Page 246 - Finite Element Analysis with ANSYS Workbench

12.2 Finite Volume Method 237

result on the left-hand-side of the equation for every cell is less
than the specified tolerance.
The Jacobian matrix is determined from,

  = R 1     
R
A
where
 1 1 1 
  c 2 0 2c 2 2c 2 
 
  u 2  n y u  c x u  c x 
2
2
1 
R
 =  c v 2c v 2c  

 c
 c
 v n y y 
 c 2 x 2c 2 2c 2 
   V c 1  V c 1 
  V n  n  
 c 2 t 2c 2 2  2c 2 2  
 V 0 0 0 
 0 n V 0 0 

 =  n  

 0 0 V n  c 0 
 0 0 0 V n  c  

   c 2  u  v  
 
R
 =  V t  n y n x 0 
  V c c  u c  v  
 n x y 

 V n c  c x  u  c y  v   

These three matrices contain coefficients which are,
c
c = n x c x ; c = n
y
y
 (  ) 1
H
c =  p  ; p = (  ) 

1
2

1
 = (u  v 2 ) ;  = 
2
V = nu x  n v y ; V =  u n  n v x
n
t
y
and H =   (   ) 1