找回密码
 注册
查看: 4076|回复: 7

泊肃叶流动

[复制链接]
发表于 2013-5-10 14:56:47 | 显示全部楼层 |阅读模式

马上注册,结交更多好友,享用更多功能,让你轻松玩转社区。

您需要 登录 才可以下载或查看,没有账号?注册

x
泊肃叶流动用D2Q9模型来模拟,关于粘度和Re数的关系,u=U*Lx/Re,它的U是怎么确定的?在演化处理过程中,我是添加的外力项。
发表于 2013-5-11 12:03:30 | 显示全部楼层
For the Poiseuille flow with a GIVEN Reynolds number Re, the following formula for the maximum velocity

U = F N^2 / 8 v

where v is the viscosity, F is the forcing magnitude, and N is the number of grid point across the channel. Also, the following condition must be satisfied for numerical stability in practice:

U / c_s < 0.3

Therefore, you will have the following upper bound for the forcing:

F = 8 U^2/(Re*N) < 0.24 / (Re*N)

where c_s^2 = 1/3 has been used. Once the forcing F is determined, so is U.
 楼主| 发表于 2013-5-11 18:02:48 | 显示全部楼层
谢谢,罗老师。
 楼主| 发表于 2013-5-11 18:03:21 | 显示全部楼层

回复 2# luo@odu.edu 的帖子

谢谢,罗老师
发表于 2013-5-11 21:24:39 | 显示全部楼层

回复 3# 1058343691 的帖子

Does my comment help at all?
 楼主| 发表于 2013-5-23 09:40:22 | 显示全部楼层

回复 5# luo@odu.edu 的帖子

我觉得还是有些不明白的。查了一些资料也没查到,罗老师你说的。按我的理解
U = F N^2 / 8 v,假设Nx=50, v=0.0135,F=密度*加速度,加速度取,0.005,密度取0.1,
代入F = 8 U^2/(Re*N) < 0.24 / (Re*N)中,不满足条件。感觉这个条件很难满足啊,我想根本问题是我理解有误。
发表于 2013-5-25 09:40:55 | 显示全部楼层
Mean density rho_0 should be set to 1.0 (because rho is NOT a meaningful
quantity in the incompressible flow), thus F = rho_0 a = a (a is the acceleration),
and you should use the following formula to pick a value of "a":

a < 0.24 / (Re*N)

That is: pick a value of viscosity nu, and a channel width N, then you can compute
Re, with Re and N, you can decide the approximated upper bound for the acceleration
(or force).

[ 本帖最后由 luo@odu.edu 于 2013-5-25 09:51 编辑 ]
发表于 2014-4-16 10:06:04 | 显示全部楼层

回复 7# luo@odu.edu 的帖子

罗老师,对于加速度就是重力加速度(重力驱动),我格子中的加速度怎么取呢?
您需要登录后才可以回帖 登录 | 注册

本版积分规则

快速回复 返回顶部 返回列表