Internal discharge in 0.5 mm air bubble

An internal (partial) discharge starts in a gas bubble when the voltage across the bubble reaches the breakdown voltage of the gas in it. The bubble and the remaining solid act as two capacitors in series.

Given

• Bubble thickness $t = 0.5$ mm $= 0.05$ cm
• Solid thickness $d = 1.0$ cm, so solid in series with the bubble $= d - t = 0.95$ cm
• $\varepsilon_r = 3.0$, applied voltage $30$ kV (rms), 50 Hz
• Breakdown strength of air $E_b = 30$ kV (peak)/cm

   +V  ======================
         solid  er = 3
        [ air bubble 0.5 mm ]
         solid
    0  ======================
          d = 1.0 cm

Equivalent capacitances (per unit area $A$)

$$C_1 = \frac{\varepsilon_0 A}{t}\ \ (\text{bubble}), \qquad C_2 = \frac{\varepsilon_0\varepsilon_r A}{d - t}\ \ (\text{solid in series})$$

Voltage across the bubble when $V$ is applied:

$$V_1 = V\,\frac{C_2}{C_1 + C_2} = \frac{V\,t}{t + \dfrac{d - t}{\varepsilon_r}}$$

Voltage across the bubble at breakdown

$$V_{1b} = E_b\,t = 30\times0.05 = 1.5\ \text{kV (peak)}$$

Applied voltage at which discharge starts

$$\begin{aligned} V_i &= V_{1b}\,\frac{t + (d - t)/\varepsilon_r}{t} = E_b\left[t + \frac{d - t}{\varepsilon_r}\right] \\ &= 30\left[0.05 + \frac{0.95}{3}\right] \\ &= 30\,(0.05 + 0.3167) = 11.0\ \text{kV (peak)} \\ V_{i,rms} &= \frac{11.0}{\sqrt2} = 7.78\ \text{kV (rms)} \end{aligned}$$

Answer: Internal discharge starts when the applied voltage reaches 11.0 kV (peak), i.e. about 7.78 kV (rms).

Since the working voltage of 30 kV (rms) is far above 7.78 kV (rms), discharges will occur in every half cycle. They slowly erode the bubble walls, form trees and finally cause breakdown of the solid. The insulation must therefore be made void-free (vacuum impregnation, careful moulding).