UDF growth rate, unsteady state
Hi!
I have a code for determining the growth rate of a crystal in terms of the supersaturation. However, I have realized that this code works only when I run the iterations in steady state. When I run iterations in unsteady state, I get an error message. Does anyone have any idea about what I need to change? Please see below the code Thanks! |
growth rate code
/************************************************** **********************UDF that computes the particle growth rate********************************************** ***************************/
#include "udf.h" #include "sg_pb.h" #include "sg_mphase.h" DEFINE_PB_GROWTH_RATE(growth_rate, cell, thread,d_1) { /* d_1 can be used if size-dependent growth is needed */ /* When using SMM, only size-independent or linear growth is allowed */ real G, S; real Kg = 2.8e-8; /* growth constant */ real Ng = 1.; /* growth law power index */ real T,solute_mass_frac,solvent_mass_frac, solute_mol_frac,solubility; real solute_mol_wt, solvent_mol_wt; Thread *tc = THREAD_SUPER_THREAD(thread); /*obtain mixture thread */ Thread **pt = THREAD_SUB_THREADS(tc); /* pointer to sub_threads */ Thread *tp = pt[P_PHASE]; /* primary phase thread */ solute_mol_wt = 74.55; /* molecular weight of potassium chloride */ solvent_mol_wt = 18.; /* molecular weight of water */ solute_mass_frac = C_YI(cell,tp,0); /* mass fraction of solute in primary phase (solvent) */ solvent_mass_frac = 1.0 - solute_mass_frac; solute_mol_frac = (solute_mass_frac/solute_mol_wt)/ ((solute_mass_frac/solute_mol_wt)+(solvent_mass_frac/solvent_mol_wt)); T = C_T(cell,tp); /* Temperature of primary phase in Kelvin */ solubility = 0.0005*T-0.0794; /* Solubility Law relating equilibrium solute mole fraction to Temperature*/ S = solute_mol_frac/solubility; /* Definition of Supersaturation */ if (S <= 1.) { G = 0.; } else { G = Kg*pow((S-1),Ng); } return G; } |
UDF cell growth
Hi,
Where do you linked your UDF code? I'm basically doing the same where i wish to determine the growth rate of cells in a reactor. However, im still in the process of learning how to do it. Anyone who can give some tips or infos related to this would is highly appreciated. |
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