Navigation

    CFD中文网

    CFD中文网

    • Login
    • Search
    • 最新

    聚并破碎的SQMOM方法

    Algorithm
    1
    1
    369
    Loading More Posts
    • Oldest to Newest
    • Newest to Oldest
    • Most Votes
    Reply
    • Reply as topic
    Log in to reply
    This topic has been deleted. Only users with topic management privileges can see it.
    • 李东岳
      李东岳 管理员 last edited by 李东岳

      对于给定的NDF,划分为$i$个$N_{pp}$,对每个$i$上定义$k$阶矩$m_k^i$,给定$m_k^i$,可以计算第$i$区间的节点$d^i_0,d^i_1$以及权重$w^i_0,w^i_1$:
      \begin{equation}
      \begin{split}
      w^i_0&=w^i_1=0.5
      \\
      d^i_0&=m_1^i-\frac{1}{\sqrt{3}}\sqrt{\frac{m_3^i}{m_1^i}-{m_1^i}^2}
      \\
      d^i_1&=m_1^i+\frac{1}{\sqrt{3}}\sqrt{\frac{m_3^i}{m_1^i}-{m_1^i}^2}
      \end{split}
      \end{equation}
      对于仅考虑破碎的PBE:
      \begin{equation}\label{pbe}
      \frac{\p n(d)}{\p t}=\int_d^{d_{max}}g(d')\beta(d|d')n(d')\rd d'-g(d)n(d)
      \end{equation}
      对方程\eqref{pbe}在$i$上取$k$阶矩:
      \begin{equation}\label{m}
      \frac{\p m_k^i}{\p t}=\int_{d_{i-1/2}}^{d_{i+1/2}}\int_d^{d_{max}}g(d')d^k\beta(d|d')n(d')\rd d'\rd d-\sum^2_{j=0} g(d_j^i)w_j^i(d_j^i)^k
      \end{equation}
      \begin{equation}
      \begin{split}
      \int_{d_{i-1/2}}^{d_{i+1/2}}\int_d^{d_{max}}g(d')\beta(d|d')n(d')\rd d'\rd d&=
      \int_{d_{i-1/2}}^{d_{max}}g(d')n(d')\left(\int_{d_{i-1/2}}^{d'}\beta(d|d')\rd d\right)\rd d'
      \\&=
      \sum_{m=i}^{N}\sum_{j=0}^2g(d_j^m)w_j^m\left(\int_{d_{i-1/2}}^{d_j^m}d^k\beta(d|d_j^m)\rd d\right)
      \end{split}
      \end{equation}
      Therefore
      \begin{equation}
      \frac{\p m_k^i}{\p t}=\sum_{m=i}^{N}\sum_{j=0}^2g(d_j^m)w_j^m\left(\int_{d_{i-1/2}}^{d_j^m}d^k\beta(d|d_j^m)\rd d\right)-\sum^2_{j=0} g(d_j^i)w_j^i(d_j^i)^k
      \end{equation}

      线上CFD课程 7月1日报名截止 http://dyfluid.com/class.html
      CFD高性能服务器 http://dyfluid.com/servers.html

      1 Reply Last reply Reply Quote
      • First post
        Last post

      CFD中文网 | 东岳流体 | 京ICP备15017992号-2
      论坛登录问题反馈可联系 li.dy@dyfluid.com