Dielectric heating of laminated wooden board

The board's thickness is missing from the question; a thickness of 2 cm is assumed (board 30 cm × 25 cm × 2 cm). The current does not depend on this assumption (see the note at the end).

Given: $A = 0.30 \times 0.25 = 0.075$ m², $d = 0.02$ m, $\Delta\theta = 170 - 20 = 150$°C, $t = 10$ min $= 600$ s, $f = 30$ MHz, $c = 0.35$ cal/g°C, density $= 0.55$ g/cc, $\varepsilon_r = 5$, pf $= 0.05$, losses 15% (efficiency 85%).

Heat and power required

$$\begin{aligned} \text{Volume} &= 30 \times 25 \times 2 = 1500\ \text{cm}^3, \quad m = 0.55 \times 1500 = 825\ \text{g} \\ H &= m c \Delta\theta = 825 \times (0.35 \times 4.186) \times 150 = 181\,306\ \text{J} \\ P_{out} &= \frac{181\,306}{600} = 302.2\ \text{W} \\ P_{in} &= \frac{302.2}{0.85} = 355.5\ \text{W} \end{aligned}$$

Capacitance of the work piece

$$C = \frac{\varepsilon_0 \varepsilon_r A}{d} = \frac{8.854\times10^{-12} \times 5 \times 0.075}{0.02} = 166.0\ \text{pF}$$

Voltage across the work

Dielectric loss: $P = V^2 \omega C \cos\phi$

$$\begin{aligned} V &= \sqrt{\frac{P}{2\pi f C \cos\phi}} = \sqrt{\frac{355.5}{2\pi \times 30\times10^6 \times 166.0\times10^{-12} \times 0.05}} \\ &= \sqrt{\frac{355.5}{1.5646 \times 10^{-3}}} = 476.7\ \text{V} \end{aligned}$$

Current

$$I = \frac{P}{V\cos\phi} = \frac{355.5}{476.7 \times 0.05} = 14.92\ \text{A}$$

Answer (for 2 cm thickness): Power input $\approx 355.5$ W, voltage $\approx 477$ V, current $\approx 14.9$ A.

Note: $V$ is proportional to the thickness (for 1 cm, $V \approx 238$ V; for 2.5 cm, $V \approx 596$ V), but the current stays 14.9 A because both the power and the capacitance scale with thickness.