Transcript Ozone Cell - Massachusetts Institute of Technology

Ozone Cell
Alan Millner 01-18-2005
Review of the Field of a Pair of
Oppositely Charged Plates in
Vacuum
 By symmetry, no field outside
 If plates have charge Q, area A, then by
Gauss’ Law D = 0E = Q/A between
plates
 If separation is d, then
 V = dE = dQ/ 0A or Q= (0A/d)
 If the capacitance C = 0A/d then
Q = C V
 Energy W = ½ QV = ½ 0E2 (dA)
 Or, W = ½ C V2
+Q
-Q
A
d
Polarization and Dielectric
Constant
 Suppose we have a polarization between the
plates P proportional to E
 Gauss’ Law says D = P+ 0E = Q/A
 Define dielectric constant in a material  by
 P = ( -0) so D =  E = Q/A
 So V = dE = dQ/ A , or Q = CV
 where capacitance C = A/d
 (substitute  for 0)
 Note W = ½ C V2 = ½  E2 (dA)
– Energy density times volume
Design Constraints
 Need 4E7 volts/meter across O2 to make O3
 Broad optimum around 30kHz
 Need insulating layer to avoid arc damage
– Alumina from Kyocera is good
 Need 1:1000 round shape factor for support
 Over 4kV becomes difficult to insulate
 Fewer larger cells are less expensive
 NOW YOU DESIGN AN OZONE CELL
Ozone Cell Construction
Ozone Cell
High Voltage
Wire
Steel shell
Alumina 3mm
ground
HV Electrode metal
90mm dia
Alumina 0.20mm
gas space 0.11mm
ground electrode
Steel shell
Ground
wire
Equivalent Circuit
Ozone Cell Equivalent Circuit
C2
R1
C1
C3
Ozone Cell Analysis
Dimensions
diameter
Th1 back
Th2 active
Airgap
mm
Dielectric const K
Mu0
Eps0
Area
Cback
Cdiel active
Cairgap
Cactive tot
Ctotal
m
90
3
0.2
0.11
0.09
0.003
0.0002
0.00011
9.5
1.25664E-06
8.85E-12
0.006361725
1.78E-10
2.68E-09
5.12E-10
4.30E-10
6.08E-10
Energy
W=  v*I dt
W=  v* (dQ/dt)* dt=  vdQ
For linear C, Q = C*V dQ= C dv
W=  C*vdv from v=0 to v=V
W = ½ C V2
Or W = ½ V*Q (area under curve of V vs Q)
Energy in a field
V= E*d = Efield/gap
Q=D*A= Dfield*area
W= (dA) * ( E*dD)
Volume V = dA
W = volume * energy per unit volume = V
*U
If linear, D = E and U =  E*dD = ½  E2
Displacement current
 I = dQ/dt
 Q = A*D = A * E
 So I = d(A* E )/dt, I/A = d(E )/dt
 We may identify d(E )/dt as displacement
current density
 Like j=I/A = current density
 Note later:
  H*dr =  {j+ d(E )/dt} dA
 so a loop around either current J or displacement
current d(E )/dt
 produces the same H field
Fringe fields
 If C = Co + Cfringe
 W = Wo + Wfringe

 Cfringe/Co = Wfringe/ Wo
 If Efringe <= Eo then
 Cfringe/Co <= Vfringe/ Vo
 Vo = dA
 Vfringe = d2 *P where P = perimeter
Fringe fields
example
Our example of a circular capacitor, radius
R
Vo = d* *R2
Vfringe = d2 * 2R
Vfringe/Vo = 2d/R
In our example, better than 1%.
Field Patterns
Consider 2 dimes, arranged on an axis, 10 meters
apart, oppositely charged to 1 coulomb
What does the field pattern look like:
1. within 1 mm of the positive dime's surface?
2. one meter from the positive dime?
3. 5meters from the axis center?
What is the field strength at each location above?