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who can help me about my udf on piston...

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Old   August 11, 2008, 05:25
Default who can help me about my udf on piston...
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i have to simulate a case about stirling engine, it is like the motion of piston,i write a udf to control the piston motion, but the result shows the pressure shakes and increases the pressure value.In normal situlation, the pressure should be shake on steady,the pressure can't always increase, who can help me to solve the problem, please!!! i had think is it my UDF wrong about the "force". can everyone help me!!!>"<I really need someone can help me !

it is my UDF #include <stdio.h> #include "udf.h" # define UDF_FILENAME "piston.dat" # define PI 3.141592654 # define rp 0.0067 # define l1 0.079 # define rpm 500. # define w 2.*PI*rpm/60. # define q w*time FILE *fp;

DEFINE_CG_MOTION(piston1, dt, cg_vel, cg_omega, time, dtime) { Thread *t=DT_THREAD (dt); face_t f; real force, loc; real velo, pressure; real v1, v2, v3; fp= fopen("piston.dat","a"); NV_S(cg_vel,=,0.0); NV_S(cg_omega,=,0.0); if (!Data_Valid_P ())

return; force=0.0;

begin_f_loop(f,t) {

real *AA;

AA = F_AREA_CACHE (f, t);

force += F_P (f, t) * AA[0];

if (rp_axi)

force *= 2.0 * PI;

pressure = force / 0.0099*0.0099*PI; } end_f_loop (f, t)

velo = rp*w*sin(PI/2.-q)+1./sqrt(2.*rp*cos(PI/2.-q)* 2.*rp*cos(PI/2.-q)-4.*(rp*rp-l1*l1))*(2.*rp*cos(PI/2.-q)) *rp*sin(PI/2.-q)*(w);

loc = 0.01945+0.02+rp*cos(PI/2.-q)+0.5*sqrt ((2.*rp*cos(PI/2.-q))*(2.*rp*cos(PI/2.-q))-4.*(rp*rp-l1*l1));





Message ("\nUDF valve:%f,%f,%f,%f,%f,%f\n",time,loc,velo,dtime

,v3,pressure); fprintf (fp, "%e %e %e %e %e %e\n", time, loc, velo, dtime,

v3,pressure); fclose (fp); }
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Old   May 28, 2014, 10:37
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I used fluently with Visual Studio in the study of Stirling Engine
I have a problem when the engine speed decreases Increase heat spread
Knowing that I used the sinusoidal motion
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