Question Bank • QB_APE_PROP_013 NAT • 2 Marks

Section 6: Agricultural Process Engineering

Engineering Properties of Agriculture Produce

Subtopic: Electrical properties

Question Statement:

A dielectric heating system operates at a microwave frequency of $f = 2.45\text{ GHz}$ to treat corn kernels. The root-mean-square electric field intensity inside the grain bed is $E = 350\text{ V/m}$. The corn has a dielectric loss factor of $\varepsilon'' = 0.85$. The permittivity of free space is $\varepsilon_0 = 8.854 \times 10^{-12}\text{ F/m}$. The volumetric power dissipation rate in the grain bed is ________ $\text{kW/m}^3$ (round off to two decimal places).

Answer Key:

14.19

Step-by-Step Solution & Explanation

Multi-Approach Solution: **Method 1: Volumetric Dielectric Heating Power Formula** $P_v = 2 \pi f \varepsilon_0 \varepsilon'' E^2$ Given: - $f = 2.45 \times 10^9\text{ Hz}$ - $\varepsilon_0 = 8.854 \times 10^{-12}\text{ F/m}$ - $\varepsilon'' = 0.85$ - $E = 350\text{ V/m} \implies E^2 = 122500\text{ V}^2/\text{m}^2$ Calculate intermediate factor: $2 \pi f \varepsilon_0 = 2 \times 3.14159265 \times 2.45 \times 10^9 \times 8.854 \times 10^{-12} \approx 0.136278\text{ S/m}$ Power dissipation per unit volume: $P_v = 0.136278 \times 0.85 \times 122500 = 14189.6\text{ W/m}^3 \approx 14.19\text{ kW/m}^3$ **Method 2: Equivalent AC Conductivity** $\sigma_{ac} = 2 \pi f \varepsilon_0 \varepsilon'' = 0.136278 \times 0.85 \approx 0.115836\text{ S/m}$ $P_v = \sigma_{ac} E^2 = 0.115836 \times 122500 = 14189.9\text{ W/m}^3 = 14.19\text{ kW/m}^3$
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