Dielectric heating: power, voltage, current (board to 160°C)

The initial temperature of the board is not given; 20°C is assumed. Calculations are done for efficiency 85% (the lower limit, which gives the largest power, used for sizing), with 90% shown for comparison.

Given: board $0.3 \times 0.15 \times 0.025$ m, final temperature 160°C, $t = 600$ s, $f = 30$ MHz, $c = 1465$ J/kg°C, density $= 575$ kg/m³, $\varepsilon_r = 5$, pf $= 0.05$.

Mass and heat required

$$\begin{aligned} m &= 575 \times (0.3 \times 0.15 \times 0.025) = 0.6469\ \text{kg} \\ H &= m c \Delta\theta = 0.6469 \times 1465 \times (160 - 20) = 132\,674\ \text{J} \\ P_{out} &= \frac{132\,674}{600} = 221.1\ \text{W} \end{aligned}$$

Power required

$$P_{in} = \frac{221.1}{0.85} = 260.1\ \text{W} \quad (\text{at } 90\%: 245.7\ \text{W})$$

Capacitance of the board

$$C = \frac{\varepsilon_0 \varepsilon_r A}{d} = \frac{8.854\times10^{-12} \times 5 \times (0.3\times0.15)}{0.025} = 79.69\ \text{pF}$$

Voltage across the work

Using $P = V^2\,2\pi f C \cos\phi$:

$$\begin{aligned} 2\pi f C \cos\phi &= 2\pi \times 30\times10^6 \times 79.69\times10^{-12} \times 0.05 = 7.510\times10^{-4} \\ V &= \sqrt{\frac{260.1}{7.510\times10^{-4}}} = 588.5\ \text{V} \end{aligned}$$

Current through the work

$$I = \frac{P}{V\cos\phi} = \frac{260.1}{588.5 \times 0.05} = 8.84\ \text{A}$$

Answer (85%): Power $\approx 260$ W, voltage $\approx 589$ V, current $\approx 8.84$ A.