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September 21, 2013, 14:21 |
udf for heat generation
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#1 |
Member
sarah
Join Date: Aug 2010
Posts: 42
Rep Power: 16 |
hi ,
I am supposed to write a udf for a heat generation that depends on time and space . here its formula: Q(t0,ts)=Q_0(X)*A*(ts^ (–a)-( t_0 + ts)^(-a)) Q_0(X)=(5000*sin(0.285*x[0])+ 1800) t_0=35424000 i try to write a udf #include "udf.h" #include "math.h" DEFINE_SOURCE(energy_source,c,t,dS,eqn) { real x[ND_ND]; real source; real t = CURRENT_TIME ; real t_0; cell_t c; t_0=35424000; begin_c_loop(c,t) { C_CENTROID(x,c,t); if((t>0.1) && (t<10)) { source = (4958000*sin(0.285*x[0])+ 1866900)*0.0603*(pow(t,-0.0639)-pow(t+t_0, -0.0639) ; dS[eqn] = 1416001.489*cos(0.285*x[0])*0.0603*(pow(t,-0.0639)-pow(t+t_0, -0.0639) ; return source; } if((t>10) && (t<150)) { source = (4958000*sin(0.285*x[0])+ 1866900)*0.0766*(pow(t,-0.181)-pow(t+t_0, -0.181) ; dS[eqn] = 1416001.489*cos(0.285*x[0])* 0.0766*(pow(t,-0.181)-pow(t+t_0, -0.181) ; return source; } if((t>150) && (t<4000000)) { source = (4958000*sin(0.285*x[0])+ 1866900)*0.13*(pow(t,-0.283)-pow(t+t_0, -0. 283) ; dS[eqn] = 1416001.489*cos(0.285*x[0])* 0.13*(pow(t,-0.283)-pow(t+t_0, -0. 283); return source; } if((t>4000000) && (t<200000000)) { source = (4958000*sin(0.285*x[0])+ 1866900)*0.266*(pow(t,-0.335)-pow(t+t_0, -0. 335) ; dS[eqn] = 1416001.489*cos(0.285*x[0])* 0.266*(pow(t,-0.335)-pow(t+t_0, -0. 335); return source; } end_c_loop(c,t) } but there are some errors ! is there any nice person who help me to correct it? Thanks |
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October 16, 2013, 18:48 |
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#2 |
New Member
David Aliaga
Join Date: Oct 2013
Posts: 4
Rep Power: 13 |
Hey! its an old post but anyway,.. im working too with source terms .
i think you dont need the cell loop in the define_source macro, cause this is already looping on the cell thread zone. another thing, the variable t that you are using could be a problem, cause its been used for time and threat pointer at the same time. i hope it helps |
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