---
title: "Dielectric heating of laminated wooden board"
subject: "Utilization of Electrical Energy"
chapter: 5
key: "semester_7/utilization_of_electrical_energy:dielectric-heating-wooden-board-30mhz"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/dielectric-heating-wooden-board-30mhz
site: BCT Engineering
---

# Dielectric heating of laminated wooden board

Utilization of Electrical Energy · Chapter 5: Electric Heating

- Asked in: 2079 Baisakh, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/dielectric-heating-wooden-board-30mhz
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-5

## Question

A laminated wooden board 30 cm × 25 cm thick is to be heated from 20°C to 170°C in 10 min by dielectric heating using 30 MHz supply source. Specific heat of wood is 0.35 calorie per gram per °C and density 0.55 gm/cc, relative permittivity = 5, pf = 0.05. Determine the voltage across the work piece and current during heating. Assume loss of energy by conduction, convection and radiation is 15%.

## Short answer

Assuming 2 cm thickness, heat required is 181,306 J and input power is 355.5 W. With C = 166 pF, voltage is about 477 V and current is about 14.9 A.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the input power found from the heat needed?
   A. P_in = H x 0.85
   B. P_in = H/t only
   C. P_in = P_out x 0.15
   D. P_out = H/t, then P_in = P_out/0.85
   Answer: D. 15% loss means efficiency 0.85: 302.2 W useful gives 355.5 W input. (Key point: Power input with losses)

2. Which relation links power to voltage across the work?
   A. P = V/ R
   B. P = I^2 ω C
   C. P = V^2 ω C cos(phi)
   D. P = V C f
   Answer: C. Loss in the dielectric is V^2 ω C cos(phi), with C = ε0 εr A/d. (Key point: Dielectric loss formula)

3. How is the current found once V is known?
   A. I = P/V^2
   B. I = P/(V cos(phi))
   C. I = V ω C
   D. I = P x V
   Answer: B. Real power P = V I cos(phi), so I = 355.5/(476.7 x 0.05) = 14.9 A. (Key point: Current)

## Related in Utilization of Electrical Energy

- [Dielectric heating: power, voltage, current (board to 160°C)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adielectric-heating-board-160c-power-voltage) — Utilization of Electrical Energy, chapter 5
- [Dielectric vs infrared vs microwave heating](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Adielectric-infrared-microwave-heating) — Utilization of Electrical Energy, chapter 5

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