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June 21, 2016, 15:28 |
heterogenous Model of a fixed bed Reactor
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#1 |
New Member
Julian Zoller
Join Date: Jun 2016
Posts: 4
Rep Power: 10 |
Hello everyone,
I try to implement a heterogeneus CO2 methanation fixed bed reactor model in Ansys Fluent 17.0. First i only try to simulate flow and heat transport without reaction. I need two energy equations one for the fluid and one for the solid phase, because i need solids temperature for calculation of reaction rate. My problem is to implement a energy equation for the solid phase of the porous media model to get its temperature. I've read some old threads on this issue and therefore tried three different methods: 1. Enable the non equilibrium button in the porous media dialoq box -> Contour plots for instance of velocity contain several rectangular areas with no velocity at all while the velocity in the other regions is quite the same as in the equilibrium. The borders between these regions are abrupt. 2. Use the Dual cell heat exchanger model for a second energy equation but i dont know what to enter in the heat exchanger data table. 3. Use an UDS as solid energy equation with solids temperature as skalar. I dont need the diffusion term of the UDS but equal it to zero causes an error. Also my source terms only lead to divergence (See udf's below). Thank you very much for every hint. DEFINE_SOURCE(fluid_heat_source,c,t,dS,eqn) { real Tm=333.15; /* Mittelwert Einlass und Wand*/ real rho_fluid=1.045; real cp_fluid=1009; real lambda_fluid=0.0286; real kin_visc_fluid=19.27*pow(10,(-6)); real Pr=kin_visc_fluid*rho_fluid*cp_fluid/(lambda_fluid); real D=0.0754; real d=0.0064; real F=M_PI*pow(D,2)/4; real Massenstrom=0.002527777; real u0=Massenstrom/(F*rho_fluid); real Re_0=u0*d/kin_visc_fluid; real por_unendlich=0.4; real por_average=por_unendlich+(1-por_unendlich)*0.526*d/D; real alpha_P=(lambda_fluid/d)*((7-10*por_average+5*pow(por_average,2))*(1+0.7*pow(Re _0,0.2)*pow(Pr,(1/3)))+(1.33-2.4*por_average+1.2*pow(por_average,2))*pow(Re_0,0 .7)*pow(Pr,(1/3))); real a=M_PI*pow(d,2)*(1-por_average)/(pow(d,3)*M_PI/6); real source; source = alpha_P*a*((C_UDSI(c,t,0))-(C_T(c,t))); dS[eqn] = -alpha_P*a; return source; } DEFINE_SOURCE(solid_heat_source,c,t,dS,eqn) { real Tm=333.15; /* Mittelwert Einlass und Wand*/ real rho_fluid=1.045; real cp_fluid=1009; real lambda_fluid=0.0286; real kin_visc_fluid=19.27*pow(10,(-6)); real Pr=kin_visc_fluid*rho_fluid*cp_fluid/(lambda_fluid); real D=0.0754; real d=0.0064; real F=M_PI*pow(D,2)/4; real Massenstrom=0.002527777; real u0=Massenstrom/(F*rho_fluid); real Re_0=u0*d/kin_visc_fluid; real por_unendlich=0.4; real por_average=por_unendlich+(1-por_unendlich)*0.526*d/D; real alpha_P=(lambda_fluid/d)*((7-10*por_average+5*pow(por_average,2))*(1+0.7*pow(Re _0,0.2)*pow(Pr,(1/3)))+(1.33-2.4*por_average+1.2*pow(por_average,2))*pow(Re_0,0 .7)*pow(Pr,(1/3))); real a=M_PI*pow(d,2)*(1-por_average)/(pow(d,3)*M_PI/6); real source; source = -alpha_P*a*((C_UDSI(c,t,0))-(C_T(c,t))); dS[eqn] = alpha_P*a; return source; } DEFINE_UDS_UNSTEADY(my_uds_unsteady,c,t,i,apu,su) { real physical_dt, vol, rho, phi_old; physical_dt = RP_Get_Real("physical-time-step"); vol = C_VOLUME(c,t); real por_unendlich=0.4; real por_average=por_unendlich+(1-por_unendlich)*0.526*d/D; real rho_solid=3000; real cp_solid=774.51; HC = rho_solid*(1-por_average)*cp_solid; *apu = -HC*vol / physical_dt;/*implicit part*/ phi_old = C_STORAGE_R(c,t,SV_UDSI_M1(i)); *su = HC*vol*phi_old/physical_dt;/*explicit part*/ } |
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June 27, 2016, 15:29 |
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#2 |
New Member
Julian Zoller
Join Date: Jun 2016
Posts: 4
Rep Power: 10 |
Hi,
i deleted UDS unsteady, returned 0 in UDS Diffusity and changed dS in solid_heat_source to dS[eqn] =-alpha_P*a; With this corrections the code work quite well. Greetings |
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November 2, 2016, 09:28 |
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#3 |
New Member
Join Date: Nov 2016
Posts: 1
Rep Power: 0 |
Hi Julian,
how did you model the heterogeneity in your simulation? I am trying to model a similar problem of an heterogenous catalyzed reaction in a packed bed. To save computation power I am trying to model it in 2D. So far we have modelled the reactor using a porous medium and a volumetric reaction, but the substance transport resistance/temperature difference between porous catalyst and gas phase is not being accounted for. Greetings, |
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Tags |
extra energy equation, heterogenous, porous media, reactor, solid temperature |
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