2D formulation for Electrostatic Problems

From KratosWiki
(Difference between revisions)
Jump to: navigation, search
Line 15: Line 15:
  
 
can be written as (see the [[General_formulation_for_Electrostatic_Problems | General formulation for Electrostatic Problems]]):
 
can be written as (see the [[General_formulation_for_Electrostatic_Problems | General formulation for Electrostatic Problems]]):
 +
 +
 +
::<math>
 +
    {
 +
    \int_{\Omega} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B} \mathbf{a}  \partial \Omega +
 +
    \oint_{\Gamma_{\infty}} \mathbf{N^T} \alpha \mathbf{N} \mathbf{a} \partial \Gamma_{\infty} =
 +
    \int_{\Omega} \mathbf{N^T} \rho_v \partial \Omega -
 +
    \oint_{\Gamma_q} \mathbf{N^T} \bar D_n \partial \Gamma_q -
 +
    \oint_{\Gamma_V} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V
 +
    }
 +
  </math>
 +
 +
 +
::<math>\mathbf{K} \mathbf{a} \,= \mathbf{f}</math>
  
  

Revision as of 15:05, 30 October 2009

The 2D Electrostatic Poisson's equation given by the governing PDE and its boundary conditions:


 A(V) = \left[ \frac{\partial}{\partial x}\cdot \left( \varepsilon_{x} \cdot \frac{\partial}{\partial x}\right) + \frac{\partial}{\partial y}\cdot \left(\varepsilon_{y} \cdot \frac{\partial}{\partial y} \right) \right]V(x,y) + \rho_S = 0  ~~ in ~ \Omega


 B(V) = 
\begin{cases} 
  \left . V - \bar V = 0 \right |_{\Gamma_{V}}  & in ~ \Gamma_{\varphi} \\
  \left . \hat n \vec{D} - \bar D_n = 0 \right |_{\Gamma_{q}}  & in ~ \Gamma_{q} \\
  \left . \frac{\partial V}{\partial r} \right |_{\Gamma_{\infty}} \approx - \frac{V}{r} & in ~ \Gamma_{\infty}
\end{cases}


can be written as (see the General formulation for Electrostatic Problems):


 
    {
    \int_{\Omega} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B} \mathbf{a}  \partial \Omega + 
    \oint_{\Gamma_{\infty}} \mathbf{N^T} \alpha \mathbf{N} \mathbf{a} \partial \Gamma_{\infty} = 
    \int_{\Omega} \mathbf{N^T} \rho_v \partial \Omega -
    \oint_{\Gamma_q} \mathbf{N^T} \bar D_n \partial \Gamma_q -
    \oint_{\Gamma_V} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V
    }


\mathbf{K} \mathbf{a} \,= \mathbf{f}


\mathbf{K}^{(e)}=
    \int_{\Omega^{(e)}} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B}  \partial \Omega^{(e)} + 
    \oint_{\Gamma_{\infty}^{(e)}} \mathbf{N^T} \alpha \mathbf{N} \partial \Gamma_{\infty}^{(e)}
\mathbf{f}^{(e)}=
    \int_{\Omega^{(e)}} \mathbf{N^T} \rho_v \partial \Omega^{(e)} -
    \oint_{\Gamma_q^{(e)}} \mathbf{N^T} \bar D_n \partial \Gamma_q^{(e)} -
    \oint_{\Gamma_V^{(e)}} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V^{(e)}



with (n is the number of nodes of the element):


 V (x,y,z) \cong \hat V (x,y,z) = \sum_{i=0}^n N_i (x,y,z) a_i = \mathbf{N}^{(e)} · \mathbf{a}^{(e)}


\mathbf{N^{(e)}} = 
   \begin{bmatrix} 
     N_1 \\ 
     \, \\
     N_2 \\ 
     \, \\
     \vdots \\
     \, \\
     N_n 
   \end{bmatrix}
\qquad
   \mathbf{a^{(e)}} = 
   \begin{bmatrix} 
     a_1 \\ 
     \, \\
     a_2 \\ 
     \, \\
     \vdots \\
     \, \\
     a_n 
   \end{bmatrix}



\mathbf{B_i}=
   \begin{bmatrix} 
     \frac{\partial N_i}{\partial x} \\ 
     \, \\
     \frac{\partial N_i}{\partial y} \\ 
     \, \\
     \frac{\partial N_i}{\partial z} 
   \end{bmatrix}


This is:


 
    {
    \int_{\Omega} W \left [ \nabla^T \mathbf{\varepsilon} \nabla \hat V + \rho_v \right ] \partial\Omega 
    + \oint_{\Gamma} \overline{W} \left [\mathbf{n}^T \mathbf{\varepsilon} \nabla \hat V + \bar \mathbf{q} + \alpha V  \right ] \partial \Gamma=0
    }


with \alpha = \frac{1}{r} the infinit condition factor and \bar \mathbf{q} the field produced whe V \, is fixed by \bar V \,.


The weak form of this expression can be obtained using the integration by parts. In addition, if      \bar W = - W \,:


 
    {
    \int_{\Omega} \nabla^T W^T \mathbf{\varepsilon} \nabla \hat V  \partial \Omega + 
    \oint_{\Gamma_{\infty}} W^T \alpha V \partial \Gamma_{\infty} = 
    \int_{\Omega} W^T \rho_v \partial \Omega -
    \oint_{\Gamma_q} W^T \bar D_n \partial \Gamma_q -
    \oint_{\Gamma_V} W^T \mathbf{q_n} \partial \Gamma_V
    }


Remembering that:


 \hat V (x,y,z) = \sum_{i=0}^n N_i (x,y,z) a_i = \mathbf{N} \mathbf{a}^{(e)}


\nabla \hat V = \nabla \mathbf{N} \mathbf{a}^{(e)} = \mathbf{B} \mathbf{a}^{(e)}


is the gradient potential with:


\mathbf{B}= \left [ \mathbf{B_1, B_2 ... B_n} \right ]


and:


\mathbf{B_i}=
   \begin{bmatrix} 
     \frac{\partial N_i}{\partial x} \\ 
     \, \\
     \frac{\partial N_i}{\partial y} \\ 
     \, \\
     \frac{\partial N_i}{\partial z} 
   \end{bmatrix}


The electric field and electric displacement field can be written as follows:


\mathbf{q} = -  \mathbf{B} \mathbf{a}^{(e)}  \qquad   \mathbf{q'} = - \mathbf{\varepsilon} \mathbf{B} \mathbf{a}^{(e)}


We will now use the Galerkin Method W_i(x) \equiv N_i(x) \,. So, finally, the integral expression ready to create the matricial system of equations is:


 
    {
    \int_{\Omega} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B} \mathbf{a}  \partial \Omega + 
    \oint_{\Gamma_{\infty}} \mathbf{N^T} \alpha \mathbf{N} \mathbf{a} \partial \Gamma_{\infty} = 
    \int_{\Omega} \mathbf{N^T} \rho_v \partial \Omega -
    \oint_{\Gamma_q} \mathbf{N^T} \bar D_n \partial \Gamma_q -
    \oint_{\Gamma_V} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V
    }


\mathbf{K} \mathbf{a} \,= \mathbf{f}


Note that K is a coefficients matrix that depends on the geometrical and physical properties of the problem, a is the vector with the n unknowns to be obtained and f is a vector that depends on the source values and boundary conditions.


\mathbf{K}^{(e)}=
    \int_{\Omega^{(e)}} \mathbf{B^T} \mathbf{\varepsilon} \mathbf{B}  \partial \Omega^{(e)} + 
    \oint_{\Gamma_{\infty}^{(e)}} \mathbf{N^T} \alpha \mathbf{N} \partial \Gamma_{\infty}^{(e)}
\mathbf{f}^{(e)}=
    \int_{\Omega^{(e)}} \mathbf{N^T} \rho_v \partial \Omega^{(e)} -
    \oint_{\Gamma_q^{(e)}} \mathbf{N^T} \bar D_n \partial \Gamma_q^{(e)} -
    \oint_{\Gamma_V^{(e)}} \mathbf{n^T} \mathbf{N^T} \mathbf{q_n} \partial \Gamma_V^{(e)}
Personal tools
Categories