium Reactor Unit file uses Method 1 and the second uses Method 3 as explained in my previous posting.

I e-mailed the three files to you. I hope that you can determine the cause(s) of the conversion discrepancies. Thanks for your help in advance.

[nrgeng]

What thermo did you use?

The thermodynamics that I used were as follows:

(1) The change in free energy can be derived from the combination of the individual heat capacities of the compounds involved in the reaction, the hydration of ethylene to ethanol.

(2) Knowing that the change in free energy is zero at equilibrium, the equilibrium constant, Keq, can be expressed as a function of the change in free energy.

(3) These two relations can be combined to provide the equilibrium constant as a function of absolute temperature.

(4) This relation can be manipulated into a usable form by regression analysis to provide the equations that I placed into the Reaction Package for the Reactor Units as Methods 1 and 3.

I hope that this explanation has answered your question.

[jasper]

I checked the results of the Gibbs reactor by making a fixed conversion reactor with the same reaction, and a measure unit to measure enthalpyF, T, entropyF and totalFlow. An information calculator calculates the total Gibbs energy from F * (H - T * S), a parametric study loops from 0.045% to 0.050% conversion (the Gibbs reactor predicts 0.047% conversion). Below image shows the results. So the Gibbs reactor seems to give the expected answer.

I can share the document if you are ok with it: your thermo is still in there, I removed your reaction package (the Gibbs reactor does not need it) and added a reaction package containing one reaction that matches that of the Gibbs reactor to set up the equilibrium reactor. For now I just attach the graph. Let me know if you want me to post the entire document.

The equilibrium reactors I cannot check: I can merely check that the reactor calculates what you specify, but I cannot check whether the equilibrium constants are correct (are you sure that this is an equilibrium constant? It looks like a rate equation) and that the units of measure are correct (see here for the units: http://www.cocosimulator.org/index_help ... ctions.htm)

Edit: note the scale of the Y-axis.

[jasper]

(1) The change in free energy can be derived from the combination of the individual heat capacities of the compounds involved in the reaction, the hydration of ethylene to ethanol.

... and the formation terms

(2) Knowing that the change in free energy is zero at equilibrium, the equilibrium constant, Keq, can be expressed as a function of the change in free energy.

yes.

(3) These two relations can be combined to provide the equilibrium constant as a function of absolute temperature.

yes

(4) This relation can be manipulated into a usable form by regression analysis to provide the equations that I placed into the Reaction Package for the Reactor Units as Methods 1 and 3.

I see a single Arrhenius term in there. I would typically expect two: one for the forward an one for the backward reaction.

[nrgeng]

Nomenclature:
Gibbs Minimization Reactor (GMR) Unit
CSTR--equilibrium mode (CSTR-eq) Unit
Equilibrium Reactor (Equilibrium) Unit

Previously submitted: (.fsd)
Ethylene (gibbs, RP) V1.1 M1 and Ethylene (equilibrium) V1.1 M1

The GMR Unit flowsheet provides the data to calculate (1) extent of reaction, (2) equilibrium constant (Keq) @ 320 C, and (3) conversion percentage. I have performed these calculations and I now agree with the two others that the GMR Unit flowsheet was/is correct. Thank you for all of your effort expended on the feedback.

The configuration of the CSTR-eq Unit was checked and found to be correct, though the units of the expression for the thermodynamically derived Keq were suspected as a possible error.

The thermodynamically derived Keq used in the CSTR-eq Unit is in agreement with the Keq calculated in Item 2; therefore the problem lies with equilibrium calculations (possibly the same subroutine) within both the CSTR-eq Unit and the Equilibrium Unit because these two Units provide identical results.

Your analyses and/or comments are solicited.

[jasper]

The results of the CSTR and equilibrium reactor are easily validated. Just calculate the product of A^n where n is the stoichoimetry of each compound and A is a property of the compound that corresponds to the basis, e.g. molar concentration (molar density times mole fraction). This should match the specified equilibrium constant, at the solution.

The CSTR and equilibrium reactor use different solvers.

But if you are in doubt regarding the solution of a particular equilibrium reactor, send me the fsd and I will check it for you.

Taking Ethylene (equilibrium) V1.1 M1.fsd that you sent last week as an example: the equilibrium constant as specified by your formula should have a value of 0.00191268 m3/mol (taking R=8.314 J/mol/K and T = 593.15 K) and the basis is molarity. The value must therefore correspond to

X[ethanol]*rho/(X[ethylene]*rho*X[water]*rho)=X[ethanol]/(X[ethylene]*X[water]*rho)=0.2796227/(0.36018865*0.36018865*1126.9096)=0.0019126

with (taken from the outlet stream, which is at reactor conditions):

rho=1126.9096 mol/m3
X[ethanol]=0.2796227
X[ethylene]=0.36018865
X[water]=0.36018865

So this solution appears correct.

(Edit: these numbers are close to the ones in the PDF, but not the same, probably because you put an equilibrium constant into the reactor that is slightly different and is the reported 0.002025? I cannot check whether this corresponds to the Gibbs reactor, but you can perform the same calculation as outlined above).

[nrgeng]

My problem, I think that it is a problem, is that I expect the Equilibrium Reactor Unit to present the same results, e.g., equilibrium conversions as presented by the Gibbs Minimization Reactor Unit.

@@@ Gibbs Minimization Reactor Unit:

Before deciding upon what to use for a financial analysis, it would be wise to understand the differences. Given reasonable thermo, I would expect the Gibbs and equilibrium results to be close to each other (if not the case, I suspect something is wrong with either the choice of thermo, or one of the two reactor configurations.

If you agree, then what I did was to calculate the Keq from the flowsheet data of Ethylene (gibbs, RP) V1.1 M1.fsd.

The GMR Unit flowsheet provides the data to calculate (1) extent of reaction, (2) equilibrium constant (Keq) @ 320 C, and (3) conversion percentage.

(Edit: these numbers are close to the ones in the PDF, but not the same, probably because you put an equilibrium constant into the reactor that is slightly different and is the reported 0.002025? I cannot check whether this corresponds to the Gibbs reactor, but you can perform the same calculation as outlined above).

The result was what you found, Keq = 2.025E-03. This Keq only applies to the operating conditions of 320 C and 50 atm. (Note: From my expression, Keq = 1.91268E-03, while from the Gibbs Minimization Reactor Unit, Keq = 2.025E-03. Perhaps too close for coincidence?)

I placed this single value into the CSTR--equilibrium mode Unit, which was shown in Posting 2.7 (Ethylene (CSTR equilibrium) V2.0.pdf).

@@@ Equilibrium Reactor Unit:

Taking Ethylene (equilibrium) V1.1 M1.fsd that you sent last week as an example: the equilibrium constant as specified by your formula should have a value of 0.00191268 m3/mol (taking R=8.314 J/mol/K and T = 593.15 K) and the basis is molarity. The value must therefore correspond to

X[ethanol]*rho/(X[ethylene]*rho*X[water]*rho)=X[ethanol]/(X[ethylene]*X[water]*rho)=0.2796227/(0.36018865*0.36018865*1126.9096)=0.0019126

Your posting for the calculation of Keq = 1.9126E-03 (for Ethylene (equilibrium) V1.1 M1.fsd) shows that the Equilibrium Reactor Unit provides the expected conversion results for the equilibrium expression that I derived and inserted in the Equilibrium Reactor Unit's GUI (via Reaction Package: Reaction M1).

@@@ Summary:
The equilibrium conversions of ethylene for the Gibbs Minimization Reactor Unit and the CSTR--equilibrium mode Unit were 0.047109 and 0.448211, respectively. This is almost a 1:10 difference that I did not expect. I am trying to determine the source of this wide discrepancy.

If the Gibbs Minimization Reactor Unit is presenting accurate conversion results and the Equilibrium Reactor Unit is also presenting accurate conversion results, then why are not the results the same?

I think that this is the essence of my problem. Do you know the answer?

[jasper]

If you calculate the equilibrium constant from the Gibbs reactor and put it in the Equilibrium Reactor they should give the same answer. So I am left to believe that you are using a different definition.

Going back to the Gibbs reactor you sent some time ago (Ethylene (gibbs, RP) V1.1 M1.fsd):

X[ethanol]*rho/(X[ethylene]*rho*X[water]*rho)=X[ethanol]/(X[ethylene]*X[water]*rho)=V*X[ethanol]/(X[ethylene]*X[water])=9.20499e-005

using (from the outlet stream)

V=0.00090895088 m3/mol
X[ethanol]=0.024111459
X[ethylene]=0.48794427
X[water]=0.48794427

putting Keq = 9.20499e-005 into the Ethylene (equilibrium) V1.1 M1.fsd document I get:

Stream 1 2 Unit
Pressure 50 50 atm
Temperature 250 320 °C
Flow rate 200 195.291 kmol / h
Mole frac Ethylene 0.5 0.487944
Mole frac Water 0.5 0.487944
Mole frac Ethanol 0 0.0241111

so the results are consistent. I hope this clears it up.