| 查看: 973 | 回复: 1 | |||
[求助]
Fluent UDF 改单机多核
|
|
最近在学习udf的并行,也看了很多资料,仍然不能掌握,所以给出一个例子请求帮助。 以下是一个简单的定义物体x方向上的速度每一个时间步后速度增加0.001,如何改成单机多核,或者说不用改就能在单机多核上运行? 谢谢! #include "udf.h" #include "mem.h" #include "math.h" #include "dynamesh_tools.h" static real init_vel=0; DEFINE_CG_MOTION(sway,dt,vel,omega,time,dtime) { FILE *fd; init_vel+=0.001;/*迭代出新的速度*/ vel[0]=init_vel; fd=fopen("vel.dat","a" ;fprintf(fd,"%f,%f\n", time, vel[0]); fclose(fd); } |
» 猜你喜欢
291求调剂
已经有13人回复
085404 22408 309分求调剂
已经有11人回复
求调剂
已经有7人回复
有没有接收比较快的sci期刊呀,最好在一个月之内的,研三孩子求毕业
已经有6人回复
307中医考研调剂
已经有9人回复
327求调剂
已经有30人回复
294求调剂
已经有15人回复
0854求调剂
已经有27人回复
300求调剂
已经有12人回复
291求调剂
已经有10人回复
|
补充一点: 我的电脑是单机4核,用的15.0,采用的串行,编译上述udf得到的速度更新结果如下:它并不是每一个时间步速度增加,而是每个时间步内速度已经叠加16次,并且我也尝试过和每一步的迭代次数无关。 不知道问题出在哪里了,请有经验的朋友指点!感谢! timestep 速度 0.001000,0.001 0.001000,0.002 0.001000,0.003 0.001000,0.004 0.001000,0.005 0.001000,0.006 0.001000,0.007 0.001000,0.008 0.001000,0.009 0.001000,0.010 0.001000,0.011 0.001000,0.012 0.001000,0.013 0.001000,0.014 0.001000,0.015 0.002000,0.016 0.002000,0.017 0.002000,0.018 0.002000,0.019 0.002000,0.020 0.002000,0.021 0.002000,0.022 0.002000,0.023 0.002000,0.024 0.002000,0.025 0.002000,0.026 0.002000,0.027 0.002000,0.028 0.002000,0.029 0.002000,0.030 0.002000,0.031 0.002000,0.032 |
2楼2015-11-14 17:23:16













;
回复此楼
50