Reference ID: MET-2B2A | Process Engineering Reference Sheets Calculation Guide
Introduction & Context
Ohmic heating is a critical process in food engineering where electrical energy is converted directly into thermal energy within a conductive fluid matrix. Unlike conventional heat exchangers that rely on thermal gradients and surface conduction, ohmic heating provides volumetric heating, which significantly reduces fouling and preserves the nutritional quality of heat-sensitive products. This calculation is essential for process engineers to determine the temperature rise in continuous flow systems, ensuring that the product reaches the required pasteurization or sterilization temperatures while maintaining flow uniformity.
Methodology & Formulas
The calculation accounts for the temperature-dependent nature of electrical conductivity, which typically increases as the fluid heats up. To determine the temperature rise, we equate the electrical power input to the sensible heat gain of the fluid, incorporating a linear correction factor for conductivity.
The fundamental constant K is defined by the geometric and electrical properties of the heater:
\[ K = \frac{V^2 \cdot A \cdot \sigma_{inlet}}{L} \]
To account for the temperature-dependent conductivity, we use the average temperature rise to determine the average conductivity. The temperature rise is derived by solving the energy balance equation:
Model invalid; non-uniform heating due to turbulent residence time distribution.
Thermal Stability
Denominator ≤ 0
System thermal runaway; power input exceeds heat removal capacity.
Current Type
DC Current
Invalidates model; causes electrolysis and electrode fouling.
To determine the temperature rise, you must account for the electrical energy input and the thermal properties of the product. Use the following approach:
Identify the electrical conductivity of the food product, which is temperature-dependent.
Apply the energy balance equation: ΔT = (V² × t) / (R × m × Cp), where V is voltage, t is time, R is resistance, m is mass, and Cp is specific heat capacity.
Adjust for heat losses to the environment if the system is not perfectly adiabatic.
Non-uniform heating is a common challenge in ohmic processing. Key contributors include:
Differences in electrical conductivity between solid particles and the liquid carrier phase.
Electrode polarization effects near the contact surfaces.
Variations in local electric field strength due to the geometry of the heating chamber.
Natural convection currents induced by density gradients as the product heats up.
Electrical conductivity is the primary driver of the heating rate in ohmic systems. Higher conductivity allows for greater current flow at a given voltage, which directly increases the rate of internal heat generation. Engineers should note:
Conductivity typically increases linearly with temperature for most food products.
If the solid phase has lower conductivity than the liquid, it will heat more slowly, potentially leading to cold spots.
Monitoring conductivity in real-time is essential for maintaining consistent process lethality.
Worked Example: Temperature Rise in an Ohmic Heater for Food Processing
A food processing plant is designing an ohmic heating system for the continuous pasteurization of a liquid fruit puree. The system must achieve a controlled temperature rise under plug flow conditions to ensure uniform thermal treatment without hot spots. The following parameters are established from system design and fluid properties.
Known Input Parameters:
Mass flow rate, \( \dot{m} \) = 0.5 kg/s
Specific heat capacity, \( C_p \) = 4180.0 J/(kg·°C)
Applied voltage, \( E \) = 230.0 V
Distance between electrodes, \( L \) = 0.2 m
Cross-sectional area of heater, \( A \) = 0.01 m²
Electrical conductivity at inlet, \( \sigma_{in} \) = 0.5 S/m
Inlet temperature, \( T_{in} \) = 20.0 °C
Reynolds number, \( Re \) = 1500.0
Temperature coefficient for conductivity, \( \alpha \) = 0.02 °C⁻¹
Reference temperature for conductivity, \( T_{ref} \) = 20.0 °C
Step-by-Step Calculation:
Flow regime validation: The Reynolds number (1500.0) is less than 2100, confirming the plug flow assumption required for the model.
Calculate the base power constant \( K \), derived from the ohmic heating formula: \( K = \frac{E^2 \cdot A \cdot \sigma_{in}}{L} \). Using the input values, \( K \) = 1322.5 W.
Account for the temperature dependence of electrical conductivity. The average conductivity is modeled as \( \sigma_{avg} = \sigma_{in} \cdot \left(1 + \alpha \cdot \frac{\Delta T}{2}\right) \).
Establish the energy balance. The electrical power input \( q \) must equal the sensible heat gain of the fluid:
\( q = \dot{m} \cdot C_p \cdot \Delta T \)
\( q = K \cdot \left(1 + \alpha \cdot \frac{\Delta T}{2}\right) \)
Solve the energy balance for the temperature rise \( \Delta T \). Combining the equations gives: \( \Delta T = \frac{K}{\dot{m} \cdot C_p - \frac{K \cdot \alpha}{2}} \).
The denominator of this equation is calculated as 2076.775 W/°C.
Therefore, the temperature rise is \( \Delta T = \frac{1322.5}{2076.775} = 0.637 °C \).
The outlet temperature is \( T_{out} = T_{in} + \Delta T = 20.0 + 0.637 = 20.637 °C \).
The final electrical power input is \( q = \dot{m} \cdot C_p \cdot \Delta T = 1330.922 W \).
The average electrical conductivity over the heating section is \( \sigma_{avg} = 0.503 S/m \).
Final Answer: For the given system, the calculated temperature rise is \( \Delta T = 0.637 \, \text{°C} \). This results in an outlet fluid temperature of \( T_{out} = 20.637 \, \text{°C} \), with a required power input of \( q = 1330.922 \, \text{W} \) and an average conductivity of \( \sigma_{avg} = 0.503 \, \text{S/m} \).
"Un projet n'est jamais trop grand s'il est bien conçu."— André Citroën
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