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Complex permittivity and loss (Cole-Cole)

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

The formula
ε*=ε∞+εs−ε∞1+jωτ1−α−jσωε0
ε
Real permittivity — the part that stores
κ
Loss permittivity — the part that heats
δ
Loss tangent
LaTeX 
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}

Work it out

Fill from a part the library

What is in the cell. Conductivity is the parameter that dominates here, and it is what most separates these.

kHz

The drive frequency.

Compare with a variation

Result

δ Loss tangent 0.1234

Dissipated over stored, per cycle. Below ~0.01 the cell behaves like a capacitor; above ~1 it is closer to a resistor.

ε Real permittivity 80.1

The storage part — this is what sets capacitance.

κ Loss permittivity 9.887

The dissipation part. At VIC frequencies this is almost entirely ionic conduction, not dielectric relaxation.

λ Dipole loss alone 5.214e-5

The relaxation part of the loss permittivity with the conduction taken out — what the water molecule itself is doing. The heating calculation takes this, and the conductivity separately.

Q Effective Q of the dielectric 8.102

One over the loss tangent — the ceiling the water alone puts on the circuit’s Q, before any copper losses.

With your numbers
80.1*=ε∞+εs−ε∞1+jωτ1−α−jσωε0
LaTeX 
80.1^{*}(\omega) = \varepsilon_\infty + \frac{\varepsilon_s - \varepsilon_\infty}{1 + (j\omega\tau)^{1-\alpha}} - j\,\frac{\sigma}{\omega\varepsilon_0} \qquad \tan0.1234 = \frac{9.887}{80.1}

Worth knowing

  • A loss tangent of 0.123 caps the circuit Q at about 8 before any copper or core losses are counted.

This result is a link — the address bar holds your numbers, so it can be pasted into a post and opened to the same answer.

What this looks like

Permittivity across frequency The storage term and the two sources of loss for distilled / deionised, from 10 Hz to 100 GHz. Note where the relaxation actually is.
Permittivity across frequencyThe real permittivity holds flat at about 80 from 10 Hz until roughly 1 GHz, then falls to 5.2 past the relaxation frequency of 19 GHz. Below about 1 MHz the loss is entirely ionic conduction, which rises as frequency falls; dielectric relaxation loss only becomes the larger of the two above about 100 MHz. Every frequency a VIC is driven at is in the flat region.1e-81e-61e-41e-211e21e41e11e31e51e71e91e11your frequencyrelaxation, 19 GHzFrequency (Hz)Relative permittivityStorage, ε′Conduction lossRelaxation loss
The formula behind the curve
ε*=ε∞+εs−ε∞1+jωτ1−α−jσωε0
ε
Real permittivity — the part that stores
κ
Loss permittivity — the part that heats
δ
Loss tangent
LaTeX 
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}
The Cole-Cole plot Loss against storage, with the conduction term left out — which is the only way the arc is visible. Each point is one frequency; the curve is traced from low frequency on the right to high on the left.
The Cole-Cole plotA depressed semicircular arc running from ε′ = 80 at low frequency down to ε′ = 5.2 at high, peaking at a loss of about 36 near the relaxation frequency. The depression below a true semicircle is the α parameter, 0.02 here: it says the water does not relax at one single time constant but across a spread of them.010203040020406080100Storage, ε′Relaxation loss, ε″
The formula behind the curve
ε*=ε∞+εs−ε∞1+jωτ1−α−jσωε0
ε
Real permittivity — the part that stores
κ
Loss permittivity — the part that heats
δ
Loss tangent
LaTeX 
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}
Loss tangent against frequency Frequency swept from 5 kHz to 15 kHz with everything else held at your numbers. The dashed lines cross where you are.
Loss tangent against frequencyLoss tangent falls from 0.247 to 0.0823 as frequency rises from 5 kHz to 15 kHz. At your frequency of 10 kHz it is 0.123.0.050.10.150.20.2557.51012.51510 kHz0.123Frequency (kHz)Loss tangent
The formula behind the curve
ε*=ε∞+εs−ε∞1+jωτ1−α−jσωε0
ε
Real permittivity — the part that stores
κ
Loss permittivity — the part that heats
δ
Loss tangent
LaTeX 
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}
What moves the answer Each input moved 10% either way, with the others held still, and the effect on loss tangent.
What moves the answerLoss tangent is most sensitive to Frequency, which moves it by about 11.1% for a 10% change. It is the only input.Change in the answer when each input moves by 10%-20%-10%10%20%Frequency±11.1
The formula behind the curve
ε*=ε∞+εs−ε∞1+jωτ1−α−jσωε0
ε
Real permittivity — the part that stores
κ
Loss permittivity — the part that heats
δ
Loss tangent
LaTeX 
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.