Introduction & Context

Ohmic (or Joule) heating is a volumetric heating method in which an alternating electric current is passed directly through a conductive food or process fluid. Because heat is generated internally, surface fouling is minimized and heating rates are rapid, making the technique attractive for continuous pasteurization, sterilization, and blanching in the food, pharmaceutical, and specialty-chemical industries. The key design task is to determine the electrical operating point—voltage and current—that will deliver the required thermal duty while respecting material, safety, and equipment constraints.

Methodology & Formulas

Energy balance
The thermal power \(q\) needed to raise the mass flow rate \(G\) from an inlet temperature \(T_{\text{in}}\) to an outlet temperature \(T_{\text{out}}\) is \[ q = G\,C_p\,(T_{\text{out}}-T_{\text{in}}) \] where \(C_p\) is the specific heat capacity (assumed constant).
Electrical power equivalence
For ohmic systems the electrical power equals the thermal power (100 % efficiency assumption): \[ q = V\,I \] where \(V\) is the voltage across the electrodes and \(I\) is the current.
Ohm’s law in resistive form
The resistance of the fluid slug between electrodes is \[ R = \frac{L}{\kappa\,A} \] with electrode gap \(L\), electrical conductivity \(\kappa\), and flow cross-section \(A=\pi\,r^{2}=\pi\,(d/2)^{2}\) for a circular tube of inner diameter \(d\).
Combine to find operating voltage
Substitute \(I=V/R\) into \(q=V\,I\) and solve for \(V\): \[ V = \sqrt{q\,R} = \sqrt{q\,\frac{L}{\kappa\,A}} \]
Current follows from power
\[ I = \frac{q}{V} \]

Typical validity & safety thresholds
Parameter	Range / Limit	Comment
Electrical conductivity \(\kappa\)	0.2–2 S m⁻¹	Food & similar fluids
Temperature rise \(\Delta T\)	< 80 °C	Minimize coagulation or quality loss
Electric field \(E = V/L\)	< 15 kV	Commercial insulation limits
Current \(I\)	< 100 A	Food-grade electrode sizing

Use \( P = V^2 / R \) where \( R = \rho L / A \).

Measure voltage \( V \) across the electrodes.
Determine electrical conductivity \( \sigma = 1/\rho \) of the food at its highest process temperature.
Calculate resistance \( R \) from the electrode gap \( L \) and cross-sectional flow area \( A \).
Multiply by duty factor if pulsed AC is used.

Result is instantaneous power in watts; multiply by residence time to get energy per kg.

Use the conductivity at the target sterilisation temperature, not the inlet value.

Measure \( \sigma \) at 90 °C or higher in a lab cell with identical formulation.
Fit \( \sigma(T) = \sigma_0[1 + \alpha(T - T_0)] \) to obtain temperature coefficient \( \alpha \).
Insert \( \sigma(T_{\text{max}}) \) into the resistance calculation to guarantee the design power is met at the coldest spot.

This prevents under-processing caused by rising resistance as the fluid heats.

Solids lower the bulk conductivity and reduce the homogeneous field strength.

Determine effective conductivity \( \sigma_{\text{eff}} = \epsilon \sigma_{\text{liquid}} + (1 - \epsilon) \sigma_{\text{solid}} \) where \( \epsilon \) is liquid volume fraction.
If particles are large (> 3 mm) treat the stream as two resistors in series; use the higher resistance path.
Increase design power by 10–20 % to compensate for non-uniform current distribution around particles.

Verify with a pilot-scale run using fibre-optic temperature loggers inside representative particles.

Yes for pure resistance, but check for additional losses.

At > 1 kHz capacitive current through the food becomes significant; measure impedance \( Z \) not just \( R \).
Skin depth \( \delta \approx 5 \) cm at 10 kHz in 1 S m⁻¹ product; if electrode gap < \( \delta \) treat as uniform resistance.
Include inverter efficiency (typically 92–96 %) in the overall power budget.

Recalculate \( P = V_{\text{rms}}^2 / \text{Re}(Z) \) and confirm the heater reaches set-point within the same residence time.

Worked Example – Ohmic Heating of Tomato Purée in a Continuous Steriliser

A small-scale food processor is designing a continuous ohmic heater to raise the temperature of tomato purée from 15 °C to 72 °C at a mass flow rate of 1.2 kg s^-1. The heater consists of a stainless-steel pipe (ID 40 mm, length 250 mm) and the purée has an electrical conductivity of 0.55 S m^-1 and a specific heat capacity of 3900 J kg^-1 K^-1. Determine the electrical power that must be supplied to achieve the target outlet temperature.

Knowns

Mass flow rate, \( \dot{m} \) = 1.2 kg s^-1
Inlet temperature, \( T_{\text{in}} \) = 15 °C
Outlet temperature, \( T_{\text{out}} \) = 72 °C
Specific heat capacity, \( C_p \) = 3900 J kg^-1 K^-1
Electrical conductivity, \( \kappa \) = 0.55 S m^-1
Heater length, \( L \) = 0.25 m
Internal diameter, \( ID \) = 0.04 m
Cross-sectional area, \( A \) = 0.00126 m²

Step-by-Step Calculation

Calculate the temperature rise required: \[ \Delta T = T_{\text{out}} - T_{\text{in}} = 72 - 15 = 57\ \text{K} \]
Compute the thermal power absorbed by the product: \[ q = \dot{m}\ C_p\ \Delta T = 1.2 \times 3900 \times 57 = 266760\ \text{W} \approx 267\ \text{kW} \]
Determine the heater volume: \[ V = A\ L = 0.00126 \times 0.25 = 0.000315\ \text{m}^3 \]
Evaluate the required volumetric power density: \[ E_V = \frac{q}{V} = \frac{266760}{0.000315} = 846857\ \text{W m}^{-3} \approx 847\ \text{kW m}^{-3} \]
Relate volumetric power density to electric field strength \( E \): \[ E_V = \kappa\ E^2 \Rightarrow E = \sqrt{\frac{E_V}{\kappa}} = \sqrt{\frac{846857}{0.55}} = 1241\ \text{V m}^{-1} \]
Finally, calculate the heater current: \[ I = \kappa\ E\ A = 0.55 \times 1241 \times 0.00126 = 0.86\ \text{A} \] (Note: The low current reflects the high conductivity and small cross-section.)

Final Answer

The ohmic heater must supply 267 kW of electrical power to raise the tomato purée temperature from 15 °C to 72 °C at the given flow rate.

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