Fluids

Fluids

Pre-mixed combustion using finite rate/eddy dissipation model

    • krany3
      Subscriber

      I'm trying to simulate pre-mixed combustion using the finite rate/eddy dissipation model in fluent. I'm not getting the results properly. Can anyone help me on this?

    • Rob
      Ansys Employee

      Please can you elaborate on what's not behaving as expected? 

    • krany3
      Subscriber

      I've a sudden expansion region with combustion occurring exactly in that region. So, I've patched a volume with a high temperature of 1700 K to initiate combustion. Once, I started the simulation, the maximum temperature in the reacting zone is reaching 5000 K. Even after 10000 iterations in steady state, the temperature doesn't decrease. I thought patching won't affect the final solution. What should I do?

    • Rob
      Ansys Employee

      If you're hitting the limiter then it's definitely running very hot; I assume you're not expecting flame temperatures in that range?  Work through the boundary conditions, chemistry etc to see if it's all set correctly and the residuals are showing it's converged. 


      Post some images of the flow field, temperature etc as ANSYS staff cannot review case files: do this with node values on & off. We may spot something. If you also add the case & data someone in the wider community may be able to help too. 

    • krany3
      Subscriber

      Thanks. I'm checking all the possibilities to see if there is any issue in the physics. I appreciate your help.

    • DrAmine
      Ansys Employee

      Please make a summary of your case: boundaries used, material EOS and models used.

    • krany3
      Subscriber

      Sorry for the late reply. 


      Methane-air mixutre


      Velocity Inlet:


      Axial velocity at inlet = 7 m/s


      Pressure and Temperature at atmospheric conditions


      Turbulent intensity = 5%


      Turbulent viscosity ratio = 10


      Pressure outlet with atmospheric conditions with same turbulence parameters as inlet


      For turbulence modeling, REALIZABLE k-epsilon model


      For combustion modeling, FINITE RATE/EDDY DISSIPATION model. To ignite the methane-air mixture, I patched the reacting zone with 1700K. I didn't include nitrogen in the reaction. It's just methane oxidation. I think because of the oxidation, I'm experiencing very high temperatures. I tried to include nitrogen to see if there's any change. This time, there's no combustion though I patched with high temperature.


       

    • klu
      Ansys Employee

      Also try to patch very small amount of products of the reaction. Sometime it may help trigger the reaction to start.

    • krany3
      Subscriber

      I did patch the products. The combustion temperature reaches 600 K and starts decreasing after few iterations, reaching the initial temperature 300 K. Is there any other way, I can simulate combustion? Do I have to include nitrogen into the reactants, even though it is inert? If not, the temperature reaches to 5000 K in presence of oxygen. What is the remedy for this? I've tried all the possible ways.

    • Rob
      Ansys Employee

      Does this mean the 5000K temperature is reached with an oxy-fuel system rather than with air?  

    • krany3
      Subscriber

      Yes. I've chosen methane-air as the mixture. But, nitrogen being inert, it's not included in the reactants. Of course, I added it later to see if there's any change. Initially there's no combustion with the same conditions that were used without nitrogen. Later, I patched a fraction of products along with temperature, and this time there is combustion. But, as I mentioned earlier, the temperature raises to 500 K and starts decreasing after a few iterations. 

    • Rob
      Ansys Employee

      Just to clarify:


      Initial very high (and converged model) no nitrogen in the system so pure oxygen fuelled flame.


      New model contains nitrogen and flame goes out.

    • krany3
      Subscriber

      Is there any solution to curb the high temperature?

    • klu
      Ansys Employee

      When you patched a small amount of products, please also patch high temperatures at reaction zones. 


      Can you attach a few screenshots of your Fluent case?


      1. Geometry or computational domain.


      2. Viscous panel.


      3. Species panel.


      4. Solution Methods.


      5. Solution Controls. Make sure the energy relaxation factor is not lower than 0.95.


      In addition, can you try to use the methane-air-2step mixture instead of the 1 step one?


      Thanks.

    • krany3
      Subscriber

      I've a small doubt regarding the species' mass fractions. How should I calculate the mass fraction of the species in multi-step reactions? What should I enter in boundary conditions for species?

    • Rob
      Ansys Employee

      The species will enter the domain, (hopefully) react and leave. You only need to specify the fractions of reactants on the inlet, and I'd suggest some care on the outlet incase of reverse flow. Unless you're expecting intermediate species (CH3 radicals for example) on the inlet they'll have zero mass (mole) fraction. Fluent will calculate all of the combustion and therefore intermediate materials assuming you set it up. 

    • krany3
      Subscriber

      In my case, methane-air mixture, I've CO, H2O and O2 in the first step of the reaction in products, and I've CO2 in the second step in the products. If I want to calculate the mass fraction of CO, should I consider the mass of all the 4 species in products? 

    • Rob
      Ansys Employee

      The mass fraction is relative to the whole (domain) or local (cell) fluid volume depending on your definition. However, O2 is a reactant, so I'm not sure why you're using it as a product; the typical two step reaction is (I may or may not have balanced the equations):


      2CH4 + 3O2 -> 2CO + 4H2O


      2CO + O2 -> 2CO2 


      You wouldn't have a primary reactant as a product.  


       

    • krany3
      Subscriber

      But, I'm using lean condition. In my case, the 2 step reaction will be,


      CH4 + 2.985 O2 -> CO + 2H2O + 0.985 O2


      CO + 0.5 O2 -> CO2


       

    • Rob
      Ansys Employee

      OK, so how much energy is required to covert O2 to O2 in the reaction scheme? A lean mixture means there is an excess of oxygen, so there is usually enough available to burn out all the fuel. There are reaction schemes where O2 is a product, but unless you want to start including OH radicals I'd steer clear. 


      I'd suggest you read "Modeling Premixed Combustion" of the Fluent User's Guide, if you want the species fractions available you need to rethink your use of the premixed combustion model. 

    • krany3
      Subscriber

      I think the activation energy was approximately 51 MJ. I've gone through that user guide. As you said, premixed combustion model doesn't calculate the species' mass fractions. I'm not concerned about the species. I'm trying to observe the flame front. The C-equation gave me a more jet spread and the Enhanced Coherent Flame Model (ECFM) didn't predict the flame front location properly. So, I moved to the Finite Rate/ Eddy Dissipation model. For this model, I should consider the species' mass/mole fractions.

    • krany3
      Subscriber

      How to calculate the boundary species' mass fractions at the inlet and at the outlet, if I have the above mentioned 2-step reaction mechanism? I'm getting different values. 

    • Gilleseggenspieler
      Ansys Employee

      Hi,


      Let me give you an example with a 1-step mechanism. Also I noticed that you only consider O2 (pure oxygen?) - I am using an example with air 


      The stochiometric mechanism is CH4 + 2 (O2 + 3.76 N2) -> CO2 + 2 H2O + 7.52 N2


      Mass fraction of CH4 (Y_CH4) = Mass of CH4 / Mass of the entire mixture - i.e.: Y_CH4 = 16 /(16 + 2*(32 + 3.76*28)


      Mass fraction of O2 (Y_O2) = Mass of O2 / Mass of the entire mixture - i.e.: Y_O2 = 2 * 32 /(16 + 2*(32 + 3.76*28)


      And of course Y_N2 = 1.0 - Y_CH4 - Y_O2


      Same operation but on the product side to compute the mass fractions for the outlet (assumes that all products are burned by the time the outlet is reached of course). 


      If you do lean combustion you will have something like


      CH4 + 2 * B (O2 + 3.76 N2) -> CO2 + 2 H2O + 7.52*B N2 + 2 (B-1) * O2 where B depends upon the equivalence ratio you use. (B>1.0 for lean combustion)


      Same operation to compute the inlet mass fractions, just include B in your computation.


      Outlet is your case (2-steps)  is trickier because it depends on how much CO was oxidized to CO2. I would start by assuming that all CO is oxidized (mass fraction of CO at the outlet = 0.0) and correct as needed.


      I hope this helps!


      Best,


      Gilles


       


       


       


       


       

    • krany3
      Subscriber

      I've no issues with the mass fraction calculations from the primary reaction or the one step reaction mechanism. As you said, it's getting trickier with the multi-step reaction mechanisms. If we assume that the CO is completely oxidized, then there will not be any CO at the outlet!


      What will be the mass fraction of oxygen at the inlet in 2-step reaction? If I calculate the mass fractions of H2O, CO2 and O2, should I just consider them as the product species for calculation? Since this is a lean combustion, I assume there will not be any CO in products. But, is there a case in lean combustion where I have to consider CO in calculating the mass fractions?

    • Gilleseggenspieler
      Ansys Employee

      What will be the mass fraction of oxygen at the inlet in 2-step reaction? Same as 1 step. The CO + 0.5 O2 -> CO2 only kicks-in in and after the flame and O2 comes from the fact that you run lean. So when computing O2 inlet mass fraction you only have to look at the 1 step.


      CO at the outlet: do not worry about it - it becomes a real problem only when you have backflow (to be avoided!) You can easily put Y_CO = 0 at the outlet and will be close enough. Lets say you run a simulation and see that Y_CO in the simulation by the exit get closer to a non-zero value. Simply adjust Y_CO @ outlet accordingly. That is the trick with Y_CO - you will not know what the exit BC should be until you run a simulation to see how much CO is left (i.e. not oxidized back in CO2). Y_CO @ outlet will strongly depend upon residence time and the post-flame temperature

    • krany3
      Subscriber

      Thanks Gilles! It makes sense. I'll try with that. 

    • krany3
      Subscriber

      Thanks @rwoolhu, @adenhadj and @klu for your help. I really appreciate it. I'm not yet done with the finite rate/eddy dissipation model. I might need help.

    • Rob
      Ansys Employee

      Not sure, other than that you may need to check you're in a combustible volume fraction range. As ANSYS staff we're unable to review cases on here, so we need to rely on the community. 

    • DrAmine
      Ansys Employee

      As my colleague said you need to check your setup. As ANSYS Staff we can only some suggestions and we do not review cases:


      1/Carry out flow without reactions at first


      2/Patch small values for product species mass fractions in the flame region (with higher temperatures). You can even at first increase B constant to very high values


      Eddy Dissipation and the extension to FR are very sensitive to the values of A and B


       

    • krany3
      Subscriber

      Thanks. I'll try to vary the values of A and B. 

    • krany3
      Subscriber

      Thanks!

    • krany3
      Subscriber

      When I select, either Eddy Dissipation or Finite Rate/Eddy Dissipation, why there is no NITROGEN in the reactions? Should I calculate the mass fractions considering nitrogen or without it? 

    • Rob
      Ansys Employee

      There probably isn't any nitrogen in the reaction as (normally) it's considered to be inert. You need to add materials based on the panels, and account for fuel, oxygen & nitrogen in your inlet mass fractions. 

    • krany3
      Subscriber

      Thank you!

    • krany3
      Subscriber

      How to write 1-step reaction mechanism into 2 or 3 step mechanisms?


      1 step reaction is CH4 + 2.985 (O2+3.76N2) -> CO2 + 2H2O + 0.985 O2 + 11.2236 N2


       

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