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Electrostatistic Precipitation 159

Above the corona threshold the interelectrode space charge becomes significant and
must be included in Eq. (2). Under these conditions, the solution of Eq. (2) yields
/
 i  2  i   12
2
E =  1 + r  0  E − 1   (4)
c
  2π Km i  r   2π Km  
i 
0
0
where m is the ionic mobility, i is current per unit length of wire, and E is the minimal
i 1 c
field intensity required to form a corona and also the field strength at the wire (r − r ).
0
This is the maximum field strength, and as r increases, E decreases until for sufficiently
large r,
 i  12 /
E =  1 
 2πK 0 m  (5)
i
The electric field in terms of the potential V can be expressed as E =−dV/dr.
Therefore, integrating the right-hand side of Eq. (4) with respect to r gives the corona
current–voltage relationship

 1 +(1 + φ) 12 / 
V = V + E r (  1 + φ) 12 / −− ln  (6)
1
0
c 0
c
  2  
where V is the corona-starting potential related to E by Eq. (3) as follows:
c 0
E = V 0 r r ) (7)
c
ln
0 0
r ( 1
φ is a dimensionless current given by
φ =  r 1   2 i 1 (8)

 Er  2 πKm i
c 0
0
Thus far, the space charge considered is the ionic space charge with no consideration
given to the presence of charged particles in the interelectrode space. Such particles do
contribute to the total space charge, and their contribution to the total field may be approx-
imated as follows. The particle space charge density σ and ionic space charge density σ
p i
are assumed to be independent of position, (i.e., r). Inclusion of these two densities into
Poisson’s equation and solving for E gives
2
2
r
E = V c + i ( σ +σ p) ( r − ) (9)
0
ln r r ) 2 K r
0 0
r ( 1
Integration of Eq. (10) gives the potential of the wire as
σ + σ 
V = V +  i p  1 2 (10)
r
0
c
 4 K 0 