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Ozone Cell

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 =  0 E = Q/A between plates If separation is d, then V = dE = dQ/  0 A or Q= (  0 A/d)

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Ozone Cell

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  1. Ozone Cell Alan Millner 01-18-2005

  2. 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/ 0Aor 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

  3. 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

  4. 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

  5. Ozone Cell Construction

  6. Equivalent Circuit

  7. Ozone Cell Analysis

  8. 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)

  9. 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

  10. 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

  11. 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

  12. Fringe fieldsexample • 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%.

  13. 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?

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