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December 24, 2022, 18:24 |
udf of MHD body force
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#1 |
Member
sofia
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Hi ,
I am trying to simulate mhd flow in horizontal pipe with udf to calculate mhd body force , bf=(sigma/rho)*B*u B is the magnetic induction,u is the velocity so i write an udf but there is an error. Does some body know this error thanck you very much |
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December 24, 2022, 18:27 |
MHD body force
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#2 | |
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sofia
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Quote:
Magnetic Bodyforce calculation ************************************************** ********************/ #include "udf.h" #define B_0 10. /* y component of magnetic field */ #define SIG 5.5E-6 /*electrical conductivity*/ #define rho 1239.1244 DEFINE_MAGNETIC-BODYFORCE(Cell_bforce, cell,thread) { real bforce,rho,veloc u; veloc u=C-U(cell,thread); bforce = -(SIG/rho)*B-0*B-0*veloc u; return bforce; } |
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December 26, 2022, 05:01 |
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#3 |
Senior Member
Alexander
Join Date: Apr 2013
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fluent is using build in macros to execute with user's code
for your case you probably wanna go with DEFINE_SOURCE macro Code:
#include "udf.h" real B_0=10.0; /* y component of magnetic field*/ real SIG=5.5E-6; /*electrical conductivity*/ real rho=1239.1244; DEFINE_SOURCE(Cell_bforce, cell, thread, dS, eqn) { real source; real rho; real veloc_u; veloc_u = C_U(cell,thread); source = -(SIG/rho)*B_0*B_0*veloc_u; dS[eqn] = -(SIG/rho)*B_0*B_0; return source; }
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December 28, 2022, 17:30 |
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#4 | |
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sofia
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Quote:
The udf works well, but inserting it in another udf it doesn't work, please i need help; /************************************************** ******************* Calculation of the properties of Fe3O4 nanoparticles ************************************************** ********************/ #include "udf.h" #define FI 0.05 #define RHO_np 1398.194 #define SI_1 0.9830 #define SI_2 12.959 #define KTC_np 7 #define TI 5.E4 #define BETA_1 8.4407 #define BETA_2 -1.07304 #define CP_w 4181.8 #define KA 1.383E-23 #define SIi_1 2.8217E-2 #define SIi_2 3.917E-3 #define SIi_3 -3.0669E-2 #define SIi_4 -3.91123E-3 #define T_0 298.15 #define D_np 59.E-9 #define CP_np 640.0 #define B_0 10.0; /* y component of magnetic field */ #define SIG 5.5E-6; /*electrical conductivity*/ DEFINE_PROPERTY(cell_conductivity,cell,thread) { real ktc,ktc_w,temp,f,beta,rho_w; temp = C_T(cell,thread); f = ((SIi_1*FI+SIi_2*temp)/T_0)+(SIi_3*FI+SIi_4); beta = BETA_1*(pow(100*FI,BETA_2)); rho_w = (-3.570*(pow(10,-3))*(pow(temp,2))+(1.88*temp+753.2)); ktc_w = (-8.354*0.000001*(pow(temp,2)))+((6.53*0.001*temp)-0.5981); ktc = ((KTC_np+(2*ktc_w)-2*(ktc_w-KTC_np)*FI)/(KTC_np+(2*ktc_w)+(ktc_w-KTC_np)*FI))+(TI*beta*FI*rho_w*CP_w*(pow(((KA*temp )/(RHO_np*D_np)),0.5))*f); return ktc; } DEFINE_PROPERTY(cell_density,cell,thread) { real temp,rho_w,rho; temp = C_T(cell,thread); rho_w = (-3.570*(pow(10,-3))*(pow(temp,2))+(1.88*temp+753.2)); rho = (FI*RHO_np)+((1-FI)*rho_w); return rho; } DEFINE_PROPERTY(cell_viscosity,cell,thread) { real mu,mu_w,temp; temp = C_T(cell,thread); mu_w = (2.591*(pow(10,-5))*(pow(10,(238.3/(temp-143.2))))); mu = (SI_1*exp(SI_2*FI)*mu_w); return mu; } DEFINE_SOURCE(Cell_bforce, cell, thread, dS, eqn) { real source,rho_w,temp,veloc_u; temp = C_T(cell,thread); rho_w = (-3.570*(pow(10,-3))*(pow(temp,2))+(1.88*temp+753.2)); veloc_u = C_U(cell,thread); source = -(SIG/rho_w)*B_0*B_0*veloc_u; dS[eqn] = -(SIG/rho_w)*B_0*B_0; return source; } |
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December 28, 2022, 19:03 |
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#5 |
Senior Member
Alexander
Join Date: Apr 2013
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was
Code:
#define B_0 10.0; /* y component of magnetic field */ #define SIG 5.5E-6; /*electrical conductivity*/ Code:
#define B_0 10.0 #define SIG 5.5E-6
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December 29, 2022, 17:54 |
udf of MHD body force
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#6 |
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sofia
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December 29, 2022, 18:03 |
udf of MHD body force
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#7 |
Member
sofia
Join Date: Apr 2010
Location: world
Posts: 31
Rep Power: 16 |
Attachment 92827Attachment 92827Thanck you very much it run finally
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