| 查看: 967 | 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
2本,初试303,0860求调剂
已经有4人回复
286求调剂
已经有26人回复
335求调剂
已经有11人回复
293求调剂
已经有9人回复
290调剂生物0860
已经有25人回复
295分求调剂
已经有11人回复
一志愿华中农微生物,288分,三年实验经历
已经有11人回复
材料相关专业344求调剂双非工科学校或课题组
已经有4人回复
求助调剂,跨调
已经有12人回复
308求调剂
已经有5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