Steam resonator

From the parts

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• w Water 0

  No water chosen; distilled assumed.

• T Starting temperature 20 °C

  Starting temperature assumed at 20 °C.

• σ Water conductivity 5 µS/cm

  Conductivity assumed at 5 µS/cm — distilled water — because the water chosen has no dissolved-solids reading.

• A Plate area 516.1 cm²

  Plate - 2x Parallel 8" x 10"'s plate area, as recorded.

• d Gap 6.35 mm

  Plate - 2x Parallel 8" x 10"'s gap, as recorded.

• No water chosen, so the cell is taken as full of distilled water at 20 °C. Pick a water to change that.
• Plate - 2x Parallel 8" x 10" is 2 cells; the heating is worked out for one pair of plates. The whole stack heats 2 times as much water with 2 times the power, so the time to the boil is the same.

The working, step by step

1. 1. Flat-plate cell capacitance and field

   C 5.764 nF

   The plates as a capacitor, and whether the field across the gap is past what water holds.

   $$5.764 \mathrm{nF}=\frac{{\epsilon }_0{\epsilon }_r·516.1 \mathrm{cm²}·1}{6.35}157.5=\frac{1000}{ }$$

   LaTeX

   5.764\,\mathrm{nF} = \frac{\varepsilon_0\varepsilon_r \cdot 516.1\,\mathrm{cm²} \cdot 1}{6.35\,\mathrm{mm}} \qquad 157.5 = \frac{1000\,\mathrm{V}}{6.35\,\mathrm{mm}}

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2. 2. Complex permittivity and loss (Cole-Cole)

   δ 0.1234

   What the water's dipoles do at this frequency — the relaxation loss the dielectric heating depends on, with the conduction kept separate so it is not counted twice.

   $${80.1}^{*}={\epsilon }_{\infty }+\frac{{\epsilon }_s-{\epsilon }_{\infty }}{1+{\left(j\omega \tau \right)}^{1-\alpha }}-j\frac{\sigma }{\omega {\epsilon }_0}$$

   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}

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

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3. 3. Heating water with a field: dielectric or conduction

   P 4.064 kW

   The heating, by dipoles and by current, and the time to the boil.

   $$1575 \mathrm{V/cm}=\frac{1000}{ }235.8 \mathrm{\mu W}=2\pi 10 \mathrm{kHz}·{\epsilon }_0·5.214e-5·{1575}^{ }·327.7 \mathrm{mL}4.064 \mathrm{kW}=5 \mathrm{\mu S/cm}·{1575}^{ }·327.7 \mathrm{mL}4.064 \mathrm{kW}=235.8 \mathrm{\mu W}+4.064 \mathrm{kW}$$

   LaTeX

   1575\,\mathrm{V/cm} = \frac{1000\,\mathrm{V}}{6.35\,\mathrm{mm}} \qquad 235.8\,\mathrm{µW} = 2\pi 10\,\mathrm{kHz} \cdot \varepsilon_0 \cdot 5.214e-5 \cdot 1575\,\mathrm{V/cm}^{2} \cdot 327.7\,\mathrm{mL} \qquad 4.064\,\mathrm{kW} = 5\,\mathrm{µS/cm} \cdot 1575\,\mathrm{V/cm}^{2} \cdot 327.7\,\mathrm{mL} \qquad 4.064\,\mathrm{kW} = 235.8\,\mathrm{µW} + 4.064\,\mathrm{kW}

   • The dipoles contribute 0.000% of the heating. At 10 kHz the water is being heated by the current through it — this is an electrode boiler, and the frequency is doing nothing the ions care about. Dipole heating overtakes conduction only in the gigahertz.
   • The conduction heating implies a current of about 4.06 A through the water at this voltage, all of which is also electrolysing it.

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