Dielectric Heating Calculator

Dielectric Heating Calculator Tutorial - Part 2

1. How the Dielectric Heating Calculator Works

The calculator is a one-dimensional, analytic model of volumetric heating in a lossy dielectric. It assumes a uniform RMS electric field inside the load, no heat loss to the surroundings and no phase change. These assumptions make it a fast sizing tool for the first stage of applicator design, before full-wave or multiphysics simulation.^[1][3]

Calculator inputs and outputs

The "First Term," "Second Term" and "Third Term" fields shown in Step 3 are intermediate values of the penetration-depth equation. They are displayed for checking and do not need to be entered.

Complex permittivity and loss tangent

A dielectric heats because its polarization lags the applied field. That lag is described by the complex permittivity, whose imaginary part ε″ converts field energy into heat.^[1][4][6]

Step 1 equation: power required for a temperature rise

The power the load must absorb follows from a simple energy balance on the heated mass:^[1][3]

Step 4 equation: power dissipated per unit volume

The fundamental dielectric heating equation gives the power converted to heat in each cubic meter of material by an RMS field E:^[1][2][3]

Step 2 equation: electric field required

Dividing Equation 2 by the heated volume V = M_a/ρ and equating it to Equation 3 gives the heating rate, and solving for E gives the field the calculator reports:^[1][2]

The calculator uses the rounded constant 5.56 × 10⁻¹¹ in Step 2 and the exact product 2πε₀ = 5.5632 × 10⁻¹¹ in Step 4. The difference is 0.058% in power and 0.029% in field, which is far below the uncertainty of any measured ε″.

Step 3 equation: power penetration depth

The field decays as it travels into a lossy material. The power penetration depth D_p is the distance at which the power density falls to e⁻¹ of its surface value. It equals half the field skin depth, 1/(2α):^[1][3][6]

For low-loss, non-magnetic materials (ε″ ≪ ε′, μ′ = 1), Equation 6 reduces to a convenient approximation:^[1][3]

Consistency check between Step 1 and Step 4

Multiplying the Step 4 result by the heated volume should return the Step 1 power. This is a quick check that the inputs are self-consistent:

2. Tutorial 1: Verify the Calculator with the Water Baseline

Problem: Before analyzing a new material, confirm that you can reproduce a known result. Heat 1 kg of water from 20 °C to 21.5 °C in 1 s at 2.45 GHz, using Metaxas' room-temperature properties ε′ = 78 and ε″ = 11.7.^[1]

1. Enter the Step 1 thermal inputs. Enter M_a = 1.0 kg, c_p = 4184 J/(kg·°C), ΔT = 1.5 °C and t = 1.0 s. These are the calculator's default values.
2. Read the absorbed power and check it by hand. The calculator returns P = 6,276 W.
   Equation 9 — Step 1 check for 1 kg of water $$P=\frac{\left(1.0\right)\left(4184\right)\left(1.5\right)}{1.0}=6,276\text{ W}$$ P = (1.0)(4184)(1.5) / 1.0 = 6,276 W
3. Enter the Step 2 inputs. Enter ρ = 997 kg/m³, f = 2.45E9 Hz and ε″ = 11.7. The calculator returns E = 1,981.42 V/m.
   Equation 10 — Step 2 check of the required field $$E=\sqrt{\frac{\left(997\right)\left(4184\right)\left(1.5\right)}{(5.56\times {10}^{-11})(2.45\times {10}^9)\left(11.7\right)\left(1.0\right)}}=1,981.4\text{ V/m}$$ E = √[(997)(4184)(1.5) / ((5.56×10^-11)(2.45×10^9)(11.7)(1.0))] = 1,981.4 V/m
4. Enter the Step 3 inputs and read the penetration depth. Enter ε′ = 78.0 and μ′ = 1.0. The calculator returns D_p = 0.014742 m (1.47 cm). With λ₀ = 0.12236 m and tan δ = 0.150, the bracketed term in Equation 6 is √(1.0225) − 1 = 0.011187. The low-loss shortcut, Equation 7, gives 1.470 cm, only 0.3% different.
5. Read the power per unit volume and close the energy balance. Step 4 returns P_v = 6,260,828 W/m³. The water volume is V = 1/997 = 0.001003 m³, so P_vV = 6,280 W, matching Step 1 to 0.06%. Substituting the field back into Equation 4 returns 1.501 °C/s.

Result: When all four checks close, the inputs and units are correct. Use the same checks on every new problem.

3. Tutorial 2: Choose a Heating Frequency for a Load

Problem: A 1 kg water-based load at 25 °C, 4 cm thick and heated from both faces, must rise 10 °C in 10 s. Compare the U.S. ISM heating frequencies of 915 MHz, 2.45 GHz and 5.8 GHz.^[11]

1. Obtain the dielectric properties at each frequency. Permittivity changes with frequency, so a single "room-temperature" value is not enough. For pure water, a validated double-Debye model gives ε′ and ε″ at any frequency and temperature:^[7][8]
   Equation 11 — Double-Debye model of pure water (Liebe, Hufford and Manabe) [7] $${\epsilon }_r\left(f\right)=\frac{{\epsilon }_s-{\epsilon }_1}{1+jf/f_1}+\frac{{\epsilon }_1-{\epsilon }_2}{1+jf/f_2}+{\epsilon }_2={\epsilon }^{\prime }-j{\epsilon }^{″}$$ ε_r(f) = (ε_s − ε_1)/(1 + jf/f_1) + (ε_1 − ε_2)/(1 + jf/f_2) + ε_2; θ = 300/(T°C + 273.15), ε_s = 77.66 + 103.3(θ − 1), ε_1 = 5.48, ε_2 = 3.51, f_1 = 20.27 − 146.5(θ − 1) + 314(θ − 1)² GHz, f_2 = 590 GHz
   At 25 °C the model gives the values in the table below. Note that it gives ε″ ≈ 9.1 at 2.45 GHz, lower than the 11.7 used in Tutorial 1; tabulated values for "room temperature" water differ because ε″ falls quickly as water warms (the model gives 12.3 at 15 °C and 10.5 at 20 °C).
2. Run the calculator at each frequency. Enter M_a = 1 kg, c_p = 4181 J/(kg·°C), ΔT = 10 °C, t = 10 s, ρ = 997 kg/m³ and μ′ = 1, then change f, ε′ and ε″ for each case.^[12]

   Calculator results: 1 kg of water at 25 °C, ΔT = 10 °C in 10 s

   Frequency	ε′	ε″	P (W)	E (V/m)	Dp (cm)	Pv (MW/m3)
   915 MHz	78.1	3.43	4,181	4,887.6	13.44	4.171
   2.45 GHz	77.2	9.07	4,181	1,836.8	1.89	4.171
   5.8 GHz	72.3	20.03	4,181	803.3	0.35	4.171

   The absorbed power and P_v do not depend on frequency; they are fixed by the thermal requirement. Frequency changes only the field needed and how deeply that power reaches.
3. Estimate heating uniformity from the penetration depth. For a slab of thickness d illuminated equally from both faces, and ignoring reflections and standing waves, the ratio of power density at the center to that at a surface is:
   Equation 12 — Center-to-surface power ratio for a slab heated from both faces (from Eq. 6) $$\frac{P_{\text{center}}}{P_{\text{surface}}}=\frac{2e^{-d/\left(2D_p\right)}}{1+e^{-d/D_p}}$$ P_center / P_surface = 2·e^(−d/(2D_p)) / (1 + e^(−d/D_p))
   For d = 4 cm the ratio is 0.99 at 915 MHz, 0.62 at 2.45 GHz and 0.007 at 5.8 GHz.
4. Select the frequency. At 5.8 GHz the center of the load would barely heat, the "cold center" familiar from reheated food. At 2.45 GHz the center heats at about 62% of the surface rate, which conduction may partly even out over a long cycle. At 915 MHz the slab heats almost uniformly, but the load is a small fraction of D_p, so the applicator must be designed to keep most of the power from passing through the load unabsorbed.

Result: For this 4 cm load, 915 MHz gives the most uniform heating. Choose 2.45 GHz when the load is no thicker than about 2 D_p (about 4 cm here, heated from both faces) and a center-to-surface ratio near 0.6 is acceptable, and 5.8 GHz only for thin films, coatings or small-diameter streams.

4. Tutorial 3: Temperature-Stepped Analysis from 20 °C to 80 °C

Problem: Heat 1 kg of water from 20 °C to 80 °C in 60 s at 2.45 GHz. Because c_p, ρ, ε′ and ε″ all change with temperature, a single calculation over 60 °C would be wrong. Break the problem into increments and use properties at each increment's mid-temperature.^[1][3]

1. Divide the heating range into increments. Use three 20 °C increments of 20 s each (20–40, 40–60 and 60–80 °C). Smaller increments reduce error; 20 °C is adequate for water, where properties change smoothly. Use smaller increments for materials whose ε″ changes sharply, such as frozen foods near thawing.
2. Tabulate the properties at each mid-temperature. Take c_p and ρ from steam-table data and ε′ and ε″ from Equation 11 (or from your own measurements):^[7][13]

   Water properties at the mid-temperature of each increment, 2.45 GHz

   Increment	Mid-T (°C)	cp (J/(kg·°C))	ρ (kg/m3)	ε′	ε″
   20–40 °C	30	4180	995.7	75.7	7.89
   40–60 °C	50	4181	988.0	69.9	4.88
   60–80 °C	70	4190	977.8	64.5	3.32

3. Run the calculator once per increment. For each row, enter M_a = 1 kg, ΔT = 20 °C, t = 20 s, f = 2.45E9 Hz, μ′ = 1 and that row's c_p, ρ, ε′ and ε″.

   Calculator results for each temperature increment

   Increment	P (W)	E (V/m)	Dp (cm)	Pv (MW/m3)
   20–40 °C	4,180	1,967.9	2.15	4.164
   40–60 °C	4,181	2,492.8	3.34	4.133
   60–80 °C	4,190	3,009.8	4.71	4.099

4. Interpret the trend in the electric field. The absorbed power is nearly constant, but the required field rises 53% because ε″ falls from 7.89 to 3.32. From Equation 5, E ∝ 1/√ε″ at fixed power:
   Equation 13 — Field scaling with loss factor at constant heating rate (from Eq. 5) $$\frac{E_2}{E_1}\approx \sqrt{\frac{{\epsilon }_1^{″}}{{\epsilon }_2^{″}}}=\sqrt{\frac{7.89}{3.32}}=1.54$$ E_2/E_1 ≈ √(ε″_1/ε″_2) = √(7.89/3.32) = 1.54 (calculator: 3,009.8/1,967.9 = 1.53)
   For water, the falling ε″ is self-limiting: hotter regions absorb less power. Materials whose ε″ rises with temperature behave the opposite way and can suffer thermal runaway, where hot spots grow hotter.^[1][3]
5. Account for frequency drift and the growing penetration depth. ε′ falls from 75.7 to 64.5 over the cycle, which shifts the resonant frequency of a cavity applicator ("frequency drift") and changes the match to the generator. D_p more than doubles, so a load that heated non-uniformly at the start heats more uniformly near the end. Record both trends; they set the tuning range and control strategy of the applicator.
6. Total the energy and check the average power. The energy absorbed is 20 × (4180 + 4181 + 4190) = 251.0 kJ, an average of 4.18 kW over 60 s.

Result: A constant-power, 4.2 kW absorbed source meets the requirement, but the field strength inside the water must rise from about 2.0 kV/m to 3.0 kV/m as it heats. A fixed-field design sized for the first increment would heat at only about 43% of the required rate in the last increment, since (1,967.9/3,009.8)² = 0.43.

5. Tutorial 4: Size a Continuous-Flow (Pipeline) Heater

Problem: Water flowing at 4 L/min through a microwave-transparent tube must be heated from 20 °C to 60 °C at 915 MHz. The heated section is 0.5 m long with a 30 mm inside diameter. Find the absorbed power, the field and whether the tube diameter suits the penetration depth.

1. Convert the flowing load into a mass and a residence time. The calculator's mass and time inputs become the mass inside the heated section and the time each fluid element spends there:
   Equation 14 — Residence time and absorbed power for a flowing load (from Eq. 2) $$V_h=\frac{\pi D^2}{4}L,t_{\text{res}}=\frac{V_h}{Q},P=\frac{M_a{c}_p\Delta T}{t_{\text{res}}}=Q\rho c_p\Delta T$$ V_h = (πD²/4)·L, t_res = V_h/Q, P = M_a·c_p·ΔT / t_res = Q·ρ·c_p·ΔT
   With D = 0.030 m and L = 0.5 m, V_h = 3.534 × 10⁻⁴ m³ (0.353 L). At Q = 4 L/min = 6.667 × 10⁻⁵ m³/s, t_res = 5.30 s. Using ρ = 992.2 kg/m³ at the 40 °C mid-temperature, M_a = 0.3507 kg.
2. Enter the inputs. Enter M_a = 0.3507 kg, c_p = 4179 J/(kg·°C), ΔT = 40 °C, t = 5.30 s, ρ = 992.2 kg/m³, f = 9.15E8 Hz, ε″ = 2.30, ε′ = 73.2 and μ′ = 1 (water at 40 °C and 915 MHz from Equation 11).
3. Read the results.

   Calculator results: 4 L/min water, 20 → 60 °C, 915 MHz, 30 mm × 0.5 m heated section

   Output	Value	Check
   Absorbed power, P	11,061 W	QρcpΔT = 11,057 W
   Electric field, E	16,354 V/m	RMS, inside the water
   Penetration depth, Dp	19.4 cm	6.5 times the tube diameter
   Power per unit volume, Pv	31.3 MW/m3	PvVh = 11,067 W

4. Judge the tube diameter against D_p. Because D_p = 19.4 cm is much larger than the 3 cm bore, the cross-section heats uniformly, but a single pass absorbs only part of the incident power. The applicator must therefore be resonant or multi-pass so that the water absorbs most of the energy. At 2.45 GHz the same water would have D_p of a few centimeters, so a smaller bore, or the 0.25 in tube suggested in the calculator's introduction, becomes the better match.
5. Size the generator. The calculator gives absorbed power only. Divide by the expected overall efficiency of the applicator, waveguide and tuning network. If that efficiency is estimated at 80% for this example, the generator must deliver 11,061 / 0.80 = 13.8 kW, so a 15 kW, 915 MHz source leaves reasonable margin. Measure the actual efficiency on the built system.
   Equation 15 — Generator power from absorbed power $$P_{\text{gen}}=\frac{P}{\eta }=\frac{11,061\text{ W}}{0.80}=13.8\text{ kW}$$ P_gen = P/η = 11,061 W / 0.80 = 13.8 kW (η = assumed overall efficiency)
6. Refine with a stepped analysis along the tube. ε″ of water at 915 MHz falls from about 4.0 at the 20 °C inlet to about 1.5 at the 60 °C outlet (Equation 11). In a uniform field the inlet end therefore heats faster than the outlet end. Split the heated length into segments, as in Tutorial 3, to see how the temperature profile develops along the tube.

Result: The pipeline heater needs about 11.1 kW absorbed and a 13.8 kW generator at an assumed 80% efficiency. The 30 mm bore heats uniformly at 915 MHz, and the applicator must be designed for efficient absorption in a load much thinner than D_p.

6. Common Input Mistakes

• Entering frequency in MHz or GHz. The frequency field expects hertz. Enter 2.45 GHz as 2.45E9; entering 2450 (the value in MHz) makes the field 1,000 times too large and D_p 10⁶ times too large.
• Entering tan δ instead of ε″. If your data gives the loss tangent, convert it first: ε″ = ε′ tan δ. For water at 2.45 GHz, entering 0.15 instead of 11.7 overstates the field by a factor of 8.8.
• Using properties at the wrong frequency or temperature. ε′, ε″ and c_p must match both the heating frequency and the mid-temperature of each increment. Use stepped analysis (Tutorial 3) over wide temperature ranges.
• Confusing the internal field with the applied field. E is the RMS field inside the material. The field in the surrounding air or in the waveguide differs, set by boundary conditions and the applicator design.^[6]
• Mixing mass and density. The calculator infers volume as M_a/ρ. Use the bulk density of the load as packed or poured, not the density of the solid material, for granular or porous loads.
• Treating absorbed power as generator power. Step 1 gives power absorbed by the load. Add reflections, wall losses and generator efficiency (Tutorial 4) before choosing a source.
• Applying it to drying or boiling. The calculator covers sensible heating only. For evaporation, add the latent heat term by hand: P = (M_ac_pΔT + m_vL_v)/t, with L_v ≈ 2,257 kJ/kg for water at 100 °C.^[13]

7. Dielectric Heating Equation Summary

Need help with a microwave or RF heating applicator, material characterization or scale-up? Contact SiberSci, LLC. Related services and tools: Microwave Heating Engineering, RF Heating Engineering, Wavelength Calculator and Skin Depth Calculator.

8. References

1. A. C. Metaxas and R. J. Meredith, Industrial Microwave Heating, IEE Power Engineering Series 4. London: Peter Peregrinus, 1983 (reprinted 1993).
2. A. C. Metaxas, Foundations of Electroheat: A Unified Approach. Chichester, UK: Wiley, 1996.
3. R. J. Meredith, Engineers' Handbook of Industrial Microwave Heating, IEE Power Series 25. London: IEE, 1998.
4. A. R. von Hippel, Dielectric Materials and Applications. Cambridge, MA: MIT Press, 1954; reprinted Boston: Artech House, 1995.
5. L. F. Chen, C. K. Ong, C. P. Neo, V. V. Varadan, and V. K. Varadan, Microwave Electronics: Measurement and Materials Characterization. Chichester, UK: Wiley, 2004.
6. D. M. Pozar, Microwave Engineering, 4th ed. Hoboken, NJ: Wiley, 2012. Ch. 1, "Electromagnetic Theory" (complex permittivity, loss tangent and plane waves in lossy media).
7. H. J. Liebe, G. A. Hufford, and T. Manabe, "A model for the complex permittivity of water at frequencies below 1 THz," International Journal of Infrared and Millimeter Waves, vol. 12, no. 7, pp. 659–675, 1991.
8. W. J. Ellison, "Permittivity of pure water, at standard atmospheric pressure, over the frequency range 0–25 THz and the temperature range 0–100 °C," Journal of Physical and Chemical Reference Data, vol. 36, no. 1, pp. 1–18, 2007.
9. National Institute of Standards and Technology, "CODATA Value: vacuum electric permittivity," ε₀ = 8.8541878188 × 10⁻¹² F/m. https://physics.nist.gov/cgi-bin/cuu/Value?ep0
10. National Institute of Standards and Technology, "CODATA Value: speed of light in vacuum," c = 299,792,458 m/s (exact). https://physics.nist.gov/cgi-bin/cuu/Value?c
11. U.S. Code of Federal Regulations, Title 47, Part 18, "Industrial, Scientific, and Medical Equipment," §18.301, "Operating frequencies." https://www.ecfr.gov/current/title-47/chapter-I/subchapter-A/part-18
12. S. Best, "Dielectric Heating Calculator," SiberSci, LLC. https://sibersci.com/dielectric-heating-calculator/
13. W. Wagner and A. Pruß, "The IAPWS formulation 1995 for the thermodynamic properties of ordinary water substance for general and scientific use," Journal of Physical and Chemical Reference Data, vol. 31, no. 2, pp. 387–535, 2002.

Tutorial by Dr. Scott Best, SiberSci, LLC. All worked-example results were reproduced in the live calculator. Water permittivities are computed from the Liebe–Hufford–Manabe model [7] and rounded as entered; measure the permittivity of your own material at the heating frequency and across the process temperature range [5].