Complex permittivity and loss (Cole-Cole)

How much of the energy put into the cell is stored, and how much just heats the water?

ε^{*}(\omega) = \varepsilon_\infty + \frac{\varepsilon_s - \varepsilon_\infty}{1 + (j\omega\tau)^{1-\alpha}} - j\,\frac{\sigma}{\omega\varepsilon_0} \qquad \tanδ = \frac{κ}{ε}

Method

1. Take the water’s Cole-Cole parameters: ε_s (permittivity at DC), ε_∞ (at optical frequencies), τ (the relaxation time, about 8.3 ps for water — a relaxation near 19 GHz), α (how spread out that relaxation is), and σ (the conductivity).
2. Work out ω = 2πf, and form (jωτ) raised to the power (1 − α). Below a gigahertz ωτ is minute, so this term is very nearly 1 and the dispersion part barely moves off ε_s.
3. Divide (ε_s − ε_∞) by 1 + that term, as complex numbers. The real part of the result adds to ε_∞ to give the storage permittivity; the imaginary part is the relaxation loss.
4. Add the conduction loss, σ ÷ (ω ε₀), to the imaginary part. At VIC frequencies this term is the whole story — it grows as frequency falls, which is why a cell is lossier at 1 kHz than at 100 kHz.
5. Divide the imaginary part by the real part to get the loss tangent, and take its reciprocal for the highest Q the dielectric will permit.

Assumptions

• The Cole-Cole parameters are for water at 20 °C. Warming shifts both the permittivity and the relaxation time; this calculation does not adjust for temperature — use the permittivity calculation for that part.
• Conductivity is taken as constant with frequency and field. In a real cell under high field it is neither: ion mobility rises with field strength, so a cell driven hard is lossier than this predicts.
• The electrodes are ideal. Real electrodes form a double layer that dominates the measured impedance below a few hundred hertz, and none of that is modelled here.
• This describes bulk water. It says nothing about what happens at the plate surface, which is where any gas actually comes from.